Resolucao Mecanica Vetorial para Engenheiros 9ed cap 6 10 estatica

PROBLEM 6.1 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression.

SOLUTION AB = 32 + 1.252 = 3.25 m BC = 32 + 42 = 5 m

Reactions: ΣM A = 0: (84 kN)(3 m) − C (5.25 m) = 0

C = 48 kN ΣFx = 0: Ax − C = 0 A x = 48 kN

ΣFy = 0: Ay = 84 kN = 0 A y = 84 kN

Joint A: ΣFx = 0: 48 kN −

12 FAB = 0 13 FAB = +52 kN

ΣFy = 0: 84 kN −

FAB = 52 kN T

5 (52 kN) − FAC = 0 13 FAC = +64.0 kN

FAC = 64.0 kN T

!

Joint C:

FBC 48 kN = 5 3

FBC = 80.0 kN C

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SOLUTION Free body: Entire truss ΣFx = 0: Bx = 0 ΣM B = 0: C (15.75 ft) − (945 lb)(12 ft) = 0 C = 720 lb

ΣFy = 0: By + 720 lb − 945 lb = 0 B y = 225 lb

Free body: Joint B: FAB FBC 225 lb = = 5 4 3

FAB = 375 lb C FBC = 300 lb T

Free body: Joint C: FAC FBC 720 lb = = 9.75 3.75 9 FBC = 300 lb T

FAC = 780 lb C

(Checks)


!


SOLUTION Free body: Entire truss ΣFx = 0: C x = 0 C x = 0 ΣM B = 0: (1.92 kN)(3 m) + C y (4.5 m) = 0 C y = −1.28 kN C y = 1.28 kN ΣFy = 0: B − 1.92 kN − 1.28 kN = 0 B = 3.20 kN

Free body: Joint B: FAB FBC 3.20 kN = = 5 3 4 FAB = 4.00 kN C FBC = 2.40 kN C

Free body: Joint C: ΣFx = 0: −

7.5 FAC + 2.40 kN = 0 8.5 FAC = +2.72 kN

!

! ΣFy =

FAC = 2.72 kN T

!

4 (2.72 kN) − 1.28 kN = 0 (Checks) 8.5



SOLUTION Reactions:

ΣM D = 0: Fy (24) − (4 + 2.4)(12) − (1)(24) = 0 Fy = 4.2 kips

ΣFx = 0: Fx = 0 ΣFy = 0: D − (1 + 4 + 1 + 2.4) + 4.2 = 0 D = 4.2 kips

Joint A: ΣFx = 0: FAB = 0

FAB = 0

ΣFy = 0 : −1 − FAD = 0 FAD = −1 kip

Joint D:

ΣFy = 0: − 1 + 4.2 +

ΣFx = 0:

8 FBD = 0 17 FBD = −6.8 kips

15 (−6.8) + FDE = 0 17 FDE = +6 kips

FAD = 1.000 kip C

FBD = 6.80 kips C

FDE = 6.00 kips T


PROBLEM 6.4 (Continued)

Joint E:

ΣFy = 0 : FBE − 2.4 = 0 FBE = +2.4 kips

FBE = 2.40 kips T

Truss and loading symmetrical about cL



SOLUTION Free body: Truss ΣFx = 0: A x = 0 ΣM A = 0: D (22.5) − (10.8 kips)(22.5) − (10.8 kips)(57.5) = 0 D = 38.4 kips

ΣFy = 0: A y = 16.8 kips

Free body: Joint A: FAB F 16.8 kips = AD = 22.5 25.5 12

FAB = 31.5 kips T FAD = 35.7 kips C

Free body: Joint B: ΣFx = 0:

FBC = 31.5 kips T

ΣFy = 0:

FBD = 10.80 kips C

Free body: Joint C: FCD FBC 10.8 kips = = 37 35 12 FBC = 31.5 kips T

FCD = 33.3 kips C

(Checks)


!


SOLUTION Free Body: Truss ΣM E = 0: F (3 m) − (900 N)(2.25 m) − (900 N)(4.5 m) = 0 F = 2025 N

ΣFx = 0: Ex + 900 N + 900 N = 0 Ex = −1800 N E x = 1800 N ΣFy = 0: E y + 2025 N = 0 E y = −2025 N E y = 2025 N FAB = FBD = 0

We note that AB and BD are zero-force members: Free body: Joint A: FAC FAD 900 N = = 2.25 3.75 3

FAC = 675 N T FAD = 1125 N C

Free body: Joint D: FCD FDE 1125 N = = 3 2.23 3.75

FCD = 900 N T FDF = 675 N C

Free body: Joint E: ΣFx = 0: FEF − 1800 N = 0

FEF = 1800 N T

ΣFy = 0: FCE − 2025 N = 0

FCE = 2025 N T



Free body: Joint F: ΣFy = 0:

2.25 FCF + 2025 N − 675 N = 0 3.75 FCF = −2250 N

ΣFx = −

FCF = 2250 N C

3 ( −2250 N) − 1800 N = 0 (Checks) 3.75



SOLUTION Free body: Truss ΣFy = 0: B y = 0 ΣM B = 0: D(4.5 m) + (8.4 kN)(4.5 m) = 0 D = −8.4 kN

D = 8.4 kN

ΣFx = 0: Bx − 8.4 kN − 8.4 kN − 8.4 kN = 0 Bx = +25.2 kN B x = 25.2 kN

Free body: Joint A: FAB FAC 8.4 kN = = 5.3 4.5 2.8

FAB = 15.90 kN C FAC = 13.50 kN T

Free body: Joint C: ΣFy = 0: 13.50 kN −

4.5 FCD = 0 5.3 FCD = +15.90 kN

ΣFx = 0: − FBC − 8.4 kN −

FCD = 15.90 kN T

2.8 (15.90 kN) = 0 5.3 FBC = −16.80 kN

FBC = 16.80 kN C



Free body: Joint D: FBD 8.4 kN = 4.5 2.8

FBD = 13.50 kN C

We can also write the proportion FBD 15.90 kN = 4.5 5.3

FBD = 13.50 kN C (Checks)



SOLUTION

AD = 52 + 122 = 13 ft BCD = 122 + 162 = 20 ft

Reactions:

ΣFx = 0: Dx = 0 ΣM E = 0: D y (21 ft) − (693 lb)(5 ft) = 0 ΣFy = 0: 165 lb − 693 lb + E = 0

Joint D:

D y = 165 lb E = 528 lb

ΣFx = 0:

5 4 FAD + FDC = 0 13 5

(1)

ΣFy = 0:

12 3 FAD + FDC + 165 lb = 0 13 5

(2)

Solving (1) and (2), simultaneously: FAD = −260 lb

FAD = 260 lb C

FDC = +125 lb

FDC = 125 lb T



Joint E: ΣFx = 0:

5 4 FBE + FCE = 0 13 5

(3)

ΣFy = 0:

12 3 FBE + FCE + 528 lb = 0 13 5

(4)

Solving (3) and (4), simultaneously: FBE = −832 lb

FBE = 832 lb C

FCE = +400 lb

FCE = 400 lb T

Joint C:

Force polygon is a parallelogram (see Fig. 6.11 p. 209) FAC = 400 lb T FBC = 125 lb T

Joint A:

ΣFx = 0:

5 4 (260 lb) + (400 lb) + FAB = 0 13 5 FAB = −420 lb

ΣFy = 0:

FAB = 420 lb C

12 3 (260 lb) − (400 lb) = 0 13 5 0 = 0 (Checks)


!

PROBLEM 6.9 Determine the force in each member of the Pratt roof truss shown. State whether each member is in tension or compression.

SOLUTION Free body: Truss

ΣFx = 0: Ax = 0

Due to symmetry of truss and load Ay = H =

1 total load = 21 kN 2

Free body: Joint A:

FAB FAC 15.3 kN = = 37 35 12 FAB = 47.175 kN FAC = 44.625 kN

FAB = 47.2 kN C FAC = 44.6 kN T

!

Free body: Joint B:

From force polygon:

FBD = 47.175 kN, FBC = 10.5 kN

FBC = 10.50 kN C FBD = 47.2 kN C



Free body: Joint C:

ΣFy = 0:

3 FCD − 10.5 = 0 5

ΣFx = 0: FCE +

FCD = 17.50 kN T

4 (17.50) − 44.625 = 0 5 FCE = 30.625 kN

Free body: Joint E:

DE is a zero-force member

FCE = 30.6 kN T

!

FDE = 0

!



PROBLEM 6.10 Determine the force in each member of the fan roof truss shown. State whether each member is in tension or compression.

SOLUTION Free body: Truss ΣFx = 0 : A x = 0

From symmetry of truss and loading: Ay = I =

1 2

Total load

A y = I = 6 kN

Free body: Joint A:

F FAB 5 kN = AC = 9.849 9 4

FAB = 12.31 kN C FAC = 11.25 kN T

Free body: Joint B: ΣFx =

9 (12.31 kN + FBD + FDC ) = 0 9.849

FBD + FBC = −12.31 kN

or ΣFy =

or

(1)

4 (12.31 kN + FBD − FBC ) − 2 kN = 0 9.849

FBD − FBC = −7.386 kN

(2)

Add (1) and (2):

2 FBD = −19.70 kN

FBD = 9.85 kN C

Subtract (2) from (1):

2 FBC = −4.924 kN

FBC = 2.46 kN C



Free body: Joint D:

FCD = 2.00 kN C

From force polygon:

FDE = 9.85 kN C

Free body: Joint C: ΣFy =

4 4 FCE − (2.46 kN) − 2 kN = 0 5 9.849

FCE = 3.75 kN T

3 9 ΣFx = 0 : FCG + (3.75 kN) + (2.46 kN) − 11.25 kN = 0 5 9.849 FCG = +6.75 kN

FCG = 6.75 kN T

From the symmetry of the truss and loading: FEF = FDE

FEF = 9.85 kN C

FEG = FCE

FEG = 3.75 kN T

FFG = FCD

FFG = 2.00 kN C

FFH = FBD

FFH = 9.85 kN C

FGH = FBC

FGH = 2.46 kN C

FGI = FAC

FGI = 11.25 kN T

FHI = FAB

FHI = 12.31 kN C


PROBLEM 6.11 Determine the force in each member of the Howe roof truss shown. State whether each member is in tension or compression.


ΣFx = 0: H x = 0

Because of the symmetry of the truss and loading: A = Hy =

1 2

Total load

A = H y = 1200 lb

Free body: Joint A: FAB FAC 900 lb = = 5 4 3

FAB = 1500 lb C FAC = 1200 lb T

Free body: Joint C: BC is a zero-force member FCE = 1200 lb T

FBC = 0


!



or

FBD + FBE = −1500 lb ΣFy = 0:

or Add Eqs. (1) and (2): Subtract (2) from (1):

4 4 4 FBD + FBC + (1500 lb) = 0 5 5 5

(1)

3 3 3 FBD − FBE + (1500 lb) − 600 lb = 0 5 5 5

FBD − FBE = −500 lb 2 FBD = −2000 lb 2 FBE = −1000 lb

(2) FBD = 1000 lb C FBE = 500 lb C

Free Body: Joint D: 4 4 (1000 lb) + FDF = 0 5 5

ΣFx = 0:

FDF = −1000 lb

FDF = 1000 lb C

3 3 (1000 lb) − ( −1000 lb) − 600 lb − FDE = 0 5 5

ΣFy = 0:

FDE = +600 lb

FDE = 600 lb T

Because of the symmetry of the truss and loading, we deduce that FEF = FBE

FEF = 500 lb C

FEG = FCE

FEG = 1200 lb T

FFG = FBC

FFG = 0

FFH = FAB

FFH = 1500 lb C

FGH = FAC

FGH = 1200 lb T


PROBLEM 6.12 Determine the force in each member of the Gambrel roof truss shown. State whether each member is in tension or compression.

SOLUTION Free body: Truss ΣFx = 0: H x = 0

Because of the symmetry of the truss and loading A = Hy =

1 2

Total load

A = H y = 1200 lb

Free body: Joint A: FAB FAC 900 lb = = 5 4 3

FAB = 1500 lb C FAC = 1200 lb T

Free body: Joint C: BC is a zero-force member FBC = 0

FCE = 1200 lb T

Free body: Joint B:

ΣFx = 0:

24 4 4 FBD + FBE + (1500 lb) = 0 25 5 5 24 FBD + 20 FBE = −30, 000 lb

or ΣFy = 0:

(1)

7 3 3 FBD − FBE + (1500) − 600 = 0 25 5 5 7 FBD − 15FBE = −7,500 lb

or

(2)



Multiply (1) by (3), (2) by 4, and add: 100 FBD = −120, 000 lb

FBD = 1200 lb C

500 FBE = −30,000 lb

FBE = 60.0 lb C

Multiply (1) by 7, (2) by –24, and add:

Free body: Joint D: 24 24 FDF = 0 (1200 lb) + 25 25

ΣFx = 0:

FDF = −1200 lb ΣFy = 0:

FDF = 1200 lb C

7 7 (1200 lb) − (−1200 lb) − 600 lb − FDE = 0 25 25 FDE = 72.0 lb

FDE = 72.0 lb T

Because of the symmetry of the truss and loading, we deduce that FEF = FBE

FEF = 60.0 lb C

FEG = FCE

FEG = 1200 lb T

FFG = FBC

FFG = 0

FFH = FAB

FFH = 1500 lb C

FGH = FAC

FGH = 1200 lb T

Note: Compare results with those of Problem 6.9.


PROBLEM 6.13 Determine the force in each member of the truss shown.

SOLUTION 12.5 kN FCD FDG = = 2.5 6 6.5

Joint D:

FCD = 30 kN T FDG = 32.5 kN C

ΣF = 0: FCG = 0

Joint G:

ΣF = 0: FFG = 32.5 kN C

Joint C:

BF =

BF 2 (2.5 m) = 1.6667 m β = ∠BCF = tan −1 = 39.81° 3 2

ΣFy = 0: − 12.5 kN − FCF sin β = 0 −12.5 kN − FCF sin 39.81° = 0 FCF = −19.526 kN

FCF = 19.53 kN C

ΣFx = 0: 30 kN − FBC − FCF cos β = 0 30 kN − FBC − (−19.526 kN) cos 39.81° = 0 FBC = +45.0 kN

Joint F:

ΣFx = 0: −

FBC = 45.0 kN T

6 6 FEF − (32.5 kN) − FCF cos β = 0 6.5 6.5 6.5 ! FEF = −32.5 kN − " # (19.526 kN) cos 39.81° $ 6 % FEF = −48.75 kN

ΣFy = 0: FBF −

FEF = 48.8 kN C

2.5 2.5 (32.5 kN) − (19.526 kN)sin 39.81° = 0 FEF − 6.5 6.5 FBF −

2.5 ( −48.75 kN) − 12.5 kN − 12.5 kN = 0 6.5

FBF = +6.25 kN

FBF = 6.25 kN T



Joint B:

tan α =

2.5 m ; γ = 51.34° 2m

ΣFy = 0: − 12.5 kN − 6.25 kN − FBE sin 51.34° = 0 FBE = −24.0 kN

FBE = 24.0 kN C

ΣFx = 0: 45.0 kN − FAB + (24.0 kN) cos 51.34° = 0 FAB = +60 kN

FAB = 60.0 kN T

Joint E:

γ = 51.34° ΣFy = 0: FAE − (24 kN) sin 51.34° − (48.75 kN) FAE = +37.5 kN

2.5 =0 6.5 FAE = 37.5 kN T


PROBLEM 6.14 Determine the force in each member of the roof truss shown. State whether each member is in tension or compression.

SOLUTION Free body: Truss ΣFx = 0: A x = 0

From symmetry of loading: Ay = E =

1 Total load 2

A y = E = 3.6 kN

We note that DF is a zero-force member and that EF is aligned with the load. Thus FDF = 0 FEF = 1.2 kN C

Free body: Joint A: FAB FAC 2.4 kN = = 13 12 5

FAB = 6.24 kN C FAC = 2.76 kN T


3 12 12 FBC + FBD + (6.24 kN) = 0 3.905 13 13

ΣFy = 0:

−

2.5 5 5 FBC + FBD + (6.24 kN) − 2.4 kN = 0 3.905 13 13

(1) (2)



Multiply (1) by 2.5, (2) by 3, and add: 45 45 FBD + (6.24 kN) − 7.2 kN = 0, FBD = −4.16 kN, 13 13

FBD = 4.16 kN C

Multiply (1) by 5, (2) by –12, and add: 45 FBC + 28.8 kN = 0, FBC = −2.50 kN, 3.905

FBC = 2.50 kN C

Free body: Joint C: ΣFy = 0:

5 2.5 (2.50 kN) = 0 FCD − 5.831 3.905 FCD = 1.867 kN T

ΣFx = 0: FCE − 5.76 kN +

3 3 (2.50 kN) + (1.867 kN) = 0 3.905 5.831 FCE = 2.88 kN T

Free body: Joint E: ΣFy = 0:

5 FDE + 3.6 kN − 1.2 kN = 0 7.81 FDE = −3.75 kN

ΣFx = 0: − FCE −

FDE = 3.75 kN C

6 ( −3.75 kN) = 0 7.81

FCE = +2.88 kN FCE = 2.88 kN T

(Checks)


PROBLEM 6.15 Determine the force in each member of the Warren bridge truss shown. State whether each member is in tension or compression.

SOLUTION ΣFx = 0: Ax = 0

Free body: Truss Due to symmetry of truss and loading Ay = G =

Free body: Joint A:

1 Total load = 6 kips 2

FAB FAC 6 kips = = 5 3 4

FAB = 7.50 kips C FAC = 4.50 kips T

!

Free body: Joint B:

FBC FBD 7.5 kips = = 5 6 5

FBC = 7.50 kips T FBD = 9.00 kips C

Free body: Joint C: ΣFy = 0:

4 4 (7.5) + FCD − 6 = 0 5 5 FCD = 0

3 ΣFx = 0: FCE − 4.5 − (7.5) = 0 5 FCE = +9 kips

FCE = 9.00 kips T



!

PROBLEM 6.16 Solve Problem 6.15 assuming that the load applied at E has been removed. PROBLEM 6.15 Determine the force in each member of the Warren bridge truss shown. State whether each member is in tension or compression.


ΣFx = 0: Ax = 0 ΣM G = 0: 6(36) − Ay (54) = 0 A y = 4 kips ΣFy = 0: 4 − 6 + G = 0 G = 2 kips

Free body: Joint A:

FAB FAC 4 kips = = 5 3 4

FAB = 5.00 kips C FAC = 3.00 kips T

Free body Joint B:

FBC FBD 5 kips = = 5 6 5

FBC = 5.00 kips T FBD = 6.00 kips C

Free body Joint C: ΣM y = 0:

4 4 (5) + FCD − 6 = 0 5 5

3 3 ΣFx = 0: FCE + (2.5) − (5) − 3 = 0 5 5

FCD = 2.50 kips T FCE = 4.50 kips T


!


Free body: Joint D: 4 4 ΣFy = 0: − (2.5) − FDE = 0 5 5 FDE = −2.5 kips

FDE = 2.50 kips C

3 3 ΣFx = 0: FDF + 6 − (2.5) − (2.5) = 0 5 5

Free body: Joint F:

FDF = −3 kips

FDF = 3.00 kips C

FEF FFG 3 kips = = 5 5 6

FEF = 2.50 kips T FFG = 2.50 kips C

!

Free body: Joint G:

FEG 2 kips = 3 4

Also:

FFG 2 kips = 5 4

FEG = 1.500 kips T

FFG = 2.50 kips C (Checks)


!

PROBLEM 6.17 Determine the force in member DE and in each of the members located to the left of DE for the inverted Howe roof truss shown. State whether each member is in tension or compression.

SOLUTION Free body: Truss:

ΣFx = 0: A x = 0 ΣM H = 0: (400 lb)(4d ) + (800 lb)(3d ) + (800 lb)(2d ) + (800 lb)d − Ay (4d ) = 0 Ay = 1600 lb

Angles:

6.72 α = 32.52° 10.54 6.72 tan β = β = 16.26° 23.04 tan α =

Free body: Joint A:

FAC FAB 1200 lb = = sin 57.48° sin106.26° sin16.26° FAB = 3613.8 lb C FAC = 4114.3 lb T

FAB = 3610 lb C , FAC = 4110 lb T



Free body: Joint B: ΣFy = 0: − FBC − (800 lb) cos16.26° = 0 FBC = −768.0 lb

FBC = 768 lb C

ΣFx = 0: FBD + 3613.8 lb + (800 lb) sin16.26° = 0 FBD = −3837.8 lb

FBD = 3840 lb C

Free body: Joint C:

ΣFy = 0: − FCE sin 32.52° + (4114.3 lb) sin 32.52° − (768 lb) cos16.26° = 0 FCE = 2742.9 lb

FCE = 2740 lb T

ΣFx = 0: FCD − (4114.3 lb) cos 32.52° + (2742.9 lb) cos 32.52° − (768 lb) sin16.26° = 0 FCD = 1371.4 lb

FCD = 1371 lb T

Free body: Joint E:

FDE 2742.9 lb = sin 32.52° sin73.74°

FDE = 1536 lb C


PROBLEM 6.18 Determine the force in each of the members located to the right of DE for the inverted Howe roof truss shown. State whether each member is in tension or compression.


ΣM A = 0: H (4d ) − (800 lb)d − (800 lb)(2d ) − (800 lb)(3d ) − (400 lb)(4d ) = 0 H = 1600 lb

Angles:

6.72 α = 32.52° 10.54 6.72 tan β = β = 16.26° 23.04 tan α =

Free body: Joint H:

FGH = (1200 lb) cot16.26° FGH = 4114.3 lb T FFH =

1200 lb = 4285.8 lb sin16.26°

FGH = 4110 lb T FFH = 4290 lb C



Free body Joint F: ΣFy = 0: − FFG − (800 lb) cos16.26° = 0 FFG = −768.0 lb

FFG = 768 lb C

ΣFx = 0: − FDF − 4285.8 lb + (800 lb) sin 16.26° = 0 FDF = −4061.8 lb

FDF = 4060 lb C

Free body: Joint G: ΣFy = 0: FDG sin 32.52° − (768.0 lb) cos16.26° = 0 FDG = +1371.4 lb FDG = 1371 lb T

! ΣFx = 0: − FEG + 4114.3 lb − (768.0 lb)sin16.26° − (1371.4 lb) cos 32.52° = 0 FEG = +2742.9 lb

FEG = 2740 lb T


!

PROBLEM 6.19 Determine the force in each of the members located to the left of FG for the scissors roof truss shown. State whether each member is in tension or compression.

SOLUTION Free Body: Truss

ΣFx = 0: A x = 0 ΣM L = 0: (1 kN)(12 m) + (2 kN)(10 m) + (2 kN)(8 m) + (1 kN)(6 m) − Ay (12 m) = 0 A y = 4.50 kN

FBC = 0

We note that BC is a zero-force member: Also:

FCE = FAC

Free body: Joint A:

ΣFx = 0:

ΣFy = 0:

(1) 1 2 1 2

FAB +

FAB +

2 5 1 5

FAC = 0

(2)

FAC + 3.50 kN = 0

(3)

Multiply (3) by –2 and add (2): −

1 2

FAB − 7 kN = 0

FAB = 9.90 kN C



Subtract (3) from (2): 1 5

FAC − 3.50 kN = 0 FAC = 7.826 kN FCE = FAC = 7.826 kN

From (1):

FAC = 7.83 kN T FCE = 7.83 kN T

Free body: Joint B: ΣFy = 0:

1 2

FBD +

1 2

(9.90 kN) − 2 kN = 0 FBD = −7.071 kN

1

ΣFx = 0: FBE +

2

(9.90 − 7.071)kN = 0 FBE = −2.000 kN

Free body: Joint E:

ΣFx = 0:

2 5

FBD = 7.07 kN C

FBE = 2.00 kN C

( FEG − 7.826 kN) + 2.00 kN = 0 FEG = 5.590 kN

ΣFy = 0: FDE −

1 5

FEG = 5.59 kN T

(7.826 − 5.590)kN = 0 FDE = 1.000 kN

FDE = 1.000 kN T

Free body: Joint D: ΣFx = 0:

2 5

1 2

(7.071 kN)

FDF + FDG = −5.590 kN

or ΣFy = 0:

or

( FDF + FDG ) +

1 5

( FDF − FDG ) +

1 2

(4)

(7.071 kN) = 2 kN − 1 kN = 0

FDE − FDG = −4.472

(5)

Add (4) and (5):

2 FDF = −10.062 kN

FDF = 5.03 kN C


2 FDG = −1.1180 kN

FDG = 0.559 kN C


PROBLEM 6.20 Determine the force in member FG and in each of the members located to the right of FG for the scissors roof truss shown. State whether each member is in tension or compression.


ΣM A = 0: L(12 m) − (2 kN)(2 m) − (2 kN)(4 m) − (1 kN)(6 m) = 0 L = 1.500 kN

Angles: tan α = 1

α = 45°

1 tan β = 2

β = 26.57°

Zero-force members: Examining successively joints K, J, and I, we note that the following members to the right of FG are zeroforce members: JK, IJ, and HI. FHI = FIJ = FJK = 0

Thus: We also note that

and

FGI = FIK = FKL

(1)

FHJ = FJL

(2)



Free body: Joint L:

FJL F 1.500 kN = KL = sin116.57° sin 45° sin18.43° FJL = 4.2436 kN

FJL = 4.24 kN C FKL = 3.35 kN T

From Eq. (1):

FGI = FIK = FKL

FGI = FIK = 3.35 kN T

From Eq. (2):

FHJ = FJL = 4.2436 kN

FHJ = 4.24 kN T

FGH FFH 4.2436 = = sin108.43° sin18.43° sin 53.14°

FFH = 5.03 kN C

Free body: Joint H:

FGH = 1.677 kN T

Free body: Joint F: ΣFx = 0: − FDF cos 26.57° − (5.03 kN) cos 26.57° = 0 FDF = −5.03 kN

ΣFy = 0: − FFG − 1 kN + (5.03 kN) sin 26.57° − ( −5.03 kN)sin 26.57° = 0 FFG = 3.500 kN

FFG = 3.50 kN T


PROBLEM 6.21 Determine the force in each of the members located to the left of line FGH for the studio roof truss shown. State whether each member is in tension or compression.


ΣFx = 0: A x = 0

Because of symmetry of loading: Ay = L =

1 Total load 2

A y = L = 1200 lb

Zero-Force Members. Examining joints C and H, we conclude that BC, EH, and GH are zero-force members. Thus FBC = FEH = 0

Also,

FCE = FAC

Free body: Joint A

FAB

(1)

FAC 1000 lb = 2 1 5 FAB = 2236 lb C =

FAB = 2240 lb C FAC = 2000 lb T FCE = 2000 lb T

From Eq. (1): Free body: Joint B ΣFx = 0:

2 5

FBD +

2 5

FBE +

2 5

(2236 lb) = 0

FBD + FBE = −2236 lb

or ΣFy = 0:

1 5

FBD −

1 5

FBE +

1

FBD − FBE = −1342 lb

or

(2)

5

(2236 lb) − 400 lb = 0

(3)



Add (2) and (3):

2 FBD = −3578 lb

FBD = 1789 lb C


2 FBE = − 894 lb

FBE = 447 lb C

Free body: Joint E 2

ΣFx = 0:

5

FEG +

2 5

(447 lb) − 2000 lb = 0

FEG = 1789 lb T ΣFy = 0: FDE +

1 5

1

(1789 lb) −

5

FDE = − 600 lb

(447 lb) = 0

FDE = 600 lb C

Free body: Joint D 2

ΣFx = 0:

5

FDF +

2 5

2

FDG +

5

(1789 lb) = 0

FDF + FDG = −1789 lb

or ΣFy = 0:

1

FDF −

1

FDG +

1

5 5 5 + 600 lb − 400 lb = 0

(4)

(1789 lb)

FDF − FDG = −2236 lb

or Add (4) and (5):

2 FDF = − 4025 lb


2 FDG = 447 lb

(5)

FDF = 2010 lb C FDG = 224 lb T

!


PROBLEM 6.22 Determine the force in member FG and in each of the members located to the right of FG for the studio roof truss shown. State whether each member is in tension or compression.

SOLUTION Reaction at L: Because of the symmetry of the loading, L=

1 Total load, 2

L = 1200 lb

(See F, B, diagram to the left for more details) Free body: Joint L 9 = 26.57° 18 3 β = tan −1 = 9.46° 18 FJL FKL 1000 lb = = sin 63.43° sin 99.46° sin17.11°

α = tan −1

FKL = 3352.7 lb C

FJL = 3040 lb T FKL = 3350 lb C

Free body: Joint K ΣFx = 0: −

2 5

FIK −

2 5

FJK −

2 5

(3352.7 lb) = 0

FIK + FJK = − 3352.7 lb

or:

ΣFy = 0:

or: Add (1) and (2):

1 5

FIK −

1 5

FJK +

(1) 1 5

FIK − FJK = − 2458.3 lb

(2)

2 FIK = − 5811.0

FJK = − 2905.5 lb


(3352.7) − 400 = 0

FIK = 2910 lb C

2 FJK = − 894.4

FJK = − 447.2 lb

FJK = 447 lb C



Free body: Joint J 2

ΣFx = 0: −

13 3

ΣFy = 0:

FIJ +

13

6

FIJ −

37 1

FGJ −

37

6

FGJ +

37 1

5

1

(3040 lb) −

37

2

(3040 lb) −

5

(447.2) = 0

(447.2) = 0

(3) (4)

Multiply (4) by 6 and add to (3): 16 13

FIJ −

8 5

(447.2) = 0

FIJ = 360.54 lb

FIJ = 361 lb T

Multiply (3) by 3, (4) by 2, and add: −

16 37

( FGJ − 3040) −

8 5

(447.2) = 0

FGJ = 2431.7 lb

FGJ = 2430 lb T

Free body: Joint I 2

ΣFx = 0: −

5

2

FFI −

5

2

FGI −

5

(2905.5) +

2 13

(360.54) = 0

FFI + FGI = − 2681.9 lb

or ΣFy = 0:

1 5

FFI −

1 5

FGI +

1 5

(2905.5) −

(5) 3 13

(360.54) − 400 = 0

FFI − FGI = −1340.3 lb

or

(6)

2 FFI = −4022.2

Add (5) and (6):


FFI = −2011.1 lb

FFI = 2010 lb C

2 FGI = −1341.6 lb

FGI = 671 lb C

Free body: Joint F From

ΣFx = 0: FDF = FFI = 2011.1 lb C

! 1 2011.1 lb # = 0 ΣFy = 0: FFG − 400 lb + 2 " $ 5 % FFG = + 1400 lb

FFG = 1400 lb T

!


PROBLEM 6.23 Determine the force in each of the members connecting joints A through F of the vaulted roof truss shown. State whether each member is in tension or compression.

SOLUTION Free body: Truss ΣFx = 0: A x = 0 ΣM K = 0: (1.2 kN)6a + (2.4 kN)5a + (2.4 kN)4a + (1.2 kN)3a − Ay (6a) = 0 A y = 5.40 kN

Free body: Joint A FAB

FAC 4.20 kN = 2 1 5 FAB = 9.3915 kN =

FAB = 9.39 kN C FAC = 8.40 kN T

Free body: Joint B 2

ΣFx = 0:

5 1

ΣFy = 0:

5

FBD + FBD −

1 2 1 2

FBC + FBC +

2 5 1 5

(9.3915) = 0

(1)

(9.3915) − 2.4 = 0

(2)

Add (1) and (2): 3 5

FBD +

3 5

(9.3915 kN) − 2.4 kN = 0

FBD = − 7.6026 kN

FBD = 7.60 kN C

Multiply (2) by –2 and add (1): 3 2

FB + 4.8 kN = 0

FBC = −2.2627 kN

FBC = 2.26 kN C



Free body: Joint C 1

ΣFx = 0:

5 2

ΣFy = 0:

5

4

FCD +

17 1

FCD +

17

FCE + FCE −

1 2 1 2

(2.2627) − 8.40 = 0

(3)

(2.2627) = 0

(4)

Multiply (4) by −4 and add (1): −

7 5

FCD +

5

(2.2627) − 8.40 = 0

2

FCD = − 0.1278 kN

FCD = 0.128 kN C

Multiply (1) by 2 and subtract (2): 7 17

FCE +

3 2

(2.2627) − 2(8.40) = 0

FCE = 7.068 kN

FCE = 7.07 kN T

Free body: Joint D ΣFx = 0:

2 5

1

+ ΣFy = 0:

5

1 5 +

FDF +

(0.1278) = 0

FDF − 2 5

1 2 FDE + (7.6026) 1.524 5

(5)

1.15 1 FDE + (7.6026) 1.524 5

(0.1278) − 2.4 = 0

(6)

Multiply (5) by 1.15 and add (6): 3.30 5

FDF +

3.30 5

(7.6026) +

3.15 5

(0.1278) − 2.4 = 0

FDF = − 6.098 kN

FDF = 6.10 kN C

Multiply (6) by –2 and add (5): 3.30 3 FDE − (0.1278) + 4.8 = 0 1.524 5

FDE = − 2.138 kN

FDE = 2.14 kN C



Free body: Joint E ΣFx = 0:

0.6 4 1 FEF + ( FEH − FCE ) + (2.138) = 0 2.04 1.524 17

(7)

ΣFy = 0:

1.95 1 1.15 FEF + ( FEH − FCE ) − (2.138) = 0 2.04 1.524 17

(8)

Multiply (8) by 4 and subtract (7): 7.2 FEF − 7.856 kN = 0 2.04

FEF = 2.23 kN T


!

PROBLEM 6.24 The portion of truss shown represents the upper part of a power transmission line tower. For the given loading, determine the force in each of the members located above HJ. State whether each member is in tension or compression.

SOLUTION Free body: Joint A F FAB 1.2 kN = AC = 2.29 2.29 1.2

FAB = 2.29 kN T FAC = 2.29 kN C

Free body: Joint F FDF F 1.2 kN = EF = 2.29 2.29 2.1

FDF = 2.29 kN T FEF = 2.29 kN C

Free body: Joint D FBD FDE 2.29 kN = = 2.21 0.6 2.29

FBD = 2.21 kN T FDE = 0.600 kN C

Free body: Joint B ΣFx = 0:

4 2.21 (2.29 kN) = 0 FBE + 2.21 kN − 5 2.29

FBE = 0 3 0.6 (2.29 kN) = 0 ΣFy = 0: − FBC − (0) − 5 2.29

FBC = − 0.600 kN

FBC = 0.600 kN C



Free body: Joint C ΣFx = 0: FCE +

2.21 (2.29 kN) = 0 2.29

FCE = − 2.21kN ΣFy = 0: − FCH − 0.600 kN −

FCE = 2.21 kN C 0.6 (2.29 kN) = 0 2.29

FCH = −1.200 kN

FCH = 1.200 kN C

Free body: Joint E ΣFx = 0: 2.21 kN −

2.21 4 (2.29 kN) − FEH = 0 2.29 5

FEH = 0 ΣFy = 0: − FEJ − 0.600 kN −

FEJ = −1.200 kN

0.6 (2.29 kN) − 0 = 0 2.29

FEJ = 1.200 kN C

!


PROBLEM 6.25 For the tower and loading of Problem 6.24 and knowing that FCH = FEJ = 1.2 kN C and FEH = 0, determine the force in member HJ and in each of the members located between HJ and NO. State whether each member is in tension or compression. PROBLEM 6.24 The portion of truss shown represents the upper part of a power transmission line tower. For the given loading, determine the force in each of the members located above HJ. State whether each member is in tension or compression.

SOLUTION Free body: Joint G FGH F 1.2 kN = GI = 3.03 3.03 1.2

FGH = 3.03 kN T FGI = 3.03 kN C

Free body: Joint L FJL F 1.2 kN = KL = 3.03 3.03 1.2

FJL = 3.03 kN T FKL = 3.03 kN C

Free body: Joint J ΣFx = 0: − FHJ +

2.97 (3.03 kN) = 0 3.03 FHJ = 2.97 kN T

0.6 (3.03 kN) = 0 3.03 = −1.800 kN FJK = 1.800 kN C

Fy = 0: − FJK − 1.2 kN − FJK

Free body: Joint H ΣFx = 0:

4 2.97 (3.03 kN) = 0 FHK + 2.97 kN − 5 3.03 FHK = 0 0.6 3 (3.03) kN − (0) = 0 3.03 5 = −1.800 kN FHI = 1.800 kN C

ΣFy = 0: − FHI − 1.2 kN −

FHI



Free body: Joint I ΣFx = 0: FIK +

2.97 (3.03 kN) = 0 3.03

FIK = −2.97 kN ΣFy = 0: − FIN − 1.800 kN −

FIN = −2.40 kN

FIK = 2.97 kN C 0.6 (3.03 kN) = 0 3.03

FIN = 2.40 kN C

Free body: Joint K ΣFx = 0: −

4 2.97 (3.03 kN) = 0 FKN + 2.97 kN − 5 3.03

FKN = 0 ΣFy = 0: − FKD −

0.6 3 (3.03 kN) − 1.800 kN − (0) = 0 3.03 5

FKD = −2.40 kN

FKD = 2.40 kN C

!


PROBLEM 6.26 Solve Problem 6.24 assuming that the cables hanging from the right side of the tower have fallen to the ground. PROBLEM 6.24 The portion of truss shown represents the upper part of a power transmission line tower. For the given loading, determine the force in each of the members located above HJ. State whether each member is in tension or compression.

SOLUTION Zero-Force Members. Considering joint F, we note that DF and EF are zero-force members: FDF = FEF = 0

Considering next joint D, we note that BD and DE are zero-force members: FBD = FDE = 0

Free body: Joint A F FAB 1.2 kN = AC = 2.29 2.29 1.2

FAB = 2.29 kN T FAC = 2.29 kN C


4 2.21 (2.29 kN) = 0 FBE − 5 2.29

FBE = 2.7625 kN ΣFy = 0: − FBC −

FBE = 2.76 kN T

0.6 3 (2.29 kN) − (2.7625 kN) = 0 2.29 5

FBC = −2.2575 kN

FBC = 2.26 kN C



Free body: Joint C ΣFx = 0: FCE +

2.21 (2.29 kN) = 0 2.29

FCE = 2.21 kN C ΣFy = 0: − FCH − 2.2575 kN −

FCH = −2.8575 kN

0.6 (2.29 kN) = 0 2.29

FCH = 2.86 kN C

Free body: joint E ΣFx = 0: −

4 4 FEH − (2.7625 kN) + 2.21 kN = 0 5 5

FEH = 0 3 3 ΣFy = 0: − FEJ + (2.7625 kN) − (0) = 0 5 5

FEJ = +1.6575 kN

FEJ = 1.658 kN T

!


PROBLEM 6.27 Determine the force in each member of the truss shown. State whether each member is in tension or compression.

SOLUTION Free body: Truss ΣM F = 0: G (20 ft) − (15 kips)(16 ft) − (40 kips)(15 ft) = 0

G = 42 kips ΣFx = 0: Fx + 15 kips = 0

Fx = 15 kips ΣFy = 0: Fy − 40 kips + 42 kips = 0

Fy = 2 kips

Free body: Joint F ΣFx = 0:

1 5

FDF − 15 kips = 0 FDF = 33.54 kips

ΣFy = 0: FBF − 2 kips +

2 5

FDF = 33.5 kips T

(33.54 kips) = 0

FBF = − 28.00 kips

FBF = 28.0 kips C

Free body: Joint B ΣFx = 0: ΣFy = 0:

5 29 2 29

FAB + FAB −

5 61 6 61

FBD + 15 kips = 0

(1)

FBD + 28 kips = 0

(2)



Multiply (1) by 6, (2) by 5, and add: 40 29

FAB + 230 kips = 0 FAB = −30.96 kips

FAB = 31.0 kips C

Multiply (1) by 2, (2) by –5, and add: 40

FBD − 110 kips = 0

61

FBD = 21.48 kips

FBD = 21.5 kips T

Free body: joint D 2

ΣFy = 0:

5

2

FAD −

5

(33.54) +

6 61

(21.48) = 0

FAD = 15.09 kips T 1

ΣFx = 0: FDE +

5

(15.09 − 33.54) −

5

(21.48) = 0

61

FDE = 22.0 kips T

Free body: joint A 5

ΣFx = 0:

29

ΣFy = 0: −

1

FAC +

2 29

5

FAC −

FAE +

2 5

5

FAE +

29

(30.36) −

2 29

1 5

(15.09) = 0

2

(30.96) −

5

(15.09) = 0

(3) (4)

Multiply (3) by 2 and add (4): 8 29

FAC +

12 29

(30.96) −

4 5

FAC = −28.27 kips,

(15.09) = 0

FAC = 28.3 kips C

Multiply (3) by 2 (4) by 5 and add: −

8 5

FAE +

20 29

(30.96) −

12 5

(15.09) = 0

FAE = 9.50 kips T



Free body: Joint C From force triangle FCE 61

=

FCG 28.27 kips = 8 29

FCE = 41.0 kips T FCG = 42.0 kips C

Free body: Joint G ΣFx = 0: ΣFy = 0: 42 kips − 42 kips = 0

FEG = 0 (Checks) !


PROBLEM 6.28 Determine the force in each member of the truss shown. State whether each member is in tension or compression.

SOLUTION Free body: Truss ΣFx = 0: H x = 0 ΣM H = 0: 48(16) − G (4) = 0 G = 192 kN ΣFy = 0: 192 − 48 + H y = 0

H y = −144 kN H y = 144 kN

Zero-Force Members: Examining successively joints C, B, E, D, and F, we note that the following are zero-force members: BC, BE, DE, DG, FG Thus,

FBC = FBE = FDE = FDG = FFG = 0

Also note:

FAB = FBD = FDF = FFH

(1)

FAC = FCE = FEG

(2)

Free body: Joint A:

FAB FAC 48 kN = = 8 3 73

FAB = 128 kN

FAB = 128.0 kN T

FAC = 136.704 kN

FAC = 136.7 kN C

From Eq. (1):

FBD = FDF = FFH = 128.0 kN T

From Eq. (2):

FCE = FEG = 136.7 kN C


!


Free body: Joint H

Also

FGH 145

=

144 kN 9

FGH = 192.7 kN C

FFH 144 kN = 8 9 FFH = 128.0 kN T

(Checks) !


PROBLEM 6.29 Determine whether the trusses of Problems 6.31a, 6.32a, and 6.33a are simple trusses.

SOLUTION Truss of Problem 6.31a Starting with triangle ABC and adding two members at a time, we obtain joints D, E, G, F, and H, but cannot go further thus, this truss is not a simple truss

Truss of Problem 6.32a Starting with triangle HDI and adding two members at a time, we obtain successively joints A, E, J, and B, but cannot go further. Thus, this truss is not a simple truss

Truss of Problem 6.33a Starting with triangle EHK and adding two members at a time, we obtain successively joints D, J, C, G, I, B, F, and A, thus completing the truss. Therefore, this truss is a simple truss


!

PROBLEM 6.30 Determine whether the trusses of Problems 6.31b, 6.32b, and 6.33b are simple trusses.

SOLUTION Truss of Problem 6.31b. Starting with triangle ABC and adding two members at a time, we obtain successively joints E, D, F, G, and H, but cannot go further. Thus, this truss is not a simple truss

Truss of Problem 6.32b. Starting with triangle CGH and adding two members at a time, we obtain successively joints B, L, F, A, K, J, then H, D, N, I, E, O, and P, thus completing the truss. Therefore, this truss is a simple truss

Truss of Problem 6.33b. Starting with triangle GLM and adding two members at a time, we obtain joints K and I but cannot continue, starting instead with triangle BCH, we obtain joint D but cannot continue, thus, this truss is not a simple truss

!


PROBLEM 6.31 For the given loading, determine the zero-force members in each of the two trusses shown.

SOLUTION Truss (a)

FB: Joint B: FBC = 0 FB: Joint C: FCD = 0 FB: Joint J : FIJ = 0 FB: Joint I : FIL = 0 FB: Joint N : FMN = 0

(a)

FB: Joint M : FLM = 0

The zero-force members, therefore, are Truss (b)

BC , CD, IJ , IL, LM , MN

FB: Joint C: FBC = 0 FB: Joint B: FBE = 0 FB: Joint G: FFG = 0 FB: Joint F : FEF = 0 FB: Joint E: FDE = 0 FB: Joint I : FIJ = 0

(b)

FB: Joint M : FMN = 0 FB: Joint N : FKN = 0

The zero-force members, therefore, are

BC , BE , DE , EF , FG , IJ , KN , MN



SOLUTION Truss (a)

FB: Joint B: FBJ = 0 FB: Joint D: FDI = 0 FB: Joint E: FEI = 0 FB: Joint I : FAI = 0 (a)

FB: Joint F : FFK = 0 FB: Joint G: FGK = 0 FB: Joint K : FCK = 0


AI , BJ , CK , DI , EI , FK , GK

FB: Joint K : FFK = 0 FB: Joint O: FIO = 0

(b)


FK and IO

All other members are either in tension or compression.



SOLUTION Truss (a)

FB: Joint F : FBF = 0 FB: Joint B: FBG = 0 FB: Joint G: FGJ = 0 FB: Joint D: FDH = 0 (a)

FB: Joint J : FHJ = 0 FB: Joint H : FEH = 0


BF , BG , DH , EH , GJ , HJ FB: Joint A: FAB = FAF = 0 FB: Joint C: FCH = 0 FB: Joint E: FDE = FEJ = 0 FB: Joint L: FGL = 0 FB: Joint N : FIN = 0


(b)

AB, AF , CH , DE , EJ , GL, IN


PROBLEM 6.34 Determine the zero-force members in the truss of (a) Problem 6.23, (b) Problem 6.28.

SOLUTION (a)

Truss of Problem 6.23 FB : Joint I : FIJ = 0 FB : Joint J : FGJ = 0 FB : Joint G: FGH = 0

The zero-force members, therefore, are (b)

GH , GJ , IJ

Truss of Problem 6.28 FB : Joint C: FBC = 0 FB : Joint B: FBE = 0 FB : Joint E: FDE = 0 FB : Joint D: FDG = 0 FB : Joint F : FFG = 0


BC , BE , DE , DG, FG


PROBLEM 6.35* The truss shown consists of six members and is supported by a short link at A, two short links at B, and a ball and socket at D. Determine the force in each of the members for the given loading.

SOLUTION Free body: Truss From symmetry: Dx = Bx

and

Dy = B y

ΣM z = 0: − A(10 ft) − (400 lb)(24 ft) = 0 A = −960 lb

ΣFx = 0: Bx + Dx + A = 0 2 Bx − 960 lb = 0, Bx = 480 lb

ΣFy = 0: B y + Dy − 400 lb = 0

2 By = 400 lb By = +200 lb B = (480 lb)i + (200 lb) j "

Thus Free body: C FCA = FAC FCB = FBC FCD = FCD

! CA FAC = ( −24i + 10 j) CA 26 ! CB FBC = (−24i + 7k ) CB 25 ! CD FCD = (−24i − 7k ) CD 25

ΣF = 0: FCA + FCB + FCD − (400 lb) j = 0


PROBLEM 6.35* (Continued)

Substituting for FCA , FCB , FCD , and equating to zero the coefficients of i, j, k : 24 24 FAC − ( FBC + FCD ) = 0 26 25

i:

−

j:

10 FAC − 400 lb = 0 26

k:

7 ( FBC − FCD ) = 0 FCD = FBC 25

(1) FAC = 1040 lb T

Substitute for FAC and FCD in Eq. (1): −

24 24 (10.40 lb) − (2 FBC ) = 0 FBC = −500 lb 26 25

FBC = FCD = 500 lb C

Free body: B FBC FBA FBD

! CB = (500 lb) = −(480 lb)i + (140 lb)k CB ! BA FAB = FAB = (10 j − 7k ) BA 12.21 = − FBD k

ΣF = 0: FBA + FBD + FBC + (480 lb)i + (200 lb) j = 0

Substituting for FBA , FBD , FBC and equating to zero the coefficients of j and k: j:

10 FAB + 200 lb = 0 FAB = −244.2 lb 12.21

k: −

7 FAB − FBD + 140 lb = 0 12.21 FBD = −

From symmetry:

7 (−244.2 lb) + 140 lb = +280 lb 12.21

FAD = FAB

FAB = 244 lb C

FBD = 280 lb T FAD = 244 lb C


PROBLEM 6.36* The truss shown consists of six members and is supported by a ball and socket at B, a short link at C, and two short links at D. Determine the force in each of the members for P = (−2184 N)j and Q = 0.

SOLUTION Free body: Truss From symmetry: Dx = Bx and Dy = By ΣFx = 0: 2 Bx = 0 Bx = Dx = 0 ΣFz = 0: Bz = 0 ΣM cz = 0: − 2 By (2.8 m) + (2184 N)(2 m) = 0 By = 780 N B = (780 N) j "

Thus Free body: A ! AB FAB FAB = FAB = ( −0.8i − 4.8 j + 2.1k ) AB 5.30 ! AC FAC = FAC = FAC (2i − 4.8 j) AC 5.20 ! AD FAD = FAD = FAD (+0.8i − 4.8 j − 2.1k ) AD 5.30 ΣF = 0: FAB + FAC + FAD − (2184 N) j = 0



Substituting for FAB , FAC , FAD , and equating to zero the coefficients of i, j, k : i:

−

0.8 2 ( FAB + FAD ) + FAC = 0 5.30 5.20

(1)

j:

−

4.8 4.8 ( FAB + FAD ) − FAC − 2184 N = 0 5.30 5.20

(2)

2.1 ( FAB − FAD ) = 0 5.30

k:

FAD = FAB

Multiply (1) by –6 and add (2): 16.8 ! −" # FAC − 2184 N = 0, FAC = −676 N $ 5.20 %

FAC = 676 N C

Substitute for FAC and FAD in (1): 0.8 ! 2 ! −" # 2 FAB + " # (−676 N) = 0, FAB = −861.25 N $ 5.30 % $ 5.20 %

FAB = FAD = 861 N C

Free body: B ! AB = −(130 N)i − (780 N) j + (341.25 N)k FAB = (861.25 N) AB 2.8i − 2.1k ! FBC = FBC " # = FBC (0.8i − 0.6k ) 3.5 $ % FBD = − FBD k ΣF = 0: FAB + FBC + FBD + (780 N) j = 0

Substituting for FAB , FBC , FBD and equating to zero the coefficients of i and k : i: k:

−130 N + 0.8 FBC = 0 FBC = +162.5 N 341.25 N − 0.6 FBC − FBD = 0 FBD = 341.25 − 0.6(162.5) = +243.75 N

From symmetry:

FBC = 162.5 N T

FCD = FBC

FBD = 244 N T FCD = 162.5 N T


PROBLEM 6.37* The truss shown consists of six members and is supported by a ball and socket at B, a short link at C, and two short links at D. Determine the force in each of the members for P = 0 and Q = (2968 N)i.

SOLUTION Free body: Truss From symmetry: Dx = Bx and Dy = By ΣFx = 0: 2 Bx + 2968 N = 0 Bx = Dx = −1484 N ΣM cz ′ = 0: − 2 By (2.8 m) − (2968 N)(4.8 m) = 0 By = −2544 N B = −(1484 N)i − (2544 N) j "

Thus Free body: A FAB = FAB

! AB AB

FAB (−0.8i − 4.8 j + 2.1k ) 5.30 ! AC FAC = FAC = (2i − 4.8 j) AC 5.20 ! AD = FAD AD FAD = (−0.8i − 4.8 j − 2.1k ) 5.30 =

FAC FAD

ΣF = 0: FAB + FAC + FAD + (2968 N)i = 0



Substituting for FAB , FAC , FAD , and equating to zero the coefficients of i, j, k : 0.8 2 FAC + 2968 N = 0 ( FAB + FAD ) + 5.30 5.20

(1)

4.8 4.8 FAC = 0 ( FAB + FAD ) − 5.30 5.20

(2)

i:

−

j:

−

k:

2.1 ( FAB − FAD ) = 0 5.30

FAD = FAB

Multiply (1) by –6 and add (2): 16.8 ! −" # FAC − 6(2968 N) = 0, FAC = −5512 N $ 5.20 %

FAC = 5510 N C

Substitute for FAC and FAD in (2): 4.8 ! 4.8 ! −" # 2 FAB − " # (−5512 N) = 0, FAB = +2809 N $ 5.30 % $ 5.20 % FAB = FAD = 2810 N T

Free body: B FAB FBC FBD

! BA = (2809 N) = (424 N)i + (2544 N) j − (1113 N)k BA 2.8 i − 2.1k ! = FBC " # = FBC (0.8i − 0.6k ) 3.5 $ % = − FBD k

ΣF = 0: FAB + FBC + FBD − (1484 N)i − (2544 N) j = 0

Substituting for FAB , FBC , FBD and equating to zero the coefficients of i and k: i: k:

From symmetry:

+24 N + 0.8FBC − 1484 N = 0, FBC = +1325 N

FBC = 1325 N T

−1113 N − 0.6 FBC − FBD = 0 FBD = −1113 N − 0.6(1325 N) = −1908 N,

FBD = 1908 N C

FCD = FBC

FCD = 1325 N T


PROBLEM 6.38* The truss shown consists of nine members and is supported by a ball and socket at A, two short links at B, and a short link at C. Determine the force in each of the members for the given loading.

SOLUTION Free body: Truss From symmetry: Az = Bz = 0 ΣFx = 0: Ax = 0 ΣM BC = 0: Ay (6 ft) + (1600 lb)(7.5 ft) = 0 A = −(2000 lb)j "

Ay = −2000 lb By = C

From symmetry:

ΣFy = 0: 2 By − 2000 lb − 1600 lb = 0 B = (1800 lb) j "

By = 1800 lb ΣF = 0: FAB + FAC + FAD − (2000 lb) j = 0

Free body: A

FAB

i+k

+ FAC

2

i−k 2

+ FAD (0.6 i + 0.8 j) − (2000 lb) j = 0

Factoring i, j, k and equating their coefficient to zero: 1 2

FAB +

1 2

FAC + 0.6 FAD = 0

(1)

0.8FAD − 2000 lb = 0 1 2

FAB −

1 2

FAC = 0

FAD = 2500 lb T FAC = FAB



Substitute for FAD and FAC into (1): 2 2

Free body: B

FAB + 0.6(2500 lb) = 0, FAB = −1060.7 lb,

FBA FBC

FAB = FAC = 1061 lb C

! BA i+k = FAB = + (1060.7 lb) = (750 lb)(i + k ) BA 2 = − FBC k

FBD = FBD (0.8 j − 0.6k ) ! BE FBE FBE = FBE = (7.5i + 8 j − 6k ) BE 12.5

ΣF = 0: FBA + FBC + FBD + FBE + (1800 lb) j = 0

Substituting for FBA , FBC , FBD + FBE and equate to zero the coefficients of i, j, k : i:

7.5 ! 750 lb + " # FBE = 0, FBE = −1250 lb $ 12.5 %

FBE = 1250 lb C

j:

8 ! 0.8 FBD + " # (−1250 lb) + 1800 lb = 0 $ 12.5 %

FBD = 1250 lb C

k:

750 lb − FBC − 0.6( −1250 lb) −

6 ( −1250 lb) = 0 12.5

FBC = 2100 lb T FBD = FCD = 1250 lb C

From symmetry: Free body: D ΣF = 0: FDA + FDB + FDC + FDE i = 0

We now substitute for FDA , FDB , FDC and equate to zero the coefficient of i. Only FDA contains i and its coefficient is −0.6 FAD = −0.6(2500 lb) = −1500 lb

i:

−1500 lb + FDE = 0

FDE = 1500 lb T


PROBLEM 6.39* The truss shown consists of nine members and is supported by a ball and socket at B, a short link at C, and two short links at D. (a) Check that this truss is a simple truss, that it is completely constrained, and that the reactions at its supports are statically determinate. (b) Determine the force in each member for P = (−1200 N)j and Q = 0.

SOLUTION Free body: Truss ΣM B = 0: 1.8i × Cj + (1.8i − 3k ) × ( D y j + Dk ) + (0.6i − 0.75k ) × (−1200 j) = 0 −1.8 Ck + 1.8 D y k − 1.8 Dz j + 3D y i − 720k − 900i = 0

Equate to zero the coefficients of i, j, k : i:

3Dy − 900 = 0, D y = 300 N

j:

Dz = 0,

D = (300 N) j "

k:

1.8C + 1.8(300) − 720 = 0

C = (100 N) j "

ΣF = 0: B + 300 j + 100 j − 1200 j = 0

B = (800 N) j "

Free body: B ΣF = 0: FBA + FBC + FBE + (800 N) j = 0, with ! BA FAB FBA = FAB (0.6i + 3j − 0.75k ) = BA 3.15 FBE = − FBE k

FBC = FBC i

Substitute and equate to zero the coefficient of j, i, k : j:

3 ! " # FAB + 800 N = 0, FAB = −840 N, $ 3.315 %

i:

0.6 ! " # (−840 N) + FBC = 0 $ 3.15 %

FBC = 160.0 N T

k:

0.75 ! "− # (−840 N) − FBE = 0 $ 3.15 %

FBE = 200 N T

FAB = 840 N C



Free body C: ΣF = 0: FCA + FCB + FCD + FCE + (100 N) j = 0, with ! F CA FCA = FAC = AC (−1.2i + 3j − 0.75 k ) CA 3.317

FCB = −(160 N) i FCD = − FCD k FCE = FCE

! F CE = CE (−1.8i − 3k ) CE 3, 499

Substitute and equate to zero the coefficient of j, i, k : 3 ! " # FAC + 100 N = 0, FAC = −110.57 N $ 3.317 %

j: −

i: k:

−

FAC = 110.6 N C

1.2 1.8 (−110.57) − 160 − FCE = 0, FCE = −233.3 3.317 3.499

FCE = 233 N C

0.75 3 (−110.57) − FCD − (−233.3) = 0 3.317 3.499

FCD = 225 N T

Free body: D ΣF = 0: FDA + FDC + FDE + (300 N) j = 0, with ! F DA FDA = FAD = AD (−1.2i + 3j + 2.25k ) DA 3.937 FDC = FCD k = (225 N)k

FDE = − FDE i

Substitute and equate to zero the coefficient of j, i, k : j:

i:

k:

3 ! " # FAD + 300 N = 0, $ 3.937 %

FAD = −393.7 N,

1.2 ! "− # ( −393.7 N) − FDE = 0 $ 3.937 %

FAD = 394 N C FDE = 1200 N T

2.25 ! " # (−393.7 N) + 225 N = 0 (Checks) $ 3.937 %

Free body: E Member AE is the only member at E which does not lie in the xz plane. Therefore, it is a zero-force member. FAE = 0


PROBLEM 6.40* Solve Problem 6.39 for P = 0 and Q = (−900 N)k. PROBLEM 6.39* The truss shown consists of nine members and is supported by a ball and socket at B, a short link at C, and two short links at D. (a) Check that this truss is a simple truss, that it is completely constrained, and that the reactions at its supports are statically determinate. (b) Determine the force in each member for P = (−1200 N)j and Q = 0.

SOLUTION Free body: Truss ΣM B = 0: 1.8i × Cj + (1.8i − 3k ) × ( D y j + Dz k ) + (0.6i + 3j − 0.75k ) × (−900N)k = 0 1.8 Ck + 1.8D y k − 1.8Dz j

+ 3D y i + 540 j − 2700i = 0

Equate to zero the coefficient of i, j, k : 3Dy − 2700 = 0 Dy = 900 N

−1.8Dz + 540 = 0 Dz = 300 N 1.8C + 1.8D y = 0 C = − Dy = −900 N

Thus:

C = −(900 N) j D = (900 N) j + (300 N)k

"

ΣF = 0: B − 900 j + 900 j + 300k − 900k = 0

B = (600 N)k "

Free body: B

Since B is aligned with member BE: FAB = FBC = 0, FBE = 600 N T

Free body: C !

ΣF = 0: FCA + FCD + FCE − (900 N) j = 0, !with



!

FCA FCD

! F CA = FAC = AC ( −1.2i + 3j − 0.75k ) ! CA 3.317 ! F CE = − FCD k FCE = FCE = CE (−1.8i − 3k ) CE 3.499

Substitute and equate to zero the coefficient of j, i, k: j:

3 ! " # FAC − 900 N = 0, FAC = 995.1 N $ 3.317 %

FAC = 995 N T

i:

−

1.2 1.8 (995.1) − FCE = 0, FCE = −699.8 N 3.317 3.499

FCE = 700 N C

k:

−

0.75 3 (995.1) − FCD − ( −699.8) = 0 3.317 3.499

FCD = 375 N T

Free body: D ΣF = 0: FDA + FDE + (375 N)k +(900 N)j + (300 N) k = 0 ! F DA FDA = FAD = AD (−1.2i + 3j + 2.25k ) DA 3.937

with

FDE = − FDE i

and

Substitute and equate to zero the coefficient j, i, k: j:

3 ! " # FAD + 900 N = 0, FAD = −1181.1 N $ 3.937 %

FAD = 1181N C

i:

1.2 ! −" # (−1181.1 N) − FDE = 0 $ 3.937 %

FDE = 360 N T

k:

2.25 ! " # (−1181.1 N + 375 N + 300 N = 0) $ 3.937 %

(Checks)

Free body: E Member AE is the only member at E which does not lie in the XZ plane. Therefore, it is a zero-force member. FAE = 0


PROBLEM 6.41* The truss shown consists of 18 members and is supported by a ball and socket at A, two short links at B, and one short link at G. (a) Check that this truss is a simple truss, that it is completely constrained, and that the reactions at its supports are statically determinate. (b) For the given loading, determine the force in each of the six members joined at E.

SOLUTION (a)

Check simple truss. (1) start with tetrahedron BEFG (2) Add members BD, ED, GD joining at D. (3) Add members BA, DA, EA joining at A. (4) Add members DH, EH, GH joining at H. (5) Add members BC, DC, GC joining at C. Truss has been completed: It is a simple truss Free body: Truss Check constraints and reactions: Six unknown reactions-ok-however supports at A and B constrain truss to rotate about AB and support at G prevents such a rotation. Thus, Truss is completely constrained and reactions are statically determinate Determination of Reactions: ΣM A = 0: 11i × ( By j + Bz k ) + (11i − 9.6k ) × G j + (10.08 j − 9.6k ) × (275i + 240k ) = 0 11By k − 11By j + 11Gk + 9.6Gi − (10.08)(275)k

+ (10.08)(240)i − (9.6)(275) j = 0

Equate to zero the coefficient of i, j, k: i : 9.6G + (10.08)(240) = 0 G = −252 lb

G = (−252 lb) j "!

j: − 11Bz − (9.6)(275) = 0 Bz = −240 lb k : 11By + 11(−252) − (10.08)(275) = 0, By = 504 lb

B = (504 lb) j − (240 lb)k "



ΣF = 0: A + (504 lb) j − (240 lb)k − (252 lb) j + (275 lb)i + (240 lb)k = 0 A = −(275 lb)i − (252 lb) j "

Zero-force members The determination of these members will facilitate our solution. FB: C. Writing

ΣFx = 0, ΣFy = 0, ΣFz = 0

Yields FBC = FCD = FCG = 0 "!

FB: F. Writing

ΣFx = 0, ΣFy = 0, ΣFz = 0

Yields FBF = FEF = FFG = 0 "

FB: A: Since FB: H: Writing

Az = 0, writing ΣFz = 0 ΣFy = 0

Yields FAD = 0 " Yields FDH = 0 "

FB: D: Since FAD = FCD = FDH = 0, we need consider only members DB, DE, and DG. Since FDE is the only force not contained in plane BDG, it must be zero. Simple reasonings show that the other two forces are also zero. FBD = FDE = FDG = 0 "

The results obtained for the reactions at the supports and for the zero-force members are shown on the figure below. Zero-force members are indicated by a zero (“0”).

(b)

Force in each of the members joined at E FDE = FEF = 0

We already found that Free body: A

ΣFy = 0

Yields FAE = 252 lb T

Free body: H

ΣFz = 0

Yields FEH = 240 lb C



ΣF = 0: FEB + FEG + (240 lb)k − (252 lb) j = 0

Free body: E

F FBE (11i − 10.08 j) + EG (11i − 9.6k ) + 240k − 252 j = 0 14.92 14.6

Equate to zero the coefficient of y and k: 10.08 ! j: − " # FBE − 252 = 0 14.92 $ %

FBE = 373 lb C

9.6 ! −" # FEG + 240 = 0 14.6 $ %

FEG = 365 lb T

k:


PROBLEM 6.42* The truss shown consists of 18 members and is supported by a ball and socket at A, two short links at B, and one short link at G. (a) Check that this truss is a simple truss, that it is completely constrained, and that the reactions at its supports are statically determinate. (b) For the given loading, determine the force in each of the six members joined at G.

SOLUTION See solution to Problem 6.41 for Part (a) and for reactions and zero-force members. (b)

Force in each of the members joined at G. We already know that FCG = FDG = FFG = 0

Free body: H

ΣFx = 0

Free body: G

ΣF = 0: FGB + FGE + (275 lb)i − (252 lb) j = 0

!

Yields FGH = 275 lb C

FBG F (−10.08 j + 9.6k ) + EG (−11i + 9.6k ) + 275i − 252 j = 0 13.92 14.6

Equate to zero the coefficient of i, j, k: 11 ! i: − " # FEG + 275 = 0 14.6 $ %

FEG = 365 lb T

10.08 ! j: − " # FBG − 252 = 0 $ 13.92 %

FBG = 348 lb C

9.6 ! 9.6 ! k: " # (−348) + " # (365) = 0 $ 13.92 % $ 14.6 %

(Checks)


!

PROBLEM 6.43 A Warren bridge truss is loaded as shown. Determine the force in members CE, DE, and DF.

SOLUTION Free body: Truss ΣFx = 0: k x = 0 ΣM A = 0: k y (62.5 ft) − (6000 lb)(12.5 ft) − (6000 lb)(25 ft) = 0 k = k y = 3600 lb "!

ΣFy = 0: A + 3600 lb − 6000 lb − 6000 lb = 0 A = 8400 lb "

We pass a section through members CE, DE, and DF and use the free body shown. ΣM D = 0: FCE (15 ft) − (8400 lb)(18.75 ft) + (6000 lb)(6.25 ft) = 0 FCE = +8000 lb ΣFy = 0: 8400 lb − 6000 lb −

FCE = 8000 lb T 15 FDE = 0 16.25

FDE = +2600 lb

FDE = 2600 lb T

ΣM E = 0: 6000 lb (12.5 ft) − (8400 lb)(25 ft) − FDF (15 ft) = 0 FDF = −9000 lb

FDF = 9000 lb C


!

PROBLEM 6.44 A Warren bridge truss is loaded as shown. Determine the force in members EG, FG, and FH.

SOLUTION See solution of Problem 6.43 for free-body diagram of truss and determination of reactions: A = 8400 lb

k = 3600 lb "

We pass a section through members EG, FG, and FH, and use the free body shown. ΣM F = 0: (3600 lb)(31.25 ft) − FEG (15 ft) = 0 FEG = +7500 lb +ΣFy = 0:

FEG = 7500 lb T

!

15 FFG + 3600 lb = 0 16.25 FFG = −3900 lb

FFG = 3900 lb C

ΣM G = 0: FFH (15 ft) + (3600 lb)(25 ft) = 0 FFH = −6000 lb

FFH = 6000 lb C


PROBLEM 6.45 Determine the force in members BD and DE of the truss shown.

SOLUTION Member BD: ΣM E = 0: FBD (4.5 m) − (135 kN)(4.8 m) − (135 kN)(2.4 m) = 0 FBD = +216 kN

FBD = 216 kN T

Member DE: ΣFx = 0: 135 kN + 135 kN − FDE = 0 FDE = +270 kN

FDE = 270 kN T


PROBLEM 6.46 Determine the force in members DG and EG of the truss shown.

SOLUTION Member DG: ΣFx = 0: 135 kN + 135 kN + 135 kN + FDG = −459 kN

15 FDG = 0 17 FDG = 459 kN C

Member EG: ΣM D = 0: (135 kN)(4.8 m) + (135 kN)(2.4 m) + FEG (4.5 m) = 0 FEG = −216 kN

FEG = 216 kN C


PROBLEM 6.47 A floor truss is loaded as shown. Determine the force in members CF, EF, and EG.

SOLUTION Free body: Truss ΣFx = 0: k x = 0 ΣM A = 0: k y (4.8 m) − (4 kN)(0.8 m) − (4 kN)(1.6 m) − (3 kN)(2.4 m) − (2 kN)(3.2 m) − (2 kN)(4 m) − (1 kN)(4.8 m) = 0 k y = 7.5 kN k = 7.5 kN "

Thus: ΣFy = 0: A + 7.5 kN − 18 kN = 0

A = 10.5 kN

A = 10.5 kN "

We pass a section through members CF, EF, and EG and use the free body shown. ΣM E = 0: FCF (0.4 m) − (10.5 kN)(1.6 m) + (2 kN)(1.6 m) + (4 kN)(0.8 m) = 0 FCF = +26.0 kN ΣFy = 0: 10.5 kN − 2 kN − 4 kN − 4 kN −

FCF = 26.0 kN T 1 5

FEF = +1.118 kN

FEF = 0 FEF = 1.118 kN T

ΣM F = 0: (2 kN)(2.4 m) + (4 kN)(1.6 m) + (4 kN)(0.8 m) − (10.5 kN)(2.4 m) − FEG (0.4 m) = 0 FEG = −27.0 kN

FEG = 27.0 kN C


PROBLEM 6.48 A floor truss is loaded as shown. Determine the force in members FI, HI, and HJ.

SOLUTION See solution of Problem 6.47 for free-body diagram of truss and determination of reactions: A = 10.5 kN , k = 7.5 kN "

We pass a section through members FI, HI, and HJ, and use the free body shown. ΣM H = 0: (7.5 kN)(1.6 m) − (2 kN)(0.8 m) − (1 kN)(1.6 m) − FFI (0.4 m) = 0 FFI = +22.0 kN ΣFy = 0:

1 5

FFI = 22.0 kN T

FHI − 2 kN − 1 kN + 7.5 kN = 0 FHI = −10.06 kN

FHI = 10.06 kN C

ΣM I = 0: FHJ (0.4 m) + (7.5 kN)(0.8 m) − (1 kN)(0.8 m) = 0 FHJ = −13.00 kN

FHJ = 13.00 kN C


PROBLEM 6.49 A pitched flat roof truss is loaded as shown. Determine the force in members CE, DE, and DF.

SOLUTION Reactions at supports: Because of the symmetry of the loading, Ax = 0 Ay = I =

1 1 (Total load) = (8 kN) 2 2

A = I = 4 kN "

We pass a section through members CD, DE, and DF, and use the free body shown. (We moved FDE to E and FDF to F) Slope

BJ =

Slope

DE =

2.16 m 9 = 9.6 m 40

−1 m −5 = 2.4 m 12 0.46 m 0.46 m a= = = 2.0444 m 9 Slope BJ 40

ΣM D = 0: FCE (1 m) + (1 kN)(2.4 m) − (4 kN)(2.4 m) = 0 FCE = 7.20 kN T

FCE = +7.20 kN ΣM K = 0: (4 kN)(2.0444 m) − (1 kN)(2.0444 m) 5 ! − (2 kN)(4.4444 m) − " FDE # (6.8444 m) = 0 $ 13 % FDE = −1.047 kN

FDE = 1.047 kN C

ΣM E = 0: (1 kN)(4.8 m) + (2kN)(2.4 m) − (4 kN)(4.8 m) 40 ! − " FDF # (1.54 m) = 0 41 $ % FDF = −6.39 kN

FDF = 6.39 kN C


PROBLEM 6.50 A pitched flat roof truss is loaded as shown. Determine the force in members EG, GH, and HJ.

SOLUTION Reactions at supports: Because of the symmetry of the loading, Ax = 0 Ay = I =

1 1 (Total load) = (8 kN) 2 2

A = I = 4 kN "

We pass a section through members EG, GH, and HJ, and use the free body shown. ΣM H = 0: (4 kN)(2.4 m) − (1 kN)(2.4 m) − FEG (2.08 m) = 0 FEG = +3.4615 kN

FEG = 3.46 kN T

ΣM J = 0: − FGH (2.4 m) − FEG (2.62 m) = 0 FGH = −

2.62 (3.4615 kN) 2.4

FGH = −3.7788 kN ΣFx = 0: − FEG −

FGH = 3.78 kN C

2.4 FHJ = 0 2.46

FHJ = −

2.46 2.46 (3.4615 kN) FEG = − 2.4 2.4

FHJ = −3.548 kN

FHJ = 3.55 kN C


PROBLEM 6.51 A Howe scissors roof truss is loaded as shown. Determine the force in members DF, DG, and EG.

SOLUTION Reactions at supports. Because of symmetry of loading. Ax = 0,

Ay = L =

1 1 (Total load) = (9.60 kips) = 4.80 kips 2 2

A = L = 4.80 kips

We pass a section through members DF, DG, and EG, and use the free body shown. We slide FDF to apply it at F: ΣM G = 0: (0.8 kip)(24 ft) + (1.6 kips)(16 ft) + (1.6 kips)(8 ft) 8FDF

− (4.8 kips)(24 ft) −

82 + 3.52

(6 ft) = 0

FDF = −10.48 kips, FDF = 10.48 kips C ΣM A = 0: − (1.6 kips)(8 ft) − (1.6 kips)(16 ft) −

2.5 FDG 2

8 + 2.5

2

(16 ft) −

8 FDG 82 + 2.52

(7 ft) = 0

FDG = −3.35 kips, FDG = 3.35 kips C ΣM D = 0: (0.8 kips)(16 ft) + (1.6 kips)(8 ft) − (4.8 kips)(16 ft) −

8 FEG 82 + 1.52

(4 ft) = 0

FEG = +13.02 kips, FEG = 13.02 kips T

!


!

PROBLEM 6.52 A Howe scissors roof truss is loaded as shown. Determine the force in members GI, HI, and HJ.

SOLUTION Reactions at supports. Because of symmetry of loading: 1 (Total load) 2 1 = (9.60 kips) 2

Ay = L =

Ax = 0,

= 4.80 kips

A = L = 4.80 kips

We pass a section through members GI, HI, and HJ, and use the free body shown.

ΣM H = 0: −

16 FGI 162 + 32

(4 ft) + (4.8 kips)(16 ft) − (0.8 kip)(16 ft) − (1.6 kips)(8 ft) = 0 FGI = +13.02 kips

FGI = 13.02 kips T

ΣM L = 0: (1.6 kips)(8 ft) − FHI (16 ft) = 0 FHI = +0.800 kips FHI = 0.800 kips T

We slide FHG to apply it at H. ΣM I = 0:

8FHJ 82 + 3.52

(4 ft) + (4.8 kips)(16 ft) − (1.6 kips)(8 ft) − (0.8 kip)(16 ft) = 0 FHJ = −13.97 kips

FHJ = 13.97 kips C


!

PROBLEM 6.53 A Pratt roof truss is loaded as shown. Determine the force in members CE, DE, and DF.

SOLUTION Free body: Entire truss ΣFx = 0: Ax = 0

Total load = 5(3 kN) + 2(1.5 kN) = 18 kN By symmetry: Ay = L =

1 (18 kN) 2

A = L = 9 kN

Free body: Portion ACD Note:

Slope of ABDF is 6.75 3 = 9.00 4

Force in CE: 2 ! ΣM D = 0: FCE " × 6.75 m # − (9 kN)(6 m) + (1.5 kN)(6 m) + (3 kN)(3 m) = 0 $3 % FCE (4.5 m) − 36 kN ⋅ m = 0 FCE = +8 kN

FCE = 8 kN T

Force in DE: ΣM A = 0: FDE (6 m) + (3 kN)(6 m) + (3 kN)(3 m) = 0 FDE = −4.5 kN

FDE = 4.5 kN C



Force in DF: Sum moments about E where FCE and FDE intersect. ΣM E = 0: (1.5 kN)(6 m) − (9 kN)(6 m) + (3 kN)(3 m) +

4 2 ! FCE " × 6.75 m # = 0 5 $3 %

4 FCE (4.5 m) − 36 kN ⋅ m 5 FCE = 10.00 kN C

FCE = −10.00 kN


!

PROBLEM 6.54 A Pratt roof truss is loaded as shown. Determine the force in members FH, FI, and GI.

SOLUTION Free body: Entire truss ΣFx = 0: Ax = 0

Total load = 5(3 kN) + 2(1.5 kN) = 18 kN By symmetry: Ay = L =

1 (18) = 9 kN 2

Free body: Portion HIL Slope of FHJL 6.75 3 = 9.00 4 tan α =

FG 6.75 m = α = 66.04° GI 3m

Force in FH: ΣM I = 0:

4 2 ! FFH " × 6.75 m # + (9 kN)(6 m) − (1.5 kN)(6 m) − (3 kN)(3 m) = 0 5 $3 % 4 FFH (4.5 m) + 36 kN ⋅ m 5 FFH = −10.00 kN

FFH = 10.00 kN C

Force in FI: ΣM L = 0: FFI sin α (6 m) − (3 kN)(6 m) − (3 kN)(3 m) = 0 FFI sin 66.04°(6 m) = 27 kN ⋅ m FFI = +4.92 kN

FFI = 4.92 kN T


!


Force in GI: ΣM H = 0: FGI (6.75 m) + (3 kN)(3 m) + (3 kN)(6 m) + (1.5 kN)(9 m) − (9 kN)(9 m) = 0 FGI (6.75 m) = +40.5 kN ⋅ m FGI = 6.00 kN T

FGI = +6.00 kN


!

PROBLEM 6.55 Determine the force in members AD, CD, and CE of the truss shown.

SOLUTION Reactions:

ΣM k = 0: 36(2.4) − B(13.5) + 20(9) + 20(4.5) = 0 ΣFx = 0: −36 + K x = 0

B = 26.4 kN

K x = 36 kN

ΣFy = 0: 26.4 − 20 − 20 + K y = 0 K y = 13.6 kN

ΣM C = 0: 36(1.2) − 26.4(2.25) − FAD (1.2) = 0

FAD = −13.5 kN 8 ! ΣM A = 0: " FCD # (4.5) = 0 17 $ %

FAD = 13.5 kN C FCD = 0

15 ! ΣM D = 0: " FCE # (2.4) − 26.4(4.5) = 0 17 $ %

FCE = +56.1 kN

FCE = 56.1 kN T


!

PROBLEM 6.56 Determine the force in members DG, FG, and FH of the truss shown.

SOLUTION See the solution to Problem 6.55 for free-body diagram and analysis to determine the reactions at the supports B and K. B = 26.4 kN ;

K x = 36.0 kN

;

K y = 13.60 kN

ΣM F = 0: 36(1.2) − 26.4(6.75) + 20(2.25) − FDG (1.2) = 0

FDG = −75 kN

FDG = 75.0 kN C

8 ! ΣM D = 0: − 26.4(4.5) + " FFG # (4.5) = 0 $ 17 %

FFG = +56.1 kN

!

15 ΣM G = 0: 20(4.5) − 26.4(9) + " FFH 17 $

! # (2.4) = 0 ! %

FFH = +69.7 kN !

!

FFG = 56.1 kN T

FFH = 69.7 kN T


!

PROBLEM 6.57 A stadium roof truss is loaded as shown. Determine the force in members AB, AG, and FG.

SOLUTION We pass a section through members AB, AG, and FG, and use the free body shown.

40 ! ΣM G = 0: " FAB # (6.3 ft) − (1.8 kips)(14 ft) − (0.9 kips)(28 ft) = 0 $ 41 %

FAB = +8.20 kips

FAB = 8.20 kips T

3 ! ΣM D = 0: − " FAG # (28 ft) + (1.8 kips)(28 ft) + (1.8 kips)(14 ft) = 0 $5 %

FAG = +4.50 kips

FAG = 4.50 kips T

M A = 0: − FFG (9 ft) − (1.8 kips)(12 ft) − (1.8 kips)(26 ft) − (0.9 kips)(40 ft) = 0 FFG = −11.60 kips

FFG = 11.60 kips C


!

PROBLEM 6.58 A stadium roof truss is loaded as shown. Determine the force in members AE, EF, and FJ.

SOLUTION We pass a section through members AE, EF, and FJ, and use the free body shown.

! 8 ΣM F = 0: " FAE # (9 ft) − (1.8 kips)(12 ft) − (1.8 kips)(26 ft) − (0.9 kips)(40 ft) = 0 " 82 + 9 2 # $ % FAE = +17.46 kips

FAE = 17.46 kips T

ΣM A = 0: − FEF (9 ft) − (1.8 kips)(12 ft) − (1.8 kips)(26 ft) − (0.9 kips)(40 ft) = 0 FEF = −11.60 kips

FEF = 11.60 kips C

ΣM E = 0: − FFJ (8 ft) − (0.9 kips)(8 ft) − (1.8 kips)(20 ft) − (1.8 kips)(34 ft) − (0.9 kips)(48 ft) = 0 FFJ = −18.45 kips

FFJ = 18.45 kips C


!

PROBLEM 6.59 A Polynesian, or duopitch, roof truss is loaded as shown. Determine the force in members DF, EF, and EG.


ΣFx = 0: N x = 0 ΣM N = 0: (200 lb)(8a ) + (400 lb)(7a + 6a + 5a)+(350 lb)(4a ) + (300 lb)(3a + 2a + a) − A(8a ) = 0 A = 1500 lb "

ΣFy = 0: 1500 lb − 200 lb − 3(400 lb) − 350 lb − 3(300 lb) − 150 lb + N y = 0 N y = 1300 lb N = 1300 lb "

We pass a section through DF, EF, and EG, and use the free body shown. (We apply FDF at F) ΣM E = 0: (200 lb)(18 ft) + (400 lb)(12 ft) + (400 lb)(6 ft) − (1500 lb)(18 ft)

! 18 F # (4.5 ft) = 0 −" " 182 + 4.52 DF # $ % FDF = −3711 lb FDF = 3710 lb C ΣM A = 0: FEF (18 ft) − (400 lb)(6 ft) − (400 lb)(12 ft) = 0 FEF = +400 lb

FEF = 400 lb T

ΣM F = 0: FEG (4.5 ft) − (1500 lb)(18 ft)+(200 lb)(18 ft) + (400 lb)(12 ft) + (400 lb)(6 ft) = 0 FEG = +3600 lb FEG = 3600 lb T


!

PROBLEM 6.60 A Polynesian, or duopitch, roof truss is loaded as shown. Determine the force in members HI, GI, and GJ.

SOLUTION See solution of Problem 6.59 for reactions: A = 1500 lb ,

N = 1300 lb "

We pass a section through HI, GI, and GJ, and use the free body shown. (We apply FHI at H.)

! 6 ΣM G = 0: " FHI # (8.5 ft) + (1300 lb)(24 ft) − (300 lb)(6 ft) " 2 # 2 $ 6 +4 % − (300 lb)(12 ft) − (300 lb)(18 ft) − (150 lb)(24 ft) = 0 FHI = −2375.4 lb FHI = 2375 lb C ΣM I = 0: (1300 lb)(18 ft) − (300 lb)(6 ft) − (300 lb)(12 ft) − (150 lb)(18 ft) − FGJ (4.5 ft) = 0 FGJ = +3400 lb FGJ = 3400 lb T ΣFx = 0: −

4 6 (−2375.4 lb) − 3400 lb = 0 FGI − 2 5 6 + 42 FGI = −1779.4 lb

FGI = 1779 lb C


!

PROBLEM 6.61 Determine the force in members AF and EJ of the truss shown when P = Q = 1.2 kN. (Hint: Use section aa.)

SOLUTION Free body: Entire truss

ΣM A = 0: K (8 m) − (1.2 kN)(6 m) − (1.2 kN)(12 m) = 0 K = +2.70 kN

K = 2.70 kN

"

Free body: Lower portion

ΣM F = 0: FEJ (12 m) + (2.70 kN)(4 m) − (1.2 kN)(6 m) − (1.2 kN)(12 m) = 0 FEJ = +0.900 kN

FEJ = 0.900 kN T

ΣFy = 0: FAF + 0.9 kN − 1.2 kN − 1.2 kN = 0 FAF = +1.500 kN

FAF = 1.500 kN T


PROBLEM 6.62 Determine the force in members AF and EJ of the truss shown when P = 1.2 kN and Q = 0. (Hint: Use section aa.)

SOLUTION Free body: Entire truss

ΣM A = 0: K (8 m) − (1.2 kN)(6 m) = 0 K = +0.900 kN

K = 0.900 kN

"

Free body: Lower portion ΣM F = 0: FEJ (12 m) + (0.900 kN)(4 m) − (1.2 kN)(6 m) = 0 FEJ = +0.300 kN

FEJ = 0.300 kN T

ΣFy = 0: FAF + 0.300 kN − 1.2 kN = 0 FAF = +0.900 kN

FAF = 1.900 kN T


PROBLEM 6.63 Determine the force in members EH and GI of the truss shown. (Hint: Use section aa.)

SOLUTION Reactions:

ΣFx = 0: Ax = 0 ΣM P = 0: 12(45) + 12(30) + 12(15) − Ay (90) = 0 A y = 12 kips

ΣFy = 0: 12 − 12 − 12 − 12 + P = 0

P = 24 kips

ΣM G = 0: − (12 kips)(30 ft) − FEH (16 ft) = 0 FEH = −22.5 kips ΣFx = 0: FGI − 22.5 kips = 0

FEH = 22.5 kips C FGI = 22.5 kips T


PROBLEM 6.64 Determine the force in members HJ and IL of the truss shown. (Hint: Use section bb.)

SOLUTION See the solution to Problem 6.63 for free body diagram and analysis to determine the reactions at supports A and P. A x = 0; A y = 12.00 kips ;

P = 24.0 kips

ΣM L = 0: FHJ (16 ft) − (12 kips)(15 ft) + (24 kips)(30 ft) = 0 FHJ = −33.75 kips

FHJ = 33.8 kips C

ΣFx = 0 : 33.75 kips − FIL = 0 FIL = +33.75 kips

FIL = 33.8 kips T


PROBLEM 6.65 The diagonal members in the center panels of the power transmission line tower shown are very slender and can act only in tension; such members are known as counters. For the given loading, determine (a) which of the two counters listed below is acting, (b) the force in that counter. Counters CJ and HE.

SOLUTION Free body: Portion ABDFEC of tower We assume that counter CJ is acting and show the forces exerted by that counter and by members CH and EJ.

ΣFx = 0:

4 FCJ − 2(1.2 kN) sin 20° = 0 FCJ = +1.026 kN 5

Since CJ is found to be in tension, our assumption was correct. Thus, the answers are (a) CJ (b) 1.026 kN T


PROBLEM 6.66 The diagonal members in the center panels of the power transmission line tower shown are very slender and can act only in tension; such members are known as counters. For the given loading, determine (a) which of the two counters listed below is acting, (b) the force in that counter. Counters IO and KN.

SOLUTION Free body: Portion of tower shown

We assume that counter IO is acting and show the forces exerted by that counter and by members IN and KO. ΣFx = 0 :

4 FIO − 4(1.2 kN)sin 20° = 0 FIO = +2.05 kN 5

Since IO is found to be in tension, our assumption was correct. Thus, the answers are (a) IO (b) 2.05 kN T


PROBLEM 6.67 The diagonal members in the center panels of the truss shown are very slender and can act only in tension; such members are known as counters. Determine the forces in the counters that are acting under the given loading.


ΣFx = 0: Fx = 0 ΣM H = 0: 4.8(3a) + 4.8(2a) + 4.8a − 2.4a − Fy (2a) = 0 Fy = +13.20 kips

F = 13.20 kips "

ΣFy = 0: H + 13.20 kips − 3(4.8 kips) − 2(2.4 kips) = 0 H = +6.00 kips

H = 6.00 kips "

Free body: ABF We assume that counter BG is acting. ΣFy = 0: −

9.6 FBG + 13.20 − 2(4.8) = 0 14.6 FBG = +5.475

FBG = 5.48 kips T

Since BG is in tension, our assumption was correct Free body: DEH We assume that counter DG is acting. +ΣFy = 0: −

9.6 FDG + 6.00 − 2(2.4) = 0 14.6 FDG = 1.825 kips T

FDG = +1.825

Since DG is in tension, O.K.


PROBLEM 6.68 The diagonal members in the center panels of the truss shown are very slender and can act only in tension; such members are known as counters. Determine the forces in the counters that are acting under the given loading.


ΣFx = 0: Fx = 0 ΣM G = 0: − Fy a + 4.8(2a) + 4.8a − 2.4a − 2.4(2a ) = 0 Fy = 7.20

F = 7.20 kips "

Free body: ABF We assume that counter CF is acting. ΣFy = 0 :

9.6 FCF + 7.20 − 2(4.8) = 0 14.6 FCF = 3.65 kips T

FCF = +3.65

Since CF is in tension, O.K. Free body: DEH We assume that counter CH is acting. +ΣFy = 0:

9.6 FCH − 2(2.4 kips) = 0 14.6 FCH = +7.30

FCH = 7.30 kips T

Since CH is in tension, O.K.


PROBLEM 6.69 Classify each of the structures shown as completely, partially, or improperly constrained; if completely constrained, further classify as determinate or indeterminate. (All members can act both in tension and in compression.)

SOLUTION Structure (a) Number of members: m = 16

Number of joints: n = 10

Reaction components: r=4 (a)

m + r = 20, 2n = 20

m + r = 2n "

Thus:

To determine whether the structure is actually completely constrained and determinate, we must try to find the reactions at the supports. We divide the structure into two simple trusses and draw the free-body diagram of each truss.

This is an improperly supported simple truss. (Reaction at C passes through B. Thus, Eq. ΣM B = 0 cannot be satisfied.)

This is a properly supported simple truss – O.K.

Structure is improperly constrained

!

Structure (b) m = 16 n = 10 r=4 m + r = 20, 2n = 20

(b)

m + r = 2n "

Thus:



We must again try to find the reactions at the supports dividing the structure as shown.

Both portions are simply supported simple trusses. Structure is completely constrained and determinate Structure (c) m = 17 n = 10 r=4 m + r = 21, 2n = 20

(c)

m + r > 2n "

Thus:

This is a simple truss with an extra support which causes reactions (and forces in members) to be indeterminate. Structure is completely constrained and indeterminate



SOLUTION Structure (a): Non-simple truss with r = 4, m = 16, n = 10 so m + r = 20 = 2n, but must examine further. FBD Sections:

FBD

I:

ΣM A = 0 &

T1

II:

ΣFx = 0 &

T2

I:

ΣFx = 0

&

Ax

I:

ΣFy = 0

&

Ay

II:

ΣM E = 0 &

Cy

II:

ΣFy = 0 &

Ey

Since each section is a simple truss with reactions determined, structure is completely constrained and determinate. Non-simple truss with r = 3, m = 16, n = 10 Structure (b): so

m + r = 19 < 2n = 20

structure is partially constrained



Structure (c):

Simple truss with r = 3, m = 17, n = 10 m + r = 20 = 2n, but the horizontal reaction forces Ax and E x are collinear and no equilibrium equation will resolve them, so the structure is improperly constrained and indeterminate



SOLUTION Structure (a):

Non-simple truss with r = 4, m = 12, n = 8 so r + m = 16 = 2n, check for determinacy: One can solve joint F for forces in EF, FG and then solve joint E for E y and force in DE. This leaves a simple truss ABCDGH with r = 3, m = 9, n = 6 so r + m = 12 = 2n

Structure is completely constrained and determinate

Structure (b):

Simple truss (start with ABC and add joints alphabetically to complete truss) with r = 4, m = 13, n = 8 so

r + m = 17 > 2n = 16

Constrained but indeterminate



Structure (c):

Non-simple truss with r = 3, m = 13, n = 8 so r + m = 16 = 2n. To further examine, follow procedure in Part (a) above to get truss at left. Since F1 ≠ 0 (from solution of joint F), ΣM A = aF1

≠ 0 and there is no equilibrium.

Structure is improperly constrained



SOLUTION Structure (a) Number of members: m = 12

Number of joints: n=8

Reaction components: r =3 m + r = 15,

Thus:

2n = 16

m + r , 2n

" Structure is partially constrained

Structure (b) m = 13, n = 8 r =3 m + r = 16, 2n = 16

Thus:

m + r = 2n

"

To verify that the structure is actually completely constrained and determinate, we observe that it is a simple truss (follow lettering to check this) and that it is simply supported by a pin-and-bracket and a roller. Thus: Structure is completely constrained and determinate



Structure (c) m = 13, r=4

n=8

m + r = 17, 2n = 16

Thus:

m + r . 2n

" Structure is completely constrained and indeterminate

This result can be verified by observing that the structure is a simple truss (follow lettering to check thus), therefore rigid, and that its supports involve 4 unknowns.



SOLUTION Structure (a):

Rigid truss with r = 3, m = 14, n = 8 so

r + m = 17 . 2n = 16

so completely constrained but indeterminate Structure (b):

Simple truss (start with ABC and add joints alphabetically), with r = 3, m = 13, n = 8 so r + m = 16 = 2n

so completely constrained and determinate

Structure (c):

Simple truss with r = 3, m = 13, n = 8 so r + m = 16 = 2n, but horizontal reactions ( Ax and Dx ) are collinear so cannot be resolved by any equilibrium equation. structure is improperly constrained



SOLUTION Structure (a): No. of members

m = 12

No. of joints

n = 8 m + r = 16 = 2n

No. of react. comps.

r = 4 unks = eqns

FBD of EH:

ΣM H = 0

FDE ; ΣFx = 0

FGH ; ΣFy = 0

Hy

Then ABCDGF is a simple truss and all forces can be determined. This example is completely constrained and determinate. Structure (b): No. of members

m = 12

No. of joints

n = 8 m + r = 15 , 2n = 16


r = 3 unks , eqns

partially constrained Note: Quadrilateral DEHG can collapse with joint D moving downward: in (a) the roller at F prevents this action. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 849


Structure (c): No. of members

m = 13

No. of joints

n = 8 m + r = 17 . 2n = 16


r = 4 unks . eqns

completely constrained but indeterminate


PROBLEM 6.75 For the frame and loading shown, determine the force acting on member ABC (a) at B, (b) at C.

SOLUTION FBD ABC:

Note: BD is two-force member

(a)

(b)

3 ! ΣM C = 0: (0.09 m)(200 N) − (2.4 m) " FBD # = 0 5 $ %

4 ΣFx = 0: 200 N − (125 N) − C x = 0 5 ΣFy = 0:

3 FBD − C y = 0 5

FBD = 125.0 N

36.9°

C = 125.0 N

36.9°

C x = 100 N

3 C y = (125 N) = 75 N 5


PROBLEM 6.76 Determine the force in member BD and the components of the reaction at C.

SOLUTION We note that BD is a two-force member. The force it exerts on ABC, therefore, is directed along the BD. Free body: ABC

Attaching FBD at D and resolving it into components, we write ΣM C = 0:

450 ! FBD # (240 mm) = 0 (400 N)(135 mm) + " 510 $ % FBD = −255 N

ΣFx = 0: C x +

FBD = 255 N C

240 (−255 N) = 0 510 C x = +120.0 N

ΣFy = 0: C y − 400 N +

C x = 120.0 N

450 (−255 N) = 0 510 C y = +625 N

C y = 625 N


PROBLEM 6.77 Rod CD is fitted with a collar at D that can be moved along rod AB, which is bent in the shape of an arc of circle. For the position when θ = 30°, determine (a) the force in rod CD, (b) the reaction at B.

SOLUTION FBD:

(a)

ΣM C = 0: (15 in.)(20 lb − By ) = 0 ΣFy = 0: − 20 lb + FCD sin 30° − 20 lb = 0

(b)

B y = 20 lb

FCD = 80.0 lb T

ΣFx = 0: (80 lb) cos 30° − Bx = 0 B x = 69.282 lb

so B = 72.1 lb

16.10°


PROBLEM 6.78 Solve Problem 6.77 when θ = 150°. PROBLEM 6.77 Rod CD is fitted with a collar at D that can be moved along rod AB, which is bent in the shape of an arc of circle. For the position when θ = 30°, determine (a) the force in rod CD, (b) the reaction at B.

SOLUTION Note that CD is a two-force member, FCD must be directed along DC.

(a)

ΣM B = 0: (20 lb)(2 R ) − ( FCD sin 30°) R = 0 FCD = 80 lb

(b)

FCD = 80.0 lb T

ΣM C = 0: (20 lb) R + ( By ) R = 0 By = −20 lb

B y = 20.0 lb

ΣFx = 0: − FCD cos 30° + Bx = 0 −(20 lb) cos 30° + Bx = 0 Bx = 69.28 lb

B x = 69.28 lb

B = 72.1 lb

16.10°


PROBLEM 6.79 Determine the components of all forces acting on member ABCD when θ = 0.

SOLUTION Free body: Entire assembly

ΣM B = 0: A(8 in.) − (60 lb)(20 in.) = 0 A = 150 lb

A = 150.0 lb

ΣFx = 0: Bx + 150.0 lb = 0 Bx = −150 lb

B x = 150.0 lb

ΣFy = 0: By − 60.0 lb = 0

Free body: Member ABCD

By = +60.0 lb

B y = 60.0 lb

We note that D is directed along DE, since DE is a two-force member. ΣM C = 0: D (12) − (60 lb)(4) + (150 lb)(8) = 0 D = −80 lb ΣFx = 0: C x + 150.0 − 150.0 = 0

Cx = 0

ΣFy = 0: C y + 60.0 − 80.0 = 0

C y = +20.0 lb

D = 80.0 lb

C = 20.0 lb


PROBLEM 6.80 Determine the components of all forces acting on member ABCD when θ = 90°.


ΣM B = 0: A(8 in.) − (60 lb)(8 in.) = 0 A = +60.0 lb

A = 60.0 lb

ΣFx = 0: Bx + 60 lb − 60 lb = 0 Bx = 0 ΣFy = 0: By = 0

Free body: Member ABCD

B=0

We note that D is directed along DE, since DE is a two-force member. ΣM C = 0: D(12 in.) + (60 lb)(8 in.) = 0 D = −40.0 lb

D = 40.0 lb

ΣFx = 0: C x + 60 lb = 0 C x = −60 lb

C x = 60.0 lb

ΣFy = 0: C y − 40 lb = 0 C y = +40 lb

C y = 40.0 lb


PROBLEM 6.81 For the frame and loading shown, determine the components of all forces acting on member ABC.

SOLUTION Free body: Entire frame ΣFx = 0: Ax + 18 kN = 0 Ax = −18 kN

A x = 18.00 kN

ΣM E = 0: −(18 kN)(4 m) − Ay (3.6 m) = 0 Ay = −20 kN

A y = 20.0 kN

ΣFy = 0: − 20 kN + F = 0 F = +20 kN F = 20 kN

Free body: Member ABC Note: BE is a two-force member. Thus B is directed along line BE. ΣM C = 0: B(4 m) − (18 kN)(6 m) + (20 kN)(3.6 m) = 0 B = 9 kN

B = 9.00 kN

ΣFx = 0: C x − 18 kN + 9 kN = 0 C x = 9 kN

C x = 9.00 kN

ΣFy = 0: C y − 20 kN = 0 C y = 20 kN

C y = 20.0 kN


PROBLEM 6.82 Solve Problem 6.81 assuming that the 18-kN load is replaced by a clockwise couple of magnitude 72 kN · m applied to member CDEF at Point D. PROBLEM 6.81 For the frame and loading shown, determine the components of all forces acting on member ABC.

SOLUTION Free body: Entire frame ΣFx = 0: Ax = 0 ΣM F = 0: − 72 kN ⋅ m − Ay (3.6 m) = 0 Ay = −20 kN A y = 20 kN

A = 20.0 kN

Free body: Member ABC Note: BE is a two-force member. Thus B is directed along line BE. ΣM C = 0: B(4 m) + (20 kN)(3.6 m) = 0 B = −18 kN

B = 18.00 kN

Fx = 0: − 18 kN + C x = 0 C x = 18 kN

C x = 18.00 kN

ΣFy = 0: C y − 20 kN = 0 C y = 20 kN

C y = 20.0 kN


PROBLEM 6.83 Determine the components of the reactions at A and E if a 750-N force directed vertically downward is applied (a) at B, (b) at D.

SOLUTION Free-body: Entire Frame The following analysis is valid for both (a) and (b) since position of load on its line of action is immaterial. ΣM E = 0: −(750 N)(240 mm) − Ax (400 mm) = 0 Ax = −450 N A x = 450 N ΣFx = 0: Ex − 450 N = 0 Ex = 450 N E x = 450 N ΣFy = 0: Ay + E y − 750 N = 0

(a)

(1)

Load applied at B. Free body: Member CE CE is a two-force member. Thus, the reaction at E must be directed along CE. Ey

=

450 N

From Eq. (1):

240 mm ; E y = 225 N 480 mm

Ay + 225 − 750 = 0; A y = 525 lb

Thus, reactions are:

(b)

A x = 450 N

,

A y = 525 lb

E x = 450N

,

E y = 225 lb

A x = 450 N

,

A y = 150.0 N

E x = 450 N

,

E y = 600 N

Load applied at D. Free body: Member AC AC is a two-force member. Thus, the reaction at A must be directed along AC. Ay 450 N

From Eq. (1):

=

160 mm 480 mm

A y = 150.0 N

Ay + E y − 750 N = 0 150 N + E y − 750 N = 0 E y = 600 N E y = 600 N



!

PROBLEM 6.84 Determine the components of the reactions at A and E if a 750-N force directed vertically downward is applied (a) at B, (b) at D.

SOLUTION Free body: Entire frame The following analysis is valid for both (a) and (b) since position of load on its line of action is immaterial. ΣM E = 0: − (750 N)(80 mm) − Ax (200 mm) = 0 Ax = − 300 N A x = 300 N ΣFx = 0: Ex − 300 N = 0 Ex = 300 N E x = 300 N ΣFy = 0: Ay + E y − 750 N = 0

(a)

(1)

Load applied at B. Free body: Member CE CE is a two-force member. Thus, the reaction at E must be directed along CE. Ey

=

300 N

From Eq. (1):

75 mm 250 mm

E y = 90 N

Ay + 90 N − 750 N = 0

Ay = 660 N

Thus reactions are:

(b)

A x = 300 N

,

A y = 660 N

E x = 300 N

,

E y = 90.0 N

Load applied at D. Free body: Member AC AC is a two-force member. Thus, the reaction at A must be directed along AC. Ay 300 N

=

125 mm 250 mm

Ay = 150 N



Ay + E y − 750 N = 0

From Eq. (1):

150 N + E y − 750 N = 0 E y = 600 N E y = 600 N


A x = 300 N

,

A y = 150.0 N

E x = 300 N

,

E y = 600 N


!

PROBLEM 6.85 Determine the components of the reactions at A and E if the frame is loaded by a clockwise couple of magnitude 36 N · m applied (a) at B, (b) at D.

SOLUTION Free body: Entire frame The following analysis is valid for both (a) and (b) since the point of application of the couple is immaterial. ΣM E = 0: −36 N ⋅ m − Ax (0.4 m) = 0 Ax = − 90 N A x = 90.0 N ΣFx = 0: −90 +Ex = 0 Ex = 90 N E x = 90.0 N ΣFy = 0: Ay + E y = 0

(a)

(1)

Couple applied at B. Free body: Member CE AC is a two-force member. Thus, the reaction at E must be directed along EC. Ey

=

90 N

From Eq. (1):

0.24 m ; E y = 45.0 N 0.48 m

Ay + 45 N = 0 Ay = − 45 N A y = 45.0 N

Thus, reactions are

(b)

A x = 90.0 N

,

A y = 45.0 N

E x = 90.0 N

,

E y = 45.0 N

Couple applied at D. Free body: Member AC AC is a two-force member. Thus, the reaction at A must be directed along AC. Ay 90 N

=

0.16 m ; A y = 30 N 0.48 m



From Eq. (1):

Ay + E y = 0 30 N + E y = 0 E y = − 30 N E y = 30 N A x = 90.0 N


E x = − 90.0 N

A y = 30.0 N

, ,

E y = 30.0 N


!

PROBLEM 6.86 Determine the components of the reactions at A and E if the frame is loaded by a clockwise couple of magnitude 36 N · m applied (a) at B, (b) at D.

SOLUTION Free body: Entire frame The following analysis is valid for both (a) and (b) since the point of application of the couple is immaterial. ΣM E = 0: −36 N ⋅ m − Ax (0.2 m) = 0 Ax = −180 N A x = 180 N ΣFx = 0: −180 N + Ex = 0 Ex = 180 N E x = 180 N ΣFy = 0: Ay + E y = 0

(a)

(1)

Couple applied at B. Free body: Member CE AC is a two-force member. Thus, the reaction at E must be directed along EC. Ey

=

180 N

From Eq. (1):

0.075 m 0.25 m

E y = 54 N

Ay + 54 N = 0 Ay = − 54 N A y = 54.0 N

Thus reactions are

(b)

A x = 180.0 N

,

A y = 54.0 N

E x = 180.0 N

,

E y = 54.0 N

Couple applied at D. Free body: Member AC AC is a two-force member. Thus, the reaction at A must be directed along EC. Ay 180 N

=

0.125 m 0.25 m

Ay = 90 N



From Eq. (1):

Ay + E y = 0 90 N + E y = 0 E y = − 90 N E y = 90 N

Thus, reactions are A x = 180.0 N E x = −180.0 N

A y = 90.0 N

, ,

E y = 90.0 N


!

PROBLEM 6.87 Determine the components of the reactions at A and B, (a) if the 500-N load is applied as shown, (b) if the 500-N load is moved along its line of action and is applied at Point F.

SOLUTION Free body: Entire frame

Analysis is valid for either (a) or (b), since position of 100-lb load on its line of action is immaterial. ΣM A = 0: By (10) − (100 lb)(6) = 0 By = + 60 lb ΣFy = 0: Ay + 60 − 100 = 0

Ay = + 40 lb

ΣFx = 0: Ax + Bx = 0

(a)

(1)

Load applied at E. Free body: Member AC

Since AC is a two-force member, the reaction at A must be directed along CA. We have Ax 40 lb = 10 in. 5 in.

From Eq. (1):

A x = 80.0 lb

,

A y = 40.0 lb

B x = 80.0 lb

,

B y = 60.0 lb

− 80 + Bx = 0 Bx = + 80 lb

Thus,



(b)

Load applied at F. Free body: Member BCD Since BCD is a two-force member (with forces applied at B and C only), the reaction at B must be directed along CB. We have therefore Bx = 0

The reaction at B is From Eq. (1): The reaction at A is

Bx = 0

Ax + 0 = 0

B y = 60.0 lb

Ax = 0 A y = 40.0 lb

Ax = 0


!

PROBLEM 6.88 The 48-lb load can be moved along the line of action shown and applied at A, D, or E. Determine the components of the reactions at B and F if the 48-lb load is applied (a) at A, (b) at D, (c) at E.

SOLUTION Free body: Entire frame The following analysis is valid for (a), (b) and (c) since the position of the load along its line of action is immaterial. ΣM F = 0: (48lb)(8 in.) − Bx (12 in.) = 0 Bx = 32 lb B x = 32 lb ΣFx = 0: 32 lb + Fx = 0 Fx = − 32 lb Fx = 32 lb ΣFy = 0: By + Fy − 48 lb = 0

(a)

(1)

Load applied at A. Free body: Member CDB

CDB is a two-force member. Thus, the reaction at B must be directed along BC. By 32 lb

From Eq. (1):

=

5 in. 16 in.

B y = 10 lb

10 lb + Fy − 48 lb = 0 Fy = 38 lb Fy = 38 lb

Thus reactions are: B x = 32.0 lb

,

B y = 10.00 lb

Fx = 32.0 lb

,

Fy = 38.0 lb


!


(b)

Load applied at D. Free body: Member ACF. ACF is a two-force member. Thus, the reaction at F must be directed along CF. Fy 32 lb

From Eq. (1):

=

7 in. 16 in.

Fy = 14 lb

By + 14 lb − 48 lb = 0 By = 34 lb By = 34 lb


(c)

B x = 32.0 lb

,

B y = 34.0 lb

Fx = 32.0 lb

,

Fy = 14.00 lb

Load applied at E. Free body: Member CDB

This is the same free body as in Part (a). Reactions are same as (a)


!

PROBLEM 6.89 The 48-lb load is removed and a 288-lb · in. clockwise couple is applied successively at A, D, and E. Determine the components of the reactions at B and F if the couple is applied (a) at A, (b) at D, (c) at E.

SOLUTION Free body: Entire frame The following analysis is valid for (a), (b), and (c), since the point of application of the couple is immaterial. ΣM F = 0: − 288 lb ⋅ in. − Bx (12 in.) = 0 Bx = − 24 lb B x = 24 lb ΣFx = 0: − 24 lb + Fx = 0 Fx = 24 lb Fx = 24 lb ΣFy = 0: By + Fy = 0

(a)

(1)

Couple applied at A. Free body: Member CDB

CDB is a two-force member. Thus, reaction at B must be directed along BC. By 24 lb

From Eq. (1):

=

5 in. 16 in.

B y = 7.5 lb

− 7.5 lb + Fy = 0 Fy = 7.5 lb Fy = 7.5 lb

Thus, reactions are: B x = 24.0 lb

,

B y = 7.50 lb

Fx = 24.0 lb

,

Fy = 7.50 lb


!


(b)

Couple applied at D. Free body: Member ACF.

ACF is a two-force member. Thus, the reaction at F must be directed along CF. Fy 24 lb

From Eq. (1):

=

7 in. 16 in.

Fy = 10.5 lb

By − 10.5 lb: By = + 10.5 lb B y = 10.5 lb


(c)

B x = 24.0 lb

,

B y = 10.50 lb

Fx = 24.0 lb

,

Fy = 10.50 lb

Couple applied at E. Free body: Member CDB

This is the same free body as in Part (a). Reactions are same as in (a)


!

PROBLEM 6.90 (a) Show that when a frame supports a pulley at A, an equivalent loading of the frame and of each of its component parts can be obtained by removing the pulley and applying at A two forces equal and parallel to the forces that the cable exerted on the pulley. (b) Show that if one end of the cable is attached to the frame at a point B, a force of magnitude equal to the tension in the cable should also be applied at B.

SOLUTION First note that, when a cable or cord passes over a frictionless, motionless pulley, the tension is unchanged. ΣM C = 0: rT1 − rT2 = 0 T1 = T2

(a)

Replace each force with an equivalent force-couple.

(b)

Cut cable and replace forces on pulley with equivalent pair of forces at A as above.

!


PROBLEM 6.91 Knowing that the pulley has a radius of 50 mm, determine the components of the reactions at B and E.

SOLUTION Free body: Entire assembly ΣM E = 0 : − (300 N)(350 mm) − Bx (150 mm) = 0 Bx = −700 N

B x = 700 N

ΣFx = 0: − 700 N + Ex = 0 Ex = 700 N

E x = 700 N

ΣFy = 0: By + E y − 300 N = 0

(1)

Free body: Member ACE

ΣM C = 0: (700 N)(150 mm) − (300 N)(50 mm) − E y (180 mm) = 0 E y = 500 N

From Eq. (1):

E y = 500 N

By + 500 N − 300 N = 0 By = −200 N

B y = 200 N

Thus, reactions are: B x = 700 N

,

B y = 200 N

E x = 700 N

,

E y = 500 N


PROBLEM 6.92 Knowing that each pulley has a radius of 250 mm, determine the components of the reactions at D and E.


ΣM E = 0: (4.8 kN)(4.25 m) − Dx (1.5 m) = 0 Dx = +13.60 kN

D x = 13.60 kN

ΣFx = 0: Ex + 13.60 kN = 0 Ex = −13.60 kN

E x = 13.60 kN

ΣFy = 0: D y + E y − 4.8 kN = 0

(1)

Free body: Member ACE

ΣM A = 0 : (4.8 kN)(2.25 m) + E y (4 m) = 0 E y = −2.70 kN

From Eq. (1):

E y = 2.70 kN

Dy − 2.70 kN − 4.80 kN = 0 Dy = +7.50 kN

D y = 7.50 kN


PROBLEM 6.93 Two 9-in.-diameter pipes (pipe 1 and pipe 2) are supported every 7.5 ft by a small frame like that shown. Knowing that the combined weight of each pipe and its contents is 30 lb/ft and assuming frictionless surfaces, determine the components of the reactions at A and G.

SOLUTION Free-body: Pipe 2 W = (30 lb/ft)(7.5 ft) = 225 lb

F D 225 lb = = 8 17 15 F = 120 lb D = 255 lb

r = 4.5 in.

Geometry of pipe 2

CF = CD

By symmetry:

(1)

Equate horizontal distance: r+

8 15 ! r = CD " # 17 $ 17 % 25 15 ! r = CD " # 17 17 $ % 25 5 CD = r= r 15 3

From Eq. (1):

5 5 CF = r = (4.5 in.) 3 3 CF = 7.5 in.

Free-body: Member CFG ΣM C = 0: (120 lb)(7.5 in.) − Gx (16 in.) = 0 Gx = 56.25 lb

G x = 56.3 lb



Free body: Frame and pipes

Note: Pipe 2 is similar to pipe 1. AE = CF = 7.5 in. E = F = 120 lb ΣM A = 0: G y (15 in.) − (56.25 lb)(24 in.) − (225 lb)(4.5 in.) −(225 lb)(19.5 in.) − (120 lb)(7.5 in.) = 0 G y = 510 lb

G y = 510 lb

!

ΣFx = 0: Ax + 120 lb + 56.25 lb = 0 Ax = 176.25 lb

A x = 176.3 lb

ΣFy = 0: Ay + 510 lb − 225 lb − 225 lb = 0 Ay = −60 lb

A y = 60.0 lb


PROBLEM 6.94 Solve Problem 6.93 assuming that pipe 1 is removed and that only pipe 2 is supported by the frames. PROBLEM 6.93 Two 9-in.-diameter pipes (pipe 1 and pipe 2) are supported every 7.5 ft by a small frame like that shown. Knowing that the combined weight of each pipe and its contents is 30 lb/ft and assuming frictionless surfaces, determine the components of the reactions at A and G.

SOLUTION Free-body: Pipe 2

W = (30 lb/ft)(7.5 ft) = 225 lb F D 225 lb = = 8 17 15 F = 120 lb D = 255 lb

r = 4.5 in.

Geometry of pipe 2

CF = CD

By symmetry:

(1)

Equate horizontal distance: r+

8 15 ! r = CD " # 17 $ 17 % 25 15 ! r = CD " # 17 17 $ % 25 5 CD = r= r 15 3

From Eq. (1):

5 5 CF = r = (4.5 in.) 3 3 CF = 7.5 in.

Free body: Member CFG ΣM C = 0: (120 lb)(7.5 in.) − Gx (16 in.) = 0 Gx = 56.25 lb

G x = 56.3 lb



Free body: Frame and pipe 2 G x = 56.3 lb

ΣM A = 0: G y (15 in.) − (56.25 lb)(24 in.) − (225 lb)(19.5 in.) = 0 G y = 382.5 lb

G y = 383 lb

!

ΣFx = 0: Ax + 56.25 lb Ax = −56.25 lb

A x = 56.3 lb

ΣFy = 0: Ay + 382.5 lb − 225 lb = 0 Ay = −157.5 lb

A y = 157.5


PROBLEM 6.95 A trailer weighing 2400 lb is attached to a 2900-lb pickup truck by a ball-and-socket truck hitch at D. Determine (a) the reactions at each of the six wheels when the truck and trailer are at rest, (b) the additional load on each of the truck wheels due to the trailer.

SOLUTION (a)

Free body: Trailer (We shall denote by A, B, C the reaction at one wheel) ΣM A = 0: − (2400 lb)(2 ft) + D(11 ft) = 0 D = 436.36 lb ΣFy = 0: 2 A − 2400 lb + 436.36 lb = 0 A = 981.82 lb

A = 982 lb

Free body: Truck ΣM B = 0: (436.36 lb)(3 ft) − (2900 lb)(5 ft) + 2C (9 ft) = 0 C = 732.83 lb

C = 733 lb

ΣFy = 0: 2 B − 436.36 lb − 2900 lb + 2(732.83 lb) = 0 B = 935.35 lb

(b)

B = 935 lb

Additional load on truck wheels Use free body diagram of truck without 2900 lb. ΣM B = 0: (436.36 lb)(3 ft) + 2C (9 ft) = 0 C = −72.73 lb

∆C = −72.7 lb

ΣFy = 0: 2 B − 436.36 lb − 2(72.73 lb) = 0 B = 290.9 lb

∆B = +291 lb


PROBLEM 6.96 In order to obtain a better weight distribution over the four wheels of the pickup truck of Problem 6.95, a compensating hitch of the type shown is used to attach the trailer to the truck. The hitch consists of two bar springs (only one is shown in the figure) that fit into bearings inside a support rigidly attached to the truck. The springs are also connected by chains to the trailer frame, and specially designed hooks make it possible to place both chains in tension. (a) Determine the tension T required in each of the two chains if the additional load due to the trailer is to be evenly distributed over the four wheels of the truck. (b) What are the resulting reactions at each of the six wheels of the trailer-truck combination? PROBLEM 6.95 A trailer weighing 2400 lb is attached to a 2900-lb pickup truck by a ball-and-socket truck hitch at D. Determine (a) the reactions at each of the six wheels when the truck and trailer are at rest, (b) the additional load on each of the truck wheels due to the trailer.

SOLUTION (a)

We small first find the additional reaction ∆ at each wheel due the trailer. Free body diagram (Same ∆ at each truck wheel) ΣM A = 0: − (2400 lb)(2 ft) + 2∆(14 ft) + 2∆(23 ft) = 0 ∆ = 64.86 lb ΣFY = 0: 2 A − 2400 lb + 4(64.86 lb) = 0; A = 1070 lb;

A = 1070 lb

Free body: Truck (Trailer loading only) ΣM D = 0: 2∆(12 ft) + 2∆(3 ft) − 2T (1.7 ft) = 0 T = 8.824∆ = 8.824(64.86 lb) T = 572.3 lb

T = 572 lb

Free body: Truck (Truck weight only) ΣM B = 0: − (2900 lb)(5 ft) + 2C ′(9 ft) = 0 C ′ = 805.6 lb

C′ = 805.6 lb


PROBLEM 6.96 (Continued) ΣFy = 0: 2 B′ − 2900 lb + 2(805.6 lb) = 0 B′ = 644.4 lb

B′ = 644.4 lb

Actual reactions B = B′ + ∆ = 644.4 lb + 64.86 = 709.2 lb

B = 709 lb

!

C = C ′ + ∆ = 805.6 lb + 64.86 = 870.46 lb

C = 870 lb

!

A = 1070 lb

(From Part a):


PROBLEM 6.97 The cab and motor units of the front-end loader shown are connected by a vertical pin located 2 m behind the cab wheels. The distance from C to D is 1 m. The center of gravity of the 300-kN motor unit is located at Gm, while the centers of gravity of the 100-kN cab and 75-kN load are located, respectively, at Gc and Gl. Knowing that the machine is at rest with its brakes released, determine (a) the reactions at each of the four wheels, (b) the forces exerted on the motor unit at C and D.

SOLUTION (a)

Free body: Entire machine

A = Reaction at each front wheel B = Reaction at each rear wheel

ΣM A = 0: 75(3.2 m) − 100(1.2 m) + 2 B(4.8 m) − 300(5.6 m) = 0 2 B = 325 kN

B = 162.5 kN

!

A = 75.0 kN

!

ΣFy = 0: 2 A + 325 − 75 − 100 − 300 = 0 2 A = 150 kN


PROBLEM 6.97 (Continued) (b)

Free body: Motor unit ΣM D = 0: C (1 m) + 2 B(2.8 m) − 300(3.6 m) = 0 C = 1080 − 5.6 B

Recalling

B = 162.5 kN, C = 1080 − 5.6(162.5) = 170 kN

ΣFx = 0: Dx − 170 = 0

!

!

(1)

! ΣFy = 0: 2(162.5) − Dy − 300 = 0

C = 170.0 kN

!

D x = 170.0 kN

!

D y = 25.0 kN


PROBLEM 6.98 Solve Problem 6.97 assuming that the 75-kN load has been removed. PROBLEM 6.97 The cab and motor units of the front-end loader shown are connected by a vertical pin located 2 m behind the cab wheels. The distance from C to D is 1 m. The center of gravity of the 300-kN motor unit is located at Gm, while the centers of gravity of the 100-kN cab and 75-kN load are located, respectively, at Gc and Gl. Knowing that the machine is at rest with its brakes released, determine (a) the reactions at each of the four wheels, (b) the forces exerted on the motor unit at C and D.

SOLUTION (a)

Free body: Entire machine A = Reaction at each front wheel B = Reaction at each rear wheel

ΣM A = 0: 2 B(4.8 m) − 100(1.2 m) − 300(5.6 m) = 0 2 B = 375 kN

B = 187.5 kN

!

A = 12.50 kN

!

ΣFy = 0: 2 A + 375 − 100 − 300 = 0 2 A = 25 kN

(b)

Free body: Motor unit See solution of Problem 6.97 for free body diagram and derivation of Eq. (1). With B = 187.5 kN, we have C = 1080 − 5.6(187.5) = 30 kN C = 30.0 kN

ΣFx = 0: Dx − 30 = 0

D x = 30.0 kN

ΣFy = 0: 2(187.5) − Dy − 300 = 0

D y = 75.0 kN


PROBLEM 6.99 For the frame and loading shown, determine the components of the forces acting on member CFE at C and F.

SOLUTION Free body: Entire frame

ΣM D = 0: (40 lb)(13 in.) + Ax (10 in.) = 0 Ax = −52 lb,

A x = 52 lb

!

Free body: Member ABF ΣM B = 0: − (52 lb)(6 in.) + Fx (4 in.) = 0 Fx = +78 lb

Free body: Member CFE Fx = 78.0 lb

From above:

!

ΣM C = 0: (40 lb)(9 in.) − (78 lb)(4 in.) − Fy (4 in.) = 0 Fy = +12 lb

Fy = 12.00 lb

ΣFx = 0: Cx − 78 lb = 0 Cx = +78 lb ΣFy = 0: − 40 lb + 12 lb + C y = 0; C y = +28 lb

C x = 78.0 lb

!

C y = 28.0 lb


PROBLEM 6.100 For the frame and loading shown, determine the components of the forces acting on member CDE at C and D.

SOLUTION Free body: Entire frame ΣM y = 0: Ay − 25 lb = 0 Ay = 25 lb

A y = 25 lb !

ΣM F = 0: Ax (6.928 + 2 × 3.464) − (25 lb)(12 in.) = 0 Ax = 21.651 lb

A y = 21.65 lb

ΣFx = 0: F − 21.651 lb = 0 F = 21.651 lb

F = 21.65 lb

Free body: Member CDE ΣM C = 0: Dy (4 in.) − (25 lb)(10 in.) = 0 Dy = +62.5 lb

D y = 62.5 lb

!

