Philpot MoM 2nd Ch07-11 ISM

7.1 For the cantilever beam and loading shown, (a) Derive equations for the shear force V and the bending moment M for any location in the beam. (Place the origin at point A.) (b) Plot the shear-force and bending-moment diagrams for the beam using the derived functions. Fig. P7.1

Solution Beam equilibrium: Fy Ay w0 L

MA

Ay

w0 L

MA

w0 L

MA Section a-a: Fy w0 L

a

L 2 w0 L2 2

w0 x V

V

Ma

0

w0 L

0

0 w0 x

w0 ( L

x)

x 2

M

w0 L2 2

w0 Lx

w0 x

M

w0 L2 2

w0 x 2 2

w0 L x

0 w0 2 (L 2

x 2 ) w0 Lx

(b) Shear-force and bending-moment diagrams

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7.2 For the simply supported beam shown, (a) Derive equations for the shear force V and the bending moment M for any location in the beam. (Place the origin at point A.) (b) Plot the shear-force and bending-moment diagrams for the beam using the derived functions. Fig. P7.2

Solution Beam equilibrium: Fy Ay C y P MA

0

Pa C y (a b) Cy

Pa a b

0

and

Section a-a: For the interval 0 ≤ x < a: Pb Fy Ay V V a b Ma

a

Ay x M

b

Pb a b

0

V

Pb x M a b

Section b-b: For the interval a ≤ x < b: Pb Fy Ay P V a b Pa V a b Mb

Ay

P V

Ay x

P( x a) M

M

Pb x a b

0

M

Pb a b Pb x a b

0

Pb x a b

P( x a) M

0

P( x a)





Solution Beam equilibrium: Fy wa a wbb C y Cy MC

0

wa a wbb

wa a b MC

a 2

b 2

wbb

MC

a 2

wa a b

wbb

0 b 2

Section a-a: For the interval 0 ≤ x < a: Fy wa x V 0

Ma

a

x wa x 2

V

M

0

Section b-b: For the interval a ≤ x < b: Fy wa a wb x a V Mb

b

wa a x M

wa a

V

wa x wa x 2 2

M

0

wb x a

a 2 wa a x

wb x a a 2

x a 2

wb x a 2

M

0

2




7.4 For the simply supported beam subjected to the loading shown, (a) Derive equations for the shear force V and the bending moment M for any location in the beam. (Place the origin at point A.) (b) Plot the shear-force and bending-moment diagrams for the beam using the derived functions. Fig. P7.4

Solution Beam equilibrium:

MC

a 2

wa a b

b 2

Ay (a b)

0

wa a(a 2b) wbb 2 2(a b)

Ay MA

wbb

a 2

wa a

wa a 2

Cy

b 2

wbb a

C y (a b)

0

wbb(2a b) 2(a b)

Section a-a: For the interval 0 ≤ x < a: Fy Ma

Ay

a

wa x V

0

Ay x

x 2

wa x wa x 2 2

M

V M Ay x

Section b-b: For the interval a ≤ x < b: Fy Ay wa a wb ( x a ) V V

Ay

wa a

wb ( x

wa a (a 2b) 2(a b) Mb M

Ay x

b

Ay x

Ay

wa a (a 2b) wbb 2 2(a b)

wa x

0 wa x 2 2

wa a (a 2b) wbb 2 x 2(a b)

0

a)

wbb 2 2(a b)

wa a x

wa a x

wa x

a 2

a 2

wa a

wb ( x a )

wb x a x a wb 2

2

x a 2

M

wa a(a 2b) 2(a b)

0 wbb 2 x 2(a b)

wa a x

a 2

x a wb 2

2





Solution Beam equilibrium: w0 L Fy By 0 2 w0 L L MB MB 2 3

By 0

MB

w0 L 2 w0 L2 6

Section a-a:

w0 x x L 2

Fy Ma

a

w0 x x L 2

V x 3

0

V M

0

M

w0 x 2 2L w0 x3 6L



7.6 For the simply supported beam shown, (a) Derive equations for the shear force V and the bending moment M for any location in the beam. (Place the origin at point A.) (b) Plot the shear-force and bending-moment diagrams for the beam using the derived functions. (c) Determine the location and the magnitude of the maximum bending moment. Fig. P7.6

Solution Beam equilibrium: MB

w0 L L 2 3

0

Ay

w0 L 6

w0 L 2 L 2 3

0

By

w0 L 3

Ay L

MA

By L

Section a-a: Fy

w0 x x L 2

Ay V

Ma

a

Ay x

w0 L 6

V

w0 L 6

w0 x 2 2L

V

0

w0 x 2 2L

w0 x x L 2

x 3

M

w0 Lx 6

w0 x 3 6L

M

0

w0 x 3 w0 Lx 6L 6 (b) Shear-force and bending-moment diagrams M


(c) Location and magnitude of maximum bending moment: The maximum bending moment is located at the location where V = 0. Therefore, the maximum bending moment occurs at: w0 L w0 x 2 V 0 6 2L w0 x 2 2L

w0 L 6

x

L2 3

L 3

0.577350 L

Substitute this value of x into the bending moment equation to determine the moment magnitude: w0 x3 w0 Lx M 6L 6 w0 (0.577350 L)3 w0 L(0.577350 L) M max 0.064150w0 L2 6L 6

Ans.

Ans.


7.7 For the simply supported beam subjected to the loading shown, (a) Derive equations for the shear force V and the bending moment M for any location in the beam. (Place the origin at point A.) (b) Plot the shear-force and bending-moment diagrams for the beam using the derived functions. (c) Report the maximum bending moment and its location. Fig. P7.7

Solution Beam equilibrium: MA (50 kN)(3 m) (75 kN)(6 m) Dy (10 m) 0 Dy Fy

Ay

60 kN

Dy

50 kN 75 kN

Ay

65 kN

Section a-a: For the interval 0 ≤ x < 3 m: Fy Ay V 65 kN V Ma

0

Ay x M

a

0

V

(65 kN)x M

0

Section b-b: For the interval 3 m ≤ x < 6 m: Fy Ay 50 kN V 65 kN 50 kN V V

Mb

b

M

65 kN

(65 kN) x

0

15 kN Ay x (50 kN)(x 3 m) M

(65 kN)x (50 kN)(x 3 m) M M

0

(15 kN) x 150 kN-m

Section c-c: For the interval 6 m ≤ x < 10 m: Fy Ay 50 kN 75 kN V 65 kN 50 kN 75 kN V V

Mc

c

0

60 kN

Ay x (50 kN)(x 3 m) (75 kN)(x 6 m) M (65 kN)x (50 kN)(x 3 m) (75 kN)(x 6 m) M M

0

(60 kN) x 600 kN-m



(c) Maximum bending moment and its location Mmax = 240 kN-m @ x = 6 m


7.8 For the simply supported beam subjected to the loading shown, (a) Derive equations for the shear force V and the bending moment M for any location in the beam. (Place the origin at point A.) (b) Plot the shear-force and bending-moment diagrams for the beam using the derived functions. (c) Report the maximum positive bending moment, the maximum negative bending moment, and their respective locations. Fig. P7.8

Solution Beam equilibrium: M B (20 kN)(2 m) (60 kN)(6 m) Dy (8 m) 0 Dy Fy

By

40 kN

Dy By

20 kN 60 kN 40 kN

Section a-a: For the interval 0 ≤ x < 2 m: Fy 20 kN V 0 Ma

(20 kN)x M

a

0

V 0

M

20 kN (20 kN) x

Section b-b: For the interval 2 m ≤ x < 8 m: Fy 20 kN By V 20 kN 40 kN V V Mb

b

0

20 kN (20 kN)x By ( x 2 m) M

(20 kN)x (40 kN)( x 2 m) M M

0

(20 kN) x 80 kN-m


Section c-c: For the interval 8 m ≤ x < 10 m: Fy 20 kN By 60 kN V 20 kN 40 kN 60 kN V V

Mc

c

0

40 kN

(20 kN)x By ( x 2 m) (60 kN)(x 8 m) M (20 kN)x (40 kN)( x 2 m) (60 kN)(x 8 m) M M

0

(40 kN) x 400 kN-m


(c) Maximum bending moment and its location Mmax-positive = 80 kN-m @ x = 8 m Mmax-negative = –40 kN-m @ x = 2 m



Solution Beam equilibrium: M C (7 kips/ft)(30 ft)(15 ft) By (21 ft) 0 By Fy

By

150 kips

Cy Cy

(7 kips/ft)(30 ft) 0 60 kips

Section a-a: For the interval 0 ≤ x < 9 ft: Fy (7 kips/ft)x V 0 Ma

(7 kips/ft)(x)

a

V

x 2

M

Section b-b: For the interval 9 ft ≤ x < 30 ft: Fy (7 kips/ft)x By V V

Mb

b

0

(7 kips/ft) 2 x 2

M

(7 kips/ft)x 150 kips V

0

(7 kips/ft) x 150 kips

(7 kips/ft)(x)

x 2

By ( x 9 ft) M

(7 kips/ft)(x)

x 2

(150 kips)( x 9 ft) M

M

(7 kips/ft) x

(7 kips/ft) 2 x 2

0

(150 kips) x 1,350 kips



(c) Maximum bending moment and its location Mmax-positive = 257.14 kip-ft @ x = 21.43 ft Mmax-negative = –283.50 kip-ft @ x = 9 ft



Solution Beam equilibrium: Fy (4 kips/ft)(8 ft) C y Cy MC

0

32 kips

(4 kips/ft)(8 ft)(12 ft) M C MC

0

384 kip-ft

Section a-a: For the interval 0 ≤ x < 8 ft: Fy (4 kips/ft)x V 0 Ma

a

(4 kips/ft)(x)

x 2

V M

0

M

(4 kips/ft) x 4 kips/ft 2 x 2

Section b-b: For the interval 8 ft ≤ x < 16 ft: Fy (4 kips/ft)(8 ft) V 0 V Mb

b

32 kips

(4 kips/ft)(8 ft)(x 4 ft) M M

0

(32 kips) x 128 kip-ft




7.11 For the simply supported beam subjected to the loading shown, (a) Derive equations for the shear force V and the bending moment M for any location in the beam. (Place the origin at point A.) (b) Plot the shear-force and bending-moment diagrams for the beam using the derived functions. (c) Report the maximum bending moment and its location. Fig. P7.11

Solution Beam equilibrium: MA (42 kips)(10 ft) (6 kips/ft)(20 ft)(20 ft) C y (30 ft) 0 Cy

Fy

Ay

Cy Ay

94 kips

42 kips (6 kips/ft)(20 ft) 0 68 kips

Section a-a: For the interval 0 ≤ x < 10 ft: Fy Ay V 68 kips V Ma

a

Ay x M

0

(68 kips) x M

V 0

M

68 kips (68 kips) x

Section b-b: For the interval 10 ft ≤ x < 30 ft: Fy Ay 42 kips (6 kips/ft)(x 10 ft) V 68 kips 42 kips (6 kips/ft)(x 10 ft) V V

Mb

b

0

(6 kips/ft) x 86 kips

Ay x (42 kips)( x 10 ft) (6 kips/ft)(x 10 ft) (68 kips) x (42 kips)( x 10 ft) M

x 10 ft 2

6 kips/ft (x 10 ft) 2 2

M M

0

3x 2 86 x 120 kip-ft



(c) Maximum bending moment and its location Mmax = 736.33 kip-ft @ x = 14.33 ft



Solution Beam equilibrium: M A 180 kN-m C y (9 m) (36 kN)(12 m) 0 Cy Fy

Ay

28 kN

Cy Ay

36 kN

0

8 kN

Section a-a: For the interval 0 ≤ x < 4 m: Fy Ay V 8 kN V Ma

Ay x M

a

0

V

(8 kN) x M

Section b-b: For the interval 4 m ≤ x < 9 m: Fy Ay V 8 kN V 0 Mx

0

M

V

8 kN (8 kN) x

8 kN

Ay x 180 kN-m M (8 kN) x 180 kN-m M M

0

(8 kN) x 180 kN-m

Section c-c: For the interval 9 m ≤ x < 12 m: Fy Ay C y V 8 kN 28 kN V V

Mc

0

36 kN

Ay x C y ( x 9 m) 180 kN-m M

c

(8 kN) x (28 kN)(x 9 m) 180 kN-m M M

0

(36 kN) x 432 kN-m



(c) Maximum bending moment and its location Mmax-positive = 32 kN-m @ x = 4 m Mmax-negative = –148 kN-m @ x = 4 m




Fy

Cy

5 kips/ft 6 ft Cy

0

30 kips

M C 120 kip-ft

5 kips/ft 6 ft 3 ft

MC

MC

0

210 kip-ft

Section a-a: For the interval 0 ≤ x < 8 ft: Fy

V

0

V

M a -a

120 kip-ft M

0

M

0 kips 120 kip-ft

Section b-b: For the interval 8 ft ≤ x < 14 ft: Fy

5 kips/ft x 8 ft V

M b -b

V

0

5 x 40 kips 120 kip-ft

M

2.5 x 2

5 kips/ft x 8 ft

x 8 ft 2

M =0

40x 280 kip-ft


(b) Shear-force and bending-moment diagrams:




Fy

1 2

Ay Ay

MA

6 kips/ft 9 ft

1 2

MA

MA

6 kips/ft 9 ft 6 ft 0

400 kip-ft

Section a-a: For the interval 0 ≤ x < 9 ft: 1 6x Fy Ay x V 2 9 V

x2 3

44 kips

MA

Ay x

1 2

a

0

44 kips

17 kips 14 ft

Ma

17 kips

400 kip-ft x3 9

M

1 2

44 kips

6x x 9

x 3 1 2

44 kips x

6x x V 9

0

M 6x x 9

x 3

M

0

44x 400 kip-ft

Section b-b: For the interval 9 ft ≤ x < 14 ft: Fy

1 6 kips/ft 9 ft 2

Ay

44 kips V

V

1 6 kips/ft 9 ft 2

V

0

17 kips


M b -b

MA

Ay x

400 kip-ft M

1 6 kips/ft 9 ft x 6 ft 2 44 kips x

M

1 6 kips/ft 9 ft x 6 ft 2

M

0

17x 238 kip-ft

(b) Shear-force and bending-moment diagrams:



Solution Beam equilibrium: MA

250 kip-ft

7 kips/ft 25 ft 12.5 ft

C y 17 ft Cy Fy

Ay

0

113.97 kips

Cy

7 kips/ft 25 ft

Ay 113.97 kips Ay

7 kips/ft 25 ft

0

61.03 kips

Section a-a: For the interval 0 ≤ x < 13 ft:

Fy

Ay

7 kips/ft x V V

Ma

7 kips/ft x

61.03 kips x M

3.5 x 2

x 2

Mb

b

Ay x

7 kips/ft x

x 2

M

0

61.03 x kip-ft

7 kips/ft x V

0

7 x 61.03 kips 7 kips/ft x

61.03 kips x M

0

M

Section b-b: For the interval 13 ft ≤ x < 17 ft: Fy Ay 7 kips/ft x V 61.03 kips

V

7 kips/ft x V

7 x 61.03 kips

Ay x

a

61.03 kips

3.5 x 2

x 2

250 kip-ft

7 kips/ft x

x 2

M

250 kip-ft

M

0

61.03x 250 kip-ft


Section c-c: For the interval 17 ft ≤ x < 25 ft: Fy Ay C y 7 kips/ft x V

61.03 kips 113.97 kips V Mc

c

0

7 x 175 kips

Ay x C y ( x 17 ft) 250 kip-ft 175 kips x

7 kips/ft x

x 2

M

113.97 kips (17 ft)

250 kip-ft M

7 kips/ft x V

M

7 kips/ft x

x 2

0

3.5 x 2 175 x 2,187.5 kip-ft

(b) Shear-force and bending-moment diagrams: (c) Maximum bending moment and its location Mmax-positive = 266.04 kip-ft @ x = 8.72 ft Mmax-nagative = –224.0 kip-ft @ x = 17 ft


7.16 Use the graphical method to construct the shear-force and bending-moment diagrams for the beam shown. Label all significant points on each diagram and identify the maximum moments (both positive and negative) along with their respective locations. Clearly differentiate straight-line and curved portions of the diagrams. Fig. P7.16


MA Fy

28 kips 4 ft

42 kips 8 ft

Dy

32.00 kips

Ay

Dy

34 kips 56 kips

Ay

32.00 kips 28 kips 42 kips Ay

Dy 14 ft

0

0

38.00 kips

Shear-force and bending-moment diagrams:




Fy

35 kN 4 m

45 kN 8 m

15 kN 14 m

Dy 10 m

Dy

71 kN

Ay

Dy

35 kN

Ay

71 kN 35 kN Ay

0

45 kN 15 kN 45 kN 15 kN

0

24 kN




Solution Beam equilibrium: Fy

15 kips 25 kips C y Cy

MC

10 kips

15 kips 9 ft MC

0

25 kips 3 ft

MC

0

60 kip-ft





MA

10 kips/ft 12 ft 6 ft Cy

Fy

0

40 kips

Ay

Cy

Ay

40 kips Ay

C y 18 ft

10 kips/ft 12 ft 10 kips/ft 12 ft

0

80 kips





MA

4.5 kips/ft 12 ft 9 ft Cy

Fy

0

40.50 kips

Ay

Cy

Ay

40.50 kips Ay

C y 12 ft

4.5 kips/ft 12 ft 4.5 kips/ft 12 ft

0

13.50 kips





Fy

Ay

40 kN/m 3 m Ay

MA

MA

50 kN

0

70 kN 40 kN/m 3 m 1.5 m

(50 kN)(3 m) 60 kN-m MA 90 kN-m

0





28 kips Cy

MC

9 kips/ft 5 ft

0

17 kips

28 kips 8 ft MC

Cy


MC

0

111.5 kip-ft





MA Fy

10 kips/ft 9 ft 4.5 ft 6 ft Ay

Dy

47.25 kips

Dy

10 kips/ft 9 ft

Ay

42.75 kips

Dy 20 ft

0

0





MA

4.5 kips/ft 7 ft 7 ft 38 kips 14.5 ft

E y 18 ft

0

E y 18.36 kip Fy

Ay

Ey

4.5 kips/ft 7 ft

Ay 18.36 kips Ay

38 kips

4.5 kips/ft 7 ft

38 kips 0

11.86 kips





MA

60 kN 2.5 m Dy 9 m Dy

Fy

Ay

Dy

45 kN/m 5 m 7.5 m

0

156.67 kN 60 kN

45 kN/m 4 m

Ay 156.67 kN 60 kN Ay

45 kN/m 4 m

0

83.33 kN





Fy

10 kips C y

0

C y 10 kips MC

10 kips 10 ft MC

60 kip-ft M C

0

40 kip-ft





Fy

Ay

2 kN 11 kN Ay

MA

0

13 kN

M A 50 kN-m 11 kN 6 m MA

2 kN 3.5 m 0

23 kN-m





MA

66 kN-m 96 kN-m Dy 12 m Dy

Fy

Ay

Dy Ay

0

13.50 kN Ay 13.50 kN

0

13.50 kN





MA

80 kN-m

25 kN/m 6 m 3 m

50 kN-m By 6 m By Fy

70 kN

Ay

By

Ay

70 kN Ay

0

25 kN/m 6 m 25 kN/m 6 m

0

80 kN





MA

25 kN-m

15 kN 8 m

7 kN/m 3 m 13.5 m Dy Fy

Ay

0

31.54 kN

Dy 15 kN

7 kN/m 3 m

Ay 31.54 kN 15 kN Ay

Dy 12 m

7 kN/m 3 m

0

4.46 kN



7.31 Draw the shear-force and bending-moment diagram for the beam shown in Fig. P7.31. Assume the upward reaction provided by the ground to be uniformly distributed. Label all significant points on each diagram. Determine the maximum value of (a) the internal shear force and (b) the internal bending moment. Fig. P7.31


Fy

2 kips/ft 8 ft

25 kips 25 kips w 16 ft

0

w 4.125 kips/ft

Shear-force and bending-moment diagrams (a) Maximum value of internal shear force: V = 16.50 kips @ x = 4 ft

Ans.

V = −16.50 kips @ x = 12 ft

Ans.

(b) Maximum value of internal bending moment: M = 33 kip-ft @ x = 4 ft, 12 ft


Ans.

7.32 Draw the shear-force and bending-moment diagram for the beam shown in Fig. P7.32. Assume the upward reaction provided by the ground to be uniformly distributed. Label all significant points on each diagram. Determine the maximum value of (a) the internal shear force and (b) the internal bending moment. Fig. P7.32

Solution Beam equilibrium: Fy 40 kN/m 1 m

50 kN

40 kN/m 1 m

w 4m

0

w 32.5 kN/m

Shear-force and bending-moment diagrams (a) Maximum value of internal shear force: V = ±25 kN @ x = 2 m

Ans.

(b) Maximum value of internal bending moment: M = −4.62 kN-m @ x = 1.23 m M = −4.62 kN-m @ x = 2.77 m Mmax = 5.00 kN-m

Ans.


7.33 Use the graphical method to construct the shear-force and bending-moment diagrams for the beam shown. Label all significant points on each diagram and identify the maximum moments along with their respective locations. Additionally: (a) Determine V and M in the beam at a point located 0.75 m to the right of B. (b) Determine V and M in the beam at a point located 1.25 m to the left of C. Fig. P7.33


MA

125 kN 3 m C y 15 m Cy

Fy

50 kN/m 12 m 9 m 0

385.00 kN

Ay

C y 125 kN

Ay

385.00 kN 125 kN Ay

50 kN/m 12 m 50 kN/m 12 m

0

340.00 kN

Shear-force and bending-moment diagrams Shear force V and bending moment M at specific locations: (a) At x = 3.75 m, V = 177.5 kN

Ans.

M = 1,167 kN-m

Ans.

(b) At x = 13.75 m, V = −323 kN

Ans.

M = 442 kN-m

Ans.


7.34 Use the graphical method to construct the shear-force and bending-moment diagrams for the beam shown. Label all significant points on each diagram and identify the maximum moments along with their respective locations. Additionally: (a) Determine V and M in the beam at a point located 0.75 m to the right of B. (b) Determine V and M in the beam at a point located 1.25 m to the left of C. Fig. P7.34


MB

15 kN 3 m

40 kN/m 6 m 3 m

18 kN 10 m Cy Fy

By

Cy 6 m

142.50 kN

C y 15 kN

40 kN/m 6 m

By 142.5 kN 15 kN By

0 18 kN

40 kN/m 6 m

18 kN

0

130.50 kN


Ans.

M = 30.4 kN-m

Ans.

(b) At x = 7.75 m, V = −74.5 kN

Ans.

M = 52.4 kN-m

Ans.


7.35 Use the graphical method to construct the shearforce and bending-moment diagrams for the beam shown. Label all significant points on each diagram and identify the maximum moments along with their respective locations. Additionally: (a) Determine V and M in the beam at a point located 0.75 m to the right of B. (b) Determine V and M in the beam at a point located 1.25 m to the left of C. Fig. P7.35


MB

25 kN/m 3 m 1.5 m 65 kN/m 5 m 2.5 m Cy

Fy

By C y

0

185.00 kN 25 kN/m 3 m

By 185.00 kN 65 kN/m 5 m By

Cy 5 m

65 kN/m 5 m

25 kN/m 3 m 0

65.00 kN


Ans.

M = 199.2 kN-m

Ans.

(b) At x = 6.75 m, V = −103.8 kN

Ans.

M = 180.5 kN-m

Ans.


7.36 Use the graphical method to construct the shearforce and bending-moment diagrams for the beam shown. Label all significant points on each diagram and identify the maximum moments along with their respective locations. Additionally: (a) Determine V and M in the beam at a point located 0.75 m to the right of B. (b) Determine V and M in the beam at a point located 1.25 m to the left of C. Fig. P7.36


75 kN 60 kN Cy

MC

35 kN/m 6 m

0

75.00 kN

75 kN 6 m

60 kN 3.5 m

35 kN/m 6 m 3 m MC

Cy

MC

120 kN-m 0

90.00 kN-m

Shear-force and bending-moment diagrams Shear force V and bending moment M at specific locations: (a) At x = 3.25 m, V = −21.3 kN

Ans.

M = 16.09 kN-m

Ans.

(b) At x = 4.75 m, V = 31.25 kN

Ans.

M = 23.6 kN-m

Ans.


7.37 Use the graphical method to construct the shearforce and bending-moment diagrams for the beam shown. Label all significant points on each diagram and identify the maximum moments along with their respective locations. Additionally: (a) Determine V and M in the beam at a point located 1.50 m to the right of B. (b) Determine V and M in the beam at a point located 1.25 m to the left of D. Fig. P7.37


52 kN 3 m 150 kN-m Dy

Fy

35 kN/m 9 m 4.5 m 36 kN 12 m

0

204.83 kN

By

Dy

By

204.83 kN 52 kN By

Dy 9 m

52 kN

35 kN/m 9 m

36 kN

35 kN/m 9 m

36 kN

0

198.17 kN


Ans.

M = 23.9 kN-m

Ans.

(b) At x = 10.75 m, V = −125.1 kN

Ans.

M = 75.7 kN-m

Ans.




MB

25 kN/m 3.5 m 1.75 m 25 kN/m 12.5 m 6.25 m

80 kN 5.5 m

20 kN 12.5 m

0

Dy Fy

By

Dy

Dy 10 m

161.00 kN 25 kN/m 16 m

By 161.00 kN

80 kN 20 kN

25 kN/m 16 m

80 kN 20 kN

0

By 179.00 kN Shear-force and bending-moment diagrams Shear force V and bending moment M at specific locations: (a) At x = 5.0 m, V = 54.0 kN

Ans.

M = −44.0 kN-m

Ans.

(b) At x = 12.25 m, V = −47.3 kN

Ans.

M = −49.5 kN-m

Ans.




MB

160 kN 2 m

50 kN/m 2 m 1 m

50 kN/m 2 m 1 m 120 kN/m 5 m 4.5 m Dy Fy

By

Dy 7 m

0

340 kN

Dy 160 kN

50 kN/m 4 m

120 kN/m 5 m By 340 kN 160 kN 120 kN/m 5 m By

50 kN/m 4 m

0

620 kN

Shear-force and bending-moment diagrams Shear force V and bending moment M at specific locations: (a) At x = 3.50 m, V = 285 kN

Ans.

M = 63.8 kN-m

Ans.

(b) At x = 7.75 m, V = −190.0 kN

Ans.

M = 331 kN-m

Ans.


7.40 Use the graphical method to construct the shear-force and bending-moment diagrams for the beam shown. Label all significant points on each diagram and identify the maximum moments along with their respective locations. Clearly differentiate straight-line and curved portions of the diagrams. Fig. P7.40


MA

225 kN-m

120 kN/m 4 m 2 m

60 kN/m 2.5 m 8.75 m Cy Fy

Ay C y Ay

0

273 kN 120 kN/m 4 m

273 kN

60 kN/m 2.5 m

120 kN/m 4 m

60 kN/m 2.5 m Ay

C y 7.5 m

0

357 kN

Shear-force and bending-moment diagrams




MB

25 kip-ft


5 kips/ft 5 ft 2.5 ft E y 15 ft

25 kips 10 ft

0

E y 17.67 kips Fy

By

Ey

5 kips/ft 8 ft

By 17.67 kips By

25 kips

5 kips/ft 8 ft

25 kips 0

47.33 kips





35 kip-ft


17 kips 12 ft Cy Fy

By C y By

0

62.56 kips 8 kips/ft 9 ft

62.56 kips By

C y 9 ft

17 kips

8 kips/ft 9 ft

17 kips 0

26.44 kips





MA

6 kips/ft 30 ft 15 ft 60 kips 20 ft

3 kips/ft 10 ft 35 ft

90 kip-ft Dy 30 ft Dy Fy

Ay

Dy

60 kips 10 ft 0

62.00 kips 6 kips/ft 30 ft

3 kips/ft 10 ft

60 kips 60 kips Ay 62.00 kips

6 kips/ft 30 ft

3 kips/ft 10 ft

60 kips 60 kips 0 Ay

28.00 kips



7.44 Use the graphical method to construct the shearforce and bending-moment diagrams for the beam shown. Label all significant points on each diagram and identify the maximum moments along with their respective locations. Clearly differentiate straight-line and curved portions of the diagrams. Fig. P7.44


5 kips 5 ft 15 kips 8 ft Dy

Fy

2 kips/ft 20 ft 10 ft 10 kips 23 ft

25 kip-ft

Dy 20 ft

0

23 kips

By

Dy 5 kips

2 kips/ft 20 ft

By

23 kips 5 kips

15 kips 10 kips

2 kips/ft 20 ft

15 kips 10 kips 0 By 17 kips





MA

50 kN/m 2 m 1 m

25 kN/m 3 m 3.5 m MA Fy

Ay

50 kN 5 m

0

47.50 kN-m

50 kN/m 2 m 25 kN/m 3 m Ay

20 kN 2 m

20 kN

50 kN 0

55 kN





MC

20 kips 15 ft

6 kips/ft 8 ft 11 ft

12 kips/ft 7 ft 3.5 ft 70 kips 7 ft MC Fy

20 kips

MC

32.00 kip-ft 6 kips/ft 8 ft

12 kips/ft 7 ft Cy

0

Cy

70 kips 0

42.00 kips




Solution Beam equilibrium: MB 4,000 lb-ft 9,000 lb-ft

800 lb/ft 4 ft 2 ft 600 lb/ft 10 ft 10 ft

3,600 lb 10 ft Ey

Fy

By

Ey

0

5,640 lb

800 lb/ft 4 ft

By 5,640 lb By

E y 15 ft

600 lb/ft 10 ft

800 lb/ft 4 ft

3,600 lb

600 lb/ft 10 ft

3,600 lb 0

7,160 lb





MB

400 kN-m 500 kN-m 600 kN-m 60 kN 2 m 1 m

120 kN/m 4 m 4 m

60 kN/m 2 m 7 m Ey Fy

By

Ey

60 kN 2 m

530 kN

0

120 kN/m 4 m

150 kN

60 kN 2 m

60 kN 2 m By

Ey 8 m

530 kN

60 kN 2 m By

150 kN 6 m

120 kN/m 4 m

150 kN

340 kN





Consider free-body diagram of DE:

MD

55 kN/m 3 m 1.5 m Ey

Fy

Dy

0

82.5 kN

Ey

55 kN/m 3 m

Dy 82.5 kN Dy

Ey 3 m

55 kN/m 3 m

0

82.5 kN

Consider free-body diagram of ABCD:

MA

60 kN-m

75 kN/m 5 m 2.5 m

100 kN 2.5 m 60 kN-m

Fy

Ay C y Ay

C y 3.5 m

75 kN/m 5 m 2.5 m

100 kN 2.5 m Cy

Dy 5 m

82.5 kN 5 m

C y 3.5 m

0

440 kN 75 kN/m 5 m

440 kN

100 kN Dy

75 kN/m 5 m

100 kN 82.5 kN 0

Ay 117.5 kN






Consider free-body diagram of ABC: MA 500 lb/ft 10 ft 5 ft C y 15 ft

0

C y 1,666.67 lb Fy

Ay C y

500 lb/ft 10 ft

Ay 1,666.67 lb Ay

500 lb/ft 10 ft

0

3,333.33 lb

Consider free-body diagram of CDE: Fy C y E y 1, 200 lb

1, 666.67 lb Ey ME

E y 1, 200 lb 0

2,866.67 lb

C y 10 ft

1, 200 lb 8 ft

1, 666.67 lb 10 ft ME

ME

1, 200 lb 8 ft

ME

0

26, 266.67 lb-ft






MA

70 kN/m 7 m (3.5 m) 1 2

70 kN/m 3 m 8 m

55 kN 10 m By Fy

Ay Ay

0

443.57 kN

By 1 2

By 7 m

70 kN/m 7 m

70 kN/m 3 m

443.57 kN

55 kN

70 kN/m 7 m

1 2

70 kN/m 3 m

Ay

206.43 kN

55 kN

0





MB

1 2

8 ft 3

6 kips/ft 8 ft

4 kips/ft 15 ft (14.5 ft) Dy 17 ft Dy Fy

By

Dy

0

47.41 kips 1 2

6 kips/ft 8 ft

4 kips/ft 15 ft By

47.41 kips

1 2

6 kips/ft 8 ft

4 kips/ft 15 ft By

0

36.59 kips





MA

(9 kips)(4 ft) 1 2

4 kips/ft 9 ft (13 ft) M A

MA Fy

0

270.00 kip-ft 1 2

Ay

(9 kips)

Ay

27.00 kips

4 kips/ft 9 ft

0





MD

(25 kN)(6 m) 1 2

30 kN/m 3 m 3 m

MD Fy

Dy

0

285.00 kN-m

25 kN Dy

MD

1 2

30 kN/m 3 m

0

70 kN





MA

(6 kips/ft)(22 ft)(11 ft) 1 2

2(8 ft) 3

(9 kips/ft)(8 ft) 22 ft

By (22 ft) 0

By 110.73 kips Fy

Ay

By (6 kips/ft)(22 ft)

1 2

(9 kips/ft)(8 ft)

Ay (110.73 kips) (6 kips/ft)(22 ft) Ay

1 2

(9 kips/ft)(8 ft) 0

57.27 kips



7.56 For the beam and loading shown, (a) Use discontinuity functions to write the expression for w(x). Include the beam reactions in this expression. (b) Integrate w(x) to twice to determine V(x) and M(x). (c) Use V(x) and M(x) to plot the shear-force and bending-moment diagrams. Fig. P7.56

Solution Beam equilibrium: MA (180 lb)(2 ft) (450 lb)(6 ft)

Dy Fy

Ay

0

340 lb

Dy 180 lb 450 lb Ay

Dy (9 ft)

0

290 lb

Load function w(x):

w( x)

1

290 lb x 0 ft

1

180 lb x 2 ft

450 lb x 6 ft

1

340 lb x 9 ft

1

Ans.

Shear-force function V(x) and bending-moment function M(x):

V ( x) M ( x)

290 lb x 0 ft 290 lb x 0 ft

0 1

180 lb x 2 ft 180 lb x 2 ft

0 1

450 lb x 6 ft 450 lb x 6 ft

0 1

340 lb x 9 ft 340 lb x 9 ft

0 1

Ans. Ans.




Solution Beam equilibrium: M B (10 kN)(2.5 m) (35 kN)(3 m)

Dy Fy

By

0

16 kN

Dy 10 kN 35 kN By

Dy (5 m)

0

29 kN

Load function w(x):

w( x)

1

10 kN x 0 m

1

29 kN x 2.5 m

35 kN x 5.5 m

1

16 kN x 7.5 m

1

Ans.


V ( x) M ( x)

10 kN x 0 m 10 kN x 0 m

0 1

29 kN x 2.5 m 29 kN x 2.5 m

0 1

35 kN x 5.5 m 35 kN x 5.5 m

0 1

16 kN x 7.5 m 16 kN x 7.5 m

0 1

Ans. Ans.




Solution Beam equilibrium: MB (30 kN)(3 m) (20 kN)(7 m) (15 kN)(15 m) Dy Fy

Ay

Dy Ay

Dy (10 m)

0

45.5 kN 30 kN

20 kN 15 kN

0

19.5 kN

Load function w(x): w( x) 19.5 kN x 0 m

1

1

30 kN x 3 m 1

45.5 kN x 10 m

20 kN x 7 m 1

15 kN x 15 m

Shear-force function V(x) and bending-moment function M(x): 0 0 V ( x) 19.5 kN x 0 m 30 kN x 3 m 20 kN x 7 m 0

45.5 kN x 10 m M ( x) 19.5 kN x 0 m

1

45.5 kN x 10 m

15 kN x 15 m 30 kN x 3 m

1

15 kN x 15 m

Ans.

0

0

1

Ans.

20 kN x 7 m 1

1

1

Ans.




Solution Beam equilibrium: Fy C y 5 kN

Cy MC

0

5 kN

(5 kN)(6 m) 20 kN-m M C MC

0

10 kN-m

Load function w(x):

w( x)

1

5 kN x 0 m

2

20 kN-m x 3 m

1

5 kN x 6 m

2

10 kN-m x 6 m

Ans.


V ( x) M ( x)

5 kN x 0 m 5 kN x 0 m

0 1

20 kN-m x 3 m 20 kN-m x 3 m

1 0

5 kN x 6 m 5 kN x 6 m

0 1

1

10 kN-m x 6 m 10 kN-m x 6 m

0

Ans. Ans.




Solution Beam equilibrium: Fy Ay (35 kN/m)(2 m)

Ay MA

0

70 kN

(35 kN/m)(2 m)(4 m) M A MA

0

280 kN-m

Load function w(x):

w( x)

1

70 kN x 0 m

2

280 kN-m x 0 m

0

35 kN/m x 3 m

35 kN/m x 5 m

0

Ans.


V ( x) M ( x)

70 kN x 0 m 70 kN x 0 m

0

1

280 kN-m x 0 m 280 kN-m x 0 m

1

0

35 kN/m x 3 m 35 kN/m x 3m 2

1

2

1

35 kN/m x 5 m 35 kN/m x 5m 2

Ans. 2

Ans.




Solution Beam equilibrium: MA (25 kN)(4 m)(2 m) (32 kN)(6 m)

Dy Fy

Ay

Dy Ay

Dy (8 m)

0

49 kN (25 kN)(4 m) 32 kN

0

83 kN

Load function w(x): w( x) 83 kN x 0 m

1 1

32 kN x 6 m

25 kN/m x 0 m

0

25 kN/m x 4 m

1

49 kN x 8 m

Ans.

Shear-force function V(x) and bending-moment function M(x): 0 1 V ( x) 83 kN x 0 m 25 kN/m x 0 m 25 kN/m x 4 m

32 kN x 6 m M ( x)

83 kN x 0 m 32 kN x 6 m

0

1

1

0

1

0

49 kN x 8 m 25 kN/m x 0m 2 49 kN x 8 m

1

Ans. 2

25 kN/m x 4m 2

2

Ans.




Solution Beam equilibrium: MA (3,000 lb)(5 ft) 8,000 lb-ft (800 lb)(7 ft)(12.5 ft) Ey Fy

Ay

Ey Ay

E y (20 ft)

0

3,850 lb 3,000 lb (800 lb)(7 ft)

0

4,750 lb

Load function w(x): w( x) 4,750 lb x 0 ft

1

1

3,000 lb x 5 ft

2

8,000 lb-ft x 5 ft

0

0

1

800 lb/ft x 9 ft 800 lb/ft x 16 ft 3,850 lb x 20 ft Shear-force function V(x) and bending-moment function M(x): 0 0 1 V ( x) 4,750 lb x 0 ft 3,000 lb x 5 ft 8,000 lb-ft x 5 ft 800 lb/ft x 9 ft M ( x)

4,750 lb x 0 ft

1

1

800 lb/ft x 16 ft 3,000 lb x 5 ft

1

1

800 lb/ft 800 lb/ft 2 x 9 ft x 16 ft 2 2 Shear-force and bending-moment diagrams:

3,850 lb x 20 ft

8,000 lb-ft x 5 ft 2

0

Ans.

Ans.

0

3,850 lb x 20 ft

1

Ans.



Solution Beam equilibrium: Fy Ay (800 lb/ft)(12 ft) (800 lb)(6 ft)

Ay MA

0

14, 400 lb

(800 lb/ft)(12 ft)(6 ft) (800 lb)(6 ft)(21 ft) M A MA

0

158, 400 lb-ft

Load function w(x): w( x) 14,400 lb x 0 ft

1

158,400 lb-ft x 0 ft 0

0

2

1

800 lb/ft x 18 ft

1

Ans.

