Mechanics of Materials 7th Edition Beer Johnson Chapter 6

CHAPTER 6

s

PROBLEM 6.1

s

Three full-size 50  100-mm boards are nailed together to form a beam that is subjected to a vertical shear of 1500 N. Knowing that the allowable shearing force in each nail is 400 N, determine the largest longitudinal spacing s that can be used between each pair of nails.

50 mm 50 mm 50 mm

100 mm

SOLUTION I 

1 3 1 (100)(150)3  28.125  106 mm 4 bh  12 12

 28.125  106 m 4

A  (100)(50)  5000 mm2 y1  50 mm

Q  A y1  250  103 mm3  250  106 m3 VQ (1500)(250  106 )   13.3333  103 N/m 6 I 28.125  10 2 Fnail (2)(400) qs  2Fnail s    60.0  103 m q 13.3333  103 q 

s  60.0 mm 

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s

PROBLEM 6.2

s

For the built-up beam of Prob. 6.1, determine the allowable shear if the spacing between each pair of nails is s  45 mm.

50 mm 50 mm

PROBLEM 6.1 Three full-size 50  100-mm boards are nailed together to form a beam that is subjected to a vertical shear of 1500 N. Knowing that the allowable shearing force in each nail is 400 N, determine the largest longitudinal spacing s that can be used between each pair of nails.

50 mm

100 mm

SOLUTION I 

1 3 1 bh  (100)(150)3  28.125  106 mm 4 12 12

 28.125  106 m 4

A  (100)(50)  5000 mm 2

y1  50 mm Q  A y1  250  103 mm3  250  106 m3 q 

Eliminating q, Solving for V,

VQ I

qs  2 Fnail

VQ 2 Fnail  I s

V 

2IFnail (2)(28.125  106 )(400)   2.00  103 N Qs (250  106 )(45  103 ) V  2.00 kN ◄


PROBLEM 6.3

s s

Three boards, each 2 in. thick, are nailed together to form a beam that is subjected to a vertical shear. Knowing that the allowable shearing force in each nail is 150 lb, determine the allowable shear if the spacing s between the nails is 3 in.

s 2 in. 4 in. 2 in. 2 in.

6 in.

SOLUTION 1 3 bh  Ad 2 12 1 (6)(2)3  (6)(2)(3) 2  112 in 4  12 1 3 1 (2)(4)3  10.6667 in 4 I2  bh  12 12 I1 

I 3  I1  112 in 4 I  I1  I 2  I 3  234.67 in 4 Q  A1 y1  (6)(2)(3)  36 in 3 qs  Fnail

(1)

VQ I

(2)

q

Dividing Eq. (2) by Eq. (1),

1 VQ  s Fnail I V 

Fnail I (150)(234.67)  (36)(3) Qs

V  326 lb 


PROBLEM 6.4

s s s

A square box beam is made of two 20  80-mm planks and two 20  120-mm planks nailed together as shown. Knowing that the spacing between the nails is s  30 mm and that the vertical shear in the beam is V  1200 N, determine (a) the shearing force in each nail, (b) the maximum shearing stress in the beam.

20 mm 80 mm 20 mm

120 mm

SOLUTION 1 1 b2 h23  b1h13 12 12 1 1  (120)(120)3  (80)(80)3  13.8667  106 mm 4 12 12 6 4  13.8667  10 m

I

(a)

A1  (120)(20)  2400 mm 2 y1  50 mm Q1  A1 y1  120  103 mm3  120  106 m3 VQ (1200)(120  106 )   10.3846  103 N/m 6 I 13.8667  10 qs  2 Fnail q

Fnail 

(b)

qs (10.3846  103 )(30  103 )  2 2

Fnail  155.8 N 

Q  Q1  (2)(20)(40)(20)  120  103  32  103  152  103 mm3  152  106 m3

 max 

VQ (1200)(152  106 )  It (13.8667  106 )(2  20  103 )

 max  329 kPa 

 329  103 Pa


16 ⫻ 200 mm

PROBLEM 6.5 The American Standard rolled-steel beam shown has been reinforced by attaching to it two 16  200-mm plates, using 18-mm-diameter bolts spaced longitudinally every 120 mm. Knowing that the average allowable shearing stress in the bolts is 90 MPa, determine the largest permissible vertical shearing force.

S310 ⫻ 52

SOLUTION Calculate moment of inertia: A (mm 2 ) 3200 6650 3200

Part Top plate S310  52 Bot. plate  *d



d (mm) *

160.5 0

*

160.5

Ad 2 (106 mm 4 ) 82.43 82.43 164.86

I (106 mm 4 ) 0.07 95.3 0.07 95.44

305 16   160.5 mm 2 2

I  Ad 2  I  260.3  106 mm 4  260.3  106 m 4 Q  Aplate d plate  (3200)(160.5)  513.6  103 mm3  513.6  106 m3 Abolt 

 4

2  d bolt

 4

(18  103 )2  254.47  106 m 2

Fbolt   all Abolt  (90  106 )(254.47  106 )  22.90  103 N qs  2 Fbolt q

VQ I

q V 

2Fbolt (2)(22.90  103 )   381.7  103 N/m 3 s 120  10

Iq (260.3  106 )(381.7  103 )   193.5  103 N Q 513.6  106

V  193.5 kN 


y

16 in. ⫻

1 2

in.

C12 ⫻ 20.7 z

C

PROBLEM 6.6 The beam shown is fabricated by connecting two channel shapes and two plates, using bolts of 34 -in. diameter spaced longitudinally every 7.5 in. Determine the average shearing stress in the bolts caused by a shearing force of 25 kips parallel to the y axis.

SOLUTION C12  20.7: d  12.00 in., I x  129 in 4 For top plate,

y

12.00  1  1       6.25 in. 2  2  2  3

1 1 1 I t  (16)    (16)   (6.25) 2  312.667 in 4 12 2 2

For bottom plate,

I b  312.667 in 4

Moment of inertia of fabricated beam: I  (2)(129)  312.667  312.667  883.33 in 4 1 Q  Aplate yplate  (16)   (6.25)  50 in 3 2 VQ (25)(50) q   1.41510 kips/in. I 883.33 1 1 Fbolt  qs    (1.41510)(7.5)  5.3066 kips 2 2

 bolt



(d bolt ) 2 

 3

2

 0.44179 in 2 4 4  4  F 5.3066  bolt   12.01 ksi Abolt 0.44179

Abolt 

 bolt  12.01 ksi 

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PROBLEM 6.7

y

A columm is fabricated by connecting the rolled-steel members shown by bolts of 34 -in. diameter spaced longitudinally every 5 in. Determine the average shearing stress in the bolts caused by a shearing force of 30 kips parallel to the y axis.

C8 ⫻ 13.7

z

C

S10 ⫻ 25.4

SOLUTION Geometry: d f     (tw )C  2 s 10.0  0.303  5.303 in. 2 x  0.534 in. 

y1  f  x  5.303  0.534  4.769 in. Determine moment of inertia. d (in.) 4.769 0 4.769

A(in 2 ) Part C8  13.7 4.04 S10  25.4 7.45 C8  13.7 4.04 

Ad 2 (in 4 ) 91.88 0 91.88 183.76

I (in 4 ) 1.52 123 1.52 126.04

I  Ad 2  I  183.76  126.04  309.8 in 4 Q  A y1  (4.04)(4.769)  19.2668 in 3 VQ (30)(19.2668)   1.86573 kip/in. I 309.8 1 1  qs    (1.86573)(5)  4.6643 kips 2 2

q Fbolt

Abolt 

 bolt 

 4

2  d bolt

 3

2

2    0.44179 in 4 4

4.6643 Fbolt   10.56 ksi Abolt 0.44179

 bolt  10.56 ksi 



PROBLEM 6.8 The composite beam shown is fabricated by connecting two W6  20 rolled-steel members, using bolts of 85 -in. diameter spaced longitudinally every 6 in. Knowing that the average allowable shearing stress in the bolts is 10.5 ksi, determine the largest allowable vertical shear in the beam.

SOLUTION W6  20: A  5.87 in 2 , d  6.20 in., I x  41.4 in 4 y 

Composite:

1 d  3.1 in. 2

I  2[41.4  (5.87)(3.1) 2 ]  195.621 in 4 Q  A y  (5.87)(3.1)  18.197 in 3

Bolts:

d 

Shear:

 5

2

2    0.30680 in 4 8   all Abolt  (10.5)(0.30680)  3.2214 kips

Abolt  Fbolt

5 in.,  all  10.5 ksi, s  6 in. 8

q

2Fbolt (2)(3.2214)   1.07380 kips/in. 6 s

q

VQ I

V 

Iq (195.621)(1.0780)  Q 18.197

V  11.54 kips 



15 15

30

PROBLEM 6.9

15 15 20

a

0.5 m

For the beam and loading shown, consider section n-n and determine (a) the largest shearing stress in that section, (b) the shearing stress at point a.

72 kN

20 n 40

120

n

20 20

0.8 m

1.5 m

90 Dimensions in mm

SOLUTION M B  0: 2.3 A  (1.5)(72)  0

A  46.957 kN  V  A  46.957 kN

At section n-n,

Calculate moment of inertia: 1  1  1 I  2  (15)(40)3   2  (15)(80)3   (30)(1203 ) 12  12  12  5.76  106 mm 4  5.76  106 m 4

ta  30 mm  0.030 m

At a,

Qa  (30  20)(50)  30  103 mm3  30  106 m3

a 

VQa (46.957  103 )(30  106 )  Ita (5.76  106 )(0.030)

 8.15  106 Pa = 8.15 MPa tb  60 mm  0.060 m

At b,

Qb  Qa  (60  20)(30)  30  103  36  103  66  103 mm3  66  106 m 4

b 

VQb (46.957  103 )(66  106 )   8.97  106 Pa  8.97 MPa Itb (5.76  106 )(0.060)

t NA  90 mm  0.090 m

At NA,

QNA  Qb  (90  20)(10)  66  103  18  103  84  103 mm3  84  106 m3

 NA  (a)

VQNA (46.957  103 )(84  106 )   7.61  106 Pa  7.61 MPa It NA (5.76  106 )(0.090)

 max occurs at b.

 max  8.97 MPa   a  8.15 MPa 

(b)



PROBLEM 6.10

0.3 m n 10 kN

40 mm

a 100 mm

12 mm 150 mm 12 mm

n 200 mm

1.5 m


SOLUTION At section n-n, V  10 kN.

I  I1  4I 2 

1  1  b1h13  4  b2h23  A2d 22  12 12  



 1   1 (100)(150)3  4   (50)(12)3  (50)(12)(69)2  12  12  

 28.125  106  4 0.0072  106  2.8566  106   39.58  106 mm 4  39.58  106 m 4

(a)

Q  A1 y1  2 A2 y2  (100)(75)(37.5)  (2)(50)(12)(69)  364.05  103 mm3  364.05  106 m3

t  100 mm = 0.100 m

 max 

(b)

VQ (10  103 )(364.05  106 )   920  103 Pa It (39.58  106 )(0.100)

 max  920 kPa 

Q  A1 y1  2 A2 y2  (100)(40)(55)  (2)(50)(12)(69)  302.8  103 mm3  302.8  106 m3 t  100 mm  0.100 m

a 

VQ (10  103 )(302.8  106 )   765  103 Pa It (39.58  106 )(0.100)

 a  765 kPa 


PROBLEM 6.11

t 3 in.

18 in.

a

t

t

n 25 kips n

3 in. 3 in.


t = 0.25 in.

25 in. 8 in.

SOLUTION

I 

1 1 2 (8 in.)(9 in.)3  (7.25 in.)(8.5 in.)3  (0.25 in.)(3 in.)3 12 12 12

I  113.8 in 4

V = 25 kips M  (25 kips)(18 in.) = 450 kip  in.

(a)

 m:

At neutral axis,

thickness  0.25 in.

Q  2(3 in.  0.25 in.)(3 in.)  (7.5 in.)(0.25 in.)(4.375 in.)  (4.25 in.)(0.25 in.)(2.125 in.)

Q  4.5 in 3  8.203 in 3  2.258 in 3  14.96 in 3 t  0.25 in. V (25 kips)(14.96 in 3 ) m  Q  It (113.8 in 4 )(0.25 in.)

 m  13.15 ksi ◄



PROBLEM 6.11 (Continued)

(b)

 a: At point a,

t = 0.25 in.

See sketch above.

Qa  4.5 in 3  8.203 in 3  12.70 in 3

a 

VQa (25 kips)(12.70 in 3 )  It (113.8 in 4 )(0.25 in.)

 a  11.16 ksi ◄



1 2

10 kips 10 kips 8 in.

PROBLEM 6.12

in.

a n

1 2

4 in.

in.

n 16 in.

12 in.


4 in.

16 in.

SOLUTION By symmetry, RA  RB . Fy  0: RA  RB  10  10  0 RA  RB  10 kips

From the shear diagram,

V  10 kips at n-n. 1 1 b2h23  b1h13 12 12 1 1  (4)(4)3  (3)(3)3  14.5833 in 4 12 12

I 

(a)

1 1 Q  A1 y1  A2 y2  (3)   (1.75)  (2)   (2)(1)  4.625 in 3 2 2 t 

 max 

(b)

1 1   1 in. 2 2 (10)(4.625) VQ  (14.5833)(1) It

 max  3.17 ksi 

1 Q  Ay  (4)   (1.75)  3.5 in 3 2 1 1 t    1 in. 2 2

 

(10)(3.5) VQ  (14.5833)(1) It

 a  2.40 ksi 



10

40

Dimensions in mm

30

10

PROBLEM 6.13 For a beam having the cross section shown, determine the largest allowable vertical shear if the shearing stress is not to exceed 60 MPa.

40

40

SOLUTION Calculate moment of inertia.

1  1 I  2  (10 mm)(120 mm)3   (30 mm)(40 mm)3 12  12  2[1.440  106 mm 4 ]  0.160  106 mm 4  3.04  106 mm 4 I  3.04  106 m 4 Assume that  m occurs at point a. t  10 mm  0.01 m Q  (10 mm  40 mm)(40 mm)  16  103 mm3 Q  16  106 m3 For

 all  60 MPa, 60  106 Pa 

 m   all  V (16  106 m3 ) (3.04  106 m 4 )(0.01m)

VQ It

V  114.0 kN 

Check  at neutral axis: t  50 mm  0.05 m Q  2[(10  60)(30)]  (30  20)(10)  42  103 m3  42  106 m3

 NA 

VQ (114.0 kN)(42  106 m3 )   31.5 MPa  60 MPa It (3.04  106 m 4 )(0.05 m)

OK



10

30

PROBLEM 6.14

10

For a beam having the cross section shown, determine the largest allowable vertical shear if the shearing stress is not to exceed 60 MPa.

10 30 40

Dimensions in mm

30 10

SOLUTION

Calculate moment of inertia. I 

1 1  (50 mm)(120 mm)3  2  (30 mm)4  (30 mm  30 mm)(35 mm)2  12 12 

I  7.2  106 mm 4  2[1.170  106 mm 4 ]  4.86  106 mm 4  4.86  106 m 4

Assume that  m occurs at point a.

t  2(10 mm)  0.02 m Q  (10 mm  50 mm)(55 mm)  2[(10 mm  30 mm)(35 mm)]  48.5  103 mm3  48.5  106 m3




For  all  60 MPa,

 m   all  60  106 Pa 

Check  at neutral axis:

VQ It

V (48.5  106 m3 ) (4.86  106 m 4 )(0.02 m)

V  120.3 kN 

t  50 mm  0.05 m Q  (50  60)(30)  (30  30)(35)  58.5  103 mm3

 

VQ (120.3 kN)(58.5  106 m3 )   29.0 MPa  60 MPa It (4.86  106 m 4 )(0.05 m)

OK



PROBLEM 6.15

1.5 in.

For a timber beam having the cross section shown, determine the largest allowable vertical shear if the shearing stress is not to exceed 150 psi.

2 in.

4 in. w = 2.5 in. 2 in.

1.5 in.

SOLUTION I 

1 (1.5  83  2(0.5)(4)3 ) 12

I  69.333 in 4

 all  150 psi

Q  (1.5 in.)(2 in.)(3 in.)  9 in 3;

At point a:

m 

VQ ; It

150 psi 

t  1.5 in.

V (9 in 3 ) ; (69.333 in 4 )(1.5 in.)

V  1733 lb 

At neutral axis:

Q  (1.5 in.)(2 in.)(3 in.)  (2.5 in.)(2 in.)(1 in.)  14 in 3, t  2.6 in.

m 

VQ ; It

150 psi 

V (14 in 3 ) ; (69.333 in 4 )(2.5 in.)

V  1857 lb  V  1733 lb ◄

We choose smaller shear.