ΣFy = 0: −C y + 62.5 lb − 25 lb = 0 C y = +37.5 lb

C y = 37.5 lb

Free body: Member ABD ΣM B = 0: Dx (3.464 in.) + (21.65 lb)(6.928 in.) −(25 lb)(4 in.) − (62.5 lb)(2 in.) Dx = +21.65 lb

Return to free body: Member CDE From above Dx = +21.65 lb

D x = 21.7 lb

ΣFx = 0: Cx − 21.65 lb Cx = +21.65 lb

C x = 21.7 lb


PROBLEM 6.101 For the frame and loading shown, determine the components of all forces acting on member ABE.

SOLUTION FBD Frame: ΣM E = 0: (1.8 m) Fy − (2.1 m)(12 kN) = 0 Fy = 14.00 kN ΣFy = 0: − E y + 14.00 kN − 12 kN = 0 E y = 2 kN E y = 2.00 kN

FBD member BCD: ΣM B = 0: (1.2 m)C y − (12 kN)(1.8 m) = 0 C y = 18.00 kN

But C is ⊥ ACF, so C x = 2C y ; C x = 36.0 kN ΣFx = 0: − Bx + C x = 0 Bx = C x = 36.0 kN Bx = 36.0 kN

on BCD

ΣFy = 0: − B y + 18.00 kN − 12 kN = 0 By = 6.00 kN on BCD B x = 36.0 kN

On ABE:

B y = 6.00 kN

FBD member ABE: ΣM A = 0: (1.2 m)(36.0 kN) − (0.6 m)(6.00 kN) + (0.9 m)(2.00 kN) − (1.8 m)( E x ) = 0 E x = 23.0 kN ΣFx = 0: − 23.0 kN + 36.0 kN − Ax = 0 ΣFy = 0: − 2.00 kN + 6.00 kN − Ay = 0

A x = 13.00 kN A y = 4.00 kN


PROBLEM 6.102 For the frame and loading shown, determine the components of all forces acting on member ABE. PROBLEM 6.101 For the frame and loading shown, determine the components of all forces acting on member ABE.

SOLUTION FBD Frame:

ΣM F = 0: (1.2 m)(2400 N) − (4.8 m) E y = 0

E y = 600 N

FBD member BC: Cy =

4.8 8 Cx = Cx 5.4 9

ΣM C = 0: (2.4 m) By − (1.2 m)(2400 N) = 0 B y = 1200 N B y = 1200 N

on ABE: ΣFy = 0: −1200 N + C y − 2400 N = 0 C y = 3600 N

so

9 Cx = C y 8 ΣFx = 0: − Bx + C x = 0 Bx = 4050 N

C x = 4050 N

on BC B x = 4050 N

on ABE:



FBD member AB0E: ΣM A = 0: a(4050 N) − 2aEx = 0 Ex = 2025 N

E x = 2025 N

ΣFx = 0 : − Ax + (4050 − 2025) N = 0

A x = 2025 N

ΣFy = 0: 600 N + 1200 N − Ay = 0

A y = 1800 N


!

PROBLEM 6.103 Knowing that P = 15 lb and Q = 65 lb, determine the components of the forces exerted (a) on member BCDF at C and D, (b) on member ACEG at E.

SOLUTION Free body: Entire frame ΣM E = 0: ( P + Q)(25 in.) − ( P + Q )(15 in.) − Dx (8 in.) = 0 10 8 10 D x = ( P + Q) 8 Dx = ( P + Q)

ΣFx = 0: − Ex + ( P + Q )

10 =0 8 Ex = ( P + Q)

10 10 ; E x = ( P + Q) 8 8

Free body: Member BCDF From above:

ΣFx = 0: −C x + Dx = 0 ΣM D = 0: − ( P + Q )

D x = ( P + Q)

10 8

v

C x = ( P + Q)

10 8

v

10 (4 in.) − P(15 in.) + Q (25 in.) + C y (10 in.) 8 Cy =

1 (− 20 P + 20 Q ) 10

C y = 2Q − 2 P

C y = 2Q − 2 P v

ΣFy = 0: D y + (2Q − 2 P ) = P − Q = 0 D y = − Q + 3P

D y = − Q + 3P v



Free body: Member ACEG

From above E x = ( P + Q)

10 8

v

Cy = 2Q − P

and

ΣFy = 0: E y − C y − P − Q = 0 E y − (2Q − 2 P ) − P − Q = 0 E y = 3Q − P

E y = 3Q − P v

P = 15 lb and Q = 65 lb C x = ( P + Q)

10 10 = (15 + 65) = +100 lb 8 8

C y = 2 Q − 2 P = 2(65) − 2(15) = +100 lb Dx = ( P + Q )

10 10 = (15 + 65) = +100 lb 8 8

Dy = − Q + 3P = − 65 + 3(15) = − 20 lb E x = ( P + Q)

10 10 = (15 + 65) = +100 lb 8 8

E y = 3Q − P = 3(65) − 15 = + 180 lb

C x = 100.0 lb C y = 100.0 lb D x = 100.0 lb D y = 20.0 lb E x = 100.0 lb E y = 180.0 lb


!

PROBLEM 6.104 Knowing that P = 25 lb and Q = 55 lb, determine the components of the forces exerted (a) on member BCDF at C and D, (b) on member ACEG at E.

SOLUTION Free body: Entire frame ΣM E = 0: ( P + Q)(25 in.) − ( P + Q )(15 in.) − Dx (8 in.) = 0 10 8 10 D x = ( P + Q) 8 Dx = ( P + Q)

ΣFx = 0: − Ex + ( P + Q )

10 =0 8 Ex = ( P + Q)

10 10 ; E x = ( P + Q) 8 8

Free body: Member BCDF From above:

ΣFx = 0: − C x + Dx = 0 ΣM D = 0: − ( P + Q )

D x = ( P + Q)

10 8

v

C x = ( P + Q)

10 8

v

10 (4 in.) − P(15 in.) + Q (25 in.) + C y (10 in.) 8 Cy =

1 (− 20 P + 20 Q ) 10

C y = 2Q − 2 P

C y = 2Q − 2 P v

ΣFy = 0: D y + (2Q − 2 P ) − P − Q = 0 D y = − Q + 3P

D y = − Q + 3P v



Free body: Member ACEG

From above: E x = ( P + Q)

10 8

v

Cy = 2Q − P

and

ΣFy = 0: E y − C y − P − Q = 0 E y − (2Q − 2 P ) − P − Q = 0 E y = 3Q − P

E y = 3Q − P v

P = 25 lb and Q = 55 lb C x = ( P + Q)

10 10 = (25 + 55) = +100 lb 8 8

C y = 2 Q − 2 P = 2(55) − 2(25) = +60 lb Dx = ( P + Q )

10 10 = (25 + 55) = +100 lb 8 8

Dy = − Q + 3P = − 55 + 3(25) = +20 lb E x = ( P + Q)

10 10 = (25 + 55) = +100 lb 8 8

E y = 3Q − P = 3(55) − 25 = +140 lb

C x = 100.0 C y = 60.0 lb D x = 100.0 lb D y = 20.0 lb E x = 100.0 lb E y = 140.0 lb


!

PROBLEM 6.105 For the frame and loading shown, determine the components of the forces acting on member DABC at B and D.

SOLUTION Free body: Entire frame ΣM G = 0: H (0.6 m) − (12 kN)(1 m) − (6 kN)(0.5 m) = 0 H = 25 kN H = 25 kN

Free body: Member BEH ΣM F = 0: Bx (0.5 m) − (25 kN)(0.2 m) = 0 Bx = + 10 kN

Free body: Member DABC B x = 10.00 kN

From above:

ΣM D = 0: − By (0.8 m) + (10 kN + 12 kN)(0.5 m) = 0 By = + 13.75 kN

B y = 13.75 kN

ΣFx = 0: − Dx + 10 kN + 12 kN = 0 Dx = + 22 kN

D x = 22.0 kN

ΣFy = 0: − D y + 13.75 kN = 0 Dy = + 13.75 kN

D y = 13.75 kN


!

PROBLEM 6.106 Solve Problem 6.105 assuming that the 6-kN load has been removed. PROBLEM 6.105 For the frame and loading shown, determine the components of the forces acting on member DABC at B and D.

SOLUTION Free body: Entire frame ΣM G = 0: H (0.6 m) − (12 kN)(1 m) = 0 H = 20 kN H = 20 kN

Free body: Member BEH ΣM E = 0: Bx (0.5 m) − (20 kN)(0.2 m) = 0 Bx = + 8 kN

Free body: Member DABC B x = −8.00 kN

From above:

ΣM D = 0: − By (0.8 m) + (8 kN + 12 kN)(0.5 m) = 0 By = + 12.5 kN

B y = 12.50 kN

ΣFx = 0: − Dx + 8 kN + 12 kN = 0 Dx = + 20 kN ΣFy = 0: − D y + 12.5 kN = 0; Dy = + 12.5 kN

D x = 20.0 kN D y = 12.50 kN


!

PROBLEM 6.107 The axis of the three-hinge arch ABC is a parabola with vertex at B. Knowing that P = 112 kN and Q = 140 kN, determine (a) the components of the reaction at A, (b) the components of the force exerted at B on segment AB.

SOLUTION

Free body: Segment AB: ΣM A = 0: Bx (3.2 m) − By (8 m) − P(5 m) = 0

(1)

Bx (2.4 m) − By (6 m) − P(3.75 m) = 0

(2)

0.75 (Eq. 1)

Free body: Segment BC: ΣM C = 0: Bx (1.8 m) + By (6 m) − Q(3 m) = 0

Add (2) and (3):

(3)

4.2 Bx − 3.75P − 3Q = 0 Bx = (3.75 P + 3Q)/4.2

Eq. (1):

(3.75P + 3Q)

(4)

3.2 − 8B y − 5 P = 0 4.2 By = (− 9P + 9.6Q )/33.6

(5)

Given that P = 112 kN and Q = 140 kN (a)

Reaction at A: Considering again AB as a free body ΣFx = 0: Ax − Bx = 0;

Ax = Bx = 200 kN

A x = 200 kN

ΣFy = 0: Ay − P − By = 0 Ay − 112 kN − 10 kN = 0 Ay = + 122 kN

A y = 122.0 kN



(b)

Force exerted at B on AB Eq. (4):

Bx = (3.75 × 112 + 3 × 140)/4.2 = 200 kN B x = 200 kN

Eq. (5):

By = (− 9 × 112 + 9.6 × 140)/33.6 = + 10 kN B y = 10.00 kN


!

PROBLEM 6.108 The axis of the three-hinge arch ABC is a parabola with vertex at B. Knowing that P = 140 kN and Q = 112 kN, determine (a) the components of the reaction at A, (b) the components of the force exerted at B on segment AB.

SOLUTION

Free body: Segment AB: ΣM A = 0: Bx (3.2 m) − By (8 m) − P(5 m) = 0

(1)

Bx (2.4 m) − By (6 m) − P(3.75 m) = 0

(2)

0.75 (Eq. 1)

Free body: Segment BC: ΣM C = 0: Bx (1.8 m) + By (6 m) − Q(3 m) = 0

(3)

4.2 Bx − 3.75P − 3Q = 0

Add (2) and (3):

Bx = (3.75 P + 3Q)/4.2 (3.75P + 3Q)

Eq. (1):

(4)

3.2 − 8B y − 5 P = 0 4.2 By = (− 9P + 9.6Q )/33.6

(5)

Given that P = 140 kN and Q = 112 kN (a)

Reaction at A: ΣFx = 0: Ax − Bx = 0;

Ax = Bx = 205 kN

A x = 205 kN

ΣFy = 0: Ay − P − By = 0 Ay − 140 kN − (− 5.5 kN) = 0 Ay = 134.5 kN

A y = 134.5 kN



(b)

Force exerted at B on AB Eq. (4):

Bx = (3.75 × 140 + 3 × 112)/4.2 = 205 kN B x = 205 kN

Eq. (5):

By = (− 9 × 140 + 9.6 × 112)/33.6 = − 5.5 kN B y = 5.5 kN


!

PROBLEM 6.109 Knowing that the surfaces at A and D are frictionless, determine the forces exerted at B and C on member BCE.

SOLUTION

Free body of Member ACD ΣM H = 0: C x (2 in.) − C y (18 in.) = 0 C x = 9C y

(1)

Free body of Member BCE ΣM B = 0: C x (6 in.) + C y (6 in.) − (50 lb)(12 in.) = 0 9C y (6) + C y (6) − 600 = 0

Substitute from (1):

C y = + 10 lb; C x = 9C y = 9(10) = + 90 lb C = 90.6 lb ΣFx = 0: Bx − 90 lb = 0

Bx = 90 lb

ΣFy = 0: By + 10 lb − 50 lb = 0

By = 40 lb B = 98.5 lb

6.34°

24.0°


!

PROBLEM 6.110 For the frame and loading shown, determine (a) the reaction at C, (b) the force in member AD.

SOLUTION Free body: Member ABC ΣM C = 0: + (100 lb)(45 in.) +

4 4 FAD (45 in.) + FBE (30 in.) = 0 5 5 3FAD + 2 FBF = − 375 lb

(1)

Free Body: Member DEF ΣM F = 0:

4 4 FAD (30 in.) + FBF (15 in.) = 0 5 5 FBE = −2 FAD

(a)

(2)

Substitute from (2) into (1) 3FAD + 2( − 2 FAD ) = − 375 lb FAD = + 375 lb

(2)

FAD = 375 lb ten.

FBE = − 2 FAD = − 2(375 lb) FBE = −750 lb FBE = 750 lb comp.



(b)

Return to free body of member ABC ΣFx = 0: C x + 100 lb + C x + 100 +

4 4 FAD + FBE = 0 5 5

4 4 (375) + (−750) = 0 5 5 C x = + 200 lb C x = 200 lb

3 3 ΣFy = 0: C y − FAD − FBF = 0 5 5 3 3 C y − (375) − (− 750) = 0 5 5 C y = − 225 lb C y = 225 lb

α = 48.37° C = 301.0 lb

C = 301 lb

48.4°


!

PROBLEM 6.111 Members ABC and CDE are pin-connected at C and supported by four links. For the loading shown, determine the force in each link.

SOLUTION Member FBDs:

I

II

From FBD I:

ΣM J = 0:

a 3a a C x + C y − P = 0 C x + 3C y = P 2 2 2

FBD II:

ΣM K = 0:

a 3a C x − C y = 0 C x − 3C y = 0 2 2

Solving:

Cx =

FBD I:

P P ; C y = as drawn 2 6

ΣM B = 0: aC y − a ΣFx = 0: −

1 2 FAG = 0 FAG = 2C y = P 6 2

FAG =

1 1 2 2 FAG + FBF − C x = 0 FBF = FAG + C x 2 = P+ P 6 2 2 2 FBF =

FBD II:

ΣM D = 0: a

2 P C 6

1 2 FEH + aC y = 0 FEH = − 2C y = − P 6 2

ΣFx = 0: C x −

2 2 P C 3

FEH =

2 P T 6

1 1 2 2 FDI + FEH = 0 FDI = FEH + C x 2 = − P+ P 6 2 2 2 FDI =

2 P C 3




I FBD I: FBD II: FBDs combined:

ΣM B = 0:

II 1

aC y − a

1

M D = 0: aC y − a ΣM G = 0:

aP − a

FAF = 0 FAF = 2C y

2

2 1 2

FEH = 0 FEH = 2C y 1

FAF − a

2

FEH = 0 P = Cy =

FBD I:

FBD II:

ΣFy = 0:

1 2

FAF +

ΣFy = 0: − C y +

1 2

1 2

P 2

FBG − P + C y = 0

FDG −

1 2

FEH = 0 −

1 2

2C y +

1

2C y

2

so FAF =

2 P C 2

FEH =

2 P T 2

!

FBG = 0

!

P 1 P + FBG − P + = 0 2 2 2

P 1 P + FDG − = 0 2 2 2 FDG = 2 P C




I

II

FBD I:

ΣM I = 0: 2aC y + aC x − aP = 0 2C y + C x = P

FBD II:

ΣM J = 0: 2aC y − aC x = 0 2C y − C x = 0

Solving: FBD I:

Cx =

P P ; C y = as shown 2 4

ΣFx = 0: −

1 2

FBG + C x = 0 FBG = C x 2

ΣFy = 0: FAF − P +

FBD II:

ΣFx = 0: − Cx +

ΣFy = 0: − C y +

1 2

2 ! P P# + = 0 "" 2 $ 2 #% 4

1

FDG = 0 FDG = Cx 2

2 ! P P P P # + FEH = 0 FEH = − = − "" # 4 2 4 2$ 2 %

1

FBG =

P 2

P 4

FAF = FDG =

P

FEH =

2

C

C

C

P T 4


PROBLEM 6.114 Members ABC and CDE are pin-connected at C and supported by the four links AF, BG, DG, and EH. For the loading shown, determine the force in each link.

SOLUTION Free body: Member ABC ΣM H = 0: Cx (a ) − C y (2a) = 0 Cx = 2C y

Note: This checks that for 3-force member the forces are concurrent. Free body: Member CDE ΣM F = 0: Cx (2a ) − C y (a) + P( a) = 0 2Cx − C y + P = 0 1 Cy = − P v 3

2(2C y ) − C y + P = 0

2 Cx = − P v 3

Cx = 2C y : ΣF = 0: Cx −

E 2

= 0; −

E=−

E 2 P− =0 3 2

2 2 P 3

FEH =

2 2 P comp. 3

ΣM E = 0: D (2a ) + C y (3a) − P(3a) = 0 D (2a) −

P (3a) − P(3a) = 0 3 D = + 2P

FDG = 2 P T



Return to free body of ABC ΣFy = 0:

A

+ Cy = 0

2 A 2

−

P =0 3 A=+

2 P 3

FAF =

2 P T 3

ΣM A = 0: B(2a) − Cx (3a) = 0 B(2a ) +

2 P(3a) = 0 3 B = −P

FBG = P C


PROBLEM 6.115 Solve Problem 6.113 assuming that the force P is replaced by a clockwise couple of moment M0 applied to member CDE at D. PROBLEM 6.113 Members ABC and CDE are pin-connected at C and supported by four links. For the loading shown, determine the force in each link.

SOLUTION Free body: Member ABC ΣM J = 0: C y (2a) + Cx ( a) = 0 Cx = − 2C y

Free body: Member CDE ΣM K = 0: C y (2a ) − Cx (a ) − M 0 = 0 C y (2a) − (− 2C y )(a) − M 0 = 0 Cx = − 2C y : D

ΣFx = 0:

2

D

+ Cx = 0; D=

2

−

M0

M0 v 4a M Cx = − 0 v 2a Cy =

M0 =0 2a FDG =

2a

M0 2a

T

ΣM D = 0: E (a) − C y ( a) + M 0 = 0 M ! E (a) − " 0 # ( a) + M 0 = 0 $ 4a % E=−

3 M0 4 a

FEH =

3 M0 4 a

C

Return to Free Body of ABC ΣFx = 0:

B 2

B

+ C x = 0; B=

2

−

M0 =0 2a

M0

FBG =

2a

ΣM B = 0: A( a) + C y (a);

A(a) +

M0 2a

T

M0 (a ) = 0 4a A=−

M0 4a

FAF =

M0 4a

C


PROBLEM 6.116 Solve Problem 6.114 assuming that the force P is replaced by a clockwise couple of moment M0 applied to member CDE at D. PROBLEM 6.114 Members ABC and CDE are pin-connected at C and supported by the four links AF, BG, DG, and EH. For the loading shown, determine the force in each link.

SOLUTION Free body: Member ABC ΣM H = 0: C x (a ) − C y (2a) = 0 C x = 2C y

Free body: Member CDE ΣM F = 0: C x (2a ) − C y (a) − M 0 = 0 (2C y )(2a ) − C y (a) − M 0 = 0 C x = 2C y : ΣFx = 0: C x −

E 2

= 0;

E= ΣFy = 0: D + D+

E 2

Cy = Cx =

M0 v 3a

2M 0 v 3a

2M 0 E − =0 3a 2

2 2 M0 3 a

FEH =

2 2 M0 3 a

+ Cy = 0

M 2 2 M0 1 + 0 =0 3 a 2 3a D=−

M0 a

FDG =

M0 a

C



Return to free body of ABC ΣFy = 0:

A 2

+ C y = 0; A=−

A 2

+

M0 =0 3a

2 M0 3 a

FAF =

2 M0 3 a

C

M0 a

T

ΣM A = 0: B(2a) − C x (3a) = 0 2 M0 ! B(2a ) − " # (3a) = 0 $3 a % B=+

M0 a

FBG =


PROBLEM 6.117 Four beams, each of length 3a, are held together by single nails at A, B, C, and D. Each beam is attached to a support located at a distance a from an end of the beam as shown. Assuming that only vertical forces are exerted at the connections, determine the vertical reactions at E, F, G, and H.

SOLUTION We shall draw the free body of each member. Force P will be applied to member AFB. Starting with member AED, we shall express all forces in terms of reaction E. Member AFB: ΣM D = 0: A(3a) + E (a ) = 0 E 3 ΣM A = 0: − D (3a ) − E (2a ) = 0 A=−

D=−

2E 3

Member DHC: 2E ! ΣM C = 0: " − # (3a) − H (a) = 0 $ 3 % H = − 2E

(1)

2E ! ΣM H = 0: " − # (2a) + C (a ) = 0 $ 3 % C=+

4E 3

Member CGB: 4E ! ΣM B = 0: + " # (3a ) − G (a ) = 0 $ 3 % G = + 4E

(2)

4E ! ΣM G = 0: + " # (2a) + B (a ) = 0 $ 3 % B=−

8E 3



Member AFB: ΣFy = 0: F − A − B − P = 0 8E ! E! F −"− # −"− #−P =0 $ 3% $ 3 % F = P − 3E

(3)

ΣM A = 0: F ( a) − B(3a) = 0 8E ! ( P − 3E )(a) − " − # (3a ) = 0 $ 3 % P − 3E + 8 E = 0; E = −

P 5

E=

P 5

Substitute E = − P5 into Eqs. (1), (2), and (3). P! H = −2 E = − 2 " − # $ 5%

H =+

2P 5

H=

2P 5

P! G = + 4E = 4 " − # $ 5%

G=−

4P 5

G=

4P 5

P! F = P − 3E = P − 3 " − # $ 5%

F =+

8P 5

F=

8P 5

!


!

PROBLEM 6.118 Four beams, each of length 2a, are nailed together at their midpoints to form the support system shown. Assuming that only vertical forces are exerted at the connections, determine the vertical reactions at A, D, E, and H.

SOLUTION Note that, if we assume P is applied to EG, each individual member FBD looks like so

2 Fleft = 2 Fright = Fmiddle

Labeling each interaction force with the letter corresponding to the joint of its application, we see that B = 2 A = 2F C = 2B = 2D G = 2C = 2 H P + F = 2G ( = 4C = 8 B = 16 F ) = 2 E

From

P + F = 16 F , F =

P 15

!

so A =

P 15

D=

2P 15

H=

4P 15

E=

8P 15


!

!

PROBLEM 6.119 Each of the frames shown consists of two L-shaped members connected by two rigid links. For each frame, determine the reactions at the supports and indicate whether the frame is rigid.

SOLUTION Note: In all three cases, the right member has only three forces acting, two of which are parallel. Thus the third force, at B, must be parallel to the link forces. (a)

FBD whole:

ΣM A = 0: − 2aP − ΣFx = 0: Ax − ΣFy = 0:

a 4 1 B + 5a B = 0 B = 2.06 P 4 17 17

4 17

Ay − P +

B=0 1 17

B = 2.06P

14.04°

A = 2.06P

14.04°

A x = 2P

B = 0 Ay =

P 2

rigid (b)

FBD whole:

Since B passes through A,

ΣM A = 2aP = 0 only if P = 0

no equilibrium if P ≠ 0

not rigid



(c)

FBD whole:

ΣM A = 0: 5a

1 17

ΣFx = 0: Ax +

B+

4 17

ΣFy = 0: Ay − P +

3a 4 17 B − 2aP = 0 B = P 4 17 4

B=0 1 17

B = 1.031P

14.04°

A = 1.250P

36.9°

Ax = − P

B=0

Ay = P −

P 3P = 4 4

System is rigid



SOLUTION (a)

Member FBDs:

I

II

FBD I:

ΣM A = 0: aF1 − 2aP = 0 F1 = 2 P; ΣFy = 0:

FBD II:

ΣM B = 0: − aF2 = 0 F2 = 0

Ay − P = 0 A y = P

ΣFx = 0: Bx + F1 = 0, Bx = − F1 = −2 P B x = 2 P ΣFy = 0: B y = 0

FBD I:

ΣFx = 0: − Ax − F1 + F2 = 0

so B = 2P Ax = F2 − F1 = 0 − 2 P A x = 2 P

so A = 2.24P

26.6°

frame is rigid (b)

FBD left:

FBD whole:

I FBD I:

ΣM E = 0:

FBD II:

ΣM B = 0:

II a a 5a P + Ax − Ay = 0 Ax − 5 Ay = − P 2 2 2 3aP + aAx − 5aAy = 0 Ax − 5 Ay = −3P

This is impossible unless P = 0

not rigid


!


(c)

Member FBDs:

I FBD I:

II

ΣFy = 0: A − P = 0

A=P

!

C=P

!

ΣM D = 0: aF1 − 2aA = 0 F1 = 2 P ΣFx = 0: F2 − F1 = 0 F2 = 2 P !

FBD II:

ΣM B = 0: 2aC − aF1 = 0 C =

F1 =P 2

ΣFx = 0: F1 − F2 + Bx = 0 Bx = P − P = 0 ! ΣFx = 0: By + C = 0 B y = − C = − P

B=P

Frame is rigid



SOLUTION (a)

Member FBDs:

I FBD II: FBD I:

ΣFy = 0: B y = 0

II ΣM B = 0: aF2 = 0

F2 = 0

ΣM A = 0: aF2 − 2aP = 0 but F2 = 0

so P = 0 (b)

not rigid for P ≠ 0

Member FBDs:

Note: 7 Unknowns ( Ax , Ay , Bx , B y , F1 , F2 , C ) but only 6 independent equations.

!

System is statically indeterminate

!

System is, however, rigid

!



(c)

FBD whole:

FBD right:

I FBD I:

II

ΣM A = 0: 5aBy − 2aP = 0 ΣFy = 0: Ay − P +

2 P=0 5

a 5a Bx − By = 0 Bx = 5B y 2 2

FBD II:

ΣM c = 0:

FBD I:

ΣFx = 0: Ax + Bx = 0

Ax = − Bx

By =

2 P ! 5

Ay =

3 P 5

B x = 2P A x = 2P A = 2.09P

16.70°

B = 2.04P

11.31°

System is rigid


PROBLEM 6.122 An 84-lb force is applied to the toggle vise at C. Knowing that θ = 90°, determine (a) the vertical force exerted on the block at D, (b) the force exerted on member ABC at B.

SOLUTION We note that BD is a two-force member. Free body: Member ABC BD = (7) 2 + (24)2 = 25 in.

We have

( FBD ) x =

7 24 FBD , ( FBD ) y = FBD 25 25

ΣM A = 0: ( FBD ) x (24) + ( FBD ) y (7) − 84(16) = 0

7 24 ! ! " FBD # (24) + " FBD # (7) = 84(16) 25 25 $ % $ % 336 FBD = 1344 25 FBD = 100 lb tan α =

FBD = 100.0 lb

(b)

Force exerted at B:

(a)

Vertical force exerted on block ( FBD ) y =

24 α = 73.7° 7

73.7°

24 24 FBD = (100 lb) = 96 lb 25 25 (FBD ) y = 96.0 lb


PROBLEM 6.123 Solve Problem 6.122 when θ = 0. PROBLEM 6.122 An 84-lb force is applied to the toggle vise at C. Knowing that θ = 90°, determine (a) the vertical force exerted on the block at D, (b) the force exerted on member ABC at B.

SOLUTION We note that BD is a two-force member. Free body: Member ABC BD = (7) 2 + (24)2 = 25 in.

We have

( FBD ) x =

7 24 FBD , ( FBD ) y = FBD 25 25

ΣM A = 0: ( FBD ) x (24) + ( FBD ) y (7) − 84(40) = 0 7 24 ! ! " FBD # (24) + " FBD # (7) = 84(40) 25 25 $ % $ % 336 FBD = 3360 25 FBD = 250 lb tan α =

FBD = 250.0 lb

(b)

Force exerted at B:

(a)

Vertical force exerted on block ( FBD ) y =

24 α = 73.7° 7

73.7°

24 24 FBD = (250 lb) = 240 lb 25 25 (FBD ) y = 240 lb


PROBLEM 6.124 The control rod CE passes through a horizontal hole in the body of the toggle system shown. Knowing that link BD is 250 mm long, determine the force Q required to hold the system in equilibrium when β = 20°.

SOLUTION We note that BD is a two-force member. Free body: Member ABC Dimensions in mm

Since

BD = 250, θ = sin −1

33.404 ; θ = 7.679° 250

ΣM C = 0: ( FBD sin θ )187.94 − ( FBD cos θ )68.404 + (100 N)328.89 = 0 FBD [187.94sin 7.679° − 68.404 cos 7.679°] = 32889 FBD = 770.6 N ΣFx = 0: (770.6 N) cos 7.679° = C x = 0 C x = + 763.7 N

Member CQ: ΣFx = 0: Q = C x = 763.7 N Q = 764 N


PROBLEM 6.125 Solve Problem 6.124 when (a) β = 0, (b) β = 6°. PROBLEM 6.124 The control rod CE passes through a horizontal hole in the body of the toggle system shown. Knowing that link BD is 250 mm long, determine the force Q required to hold the system in equilibrium when β = 20°.

SOLUTION We note that BD is a two-force member. (a)

β = 0:

Free body: Member ABC BD = 250 mm, sinθ =

Since

35 mm ; θ = 8.048° 250 mm

ΣM C = 0: (100 N)(350 mm) − FBD sin θ (200 mm) = 0 FBD = 1250 N ΣFx = 0: FBD cos θ − Cx = 0 (1250 N)(cos 8.048°) − C x = 0

C x = 1237.7 N

Member CE: ΣFx = 0: (1237.7 N) − Q = 0 Q = 1237.7 N

(b)

β = 6°

Q = 1238 N

Free body: Member ABC

Dimensions in mm

Since

14.094 mm 250 mm θ = 3.232°

BD = 250 mm, θ = sin −1

ΣM C = 0: ( FBD sin θ )198.90 + ( FBD cos θ )20.906 − (100 N)348.08 = 0 FBD [198.90sin 3.232° + 20.906 cos 3.232°] = 34808 FBD = 1084.8 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 923


ΣFx = 0: FBD cos θ − Cx = 0 (1084.8 N) cos 3.232° − C x = 0 C x = + 1083.1 N

Member DE:

ΣFx = 0: Q = C x Q = 1083.1 N

Q = 1083 N


!

PROBLEM 6.126 The press shown is used to emboss a small seal at E. Knowing that P = 250 N, determine (a) the vertical component of the force exerted on the seal, (b) the reaction at A.

SOLUTION FBD Stamp D: ΣFy = 0: E − FBD cos 20° = 0, E = FBD cos 20°

FBD ABC:

ΣM A = 0: (0.2 m)(sin 30°)( FBD cos 20°) + (0.2 m)(cos 30°)( FBD sin 20°) − [(0.2 m)sin 30° + (0.4 m) cos15°](250 N) = 0 FBD = 793.64 N C

and, from above, E = (793.64 N) cos 20°

(a)

E = 746 N

ΣFx = 0: Ax − (793.64 N)sin 20° = 0 A x = 271.44 N

ΣFy = 0: Ay + (793.64 N) cos 20° − 250 N = 0 A y = 495.78 N

so

(b)

A = 565 N

61.3°


PROBLEM 6.127 The press shown is used to emboss a small seal at E. Knowing that the vertical component of the force exerted on the seal must be 900 N, determine (a) the required vertical force P, (b) the corresponding reaction at A.

SOLUTION FBD Stamp D: ΣFy = 0: 900 N − FBD cos 20° = 0, FBD = 957.76 N C

(a) FBD ABC:

ΣM A = 0: [(0.2 m)(sin 30°)](957.76 N) cos 20° + [(0.2 m)(cos 30°)](957.76 N) sin 20° − [(0.2 m)sin 30° + (0.4 m) cos15°]P = 0 P = 301.70 N,

(b)

P = 302 N

ΣFx = 0: Ax − (957.76 N)sin 20° = 0 A x = 327.57 N

ΣFy = 0: −Ay + (957.76 N) cos 20° − 301.70 N = 0 A y = 598.30 N

so

A = 682 N

61.3°


PROBLEM 6.128 Water pressure in the supply system exerts a downward force of 135 N on the vertical plug at A. Determine the tension in the fusible link DE and the force exerted on member BCE at B.

SOLUTION Free body: Entire linkage + ΣFy = 0: C − 135 = 0 C = + 135 N

Free body: Member BCE ΣFx = 0: Bx = 0 ΣM B = 0: (135 N)(6 mm) − TDE (10 mm) = 0 TDE = 81.0 N ΣFy = 0: 135 + 81 − By = 0 By = + 216 N

B = 216 N


!

PROBLEM 6.129 A couple M of magnitude 1.5 kN ⋅ m is applied to the crank of the engine system shown. For each of the two positions shown, determine the force P required to hold the system in equilibrium.

SOLUTION (a)

FBDs:

50 mm 175 mm 2 = 7

Dimensions in mm

Note:

tan θ =

FBD whole:

ΣM A = 0: (0.250 m)C y − 1.5 kN ⋅ m = 0 C y = 6.00 kN

FBD piston:

ΣFy = 0: C y − FBC sin θ = 0 FBC =

Cy

sin θ

=

6.00 kN sinθ

ΣFx = 0: FBC cos θ − P = 0 P = FBC cos θ =

6.00 kN = 7 kips tan θ P = 21.0 kN



(b)

FBDs:

Dimensions in mm

2 as above 7

Note:

tan θ =

FBD whole:

ΣM A = 0: (0.100 m)C y − 1.5 kN ⋅ m = 0 C y = 15 kN ΣFy = 0: C y − FBC sin θ = 0 FBC = ΣFx = 0:

Cy

sin θ

FBC cos θ − P = 0 P = FBC cos θ =

Cy

tan θ

=

15 kN 2/7

P = 52.5 kN


PROBLEM 6.130 A force P of magnitude 16 kN is applied to the piston of the engine system shown. For each of the two positions shown, determine the couple M required to hold the system in equilibrium.

SOLUTION (a)

FBDs:

Note:

FBD piston:

50 mm tan θ = 175 mm 2 = 7

Dimensions in mm

ΣFx = 0: FBC cos θ − P = 0 FBC =

P cos θ

ΣFy = 0: C y − FBC sin θ = 0 C y = FBC sin θ = P tan θ =

FBD whole:

2 P 7

ΣM A = 0: (0.250 m)C y − M = 0

2! M = (0.250 m) " # (16 kN) 7 $ % = 1.14286 kN ⋅ m

M = 1143 N ⋅ m



(b)

FBDs:

Dimensions in mm

tan θ =

Note: FBD piston: as above FBD whole:

2 as above 7

C y = P tan θ =

2 P 7

2 ΣM A = 0: (0.100 m)C y − M = 0 M = (0.100 m) (16 kN) 7 M = 0.45714 kN ⋅ m

M = 457 N ⋅ m


PROBLEM 6.131 The pin at B is attached to member ABC and can slide freely along the slot cut in the fixed plate. Neglecting the effect of friction, determine the couple M required to hold the system in equilibrium when θ = 30°.

SOLUTION Free body: Member ABC ΣM C = 0: (25 lb)(13.856 in.) − B(3 in.) = 0 B = +115.47 lb ΣFy = 0: − 25 lb + C y = 0 C y = +25 lb ΣFx = 0: 115.47 lb − Cx = 0 Cx = +115.47 lb

Free body: Member CD

β = sin −1

5.196 ; β = 40.505° 8

CD cos β = (8 in.) cos 40.505° = 6.083 in. ΣM D = 0: M − (25 lb)(5.196 in.) − (115.47 lb)(6.083 in.) = 0 M = +832.3 lb ⋅ in. M = 832 lb ⋅ in.


PROBLEM 6.132 The pin at B is attached to member ABC and can slide freely along the slot cut in the fixed plate. Neglecting the effect of friction, determine the couple M required to hold the system in equilibrium when θ = 60°.

SOLUTION Free body: Member ABC ΣM C = 0: (25 lb)(8 in.) − B(5.196 in.) = 0 B = +38.49 lb ΣFx = 0: 38.49 lb − C x = 0 Cx = +38.49 lb ΣFy = 0: − 25 lb + C y = 0 C y = +25 lb

Free body: Member CD 3 8

β = sin −1 ; β = 22.024° CD cos β = (8 in.) cos 22.024° = 7.416 in. ΣM D = 0: M − (25 lb)(3 in.) − (38.49 lb)(7.416 in.) = 0 M = +360.4 lb ⋅ in.

M = 360 lb ⋅ in.


PROBLEM 6.133 Arm ABC is connected by pins to a collar at B and to crank CD at C. Neglecting the effect of friction, determine the couple M required to hold the system in equilibrium when θ = 0.

SOLUTION Free body: Member ABC ΣFx = 0: Cx − 240 N = 0 Cx = +240 N ΣM C = 0: (240 N)(500 mm) − B(160 mm) = 0 B = +750 N ΣFy = 0: C y − 750 N = 0 C y = +750 N

Free body: Member CD ΣM D = 0: M + (750 N)(300 mm) − (240 N)(125 mm) = 0 M = −195 × 103 N ⋅ mm M = 195.0 kN ⋅ m


PROBLEM 6.134 Arm ABC is connected by pins to a collar at B and to crank CD at C. Neglecting the effect of friction, determine the couple M required to hold the system in equilibrium when θ = 90°.

SOLUTION Free body: Member ABC ΣFx = 0: Cx = 0 ΣM B = 0: C y (160 mm) − (240 N)(90 mm) = 0 C y = +135 N

Free body: Member CD ΣM D = 0: M − (135 N)(300 mm) = 0 M = +40.5 × 103 N ⋅ mm M = 40.5 kN ⋅ m


PROBLEM 6.135 Two rods are connected by a slider block as shown. Neglecting the effect of friction, determine the couple MA required to hold the system in equilibrium.

SOLUTION D ⊥ AC

Note: Member FBDs:

ΣM B = 0:

lD sin115° − 250 lb ⋅ in. = 0 D=

250 lb ⋅ in. 2(15 in.) cos 25° sin115°

= 10.1454 lb ΣM A = 0:

M A − (15 in.)(10.1454 lb) = 0 M A = 152.2 lb ⋅ in.

!

! ! !


PROBLEM 6.136 Two rods are connected by a slider block as shown. Neglecting the effect of friction, determine the couple MA required to hold the system in equilibrium.

SOLUTION C ⊥ BD

Note: Member FBD’s:

ΣM B = 0: lC − 250 lb ⋅ in. = 0 C=

250 lb ⋅ in. 2(15 in.) cos 25°

C = 9.1948 lb ΣM A = 0: M A − (15 in.)C sin 65° = 0 M A = (15 in.)(9.1948 lb) sin 65° M A = 125.0 lb ⋅ in.

!

!


PROBLEM 6.137 Rod CD is attached to the collar D and passes through a collar welded to end B of lever AB. Neglecting the effect of friction, determine the couple M required to hold the system in equilibrium when θ = 30°.

SOLUTION B ⊥ CD

Note: FBD DC:

ΣFx′ = 0: D y sin 30° − (300 N) cos 30° = 0 Dy =

300 N = 519.62 N tan 30°

FBD machine:

ΣM A = 0:

0.200 m 519.62 N − M = 0 sin 30° M = 207.85 N ⋅ m

M = 208 N ⋅ m


PROBLEM 6.138 Rod CD is attached to the collar D and passes through a collar welded to end B of lever AB. Neglecting the effect of friction, determine the couple M required to hold the system in equilibrium when θ = 30°.

SOLUTION FBD DC: ΣFx′ = 0:

D y sin 30° − (150 N) cos 30° = 0 Dy = (150 N) ctn 30° = 259.81 N

FBD machine: ΣM A = 0: (0.100 m)(150 N) + d (259.81 N) − M = 0 d = b − 0.040 m b= so

0.030718 m tan 30

b = 0.053210 m d = 0.0132100 m M = 18.4321 N ⋅ m M = 18.43 N ⋅ m.


PROBLEM 6.139 Two hydraulic cylinders control the position of the robotic arm ABC. Knowing that in the position shown the cylinders are parallel, determine the force exerted by each cylinder when P = 160 N and Q = 80 N.

SOLUTION Free body: Member ABC

ΣM B = 0:

4 FAE (150 mm) − (160 N)(600 mm) = 0 5

FAE = +800 N

FAE = 800 N T

3 ΣFx = 0: − (800 N) + Bx − 80 N = 0 5

Bx = +560 N 4 ΣFy = 0: − (800 N) + By − 160 N = 0 5

By = +800 N

Free body: Member BDF ΣM F = 0: (560 N)(400 mm) − (800 N)(300 mm) − FDG = −100 N

4 FDG (200 mm) = 0 5

FDG = 100.0 N C


PROBLEM 6.140 Two hydraulic cylinders control the position of the robotic arm ABC. In the position shown, the cylinders are parallel and both are in tension. Knowing the FAE = 600 N and FDG = 50 N, determine the forces P and Q applied at C to arm ABC.

SOLUTION Free body: Member ABC ΣM B = 0:

4 (600 N)(150 mm) − P(600 mm) = 0 5

P = +120 N ΣM C = 0:

P = 120.0 N

4 (600)(750 mm) − By (600 mm) = 0 5

By = +600 N

Free body: Member BDF ΣM F = 0: Bx (400 mm) − (600 N)(300 mm) 4 − (50 N)(200 mm) = 0 5

Bx = +470 N

Return to free body: Member ABC 3 ΣFx = 0: − (600 N) + 470 N − Q = 0 5

Q = +110 N

Q = 110.0 N


!

PROBLEM 6.141 A log weighing 800 lb is lifted by a pair of tongs as shown. Determine the forces exerted at E and F on tong DEF.

SOLUTION FBD AB:

Ay = By = 400 lb

By symmetry:

Ax = Bx

and

6 (400 lb) 5 = 480 lb =

D = −B

Note:

Dx = 480 lb

so FBD DEF:

Dy = 400 lb ΣM F = (10.5 in.)(400 lb) + (15.5 in.)(480 lb) − (12 in.) Ex = 0 Ex = 970 lb

E = 970 lb

ΣFx = 0: − 480 lb + 970 lb − Fx = 0 Fx = 490 lb ΣFy = 0: 400 lb − Fy = 0 Fy = 400 lb F = 633 lb

39.2°


PROBLEM 6.142 A 39-ft length of railroad rail of weight 44 lb/ft is lifted by the tongs shown. Determine the forces exerted at D and F on tong BDF.

SOLUTION Free body: Rail W = (39 ft)(44 lb/ft) = 1716 lb

Free body: Upper link

By symmetry

1 E y = Fy = W = 858 lb 2

By symmetry,

1 ( FAB ) y = ( FAC ) y = W = 858 lb 2

Since AB is a two-force member, ( FAB ) x ( FAB ) y = 9.6 6

Free Body: Tong BDF

( FAB ) x =

9.6 (858) = 1372.8 lb 6

ΣM D = 0: (Attach FAB at A) Fx (8) − ( FAB ) x (18) − Fy (0.8) = 0 Fx (8) − (1372.8 lb)(18) − (858 lb)(0.8) = 0 Fx = +3174.6 lb

F = 3290 lb

15.12°

ΣFx = 0: − Dx + ( FAB ) x + Fx = 0 Dx = ( FAB ) x + Fx = 1372.8 + 3174.6 = 4547.4 lb ΣFy = 0: D y + ( FAB ) y − Fy = 0 Dy = 0

D = 4550 lb


!

PROBLEM 6.143 The tongs shown are used to apply a total upward force of 45 kN on a pipe cap. Determine the forces exerted at D and F on tong ADF.

SOLUTION FBD whole:

By symmetry

FBD ADF:

A = B = 22.5 kN

ΣM F = 0: (75 mm)CD − (100 mm)(22.5 kN) = 0 CD = 30.0 kN

ΣFx = 0: Fx − CD = 0 Fx = CD = 30 kN ΣFy = 0: 22.5 kN − Fy = 0 Fy = 22.5 kN F = 37.5 kN

so

36.9°


!

PROBLEM 6.144 If the toggle shown is added to the tongs of Problem 6.143 and a single vertical force is applied at G, determine the forces exerted at D and F on tong ADF.

SOLUTION Free body: Toggle By symmetry

Ay =

1 (45 kN) = 22.5 kN 2

AG is a two-force member Ax 22.5 kN = 22 mm 55 mm Ax = 56.25 kN

Free body: Tong ADF ΣFy = 0: 22.5 kN − Fy = 0 Fy = +22.5 kN ΣM F = 0 : D (75 mm) − (22.5 kN)(100 mm) − (56.25 kN)(160 mm) = 0 D = +150 kN

D = 150.0 kN

!

ΣFx = 0: 56.25 kN − 150 kN + Fx = 0 Fx = 93.75 kN

F = 96.4 kN

13.50°


PROBLEM 6.145 The pliers shown are used to grip a 0.3-in.-diameter rod. Knowing that two 60-lb forces are applied to the handles, determine (a) the magnitude of the forces exerted on the rod, (b) the force exerted by the pin at A on portion AB of the pliers.

SOLUTION Free body: Portion AB (a)

ΣM A = 0: Q (1.2 in.) − (60 lb)(9.5 in.) = 0 Q = 475 lb

(b)

!

ΣFx = 0: Q(sin 30°) + Ax = 0 (475 lb)(sin 30°) + Ax = 0 Ax = −237.5 lb

A y = 237.5 lb

ΣFy = 0: − Q (cos 30°) + Ay − 60 lb = 0 −(475 lb)(cos 30°) + Ay − 60 lb = 0 Ay = +471.4 lb

A y = 471.4 lb A = 528 lb

63.3°


PROBLEM 6.146 In using the bolt cutter shown, a worker applies two 300-N forces to the handles. Determine the magnitude of the forces exerted by the cutter on the bolt.

SOLUTION FBD cutter AB:

I FBD I:

FBD handle BC:

II

Dimensions in mm

ΣFx = 0: Bx = 0

FBD II:

ΣM C = 0: (12 mm)By − (448 mm)300 N = 0 By = 11, 200.0 N

Then

FBD I:

ΣM A = 0: (96 mm) By − (24 mm) F = 0

F = 4 By F = 44,800 N = 44.8 kN


PROBLEM 6.147 Determine the magnitude of the gripping forces exerted along line aa on the nut when two 50-lb forces are applied to the handles as shown. Assume that pins A and D slide freely in slots cut in the jaws.

SOLUTION FBD jaw AB:

ΣFx = 0: Bx = 0 ΣM B = 0: (0.5 in.)Q − (1.5 in.) A = 0 A=

Q 3

ΣFy = 0: A + Q − B y = 0 By = A + Q =

FBD handle ACE:

4Q 3

By symmetry and FBD jaw DE: Q 3 E x = Bx = 0 D= A=

E y = By =

4Q 3

ΣM C = 0: (5.25 in.)(50 lb) + (0.75 in.)

Q 4Q − (0.75 in.) =0 3 3 Q = 350 lb


PROBLEM 6.148 Determine the magnitude of the gripping forces produced when two 300-N forces are applied as shown.

SOLUTION We note that AC is a two-force member FBD handle CD: ΣM D = 0: − (126 mm)(300 N) − (6 mm)

2.8 8.84

A

! 1 + (30 mm) " A# = 0 $ 8.84 %

A = 2863.6 8.84 N

Dimensions in mm

FBD handle AB: ΣM B = 0: (132 mm)(300 N) − (120 mm)

1 8.84

(2863.6 8.84 N)

+ (36 mm) F = 0 F = 8.45 kN

!

Dimensions in mm


PROBLEM 6.149 Knowing that the frame shown has a sag at B of a = 1 in., determine the force P required to maintain equilibrium in the position shown.

SOLUTION We note that AB and BC are two-force members Free body: Toggle Cy =

By symmetry:

P 2 Cy

Cx = 10 in. a 10 10 P 5 P Cx = C y = ⋅ = a a 2 a

Free body: Member CDE ΣM E = 0: C x (6 in.) − C y (20 in.) − (50 lb)(10 in.) = 0 P 5P (b) − (20) = 500 a 2 30 ! P " − 10 # = 500 $ a %

For

(1)

a = 1.0 in. 30 ! P " − 10 # = 500 $ 1 % 20 P = 500

P = 25.0 lb


PROBLEM 6.150 Knowing that the frame shown has a sag at B of a = 0.5 in., determine the force P required to maintain equilibrium in the position shown.

SOLUTION We note that AB and BC are two-force members Free body: Toggle Cy =

By symmetry:

P 2 Cy

Cx = 10 in. a 10 10 P 5 P Cx = C y = ⋅ = a a 2 a

Free body: Member CDE ΣM E = 0: C x (6 in.) − C y (20 in.) − (50 lb)(10 in.) = 0 5P P (6) − (20) = 500 a 2 30 ! P " − 10 # = 500 $ a %

For

(1)

a = 0.5 in. 30 ! − 10 # = 500 P" $ 0.5 % 50 P = 500

P = 10.00 lb


PROBLEM 6.151 The garden shears shown consist of two blades and two handles. The two handles are connected by pin C and the two blades are connected by pin D. The left blade and the right handle are connected by pin A; the right blade and the left handle are connected by pin B. Determine the magnitude of the forces exerted on the small branch at E when two 80-N forces are applied to the handles as shown.

SOLUTION By symmetry vertical components Cy, Dy, Ey are 0. Then by considering ΣFy = 0 on the blades or handles, we find that Ay and By are 0. Thus forces at A, B, C, D, and E are horizontal. Free body: Right handle ΣM C = 0: A(30 mm) − (80 N)(270 mm) = 0 A = +720 N ΣFx = 0: C − 720 N − 80 N = 0 C = +800 N

Free body: Left blade ΣM D = 0: E (180 mm) − (720 N)(60 mm) = 0 E = 240 N !


PROBLEM 6.152 The telescoping arm ABC is used to provide an elevated platform for construction workers. The workers and the platform together have a mass of 200 kg and have a combined center of gravity located directly above C. For the position when θ = 20°, determine (a) the force exerted at B by the single hydraulic cylinder BD, (b) the force exerted on the supporting carriage at A.

SOLUTION a = (5 m) cos 20° = 4.6985 m

Geometry:

b = (2.4 m) cos 20° = 2.2553 m c = (2.4 m)sin 20° = 0.8208 m d = b − 0.5 = 1.7553 m e = c + 0.9 = 1.7208 m tan β =

e 1.7208 ; β = 44.43° = d 1.7553

Free body: Arm ABC We note that BD is a two-force member W = (200 kg)(9.81 m/s 2 ) = 1.962 kN

(a)

ΣM A = 0: (1.962 kN)(4.6985 m) − FBD sin 44.43°(2.2553 m) + FBD cos 44.43(0.8208 m) = 0 9.2185 − FBD (0.9927) = 0: FBD = 9.2867 kN FBD = 9.29 kN

(b)

44.4° !

ΣFx = 0: Ax − FBD cos β = 0 Ax = (9.2867 kN) cos 44.43° = 6.632 kN

A x = 6.632 kN

ΣFy = 0: Ay − 1.962 kN + FBD sin β = 0 Ay = 1.962 kN − (9.2867 kN)sin 44.43° = −4.539 kN

A y = 4.539 kN A = 8.04 kN

34.4° !


PROBLEM 6.153 The telescoping arm ABC can be lowered until end C is close to the ground, so that workers can easily board the platform. For the position when θ = −20°, determine (a) the force exerted at B by the single hydraulic cylinder BD, (b) the force exerted on the supporting carriage at A.

SOLUTION Geometry:

a = (5 m) cos 20° = 4.6985 m b = (2.4 m) cos 20° = 2.2552 m c = (2.4 m)sin 20° = 0.8208 m d = b − 0.5 = 1.7553 m e = 0.9 − c = 0.0792 m tan β =

e 0.0792 ; β = 2.584° = d 1.7552

Free body: Arm ABC We note that BD is a two-force member W = (200 kg)(9.81 m/s 2 ) W = 1962 N = 1.962 kN

(a)

ΣM A = 0: (1.962 kN)(4.6985 m) − FBD sin 2.584°(2.2553 m) − FBD cos 2.584°(0.8208 m) = 0 9.2185 − FBD (0.9216) = 0 FBD = 10.003 kN FBD = 10.00 kN

(b)

2.58° !

ΣFx = 0: Ax − FBD cos β = 0 Ax = (10.003 kN) cos 2.583° = 9.993 kN

A x = 9.993 kN

ΣFy = 0: Ay − 1.962 kN + FBD sin β = 0 Ay = 1.962 kN − (10.003 kN)sin 2.583° = −1.5112 kN A y = 1.5112 kN A = 10.11 kN

8.60° !


PROBLEM 6.154 The position of member ABC is controlled by the hydraulic cylinder CD. Knowing that θ = 30°, determine for the loading shown (a) the force exerted by the hydraulic cylinder on pin C, (b) the reaction at B.

SOLUTION Geometry: In DBCD (CD ) 2 = (0.5)2 + (1.5) 2 − 2(0.5)(1.5) cos 60° CD = 1.3229 m

Law of cosines

sin β sin 60° = 0.5 m 1.3229 m sin β = 0.3273 β = 19.107°

Law of sines

Free body: Entire system Move force FCD along its line of action so it acts at D. (a)

ΣM B = 0: (10 kN)(0.69282 m) − FCD sin β (1.5 m) = 0 6.9282 kN ⋅ m − FCD sin 19.107°(1.5 m) = 0 FCD = 14.111 kN FCD = 14.11 kN

(b)

19.11° !

ΣFx = 0: Bx + FCD cos β = 0 Bx + (14.111 kN) cos 19.107° = 0 B x = 13.333 kN

Bx = −13.333 kN

ΣFy = 0: By − 10 kN − FCD sin 19.107° = 0

By − 10 kN − (14.111 kN)sin 19.107° = 0 By = +14.619 kN

B y = 14.619 kN B = 19.79 kN

47.6° !


PROBLEM 6.155 The motion of the bucket of the front-end loader shown is controlled by two arms and a linkage that are pin-connected at D. The arms are located symmetrically with respect to the central, vertical, and longitudinal plane of the loader; one arm AFJ and its control cylinder EF are shown. The single linkage GHDB and its control cylinder BC are located in the plane of symmetry. For the position and loading shown, determine the force exerted (a) by cylinder BC, (b) by cylinder EF.

SOLUTION Free body: Bucket ΣM J = 0: (4500 lb)(20 in.) − FGH (22 in.) = 0 FGH = 4091 lb

Free body: Arm BDH ΣM D = 0: − (4091 lb)(24 in.) − FBC (20 in.) = 0 FBC = −4909 lb FBC = 4.91 kips C !

Free body: Entire mechanism (Two arms and cylinders AFJE)

Note: Two arms thus 2 FEF 18 in. 65 in. β = 15.48°

tan β =

ΣM A = 0: (4500 lb)(123 in.) + FBC (12 in.) + 2 FEF cos β (24 in.) = 0 (4500 lb)(123 in.) − (4909 lb)(12 in.) + 2 FEF cos 15.48°(24 in.) = 0

FEF = −10.690 lb

FEF = 10.69 kips C !


PROBLEM 6.156 The bucket of the front-end loader shown carries a 3200-lb load. The motion of the bucket is controlled by two identical mechanisms, only one of which is shown. Knowing that the mechanism shown supports one-half of the 3200-lb load, determine the force exerted (a) by cylinder CD, (b) by cylinder FH.

SOLUTION Free body: Bucket (One mechanism) ΣM D = 0: (1600 lb)(15 in.) − FAB (16 in.) = 0

FAB = 1500 lb

Note: There are 2 identical support mechanisms. Free body: One arm BCE 8 20 β = 21.8°

tan β =

ΣM E = 0: (1500 lb)(23 in.) + FCD cos 21.8°(15 in.) − FCD sin 21.8°(5 in.) = 0

FCD = −2858 lb

FCD = 2.86 kips C !

Free body: Arm DFG

ΣM G = 0: (1600 lb)(75 in.) + FFH sin 45°(24 in.) − FFH cos 45°(6 in.) = 0

FFH = −9.428 kips

FFH = 9.43 kips C !


PROBLEM 6.157 The motion of the backhoe bucket shown is controlled by the hydraulic cylinders AD, CG, and EF. As a result of an attempt to dislodge a portion of a slab, a 2-kip force P is exerted on the bucket teeth at J. Knowing that θ = 45°, determine the force exerted by each cylinder.

SOLUTION Free body: Bucket ΣM H = 0

(Dimensions in inches) 4 3 FCG (10) + FCG (10) + P cos θ (16) + P sin θ (8) = 0 5 5

FCG = −

P (16 cos θ + 8 sin θ ) 14

(1)

Free body: Arm ABH and bucket (Dimensions in inches)

ΣM B = 0:

4 3 FAD (12) + FAD (10) + P cos θ (86) − P sin θ (42) = 0 5 5

FAD = −

P (86 cos θ − 42 sin θ ) 15.6

(2)

Free body: Bucket and arms IEB + ABH Geometry of cylinder EF

16 in. 40 in. β = 21.801°

tan β =



ΣM I = 0: FEF cos β (18 in.) + P cos θ (28 in.) − P sin θ (120 in.) = 0

FEF = =

P(120 sin θ − 28 cos θ ) cos 21.8°(18) P (120 sin θ − 28 cos θ ) 16.7126

(3)

For P = 2 kips, θ = 45° Eq. (1):

FCG = −

2 (16 cos 45° + 8 sin 45°) = −2.42 kips 14

FCG = 2.42 kips C !

Eq. (2):

FAD = −

2 (86 cos 45° − 42 sin 45°) = −3.99 kips 15.6

FAD = 3.99 kips C !

Eq. (3):

FEF =

2 (120 sin 45° − 28 cos 45°) = +7.79 kips 16.7126

FEF = 7.79 kips T !


PROBLEM 6.158 Solve Problem 6.157 assuming that the 2-kip force P acts horizontally to the right (θ = 0). PROBLEM 6.157 The motion of the backhoe bucket shown is controlled by the hydraulic cylinders AD, CG, and EF. As a result of an attempt to dislodge a portion of a slab, a 2-kip force P is exerted on the bucket teeth at J. Knowing that θ = 45°, determine the force exerted by each cylinder.

SOLUTION Free body: Bucket ΣM H = 0

(Dimensions in inches) 4 3 FCG (10) + FCG (10) + P cos θ (16) + P sin θ (8) = 0 5 5 FCG = −

P (16 cos θ + 8 sin θ ) 14

(1)

Free body: Arm ABH and bucket (Dimensions in inches)

ΣM B = 0:

4 3 FAD (12) + FAD (10) + P cos θ (86) − P sin θ (42) = 0 5 5 FAD = −

P (86 cos θ − 42 sin θ ) 15.6

(2)

Free body: Bucket and arms IEB + ABH Geometry of cylinder EF

16 in. 40 in. β = 21.801°

tan β =



ΣM I = 0: FEF cos β (18 in.) + P cos θ (28 in.) − P sin θ (120 in.) = 0 FEF = =

P(120 sin θ − 28 cos θ ) cos 21.8°(18) P (120 sin θ − 28 cos θ ) 16.7126

(3)

For P = 2 kips, θ = 0 Eq. (1):

FCG = −

2 (16 cos 0 + 8 sin 0) = −2.29 kips 14

Eq. (2):

FAD = −

2 (86 cos 0 − 42 sin 0) = −11.03 kips 15.6

FAD = 11.03 kips C !

Eq. (3):

FEF =

2 (120 sin 0 − 28 cos 0) = −3.35 kips 16.7126

FEF = 3.35 kips C !

FCG = 2.29 kips C !


PROBLEM 6.159 The gears D and G are rigidly attached to shafts that are held by frictionless bearings. If rD = 90 mm and rG = 30 mm, determine (a) the couple M0 that must be applied for equilibrium, (b) the reactions at A and B.

SOLUTION (a)

Projections on yz plane Free body: Gear G ΣM G = 0: 30 N ⋅ m − J (0.03 m) = 0; J = 1000 N

Free body: Gear D ΣM D = 0: M 0 − (1000 N)(0.09 m) = 0 M 0 = 90 N ⋅ m

(b)

M 0 = (90.0 N ⋅ m)i !

Gear G and axle FH ΣM F = 0: H (0.3 m) − (1000 N)(0.18 m) = 0 H = 600 N ΣFy = 0: F + 600 − 1000 = 0 F = 400 N

Gear D and axle CE ΣM C = 0: (1000 N)(0.18 m) − E (0.3 m) = 0 E = 600 N ΣFy = 0: 1000 − C − 600 = 0 C = 400 N



Free body: Bracket AE ΣFy = 0: A − 400 + 400 = 0

A=0 !

ΣM A = 0: M A + (400 N)(0.32 m) − (400 N)(0.2 m) = 0 M A = −48 N ⋅ m

M A = −(48.0 N ⋅ m)i !

Free body: Bracket BH ΣFy = 0: B − 600 + 600 = 0

B=0 !

ΣM B = 0: M B + (600 N)(0.32 m) − (600 N)(0.2 m) = 0 M B = −72 N ⋅ m

M B = −(72.0 N ⋅ m)i !


PROBLEM 6.160 In the planetary gear system shown, the radius of the central gear A is a = 18 mm, the radius of each planetary gear is b, and the radius of the outer gear E is (a + 2b). A clockwise couple of magnitude MA = 10 N ⋅ m is applied to the central gear A and a counterclockwise couple of magnitude MS = 50 N ⋅ m is applied to the spider BCD. If the system is to be in equilibrium, determine (a) the required radius b of the planetary gears, (b) the magnitude ME of the couple that must be applied to the outer gear E.

SOLUTION FBD Central Gear:

By symmetry:

F1 = F2 = F3 = F

ΣM A = 0: 3(rA F ) − 10 N ⋅ m = 0,

ΣM C = 0: rB ( F − F4 ) = 0,

FBD Gear C:

F=

10 N⋅m 3rA

F4 = F

ΣFx′ = 0: Cx′ = 0 ΣFy ′ = 0: C y′ − 2 F = 0,

C y′ = 2 F

Gears B and D are analogous, each having a central force of 2F

ΣM A = 0: 50 N ⋅ m − 3(rA + rB )2 F = 0

FBD Spider:

50 N ⋅ m − 3(rA + rB )

20 N⋅m = 0 rA

rA + rB r = 2.5 = 1 + B , rA rA

rB = 1.5rA

Since rA = 18 mm, FBD Outer Gear:

rB = 27.0 mm !

(a) ΣM A = 0: 3(rA + 2rB ) F − M E = 0 3(18 mm + 54 mm)

10 N ⋅ m − ME = 0 54 mm M E = 40.0 N ⋅ m

(b)

!


PROBLEM 6.161* Two shafts AC and CF, which lie in the vertical xy plane, are connected by a universal joint at C. The bearings at B and D do not exert any axial force. A couple of magnitude 500 lb ⋅ in. (clockwise when viewed from the positive x axis) is applied to shaft CF at F. At a time when the arm of the crosspiece attached to shaft CF is horizontal, determine (a) the magnitude of the couple that must be applied to shaft AC at A to maintain equilibrium, (b) the reactions at B, D, and E. (Hint: The sum of the couples exerted on the crosspiece must be zero.)

SOLUTION We recall from Figure 4.10, that a universal joint exerts on members it connects a force of unknown direction and a couple about an axis perpendicular to the crosspiece. Free body: Shaft DF

ΣM x = 0: M C cos 30° − 500 lb ⋅ in. = 0 M C = 577.35 lb ⋅ in.

Free body: Shaft BC We use here x′, y ′, z with x′ along BC

ΣM C = 0: − M R i ′ − (577.35 lb ⋅ in)i ′ + (−5 in)i ′ × ( By j ′ + Bz k ) = 0



Equate coefficients of unit vectors to zero: i:

M A − 577.35 lb ⋅ in = 0

M A = 577.35 lb ⋅ in.

j:

Bz = 0

k:

By = 0

M A = 577 lb ⋅ in. ! B=0

ΣF = 0:

B + C = 0,

B=0

since B = 0,

C=0

Return to free body of shaft DF ΣM D = 0

(Note that C = 0 and M C = 577.35 lb ⋅ in. )

(577.35 lb ⋅ in.)(cos 30°i + sin 30° j) − (500 lb ⋅ in.)i + (6 in.)i × ( Ex i + E y j + Ez k ) = 0 (500 lb ⋅ in.)i + (288.68 lb ⋅ in.) j − (500 lb ⋅ in.)i + (6 in.) E y k − (6 in.) Ez j = 0

Equate coefficients of unit vectors to zero: j:

288.68 lb ⋅ in. − (6 in.)Ez = 0 Ez = 48.1 lb

k:

Ey = 0 ΣF = 0: C + D + E = 0 0 + Dy j + Dz k + Ex i + (48.1 lb)k = 0

i:

Ex = 0

j:

Dy = 0

k:

Dz + 48.1 lb = 0

Dz = −48.1 lb B=0 !

Reactions are:

D = −(48.1 lb)k ! E = (48.1 lb)k !


PROBLEM 6.162* Solve Problem 6.161 assuming that the arm of the crosspiece attached to shaft CF is vertical. PROBLEM 6.161 Two shafts AC and CF, which lie in the vertical xy plane, are connected by a universal joint at C. The bearings at B and D do not exert any axial force. A couple of magnitude 500 lb · in. (clockwise when viewed from the positive x axis) is applied to shaft CF at F. At a time when the arm of the crosspiece attached to shaft CF is horizontal, determine (a) the magnitude of the couple that must be applied to shaft AC at A to maintain equilibrium, (b) the reactions at B, D, and E. (Hint: The sum of the couples exerted on the crosspiece must be zero.)

SOLUTION Free body: Shaft DF

ΣM x = 0: M C − 500 lb ⋅ in. = 0 M C = 500 lb ⋅ in.

Free body: Shaft BC

We resolve −(520 lb ⋅ in.)i into components along x′ and y ′ axes: −M C = −(500 lb ⋅ in.)(cos 30°i ′ + sin 30° j′) ΣM C = 0: M Ai ′ − (500 lb ⋅ in.)(cos 30°i ′ + sin 30° j′) + (5 in.)i ′ × ( By ′ j′ + Bz k ) = 0 M A i ′ − (433 lb ⋅ in.)i ′ − (250 lb ⋅ in.) j + (5 in.) By′k − (5 in.) Bz j′ = 0 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 967


Equate to zero coefficients of unit vectors: i ′: M A − 433 lb ⋅ in. = 0

M A = 433 lb ⋅ in. !

j′: − 250 lb ⋅ in. − (5 in.) Bz = 0

Bz = −50 lb

k : B y′ = 0 B = −(50 lb)k

Reactions at B: ΣF = 0: B − C = 0 −(50 lb)k − C = 0

C = −(50 lb)k

Return to free body of shaft DF: ΣM D = 0: (6 in.)i × ( Ex i + E y j + Ez k ) − (4 in.)i × (−50 lb)k − (500 lb ⋅ in.)i + (500 lb ⋅ in.)i = 0 (6 in.) E y k − (6 in.) Ez j − (200 lb ⋅ in.) j = 0 k : Ey = 0 j: −(6 in.) Ez − 200 lb ⋅ in. = 0

Ez = −33.3 lb

ΣF = 0: C + D + E = 0 −(50 lb)k + D y j + Dz k + Ex i − (33.3 lb)k = 0 i : Ex = 0 k : −50 lb − 33.3 lb + Dz = 0

Dz = 83.3 lb B = −(50 lb)k !

Reactions are:

D = (83.3 lb)k ! E = −(33.3 lb)k !


PROBLEM 6.163* The large mechanical tongs shown are used to grab and lift a thick 7500-kg steel slab HJ. Knowing that slipping does not occur between the tong grips and the slab at H and J, determine the components of all forces acting on member EFH. (Hint: Consider the symmetry of the tongs to establish relationships between the components of the force acting at E on EFH and the components of the force acting at D on CDF.)

SOLUTION Free body: Pin A

T = W = mg = (7500 kg)(9.81 m/s 2 ) = 73.575 kN

ΣFx = 0: ( FAB ) x = ( FAC ) x 1 ΣFy = 0: ( FAB ) y = ( FAC ) y = W 2

Also:

( FAC ) x = 2( FAC ) y = W

Free body: Member CDF 1 ΣM D = 0: W (0.3) + W (2.3) − Fx (1.8) − Fy (0.5 m) = 0 2

or

1.8 Fx + 0.5Fy = 1.45W

(1)



ΣFx = 0: Dx − Fx − W = 0 Ex − Fx = W

or

(2)

1 ΣFy = 0: Fy − Dy + W = 0 2 1 E y − Fy = W 2

or

(3)

Free body: Member EFH 1 ΣM E = 0: Fx (1.8) + Fy (1.5) − H x (2.3) + W (1.8 m) = 0 2 1.8 Fx + 1.5Fy = 2.3H x − 0.9W

or:

(4)

ΣFx = 0: Ex + Fx − H x = 0 Ex + Fx = H x

or

2 Fx = H x − W

Subtract (2) from (5): Subtract (4) from 3 × (1):

3.6 Fx = 5.25W − 2.3H x

Add (7) to 2.3 × (6):

8.2 Fx = 2.95W Fx = 0.35976W

(5) (6) (7)

(8)

Substitute from (8) into (1): (1.8)(0.35976W ) + 0.5Fy = 1.45W 0.5Fy = 1.45W − 0.64756W = 0.80244W Fy = 1.6049W

(9)

Substitute from (8) into (2):

Ex − 0.35976W = W ; Ex = 1.35976W

Substitute from (9) into (3):

1 E y − 1.6049W = W 2

From (5):

H x = Ex + Fx = 1.35976W + 0.35976W = 1.71952W

Recall that:

1 Hy = W 2

E y = 2.1049W



Since all expressions obtained are positive, all forces are directed as shown on the free-body diagrams. Substitute

W = 73.575 kN E x = 100.0 kN

E y = 154.9 kN !

Fx = 26.5 kN

Fy = 118.1 kN !

H x = 126.5 kN

H y = 36.8 kN !



SOLUTION Free body: Truss. C = D = 600 lb

From the symmetry of the truss and loading, we find

Free body: Joint B

FAB 5

=

FBC 300 lb = 2 1 FAB = 671 lb T

FBC = 600 lb C !

Free body: Joint C ΣFy = 0:

3 FAC + 600 lb = 0 5 FAC = −1000 lb

ΣFx = 0:

4 ( −1000 lb) + 600 lb + FCD = 0 5

FAC = 1000 lb C ! FCD = 200 lb T !

From symmetry: FAD = FAC = 1000 lb C , FAE = FAB = 671 lb T , FDE = FBC = 600 lb C !


PROBLEM 6.165 Using the method of joints, determine the force in each member of the double-pitch roof truss shown. State whether each member is in tension or compression.


ΣM A = 0: H (18 m) − (2 kN)(4 m) − (2 kN)(8 m) − (1.75 kN)(12 m) − (1.5 kN)(15 m) − (0.75 kN)(18 m) = 0 H = 4.50 kN

ΣFx = 0: Ax = 0 ΣFy = 0: Ay + H − 9 = 0 Ay = 9 − 4.50,

Ay = 4.50 kN

Free body: Joint A

FAB

5 FAB

FAC 3.50 kN = 2 1 = 7.8262 kN C

=

FAB = 7.83 kN C ! FAC = 7.00 kN T !




2 5

FBD +

2 5

(7.8262 kN) +

1 2

FBC = 0

FBD + 0.79057 FBC = −7.8262 kN

or ΣFy = 0:

1 5

FBD +

1 5

(7.8262 kN) −

1 2

(1)

FBC − 2 kN = 0

FBD − 1.58114 FBC = −3.3541

or

(2)

Multiply (1) by 2 and add (2): 3FBD = −19.0065 FBD = 6.34 kN C !

FBD = −6.3355 kN

Subtract (2) from (1): 2.37111FBC = −4.4721 FBC = −1.8861 kN

FBC = 1.886 kN C !

Free body: Joint C 2

ΣFy = 0:

FCD −

5

1 2

(1.8861 kN) = 0

FCD = +1.4911 kN

ΣFx = 0: FCE − 7.00 kN +

1 2

FCD = 1.491 kN T !

(1.8861 kN) +

1 5

(1.4911 kN) = 0

FCE = 5.000 kN

FCE = 5.00 kN T !

Free body: Joint D ΣFx = 0:

2 5

FDF + FDF

or ΣFy = 0:

or Add (1) and (2):

1 5

1

FDE +

2

(6.3355 kN) −

2 5 + 0.79057 FDE = −5.5900 kN

FDF −

1 2

FDE +

1 5

(6.3355 kN) −

5

(1.4911 kN) = 0

(1) 2 5

(1.4911 kN) − 2 kN = 0

FDF − 0.79057 FDE = −1.1188 kN (2)

2 FDF = −6.7088 kN FDF = −3.3544 kN


1

FDF = 3.35 kN C !

1.58114 FDE = −4.4712 kN FDE = −2.8278 kN

FDE = 2.83 kN C !



Free body: Joint F ΣFx = 0:

1

FFG +

2

2

(3.3544 kN) = 0

5

FFG = −4.243 kN

ΣFy = 0: −FEF − 1.75 kN +

1 5

FFG = 4.24 kN C !

(3.3544 kN) −

1 2

FEF = 2.750 kN

(−4.243 kN) = 0 FEF = 2.75 kN T !

Free body: Joint G ΣFx = 0:

1

FGH −

2

ΣFy = 0: −

1 2

1 2

(4.243 kN) = 0

FGH −

1 2

FEG −

1 2

(1)

(4.243 kN) − 1.5 kN = 0

FGH + FEG = −6.364 kN

or:

(2)

2 FGH = −10.607 FGH = −5.303


2

FEG +

FGH − FEG = −4.243 kN

or:

Add (1) and (2):

1

FGH = 5.30 kN C !

2 FEG = −2.121 kN FEG = −1.0605 kN

FEG = 1.061 kN C !

Free body: Joint H

FEH 3.75 kN = 1 1

We can also write:

FGH

2

=

FEH = 3.75 kN T !

3.75 kN 1

FGH = 5.30 kN C (Checks)


PROBLEM 6.166 The truss shown was designed to support the roof of a food market. For the given loading, determine the force in members FG, EG, and EH.

SOLUTION Reactions at supports. Because of the symmetry of the loading Ax = 0,

Ay = O =

1 (Total load) 2

A = O = 4.48 kN !

We pass a section through members FG, EG, and EH, and use the Free body shown. Slope FG = Slope FI = Slope EG =

1.75 m 6m

5.50 m 2.4 m

ΣM E = 0: (0.6 kN)(7.44 m) + (1.24 kN)(3.84 m) − (4.48 kN)(7.44 m) 6 ! FFG # (4.80 m) = 0 −" 6.25 $ % FFG = −5.231 kN

FFG = 5.23 kN C !

ΣM G = 0: FEH (5.50 m) + (0.6 kN)(9.84 m) + (1.24 kN)(6.24 m) + (1.04 kN)(2.4 m) −(4.48 kN)(9.84 m) = 0 ΣFy = 0:

FEH = 5.08 kN T !

5.50 1.75 FEG + (−5.231 kN) + 4.48 kN − 0.6 kN − 1.24 kN − 1.04 kN = 0 6.001 6.25 FEG = −0.1476 kN

FEG = 0.1476 kN C !


PROBLEM 6.167 The truss shown was designed to support the roof of a food market. For the given loading, determine the force in members KM, LM, and LN.

SOLUTION Because of symmetry of loading,

O=

1 (Load) 2

O = 4.48 kN

We pass a section through KM, LM, LN, and use free body shown 3.84 ! ΣM M = 0: " FLN # (3.68 m) 4 $ % + (4.48 kN − 0.6 kN)(3.6 m) = 0 FLN = −3.954 kN

FLN = 3.95 kN C !

ΣM L = 0: − FKM (4.80 m) − (1.24 kN)(3.84 m) + (4.48 kN − 0.6 kN)(7.44 m) = 0 FKM = +5.022 kN +ΣFy = 0:

FKM = 5.02 kN T !

4.80 1.12 (−3.954 kN) − 1.24 kN − 0.6 kN + 4.48 kN = 0 FLM + 6.147 4 FLM = −1.963 kN

FLM = 1.963 kN C !"


PROBLEM 6.168 For the frame and loading shown, determine the components of all forces acting on member ABC.

SOLUTION Free body: Entire frame ΣM E = 0: − Ax (4) − (20 kips)(5) = 0 Ax = −25 kips,

A x = 25.0 kips

!

ΣFy = 0: Ay − 20 kips = 0 Ay = 20 kips

A y = 20.0 kips !

Free body: Member ABC Note: BE is a two-force member, thus B is directed along line BE and By =

2 Bx 5

ΣM C = 0: (25 kips)(4 ft) − (20 kips)(10 ft) + Bx (2 ft) + By (5 ft) = 0 −100 kip ⋅ ft + Bx (2 ft) +

2 Bx (5 ft) = 0 5

Bx = 25 kips By =

B x = 25.0 kips

2 2 ( Bx ) = (25) = 10 kips 5 5

!

B y = 10.00 kips !

ΣFx = 0: C x − 25 kips − 25 kips = 0 Cx = 50 kips

C x = 50.0 kips

!

ΣFy = 0: C y + 20 kips − 10 kips = 0 C y = −10 kips

C y = 10.00 kips !"


PROBLEM 6.169 Solve Problem 6.168 assuming that the 20-kip load is replaced by a clockwise couple of magnitude 100 kip ⋅ ft applied to member EDC at Point D. PROBLEM 6.168 For the frame and loading shown, determine the components of all forces acting on member ABC.

SOLUTION Free body: Entire frame ΣFy = 0: Ay = 0 ΣM E = 0: − Ax (4 ft) − 100 kip ⋅ ft = 0 Ax = −25 kips

A x = 25.0 kips A = 25.0 kips

!

B x = 25.0 kips

!

Free Body: Member ABC Note: BE is a two-force member, thus B is directed along line BE and By =

2 Bx 5

ΣM C = 0: (25 kips)(4 ft) + Bx (2 ft) + By (5 ft) = 0 100 kip ⋅ ft + Bx (2 ft) +

2 Bx (5 ft) = 0 5

Bx = −25 kips By =

2 2 Bx = (−25) = −10 kips; 5 5

ΣFx = 0: − 25 kips + 25 kips + Cx = 0

B y = 10.00 kips !

Cx = 0

ΣFy = 0: +10 kips + C y = 0 C y = −10 kips

C y = 10 kips C = 10.00 kips !"


PROBLEM 6.170 Knowing that the pulley has a radius of 0.5 m, determine the components of the reactions at A and E.

SOLUTION FBD Frame: ΣM A = 0: (7 m) E y − (4.5 m)(700 N) = 0 ΣFy = 0: Ay − 700 N + 450 N = 0 ΣFx = 0: Ax − Ex = 0

E y = 450 N ! A y = 250 N !

Ax = Ex

Dimensions in m

FBD Member ABC: ΣM C = 0: (1 m)(700 N) − (1 m)(250 N) − (3 m) Ax = 0 A x = 150.0 N

so E x = 150.0 N

! !"

"


PROBLEM 6.171 For the frame and loading shown, determine the reactions at A, B, D, and E. Assume that the surface at each support is frictionless.

SOLUTION Free body: Entire frame ΣFx = 0: A − B + (1000 lb) sin 30° = 0 A − B + 500 = 0

(1)

ΣFy = 0: D + E − (1000 lb) cos 30° = 0 D + E − 866.03 = 0

(2)

Free body: Member ACE ΣM C = 0: − A(6 in.) + E (8 in.) = 0 E=

3 A 4

(3)

Free body: Member BCD ΣM C = 0: − D(8 in.) + B(6 in.) = 0 D=

3 B 4

(4)

Substitute E and D from (3) and (4) into (2): −

3 3 A + B − 866.06 = 0 4 4 A + B − 1154.71 = 0

(5)

(1)

A − B + 500 = 0

(5) + (6)

2 A − 654.71 = 0

A = 327.4 lb

A = 327 lb

!

(5) − (6)

2 B − 1654.71 = 0

B = 827.4 lb

B = 827 lb

!

(4)

D=

3 (827.4) 4

D = 620.5 lb

D = 621 lb !

(3)

E=

3 (327.4) 4

E = 245.5 lb

E = 246 lb !

(6)


PROBLEM 6.172 For the system and loading shown, determine (a) the force P required for equilibrium, (b) the corresponding force in member BD, (c) the corresponding reaction at C.


FBD I:

I: ΣM C = 0: R( FBD sin 30°) − [ R(1 − cos 30°)](100 N) − R(50 N) = 0 FBD = 126.795 N ΣFx = 0: −Cx + (126.795 N) cos 30° = 0

(b) FBD = 126.8 N T !" C x = 109.808 N

ΣFy = 0: C y + (126.795 N) sin 30° − 100 N − 50 N = 0 C y = 86.603 N

II:

(c) so C = 139.8 N

38.3° !"

FBD II: ΣM A = 0: aP − a[(126.795 N) cos 30°] = 0

(a ) P = 109.8 N

!


PROBLEM 6.173 A small barrel weighing 60 lb is lifted by a pair of tongs as shown. Knowing that a = 5 in., determine the forces exerted at B and D on tong ABD.

SOLUTION We note that BC is a two-force member. Free body: Tong ABD Bx By = 15 5

Bx = 3By

ΣM D = 0: By (3 in.) + 3By (5 in.) − (60 lb)(9 in.) = 0 By = 30 lb Bx = 3By : Bx = 90 lb ΣFx = 0: −90 lb + Dx = 0 ΣFy = 0: 60 lb − 30 lb − D y = 0

D x = 90 lb D y = 30 lb B = 94.9 lb

18.43° !"

D = 94.9 lb

18.43° !


PROBLEM 6.174 A 20-kg shelf is held horizontally by a self-locking brace that consists of two Parts EDC and CDB hinged at C and bearing against each other at D. Determine the force P required to release the brace.

SOLUTION Free body: Shelf W = (20 kg)(9.81 m/s 2 ) = 196.2 N ΣM A = 0: By (200 mm) − (196.2 N)(125 mm) = 0 By = 122.63 N

Free body: Portion ACB ΣM A = 0: − Bx (150 mm) − P(150 mm) − (122.63 N)(200 mm) = 0 Bx = −163.5 − P

(1)

ΣM C = 0: + (122.63 N)(92 mm) + Bx (94 mm) = 0 + (122.63 N)(92 mm) + (−163.5 − P)(44 mm) = 0 P = 92.9 N

P = 92.9 N

!


PROBLEM 6.175 The specialized plumbing wrench shown is used in confined areas (e.g., under a basin or sink). It consists essentially of a jaw BC pinned at B to a long rod. Knowing that the forces exerted on the nut are equivalent to a clockwise (when viewed from above) couple of magnitude 135 lb ⋅ in., determine (a) the magnitude of the force exerted by pin B on jaw BC, (b) the couple M0 that is applied to the wrench.

SOLUTION Free body: Jaw BC This is a two-force member Cy 1.5 in.

Free body: Nut

=

Cx C y = 2.4 C x in.

5 8

ΣFx = 0: Bx = C x

(1)

ΣFy = 0: By = C y = 2.4 Cx

(2)

ΣFx = 0: Cx = Dx ΣM = 135 lb ⋅ in.

Cx (1.125 in.) = 135 lb ⋅ in. Cx = 120 lb

(a)

Eq. (1):

Bx = Cx = 120 lb

Eq. (2):

By = C y = 2.4(120 lb) = 288 lb

(

B = Bx2 + By2

(b)

1/ 2

)

= (1202 + 2882 )1/ 2

B = 312 lb !"

Free body: Rod ΣM D = 0: − M 0 + By (0.625 in.) − Bx (0.375 in.) = 0 − M 0 + (288)(0.625) − (120)(0.375) = 0 M 0 = 135.0 lb ⋅ in.

!


CHAPTER 7

PROBLEM 7.1 Determine the internal forces (axial force, shearing force, and bending moment) at Point J of the structure indicated. Frame and loading of Problem 6.79.

SOLUTION From Problem 6.79: On JD

D = 80 lb

FBD of JD:

ΣFy = 0: V − 80 lb = 0 ΣFx = 0: F = 0

V = 80.0 lb F =0

ΣM J = 0: M − (80 lb)(6 in.) = 0

M = 480 lb ⋅ in.


PROBLEM 7.2 Determine the internal forces (axial force, shearing force, and bending moment) at Point J of the structure indicated. Frame and loading of Problem 6.80.

SOLUTION From Problem 6.80: On JD

D = 40 lb

FBD of JD:

ΣFy = 0: V − 40 lb = 0 ΣFx = 0: F = 0

V = 40.0 lb F =0

ΣM J = 0: M − (40 lb)(6 in.) = 0

M = 240 lb ⋅ in.


PROBLEM 7.3 Determine the internal forces at Point J when α = 90°.

SOLUTION Free body: Entire bracket

ΣM D = 0: (1.4 kN)(0.6 m) − A(0.175 m) = 0 A = + 4.8 kN A = 4.8 kN ΣFx = 0: Dx − 4.8 = 0

D x = 4.8 kN

ΣFy = 0: D y − 1.4 = 0

D y = 1.4 kN

Free body: JCD

ΣFx = 0: 4.8 kN − F = 0

F = 4.80 kN

ΣFy = 0: 1.4 kN − V = 0

V = 1.400 kN

ΣM J = 0: (4.8 kN)(0.2 m) + (1.4 kN)(0.3 m) − M = 0 M = + 1.38 kN ⋅ m

M = 1.380 kN ⋅ m


PROBLEM 7.4 Determine the internal forces at Point J when α = 0.

SOLUTION Free body: Entire bracket

ΣFy = 0: D y = 0

Dy = 0

ΣM A = 0: (1.4 kN)(0.375 m) − Dx (0.175 m) = 0 Dx = + 3 kN

D x = 3 kN

Free body: JCD

ΣFx = 0: 3 kN − F = 0

F = 3.00 kN

ΣFy = 0: − V = 0

V =0

ΣM J = 0: (3 kN)(0.2 m) − M = 0 M = + 0.6 kN ⋅ m

M = 0.600 kN ⋅ m


PROBLEM 7.5 Determine the internal forces at Point J of the structure shown.

SOLUTION FBD ABC:

ΣM D = 0: (0.375 m)(400 N) − (0.24 m)C y = 0

C y = 625 N ΣM B = 0: − (0.45 m)C x + (0.135 m)(400 N) = 0

C x = 120 N

FBD CJ:

ΣFy = 0: 625 N − F = 0 ΣFx = 0: 120 N − V = 0

F = 625 N V = 120.0 N

! !

ΣM J = 0: M − (0.225 m)(120 N) = 0 M = 27.0 N ⋅ m


PROBLEM 7.6 Determine the internal forces at Point K of the structure shown.

SOLUTION FBD AK: ΣFx = 0: V = 0 V=0 ΣFy = 0: F − 400 N = 0 F = 400 N ΣM K = 0: (0.135 m)(400 N) − M = 0 M = 54.0 N ⋅ m


PROBLEM 7.7 A semicircular rod is loaded as shown. Determine the internal forces at Point J.

SOLUTION FBD Rod: ΣM B = 0: Ax (2r ) = 0 Ax = 0 ΣFx′ = 0: V − (120 N) cos 60° = 0 V = 60.0 N

! FBD AJ: ΣFy′ = 0: F + (120 N)sin 60° = 0 F = −103.923 N F = 103.9 N ΣM J = 0: M − [(0.180 m)sin 60°](120 N) = 0 M = 18.7061 M = 18.71


!

PROBLEM 7.8 A semicircular rod is loaded as shown. Determine the internal forces at Point K.

SOLUTION FBD Rod: ΣFy = 0: B y − 120 N = 0 B y = 120 N ΣM A = 0: 2rBx = 0 B x = 0 ΣFx′ = 0: V − (120 N) cos 30° = 0 V = 103.923 N V = 103.9 N

FBD BK: ΣFy′ = 0: F + (120 N)sin 30° = 0 F = − 60 N F = 60.0 N ΣM K = 0: M − [(0.180 m)sin 30°](120 N) = 0 M = 10.80 N ⋅ m


PROBLEM 7.9 An archer aiming at a target is pulling with a 45-lb force on the bowstring. Assuming that the shape of the bow can be approximated by a parabola, determine the internal forces at Point J.

SOLUTION FBD Point A: By symmetry

T1 = T2

3 ! ΣFx = 0: 2 " T1 # − 45 lb = 0 T1 = T2 = 37.5 lb $5 %

Curve CJB is parabolic: x = ay 2 FBD BJ: At B : x = 8 in.

y = 32 in. 8 in. 1 a= = (32 in.) 2 128 in. x=

y2 128

Slope of parabola = tan θ =

At J: So

dx 2 y y = = dy 128 64

& 16 ' ) = 14.036° * 64 +

θ J = tan −1 ( α = tan −1

4 − 14.036° = 39.094° 3

ΣFx′ = 0: V − (37.5 lb) cos(39.094°) = 0

V = 29.1 lb


!


ΣFy ′ = 0: F + (37.5 lb) sin (39.094°) = 0

F = − 23.647

F = 23.6 lb

&3 ' &4 ' ΣMJ = 0: M + (16 in.) ( (37.5 lb) ) + [(8 − 2) in.] ( (37.5 lb) ) = 0 5 5 * + * + M = 540 lb ⋅ in.


!

PROBLEM 7.10 For the bow of Problem 7.9, determine the magnitude and location of the maximum (a) axial force, (b) shearing force, (c) bending moment. PROBLEM 7.9 An archer aiming at a target is pulling with a 45-lb force on the bowstring. Assuming that the shape of the bow can be approximated by a parabola, determine the internal forces at Point J.

SOLUTION Free body: Point A

3 ! ΣFx = 0: 2 " T # − 45 lb = 0 $5 %

T = 37.5 lb v

Free body: Portion of bow BC ΣFy = 0: FC − 30 lb = 0

FC = 30 lb v

ΣFx = 0: VC − 22.5 lb = 0

VC = 22.5 lb

v

M C = 960 lb ⋅ in.

v

ΣM C = 0: (22.5 lb)(32 in.) + (30 lb)(8 in.) − M C = 0

Equation of parabola x = ky 2

At B:

8 = k (32) 2

Therefore, equation is

x=

k=

1 128

y2 128

(1)

The slope at J is obtained by differentiating (1): dx =

2 y dy dx y = , tan θ = 128 dy 64

(2)



(a)

Maximum axial force ΣFV = 0: − F + (30 lb) cos θ − (22.5 lb)sin θ = 0

Free body: Portion bow CK F = 30 cos θ − 22.5sin θ

F is largest at C (θ = 0) Fm = 30.0 lb at C

(b)

Maximum shearing force ΣFV = 0: −V + (30 lb)sin θ + (22.5 lb) cos θ = 0 V = 30sin θ + 22.5cos θ

V is largest at B (and D) Where

1 θ = θ max = tan −1 " !# = 26.56°

$2%

Vm = 30sin 26.56° + 22.5cos 26.56°

(c)

Vm = 33.5 lb at B and D

Maximum bending moment ΣM K = 0: M − 960 lb ⋅ in. + (30 lb) x + (22.5 lb) y = 0 M = 960 − 30 x − 22.5 y

M is largest at C, where x = y = 0.

M m = 960 lb ⋅ in. at C


PROBLEM 7.11 Two members, each consisting of a straight and a quarter-circular portion of rod, are connected as shown and support a 75-lb load at A. Determine the internal forces at Point J.

SOLUTION Free body: Entire frame ΣM C = 0: (75 lb)(12 in.) − F (9 in.) = 0 F = 100 lb v

ΣFx = 0: C x = 0 ΣFy = 0: C y − 75 lb − 100 lb = 0 C y = + 175 lb

C = 175 lb v

Free body: Member BEDF ΣM B = 0: D (12 in.) − (100 lb)(15 in.) = 0 D = 125 lb v

ΣFx = 0: Bx = 0 ΣFy = 0: By + 125 lb − 100 lb = 0 By = − 25 lb

B = 25 lb v

Free body: BJ ΣFx = 0: F − (25 lb)sin 30° = 0 F = 12.50 lb

30.0°

V = 21.7 lb

60.0°

ΣFy = 0: V − (25 lb) cos 30° = 0

ΣM J = 0: − M + (25 lb)(3 in.) = 0 M = 75.0 lb ⋅ in.


PROBLEM 7.12 Two members, each consisting of a straight and a quartercircular portion of rod, are connected as shown and support a 75-lb load at A. Determine the internal forces at Point K.

SOLUTION Free body: Entire frame ΣM C = 0: (75 lb)(12 in.) − F (9 in.) = 0 F = 100 lb v

ΣFx = 0: C x = 0 ΣFy = 0: C y − 75 lb − 100 lb = 0 C y = + 175 lb

C = 175 lb v

Free body: Member BEDF ΣM B = 0: D (12 in.) − (100 lb)(15 in.) = 0 D = 125 lb v

ΣFx = 0: Bx = 0 ΣFy = 0: By + 125 lb − 100 lb = 0 By = − 25 lb

B = 25 lb v

Free body: DK We found in Problem 7.11 that D = 125 lb on BEDF. D = 125 lb on DK. v

Thus ΣFx = 0: F − (125 lb) cos 30° = 0

F = 108.3 lb

60.0°



ΣFy = 0: V − (125 lb)sin 30° = 0 V = 62.5 lb

30.0°

ΣM K = 0: M − (125 lb)d = 0 M = (125 lb)d = (125 lb)(0.8038 in.) = 100.5 lb ⋅ in. M = 100.5 lb ⋅ in.


PROBLEM 7.13 A semicircular rod is loaded as shown. Determine the internal forces at Point J knowing that θ = 30°.

SOLUTION FBD AB:

4 ! 3 ! ΣM A = 0: r " C # + r " C # − 2r (280 N) = 0 $5 % $5 % C = 400 N

ΣFx = 0: − Ax +

4 (400 N) = 0 5 A x = 320 N

3 ΣFy = 0: Ay + (400 N) − 280 N = 0 5 A y = 40.0 N

FBD AJ:

ΣFx′ = 0: F − (320 N) sin 30° − (40.0 N) cos 30° = 0 F = 194.641 N F = 194.6 N

60.0°

!

30.0°

!

ΣFy ′ = 0: V − (320 N) cos 30° + (40 N) sin 30° = 0 V = 257.13 N V = 257 N

ΣM 0 = 0: (0.160 m)(194.641 N) − (0.160 m)(40.0 N) − M = 0 M = 24.743 M = 24.7 N ⋅ m


PROBLEM 7.14 A semicircular rod is loaded as shown. Determine the magnitude and location of the maximum bending moment in the rod.

SOLUTION Free body: Rod ACB 4 3 ! ! ΣM A = 0: " FCD # (0.16 m) + " FCD # (0.16 m) $5 % $5 % − (280 N)(0.32 m) = 0

ΣFx = 0: Ax +

FCD = 400 N

v

A x = 320 N

v

4 (400 N) = 0 5

Ax = − 320 N 3 + ΣFy = 0: Ay + (400 N) − 280 N = 0 5 Ay = + 40.0 N

A y = 40.0 N v

Free body: AJ (For θ , 90°) ΣM J = 0: (320 N)(0.16 m) sin θ − (40.0 N)(0.16 m)(1 − cos θ ) − M = 0 M = 51.2sin θ + 6.4 cos θ − 6.4

(1)

For maximum value between A and C: dM = 0: 51.2 cos θ − 6.4sin θ = 0 dθ tan θ =

51.2 =8 6.4

θ = 82.87° v

Carrying into (1): M = 51.2sin 82.87° + 6.4cos82.87° − 6.4 = + 45.20 N ⋅ m

v



Free body: BJ (For θ . 90°) ΣM J = 0: M − (280 N)(0.16 m)(1 − cos φ ) = 0 M = (44.8 N ⋅ m)(1 − cos φ )

Largest value occurs for φ = 90°, that is, at C, and is M C = 44.8 N ⋅ m v

We conclude that M max = 45.2 N ⋅ m

for

θ = 82.9°


PROBLEM 7.15 Knowing that the radius of each pulley is 150 mm, that α = 20°, and neglecting friction, determine the internal forces at (a) Point J, (b) Point K.

SOLUTION Tension in cable = 500 N. Replace cable tension by forces at pins A and B. Radius does not enter computations: (cf. Problem 6.90) (a)

Free body: AJ

ΣFx = 0: 500 N − F = 0 F = 500 N

F = 500 N

ΣFy = 0: V − 500 N = 0 V = 500 N

V = 500 N

ΣM J = 0: (500 N)(0.6 m) = 0 M = 300 N ⋅ m

(b)

M = 300 N ⋅ m

Free body: ABK

ΣFx = 0: 500 N − 500 N + (500 N)sin 20° − V = 0 V = 171.01 N

V = 171.0 N



ΣFy = 0: − 500 N − (500 N) cos 20° + F = 0 F = 969.8 N

F = 970 N

ΣM K = 0: (500 N)(1.2 m) − (500 N)sin 20°(0.9 m) − M = 0 M = 446.1 N ⋅ m M = 446 N ⋅ m


PROBLEM 7.16 Knowing that the radius of each pulley is 150 mm, that α = 30°, and neglecting friction, determine the internal forces at (a) Point J, (b) Point K.

SOLUTION Tension in cable = 500 N. Replace cable tension by forces at pins A and B. Radius does not enter computations: (cf. Problem 6.90) (a)

Free body: AJ:

ΣFx = 0: 500 N − F = 0 F = 500 N

F = 500 N

ΣFy = 0: V − 500 N = 0 V = 500 N

V = 500 N

ΣM J = 0: (500 N)(0.6 m) = 0 M = 300 N ⋅ m

(b)

M = 300 N ⋅ m

FBD: Portion ABK

ΣFx = 0: 500 N − 500 N + (500 N)sin 30° − V ΣFy = 0: − 500 N − (500 N) cos 30° + F = 0 ΣM K = 0: (500 N)(1.2 m) − (500 N)sin 30°(0.9 m) − M = 0

V = 250 N F = 933 N M = 375 N ⋅ m


PROBLEM 7.17 Knowing that the radius of each pulley is 200 mm and neglecting friction, determine the internal forces at Point J of the frame shown.

SOLUTION Free body: Frame and pulleys ΣMA = 0: − Bx (1.8 m) − (360 N)(2.6 m) = 0 Bx = − 520 N

B x = 520 N

v

A x = 520 N

v

ΣFx = 0: Ax − 520 N = 0 Ax = + 520 N ΣFy = 0: Ay + By − 360 N = 0 Ay + By = 360 N

(1)

Free body: Member AE ΣME = 0: − Ay (2.4 m) − (360 N)(1.6 m) = 0 Ay = − 240 N

From (1):

A y = 240 N v

By = 360 N + 240 N By = + 600 N

B y = 600 N v

Free body: BJ We recall that the forces applied to a pulley may be applied directly to its axle. ΣFy = 0:

3 4 (600 N) + (520 N) 5 5 3 − 360 N − (360 N) − F = 0 5 F = + 200 N

F = 200 N



ΣFx = 0:

4 3 4 (600 N) − (520 N) − (360 N) + V = 0 5 5 5 V = + 120.0 N

V = 120.0 N

ΣM J = 0: (520 N)(1.2 m) − (600 N)(1.6 m) + (360 N)(0.6 m) + M = 0 M = + 120.0 N ⋅ m

M = 120.0 N ⋅ m


PROBLEM 7.18 Knowing that the radius of each pulley is 200 mm and neglecting friction, determine the internal forces at Point K of the frame shown.

SOLUTION Free body: Frame and pulleys ΣMA = 0: − Bx (1.8 m) − (360 N)(2.6 m) = 0 Bx = − 520 N

B x = 520 N

v

A x = 520 N

v

ΣFx = 0: Ax − 520 N = 0 Ax = + 520 N ΣFy = 0: Ay + By − 360 N = 0 Ay + By = 360 N

(1)

Free body: Member AE ΣME = 0: − Ay (2.4 m) − (360 N)(1.6 m) = 0 Ay = − 240 N

From (1):

A y = 240 N v

By = 360 N + 240 N By = + 600 N

B y = 600 N v

Free body: AK ΣFx = 0: 520 N − F = 0 F = + 520 N

F = 520 N

ΣFy = 0: 360 N − 240 N − V = 0 V = + 120.0 N

V = 120.0 N

ΣMK = 0: (240 N)(1.6 m) − (360 N)(0.8 m) − M = 0 M = + 96.0 N ⋅ m

M = 96.0 N ⋅ m


PROBLEM 7.19 A 5-in.-diameter pipe is supported every 9 ft by a small frame consisting of two members as shown. Knowing that the combined weight of the pipe and its contents is 10 lb/ft and neglecting the effect of friction, determine the magnitude and location of the maximum bending moment in member AC.

SOLUTION Free body: 10-ft section of pipe 4 ΣFx = 0: D − (90 lb) = 0 5

D = 72 lb

v

3 ΣFy = 0: E − (90 lb) = 0 5

E = 54 lb

v

Free body: Frame ΣM B = 0: − Ay (18.75 in.) + (72 lb)(2.5 in.) + (54 lb)(8.75 in.) = 0 Ay = + 34.8 lb

A y = 34.8 lb v

4 3 ΣFy = 0: By + 34.8 lb − (72 lb) − (54 lb) = 0 5 5 By = + 55.2 lb

B y = 55.2 lb v

3 4 ΣFx = 0: Ax + Bx − (72 lb) + (54 lb) = 0 5 5 Ax + Bx = 0

(1)

Free body: Member AC ΣM C = 0: (72 lb)(2.5 in.) − (34.8 lb)(12 in.) − Ax (9 in.) = 0 Ax = − 26.4 lb

A x = 26.4 lb

v

B x = 26.4 lb

v

Bx = − Ax = + 26.4 lb

From (1):



Free body: Portion AJ For x # 12.5 in. ( AJ # AD ) : 3 4 ΣM J = 0: (26.4 lb) x − (34.8 lb) x + M = 0 5 5 M = 12 x M max = 150 lb ⋅ in. for x = 12.5 in. M max = 150.0 lb ⋅ in. at D v For x . 12.5 in.( AJ . AD): 3 4 ΣM J = 0: (26.4 lb) x − (34.8 lb) x + (72 lb)( x − 12.5) + M = 0 5 5 M = 900 − 60 x M max = 150 lb ⋅ in. for x = 12.5 in. M max = 150.0 lb ⋅ in. at D

Thus:


PROBLEM 7.20 For the frame of Problem 7.19, determine the magnitude and location of the maximum bending moment in member BC. PROBLEM 7.19 A 5-in.-diameter pipe is supported every 9 ft by a small frame consisting of two members as shown. Knowing that the combined weight of the pipe and its contents is 10 lb/ft and neglecting the effect of friction, determine the magnitude and location of the maximum bending moment in member AC.

SOLUTION Free body: 10-ft section of pipe 4 ΣFx = 0: D − (90 lb) = 0 5

D = 72 lb

v

3 ΣFy = 0: E − (90 lb) = 0 5

E = 54 lb

v

Free body: Frame ΣM B = 0: − Ay (18.75 in.) + (72 lb)(2.5 in.) + (54 lb)(8.75 in.) = 0 Ay = + 34.8 lb

A y = 34.8 lb v

4 3 ΣFy = 0: By + 34.8 lb − (72 lb) − (54 lb) = 0 5 5 By = + 55.2 lb

B y = 55.2 lb v

3 4 ΣFx = 0: Ax + Bx − (72 lb) + (54 lb) = 0 5 5 Ax + Bx = 0

(1)

Free body: Member AC ΣM C = 0: (72 lb)(2.5 in.) − (34.8 lb)(12 in.) − Ax (9 in.) = 0 Ax = − 26.4 lb

From (1):

A x = 26.4 lb

v

B x = 26.4 lb

v

Bx = − Ax = + 26.4 lb



Free body: Portion BK For x # 8.75 in.( BK # BE ): 3 4 ΣM K = 0: (55.2 lb) x − (26.4 lb) x − M = 0 5 5 M = 12 x M max = 105.0 lb ⋅ in. for

x = 8.75 in.

M max = 105.0 lb ⋅ in. at E v For x . 8.75 in.( BK . BE ): 3 4 ΣM K = 0: (55.2 lb) x − (26.4 lb) x − (54 lb)( x − 8.75 in.) − M = 0 5 5 M = 472.5 − 42 x M max = 105.0 lb ⋅ in. for

x = 8.75 in.

M max = 105.0 lb ⋅ in. at E

Thus


PROBLEM 7.21 A force P is applied to a bent rod that is supported by a roller and a pin and bracket. For each of the three cases shown, determine the internal forces at Point J.

SOLUTION (a)

FBD Rod: ΣFx = 0:

Ax = 0

ΣM D = 0: aP − 2aAy = 0

Ay =

P 2

ΣFx = 0: V = 0

FBD AJ: ΣFy = 0:

P −F =0 2

F=

P 2

ΣM J = 0: M = 0

(b)

FBD Rod:

4 ! 3 ! ΣM A = 0: 2a " D # + 2a " D # − aP = 0 5 5 $ % $ % ΣFx = 0: Ax −

4 5 P=0 5 14

ΣFy = 0: Ay − P +

3 5 P=0 5 14

D=

5P 14

Ax =

2P 7

Ay =

11P 14



FBD AJ:

!

! ! (c)

ΣFx = 0:

2 P −V = 0 7

ΣFy = 0:

11P −F =0 14

ΣM J = 0: a

2P −M =0 7

ΣM A = 0:

a 4D ! − aP = 0 2 "$ 5 #%

V=

2P 7

F= M=

11P 14

2 aP 7

FBD Rod: D=

5P 2

Ax −

4 5P =0 5 2

Ax = 2 P

ΣFy = 0: Ay − P −

3 5P =0 5 2

Ay =

ΣFx = 0:

5P 2

FBD AJ: ΣFx = 0:

2P − V = 0

ΣFy = 0:

5P −F =0 2

ΣM J = 0: a(2 P) − M = 0

V = 2P F=

5P 2

M = 2aP

!


!

PROBLEM 7.22 A force P is applied to a bent rod that is supported by a roller and a pin and bracket. For each of the three cases shown, determine the internal forces at Point J.

SOLUTION (a)

FBD Rod:

ΣM D = 0: aP − 2aA = 0 A=

ΣFx = 0: V −

FBD AJ:

P =0 2

V=

ΣFy = 0:

P 2

F=0

ΣM J = 0: M − a

(b)

P 2

P =0 2

M=

aP 2

FBD Rod: ΣM D = 0: aP −

a 4 ! A =0 2 "$ 5 #% A=

5P 2



FBD AJ:

(c)

ΣFx = 0:

3 5P −V = 0 5 2

ΣFy = 0:

4 5P −F =0 5 2

V=

3P 2

F = 2P M=

3 aP 2

V=

3P 14

!

FBD Rod: 3 ! 4 ! ΣM D = 0: aP − 2a " A # − 2a " A # = 0 $5 % $5 % A= 3 5P ! ΣFx = 0: V − " #=0 $ 5 14 % ΣFy = 0:

4 5P −F =0 5 14

3 5P ! ΣM J = 0: M − a " #=0 5 $ 14 %

5P 14

F= M=

2P 7

3 aP 14


!

PROBLEM 7.23 A semicircular rod of weight W and uniform cross section is supported as shown. Determine the bending moment at Point J when θ = 60°.

SOLUTION FBD Rod: ΣM A = 0:

2r

π

W − 2rB = 0 B=

ΣFy′ = 0: F +

W

π

W W sin 60° − cos 60° = 0 3 π

F = − 0.12952W

FBD BJ:

W ! 3r W ! ΣM 0 = 0: r " F − # + +M =0 π % 2π "$ 3 #% $ 1 1 ! = 0.28868Wr M = Wr " 0.12952 + − 2 π π #% $

On BJ

M J = 0.289Wr


!

PROBLEM 7.24 A semicircular rod of weight W and uniform cross section is supported as shown. Determine the bending moment at Point J when θ = 150°.

SOLUTION FBD Rod: ΣFy = 0: Ay − W = 0 ΣM B = 0:

2r

π

Ay = W

W − 2rAx = 0 Ax =

W

π

FBD AJ:

ΣFx′ = 0:

W

π

cos 30° +

5W sin 30° − F = 0 F = 0.69233W 6

W! W! ΣM 0 = 0: 0.25587 r " # + r " F − # − M = 0 π % 6 $ % $ 1' & 0.25587 M = Wr ( + 0.69233 − ) 6 π+ * M = Wr (0.4166)

On AJ

M = 0.417Wr


!

PROBLEM 7.25 A quarter-circular rod of weight W and uniform cross section is supported as shown. Determine the bending moment at Point J when θ = 30°.

SOLUTION FBD Rod: ΣFx = 0: A x = 0 2r

ΣM B = 0:

FBD AJ:

π

W − rAy = 0

α = 15°, weight of segment = W r=

r

α

ΣFy ′ = 0:

sin α =

r π

Ay =

2W

π

30° W = 90° 3

sin15° = 0.9886r

12

2W

π

cos 30° −

W cos 30° − F = 0 3 F=

W 3 2 1! " − # 2 $π 3%

W 2W ! ΣM 0 = M + r " F − # + r cos15° 3 = 0 π $ %

M = 0.0557 Wr


PROBLEM 7.26 A quarter-circular rod of weight W and uniform cross section is supported as shown. Determine the bending moment at Point J when θ = 30°.

SOLUTION FBD Rod: ΣM A = 0: rB −

2r

π

W =0 B=

α = 15° =

FBD BJ:

r=

r π

2W

π

π 12

sin15° = 0.98862r

12

Weight of segment = W

30° W = 90° 3

ΣFy′ = 0: F −

W 2W cos 30° − sin 30° = 0 3 π

3 1! W + F=" " 6 π ## $ % ΣM 0 = 0: rF − ( r cos15°)

W −M =0 3 3 1! cos15° ! M = rW " + − 0.98862 #Wr " 6 π ## "$ 3 % $ %

M = 0.289Wr


PROBLEM 7.27 For the rod of Problem 7.25, determine the magnitude and location of the maximum bending moment. PROBLEM 7.25 A quarter-circular rod of weight W and uniform cross section is supported as shown. Determine the bending moment at Point J when θ = 30°.

SOLUTION ΣFx = 0: Ax = 0

FBD Rod:

ΣM B = 0:

2r

π

W − rAy = 0

α=

θ 2

Ay = r=

,


2α π

=

F=

FBD AJ:

2W ΣM 0 = 0: M + " F − π $ M=

But, so

sin α cos α = M=

2W

π

4α

π

W cos 2α +

2W

π

π r

α

4α

2

ΣFx′ = 0: − F −

2W

π

2W

π

sin α

W

cos 2α = 0

(1 − 2α ) cos 2α =

2W

π

(1 − θ ) cos θ

4α ! # r + (r cos α ) π W = 0 %

(1 + θ cos θ − cos θ ) r −

4αW r

π

α

sin α cos α

1 1 sin 2α = sin θ 2 2 2r

π

W (1 − cos θ + θ cos θ − sin θ )

dM 2rW = (sin θ − θ sin θ + cos θ − cos θ ) = 0 π dθ

for

(1 − θ )sin θ = 0 dM = 0 for θ = 0, 1, nπ ( n = 1, 2, ) dθ

Only 0 and 1 in valid range At

θ = 0 M = 0, at θ = 1 rad at

θ = 57.3°

M = M max = 0.1009Wr


PROBLEM 7.28 For the rod of Problem 7.26, determine the magnitude and location of the maximum bending moment. PROBLEM 7.26 A quarter-circular rod of weight W and uniform cross section is supported as shown. Determine the bending moment at Point J when θ = 30°.

SOLUTION FBD Bar: ΣM A = 0: rB −

2r

π

W =0

α=

B=

θ

r=

α

π 0#α #

so

2 r

2W

π 4

sin α


2α π 2

= ΣFx′ = 0: F − F= =

4α

π

π

W

W cos 2α −

2W

π 2W

π

4α

2W

π

sin 2α = 0

(sin 2α + 2α cos 2α ) (sin θ + θ cos θ )

FBD BJ: ΣM 0 = 0: rF − (r cos α ) M=

But,

sin α cos α =

2

π

4α

π

W −M =0

r ! 4α Wr (sin θ + θ cos θ ) − " sin α cos α # W α $ % π

1 1 sin 2α = sin θ 2 2



so

M=

or

M=

2Wr

π 2

π

(sin θ + θ cos θ − sin θ )

Wrθ cos θ

dM 2 = Wr (cos θ − θ sin θ ) = 0 at θ tan θ = 1 dθ π

Solving numerically

θ = 0.8603 rad M = 0.357Wr

and

at θ = 49.3° (Since M = 0 at both limits, this is the maximum)!


!

PROBLEM 7.29 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.

SOLUTION (a)

From A to B: ΣFy = 0:

V = −P

ΣM1 = 0: M = − Px

From B to C: ΣFy = 0: − P − P − V = 0

V = −2 P

ΣM 2 = 0: Px + P( x − a) + M = 0

(b)

|V | max = 2 P;

M = −2 Px + Pa

| M | max = 3Pa


!

PROBLEM 7.30 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.

SOLUTION (a)

Reactions: A=

2P 3

C=

P 3

From A to B: ΣFy = 0: V = +

2P 3

ΣM1 = 0: M = +

2P x 3

From B to C: ΣFy = 0: V = −

P 3

ΣM 2 = 0: M = +

(b)

|V | max =

2P ; 3

P ( L − x) 3

|M | max =

2 PL 9



SOLUTION FBD beam: (a)

Ay = D =

By symmetry:

1 L ( w) 2 2

Ay = D =

wL 4

Along AB: ΣFy = 0:

wL −V = 0 4

ΣM J = 0: M − x

V=

wL 4

wL =0 4 wL M= x (straight) 4

Along BC: ΣFy = 0:

wL − wx1 − V = 0 4 wL V= − wx1 4

Straight with ΣM k = 0: M +

Parabola with

V = 0 at

x1 =

L 4

x1 L ! wL wx1 − + x1 ! =0 2 "4 # 4 M=

% w $ L2 L + x1 − x12 !! 2" 8 2 #

M=

3 L wL2 at x1 = 32 4

Section CD by symmetry (b)

|V |max =

From diagrams:

wL on AB and CD 4

| M |max =

3wL2 at center 32



SOLUTION (a)

Along AB: ΣFy = 0: − wx − V = 0

Straight with

V =−

wL 2

ΣM J = 0: M +

Parabola with

x=

at

V = − wx

L 2

x 1 wx = 0 M = − wx 2 2 2

M =−

wL2 8

at

x=

L 2

Along BC: ΣFy = 0: − w

L −V = 0 2

1 V = − wL 2

L% L $ ΣM k = 0: M + x1 + ! w = 0 4# 2 " M =−

Straight with (b)

wL $ L % + x1 ! 2 "4 #

3 M = − wL2 8

From diagrams:

at

x1 =

L 2 |V |max =

wL on BC 2

|M |max =

3wL2 at C 8



SOLUTION Free body: Portion AJ ΣFy = 0: − P − V = 0 ΣM J = 0: M + Px − PL = 0

(a)

V = −P ! M = P( L − x) !

The V and M diagrams are obtained by plotting the functions V and M.

|V |max = P

(b)

| M |max = PL



SOLUTION (a)

FBD Beam:

ΣM C = 0: LAy − M 0 = 0 Ay =

M0 L

ΣFy = 0: − Ay + C = 0 C=

M0 L

Along AB: ΣFy = 0: −

M0 −V = 0 L V =−

ΣM J = 0: x

M0 L

M0 +M =0 L

Straight with

M =−

M0 x L

M =−

M0 at B 2

Along BC: ΣFy = 0: −

M0 −V = 0 L

ΣM K = 0: M + x

Straight with (b)

M=

V =−

M0 − M0 = 0 L

M0 at B 2

M0 L x% $ M = M0 1− ! L# "

M = 0 at C |V |max = P everywhere

From diagrams:

| M |max =

#

M0 at B 2



SOLUTION (a)

Just to the right of A: ΣFy = 0 V1 = +15 kN M1 = 0

Just to the left of C: V2 = +15 kN M 2 = +15 kN ⋅ m

Just to the right of C: V3 = +15 kN M 3 = +5 kN ⋅ m

Just to the right of D: V4 = −15 kN M 4 = +12.5 kN ⋅ m

Just to the right of E: V5 = −35 kN M 5 = +5 kN ⋅ m

At B:

(b)

M B = −12.5 kN ⋅ m

|V |max = 35.0 kN

| M |max = 12.50 kN ⋅ m



SOLUTION Free body: Entire beam

ΣM A = 0: B(3.2 m) − (40 kN)(0.6 m) − (32 kN)(1.5 m) − (16 kN)(3 m) = 0 B = +37.5 kN

B = 37.5 kN !#

ΣFx = 0: Ax = 0 ΣFy = 0: Ay + 37.5 kN − 40 kN − 32 kN − 16 kN = 0 Ay = +50.5 kN

(a)

A = 50.5 kN !

Shear and bending moment.

Just to the right of A:

V1 = 50.5 kN

M1 = 0 !

Just to the right of C: ΣFy = 0: 50.5 kN − 40 kN − V2 = 0 ΣM 2 = 0: M 2 − (50.5 kN)(0.6 m) = 0

V2 = +10.5 kN ! M 2 = +30.3 kN ⋅ m !#

Just to the right of D: ΣFy = 0: 50.5 − 40 − 32 − V3 = 0 ΣM 3 = 0: M 3 − (50.5)(1.5) + (40)(0.9) = 0

V3 = −21.5 kN ! M 3 = +39.8 kN ⋅ m !



Just to the right of E: ΣFy = 0: V4 + 37.5 = 0

V4 = −37.5 kN !

ΣM 4 = 0: − M 4 + (37.5)(0.2) = 0

M 4 = +7.50 kN ⋅ m !

VB = M B = 0

At B:

!

|V |max = 50.5 kN

(b)

|M |max = 39.8 kN ⋅ m




ΣM A = 0: E (6 ft) − (6 kips)(2 ft) − (12 kips)(4 ft) − (4.5 kips)(8 ft) = 0 E = +16 kips

E = 16 kips !#

ΣFx = 0: Ax = 0 ΣFy = 0: Ay + 16 kips − 6 kips − 12 kips − 4.5 kips = 0 Ay = +6.50 kips

(a)

A = 6.50 kips !

Shear and bending moment Just to the right of A: V1 = +6.50 kips M1 = 0

!

Just to the right of C: ΣFy = 0: 6.50 kips − 6 kips − V2 = 0 ΣM 2 = 0: M 2 − (6.50 kips)(2 ft) = 0

V2 = +0.50 kips ! M 2 = +13 kip ⋅ ft !#

Just to the right of D: ΣFy = 0: 6.50 − 6 − 12 − V3 = 0 ΣM 3 = 0: M 3 − (6.50)(4) − (6)(2) = 0

V3 = +11.5 kips ! M 3 = +14 kip ⋅ ft !



Just to the right of E: ΣFy = 0: V4 − 4.5 = 0 ΣM 4 = 0: − M 4 − (4.5)2 = 0

V4 = +4.5 kips ! M 4 = −9 kip ⋅ ft !

VB = M B = 0

At B:

!

|V |max = 11.50 kips

(b)

| M |max = 14.00 kip ⋅ ft




ΣM C = 0: (120 lb)(10 in.) − (300 lb)(25 in.) + E (45 in.) − (120 lb)(60 in.) = 0 E = +300 lb

E = 300 lb !#

ΣFx = 0: Cx = 0 ΣFy = 0: C y + 300 lb − 120 lb − 300 lb − 120 lb = 0 C y = +240 lb

(a)

C = 240 lb !

Shear and bending moment Just to the right of A: ΣFy = 0: − 120 lb − V1 = 0

V1 = −120 lb, M 1 = 0 !

Just to the right of C: ΣFy = 0: 240 lb − 120 lb − V2 = 0 ΣM C = 0: M 2 + (120 lb)(10 in.) = 0

V2 = +120 lb ! M 2 = −1200 lb ⋅ in. !#

Just to the right of D: ΣFy = 0: 240 − 120 − 300 − V3 = 0 ΣM 3 = 0: M 3 + (120)(35) − (240)(25) = 0,

V3 = −180 lb ! M 3 = +1800 lb ⋅ in. !



Just to the right of E: +ΣFy = 0: V4 − 120 lb = 0 ΣM 4 = 0: − M 4 − (120 lb)(15 in.) = 0

V4 = +120 lb ! M 4 = −1800 lb ⋅ in. ! VB = M B = 0 !

At B:

|V |max = 180.0 lb

(b)

| M |max = 1800 lb ⋅ in.




ΣM A = 0: B(5 m) − (60 kN)(2 m) − (50 kN)(4 m) = 0 B = +64.0 kN

B = 64.0 kN !#

ΣFx = 0: Ax = 0 ΣFy = 0: Ay + 64.0 kN − 6.0 kN − 50 kN = 0 Ay = +46.0 kN

(a)

A = 46.0 kN !

Shear and bending-moment diagrams. From A to C: ΣFy = 0: 46 − V = 0 ΣM y = 0: M − 46 x = 0

V = +46 kN ! M = (46 x)kN ⋅ m !

From C to D: ΣFy = 0: 46 − 60 − V = 0

V = −14 kN !

ΣM j = 0: M − 46 x + 60( x − 2) = 0 M = (120 − 14 x)kN ⋅ m

For

x = 2 m: M C = +92.0 kN ⋅ m

!#

For

x = 3 m: M D = +78.0 kN ⋅ m

!



From D to B: ΣFy = 0: V + 64 − 25µ = 0 V = (25µ − 64)kN ΣM j = 0: 64 µ − (25µ )

$µ% !−M =0 "2# M = (64µ − 12.5µ 2 )kN ⋅ m

For

µ = 0: VB = −64 kN

MB = 0 !

|V |max = 64.0 kN

(b)

| M |max = 92.0 kN ⋅ m

#




ΣM B = 0: (60 kN)(6 m) − C (5 m) + (80 kN)(2 m) = 0 C = +104 kN

From A to C:

C = 104 kN

ΣFy = 0: 104 − 60 − 80 + B = 0

B = 36 kN

ΣFy = 0: − 30 x − V = 0

V = −30 x

ΣM1 = 0: (30 x)

$x% !+M =0 "2#

M = −15 x 2

From C to D:

ΣFy = 0: 104 − 60 − V = 0 ΣM 2 = 0: (60)( x − 1) − (104)( x − 2) + M = 0

V = +44 kN M = 44 x − 148



From D to B:

ΣFy = 0: V = −36 kN ΣM 3 = 0: (36)(7 − x) − M = 0 M = −36 x + 252

(b)

|V |max = 60.0 kN

| M |max = 72.0 kN ⋅ m.