0

800 lb/ft x 12 ft 800 lb/ft x 18 ft 800 lb/ft x 24 ft Shear-force function V(x) and bending-moment function M(x): 0 1 V ( x) 14,400 lb x 0 ft 158,400 lb-ft x 0 ft 800 lb-ft x 0 ft 800 lb/ft x 12 ft

0

800 lb-ft x 0 ft

1

1

800 lb/ft x 24 ft 800 lb-ft 1 0 M ( x) 14, 400 lb x 0 ft 158, 400 lb-ft x 0 ft x 0 ft 2 800 lb/ft 800 lb/ft 800 lb/ft 2 2 x 12 ft x 18 ft x 24 ft 2 2 2 Shear-force and bending-moment diagrams:

Ans. 2

2

Ans.



Solution Beam equilibrium: MA 12 kN-m (18 kN/m)(2 m)(2 m)

Dy Fy

Ay

0

12 kN

Dy Ay

Dy (5 m)

(18 kN/m)(2 m)

0

24 kN


1

2

12 kN-m x 0 m

0

1

18 kN/m x 3 m 12 kN x 5 m Shear-force function V(x) and bending-moment function M(x): 0 1 V ( x) 24 kN x 0 m 12 kN-m x 0 m 18 kN/m x 1 m 18 kN/m x 3 m M ( x)

24 kN x 0 m

1

1

0

18 kN/m x 1 m

12 kN x 5 m

Ans. 1

0

Ans.

12 kN-m x 0 m

18 kN/m 2 x 3m 12 kN x 5 m 2 Shear-force and bending-moment diagrams:

1

0

18 kN/m x 1m 2

2

Ans.




MA

(6 kips/ft)(22 ft)(11 ft) 1 2

2(8 ft) 3

(9 kips/ft)(8 ft) 22 ft

By (22 ft) 0

By 110.73 kips Fy

Ay

1 2

By (6 kips/ft)(22 ft)

(9 kips/ft)(8 ft)

Ay (110.73 kips) (6 kips/ft)(22 ft) Ay

1 2

(9 kips/ft)(8 ft) 0

57.27 kips

Load function w(x): w( x) 57.27 kips x 0 ft

1

9 kips/ft x 22 ft 8 ft

6 kips/ft x 0 ft

1

0


1

9 kips/ft x 30 ft

Shear-force function V(x) and bending-moment function M(x): 0 1 V ( x) 57.27 kips x 0 ft 6 kips/ft x 0 ft 110.73 kips x 22 ft 9 kips/ft x 22 ft 2(8 ft) M ( x)

57.27 kips x 0 ft 9 kips/ft x 22 ft 6(8 ft)

2

1

3

1

110.73 kips x 22 ft

6 kips/ft x 22 ft

0

0

0

9 kips/ft 2 1 x 30 ft 9 kips/ft x 30 ft 2(8 ft) 6 kips/ft 2 1 x 0 ft 110.73 kips x 22 ft 2 9 kips/ft 9 kips/ft 3 2 x 30 ft x 30 ft 6(8 ft) 2

Ans.

6 kips/ft x 22 ft

1

Ans. 6 kips/ft x 22 ft 2

2

Ans.





Solution Beam equilibrium: Fy C y (20 kN/m)(3 m) Cy MC

1 2

(30 kN/m)(3 m)

0

105 kN

(20 kN/m)(3 m)(2.5 m) 1 2

(30 kN/m)(3 m)(2 m)

MC

MC

0

240 kN-m

Load function w(x):

w( x)

20 kN/m x 0 m

0

30 kN/m x 0m 3m

0

1

20 kN/m x 3 m

1

30 kN/m x 3 m 105 kN x 4 m 240 kN-m x 4 m Shear-force function V(x) and bending-moment function M(x): 30 kN/m 1 2 V ( x) 20 kN/m x 0 m x 0m 20 kN/m x 3 m 2(3 m) 30 kN/m x 3 m

1

105 kN x 4 m

0

240 kN-m x 4 m

1

2

Ans. 30 kN/m x 3m 2(3 m)

1

2

1

20 kN/m 30 kN/m 20 kN/m 2 3 x 0m x 0m x 3m 2 6(3 m) 2 30 kN/m 2 1 0 x 3m 105 kN x 4 m 240 kN-m x 4 m 2 Shear-force and bending-moment diagrams: M ( x)

30 kN/m x 3m 3m

0

Ans. 2

30 kN/m x 3m 6(3 m)

3

Ans.


7.67 For the beam and loading shown, (a) Use discontinuity functions to write the expression for w(x). Include the beam reactions in this expression. (b) Integrate w(x) to twice to determine V(x) and M(x). (c) Determine the maximum bending moment in the beam between the two simple supports. Fig. P7.67

Solution Beam equilibrium: M B 9 kN-m 12 (18 kN/m)(3 m)(1 m) C y (3 m) Cy Fy

By

6 kN

Cy By

0

1 2

(18 kN/m)(3 m)

0

21 kN

Load function w(x): w( x) 9 kN-m x 0 m

18 kN/m x 1m 3m

2

1

21 kN x 1 m

18 kN/m x 4m 3m

1

1

1

6 kN x 4 m

Shear-force function V(x) and bending-moment function M(x): 1 0 V ( x) 9 kN-m x 0 m 21 kN x 1 m 18 kN/m x 1 m 18 kN/m x 1m 2(3 m) M ( x)

9 kN-m x 0 m 18 kN/m x 1m 6(3 m)

18 kN/m x 4m 2(3 m)

2

0

3

21 kN x 1 m

0

18 kN/m x 1 m

1

18 kN/m x 4m 6(3 m)

2

6 kN x 4 m

1

0

18 kN/m x 1m 2 3

6 kN x 4 m

Ans.

Ans. 2

1

Ans.


Maximum bending moment: Mmax = 5.66 kN-m at x = 2.59 m

Ans.




Solution Beam equilibrium: 1 2

MA

(5 kips/ft)(9 ft)(6 ft) C y (14 ft) 0

Cy Fy

Ay

9.64 kips

Cy Ay

1 2

(5 kips/ft)(9 ft) 0

12.86 kips

Load function w(x): 1

w( x) 12.86 kips x 0 ft 5 kips/ft x 9 ft

0



1

1

9.64 kips x 14 ft

Ans.

Shear-force function V(x) and bending-moment function M(x): 5 kips/ft 5 kips/ft 0 2 V ( x) 12.86 kips x 0 ft x 0 ft x 9 ft 2(9 ft) 2(9 ft) 5 kips/ft x 9 ft

1

M ( x) 12.86 kips x 0 ft 5 kips/ft x 9 ft 2

2

5 kips/ft x 0 ft 6(9 ft)

2

0

9.64 kips x 14 ft 1

1

Ans.


3

9.64 kips x 14 ft

1

3

Ans.


Maximum bending moment: Mmax = 58.3 kip-ft at x = 6.80 ft

Ans.





MA

(5 kips/ft)(6 ft)(3 ft) 1 2

(9 kips/ft)(21 ft) 6 ft Cy

Fy

C y (16 ft) 0

82.41 kips

Ay C y Ay

21 ft 3 1 2

(5 kips/ft)(6 ft)

(9 kips/ft)(21 ft) 0

42.09 kips



1

1

5 kips/ft x 0 ft

0

82.41 kips x 16 ft

1

M ( x)

42.09 kips x 0 ft 9 kips/ft x 6 ft 6(21 ft)

2

1

3

9 kips/ft x 6 ft


Shear-force function V(x) and bending-moment function M(x): 0 1 V ( x) 42.09 kips x 0 ft 5 kips/ft x 0 ft 5 kips/ft x 6 ft 9 kips/ft x 6 ft 2(21 ft)

0

5 kips/ft x 6 ft

1

0

1

9 kips/ft x 6 ft

Ans.

1

9 kips/ft 2 x 27 ft 2(21 ft) 5 kips/ft 5 kips/ft 9 kips/ft 2 2 x 0 ft x 6 ft x 6 ft 2 2 2 9 kips/ft 1 3 82.41 kips x 16 ft x 27 ft 6(21 ft) 82.41 kips x 16 ft

0

Ans. 2

Ans.


Maximum bending moment: Mmax = 170.9 kip-ft at x = 7.39 ft

Ans.




Solution Beam equilibrium: MA (25 kN/m)(4.0 m)(4.5 m) 1 2

Dy Fy

Ay

2(4.0 m) 3

(45 kN/m)(4.0 m) 2.5 m

0

114.38 kN

Dy Ay

Dy (8 m)

(25 kN/m)(4.0 m)

1 2

(45 kN/m)(4.0 m)

0

75.63 kN

Load function w(x): w( x)

1

75.63 kN x 0 m

45 kN/m x 6.5 m 4.0 m

0

25 kN/m x 2.5 m 1

45 kN/m x 6.5 m

0

45 kN/m x 2.5 m 4.0 m

0

25 kN/m x 6.5 m

114.38 kN x 8 m

1

1

Ans.


V ( x)

75.63 kN x 0 m

0

45 kN/m x 6.5 m 2(4.0 m) M ( x)

25 kN/m x 2.5 m 2

1

45 kN/m x 6.5 m

25 kN/m x 6.5 m 1

1

114.38 kN x 8 m

45 kN/m x 2.5 m 2(4.0 m) 0

25 kN/m 25 kN/m 45 kN/m 2 2 x 2.5 m x 6.5 m x 2.5 m 2 2 6(4.0 m) 45 kN/m 45 kN/m 3 2 1 x 6.5 m x 6.5 m 114.38 kN x 8 m 6(4.0 m) 2

75.63 kN x 0 m

1

2

Ans.

3

Ans.


Maximum bending moment: Mmax = 275 kN-m at x = 4.57 m

Ans.





MC

1 2

(30 kN/m)(7.0 m)(3.5 m) (50 kN/m)(2.0 m)(1.0 m) By

Fy

By

(40 kN/m)(7.0 m) By (5.5 m)

2(7.0 m) 3

0

234.24 kN

Cy

1 2

(30 kN/m)(7.0 m)

(50 kN/m)(2.0 m) C y 215.76 kN

(40 kN/m)(7.0 m)

0


30 kN/m x 0 m

0

40 kN/m 1 x 0m 7.0 m 40 kN/m 0 30 kN/m x 7 m x 7m 7.0 m

40 kN/m x 0 m 1

234.24 kN x 1.5 m 1

215.76 kN x 7 m

0

0

50 kN/m x 7.0 m

50 kN/m x 9.0 m

Shear-force function V(x) and bending-moment function M(x): 40 kN/m 1 1 2 V ( x) 30 kN/m x 0 m 40 kN/m x 0 m x 0m 2(7.0 m) 40 kN/m 0 1 234.24 kN x 1.5 m 30 kN/m x 7 m x 7m 2(7.0 m) 0

M ( x)

215.76 kN x 7 m 30 kN/m 2 x 0m 2

1

1

0

Ans.

2

1

50 kN/m x 7.0 m 50 kN/m x 9.0 m 40 kN/m 40 kN/m 2 3 x 0m x 0m 2 6(7.0 m) 30 kN/m 40 kN/m 1 2 3 234.24 kN x 1.5 m x 7m x 7m 2 6(7.0 m) 50 kN/m 50 kN/m 1 2 215.76 kN x 7 m x 7.0 m x 9.0 m 2 2

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2

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(60 kN)(1.5 m) Dy (6.5 m) Dy

Fy

By

Dy By

1 2

(90 kN/m)(4.5 m)

1 2

(90 kN/m)(4.5 m)

2(4.5 m) 3

0

79.62 kN 60 kN

0

182.88 kN


60 kN x 0 m

1

90 kN/m x 6m 4.5 m

182.88 kN x 1.5 m 1

90 kN/m x 6 m

0

1

90 kN/m x 1.5 m 4.5 m

79.62 kN x 8 m

1

Shear-force function V(x) and bending-moment function M(x): 90 kN/m 0 0 2 V ( x) 60 kN x 0 m 182.88 kN x 1.5 m x 1.5 m 2(4.5 m) 90 kN/m 2 1 0 x 6m 90 kN/m x 6 m 79.62 kN x 8 m 2(4.5 m) 90 kN/m 1 1 3 M ( x) 60 kN x 0 m 182.88 kN x 1.5 m x 1.5 m 6(4.5 m) 90 kN/m 90 kN/m 3 2 1 x 6m x 6m 79.62 kN x 8 m 6(4.5 m) 2

1

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8.1 During fabrication of a laminated timber arch, one of the 10 in. wide by 1 in. thick Douglas fir [E = 1,900 ksi] planks is bent to a radius of curvature of 40 ft. Determine the maximum bending stress developed in the plank.

Solution From Eq. (8.3): E 1,900 ksi x   y   (0.5 in.)  1.979 ksi  1.979 ksi  (40 ft)(12 in./ft)

Ans.


8.2 A high-strength steel [E = 200 GPa] tube having an outside diameter of 80 mm and a wall thickness of 3 mm is bent into a circular curve having a 52-m radius of curvature. Determine the maximum bending stress developed in the tube.

Solution From Eq. (8.3): E 200,000 MPa x   y   (80 mm / 2)  153.846 MPa  153.8 MPa  (52 m)(1,000 mm/m)

Ans.


8.3 A high-strength steel [E = 200 GPa] band saw blade wraps around a pulley that has a diameter of 450 mm. Determine the maximum bending stress developed in the blade. The blade is 12-mm wide and 1-mm thick.

Solution The radius of curvature of the band saw blade is: 450 mm 1 mm    225.5 mm 2 2 From Eq. (8.3): E 200,000 MPa x   y   (0.5 mm)  443.459 MPa  443 MPa  225.5 mm

Ans.


8.4 The boards for a concrete form are to be bent into a circular shape having an inside radius of 10 m. What maximum thickness can be used for the boards if the normal stress is not to exceed 7 MPa? Assume that the modulus of elasticity for the wood is 12 GPa.

Solution The radius of curvature of the concrete form is dependent on the board thickness: t   10,000 mm  2 From Eq. (8.3): E 12,000 MPa  t  x   y    7 MPa t   2    10,000 mm   2 Solve for t: t t  12,000 MPa    7 MPa 10,000 mm    2  2 6,000t  70,000  3.5t

5,996.5t  70,000  t  11.67 mm

Ans.


8.5 A beam having a tee-shaped cross section is subjected to equal 12 kN-m bending moments, as shown in Fig. P8.5a. The cross-sectional dimensions of the beam are shown in Fig. P8.5b. Determine: (a) the centroid location, the moment of inertia about the z axis, and the controlling section modulus about the z axis. (b) the bending stress at point H. State whether the normal stress at H is tension or compression. (c) the maximum bending stress produced in the cross section. State whether the stress is tension or compression.

Fig. P8.5a

Fig. P8.5b

Solution (a) Centroid location in y direction: (reference axis at bottom of tee shape) yi Shape Area Ai (from bottom) yi Ai 2 (mm ) (mm) (mm3) top flange 2,500.0 162.5 406,250.0 stem 3,750.0 75.0 281,250.0 2 6,250.0 mm 687,500.0 mm3

y

yi Ai Ai



687,500.0 mm3  110.0 mm (measured upward from bottom edge of stem) 6,250.0 mm2

Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (mm ) (mm) (mm4) top flange 130,208.33 52.50 6,890,625.00 stem 7,031,250.00 −35.00 4,593,750.00 Moment of inertia about the z axis (mm4) =

Ans.

IC + d²A (mm4) 7,020,833.33 11,625,000.00 18,645,833.33

I z  18,656,000 mm4

Ans.


Section moduli: Iz 18,645,833.33 mm 4 S top    286,858.974 mm3 ctop (175 mm  110 mm)

Sbot 

I z 18,645,833.33 mm 4   169,507.576 mm3 cbot 110 mm

 S  169,500 mm3

Ans.

(b) Bending stress at point H: (y = 175 mm − 25 mm − 110 mm = 40 mm) My x   Iz



(12 kN-m)(40 mm)(1,000 N/kN)(1,000 mm/m) 18,654,833.33 mm4

 25.743 MPa  25.7 MPa (C)

Ans.

(c) Maximum bending stress: The maximum bending stress occurs at either the top or the bottom surface of the beam. The top of the cross section is at y = +65 mm, and the bottom of the cross section is at y = −110 mm. The larger bending stress magnitude occurs at the larger magnitude of these two values; in this case, at the bottom of the cross section. My x   Iz



(12 kN-m)(  110 mm)(1,000 N/kN)(1,000 mm/m) 18,654,833.33 mm 4

 70.793 MPa  70.8 MPa (T)

Ans.


8.6 A beam is subjected to equal 6.5 kip-ft bending moments, as shown in Fig. P8.6a. The crosssectional dimensions of the beam are shown in Fig. P8.6b. Determine: (a) the centroid location, the moment of inertia about the z axis, and the controlling section modulus about the z axis. (b) the bending stress at point H, which is located 2 in. below the z centroidal axis. State whether the normal stress at H is tension or compression. (c) the maximum bending stress produced in the cross section. State whether the stress is tension or compression.

Fig. P8.6a

Fig. P8.6b

Solution (a) Centroid location in y direction: (reference axis at bottom of shape) yi Shape Area Ai (from bottom) yi Ai (in.2) (in.) (in.3) left side 8.0 4.0 32.0 top flange 4.0 7.5 30.0 right side 8.0 4.0 32.0 20.0 in.2 94.0 in.3

y

yi Ai Ai



94.0 in.3  4.70 in. (measured upward from bottom edge of section) 20.0 in.2

Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (in. ) (in.) (in.4) left side 42.667 −0.700 3.920 top flange 0.333 2.800 31.360 right side 42.667 −0.700 3.920 Moment of inertia about the z axis (in.4) =

Ans.

IC + d²A (in.4) 46.587 31.693 46.587 124.867

I z  124.9 in.4

Ans.


Section moduli: Iz 124.867 in.4 S top    37.8384 in.3 ctop (8 in.  4.7 in.) S bot 

I z 124.867 in.4   26.5674 in.3 cbot 4.7 in.

 S  26.6 in.3

Ans.

(b) Bending stress at point H: (y = −2 in.) My x   Iz



( 6.5 kip-ft)(  2 in.)(12 in./ft) 124.867 in.4

 1, 249 psi  1, 249 psi (C)

Ans.

(c) Maximum bending stress: The maximum bending stress occurs at either the top or the bottom surface of the beam. The top of the cross section is at y = +3.30 in., and the bottom of the cross section is at y = −4.7 in. The larger bending stress magnitude occurs at the larger magnitude of these two values; in this case, at the bottom of the cross section. My x   Iz



( 6.5 kip-ft)(  4.7 in.)(12 in./ft) 124.867 in.4

 2,935.9 psi  2,940 psi (C)

Ans.


8.7 A beam is subjected to equal 470 N-m bending moments, as shown in Fig. P8.7a. The crosssectional dimensions of the beam are shown in Fig. P8.7b. Determine: (a) the centroid location, the moment of inertia about the z axis, and the controlling section modulus about the z axis. (b) the bending stress at point H. State whether the normal stress at H is tension or compression. (c) the maximum bending stress produced in the cross section. State whether the stress is tension or compression.

Fig. P8.7a

Fig. P8.7b

Solution (a) Centroid location in y direction: (reference axis at bottom of U shape) yi Shape Area Ai (from bottom) yi Ai (mm2) (mm) (mm3) left side 400.0 25.0 10,000.0 bottom flange 272.0 4.0 1,088.0 right side 400.0 25.0 10,000.0 1,072.0 mm2 21,088.0 mm3

yi Ai

21,088.0 mm3 y   19.67 mm (measured upward from bottom edge of section) Ai 1,072.0 mm2 Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (mm ) (mm) (mm4) left side 83,333.33 5.33 11,356.56 bottom flange 1,450.67 −15.67 66,803.30 right side 83,333.33 5.33 11,356.56 Moment of inertia about the z axis (mm4) =

Ans.

IC + d²A (mm4) 94,689.89 68,253.96 94,689.89 257,633.75

I z  257,600 mm4

Ans.


Section moduli: Iz 257,633.75 mm 4 S top    8,494.814 mm3 ctop (50 mm  19.672 mm)

Sbot 

Iz 257,633.75 mm 4   13,096.708 mm3 cbot 19.672 mm

 S  8,495 mm3

Ans.

(b) Bending stress at point H: (y = 8 mm − 19.672 mm = −11.672 mm) My x   Iz



(470 N-m)(  11.672 mm)(1,000 mm/m) 257,633.75 mm 4

 21.293 MPa  21.3 MPa (T)

Ans.

(c) Maximum bending stress: The maximum bending stress occurs at either the top or the bottom surface of the beam. The top of the cross section is at y = +30.328 mm, and the bottom of the cross section is at y = −19.672 mm. The larger bending stress magnitude occurs at the larger magnitude of these two values; in this case, at the top of the cross section. My x   Iz



(470 N-m)(30.328 mm)(1,000 mm/m) 257,633.75 mm 4

 55.328 MPa  55.3 MPa (C)

Ans.


8.8 A beam is subjected to equal 17.5 kip-ft bending moments, as shown in Fig. P8.8a. The crosssectional dimensions of the beam are shown in Fig. P8.8b. Determine: (a) the centroid location, the moment of inertia about the z axis, and the controlling section modulus about the z axis. (b) the bending stress at point H. State whether the normal stress at H is tension or compression. (c) the bending stress at point K. State whether the normal stress at K is tension or compression. (d) the maximum bending stress produced in the cross section. State whether the stress is tension or compression.

Fig. P8.8a

Fig. P8.8b

Solution (a) Centroid location in y direction: (reference axis at bottom of shape) yi Shape Area Ai (from bottom) yi Ai (in.2) (in.) (in.3) top flange 12.0000 13.0000 156.0000 web 20.0000 7.0000 140.0000 bottom flange 20.0000 1.0000 20.0000 2 52.0000 in. 316.0000 in.3

yi Ai

316.0 in.3 y   6.077 in.  6.08 in. Ai 52.0 in.2

(measured upward from bottom edge of bottom

flange) Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (in. ) (in.) (in.4) top flange 4.000 6.923 575.148 web 166.667 0.923 17.041 bottom flange 6.667 -5.077 515.503 Moment of inertia about the z axis (in.4) =

Ans.

IC + d²A (in.4) 579.148 183.708 522.170 1,285.026 Ans.


Section Moduli cbot  6.0769 in.

ctop  14 in.  6.0769 in.  7.9231 in. Sbot

I z 1, 285.026 in.4    211.460 in.3 cbot 6.0769 in.

Stop 

I z 1, 285.026 in.4   162.188 in.3 ctop 7.9231 in.

The controlling section modulus is the smaller of the two values; therefore, S  162.2 in.3 Bending stress at point H: From the flexure formula: My ( 17.5 kip-ft)(7.9231 in.  2 in.)(12 in./ft) x     967.9544 psi  968 psi (T) Iz 1,285.0256 in.4 Bending stress at point K: From the flexure formula: My ( 17.5 kip-ft)(  6.0769 in.  2 in.)(12 in./ft) x     666.2543 psi  666 psi (C) Iz 1,285.026 in.4

Ans.

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Maximum bending stress Since ctop > cbot, the maximum bending stress occurs at the top of the flanged shape. From the flexure formula: My ( 17.5 kip-ft)(7.9231 in.)(12 in./ft) x     1, 294.8 psi  1, 295 psi (T) Ans. Iz 1,285.026 in.4 Also, note that the same maximum bending stress magnitude can be calculated with the section modulus: M (17.5 kip-ft)(12 in./ft) x    1,294.8 psi  1,295 psi Ans. S 162.1877 in.3 The sense of the stress (either tension or compression) would be determined by inspection.


8.9 The cross-sectional dimensions of a beam are shown in Fig. P8.9. (a) If the bending stress at point K is 43 MPa (C), determine the internal bending moment Mz acting about the z centroidal axis of the beam. (b) Determine the bending stress at point H. State whether the normal stress at H is tension or compression.

Fig. P8.9

Solution Centroid location in y direction: (reference axis at bottom of double-tee shape) yi Shape Area Ai (from bottom) yi Ai 2 (mm ) (mm) (mm3) top flange 375.0 47.5 17,812.5 left stem 225.0 22.5 5,062.5 right stem 225.0 22.5 5,062.5 2 825.0 mm 27,937.5 mm3 yi Ai 27,937.5 mm3 y   33.864 mm  33.9 mm (measured upward from bottom of section) Ai 825.0 mm2 Moment of inertia about the z axis: d = yi – y Shape IC d²A IC + d²A 4 4 (mm ) (mm) (mm ) (mm4) top flange 781.250 13.636 69,731.405 70,512.655 left stem 37,968.750 −11.364 29,054.752 67,023.502 right stem 37,968.750 −11.364 29,054.752 67,023.502 4 Moment of inertia about the z axis (mm ) = 204,559.659 (a) Determine bending moment: At point K, y = 50 mm − 5 mm − 33.864 mm = 11.136 mm. The bending stress at K is x = −43 MPa; therefore, the bending moment magnitude can be determined from the flexure formula: My x   Iz

 xIz

( 43 N/mm 2 )(204,559.659 mm 4 ) M    y 11.136 mm  789,850.765 N-mm  790 N-m (b) Bending stress at point H: At point H, y = −33.864 mm. The bending stress is computed with the flexure formula: My (789,850.765 N-mm)(  33.864 mm) x     130.755 MPa  130.8 MPa (T) Iz 204,559.659 mm 4

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8.10 The cross-sectional dimensions of a beam are shown in Fig. P8.10. (a) If the bending stress at point K is 2,600 psi (T), determine the internal bending moment Mz acting about the z centroidal axis of the beam. (b) Determine the bending stress at point H. State whether the normal stress at H is tension or compression.

Fig. P8.10

Solution Centroid location in y direction: (reference axis at bottom of inverted-tee shape) yi Shape Area Ai (from bottom) yi Ai 2 (in. ) (in.) (in.3) bottom flange 0.56250 0.12500 0.07031 stem 0.56250 1.37500 0.77344 2 1.12500 in. 0.84375 in.3 yi Ai 0.84375 in.3 (measured upward from bottom edge of section) y   0.750 in. Ai 1.1250 in.2 Moment of inertia about the z axis: d = yi – y Shape IC d²A IC + d²A 4 (in. ) (in.) (in.4) (in.4) bottom flange 0.00293 −0.62500 0.21973 0.22266 stem 0.23730 0.62500 0.21973 0.45703 4 Moment of inertia about the z axis (in. ) = 0.67969 (a) Determine bending moment: At point K, y = 2.50 in. − 0.75 in. = 1.750 in. The bending stress at K is x = +2,600 psi; therefore, the bending moment magnitude can be determined from the flexure formula: My x   Iz

 xIz

(2,600 psi)(0.67967 in.4 ) M    y 1.750 in.  1,009.820 lb-in.  1,010 lb-in.  84.2 lb-ft (b) Bending stress at point H: At point H, y = −0.75 in. The bending stress is computed with the flexure formula: My ( 1,009.820 lb-in.)(  0.75 in.) x     1,114.286 psi  1,114 psi (C) Iz 0.67969 in.4

Ans.

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8.11 The cross-sectional dimensions of a boxshaped beam are shown in Fig. P8.11. If the maximum allowable bending stress is b = 15,000 psi, determine the maximum internal bending moment Mz magnitude that can be applied to the beam.

Fig. P8.11

Solution Moment of inertia about z axis: (3 in.)(2 in.)3 (2.5 in.)(1 in.)3 Iz    1.791667 in.4 12 12 Maximum internal bending moment Mz: Mc x  z Iz

M 

 xIz c



(15,000 psi)(1.791667 in.4 )  26,875 lb-in.  2,240 lb-ft 1 in.

Ans.


8.12 The cross-sectional dimensions of a beam are shown in Fig. P8.12. The internal bending moment about the z centroidal axis is Mz = +2.70 kip-ft. Determine: (a) the maximum tension bending stress in the beam. (b) the maximum compression bending stress in the beam. Fig. P8.12

Solution Centroid location in y direction: (reference axis at bottom of shape) yi Shape Area Ai (from bottom) yi Ai 2 (in. ) (in.) (in.3) left stem 2.000 2.000 4.000 top flange 2.500 3.750 9.375 right stem 2.000 2.000 4.000 2 6.500 in. 17.375 in.3 yi Ai 17.375 in.3 y   2.673 in. Ai 6.500 in.2 (measured upward from bottom edge of section) Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (in. ) (in.) (in.4) left stem 2.66667 −0.67308 0.90607 top flange 0.05208 1.07692 2.89941 right stem 2.66667 −0.67308 0.90607 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 3.57273 2.95149 3.57273 10.09696

(a) Determine maximum tension bending stress: For a positive bending moment, tension bending stresses will be created below the neutral axis. Therefore, the maximum tension bending stress will occur at point K (i.e., y = −2.673 in.): My (2.70 kip-ft)(  2.673 in.)(12 in./ft) x     8.578 ksi  8.58 ksi (T) Ans. Iz 10.09696 in.4 (b) Determine maximum compression bending stress: For a positive bending moment, compression bending stresses will be created above the neutral axis. Therefore, the maximum compression bending stress will occur at point H (i.e., y = 4 in. − 2.673 in. = 1.327 in.): My (2.70 kip-ft)(1.327 in.)(12 in./ft) x     4.258 ksi  4.26 ksi (C) Ans. Iz 10.09696 in.4


8.13 The cross-sectional dimensions of a beam are shown in Fig. P8.13. (a) If the bending stress at point K is 35.0 MPa (T), determine the bending stress at point H. State whether the normal stress at H is tension or compression. (b) If the allowable bending stress is b = 165 MPa, determine the magnitude of the maximum bending moment Mz that can be supported by the beam.

Fig. P8.13

Solution Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (mm ) (mm) (mm4) top flange 540,000.000 160.000 184,320,000.000 web 32,518,666.667 0.000 0.000 bottom flange 540,000.000 −160.000 184,320,000.000 4 Moment of inertia about the z axis (mm ) =

IC + d²A (mm4) 184,860,000.000 32,518,666.667 184,860,000.000 402,238,666.667

(a) At point K, y = −90 mm, and at point H, y = −175 mm. The bending stress at K is x = +35 MPa, and the bending stress is distributed linearly over the depth of the cross section. Therefore, the bending stress at H can be found from the ratio:

H yH



K yK

 H   K

yH 175 mm  (35.0 MPa)  68.056 MPa  68.1 MPa (T) yK 90 mm

Ans.

(b) Maximum internal bending moment Mz: Mc x  z Iz

 xIz

(165 N/mm 2 )(402,238,667 mm 4 ) Mz    379,253,600 N-mm  379 kN-m c 175 mm

Ans.


8.14 The cross-sectional dimensions of a beam are shown in Fig. P8.14. (a) If the bending stress at point K is 9.0 MPa (T), determine the bending stress at point H. State whether the normal stress at H is tension or compression. (b) If the allowable bending stress is b = 165 MPa, determine the magnitude of the maximum bending moment Mz that can be supported by the beam. Fig. P8.14

Solution Moment of inertia about the z axis: d = yi – y Shape IC 4 (mm ) (mm) left flange 9,720,000 0 web 31,680 0 right flange 9,720,000 0 Moment of inertia about the z axis (mm4) =

d²A (mm4) 0 0 0

IC + d²A (mm4) 9,720,000 31,680 9,720,000 19,471,680

(a) At point K, y = −60 mm, and at point H, y = +90 mm. The bending stress at K is x = +9.0 MPa, and the bending stress is distributed linearly over the depth of the cross section. Therefore, the bending stress at H can be found from the ratio:

H yH



K yK

 H   K

yH 90 mm  (9.0 MPa)  13.50 MPa  13.50 MPa (C) yK 60 mm

Ans.

(b) Maximum bending moment Mz: Mc x  z Iz

 xIz

(165 N/mm 2 )(19,471,680 mm 4 ) M    35,698,080 N-mm  35.7 kN-m c 90 mm

Ans.


8.15 The cross-sectional dimensions of a beam are shown in Fig. P8.15. The internal bending moment about the z centroidal axis is Mz = −1.55 kip-ft. Determine: (a) the maximum tension bending stress in the beam. (b) the maximum compression bending stress in the beam. Fig. P8.15

Solution Centroid location in y direction: Shape top flange left web left bottom flange right web right bottom flange

y

yi Ai Ai



Area Ai (in.2) 8.0 3.0 3.0 3.0 3.0 20.0 in.2

yi (from bottom) (in.) 4.5 2.5 0.5 2.5 0.5

yi Ai (in.3) 36.0 7.5 1.5 7.5 1.5 54.0 in.3

54.0 in.3  2.70 in. (measured upward from bottom edge of bottom flange) 20.0 in.2

Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (in. ) (in.) (in.4) top flange 0.6667 1.8000 25.9200 left web 2.2500 −0.2000 0.1200 left bottom flange 0.2500 −2.2000 14.5200 right web 2.2500 −0.2000 0.1200 right bottom flange 0.2500 −2.2000 14.5200 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 26.5867 2.3700 14.7700 2.3700 14.7700 60.8667

(a) Maximum tension bending stress: For a negative bending moment, the maximum tension bending stress will occur at the top surface of the cross section. From the flexure formula, the bending stress at the top surface is: My ( 1.55 kip-ft)(5.0 in.  2.70 in.)(12 in./ft) x     0.7028 ksi  703 psi (T) Ans. Iz 60.8667 in.4 (b) Maximum compression bending stress: The maximum compression bending stress will occur at the bottom surface of the cross section. From the flexure formula, the bending stress at the bottom surface is: My ( 1.55 kip-ft)(  2.70 in.)(12 in./ft) x     0.8251 ksi  825 psi (C) Ans. Iz 60.8667 in.4 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

8.16 The cross-sectional dimensions of a beam are shown in Fig. P8.16. The internal bending moment about the z centroidal axis is Mz = +270 lb-ft. Determine: (a) the maximum tension bending stress in the beam. (b) the maximum compression bending stress in the beam.

Solution

Fig. P8.16

Centroid location in y direction: yi Area Ai (from bottom) yi Ai 2 (in. ) (in.) (in.3) bottom flange 0.40625 0.06250 0.02539 left web 0.28125 1.25000 0.35156 left top flange 0.09375 2.43750 0.22852 right web 0.28125 1.25000 0.35156 right top flange 0.09375 2.43750 0.22852 2 1.15625 in. 1.18555 in.3 yi Ai 1.18555 in.3 y   1.0253 in. (measured upward from bottom edge of bottom flange) Ai 1.15625 in.2 Shape

Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (in. ) (in.) (in.4) bottom flange 0.000529 −0.962838 0.376617 left web 0.118652 0.224662 0.014196 left top flange 0.000122 1.412162 0.186956 right web 0.118652 0.224662 0.014196 right top flange 0.000122 1.412162 0.186956 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 0.377146 0.132848 0.187079 0.132848 0.187079 1.016999

(a) Maximum tension bending stress: For a positive bending moment of Mz = +270 lb-ft, the maximum tension bending stress will occur at the bottom surface of the cross section (i.e., y = −1.0253 in.). From the flexure formula, the bending stress at the bottom of the cross section is: My (270 lb-ft)(  1.0253 in.)(12 in./ft) x     3, 266.446 psi  3, 270 psi (T) Ans. Iz 1.016999 in.4 (b) Maximum compression bending stress: The maximum compression bending stress will occur at the top surface of the cross section (i.e., y = 2.50 in. − 1.0253 in. = 1.4747 in.). From the flexure formula, the bending stress at the top of the cross section is: My (270 lb-ft)(1.4747 in.)(12 in./ft) x     4,698.164 psi  4,700 psi (C) Ans. Iz 1.016999 in.4 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

8.17 Two vertical forces are applied to a simply supported beam (Fig. P8.17a) having the cross section shown in Fig. P8.17b. Determine the maximum tension and compression bending stresses produced in segment BC of the beam.

Fig. P8.17a

Fig. P8.17b

Solution Centroid location in y direction: Shape top flange stem

y

yi Ai Ai



Area Ai (mm2) 3,000.0 1,440.0 4,440 mm2

yi (from bottom) (mm) 167.5 80.0

yi Ai (mm3) 502,500.0 115,200.0 617,700 mm3

617, 700 mm3  139.1216 mm (measured upward from bottom edge of stem) 4,440 mm2

Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (mm ) (mm) (mm4) top flange 56,250.00 28.38 2,415,997.08 stem 3,072,000.00 −59.12 5,033,327.25 Moment of inertia about the z axis (mm4) =

IC + d²A (mm4) 2,472,247.08 8,105,327.25 10,577,574.32



The maximum moment occurs between B and C. The moment magnitude is 12 kN-m. Maximum tension bending stress: For a positive bending moment, the maximum tension bending stress will occur at the bottom surface of this cross section. From the flexure formula, the bending stress at the bottom of the tee stem is: My (12 kN-m)(  139.1216 mm)(1,000 N/kN)(1,000 mm/m) x     157.8 MPa (T) Ans. Iz 10.5776  106 mm 4 Maximum compression bending stress: The maximum compression bending stress will occur at the top of the flange: My x   Iz



(12 kN-m)(175 mm  139.1216 mm)(1,000 N/kN)(1,000 mm/m) 10.5776  106 mm 4

 40.7 MPa  40.7 MPa (C)

Ans.


8.18 Two vertical forces are applied to a simply supported beam (Fig. P8.18a) having the cross section shown in Fig. P8.18b. Determine the maximum tension and compression bending stresses produced in segment BC of the beam.

Fig. P8.18a

Fig. P8.18b

Solution Centroid location in y direction: Shape left stem bottom flange right stem

Area Ai (in.2) 0.7500 0.5000 0.7500 2.000 in.2

yi (from bottom) (in.) 1.5000 0.1250 1.5000

yi Ai (in.3) 1.1250 0.0625 1.1250 2.3125 in.3

yi Ai

2.3125 in.3 y   1.1563 in. (measured upward from bottom edge of stem) Ai 2.000 in.2 Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (in. ) (in.) (in.4) left stem 0.56250 0.34375 0.08862 bottom flange 0.00260 −1.03125 0.53174 right stem 0.56250 0.34375 0.08862 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 0.65112 0.53434 0.65112 1.83659



The maximum moment occurs between B and C. The moment magnitude is 600 lb-ft. Maximum tension bending stress: For a positive bending moment, the maximum tension bending stress will occur at the bottom surface of this cross section at y = −1.1563 in. From the flexure formula, the bending stress at the bottom of the U shape is: My (600 lb-ft)(  1.1563 in.)(12 in./ft) x     4,533.053 psi  4,530 psi (T) Ans. Iz 1.83659 in.4 Maximum compression bending stress: The maximum compression bending stress will occur at the top of the U shape, where y = 1.8438 in.: My (600 lb-ft)(1.8438 in.)(12 in./ft) x     7, 228.265 psi  7, 230 psi (C) Ans. Iz 1.83659 in.4


8.19 A WT230 × 26 standard steel shape is used to support the loads shown on the beam in Fig. P8.19a. The dimensions from the top and bottom of the shape to the centroidal axis are shown on the sketch of the cross section (Fig. P8.19b). Consider the entire 4-m length of the beam and determine: (a) the maximum tension bending stress at any location along the beam, and (b) the maximum compression bending stress at any location along the beam.