PROBLEM 6.16

220 mm 12 mm

W250 3 58

Two steel plates of 12  220-mm rectangular cross section are welded to the W250  58 beam as shown. Determine the largest allowable vertical shear if the shearing stress in the beam is not to exceed 90 MPa.

252 mm

12 mm

SOLUTION Calculate moment of inertia. Part Top plate

A(mm 2 ) 2640

W250 × 58

7420

0

Bot. plate

2640

*132

Ad 2 (106 mm 4 ) 45.999

d (mm) *132

0



*d 

I (106 mm 4 ) 0.032

87.3

45.999

0.032

91.999

87.364

Ay (103 mm3 )

252 12  2 2

I  Ad 2  I  179.363  106 mm 4  179.363  106 m 4

 max occurs at neutral axis. t  8.0 mm  8.0  103 m

Part  Top plate

A(mm 2 )

y (mm)

2640

132

348.48

 Top flange

2740.5

119.25

326.805

56.25

50.625

 Half web

900

725.91



Dimensions in mm:  12  220 ;  13.5  203 ; 

8.0  112.5

Q  A y  725.91  103 mm3  725.91  106 m3

 

VQ It

V 

It (179.363  106 )(8.0  103 )(90  106 )  Q 725.91  106  177.9  103 N

V  177.9 kN 



PROBLEM 6.17

A

B

A

(a)

B

Two W8  31 rolled-steel sections may be welded at A and B in either of the two ways shown in order to form a composite beam. Knowing that for each weld the allowable shearing force is 3000 lb per inch of weld, determine for each arrangement the maximum allowable vertical shear in the composite beam.

(b)

SOLUTION

A  9.12 in 2 I x  110 in 4

W8  31

I y  37.1 in 4 I  2[ I x  Ad 2 ]

(a)

 2[110 in 4  (9.12 in 2 )(4 in.)2 ]

I  511.84 in 4 Q  Ay  (9.12 in 2 )(4 in.)  36.4 in 3

For two welds each with allowable shearing force of 3 kips/in., q  2(3 kips/in.)  6 kips/in.

q

VQ ; I

6 kips/in. 

V (36.4 in 3 ) (511.84 in 4 )

V  84.2 kips 

(b) I  2[ I y  Ad 2 ]  2[37.1 in 4  (9.12 in 2 )(4 in.) 2 ]

I  366.04 in 4 Q  Ay  (9.12 in 2 )(4 in.)  36.4 in 3 q  6 kips/in. (same as in part a)

q

V (36.4 in 3 ) VQ ; 6 kips/in.  I (366.04 in 4 )

V  60.2 kips 



2.4 kN

4.8 kN

PROBLEM 6.18

7.2 kN b

B

C

D

A

150 mm

E

1m

1m

1m

For the beam and loading shown, determine the minimum required width b, knowing that for the and grade of timber used,  all  12 MPa  all  825 kPa.

0.5 m

SOLUTION M D  0: 3 A  (2)(2.4)  (1)(4.8)  (0.5)(7.2)  0 A  2 kN 

Draw the shear and bending moment diagrams. |V |max  7.2 kN  7.2  103 N |M |max  3.6 kN  m  3.6  103 N  m



Bending:

S min 

M S |M |max



3.6  103  12  106  300  106 m3  300  103 mm3 1 S  bh 2 6 6 S (6)(300  103 ) b 2   80 mm h (150)2

For a rectangular section,

Shear: Maximum shearing stress occurs at the neutral axis of bending for a rectangular section. 1 1 1 bh, y  h, Q  Ay  bh 2 2 4 8 1 3 I  bh t  b 12 V ( 18 bh 2 ) 3V VQ    3 1 It ( 12 bh )(b) 2 bh

A

b

3V (3)(7.2  103 )   87.3  103 m 2h (2)(150  103 )(825  103 )

b  87.3 mm 

The required value of b is the larger one.



PROBLEM 6.19 P L/2 A

b

L/2

C

B

h

A timber beam AB of length L and rectangular cross section carries a single concentrated load P at its midpoint C. (a) Show that the ratio  m / m of the maximum values of the shearing and normal stresses in the beam is equal to 2h/L, where h and L are, respectively, the depth and the length of the beam. (b) Determine the depth h and the width b of the beam, knowing that L  2 m, P  40 kN,  m  960 kPa, and  m  12 MPa.

SOLUTION RA  RB  P/2 

Reactions: P 2

(1)

Vmax  RA 

(2)

A  bh for rectangular section.

(3)

m 

(4)

M max 

(5)

S 

(6)

m 

3 Vmax 3P  for rectangular section. 2 A 4bh PL 4

1 2 bh for rectangular section. 6 M max 3PL  S 2bh 2

m h    m 2L

(a)

(b)

Solving for h,

h

2 L m

(2)(2)(960  103 )  320  103 m 6 12  10

h  320 mm 

3P (3)(40  103 )   97.7  103 m 4h m (4)(320  103 )(960  103 )

b  97.7 mm 

m



Solving Eq. (3) for b, b



PROBLEM 6.20 w

b

A

B C L/4

D L/2

L/4

h

A timber beam AB of length L and rectangular cross section carries a uniformly distributed load w and is supported as shown. (a) Show that the ratio  m / m of the maximum values of the shearing and normal stresses in the beam is equal to 2h/L, where h and L are, respectively, the depth and the length of the beam. (b) Determine the depth h and the width b of the beam, knowing that L  5 m, w  8 kN/m,  m  1.08 MPa, and  m  12 MPa.

SOLUTION RA  RB  From shear diagram, For rectangular section,

|V |m 

wL 4

(1)

A  bh

m  From bending moment diagram,

wL 2

(2)

3 Vm 3wL  2 A 8bh

|M |m 

(3)

wL2 32

(4)

1 2 bh 6

(5)

|M |m 3wL2  S 16 bh 2

(6)

For a rectangular cross section, S 

m  (a)

Dividing Eq. (3) by Eq. (6),

(b)

Solving for h, h

L m (5)(1.08  106 )   225  103 m 6 2 m (2)(12  10 )

m 2h   L m

h  225 mm 

Solving Eq. (3) for b, b

3wL (3)(8  103 )(5)  8h m (8)(225  103 )(1.08  106 )

 61.7  103 m

b  61.7 mm 



25 kips

PROBLEM 6.21

25 kips

n

3 4

7.25 in.

in. b a

B

A n 20 in.

10 in.

3 4

20 in.

1.5 in. 1.5 in. 3 4

in.

For the beam and loading shown, consider section n-n and determine the shearing stress at (a) point a, (b) point b.

in.

8 in.

SOLUTION RA  RB  25 kips V  25 kips

At section n-n,

Locate centroid and calculate moment of inertia. Part

A(in 2 )

y (in.)

A y (in 3 )

d(in.)

Ad 2 (in 4 )

I (in 4 )



4.875

6.875

33.52

2.244

24.55

0.23



10.875

3.625

39.42

1.006

11.01

47.68



15.75

35.56

47.91

72.94

Ay 72.94   4.631 in. A 15.75 I  Ad 2  I  35.56  47.91  83.47 in 4

Y 

(a)

3 Qa  Ay    (1.5)(4.631  0.75)  4.366 in 3 4 3 t   0.75 in. 4

a  (b)

VQ (25)(4.366)  It (83.47)(0.75)

 a  1.744 ksi 

3 Qb  Ay    (3)(4.631  1.5)  7.045 in 3 4 t  0.75 in.

b 

VQ (25)(7.045)  It (83.47)(0.75)

 b  2.81 ksi 



PROBLEM 6.22

160 mm

180 kN

a

n A

20 mm

100 mm

B

b

n 500 mm

30 mm


500 mm 30 mm

30 mm 20 mm

SOLUTION |V |max  90 kN

Draw the shear diagram. Part

A(mm 2 )

y (mm)

A y (103 mm3 )



3200

90

288



1600

40



1600

40



6400

Ad 2 (106 mm 4 )

I (106 mm 4 )

25

2.000

0.1067

64

25

1.000

0.8533

64

25

1.000

0.8533

4.000

1.8133

d(mm)

416 Ay 416  103   65 mm A 6400

Y 

I  Ad 2  I  (4.000  1.8133)  106 mm 4  5.8133  106 mm 4  5.8133  106 m 4 (a)

A  (80)(20)  1600 mm 2 y  25 mm Qa  Ay  40  103 mm3  40  106 m3

a 

VQa (90  103 )(40  106 )   31.0  106 Pa It (5.8133  106 )(20  103 )

 a  31.0 MPa  (b)

A  (30)(20)  600 mm 2

y  65  15  50 mm

Qb  Ay  30  103 mm3  30  106 m3

b 

VQb (90  103 )(30  106 )   23.2  106 Pa It (5.8133  106 )(20  103 )

 b  23.2 MPa 



25 kips

PROBLEM 6.23

25 kips

n

7.25 in.

3 4

in. b a

B

A n 20 in.

10 in.

3 4

20 in.

1.5 in. 1.5 in. 3 4

in.

For the beam and loading shown, determine the largest shearing stress in section n-n.

in.

8 in.

SOLUTION RA  RB  25 kips V  25 kips

At section n-n,

Locate centroid and calculate moment of inertia.

Part

A(in 2 )

y (in.)

Ay (in 3 )

d(in.)

Ad 2 (in 4 )



4.875

6.875

33.52

2.244

24.55

0.23



10.875

3.625

39.42

1.006

11.01

47.68



15.75

35.56

47.91

72.94

I (in 4 )

A y 72.94   4.631 in. A 15.75 I  Ad 2  I  35.56  47.91  83.47 in 4

Y 

Largest shearing stress occurs on section through centroid of entire cross section.

3  4.631   8.042 in 3 Q  Ay    (4.631)   4  2  3 t   0.75 in. 4 VQ (25)(8.042)   It (83.47)(0.75)

 m  3.21 ksi 



PROBLEM 6.24

160 mm

180 kN

a

n A

For the beam and loading shown, determine the largest shearing stress in section n-n.

100 mm

B

b

n 500 mm

20 mm

30 mm

500 mm 30 mm

30 mm 20 mm

SOLUTION |V |max  90 kN

Draw the shear diagram. Part

A (mm 2 )

x



3200

90

288

25

2.000

0.1067



1600

40

64

25

1.000

0.8533



1600

40

64

25

1.000

0.8533



6400

4.000

1.8133

Ay (103 mm3 ) d(mm) Ad 2 (106 mm 4 ) I (106 mm 4 )

416 Y 

Ay 416  103   65 mm A 6400

I  Ad 2  I  (4.000  1.8133)  106 mm 4  5.8133  106 mm 4  5.8133  106 m 4 Part

A(mm 2 )



3200



300

7.5

2.25



300

7.5

2.25

y (mm) 25



Ay (103 mm3 )

80

84.5 Q  Ay  84.5  103 mm3  84.5  106 m3 t  (2)(20)  40 mm  40  103 m

 max 

VQ (90  103 )(84.5  106 )  It (5.8133  106 )(40  103 )

 32.7  106 Pa

 m  32.7 MPa 



PROBLEM 6.25

c

A beam having the cross section shown is subjected to a vertical shear V. Determine (a) the horizontal line along which the shearing stress is maximum, (b) the constant k in the following expression for the maximum shearing stress

 max  k

V A

where A is the cross-sectional area of the beam.

SOLUTION I 

For semicircle,

As 

 4

 2

c4

c2

Q  As y  (a)

 max occurs at center where

(b)

 max 

and

A   c2

y 

 2

c2 

4c 3 4c 2  c3 3 3 t  2c 

V  2 c3 VQ 4V 4V   43   2 It 3A c  2c 3 c 4

k 

4  1.333  3



PROBLEM 6.26 tm

rm


 max  k

V A


SOLUTION A  2 rmtm

For a thin-walled circular section,

J  Arm2  2 rm3tm ,

For a semicircular arc,

y 

I 

1 J   rm3tm 2

2rm



As   rmtm Q  As y   rmtm

(a)

t  2tm

(b)

 max 

2rm



 2rm2tm

at neutral axis where maximum occurs.  VQ V (2rm2tm ) V 2V    3 It A ( rmtm )(2tm )  rmtm

k  2.00 



PROBLEM 6.27 h


h

 max  k

b

V A


SOLUTION 1  A  2  bh   bh 2 

 1  1 I  2  bh3   bh3  12  6

For a cut at location y, where y  h, A( y) 

1  by  by 2  y  2 h  2h

y ( y)  h 

2 y 3

Q( y)  Ay  t ( y) 

 ( y) 

(a)

by 2 by 3  2 3h

by h

2 VQ 6 h by 2 by 3 V   y  y  V 3     3    2    It 3h bh   h  bh by 2  h  

To find location of maximum of  , set

d  0. dy

d V y  2 [3  4 m ]  0 dy h bh (b)

 ( ym ) 

2 V  3 V 3  9 V  1.125 3    2     bh   4  A  4   8 bh

ym 3 1  , i.e.,  h from neutral axis.  h 4 4 k  1.125 



PROBLEM 6.28 A beam having the cross section shown is subjected to a vertical shear V. Determine (a) the horizontal line along which the shearing stress is maximum, (b) the constant k in the following expression for the maximum shearing stress

b

h

 max  k

V A


SOLUTION A

1 bh 2

I 

1 3 bh 36

For a cut at location y, A( y) 

1  by  by 2  y   2 h  2h

y ( y) 

2 2 h y 3 3

Q( y)  Ay  t ( y) 

by 2 (h  y ) 3

by h

2

by VQ V 3 (h  y) 12Vy(h  y) 12V    (hy  y 2 )  ( y)  3 3 1 bh3 ) by It bh bh ( 36 h

(a)

To find location of maximum of  , set

d  0. dy

d 12V ( h  2 ym )  0  dy bh3 (b)

m

2 12V 12V  1 2  1   3V 3V 2 (hym  ym )     h   h   3 3 2A bh bh  2  2   bh

ym 

1 h, i.e., at mid-height  2 k 

3  1.500  2



PROBLEM 6.29

2 in.

The built-up timber beam shown is subjected to a vertical shear of 1200 lb. Knowing that the allowable shearing force in the nails is 75 lb, determine the largest permissible spacing s of the nails.

10 in.

2 in. s

s

s

2 in.

SOLUTION 1 b1h13  A1d12 12 1 (2)(2)3  (2)(2)(4)2  65.333 in 4  12 1 1 (2)(10)3  166.67 in 4 I2  b2h23  12 12 I1 

I  4 I1  I 2  428 in 4 Q  Q1  A1 y1  (2)(2)(4)  16 in 3

q

(1200)(16) VQ   44.86 lb/in. I 428

Fnail  qs s 

75 Fnail   1.672 in. q 44.86





PROBLEM 6.30

50 mm

The built-up beam shown is made by gluing together two 20  250-mm plywood strips and two 50  100-mm planks. Knowing that the allowable average shearing stress in the glued joints is 350 kPa, determine the largest permissible vertical shear in the beam.

150 mm

50 mm 20 mm

20 mm

100 mm

SOLUTION I 

1 1 (140)(250)3  (100)(150)3  154.167  106 mm 4 12 12  154.167  106 m 4

Q  Ay  (100)(50)(100)  500  103 mm3  500  106 m3 t  50 mm  50 mm  100 mm  100  103 m

 

VQ It

V 

It (154.167  106 )(100  103 )(350  103 )  Q 500  106

 10.79  103 N

V  10.79 kN 



1.5

0.8

0.8 4

PROBLEM 6.31

1.5 0.8

A

B 3.2

The built-up beam was made by gluing together several wooden planks. Knowing that the beam is subjected to a 1200-lb vertical shear, determine the average shearing stress in the glued joint (a) at A, (b) at B.

0.8 Dimensions in inches

SOLUTION 1 1  I  2  (0.8)(4.8)3  (7)(0.8)3  (7)(0.8)(2.0) 2  12 12   60.143 in 4

(a)

Aa  (1.5)(0.8)  1.2 in 2 Qa  Aa ya  2.4 in

ya  2.0 in.

3

ta  0.8 in.

a  (b)

VQa (1200)(2.4)  Ita (60.143)(0.8)

Ab  (4)(0.8)  3.2 in 2

 a  59.9 psi  yb  2.0 in.

Qb  Ab yb  (3.2)(2.0)  6.4 in 3

tb  (2)(0.8)  1.6 in.

b 

VQb (1200)(6.4)   Itb (60.143)(1.6)

 b  79.8 psi 



PROBLEM 6.32

20 mm 60 mm 20 mm A 20 mm 30 mm

B

Several wooden planks are glued together to form the box beam shown. Knowing that the beam is subjected to a vertical shear of 3 kN, determine the average shearing stress in the glued joint (a) at A, (b) at B.