SOLUTION (a)

By symmetry: Ay = B = 8 kips +

1 (4 kips)(5 ft) A y = B = 18 kips 2

Along AC: ΣFy = 0 : 18 kips − V = 0 V = 18 kips ΣM J = 0: M − x(18 kips) M = (18 kips) x M = 36 kip ⋅ ft at C ( x = 2 ft)

Along CD: ΣFy = 0: 18 kips − 8 kips − (4 kips/ft) x1 − V = 0 V = 10 kips − (4 kips/ft) x1 V = 0 at x1 = 2.5 ft (at center) ΣM K = 0: M +

x1 (4 kips/ft) x1 + (8 kips) x1 − (2 ft + x1 )(18 kips) = 0 2

M = 36 kip ⋅ ft + (10 kips/ft) x1 − (2 kips/ft) x12 M = 48.5 kip ⋅ ft at x1 = 2.5 ft

Complete diagram by symmetry (b)

|V |max = 18.00 kips on AC and DB

From diagrams:

|M |max = 48.5 kip ⋅ ft at center




ΣM A = 0: B (10 ft) − (15 kips)(3 ft) − (12 kips)(6 ft) = 0 B = +11.70 kips

B = 11.70 kips !

ΣFx = 0: Ax = 0 ΣFy = 0: Ay − 15 − 12 + 11.70 = 0 Ay = +15.30 kips

(a)

A = 15.30 kips !

Shear and bending-moment diagrams From A to C: ΣFy = 0: 15.30 − 2.5 x − V = 0 V = (15.30 − 2.5 x) kips

x! ΣM J = 0: M + (2.5 x) " # − 15.30 x = 0 2 $ % M = 15.30 x − 1.25 x 2

For x = 0:

VA = +15.30 kips

For x = 6 ft:

VC = +0.300 kip

MA = 0 ! M C = +46.8 kip ⋅ ft !



From C to B: ΣFy = 0: V + 11.70 = 0

V = −11.70 kips !

ΣM J = 0: 11.70µ − M = 0 M = (11.70µ ) kip ⋅ ft M C = +46.8 kip ⋅ ft !

For µ = 4 ft: For µ = 0:

MB = 0 !

(b)

|V |max = 15.30 kips

"

" " " " " " |M |max = 46.8 kip ⋅ ft


PROBLEM 7.43 Assuming the upward reaction of the ground on beam AB to be uniformly distributed and knowing that a = 0.3 m, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.

SOLUTION (a)

FBD Beam:

ΣFy = 0: w(1.5 m) − 2(3.0 kN) = 0 w = 4.0 kN/m

Along AC: ΣFy = 0: (4.0 kN/m) x − V = 0 V = (4.0 kN/m) x

x (4.0 kN/m) x = 0 2 M = (2.0 kN/m) x 2

ΣM J = 0: M −

Along CD: ΣFy = 0: (4.0 kN/m) x − 3.0 kN − V = 0 V = (4.0 kN/m) x − 3.0 kN ΣM K = 0: M + ( x − 0.3 m)(3.0 kN) −

x (4.0 kN/m) x = 0 2

M = 0.9 kN ⋅ m − (3.0 kN) x + (2.0 kN/m) x 2

Note: V = 0 at x = 0.75 m, where M = −0.225 kN ⋅ m Complete diagrams using symmetry. |V |max = 1.800 kN at C and D

(b)

|M |max = 0.225 kN ⋅ m at center


PROBLEM 7.44 Solve Problem 7.43 knowing that a = 0.5 m. PROBLEM 7.43 Assuming the upward reaction of the ground on beam AB to be uniformly distributed and knowing that a = 0.3 m, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.


ΣFy = 0: wg (1.5 m) − 3 kN − 3 kN = 0

(a)

wg = 4 kN/m !

Shear and bending moment From A to C: ΣFy = 0: 4 x − V = 0 V = (4 x) kN ΣM J = 0: M − (4 x)

x = 0, M = (2 x 2 ) kN ⋅ m 2

For x = 0: For x = 0.5 m: From C to D:

VA = M A = 0 ! VC = 2 kN,

M C = 0.500 kN ⋅ m !"

ΣFy = 0: 4 x − 3 kN − V = 0 V = (4 x − 3) kN ΣM J = 0: M + (3 kN)( x − 0.5) − (4 x)

x =0 2

M = (2 x 2 − 3 x + 1.5) kN ⋅ m

For x = 0.5 m:

VC = −1.00 kN,

For x = 0.75 m:

VC = 0,

For x = 1.0 m:

VC = 1.00 kN,

M C = 0.500 kN ⋅ m ! M C = 0.375 kN ⋅ m !" M C = 0.500 kN ⋅ m !



From D to B: ΣFy = 0: V + 4µ = 0 V = −(4µ ) kN ΣM J = 0: (4 µ )

µ 2

− M = 0, M = 2µ 2

For µ = 0: For µ = 0.5 m:

VB = M B = 0 ! VD = −2 kN,

M D = 0.500 kN ⋅ m !

(b)

|V |max = 2.00 kN

"

"

|M |max = 0.500 kN ⋅ m

"

" " " " " " " " "


PROBLEM 7.45 Assuming the upward reaction of the ground on beam AB to be uniformly distributed, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.


ΣFy = 0: wg (12 ft) − (3 kips/ft)(6 ft) = 0

(a)

wg = 1.5 kips/ft !

Shear and bending-moment diagrams. From A to C: ΣFy = 0: 1.5 x − 3x − V = 0 V = (−1.5 x) kips x x − (1.5 x) = 0 2 2 2 M = ( −0.75 x ) kip ⋅ ft

ΣM J = 0: M + (3x)

For x = 0:

VA = M A = 0 !

For x = 3 ft:

VC = −4.5 kips M C = −6.75 kip ⋅ ft !"

From C to D: ΣFy = 0: 1.5 x = 9 − V = 0, V = (1.5 x − 9) kips ΣM J = 0: M + 9( x − 1.5) − (1.5 x)

x =0 2

M = 0.75 x 2 − 9 x + 13.5

For x = 3 ft:

VC = −4.5 kips,

For x = 6 ft:

VC = 0,

M C = −6.75 kip ⋅ ft ! M C = −13.50 kip ⋅ ft !


PROBLEM 7.45 (Continued) For x = 9 ft :

VD = 4.5 kips, M D = −6.75 kip ⋅ ft ! VB = M B = 0 !

At B: (b)

|V |max = 4.50 kips

"

|M |max = 13.50 kip ⋅ ft

"

" " " Bending moment diagram consists of three distinct arcs of parabola.

Since entire diagram is below the x axis:

M # 0 everywhere


PROBLEM 7.46 Assuming the upward reaction of the ground on beam AB to be uniformly distributed, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.

SOLUTION Free body: Entire beam ΣFy = 0: wg (12 ft) − (3 kips/ft)(6 ft) = 0 wg = 1.5 kips/ft !

(a)

Shear and bending-moment diagrams from A to C: ΣFy = 0: 1.5 x − V = 0 ΣM J = 0: M − (1.5 x)

x 2

V = (1.5 x)kips M = (0.75 x 2 )kip ⋅ ft

For x = 0:

VA = M A = 0 !

For x = 3 ft: VC = 4.5 kips,

M C = 6.75 kip ⋅ ft !

From C to D: ΣFy = 0: 1.5 x − 3( x − 3) − V = 0 V = (9 − 1.5 x)kips ΣM J = 0: M + 3( x − 3)

x−3 x − (1.5 x) = 0 2 2

M = [0.75 x 2 − 1.5( x − 3)2 ]kip ⋅ ft

For x = 3 ft: VC = 4.5 kips,

M C = 6.75 kip ⋅ ft !

For x = 6 ft: VcL = 0,

McL= 13.50 kip ⋅ ft !

For x = 9 ft: VD = −4.5 kips,

M D = 6.75 kip ⋅ ft ! VB = M B = 0 !

At B:

|V |max = 4.50 kips

(b)

| M | max = 13.50 kip ⋅ ft

Bending-moment diagram consists or three distinct arcs of parabola, all located above the x axis. Thus: M $ 0 everywhere "


PROBLEM 7.47 Assuming the upward reaction of the ground on beam AB to be uniformly distributed and knowing that P = wa, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.

SOLUTION Free body: Entire beam ΣFy = 0: wg (4a) − 2wa − wa = 0 wg =

(a)

3 w ! 4

Shear and bending-moment diagrams From A to C: ΣFy = 0:

3 wx − wx − V = 0 4 1 V = − wx 4

ΣM J = 0 : M + ( wx)

x 3 !x − " wx # = 0 2 $4 %2

1 M = − wx 2 8

For x = 0:

VA = M A = 0 ! 1 VC = − wa 4

For x = a :

1 M C = − wa 2 ! 8

From C to D: ΣFy = 0:

3 wx − wa − V = 0 4 3 ! V = " x − a#w 4 $ %

a! 3 x! ΣM J = 0: M + wa " x − # − wx " # = 0 2 4 2 $ % $ % 3 a! M = wx 2 − wa " x − # 8 2% $

(1)



For x = a :

1 VC = − wa 4

1 M C = − wa 2 ! 8

For x = 2a :

1 VD = + wa 2

MD = 0 !

Because of the symmetry of the loading, we can deduce the values of V and M for the right-hand half of the beam from the values obtained for its left-hand half. |V |max =

(b)

To find | M |max , we differentiate Eq. (1) and set

dM dx

1 wa 2

= 0:

dM 3 4 = wx − wa = 0, x = a dx 4 3 2

wa 2 3 4 ! 4 1! M = w " a # − wa 2 " − # = − 8 $3 % 6 $3 2% |M |max =

1 2 wa 6

Bending-moment diagram consists of four distinct arcs of parabola."


PROBLEM 7.48 Solve Problem 7.47 knowing that P = 3wa. PROBLEM 7.47 Assuming the upward reaction of the ground on beam AB to be uniformly distributed and knowing that P = wa, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.

SOLUTION Free body: Entire beam ΣFy = 0: wg (4a ) − 2wa − 3wa = 0 wg =

(a)

5 w ! 4

Shear and bending-moment diagrams From A to C: ΣFy = 0:

5 wx − wx − V = 0 4 1 V = + wx 4

ΣM J = 0 : M + ( wx)

x 5 !x − " wx # = 0 2 $4 %2

1 M = + wx 2 8

For x = 0:

VA = M A = 0 ! 1 VC = + wa 4

For x = a :

1 M C = + wa 2 ! 8

From C to D: ΣFy = 0:

5 wx − wa − V = 0 4 5 ! V = " x − a#w $4 %

a! 5 x! ΣM J = 0: M + wa " x − # − wx " # = 0 2% 4 $2% $ M=

5 2 a! wx − wa " x − # 8 2% $

(1)



1 1 VC = + wa, M C = + wa 2 ! 8 4

For x = a : For x = 2a :

3 VD = + wa, M D = + wa 2 ! 2

Because of the symmetry of the loading, we can deduce the values of V and M for the right-hand half of the beam from the values obtained for its left-hand half. |V |max =

(b) To find | M |max , we differentiate Eq. (1) and set

dM dx

3 wa 2

= 0:

dM 5 = wx − wa = 0 dx 4 4 x = a , a (outside portion CD ) 5

The maximum value of | M | occurs at D: | M |max = wa 2

Bending-moment diagram consists of four distinct arcs of parabola."


PROBLEM 7.49 Draw the shear and bending-moment diagrams for the beam AB, and determine the shear and bending moment (a) just to the left of C, (b) just to the right of C.

SOLUTION Free body: Entire beam ΣM A = 0: B(0.6 m) − (600 N)(0.2 m) = 0 B = +200 N

B = 200 N !

ΣFx = 0: Ax = 0 ΣFy = 0: Ay − 600 N + 200 N = 0 Ay = +400 N

A = 400 N !

We replace the 600-N load by an equivalent force-couple system at C Just to the right of A: V1 = +400 N , M1 = 0 !

(a)

Just to the left of C: V2 = +400 N M 2 = (400 N)(0.4 m)

(b)

Just to the right of C: M 3 = (200 N)(0.2 m)

Just to the left of B:

M 2 = +160.0 N ⋅ m V3 = −200 N M 3 = + 40.0 N ⋅ m V4 = −200 N , M 4 = 0 !

"


PROBLEM 7.50 Two small channel sections DF and EH have been welded to the uniform beam AB of weight W = 3 kN to form the rigid structural member shown. This member is being lifted by two cables attached at D and E. Knowing that θ = 30° and neglecting the weight of the channel sections, (a) draw the shear and bending-moment diagrams for beam AB, (b) determine the maximum absolute values of the shear and bending moment in the beam.

SOLUTION FBD Beam + channels: (a)

T1 = T2 = T

By symmetry:

ΣFy = 0: 2T sin 60° − 3 kN = 0 T= T1x =

3 3 3

kN

2 3 3 T1 y = kN 2 M = (0.5 m)

FBD Beam:

3

2 3 = 0.433 kN ⋅ m

kN

With cable force replaced by equivalent force-couple system at F and G Shear Diagram:

V is piecewise linear dV ! " dx = −0.6 kN/m # with 1.5 kN $ %

discontinuities at F and H. VF − = −(0.6 kN/m)(1.5 m) = 0.9 kN + V increases by 1.5 kN to +0.6 kN at F

VG = 0.6 kN − (0.6 kN/m)(1 m) = 0

Finish by invoking symmetry



Moment diagram: M is piecewise parabolic dM ! " dx decreasing with V # $ %

with discontinuities of .433 kN at F and H. 1 M F − = − (0.9 kN)(1.5 m) 2 = −0.675 kN ⋅ m

M increases by 0.433 kN m to –0.242 kN ⋅ m at F+ M G = −0.242 kN ⋅ m +

1 (0.6 kN)(1 m) 2

= 0.058 kN ⋅ m

Finish by invoking symmetry |V |max = 900 N

(b)

+ − at F and G

M

max

= 675 N ⋅ m

at F and G


PROBLEM 7.51 Solve Problem 7.50 when θ = 60°. PROBLEM 7.50 Two small channel sections DF and EH have been welded to the uniform beam AB of weight W = 3 kN to form the rigid structural member shown. This member is being lifted by two cables attached at D and E. Knowing that θ = 30° and neglecting the weight of the channel sections, (a) draw the shear and bending-moment diagrams for beam AB, (b) determine the maximum absolute values of the shear and bending moment in the beam.

SOLUTION Free body: Beam and channels From symmetry: E y = Dy

E x = Dx = Dy tan θ

Thus:

(1)

ΣFy = 0: D y + E y − 3 kN = 0

D y = E y = 1.5 kN !

D x = (1.5 kN) tan θ

From (1):

E = (1.5 kN) tan θ

!

We replace the forces at D and E by equivalent force-couple systems at F and H, where M 0 = (1.5 kN tan θ )(0.5 m) = (750 N ⋅ m) tan θ

(2)

We note that the weight of the beam per unit length is w=

(a)

W 3 kN = = 0.6 kN/m = 600 N/m L 5m

Shear and bending moment diagrams From A to F: ΣFy = 0: − V − 600 x = 0 V = (−600 x)N ΣM J = 0: M + (600 x)

x = 0, M = ( −300 x 2 )N ⋅ m 2 VA = M A = 0 !

For x = 0: For x = 1.5 m:

VF = −900 N, M F = −675 N ⋅ m !



From F to H: ΣFy = 0: 1500 − 600 x − V = 0 "

"

V = (1500 − 600 x)N "

" ΣM J = 0: M + (600 x)

"

x − 1500( x − 1.5) − M 0 = 0 2

M = M 0 − 300 x 2 + 1500( x − 1.5)N ⋅ m "

For x = 1.5 m:

VF = +600 N, M F = ( M 0 − 675) N ⋅ m !

For x = 2.5 m:

VG = 0, M G = ( M 0 − 375) N ⋅ m !

From G To B, The V and M diagrams will be obtained by symmetry, (b)

Making θ = 60° in Eq. (2):

|V |max = 900 N

M 0 = 750 tan 60° = 1299 N ⋅ m M = 1299 − 675 = 624 N ⋅ m !

Thus, just to the right of F:

M G = 1299 − 375 = 924 N ⋅ m !

and

(b)

|V |max = 900 N

"

| M |max = 924 N ⋅ m

"

" " "


PROBLEM 7.52 Draw the shear and bending-moment diagrams for the beam AB, and determine the maximum absolute values of the shear and bending moment.

SOLUTION FBD whole: ΣM D = 0: (1.2 ft)(1.5 kips) − (1.2 ft)(6 kips) − (3.6 ft)(1.5 kips) + (1.6 ft)G = 0 G = 6.75 kips

(Dimensions in ft., loads in kips, moments in kips ⋅ ft) ΣFx = 0: − Dx + G = 0

D x = 6.75 kips

ΣFy = 0: D y − 1.5 kips − 6 kips − 1.5 kips = 0

D y = 9 kips

Beam AB, with forces D and G replaced by equivalent force/couples at C and F Along AD: ΣFy = 0: − 1.5 kips − V = 0 V = −1.5 kips ΣM J = 0: x(1.5 kips) + M = 0 M = −(1.5 kips) x M = −1.8 kips at x = 1.2 ft

Along DE: ΣFy = 0: −1.5 kips + 9 kips − V = 0 V = 7.5 kips ΣM K = 0: M + 5.4 kip ⋅ ft − x1 (9 kips) + (1.2 ft + x1 )(1.5 kips) = 0 M = 7.2 kip ⋅ ft + (7.5 kips) x1 M = 1.8 kip ⋅ ft at x1 = 1.2 ft



Along EF:

ΣFy = 0: V − 1.5 kips = 0 V = 1.5 kips ΣM N = 0: − M + 5.4 kip ⋅ ft − ( x4 + 1.2 ft)(1.5 kips) M = 3.6 kip ⋅ ft − (1.5 kips) x4 M = 1.8 kip ⋅ ft at x4 = 1.2 ft M = 3.6 kip ⋅ ft at x4 = 0

Along FB: ΣFy = 0: V − 1.5 kips = 0 V = 1.5 kips ΣM L = 0: − M − x3 (1.5 kips) = 0 M = (−1.5 kips) x3 M = −1.8 kip ⋅ ft at x3 = 1.2 ft |V |max = 7.50 kips on DE

From diagrams:

| M |max = 7.20 kip ⋅ ft at D +



SOLUTION

ΣFG = 0: T (9 in.) − (45 lb)(9 in.) − (120 lb)(21 in.) = 0

T = 325 lb

ΣFx = 0: − 325 lb + Gx = 0

G x = 325 lb

ΣFy = 0: G y − 45 lb − 120 lb = 0

G y = 165 lb

Equivalent loading on straight part of beam AB

From A to E: ΣFy = 0: V = +165 lb ΣM1 = 0: + 1625 lb ⋅ in. − (165 lb) x + M = 0 M = −1625 + 165 x

From E to F: ΣFy = 0: 165 − 45 − V = 0 V = +120 lb ΣM 2 = 0 : + 1625 lb ⋅ in. − (165 lb) x + (45 lb)( x − 9) + M = 0 M = −1220 + 120 x



From F to B:

ΣFy = 0 V = +120 lb ΣM 3 = 0: − (120)(21 − x) − M = 0 M = −2520 + 120 x

|V |max = 165.0 lb

| M |max = 1625lb ⋅ in.

"

"



SOLUTION Free body: Entire beam ΣM A = 0: B(0.9 m) − (400 N)(0.3 m) − (400 N)(0.6 m) − (400 N)(0.9 m) = 0 B = +800 N

B = 800 N !

ΣFx = 0: Ax = 0 ΣFy = 0: Ay + 800 N − 3(400 N) = 0 Ay = +400 N

A = 400 N !

We replace the loads by equivalent force-couple systems at C, D, and E. We consider successively the following F-B diagrams. V5 = −400 N

V1 = +400 N M1 = 0

M 5 = +180 N ⋅ m

V2 = +400 N

V6 = −400 N

M 2 = +60 N ⋅ m

M 6 = +60 N ⋅ m

V3 = 0

V7 = −800 N

M 3 = +120 N ⋅ m

M 7 = +120 N ⋅ m V8 = −800 N

V4 = 0 M 4 = +120 N ⋅ m

M8 = 0



(b)

|V |max = 800 N

|M |max = 180.0 N ⋅ m

"

" " "


PROBLEM 7.55 For the structural member of Problem 7.50, determine (a) the angle θ for which the maximum absolute value of the bending moment in beam AB is as small as possible, (b) the corresponding value of | M |max. (Hint: Draw the bending-moment diagram and then equate the absolute values of the largest positive and negative bending moments obtained.) PROBLEM 7.50 Two small channel sections DF and EH have been welded to the uniform beam AB of weight W = 3 kN to form the rigid structural member shown. This member is being lifted by two cables attached at D and E. Knowing that θ = 30° and neglecting the weight of the channel sections, (a) draw the shear and bendingmoment diagrams for beam AB, (b) determine the maximum absolute values of the shear and bending moment in the beam.

SOLUTION See solution of Problem 7.50 for reduction of loading or beam AB to the following:

M 0 = (750 N ⋅ m) tan θ !

where [Equation (2)] The largest negative bending moment occurs Just to the left of F: ΣM1 = 0: M 1 + (900 N) " $

1.5 m ! =0 2 #%

M1 = −675 N ⋅ m !

The largest positive bending moment occurs At G: ΣM 2 = 0: M 2 − M 0 + (1500 N)(1.25 m − 1 m) = 0 M 2 = M 0 − 375 N ⋅ m !

Equating M 2 and − M 1 : M 0 − 375 = +675 M 0 = 1050 N ⋅ m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1069


(a)

From Equation (2):

tan θ =

1050 = 1.400 750

θ = 54.5°

|M |max = 675 N ⋅ m

(b)


PROBLEM 7.56 For the beam of Problem 7.43, determine (a) the distance a for which the maximum absolute value of the bending moment in the beam is as small as possible, (b) the corresponding value of | M |max. (See hint for Problem 7.55.) PROBLEM 7.43 Assuming the upward reaction of the ground on beam AB to be uniformly distributed and knowing that a = 0.3 m, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.

SOLUTION Force per unit length exerted by ground: wg =

6 kN = 4 kN/m 1.5 m

The largest positive bending moment occurs Just to the left of C: ΣM1 = 0: M 1 = (4a)

a 2

M 1 = 2a 2 !

The largest negative bending moment occurs

At the center line: ΣM 2 = 0: M 2 + 3(0.75 − a ) − 3(0.375) = 0

Equating M1 and − M 2 :

M 2 = −(1.125 − 3a) !

2a 2 = 1.125 − 3a a 2 + 1.5a − 0.5625 = 0

(a)

Solving the quadratic equation:

(b)

Substituting:

a = 0.31066, | M |max = M1 = 2(0.31066) 2

a = 0.311 m | M |max = 193.0 N ⋅ m


"

PROBLEM 7.57 For the beam of Problem 7.47, determine (a) the ratio k = P/wa for which the maximum absolute value of the bending moment in the beam is as small as possible, (b) the corresponding value of Mmax. (See hint for Problem 7.55.) PROBLEM 7.47 Assuming the upward reaction of the ground on beam AB to be uniformly distributed and knowing that P = wa, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.

SOLUTION Free body: Entire beam ΣFy = 0: wg (4a) − 2wa − kwa = 0 wg = wg

Setting

w

w (2 + k ) 4

=α

(1)

k = 4α − 2

We have

(2)

Minimum value of B.M. For M to have negative values, we must have wg , w. We verify that M will then be negative and keep decreasing in the portion AC of the beam. Therefore, M min will occur between C and D. From C to D:

a! x! ΣM J = 0: M + wa " x − # − α wx " # = 0 2% $ $2% M=

We differentiate and set

dM = 0: dx

1 w(α x 2 − 2ax + a 2 ) 2

αx − a = 0

xmin =

a

α

(3) (4)

Substituting in (3): 1 2 1 2 ! wa " − + 1# 2 $α α % 1−α = − wa 2 2α

M min = M min

(5)



Maximum value of bending moment occurs at D

3a ! ΣM D = 0: M D + wa " # − (2α wa)a = 0 $ 2 % 3! M max = M D = wa 2 " 2α − # 2% $

(6)

Equating − M min and M max : wa 2

1−α 3! = wa 2 " 2α − # 2α 2% $

4α 2 − 2α − 1 = 0

α= (a)

Substitute in (2):

(b)

Substitute for α in (5):

2 + 20 8

α=

1+ 5 = 0.809 4

k = 4(0.809) − 2

|M |max = − M min = − wa 2

Substitute for α in (4):

k = 1.236

1 − 0.809 2(0.809)

|M |max = 0.1180wa 2 xmin =

a 1.236a ! 0.809


PROBLEM 7.58 A uniform beam is to be picked up by crane cables attached at A and B. Determine the distance a from the ends of the beam to the points where the cables should be attached if the maximum absolute value of the bending moment in the beam is to be as small as possible. (Hint: Draw the bending-moment diagram in terms of a, L, and the weight w per unit length, and then equate the absolute values of the largest positive and negative bending moments obtained.)

SOLUTION w = weight per unit length

To the left of A:

x! ΣM1 = 0: M + wx " # = 0 $2% 1 M = − wx 2 2 1 2 M A = − wa 2

Between A and B: 1 1 ! ! ΣM 2 = 0: M − " wL # ( x − a) + ( wx) " x # = 0 2 $ % $2 % 1 1 1 M = − wx 2 + wLx − wLa 2 2 2

At center C:

x=

L 2 2

L! L! 1 1 1 M C = − w " # + wL " # − wLa 2 $2% 2 $2% 2



We set

| M A| = | M C | :

1 1 1 1 1 1 − wa 2 = wL2 − wLa + wa 2 = wL2 − wLa 2 8 2 2 8 2 a 2 + La − 0.25L2 = 0 1 1 ( L ± L2 + L2 ) = ( 2 − 1) L 2 2 1 = w(0.207 L)2 = 0.0214 wL2 2

a= M max

a = 0.207 L


PROBLEM 7.59 For the beam shown, determine (a) the magnitude P of the two upward forces for which the maximum absolute value of the bending moment in the beam is as small as possible, (b) the corresponding value of |M |max . (See hint for Problem 7.55.)

SOLUTION Ay = B = 60 kips − P

By symmetry: Along AC:

ΣM J = 0: M − x(60 kips − P) = 0 M = (60 kips − P) x M = 120 kips ⋅ ft − (2 ft) P at

x = 2 ft

Along CD: ΣM K = 0: M + ( x − 2 ft)(60 kips) − x(60 kips − P) = 0 M = 120 kip ⋅ ft − Px M = 120 kip ⋅ ft − (4 ft) P at

x = 4 ft

Along DE: ΣM L = 0: M − ( x − 4 ft) P + ( x − 2 ft)(60 kips) − x(60 kips − P) = 0 M = 120 kip ⋅ ft − (4 ft) P (const)

Complete diagram by symmetry For minimum |M |max , set M max = − M min 120 kip ⋅ ft − (2 ft) P = − [120 kip ⋅ ft − (4 ft) P]

M min = 120 kip ⋅ ft − (4 ft) P

(a)

P = 40.0 kips

(b)

| M |max = 40.0 kip ⋅ ft


PROBLEM 7.60 Knowing that P = Q = 150 lb, determine (a) the distance a for which the maximum absolute value of the bending moment in beam AB is as small as possible, (b) the corresponding value of |M |max . (See hint for Problem 7.55.)

SOLUTION Free body: Entire beam ΣM A = 0: Da − (150)(30) − (150)(60) = 0 D=

13,500 a

!

Free body: CB ΣM C = 0: − M C − (150)(30) +

13,500 (a − 30) = 0 a

45 ! M C = 9000 "1 − # a % $

!

Free body: DB ΣM D = 0: − M D − (150)(60 − a) = 0 M D = −150(60 − a)

(a)

!"

We set 45 ! M max = | M min | or M C = − M D : 9000 "1 − # = 150(60 − a) a % $ 60 −

2700 = 60 − a a

a 2 = 2700 a = 51.96 in.

(b)

a = 52.0 in.

| M |max = − M D = 150(60 − 51.96) | M | max = 1206 lb ⋅ in.


PROBLEM 7.61 Solve Problem 7.60 assuming that P = 300 lb and Q = 150 lb. PROBLEM 7.60 Knowing that P = Q = 150 lb, determine (a) the distance a for which the maximum absolute value of the bending moment in beam AB is as small as possible, (b) the corresponding value of |M |max . (See hint for Problem 7.55.)

SOLUTION Free body: Entire beam ΣM A = 0: Da − (300)(30) − (150)(60) = 0 D=

18,000 ! a

Free body: CB ΣM C = 0: − M C − (150)(30) +

18,000 (a − 30) = 0 a 40 ! M C = 13,500 "1 − # ! a % $

Free body: DB ΣM D = 0: − M D − (150)(60 − a) = 0 M D = −150(60 − a) !

(a)

We set 40 ! M max = | M min | or M C = − M D : 13,500 "1 − # = 150(60 − a ) a % $ 90 −

3600 = 60 − a a

a 2 + 30a − 3600 = 0 a=

−30 + 15.300 = 46.847 2 a = 46.8 in.

(b)

| M |max = − M D = 150(60 − 46.847) |M | max = 1973 lb ⋅ in.


PROBLEM 7.62* In order to reduce the bending moment in the cantilever beam AB, a cable and counterweight are permanently attached at end B. Determine the magnitude of the counterweight for which the maximum absolute value of the bending moment in the beam is as small as possible and the corresponding value of Mmax. Consider (a) the case when the distributed load is permanently applied to the beam, (b) the more general case when the distributed load may either be applied or removed.

SOLUTION M due to distributed load: ΣM J = 0: − M −

x wx = 0 2 1 M = − wx 2 2

M due to counter weight: ΣM J = 0: − M + xw = 0 M = wx

(a)

Both applied: M = Wx −

w 2 x 2

dM W = W − wx = 0 at x = dx w

And here M =

W2 > 0 so Mmax; Mmin must be at x = L 2w

So M min = WL − so

1 2 wL . For minimum | M |max set M max = − M min , 2 W2 1 = − WL + wL2 2w 2

or

W 2 + 2 wLW − w2 L2 = 0

W = − wL ± 2w2 L2 (need + )

W = ( 2 − 1) wL = 0.414 wL



(b)

w may be removed M max =

Without w,

With w (see Part a)

W 2 ( 2 − 1) 2 2 = wL 2w 2

M max = 0.858 wL2

M = Wx M max = WL at A M = Wx −

w 2 x 2

W2 W at x = 2w w 1 2 = WL − wL at x = L 2

M max = M min

For minimum M max , set M max (no w) = − M min (with w) WL = − WL +

1 2 1 wL → W = wL → 2 4

With

M max =

1 2 wL 4

W=

1 wL 4


PROBLEM 7.63 Using the method of Section 7.6, solve Problem 7.29. PROBLEM 7.29 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.

SOLUTION Free body: Entire beam ΣFy = 0: C − P − P = 0 C = 2P

ΣM C = 0: P(2a) + P(a) − M C = 0 M C = 3Pa

Shear diagram VA = − P

At A:

|V | max = 2 P

Bending-moment diagram MA = 0

At A:

| M | max = 3Pa



SOLUTION Free body: Entire beam 2L ! ΣM C = 0: P " # − A( L) = 0 $ 3 % A=

2 P 3

2 P−P+C =0 3

ΣFY = 0:

C=

1 P 3

Shear diagram VA =

At A:

2 P 3 |V | max =

2 P 3


At A:

|M | max =

2 PL 9



SOLUTION Reactions at A and D Because of the symmetry of the supports and loading. A=D=

1 L! 1 w # = wL " 2$ 2 % 4 A=D=

1 wL v 4

|V | max =

1 wL 4

Shear diagram 1 VA = + wL 4

At A: From B to C: Oblique straight line


At A:

From B to C: ARC of parabola

|M | max =

3 wL2 32

Since V has no discontinuity at B nor C, the slope of the parabola at these points is the same as the slope of the adjoining straight line segment.



SOLUTION Free body: Entire beam ΣFy = 0: C − w

L =0 2 C=

1 wL 2

1 ! 3L ! ΣM C = 0: " wL #" # = M C = 0 $2 %$ 4 % 3 M C = wL2 8

Shear diagram VA = 0

At A:

|V | max =

1 wL 2

Bending-moment diagram M =0

At A:

dM =V = 0 dx 3 |M | max = wL2 8

From A to B: ARC of parabola Since V has no discontinuity at B, the slope of the parabola at B is equal to the slope of the straightline segment.



SOLUTION Free body: Entire beam ΣFy = 0: B − P = 0 B=P

Σ M B = 0: M B − M 0 + PL = 0 MB = 0

Shear diagram VA = − P

At A:

|V | max = P

Bending-moment diagram M A = M 0 = PL

At A:

|M | max = PL



SOLUTION Free body: Entire beam ΣFy = 0: A = C ΣM C = 0: Al − M 0 = 0 A=C =

M0 L

Shear diagram VA = −

At A:

M0 L |V | max =

M0 L


At A:

At B, M increases by M0 on account of applied couple. |M | max = M 0 /2



SOLUTION ΣM A = 0: (8)(2) + (4)(3.2) − 4C = 0 C = 7.2 kN

ΣFy = 0: A = 4.8 kN

(a)

Shear diagram

Similar Triangles:

x 3.2 − x 3.2 = = ; x = 2.4 m 4.8 1.6 6.4 ↑ Add num. & den.

Bending-moment diagram

(b)

|V | max = 7.20 kN

!

|M | max = 5.76 kN ⋅ m




ΣM D = 0: (500 N)(0.8 m) + (500 N)(0.4 m) − (2400 N/m)(0.3 m)(0.15 m) − A(1.2 m) = 0 A = 410 N

ΣFy = 0: 410 − 2(500) − 2400(0.3) + D = 0 D = 1310 N

Shear diagram

At A:

VA = + 410 N

|V | max = 720 N


At A:

MA = 0

| M | max = 164.0 N ⋅ m



SOLUTION Free body: Beam ΣFx = 0: Ax = 0 ΣM B = 0: (60 kN)(3 m) + (50 kN)(1 m) − Ay (5 m) = 0 Ay = + 46.0 kN v ΣFy = 0: B + 46.0 kN − 60 kN − 50 kN = 0 B = + 64.0 kN v

Shear diagram At A:

VA = Ay = + 46.0 kN |V |max = 64.0 kN

Bending-moment diagram At A:

MA = 0 | M |max = 92.0 kN ⋅ m

Parabola from D to B. Its slope at D is same as that of straight-line segment CD since V has no discontinuity at D.



SOLUTION Free body: Entire beam ΣM B = 0: (30 kN/m)(2 m)(6 m) − C (5 m) + (80 kN)(2 m) = 0 C = 104 kN

ΣFy = 0: 104 − 30(2) − 80 + B = 0 B = 36 kN


At A:

|V |max = 60.0 kN


At A:

| M |max = 72.0 kN ⋅ m



SOLUTION Reactions at supports. Because of the symmetry: A= B=

1 (8 + 8 + 4 × 5) kips 2 A = B = 18 kips v

Shear diagram At A:

VA = + 18 kips |V |max = 18.00 kips


MA = 0 | M |max = 48.5 kip ⋅ ft

Discontinuities in slope at C and D, due to the discontinuities of V.



SOLUTION Free body: Beam ΣFx = 0: Ax = 0 ΣM B = 0: (12 kips)(4ft) + (15 kips)(7 ft) − Ay (10 ft) = 0 Ay = + 15.3 kips v ΣFy = 0: B + 15.3 − 15 − 12 = 0 B = + 11.7 kips v

Shear diagram At A:

VA = Ay = 15.3 kips |V |max = 15.30 kips


MA = 0 | M |max = 46.8 kip ⋅ ft



SOLUTION Free body: Beam ΣM B = 0: 5 kip ⋅ ft + (1 kips)(10 ft) + (4 kips)(5 ft) − Ay (15 ft) = 0 Ay = + 3.00 kips v ΣFx = 0: Ax = 0

Shear diagram At A:

VA = Ay = + 3.00 kips |V | max = 3.00 kips


M A = − 5 kip ⋅ ft | M |max = 15.00 kip ⋅ ft



SOLUTION Free body: Beam ΣM B = 0: 6 kip ⋅ ft + 12 kip ⋅ ft + (2 kips)(18 ft) + (3 kips)(12 ft) + (4 kips)(6 ft) − Ay (24 ft) = 0 Ay = + 4.75 kips v ΣFx = 0: Ax = 0

Shear diagram At A:

VA = Ay = + 4.75 kips |V |max = 4.75 kips


M A = − 6 kip ⋅ ft | M |max = 39.0 kip ⋅ ft



SOLUTION Free body: Beam ΣFx = 0: Ax = 0 ΣM B = 0: (19.2 kN)(1.2 m) − Ay (3.2 m) = 0 Ay = + 7.20 kN v

Shear diagram VA = VC = Ay = + 7.20 kN

To determine Point D where V = 0, we write VD − VC = wu 0 − 7.20 kN = − (8 kN/m)u

u = 0.9 m v

We next compute all areas Bending-moment diagram At A:

MA = 0

Largest value occurs at D with AD = 0.8 + 0.9 = 1.700 m | M |max = 9.00 kN ⋅ m

!

!

1.700 m from A


PROBLEM 7.78 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the magnitude and location of the maximum bending moment.

SOLUTION Free body: Beam ΣFx = 0: Ax = 0 ΣM B = 0: (40 kN)(1.5 m) − Ay (3.75 m) = 0 Ay = + 16.00 kN v

Shear diagram VA = VC = Ay = + 16.00 kN

To determine Point E where V = 0, we write VE − VC = − wu 0 − 16 kN = − (20 kN/m)u

u = 0.800 m v

We next compute all areas Bending-moment diagram At A:

MA = 0

Largest value occurs at E with AE = 1.25 + 0.8 = 2.05 m

| M |max = 26.4 kN ⋅ m 2.05 m from A

From A to C and D to B: Straight line segments. From C to D: Parabola


PROBLEM 7.79 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the magnitude and location of the maximum absolute value of the bending moment.

SOLUTION Free body: Entire beam ΣM C = 0: − (27 kip ⋅ ft) + A(22.5 ft) − (1 kip/ft)(18 ft)(13.5 ft) = 0 A = 12 kips

ΣFy = 0: 12 − 1(18) + C = 0 C = 6 kips

Shear diagram VA = + 12 kips

At A:

To locate Point D (where V = 0) VD − VA = − wu 0 − 12 kips = − (1 kip/ft)u u = 12 ft


At A:

| M |max = 45.0 kip ⋅ ft 12.00 ft from A


PROBLEM 7.80 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the magnitude and location of the maximum absolute value of the bending moment.

SOLUTION Free body: Entire beam ΣM A = 0: (6 kips)(10 ft) − (9 kips)(7 ft) + 8(4 ft) = 0 B = + 0.75 kips B = 0.75 kips

ΣFy = 0: A + 0.75 kips − 9 kips + 6 kips = 0 A = + 2.25 kips A = 2.25 kips

M max = 12.00 kip ⋅ ft 6.00 ft from A


PROBLEM 7.81 (a) Draw the shear and bending-moment diagrams for beam AB, (b) determine the magnitude and location of the maximum absolute value of the bending moment.

SOLUTION Reactions at supports Because of symmetry of load A= B=

1 (300 × 8 + 300) 2

A = B = 1350 lb v

Load diagram for AB The 300-lb force at D is replaced by an equivalent force-couple system at C.

Shear diagram VA = A = 1350 lb

At A:

To determine Point E where V = 0: VE − VA = − wx 0 − 1350 lb = − (300 lb/ft) x x = 4.50 ft v

We compute all areas Bending-moment diagram At A:

MA = 0

Note 600 − lb ⋅ ft drop at C due to couple | M |max = 3040 lb ⋅ ft

!

!

4.50 ft from A


PROBLEM 7.82 Solve Problem 7.81 assuming that the 300-lb force applied at D is directed upward. PROBLEM 7.81 (a) Draw the shear and bending-moment diagrams for beam AB, (b) determine the magnitude and location of the maximum absolute value of the bending moment.

SOLUTION Reactions at supports Because of symmetry of load: A= B=

1 (300 × 8 − 300) 2

A = B = 1050 lb v

Load diagram The 300-lb force at D is replaced by an equivalent force-couple system at C.

Shear diagram VA = A = 1050 lb

At A:

To determine Point E where V = 0: VE − VA = − wx 0 − 1050 lb = − (300 lb/ft) x x = 3.50 ft v

We compute all areas Bending-moment diagram MA = 0

At A:

Note 600 − lb ⋅ ft increase at C due to couple | M |max = 1838 lb ⋅ ft

!

!

3.50 ft from A


PROBLEM 7.83 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the magnitude and location of the maximum absolute value of the bending moment.

SOLUTION Free body: Beam ΣM A = 0: B (4 m) − (100 kN)(2 m) − 20 kN ⋅ m = 0 B = + 55 kN v ΣFx = 0: Ax = 0 ΣFy = 0: Ay + 55 − 100 = 0 Ay = + 45 kN v

Shear diagram VA = Ay = + 45 kN

At A:

To determine Point C where V = 0: VC − VA = − wx 0 − 45 kN = − (25 kN ⋅ m) x x = 1.8 m v

We compute all areas bending-moment Bending-moment diagram At A:

MA = 0

At B:

M B = − 20 kN ⋅ m | M |max = 40.5 kN ⋅ m

!

!

1.800 m from A

!

Single arc of parabola


PROBLEM 7.84 Solve Problem 7.83 assuming that the 20-kN ⋅ m couple applied at B is counterclockwise. PROBLEM 7.83 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the magnitude and location of the maximum absolute value of the bending moment.

SOLUTION Free body: Beam ΣM A = 0: B (4 m) − (100 kN)(2 m) − 20 kN ⋅ m = 0 B = + 45 kN v ΣFx = 0: Ax = 0 ΣFy = 0: Ay + 45 − 100 = 0 Ay = + 55 kN v

Shear diagram VA = Ay = + 55 kN

At A:

To determine Point C where V = 0 : VC − VA = − wx 0 − 55 kN = − (25 kN/m) x x = 2.20 m v

We compute all areas bending-moment Bending-moment diagram At A:

MA = 0

At B:

M B = + 20 kN ⋅ m | M |max = 60.5 kN ⋅ m

!

!

2.20 m from A

!

Single arc of parabola


PROBLEM 7.85 For the beam and loading shown, (a) write the equations of the shear and bending-moment curves, (b) determine the magnitude and location of the maximum bending moment.

SOLUTION Distributed load 1 x! ! w = w0 "1 − # " Total = w0 L # 2 L% $ $ % L 1 ! w0 L # − LB = 0 " 3 $2 %

ΣM A = 0:

ΣFy = 0: Ay −

B=

w0 L 6

wL wL 1 w0 L + 0 = 0 A y = 0 2 6 3

w0 L 3

Shear:

VA = Ay =

Then

dV = −w → V dx x! w0 "1 − # dx L% $ w0 L ! 1 w0 2 V =" x − w0 x + # 2 L $ 3 % = VA −

,

x

0

& 1 x 1 x !2 ' = w0 L ( − + " # ) (* 3 L 2 $ L % )+

Note: At

x=L V =−

w0 L 6 2

x! x! 2 V = 0 at " # − 2 " # + $L% $L% 3 =0→

x 1 =1− L 3



Moment: Then

MA = 0 x x/ L dM ! x! x! " # = V → M = 0 Vdx = L 0 V " # d " # $ dx % $ L% $ L% 2 x/ L & 1 x 1 x! ' x! M = w0 L2 ( − + " # )d" # 0 (3 L L L 2 $ % $ % * +)

,

,

,

& 1 x ! 1 x ! 2 1 x !3 ' M = w0 L ( " # − " # + " # ) 6 $ L % )+ (* 3 $ L % 2 $ L % 1! x 2 M max " at = 1 − ## = 0.06415w0 L " L 3 $ % 2

& 1 x 1 x !2 ' V = w0 L ( − + " # ) (* 3 L 2 $ L % )+

!

& 1 x ! 1 x ! 2 1 x !3 ' M = w0 L ( " # − " # + " # ) 6 $ L % )+ (* 3 $ L % 2 $ L %

!

M max = 0.0642 w0 L2

!

(a)

2

(c)

at x = 0.423L



SOLUTION (a)

We note that at Load:

B( x = L): VB = 0, M B = 0

(1)

4x ! x! 1 x! w( x) = w0 "1 − # − w0 " # = w0 "1 − # L% 3 $L% $ $ 3L %

Shear: We use Eq. (7.2) between C ( x = x) and B( x = L): VB − VC = −

,

V ( x) = w0

L x

,

w( x) dx 0 − V ( x) = − L

x

,

L x

w( x)dx

4x ! "1 − 3L # dx $ % L

& 2x2 ' 2L 2x2 ! = w0 ( x − −x+ # ) = w0 "" L − 3L + x 3 3L %# * $ V ( x) =

w0 (2 x 2 − 3Lx + L2 ) 3L

(2)

Bending moment: We use to Eq. (7.4) between C ( x = x) and B( x = L) : M B − MC = =

,

L x

w0 3L

V ( x)dx 0 − M ( x)

,

L x

(2 x 2 − 3Lx + L2 )dx L

w0 & 2 3 3 2 ' x − Lx + L2 x ) ( 3L * 3 2 + w = − 0 [4 x3 − 9 Lx 2 + 6 L2 x]Lx 18L w = − 0 [(4 L3 − 9 L3 + 6 L3 ) − (4 x3 − 9 Lx 2 + 6 L2 x)] 18L

M ( x) = −

M ( x) =

w0 (4 x3 − 9 Lx 2 + 6 L2 x − L3 ) 18L

(3)



(b)

Maximum bending moment dM =V = 0 dx

Eq. (2):

2 x 2 − 3Lx + L2 = 0 x=

Carrying into (3): Note:

M max = | M |max =

3− 9−8 L L= 4 2 w0 L2 , At 72 w0 L2 18

At

x=

L 2

x =0!



SOLUTION

π x dv = − w = w0 cos 2 L dx 2L ! π x V = − wdx = − w0 " # sin 2 L + C1 $ π %

,

(1)

2L ! π x dM = V = − w0 " # sin 2 L + C1 dx $ π % 2

2L ! πx cos M = Vdx = + w0 " + C1 x + C2 # 2L $ π %

,

(2)

Boundary conditions At x = 0:

V = C1 = 0 C1 = 0

At x = 0:

2L ! M = + w0 " # cos(0) + C2 = 0 $ π %

2

2L ! C2 = − w0 " # $ π %

2

2L ! π x V = − w0 " # sin 2 L $ π %

Eq. (1)

2

πx! 2L ! −1 + cos M = w0 " # " 2 L #% $ π % $ 2

2L ! 4 | M max | = w0 " | −1 + 0|= 2 w0 L2 # π $ π %

M max at x = L :


!

PROBLEM 7.88 The beam AB, which lies on the ground, supports the parabolic load shown. Assuming the upward reaction of the ground to be uniformly distributed, (a) write the equations of the shear and bending-moment curves, (b) determine the maximum bending moment.

SOLUTION ΣFy = 0: wg L −

(a)

,

L 0

4 w0 L2

( Lx − x 2 ) dx = 0

4 w0 1 2 1 3 ! 2 " LL − 3 L # = 3 w0 L L2 $ 2 % 2 w0 wg = 3

wg L =

2 &x x! ' 2 net load w = 4 w0 ( − " # ) − w0 (* L $ L % )+ 3

Define

x dx ξ = so d ξ = L L

or

1 ! w = 4 w0 " − + ξ − ξ 2 # 6 $ % 1 ! 4w0 L " − + ξ − ξ 2 # d ξ 6 $ % 1 1 2 1 3! 2 V = w0 L (ξ − 3ξ 2 + 2ξ 3 ) = 0 + 4w0 L " ξ + ξ − ξ # 2 3 % 3 $6 x ξ 2 M = M 0 + Vdx = 0 + w0 L2 (ξ − 3ξ 2 + 2ξ 3 )d ξ 0 0 3 2 1 1 ! 1 = w0 L2 " ξ 2 − ξ 3 + ξ 4 # = w0 L2 (ξ 2 − 2ξ 3 + ξ 4 ) 3 2 % 3 $2 V = V (0) −

,

ξ

0

,

(b)

Max M occurs where

V =0

,

1 − 3ξ + 2ξ 2 = 0

ξ=

1 2

1! 1 1 2 1 ! w L2 M " ξ = # = w0 L2 " − + # = 0 2% 3 48 $ $ 4 8 16 % M max =

w0 L2 at center of beam 48


!

PROBLEM 7.89 The beam AB is subjected to the uniformly distributed load shown and to two unknown forces P and Q. Knowing that it has been experimentally determined that the bending moment is +800 N ⋅ m at D and +1300 N ⋅ m at E, (a) determine P and Q, (b) draw the shear and bending-moment diagrams for the beam.

SOLUTION (a)

Free body: Portion AD ΣFx = 0: Cx = 0 ΣM D = 0: −C y (0.3 m) + 0.800 kN ⋅ m + (6 kN)(0.45 m) = 0 C y = +11.667 kN

C = 11.667 kN "

Free body: Portion EB ΣM E = 0: B(0.3 m) − 1.300 kN ⋅ m = 0 B = 4.333 kN "

Free body: Entire beam ΣM D = 0: (6 kN)(0.45 m) − (11.667 kN)(0.3 m) − Q (0.3 m) + (4.333 kN)(0.6 m) = 0 Q = 6.00 kN ΣM y = 0: 11.667 kN + 4.333 kN −6 kN − P − 6 kN = 0 P = 4.00 kN

Load diagram

(b)

Shear diagram At A: VA = 0 |V |max = 6 kN "!




At A:

| M |max = 1300 N ⋅ m "!

We check that M D = +800 N ⋅ m and M E = +1300 N ⋅ m

As given: M C = −900 N ⋅ m

At C:

!


PROBLEM 7.90 Solve Problem 7.89 assuming that the bending moment was found to be +650 N ⋅ m at D and +1450 N ⋅ m at E. PROBLEM 7.89 The beam AB is subjected to the uniformly distributed load shown and to two unknown forces P and Q. Knowing that it has been experimentally determined that the bending moment is +800 N ⋅ m at D and +1300 N ⋅ m at E, (a) determine P and Q, (b) draw the shear and bending-moment diagrams for the beam.

SOLUTION !

(a)

Free body: Portion AD ΣFx = 0: Cx = 0 ΣM D = 0: −C (0.3 m) + 0.650 kN ⋅ m + (6 kN)(0.45 m) = 0

C y = +11.167 kN

C = 11.167 kN "

Free body: Portion EB ΣM E = 0: B(0.3 m) − 1.450 kN ⋅ m = 0

! !

B = 4.833 kN "

!

Free body: Entire beam ΣM D = 0: (6 kN)(0.45 m) − (11.167 kN)(0.3 m) −Q (0.3 m) + (4.833 kN)(0.6 m) = 0 Q = 7.50 kN

! !

ΣM y = 0: 11.167 kN + 4.833 kN − 6 kN − P − 7.50 kN = 0 P = 2.50 kN

Load diagram ! ! (b)


At A:

|V |max = 6 kN "!

! ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1111


! !


At A:

| M |max = 1450 N ⋅ m "!

We check that M D = +650 N ⋅ m and M E = +1450 N ⋅ m

As given: !

M C = −900 N ⋅ m

At C:

!


PROBLEM 7.91* The beam AB is subjected to the uniformly distributed load shown and to two unknown forces P and Q. Knowing that it has been experimentally determined that the bending moment is +6.10 kip ⋅ ft at D and +5.50 kip ⋅ ft at E, (a) determine P and Q, (b) draw the shear and bending-moment diagrams for the beam.

SOLUTION (a)

Free body: Portion DE ΣM E = 0: 5.50 kip ⋅ ft − 6.10 kip ⋅ ft + (1 kip)(2 ft) − VD (4 ft) = 0

VD = +0.350 kip ΣFy = 0: 0.350 kip − 1kip − VE = 0

VE = −0.650 kip

Free body: Portion AD ΣM A = 0: 6.10 kip ⋅ ft − P(2ft) − (1 kip)(2 ft) − (0.350 kip)(4 ft) = 0 P = 1.350 kips ΣFx = 0: Ax = 0 ΣFy = 0: Ay − 1 kip − 1.350 kip − 0.350 kip = 0

Ay = +2.70 kips

A = 2.70 kips

Free body: Portion EB ΣM B = 0: (0.650 kip)(4 ft) + (1 kip)(2 ft) + Q(2 ft) − 5.50 kip ⋅ ft = 0 Q = 0.450 kip ΣFy = 0: B − 0.450 − 1 − 0.650 = 0 B = 2.10 kips


PROBLEM 7.91* (Continued) (b)

Load diagram

Shear diagram VA = A = +2.70 kips

At A:

To determine Point G where V = 0, we write

VG − VC = − wµ 0 − 0.85 kips = −(0.25 kip/ft)µ

µ = 3.40 ft |V |max = 2.70 kips at A

We next compute all areas

Bending-moment diagram At A: M A = 0 Largest value occurs at G with AG = 2 + 3.40 = 5.40 ft | M |max = 6.345 kip ⋅ ft

5.40 ft from A Bending-moment diagram consists of 3 distinct arcs of parabolas.!


PROBLEM 7.92* Solve Problem 7.91 assuming that the bending moment was found to be +5.96 kip ⋅ ft at D and +6.84 kip ⋅ ft at E. PROBLEM 7.91* The beam AB is subjected to the uniformly distributed load shown and to two unknown forces P and Q. Knowing that it has been experimentally determined that the bending moment is +6.10 kip ⋅ ft at D and +5.50 kip ⋅ ft at E, (a) determine P and Q, (b) draw the shear and bendingmoment diagrams for the beam.

SOLUTION (a)

Free body: Portion DE ΣM E = 0: 6.84 kip ⋅ ft − 5.96 kip ⋅ ft + (1 kip)(2 ft) − VD (4 ft) = 0 VD = +0.720 kip ΣFy = 0: 0.720 kip − 1kip − VE = 0 VE = −0.280 kip

Free body: Portion AD ΣM A = 0: 5.96 kip ⋅ ft − P (2 ft) − (1 kip)(2 ft) − (0.720 kip)(4 ft) = 0 P = 0.540 kip ΣFx = 0: Ax = 0 ΣFy = 0: Ay − 1 kip − 0.540 kip − 0.720 kip = 0 Ay = +2.26 kips

A = 2.26 kips "

Free body: Portion EB ΣM B = 0: (0.280 kip)(4 ft) + (1 kip)(2 ft) + Q(2 ft) − 6.84 kip ⋅ ft = 0 Q = 1.860 kips ΣFy = 0: B − 1.860 − 1 − 0.280 = 0 B = 3.14 kips



(b)

Load diagram

Shear diagram VA = A = +2.26 kips

At A:

To determine Point G where V = 0, we write

VG − VC = − wµ 0 − (1.22 kips) = −(0.25 kip/ft)µ

µ = 4.88 ft " |V |max = 3.14 kips at B

We next compute all areas Bending-moment diagram At A: M A = 0 Largest value occurs at G with AG = 2 + 4.88 = 6.88 ft

| M |max = 6.997 kip ⋅ ft

6.88 ft from A Bending-moment diagram consists of 3 distinct arcs of parabolas.!


PROBLEM 7.93 Two loads are suspended as shown from the cable ABCD. Knowing that hB = 1.8 m, determine (a) the distance hC , (b) the components of the reaction at D, (c) the maximum tension in the cable.

SOLUTION ΣFx = 0: − Ax + Dx = 0

FBD Cable:

Ax = Dx

ΣM A = 0: (10 m)Dy − (6 m)(10 kN) − (3 m)(6 kN) = 0 D y = 7.8 kN ΣFy = 0: Ay − 6 kN − 10 kN + 7.8 kN = 0 A y = 8.2 kN

FDB AB:

ΣM B = 0: (1.8 m) Ax − (3 m)(8.2 kN) = 0 Ax =

From above

Dx = Ax =

41 kN 3

41 kN 3

FBD CD: 41 ! ΣM C = 0: (4 m)(7.8 kN) − hC " kN # = 0 $ 3 % hC = 2.283 m

hC = 2.28 m

(a)

D x = 13.67 kN

(b)

D y = 7.80 kN

Since Ax = Bx and Ay . B y , max T is TAB 2

TAB =

Ax2

+

Ay2

41 ! kN # + (8.2 kN)2 = " $ 3 % Tmax = 15.94 kN

(c)


!

PROBLEM 7.94 Knowing that the maximum tension in cable ABCD is 15 kN, determine (a) the distance hB , (b) the distance hC .

SOLUTION ΣFx = 0: − Ax + Dx = 0

FBD Cable:

Ax = Dx

ΣM A = 0: (10 m) Dy − (6 m)(10 kN) − (3 m)(6 kN) = 0 D y = 7.8 kN

ΣFy = 0: Ay − 6 kN − 10 kN + 7.8 kN = 0 A y = 8.2 kN

Since

Ax = Dx

and

Ay > Dy , Tmax = TAB

FBD Pt A: ΣFy = 0: 8.2 kN − (15 kN)sin θ A = 0

θ A = sin −1

8.2 kN = 33.139° 15 kN

ΣFx = 0: − Ax + (15 kN) cos θ A = 0 Ax = (15 kN) cos(33.139°) = 12.56 kN

FBD CD: From FBD cable: hB = (3 m) tan θ A = (3 m) tan(33.139°) hB = 1.959 m

(a) ΣM C = 0: (4 m)(7.8 kN) − hC (12.56 kN) = 0

hC = 2.48 m

(b)


!

PROBLEM 7.95 If dC = 8 ft, determine (a) the reaction at A, (b) the reaction at E.

SOLUTION Free body: Portion ABC ΣM C = 0 2 Ax − 16 Ay + 300(8) = 0 Ax = 8 Ay − 1200

(1)

Free body: Entire cable

ΣM E = 0: + 6 Ax + 32 Ay − (300 lb + 200 lb + 300 lb)16 ft = 0 3 Ax + 16 Ay − 6400 = 0

Substitute from Eq. (1): 3(8 Ay − 1200) + 16 Ay − 6400 = 0

Eq. (1)

Ax = 8(250) − 1200

A y = 250 lb A x = 800 lb

ΣFx = 0: − Ax + Ex = 0 − 800 lb + Ex = 0

E x = 800 lb

ΣFy = 0: 250 + E y − 300 − 200 − 300 = 0

E y = 550 lb

A = 838 lb E = 971 lb

17.4° 34.5°


!

PROBLEM 7.96 If dC = 4.5 ft, determine (a) the reaction at A, (b) the reaction at E.

SOLUTION Free body: Portion ABC

ΣM C = 0: − 1.5 Ax − 16 Ay + 300 × 8 = 0 Ax =


(2400 − 16 Ay )

(1)

1.5

ΣM E = 0: + 6 Ax + 32 Ay − (300 lb + 200 lb + 300 lb)16 ft = 0 3 Ax + 16 Ay − 6400 = 0

Substitute from Eq. (1):

3(2400 − 16 Ay ) 1.5

+ 16 Ay − 6400 = 0 Ay = −100 lb A y = 100 lb

Thus Ay acts downward Eq. (1)

(2400 − 16( −100)) = 2667 lb 1.5

A x = 2667 lb

ΣFx = 0: − Ax + Ex = 0 − 2667 + Ex = 0

E x = 2667 lb

Ax =

ΣFy = 0: Ay + E y − 300 − 200 − 300 = 0 −100 lb + E y − 800 lb = 0

E y = 900 lb

A = 2670 lb

2.10°

E = 2810 lb

18.6°


!

PROBLEM 7.97 Knowing that dC = 3 m, determine (a) the distances dB and dD (b) the reaction at E.

SOLUTION Free body: Portion ABC ΣM C = 0: 3 Ax − 4 Ay + (5 kN)(2 m) = 0 Ax =

4 10 Ay − 3 3

(1)


ΣM E = 0: 4 Ax − 10 Ay + (5 kN)(8 m) + (5 kN)(6 m) + (10 kN)(3 m) = 0 4 Ax − 10 Ay + 100 = 0

Substitute from Eq. (1): 4 10 ! 4 " Ay − # − 10 Ay + 100 = 0 3 3% $ Ay = +18.571 kN

Eq. (1)

A y = 18.571 kN

4 10 (18.511) − = +21.429 kN 3 3

A x = 21.429 kN

ΣFx = 0: − Ax + Ex = 0 − 21.429 + Ex = 0

E x = 21.429 kN

ΣFy = 0: 18.571 kN + E y + 5 kN + 5 kN + 10 kN = 0

E y = 1.429 kN

Ax =

E = 21.5 kN

3.81°



Portion AB ΣM B = 0: (18.571 kN)(2 m) − (21.429 kN)d B = 0 d B = 1.733 m

Portion DE Geometry h = (3 m) tan 3.8° = 0.199 m d D = 4 m + 0.199 m d D = 4.20 m

!


!

PROBLEM 7.98 Determine (a) distance dC for which portion DE of the cable is horizontal, (b) the corresponding reactions at A and E.

SOLUTION Free body: Entire cable

ΣFy = 0: Ay − 5 kN − 5 kN − 10 kN = 0

A y = 20 kN

ΣM A = 0: E (4 m) − (5 kN)(2 m) − (5 kN)(4 m) − (10 kN)(7m) = 0 E = +25 kN

E = 25.0 kN

ΣFx = 0: − Ax + 25 kN = 0

A x = 25 kN A = 32.0 kN

38.7°

!

Free body: Portion ABC

ΣM C = 0: (25 kN)dC − (20 kN)(4 m) + (5 kN)(2 m) = 0 25dC − 70 = 0

dC = 2.80 m


PROBLEM 7.99 If dC = 15 ft, determine (a) the distances dB and dD, (b) the maximum tension in the cable.


ΣM A = 0: E y (30 ft) − Ex (7.5 ft) − (2 kips)(6 ft) − (2 kips)(15 ft) − (2 kips)(21 ft) = 0 7.5Ex − 30 E y + 84 = 0

(1)

Free body: Portion CDE ΣM C = 0: E y (15 ft) − Ex (15 ft) − (2 kips)(6 ft) = 0 15Ex − 15E y + 12 = 0

(2)

1 Eq. (1) × : 3.75 Ex − 15E y + 42 = 0 2

(2) – (3):

(3)

11.25Ex − 30 = 0 Ex = 2.6667 kips

Eq. (1):

7.5(2.6667) − 30 E y + 84 = 0 E y = 3.4667 kips

Tm = Ex2 + E y2 = (2.6667) 2 + (3.4667)2

Tm = 4.37 kips

!



Free body: Portion DE

ΣM D = 0: (3.4667 kips)(9 ft) − (2.6667 kips) d D = 0

Return to free body of entire cable

d D = 11.70 ft

(with Ex = 2.6667 kips, E y = 3.4667 kips)

ΣFy = 0: Ay − 3(2 kips) + 3.4667 kips = 0

Ay = 2.5333 kips

ΣFx = 0: 2.6667 kips − Ax = 0

Ax = 2.6667 kips

Free body: Portion AB

ΣM B = 0: Ax ( d B − 7.5) − Ay (6) = 0 (2.6667)(d B − 7.5) − (2.5333)(6) = 0

d B = 13.20 ft


PROBLEM 7.100 Determine (a) the distance dC for which portion BC of the cable is horizontal, (b) the corresponding components of the reaction at E.

SOLUTION Free body: Portion CDE

ΣFy = 0: E y − 2(2 kips) = 0 E y = 4 kips ΣM C = 0: (4 kips)(15 ft) − Ex dC − (2 kips)(6 ft) = 0 Ex dC = 48 kip ⋅ ft

(1)


ΣM A = 0: (4 kips)(30 ft) − Ex (7.5 ft) − (2 kips)(6 ft) − (2 kips)(15 ft) − (2 kips)(21 ft) = 0 Ex = 4.8 kips

From Eq. (1):

dC =

48 48 = Ex 4.8

dC = 10.00 ft E x = 4.80 kips

Components of reaction at E:

;

E y = 4.00 kips


PROBLEM 7.101 Cable ABC supports two loads as shown. Knowing that b = 4 ft, determine (a) the required magnitude of the horizontal force P, (b) the corresponding distance a.

SOLUTION FBD ABC: ΣFy = 0: − 40 lb − 80 lb + C y = 0 C y = 120 lb

FBD BC:

ΣM B = 0: (4 ft)(120 lb) − (10 ft)Cx = 0 C x = 48 lb

From ABC:

ΣFx = 0: − P + Cx = 0 P = Cx = 48 lb

(a)

P = 48.0 lb

(b)

a = 10.00 ft

!

ΣM C = 0: (4 ft)(80 lb) + a(40 lb) − (15 ft)(48 lb) = 0


PROBLEM 7.102 Cable ABC supports two loads as shown. Determine the distances a and b when a horizontal force P of magnitude 60 lb is applied at A.

SOLUTION FBD ABC: ΣFx = 0: Cx − P = 0

C x = 60 lb

ΣFy = 0: C y − 40 lb − 80 lb = 0 C y = 120 lb

FBD BC:

ΣM B = 0: b(120 lb) − (10 ft)(60 lb) = 0

b = 5.00 ft

FBD AB: ΣM B = 0: (a − b)(40 lb) − (5 ft)60 lb = 0 a − b = 7.5 ft a = b + 7.5 ft = 5 ft + 7.5 ft a = 12.50 ft


PROBLEM 7.103 Knowing that mB = 70 kg and mC = 25 kg, determine the magnitude of the force P required to maintain equilibrium.

SOLUTION Free body: Portion CD ΣM C = 0: D y (4 m) − Dx (3 m) = 0 Dy =

3 Dx 4


ΣM A = 0:

3 Dx (14 m) − WB (4 m) − WC (10 m) − P(5 m) = 0 4

ΣM B = 0:

3 Dx (10 m) − Dx (5 m) − WC (6 m) = 0 4

(1)

Free body: Portion BCD

Dx = 2.4WC

For

mB = 70 kg mC = 25 kg

g = 9.81 m/s 2 :

WB = 70 g

Eq. (2):

Dx = 2.4WC = 2.4(25g ) = 60 g

Eq. (1):

(2)

WC = 25 g

3 60 g (14) − 70 g (4) − 25 g (10) − 5P = 0 4 100 g − 5 P = 0: P = 20 g P = 20(9.81) = 196.2 N

P = 196.2 N


PROBLEM 7.104 Knowing that mB = 18 kg and mC = 10 kg, determine the magnitude of the force P required to maintain equilibrium.

SOLUTION Free body: Portion CD ΣM C = 0: D y (4 m) − Dx (3 m) = 0 Dy =

3 Dx 4


ΣM A = 0:

3 Dx (14 m) − WB (4 m) − WC (10 m) − P(5 m) = 0 4

ΣM B = 0:

3 Dx (10 m) − Dx (5 m) − WC (6 m) = 0 4

(1)

Free body: Portion BCD

Dx = 2.4WC

For

mB = 18 kg mC = 10 kg

g = 9.81 m/s 2 :

WB = 18 g

Eq. (2):

Dx = 2.4WC = 2.4(10 g ) = 24 g

Eq. (1):

3 24 g (14) − (18 g )(4) − (10 g )(10) − 5P = 0 4

(2)

WC = 10 g

80 g − 5P : P = 16 g P = 16(9.81) = 156.96 N

P = 157.0 N


PROBLEM 7.105 If a = 3 m, determine the magnitudes of P and Q required to maintain the cable in the shape shown.

SOLUTION Free body: Portion DE ΣM D = 0: E y (4 m) − Ex (5 m) = 0 Ey =

5 Ex 4

Free body: Portion CDE ΣM C = 0:

5 Ex (8 m) − Ex (7 m) − P(2 m) = 0 4 Ex =

2 P 3

(1)

Free body: Portion BCDE

ΣM B = 0:

5 Ex (12 m) − Ex (5 m) − (120 kN)(4 m) = 0 4 10 Ex − 480 = 0; Ex = 48 kN

Eq. (1):

48 kN =

2 P 3

P = 72.0 kN




ΣM A = 0:

5 Ex (16 m) − Ex (2 m) + P(3 m) − Q (4 m) − (120 kN)(8 m) = 0 4 (48 kN)(20 m − 2 m) + (72 kN)(3 m) − Q(4 m) − 960 kN ⋅ m = 0 4Q = 120

Q = 30.0 kN


PROBLEM 7.106 If a = 4 m, determine the magnitudes of P and Q required to maintain the cable in the shape shown.

SOLUTION Free body: Portion DE ΣM D = 0: E y (4 m) − Ex (6 m) = 0 Ey =

3 Ex 2

Free body: Portion CDE ΣM C = 0:

3 Ex (8 m) − Ex (8 m) − P(2 m) = 0 2 Ex =

1 P 2

(1)

Free body: Portion BCDE

ΣM B = 0:

3 Ex (12 m) − Ex (6 m) + (120 kN)(4 m) = 0 2 12 Ex = 480

Eq (1):

Ex =

1 P; 2

Ex = 40 kN 40 kN =

1 P 2

P = 80.0 kN




ΣM A = 0:

3 Ex (16 m) − Ex (2 m) + P(4 m) − Q(4 m) − (120 kN)(8 m) = 0 2 (40 kN)(24 m − 2 m) + (80 kN)(4 m) − Q (4 m) − 960 kN ⋅ m = 0 4Q = 240

Q = 60.0 kN


PROBLEM 7.107 A wire having a mass per unit length of 0.65 kg/m is suspended from two supports at the same elevation that are 120 m apart. If the sag is 30 m, determine (a) the total length of the wire, (b) the maximum tension in the wire.

SOLUTION

Eq. 7.16:

Solve by trial and error: Eq. 7.15:

xB c 60 30m + c = c cosh c yB = c cosh

c = 64.459 m sB = c sin h

xB c

sB = (64.456 m) sinh

60 m 64.459 m

sB = 69.0478 m

Length = 2 sB = 2(69.0478 m) = 138.0956 m Eq. 7.18:

L = 138.1 m

Tm = wyB = w(h + c) = (0.65 kg/m)(9.81 m/s 2 )(30 m + 64.459 m) Tm = 602.32 N

Tm = 602 N


PROBLEM 7.108 Two cables of the same gauge are attached to a transmission tower at B. Since the tower is slender, the horizontal component of the resultant of the forces exerted by the cables at B is to be zero. Knowing that the mass per unit length of the cables is 0.4 kg/m, determine (a) the required sag h, (b) the maximum tension in each cable.

SOLUTION W = wxB ΣM B = 0: T0 yB − ( wxB )

Horiz. comp. = T0 =

(a) Cable AB

w(45 m) 2 2h

xB = 30 m, T0 =

Equate T0 = T0

wxB2 2 yB

xB = 45 m T0 =

Cable BC

yB =0 2

yB = 3 m

w(30 m) 2 2(3 m)

w (45 m) 2 w (30 m) 2 = 2h 2(3 m)

h = 6.75 m

Tm2 = T02 + W 2

(b) Cable AB:

w = (0.4 kg/m)(9.81 m/s) = 3.924 N/m xB = 45 m, T0 =

yB = h = 6.75 m

wxB2 (3.924 N/m)(45 m)2 = = 588.6 N 2 yB 2(6.75 m)

W = wxB = (3.924 N/m)(45 m) = 176.58 N Tm2 = (588.6 N) 2 + (176.58 N)2 Tm = 615 N

For AB:



Cable BC

xB = 30 m, T0 =

yB = 3 m

wxB2 (3.924 N/m)(30 m) 2 = = 588.6 N (Checks) 2 yB 2(3 m)

W = wxB = (3.924 N/m)(30 m) = 117.72 N Tm2 = (588.6 N) 2 + (117.72 N) 2 Tm = 600 N

For BC


PROBLEM 7.109 Each cable of the Golden Gate Bridge supports a load w = 11.1 kips/ft along the horizontal. Knowing that the span L is 4150 ft and that the sag h is 464 ft, determine (a) the maximum tension in each cable, (b) the length of each cable.

SOLUTION Eq. (7.8) Page 386: At B:

(a)

yB =

wxB2 2T0

T0 =

wxB2 (11.1 kip/ft)(2075 ft)2 = 2 yB 2(464 ft)

T0 = 51.500 kips W = wxB = (11.1 kips/ft)(2075 ft) = 23.033 kips Tm = T02 + W 2 = (51.500 kips)2 + (23.033 kips) 2 Tm = 56, 400 kips

(b)

& sB = xB (1 + ( *

2

4

2 yB ! 2 yB ! " # − " # + 3 $ xB % 5 $ yB %

' ) ) +

yB 464 ft = = 0.22361 xB 2075 ft

2 & 2 sB = (2075 ft) (1 + (0.22361) 2 − (0.22361) 4 + 3 5 *

' ) = 2142.1 ft +

Length = 4280 ft

Length = 2sB = 2(2142.1 ft)


PROBLEM 7.110 The center span of the George Washington Bridge, as originally constructed, consisted of a uniform roadway suspended from four cables. The uniform load supported by each cable was w = 9.75 kips/ft along the horizontal. Knowing that the span L is 3500 ft and that the sag h is 316 ft, determine for the original configuration (a) the maximum tension in each cable, (b) the length of each cable.

SOLUTION W = wxB = (9.75 kips/ft) (1750 ft) W = 17,063 kips ΣM B = 0: T0 (316 ft) − (17063 kips) (875 ft) = c T0 = 47, 247 kips

(a)

Tm = T02 + W 2 = (47, 247 kips) 2 + (17, 063 kips) 2 Tm = 50, 200 kips

(b)

& 2 y !2 2 y !4 sB = xB (1 + " B # − " B # + 5 $ xB % ( 3 $ xB % * yB 316 ft = = 0.18057 xB 1750 ft

!

' ) ) +

2 & 2 = (1750 ft) (1+ (0.18057)2 − (0.18057) 4 + 3 5 * sB = 1787.3 ft; Length = 2 sB = 3579.6 ft

' ) + Length = 3580 ft


PROBLEM 7.111 The total mass of cable AC is 25 kg. Assuming that the mass of the cable is distributed uniformly along the horizontal, determine the sag h and the slope of the cable at A and C.

SOLUTION Cable:

m = 25 kg W = 25 (9.81) = 245.25 N

Block:

m = 450 kg W = 4414.5 N ΣM B = 0: (245.25) (2.5) + (4414.5)(3) − C x (2.5) = 0 C x = 5543 N ΣFx = 0: Ax = C x = 5543 N ΣM A = 0: C y (5) − (5543) (2.5) − (245.25) (2.5) = 0 C y = 2894 N + ΣFy = 0: C y − Ay − 245.25N = 0

2894 N − Ay − 245.25 N = 0

Point A:

tan φ A =

Point C:

tan φC =

Free body: Half cable

Ay Ax

Cy Cx

A y = 2649 N

=

2649 = 0.4779; 5543

φ A = 25.5°

=

2894 = 0.5221; 5543

φC = 27.6°

!

W = (12.5 kg) g = 122.6 ΣM 0 = 0: (122.6 N) (1.25 M) + (2649 N) (2.5 m) − (5543 N) yd = 0

yd = 1.2224 m; sag = h = 1.25 m − 1.2224 m h = 0.0276 m = 27.6 mm PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1140

PROBLEM 7.112 A 50.5-m length of wire having a mass per unit length of 0.75 kg/m is used to span a horizontal distance of 50 m. Determine (a) the approximate sag of the wire, (b) the maximum tension in the wire. [Hint: Use only the first two terms of Eq. (7.10).]

SOLUTION First two terms of Eq. 7.10 1 (50.5 m) = 25.25 m, 2 1 xB = (50 m) = 25 m 2 yB = h sB =

(a)

& sB = xB (1 + ( *

2 2 yB ! ' " # ) 3 $ xB % ) +

& 2 y !2 ' 25.25 m = 25 m (1 − " B # ) ( 3 $ xB % ) * + 2

2

yB ! 3! " # = 0.01" # = 0.015 $2% $ xB % yB = 0.12247 xB h = 0.12247 25m h = 3.0619 m

(b)

h = 3.06 m "

Free body: Portion CB w = (0.75 kg/m) (9.81m) = 7.3575 N/m W = sB w = (25.25 m) (7.3575 N/m) W = 185.78 N ΣM 0 = 0: T0 (3.0619 m) − (185.78 N) (12.5 m) = 0 T0 = 758.4 N Bx = T0 = 758.4 N + ΣFy = 0: By − 185.78 N = 0 By = 185.78 N Tm = Bx2 + By2 = (758.4 N) 2 + (185.78 N) 2

Tm = 781 N "


PROBLEM 7.113 A cable of length L + ∆ is suspended between two points that are at the same elevation and a distance L apart. (a) Assuming that ∆ is small compared to L and that the cable is parabolic, determine the approximate sag in terms of L and ∆. (b) If L = 100 ft and ∆ = 4 ft, determine the approximate sag. [Hint: Use only the first two terms of Eq. (7.10).

SOLUTION Eq. 7.10

(First two terms)

(a)

& sB = xB (1 + ( *

2 2 yB ! ' " # ) 3 $ xB % ) +

xB = L/2 1 ( L + ∆) 2 yB = h sB =

2' & 1 L( 2 h! ) ( L + ∆) = 1 + " L # 2 2 ( 3 "$ 2 #% ) * +

∆ 4 h2 3 = ; h 2 = L∆; 2 3 L 8

(b)

L = 100 ft, h = 4 ft.

h=

h=

3 (100)(4); 8

3 L∆ 8

h = 12.25 ft


PROBLEM 7.114 The center span of the Verrazano-Narrows Bridge consists of two uniform roadways suspended from four cables. The design of the bridge allows for the effect of extreme temperature changes that cause the sag of the center span to vary from hw = 386 ft in winter to hs = 394 ft in summer. Knowing that the span is L = 4260 ft, determine the change in length of the cables due to extreme temperature changes.

SOLUTION Eq. 7.10.

& sB = xB (1 + ( *

Winter:

yB = h = 386 ft, xB =

2

4

2 yB ! 2 yB ! " # − " # + 3 $ xB % 5 $ xB %

' ) ) +

1 L = 2130 ft 2

& 2 386 !2 2 386 !4 sB = (2130) (1 + " # − 5 " 2130 # + $ % (* 3 $ 2130 %

Summer:

yB = h = 394 ft, xB = & sB = (2130) (1 + (*

' ) = 2175.715 ft )+

1 L = 2130 ft 2 2

4

2 394 ! 2 394 ! − " + 3 "$ 2130 #% 5 $ 2130 #%

' ) = 2177.59 ft )+

∆ = 2( ∆ sB ) = 2(2177.59 ft − 2175.715 ft) = 2(1.875 ft)

Change in length = 3.75 ft


PROBLEM 7.115 Each cable of the side spans of the Golden Gate Bridge supports a load w = 10.2 kips/ft along the horizontal. Knowing that for the side spans the maximum vertical distance h from each cable to the chord AB is 30 ft and occurs at midspan, determine (a) the maximum tension in each cable, (b) the slope at B.

SOLUTION FBD AB: ΣM A = 0: (1100 ft)TBy − (496 ft)TBx − (550 ft)W = 0 ! 11TBy − 4.96TBx = 5.5W

(1)!

FBD CB:

ΣM C = 0: (550 ft)TBy − (278 ft)TBx − (275 ft)

W =0 2

11TBy − 5.56TBx = 2.75W

Solving (1) and (2)

TBy = 28, 798 kips

Solving (1) and (2

TBx = 51, 425 kips ! Tmax = TB = TB2x + TB2y

So that

tan θ B =

TBy

(2)

!

TBx

(a)

Tmax = 58,940 kips

(b)

θ B = 29.2°

!


PROBLEM 7.116 A steam pipe weighting 45 lb/ft that passes between two buildings 40 ft apart is supported by a system of cables as shown. Assuming that the weight of the cable system is equivalent to a uniformly distributed loading of 5 lb/ft, determine (a) the location of the lowest Point C of the cable, (b) the maximum tension in the cable.

SOLUTION Note:

xB − x A = 40 ft

or

x A = xB − 40 ft

(a)

Use Eq. 7.8 Point A:

yA =

wx A2 w ( xB − 40) 2 ; 9= 2T0 2T0

(1)

Point B:

yB =

wxB2 wx 2 ; 4= B 2T0 2T0

(2)

Dividing (1) by (2): (b)

9 ( xB − 40) 2 = ; xB = 16 ft 4 xB2

Point C is 16 ft to left of B

Maximum slope and thus Tmax is at A x A = xB − 40 = 16 − 40 = −24 ft yA =

wx A2 (50 lb/ft)(−24 ft) 2 ; 9 ft = ; T0 = 1600 lb 2T0 2T0

WAC = (50 lb/ft)(24 ft) = 1200 lb

Tmax = 2000 lb


PROBLEM 7.117 Cable AB supports a load uniformly distributed along the horizontal as shown. Knowing that at B the cable forms an angle θB = 35° with the horizontal, determine (a) the maximum tension in the cable, (b) the vertical distance a from A to the lowest point of the cable.


By = Bx tan 35° W = (45 kg/m)(12 m)(9.81 m/s 2 ) W = 5297.4 N ΣM A = 0 : W (6 m) + Bx (1.8 m) − By (12 m) = 0 (5297.4)(6) + 1.8Bx − Bx tan 35°(12) = 0 Bx = 4814 N By = (4814 N) tan 35° = 3370.8 N

Free body: Portion CB ΣFy = 0: By − WBC = 0 WBC = By = 3370.8 N WBC = (45 kg/m)(9.81 m/s 2 )b 3370.8 N = (441.45 N/m)b b = 7.6357 m 1 ! ΣM B = 0: T0 dC − WBC " b # = 0 $2 % 1 (4814 N)dC − (3370.8 N) (7.6357 m) = 0 2 dC = 2.6733 m

(a)

dC = 1.8 m + a; 2.6733 m = 1.8 m + a;

(b)

Tm = B = Bx2 + By2 = (4814 N) 2 + (3370.8 N) Tm = 5877 N

a = 0.873 m

Tm = 5880 N


PROBLEM 7.118 Cable AB supports a load uniformly distributed along the horizontal as shown. Knowing that the lowest point of the cable is located at a distance a = 0.6 m below A, determine (a) the maximum tension in the cable, (b) the angle θB that the cable forms with the horizontal at B.

SOLUTION Note:

xB − x A = 12 m

or

x A = xB − 12 m

Point A:

yA =

wx A2 w( xB − 12) 2 ; 0.6 = 2T0 2T0

(1)

Point B:

yB =

wxB2 wx 2 ; 2.4 = B 2T0 2T0

(2)

Dividing (1) by (2):

0.6 ( xB − 12) 2 ; xB = 8 m = 2.4 xB2

(a)

2.4 =

Eq. (2):

w(8) 2 ; T0 = 13.333w 2T0

Free body: Portion CB ΣFy = 0 By = wxB B y = 8w

Tm2 = Bx2 + By2 ; Tm2 = (13.333w) 2 + (8w) 2 Tm = 15.549w = 15.549(45)(9.81)

θ B = tan −1 By /Bx = tan −1 8w/13.333w

Tm = 6860 N

θ B = 31.0°


PROBLEM 7.119* A cable AB of span L and a simple beam A′B′ of the same span are subjected to identical vertical loadings as shown. Show that the magnitude of the bending moment at a point C′ in the beam is equal to the product T0h, where T0 is the magnitude of the horizontal component of the tension force in the cable and h is the vertical distance between Point C and the chord joining the points of support A and B.

SOLUTION ΣM B = 0: LACy + aT0 − ΣM B loads = 0

(1)

FBD Cable: (Where ΣM B loads includes all applied loads) x! ΣM C = 0: xACy − " h − a # T0 − ΣM C left = 0 L% $

(2)

FBD AC: (Where ΣM C left includes all loads left of C) x x (1) − (2): hT0 − ΣM B loads + ΣM C left = 0 L L

(3)

ΣM B = 0: LABy − ΣM B loads = 0

(4)

ΣM C = 0: xABy − ΣM C left − M C = 0

(5)

FBD Beam:

FBD AC:

Comparing (3) and (6)

x x (4) − (5): − ΣM B loads + ΣM C left + M C = 0 L L M C = hT0

(6) Q.E.D.


PROBLEM 7.120 Making use of the property established in Problem 7.119, solve the problem indicated by first solving the corresponding beam problem. PROBLEM 7.94 (a) Knowing that the maximum tension in cable ABCD is 15 kN, determine the distance hB .

SOLUTION ΣM B = 0: A(10 m) − (6 kN)(7 m) − (10 kN)(4 m) = 0 A = 8.2 kN ΣFy = 0: 8.2 kN − 6 kN − 10 kN + B = 0 B = 7.8 kN

At A:

Tm2 = T02 + A2 152 = T02 + 8.2 T0 = 12.56 kN

At B:

M B = T0 hB ; 24.6 kN ⋅ m = (12.56 kN)hB ;

hB = 1.959 m

At C:

M C = T0 hC ; 31.2 kN ⋅ m = (12.56 kN)hC ;

hC = 2.48 m

!


PROBLEM 7.121 Making use of the property established in Problem 7.119, solve the problem indicated by first solving the corresponding beam problem. PROBLEM 7.97 (a) Knowing that dC = 3 m, determine the distances dB and dD .

SOLUTION

ΣM B = 0: A(10 m) − (5 kN)(8 m) − (5 kN)(6 m) − (10 kN)(3 m) = 0

A = 10 kN

Geometry: dC = 1.6 m + hC 3 m = 1.6 m + hC hC = 1.4 m

Since M = T0h, h is proportional to M, thus h hB h = C = D ; M B MC M D

hB hD 1.4 m = = 20 kN ⋅ m 30 kN ⋅ m 30 kN ⋅ m

20 ! hB = 1.4 " # = 0.9333 m 30 $ %

30 ! hD = 1.4 " # = 1.4 m $ 30 %

d B = 0.8 m + 0.9333 m

d D = 2.8 m + 1.4 m

d B = 1.733 m

!

d D = 4.20 m


PROBLEM 7.122 Making use of the property established in Problem 7.119, solve the problem indicated by first solving the corresponding beam problem. PROBLEM 7.99 (a) If dC = 15 ft, determine the distances dB and dD.

SOLUTION Free body: Beam AE ΣM E = 0: − A(30) + 2(24) + 2(15) + 2(9) = 0

A = 3.2 kips +ΣFy = 0: 3.2 − 3(2) + B = 0

B = 2.8 kips

Geometry: Given:

dC = 15 ft

Then,

hC = dC − 3.75 ft = 11.25 ft

Since M = T0 h, h is proportional to M. Thus, h hB h = C = D M B MC M D

Then,

or,

hB h 11.25 ft = = D 19.2 30 25.2

or,

hB = 7.2 ft hD = 9.45 ft

d B = 6 + hB = 6 + 7.2

d B = 13.20 ft

d D = 2.25 + hD = 2.25 + 9.45

d D = 11.70 ft


PROBLEM 7.123 Making use of the property established in Problem 7.119, solve the problem indicated by first solving the corresponding beam problem. PROBLEM 7.100 (a) Determine the distance dC for which portion BC of the cable is horizontal.

SOLUTION Free body: Beam AE ΣM E = 0: − A(30) + 2(24) + 2(15) + 2(9) = 0 A = 3.2 kips

ΣFy = 0: 3.2 − 3(2) + B = 0 B = 2.8 kips

Geometry: Given: Then,

dC = d B 3.75 + hC = 6 + hB hC = 2.25 + hB

(1)

Since M = T0h, h is proportional to M. Thus, h hB = C M B MC

or,

h hB = C 19.2 30

hB = 0.64hC

(2)

Substituting (2) into (1): hC = 2.25 + 0.64hC

Then,

hC = 6.25 ft

dC = 3.75 + hC = 3.75 + 6.25

dC = 10.00 ft


PROBLEM 7.124* Show that the curve assumed by a cable that carries a distributed load w(x) is defined by the differential equation d2y/dx2 = w(x)/T0, where T0 is the tension at the lowest point.

SOLUTION FBD Elemental segment: ΣFy = 0: Ty ( x + ∆ x) − Ty ( x) − w( x)∆ x = 0

So

Ty ( x + ∆ x ) T0

−

Ty ( x )

Ty

But

So

In lim : ∆ x →0

T0

T0

dy dx

− x +∆x

∆x

dy dx

x

=

w( x) ∆x T0

=

dy dx

=

w( x) T0

d 2 y w( x) = T0 dx 2

Q.E.D.


PROBLEM 7.125* Using the property indicated in Problem 7.124, determine the curve assumed by a cable of span L and sag h carrying a distributed load w = w0 cos (πx/L), where x is measured from mid-span. Also determine the maximum and minimum values of the tension in the cable. PROBLEM 7.124 Show that the curve assumed by a cable that carries a distributed load w(x) is defined by the differential equation d2y/dx2 = w(x)/T0, where T0 is the tension at the lowest point.

SOLUTION

w( x) = w0 cos

πx L

From Problem 7.124 d 2 y w( x) w0 πx cos = = 2 T0 T0 L dx

So

! dy = 0# " using dx 0 $ %

πx dy W0 L = sin dx T0π L

y=

But

And

w0 L2

πx! "1 − cos L # [using y (0) = 0] T0π $ % 2

!

w L2 w L2 π! L! y " # = h = 0 2 "1 − cos # so T0 = 02 2% π h T0π $ $2% T0 = Tmin

Tmax = TA = TB :

TBy =

Tmin =

so TBy T0

=

w0 L

dy dx

= x = L/2

π 2h

w0 L T0π 2 TB = TBy + T02 =

π

w0 L2

w0 L

π

L ! 1+ " # π $ h%

2


!

PROBLEM 7.126 If the weight per unit length of the cable AB is w0/cos2 θ, prove that the curve formed by the cable is a circular arc. (Hint: Use the property indicated in Problem 7.124.) PROBLEM 7.124 Show that the curve assumed by a cable that carries a distributed load w(x) is defined by the differential equation d2y/dx2 = w(x)/T0, where T0 is the tension at the lowest point.

SOLUTION Elemental Segment: Load on segment*

w ( x)dx =

w0 cos 2 θ

ds

dx = cos θ ds, so w( x) =

But

w0 cos3 θ

From Problem 7.119

w0 d 2 y w( x) = = T0 dx 2 T0 cos3 θ

In general

d 2 y d dy ! d dθ = " # = (tan θ ) = sec2 θ 2 dx $ dx % dx dx dx

So

or

Giving r =

w0 w0 dθ = = dx T0 cos3 θ sec 2 θ T0 cos θ T0 cos θ dθ = dx = rdθ cos θ w0 T0 = constant. So curve is circular arc w0

Q.E.D.!

*For large sag, it is not appropriate to approximate ds by dx.


PROBLEM 7.127 A 30-m cable is strung as shown between two buildings. The maximum tension is found to be 500 N, and the lowest point of the cable is observed to be 4 m above the ground. Determine (a) the horizontal distance between the buildings, (b) the total mass of the cable.

SOLUTION

sB = 15 m Tm = 500 N

Eq. 7.17:

yB2 − sB2 = c 2 ; (6 + c) 2 − 152 = c 2 36 + 12c + c 2 − 225 = c 2 12c = 189 c = 15.75 m

Eq. 7.15:

xB = 0.8473(15.75) = 13.345 m; L = 2 xB

(a) (b)

xB x ; 15 = (15.75) sinh B c c xB xB sinh = 0.95238 = 0.8473 c c sB = c sinh

Eq. 7.18:

L = 26.7 m

Tm = wyB ; 500 N = w(6 + 15.75) w = 22.99 N/m W = 2sB w = (30 m)(22.99 N/m) = 689.7 N 689.7 N W = m= g 9.81 m/s 2

Total mass = 70.3 kg


PROBLEM 7.128 A 200-ft steel surveying tape weighs 4 lb. If the tape is stretched between two points at the same elevation and pulled until the tension at each end is 16 lb, determine the horizontal distance between the ends of the tape. Neglect the elongation of the tape due to the tension.

SOLUTION

sB = 100 ft 4 lb ! w=" = 0.02 lb/ft Tm = 16 m 200 ft #% $

Eq. 7.18: Eq. 7.17: Eq. 7.15:

Tm = wyB ; 16 lb = (0.02 lb/ft) yB ;

yB = 800 ft

yB2 − sB2 = c 2 ; (800)2 − (100) 2 = c 2 ; c = 793.73 ft sB = c sinh

xB x ; 100 = 793.73 sinh B c c

xB = 0.12566; xB = 99.737 ft c L = 2 xB = 2(99.737 ft)

L = 199.5 ft


PROBLEM 7.129 A 200-m-long aerial tramway cable having a mass per unit length of 3.5 kg/m is suspended between two points at the same elevation. Knowing that the sag is 50 m, find (a) the horizontal distance between the supports, (b) the maximum tension in the cable.

SOLUTION

Given: Length = 200 m Unit mass = 3.5 kg/m h = 50 m w = (3.5 kg/m)(9.81 m/s 2 ) = 34.335 N/m

Then,

sB = 100 m yB = h + c = 50 m + c yB2

− sB2 = c 2 ; (50 + c)2 − (100) 2 = c 2

502 + 100c + c 2 − 1002 = c 2 c = 75 m sB = c sinh

xB x ; 100 = 75sinh B c 75 xB = 82.396 m

span = L = 2 xB = 2(82.396 m) Tm = wyB = (34.335 N/m)(50 m + 75 m)

L = 164.8 m Tm = 4290 N


PROBLEM 7.130 An electric transmission cable of length 400 ft weighing 2.5 lb/ft is suspended between two points at the same elevation. Knowing that the sag is 100 ft, determine the horizontal distance between the supports and the maximum tension.

SOLUTION

sB = 200 ft

Eq. 7.17:

yB2 − sB2 = c 2 ; (100 + c) 2 − 2002 = c 2 10000 + 200c + c 2 − 40000 = c 2 ; c = 150 ft sB = c sinh

Eq. 7.15: sinh

xB x ; 200 = 150sinh B c c

xB 4 xB = ; = 1.0986 3 c c xB = (150)(1.0986) = 164.79 ft

L = 2 xB = 2(164.79 ft) = 329.58 ft

Eq. 7.18:

Tm = wyB = (2.5 lb/ft)(100 ft + 150 ft)

L = 330 ft Tm = 625 lb


PROBLEM 7.131 A 20-m length of wire having a mass per unit length of 0.2 kg/m is attached to a fixed support at A and to a collar at B. Neglecting the effect of friction, determine (a) the force P for which h = 8 m, (b) the corresponding span L.

SOLUTION FBD Cable:

sT = 20 m " so sB = $

20 m ! = 10 m # 2 %

w = (0.2 kg/m)(9.81 m/s 2 ) = 1.96200 N/m hB = 8 m yB2 = (c + hB )2 = c 2 + sB2

So

c=

sB2 − hB2 2hB

(10 m)2 − (8 m)2 2(8 m) = 2.250 m

c=

Now

xB s → xB = c sinh −1 B c c 10 m ! = (2.250 m)sinh −1 " # 2.250 m% $

sB = c sinh

xB = 4.9438 m P = T0 = wc = (1.96200 N/m)(2.250 m) L = 2 xB = 2(4.9438 m)

(a) (b)

P = 4.41 N

L = 9.89 m


!

PROBLEM 7.132 A 20-m length of wire having a mass per unit length of 0.2 kg/m is attached to a fixed support at A and to a collar at B. Knowing that the magnitude of the horizontal force applied to the collar is P = 20 N, determine (a) the sag h, (b) the span L.

SOLUTION FBD Cable:

sT = 20 m, w = (0.2 kg/m)(9.81 m/s 2 ) = 1.96200 N/m P = T0 = wc c = c=

P w

20 N = 10.1937 m 1.9620 N/m

yB2 = (hB + c) 2 = c 2 + sB2 h 2 + 2ch − sB2 = 0 sB =

20 m = 10 m 2

h 2 + 2(10.1937 m)h − 100 m 2 = 0 h = 4.0861 m sB = c sinh

(a)

h = 4.09 m

xA s 10 m ! → xB = c sinh −1 B = (10.1937 m)sinh −1 " # c c $ 10.1937 m % = 8.8468 m L = 2 xB = 2(8.8468 m)

(b)

L = 17.69 m


PROBLEM 7.133 A 20-m length of wire having a mass per unit length of 0.2 kg/m is attached to a fixed support at A and to a collar at B. Neglecting the effect of friction, determine (a) the sag h for which L = 15 m, (b) the corresponding force P.

SOLUTION FBD Cable:

20 m = 10 m 2 w = (0.2 kg/m)(9.81 m/s 2 ) = 1.96200 N/m L = 15 m

sT = 20 m → sB =

L xB = c sinh 2 c c 7.5 m 10 m = c sinh c

sB = c sinh

Solving numerically:

c = 5.5504 m x yB = c cosh " B $ c yB = 11.4371 m

7.5 ! ! # = (5.5504) cosh " 5.5504 # $ % %

hB = yB − c = 11.4371 m − 5.5504 m

(a) P = wc = (1.96200 N/m)(5.5504 m)

(b)

hB = 5.89 m

P = 10.89 N


PROBLEM 7.134 Determine the sag of a 30-ft chain that is attached to two points at the same elevation that are 20 ft apart.

SOLUTION

30 ft = 15 ft 2 L xB = = 10 ft 2 x sB = c sinh B c 10 ft 15 ft = c sinh c sB =


L = 20 ft

c = 6.1647 ft yB = c cosh

xB c

= (6.1647 ft)cosh

10 ft 6.1647 ft

= 16.2174 ft hB = yB − c = 16.2174 ft − 6.1647 ft

hB = 10.05 ft


PROBLEM 7.135 A 90-m wire is suspended between two points at the same elevation that are 60 m apart. Knowing that the maximum tension is 300 N, determine (a) the sag of the wire, (b) the total mass of the wire.

SOLUTION

sB = 45 m

Eq. 7.17:

Eq. 7.16:

xB c 30 45 = c sinh ; c = 18.494 m c sB = c sinh

yB = c cosh

xB c

yB = (18.494) cosh

30 18.494

yB = 48.652 m yB = h + c 48.652 = h + 18.494 h = 30.158 m

Eq. 7.18:

h = 30.2 m

Tm = wyB 300 N = w(48.652 m) w = 6.166 N/m

Total weight of cable

W = w(Length) = (6.166 N/m)(90 m) = 554.96 N

Total mass of cable

m=

W 554.96 N = = 56.57 kg 9.81 m/s g

m = 56.6 kg


PROBLEM 7.136 A counterweight D is attached to a cable that passes over a small pulley at A and is attached to a support at B. Knowing that L = 45 ft and h = 15 ft, determine (a) the length of the cable from A to B, (b) the weight per unit length of the cable. Neglect the weight of the cable from A to D.

SOLUTION Given:

L = 45 ft h = 15 ft TA = 80 lb xB = 22.5 ft

By symmetry:

TB = TA = Tm = 80 lb

We have

yB = c cosh

and

yB = h + c = 15 + c

Then or Solve by trial for c: (a)

xB 22.5 = c cosh c c

c cosh

22.5 = 15 + c c

cosh

22.5 15 = +1 c c c = 18.9525 ft sB = c sinh

xB c

= (18.9525 ft) sinh

22.5 18.9525

= 28.170 ft

Length = 2sB = 2(28.170 ft) = 56.3 ft (b)

Tm = wyB = w(h + c) 80 lb = w(15 ft + 18.9525 ft)

w = 2.36 lb/ft


PROBLEM 7.137 A uniform cord 50 in. long passes over a pulley at B and is attached to a pin support at A. Knowing that L = 20 in. and neglecting the effect of friction, determine the smaller of the two values of h for which the cord is in equilibrium.

SOLUTION

b = 50 in. − 2sB

Length of overhang: Weight of overhang equals max. tension

Tm = TB = wb = w(50 in. − 2sB )

Eq. 7.15:

sB = c sinh

xB c

Eq. 7.16:

yB = c cosh

xB c

Eq. 7.18:

Tm = wyB w(50 in. − 2sB ) = wyB x ! x w " 50 in. − 2c sinh B # = wc cosh B c c $ % xB = 10: 50 − 2c sinh

Solve by trial and error: For c = 5.549 in.

c = 5.549 in.

and

10 10 = c cosh c c

c = 27.742 in.

10 in. = 17.277 in. 5.549 in. yB = h + c; 17.277 in. = h + 5.549 in. yB = (5.549 in.) cosh

h = 11.728 in.

For c = 27.742 in.

h = 11.73 in.

10 in. = 29.564 in. 27.742 in. yB = h + c; 29.564 in. = h + 27.742 in. yB = (27.742 in.) cosh

h = 1.8219 in.

h = 1.822 in.


PROBLEM 7.138 A cable weighing 2 lb/ft is suspended between two points at the same elevation that are 160 ft apart. Determine the smallest allowable sag of the cable if the maximum tension is not to exceed 400 lb.

SOLUTION

Eq. 7.18: Eq. 7.16:

Tm = wyB ; 400 lb = (2 lb/ft) yB ;

yB = 200 ft

xB c 80 ft 200 ft = c cosh c yB = c cosh

Solve for c: c = 182.148 ft and c = 31.592 ft yB = h + c; h = yB − c

For

c = 182.148 ft;

h = 200 − 182.147 = 17.852 ft "

For

c = 31.592 ft;

h = 200 − 31.592 = 168.408 ft "

For

smallest h = 17.85 ft

Tm # 400 lb:


PROBLEM 7.139 A motor M is used to slowly reel in the cable shown. Knowing that the mass per unit length of the cable is 0.4 kg/m, determine the maximum tension in the cable when h = 5 m.

SOLUTION

w = 0.4 kg/m L = 10 m hB = 5 m xB c L hB + c = c cosh 2c 5m ! 5 m = c " cosh − 1# c $ % yB = c cosh


c = 3.0938 m yB = hB + c = 5 m + 3.0938 m Tmax

= 8.0938 m = TB = wyB = (0.4 kg/m)(9.81 m/s 2 )(8.0938 m) Tmax = 31.8 N


PROBLEM 7.140 A motor M is used to slowly reel in the cable shown. Knowing that the mass per unit length of the cable is 0.4 kg/m, determine the maximum tension in the cable when h = 3 m.

SOLUTION

w = 0.4 kg/m, L = 10 m, hB = 3 m xB L = c cosh c 2c 5m ! − 1# 3 m = c " c cosh c $ % yB = hB + c = c cosh


c = 4.5945 m yB = hB + c = 3 m + 4.5945 m Tmax

= 7.5945 m = TB = wyB = (0.4 kg/m)(9.81 m/s 2 )(7.5945 m) Tmax = 29.8 N


PROBLEM 7.141 A uniform cable weighing 3 lb/ft is held in the position shown by a horizontal force P applied at B. Knowing that P = 180 lb and θA = 60°, determine (a) the location of Point B, (b) the length of the cable.

SOLUTION T0 = P = cw

Eq. 7.18:

c=

c = 60 ft

P cos 60° cw = = 2cw 0.5

Tm =

At A:

(a)

P 180 lb = w 3 lb/ft

Tm = w(h + c)

Eq. 7.18:

2cw = w(h + c) 2c = h + c h = b = c

b = 60.0 ft

xA c x h + c = c cosh A c y A = c cosh

Eq. 7.16:

(60 ft + 60 ft) = (60 ft) cosh cosh

xA =2 60 m

xA 60

xA = 1.3170 60 m

x A = 79.02 ft

(b)

Eq. 7.15:

a = 79.0 ft

xB 79.02 ft = (60 ft)sinh c 60 ft s A = 103.92 ft s A = c sinh

length = s A

s A = 103.9 ft


PROBLEM 7.142 A uniform cable weighing 3 lb/ft is held in the position shown by a horizontal force P applied at B. Knowing that P = 150 lb and θA = 60°, determine (a) the location of Point B, (b) the length of the cable.

SOLUTION Eq. 7.18:

T0 = P = cw c=

P cos 60° cw = = 2 cw 0.5

Tm =

At A:

(a)

P 150 lb = = 50 ft w 3 lb/ft

Tm = w(h + c)

Eq. 7.18:

2cw = w(h + c) 2c = h + c h = c = b

Eq. 7.16:

xA c xA h + c = c cosh c y A = c cosh

(50 ft + 50 ft) = (50 ft) cosh cosh

xA =2 c

xA c

xA = 1.3170 c

x A = 1.3170(50 ft) = 65.85 ft

(b)

Eq. 7.15:

b = 50.0 ft

s A = c sinh

a = 65.8 ft

xA 65.85 ft = (50 ft)sinh c 50 ft

s A = 86.6 ft

length = s A = 86.6 ft


PROBLEM 7.143 To the left of Point B the long cable ABDE rests on the rough horizontal surface shown. Knowing that the mass per unit length of the cable is 2 kg/m, determine the force F when a = 3.6 m.

SOLUTION

xD = a = 3.6 m h = 4 m xD c a h + c = c cosh c 3.6 m ! 4 m = c " cosh − 1# c $ % yD = c cosh

Solving numerically Then

c = 2.0712 m yB = h + c = 4 m + 2.0712 m = 6.0712 m F = Tmax = wyB = (2 kg/m)(9.81 m/s 2 )(6.0712 m)

F = 119.1 N


PROBLEM 7.144 To the left of Point B the long cable ABDE rests on the rough horizontal surface shown. Knowing that the mass per unit length of the cable is 2 kg/m, determine the force F when a = 6 m.

SOLUTION

xD = a = 6 m h = 4 m xD c a h + c = c cosh c 6m ! 4 m = c " cosh − 1# c $ % y D = c cosh

Solving numerically

c = 5.054 m yB = h + c = 4 m + 5.054 m = 9.054 m F = TD = wyD = (2 kg/m)(9.81 m/s 2 )(9.054 m)

F = 177.6 N


PROBLEM 7.145 The cable ACB has a mass per unit length of 0.45 kg/m. Knowing that the lowest point of the cable is located at a distance a = 0.6 m below the support A, determine (a) the location of the lowest Point C, (b) the maximum tension in the cable.

SOLUTION xB − x A = 12 m

Note:

− x A = 12 m − xB

or, Point A:

y A = c cosh

12 − xB − xA ; c + 0.6 = c cosh c c

(1)

Point B:

yB = c cosh

xB x ; c + 2.4 = c cosh B c c

(2)

From (1):

c + 0.6 ! 12 xB − = cosh −1 " # c c $ c %

(3)

From (2):

xB c + 2.4 ! = cosh −1 " # c $ c %

(4)

Add (3) + (4):

12 c + 0.6 ! c + 2.4 ! = cosh −1 " + cosh −1 " # # c $ c % $ c % c = 13.6214 m

Solve by trial and error: Eq. (2)

13.6214 + 2.4 = 13.6214 cosh cosh

xB = 1.1762; c

xB c

xB = 0.58523 c

xB = 0.58523(13.6214 m) = 7.9717 m

Point C is 7.97 m to left of B yB = c + 2.4 = 13.6214 + 2.4 = 16.0214 m

Eq. 7.18:

Tm = wyB = (0.45 kg/m)(9.81 m/s 2 )(16.0214 m) Tm = 70.726 N

Tm = 70.7 N


PROBLEM 7.146 The cable ACB has a mass per unit length of 0.45 kg/m. Knowing that the lowest point of the cable is located at a distance a = 2 m below the support A, determine (a) the location of the lowest Point C, (b) the maximum tension in the cable.

SOLUTION Note: xB − x A = 12 m

or − x A = 12 m − xB

Point A:

y A = c cosh

12 − xB − xA ; c + 2 = c cosh c c

(1)

Point B:

yB = c cosh

xB x ; c + 3.8 = c cosh B c c

(2)

c+2! 12 xB − = cosh −1 " # c c $ c %

From (1):

(3)

From (2):

xB c + 3.8 ! = cosh −1 " # c $ c %

Add (3) + (4):

12 c+2! c + 3.8 ! = cosh −1 " + cosh −1 " # # c $ c % $ c % c = 6.8154 m

Solve by trial and error: Eq. (2):

(4)

6.8154 m + 3.8 m = (6.8154 m) cosh cosh

xB c

xB xB = 1.5576 = 1.0122 c c xB = 1.0122(6.8154 m) = 6.899 m

Point C is 6.90 m to left of B yB = c + 3.8 = 6.8154 + 3.8 = 10.6154 m

Eq. (7.18):

Tm = wyB = (0.45 kg/m)(9.81 m/s 2 )(10.6154 m) Tm = 46.86 N

Tm = 46.9 N


PROBLEM 7.147* The 10-ft cable AB is attached to two collars as shown. The collar at A can slide freely along the rod; a stop attached to the rod prevents the collar at B from moving on the rod. Neglecting the effect of friction and the weight of the collars, determine the distance a.

SOLUTION Collar at A: Since µ = 0, cable ' rod

Point A:

x dy x = sinh y = c cosh ; c dx c xA dy tan θ = = sinh c dx A xA = sinh(tan (90° − θ )) c x A = c sinh(tan (90° − θ ))

Length of cable = 10 ft

(1)

10 ft = AC + CB xA x + c sinh B c c xB 10 xA = − sinh sinh c c c 10 = c sinh

x ' &10 xB = c sinh −1 ( − sinh A ) c c + * x x y A = c cosh A yB = c cosh B c c

In ∆ ABD:

tan θ =

yB − y A xB + x A

(2) (3) (4)



Method of solution: For given value of θ, choose trial value of c and calculate: From Eq. (1): xA Using value of xA and c, calculate: From Eq. (2): xB From Eq. (3): yA and yB Substitute values obtained for xA, xB, yA, yB into Eq. (4) and calculate θ Choose new trial value of θ and repeat above procedure until calculated value of θ is equal to given value of θ. For θ = 30° Result of trial and error procedure: c = 1.803 ft x A = 2.3745 ft xB = 3.6937 ft y A = 3.606 ft yB = 7.109 ft a = yB − y A = 7.109 ft − 3.606 ft = 3.503 ft

a = 3.50 ft


PROBLEM 7.148* Solve Problem 7.147 assuming that the angle θ formed by the rod and the horizontal is 45°. PROBLEM 7.147 The 10-ft cable AB is attached to two collars as shown. The collar at A can slide freely along the rod; a stop attached to the rod prevents the collar at B from moving on the rod. Neglecting the effect of friction and the weight of the collars, determine the distance a.

SOLUTION Collar at A: Since µ = 0, cable ' rod

Point A:

x dy x = sinh y = c cosh ; c dx c x dy tan θ = = sinh A c dx A xA = sinh(tan (90° − θ )) c x A = c sinh(tan (90° − θ ))

Length of cable = 10 ft

(1)

10 ft = AC + CB x x 10 = c sinh A + c sinh B c c xB 10 xA sinh = − sinh c c c x ' &10 xB = c sinh −1 ( − sinh A ) c + *c y A = c cosh

xA c

yB = c cosh

(2) xB c

(3)



In ∆ ABD:

tan θ =

yB − y A xB + x A

(4)

Method of solution: For given value of θ, choose trial value of c and calculate: From Eq. (1): xA Using value of xA and c, calculate: From Eq. (2): xB From Eq. (3): yA and yB Substitute values obtained for xA, xB, yA, yB into Eq. (4) and calculate θ Choose new trial value of θ and repeat above procedure until calculated value of θ is equal to given value of θ. For θ = 45° Result of trial and error procedure: c = 1.8652 ft x A = 1.644 ft xB = 4.064 ft y A = 2.638 ft yB = 8.346 ft a = yB − y A = 8.346 ft − 2.638 ft = 5.708 ft

a = 5.71 ft


PROBLEM 7.149 Denoting by θ the angle formed by a uniform cable and the horizontal, show that at any point (a) s = c tan θ, (b) y = c sec θ.

SOLUTION dy x = sinh dx c x s = c sinh = c tan θ c

tan θ =

(a)

(b)

Q.E.D.

Also

y 2 = s 2 + c 2 (cosh 2 x = sinh 2 x + 1)

so

y 2 = c 2 (tan 2 θ + 1)c 2 sec2 θ

and

y = c secθ

Q.E.D.


PROBLEM 7.150* (a) Determine the maximum allowable horizontal span for a uniform cable of weight per unit length w if the tension in the cable is not to exceed a given value Tm. (b) Using the result of part a, determine the maximum span of a steel wire for which w = 0.25 lb/ft and Tm = 8000 lb.

SOLUTION

Tm = wyB

(a)

xB c ! x 1 # # cosh B xB # c # c %

= wc cosh " = wxB " "" $

We shall find ratio

( ) for when T xB c

m

is minimum

2 & ' ! ( " # d Tm x xB ) 1 1 = wxB ( sinh B − " # cosh ) = 0 ( xB x ! c " xB # c ) d" B # " # ( ) $ c % $ c % * c + x sinh B c = 1 xB xB cosh c c xB c = tanh c xB

Solve by trial and error for: sB = c sinh

Eq. 7.17:

xB = c sinh(1.200): c

xB = 1.200 c

(1)

sB = 1.509 c yB2 − sB2 = c 2 2 & s ! ' yB2 = c 2 (1 + " B # ) = c 2 (1 + 1.5092 ) (* $ c % )+ yB = 1.810c



Eq. 7.18:

Tm = wyB = 1.810 wc Tm c= 1.810 w

Eq. (1): Span: (b)

xB = 1.200c = 1.200

Tm T = 0.6630 m w 1.810 w

L = 2 xB = 2(0.6630)

Tm w

L = 1.326

Tm w

For w = 0.25lb/ft and Tm = 8000 lb 8000 lb 0.25lb/ft = 42, 432ft

L = 1.326

L = 8.04 miles


PROBLEM 7.151* A cable has a mass per unit length of 3 kg/m and is supported as shown. Knowing that the span L is 6 m, determine the two values of the sag h for which the maximum tension is 350 N.

SOLUTION

L =h+c 2c w = (3 kg/m)(9.81 m/s 2 ) = 29.43 N/m

ymax = c cosh

Tmax = wymax Tmax w 350 N = = 11.893 m 29.43 N/m

ymax = ymax c cosh

Solving numerically

3m = 11.893 m c c1 = 0.9241 m c2 = 11.499 m h = ymax − c h1 = 11.893 m − 0.9241 m

h1 = 10.97 m

h2 = 11.893 m − 11.499 m

h2 = 0.394 m


PROBLEM 7.152* Determine the sag-to-span ratio for which the maximum tension in the cable is equal to the total weight of the entire cable AB.

SOLUTION

Tmax = wyB = 2wsB y B = 2 sB L 2c L tanh 2c L 2c hB c

c cosh

= 2c sinh =

L 2c

1 2

1 = 0.549306 2 y −c L = B = cosh − 1 = 0.154701 2c c = tanh −1

h

B hB 0.5(0.154701) = cL = = 0.14081 0.549306 L 2 ( 2c )

hB = 0.1408 L


PROBLEM 7.153* A cable of weight w per unit length is suspended between two points at the same elevation that are a distance L apart. Determine (a) the sag-to-span ratio for which the maximum tension is as small as possible, (b) the corresponding values of θB and Tm.

SOLUTION

L 2c L L L! = w " cosh − sinh # 2c 2c 2c % $

Tmax = wyB = wc cosh

(a)

dTmax dc

For

min Tmax ,

dTmax =0 dc tanh

L 2c = 2c L

L = 1.1997 2c

yB L = cosh = 1.8102 2c c h yB = − 1 = 0.8102 c c 0.8102 h & 1 h 2c ! ' =( = 0.3375 )= L * 2 c "$ L #% + 2(1.1997) T0 = wc Tmax = wc cosh

(b)

But

T0 = Tmax cos θ B

So

θ B = sec −1 "

L 2c

!

θ B = 56.5°

!

Tmax L yB = cosh = 2c T0 c

Tmax = sec θ B T0

yB ! −1 # = sec (1.8102) = 56.46° $ c %

Tmax = wyB = w

h = 0.338 L

y B 2c ! L ! L = w(1.8102) " #" # c $ L %$ 2 % 2(1.1997)

Tmax = 0.755wL


PROBLEM 7.154 It has been experimentally determined that the bending moment at Point K of the frame shown is 300 N ⋅ m. Determine (a) the tension in rods AE and FD, (b) the corresponding internal forces at Point J.

SOLUTION Free body: ABK

(a)

ΣM K = 0: 300 N ⋅ m −

8 15 T (0.2 m) − T (0.12 m) = 0 17 17

T = 1500 N

Free body: AJ 8 8 T = (1500 N) 17 17 = 705.88 N

Tx =

15 15 T = (1500 N) 17 17 = 1323.53 N

Ty =

Internal forces on ABJ (b)

ΣFx = 0: 705.88 N − V = 0 V = + 705.88 N

V = 706 N

ΣFy = 0: F − 1323.53 N = 0 F = + 1323.53 N

F = 1324 N

ΣM J = 0: M − (705.88 N)(0.1m) − (1323.53 N)(0.12 m) = 0 M = + 229.4 N ⋅ m

M = 229 N ⋅ m


PROBLEM 7.155 Knowing that the radius of each pulley is 200 mm and neglecting friction, determine the internal forces at Point J of the frame shown.

SOLUTION FBD Frame with pulley and cord:

ΣM A = 0: (1.8 m) Bx − (2.6 m)(360 N) − (0.2 m)(360 N) = 0 B x = 560 N

FBD BE: Note: Cord forces have been moved to pulley hub as per Problem 6.91. ΣM E = 0: (1.4 m)(360 N) + (1.8 m)(560 N) − (2.4 m) By = 0 B y = 630 N

FBD BJ: 3 ΣFx′ = 0: F + 360 N − (630 N − 360 N) 5 4 − (560 N) = 0 5 F = 250 N

ΣFy′ = 0: V +

4 3 (630 N − 360 N) − (560 N) = 0 5 5 V = 120.0 N

ΣM J = 0: M + (0.6 m)(360 N) + (1.2 m)(560 N) − (1.6 m)(630 N) = 0 M = 120.0 N ⋅ m


PROBLEM 7.156 A steel channel of weight per unit length w = 20 lb/ft forms one side of a flight of stairs. Determine the internal forces at the center C of the channel due to its own weight for each of the support conditions shown.

SOLUTION (a)

Free body: AB AB = 92 + 122 = 15 ft W = (20 lb/ft)(15 ft) = 300 lb ΣM B = 0: A(12 ft) − (300 lb)(6 ft) = 0 A = +150 lb

A = 150 lb

Free body: AC (150-lb Forces form a couple) ΣF = 0

F=0

ΣF

V=0

ΣM C = 0: M − (150 lb)(3 ft) = 0 M = +450 lb ⋅ ft M = 450 lb ⋅ ft

(b)

Free body: AB ΣM A = 0: B(9 ft) − (300 lb)(6 ft) = 0 B = +200 lb B = 200 lb



Free body: CB ΣM C = 0: (200 lb)(4.5 ft) − (150 lb)(3 ft) − M = 0 M = +450 lb ⋅ ft M = 450 lb ⋅ ft

tan θ =

9 3 3 4 = ; sin θ = ; cos θ = 12 4 5 5

3 4 ΣF = 0: F − (150 lb) − (200 lb) = 0 5 5 F = +250 lb F = 250 lb 4 3 ΣF = 0: V − (150 lb) + (200 lb) = 0 5 5 V =0 V=0 M = 450 lb ⋅ ft , F = 250 lb

On portion AC internal forces are

, V=0


!

PROBLEM 7.157 For the beam shown, determine (a) the magnitude P of the two concentrated loads for which the maximum absolute value of the bending moment is as small as possible, (b) the corresponding value of | M |max .

SOLUTION Free body: Entire beam By symmetry 1 (16 lb/in.)(30 in.) + P 2 D = 240 lb + P D=E=

Free body: Portion AD ΣM D = 0: M D = −10 P

Free body: Portion ADC ΣM C = 0: P(25 in.) − (240 lb + P)(15 in.) + (240 lb)(7.5 in.) + M C = 0 M C = 1800 lb ⋅ in. − (10 in.)P

(a)

We equate:

|M D | = | MC | | − 10 P | = |1800 − 10 P | 10 P = 1800 − 10 P

(b)

For P = 90 lb:

M D = −10 lb (90 lb)

P = 90.0 lb

!

| M |max = 900 lb ⋅ in.

!


PROBLEM 7.158 Knowing that the magnitude of the concentrated loads P is 75 lb, (a) draw the shear and bending-moment diagrams for beam AB, (b) determine the maximum absolute values of the shear and bending moment.

SOLUTION By symmetry:

D=E 1 (16 lb/in.)(30 in.) + 75 lb 2 = 315 lb =

!



SOLUTION Free body: Entire beam ΣM B = 0: (24 kN)(3 m) + (24 kN)(2.4 m) + (21.6 kN)(0.9 m) − Ay (3.6 m) = 0 Ay = +91.4 kN ΣFx = 0: Ax = 0

Shear diagram

At A:

VA = Ay = +41.4 kN

|V |max = 41.4 kN


At A:

| M |max = 35.3 kN ⋅ m

MA = 0

The slope of the parabola at E is the same as that of the segment DE!


PROBLEM 7.160 For the beam shown, draw the shear and bending-moment diagrams, and determine the magnitude and location of the maximum absolute value of the bending moment, knowing that (a) P = 6 kips, (b) P = 3 kips.

SOLUTION Free body: Beam ΣFx = 0: Ax = 0 ΣM A = 0: C (6 ft) − (12 kips)(3 ft) − P(8 ft) = 0 C = 6 kips +

4 P 3

(1) "

4 ! ΣFy = 0: Ay + " 6 + P # − 12 − P = 0 3 % $ 1 Ay = 6 kips − P 3

(a)

(2) "

P = 6 kips.

Load diagram Substituting for P in Eqs. (2) and (1): 1 Ay = 6 − (6) = 4 kips 3 4 C = 6 + (6) = 14 kips 3

Shear diagram

VA = Ay = +4 kips

To determine Point D where V = 0: VD − VA = − wx 0 − 4 kips = (2 kips/ft)x

x = 2 ft "


MA = 0 | M |max = 12.00 kip ⋅ ft, at C

Parabola from A to C PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1193


(b)

P = 3 kips

Load diagram Substituting for P in Eqs. (2) and (1): 1 A = 6 − (3) = 5 kips 3 4 C = 6 + (3) = 10 kips 3

Shear diagram

VA = Ay = +5 kips

To determine D where V = 0: VD − VA = − wx 0 − (5 kips) = −(2 kips/ft) x

x = 2.5 ft "


MA = 0 | M |max = 6.25 kip ⋅ ft 2.50 ft from A

!

Parabola from A to C.


PROBLEM 7.161 For the beam and loading shown, (a) write the equations of the shear and bending-moment curves, (b) determine the maximum bending moment.

SOLUTION (a)

Reactions at supports:

1 W A = B = W , where = Total load 2 L W=

-

L

wdx = w0

0

-

L

0

πx! "1 − sin # dx L % $ L

π x' L & = w0 ( x + cos ) x L +0 * 2! = w0 L "1 − # $ π%

1 1 2! VA = A = W = w0 L "1 − # 2 2 $ π%

Thus

MA = 0

(1)

πx! w( x) = w0 "1 − sin L #% $

Load: Shear: From Eq. (7.2):

V ( x ) − VA = −

-

x

w( x) dx

0

= − w0

-

x

0

πx! "1 − sin L # dx $ %

Integrating and recalling first of Eqs. (1), x

L 1 2! π x' & V ( x) − w0 L "1 − # = − w0 ( x + cos ) L +0 2 π $ π% * V ( x) =

1 2! L L πx! w0 L "1 − # − w0 " 2 + cos # + w0 2 L π π π $ % $ %

L L πx! V ( x) = w0 " − x − cos L #% π $2

(2)



Bending moment: From Eq. (7.4) and recalling that M A = 0. M ( x) − M A =

-

x

V ( x) dx

0 x

2 &L L! π x' 1 = w0 ( x − x 2 − " # sin ) L )+ 2 $π % (* 2 0

M ( x) =

(b)

πx! 1 2 L2 w0 "" Lx − x 2 − 2 sin # L #% 2 $ π

(3)

Maximum bending moment dM = V = 0. dx

This occurs at x =

L 2

as we may check from (2): L! L L L π! V " # = w0 " − − cos # = 0 2% $2% $2 2 π

From (3):

π! L! 1 L2 L2 2 L2 M " # = w0 "" − − 2 sin ## 4 π 2% $2% 2 $ 2 1 8 ! = w0 L2 "1 − 2 # 8 $ π % = 0.0237 w0 L2 M max = 0.0237 w0 L2 , at

x=

L 2


!