Fig. P8.19a

Fig. P8.19b

Solution Section properties From Appendix B:

Iz

16.7 106 mm 4

Shear-force and bending-moment diagrams Maximum bending moments positive M = 13.61 kN-m negative M = −20.00 kN-m Bending stresses at max positive moment (13.61 kN-m)(60.7 mm)(1,000) 2 x 16.7 106 mm 4 49.5 MPa (C) x

(13.61 kN-m)( 164.3 mm)(1,000) 2 16.7 106 mm 4 133.9 MPa (T)

Bending stresses at max negative moment ( 20 kN-m)(60.7 mm)(1,000)2 x 16.7 106 mm 4 72.7 MPa (T) x

(a) Maximum tension bending stress

133.9 MPa (T)

(b) Maximum compression bending stress

196.8 MPa (C)

( 20 kN-m)( 164.3 mm)(1,000) 2 16.7 106 mm 4 196.8 MPa (C)

Ans. Ans.


8.20 A WT305 × 41 standard steel shape is used to support the loads shown on the beam in Fig. P8.20a. The dimensions from the top and bottom of the shape to the centroidal axis are shown on the sketch of the cross section (Fig. P8.19b). Consider the entire 10-m length of the beam and determine: (a) the maximum tension bending stress at any location along the beam, and (b) the maximum compression bending stress at any location along the beam.

Fig. P8.20a

Fig. P8.20b


Iz

48.7 106 mm 4

Shear-force and bending-moment diagrams Maximum bending moments positive M = 45.84 kN-m negative M = −24.00 kN-m Bending stresses at max positive moment (45.84 kN-m)(88.9 mm)(1,000) 2 x 48.7 106 mm 4 83.7 MPa (C) x

(45.84 kN-m)( 211.1 mm)(1,000) 2 48.7 106 mm 4 198.7 MPa (T)

Bending stresses at max negative moment ( 24 kN-m)(88.9 mm)(1,000) 2 x 48.7 106 mm 4 43.8 MPa (T) x


198.7 MPa (T)


104.0 MPa (C)

( 24 kN-m)( 211.1 mm)(1,000) 2 48.7 106 mm 4 104.0 MPa (C)

Ans. Ans.


8.21 A steel tee shape is used to support the loads shown on the beam in Fig. P8.21a. The dimensions of the shape are shown in Fig. P8.21b. Consider the entire 24-ft length of the beam and determine: (a) the maximum tension bending stress at any location along the beam, and (b) the maximum compression bending stress at any location along the beam.

Fig. P8.21a

Fig. P8.21b


y

yi Ai Ai

Area Ai (in.2) 24.0000 13.8750 37.875 in.2 590.3438 in.3 37.8750 in.2

yi (from bottom) (in.) 19.2500 9.2500

yi Ai (in.3) 462.0000 128.3438 590.3438 in.3

15.5866 in.

(from bottom of shape to centroid)

4.4134 in.

(from top of shape to centroid)

Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (in. ) (in.) (in.4) top flange 4.5000 3.6634 322.0861 stem 395.7266 −6.3366 557.1219 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 326.5861 952.8484 1,279.4345


Shear-force and bending-moment diagrams Maximum bending moments positive M = 100.75 kip-ft negative M = −68.00 kip-ft Bending stresses at max positive moment (100.75 kip-ft)(4.4134 in.)(12 in./ft) x 1, 279.4345 in.4 4.17 ksi (C) (100.75 kip-ft)( 15.5866 in.)(12 in./ft) x 1, 279.4345 in.4 14.73 ksi (T) Bending stresses at max negative moment ( 68 kip-ft)(4.4134 in.)(12 in./ft) x 1, 279.4345 in.4 2.81 ksi (T) ( 68 kip-ft)( 15.5866 in.)(12 in./ft) x 1, 279.4345 in.4 9.94 ksi (C)


14.73 ksi (T)


9.94 ksi (C)

Ans. Ans.


8.22 A flanged wooden shape is used to support the loads shown on the beam in Fig. P8.22a. The dimensions of the shape are shown in Fig. P8.22b. Consider the entire 18-ft length of the beam and determine: (a) the maximum tension bending stress at any location along the beam, and (b) the maximum compression bending stress at any location along the beam.

Fig. P8.22a

Fig. P8.22b

Solution Centroid location in y direction: Shape top flange web bottom flange

y

yi Ai Ai

Area Ai (in.2) 20.0 16.0 12.0 48.0 in.2 328.0 in.3 48.0 in.2

yi (from bottom) (in.) 11.0 6.0 1.0

yi Ai (in.3) 220.0 96.0 12.0 328.0 in.3

6.8333 in. (from bottom of shape to centroid) 5.1667 in.


Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (in. ) (in.) (in.4) top flange 6.667 4.167 347.222 web 85.333 –0.833 11.111 bottom flange 4.000 –5.833 408.333 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 353.889 96.444 412.333 862.667


Shear-force and bending-moment diagrams Maximum bending moments positive M = 10,580 lb-ft negative M = −8,400 lb-ft Bending stresses at max positive moment (10,580 lb-ft)(5.1667 in.)(12 in./ft) x 862.667 in.4 760.4 psi (C) x

(10,580 lb-ft)( 6.8333 in.)(12 in./ft) 862.667 in.4 1,005.6 psi (T)

Bending stresses at max negative moment ( 8, 400 lb-ft)(5.1667 in.)(12 in./ft) x 862.667 in.4 603.7 psi (T) x


1,006 psi (T)


799 psi (C)

( 8, 400 lb-ft)( 6.8333 in.)(12 in./ft) 862.667 in.4 798.5 psi (C)

Ans. Ans.


8.23 A channel shape is used to support the loads shown on the beam in Fig. P8.23a. The dimensions of the shape are shown in Fig. P8.23b. Consider the entire 12-ft length of the beam and determine: (a) the maximum tension bending stress at any location along the beam, and (b) the maximum compression bending stress at any location along the beam.

Fig. P8.23a

Fig. P8.23b

Solution Centroid location in y direction: Shape left stem top flange right stem

y

yi Ai Ai

Area Ai (in.2) 3.000 5.500 3.000 11.500 in.2 49.625 in.3 11.500 in.2

yi (from bottom) (in.) 3.000 5.750 3.000

4.3152 in. 1.6848 in.

yi Ai (in.3) 9.000 31.625 9.000 49.625 in.3

(from bottom of shape to centroid) (from top of shape to centroid)

Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (in. ) (in.) (in.4) left stem 9.0000 −1.3152 5.1894 top flange 0.1146 1.4348 11.3223 right stem 9.0000 −1.3152 5.1894 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 14.1894 11.4369 14.1894 39.8157


Shear-force and bending-moment diagrams Maximum bending moments positive M = 8,850 lb-ft negative M = −9,839 lb-ft Bending stresses at max positive moment (8,850 lb-ft)(1.6848 in.)(12 in./ft) x 39.8157 in.4 4, 494 psi (C) 4.49 ksi (C) x

(8,850 lb-ft)( 4.3152 in.)(12 in./ft) 39.8157 in.4 11,510 psi (T) 11.51 ksi (T)

Bending stresses at max negative moment ( 9,839 lb-ft)(1.6848 in.)(12 in./ft) x 39.8157 in.4 4,996 psi (T) 5.00 ksi (T) x


11.51 ksi (T)


12.80 ksi (C)

( 9,839 lb-ft)( 4.3152 in.)(12 in./ft) 39.8157 in.4 12, 796 psi (C) 12.80 ksi (C)

Ans. Ans.


8.24 A W360 × 72 standard steel shape is used to support the loads shown on the beam in Fig. P8.24a. The shape is oriented so that bending occurs about the weak axis as shown in Fig. P8.24b. Consider the entire 6-m length of the beam and determine: (a) the maximum tension bending stress at any location along the beam, and (b) the maximum compression bending stress at any location along the beam.

Fig. P8.24a

Fig. P8.24b


Iz

21.4 106 mm4

bf

204 mm

Shear-force and bending-moment diagrams Maximum bending moments positive M = 31.50 kN-m negative M = −25.87 kN-m Since the shape is symmetric about the z axis, the largest bending stresses will occur at the location of the largest moment magnitude – either positive or negative. In this case, the largest bending stresses will occur where the moment magnitude is 31.50 kN-m. Bending stresses at maximum moment (31.50 kN-m)( 204 mm/2)(1,000)2 x 21.4 106 mm4 150.1 MPa (T) and (C)


150.1 MPa (T)


150.1 MPa (C)

Ans. Ans.


8.25 A 1.00-in.-diameter solid steel shaft supports loads PA = 180 lb and PC = 240 lb as shown in Fig. P8.25. Assume L1 = 5 in., L2 = 16 in., and L3 = 8 in. The bearing at B can be idealized as a roller support and the bearing at D can be idealized as a pin support. Determine the magnitude and location of the maximum bending stress in the shaft. Fig. P8.25

Solution Section properties I

64

D4

64

(1.00 in.)4

0.049087 in.4

Shear-force and bending-moment diagrams Maximum bending moments positive M = 980 lb-in. negative M = −900 lb-in. Since the circular cross section is symmetric about the z axis, the largest bending stresses will occur at the location of the largest moment magnitude – either positive or negative. In this case, the largest bending stresses will occur at C, where the moment magnitude is 980 lb-in. Bending stresses at maximum moment (980 lb-in.)( 1.00 in./2) x 0.049087 in.4 9,980 psi Ans.


8.26 A 30-mm-diameter solid steel shaft supports loads PA = 1,400 N and PC = 2,600 N as shown in Fig. P8.26. Assume L1 = 100 mm, L2 = 200 mm, and L3 = 150 mm. The bearing at B can be idealized as a roller support and the bearing at D can be idealized as a pin support. Determine the magnitude and location of the maximum bending stress in the shaft. Fig. P8.26


64

D4

64

(30 mm)4

39, 760.8 mm4

Shear-force and bending-moment diagrams Maximum bending moments positive M = 162,857 N-mm negative M = −140,000 N-mm Since the circular cross section is symmetric about the z axis, the largest bending stresses will occur at the location of the largest moment magnitude – either positive or negative. In this case, the largest bending stresses will occur where the moment magnitude is 162,857 Nmm. Bending stresses at maximum moment (162,857 N-mm)( 30 mm/2) x 39,760.8 mm 4

61.4 MPa

Ans.


8.27 A 20-mm-diameter solid steel shaft supports loads PA = 500 N, PC = 1,750 N, and PE = 500 N as shown in Fig. P8.27. Assume L1 = 90 mm, L2 = 260 mm, L3 = 140 mm, and L4 = 160 mm. The bearing at B can be idealized as a roller support and the bearing at D can be idealized as a pin support. Determine the magnitude and location of the maximum bending stress in the shaft. Fig. P8.27

Solution Section properties Iz

64

D4

64

(20 mm)4

7,853.9816 mm4

Shear-force and bending-moment diagrams Maximum bending moments positive M = 91,500 N-mm negative M = −80,000 N-mm Since the circular cross section is symmetric about the z axis, the largest bending stresses will occur at the location of the largest moment magnitude – either positive or negative. In this case, the largest bending stresses will occur at C, where the moment magnitude is 91,500 Nmm. Bending stresses at maximum moment (91,500 N-mm)( 20 mm/2) x 7,853.9816 mm 4

116.5 MPa

Ans.


8.28 A 1.75-in.-diameter solid steel shaft supports loads PA = 250 lb, PC = 600 lb, and PE = 250 lb as shown in Fig. P8.28. Assume L1 = 9 in., L2 = 24 in., L3 = 12 in., and L4 = 15 in. The bearing at B can be idealized as a roller support and the bearing at D can be idealized as a pin support. Determine the magnitude and location of the maximum bending stress in the shaft. Fig. P8.28


64

D4

64

(1.75 in.)4

0.460386 in.4

Shear-force and bending-moment diagrams Maximum bending moments positive M = 1,550 lb-in. negative M = −3,750 lb-in. Since the circular cross section is symmetric about the z axis, the largest bending stresses will occur at the location of the largest moment magnitude – either positive or negative. In this case, the largest bending stresses will occur at support D, where the moment magnitude is 3,750 lb-in. Bending stresses at maximum moment ( 3,750 lb-in.)( 1.75 in./2) x 0.460386 in.4 7,130 psi Ans.


8.29 A HSS12 × 8 × 1/2 standard steel shape is used to support the loads shown on the beam in Fig. P8.29. The shape is oriented so that bending occurs about the strong axis. Determine the magnitude and location of the maximum bending stress in the beam.

Fig. P8.29


Iz

333 in.4

d

12 in.

Shear-force and bending-moment diagrams Maximum bending moments positive M = 124.59 kip-ft negative M = −72.00 kip-ft Since the shape is symmetric about the z axis, the largest bending stresses will occur at the location of the largest moment magnitude – either positive or negative. In this case, the largest bending stresses will occur at C, where the moment magnitude is 124.59 kip-ft.

Bending stresses at max moment magnitude (124.59 kip-ft)( 12 in./2)(12 in./ft) x 333 in.4

26.9 ksi

Ans.


8.30 A W410 × 60 standard steel shape is used to support the loads shown on the beam in Fig. P8.30. The shape is oriented so that bending occurs about the strong axis. Determine the magnitude and location of the maximum bending stress in the beam. Fig. P8.30


Iz

216 106 mm 4

d

406 mm

Shear-force and bending-moment diagrams Maximum bending moments positive M = 50 kN-m negative M = −70 kN-m Since the shape is symmetric about the z axis, the largest bending stresses will occur at the location of the largest moment magnitude – either positive or negative. In this case, the largest bending stresses will occur between B and C, where the moment magnitude is 70 kNm.

Bending stresses at max moment magnitude (70 kN-m)( 406 mm/2)(1,000)2 65.8 MPa x 216 106 mm 4

Ans.


8.31 A solid steel shaft supports loads PA = 200 lb and PD = 300 lb as shown in Fig. P8.31. Assume L1 = 6 in., L2 = 20 in., and L3 = 10 in. The bearing at B can be idealized as a roller support and the bearing at C can be idealized as a pin support. If the allowable bending stress is 8 ksi, determine the minimum diameter that can be used for the shaft. Fig. P8.31

Solution Shear-force and bending-moment diagrams Maximum bending moment magnitude M = 3,000 lb-in. Minimum required section modulus M x  S M S 

x



3, 000 lb-in.  0.375 in.3 8, 000 psi

Section modulus for solid circular section d3 S 32 Minimum shaft diameter d 3  0.375 in.3 32

 d  1.563 in.

Ans.


8.32 A solid steel shaft supports loads PA = 500 N and PD = 400 N as shown in Fig. P8.32. Assume L1 = 200 mm, L2 = 660 mm, and L3 = 340 mm. The bearing at B can be idealized as a roller support and the bearing at C can be idealized as a pin support. If the allowable bending stress is 25 MPa, determine the minimum diameter that can be used for the shaft. Fig. P8.32

Solution Shear-force and bending-moment diagrams Maximum bending moment magnitude M = 136,000 N-mm Minimum required section modulus M x  S M S 

x



136, 000 N-mm  5, 440 mm3 25 N/mm 2

Section modulus for solid circular section d3 S 32 Minimum shaft diameter d 3  5,440 mm3 32

 d  38.1 mm

Ans.


8.33 A simply supported wood beam (Fig. P8.33a) with a span of L = 20 ft supports a uniformly distributed load of w = 800 lb/ft. The allowable bending stress of the wood is 1,400 psi. If the aspect ratio of the solid rectangular wood beam is specified as h/b = 1.5 (Fig. P8.33b), determine the minimum width b that can be used for the beam.

Fig. P8.33a

Fig. P8.33b

Solution Shear-force and bending-moment diagrams Also, see Example 7-3 for shear-force and bending-moment diagram development. Maximum bending moment wL2 (800 lb/ft)(20 ft) 2 M max   8 8  40, 000 lb-ft  480, 000 lb-in. Minimum required section modulus M x  S M S 

x



480, 000 lb-in.  342.8571 in.3 1, 400 psi

Section modulus for solid rectangular section I bh3 /12 bh 2 S   c h/2 6

The aspect ratio of the solid rectangular wood beam is specified as h/b = 1.5; therefore, the section modulus can be expressed as: bh 2 b(1.5b) 2 2.25b3 S    0.3750b3 6 6 6 Minimum allowable beam width 0.3750b3  342.8571 in.3  b  9.71 in.

Ans.

The corresponding beam height h is h / b  1.5  h  1.5b  1.5(9.71 in.)  14.57 in.


8.34 A simply supported wood beam (Fig. P8.34a) with a span of L = 14 ft supports a uniformly distributed load of w. The beam width is b = 6 in. and the beam height is h = 10 in. (Fig. P8.34b). The allowable bending stress of the wood is 900 psi. Determine the magnitude of the maximum load w that may be carried by the beam.

Fig. P8.34a

Fig. P8.34b

Solution Moment of inertia for rectangular cross section about horizontal centroidal axis bh3 (6 in.)(10 in.)3 I   500 in.4 12 12 Maximum allowable moment  x I (900 psi)(500 in.4 ) Mc x  M    90, 000 lb-in.  7,500 lb-ft I c 5 in. Shear-force and bending-moment diagrams Also, see Example 7-3 for shear-force and bending-moment diagram development. Determine distributed load intensity Equate the moment expression from the bendingmoment diagram to the maximum allowable moment that can be applied to the rectangular cross section: wL2 M max   7,500 lb-ft 8 Solve for the maximum distributed load w that can be applied to the 14-ft simple span: wL2 w(14 ft) 2   7,500 lb-ft 8 8 8(7,500 lb-ft) Ans. w   306 lb/ft (14 ft) 2


8.35 A cantilever timber beam (Fig. P8.35a) with a span of L = 2.5 m supports a uniformly distributed load of w = 4 kN/m. The allowable bending stress of the wood is 9 MPa. If the aspect ratio of the solid rectangular timber is specified as h/b = 0.5 (Fig. P8.35b), determine the minimum width b that can be used for the beam.

Fig. P8.35a

Fig. P8.35b

Solution Maximum moment magnitude: The maximum bending moment magnitude in the cantilever beam occurs at support A: wL2 (4 kN/m)(2.5 m) 2 M max    12.5 kN-m  12.5 106 N-mm 2 2 Minimum required section modulus M x  S M S 

x



12.5 106 N-mm  1.3889 106 mm3 2 9 N/mm

Section modulus for solid rectangular section I bh3 /12 bh 2 S   c h/2 6 The aspect ratio of the solid rectangular wood beam is specified as h/b = 0.5; therefore, the section modulus can be expressed as: bh 2 b(0.5b) 2 0.25b3 S    0.0416667b3 6 6 6 Minimum allowable beam width 0.0416667b3  1.3889  106 mm3  b  321.83 mm  322 mm

Ans.

The corresponding beam height h is h / b  0.5

 h  0.5b  0.5(321.83 mm)  161 mm


8.36 A cantilever timber beam (Fig. P8.36a) with a span of L = 3 m supports a uniformly distributed load of w. The beam width is b = 300 mm and the beam height is h = 200 mm (Fig. P8.36b). The allowable bending stress of the wood is 6 MPa. Determine the magnitude of the maximum load w that may be carried by the beam.

Fig. P8.36a

Fig. P8.36b

Solution Section modulus for solid rectangular section I bh3 /12 bh 2 (300 mm)(200 mm) 2 S     2 106 mm3 c h/2 6 6 Maximum allowable bending moment: M x   M allow   x S  (6 N/mm2 )(2 106 mm3 )  12 106 N-mm S Maximum bending moment in cantilever span: The maximum bending moment magnitude in the cantilever beam occurs at support A: wL2 M max  2 Maximum distributed load: wL2  M allow 2 2M allow 2(12  106 N-mm)  wallow    2.67 N/mm  2.67 kN/m L2 [(3 m)(1,000 mm/m)]2

Ans.


8.37 The beam shown in Fig. P8.37 will be constructed from a standard steel W-shape using an allowable bending stress of 24 ksi. (a) Develop a list of five acceptable shapes that could be used for this beam. On this list, include the most economical W10, W12, W14, W16, and W18 shapes. (b) Select the most economical W shape for this beam. Fig. P8.37

Solution Shear-force and bending-moment diagrams Maximum bending moment magnitude M = 90 kip-ft Minimum required section modulus M x  S M S 

x



(90 kip-ft)(12 in./ft)  45 in.3 24 ksi

(a) Acceptable steel W-shapes W10  45, S  49.1 in.3 W12  40,

S  51.5 in.3

W14  34,

S  48.6 in.3

W16  31,

S  47.2 in.3

W18  35,

S  57.6 in.3

(b) Most economical W-shape W16  31

Ans.


8.38 The beam shown in Fig. P8.38 will be constructed from a standard steel W-shape using an allowable bending stress of 165 MPa. (a) Develop a list of four acceptable shapes that could be used for this beam. Include the most economical W360, W410, W460, and W530 shapes on the list of possibilities. (b) Select the most economical W shape for this beam. Fig. P8.38

Solution Shear-force and bending-moment diagrams Maximum bending moment magnitude M = 206.630 kN-m Minimum required section modulus M x  S M S 

x



(206.63 kN-m)(1,000) 2  1, 252 103 mm3 2 165 N/mm

(a) Acceptable steel W-shapes W360  79, S  1, 270 103 mm3

W410  75,

S  1,330 103 mm3

W460  74,

S  1, 460 103 mm3

W530  66,

S  1,340 103 mm3


Ans.


8.39 The beam shown in Fig. P8.39 will be constructed from a standard steel W-shape using an allowable bending stress of 165 MPa. (a) Develop a list of four acceptable shapes that could be used for this beam. Include the most economical W360, W410, W460, and W530 shapes on the list of possibilities. (b) Select the most economical W shape for this beam. Fig. P8.39

Solution Shear-force and bending-moment diagrams Maximum bending moment magnitude M = 238.57 kN-m Minimum required section modulus M x  S M S 

x



(238.57 kN-m)(1,000)2  1, 446 103 mm3 2 165 N/mm

(a) Acceptable steel W-shapes W360 101, S  1, 690 103 mm3

W410  85,

S  1,510 103 mm3

W460  74,

S  1, 460 103 mm3

W530  74,

S  1,550 103 mm3

(b) Most economical W-shape W460  74 or W530  74

Ans.


8.40 The beam shown in Fig. P8.40 will be constructed from a standard steel W-shape using an allowable bending stress of 165 MPa. (a) Develop a list of four acceptable shapes that could be used for this beam. Include the most economical W310, W360, W410, and W460 shapes on the list of possibilities. (b) Select the most economical W shape for this beam.

Fig. P8.40

Solution Maximum moment magnitude: The maximum bending moment magnitude occurs at the base of the cantilever beam: 1 1 M max  (15 kN)(3.0 m)  (40 kN/m)(3.0 m) (3.0 m) 2 3 6  105.0 kN-m  105.0 10 N-mm Minimum required section modulus M x  S M (105.0 kN-m)(1,000) 2 S    636 103 mm3 2 x 165 N/mm (a) Acceptable steel W-shapes W310  60, S  844 103 mm3

W360  44,

S  688 103 mm3

W410  46.1, S  773 103 mm3 W460  52,

S  944 103 mm3


Ans.


8.41 The beam shown in Fig. P8.41 will be constructed from a standard steel HSS-shape using an allowable bending stress of 30 ksi. (a) Develop a list of three acceptable shapes that could be used for this beam. On this list, include the most economical HSS8, HSS10, and HSS12 shapes. (b) Select the most economical HSS-shape for this beam. Fig. P8.41

Solution Shear-force and bending-moment diagrams Maximum bending moment magnitude M = 45.56 kip-ft Minimum required section modulus M x  S M S 

x



(45.56 kip-ft)(12 in./ft)  18.22 in.3 30 ksi

(a) Acceptable steel HSS shapes HSS8 none are acceptable

HSS10  4  3 / 8,

S  20.8 in.3

HSS10  6  3 / 8,

S  27.4 in.3

HSS12  6  3 / 8,

S  35.9 in.3

HSS12  8  3 / 8,

S  43.7 in.3

(b) Most economical HSS shape HSS10  4  3 / 8

Ans.


8.42 A composite beam is fabricated by bolting two 3 in. wide × 12 in. deep timber planks to the sides of a 0.50 in. × 12 in. steel plate (Fig. P8.42b). The moduli of elasticity of the timber and the steel are 1,800 ksi and 30,000 ksi, respectively. The simply supported beam spans a distance of 20 ft and carries two concentrated loads P, which are applied at the quarter points of the span (Fig. P8.42a). (a) Determine the maximum bending stresses produced in the timber planks and the steel plate if P = 3 kips. (b) Assume that the allowable bending stresses of the timber and the steel are 1,200 psi and 24,000 psi, respectively. Determine the largest acceptable magnitude for concentrated loads P. (You may neglect the weight of the beam in your calculations.)

Fig. P8.42a Fig. P8.42b

Solution Let the timber be denoted as material (1) and the steel plate as material (2). The modular ratio is: E2 30, 000 ksi n 16.6667 E1 1,800 ksi Transform the steel plate (2) into an equivalent amount of wood (1) by multiplying its width by the modular ratio: b2, trans = 16.6667(0.50 in.) = 8.3333 in. Thus, for calculation purposes, the 12 in. × 0.50 in. steel plate is replaced by a wood board that is 12 in. deep and 8.3333-in. thick. Moment of inertia about the horizontal centroidal axis d = yi – y Shape IC 4 (in. ) (in.) timber (1) 864 0 transformed steel plate (2) 1,200 0 Moment of inertia about the z axis =

d²A (in.4) 0 0

IC + d²A (in.4) 864 1,200 2,064 in.4

Maximum bending moment in beam for P = 3 kips The maximum bending moment in the simply supported beam with two 3-kip concentrated loads is: M max (3 kips)(5 ft) 15 kip-ft 180 kip-in. Bending stress in timber (1) From the flexure formula, the maximum bending stress in timber (1) is: My (180 kip-in.)( 6 in.) 0.5233 ksi 523 psi 1 I 2, 064 in.4

Ans.


Bending stress in steel plate (2) The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the maximum bending stress in steel plate (2) is: My (180 kip-in.)( 6 in.) n (16.6667) 8.7209 ksi 8,720 psi Ans. 2 I 2,064 in.4 Determine maximum P If the allowable bending stress in the timber is 1,200 psi, then the maximum bending moment that may be supported by the beam is: My (1.200 ksi)(2, 064 in.4 ) 1I M 412.80 kip-in. 1 max I y 6 in. If the allowable bending stress in the steel is 165 MPa, then the maximum bending moment that may be supported by the beam is: My (24.00 ksi)(2,064 in.4 ) 2I n M 495.36 kip-in. 2 max I ny (16.667)(6 in.) Note: The negative signs were omitted in the previous two equations because only the moment magnitude is of interest here. From these two results, the maximum moment that the beam can support is 412.80 kip-in. The maximum concentrated load magnitude P that can be supported is found from: M max (5 ft)P

P

M max 5 ft

412.80 kip-in. (5 ft)(12 in./ft)

6.88 kips

Ans.


8.43 The cross section of a composite beam that consists of 4-mm-thick fiberglass faces bonded to a 20mm-thick particleboard core is shown in Fig. P8.43. The beam is subjected to a bending moment of 55 N-m acting about the z axis. The elastic moduli for the fiberglass and the particleboard are 30 GPa and 10 GPa, respectively. Determine: (a) the maximum bending stresses in the fiberglass faces and the particleboard core. (b) the stress in the fiberglass at the joint where the two materials are bonded together. Fig. P8.43

Solution Let the particleboard be denoted as material (1) and the fiberglass as material (2). The modular ratio is: E2 30 GPa n 3 E1 10 GPa Transform the fiberglass faces into an equivalent amount of particleboard by multiplying their width by the modular ratio: b2, trans = 3(50 mm) = 150 mm. Thus, for calculation purposes, the 50 mm × 4 mm fiberglass faces are replaced by particleboard faces that are 150-mm wide and 4-mm thick. Moment of inertia about the horizontal centroidal axis d = yi – y Shape IC 4 (mm ) (mm) transformed fiberglass top face 800.00 12.00 particleboard core 33,333.33 0 transformed fiberglass bot face 800.00 12.00 Moment of inertia about the z axis =

d²A (mm4) 86,400.00 0 86,400.00

Bending stress in particleboard core (1) From the flexure formula, the maximum bending stress in the particleboard core is: My (55 N-m)( 10 mm)(1,000 mm/m) 2.65 MPa 1 I 207,733.33 mm 4

IC + d²A (mm4) 87,200.00 33,333.33 87,200.00 207,733.33 mm4

Ans.

Bending stress in fiberglass faces (2) The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the maximum bending stress in the fiberglass faces (2) is: My (55 N-m)( 14 mm)(1,000 mm/m) n (3) 11.12 MPa Ans. 2 I 207,733.33 mm 4 Bending stress in fiberglass (2) at interface At the interface between the particleboard and the fiberglass, y = ±10 mm: My (55 N-m)( 10 mm)(1,000 mm/m) n (3) 7.94 MPa 2 I 207,733.33 mm 4

Ans.


8.44 A composite beam is made of two brass [E = 100 GPa] plates bonded to an aluminum [E = 75 GPa] bar, as shown in Fig. P8.44. The beam is subjected to a bending moment of 1,750 N-m acting about the z axis. Determine: (a) the maximum bending stresses in the brass plates and the aluminum bar. (b) the stress in the brass at the joints where the two materials are bonded together.

Fig. P8.44

Solution Let the aluminum be denoted as material (1) and the brass as material (2). The modular ratio is: E2 100 GPa n 1.3333 E1 75 GPa Transform the brass plates into an equivalent amount of aluminum by multiplying their width by the modular ratio: b2, trans = 1.3333(50 mm) = 66.6666 mm. Thus, for calculation purposes, the 50 mm × 10 mm brass plates are replaced by aluminum plates that are 66.6666-mm wide and 10-mm thick. Moment of inertia about the horizontal centroidal axis d = yi – y Shape IC 4 (mm ) (mm) transformed top brass plate 5,555.55 20 aluminum bar 112,500.00 0 transformed bot brass plate 5,555.55 –20 Moment of inertia about the z axis =

d²A (mm4) 266,666.40 0 266,666.40

Bending stress in aluminum bar (1) From the flexure formula, the maximum bending stress in the aluminum bar is: My (1,750 N-m)( 15 mm)(1,000 mm/m) 40.0 MPa 1 I 656,943.90 mm 4

IC + d²A (mm4) 272,221.95 112,500.00 272,221.95 656,943.90 mm4

Ans.

Maximum bending stress in brass plates (2) The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the maximum bending stress in the brass plates (2) is: My (1,750 N-m)( 25 mm)(1,000 mm/m) n (1.3333) 88.8 MPa Ans. 2 I 656,943.90 mm 4 Bending stress in brass plates (2) at interface At the interface between the brass plates and the aluminum bar, y = ±15 mm: My (1,750 N-m)( 15 mm)(1,000 mm/m) n (1.3333) 53.3 MPa 2 I 656,943.90 mm 4

Ans.


8.45 An aluminum [E = 10,000 ksi] bar is bonded to a steel [E = 30,000 ksi] bar to form a composite beam (Fig. P8.45b). The composite beam is subjected to a bending moment of M = +300 lb-ft about the z axis (Fig. P8.45a). Determine: (a) the maximum bending stresses in the aluminum and steel bars. (b) the stress in the two materials at the joint where they are bonded together.

Fig. P8.45a

Fig. P8.45b

Solution Denote the aluminum as material (1) and denote the steel as material (2). The modular ratio is: E2 30, 000 ksi n 3 E1 10,000 ksi Transform the steel bar (2) into an equivalent amount of aluminum (1) by multiplying its width by the modular ratio: b2, trans = 3(2.00 in.) = 6.00 in. Thus, for calculation purposes, the 2.00 in. × 0.75 in. steel bar is replaced by an aluminum bar that is 6.00-in. wide and 0.75-in. thick. Centroid location of the transformed section in the vertical direction Shape aluminum bar (1) transformed steel bar (2)

y

yi Ai Ai

4.1875 in.3 5.50 in.2

Width b (in.) 2.00 6.00

Height h (in.) 0.50 0.75

Area Ai (in.2) 1.00 4.50 5.50

yi (from bottom) (in.) 0.25 0.875

yi Ai (in.3) 0.2500 3.9375 4.1875

0.7614 in. (measured upward from bottom edge of section)

Moment of inertia about the horizontal centroidal axis d = yi – y Shape IC 4 (in. ) (in.) aluminum bar (1) 0.02083 –0.5114 transformed steel bar (2) 0.2109 0.1136 Moment of inertia about the z axis =

d²A (in.4) 0.2615 0.05811

IC + d²A (in.4) 0.2823 0.2690 0.5514 in.4


(a) Maximum bending stress in aluminum bar (1) From the flexure formula, the maximum bending stress in aluminum bar (1) is: My (300 lb-ft)( 0.7614 in.)(12 in./ft) 4,970 psi (T) 1 I 0.5514 in.4

Ans.

(a) Maximum bending stress in steel bar (2) The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the maximum bending stress in steel bar (2) is: My (300 lb-ft)(1.250 in. 0.7614 in.)(12 in./ft) Ans. (3) 9,570 psi (C) 2 I 0.5514 in.4 (b) Bending stress in aluminum bar (1) at interface My (300 lb-ft)(0.50 in. 0.7614 in.)(12 in./ft) 1 I 0.5514 in.4

1,706 psi (T)

(b) Bending stress in steel bar (2) at interface My (300 lb-ft)(0.50 in. 0.7614 in.)(12 in./ft) (3) 2 I 0.5514 in.4

5,120 psi (T)

Ans.

Ans.


8.46 An aluminum [E = 10,000 ksi] bar is bonded to a steel [E = 30,000 ksi] bar to form a composite beam (Fig. P8.46b). The allowable bending stresses for the aluminum and steel bars are 20 ksi and 30 ksi, respectively. Determine the maximum bending moment M that can be applied to the beam.

Fig. P8.46a

Fig. P8.46b

Solution Denote the aluminum as material (1) and denote the steel as material (2). The modular ratio is: E2 30, 000 ksi n 3 E1 10,000 ksi Transform the steel bar (2) into an equivalent amount of aluminum (1) by multiplying its width by the modular ratio: b2, trans = 3(2.00 in.) = 6.00 in. Thus, for calculation purposes, the 2.00 in. × 0.75 in. steel bar is replaced by an aluminum bar that is 6.00-in. wide and 0.75-in. thick. Centroid location of the transformed section in the vertical direction Shape aluminum bar (1) transformed steel bar (2)

y

yi Ai Ai

4.1875 in.3 5.50 in.2

Width b (in.) 2.00 6.00

Height h (in.) 0.50 0.75

Area Ai (in.2) 1.00 4.50 5.50


yi Ai (in.3) 0.2500 3.9375 4.1875


Moment of inertia about the horizontal centroidal axis d = yi – y Shape IC 4 (in. ) (in.) aluminum bar (1) 0.02083 –0.5114 transformed steel bar (2) 0.2109 0.1136 Moment of inertia about the z axis =

d²A (in.4) 0.2615 0.05811

IC + d²A (in.4) 0.2823 0.2690 0.5514 in.4


(a) Maximum bending moment magnitude based on allowable aluminum stress Based on an allowable bending stress of 20 ksi for the aluminum, the maximum bending moment magnitude that be applied to the cross section is: My (20 ksi)(0.5514 in.4 ) 1I (a) M 14.484 kip-in. 1 I y 0.7614 in. Maximum bending moment magnitude based on allowable steel stress Based on an allowable bending stress of 30 ksi for the steel, the maximum bending moment magnitude that be applied to the cross section is: My (30 ksi)(0.5514 in.4 ) 2I (b) n M 11.285 kip-in. 2 I ny (3)(1.25 in. 0.7614 in.) Maximum bending moment magnitude From the values obtained in Eqs. (a) and (b), the maximum bending moment that can be applied to the cross section is Ans. M max 11.285 kip-in. 940 lb-ft


8.47 Two steel [E = 30,000 ksi] plates are securely attached to a Southern pine [E = 1,800 ksi] timber to form a composite beam (Fig. P8.47). The allowable bending stress for the steel plates is 24,000 psi and the allowable bending stress for the Southern pine is 1,200 psi. Determine the maximum bending moment that can be applied about the horizontal axis of the beam.

Fig. P8.47

Solution Denote the timber as material (1) and denote the steel as material (2). The modular ratio is: E2 30, 000 ksi n 16.6667 E1 1,800 ksi Transform the steel plates into an equivalent amount of timber by multiplying their width by the modular ratio: b2, trans = 16.6667(8 in.) = 133.3333 in. Thus, for calculation purposes, the 8 in. × 0.25 in. steel plates can be replaced by wood plates that are 133.3333-in. wide and 0.25-in. thick. Moment of inertia about the horizontal centroidal axis d = yi – y Shape IC 4 (in. ) (in.) transformed steel plate at top 0.1736 8.125 timber (1) 3,413.3333 0 transformed steel plate at bottom 0.1736 –8.125 Moment of inertia about the z axis =

d²A (in.4) 2,200.52 0 2,200.52

IC + d²A (in.4) 2,200.694 3,413.333 2,200.694 7,814.72 in.4

(a) Maximum bending moment magnitude based on allowable Southern pine stress Based on an allowable bending stress of 1,200 psi for the Southern pine timber, the maximum bending moment magnitude that be applied to the cross section is: My (1.200 ksi)(7,814.72 in.4 ) 1I (a) M 1,172.208 kip-in. 1 I y 8 in. Maximum bending moment magnitude based on allowable steel stress Based on an allowable bending stress of 24,000 psi for the steel plates, the maximum bending moment magnitude that be applied to the cross section is: My (24 ksi)(7,814.72 in.4 ) 2I (b) n M 1,364.021 kip-in. 2 I ny (16.6667)( 8.25 in.) Maximum bending moment magnitude From the values obtained in Eqs. (a) and (b), the maximum bending moment that can be applied to the cross section is Ans. M max 1,172.208 kip-in. 97.7 kip-ft


8.48 A simply supported composite beam 5 m long carries a uniformly distributed load w (Fig. P8.48a). The beam is constructed of a Southern pine [E = 12 GPa] timber, 200 mm wide by 360 mm deep, that is reinforced on its lower surface by a steel [E = 200 GPa] plate that is 150 mm wide by 12 mm thick (Fig. P8.48b). (a) Determine the maximum bending stresses produced in the timber and the steel if w = 12 kN/m. (b) Assume that the allowable bending stresses of the timber and the steel are 9 MPa and 165 MPa, respectively. Determine the largest acceptable magnitude for distributed load w. (You may neglect the weight of the beam in your calculations.)

Fig. P8.48a

Fig. P8.48b

Solution Let the timber be denoted as material (1) and the steel plate as material (2). The modular ratio is: E2 200 GPa n 16.6667 E1 12 GPa Transform the steel plate (2) into an equivalent amount of wood (1) by multiplying its width by the modular ratio: b2, trans = 16.6667(150 mm) = 2,500 mm. Thus, for calculation purposes, the 150 mm × 12 mm steel plate is replaced by a wood board that is 2,500-mm wide and 12-mm thick. Centroid location of the transformed section in the vertical direction Shape timber (1) transformed steel plate (2)

y

yi Ai Ai

Width b (mm) 200 2,500

14,004,000 mm3 102,000 mm2

Height h (mm) 360 12

Area Ai (mm2) 72,000 30,000 102,000

yi (from bottom) (mm) 192 6

yi Ai (mm3) 13,824,000 180,000 14,004,000

137.294 mm (measured upward from bottom edge of section)

Moment of inertia about the horizontal centroidal axis d = yi – y Shape IC 4 (mm ) (mm) timber (1) 777,600,000 54.71 transformed steel plate (2) 360,000 –131.29 Moment of inertia about the z axis =

d²A (mm4) 215,476,817 517,144,360

IC + d²A (mm4) 993,076,817 517,504,360 1,510,581,176 mm4 = 1.5106 ×109 mm4


Bending moment in beam for w = 12 kN/m The bending moment in the simply supported beam with a uniformly distributed load of 12 kN/m is: wL2 (12 kN/m)(5 m) 2 M max 37.5 kN-m 37.5 106 N-mm 8 8 Bending stress in timber (1) From the flexure formula, the maximum bending stress in timber (1) is: My (37.5 106 N-mm)(372 mm 137.294 mm) 5.83 MPa (C) 1 I 1.5106 109 mm4

Ans.