20 mm 30 mm 20 mm

SOLUTION IA 

1 3 1 (60)(20)3  (60)(20)(50)2 bh  Ad 2  12 12

 3.04  106 mm 4 1 3 1 (60)(20)3  0.04  106 mm 4 bh  12 12 1 3 1 (20)(120)3  2.88  106 mm 4 IC  bh  12 12 IB 

I  2 I A  I B  2 I C  11.88  106 mm 4  11.88  106 m 4 QA  Ay  (60)(20)(50)  60  103 mm3  60  106 m3 t  20 mm  20 mm  40 mm  40  103 m (a)

A 

VQA (3  103 )(60  106 )   379  103 Pa It (11.88  10 6 )(40  103 )

 A  379 kPa 

QB  0 (b)

B 

VQB 0 It

B  0 



PROBLEM 6.33

50

300

50

B A 100

A 50 C

400

x

The built-up wooden beam shown is subjected to a vertical shear of 8 kN. Knowing that the nails are spaced longitudinally every 60 mm at A and every 25 mm at B, determine the shearing force in the nails (a) at A, (b) at B. (Given: I x  1.504  109 mm 4.)

50 A

A

200

B Dimensions in mm

SOLUTION I x  1.504  109 mm 4  1504  106 m 4 s A  60 mm  0.060 m sB  25 mm  0.025 m (a)

QA  Q1  A1 y1  (50)(100)(150)  750  103 mm3  750  106 m3 FA  q As A 

VQ1s A (8  103 )(750  106 )(0.060)  I 1504  106

FA  239 N  (b)

Q2  A2 y2  (300)(50)(175)  2625  103 mm3 QB  2Q1  Q2  4125  103 mm3  4125  106 m3 FB  qB sB 

VQB sB (8  103 )(4125  106 )(0.025)  I 1504  106

FB  549 N 



105 mm

PROBLEM 6.34

A

Knowing that a W360  122 rolled-steel beam is subjected to a 250-kN vertical shear, determine the shearing stress (a) at point A, (b) at the centroid C of the section.

C

SOLUTION For W360  122, d  363 mm, bF  257 mm, t F  21.70 mm, tw  13.0 mm I  367  106 mm 4  367  106 m 4

(a)

Aa  (105)(21.70)  2278.5 mm 2

ya 

d tF 363 21.70     170.65 mm 2 2 2 2

Qa  Aa ya  388.8  103 mm3  388.8  106 m3 ta  t F  21.70 mm  21.7  103 m

a  (b)

VQa (250  103 )(388.8  106 )   12.21  106 Pa Ita (367  106 )(21.7  103 )

 a  12.21 MPa 

A1  bF t F  (257)(21.70)  5577 mm 2

y1 

d tF 363 21.70     170.65 mm 2 2 2 2

d  A2  tw   t F   (13.0)(159.8)  2077 mm 2 2  y2 

1d    t F   79.9 mm 2 2 

Qc  Ay  (5577)(170.65)  (2077)(79.9)  1117.7  103 mm3  1117.7  106 m3 tc  t w  13.0 mm  13  103 m

c 

VQc (250  103 )(1117.7  106 )   58.6  106 Pa 6 3 Itc (367  10 )(13  10 )

 c  58.6 MPa 



PROBLEM 6.35

b 12

12

6 a

80 6

40

An extruded aluminum beam has the cross section shown. Knowing that the vertical shear in the beam is 150 kN, determine the shearing stress at (a) point a, (b) point b.

150 Dimensions in mm

SOLUTION I 

1 1 (150)(80)3  (126)(68)3 12 12

 3.098  106 mm 4  3.0985  106 m 4

(a)

Qa  A1 y1  2 A2 y2  (126)(6)(37)  (2)(12)(40)(20)  47.172  103 mm3  47.172  106 m3

ta  (2)(12)  24 mm  0.024 m

a 

VQa (150  103 )(47.172  106 )   95.2  106 Pa 6 Ita (3.0985  10 )(0.024)

 a  95.2 MPa  (b)

Qb  A1 y1  (126)(6)(37)  27.972  103 mm3  27.972  106 m3 tb  (2)(6)  12 mm  0.012 m

b 

VQb (150  103 )(27.972  106 )   112.8  106 Pa Itb (3.0985  106 )(0.012)

 b  112.8 MPa 



6

PROBLEM 6.36

b

12

12

80

a 6

An extruded aluminum beam has the cross section shown. Knowing that the vertical shear in the beam is 150 kN, determine the shearing stress at (a) point a, (b) point b.

40

80 Dimensions in mm

SOLUTION I

1 1 (80)(80)3  (56)(68)3  1.9460  106 mm 4 12 12

 1.946  106 m 4

(a)

Qa  A1 y1  2 A2 y2  (56)(6)(37)  (2)(12)(40)(20)  31.632  103 mm3  31.632  106 m3 ta  (2)(12)  24 mm  0.024 m

a 

VQa (150  103 )(31.632  106 )   101.6  106 Pa 6 Ita (1.946  10 )(0.024)

 a  101.6 MPa  (b)

Qb  A1 y1  (56)(6)(37)  12.432  103 mm3  12.432  106 m3 tb  (2)(6)  12 mm  0.012 m

b 

VQb (150  103 )(12.432  106 )   79.9  106 Pa Itb (1.946  106 )(0.012)

 b  79.9 MPa 



PROBLEM 6.37

120 50

50

Knowing that a given vertical shear V causes a maximum shearing stress of 75 MPa in an extruded beam having the cross section shown, determine the shearing stress at the three points indicated.

10

c b

40 30

a

160 30 40 10

20

20

Dimensions in mm

SOLUTION

 

VQ It

τ is proportional to Q/t.

Qc  (30)(10)(75)

Point c:

 22.5  103 mm3 tc  10 mm Qc /tc  2250 mm 2 Qb  Qc  (20)(50)(55)

Point b:

 77.5  103 mm3 tb  20 mm Qb /tb  3875 mm 2

Qa  2Qb  (120)(30)(15)

Point a:

 209  103 mm3 ta  120 mm Qa /ta  1741.67 mm 2

 m   b  75 MPa

(Q/t )m occurs at b.

a Qa /ta

a 1741.67 mm

2

 

b Qb /tb



c Qc /tc

75 MPa a  2 3875 mm 2250 mm 2

 a  33.7 MPa   b  75.0 MPa   c  43.5 MPa  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.


0.5 in.

d

5 in.

PROBLEM 6.38

d

The vertical shear is 1200 lb in a beam having the cross section shown. Knowing that d  4 in., determine the shearing stress at (a) point a, (b) point b.

b 8 in.

a 4 in.

0.5 in.

SOLUTION

1 (4)(0.5)3  (4)(0.5)(3.75) 2  28.167 in 4 12 1 I 2  (5)(4)3  106.67 in 4 3 I1 

I  4 I1  2I 2  326 in 4

(a)

Qa  2 A1 y1  A2 y2  (2)(4)(0.5)(3.75)  (5)(4)(2)  55 in 3

ta  5 in.

a  (b)

VQa (1200)(55)   40.5 psi Ita (326)(5)



Qb  A1 y1  (4)(0.5)(3.75)  7.5 in 4

tb  0.5 in.

b 

VQb (1200)(7.5)   55.2 psi  Itb (326)(0.5)





d

0.5 in.

5 in.

PROBLEM 6.39

d

The vertical shear is 1200 lb in a beam having the cross section shown. Determine (a) the distance d for which a  b, (b) the corresponding shearing stress at points a and b.

b a

8 in.

4 in.

0.5 in.

SOLUTION A1  0.5d in 2 ,

y1  3.75 in., tb  0.5 in.

A2  (5)(4)  20 in 2 ,

y2  2 in., ta  5 in.

Qb  A1 y1  1.875d in 3

b 

VQb V 1.875d Vd   3.75 Itb I 0.5 I

Qa  A2 y2  2Qb  (20)(2)  (2)(1.875d )  40  3.75d ta  5 in.

(a)

a 

VQa V (40  3.75d ) V Vd Vd   8  0.75   b  3.75 Ita I (5) I I I

8  0.75d  3.75d

d 

8  2.6667 in. 3 d  2.67 in .

(b)

1 (2.6667)(0.5)3  (2.6667)(0.5)(3.75) 2  18.780 in 4 12 1 I 2  (0.5)(4)3  106.667 in 4 3 I1 

I  4 I1  2 I 2  288.46 in 4

 a   b  3.75

Vd (3.75)(1200)(2.6667)  288.46 I

 a  41.6 psi 



c

PROBLEM 6.40

b

d

1.25 in.

a

e 1.25 in.

1.25 in.

The extruded aluminum beam has a uniform wall thickness of 18 in. Knowing that the vertical shear in the beam is 2 kips, determine the corresponding shearing stress at each of the five points indicated.

1.25 in.

SOLUTION I

1 1 (2.50)(2.50)3  (2.125)(2.25)3  1.23812 in 4  12 12

t  0.125 in.at all sections. V  2 kips 

Qa  0

a 

VQa It

 a  0 

 1.25  3 Qb  (0.125)(1.25)    0.097656 in   2 

b 

VQb (2)(0.097656)  It (1.23812)(0.125)

 b  1.262 ksi 

Qc  Qb  (1.0625)(0.125)(1.1875)  0.25537 in.2 

c 

VQc (2)(0.25537)   It (1.23812)(0.125)

 c  3.30 ksi 

Qd  2Qc  (0.125)2 (1.1875)  0.52929 

d 

VQd (2)(0.52929)  It (1.23812)(0.125)

  d  6.84 ksi 

 1.125  Qe  Qd  (0.125)(1.125)    0.60839   2 

e 

VQ (2)(0.60839)  It (1.23812)(0.125)

 e  7.86 ksi 



c

PROBLEM 6.41

b

d

1.25 in.

e 1.25 in.

1.25 in.

a

The extruded aluminum beam has a uniform wall thickness of 18 in. Knowing that the vertical shear in the beam is 2 kips, determine the corresponding shearing stress at each of the five points indicated.

1.25 in.

SOLUTION I

1 1 (2.50)(2.50)3  (2.125)(2.25)3  1.23812 in 4 12 12

Add symmetric points c’, b’, and a’. Qe  0   1.125  3 Qd  (0.125)(1.125)    0.079102 in  2 

td  0.125 in. 

Qc  Qd  (0.125)2 (1.1875)  0.097657 in 4

tc  0.25 in. 

Qb  Qc  (2)(1.0625)(0.125)(1.1875)  0.41309 in 3

tb  0.25 in. 

 1.25  3 Qa  Qb  (2)(0.125)(1.25)    0.60840 in 2  

ta  0.25 in. 

a 

VQa (2)(0.60840)  Ita (1.23812)(0.25)

 a  3.93 ksi 

b 

VQb (2)(0.41309)  Itb (1.23812)(0.25)

 b  2.67 ksi 

c 

VQc (2)(0.097657)  Itc (1.23812)(0.25)

 c  0.631 ksi 

d 

VQd (2)(0.079102)  Itd (1.23812)(0.125)

 d  1.02 ksi 

e 

VQe Ite

 e  0 



PROBLEM 6.42

40 mm 12 mm 40 mm

30 mm

a

10 mm

b

Knowing that a given vertical shear V causes a maximum shearing stress of 50 MPa in a thin-walled member having the cross section shown, determine the corresponding shearing stress (a) at point a, (b) at point b, (c) at point c.

c 50 mm

10 mm 30 mm

SOLUTION Qa  (12)(30)(25  10  15)  18  103 mm3 Qb  (40)(10)(25  5)  12  103 mm3 Qc  Qa  2Qb  (12)(10)(25  5)  45.6  103 mm3  25  Qm  Qc  (12)(25)    49.35  103 mm3  2  ta  tc  tm  12 mm tb  10 mm

 m  50 MPa (a)

 a Qa tm 18 12      0.36474 49.35 12  m Qm ta

 a  18.23 MPa 

(b)

 b Qb tm 12 12      0.29179 49.35 10  m Qm tb

 b  14.59 MPa 

(c)

c Q t 45.6 12  c  m    0.92401 49.35 12  m Qm tc

 c  46.2 MPa 



2 in.

PROBLEM 6.43

2 in. 10 in. 4 in. 10 in.

Three planks are connected as shown by bolts of 83 -in. diameter spaced every 6 in. along the longitudinal axis of the beam. For a vertical shear of 2.5 kips, determine the average shearing stress in the bolts.

SOLUTION Locate neutral axis. A  (2)(2)(10)  (10)(4)  80 in 2 Ay  (2)(2)(10)(5)  (10)(4)(8)  520 in 3

Y 

Ay 520   6.5 in. A 80

1  I  2  (2)(10)3  (2)(10)(1.5) 2  12   1  (10)(4)3  (10)(4)(1.5) 2  566.67 in 4 12 Q  (2)(10)(1.5)  30 in 3

F  qs  Abolt 

 4

VQs (2.5)(30)(6)   0.79411 kips 566.67 I

2 d bolt 

  3

2

2    0.110447 in 4 8

 bolt 

F 0.79411   7.19 ksi  Abolt 0.110447



PROBLEM 6.44 100 mm

A beam consists of three planks connected as shown by steel bolts with a longitudinal spacing of 225 mm. Knowing that the shear in the beam is vertical and equal to 6 kN and that the allowable average shearing stress in each bolt is 60 MPa, determine the smallest permissible bolt diameter that can be used.

25 mm 25 mm 100 mm

50 mm 100 mm 50 mm

SOLUTION Part

A(mm 2 )

y (mm)

Ay 2 (106 mm 4 )

I (106 mm 4 )



7500

50

18.75

14.06



7500

50

18.75

14.06



15,000

50

37.50

28.12

75.00

56.24

Σ

I   Ay 2   I  131.25  106 mm 4  131.25  106 m 4 Q  A1 y1  (7500)(50)  375  103 mm3  375  106 m3 Fbolt   bolt Abolt  qs  Abolt 

VQs I

VQs (6  103 )(375  106 )(0.225)   64.286  106 m 2  bolt I (6  106 )(131.25  106 )  64.286 mm 2

d bolt 

4 Abolt





(4)(64.286)

d bolt  9.05 mm 





PROBLEM 6.45 6 in.

1 in. 1 in.

A beam consists of five planks of 1.5  6-in. cross section connected by steel bolts with a longitudinal spacing of 9 in. Knowing that the shear in the beam is vertical and equal to 2000 lb and that the allowable average shearing stress in each bolt is 7500 psi, determine the smallest permissible bolt diameter that can be used.

SOLUTION Part

A(in 2 )

y0 (in.)

Ay0 (in 3 )

y (in.)

Ay 2 (in 4 )

I (in 4 )



9

5

45

0.8

5.76

27



9

4

36

0.2

0.36

27



9

3

27

1.2

12.96

27



9

4

36

0.2

0.36

27



9

5

45

0.8

5.76

27



45

25.20

135

189 Y0 

Ay 189   4.2 in. A 45

I  Ad 2  I  160.2 in 4

Between  and :

Q12  Q1  Ay1  (9)(0.8)  7.2 in 3

Between  and :

Q23  Q1  Ay2  7.2  (9)(0.2)  5.4 in 3 q

VQ I

Maximum q is based on Q12  7.2 in 3. (2000)(7.2)  89.888 lb/in. 160.2  qs  (89.888)(9)  809 lb

q Fbolt

 bolt  Abolt 

Fbolt Abolt

 4

2 d bolt

Abolt 

Fbolt

 bolt

d bolt 



4 Abolt



809  0.1079 in 2 7500 

(4)(0.1079)



d bolt  0.371 in. 



PROBLEM 6.46 400 mm

Four L102  102  9.5 steel angle shapes and a 12  400-mm plate are bolted together to form a beam with the cross section shown. The bolts are of 22-mm diameter and are spaced longitudinally every 120 mm. Knowing that the beam is subjected to a vertical shear of 240 kN, determine the average shearing stress in each bolt.

12 mm

SOLUTION For one L102  102  9.5, A  1845 mm 2 I  1.815  106 mm 2 Q  (1845 mm 2 )(171 mm)  315.5  103 mm3  315.5  106 m 4

For 12-mm  400-mm plate and four angle, I 

1 (12 mm)(400 mm)3  4[1.815  106 mm 4  (1845 mm 2 )(171 mm) 2 ] 12

 287.06  106 mm 4  287.06  106 m 4

One angle:

q

VQ (240 kN)(315.5  106 m3 )   263.78 kN/m I 287.06  106 m 4

F  qs  (263.78 kN/m)(0.120 m)  31.65 kN

Diam. bolt  22 mm

 

F 31.65 kN  ;  A (0.022 m) 2 4

  83.3 MPa ◄



PROBLEM 6.47

D 1.6 in. A

B 2 in.

E

1.2 in. 1.2 in.

F 2 in.

A plate of 14 -in. thickness is corrugated as shown and then used as a beam. For a vertical shear of 1.2 kips, determine (a) the maximum shearing stress in the section, (b) the shearing stress at point B. Also, sketch the shear flow in the cross section.

SOLUTION LBD 

(1.2) 2  (1.6) 2  2.0 in.