PROBLEM 7.162 An oil pipeline is supported at 6-ft intervals by vertical hangers attached to the cable shown. Due to the combined weight of the pipe and its contents the tension in each hanger is 400 lb. Knowing that dC = 12 ft, determine (a) the maximum tension in the cable, (b) the distance dD.

SOLUTION FBD Cable: Hanger forces at A and F act on the supports, so A y and Fy act on the cable. ΣM F = 0: (6 ft + 12 ft + 18 ft + 24 ft)(400 lb) − (30 ft)Ay − (5 ft) Ax = 0 Ax + 6 Ay = 4800 lb

FBD ABC:

(1)

ΣM C = 0: (7 ft)Ax − (12 ft)Ay + (6 ft)(400 lb) = 0

(2)

Solving (1) and (2) A x = 800 lb

Ay =

From FBD Cable:

2000 lb 3

ΣFx = 0: − 800 lb + Fx = 0

Fx = 800 lb

FBD DEF: ΣFy = 0:

200 lb − 4 (400 lb) + Fy = 0 3

Fy =

Since Ax = Fx and Fy . Ay ,

2800 ! Tmax = TEF = (800 lb) 2 + " lb # $ 3 %

2800 lb 3

2



(a)

Tmax = 1229.27 lb,

Tmax = 1229 lb 2800 ! lb # − d D (800 lb) − (6 ft)(400 lb) = 0 ΣM D = 0: (12 ft) " $ 3 % d D = 11.00 ft

(b)


!

PROBLEM 7.163 Solve Problem 7.162 assuming that dC = 9 ft. PROBLEM 7.162 An oil pipeline is supported at 6-ft intervals by vertical hangers attached to the cable shown. Due to the combined weight of the pipe and its contents the tension in each hanger is 400 lb. Knowing that dC = 12 ft, determine (a) the maximum tension in the cable, (b) the distance dD.

SOLUTION FBD CDEF:

ΣM C = 0: (18 ft)Fy − (9 ft)Fy − (6 ft + 12 ft)(400 lb) = 0 Fx − 2 Fy = −800 lb

(1)

EM A = 0: (30 ft)Fy − (5 ft)Fx

FBD Cable:

− (6 ft)(1 + 2 + 3 + 4)(400 lb) = 0 Fx − 6 Fy = −4800 lb

(2)

Solving (1) and (2), Fx = 1200 lb

, Fy = 1000 lb

ΣFx = 0: − Ax + 1200 lb = 0, A x = 1200 lb

Point F:

ΣFy = 0: Ay + 1000 lb − 4(400 lb) = 0,

A y = 600 lb



Since Ax = Ay

and Fy . Ay , Tmax = TEF

Tmax = (1 kip)2 + (1.2 kips) 2 Tmax = 1.562 kips

(a) FBD DEF:

ΣM D = 0: (12 ft)(1000 lb) − d D (1200 lb) − (6 ft)(400 lb) = 0

d D = 8.00 ft

(b)


!

PROBLEM 7.164 A transmission cable having a mass per unit length of 0.8 kg/m is strung between two insulators at the same elevation that are 75 m apart. Knowing that the sag of the cable is 2 m, determine (a) the maximum tension in the cable, (b) the length of the cable.

SOLUTION w = (0.8 kg/m)(9.81 m/s 2 ) = 7.848 N/m W = (7.848 N/m)(37.5 m) W = 294.3 N

(a)

1 ! ΣM B = 0: T0 (2 m) − W " 37.5 m # = 0 $2 % 1 T0 (2 m) − (294.3 N) (37.5 m) = 0 2 T0 = 2759 N Tm2 = (294.3 N) 2 + (2759 N)2

(b)

Tm = 2770 N

& 2 y !2 ' sB = xB (1 + " B # + !) ( 3 $ xB % ) * + & 2 2 m !2 ' = 37.5 m (1+ " + !) # (* 3 $ 37.5 m % )+ = 37.57 m

Length = 2sB = 2(37.57 m)

Length = 75.14 m


!

PROBLEM 7.165 Cable ACB supports a load uniformly distributed along the horizontal as shown. The lowest Point C is located 9 m to the right of A. Determine (a) the vertical distance a, (b) the length of the cable, (c) the components of the reaction at A.

SOLUTION Free body: Portion AC ΣFy = 0: Ay − 9 w = 0 A y = 9w

ΣM A = 0: T0 a − (9 w)(4.5 m) = 0

(1)

Free body: Portion CB

ΣFy = 0: By − 6w = 0 B y = 6w


ΣM A = 0: 15w (7.5 m) − 6 w (15 m) − T0 (2.25 m) = 0 T0 = 10 w

(a) Eq. (1):

T0 a − (9w)(4.5 m) = 0 10wa = (9w)(4.5) = 0

a = 4.05 m



(b)

Length = AC + CB Portion AC:

x A = 9 m, s AC s AC

Portion CB:

y A = a = 4.05 m;

y A 4.05 = = 0.45 9 xA

& 2 y !2 2 y !4 ' = xB (1 + " A # − " B # + !) 5 $ xA % ( 3 $ xA % ) * + 2 2 ! = 9 m "1+ 0.452 − 0.454 + ! # = 10.067 m 5 $ 3 %

xB = 6 m,

yB = 4.05 − 2.25 = 1.8 m;

yB = 0.3 xB

2 2 ! sCB = 6 m "1 + 0.32 − 0.34 + ! # = 6.341 m 3 5 $ % Total length = 10.067 m + 6.341 m

(c)

Total length = 16.41 m

Components of reaction at A. Ay = 9 w = 9(60 kg/m)(9.81 m/s 2 ) = 5297.4 N Ax = T0 = 10 w = 10(60 kg/m)(9.81 m/s 2 ) = 5886 N

A x = 5890 N A y = 5300 N


!

CHAPTER 8

PROBLEM 8.1 Determine whether the block shown is in equilibrium and find the magnitude and direction of the friction force when θ = 25° and P = 150 lb.

SOLUTION Assume equilibrium: ΣFx = 0: F + (240 lb) sin 25° − (150 lb) cos 25° = 0 F = +34.518 lb

F = 34.518 lb

ΣFy = 0: N − (240 lb) cos 25° − (150 lb)sin 25° = 0 N = +280.91 lb

Maximum friction force:

N = 280.91 lb

Fm = µ s N = 0.35(280.91 lb) = 98.319 lb

Block is in equilibrium F = 34.5 lb

Since F , Fm ,


PROBLEM 8.2 Determine whether the block shown is in equilibrium and find the magnitude and direction of the friction force when θ = 30° and P = 30 lb.

SOLUTION Assume equilibrium:

ΣFx = 0: F + (240 lb) sin 30° − (30 lb) cos 30° = 0 F = −94.019 lb

F = 94.019 lb

ΣFy = 0: N − (240 lb) cos 30° − (30 lb) sin 30° = 0 N = +222.85 lb

Maximum friction force:

N = 222.85 lb

Fm = µ s N = 0.35(222.85 lb) = 77.998 lb

Since F is

and F . Fm ,

Actual friction force:

Block moves down F = Fk = µ k N = 0.25(222.85 lb)

F = 55.7 lb


!

PROBLEM 8.3 Determine whether the block shown is in equilibrium and find the magnitude and direction of the friction force when θ = 40° and P = 400 N.

SOLUTION Assume equilibrium: ΣFy = 0: N − (800 N) cos 25° + (400 N) sin 15° = 0 N = +621.5 N

N = 621.5 N

ΣFx = 0: − F + (800 N) sin 25° − (400 N) cos 15° = 0 F = +48.28 N

F = 48.28 N

Maximum friction force: Fm = µ s N = 0.20(621.5 N) = 124.3 N

Block is in equilibrium F = 48.3 N

Since F , Fm , !


!

PROBLEM 8.4 Determine whether the block shown is in equilibrium and find the magnitude and direction of the friction force when θ = 35° and P = 200 N.

SOLUTION Assume equilibrium: ΣFy = 0: N − (800 N) cos 25° + (200 N) sin 10° = 0 N = 690.3 N

N = 690.3 N

ΣFx = 0: − F + (800 N) sin 25° − (200 N) cos 10° = 0 F = 141.13 N

F = 141.13 N

Maximum friction force: Fm = µs N = (0.20)(690.3 N) = 138.06 N

Since F . Fm ,

Block moves down

Friction force: F = µk N

!

= (0.15)(690.3 N) = 103.547 N !

F = 103.5 N


!

PROBLEM 8.5 Knowing that θ = 45°, determine the range of values of P for which equilibrium is maintained.

SOLUTION To start block up the incline:

µs = 0.20 φs = tan −1 0.20 = 11.31°

Force triangle: P 800 N = sin 36.31° sin 98.69°

P = 479.2 N "!

To prevent block from moving down:

Force triangle: P 800 N = sin 13.69° sin 121.31°

P = 221.61 N " 222 N # P # 479 N

Equilibrium is maintained for!


PROBLEM 8.6 Determine the range of values of P for which equilibrium of the block shown is maintained.

SOLUTION FBD block: (Impending motion down):

φs = tan −1 µs = tan −1 0.25 P = (500 lb) tan (30° − tan −1 0.25) = 143.03 lb

(Impending motion up):

P = (500 lb) tan (30° + tan −1 0.25) = 483.46 lb 143.0 lb # P # 483 lb

Equilibrium is maintained for


PROBLEM 8.7 Knowing that the coefficient of friction between the 15-kg block and the incline is µ s = 0.25, determine (a) the smallest value of P required to maintain the block in equilibrium, (b) the corresponding value of β.

SOLUTION FBD block (Impending motion downward): W = (15 kg)(9.81 m/s 2 ) = 147.150 N

φs = tan −1 µ s = tan −1 (0.25) = 14.036°

(a)

Note: For minimum P,

P

So

β =α

R

= 90° − (30° + 14.036°) = 45.964°

and

P = (147.150 N)sin α = (147.150 N)sin (45.964°) P = 108.8 N

β = 46.0°

(b)


PROBLEM 8.8 Considering only values of θ less than 90°, determine the smallest value of θ required to start the block moving to the right when (a) W = 75 lb, (b) W = 100 lb.

SOLUTION FBD block (Motion impending):

φs = tan −1 µs = 14.036° W 30 lb = sin φs sin(θ − φs ) sin(θ − φs ) =

or

sin(θ − 14.036°) =

W sin 14.036° 30 lb W 123.695 lb

(a)

W = 75 lb:

θ = 14.036° + sin −1

75 lb 123.695 lb

θ = 51.4°

!

(b)

W = 100 lb: θ = 14.036° + sin −1

100 lb 123.695 lb

θ = 68.0°

!


PROBLEM 8.9 The coefficients of friction between the block and the rail are µ s = 0.30 and µk = 0.25. Knowing that θ = 65°, determine the smallest value of P required (a) to start the block moving up the rail, (b) to keep it from moving down.

SOLUTION (a)

To start block up the rail:

µs = 0.30 φs = tan −1 0.30 = 16.70°

Force triangle: P 500 N = sin 51.70° sin(180° − 25° − 51.70°)

(b)

P = 403 N

To prevent block from moving down:

Force triangle: P 500 N = sin 18.30° sin(180° − 25° − 18.30°)

P = 229 N


!

PROBLEM 8.10 The 80-lb block is attached to link AB and rests on a moving belt. Knowing that µ s = 0.25 and µk = 0.20, determine the magnitude of the horizontal force P that should be applied to the belt to maintain its motion (a) to the right, (b) to the left.

SOLUTION We note that link AB is a two-force member, since there is motion between belt and block µk = 0.20 and φk = tan −1 0.20 = 11.31° (a)

Belt moves to right

Free body: Block

Force triangle: R 80 lb = sin 120° sin 48.69° R = 92.23 lb

Free body: Belt ΣFx = 0: P − (92.23 lb) sin 11.31°

P = 18.089 lb P = 18.09 lb

(b)

Belt moves to left

Free body: Block

Force triangle: R 80 lb = sin 60° sin 108.69° R = 73.139 lb

Free body: Belt ΣFx = 0: (73.139 lb)sin 11.31° − P = 0

P = 14.344 lb P = 14.34 lb


PROBLEM 8.11 The coefficients of friction are µ s = 0.40 and µk = 0.30 between all surfaces of contact. Determine the smallest force P required to start the 30-kg block moving if cable AB (a) is attached as shown, (b) is removed.

SOLUTION (a)

Free body: 20-kg block W1 = (20 kg)(9.81 m/s 2 ) = 196.2 N F1 = µ s N1 = 0.4(196.2 N) = 78.48 N ΣF = 0: T − F1 = 0 T = F1 = 78.48 N

Free body: 30-kg block W2 = (30 kg)(9.81 m/s 2 ) = 294.3 N N 2 = 196.2 N + 294.3 N = 490.5 N F2 = µ s N 2 = 0.4(490.5 N) = 196.2 N ΣF = 0: P − F1 − F2 − T = 0

P = 78.48 N + 196.2 N + 78.48 N = 353.2 N

P = 353 N

(b)

Free body: Both blocks Blocks move together W = (50 kg)(9.81 m/s 2 ) = 490.5 N ΣF = 0: P − F = 0

P = µ s N = 0.4(490.5 N) = 196.2 N

P = 196.2 N


PROBLEM 8.12 The coefficients of friction are µ s = 0.40 and µk = 0.30 between all surfaces of contact. Determine the smallest force P required to start the 30-kg block moving if cable AB (a) is attached as shown, (b) is removed.

SOLUTION (a)

Free body: 20-kg block W1 = (20 kg)(9.81 m/s 2 ) = 196.2 N F1 = µ s N1 = 0.4(196.2 N) = 78.48 N ΣF = 0: T − F1 = 0 T = F1 = 78.48 N

Free body: 30-kg block W2 = (30 kg)(9.81 m/s 2 ) = 294.3 N N 2 = 196.2 N + 294.3 N = 490.5 N F2 = µ s N 2 = 0.4(490.5 N) = 196.2 N ΣF = 0: P − F1 − F2 = 0

P = 78.48 N + 196.2 N = 274.7 N

P = 275 N

(b)

Free body: Both blocks Blocks move together W = (50 kg)(9.81 m/s 2 ) = 490.5 N ΣF = 0: P − F = 0 P = µ s N = 0.4(490.5 N) = 196.2 N

P = 196.2 N


PROBLEM 8.13 Three 4-kg packages A, B, and C are placed on a conveyor belt that is at rest. Between the belt and both packages A and C the coefficients of friction are µ s = 0.30 and µk = 0.20; between package B and the belt the coefficients are µ s = 0.10 and µk = 0.08. The packages are placed on the belt so that they are in contact with each other and at rest. Determine which, if any, of the packages will move and the friction force acting on each package.

SOLUTION Consider C by itself: Assume equilibrium ΣFy = 0: N C − W cos 15° = 0

N C = W cos 15° = 0.966W ΣFx = 0: FC − W sin 15° = 0

FC = W sin 15° = 0.259W

But

Fm = µ s NC = 0.30(0.966W ) = 0.290W

Package C does not move

Thus, FC , Fm FC = 0.259W = 0.259(4 kg)(9.81 m/s 2 ) = 10.16 N

FC = 10.16 N

Consider B by itself: Assume equilibrium. We find, FB = 0.259W N B = 0.966W

But

Fm = µ s N B = 0.10(0.966W ) = 0.0966W

Package B would move if alone

Thus, FB . Fm .



Consider A and B together: Assume equilibrium FA = FB = 0.259W N A = N B = 0.966W FA + FB = 2(0.259W ) = 0.518W ( FA ) m + ( FB ) m = 0.3 N A + 0.1N B = 0.386W

Thus,

FA + FB . ( FA ) m + ( FB )m

!

FA = µk N A = 0.2(0.966)(4)(9.81) !

!

!

A and B move

FA = 7.58 N

FB = µk N B = 0.08(0.966)(4)(9.81) FB = 3.03 N


!

PROBLEM 8.14 Solve Problem 8.13 assuming that package B is placed to the right of both packages A and C. PROBLEM 8.13 Three 4-kg packages A, B, and C are placed on a conveyor belt that is at rest. Between the belt and both packages A and C the coefficients of friction are µ s = 0.30 and µk = 0.20; between package B and the belt the coefficients are µ s = 0.10 and µk = 0.08. The packages are placed on the belt so that they are in contact with each other and at rest. Determine which, if any, of the packages will move and the friction force acting on each package.

SOLUTION Consider package B by itself: Assume equilibrium ΣFy = 0: N B − W cos 15° = 0 N B = W cos 15° = 0.966W ΣFx = 0: FB − W sin 15° = 0 FB = W sin 15° = 0.259W

But

Fm = µ s N B = 0.10(0.966W ) = 0.0966W

Thus, FB . Fm . Package B would move if alone. Consider all packages together: Assume equilibrium. In a manner similar to above, we find N A = N B = N C = 0.966W FA = FB = FC = 0.259W FA + FB + FC = 3(0.259W ) = 0.777 W

But

( FA ) m = ( FC ) m = µ s N = 0.30(0.966W ) = 0.290W

and

( FB ) m = 0.10(0.966W ) = 0.0966W



Thus,

( FA ) m + ( FC ) m + ( FB ) m = 2(0.290W ) + 0.0966W = 0.677 W

and we note that

FA + FB + FC . ( FA ) m + ( FC ) m + ( FB ) m

All packages move FA = FC = µk N = 0.20(0.966)(4 kg)(9.81 m/s 2 ) = 7.58 N FB = µk N = 0.08(0.966)(4 kg)(9.81 m/s 2 ) = 3.03 N FA = FC = 7.58 N

;

FB = 3.03 N


PROBLEM 8.15 A 120-lb cabinet is mounted on casters that can be locked to prevent their rotation. The coefficient of static friction between the floor and each caster is 0.30. If h = 32 in., determine the magnitude of the force P required to move the cabinet to the right (a) if all casters are locked, (b) if the casters at B are locked and the casters at A are free to rotate, (c) if the casters at A are locked and the casters at B are free to rotate.

SOLUTION FBD cabinet: Note: for tipping, N A = FA = 0 ΣM B = 0: (12 in.)W − (32 in.)Ptip = 0 Ptip = 2.66667

(a)

All casters locked. Impending slip: FA = µ s N A FB = µ s N B ΣFy = 0 : N A + N B − W = 0 W = 120 lb

N A + NB = W

So

FA + FB = µ sW

µs = 0.3

ΣFx = 0: P − FA − FB = 0 P = FA + FB = µ sW P = 0.3(120 lb) or P = 36.0 lb ( P = 0.3W , Ptip

(b)

Casters at A free, so

FA = 0

Impending slip:

FB = µ s N B

OK)

ΣFx = 0: P − FB = 0 P = FB = µ s N B

NB =

P

µs



ΣM A = 0: (32 in.)P + (12 in.)W − (24 in.)N B = 0 8 P + 3W − 6

P = 0 P = 0.25W 0.3

( P = 0.25W , Ptip

OK)

P = 0.25(120 lb)

(c)

Casters at B free, so

FB = 0

Impending slip:

FA = µ s N A

or P = 30.0 lb

ΣFx = 0: P − FA = 0 P = FA = µ s N A NA =

P

µs

=

P 0.3

ΣM B = 0: (12 in.)W − (32 in.)P − (24 in.)N A = 0

3W − 8P − 6

( P , Ptip

P =0 0.3 P = 0.107143W = 12.8572

OK) P = 12.86 lb

!


PROBLEM 8.16 A 120-lb cabinet is mounted on casters that can be locked to prevent their rotation. The coefficient of static friction between the floor and each caster is 0.30. Assuming that the casters at both A and B are locked, determine (a) the force P required to move the cabinet to the right, (b) the largest allowable value of h if the cabinet is not to tip over.

SOLUTION FBD cabinet: (a)

ΣFy = 0: N A + N B − W = 0 N A + NB = W

Impending slip: FA = µ s N A FB = µ s N B

So FA + FB = µ sW

W = 120 lb µs = 0.3

ΣFx = 0: P − FA − FB = 0 P = FA + FB = µ sW

P = 0.3(120 lb) = 141.26 N P = 36.0 lb

(b)

For tipping,

N A = FA = 0 ΣM B = 0: hP − (12 in.)W = 0 hmax = (12 in.)

W 1 12 in. = (12 in.) = P 0.3 µs hmax = 40.0 in.


!

PROBLEM 8.17 The cylinder shown is of weight W and radius r, and the coefficient of static friction µ s is the same at A and B. Determine the magnitude of the largest couple M that can be applied to the cylinder if it is not to rotate.

SOLUTION FBD cylinder: For maximum M, motion impends at both A and B FA = µ s N A FB = µ s N B ΣFx = 0: N A − FB = 0 N A = FB = µ s N B FA = µ s N A = µ s2 N B ΣFy = 0: N B + FA − W = 0 N B + µ s2 N B = W

or

NB =

and

FB = FA =

W 1 + µ s2

µsW 1 + µs2

µs2W 1+ µ2

ΣM C = 0: M − r ( FA + FB ) = 0

(

M = r µs + µs2

) 1 +Wµ

2 s

M max = Wr µ s

1 + µs 1 + µ s2


!

PROBLEM 8.18 The cylinder shown is of weight W and radius r. Express in terms W and r the magnitude of the largest couple M that can be applied to the cylinder if it is not to rotate, assuming the coefficient of static friction to be (a) zero at A and 0.30 at B, (b) 0.25 at A and 0.30 at B.

SOLUTION FBD cylinder: For maximum M, motion impends at both A and B FA = µ A N A FB = µ B N B ΣFx = 0: N A − FB = 0 N A = FB = µ B N B FA = µ A N A = µ A µ B N B ΣFy = 0: N B + FA − W = 0 N B (1 + µ A µ B ) = W

or and

NB =

1 1 + µ A µB

W

µB W 1 + µ AµB µ A µB FA = µ A µ B N B = W 1 + µ A µB FB = µ B N B =

ΣM C = 0: M − r ( FA + FB ) = 0 M = Wr µ B

(a)

1 + µA 1 + µ A µB

For µ A = 0 and µ B = 0.30: M = 0.300Wr

(b)

For µ A = 0.25 and µ B = 0.30: M = 0.349Wr


!

PROBLEM 8.19 The hydraulic cylinder shown exerts a force of 3 kN directed to the right on Point B and to the left on Point E. Determine the magnitude of the couple M required to rotate the drum clockwise at a constant speed.

SOLUTION Free body: Drum ΣM C = 0: M − (0.25 m)( FL + FR ) = 0 M = (0.25 m)( FL + FR )

(1)

Since drum is rotating FL = µk N L = 0.3N L FR = µk N R = 0.3 N R

Free body: Left arm ABL

ΣM A = 0: (3 kN)(0.15 m) + FL (0.15 m) − N L (0.45 m) = 0 0.45 kN ⋅ m + (0.3N L )(0.15 m) − N L (0.45 m) = 0 0.405 N L = 0.45 N L = 1.111 kN FL = 0.3N L = 0.3(1.111 kN) = 0.3333 kN

Free body: Right arm DER

(2)

ΣM D = 0: (3 kN)(0.15 m) − FR (0.15 m) − N R (0.45 m) = 0 0.45 kN ⋅ m − (0.3N R )(0.15 m) − N R (0.45 m) = 0 0.495 N R = 0.45 N R = 0.9091 kN FR = µk N R = 0.3(0.9091 kN) = 0.2727 kN

Substitute for FL and FR into (1):

(3)

M = (0.25 m)(0.333 kN + 0.2727 kN) M = 0.1515 kN ⋅ m M = 151.5 N ⋅ m

!


PROBLEM 8.20 A couple M of magnitude 100 N ⋅ m is applied to the drum as shown. Determine the smallest force that must be exerted by the hydraulic cylinder on joints B and E if the drum is not to rotate.

SOLUTION Free body: Drum ΣM C = 0: 100 N ⋅ m − (0.25 m)( FL + FR ) = 0 FL + FR = 400 N

(1)

Since motion impends FL = µs N L = 0.4 N L FR = µs N R = 0.4 N R

Free body: Left arm ABL ΣM A = 0: T (0.15 m) + FL (0.15 m) − N L (0.45 m) = 0 0.15T + (0.4 N L )(0.15 m) − N L (0.45 m) = 0 0.39 N L = 0.15T ; N L = 0.38462T FL = 0.4 N L = 0.4(0.38462T ) FL = 0.15385T

(2)

Free body: Right arm DER ΣM D = 0: T (0.15 m) − FR (0.15 m) − N R (0.45 m) = 0 0.15T − (0.4 N R )(0.15 m) − N R (0.45 m) = 0 0.51N R = 0.15T ; N R = 0.29412T FR = 0.4 N R = 0.4(0.29412T ) FR = 0.11765T

(3)

Substitute for FL and FR into Eq. (1): 0.15385T + 0.11765T = 400 T = 1473.3 N T = 1.473 kN PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1229

!

PROBLEM 8.21 A 6.5-m ladder AB leans against a wall as shown. Assuming that the coefficient of static friction µs is zero at B, determine the smallest value of µs at A for which equilibrium is maintained.

SOLUTION Free body: Ladder Three-force body. Line of action of A must pass through D, where W and B intersect.

At A:

µs = tan φs =

1.25 m = 0.2083 6m

µs = 0.208


!

PROBLEM 8.22 A 6.5-m ladder AB leans against a wall as shown. Assuming that the coefficient of static friction µs is the same at A and B, determine the smallest value of µs for which equilibrium is maintained.

SOLUTION Free body: Ladder Motion impending: FA = µs N A FB = µ s N B ΣM A = 0: W (1.25 m) − N B (6 m) − µ s N B (2.5 m) = 0 NB =

1.25W 6 + 2.5µs

(1)

ΣFy = 0: N A + µs N B − W = 0 N A = W − µs N B NA = W −

1.25µ sW 6 + 2.5µ s

(2)

ΣFx = 0: µ s N A − N B = 0

Substitute for NA and NB from Eqs. (1) and (2):

µsW −

1.25µ s2W 1.25W = 6 + 2.5µ s 6 + 2.5µ s

6µ s + 2.5µ s2 − 1.25µ s2 = 1.25 1.25µ s2 + 6µ s − 1.25 = 0

µ s = 0.2 and

µs = −5 (Discard) µs = 0.200


!

PROBLEM 8.23 End A of a slender, uniform rod of length L and weight W bears on a surface as shown, while end B is supported by a cord BC. Knowing that the coefficients of friction are µs = 0.40 and µk = 0.30, determine (a) the largest value of θ for which motion is impending, (b) the corresponding value of the tension in the cord.

SOLUTION Free-body diagram Three-force body. Line of action of R must pass through D, where T and R intersect. Motion impends: tan φs = 0.4

φs = 21.80° (a)

Since BG = GA, it follows that BD = DC and AD bisects ∠BAC

θ 2

θ 2

+ φs = 90°

+ 21.8° = 90°

θ = 136.4°

(b)

Force triangle (right triangle): T = W cos 21.8° T = 0.928W


!

PROBLEM 8.24 End A of a slender, uniform rod of length L and weight W bears on a surface as shown, while end B is supported by a cord BC. Knowing that the coefficients of friction are µs = 0.40 and µk = 0.30, determine (a) the largest value of θ for which motion is impending, (b) the corresponding value of the tension in the cord.

SOLUTION Free-body diagram Rod AB is a three-force body. Thus, line of action of R must pass through D, where W and T intersect. Since AG = GB, CD = DB and the median AD of the isosceles triangle ABC bisects the angle θ. (a)

Thus,

1 2

φs = θ

Since motion impends,

φs = tan −1 0.40 = 21.80° θ = 2φs = 2(21.8°) θ = 43.6° (b)

Force triangle: This is a right triangle. T = W sin φs

= W sin 21.8° T = 0.371W


!

PROBLEM 8.25 A window sash weighing 10 lb is normally supported by two 5-lb sash weights. Knowing that the window remains open after one sash cord has broken, determine the smallest possible value of the coefficient of static friction. (Assume that the sash is slightly smaller than the frame and will bind only at Points A and D.)

SOLUTION FBD window:

T = 5 lb

ΣFx = 0: N A − N D = 0 N A = ND

Impending motion:

FA = µs N A FD = µs N D

ΣM D = 0: (18 in.)W − (27 in.) NA − (36 in.) FA = 0

W = 10 lb

3 N A + 2µs N A 2 2W NA = 3 + 4µs W=

ΣFy = 0: FA − W + T + FD = 0 FA + FD = W − T =

Now

W 2

FA + FD = µs ( N A + N D ) = 2µs N A

Then

W 2W = 2µs 2 3 + 4µs

µs = 0.750

or


!

PROBLEM 8.26 A 500-N concrete block is to be lifted by the pair of tongs shown. Determine the smallest allowable value of the coefficient of static friction between the block and the tongs at F and G.

SOLUTION Free body: Members CA, AB, BD C y = Dy =

By symmetry:

1 (500) = 250 N 2

Since CA is a two-force member, Cy Cx 250 N = = 90 mm 75 mm 75 mm Cx = 300 N ΣFx = 0: Dx = C x Dx = 300 N

Free body: Tong DEF ΣM E = 0: (300 N)(105 mm) + (250 N)(135 mm) + (250 N)(157.5 mm) − Fx (360 mm) = 0 Fx = +290.625 N

Minimum value of µs :

µs =

Fy Fx

=

250 N 290.625 N

µs = 0.860


!

PROBLEM 8.27 The press shown is used to emboss a small seal at E. Knowing that the coefficient of static friction between the vertical guide and the embossing die D is 0.30, determine the force exerted by the die on the seal.

SOLUTION Free body: Member ABC

ΣM A = 0: FBD cos 20°(4) + FBD sin 20°(6.9282) − (50 lb)(4 + 15.4548) = 0 FBD = 158.728 lb

Free body: Die D

φs = tan −1 µs = tan −1 0.3 = 16.6992°

Force triangle: 158.728 lb D = sin 53.301° sin 106.6992° D = 132.869 lb

D = 132.9 lb

On seal:

!


PROBLEM 8.28 The 100-mm-radius cam shown is used to control the motion of the plate CD. Knowing that the coefficient of static friction between the cam and the plate is 0.45 and neglecting friction at the roller supports, determine (a) the force P required to maintain the motion of the plate, knowing that the plate is 20 mm thick, (b) the largest thickness of the plate for which the mechanism is self locking (i.e., for which the plate cannot be moved however large the force P may be).

SOLUTION Free body: Cam

Impending motion:

F = µs N ΣM A = 0: QR − NR sin θ + ( µs N ) R cos θ = 0 N=

Free body: Plate

ΣFx = 0 P = µs N

Q sin θ − µs cos θ

(1) (2)

Geometry in ∆ABD with R = 100 mm and d = 20 mm R−d R 80 mm = 100 mm = 0.8

cos θ =

sin θ = 1 − cos 2 θ = 0.6



(a)

Eq. (1) using

Q = 60 N and µs = 0.45 60 N 0.6 − (0.45)(0.8) 60 = = 250 N 0.24

N=

Eq. (2)

P = µs N = (0.45)(250 N) P = 112.5 N

(b)

For P = ∞, N = ∞. Denominator is zero in Eq. (1). sin θ − µ s cos θ = 0 tan θ = µ s = 0.45

θ = 24.23° R−d R 100 − d cos 24.23 = 100 cos θ =

d = 8.81 mm


!

PROBLEM 8.29 A slender rod of length L is lodged between peg C and the vertical wall and supports a load P at end A. Knowing that the coefficient of static friction is 0.20 at both B and C, find the range of values of the ratio L/a for which equilibrium is maintained.

SOLUTION We shall first assume that the motion of end B is impending upward. The friction forces at B and C will have the values and directions indicated in the FB diagram.

a ! ΣM B = 0: PL sin θ − N C " #=0 $ sin θ % NC =

PL 2 sin θ a

(1)

ΣFx = 0: N C cos θ + µ N C sin θ − N B = 0

(2)

ΣFy = 0: N C sin θ − µ N C cos θ − µ N B − P = 0

(3)

Multiply Eq. (2) by µ and subtract from Eq. (3): N C (sin θ − µ cos θ ) − µ N C (cos θ + µ sin θ ) − P = 0 P = N C [sin θ (1 − µ 2 ) − 2 µ cos θ ]

Substitute for N C from Eq. (1) and solve for a/L: a = sin 2 θ [(1 − µ 2 )sin θ − 2µ cos θ ] L

(4)

Making θ = 35° and µ = 0.20 in Eq. (4): a = sin 2 35°[(1 − 0.04)sin 35° − 2(0.20) cos 35°] L = 0.07336 L = 13.63 # a



Assuming now that the motion at B is impending downward, we should reverse the direction of FB and FC in. the FB diagram. The same result may be obtained by making θ = 35° and µ = −0.20 in Eq. (4): a = sin 2 35°[(1 − 0.04)sin 35° − 2(−0.20) cos 35°] L = 0.2889 L = 3.461 # a

Thus, the range of values of L/a for which equilibrium is maintained is

3.46 #

L # 13.63 a


!

PROBLEM 8.30 The 50-lb plate ABCD is attached at A and D to collars that can slide on the vertical rod. Knowing that the coefficient of static friction is 0.40 between both collars and the rod, determine whether the plate is in equilibrium in the position shown when the magnitude of the vertical force applied at E is (a) P = 0, (b) P = 20 lb.

SOLUTION (a)

P=0

ΣM D = 0: N A (2 ft) − (50 lb)(3 ft) = 0 N A = 75 lb ΣFx = 0: N D = N A = 75 lb ΣFy = 0: FA + FD − 50 lb = 0 FA + FD = 50 lb

But:

( FA ) m = µs N A = 0.40(75 lb) = 30 lb (FD ) m = µ s N D = 0.40(75 lb) = 30 lb

(b)

Thus:

( FA ) m + ( FD )m = 60 lb

and

( FA ) m + ( FD )m . FA + FD

P = 20 lb

Plate is in equilibrium

ΣM D = 0: N A (2 ft) − (50 lb)(3 ft) + (20 lb)(5 ft) = 0 N A = 25 lb ΣFx = 0: N D = N A = 25 lb ΣFy = 0: FA + FD − 50 lb + 20 lb = 0 FA + FD = 30 lb

But:

( FA ) m = µs N A = 0.4(25 lb) = 10 lb (FD ) m = µ s N D = 0.4(25 lb) = 10 lb

Thus: and

( FA ) m + ( FD )m = 20 lb FA + FD . ( FA ) m + ( FD ) m

Plate moves downward


!

PROBLEM 8.31 In Problem 8.30, determine the range of values of the magnitude P of the vertical force applied at E for which the plate will move downward. PROBLEM 8.30 The 50-lb plate ABCD is attached at A and D to collars that can slide on the vertical rod. Knowing that the coefficient of static friction is 0.40 between both collars and the rod, determine whether the plate is in equilibrium in the position shown when the magnitude of the vertical force applied at E is (a) P = 0, (b) P = 20 lb.

SOLUTION We shall consider the following two cases: (1)

0 , P , 30 lb

ΣM D = 0: N A (2 ft) − (50 lb)(3 ft) + P (5 ft) = 0 N A = 75 lb − 2.5P

(Note: N A $ 0 and directed

for P # 30 lb as assumed here) ΣFx = 0: N A = N D

ΣFy = 0: FA + FD + P − 50 = 0 FA + FD = 50 − P

But:

( FA ) m = ( F0 ) m = µ s N A = 0.40(75 − 2.5P) = 30 − P

Plate moves if: or (2)

FA + FD . ( FA ) m + ( FD ) m

50 − P . (30 − P) + (30 − P)

P . 10 lb

30 lb , P , 50 lb ΣM D = 0: − N A (2 ft) − (50 lb)(3 ft) + P (5 ft) = 0 N A = 2.5P − 75

(Note: NA . and directed

for P . 30 lb as assumed) ΣFx = 0: N A = N D ΣFy = 0: FA + FD + P − 50 = 0 FA + FD = 50 − P



But:

( FA ) m = ( FD ) m = µ s N A = 0.40(2.5 P − 75) = P − 30 lb

Plate moves if:

FA + FD . ( FA ) m + ( FD ) m 50 − P . ( P − 30) + ( P − 30)

P,

110 = 36.7 lb 3

10.00 lb , P , 36.7 lb !

Thus, plate moves downward for: (Note: For P . 50 lb, plate is in equilibrium)


PROBLEM 8.32 A pipe of diameter 60 mm is gripped by the stillson wrench shown. Portions AB and DE of the wrench are rigidly attached to each other, and portion CF is connected by a pin at D. If the wrench is to grip the pipe and be self-locking, determine the required minimum coefficients of friction at A and C.

SOLUTION FBD ABD: ΣM D = 0: (15 mm)N A − (110 mm)FA = 0 FA = µ A N A

Impending motion: 15 − 110µ A = 0

Then

µ A = 0.13636

or

µ A = 0.1364 !

ΣFx = 0: FA − Dx = 0 Dx = FA

FBD pipe: ΣFy = 0: N C − N A = 0 NC = N A

FBD DF: ΣM F = 0: (550 mm)FC − (15 mm)N C − (500 mm)Dx = 0

Impending motion: Then

But So

FC = µC N C

550 µC − 15 = 500 NC = N A

FA NC and

FA = µ A = 0.13636 NA

550 µC = 15 + 500(0.13636)

µC = 0.1512 ! "


PROBLEM 8.33 Solve Problem 8.32 assuming that the diameter of the pipe is 30 mm. PROBLEM 8.32 A pipe of diameter 60 mm is gripped by the stillson wrench shown. Portions AB and DE of the wrench are rigidly attached to each other, and portion CF is connected by a pin at D. If the wrench is to grip the pipe and be self-locking, determine the required minimum coefficients of friction at A and C.

SOLUTION FBD ABD: ΣM D = 0: (15 mm)N A − (80 mm)FA = 0

FA = µ A N A

Impending motion: Then

15 mm − (80 mm)µ A = 0

µ A = 0.1875 !

ΣFx = 0: FA − Dx = 0 Dx = FA = 0.1875 N A

So that FBD pipe:

ΣFy = 0: N C − N A = 0 NC = N A

FBD DF:

ΣM F = 0: (550 mm)FC − (15 mm)N C − (500 mm)Dx = 0

Impending motion:

FC = µC N C 550 µC − 15 = 500(0.1875)

But N A = N C (from pipe FBD) so

NA NC

NA =1 NC

µC = 0.1977 !"

and


PROBLEM 8.34 A 10-ft beam, weighing 1200 lb, is to be moved to the left onto the platform. A horizontal force P is applied to the dolly, which is mounted on frictionless wheels. The coefficients of friction between all surfaces are µs = 0.30 and µk = 0.25, and initially x = 2 ft. Knowing that the top surface of the dolly is slightly higher than the platform, determine the force P required to start moving the beam. (Hint: The beam is supported at A and D.)

SOLUTION FBD beam:

ΣM A = 0: N D (8ft) − (1200 lb)(5ft) = 0 N D = 750 lb

ΣFy = 0: N A − 1200 + 750 = 0 N A = 450 lb

( FA ) m = µ s N A = 0.3(450) = 135.0 lb ( FD ) m = µ s N D = 0.3(750) = 225 lb

Since ( FA ) m , ( FD )m , sliding first impends at A with FA = ( FA ) m = 135 lb ΣFx = 0: FA − FD = 0 FD = FA = 135.0 lb

FBD dolly: From FBD of dolly: ΣFx = 0: FD − P = 0 P = FD = 135.0 lb

P = 135.0 lb !"


PROBLEM 8.35 (a) Show that the beam of Problem 8.34 cannot be moved if the top surface of the dolly is slightly lower than the platform. (b) Show that the beam can be moved if two 175-lb workers stand on the beam at B and determine how far to the left the beam can be moved. PROBLEM 8.34 A 10-ft beam, weighing 1200 lb, is to be moved to the left onto the platform. A horizontal force P is applied to the dolly, which is mounted on frictionless wheels. The coefficients of friction between all surfaces are µs = 0.30 and µk = 0.25, and initially x = 2 ft. Knowing that the top surface of the dolly is slightly higher than the platform, determine the force P required to start moving the beam. (Hint: The beam is supported at A and D.)

SOLUTION (a)

Beam alone ΣM C = 0: N B (8 ft) − (1200 lb)(3 ft) = 0 N B = 450 lb

ΣFy = 0: N C + 450 − 1200 = 0 N C = 750 lb

( FC ) m = µs N C = 0.3(750) = 225 lb (FB ) m = µs N B = 0.3(450) = 135 lb

Beam cannot be moved !

Since ( FB ) m , ( FC ) m , sliding first impends at B, and

(b)

Beam with workers standing at B

ΣM C = 0: N B (10 − x) − (1200)(5 − x) − 350(10 − x) = 0 NB =

9500 − 1550 x 10 − x

ΣM B = 0: (1200)(5) − N C (10 − x) = 0 NC =

6000 10 − x



Check that beam starts moving for x = 2 ft: For x = 2 ft:

9500 − 1550(2) = 800 lb 10 − 2 6000 NC = = 750 lb 10 − 2 ( FC ) m = µs NC = 0.3(750) = 225 lb NB =

( FB ) m = µs N B = 0.3(800) = 240 lb

Since ( FC ) m , ( FB ) m , sliding first impends at C,

Beam moves !

How far does beam move? Beam will stop moving when FC = ( FB ) m

But

FC = µk NC = 0.25

and

( FB ) m = µ s N B = 0.30

Setting FC = ( FB ) m :

6000 1500 = 10 − x 10 − x 9500 − 1550 x 2850 − 465 x = 10 − x 10 − x

1500 = 2850 − 465 x

x = 2.90ft !"

(Note: We have assumed that, once started, motion is continuous and uniform (no acceleration).)


PROBLEM 8.36 Knowing that the coefficient of static friction between the collar and the rod is 0.35, determine the range of values of P for which equilibrium is maintained when θ = 50° and M = 20 N ⋅ m.

SOLUTION Free body member AB: BC is a two-force member. ΣM A = 0: 20 N ⋅ m − FBC cos 50°(0.1 m) = 0 FBC = 311.145 N

Motion of C impending upward: ΣFx = 0: (311.145 N) cos 50° − N = 0 N = 200 N ΣFy = 0: (311.145 N) sin 50° − P − (0.35)(200 N) = 0 P = 168.351 N

Motion of C impending downward: ΣFx = 0: (311.145 N) cos 50° − N = 0 N = 200 N ΣFy = 0: (311.145 N) sin 50° − P + (0.35)(200 N) = 0 P = 308.35 N 168.4 N # P # 308 N !

Range of P:


PROBLEM 8.37 Knowing that the coefficient of static friction between the collar and the rod is 0.40, determine the range of values of M for which equilibrium is maintained when θ = 60° and P = 200 N.

SOLUTION Free body member AB: BC is a two-force member. ΣM A = 0: M − FBC cos 60°(0.1 m) = 0 M = 0.05FBC

(1)

Motion of C impending upward: ΣFx = 0: FBC cos 60° − N = 0 N = 0.5FBC ΣFy = 0: FBC sin 60° − 200 N − (0.40)(0.5FBC ) = 0 FBC = 300.29 N

Eq. (1):

M = 0.05(300.29) M = 15.014 N ⋅ m

Motion of C impending downward: ΣFx = 0: FBC cos 60° − N = 0 N = 0.5FBC ΣFy = 0: FBC sin 60° − 200 N + (0.40)(0.5 FBC ) = 0 FBC = 187.613 N

Eq. (1):

M = 0.05(187.613) M = 9.381 N ⋅ m 9.38 N ⋅ m # M # 15.01 N ⋅ m !

Range of M:


PROBLEM 8.38 The slender rod AB of length l = 600 mm is attached to a collar at B and rests on a small wheel located at a horizontal distance a = 80 mm from the vertical rod on which the collar slides. Knowing that the coefficient of static friction between the collar and the vertical rod is 0.25 and neglecting the radius of the wheel, determine the range of values of P for which equilibrium is maintained when Q = 100 N and θ = 30°.

SOLUTION For motion of collar at B impending upward: F = µs N

ΣM B = 0: Ql sin θ −

Ca =0 sin θ

l! C = Q " # sin 2 θ a $ % l! ΣFx = 0: N = C cos θ = Q " # sin 2 θ cos θ $a% ΣFy = 0: P + Q − C sin θ − µ s N = 0 l! l! P + Q − Q " # sin 3 θ − µ s Q " # sin 2 θ cos θ = 0 a a $ % $ % l & ' P = Q ( sin 2 θ (sin θ − µ s cosθ ) − 1) *a +

Substitute data:

(1)

& 600 mm 2 ' P = (100 N) ( sin 30°(sin 30° − 0.25cos 30°) − 1) * 80 mm +

P = −46.84 N (P is directed ) P = −46.8 N



For motion of collar, impending downward: F = µs N

In Eq. (1) we substitute − µ s for µ s . &l ' P = Q ( sin 2 θ (sin θ + µ s cos θ ) − 1) *a + & 600 mm 2 ' P = (100 N) ( sin 30°(sin 30° + 0.25cos θ ) − 1) 80 mm * + P = + 34.34 N −46.8 N # P # 34.3 N !"

For equilibrium:


PROBLEM 8.39 Two 10-lb blocks A and B are connected by a slender rod of negligible weight. The coefficient of static friction is 0.30 between all surfaces of contact, and the rod forms an angle θ = 30°. with the vertical. (a) Show that the system is in equilibrium when P = 0. (b) Determine the largest value of P for which equilibrium is maintained.

SOLUTION FBD block B: (a)

Since P = 2.69 lb to initiate motion,

equilibrium exists with P = 0 !

(b)

For Pmax , motion impends at both surfaces:

ΣFy = 0: N B − 10 lb − FAB cos 30° = 0

Block B:

N B = 10 lb +

Impending motion:

3 FAB 2

(1)

FB = µs N B = 0.3N B ΣFx = 0: FB − FAB sin 30° = 0 FAB = 2 FB = 0.6 N B

Solving Eqs. (1) and (2):

N B = 10 lb +

(2)

3 (0.6 N B ) = 20.8166 lb 2

FBD block A: FAB = 0.6 N B = 12.4900 lb

Then Block A:

ΣFx = 0: FAB sin 30° − N A = 0 NA =

Impending motion:

1 1 FAB = (12.4900 lb) = 6.2450 lb 2 2

FA = µs N A = 0.3(6.2450 lb) = 1.8735 lb ΣFy = 0: FA + FAB cos 30° − P − 10 lb = 0 3 FAB − 10 lb 2 3 (12.4900 lb) − 10 lb = 1.8735 lb + 2 = 2.69 lb

P = FA +

P = 2.69 lb !"


PROBLEM 8.40 Two identical uniform boards, each of weight 40 lb, are temporarily leaned against each other as shown. Knowing that the coefficient of static friction between all surfaces is 0.40, determine (a) the largest magnitude of the force P for which equilibrium will be maintained, (b) the surface at which motion will impend.

SOLUTION Board FBDs:

Assume impending motion at C, so FC = µs N C = 0.4 N C

FBD II:

ΣM B = 0: (6 ft)N C − (8 ft)FC − (3 ft)(40 lb) = 0 [6 ft − 0.4(8 ft)]N C = (3 ft)(40 lb)

or

N C = 42.857 lb

and

FC = 0.4 N C = 17.143 lb ΣFx = 0: N B − FC = 0 N B = FC = 17.143 lb ΣM y = 0: − FB − 40 lb + N C = 0 FB = N C − 40 lb = 2.857 lb

Check for motion at B: FBD I:

FB 2.857 lb = = 0.167 , µs , OK, no motion. N B 17.143 lb ΣM A = 0: (8 ft)N B + (6 ft)FB − (3 ft)(P + 40 lb) = 0

(8 ft)(17.143 lb) + (6 ft)(2.857 lb) − 40 lb 3 ft = 11.429 lb

P=



Check for slip at A (unlikely because of P): ΣFx = 0: FA − N B = 0 or

FA = N B = 17.143 lb

ΣFy = 0: N A − P − 40 lb + FB = 0

or

NA = 11.429 lb + 40 lb − 2.857 lb = 48.572 lb

Then

FA 17.143 lb = = 0.353 , µ s N A 48.572 lb

OK, no slip , assumption is correct. Therefore (a)

Pmax = 11.43 lb !

(b)

Motion impends at C !


PROBLEM 8.41 Two identical 5-ft-long rods connected by a pin at B are placed between two walls and a horizontal surface as shown. Denoting by µs the coefficient of static friction at A, B, and C, determine the smallest value of µs for which equilibrium is maintained.

SOLUTION Sense of impending motion:

ΣM B = 0: 2W − 3N A − 4 µs N A = 0

ΣM B = 0: 1.5W − 4 NC + 3µ s N C = 0

2W (3 + 4µs )

(1)

ΣFy : N AB = W − µs N A

(3)

NA =

ΣFx = 0: Bx + µs N BA − N A = 0 Bx = N A − µs N BA

Equate (5) and (6): Substitute from Eqs. (3) and (4):

1.5W (4 − 3µs )

(2)

ΣFy : N BC = W + µs NC

(4)

NC =

ΣFx = 0: Bx − NC − µs N BC = 0

(5)

Bx = NC + µs N BC

(6)

N A − µs N BA = NC + µs N BC N A − µ s (W − µ s N A ) = NC + µ s (W + µ s NC ) N A (1 + µ s2 ) − µ sW = NC (1 + µ s2 ) + µ sW

Substitute from Eqs. (1) and (2):

2W 1.5W (1 + µ s2 ) − µ sW = (1 + µ s2 ) + µsW 3 + 4µs 4 − 3µ s 2µs 2 1.5 − = 3 + 4µ s 4 − 3µ s 1 + µ s2

Solve for µ s :

µs = 0.0949 !


PROBLEM 8.42 Two 8-kg blocks A and B resting on shelves are connected by a rod of negligible mass. Knowing that the magnitude of a horizontal force P applied at C is slowly increased from zero, determine the value of P for which motion occurs, and what that motion is, when the coefficient of static friction between all surfaces is (a) µ s = 0.40, (b) µ s = 0.50.

SOLUTION (a)

µs = 0.40:

Assume blocks slide to right. W = mg = (8 kg)(9.81 m/s 2 ) = 78.48 N FA = µ s N A FB = µ s N B ΣFy = 0: N A + N B − 2W = 0 N A + N B = 2W ΣFx = 0: P − FA − FB = 0 P = FA + FB = µ s ( N A + N B ) = µ s (2W )

(1)

P = 0.40(2)(78.48 N) = 62.78 N ΣM B = 0: P (0.1 m) − ( N A − W )(0.09326 m) + FA (0.2 m) = 0 (62.78)(0.1) − ( N A − 78.48)(0.09326) + (0.4)( N B )(0.2) = 0 0.17326 N A = 1.041 N A = 6.01 N . 0 OK

System slides: P = 62.8 N ! (b)

µs = 0.50: See part a. Eq. (1):

P = 0.5(2)(78.48 N) = 78.48 N ΣM B = 0: P(0.1 m) + ( N A − W )(0.09326 m) + FA (0.2 m) = 0 (78.48)(0.1) + ( N A − 78.48)(0.09326) + (0.5) N A (0.2) = 0 0.19326 N A = −0.529 N A = −2.73 N , 0 uplift, rotation about B



For N A = 0:

ΣM B = 0: P(0.1 m) − W (0.09326 m) = 0 P = (78.48 N)(0.09326 m)/(0.1) = 73.19

System rotates about B: P = 73.2 N !


PROBLEM 8.43 A slender steel rod of length 225 mm is placed inside a pipe as shown. Knowing that the coefficient of static friction between the rod and the pipe is 0.20, determine the largest value of θ for which the rod will not fall into the pipe.

SOLUTION Motion of rod impends down at A and to left at B. FA = µ s N A FB = µ s N B ΣFx = 0: N A − N B sin θ + FB cos θ = 0 N A − N B sin θ + µ s N B cos θ = 0 N A = N B (sin θ − µ s cos θ )

(1)

ΣFy = 0: FA + N B cos θ + FB sin θ − W = 0

µs N A + N B cos θ + µs N B sin θ − W = 0

(2)

Substitute for NA from Eq. (1) into Eq. (2):

µs N B (sin θ − µ s cos θ ) + N B cos θ + µs N B sin θ − W = 0 NB =

W (1 −

µs2 ) cos θ

+ 2µ s sin θ

=

W (1 − 0.2 ) cos θ + 2(0.2) sin θ 2

(3)

75 ! ΣM A = 0: N B " # − W (112.5cos θ ) = 0 $ cos θ %

Substitute for N B from Eq. (3), cancel W, and simplify to find 9.6 cos3 θ + 4sin θ cos 2 θ − 6.6667 = 0 cos3 θ (2.4 + tan θ ) = 1.6667

θ = 35.8° !

Solve by trial & error:


PROBLEM 8.44 In Problem 8.43, determine the smallest value of θ for which the rod will not fall out the pipe. PROBLEM 8.43 A slender steel rod of length 225 mm is placed inside a pipe as shown. Knowing that the coefficient of static friction between the rod and the pipe is 0.20, determine the largest value of θ for which the rod will not fall into the pipe.

SOLUTION Motion of rod impends up at A and right at B. FA = µs N A

FB = µ s N B

ΣFx = 0: N A − N B sin θ − FB cos θ = 0 N A − N B sin θ − µ s N B cos θ = 0 N A = N B (sin θ + µ s cos θ )

(1)

ΣFy = 0: −FA + N B cos θ − FB sin θ − W = 0 − µ s N A + N B cos θ − µs N B sin θ − W = 0

(2)

Substitute for N A from Eq. (1) into Eq. (2): − µ s N B (sin θ + µ s cos θ ) + N B cos θ − µ s N B sin θ − W = 0 NB =

W

(1 −

µs2 ) cos θ

− 2µs sin θ

=

W (1 − 0.2 ) cos θ − 2(0.2)sin θ

(3)

2

75 ! ΣM A = 0: N B " − W (112.5cos θ ) = 0 cos θ #% $

Substitute for NB from Eq. (3), cancel W, and simplify to find 9.6 cos3 θ − 4sin θ cos 2 θ − 6.6667 = 0 cos3 θ (2.4 − tan θ ) = 1.6667

θ = 20.5°

Solve by trial + error:


PROBLEM 8.45 Two slender rods of negligible weight are pin-connected at C and attached to blocks A and B, each of weight W. Knowing that θ = 80° and that the coefficient of static friction between the blocks and the horizontal surface is 0.30, determine the largest value of P for which equilibrium is maintained.

SOLUTION FBD pin C:

FAC = P sin 20° = 0.34202 P FBC = P cos 20° = 0.93969 P

ΣFy = 0: N A − W − FAC sin 30° = 0 N A = W + 0.34202 P sin 30° = W + 0.171010 P

or FBD block A:

ΣFx = 0: FA − FAC cos 30° = 0

or

FA = 0.34202 P cos 30° = 0.29620 P

For impending motion at A:

FA = µs N A

Then

NA =

FA

µs

: W + 0.171010 P =

0.29620 P 0.3

P = 1.22500W

or

ΣFy = 0: N B − W − FBC cos 30° = 0 N B = W + 0.93969 P cos 30° = W + 0.81380 P

ΣFx = 0: FBC sin 30° − FB = 0 FB = 0.93969 P sin 30° = 0.46985P



FBD block B: For impending motion at B:

FB = µs N B

Then

NB =

FB

µs

: W + 0.81380 P =

0.46985P 0.3

P = 1.32914W

or

Pmax = 1.225W

Thus, maximum P for equilibrium


PROBLEM 8.46 The machine part ABC is supported by a frictionless hinge at B and a 10° wedge at C. Knowing that the coefficient of static friction is 0.20 at both surfaces of the wedge, determine (a) the force P required to move the wedge to the left, (b) the components of the corresponding reaction at B.

SOLUTION

µs = 0.20 φs = tan −1 µs = tan −1 0.20 = 11.3099° Free body: ABC

10° + 11.3099° = 21.3099° ΣM B = 0: ( RC cos 21.3099°)(10) − (120 lb)(8) = 0 RC = 103.045 lb

Free body: Wedge

Force triangle:



Law of sines: ( R = 103.045 lb) P = C sin 32.6198° sin 78.690° P = 56.6 lb

(a) (b)

Returning to free body of ABC: ΣFx = 0: Bx + 120 − (103.045) sin 21.3099° = 0 Bx = −82.552 lb

B x = 82.6 lb

ΣFy = 0: By + (103.045) cos 21.3099° = 0 By = −96.000 lb

B y = 96.0 lb


!

PROBLEM 8.47 Solve Problem 8.46 assuming that the wedge is to be moved to the right. PROBLEM 8.46 The machine part ABC is supported by a frictionless hinge at B and a 10° wedge at C. Knowing that the coefficient of static friction is 0.20 at both surfaces of the wedge, determine (a) the force P required to move the wedge to the left, (b) the components of the corresponding reaction at B.

SOLUTION

µs = 0.20 φs = tan −1 µ s = tan −1 0.20 = 11.30993° Free body: ABC

11.30993° − 10° = 1.30993° ΣM B = 0: ( RC cos 1.30993°)(10) − (120 lb)(8) = 0 RC = 96.025 lb

Free body: Wedge

Force triangle:



Law of sines: ( R = 96.025 lb) P = C sin 12.6198° sin 78.690° P = 21.4 lb

(a) (b)

Returning to free body of ABC: ΣFx = 0: Bx + 120 + (96.025)sin 1.30993° = 0 Bx = −122.195 lb

B x = 122.2 lb

ΣFy = 0: By + (96.025) cos 1.30993° = 0 By = −96.000 lb

B y = 96.0 lb


!

PROBLEM 8.48 Two 8° wedges of negligible weight are used to move and position the 800-kg block. Knowing that the coefficient of static friction is 0.30 at all surfaces of contact, determine the smallest force P that should be applied as shown to one of the wedges.

SOLUTION

µs = 0.30 φs = tan −1 0.30 = 16.70° Free body: 800-kg block and right-hand wedge W = (800 kg)(9.81 m/s2 ) = 7848 N

Force triangle:

α = 90° − 16.70° − 24.70° = 48.60°

Law of sines:

R1 7848 N = sin 16.70° sin 48.60° R1 = 3006.5 N

Free body: Left-hand wedge

Force triangle:

β = 16.70° + 24.70° = 41.40°

Law of sines:

R1 P = sin 41.40° sin(90° − 16.70°) P 3006.5 N = sin 41.40° cos 16.70°

P = 2080 N


!

PROBLEM 8.49 Two 8° wedges of negligible weight are used to move and position the 800-kg block. Knowing that the coefficient of static friction is 0.30 at all surfaces of contact, determine the smallest force P that should be applied as shown to one of the wedges.

SOLUTION

µs = 0.30 φs = tan −1 0.30 = 16.70° Free body: 800-kg block

Force triangle:

W = (800 kg)(9.81 m/s 2 ) = 7848 N

α = 90° − 2φs = 90° − 2(16.70°) = 56.60°

Law of sines:

R1 7848 N = sin 16.70° sin 56.60° R1 = 2701 N

Free body: Right-hand wedge Force triangle:

β = 16.70° + 24.70° = 41.40°

Law of sines:

R1 P = sin 41.40° sin(90° − 24.70°) P 2701 N = sin 41.40° cos 24.70°

P = 1966 N


!

PROBLEM 8.50 The elevation of the end of the steel beam supported by a concrete floor is adjusted by means of the steel wedges E and F. The base plate CD has been welded to the lower flange of the beam, and the end reaction of the beam is known to be 100 kN. The coefficient of static friction is 0.30 between two steel surfaces and 0.60 between steel and concrete. If the horizontal motion of the beam is prevented by the force Q, determine (a) the force P required to raise the beam, (b) the corresponding force Q.

SOLUTION Free body: Beam and plate CD (100 kN) cos16.7° R1 = 104.4 kN R1 =

Free body: Wedge E

P 104.4 kN = sin 43.4° sin 63.3°

(a)

P = 80.3 kN

!

φs = tan −1 0.3 = 16.7°

(b)

Q = (100 kN) tan16.7°

Q = 30 kN

Free body: Wedge F (To check that it does not move.) Since wedge F is a two-force body, R2 and R3 are colinear Thus But

θ = 26.7° φconcrete = tan −1 0.6 = 31.0° . θ

OK!


PROBLEM 8.51 The elevation of the end of the steel beam supported by a concrete floor is adjusted by means of the steel wedges E and F. The base plate CD has been welded to the lower flange of the beam, and the end reaction of the beam is known to be 100 kN. The coefficient of static friction is 0.30 between two steel surfaces and 0.60 between steel and concrete. If the horizontal motion of the beam is prevented by the force Q, determine (a) the force P required to raise the beam, (b) the corresponding force Q.

SOLUTION Free body: Wedge F

φs = tan −1 0.30 = 16.7°

(a)

P = (100 kN) tan 26.7° + (100 kN) tan φs P = 50.29 kN + 30 kN P = 80.29 kN R1 =

P = 80.3 kN

(100 kN) = 111.94 kN cos 26.7°

Free body: Beam, plate, and wedge E

(b)

Q = W tan 26.7° = (100 kN) tan 26.7° Q = 50.29 kN

Q = 50.3 kN


!

PROBLEM 8.52 A wedge A of negligible weight is to be driven between two 100-lb plates B and C. The coefficient of static friction between all surfaces of contact is 0.35. Determine the magnitude of the force P required to start moving the wedge (a) if the plates are equally free to move, (b) if plate C is securely bolted to the surface.

SOLUTION (a)

With plates equally free to move Free body: Plate B

φs = tan −1 µs = tan −1 0.35 = 19.2900°

Force triangle:

α = 180° − 124.29° − 19.29° = 36.42° Law of sines:

R1 100 lb = sin 19.29° sin 36.42° R1 = 55.643 lb

Free body: Wedge A Force triangle:

By symmetry,

R3 = R1 = 55.643 lb

β = 19.29° + 15° = 34.29°

(b)

Then

P = 2 R1 sin β

or

P = 2(55.643) sin 34.29°

P = 62.7 lb

With plate C bolted The free body diagrams of plate B and wedge A (the only members to move) are same as above. Answer is thus the same. P = 62.7 lb


!

PROBLEM 8.53 Block A supports a pipe column and rests as shown on wedge B. Knowing that the coefficient of static friction at all surfaces of contact is 0.25 and that θ = 45°, determine the smallest force P required to raise block A.

SOLUTION φs = tan −1 µ s = tan −1 0.25 = 14.036° FBD block A:

R2 3 kN = sin 104.036° sin 16.928° R2 = 10.0000 kN

FBD wedge B:

P 10.0000 kN = sin 73.072° sin 75.964° P = 9.8611 kN

P = 9.86 kN


!

PROBLEM 8.54 Block A supports a pipe column and rests as shown on wedge B. Knowing that the coefficient of static friction at all surfaces of contact is 0.25 and that θ = 45°, determine the smallest force P for which equilibrium is maintained.

SOLUTION φs = tan −1 µ s = tan −1 0.25 = 14.036° FBD block A:

R2 3 kN = sin(75.964°) sin(73.072°) R2 = 3.0420 kN

FBD wedge B:

P 3.0420 kN = sin 16.928° sin 104.036° P = 0.91300 kN

P = 913 N


!

PROBLEM 8.55 Block A supports a pipe column and rests as shown on wedge B. The coefficient of static friction at all surfaces of contact is 0.25. If P = 0, determine (a) the angle θ for which sliding is impending, (b) the corresponding force exerted on the block by the vertical wall.

SOLUTION Free body: Wedge B (a)

φs = tan −1 0.25 = 14.04°

Since wedge is a two-force body, R 2 and R 3 must be equal and opposite. Therefore, they form equal angles with vertical

β = φs and

θ − φs = φs θ = 2φs = 2(14.04°) θ = 28.1° Free body: Block A

R1 = (3 kN)sin 14.04° = 0.7278 kN

(b)

R1 = 728 N

Force exerted by wall:

14.04°


!

PROBLEM 8.56 A 12° wedge is used to spread a split ring. The coefficient of static friction between the wedge and the ring is 0.30. Knowing that a force P of magnitude 25 lb was required to insert the wedge, determine the magnitude of the forces exerted on the ring by the wedge after insertion.

SOLUTION Free body: Wedge

φs = tan −1 µ s = tan −1 0.30 = 16.6992°

Force triangle:

α = 6° + φ s = 6° + 16.6992° = 22.6992° Q = Horizontal component of R Q=

1 2

(25 lb)

tan 22.6992°

= 29.9 lb

Free body: After wedge has been inserted

Wedge is now a two-force body with forces shown. Q = 29.9 lb

(Note: Since angles between force Q and normal to wedge is 6° , φs , wedge stays in place.)!


PROBLEM 8.57 A 10° wedge is to be forced under end B of the 5-kg rod AB. Knowing that the coefficient of static friction is 0.40 between the wedge and the rod and 0.20 between the wedge and the floor, determine the smallest force P required to raise end B of the rod.

SOLUTION FBD AB: W = mg W = (5 kg)(9.81 m/s 2 ) W = 49.050 N

φs1 = tan −1 ( µ s )1 = tan −1 0.40 = 21.801° ΣM A = 0: rR1 cos(10° + 21.801°) − rR1 sin(10° + 21.801°) −

2r

π

(49.050 N) = 0

R1 = 96.678 N

FBD wedge:

φs 2 = tan −1 ( µ s ) 2 = tan −1 0.20 = 11.3099 P 96.678 N = sin(43.111°) sin 78.690°

P = 67.4 N


!

PROBLEM 8.58 A 10° wedge is used to split a section of a log. The coefficient of static friction between the wedge and the log is 0.35. Knowing that a force P of magnitude 600 lb was required to insert the wedge, determine the magnitude of the forces exerted on the wood by the wedge after insertion.

SOLUTION FBD wedge (impending motion ):

φs = tan −1 µs = tan −1 0.35 = 19.29°

By symmetry:

R1 = R2 ΣFy = 0: 2 R1 sin(5° + φs ) − 600 lb = 0

or

R1 = R2 =

300 lb = 729.30 lb sin (5°+19.29°)

When P is removed, the vertical components of R1 and R2 vanish, leaving the horizontal components R1x = R2 x = R1 cos(5° + φs ) = (729.30 lb) cos(5° + 19.29°)

R1x = R2 x = 665 lb

(Note that φs . 5°, so wedge is self-locking.)!


PROBLEM 8.59 A conical wedge is placed between two horizontal plates that are then slowly moved toward each other. Indicate what will happen to the wedge (a) if µs = 0.20, (b) if µs = 0.30.

SOLUTION

As the plates are moved, the angle θ will decrease. (a)

φs = tan −1 µs = tan −1 0.2 = 11.31°. As θ decrease, the minimum angle at the contact approaches 12.5° . φs = 11.31°, so the wedge will slide up and out from the slot.

(b)

φs = tan −1 µs = tan −1 0.3 = 16.70°. As θ decreases, the angle at one contact reaches 16.7°. (At this time the angle at the other contact is 25° − 16.7° = 8.3° , φs ). The wedge binds in the slot.


!

PROBLEM 8.60 A 15° wedge is forced under a 50-kg pipe as shown. The coefficient of static friction at all surfaces is 0.20. (a) Show that slipping will occur between the pipe and the vertical wall. (b) Determine the force P required to move the wedge.

SOLUTION Free body: Pipe ΣM B = 0: Wr sin θ + FA r (1 + sin θ ) − N A r cos θ = 0

Assume slipping at A: FA = µs NA NA cos θ − µs NA (1 + sin θ ) = W sin θ NA =

W sin θ cos θ − µs (1 + sin θ )

W sin 15° cos 15° − (0.20)(1 + sin 15°) = 0.36241W

NA =

ΣFx = 0: − FB − W sin θ − FA sin θ + N A cos θ = 0 FB = N A cos θ − µs N A sin θ − W sin θ FB = 0.36241W cos 15° − 0.20(0.36241W )sin 15° − W sin 15° FB = 0.072482W ΣFy = 0: N B − W cos θ − FA cos θ − NA sin θ = 0 N B = N A sin θ + µ s N A cos θ + W cos θ N B = (0.36241W )sin 15° + 0.20(0.36241W ) cos 15° + W cos 15° N B = 1.12974W

Maximum available: (a)

FB = µs N B = 0.22595W

No slip at B

We note that FB , Fmax



(b)

Free body: Wedge

ΣFy = 0: N 2 − N B cos θ + FB sin θ = 0 N 2 = N B cos θ − FB sin θ N 2 = (1.12974W ) cos15° − (0.07248W ) sin 15° N 2 = 1.07249W ΣFx = 0: FB cos θ + N B sin θ + µs N 2 − P = 0 P = FB cos θ + N B sin θ + µ s N 2 P = (0.07248W ) cos 15° + (1.12974W ) sin 15° + 0.2(1.07249W ) P = 0.5769W W = mg : P = 0.5769(50 kg)(9.81 m/s 2 )

P = 283 N


!

PROBLEM 8.61 A 15° wedge is forced under a 50-kg pipe as shown. Knowing that the coefficient of static friction at both surfaces of the wedge is 0.20, determine the largest coefficient of static friction between the pipe and the vertical wall for which slipping will occur at A.

SOLUTION Free body: Pipe ΣM A = 0: N B r cos θ − µ B N B r − ( µ B N B sin θ )r − Wr = 0

NB =

W cos θ − µ B (1 + sin θ )

W cos15° − 0.2(1 + sin15°) N B = 1.4002W NB =

ΣFx = 0: N A − N B sin θ − µ B N B cos θ = 0 N A = N B (sin θ + µ B cos θ ) = (1.4002W )(sin15° + 0.2 × cos15°) N A = 0.63293W ΣFy = 0: − FA − W + N B cos θ − µ B N B sin θ = 0 FA = N B (cos θ − µ B sin θ ) − W FA = (1.4002W )(cos15° − 0.2 × sin θ ) − W FA = 0.28001W

For slipping at A:

FA = µ A N A

µA =

FA 0.28001W = N A 0.63293W

µ A = 0.442


PROBLEM 8.62 An 8° wedge is to be forced under a machine base at B. Knowing that the coefficient of static friction at all surfaces of contact is 0.15, (a) determine the force P required to move the wedge, (b) indicate whether the machine base will slide on the floor.

SOLUTION Free body: Machine base ΣM B = 0: (200 lb)(3 ft) + (400 lb)(1.5 ft) − Ay (6 ft) = 0 Ay = 200 lb ΣFy = 0: Ay + By − 200 lb − 400 lb = 0 200 lb + By − 200 lb − 400 lb = 0 By = 400 lb

Free body: Wedge (Assume machine base will not move)

µs = 0.15, φs = tan −1 0.15 = 8.53° We know that Force triangle:

By = 400 lb 8° + φs = 8° + 8.53° = 16.53°

P = (400 lb) tan16.53° + (400 lb) tan 8.53° P = 178.7 lb

Total maximum friction force at A and B: Fm = µ sW = 0.15(200 lb + 400 lb) = 90 lb

If machine moves with wedge:

P = Fm = 90 lb

Using smaller P, we have (a)

P = 90.0 lb

(b)

Machine base moves


PROBLEM 8.63 Solve Problem 8.62 assuming that the wedge is to be forced under the machine base at A instead of B. PROBLEM 8.62 An 8° wedge is to be forced under a machine base at B. Knowing that the coefficient of static friction at all surfaces of contact is 0.15, (a) determine the force P required to move the wedge, (b) indicate whether the machine base will slide on the floor.

SOLUTION FBD: Machine base ΣM B = 0: (200 lb)(3 ft) + (400 lb)(1.5 ft) − Ay (6 ft) = 0

A y = 200 lb ΣFy = 0: Ay + B y − 200 lb − 400 lb = 0

B y = 400lb

φs = tan −1 0.15 = 8.53° We known that

Ay = 200 lb

FBD: Wedge Force triangle:

P = (200 lb) tan 8.53° + (200 lb) tan 16.53°

(a) (b)

P = 89.4 lb

Total maximum friction force at A and B: Fm = µs (W ) = 0.15(200 lb + 400 lb) = 90 lb

Since P , Fm ,

machine base will not move


PROBLEM 8.64* A 200-N block rests as shown on a wedge of negligible weight. The coefficient of static friction µs is the same at both surfaces of the wedge, and friction between the block and the vertical wall may be neglected. For P = 100 N, determine the value of µs for which motion is impending. (Hint: Solve the equation obtained by trial and error.)

SOLUTION Free body: Wedge Force triangle:

Law of sines:

R2 P = sin(90° − φs ) sin(15° + 2φs ) R2 = P

sin(90° − φs ) sin(15° + 2φs )

(1)

Free body: Block Fy = 0

Vertical component of R2 is 200 N Return to force triangle of wedge. Note P = 100 N 100 N = (200 N) tan φ + (200 N) tan(15° + φs ) 0.5 = tan φ + tan(15° + φs )

Solve by trial and error

φs = 6.292 µs = tan φs = tan 6.292°

µs = 0.1103


PROBLEM 8.65* Solve Problem 8.64 assuming that the rollers are removed and that µs is the coefficient of friction at all surfaces of contact. PROBLEM 8.64* A 200-N block rests as shown on a wedge of negligible weight. The coefficient of static friction µs is the same at both surfaces of the wedge, and friction between the block and the vertical wall may be neglected. For P = 100 N, determine the value of µs for which motion is impending. (Hint: Solve the equation obtained by trial and error.)

SOLUTION Free body: Wedge Force triangle:

Law of sines:

R2 P = sin (90° − φs ) sin (15° + 2φs ) R2 = P

sin (90° − φs ) sin (15° + 2φs )

(1)

Free body: Block (Rollers removed) Force triangle:

Law of sines:

R2 W = sin (90° + φs ) sin (75° − 2φs ) R2 = W

sin (90° + φs ) sin (75° − 2θ s )

(2)



Equate R2 from Eq. (1) and Eq. (2): P

sin (90° − φs ) sin (90° + φs ) =W sin (15° + 2φs ) sin (75° − 2φs ) P = 100 lb W = 200 N sin (90° + φs )sin (15° + 2φs ) 0.5 = sin (75° − 2φs )sin (90° − φs )

Solve by trial and error:

φs = 5.784° µs = tan φs = tan 5.784°

µs = 0.1013


PROBLEM 8.66 Derive the following formulas relating the load W and the force P exerted on the handle of the jack discussed in Section 8.6. (a) P = (Wr/a) tan (θ + φs), to raise the load; (b) P = (Wr/a) tan(φs − θ ), to lower the load if the screw is self-locking; (c) P = (Wr/a) tan (θ − φs), to hold the load if the screw is not self-locking.

SOLUTION FBD jack handle: See Section 8.6.

ΣM C = 0: aP − rQ = 0 or

P=

r Q a

FBD block on incline: (a)

Raising load

Q = W tan (θ + φs )

(b)

Lowering load if screw is self-locking (i.e.,: if φs > θ )

Q = W tan (φs − θ )

(c)

r P = W tan (θ + φs ) a

r P = W tan (φs − θ ) a

Holding load is screw is not self-locking (i.e., if φs , θ )

Q = W tan (θ − φs )

r P = W tan (θ − φs ) a


PROBLEM 8.67 The square-threaded worm gear shown has a mean radius of 1.5 in. and a lead of 0.375 in. The large gear is subjected to a constant clockwise couple of 7.2 kip ⋅ in. Knowing that the coefficient of static friction between the two gears is 0.12, determine the couple that must be applied to shaft AB in order to rotate the large gear counterclockwise. Neglect friction in the bearings at A, B, and C.

SOLUTION FBD large gear: ΣM C = 0: (12 in.)W − 7.2 kip ⋅ in. = 0 W = 0.600 kips = 600 lb

Block on incline:

θ = tan −1

0.375 in. = 2.2785° 2π (1.5 in.)

φs = tan −1 µs = tan −1 0.12 = 6.8428° Q = W tan (θ + φs ) = (600 lb) tan 9.1213° = 96.333lb

FBD worm gear: r = 1.5 in. ΣM B = 0: (1.5in.)(96.333lb) − M = 0 M = 144.5 lb ⋅ in.


PROBLEM 8.68 In Problem 8.67, determine the couple that must be applied to shaft AB in order to rotate the large gear clockwise. PROBLEM 8.67 The square-threaded worm gear shown has a mean radius of 1.5 in. and a lead of 0.375 in. The large gear is subjected to a constant clockwise couple of 7.2 kip ⋅ in. Knowing that the coefficient of static friction between the two gears is 0.12, determine the couple that must be applied to shaft AB in order to rotate the large gear counterclockwise. Neglect friction in the bearings at A, B, and C.

SOLUTION FBD large gear: ΣM C = 0: (12 in.)W − 7.2 kip ⋅ in. = 0 W = 0.600 kips = 600 lb

Block on incline:

θ = tan −1

0.375 in. 2π (1.5 in.)

= 2.2785°

φs = tan −1 µs = tan −1 0.12 = 6.8428° Q = W tan (φs − θ ) = (600 lb) tan 4.5643° = 47.898 lb

FBD worm gear: r = 1.5 in. ΣM B = 0: M − (1.5in.)(47.898 lb) = 0 M = 71.8lb ⋅ in.


PROBLEM 8.69 High-strength bolts are used in the construction of many steel structures. For a 24-mmnominal-diameter bolt the required minimum bolt tension is 210 kN. Assuming the coefficient of friction to be 0.40, determine the required couple that should be applied to the bolt and nut. The mean diameter of the thread is 22.6 mm, and the lead is 3 mm. Neglect friction between the nut and washer, and assume the bolt to be square-threaded.

SOLUTION FBD block on incline:

3 mm (22.6 mm)π = 2.4195°

θ = tan −1

φs = tan −1 µ s = tan −1 0.40 φs = 21.801° Q = (210 kN) tan (21.801° + 2.4195°) Q = 94.468 kN d Torque = Q 2 22.6 mm = (94.468 kN) 2 = 1067.49 N ⋅ m

Torque = 1068 N ⋅ m


PROBLEM 8.70 The ends of two fixed rods A and B are each made in the form of a single-threaded screw of mean radius 6 mm and pitch 2 mm. Rod A has a right-handed thread and rod B has a left-handed thread. The coefficient of static friction between the rods and the threaded sleeve is 0.12. Determine the magnitude of the couple that must be applied to the sleeve in order to draw the rods closer together.

SOLUTION To draw rods together: Screw at A tan θ =

2 mm 2π (6 mm)

θ = 3.037° φs = tan −1 0.12 = 6.843° Q = (2 kN) tan 9.88° = 348.3 N Torque at A = Qr = (348.3 N)(6 mm) = 2.09 N ⋅ m

Total torque = 4.18 N ⋅ m

Same torque required at B


PROBLEM 8.71 Assuming that in Problem 8.70 a right-handed thread is used on both rods A and B, determine the magnitude of the couple that must be applied to the sleeve in order to rotate it. PROBLEM 8.70 The ends of two fixed rods A and B are each made in the form of a single-threaded screw of mean radius 6 mm and pitch 2 mm. Rod A has a right-handed thread and rod B has a left-handed thread. The coefficient of static friction between the rods and the threaded sleeve is 0.12. Determine the magnitude of the couple that must be applied to the sleeve in order to draw the rods closer together.

SOLUTION From the solution to Problem 8.70, Torque at A = 2.09 N ⋅ m Screw at B: Loosening

θ = 3.037° φs = 6.843° Q = (2 kN) tan 3.806° = 133.1 N

Torque at B = Qr = (133.1 N)(6 mm) = 0.798 N ⋅ m

Total torque = 2.09 N ⋅ m + 0.798 N ⋅ m

Total torque = 2.89 N ⋅ m


PROBLEM 8.72 In the machinist’s vise shown, the movable jaw D is rigidly attached to the tongue AB that fits loosely into the fixed body of the vise. The screw is single-threaded into the fixed base and has a mean diameter of 0.75 in. and a pitch of 0.25 in. The coefficient of static friction is 0.25 between the threads and also between the tongue and the body. Neglecting bearing friction between the screw and the movable head, determine the couple that must be applied to the handle in order to produce a clamping force of 1 kip.

SOLUTION Free body: Jaw D and tongue AB P is due to elastic forces in clamped object. W is force exerted by screw. ΣFy = 0: N H − N J = 0 N J = N H = N

For final tightening, FH = FJ = µs N = 0.25 N ΣFx = 0: W − P − 2(0.25 N) = 0 N = 2(W − P)

(1)

ΣM H = 0: P(3.75) − W (2) − N (3) + (0.25 N)(1.25) = 0 3.75P − 2W − 2.6875 N = 0

(2)

3.75 P − 2W − 2.6875[2(W − P )] = 0

Substitute Eq. (1) into Eq. (2):

7.375W = 9.125P = 9.125(1 kip) W = 1.23729 kips

Block-and-incline analysis of screw: tan φs = µ s = 0.25

φs = 14.0362° tan θ =

0.25 in.

π (0.75 in.)

θ = 6.0566° θ + φs = 20.093° Q = (1.23729 kips) tan 20.093° = 0.45261 kip ! 0.75 in. " T = Qr = (452.61 lb) # $ % 2 &

T = 169.7 lb ⋅ in.


PROBLEM 8.73 In Problem 8.72, a clamping force of 1 kip was obtained by tightening the vise. Determine the couple that must be applied to the screw to loosen the vise. PROBLEM 8.72 In the machinist’s vise shown, the movable jaw D is rigidly attached to the tongue AB that fits loosely into the fixed body of the vise. The screw is single-threaded into the fixed base and has a mean diameter of 0.75 in. and a pitch of 0.25 in. The coefficient of static friction is 0.25 between the threads and also between the tongue and the body. Neglecting bearing friction between the screw and the movable head, determine the couple that must be applied to the handle in order to produce a clamping force of 1 kip.

SOLUTION Free body: Jaw D and tongue AB P is due to elastic forces in clamped object. W is force exerted by screw. ΣFy = 0: N H − N J = 0 N J = N H = N

As vise is just about to loosen,

FH = FJ = µs N = 0.25 N

ΣFx = 0: W − P + 2(0.25 N) = 0 N = 2( P − W )

(1)

ΣM H = 0: P(3.75) − W (2) − N (3) − (0.25 N)(1.25) = 0 3.75P − 2W − 3.3125 N = 0

Substitute Eq. (1) into Eq. (2):

(2)

3.75 P − 2W − 3.3125[2( P − W )] = 0 4.625W = 2.875P = 2.875(1 kip) W = 0.62162 kip

Block-and-incline analysis of screw: tan φs = µs = 0.25 0.25in. π (0.75in.) φs − θ = 7.9796° tan θ =

φs = 14.0362° θ = 6.0566°

Q = (621.62 lb) tan 7.9796° = 87.137 lb ! 0.75 in. " M = Qr = (87.137 lb) # $ % 2 &

M = 32.7 lb ⋅ in.


PROBLEM 8.74 In the gear-pulling assembly shown the square-threaded screw AB has a mean radius of 15 mm and a lead of 4 mm. Knowing that the coefficient of static friction is 0.10, determine the couple that must be applied to the screw in order to produce a force of 3 kN on the gear. Neglect friction at end A of the screw.

SOLUTION Block/Incline: 0.25 in. 1.875π in. = 2.4302°

θ = tan −1

φs = tan −1 µs = tan −1 (0.10) = 5.7106° Q = (3000 N) tan (8.1408°) = 429.14 N

Couple = rQ = (0.015 m)(429.14 N) = 6.4371 N ⋅ m

M = 6.44 N ⋅ m


PROBLEM 8.75 A 6-in.-radius pulley of weight 5 lb is attached to a 1.5-in.-radius shaft that fits loosely in a fixed bearing. It is observed that the pulley will just start rotating if a 0.5-lb weight is added to block A. Determine the coefficient of static friction between the shaft and the bearing.

SOLUTION

ΣM D = 0: (10.5 lb)(6 in. − rf ) − (10 lb)(6 in. + rf ) = 0 (0.5 lb)(6 in.) = (20.5 lb)r f rf = 0.146341 in. r f = r sin φs 0.146341 in. 1.5 in. = 0.097561

sin φs =

φs = 5.5987° µs = tan φs µs = 0.0980

= tan 5.5987°


PROBLEM 8.76 The double pulley shown is attached to a 10-mm-radius shaft that fits loosely in a fixed bearing. Knowing that the coefficient of static friction between the shaft and the poorly lubricated bearing is 0.40, determine the magnitude of the force P required to start raising the load.

SOLUTION ΣM D = 0: P(45 − r f ) − W (90 + rf ) = 0 P =W

90 + rf 45 − rf

= (196.2 N) P = 449.82 N

90 mm + 4 mm 45 mm − 4 mm P = 450 N


!

PROBLEM 8.77 The double pulley shown is attached to a 10-mm-radius shaft that fits loosely in a fixed bearing. Knowing that the coefficient of static friction between the shaft and the poorly lubricated bearing is 0.40, determine the magnitude of the force P required to start raising the load.

SOLUTION Find P required to start raising load ΣM D = 0: P(45 − r f ) − W (90 − rf ) = 0 P =W

90 − r f 45 − rf

= (196.2 N) P = 411.54 N

90 mm − 4 mm 45 mm − 4 mm P = 412 N


!

PROBLEM 8.78 The double pulley shown is attached to a 10-mm-radius shaft that fits loosely in a fixed bearing. Knowing that the coefficient of static friction between the shaft and the poorly lubricated bearing is 0.40, determine the magnitude of the smallest force P required to maintain equilibrium.

SOLUTION Find smallest P to maintain equilibrium ΣM D = 0: P(45 + rf ) − W (90 − rf ) = 0 P =W

90 − rf 45 + rf

= (196.2 N) P = 344.35 N

90 mm − 4 mm 45 mm + 4 mm P = 344 N


!

PROBLEM 8.79 The double pulley shown is attached to a 10-mm-radius shaft that fits loosely in a fixed bearing. Knowing that the coefficient of static friction between the shaft and the poorly lubricated bearing is 0.40, determine the magnitude of the smallest force P required to maintain equilibrium.

SOLUTION Find smallest P to maintain equilibrium ΣM D = 0: P(45 + rf ) − W (90 + r f ) = 0 P =W

90 + rf 45 + rf

= (196.2 N) P = 376.38 N

90 mm + 4 mm 45 mm + 4 mm P = 376 N


!

PROBLEM 8.80 A lever of negligible weight is loosely fitted onto a 75-mmdiameter fixed shaft. It is observed that the lever will just start rotating if a 3-kg mass is added at C. Determine the coefficient of static friction between the shaft and the lever.

SOLUTION ΣM O = 0: WC (150) − WD (100) − Rr f = 0 WC = (23 kg)(9.81 m/s 2 )

But

WD = (30 kg)(9.81 m/s 2 ) R = WC + WD = (53 kg)(9.81)

Thus, after dividing by 9.81, 23(150) − 30(100) − 53 rf = 0 rf = 8.49 mm

But

µs ≈

rf r

=

8.49 mm 37.5 mm

µs ≈ 0.226


!

PROBLEM 8.81 The block and tackle shown are used to raise a 150-lb load. Each of the 3-in.-diameter pulleys rotates on a 0.5-in.-diameter axle. Knowing that the coefficient of static friction is 0.20, determine the tension in each portion of the rope as the load is slowly raised.

SOLUTION For each pulley:

Axle diameter = 0.5 in. ! 0.5 in. " r f = r sin φs ≈ µs r = 0.20 # $ = 0.05 in. % 2 &

Pulley BC:

ΣM B = 0: TCD (3 in.) − (150 lb)(1.5 in. + rf ) = 0 1 TCD = (150 lb)(1.5 in. + 0.05 in.) 3 ΣFy = 0: TAB + 77.5 lb − 150 lb = 0

Pulley DE:

TCD = 77.5 lb TAB = 72.5 lb

ΣM B = 0: TCD (1.5 + rf ) − TEF (1.5 − rf ) = 0

TEF = TCD

1.5 + rf 1.5 − rf

= (77.5 lb)

1.5 in. + 0.05 in. 1.5 in. − 0.05 in.

TEF = 82.8 lb


!

PROBLEM 8.82 The block and tackle shown are used to lower a 150-lb load. Each of the 3-in.-diameter pulleys rotates on a 0.5-in.-diameter axle. Knowing that the coefficient of static friction is 0.20, determine the tension in each portion of the rope as the load is slowly lowered.

SOLUTION For each pulley:

Pulley BC:

! 0.5 in. " rf = r µs = # $ 0.2 = 0.05 in. % 2 &

ΣM B = 0: TCD (3 in.) − (150 lb)(1.5 in. − rf ) = 0 TCD =

(150 lb)(1.5 in. − 0.05 in.) 3 in.

ΣFy = 0: TAB + 72.5 lb − 150 lb = 0

Pulley DE:

TCD = 72.5 lb TAB = 77.5 lb

TCD (1.5 in. − rf ) − TEF (1.5 in. + rf ) = 0 TEF = TCD

1.5 in. − rf 1.5 in. + r f

= (72.5 lb)

1.5 in. − 0.05 in. 1.5 in. + 0.05 in.

TEF = 67.8 lb


!

PROBLEM 8.83 A loaded railroad car has a mass of 30 Mg and is supported by eight 800-mm-diameter wheels with 125-mmdiameter axles. Knowing that the coefficients of friction are µs = 0.020 and µk = 0.015, determine the horizontal force required (a) to start the car moving, (b) to keep the car moving at a constant speed. Neglect rolling resistance between the wheels and the track.

SOLUTION rf = µ r ; R = 400 mm sin θ = tan θ =

rf

=

µr

R µr P = W tan θ = W R 62.5 mm P =Wµ 400 mm = 0.15625Wµ

For one wheel:

For eight wheels of rail road car:

(a)

To start motion:

R

1 W = (30 mg)(9.81 m/s 2 ) 8 1 = (294.3 kN) 8 1 ΣP = 8(0.15625) (294.3 kN) µ 8 = (45.984µ ) kN

µs = 0.020 ΣP = (45.984)(0.020) = 0.9197 kN

(b)

To maintain motion:

ΣP = 920 N

µk = 0.015 ΣP = (45.984)(0.015) = 0.6897 kN

ΣP = 690 N


!

PROBLEM 8.84 A lever AB of negligible weight is loosely fitted onto a 2.5-in.-diameter fixed shaft. Knowing that the coefficient of static friction between the fixed shaft and the lever is 0.15, determine the force P required to start the lever rotating counterclockwise.

SOLUTION rf = µs r = 0.15(1.25 in.) = 0.1875 in. ΣM D = 0: P(5 in. + rf ) − (50 lb) rf = 0 50(0.1875) 5.1875 = 1.807 lb

P=

P = 1.807 lb


!

PROBLEM 8.85 A lever AB of negligible weight is loosely fitted onto a 2.5-in.-diameter fixed shaft. Knowing that the coefficient of static friction between the fixed shaft and the lever is 0.15, determine the force P required to start the lever rotating counterclockwise.

SOLUTION rf = µ s r = 0.15(1.25 in.) rf = 0.1875 in. 2 in. 5 in . γ = 21.801°

tan γ =

In ∆EOD: OD = (2 in.) 2 + (5 in.) 2 = 5.3852 in. rf OE sin θ = = OD OD 0.1875 in. = 5.3852 in. θ = 1.99531°

Force triangle: P = (50 lb) tan (γ + θ ) = (50 lb) tan 23.796° = 22.049 lb

P = 22.0 lb


!

PROBLEM 8.86 A lever AB of negligible weight is loosely fitted onto a 2.5-in.-diameter fixed shaft. Knowing that the coefficient of static friction between the fixed shaft and the lever is 0.15, determine the force P required to start the lever rotating clockwise.

SOLUTION rf = µs r = 0.15(1.25 in.) r f = 0.1875 in. ΣM D = 0: P (5 in. − rf ) − (50 lb)r f = 0 50(0.1875) 5 − 0.1875 = 1.948 lb

P=

P = 1.948 lb


!

PROBLEM 8.87 A lever AB of negligible weight is loosely fitted onto a 2.5-in.-diameter fixed shaft. Knowing that the coefficient of static friction between the fixed shaft and the lever is 0.15, determine the force P required to start the lever rotating clockwise.

SOLUTION rf = µ s r = 0.15(1.25 in.) = 0.1875 in. 5 in. tan β = 2 in.

β = 68.198° In ∆EOD: OD = (2 in.) 2 + (5 in.) 2 OD = 5.3852 in. OE 0.1875 in. sin θ = = OD 5.3852 in. θ = 1.99531°

Force triangle:

P=

50 50 lb = tan ( β + θ ) tan 70.193°

P = 18.01 lb


!

PROBLEM 8.88 The link arrangement shown is frequently used in highway bridge construction to allow for expansion due to changes in temperature. At each of the 60-mmdiameter pins A and B the coefficient of static friction is 0.20. Knowing that the vertical component of the force exerted by BC on the link is 200 kN, determine (a) the horizontal force that should be exerted on beam BC to just move the link, (b) the angle that the resulting force exerted by beam BC on the link will form with the vertical.

SOLUTION Bearing: r = 30 mm rf = µs r = 0.20(30 mm) = 6 mm

Resultant forces R must be tangent to friction circles at Points C and D. (a) Ry = Vertical component = 200 kN Rx = Ry tan θ = (200 kN) tan 1.375° = 4.80 kN Horizontal force = 4.80 kN

(b) 6 mm 250 mm sin θ = 0.024 sin θ =

θ = 1.375°

!


PROBLEM 8.89 A scooter is to be designed to roll down a 2 percent slope at a constant speed. Assuming that the coefficient of kinetic friction between the 25-mm-diameter axles and the bearings is 0.10, determine the required diameter of the wheels. Neglect the rolling resistance between the wheels and the ground.

SOLUTION

tan θ =

2 = 0.02 100

Since a scooter rolls at constant speed, each wheel is in equilibrium. Thus, W and R must have a common line of action tangent to the friction circle. rf = µ k r = (0.10)(12.5 mm) = 1.25 mm rf OB 1.25 mm = = tan θ tan θ 0.02 = 62.5 mm

OA =

Diameter of wheel = 2(OA) = 125.0 mm


!

PROBLEM 8.90 A 50-lb electric floor polisher is operated on a surface for which the coefficient of kinetic friction is 0.25. Assuming that the normal force per unit area between the disk and the floor is uniformly distributed, determine the magnitude Q of the horizontal forces required to prevent motion of the machine.

SOLUTION See Figure 8.12 and Eq. (8.9). Using: R = 9 in. P = 50 lb

and

µk = 0.25 2 2 µk PR = (0.25)(50 lb)(9 in.) 3 3 = 75 lb ⋅ in.

M =

ΣM y = 0 yields:

M = Q(20 in.) 75 lb ⋅ in. = Q(20 in.) Q = 3.75 lb


!

PROBLEM 8.91 Knowing that a couple of magnitude 30 N ⋅ m is required to start the vertical shaft rotating, determine the coefficient of static friction between the annular surfaces of contact.

SOLUTION For annular contact regions, use Equation 8.8 with impending slipping: M=

So,

30 N ⋅ m =

R3 − R13 2 µ s N 22 3 R2 − R12 2 (0.06 m)3 − (0.025 m)3 µ s (4000 N) 3 (0.06 m) 2 − (0.025 m) 2

µs = 0.1670


PROBLEM 8.92* The frictional resistance of a thrust bearing decreases as the shaft and bearing surfaces wear out. It is generally assumed that the wear is directly proportional to the distance traveled by any given point of the shaft and thus to the distance r from the point to the axis of the shaft. Assuming, then, that the normal force per unit area is inversely proportional to r, show that the magnitude M of the couple required to overcome the frictional resistance of a worn-out end bearing (with contact over the full circular area) is equal to 75 percent of the value given by Eq. (8.9) for a new bearing.

SOLUTION Using Figure 8.12, we assume ∆N =

k ∆A: ∆A = r ∆θ ∆ r r ∆N =

P = Σ∆N

We write

P= ∆N =

2π 0

or R 0

k r ∆θ ∆r = k ∆θ ∆r r

P = dN

k ∆θ ∆r = 2π Rk ; k =

P 2π R

P∆θ ∆r 2π R P∆θ ∆r 2π R R µ P 2πµk P R 2 1 k rdrdθ = ⋅ = µk PR 0 2π R 2π R 2 2

∆M = r ∆F = r µ k ∆N = r µk M=

2π 0

From Eq. (8.9) for a new bearing,

Thus,

M New =

2 µk PR 3

M = M New

1 2 2 3

=

3 4

M = 0.75M New


!

PROBLEM 8.93* Assuming that bearings wear out as indicated in Problem 8.92, show that the magnitude M of the couple required to overcome the frictional resistance of a worn-out collar bearing is M=

1 µk P( R1 + R2 ) 2

where P = magnitude of the total axial force R1, R2 = inner and outer radii of collar

SOLUTION Let normal force on ∆A be ∆N , and

∆N k = . ∆A r

As in the text,

∆F = µ∆N

∆M = r ∆F

The total normal force P is P = lim Σ∆N = ∆A→ 0

P = 2π

Total couple:

R2 R1

! # %

k " rdr $ dθ r &

R2 R1

kdr = 2π k ( R2 − R1 ) or k =

M worn = lim Σ∆M = ∆A→ 0

M worn = 2πµ k

2π 0

R2 R1

2π 0

! # %

R2 R1

(rdr ) = πµ k

(

rµ

R22

P 2π ( R2 − R1 )

k " rdr $ dθ r &

−

R12

)=

(

πµ P R22 − R12

)

2π ( R2 − R1 ) M worn =

1 µ P( R2 + R1 ) 2


!

PROBLEM 8.94* Assuming that the pressure between the surfaces of contact is uniform, show that the magnitude M of the couple required to overcome frictional resistance for the conical bearing shown is 2 µ k P R23 − R13 3 sin θ R22 − R12

M=

SOLUTION Let normal force on ∆A be ∆N and So

∆N = k ∆A

∆N = k. ∆A ∆A = r ∆s ∆φ

∆s =

∆r sin θ

where φ is the azimuthal angle around the symmetry axis of rotation. ∆Fy = ∆N sin θ = kr ∆r ∆φ

Total vertical force:

P = lim Σ∆Fy ∆A→0

P=

2π 0

! # %

R2 R1

" krdr $ dφ = 2π k &

(

P = π k R22 − R12

Friction force: Moment: Total couple:

)

or k =

R2 R1

rdr

P

π

(

R22

− R12

)

∆F = µ∆N = µ k ∆A ∆M = r ∆F = r µ kr M = lim Σ∆M = ∆A→0

M = 2π

µk sin θ

R2 R1

∆r ∆φ sin θ 2π !

R2

0

R1

# %

r 2 dr =

µk 2 " r dr $ dφ sin θ &

2 πµ P 3 sin θ π R22 − R32

(

)

(R

3 2

− R33

)

M=

2 µ P R23 − R13 3 sin θ R22 − R12


!

PROBLEM 8.95 Solve Problem 8.90 assuming that the normal force per unit area between the disk and the floor varies linearly from a maximum at the center to zero at the circumference of the disk. PROBLEM 8.90 A 50-lb electric floor polisher is operated on a surface for which the coefficient of kinetic friction is 0.25. Assuming that the normal force per unit area between the disk and the floor is uniformly distributed, determine the magnitude Q of the horizontal forces required to prevent motion of the machine.

SOLUTION r" ∆N ! = k #1 − $ . R ∆A % &

Let normal force on ∆A be ∆N and

r" r" ! ! ∆F = µ∆N = µ k #1 − $ ∆A = µ k #1 − $ r ∆r ∆θ R R % & % & P = lim Σ∆N = ∆A→0

P = 2π k

0

' ) +

R 0

( r" ! k #1 − $ rdr * dθ R& % ,

! R 2 R3 " r" rdr k π 1 2 − = # $ ## 2 − 3R $$ R& % % &

R! 0

1 P = π kR 2 3

or k =

3P π R2

( r" ! r µ k #1 − $ rdr * dθ R % & , 3 R! r " = 2πµ k ## r 2 − $$ dr 0 R& %

M = lim Σr ∆F = ∆A→0

2π

2π

0

' ) +

R

0

! R3 R 4 " 1 3 = 2πµ k ## − $$ = πµ kR R 3 4 6 % & πµ 3P 3 1 R = µ PR = 6 π R2 2

where

µ = µk = 0.25 R = 9 in.



P = W = 50 lb 1 (0.25)(50 lb)(9 in.) 2 = 56.250 lb ⋅ in.

Then

M=

Finally

Q=

M 56.250 lb ⋅ in. = d 20 in.

Q = 2.81 lb


PROBLEM 8.96 A 900-kg machine base is rolled along a concrete floor using a series of steel pipes with outside diameters of 100 mm. Knowing that the coefficient of rolling resistance is 0.5 mm between the pipes and the base and 1.25 mm between the pipes and the concrete floor, determine the magnitude of the force P required to slowly move the base along the floor.

SOLUTION FBD pipe:

W = mg = (900 kg)(9.81 m/s 2 ) = 8829.0 N .5 mm + 1.25 mm θ = sin −1 100 mm = 1.00273° P = W tan θ for each pipe, so also for total P = (8829.0 N) tan (1.00273°)

P = 154.4 N


PROBLEM 8.97 Knowing that a 6-in.-diameter disk rolls at a constant velocity down a 2 percent incline, determine the coefficient of rolling resistance between the disk and the incline.

SOLUTION FBD disk:

tan θ = slope = 0.02 b = r tan θ = (3 in.)(0.02)

b = 0.0600 in.


PROBLEM 8.98 Determine the horizontal force required to move a 2500-lb automobile with 23-in.-diameter tires along a horizontal road at a constant speed. Neglect all forms of friction except rolling resistance, and assume the coefficient of rolling resistance to be 0.05 in.

SOLUTION FBD wheel:

r = 11.5 in. b = 0.05 in. b θ = sin −1 r b" ! P = W tan θ = W tan # sin −1 $ for each wheel, so for total r& % 0.05 " ! P = 2500 lb tan # sin −1 11.5 $& %

P = 10.87 lb


!

PROBLEM 8.99 Solve Problem 8.83 including the effect of a coefficient of rolling resistance of 0.5 mm. PROBLEM 8.83 A loaded railroad car has a mass of 30 Mg and is supported by eight 800-mm-diameter wheels with 125-mm-diameter axles. Knowing that the coefficients of friction are µs = 0.020 and µk = 0.015, determine the horizontal force required (a) to start the car moving, (b) to keep the car moving at a constant speed. Neglect rolling resistance between the wheels and the track.

SOLUTION For one wheel: rf = µ r tan θ ≈ sin θ ≈ tan θ =

rf + b a

µr + b

a W W µr + b Q = tan θ = a 8 8

For eight wheels of car:

P =W

µr + b

a W = mg = (30 Mg)(9.81 m/s 2 ) = 294.3 kN a = 400 mm, r = 62.5 mm, b = 0.5 mm

(a)

To start motion:

µ = µs = 0.02 P = (294.3 kN)

(b)

To maintain constant speed

(0.020)(62.5 mm) + 0.5 mm 400 mm P = 1.288 kN

!

P = 1.058 kN

!

µ = µk = 0.015 P = (294.3 kN)

(0.015)(62.5 mm) + 0.5 mm 400 mm


PROBLEM 8.100 Solve Problem 8.89 including the effect of a coefficient of rolling resistance of 1.75 mm. PROBLEM 8.89 A scooter is to be designed to roll down a 2 percent slope at a constant speed. Assuming that the coefficient of kinetic friction between the 25-mm-diameter axles and the bearings is 0.10, determine the required diameter of the wheels. Neglect the rolling resistance between the wheels and the ground.

SOLUTION Since the scooter rolls at a constant speed, each wheel is in equilibrium. Thus, W and R must have a common line of action tangent to the friction circle. a = Radius of wheel tan θ =

2 = 0.02 100

Since b and r f are small compared to a, tan θ ≈

Data:

rf + b a

=

µk r + b a

= 0.02

µk = 0.10, b = 1.75 mm, r = 12.5 mm (0.10)(12.5 mm) + 1.75 mm = 0.02 a a = 150 mm

Diameter = 2a = 300 mm


PROBLEM 8.101 A hawser is wrapped two full turns around a bollard. By exerting an 80-lb force on the free end of the hawser, a dockworker can resist a force of 5000 lb on the other end of the hawser. Determine (a) the coefficient of static friction between the hawser and the bollard, (b) the number of times the hawser should be wrapped around the bollard if a 20,000-lb force is to be resisted by the same 80-lb force.

SOLUTION

β = 2 turns = 2(2π ) = 4π

(a)

T1 = 80 lb, ln

T2 = 5000 lb

T2 = µs β T1

µs = µs =

1

β

ln

T2 1 5000 lb ln = T1 4π 80 lb

1 4.1351 ln 62.5 = 4π 4π

µs = 0.329

T1 = 80 lb, T2 = 20,000 lb, µs = 0.329

(b) ln

T2 = µs β T1

β= β=

1

µ

ln

T2 1 20, 000 lb = ln T1 0.329 80 lb

1 5.5215 = 16.783 ln(250) = 0.329 0.329

Number of turns =

16.783 2π

Number of turns = 2.67


PROBLEM 8.102 A rope ABCD is looped over two pipes as shown. Knowing that the coefficient of static friction is 0.25, determine (a) the smallest value of the mass m for which equilibrium is possible, (b) the corresponding tension in portion BC of the rope.

SOLUTION We apply Eq. (8.14) to pipe B and pipe C. T2 = eµs β T1

Pipe B:

(8.14)

T2 = WA , T1 = TBC

µs = 0.25, β =

2π 3

WA = e0.25(2π /3) = eπ /6 TBC

Pipe C:

(1)

T2 = TBC , T1 = WD , µ s = 0.25, β =

π 3

TBC = e0.025(π /3) = eπ /12 WD

(a)

(2)

Multiplying Eq. (1) by Eq. (2): WA = eπ /6 ⋅ eπ /12 = eπ /6 + π /12 = eπ / 4 = 2.193 WD WA

WD =

WA W mA 50 kg g m= D = = = g 2.193 2.193 2.193 2.193 m = 22.8 kg

(b)

TBC =

From Eq. (1):

WA eπ /6

=

(50 kg)(9.81 m/s 2 ) = 291 N 1.688


PROBLEM 8.103 A rope ABCD is looped over two pipes as shown. Knowing that the coefficient of static friction is 0.25, determine (a) the largest value of the mass m for which equilibrium is possible, (b) the corresponding tension in portion BC of the rope.

SOLUTION See FB diagrams of Problem 8.102. We apply Eq. (8.14) to pipes B and C. Pipe B:

T1 = WA , T2 = TBC , µ s = 0.25, β =

2π 3

T T2 = e µs β : BC = e0.25(2π /3) = eπ /6 T1 WA

Pipe C:

T1 = TBC , T2 = WD , µ s = 0.25, β =

(1)

π 3

T2 WD = e µs β : = e0.25(π /3) = eπ /12 T1 TBC

(a)

(2)

Multiply Eq. (1) by Eq. (2): WD = eπ /6 ⋅ eπ /12 = eπ /6 + π /12 = eπ /4 = 2.193 WA W0 = 2.193WA

(b)

From Eq. (1):

m = 2.193mA = 2.193(50 kg)

m = 109.7 kg

TBC = WAeπ /6 = (50 kg)(9.81 m/s 2 )(1.688) = 828 N


PROBLEM 8.104 A 300-lb block is supported by a rope that is wrapped 1 12 times around a horizontal rod. Knowing that the coefficient of static friction between the rope and the rod is 0.15, determine the range of values of P for which equilibrium is maintained.

SOLUTION

β = 1.5 turns = 3π rad For impending motion of W up, P = We µs β = (300 lb)e(0.15)3π = 1233.36 lb

For impending motion of W down, P = We− µs β = (300 lb)e− (0.15)3π = 72.971 lb 73.0 lb # P # 1233 lb

For equilibrium,


PROBLEM 8.105 The coefficient of static friction between block B and the horizontal surface and between the rope and support C is 0.40. Knowing that mA = 12 kg, determine the smallest mass of block B for which equilibrium is maintained.

SOLUTION Support at C:

FBD block B:

WA = mg = (12 kg) g ΣFy = 0: N B − WB = 0 or

N B = WB

FB = µs N B = 0.4 N B = 0.4WB

Impending motion:

ΣFx = 0: FB − TB = 0 or TB = FB = 0.4WB

At support, for impending motion of WA down: WA = TB e µs β

so

TB = WAe − µs β = (12 kg) g − (0.4)π /2 = (6.4019 kg) g

Now

WB =

TB ! 6.4019 kg " = $ g = 16.0048 g 0.4 #% 0.4 &

so that

mB =

WB 16.0048 g = g g

mB = 16.00 kg


PROBLEM 8.106 The coefficient of static friction µs is the same between block B and the horizontal surface and between the rope and support C. Knowing that mA = mB, determine the smallest value of µs for which equilibrium is maintained.

SOLUTION Support at C:

FBD B:

ΣFy = 0: N B − W = 0 or

FB = µs N B = µsW

Impending motion:

ΣFx = 0: FB − TB = 0 or

Impending motion of rope on support: or or

NB = W

TB = FB = µ sW

W = TB e µs β = µsWe µs β 1 = µ s e µs β

µs eπ /2 µs = 1 µs = 0.475



PROBLEM 8.107 A flat belt is used to transmit a couple from drum B to drum A. Knowing that the coefficient of static friction is 0.40 and that the allowable belt tension is 450 N, determine the largest couple that can be exerted on drum A.

SOLUTION FBD’s drums: 7π 6 π 5π β B = 180° − 30° = π − = 6 6

β A = 180° + 30° = π +

π

6

=

Since β B , β A , slipping will impend first on B (friction coefficients being equal) So

T2 = Tmax = T1e µs β B

450 N = T1e(0.4)5π /6

or T1 = 157.914 N

ΣM A = 0: M A + (0.12 m)(T1 − T2 ) = 0 M A = (0.12 m)(450 N − 157.914 N) = 35.05 N ⋅ m M A = 35.1 N ⋅ m


!

PROBLEM 8.108 A flat belt is used to transmit a couple from pulley A to pulley B. The radius of each pulley is 60 mm, and a force of magnitude P = 900 N is applied as shown to the axle of pulley A. Knowing that the coefficient of static friction is 0.35, determine (a) the largest couple that can be transmitted, (b) the corresponding maximum value of the tension in the belt.

SOLUTION Drum A: T2 = e µsπ = e(0.35)π T1 T2 = 3.0028 T1

β = 180° = π radians (a)

Torque:

ΣM A = 0: M − (675.15 N)(0.06 m) + (224.84 N)(0.06 m) M = 27.0 N ⋅ m

(b)

ΣFx = 0: T1 + T2 − 900 N = 0 T1 + 3.0028 T1 − 900 N = 0 4.00282 T1 = 900 T1 = 224.841 N T2 = 3.0028(224.841 N) = 675.15 N Tmax = 675 N


!

PROBLEM 8.109 Solve Problem 8.108 assuming that the belt is looped around the pulleys in a figure eight. PROBLEM 8.108 A flat belt is used to transmit a couple from pulley A to pulley B. The radius of each pulley is 60 mm, and a force of magnitude P = 900 N is applied as shown to the axle of pulley A. Knowing that the coefficient of static friction is 0.35, determine (a) the largest couple that can be transmitted, (b) the corresponding maximum value of the tension in the belt.

SOLUTION Drum A:

β = 240° = 240°

60 1 = 120 2 θ = 30°

π

sin θ =

4 = π 180° 3 T2 = e µs β = e0.35(4/3π ) T1 T2 = 4.3322 T1

(a)

Torque:

ΣM B = 0: M − (844.3 N)(0.06 m) + (194.9 N)(0.06 m) = 0 M = 39.0 N ⋅ m

(b)

ΣFx = 0: (T1 + T2 ) cos30° − 900 N (T1 + 4.3322 T1 ) cos 30° = 900 T1 = 194.90 N T2 = 4.3322(194.90 N) = 844.3 N Tmax = 844 N


PROBLEM 8.110 In the pivoted motor mount shown the weight W of the 175-lb motor is used to maintain tension in the drive belt. Knowing that the coefficient of static friction between the flat belt and drums A and B is 0.40, and neglecting the weight of platform CD, determine the largest couple that can be transmitted to drum B when the drive drum A is rotating clockwise.

SOLUTION FBD motor and mount:

Impending belt slip: cw rotation T2 = T1e µs β = T1e0.40π = 3.5136T1

ΣM D = 0: (12 in.)(175 lb) − (7 in.)T2 − (13 in.)T1 = 0 2100 lb = [(7 in.)(3.5136) + 13 in.]T1 T1 = 55.858 lb, T2 = 3.5136 T1 = 196.263 lb

FBD drum at B: ΣM B = 0: M B − (3 in.)(196.263 lb − 55.858 lb) = 0

r = 3 in. M B = 421 lb ⋅ in.

(Compare to 857 lb ⋅ in. using V-belt, Problem 8.130)


PROBLEM 8.111 Solve Problem 8.110 assuming that the drive drum A is rotating counterclockwise. PROBLEM 8.110 In the pivoted motor mount shown the weight W of the 175-lb motor is used to maintain tension in the drive belt. Knowing that the coefficient of static friction between the flat belt and drums A and B is 0.40, and neglecting the weight of platform CD, determine the largest couple that can be transmitted to drum B when the drive drum A is rotating clockwise.

SOLUTION FBD motor and mount:

Impending belt slip: ccw rotation T1 = T2 e µs β = T2 e0.40π = 3.5136T2

ΣM D = 0: (12 in.)(175 lb) − (13 in.)T1 − (7 in.)T2 = 0 2100 lb = [(13 in.)(3.5136) + 7 in.]T2 = 0 T2 = 39.866 lb, T1 = 3.5136 T2 = 140.072 lb

FBD drum at B:

ΣM B = 0: (3 in.)(140.072 lb − 39.866 lb) − M B = 0 M B = 301 lb ⋅ in.


PROBLEM 8.112 A band brake is used to control the speed of a flywheel as shown. The coefficients of friction are µs = 0.30 and µk = 0.25. Determine the magnitude of the couple being applied to the flywheel, knowing that P = 45 N and that the flywheel is rotating counterclockwise at a constant speed.

SOLUTION Free body: Cylinder

Since slipping of band relative to cylinder is clockwise, T1 and T2 are located as shown. From free body: Lever ABC ΣM C = 0: (45 N)(0.48 m) − T2 (0.12 m) = 0 T2 = 180 N

Free body: Lever ABC

From free body: Cylinder Using Eq. (8.14) with µk = 0.25 and β = 270° =

3π rad: 2

T2 = e µs β = e(0.25)(3π / 2) = e3π /8 T1 T1 =

T2 e

3π /8

=

180 N = 55.415 N 3.2482

ΣM D = 0: (55.415 N)(0.36 m) − (180 N)(0.36 m) + M = 0

M = 44.9 N ⋅ m


PROBLEM 8.113 The speed of the brake drum shown is controlled by a belt attached to the control bar AD. A force P of magnitude 25 lb is applied to the control bar at A. Determine the magnitude of the couple being applied to the drum, knowing that the coefficient of kinetic friction between the belt and the drum is 0.25, that a = 4 in., and that the drum is rotating at a constant speed (a) counterclockwise, (b) clockwise.

SOLUTION (a)

Counterclockwise rotation Free body: Drum r = 8 in. β = 180° = π radians T2 = e µk β = e0.25π = 2.1933 T1 T2 = 2.1933T1

Free body: Control bar ΣM C = 0: T1 (12 in.) − T2 (4 in.) − (25 lb)(28 in.) = 0 T1 (12) − 2.1933T1 (4) − 700 = 0 T1 = 216.93 lb T2 = 2.1933(216.93 lb) = 475.80 lb

Return to free body of drum ΣM E = 0: M + T1 (8 in.) − T2 (8 in.) = 0 M + (216.96 lb)(8 in.) − (475.80 lb)(8 in.) = 0 M = 2070.9 lb ⋅ in.

(b)

M = 2070 lb ⋅ in.

Clockwise rotation r = 8 in. β = π rad T2 = e µk β = e0.25π = 2.1933 T1 T2 = 2.1933T1


PROBLEM 8.113 (Continued) Free body: Control rod ΣM C = 0: T2 (12 in.) − T1 (4 in.) − (25 lb)(28 in.) = 0 2.1933T1 (12) − T1 (4) − 700 = 0 T1 = 31.363 lb T2 = 2.1933(31.363 lb) T2 = 68.788 lb

Return to free body of drum ΣM E = 0: M + T1 (8 in.) − T2 (8 in.) = 0 M + (31.363 lb)(8 in.) − (68.788 lb)(8 in.) = 0 M = 299.4 lb ⋅ in.

M = 299 lb ⋅ in.


PROBLEM 8.114 Knowing that a = 4 in., determine the maximum value of the coefficient of static friction for which the brake is not self-locking when the drum rotates counterclockwise.

SOLUTION r = 8 in., β = 180° = π radians T2 = e µ s β = e µsπ T1 T2 = e µsπ T1

Free body: Control rod

ΣM C = 0: P(28 in.) − T1 (12 in.) + T2 (4 in.) = 0 28 P − 12T1 + e µπ T1 (4) = 0 P=0

For self-locking brake: 12 T1 = 4 T1e µsπ e µs π = 3 µ sπ = ln 3 = 1.0986

µs =

1.0986

π

= 0.3497

µs = 0.350


PROBLEM 8.115 Knowing that the coefficient of static friction is 0.30 and that the brake drum is rotating counterclockwise, determine the minimum value of a for which the brake is not self-locking.

SOLUTION r = 8 in., β = π radians T2 = e µs β = e0.30π = 2.5663 T1 T2 = 2.5663T1

Free body: Control rod

b = 16 in. − a ΣM C = 0: P(16 in. + b) − T1b + T2 a = 0

For brake to be self-locking,

P=0 T2 a = T1b ; 2.5663T1a = T1 (16 − a)

2.5663a = 16 − a 3.5663a = 16

a = 4.49 in.


PROBLEM 8.116 Bucket A and block C are connected by a cable that passes over drum B. Knowing that drum B rotates slowly counterclockwise and that the coefficients of friction at all surfaces are µs = 0.35 and µk = 0.25, determine the smallest combined mass m of the bucket and its contents for which block C will (a) remain at rest, (b) start moving up the incline, (c) continue moving up the incline at a constant speed.

SOLUTION Free body: Drum 2

µ π T2 =e 3 mg

T2 = mge 2 µπ /3

(a)

(1)

Smallest m for block C to remain at rest Cable slips on drum. Eq. (1) with µk = 0.25; T2 = mge2(0.25)π /3 = 1.6881mg Block C: At rest, motion impending ΣF = 0: N − mC g cos 30° N = mC g cos 30° F = µ s N = 0.35 mC g cos 30° mC = 100 kg ΣF = 0: T2 + F − mC g sin 30° = C 1.6881 mg + 0.35mC g cos 30° − mC g sin 30° = 0 1.6881m = 0.19689mC m = 0.11663mC = 0.11663(100 kg);

(b)

m = 11.66 kg

Smallest m to start block moving up No slipping at both drum and block:

µs = 0.35

Eq. (1):

T2 = mge 2(0.35)π /3 = 2.0814 mg



Block C: Motion impending mC = 100 kg ΣF = 0: N − mg cos 30° N = mC g cos30° F = µs N = 0.35mC g cos 30° ΣF = 0: T2 − F − mC g sin 30° = 0 2.0814mg − 0.35mC g cos 30° − mC g sin 30° = 0 2.0814m = 0.80311mC m = 0.38585mC = 0.38585(100 kg) m = 38.6 kg

(c)

Smallest m to keep block moving up drum: No slipping: µs = 0.35 Eq. (1) with µs = 0.35 T2 = mg 2 µsπ /3 = mge2(0.35)π /3 T2 = 2.0814mg

Block C: Moving up plane, thus µk = 0.25 Motion up ΣF = 0: N − mC g cos 30° = 0 N = mC g cos 30° F = µk N = 0.25 mC g cos 30° ΣF = 0: T2 − F − mC g sin 30° = 0 2.0814mg − 0.25mC g cos 30° − mC g sin 30° = 0 2.0814m = 0.71651mC m = 0.34424mC = 0.34424 (100 kg) m = 34.4 kg


PROBLEM 8.117 Solve Problem 8.116 assuming that drum B is frozen and cannot rotate. PROBLEM 8.116 Bucket A and block C are connected by a cable that passes over drum B. Knowing that drum B rotates slowly counterclockwise and that the coefficients of friction at all surfaces are µs = 0.35 and µk = 0.25, determine the smallest combined mass m of the bucket and its contents for which block C will (a) remain at rest, (b) start moving up the incline, (c) continue moving up the incline at a constant speed.

SOLUTION (a)

Block C remains at rest: Motion impends T2 = e µk β = e0.35(2π /3) mg T2 = 2.0814mg

Drum:

Block C:

Motion impends ΣF = 0: N − mC g cos 30° = 0 N = mC g cos30° F = µs N = 0.35mC g cos 30° ΣF = 0: T2 + F − mC g sin 30° = 0 2.0814mg + 0.35mC g cos 30° − mC g sin 30° = 0 2.0814mg = 0.19689mC m = 0.09459mC = 0.09459(100 kg)

(b)

Block C: Starts moving up

m = 9.46 kg

µs = 0.35

Drum: Impending motion of cable T2 = e µs β T1 mg = e0.35(2/3π ) T1 mg 2.0814 = 0.48045mg

T1 =



Block C: Motion impends ΣF = 0: N − mC g cos 30° N = mC g cos30° F = µs N = 0.35mC g cos 30° ΣF = 0: T1 − F − mC g sin 30° = 0 0.48045mg − 0.35mC g cos 30° − 0.5mC g = 0 0.48045m = 0.80311mC m = 1.67158mC = 1.67158(100 kg)

(c)

m = 167.2 kg

Smallest m to keep block moving Drum: Motion of cable

µk = 0.25 T2 = e µk β = e0.25(2/3π ) T1 mg = 1.6881 T1 T1 =

mg = 0.59238mg 1.6881

Block C: Block moves ΣF = 0: N − mC g cos 30° = 0 N = mC g cos30° F = µk N = 0.25mC g cos 30° ΣF = 0: T1 − F − mC g sin 30° = 0 0.59238mg − 0.25mC g cos 30° − 0.5mC g = 0 0.59238m = 0.71651mC m = 1.20954mC = 1.20954(100 kg)

m = 121.0 kg


!

PROBLEM 8.118 A cable is placed around three parallel pipes. Knowing that the coefficients of friction are µs = 0.25 and µk = 0.20, determine (a) the smallest weight W for which equilibrium is maintained, (b) the largest weight W that can be raised if pipe B is slowly rotated counterclockwise while pipes A and C remain fixed.