Bending stress in steel plate (2) The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the maximum bending stress in steel plate (2) is: My (37.5 106 N-mm)( 137.294 mm) (16.6667) 56.8 MPa (T) Ans. 2 I 1.5106 109 mm 4 Determine maximum w If the allowable bending stress in the timber is 9 MPa, then the maximum bending moment that may be supported by the beam is: My (9 N/mm 2 )(1.5106 109 mm 4 ) 1I M 57.925 106 N-mm 1 max I y (372 mm 137.294 mm) If the allowable bending stress in the steel is 165 MPa, then the maximum bending moment that may be supported by the beam is: My (165 N/mm2 )(1.5106 109 mm4 ) 2I n M 108.926 106 N-mm 2 max I ny (16.6667)(137.294 mm) Note: The negative signs were omitted in the previous two equations because only the moment magnitude is of interest here. From these two results, the maximum moment that the beam can support is 57.925×106 N-mm. The maximum distributed load magnitude w that can be supported is found from: wL2 M max 8 8M max 8(57.925 106 N-mm)(1 m/1000 mm) Ans. w 18,536 N/m 18.54 kN/m L2 (5 m)2


8.49 A glue-laminated timber beam is reinforced by carbon fiber reinforced plastic (CFRP) material bonded to its bottom surface. The cross section of the composite beam is shown in Fig. P8.49b. The elastic modulus of the wood is E = 12 GPa and the elastic modulus of the CFRP is 112 GPa. The simply supported beam spans 6 m and carries a concentrated load P at midspan (Fig. P8.49a). (a) Determine the maximum bending stresses produced in the timber and the CFRP if P = 4 kN. (b) Assume that the allowable bending stresses of the timber and the CFRP are 9 MPa and 1,500 MPa, respectively. Determine the largest acceptable magnitude for concentrated load P. (You may neglect the weight of the beam in your calculations.)


Solution Denoted the timber as material (1) and denote the CFRP as material (2). The modular ratio is: E2 112 GPa n 9.3333 E1 12 GPa Transform the CFRP into an equivalent amount of wood by multiplying its width by the modular ratio: b2, trans = 9.3333(40 mm) = 373.33 mm. Thus, for calculation purposes, the 40 mm × 3 mm CFRP is replaced by a wood board that is 373.33-mm wide and 3-mm thick. Centroid location of the transformed section in the vertical direction Shape timber (1) transformed CFRP (2)

y

yi Ai Ai

Width b (mm) 90 373.33

2,881,680 mm3 23,620 mm2

Height h (mm) 250 3

Area Ai (mm2) 22,500 1,120 23,620

yi (from bottom) (mm) 128 1.5

yi Ai (mm3) 2,880,000 1,680 2,881,680

122.00 mm (measured upward from bottom edge of section)

Moment of inertia about the horizontal centroidal axis d = yi – y Shape IC 4 (mm ) (mm) timber (1) 117,187,500 6.00 transformed CFRP (2) 840 –120.50 Moment of inertia about the z axis =

d²A (mm4) 810,000 16,262,680

IC + d²A (mm4) 117,997,500 16,263,520 134,261,020 mm4 = 134.261 ×106 mm4


Maximum bending moment in beam for P = 4 kN The maximum bending moment in the simply supported beam with a concentrated load of 4 kN at midspan is: PL (4 kN)(6 m) M max 6 kN-m 6 106 N-mm 4 4 (a) Bending stress in timber (1) From the flexure formula, the maximum bending stress in timber (1) is: My (6 106 N-mm)(253 mm 122.00 mm) 5.85 MPa (C) 1 I 134.261 106 mm 4

Ans.

(a) Bending stress in CFRP (2) The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the maximum bending stress in the CFRP is: My (6 106 N-mm)( 122.00 mm) (9.3333) 50.9 MPa (T) Ans. 2 I 134.261 106 mm 4 (b) Determine maximum P If the allowable bending stress in the timber is 9 MPa, then the maximum bending moment that may be supported by the beam is: My (9 N/mm 2 )(134.261 106 mm 4 ) 1I M 9.224 106 N-mm 1 max I y (253 mm 122.00 mm) If the allowable bending stress in the CFRP is 1,500 MPa, then the maximum bending moment that may be supported by the beam is: My (1,500 N/mm2 )(134.261 106 mm4 ) 2I n M 176.867 106 N-mm 2 max I ny (9.3333)(122.00 mm) Note: The negative signs were omitted in the previous two equations because only the moment magnitude is of interest here. From these two results, the maximum moment that the beam can support is 9.224×106 N-mm. The maximum concentrated load magnitude P that can be supported is found from: PL M max 4 4M max 4(9.224 106 N-mm)(1 m/1000 mm) Ans. P 6,149 N 6.15 kN L (6 m)


8.50 Two steel plates, each 4 in. wide and 0.25 in. thick, reinforce a wood beam that is 3 in. wide and 8 in. deep. The steel plates are attached to the vertical sides of the wood beam in a position such that the composite shape is symmetric about the z axis, as shown in the sketch of the beam cross section (Fig. P8.50). Determine the maximum bending stresses produced in both the wood and the steel if a bending moment of Mz = +50 kip-in is applied about the z axis. Assume Ewood = 2,000 ksi and Esteel = 30,000 ksi. Fig. P8.50

Solution Let the wood be denoted as material (1) and the steel plates as material (2). The modular ratio is: E2 30, 000 ksi n 15 E1 2,000 ksi Transform the steel plates (2) into an equivalent amount of wood (1) by multiplying the plate thicknesses by the modular ratio: b2, trans = 15(0.25 in.) = 3.75 in. (each). Thus, for calculation purposes, each 4 in. × 0.25 in. steel plate is replaced by a wood board that is 4-in. tall and 3.75-in. wide. Centroid location: Since the transformed section is doubly symmetric, the centroid location is found from symmetry. Moment of inertia about the z centroidal axis Shape IC (in.4) wood beam (1) 128 two transformed steel plates (2) 40 Moment of inertia about the z axis =

d = yi – y (in.) 0 0

d²A (in.4) 0 0

Bending stress in wood beam (1) From the flexure formula, the maximum bending stress in wood beam (1) is: M z c (50 kip-in.)(4 in.) 1.190 ksi 1,190 psi 1 Iz 168 in.4

IC + d²A (in.4) 128 40 168 in.4

Ans.

Bending stress in steel plates (2) The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the maximum bending stress in the steel plates (2) is: M c (50 kip-in.)(2 in.) n z (15) 8.93 ksi 8,930 psi Ans. 2 Iz 168 in.4


8.51 A glue-laminated timber beam is reinforced by carbon fiber reinforced plastic (CFRP) material bonded to its bottom surface. The cross section of the composite beam is shown in Fig. P8.51b. The elastic modulus of the wood is 1,700 ksi and the elastic modulus of the CFRP is 23,800 ksi. The simply supported beam spans 24 ft and carries two concentrated loads P, which act at the quarter-points of the span (Fig. P8.51a). The allowable bending stresses of the timber and the CFRP are 2,400 psi and 175,000 psi, respectively. Determine the largest acceptable magnitude for the concentrated loads P. (You may neglect the weight of the beam in your calculations.)


Solution Denoted the timber as material (1) and denote the CFRP as material (2). The modular ratio is: E2 23,800 ksi n 14 E1 1,700 ksi Transform the CFRP into an equivalent amount of wood by multiplying its width by the modular ratio: b2, trans = 14(3 in.) = 42 in. Thus, for calculation purposes, the 3 in. × 0.125 in. CFRP is replaced by a wood board that is 42-in. wide and 0.125-in. thick. Centroid location of the transformed section in the vertical direction Shape timber (1) transformed CFRP (2)

y

yi Ai Ai

404.5781 in.3 71.25 in.2

Width b (in.) 5.5 42.0

Height h (in.) 12 0.125

Area Ai (in.2) 66 5.25 71.25


yi Ai (in.3) 404.25 0.3281 404.5781


Moment of inertia about the horizontal centroidal axis d = yi – y Shape IC 4 (in. ) (in.) timber (1) 792 0.4467 transformed CFRP (2) 0.00684 –5.6158 Moment of inertia about the z axis =

d²A (in.4) 13.1703 165.5697

IC + d²A (in.4) 805.170 165.577 970.747 in.4


Determine maximum P If the allowable bending stress in the timber is 2,400 psi, then the maximum bending moment that may be supported by the beam is: My (2.40 ksi)(970.747 in.4 ) 1I M 361.393 kip-in. 1 max I y (12.125 in. 5.6783 in.) If the allowable bending stress in the CFRP is 175,000 psi, then the maximum bending moment that may be supported by the beam is: My (175 ksi)(970.747 in.4 ) 2I n M 2,137 kip-in. 2 max I ny (14)(5.6783 in.) Note: The negative signs were omitted in the previous two equations because only the moment magnitude is of interest here. From these two results, the maximum moment that the beam can support is 351.393 kip-in. = 30.116 kip-ft. The maximum concentrated load magnitude P that can be supported is found from: M max (6 ft)P P

M max 6 ft

30.116 kip-ft 6 ft

5.02 kips

Ans.


8.52 A steel pipe assembly supports a concentrated load of P = 22 kN as shown in Fig. P8.52. The outside diameter of the pipe is 142 mm and the wall thickness is 6.5 mm. Determine the normal stresses produced at points H and K.

Fig. P8.52

Solution Section properties d  D  2t  142 mm  2(6.5 mm)  129 mm A



Iz 



 D 2  d 2   (142 mm)2  (129 mm)2   2, 766.958 mm 2 4 4





 D 4  d 4   (142 mm) 4  (129 mm) 4   6,364,867 mm 4 64  64 

Internal forces and moments F  22 kN  22,000 N M z  (22,000 N)(370 mm)  8,140,000 N-mm Stresses

 axial   bending

F 22,000 N   7.951 MPa (C) A 2,766.958 mm 2 M c (8,140,000 N-mm)(142 mm/2)  z   90.802 MPa Iz 6,364,867 mm 4

Normal stress at H By inspection, the bending stress at H will be compression; therefore, the normal stress at H is:  H  7.951 MPa  90.802 MPa  98.753 MPa  98.8 MPa (C)

Ans.

Normal stress at K By inspection, the bending stress at K will be tension; therefore, the normal stress at K is:  K  7.951 MPa  90.802 MPa  82.851 MPa  82.9 MPa (T)

Ans.


8.53 The screw of a clamp exerts a compressive force of 350 lb on the wood blocks. Determine the normal stresses produced at points H and K. The clamp cross-sectional dimensions at the section of interest are 1.25 in. by 0.375 in. thick.

Fig. P8.53

Solution Section properties A  (0.375 in.)(1.250 in.)  0.468750 in.2

(0.375 in.)(1.250 in.)3 Iz   0.061035 in.4 12 Internal forces and moments F  350 lb M z  (350 lb)(3.75 in.  1.25 in./2)  1,531.25 lb-in. Stresses


F 350 lb   746.667 psi (T) A 0.468750 in.2 M c (1,531.25 lb-in.)(1.250 in./2)  z   15,680.0 psi Iz 0.061035 in.4

Normal stress at H By inspection, the bending stress at H will be tension; therefore, the normal stress at H is:  H  746.667 psi  15,680 psi  16,426.667 psi  16,430 psi (T)

Ans.

Normal stress at K By inspection, the bending stress at K will be compression; therefore, the normal stress at K is:  K  746.667 psi  15,680 psi  14,933.333 psi  14,930 psi (C)

Ans.


8.54 Determine the normal stresses produced at points H and K of the pier support shown in Fig. P8.54a.

Fig. P8.54a

Fig. P8.54b Cross section a–a

Solution Section properties A  (250 mm)(500 mm)  125, 000 mm 2

Iz 

(250 mm)(500 mm)3  2.60417 109 mm 4 12

Internal forces and moments F  250 kN  400 kN  650 kN M z  (250 kN)(3.25 m)  (400 kN)(2.25 m)  87.50 kN-m Stresses

 axial 

F 650, 000 N   5.20 MPa (C) A 125, 000 mm 2

 bending 

M z c (87.5 kN-m)(500 mm/2)(1,000)2   8.40 MPa Iz 2.60417 109 mm 4

Normal stress at H By inspection, the bending stress at H will be tension; therefore, the normal stress at H is:  H  5.20 MPa  8.40 MPa  3.20 MPa  3.20 MPa (T)

Ans.

Normal stress at K By inspection, the bending stress at K will be compression; therefore, the normal stress at K is:  K  5.20 MPa  8.40 MPa  13.60 MPa  13.60 MPa (C)

Ans.


8.55 A tubular steel column CD supports horizontal cantilever arm ABC, as shown in Fig. P8.55. Column CD has an outside diameter of 10.75 in. and a wall thickness of 0.365 in. Determine the maximum compression stress at the base of column CD.

Fig. P8.55

Solution Section properties d  D  2t  10.750 in.  2(0.365 in.)  10.020 in. A



Iz 



 D 2  d 2   (10.750 in.) 2  (10.020 in.) 2   11.908 in.2 4 4





 D 4  d 4   (10.750 in.) 4  (10.020 in.) 4   160.734 in.4 64 64

Internal forces and moments F  700 lb  900 lb  1, 600 lb M  (700 lb)(13 ft)  (900 lb)(23 ft)  29,800 lb-ft  357, 600 lb-in. Stresses


F 1, 600 lb   134.36 psi (C) A 11.908 in.2 M c (357, 600 lb-in.)(10.75 in./2)    11,958.27 psi I 160.734 in.4

Maximum compression stress at base of column  compression  134.36 psi  11,958.27 psi  12,092.63 psi  12.09 ksi (C)

Ans.


8.56 Determine the normal stresses acting at points H and K for the structure shown in Fig. P8.56a. The cross-sectional dimensions of the vertical member are shown in Fig. P8.56b.

Fig. P8.56b Cross section Fig. P8.56a

Solution Section properties A  (4 in.)(8 in.)  32 in.2

(4 in.)(8 in.)3 Iz   170.6667 in.4 12 Internal forces and moments F  1, 200 lb  2,800 lb  4,000 lb M z  (1, 200 lb)(12 in.  8 in./2)  19, 200 lb-in. Stresses


F 4,000 lb   125 psi (C) A 32 in.2 M c (19,200 lb-in.)(8 in./2)  z   450 psi Iz 170.6667 in.4

Normal stress at H By inspection, the bending stress at H will be compression; therefore, the normal stress at H is:  H  125 psi  450 psi  575 psi  575 psi (C)

Ans.

Normal stress at K By inspection, the bending stress at K will be tension; therefore, the normal stress at K is:  K  125 psi  450 psi  325 psi  325 psi (T)

Ans.


8.57 A W18 × 35 standard steel shape is subjected to a tension force P that is applied 15 in. above the bottom surface of the wideflange shape as shown in Fig. P8.57. If the tension normal stress of the upper surface of the W-shape must be limited to 18 ksi, determine the allowable force P that may be applied to the member.

Fig. P8.57

Solution Section properties (from Appendix B) Depth d  17.7 in. A  10.3 in.2 I z  510 in.4

Stresses


F P  A 10.3 in.2 M c P(15 in.  17.7 in./2)(17.7 in./2) P(6.15 in.)(8.85 in.) P(54.4275 in.2 )  z    Iz 510 in.4 510 in.4 510 in.4

Normal stress on the upper surface of the W-shape The tension normal stress on the upper surface is equal to the sum of the axial and bending stresses. Since these stresses are expressed in terms of the unknown force P, the tension normal stress is given by: P P(54.4275 in.2 )  upper surface   10.3 in.2 510 in.4  P(0.097087 in.2  0.106721 in.2 )  (0.203808 in.2 ) P The normal stress on the upper surface of the W-shape must be limited to 18 ksi; therefore, (0.203808 in.2 ) P  18 ksi

P 

18 ksi  88.3 kips 0.203808 in.2

Ans.


8.58 A WT305 × 41 standard steel shape is subjected to a tension force P that is applied 250 mm above the bottom surface of the tee shape, as shown in Fig. P8.58. If the tension normal stress of the upper surface of the WT-shape must be limited to 150 MPa, determine the allowable force P that may be applied to the member.

Fig. P8.58

Solution Section properties (from Appendix B) Depth d  300 mm

Centroid y  88.9 mm (from flange to centroid) A  5, 230 mm 2 I z  48.7 106 mm 4 Stresses

 axial 

 bending

F P   P(1.9120 104 mm 2 ) A 5, 230 mm 2 M c P(250 mm  88.9 mm)(300 mm  88.9 mm)  z  Iz 48.7 106 mm 4

P(161.1 mm)(211.1 mm) 48.7 106 mm 4  P(6.9832 104 mm 2 ) 

Normal stress on the upper surface of the WT-shape The tension normal stress on the upper surface is equal to the sum of the axial and bending stresses. Since these stresses are expressed in terms of the unknown force P, the tension normal stress is given by:  upper surface  P(1.9120 104 mm2 )  P(6.9832 104 mm2 )  (8.8953 104 mm2 ) P The normal stress on the upper surface of the WT-shape must be limited to 150 MPa; therefore, (8.8953  104 mm2 ) P  150 MPa

150 N/mm2 P   168,629 N  168.6 kN 8.8953  104 mm2

Ans.


8.59 A pin support consists of a vertical plate 60 mm wide by 10 mm thick. The pin carries a load of 1,200 N. Determine the normal stresses acting at points H and K for the structure shown in Fig. P8.59.

Fig. P8.59

Solution Section properties A  (60 mm)(10 mm)  600 mm 2

I

(60 mm)(10 mm)3  5, 000 mm4 12

Internal forces and moments F  1, 200 N M  (1, 200 N)(30 mm  10 mm/2)  42,000 N-mm Stresses


F 1, 200 N   2.00 MPa (T) A 600 mm 2 M c (42,000 N-mm)(10 mm/2)    42.00 MPa I 5,000 mm 4

Normal stress at H By inspection, the bending stress at H will be compression; therefore, the normal stress at H is:  H  2.00 MPa  42.00 MPa  40.00 MPa  40.0 MPa (C)

Ans.

Normal stress at K By inspection, the bending stress at K will be tension; therefore, the normal stress at K is:  K  2.00 MPa  42.00 MPa  44.00 MPa  44.0 MPa (T)

Ans.


8.60 The tee shape shown in Fig. P8.60b is used as a short post to support a load of P = 4,600 lb. The load P is applied at a distance of 5 in. from the surface of the flange, as shown in Fig. P8.60a. Determine the normal stresses at points H and K, which are located on section a–a.

Fig. P8.60b Cross-sectional dimensions Fig. P8.60a

Solution Centroid location in x direction: Shape

width b (in.) 12 2

flange stem

x

xi Ai Ai



height h (in.) 2 10

164 in.3  3.7273 in. 44 in.2  8.2727 in.

Area Ai (in.2) 24 20 44 in.2

xi (from left) (in.) 1 7

xi Ai (in.3) 24 140 164 in.3

(from left side to centroid) (from right side to centroid)

Moment of inertia about the z axis: Shape IC d = xi – x d²A (in.4) (in.) (in.4) flange 8 −2.7273 178.5160 stem 166.6667 3.2727 214.2113 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 186.5160 380.8790 567.3940

Internal forces and moments F  4,600 lb M z  (4,600 lb)(5 in.  3.7273 in.)  40,145.455 lb-in.


Stresses

F 4,600 lb   104.545 psi A 44 in.2 M x (40,145.455 lb-in.)(  3.7273 in.)  H ,bending  z   263.720 psi Iz 567.3940 in.4

 axial 

 K ,bending 

M z x (40,145.455 lb-in.)(8.2727 in.)   585.329 psi Iz 567.3940 in.4

Normal stress at H  H  104.545 psi  263.720 psi  368.265 psi  368 psi (C)

Ans.

Normal stress at K  K  104.545 psi  585.329 psi  480.784 psi  481 psi (T)

Ans.


8.61 The tee shape shown in Fig. P8.61b is used as a short post to support a load of P. The load P is applied at a distance of 5 in. from the surface of the flange, as shown in Fig. P8.61a. The tension and compression normal stresses in the post must be limited to 1,000 psi and 800 psi, respectively. Determine the maximum magnitude of load P that satisfies both the tension and compression stress limits.

Fig. P8.61b Cross-sectional dimensions Fig. P8.61a


width b (in.) 12 2

flange stem

x

xi Ai Ai



height h (in.) 2 10

164 in.3  3.7273 in. 44 in.2  8.2727 in.

Area Ai (in.2) 24 20 44 in.2

xi (from left) (in.) 1 7

xi Ai (in.3) 24 140 164 in.3


Moment of inertia about the z axis: Shape IC d = xi – x d²A (in.4) (in.) (in.4) flange 8 −2.7273 178.5160 stem 166.6667 3.2727 214.2113 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 186.5160 380.8790 567.3940

Internal forces and moments FP M z  P(5 in.  3.7273 in.)  (8.7273 in.)P


Stresses F P   (0.022727 in.2 ) P A 44 in.2 M x (8.7273 in.)P(  3.7273 in.)  H ,bending  z   (0.057331 in.2 ) P Iz 567.3940 in.4

 axial 

 K ,bending 

M z x (8.7273 in.)P(8.2727 in.)   (0.127246 in.2 ) P 4 Iz 567.3940 in.

Compression stress limit (at H)  H  (0.022727 in.2 ) P  (0.057331 in.2 ) P  (0.080058 in.2 ) P (0.080058 in.2 ) P  800 psi  P  9,992.76 lb

Tension stress limit (at K)  K  (0.022727 in.2 ) P  (0.127246 in.2 ) P  (0.104519 in.2 ) P (0.104519 in.2 ) P  1, 000 psi  P  9,567.64 lb

Maximum magnitude of load P Pmax  9,570 lb

Ans.


8.62 The tee shape shown in Fig. P8.62b is used as a post that supports a load of P = 25 kN. Note that the load P is applied 400 mm from the flange of the tee shape, as shown in Fig. P8.62a. Determine the normal stresses at points H and K.

Fig. P8.62a

Fig. P8.62b Cross-sectional dimensions


width b (mm) 20 120

stem flange

x

xi Ai Ai



height h (mm) 130 20

505, 000 mm3  101.0 mm 5,000 mm 2  49.0 mm

Area Ai (mm2) 2,600 2,400 5,000

xi (from left) (mm) 65 140

xi Ai (mm3) 169,000 336,000 505,000


Moment of inertia about the z axis: Shape IC d = xi – x d²A IC + d²A (mm4) (mm) (mm4) (mm4) stem 3,661,666.67 −36.0 3,369,600.00 7,031,266.67 flange 80,000.00 39.0 3,650,400.00 3,730,400.00 4 Moment of inertia about the z axis (mm ) = 10,761,666.67 Internal forces and moments F  25 kN  25, 000 N M z  (25, 000 N)(400 mm  49.0 mm)  11, 225, 000 N-mm


Stresses

F 25, 000 N   5 MPa A 5, 000 mm 2 M x (11, 225, 000 N-mm)(  101.0 mm)  H ,bending  z   105.35 MPa Iz 10,761,666.67 mm 4

 axial 

 K ,bending 

M z x (11, 225, 000 N-mm)(49.0 mm)   51.11 MPa Iz 10,761,666.67 mm 4

Normal stress at H  H  5 MPa  105.35 MPa  100.4 MPa (T)

Ans.

Normal stress at K  K  5 MPa  51.11 MPa  56.1 MPa (C)

Ans.


8.63 The tee shape shown in Fig. P8.63b is used as a post that supports a load of P, which is applied 400 mm from the flange of the tee shape, as shown in Fig. P8.63a. The tension and compression normal stresses in the post must be limited to 165 MPa and 80 MPa, respectively. Determine the maximum magnitude of load P that satisfies both the tension and compression stress limits.

Fig. P8.63a


Solution Centroid location in x direction: Shape stem flange

width b (mm) 20 120


xi Ai

505, 000 mm3 x   101.0 mm Ai 5,000 mm 2  49.0 mm

Area Ai (mm2) 2,600 2,400 5,000


xi Ai (mm3) 169,000 336,000 505,000


Moment of inertia about the z axis: Shape IC d = xi – x d²A IC + d²A 4 4 (mm ) (mm) (mm ) (mm4) stem 3,661,666.67 −36.0 3,369,600.00 7,031,266.67 flange 80,000.00 39.0 3,650,400.00 3,730,400.00 4 Moment of inertia about the z axis (mm ) = 10,761,666.67 Internal forces and moments FP M z   P(400 mm  49.0 mm)  (449.0 mm)P


Stresses F P   (2 104 mm 2 ) P 2 A 5, 000 mm M x (449 mm)P(  101.0 mm)  H ,bending  z   (4.21394  103 mm 2 ) P 4 Iz 10,761,666.67 mm

 axial 

 K ,bending 

M z x (449 mm)P(49.0 mm)   (2.04439  103 mm 2 ) P Iz 10,761,666.67 mm 4

Tension stress limit (at H)

 H  (2 104 mm 2 ) P  (4.21394 103 mm 2 ) P  (4.01394 103 mm 2 ) P (4.01394 103 mm 2 ) P  165 N/mm 2  P  41,106.7 N Compression stress limit (at K)  K  (2 104 mm 2 ) P  (2.04439 103 mm 2 ) P  (2.24439 10 3 mm 2 ) P (2.24439 103 mm 2 ) P  80 N/mm 2  P  35, 644.43 N

Maximum magnitude of load P Pmax  35.6 kN

Ans.


8.64 The tee shape shown in Fig. P8.64b is used as a post that supports a load of P = 25 kN, which is applied 400 mm from the flange of the tee shape, as shown in Fig. P8.64a. Determine the magnitudes and locations of the maximum tension and compression normal stresses within the vertical portion BC of the post.

Fig. P8.64a


Solution Centroid location in x direction: Shape stem flange

width b (mm) 20 120


xi Ai

505, 000 mm3 x   101.0 mm Ai 5,000 mm 2  49.0 mm

Area Ai (mm2) 2,600 2,400 5,000


xi Ai (mm3) 169,000 336,000 505,000


Moment of inertia about the z axis: Shape IC d = xi – x d²A IC + d²A 4 4 (mm ) (mm) (mm ) (mm4) stem 3,661,666.67 −36.0 3,369,600.00 7,031,266.67 flange 80,000.00 39.0 3,650,400.00 3,730,400.00 4 Moment of inertia about the z axis (mm ) = 10,761,666.67 Internal forces and moments F  (25 kN) cos 35  20.4788 kN  20, 478.8 N (vertical component) V  (25 kN) sin 35  14.3394 kN  14,339.4 N (horizontal component) at B M z  (20, 478.8 N)(400 mm  49.0 mm)  9,194,981.2 N-mm at C M z  (20, 478.8 N)(400 mm  49.0 mm)  (14,339.4 N)(1,200 mm)  8, 012, 298.8 N-mm Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Normal stress at H at location B F 20, 478.8 N  axial    4.0958 MPa A 5, 000 mm 2 M x (9,194,981.2 N-mm)(  101.0 mm)  H ,bending  z   86.2964 MPa Iz 10,761,666.67 mm 4

 H  4.0958 MPa  86.2964 MPa  82.2 MPa Normal stress at H at location C M x (8, 012, 298.8 N-mm)(  101.0 mm)  H ,bending  z   75.1967 MPa Iz 10,761,666.67 mm4

 H  4.0958 MPa  75.1967 MPa  79.3 MPa Normal stress at K at location B M x (9,194,981.2 N-mm)(49.0 mm)  K ,bending  z   41.8666 MPa Iz 10,761,666.67 mm4

 K  4.0958 MPa  41.8666 MPa  46.0 MPa Normal stress at K at location C M x (8, 012, 298.8 N-mm)(49.0 mm)  K ,bending  z   36.4816 MPa Iz 10,761,666.67 mm4

 K  4.0958 MPa  36.4816 MPa  32.4 MPa Maximum tension stress  max tension  82.2 MPa (T)

at location B

Ans.

Maximum compression stress  max compression  79.3 MPa (C)

at location C

Ans.


8.65 A beam with a box cross section is subjected to a resultant moment magnitude of 2,100 N-m acting at the angle shown in Fig. P8.65. Determine: (a) the maximum tension and the maximum compression bending stresses in the beam. (b) the orientation of the neutral axis relative to the +z axis. Show its location on a sketch of the cross section.

Fig. P8.65

Solution Section properties (90 mm)(55 mm)3 Iy 12 (55 mm)(90 mm)3 Iz 12

(80 mm)(45 mm)3 12 (45 mm)(80 mm)3 12

Moment components M y (2,100 N-m)sin 30

Mz

(2,100 N-m) cos30

640,312.5 mm 4 1, 421, 250.0 mm 4

1, 050 N-m 1,818.65 N-m

(a) Maximum bending stresses For a shape having at least one axis of symmetry, Eq. (8-24) can be used to determine bending stresses. Compute normal stress at y = 45 mm, z = 27.5 mm: M yz Mz y x Iy Iz

(1,050 N-m)(27.5 mm)(1,000 mm/m) 640,312.5 mm 4 45.0952 MPa 57.5827 MPa 102.6779 MPa

( 1,818.65 N-m)(45 mm)(1,000 mm/m) 1,421,250.0 mm 4 Ans.

102.7 MPa (T)

Compute normal stress at y = −45 mm, z = −27.5 mm: M yz Mz y x Iy Iz

(1,050 N-m)( 27.5 mm)(1,000 mm/m) 640,312.5 mm 4 45.0952 MPa 57.5827 MPa 102.6779 MPa

102.7 MPa (C)

( 1,818.65 N-m)( 45 mm)(1,000 mm/m) 1,421,250.0 mm 4 Ans.

(b) Orientation of neutral axis For a shape having at least one axis of symmetry, Eq. (8-25) can be used to determine the orientation of the neutral axis: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

tan

M yIz M zIy

(1,050 N-m)(1, 421, 250.0 mm 4 ) ( 1,818.65 N-m)(640,312.5 mm 4 ) 52.03

1.2815

(i.e., 52.03 CCW from

z axis)

Ans.


8.66 The moment acting on the cross section of the T-beam has a magnitude of 22 kip-ft and is oriented as shown in Fig. P8.66. Determine: (a) the bending stress at point H. (b) the bending stress at point K. (c) the orientation of the neutral axis relative to the +z axis. Show its location on a sketch of the cross section.

Fig. P8.66

Solution Section properties Centroid location in y direction: Shape top flange stem

y

yi Ai Ai

Width b (in.) 7.00 0.75

95.80469 in.3 14.5625 in.2

Height h (in.) 1.25 7.75

Area Ai (in.2) 8.7500 5.8125 14.5625


6.5789 in.


2.4211 in.


Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (in. ) (in.) (in.4) top flange 1.1393 1.7961 28.2273 stem 29.0928 −2.7039 42.4956 Moment of inertia about the z axis (in.4) = Moment of inertia about the y axis: (1.25 in.)(7.00 in.)3 (7.75 in.)(0.75 in.) 3 Iy 12 12 Moment components My (22 kip-ft) cos55

Mz

(22 kip-ft)sin 55

yi Ai (in.3) 73.28125 22.52344 95.80469

IC + d²A (in.4) 29.3666 71.5884 100.9550

36.0016 in.4

12.6187 kip-ft

151.4242 kip-in.

18.0213 kip-ft

216.2561 kip-in.


(a) Bending stress at H For a shape having at least one axis of symmetry, Eq. (8-24) can be used to determine bending stresses. To compute the normal stress at H, use the (y, z) coordinates y = 2.4211 in. and z = −3.5 in.: M yz Mz y x Iy Iz ( 151.4242 kip-in.)( 3.50 in.) 36.0016 in.4 14.7211 ksi 5.1862 ksi 19.9074 ksi

( 216.2561 kip-in.)(2.4211 in.) 100.9550 in.4

19.91 ksi (T)

Ans.

(b) Bending stress at K To compute the normal stress at K, use the (y, z) coordinates y = −6.5789 in. and z = 0.375 in.: M yz Mz y x Iy Iz ( 151.4242 kip-in.)(0.375 in.) 36.0016 in.4 1.5773 ksi 14.0927 ksi 15.6700 ksi

( 216.2561 kip-in.)( 6.5789 in.) 100.9550 in.4

15.67 ksi (C)

Ans.

(c) Orientation of neutral axis For a shape having at least one axis of symmetry, Eq. (8-25) can be used to determine the orientation of the neutral axis: M y I z ( 151.4242 kip-in.)(100.9550 in.4 ) tan 1.9635 M zIy ( 216.2561 kip-in.)(36.0016 in.4 ) 63.01

(i.e., 63.01 CW from

z axis)

Ans.


8.67 A beam with a box cross section is subjected to a resultant moment magnitude of 75 kip-in. acting at the angle shown in Fig. P8.67. Determine: (a) the bending stress at point H. (b) the bending stress at point K. (c) the maximum tension and the maximum compression bending stresses in the beam. (d) the orientation of the neutral axis relative to the +z axis. Show its location on a sketch of the cross section.

Fig. P8.67

Solution Section properties (4 in.)(6 in.)3 Iy 12 (6 in.)(4 in.)3 Iz 12

(3.25 in.)(5.25 in.)3 12 (5.25 in.)(3.25 in.)3 12

Moment components M y (75 kip-in.) cos 20

Mz

(75 kip-in.)sin 20

32.8096 in.4 16.9814 in.4

70.4769 kip-in. 25.6515 kip-in.

(a) Bending stress at H For a shape having at least one axis of symmetry, Eq. (8-24) can be used to determine bending stresses. To compute the normal stress at H, use the (y, z) coordinates y = −2.0 in. and z = −3.0 in.: M yz Mz y x Iy Iz (70.4769 kip-in.)( 3.0 in.) 32.8096 in.4 6.4442 ksi 3.0211 ksi 3.4231 ksi

(25.6515 kip-in.)( 2.0 in.) 16.9814 in.4

3.42 ksi (C)

Ans.

(b) Bending stress at K To compute the normal stress at K, use the (y, z) coordinates y = 2.0 in. and z = 3.0 in.: M yz Mz y x Iy Iz (70.4769 kip-in.)(3.0 in.) 32.8096 in.4 6.4442 ksi 3.0211 ksi 3.4231 ksi

3.42 ksi (T)

(25.6515 kip-in.)(2.0 in.) 16.9814 in.4

Ans.


(c) Maximum bending stresses The maximum tension normal stress occurs at the (y, z) coordinates y = −2.0 in. and z = 3.0 in.: M yz Mz y x Iy Iz (70.4769 kip-in.)(3.0 in.) 32.8096 in.4 6.4442 ksi 3.0211 ksi 9.4653 ksi

(25.6515 kip-in.)( 2.0 in.) 16.9814 in.4

9.47 ksi (T)

Ans.

The maximum compression normal stress occurs at the (y, z) coordinates y = 2.0 in. and z = −3.0 in.: M yz Mz y x Iy Iz (70.4769 kip-in.)( 3.0 in.) 32.8096 in.4 6.4442 ksi 3.0211 ksi 9.4653 ksi

(25.6515 kip-in.)(2.0 in.) 16.9814 in.4

9.47 ksi (C)

Ans.

(d) Orientation of neutral axis For a shape having at least one axis of symmetry, Eq. (8-25) can be used to determine the orientation of the neutral axis: M y I z (70.4769 kip-in.)(16.9814 in.4 ) tan 1.4220 M z I y (25.6515 kip-in.)(32.8096 in.4 ) 54.88


z axis)

Ans.


8.68 The moment acting on the cross section of the wide-flange beam has a magnitude of M = 12 kN-m and is oriented as shown in Fig. P8.68. Determine: (a) the bending stress at point H. (b) the bending stress at point K. (c) the orientation of the neutral axis relative to the +z axis. Show its location on a sketch of the cross section.

Fig. P8.68

Solution Section properties Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (mm ) (mm) (mm4) top flange 59,062.5 97.5 29,944,687.5 web 4,860,000 0 0 bottom flange 59,062.5 −97.5 29,944,687.5 Moment of inertia about the z axis (mm4) = Moment of inertia about the y axis: (15 mm)(210 mm)3 (180 mm)(10 mm) 3 Iy 2 12 12 Moment components M y (12 kN-m)sin 35 Mz

(12 kN-m) cos 35

IC + d²A (mm4) 30,003,750 4,860,000 30,003,750 64,867,500

23,167,500 mm 4

6.8829 kN-m 6.8829 106 N-mm 9.8298 kN-m 9.8298 106 N-mm

(a) Bending stress at H For a shape having at least one axis of symmetry, Eq. (8-24) can be used to determine bending stresses. To compute the normal stress at H, use the (y, z) coordinates y = 105 mm and z = −105 mm: M yz Mz y x Iy Iz

(6.8829 106 N-mm)( 105 mm) 23,167,500 mm 4 31.1948 MPa 15.9114 MPa 47.1062 MPa

47.1 MPa (C)

(9.8298 106 N-mm)(105 mm) 64, 867,500 mm 4 Ans.


(b) Bending stress at K To compute the normal stress at K, use the (y, z) coordinates y = −105 mm and z = 105 mm: M yz Mz y x Iy Iz

(6.8829 106 N-mm)(105 mm) 23,167,500 mm 4 31.1948 MPa 15.9114 MPa 47.1062 MPa

(9.8298 106 N-mm)( 105 mm) 64,867,500 mm 4 Ans.

47.1 MPa (T)

(b) Orientation of neutral axis For a shape having at least one axis of symmetry, Eq. (8-25) can be used to determine the orientation of the neutral axis: M y I z (6.8829 kN-m)(64,867,500 mm 4 ) tan 1.9605 M z I y (9.8298 kN-m)(23,167,500 mm 4 ) 62.98


z axis)

Ans.


8.69 For the cross section shown in Fig. P8.69, determine the maximum magnitude of the bending moment M so that the bending stress in the wideflange shape does not exceed 165 MPa.

Fig. P8.69

Solution Section properties Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (mm ) (mm) (mm4) top flange 59,062.5 97.5 29,944,687.5 web 4,860,000 0 0 bottom flange 59,062.5 −97.5 29,944,687.5 Moment of inertia about the z axis (mm4) = Moment of inertia about the y axis: (15 mm)(210 mm)3 (180 mm)(10 mm) 3 Iy 2 12 12 Moment components M y M sin 35

Mz

IC + d²A (mm4) 30,003,750 4,860,000 30,003,750 64,867,500

23,167,500 mm 4

M cos 35

Maximum bending moment magnitude The maximum tension bending stress should occur at point K, which has the (y, z) coordinates y = −105 mm and z = 105 mm: M y z M z y M sin 35 (105 mm) M cos35 ( 105 mm) 165 MPa x Iy Iz 23,167,500 mm4 64,867,500 mm 4 M 2.59957 10 M

6

mm

3

1.32595 10

42.0327 106 N-mm

6

mm

3

42.0 kN-m

165 N/mm 2

Ans.


8.70 The unequal-leg angle is subjected to a bending moment of Mz = 20 kip-in. that acts at the orientation shown in Fig. P8.70. Determine: (a) the bending stress at point H. (b) the bending stress at point K. (c) the maximum tension and the maximum compression bending stresses in the cross section. (d) the orientation of the neutral axis relative to the +z axis. Show its location on a sketch of the cross section.