ABD  (0.25)(2.0)  0.5 in 2

Locate neutral axis and compute moment of inertia. Part

A(in 2 )

d (in.)

Ad 2 (in 4 )

I (in 4 )

AB

0.5

0

0

0.4

0.080

neglect

BD

0.5

0.8

0.4

0.4

0.080

*0.1067

DE

0.5

0.8

0.4

0.4

0.080

*0.1067

EF

0.5

0

0

0.4

0.080

neglect

Σ

2.0

0.320

0.2134

*

y (in.)

Ay (in 3 )

0.8

1 1 ABD h 2  (0.5)(1.6)2  0.1067 in 4 12 12

Y 

Ay 0.8   0.4 in. 2.0 A

I  Ad 2  I  0.5334 in 4

(a)

Qm  QAB  QBC QAB  (2)(0.25)(0.4)  0.2 in 3 QBC  (0.5)(0.25)(0.2)  0.025 in 3 Qm  0.225 in 3

m  (b)

VQm (1.2)(0.225)  It (0.5334)(0.25)

 m  2.02 ksi 

QB  QAB  0.2 in 3

B 

VQB (1.2)(0.2)  It (0.5334)(0.25)

 B  1.800 ksi 

D  0 


22 mm

PROBLEM 6.48

e

A plate of 2-mm thickness is bent as shown and then used as a beam. For a vertical shear of 5 kN, determine the shearing stress at the five points indicated and sketch the shear flow in the cross section. a

d 50 mm

b c

10 mm 10 mm

SOLUTION 1 1 1  I  2  (2)(48)3  (2)(52)3  (20)(2)3  (20)(2)(25) 2  12 12 12   133.75  103 mm 4  133.75  109 mm 4 Qa  (2)(24)(12)  576 mm3  576  109 mm3 Qa  0 Qc  Qb  (12)(2)(25)  600 mm3  600  109 m3 Qd  Qc  (2)(24)(12)  1.176  103 mm3  1.176  106 m3 Qe  Qd  (2)(26)(13)  600 mm3  500  109 m3

a 

VQa (5  103 )(576  109 )   10.77  106 Pa It (133.75  109 )(2  103 )

b 

VQb It

c 

VQc (5  103 )(600  109 )   11.21  106 Pa It (133.75  109 )(2  103 )

 c  11.21 MPa 

d 

VQd (5  103 )(1.176  106 )   22.0  106 Pa 9 3 It (133.75  10 )(2  10 )

 d  22.0 MPa 

e 

VQe (5  103 )(500  109 )   9.35  106 Pa It (133.75  109 )(2  103 )

 e  9.35 MPa 

 a  10.76 MPa  b  0 





60 mm

PROBLEM 6.49

A

An extruded beam has the cross section shown and a uniform wall thickness of 3 mm. For a vertical shear of 10 kN, determine (a) the shearing stress at point A, (b) the maximum shearing stress in the beam. Also, sketch the shear flow in the cross section.

30 mm

16 mm

28 mm

16 mm

SOLUTION tan  

16 30

  28.07

A  (3 sec  )(30)  102 mm 2 1 I  (3 sec  )(30)3  7.6498  103 mm 4 12

Side:

A y (103 mm3 )

Ad 2 (103 mm 4 )

I (103 mm 4 )

Part

A (mm 2 )

y0 (mm)

Top

180

30

5.4

11.932

25.627

neglect

Side

102

15

1.53

3.077

0.966

7.6498

Side

102

15

1.53

3.077

0.966

7.6498

Bot.

84

0

18.077

27.449

neglect

55.008

15.2996

Σ

468

0

d (mm)

8.46

A y 8.46  103   18.077 mm 468 A I  Ad 2  I  70.31  103 mm 4  70.31  109 m 4

Y0 

(a)

QA  (180)(11.932)  2.14776  103 mm3  2.14776  10 6 m3 t  (2)(3  103 )  6  103 m

A  (b)

VQ (10  103 )(2.14776  106 )   50.9  106 Pa 9 6 It (70.31  10 )(6  10 )

 A  50.9 MPa 

1  Qm  QA  (2)(3 sec  )(11.932)   11.932  2   2.14776  103  484.06  2.6318  103 mm3  2.6318  106 m3 t  6  103 m

m 

VQm (10  103 )(2.6318  106 )   62.4  106 Pa It (70.31  109 )(6  103 )

 m  62.4 MPa 



QB  (28)(3)(18.077)  1.51847  103 mm3 QB 1.51847  103 A  (50.9) QA 2.14776  103  36.0 MPa

B 

Multiply shearing stresses by t (3 mm  0.003 m) to get shear flow.



PROBLEM 6.50

6 in. E

D

4.8 in. A

G

B F 3 in.

2 in.

A plate of thickness t is bent as shown and then used as a beam. For a vertical shear of 600 lb, determine (a) the thickness t for which the maximum shearing stress is 300 psi, (b) the corresponding shearing stress at point E. Also, sketch the shear flow in the cross section.

3 in.

SOLUTION LBD  LEF 

4.82  22  5.2 in.

Neutral axis lies at 2.4 in. above AB. Calculate I. I AB  (3t )(2.4) 2  17.28t I BD 

1 (5.2t )(4.8)2  9.984t 12

I DE  (6t )(2.4)2  34.56t I EF  I DB  9.984t I FG  I AB  17.28t I  I  89.09t

(a)

At point C,

QC  QAB  QBC  (3t )(2.4)  (2.6t )(1.2)  10.32t

 

(b)

VQC It

 t 

VQ (600)(10.32t )   0.23168 in. (300)(89.09t ) I

t  0.232 in. 

I  (89.09)(0.23168)  20.64 in 4 QE  QEF  QFG  0  (3)(0.23168)(2.4)  1.668 in 3

E 

VQE (600)(1.668)  It (20.64)(0.23168)

 E  209 psi 



3 8

3 8

in.

PROBLEM 6.51

in.

2 in.

2 in. 1 2

in.

a

The design of a beam calls for connecting two vertical rectangular 83  4 -in. plates by welding them to horizontal 12  2 -in. plates as shown. For a vertical shear V, determine the dimension a for which the shear flow through the welded surface is maximum.

a

2 in.

1 2

in.

SOLUTION 3

 1  3   1  1 1 I  (2)    (4)3  (2)   (2)    (2)(2)   a 2  12  8   12   2  2  4.041667  2a 2 in 4 1 Q  (2)   a  a in 3 2 VQ Va q  I 4.041667  2a 2

Set

dq  0. da

dq  (4.041667  2a 2 )  (a)(4a)   V  0 da  (4.041667  2a 2 )2 

2a 2  4.041667 a  1.422 in.  



PROBLEM 6.52 The cross section of an extruded beam is a hollow square of side a  3 in. and thickness t  0.25 in. For a vertical shear of 15 kips, determine the maximum shearing stress in the beam and sketch the shear flow in the cross section. a

a

SOLUTION Iu  I v 

1 (3.254  2.754 ) 12

 4.53125 in 4

Since products of inertia  0, I x  I y  Iu  I v I x  4.53125 in 4

V  15 kips

QNA  2[(3 in.  0.25 in.)(1.0607 in.)]  1.59105 in 3

 m   NA 

VQNA (15 kips)(1.59105 in 3 )  I (2t ) (4.53125 in 4 )(2)(0.25 in.)

 m  10.53 ksi ◄ 



PROBLEM 6.53

a a

(a)

An extruded beam has a uniform wall thickness t. Denoting by V the vertical shear and by A the cross-sectional area of the beam, express the maximum shearing stress as  max  k (V/A) and determine the constant k for each of the two orientations shown.

(b)

SOLUTION

(a)

3 a 2 A1  A2  at h

(b)

3 3 at 4 1 1 3 1 I 2  A2 h 2  at a 2  a 3t 3 3 4 4 5 3 I  2 I1  4 I 2  a t 2

A2 

I1  A1h 2  ath 2 

2

1 a h  ath 2  at    12 2 2 1 3 9 7  a t  a 3t  a 3t 48 16 12

3 2 a t 2 3 2 h Q2  A2  a t 2 4 Qm  Q1  2Q2  3a 2 t

 k

3

1 a 1 3 I2  t    at 3 2 24 5 I  4 I1  4 I 2  a3t 2 a h 3 2 Q1  at     a t 2 2 4

VQ V 3a 2 t 3V   3 5 I (2t ) 5 at a t 2t 2





 1  a  1 Q2   at    a 2t  2  4  8 7 Q  2Q1  2Q2  a 3t 4 V  74 a3t VQ  m  5 3 I (2t ) a t (2t ) 2

V 6 3 V 6 3V  k A 5 6at 5 A 6 3 5

1 at 2

I1  I1  A1d 2

Q1  A1h 

m 

a 2 A1  at h

k  2.08 







7 V 42 V 21 V   20 at 20 6at 10 A

k

V A

k

21  2.10  10


PROBLEM 6.54

u P C rm

t

(a) Determine the shearing stress at point P of a thin-walled pipe of the cross section shown caused by a vertical shear V. (b) Show that the maximum shearing stress occurs for   90 and is equal to 2V/A, where A is the cross-sectional area of the pipe.

SOLUTION A  2 rm t r

sin 



J  Arm2  2 rm3t

I

1 J   rm3t 2

for a circular arc.

AP  2r t QP  AP r  2rt sin  (a)

P 

(b)

m 

VQP (V )(2rt sin  )  I (2t )  rm3t (2t )



P 



2V sin 2

V sin    rm t

m 

2 rm t

2V  A



PROBLEM 6.55 For a beam made of two or more materials with different moduli of elasticity, show that Eq. (6.6)

 ave 

VQ It

remains valid provided that both Q and I are computed by using the transformed section of the beam (see Sec. 4.4), and provided further that t is the actual width of the beam where  ave is computed.

SOLUTION Let Eref be a reference modulus of elasticity. n1 

E1 E , n2  2 , etc. Eref Eref

Widths b of actual section are multiplied by n’s to obtain the transformed section. The bending stress distribution in the cross section is given by

x  

nMy I

where I is the moment of inertia of the transformed cross section and y is measured from the centroid of the transformed section. The horizontal shearing force over length  x is



H   ( x ) dA 



n(M ) y (M ) Q(M ) dA  ny dA  I I I





Q  ny dA  first moment of transformed section.

Shear flow:

q

 H M Q VQ   x x I I

q is distributed over actual width t, thus

q t VQ  It

 .



PROBLEM 6.56

90 mm

A composite beam is made by attaching the timber and steel portions shown with bolts of 12-mm diameter spaced longitudinally every 200 mm. The modulus of elasticity is 10 GPa for the wood and 200 GPa for the steel. For a vertical shear of 4 kN, determine (a) the average shearing stress in the bolts, (b) the shearing stress at the center of the cross section. (Hint: Use the method indicated in Prob. 6.55.)

84 mm 90 mm 6 mm

140 mm

6 mm

SOLUTION Let steel be the reference material. ns  1.0

Ew 10 GPa   0.05 Es 200 GPa

nw 

Depth of section:

d  90  84  90  264 mm

For steel portion,

Is  2

For the wooden portion,

Iw 

For the transformed section,

1 3  1  bd  (2)   (6)(264)3  18.400  106 mm 4 12  12 





1 1 (140)(2643  843 )  207.75  106 mm 4 b d13  d 23  12 12

I  ns I s  nw I w

I  (1.0)(18.400  106 )  (0.05)(207.75  106 )  28.787  106 mm 4  28.787  106 m 4 (a)

Shearing stress in the bolts.

For the upper wooden portion,

Qw  (90)(140)(42  45)  1.0962  106 mm3

For the transformed wooden portion, Q  nwQw  (0.05)(1.0962  106 )  54.81  103 mm3  54.81  106 m3 Shear flow on upper wooden portion: q

VQ (4000)(54.81  106 )   7616 N/m I 28.787  106

Fbolt  qs  (7616)(0.200)  1523.2 N Abolt  Double shear:

 4

2 d bolt 



 bolt 

4

(12) 2  113.1 mm 2  113.1  106 m 2

Fbolt 1523.2  2 Abolt (2)(113.1  106 )

 bolt  6.73 MPa 

 6.73  106 Pa  




(b)

Shearing stress at the center of the cross section.

For two steel plates,

Qs  (2)(6)(90  42)(90  42)  76.032  103 mm3  76.032  106 m3

For the neutral axis,

Q  54.81  106  76.032  106  130.842  106 m3

Shear flow across the neutral axis: q

VQ (4000)(130.842  106 )   18.181  103 N/m 6 I 28.787  10

Double thickness:

2t  12 mm  0.012 m

Shearing stress:

 

q 18.181  103   1.515  106 Pa 2t 0.012

  1.515 MPa 



PROBLEM 6.57

150 mm 12 mm

A composite beam is made by attaching the timber and steel portions shown with bolts of 12-mm diameter spaced longitudinally every 200 mm. The modulus of elasticity is 10 GPa for the wood and 200 GPa for the steel. For a vertical shear of 4 kN, determine (a) the average shearing stress in the bolts, (b) the shearing stress at the center of the cross section. (Hint: Use the method indicated in Prob. 6.55.)

250 mm

12 mm

SOLUTION Eref  Es  200 GPa

Let

ns  1

nw 

Ew 10 GPa 1   Es 200 GPa 20

Widths of transformed section: bs  150 mm

 1  bw    (150)  7.5 mm  20 

1  1 I  2  (150)(12)3  (150)(12)(125  6)2   (7.5)(250)3 12  12  2[0.0216  106  30.890  106 ]  9.766  106  71.589  106 mm 4  71.589  106 m 4 (a)

Q1  (150)(12)(125  6)  235.8  103 mm3  235.8  106 m3 VQ1 (4  103 )(235.8  106 )   13.175  103 N/m I 71.589  106  qs  (23.187  103 )(200  103 )  2.635  103 N

q Fbolt

Abolt 

 bolt  (b)



  2    (12)2  113.1 mm 2  113.1  106 m 2 d bolt 4 4

Fbolt 2.635  103   23.3  106 Pa Abolt 113.1  106

 bolt  23.3 MPa 

Q2  Q1  (7.5)(125)(62.5)  235.8  103  58.594  103  294.4  103 mm3  294.4  106 m3 t  150 mm  150  103 m

c 

VQ2 (4  103 )(294.4  106 )   109.7  103 Pa It (71.589  106 )(150  103 )

 c  109.7 kPa 



PROBLEM 6.58 2 in.

Aluminum

1 in.

Steel 1.5 in.

A steel bar and an aluminum bar are bonded together as shown to form a composite beam. Knowing that the vertical shear in the beam is 4 kips and that the modulus of elasticity is 29  106 psi for the steel and 10.6  106 psi for the aluminum, determine (a) the average stress at the bonded surface, (b) the maximum shearing stress in the beam. (Hint: Use the method indicated in Prob. 6.55.)

SOLUTION n  1 in aluminum. n

29  106 psi  2.7358 in steel. 10.6  106 psi

Part

nA (in 2 )

Alum.

3.0

2.0

Steel

4.1038

0.5

Σ

7.1038

y (in.)

nA y (in 3 )

d (in.)

nAd 2 (in 2 )

6.0

0.8665

2.2525

1.0

2.0519

0.6335

1.6469

0.3420

3.8994

1.3420

8.0519

Y 

nI (in 4 )

nA y 8.0519   1.1335 in. nA 7.1038

I  nAd 2  nI  5.2414 in 4

(a)

At the bonded surface,

Q  (1.5)(2)(0.8665)  2.5995 in 3

  (b)

At the neutral axis,

VQ (4)(2.5995)  It (5.2414)(1.5)

  1.323 ksi 

 1.8665  3 Q  (1.5)(1.8665)    2.6129 in 2  

 max 

VQ (4)(2.6129)  It (5.2814)(1.5)

 max  1.329 ksi 



PROBLEM 6.59 2 in.

Steel

1 in.

Aluminum 1.5 in.

A steel bar and an aluminum bar are bonded together as shown to form a composite beam. Knowing that the vertical shear in the beam is 4 kips and that the modulus of elasticity is 29  106 psi for the steel and 10.6  106 psi for the aluminum, determine (a) the average stress at the bonded surface, (b) the maximum shearing stress in the beam. (Hint: Use the method indicated in Prob. 6.55.)

SOLUTION n  1 in aluminum. n

29  106 psi  2.7358 in steel. 10.6  106 psi

Part

nA (in 2 )

y (in.)

nA y (in 3 )

d (in.)

nAd 2 (in 2 )

nI (in 4 )

Steel

8.2074

2.0

16.4148

0.2318

0.4410

2.7358

Alum.