SOLUTION (a)

Smallest W for equilibrium

B = π , µ = µs

TAC = e0.25π 50 lb

TBC = e0.25π TAC

W = e0.25π TBC

TAC TBC W ⋅ ⋅ = eπ /4 ⋅ eπ / 4 ⋅ eπ / 4 = e3π / 4 = 10.551 50 lb TAC TBC W = 10.551; 50 lb

(b)

W = 4.739 lb

W = 4.74 lb

Largest W which can be raised by pipe B rotated

β = π , µ = µk

β = π , µ = µk

β = π , µ = µs

50 lb = e0.2π TAC

TAC = e0.2π TBC

W = e0.25π TBC

50 lb TAC TBC ⋅ ⋅ = eπ /5 ⋅ eπ /5 ⋅ e−π /4 = eπ (1/5+1/5−1/4) TAC TBC W = e3π / 20 = 1.602 50 lb = 1.602; W

W=

50 lb = 31.21 lb 1.602

W = 31.2 lb


!

PROBLEM 8.119 A cable is placed around three parallel pipes. Two of the pipes are fixed and do not rotate; the third pipe is slowly rotated. Knowing that the coefficients of friction are µs = 0.25 and µk = 0.20, determine the largest weight W that can be raised (a) if only pipe A is rotated counterclockwise, (b) if only pipe C is rotated clockwise.

SOLUTION (a)

Pipe a rotates

β = π , µ = µs

β = π , µ = µk

TAC = e0.25π 50 lb

TAC = e0.2π TBC

β = π , µ = µk

TBC = e0.2π W

TAC TBC W ⋅ ⋅ = eπ /4 ⋅ e−π /5 ⋅ e−π /5 50 lb TAC TBC = eπ (1/ 4−1/5−1/5) = e−3π / 20 = 0.62423 W = 0.62423; W = 31.21 lb 50 lb

(b)

Pipe C rotates

β = π , µ = µk

50 lb = e0.2π TAC

β = π , µ = µs

TBC = e0.25π TAC

W = 31.2 lb

β = π , µ = µk

TBC = e0.2π W

50 lb TAC TBC ⋅ ⋅ = eπ /5 ⋅ e −π / 4 ⋅ eπ /5 = eπ (1/5−1/4 +1/5) = e3π / 20 TAC TBC W 50 lb = e3π /20 = 1.602 W 50 lb W= = 31.21 lb 1.602

W = 31.2 lb


!

PROBLEM 8.120 A cable is placed around three parallel pipes. Knowing that the coefficients of friction are µs = 0.25 and µk = 0.20, determine (a) the smallest weight W for which equilibrium is maintained, (b) the largest weight W that can be raised if pipe B is slowly rotated counterclockwise while pipes A and C remain fixed.

SOLUTION (a)

µ = µ s = 0.25 at all pipes.

TAC = e0.25π TBC

50 lb = e0.25π /2 TAB

TBC = e0.25π / 2 W

50 lb TAB TBC ⋅ ⋅ = eπ /8 ⋅ eπ /4 ⋅ eπ /8 = eπ /8+π / 4+π /8 = eπ / 2 = 4.8105 TAB TBC W 50 lb = 4.8015; W = 10.394 lb W

(b)

Pipe B rotated

β=

π 2

; µ = µk

50 lb = e0.2π /2 TAB

β = π ; µ = µs

TBC = e0.25π TAB

W = 10.39 lb

β=

π 2

; µ = µk

TBC = e0.2π / 2 W

50 lb TAB TBC ⋅ ⋅ = eπ /10 ⋅ e −π /4 ⋅ eπ /10 TAB TBC W = eπ /10 −π /4 +π /10 = e −π / 20 = 0.85464 50 lb = 0.85464 W 50 lb W= = 58.504 lb 0.85464

W = 58.5 lb


!

PROBLEM 8.121 A cable is placed around three parallel pipes. Two of the pipes are fixed and do not rotate; the third pipe is slowly rotated. Knowing that the coefficients of friction are µs = 0.25 and µk = 0.20, determine the largest weight W that can be raised (a) if only pipe A is rotated counterclockwise, (b) if only pipe C is rotated clockwise.

SOLUTION (a)

Pipe a rotates

β=

π 2

; µ = µs

β = π , µ = µk

TAB = e0.25π /2 50 lb

β=

π 2

; µ = µk

TBC = e0.2π /2 W

TAB = e0.2π TBC

TAB TBC W ⋅ ⋅ = eπ /8 ⋅ e−π /5 ⋅ e−π /10 50 lb TAB TBC = eπ (1/8−1/5−1/10) = e −7π / 40 = 0.57708 W = 0.57708; W = 28.854 lb 50 lb

(b)

Pipe C rotates

β=

π

; µ = µk

β = π ; µ = µk

50 lb = e0.2π / 2 TAB

TAB = e0.2π TBC

2

β=

π 2

W = 28.9 lb

, µ = µs

W = e0.25π /2 TBC

50 lb TAB TBC ⋅ ⋅ = eπ /10 ⋅ eπ /5 ⋅ e −π /8 = e7π /40 = 0.57708 TAB TBC W 50 lb = 0.57708 W W = 28.854 lb

W = 28.9 lb


!

PROBLEM 8.122 A recording tape passes over the 20-mm-radius drive drum B and under the idler drum C. Knowing that the coefficients of friction between the tape and the drums are µs = 0.40 and µk = 0.30 and that drum C is free to rotate, determine the smallest allowable value of P if slipping of the tape on drum B is not to occur.

SOLUTION FBD drive drum:

ΣM B = 0: r (TA − T ) − M = 0 TA − T =

Impending slipping:

M 300 N ⋅ mm = = 15.0000 N r 20 mm

TA = Te µs β = Te0.4π

So

T (e0.4π − 1) = 15.0000 N

or

T = 5.9676 N

If C is free to rotate, P = T

P = 5.97 N


!

PROBLEM 8.123 Solve Problem 8.122 assuming that the idler drum C is frozen and cannot rotate. PROBLEM 8.122 A recording tape passes over the 20-mm-radius drive drum B and under the idler drum C. Knowing that the coefficients of friction between the tape and the drums are µs = 0.40 and µk = 0.30 and that drum C is free to rotate, determine the smallest allowable value of P if slipping of the tape on drum B is not to occur.

SOLUTION FBD drive drum:

ΣM B = 0: r (TA − T ) − M = 0 TA − T =

Impending slipping:

M = 300 N ⋅ mm = 15.0000 N r

TA = Te µs β = Te0.4π

So

(e0.4π − 1)T = 15.000 N

or

T = 5.9676 N

If C is fixed, the tape must slip So

P = Te µk βC = (5.9676 N)e0.3π / 2 = 9.5600 N

P = 9.56 N


!

PROBLEM 8.124 The 10-lb bar AE is suspended by a cable that passes over a 5-in.-radius drum. Vertical motion of end E of the bar is prevented by the two stops shown. Knowing that µs = 0.30 between the cable and the drum, determine (a) the largest counterclockwise couple M 0 that can be applied to the drum if slipping is not to occur, (b) the corresponding force exerted on end E of the bar.

SOLUTION Drum: Slipping impends

µs = 0.30 T2 T = e µβ : D = e0.30π = 2.5663 T1 TB TD = 2.5663TB

(a)

Free-body: Drum and bar

ΣM C = 0: M 0 − E (8 in.) = 0 M 0 = (3.78649 lb)(8 in.) = 30.27 lb ⋅ in.

(b)

Bar AE:

M 0 = 30.3 lb ⋅ in.

ΣFy = 0: TB + TD − E − 10 lb = 0 TB + 2.5663TB − E − 10 lb = 0 3.5663TB − E − 10 lb = 0 E = 3.5663TB − 10 lb

(1)

ΣM D = 0: E (3 in.) − (10 lb)(5 in.) + TB (10 in.) = 0 (3.5663TB − 10 lb)(3 in.) − 50 lb ⋅ in. + TB (10 in.) = 0 20.699TB = 80 TB = 3.8649 lb

Eq. (1):

E = 3.5663(3.8649 lb) − 10 lb E = +3.78347 lb

E = 3.78 lb


!

PROBLEM 8.125 Solve Problem 8.124 assuming that a clockwise couple M 0 is applied to the drum. PROBLEM 8.124 The 10-lb bar AE is suspended by a cable that passes over a 5-in.-radius drum. Vertical motion of end E of the bar is prevented by the two stops shown. Knowing that µs = 0.30 between the cable and the drum, determine (a) the largest counterclockwise couple M 0 that can be applied to the drum if slipping is not to occur, (b) the corresponding force exerted on end E of the bar.

SOLUTION Drum: Slipping impends

µs = 0.30 T2 = e µβ T1 TB = e0.30π = 2.5663 TD TB = 2.5663TD

(a)

Free body: Drum and bar

ΣM C = 0: M 0 − E (8 in.) = 0 M 0 = (2.1538 lb)(8 in.) M 0 = 17.23 lb ⋅ in.

(b)

Bar AE: ΣFy = 0: TB + TD + E − 10 lb = 0 = 2.5663TD + TD + E − 10 lb E = −3.5663TD + 10 lb

(1)

ΣM B = 0: TD (10 in.) − (10 lb)(5 in.) + E (13 in.) = 0 TD (10 in.) − 50 lb ⋅ in. + ( − 3.5663TD + 10 lb)(13 in.) = 0 −36.362TD + 80 lb ⋅ in. = 0; TD = 2.200 lb E = −3.5633(2.200 lb) + 10 lb

Eq. (1):

E = + 2.1538 lb

E = 2.15 lb


!

PROBLEM 8.126 The strap wrench shown is used to grip the pipe firmly without marring the external surface of the pipe. Knowing that the coefficient of static friction is the same for all surfaces of contact, determine the smallest value of µs for which the wrench will be self-locking when a = 200 mm, r = 30 mm, and θ = 65°.

SOLUTION For wrench to be self-locking ( P = 0), the value of µs must prevent slipping of strap which is in contact with the pipe from Point A to Point B and must be large enough so that at Point A the strap tension can increase from zero to the minimum tension required to develop “belt friction” between strap and pipe. Free body: Wrench handle

Geometry

In ∆CDH:

On wrench handle

a tan θ a CD = sin θ DE = BH = CH − BC a −r DE = tan θ a −r AD = CD − CA = sin θ

CH =

ΣM D = 0: TB ( DE ) − F ( AD ) = 0

a −r TB AD sin θ = = a F DE −r tan θ

(1)



Free body: Strap at Point A

ΣF = 0: T1 − 2 F = 0

T1 = 2 F

(2)

Pipe and strap

β = (2π − θ ) radians µs β = ln

Eq. (8.13):

µs =

1

β

T2 T1 ln

TB 2F

(3)

Return to free body of wrench handle ΣFx = 0: N sin θ + F cos θ − TB = 0

T N sin θ = B − cos θ F F

Since F = µ s N , we have or

1

µs

sin θ =

µs =

TB − cos θ F sin θ

(4)

TB − cos θ F

(Note: For a given set of data, we seek the larger of the values of µs from Eqs. (3) and (4).) For

Eq. (1):

a = 200 mm, r = 30 mm, θ = 65° 200 mm − 30 mm TB = sin 65° 200 mm F − 30 mm tan 65° 190.676 mm = = 3.0141 63.262 mm

β = 2π − θ = 2π − 65°

π 180°

= 5.1487 radians



Eq. (3):

1 3.0141 ln 5.1487 rad 2 0.41015 = 5.1487

µs =

= 0.0797

Eq. (4):

#

sin 65° 3.0141 − cos 65° 0.90631 = 2.1595

µs =

= 0.3497

#

µs = 0.350

We choose the larger value:


!

PROBLEM 8.127 Solve Problem 8.126 assuming that θ = 75°. PROBLEM 8.126 The strap wrench shown is used to grip the pipe firmly without marring the external surface of the pipe. Knowing that the coefficient of static friction is the same for all surfaces of contact, determine the smallest value of µs for which the wrench will be self-locking when a = 200 mm, r = 30 mm, and θ = 65°.

SOLUTION For wrench to be self-locking ( P = 0), the value of µs must prevent slipping of strap which is in contact with the pipe from Point A to Point B and must be large enough so that at Point A the strap tension can increase from zero to the minimum tension required to develop “belt friction” between strap and pipe. Free body: Wrench handle

Geometry

In ∆CDH:

On wrench handle

a tan θ a CD = sin θ DE = BH = CH − BC a DE = −r tan θ a AD = CD − CA = −r sin θ

CH =

ΣM D = 0: TB ( DE ) − F ( AD ) = 0

a −r TB AD sin θ = = a F DE −r tan θ

(1)



Free body: Strap at Point A

ΣF = 0: T1 − 2 F = 0

T1 = 2 F

(2)

Pipe and strap

β = (2π − θ ) radians µs β = ln

Eq. (8.13):

µs =

1

β

T2 T1 ln

TB 2F

(3)

Return to free body of wrench handle ΣFx = 0: N sin θ + F cos θ − TB = 0

T N sin θ = B − cos θ F F

Since F = µ s N , we have or

1

µs

sin θ =

µs =

TB − cos θ F sin θ TB − cos θ F

(4)

(Note: For a given set of data, we seek the larger of the values of µs from Eqs. (3) and (4).) For

Eq. (1):

a = 200 mm, r = 30 mm, θ = 75° 200 mm − 30 mm TB = sin 75° 200 mm F − 30 mm tan 75° 177.055 mm = = 7.5056 23.590 mm

β = 2π − θ = 2π − 75°

π 180°

= 4.9742



Eq. (3):

1 7.5056 ln 4.9742 rad 2 1.3225 = 4.9742

µs =

= 0.2659

Eq. (4):

#

sin 75° 7.5056 − cos 75° 0.96953 = 7.2468

µs =

= 0.1333

#

µs = 0.266

We choose the larger value:


!

PROBLEM 8.128 Prove that Eqs. (8.13) and (8.14) are valid for any shape of surface provided that the coefficient of friction is the same at all points of contact.

SOLUTION ΣFn = 0: ∆N − [T + (T + ∆T )]sin ∆N = (2T + ∆T ) sin

or

ΣFt = 0: [(T + ∆T ) − T ]cos ∆F = ∆T cos

or

∆T cos

So

So

∆θ − ∆F = 0 2

∆θ 2

∆θ

∆θ ∆θ sin ∆θ = µ s 2T sin + µs ∆T 2 2 2

0: dT = µsTdθ T2 T1

and

∆θ 2

∆F = µ s ∆N

Impending slipping:

In limit as

∆θ =0 2

ln

dT = T

β 0

or

dT = µ s dθ T

µ s dθ

T2 = µs β T1

or T2 = T1e µs β

(Note: Nothing above depends on the shape of the surface, except it is assumed to be a smooth curve.)!


PROBLEM 8.129 Complete the derivation of Eq. (8.15), which relates the tension in both parts of a V belt.

SOLUTION Small belt section: Side view:

End view:

ΣFy = 0: 2

∆N α ∆θ sin − [T + (T + ∆T )]sin =0 2 2 2

ΣFx = 0: [(T + ∆T ) − T ]cos ∆F = µs ∆N . ∆T cos

Impending slipping:

In limit as

∆θ

0: dT =

µ sTdθ α sin

so

or

T2 T1

2

µs dT = α T sin 2

ln

or

β 0

∆θ − ∆F = 0 2

∆θ 2T + ∆T ∆θ sin = µs α 2 2 sin 2

µs dT dθ = α T sin 2

dθ

µβ T2 = s T1 sin α 2 T2 = T1e

or

µ s β / sin α2


!

PROBLEM 8.130 Solve Problem 8.107 assuming that the flat belt and drums are replaced by a V belt and V pulleys with α = 36°. (The angle α is as shown in Figure 8.15a.) PROBLEM 8.107 A flat belt is used to transmit a couple from drum B to drum A. Knowing that the coefficient of static friction is 0.40 and that the allowable belt tension is 450 N, determine the largest couple that can be exerted on drum A.

SOLUTION Since β is smaller for pulley B. The belt will slip first at B. ! π rad " 5 $ = π rad % 180° & 6

β = 150° #

T2 µ β / sin α2 =e s T1

450 N (0.4) 5 π / sin18° =e 6 = e3.389 T1 450 N = 29.63; T1 = 15.187 N T1

Torque on pulley A:

ΣM B = 0: M − (Tmax − T1 )(0.12 m) = 0 M − (450 N − 15.187 N)(0.12 m) = 0 M = 52.18 N ⋅ m

M = 52.2 N ⋅ m


!

PROBLEM 8.131 Solve Problem 8.108 assuming that the flat belt and pulleys are replaced by a V belt and V pulleys with α = 36°. (The angle α is as shown in Figure 8.15a.) PROBLEM 8.108 A flat belt is used to transmit a couple from pulley A to pulley B. The radius of each pulley is 60 mm, and a force of magnitude P = 900 N is applied as shown to the axle of pulley A. Knowing that the coefficient of static friction is 0.35, determine (a) the largest couple that can be transmitted, (b) the corresponding maximum value of the tension in the belt.

SOLUTION Pulley A:

β = π rad T2 µ β / sin α2 =e s T1 T2 = e0.35π / sin18° T1 T2 = e3.558 = 35.1 T1 T2 = 35.1T1 ΣFx = 0: T1 + T2 − 900 N = 0 T1 + 35.1T1 − 900 N = 0 T1 = 24.93 N T2 = 35.1(24.93 N) = 875.03 N ΣM A = 0: M − T2 (0.06 m) + T1 (0.06 m) = 0 M − (875.03 N)(0.06 m) + (24.93 N)(0.06 m) = 0 M = 51.0 N ⋅ m Tmax = T2

Tmax = 875 N


!

PROBLEM 8.132 Knowing that the coefficient of friction between the 25-kg block and the incline is µs = 0.25, determine (a) the smallest value of P required to start the block moving up the incline, (b) the corresponding value of β.

SOLUTION FBD block (Impending motion up) W = mg = (25 kg)(9.81 m/s 2 ) = 245.25 N

φs = tan −1 µs = tan −1 (0.25) = 14.04°

(a)

(Note: For minimum P, P

R so β = φs .)

Then P = W sin (30° + φs ) = (245.25 N)sin 44.04° Pmin = 170.5 N

(b)

We have β = φs

β = 14.04°


PROBLEM 8.133 The 20-lb block A and the 30-lb block B are supported by an incline that is held in the position shown. Knowing that the coefficient of static friction is 0.15 between all surfaces of contact, determine the value of θ for which motion is impending.

SOLUTION Since motion impends, F = µs N at all surfaces. Free body: Block A

Impending motion: ΣFy = 0: N1 = 20cos θ ΣFx = 0: T − 20sin θ − µ s N1 = 0 T = 20sin θ + 0.15(20 cos θ ) T = 20sin θ + 3cos θ

(1)

Free body: Block B

Impending motion: ΣFy = 0: N 2 − 30 cos θ − N1 = 0 N 2 = 30 cos θ + 20cos θ = 50 cos θ F2 = µ s N 2 = 0.15(50 cos θ ) = 6cos θ ΣFx = 0: T − 30sin θ + µ s N1 + µs N 2 = 0 T = 30sin θ − 0.15(20 cos θ ) − 0.15(50cos θ ) T = 30sin θ − 3cos θ − 7.5cos θ

Eq. (1) subtracted by Eq. (2):

(2)

20sin θ + 3cos θ − 30sin θ + 3cos θ + 7.5cos θ = 0 13.5cos θ = 10sin θ , tan θ =

13.5° 10

θ = 53.5°


!

PROBLEM 8.134 A worker slowly moves a 50-kg crate to the left along a loading dock by applying a force P at corner B as shown. Knowing that the crate starts to tip about the edge E of the loading dock when a = 200 mm, determine (a) the coefficient of kinetic friction between the crate and the loading dock, (b) the corresponding magnitude P of the force.

SOLUTION Free body: Crate Three-force body. Reaction E must pass through K where P and W intersect. Geometry: HK = (0.6 m) tan15° = 0.16077 m

(a)

JK = 0.9 m + HK = 1.06077 m 0.4 m = 0.37708 tan φs = 1.06077 m

µs = tan φs = 0.377

φs = 20.66° Force triangle: W = (50 kg)(9.81 m/s) = 490.5 N

(b) Law of sines:

P 490.5 N = sin 20.66° sin 84.34° P = 173.91 N

P = 173.9 N


!

PROBLEM 8.135 A slender rod of length L is lodged between peg C and the vertical wall and supports a load P at end A. Knowing that the coefficient of static friction between the peg and the rod is 0.15 and neglecting friction at the roller, determine the range of values of the ratio L/a for which equilibrium is maintained.

SOLUTION FBD rod: Free-body diagram: For motion of B impending upward: ! a ΣM B = 0: PL sin θ − N C # % sin θ

" $=0 &

NC =

PL 2 sin θ a

(1)

ΣFy = 0: NC sin θ − µ s N C cos θ − P = 0 NC (sin θ − µ cos θ ) = P

Substitute for NC from Eq. (1), and solve for a/L. a = sin 2 θ (sin θ − µ s cos θ ) L

For θ = 30° and µ s = 0.15:

(2)

a = sin 2 30°(sin 30° − 0.15cos 30°) L a L = 0.092524 = 10.808 L a

For motion of B impending downward, reverse sense of friction force FC . To do this we make µs = −0.15 in. Eq. (2). Eq. (2):

a = sin 2 30°(sin 30° − (−0.15) cos 30°) L a L = 0.15748 = 6.350 L a

Range of values of L/a for equilibrium:

6.35 #

L # 10.81 a


!

PROBLEM 8.136 A safety device used by workers climbing ladders fixed to high structures consists of a rail attached to the ladder and a sleeve that can slide on the flange of the rail. A chain connects the worker’s belt to the end of an eccentric cam that can be rotated about an axle attached to the sleeve at C. Determine the smallest allowable common value of the coefficient of static friction between the flange of the rail, the pins at A and B, and the eccentric cam if the sleeve is not to slide down when the chain is pulled vertically downward.

SOLUTION Free body: Cam

ΣM C = 0: N D (0.8 in.) − µ s N D (3in.) − P (6 in.) = 0 ND =

6P 0.8 − 3µ s

(1)

Free body: Sleeve and cam

ΣFx = 0: N D − N A − N B = 0 N A + NB = ND

(2)

ΣFy = 0: FA + FB + FD − P = 0

µs ( N A + N B + N D ) = P

or

µs (2 N D ) = P

Substitute from Eq. (2) into Eq. (3):

ND =

(3)

P 2µ s

(4)

Equate expressions for N D from Eq. (1) and Eq. (4): P 6P ; 0.8 − 3µ s = 12 µ s = 2µ s 0.8 − 3µ s

µs =

0.8 15

µs = 0.0533

(Note: To verify that contact at pins A and B takes places as assumed, we shall check that N A . 0 and N B = 0.)



From Eq. (4):

ND =

P P = = 9.375P 2µ s 2(0.0533)

From free body of cam and sleeve: ΣM B = 0: N A (8 in.) − N D (4 in.) − P(9 in.) = 0 8 N A = (9.375 P)(4) + 9 P N A = 5.8125P . 0 OK

From Eq. (2):

N A + NB = ND 5.8125P + N B = 9.375P N B = 3.5625P . 0 OK


PROBLEM 8.137 To be of practical use, the safety sleeve described in the preceding problem must be free to slide along the rail when pulled upward. Determine the largest allowable value of the coefficient of static friction between the flange of the rail and the pins at A and B if the sleeve is to be free to slide when pulled as shown in the figure, assuming (a) θ = 60°, (b) θ = 50°, (c) θ = 40°.

SOLUTION Note the cam is a two-force member. Free body: Sleeve We assume contact between rail and pins as shown. ΣM C = 0: FA (3 in.) + FB (3 in.) − N A (4 in.) − N B (4 in.) = 0

But

FA = µ s N A FB = µ s N B

We find

3µ s ( N A + N B ) − 4( N A + N B ) = 0

µs =

4 = 1.33333 3

We now verify that our assumption was correct. ΣFx = 0: N A − N B + P cos θ = 0 N B − N A = P cos θ

(1)

ΣFy = 0: − FA − FB + P sin θ = 0

µ s N A + µ s N B = P sin θ N A + NB =

Add Eqs. (1) and (2):

P sin θ µs

(2)

! sin θ " 2 N B = P # cos θ + $ . 0 OK µs & %



Subtract Eq. (1) from Eq. (2):

NA . 0 only if

! sin θ " 2N A = P # − cos θ $ % µs &

sin θ − cos θ . 0 µs tan θ . µ s = 1.33333

θ = 53.130° (a)

For case (a): Condition is satisfied, contact takes place as shown. Answer is correct.

µ s = 1.333 But for (b) and (c): θ , 53.130° and our assumption is wrong, N A is directed to left. ΣFx = 0: − NA − N B + P cos θ = 0 N A + N B = P cos θ

(3)

ΣFy = 0: − FA − FA + P sin θ = 0

µ s ( N A + N B ) = P sin θ

(4)

Divide Eq. (4) by Eq. (3):

µ s = tan θ (b)

(c)

(5)

We make θ = 50° in Eq. (5):

µ s = tan 50°

µ s = 1.192

µ s = tan 40°

µ s = 0.839

We make θ = 40° in Eq. (5):


PROBLEM 8.138 Bar AB is attached to collars that can slide on the inclined rods shown. A force P is applied at Point D located at a distance a from end A. Knowing that the coefficient of static friction µs between each collar and the rod upon which it slides is 0.30 and neglecting the weights of the bar and of the collars, determine the smallest value of the ratio a/L for which equilibrium is maintained.

SOLUTION FBD bar and collars: Impending motion:

φs = tan −1 µs = tan −1 0.3 = 16.6992°

Neglect weights: 3-force FBD and " ACB = 90° so

AC =

a cos (45° + φs )

= l sin (45° − φs )

a = sin (45° − 16.6992°) cos (45° + 16.6992°) l a = 0.225 l


PROBLEM 8.139 The machine part ABC is supported by a frictionless hinge at B and a 10° wedge at C. Knowing that the coefficient of static friction at both surfaces of the wedge is 0.20, determine (a) the force P required to move the wedge, (b) the components of the corresponding reaction at B.

SOLUTION

φs = tan −1 0.20 = 11.31° Free body: Part ABC

ΣM B = 0 (1800 N)(0.35 m) − R cos 21.31°(0.6 m) = 0

RC = 1127.1 N

Force triangle:

Free body: Wedge

(a)

Law of sines:

P 1127.1 N = sin(11.31° + 21.31°) sin 78.69° P = 619.6 N

(b)

P = 620 N

Return to part ABC: ΣFx = 0: Bx + 1800 N − RC sin 21.31° = 0

Bx + 1800 N − (1127.1 N) sin 21.31° Bx = −1390.4 N

B x = 1390 N

ΣFy = 0: By + RC cos 21.31° = 0

By + (1127.1 N) cos 21.31° = 0 By = −1050 N

B y = 1050 N


!

PROBLEM 8.140 A wedge A of negligible weight is to be driven between two 100-lb blocks B and C resting on a horizontal surface. Knowing that the coefficient of static friction at all surfaces of contact is 0.35, determine the smallest force P required to start moving the wedge (a) if the blocks are equally free to move, (b) if block C is securely bolted to the horizontal surface.

SOLUTION Wedge angle θ :

(a)

0.75 in. 4 in. θ = 10.62°

θ = tan −1

Free body: Block B

φs = tan −1 0.35 = 19.29° R1 100 lb = sin19.29° sin 40.80° R1 = 50.56 lb

Free body: Wedge By symmetry:

R3 = R1

P = 2 R1 sin (θ + φs ) = 2(50.56)sin 29.91° P = 50.42 lb

(b)

P = 50.4 lb P = 50.4 lb

Free bodies unchanged: Same result.


!

PROBLEM 8.141 The position of the automobile jack shown is controlled by a screw ABC that is single-threaded at each end (right-handed thread at A, left-handed thread at C). Each thread has a pitch of 0.1 in. and a mean diameter of 0.375 in. If the coefficient of static friction is 0.15, determine the magnitude of the couple M that must be applied to raise the automobile.

SOLUTION Free body: Parts A, D, C, E Two-force members Joint D: FAD = FCD

Symmetry:

ΣFy = 0: 2 FCD sin 25° − 800 lb = 0

FCD = 946.5 lb

Joint C: FCE = FCD

Symmetry:

ΣFx = 0: 2 FCD cos 25° − FAC = 0

FAC = 2(946.5 lb) cos 25° FAC = 1715.6 lb

Block-and-incline analysis of one screw:

tan θ =

0.1 m

π (0.375 in.)

θ = 4.852° φs = tan −1 0.15 = 8.531°

Q = (1715.6 lb) tan13.383° Q = 408.2 lb

But, we have two screws:

! 0.375 in. " Torque = 2Qr = 2(408.2 lb) # $ 2 % &

Torque = 153.1 lb ⋅ in.


PROBLEM 8.142 A lever of negligible weight is loosely fitted onto a 30-mmradius fixed shaft as shown. Knowing that a force P of magnitude 275 N will just start the lever rotating clockwise, determine (a) the coefficient of static friction between the shaft and the lever, (b) the smallest force P for which the lever does not start rotating counterclockwise.

SOLUTION (a)

Impending motion W = (40 kg)(9.81 m/s 2 ) = 392.4 N ΣM D = 0: P(160 − rf ) − W (100 + rf ) = 0 160 P − 100W P +W (160 mm)(275 N) − (100 mm)(392.4 N) rf = 275 N + 392.4 N r f = 7.132 mm

rf =

r f = r sin φs = r µ s

µs = (b)

Impending motion

rf r

=

7.132 mm = 0.2377 30 mm

µs = 0.238

r f = r sin φs = r µ s = (30 mm)(0.2377)

r f = 7.132 mm ΣM D = 0: P(160 + rf ) − W (100 − rf ) = 0

P =W

100 − r f 160 + rf

P = (392.4 N)

100 mm − 7.132 mm 160 mm + 7.132 mm P = 218 N

P = 218.04 N


PROBLEM 8.143 A couple MB is applied to the drive drum B to maintain a constant speed in the polishing belt shown. Knowing that µk = 0.45 between the belt and the 15-kg block being polished and µ s = 0.30 between the belt and the drive drum B, determine (a) the couple MB, (b) the minimum tension in the lower portion of the belt if no slipping is to occur between the belt and the drive drum.

SOLUTION Block:

Portion of belt located under block: ΣFx = 0: T2 − T1 − 66.217 N = 0

(1)

Drum B:

T2 = e µsπ = e0.3π = 2.5663 T1 T2 = 2.5663T1

Eq. (1):

(2)

2.5663T1 − T1 − 66.217 N = 0 1.5663T1 = 66.217 N

T1 = 42.276 N

Eq. (2):

Tmin = 42.3 N

T2 = 2.5663(42.276 N) = 108.493 N ΣM B = 0: M B − (108.493 N)(0.075 m) + (42.276 N)(0.075 m) = 0

M B = 4.966 N ⋅ m

M B = 4.97 N ⋅ m


CHAPTER 9

PROBLEM 9.1 Determine by direct integration the moment of inertia of the shaded area with respect to the y axis.

SOLUTION By observation Now

y=

h x b

dI y = x 2

dA = x 2 [(h − y ) dx]

x! = hx 2 "1 − # dx $ b%

Then

&

I y = dI y =

&

x! hx 2 "1 − # dx 0 $ b% b

'1

b

x4 ( =h x − ! 4b # 0 "3 3

or

Iy =

1 3 bh 12


!


SOLUTION a=

k a

Then

y=

a2 x

Now

dIy = x 2

At x = a,

y = a:

or k = a 2

dA = x 2 ( y dx)

$ a2 % = x 2 && dx '' = a 2 x dx ( x )

Then

,

Iy = dI y =

,

2a a

2a

a 2 x dx = a 2

a2 *1 2+ x ! = [(2a) 2 − (a) 2 ] 2 "2 #a or

Iy =

3 4 a 2


!


SOLUTION y = kx 2

For x = a :

Thus:

b = ka 2

k=

b a2

y=

b 2 x a2

dA = (b − y )dx b $ % dIy = x 2 dA = x 2 (b − y )dx = x 2 & b − 2 x 2 ' dx a ( ) a$ 1 1 b % Iy = dI y = & bx 2 − 2 x 4 ' dx = a 3b − a 3b 0 ( 3 5 a )

,

,

Iy =

2 3 ab 15


!


SOLUTION $ x x2 % y = 4h && − 2 '' (a a ) dA = y dx $ x x2 dI y = x 2 dA = 4hx 2 && − 2 (a a a $ x3 x4 % I y = 4h && − 2 '' dx 0 ( a a )

% '' dx )

,

a

* x4

$ a3 a3 % x5 + − 2 ! = 4h && − '' I y = 4h 5 ) " 4 a 5a # 0 ( 4

1 I y = ha3 5


!

PROBLEM 9.5 Determine by direct integration the moment of inertia of the shaded area with respect to the x axis.

SOLUTION By observation

y=

h x b

or

x=

b y h

Now

dI x = y 2 dA = y 2 ( x dy ) $b % = y 2 & y dy ' (h ) b = y 3 dy h

Then

,

I x = dI x =

,

h 0

b 3 y dy h h

=

b *1 4 + y h " 4 !# 0

or

Ix =

1 3 bh 4


!


SOLUTION At x = a, Then

Now

y = a:

a=

k a

y=

a2 x

1 1 $ a2 dI x = y 3 dx = && 3 3( x =

Then

or k = a 2

3

% '' dx )

1 a6 dx 3 x3

,

I x = dI x =

,

a a

2a

1 a6 1 * 1 1+ dx = a 6 − 3 x3 3 " 2 x 2 !# a

1 * 1 1 + = − a6 − ! 2 6 " (2a) (a) 2 #

or

1 I x = a4 8


!


SOLUTION See figure of solution of Problem 9.3. y=

b x a2

1 1 1 1 b3 6 dI x = b3 dx − y 3 dx = b3 dx − x dx 3 3 3 3 a6

,

I x = dI x =

,

a$ 1 0

1 b3 6 % 3 x ' dx && b − 3 a 6 ') (3

1 1 b3 a 7 $ 1 1 % 3 = b3 a − = & − ' ab 3 3 a 6 7 ( 3 21 ) 1 % 6 $ 7 = & − ' ab3 = ab3 21 ( 21 21 )

Ix =

2 3 ab 7


!


SOLUTION See figure of solution of Problem 9.4. y = h1 + ( h2 − h1 ) dI x =

x a

1 3 y dx 3

,

I x = dI y =

1 3

,

3

a* 0

x+ h1 + ( h2 − h1 ) ! dx a# " 4

a

x+ $ a % 1 * h1 + ( h2 − h1 ) ! & = ' a # ( h2 − h1 ) 12 " 0 =

2 2 a a ( h + h1 ) (h2 + h1 )(h2 − h1 ) h24 − h14 = ⋅ 2 12(h2 − h1 ) 12 h2 − h1

(

)

Ix =

a 2 h1 + h22 (h1 + h2 ) 12

(

)


!


SOLUTION

At x = 0, y = b :

b = c(1 − 0)

or

c=b

At x = a, y = 0:

0 = b(1 − ka1/2 )

or k =

Then

$ x1/2 % y = b &&1 − 1/ 2 '' ( a )

1 a

1/2

3

Now

1 1* $ x1/ 2 dI x = y 3 dx = b &&1 − 1/ 2 3 3" ( a

or

1 $ x1/ 2 x x3/ 2 dI x = b3 &&1 − 3 1/2 + 3 − 3/2 3 ( a a a

Then

,

I x = dI x = 2

,

%+ '' ! dx ) #! % '' dx )

$ x1/2 x x3/2 b3 &&1 − 3 1/ 2 + 3 − 3/2 a a 3 ( a

a1 0

% '' dx )

a

2 * x3/2 3 x 2 2 x5/2 + = b3 x − 2 1/ 2 + − ! 3 " 2 a 5 a3/ 2 # 0 a

or

Ix =

1 3 ab 15


!


SOLUTION

At x = a, y = b :

b = ke a/a

Then

y=

Now

dI x =

Then

or k =

b e

b x/a e = be x/a −1 e

1 3 1 y dx = (be x/a −1 )3 dx 3 3 1 3 3( x/a −1) = be dx 3

,

I x = dI x =

,

a 0

a

1 3 3( x/a −1) b3 * a 3( x/a −1) + be dx = e ! 3 3 "3 #0

1 = ab3 (1 − e−3 ) 9

or

I x = 0.1056ab3


!


SOLUTION At x = 3a,

y = b:

b = k (3a − a)3

or

k=

b 8a 3

Then

y=

b ( x − a )3 8a 3

Now

dI x =

1 3 y dx 3 3

Then

=

1* b + ( x − a)3 ! dx 3 " 8a 3 #

=

b3 ( x − a )9 dx 1536a9

,

I x = dI x = =

,

3a a

3a

b3 b3 * 1 + 9 − = ( x a ) dx ( x − a )10 ! 9 9 1536a 1536a "10 #a

b3 [(3a − a )10 − 0] 15,360a9 or

Ix =

1 3 ab 15


!


SOLUTION At x = 0,

y = b:

b = c(1 − 0)

or c = b

x = a,

y = 0:

0 = c(1 − ka1/ 2 )

or k =

1 a1/2

$ x1/2 y = b &&1 − 1/ 2 ( a

Then

% '' )


!


SOLUTION At x = a,

b = kea/a

y = b:

or

k=

b e

Then

y=

b x/a e = be x/a −1 e

dI y = x 2 dA = x 2 ( y dx)

Now

= x 2 (be x/a −1dx)

,

I y = dI y =

Then

,

a

bx 2 e x/a −1dx

0

Now use integration by parts with dv = e x/a −1dx

u = x2

v = ae x/a −1

du = 2 x dx

Then

,

a 0

a

x 2 e x/a −1dx = *" x 2 ae x/a −1 +# − 0

= a 3 − 2a

,

a 0

,

a 0

(ae x/a −1 )2 x dx

xe x/a −1dx

Using integration by parts with

Then

u=x

dv = e x/a −1dx

du = dx

v = ae x/a −1

* I y = b / a3 − 2a ( xae x/a −1 ) |0a − " 1

{ = b {a

,

a 0

+. ( ae x/a −1 ) dx ! 0 #2

} ) +#}

= b a3 − 2a *" a 2 − ( a 2 e x/a −1 ) |0a +# 3

− 2a *" a 2 − ( a 2 − a 2 e−1

or

I y = 0.264a3b


!


SOLUTION At x = 3a,

y = b:

b = k (3a − a)3

or

k=

b 8a 3

Then

y=

b ( x − a )3 8a 3

Now

dI y = x 2 dA = x 2 ( y dx) = x 2 =

Then

* b + ( x − a)3 dx ! 3 " 8a #

b 2 3 x ( x − 3x 2 a + 3xa 2 − a 3 )dx 8a 3

,

I y = dI y =

,

3a a

b 5 ( x − 3x 4 a + 3x3 a 2 − a3 x 2 )dx 3 8a 3a

b *1 6 3 5 3 2 4 1 3 3 + = 3 x − ax + a x − a x ! 5 4 3 8a " 6 #a =

3 3 1 b -* 1 + (3a)6 − a(3a)5 + a 2 (3a) 4 − a3 (3a )3 ! 3 / 5 4 3 8a 1 " 6 # 3 3 1 *1 +. − ( a )6 − a ( a )5 + a 2 ( a ) 4 − a 3 ( a )3 ! 0 5 4 3 "6 #2

or I y = 3.43a3b


!

PROBLEM 9.15 Determine the moment of inertia and the radius of gyration of the shaded area shown with respect to the x axis.

SOLUTION At x = a,

y1 = y2 = b :

b a b or k = 3 a

y1 : b = ma or m = y2 : b = ka3

Then

y1 =

b x a

or

x1 =

a y b

and

y2 =

b 3 x a3

or

x2 =

a 1/3 y b

Now

dA = ( x2 − x1 ) dy

1/3

a % $ a = & 1/3 y1/3 − y ' dy b ) (b

Then

,

A = dA = 2

,

b 0

$ y1/3 1 % a && 1/3 − y '' dy b ) (b b

= 2a

Now

1 * 3 1 4/3 1 2 + y − y ! = ab 1/3 2b # 0 2 "4 b

*$ a a % + dI x = y 2 dA = y 2 & 1/3 y1/3 − y ' dy ! b ) # "( b



Then

Ix = 2

,

b 0

1 % $ 1 a & 1/3 y 7/3 − y 3 ' dy b b ( ) b

= 2a

and

k x2 =

* 3 1 10/3 1 4 + y y − 1/3 4b !# 0 "10 b

Ix = A

1 ab3 10 1 ab 2

1 = b2 5

or

or

Ix =

1 3 ab 10

kx =

b 5



SOLUTION y1 = k1 x 2

For

y2 = k2 x1/ 2

x = 0 and

y1 = y2 = b

b = k1a 2

b = k2 a1/ 2

k1 =

b a2

y1 =

b 2 x a2

k2 =

b a

1/ 2

Thus, y2 =

b 1/ 2 x a 1/2

dA = ( y2 − y1 )dx A=

,

a*

0

b 1/2 b 2 + x − 2 x ! dx 1/ 2 a a " #

2 ba 3/2 ba 3 − 3 a1/2 3a 2 1 A = ab 3

A=

1 3 1 y2 dx − y13 dx 3 3 3 1 b 1 b3 6 3/ 2 x dx x dx = − 3 a3/ 2 3 a6

dI x =

,

b3 a 3/ 2 b3 a 6 − x dx x dx 3a 3/2 0 3a 6 0 1 % b3 a 5/2 b3 a 7 $ 2 = 3/ 2 5 − 6 = & − ' ab3 7 15 21 3a 3 a ( ) (2)

,

I x = dI x =

k x2 =

Ix = A

(

3 35

ab3 ab b

)

,

Ix =

3 3 ab 35

kx = b

9 35


PROBLEM 9.17 Determine the moment of inertia and the radius of gyration of the shaded area shown with respect to the y axis.

SOLUTION At x = a,

Then

Now

Then

y1 = y2 = b :

b a3 b y2 : b = ma or m = a y1 : b = ka3

or k =

b 3 x a3 b y2 = x a y1 =

b $b % dA = ( y2 − y1 )dx = & x − 3 x3 ' dx a a ( )

,

A = dA = 2

,

a 0

1 3% b$ & x − 2 x ' dx a( a ) a

=2

Now

Then

1 1 b *1 2 + x − 2 x 4 ! = ab a "2 4a #0 2

*$ b b % + dI y = x 2 dA = x 2 & x − 3 x3 ' dx ! a ) # "( a

,

I y = dI y = 2

,

a 0

b 2$ 1 % x x − 2 x3 ' dx a &( a ) a

=2

and

k y2 =

Iy A

=

b *1 4 1 1 6+ x − x a "4 6 a 2 !# 0

1 3 ab 6 1 ab 2

1 = a2 3

or

or

Iy =

1 3 ab 6

ky =

a 3



SOLUTION See figure of solution on Problem 9.16. 1 A = ab 3 a

dI y = x 2 dA = x 2 ( y2 − y1 )dx

b b $ b % x 2 & 1/2 x1/ 2 − 2 x 2 ' dx = 1/2 a a (a )

Iy =

,

Iy =

b b7/2 b a5 $ 2 1 % 3 ⋅ − ⋅ = & − 'a b a1/ 2 ( 72 ) a 2 5 ( 7 5 )

k y2

=

0

Iy A

( =

3 35

a 3b

)

,

a 0

x5/ 2 dx −

b a2

,

a 0

x 4 dx

Iy =

3 3 ab 35

ky = a

ab 3

9 35



SOLUTION y1 :

At x = 2a,

y = 0:

0 = c sin k (2a) 2ak = π

y2 :

At x = a,

y = h:

At x = a,

y = 2h :

At x = 2a,

y = 0:

or k =

h = c sin

π 2a

π 2a

( a) or

2h = ma − b 0 = m(2a) + b 2h , b = 4h a

Solving yields

m=−

Then

y1 = h sin

Now

dA = ( y2 − y1 )dx =

Then

c=h

,

π 2a

A = dA =

,

2h x + 4h a 2h ( − x + 2a ) = a

y2 = −

x

2a a

h

π + * 2h (− x + 2a) − h sin x dx 2a !# "a

π + *2 (− x + 2a) − sin x dx 2a !# "a 2a

2a π + * 1 cos x = h − ( − x + 2a ) 2 + 2a !# a π " a

*$ 2 a % 1 + 2% $ = h & − ' + (− a + 2a ) 2 ! = ah &1 − ' ( π) "( π ) a # = 0.36338ah



Find: We have

I x and k x 3 3 1 % 1 -3 * 2h π % .3 $1 + $ dI x = & y2 − y1 ' dx = / (− x + 2a) ! − & h sin x 0 dx 3 ) 3 13 " a 2a )' 23 (3 # (

=

Then Now Then

h3 * 8 π + ( − x + 2a )3 − sin 3 x 3 " a3 2a !#

,

I x = dI x =

,

2a a

h3 * 8 π + (− x + 2a)3 − sin 3 x dx 3 3 "a 2a !#

sin 3 θ = sin θ (1 − cos 2 θ ) = sin θ − sin θ cos 2 θ Ix =

h3 3

,

2a * a

8 π π π %+ $ ( − x + 2a )3 − & sin x − sin x cos 2 x ! dx 3 2a 2a 2a ') # ( "a 2a

=

2a 2a π π + h3 * 2 cos cos3 − ( − x + 2a ) 4 + x− x 3 " a3 2a 3π 2a !# a π

=

h3 3

=

2 3$ 2 % ah &1 − ' 3 ( 3π )

*$ 2a 2a % 2 4+ & − π + 3π ' + 3 (− a + 2a) ! ) a "( #

I x = 0.52520ah3

and

k x2 =

I x 0.52520ah3 = 0.36338ah A

or

I x = 0.525ah3

or

k x = 1.202h


!


SOLUTION y1 :

At x = 2a,

y = 0:

0 = c sin k (2a) 2ak = π

y2 :

At x = a,

y = h:

At x = a,

y = 2h :

At x = 2a,

y = 0:

or k =

h = c sin

π 2a

π 2a

( a) or

2h = ma − b 0 = m(2a) + b 2h , b = 4h a

Solving yields

m=−

Then

y1 = h sin

Now

dA = ( y2 − y1 )dx =

Then

c=h

,

π 2a

A = dA =

,

2h x + 4h a 2h ( − x + 2a ) = a

y2 = −

x

2a a

h

π + * 2h (− x + 2a) − h sin x dx 2 a a !# "

π + *2 x dx (− x + 2a) − sin a a !# 2 " 2a

2a π + * 1 cos x = h − ( − x + 2a ) 2 + 2a !# a π " a

*$ 2 a % 1 + 2% $ = h & − ' + (− a + 2a ) 2 ! = ah &1 − ' π π a ( ) ( ) " # = 0.36338ah


PROBLEM 9.20 (Continued) Find: I y and k y - * 2h π + . dI y = x 2 dA = x 2 / (− x + 2a) − h sin x dx 0 2a !# 2 1" a π + *2 = h ( − x3 + 2ax 2 ) − x 2 sin x dx 2a !# "a

We have

,

I y = dI y =

Then

,

2a

h

a

π + *2 x dx ( − x3 + 2ax 2 ) − x 2 sin a a !# 2 "

Now using integration by parts with u = x2

dv = sin

du = 2 x dx

,x

Then

2

sin

π 2a

,x

2

sin

π 2a

2a

π

x dx

cos

π 2a

x

,

dv = cos

du = dx

Then

2a

π % $ 2a π % $ 2a x dx = x 2 & − cos x ' − & − cos x (2 x dx) a a ') π π 2 2 ( ) ( u=x

Now let

v=−

π

x dx = −

Iy = h

2a

π

x 2 cos

v=

π 2a

2a

π

x+

π 2a sin

x dx

π 2a

x

4a * $ 2a π % $ 2a π % + x & sin x ' − & sin x dx ! 2a ) 2a ') # π " (π (π

,

*2$ 1 4 2 3% & − x + 3 ax ' ) "a ( 4 2a

$ 2a 2 π π π %+ 8a 2 16a 3 − && − x cos x + 2 x sin x + 3 cos x '! 2a 2a 2a ') !# π π ( π a

-3 2 * 1 2 16a3 .3 + 2a (2a )2 + 3 0 = h / − (2a )4 + a (2a)3 ! − 3 π 23 # π 31 a " 4 2 2 3- 2 * 1 + 8a 3. − h / − ( a ) 4 + a ( a )3 ! − 2 ( a ) 0 3 # π 31 a " 4 32 = 0.61345a3 h

and

k y2 =

Iy A

=

0.61345a 3 h 0.36338ah

or

I y = 0.613a3 h

or

k y = 1.299a


!

PROBLEM 9.21 Determine the polar moment of inertia and the polar radius of gyration of the shaded area shown with respect to Point P.

SOLUTION

We have Then

dI x = y 2 dA = y 2 [( x2 − x1 )dy ] 2a * a 2 + y (2a − a) dy + y 2 (2a − 0)dy ! a 2 " # 2a . -3 * 1 3 + a *1 + 3 = 2 / a y ! + 2a y 3 ! 0 " 3 # a 23 13 " 3 # 0

,

Ix = 2

,

- *1 . + 2 = 2 / a ( a)3 ! + a *"(2a )3 − ( a)3 +# 0 3 3 # 1 " 2 = 10a 4

Also Then

dI y = x 2 dA = x 2 [( y2 − y1 )dx] Iy = 2

* "

,

a 0

x 2 (2a − a)dx +

,

2a a

+ x 2 (2a − 0)dx ! #

= 10a 4

Now

JP = I x + I y = 10a 4 + 10a 4

and

k P2 = =

or

J P = 20a 4

or

k P = 1.826a

JP 20a 4 = A (4a)(2a ) − (2a)(a) 10 2 a 3


!


SOLUTION

x=

a a y 1 + = (a + y ) 2 2a 2

dA = x dy = 1 A= 2

,

a 0

dA =

,

1 I x = y 2 dA = 2

,

a 0

,

1 1 y2 (a + y )dy = ay + 2 2 2 a

0

1 y3 y4 = a + 2 3 4 1 Iy = 2

,

a 0

1 1 Iy = 2 24

,

0

1 1 (a + y )dy = 2 2

y2 a

= 0

1 3 1 x dy = 3 3 a

1 (a + y )dy 2

,

a 0

= 0

1 $ 2 a2 % 3 2 & a + '' = a 2 &( 2 ) 4

A=

Ix =

7 4 a " 12

Iy =

5 4 a " 16

3

1 % & 2 (a + y ) ' dy ( ) a

(a + y )3 dy =

1 1 1 * 15 4 4 4+ (a + y ) 4 = "(2a) − a # = 96 a 24 4 96 0

1 5 4 Iy = a 2 32

From Eq. (9.4):

3 2 a " 2

(ay 2 + y 3 )dy

1$1 1% 4 1 7 4 a = a + 2 &( 3 4 ') 2 12

a$ 0

,

a

JO = I x + I y = J O = kO2 A

7 4 5 4 $ 28 + 15 % 4 a + a =& 'a 12 16 ( 48 )

kO2 =

JO = A

43 4 a 48 3 2 a 2

=

43 2 a 72

JO =

kO = a

43 72

43 4 a 48

kO = 0.773a


!

PROBLEM 9.23 (a) Determine by direct integration the polar moment of inertia of the annular area shown with respect to Point O. (b) Using the result of Part a, determine the moment of inertia of the given area with respect to the x axis.

SOLUTION dA = 2π u du

(a)

dJ O = u 2 dA = u 2 (2π u du ) = 2π u 3 du

,

J O = dJ O = 2π = 2π

(b)

From Eq. (9.4):

(Note by symmetry.)

,

R2

R1

u 3 du

1 4 u 4

R2

JO =

π* 4 R2 − R14 +

Ix =

π* 4 R2 − R14 +

R1

2"

#

Ix = Iy JO = I x + I y = 2I x Ix =

1 π J O = *" R24 − R14 +# 2 4

4"

#


PROBLEM 9.24 (a) Show that the polar radius of gyration kO of the annular area shown is approximately equal to the mean radius Rm = ( R1 + R2 ) 2 for small values of the thickness t = R2 − R1. (b) Determine the percentage error introduced by using Rm in place of kO for the following values of t/Rm: 1, 12 , and 101 .

SOLUTION (a)

From Problem 9.23: JO =

kO2

π* 4 R2 − R14 + 2"

#

4 4 π J O 2 *" R2 − R1 +# = = A π R22 − R12

(

)

kO2 =

1 2 * R2 + R12 + # 2"

Mean radius:

Rm =

1 ( R1 + R2 ) 2

Thus,

1 1 R1 = Rm − t and R2 = Rm + t 2 2

Thickness: t = R2 − R1

kO2 =

2 2 1 *$ 1 % $ 1 % + 1 R + t + R − t ! = Rm2 + t 2 m m & ' & ' 2 "( 2 ) ( 2 ) #! 4

For t small compared to Rm : kO2 ≈ Rm2 (b)

Percentage error =

kO ≈ Rm

(exact value) − (approximation value) 100 (exact value)

P.E. = −

Rm2 + 14 t 2 − Rm Rm2 + 14 t 2

(100) = −

2

( ) − 1100 1+ ( )

1+

t Rm

1 4

1 4

t Rm

2



For

t = 1: Rm

t 1 For = : Rm 2

1 t For = : Rm 10

P.E. =

1 + 14 − 1 1+

P.E. = −

P.E. =

1 4

(100) = −10.56% 2

( 12 ) − 1 (100) = −2.99% 1 1 2 1+ 4 ( 2 )

1+

1 4

2

( ) −1 (100) = −0.125% 2 1 + 14 ( 101 )

1+

1 1 4 10



SOLUTION y1 :

At x = 2a,

y = 2a : 2a = k1 (2a) 2

y2 :

At x = 0, At x = 2a,

y = a:

or k1 =

a=c

y = 2a :

2a = a + k2 (2a )2

Then

y1 =

1 2 x 2a

Then

Now

k2 =

or

y2 = a + =

Now

1 2a

1 4a

1 2 x 4a

1 (4a 2 + x 2 ) 4a

1 2+ *1 (4a 2 + x 2 ) − x dx 2a !# " 4a 1 (4a 2 − x 2 ) dx = 4a

dA = ( y2 − y1 )dx =

,

A = dA = 2

,

2a 0

2a

1 1 * 2 1 + 8 (4a 2 − x 2 )dx = 4a x − x 3 ! = a 2 4a 2a " 3 #0 3

3 3 1 % 1 -3 * 1 $1 + * 1 2 + .3 (4a 2 + x 2 ) ! − dI x = & y23 − y13 ' dx = / x ! 0 dx 3 ) 3 13 " 4a (3 # " 2a # 23

1* 1 1 + (64a 6 + 48a 4 x 2 + 12a 2 x 4 + x6 ) − 3 x6 ! dx 3 3 " 64a 8a # 1 (64a 6 + 48a 4 x 2 + 12a 2 x 4 − 7 x 6 )dx = 3 192a

=



Then

,

I x = dI x = 2

,

2a 0

1 (64a 6 + 48a 4 x 2 + 12a 2 x 4 − 7 x 6 )dx 192a3 2a

=

1 * 12 + 64a 6 x + 16a 4 x3 + a 2 x5 − x 7 ! 3 5 96a " #0

1 * 12 + 64a 6 (2a) + 16a 4 (2a )3 + a 2 (2a)5 − (2a )7 ! 5 96a3 " # 1 4$ 12 % 32 4 a &128 + 128 + × 32 − 128 ' = a = 96 ( 5 ) 15 =

Also

Then

*1 + (4a 2 − x 2 )dx ! " 4a #

dI y = x 2 dA = x 2

,

I y = dI y = 2 =

,

2a 0

2a

1 2 1 *4 2 3 1 5+ x (4a 2 − x 2 )dx = a x − x ! 4a 2a " 3 5 #0

1 *4 2 1 + 32 4 $ 1 1 % 32 4 a (2a )3 − (2a )5 ! = a & − '= a 2a " 3 5 # 2 ( 3 5 ) 15

Now

JP = I x + I y =

and

k P2 =

JP = A

32 4 32 4 a + a 15 15

64 4 a 15 8 2 a 3

=

8 2 a 5

or

JP =

64 4 a 15

or k P = 1.265a



SOLUTION The equation of the circle is x2 + y 2 = r 2

So that Now

x = r 2 − y2 dA = x dy = r 2 − y 2 dy

,

,

r

r 2 − y 2 dy

Then

A = dA = 2

Let

y = r sin θ ; dy = r cos θ dθ

Then

A=2 =2

π /2

,π

− /6

π /2

,π

− /6

= 2r 2

− r/2

r 2 − (r sin θ ) 2 r cos dθ π /2

r 2 cos 2 θ dθ = 2r 2

* π2

$ − π sin − π3 − && 6 + 4 "2 ( 2

* θ sin 2θ + + 4 !# −π /6 "2

$ %+ 3% 2 π ' '' ! = 2r && + 8 )' ) #! (3

= 2.5274r 2

Now

dI x = y 2 dA = y 2

Then

I x = dI x = 2

Let Then

,

r

− r/2

r 2 − y 2 dy

)

y 2 r 2 − y 2 dy

y = r sin θ ; dy = r cos θ dθ Ix = 2 =2

Now

,

(

π /2

,π

− /6

π /2

,π

− /6

(r sin θ )2 r 2 − (r sin θ ) 2 r cos θ dθ r 2 sin 2 θ (r cos θ )r cos θ dθ

sin 2θ = 2sin θ cos θ 4 sin 2 θ cos 2 θ =

1 sin 2θ 4



Then

Ix = 2

,

π /2

r 4 * θ sin 4θ + $1 % r 4 & sin 2 2θ ' dθ = − −π /6 2 "2 8 #! −π /6 (4 ) π /2

r 4 * π2 $ π6 sin − 23π −& − 2 " 2 &( 2 8 3% r4 $ π = & − ' & 2 ( 3 16 ') =

Also

dI y =

1 3 1 x dy = 3 3

(

r 2 − y2

,

r

1

,

(r 2 − y 2 )3/2 dy

I y = dI y = 2

Let

y = r sin θ ; dy = r cos θ dθ

Then

Iy =

2 3

,π

Iy =

2 3

,π

Now Then

π /2

− /6

π /2

− /6

3

) dy

Then

− r/ 2 3

%+ '' ! ) !#

[r 2 − (r sin θ ) 2 ]3/ 2 r cos θ dθ ( r 3 cos3 θ )r cos θ dθ

1 cos 4 θ = cos 2 θ (1 − sin 2 θ ) = cos 2 θ − sin 2 2θ 4 Iy = =

2 3

π /2

,π

− /6

1 $ % r 4 & cos 2 θ − sin 2 2θ ' dθ 4 ( )

2 4 *$ θ sin 2θ r & + 3 "( 2 4

% 1 $ θ sin 4θ '− & − 8 ) 4( 2

π /2

%+ '! ) # −π / 6

2 4 -3 * π2 1 $ π2 % + * − π6 sin − π3 1 $ − π6 sin − 23π + − & − r / − & '! − 3 3 " 2 4 &( 2 ') #! " 2 4 4 (& 2 8 1 2 *π π π 1 $ 3 % π 1 $ 3 %+ = r4 − + + & + & '' − '! & 3 " 4 16 12 4 ( 2 ) 48 32 &( 2 ') #! 2 $π 9 3 % = r4 & + ' 3 &( 4 64 ')

=

% + .3 '' ! 0 ) !# 32



Now

JP = Ix + Iy =

r4 2

$π 3 % 2 4$π 9 3 % && − '' + r && + ' 64 ') ( 3 16 ) 3 ( 4

$π 3% = r4 & + = 1.15545r 4 & 3 16 '' ( )

and

k P2 =

J P 1.15545r 4 = A 2.5274r 2

or

J P = 1.155r 4

or k P = 0.676r


!

PROBLEM 9.27 Determine the polar moment of inertia and the polar radius of gyration of the shaded area shown with respect to Point O.

SOLUTION At θ = π , R = 2a :

2a = a + k (π ) a

or

k=

Then

R=a+

π a

π

$ θ% θ = a &1 + ' π (

)

dA = (dr )(r dθ ) = rdr dθ

Now

,

A = dA =

Then

π

a (1+θ /π )

0

0

, ,

rdr dθ =

,

π 0

a (1+θ /π )

*1 2 + r ! " 2 #0

dθ π

A=

,

π 0

2 3 1 2$ θ % 1 *π $ θ % + a & 1 + ' dθ = a 2 1 + ! 2 ( π) 2 " 3 &( π ') !# 0

*$ π % 3 + 7 1 = π a 2 &1 + ' − (1)3 ! = π a 2 6 "( π ) #! 6

Now Then

dJ O = r 2 dA = r 2 (rdr dθ )

,

J O = dJ O = =

,

π 0

π

a (1+θ /π )

0

0

, ,

r 3 dr dθ

a (1+θ /π )

*1 4 + r ! " 4 #0

dθ =

,

π 0

4

1 4$ θ % a & 1 + ' dθ 4 ( π)

π 5+

=

and

kO2 =

5 * + π% 1 4 *π $ θ % 1 4 $ a a π 1 1 + = + − (1)5 ! ! & ' & ' 4 " 5 ( π ) #! 20 "( π ) #! 0

JO = A

31 π a4 20 7 π a2 6

=

93 2 a 70

31 4 πa 20

or

JO =

or

kO = 1.153a


!

PROBLEM 9.28 Determine the polar moment of inertia and the polar radius of gyration of the isosceles triangle shown with respect to Point O.

SOLUTION By observation:

y=

or

x=

h b 2

x

b y 2h

Now

$ b dA = xdy = & ( 2h

and

dI x = y 2 dA =

Then

,

% y ' dy )

b 3 y dy 2h

I x = dI x = 2

,

h 0

b 3 y dy 2h

b y4 = h 4

From above: Now

y=

= 0

1 3 bh 4

2h x b

2h % $ dA = (h − y )dx = & h − x dx b ') ( =

and

h

dI y = x 2 dA = x 2

h (b − 2 x)dx b

h (b − 2 x)dx b



Then

,

I y = dI y = 2

,

b/2 0

h 2 x (b − 2 x)dx b b/2

=2

h *1 3 1 4 + bx − x ! b "3 2 #0

h *b $ b % 1 $ b % =2 − b " 3 &( 2 ') 2 &( 2 ') 3

Now

JO = I x + I y =

and

kO2 =

JO = A

bh 48

4+

1 ! = b3 h 48 !#

1 3 1 3 bh + b h 4 48

(12h 2 + b 2 ) 1 2

bh

=

1 (12h 2 + b 2 ) 24

or

or

JO =

bh (12h 2 + b 2 ) 48

kO =

12h 2 + b 2 24


PROBLEM 9.29* Using the polar moment of inertia of the isosceles triangle of Problem 9.28, show that the centroidal polar moment of inertia of a circular area of radius r is π r 4 /2. (Hint: As a circular area is divided into an increasing number of equal circular sectors, what is the approximate shape of each circular sector?) PROBLEM 9.28 Determine the polar moment of inertia and the polar radius of gyration of the isosceles triangle shown with respect to Point O.

SOLUTION First the circular area is divided into an increasing number of identical circular sectors. The sectors can be approximated by isosceles triangles. For a large number of sectors the approximate dimensions of one of the isosceles triangles are as shown. For an isosceles triangle (see Problem 9.28): JO =

b = r ∆θ

Then with

= dJ O sector dθ

and h = r

1 (r ∆θ )(r ) *"12r 2 + (r ∆θ ) 2 +# 48

(∆ J O )sector

Now

bh (12h 2 + b 2 ) 48

1 4 r ∆θ *"(12 + ∆θ 2 ) +# 48

$ ∆ J O sector = lim & ∆θ →0 ( ∆θ

% -1 4 2 . ' = lim / r *"12 + ( ∆θ ) +# 0 ∆θ →0 1 48 2 ) =

Then

,

( J O )circle = dJ O sector =

,

2π 0

1 4 r 4

1 4 1 2π r dθ = r 4 [θ ]0 4 4

or

( J O )circle =

π 2

r4


!

PROBLEM 9.30* Prove that the centroidal polar moment of inertia of a given area A cannot be smaller than A2 /2π . (Hint: Compare the moment of inertia of the given area with the moment of inertia of a circle that has the same area and the same centroid.)

SOLUTION From the solution to sample Problem 9.2, the centroidal polar moment of inertia of a circular area is ( J C )cir =

π 2

r4

The area of the circle is Acir = π r 2

So that

[ J C ( A)]cir =

A2 2π

Two methods of solution will be presented. However, both methods depend upon the observation that as a given element of area dA is moved closer to some Point C, The value of J C will be decreased ( J C = , r 2 dA; as r decreases, so must J C ). Solution 1 Imagine taking the area A and drawing it into a thin strip of negligible width and of sufficient length so that its area is equal to A. To minimize the value of ( J C ) A, the area would have to be distributed as closely as possible about C. This is accomplished by winding the strip into a tightly wound roll with C as its center; any voids in the roll would place the corresponding area farther from C than is necessary, thus increasing the value of ( J C ) A. (The process is analogous to rewinding a length of tape back into a roll.) Since the shape of the roll is circular, with the centroid of its area at C, it follows that ( JC ) A ≥

A2 Q.E.D. 2π

! ! ! where the equality applies when the original area is circular.



Solution 2 Consider an area A with its centroid at Point C and a circular area of area A with its center (and centroid) at Point C. Without loss of generality, assume that A1 = A2

A3 = A4

It then follows that ( J C ) A = ( J C )cir + [ J C ( A1 ) − J C ( A2 ) + J C ( A3 ) − J C ( A4 )]

Now observe that J C ( A1 ) − J C ( A2 ) $ 0 J C ( A3 ) − J C ( A4 ) $ 0

since as a given area is moved farther away from C its polar moment of inertia with respect to C must increase. ( J C ) A $ ( J C )cir

or ( J C ) A $

A2 Q.E.D. 2π


!

PROBLEM 9.31 Determine the moment of inertia and the radius of gyration of the shaded area with respect to the x axis.

SOLUTION First note that A = A1 + A2 + A3 = [(24)(6) + (8)(48) + (48)(6)] mm 2 = (144 + 384 + 288) mm 2 = 816 mm 2

Now

I x = ( I x )1 + ( I x )2 + ( I x )3

where 1 (24 mm)(6 mm)3 + (144 mm 2 )(27 mm)2 12 = (432 + 104,976) mm 4

( I x )1 =

= 105, 408 mm 4 1 (8 mm)(48 mm)3 = 73,728 mm 4 12 1 ( I x )3 = (48 mm)(6 mm)3 + (288 mm 2 )(27 mm)2 12 = (864 + 209,952) mm 4 = 210,816 mm 4

( I x )2 =

Then

I x = (105, 408 + 73, 728 + 210,816) mm 4

= 389,952 mm 4

and

k x2 =

I x 389,952 mm 4 = A 816 mm 2

or

I x = 390 × 103 mm 4

or

k x = 21.9 mm


!

PROBLEM 9.32 Determine the moment of inertia and the radius of gyration of the shaded area with respect to the x axis.

SOLUTION First note that A = A1 − A2 − A3

= [(5)(6) − (4)(2) − (4)(1)] in.2 = (30 − 8 − 4) in.2 = 18 in.2

Now where

I x = ( I x )1 − ( I x ) 2 − ( I x )3

( I x )1 =

1 (5 in.)(6 in.)3 = 90 in.4 12

1 (4 in.)(2 in.)3 + (8 in.2 )(2 in.) 2 12 2 = 34 in.4 3

( I x )2 =

1 $3 % (4 in.)(1 in.)3 + (4 in.2 ) & in. ' 12 2 ( ) 1 4 = 9 in. 3

2

( I x )3 =

Then

2 1% $ I x = & 90 − 34 − 9 ' in.4 3 3) (

and

k x2 =

I x 46.0 in.4 = A 18 in.4

or

I x = 46.0 in.4

or k x = 1.599 in.


PROBLEM 9.33 Determine the moment of inertia and the radius of gyration of the shaded area with respect to the y axis.

SOLUTION First note that A = A1 + A2 + A3 = [(24 × 6) + (8)(48) + (48)(6)] mm 2 = (144 + 384 + 288) mm 2 = 816 mm 2

Now

I y = ( I y )1 + ( I y ) 2 + ( I y )3

where 1 (6 mm)(24 mm)3 = 6912 mm 4 12 1 ( I y ) 2 = (48 mm)(8 mm)3 = 2048 mm 4 12 1 ( I y )3 = (6 mm)(48 mm)3 = 55, 296 mm 4 12 ( I y )1 =

Then

and

I y = (6912 + 2048 + 55, 296) mm 4 = 64, 256 mm 4

k y2 =

Iy A

=

64, 256 mm 4 816 mm 2

or

I y = 64.3 × 103 mm 4

or

k y = 8.87 mm


PROBLEM 9.34 Determine the moment of inertia and the radius of gyration of the shaded area with respect to the y axis.

SOLUTION First note that A = A1 − A2 − A3 = [(5)(6) − (4)(2) − (4)(1)] in.2 = (30 − 8 − 4) in.2 = 18 in.2

Now

I y = ( I y )1 − ( I y ) 2 − ( I y )3

where ( I y )1 =

1 (6 in.)(5 in.)3 = 62.5 in.4 12

( I y )2 =

1 2 (2 in.)(4 in.)3 = 10 in.4 12 3

( I y )3 =

1 1 (1 in.)(4 in.)3 = 5 in.4 12 3

Then

2 1% $ I y = & 62.5 − 10 − 5 ' in.4 3 3) (

and

k y2 =

Iy A

=

46.5 in.4 18 in.2

or

I y = 46.5 in.4

or k y = 1.607 in.


PROBLEM 9.35 Determine the moments of inertia of the shaded area shown with respect to the x and y axes when a = 20 mm.

SOLUTION We have where

I x = ( I x )1 + 2( I x ) 2 1 (40 mm)(40 mm)3 12 = 213.33 × 103 mm 4

( I x )1 =

( I x )2 =

*π "8

(20 mm)4 −

2 $ 4 × 20 % + (20 mm) 2 & mm ' ! 2 ( 3π ) !#

π

*$ 4 × 20 + % + (20 mm) & + 20 ' mm ! 2 ) "( 3π #

π

2

2

= 527.49 × 103 mm 4

Then

I x = [213.33 + 2(527.49)] × 103 mm 4

or Also where

I y = ( I y )1 + 2( I y )2 ( I y )1 = ( I y )2 =

Then

I x = 1.268 × 106 mm 4

1 (40 mm)(40 mm)3 = 213.33 × 103 mm 4 12

π 8

(20 mm)4 = 62.83 × 103 mm 4

I y = [213.33 + 2(62.83)] × 103 mm 4

or

I y = 339 × 103 mm 4


!

PROBLEM 9.36 Determine the moments of inertia of the shaded area shown with respect to the x and y axes when a = 20 mm.

SOLUTION Area = Square − 2(Semicircles)

Given: For a = 20 mm, we have

Ix = Iy =

Square: Semicircle

1 (60) 4 = 1080 × 103 mm 4 12

: Ix =

π 8

(20) 4 = 62.83 × 103 mm 4

I AA′ = I y′ + Ad 2 ;

π 8

$π % (20)4 = I y′ + & ' (20) 2 (8.488) 2 (2)

I y′ = 62.83 × 103 − 45.27 × 103 I y′ = 17.56 × 103 mm 4 I y = I y′ + A(30 − 8.488)2 = 17.56 × 103 +

π 2

(20) 2 (21.512)2

I y = 308.3 × 103 mm 4



Semicircle !: Same as semicircle

.

Entire Area: I x = 1080 × 103 − 2(62.83 × 103 ) = 954.3 × 103 mm 4

I x = 954 × 103 mm 4

I y = 1080 × 103 − 2(308.3 × 103 ) = 463.3 × 103 mm 4

I y = 463 × 103 mm 4


!

PROBLEM 9.37 For the 4000-mm 2 shaded area shown, determine the distance d2 and the moment of inertia with respect to the centroidal axis parallel to AA′ knowing that the moments of inertia with respect to AA′ and BB′ are 12 × 106 mm 4 and 23.9 × 106 mm4, respectively, and that d1 = 25 mm.

SOLUTION We have

I AA′ = I + Ad12

and

I BB′ = I + Ad 22

Subtracting or

(

I AA′ − I BB′ = A d12 − d 22 d 22 = d12 −

(1)

)

I AA′ − I BB′ A

= (25 mm)2 −

(12 − 23.9)106 mm 4 4000 mm 2

= 3600 mm 2

Using Eq. (1):

or d 2 = 60.0 mm

I = 12 × 106 mm 4 − (4000 mm 2 )(25 mm) 2

or

I = 9.50 × 106 mm 4


!

PROBLEM 9.38 Determine for the shaded region the area and the moment of inertia with respect to the centroidal axis parallel to BB′, knowing that d1 = 25 mm and d 2 = 15 mm and that the moments of inertia with respect to AA′ and BB′ are 7.84 × 106 mm 4 and 5.20 × 106 mm4, respectively.

SOLUTION We have

I AA′ = I + Ad12

and

I BB′ = I + Ad 22

Subtracting or

(

I AA′ − I BB′ = A d12 − d 22 A=

I AA′ − I BB′ d12 − d 22

(1)

) =

(7.84 − 5.20)106 mm 4 (25 mm) 2 − (15 mm)2

or Using Eq. (1):

A = 6600 mm 2

I = 7.84 × 106 mm 4 − (6600 mm 2 )(25 mm)2 = 3.715 × 106 mm 4

or I = 3.72 × 106 mm 4


!

PROBLEM 9.39 The shaded area is equal to 50 in.2 . Determine its centroidal moments of inertia I x and I y , knowing that I y = 2 I x and that the polar moment of inertia of the area about Point A is JA = 2250 in.4 .

SOLUTION Given:

A = 50 in.2

I y = 2 I x , J A = 2250 in.4 J A = J C + A(6 in.) 2

2250 in.4 = J C + (50 in.2 )(6 in.) 2 J C = 450 in.4 JC = I x + I y

with

450 in.4 = I x + 2 I x

I y = 2I x I x = 150.0 in.4 I y = 2 I x = 300 in.4


!

PROBLEM 9.40 The polar moments of inertia of the shaded area with respect to Points A, B, and D are, respectively, JA = 2880 in.4, JB = 6720 in.4, and JD = 4560 in.4. Determine the shaded area, its centroidal moment of inertia JC , and the distance d from C to D.

SOLUTION See figure at solution of Problem 9.39. Given:

J A = 2880 in.4 , J B = 6720 in.4 , J D = 4560 in.4 J B = J C + A(CB ) 2 ; 6720 in.4 = J C + A(62 + d 2 )

(1)

J D = J C + A(CD) 2 ; 4560 in.4 = J C + Ad 2

(2)

Eq. (1) subtracted by Eq. (2): J B − J D = 2160 in.4 = A(6) 2 J A = J C + A( AC )2 ; 2880 in.4 = J C + (60 in.2 )(6 in.) 2

Eq. (2):

4560 in.4 = 720 in.4 + (60 in.2 )d 2

A = 60.0 in.2 J C = 720 in.4 d = 8.00 in.


!

PROBLEM 9.41 Determine the moments of inertia I x and I y of the area shown with respect to centroidal axes respectively parallel and perpendicular to side AB.

SOLUTION

Dimensions in mm

First locate centroid C of the area. Symmetry implies Y = 30 mm.

1

Then

A, mm 2

x , mm

108 × 60 = 6480

54

349,920

46

–59,616

2

1 − × 72 × 36 = −1296 2

Σ

5184

290,304

X ΣA = Σ xA : X (5184 mm 2 ) = 290,304 mm3

or

X = 56.0 mm

Now

I x = ( I x )1 − ( I x )2

where

xA, mm3

1 (108 mm)(60 mm)3 = 1.944 × 106 mm 4 12 *1 + $1 % ( I x )2 = 2 (72 mm)(18 mm)3 + & × 72 mm × 18 mm ' (6 mm) 2 ! (2 ) " 36 # ( I x )1 =

= 2(11, 664 + 23,328) mm 4 = 69.984 × 103 mm 4 [( I x ) 2 is obtained by dividing A2 into

Then

]

I x = (1.944 − 0.069984) × 106 mm 4

or

I x = 1.874 × 106 mm 4



Also where

I y = ( I y )1 − ( I y ) 2 1 (60 mm)(108 mm)3 + (6480 mm 2 )[(56.54) mm]2 12 = (6, 298,560 + 25,920) mm 4 = 6.324 × 106 mm 4

( I y )1 =

1 (36 mm)(72 mm)3 + (1296 mm 2 )[(56 − 46) mm]2 36 = (373, 248 + 129, 600) mm 4 = 0.502 × 106 mm 4

( I y )2 =

Then

I y = (6.324 − 0.502)106 mm 4

or I y = 5.82 × 106 mm 4


!


SOLUTION First locate C of the area: Symmetry implies X = 12 mm.

Then

A, mm 2

y , mm

yA, mm3

1

12 × 22 = 264

11

2904

2

1 (24)(18) = 216 2

28

6048

Σ

480

8952

Y ΣA = Σ yA: Y (480 mm 2 ) = 8952 mm3 Y = 18.65 mm

Now where

I x = ( I x )1 + ( I x ) 2

1 (12 mm)(22 mm)3 + (264 mm 2 )[(18.65 − 11) mm]2 12 = 26, 098 mm 4

( I x )1 =

1 (24 mm)(18 mm)3 + (216 mm 2 )[(28 − 18.65) mm]2 36 = 22, 771 mm 4

( I x )2 =

Then

I x = (26.098 + 22.771) × 103 mm 4

or

I x = 48.9 × 103 mm 4



Also where

I y = ( I y )1 + ( I y ) 2 1 (22 mm)(12 mm)3 = 3168 mm 4 12 *1 + $1 % ( I y )2 = 2 (18 mm)(12 mm)3 + & × 18 mm × 12 mm ' (4 mm)2 ! 36 2 ( ) " # ( I y )1 =

= 5184 mm 4 [( I y )2 is obtained by dividing A2 into

Then

]

I y = (3168 + 5184) mm 4

or I y = 8.35 × 103 mm 4


!


SOLUTION

First locate centroid C of the area.

Then

A, in.2

x , in.

1

5 × 8 = 40

2.5

2

−2 × 5 = −10

1.9

Σ

30

where

4

100

160

4.3

–19

–43

81

117

X = 2.70 in. Y ΣA = Σ yA: Y (30 in.2 ) = 117 in.3 Y = 3.90 in.

or Now

yA, in.3

X ΣA = Σ xA: X (30 in.2 ) = 81 in.3

or and

xA, in.3

y , in.

I x = ( I x )1 − ( I x )2 1 (5 in.)(8 in.)3 + (40 in.2 )[(4 − 3.9) in.]2 12 = (213.33 + 0.4) in.4 = 213.73 in.4

( I x )1 =

1 (2 in.)(5 in.)3 + (10 in.2 )[(4.3 − 3.9) in.]2 12 = (20.83 + 1.60) = 22.43 in.4

( I x )2 =



Then

I x = (213.73 − 22.43) in.4

Also

I y = ( I y )1 − ( I y ) 2

where

or I x = 191.3 in.4

1 (8 in.)(5 in.)3 + (40 in.2 )[(2.7 − 2.5) in.]2 12 = (83.333 + 1.6)in.4 = 84.933 in.4

( I y )1 =

1 (5 in.)(2 in.)3 + (10 in.2 )[(2.7 − 1.9) in.]2 12 = (3.333 + 6.4)in.4 = 9.733 in.4

( I y )2 =

Then

I y = (84.933 − 9.733)in.4

or

I y = 75.2 in.4


!


SOLUTION

First locate centroid C of the area. A, in.2

Then or and or

x , in.

y , in.

xA, in.3

yA, in.3

1

3.6 × 0.5 = 1.8

1.8

0.25

3.24

0.45

2

0.5 × 3.8 = 1.9

0.25

2.4

0.475

4.56

3

1.3 × 1 = 1.3

0.65

4.8

0.845

6.24

Σ

5.0

4.560

11.25

X ΣA = Σ xA: X (5 in.2 ) = 4.560 in.3 X = 0.912 in. Y ΣA = Σ yA: Y (5 in.2 ) = 11.25 in.3 Y = 2.25 in.



Now where

I x = ( I x )1 + ( I x )2 + ( I x )3 1 (3.6 in.)(0.5 in.)3 + (1.8 in.2 )[(2.25 − 0.25)in.]2 12 = (0.0375 + 7.20)in.4 = 7.2375 in.4

( I x )1 =

1 (0.5 in.)(3.8 in.)3 + (1.9 in.2 )[(2.4 − 2.25)in.]2 12 = (2.2863 + 0.0428)in.4 = 2.3291 in.4

( I x )2 =

1 (1.3 in.)(1 in.)3 + (1.3 in.2 )[(4.8 − 2.25 in.)]2 12 = (0.1083) + 8.4533) in.4 = 8.5616 in.4

( I x )3 =

Then

I x = (7.2375 + 2.3291 + 8.5616) in.4 = 18.1282 in.4

or Also where

I x = 18.13 in.4

I y = ( I y )1 + ( I y ) 2 + ( I y )3 1 (0.5 in.)(3.6 in.)3 + (1.8 in.2 )[(1.8 − 0.912)in.]2 12 = (1.9440 + 1.4194)in.4 = 3.3634 in.4

( I y )1 =

1 (3.8 in.)(0.5 in.)3 + (1.9 in.2 )[(0.912 − 0.25) in.]2 12 = (0.0396 + 0.8327) in.4 = 0.8723 in.4

( I y )2 =

1 (1 in.)(1.3 in.)3 + (1.3 in.2 )[(0.912 − 0.65) in.]2 12 = (0.1831 + 0.0892) in.4 = 0.2723 in.4

( I y )3 =

Then

I y = (3.3634 + 0.8723 + 0.2723) in.4 = 4.5080 in.4

or

I y = 4.51 in.4


!

PROBLEM 9.45 Determine the polar moment of inertia of the area shown with respect to (a) Point O, (b) the centroid of the area.

SOLUTION Symmetry: X = Y

Dimensions in mm

Determination of centroid, C, of entire section. Area, mm 2

Section

π

1

4

(100) 2 = 7.854 × 103

y , mm

yA, mm3

42.44

333.3 × 103

2

(50)(100) = 5 × 103

50

250 × 103

3

(100)(50) = 5 × 103

–25

−125 × 103

Σ

17.854 × 103

458.3 × 103

Y ΣA = Σ yA: Y (17.854 × 103 mm 2 ) = 458.3 × 103 mm3 Y = 25.67 mm

Distance O to C: (a)

X = Y = 25.67 mm

OC = 2Y = 2(25.67) = 36.30 mm

π

(100) 4 = 39.27 × 106 mm 4

Section 1:

JO =

Section 2:

JO = J + A(OD) 2 =

8

*$ 50 %2 $ 100 %2 + 1 (50)(100)[502 + 1002 ] + (50)(100) & ' + & ' ! 12 "( 2 ) ( 2 ) #!

JO = 5.208 × 106 + 15.625 × 106 = 20.83 × 106 mm 4



Section 3: Same as Section 2;

J O = 20.83 × 106 mm 4

Entire section: J O = 39.27 × 106 + 2(20.83 × 106 ) JO = 80.9 × 106 mm 4

= 80.94 × 106

(b)

Recall that,

OC = 36.30 mm and

A = 17.854 × 103 mm 2

J O = J C + A(OC ) 2 80.94 × 106 mm 4 = J C + (17.854 × 103 mm 2 )(36.30 mm) 2 J C = 57.41 × 106 mm 4

J C = 57.4 × 106 mm 4


!


SOLUTION First locate centroid C of the figure. Note that symmetry implies Y = 0.

Dimensions in mm

A, mm 2

π

1

2

(84)(42) = 5541.77

π

2

2 −

3

π 2 −

4

X , mm −

2

π 56

(42)2 = 2770.88

π

(54)(27) = −2290.22

π

112

−

72

π 36

(27) 2 = −1145.11

π

= −35.6507

−197,568

= 17.8254

49,392

= −22.9183

52,488

= 11.4592

4877.32

Σ

XA, mm3

–13,122 –108,810

X ΣA = Σ xA: X (4877.32 mm 2 ) = −108,810 mm3

Then

X = −22.3094

or

J O = ( J O )1 + ( J O ) 2 − ( J O )3 − ( J O )4

(a) where

π

(84 mm)(42 mm)[(84 mm) 2 + (42 mm)2 ] 8 = 12.21960 × 106 mm 4

( J O )1 =



π

(42 mm) 4 4 = 2.44392 × 106 mm 4

( J O )2 =

π

(54 mm)(27 mm)[(54 mm)2 + (27 mm)2 ] 8 = 2.08696 × 106 mm 4

( J O )3 =

π

(27 mm) 4 4 = 0.41739 × 106 mm 4

( J O )4 =

Then

J O = (12.21960 + 2.44392 − 2.08696 − 0.41739) × 106 mm 4 = 12.15917 × 106 mm 4

or

J O = 12.16 × 106 mm 4

J O = J C + AX 2

(b) or

J C = 12.15917 × 106 mm 4 − (4877.32 mm 2 )(−22.3094 mm) 2

or

J C = 9.73 × 106 mm 4


!


SOLUTION

Area, in.2

Section

π

1 2

(4.5) 2 = 15.904

4 −

π 4

xA, in.3

1.9099

30.375

1.2732

–9.00

8.835

Σ

Then

(3) 2 = −7.069

x , in.

21.375

XA = Σ xA: X (8.835 in.2 ) = 21.375 in.3 X = 2.419 in.

Then

JO =

π 8

(4.5 in.)4 −

π 8

(3 in.)4 = 129.22 in.4

J O = 129.2 in.4

OC = 2 X = 2(2.419 in.) = 3.421 in. J O = J C + A(OC ) 2 : 129.22 in.4 = J C + (8.835 in.)(3.421 in.)2

J C = 25.8 in.4


!


SOLUTION First locate centroid C of the figure.

A, in.2

1

π 2

y , in. 8

(6)2 = 56.5487

2

1 − (12)(4.5) = −27 2

Σ

29.5487

π

= 2.5465

1.5

yA, in.3

144 –40.5 103.5

Y ΣA = Σ yA: Y (29.5487 in.2 ) = 103.5 in.3

Then

Y = 3.5027 in.

or J O = ( J O )1 − ( J O ) 2

(a)

π

where

(6 in.) 4 = 107.876 in.4 4 ( J O ) 2 = ( I x′ ) 2 + ( I y ′ ) 2

Now

( I x′ ) 2 =

and

( I y′ ) 2 = 2

[Note: ( I y ′ ) 2 is obtained using

( J O )1 =

1 (12 in.)(4.5 in.)3 = 91.125 in.4 12

*1 + (4.5 in.)(6 in.)3 ! = 162.0 in.4 12 " #

]



Then Finally

( J O ) 2 = (91.125 + 162.0) in.4 = 253.125 in.4 J O = (1017.876 − 253.125) in.4 = 764.751 in.4

or

J O = 765 in.4

J O = J C + AY 2

(b) or

J C = 764.751 in.4 − (29.5487 in.2 )(3.5027 in.) 2

or

J C = 402 in.4


!

PROBLEM 9.49 Two 20-mm steel plates are welded to a rolled S section as shown. Determine the moments of inertia and the radii of gyration of the combined section with respect to the centroidal x and y axes.

SOLUTION S section:

A = 6010 mm 2 I x = 90.3 × 106 mm 4 I y = 3.88 × 106 mm 4

Note:

Atotal = AS + 2 Aplate = 6010 mm 2 + 2(160 mm)(20 mm) = 12, 410 mm 2

Now where

I x = ( I x )S + 2( I x ) plate ( I x )plate = I xplate + Ad 2 1 (160 mm)(20 mm)3 + (3200 mm 2 )[(152.5 + 10) mm]2 12 = 84.6067 × 106 mm 4 =

Then

I x = (90.3 + 2 × 84.6067) × 106 mm 4 = 259.5134 × 106 mm 4

and

k x2 =

Ix 259.5134 × 106 mm 4 = Atotal 12410 mm 2

or I x = 260 × 106 mm 4 or

k x = 144.6 mm

! ! ! ! ! !


!


Also

I y = ( I y )S + 2( I y ) plate = 3.88 × 106 mm 4 + 2

*1 + (20 mm)(160 mm)3 ! 12 " #

or I y = 17.53 × 106 mm 4

= 17.5333 × 106 mm 4

and

k y2 =

Iy Atotal

=

17.5333 × 106 mm 4 12, 410 mm 2

or

k y = 37.6 mm


!

PROBLEM 9.50 Two channels are welded to a rolled W section as shown. Determine the moments of inertia and the radii of gyration of the combined section with respect to the centroidal x and y axes.

SOLUTION W section:

A = 9.12 in.2 I x = 110 in.4 I y = 37.1 in.4

Channel:

A = 3.37 in.2 I x = 1.31 in.4 I y = 32.5 in.4 Atotal = AW + 2Achan = 9.12 + 2(3.37) = 15.86 in.2

Now where

I x = ( I x ) W + 2( I x )chan ( I x )chan = I xchan + Ad 2 = 1.31 in.4 + (3.37 in.2 )(4.572 in.)2 = 71.754 in.4

Then

I x = (110 + 2 × 71.754) in.4 = 253.51 in.4

and

k x2 =

Also

I y = ( I y ) W + 2( I y )chan

Ix 253.51 in.4 = Atotal 15.86 in.2

= (37.1 + 2 × 32.5) in.4 = 102.1 in.4

and

k y2 =

Iy Atotal

=

102.1 in.4 15.86 in.2

I x = 254 in.4 k x = 4.00 in.

I y = 102.1 in.4 k y = 2.54 in.


!

PROBLEM 9.51 To form a reinforced box section, two rolled W sections and two plates are welded together. Determine the moments of inertia and the radii of gyration of the combined section with respect to the centroidal axes shown.

SOLUTION W section:

A = 5880 mm 2 I x = 15.4 × 106 mm 4 I y = 45.8 × 106 mm 4

Note:

Atotal = 2 AW + 2 Aplate = 2(5880 mm 2 ) + 2(203 mm)(6 mm) = 14,196 mm 2

Now where

I x = 2( I x ) W + 2( I x )plate

(I x )W

203 ! mm ! = I x + Ad = 15.4 × 10 mm + (5880 mm ) " 2 # 2

6

4

2

2

= 75.9772 × 106 mm 4 ( I x )plate = I xplate + Ad 2 1 (203 mm)(6 mm)3 + (1218 mm 2 )[(203 + 3) mm]2 12 = 51.6907 × 106 mm 4 =

Then

I x = [2(75.9772) + 2(51.6907)] × 106 mm 4 = 255.336 × 106 mm 4

and

k x2 =

or

I x = 255 × 106 mm 4

or

k x = 134.1 mm

Ix 255.336 × 106 mm 4 = Atotal 14,196 mm 2



also where

I y = 2( I y ) W + 2( I y ) plate (I y )W = I y ( I y ) plate =

Then

1 (6 mm)(203 mm)3 = 4.1827 × 106 mm 4 12

I y = [2(45.8) + 2(4.1827)] × 106 mm 4 = 99.9654 × 106 mm 4

or I y = 100.0 × 106 mm 4 and

k y2 =

Iy Atotal

=

99.9654 × 106 mm 4 14,196 mm 2

or

k y = 83.9 mm


PROBLEM 9.52 Two channels are welded to a d × 12-in. steel plate as shown. Determine the width d for which the ratio I x /I y of the centroidal moments of inertia of the section is 16.

SOLUTION Channel:

A = 4.48 in.2 I x = 67.3 in.4 I y = 2.27 in.4

Now

I x = 2( I x )C + ( I x )plate 1 (d in.)(12 in.)3 12 = (134.6 + 144d )in.4 = 2(67.3 in.4 ) +

and where

I y = 2( I y )C + ( I y ) plate ( I y )C = I y + Ad 2 $& d ' % + 0.634 ! in.) ( I y )C = 2.27 in. + (4.48 in. ) ( # + *" 2 4

( I y ) plate

Then

2

2

= (1.1200d 2 + 2.84032d + 4.07076) in.4 1 = (12 in.)(d in.)3 = d 3 in.4 12

I y = [2(1.1200d 2 + 2.84032d + 4.07076) + d 3 ] in.4 = ( d 3 + 2.240d 2 + 5.68064d + 8.14152) in.4

Now or

Ix = 16: (134.6 + 144d ) = 16( d 3 + 2.240d 2 + 5.68064d + 8.14152) Iy d 3 + 2.240d 2 − 3.31936d − 0.27098 = 0

Solving yields d = 1.07669 in. (the other roots are negative). d = 1.077 in.


PROBLEM 9.53 Two L76 × 76 × 6.4-mm angles are welded to a C250 × 22.8 channel. Determine the moments of inertia of the combined section with respect to centroidal axes respectively parallel and perpendicular to the web of the channel.

SOLUTION A = 929 mm 2

Angle:

I x = I y = 0.512 × 106 mm 4 A = 2890 mm 2

Channel:

I x = 0.945 × 106 mm 4 I y = 28.0 × 106 mm 4

First locate centroid C of the section

Then

A, mm 2

y , mm

yA, mm3

Angle

2(929) = 1858

21.2

39,389.6

Channel

2890

−16.1

−46,529

Σ

4748

−7139.4

Y Σ A = Σ y A: Y (4748 mm 2 ) = −7139.4 mm3 Y = −1.50366 mm

or Now where

I x = 2( I x ) L + ( I x )C ( I x ) L = I x + Ad 2 = 0.512 × 106 mm 4 + (929 mm 2 )[(21.2 + 1.50366) mm]2

= 0.990859 × 106 mm 4 ( I x )C = I x + Ad 2 = 0.949 × 106 mm 4 + (2890 mm 2 )[(16.1 − 1.50366) mm]2

= 1.56472 × 106 mm 4

Then

I x = [2(0.990859) + 1.56472 × 106 mm 4

or

I x = 3.55 × 106 mm 4



Also where

I y = 2( I y ) L + ( I y )C ( I y ) L = I y + Ad 2 = 0.512 × 106 mm 4 + (929 mm 2 )[(127 − 21.2) mm]2

= 10.9109 × 106 mm 4 ( I y )C = I y

Then

I y = [2(10.9109) + 28.0] × 106 mm 4

or

I y = 49.8 × 106 mm 4


PROBLEM 9.54 Two L4 × 4 × 12 -in. angles are welded to a steel plate as shown. Determine the moments of inertia of the combined section with respect to centroidal axes respectively parallel and perpendicular to the plate.

SOLUTION

1 For 4 × 4 × -in. angle: 2

A = 3.75 in.2 , I x = I y = 5.52 in.4 YA = Σ y A Y (12.5 in.2 ) = 33.85 in.3 Y = 2.708 in.

Area, in.2

y in.

yA, in.3

Plate

(0.5)(10) = 5

5

25

Two angles

2(3.75) = 7.5

1.18

8.85

Section

Σ

12.5

33.85



Entire section: $1 % I x = Σ( I x′ + Ad 2 ) = ( (0.5)(10)3 + (0.5)(10)(2.292) 2 ) + 2[5.52 + (3.75)(1.528) 2 ] *12 + = 41.667 + 26.266 + 1604 + 17.511 = 96.48 in.4 Iy =

I x = 96.5 in.4

1 (10)(0.5)3 + 2[5.52 + (3.75)(1.43) 2 ] 12

= 0.104 + 11.04 + 15.367 = 26.51 in.4

I y = 26.5 in.4


PROBLEM 9.55 The strength of the rolled W section shown is increased by welding a channel to its upper flange. Determine the moments of inertia of the combined section with respect to its centroidal x and y axes.

SOLUTION A = 14, 400 mm 2

W Section:

I x = 554 × 106 mm 4 I y = 63.3 × 106 mm 4 A = 2890 mm 2

Channel:

I x = 0.945 × 106 mm 4 I y = 28.0 × 106 mm 4

First locate centroid C of the section.

W Section Channel Σ

Then

A, mm2

y , mm

yA, mm3

14400

−231

−3326400

2890

49.9

144211

17290

−3182189

Y Σ A = Σ y A: Y (17, 290 mm 2 ) = −3,182,189 mm3 Y = −184.047 mm

or Now where

I x = ( I x ) W + ( I x )C

( I x ) W = I x + Ad 2 = 554 × 106 mm 4 + (14, 400 mm 2 )(231 − 184.047) 2 mm 2 = 585.75 × 106 mm 4 ( I x )C = I x − Ad 2 = 0.945 × 106 mm 4 + (2890 mm 2 )(49.9 + 184.047) 2 mm 2 = 159.12 × 106 mm 4



Then

also

I x = (585.75 + 159.12) × 106 mm 4

or

I x = 745 × 106 mm 4

or

I y = 91.3 × 106 mm 4

I y = ( I y ) W + ( I y )C = (63.3 + 28.0) × 106 mm 4


PROBLEM 9.56 Two L5 × 3 × 12 -in. angles are welded to a 12 -in. steel plate. Determine the distance b and the centroidal moments of inertia I x and I y of the combined section, knowing that I y = 4 I x .

SOLUTION A = 3.75in.2

Angle:

I x = 9.43 in.4 I y = 2.55 in.4

First locate centroid C of the section.

Area, in.2

y , in.

yA, in.3

Angle

2(3.75) = 7.50

1.74

13.05

Plate

(10)(0.5) = 5

−0.25

−1.25

Σ

Then

12.50

11.80

Y Σ A = Σ yA: Y (12.50 in.2 ) = 11.80 in.3 Y = 0.944 in.

or Now where

I x = 2( I x )angle + ( I x ) plate ( I x )angle = I x + Ad 2 = 9.43 in.4 + (3.75 in.2 )[(1.74 − 0.944) in.]2

( I x ) plate

Then

= 11.8061 in.4 1 = I x + Ad 2 = (10 in.)(0.5 in.)3 + (5 in.2 )[(0.25 + 0.944) in.]2 12 = 7.2323 in.4

I x = [2(11.8061) + 7.2323] in.4 = 30.8445 in.4

or

I x = 30.8 in.4



We have

I y = 4 I x = 4(30.8445 in.4 ) = 123.378 in.4

or Now where

I y = 2( I y )angle + ( I y )plate ( I y )angle = I y + Ad 2 = 2.55 in.4 + (3.75 in.2 )[(b − 0.746) in.]2 ( I y ) plate =

Then

I y = 123.4 in.4

1 (0.5 in.)(10 in.)3 = 41.6667 in.4 12

123.378 in.4 = 2[2.55 + 3.75(b − 0.746)2 ] in.4 + 41.6667 in.4

or

b = 3.94 in.


PROBLEM 9.57 The panel shown forms the end of a trough that is filled with water to the line AA′. Referring to Section 9.2, determine the depth of the point of application of the resultant of the hydrostatic forces acting on the panel (the center of pressure).

SOLUTION From Section 9.2:

,

,

R = γ y dA, M AA′ = γ y 2 dA

Let yP = distance of center of pressure from AA′. We must have

, ,

2 M AA′ γ y dA I AA′ RyP = M AA′ : yP = = = R γ y dA y A

(1)

For semicircular panel: I AA′ =

π 8

r4

y=

4r 3π

A=

π 2

r2

π

r4 I AA′ 8 yP = = yA & 4r ' π 2 ! r " 3π # 2

yP =

3π r 16



SOLUTION See solution of Problem 9.57 for derivation of Eq. (1): yP =

I AA′ yA

(1)

For quarter ellipse panel: I AA′ =

π 16

ab3

y=

4b 3π

A=

π 4

ab

π

ab3 I AA′ 16 yP = = yA & 4b '& π ' ab ! ! " 3π #" 4 #

yP =

3π b 16



SOLUTION Using the equation developed on page 491 of the text: yP =

Now

I AA′ yA

YA = Σ yA 4 &1 h ' (2b × h) + h × 2b × h ! 2 3 "2 # 7 2 = bh 3 =

and

I AA′ = ( I AA′ )1 + ( I AA′ ) 2

where 1 2 ( I AA′ )1 = (2b)(h)3 = bh3 3 3 1 &1 '& 4 ' (2b)( h)3 + × 2b × h ! h ! 36 "2 #" 3 # 11 = bh3 6

2

( I AA′ ) 2 = I x + Ad 2 =

Then

Finally,

I AA′ =

2 3 11 3 5 3 bh + bh = bh 3 6 2

5 3 bh yP = 2 7 2 bh 3

or

yP =

15 h 14


PROBLEM 9.60* The panel shown forms the end of a trough that is filled with water to the line AA′. Referring to Section 9.2, determine the depth of the point of application of the resultant of the hydrostatic forces acting on the panel (the center of pressure).

SOLUTION Using the equation developed on page 491 of the text:

For a parabola: Now By observation

yP =

I AA′ yA

y=

2 h 5

A=

4 ah 3

1 dI AA′ = (h − y )3 dx 3

y=

h 2 x a2 3

So that

Then

dI AA′ =

1& 1 h3 2 h ' (a − x 2 )3 dx h − 2 x 2 ! dx = 6 3" 3 a a # 1 h 6 (a − 3a 4 x 2 + 3a 2 x 4 − x 6 )dx = 6 3a

I AA′ = 2

,

a

0

1 h3 6 (a − 3a 4 x 2 + 3a 2 x 4 − x 6 )dx 6 3a a

2 h3 $ 6 3 1 % a x − a 4 x3 + a 2 x5 − x 7 ) 3 a 6 (* 5 7 +0 32 3 = ah 105 =

Finally,

32 3 ah 105 yp = 2 4 h × ah 5 3

or

yP =

4 h 7


PROBLEM 9.61 The cover for a 0.5-m-diameter access hole in a water storage tank is attached to the tank with four equally spaced bolts as shown. Determine the additional force on each bolt due to the water pressure when the center of the cover is located 1.4 m below the water surface.

SOLUTION From Section 9.2: R = γ yA

yP =

I AA′ yA

where R is the resultant of the hydrostatic forces acting on the cover and yP is the depth to the point of application of R. Recalling that γ = p ⋅ y , we have R = (103 kg/m3 × 9.81 m/s 2 )(1.4 m)[π (0.25 m) 2 ] = 2696.67 N I AA′ = I x + Ay 2 =

Also

π

(0.25 m) 4 + [π (0.25 m)2 ](1.4 m) 2

4 = 0.387913 m 4

yP =

Then

0.387913 m 4 = 1.41116 m (1.4 m)[π (0.25 m) 2 ]

Now note that symmetry implies FA = FB

FC = FD

Next consider the free-body of the cover. We have

ΣM CD = 0: [2(0.32 m) sin 45°](2 FA ) − [0.32 sin 45° − (1.41116 − 1.4)] m × (2696.67 N) = 0

or Then or

FA = 640.92 N ΣFz = 0: 2(640.92 N) + 2 FC − 2696.67 N = 0 FC = 707.42 N FA = FB = 641 N FC = FD = 707 N

and


PROBLEM 9.62 A vertical trapezoidal gate that is used as an automatic valve is held shut by two springs attached to hinges located along edge AB. Knowing that each spring exerts a couple of magnitude 1470 N · m, determine the depth d of water for which the gate will open.

SOLUTION From Section 9.2: R = γ yA

yP =

I SS ′ yA

where R is the resultant of the hydrostatic forces acting on the gate and yP is the depth to the point of application of R. Now &1 ' &1 ' × 1.2 m × 0.51 m ! + [( h + 0.34) m] × 0.84 × 0.51 m ! "2 # "2 # 3 = (0.5202h + 0.124848) m

yA = Σ yA = [(h + 0.17) m]

Recalling that γ = py, we have R = (103 kg/m3 × 9.81 m/s 2 )(0.5202h + 0.124848) m3 = 5103.162(h + 0.24) N

Also where

I SS ′ = ( I SS ′ )1 + ( I SS ′ ) 2 ( I SS ′ )1 = I x + Ad 2 =

1 &1 ' (1.2 m)(0.51 m)3 + × 1.2 m × 0.51 m ! [(h + 0.17) m]2 36 2 " #

= [0.0044217 + 0.306(h + 0.17)2 ] m 4 = (0.306h2 + 0.10404h + 0.0132651) m 4 ( I SS ′ ) 2 = I X 2 + Ad 2 =

1 &1 ' (0.84 m)(0.51 m)3 + × 0.84 m × 0.51 m ! [(h + 0.34) m]2 36 "2 #

= [0.0030952 + 0.2142( h + 0.34) 2 ] m 4 = (0.2142h 2 + 0.145656h + 0.0278567) m 4



I SS ′ = ( I SS ′ )1 + ( I SS ′ ) 2

Then

= (0.5202h 2 + 0.249696h + 0.0411218) m 4 yP =

and

=

(0.5202h2 + 0.244696h + 0.0411218) m 4 (0.5202h + 0.124848) m3 h 2 + 0.48h + 0.07905 m h + 0.24

For the gate to open, require that ΣM AB : M open = ( yP − h) R

Substituting or

2940 N ⋅ m =

& h 2 + 0.48h + 0.07905 ' − h !! m × 5103.162( h + 0.24) N h + 0.24 " #

5103.162(0.24h + 0.07905) = 2940

or

h = 2.0711 m

Then

d = (2.0711 + 0.79) m

or

d = 2.86 m


PROBLEM 9.63* Determine the x coordinate of the centroid of the volume shown. (Hint: The height y of the volume is proportional to the x coordinate; consider an analogy between this height and the water pressure on a submerged surface.)

SOLUTION

y=

First note that Now where

Then

,

h x a

,

x dV = xEL dV xEL = x dV = y dA =

&h ' x ! dA "a #

, x ( x dA) = , x dA = ( I ) x= , x dA , x dA ( xA) 2

h a

z A

h a

A

where ( I z ) A and ( x A ) A pertain to area. OABC: ( I z ) A is the moment of inertia of the area with respect to the z axis, x A is the x coordinate of the centroid of the area, and A is the area of OABC. Then ( I z ) A = ( I z ) A1 + ( I z ) A2 1 1 = (b)(a)3 + (b)( a)3 3 12 5 3 = ab 12

and

( xA) A = Σ xA $& a ' % $& a '& 1 '% = ( ! ( a × b) ) + ( ! × a × b ! ) #+ *" 2 # + *" 3 #" 2 2 = a 2b 3



Finally,

x=

5 3 ab 12 2 a 2b 3

or

5 x= a 8

Analogy with hydrostatic pressure on a submerged plate: Recalling that P = γ y, it follows that the following analogies can be established. Height y ~ P dV = y dA ~ p dA = dF xdV = x( y dA) ~ y dF = dM

Recalling that

It can then be concluded that

yP =

& dM ' Mx ! = R dF !! " #

, ,

x ~ yP


!

PROBLEM 9.64* Determine the x coordinate of the centroid of the volume shown; this volume was obtained by intersecting an elliptic cylinder with an oblique plane. (Hint: The height y of the volume is proportional to the x coordinate; consider an analogy between this height and the water pressure on a submerged surface.)

SOLUTION Following the “Hint,” it can be shown that (see solution to Problem 9.63) x=

(I z ) A x ( xA) A

where ( I z ) A and ( xA) A are the moment of inertia and the first moment of the area, respectively, of the elliptical area of the base of the volume with respect to the z axis. Then ( I z ) A = I z + Ad 2

π

(39 mm)(64 mm)3 + [π (64 mm)(39 mm)](64 mm) 2 4 = 12.779520π × 106 mm 4 =

( xA) A = (64 mm)[π (64 mm)(39 mm)] = 0.159744π × 106 mm3

Finally

x=

12.779520π × 106 mm 4 0.159744π × 106 mm 4

or

x = 80.0 mm


!

PROBLEM 9.65* Show that the system of hydrostatic forces acting on a submerged plane area A can be reduced to a force P at the centroid C of the area and two couples. The force P is perpendicular to the area and is of magnitude P = γ Ay sin θ , where γ is the specific weight of the liquid, and the couples are M x′ = (γ I x′ sin θ )i and M y′ = (γ I x′y ′ sin θ ) j, where I x′y′ = , x′y ′ dA (see Section 9.8). Note that the couples are independent of the depth at which the area is submerged.

SOLUTION The pressure p at an arbitrary depth ( y sin θ ) is p = γ ( y sin θ )

so that the hydrostatic force dF exerted on an infinitesimal area dA is dF = (γ y sin θ )dA

Equivalence of the force P and the system of infinitesimal forces dF requires

,

,

,

ΣF : P = dF = γ y sin θ dA = γ sin θ y dA

or

P = γ Ay sin θ

Equivalence of the force and couple (P, M x′ + M y′ ) and the system of infinitesimal hydrostatic forces requires

,

ΣM x : − yP − M x′ = (− y dF )

Now

,

,

,

− y dF = − y (γ y sin θ ) dA = −γ sin θ y 2 dA = −(γ sin θ ) I x

Then or

− yP − M x′ = −(γ sin θ ) I x M x′ = (γ sin θ ) I x − y (γ Ay sin θ ) = γ sin θ ( I x − Ay 2 )

or

M x′ = γ I x′ sin θ



,

ΣM y : xP + M y′ = x dF

Now

, x dF = , x(γ y sin θ )dA = γ sin θ , xy dA = (γ sin θ ) I xy

Then or

(Equation 9.12)

xP + M y′ = (γ sin θ ) I xy M y′ = (γ sin θ ) I xy − x (γ Ay sin θ ) = γ sin θ ( I xy − Ax y )

or

M y′ = γ I x′y′ sin θ


!

PROBLEM 9.66* Show that the resultant of the hydrostatic forces acting on a submerged plane area A is a force P perpendicular to the area and of magnitude P = γ Ay sin θ = pA, where γ is the specific weight of the liquid and p is the pressure at the centroid C of the area. Show that P is applied at a Point CP, called the center of pressure, whose coordinates are xP = I xy /Ay and yP = I x /Ay , where I xy = , xy dA (see Section 9.8). Show also that the difference of ordinates yP − y is equal to k x2′ / y and thus depends upon the depth at which the area is submerged.

SOLUTION The pressure P at an arbitrary depth ( y sin θ ) is P = γ ( y sin θ )

so that the hydrostatic force dP exerted on an infinitesimal area dA is dP = (γ y sin θ )dA

The magnitude P of the resultant force acting on the plane area is then

,

,

,

P = dP = γ y sin θ dA = γ sin θ y dA = γ sin θ ( yA)

Now

p = γ y sin θ

P = pA

Next observe that the resultant P is equivalent to the system of infinitesimal forces dP. Equivalence then requires

,

ΣM x : − yP P = − y dP

Now

, y dP = , y(γ y sin θ )dA = γ sin θ , y dA 2

= (γ sin θ ) I x

Then or

yP P = (γ sin θ ) I x yP =

(γ sin θ ) I x γ sin θ ( yA)

or

yP =

Ix Ay



,

ΣM y : xP P = x dP

Now

, x dP = , x(γ y sin θ )dA = γ sin θ , xy dA = (γ sin θ ) I xy

Then or

(Equation 9.12)

xP P = (γ sin θ ) I xy xP =

(γ sin θ ) I xy

γ sin θ ( yA) or

Now

I x = I x′ + A y 2

From above

I x = ( A y ) yP

By definition

I x′ = k x′ A

Substituting

( Ay ) yP = k x2′ A + Ay 2

xP =

I xy

!

Ay

2

yP − y =

Rearranging yields

k x2′ y

Although k x′ is not a function of the depth of the area (it depends only on the shape of A), y is dependent on the depth. ( yP − y ) = f (depth) !


!

PROBLEM 9.67 Determine by direct integration the product of inertia of the given area with respect to the x and y axes.

SOLUTION y = a 1−

First note

=

x2 4a 2

1 4a 2 − x 2 2

We have

dI xy = dI x′y′ + xEL yEL dA

where

dI x′y′ = 0 (symmetry) xEL = x yEL =

1 1 4a 2 − x 2 y= 2 4

dA = ydx =

Then

1 4a 2 − x 2 dx 2

,

2a

I xy = dI xy =

,

=

1 8

=

a4 8

0

,

x

&1 '& 1 4a 2 − x 2 ! 4a 2 − x 2 4 2 " #"

2a 0

' ! dx # 2a

1$ 1 % (4a 2 x − x3 )dx = ( 2a 2 x 2 − x 4 ) 8* 4 +0

1 4% $ 2 ( 2(2) − 4 (2) ) * +

or

I xy =

1 4 a 2



SOLUTION At

x = a, y = b: a = kb 2

or

k=

a b2

Then

x=

a 2 y b2

We have


where

dI x′y′ = 0 (symmetry) xEL = yEL = y dA = x dy =

Then

,

I xy = dI xy =

=

,

1 a x = 2 y2 2 2b

a 2 y dy b2

b& 0

a 2' & a 2 ' y ! ( y ) 2 y dy ! 2 " 2b # "b #

a2 2b 4

,

b 0

y 5 dy =

a2 2b4

b

$1 6 % (6 y ) * +0

or

I xy =

1 2 2 a b 12



SOLUTION First note that h y=− x b

Now


where

dI x′y′ = 0 (symmetry) xEL = x

yEL =

1 1h y=− x 2 2b

h dA = y dx = − x dx b

Then

,

I xy = dI xy =

=

,

& 1 h '& h ' x − x ! − x dx ! −b " 2 b #" b # 0

1 h2 2 b2

,

0

−b

0

x3 dx =

1 h2 $ 1 4 % x 2 b 2 (* 4 )+ −b

or

1 I xy = − b 2 h 2 8



SOLUTION y = h1 + ( h2 − h1 )

x a

dI xy = dI x′y′ + xEL yEL dA dI x′y′ = 0 (by symmetry) xEL = x

yEL =

,

I xy = dI xy − =

1 2

=

1 2

, ,

a 0

,

a 0

x

1 y 2

dA = y dx

1 &1 ' y ! y dx = 2 "2 #

,

a 0

xy 2 dx

2

x% $ x ( h1 + (h2 − h1 ) ) dx a+ *

a$ 0

3 x2 2 2 x % ( h1 x + 2h1 (h2 − h1 ) + (h2 − h1 ) 2 ) dx a a + *

1 $ 2 a2 a3 a4 % + 2h1 ( h2 − h1 ) + (h2 − h1 ) 2 2 ) ( h1 2* 2 3a 4a + % 1 $ a2 2 2 1 1 1 = ( h12 + h1h2 a 2 − h12 a 2 + h22 a 2 − h2 h1a + h12 a 2 ) 2* 2 3 3 4 2 4 + 2 a $ 2&1 2 1' & 2 1 ' 2 & 1 '% = − + ! + h1h2 − ! + h2 ( h1 !) 2 * "2 3 4# "3 2# " 4 #+

=

=

a2 2

$ 2& 1 ' &1' & 1 '% + h1h2 ! + h22 ( h1 ! !) "6# " 4 #+ * " 12 # I xy =

a2 2 h1 + 2h1h2 + 3h22 24

(

)



The following table is provided for the convenience of the instructor, as many problems in this and the next lesson are related. Type of Problem Compute I x and I y

Figure 9.12

Compute I xy

9.67

9.72

9.73

9.74

9.75

9.78

9.79

9.80

9.81

9.83

9.82

9.84

9.85

9.86

9.87

9.89

9.88

9.90

9.91

9.92

9.93

9.95

9.94

9.96

9.97

9.98

9.100

9.101

9.103

9.106

I x′, I y′, I x′y′ by equations

Principal axes by equations

Figure 9.13B

Figure 9.13A

I x′, I y′, I x′y′ by Mohr's circle

Principal axes by Mohr’s circle !


PROBLEM 9.71 Using the parallel-axis theorem, determine the product of inertia of the area shown with respect to the centroidal x and y axes.

SOLUTION Dimensions in mm

I xy = ( I xy )1 + ( I xy ) 2 + ( I xy )3

We have Now symmetry implies

( I xy ) 2 = 0

and for the other rectangles I xy = I x′y′ + x yA I x′y′ = 0 (symmetry)

where

I xy = ( x yA)1 + ( x yA)3

Thus A, mm 2

x , mm

y , mm

x yA, mm 4

1

10 × 80 = 800

−55

20

−880, 000

3

10 × 80 = 800

55

−20

−880, 000 −1, 760,000

Σ

I xy = −1.760 × 106 mm 4


!


SOLUTION

Dimensions in mm

I xy = ( I xy )1 − ( I xy ) 2 − ( I xy )3

We have Now symmetry implies and for each triangle

( I xy )1 = 0 I xy = I x′y′ + x yA

1 2 2 b h for both triangles. Note that the sign of I x′y′ where, using the results of Sample Problem 9.6, I x′y′ = − 72

is unchanged because the angles of rotation are 0° and 180° for triangles 2 and 3, respectively. Now A, mm 2

x , mm

x yA, mm 4

2

1 (120)(60) = 3600 2

−80

−40

11.520 × 106

3

1 (120)(60) = 3600 2

80

40

11.520 × 106 23.040 × 106

Σ

Then

y , mm

- $ 1 . % I xy = − /2 ( − (120 mm) 2 (60 mm) 2 ) + 23.040 × 106 mm 4 0 72 + 1 * 2

or

I xy = −21.6 × 106 mm 4


!


SOLUTION We have

I xy = ( I xy )1 + ( I xy ) 2

For each semicircle

I xy = I x′y′ + x yA

and

I x′y′ = 0 (symmetry)

Thus

I xy = Σ x yA A, in.2

1

2

π 2

π 2

x , in.

(6) 2 = 18π

−3

(6) 2 = 18π

3

Σ

y , in. −

8

π

8

π

xyA

432

432 864 I xy = 864 in.4


!


SOLUTION

I xy = ( I xy )1 + ( I xy ) 2

We have For each rectangle

I xy = I x′y′ + x yA I x′y′ = 0 (symmetry)

and

I xy = Σ x yA

Thus

A, in.2

x , in.

y , in.

x yA, in.4

1

3 × 0.25 = 0.75

−0.520

0.362

−0.141180

2

0.25 × 1.75 = 0.4375

0.855

−0.638

−0.238652 −0.379832

Σ

I xy = −0.380 in.4

!


!


SOLUTION I xy = ( I xy )1 + ( I xy ) 2 + ( I xy )3

We have

( I xy )1 = 0

Now symmetry implies and for the other rectangles


where


Thus

I xy = ( x yA) 2 + ( x yA)3 A, mm 2

X , mm

y , mm

x yA, mm 4

2

8 × 32 = 256

–46

–20

235,520

3

8 × 32 = 256

46

20

235,520

Σ

471,040 I xy = 471 × 103 mm 4


!


SOLUTION I xy = ( I xy )1 + ( I xy ) 2 + ( I xy )3

We have Now, symmetry implies

( I xy )1 = 0 I xy = I x′y′ + x yA

and for each triangle

1 2 2 where, using the results of Sample Problem 9.6, I x′y′ = − 72 b h for both triangles. Note that the sign of I x′y′

is unchanged because the angles of rotation are 0° and 180° for triangles 2 and 3, respectively. Now A, in.2

x , in.

y , in.

x yA, in.4

2

1 (9)(15) = 67.5 2

–9

7

–4252.5

3

1 (9)(15) = 67.5 2

9

–7

–4252.5

Σ

Then

–8505 $ 1 % I xy = 2 ( − (9 in.) 2 (15 in.) 2 ) − 8505 in.4 72 * + = −9011.25 in.4

or

I xy = −9010 in.4


!


SOLUTION We have

I xy = ( I xy )1 + ( I xy ) 2 + ( I xy )3

For each rectangle


and

I x′y′ = 0 (symmetry) I xy = Σ x yA

Thus

A, in.2

x , in.

y , in.

1

3.6 × 0.5 = 1.8

0.888

–2.00

2

0.5 × 3.8 = 1.9

–0.662

0.15

–0.18867

3

1.3 × 1.0 = 1.3

–0.262

2.55

–0.86853

Σ

x yA, in.4

–3.196

–4.25320 I xy = −4.25 in.4


!


SOLUTION We have

I xy = ( I xy )1 + ( I xy ) 2

For each rectangle


and


Thus

I xy = Σ x yA

x yA, mm 4

A, mm 2

x , mm

y , mm

1

76 × 12.7 = 965.2

–19.1

–37.85

697,777

2

12.7 × (127 − 12.7) = 1451.61

12.55

25.65

467,284

Σ

1,165,061 I xy = 1.165 × 106 mm 4


!

PROBLEM 9.79 Determine for the quarter ellipse of Problem 9.67 the moments of inertia and the product of inertia with respect to new axes obtained by rotating the x and y axes about O (a) through 45° counterclockwise, (b) through 30° clockwise.

SOLUTION From Figure 9.12:

Ix = = Iy = =

From Problem 9.67: First note

π 16

π 8

a4

π 16

π 2

I xy =

(2a )( a)3

(2a )3 (a)

a4

1 4 a 2

1 1&π 4 π 4 ' 5 (I x + I y ) = a + a ! = π a4 2 2" 8 2 # 16 1 1&π 4 π 4 ' 3 (I x − I y ) = a − a ! = − π a4 2 2" 8 2 # 16

Now use Equations (9.18), (9.19), and (9.20). Equation (9.18):

I x′ = =

Equation (9.19):

I y′ = =

Equation (9.20):

I x′y′ =

1 1 ( I x + I y ) + ( I x − I y ) cos 2θ − I xy sin 2θ 2 2 5 3 1 π a 4 − π a 4 cos 2θ − a 4 sin 2θ 16 16 2 1 1 ( I x + I y ) − ( I x − I y ) cos 2θ + I xy sin 2θ 2 2 5 3 1 π a 4 + π a 4 cos 2θ + a 4 sin 2θ 16 16 2 1 ( I x − I y ) sin 2θ + I xy cos 2θ 2

=−

3 1 π a 4 sin 2θ + a 4 cos 2θ 16 2



(a)

θ = +45°:

I x′ =

I y′ =

5 3 1 π a 4 − π a 4 cos 90° − a 4 sin 90° 16 16 2

or

I x′ = 0.482a 4

or

I y′ = 1.482a 4

5 3 1 π + π a 4 cos 90° + a 4 16 16 2

I x′y′ = −

3 1 π a 4 sin 90° + a 4 cos 90° 16 2 or I x′y′ = −0.589a 4

(b)

θ = −30° :

I x′ =

I y′ =

5 3 1 π a 4 − π a 4 cos(−60°) − a 4 sin( −60°) 16 16 2

or

I x′ = 1.120a 4

or

I y′ = 0.843a 4

or

I x′y′ = 0.760a 4

5 3 1 π a 4 + π a 4 cos(−60°) + a 4 sin( −60°) 16 16 2

I x′y′ = −

3 1 π a 4 sin(−60°) + a 4 cos(−60°) 16 2


!

PROBLEM 9.80 Determine the moments of inertia and the product of inertia of the area of Problem 9.72 with respect to new centroidal axes obtained by rotating the x and y axes 30° counterclockwise.