Fig. P8.70

Solution Section properties Centroid location in y direction: Shape upright leg bottom leg y

yi Ai Ai

Width b (in.) 0.375 2.625 3.18457 in.3 2.4844 in.2

Height h (in.) 4.000 0.375

Area Ai (in.2) 1.5000 0.9844 2.4844


yi Ai (in.3) 3.00 0.18457 3.18457

1.2818 in.(from bottom of shape to centroid) 2.7182 in.


Centroid location in z direction: Shape upright leg bottom leg z

zi Ai Ai

Area Ai (in.2) 1.5000 0.9844 2.4844 1.94243 in.3 2.4844 in.2

zi (from right edge) (in.) 0.1875 1.6875

zi Ai (in.3) 0.2813 1.6612 1.94243

0.7818 in.

(from right edge of shape to centroid)

2.2182 in.

(from left edge of shape to centroid)

Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (in. ) (in.) (in.4) upright leg 2.000 0.7182 0.77372 bottom leg 0.011536 −1.0943 1.17881 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 2.7737 1.1903 3.9640


Moment of inertia about the y axis: Shape IC d = zi – z d²A (in.4) (in.) (in.4) upright leg 0.017578 −0.5943 0.52979 bottom leg 0.565247 0.9057 0.80750 Moment of inertia about the y axis (in.4) = Product of inertia about the centroidal axes: Shape Iy’z’ yc zc 4 (in. ) (in.) (in.) upright leg 0 0.7182 −0.5943 bottom leg 0 −1.0943 0.9057

IC + d²A (in.4) 0.5474 1.3727 1.9201

Area Ai yc zc Ai 2 (in. ) (in.4) 1.5000 −0.6402 0.9844 −0.9757 Product of inertia (in.4) =

Iyz (in.4) −0.6402 −0.9757 −1.6159

(a) Bending stress at H Since the angle shape has no axis of symmetry, Eq. (8.21) or Eq. (8.22) must be used to determine the bending stresses. Equation (8.22) will be used here. Note that the bending moment component about the y axis is zero (i.e., My = 0); therefore, the first term in Eq. (8.22) is eliminated. To compute the normal stress at H, use (y, z) coordinates of y = 2.7182 in. and z = −0.4068 in.: I y y I yz z Mz x I y I z I yz2 (1.9201 in.4 )(2.7182 in.) ( 1.6159 in.4 )( 0.4068 in.) (20 kip-in.) (1.9201 in.4 )(3.9640 in.4 ) ( 1.6159 in.4 )2 4.5619 in.5 (20 kip-in.) 5.0001 in.8 18.2469 ksi

18.25 ksi (C)

Ans.

(b) Bending stress at K To compute the normal stress at K, use (y, z) coordinates of y = −0.9068 in. and z = 2.2182 in.: I y y I yz z Mz x I y I z I yz2 (1.9201 in.4 )( 0.9068 in.) ( 1.6159 in.4 )(2.2182 in.) (20 kip-in.) (1.9201 in.4 )(3.9640 in.4 ) ( 1.6159 in.4 ) 2 1.8432 in.5 (20 kip-in.) 5.0001 in.8 7.3728 ksi

7.37 ksi (C)

Ans.


(d) Orientation of neutral axis Since the angle shape has no axis of symmetry, Eq. (8.23) must be used to determine the orientation of the neutral axis: M y I z M z I yz (20 kip-in.)( 1.6159 in.4 ) tan 0.8416 M z I y M y I yz (20 kip-in.)(1.9201 in.4 ) 40.08


z axis)

Ans.

(c) Maximum bending stresses Sketch the orientation of the neutral axis. By inspection, the points on the angle cross section that are farthest from the neutral axis are point H and the corner of the angle. The bending stress at H has already been computed. To compute the normal stress at the corner of the angle, use (y, z) coordinates of y = −1.2818 in. and z = −0.7818 in. I y y I yz z Mz x I y I z I yz2 (1.9201 in.4 )( 1.2818 in.) ( 1.6159 in.4 )( 0.7818 in.) (20 kip-in.) (1.9201 in.4 )(3.9640 in.4 ) ( 1.6159 in.4 ) 2 3.7245 in.5 (20 kip-in.) 5.0001 in.8 14.8977 ksi

14.90 ksi (T)

Therefore, the maximum compression bending stress is: 18.25 ksi (C) x

Ans.

and the maximum tension bending stress is: 14.90 ksi (T) x

Ans.


8.71 For the cross section shown in Fig. P8.71, determine the maximum magnitude of the bending moment M so that the bending stress in the unequalleg angle shape does not exceed 24 ksi.

Fig. P8.71

Solution Section properties Centroid location in y direction: Shape upright leg bottom leg y

yi Ai Ai

Width b (in.) 0.375 2.625 3.18457 in.3 2.4844 in.2

Height h (in.) 4.000 0.375

Area Ai (in.2) 1.5000 0.9844 2.4844


yi Ai (in.3) 3.00 0.18457 3.18457

1.2818 in.(from bottom of shape to centroid) 2.7182 in.


Centroid location in z direction: Shape upright leg bottom leg z

zi Ai Ai

Area Ai (in.2) 1.5000 0.9844 2.4844 1.94243 in.3 2.4844 in.2

zi (from right edge) (in.) 0.1875 1.6875

zi Ai (in.3) 0.2813 1.6612 1.94243

0.7818 in.

(from right edge of shape to centroid)

2.2182 in.

(from left edge of shape to centroid)

Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (in. ) (in.) (in.4) upright leg 2.000 0.7182 0.77372 bottom leg 0.011536 −1.0943 1.17881 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 2.7737 1.1903 3.9640


Moment of inertia about the y axis: Shape IC d = zi – z d²A (in.4) (in.) (in.4) upright leg 0.017578 −0.5943 0.52979 bottom leg 0.565247 0.9057 0.80750 Moment of inertia about the y axis (in.4) = Product of inertia about the centroidal axes: Shape Iy’z’ yc zc 4 (in. ) (in.) (in.) upright leg 0 0.7182 −0.5943 bottom leg 0 −1.0943 0.9057

IC + d²A (in.4) 0.5474 1.3727 1.9201

Area Ai yc zc Ai 2 (in. ) (in.4) 1.5000 −0.6402 0.9844 −0.9757 Product of inertia (in.4) =

Iyz (in.4) −0.6402 −0.9757 −1.6159

Orientation of neutral axis Since the angle shape has no axis of symmetry, it is helpful to determine the orientation of the neutral axis from Eq. (8.23) before beginning the stress calculations: M y I z M z I yz (20 kip-in.)( 1.6159 in.4 ) tan 0.8416 M z I y M y I yz (20 kip-in.)(1.9201 in.4 ) 40.08


z axis)

Allowable moments based on maximum tension and compression bending stresses Sketch the orientation of the neutral axis. By inspection, the points on the angle cross section that are farthest from the neutral axis are point H and the corner of the angle. To compute the normal stress at H, use (y, z) coordinates of y = 2.7182 in. and z = −0.4068 in.: I y y I yz z (1.9201 in.4 )(2.7182 in.) ( 1.6159 in.4 )( 0.4068 in.) Mz Mz x I y I z I yz2 (1.9201 in.4 )(3.9640 in.4 ) ( 1.6159 in.4 )2 4.5619 in.5 M z ( 0.9124 in. 3 )M z 8 5.0001 in. Therefore, based on an allowable bending stress of 24 ksi at H, the maximum magnitude of Mz is: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

(0.9124 in. 3 )M z Mz

24 ksi (a)

26.3054 kip-in.

To compute the normal stress at the corner of the angle, use (y, z) coordinates of y = −1.2818 in. and z = −0.7818 in. I y y I yz z (1.9201 in.4 )( 1.2818 in.) ( 1.6159 in.4 )( 0.7818 in.) M Mz x z I y I z I yz2 (1.9201 in.4 )(3.9640 in.4 ) ( 1.6159 in.4 ) 2 3.7245 in.5 Mz 5.0001 in.8

(0.7449 in. 3 ) M z

Therefore, based on the bending stress at the corner of the angle, the maximum magnitude of Mz is: (0.7449 in. 3 )M z 24 ksi

Mz

32.2197 kip-in.

(b)

Maximum bending moment Mz Compare the results in Eqs. (a) and (b) to find that the maximum bending moment that can be applied to the angle shape is: Ans. M z 26.3 kip-in.


8.72 The moment acting on the cross section of the unequal-leg angle has a magnitude of M = 20 kip-in. and is oriented as shown in Fig. P8.72. Determine: (a) the bending stress at point H. (b) the bending stress at point K. (c) the maximum tension and the maximum compression bending stresses in the cross section. (d) the orientation of the neutral axis relative to the +z axis. Show its location on a sketch of the cross section.

Fig. P8.72

Solution Moment of inertia about the z axis: d = yi – y Shape IC Area Ai d²A 4 2 (mm ) (mm) (mm ) (mm4) top flange 130,208.3 112.5 2,500 31,640,625.0 web 10,666,666.7 0 3,200 0 bottom flange 130,208.3 −112.5 2,500 31,640,625.0 Moment of inertia about the z axis (mm4) =

IC + d²A (mm4) 31,770,883.3 10,666,666.7 31,770,883.3 74,208,333.3

Moment of inertia about the y axis: Shape IC d = zi – z Area Ai d²A IC + d²A 4 2 4 (mm ) (mm) (mm ) (mm ) (mm4) top flange 2,083,333.3 −42.0 2,500 4,410,000 6,493,333.3 web 68,266.7 0 3,200 0 68,266.7 bottom flange 2,083,333.3 42.0 2,500 4,410,000 6,493,333.3 Moment of inertia about the y axis (mm4) = 13,054,933.3 Product of inertia about the centroidal axes: Shape yc zc (mm) (mm) top flange 112.5 −42.0 web 0 0 bottom flange −112.5 42.0

Moment components My (40 kN-m)sin15 Mz

(40 kN-m) cos15

Area Ai yc zc Ai Iyz 2 4 (mm ) (mm ) (mm4) 2,500 −11,812,500 −11,812,500 3,200 0 0 2,500 −11,812,500 −11,812,500 Product of inertia (mm4) = −23,625,000

10.3528 kN-m

10.3528 106 N-mm

38.6370 kN-m

38.6370 106 N-mm


(a) Bending stress at H Since the zee shape has no axis of symmetry, Eq. (8.21) or Eq. (8.22) must be used to determine the bending stresses. Equation (8.21) will be used here. M z I y M y I yz y M y I z M z I yz z x

I yIz

I yz2

I yIz

I yz2

( 38.6370 106 N-mm)(13,054,933.3 mm4 ) ( 10.3528 106 N-mm)( 23,625,000 mm4 ) y (13,054,933.3 mm4 )(74,208,333.3 mm4 ) ( 23,625,000 mm 4 )2 ( 10.3528 106 N-mm)(74,208,333.3 mm 4 ) ( 38.6370 106 N-mm)( 23,625,000 mm4 ) z (13,054,933.3 mm4 )(74,208,333.3 mm4 ) ( 23,625,000 mm 4 )2 (0.63271 N/mm3 ) y (0.35197 N/mm3 )z

To compute the normal stress at H, use (y, z) coordinates of y = 125 mm and z = −92 mm: (0.63271 N/mm3 )(125 mm) (0.35197 N/mm3 )( 92 mm) x 46.7073 MPa

46.7 MPa (T)

Ans.

(b) Bending stress at K To compute the normal stress at K, use (y, z) coordinates of y = −125 mm and z = 92 mm: (0.63271 N/mm3 )( 125 mm) (0.35197 N/mm 3 )(92 mm) x 46.7073 MPa

46.7 MPa (C)

Ans.

(d) Orientation of neutral axis Since the zee shape has no axis of symmetry, it is helpful to determine the orientation of the neutral axis from Eq. (8.23) to help identify points of maximum stress. M y I z M z I yz tan M z I y M y I yz ( 10.3528 kN-m)(74,208,333.3 mm 4 ) ( 38.6370 kN-m)( 23,625,000 mm 4 ) ( 38.6370 kN-m)(13,054,933.3 mm 4 ) ( 10.3528 kN-m)( 23,625,000 mm 4 ) 0.55629 29.09


z axis)


(c) Maximum tension and compression bending stresses Sketch the orientation of the neutral axis. By inspection, the points on the zee cross section that are farthest from the neutral axis are on the top and bottom surfaces at the outside corners of the web. To compute bending stresses at the upper point, use (y, z) coordinates of y = 125 mm and z = 8 mm: (0.63271 N/mm3 )(125 mm) (0.35197 N/mm 3 )(8 mm) x 81.9045 MPa

81.9 MPa (T)

Maximum tension bending stress

Ans.

To compute bending stresses at the lower point, use (y, z) coordinates of y = −125 mm and z = −8 mm: (0.63271 N/mm3 )( 125 mm) (0.35197 N/mm3 )( 8 mm) x 81.9045 MPa

81.9 MPa (C)

Maximum compression bending stress

Ans.


8.73 The moment acting on the cross section of the unequal-leg angle has a magnitude of 14 kN-m and is oriented as shown in Fig. P8.73. Determine: (a) the bending stress at point H. (b) the bending stress at point K. (c) the maximum tension and the maximum compression bending stresses in the cross section. (d) the orientation of the neutral axis relative to the +z axis. Show its location on a sketch of the cross section. Fig. P8.73

Solution Section properties Centroid location in y direction: Shape horizontal leg vertical leg

y

yi Ai Ai

Width b (mm) 150 19

Height h (mm) 19 181

854,154.5 mm3 6,289 mm 2

Area Ai (mm2) 2,850 3,439 6,289

yi (from bottom) (mm) 190.50 90.50

yi Ai (mm3) 542,925.0 311,229.5 854,154.5

135.82 mm


64.18 mm


Centroid location in z direction: Shape horizontal leg vertical leg

z

zi Ai Ai

Area Ai (mm2) 2,850 3,439 6,289 246, 420.5 mm3 6,289 mm 2

zi (from right edge) (mm) 75.0 9.5

39.18 mm 110.82 mm

zi Ai (mm3) 213,750.0 32,670.5 246,420.5 (from right edge of shape to centroid) (from left edge of shape to centroid)

Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (mm ) (mm) (mm4) horizontal leg 85,737.50 54.68 8,522,088.15 vertical leg 9,388,756.58 −45.32 7,062,503.99 Moment of inertia about the z axis (mm4) =

IC + d²A (mm4) 8,607,825.65 16,451,260.58 25,059,086.23


Moment of inertia about the y axis: Shape IC d = zi – z d²A (mm4) (mm) (mm4) horizontal leg 5,343,750.00 35.82 3,656,188.87 vertical leg 103,456.58 −29.68 3,029,990.78 Moment of inertia about the y axis (mm4) = Product of inertia about the centroidal axes: Shape Iy’z’ yc zc 4 (mm ) (mm) (mm) horizontal leg 0 54.68 35.82 vertical leg 0 −45.32 −29.68

IC + d²A (mm4) 8,999,938.87 3,133,447.36 12,133,386.23

Area Ai yc zc Ai Iyz 2 4 (mm ) (mm ) (mm4) 2,850 5,582,117.16 5,582,117.16 3,439 4,625,790.65 4,625,790.65 4 Product of inertia (mm ) = 10,207,907.81

Since the angle shape has no axis of symmetry, Eq. (8.21) or Eq. (8.22) must be used to determine the bending stresses. Equation (8.21) will be used here. M z I y M y I yz y M y I z M z I yz z x

I y Iz

I yz2

I y Iz

I yz2

(14 106 N-mm)(12,133,386.23 mm4 ) y (12,133,386.23 mm 4 )(25,059,086.23 mm 4 ) (10, 207,907.81 mm 4 )2 (14 106 N-mm)(10, 207,907.81 mm 4 ) z (12,133,386.23 mm 4 )(25,059,086.23 mm 4 ) (10, 207,907.81 mm 4 ) 2 ( 0.84997 N/mm3 ) y (0.71509 N/mm3 )z

To compute the normal stress at H, use (y, z) coordinates of y = 45.18 mm and z = 110.82 mm: ( 0.84997 N/mm3 )(45.18 mm) (0.71509 N/mm 3 )(110.82 mm) x 40.8444 MPa

40.8 MPa (T)

Ans.

(b) Bending stress at K To compute the normal stress at K, use (y, z) coordinates of y = 64.18 mm and z = −39.18 mm: ( 0.84997 N/mm3 )(64.18 mm) (0.71509 N/mm 3 )( 39.18 mm) x 82.5685 MPa

82.6 MPa (C)

Ans.

(d) Orientation of neutral axis Since the angle shape has no axis of symmetry, it is helpful to determine the orientation of the neutral axis from Eq. (8.23) to help identify points of maximum stress. M y I z M z I yz tan M z I y M y I yz (14 kN-m)(10, 207,907.81 mm 4 ) (14 kN-m)(12,133,386.23 mm 4 ) 0.84131 40.07


z axis)


(c) Maximum tension and compression bending stresses Sketch the orientation of the neutral axis. By inspection, the points on the angle cross section that are farthest from the neutral axis are on the top corner (at K) and on the inside corner of the vertical leg. To compute bending stresses at the lower point, use (y, z) coordinates of y = −135.82 mm and z = −20.18 mm: ( 0.84997 N/mm 3 )( 135.82 mm) (0.71509 N/mm 3 )( 20.18 mm) x 101.0129 MPa

101.0 MPa (T)


Ans.

The maximum compression bending stress is ( 0.84997 N/mm3 )(64.18 mm) (0.71509 N/mm 3 )( 39.18 mm) x 82.5685 MPa

82.6 MPa (C)


Ans.


8.74 The moment acting on the cross section of the zee shape has a magnitude of M = 4.75 kip-ft and is oriented as shown in Fig. P8.74. Determine: (a) the bending stress at point H. (b) the bending stress at point K. (c) the maximum tension and the maximum compression bending stresses in the cross section. (d) the orientation of the neutral axis relative to the +z axis. Show its location on a sketch of the cross section.

Fig. P8.74

Solution Moment of inertia about the z axis: d = yi – y Shape IC Area Ai d²A 4 (in. ) (in.) (in.2) (in.4) top flange 0.0260 2.75 1.25 9.4531 web 3.6458 0 1.75 0 bottom flange 0.0260 −2.75 1.25 9.4531 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 9.4792 3.6458 9.4792 22.6042

Moment of inertia about the y axis: Shape IC d = zi – z Area Ai d²A (in.4) (in.) (in.2) (in.4) top flange 0.6510 1.075 1.25 1.4445 web 68,266.7 0 1.75 0 bottom flange 0.6510 −1.075 1.25 1.4445 Moment of inertia about the y axis (in.4) =

IC + d²A (in.4) 2.0956 0.0179 2.0956 4.2091

Product of inertia about the centroidal axes: Shape yc zc (in.) (in.) top flange 2.75 1.075 web 0 0 bottom flange −2.75 −1.075

Area Ai yc zc Ai 2 (in. ) (in.4) 1.25 3.6953 1.75 0 1.25 3.6953 Product of inertia (in.4) =

Iyz (in.4) 3.6953 0 3.6953 7.3906

(a) Bending stress at H Since the zee shape has no axis of symmetry, Eq. (8.21) or Eq. (8.22) must be used to determine the bending stresses. Equation (8.21) will be used here.


MzIy x

I yIz

M y I yz y I

2 yz

M yIz

M z I yz z

I yIz

I yz2

( 4.75 kip-ft)(12 in./ft)(4.2091 in.4 ) y (4.2091 in.4 )(22.6042 in.4 ) (7.3906 in.4 ) 2

( 4.75 kip-ft)(12 in./ft)(7.3906 in.4 ) z (4.2091 in.4 )(22.6042 in.4 ) (7.3906 in.4 ) 2

(5.92065 kips/in.3 ) y (10.39584 kips/in.3 )z

To compute the normal stress at H, use (y, z) coordinates of y = 3 in. and z = 2.325 in.: (5.92065 kips/in.3 )(3 in.) (10.39584 kips/in.3 )(2.325 in.) x 6.4084 ksi

6.41 ksi (C)

Ans.

(b) Bending stress at K To compute the normal stress at K, use (y, z) coordinates of y = −2.50 in. and z = −2.325 in.: (5.92065 kips/in.3 )( 2.50 in.) (10.39584 kips/in.3 )( 2.325 in.) x 9.3687 ksi

9.37 ksi (T)

Ans.

(d) Orientation of neutral axis Since the zee shape has no axis of symmetry, it is helpful to determine the orientation of the neutral axis from Eq. (8.23) to help identify points of maximum stress. M y I z M z I yz tan M z I y M y I yz ( 4.75 kip-ft)(7.3906 in.4 ) ( 4.75 kip-ft)(4.2091 in.4 ) 1.7559 60.34


z axis)


(c) Maximum tension and compression bending stresses Sketch the orientation of the neutral axis. By inspection, the points on the zee cross section that are farthest from the neutral axis are on the top and bottom surfaces at the outside corners of the web. To compute bending stresses at the upper point, use (y, z) coordinates of y = 3 in. and z = −0.175 in.: (5.92065 kips/in.3 )(3 in.) (10.39584 kips/in.3 )( 0.175 in.) x 19.5812 ksi

19.58 ksi (T)


Ans.

To compute bending stresses at the lower point, use (y, z) coordinates of y = −3 in. and z = 0.175 in.: (5.92065 kips/in.3 )( 3 in.) (10.39584 kips/in.3 )(0.175 in.) x 19.5812 ksi

19.58 ksi (C)


Ans.


8.75 For the cross section shown in Fig. P8.75, determine the maximum magnitude of the bending moment M so that the bending stress in the zee shape does not exceed 24 ksi.

Fig. P8.75

Solution Moment of inertia about the z axis: d = yi – y Shape IC Area Ai d²A 4 (in. ) (in.) (in.2) (in.4) top flange 0.0260 2.75 1.25 9.4531 web 3.6458 0 1.75 0 bottom flange 0.0260 −2.75 1.25 9.4531 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 9.4792 3.6458 9.4792 22.6042

Moment of inertia about the y axis: Shape IC d = zi – z Area Ai d²A (in.4) (in.) (in.2) (in.4) top flange 0.6510 1.075 1.25 1.4445 web 68,266.7 0 1.75 0 bottom flange 0.6510 −1.075 1.25 1.4445 Moment of inertia about the y axis (in.4) =

IC + d²A (in.4) 2.0956 0.0179 2.0956 4.2091

Product of inertia about the centroidal axes: Shape yc zc (in.) (in.) top flange 2.75 1.075 web 0 0 bottom flange −2.75 −1.075

Area Ai yc zc Ai 2 (in. ) (in.4) 1.25 3.6953 1.75 0 1.25 3.6953 Product of inertia (in.4) =

Iyz (in.4) 3.6953 0 3.6953 7.3906

Bending stresses in the section Since the zee shape has no axis of symmetry, Eq. (8.21) or Eq. (8.22) must be used to determine the bending stresses. Equation (8.21) will be used here. For this problem, My = 0 and from the sketch, Mz is observed to be negative. The bending stress in the zee cross section is described by:


MzIy x

I yIz

M y I yz y I

M yIz

2 yz

I yIz

M z I yz z I yz2

M z (4.2091 in.4 ) y (4.2091 in.4 )(22.6042 in.4 ) (7.3906 in.4 ) 2

M z (7.3906 in.4 ) z (4.2091 in.4 )(22.6042 in.4 ) (7.3906 in.4 ) 2

(0.103871 in. 4 ) M z y (0.182383 in. 4 )M z z M z (0.103871 in. 4 ) y (0.182383 in. 4 )z

Orientation of neutral axis Since the angle shape has no axis of symmetry, it is helpful to determine the orientation of the neutral axis from Eq. (8.23) before beginning the stress calculations: M y I z M z I yz M z (7.3906 in.4 ) tan 1.7559 M z I y M y I yz M z (4.2091 in.4 ) 60.34


z axis)

Allowable moments based on maximum tension and compression bending stresses Sketch the orientation of the neutral axis. By inspection, the points on the zee cross section that are farthest from the neutral axis are on the top and bottom surfaces at the outside corners of the web. To compute bending stresses at the upper point, coordinates of y = 3 in. and z = −0.175 in. are used. Set the bending stress at this point to the 24-ksi allowable bending stress and solve for the moment magnitude: M z (0.103871 in. 4 )(3 in.) (0.182383 in. 4 )( 0.175 in.) 24 ksi x

Mz

24 ksi 0.343530 in. 3

69.86287 kip-in.

5.82 kip-ft

Ans.


8.76 A stainless-steel spring (shown in Fig. P8.76) has a thickness of ¾ in. and a change in depth at section B from D = 1.50 in. to d = 1.25 in. The radius of the fillet between the two sections is r = 0.125 in. If the bending moment applied to the spring is M = 2,000 lb-in., determine the maximum normal stress in the spring. Fig. P8.76

Solution From Figure 8.18 r 0.125 in. d 1.25 in.

0.10

D d

1.50 in. 1.20 1.25 in.

K

1.69

Moment of inertia at minimum depth section: (0.75 in.)(1.25 in.)3 I 0.122070 in.4 12 Nominal bending stress at minimum depth section: My (2,000 lb-in.)(1.25 in./2) 10.2400 ksi nom I 0.122070 in.4 Maximum bending stress: K nom 1.69(10.2400 ksi) 17.3056 ksi max

17.31 ksi

Ans.


8.77 An alloy-steel spring (shown in Fig. P8.77) has a thickness of 25 mm and a change in depth at section B from D = 75 mm to d = 50 mm. If the radius of the fillet between the two sections is r = 8 mm, determine the maximum moment that the spring can resist if the maximum bending stress in the spring must not exceed 120 MPa. Fig. P8.77

Solution From Figure 8.18 r 8 mm 0.16 d 50 mm

D d

75 mm 50 mm

1.50

K

1.57

Determine maximum nominal bending stress: 120 MPa max 76.4331 MPa nom K 1.57 Moment of inertia at minimum depth section: (25 mm)(50 mm)3 I 260, 416.67 mm 4 12 Maximum bending moment: (76.4331 N/mm 2 )(260,416.67 mm 4 ) nom I M max y 50 mm/2

796,178.3 N-mm

796 N-m

Ans.


8.78 The notched bar shown in Fig. P8.78 is subjected to a bending moment of M = 300 N-m. The major bar width is D = 75 mm, the minor bar width at the notches is d = 50 mm, and the radius of each notch is r = 10 mm. If the maximum bending stress in the bar must not exceed 90 MPa, determine the minimum required bar thickness b. Fig. P8.78


D d

75 mm 50 mm

1.50

K

1.76

Determine maximum nominal bending stress: 90 MPa max 51.1364 MPa nom K 1.76 Minimum bar thickness b: M y M (d /2) 6M nom I bd 3 /12 bd 2 6M 6(300 N-m)(1,000 mm/m) b 2 (51.1364 N/mm 2 )(50 mm) 2 nom d

14.08 mm

Ans.


8.79 The machine part shown in Fig. P8.79 is made of cold-rolled 18-8 stainless steel (see Appendix D for properties). The major bar width is D = 1.50 in., the minor bar width at the notches is d = 1.00 in., the radius of each notch is r = 0.125 in., and the bar thickness is b = 0.25 in. Determine the maximum safe moment M that may be applied to the bar if a factor of safety of 2.5 with respect to failure by yield is specified. Fig. P8.79


0.125

D d

1.50 in. 1.50 1.00 in.

K

2.05

Moment of inertia at minimum depth section: (0.25 in.)(1.00 in.)3 I 0.020833 in.4 12 Maximum allowable bending moment: From the specified factor of safety and the yield stress of the material, the allowable bending stress is: 165 ksi Y 66 ksi allow FS 2.5 Thus, the maximum allowable bending moment can be determined from: My K allow I (66 ksi)(0.020833 in.4 ) allow I M max 1.3415 kip-in. 111.8 lb-ft Ky (2.05)(1.00 in./2)

Ans.


8.80 The shaft shown in Fig. P8.80 is supported at each end by self-aligning bearings. The major shaft diameter is D = 2.00 in., the minor shaft diameter is d = 1.50 in., and the radius of the fillet between the major and minor diameter sections is r = 0.125 in. The shaft length is L = 24 in. and the fillets are located at x = 8 in. and x = 16 in. Determine the maximum load P that may be applied to the shaft if the maximum normal stress must be limited to 24,000 psi. Fig. P8.80


D d

0.083

2.00 in. 1.33 1.50 in.

K

1.78

Moment of inertia at minimum diameter section: I

64

(1.50 in.)4

0.248505 in.4

Maximum allowable bending moment: My K allow I (24,000 psi)(0.248505 in.4 ) allow I M max Ky (1.78)(1.50 in./2)

4,467.50 lb-in.

Bending moment at x = 8 in.: P P M x (8 in.) P(4 in.) 2 2 Maximum load P: P(4 in.) 4,467.50 lb-in. P 1,116.88 lb

1,117 lb

Check stress at midspan: PL (1,116.88 lb)(24 in.) M midspan 4 4 I midspan

(2.00 in.) 4

64 My I

Ans.

6,701.28 lb-in.

0.785398 in.4

(6,701.28 lb-in.)(2.00 in./2) 0.785398 in.4

8,532 psi

24,000 psi

OK


8.81 The C86100 bronze (see Appendix D for properties) shaft shown in Fig. P8.81 is supported at each end by self-aligning bearings. The major shaft diameter is D = 40 mm, the minor shaft diameter is d = 25 mm, and the radius of the fillet between the major and minor diameter sections is r = 5 mm. The shaft length is L = 500 mm and the fillets are located at x = 150 mm and x = 350 mm. Determine the maximum load P that may be applied to the shaft if a factor of safety of 3.0 with respect to failure by yield is specified. Fig. P8.81


D d

40 mm 25 mm

1.60

K

1.48


64

(25 mm)4

19,174.76 mm 4

Maximum allowable bending moment: 331 MPa yield 110.33 MPa allow FS 3.0 My K allow I (110.33 N/mm 2 )(19,174.76 mm 4 ) allow I M max Ky (1.48)(25 mm/2)

114,357.58 N-mm

Bending moment at x = 150 mm: P P M x (150 mm) P(75 mm) 2 2 Maximum load P: P(75 mm) 114,357.58 N-mm P 1,524.77 N

1,525 N

Check stress at midspan: PL (1,524.77 N)(500 mm) M midspan 4 4

I midspan

(40 mm)4

64 My I

Ans.

190,596.25 N-mm

125,663.71 mm 4

(190,596.25 N-mm)(40 mm/2) 125,663.71 mm4

30.33 MPa 110.33 MPa

OK


8.82 The machine shaft shown in Fig. P8.82 is made of 1020 cold-rolled steel (see Appendix D for properties). The major shaft diameter is D = 1.000 in., the minor shaft diameter is d = 0.625 in., and the radius of the fillet between the major and minor diameter sections is r = 0.0625 in. The fillet is located at x = 4 in. from C. If a load of P = 125 lb is applied at C, determine the factor of safety with respect to failure by yield in the fillet at B. Fig. P8.82

Solution For 1020 cold-rolled steel: 62,000 psi Y From Figure 8.20 r 0.0625 in. d 0.625 in.

0.10

D d

1.000 in. 1.6 0.625 in.

K

1.74


64

(0.625 in.)4

0.0074901 in.4

Bending moment at x = 4 in.: M Px (125 lb)(4 in.) 500 lb-in. Maximum bending stress: My (500 lb-in.)(0.625 in./2) K (1.74) max I 0.0074901 in.4 Factor of safety: 62,000 psi Y FS 36, 297.7 psi max

1.708

36,297.7 psi

Ans.


8.83 The machine shaft shown in Fig. P8.83 is made of 1020 cold-rolled steel (see Appendix D for properties). The major shaft diameter is D = 30 mm, the minor shaft diameter is d = 20 mm, and the radius of the fillet between the major and minor diameter sections is r = 3 mm. The fillet is located at x = 90 mm from C. Determine the maximum load P that can be applied to the shaft at C if a factor of safety of 1.5 with respect to failure by yield is specified for the fillet at B. Fig. P8.83


D d

30 mm 20 mm

1.5

K

1.58


64

(20 mm)4

7,853.98 mm 4

Maximum allowable bending moment: 427 MPa Y 284.6667 MPa allow FS 1.5 My K allow I (284.6667 N/mm 2 )(7,853.98 mm 4 ) allow I M max Ky (1.58)(20 mm/2)

141,504.2261 N-mm

Bending moment at x = 90 mm: M Px P(90 mm) Maximum load P: P(90 mm) 141,504.2261 N-mm P 1,572.3 N

1,572 N

Ans.


8.84 The grooved shaft shown in Fig. P8.84 is made of C86100 bronze (see Appendix D for properties). The major shaft diameter is D = 50 mm, the minor shaft diameter at the groove is d = 34 mm, and the radius of the groove is r = 4 mm. Determine the maximum allowable moment M that may be applied to the shaft if a factor of safety of 1.5 with respect to failure by yield is specified. Fig. P8.84


D d

50 mm 34 mm

1.471

K

1.96


64

(34 mm)4

65,597.24 mm 4

Maximum allowable bending moment: 331 MPa Y 220.6667 MPa allow FS 1.5 My K allow I (220.6667 N/mm 2 )(65,597.24 mm 4 ) allow I M max Ky (1.96)(34 mm/2) 434, 427.5 N-mm

434 N-m

Ans.


9.1 For the following problems, a beam segment subjected to internal bending moments at sections A and B is shown along with a sketch of the cross-sectional dimensions. For each problem: (a) Sketch a side view of the beam segment and plot the distribution of bending stresses acting at sections A and B. Indicate the magnitude of key bending stresses on the sketch. (b) Determine the resultant forces acting in the x direction on the specified area at sections A and B and show these resultant forces on the sketch. (c) Is the specified area in equilibrium with respect to forces acting in the x direction? If not, determine the horizontal force required to satisfy equilibrium for the specified area and show the location and direction of this force on the sketch. Consider area (1) of the 20-in.-long beam segment, which is subjected to internal bending moments of MA = 24 kip-ft and MB = 28 kip-ft.

Fig. P9.1a Beam segment


Solution Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (in. ) (in.) (in.4) left web 864.000 0.000 0.000 top flange 12.505 10.250 1,287.016 bottom flange 12.505 –10.250 1,287.016 right web 864.000 0.000 0.000 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 864.000 1,299.521 1,299.521 864.000 4,327.042


(a) Bending stress distribution

(b) Resultant forces acting on area (1) On section A, the resultant force on area (1) in the x direction is 1 FA (798.699 psi 565.745 psi)(3.5 in.)(3.5 in.) 8,357.227 lb 2 and on section B, the horizontal resultant force on area (1) is 1 FB (931.816 psi 660.036 psi)(3.5 in.)(3.5 in.) 9,750.098 lb 2 (c) Equilibrium of area (1) Fx 8,357.227 lb 9,750.098 lb FH

1.393 kips

1,392.871 lb

8.36 kips (C)

Ans.

9.75 kips (C)

Ans.

0

Ans.

The horizontal shear force is directed from section A toward section B at the interface between area (1) and the web elements.


9.2 For the following problems, a beam segment subjected to internal bending moments at sections A and B is shown along with a sketch of the cross-sectional dimensions. For each problem: (a) Sketch a side view of the beam segment and plot the distribution of bending stresses acting at sections A and B. Indicate the magnitude of key bending stresses on the sketch. (b) Determine the resultant forces acting in the x direction on the specified area at sections A and B and show these resultant forces on the sketch. (c) Is the specified area in equilibrium with respect to forces acting in the x direction? If not, determine the horizontal force required to satisfy equilibrium for the specified area and show the location and direction of this force on the sketch. Consider area (1) of the 12-in.-long beam segment, which is subjected to internal bending moments of MA = 700 lb-ft and MB = 400 lb-ft.



Solution Centroid location in y direction: (reference axis at bottom of tee shape) Shape top flange stem

y

yi Ai Ai

Width b (in.) 4.5 1.0

47.25 in.3 10.50 in.2

Height h (in.) 1.0 6.0

Area Ai (in.2) 4.50 6.00 10.50


yi Ai (in.3) 29.25 18.00 47.25

4.50 in. (measured upward from bottom edge of shape)

Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (in. ) (in.) (in.4) top flange 0.375 2.000 18.000 stem 18.000 –1.500 13.500 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 18.375 31.500 49.875



(b) Resultant forces acting on area (1) On section A, the resultant force on area (1) in the x direction is 1 FA (421.053 psi 252.632 psi)(4.5 in.)(1 in.) 1,515.792 lb 1,516 lb (C) 2 and on section B, the horizontal resultant force on area (1) is 1 FB (240.602 psi 144.361 psi)(4.5 in.)(1 in.) 866.167 lb 866 lb (C) 2 (c) Equilibrium of area (1) Fx 1,515.792 lb 866.167 lb

649.625 lb

Ans.

Ans.

0

Ans. FH 650 lb The horizontal shear force is directed from section B toward section A at the interface between area (1) and the stem.


9.3 For the following problems, a beam segment subjected to internal bending moments at sections A and B is shown along with a sketch of the cross-sectional dimensions. For each problem: (a) Sketch a side view of the beam segment and plot the distribution of bending stresses acting at sections A and B. Indicate the magnitude of key bending stresses on the sketch. (b) Determine the resultant forces acting in the x direction on the specified area at sections A and B and show these resultant forces on the sketch. (c) Is the specified area in equilibrium with respect to forces acting in the x direction? If not, determine the horizontal force required to satisfy equilibrium for the specified area and show the location and direction of this force on the sketch. Consider area (1) of the 500-mm-long beam segment, which is subjected to internal bending moments of MA = −5.8 kN-m and MB = −3.2 kN-m.



Solution Centroid location in y direction: (reference axis at bottom of shape) Shape top flange left stem right stem

y

yi Ai Ai

Width b (mm) 160 20 20

2,826,000 mm3 15,600 mm2

Height h (mm) 30 270 270

Area Ai (mm2) 4,800 5,400 5,400 15,600

yi (from bottom) (mm) 285 135 135

yi Ai (mm3) 1,368,000 729,000 729,000 2,826,000

181.154 mm (measured upward from bottom edge of shape)

Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (mm ) (mm) (mm4) top flange 360,000 103.846 51,763,160.3 left stem 32,805,000 –46.154 11,503,035.3 right stem 32,805,000 –46.154 11,503,035.3 Moment of inertia about the z axis (mm4) =

IC + d²A (mm4) 52,123,160.3 44,308,035.3 44,308,035.3 140,739,231



(b) Resultant forces acting on area (1) On section A, the resultant force on area (1) in the x direction is 1 FA (4.898 MPa 3.661 MPa)(160 mm)(30 mm) 20,542 N 2 and on section B, the horizontal resultant force on area (1) is 1 FB (2.702 MPa 2.020 MPa)(160 mm)(30 mm) 11,334 N 2 (c) Equilibrium of area (1) Fx 20,542 N 11,334 N

9,209 N

20.5 kN (T)

Ans.

11.33 kN (T)

Ans.

0

Ans. FH 9.21 kN The horizontal shear force is directed from section A toward section B at the interface between area (1) and the stems.


9.4 For the following problems, a beam segment subjected to internal bending moments at sections A and B is shown along with a sketch of the cross-sectional dimensions. For each problem: (a) Sketch a side view of the beam segment and plot the distribution of bending stresses acting at sections A and B. Indicate the magnitude of key bending stresses on the sketch. (b) Determine the resultant forces acting in the x direction on the specified area at sections A and B and show these resultant forces on the sketch. (c) Is the specified area in equilibrium with respect to forces acting in the x direction? If not, determine the horizontal force required to satisfy equilibrium for the specified area and show the location and direction of this force on the sketch. Consider area (1) of the 16-in.-long beam segment, which is subjected to internal bending moments of MA = −3,300 lb-ft and MB = −4,700 lb-ft.