1.5

0.5

0.75

1.2682

2.4125

0.1250

Σ

9.7074

2.8535

2.8608

17.1648 nA y 17.1648   1.7682 in. 9.7074 A I  nAd 2  nI  5.7143 in 4

Y 

(a)

At the bonded surface,

Q  (1.5)(1.2682)  1.9023 in 3

 (b)

At the neutral axis,

VQ (4)(1.9023)  It (5.7143)(1.5)

  0.888 ksi 

 1.2318  Q  (2.7358)(1.5)(1.2318)   3.1133 in 3   2 

 max 

VQ (4)(3.1133)  It (5.7143)(1.5)

 max  1.453 ksi 



PROBLEM 6.60

P

Consider the cantilever beam AB discussed in Sec. 6.5 and the portion ACKJ of the beam that is located to the left of the transverse section CC and above the horizontal plane JK, where K is a point at a distance y  yY above the neutral axis. (See figure.) (a) Recalling that  x   Y between C and E and  x  ( Y /yY ) y between E and K, show that the magnitude of the horizontal shearing force H exerted on the lower face of the portion of beam ACKJ is

Plastic C

A J

E yY

K B C'

E'

H

y

 1 y2  b Y  2c  yY   2 yY  

x Neutral axis

(b) Observing that the shearing stress at K is H 1 H 1  H  lim   A 0  A  x 0 b  x b x

 xy  lim

and recalling that yY is a function of x defined by Eq. (6.14), derive Eq. (6.15).

SOLUTION Point K is located a distance y above the neutral axis. The stress distribution is given by y   Y for 0  y  yY and    Y yY

for

yY  y  c.




For equilibrium of horizontal forces acting on ACKJ,



H   dA 



yY

 Y yb



c

 Y b dy yY yY  b  y2  y2   Y  Y    y b(c  yY ) yY  2   1 y2  H  b  Y  2c  yY   yY  2  y

dy 

(a)

Note that yY is a function of x.

y 2 dyY  1  H 1   yY  Y     b  x 2   x yY 2 dx  1  y 2  dyY    Y 1   2  yY 2  dx

 xy 

But

Differentiating,

M  Px 

 1 yY2  3 M y 1   3 c 2  2  

dM 3  2 yY dyY   P  MY    2 dx 2  3 c dx  dyY Pc 2 Pc 2 3 P    2 2 dx yY M Y 2  Y byY yY 3  Y bc

Then

1 2

  

 xy   Y 1 

y2 yY 2

3 P 3P    2  Y b Y 4byY 

 y2  1    yY 2  

(b)


a

A

PROBLEM 6.61

B a

D

E

Determine the location of the shear center O of a thin-walled beam of uniform thickness having the cross section shown.

a

O e F

G a

H

J

SOLUTION 2

1 9  3a  I AB  I HJ  at    at 3  ta 3 2 12 4   2

1 1 a I DE  I FG  at    at 3  ta 2 4  2  12 1 9 I AH  t (3a )3  ta 3 12 4 29 3 I  I  ta 4 Part AB:

3a 3 Q  atx 2 2 3 VQ V  2 atx 6Vx    3 29 It ta t 29a 2 t 4 A  tx

y





F1   dA  Part DE:

a 0

6Vx 6V t dx  2 29a t 29a 2



x dx 

3 V 29

x dx 

1 V 29

a 0

a 1 Q  atx 2 2 1 2Vx VQ V  2 atx    3 29 It 29 a 2t ta t 4 A  tx



y

F2   dA 

M K 



a 0

2Vx 2V t dx  2 29a t 29a 2



a 0

M K : Ve  F1 (3a )  F2 (a )  e

9 1 10 Va  Va  Va 29 29 29 10 a 29

e  0.345a 



PROBLEM 6.62

A D

a

B

a


O e

a F

E a G 2a

SOLUTION 2

1 3 7  3a  ta  (ta)    ta 3 12 2 3   1  I EF  (2at ) a 2  (2a ) t 3  2a3t 12 1 2 28 3 I  I  ta  t (2a )3  ta 3 12 3 3

I AB  I FG  I DB I DE Part AB:

A  t (2a  y ); Q  Ay 

y

2a  y 3

1 t (2a  y )(2a  y ) 2

1 t (4a 2  y 2 ) 2 VQ V (4a 2  y 2 )   2I It 2a V F1   dA  (4a 2  y 2 ) t dy a 2I Vt  2 y 3  2 a Vta 3  (2)3  1    (4)(1)      4a y    (4)(2)  2I  3 a 2I  3  3  







5 Vta 3 5  V 6 I 56




3a  txa 2  3a   ta   x  2  

Q  (ta )

Part DB:



VQ Va  3a    x It I  2 



F2   dA  

Vta  3a   x  t dx   I  2 I 

2 a Va 0

Vta  3ax x 2   I  2 2

5 MH 



  

2a 0





2a  0

3a   2  x  dx  

Vta 3  (3)(2) (2)2     I  2 2 

3

Vta 15  V I 28  M H:

Ve  F2 (2a )  2 F1 (2a ) 

30 20 5 Va  Va  Va 28 56 7 e

5 a  0.714a  7



D

PROBLEM 6.63

B a A G

O e


a E

2a

F

SOLUTION I AB  I FG 

1 3 ta 3

1 2att 3  2ta3 12

I DB  I CP  2ata3 

1 2 16 3 t (2a)3  ta 3 I  I  ta 12 3 3 y 1 A  ty y  Q  ty 2 2 2

I DE  Part AB:

2 VQ V  12 ty 3Vy 2    16 3  It ta t 32a3t 3 a

F1    dA   0  t dy  Part BD:

Q  QB  txa 

 

VQ  It

3V a 2 1 y dt  V 3 0 32 32a

1 2 ta  tax 2

Vt  1 2   a  ax  a3t  2 

16 3

3 V ( a  2 x) 32 a 2 2 a 3V F2    dA   0 (a  2 x)dx 32a 2 2a 3V  (ax  x 2 ) 2 0 32a 3V 9  (2a 2  4a 2 )  V 16 32a 2 

M H 

M H :

Ve  (2a)(2 F1)  (2a)( F2 ) 

1 9 5 Va  Va  Va 8 8 4

e

5 a  1.250a  4



a

PROBLEM 6.64

b D

A

B

h

O


e E

G

F

SOLUTION 2

Part AD:

1 1 h I AB  I EF  (a  b)t    (a  b)t 3  t (a  b)h 2 4  2  12 1 1 I DG  th3 I  I  t (6a  6b  h)h 2 12 12 h 1 Q  tx  thx 2 2 VQ Vhx   It 2I a Vhx Vht a F1   dA  t dx  x dx 0 2I 2I 0



 Part BD:



Vht x 2 2I 2



a

 0

Vhta 2 4I

h 1  thx 2 2 VQ Vhx   2I It b Vhx Vht F2   dA  t dx  0 2I 2I Q  tx





Vht x 2  2I 2 M H 

b

 0



b

0

x dx

Vhtb 2 4I

M H :

Vh 2 t (b 2  a 2 ) 4I 2 2 2 3V (b 2  a 2 ) Vh t (b  a )   2 6a  6b  h 4  121 t (6a  6b  h)h

Ve  F2 h  F1h 

e

3(b 2  a 2 )  6(a  b)  h



PROBLEM 6.65

6 mm B A

An extruded beam has the cross section shown. Determine (a) the location of the shear center O, (b) the distribution of the shearing stresses caused by the vertical shearing force V shown applied at O.

12 mm O

C

192 mm

e V ⫽ 110 kN

6 mm E

D 72 mm

SOLUTION 2  1   192   1 3 I  2   (72)(6)3  (72)(6)     12 (12)(192) 12 2     

 15.0431  106 mm 4  15.0431  106 m 4

Part AB:

 192  Q  Ay  (6 x)    576 x  2  VQ 576Vx  q I I

A  6x

x  0 at point A. x  l AB  72 mm at point B. F1   MC  (a) (b)



xB xA

q dx 



72 0

576Vx 576V (72)2 dx  2 I I

2

(288)(72) V  0.099247 V 15.0431  106 M C : Ve  (0.099247 ) V (192)

e  19.0555 mm Point A: x  0,

e  19.06 mm  Q  0,

Point B in part AB:

A  0 

q0

x  72 mm Q  (576)(72)  41.472  103 mm3  41.472  106 m3 t  6 mm  0.006 m

B 

VQ (110  103 )(41.472  106 )  It (15.0431  106 )(0.006)

 B  50.5 MPa 

 50.5  106 Pa




Part BD: Point B:

y  96 mm

Q  41.472  103 mm3  41.472  106 m3

t  12 mm  0.012 m

B 

VQ (110  103 )(41.472  106 )  It (15.0431  106 )(0.012)

 B  25.3 MPa 

 25.271  106 Pa Point C:

y  0, t  0.012 m  96  Q  41.472  103  (12)(96)    96.768  103 mm3  96.768  106 m3  2 



VQ (110  103 )(96.768  106 )   58.967  106 Pa It (15.0431  106 )(0.012)

 C  59.0 MPa 



PROBLEM 6.66

4.0 in. D

B

O

A G

e

6.0 in.


V ⫽ 2.75 kips E

F t⫽

1 8

in.

SOLUTION 1 I AB  (0.125)(3)3  1.125 in 4 3 1 I BD  (4)(0.125)3  (4)(0.125)(3)2  4.50065 in 4 12 1 I DE  (0.125)(6)3  2.25 in 4 12 I EF  I BD  4.50065 in 4 I FG  I AB  1.125 in 4 I   I  13.50 in 4 (a)

Part AB:

y  0.5ty 2 2 VQ ( y ) 0.5Vt 2 q( y )  y  I I 3 0.5Vt FAB  q ( y ) dy  0 I

Q ( y )  ty



Its moment about H is

4 FAB  18



3 0

y 2 dy  4.5

Vt  I

Vt I

QB  (0.5)(t )(3) 2  4.5 t

Part BD:

Q( x)  QB  xt (3)  (4.5  3x) t Vq ( x) Vt  (4.5  3 x) I I 4 t 4 Vt  q ( x) dx  (4.5  3x) dx  42  0 I 0 I

q( x)  FBD Its moment about H : 3FBD  126





Vt I

QD  [4.5  (3)(4)] t  16.5 t




By symmetry with part BD, FEF  42

Part EF:

Its moment about H is 3FEF  126

Vt  I

Vt I

By symmetry with part AB, FFG  4.5

Part FG:

Its moment about H is 4 FFG  18

VT  I

Vt I

Moment about H of force in part DE is zero. Vt 144Vt (18  126  0  126  18)  I I 144 t (2.88)(0.125)  e 13.50 I

Ve   M H 

(b)

e  2.67 in. 

 A  G  0 

QA  QG  0 QB  QF  4.5t

B F 

VQB (2.75)(4.5 t )  13.50 t It



QD  QE  16.5t 



D E 

 B   F  0.917 ksi 

VQ0 (2.75)(16.5t )   13.50 t It

At H (neutral axis),

 D   E  3.36 ksi 

QH  QD  t (3)(1.5)  21t

H 

VQH (2.75)(21t )  13.50t It

 H  4.28 ksi 


PROBLEM 6.67

A 2 in. B

D

O

6 in.


e V 5 2.75 kips

F

E

2 in. G 4 in. t5

1 8

in.

SOLUTION Part

A (in2)

d (in.)

Ad 2 (in 4 )

I (in 4 )

BD

0.50

3

4.50

0

ABEG

1.25

0

0

EF

0.50

3

4.50

0



2.25

9.00

10.417

10.417

I   Ad 2   I  19.417 in 4 (a)

Q( x)  3 tx VQ ( x) V q( x)   (3tx) I I

Part BD:

3Vt 24Vt (8)  0 I I 72Vt ( M BD ) H  3FBD  I

FBD  Its moment about H: Part EF:

By same method,

3Vt I



4

x dx 

FEF 

24Vt I

( M EF ) H 

72Vt I

Moments of FAB , FBE , and FEG about H are zero. Ve   M H  e

72Vt 72Vt 144Vt   I I I

144 t (144)(0.125)  19.417 I

e  0.927 in. 




(b)

 A   D   F  G  0 

At A, D, F , and G,

Q0

Just above B :

Q1  QAB  (2t )(4)  8 t

1  Just to the right of B:

VQ2 (2.75)(12t )  (19.417) t It

 2  1.700 ksi 

Q3  Q1  Q2  20t

3  At H (neutral axis),

1  1.133 ksi 

Q2  QBD  (3) t (4)  12t

2  Just below B:

VQ1 (2.75)(8t )  (19.417) t It

VQ3 (2.75)(20t )  (19.417) t It

 3  2.83 ksi 

QH  Q3  QBH  20t  t (3)(1.5)  24.5 t

H 

VQH (2.75)(24.5t )  (19.417) t It

 H  3.47 ksi   4   3  2.83 ksi 

By symmetry,

 5   2  1.700 ksi 



 6  1  1.133 ksi 


PROBLEM 6.68

6 mm A B 6 mm

z

30 mm

4 mm D

O

E


30 mm 4 mm

e F N = 35 kN

G 30 mm

6 mm H

J

30 mm Iz = 1.149 × 106 mm4

SOLUTION 1 (30)(6)3  (30)(6)(45)2  0.365  106 mm 4 12 1  (30)(4)3  (30)(4)(15) 2  0.02716  106 mm 4 12

I AB  I HJ  I DE  I FG

1 (6)(90)3  0.3645  106 mm 4 12 I  I  1.14882  106 mm 4

I AH 

(a)

For a typical flange,

A( s )  ts Q( s )  yts VQ( s ) Vyts  I I b Vytb 2 F  q( s )ds  0 2I

q(s) 



Flange AB:

FAB 

V (45)(6)(302 )  0.10576V  (2)(1.14882  106 )

Flange DE:

FDE 

V (15)(4)(30)2  0.023502V  (2)(1.14882  106 )

Flange FG:

FFG  0.023502V 

Flange HJ:

FHJ  0.10576V  M K 

M K : Ve  45 FAB  15FDE  15 FFG  45 FHJ  10.223V e  10.22 mm 

Dividing by V,




(b)

Calculation of shearing stresses. I  1.14882  106 m 4

V  35  103 N

 0 

At B, E, G, and J, At A and H, Q  (30)(6)(45)  8.1  103 mm3  8.1  106 m3 t  6  103 m



VQ (35  103 )(8.1  106 )   41.1  106 Pa It (1.14882  106 )(6  103 )

  41.1 MPa 

Just above D and just below F: Q  8.1  103  (6)(30)(30)  13.5  103 mm3  13.5  106 m3 t  6  103 m



VQ (35  103 )(13.5  106 )   68.5  106 Pa It (1.14882  106 )(6  103 )

  68.5 MPa 

Just to right of D and just to the right of F: Q  (30)(4)(15)  1.8  103 mm3  1.8  106 m3



t  4  103 m

VQ (35  103 )(1.8  106 )   13.71  106 Pa It (1.14882  106 )(4  103 )

  13.71 MPa 

Just below D and just above F: Q  13.5  103  1.8  103  15.3  103 mm3  15.3  106 m3 t  6  103 m

 At K,

VQ (35  103 )(15.3  106 )   77.7  106 Pa It (1.14882  106 )(6  103 )

  77.7 MPa 

Q  15.3  103  (6)(15)(7.5)  15.975  103 mm3  15.975  106 m3



VQ (35  103 )(15.975  106 )   81.1  106 Pa It (1.14882  106 )(6  103 )

  81.1 MPa 



PROBLEM 6.69

A 1 4

2 in.

in. B O

D

3 in.


3 in. r

E 2 in. F

4 in.

SOLUTION LDB 

42  32  5 in.

1 ADB  LDBt  (5)    1.25 in 2 4

I DB 

1 1 AAB h 2    (1.25)(3)2  3.75 in 4 3 3

I AB 

1 1 3 1 2 4   (2)    (2)(4)  8.1667 in 12  4  4

I  (2)(3.75)  (2)(8.1667)  23.833 in 4

Part AB:

1 (5  y) in 2 4 1 y  (5  y) in. 2 1 1 Q  Ay  (5  y)(5  y)  (25  y 2 ) 8 8 A

 

VQ V (25  y 2 ) V (25  y 2 )   It (8)(23.833)(0.25) 47.667  y2) 1  dy (47.667) 4

5 V (25

F1    dA   3

5

1  V  25 y  y 3    190.667  3 3  MD 

V 1 1   (25)(5)  (5)3  (25)(3)  (3)3   0.09091V 190.667  3 3  M D : Ve  2F1(4)  0.7273V

e  0.727 in. 



PROBLEM 6.70

B 6 mm 35 mm

60⬚

O 35 mm

D e


A F

60⬚

E

SOLUTION 1 I DB  (6)(35)3  85.75  103 mm 4 3 LAB  70 mm AAB  (70)(6)  420 mm 2 I AB 

1 1 AAB h 2    (420)(35)2  171.5  103 mm 4 3  3

I  (2)(85.75  103 )  (2)(171.5  103 )  514.5  103 mm 4 A  ts  6 s 1 1 y  s sin 30  s 2 4 3 2 Q  Ay  s 2 VQ 3Vs 2   It It

Part AB:



F1   dA  

M D 



70 3Vs 2 0

2 It

t ds 

3V I



70 0

s 2 ds

(3)(70)3 1 V V (2)(3) I 3

 M D:

Ve  2[( F1 cos 60°)(70 sin 60°)]  20.2V e  20.2 mm 

Dividing by V,



PROBLEM 6.71

4 in. A


3 in. B O

5 in.