SOLUTION From Problem 9.72:

I xy = −21.6 × 106 mm 4

Now

I x = ( I x )1 − ( I x ) 2 − ( I x )3

where 1 (240 mm)(160 mm)3 12 = 81.920 × 106 mm 4

( I x )1 =

( I x ) 2 = ( I x )3 =

1 (120 mm)(60 mm)3 36 $1 % + ( (120 mm)(60 mm) ) (40 mm) 2 *2 +

= 6.480 × 106 mm 4

Then

I x = [81.920 − 2(6.480)] × 106 mm 4 = 68.96 × 106 mm 4

Also

I y = ( I y )1 − ( I y ) 2 − ( I y )3

where

1 (160 mm)(240 mm)3 = 184.320 × 106 mm 4 12 1 ( I y ) 2 = ( I y )3 = (60 mm)(120 mm)3 36 $1 % + ( (120 mm)(60 mm) ) (80 mm) 2 2 * + ( I y )1 =

= 25.920 × 106 mm 4

Then

I y = [184.320 − 2(25.920)] × 106 mm 4 = 132.48 × 106 mm 4



Now

1 (I x + I y ) = 2 1 (I x − I y ) = 2

1 (68.96 + 132.48) × 106 = 100.72 × 106 mm 4 2 1 (68.96 − 132.48) × 106 = −31.76 × 106 mm 4 2

Using Eqs. (9.18), (9.19), and (9.20): Eq. (9.18):

1 1 ( I x + I y ) + ( I x − I y ) cos 2θ − I xy sin 2θ 2 2 = [100.72 + (−31.76) cos 60° − (−21.6)sin 60°] × 106 mm 4

Ix =

or Eq. (9.19):

1 1 ( I x + I y ) − ( I x − I y ) cos 2θ + I xy sin 2θ 2 2 = [100.72 − ( −31.76) cos 60° + (−21.6) sin 60°] × 106 mm 4

I y′ =

or Eq. (9.20):

I x′ = 103.5 × 106 mm 4

I y′ = 97.9 × 106 mm 4

1 ( I x − I y ) sin 2θ + I xy sin 2θ 2 = [−31.76sin 60° + (−21.6) cos 60°] × 106 mm 4

I x′y′ =

or

I x′y′ = −38.3 × 106 mm 4


!

PROBLEM 9.81 Determine the moments of inertia and the product of inertia of the area of Problem 9.73 with respect to new centroidal axes obtained by rotating the x and y axes 60° counterclockwise.

SOLUTION From Problem 9.73: I xy = 864 in.4 I x = ( I x )1 + ( I x ) 2

Now

π

(6 in.) 4 8 = 162π in.4

( I x )1 = ( I x ) 2 =

where

Then

I x = 2(162π in.4 ) = 324π in.4

Also

I y = ( I y )1 + ( I y ) 2 ( I y )1 = ( I y ) 2 =

where

8

$π % (6 in.) 4 + ( (6 in.) 2 ) (3 in.) 2 = 324π in.4 *2 +

I y = 2(324π in.4 ) = 648π in.4

Then Now

π

1 (I x + I y ) = 2 1 (I x − I y ) = 2

1 (324π + 648π ) = 486π in.4 2 1 (324π − 648π ) = −162π in.4 2

Using Eqs. (9.18), (9.19), and (9.20): Eq. (9.18):

1 1 ( I x + I y ) + ( I x − I y ) cos 2θ − I xy sin 2θ 2 2 = [486π + (−162π ) cos120° − 864sin120°] in.4

I x′ =

or

I x′ = 1033 in.4



Eq. (9.19):

Eq. (9.20):

1 1 ( I x + I y ) − ( I x − I y ) cos 2θ + I xy sin 2θ 2 2 = [486π − (−162π ) cos120° + 864sin120°] in.4

I y′ =

or

I y′ = 2020 in.4

or

I x′y′ = −873 in.4

1 ( I x − I y )sin 2θ + I xy cos 2θ 2 = [(−162π ) sin120° + 864cos120°] in.4

I x′y′ =


!

PROBLEM 9.82 Determine the moments of inertia and the product of inertia of the area of Problem 9.75 with respect to new centroidal axes obtained by rotating the x and y axes 45° clockwise.

SOLUTION From Problem 9.75: I xy = 471, 040 mm 4

Now where

I x = ( I x )1 + ( I x )2 + ( I x )3 1 (100 mm)(8 mm)3 12 = 4266.67 mm 4

( I x )1 =

1 (8 mm)(32 mm)3 + [(8 mm)(32 mm)](20 mm) 2 12 = 124, 245.33

( I x ) 2 = ( I x )3 =

Then

I x = [4266.67 + 2(124, 245.33)] mm 4 = 252, 757 mm 4

Also

I y = ( I y )1 + ( I y ) 2 + ( I y )3

where

Then Now

1 (8 mm)(100 mm)3 = 666, 666.7 mm 4 12 1 ( I y ) 2 = ( I y )3 = (32 mm)(8 mm)3 + [(8 mm)(32 mm)](46 mm) 2 12 = 543,061.3 mm 4 ( I y )1 =

I y = [666, 666.7 + 2(543,061.3)] mm 4 = 1, 752, 789 mm 4

1 (I x + I y ) = 2 1 (I x − I y ) = 2

1 (252, 757 + 1, 752, 789) mm 4 = 1, 002,773 mm 4 2 1 (252, 757 − 1, 752, 789) mm 4 = −750, 016 mm 4 2

Using Eqs. (9.18), (9.19), and (9.20): Eq. (9.18):

1 1 ( I x + I y ) + ( I x − I y ) cos 2θ − I xy sin 2θ 2 2 = [1, 002,773 + (−750,016) cos(−90°) − 471, 040sin( −90°)]

I x′ =

or

I x′ = 1.474 × 106 mm 4


!


Eq. (9.19):

1 1 ( I x + I y ) − ( I x − I y ) cos 2θ + I xy sin 2θ 2 2 = [1,002, 773 − (−750, 016) cos(−90°) + 471,040sin(−90°)]

I y′ =

or Eq. (9.20):

I y′ = 0.532 × 106 mm 4

1 ( I x − I y ) sin 2θ + I xy cos 2θ 2 = [( −750, 016) sin(−90°) + 471,040cos( −90°)]

I x′y′ =

or

I x′y′ = 0.750 × 106 mm 4


!

PROBLEM 9.83 Determine the moments of inertia and the product of inertia of the L3 × 2 × 14 -in. angle cross section of Problem 9.74 with respect to new centroidal axes obtained by rotating the x and y axes 30° clockwise.

SOLUTION From Figure 9.13: I x = 0.390 in.4 I y = 1.09 in.4

From Problem 9.74: I xy = −0.37983 in.4

Now

1 (I x + I y ) = 2 1 (I x − I y ) = 2

1 (0.390 + 1.09) in.4 = 0.740 in.4 2 1 (0.390 − 1.09) in.4 = −0.350 in.4 2

Using Eqs. (9.18), (9.19), and (9.20): Eq. (9.18):

1 1 ( I x + I y ) + ( I x − I y ) cos 2θ − I xy sin 2θ 2 2 = [0.740 + (−0.350) cos(−60°) − (−0.37983) sin( −60°)]

I x′ =

or Eq. (9.19):

1 1 ( I x + I y ) − ( I x − I y ) cos 2θ + I xy sin 2θ 2 2 = [0.740 − (−0.350) cos(−60°) + ( −0.37983)sin(−60°)]

I y′ =

or Eq. (9.20):

I x′ = 0.236 in.4

I y′ = 1.244 in.4

1 ( I x − I y )sin 2θ + I xy cos 2θ 2 = [(−0.350) sin(−60°) + (−0.37983) cos(−60°)]

I x′y′ =

or I x′y′ = 0.1132 in.4


!

PROBLEM 9.84 Determine the moments of inertia and the product of inertia of the L127 ×76 × 12.7-mm angle cross section of Problem 9.78 with respect to new centroidal axes obtained by rotating the x and y axes 45° counterclockwise.

SOLUTION From Figure 9.13: I x = 3.93 × 106 mm 4 I y = 1.06 × 106 mm 4

From Problem 9.78: I xy = 1.165061 × 106 mm 4

Now

1 1 ( I x + I y ) = (3.93 + 1.06) × 106 mm 4 2 2 = 2.495 × 106 mm 4 1 1 ( I x − I y ) = (3.93 − 1.06) × 106 mm 4 = 1.435 × 106 mm 4 2 2

Using Eqs. (9.18), (9.19), and (9.20): Eq. (9.18):

1 1 ( I x + I y ) + ( I x − I y ) cos 2θ − I xy sin 2θ 2 2 = [2.495 + 1.435cos 90° − 1.165061sin 90°] × 106 mm 4

I x′ =

or Eq. (9.19):

1 1 ( I x + I y ) − ( I x − I y ) cos 2θ + I xy sin 2θ 2 2 = [2.495 − 1.435cos 90° + 1.165061sin 90°] × 106 mm 4

I y′ =

or Eq. (9.20):

I x′ = 1.330 × 106 mm 4

I y′ = 3.66 × 106 mm 4

1 ( I x − I y ) sin 2θ + I xy cos 2θ 2 = [(1.435sin 90° + 1.165061cos 90°] × 106 mm 4

I x′y′ =

or

I x′y′ = 1.435 × 106 mm 4


!

PROBLEM 9.85 For the quarter ellipse of Problem 9.67, determine the orientation of the principal axes at the origin and the corresponding values of the moments of inertia.


Ix =

Problem 9.67:

I xy =

Now Eq. (9.25):

π 8

a4

Iy =

2

a4

1 4 a 2

tan2θ m = −

2 I xy Ix − I y

=−

2 π 8

=

Then

π

( 4

1 2

a4

)

a − 2 a4 π

8 = 0.84883 3π

2θ m = 40.326° and 220.326°

or Also Eq. (9.27):

I max, min = =

Ix + I y 2

θ m = 20.2° and 110.2°

2

±

& Ix − Iy ' 2 ! + I xy 2 " #

1&π 4 π 4 ' a + a ! 2" 8 2 # 2

$1 & π 4 π 4 % & 1 4 ' a − a ) + a ! ± ( 2 + "2 # *2" 8

2

= (0.981, 748 ± 0.772,644)a 4

or

I max = 1.754a 4

and

I min = 0.209a 4

By inspection, the a axis corresponds to I min and the b axis corresponds to I max .

!


!

PROBLEM 9.86 For the area indicated, determine the orientation of the principal axes at the origin and the corresponding values of the moments of inertia. Area of Problem 9.72.


I x = 68.96 × 106 mm 4 I y = 132.48 × 106 mm 4

Problem 9.72: Now Eq. (9.25):

I xy = −21.6 × 106 mm 4 tan2θ m = −

2 I xy Ix − I y

=−

2( −21.6 × 106 ) (68.96 − 132.48) × 106

= −0.68010

Then

2θ m = −34.220° and 145.780°

or Ix + Iy

θ m = −17.11° and 72.9°

2

& Ix − I y ' 2 ± ! + I xy ! 2 " #

Also Eq. (9.27):

I max, min =

Then

2 $ 68.96 + 132.48 % & 68.96 − 132.48 ' 2) ( 21.6) I max, min = ( ± + − × 106 ! ( ) 2 2 " # * +

2

= (100.72 ± 38.409) × 106 mm 4

or and

I max = 139.1 × 106 mm 4 I min = 62.3 × 106 mm 4


!


!



I x = 324π in.4

Problem 9.73:

I xy = 864 in.4

Now Eq. (9.25):

tan2θ m = −

2 I xy Ix − I y

I y = 648π in.4

=−

2(864) 324π − 648π

= 1.69765

Then

2θ m = 59.500° and 239.500°

θ m = 29.7° and 119.7°

or

Also Eq. (9.27):

I max, min =

Then

I max, min =

Ix + Iy 2

2

& Ix − I y ' ± ! + I xy2 ! 2 " # 2

324π + 648π & 324π − 648π ' 2 ± ! + 864 2 2 " #

= (1526.81 ± 1002.75) in.4

or

I max = 2530 in.4

and

I min = 524 in.4


!




I x = 252, 757 mm 4 I y = 1, 752,789 mm 4


I xy = 471, 040 mm 4 tan 2θ m = −

2 I xy Ix − Iy

2(471, 040) 252, 757 − 1, 752, 789 = 0.62804 =−

Then

2θ m = 32.130° and 212.130°

or Also Eq. (9.27):

I max, min =

Then

I max, min =

Ix − I y 2

θ m = 16.07° and 106.1°

2

& Ix − Iy ' ± ! + I xy2 ! 2 " # 2

252, 757 + 1,752, 789 & 252, 757 − 1,752, 789 ' 2 ± ! + 471040 2 2 " #

= (1,002, 773 ± 885, 665) mm 4

or

I max = 1.888 × 106 mm 4

and

I min = 0.1171 × 106 mm 4



PROBLEM 9.89 For the angle cross section indicated, determine the orientation of the principal axes at the origin and the corresponding values of the moments of inertia. The L3 × 2 × 14 -in. angle cross section of Problem 9.74.


I x = 0.390 in.4

Problem 9.74:

I xy = −0.37983 in.4

Now Eq. (9.25):

tan 2θ m = −

2 I xy Ix − I y

I y = 1.09 in.4

=−

2(−0.37983) 0.390 − 1.09

= −1.08523

Then

2θ m = −47.341° and 132.659°

or

Also Eq. (9.27):

I max, min =

Then

I max, min =

Ix + I y 2

θ m = −23.7° and 66.3°

2

& Ix − I y ' ± ! + I xy2 ! 2 " # 2

0.390 + 1.09 & 0.390 − 1.09 ' 2 ± ! + (−0.37983) 2 2 " #

= (0.740 ± 0.51650)2 in.4

or

I max = 1.257 in.4

and

I min = 0.224 in.4


!


!

PROBLEM 9.90 For the angle cross section indicated, determine the orientation of the principal axes at the origin and the corresponding values of the moments of inertia. The L127 × 76 × 12.7-mm angle cross section of Problem 9.78.


I x = 3.93 × 106 mm 4 I y = 1.06 × 106 mm 4


I xy = 1.165061 × 106 mm 4 tan 2θ m = −

2 I xy Ix − Iy

=−

2(1.165061 × 106 ) (3.93 − 1.06) × 106

= −0.81189

Then

2θ m = −39.073° and 140.927°

or θ m = −19.54° and 70.5° Also Eq. (9.27):

Then

I max, min =

I max, min

Ix + I y 2

2

& Ix − I y ' ± ! + I xy2 ! 2 " #

2 $ 3.93 + 1.06 % & 3.93 − 1.06 ' ( = ± + 1.1650612 ) × 106 mm 4 ! ( ) 2 2 " # * +

= (2.495 ± 1.84840) × 106 mm 4

or

I max = 4.34 × 106 mm 4

and

I min = 0.647 × 106 mm 4

By inspection, the a axis corresponds to I max and the b axis corresponds to I min .


!

PROBLEM 9.91 Using Mohr’s circle, determine for the quarter ellipse of Problem 9.67 the moments of inertia and the product of inertia with respect to new axes obtained by rotating the x and y axes about O (a) through 45° counterclockwise, (b) through 30° clockwise.

SOLUTION Ix =

From Problem 9.79:

π 8

π

Iy =

2

a4

1 4 a 2

I xy =

Problem 9.67:

a4

The Mohr’s circle is defined by the diameter XY, where X

Now

&π 4 1 4 ' &π 4 1 4' a , a ! and Y a ,− a ! 2 # 2 # "8 "2

I ave =

1 1&π 4 π 4 ' 5 (I x + I y ) = a + a ! = π a 4 = 0.98175a 4 2 2" 8 2 # 16

R=

2 & Ix − Iy ' $ 1 & π 4 π 4 '% & 1 4 ' 2 + = − + I a a a ! xy ( ) 2 !# + " 2 !# *2 " 8 " 2 #

2

and

2

= 0.77264a 4

The Mohr’s circle is then drawn as shown. tan 2θ m = −

=−

2 I xy Ix − I y 2 π 8

(

1 2

a4

4

)

a − 2 a4 π

= 0.84883

or

2θ m = 40.326°



α = 90° − 40.326°

Then

= 49.674°

β = 180° − (40.326° + 60°) = 79.674°

(a)

θ = +45°:

I x′ = I ave − R cos α = 0.98175a 4 − 0.77264a 4 cos 49.674°

or

I x′ = 0.482a 4

I y′ = I ave + R cos α = 0.98175a 4 + 0.77264a 4 cos 49.674°

or

I y′ = 1.482a 4

or

I x′y′ = −0.589a 4

I x′y′ = − R sin α = −0.77264a 4 sin 49.674°

(b)

θ = −30°:

!

I x′ = I ave + R cos β = 0.98175a 4 + 0.77264a 4 cos 79.674°

or

I x′ = 1.120a 4

I y′ = I ave − R cos β = 0.98175a 4 − 0.77264a 4 cos 79.674°

or

I y′ = 0.843a 4

or

I x′y′ = 0.760a 4

I x′y′ = R sin β = 0.77264a 4 sin 79.674°


!

PROBLEM 9.92 Using Mohr’s circle, determine the moments of inertia and the product of inertia of the area of Problem 9.72 with respect to new centroidal axes obtained by rotating the x and y axes 30° counterclockwise.

SOLUTION I x = 68.96 × 106 mm 4

From Problem 9.80:

I y = 132.48 × 106 mm 4 I xy = −21.6 × 106 mm 4

Problem 9.72:

The Mohr’s circle is defined by the diameter XY, where X (68.96 × 106 , −21.6 × 106 ) and Y (132.48 × 106 , 21.6 × 106 ). Now

and

I ave =

1 1 ( I x − I y ) = (68.96 + 132.48) × 106 = 100.72 × 106 mm 4 2 2

2 2 - 1 . 4 $ 4 $1 % % 2 R = ( ( I x − I y ) ) + I xy = / ( (68.96 − 132.48) ) + (−21.6) 2 0 × 106 mm 4 *2 + + 41 * 2 42

= 38.409 × 106 mm 4


2 I xy Ix − Iy

2(−21.6 × 106 ) (68.96 − 132.48) × 106 = −0.68010 =−

or Then

2θ m = −34.220°

α = 180° − (34.220° + 60°) = 85.780°



Then

I x′ = I ave + R cos α = (100.72 + 38.409 cos85.780°) × 106

or

I x′ = 103.5 × 106 mm 4

I y′ = I ave − R cos α = (100.72 − 38.409 cos85.780°) × 106

or

I y′ = 97.9 × 106 mm 4

or

I x′y′ = −38.3 × 106 mm 4

I x′y′ = − R sin α = −(38.409 × 106 ) sin 85.780°


PROBLEM 9.93 Using Mohr’s circle, determine the moments of inertia and the product of inertia of the area of Problem 9.73 with respect to new centroidal axes obtained by rotating the x and y axes 60° counterclockwise.

SOLUTION I x = 324π in.4

From Problem 9.81:

I y = 648π in.4 I xy = 864 in.4

Problem 9.73:

The Mohr’s circle is defined by the diameter XY, where X (324π , 864) and Y (648π , − 864). Now

I ave =

1 1 ( I x + I y ) = (324π + 648π ) = 1526.81 in.4 2 2 2

and

$1 $1 % R = ( ( I x − I y ) 2 + I xy2 = ( (324π + 648π ) ) + 8642 2 2 * * + = 1002.75 in.4


2 I xy Ix − Iy

2(864) 324π − 648π = 1.69765 =−

or Then

2θ m = 59.500°

α = 120° − 59.500° = 60.500°



Then

!

I x′ = I ave − R cos α = 1526.81 − 1002.75cos 60.500°

or

I x′ = 1033 in.4

or

I y′ = 2020 in.4

or

I x′y′ = −873 in.4

I y = I ave + R cos α = 1526.81 + 1002.75cos 60.500° !

I x′y′ = − R sin α = −1002.75sin 60.500°


!

PROBLEM 9.94 Using Mohr’s circle, determine the moments of inertia and the product of inertia of the area of Problem 9.75 with respect to new centroidal axes obtained by rotating the x and y axes 45° clockwise.


I x = 252, 757 mm 4 I y = 1, 752,789 mm 4

Problem 9.75:

I xy = 471, 040 mm 4

The Mohr’s circle is defined by the diameter XY, where X (252,757; 471,040) and Y (1,752,789; −471, 040). Now

1 1 ( I x + I y ) = (252, 757 + 1, 752, 789) 2 2 = 1, 002, 773 mm 4

I ave =

2

and

$1 % R = ( ( I x − I y ) ) + I xy2 *2 + $1 % = ( (252, 757 − 1,752, 789) 2 ) + 471, 0402 2 * + = 885,665 mm 4


2 I xy Ix − Iy

2(471, 040) 252, 757 − 1, 752, 789 = 0.62804 =−

or Then

2θ m = 32.130°

α = 180° − (32.130 + 90°) = 57.870°



Then

I x′ = I ave + R cos α = 1,002, 773 + 885, 665cos 57.870° I x′ = 1.474 × 106 mm 4

or I y′ = I ave − R cos α = 1, 002,773 − 885, 665cos 57.870°

or

I y′ = 0.532 × 106 mm 4

or

I x′y′ = 0.750 × 106 mm 4

I x′y′ = R sin α = 885, 665sin 57.870°


PROBLEM 9.95 Using Mohr’s circle, determine the moments of inertia and the product of inertia of the L3 × 2 × 14 -in. angle cross section of Problem 9.74 with respect to new centroidal axes obtained by rotating the x and y axes 30° clockwise.


I x = 0.390 in.4 I y = 1.09 in.4

Problem 9.74: I xy = −0.37983 in.4

The Mohr’s circle is defined by the diameter XY, where X (0.390, − 0.37983) and Y (1.09, 0.37983). Now

1 (I x + I y ) 2 1 = (0.390 + 1.09) 2 = 0.740 in.4

I ave =

2

$1 % R = ( ( I x − I y ) ) + I xy2 *2 +

and

2

$1 % = ( (0.390 − 1.09) ) + (−0.37983) 2 *2 + = 0.51650 in.4


2 I xy Ix − Iy

2( −0.37983) 0.390 − 1.09 = −1.08523 =−

or Then

2θ m = −47.341°

α = 60° − 47.341° = 12.659°



Then

I x′ = I ave − R cos α = 0.740 − 0.51650 cos 12.659°

or

I x′ = 0.236 in.4

I y′ = I ave + R cos α = 0.740 + 0.51650cos 12.659°

or

I y = 1.244 in.4

or

I x′y′ = 0.1132 in.4

I x′y′ = R sin α = 0.51650 sin 12.659°


!

PROBLEM 9.96 Using Mohr’s circle, determine the moments of inertia and the product of inertia of the L127 × 76 × 12.7-mm angle cross section of Problem 9.78 with respect to new centroidal axes obtained by rotating the x and y axes 45° counterclockwise.


I x = 3.93 × 106 mm 4 I y = 1.06 × 106 mm 4

Problem 9.78:

I xy = 1.165061 × 106 mm 4

The Mohr’s circle is defined by the diameter XY, where X (3.93 × 106 , 1.165061 × 106 ), Y (1.06 × 106 , −1.165061 × 106 ). Now

I ave =

1 1 ( I x + I y ) = (3.93 + 1.06) × 106 = 2.495 × 106 mm 4 2 2 2

and

$1 % R = ( ( I x − I y ) ) + I xy2 *2 + 2 - 1 . 4 $ % 4 = / ( (3.93 − 1.06) ) + 1.1650612 0 × 106 mm 4 + 41 * 2 42

= 1.84840 × 106 mm 4


2 I xy Ix − Iy

2(1.165061 × 106 ) (3.93 − 1.06) × 106 = −0.81189 =−

or

2θ m = −39.073°



Then

α = 180° − (39.073° + 90°) = 50.927°

Then

I x′ = I ave − R cos α = (2.495 − 1.84840 cos 50.927°) × 106

or

I x′ = 1.330 × 106 mm 4

I y′ = I ave + R cos α = (2.495 + 1.84840 cos 50.927°) × 106

or

I y′ = 3.66 × 106 mm 4

or

I x′y′ = 1.435 × 106 mm 4

I x′y′ = R sin α = (1.84840 × 106 ) sin 50.927°


!

PROBLEM 9.97 For the quarter ellipse of Problem 9.67, use Mohr’s circle to determine the orientation the principal axes at the origin and the corresponding values of the moments of inertia.


Ix =

Problem 9.67:

I xy =

π 8

a4

Iy =

π 2

a4

1 4 a 2

The Mohr’s circle is defined by the diameter XY, where X

Now

1 ' &π 4 1 4 ' &π 4 a , a ! and Y a , − a4 ! 8 2 2 2 # " # " I ave =

1 1&π 4 π 4 ' a + a ! = 0.98175a 4 (I x + I y ) = 2 2" 8 2 #

and 2

$1 % 2 R = ( ( I x − I y ) ) + I xy *2 + 2

$ 1 & π 4 π 4 '% & 1 4 ' = ( a − a !) + a ! 2 #+ " 2 # *2 " 8

2

= 0.77264a 4


=−

2 I xy Ix − I y 2 π 8

(

1 2

a4

4

)

a − 2 a4 π

= 0.84883

or and

2θ m = 40.326°

θ m = 20.2°



The Principal axes are obtained by rotating the xy axes through 20.2° counterclockwise

!

!

about O.

! Now

I max, min = I ave ± R = 0.98175a 4 ± 0.77264a 4

or

I max = 1.754 a 4

!

and

I min = 0.209 a 4

!

From the Mohr’s circle it is seen that the a axis corresponds to I min and the b axis corresponds to I max . !


PROBLEM 9.98 Using Mohr’s circle, determine for the area indicated the orientation of the principal centroidal axes and the corresponding values of the moments of inertia. Area of Problem 9.72.

SOLUTION I x = 68.96 × 106 mm 4

From Problem 9.80:

I y = 132.48 × 106 mm 4 I xy = −21.6 × 106 mm 4

Problem 9.72:

The Mohr’s circle is defined by the diameter XY, where X (68.96 × 106 , − 21.6 × 106 ) and Y (132.48 × 106 , 21.6 × 106 ) Now

1 (I x + I y ) 2 1 = (68.96 + 132.48) × 106 2 = 100.72 × 106 mm 4

I ave =

2

and

$1 % R = ( ( I x − I y ) ) + I xy2 *2 + 2 - 1 . 4 $ % 4 = / ( (68.96 − 132.48) ) + (−21.6) 2 0 × 106 + 41 * 2 42

= 38.409 × 106 mm 4


2 I xy Ix − Iy

2(−21.6 × 106 ) (68.96 − 132.48) × 106 = −0.68010 =−

or and

2θ m = −34.220°

θ m = −17.11°



The principal axes are obtained by rotating the xy axes through 17.11° clockwise about C.

Now

I max, min = I ave ± R = (100.72 + 38.409) × 106

or

I max = 139.1 × 106 mm 4

and

I min = 62.3 × 106 mm 4



!


SOLUTION From Problem 9.76: I xy = −9011.25 in.4

Now where

I x = ( I x )1 + ( I x )2 + ( I x )3 ( I x )1 =

1 (24 in.)(4 in.)3 = 128 in.4 12

( I x ) 2 = ( I x )3 =

1 $1 % (9 in.)(15 in.)3 + ( (9 in.)(15 in.) ) (7 in.)2 36 *2 +

= 4151.25 in.4

Then

I x = [128 + 2(4151.25)] in.4 = 8430.5 in.4

Also where

I y = ( I y )1 + ( I y ) 2 + ( I y )3 ( I y )1 =

1 (4 in.)(24 in.) = 4608 in.4 12

( I y ) 2 = ( I y )3 =

1 $1 % (15 in.)(9 in.)3 + ( (9 in.)(15 in.) ) (9 in.) 2 36 2 * +

= 5771.25 in.4

Then

I y = [4608 + 2(5771.25)] in.4 = 16150.5 in.4




I ave =

1 1 ( I x + I y ) = (8430.5 + 16150.5) = 12290.5 in.4 2 2 2

and

$1 % R = ( ( I x − I y ) ) + I xy2 *2 + 2

$1 % = ( (8430.5 − 16150.5) ) + (−9011.25) 2 *2 + = 9803.17 in.4


2 I xy Ix − I y

2( −9011.25) 8430.5 − 16150.5 = −2.33452 =−

or and

2θ m = −66.812°

θ m = −33.4° The principal axes are obtained by rotating the xy axes through 33.4° clockwise

!

Now

about C.

I max, min = I ave ± R = 12290.5 ± 9803.17

or

I max = 22.1 × 103 in.4

and

I min = 2490 in.4



!

PROBLEM 9.100 Using Mohr’s circle, determine for the area indicated the orientation of the principal centroidal axes and the corresponding values of the moments of inertia. Area of Problem 9.73


I x = 324π in.4

Problem 9.73:

I xy = 864 in.4

I y = 648π in.4

The Mohr’s circle is defined by the diameter XY, where X (324π , 864) and Y (648π , − 864). Now

and

I ave =

1 1 ( I x + I y ) = (324π + 648π ) = 1526.81 in.4 2 2

$1 % 2 R = ( ( I x − I y ) 2 ) + I xy *2 + 2

$1 % = ( (324π − 648π ) ) + 8642 *2 + = 1002.75 in.4


2 I xy Ix − Iy

2(864) 324π − 648π = 1.69765 =−

or

2θ m = 59.4998°

and

θ m = 29.7° The principal axes are obtained by rotating the xy axes through 29.7° counterclockwise about C.


!


Now

I max, min = I ave ± R = 1526.81 ± 1002.75

or

I max = 2530 in.4

and

I min = 524 in.4




SOLUTION I x = 0.390 in.4

From Problem 9.83:

I y = 1.09 in.4

I xy = −0.37983 in.4

Problem 9.74:


I ave =

1 1 ( I x + I y ) = (0.390 + 1.09) = 0.740 in.4 2 2 2

and

1 ! R = " ( I x − I y ) # + I xy2 2 $ % 2

1 ! = " (0.390 − 1.09) # + (−0.37983) 2 2 $ % = 0.51650 in.4


2 I xy Ix − Iy

2( −0.37983) 0.390 − 1.09 = −1.08523 =−

Then and

2θ m = −47.341°

θ m = −23.7° The principal axes are obtained by rotating the xy axes through 23.7° clockwise about C.


!


Now

I max, min = I ave ± R = 0.740 ± 0.51650

or

I max = 1.257 in.4

and

I min = 0.224 in.4



PROBLEM 9.102 Using Mohr’s circle, determine for the area indicated the orientation of the principal centroidal axes and the corresponding values of the moments of inertia. Area of Problem 9.77 (The moments of inertia I x and I y of the area of Problem 9.102 were determined in Problem 9.44).


I x = 18.1282 in.4 I y = 4.5080 in.4

Problem 9.77:

I xy = −4.25320 in.4


I ave =

1 1 ( I x + I y ) = (18.1282 + 4.5080) = 11.3181 in.4 2 2 2

1 ! R = " ( I x − I y ) # + I xy2 $2 %

and

2

1 ! = " (18.1282 − 4.5080) # + ( −4.25320)2 $2 % = 8.02915 in.4


2 I xy

Ix − Iy

2(−4.25320) 18.1282 − 4.5080 = 0.62454 =−

or and

2θ m = 31.986°

θ m = 15.99° The principal axes are obtained by rotating the xy axes through 15.99° counterclockwise ! about C.



Now

I max, min = I ave ± R = 11.3181 ± 8.02915 I max = 19.35 in.4

or

I min = 3.29 in.4

and

From the Mohr’s circle it is seen that the a axis corresponds to I max and the b axis corresponds to I min . !


PROBLEM 9.103 The moments and product of inertia of an L4 × 3 × 14 -in. angle cross section with respect to two rectangular axes x and y through C are, respectively, I x = 1.33 in.4, I y = 2.75 in.4, and I xy , 0, with the minimum value of the moment of inertia of the area with respect to any axis through C being I min = 0.692 in.4. Using Mohr’s circle, determine (a) the product of inertia I xy of the area, (b) the orientation of the principal axes, (c) the value of I max .

SOLUTION (Note: A review of a table of rolled-steel shapes reveals that the given values of I x and I y are obtained when the 4-in. leg of the angle is parallel to the x axis. Further, for I xy , 0, the angle must be oriented as shown.) 1 1 ( I x + I y ) = (1.33 + 2.75) = 2.040 in.4 2 2

Now

I ave =

and

I min = I ave − R or

R = 2.040 − 0.692 = 1.348 in.4

Using I ave and R, the Mohr’s circle is then drawn as shown; note that for the diameter XY, X (1.33, I xy ) and Y (2.75, |I xy |). 2

(a)

We have

1 ! R 2 = " ( I x − I y ) # + I xy2 $2 %

or

I xy2

1 ! = 1.348 − " (1.33 − 2.75) # $2 %

2

2

Solving for I xy and taking the negative root (since I xy , 0) yields I xy = −1.14586 in.4 . I xy = −1.146 in.4

(b)

We have

tan 2θ m = −

2 I xy Ix − Iy

=−

2(−1.14586) 1.33 − 2.75

= −1.61389

or

2θ m = −58.217°

θ m = −29.1°

The principal axes are obtained by rotating the xy axes through 29.1° clockwise about C. (c)

We have

I max = I ave + R = 2.040 + 1.348

or

I max = 3.39 in.4


!

PROBLEM 9.104 Using Mohr’s circle, determine for the cross section of the rolledsteel angle shown the orientation of the principal centroidal axes and the corresponding values of the moments of inertia. (Properties of the cross sections are given in Figure 9.13.)

SOLUTION I x = 0.162 × 106 mm 4

From Figure 9.13B:

I y = 0.454 × 106 mm 4

We have

I xy = ( I xy )1 + ( I xy ) 2

For each rectangle


and

I x′y′ = 0

I xy = Σ x yA

(symmetry)

A, mm 2

x , mm

y , mm

x yA, mm 4

1

76 × 6.4 = 486.4

−13.1

9.2

−58620.93

2

6.4 × (51 − 6.4) = 285.44

21.7

–16.3

–100962.98

Σ

–159583.91 I xy = −159584 mm 4

The Mohr’s circle is defined by the diameter XY where X (0.162 × 106 , − 0.159584 × 106 ) and Y (0.454 × 106 , 0.159584 × 106 ) Now

1 1 ( I x + I y ) = (0.162 + 0.454) × 106 2 2 = 0.3080 × 106 mm 4

I ave =



and

2 2 & 1 ' 1 ( ( ! ! 2 = " (0.162 − 0.454) # + ( −0.159584) 2 ! × 106 R = " ( I x − I y ) # + I xy $2 % % &' $ 2 &(

= 0.21629 × 106 mm 4


2 I xy Ix − Iy

2(−0.159584 × 106 ) (0.162 − 0.454) × 106 = −1.09304 =−

or and

2θ m = −47.545

θ m = −23.8° The principal axes are obtained by rotating the xy axes through 23.8° clockwise About C.

Now

I max, min = I ave ± R = (0.3080 ± 0.21629) × 106 I max = 0.524 × 106 mm 4

or

I min = 0.0917 × 106 mm 4

and



PROBLEM 9.105 Using Mohr’s circle, determine for the cross section of the rolledsteel angle shown the orientation of the principal centroidal axes and the corresponding values of the moments of inertia. (Properties of the cross sections are given in Figure 9.13.)

SOLUTION I x = 3.93 × 106 mm 4

From Figure 9.13B:

I y = 1.06 × 106 mm 4 I xy = −1.165061 × 106 mm 4

Problem 9.7B:

(Note that the figure of Problem 9.105 is obtained by replacing x with –x in the figure of Problem 9.78; thus the change in sign of I xy .) 6 6 The Mohr’s circle is defined by the diameter XY, where X (3.93 × 10 , − 1.165061 × 10 ) 6 6 and Y (1.06 × 10 , 1.165061 × 10 ).

Now

1 (I x + I y ) 2 1 = (3.93 + 1.06) × 106 2 = 2.495 × 106 mm 4

I ave =

2

and

)1 * R = " ( I x − I y ) # + I xy2 2 $ % =

2 + 1 , & ) * 2& − + − × 106 (3.93 1.06) ( 1.165061) ! "2 # $ % &' (&

= 1.84840 × 106 mm 4




2 I xy Ix − Iy

2( −1.165061 × 106 ) (3.93 − 1.06) × 106 = 0.81189 =−

or and

2θ m = 39.073°

θ m = 19.54° The principal axes are obtained by rotating the xy axes through 19.54° counterclockwise about C.

Now

I max, min = I ave ± R = (2.495 ± 1.84840) × 106

or

I max = 4.34 × 106 mm 4

and

I min = 0.647 × 106 mm 4

From the Mohr’s circle it is seen that the a axis corresponds to I max and the b axis corresponds to I min .


PROBLEM 9.106* For a given area the moments of inertia with respect to two rectangular centroidal x and y axes are I x = 1200 in.4 and I y = 300 in.4, respectively. Knowing that after rotating the x and y axes about the centroid 30° counterclockwise, the moment of inertia relative to the rotated x axis is 1450 in.4, use Mohr’s circle to determine (a) the orientation of the principal axes, (b) the principal centroidal moments of inertia.

SOLUTION We have

I ave =

1 1 ( I x + I y ) = (1200 + 300) = 750 in.4 2 2

Now observe that I x . I ave , I x′ . I x , and 2θ = +60°. This is possible only if I xy < 0. Therefore, assume I xy , 0 and (for convenience) I x′y′ . 0. Mohr’s circle is then drawn as shown.

We have Now using DABD:

2θ m + α = 60° R=

I x − I ave 1200 − 750 = cos 2θ m cos 2θ m =

Using DAEF:

R=

450 cos 2θ m

(in.4 )

I x′ − I ave 1450 − 750 = cos α cos α 700 (in.4 ) = cos α

Then

450 700 = cos 2θ m cos α

or

9cos (60° − 2θ m ) = 14 cos 2θ m

Expanding:

9(cos 60° cos 2θ m + sin 60° sin 2θ m ) = 14 cos 2θ m tan 2θ m =

or

α = 60° − 2θ m

14 − 9 cos 60° = 1.21885 9 sin 60°

2θ m = 50.633° and θ m = 25.3°

or

(Note: 2θ m , 60° implies assumption I x′y′ . 0 is correct.) Finally, (a)

R=

450 = 709.46 in.4 cos 50.633°

From the Mohr’s circle it is seen that the principal axes are obtained by rotating the given centroidal x and y axes through θ m about the centroid C or 25.3° counterclockwise


!


(b)

We have

I max, min = I ave ± R = 750 ± 709.46

or

I max = 1459 in.4

and

I min = 40.5 in.4



PROBLEM 9.107 It is known that for a given area I y = 48 × 106 mm4 and I xy = –20 × 106 mm4, where the x and y axes are rectangular centroidal axes. If the axis corresponding to the maximum product of inertia is obtained by rotating the x axis 67.5° counterclockwise about C, use Mohr’s circle to determine (a) the moment of inertia I x of the area, (b) the principal centroidal moments of inertia.

SOLUTION First assume I x . I y and then draw the Mohr’s circle as shown. (Note: Assuming I x , I y is not consistent with the requirement that the axis corresponding to ( I xy ) max is obtained after rotating the x axis through 67.5° CCW.)

From the Mohr’s circle we have 2θ m = 2(67.5°) − 90° = 45°

(a)

From the Mohr’s circle we have Ix = Iy + 2

(b)

We have

and Now

| I xy | tan 2θ m

= 48 × 106 + 2

20 × 106 tan 45°

or

I x = 88.0 × 106 mm 4

or

I max = 96.3 × 106 mm 4

and

I min = 39.7 × 106 mm 4

1 1 ( I x + I y ) = (88.0 + 48) × 106 2 2 6 = 68.0 × 10 mm 4

I ave =

R=

| I xy | sin 2θ m

=

20 × 106 = 28.284 × 106 mm 4 sin 45°

I max, min = I ave ± R = (68.0 ± 28.284) × 106


PROBLEM 9.108 Using Mohr’s circle, show that for any regular polygon (such as a pentagon) (a) the moment of inertia with respect to every axis through the centroid is the same, (b) the product of inertia with respect to every pair of rectangular axes through the centroid is zero.

SOLUTION Consider the regular pentagon shown, with centroidal axes x and y.

Because the y axis is an axis of symmetry, it follows that I xy = 0. Since I xy = 0, the x and y axes must be principal axes. Assuming I x = I max and I y = I min , the Mohr’s circle is then drawn as shown.

Now rotate the coordinate axes through an angle α as shown; the resulting moments of inertia, I x′ and I y′ , and product of inertia, I x′y′ , are indicated on the Mohr’s circle. However, the x′ axis is an axis of symmetry, which implies I x′y′ = 0. For this to be possible on the Mohr’s circle, the radius R must be equal to zero (thus, the circle degenerates into a point). With R = 0, it immediately follows that (a)

I x = I y = I x′ = I y ′ = I ave (for all moments of inertia with respect to an axis through C)

(b)

I xy = I x′y′ = 0 (for all products of inertia with respect to all pairs of rectangular axes with origin at C)

!


PROBLEM 9.109 Using Mohr’s circle, prove that the expression I x′ I y′ − I x2′y′ is independent of the orientation of the x′ and y′ axes, where Ix′, Iy′, and Ix′y′ represent the moments and product of inertia, respectively, of a given area with respect to a pair of rectangular axes x′ and y′ through a given Point O. Also show that the given expression is equal to the square of the length of the tangent drawn from the origin of the coordinate system to Mohr’s circle.

SOLUTION First observe that for a given area A and origin O of a rectangular coordinate system, the values of I ave and R are the same for all orientations of the coordinate axes. Shown below is a Mohr’s circle, with the moments of inertia, I x′ and I y′ , and the product of inertia, I x′y′ , having been computed for an arbitrary orientation of the x′y ′ axes.

From the Mohr’s circle I x′ = I ave + R cos 2θ I y′ = I ave − R cos 2θ I x′y′ = R sin 2θ

Then, forming the expression

I x′ I y′ − I x2′y′

I x′ I y′ − I x2′y′ = ( I ave + R cos 2θ )( I ave − R cos 2θ ) − ( R sin 2θ ) 2

(

)

2 = I ave − R 2 cos 2 2θ − ( R 2 sin 2 2θ ) 2 = I ave − R2

which is a constant

I x′ I y′ − I x2′y′ is independent of the orientation of the coordinate axes Q.E.D.

Shown is a Mohr’s circle, with line OA, of length L, the required tangent.

Noting that

OAC is a right angle, it follows that 2 L2 = I ave − R2

or

L2 = I x′ I y′ − I x2′y′ Q.E.D.


!

PROBLEM 9.110 Using the invariance property established in the preceding problem, express the product of inertia Ixy of an area A with respect to a pair of rectangular axes through O in terms of the moments of inertia Ix and Iy of A and the principal moments of inertia Imin and Imax of A about O. Use the formula obtained to calculate the product of inertia Ixy of the L3 × 2 × 14 -in. angle cross section shown in Figure 9.13A, knowing that its maximum moment of inertia is 1.257 in4.

SOLUTION Consider the following two sets of moments and products of inertia, which correspond to two different orientations of the coordinate axes whole origin is at Point O. Case 1:

I x′ = I x , I y ′ = I y , I x′y′ = I xy

Case 2:

I x′ = I max , I y′ = I min , I x′y′ = 0

The invariance property then requires 2 I x I y − I xy = I max I min

From Figure 9.13A:

or I xy = ± I x I y − I max I min

I x = 1.09 in.4 I y = 0.390 in.4

Using Eq. (9.21): Substituting or Then

I x + I y = I max + I min 1.09 + 0.390 = 1.257 + I min I min = 0.223 in.4

I xy = (1.09)(0.390) − (1.257)(0.223) = ±0.381 in.4

The two roots correspond to the following two orientations of the cross section. I xy = −0.381 in.4

For

I xy = 0.381 in.4

and for


!

PROBLEM 9.111 Determine the mass moment of inertia of a ring of mass m, cut from a thin uniform plate, with respect to (a) the axis AA′, (b) the centroidal axis CC′ that is perpendicular to the plane of the ring.

SOLUTION Mass = m = ρV = ρ tA I mass = ρ t I area =

m I area A

We first determine

(

A = π r22 − π r12 = π r22 − r12 I AA′,area = I AA′, mass =

(a)

(b)

π 4

r24 −

π 4

r14 =

π 4

(r

4 2

)

− r14

)

m m π 4 4 1 I AA′, area = r2 − r1 = m r22 + r12 2 2 4 A π r2 − r1 4

(

)

(

By symmetry:

I BB′ = IAA′

Eq. (9.38):

I CC ′ = IAA′ + I BB′ = 2 IAA′

)

(

) I AA′ =

1 m r12 + r22 4

)

I CC ′ =

1 m r12 + r22 2

)

(

(


PROBLEM 9.112 A thin semicircular plate has a radius a and a mass m. Determine the mass moment of inertia of the plate with respect to (a) the centroidal axis BB′, (b) the centroidal axis CC′ that is perpendicular to the plate.

SOLUTION mass = m = ρ tA I mass = ρ t I area =

m I area A

1 A = π a2 2

Area:

I AA′,area = I DD′,area =

1-π 4. 1 4 / a 0 = πa 21 4 2 8

I AA′, mass = I DD′,mass =

m m -1 4. 1 2 I AA′,area = / π a 0 = ma 1 A π a2 1 8 2 4 2

I BB′ = I DD′ − m( AC ) 2 =

(a)

1 2 - 4a . ma − m / 0 4 1 3π 2

= (0.25 − 0.1801) ma 2

(b)

Eq. (9.38):

I CC ′ = I AA′ + I BB′ =

2

I BB′ = 0.0699ma 2

1 ma 2 + 0.0699ma 2 4 I CC ′ = 0.320ma 2


PROBLEM 9.113 The quarter ring shown has a mass m and was cut from a thin, uniform plate. Knowing that r1 = 34 r2, determine the mass moment of inertia of the quarter ring with respect to (a) the axis AA′, (b) the centroidal axis CC′ that is perpendicular to the plane of the quarter ring.

SOLUTION First note

mass = m = ρV = ρ tA = ρt

(r

2 2

4

− r12

=

m

π 4

Using Figure 9.12,

I AA′,area =

Then

I AA′, mass =

(

π

r22 − r12

(r 16

4 2

− r14

(

r22

−

r12

(

=

)

I area

)

m

π

4 m 2 r2 + r12 = 4

(b)

)

I mass = ρ t I area

Also

(a)

π

)

×

π 16

(r

4 2

− r14

)

)

2 m) 2 -3 . * " r2 + / r2 0 # 4 $" 1 4 2 %#

or

I AA′ =

25 2 mr2 64

Symmetry implies I BB′, mass = I AA′,mass

Then

I DD′ = I AA′ + I BB′ - 25 . = 2 / mr22 0 64 1 2 25 2 mr2 = 32



Now locate centroid C. X Σ A = Σx A

or

π . 4r - π . 4r - π . -π X / r22 − r12 0 = 2 / r22 0 − 1 / r12 0 4 4 2 3π 1 4 2 3π 1 4 2 1 4 r23 − r13 3π r22 − r12

or

X=

Now

r=X 2 3

-3 . r23 − / r2 0 4 2 14 2 = 3π 2 - 3 .2 r2 − / r2 0 14 2 =

Finally, or

37 2 r2 21π

I DD′ = I CC ′ + mr 2 - 37 2 . 25 2 mr2 = I CC ′ + m / r / 21π 2 00 32 1 2

2

or

I CC ′ = 0.1522mr22


PROBLEM 9.114 The parabolic spandrel shown was cut from a thin, uniform plate. Denoting the mass of the spandrel by m, determine its mass moment of inertia with respect to (a) the axis BB′, (b) the axis DD′ that is perpendicular to the spandrel. (Hint: See Sample Problem 9.3.)

SOLUTION First note

mass = m = ρV = ρ tA -π . = ρ t / ab 0 12 2

I mass = ρ t I area

Also

=

(a)

2m I π ab area

We have I x,area = I BB′,area + Ay 2

or (a)

I BB′,area =

- π .- 4b . ab3 − / ab 0/ 0 8 1 2 21 3π 2

π

2

From Sample Problem 9.3: 1 3 ab 21 3m 1 3 = × ab ab 21

I BB′,area =

Then

I BB′, mass

or (b)

I BB′ =

1 2 mb 7

From Sample Problem 9.3: I AA′,area =

Now

1 3 ab 5

-3 . I AA′,area = I11′,area + A / a 0 14 2

2



and

-1 . I 22′,area = I11′,area + A / a 0 14 2

2

)- 1 . - 3 . I 22′,area = I AA′,area + A "/ a 0 − / a 0 "$1 4 2 1 4 2 1 1 - 1 9 . = a3b + ab / a 2 − a 2 0 5 3 1 16 16 2 1 3 = ab 30 2

Then

and

Finally,

2*

# #%

3m 1 3 × ab ab 30 1 = ma 2 10

I 22′, mass =

I DD′,mass = I BB′,mass + I 22′, mass =

1 2 1 mb + ma 2 7 10

or

I DD′ =

1 m(7a 2 + 10b 2 ) 70


PROBLEM 9.115 A thin plate of mass m was cut in the shape of a parallelogram as shown. Determine the mass moment of inertia of the plate with respect to (a) the x axis, (b) the axis BB′, which is perpendicular to the plate.

SOLUTION mass = m = ρ t A I mass = ρ t I area =

(a)

m I area A

Consider parallelogram as made of horizontal strips and slide strips to form a square since distance from each strip to x axis is unchanged.

1 I x,area = a 4 3 I x, mass =

(b)

m m I x,area = 2 A a

-1 4. /3a 0 1 2

1 I x = ma 2 3

For centroidal axis y′: )1 * 1 I y′,area = 2 " a 4 # = a 4 $12 % 6 m m -1 . 1 I y′,mass = I y′,area = 2 / a 4 0 = ma 2 A a 16 2 6 1 7 I y′′ = I y′ + ma 2 = ma 2 + ma 2 = ma 2 6 6

For axis BB′ ⊥ to plate, Eq. (9.38): 1 7 I BB′ = I x + I y′′ = ma 2 + ma 2 3 6

I BB′ =

3 2 ma 2


PROBLEM 9.116 A thin plate of mass m was cut in the shape of a parallelogram as shown. Determine the mass moment of inertia of the plate with respect to (a) the y axis, (b) the axis AA′, which is perpendicular to the plate.

SOLUTION See sketch of solution of Problem 9.115. (a)

From Part b of solution of Problem 9.115: I y′ =

1 2 ma 6

I y = I y′ + ma 2 =

(b)

1 2 ma + ma 2 6

Iy =

7 ma 2 6

I AA′ =

1 ma 2 2

From solution of Problem 9.115: I y′ =

Eq. (9.38):

1 2 ma 6

1 and I x = ma 2 3

I AA′ = I y′ + I x =

1 2 1 2 ma + ma 6 3


PROBLEM 9.117 A thin plate of mass m has the trapezoidal shape shown. Determine the mass moment of inertia of the plate with respect to (a) the x axis, (b) the y axis.

SOLUTION First note

mass = m = ρV = ρ tA 1 ) * = ρ t "(2a )( a) + (2a)(a) # = 3ρ ta 2 2 $ %

I mass = ρ tI area =

Also (a)

Now

m I area 3a 2

I x,area = ( I x )1,area + ( I x ) 2,area 1 1 = (2a)(a)3 + (2a )(a )3 3 12 5 4 = a 6

Then (b)

I x, mass =

m 5 4 × a 3a 2 6

or

I x , mass =

5 ma 2 18

We have I z ,area = ( I z )1,area + ( I z )2, area 2 )1 1 1 )1 . * 3* 3 = " (a )(2a) # + " ( a)(2a) + (2a)(a ) / 2a + × 2a 0 # = 10a 4 2 3 $3 % "$ 36 1 2 #%

m 10 × 10a 4 = ma 2 2 3 3a

Then

I x, mass =

Finally,

I y ,mass = I x ,mass + I z ,mass =

5 10 ma 2 + ma 2 18 3

=

65 2 ma 18

or

I y , mass = 3.61ma 2


PROBLEM 9.118 A thin plate of mass m has the trapezoidal shape shown. Determine the mass moment of inertia of the plate with respect to (a) the centroidal axis CC′ that is perpendicular to the plate, (b) the axis AA′ that is parallel to the x axis and is located at a distance 1.5a from the plate.

SOLUTION First locate the centroid C. 1 . X ΣA = Σ xA: X (2a 2 + a 2 ) = a(2a 2 ) + / 2a + × 2a 0 ( a 2 ) 3 1 2 X=

or

14 a 9

-1 . -1 . Z ΣA = Σ zA: Z (2a 2 + a 2 ) = / a 0 (2a 2 ) + / a 0 (a 2 ) 2 1 2 13 2

Z=

or (a)

We have

4 a 9

I y , mass = I CC ′, mass + m( X 2 + Z 2 )

From the solution to Problem 9.117: I y ,mass =

Then

I cc′,mass =

65 2 ma 18 )- 14 .2 - 4 . 2 * 65 2 ma − m "/ a 0 + / a 0 # 18 "$1 9 2 1 9 2 #%

or (b)

We have and Then

I cc′ = 0.994ma 2

I x, mass = I BB′,mass + m( Z ) 2 I AA′, mass = I BB′, mass + m(1.5a) 2 I AA′, mass = I x, mass

2 ) -4 . * 2 + m "(1.5a) − / a 0 # 1 9 2 #% "$


!


From the solution to Problem 9.117: I x, mass =

Then

I AA′, mass =

5 ma 2 18 2 ) 5 -4 . * ma 2 + m "(1.5a) 2 − / a 0 # 18 1 9 2 #% "$

or

I AA′ = 2.33ma 2


PROBLEM 9.119 The area shown is revolved about the x axis to form a homogeneous solid of revolution of mass m. Using direct integration, express the mass moment of inertia of the solid with respect to the x axis in terms of m and h.

SOLUTION We have

y=

2h − h x+h a

so that

r=

h ( x + a) a

For the element shown: dm = ρπ r 2 dx dI x =

1 2 r dm 2

2

)h * = ρπ " ( x + a) # dx $a %

h2 h2 1 ( x + a ) 2 dx = ρπ 2 [( x + a)3 ]0a 2 0 3 a a 2 h 1 7 = ρπ 2 (8a 3 − a3 ) = ρπ ah 2 3 3 a

3

Then

m = dm =

Now

I x = dI x =

3

3

a

3

1 2 1 r ( ρπ r 2 dx) = ρπ 2 2

ρπ

3

a)h 0

4

* " a ( x + a ) # dx $ %

h4 1 1 h4 1 [( x + a)5 ]0a = ρπ 4 (32a5 − a 5 ) ρπ × 4 2 5a 10 a 31 4 = ρπ ah 10 =

From above: Then

3 7

ρπ ah 2 = m Ix =

31 - 3 . 2 93 2 m h = mh 10 /1 7 02 70

or

I x = 1.329 mh 2


PROBLEM 9.120 Determine by direct integration the mass moment of inertia with respect to the y axis of the right circular cylinder shown, assuming that it has a uniform density and a mass m.

SOLUTION For element shown:

dm = ρ dV = ρπ a 2 dx dI y = dI y + x 2 dm 1 2 a dm + x 2 dm 4 -1 . = / a 2 + x 2 0 ρπ a 2 dx 14 2 + L/2 -1 . ρπ a 2 / a 2 + x 2 0 dx I y = dI y = − L/2 14 2 =

3

= ρπ a 2

3

1 2 x3 a x+ 4 3

+ L/2

− L/2

) 1 L 1 - L .3 * = 2 ρπ a 2 " a 2 + / 0 # "$ 4 2 3 1 2 2 #%

Iy =

But for whole cylinder:

1 ρπ a 2 L(3a 2 + L2 ) " 12

m = ρV = ρπ a 2 L Iy =

Thus,

1 m(3a 2 + L2 ) 12


PROBLEM 9.121 The area shown is revolved about the x axis to form a homogeneous solid of revolution of mass m. Determine by direct integration the mass moment of inertia of the solid with respect to (a) the x axis, (b) the y axis. Express your answers in terms of m and the dimensions of the solid.

SOLUTION We have at

(a, h): h =

k a

k = ah

or For the element shown:

r=4 dm = ρπ r 2 dx 2

- ah . = ρπ / 0 dx 1 x 2

3

m = dm =

Then

3

3a a

2

- ah . ρπ / 0 dx 1 x 2 3a

) 1* = ρπ a 2 h2 " − # $ x %a

) 1 - 1 .* 2 = ρπ a 2 h2 " − − / − 0 # = ρπ ah 2 $ 3a 1 a 2 % 3

(a)

For the element:

Then

dI x =

1 2 1 r dm = ρπ r 4 dx 2 2

3

I x = dI x =

3

3a a

3a

4

1 1 - ah . ) 1 1* ρπ / 0 dx = ρπ a 4 h 4 " − 3 # 2 2 1 x 2 $ 3 x %a

)- 1 .3 - 1 .3 * 1 26 1 = − ρπ a 4 h 4 "/ 0 − / 0 # = × ρπ ah4 6 3 6 27 a a $"1 2 1 2 %# =

1 2 13 13 2 mh × ρπ ah 2 × h 2 = 6 3 9 54

or

I x = 0.241 mh 2



(b)

For the element:

dI y = dI y + x 2 dm =

Then

1 2 r dm + x 2 dm 4

3

I y = dI y =

3

3a ) 1 a

2 2 * ah . - ah . " / 0 + x 2 # ρπ / 0 dx 1 x 2 "$ 4 1 x 2 #% 2 2

= ρπ a h

3

3a

3a a

. * 1 a 2 h2 1 a 2 h2 2 2) + = − + x# ρπ dx a h 1 // 0 " 4 3 0 14 x 2 $ 12 x %a

2 2 * ) 1 a 2 h2 * ,& - 3 . +& ) 1 a h = / m 0 a "− + + a# ! a 3 # − "− 3 3 1 2 2 '& $ 12 (3a) % $ 12 (a ) % &( 2 - 1 26 h . 3 - 13 2 . h + 3a 2 0 = ma // × + 2a 00 = m / 2 1 108 2 1 12 27 a 2

or

I y = m(3a 2 + 0.1204h 2 )


PROBLEM 9.122 Determine by direct integration the mass moment of inertia with respect to the x axis of the tetrahedron shown, assuming that it has a uniform density and a mass m.

SOLUTION We have

x=

a y. y + a = a /1 + 0 h h2 1

and

z=

b y. y + b = b /1 + 0 h h2 1 2

For the element shown:

Then

y. -1 . 1 dm = ρ / xz dy 0 = ρ ab /1 + 0 dy h 2 2 1 2 1 2

3

m = dm =

3

2

y. 1 ρ ab /1 + 0 dy −h 2 h2 1 0

0

3 h )y. * 1 = ρ ab × "/1 + 0 # h 2 %# 2 3 $"1 −h

1 ρ abh )$(1)3 − (1 − 1)3 *% 6 1 = ρ abh 6 =

Now, for the element:

Then

I AA′,area =

y. 1 3 1 3xz = ab /1 + 0 36 36 h2 1

4

4 )1 y. * dI AA′,mass = ρ tI AA′,area = ρ ( dy ) " ab3 /1 + 0 # h 2 #% 1 "$ 3



Now 2 ) -1 . * dI x = dI AA′, mass + " y 2 + / z 0 # dm 1 3 2 %# $" 4

=

1 y. ρ ab3 /1 + 0 dy 36 h 1 2 2 2 * )1 y . * &, ) 1 y. &+ + y 2 + " b /1 + 0 # ! " ρ ab /1 + 0 dy # h 2 % & $" 2 h2 1 $3 1 %# '& (

=

Now

Then

and

m=

4 1 1 y. y3 y4 ρ ab3 /1 + 0 dy + ρ ab // y 2 + 2 + 2 12 2 h2 h h 1 1

. 00 dy 2

1 ρ abh 6

) 1 b2 dI x = " m $" 2 h

3

I x = dI x =

4 y. 3m - 2 y3 y 4 .* + + + + 1 y 2 / 0 # dy / h 02 h 1/ h h2 20 %# 1

3

4 m ) 2y. y3 y 4 .* + "b /1 + 0 + 6 // y 2 + 2 0 #dy − h 2h " h2 h h 2 20 #% 1 $ 1 0

5 -1 m ) 2 hy. y5 1 y4 = + 2 "b × /1 + 0 + 6 // y 3 + h2 2h $" 51 2 h 5h 13

=

0

.* 00 # 2 %# − h

m +1 2 1 1 )1 *, b h(1)5 − 6 " (− h)3 + ( − h ) 4 + 2 ( − h )5 # ! 2h ' 5 2h 5h $3 %(

or

Ix =

1 m(b 2 + h 2 ) 10


PROBLEM 9.123 Determine by direct integration the mass moment of inertia with respect to the y axis of the tetrahedron shown, assuming that it has a uniform density and a mass m.

SOLUTION We have

x=

a y. y + a = a /1 + 0 h h2 1

and

z=

b y. y + b = b /1 + 0 h h2 1 2


Then

y. -1 . 1 dm = ρ / xz dy 0 = ρ ab /1 + 0 dy 2 2 h 1 2 1 2

3

m = dm =

3

2

1 y. ρ ab /1 + 0 dy −h 2 h2 1 0

0

3 1 h )y. * = ρ ab × "/1 + 0 # 2 3 $"1 h 2 %# −h

1 ρ abh )$(1)3 − (1 − 1)3 *% 6 1 = ρ abh 6 =

Also Then, using we have

I BB′,area =

1 3 xz 12

I DD′,area =

1 3 zx 12

I mass = ρ tI area - 1 . dI BB′,mass = ρ (dy ) / xz 3 0 1 12 2

- 1 . dI DD′,mass = ρ (dy ) / zx3 0 1 12 2



Now dI y = dI BB′, mass + dI DD′, mass 1 ρ xz ( x 2 + z 2 )dy 12 2 2 y. ) y. * 1 = ρ ab /1 + 0 "(a 2 + b2 ) /1 + 0 # dy h 2 "$ h 2 #% 12 1 1 =

4

We have

m=

1 m y. ρ abh 5 dI y = (a 2 + b 2 ) /1 + 0 dy 6 2h h 1 2

Then

3

I y = dI y =

3

4

m 2 y. (a + b 2 ) /1 + 0 dy − h 2h h 1 2 0

0

5 m 2 h )y. * a + b 2 × "/ 1 + 0 # = h 2 #% 2h 5 "$1 −h m 2 a + b 2 )$(1)5 − (1 − 1)5 *% = 10

(

)

(

)

or

Iy =

1 m(a 2 + b 2 ) 10


!

PROBLEM 9.124* Determine by direct integration the mass moment of inertia with respect to the z axis of the semiellipsoid shown, assuming that it has a uniform density and a mass m.

SOLUTION First note that when


Then

1/2

x2 z = 0: y = b //1 − 2 1 a

. 00 2

x2 y = 0: z = c //1 − 2 1 a

. 00 2

1/2

x2 dm = ρ (π yzdx) = πρ bc //1 − 2 1 a

3

m = dm =

3

a 0

-

πρ bc //1 − 1

x2 a2

. 00 dx 2

. 00 dx 2 a

1 2 ) * = πρ bc " x − 2 x3 # = πρ abc 3a $ %0 3

For the element: Then Now

I AA′,area =

π 4

zy 3

-π . dI AA′,mass = ρ tI AA′,area = ρ (dx) / zy 3 0 14 2 dI z = dI AA′, mass + x 2 dm x2 = ρ b c //1 − 2 4 1 a

π

=

3

3m ) b 2 " 2a "$ 4

2

) . x2 2 00 dx + x "πρ bc //1 − 2 "$ 2 1 a

. * 00 dx # 2 #%

x2 x4 . - 2 x4 . * //1 − 2 2 + 4 00 + // x − 2 00 # dx a a 2 1 a 2 #% 1



Finally,

3

I z = dI z a ) b2

x2 x4 . - 2 x4 .* " //1 − 2 2 + 4 00 + // x − 2 00 # dx a a 2 1 a 2 #% "$ 4 1

=

3m 2a

=

3m ) b 2 " 2a "$ 4

2 x3 1 x5 + // x − 3 a2 5 a4 1

=

3 ) b2 m" 2 $4

1 .* - 2 1. 2-1 /1 − + 0 + a / − 0 # 1 3 52 1 3 5 2%

3

0

. - 1 3 1 x5 00 + // x − 5 a2 2 13

a

.* 00 # 2 #% 0

or

Iz =

1 m a 2 + b2 5

(

)


!

PROBLEM 9.125* A thin steel wire is bent into the shape shown. Denoting the mass per unit length of the wire by m′, determine by direct integration the mass moment of inertia of the wire with respect to each of the coordinate axes.

SOLUTION dy = − x −1/3 (a 2/3 − x 2/3 )1/2 dx

First note

2

Then

- dy . 1 + / 0 = 1 + x −2/3 (a 2/3 − x 2/3 ) 1 dx 2 -a. =/ 0 1 x2

2/3

2

- dy . dm = m′dL = m′ 1 + / 0 dx 1 dx 2


1/3

-a. = m′ / 0 1 x2

3

3

a

Then

m = dm =

Now

I x = y 2 dm =

3

0

m′

dx

a a1/3 3 3 dx = m′a1/3 $) x 2/3 %* = m′a 1/3 0 2 2 x

- a1/3 . (a 2/3 − x 2/3 )3 // m′ 1/3 dx 00 0 1 x 2 2 a- a . = m′a1/3 // 1/3 − 3a 4/3 x1/3 + 3a 2/3 x − x5/3 00 dx 0 x 1 2

3

a

3

a

9 3 3 )3 * = m′a1/3 " a 2 x 2/3 − a 4/3 x 4/3 + a 2/3 x 2 − x8/3 # 2 4 2 8 $ %0 3 9 3 3 3 . = m′a3 / − + − 0 = m′a3 1 2 4 2 82 8

or Symmetry implies

Ix =

1 2 ma 4

Iy =

1 2 ma 4


!


Alternative solution: - a1/3 . x 2 // m′ 1/3 dx 00 = m′a1/3 0 1 x 2 a 3 3 = m′a1/3 × )$ x8/3 *% = m′a3 0 8 8 1 = ma 2 4

3

Also

I y = x 2 dm =

3

3(x

)

Iz =

2

a

3

a

x5/3 dx

0

+ y 2 dm = I y + I x

or

Iz =

1 2 ma 2


!

PROBLEM 9.126 A thin triangular plate of mass m is welded along its base AB to a block as shown. Knowing that the plate forms an angle θ with the y axis, determine by direct integration the mass moment of inertia of the plate with respect to (a) the x axis, (b) the y axis, (c) the z axis.

SOLUTION For line BC:

ζ =−

=

h x+h a 2

h (a − 2 x) a

-1 . m = ρV = ρ t / ah 0 12 2 1 = ρ tah 2

Also

2

(a)

We have

where Then

1 -ζ . dI x = ζ 2 dm′ + / 0 dm′ 12 122 1 2 = ζ dm′ 3

dm′ = ρ tζ dx

3

I x = dI x = 2

3

a/2 1

3

0

ζ 2 ( ρ tζ dx)

a/2 ) h

3

* " a (a − 2 x) # dx $ %

=

2 ρt 3

=

a/2 2 h3 1 - 1 . ρ t 3 × / − 0 )$(a − 2 x) 4 *% 0 3 a 41 22

3

0

1 h3 ρ t 3 )$(a − a ) 4 − (a )4 *% 12 a 1 = ρ tah3 12

=−

or

Ix =

1 2 mh 6



3

Now

Iζ = x 2 dm

and

Iζ = x 2 dm′ = 2

3

3

a/2 0

x 2 ( ρ tζ dx) a/2

)h * x 2 " (a − 2 x) # dx $a %

= 2ρt

3

= 2ρt

h )a 3 1 4* x − x # a "$ 3 4 %0

0

a/2

h )a - a . 1 - a . = 2ρt " / 0 − / 0 a "$ 3 1 2 2 4 1 2 2 3

=

(b)

We have

3

4*

# #%

1 1 ρ ta3 h = ma 2 48 24

3

I y = ry2 dm = )$ x 2 + (ζ sin θ )2 *% dm

3

3

= x 2 dm + sin 2 θ ζ 2 dm

Now

3

I x = ζ 2 dm 5 I y = Iζ + I x sin 2 θ =

(c)

We have

3

1 1 ma 2 + mh 2 sin 2 θ 24 6

or

Iy =

m 2 (a + 4h2 sin 2 θ ) 24

or

Iz =

m 2 ( a + 4h 2 cos 2 θ ) 24

3

I z = rz2 dm = ( x 2 + y 2 )dm

3 = 3 x dm + cos θ 3 ζ

= )$ x 2 + (ζ cos θ ) 2 *% dm 2

2

2

dm

= Iζ + I x cos 2 θ =

1 1 ma 2 + mh2 cos 2 θ 24 6


!

PROBLEM 9.127 Shown is the cross section of a molded flat-belt pulley. Determine its mass moment of inertia and its radius of gyration with respect to the axis AA′. (The density of brass is 8650 kg/m3 and the density of the fiber-reinforced polycarbonate used is 1250 kg/m3.)

SOLUTION First note for the cylindrical ring shown that m = ρV = ρ t ×

π 4

(d

2 2

− d12

)

and, using Figure 9.28, that 2

2

1 - d2 . 1 -d . m2 − m1 / 1 0 2 /1 2 02 2 1 22 π . π . * 1 )= "/ ρ t × d 22 0 d 22 − / ρ t × d12 0 d12 # 8 $1 4 2 4 2 % 1 1-π . = / ρ t 0 d 24 − d14 81 4 2 1-π . = / ρ t 0 d 22 − d12 d 22 + d12 81 4 2 1 = m d12 + d 22 8

I AA′ =

(

)

(

)(

(

)

)

Now treat the pulley as four concentric rings and, working from the brass outward, we have m=

π

{

8650 kg/m3 × (0.0175 m) × (0.0112 − 0.0052 ) m 2 4 + 1250 kg/m3 )$(0.0175 m) × (0.017 2 − 0.0112 ) m 2 + (0.002 m) × (0.0222 − 0.017 2 ) m 2

}

+ (0.0095 m) × (0.0282 − 0.0222 ) m 2 *%

= (11.4134 + 2.8863 + 0.38288 + 2.7980) × 10−3 kg = 17.4806 × 10−3 kg PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1559


and

I AA′ =

1 )(11.4134)(0.0052 + 0.0112 ) + (2.8863)(0.0112 + 0.017 2 ) 8$ + (0.38288)(0.017 2 + 0.0222 ) +(2.7980)(0.0222 + 0.0282 ) *% × 10−3 kg × m 2

= (208.29 + 147.92 + 37.00 + 443.48) × 10−9 kg ⋅ m 2 = 836.69 × 10−9 kg ⋅ m 2

or Now

2 k AA ′ =

I AA′ = 837 × 10−9 kg ⋅ m 2

I AA′ 836.69 × 10−9 kg ⋅ m 2 = = 47.864 × 10−6 m 2 −3 m 17.4806 × 10 kg

or

k AA′ = 6.92 mm


!

PROBLEM 9.128 Shown is the cross section of an idler roller. Determine its mass moment of inertia and its radius of gyration with respect to the axis AA′. (The specific weight of bronze is 0.310 lb/in3; of aluminum, 0.100 lb/in3; and of neoprene, 0.0452 lb/in3.)

SOLUTION First note for the cylindrical ring shown that m = ρV = ρ t ×

π

(d 4

2 2

)

− d12 =

π 4

(

ρ t d 22 − d12

)

and, using Figure 9.28, that 2

2

1 - d2 . 1 -d . m2 − m1 / 1 0 2 /1 2 02 2 1 22 1 )π . π . * = "/ ρ t × d 22 0 d 22 − / ρ t × d12 0 d12 # 8 $1 4 2 4 2 % 1

I AA′ =

1-π . = / ρ t 0 d 24 − d14 81 4 2 1-π . = / ρ t 0 d 22 − d12 81 4 2 1 = m d12 + d 22 8

(

)

(

)( d

(

2 2

+ d12

)

)

Now treat the roller as three concentric rings and, working from the bronze outward, we have m=

2 2 +& 1 - 13 . )- 3 . - 1 . * ft/s 2 (0.310 lb/in.3 ) / in. 0 "/ 0 − / 0 # in.2 4 32.2 1 16 2 $"1 8 2 1 4 2 %# &' 2 2 - 11 . )- 1 . - 3 . * + 0.100 lb/in.3 / in. 0 "/ 0 − / 0 # in.2 1 16 2 "$1 2 2 1 8 2 #% 2 2 ,& - 11 . )- 1 . - 1 . * + 0.0452 lb/in.3 / in. 0 "/1 0 − / 0 # in.2 ! 1 16 2 "$1 8 2 1 2 2 #% &(

π

×

( (

)

)

= ( 479.96 + 183.41 + 769.80 ) × 10 −6 lb ⋅ s 2 /ft = 1.4332 × 10−3 lb ⋅ s 2 /ft



and

)- 1 . 2 - 3 . 2 * )- 3 . 2 - 1 . 2 * 1 +& I AA′ = (479.96) "/ 0 + / 0 # + (183.41) "/ 0 + / 0 # 8& "$1 4 2 1 8 2 #% "$1 8 2 1 2 2 #% ' )- 1 .2 - 1 . 2 * ,& 1 ft 2 + (769.80) "/1 0 + / 0 # ! × 10−6 lb ⋅ s 2 /ft × in.2 × 144 in.2 "$1 8 2 1 2 2 #% &( = (84.628 + 62.191 + 1012.78) × 10−9 lb ⋅ ft ⋅ s 2 = 1.15960 × 10−6 lb ⋅ ft ⋅ s 2

or

I AA′ = 1.160 × 10−6 lb ⋅ ft ⋅ s 2

I AA′ 1.15960 × 10 −6 lb ⋅ ft ⋅ s 2 = = 809.09 × 10−6 ft 2 m 1.4332 × 10 −3 lb ⋅ s 2 /ft

Now

2 k AA ′ =

Then

k AA′ = 28.445 × 10−3 ft

or

k AA′ = 0.341 in.


!

PROBLEM 9.129 Knowing that the thin cylindrical shell shown is of mass m, thickness t, and height h, determine the mass moment of inertia of the shell with respect to the x axis. (Hint: Consider the shell as formed by removing a cylinder of radius a and height h from a cylinder of radius a + t and height h; then neglect terms containing t2 and t3 and keep those terms containing t.)

SOLUTION From Figure 9.28 for a solid cylinder (change orientation of axes): I x′ =

1 m(3a 2 + h 2 ) 12

-h. I x = I x′ + m / 0 122

2

1 h2 m(3a 2 + h 2 ) + m 12 4 - 3 2 h2 . = m // a + 00 3 2 1 12 =

Shell = Cylinder

Ix =

1 m(3a 2 + 4h2 ) " 12

– Cylinder !

1 1 m1[3(a + t ) 2 + 4h 2 ] − m2 (3a 2 + 4h 2 ) 12 12 1 1 2 = [ ρπ (a + t ) h][3(a + t ) 2 + 4h 2 ] − ( ρπ a 2 h)(3a 2 + 4h 2 ) 12 12

Ix =



Expand (a + t ) factors and neglecting terms in t 2, t 3, and t 4 :

ρπ h )

(

)(

)

a 2 + 2at + t 2 3a 2 + 6at + 3t 2 + 4h 2 * − 3a 4 − 4a 2 h2 % 12 $ ρπ h 3a 4 + 6a3t + 3a 2 t 2 + 4a 2 h2 + 6a 3t + 8ath 2 − 3a 4 − 4a 2 h 2 = 12 ρπ h 12a 3t + 8ath 2 = 12 ρπ hat 12a 2 + 8h 2 = 12

Ix =

( (

)

(

Mass of shell:

)

)

m = ρ (2π ath) = 2 ρπ hat 2 ρπ hat 12a 2 + 8h 2 24 4m 3a 2 + 2h 2 = 24

(

Ix =

(

)

)

or

Ix =

m 3a 2 + 2h2 6

(

)

Checks: 1 , a = 0; I x = mh 2 − slender rod & & 3 ! OK ! 1 2 h = 0; I x = ma − thin ring & &( 2


!

PROBLEM 9.130 The machine part shown is formed by machining a conical surface into a circular cylinder. For b = 12 h, determine the mass moment of inertia and the radius of gyration of the machine part with respect to the y axis.

SOLUTION

Mass:

mcyl = ρπ a 2 h 1 mcyl a 2 2 1 = ρπ a 4 h 2

I y : I cyl =

mcone =

h 1 1 ρπ a 2 = ρπ a 2 h 3 2 6

3 mcone a 2 10 1 = ρπ a 4 h 20

I cone =

For entire machine part: 1 5 ρπ a 2 h = ρπ a 2 h 6 6 1 1 9 ρπ a 4 h = ρπ a 4 h = ρπ a 4 h − 2 20 20

m = mcyl − mcone = ρπ a 2 h − I y = I cyl − I cone

or

-5 .- 6 .- 9 . I y = / ρπ a 2 h 0/ 0/ 0 a 2 16 21 5 21 20 2

Then

k y2 =

I 27 2 a = m 50

Iy =

27 2 ma 50

k y = 0.735a


!

PROBLEM 9.131 After a period of use, one of the blades of a shredder has been worn to the shape shown and is of mass 0.18 kg. Knowing that the mass moments of inertia of the blade with respect to the AA′ and BB′ axes are 0.320 g ⋅ m 2 and 0.680 g ⋅ m 2 , respectively, determine (a) the location of the centroidal axis GG′, (b) the radius of gyration with respect to axis GG′.

SOLUTION (a)

d B = (0.08 − d A ) m

We have and, using the parallel axis

I AA′ = I GG′ + md A2 I BB′ = I GG′ + md B2

Then

(

I BB′ − I AA′ = m d B2 − d A2

)

= m "(0.08 − d A ) 2 − d A2 !# = m(0.0064 − 0.16d A )

Substituting:

(0.68 − 0.32) × 10−3 kg ⋅ m 2 = 0.18 kg(0.0064 − 0.16d A ) m 2

or (b)

We have

I AA′ = I GG′ + md A2

or

I GG ′ = 0.32 × 10−3 kg ⋅ m 2 − 0.18 kg ⋅ (0.0275 m)2

d A = 27.5 mm

= 0.183875 × 10−3 kg ⋅ m 2

Then

2 kGG ′ =

I GG′ 0.183875 × 10−3 kg ⋅ m 2 = = 1.02153 × 10−3 m 2 m 0.18 kg

or

kGG ′ = 32.0 mm


!

PROBLEM 9.132 Determine the mass moment of inertia of the 0.9-lb machine component shown with respect to the axis AA′.

SOLUTION First note that the given shape can be formed adding a small cone to a cylinder and then removing a larger cone as indicated.

h h + 2.4 = 0.4 1.2

Now

or h = 1.2 in.

The weight of the body is given by W = mg = g (m1 + m2 − m3 ) = ρ g (V1 + V2 − V3 )

or

0.9 lb = ρ × 32.2 ft/s 2 2 (π (0.8) (2.4) +

"

π 3

(0.2) 2 (1.2) −

! $ 1 ft % (0.6)2 (3.6) ) in 3 × & ' 3 # * 12 in. +

π

3

= ρ × 32.2 ft/s 2 (2.79253 + 0.02909 − 0.78540) × 10−3 ft 3

or Then

ρ = 13.7266 lb ⋅ s 2 /ft 4 m1 = (13.7266)(2.79253 × 10 −3 ) = 0.038332 lb ⋅ s 2 /ft m2 = (13.7266)(0.02909 × 10 −3 ) = 0.000399 lb ⋅ s 2 /ft m3 = (13.7266)(0.78540 × 10−3 ) = 0.010781 lb ⋅ s 2 /ft



Finally, using Figure 9.28, we have I AA′ = ( I AA′ )1 + ( I AA′ ) 2 − ( I AA′ )3 =

1 3 3 m1a12 + m2 a22 − m3 a32 2 10 10

1 3 3 ! $ 1 ft % = ( (0.038332)(0.8)2 + (0.000399)(0.2)2 − (0.010781)(0.6)2 ) (lb ⋅ s2 /ft) × in 2 × & ' 10 10 "2 # * 12 in. +

2

= (85.1822 + 0.0333 − 8.0858) × 10−6 lb ⋅ ft ⋅ s 2

or

I AA′ = 77.1 × 10−6 lb ⋅ ft ⋅ s 2


!

PROBLEM 9.133 A square hole is centered in and extends through the aluminum machine component shown. Determine (a) the value of a for which the mass moment of inertia of the component with respect to the axis AA′, which bisects the top surface of the hole, is maximum, (b) the corresponding values of the mass moment of inertia and the radius of gyration with respect to the axis AA′. (The specific weight of aluminum is 0.100 lb/in.3.)

SOLUTION First note

m1 = ρV1 = ρ b 2 L

and

m2 = ρV2 = ρ a 2 L

(a)

Using Figure 9.28 and the parallel-axis theorem, we have I AA′ = ( I AA′ )1 − ( I AA′ ) 2 1 $a% = ( m1 (b 2 + b 2 ) + m1 & ' 12 *2+ ("

2!

) )#

2 1 $a% ! − ( m2 (a 2 + a 2 ) + m2 & ' ) * 2 + )# ("12 1 % $1 $ 5 % = ( ρ b 2 L) & b 2 + a 2 ' − ( ρ a 2 L) & a 2 ' 4 + *6 * 12 + ρL (2b 4 + 3b 2 a 2 − 5a 4 ) = 12

Then

dI AA′ ρ L (6b 2 a − 20a 3 ) = 0 = da 12

or

a=0 d 2 I AA′

Also

da 2

and for a = b

=

ρL 12

d 2 I AA′

Now for a = 0,

da 2 3 , 10

a=b

and

d 2 I AA′ da 2

3 10

(6b 2 − 60a 2 ) =

1 ρ L(b2 − 10a 2 ) 2

.0 ,0



( I AA′ )max occurs when

Then

a=b

3 10

a = (4.2 in.)

3 10

or (b)

a = 2.30 in.

From Part a: ( I AA′ ) max =

=

2 4! $ $ 3 % 3 % ) 49 − 5&b = 2b 4 + 3b 2 & b ρ Lb 4 ' ' & 10 ' & 10 ' ) 240 12 ( * + * + # "

ρL (

$ 1 ft % 49 γ AL 4 49 0.100 lb/in.3 × × (15 in.)(4.2 in.)4 & Lb = ' 2 240 g 240 32.2 ft/s * 12 in. +

or Now where

2 k AA ′ =

2

(I AA′ )max = 20.6 × 10−3 lb ⋅ ft ⋅ s 2

( I AA′ ) max m

m = m1 − m2 = ρ L(b 2 − a 2 ) 2 $ 3 % !) 2 ( = ρL b − &b & 10 '' ) ( * + # " 7 = ρ Lb2 10

Then

2 k AA ′ =

49 ρ Lb 4 240 7 ρ Lb 2 10

=

7 2 7 b = (4.2 in.)2 24 24

or

k AA′ = 2.27 in.

!


PROBLEM 9.134 The cups and the arms of an anemometer are fabricated from a material of density ρ. Knowing that the mass moment of inertia of a thin, hemispherical shell of mass m and thickness t with respect to its centroidal axis GG′ is 5ma 2 /12, determine (a) the mass moment of inertia of the anemometer with respect to the axis AA′, (b) the ratio of a to l for which the centroidal moment of inertia of the cups is equal to 1 percent of the moment of inertia of the cups with respect to the axis AA′.

SOLUTION (a)

First note and

marm = ρVarm = ρ ×

π 4

d 2l

dmcup = ρ dVcup = ρ[(2π a cos θ )(t )(adθ )]

Then

,

mcup = dmcup =

,

π /2 0

2πρ a 2 t cos θ dθ

= 2πρ a 2 t[sin θ ]π0 /2 = 2πρ a 2 t

Now

( I AA′ )anem = ( I AA′ )cups + ( I AA′ )arms

Using the parallel-axis theorem and assuming the arms are slender rods, we have 2 ! ! ( I AA′ )anem = 3 "( I GG ′ )cup + mcup d AG # + 3 " I arm + marm d AGarm # 2 2 1 /- 5 $ a % ! /. $l% ! 2 2 2 = 3 0 mcup a + mcup ((l + a) + & ' ) 1 + 3 ( marm l + marm & ' ) * 2 + )# /3 * 2 + )# (" (" 2 /212 $5 % = 3mcup & a 2 + 2la + l 2 ' + marm l 2 3 * +

% $5 % $π = 3(2πρ a 2 t ) & a 2 + 2la + l 2 ' + & ρ d 2 l ' (l 2 ) *3 + *4 +

or

$ 5 a2 a % d 2l ! ( I AA′ )anem = πρ l 2 (6a 2 t && 2 + 2 + 1'' + ) l *3 l + 4 )# "(


!


(b)

( I GG ′ )cup

We have

( I AA′ )cup

5 $5 % mcup a 2 = 0.01mcup & a 2 + 2la + l 2 ' (from Part a ) 12 *3 +

or Now let ζ =

= 0.01

a . l

Then

$5 % 5ζ 2 = 0.12 & ζ 2 + 2ζ + 1' 3 * +

or

40ζ 2 − 2ζ − 1 = 0 2 ± (−2) 2 − 4(40)(−1) 2(40)

Then

ζ =

or

ζ = 0.1851 and ζ = − 0.1351 a = 0.1851 l

To the instructor: The following formulas for the mass moment of inertia of thin plates and a half cylindrical shell are derived at this time for use in the solutions of Problems 9.135 through 9.140. Thin rectangular plate ( I x )m = ( I x′ ) m + md 2 2

=

1 $b% $h% m(b 2 + h 2 ) + m (& ' + & ' 12 ("* 2 + * 2 +

2!

) )#

1 = m(b2 + h 2 ) 3 ( I y ) m = ( I y′ ) m + md 2 2

=

1 1 $b% mb 2 + m & ' = mb 2 12 2 3 * +

I z = ( I z′ ) m + md 2 2

=

1 1 $h% mh 2 + m & ' = mh 2 12 2 3 * +



Thin triangular plate We have

$1 % m = ρV = ρ & bht ' *2 + 1 3 bh 36

and

I z ,area =

Then

I z ,mass = ρ tI z ,area

= ρt ×

1 3 bh 36

=

1 mh 2 18

Similarly,

I y ,mass =

1 mb2 18

Now

I x, mass = I y ,mass + I z ,mass =

1 m(b2 + h 2 ) 18

Thin semicircular plate We have

$π % m = ρV = ρ & a 2 t ' *2 +

π

a4

and

I y ,area = I z ,area =

Then

I y ,mass = I z ,mass = ρ tI y ,area

8

= ρt × =

π 8

a4

1 2 ma 4 1 ma 2 2

Now


Also

I x, mass = I x′,mass + my 2

or

$ 1 16 I x′,mass = m & − 2 * 2 9π

% 2 'a +

and

I z ,mass = I z′,mass + my 2

or

$ 1 16 I z ′,mass = m & − 2 * 4 9π

% 2 'a +



y=z =

4a 3π

Thin Quarter-Circular Plate We have

$π % m = ρV = ρ & a 2 t ' *4 +

π

a4

and

I y ,area = I z ,area =

Then

I y ,mass = I z ,mass = ρ tI y ,area

16

= ρt × =

π 16

a4

1 ma 2 4 1 ma 2 2

Now


Also

I x, mass = I x′,mass + m( y 2 + z 2 )

or

$ 1 32 I x′,mass = m & − 2 * 2 9π

and

I y ,mass = I y′,mass + mz 2

or

$ 1 16 % I y′,mass = m & − 2 ' a 2 * 4 9π +

% 2 'a +


PROBLEM 9.135 A 2-mm thick piece of sheet steel is cut and bent into the machine component shown. Knowing that the density of steel is 7850 kg/m3, determine the mass moment of inertia of the component with respect to each of the coordinate axes.

SOLUTION First compute the mass of each component. We have m = ρSTV = ρST tA

Then m1 = (7850 kg/m3 )(0.002 m)(0.3 m) 2 = 1.413 kg m2 = m3 = (7850 kg/m3 )(0.002 m) × (0.15 × 0.12) m 2 = 0.2826 kg

Using Figure 9.28 and the parallel-axis theorem, we have I x = ( I x )1 + 2( I x )2 1 ! = ( (1.413 kg)(0.3 m) 2 ) "12 # 1 ! + 2 ( (0.2826 kg)(0.12 m) 2 + (0.2828 kg)(0.152 + 0.06 2 )m 2 ) 12 " # = [(0.0105975) + 2(0.0003391 + 0.0073759)] kg ⋅ m 2 = [(0.0105975) + 2(0.0077150)] kg ⋅ m 2

or

I x = 26.0 × 10−3 kg ⋅ m 2

I y = ( I y )1 + 2( I y )2 1 ! = ( (1.413 kg)(0.32 + 0.32 ) m 2 ) 12 " # 1 ! 2 + 2 ( (0.2826 kg)(0.15 m) + (0.2826 kg)(0.0752 + 0.152 ) m 2 ) "12 # = [(0.0211950) + 2(0.0005299 + 0.0079481)] kg ⋅ m 2 = [(0.0211950) + 2(0.0084780)] kg ⋅ m 2

or

I y = 38.2 × 10−3 kg ⋅ m 2



I z = ( I z )1 + 2( I z ) 2 1 ! = ( (1.413 kg)(0.3 m) 2 ) 12 " # 1 ! + 2 ( (0.2826 kg)(0.152 + 0.122 ) m 2 + (0.2826 kg)(0.0752 + 0.062 ) m 2 ) "12 # = [(0.0105975) + 2(0.0008690 + 0.0026070)] kg ⋅ m 2 = [(0.0105975) + 2(0.0034760)] kg ⋅ m 2

or

I z = 17.55 × 10−3 kg ⋅ m 2


!



Then m1 = (7850 kg/m3 )(0.002 m)(0.35 × 0.39) m 2 = 2.14305 kg $π % m2 = (7850 kg/m3 )(0.002 m) & × 0.1952 ' m 2 = 0.93775 kg 2 * + $1 % m3 = (7850 kg/m3 )(0.002 m) & × 0.39 × 0.15 ' m 2 = 0.45923 kg 2 * +

Using Figure 9.28 for component 1 and the equations derived above for components 2 and 3, we have I x = ( I x )1 + ( I x )2 + ( I x )3 2 1 $ 0.39 % ! 2 m' ) = ( (2.14305 kg)(0.39 m) + (2.14305 kg) & * 2 + )# ("12 2 -/$ 1 16 % ! ./ $ 4 × 0.195 % + 0& − 2 ' (0.93775 kg)(0.195 m) 2 + (0.93775 kg) (& + (0.195) 2 ) m 2 1 ' + /2* 2 9π + "(* 3π #) /3 2 2 -/ 1 $ 0.39 % $ 0.15 % ! 2 ./ + 0 (0.45923 kg)[(0.39) 2 + (0.15)2 ] m 2 + (0.45923 kg) (& + ' & ' )m 1 ("* 3 + * 3 + #) /3 /218

= [(0.027163 + 0.081489) + (0.011406 + 0.042081) + (0.004455 + 0.008909)] kg ⋅ m 2 = (0.108652 + 0.053487 + 0.013364) kg ⋅ m 2 = 0.175503 kg ⋅ m 2

or

I x = 175.5 × 10−3 kg ⋅ m 2



I y = ( I y )1 + ( I y )2 + ( I y )3 2 2 -/ 1 $ 0.35 % $ 0.39 % ! 2 ./ = 0 (2.14305 kg)[(0.35) 2 + (0.39) 2 ] m 2 + (2.14305 kg) (& + ' & ' )m 1 ("* 2 + * 2 + )# 3/ /212

1 ! + ( (0.93775 kg)(0.195 m)2 + (0.93775 kg)(0.195 m) 2 ) "4 # 2 -/ 1 $ 0.39 % ! 2 ./ 2 2 + 0 (0.45923 kg)(0.39 m) + (0.45923 kg) ((0.35) + & ' )m 1 * 3 + )# /3 (" /218 = [(0.049040 + 0.147120) + (0.008914 + 0.035658) + (0.003880 + 0.064017)] kg ⋅ m 2 = (0.196160 + 0.044572 + 0.067897) kg ⋅ m 2 = 0.308629 kg ⋅ m 2

or

I y = 309 × 10−3 kg ⋅ m 2

I z = ( I z )1 + ( I z ) 2 + ( I z )3 2 1 $ 0.35 % ! m' ) = ( (2.14305 kg)(0.35 m) 2 + (2.14305 kg) & * 2 + )# ("12

1 ! + ( (0.93775 kg)(0.195 m)2 ) "4 # 2 /- 1 $ 0.15 % ! 2 /. + 0 (0.45923 kg)(0.15 m) 2 + (0.45923 kg) ((0.35) 2 + & ' )m 1 * 3 + #) 3/ (" /218

= [(0.021877 + 0.065631) + 0.008914) + (0.000574 + 0.057404)] kg ⋅ m 2 = (0.087508 + 0.008914 + 0.057978) kg ⋅ m 2 = 0.154400 kg ⋅ m 2

or

I z = 154.4 × 10−3 kg ⋅ m 2


!

PROBLEM 9.137 The cover for an electronic device is formed from sheet aluminum that is 0.05 in. thick. Determine the mass moment of inertia of the cover with respect to each of the coordinate axes. (The specific weight of aluminum is 0.100 lb/in.3 .)

SOLUTION First compute the mass of each component. We have m = ρV =

Then

γ g

tA

m1 =

0.100 lb/in.3 × 0.05 in. × (3 × 2.4) in.2 = 1.11801 × 10 −3 lb ⋅ s 2 /ft 2 32.2 ft/s

m2 =

0.100 lb/in.3 × 0.05 in. × (3 × 6.2) in.2 = 2.88820 × 10−3 lb ⋅ s 2 /ft 32.2 ft/s 2

m3 = m4 =

0.100 lb/in.3 × 0.05 in. × (2.4 × 6.2) in.2 = 2.31056 × 10−3 lb ⋅ s 2 /ft 2 32.2 ft/s

Using Figure 9.28 and the parallel-axis theorem, we have ( I x )3 = ( I x ) 4 I x = ( I x )1 + ( I x )2 + ( I x )3 + ( I x )4 1 = ( (1.11801 × 10 −3 lb ⋅ s 2 /ft)(2.4 in.) 2 "12 2 2 $ 2.4 % ! $ 1 ft % in. ' ) & + (1.11801 × 10−3 lb ⋅s 2 /ft) & ' * 2 + )# * 12 in. + 1 + ( (2.88820 × 10−3 lb ⋅ s 2 /ft)(6.2 in.) 2 12 " 2 2 $ 6.2 % ! $ 1 ft % in. ' ) & + (2.88820 × 10 lb ⋅ s /ft) & ' * 2 + )# * 12 in. + -1 + 2 0 (2.31056 × 10−3 lb ⋅ s 2 /ft) "(2.4)2 + (6.2) 2 #! in.2 212

−3

2

2 2 2 $ 2.4 % $ 6.2 % ! 2 /. $ 1 ft % + (2.31056 × 10−3 lb ⋅ s 2 /ft) (& + in. 1& ' ' & ' ) /3 * 12 in. + "(* 2 + * 2 + #)



= [(3.7267 + 11.1801) + (64.2491 + 192.7472) + 2(59.1011 + 177.3034)] × 10 −6 lb ⋅ ft ⋅ s 2 = [14.9068 + 256.9963 + 2(236.4045)] × 10 −6 lb ⋅ ft ⋅ s 2

or

I x = 745 × 10−6 lb ⋅ ft ⋅ s 2

I y = ( I y )1 + ( I y ) 2 + ( I y )3 + ( I y ) 4 1 = ( (1.11801 × 10−3 lb ⋅ s 2 /ft)(3 in.) 2 "12 2 2 $3 % ! $ 1 ft % 2 −3 + (1.11801 × 10 lb ⋅ s /ft) & in. ' ) & ' *2 + )# * 12 in. +

-1 + 0 (2.88820 × 10−3 lb ⋅ s 2 /ft)[(3)2 + (6.2) 2 ] in.2 212 2 2 2 $ 3 % $ 6.2 % ! 2 ./ $ 1 ft % + (2.88820 × 10−3 lb ⋅ s 2 /ft) (& ' + & in. ) 1 ' & ' /3 * 12 in. + "(* 2 + * 2 + #)

1 + ( (2.31056 × 10−3 lb ⋅ s 2 /ft)(6.2 in.)2 "12 2 2 $ 6.2 % ! $ 1 ft % −3 2 + (2.31056 × 10 lb ⋅ s /ft) & in. ' ) & ' * 2 + )# * 12 in. +

-1 + 0 (2.31056 × 10−3 lb ⋅ s 2 /ft)(6.2 in.)2 212 2 2 $ 6.2 % ! 2 ./ $ 1 ft % + (2.31056 × 10−3 lb ⋅ s 2 /ft) ((3)2 + & in. ) 1 ' & ' * 2 + #) /3 * 12 in. + "( = [(5.8230 + 17.4689) + (79.2918 + 237.8754) + (51.3993 + 154.1978)

+ (51.3993 + 298.6078)] × 10−6 lb ⋅ ft ⋅ s 2 = (23.2919 + 317.1672 + 205.5971 + 350.0071) × 10−6 lb ⋅ ft ⋅ s 2

or

I y = 896 × 10−6 lb ⋅ ft ⋅ s 2



I z = ( I z )1 + ( I z ) 2 + ( I z )3 + ( I z ) 4

-1 = 0 (1.11801 × 10 −3 lb ⋅ s 2 /ft)[(3) 2 + (2.4)2 ] in.2 212 2

$ 3 % $ 2.4 % + (1.11801 × 10−3 lb ⋅ s 2 /ft) (& ' + & ' ("* 2 + * 2 + 1 + ( (2.88820 × 10 −3 lb ⋅ s 2 /ft)(3 in.) 2 12 "

./ $ 1 ft % 2 ) in.2 1 & ' )# /3 * 12 in. +

2!

2 2 $3 % ! $ 1 ft % + (2.88820 × 10−3 lb ⋅ s 2 /ft) & in. ' ) & ' *2 + )# * 12 in. +

1 + ( (2.31056 × 10−3 lb ⋅ s 2 /ft)(2.4 in.) 2 "12 2 2 $ 2.4 % ! $ 1 ft % + (2.31056 × 10−3 lb ⋅ s 2 / ft) & in. ' ) & ' * 2 + )# * 12 in. +

-1 + 0 (2.31056 × 10−3 lb ⋅ s 2 /ft)(2.4 in.) 2 212 $ 2.4 % + (2.31056 × 10−3 lb ⋅ s 2 /ft) ( (3) 2 + & ' * 2 + ("

2

! 2 ./ $ 1 ft % 2 ) in. 1 & ' 12 in. + )# 3/ *

= [(9.5497 + 28.6490) + (15.0427 + 45.1281) + (7.7019 + 23.1056) + (7.7019 + 167.5156)] × 10−6 lb ⋅ ft ⋅ s 2 = (38.1987 + 60.1708 + 30.8075 + 175.2175) × 10−6 lb ⋅ ft ⋅ s 2

or

I z = 304 × 10−6 lb ⋅ ft ⋅ s 2


!

PROBLEM 9.138 A framing anchor is formed of 0.05-in.-thick galvanized steel. Determine the mass moment of inertia of the anchor with respect to each of the coordinate axes. (The specific weight of galvanized steel is 470 lb/ft 3 .)

SOLUTION First compute the mass of each component. We have m = ρV =

γ G.S. g

tA

Then 470 lb/ft 3 $ 1 ft % × 0.05 in. × (2.25 × 3.5) in.2 × & ' 2 32.2 ft/s * 12 in. + = 3325.97 × 10−6 lb ⋅ s 2 /ft

3

m1 =

470 lb/ft 3 $ 1 ft % m2 = × 0.05 in. × (2.25 × 1) in.2 × & ' 2 32.2 ft/s * 12 in. +

3

= 950.28 × 10−6 lb ⋅ s 2 /ft 470 lb/ft 3 $1 % $ 1 ft % × 0.05 in. × & × 2 × 4.75 ' in.2 × & ' 2 32.2 ft/s *2 + * 12 in. + = 2006.14 × 10−6 lb ⋅ s 2 /ft

3

m3 =

Using Figure 9.28 for components 1 and 2 and the equations derived above for component 3, we have I x = ( I x )1 + ( I x ) 2 + ( I x )3 1 = ( (3325.97 × 10−6 lb ⋅ s 2 /ft)(3.5 in.) 2 "12 $ 3.5 2 % ! $ 1 ft % in. ' ) & + (3325.97 × 10−6 lb ⋅ s 2 /ft) & ' * 2 + # * 12 in. + -1 + 0 (950.28 × 10−6 lb ⋅ s 2 /ft)(1 in.) 2 212

2

2 ./ $ 1 ft %2 $1% ! + (950.28 × 10−6 lb ⋅ s 2 /ft) ((3.5) 2 + & ' ) in.2 1 & ' * 2 + )# /3 * 12 in. + "(



-1 + 0 (2006.14 × 10−6 lb ⋅ s 2 /ft)[(4.75) 2 + (2)2 ] in.2 218 2

$2 % $1 % + (2006.14 × 10−6 lb ⋅ s 2 /ft) (& × 4.75 ' + & × 2 ' + *3 + ("* 3

./ $ 1 ft % 2 ) in.2 1 & ' )# /3 * 12 in. +

2!

= [(23.578 + 70.735) + (0.550 + 82.490) + (20.559 + 145.894)] × 10−6 lb ⋅ ft ⋅ s 2 = (94.313 + 83.040 + 166.453) × 10−6 lb ⋅ ft ⋅ s 2 I x = 344 × 10−6 lb ⋅ ft ⋅ s 2

or I y = ( I y )1 + ( I y ) 2 + ( I y )3 1 = ( (3325.97 × 10−6 lb ⋅ s 2 /ft)(2.25 in.)2 "12 2 2 $ 2.25 % ! $ 1 ft % in. ' ) & + (3325.97 × 10 lb ⋅ s /ft) & ' * 2 + )# * 12 in. +

−6

2

-1 + 0 (950.28 × 10 −6 lb ⋅ s 2 /ft)[(2.25) 2 + (1) 2 ] in.2 212 2

$ 2.25 % $ 1 % + (950.28 × 10−6 lb ⋅ s 2 /ft) (& ' +& ' ("* 2 + * 2 +

./ $ 1 ft % 2 ) in.2 1 & ' )# /3 * 12 in. +

2!

-1 + 0 (2006.14 × 10−6 lb ⋅ s 2 /ft)(2 in.)2 218 $1 % + (2006.14 × 10−6 lb ⋅ s 2 /ft) ((2.25) 2 + & × 2 ' *3 + ("

./ $ 1 ft %2 ) in.2 1 & ' )# /3 * 12 in. +

2!

= [(9.744 + 29.232) + (3.334 + 10.002) + (3.096 + 76.720)] × 10−6 lb ⋅ ft ⋅ s 2 = (38.976 + 13.336 + 79.816) × 10−6 lb ⋅ ft ⋅ s 2

or

I y = 132.1 × 10−6 lb ⋅ ft ⋅ s 2



I z = ( I z )1 + ( I z ) 2 + ( I z )3 -1 = 0 (3325.97 × 10 −6 lb ⋅ s 2 /ft)[(2.25) 2 + (3.5)2 ] in.2 212 2

$ 2.25 % $ 3.5 % + (3325.97 × 10−6 lb ⋅ s 2 /ft) (& ' +& ' ("* 2 + * 2 +

./ $ 1 ft % 2 ) in.2 1 & ' )# /3 * 12 in. +

2!

-1 + 0 (950.28 × 10−6 lb ⋅ s 2 /ft)(2.25 in.) 2 212 2 ! 2 ./ $ 1 ft % 2 $ 2.25 % 2 + (950.28 × 10 lb ⋅ s /ft) (& ' + (3.5) ) in. 1 & ' ("* 2 + )# /3 * 12 in. + -1 + 0 (2006.14 × 10−6 lb ⋅ s 2 /ft)(4.75 in.) 2 218

−6

2

2 ./ $ 1 ft %2 $2 % ! + (2006.14 × 10−6 lb ⋅ s 2 /ft) ((2.25) 2 + & × 4.75 ' ) in.2 1 & ' *3 + )# /3 * 12 in. + "( = [(33.322 + 99.967) + (2.784 + 89.192)

+ (17.463 + 210.231)] × 10−6 lb ⋅ ft ⋅ s 2 = (133.289 + 91.976 + 227.694) × 10−6 lb ⋅ ft ⋅ s 2

or

I z = 453 × 10−6 lb ⋅ ft ⋅ s 2


PROBLEM 9.139 A subassembly for a model airplane is fabricated from three pieces of 1.5-mm plywood. Neglecting the mass of the adhesive used to assemble the three pieces, determine the mass moment of inertia of the subassembly with respect to each of the coordinate axes. (The density of the plywood is 780 kg/m3 .)

SOLUTION First compute the mass of each component. We have m = ρV = ρ tA

Then $1 % m1 = m2 = (780 kg/m3 )(0.0015 m) & × 0.3 × 0.12 ' m 2 *2 + = 21.0600 × 10−3 kg $π % m3 = (780 kg/m3 )(0.0015 m) & × 0.122 ' m 2 = 13.2324 × 10−3 kg *4 +

Using the equations derived above and the parallel-axis theorem, we have ( I x )1 = ( I x ) 2 I x = ( I x )1 + ( I x ) 2 + ( I x )3 2 1 $ 0.12 % ! m' ) = 2 ( (21.0600 × 10−3 kg)(0.12 m)2 + (21.0600 × 10−3 kg) & * 3 + )# ("18 1 ! + ( (13.2324 × 10 −3 kg)(0.12 m) 2 ) "2 #

= [2(16.8480 + 33.6960) + (95.2733)] × 10−6 kg ⋅ m 2 = [2(50.5440) + (95.2733)] × 10−6 kg ⋅ m 2

or

I x = 196.4 × 10−6 kg ⋅ m 2

! ! ! ! ! ! ! ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1585

!

PROBLEM 9.139 (Continued) ! !

I y = ( I y )1 + ( I y )2 + ( I y )3

! 2!

1 $ 0.3 % = ( (21.0600 × 10−3 kg)(0.3 m) 2 + (21.0600 kg) & m' ) * 3 + )# ("18

!

-1 + 0 (21.0600 × 10−3 kg)[(0.3)2 + (0.12) 2 ] m 2 ! 218 2 2 $ 0.3 % $ 0.12 % ! 2 ./ + (21.0600 × 10−3 kg) (& + ' & ' )m 1 /3 "(* 3 + * 3 + #)

! !

1 ! + ( (13.2324 × 10−3 kg)(0.12 m) 2 ) ! "4 # = [(105.300 + 210.600) + (122.148 + 244.296) ! + (47.637)] × 10−6 kg ⋅ m 2

!

= (315.900 + 366.444 + 47.637) × 10−6 kg ⋅ m 2 !

or

I y = 730 × 10−6 kg ⋅ m 2

Symmetry implies I y = I z I z = 730 × 10−6 kg ⋅ m 2


PROBLEM 9.140* A farmer constructs a trough by welding a rectangular piece of 2-mm-thick sheet steel to half of a steel drum. Knowing that the density of steel is 7850 kg/m3 and that the thickness of the walls of the drum is 1.8 mm, determine the mass moment of inertia of the trough with respect to each of the coordinate axes. Neglect the mass of the welds.


Then m1 = (7850 kg/m3 )(0.002 m)(0.84 × 0.21) m 2 = 2.76948 kg m2 = (7850 kg/m3 )(0.0018 m)(π × 0.285 × 0.84) m 2 = 10.62713 kg $π % m3 = m4 = (7850 kg/m3 )(0.0018 m) & × 0.2852 ' m 2 2 * + = 1.80282 kg

Using Figure 9.28 for component 1 and the equations derived above for components 2 through 4, we have ( I x )3 = ( I x ) 4 I x = ( I x )1 + ( I x ) 2 + ( I x )3 + ( I x ) 4 2 1 0.21 % 2 ! $ 2 = ( (2.76948 kg)(0.21 m) + (2.76948 kg) & 0.285 − m ) 2 '+ * ("12 )# 1 ! + [(10.62713 kg)(0.285 m)2 ] + 2 ( (1.80282 kg)(0.285 m) 2 ) "2 #

= [(0.01018 + 0.08973) + (0.86319) + 2(0.07322)] kg ⋅ m 2 = [(0.09991 + 0.86319 + 2(0.07322)] kg ⋅ m 2

or

I x = 1.110 kg ⋅ m 2

! ! !


!


I y = ( I y )1 + ( I y )2 + ( I y )3 + ( I y ) 4 -1 = 0 (2.76948 kg)[(0.84) 2 + (0.21) 2 ] m 2 212 2

0.21 % $ 0.84 % $ + (2.76948 kg) (& + & 0.285 − ' 2 '+ ("* 2 + *

./ ) m2 1 )# /3

2!

2 /- 1 $ 0.84 % /. + 0 (10.62713 kg)[(0.84) 2 + 6(0.285) 2 ] m 2 + (10.62713 kg) & m' 1 * 2 + /3 2/12

1 ! + ( (1.80282 kg)(0.285 m) 2 ) "4 # 1 ! + ( (1.80282 kg)(0.285 m) 2 + (1.80282 kg)(0.84 m) 2 ) "4 # = [(0.17302 + 0.57827) + (1.05647 + 1.87463) + (0.03661) + (0.03661 + 1.27207)] kg ⋅ m 2 = (0.75129 + 2.93110 + 0.03661 + 1.30868) kg ⋅ m 2

or

I y = 5.03 kg ⋅ m 2

I z = ( I z )1 + ( I z ) 2 + ( I z )3 + ( I z ) 4 2 1 $ 0.84 % ! m' ) = ( (2.76948 kg)(0.84 m) 2 + (2.76948 kg) & * 2 + )# ("12 2 -/ 1 $ 0.84 % ./ m' 1 + 0 (10.62713 kg)[(0.84)2 + 6(0.285) 2 ] m 2 + (10.62713 kg) & * 2 + /3 2/12 1 ! + ( (1.80282 kg)(0.285 m) 2 ) "4 #

1 ! + ( (1.80282 kg)(0.285 m)2 + (1.80282 kg)(0.84 m)2 ) 4 " # = [(0.16285 + 0.48854) + (1.05647 + 1.87463) + (0.03661) + (0.03661 + 1.27207)] kg ⋅ m 2 = (0.65139 + 2.93110 + 0.03661 + 1.30868) kg ⋅ m 2

or

I z = 4.93 kg ⋅ m 2


PROBLEM 9.141 The machine element shown is fabricated from steel. Determine the mass moment of inertia of the assembly with respect to (a) the x axis, (b) the y axis, (c) the z axis. (The density of steel is 7850 kg/m3 .)

SOLUTION First compute the mass of each component. We have m = ρSTV

Then m1 = (7850 kg/m3 )(π (0.08 m) 2 (0.04 m)]

= 6.31334 kg m2 = (7850 kg/m3 )[π (0.02 m) 2 (0.06 m)] = 0.59188 kg m3 = (7850 kg/m3 )[π (0.02 m)2 (0.04 m)] = 0.39458 kg

Using Figure 9.28 and the parallel-axis theorem, we have (a)

I x = ( I x )1 + ( I x )2 − ( I x )3

-1 . = 0 (6.31334 kg)[3(0.08)2 + (0.04)2 ] m 2 + (6.31334 kg)(0.02 m) 2 1 212 3 -1 . + 0 (0.59188 kg)[3(0.02) 2 + (0.06)2 ] m 2 + (0.59188 kg)(0.03 m) 2 1 212 3 -1 . − 0 (0.39458 kg)[3(0.02)2 + (0.04) 2 ] m 2 + (0.39458 kg)(0.02 m) 2 1 212 3 = [(10.94312 + 2.52534) + (0.23675 + 0.53269) − (0.09207 + 0.15783)] × 10−3 kg ⋅ m 2 = (13.46846 + 0.76944 − 0.24990) × 10−3 kg ⋅ m 2 = 13.98800 × 10−3 kg ⋅ m 2

or I x = 13.99 × 10−3 kg ⋅ m 2


!


(b)

I y = ( I y )1 + ( I y )2 − ( I y )3

1 ! = ( (6.31334 kg)(0.08 m) 2 ) "2 # 1 ! + ( (0.59188 kg)(0.02 m) 2 + (0.59188 kg)(0.04 m) 2 ) 2 " # 1 ! − ( (0.39458 kg)(0.02 m 2 ) + (0.39458 kg)(0.04 m) 2 ) "2 # = [(20.20269) + (0.11838 + 0.94701) − (0.07892 + 0.63133)] × 10−3 kg ⋅ m 2 = (20.20269 + 1.06539 − 0.71025) × 10−3 kg ⋅ m 2 = 20.55783 × 10−3 kg ⋅ m 2

or (c)

I y = 20.6 × 10−3 kg ⋅ m 2

I z = ( I z )1 + ( I z ) 2 − ( I z )3

-1 . = 0 (6.31334 kg)[3(0.08)2 + (0.04)2 ] m 2 + (6.31334 kg)(0.02 m) 2 1 212 3 -1 . + 0 (0.59188 kg)[3(0.02) 2 + (0.06)2 ] m 2 + (0.59188 kg)[(0.04) 2 + (0.03) 2 ] m 2 1 212 3 -1 . − 0 (0.39458 kg)[3(0.02) 2 + (0.04) 2 ] m 2 + (0.03958 kg)[(0.04)2 + (0.02) 2 ] m 2 1 12 2 3 = [(10.94312 + 2.52534) + (0.23675 + 1.47970) − (0.09207 + 0.78916)] × 10−3 kg ⋅ m 2 = (13.46846 + 1.71645 − 0.88123) × 10−3 kg ⋅ m 2 = 14.30368 × 10−3 kg ⋅ m 2

or

I z = 14.30 × 10−3 kg ⋅ m 2

To the Instructor: The following formulas for the mass of inertia of a semicylinder are derived at this time for use in the solutions of Problems 9.142 through 9.145. From Figure 9.28: Cylinder

( I x ) cyl =

1 mcyl a 2 2

( I y )cyl = ( I z )cyl =

1 mcyl (3a 2 + L2 ) 12



Symmetry and the definition of the mass moment of inertia ( I = , r 2 dm) imply ( I )semicylinder = ( I x )sc =

and However,

1 ( I )cylinder 2 1$1 % mcyl a 2 ' 2 &* 2 +

( I y )sc = ( I z )sc = msc =

1 1 ! mcyl (3a 2 + L2 ) ) 2 ("12 #

1 mcyl 2 1 msc a 2 2

Thus,

( I x )sc =

and

( I y )sc = ( I z )sc =

1 msc (3a 2 + L2 ) 12

Also, using the parallel axis theorem find $ 1 16 % I x′ = msc & − 2 ' a 2 * 2 9π + $ 1 16 I z′ = msc (& − 2 "* 4 9π

% 2 1 2! ' a + 12 L ) + #

where x′ and z ′ are centroidal axes.!


PROBLEM 9.142 Determine the mass moment of inertia of the steel machine element shown with respect to the y axis. (The specific weight of steel is 490 lb/ft 3 .)

SOLUTION First compute the mass of each component. We have m = ρSTV =

Then

m1 =

γ ST g

V

490 lb/ft 3 $ 1 ft % × (2.7 × 3.7 × 9) in.3 × & ' 32.2 ft/s 2 * 12 in. +

3

= 791.780 × 10−3 lb ⋅ s 2 /ft

490 lb/ft 3 $ π % $ 1 ft % m2 = × & × 1.352 × 1.4 ' in.3 × & ' 2 32.2 ft/s * 2 + * 12 in. +

3

= 35.295 × 10−3 lb ⋅ s 2 /ft m3 =

490 lb/ft 3 $ 1 ft % × (π × 0.62 × 1.2) in.3 × & ' 2 32.2 ft/s * 12 in. +

3

= 11.952 × 10−3 lb ⋅ s 2 /ft 3

m4 =

490 lb/ft 3 $ 1 ft % 2 −3 × (0.9 × 0.6 × 9) in.3 × & ' = 42.799 × 10 lb ⋅ s /ft 2 32.2 ft/s * 12 in. +

The mass moments of inertia are now computed using Figure 9.28 (components 1, 3, and 4) and the equations derived above (component 2).



Find: I y We have

I y = ( I y )1 + ( I y ) 2 + ( I y )3 − ( I y ) 4

-1 = 0 (791.780 × 10−3 lb ⋅ s 2 /ft)[(2.7) 2 + (9) 2 ] in.2 212 2 2 2 $ 2.7 % $ 9 % ! 2 ./ $ 1 ft % + (791.780 × 10−3 lb ⋅ s 2 /ft) (& + × in. ) 1 & ' & ' ' ("* 2 + * 2 + )# /3 * 12 in. +

-1 + 0 (35.295 × 10 −3 lb ⋅ s 2 /ft)[3(1.35) 2 + (1.4) 2 ] in.2 212 1.4 % $ + (35.295 × 10−3 lb ⋅ s 2 /ft) ( (1.35) 2 + & 9 − ' 2 + * ("

./ $ 1 ft %2 ) in.2 1 × & ' )# /3 * 12 in. +

2!

-1 + 0 (11.952 × 10 −3 lb ⋅ s 2 /ft)[3(0.6) 2 + (1.2) 2 ] in.2 212 2 2 . 1.2 % $ 2 / $ 1 ft % in. + (11.952 × 10 −3 lb ⋅ s 2 /ft)[(1.35) 2 + & 9 + × 1 & ' 2 '+ * /3 * 12 in. + -1 − 0 (42.799 × 10−3 lb ⋅ s 2 /ft)[(0.9) 2 + (9) 2 ] in.2 212 2 2 $ 9 % ! 2 ./ $ 1 ft % + (42.799 × 10 lb ⋅ s /ft) ((1.35) + & ' ) in. 1 × & ' * 2 + )# (" /3 * 12 in. + = [(40.4550 + 121.3650) + (0.1517 + 17.3319)

−3

2

2

+ (0.0174 + 7.8005) − (2.0263 + 6.5603)] × 10−3 lb ⋅ ft ⋅ s 2 = (161.8200 + 17.4836 + 7.8179 − 8.5866) × 10−3 lb ⋅ ft ⋅ s 2

or

I y = 0.1785 lb ⋅ ft ⋅ s 2


!

PROBLEM 9.143 Determine the mass moment of inertia of the steel machine element shown with respect to the z axis. (The specific weight of steel is 490 lb/ft 3 .)


γ ST g

V

Then m1 =

490 lb/ft 3 $ 1 ft % × (2.7 × 3.7 × 9) in.3 × & ' 2 32.2 ft/s * 12 in. +

3

= 791.780 × 10−3 lb ⋅ s 2 /ft m2 =

490 lb/ft 3 $ π % $ 1 ft % × & × 1.352 × 1.4 ' in.3 × & ' 2 32.2 ft/s * 2 + * 12 in. +

3

= 35.295 × 10−3 lb ⋅ s 2 /ft m3 =

490 lb/ft 3 $ 1 ft % × (π × 0.62 × 1.2) in.3 × & ' 2 32.2 ft/s * 12 in. +

3

= 11.952 × 10−3 lb ⋅ s 2 /ft 3

m4 =

490 lb/ft 3 $ 1 ft % 2 −3 × (0.9 × 0.6 × 9) in.3 × & ' = 42.799 × 10 lb ⋅ s /ft 2 32.2 ft/s * 12 in. +

The mass moments of inertia are now computed using Figure 9.28 (components 1, 3, and 4) and the equations derived above (component 2).



Find: I z We have

I z = ( I z )1 + ( I z ) 2 + ( I z )3 − ( I z ) 4

-/ 1 = 0 (791.780 × 10 −3 lb ⋅ s 2 /ft)[(2.7)2 + (3.7) 2 ] in.2 /2 12 2

$ 2.7 % $ 3.7 % + (791.780 lb ⋅ s /ft) (& ' +& ' ("* 2 + * 2 + 2

-/ $ 1 16 + 0 (35.295 × 10−3 lb ⋅ s 2 /ft) & − 2 * 2 9π /2

. $ 1 ft % 2 ) in. 1 × & ' )# /3 * 12 in. +

2!

2/

% 2 ' (1.35 in.) +

4 × 1.35 % $ + (35.295 × 10−3 lb ⋅ s 2 /ft) ((1.35) 2 + & 3.7 + ' 3π + * ("

2 . 2 / $ 1 ft % × in. ) 1 & ' )# /3 * 12 in. +

2!

-1 + 0 (11.952 × 10−3 lb ⋅ s 2 /ft)(0.6 in.) 2 212 ./ $ 1 ft %2 + (11.952 × 10−3 lb ⋅ s 2 /ft)[(1.35)2 + (3.7)2 ] in.2 1 × & ' /3 * 12 in. + -/ 1 − 0 (42.799 × 10−3 lb ⋅ s 2 /ft)[(0.9) 2 + (0.6) 2 ] in.2 /2 12 $ 0.6 % + (42.799 × 10−3 lb ⋅ s 2 /ft) ((1.35) 2 + & ' * 2 + "( I z = [(9.6132 + 28.8395) + (0.1429 + 4.9219)

2

./ $ 1 ft %2 ! ) in.2 1 × & ' /3 * 12 in. + #)

+ (0.0149 + 1.2875) − (0.0290 + 0.5684)] × 10−3 lb ⋅ ft ⋅ s 2 = (38.4527 + 5.0648 + 1.3024 − 0.5974) × 10−3 lb ⋅ ft ⋅ s 2

or

I z = 0.0442 lb ⋅ ft ⋅ s 2


!

PROBLEM 9.144 Determine the mass moment of inertia and the radius of gyration of the steel machine element shown with respect to the x axis. (The density of steel is 7850 kg/m3 .)


Then

m1 = (7850 kg/m3 )(0.09 × 0.03 × 0.15) m3 = 3.17925 kg

π ! m2 = (7850 kg/m3 ) ( (0.045) 2 × 0.04 ) m3 "2 # = 0.998791 kg m3 = (7850 kg/m3 )[π (0.038) 2 × 0.03] m3 = 1.06834 kg

Using Figure 9.28 for components 1 and 3 and the equation derived above (before the solution to Problem 9.142) for a semicylinder, we have I x = ( I x )1 + ( I x )2 − ( I x )3 1 ! = ( (3.17925 kg)(0.032 + 0.152 ) m 2 + (3.17925 kg)(0.075 m)2 ) "12 # -/ $ 1 16 + 0(0.998791 kg) (& − 2 "* 4 9π 2/

1 % 2 2! ' (0.045 m) + 12 (0.04 m) ) + #

$ 4 × 0.045 % + (0.998791 kg) ((0.13) + & + 0.015 ' * 3π + (" 2

./ ) m2 1 )# /3

2!

-1 . − 0 (1.06834 kg)[3(0.038 m) 2 + (0.03 m)2 ] + (1.06834 kg)(0.06 m)2 1 212 3



= [(0.0061995 + 0.0178833) + (0.0002745 + 0.0180409) − (0.0004658 + 0.0038460)] kg ⋅ m 2 = (0.0240828 + 0.0183154 − 0.0043118) kg ⋅ m 2 = 0.0380864 kg ⋅ m 2

or Now

and

I x = 38.1 × 10−3 kg ⋅ m 2

m = m1 + m2 − m3 = (3.17925 + 0.998791 − 1.06834) kg = 3.10970 kg k x2 =

I x 0.0380864 kg ⋅ m 2 = m 3.10970 kg

or

k x = 110.7 mm


!

PROBLEM 9.145 Determine the mass moment of inertia of the steel fixture shown with respect to (a) the x axis, (b) the y axis, (c) the z axis. (The density of steel is 7850 kg/m3.)