Solution Centroid location in y direction: (reference axis at bottom of shape) Shape left flange (1) right flange (2) central stem

y

yi Ai Ai

Width b (in.) 1.50 1.50 1.50

215.625 in.3 28.50 in.2

Height h (in.) 3.50 3.50 12.00

Area Ai (in.2) 5.25 5.25 18.00 28.50

yi (from bottom) (in.) 10.25 10.25 6.00

yi Ai (in.3) 53.8125 53.8125 108.0000 215.6250


Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (in. ) (in.) (in.4) left flange (1) 5.3594 2.6842 37.8262 right flange (2) 5.3594 2.6842 37.8262 central stem 216.0000 –1.5658 44.1305 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 43.1856 43.1856 260.1305 346.5016



(b) Resultant forces acting on area (1) On section A, the resultant force on area (1) in the x direction is 1 FA (506.765 psi 106.767 psi)(1.5 in.)(3.5 in.) 1,610.522 lb 2 and on section B, the horizontal resultant force on area (1) is 1 FB (721.756 psi 152.061 psi)(1.5 in.)(3.5 in.) 2,293.773 lb 2 (c) Equilibrium of area (1) Fx 1,610.522 lb 2,293.773 lb

683.252 lb

1,611 lb (T)

Ans.

2,290 lb (T)

Ans.

0

Ans. FH 683 lb The horizontal shear force is directed from section B toward section A at the interface between area (1) and the stem.


9.5 For the following problems, a beam segment subjected to internal bending moments at sections A and B is shown along with a sketch of the cross-sectional dimensions. For each problem: (a) Sketch a side view of the beam segment and plot the distribution of bending stresses acting at sections A and B. Indicate the magnitude of key bending stresses on the sketch. (b) Determine the resultant forces acting in the x direction on the specified area at sections A and B and show these resultant forces on the sketch. (c) Is the specified area in equilibrium with respect to forces acting in the x direction? If not, determine the horizontal force required to satisfy equilibrium for the specified area and show the location and direction of this force on the sketch. Consider area (1) of the 18-in.-long beam segment, which is subjected to internal bending moments of MA = −42 kip-in. and MB = −36 kip-in.



Solution Centroid location in y direction: (reference axis at bottom of shape) Shape top flange (1) bottom flange (2) web

y

yi Ai Ai

Width b (in.) 6 10 2

248 in.3 48 in.2

Height h (in.) 2 2 8

Area Ai (in.2) 12 20 16 48

yi (from bottom) (in.) 11 1 6

yi Ai (in.3) 132 20 96 248


Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (in. ) (in.) (in.4) top flange (1) 4.0000 5.8333 408.3333 bottom flange (2) 6.6667 –4.1667 347.2222 web 85.3333 0.8333 11.1111 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 412.3333 353.8889 96.4444 862.6667



(b) Resultant forces acting on area (1) On section A, the resultant force on area (1) in the x direction is 1 FA (332.690 psi 235.317 psi)(6 in.)(2 in.) 3, 408.043 lb 2 and on section B, the horizontal resultant force on area (1) is 1 FB (251.546 psi 154.173 psi)(6 in.)(2 in.) 2,921.180 lb 2 (c) Equilibrium of area (1) Fx 3, 408.043 lb 2,921.180 lb FH

0.487 kips

486.863 lb

3.41 kips (T)

Ans.

2.92 kips (T)

Ans.

0

Ans.

The horizontal shear force is directed from section A toward section B at the interface between area (1) and the web.


9.6 For the following problems, a beam segment subjected to internal bending moments at sections A and B is shown along with a sketch of the cross-sectional dimensions. For each problem: (a) Sketch a side view of the beam segment and plot the distribution of bending stresses acting at sections A and B. Indicate the magnitude of key bending stresses on the sketch. (b) Determine the resultant forces acting in the x direction on the specified area at sections A and B and show these resultant forces on the sketch. (c) Is the specified area in equilibrium with respect to forces acting in the x direction? If not, determine the horizontal force required to satisfy equilibrium for the specified area and show the location and direction of this force on the sketch. Consider area (2) of the beam segment shown in Problem 9.5.



Solution Centroid location in y direction: (reference axis at bottom of shape) Shape top flange (1) bottom flange (2) web

y

yi Ai Ai

Width b (in.) 6 10 2

248 in.3 48 in.2


Area Ai (in.2) 12 20 16 48


yi Ai (in.3) 132 20 96 248


Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (in. ) (in.) (in.4) top flange (1) 4.0000 5.8333 408.3333 bottom flange (2) 6.6667 –4.1667 347.2222 web 85.3333 0.8333 11.1111 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 412.3333 353.8889 96.4444 862.6667



(b) Resultant forces acting on area (2) On section A, the resultant force on area (2) in the x direction is 1 FA (251.546 psi 154.173 psi)(10 in.)(2 in.) 4,057.195 lb 2 and on section B, the horizontal resultant force on area (2) is 1 FB (215.611 psi 132.149 psi)(10 in.)(2 in.) 3,477.595 lb 2 (c) Equilibrium of area (2) Fx 4,057.195 lb 3, 477.595 lb FH

0.580 kips

579.599 lb

4.06 kips (C)

Ans.

3.48 kips (C)

Ans.

0

Ans.

The horizontal shear force is directed from section B toward section A at the interface between area (2) and the web.


9.7 For the following problems, a beam segment subjected to internal bending moments at sections A and B is shown along with a sketch of the cross-sectional dimensions. For each problem: (a) Sketch a side view of the beam segment and plot the distribution of bending stresses acting at sections A and B. Indicate the magnitude of key bending stresses on the sketch. (b) Determine the resultant forces acting in the x direction on the specified area at sections A and B and show these resultant forces on the sketch. (c) Is the specified area in equilibrium with respect to forces acting in the x direction? If not, determine the horizontal force required to satisfy equilibrium for the specified area and show the location and direction of this force on the sketch. Consider area (1) of the 300-mm-long beam segment, which is subjected to internal bending moments of MA = 7.5 kN-m and MB = 8.0 kN-m.



Solution Centroid location in y direction: (reference axis at bottom of shape) Shape left stiff (1) flange (2) right stiff (3) stem

y

yi Ai Ai

Width b (mm) 40 150 40 40

5,348,000 mm3 24,400 mm2

Height h (mm) 90 40 90 280

Area Ai (mm2) 3,600 6,000 3,600 11,200 24,400

yi (from bottom) (mm) 275 300 275 140

yi Ai (mm3) 990,000 1,800,000 990,000 1,568,000 5,348,000



Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (mm ) (mm) (mm4) left stiff (1) 2,430,000.00 55.8197 11,217,008.87 flange (2) 800,000.00 80.8197 39,190,916.42 right stiff (3) 2,430,000.00 55.8197 11,217,008.87 stem 73,173,333.33 –79.1803 70,218,672.40 Moment of inertia about the z axis (mm4) =

IC + d²A (mm4) 13,647,008.87 39,990,916.42 13,647,008.87 143,392,005.73 210,676,939.89


(b) Resultant forces acting on area (1) On section A, the resultant force on area (1) in the x direction is 1 FA (3.589 MPa 0.385 MPa)(40 mm)(90 mm) 7,153.755 N 2 and on section B, the horizontal resultant force on area (1) is 1 FB (3.828 MPa 0.411 MPa)(40 mm)(90 mm) 7,630.672 N 2 (c) Equilibrium of area (1) Fx 7,153.755 N 7,630.672 N

476.917 N

7.15 kN (C)

Ans.

7.63 kN (C)

Ans.

0

FH 0.477 kN Ans. The horizontal shear force is directed from section A toward section B at the interface between area (1) and area (2).


9.8 For the following problems, a beam segment subjected to internal bending moments at sections A and B is shown along with a sketch of the cross-sectional dimensions. For each problem: (a) Sketch a side view of the beam segment and plot the distribution of bending stresses acting at sections A and B. Indicate the magnitude of key bending stresses on the sketch. (b) Determine the resultant forces acting in the x direction on the specified area at sections A and B and show these resultant forces on the sketch. (c) Is the specified area in equilibrium with respect to forces acting in the x direction? If not, determine the horizontal force required to satisfy equilibrium for the specified area and show the location and direction of this force on the sketch. Combine areas (1), (2), and (3) of the beam segment shown in Problem 9.7.



Solution (a) Centroid location in y direction: (reference axis at bottom of shape) Shape left stiff (1) flange (2) right stiff (3) stem

y

yi Ai Ai

Width b (mm) 40 150 40 40

5,348,000 mm3 24,400 mm2

Height h (mm) 90 40 90 280

Area Ai (mm2) 3,600 6,000 3,600 11,200 24,400

yi (from bottom) (mm) 275 300 275 140

yi Ai (mm3) 990,000 1,800,000 990,000 1,568,000 5,348,000


Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (mm ) (mm) (mm4) left stiff (1) 2,430,000.00 55.8197 11,217,008.87 flange (2) 800,000.00 80.8197 39,190,916.42 right stiff (3) 2,430,000.00 55.8197 11,217,008.87 stem 73,173,333.33 -79.1803 70,218,672.40 Moment of inertia about the z axis (mm4) =

IC + d²A (mm4) 13,647,008.87 39,990,916.42 13,647,008.87 143,392,005.73 210,676,939.89



(b) Resultant forces acting on area (1) On section A, the resultant force on area (1) in the x direction is 1 FA (3.589 MPa 0.385 MPa)(40 mm)(90 mm) 7,153.755 N 2 and on section B, the horizontal resultant force on area (1) is 1 FB (3.828 MPa 0.411 MPa)(40 mm)(90 mm) 7,630.672 N 2 Resultant forces acting on area (3) The forces acting on area (3) are identical to those acting on area (1). Resultant forces acting on area (2) On section A, the resultant force on area (2) in the x direction is 1 FA (3.589 MPa 2.165 MPa)(150 mm)(40 mm) 17,262.855 N 2 and on section B, the horizontal resultant force on area (2) is 1 FB (3.828 MPa 2.309 MPa)(150 mm)(40 mm) 18,413.697 N 2 Resultant forces acting on combined areas (1), (2), and (3) On section A, the resultant force on combined areas (1), (2), and (3) is FA 2(7,153.755 N) 17,262.855 N 31,570.363 N 31.6 kN (C)

Ans.

and on section B, the horizontal resultant force on area (2) is FB 2(7,630.672 N) 18,413.697 N 33,675.054 N 33.7 kN (C)

Ans.

(c) Equilibrium of combined areas (1), (2), and (3) Fx 31,570.363 N 33,675.054 N 2,104.691 N

0

FH 2.10 kN Ans. The horizontal shear force is directed from section A toward section B at the interface between area (2) and the stem of the tee.


9.9 A 1.6-m long cantilever beam supports a concentrated load of 7.2 kN, as shown below. The beam is made of a rectangular timber having a width of 120 mm and a depth of 280 mm. Calculate the maximum horizontal shear stresses at points located 35 mm, 70 mm, 105 mm, and 140 mm below the top surface of the beam. From these results, plot a graph showing the distribution of shear stresses from top to bottom of the beam.

Fig. P9.9a Cantilever beam


Solution Shear force in cantilever beam: V = 7.2 kN = 7,200 N Shear stress formula: VQ  It Section properties: (120 mm)(280 mm)3 I  219.52 106 mm 4 12 t = 120 mm

Distance below top surface of beam

y

Q



35 mm

105 mm

514,500 mm3

140.6 kPa

70 mm

70 mm

882,000 mm3

241 kPa 3

301 kPa 321 kPa

105 mm

35 mm

1,102,500 mm

140 mm

0 mm

1,176,000 mm3


9.10 A 14-ft long simply supported timber beam carries a 6-kip concentrated load at midspan, as shown in Fig. P9.10a. The cross-sectional dimensions of the timber are shown in Fig. P9.10b. (a) At section a–a, determine the magnitude of the shear stress in the beam at point H. (b) At section a–a, determine the magnitude of the shear stress in the beam at point K. (c) Determine the maximum horizontal shear stress that occurs in the beam at any location within the 14-ft span length. (d) Determine the maximum tension bending stress that occurs in the beam at any location within the 14-ft span length.


Fig. P9.10a Simply supported timber beam

Solution Section properties: (6 in.)(15 in.)3 I  1, 687.5 in.4 12

t  6 in.

(a) Shear stress at H: Q  (6 in.)(3 in.)(6 in.)  108 in.3

 

VQ It (3,000 lb)(108 in.3 )  32.0 psi (1,687.500 in.4 )(6 in.)

Ans.

(b) Shear stress at K: Q  (6 in.)(1 in.)(7 in.)  42 in.3

 

VQ It (3,000 lb)(42 in.3 )  12.44 psi (1,687.500 in.4 )(6 in.)

Ans.


(c) Maximum shear stress at any location: Qmax  (6 in.)(7.5 in.)(3.75 in.)  168.75 in.3

V Q (3,000 lb)(168.75 in.3 )    50.0 psi It (1,687.500 in.4 )(6 in.)

Ans.

(d) Maximum bending stress at any location: M max  21 kip-ft  21,000 lb-ft

x 

M c (21,000 lb-ft)(7.5 in.)(12 in./ft)   1,120 psi (T) and (C) I 1,687.500 in.4

Ans.


9.11 A 5-m long simply supported timber beam carries a uniformly distributed load of 12 kN/m, as shown in Fig. P9.11a. The cross-sectional dimensions of the beam are shown in Fig. P9.11b. (a) At section a–a, determine the magnitude of the shear stress in the beam at point H. (b) At section a–a, determine the magnitude of the shear stress in the beam at point K. (c) Determine the maximum horizontal shear stress that occurs in the beam at any location within the 5-m span length. (d) Determine the maximum compression bending stress that occurs in the beam at any location within the 5-m span length.



Solution Section properties: (100 mm)(300 mm)3 I  225 106 mm 4 12

t  100 mm

(a) Shear stress magnitude at H: Q  (100 mm)(90 mm)(105 mm)  945,000 mm 3 VQ  It 

(18,000 N)(945,000 mm 3 ) (225  106 mm 4 )(100 mm)

 756 kPa

Ans.

(b) Shear stress magnitude at K: Q  (100 mm)(40 mm)(130 mm)  520,000 mm 3 VQ  It 

(18,000 N)(520,000 mm 3 ) (225  106 mm 4 )(100 mm)

 416 kPa

Ans.


(c) Maximum shear stress at any location: Qmax  (100 mm)(150 mm)(75 mm)  1,125,000 mm3

V Q (30,000 N)(1,125,000 mm3 )    1,500 kPa It (225  106 mm4 )(100 mm)

Ans.

(d) Maximum compression bending stress at any location: M max  37.5 kN-m

My (37.5 kN-m)(150 mm)(1,000 N/kN)(1,000 mm/m)  I 225  106 mm 4  25.0 MPa  25.0 MPa (C)

x  

Ans.


9.12 A 5-m long simply supported timber beam carries two concentrated loads, as shown in Fig. P9.12a. The cross-sectional dimensions of the beam are shown in Fig. P9.12b. (a) At section a–a, determine the magnitude of the shear stress in the beam at point H. (b) At section a–a, determine the magnitude of the shear stress in the beam at point K. (c) Determine the maximum horizontal shear stress that occurs in the beam at any location within the 5m span length. (d) Determine the maximum compression bending stress that occurs in the beam at any location within the 5-m span length.



Solution Section properties: (150 mm)(450 mm)3 I  1,139.1106 mm 4 12

t  150 mm

(a) Shear stress magnitude at H: Q  (150 mm)(150 mm)(150 mm)  3,375,000 mm 3 VQ  It 

(39, 200 N)(3,375,000 mm 3 ) (1,139.1  106 mm 4 )(150 mm)

 774 kPa

Ans.

(b) Shear stress magnitude at K: Q  (150 mm)(100 mm)(175 mm)  2,625,000 mm 3 VQ  It 

(39, 200 N)(2,625,000 mm 3 ) (1,139.1  106 mm 4 )(150 mm)

 602 kPa

Ans.


(c) Maximum shear stress at any location: Qmax  (150 mm)(225 mm)(112.5 mm)  3,796,875 mm3

V Q (39,200 N)(3,796,875 mm3 )    871 kPa I t (1,139.1  106 mm 4 )(150 mm)

Ans.

(d) Maximum bending stress at any location: M max  39.2 kN-m

x  

My (39.2 kN-m)(225 mm)(1,000 N/kN)(1,000 mm/m)  I 1,139.1  106 mm 4

 7.74296 MPa  7,740 kPa (C)

Ans.


9.13 A laminated wood beam consists of eight 2 in. × 6-in. planks glued together to form a section 6 in. wide by 16 in. deep, as shown in Fig. P9.13a. If the allowable strength of the glue in shear is 160 psi, determine: (a) the maximum uniformly distributed load w that can be applied over the full length of the beam if the beam is simply supported and has a span of 20 ft. (b) the shear stress in the glue joint at H, which is located 4 in. above the bottom of the beam and 3 ft from the left support. Assume the beam is subjected to the load w determined in part (a). (c) the maximum tension bending stress in the beam when the load of part (a) is applied.



Solution Section properties: (6 in.)(16 in.)3 I  2, 048 in.4 12

t  6 in.

(a) Maximum Q: Q  (6 in.)(8 in.)(4 in.)  192 in.3 Maximum shear force V: VQ   160 psi It

V 

(160 psi)(2,048 in.4 )(6 in.)  10, 240 lb 192 in.3

Maximum distributed load w: wL Vmax   10, 240 lb 2 2(10, 240 lb)  wmax   1,024 lb/ft 20 ft

Ans.


(b) Shear force at x = 3 ft: V  10, 240 lb  (1,024 lb/ft)(3 ft)  7,168 lb Q  (6 in.)(4 in.)(6 in.)  144 in.3



V Q (7,168 lb)(144 in.3 )   84.0 psi It (2,048 in.4 )(6 in.)

Ans.

(c) Maximum tension bending stress at any location: wL2 (1,024 lb/ft)(20 ft) 2 M max    51, 200 lb-ft 8 8

x  

My (51, 200 lb-ft)(  8 in.)(12 in./ft)   2, 400 psi (T) I 2,048 in.4

Ans.


9.14 A 5-ft long simply supported wood beam carries a concentrated load P at midspan, as shown in Fig. P9.14a. The cross-sectional dimensions of the beam are shown in Fig. P9.14b. If the allowable shear strength of the wood is 80 psi, determine the maximum load P that may be applied at midspan. Neglect the effects of the beam’s self weight.



Solution Section properties: (6 in.)(10 in.)3 I  500 in.4 12

t  6 in.

Maximum Q: Q  (6 in.)(5 in.)(2.5 in.)  75 in.3 Maximum shear force V: VQ   80 psi It

V 

(80 psi)(500 in.4 )(6 in.)  3, 200 lb 75 in.3

Maximum concentrated load P: P Vmax   3,200 lb 2  Pmax  2(3,200 lb)  6,400 lb

Ans.


9.15 A wood beam supports the loads shown in Fig. P9.15a. The cross-sectional dimensions of the beam are shown in Fig. P9.15b. Determine the magnitude and location of: (a) the maximum horizontal shear stress in the beam. (b) the maximum tension bending stress in the beam.



Solution Section properties: (75 mm)(240 mm)3 (20 mm)(100 mm)3 I 2  89, 733,333 mm 4 12 12 (a) Maximum shear force: Vmax = 9.54 kN = 9,540 N @ support A Check shear stress at neutral axis: Q  (75 mm)(120 mm)(60 mm) 2(20 mm)(50 mm)(25 mm)  590, 000 mm3



VQ (9,540 N)(590,000 mm3 )   545 kPa I t (89,733,333 mm4 )(115 mm)

Check shear stress at top edge of cover plates: Q  (75 mm)(70 mm)(85 mm)  446, 250 mm3



VQ (9,540 N)(446,250 mm3 )   633 kPa I t (89,733,333 mm4 )(75 mm)

Maximum shear stress in beam:  H ,max  633 kPa

Ans.

(b) Maximum bending moment: Mmax = 6.49 kN-m (between support A and point B) Maximum tension bending stress: My (6.49 kN-m)(  120 mm)(1,000 N/kN)(1,000 mm/m) x    I 89,733,333 mm 4  8.67905 MPa  8,680 kPa (T)

Ans.


9.16 A 50-mm-diameter solid steel shaft supports loads PA = 1.5 kN and PC = 3.0 kN, as shown in Fig. P9.16. Assume L1 = 150 mm, L2 = 300 mm, and L3 = 225 mm. The bearing at B can be idealized as a roller support and the bearing at D can be idealized as a pin support. Determine the magnitude and location of: (a) the maximum horizontal shear stress in the shaft. (b) the maximum tension bending stress in the shaft.

Fig. P9.16

Solution Section properties: I



D4 



(50 mm) 4

64 64  306, 796.158 mm 4 D 3 (50 mm)3  12 12  10, 416.667 mm 3

Q

Maximum shear force magnitude: Vmax = 1.71 kN (between B and C)

Maximum bending moment magnitude: Mmax = 289.29 kN-mm (at C)

(a) Maximum horizontal shear stress: V Q (1,710 N)(10,416.667 mm3 )   It (306,796.158 mm 4 )(50 mm)  1.161 MPa

(at neutral axis between B and C )

Ans.

(b) Maximum tension bending stress: My (289.29 kN-mm)(  50 mm/2)(1,000 N/kN) x    I 306,796.158 mm 4  23.574 MPa  23.6 MPa (T)

(on bottom of shaft at C )

Ans.


9.17 A 1.25-in.-diameter solid steel shaft supports loads PA = 400 lb and PC = 900 lb, as shown in Fig. P9.17. Assume L1 = 6 in., L2 = 12 in., and L3 = 8 in. The bearing at B can be idealized as a roller support and the bearing at D can be idealized as a pin support. Determine the magnitude and location of: (a) the maximum horizontal shear stress in the shaft. (b) the maximum tension bending stress in the shaft.

Fig. P9.17




D4 



(1.25 in.) 4

64 64  0.119842 in.4

D 3 (1.25 in.)3  12 12  0.162760 in.3

Q

Maximum shear force magnitude: Vmax = 480 lb (between B and C)

Maximum bending moment magnitude: Mmax = 3,360 lb-in. (at C)

(a) Maximum horizontal shear stress: V Q (480 lb)(0.162760 in.3 )   I t (0.119842 in.4 )(1.25 in.)

 521.519 psi  522 psi


(b) Maximum tension bending stress: My (3,360 lb-in.)(  1.25 in./2) x    I 0.119842 in.4  17, 523.022 psi  17,520 psi (T)


Ans.

Ans.


9.18 A 1.00-in.-diameter solid steel shaft supports loads PA = 200 lb and PD = 240 lb, as shown in Fig. P9.18. Assume L1 = 2 in., L2 = 5 in., and L3 = 4 in. The bearing at B can be idealized as a pin support and the bearing at C can be idealized as a roller support. Determine the magnitude and location of: (a) the maximum horizontal shear stress in the shaft. (b) the maximum tension bending stress in the shaft. Fig. P9.18




D4 



(1.00 in.) 4 64 64  0.049087 in.4 D 3 (1.00 in.)3  12 12  0.083333 in.3

Q

Maximum shear force magnitude: Vmax = 272 lb (between B and C)

Maximum bending moment magnitude: Mmax = 960 lb-in. (at C)

(a) Maximum horizontal shear stress: V Q (272 lb)(0.083333 in.3 )   I t (0.049087 in.4 )(1.00 in.)

 461.762 psi  462 psi


(b) Maximum tension bending stress: My (960 lb-in.)(  1.00 in./2) x    I 0.049087 in.4  9,778.480 psi  9,780 psi (T)


Ans.

Ans.


9.19 A 20-mm-diameter solid steel shaft supports loads PA = 900 N and PD = 1,200 N, as shown in Fig. P9.19. Assume L1 = 50 mm, L2 = 120 mm, and L3 = 90 mm. The bearing at B can be idealized as a pin support and the bearing at C can be idealized as a roller support. Determine the magnitude and location of: (a) the maximum horizontal shear stress in the shaft. (b) the maximum compression bending stress in the shaft. Fig. P9.19




D4 



(20 mm) 4 64 64  7,853.982 mm 4 D3 (20 mm)3  12 12  666.667 mm3

Q

Maximum shear force magnitude: Vmax = 1,275 N (between B and C)

Maximum bending moment magnitude: Mmax = 108,000 N-mm (at C)

(a) Maximum horizontal shear stress: VQ (1, 275 N)(666.667 mm3 )   It (7,853.982 mm 4 )(20 mm)  5.411 MPa  5.41 MPa


Ans.

(b) Maximum compression bending stress: My (108,000 N-mm)(20 mm/2) x    I 7,853.982 mm 4  137.510 MPa  137.5 MPa (C)

(on top of shaft at C )

Ans.


9.20 A 1.25-in.-diameter solid steel shaft supports loads PA = 600 lb, PC = 1,600 lb, and PE = 400 lb, as shown in Fig. P9.20. Assume L1 = 6 in., L2 = 15 in., L3 = 8 in., and L4 = 10 in. The bearing at B can be idealized as a roller support and the bearing at D can be idealized as a pin support. Determine the magnitude and location of: (a) the maximum horizontal shear stress in the shaft. (b) the maximum tension bending stress in the shaft.

Fig. P9.20




D4 



(1.25 in.) 4

64 64  0.119842 in.4

D 3 (1.25 in.)3  12 12  0.162760 in.3

Q

Maximum shear force magnitude: Vmax = 1,060.9 lb (between C and D)

Maximum bending moment magnitude: Mmax = 4,487 lb-in. (at C)

(a) Maximum horizontal shear stress: V Q (1,060.9 lb)(0.162760 in.3 )   It (0.119842 in.4 )(1.25 in.)

 1,152.632 psi  1,153 psi

(at neutral axis between C and D)

(b) Maximum tension bending stress: My (4, 487 lb-in.)(  1.25 in./2) x    I 0.119842 in.4  23, 400.309 psi  23, 400 psi (T)


Ans.

Ans.


9.21 A 25-mm-diameter solid steel shaft supports loads PA = 1,000 N, PC = 3,200 N, and PE = 800 N, as shown in Fig. P9.21. Assume L1 = 80 mm, L2 = 200 mm, L3 = 100 mm, and L4 = 125 mm. The bearing at B can be idealized as a roller support and the bearing at D can be idealized as a pin support. Determine the magnitude and location of: (a) the maximum horizontal shear stress in the shaft. (b) the maximum tension bending stress in the shaft.

Fig. P9.21




D4 



(25 mm) 4

64 64  19,174.760 mm 4 D 3 (25 mm)3  12 12  1,302.083 mm 3

Q

Maximum shear force magnitude: Vmax = 2,200 N (between C and D)

Maximum bending moment magnitude: Mmax = 120,000 N-mm (at C)

(a) Maximum horizontal shear stress: V Q (2, 200 N)(1,302.083 mm3 )   It (19,174.760 mm 4 )(25 mm)  5.98 MPa

(at neutral axis between C and D)

Ans.

(b) Maximum tension bending stress: My (120,000 N-mm)(  25 mm/2) x    I 19,174.760 mm 4  78.228 MPa  78.2 MPa (T)


Ans.


9.22 A 3-in. standard steel pipe (D = 3.500 in.; d = 3.068 in.) supports a concentrated load of P = 900 lb, as shown in Fig. P9.22a. The span length of the cantilever beam is L = 3 ft. Determine the magnitude of: (a) the maximum horizontal shear stress in the pipe. (b) the maximum tension bending stress in the pipe.


Fig. P9.22b Pipe cross section




[D4  d 4 ] 



[(3.500 in.) 4  (3.068 in.) 4 ]  3.017157 in.4

64 64 1 1 Q  [ D 3  d 3 ]  [(3.500 in.)3  (3.068 in.)3 ]  1.166422 in.3 12 12

Maximum shear force magnitude: Vmax = 900 lb Maximum bending moment magnitude: Mmax = (900 lb)(3 ft)(12 in./ft) = 32,400 lb-in. (a) Maximum horizontal shear stress: VQ (900 lb)(1.166422 in.3 )    805.410 psi  805 psi I t (3.017157 in.4 )(3.500 in.  3.068 in.) (b) Maximum tension bending stress: My ( 32,400 lb-in.)(3.500 in./2) x     18,792.529 psi  18,790 psi (T) I 3.017157 in.4

Ans.

Ans.


9.23 A steel pipe (D = 170 mm; d = 150 mm) supports a concentrated load of P as shown in Fig. P9.23a. The span length of the cantilever beam is L = 1.2 m. (a) Compute the value of Q for the pipe. (b) If the allowable shear stress for the pipe shape is 75 MPa, determine the maximum load P than can be applied to the cantilever beam.


Fig. P9.23b Pipe cross section

Solution (a) Section properties: I



[D4  d 4 ] 



[(170 mm) 4  (150 mm) 4 ]  16,147,786.239 mm 4 64 64 1 1 Q  [ D 3  d 3 ]  [(170 mm)3  (150 mm)3 ]  128,166.667 mm3  128,170 mm3 12 12

Ans.

(b) Maximum load P: VQ   75 MPa It Vmax 

(75 N/mm 2 )(16,147,786.239 mm 4 )(170 mm  150 mm)  188,986 N 128,166.667 mm3

For the cantilever beam shown here, V = P; therefore, Pmax  Vmax  189.0 kN

Ans.


9.24 A concentrated load P is applied to the upper end of a 1-m long pipe, as shown in Fig. P9.24. The outside diameter of the pipe is D = 114 mm and the inside diameter is d = 102 mm. (a) Compute the value of Q for the pipe. (b) If the allowable shear stress for the pipe shape is 75 MPa, determine the maximum load P than can be applied to the cantilever beam.

Fig. P9.24

Solution (a) Section properties: I



[D4  d 4 ] 



[(114 mm) 4  (102 mm) 4 ]  2,977, 287 mm 4

64 64 1 1 Q  [ D 3  d 3 ]  [(114 mm)3  (102 mm)3 ]  35,028 mm3 12 12

Ans.

(b) Maximum load P: VQ   75 MPa It Vmax

(75 N/mm 2 )(2,977, 287 mm 4 )(114 mm  102 mm)   76, 498 N 35,028 mm3

For the cantilever beam shown here, V = P; therefore, Pmax  Vmax  76.5 kN

Ans.


9.25 A concentrated load of P = 6 kips is applied to the upper end of a 4-ft long pipe, as shown in Fig. P9.25. The pipe is an 8 in. standard steel pipe, which has an outside diameter of D = 8.625 in. and an inside diameter of d = 7.981 in. Determine the magnitude of: (a) the maximum vertical shear stress in the pipe. (b) the maximum tension bending stress in the pipe.

Fig. P9.25




[D4  d 4 ] 



[(8.625 in.) 4  (7.981 in.) 4 ]  72.489241 in.4

64 64 1 1 Q  [ D3  d 3 ]  [(8.625 in.)3  (7.981 in.)3 ]  11.104874 in.3 12 12

Maximum shear force magnitude: Vmax = 6 kips = 6,000 lb Maximum bending moment magnitude: Mmax = (6,000 lb)(4 ft)(12 in./ft) = 288,000 lb-in. (a) Maximum vertical shear stress: VQ (6,000 lb)(11.104874 in.3 )    1,427.268 psi  1,427 psi I t (72.489241 in.4 )(8.625 in.  7.981 in.) (b) Maximum tension bending stress: M c (288,000 lb-in.)(8.625 in./2) x    17,133.578 psi  17,130 psi (T) I 72.489241 in.4

Ans.

Ans.


9.26 The cantilever beam shown in Fig. P9.26a is subjected to a concentrated load of P = 38 kips. The cross-sectional dimensions of the wide-flange shape are shown in Fig. P9.26b. Determine: (a) the shear stress at point H, which is located 4 in. below the centroid of the wide-flange shape. (b) the maximum horizontal shear stress in the wide-flange shape.

Fig. P9.26a

Fig. P9.26b

Solution Moment of inertia about the z axis: Shape Width b Height h (in.) (in.) flange 6.75 0.455 web 0.285 13.090 flange 6.75 0.455

d = yi – y IC d²A 4 (in. ) (in.) (in.4) 0.0530 6.7725 140.8683 53.2700 0.0000 0.0000 0.0530 –6.7725 140.8683 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 140.9213 53.2700 140.9213 335.1125

(a) Shear stress at H:

0.455 in.  QH  (6.75 in.)(0.455 in.)  7 in.    2 7 in.  0.455 in.  4 in.  3  (0.285 in.)(7 in.  0.455 in.  4 in.)  4 in.    24.6243 in.  2

H 

(38 kips)(24.6243 in.3 )  9.7974 ksi  9.80 ksi (335.1125 in.4 )(0.285 in.)

Ans.

(b) Maximum horizontal shear stress: 0.455 in.  Qmax  (6.75 in.)(0.455 in.)  7 in.    2

 7 in.  0.455 in. 3  (0.285 in.)(7 in.  0.455 in.)    26.9043 in.  2

 max 

(38 kips)(26.9043 in.3 )  10.7046 ksi  10.70 ksi (335.1125 in.4 )(0.285 in.)

Ans.


9.27 The cantilever beam shown in Fig. P9.27a is subjected to a concentrated load of P. The crosssectional dimensions of the wide-flange shape are shown in Fig. P9.27b. (a) Compute the value of Q that is associated with point K, which is located 2 in. above the centroid of the wide-flange shape. (b) If the allowable shear stress for the wide-flange shape is 14 ksi, determine the maximum concentrated load P than can be applied to the cantilever beam.

Fig. P9.27a

Fig. P9.27b

Solution Moment of inertia about the z axis: Shape Width b Height h (in.) (in.) flange 6.75 0.455 web 0.285 13.090 flange 6.75 0.455

d = yi – y IC d²A 4 (in. ) (in.) (in.4) 0.0530 6.7725 140.8683 53.2700 0.0000 0.0000 0.0530 –6.7725 140.8683 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 140.9213 53.2700 140.9213 335.1125

(a) Q associated with point K: 0.455 in.  QK  (6.75 in.)(0.455 in.)  7 in.    2 7 in.  0.455 in.  2 in.   (0.285 in.)(7 in.  0.455 in.  2 in.)  2 in.   26.3343 in.3    2

Ans.

(b) Maximum load P: 0.455 in.  Qmax  (6.75 in.)(0.455 in.)  7 in.    2

 max

 7 in.  0.455 in. 3  (0.285 in.)(7 in.  0.455 in.)    26.9043 in.  2 VQmax   14 ksi It

Vmax

(14 ksi)(335.1125 in.4 )(0.285 in.)   49.6983 kips 26.9043 in.3

For the cantilever beam shown here, V = P; therefore, Pmax  Vmax  49.7 kips

Ans.


9.28 The cantilever beam shown in Fig. P9.28a is subjected to a concentrated load of P. The crosssectional dimensions of the rectangular tube shape are shown in Fig. P9.28b. (a) Compute the value of Q that is associated with point H, which is located 90 mm above the centroid of the rectangular tube shape. (b) If the allowable shear stress for the rectangular tube shape is 125 MPa, determine the maximum concentrated load P than can be applied to the cantilever beam.

Fig. P9.28a

Fig. P9.28b

Solution Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (mm ) (mm) (mm4) outer rectangle 195,312,500 0.000 0.000 inner rectangle −143,077,428 0.000 0.000 Moment of inertia about the z axis (mm4) = (a) Q associated with point H:  250 mm 8 mm  QH  (150 mm)(8 mm)     2 2 

IC + d²A (mm4) 195,312,500 −143,077,428 52,235,072

250 mm    8 mm  90 mm 250 mm    2 2(8 mm)   8 mm  90 mm  90 mm     2 2    189,912 mm3 (b) Maximum load P:

Ans.

 250 mm 8 mm  Qmax  (150 mm)(8 mm)     2 2 

 max

 250 mm   8 mm  250 mm   2 2(8 mm)   8 mm   254,712 mm3    2 2   VQmax   125 MPa It Vmax 

(125 N/mm 2 )(52, 235,072 mm 4 )(2  8 mm)  410,150 N  410.15 kN 254,712 mm3

For the cantilever beam shown here, V = P; therefore, Pmax  Vmax  410 kN

Ans.


9.29 The cantilever beam shown in Fig. P9.29a is subjected to a concentrated load of P = 175 kN. The cross-sectional dimensions of the rectangular tube shape are shown in Fig. P9.29b. Determine: (a) the shear stress at point K, which is located 50 mm below the centroid of the rectangular tube shape. (b) the maximum horizontal shear stress in the rectangular tube shape.

Fig. P9.29a

Fig. P9.29b

Solution Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (mm ) (mm) (mm4) outer rectangle 195,312,500 0.000 0.000 inner rectangle −143,077,428 0.000 0.000 Moment of inertia about the z axis (mm4) =

IC + d²A (mm4) 195,312,500 −143,077,428 52,235,072

(a) Shear stress at K:

 250 mm 8 mm  QK  (150 mm)(8 mm)     2 2  250 mm    8 mm  50 mm  250 mm   2 2(8 mm)   8 mm  50 mm  50 mm     2 2    234,712 mm3

K 

(175,000 N)(234,712 mm3 )  49.1463 MPa  49.1 MPa (52, 235,072 mm 4 )(2  8 mm)

Ans.

(b) Maximum horizontal shear stress:  250 mm 8 mm  Qmax  (150 mm)(8 mm)     2 2 

 250 mm   8 mm  250 mm   3 2 2(8 mm)   8 mm    254,712 mm   2 2  

 max 

(175,000 N)(254,712 mm 3 )  53.3341 MPa  53.3 MPa (52, 235,072 mm 4 )(2  8 mm)

Ans.


9.30 The internal shear force V at a certain section of an aluminum beam is 8 kN. If the beam has a cross section shown in Fig. P9.30, determine: (a) the shear stress at point H, which is located 30 mm above the bottom surface of the tee shape. (b) the maximum horizontal shear stress in the tee shape.

Fig. P9.30


yi Area Ai (from bottom) yi Ai 2 (mm ) (mm) (mm3) 375.0 72.5 27,187.5 350.0 35.0 12,250.0 2 725.0 mm 39,437.5 mm3

yi Ai

39, 437.5 mm3 y   54.397 mm Ai 725.0 mm 2  20.603 mm


Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (mm ) (mm) (mm4) top flange 781.250 18.103 122,900.565 stem 142,916.667 −19.397 131,679.177 Moment of inertia about the z axis (mm4) =

IC + d²A (mm4) 123,681.815 274,595.843 398,277.658

(a) Shear stress at H: QH  (5 mm)(30 mm)(39.397 mm)  5,909.550 mm 3

H 

(8,000 N)(5,909.550 mm3 )  23.7 MPa (398, 277.658 mm 4 )(5 mm)

Ans.

(b) Maximum horizontal shear stress: At neutral axis: Qmax  (5 mm)(54.397 mm)(27.199 mm)  7,397.720 mm3

 max 

(8,000 N)(7,397.720 mm3 )  29.7 MPa (398, 277.658 mm 4 )(5 mm)

Ans.


9.31 The internal shear force V at a certain section of a steel beam is 80 kN. If the beam has a cross section shown in Fig. P9.31, determine: (a) the shear stress at point H, which is located 30 mm below the centroid of the wide-flange shape. (b) the maximum horizontal shear stress in the wideflange shape.