D 3 in. e E

SOLUTION

LAB  42  32  5 in.

AAB  5t

1 1 AAB h 2  AAB d 2  (5t )(3)2  (5t )(4)2  83.75t in 4 12 12 1 I BD  (t )(5)3  10.417t in 4 12 I  2 I AB  I BD  177.917t in 4 I AB 

Q  QAB  QBY

In part BD,

1 Q  (5t )(4)  (2.5  y )t   (2.5  y ) 2 1 1    20t  3.125t  ty 2   23.125  y 2  t 2 2  



VQ It



FBD   dA 

2.5

V (23.125  12 y 2 )t

2.5

It



 t dy 2.5

1  Vt  1   23.125  y 2  dy   23.125 y  y 3    2.5  2  I  6  2.5



Vt I



Vt  (2.5)3  Vt (110.417)  2 (23.125)(2.5)   0.62061V  I 6  177.917t 



2.5

M K  e

10  10  M K : V   e    (0.62061V ) 3  3  10 [1  0.62061] 3

e  1.265 in. 

Note that the lines of action of FAB and FDE pass through point K. Thus, these forces have zero moment about point K.



PROBLEM 6.72

A

B

60 mm O

D


60 mm

e

E

F

80 mm 40 mm

SOLUTION I AB  (40t )(60)2  144  103t LDB  802  602  100 mm

ADB  100t

1 1 ADB h 2  (100t )(60)2  120  103t 3 3 I  2 I AB  2 I DB  528  103t

I DB 

A  tx y  60 mm

Part AB:

Q  A y  60tx mm3 VQ V (60tx) 60Vx    It It I 40 60Vx 60Vt F1   dA  t dx  0 I I



 M D 



60Vt x 2 I 2  M D:

30 0





40

0

x dx

(60)(30)2 Vt  0.051136V (2)(528  103 ) t

Ve  (0.051136V )(120)

e  6.14 mm 



PROBLEM 6.73 a

O

t

A B


e

SOLUTION For whole cross section,

A  2 at J  Aa 2  2 a 3t

I

1 J   a 3t 2

Use polar coordinate  for partial cross section. A  st  a t s  arc length sin  1 r a where     2 2 sin  y  r sin   a



sin 2 

Q  Ay  a t a  a 2t 2sin 2





But

M C  Ve,

2

 a 2 t 2 sin 2 

 a 2 t (1  cos  )

VQ Va 2  (1  cos  ) It I

M C  a  dA  







2 0

Va3 Va 4t (1  cos  ) tad  (  sin  ) I I

2 0

2 Va t  2aV  a 3t 4

e  2a 

hence



A a

O

PROBLEM 6.74 Determine the location of the shear center O of a thin-walled beam of uniform thickness having the cross section shown.

t B e

SOLUTION For a thin-walled hollow circular cross section, A  2 at J  a 2 A  2 a 3t I

For the half-pipe section,

 2

I

1 J   a 3t 2

a 3t

Use polar coordinate  for partial cross section. A  st  a  t r a

sin 



s  arc length where  

y  r cos   a Q  Ay  a t a

 2

sin  cos 



sin  cos 



 a 2t (2 sin  cos  )

 a 2t sin 2  a 2t sin 



VQ Va 2  sin  It I



M H  a  dA  2



 0

Va 2 Va 4 t sin  ta d   cos  a I I



0

4

Va t 4  Va I 

But M H  Ve, hence

e

4



a  1.273a 



3 4

PROBLEM 6.75

in. 3 4

in.

A thin-walled beam has the cross section shown. Determine the location of the shear center O of the cross section.

1 2 in.

O

8 in.

6 in.

e 8 in.

SOLUTION I

1 3 1 t1h1  t2 h23 12 12

1  A   h2  y  t2 2   11  y   h2  y  t2 22  Q  Ay

Right flange:

11  1  h2  y  h2  y  t2  22  2  11    h22  y 2  t2 24  VQ V 1 2    h2  y 2  t2  It2 2 It2  4  

Vt2 Vt2  1 2 2 F2   dA   h2  y  t2 dy  2 I  h2 / 2 2 I t  4  2







Vt2 2I

h2 / 2

1 2 y3   h2 y   3  4

h2 / 2 h2 / 2

 1 2 h 1  h 3 1 2 h 1  h 3  Vt h3 Vt2 h23 2 2 2 2 2 2      h h  2  2 2 3  2  4 2 3  2   12 I t1h13  t2 h23  4

MH 

M H:

 Ve   F2 b  V

e

t2 h23b



t1h13  t2 h23

t2 h23b t1h13  t2 h23

(0.75)(6)3(8) (0.75)(8)3  (0.75)(6)3

e  2.37 in. 



50 mm

PROBLEM 6.76

50 mm

A 6 mm

D

O

80 mm

A thin-walled beam has the cross section shown. Determine the location of the shear center O of the cross section.

F 60 mm 40 mm G E

B

e

SOLUTION Let

h1  AB  h, h2  DE ,

I  Part AB:



1 t h13  h23  h33 12

h3  FG



1  A   h1  y  t 2   y 

11   h1  y  22 

Q  Ay  

 

1 1  1  t  h1  y  h1  y  2 2  2  1 1 2  t  h1  y 2  2 4 

VQ V 1 2 2   h1  y  It 2I  4  1h

F1    dA   2 1 h1 1 2

V 2I

1 2 2  h1  y  t dy 4   1

h1

Vt  1 2 y3  2   h1 y   2I  4 3  1  h1 2



Vt  1 2 1 1  h1  h1 h1    3 2  I 4 2 



h13V h13  h23  h33

3

Vth13   12 I 




Likewise, for Part DE,

F2 

h23V h13  h23  h33

and for Part FG,

F3 

h33V h13  h23  h33

M H  e

M H : Ve  bF2  2bF3 

bh23  2bh33 V h13  h23  h33

h23  2h33 (60)3  (2)(40)3 b (50) 3 3  h2  h3 (80)3  (60)3  (40)3

h13

 21.7 mm

e  21.7 mm ◄



PROBLEM 6.77

A 60 mm

B

60 mm

60 mm

D O

F

A thin-walled beam of uniform thickness has the cross section shown. Determine the dimension b for which the shear center O of the cross section is located at the point indicated.

E

G b

SOLUTION Part AB:

1 s 2 1 Q( s )  A( s ) y ( s)  ty A s  ts 2 2 VQ( s ) Vt  1  q(s)    yAs  s2  I I  2  A( s )  ts

FAB 



y ( s)  y A 

l AB

0

q( s ) ds

2 l3  Vt  y Al AB  AB    6  I  2 1 2 QB  ty Al AB  tl AB 2 FFG  FAB



At B, By symmetry, Part BD:

A( x)  tx Q( x)  QB  yB A( x)  QB  tyB x VQ( x) V  (QB  tyB x) I I b V 1   q ( x) dx   QB b  tyB b 2   0 I 2 

q( x)  FBD By symmetry,



FEF  FBD

FDE is not required, since its moment about O is zero.




 M O  0 : b( FAB  FFG )  yB FBD  yF FEF  0 2b FAB  2 yB FBD  0 2b 

2 l3  Vt  y Al AB V  AB   2 yB   I  2 I 6 

1  2  QB b  2 tyB b   0  

2Vt  1 1 3  2Vt  1 2  1 2 2 2  y A l AB  l AB  b   y Al AB  l AB  yB b  yB b  0 I 2 I  6 2 2  

Dividing by

2Vt and substituting numerical data, I 1 1 1 1  2 3 2 2 2  (90)(60)  (60)  b  (90)(60)  (60)  (30)b  (30) b  0 6 2 2 2    126  103b  108  103b  450b 2  0 18  103b  450b 2  0

b  0 and b  40.0 mm 



A

PROBLEM 6.78

B

1 in. D


E

8 in.

10 in.

O

F

G

1 in. H b

J 3 in.

SOLUTION Part AB:

A  tx

 

y  5 in.

VQ V .5 tx 5Vx   It It I 3 5Vx

F1    dA   0 

Part DE:

Q  Ay  5tx

I

tdx 

5Vt 3  x dx I 0

2

(5)(3) Vt Vt  22.5 2 I I

A  tx

y  4 in.

Q  Ay  4tx

VQ V 4tx 4Vx   It It I a 4Vx F2    dA   0 t dx I 4Vt a   x dx I 0

 



M O  a2 

M O :

2Vta 2 I

Vt  2Vta 2  O  (10)  22.5   (8) I  I 

(10)(22.5)  14.0625 in 2 (8)(2)

a  3.75 in. 



PROBLEM 6.79



For the angle shape and loading of Sample Prob. 6.6, check that q dz  0 along the horizontal leg of the angle and q dy  P along its vertical leg.



SOLUTION Refer to Sample Prob. 6.6. 3P(a  z )(a  3 z ) 3P 2  (a  4az  3z 2 ) 4ta 3 4ta3 a 3P a q dz   f t dz  3 (a 2  4az  3z 2 )dz 0 4a 0 a 3P  z2 z2   3  a 2 z  4a  3  2 3 0 4a 

f 

Along horizontal leg:







3P 3 ( a  2a 3  a 3 )  0 4a 3 3P(a  y )(a  5 y ) 3P 2  (a  4ay  5 y 2 ) e  3 3 4ta 4ta a a 3P q dy   e t dy  3 (a 2  4ay  5y 2 )dy 0 4a 0 

Along vertical leg:





 





3P  2 y2 y3  a y a   4 5   2 3  4a3 

a

0

3P  3 5 3  3P 4 3 3  a  2a  3 a   3  3 a  P 4a 3   4a





PROBLEM 6.80 For the angle shape and loading of Sample Prob. 6.6, (a) determine the points where the shearing stress is maximum and the corresponding values of the stress, (b) verify that the points obtained are located on the neutral axis corresponding to the given loading.

SOLUTION Refer to Sample Prob. 6.6. (a)

e 

Along vertical leg:

3P(a  y )(a  5 y ) 3P 2  (a  4ay  5 y 2 ) 3 4ta 4ta 3

d e 3P (4a  10 y )  0  dy 4ta 3 2  2   2   3P  9 2  a   a 2  (4a )  a   (5)  a    3  5   5   4ta  5   3P(a  z )(a  3 z ) 3P 2  (a  4az  3z 2 ) f  3 4ta 4ta3

m 

Along horizontal leg:

d f dz



m  

At the corner:

(b)

1 I y  ta 3 3

3P 4ta3

3P (4a  6 z )  0 4ta 3 3P 4ta 3

2  2 2   2   3P  5 2  a a a a    a  (4 ) (3)  3  3   3       4ta  3  

y  0, z  0,

I z  tan  

y

m 

2 a  5

27 P  20 ta

z

2 a  3

m  

1 P  4 ta



3P  4 ta

1 3 ta   45 12 I z 1 tan     14.036 4 I y

    45  14.036  30.964 Ay at (a/2) 1   a A 2at 4 Az at (a/2) 1   a z A 2at 4

y




Neutral axis intersects vertical leg at

y  y  z tan 30.964 1 1     tan 30.964  a  0.400a 4 4 

y

2 a  5

z

2 a  3

Neutral axis intersects horizontal leg at

z  z  y tan (45 +  ) 1 1     tan 59.036  a  0.667a 4 4 



PROBLEM 6.81

P D'

Determine the distribution of the shearing stresses along line DB in the horizontal leg of the angle shape for the loading shown. The x and y  axes are the principal centroidal axes of the cross section.

D a

A'

B'

B

A 2a

0.596a

y'

D'

B'

0.342a

C' 2 3

y

a 6

a

A' 15.8⬚

x'

Ix' ⫽ 1.428ta3 Iy' ⫽ 0.1557ta3

x

SOLUTION

  15.8

Vx  P cos 

A( y )  (2a  y )t Coordinate transformation.

In particular,

y

Vy   P sin 

1 (2a  y ), 2

x 0

2  1    y    y  a  cos    x  a  sin  3  6    1  2    x   x  a  cos    y  a  sin  6  3   

2  1    y    y  a  cos    x  a  sin  3  6    1  1  1    y  a  cos     a  sin  3  2  6   0.48111y  0.36612a 1  2    x   x  a  cos    y  a  sin  6 3     1   1  1    a  cos    y  a  sin  3   6  2  0.13614 y  0.06961a


PROBLEM 6.81 (Continued)

  

Vx Ax Vy A y   I yt I x t

( P cos  )(2a  y )(t )(0.13614 y  0.06961a) (0.1557ta 3 )(t ) ( P sin  )(2a  y )(0.48111 y  0.36612a)  (1.428 a 3t )(t ) P(2a  y )(0.750 y  0.500a)  ta3 

y (a )

P     at  

0

1 3

2 3

1

4 3

5 3

2

1.000

0.417

0

0.250

0.333

0.250

0




PROBLEM 6.82

P D'

For the angle shape and loading of Prob. 6.81, determine the distribution of the shearing stresses along line DA in the vertical leg.

D a

A'

B'

B

A 2a

0.596a

PROBLEM 6.81* Determine the distribution of the shearing stresses along line DB in the horizontal leg of the angle shape for the loading shown. The x and y  axes are the principal centroidal axes of the cross section.

y'

D'

B'

0.342a

C' 2 3

y

a 6

a

A' 15.8⬚

x'

Ix' ⫽ 1.428ta3 Iy' ⫽ 0.1557ta3

x

SOLUTION

  15.8

Vx  P cos 

Vx   P sin  x

1 (a  x), 2

A( x)  (a  x)t y 0

Coordinate transformation. 2  1    y    y  a  cos    x  a  sin  3  6    1  2    x   x  a  cos    y  a  sin  6 3    

In particular, 2  1    y    y  a  cos    x  a  sin  3  6    1   2  1    a  cos    x  a  sin  3   3  2  0.13614 x  0.73224a 1  2    x   x  a  cos    y  a  sin  6  3    1  1  2    x  a  cos     a  sin  3  2  3   0.48111x  0.13922a


PROBLEM 6.82* (Continued)



Vx A( x) x Vy A( x) y   I y t I x t

( P cos  )(a  x)(t )(0.48111x  0.13922a) (0.1557ta3 )(t ) ( P sin  )(a  x)(t )(0.13614 x  0.73224a)  (1.428ta 3 )(t ) P(a  x)(3.00 x  1.000a )  ta 3 

x( a )

P

    at 

0

1 6

1 3

1 2

2 3

5 6

1

1.000

1.250

1.333

1.250

1.000

0.583

0






PROBLEM 6.83* A steel plate, 160 mm wide and 8 mm thick, is bent to form the channel shown. Knowing that the vertical load P acts at a point in the midplane of the web of the channel, determine (a) the torque T that would cause the channel to twist in the same way that it does under the load P, (b) the maximum shearing stress in the channel caused by the load P.

B 100 mm

A

D E P ⫽ 15 kN 30 mm

SOLUTION Use results of Example 6.06 with

e

b  30 mm,

h  100 mm, and t  8 mm.

b 30   9.6429 mm  9.6429  103 m 100 h 2  3b 2  (3)(30)

1 2 1 th (6b  h)  (8)(100)2 [(16)(30)  100]  1.86667  106 mm 4  1.86667  106 m 4 12 12 3 V  15  10 N

I

(a)

T  Ve  (15  103 )(9.6429  103 )

T  144.6 N  m 

Stress at neutral axis due to V:

Q  bt

h  h  h  1 t   th(h  4b) 2  2   4  8

1  (8)(100) 100  (4)(30)  22  103 mm3  22  106 m3 8 t  8  103 m

V 

VQ (15  103 )(22  106 )   22.10  106 Pa  22.10 MPa 6 3 It (1.86667  10 )(8  10 )

Stress due to T :

a  2b  h  160 mm  0.160 m 1 t  1 8   0.3228 c1  1  0.630   1  (0.630) 3 a 3 160  T 144.64 V    43.76  106 Pa  43.76 MPa 3 2 2 c1at (0.3228)(0.160)(8  10 )

(b)

By superposition,

 max   V   T

 max  65.9 MPa 



PROBLEM 6.84* Solve Prob. 6.83, assuming that a 6-mm-thick plate is bent to form the channel shown.

B 100 mm

A

D E

PROBLEM 6.83* A steel plate, 160 mm wide and 8 mm thick, is bent to form the channel shown. Knowing that the vertical load P acts at a point in the midplane of the web of the channel, determine (a) the torque T that would cause the channel to twist in the same way that it does under the load P, (b) the maximum shearing stress in the channel caused by the load P.