Then

m1 = 7850 kg/m3 × (0.08 × 0.05 × 0.160) m3 = 5.02400 kg m2 = 7850 kg/m3 × (0.08 × 0.038 × 0.07) m3 = 1.67048 kg m3 = 7850 kg/m3 ×

$π % × 0.0242 × 0.04 ! m3 = 0.28410 kg "2 #

Using Figure 9.28 for components 1 and 2 and the equations derived above for component 3, we have (a)

I x = ( I x )1 − ( I x ) 2 − ( I x )3 &) 0.05 *2 ) 0.16 *2 ' 2 %( ($ 1 = - (5.02400 kg)[(0.05)2 + (0.16) 2 ] m 2 + (5.02400 kg) + ! + ! ,m . (112 /+" 2 # " 2 # 0, 2( 2 2 &) 0.038 * ) 0.07 * ' 2 (% ($ 1 − - (1.67048 kg)[(0.038)2 + (0.07) 2 ] m 2 + (1.67048 kg) + 0.05 − + + 0.05 ,m . 2 !# " 2 !# ,0 +/" (112 (2 &) 1 16 * ' 1 ($ − -(0.28410 kg) + − 2 ! (0.024) 2 + (0.04) 2 , m 2 12 /" 4 9π # 0 1( 2 2 &) 4 × 0.024 * ) 0.04 * ' 2 %( 0.16 + (0.28410 kg) + 0.05 − + − ,m . ! 3π 2 !# ,0 # " +/" 2(

= [(11.7645 + 35.2936) − (0.8831 + 13.6745) − (0.0493 + 6.0187)] × 10−3 kg ⋅ m 2 = (47.0581 − 14.5576 − 6.0680) × 10−3 kg ⋅ m 2 = 26.4325 × 10−3 kg ⋅ m 2

or

I x = 26.4 × 10−3 kg ⋅ m 2



(b)

I y = ( I y )1 − ( I y ) 2 − ( I y )3

&) 0.08 *2 ) 0.16 * 2 ' 2 %( ($ 1 2 2 2 = - (5.02400 kg)[(0.08) + (0.16) ] m + (5.02400 kg) + ! + ! ,m . 12 (2 /+" 2 # " 2 # 0, 1( 2 &) 0.08 *2 ) 0.07 * ' 2 (% ($ 1 2 2 2 − - (1.67048 kg)[(0.08) + (0.07) ] m + (1.67048 kg) + ! + 0.05 + 2 ! , m . 12 # ,0 +/" 2 # " 1( 2( 2 $( 1 &) 0.08 *2 ) 0.04 * ' 2 %( 2 2 2 − - (0.28410 kg)[3(0.024) + (0.04) ] m + (0.28410 kg) + ! + 0.16 − 2 ! , m . # ,0 +/" 2 # " (112 (2

= [(13.3973 + 40.1920) − (1.5730 + 14.7420) − (0.0788 + 6.0229)] × 10−3 kg ⋅ m 2 = (53.5893 − 16.3150 − 6.1017) × 10−3 kg ⋅ m 2 = 31.1726 × 10−3 kg ⋅ m 2

(c)

or

I y = 31.2 × 10−3 kg ⋅ m 2

I z = ( I z )1 − ( I z ) 2 − ( I z )3

$( 1 &) 0.08 *2 ) 0.05 *2 ' 2 %( = - (5.02400 kg)[(0.08) 2 + (0.05) 2 ] m 2 + (5.02400 kg) + ! + ! ,m . +/" 2 # " 2 # 0, (112 2( 2 &) 0.08 *2 ) 0.038 * ' 2 (% ($ 1 0.05 − - (1.67048 kg)[(0.08) 2 + (0.038) 2 ] m 2 + (1.67048 kg) + + − ,m . ! 2 !# 0, +/" 2 # " (112 2(

$( ) 1 16 − -(0.28410 kg) − 2 " 2 9π 1(

2 &) 0.08 *2 ) 4 × 0.024 * ' 2 %( * 2 (0.024 m) (0.28410 kg) 0.05 + + − + ! ! ! ,m . 3π # # ,0 +/" 2 # " 2(

= [(3.7261 + 11.1784) − (1.0919 + 4.2781) − (0.0523 + 0.9049)] × 10−3 kg ⋅ m 2 = (14.9045 − 5.3700 − 0.9572) × 10−3 kg ⋅ m 2 = 8.5773 × 10−3 kg ⋅ m 2

or

I z = 8.58 × 10−3 kg ⋅ m 2

To the instructor: The following formulas for the mass moment of inertia of wires are derived or summarized at this time for use in the solutions of Problems 9.146 through 9.148. Slender Rod Ix = 0

I y′ = I z′ =

1 mL2 (Figure 9.28) 12

1 I y = I z = mL2 (Sample Problem 9.9) 3



Circle

3

We have

I y = r 2 dm = ma 2

Now

Iy = Ix + Iz

And symmetry implies

Ix = Iz Ix = Iz =

1 2 ma 2

Semicircle Following the above arguments for a circle, We have Ix = Iz =

1 2 ma 2

I y = ma 2

Using the parallel-axis theorem I z = I z ′ + mx 2

or

I z′ = m

)1 4 − 2 "2 π

x=

2a

π

* 2 !a ! #


PROBLEM 9.146 Aluminum wire with a weight per unit length of 0.033 lb/ft is used to form the circle and the straight members of the figure shown. Determine the mass moment of inertia of the assembly with respect to each of the coordinate axes.

SOLUTION First compute the mass of each component. We have m=

Then

W 1 )W * L = g g " L !#AL

1 1 ft × 0.033 lb/ft × (2π × 16 in.) × 12 in. 32.2 ft/s 2 = 8.5857 × 10−3 lb ⋅ s 2 /ft 1 1 ft m2 = m3 = m4 = m5 = × 0.033 lb/ft × 8 in. × 2 12 in. 32.2 ft/s 2 = 0.6832 lb ⋅ s /ft m1 =

Using the equations given above and the parallel-axis theorem, we have I x = ( I x )1 + ( I x )2 + ( I x )3 + ( I x ) 4 + ( I x )5 &1 ' ) 1 ft * = + (8.5857 × 10−3 lb ⋅ s 2 /ft)(16 in.) 2 , × ! 2 / 0 " 12 in. # &1 ' ) 1 ft * + + (0.6832 lb ⋅ s 2 /ft)(8 in.) 2 , × ! /3 0 " 12 in. #

+ [0 + (0.6832 lb ⋅ s 2 /ft)(8 in.) 2 ] ×

2

2

) 1 ft * ! " 12 in. #

2

&1 ' ) 1 ft * + + (0.6832 lb ⋅ s 2 /ft)(8 in.) 2 + (0.6832 lb ⋅ s 2 /ft)(42 + 162 ) in.2 , × ! 12 / 0 " 12 in. # &1 ' ) 1 ft * + + (0.6832 lb ⋅ s 2 /ft)(8 in.) 2 + (0.6832 lb ⋅ s 2 /ft)(82 + 122 ) in.2 , × ! 12 / 0 " 12 in. #

2

2

= [(7.6315) + (0.1012) + (0.3036) + (0.0253 + 1.2905) + (0.0253 + 0.9868)] × 10−3 lb ⋅ ft ⋅ s 2 = (7.6315 + 0.1012 + 0.3036) + 1.3158 + 1.0121) × 10−3 lb ⋅ ft ⋅ s 2 = 10.3642 × 10−3 lb ⋅ ft ⋅ s 2

or

I x = 10.36 × 10−3 lb ⋅ ft ⋅ s 2



( I y )2 = ( I y )4 ,

( I y )3 = ( I y )5

I y = ( I y )1 + ( I y ) 2 + ( I y )3 + ( I y ) 4 + ( I y )5

= /&(8.5857 × 10−3 lb ⋅ s 2 /ft)(16 in.) 2 0' ×

) 1 ft * ! " 12 in. #

+2[0 + (0.6832 × 10−3 lb ⋅ s 2 /ft)(16 in.) 2 ] ×

2

) 1 ft * ! " 12 in. #

2

&1 ' ) 1 ft * +2 + (0.6832 × 10−3 lb ⋅ s 2 /ft)(8 in.) 2 + (0.6832 lb ⋅ s 2 /ft)(12 in.) 2 , × ! 12 / 0 " 12 in. #

2

= [(15.2635) + 2(1.2146) + 2(0.0253 + 0.6832)] × 10−3 lb ⋅ ft ⋅ s 2 = [15.2635 + 2(1.2146) + 2(0.7085)] × 10−3 lb ⋅ ft ⋅ s 2 = 19.1097 × 10−3 lb ⋅ ft ⋅ s 2

Symmetry implies

or

I y = 19.11 × 10−3 lb ⋅ ft ⋅ s 2 I z = 10.36 × 10−3 lb ⋅ ft ⋅ s 2

Ix = Iz


!

PROBLEM 9.147 The figure shown is formed of 18 -in.-diameter steel wire. Knowing that the specific weight of the steel is 490 lb/ft3, determine the mass moment of inertia of the wire with respect to each of the coordinate axes.


γ ST g

AL

Then 2 3 490 lb/ft 3 & π ) 1 * ' ) 1 ft * in. ! , × (π × 18 in.) × ×+ m1 = m2 = ! 32.2 ft/s 2 +/ 4 " 8 # ,0 " 12 in. #

= 6.1112 × 10−3 lb ⋅ s 2 /ft

m3 = m4 =

2 3 490 lb/ft 3 & π ) 1 * ' ) 1 ft * in. 18 in. = × × + , ! ! 32.2 ft/s 2 /+ 4 " 8 # 0, " 12 in. #

= 1.9453 × 10−3 lb ⋅ s 2 /ft

Using the equations given above and the parallel-axis theorem, we have ( I x )3 = ( I x ) 4 I x = ( I x )1 + ( I x )2 + ( I x )3 + ( I x ) 4 &1 ' ) 1 ft * = + (6.1112 × 10−3 lb ⋅ s 2 /ft)(18 in.) 2 , × ! /2 0 " 12 in. #

2

&1 ' ) 1 ft * + + (6.1112 × 10 −3 lb ⋅ s 2 /ft)(18 in.) 2 + (6.1112 × 10 −3 lb ⋅ s 2 /ft)(18 in.) 2 , × ! /2 0 " 12 in. #

2

&1 ' ) 1 ft * + 2 + (1.9453 × 10−3 lb ⋅ s 2 /ft)(18 in.) 2 + (6.1112 × 10 −3 lb ⋅ s 2 /ft)(92 + 182 ) in.2 , × ! /12 0 " 12 in. #

2

= [(6.8751) + (6.8751 + 13.1502) + 2(0.3647 + 5.4712)] × 10 −3 lb ⋅ ft ⋅ s 2 = [6.8751 + 20.6252 + 2(5.8359)] × 10−3 lb ⋅ ft ⋅ s 2 = 39.1721 × 10−3 lb ⋅ ft ⋅ s 2

or

I x = 0.0392 lb ⋅ ft ⋅ s 2

! ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1603

!


( I y )1 = ( I y ) 2 , ( I y )3 = ( I y ) 4 I y = ( I y )1 + ( I y ) 2 + ( I y )3 + ( I y ) 4 ) 1 ft * = 2 &/(6.1112 × 10−3 lb ⋅ s 2 /ft)(18 in.)2 '0 × ! " 12 in. # + 2[(0 + 1.9453 × 10−3 lb ⋅ s 2 /ft)(18 in.) 2 ] ×

2

) 1 ft * ! " 12 in. #

2

= [2(13.7502) + 2(4.3769)] × 10−3 lb ⋅ ft ⋅ s 2 = 36.2542 × 10−3 lb ⋅ ft ⋅ s 2

or

I y = 0.0363 lb ⋅ ft ⋅ s 2

or

I z = 0.0304 lb ⋅ ft ⋅ s 2

( I z )3 = ( I z ) 4 I z = ( I z )1 + ( I z ) 2 + ( I z )3 + ( I z ) 4 &1 ' ) 1 ft * = + (6.1112 × 10−3 lb ⋅ s 2 /ft)(18 in.)2 , × ! /2 0 " 12 in. # $ )1 4 + -(6.1112 × 10−3 lb ⋅ s 2 /ft) − 2 "2 π 1

2

* 2 ! (18 in.) #

&) 2 × 18 *2 ' 2 (% ) 1 ft *2 2 + (6.1112 × 10−3 lb ⋅ s 2 /ft) + + (18) in. . × , ! 12 in. !# ,0 /+" π # 2( " &1 ' ) 1 ft * + 2 + (1.9453 × 10−3 lb ⋅ s 2 /ft)(18 in.) 2 , × ! 3 / 0 " 12 in. #

2

= [(6.8751) + (1.3024 + 19.3229) + 2(1.4590)] × 10−3 lb ⋅ ft ⋅ s 2 = [6.8751 + 20.6253 + 2(1.4590)] × 10−3 lb ⋅ ft ⋅ s2 = 30.4184 × 10−3 lb ⋅ ft ⋅ s 2


!

PROBLEM 9.148 A homogeneous wire with a mass per unit length of 0.056 kg/m is used to form the figure shown. Determine the mass moment of inertia of the wire with respect to each of the coordinate axes.

SOLUTION First compute the mass m of each component. We have m = (m/L) L = 0.056 kg/m × 1.2 m = 0.0672 kg

Using the equations given above and the parallel-axis theorem, we have I x = ( I x )1 + ( I x ) 2 + ( I x )3 + ( I x ) 4 + ( I x )5 + ( I x )6

Now Then

( I x )1 = ( I x )3 = ( I x ) 4 = ( I x )6

and ( I x )2 = ( I x )5

&1 ' I x = 4 + (0.0672 kg)(1.2 m) 2 , + 2[0 + (0.0672 kg)(1.2 m) 2 ] /3 0 = [4(0.03226) + 2(0.09677)] kg ⋅ m 2 = 0.32258 kg ⋅ m 2

or

I x = 0.323 kg ⋅ m 2

I y = ( I y )1 + ( I y ) 2 + ( I y )3 + ( I y ) 4 + ( I y )5 + ( I y )6

Now Then

( I y )1 = 0, ( I y ) 2 = ( I y )6 , and ( I y )4 = ( I y )5 &1 ' I y = 2 + (0.0672 kg)(1.2 m) 2 , + [0 + (0.0672 kg)(1.2 m) 2 ] /3 0 &1 ' + 2 + (0.0672 kg)(1.2 m) 2 + (0.0672 kg)(1.22 + 0.62 ) m 2 , 12 / 0 = [2(0.03226) + (0.09677) + 2(0.00806 + 0.12096)] kg ⋅ m 2 = [2(0.03226) + (0.09677) + 2(0.12902)] kg ⋅ m 2 = 0.41933 kg ⋅ m 2

Symmetry implies

or

I y = 0.419 kg ⋅ m 2 I z = 0.419 kg ⋅ m 2

I y = Iz

!


PROBLEM 9.149 Determine the mass products of inertia Ixy, Iyz, and Izx of the steel fixture shown. (The density of steel is 7850 kg/m3.)

SOLUTION First compute the mass of each component. We have m = ρV m1 = 7850 kg/m3 × (0.08 × 0.05 × 0.16) m3 = 5.02400 kg

Then

m2 = 7850 kg/m3 × (0.08 × 0.038 × 0.07) m3 = 1.67048 kg m3 = 7850 kg/m3 ×

)π * × 0.0242 × 0.04 ! m3 = 0.28410 kg 2 " #

Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of each component are zero because of symmetry. Now 0.038 * ) m = 0.031 m y2 = 0.05 − 2 !# " 4 × 0.024 * ) y3 = 0.05 − ! m = 0.039814 m 3π " # and then mx y , kg ⋅ m 2

my z , kg ⋅ m 2

mz x , kg ⋅ m 2

0.08

5.0240 × 10−3

10.0480 × 10−3

16.0768 × 10−3

0.031

0.085

2.0714 × 10−3

4.4017 × 10−3

5.6796 × 10−3

0.039814

0.14

0.4524 × 10−3

1.5836 × 10−3

1.5910 × 10−3

m, kg

x, m

y, m

1

5.02400

0.04

0.025

2

1.67048

0.04

3

0.28410

0.04

z, m

Finally, I xy = ( I xy )1 − ( I xy ) 2 − ( I xy )3 = [(0 + 5.0240) − (0 + 2.0714) − (0 + 0.4524)] × 10−3 = 2.5002 × 10−3 kg ⋅ m 2

or

I xy = 2.50 × 10−3 kg ⋅ m 2



I yz = ( I yz )1 − ( I yz ) 2 − ( I yz )3 = [(0 + 10.0480) − (0 + 4.4017) − (0 + 1.5836)] × 10−3 = 4.0627 × 10−3 kg ⋅ m 2

or

I yz = 4.06 × 10−3 kg ⋅ m 2

I zx = ( I zx )1 − ( I zx ) 2 − ( I zx )3 = [(0 + 16.0768) − (0 + 5.6796) − (0 + 1.5910)] × 10−3 = 8.8062 × 10−3 kg ⋅ m 2

or

I zx = 8.81 × 10−3 kg ⋅ m 2


PROBLEM 9.150 Determine the mass products of inertia Ixy, Iyz, and Izx of the steel machine element shown. (The density of steel is 7850 kg/m3.)

SOLUTION First compute the mass of each component. We have m = ρSTV m1 = 7850 kg/m3 × (0.1 × 0.018 × 0.09) m3 = 1.27170 kg

Then

m2 = 7850 kg/m3 × (0.016 × 0.06 × 0.05) m3 = 0.37680 kg m3 = 7850 kg/m3 × (π × 0.0122 × 0.01) m3 = 0.03551 kg

Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of each component are zero because of symmetry. Now m, kg

x, m

y, m

z, m

mx y, kg ⋅ m 2

my z , kg ⋅ m 2

mz x , kg ⋅ m 2

1

1.27170

0.05

0.009

0.045

0.57227 × 10−3

0.51504 × 10−3

2.86133 × 10−3

2

0.37680

0.092

0.048

0.045

1.66395 × 10−3

0.81389 × 10−3

1.55995 × 10−3

3

0.03551

0.105

0.054

0.045

0.20134 × 10−3

0.08629 × 10−3

0.16778 × 10−3

2.43756 × 10−3

1.41522 × 10−3

4.58906 × 10−3

Σ

Then

0 + mx y )

or

I xy = 2.44 × 10−3 kg ⋅ m 2

0 = Σ( I y′z ′ + my z )

or

I yz = 1.415 × 10−3 kg ⋅ m 2

or

I zx = 4.59 × 10−3 kg ⋅ m 2

I xy = Σ( I x′y′

I yz

0 I zx = Σ( I z′x′ + mz x )


PROBLEM 9.151 Determine the mass products of inertia Ixy, Iyz, and Izx of the cast aluminum machine component shown. (The specific weight of aluminum is 0.100 lb/in.3)

SOLUTION First compute the mass of each component. We have m = ρ ALV =

Then

m1 =

γ AL g

V

0.100 lb/in.3 × (7.8 × 0.8 × 1.6)in.3 32.2 ft/s 2

= 31.0062 × 10−3 lb ⋅ s 2 /ft m2 =

0.100 lb/in.3 × (2.4 × 0.8 × 2) in.3 2 32.2 ft/s

= 11.9255 × 10−3 lb ⋅ s 2 /ft m3 =

0.100 lb/in.3 π ! × " (0.8)2 × 0.8# in.3 = 2.4977 lb ⋅ s 2 /ft 2 32.2 ft/s 2 $ %

m4 =

0.100 lb/in.3 × [π (0.55)2 × 0.6]in.3 = 1.7708 lb ⋅ s 2 /ft 2 32.2 ft/s

Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of each component are zero because of symmetry. Now 4 × 0.8 ' & x3 = − ( 7.8 + in. = −8.13953 in. 3π )+ *


PROBLEM 9.151 (Continued) and then m, lb ⋅ s 2 /ft

x , ft

y , ft

z , ft

mx y , lb ⋅ ft ⋅ s 2

my z , lb ⋅ ft ⋅ s 2

mz x , lb ⋅ ft ⋅ s 2

1

31.0062 × 10−3

−

3.9 12

0.4 12

−

0.8 12

− 335.901 × 10−6

− 68.903 × 10−6

671.801 × 10−6

2

11.9255 × 10−3

−

1.2 12

0.4 12

−

2.6 12

− 39.752 × 10−6

−86.129 × 10−6

258.386 × 10−6

3

2.4977 × 10−3

8.13953 12

0.4 12

−

0.8 12

− 56.473 × 10−6

− 5.550 × 10−6

112.945 × 10−6

4

1.7708 × 10−3

7.8 12

1.1 12

−

0.8 12

−105.511 × 10−6

−10.822 × 10−6

76.735 × 10−6

− 537.637 × 10−6

−171.404 × 10−6

1119.867 × 10−6

Σ

Then

−

−

0 I xy = Σ( I x′y′ + mx y ) 0 I yz = Σ( I y ′z ′ + m y z )

or

I xy = −538 × 10−6 lb ⋅ ft ⋅ s 2

or

I yz = −171.4 × 10−6 lb ⋅ ft ⋅ s 2

0 I zx = Σ( I z ′x′ + mz x )

or

I zx = 1120 × 10−6 lb ⋅ ft ⋅ s 2


PROBLEM 9.152 Determine the mass products of inertia Ixy, Iyz, and Izx of the cast aluminum machine component shown. (The specific weight of aluminum is 0.100 lb/in.3)

SOLUTION First compute the mass of each component. We have m = ρ ALV =

Then

m1 =

γ AL g

V

0.100 lb/in.3 × (5.9 × 1.8 × 2.2) in.3 32.2 ft/s 2

= 72.5590 × 10−3 lb ⋅ s 2 /ft m2 =

0.100 lb/in.3 π ! × " (1.1)2 × 1.8# in.3 2 32.2 ft/s $2 %

= 10.6248 × 10−3 lb ⋅ s 2 /ft m3 =

0.100 lb/in.3 × (1.4 × 1.1 × 1.2) in.3 32.2 ft/s 2

= 5.7391 × 10−3 lb ⋅ s 2 /ft

Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of each component are zero because of symmetry. Now 4 × 1.1 ' & in. x2 = − ( 5.9 + 3π )+ * = − 6.36685 in. 1.1 ' & in. y3 = (1.8 − 2 )+ * = 1.25 in.



and then m, lb ⋅ s 2 /ft

1

72.5590 × 10−3

2

10.6248 × 10−3

3

5.7391 × 10−3

Finally,

x , ft

y , ft

z , ft

mx y , lb ⋅ ft ⋅ s 2

my z , lb ⋅ ft ⋅ s 2

mz x , lb ⋅ ft ⋅ s 2

2.95 12

0.9 12

1.1 12

−1.33781 × 10−3

0.49884 × 10−3

−1.63510 × 10−3

6.36685 12

0.9 12

1.1 12

− 0.42279 × 10−3

0.07305 × 10−3

− 0.51674 × 10−3

0.7 12

1.25 12

1.3 12

− 0.03487 × 10−3

0.06476 × 10−3

− 0.03627 × 10−3

− −

−

I xy = ( I xy )1 + ( I xy ) 2 − ( I xy )3 = [(0 − 1.33781) + (0 − 0.42279) − (0 − 0.03487)] × 10−3

or

I xy = −1.726 × 10−3 lb ⋅ ft ⋅ s 2

I yz = ( I yz )1 + ( I yz ) 2 − ( I yz )3 = [(0 + 0.49884) + (0 + 0.07305) − (0 + 0.06476)] × 10−3

or

I yz = 0.507 × 10−3 lb ⋅ ft ⋅ s 2

I zx = ( I zx )1 + ( I zx )2 − ( I zx )3 = [(0 − 1.63510) + (0 − 0.51674) − (0 − 0.3627)] × 10 −3

or

I zx = −2.12 × 10 −3 lb ⋅ ft ⋅ s 2


PROBLEM 9.153 A section of sheet steel 2 mm thick is cut and bent into the machine component shown. Knowing that the density of steel is 7850 kg/m3, determine the mass products of inertia Ixy, Iyz, and Izx of the component.

SOLUTION

m1 = ρV = 7850 kg/m3 (0.2 m)(0.3 m)(0.002 m) = 0.942 kg m2 = ρV = 7850 kg/m3 (0.4 m)(0.3 m)(0.002 m) = 1.884 kg

For each panel the centroidal product of inertia is zero with respect to each pair of coordinate axes.

! Σ

m, kg

x, m

y, m

z, m

mx y

my z

mz x

0.942

0.15

0.1

0.4

+ 14.13 × 10−3

+ 37.68 × 10−3

+ 56.52 × 10−3

1.884

0.15

0

0.2

0

0

+ 56.52 × 10−3

+ 14.13 × 10−3

+ 37.68 × 10−3

+ 113.02 × 10−3

I xy = + 14.13 × 10−3 kg ⋅ m 2

I xy = + 14.13 g ⋅ m 2

I yz = + 37.68 × 10−3 kg ⋅ m 2

I yz = + 37.7 g ⋅ m 2

I zx = + 113.02 × 10−3 kg ⋅ m 2

I zx = + 113.0 g ⋅ m 2



SOLUTION

First compute the mass of each component. We have Then

m = ρSTV = ρST tA

m1 = (7850 kg/m3 )[(0.002)(0.4)(0.45)]m3 = 2.8260 kg &1 '! m2 = 7850 kg/m3 "(0.002) ( × 0.45 × 0.18 ) # m3 2 * +% $ = 0.63585 kg

Now observe that ( I x′y′ )1 = ( I y′z ′ )1 = ( I z′x′ )1 = 0 ( I y′z′ ) 2 = ( I z′x′ ) 2 = 0

From Sample Problem 9.6:

( I x′y′ )2,area = −

1 2 2 b2 h2 72

Then

1 & 1 ' ( I x′y′ ) 2 = ρST t ( I x′y′ ) 2,area = ρST t ( − b22 h22 ) = − m2 b2 h2 72 36 * +

Also

x1 = y1 = z2 = 0

0.45 ' & x2 = ( − 0.225 + m = − 0.075 m 3 )+ *



Finally,

1 I xy = Σ( I xy + mx y ) = (0 + 0) + " − (0.63585 kg)(0.45 m)(0.18 m) $ 36 & 0.18 m ' ! + (0.63585 kg)(− 0.075 m) ( )# * 3 +% = (−1.43066 × 10−3 − 2.8613 × 10−3 ) kg ⋅ m 2

and

or

I xy = − 4.29 × 10−3 kg ⋅ m 2

or

I yz = 0

or

I zx = 0

I yz = Σ( I y′z′ + m y z ) = (0 + 0) + (0 + 0) = 0

I zx = Σ( I z ′x′ + m z x ) = (0 + 0) + (0 + 0) = 0




Then

m1 = (7850 kg/m3 )(0.002 m)(0.35 × 0.39) m 2 = 2.14305 kg &π ' m2 = (7850 kg/m3 )(0.002 m) ( × 0.1952 ) m 2 *2 + = 0.93775 kg &1 ' m3 = (7850 kg/m3 )(0.002 m) ( × 0.39 × 0.15 ) m 2 *2 + = 0.45923 kg

Now observe that because of symmetry the centroidal products of inertia of components 1 and 2 are zero and ( I x′y′ )3 = ( I z′x′ )3 = 0

Also

( I y′z ′ )3,mass = ρST t ( I y′z′ )3,area

Using the results of Sample Problem 9.6 and noting that the orientation of the axes corresponds to a 90° rotation, we have ( I y′z ′ )3,area =

Then Also

1 2 2 b3 h3 72

& 1 ' 1 m3b3 h3 ( I y′z ′ )3 = ρST t ( b32 h32 ) = 72 * + 36 y1 = x2 = 0 y2 =

4 × 0.195 m = 0.082761 m 3π



Finally, I xy = Σ( I x′y′ + mx y ) & 0.15 ' ! = (0 + 0) + (0 + 0) + "0 + (0.45923 kg)(0.35 m) ( − m )# 3 * +% $ = − 8.0365 × 10−3 kg ⋅ m 2

or I xy = − 8.04 × 10−3 kg ⋅ m 2

I yz = Σ( I y′z′ + my z ) = (0 + 0) + [0 + (0.93775 kg)(0.082761 m)(0.195 m)] 1 & 0.15 '& 0.39 ' ! m )( m )# + " (0.45923 kg)(0.39 m)(0.15 m) + (0.45923 kg) ( − 3 * +* 3 +% $ 36 = [(15.1338) + (0.7462 − 2.9850)] × 10−3 kg ⋅ m 2 = (15.1338 − 2.2388) × 10−3 kg ⋅ m 2 = 12.8950 × 10−3 kg ⋅ m 2

or I yz = 12.90 × 10−3 kg ⋅ m 2

I zx = Σ( I z ′x′ + mz x ) = [0 + (2.14305 kg)(0.175 m)(0.195 m)] + (0 + 0) ! & 0.39 ' m ) (0.35 m) # + "0 + (0.45923 kg) ( 3 * + $ % = (73.1316 + 20.8950) × 10−3 kg ⋅ m 2 = 94.0266 × 10−3 kg ⋅ m 2

or I zx = 94.0 × 10−3 kg ⋅ m 2


!



Then &1 ' m1 = m3 = (7850 kg/m3 )(0.002 m) ( × 0.225 × 0.135 ) m 2 = 0.23844 kg *2 + &π ' m2 = (7850 kg/m3 )(0.002 m) ( × 0.2252 ) m 2 = 0.62424 kg *4 + π & ' m4 = (7850 kg/m3 )(0.002 m) ( × 0.1352 ) m 2 = 0.22473 kg *4 +

Now observe that the following centroidal products of inertia are zero because of symmetry.

Also Now

( I x′y′ )1 = ( I y′z′ )1 = 0

( I x′y′ ) 2 = ( I y′z′ )2 = 0

( I x′y′ )3 = ( I z′x′ )3 = 0

( I x′y′ ) 4 = ( I z′x′ ) = 0

y1 = y2 = 0

x3 = x4 = 0

I xy = Σ( I x′y′ + mx y ) I xy = 0

so that Using the results of Sample Problem 9.6, we have I uv ,area =

Now

1 2 2 b h 24

I uv ,mass = ρST t I uv ,area & 1 ' = ρST t ( b2 h 2 ) 24 * + 1 = mbh 12

Thus,

( I zx )1 =

1 m1b1h1 12



( I yz )3 = −

While

1 m3b3 h3 12

because of a 90° rotation of the coordinate axes. To determine I uv for a quarter circle, we have 0 dI uv = d I u ′v′ + uEL vEL dm 1 1 2 v= a − u2 2 2

uEL = u vEL =

Where

dm = ρST t dA = ρST tv du = ρST t a 2 − u 2 du

,

I uv = dI uv =

Then

=

1 ρST t 2

,

, a

0

a

0

&1 2 ' (u ) ( a − u 2 ) ρST t a 2 − u 2 du 2 * +

(

)

u (a 2 − u 2 ) du a

1 1 1 ! 1 1 = ρST t " a 2 u 2 − u 4 # = ρST t a 4 = ma 4 2 4 %0 8 2π $2 ( I zx ) 2 = −

Thus

1 m2 a22 2π

because of a 90° rotation of the coordinate axes. Also ( I yz ) 4 =

Finally, I yz

1 m4 a42 2π

0 0 = Σ( I yz ) = [( I y′z′ ) + m1 y1 z1 ] + [( I y′z′ ) + m2 y2 z2 ] + ( I yz )3 + ( I yz ) 4 1 1 ! ! = " − (0.23844 kg)(0.225 m)(0.135 m) # + " (0.22473 kg)(0.135 m) 2 # $ 12 % $ 2π % = (− 0.60355 + 0.65185) × 10−3 kg ⋅ m 2

or

I yz = 48.3 × 10−6 kg ⋅ m 2

0 0 I zx = Σ( I zx ) = ( I zx )1 + ( I zx ) 2 + [( I z ′x′ )3 + m3 z3 x3 ] + [( I z′x′ ) 4 + m4 z4 x4 ] 1 1 ! ! = " (0.23844 kg)(0.225 m)(0.135 m) # + " − (0.62424 kg)(0.225 m)2 # $12 % $ 2π % = (0.60355 − 5.02964) × 10−3 kg ⋅ m 2

or

I zx = − 4.43 × 10−3 kg ⋅ m 2


PROBLEM 9.157 Brass wire with a weight per unit length w is used to form the figure shown. Determine the mass products of inertia Ixy, Iyz, and Izx of the wire figure.


W 1 = wL g g

W w (2π × a ) = 2π a g g w w m2 = (a) = a g g w w m3 = (2a) = 2 a g g 3 ' w& w m4 = ( 2π × a ) = 3π a 2 + g* g m1 =

Then

Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of each component are zero because of symmetry. m 1 2

2π

w a g

3

2

4

3π

Σ

w a g

w a g w a g

x

y

z

2a

a

−a

2a

1 a 2

0

w 3 a g

0

2a

0

a

0

0

2a

3 − a 2

2a

mx y w 3 a g

4π

− 9π

w 3 a g

w (1 − 5π )a 3 g

my z

mz x

w 3 a g

−2π

− 4π

w 3 a g

0 4

− 9π

w 3 a g

−11π

w 3 a g

w 3 a g

12π 4

w 3 a g

w (1 + 2π )a3 g



Then

0 I xy = Σ( I x′y′ + mx y )

or

0 = Σ( I y′z′ + my z )

or

I yz

I xy =

I yz = −11π

0 I zx = Σ( I z′x′ + mz x )

or

w 3 a (1 − 5π ) g

I zx = 4

w 3 a g

w 3 a (1 + 2π ) g


PROBLEM 9.158 Brass wire with a weight per unit length w is used to form the figure shown. Determine the mass products of inertia Ixy, Iyz, and Izx of the wire figure.


W 1 = wL g g

W& 3 ' 3 w (π × a ) = π a 2 + 2 g g* w w m2 = (3a ) = 3 a g g w w m3 = (π × a ) = π a g g m1 =

Then

Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of each component are zero because of symmetry. m 1

3 w π a 2 g

2

3

w a g

3

π

w a g

x −

y

z

2a

1 a 2

0

1 a 2

2a

(a )

−a

a

2&3 ' a π (* 2 )+

2

π

mx y w 3 a g

−9

0 −2

w 3 a g

−11

Σ

w 3 a g

my z 3 w 3 π a 2 g 3

mz x −

9w 3 a 4g

w 3 a g

−π

w 3 a g

w&π ' + 3 ) a3 ( g*2 +

0 2 −

w 3 a g

1w 3 a 4g



Then

0 I xy = Σ( I x′y′ + mx y )

I xy = −11

or

0 I yz = Σ( I y′z′ + my z )

or

0 + mz x )

or

I zx = Σ( I z′x′

I yz =

w 3 a g

1w 3 a (π + b) 2g

I zx = −

1w 3 a 4g


PROBLEM 9.159 The figure shown is formed of 1.5-mm-diameter aluminum wire. Knowing that the density of aluminum is 2800 kg/m3, determine the mass products of inertia Ixy, Iyz, and Izx of the wire figure.

SOLUTION

First compute the mass of each component. We have m = ρ ALV = ρ AL AL

Then

π ! m1 = m4 = (2800 kg/m3 ) " (0.0015 m) 2 # (0.25 m) $4 % = 1.23700 × 10−3 kg

π ! m2 = m5 = (2800 kg/m3 ) " (0.0015 m)2 # (0.18 m) $4 % = 0.89064 × 10−3 kg

π ! m3 = m6 = (2800 kg/m3 ) " (0.0015 m) 2 # (0.3 m) $4 % = 1.48440 × 10−3 kg



Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of each component are zero because of symmetry. m, kg

x, m

y, m

z, m

mx y , kg ⋅ m 2

my z , kg ⋅ m 2

mz x , kg ⋅ m 2

1

1.23700 × 10−3

0.18

0.125

0

27.8325 × 10−6

0

0

2

0.89064 × 10−3

0.09

0.25

0

20.0394 × 10−6

0

0

3

1.48440 × 10−3

0

0.25

0.15

0

55.6650 × 10−6

0

4

1.23700 × 10−3

0

0.125

0.3

0

46.3875 × 10−6

0

5

0.89064 × 10−3

0.09

0

0.3

0

0

24.0473 × 10−6

6

1.48440 × 10−3

0.18

0

0.15

0

0

40.0788 × 10−6

47.8719 × 10−6

102.0525 × 10−6

64.1261 × 10−6

Σ

Then

0 I xy = Σ( I x′y′ + mx y ) 0 I yz = Σ( I y′z′ + my z ) 0 I zx = Σ( I z′x′ + mz x )

or

I xy = 47.9 × 10−6 kg ⋅ m 2

or

I yz = 102.1 × 10−6 kg ⋅ m 2

or

I zx = 64.1 × 10−6 kg ⋅ m 2


PROBLEM 9.160 Thin aluminum wire of uniform diameter is used to form the figure shown. Denoting by m′ the mass per unit length of the wire, determine the mass products of inertia Ixy, Iyz, and Izx of the wire figure.

SOLUTION First compute the mass of each component. We have &m' m = ( ) L = m′L *L+

Then

&π ' π m1 = m5 = m′ ( R1 ) = m′R1 *2 + 2 m2 = m4 = m′( R2 − R1 ) &π ' π m3 = m′ ( R2 ) = m′R2 *2 + 2

Now observe that because of symmetry the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of components 2 and 4 are zero and ( I x′y′ )1 = ( I z′x′ )1 = 0

( I x′y′ )3 = ( I y′z′ )3 = 0

( I y′z ′ )5 = ( I z′x′ )5 = 0

Also

x1 = x2 = 0

y2 = y3 = y4 = 0

z4 = z5 = 0

Using the parallel-axis theorem [Equations (9.47)], it follows that I xy = I yz = I zx for components 2 and 4. To determine I uv for one quarter of a circular arc, we have dI uv = uvdm where and

u = a cos θ

v = a sin θ

dm = ρ dV = ρ [ A( adθ )]

where A is the cross-sectional area of the wire. Now &π ' &π ' m = m′ ( a ) = ρ A ( a ) 2 * + *2 +

so that and

dm = m′adθ dI uv = (a cos θ )(a sin θ )(m′adθ ) = m′a3 sin θ cos θ dθ



,

I uv = dI uv =

Then

,

π /2 0

m′a3 sin θ cos θ dθ π /2

1 ! = m′a3 " sin 2 θ # $2 %0

=

1 m′a3 2

1 m′R13 2

Thus,

( I yz )1 =

and

1 ( I zx )3 = − m′R23 2

1 ( I xy )5 = − m′R13 2

because of 90° rotations of the coordinate axes. Finally,

I xy

0 0 = Σ( I xy ) = [( I x′y ′ )1 + m1 x1 y1 ] + [( I x′y′ )3 + m3 x3 y3 ] + ( I xy )5 or 0

0

1 I xy = − m′R13 2

I yz = Σ( I yz ) = ( I yz )1 + [( I y′z′ )3 + m3 y3 z3 ] + [( I y′z′ )5 + m5 y5 z5 ]

or 0 0 I zx = Σ( I zx ) = [( I z′x′ )1 + m1 z1 x1 ] + ( I zx )3 + [( I z ′x′ )5 + m5 z5 x5 ] or

I yz =

1 m′R13 2

1 I zx = − m′R23 2


PROBLEM 9.161 Complete the derivation of Eqs. (9.47), which express the parallel-axis theorem for mass products of inertia.

SOLUTION We have and Consider

,

,

,

I xy = xydm I yz = yzdm I zx = zxdm x = x′ + x

y = y′ + y

(9.45)

z = z′ + z

(9.31)

,

I xy = xydm

Substituting for x and for y

, = , x′y ′dm + y , x′dm + x , y ′dm + x y , dm

I xy = ( x′ + x )( y ′ + y ) dm

By definition and

,

I x′y′ = x′y ′dm

, x′dm = mx ′ , y′dm = my ′

However, the origin of the primed coordinate system coincides with the mass center G, so that x′ = y′ = 0 I xy = I x′y′ + mx y Q.E.D.

The expressions for I yz and I zx are obtained in a similar manner.


PROBLEM 9.162 For the homogeneous tetrahedron of mass m shown, (a) determine by direct integration the mass product of inertia Izx, (b) deduce Iyz and Ixy from the result obtained in Part a.

SOLUTION (a)

First divide the tetrahedron into a series of thin vertical slices of thickness dz as shown. a z' & z + a = a (1 − ) c c * +

Now

x=−

and

b z' & y = − z + b = b (1 − ) c * c+

The mass dm of the slab is 2

z' &1 ' 1 & dm = ρ dV = ρ ( xydz ) = ρ ab (1 − ) dz 2 2 c * + * +

Then

,

m = dm =

,

c 0

2

z' 1 & ρ ab (1 − ) dz c 2 * +

z' 1 & c '& = ρ ab "( − )(1 − ) 2 3 c +* + "$*

Now where

3 !c

1 # = ρ abc #% 0 6

dI zx = dI z′x′ + zEL xEL dm dI z′x′ = 0 (symmetry)

and

zEL = z

Then

I zx = dI zx =

,

xEL =

1 1 & z' x = a (1 − ) 3 3 * c+

2 ! z '! 1 z' 1 & & z " a (1 − ) # " ρ ab (1 − ) dz # 0 * c+ #% $ 3 * c + % $" 2 2 3 4 c& 1 z z z ' = ρ a 2 b (( z − 3 + 3 2 − 3 )) dz 0 6 c c c + *

,

c

,

c

m 1 z3 3 z 4 1 z5 ! = a " z2 − + − # c $2 c 4 c 2 5 c3 % 0

or

I zx =

1 mac 20



(b)

Because of the symmetry of the body, I xy and I yz can be deduced by considering the circular permutation of ( x, y, z ) and (a, b, c) . Thus,

!

I xy =

1 mab 20

!

I yz =

1 mbc 20

!

Alternative solution for Part a: First divide the tetrahedron into a series of thin horizontal slices of thickness dy as shown. Now

x=−

a y' & y + a = a (1 − ) b b+ *

and

z=−

c y' & y + c = c (1 − ) b * b+

The mass dm of the slab is 2

y' &1 ' 1 & dm = ρ dV = ρ ( xzdy ) = ρ ac (1 − ) dy *2 + 2 * b+

Now where and dI zx,area =

Then

dI zx = ρ tdI zx ,area t = dy

1 2 2 x z from the results of Sample Problem 9.6. 24 2 2 -/ 1 y' ! & y ' ! ./ & dIzx = ρ (dy ) 0 " a (1 − ) # "c (1 − ) # 1 b + % $ * b +% / /2 24 $ * 3 4

=

Finally,

4

1 1m & y' y' & ac (1 − ) dy ρ a 2 c 2 (1 − ) dy = 24 4b * b+ b+ *

,

I zx = dI zx =

,

b 0

4

1m & y' ac (1 − ) dy 4 b * b+ b

5 y' ! 1m & b '& ac "( − )(1 − ) # = b + %0 4 b $* 5 +*

or

I zx =

1 mac 20



Alternative solution for Part a: The equation of the included face of the tetrahedron is x y z + + =1 a b c x z' & y = b (1 − − ) a c+ *

so that

For an infinitesimal element of sides dx, dy, and dz : dm = ρ dV = ρ dydxdz z' & x = a (1 − ) * c+

From Part a Now

,

I zx = zxdm =

c

a (1− z/c )

0

0

,,

=ρ

= ρb

= ρb =

b (1− x/a − z/c ) 0

c

a (1− z/c )

0

0

,,

= ρb

,

, , ,

c 0 c 0 c 0

zx( ρ dydxdz )

zx $b (1 − ax −

z c

)!% dxdz a (1− z/c )

1 1 x3 1 z 2 ! z " x2 − x # − 3 a 2 c %0 $2 2

dz 3

1 & 1 z' z' 1 z 2& z' & z " a 2 (1 − ) − a3 (1 − ) − a (1 − ) 2 3 2 c a c c c * + * + * + "$

2!

# dz #%

3

1 2 & z' a z (1 − ) dz 6 * c+

1 2 ρa b 6

,

c&

z2 z3 z4 ' (( z − 3 + 3 2 − 3 )) dz 0 c c c + * c

m 1 z3 3 z4 1 z5 ! = a " z2 − + − # c $2 c 4 c 2 5 c3 % 0

or I zx =

1 mac 20


PROBLEM 9.163 The homogeneous circular cylinder shown has a mass m. Determine the mass moment of inertia of the cylinder with respect to the line joining the origin O and Point A that is located on the perimeter of the top surface of the cylinder.

SOLUTION Iy =

From Figure 9.28:

1 ma 2 2

and using the parallel-axis theorem 2

Ix = Iz =

Symmetry implies

1 1 &h' m(3a 2 + h 2 ) + m ( ) = m(3a 2 + 4h 2 ) 12 2 12 * +

I xy = I yz = I zx = 0

For convenience, let Point A lie in the yz plane. Then

λOA =

1 2

h + a2

( hj + ak )

With the mass products of inertia equal to zero, Equation (9.46) reduces to 0 I OA = I x λx2 + I y λ y2 + I z λz2 & 1 h = ma 2 ( ( h2 + a 2 2 *

2

' & 1 a ) + m(3a 2 + 4h2 ) ( ) 12 ( h2 + a 2 + *

or

' ) ) +

2

I OA =

1 10h 2 + 3a 2 ma 2 12 h2 + a 2

Note: For Point A located at an arbitrary point on the perimeter of the top surface, λOA is given by

λOA =

1 2

h + a2

( a cos φ i + hj + a sin φ k )

which results in the same expression for I OA . !


PROBLEM 9.164 The homogeneous circular cone shown has a mass m. Determine the mass moment of inertia of the cone with respect to the line joining the origin O and Point A.

SOLUTION 2

9 &3 ' = ( a ) + (−3a) 2 + (3a) 2 = a 2 *2 +

First note that

dOA

Then

λOA =

1 &3 ' 1 ai − 3aj + 3ak ) = (i − 2 j + 2k ) ( a*2 + 3

9 2

For a rectangular coordinate system with origin at Point A and axes aligned with the given x, y , z axes, we have (using Figure 9.28) 3 1 ! I x = I z = m " a 2 + (3a) 2 # 5 $4 % =

Also, symmetry implies

Iy =

3 ma 2 10

111 2 ma 20


With the mass products of inertia equal to zero, Equation (9.46) reduces to I OA = I x λx2 + I y λ y2 + I z λz2 2

2

111 2 & 1 ' 3 & 2 ' 111 2 & 2 ' = ma ( ) + ma 2 ( − ) + ma ( ) 20 20 * 3 + 10 * 3+ *3+ =

193 2 ma 60

or

2

I OA = 3.22ma 2


!

PROBLEM 9.165 Shown is the machine element of Problem 9.141. Determine its mass moment of inertia with respect to the line joining the origin O and Point A.

SOLUTION

First compute the mass of each component. We have

m = ρSTV =

0.284 lb/in.3 V = (0.008819 lb ⋅ s 2 /ft ⋅ in.3 )V 2 32.2 ft/s

Then m1 = (7850 kg/m3 )[π (0.08 m)2 (0.04 m)] = 6.31334 kg m2 = (7850 kg/m3 )[π (0.02 m) 2 (0.06 m)] = 0.59188 kg m3 = (7850 kg/m3 )[π (0.02 m) 2 (0.04 m)] = 0.39458 kg

Symmetry implies and Now

I yz = I zx = 0

( I xy )1 = 0

( I x′y′ ) 2 = ( I x′y′ )3 = 0 I xy = Σ( I x′y′ + mx y ) = m2 x2 y2 − m3 x3 y3 = [0.59188 kg (0.04 m)(0.03 m)] − [0.39458 kg (− 0.04 m)(− 0.02 m)] = (0.71026 − 0.31566) × 10−3 kg ⋅ m 2 = 0.39460 × 10−3 kg ⋅ m 2



From the solution to Problem 9.141, we have I x = 13.98800 × 10−3 kg ⋅ m 2 I y = 20.55783 × 10−3 kg ⋅ m 2 I z = 14.30368 × 10−3 kg ⋅ m 2

By observation

OA

=

1 13

(2i + 3j)

Substituting into Eq. (9.46) I OA =

I x λx2

+

I y λ y2

+

0 0 0 − 2 I xy λx λ y − 2 I yz λ y λz − 2 I zx λz λx

I z λz2

2

2

& 2 ' & 3 ' = "(13.98800) ( ) + (20.55783) ( ) 13 "$ * + * 13 + & 2 '& 2 ' ! 2 −3 − 2(0.39460) ( )( ) # × 10 kg ⋅ m * 13 +* 13 + %

= (4.30400 + 14.23234 − 0.36425) × 10−3 kg ⋅ m 2 or

I OA = 18.17 × 10−3 kg ⋅ m 2


!

PROBLEM 9.166 Determine the mass moment of inertia of the steel fixture of Problems 9.145 and 9.149 with respect to the axis through the origin that forms equal angles with the x, y, and z axes.

SOLUTION From the solutions to Problems 9.145 and 9.149, we have Problem 9.145:

I x = 26.4325 × 10−3 kg ⋅ m 2 I y = 31.1726 × 10−3 kg ⋅ m 2 I z = 8.5773 × 10−3 kg ⋅ m 2

Problem 9.149:

I xy = 2.5002 × 10−3 kg ⋅ m 2 I yz = 4.0627 × 10−3 kg ⋅ m 2 I zx = 8.8062 × 10−3 kg ⋅ m 2

From the problem statement it follows that

λx = λ y = λz Now or

λx2 + λ y2 + λz2 = 1 4 3λx2 = 1 λx = λ y = λz =

1 3

Substituting into Eq. (9.46) I OL = I x λx2 + I y λ y2 + I z λz2 − 2 I xy λx λ y − 2 I yz λ y λz − 2 I zx λz λx

Noting that We have

λx2 = λ y2 = λz2 = λx λ y = λ y λz = λz λx =

1 3

1 I OL = [26.4325 + 31.1726 + 8.5773 3 − 2(2.5002 + 4.0627 + 8.8062)] × 10−3 kg ⋅ m 2 or

I OL = 11.81 × 10−3 kg ⋅ m 2


!

PROBLEM 9.167 The thin bent plate shown is of uniform density and weight W. Determine its mass moment of inertia with respect to the line joining the origin O and Point A.

SOLUTION First note that and that

m1 = m2 =

λOA =

1 3

1W 2 g

(i + j + k )

Using Figure 9.28 and the parallel-axis theorem, we have I x = ( I x )1 + ( I x )2 1 &1W =" ( 12 $" * 2 g

' 2 1W )a + 2 g +

-/ 1 & 1 W +0 ( /212 * 2 g

2 &a' ! ( ) # * 2 + %# 2

' 2 1W 2 ) (a + a ) + 2 g +

&a' &a' "( ) + ( ) "$* 2 + * 2 + 1 W & 1 1 ' 2 & 1 1 ' 2! 1 W 2 a = "( + ) a + ( + ) a # = 2 g $* 12 4 + *6 2+ % 2 g

2 !.

/ #1 #% /3

I y = ( I y )1 + ( I y )2 -/ 1 & 1 W ' 1W 2 2 =0 ( ) (a + a ) + 12 2 g + 2 g 2/ *

2

&a' &a' "( ) + ( ) "$* 2 + * 2 + 2 -/ 1 & 1 W ' 2 1 W & a ' ! /. 2 +0 ( + + ( ) a a " ) ( ) #1 2 g "$ * 2 + #% /3 /212 * 2 g + 1 W & 1 1 ' 2 & 1 5' 2! W 2 = a + ( + )a # = a " + 2 g $(* 6 2 )+ * 12 4 + % g

2 !.

/ #1 #% 3/

I z = ( I z )1 + ( I z ) 2 1 &1W =" ( 12 $" * 2 g

' 2 1W )a + 2 g +

-/ 1 & 1 W +0 ( 2/12 * 2 g

2 &a' ! ( ) # * 2 + %# 2 !.

' 2 1W &a' " (a ) 2 + ( ) )a + 2 g *2+ + $" 1 W & 1 1 ' 2 & 1 5' 2! = + a + ( + )a # = " 2 g $(* 12 4 )+ * 12 4 + %

/ #1 %# /3 5W 2 a 6 g



Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of both components are zero because of symmetry. Also, y1 = 0 0 1W &a' 1W 2 (a) ( ) = I xy = Σ( I x′y ′ + mx y ) = m2 x2 y2 = a 2 g *2+ 4 g

Then

0 1 W & a '& a ' 1 W 2 I yz = Σ( I y′z′ + my z ) = m2 y2 z2 = a )= 2 g (* 2 )( +* 2 + 8 g 0 I zx = Σ( I z ′x′ + mz x ) = m1z1 x1 + m2 z2 x2 =

1 W & a '& a ' 1 W & a ' 3W 2 a + (a) = 2 g (* 2 )+ (* 2 )+ 2 g (* 2 )+ 8 g

Substituting into Equation (9.46) I OA = I x λx2 + I y λ y2 + I z λz2 − 2 I xy λx λ y − 2 I yz λ y λz − 2 I zx λz λx

Noting that

λx2 = λ y2 = λz2 = λx λ y = λ y λz = λz λx =

1 3

We have & 1 W 2 1 W 2 3 W 2 '! 1 1W 2 W 2 5W 2 I OA = " a + a + a − 2( a + a + a )# 3 $2 g 6 g 8 g 8 g g *4 g +% 1 14 & 3 '! W = " − 2 ( )# a 2 3$ 6 * 4 +% g

or

I OA =

5 W 2 a 18 g


!

PROBLEM 9.168 A piece of sheet steel of thickness t and specific weight γ is cut and bent into the machine component shown. Determine the mass moment of inertia of the component with respect to the joining the origin O and Point A.

SOLUTION First note that OA

1

=

6

(i + 2 j + k )

Next compute the mass of each component. We have m = ρV =

Then

m1 = m2 =

γ g

γ g

(tA)

t (2a × 2a) = 4

γt g

a2

γ &π

' π γt 2 × a2 ) = t a g (* 2 + 2 g

Using Figure 9.28 for component 1 and the equations derived above (following the solution to Problem 9.134) for a semicircular plate for component 2, we have I x = ( I x )1 + ( I x )2 -/ 1 & γ t ' ./ γt = 0 ( 4 a 2 ) [(2a) 2 + (2a )2 ] + 4 a 2 (a 2 + a 2 ) 1 g /212 * g /3 + /- 1 & π γ t 2 ' 2 π γ t 2 /. a )a + a [(2a) 2 + (a) 2 ]1 +0 ( 2 g /2 4 * 2 g + /3

π γt 2 & 1 &2 ' ' a2 ( + 2 ) a2 + a + 5 ) a2 g 2 g (* 4 *3 + + γt = 18.91335 a 4 g =4

γt



I y = ( I y )1 + ( I y ) 2 ! 1 & γt ' γt = " ( 4 a 2 ) (2a )2 + 4 a 2 ( a 2 ) # g $12 * g + % 2 -/ π γ t & 1 16 ' ! ./ π γ t 2 & 4a ' a2 ( − 2 ) a2 + a "( +0 + (a) 2 # 1 ) 2 g "$* 3π + #% /3 /2 2 g * 2 9π + 16 γt & 1 ' π γ t 2 & 1 16 ' a = 4 a 2 ( + 1) a 2 + − + + 1) a 2 g *3 + 2 g (* 2 9π 2 9π 2 + γt = 7.68953 a 4 g I z = ( I z )1 + ( I z ) 2 ! γt 1 & γt ' = " ( 4 a 2 ) (2a) 2 + 4 a 2 ( a 2 ) # g $12 * g + % 2 -/ π γ t & 1 16 ' ! ./ π γ t 2 & 4a ' +0 + (2a) 2 # 1 a2 ( − 2 ) a2 + a "( ) 2 g * 4 9π + "$* 3π + #% /3 /2 2 g γt &1 ' π γ t 2 & 1 16 16 ' = 4 a 2 ( + 1) a 2 + a ( − 2 + 2 + 4 ) a2 2 g * 4 9π g 9π *3 + + γt = 12.00922 a 4 g

Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of both components are zero because of symmetry. Also x1 = 0. 0 π γ t 2 & 4a ' I xy = Σ( I x′y′ + mx y ) = m2 x2 y2 = a ( Then ) (2a) 2 g * 3π + γt = 1.33333 a 4 g I yz = Σ( I y′z′ =4

γt g

0 + my z ) = m1 y1 z1 + m2 y2 z2

a 2 ( a)(a) +

= 7.14159

γt g

π γt 2 g

a 2 (2a)(a)

a4

0 π γ t 2 & 4a ' I zx = Σ( I z ′x′ + mz x ) = m2 z2 x2 = a (a) ( ) 2 g * 3π + γt = 0.66667 a 4 g PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1640


Substituting into Eq. (9.46) I OA = I x λx2 + I y λ y2 + I z λz2 − 2 I xy λx λ y − 2 I yz λ y λz − 2 I zx λz λx = 18.91335

2

& 1 ' γt 4 & 2 ' a4 ( ) + 7.68953 a ( ) g g * 6+ * 6+

γt

2

2

& & 1 ' γ t 4 ' & 1 '& 2 ' a4 ( ) − 2 (1.33333 a ) ( )( ) g * 6+ g * + * 6 +* 6 + & γ t ' & 2 '& 2 ' & γ t 4 ' & 1 '& 1 ' − 2 ( 7.14159 a 4 ) ( )( ) − 2 ( 0.66667 a ) ( )( ) g g + * 6 +* 6 + * + * 6 +* 6 + * γt = (3.15223 + 5.12635 + 2.00154 − 0.88889 − 4.76106 − 0.22222) a 4 g + 12.00922

γt

or

I OA = 4.41

γt g

a4


!

PROBLEM 9.169 Determine the mass moment of inertia of the machine component of Problems 9.136 and 9.155 with respect to the axis through the origin characterized by the unit vector λ = (− 4i + 8j + k )/9.

SOLUTION From the solutions to Problems 9.136 and 9.155. We have Problem 9.136:

I x = 175.503 × 10−3 kg ⋅ m 2 I y = 308.629 × 10−3 kg ⋅ m 2 I z = 154.400 × 10−3 kg ⋅ m 2

Problem 9.155:

I xy = −8.0365 × 10−3 kg ⋅ m 2 I yz = 12.8950 × 10−3 kg ⋅ m 2 I zx = 94.0266 × 10−3 kg ⋅ m 2

Substituting into Eq. (9.46) I OL = I x λx2 + I y λ y2 + I z λz2 − 2 I xy λx λ y − 2 I yz λ y λz − 2 I zx λz λx 2

2

& 4' &8' &1' = "175.503 ( − ) + 308.629 ( ) + 154.400 ( ) * 9+ *9+ *9+ "$ & 4 '& 8 ' & 8 '& 1 ' − 2(−8.0365) ( − )( ) − 2(12.8950) ( )( ) 9 9 * +* + * 9 +* 9 + & 1 '& 4 ' ! − 2(94.0266) ( )( − ) # × 10−3 kg ⋅ m 2 * 9 +* 9 + %

2

= (34.6673 + 243.855 + 1.906 − 6.350 − 2.547 + 9.287) × 10−3 kg ⋅ m 2

or

I OL = 281 × 10−3 kg ⋅ m 2


!

PROBLEM 9.170 For the wire figure of Problem 9.148, determine the mass moment of inertia of the figure with respect to the axis through the origin characterized by the unit vector λ = (−3i − 6j + 2k )/7.

SOLUTION First compute the mass of each component. We have &m' m = ( ) L = 0.056 kg/m × 1.2 m *L+ = 0.0672 kg

Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , for each component are zero because of symmetry. Also x1 = x6 = 0

Then

y4 = y5 = y6 = 0

z1 = z2 = z3 = 0

0 I xy = Σ( I x′y ′ + mx y ) = m2 x2 y2 + m3 x3 y3 = (0.0672 kg)(0.6 m)(1.2 m) + (0.0672 kg)(1.2 m)(0.6 m) = 0.096768 kg ⋅ m 2

0 I yz = Σ( I y′z′ + my z ) = 0 0 I zx = Σ( I z ′x′ + mz x ) = m4 z4 x4 + m5 z5 x5 = (0.0672 kg)(0.6 m)(1.2 m) + (0.0672 kg)(1.2 m)(0.6 m) = 0.096768 kg ⋅ m 2 From the solution to Problem 9.148, we have I x = 0.32258 kg ⋅ m 2 I y = I z = 0.41933 kg ⋅ m 2



Substituting into Eq. (9.46)

0 I OL = I x λx2 + I y λ y2 + I z λz2 − 2 I xy λx λ y − 2 I yz λ y λz − 2 I zx λz λx 2

2

& 3' & 6' &2' = "0.32258 ( − ) + 0.41933 ( − ) + 0.41933 ( ) 7 7 * + * + *7+ "$

2

& 3 '& 6 ' & 2 '& 3 ' ! − 2(0.096768) ( − )( − ) − 2(0.096768) ( )( − ) # kg ⋅ m 2 * 7 +* 7 + * 7 +* 7 + % I OL = (0.059249 + 0.30808 + 0.034231 − 0.071095 + 0.023698) kg ⋅ m 2

or

I OL = 0.354 kg ⋅ m 2


!



γ ST g

AL

Then m1 = m2 =

3 2 & 1 ft ' π &1 ' ! 490 lb/ft 3 × × × × π in. ( 18 in.) " # ( ) ( ) 32.2 ft/s 2 $" 4 * 8 + %# * 12 in. +

= 6.1112 × 10−3 lb ⋅ s 2 /ft

π &1 ' 490 lb/ft 3 × " ( in. ) m3 = m4 = 2 32.2 ft/s $" 4 * 8 +

2!

& 1 ft ' # × 18 in. × ( ) * 12 in. + %#

3

= 1.9453 lb ⋅ s 2 /ft

Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , for each component are zero because of symmetry. Also

x3 = x4 = 0

y1 = 0

z1 = z2 = 0

Then

0 I xy = Σ( I x′y′ + mx y ) = m2 x2 y2 & 1 ft ' & 2 × 18 ' in. ) (18 in.) ( = (6.1112 × 10−3 lb ⋅ s 2 /ft) ( − ) π * + * 12 in. +

2

= −8.75480 × 10−3 lb ⋅ ft ⋅ s 2

0 I yz = Σ( I y ′z′ + my z ) = m3 y3 z3 + m4 y4 z4 Now

m3 = m4 ,

y3 = y4 ,

z 4 = − z3

0 I zx = Σ( I z′x′ + mz x ) or

I yz = 0

I zx = 0



From the solution to Problem 9.147, we have I x = 39.1721 × 10 −3 lb ⋅ ft ⋅ s 2 I y = 36.2542 × 10−3 lb ⋅ ft ⋅ s 2 I z = 30.4184 × 10−3 lb ⋅ ft ⋅ s 2


0 0 I OL = I x λx2 + I y λ y2 + I z λz2 − 2 I xy λx λ y − 2 I yz λ y λz − 2 I zx λz λx 2

2

& 3' & 6' &2' = "39.1721( − ) + 36.2542 ( − ) + 30.4184 ( ) * 7+ * 7+ *7+ "$ & 3 '& 6 ' ! − 2(−8.75480) ( − )( − ) # × 10−3 lb ⋅ ft ⋅ s 2 * 7 +* 7 + %

2

= (7.19488 + 26.6357 + 2.48313 + 6.43210) × 10−3 lb ⋅ ft ⋅ s 2

or

I OL = 0.0427 lb ⋅ ft ⋅ s 2


!



W 1 = (W/L) AL L g g

m1 =

1 1 ft (0.033 lb/ft)(2π × 16 in.) × 2 12 in. 32.2 ft/s

Then

= 8.5857 × 10 −3 lb ⋅ s 2 /ft m2 = m3 = m4 = m5 =

1 1 ft (0.033 lb/ft)(8 in.) × 12 in. 32.2 ft/s 2

= 0.6832 × 10−3 lb ⋅ ft/s 2

Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of each component are zero because of symmetry. Also

Then

x1 = x4 = x5 = 0 y1 = 0 z1 = z2 = z3 = 0 0 I xy = Σ( I x′y ′ + mx y ) = m2 x2 y2 + m3 x3 y3

& 16 '& 4 ' = 0.6832 × 10 −3 lb ⋅ s 2 /ft ( ft )( ft ) * 12 +* 12 + & 12 '& 8 ' + 0.6832 × 10−3 lb ⋅ s 2 /ft ( ft )( ft ) * 12 +* 12 + = (0.30364 + 0.45547) × 10−3 lb ⋅ ft ⋅ s 2 = 0.75911 × 10−3 lb ⋅ ft ⋅ s 2

Symmetry implies

I yz = I xy

I yz = 0.75911 × 10−3 lb ⋅ ft ⋅ s 2

0 I zx = Σ( I z ′x′ + mz x ) = 0



From the solution to Problem 9.146, we have I x = I z = 10.3642 × 10−3 lb ⋅ ft ⋅ s 2 I y = 19.1097 × 10−3 lb ⋅ ft ⋅ s 2


0 I OL = I x λx2 + I y λ y2 + I z λz2 − 2 I xy λx λ y − 2 I yz λ y λz − 2 I zx λz λx 2

2

& 3' & 6' &2' = "10.3642 ( − ) + 19.1097 ( − ) + 10.3642 ( ) 7 7 * + * + *7+ "$

2

& 3 '& 6 ' & 6 '& 2 ' ! − 2(0.75911) ( − )( − ) − 2(0.75911) ( − )( ) # × 10−3 lb ⋅ ft ⋅ s 2 * 7 +* 7 + * 7 +* 7 + % = (1.90663 + 14.03978 + 0.84606 − 0.55771 + 0.37181) × 10−3 lb ⋅ ft ⋅ s 2

or

I OL = 16.61 × 10−3 lb ⋅ ft ⋅ s 2


!

PROBLEM 9.173 For the rectangular prism shown, determine the values of the ratios b/a and c/a so that the ellipsoid of inertia of the prism is a sphere when computed (a) at Point A, (b) at Point B.

SOLUTION (a)

Using Figure 9.28 and the parallel-axis theorem, we have at Point A 1 I x′ = m(b 2 + c 2 ) 12 1 &a' I y ′ = m( a 2 + c 2 ) + m ( ) 12 * 2+ 1 = m(4a 2 + c 2 ) 12

2

2

I z′ =

1 1 &a' m(a 2 + b 2 ) + m ( ) = m(4a 2 + b 2 ) 12 * 2 + 12

Now observe that symmetry implies I x′y′ = I y′z′ = I z′x′ = 0

Using Eq. (9.48), the equation of the ellipsoid of inertia is then I x′ x 2 + I y′ y 2 + I z′ z 2 = 1:

1 1 1 m(b 2 + c 2 ) x 2 + m(4a 2 + c 2 ) y 2 + m(4a 2 + b 2 ) z 2 = 1 12 12 12

For the ellipsoid to be a sphere, the coefficients must be equal. Therefore 1 1 m(b 2 + c 2 ) = m(4a 2 + c 2 ) 12 12 1 = m(4a 2 + b 2 ) 12

Then

b 2 + c 2 = 4a 2 + c 2

or

b =2 a

and

b 2 + c 2 = 4a 2 + b 2

or

c =2 a



(b)

Using Figure 9.28 and the parallel-axis theorem, we have at Point B 2

I x′′ =

1 1 &c' m(b 2 + c 2 ) + m ( ) = m(b 2 + 4c 2 ) 12 12 *2+ 2

1 1 &c' m ( a 2 + c 2 ) + m ( ) = m ( a 2 + 4c 2 ) 12 * 2 + 12 1 I z′′ = m(a 2 + b 2 ) 12

I y′′ =

Now observe that symmetry implies I x′′y′′ = I y′′z′′ = I z′′x′′ = 0

From Part a it then immediately follows that 1 1 1 m(b 2 + 4c 2 ) = m(a 2 + 4c 2 ) = m(a 2 + b 2 ) 12 12 12

Then

b 2 + 4c 2 = a 2 + 4c 2

or

b =1 a

and

b 2 + 4c 2 = a 2 + b 2

or

c 1 = a 2


!

PROBLEM 9.174 For the right circular cone of Sample Problem 9.11, determine the value of the ratio a/h for which the ellipsoid of inertia of the cone is a sphere when computed (a) at the apex of the cone, (b) at the center of the base of the cone.

SOLUTION (a)

From Sample Problem 9.11, we have at the apex A 3 ma 2 10 3 &1 ' I y = I z = m ( a 2 + h2 ) 5 *4 + Ix =


Now observe that symmetry implies

Using Eq. (9.48), the equation of the ellipsoid of inertia is then

I x x 2 + I y y 2 + I z z 2 = 1:

3 3 &1 3 &1 ' ' ma 2 x 2 + m ( a 2 + h 2 ) y 2 + m ( a 2 + h 2 ) z 2 = 1 10 5 *4 5 4 + * +

For the ellipsoid to be a sphere, the coefficients must be equal. Therefore, 3 3 &1 ' ma 2 = m ( a 2 + h 2 ) 10 5 *4 +

(b)

or

a =2 h

From Sample Problem 9.11, we have

and at the centroid C

I x′ =

3 ma 2 10

I y′′ =

3 & 2 1 2' m a + h ) 20 (* 4 + 2

Then

I y′ = I z′ =

3 & 2 1 2' 1 &h' m( a + h ) + m( ) = m(3a 2 + 2h 2 ) 20 * 4 + 20 *4+


From Part a it then immediately follows that 3 1 ma 2 = m(3a 2 + 2h 2 ) 10 20

or

a = h

2 3


!

PROBLEM 9.175 For the homogeneous circular cylinder shown, of radius a and length L, determine the value of the ratio a/L for which the ellipsoid of inertia of the cylinder is a sphere when computed (a) at the centroid of the cylinder, (b) at Point A.

SOLUTION (a)

From Figure 9.28: 1 2 ma 2 1 I y = I z = m(3a 2 + L2 ) 12 Ix =

Now observe that symmetry implies


Using Eq. (9.48), the equation of the ellipsoid of inertia is then I x x 2 + I y y 2 + I z z 2 = 1:

1 2 2 1 1 ma x + m(3a 2 + L2 ) y 2 + m(3a 2 + L2 ) = 1 2 12 12

For the ellipsoid to be a sphere, the coefficients must be equal. Therefore 1 2 1 ma = m(3a 2 + L2 ) 2 12

(b)

or

a 1 = L 3

Using Figure 9.28 and the parallel-axis theorem, we have I x′ =

1 2 ma 2 2

1 7 2' &L' &1 I y′ = I z′ = m(3a 2 + L2 ) + m ( ) = m ( a 2 + L ) 12 48 + *4+ *4


From Part a it then immediately follows that 1 2 7 2' &1 ma = m ( a 2 + L ) 2 4 48 * +

or

a 7 = L 12


!

PROBLEM 9.176 Given an arbitrary body and three rectangular axes x, y, and z, prove that the mass moment of inertia of the body with respect to any one of the three axes cannot be larger than the sum of the mass moments of inertia of the body with respect to the other two axes. That is, prove that the inequality I x # I y + I z and the two similar inequalities are satisfied. Further, prove that I y $ 12 I x if the body is a homogeneous solid of revolution, where x is the axis of revolution and y is a transverse axis.

SOLUTION (i)

Iy + Iz $ Ix

To prove

, = , (x

I y = ( z 2 + x 2 )dm

By definition

Iz

2

+ y 2 ) dm

,

,

I y + I z = ( z 2 + x 2 ) dm + ( x 2 + y 2 ) dm

Then

,

,

= ( y 2 + z 2 )dm + 2 x 2 dm

Now

,(y

2

+ z 2 )dm = I x

, x dm $ 0 2

and

Iy + Iz $ Ix

Q.E.D.

The proofs of the other two inequalities follow similar steps. (ii)

If the x axis is the axis of revolution, then I y = Iz

and from Part (i) or

Iy + Iz $ Ix

2I y $ I x Iy $

or

1 Ix 2

Q.E.D. !


PROBLEM 9.177 Consider a cube of mass m and side a. (a) Show that the ellipsoid of inertia at the center of the cube is a sphere, and use this property to determine the moment of inertia of the cube with respect to one of its diagonals. (b) Show that the ellipsoid of inertia at one of the corners of the cube is an ellipsoid of revolution, and determine the principal moments of inertia of the cube at that point.

SOLUTION (a)

At the center of the cube have (using Figure 9.28) Ix = I y = Iz =

1 1 m(a 2 + a 2 ) = ma 2 12 6

Now observe that symmetry implies I xy = I yz = I zx = 0

Using Equation (9.48), the equation of the ellipsoid of inertia is &1 2' 2 &1 2' 2 &1 2' 2 ( 6 ma ) x + ( 6 ma ) y + ( 6 ma ) z = 1 * + * + * +

or

x2 + y2 + z 2 =

6 (= R 2 ) ma 2

which is the equation of a sphere. Since the ellipsoid of inertia is a sphere, the moment of inertia with respect to any axis OL through the center O of the cube must always & 1 ' 1 ). be the same ( R = I OL = ma 2 ( 6 I OL )+ * (b)

The above sketch of the cube is the view seen if the line of sight is along the diagonal that passes through corner A. For a rectangular coordinate system at A and with one of the coordinate axes aligned with the diagonal, an ellipsoid of inertia at A could be constructed. If the cube is then rotated 120° about the diagonal, the mass distribution will remain unchanged. Thus, the ellipsoid will also remain unchanged after it is rotated. As noted at the end of Section 9.17, this is possible only if the ellipsoid is an ellipsoid of revolution, where the diagonal is both the axis of revolution and a principal axis.! 1 It then follows that I x′ = I OL = ma 2 ! 6 In addition, for an ellipsoid of revolution, the two transverse principal moments of inertia are equal and any axis perpendicular to the axis of revolution is a principal axis. Then, applying the parallel-axis theorem between the center of the cube and corner A for any perpendicular axis I y′ = I z′ =

& 3 ' 1 2 ma + m ( a ( 2 )) 6 * +

2

or

I y′ = I z′ =

11 2 ma 12

(Note: Part b can also be solved using the method of Section 9.18.) PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1654


First note that at corner A 2 2 ma 3 1 = ma 2 4

Ix = I y = Iz = I xy = I yz = I zx

Substituting into Equation (9.56) yields k 3 − 2ma 2 k 2 +

55 2 6 121 3 9 m a k− m a =0 48 864

For which the roots are 1 2 ma 6 ! 11 k2 = k3 = ma 2 12 k1 =


PROBLEM 9.178 Given a homogeneous body of mass m and of arbitrary shape and three rectangular axes x, y, and z with origin at O, prove that the sum Ix + Iy + Iz of the mass moments of inertia of the body cannot be smaller than the similar sum computed for a sphere of the same mass and the same material centered at O. Further, using the result of Problem 9.176, prove that if the body is a solid of revolution, where x is the axis of revolution, its mass moment of inertia Iy about a transverse axis y cannot be smaller than 3ma2/10, where a is the radius of the sphere of the same mass and the same material.

SOLUTION (i)

Using Equation (9.30), we have

,

,

,

I x + I y + I z = ( y 2 + z 2 )dm + ( z 2 + x 2 ) dm + ( x 2 + y 2 )dm

,

= 2 ( x 2 + y 2 + z 2 )dm

,

= 2 r 2 dm

where r is the distance from the origin O to the element of mass dm. Now assume that the given body can be formed by adding and subtracting appropriate volumes V1 and V2 from a sphere of mass m and radius a which is centered at O; it then follows that m1 = m2 (mbody = msphere = m)

Then ( I x + I y + I z ) body = ( I x + I y + I z )sphere + ( I x + I y + I z )V1 − ( I x + I y + I z )V2

or

( I x + I y + I z ) body = ( I x + I y + I z )sphere + 2

,

m1

r 2 dm − 2

,

m2

r 2 dm

Now, m1 = m2 and r1 $ r2 for all elements of mass dm in volumes 1 and 2.

,

m1

r 2 dm −

,

m2

r 2 dm $ 0 ( I x + I y + I z ) body $ ( I x + I y + I z )sphere

so that

Q.E.D.



(ii)

First note from Figure 9.28 that for a sphere Ix = Iy = Iz =

Thus

( I x + I y + I z )sphere =

2 2 ma 5

6 2 ma 5

For a solid of revolution, where the x axis is the axis of revolution, we have I y = Iz

Then, using the results of Part (i) 6 ( I x + 2 I y ) body $ ma 2 5

From Problem 9.178 we have or

Iy $

1 Ix 2

(2 I y − I x ) body $ 0

Adding the last two inequalities yields 6 (4 I y ) body $ ma 2 5

or

( I y ) body $

3 ma 2 10

Q.E.D.


PROBLEM 9.179* The homogeneous circular cylinder shown has a mass m, and the diameter OB of its top surface forms 45° angles with the x and z axes. (a) Determine the principal mass moments of inertia of the cylinder at the origin O. (b) Compute the angles that the principal axes of inertia at O form with the coordinate axes. (c) Sketch the cylinder, and show the orientation of the principal axes of inertia relative to the x, y, and z axes.

SOLUTION (a)

First compute the moments of inertia using Figure 9.28 and the parallel-axis theorem. 2 2 " a # " a # ! 13 2 1 + m(3a 2 + a 2 ) + m $& ' & ' % = ma 12 $*( 2 ) ( 2 ) %+ 12 1 3 I y = ma 2 + m(a) 2 = ma 2 2 2

Ix = Iz =

Next observe that the centroidal products of inertia are zero because of symmetry. Then 0 " a #" a # 1 I xy = I x′y′ + mx y = m & ma 2 '& − ' = − 2 2 2 2 ( ) ( ) 0 1 " a #" a # I yz = I y′z ′ + my z = m & − ' & ma 2 '=− 2 2 ( 2 )( 2 ) 0 " a #" a # 1 2 I zx = I z′x′ + mz x = m & '& ' = ma ( 2 )( 2 ) 2 Substituting into Equation (9.56) " 13 3 13 # K 3 − & + + ' ma 2 K 2 ( 12 2 12 ) 2

2

1 # " 1 # "1# " 13 3 # " 3 13 # " 13 13 # " + $& × ' + & × ' + & × ' − & − ' −&− ' −& ' *( 12 2 ) ( 2 12 ) ( 12 12 ) ( 2 2 ) ( 2 2 ) ( 2 ) 2

1 # " 3 #" 1 # " 13 3 13 # " 13 # " − $& × × '− & ' & − ' − & '& ' *( 12 2 12 ) ( 12 ) ( 2 2 ) ( 2 )( 2 )

2!

% (ma 2 ) 2 K %+

2

2

" 1 # 1 #" 1 # " 1 #! " 13 # " 2 3 − & '& − ' − 2& − '& − ' & ' % ( ma ) = 0 12 2 ( )( 2 2 ) ( 2 2 )( 2 2 ) ( ) +



Simplifying and letting K = ma 2ζ yields

ζ3−

11 2 565 95 ζ + ζ− =0 3 144 96

Solving yields

ζ 1 = 0.363383

ζ2 =

19 12

ζ 3 = 1.71995

The principal moments of inertia are then K1 = 0.363ma 2 K 2 = 1.583ma 2 K 3 = 1.720ma 2

(b)

To determine the direction cosines λx , λ y , λz of each principal axis, we use two of the equations of Equations (9.54) and (9.57). Thus, ( I x − K )λx − I xy λ y − I zx λz = 0

(9.54a)

− I zx λx − I yz λ y + ( I z − K )λz = 0

(9.54c)

λx2 + λ y2 + λz2 = 1

(9.57)

(Note: Since I xy = I yz , Equations (9.54a) and (9.54c) were chosen to simplify the “elimination” of λ y during the solution process.) Substituting for the moments and products of inertia in Equations (9.54a) and (9.54c) " # 1 " 13 2 # "1 # ma 2 ' λ y − & ma 2 ' λz = 0 & 12 ma − K ' λx − & − ( ) (2 ) ( 2 2 ) " # 1 "1 # " 13 # ma 2 ' λ y + & ma 2 − K ' λz = 0 − & ma 2 ' λx − & − 12 2 2 (2 ) ( ) ( )

or

and

" 13 & 12 − ζ (

1 1 # λ y − λz = 0 ' λx + 2 2 2 )

1 1 " 13 λy + & − ζ − λx + 2 2 2 ( 12

# ' λz = 0 )

(i)

(ii)

Observe that these equations will be identical, so that one will need to be replaced, if 13 1 −ζ = − 12 2

ζ =

or

19 12



Thus, a third independent equation will be needed when the direction cosines associated with K 2 are determined. Then for K1 and K 3 Eq. (i) through Eq. (ii):

13 1 " 13 " 1 #! #! $ − ζ − & − ' % λx + $ − − & − ζ ' % λz = 0 ( 2 )+ )+ *12 * 2 ( 12

λz = λx

or Substituting into Eq. (i):

" 13 & 12 − ζ (

1 1 # λ y − λx = 0 ' λx + 2 2 2 )

"

7 #

λ y = 2 2 & ζ − ' λx 12 ) (

or Substituting into Equation (9.57): "

!

7 #

2

λx2 + $ 2 2 & ζ − ' λx % + (λx ) 2 = 1 12 ) + ( * or

2 7# ! " $ 2 + 8 & ζ − ' % λx2 = 1 12 ) %+ ( *$

(iii)

K1 : Substituting the value of ζ 1 into Eq. (iii): 2 7 # ! 2 " $ 2 + 8 & 0.363383 − ' % ( λx )1 = 1 12 ( ) %+ *$

or and then

(λx )1 = (λz )1 = 0.647249 7 # " (λ y )1 = 2 2 & 0.363383 − ' (0.647249) 12 ( ) = −0.402662 (θ x )1 = (θ z )1 = 49.7°

(θ y )1 = 113.7°

K 3 : Substituting the value of ζ 3 into Eq. (iii): 2 7 # ! " $ 2 + 8 &1.71995 − ' % (λx )32 = 1 12 ) %+ ( *$

or and then

(λx )3 = (λz )3 = 0.284726 7 # " (λ y )3 = 2 2 &1.71995 − ' (0.284726) 12 ) ( = 0.915348 (θ x )3 = (θ z )3 = 73.5°

(θ y )3 = 23.7°

! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1660

!

PROBLEM 9.179* (Continued) ! K2: For this case, the set of equations to be solved consists of Equations (9.54a), (9.54b), and (9.57). Now − I xy λx + ( I y − K )λ y − I yz λz = 0

(9.54b)

Substituting for the moments and products of inertia. " # " # 1 1 "3 # ma 2 ' λx + & ma 2 − K ' λ y − & − ma 2 ' λz = 0 −& − (2 ) ( 2 2 ) ( 2 2 ) "3

1

#

1

λx + & − ξ ' λ y + λz = 0 2 2 2 2 (2 )

or

(iv)

Substituting the value of ξ 2 into Eqs. (i) and (iv): 1 1 " 13 19 # (λ y ) 2 − (λz ) 2 = 0 & 12 − 12 ' (λx ) 2 + 2 2 2 ( ) 1 1 " 3 19 # (λx ) 2 + & − ' (λ y ) 2 + (λ z ) 2 = 0 2 2 2 2 ( 2 12 )

or

− (λx ) 2 +

and

1 2

(λx ) 2 −

(λ y ) 2 − (λ z ) 2 = 0

2 (λ y ) 2 + (λz ) 2 = 0 6

Adding yields

(λ y ) 2 = 0

and then

(λ y ) 2 = −(λx ) 2

Substituting into Equation (9.57)

or

0 (λx ) 22 + (λ y )22 + (−λx ) 22 = 1 (λx ) 2 =

1 2

and (λz ) 2 = −

1 2

(θ x ) 2 = 45.0° (θ y )2 = 90.0° (θ z ) 2 = 135.0°

(c)

Principal axes 1 and 3 lie in the vertical plane of symmetry passing through Points O and B. Principal axis 2 lies in the xz plane.


PROBLEM 9.180 For the component described in Problem 9.165, determine (a) the principal mass moments of inertia at the origin, (b) the principal axes of inertia at the origin. Sketch the body and show the orientation of the principal axes of inertia relative to the x, y, and z axes.

SOLUTION (a)

From the solutions to Problems 9.141 and 9.165 we have I x = 13.98800 × 10−3 kg ⋅ m 2

Problem 9.141:

I y = 20.55783 × 10−3 kg ⋅ m 2 I z = 14.30368 × 10−3 kg ⋅ m 2 I yz = I zx = 0

Problem 9.165:

I xy = 0.39460 × 10−3 kg ⋅ m 2

Eq. (9.55) then becomes

Thus Substituting:

Ix − K

− I xy

0

− I xy

Iy − K

0

0

0

Iz − K

2 = 0 or ( I x − K )( I y − K )( I z − K ) − ( I z − K ) I xy =0

Iz − K = 0

2 I x I y − ( I x + I y ) K + K 2 − I xy =0

K1 = 14.30368 × 10−3 kg ⋅ m 2

or

K1 = 14.30 × 10−3 kg ⋅ m 2

−3 −3 −3 2 −3 2 and (13.98800 × 10 )(20.55783 × 10 ) − (13.98800 + 20.55783)(10 ) K + K − (0.39460 × 10 ) = 0

K 2 − (34.54583 × 10−3 ) K + 287.4072 × 10−6 = 0

or Solving yields

and

K 2 = 13.96438 × 10−3 kg ⋅ m 2

K 3 = 20.58145 × 10−3 kg ⋅ m 2

or

K 2 = 13.96 × 10−3 kg ⋅ m 2

or

K 3 = 20.6 × 10−3 kg ⋅ m 2



(b)

To determine the direction cosines λx , λ y , λz of each principal axis, use two of the equations of Eqs. (9.54) and Eq. (9.57). Then K1:

Begin with Eqs. (9.54a) and (9.54b) with I yz = I zx = 0. ( I x − K1 )(λx )1 − I xy (λ y )1 = 0 − I xy (λx )1 + ( I y − K1 )(λ y )1 = 0

Substituting: [(13.98800 − 14.30368) × 10 −3 ](λx )1 − (0.39460 × 10 −3 )(λ y )1 = 0 −(0.39460 × 10−3 )(λx )1 + [(20.55783 − 14.30368) × 10−3 ](λ y )1 = 0 (λx )1 = (λ y )1 = 0

Adding yields Then using Eq. (9.57)

0 0 (λx )12 + (λ y )12 + (λz )12 = 1 (λz )1 = 1

or

(θ x )1 = 90.0° (θ y )1 = 90.0° (θ z )1 = 0°

K2: Begin with Eqs. (9.54b) and (9.54c) with I yz = I zx = 0. − I xy (λx ) 2 + ( I y − K 2 )(λ y ) 2 = 0

(i)

( I z − K 2 )(λz ) 2 = 0 I z ≠ K 2 , (λz ) 2 = 0

Now Substituting into Eq. (i):

−(0.39460 × 10−3 )(λx ) z + [(20.55783 − 13.96438) × 10−3 ](λ y ) 2 = 0

or

(λ y ) 2 = 0.059847(λx ) 2

Using Eq. (9.57):

(λx ) 22

or

(λx ) 2 = 0.998214

and

(λ y ) 2 = 0.059740

2

+ [0.05984](λx ) 2 ] +

(λz ) 22

(θ x ) 2 = 3.4°

0 =1

(θ y ) 2 = 86.6°

(θ z ) 2 = 90.0°

K3: Begin with Eqs. (9.54b) and (9.54c) with I yz = I zx = 0. − I xy (λx )3 + ( I y − K 3 )(λ y )3 = 0

(ii)

( I z − K 3 )(λz )3 = 0



I z ≠ K 3 , (λz )3 = 0

Now Substituting into Eq. (ii):

−(0.39460 × 10−3 )(λx )3 + [(20.55783 − 20.58145) × 10−3 ](λ y )3 = 0 (λ y )3 = −16.70618(λx )3

or Using Eq. (9.57):

(λx )32

2

+ [−16.70618(λx )3 ]

+ (λz )32

0 =1

or

(λx )3 = −0.059751 (axes right-handed set , "_")

and

(λ y )3 = 0.998211 (θ x )3 = 93.4°

(θ y )3 = 3.43°

(θ z )3 = 90.0°

Note: Since the principal axes are orthogonal and (θ z ) 2 = (θ z )3 = 90°, it follows that |(λx ) 2 | = |(λ y )3 |

|(λ y ) 2 | = | (λz )3 |

The differences in the above values are due to round-off errors. (c)

Principal axis 1 coincides with the z axis, while principal axes 2 and 3 lie in the xy plane.


PROBLEM 9.181* For the component described in Problems 9.145 and 9.149, determine (a) the principal mass moments of inertia at the origin, (b) the principal axes of inertia at the origin. Sketch the body and show the orientation of the principal axes of inertia relative to the x, y, and z axes.

SOLUTION (a)

From the solutions to Problems 9.145 and 9.149, we have Problem 9.145:

I x = 26.4325 × 10−3 kg ⋅ m 2

Problem 9.149:

I xy = 2.5002 × 10−3 kg ⋅ m 2

I y = 31.1726 × 10−3 kg ⋅ m 2

I yz = 4.0627 × 10−3 kg ⋅ m 2

I z = 8.5773 × 10−3 kg ⋅ m 2

I zx = 8.8062 × 10−3 kg ⋅ m 2

Substituting into Eq. (9.56): K 3 − [(26.4325 + 31.1726 + 8.5773)(10−3 )]K 2 + [(26.4325)(31.1726) + (31.1726)(8.5773) + (8.5773)(26.4325) − (2.5002) 2 − (4.0627)2 − (8.8062)2 ](10−6 ) K − [(26.4325)(31.1726)(8.5773) − (26.4325)(4.0627)2 − (31.1726)(8.8062)2 − (8.5773)(2.5002) 2 − 2(2.5002)(4.0627)(8.8062)](10−9 ) = 0

or

K 3 − (66.1824 × 10−3 ) K 2 + (1217.76 × 10−6 ) K − (3981.23 × 10−9 ) = 0

Solving yields

!

K1 = 4.1443 × 10−3 kg ⋅ m 2

or

K1 = 4.14 × 10−3 kg ⋅ m 2

K 2 = 29.7840 × 10−3 kg ⋅ m 2

or

K 2 = 29.8 × 10−3 kg ⋅ m 2

K3 = 32.2541 × 10−3 kg ⋅ m 2 !

or

K3 = 32.3 × 10−3 kg ⋅ m 2


!


(b)

To determine the direction cosines λx , λ y , λz of each principal axis, use two of the equations of Eqs. (9.54) and Eq. (9.57). Then K 1: Begin with Eqs. (9.54a) and (9.54b).

( I x − K1 )(λx )1 − I xy (λ y )1 − I zx (λz )1 = 0 I xy (λx )1 + ( I y − K1 )(λ y )1 − I yz (λz )1 = 0

Substituting: [(26.4325 − 4.1443)(10−3 )](λx )1 − (2.5002 × 10−3 )(λ y )1 − (8.8062 × 10−3 )(λz )1 = 0 −(2.5002 × 10−3 )(λx )1 + [(31.1726 − 4.1443)(10 −3 )](λ y )1 − (4.0627 × 10−3 )(λz )1 = 0

Simplifying 8.9146(λx )1 − (λ y )1 − 3.5222(λz )1 = 0 −0.0925(λx )1 + (λ y )1 − 0.1503(λz )1 = 0

Adding and solving for (λz )1: (λz )1 = 2.4022(λx )1 (λ y )1 = [8.9146 − 3.5222(2.4022)](λx )1

and then

= 0.45357(λx )1

Now substitute into Eq. (9.57): (λx )12 + [0.45357(λx )1 ]2 + [2.4022(λx )1 ]2 = 1 (λx )1 = 0.37861

or and

(λ y )1 = 0.17173

(λz )1 = 0.90950 (θ x )1 = 67.8° (θ y )1 = 80.1° (θ z )1 = 24.6°

K 2 : Begin with Eqs. (9.54a) and (9.54b). ( I x − K 2 )(λx )2 − I xy (λ y ) 2 − I zx (λz ) 2 = 0 − I xy (λx ) 2 + ( I y − K 2 )(λ y ) 2 − I yz (λz ) 2 = 0

Substituting: [(26.4325 − 29.7840)(10−3 )](λx )2 − (2.5002 × 10−3 )(λ y ) 2 − (8.8062 × 10−3 )(λz )2 = 0 −(2.5002 × 10 −3 )(λx ) 2 + [(31.1726 − 29.7840)(10−3 )](λ y ) 2 − (4.0627 × 10−3 )( λz ) 2 = 0



Simplifying −1.3405(λx )2 − (λ y ) 2 − 3.5222(λz ) 2 = 0 −1.8005(λx )2 + (λ y ) 2 − 2.9258(λz ) 2 = 0

Adding and solving for (λz ) 2 : (λz )2 = −0.48713(λx ) 2 (λ y ) 2 = [−1.3405 − 3.5222(−0.48713)](λx )2

and then

= 0.37527(λx ) 2

Now substitute into Eq. (9.57): (λx ) 22 + [0.37527(λx ) 2 ]2 + [−0.48713(λx ) 2 ]2 = 1 (λx ) 2 = 0.85184

or (λ y ) 2 = 0.31967

and

(λz )2 = −0.41496 (θ x )1 = 31.6° (θ y ) 2 = 71.4° (θ z ) 2 = 114.5°

K 3 : Begin with Eqs. (9.54a) and (9.54b). ( I x − K3 )(λx )3 − I xy (λ y )3 − I zx (λz )3 = 0 − I xy (λx )3 + ( I y − K3 )(λ y )3 − I yz (λz )3 = 0

Substituting: [(26.4325 − 32.2541)(10−3 )](λx )3 − (2.5002 × 10−3 )(λ y )3 − (8.8062 × 10−3 )(λz )3 = 0 −(2.5002 × 10−3 )(λx )3 + [(31.1726 − 32.2541)(10−3 )](λ y )3 − (4.0627 × 10−3 )(λz )3 = 0

Simplifying −2.3285(λx )3 − (λ y )3 − 3.5222(λz )3 = 0

2.3118(λx )3 + (λ y )3 + 3.7565(λz )3 = 0

Adding and solving for (λz )3 : (λz )3 = 0.071276(λx )3

and then

(λ y )3 = [ −2.3285 − 3.5222(0.071276)](λx )3 = −2.5795(λx )3



Now substitute into Eq. (9.57): (λx )32 + [−2.5795(λx )3 ]2 + [0.071276(λx )3 ]2 = 1 (λx )3 = 0.36134

or and

(i)

(λ y )3 = 0.93208

(λz )3 = 0.025755 (θ x )3 = 68.8° (θ y )3 = 158.8° (θ z )3 = 88.5°

(c)

Note: Principal axis 3 has been labeled so that the principal axes form a right-handed set. To obtain the direction cosines corresponding to the labeled axis, the negative root of Eq. (i) must be chosen; that is, (λx )3 = −0.36134.

(θ x )3 = 111.2° (θ y )3 = 21.2° (θ z )3 = 91.5°

Then


PROBLEM 9.182* For the component described in Problem 9.167, determine (a) the principal mass moments of inertia at the origin, (b) the principal axes of inertia at the origin. Sketch the body and show the orientation of the principal axes of inertia relative to the x, y, and z axes.

SOLUTION (a)

From the solution of Problem 9.167, we have 1W 2 a 2 g W I y = a2 g 5W 2 Iz = a 6 g Ix =

1W 2 a 4 g 1W 2 I yz = a 8 g 3W 2 I zx = a 8 g

I xy =

Substituting into Eq. (9.56): 5 # " W #! "1 K 3 − $& + 1 + ' & a 2 ' % K 2 6 )( g )+ *( 2 2

2

"1# " 5 # " 5 #" 1 # " 1 # " 1 # " 3 # + $& ' (1) + (1) & ' + & '& ' − & ' − & ' − & ' ( 6 ) ( 6 )( 2 ) ( 4 ) ( 8 ) ( 8 ) *$( 2 )

2!

2

"W # % & a2 ' K ) +% ( g

3 2 2 2 " 1 # " 5 # " 1 #" 1 # " 3 # " 5 #" 1 # " 1 #" 1 #" 3 # ! " W 2 # − $& ' (1) & ' − & '& ' − (1) & ' − & '& ' − 2 & '& '& ' % & a ' = 0 ( 8 ) ( 6 )( 4 ) ( 4 )( 8 )( 8 ) %+ ( g $*( 2 ) ( 6 ) ( 2 )( 8 ) )

Simplifying and letting K =

W 2 a K yields g K 3 − 2.33333K 2 + 1.53125K − 0.192708 = 0

Solving yields K1 = 0.163917 K 2 = 1.05402 K3 = 1.11539

The principal moments of inertia are then

K1 = 0.1639

W 2 a g

K 2 = 1.054

W 2 a g

K 3 = 1.115

W 2 a g


!


(b)

To determine the direction cosines λx , λ y , λz of each principal axis, use two of the equations of Eqs. (9.54) and Eq. (9.57). Then K 1:

Begin with Eqs. (9.54a) and (9.54b). ( I x − K1 )(λx )1 − I xy (λ y )1 − I zx (λz )1 = 0 − I xy (λx )1 + ( I y − K 2 )(λ y )1 − I yz (λz )1 = 0

Substituting "1W 2# "3W 2# "1 # " W 2 #! a ' (λ y )1 − & a ' (λz )1 = 0 $& − 0.163917 ' & a ' % (λx )1 − & )( g )+ (4 g ) (8 g ) *( 2 "1W 2# " W #! "1W 2# −& a ' (λx )1 + $(1 − 0.163917) & a 2 ' % (λ y )1 − & a ' (λz )1 = 0 (4 g ) ( g )+ (8 g ) *


Adding and solving for (λz )1: (λz )1 = 0.633715(λx )1 (λ y )1 = [1.34433 − 1.5(0.633715)](λx )1

and then

= 0.393758(λx )1

Now substitute into Eq. (9.57): (λx )12 + [0.393758(λx )1 ]2 + (0.633715(λx )1 ]2 = 1 (λx )1 = 0.801504 !

or and

(λ y )1 = 0.315599

(λz )1 = 0.507925 (θ x )1 = 36.7° (θ y )1 = 71.6° (θ z )1 = 59.5°

K 2:

Begin with Eqs. (9.54a) and (9.54b): ( I x − k2 )(λx ) 2 − I xy (λ y ) 2 − I zx (λz ) 2 = 0 − I xy (λx ) 2 + ( I y − k2 )(λ y )2 − I yz (λz ) 2 = 0



Substituting "1W 2# "3W 2# "1 # " W 2 #! a ' (λ y ) 2 − & a ' (λz ) 2 = 0 $& − 1.05402 ' & a ' % (λx ) 2 − & )( g )+ (4 g ) (8 g ) *( 2 "1W 2# " W #! "1W 2# −& a ' (λx )2 + $(1 − 1.05402) & a 2 ' % (λ y ) 2 − & a ' (λz ) 2 = 0 (4 g ) ( g )+ (8 g ) *

Simplifying −2.21608(λx ) 2 − (λ y ) 2 − 1.5(λz ) 2 = 0 4.62792(λx ) 2 + (λ y ) 2 + 2.31396(λz )2 = 0

Adding and solving for (λz ) 2 (λz )2 = −2.96309(λx ) 2 (λ y ) 2 = [−2.21608 − 1.5(−2.96309)](λx ) 2

and then

= 2.22856(λx ) 2

Now substitute into Eq. (9.57): (λx ) 22 + [2.22856(λx ) 2 ]2 + [−2.96309(λx ) 2 ]2 = 1 (λx ) 2 = 0.260410

or and

(λ y ) 2 = 0.580339

(λz ) 2 = −0.771618 (θ x ) 2 = 74.9° (θ y ) 2 = 54.5° (θ z )2 = 140.5°

! K 3 : Begin with Eqs. (9.54a) and (9.54b):

( I x − K3 )(λx )3 − I xy (λ y )3 − I zx (λz )3 = 0 − I xy (λx )3 + ( I y − K3 )(λ y )3 − I yz (λz )3 = 0

Substituting "1W 2# "3W 2# "1 # " W 2 #! a ' (λ y )3 − & a ' (λz )3 = 0 $& − 1.11539 ' & a ' % (λx )3 − & )( g )+ (4 g ) (8 g ) *( 2 "1W 2# " W #! "1W 2# −& a ' (λx )3 + $(1 − 1.11539) & a 2 ' % (λ y )3 − & a ' (λ z )3 = 0 (4 g ) ( g )+ (8 g ) *

Simplifying −2.46156(λx )3 − (λ y )3 − 1.5(λz )3 = 0 2.16657(λx )3 + (λ y )3 + 1.08328(λz )3 = 0 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1671


Adding and solving for (λz )3 (λz )3 = −0.707885(λx )3

and then

(λ y )3 = [ −2.46156 − 1.5(−0.707885)](λx )3 = −1.39973(λx )3

Now substitute into Eq. (9.57): (λx )32 + [ −1.39973(λx )3 ]2 + [−0.707885(λx )3 ]2 = 1

or

(λx )3 = 0.537577

and

(λ y )3 = −0.752463

(i)

(λz )3 = −0.380543 (θ x )3 = 57.5° (θ y )3 = 138.8° (θ z )3 = 112.4°

(c)

Note: Principal axis 3 has been labeled so that the principal axes form a right-handed set. To obtain the direction cosines corresponding to the labeled axis, the negative root of Eq. (i) must be chosen; that is, (λx )3 = −0.537577

(θ x )3 = 122.5° (θ y )3 = 41.2° (θ z )3 = 67.6°

Then


!

PROBLEM 9.183* For the component described in Problem 9.168, determine (a) the principal mass moments of inertia at the origin, (b) the principal axes of inertia at the origin. Sketch the body and show the orientation of the principal axes of inertia relative to the x, y, and z axes.

SOLUTION (a)

From the solution to Problem 9.168, we have I x = 18.91335

γt

a4

g γt I y = 7.68953 a 4 g γt I z = 12.00922 a 4 g

I yz I zx

γt

a4 g γt = 7.14159 a 4 g γt = 0.66667 a 4 g

I xy = 1.33333

Substituting into Eq. (9.56): " γ t #! K 3 − $(18.91335 + 7.68953 + 12.00922) & a 4 ' % K 2 (g )+ * + [(18.91335)(7.68953) + (7.68953)(12.00922) + (12.00922)(18.91335) 2

" γt # − (1.33333) − (7.14159) + (0.66667) ] & a 4 ' K (g ) 2

2

2

− [(18.91335)(7.68953)(12.00922) − (18.91335)(7.14159) 2 − (7.68953)(0.66667)2 − (12.00922)(1.33333) 2 3

"γt # − 2(1.33333)(7.14159)(0.66667)] & a 4 ' = 0 (g )

Simplifying and letting K=

γt g

a4 K

yields

K 3 − 38.61210 K 2 + 411.69009 K − 744.47027 = 0



Solving yields K1 = 2.25890 K 2 = 17.27274 K3 = 19.08046 K1 = 2.26

The principal moments of inertia are then

K 2 = 17.27 K 3 = 19.08

(b)

γt g

γt g

γt g

a4 a4 a4

To determine the direction cosines λx , λ y , λz of each principal axis, use two of the equations of Eqs. (9.54) and Eq. (9.57). Then K 1 : Begin with Eqs. (9.54a) and (9.54b): ( I x − K1 )(λx )1 − I xy (λ y )1 − I zx (λz )1 = 0 − I xy (λx )1 + ( I y − K 2 )(λ y )1 − I yz (λz )1 = 0

Substituting " γ t 4 #! " γt 4 # γt 4 # " $(18.91335 − 2.25890) & a ' % (λx )1 − &1.33333 a ' (λ y )1 − & 0.66667 a ' (λz )1 = 0 a ) g ( ( g )+ ( ) * " " γ t #! " γt # γt # − &1.33333 a 4 ' (λx )1 + $(7.68953 − 2.25890) & a 4 ' % (λ y )1 − & 7.14159 a 4 ' (λz )1 = 0 g ) g ) ( ( g )+ ( *


Adding and solving for (λz )1 (λz )1 = 6.74653(λx )1

and then

(λ y )1 = [12.49087 − (0.5)(6.74653)](λx )1 = 9.11761(λx )1

Now substitute into Eq. (9.57):

(λx )12 + [9.11761(λx )1 ]2 + [6.74653(λx )1 ]2 = 1

or

(λx )1 = 0.087825

and

(λ y )1 = 0.80075

(λz )1 = 0.59251 (θ x )1 = 85.0° (θ y )1 = 36.8° (θ z )1 = 53.7°



K 2 : Begin with Eqs. (9.54a) and (9.54b): ( I x − K 2 )(λx )2 − I xy (λ y ) 2 − I zx (λz ) 2 = 0 − I xy (λx ) 2 + ( I y − K 2 )(λ y ) 2 − I yz (λz ) 2 = 0

Substituting "γt # " " γt # γt # [(18.91335 − 17.27274) & a 4 ' (λx )2 − & 1.33333 a 4 ' (λ y ) 2 − & 0.66667 a 4 ' (λz ) 2 = 0 g g ) (g ) ( ) ( " " γ t #! " γt # γt # − &1.33333 a 4 ' (λx ) 2 + $(7.68953 − 17.27274) & a 4 ' % (λ y )2 − & 7.14159 a 4 ' (λz ) 2 = 0 g ) g ) ( (g )+ ( *

Simplifying 1.23046(λx )2 − (λ y )2 − 0.5(λz ) 2 = 0 0.13913(λx )2 + (λ y ) 2 + 0.74522(λz ) 2 = 0

Adding and solving for (λz ) 2 (λz )2 = −5.58515(λx )2

and then

(λ y ) 2 = [1.23046 − (0.5)(−5.58515)](λx )2 = 4.02304 (λx ) 2

Now substitute into Eq. (9.57): (λx ) 22 + [4.02304(λx ) 2 ]2 + [−5.58515(λx ) 2 ]2 = 1

or

(λx ) 2 = 0.14377

and

(λ y ) 2 = 0.57839

(λz )2 = −0.80298 (θ x ) 2 = 81.7° (θ y ) 2 = 54.7° (θ z ) 2 = 143.4°

K3 :

Begin with Eqs. (9.54a) and (9.54b): ( I x − K3 )(λx )3 − I xy (λ y )3 − I zx (λz )3 = 0 − I xy (λx )3 + ( I y − K 3 )(λ y )3 − I yz (λ y )3 = 0

Substituting "γt # " " γt # γt # [(18.91335 − 19.08046) & a 4 ' (λx )3 − & 1.33333 a 4 ' (λ y )3 − & 0.66667 a 4 ' (λz )3 = 0 g ) g ) (g ) ( ( " " γ t #! " γt # γt # − &1.33333 a 4 ' (λx )3 + $(7.68953 − 19.08046) & a 4 ' % (λ y )3 − & 7.14159 a 4 ' (λz )3 = 0 g g g ) ( ) ( )+ ( * PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1675


Simplifying −0.12533(λx )3 − (λ y )3 − 0.5(λz )3 = 0 0.11705(λx )3 + (λ y )3 + 0.62695(λz )3 = 0

Adding and solving for (λz )3 (λz )3 = 0.06522(λx )3

and then

(λ y )3 = [ −0.12533 − (0.5)(0.06522)](λx )3 = −0.15794(λx )3

Now substitute into Eq. (9.57): (λx )32 + [−0.15794(λx )3 ]2 + [0.06522(λx ]3 ]2 = 1

or

(λx )3 = 0.98571

and

(λ y )3 = −0.15568

(i)

(λz )3 = 0.06429 (θ x )3 = 9.7° (θ y )3 = 99.0° (θ z )3 = 86.3°

(c)

Note: Principal axis 3 has been labeled so that the principal axes form a right-handed set. To obtain the direction cosines corresponding to the labeled axis, the negative root of Eq. (i) must be chosen; that is, (λx )3 = −0.98571.

(θ x )3 = 170.3° (θ y )3 = 81.0° (θ z )3 = 93.7°

Then


PROBLEM 9.184* For the component described in Problems 9.148 and 9.170, determine (a) the principal mass moments of inertia at the origin, (b) the principal axes of inertia at the origin. Sketch the body and show the orientation of the principal axes of inertia relative to the x, y, and z axes.

SOLUTION (a)

From the solutions to Problems 9.148 and 9.170. We have I x = 0.32258 kg ⋅ m 2

I y = I z = 0.41933 kg ⋅ m 2

I xy = I zx = 0.096768 kg ⋅ m 2

I yz = 0

Substituting into Eq. (9.56) and using Iy = Iz

I xy = I zx

I yz = 0

K 3 − [0.32258 + 2(0.41933)]K 2 + [2(0.32258)(0.41933) + (0.41933)2 − 2(0.096768)2 ]K − [(0.32258)(0.41933) 2 − 2(0.41933)(0.096768)2 ] = 0

Simplifying K 3 − 1.16124 K 2 + 0.42764 K − 0.048869 = 0

Solving yields

(b)

K1 = 0.22583 kg ⋅ m 2

or

K1 = 0.226 kg ⋅ m 2

K 2 = 0.41920 kg ⋅ m 2

or

K 2 = 0.419 kg ⋅ m 2

K3 = 0.51621 kg ⋅ m 2

or

K3 = 0.516 kg ⋅ m 2

To determine the direction cosines λx , λ y , λz of each principal axis, use two of the equations of Eqs. (9.54) and (9.57). Then K1 : Begin with Eqs. (9.54b) and (9.54c): 0 − I xy (λx )1 + ( I y − K1 )(λ y )1 − I yz (λz )3 = 0 0 − I zx (λx )1 − I yz (λ y )1 + ( I z − K1 )(λz )3 = 0



Substituting −(0.096768)(λx )1 + (0.41933 − 0.22583)(λ y )1 = 0 −(0.096768)(λx )1 + (0.41933 − 0.22583)(λz )1 = 0

Simplifying yields (λ y )1 = (λz )1 = 0.50009(λx )1

Now substitute into Eq. (9.54): (λx ) 2 + 2[0.50009(λx )1 ]2 = 1

or

(λx )1 = 0.81645

and

(λ y )1 = (λz )1 = 0.40830 (θ x )1 = 35.3° (θ y )1 = (θ z )1 = 65.9°

K 2 : Begin with Eqs. (9.54a) and (9.54b) ( I x − K 2 )(λx ) 2 − I xy (λ y ) 2 − I zx (λz ) 2 = 0 0 − I xy (λx ) 2 + ( I y − K 2 )(λ y ) 2 − I yz (λz ) 2 = 0

Substituting (0.32258 − 0.41920)(λx ) 2 − (0.096768)(λ y ) 2 − (0.096768)(λz ) 2 = 0

(i)

−(0.96768)(λx ) 2 + (0.41933 − 0.41920)(λ y ) 2 = 0

(ii)

Eq. (ii) , (λx ) 2 = 0 and then Eq. (i) , (λz )2 = −(λ y )2 Now substitute into Eq. (9.57):

0 (λx ) 22 + (λ y ) 22 + [−(λ y ) 2 ]2 = 1 1

or

(λ y ) 2 =

and

(λz ) 2 = −

2 1 2

(θ x ) 2 = 90° (θ y ) 2 = 45° (θ z )2 = 135°



K3 :

Begin with Eqs. (9.54b) and (9.54c):

0 I xy (λx )3 + ( I y − K3 )(λ y )3 + I yz (λz )3 = 0

0 − I zx (λx )3 − I yz (λ y )3 + ( I z − K3 )(λz )3 = 0 Substituting −(0.096768)(λx )3 + (0.41933 − 0.51621)(λ y )3 = 0 −(0.096768)(λx )3 + (0.41933 − 0.51621)(λz )3 = 0

Simplifying yields (λ y )3 = (λz )3 = − (λx )3

Now substitute into Eq. (9.57): (λx )32 + 2[−(λx )3 ]2 = 1

(i)

1

or

(λx )3 =

and

(λ y )3 = (λz ) = −

3 1 3 (θ x )3 = 54.7° (θ y )3 = (θ z )3 = 125.3°

!

(c)

Note: Principal axis 3 has been labeled so that the principal axes form a right-handed set. To obtain the direction cosines corresponding to the labeled axis, the negative root of Eq. (i) must be chosen; That is,

(λx )3 = −

1 3 (θ x )3 = 125.3°

Then

(θ y )3 = (θ z )3 = 54.7°


!

PROBLEM 9.185 Determine by direct integration the moments of inertia of the shaded area with respect to the x and y axes.

SOLUTION x1 = a,

At

Then

Now

Then

y1 = y2 = b

y1 :

b = ka 2

y2 :

b = ma or m = y1 =

b 2 x a2

y2 =

b x a

or k =

b a2 b a

1 3 y2 − y13 dx 3 b3 # 1 " b3 = && 3 x3 − 6 x6 '' dx 3( a a )

(

dI x =

-

I x = dI x =

)

-

a 0

b3 " 1 3 1 6 # x − 6 x ' dx 3 &( a3 a ) a

3

=

Also

Then

1 4 1 b ! x − 6 x7 % 3 $* 4a 3 7a +0

or

Ix =

1 3 ab 28

or

Iy =

1 3 ab 20

b "b # ! dI y = x 2 dA = x 2 [( y2 − y1 )dx = x 2 $& x − 2 x 2 ' dx % a a ) + *(

-

I y = dI y =

-

a

0

1 "1 # b & x3 − 2 x 4 ' dx a (a ) a

1 1 ! = b $ x 4 − 2 x5 % 5a * 4a +0


PROBLEM 9.186 Determine the moments of inertia and the radii of gyration of the shaded area shown with respect to the x and y axes.

SOLUTION 1/2 !

We have

Then

" x# dA = ydx = b $1 − & ' $* ( a )

-

A = dA =

-

a

0

% dx %+ 1/ 2 !

" x# b $1 − & ' $* ( a )

% dx %+

a

2 3/ 2 ! 1 = b $x − x % = ab 3 a * +0 3 3

Now

1/2 1 1 .0 " x # ! /0 dI x = y 3 dx = 1b $1 − & ' % 2 dx 3 3 0 $* ( a ) %+ 0 3 4 1/ 2

b3 " x# = $1 − 3 & ' 3 $* (a)

Then

-

I x = dI x =

-

a

0

"x# " x# + 3& ' − & ' (a) (a) 1/ 2

b3 " x# $1 − 3 & ' 3 $* (a)

3/2 !

% dx %+

" x# " x# + 3& ' − & ' (a) (a)

3/ 2 !

% dx %+

a

! 2 3/ 2 3 2 2 b3 = $x − x + x + 3/ 2 x5/2 % 3 * 2a 5a a +0

or

and

k x2 =

Ix = A

1 3 ab 30

1 ab3 30 1 ab 3

or

Also

Ix =

kx =

b

10

1/2 .0 " x # ! /0 dI y = x 2 dA = x 2 ( ydx) = x 2 1b $1 − & ' % dx 2 03 *$ ( a ) +% 04



Then

-

I y = dI y =

-

a 0

a

" 1 5/2 # 1 2 7/ 2 ! b & x2 − x ' dx = b $ x3 − x % 3 7 a a ( ) * +0

or

and

k y2

=

Iy A

=

Iy =

1 3 ab 21

1 3 ab 21 1 ab 3

or

ky =

a 7



SOLUTION At x = a,

y1 = y2 = b :

y1 : b = ka 2

or k =

y2 : b = 2b − ca 2

Then

Now

b a2

or c =

b 2 x a2 " x2 y2 = b && 2 − 2 a (

b a2

y1 =

# '' )

dA = ( y2 − y1 ) dx ! " x2 # b = $b && 2 − 2 '' − 2 x 2 % dx a ) a $* ( %+ 2b = 2 (a 2 − x 2 ) dx a

Then

-

A = dA =

-

a

0

2b 2 (a − x 2 )dx a2 a

2b 2 1 ! a x − x3 % 2 $ 3 +0 a * 4 = ab 3 =

Now

2b ! dI y = x 2 dA = x 2 $ 2 (a 2 − x 2 ) dx % a * +



Then

-

I y = dI y =

-

a 0

2b 2 2 x (a − x 2 ) dx a2 a

=

2b 1 2 3 1 5 ! a x − x % 5 +0 a 2 $* 3

or

and

k y2 =

Iy A

=

4 3 ab 15 4 ab 3

=

Iy =

4 3 ab 15

1 2 a 5

or

ky =

a 5


PROBLEM 9.188 Determine the moments of inertia of the shaded area shown with respect to the x and y axes.

SOLUTION We have where

I x = ( I x )1 + ( I x )2 − ( I x )3 ( I x )1 =

1 4 (2a )(2a)3 = a 4 12 3

π

a4

Now

( I AA ) 2 = ( I BB )3 =

and

( I AA ) 2 = ( I x2 ) 2 + Ad 2

or

( I x2 ) 2 = ( I x3 )3 =

Then

8

2

8 " π #" 4a # " π a 4 − & a 2 '& ' =& − 8 ( 2 )( 3π ) ( 8 9π

π

8 "π ( I x )2 = ( I x2 )2 + Ad 22 = & − ( 8 9π " 4 5π # 4 =& + 'a (3 8 )

# 4 ' a1 )

4a # # 4 " π 2 #" ' a + & a '& a + ' 3π ) ) ( 2 )(

8 # 4 " π 2 #" 4a # "π ( I x )3 = ( I x3 )3 + Ad32 = & − ' a + & a '& a − ' 3π ) ( 8 9π ) ( 2 )( " 4 5π # 4 = &− + 'a ( 3 8 ) 4 4 " 4 5π # 4 ! " 4 5π # 4 ! a + $& + ' a % − $& − + 'a % 3 * ( 3 8 ) + *( 3 8 ) +

Finally

Ix =

Also

I y = ( I y )1 + ( I y ) 2 − ( I y )3

Where

( I y )1 =

2

2

I x = 4a 4

1 16 (2a)(2a)3 + (2a) 2 (a) 2 = a 4 12 3

! π "π # ( I y ) 2 = ( I y )3 $ = a 4 + & a 2 ' ( a ) 2 % 8 2 ( ) * + Iy =

16 4 a 3



SOLUTION First locate centroid C of the area.

A, mm 2

1

2 Σ

π 2

−

48

(54)(36) = 3053.6

π 2

π 24

(18) 2 = −508.9

π

y , mm

yA, mm3

= 15.2789

46,656

= 7.6394

−3888 42,768

2544.7

Y Σ A = Σ y A: Y (2544.7 mm 2 ) = 42, 768 mm3

Then

Y = 16.8067 mm "

or J O = ( J O )1 − ( J O ) 2

(a)

π

π

(54 mm)(36 mm)[(54 mm)2 + (36 mm)2 ] − (18 mm) 4 8 4 6 6 4 = (3.2155 × 10 − 0.0824 × 10 ) mm =

= 3.1331 × 106 mm 4

or

J O = 3.13 × 106 mm 4

J O = J C + A(Y )2

(b) or

J C = 3.1331 × 106 mm 4 − (2544.7 mm 2 )(16.8067 mm) 2

or

J C = 2.41 × 106 mm 4


PROBLEM 9.190 To form an unsymmetrical girder, two L76 × 76 × 6.4-mm angles and two L152 × 102 × 12.7-mm angles are welded to a 16-mm steel plate as shown. Determine the moments of inertia of the combined section with respect to its centroidal x and y axes.

SOLUTION A = 929 mm 2

L76 × 76 × 6.4:

I x × I y = 0.512 × 106 mm 4 A = 3060 mm 2

L152 × 102 × 12.7:

I x = 2.59 × 106 mm 4 I y = 7.20 × 106 mm 4

First locate centroid C of the section:

Then or

A, mm 2

y

yA, mm3

1

2(3060) = 6120

24.9

152,388

2

(16)(540) = 8640

270

2,332,800

3

2(929) = 1858

518.8

963,930

Σ

16,618

3,449,118

YΣA = Σy A Y (16, 618 mm 2 ) = 3, 449,118

or

Y = 207.553 mm

Now

I x = 2( I x )1 + ( I x ) 2 + 2( I x )3



where

( I x )1 = I x + Ad 2 = 2.59 × 106 mm 4 + (3060 mm 2 )[(207.553 − 24.9) mm]2 = 104.678 × 106 mm 4 1 ( I x )2 = I x + Ad 2 = (16 mm)(540 mm)3 + (16 mm)(540 mm)[(270 − 207.553) mm]2 12 = 243.645 × 106 mm 4 ( I x )3 = I x + Ad 2 = 0.512 × 106 mm 4 + (929 mm 2 [(518.8 − 207.553) mm]2 = 90.5086 × 106 mm 4

Then

I x = [2(104.678) + 243.645 + 2(90.5086)] × 106 mm 4

or Also where

I x = 634 × 106 mm 4

I y = 2( I y )1 + ( I y ) 2 + 2( I y )3 ( I y )1 = I y + Ad 2 = 7.20 × 106 mm 4 + (3060 mm 2 )[(8 + 50.3) mm]2 = 17.6006 × 106 mm 4 1 ( I y ) 2 = (540 mm)(16 mm)3 12 = 0.1843 × 106 mm 4 ( I y )3 = I y + Ad 2 = 0.512 × 106 mm 4 + (929 mm 2 )[(8 + 21.2) mm]2 = 1.3041 × 106 mm 4

Then

I y = [2(17.6006) + 0.1843 + 2(1.3041)] × 106 mm 4

or

I y = 38.0 × 106 mm 4


PROBLEM 9.191 Using the parallel-axis theorem, determine the product of inertia of the L5 × 3 × 12 -in. angle cross section shown with respect to the centroidal x and y axes.

SOLUTION We have I xy = ( I xy )1 + ( I xy ) 2

For each rectangle: I xy = I x′y′ + x yA I x′y′ = 0 (symmetry)

and

I xy = Σ x yA

Thus

A, in.2

1

3×

2

4.5 ×

1 = 1.5 2 1 = 2.25 2

x , in.

y , in.

x yA, in.4

0.754

−1.49

−1.68519

− 0.496

1.01

−1.12716 − 2.81235

Σ

I xy = − 2.81 in.4


PROBLEM 9.192 For the L5 × 3 × 12 -in. angle cross section shown, use Mohr’s circle to determine (a) the moments of inertia and the product of inertia with respect to new centroidal axes obtained by rotating the x and y axes 30° clockwise, (b) the orientation of the principal axes through the centroid and the corresponding values of the moments of inertia.

SOLUTION From Figure 9.13a: I x = 9.43 in.4

I y = 2.55 in.4

From the solution to Problem 9.191 I xy = −2.81235 in.4

The Mohr’s circle is defined by the diameter XY, where X (9.43 − 2.81235),

Now

I ave =

Y (2.55, 2.81235)

1 1 ( I x + I y ) = (9.43 + 2.55) = 5.99 in.4 2 2 2

and

2

1 1 ! ! R = $ ( I x − I y ) % + I xy2 = $ (9.43 − 2.55) % + (−2.81235) 2 2 2 * + * + = 4.4433 in.4


2 I xy Ix − Iy

2( −2.81235) 9.43 − 2.55 = 0.81754 =−

or and Then

2θ m = 39.267°

θ m = 19.6335° α = 180° − (39.267° + 60°) = 80.733°



(a)

We have

I x′ = I ave − R cos α = 5.99 − 4.4433cos80.733°

or

I x′ = 5.27 in.4

or

I y ′ = 6.71 in.4

or

I x′y′ = − 4.39 in.4

I y′ = I ave + R cos α = 5.99 + 4.4433cos80.733°

I x′y′ = − R sin α = −4.4433sin 80.733°

First observe that the principal axes are obtained by rotating the xy axes through

(b)

19.63° counterclockwise about C. Now

I max, min = I ave ± R = 5.99 ± 4.4433

or

I max = 10.43 in.4 I min = 1.547 in.4



PROBLEM 9.193 A piece of thin, uniform sheet metal is cut to form the machine component shown. Denoting the mass of the component by m, determine its mass moment of inertia with respect to (a) the x axis, (b) the y axis.