Fig. P9.31

Solution Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (mm ) (mm) (mm4) top flange 59,062.5 97.5 29,944,687.5 web 4,860,000.0 0.0 0.0 bottom flange 59,062.5 −97.5 29,944,687.5 Moment of inertia about the z axis (mm4) =

IC + d²A (mm4) 30,003,750.0 4,860,000.0 30,003,750.0 64,867,500.0

(a) Shear stress at H: QH  (210 mm)(15 mm)(97.5 mm)  (10 mm)(60 mm)(60 mm)  343,125 mm3

H 

(80,000 N)(343,125 mm3 )  42.3 MPa (64,867,500 mm 4 )(10 mm)

Ans.

(b) Maximum horizontal shear stress: At neutral axis: Qmax  (210 mm)(15 mm)(97.5 mm)  (10 mm)(90 mm)(45 mm)  347,625 mm3

 max 

(80,000 N)(347,625 mm3 )  42.9 MPa (64,867,500 mm 4 )(10 mm)

Ans.


9.32 The internal shear force V at a certain section of a steel beam is 110 kips. If the beam has a cross section shown in Fig. P9.32, determine: (a) the value of Q associated with point H, which is located 2 in. below the top surface of the flanged shape. (b) the maximum horizontal shear stress in the flanged shape.

Fig. P9.32

Solution Centroid location in y direction: Shape

Area Ai (in.2) 5.0 10.0 8.0 23.0 in.2

top flange web bottom flange

y

yi Ai Ai



yi (from bottom) (in.) 11.5 6.0 0.5

yi Ai (in.3) 57.5 60.0 4.0 121.5 in.3

121.5 in.3  5.2826 in. 23.0 in.2


 6.7174 in.


Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (in. ) (in.) (in.4) top flange 0.4167 6.2174 193.2798 web 83.3333 0.7174 5.1465 bottom flange 0.6667 −4.7826 182.9868 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 193.6964 88.4798 183.6534 465.8297

(a) Q at point H:

Q  (5 in.)(1 in.)(6.2174 in.)  (1 in.)(1 in.)(5.2174 in.)  36.3044 in.3

Ans.

(b) Maximum horizontal shear stress: At neutral axis: Qmax  (5 in.)(1 in.)(6.2174 in.)  (1 in.)(5.7174 in.)(2.8587 in.)  47.4313 in.3

 max 

(110 kips)(47.4313 in.3 )  11.20 ksi (465.8297 in.4 )(1 in.)

Ans.


9.33 The internal shear force V at a certain section of a steel beam is 75 kips. If the beam has a cross section shown in Fig. P9.33, determine: (a) the shear stress at point H, which is located 2 in. above the bottom surface of the flanged shape. (b) the shear stress at point K, which is located 4.5 in. below the top surface of the flanged shape. Fig. P9.33

Solution Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (in. ) (in.) (in.4) left flange 62.5000 0.0000 0.0000 web 0.0885 0.0000 0.0000 right flange 62.5000 0.0000 0.0000 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 62.5000 0.0885 62.5000 125.0885

(a) Shear stress at H: QH  2(0.75 in.)(2 in.)(4 in.)  12 in.3

H 

(75 kips)(12 in.3 )  4.80 ksi (125.0885 in.4 )(2  0.75 in.)

Ans.

(b) Shear stress at K: QK  2(0.75 in.)(4.5 in.)(2.75 in.)  18.5625 in.3

K 

(75 kips)(18.5625 in.3 )  7.42 ksi (125.0885 in.4 )(2  0.75 in.)

Ans.


9.34 Consider a 100-mm-long segment of a simply supported beam (Fig. P9.34a). The internal bending moments on the left and right sides of the segment are 75 kN-m and 80 kN-m, respectively. The crosssectional dimensions of the flanged shape are shown in Fig. P9.34b. Determine the maximum horizontal shear stress in the beam at this location.


Fig. P9.34a Beam segment (side view)

Solution Centroid location in y direction: yi Area Ai (from bottom) yi Ai 2 (mm ) (mm) (mm3) top flange 9,000 300 2,700,000 web 8,400 165 1,386,000 bottom flange 15,000 30 450,000 2 32,400 mm 4,536,000 mm3 yi Ai 4,536, 000 mm3 y   140 mm (from bottom of shape to centroid) Ai 32,400 mm 2 Shape

 190 mm (from top of shape to centroid) Moment of inertia about the z axis: d = yi – y Shape IC d²A IC + d²A 4 4 (mm ) (mm) (mm ) (mm4) top flange 2,700,000 160 230,400,000 233,100,000 web 30,870,000 25 5,250,000 36,120,000 bottom flange 4,500,000 −110 181,500,000 186,000,000 4 Moment of inertia about the z axis (mm ) = 455,220,000

Shear force in beam: M 80 kN-m  75 kN-m 5 kN-m V    50 kN x 100 mm 0.1 m Maximum horizontal shear stress: At neutral axis: Qmax  (250 mm)(60 mm)(110 mm)  (40 mm)(80 mm)(40 mm)  1,778,000 mm3

 max 

(50,000 N)(1,778,000 mm3 )  4.88 MPa (455, 220,000 mm 4 )(40 mm)

Ans.


9.35 A simply supported beam supports the loads shown in Fig. P9.35a. The cross-sectional dimensions of the wide-flange shape are shown in Fig. P9.35b. (a) Determine the maximum shear force in the beam. (b) At the section of maximum shear force, determine the shear stress in the cross section at point H, which is located 100 mm below the neutral axis of the wide-flange shape. (c) At the section of maximum shear force, determine the maximum horizontal shear stress in the cross section. (d) Determine the magnitude of the maximum bending stress in the beam.

Fig. P9.35a

Fig. P9.35b

Solution (a) Maximum shear force magnitude: Vmax = 175 kN (just to the right of B)

Maximum bending moment magnitude: Mmax = 156.25 kN-m (between B and C)


Section properties: Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (mm ) (mm) (mm4) top flange 56,250 142.5 60,918,750 web 16,402,500 0 0 bottom flange 56,250 −142.5 60,918,750 Moment of inertia about the z axis (mm4) =

IC + d²A (mm4) 60,975,000 16,402,500 60,975,000 138,352,500

(b) Shear stress at H: QH  (200 mm)(15 mm)(142.5 mm)  (10 mm)(35 mm)(117.5 mm)  468,625 mm3

H 

(175,000 N)(468,625 mm3 )  59.3 MPa (138,352,500 mm 4 )(10 mm)

Ans.

(c) Maximum horizontal shear stress: At neutral axis: Qmax  (200 mm)(15 mm)(142.5 mm)  (10 mm)(135 mm)(67.5 mm)  518,625 mm3

 max 

(175,000 N)(518,625 mm3 )  65.6 MPa (138,352,500 mm4 )(10 mm)

Ans.

(d) Maximum tension bending stress: Mc (156.25  106 N-mm)(300 mm/2) x   I 138,352,500 mm 4  169.404 MPa  169.4 MPa

Ans.


9.36 A simply supported beam supports the loads shown in Fig. P9.36a. The cross-sectional dimensions of the structural tube shape are shown in Fig. P9.36b. (a) At section a–a, which is located 4 ft to the right of pin support B, determine the bending stress and the shear stress at point H, which is located 3 in. below the top surface of the tube shape. (b) Determine the magnitude and the location of the maximum horizontal shear stress in the tube shape at section a–a.

Fig. P9.36a

Fig. P9.36b

Solution Shear force magnitude at a–a: V = 27.60 kips Bending moment at a–a: M = 60.90 kip-ft Section properties: (12 in.)(16 in.)3 (11.25 in.)(15.25 in.)3 I  12 12 4  771.0830 in. (a) Bending stress at H: My H   I (60,900 lb-ft)(5 in.)(12 in./ft)  771.0830 in.4  4,738.79 psi  4,740 psi (C)

Ans.

Shear stress at H: QH  (12 in.)(0.375 in.)(7.8125 in.)  2(0.375 in.)(2.625 in.)(6.3125 in.)  47.5840 in.3

H 

(27,600 lb)(47.5840 in.3 )  2, 270 psi (771.0830 in.4 )(2  0.375 in.)

Ans.

(b) Maximum shear force magnitude: V = 39.60 kips (at pin B)


Maximum horizontal shear stress: At neutral axis: Qmax  (12 in.)(0.375 in.)(7.8125 in.)  2(0.375 in.)(7.625 in.)(3.8125 in.)  56.9590 in.3

 max 

(39,600 lb)(56.9590 in.3 )  3,900 psi (771.0830 in.4 )(2  0.375 in.)

Ans.


9.37 A cantilever beam supports the loads shown in Fig. P9.37a. The cross-sectional dimensions of the shape are shown in Fig. P9.37b. Determine: (a) the maximum horizontal shear stress. (b) the maximum compression bending stress. (c) the maximum tension bending stress.

Fig. P9.37a

Fig. P9.37b

Solution Centroid location in y direction: Shape top flange left stem right stem

Width b (in.) 12.0 0.5 0.5

Height h (in.) 0.5 5.5 5.5

yi Ai

49.6250 in.3 y   4.3152 in. Ai 11.50 in.2  1.6848 in.

Area Ai (in.2) 6.00 2.75 2.75 11.50

yi (from bottom) (in.) 5.75 2.75 2.75

yi Ai (in.3) 34.5000 7.5625 7.5625 49.6250


Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (in. ) (in.) (in.4) top flange 0.1250 1.4348 12.3519 left stem 6.9323 −1.5652 6.7371 right stem 6.9323 −1.5652 6.7371 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 12.4769 13.6694 13.6694 39.8157


Maximum shear force magnitude: V = 5,800 lb Maximum positive bending moment: Mpos = 8,850 lb-ft Maximum negative bending moment: Mneg = −9,839 lb-ft (a) Maximum shear stress: Qmax  2(0.5 in.)(4.3152 in.)(4.3152 in./2)  9.3105 in.3

 max

(5,800 lb)(9.3105 in.3 )  (39.8157 in.4 )(2  0.5 in.)  1,356 psi

Ans.

(b) Maximum compression bending stress: Check two possibilities. First, check the bending stress created by the largest positive moment at the top of the cross section: M y (8,850 lb-ft)(1.6848 in.)(12 in./ft)  x   pos top    4,494 psi Iz 39.8157 in.4 Next, for the largest negative moment, compute the bending stress at the bottom of the cross section: M y (9,839 lb-ft)(  4.3152 in.)(12 in./ft)  x   neg bot    12,796 psi Iz 39.8157 in.4 Therefore, the maximum compression bending stress is: Ans.  comp  12,800 psi (C) (c) Maximum tension bending stress: Check two possibilities. First, check the bending stress created by the largest positive moment at the bottom of the cross section: M y (8,850 lb-ft)(  4.3152 in.)(12 in./ft)  x   pos bot    11,510 psi Iz 39.8157 in.4 Next, for the largest negative moment, compute the bending stress at the top of the cross section: M y (9,839 lb-ft)(1.6848 in.)(12 in./ft)  x   neg top    4,996 psi Iz 39.8157 in.4 Therefore, the maximum tension bending stress is: Ans.  tens  11,510 psi (T)


9.38 A cantilever beam supports the loads shown in Fig. P9.38a. The cross-sectional dimensions of the shape are shown in Fig. P9.38b. Determine: (a) the maximum vertical shear stress. (b) the maximum compression bending stress. (c) the maximum tension bending stress.

Fig. P9.38a

Fig. P9.38b

Solution Maximum shear force magnitude: Vmax = 5 kN Maximum positive bending moment: Mpos = 2.00 kN-m Maximum negative bending moment: Mneg = −1.50 kN-m

Centroid location in y direction: Shape flange stem

Width b (mm) 100 6

Height h (mm) 8 92

Area Ai (mm2) 800 552 1,352


yi Ai (mm3) 76,800 25,392 102,192


y

yi Ai Ai



102,192 mm3  75.5858 mm 1,352 mm 2  24.4142 mm


Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (mm ) (mm) (mm4) flange 4,266.67 20.4142 333,391.69 stem 389,344.00 -29.5858 483,176.36 Moment of inertia about the z axis (mm4) =

IC + d²A (mm4) 337,658.35 872,520.36 1,210,178.71

(a) Maximum vertical shear stress: At neutral axis: Qmax  (6 mm)(75.5858 mm)(75.5858 mm/2)  17,139.640 mm3

 max 

(5,000 N)(17,139.640 mm3 )  11.80 MPa (1,210,178.71 mm4 )(6 mm)

Ans.

(b) Maximum compression bending stress: Check two possibilities. First, check the bending stress created by the largest positive moment at the top of the cross section: M pos ytop (2.00  106 N-mm)(24.4142 mm) x     40.348 MPa Iz 1,210,178.71 mm4 Next, for the largest negative moment, compute the bending stress at the bottom of the cross section: M y (1.50  106 N-mm)(  75.5858 mm)  x   neg bot    93.688 MPa Iz 1,210,178.71 mm4 Therefore, the maximum compression bending stress is: Ans.  comp  93.7 MPa (C) (c) Maximum tension bending stress: Check two possibilities. First, check the bending stress created by the largest positive moment at the bottom of the cross section: M pos ybot (2.00  106 N-mm)(  75.5858 mm) x     124.917 MPa Iz 1,210,178.71 mm4 Next, for the largest negative moment, compute the bending stress at the top of the cross section: M y (1.50  106 N-mm)(24.4142 mm)  x   neg top    30.261 MPa Iz 1,210,178.71 mm4 Therefore, the maximum tension bending stress is: Ans.  tens  124.9 MPa (T)


9.39 A simply supported beam fabricated from pultruded reinforced plastic supports the loads shown in Fig. P9.39a. The cross-sectional dimensions of the plastic wide-flange shape are shown in Fig. P9.39b. (a) Determine the magnitude of the maximum shear force in the beam. (b) At the section of maximum shear force, determine the shear stress magnitude in the cross section at point H, which is located 2 in. above the bottom surface of the wide-flange shape. (c) At the section of maximum shear force, determine the magnitude of the maximum horizontal shear stress in the cross section. (d) Determine the magnitude of the maximum compression bending stress in the beam. Where along the span does this stress occur?

Fig. P9.39a

Fig. P9.39b

Solution Section properties: (4 in.)(8 in.)3 (3.625 in.)(7.25 in.) 3 Iz   12 12 4  55.5493 in. (a) Maximum shear force magnitude: V = 3,664 lb

Ans.

(b) Shear stress magnitude at H: QH  (4 in.)(0.375 in.)(3.8125 in.)

 (0.375 in.)(1.625 in.)(2.8125 in.)  7.4326 in.3

H 

(3,664 lb)(7.4326 in.3 ) (55.5493 in.4 )(0.375 in.)

 1,307 psi

Ans.


(c) Maximum horizontal shear stress magnitude: At neutral axis: Qmax  (4 in.)(0.375 in.)(3.8125 in.)  (0.375 in.)(3.625 in.)(1.8125 in.)  8.1826 in.3

 max

(3,664 lb)(8.1826 in.3 )   1, 439 psi (55.5493 in.4 )(0.375 in.)

(d) Maximum compression bending stress: My (7,719 lb-ft)(4 in.)(12 in./ft) H     6,669.965 psi  6,670 psi (C) I 55.5493 in.4

Ans.

Ans.

This stress occurs at 5.86 ft to the right of A.


9.40 A wooden beam is fabricated from one 2 × 10 and two 2 × 4 pieces of dimension lumber to form the I-beam cross section shown in Fig. P9.40. The flanges of the beam are fastened to the web with nails that can safely transmit a force of 120 lb in direct shear. If the beam is simply supported and carries a 1,000-lb load at the center of a 12-ft span, determine: (a) the horizontal force transferred from each flange to the web in a 12in. long segment of the beam. (b) the maximum spacing s (along the length of the beam) required for the nails. (c) the maximum horizontal shear stress in the I-beam. Fig. P9.40

Solution Moment of inertia about the z axis: Shape Width b Height h (in.) (in.) top flange 4 2 web 2 10 bottom flange 4 2

d = yi – y (in.) 6.000 0.000 –6.000

IC (in.4) 2.667 166.667 2.667

d²A (in.4) 288.000 0.000 288.000

IC + d²A (in.4) 290.667 166.667 290.667

Moment of inertia about the z axis (in.4) =

748.000

Maximum shear force For P = 1,000 lb, V = P/2 = 500 lb (a) Horizontal force transferred from each flange (in a 12-in. length): Q (4 in.)(2 in.)(6 in.) 48 in.3

(500 lb)(48 in.3 ) 748 in.4

VQ I

q FH

(32.086 lb/in.)(12 in.)

32.086 lb/in. 385 lb

Ans.

(b) Maximum nail spacing: q s nfVf s

(c) Maximum horizontal shear stress: Qmax (4 in.)(2 in.)(6 in.) (2 in.)(5 in.)(2.5 in.) max

(500 lb)(73 in.3 ) (748 in.4 )(2 in.)

24.4 psi

nfVf q

(1 nail)(120 lb/nail) 32.086 lb/in.

3.74 in.

Ans.

73 in.3

Ans.


9.41 A wooden beam is fabricated from one 2 × 10 and two 2 × 4 pieces of dimension lumber to form the I-beam cross section shown in Fig. P9.41. The I-beam will be used as a simply supported beam to carry a concentrated load P at the center of a 20-ft span. The wood has an allowable bending stress of 1,200 psi and an allowable shear stress of 90 psi. The flanges of the beam are fastened to the web with nails that can safely transmit a force of 120 lb in direct shear. (a) If the nails are uniformly spaced at an interval of s = 4.5 in. along the span, what is the maximum concentrated load P that can be supported by the beam? Demonstrate that the maximum bending and shear stresses produced by P are acceptable. (b) Determine the magnitude of load P that produces the allowable bending stress in the span (i.e., b = 1,200 psi). What nail spacing s is required to support this load magnitude? Demonstrate that the maximum horizontal shear stresses produced by P are acceptable. Fig. P9.41

Solution Moment of inertia about the z axis: Shape Width b Height h (in.) (in.) top flange 4 2 web 2 10 bottom flange 4 2

d = yi – y (in.) 6.000 0.000 –6.000

IC (in.4) 2.667 166.667 2.667

d²A (in.4) 288.000 0.000 288.000

Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 290.667 166.667 290.667 748.000

(a) Maximum concentrated load P: Q (4 in.)(2 in.)(6 in.) 48 in.3 q s nfVf q

q

s

(1 nail)(120 lb/nail) 4.5 in.

26.667 lb/in.

VQ I V

V

nfVf

P 2

qI Q

(26.667 lb/in.)(748 in.4 ) 48 in.3 Pmax

831 lb

415.556 lb Ans.


Check maximum bending and shear stresses: (831.1 lb)(240 in.)/4 (7 in.) Mc ( PL / 4)c max Iz Iz 748 in.4

Qmax max

(2 in.)(5 in.)(2.5 in.) (4 in.)(2 in.)(6 in.) VQ I zt

( P / 2)Qmax I zt

(831.1 lb / 2)(73 in.3 ) (748 in.4 )(2 in.)

466.7 psi 1,200 psi

OK

73 in.3 20.3 psi

90 psi

OK

(b) Magnitude of load P that produces the allowable bending stress in the span: Mc 1, 200 psi x I (1, 200 psi)(748 in.4 ) M 128, 228.566 lb-in. 7 in. PL M max 4 4M max 4(128, 228.566 lb-in.) Pmax 2,137.143 lb 2,140 lb L (20 ft)(12 in./ft)

Ans.

Required nail spacing s: Pmax 2,137.143 lb Vmax 1,068.571 lb 2 2 VQ (1,068.571 lb)(48 in.3 ) q 68.571 lb/in. I 748 in.4 q s nfVf

s

nfVf q

(1 nail)(120 lb) 68.571 lb/in.

Ans.

1.750 in.

Check maximum shear stresses: VQ ( P / 2)Qmax (2,137.143 lb / 2)(73 in.3 ) max I zt I zt (748 in.4 )(2 in.)

52.143 psi 90 psi

OK


9.42 A wooden box beam is fabricated from four boards, which are fastened together with nails, as shown in Fig. P9.42b. The nails are installed at a spacing of s = 125 mm (Fig. P9.42a), and each nail can provide a resistance of Vf = 500 N. In service, the box beam will be installed so that bending occurs about the z axis. Determine the maximum shear force V that can be supported by the box beam based on the shear capacity of the nailed connections.

Fig. P9.42a

Fig. P9.42b

Solution Moment of inertia Iz: (200 mm)(300 mm)3 Iz 12

(120 mm)(250 mm)3 12

First moment of area Q: Q (200 mm)(25 mm)(137.5 mm)

293,750,000 mm 4

687,500 mm3

Shear flow q based on nail shear force: q s nfVf

q

nfVf s

(2 nails)(500 N/nail) 125 mm

Maximum shear force V: VQ q Iz q V Iz Q

8 N/mm

(8 N/mm)(293,750,000 mm4 ) 687,500 mm3

3,418 N

3.42 kN

Ans.


9.43 A wooden box beam is fabricated from four boards, which are fastened together with screws, as shown in Fig. P9.43b. Each screw can provide a resistance of 800 N. In service, the box beam will be installed so that bending occurs about the z axis, and the maximum shear force in the beam will be 9 kN. Determine the maximum permissible spacing interval s for the screws (see Fig. P9.43a).

Fig. P9.43a

Fig. P9.43b


(140 mm)(150 mm)3 12

First moment of area Q: Q (140 mm)(50 mm)(100 mm)

208,020,833 mm 4

700,000 mm3

Shear flow q based on beam shear force V: VQ (9,000 N)(750,000 mm3 ) q 30.285 N/mm Iz 208,020,833 mm4 Maximum spacing interval s: q s nfVf s

nfVf q

(2 screws)(800 N/screw) 30.285 N/mm

52.8 mm

Ans.


9.44 A wooden beam is fabricated by nailing together three pieces of dimension lumber, as shown in Fig. P9.44a. The cross-sectional dimensions of the beam are shown in Fig. P9.44b. The beam must support an internal shear force of V = 750 lb. (a) Determine the maximum horizontal shear stress in the cross section for V = 750 lb. (b) If each nail can provide 100 lb of horizontal resistance, determine the maximum spacing s for the nails. (c) If the three boards were connected by glue instead of nails, what minimum shear strength would be necessary for the glued joints?

Fig. P9.44a

Fig. P9.44b

Solution Centroid location in y direction: Shape left board flange board right board y

yi Ai Ai

Width b (in.) 2 4 2 184 in.3 40 in.2

4.6 in.


Area Ai (in.2) 16 8 16 40


yi Ai (in.3) 64 56 64 184


3.4 in. (from top of shape to centroid) Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (in. ) (in.) (in.4) left board 85.3333 –0.60 5.7600 flange board 2.6667 2.40 46.0800 right board 85.3333 –0.60 5.7600 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 91.0933 48.7467 91.0933 230.9333

(a) Maximum horizontal shear stress: At neutral axis: Qmax 2(2 in.)(4.6 in.)(4.6 in./2) 42.32 in.3 max

(750 lb)(42.32 in.3 ) (230.9333 in.4 )(4 in.)

34.4 psi

Ans.

(b) Shear flow q based on beam shear force V: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Q

(4 in.)(2 in.)(3.4 in. 1 in.) 19.20 in.3

q

VQ Iz

(750 lb)(19.20 in.3 ) 230.9333 in.4

62.356 lb/in.

Maximum spacing interval s: q s nfVf s

nfVf q

(2 nails)(100 lb/nail) 62.356 lb/in.

(c) Glue joint shear stress: 62.356 lb/in. 15.59 psi 2(2 in.)

3.21 in.

Ans.

Ans.


9.45 A wooden beam is fabricated by gluing four dimension lumber boards, each 40-mm wide and 90-mm deep, to a 32 × 400 plywood web, as shown in Fig. P9.45. Determine the maximum allowable shear force and the maximum allowable bending moment that this section can carry if the allowable bending stress is 6 MPa, the allowable shear stress in the plywood is 640 kPa, and the allowable shear stress in the glued joints is 250 kPa.

Fig. P9.45


(80 mm)(220 mm)3 12

Maximum allowable bending moment: Mc I I (6 N/mm 2 )(526,346,667 mm 4 ) M max c 200 mm

526,346,667 mm 4

15,790, 400 N-mm

15.79 kN-m

Ans.

Maximum allowable shear force: Consider maximum shear stress, which occurs at the neutral axis: Q (32 mm)(200 mm)(100 mm) 2(40 mm)(90 mm)(200 mm 90 mm/2) 1,756,000 mm3 VQ It

V

It Q

(0.640 N/mm)(526,346,667 mm 4 )(32 mm) 1,756,000 mm3

6,138.7 N

6.14 kN

(a)

Consider shear stress in glue joints: Q (40 mm)(90 mm)(200 mm 90 mm/2) 558,000 mm3 The shear stress in the glue joints can be found from the shear flow across the glue joint divided by the width of the glue joint; thus, q VQ / I glue tglue tglue (0.250 N/mm)(526,346,667 mm 4 )(90 mm) 21, 224 N 21.2 kN Q 558,000 mm3 Compare results (a) and (b) to find that the maximum allowable shear force for the section is: Vmax 6.14 kN V

glue

I tglue

(b)

Ans.


9.46 A wooden beam is fabricated from one 2 × 12 and two 2 × 10 dimension lumber boards to form the double-tee cross section shown in Fig. P9.46. The beam flange is fastened to the stem with nails. Each nail can safely transmit a force of 175 lb in direct shear. The allowable shear stress of the wood is 70 psi. (a) If the nails are uniformly spaced at an interval of s = 4 in. along the span, what is the maximum internal shear force V that can be supported by the double-tee cross section? (b) What nail spacing s would be necessary to develop the full strength of the double-tee shape in shear? (Full strength means that the maximum horizontal shear stress in the double-tee shape equals the allowable shear stress of the wood.) Fig. P9.46

Solution Centroid location in y direction: Shape top flange left stem right stem y

yi Ai Ai

Width b (in.) 12 2 2 464 in.3 64 in.2


7.25 in.

Area Ai (in.2) 24 20 20 64


yi Ai (in.3) 264 100 100 464


4.75 in. (from top of shape to centroid) Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (in. ) (in.) (in.4) left board 8.000 3.750 337.500 flange board 166.667 –2.250 101.250 right board 166.667 –2.250 101.250 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 345.500 267.917 267.917 881.333

(a) Maximum shear force based on capacity of nails at s = 4 in.: Q (12 in.)(2 in.)(4.75 in. 2 in./2) 90.000 in.3 (2 nails)(175 lb/nail) q 87.5 lb/in. 4 in. VQ q I (87.5 lb/in.)(881.333 in.4 ) q Vmax 856.852 lb I Q 90.000 in.3 Maximum shear force based on full shear strength of double tee shape: At neutral axis: Qmax 2(2 in.)(7.25 in.)(7.25 in./2) 105.125 in.3 max

VQ It

Vmax

max

Q

It

(70 psi)(881.333 in.4 )(2 2 in.) 90.000 in.3

2,347.428 lb


Comparison of two Vmax values: The smaller of the two values computed above for Vmax gives the limiting shear force: Vmax 857 lb

Ans.

(b) Nail spacing necessary to develop the full shear strength of the section: Consider the nailed portion of the beam (i.e., only the top flange) to establish the minimum required nail spacing for Vmax = 2,347.428 lb, which is the shear force that produces a maximum horizontal shear stress of 70 psi: V Q (2,347.428 lb)(90.000 in.3 ) q 239.715 lb/in. I 881.333 in.4 q s nfVf smax

nfVf q

(2 nails)(175 lb/nail) 239.715 lb/in.

1.460 in.

Ans.


9.47 A box beam is fabricated from two plywood webs that are secured to dimension lumber boards at its top and bottom flanges (Fig. P9.47b). The beam supports a concentrated load of P = 5,000 lb at the center of a 15-ft span (Fig. P9.47a). Bolts (⅜-in. diameter) connect the plywood webs and the lumber flanges at a spacing of s = 12 in. along the span. Supports A and C can be idealized as a pin and a roller, respectively. Determine: (a) the maximum horizontal shear stress in the plywood webs. (b) the average shear stress in the bolts. (c) the maximum bending stress in the lumber flanges.

Fig. P9.47a

Fig. P9.47b

Solution Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (in. ) (in.) (in.4) left web 576 0 0 top flange 16 10 1,200 bottom flange 16 –10 1,200 right web 576 0 0 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 576 1,216 1,216 576 3,584

Maximum shear force: For P = 5,000 lb, V = P/2 = 2,500 lb Maximum bending moment: For P = 5,000 lb, M = PL/4 = 18,750 lb-ft = 225,000 lb-in. (a) Maximum horizontal shear stress (in plywood webs): Qmax (3 in.)(4 in.)(10 in.)

2(0.5 in.)(12 in.)(6 in.) 192 in.3 max

VQ It

(2,500 lb)(192 in.3 ) (3,584 in.4 )(2 0.5 in.)

133.9 psi

Ans.


(b) Bolt shear stress: Consider the dimension lumber boards that comprise the top flange. Qflange (3 in.)(4 in.)(10 in.) 120 in.3 (2,500 lb)(120 in.3 ) 83.7054 lb/in. I 3,584 in.4 Determine the force carried by one bolt: q s nfVf VQflange

q

Vf

qs nf

(83.7054 lb/in.)(12 in.) 1 bolt

1,004.4643 lb/bolt

The bolt cross-sectional area is: (0.375 in.)2 0.110447 in.2 4 Each bolt acts in double shear; therefore, the shear stress in each bolt is: 1,004.4643 lb/bolt 4,547.284 psi 4,550 psi bolt 2(0.110447 in.2 ) Abolt

(c) Maximum bending stress in lumber flanges: M c (225,000 lb-in.)(12 in.) 753 psi I 3,584 in.4

Ans.

Ans.


9.48 A box beam is fabricated from two plywood webs that are secured to dimension lumber boards at its top and bottom flanges (Fig. P9.48b). The lumber has an allowable bending stress of 1,500 psi. The plywood has an allowable shear stress of 300 psi. The ⅜-in. diameter bolts have an allowable shear stress of 6,000 psi, and they are spaced at intervals of s = 9 in. The beam span is L = 15 ft (Fig. P9.48a). Support A can be assumed to be pinned and support C can be idealized as a roller. (a) Determine the maximum load P that can be applied to the beam at midspan. (b) Report the bending stress in the lumber, the shear stress in the plywood, and the average shear stress in the bolts at the load P determined in part (a).

Fig. P9.48a

Fig. P9.48b

Solution Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (in. ) (in.) (in.4) left web 576 0 0 top flange 16 10 1,200 bottom flange 16 –10 1,200 right web 576 0 0 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 576 1,216 1,216 576 3,584

Maximum shear force: V = P/2 Maximum bending moment: M = PL/4 (a) Determine maximum load P: Consider maximum bending stress: Mc 1,500 psi I (1,500 psi)(3,584 in.4 ) M 448,000 lb-in. 12 in. PL M 448,000 lb-in. 4 4(448,000 lb-in.) Pmax 9,956 lb (15 ft)(12 in./ft)

(a)


Consider maximum horizontal shear stress (in plywood webs): Q (3 in.)(4 in.)(10 in.) 2(0.5 in.)(12 in.)(6 in.) 192 in.3 VQ 300 psi It (300 psi)(3,584 in.4 )(2 0.5 in.) 192 in.3

Vmax

5,600 lb

P 2

Vmax Pmax

2Vmax

(b)

2(5,600 lb) 11, 200 lb

Consider bolt shear stress: The bolt cross-sectional area is: (0.375 in.)2

Abolt

0.110447 in.2

4 Each bolt acts in double shear; therefore, the maximum shear force that can be carried by one bolt is: Vbolt 2(0.110447 in.2 )(6,000 psi) 1,325.364 lb Determine the shear flow that can be allowed based on the bolt shear stress: q s nfVf

nf Vf

(1 bolt)(1,325.364 lb/bolt) 147.263 lb/in. s 9 in. Consider the dimension lumber boards that comprise the top flange. Qflange (3 in.)(4 in.)(10 in.) 120 in.3 q

VQflange

q

I qI Qflange

Vmax Vmax Pmax

(147.263 lb/in.)(3,584 in.4 ) 120 in.3

4,398.245 lb

P 2 2Vmax

2(4,398.245 lb) 8,796 lb

Compare the three values obtained for Pmax in Eqs. (a), (b), and (c) to find Pmax 8,796 lb 8.80 kips

(c)

Ans.


(b) Bending stress in lumber flanges for Pmax: PL (8,796 lb)(15 ft) M 32,985 lb-ft 4 4 M c (32,985 lb-ft)(12 in.)(12 in./ft) 1,325 psi I 3,584 in.4 Maximum shear stress in plywood webs: P 8,796 lb V 4,398 lb 2 2 VQ (4,398 lb)(192 in.3 ) 236 psi I t (3,584 in.4 )(2 0.5 in.) Bolt shear stress: 6,000 psi bolt

Ans.

Ans.

Ans.


9.49 A wooden beam is fabricated from three boards, which are fastened together with screws, as shown in Fig. P9.49b. The screws are uniformly spaced along the span of the beam at intervals of 150 mm (see Fig. P9.49a). In service, the beam will be positioned so that bending occurs about the z axis. The maximum bending moment in the beam is Mz = −4.50 kN-m, and the maximum shear force in the beam is Vy = −2.25 kN. Determine: (a) the magnitude of the maximum horizontal shear stress in the beam. (b) the shear force in each screw. (c) the magnitude of the maximum bending stress in the beam.

Fig. P9.49a

Fig. P9.49b

Solution Centroid location in y direction: Shape left board bottom board right board y

yi Ai Ai

Width b (mm) 40 140 40 1, 408,000 mm3 20,000 mm 2

Height h (mm) 180 40 180 70.4 mm

Area Ai (mm2) 7,200 5,600 7,200 20,000



109.6 mm (from top of shape to centroid) Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (mm ) (mm) (mm4) left board 19,440,000.00 19.60 2,765,952.00 bottom board 746,666.67 –50.40 14,224,896.00 right board 19,440,000.00 19.60 2,765,952.00 Moment of inertia about the z axis (mm4) =

(a) Maximum horizontal shear stress: At neutral axis: Qmax 2(40 mm)(109.6 mm)(109.6 mm/2) max

VQ It

yi Ai (mm3) 648,000 112,000 648,000 1,408,000

IC + d²A (mm4) 22,205,952.00 14,971,562.67 22,205,952.00 59,383,466.67

480,486.4 mm3

(2, 250 N)(480,486.4 mm3 ) (59,383,466.67 mm 4 )(2 40 mm)

0.2276 MPa

228 kPa

Ans.


(b) Shear force in each screw Consider bottom board, which is held in place by two screws: Q (140 mm)(40 mm)(70.4 mm 40 mm/2) 282, 240 mm3 V Q (2, 250 N)(282, 240 mm 3 ) I 59,383,466.67 mm 4 q s nfVf q

Vf

qs nf

10.6939 N/mm

(10.6939 N/mm)(150 mm) 2 screws

802 N per screw

(c) Maximum bending stress: Mz y ( 4.50 106 N-mm)(109.6 mm) x Iz 59,383,466.67 mm4

8.31 MPa (T)

Ans.

Ans.


9.50 A wooden beam is fabricated by bolting together three members, as shown in Fig. P9.50a. The cross-sectional dimensions are shown in Fig. P9.50b. The 8-mm-diameter bolts are spaced at intervals of s = 200 mm along the x axis of the beam. If the internal shear force in the beam is V = 7 kN, determine the shear stress in each bolt.

Fig. P9.50a

Fig. P9.50b

Solution Centroid location in y direction: Shape left board center board right board

y

yi Ai Ai

Width b (mm) 40 40 40

3,636,000 mm3 19,200 mm 2

Height h (mm) 90 300 90

Area Ai (mm2) 3,600 12,000 3,600 19,200


189.375 mm


110.625 mm


Moment of inertia about the z axis: Shape IC (mm4) left board 2,430,000 center board 90,000,000 right board 2,430,000

yi Ai (mm3) 918,000 1,800,000 918,000 3,636,000

d = yi – y d²A IC + d²A 4 (mm) (mm ) (mm4) 65.625 15,503,906.25 17,933,906.25 -39.375 18,604,687.50 108,604,687.50 65.625 15,503,906.25 17,933,906.25 4 Moment of inertia about the z axis (mm ) = 144,472,500.00


Shear force in each bolt Consider left board, which is held in place by the bolt: Q (40 mm)(90 mm)(110.625 mm 90 mm/2) 236, 250 mm3

V Q (7,000 N)(236, 250 mm3 ) 11.4468 N/mm I 144,472,500 mm 4 Note that this value of q is the shear flow that must be transmitted by one surface of the bolt cross section. The cross-sectional area of the bolt is: q

Abolt

(8 mm)2

50.2655 mm2

4 Relate the shear flow and the bolt shear stress with Eq. (9.14): q s n f f Af f

qs n f Af

(11.4468 N/mm)(200 mm) (1 bolt surface)(50.2655 mm 2 )

45.5 MPa

Ans.


9.51 A wooden beam is fabricated by bolting together three members, as shown in Fig. P9.51a. The cross-sectional dimensions are shown in Fig. P9.51b. The allowable shear stress of the wood is 850 kPa, and the allowable shear stress of the 10mm-diameter bolts is 40 MPa. Determine: (a) the maximum internal shear force V that the cross section can withstand based on the allowable shear stress in the wood. (b) the maximum bolt spacing s required to develop the internal shear force computed in part (a).

Fig. P9.51a

Fig. P9.51b

Solution Centroid location in y direction: Shape left board center board right board

y

yi Ai Ai

Width b (mm) 40 40 40

3,636,000 mm3 19,200 mm 2

Height h (mm) 90 300 90

Area Ai (mm2) 3,600 12,000 3,600 19,200


189.375 mm


110.625 mm


Moment of inertia about the z axis: Shape IC (mm4) left board 2,430,000 center board 90,000,000 right board 2,430,000

yi Ai (mm3) 918,000 1,800,000 918,000 3,636,000

d = yi – y d²A IC + d²A (mm) (mm4) (mm4) 65.625 15,503,906.25 17,933,906.25 -39.375 18,604,687.50 108,604,687.50 65.625 15,503,906.25 17,933,906.25 4 Moment of inertia about the z axis (mm ) = 144,472,500.00


Consider maximum horizontal shear stress: Q (40 mm)(189.375 mm)(189.375 mm/2)

VQ It

850 kPa

717,257.813 mm3

0.850 MPa

(0.850 N/mm 2 )(144,472,500 mm4 )(40 mm) 717,257.813 mm3

Vmax

6,848.4 N

6.85 kN

Ans.

Maximum bolt spacing Consider left board, which is held in place by the bolt: Q (40 mm)(90 mm)(110.625 mm 90 mm/2) 236, 250 mm3

V Q (6,848.4 N)(236, 250 mm3 ) 11.1989 N/mm I 144,472,500 mm 4 Note that this value of q is the shear flow that must be transmitted by one surface of the bolt cross section. The cross-sectional area of the bolt is: q

(10 mm)2

Abolt

78.5398 mm2

4 Relate the shear flow and the bolt shear stress with Eq. (9.14): q s n f f Af s

nf

f

q

Af

(1 bolt surface)(40 N/mm 2 )(78.5398 mm 2 ) 11.1989 N/mm

281 mm

Ans.