P ⫽ 15 kN 30 mm

SOLUTION Use results of Example 6.06 with b  30 mm, h  100 mm, and t  6 mm.

e

b 30   9.6429 mm  9.6429  103 m 100 h 2  3b 2  (3)(30)

1 2 1 th (6b  h)  (6)(100) 2 [(6)(30)  100]  1.400  106 mm 4  1.400  106 m 4 12 12 3 V  15  10 N

I

(a)

T  Ve  (15  103 )(9.6429  103 )

T  144.6 N  m 

Stress at neutral axis due to V:

Q  bt

h  h  h  1 t   th(h  4b) 2  2   4  8

1  (6)(100) 100  (4)(30)   16.5  103 mm3  16.5  106 m3 8 t  6  103 m

V 

VQ (15  103 )(16.5  106 )   29.46  106 Pa  29.46 MPa It (1.400  106 )(6  106 )

Stress due to T :

a  2b  h  160 mm  0.160 m 1 t  1  6  c1  1  0.630   1  (0.630)     0.32546 3 a 3  160   T 144.64   77.16  106 Pa  77.16 MPa V  2 c1a t (0.32546)(0.160)(6  103 ) 2

(b)

By superposition,

 max   V   T

 max  106.6 MPa 



PROBLEM 6.85

B

The cantilever beam AB, consisting of half of a thinwalled pipe of 1.25-in. mean radius and 83 -in. wall thickness, is subjected to a 500-lb vertical load. Knowing that the line of action of the load passes through the centroid C of the cross section of the beam, determine (a) the equivalent force-couple system at the shear center of the cross section, (b) the maximum shearing stress in the beam. (Hint: The shear center O of this cross section was shown in Prob. 6.74 to be located twice as far from its vertical diameter as its centroid C.)

1.25 in.

A

A C

a

O t

B 500 lb

e

SOLUTION

From the solution to Prob. 6.74,



I

 

e



 2 4



Q  a 2t sin 

a 3t

Qmax  a 2 t

a

For a half-pipe section, the distance from the center of the semi-circle to the centroid is

 

x



2



a

At each section of the beam, the shearing force V is equal to P. Its line of action passes through the centroid C. The moment arm of its moment about the shear center O is

d ex  (a)

4



a

2



a

2



a

Equivalent force-couple system at O.

V P

M O  Vd 

Data: P  500 lb

a = 1.25 in.

2



Pa

V  500 lb  2 M O    (500)(1.25)  



M O  398 lb  in. 








(b)

Shearing stresses. (1)

Due to V :  V 

V 





(2)

VQmax  It ( P)(a 2 t ) 2P (2)(500)    679 psi   at  (1.25)(0.375)  3  a t t ( ) 2   

Due to the torque M O : For a long rectangular section of length l and width t, the shearing stress due to torque M O is

M  Data:

c1lt 2

1 t where c1   1  0.630  l 3

l   a   (1.25)  3.927 in. t  0.375 in. c1  0.31328

M  By superposition,

MO

397.9  2300 psi (0.31328)(3.927)(0.375) 2

   V   M  679 psi  2300 psi

  2980 psi 



PROBLEM 6.86

B

Solve Prob. 6.85, assuming that the thickness of the beam is reduced to 14 in.

1.25 in.

A

A C

a

O t

B

500 lb

e

PROBLEM 6.85 The cantilever beam AB, consisting of half of a thin-walled pipe of 1.25-in. mean radius and 83 -in. wall thickness, is subjected to a 500-lb vertical load. Knowing that the line of action of the load passes through the centroid C of the cross section of the beam, determine (a) the equivalent force-couple system at the shear center of the cross section, (b) the maximum shearing stress in the beam. (Hint: The shear center O of this cross section was shown in Prob. 6.74 to be located twice as far from its vertical diameter as its centroid C.)

SOLUTION 

From the solution to Prob. 6.74,



I   a 3t 4 e a







Q  a 2 t sin  Qmax  a 2t

For a half-pipe section, the distance from the center of the semi-circle to the centroid is



x

2



a

At each section of the beam, the shearing force V is equal to P. Its line of action passes through the centroid C. The moment arm of its moment about the shear center O is

d ex  (a)



a

2



a

2



a

Equivalent force-couple system at O.

V P 

4

M O  Vd 

P  500 lb

Data:



2



Pa

a  1.25 in. V  500 lb  M O  398 lb  in. 






(b)

Shearing stresses. (1)

Due to V ,  V 

V 

(2)

VQmax It ( P)(a 2t )  3   2 a t (t )   



2P (2)(500)   1019 psi  at  (1.25)(0.250)

Due to the torque M O : For a long rectangular section of length l and width t, the shearing stress due to torque M O is

M  Data:

MO c1lt 2

1 t where c1 = 1  0.630  3 l

l   a   (1.25)  3.927 in. t  0.250 in. c1  0.31996

M 

397.9  5067 psi (0.31996)(3.927)(0.250) 2

By superposition,    V   M  1019 psi  5067 psi

  6090 psi 


3 kips

PROBLEM 6.87

y

y' A'

B' x'

A' B'

A

C'

22.5⬚

D'

E'

B

12 in.

D' D

E'

E

6 in. 6 in. (a)

x

The cantilever beam shown consists of a Z shape of 14 -in. thickness. For the given loading, determine the distribution of the shearing stresses along line AB in the upper horizontal leg of the Z shape. The x and y  axes are the principal centroidal axes of the cross section and the corresponding moments of inertia are I x  166.3 in 4 and I y  13.61 in 4 .

(b)

SOLUTION V  3 kips   22.5 Vx  V sin  Vy  V cos  In upper horizontal leg, use coordinate x : (6 in ⭐ x ⭐ 0)

1 (6  x) in. 4 1 x  (6  x) in. 2 y  6 in. x  x cos  + y sin  y   y cos   x sin  A

1 

Due to Vx :

Vx Ax I yt

1 1  (V sin  )   (6  x)  (6  x) cos   6sin   4 2     1  1 (13.61)   4  0.084353(6  x)(0.47554  0.46194 x)

1 1   (V cos  )   (6  x) 6 cos   (6  x)sin   2 4   2   1 I xt   (166.3)   4  0.0166665(6  x)[6.69132  0.19134 x] Vy A y 

Due to Vy :

1   2  (6  x)[0.07141  0.035396 x]

Total: x (in.)

6

5

4

3

2

1

0

 (ksi)

0

0.105

0.140

0.104

0.003

0.180

0.428

 


PROBLEM 6.88 y'

3 kips

y

A'

B' x'

A' B'

A

C'

22.5⬚

D'

E'

x

B

12 in.

D' D

E'

E

6 in. 6 in. (a)

(b)

For the cantilever beam and loading of Prob. 6.87, determine the distribution of the shearing stress along line BD in the vertical web of the Z shape. PROBLEM 6.87 The cantilever beam shown consists of a Z shape of 14 -in. thickness. For the given loading, determine the distribution of the shearing stresses along line AB in the upper horizontal leg of the Z shape. The x and y  axes are the principal centroidal axes of the cross section and the corresponding moments of inertia are I x  166.3 in 4 and I y  13.61 in 4 .

SOLUTION V  3 kips   22.5 Vx  V sin  Vy  V cos  For part AB’,

For part B′Y,

Due to Vx :

1 A    (6)  1.5 in 2 4 x  3 in., y  6 in.

1 (6  y ) 4 1 x  0 y  (6  y ) 2 x  x cos   y sin  y   y cos   x sin 

A

1  1  

 ) Vx ( AAB x AB  ABY xBY I yt (V sin  )[(1.5)(3cos   6sin  )  14 (6  y ) 12 (6  y ) sin  ] (13.61) 14 

(V sin  )[0.7133  1.7221  0.047835 y 2 ]  0.3404  0.01614 y 2 3.4025




2 

Due to Vy :

2  

Vy ( AAB y AB  ABY y ) I x t (V cos  )[(1.5)(6 cos   3 sin  )  14 (6  y ) 12 (6  y ) cos  ] (166.3) 14 

(V cos  )[10.037  4.1575  0.11548y 2 ]  0.9463  0.00770 y 2 (166.3) 14 

1   2  1.2867  0.02384 y 2

Total: y (in.)

0

2

4

6

 (ksi)

1.287

1.191

0.905

0.428








s s s

PROBLEM 6.89 Three boards are nailed together to form a beam shown, which is subjected to a vertical shear. Knowing that the spacing between the nails is s  75mm and that the allowable shearing force in each nail is 400 N, determine the allowable shear when w  120 mm.

60 mm 60 mm 60 mm

w 200 mm

SOLUTION Part

A (mm 2 )

d (mm)

7200

Middle Plank Bottom Plank

Top Plank

Ad 2 (106 mm 4 )

I (106 mm 4 )

60

25.92

2.16

12,000

0

0

3.60

7200

60

25.92

2.16

51.84

7.92



I  Ad 2   I  59.76  106 mm 4  59.76  106 m 4 Q  (7200)(60)  432  103 mm3  432  106 m3 VQ I F q  nail s

q

V

Fnail  qs V 

Iq IFnail  Q Qs

(59.76  106 )(400) (432  106 )(75  103 )

V  738 N 



PROBLEM 6.90

180 160 kN 16

12 a

0.6 m

80 100

16


n

n

80

0.9 m

0.9 m

Dimensions in mm

SOLUTION

V  80 kN

At section n-n,

Consider cross section as composed of rectangles of types , , and . 1 (12)(80)3  (12)(80)(90)2  8.288  106 mm 4 12 1 (180)(16)3  (180)(16)(42)2  5.14176  106 mm 4 I2  12 1 (16)(68)3  419.24  103 mm 4 I3  12

I1 

I  4I1  2 I 2  2 I 3  44.274  106 mm 4  44.274  106 m 4 (a)

Calculate Q at neutral axis. Q1  (12)(80)(90)  86.4  103 mm 4 Q2  (180)(16)(42)  120.96  103 mm 4 Q3  (16)(34)(17)  9.248  103 mm 4

Q  2Q1  Q2  2Q3  312.256  103 mm3  312.256  106 m3

  (b)

At point a,

VQ (80  103 )(312.256  106 )   17.63  106 Pa It (44.274  106 )(2  16  103 )

  17.63 MPa 

Q  Q1  86.4  103 mm 4  86.4  106 m 4

 

VQ (80  103 )(86.4  106 )   13.01  106 Pa It (44.274  106 )(12  103 )

  13.01 MPa 



PROBLEM 6.91

P W24 × 104 A

C B 6 ft

9 ft

For the wide-flange beam with the loading shown, determine the largest load P that can be applied, knowing that the maximum normal stress is 24 ksi and the largest shearing stress using the approximation  m  V/Aweb is 14.5 ksi.

SOLUTION

 M C  0:  15RA  qP  0 RA  0.6 P

Draw shear and bending moment diagrams. V

 0.6P

max

M

max

 0.6PLAB

LAB  6 ft  72 in.

S  258 in 3

For W24  104,

Bending.

S  P

M

max

 all



0.6 PLAB

 all

(24)(258)  all S   143.3 kips 0.6LAB (0.6)(72)

Aweb  dtw

Shear.

 (24.1)(0.500)  12.05 in 2

  P

V

max

Aweb

 Aweb 0.6



0.6P Aweb



(14.5)(12.05)  291 kips 0.6

The smaller value of P is the allowable value.

P  143.3 kips 



12 kips n A

PROBLEM 6.92

1 in.

12 kips

4 in.

B

1 in. 1 in.

a b

n 16 in.

10 in.


2 in.

16 in.

4 in.

SOLUTION RA  RB  12 kips Draw shear diagram.

V  12 kips Determine section properties. Part

A(in 2 )

y (in.)

Ay (in 3 )

d(in.)

Ad 2 (in 4 )

I (in 4 )



4

4

16

2

16

5.333



8

1

8

1

8

2.667



12

24

8.000

24 Y 

24 Ay   2 in. 12 A

I  Ad 2  I  32 in 4

(a)

A  1 in 2

y  3.5 in. Qa  Ay  3.5 in 3

t  1 in.

a  (b)

VQa (12)(3.5)  (32)(1) It

A  2 in 2

 a  1.313 ksi 

y  3 in. Qb  A y  6 in 3

t  1 in.

b 

VQb (12)(6)  (32)(1) It

 b  2.25 ksi 



PROBLEM 6.93

2 in. 4 in. 6 in.

The built-up timber beam is subjected to a 1500-lb vertical shear. Knowing that the longitudinal spacing of the nails is s  2.5 in. and that each nail is 3.5 in. long, determine the shearing force in each nail.

4 in.

4 in. 2 in.

2 in.

2 in. 2 in.

SOLUTION I1 

1 (2)(4)3  (2)(4)(3) 2 12

 82.6667 in 4 1 (2)(6)3  36 in 4 12 I  2 I1  2I 2

I2 

 237.333 in 4 Q  A1 y1  (2)(4)(3)  24 in 3 q

VQ (1500)(24)   151.685 lb/in. I 237.333

2Fnail  qs

Fnail 

1 1 qs    (151.685)(2.5) 2 2

Fnail  189.6 lb 



PROBLEM 6.94

40 mm b

6 mm 60 mm

4 mm 6 mm

Knowing that a given vertical shear V causes a maximum shearing stress of 75 MPa in the hat-shaped extrusion shown, determine the corresponding shearing stress at (a) point a, (b) point b.

14 mm a

4 mm

20 mm 28 mm 20 mm

SOLUTION Neutral axis lies 30 mm above bottom.

c 

VQc It

 a Qatc   c Qcta

a 

VQa Ita

b 

VQb Itb

 b Qbtc   c Qctb

Qc  (6)(30)(15)  (14)(4)(28)  4260 mm3 tc  6 mm Qa  (14)(4)(28)  1568 mm3 ta  4 mm Qb  (14)(4)(28)  1568 mm3 tb  4 mm

 c  75 MPa (a)

a 

Qa tc 1568 6  c    75 Qc ta 4260 4

(b)

b 

Qb tc 1568 6  c    75 Qc tb 4260 4

 a  41.4 MPa   b  41.4 MPa 



125 mm

100 mm 125 mm

PROBLEM 6.95 Three planks are connected as shown by bolts of 14-mm diameter spaced every 150 mm along the longitudinal axis of the beam. For a vertical shear of 10 kN, determine the average shearing stress in the bolts.

100 mm 250 mm

SOLUTION Locate neutral axis and compute moment of inertia. Part

A(mm 2 )

y (mm)



12,500

200



25,000



12,500

Σ

50,000

Ay (mm3 )

d (mm)

Ad 2 (mm 4 )

I (mm 4 )

2.5  106

37.5

17.5781  106

10.4167  106

125

3.125  106

37.5

35.156  106

200

2.5  106

37.5

17.5781  106

8.125  106

Y 

70.312  106

130.208  106 10.4167  106 151.041  106

Ay 8.125  106   162.5 mm A 50  103

I  Ad 2  I  221.35  106 mm 4  221.35  106 m 4 Q  A1 y1  (12,500)(37.5)  468.75  103 mm3  468.75  106 m3 q

VQ (10  103 )(468.75  106 )  I 221.35  106

 21.177  103 N/m Fbolt  qs  (21.177  103 )(150  103 )  3.1765  103 N Abolt 

 bolt 

 4

(14) 2  153.938 mm 2  153.938  106 m 2

Fbolt 3.1765  103   20.6  106 Pa Abolt 153.938  106

 bolt  20.6 MPa 



PROBLEM 6.96 1 in. 1 in. C

x

18 in. 1 in.

Three 1  18-in. steel plates are bolted to four L6  6  1 angles to form a beam with the cross section shown. The bolts have a 78 -in. diameter and are spaced longitudinally every 5 in. Knowing that the allowable average shearing stress in the bolts is 12 ksi, determine the largest permissible vertical shear in the beam. (Given:I x  6123 in 4.)

18 in.

SOLUTION Flange:

If 

1 (18)(1)3  (18)(1)(9.5) 2  1626 in 4 12

Web:

Iw 

1 (1)(18)3  486 in 4 12

I  35.5 in 4 ,

Angle:

y  1.86 in.

A  11.0 in 2 d  9  1.86  7.14 in.

I a  I  Ad 2  596.18 in 4 I  2I f  I w  4I a  6123 in 4 , which agrees with the given value.

Flange:

Q f  (18)(1)(9.5)  171 in 3

Angle:

Qa  Ad  (11.0)(7.14)  78.54 in 3 Q  Q f  2Qa  328.08 in 3

Fbolt

qall 

q

 7

2

2    0.60132 in 48  2 bolt Abolt  (2)(12)(0.60132)  14.4317 kips

Abolt 

Fbolt 14.4317   2.8863 kip/s s 5

VQ I

Vall 

qall I (2.8863)(6123)  Q 328.08

Vall  53.9 kips 



PROBLEM 6.97

a 112 mm

The composite beam shown is made by welding C 200  17.1 rolled-steel channels to the flanges of a W250  80 wide-flange rolled-steel shape. Knowing that the beam is subjected to a vertical shear of 200 kN, determine (a) the horizontal shearing force per meter at each weld, (b) the shearing stress at point a of the flange of the wide-flange shape.