SOLUTION First note

mass = m = ρV = ρ tA 1 " a #! = ρ t $(2a )( a) − (2a) & ' % 2 ( 2 )+ * 3 = ρ ta 2 2 I mass = ρ tI area

Also

=

(a)

Now

Then

2m I area 3a 2

I x,area = ( I x )1,area − 2( I x ) 2,area =

! 1 1 "a# ( a)(2a)3 − 2 $ & ' (a)3 % 12 *12 ( 2 ) +

=

7 4 a 12

I x, mass =

2m 7 4 × a 3a 2 12

or (b)

We have

Ix =

7 ma 2 18

I z ,area = ( I z )1,area − 2( I z )2,area 3 1 1 "a# ! = (2a)(a )3 − 2 $ (a) & ' % 3 ( 2 ) +% *$12

=

Then

31 4 a 48

2m 31 4 × a 3a 2 48 31 2 = ma 72

I z ,mass =


!


Finally,

I y ,mass = I x ,mass + I z ,mass 7 31 2 ma 2 + ma 18 72 59 2 = ma 72 =

or

I y = 0.819ma 2


PROBLEM 9.194 A piece of thin, uniform sheet metal is cut to form the machine component shown. Denoting the mass of the component by m, determine its mass moment of inertia with respect to (a) the axis AA′, (b) the axis BB′, where the AA′ and BB′ axes are parallel to the x axis and lie in a plane parallel to and at a distance a above the xz plane.

SOLUTION First note that the x axis is a centroidal axis so that I = I x ,mass + md 2

and that from the solution to Problem 9.115, I x, mass =

(a)

(b)

We have

We have

I AA′, mass =

I BB′, mass =

7 ma 2 18 7 ma 2 + m(a) 2 18

or

I AA′ = 1.389ma 2

or

I BB′ = 2.39ma 2

7 ma 2 + m(a 2) 2 18



SOLUTION First compute the mass of each component. We have m = ρSTV = ρST t A

Then

m1 = (7850 kg/m3 )(0.002 m)(0.76 × 0.48) m 2 = 5.72736 kg "1 # m2 = (7850 kg/m3 )(0.002 m) & × 0.76 × 0.48 ' m 2 2 ( ) = 2.86368 kg "π # m3 = (7850 kg/m3 )(0.002 m) & × 0.482 ' m 2 (4 ) = 2.84101 kg

Using Figure 9.28 for component 1 and the equations derived above for components 2 and 3, we have I x = ( I x )1 + ( I x )2 + ( I x )3 2 1 " 0.48 # ! m' % = $ (5.72736 kg)(0.48 m) 2 + (5.72736 kg) & ( 2 ) %+ $*12

1 " 0.48 # m' + $ (2.86368 kg)(0.48 m) 2 + (2.86368 kg) & ( 3 ) $*18

2!

1 ! % + $ (2.84101 kg)(0.48 m) 2 % + %+ * 2

= [(0.109965 + 0.329896) + (0.036655 + 0.073310) + (0.327284)] kg ⋅ m 2 = (0.439861 + 0.109965 + 0.327284) kg ⋅ m 2

or

I x = 0.877 kg ⋅ m 2

! ! ! ! ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1695

!


I y = ( I y )1 + ( I y )2 + ( I y )3 2 2 0. 1 " 0.76 # " 0.48 # ! 2 /0 = 1 (5.72736 kg)(0.762 + 0.482 ) m 2 + (5.72736 kg) $& + ' & ' %m 2 12 *$( 2 ) ( 2 ) +% 30 40 2 1 1 " 0.76 # ! ! m ' % + $ (2.84101 kg)(0.48 m) 2 % + $ (2.86368 kg)(0.76 m) 2 + (2.86368 kg) & 18 3 4 ( ) %+ * + $*

= [(0.385642 + 1.156927) + (0.091892 + 0.183785) + (0.163642)] kg ⋅ m2 = (1.542590 + 0.275677 + 0.163642) kg ⋅ m 2

or

I y = 1.982 kg ⋅ m 2

I z = ( I z )1 + ( I z ) 2 + ( I z )3 2 1 " 0.76 # ! m' % = $ (5.72736 kg)(0.76 m)2 + (5.72736 kg) & ( 2 ) %+ *$12 2 2 0. 1 " 0.76 # " 0.48 # ! 2 0/ + 1 (2.86368 kg)(0.762 + 0.482 ) m 2 + (2.86368 kg) $& +& %m 2 ' ' 18 *$( 3 ) ( 3 ) %+ 30 40

1 ! + $ (2.84101 kg)(0.48 m) 2 % 4 * + I z = [(0.275677 + 0.827031) + (0.128548 + 0.257095) + (0.163642)] kg ⋅ m 2 = (1.102708 + 0.385643 + 0.163642) kg ⋅ m 2

or

I z = 1.652 kg ⋅ m 2


PROBLEM 9.196 Determine the mass moments of inertia and the radii of gyration of the steel machine element shown with respect to the x and y axes. (The density of steel is 7850 kg/m3.)

SOLUTION

First compute the mass of each component. We have m = ρSTV

Then

m1 = (7850 kg/m3 )(0.24 × 0.04 × 0.14) m3 = 10.5504 kg

π ! m2 = m3 = (7850 kg/m3 ) $ (0.07)2 × 0.04 % m3 = 2.41683 kg *2 + m4 = m5 = (7850 kg/m3 )[π (0.044) 2 × (0.04)] m3 = 1.90979 kg

Using Figure 9.28 for components 1, 4, and 5 and the equations derived above (before the solution to Problem 9.144) for a semicylinder, we have I x = ( I x )1 + ( I x )2 + ( I x )3 + ( I x )4 − ( I x )5

where ( I x ) 2 = ( I x )3

1 ! .1 / = $ (10.5504 kg)(0.042 + 0.142 ) m 2 % + 2 1 (2.41683 kg) *3(0.07 m)2 + (0.04 m)2 !+ 2 12 12 * + 3 4 .1 / + 1 (1.90979 kg) *3(0.044 m) 2 + (0.04 m) 2 !+ + (1.90979 kg)(0.04 m) 2 2 4 312 .1 / − 1 (1.90979 kg) *3(0.044 m)2 + (0.04 m) 2 +! 2 12 3 4 = [(0.0186390) + 2(0.0032829) + (0.0011790 + 0.0030557) − (0.0011790)] kg ⋅ m 2 = 0.0282605 kg ⋅ m 2

or

I x = 28.3 × 10−3 kg ⋅ m 2



I y = ( I y )1 + ( I y )2 + ( I y )3 + ( I y ) 4 − ( I y )5 ( I y ) 2 = ( I y )3

where Then

(I y )4 = |( I y )5 |

1 ! I y = $ (10.5504 kg)(0.242 + 0.142 ) m 2 % *12 +

" 1 16 + 2 $(2.41683 kg) & − 2 ( 2 9π *

2 4 × 0.07 # 2 ! # " 2 (0.07 m ) (2.41683 kg) 0.12 m % + + ' & 3π ') ) ( %+

= [(0.0678742) + 2(0.0037881 + 0.0541678)] kg ⋅ m 2 = 0.1837860 kg ⋅ m 2

or Also

I y = 183.8 × 10−3 kg ⋅ m 2

m = m1 + m2 + m3 + m4 − m5 where m2 = m3 , m4 = m5 = (10.5504 + 2 × 2.41683) kg = 15.38406 kg

Then

and

k x2 =

k y2 =

I x 0.0282605 kg ⋅ m 2 = m 15.38406 kg

Iy m

=

or

k x = 42.9 mm

or

k y = 109.3 mm

0.1837860 kg ⋅ m 2 15.38406 kg


CHAPTER 10

PROBLEM 10.1 Determine the vertical force P that must be applied at C to maintain the equilibrium of the linkage.

SOLUTION Assume δ y A

δ yC = 2δ yA δ yD = δ yA 1 2

1 2

δ yE = δ yD = δ yA 1 2

δ yG = δ yE = δ yA Also

δθ =

δ yA a

Virtual Work: We apply both P and M to member ABC.

δ U = 100δ y A − Pδ yC − M δθ + 60δ yD + 50δ yE − 40δ yG = 0 100δ y A − P(2δ y A ) − M 100 − 2 P −

δ yA !

1 1 ! ! + 60δ y A + 50 δ y A ! − 40 δ y A ! = 0 ! "2 # "2 # " a #

M + 60 + 25 − 20 = 0 a M = 165 N a

2P +

Now from Eq. (1) for M = 0

2 P = 165 N

(1) P = 82.5 N


!

PROBLEM 10.2 Determine the horizontal force P that must be applied at A to maintain the equilibrium of the linkage.

SOLUTION Assume δθ

δ x A = 10δθ δ yC = 4δθ δ yD = δ yC = 4δθ δφ =

δ yD

2 = δθ 6 3

δ xG = 15δφ = 15

$2 % δθ ! "3 #

= 10δθ

Virtual Work: We shall assume that a force P and a couple M are applied to member ABC as shown.

δ U = − Pδ x A − M δθ + 30δ yC + 40δ yD + 180δφ + 80δ xG = 0 − P (10δθ ) − M δθ + 30(4δθ ) + 40(4δθ ) + 180

$2 % δθ ! + 80(10δθ ) = 0 "3 #

−10 P − M + 120 + 160 + 120 + 800 = 0 (10 in.) P + M = 1200 lb ⋅ in.

Making M = 0 in Eq. (1):

P = +120.0 lb

(1) P = 120.0 lb


!

PROBLEM 10.3 Determine the couple M that must be applied to member ABC to maintain the equilibrium of the linkage.

SOLUTION Assume δ y A

δ yC = 2δ y A δ yD = δ y A 1 2

1 2

δ yE = δ yD = δ y A 1 2

δ yG = δ yE = δ y A Also

δθ =

δ yA a

Virtual Work: We apply both P and M to member ABC.

δ U = 100δ y A − Pδ yC − M δθ + 60δ yD + 50δ yE − 40δ yG = 0 100δ y A − P (2δ y A ) − M 100 − 2 P −

$ δ yA " a

M + 60 + 25 − 20 = 0 a 2P +

Now from Eq. (1) for

% $1 % $1 % ! + 60δ y A + 50 2 δ y A ! − 40 2 δ y A ! = 0 " # " # #

M = 165 N a

(1)

P = 0 and a = 0.3 M = 165 N 0.3 m

M = 49.5 N ⋅ m


!

PROBLEM 10.4 Determine the couple M that must be applied to member ABC to maintain the equilibrium of the linkage.

SOLUTION Assume δθ

δ x A = 10δθ δ yC = 4δθ δ yD = δ yC = 4δθ δφ =

δ yD

2 = δθ 6 3

δ xG = 15δ φ = 15

$2 % δθ ! "3 #

= 10δθ

Virtual Work: We shall assume that a force P and a couple M are applied to member ABC as shown.

δ U = − Pδ x A − M δθ + 30δ yC + 40δ yD + 180δφ + 80δ xG = 0 − P (10δθ ) − M δθ + 30(4δθ ) + 40(4δθ ) + 180

$2 % δθ ! + 80(10δθ ) = 0 "3 #

−10 P − M + 120 + 160 + 120 + 800 = 0 (10 in.) P + M = 1200 lb ⋅ in. P=0

Now from Eq. (1) for

(1) M = 1200 lb ⋅ in.


!

PROBLEM 10.5 Knowing that the maximum friction force exerted by the bottle on the cork is 60 lb, determine (a) the force P that must be applied to the corkscrew to open the bottle, (b) the maximum force exerted by the base of the corkscrew on the top of the bottle.

SOLUTION From sketch

δ y A = 4δ yC

Thus (a)

y A = 4 yC

Virtual Work:

δ U = 0: Pδ y A − Fδ yC = 0 P (4δ yC ) − F δ yC = 0

P= F = 60 lb: P =

1 F 4

1 (60 lb) = 15 lb 4

P = 15.00 lb

(b)

Free body: Corkscrew ΣFy = 0 R + P − F = 0; R + 15 lb − 60 lb = 0

On corkscrew:

R = 45 lb ,

On battle:

R = 45.0 lb


!

PROBLEM 10.6 The two-bar linkage shown is supported by a pin and bracket at B and a collar at D that slides freely on a vertical rod. Determine the force P required to maintain the equilibrium of the linkage.

SOLUTION

Assume δ y A :

δ yC =

8 in. 1 δ y A ; δ yC = δ y A 16 in. 2

Since bar CD move in translation

δ yE = δ yF = δ yC or Virtual Work:

δ yE = δ y F =

1 yA 2

δ U = 0: − Pδ y A + (100 lb)δ yE + (150 lb)δ yF = 0 − Pδ y A + 100

$1 % $1 % δ y A ! + 150 δ y A ! = 0 2 2 " # " # P = 125 lb P = 125.0 lb


!

PROBLEM 10.7 A spring of constant 15 kN/m connects Points C and F of the linkage shown. Neglecting the weight of the spring and linkage, determine the force in the spring and the vertical motion of Point G when a vertical downward 120-N force is applied (a) at Point C, (b) at Points C and H.

SOLUTION yG = 4 yC yH = 4 yC yF = 3 yC yE = 2 yC

δ yH = 4δ yC δ yF = 3δ yC δ yE = 2δ yC

For spring: ∆ = yF − yC

Q = Force in spring (assumed in tension) Q = + k ∆ = k ( yF − yC ) = k (3 yC − yC ) = 2kyC

(1)

C = 120 N, E = F = H = 0

(a) Virtual Work:

δ U = 0: − (120 N)δ yC + Qδ yC − Qδ yF = 0 −120δ yC + Qδ yC − Q (3δ yC ) = 0 Q = −60 N

Eq. (1):

Q = 2kyC , − 60 N = 2(15 kN/m) yC ,

Q = 60.0 N C

yC = −2 mm

yG = 4 yC = 4(−2 mm) = −8 mm

At Point G:

y G = 8.00 mm

C = H = 120 N, E = F = 0

(b) Virtual Work:

δ U = 0: − (120 N)δ yC − (120 N) yH + Qδ yC − Qδ yF = 0 −120δ yC − 120(4δ yC ) + Qδ yC − Q(3δ yC ) = 0 Q = −300 N

Eq. (1): At Point G:

Q = 2kyC

Q = 300 N C

−300 N = 2(15 kN/m)yC , yC = −10 mm yG = 4 yC = 4(−10 mm) = − 40 mm

y G = 40.0 mm


!

PROBLEM 10.8 A spring of constant 15 kN/m connects Points C and F of the linkage shown. Neglecting the weight of the spring and linkage, determine the force in the spring and the vertical motion of Point G when a vertical downward 120-N force is applied (a) at Point E, (b) at Points E and F.

SOLUTION yG = 4 yC yH = 4 yC yF = 3 yC yE = 2 yC

δ yH = 4δ yC δ yF = 3δ yC δ yE = 2δ yC

∆ = yF − yC

For spring:

Q = Force in spring (assumed in tension) Q = + k ∆ = k ( yF − yC ) = k (3 yC − yC ) = 2kyC

(1)

E = 120 N, C = F = H = 0

(a) Virtual Work:

δ U = 0: − (120 N)δ yE + Qδ yC − Qδ yF = 0 −120(2δ yC ) + Qδ yC − Q (3δ yC ) = 0 Q = −120 N Q = 2kyC , −120 N = 2(15 kN/m) yC ,

Eq. (1):

yC = −4 mm

yG = 4 yC = 4(−4 mm) = 16 mm

At Point G:

Q = 120.0 N C

y G = 16.00 mm

E = F = 120 N, C = H = 0

(b) Virtual Work:

δ U = 0: − (120 N)δ yE − (120 N)δ yF + Qδ yC − Qδ yF = 0 −120(2δ yC ) − 120(3δ yC ) + Qδ yC − Q (3δ yC ) = 0 Q = −300 N

Eq. (1): At Point G:

Q = 2kyC , − 300 N = 2(15 kN/m)yC ,

Q = 300 N C

yC = −10 mm

yG = 4 yC = 4(−10 mm) = − 40 mm

y G = 40.0 mm


!

PROBLEM 10.9 Knowing that the line of action of the force Q passes through Point C, derive an expression for the magnitude of Q required to maintain equilibrium.

SOLUTION

We have

y A = 2l cos θ ; δ y A = −2l sin θ δθ

θ

θ

CD = 2l sin ; δ (CD ) = l cos δθ 2 2

Virtual Work:

δ U = 0: − Pδ y A − Qδ (CD ) = 0 θ % $ − P( −2l sin θ δθ ) − Q l cos δθ ! = 0 2 # "

Q = 2P

sin θ cos(θ /2)


!

PROBLEM 10.10 Solve Problem 10.9 assuming that the force P applied at Point A acts horizontally to the left. PROBLEM 10.9 Knowing that the line of action of the force Q passes through Point C, derive an expression for the magnitude of Q required to maintain equilibrium.

SOLUTION

We have

x A = 2l cos θ ; δ x A = 2l cos θ δθ

θ

θ

CD = 2l sin ; δ (CD ) = l cos δθ 2 2

Virtual Work:

δ U = 0: Pδ x A − Qδ (CD) = 0 θ % $ P (2l cos θ δθ ) − Q l cos δθ ! = 0 2 # "

Q = 2P

cos θ cos(θ /2)


!

PROBLEM 10.11 The mechanism shown is acted upon by the force P derive an expression for the magnitude of the force Q required to maintain equilibrium.

SOLUTION Virtual Work: We have

x A = 2l sin θ

δ x A = 2l cos θδθ and

yF = 3l cos θ

δ yF = −3l sin θδθ Virtual Work:

δ U = 0: Qδ x A + Pδ yF = 0 Q(2l cos θδθ ) + P(−3l sin θδθ ) = 0

Q=

3 P tan θ 2


!

PROBLEM 10.12 The slender rod AB is attached to a collar A and rests on a small wheel at C. Neglecting the radius of the wheel and the effect of friction, derive an expression for the magnitude of the force Q required to maintain the equilibrium of the rod.

SOLUTION For ∆AA′C : A′C = a tan θ y A = −( A′C ) = − a tan θ

δ yA = −

a δθ cos 2 θ

For ∆CC ′B : BC ′ = l sin θ − A′C = l sin θ − a tan θ yB = BC ′ = l sin θ − a tan θ

δ yB = l cos θδθ − Virtual Work:

a δθ cos 2 θ

δ U = 0: Qδ y A − Pδ yB = 0 a % a $ $ δθ − P l cos θ − −Q − 2 ! 2 θ θ cos cos " # " a % $ a % $ = P l cos θ − Q ! 2 ! cos 2 θ # " cos θ # " Q=P

% ! δθ = 0 #

$l % cos3 θ − 1! "a #


!

PROBLEM 10.13 The slender rod AB is attached to a collar A and rests on a small wheel at C. Neglecting the radius of the wheel and the effect of friction, derive an expression for the magnitude of the force Q required to maintain the equilibrium of the rod.

SOLUTION For ∆ AA′C : A′C = a tan θ y A = −( A′C ) = − a tan θ

δ yA = −

a δθ cos 2 θ

For ∆ BB′C : B′C = l sin θ − A′C = l sin θ − a tan θ B′C l sin θ − a tan θ = BB′ = tan θ tan θ xB = BB′ = l cos θ − a

δ xB = −l sin θδθ Virtual Work:

δ U = 0: Pδ xB − Qδ y A = 0 a $ % δθ ! = 0 P(−l sin θδθ ) − Q − 2 " cos θ #

or

Pl sin θ cos 2 θ = Qa

1 Q = P sin θ cos 2 θ a


!

PROBLEM 10.14 Derive an expression for the magnitude of the force Q required to maintain the equilibrium of the mechanism shown.

SOLUTION

We have

xD = 2l cosθ

so that δ xD = −2l sin θδθ

δ A = 2lδθ δ B = lδθ Virtual Work:

δ U = 0: − Qδ xD − Pδ A − Pδ B = 0 −Q (−2l sin θδθ ) − P(2lδθ ) − P(lδθ ) = 0 2Ql sin θ − 3Pl = 0

Q=

3 P 2 sin θ


!

PROBLEM 10.15 A uniform rod AB of length l and weight W is suspended from two cords AC and BC of equal length. Derive an expression for the magnitude of the couple M required to maintain equilibrium of the rod in the position shown.

SOLUTION

yG = h cos θ =

1 l tan α cos θ 2

1 2

δ yG = − l tan α sin θδθ Virtual Work:

δ U = Wδ yG + M δθ = 0 $ 1 % W − l tan α sin θδθ ! + M δθ = 0 " 2 # 1 M = Wl tan α sin θ 2


!

PROBLEM 10.16 Derive an expression for the magnitude of the couple M required to maintain the equilibrium of the linkage shown.

SOLUTION

We have

xB = l cos θ

δ xB = −l sin θδθ

(1)

yC = l sin θ

δ yC = l cos θδθ

Now

1 2

δ xB = lδθ

Substituting from Equation (1) −l sin θδθ =

or Virtual Work:

1 lδφ 2

δφ = −2sin θδθ δ U = 0:

M δϕ + Pδ yC = 0

M (−2sin θδθ ) + P(l cos θδθ ) = 0 M=

or

1 cos θ Pl 2 sin θ

M=

Pl 2 tan θ



SOLUTION

yE = 3a sin θ

δ yE = 3a cos θδθ yF = 4a sin θ δ yF = 4a cos θδθ Virtual Work:

δ U = 0:

− M δθ + Pδ yE + Pδ yF = 0

− M δθ + P(3a cos θδθ ) + P(4a cos θδθ ) = 0

M = 7Pa cos θ


!

PROBLEM 10.18 The pin at C is attached to member BCD and can slide along a slot cut in the fixed plate shown. Neglecting the effect of friction, derive an expression for the magnitude of the couple M required to maintain equilibrium when the force P that acts at D is directed (a) as shown, (b) vertically downward, (c) horizontally to the right.

SOLUTION We have

xD = l cos θ

δ xD = −l sin θδθ yD = 3l sin θ

δ yD = 3l cos θδθ Virtual Work:

δ U = 0: M δθ − ( P cos β )δ xD − ( P sin β )δ yD = 0 M δθ − ( P cos β )(−l sin θδθ ) − ( P sin β )(3l cos θδθ ) = 0 M = Pl (3sin β cos θ − cos β sin θ )

(a)

For P directed along BCD, β = θ M = Pl (3sin θ cos θ − cos θ sin θ )

Equation (1):

(b)

M = Pl (2sin θ cos θ )

M = Pl sin 2θ

!

M = Pl (3sin 90° cos θ − cos 90° sin θ )

M = 3Pl cos θ

!

M = Pl sin θ

!

For P directed , β = 90° Equation (1):

(c)

(1)

For P directed Equation (1):

, β = 180° M = Pl (3sin180° cos θ − cos180° sin θ )


PROBLEM 10.19 A 4-kN force P is applied as shown to the piston of the engine system. Knowing that AB = 50 mm and BC = 200 mm, determine the couple M required to maintain the equilibrium of the system when (a) θ = 30°, (b) θ = 150°.

SOLUTION Analysis of the geometry:

Law of sines sin φ sin θ = AB BC sin φ =

AB sin θ BC

(1)

Now xC = AB cos θ + BC cos φ

δ xC = − AB sin θδθ − BC sin φδφ Now, from Equation (1) or

cos φδφ =

δφ =

(2)

AB cos θδθ BC AB cos θ δθ BC cos φ

(3)

From Equation (2)

δ xC = − AB sin θδθ − BC sin φ or

δ xC = −

$ AB cos θ % δθ ! BC cos φ " #

AB (sin θ cos φ + sin φ cos θ )δθ cos φ



AB sin(θ + φ ) δθ cos φ

Then

δ xC = −

Virtual Work:

δ U = 0: − Pδ xC − M δθ = 0 & AB sin(θ + φ ) ' − P (− δθ ) − M δθ = 0 cos φ * +

M = AB

Thus,

sin(θ + φ ) P cos φ

(4)

P = 4 kN, θ = 30°

(a) Eq. (1):

Eq. (4):

sin φ =

50 mm sin 30° 200 mm

M = (0.05 m)

φ = 7.181°

sin (30°+7.181°) (4 kN) cos 7.818°

M = 121.8 N ⋅ m

P = 4 kN, θ = 150°

(b) Eq. (1): Eq. (4):

sin φ =

50 mm sin160° 200 mm

M = (0.05 m)

φ = 7.181°

sin (150°+7.181°) (4 kN) cos 7.181°

M = 78.2 N ⋅ m

!


PROBLEM 10.20 A couple M of magnitude 100 N ⋅ m is applied as shown to the crank of the engine system. Knowing that AB = 50 mm and BC = 200 mm, determine the force P required to maintain the equilibrium of the system when (a) θ = 60°, (b) θ = 120°.

SOLUTION Analysis of the geometry:

Law of sines sin φ sin θ = AB BC sin φ =

AB sin θ BC

(1)

Now xC = AB cos θ + BC cos φ

δ xC = − AB sin θδθ − BC sin φδφ Now, from Equation (1) or

cos φδφ =

δφ =

(2)

AB cos θδθ BC AB cos θ δθ BC cos φ

(3)

From Equation (2)

δ xC = − AB sin θδθ − BC sin φ or

δ xC = −

$ AB cos θ % δθ ! BC cos φ " #

AB (sin θ cos φ + sin φ cos θ )δθ cos φ



AB sin(θ + φ ) δθ cos φ

Then

δ xC = −

Virtual Work:

δ U = 0: − Pδ xC − M δθ = 0 & AB sin(θ + φ ) ' − P (− δθ ) − M δθ = 0 cos φ * + M = AB

Thus,

sin(θ + φ ) P cos φ

(4)

M = 100 N ⋅ m, θ = 60°

(a) Eq. (1): Eq. (4):

sin φ =

50 mm sin 60° 200 mm

100 N ⋅ m = (0.05 m)

φ = 12.504°

sin (60° + 12.504°) P cos 12.504°

P = 2047 N

P = 2.05 kN

M = 100 N ⋅ m, θ = 120°

(b) Eq. (1): Eq. (4):

sin φ =

50 mm sin120° φ = 12.504° 200 mm

100 N ⋅ m = (0.05 m)

sin (120° + 12.504°) P cos12.504°

P = 2649 N

P = 2.65 kN


PROBLEM 10.21 For the linkage shown, determine the couple M required for equilibrium when l = 1.8 ft, Q = 40 lb, and θ = 65°.

SOLUTION

δC =

1 2

lδφ

cos θ

Virtual Work:

δU = 0 :

M δφ − Qδ C = 0

M δφ − Q

$1 l " 2 cos θ

% ! δφ = 0 # M=

Data:

M=

1 Ql 2 cos θ

1 (40 lb)(1.8 ft) = 85.18 lb ⋅ ft 2 cos 65°

M = 85.2 lb ⋅ ft


PROBLEM 10.22 For the linkage shown, determine the force Q required for equilibrium when l = 18 in., M = 600 lb ⋅ in., and θ = 70°.

SOLUTION

δC =

1 lδφ 2 cos θ

Virtual Work:

δ U = 0: M δφ − Q

M δφ − Qδ C = 0

$1 l % ! δφ = 0 " 2 cos θ # Q=

Data:

Q=

2 M cos θ l

2(600 lb ⋅ in.)cos70° = 22.801 lb 18 in.

Q = 22.8 lb

70.0°


PROBLEM 10.23 Determine the value of θ corresponding to the equilibrium position of the mechanism of Problem 10.11 when P = 45 lb and Q = 160 lb. PROBLEM 10.11 The mechanism shown is acted upon by the force P derive an expression for the magnitude of the force Q required to maintain equilibrium.

SOLUTION Virtual Work: x A = 2l sin θ

δ x A = 2l cos θ δθ yF = 3l cosθ

δ yF = −3l sin θ δθ δ U = 0:

Qδ x A + Pδ yF = 0

Q(2l cosθ δθ ) + P (−3l sin θ δθ ) = 0 Q=

Data:

3 P tan θ 2

P = 45 lb Q = 160 lb (160 lb) =

3 (45 lb) tanθ 2

θ = 67.1°


PROBLEM 10.24 Determine the value of θ corresponding to the equilibrium position of the mechanism of Problem 10.9 when P = 80 N and Q = 100 N.

SOLUTION From geometry y A = 2l cos θ , δ y A = −2l sin θ δθ

θ

θ

CD = 2l sin , δ (CD ) = l cos δθ 2 2

Virtual Work:

δ U = 0:

− Pδ y A − Qδ (CD) = 0

θ % $ − P(−2l sin θ δθ ) − Q l cos δθ ! = 0 2 # " Q = 2P

or

sin θ cos ( θ2 )

P = 80 N, Q = 100 N

With

(100 N) = 2(80 N)

sinθ cos ( θ2 )

sin θ = 0.625 cos ( θ2 )

or

2sin ( θ2 ) cos ( θ2 ) cos ( θ2 )

= 0.625

θ = 36.42°

θ = 36.4°

(Additional solutions discarded as not applicable are θ = ±180°)


!

PROBLEM 10.25 Rod AB is attached to a block at A that can slide freely in the vertical slot shown. Neglecting the effect of friction and the weights of the rods, determine the value of θ corresponding to equilibrium.

SOLUTION Dimensions in mm.

y A = 400sin θ

δ y A = 400 cos θ δθ xD = 100 cos θ

δ xD = −100sin θ δθ Virtual Work:

δ U = −(160 N)δ yA − (800 N)δ xD = 0 −(160)(400 cosθ δθ ) − (800)(−100sin θ δθ ) = 0 sin θ = 0.8; tan θ = 0.8 cos θ

θ = 38.7°


PROBLEM 10.26 Solve Problem 10.25 assuming that the 800-N force is replaced by a 24-N ⋅ m clockwise couple applied at D. PROBLEM 10.25 Rod AB is attached to a block at A that can slide freely in the vertical slot shown. Neglecting the effect of friction and the weights of the rods, determine the value of θ corresponding to equilibrium.

SOLUTION

y A = (0.4 m) sin θ

δ y A = (0.4 m) cos θ δθ Virtual Work:

δ U = −(160 N)δ y A + (24 N ⋅ m)δθ = 0 −(160 N)(0.4 m)cosθ δθ + (24 N ⋅ m)δθ = 0 cos θ = 0.375

θ = 68.0°


PROBLEM 10.27 Determine the value of θ corresponding to the equilibrium position of the rod of Problem 10.12 when l = 30 in., a = 5 in., P = 25 lb, and Q = 40 lb. PROBLEM 10.12 The slender rod AB is attached to a collar A and rests on a small wheel at C. Neglecting the radius of the wheel and the effect of friction, derive an expression for the magnitude of the force Q required to maintain the equilibrium of the rod.

SOLUTION For ∆ AA′C :

A′C = a tan θ y A = −( A′C ) = − a tan θ

δ yA = −

a δθ cos 2 θ

For ∆CC ′B : BC ′ = l sin θ − A′C = l sin θ − a tan θ yB = BC ′ = l sin θ − a tan θ

δ yB = l cos θ δθ −

a δθ cos 2 θ

Virtual Work:

δ U = 0: − Qδ y A − Pδ yB = 0 a $ −Q − 2 " cos θ Q

or

Q=P

$ a 2 " cos θ

a % $ ! δθ − P l cos θ − cos 2 θ # "

% ! δθ = 0 #

a % % $ ! = P l cos θ − ! cos 2 θ # # "

$l % cos3 θ − 1! "a #



with

l = 30 in., a = 5 in., P = 25 lb, and Q = 40 lb (40 lb) = (25 lb)

or

$ 30 in. 3 % cos θ − 1! 5 in. " #

cos3 θ = 0.4333

θ = 40.8°


PROBLEM 10.28 Determine the values of θ corresponding to the equilibrium position of the rod of Problem 10.13 when l = 600 mm, a = 100 mm, P = 50 N, and Q = 90 N. PROBLEM 10.13 The slender rod AB is attached to a collar A and rests on a small wheel at C. Neglecting the radius of the wheel and the effect of friction, derive an expression for the magnitude of the force Q required to maintain the equilibrium of the rod.

SOLUTION For ∆ AA′C :

A′C = a tan θ y A = −( A′C ) = − a tan θ

δ yA = −

a δθ cos 2 θ

For ∆BB′C : B′C = l sin θ − A′C = l sin θ − a tan θ BB′ =

B′C l sin θ − a tan θ = tan θ tan θ

xB = BB′ = l cos θ − a

δ xB = −l sin θ δθ Virtual Work:

δ U = 0: Pδ xB − Qδ y A = 0 a $ % P(−l sin θ δθ ) − Q − δθ ! = 0 2 " cos θ # Pl sin θ cos 2 θ = Qa

or

l Q = P sin θ cos 2 θ a



with

l = 600 mm, a = 100 mm, P = 50 N, and Q = 90 N 90 N = (50 N)

or

600 mm sin θ cos 2 θ 100 mm

sin θ cos 2 θ = 0.300

θ = 19.81° and 51.9°

Solving numerically


PROBLEM 10.29 Two rods AC and CE are connected by a pin at C and by a spring AE. The constant of the spring is k, and the spring is unstretched when θ = 30°. For the loading shown, derive an equation in P, θ, l, and k that must be satisfied when the system is in equilibrium.

SOLUTION yE = l cos θ

δ yE = −l sin θ δθ Spring: Unstretched length = 2l x = 2(2l sin θ ) = 4l sin θ

δ x = 4l cos θ δθ F = k ( x − 2l ) F = k ( 4l sin θ − 2l )

Virtual Work:

δ U = 0:

Pδ yE − F δ x = 0 P(−l sin θ δθ ) − k (4l sin θ − 2l )(4l cos θ δθ ) = 0 − P sin θ − 8kl (2sin θ − 1) cos θ = 0

or

cos θ P = (1 − 2sin θ ) 8kl sin θ

P 1 − 2sin θ = 8kl tan θ


PROBLEM 10.30 Two rods AC and CE are connected by a pin at C and by a spring AE. The constant of the spring is 1.5 lb/in., and the spring is unstretched when θ = 30°. Knowing that l = 10 in. and neglecting the weight of the rods, determine the value of θ corresponding to equilibrium when P = 40 lb.

SOLUTION From the analysis of Problem 10.29 Then with

P 1 − 2sin θ = 8kl tan θ P = 160 N, l = 0.2 m, and k = 300 N/m

160 N 1 − 2sin θ = 8(300 N/m)(0.2 m) tan θ

or

1 − 2sin θ 1 = = 0.3333 tan θ 3

θ = 25.0°

Solving numerically


PROBLEM 10.31 Solve Problem 10.30 assuming that force P is moved to C and acts vertically downward. PROBLEM 10.30 Two rods AC and CE are connected by a pin at C and by a spring AE. The constant of the spring is 1.5 lb/in., and the spring is unstretched when θ = 30°. Knowing that l = 10 in. and neglecting the weight of the rods, determine the value of θ corresponding to equilibrium when P = 40 lb.

SOLUTION yC = l cos θ , δ yC = −l sin θδθ

Spring: Unstretched length = 2l x = 2(2l sin θ ) = 4l sin θ δ x = 4l cos θδθ F = k ( x − 2l ) F = k (4l sin θ − 2l )

Virtual Work:

δ U = 0: − Pδ yC − Fδ x − P(−l sin θδθ ) − k (4l sin θ − 2l )(4l cos θδθ ) = 0 P sin θ − 8kl (2sin θ − 1) cosθ = 0 cos θ P = (2sin θ − 1) 8kl sin θ

or

l = 200 mm, k = 300 N/m, and P = 160 N

with

(160 N) cos θ = (2sin θ − 1) 8(300 N/m)(0.2) sin θ

or

(2sin θ − 1)

cos θ 1 = sin θ 3

Solving numerically

θ = 39.65°

and

θ = 68.96°

θ = 39.7° ! θ = 69.0°

and


PROBLEM 10.32 Rod ABC is attached to blocks A and B that can move freely in the guides shown. The constant of the spring attached at A is k = 3 kN/m, and the spring is unstretched when the rod is vertical. For the loading shown, determine the value of θ corresponding to equilibrium.

SOLUTION xC = (0.4 m)sin θ

δ xC = 0.4cos θδθ y A = (0.2 m) cos θ

δ y A = −0.2sin θδθ Spring: Unstretched length = 0.2 m F = k (0.2 m − y A ) = k (0.2 − 0.2 cos θ ) = (300 N/m)(0.2)(1 − cos θ ) F = 600(1 − cos θ )

Virtual Work:

δ U = 0: (150 N)δ xC + Fδ y A = 0 150(0.4cos θδθ ) + 600(1 − cos θ )(−0.2sin θδθ ) = 0 150(0.4) 1 1 = : = (1 − cos θ ) tan θ 600(0.2) 2 2

θ = 57.2°



!

PROBLEM 10.33 A load W of magnitude 600 N is applied to the linkage at B. The constant of the spring is k = 2.5 kN/m, and the spring is unstretched when AB and BC are horizontal. Neglecting the weight of the linkage and knowing that l = 300 mm, determine the value of θ corresponding to equilibrium.

SOLUTION

xC = 2l cos θ

δ xC = −2l sin θδθ yB = l sin θ δ yB = l cos θδθ F = ks = k (2l − xC ) = 2kl (1 − cos θ )

Virtual Work:

δ U = 0: F δ xC + W δ yB = 0 2kl (1 − cos θ )( −2l sin θδθ ) + W (l cos θδθ ) = 0 4kl 2 (1 − cos θ )sin θ = Wl cos θ

or Given:

(1 − cos θ ) tan θ =

l = 0.3 m, W = 600 N, k = 2500 N/m 600 N 4(2500 N/m)(0.3 m)

Then


or

(1 − cos θ ) tan θ = 0.2

Solving numerically

W 4kl

θ = 40.22°

θ = 40.2°


!

PROBLEM 10.34 A vertical load W is applied to the linkage at B. The constant of the spring is k, and the spring is unstretched when AB and BC are horizontal. Neglecting the weight of the linkage, derive an equation in θ, W, l, and k that must be satisfied when the linkage is in equilibrium.

SOLUTION

xC = 2l cos θ

δ xC = −2l sin θδθ yB = l sin θ δ yB = l cos θδθ F = ks = k (2l − xC ) = 2kl (1 − cos θ )

Virtual Work:

δ U = 0: F δ xC + W δ yB = 0 2kl (1 − cos θ )( −2l sin θδθ ) + W (l cos θδθ ) = 0 4kl 2 (1 − cos θ )sin θ = Wl cos θ

or


W 4kl (1 − cos θ ) tan θ =

From above

W 4kl


!

PROBLEM 10.35 Knowing that the constant of spring CD is k and that the spring is unstretched when rod ABC is horizontal, determine the value of θ corresponding to equilibrium for the data indicated. P = 300 N, l = 400 mm, k = 5 kN/m.

SOLUTION y A = l sin θ

δ y A = l cos θδθ Spring:

v = CD

Unstretched when

θ =0

so that

v0 = 2l

For θ :

v = 2l sin " $

90° + θ ! # 2 %

δ v = l cos " 45° + $

θ!

# δθ 2%

Stretched length:

θ! s = v − v0 = 2l sin " 45° + # − 2l 2% $

Then

& ' θ! F = ks = kl ( 2sin " 45° + # − 2 ) 2% $ * +

Virtual Work:

δ U = 0: Pδ y A − F δ v = 0 & ' θ! θ! Pl cos θδθ − kl ( 2sin " 45° + # − 2 ) l cos " 45° + # δθ = 0 2 2% $ % $ * +

or

1 & θ! θ! θ !' P = ( 2sin " 45° + # cos " 45° + # − 2 cos " 45° + # ) kl cos θ * 2% 2% 2 %+ $ $ $ =

1 cos θ

& θ! θ! θ !' ( 2sin " 45° + # cos " 45° + # cos θ − 2 cos " 45° + # ) 2% 2% 2 %+ $ $ $ *

= 1− 2

cos ( 45° θ2 ) cos θ



Now, with

P = 300 N, l = 400 mm, and k = 5 kN/m cos ( 45° + (300 N) =1− 2 (5000 N/m)(0.4 m) cos θ

or

cos ( 45° + θ2 ) cos θ

θ 2

)

= 0.60104

θ = 22.6°

Solving numerically


!

PROBLEM 10.36 Knowing that the constant of spring CD is k and that the spring is unstretched when rod ABC is horizontal, determine the value of θ corresponding to equilibrium for the data indicated. P = 75 lb, l = 15 in., k = 20 lb/in.

SOLUTION From the analysis of Problem 10.35, we have cos ( 45° + P =1− 2 kl cos θ

with

θ 2

)

P = 75 lb, l = 15 in. and k = 20 lb/in. cos ( 45° + θ2 ) (75 lb) =1− 2 (20 lb/in.)(15 in.) cos θ

or

cos ( 45° + θ2 ) cos θ

= 0.53033

θ = 51.1°

Solving numerically


!

PROBLEM 10.37 A load W of magnitude 72 lb is applied to the mechanism at C. Neglecting the weight of the mechanism, determine the value of θ corresponding to equilibrium. The constant of the spring is k = 20 lb/in., and the spring is unstretched when θ = 0.

SOLUTION s = rθ δ s = rδθ F = ks = krθ yC = l sin θ

δ yC = l cos θδθ Virtual Work:

δ U = − Fδ s + Wδ yC = 0 − krθ (rδθ ) + W (l cos θδθ ) = 0

θC cos θ

Solving by trial and error:

=

Wl (72 lb)(15 in.) = = 1.500 2 kr (20 lb/in.)(6 in.) 2

θ = 0.91486 rad

θ = 52.4°


!

PROBLEM 10.38 A force P of magnitude 240 N is applied to end E of cable CDE, which passes under pulley D and is attached to the mechanism at C. Neglecting the weight of the mechanism and the radius of the pulley, determine the value of θ corresponding to equilibrium. The constant of the spring is k = 4 kN/m, and the spring is unstretched when θ = 90°.

SOLUTION

π ! s = r " −θ # $2 % δ s = − rδθ π ! F = ks = kr " − θ # 2 $ % CD = 2l sin

θ 2

θ 1 ! δ (CD) = 2l cos " δθ # 2$2 % θ = l cos δθ 2

Virtual Work: Since F tends to decrease s and P tends to decrease CD, we have

δ U = − F δ s − Pδ (CD ) = 0 π θ ! ! − kr " − θ # (− rδθ ) − P " l cos δθ # = 0 2 % $2 % $ π 2

−θ

cos θ2


=

pl (240 N)(0.3 m) = = 1.25 kr 2 (4000 N/m)(0.12 m)2

θ = 0.33868 rad

θ = 19.40°


!

PROBLEM 10.39 The lever AB is attached to the horizontal shaft BC that passes through a bearing and is welded to a fixed support at C. The torsional spring constant of the shaft BC is K; that is, a couple of magnitude K is required to rotate end B through 1 rad. Knowing that the shaft is untwisted when AB is horizontal, determine the value of θ corresponding to the position of equilibrium when P = 100 N, l = 250 mm, and K = 12.5 N · m/rad.

SOLUTION y A = l sin θ

We have

δ y A = l cos θδθ Virtual Work:

δ U = 0: Pδ y A − M δθ = 0 Pl cos θδθ − Kθδθ = 0

or with

θ cos θ

Pl K

(1)

P = 100 N, l = 250 mm and K = 12.5 N ⋅ m/rad

θ cos θ

or

=

θ cos θ

=

(100 N)(0.250 m) 12.5 N ⋅ m/rad

= 2.0000

θ = 59.0°

Solving numerically


!

PROBLEM 10.40 Solve Problem 10.39 assuming that P = 350 N, l = 250 mm, and K = 12.5 N · m/rad. Obtain answers in each of the following quadrants: 0 , θ , 90°, 270° , θ , 360°, 360° , θ , 450°.

PROBLEM 10.39 The lever AB is attached to the horizontal shaft BC that passes through a bearing and is welded to a fixed support at C. The torsional spring constant of the shaft BC is K; that is, a couple of magnitude K is required to rotate end B through 1 rad. Knowing that the shaft is untwisted when AB is horizontal, determine the value of θ corresponding to the position of equilibrium when P = 100 N, l = 250 mm, and K = 12.5 N · m/rad.

SOLUTION Using Equation (1) of Problem 10.39 and P = 350 N, l = 250 mm and K = 12.5 N ⋅ m/rad

We have or

θ cos θ

θ cos θ

=

(350 N)(0.250 m) 12.5 N ⋅ m/rad

= 7 or θ = 7 cos θ

(1)

The solutions to this equation can be shown graphically using any appropriate graphing tool, such as Maple, with the command: plot ({theta, 7 * cos(theta)}, t = 0.5 * Pi/2); Thus, we plot y = θ and y = 7 cos θ in the range 0 #θ #

5π 2



We observe that there are three points of intersection, which implies that Equation (1) has three roots in the specified range of θ .

π! 0 # θ # 90° " # ; $2%

θ = 1.37333 rad, θ = 78.69°

!

3π ! # θ # 2π # ; θ = 5.65222 rad, θ = 323.85° ! 270 # θ # 360° " $ 2 %

!

5π 360 # θ # 450° " 2π # θ # 2 $

! #; %

θ = 6.61597 rad, θ = 379.07 !

θ = 78.7°

!

θ = 324°

!

θ = 379°

!


PROBLEM 10.41 The position of boom ABC is controlled by the hydraulic cylinder BD. For the loading shown, determine the force exerted by the hydraulic cylinder on pin B when θ = 65°.

SOLUTION We have yA = (72 in.) cos θ

δ y A = −72sin θδθ

( BD )2 = ( BC )2 + (CD ) 2 − 2( BC )(CD ) cos(180° − θ ) = (45)2 + (24) 2 + 2(45)(24) cos θ ( BD )2 = 2601 + 2160 cos θ

Differentiating:

(1)

2( BD )δ ( BD ) = −2160sin θδθ

δ ( BD) = −

1080 sin θδθ BD

(2)

Virtual Work: Noting that P tends to decrease yA and FBD tends to increase BD, write

δ U = − Pδ y A + FBDδ ( BD) = 0 1080 ! sin θδθ # = 0 − P(−72sin θδθ ) + FBD " − BD $ % FBD =

or, since

1 ( BD) P 15

P = 600 lb: FBD =

600 ( BD ) = (40 lb)( BD ) 15

(3)

Making θ = 65° in Eq. (1), we have ( BD )2 = 2601 + 2160 cos 65° = 3513.9 BD = 59.278

Carrying into Eq. (3):

FBD = (40 lb)(59.278) = 2371.1 lb

FBD = 2370 lb


!

PROBLEM 10.42 The position of boom ABC is controlled by the hydraulic cylinder BD. For the loading shown, (a) express the force exerted by the hydraulic cylinder on pin B as a function of the length BD, (b) determine the smallest possible value of the angle θ if the maximum force that the cylinder can exert on pin B is 2.5 kips.

SOLUTION (a)

See solution of Problem 10.41 for the derivation of Eq. (3): FBD = (40 lb)( BD)

(b)

For ( FBD )max = 2.5 kips = 2500 lb, we have 2500 lb = (40 lb)( BD) BD = 62.5

Carrying this value into Eq. (1) of Problem 10.41, write ( BD ) 2 = 2601 + 2160 cos θ (62.5) 2 = 2601 + 2160 cos θ cos θ = 0.60428

θ = 52.8°


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PROBLEM 10.43 The position of member ABC is controlled by the hydraulic cylinder CD. For the loading shown, determine the force exerted by the hydraulic cylinder on pin C when θ = 55°.

SOLUTION

y A = (0.8 m)sin θ

δ y A = 0.8cos θδθ CD 2 = BC 2 + BD 2 − 2( BC )( BD) cos(90° − θ ) CD 2 = 0.52 + 1.52 − 2(0.5)(1.5)sin θ CD 2 = 2.5 − 1.5sin θ 2(CD )(δ CD ) = −1.5cos θδθ

Virtual Work:

(1)

δ CD = −

3cos θ δθ 4CD

δ U = 0: − (10 kN)δ y A − FCDδ CD = 0 3cos θ ! −10(0.8cos θδθ ) − FCD " − δθ # = 0 CD 4 $ % FCD =

32 CD 3

(2)

For θ = 55°: Eq. (1): Eq. (2):

CD 2 = 2.5 − 1.5sin 55° = 1.2713; CD = 1.1275 m FCD =

32 32 CD = (1.1275) = 12.027 kN 3 3

FCD = 12.03 kN


!

PROBLEM 10.44 The position of member ABC is controlled by the hydraulic cylinder CD. Determine the angle θ knowing that the hydraulic cylinder exerts a 15-kN force on pin C.

SOLUTION

y A = (0.8 m)sin θ

δ y A = 0.8cos θδθ CD 2 = BC 2 + BD 2 − 2( BC )( BD) cos(90° − θ ) CD 2 = 0.52 + 1.52 − 2(0.5)(1.5)sin θ CD 2 = 2.5 − 1.5sin θ 2(CD )(δ CD ) = −1.5cos θδθ ;

Virtual Work:

(1)

δ CD = −

3cos θ δθ 4CD

δ U = 0: − (10 kN)δ y A − FCDδ CD = 0 3cos θ ! −10(0.8cos θδθ ) − FCD " − δθ # = 0 $ 4CD % FCD =

32 CD 3

(2)

For FCD = 15 kN: Eq. (2): Eq. (1):

15 kN =

32 CD : 3

CD =

(1.40625) 2 = 2.5 − 1.5sin θ :

45 = 1.40625 m 32

sin θ = 0.34831

θ = 20.38°

θ = 20.4°


!

PROBLEM 10.45 The telescoping arm ABC is used to provide an elevated platform for construction workers. The workers and the platform together weigh 500 lb and their combined center of gravity is located directly above C. For the position when θ = 20°, determine the force exerted on pin B by the single hydraulic cylinder BD.

SOLUTION

In ∆ ADE :

From the geometry:

AE 2.7 ft = DE 1.5 ft α = 60.945° 2.7 ft AD = = 3.0887 m sin 60.945°

tan α =

yC = (15 ft) sin θ

δ yC = (15 ft) cos θδθ Then, in triangle BAD:

Angle BAD = α + θ

Law of cosines: BD 2 = AB 2 + AD 2 − 2( AB )( AD ) cos(α + θ )

or

BD 2 = (7.2 ft)2 + (3.0887 ft) 2 − 2(7.2 ft)(3.0887 ft) cos(α + θ ) BD 2 = 61.380 ft 2 − (44.477 cos(α + θ )) ft 2

(1)



And then 2( BD )(δ BD) = (44.477 sin(α + θ ))δθ 44.477sin(α + θ ) δ BD = δθ 2( BD)

Virtual Work:

δ U = 0: − Pδ yC + FBDδ BD = 0 & (44.477 ft 2 )sin(α + θ ) ' −(500 lb)(15 ft) cos θδθ + FBD ( δθ ) = 0 2( BD ) * +

Substituting

or

& ' cos θ FBD = (337.25 BD ) lb/ft sin(α + θ ) * +

(2)

Now, with θ = 20° and α = 60.945° Equation (1): BD 2 = 61.380 − 44.477 cos(60.945° + 20°) BD 2 = 54.380 BD = 7.3743 ft

Equation (2): & ' cos 20° (7.3743 ft) ) lb/ft FBD = (337.25 sin(60.945° + 20°) * +

or

FBD = 2366 lb

FBD = 2.370 lb


!

PROBLEM 10.46 Solve Problem 10.45 assuming that the workers are lowered to a point near the ground so that θ = −20°. PROBLEM 10.45 The telescoping arm ABC is used to provide an elevated platform for construction workers. The workers and the platform together weigh 500 lb and their combined center of gravity is located directly above C. For the position when θ = 20°, determine the force exerted on pin B by the single hydraulic cylinder BD.

SOLUTION Using the figure and analysis of Problem 10.45, including Equations (1) and (2), and with θ = −20°, we have Equation (1):

BD 2 = 61.380 − 44.477 cos(60.945° − 20°) BD 2 = 27.785 BD = 5.2711 ft

Equation (2):

FBD = 337.25

cos(−20°) (5.2711) sin(60.945° − 20°)

FBD = 2549 lb

or FBD = 2550 lb


!

PROBLEM 10.47 A block of weight W is pulled up a plane forming an angle α with the horizontal by a force P directed along the plane. If µ is the coefficient of friction between the block and the plane, derive an expression for the mechanical efficiency of the system. Show that the mechanical efficiency cannot exceed 12 if the block is to remain in place when the force P is removed.

SOLUTION Input work = Pδ x Output work = (W sin α )δ x

Efficiency:

η=

W sin αδ x Pδ x

or η =

W sin α P

ΣFx = 0: P − F − W sin α = 0 or ΣFy = 0: N − W cos α = 0 or

(1) P = W sin α + F

(2)

N = W cos α

F = µ N = µW cos α

Equation (2):

P = W sin α + µW cos α = W (sin α + µ cos α )

Equation (1):

η=

W sin α W (sin α + µ cos α )

or η =

1 1 + µ cot α

!

If block is to remain in place when P = 0, we know (see Chapter 8) that φs $ α or, since

µ = tan φs,

µ cot α $ tan α cot α = 1

Multiply by cot α : Add 1 to each side:

µ $ tan α

1 + µ cot α $ 2

Recalling the expression for η , we find

η#

1 2


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PROBLEM 10.48 Denoting by µ s the coefficient of static friction between the block attached to rod ACE and the horizontal surface, derive expressions in terms of P, µ s, and θ for the largest and smallest magnitude of the force Q for which equilibrium is maintained.

SOLUTION For the linkage: ΣM B = 0: − x A +

xA P P = 0 or A = 2 2

Then:

F = µ s A = µs

Now

x A = 2l sin θ

P 1 = µs P 2 2

δ x A = 2l cos θ δθ and

yF = 3l cos θ

δ yF = −3l sin θ δθ Virtual Work:

δ U = 0: (Qmax − F )δ x A + Pδ yF = 0 1 ! " Qmax − 2 µ s P # (2l cos θ δθ ) + P(−3l sin θ δθ ) = 0 $ %

or

Qmax =

3 1 P tan θ + µ s P 2 2 Qmax =

P (3 tan θ + µ s ) 2

For Qmin, motion of A impends to the right and F acts to the left. We change µ s to − µ s and find Qmin =

P (3tan θ − µ s ) 2


!

PROBLEM 10.49 Knowing that the coefficient of static friction between the block attached to rod ACE and the horizontal surface is 0.15, determine the magnitude of the largest and smallest force Q for which equilibrium is maintained when θ = 30°, l = 0.2 m, and P = 40 N.

SOLUTION Using the results of Problem 10.48 with

θ = 30° l = 0.2 m P = 40 N, and µs = 0.15

We have

P (3 tan θ + µ s ) 2 (40 N) = (3tan 30° + 0.15) 2 = 37.64 N

Qmax =

Qmax = 37.6 N

and

P (3 tan θ − µ s ) 2 (40 N) = (3tan 30° − 0.15) 2 = 31.64 N

Qmin =

Qmin = 31.6 N


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PROBLEM 10.50 Denoting by µ s the coefficient of static friction between collar C and the vertical rod, derive an expression for the magnitude of the largest couple M for which equilibrium is maintained in the position shown. Explain what happens if µs $ tan θ .

SOLUTION Member BC: We have xB = l cos θ

δ xB = −l sin θδθ

(1)

and yC = l sin θ

δ yC = l cos θδθ

(2)

Member AB: We have 1 2

δ xB = lδφ Substituting from Equation (1), −l sin θδθ =

1 lδφ 2

δφ = −2sin θδθ

or

(3)

Free body of rod BC For M max, motion of collar C impends upward ΣM B = 0: Nl sin θ − ( P + µ s N )(l cos θ ) = 0

N tan θ − µ s N = P N=

! P tan θ − µ s



Virtual Work:

δ U = 0: M δφ + ( P + µs N )δ yC = 0 M (−2sin θδθ ) + ( P + µs N )l cos θδθ = 0 M max

P + µ s tan θP− µ ( P + µs N ) s = l= l 2 tan θ 2 tan θ

or M max =

Pl 2(tan θ − µs )

If µs = tan θ , M = ∞, system becomes self-locking!


PROBLEM 10.51 Knowing that the coefficient of static friction between collar C and the vertical rod is 0.40, determine the magnitude of the largest and smallest couple M for which equilibrium is maintained in the position shown, when θ = 35°, l = 600 mm, and P = 300 N.

SOLUTION From the analysis of Problem 10.50, we have M max =

With

Pl 2(tan θ + µ s )

θ = 35°, l = 0.6 m, P = 300 N (300 N)(0.6 m) 2(tan 35° − 0.4) = 299.80 N ⋅ m

M max =

M max = 300 N ⋅ m

!

For M min , motion of C impends downward and F acts upward. The equations of Problem 10.50 can still be used if we replace µs by − µs . Then M min =

Pl 2(tan θ + µ s )

Substituting, (300 N)(0.6 m) 2(tan 35° + 0.4) = 81.803 N ⋅ m

M min =

M min = 81.8 N ⋅ m


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PROBLEM 10.52 Derive an expression for the mechanical efficiency of the jack discussed in Section 8.6. Show that if the jack is to be self-locking, the mechanical efficiency cannot exceed 12 .

SOLUTION Recall Figure 8.9a. Draw force triangle

Q = W tan(θ + φs ) y = x tan θ so that δ y = δ x tan θ Input work = Qδ x = W tan(θ + φs )δ x Output work = W δ y = W (δ x) tan θ

Efficiency:

η=

W tan θδ x ; W tan(θ + φs )δ x

η=

tan θ tan(θ + φs )

!

1 2

!

From Page 432, we know the jack is self-locking if

φs $ θ Then so that

θ + φs $ 2θ tan(θ + φs ) $ tan 2θ !

From above

η=

tan θ tan(θ + φs )

It then follows that

η#

tan θ tan 2θ

But

Then

tan 2θ =

2 tan θ 1 − tan 2 θ

η#

tan θ (1 − tan 2 θ ) 1 − tan 2 θ = 2 tan θ 2

η#


PROBLEM 10.53 Using the method of virtual work, determine the reaction at E.

SOLUTION We release the support at E and assume a virtual displacement δ yE for Point E.

From similar triangles: 2.1 δ yE = 1.75δ yE 1.2 2.7 δ yC = δ yE = 2.25δ yE 1.2 0.5 0.5 (2.25δ yE ) = 1.25δ yE δ yB = δ yC = 0.9 0.9

δ yD =

Virtual Work:

δ U = (2 kN)δ yB + (3 kN)δ yD − Eδ yE = 0 2(1.25δ yE ) + 3(1.75δ yE ) − Eδ yE = 0 E = +7.75 kN

E = 7.75 kN


!

PROBLEM 10.54 Using the method of virtual work, determine separately the force and couple representing the reaction at H.

SOLUTION Force at H. We give a vertical virtual displacement δ yH to Point H, keeping member FH horizontal.

From the geometry of the diagram:

δ yF = δ yG = δ yH 0.9 0.9 δ yF = δ yH = 0.75δ yH 1.2 1.2 1.5 1.5 (0.75δ yH ) = 1.25δ yH δ yC = δ yD = 0.9 0.9 0.5 0.5 (1.25δ yH ) = 0.69444δ yH δ yB = δ yC = 0.9 0.9

δ yD =

Virtual Work:

δ U = (2 kN)δ yB + (3 kN)δ yD − (5 kN)δ yG + H δ yH = 0 2(0.69444δ yH ) + 3(0.75δ yH ) − 5δ yH + H δ yH = 0 H = +1.3611 kN H = 1.361 kN

Couple at H. We rotate beam FH through δθ about Point H.



From the geometry of the diagram:

δ yG = 1.2δθ

δ yF = 1.8δθ

0.9 0.9 (1.8δθ ) = 1.35δθ δ yF = 1.2 1.2 1.5 1.5 (1.35δθ ) = 2.25δθ δ yC = δ yD = 0.9 0.9 0.5 0.5 (2.25δθ ) = 1.25δθ δ yB = δ yC = 0.9 0.9

δ yD =

Virtual Work:

δ U = (2 kN)δ yB + (3 kN)δ yD − (5 kN)δ yG + M H δθ = 0 2(1.25δθ ) + 3(1.35δθ ) − 5(1.2δθ ) + M H δθ = 0 MH = −0.550 kN ⋅ m

M H = 550 N ⋅ m


PROBLEM 10.55 Referring to Problem 10.43 and using the value found for the force exerted by the hydraulic cylinder CD, determine the change in the length of CD required to raise the 10-kN load by 15 mm. PROBLEM 10.43 The position of member ABC is controlled by the hydraulic cylinder CD. For the loading shown, determine the force exerted by the hydraulic cylinder on pin C when θ = 55°.

SOLUTION

Virtual Work: Assume both δ yA and δ CD increase

δ U = 0: − (10 kN)δ y A − FCDδ CD = 0 Substitute:

δ y A = 15 mm and FCD = 12.03 kN −(10 kN)(15 mm) − (12.03 kN)δ CD = 0

δ CD = −12.47 mm δ CD = 12.47 mm shorter

The negative sign indicates that CD shortened


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PROBLEM 10.56 Referring to Problem 10.45 and using the value found for the force exerted by the hydraulic cylinder BD, determine the change in the length of BD required to raise the platform attached at C by 2.5 in. PROBLEM 10.45 The telescoping arm ABC is used to provide an elevated platform for construction workers. The workers and the platform together weigh 500 lb and their combined center of gravity is located directly above C. For the position when θ = 20°, determine the force exerted on pin B by the single hydraulic cylinder BD.

SOLUTION

Virtual Work: Assume both δ yC and δ BD increase

δ U = 0: − (500 lb)δ yC + FBDδ BD = 0 Substitute:

δ yC = 2.5 in. and FBD = 2370 lb −(500 lb)(2.5 in.) + (2370 lb)δ BD = 0

δ BD = +0.527 in. The positive sign indicates that cylinder BD increases in length

δ BD = 0.527 in. longer


!

PROBLEM 10.57 Determine the vertical movement of joint D if the length of member BF is increased by 1.5 in. (Hint: Apply a vertical load at joint D, and, using the methods of Chapter 6, compute the force exerted by member BF on joints B and F. Then apply the method of virtual work for a virtual displacement resulting in the specified increase in length of member BF. This method should be used only for small changes in the lengths of members.)

SOLUTION Apply vertical load P at D. ΣM H = 0: − P(40 ft) + E (120 ft) = 0 E=

ΣFy = 0:

P 3

P 3 FBF − = 0 5 3 FBF =

5 P 9

Virtual Work: We remove member BF and replace it with forces FBF and −FBF at pins F and B, respectively. Denoting the ! virtual displacements of Points B and F as δ rB and δ rF , respectively, and noting that P and δ D have the same direction, we have Virtual Work:

δ U = 0:

Pδ D + FBF ⋅ δ rF + (−FBF ) ⋅ δ rB = 0 Pδ D + FBF δ rF cos θ F − FBF δ rB cos θ B = 0 Pδ D − FBF (δ rB cos θ B − δ rF cos θ F ) = 0

where (δ rB cos θ B − δ rF cos θ F ) = δ BF , which is the change in length of member BF. Thus, Pδ D − FBF δ BF = 0 5 ! Pδ D − " P # (1.5 in.) = 0 9 $ % δ D = 0.833 in.

δ D = 0.833 in.


!

PROBLEM 10.58 Determine the horizontal movement of joint D if the length of member BF is increased by 1.5 in. (See the hint for Problem 10.57.)

SOLUTION Apply horizontal load P at D.

ΣM H = 0: P(30 ft) − E y (120 ft) = 0 Ey =

ΣFy = 0:

P 4

3 P FBF − = 0 5 4 FBF =

5 P 12

the We remove member BF and replace it with forces FBF and −FBF at pins F and B, respectively. Denoting ! virtual displacements of Points B and F as δ rB and δ rF , respectively, and noting that P and δ D have the same direction, we have Virtual Work:

δ U = 0:

Pδ D + FBF ⋅ δ rF + (−FBF ) ⋅ δ rB = 0

Pδ D + FBF δ rF cos θ F − FBF δ rB cos θ B = 0 Pδ D − FBF (δ rB cos θ B − δ rF cos θ F ) = 0

where (δ rB cos θ B − δ rF cos θ F ) = δ BF , which is the change in length of member BF. Thus, Pδ D − FBF δ BF = 0 5 ! Pδ D − " P # (1.5 in.) = 0 $9 % δ D = 0.625 in.

δ D = 0.625 in.


PROBLEM 10.59 Using the method of Section 10.8, solve Problem 10.29. PROBLEM 10.29 Two rods AC and CE are connected by a pin at C and by a spring AE. The constant of the spring is k, and the spring is unstretched when θ = 30°. For the loading shown, derive an equation in P, θ, l, and k that must be satisfied when the system is in equilibrium.

SOLUTION AE = x = 2(2l sin θ ) = 4l sin θ

Spring:

x0 = 4l sin 30° = 2l

Unstretched length: Deflection of spring s = x − x0

s = 2l (2sin θ − 1) 1 V = ks 2 + PyE 2 1 2 V = k [ 2l (2sin θ − 1) ] + P(−l cos θ ) 2 dV = 4kl 2 (2sin θ − 1)2 cos θ + Pl sin θ = 0 dθ (2sin θ − 1)

P cos θ + =0 sin θ 8kl

P 1 − 2sin θ = 8kl tan θ


PROBLEM 10.60 Using the method of Section 10.8, solve Problem 10.30. PROBLEM 10.30 Two rods AC and CE are connected by a pin at C and by a spring AE. The constant of the spring is 1.5 lb/in., and the spring is unstretched when θ = 30°. Knowing that l = 10 in. and neglecting the weight of the rods, determine the value of θ corresponding to equilibrium when P = 40 lb.

SOLUTION Using the result of Problem 10.59, with P = 40 lb l = 10 in. and k = 1.5 lb/in. P 1 − 2sin θ = 8kl tan θ

or

1 − 2sin θ 40 lb = tan θ 8(1.5 lb/in.)(10 in.) 1 = 3

θ = 25.0°

Solving numerically,


PROBLEM 10.61 Using the method of Section 10.8, solve Problem 10.33. PROBLEM 10.33 A load W of magnitude 600 N is applied to the linkage at B. The constant of the spring is k = 2.5 kN/m, and the spring is unstretched when AB and BC are horizontal. Neglecting the weight of the linkage and knowing that l = 300 mm, determine the value of θ corresponding to equilibrium.

SOLUTION

P = 600 N l = 0.3 m, and k = 2500 N ⋅ m

We have

600 N 4(2500 N/m)(0.3 m) = 0.2

(l − cos θ )tan θ =

θ = 40.2°

Solving numerically


PROBLEM 10.62 Using the method of Section 10.8, solve Problem 10.34. PROBLEM 10.34 A vertical load W is applied to the linkage at B. The constant of the spring is k, and the spring is unstretched when AB and BC are horizontal. Neglecting the weight of the linkage, derive an equation in θ, W, l, and k that must be satisfied when the linkage is in equilibrium.

SOLUTION

1 2 ks + PyB 2 1 V = k (2i − xC )2 + PyB 2 xC = 2l cos θ and yB = −l sin θ V=

Thus

1 k (2l − 2l cos θ ) 2 − Pl sin θ 2 = 2kl 2 (1 − cos θ ) 2 − Pl sin θ

V=

dV = 2kl 2 2(1 − cos θ ) sin θ − Pl cos θ = 0 dθ

(1 − cos θ )tan θ =

P 4kl


PROBLEM 10.63 Using the method of Section 10.8, solve Problem 10.35. PROBLEM 10.35 Knowing that the constant of spring CD is k and that the spring is unstretched when rod ABC is horizontal, determine the value of θ corresponding to equilibrium for the data indicated. P = 300 N, l = 400 mm, k = 5 kN/m.

SOLUTION Spring

v = 2l sin " $

90° + θ ! # 2 %

θ! v = 2l sin " 45° + # 2% $ Unstretched (θ = 0) Deflection of spring

v0 = 2l sin 45° = 2l

θ! s = v − v0 = 2l sin " 45° + # − 2l 2% $ 2

& ' 1 2 1 θ! ks + PyA = kl 2 ( 2sin " 45° + # − 2 ) + P(−l sin θ ) 2 2 2% $ * + & ' dV θ θ ! ! = kl 2 ( 2sin " 45° + # − 2 ) cos " 45° + # − Pl cos θ = 0 dθ 2 2 $ % $ % * + V=

& θ! θ! θ !' ( 2sin " 45° + # cos " 45° + # − 2 cos " 45° + # ) = 2 2 2 %+ $ % $ % $ * θ! cos θ − 2 cos " 45° + # = 2% $

Divide each member by cos θ

1− 2

cos ( 45° + θ2 ) cos θ

=

P cos θ kl P cos θ kl P kl

Then with P = 300 N, l = 0.4 m and k = 5000 N/m 1− 2

or

cos ( 45° + θ2 ) cos θ cos ( 45° + θ2 ) cos θ

300 N (5000 N/m)(0.4 m) = 0.15 =

= 0.60104

θ = 22.6°

Solving numerically


PROBLEM 10.64 Using the method of Section 10.8, solve Problem 10.36. PROBLEM 10.36 Knowing that the constant of spring CD is k and that the spring is unstretched when rod ABC is horizontal, determine the value of θ corresponding to equilibrium for the data indicated. P = 75 lb, l = 15 in., k = 20 lb/in.

SOLUTION Using the results of Problem 10.63 with P = 75 lb, l = 15 in. and k = 20 lb/in., we have 1− 2

cos ( 45° + θ2 ) cos θ

=

P kl

75 lb (20 lb/in.)(15 in.) = 0.25 =

or Solving numerically

cos ( 45° + θ2 ) cos θ

= 0.53033

θ = 51.058°

θ = 51.1°


PROBLEM 10.65 Using the method of Section 10.8, solve Problem 10.31. PROBLEM 10.31 Solve Problem 10.30 assuming that force P is moved to C and acts vertically downward.

SOLUTION AE = x = 2(2l sin θ ) = 4l sin θ

Spring: Unstretched length: Deflection of spring

x0 = 4l sin 30° = 2l

s = x − x0 s = 2l (2sin θ − 1) 1 V = ks 2 + PyC 2 1 = k[2l (2sin θ − 1)]2 + P (l cos θ ) 2

V dV dθ P cos θ + (1 − 2sin θ ) sin θ 8kl P 8kl

with We have or Solving numerically

= 2kl 2 (2sin θ − 1) 2 + Pl cos θ = 4kl 2 (2sin θ − 1)2 cos θ − Pl sin θ = 0 =0 =

2sin θ − 1 tan θ

P = 40 lb, l = 10 in. and k = 1.5 lb/in. (40 lb) 2sin θ − 1 = 8(1.5 lb/in.)(10 in.) tan θ 2sin − 1 1 = tan θ 3

θ = 39.65° and 68.96°

θ = 39.7° θ = 69.0°


PROBLEM 10.66 Using the method of Section 10.8, solve Problem 10.38. PROBLEM 10.38 A force P of magnitude 240 N is applied to end E of cable CDE, which passes under pulley D and is attached to the mechanism at C. Neglecting the weight of the mechanism and the radius of the pulley, determine the value of θ corresponding to equilibrium. The constant of the spring is k = 4 kN/m, and the spring is unstretched when θ = 90°.

SOLUTION

π ! s = r " −θ # $2 %

For spring:

Ve =

1 2 1 2 π ! ks = kr " − θ # 2 2 $2 %

x = 2l sin

For force P:

2

θ 2

V p = Px = 2 Pl sin

θ 2 2

V = Ve + V p =

Then

1 2 π θ ! kr " − θ # + 2 Pl sin 2 2 $2 %

π θ dV ! = −kr 2 " − θ # + Pl cos 2 2 dθ $ % π θ ! = −(4000 N/m)(0.12 m) 2 " − θ # + (240 N)(0.3 m) cos = 0 2 $2 % π 2


− θ = 1.25cos

θ 2

θ = 0.33868 rad

θ = 19.40°


PROBLEM 10.67 Show that equilibrium is neutral in Problem 10.1. PROBLEM 10.1 Determine the vertical force P that must be applied at C to maintain the equilibrium of the linkage.

SOLUTION We have

yA = b − u yC = b + 2u yD = −u 1 yE = − u 2 1 yG = + u 2 V = (100 N)y A + PyC + (60 N)yD + (50 N)yE + (40 N)yG 1 ! 1 ! = 100(b − u ) + P(b + 2u ) + 60(−u ) + 50 " − u # + 40 " u # 2 $ % $2 % dV = −100 + 2 P − 60 − 25 + 20 = 0 du P = 82.5 N

Substituting P = 82.5 N in the expression obtained for V : V = (100 + 82.5)b + (−100 + 165 − 60 − 25 + 20)u V = 182.5 b = constant

Since V is constant, the equilibrium is neutral


!

PROBLEM 10.68 Show that equilibrium is neutral in Problem 10.6. PROBLEM 10.6 The two-bar linkage shown is supported by a pin and bracket at B and a collar at D that slides freely on a vertical rod. Determine the force P required to maintain the equilibrium of the linkage.

SOLUTION

y A = 2u yE = −u yF = −u V = PyA + (100 lb)yE + (150 lb)yF V = P(2u ) + (100 lb)(−u ) + (150 lb)( −u ) dV = 2 P − 100 − 150 = 0 du P = 125 lb

Now, substitute P = 125 lb in expression for V, making V = 0. Thus, V is constant and equilibrium is neutral.


!

PROBLEM 10.69 Two uniform rods, each of mass m and length l, are attached to drums that are connected by a belt as shown. Assuming that no slipping occurs between the belt and the drums, determine the positions of equilibrium of the system and state in each case whether the equilibrium is stable, unstable, or neutral.

SOLUTION W = mg l l ! ! V = W " cos 2θ # − W " cos θ # $2 % $2 % dV l = W ( −2sin 2 + sin θ ) dθ 2 d 2V 1 = W ( −4 cos 2θ − cos θ ) 2 2 dθ

Equilibrium: or Solving

dV Wl = 0: (−2sin 2θ + sin θ ) = 0 2 dθ sin θ (−4cos θ + 1) = 0

θ = 0, 75.5°, 180°, and 284°

Stability:

d 2V l = W (−4 cos 2θ − cos θ ) 2 2 dθ

At θ = 0:

d 2V l = W ( −4 − 1) , 0 2 2 dθ

At θ = 75.5°:

d 2V l = W (−4(−.874) − .25) . 0 2 2 dθ

At θ = 180°:

d 2V l = W ( −4 + 1) , 0 2 2 dθ

At θ = 284°:

d 2V l = W (−4(−.874) − .25) . 0 2 2 dθ

θ = 0, Unstable θ = 75.5°, Stable θ = 180.0°, Unstable θ = 284°, Stable


!

PROBLEM 10.70 Two uniform rods AB and CD, of the same length l, are attached to gears as shown. Knowing that rod AB weighs 3 lb and that rod CD weighs 2 lb, determine the positions of equilibrium of the system and state in each case whether the equilibrium is stable, unstable, or neutral.

SOLUTION l l ! ! V = −WAB " sin θ # − WCD " cos 2θ # 2 2 $ % $ % dV 1 = − WAB l cos θ + WCD l sin 2θ dθ 2 dV =0 dθ

Equilibrium:

1 − WAB l cos θ + WCD l sin 2θ = 0 2 −WAB cos θ + 4WCD sin θ cos θ = 0 ! W −WAB cos θ "1 − CD sin θ # = 0 $ WAB % cos θ = 0 and sin θ =

WAB 4WCD

θ = 90°, θ = 270°, θ = sin −1

WAB 4WCD

(1)

WAB = 3 lb and WCD = 2 lb, We have

For

3 8 θ = 90°, θ = 270°, θ = 22.0°, θ = 158.0°

θ = 90°, θ = 270°, θ = sin −1

1 d 2V ! = " WAB sin θ + 2WCD cos 2θ ) # l 2 dθ $2 %

Stability:

θ = 270° :

d 2V & 1 ' = ( (3) sin(270°) + 2(2) cos(540°) ) l = −5.5l , 0, unstable 2 dθ *2 +

θ = 22.0°:

d 2V & 1 ' = ( (3) sin 220° + 2(2) cos 44.0°) l . 0 Stable 2 dθ *2 +

(2)


!


θ = 90°:

d 2V & 1 ' = ( (3)sin 90° + 2(2) cos180°) l = 2.5l , 0, Unstable 2 dθ *2 +

θ = 150.0° :

d 2V & 1 ' = ( (3) sin158° + 2(2) cos 316° ) l = 3.44l . 0, Stable 2 dθ *2 +


PROBLEM 10.71 Two uniform rods, each of mass m, are attached to gears of equal radii as shown. Determine the positions of equilibrium of the system and state in each case whether the equilibrium is stable, unstable, or neutral.

SOLUTION Potential energy l l ! ! V = W " − sin θ # + W " cos θ # W = mg $ 2 % $2 % l = W (cos θ − sin θ ) 2 dV Wl ( − sin θ − cos θ ) = 2 dθ d 2V Wl (sin θ − cos θ ) = 2 dθ 2

For equilibrium: or Thus

dV = 0: sin θ = − cos θ dθ

tan θ = −1

θ = −45.0° and θ = 135.0°

Stability: At θ = −45.0°:

d 2V Wl = [sin( −45°) − cos 45°] 2 dθ 2 Wl 2 2! = − "" − #,0 2 $ 2 2 #%

θ = −45.0°, Unstable At θ = 135.0°:

d 2V Wl = (sin135° − cos135°) 2 dθ 2 2 2! Wl = + "" #.0 2 $ 2 2 #%

θ = 135.0°, Stable PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1781

PROBLEM 10.72 Two uniform rods, AB and CD, are attached to gears of equal radii as shown. Knowing that WAB = 8 lb and WCD = 4 lb, determine the positions of equilibrium of the system and state in each case whether the equilibrium is stable, unstable, or neutral.

SOLUTION Potential energy l l ! ! V = (3.5 kg × 9.81 m/s 2 ) " − sin θ # + (1.75 kg × 9.81 m/s 2 ) " cos θ # $ 2 % $2 % = (8.5838 N)l (−2sin θ + cos θ ) dV = (8.5838 N)l ( −2 cos θ − sin θ ) dθ d 2V = (8.5838 N)l (2sin θ − cos θ ) dθ 2

Equilibrium: or Thus

dV = 0: − 2 cos θ − sin θ = 0 dθ tan θ = −2

θ = −63.4° and 116.6°

Stability At θ = −63.4°:

d 2V = (8.5838 N)l[2sin(−63.4°) − cos( −63.4°)] dθ 2 = (8.5838 N)l (−1.788 − 0.448) , 0

θ = −63.4°, Unstable At θ = 116.6°:

d 2V = (8.5838 N)l[2sin(116.6°) − cos(116.6°)] dθ 2 = (8.5838 N)l (1.788 + 0.447) . 0

θ = 116.6°, Stable


PROBLEM 10.73 Using the method of Section 10.8, solve Problem 10.39. Determine whether the equilibrium is stable, unstable, or neutral. (Hint: The potential energy corresponding to the couple exerted by a torsion spring is 12 Kθ 2 , where K is the torsional spring constant and θ is the angle of twist.) PROBLEM 10.39 The lever AB is attached to the horizontal shaft BC that passes through a bearing and is welded to a fixed support at C. The torsional spring constant of the shaft BC is K; that is, a couple of magnitude K is required to rotate end B through 1 rad. Knowing that the shaft is untwisted when AB is horizontal, determine the value of θ corresponding to the position of equilibrium when P = 100 N, l = 250 mm, and K = 12.5 N ⋅ m/rad.

SOLUTION Potential energy

V=

1 Kθ 2 − Pl sin θ 2

dV = Kθ − Pl cos θ dθ d 2V = K + Pl sin θ dθ 2

Equilibrium: For

dV K = 0: cos θ = θ dθ Pl P = 100 N, l = 0.25 m., K = 12.5 N ⋅ m/rad 12.5 N ⋅ m/rad θ (100)(0.25 m) = 0.500θ

cos θ =

Solving numerically, we obtain

θ = 1.02967 rad = 59.000°

θ = 59.0°

!

Stability d 2V = (12.5 N ⋅ m/rad) + (100 N)(0.25 m)sin 59.0° . 0 dθ 2

Stable


PROBLEM 10.74 In Problem 10.40, determine whether each of the positions of equilibrium is stable, unstable, or neutral. (See hint for Problem 10.73.) PROBLEM 10.40 Solve Problem 10.39 assuming that P = 350 N, l = 250 mm, and K = 12.5 N ⋅ m/rad. Obtain answers in each of the following quadrants: 0 ,θ , 90°, 270°,θ , 360°, 360° ,θ , 450°.

SOLUTION Potential energy

V=

1 Kθ 2 − Pl sin θ 2

dV = Kθ − Pl cos θ dθ d 2V = K + Pl sin θ dθ 2

Equilibrium

dV K = 0: cos θ = θ dθ Pl

For

P = 350 N, l = 0.250 m and 12.5 N ⋅ m/rad cos θ = θ (350 N)(0.250 m)

or

cos θ =

K = 12.5 N ⋅ m/rad

θ 7

Solving numerically

θ = 1.37333 rad, 5.652 rad, and 6.616 rad

or

θ = 78.7°, 323.8°, 379.1° !

Stability at θ = 78.7°:

d 2V = (12.5 N ⋅ m/rad) + (350 N)(0.250 m) sin 78.7° dθ 2

θ = 78.7°, Stable

!

θ = 324°, Unstable

!

= 98.304 . 0 At θ = 323.8°:

d 2V = (12.5 N ⋅ m/rad) + (350 N)(0.250 m) sin 323.8° dθ 2

= −39.178 N ⋅ m , 0

At At θ = 379.1°:

d 2V = (12.5 N ⋅ m/rad) + (350 N)(0.250 m)sin 379.1° dθ 2

= 44.132 N ⋅ m . 0

θ = 379°, Stable


PROBLEM 10.75 A load W of magnitude 100 lb is applied to the mechanism at C. Knowing that the spring is unstretched when θ = 15°, determine that value of θ corresponding to equilibrium and check that the equilibrium is stable.

SOLUTION We have

yC = l cos θ

π 1 k[ r (θ − θ0 )]2 + WyC θ0 = 15° = rad 2 12 1 = kr 2 (θ − θ0 )2 + Wl cos θ 2

V=

dV = kr 2 (θ − θ0 ) − Wl sin θ dθ

Equilibrium

dV = 0: kr 2 (θ − θ0 ) − wl sin θ = 0 dθ

with

W = 100 lb, R = 50 lb./in., l = 20 in., and r = 5 in.

(1)

π ! (50 lb./in.)(25 in.2 ) " θ − # − (100 lb)(20 in.)sin θ = 0 12 % $ or Solving numerically

0.625θ − sin θ = 0.16362

θ = 1.8145 rad = 103.97° θ = 104.0°

Stability

d 2V = kr 2 − Wl cos θ dθ 2

or

= 1250 − 2000cos θ

For θ = 104.0°:

= 1734 in. ⋅ lb . 0

! (2)

Stable


PROBLEM 10.76 A load W of magnitude 100 lb is applied to the mechanism at C. Knowing that the spring is unstretched when θ = 30°, determine that value of θ corresponding to equilibrium and check that the equilibrium is stable.

SOLUTION

π ! Using the solution of Problem 10.111, particularly Equations (1), with 15° replace by 30° " rad # : $6 % For equilibrium

π! kr 2 " θ − # − Wl sin θ = 0 6% $ k = 50 lb/in., W = 100 lb, r = 5 in., and l = 20 in.

With

π! (50 lb/in.)(25 in.2 ) " θ − # − (100 lb)(20 in.)sin θ = 0 6% $ or

1250θ − 654.5 − 200sin θ = 0

θ = 1.9870 rad = 113.8°

Solving numerically,

θ = 113.8° Stability: Equation (2), Problem 111: d 2V = kr 2 − Wl cos θ dθ 2

or

= 1250 − 2000cos θ

For θ = 113.8°:

= 2057 in. ⋅ lb . 0

Stable


PROBLEM 10.77 A slender rod AB, of weight W, is attached to two blocks A and B that can move freely in the guides shown. The constant of the spring is k, and the spring is unstretched when AB is horizontal. Neglecting the weight of the blocks, derive an equation in θ, W, l, and k that must be satisfied when the rod is in equilibrium.

SOLUTION

Elongation of spring:

Potential energy:

s = l sin θ + l cos θ − l s = l (sin θ + cos θ − 1) 1 2 1 ks − W sin θ W = mg 2 2 l 1 2 = kl (sin θ + cos θ − 1)2 − mg sin θ 2 2

V=

1 dV = kl 2 (sin θ + cos θ − 1)(cos θ − sin θ ) − mgl cos θ dθ 2

Equilibrium:

(1)

dV mg = 0: (sin θ + cos θ − 1)(cos θ − sin θ ) − cos θ = 0 dθ 2kl mg ' & or cos θ ((sin θ + cos θ − 1)(1 − tan θ ) − =0 2kl )+ *


PROBLEM 10.78 A slender rod AB, of weight W, is attached to two blocks A and B that can move freely in the guides shown. Knowing that the spring is unstretched when AB is horizontal, determine three values of θ corresponding to equilibrium when W = 300 lb, l = 16 in., and k = 75 lb/in. State in each case whether the equilibrium is stable, unstable, or neutral.

SOLUTION Using the results of Problem 10.77, particularly the condition of equilibrium mg ' & cos θ ((sin θ + cos θ − 1)(1 − tan θ ) − =0 2kl )+ *

Now, with W = 300 lb, l = 16 in., and k = 75 lb/in. W 300 lb = = 0.25 2kl (16 in.)(75 lb/in.) cos θ [ (sin θ + cos θ − 1)(1 − tan θ ) − 0.25] = 0

Thus:

cos θ = 0 and (sinθ + cos θ − 1)(1 − tan θ ) = 0.25

First equation yields θ = 90°. Solving the second equation by trial, we find θ = 9.39° and 34.16° Values of θ for equilibrium are

θ = 9.39°, 34.2°, and 90.0° Stability we differentiate Eq. (1). d2y 1 = kl 2 [(cos θ − sin θ )(cos θ − sin θ ) + (sin θ + cos θ − 1)(− sin θ − cos θ )] + wl sin θ 2 2 ds W & ' = kl 2 (cos 2 θ + sin 2 θ − 2 cos θ sin θ − sin 2 θ − cos 2 θ − 2 cos θ sin θ + sin θ + cos θ + sin θ ) 2kl * + & ' W ! = kl 2 ("1 + # sin θ + cos θ − 2sin 2θ ) *$ 2kl % + d 2V = kl 2 (1.25sin θ + cos θ − 2sin 2θ ) dθ 2

θ = 9.39° :

d 2V = kl 2 (1.25sin 9.4 + cos 9.4 − 2sin18.8) dθ 2 = kl 2 (+0.55) , 0

Stable



θ = 34.2° :

d 2v = kl 2 (1.25sin 34.2° + cos 34.3° − 2sin 68.4°) 2 dθ = kl 2 (−0.33) , 0

θ = 90.0° :

Unstable

d 2V = kl 2 (1.25sin 90° + cos 90° − 2sin180°) 2 dθ = kl 2 (1.25) . 0

Stable


!

PROBLEM 10.79 A slender rod AB, of weight W, is attached to two blocks A and B that can move freely in the guides shown. Knowing that the spring is unstretched when y = 0, determine the value of y corresponding to equilibrium when W = 80 N, l = 500 mm, and k = 600 N/m.

SOLUTION s = l2 + y2 − l

Deflection of spring = s, where

ds y = 2 dy l − y2

Potential energy:

y 1 2 ks − W 2 2 dV ds 1 = ks − W dy dy 2 dV = k l2 + y2 − l dy V=

(

)

y

1 − W 2 l +y 2

2

! l # y − 1W = k "1 − 2 2 # " 2 l +y % $

dV l = 0: "1 − 2 " dy l + y2 $

Equilibrium

W = 80 N, l = 0.500 m, and k = 600 N/m

Now Then

or Solving numerically,

! #y= 1W # 2 k %

! 0.500 m "1 − # y = 1 (80 N) 2 2 # " 2 (600 N/m) (0.500) + y % $ "1 − 0.500 " 0.25 + y 2 $

! # y = 0.066667 # % y = 0.357 m

y = 357 mm


PROBLEM 10.80 Knowing that both springs are unstretched when y = 0, determine the value of y corresponding to equilibrium when W = 80 N, l = 550 mm, and k = 600 N/m.

SOLUTION Spring deflections S AD = l − l 2 − y 2 S BC = l 2 + y 2 − l 1 2 1 2 y −W kS AD + kS BC 2 2 2 2 1 1 V = k l − l 2 − y2 + k 2 2 V=

)

(

(

)

(

dV y = k l − l 2 − y2 " " l 2 − y2 dy $

2

) − W 2y ! # + k ( l + y − l )" # " % $ l 2 + y2 − l 2

2

! W #− l 2 + y 2 #% 2 y

& ! !' dV l l #) y = W = 0: (" − 1# + " 1 − (" l 2 − y 2 # " dy 2k l 2 + y 2 #% )+ % $ *$

Data: W = 80 N, l = 0.5 m, k = 600 N/m & ' 0.5 0.5 80 ( )y= − = 0.066667 2 2 2 2 2(1200) ( (0.5) − y ) (0.5) y + * +

y = 0.252 m


y = 252 mm


PROBLEM 10.81 A spring AB of constant k is attached to two identical gears as shown. Knowing that the spring is undeformed when θ = 0, determine two values of the angle θ corresponding to equilibrium when P = 30 lb, a = 4 in., b = 3 in., r = 6 in., and k = 5 lb/in. State in each case whether the equilibrium is stable, unstable, or neutral.

SOLUTION Elongation of spring s = 2(a sin θ ) = 2a sin θ 1 2 ks − Pbθ 2 1 = k (2a sin θ ) 2 − Pbθ 2

V=

dV = 4ka 2 sin θ cos θ − Pb dθ = 2ka 2 sin 2θ − Pb

Equilibrium

(1)

dV Pb = 0: sin 2θ = dθ 2ka 2 sin 2θ =

(30 lb)(3 in.) ; sin 2θ = 0.5625 2(5 lb/in.)(4 in.) 2 2θ = 34.229° and 145.771°

θ = 17.11° and 72.9° Stability: We differentiate Eq. (1) d 2V = 4ka 2 cos 2θ dθ 2

θ = 17.11° :

d 2v = 4ka 2 cos 34.2° = 4ka 2 (0.83) . 0 dθ 2

θ = 72.9°:

d 2v = 4ka 2 cos145.8° = 4ka 2 (−0.83) , 0 dθ 2

Stable

Unstable


!

PROBLEM 10.82 A spring AB of constant k is attached to two identical gears as shown. Knowing that the spring is undeformed when θ = 0, and given that a = 60 mm, b = 45 mm, r = 90 mm, and k = 6 kN/m, determine (a) the range of values of P for which a position of equilibrium exists, (b) two values of θ corresponding to equilibrium if the value of P is equal to half the upper limit of the range found in Part a.

SOLUTION Elongation of spring s = 2(a sin θ ) = 2a sin θ

Potential energy 1 2 1 ks − Pbθ = k (2a sin θ )2 − Pbθ 2 2 V = 2ka 2 sin 2 θ − Pbθ V=

Equilibrium dV dV = 0: = 4ka 2 sin θ cos θ − Pb dθ dθ = 2ka 2 sin 2θ − Pb = 0 sin 2θ =

Pmax ;

Pmax b 2ka 2

Pmax (0.045 m)

(a)

(b)

Pb ; For 2ka 2

2(6000 N/m)(0.06 m) 2

For P =

1 Pmax , 2

(1)

=1 =1

Pmax = 960 N

1 sin 2θ = ; 2θ = 30° and 150° 2

θ = 15.00° and 75.0°


!

PROBLEM 10.83 A slender rod AB is attached to two collars A and B that can move freely along the guide rods shown. Knowing that β = 30° and P = Q = 400 N, determine the value of the angle θ corresponding to equilibrium.

SOLUTION

Law of Sines

yA L = sin(90° + β − θ ) sin(90 − β ) yA L = cos(θ − β ) cos β

or

yA = L

cos(θ − β ) cos β

From the figure:

yB = L

cos(θ − β ) − L cos θ cos β

Potential Energy:

& cos(θ − β ) ' cos(θ − β ) V = − PyB − Qy A = − P ( L − L cos θ ) − QL cos cos β β * + & sin(θ − β ) ' dV sin(θ − β ) = − PL ( − + sin θ ) + QL dθ cos β cos β * +

= L( P + Q )

sin(θ − β ) − PL sin θ cos β



dV sin(θ − β ) = 0: L( P + Q ) − PL sin θ = 0 dθ cos β

Equilibrium or

( P + Q) sin(θ − β ) = P sin θ cos β ( P + Q)(sin θ cos β − cos θ sin β ) = P sin θ cos β

or

−( P + Q ) cos θ sin β + Q sin θ cos β = 0 −

P + Q sin β sin θ + =0 Q cos β cos θ

tan θ =

With

P+Q tan β Q

(2)

P = Q = 400 N, β = 30° tan θ =

800 N tan 30° = 1.1547 400 N

θ = 49.1°


!

PROBLEM 10.84 A slender rod AB is attached to two collars A and B that can move freely along the guide rods shown. Knowing that β = 30°, P = 100 N, and Q = 25 N, determine the value of the angle θ corresponding to equilibrium.

SOLUTION Using Equation (2) of Problem 10.83, with P = 100 N, Q = 25 N, and β = 30°, we have (100 N)(25 N) tan 30° (25 N) = 57.735 θ = 89.007°

tan θ =

θ = 89.0°


!

PROBLEM 10.85 Collar A can slide freely on the semicircular rod shown. Knowing that the constant of the spring is k and that the unstretched length of the spring is equal to the radius r, determine the value of θ corresponding to equilibrium when W = 50 lb, r = 9 in., and k = 15 lb/in.

SOLUTION Stretch of Spring

s = AB − r s = 2(r cos θ ) − r s = r (2cos θ − 1)

Potential Energy:

Equilibrium

1 2 ks − Wr sin 2θ W = mg 2 1 V = kr 2 (2 cos θ − 1) 2 − Wr sin 2θ 2 dV = − kr 2 (2 cos θ − 1)2sin θ − 2Wr cos 2θ dθ V=

dV = 0: − kr 2 (2 cos θ − 1) sin θ − Wr cos 2θ = 0 dθ (2cos θ − 1) sin θ W =− cos 2θ kr

Now Then Solving numerically,

W (50 lb) = = 0.37037 kr (15 lb/in.)(9 in.) (2cos θ − 1)sin θ = −0.37037 cos 2θ

θ = 0.95637 rad = 54.8°

θ = 54.8°


!

PROBLEM 10.86 Collar A can slide freely on the semicircular rod shown. Knowing that the constant of the spring is k and that the unstretched length of the spring is equal to the radius r, determine the value of θ corresponding to equilibrium when W = 50 lb, r = 9 in., and k = 15 lb/in.

SOLUTION Stretch of spring s = AB − r = 2( r cos θ ) − r s = r (2cos θ − 1) 1 V = ks 2 − Wr cos 2θ 2 1 2 = kr (2 cos θ − 1) 2 − Wr cos 2θ 2 dV = −kr 2 (2 cos θ − 1)2sin θ + 2Wr sin 2θ dθ

Equilibrium dV = 0: − kr 2 (2 cos θ − 1) sin θ + Wr sin 2θ = 0 dθ − kr 2 (2cos θ − 1) sin θ + Wr (2sin θ cos θ ) = 0 (2cos θ − 1) sin θ W = 2 cos θ kr

or Now Then

W (50 lb) = = 0.37037 kr (15 lb/in.)(9 in.) 2cos θ − 1 = 0.37037 2 cos θ

θ = 37.4°

Solving


!

PROBLEM 10.87 Cart B, which weighs 75 kN, rolls along a sloping track that forms an angle β with the horizontal. The spring constant is 5 kN/m, and the spring is unstretched when x = 0. Determine the distance x corresponding to equilibrium for the angle β indicated. Angle β = 30°.

SOLUTION

x = (4 m) tan θ

(1)

yB = x sin β = 4 tan θ sin β AC = (4 m) cos θ

For x = 0, Stretch of spring.

( AC )0 = 4 m s = AC − ( AC )0 = V=

4 1 ! − 4 = 4" − 1# cos θ $ cos θ %

1 2 ks − (75 kN) yB 2 2

=

1 1 ! (5 kN/m) 16 " − 1# − (75 kN)4 tan θ sin β 2 $ cos θ %

1 sin β dV ! sin θ = 80 " − 1# − 300 2 dθ cos 2 θ $ cos θ % cos θ

Equilibrium Given: Eq. (2): Solve by trial and error: Eq. (1):

1 dV ! = 0: " − 1# sin θ = 3.75sin β dθ $ cos θ %

(2)

β = 30°, sin θ = 0.5 1 ! " cos θ − 1# sin θ = 3.75(0.5) = 1.875 $ %

θ = 70.46° x = (4 m) tan 70.46°

x = 11.27 m


!

PROBLEM 10.88 Cart B, which weighs 75 kN, rolls along a sloping track that forms an angle β with the horizontal. The spring constant is 5 kN/m, and the spring is unstretched when x = 0. Determine the distance x corresponding to equilibrium for the angle β indicated. Angle β = 60°.

SOLUTION

x = (4 m) tan θ

(1)

yB = x sin β = 4 tan θ sin β AC = (4 m) cos θ

For x = 0, Stretch of spring.

( AC )0 = 4 m s = AC − ( AC )0 = V=

4 1 ! − 4 = 4" − 1# cos θ $ cos θ %

1 2 ks − (75 kN) yB 2 2

=

1 1 ! (5 kN/m) 16 " − 1# − (75 kN)4 tan θ sin β 2 $ cos θ %

1 sin β dV ! sin θ = 80 " − 1# − 300 2 dθ cos 2 θ $ cos θ % cos θ

Equilibrium Given: Eq. (2):

1 dV ! = 0: " − 1# sin θ = 3.75sin β dθ $ cos θ %

(2)

β = 60°, sin θ = 0.86603 1 ! " cos θ − 1# sin θ = 3.75(0.86603) = 3.2476 $ %


θ = 76.67°

Eq. (1):

x = (4 m) tan 26.67°

x = 16.88 m


!

PROBLEM 10.89 A vertical bar AD is attached to two springs of constant k and is in equilibrium in the position shown. Determine the range of values of the magnitude P of two equal and opposite vertical forces P and −P for which the equilibrium position is stable if (a) AB = CD, (b) AB = 2CD.

SOLUTION For both (a) and (b): Since P and − P are vertical, they form a couple of moment M P = + Pl sin θ

The forces F and −F exerted by springs must, therefore, also form a couple, with moment M F = − Fa cos θ

We have

dU = M P dθ + M F dθ = ( Pl sin θ − Fa cos θ )dθ

but

Thus,

1 ! F = ks = k " a sin θ # 2 $ % 1 ! dU = " Pl sin θ − ka 2 sin θ cos θ # dθ 2 $ %

From Equation (10.19), page 580, we have dV = − dU = − Pl sin θ dθ +

or

1 2 ka sin 2θ dθ 4

dV 1 = − Pl sin θ + ka 2 sin 2θ dθ 4

and

d 2V 1 = − Pl cos θ + ka 2 cos 2θ 2 2 dθ

For θ = 0:

d 2V 1 = − Pl + ka 2 2 2 dθ

(1)



For Stability:

d 2V 1 . 0, − Pl + ka 2 . 0 2 2 dθ

or (for Parts a and b)

P,

Note: To check that equilibrium is unstable for P =

ka 2 2l

, we differentiate (1) twice:

d 3V = + Pl sin θ − ka 2 sin 2θ = 0, for 3 dθ d 4V = Pl cos θ − 2ka 2 cos 2θ dθ 4

For θ = 0

Thus, equilibrium is unstable when

ka 2 2l

θ = 0,

d 4V ka 2 2 = − = − 2ka 2 , 0 2 Pl ka 2 dθ 4 P=

ka 2 ! 2l


!

PROBLEM 10.90 Rod AB is attached to a hinge at A and to two springs, each of constant k. If h = 25 in., d = 12 in., and W = 80 lb, determine the range of values of k for which the equilibrium of the rod is stable in the position shown. Each spring can act in either tension or compression.

SOLUTION We have

xC = d sin θ

yB = h cos θ

Potential Energy:

1 ! V = 2 " kxC2 + WyB # $2 % = kd 2 sin 2 θ + Wh cos θ

dV = 2kd 2 sin θ cos θ − Wh sin θ dθ = kd 2 sin 2θ − Wh sin θ

Then

d 2V = 2kd 2 cos 2θ − Wh cos θ dθ 2

and

(1)

For equilibrium position θ = 0 to be stable, we must have d 2V = 2kd 2 − Wh . 0 2 dθ 1 kd 2 . Wh 2

or Note: For kd 2 = 12 Wh, we have

d 2V dθ 2

(2)

= 0, so that we must determine which is the first derivative that is not

equal to zero. Differentiating Equation (1), we write d 3V = −4kd 2 sin 2θ + Wh sin θ = 0 dθ 3 d 4V = −8kd 2 cos 2θ + Wh cos θ dθ 2

For θ = 0:

for θ = 0

d 4V = −8kd 2 + Wh dθ 4



Since kd 2 = 12 Wh,

d 4V dθ 4

= −4Wh + Wh , 0, we conclude that the equilibrium is unstable for kd 2 = 12 Wh and

the . sign in Equation (2) is correct. With Equation (2) gives or

W = 80 lb, h = 25 in., and d = 12 in. 1 k (12 in.) 2 . (80 lb)(25 in.) 2 k . 6.944 lb/in.

k . 6.94 lb/in.


!

PROBLEM 10.91 Rod AB is attached to a hinge at A and to two springs, each of constant k. If h = 45 in., k = 6 lb/in., and W = 60 lb, determine the smallest distance d for which the equilibrium of the rod is stable in the position shown. Each spring can act in either tension or compression.

SOLUTION Using Equation (2) of Problem 10.90 with h = 45 in., k = 6 lb/in., and W = 60 lb 1 (6 lb/in.)d 2 . (60 lb)(45 in.) 2

or

d 2 . 225 in.2 d . 15.0000 in.

smallest d = 15.00 in.

!


PROBLEM 10.92 Two bars are attached to a single spring of constant k that is unstretched when the bars are vertical. Determine the range of values of P for which the equilibrium of the system is stable in the position shown.

SOLUTION s=

2L L sin φ = sin θ 3 3

For small values of φ and θ

φ = 2θ 2L L ! 1 cos θ # + ks 2 V = P " cos φ + 3 3 $ % 2 2

V dV dθ

d 2V dθ 2

1 2L PL ! (cos 2θ + 2cos θ ) + k " sin θ # = 3 2 $ 3 % 2 2 PL (−2sin 2θ − 2sin θ ) + kL sin θ cos θ = 3 9 2 PL (2sin 2θ + 2sin θ ) + kL2 sin 2θ =− 3 9 4 PL (4 cos 2θ + 2 cos θ ) + kL2 cos 2θ =− 3 9

when θ = 0:

d 2V 6 PL 4 2 =− + kL 2 3 9 dθ

For stability:

4 d 2V . 0, − 2 PL + kL2 . 0 2 9 dθ

2 P , kL 9

!


PROBLEM 10.93 Two bars are attached to a single spring of constant k that is unstretched when the bars are vertical. Determine the range of values of P for which the equilibrium of the system is stable in the position shown.

SOLUTION a=

2L L sin θ = sin φ 3 3

For small values of φ and θ

φ = 2θ s=

L sin θ 3

2L L ! 1 cos θ + cos φ # + ks 2 V = P" 3 $ 3 % 2 =

1 PL L ! (2 cos θ + cos 2θ ) + k " sin θ # 3 2 $3 %

2

dV PL kL2 (−2sin θ − 2sin 2θ ) + sin θ cos θ = dθ 3 9 dV kL2 2 PL (sin θ + sin 2θ ) + sin 2θ =− dθ 3 18 d 2V kL2 2 PL (cos 2 cos 2 ) cos 2θ θ θ = − + + 3 9 dθ 2

when θ = 0:

dV 2 kL2 2 = − PL + 9 dθ 2

For stability:

d 2V kL2 0, 2 . − + .0 PL 9 dθ 2

P,

1 kL 18

!


PROBLEM 10.94 Two bars AB and BC are attached to a single spring of constant k that is unstretched when the bars are vertical. Determine the range of values of P for which the equilibrium of the system is stable in the position shown.

SOLUTION s = (l − a) sin θ V = P (2l cos θ ) + = 2 Pl cos θ +

1 2 ks 2

1 k (l − a) 2 sin 2 θ 2

dV = −2 Pl sin θ + k (l − a) 2 sin θ cos θ dθ 1 = −2 Pl sin θ + k (l − a )2 sin 2θ 2 d 2V = −2 Pl cos θ + k (l − a) 2 cos 2θ dθ 2

when

Stability:

θ = 0:

(1)

d 2V = −2 Pl + k (l − a) 2 dθ 2

d 2V . 0: − 2 Pl + k (l − a) 2 > 0 dθ 2

To check whether equilibrium is unstable for P =

k (l − a )2 2l

P,

k (l − a) 2 2l

, we differentiate

Eq. (1) twice: d 3V = 2 Pl sin θ − 2k (l − a) 2 sin 2θ = 0, For θ = 0 dθ 3 d 4V = 2 Pl cos θ − 4k (l − a) 2 cos 2θ dθ 4



For G = 0 and

P=

k (l − a) 2 2l

d 4V = 2 Pl − 4k (l − a )2 dθ 4 = k (l − a)3 − 4k (l − a) 2 , 0

Thus equil is unstable for

P=

k (l − a) 2 ! 2l


PROBLEM 10.95 The horizontal bar BEH is connected to three vertical bars. The collar at E can slide freely on bar DF. Determine the range of values of Q for which the equilibrium of the system is stable in the position shown when a = 24 in., b = 20 in., and P = 150 lb.

SOLUTION

First note For small values of θ and φ : or

A = a sin θ = b sin φ aθ = bφ a b

φ= θ V = P( a + b) cos φ − 2Q (a + b) cos θ

& ' a ! = ( a + b) ( P cos " θ # − 2Q cos θ ) b $ % * + & a ' dV a ! = ( a + b) ( − P sin " θ # + 2Q sin θ ) dθ b b $ % * + & a2 ' d 2V a ! a b ( ) = + ( − 2 P cos " θ # + 2Q cos θ ) 2 dθ $b % * b +

when θ = 0:

! d 2V a2 ( a b ) P + 2Q ## = + − " 2 2 " dθ $ b %



Stability:

d 2V a2 . − 0: P + 2Q . 0 dθ 2 b2 b2 Q a2

(1)!

a2 P 2b 2

(2)!

P,2 Q.

or with

P = 150 lb, a = 24 in., and b = 20 in.

Equation (1):

Q.

(24 in.) 2 (150 lb) = 108.000 lb 2(20 in.) 2 Q . 108.0 lb

For stability


PROBLEM 10.96 The horizontal bar BEH is connected to three vertical bars. The collar at E can slide freely on bar DF. Determine the range of values of P for which the equilibrium of the system is stable in the position shown when a = 150 mm, b = 200 mm, and Q = 45 N.

SOLUTION Using Equation (2) of Problem 10.95 with Q = 45 N, a = 150 mm, and b = 200 mm

Equation (2)

(200 mm) 2 (45 N) (150 mm) 2 = 160.000 N

P,2

P , 160.0 N

For stability


PROBLEM 10.97* Bars AB and BC, each of length l and of negligible weight, are attached to two springs, each of constant k. The springs are undeformed, and the system is in equilibrium when θ1 = θ2 = 0. Determine the range of values of P for which the equilibrium position is stable.

SOLUTION We have

xB = l sin θ xC = l sin θ1 + l sin θ 2 yC = l cos θ1 + l cos θ 2 V = PyC +

or

V = Pl (cos θ1 + cos θ 2 ) +

1 2 1 2 kxB + kxC 2 2 1 2 kl &*sin 2 θ1 + (sin θ1 + sin θ 2 ) 2 '+ 2

For small values of θ1 and θ 2 : 1 1 sin θ1 ≈ θ1 , sin θ 2 ≈ θ 2 , cosθ1 ≈ 1 − θ12 , cos θ 2 ≈ 1 − θ 22 2 2

Then

and

θ2 θ2 ! 1 2 V = Pl "1 − 1 + 1 − 2 # + kl 2 &θ12 + (θ1 + θ 2 ) ' " # * + 2 2 % 2 $

∂V = − Plθ1 + kl 2 [θ1 + (θ1 + θ 2 )] ∂ θ1 ∂V = − Plθ 2 + kl 2 (θ1 + θ 2 ) ∂ θ2 ∂ 2V = − Pl + 2kl 2 2 ∂ θ1

∂ 2V = − Pl + kl 2 2 ∂ θ2

∂ 2V = kl 2 ∂ θ1∂ θ 2 Stability: Conditions for stability (see Page 583).



For

θ1 = θ 2 = 0:

∂V ∂V = = 0 (condition satisfied) ∂ θ1 ∂ θ 2 2

∂ 2V ! ∂ 2V ∂ 2V "" ∂ θ ∂ θ ## − ∂ θ 2 ∂ θ 2 , 0 1 2 $ 1 2% Substituting

(kl 2 ) 2 − (− Pl + 2kl 2 )(− Pl + kl ) , 0 k 2 l 4 − P 2 l 2 + 3Pkl 3 − 2k 2 l 4 , 0 P 2 − 3klP + k 2 l 2 . 0 3− 5 kl or 2

Solving

P,

or

P , 0.382kl or

P.

3+ 5 kl 2

P . 2.62kl

∂ 2V . 0: − Pl + 2kl 2 . 0 ∂ θ12 or

1 P , kl 2

∂ 2V . 0: − Pl + kl 2 . 0 2 ∂ θ2 or

P , kl

Therefore, all conditions for stable equilibrium are satisfied when

0 # P , 0.382kl


PROBLEM 10.98* Solve Problem 10.97 knowing that l = 800 mm and k = 2.5 kN/m. PROBLEM 10.97* Bars AB and BC, each of length l and of negligible weight, are attached to two springs, each of constant k. The springs are undeformed, and the system is in equilibrium when θ1 = θ 2 = 0. Determine the range of values of P for which the equilibrium position is stable.

SOLUTION From the analysis of Problem 10.97 with l = 800 mm and k = 2.5 kN/m P , 0.382kl = 0.382(2500 N/m)(0.8 m) = 764 N

P , 764 N


PROBLEM 10.99* Two rods of negligible weight are attached to drums of radius r that are connected by a belt and spring of constant k. Knowing that the spring is undeformed when the rods are vertical, determine the range of values of P for which the equilibrium position θ1 = θ2 = 0 is stable.

SOLUTION

Left end of spring moves from a to b. Right end of spring moves from a′ to b′. Elongation of spring s = a′b′ − ab = rθ1 − rθ 2 = r (θ1 − θ 2 ) 1 2 ks + pl cos θ1 − wl cos θ 2 2 1 = kr 2 (θ1 − θ 2 ) 2 + pl cos θ1 − wl cos θ 2 2

V=

∂v = kr 2 (θ1 − θ 2 ) − pl sin θ1 ∂ θ1 ∂v = − kr 2 (θ1 − θ 2 ) + wl sin θ 2 ∂ θ2 ∂ 2v = kr 2 − pl cos θ1 ∂ θ12 ∂ 2v = kr 2 + wl cos θ 2 ∂ θ 22 ∂ 2v = − kr 2 ∂ θ1∂ θ 2



For

θ1 = θ 2 = 0:

∂ 2v ∂ 2v r 2v 2 2 , , = − = + + = − kr 2 kr pl kr wl 2 2 ∂ θ , ∂ θ2 ∂ θ1 ∂ θ2

Stability conditions for stability (see Page 583) 2

∂ 2v ! ∂ 2v ∂ 2v ,0 ⋅ "" ## − 2 2 $ ∂ θ1∂ θ 2 % ∂ θ1 ∂ θ 2 (kr 2 ) 2 − (kr 2 − pl )(kr 2 + wl ) , 0 pl (kr 2 + wl ) − kr 2 wl , 0 P,

wkr 2 kr 2 " P ; , l "$ kr 2 + wl

! # + w #%

W kr l

2

v

∂ 2v kr 2 2 . : kr − pl . 0; P , l ∂ θ12

v

P,

We choose:

kr 2 " l "$

! # + W #%

W kr 2 l


PROBLEM 10.100* Solve Problem 10.99 knowing that k = 20 lb/in., r = 3 in., l = 6 in., and (a) W = 15 lb, (b) W = 60 lb. PROBLEM 10.99* Two rods of negligible weight are attached to drums of radius r that are connected by a belt and spring of constant k. Knowing that the spring is undeformed when the rods are vertical, determine the range of values of P for which the equilibrium position θ1 = θ 2 = 0 is stable.

SOLUTION k = 20 lb/in. r = 3 in. l = 6 in. 2

kr (20 lb/in.)(3 in.) 2 = = 30 lb l 6 in.

(a)

W = 15 lb: P , (30 lb)

15 lb (30 lb) + (15 lb)

P , 10.00 lb

(b)

W = 60 lb: P , (30 lb)

60 lb (30 lb) + (60 lb)

P , 20.0 lb


PROBLEM 10.101 Determine the vertical force P that must be applied at G to maintain the equilibrium of the linkage.

SOLUTION

Assuming δ y A it follows

δ yC =

12 δ y A = 1.5δ y A 8

δ yE = δ yC = 1.5δ y A δ yD =

18 δ y A = 3(1.5δ y A ) = 4.5δ y A 6

δ yG =

10 10 δ y A = (1.5δ y A ) = 2.5δ y A 6 6

Then, by virtual work

δ U = 0: (300 lb)δ y A − (100 lb)δ yD + Pδ yG = 0 300δ y A − 100(4.5δ y A ) + P(2.5δ y A ) = 0 300 − 450 + 2.5 P = 0 P = 60.0 lb

P = 60.0 lb


PROBLEM 10.102 Determine the couple M that must be applied to member DEFG to maintain the equilibrium of the linkage.

SOLUTION

Assume δ y A :

δ yC =

12 δ y A = 1.5δ y A , 8

δ yD =

18 δ yE = 3(1.5δ y A ) = 4.5δ y A 6

δφ =

δ yE 6

=

δ yE = δ yC = 1.5δ y A

1.5δ y A 1 = δ yA 6 4

Virtual Work:

δ U = 0:

(300 lb)δ y A − (100 lb)δ yD + M δφ = 0 1 ! 300δ y A − 100(1.5δ y A ) + M " δ y A # = 0 $4 % 1 300 − 450 + M = 0 4 M = + 600 lb ⋅ in.

M = 600 lb ⋅ in.



SOLUTION We have xB = l sin θ

δ xB = l cos θδθ y A = l cos θ δ y A = −l sin θδθ Virtual Work:

δ U = 0: M δθ − Pδ xB + Pδ y A = 0 M δθ − P(l cos θδθ ) + P(−l sin θδθ ) = 0 M = Pl (sin θ + cos θ )


PROBLEM 10.104 Collars A and B are connected by the wire AB and can slide freely on the rods shown. Knowing that the length of the wire is 440 mm and that the weight W of collar A is 90 N, determine the magnitude of the force P required to maintain equilibrium of the system when (a) c = 80 mm, (b) c = 280 mm.

SOLUTION Since AB = 440 mm, we have (440 mm) 2 = (240 mm) 2 + b2 + c 2 b 2 = (440) 2 − (240) 2 − c 2 = 136 × 103 − c 2 2b δ b = −2c δ c

Differentiate:

c b

δb = − δc δb = −

cδ c 136 × 103 − c 2

Virtual Work:

δ U = 0: Pδ c + W δ b = 0 Pδ c = W

For W = 90 N: (a)

136 × 103 − c 2 c

(1)

136 × 103 − c 2

When c = 80 mm: Eq. (1):

(b)

P = 90

cδ c

P = 90

80 136 × 103 − (80) 2

P = 20.0 N

When c = 280 mm: Eq. (1)

P = 90

280 136 × 103 − (280) 2

P = 105.0 N


PROBLEM 10.105 Collar B can slide along rod AC and is attached by a pin to a block that can slide in the vertical slot shown. Derive an expression for the magnitude of the couple M required to maintain equilibrium.

SOLUTION R tan(90° − θ ) − Rδθ δ yB = 2 cos (90° − θ ) − Rδθ δ yB = 2 sin θ yB =

Virtual Work:

δ U = 0:

δ U = − M δθ − Pδ yB = 0 − M δθ + PR

1 δθ = 0 sin 2 θ M=

PR sin 2 θ

M = PR csc2 θ


PROBLEM 10.106 A slender rod of length l is attached to a collar at B and rests on a portion of a circular cylinder of radius r. Neglecting the effect of friction, determine the value of θ corresponding to the equilibrium position of the mechanism when l = 200 mm, r = 60 mm, P = 40 N, and Q = 80 N.

SOLUTION Geometry

OC = r OC r = cos θ = OB xB

r cos θ r sin θ δ xB = δθ cos 2 θ y A = l cos θ xB =

δ y A = −l sin θδθ Virtual Work:

δ U = 0: P( −δ y A ) − Qδ xB = 0 Pl sin θδθ − Q

r sin θ δθ = 0 cos 2 θ cos 2 θ =

Qr Pl

(1)

Then, with l = 200 mm, r = 60 mm, P = 40 N, and Q = 80 N cos 2 θ =

or

(80 N)(60 mm) = 0.6 (40 N)(200 mm)

θ = 39.231°

θ = 39.2°


PROBLEM 10.107 A horizontal force P of magnitude 40 lb is applied to the mechanism at C. The constant of the spring is k = 9 lb/in., and the spring is unstretched when θ = 0. Neglecting the weight of the mechanism, determine the value of θ corresponding to equilibrium.

SOLUTION s = rθ δ s = rδθ

Spring is unstretched at θ = 0° FSP = ks = krθ xC = l sin θ

δ xC = l cos θδθ Virtual Work:

δ U = 0: Pδ xC − FSPδ s = 0 P(l cos θδθ ) − krθ (rδθ ) = 0 Pl θ = 2 cos θ kr

or Thus

or

(40 lb)(12 in.) θ = 2 cos θ (9 lb/in.)(5 in.)

θ cos θ

= 2.1333

θ = 1.054 rad = 60.39°

θ = 60.4°


!

PROBLEM 10.108 Two identical rods ABC and DBE are connected by a pin at B and by a spring CE. Knowing that the spring is 4 in. long when unstretched and that the constant of the spring is 8 lb/in., determine the distance x corresponding to equilibrium when a 24-lb load is applied at E as shown.

SOLUTION

Deformation of spring s = EC − 4 in. =

2x −4 3 2

x 1 1 2 2x ! ks − (24 lb) = (8 lb/in.) " − 4 # − 4 x 2 6 2 3 $ % dV 2x !2 = 8" − 4# − 4 dx $ 3 %3 V=

Equilibrium:

dV =0 dx

16 2 x ! − 4# − 4 = 0 " 3$ 3 % 2x 3! − 4 = 4" # 3 $ 16 % 2x 3 =4+ 3 4

x = 7.13 in.


!

PROBLEM 10.109 Solve Problem 10.108 assuming that the 24-lb load is applied at C instead of E. PROBLEM 10.108 Two identical rods ABC and DBE are connected by a pin at B and by a spring CE. Knowing that the spring is 4 in. long when unstretched and that the constant of the spring is 8 lb/in., determine the distance x corresponding to equilibrium when a 24-lb load is applied at E as shown.

SOLUTION

Deformation of spring s = EC − 4 in. =

2x −4 3 2

1 5x 1 2x ! = (8 lb/in.) " − 4 # − 20 x V = ks 2 − (24 lb) 2 6 2 $ 3 % 2x dV !2 = 8" − 4 # − 20 3 dx $ %3

Equilibrium:

dV =0 dx

16 2 x ! − 4 # − 20 = 0 3 "$ 3 % 2x 3! − 4 = 20 " # 3 16 $ % 2x = 4 + 3.75 3

x = 11.63 in.


!

PROBLEM 10.110 Two uniform rods, each of mass m and length l, are attached to gears as shown. For the range 0 # θ # 180°, determine the positions of equilibrium of the system and state in each case whether the equilibrium is stable, unstable, or neutral.

SOLUTION

Potential energy

l l ! ! V = W " cos1.5θ # + W " cos θ # W = mg $2 % $2 % dV Wl Wl = ( −1.5sin1.5θ ) + (− sin θ ) dθ 2 2 Wl = − (1.5sin1.5θ + sin θ ) 2 2 d V Wl = − (2.25cos1.5θ + cos θ ) 2 2 dθ

For equilibrium

dV = 0: 1.5sin1.5θ + sin θ = 0 dθ

Solutions: One solution, by inspection, is θ = 0, and a second angle less than 180° can be found numerically:

θ = 2.4042 rad = 137.8° Now

d 2V Wl = − (2.25cos1.5θ + cos θ ) 2 2 dθ

At θ = 0:

d 2V Wl = − (2.25cos 0° + cos 0°) 2 2 dθ =−

Wl (3.25)(, 0) 2

θ = 0, Unstable

! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1828

!


At θ = 137.8°:

d 2V Wl = − [2.25cos(1.5 × 137.8°) + cos137.8°] 2 2 dθ =

Wl (2.75)(. 0) 2

θ = 137.8°, Stable


PROBLEM 10.111 A homogeneous hemisphere of radius r is placed on an incline as shown. Assuming that friction is sufficient to prevent slipping between the hemisphere and the incline, determine the angle θ corresponding to equilibrium when β = 10°.

SOLUTION

Detail 3 CG = r 8 V = W (− rθ sin β − (CG ) cos θ ) 3 dV ! = W " − r sin β + r sin θ # 8 dθ $ %

Equilibrium:

dV 3 = 0, − sin β + sin θ = 0 dθ 8 3 sin β = sin θ 8

For β = 10°

3 sin10° = sin θ 8

(1) sin θ = 0.46306, θ = 27.6°


PROBLEM 10.112 A homogeneous hemisphere of radius r is placed on an incline as shown. Assuming that friction is sufficient to prevent slipping between the hemisphere and the incline, determine (a) the largest angle β for which a position of equilibrium exists, (b) the angle θ corresponding to equilibrium when the angle β is equal to half the value found in Part a.

SOLUTION

Detail 3 CG = r 8 V = W (− rθ sin β − (CG ) cos θ ) 3 dV ! = W " − r sin β + r sin θ # 8 dθ $ % 3 dV = 0, − sin β + sin θ = 0 dθ 8

Equilibrium:

3 sin β = sin θ 8

(a)

For β max , θ = 90° Eq. (1)

(b)

(1)

3 3 sin β max = sin 90°, sin β max = = 22.02° 8 8

β max = 22.0°

When β = 12 β max = 11.01° Eq. (1)

3 sin11.01° = sin θ ; sin θ = 0.5093 8

θ = 30.6°

Note: We can also use ∆CGD and law of sines to derive Eq. (1). CG sin β sin θ 3 ; sin β = sin θ ; sin β = sin θ = CG CD CD 8


!

Resolucao Mecanica Vetorial para Engenheiros 9ed cap 6 10 estatica

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