SOLUTION For W250  80, d  257 mm, t f  15.6 mm, I x  126  106 mm 4 For C200  17.1,

A  2170 mm 2 , b f  57.4 mm, t f  9.91 mm

I y  0.545  106 mm 4 , x  14.5 mm For the channel in the composite beam,

yc 

257  57.4  14.5  171.4 mm 2

For the composite beam, I  126  106 +2 0.545  106  (2170)(171.4) 2   254.59  106 mm 4  254.59  106 m 4

(a) For the two welds,

Qw  Ayc  (2170) (171.4)  371.94  103 mm3  371.94  106 m3  q

VQ (200  103 ) (371.94  106 )   292.2  103 N/m I 254.59  106

For one weld,

q  146.1  103 N m 2

Shearing force per meter of weld: (b)

146.1 kN m 

For cuts at a and a ' together, Aa  2(112)(15.6)  3494.4 mm 2

ya 

257 15.6   120.7 mm 2 2

Qa  371.94  103  (3494.4)(120.7)  793.71  103 mm3  793.71  106 m3

Since there are cuts at a and a ', t  2t f  (2)(15.6)  31.2 mm  0.0312 m.

a 

VQa (200  103 )(793.71  106 )   19.99  106 Pa 6 It (254.59  10 )(0.0312)

 a  19.99 MPa 



PROBLEM 6.98

0.5 in. 2.5 in.

The design of a beam requires welding four horizontal plates to a vertical 0.5  5-in. plate as shown. For a vertical shear V, determine the dimension h for which the shear flow through the welded surface is maximum.

h 0.5 in.

2.5 in.

h

4.5 in.

4.5 in. 0.5 in.

SOLUTION Horizontal plate: 1 (4.5)(0.5)3  (4.5)(0.5)h 2 12  0.046875  2.25h 2

Ih 

Vertical plate:

Iv 

1 (0.5)(5)3  5.2083 in 4 12

Whole section:

I  4 I h  I v  9h 2  5.39583 in 4

For one horizontal plate,

Q  (4.5)(0.5)h  2.25 h in 3 q

To maximize q, set

VQ 2.25Vh  2 I 9h  5.39583

dq  0. dh 2.25V

(9h 2  5.39583)  18h 2 0 (9h 2  5.39583) 2

h  0.774 in. 



A

PROBLEM 6.99

B D E

60 mm

45 mm F

O


45 mm

60 mm

H J

G K

30 mm

b

SOLUTION Part AB:

A  tx

y  60 mm

Q  Ay  60tx mm3

 

VQ 60Vx  It I



F1   dA  60Vt x 2  I 2 Part DE:

A  tx



60 Vx 60Vt t dx  I I

30 0

30

 0



30

0

x dx

(60)(30) 2 Vt Vt  27  103 I I 2

y  45 mm

Q  Ay  45tx

 

VQ 45Vx  It I b

F2    dA   O

 MO 

45Vx 45Vt b 45b 2Vt t dx  x dx   2I I I O

 M O : 0  (2)(45) F2  (2)(60) F1

(45) 2 b 2  (2)(60)(27  103 )  Vt  0   I b2 

(2)(60)(27  103 )  1600 mm 2 452

b  40 mm 

Note that the pair of F1 forces form a couple. Likewise, the pair of F2 forces form a couple. The lines of action of the forces in BDOGK pass through point O.


PROBLEM 6.100

B 1 4

in. 60⬚

O

D e

1.5 in. A F

60⬚


1.5 in. E

SOLUTION 11 I AB    (1.5)3  0.28125 in 4 3 4  1 LBD  3 in. ABD  (3)    0.75 in 2 4 1 1 I BD  ABD h 2  (0.75)(1.5) 2  0.5625 in 4 3 3 I  (2)(0.28125)  (2)(0.5625)  1.6875 in 4 Part AB:

1 1 1 y y  y Q  Ay  y 2 4 2 8 2 VQ Vy Vy 2    (8)(1.6875)(0.25) 3.375 It

A



F1   dA  



1.5

0

(0.25)V y 3 3.375 3

Vy 2 (0.25dy ) 3.375 1.5 0



(0.25)(1.5)3 (3.375)(3)

 0.083333V MD 

M D : Ve  2 F1 (3 sin 60) Ve  (2)(0.083333) V (3 sin 60) e  (2)(0.083333)(3 sin 60)

e  0.433 in. 



PROBLEM 6.C1

x4 x3 x1

x2 w

P1

P2 t h

A

B L

a

b

A timber beam is to be designed to support a distributed load and up to two concentrated loads as shown. One of the dimensions of its uniform rectangular cross section has been specified and the other is to be determined so that the maximum normal stress and the maximum shearing stress in the beam will not exceed given allowable values  all and  all . Measuring x from end A and using either SI or U.S. customary units, write a computer program to calculate for successive cross sections, from x  0 to x  L and using given increments x, the shear, the bending moment, and the smallest value of the unknown dimension that satisfies in that section (1) the allowable normal stress requirement and (2) the allowable shearing stress requirement. Use this program to solve Prob. 5.65, assuming  all  12 MPa and  all  825 kPa and using x  0.1 m.

SOLUTION See solution of Prob. 5.C2 for the determination of RA , RB , V ( x), and M ( x) We recall that

V ( x)  RA STPA  RB STPB  P1 STP1  P2 STP2 w( x  x3 ) STP3  w( x  x4 ) STP4 M ( x)  RA ( x  a ) STPA  RB ( x  a  L) STPB  P1 ( x  x1 ) STP1 1 1 w( x  x3 ) 2 STP3  w( x  x4 )2 STP4 2 2 where STPA, STPB, STP1, STP2, STP3, and STP4 are step functions defined in Problem 5.C2.

P2 ( x  x2 ) STP2 

(1)

To satisfy the allowable normal stress requirement: If unknown dimension is h:

S min  |M |/ all . From

1 S  th 2 , we have h  h  6 S /t 6

If unknown dimension is t:

S min  |M |/ all . From (2)

1 S  th 2 , we have t  t  6S /h 2 6

To satisfy the allowable shearing stress requirement: We use Equation (6.10), Page 378: If unknown dimension is h: If unknown dimension is t:

3 |V | 3 |V |  2 A 2 th 3|V | h  h  2t all 3M t  t  2h all

 max 



PROBLEM 6.C1 (Continued) Program Outputs

Problem 5.65

RA  2.40 kN RB  3.00 kN X m

V kN

M kN  m

HSIG mm

HTAU mm

0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 2.10 2.20 2.30 2.40

2.40 2.40 2.40 2.40 2.40 2.40 2.40 2.40 0.60 0.60 0.60 0.60 0.60 0.60 0.60 0.60 –3.00 –3.00 –3.00 –3.00 –3.00 –3.00 –3.00 –3.00 0.00

0.000 0.240 0.480 0.720 0.960 1.200 1.440 1.680 1.920 1.980 2.040 2.100 2.160 2.220 2.280 2.340 2.400 2.100 1.800 1.500 1.200 0.900 0.600 0.300 0.000

0.00 54.77 77.46 94.87 109.54 122.47 134.16 144.91 154.92 157.32 159.69 162.02 164.32 166.58 168.82 171.03 173.21 162.02 150.00 136.93 122.47 106.07 86.60 61.24 0.05

109.09 109.09 109.09 109.09 109.09 109.09 109.09 109.09 27.27 27.27 27.27 27.27 27.27 27.27 27.27 27.27 136.36 136.36 136.36 136.36 136.36 136.36 136.36 136.36 0.00

        

The smallest allowable value of h is the largest of the values shown in the last two columns.

h  h  173.2 mm 

For Problem 5.65,



PROBLEM 6.C2 P

b

w B

A

8b

L

A cantilever timber beam AB of length L and of uniform rectangular section shown supports a concentrated load P at its free end and a uniformly distributed load w along its entire length. Write a computer program to determine the length L and the width b of the beam for which both the maximum normal stress and the maximum shearing stress in the beam reach their largest allowable values. Assuming  all  1.8 ksi and  all  120 psi, use this program to determine the dimensions L and b when (a) P  1000 lb and w  0, (b) P  0 and w  12.5 lb/in., (c) P  500 lb and w  12.5 lb/in.

SOLUTION Both the maximum shear and the maximum bending moment occur at A. We have VA  P  wL M A  PL 

1 2 wL 2

To satisfy the allowable normal stress requirement:

 all 

MA MA 3M A   2 1 S (8 ) 32b3 b b 6

 3 MA  b  b     32  all 

1/3

To satisfy the allowable shearing stress requirement: We use Equation (6.10), Page 378.

 all 

3V 3V 3 VA   A2 2 A 2 b(8b) 16b

 3 VA  b  b    16  all 

1/ 2

Program For L  0, VA  P and b > 0, while M A  0 and b  0. Starting with L  0 and using increments L  0.001 in., we increase L until b and b become equal. We then Print L and b.



PROBLEM 6.C2 (Continued)

Program Outputs

For P  1000 lb, w  0.0 lb/in.

For P  0 lb, w  12.5 lb/in.

Increment  0.0010 in.

Increment  0.0010 in.

L  37.5 in., b  1.250 in.

L  70.3 in., b  1.172 in.

For P  500 lb, w  12.5 lb/in. Increment  0.0010 in.

L  59.8 in., b  1.396 in.



bn

PROBLEM 6.C3

b2

A beam having the cross section shown is subjected to a vertical shear V. Write a computer program that, for loads and dimensions expressed in either SI or U.S. customary units, can be used to calculate the shearing stress along the line between any two adjacent rectangular areas forming the cross section. Use this program to solve (a) Prob. 6.10, (b) Prob. 6.12, (c) Prob. 6.22.

hn h2

V

h1

b1

SOLUTION 1.

Enter V and the number n of rectangles.

2.

For i  1 to n, enter the dimensions bi and hi .

3.

Determine the area Ai  bi hi of each rectangle.

4.

Determine the elevation of the centroid of each rectangle:

yi 

i

h

k

 0.5hi

k 1

and the elevation y of the centroid of the entire section:

 y    5.





i i

i

i

i

Determine the centroidal moment of inertia of the entire section: I

1

 12 b h

3 i i

i

6.



 A y    A    A i ( yi  y ) 2  

For each surface separating two rectangles i and i  1, determine Qi of the area below that surface:

Qi 

i

 A (y k

k

 y)

k 1

7.

Select for ti the smaller of bi and bi 1. The shearing stress on the surface between the rectangles i and i  1 is

i 

VQi Iti



    PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.



Program Outputs

Problem 6.10 Shearing force  10 kN y  75.000 mm above base

I  39.580  106 mm 4 Between Elements 1 and 2:

  418.39 kPA Between Elements 2 and 3:

  919.78 kPA

(a)

  765.03 kPA

(b)

Between Elements 3 and 4:


  418.39 kPA Problem 6.12 Shearing force  10 kips y  2.000 in. I  14.58 in 4


  2.400 ksi Between Elements 2 and 3:

  3.171 ksi

(a)

  2.400 ksi

(b)





Program Outputs (Continued)

Problem 6.22 Shearing force  90 kN y  65.000 mm

I  58.133  106 mm 4 Between Elements 1 and 2:

  23.222 MPA

(b)

  30.963 MPA

(a)




PROBLEM 6.C4

y xn x y2 y1 x2 x1

A plate of uniform thickness t is bent as shown into a shape with a vertical plane of symmetry and is then used as a beam. Write a computer program that, for loads and dimensions expressed in either SI or U.S. customary units, can be used to determine the distribution of shearing stresses caused by a vertical shear V. Use this program (a) to solve Prob. 6.47, (b) to find the shearing stress at a Point E for the shape and load of Prob. 6.50, assuming a thickness t  14 in.

SOLUTION For each element on the right-hand side, we compute (for i  1 to n): Length of element  Li  ( xi  xi 1 ) 2  ( yi  yi 1 )2 Area of element  Ai  tLi where

t

1 in. 4

1 ( yi  yi 1 ) 2

Distance from x axis to centroid of element  yi  Distance from x axis to centroid of section:

y    Ai yi   Ai Note that yn  0 and that xn 1  yn 1  0. Moment of inertia of section about centroidal axis:

1  I  2 Ai  ( yi  yi 1 ) 2  ( yi  y )2  12  Computation of Q at Point P where stress is desired:

Q   Ai ( yi  y ) where sum extends to the areas located between one end of section and Point P. Shearing stress at P:



VQ It

Note:  max occurs on neutral axis, i.e., for yP  y .



PROBLEM 6.C4 (Continued) Program Outputs

Part (a): I  0.5333 in 4

 max  2.02 ksi



 B  1.800 ksi



Part (b): I  22.27 in 4

 E  194.0 psi





PROBLEM 6.C5

y x1 x2

yn

y1

tn

O

y2 x

e

t2 t 1

The cross section of an extruded beam is symmetric with respect to the x axis and consists of several straight segments as shown. Write a computer program that, for loads and dimensions expressed in either SI or U.S. customary units, can be used to determine (a) the location of the shear center O, (b) the distribution of shearing stresses caused by a vertical force applied at O. Use this program to solve Prob. 6.70.

V

SOLUTION Since section is symmetric with x axis, computations will be done for top half. For i  1 to n  1: (Note: n  1 is the origin)

Enter ti , xi ,

yi

Compute length of each segment. For i  1 to n:  xi  xi 1  xi yi  yi 1  yi L  ( xi2  yi2 )1/ 2

Calculate moment of inertia I x . Consider each segment as made of 100 equal parts. For i  1 to n:  Area  Li ti /100 For j  1 to 100: y  yi  yi ( j  0.5)/100 I  ( Area) y 2 Ix  Ix  I



Since only top half was used,

I x  2I x Calculate shearing stress at ends of segments and shear forces in segments. For i  1 to n:

 Area  Li ti /100,

 new   next

For j  1 to 100: y  yi  yi ( j  0.5)/100 Q  ( Area) y  old   new , Q  Q  Q

 new  VQ /I x ti  ave  0.5( old   new )      ave Force i   (∆Area)

 i  VQ/I x ti ( adjacent )i  VQ/I x ti 1

Qi  Q

 next  ( adjacent )i Location of shear center. Calculate moment of shear forces about origin. For L  1 to n, ( Fx )i  Forcei ( xi )/Li ( Fy )i  Forcei (yi )/Li Moment i  ( Fx )i yL  ( Fy )i xi Moment  Moment  Moment i For whole section, moment  2(moment), Shear center is at

e  Moment/V




Program Output

Problem 6.70 T(K) mm

X(K) mm

Y(K) mm

L(K) mm

1

6.00

60.62

0.00

70.00

2

6.00

0.00

35.00

35.00

3

6.00

0.00

0.00

Moment of inertia: I x  514487 mm 4 Junction of Segments

Q mm3

Shear  1000.000 N Tau Before MPa

Tau After MPa

Force in Segment kN

1 and 2

7350.00

2.38

2.38

335.01

2 and 3

11,025.00

3.57

3.57

666.27

Moment of shear forces about origin: M  20.309 N  m  counterclockwise Distance from origin to shear center: e  20.309 mm



tn

t2

t1

ti a1

t0

an

a2

ai

O a1

PROBLEM 6.C6 A thin-walled beam has the cross section shown. Write a computer program that, for loads and dimensions expressed in either SI or U.S. customary units, can be used to determine the location of the shear center O of the cross section. Use the program to solve Prob. 6.75.

ai a2

an

b2 e bi bn

SOLUTION Distribution of shearing stresses in element i. Let

V  Shear in cross section I  Centroidal moment of inertia of section

We have for shaded area

Q  Ay  ti (ai  y )

ai  y 2

1 ti (ai2  y 2 ) 2 QV V 2   (ai  y 2 ) Iti 2I



Force exerted on element i.

Fi 



ai ai

Vti 2I Vti  I Vti  I 

 (ti dy )

 a  a ai

ai ai

0



2 i

 y 2 dy

2 i

 y 2 dy



 3 1 3 2V 3  ai  3 ai   3 I ti ai  




The system of the forces Fi must be equivalent to V at shear center. F  F :

M A  M A :

Divide (2) by (1):

e

2V ti ai3  V 3 I 2V ti ai3bi  eV 3 I

(1) (2)

ti ai3bi



ti ai3

Program Output

Problem 6.75 For Element 1:

t  0.75 in., a  4 in., b  0 For Element 2:

t  0.75 in., a  3 in., b  8 in. Answer:

e  2.37 in.





Mechanics of Materials 7th Edition Beer Johnson Chapter 6

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