9.52 A cantilever flexural member is fabricated by bolting two identical coldrolled steel channels back-to-back, as shown in Fig. P9.52a. The cantilever beam has a span of L = 1,600 mm and supports a concentrated load of P = 600 N. The cross-sectional dimensions of the built-up shape are shown in Fig. P9.52b. The effect of the rounded corners can be neglected in determining the section properties for the built-up shape. (a) If 4-mm-diameter bolts are installed at intervals of s = 75 mm, determine the shear stress produced in the bolts. (b) If the allowable average shear stress in the bolts is 96 MPa, determine the minimum bolt diameter required if a spacing of s = 400 mm is used.


Solution Centroid location in y direction for the upper channel shown in Figure 9.52b: yi Shape Width b Height h Area Ai (from z axis) (mm) (mm) (mm2) (mm) left element 3 40 120 20 center element 59 3 177 1.5 right element 3 40 120 20 417 3 yi Ai 5,065.5 mm y 12.1475 mm Ai 417 mm2

yi Ai (mm3) 2400 265.5 2400 5065.5

Note: y is measured from the z axis to the centroid of the upper channel shown in Figure 9.52b. Moment of inertia (both channels): (3 mm)(40 mm)3 (65 mm 2(3 mm))(3 mm)3 I 2 2 3 3

257,062 mm4

Shear flow: Q (12.1475 mm)(417 mm 2 ) 5,065.51 mm3

q

VQ I

(600 N)(5,065.51 mm3 ) 11.8232 N/mm 257,062 mm4

Bolt area: Abolt

4

(4 mm)2

12.5664 mm 2


(a) Bolt shear stress: q s n f f Af f

qs n f Af

(11.8232 N/mm)(75 mm) (1 bolt surface)(12.5664 mm 2 )

70.6 MPa

Ans.

(b) Minimum bolt diameter for s = 400 mm: q s n f f Af Af

qs nf f

Abolt

D2

Dbolt

4

(11.8232 N/mm)(400 mm) (1 bolt surface)(96 N/mm 2 )

49.2633 mm 2

49.2633 mm 2

4(49.2633 mm 2 )

7.92 mm

Ans.


9.53 A W310 × 60 steel beam (see Appendix B) in an existing structure is to be strengthened by adding a 250 mm wide by 16 mm thick cover plate to its lower flange, as shown in Fig. P9.53. The cover plate is attached to the lower flange by pairs of 24-mm-diameter bolts spaced at intervals of s along the beam span. Bending occurs about the z centroidal axis. (a) If the allowable bolt shear stress is 96 MPa, determine the maximum bolt spacing interval s required to support an internal shear force in the beam of V = 50 kN. (b) If the allowable bending stress is 150 MPa, determine the allowable bending moment for the existing W310 × 60 shape, the allowable bending moment for the W310 × 60 with the added cover plate, and the percentage increase in moment capacity that is gained by adding the cover plate.

Fig. P9.53

Solution Centroid location in y direction: Shape

Width b (mm) 250

W310 × 60 cover plate

1, 292,850 mm3 11,550 mm 2

yi Ai

y

Ai

Height h (mm) 16 111.935 mm

Area Ai (mm2) 7,550 4,000 11,550


yi Ai (mm3) 1,260,850 32,000 1,292,850


206.065 mm (from top of shape to centroid) Moment of inertia about the z axis: d = yi – y Shape IC d²A IC + d²A 4 4 (mm ) (mm) (mm ) (mm4) W310 × 60 128,000,000 55.065 22,892,764 150,892,764 cover plate 85,333.33 –103.935 43,209,937 43,295,270 4 Moment of inertia about the z axis (mm ) = 194,188,035

(a) Maximum bolt spacing Consider the cover plate, which is connected to the W310 × 60 shape with two bolts: Q (250 mm)(16 mm)(111.935 mm 16 mm/2) 415,740 mm3

V Q (50,000 N)(415,740 mm3 ) 107.0457 N/mm I 194,188,035 mm4 The cross-sectional area of a 24-mm-diameter bolt is: q

(24 mm)2 452.389 mm2 4 Relate the shear flow and the bolt shear stress with Eq. (9.14): q s n f f Af Abolt

s

nf

f

q

Af

(2 bolts)(96 N/mm 2 )(452.389 mm 2 ) 107.0457 N/mm

811 mm

Ans.


(b) Allowable bending moment for W310 × 60 shape (without cover plate): Mc 150 MPa I (150 N/mm 2 )(128,000,000 mm 4 ) M allow 127,152, 318 N-mm 127.2 kN-m (302 mm/2) Allowable bending moment for W310 × 60 shape (with cover plate): Mc 150 MPa I (150 N/mm 2 )(194,188,035 mm 4 ) M allow 141,354,452 N-mm 141.4 kN-m (206.065 mm) Percentage increase in moment capacity: 141,354, 452 N-mm 127,152,318 N-mm % increase (100%) 127,152,318 N-mm

11.17%

Ans.

Ans.

Ans.


9.54 A W410 × 60 steel beam (see Appendix B) is simply supported at its ends and carries a concentrated load P at the center of a 7-m span. The W410 × 60 shape will be strengthened by adding two 250 mm wide by 16 mm thick cover plate to its flanges, as shown in Fig. P9.54. Each cover plate is attached to its flange by pairs of 20-mm-diameter bolts spaced at intervals of s along the beam span. The allowable bending stress is 150 MPa, the allowable average shear stress in the bolts is 96 MPa, and bending occurs about the z centroidal axis. (a) Based on the 150 MPa allowable bending stress, determine the maximum concentrated load P that may be applied at the center of a 7-m span for a W410 × 60 steel beam with two cover plates. (b) For the internal shear force V associated with the concentrated load P determined in part (a), compute the maximum spacing interval s required for the bolts that attach the cover plates to the flanges.

Fig. P9.54

Solution Moment of inertia about the z axis (with cover plates): d = yi – y Shape IC d²A 4 (mm ) (mm) (mm4) top cover plate 85,333.33 211 178,084,000 W410 × 60 216,000,000 0 0 bottom cover plate 85,333.33 –211 178,084,000 Moment of inertia about the z axis (mm4) =

IC + d²A (mm4) 178,169,333 216,000,000 178,169,333 572,338,666

Maximum shear force: V = P/2 Maximum bending moment: M = PL/4


Consider W410 × 60 with two cover plates: Mc 150 MPa I (150 N/mm 2 )(572,338,666 mm 3 ) M allow (438 mm/2) PL M 392.013 kN-m 4 4(392.013 kN-m) Pmax 224.007 kN 7m

392,012,785 N-mm

392.013 kN-m

224 kN

Ans.

(b) Maximum bolt spacing Consider top cover plate, which is held in place by two bolts: Q (250 mm)(16 mm)(406 mm/2 16 mm/2) 844,000 mm3

(223,418 N/2)(844,000 mm3 ) 572,338,666 mm4

VQ I

q

165.166 N/mm

Note that this value of q is the shear flow that must be transmitted across two bolt surfaces. The crosssectional area of the bolt is: Abolt

4

(20 mm) 2

314.159 mm 2

Relate the shear flow and the bolt shear stress with Eq. (9.14): q s n f f Af s

nf

f

q

Af

(2 bolt surfaces)(96 N/mm 2 )(314.159 mm 2 ) 165.166 N/mm

365 mm

Ans.


9.55 A W410 × 60 steel beam (see Appendix B) is simply supported at its ends and carries a concentrated load of P = 420 kN at the center of a 7-m span. The W410 × 60 shape will be strengthened by adding two 250 mm wide by 16 mm thick cover plate to its flanges, as shown in Fig. P9.55. Each cover plate is attached to its flange by pairs of bolts spaced at intervals of s = 250 mm along the beam span. The allowable average shear stress in the bolts is 96 MPa, and bending occurs about the z centroidal axis. Determine the minimum required diameter for the bolts.

Fig. P9.55

Solution Moment of inertia about the z axis (with cover plates): d = yi – y Shape IC d²A 4 (mm ) (mm) (mm4) top cover plate 85,333.33 211 178,084,000 W410 × 60 216,000,000 0 0 bottom cover plate 85,333.33 –211 178,084,000 Moment of inertia about the z axis (mm4) =

IC + d²A (mm4) 178,169,333 216,000,000 178,169,333 565,618,666

Maximum shear force: V = P/2 = 420 kN/2 = 210 kN

Minimum bolt diameter Consider top cover plate, which is held in place by two bolts: Q (250 mm)(16 mm)(406 mm/2 16 mm/2) 844,000 mm3

q

VQ I

(210,000 N)(844,000 mm3 ) 572,338,666 mm4

309.677 N/mm


Relate the shear flow and the required bolt area with Eq. (9.14). Note that the shear flow will be transmitted by means of two fasteners. q s n f f Af Af

qs nf f

(309.677 N/mm)(250 mm) 2(96 N/mm 2 )

403.225 mm 2

Use the minimum required cross-sectional area of the bolt to calculate the minimum bolt diameter:

Abolt

D2

4 Dmin

403.225 mm2 22.658 mm

22.7 mm

Ans.


9.56 A W310 × 60 steel beam (see Appendix B) has a C250 × 45 channel bolted to the top flange, as shown in Fig. P9.56. The beam is simply supported at its ends and carries a concentrated load of 100 kN at the center of a 6-m span. Pairs of 24-mm-diameter bolts are spaced at intervals of s along the beam. If the allowable average shear stress in the bolts must be limited to 125 MPa, determine the maximum spacing interval s for the bolts. Fig. P9.56

Solution Centroid location in y direction: Shape W310 × 60 C250 × 45

y

yi Ai Ai

yi (from bottom) (mm) 151 302 + 17.1 – 16.5 = 302.6

Area Ai (mm2) 7,550 5,680 13,230 2,858,818 mm3 13,230 mm 2

216.086 mm 103.014 mm

Moment of inertia about the z axis: Shape IC (mm4) W310 × 60 128,000,000 C250 × 45 1,640,000

yi Ai (mm3) 1,140,050 1,718,768 2,858,818


d = yi – y d²A (mm) (mm4) –65.086 31,983,215 86.5140 42,512,938 Moment of inertia about the z axis (mm4) =

IC + d²A (mm4) 159,983,315 44,152,938 204,136,153

Maximum shear force: V = P/2 = 100 kN/2 = 50 kN Shear flow through the bolts Consider the C250 × 45 shape, which is connected to the W310 × 60 shape with two bolts: Q (5,680 mm 2 )(302.6 mm 216.086 mm)

491,399.5 mm3 q

VQ I (50,000 N)(491,399.5 mm3 ) 204,136,153 mm 4 120.361 N/mm

The cross-sectional area of a single 24-mm-diameter bolt is:


Abolt

4

(24 mm) 2

452.389 mm 2

Use Eq. (9.13) to determine the maximum spacing s: q s n f f Af s

nf

f

q

Af

(2 bolts)(125 N/mm2 )(452.389 mm 2 ) 120.361 N/mm

939.653 mm

940 mm

Ans.


9.57 A W310 × 60 steel beam (see Appendix B) has a C250 × 45 channel bolted to the top flange, as shown in Fig. P9.57. The beam is simply supported at its ends and carries a concentrated load of 90 kN at the center of an 8-m span. If pairs of bolts are spaced at 600-mm intervals along the beam, determine: (a) the shear force carried by each of the bolts. (b) the bolt diameter required if the average shear stress in the bolts must be limited to 75 MPa. Fig. P9.57

Solution Centroid location in y direction: Shape W310 × 60 C250 × 45

y

yi Ai Ai

yi (from bottom) (mm) 151 302 + 17.1 – 16.5 = 302.6

Area Ai (mm2) 7,550 5,680 13,230 2,858,818 mm3 13,230 mm 2

216.086 mm 103.014 mm

Moment of inertia about the z axis: Shape IC (mm4) W310 × 60 128,000,000 C250 × 45 1,640,000

yi Ai (mm3) 1,140,050 1,718,768 2,858,818


d = yi – y d²A (mm) (mm4) –65.086 31,983,215 86.5140 42,512,938 Moment of inertia about the z axis (mm4) =

IC + d²A (mm4) 159,983,315 44,152,938 204,136,153

Maximum shear force: V = P/2 = 90 kN/2 = 45 kN (a) Shear force in each bolt Consider the C250 × 45 shape, which is connected to the W310 × 60 shape with two bolts: Q (5,680 mm 2 )(302.6 mm 216.086 mm)

491,399.5 mm3 q

VQ I (45,000 N)(491,399.5 mm3 ) 204,136,153 mm 4 108.325 N/mm

Relate the shear flow and the bolt shear force with Eq. (9.13): Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

qs Vf

nfVf qs nf

(108.325 N/mm)(600 mm) 2 bolts

(b) Required bolt diameter Vbolt Vbolt Abolt Abolt allow

Abolt

4

2 Dbolt

Dbolt

32, 497 N 75 N/mm 2

32, 497 N

32.5 kN per bolt

Ans.

433.299 mm 2

433.299 mm 2 4(433.299 mm 2 )

23.5 mm

Ans.


10.1 For the loading shown, use the doubleintegration method to determine (a) the equation of the elastic curve for the cantilever beam, (b) the deflection at the free end, and (c) the slope at the free end. Assume that EI is constant for each beam. Fig. P10.1

Solution Integration of moment equation: d 2v EI 2 M ( x) M0 dx dv EI M 0 x C1 dx M 0 x2 EI v C1 x C2 2

(a) (b)

Boundary conditions: dv 0 at x 0 dx v 0 at x 0 Evaluate constants: From Eq. (a), C1 = 0. From Eq. (b), C2 = 0 (a) Elastic curve equation:

EI v

M 0 x2 2

M 0 x2 2 EI

v

Ans.

(b) Deflection at the free end:

vB

M 0 ( L) 2 2 EI

M 0 L2 2 EI

(c) Slope at the free end: dv M 0 ( L) B dx B EI

Ans.

M0L EI

Ans.



Solution Integration of moment equation: d 2v wx 2 EI 2 M ( x) dx 2 3 dv wx EI C1 dx 6 wx 4 EI v C1 x C2 24

(a) (b)

Boundary conditions: dv 0 at x L dx v 0 at x L Evaluate constants: Substitute x = L and dv/dx = 0 into Eq. (a) to determine C1: w( L)3 wL3 EI (0) C1 C1 6 6 Substitute x = L and v = 0 into Eq. (b) to determine C2: w( L) 4 wL4 wL4 EI (0) C1 ( L) C2 C2 24 24 6 (a) Elastic curve equation: wx 4 wL3 x wL4 w EI v v x4 24 6 8 24EI

C2

4L3 x 3L4

wL4 8

Ans.

(b) Deflection at the free end: w 24 EI

vA

(0)4

4 L3 (0) 3L4

3wL4 24 EI

wL4 8EI

Ans.

(c) Slope at the free end: dv dx

A A

w(0)3 6 EI

wL3 6 EI

wL3 6 EI

Ans.



Solution Integration of moment equation: d 2v w0 x3 EI 2 M ( x) dx 6L 4 dv w0 x EI C1 dx 24 L w0 x 5 EI v C1 x C2 120 L

(a) (b)

Boundary conditions: dv 0 at x L dx v 0 at x L Evaluate constants: Substitute x = L and dv/dx = 0 into Eq. (a) to determine C1: w0 ( L)4 w0 L3 EI (0) C1 C1 24 L 24 Substitute x = L and v = 0 into Eq. (b) to determine C2: w0 ( L)5 w0 L5 w0 L3 EI (0) C1 ( L) C2 ( L) C2 120 L 120 L 24 w0 L4 w0 L4 w0 L4 C2 120 24 30 (a) Elastic curve equation: w0 x5 w0 L3 w0 L4 EI v x 120 L 24 30

v

w0 x5 120 L EI

5L4 x 4 L5

Ans.

(b) Deflection at the free end: w0 (0)5 120 L EI

vA

4

5

5L (0) 4 L

w0 L4 30 EI

Ans.


A A

w0 (0)4 24 L EI

w0 L3 24 EI

w0 L3 24 EI

Ans.


10.4 For the beam and loading shown in Fig. P10.4, use the double-integration method to determine (a) the equation of the elastic curve for segment AB of the beam, (b) the deflection at B, and (c) the slope at A. Assume that EI is constant for the beam. Fig. P10.4

Solution Integration of moment equation: d 2v P EI 2 M ( x) x dx 2 dv Px 2 EI C1 dx 4 EI v

Px3 12

(a)

C1 x C2

(b)

Boundary conditions: v 0 at x 0 dv L 0 at x dx 2 Evaluate constants: Substitute x = L/2 and dv/dx = 0 into Eq. (a) to determine C1: P ( L / 2) 2 PL2 EI (0) C1 C1 4 16 Substitute x = 0 and v = 0 into Eq. (b) to determine C2: P (0)3 PL2 (0) EI (0) C2 C2 0 12 16 (a) Elastic curve equation: P x3 PL2 x EI v 12 16

Px 3L2 48EI

v

4 x2

(0

x

L ) 2

Ans.

(b) Deflection at B: P( L / 2) 3L2 48EI

vB

4

L 2

2

PL3 48EI

Ans.

(c) Slope at A: dv dx

A A

P(0) 2 4 EI

PL2 16 EI

PL2 16 EI

Ans.


10.5 For the beam and loading shown in Fig. P10.5, use the double-integration method to determine (a) the equation of the elastic curve for the beam, (b) the slope at A, (c) the slope at B, and (d) the deflection at midspan. Assume that EI is constant for the beam. Fig. P10.5

Solution Beam FBD: Fy Ay

Ay MA

By By

By L M 0 By

0 0

M0 L

and

Ay

M0 L

Moment equation: Ma

a

M ( x)

Ay x M 0

M ( x)

M0

M ( x)

M0 x M0 L

0

M0x L

Integration of moment equation: d 2v M0x EI 2 M ( x) M 0 dx L 2 dv M0x EI M0x C1 dx 2L M 0 x 2 M 0 x3 EI v C1 x C2 2 6L

(a) (b)

Boundary conditions: v 0 at x 0 v 0 at x L Evaluate constants: Substitute x = 0 and v = 0 into Eq. (b) to determine C2: M 0 (0) 2 M 0 (0)3 EI (0) C1 (0) C2 2 6L Substitute x = L and v = 0 into Eq. (b) to determine C1:

C2

0


M 0 ( L) 2 M 0 ( L)3 EI (0) C1 ( L) 2 6L M 0L M 0L M 0L C1 6 2 3 (a) Elastic curve equation: M 0 x 2 M 0 x3 M 0 Lx EI v 2 6L 3

(b) Slope at A: dv M 0 (0) A dx A (c) Slope at B: dv M 0 ( L) B dx B EI

M 0 (0)2 2L EI

M0L 3EI

M 0 ( L) 2 2 L EI

M 0L 3EI

M0x 2 x 6 L EI

v

3Lx 2 L2

M0L 3EI

Ans.

Ans.

M0 6 L 3L 2L 6 EI

M0L 6 EI

Ans.

(d) Deflection at midspan: vx

L/2

M 0 ( L / 2) 6 L EI

L 2

2

3L

L 2

2 L2

M 0 L2 16 EI

Ans.


10.6 For the beam and loading shown in Fig. P10.6, use the double-integration method to determine (a) the equation of the elastic curve for the beam, (b) the maximum deflection, and (c) the slope at A. Assume that EI is constant for the beam. Fig. P10.6

Solution Moment equation:

wLx wx 2 0 a 2 2 wx 2 wLx M ( x) 2 2 Integration of moment equation: d 2v wx 2 wLx EI 2 M ( x) dx 2 2 3 2 dv wx wLx EI C1 dx 6 4 wx 4 wLx3 EI v C1 x C2 24 12 Ma

M ( x)

(a) (b)

Boundary conditions: v 0 at x 0 v 0 at x L Evaluate constants: Substitute x = 0 and v = 0 into Eq. (b) to determine C2: w(0) 4 wL(0)3 EI (0) C1 (0) C2 C2 0 24 12 Substitute x = L and v = 0 into Eq. (b) to determine C1: w( L) 4 wL( L)3 w( L)4 w( L)4 EI (0) C1 ( L) C1 24 12 24 L 12 L (a) Elastic curve equation: wx 4 wLx3 wL3 x wx EI v v x3 2Lx 2 L3 24 12 24 24EI (b) Maximum deflection: At x = L/2: vmax

w( L / 2) 24 EI

L 2

3

L 2L 2

2 2

L

wL L3 48 EI 8

L3 2

2

L

5wL4 384 EI

wL3 24

Ans.

Ans.


A A

w(0)3 6 EI

wL(0) 2 4 EI

wL3 24 EI

wL3 24 EI

Ans.


10.7 For the beam and loading shown in Fig. P10.7, use the double-integration method to determine (a) the equation of the elastic curve for segment AB of the beam, (b) the deflection midway between the two supports, (c) the slope at A, and (d) the slope at B. Assume that EI is constant for the beam. Fig. P10.7


MA

By L By

By

P

0

3L P 0 2 3P and 2

Ay

P 2

Moment equation: P x 0 2 Integration of moment equation: d 2v Px EI 2 M ( x) dx 2 2 dv Px EI C1 dx 4 Px 3 EI v C1 x C2 12 Ma

a

M ( x)

M ( x)

Px 2

(a) (b)

Boundary conditions: v 0 at x 0 v 0 at x L Evaluate constants: Substitute x = 0 and v = 0 into Eq. (b) to determine C2: P(0)3 EI (0) C1 (0) C2 C2 0 12 Substitute x = L and v = 0 into Eq. (b) to determine C1: P ( L )3 PL2 EI (0) C1 ( L) C1 12 12


(a) Elastic curve equation for segment AB of the beam: Px3 PL2 x Px EI v v L2 x 2 12 12 12 EI

Ans.

(b) Deflection at midspan: vx

L/2

P( L / 2) 2 L 12 EI

L 2

2

PL 3L 24 EI 4

PL3 32 EI

Ans.


A A

P(0)2 4 EI

PL2 12 EI

P( L) 2 4 EI

PL2 12 EI

PL2 12 EI

Ans.

(d) Slope at B: dv dx

B B

PL2 6 EI

Ans.


10.8 For the beam and loading shown in Fig. P10.8, use the double-integration method to determine (a) the equation of the elastic curve for segment BC of the beam, (b) the deflection midway between B and C, and (c) the slope at C. Assume that EI is constant for the beam. Fig. P10.8

Solution Beam FBD: M B PL C y (4 L) P (5L)

Fy

Cy

P

By

Cy

By

P

2P

0

0

Moment equation: M a a M ( x) By x P( L x)

M ( x)

M ( x) Px P( L x) 0

PL

Integration of moment equation: d 2v EI 2 M ( x) PL dx dv EI PLx C1 dx PLx 2 EI v C1 x C2 2

(a) (b)

Boundary conditions: v 0 at x 0 v 0 at x 4L Evaluate constants: Substitute x = 0 and v = 0 into Eq. (b) to determine C2: PL(0) 2 EI (0) C1 (0) C2 C2 0 2 Substitute x = 4L and v = 0 into Eq. (b) to determine C1: PL(4 L) 2 EI (0) C1 (4 L) C1 2 PL2 2


(a) Elastic curve equation for segment BC of the beam: PLx 2 PLx EI v 2PL2 x v 4L x 2 2EI

Ans.

(b) Deflection at midspan:

vx

2L

PL(2 L) 4 L ( 2 L) 2 EI

2 PL3 EI

Ans.

(c) Slope at C: dv dx C

C

PL(4 L) EI

2 PL2 EI

2 PL2 EI

Ans.


10.9 For the beam and loading shown in Fig. P10.9, use the double-integration method to determine (a) the equation of the elastic curve for segment AB of the beam, (b) the deflection midway between A and B, and (c) the slope at B. Assume that EI is constant for the beam. Fig. P10.9

Solution Beam FBD: MA

Fy

wL2 5L P By L 2 4 wL 5 P By 2 4 Ay By wL P 0 Ay

wL 2

0

P 4

Moment equation:

wx 2 wx 2 A x M ( x ) a y 2 2 2 wx wLx Px M ( x) 2 2 4 Integration of moment equation: d 2v wx 2 wLx Px EI 2 M ( x) dx 2 2 4 3 2 2 dv wx wLx Px EI C1 dx 6 4 8 wx 4 wLx3 Px3 EI v C1 x C2 24 12 24 Ma

M ( x)

wLx 2

Px 4

0

(a) (b)

Boundary conditions: v 0 at x 0 v 0 at x L Evaluate constants: Substitute x = 0 and v = 0 into Eq. (b) to determine C2: w(0) 4 wL(0)3 P(0)3 EI (0) C1 (0) C2 24 12 24 Substitute x = L and v = 0 into Eq. (b) to determine C1: w( L) 4 wL( L)3 P( L)3 EI (0) C1 ( L) 24 12 24

C2

C1

0 wL3 24

PL2 24


(a) Elastic curve equation for segment AB of the beam: wx 4 wLx3 Px3 wL3 x PL2 x EI v 24 12 24 24 24 wx Px v x3 2 Lx 2 L3 x 2 L2 24 EI 24 EI

Ans.

(b) Deflection at midspan: vx

L/2

3

w( L / 2) 24 EI

L 2

5wL4 384 EI

PL3 64 EI

2L

L 2

2

L3

P ( L / 2) 24 EI

L 2

2

L2

Ans.

(c) Slope at B: dv dx

B B

w( L)3 6 EI

wL( L) 2 4 EI

P ( L) 2 8EI

wL3 24 EI

PL2 24 EI

wL3 24 EI

PL2 12 EI

Ans.


10.10 For the beam and loading shown in Fig. P10.10, use the double-integration method to determine (a) the equation of the elastic curve for segment AC of the beam, (b) the deflection at B, and (c) the slope at A. Assume that EI is constant for the beam. Fig. P10.10

Solution Beam FBD:

MA

Ay

3L C y (2 L) 2 9wL 4 C y w(3L) 0

Ay

3wL

w(3L) Cy

Fy

0

9wL 4

3wL 4

M ( x)

Ay x wx

x 2

M ( x)

wx 2 2

Moment equation:

Ma

a

M ( x)

3wL x x wx 4 2

0

3wLx 4

Integration of moment equation: d 2v wx 2 3wLx EI 2 M ( x) dx 2 4 3 2 dv wx 3wLx EI C1 dx 6 8 wx 4 3wLx3 EI v C1 x C2 24 24

(a) (b)

Boundary conditions: v 0 at x 0 v 0 at x 2L Evaluate constants: Substitute x = 0 and v = 0 into Eq. (b) to determine C2: w(0) 4 3wL(0)3 EI (0) C1 (0) C2 24 24 Substitute x = 2L and v = 0 into Eq. (b) to determine C1: w(2 L) 4 3wL(2 L)3 EI (0) C1 (2 L) 24 24 8wL3 12wL3 wL3 C1 24 24 6

C2

0


(a) Elastic curve equation for segment AC of the beam: wx 4 3wLx 3 wL3 x wx 3 EI v x 3Lx 2 4 L3 24 24 6 24 wx v x3 3Lx 2 4 L3 24 EI

Ans.

(b) Deflection at B:

w( L) ( L) 3 24 EI

vB

3L( L)

2

wL4 12 EI

3

4L

Ans.


A A

w(0)3 6 EI

3wL(0) 2 8EI

wL3 6 EI

wL3 6 EI

Ans.


10.11 For the simply supported steel beam [E = 200 GPa; I = 129 × 106 mm4] shown in Fig. P10.11, use the double-integration method to determine the deflection at B. Assume L = 4 m, P = 60 kN, and w = 40 kN/m.

Fig. P10.11


wL Cy

Fy

Ay Ay

L L P C y ( L) 2 2 wL P 2 2 C y w( L) P 0 wL 2

0

P 2

Moment equation: wx 2 M a a M ( x) Ay x M ( x) 2 wx 2 wLx Px M ( x) 2 2 2

wx 2 2

Integration of moment equation: d 2v wx 2 wLx Px EI 2 M ( x) dx 2 2 2 3 2 2 dv wx wLx Px EI C1 dx 6 4 4 wx 4 wLx3 Px3 EI v C1 x C2 24 12 12

wLx 2

Px 2

0

(a) (b)

Boundary conditions: v 0 at x 0 dv L 0 at x dx 2 Evaluate constants: Substitute x = 0 and v = 0 into Eq. (b) to determine C2 = 0. Next, substitute x = L/2 and dv/dx = 0 into Eq. (b) to determine C1: w( L / 2)3 wL( L / 2) 2 P( L / 2) 2 EI (0) C1 6 4 4 wL3 wL3 PL2 wL3 PL2 C1 48 16 16 24 16 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Elastic curve equation: wx 4 wLx 3 Px 3 wL3 x PL2 x EI v 24 12 12 24 16 wx Px v x 3 2 Lx 2 L3 3L2 24 EI 48 EI

4x2

Deflection at B: At x = L/2: 5wL4 PL3 vB 384 EI 48 EI Let E = 200 GPa, I = 129 × 106 mm4, w = 40 kN/m, P = 60 kN, and L = 4 m. 5(40 N/mm)(4,000 mm) 4 (60,000 N)(4,000 mm)3 vB 384(200,000 N/mm 2 )(129 106 mm 4 ) 48(200,000 N/mm2 )(129 106 mm4 ) 5.1680 mm 3.1008 mm

8.27 mm

Ans.


10.12 For the cantilever steel beam [E = 200 GPa; I = 129 × 106 mm4] shown in Fig. P10.12, use the double-integration method to determine the deflection at A. Assume L = 2.5 m, P = 50 kN, and w = 30 kN/m. Fig. P10.12

Solution Moment equation: Ma

a

M ( x)

wx 2 2

Px

0

M ( x)

wx 2 2

Px

Integration of moment equation: d 2v wx 2 EI 2 M ( x) Px dx 2 dv wx3 Px 2 EI C1 dx 6 2 wx 4 Px 3 EI v C1 x C2 24 6

(a) (b)

Boundary conditions: v 0 at x L dv 0 at x L dx Evaluate constants: Substitute x = L and dv/dx = 0 into Eq. (a) to determine C1: w( L)3 P( L) 2 wL3 PL2 EI (0) C1 C1 6 2 6 2 Substitute x = L and v = 0 into Eq. (b) to determine C2: w( L)4 P( L)3 wL3 PL2 wL4 EI (0) ( L) ( L) C2 24 6 6 2 24 4 3 wL PL C2 8 3

wL4 6

PL3 6

PL3 2

C2

Elastic curve equation: wx 4 wL3 x wL4 Px 3 PL2 x PL3 EI v 24 6 8 6 2 3 w P v x 4 4 L3 x 3L4 x 3 3L2 x 2 L3 24 EI 6 EI Deflection at A: w vA (0) 4 24 EI

4 L3 (0) 3L4

P (0)3 3L2 (0) 2 L3 6 EI

3wL4 24 EI

PL3 3EI


Let E = 200 GPa, I = 129 × 106 mm4, w = 30 kN/m, P = 50 kN, and L = 2.5 m. 3(30 N/mm)(2,500 mm)4 (50,000 N)(2,500 mm)3 vA 24(200,000 N/mm2 )(129 106 mm4 ) 3(200,000 N/mm2 )(129 106 mm4 ) 5.6777 mm 10.0937 mm

= 15.77 mm

Ans.


10.13 For the cantilever steel beam [E = 200 GPa; I = 129 × 106 mm4] shown in Fig. P10.13, use the double-integration method to determine the deflection at B. Assume L = 3 m, M0 = 70 kN-m, and w = 15 kN/m. Fig. P10.13


Ma

a

M ( x) M ( x)

w( L x)2 M0 0 2 w( L x)2 M0 2

Integration of moment equation: d 2v w( L x)2 w 2 EI 2 M ( x) M0 L 2 Lx dx 2 2 dv wL2 x wLx 2 wx3 EI M 0 x C1 dx 2 2 6 wL2 x 2 wLx3 wx 4 M 0 x 2 EI v C1 x C2 4 6 24 2

x2

M0

wL2 2

wLx

wx 2 2

M0

(a) (b)

Boundary conditions: v 0 at x 0 dv 0 at x 0 dx Evaluate constants: Substitute x = 0 and dv/dx = 0 into Eq. (a) to determine C1 = 0. Next, substitute x = 0 and v = 0 into Eq. (b) to determine C2 = 0. Elastic curve equation: wL2 x 2 wLx3 wx 4 M 0 x 2 EI v 4 6 24 2 w v x 4 4 Lx3 6 L2 x 2 24 EI

M 0 x2 2 EI

Deflection at B: w vB ( L) 4 24 EI

M 0 ( L) 2 2 EI

4 L( L)3

6 L2 ( L) 2

wL4 8EI

M 0 L2 2 EI


Let E = 200 GPa, I = 129 × 106 mm4, w = 15 kN/m, M0 = 70 kN-m, and L = 3 m. (15 N/mm)(3,000 mm)4 (70 kN-m)(1,000 N/kN)(1,000 mm/m)(3,000 mm)2 vB 8(200,000 N/mm 2 )(129 106 mm 4 ) 2(200,000 N/mm 2 )(129 106 mm4 ) 5.8866 mm 12.2093 mm

= 18.10 mm

Ans.


10.14 For the cantilever steel beam [E = 200 GPa; I = 129 × 106 mm4] shown in Fig. P10.14, use the double-integration method to determine the deflection at A. Assume L = 2.5 m, P = 50 kN-m, and w0 = 90 kN/m. Fig. P10.14


Ma

a

M ( x) M ( x)

w0 x x ( x) 2L 3 w0 x3 6L

Px

0

Px

Integration of moment equation: d 2v w0 x 3 EI 2 M ( x) Px dx 6L dv w0 x 4 Px 2 EI C1 dx 24 L 2 w0 x5 Px3 EI v C1 x C2 120 L 6

(a) (b)

Boundary conditions: v 0 at x L dv 0 at x L dx Evaluate constants: Substitute x = L and dv/dx = 0 into Eq. (a) to determine C1: w0 ( L) 4 P( L) 2 w0 L3 PL2 EI (0) C1 C1 24 L 2 24 2 Substitute x = L and v = 0 into Eq. (b) to determine C2: w0 ( L)5 P( L)3 w0 L3 PL2 w0 L4 w0 L4 EI (0) ( L) ( L) C2 120 L 6 24 2 120 24 4 3 w0 L PL C2 30 3 Elastic curve equation: w0 x5 Px3 w0 L3 PL2 EI v x x 120 L 6 24 2 w0 v x5 5L4 x 4 L5 120 L EI

PL3 6

PL3 2

C2

w0 L4 PL3 30 3 P x3 3L2 x 2 L3 6 EI


Deflection at A: Let E = 200 GPa, I = 129 × 106 mm4, w0 = 90 kN/m, P = 50 kN, and L = 2.5 m. w0 P vA (0)5 5 L4 (0) 4 L5 (0)3 3L2 (0) 2 L3 120 L EI 6 EI w0 L4 30 EI

PL3 3EI (90 N/mm)(2,500 mm) 4 30(200,000 N/mm 2 )(129 106 mm 4 ) 4.5422 mm 10.0937 mm = 14.64 mm

(50,000 N)(2,500 mm)3 3(200,000 N/mm 2 )(129 106 mm 4 )

Ans.


10.15 For the beam and loading shown in Fig. P10.15, use the double-integration method to determine (a) the equation of the elastic curve for the cantilever beam AB, (b) the deflection at the free end, and (c) the slope at the free end. Assume that EI is constant for each beam. Fig. P10.15


w0 L 2 L 2 3

MA MA

Fy

Ay Ay

0

w0 L2 3 w0 L 0 2 w0 L 2

Moment equation:

Ma

a

M ( x) M A M ( x)

w0 L2 3

w0 x x ( x) 2L 3 w0 x x ( x) 2L 3

Ay x w0 L ( x) 2

w0 x3 w0 Lx w0 L2 M ( x) 6L 2 3 Integration of moment equation: d 2v w0 x3 w0 Lx w0 L2 EI 2 M ( x) dx 6L 2 3 4 2 2 dv w0 x w0 Lx w0 L x EI C1 dx 24 L 4 3 w0 x5 w0 Lx 3 w0 L2 x 2 EI v C1 x C2 120 L 12 6

0

(a) (b)

Boundary conditions: v 0 at x 0 dv 0 at x 0 dx Evaluate constants: Substitute x = 0 and v = 0 into Eq. (b) to determine C2 = 0. Next, substitute x = 0 and dv/dx = 0 into Eq. (b) to determine C1 = 0.


(a) Elastic curve equation: w0 x5 w0 Lx3 w0 L2 x 2 EI v 120 L 12 6

w0 x 2 x3 10 L2 x 20 L3 120 L EI

v

Ans.

(b) Deflection at the free end: w0 ( L)5 10 L2 ( L)3 120 L EI

vB

3

20 L ( L)

2

11w0 L4 120 EI

Ans.


B B

w0 ( L)4 24 L

w0 L( L)2 4

w0 L2 ( L) 3

w0 L3 24

6w0 L3 24

8w0 L3 24

w0 L3 8EI

Ans.


10.16 For the beam and loading shown in Fig. P10.16, use the double-integration method to determine (a) the equation of the elastic curve for the cantilever beam AB, (b) the deflection at the free end, and (c) the slope at the free end. Assume that EI is constant for each beam. Fig. P10.16


MA MA

Fy

Ay Ay

w0 L L 2 3

0

w0 L2 6 w0 L 0 2 w0 L 2

Moment equation: Ma

M ( x)

a

M ( x)

w0 ( L x)3 2L 3

w0 ( L x )3 6L w0 3 ( L 3L2 x 3Lx 2 6L w0 L2 w0 Lx w0 x 2 6 2 2

0

x3 ) w0 x 3 6L

Integration of moment equation: d 2v w0 x3 w0 x 2 w0 Lx w0 L2 EI 2 M ( x) dx 6L 2 2 6 4 3 2 2 dv w0 x w0 x w0 Lx w0 L x EI C1 dx 24 L 6 4 6 w0 x5 w0 x 4 w0 Lx3 w0 L2 x 2 EI v C1 x C2 120 L 24 12 12

(a) (b)

Boundary conditions: v 0 at x 0 dv 0 at x 0 dx


Evaluate constants: Substitute x = 0 and v = 0 into Eq. (b) to determine C2 = 0. Next, substitute x = 0 and dv/dx = 0 into Eq. (b) to determine C1 = 0. (a) Elastic curve equation: w0 x5 w0 x 4 w0 Lx3 EI v 120 L 24 12

w0 x 2 120 L EI

v

x3

w0 L2 x 2 12 5Lx 2 10 L2 x 10 L3

Ans.

(b) Deflection at the free end: w0 ( L)5 120 L EI

vB

5L( L)4 10 L2 ( L)3 10 L3 ( L) 2

4w0 L5 120 L EI

w0 L4 30 EI

Ans.


B B

w0 ( L)4 24 L EI

w0 ( L)3 6 EI

w0 L( L) 2 4 EI

w0 L2 ( L) 6 EI

w0 L3 24 EI

Ans.


10.17 For the beam and loading shown in Fig. P10.17, use the double-integration method to determine (a) the equation of the elastic curve for the cantilever beam, (b) the deflection at B, (c) the deflection at the free end, and (d) the slope at the free end. Assume that EI is constant for the beam. Fig. P10.17


wL L 2 2

MA MA

Fy

Ay Ay

L 4

0

3wL2 8 wL 0 2 wL 2

Consider beam segment AB (0 ≤ x ≤ L/2) Moment equation:

Ma

a

M ( x) M A M ( x)

Ay x

3wL2 8

M ( x)

3wL2 8

wL x 2

0

wLx 2

Integration of moment equation: d 2v 3wL2 wLx EI 2 M ( x) dx 8 2 2 2 dv 3wL x wLx EI C1 dx 8 4 3wL2 x 2 wLx3 EI v C1 x C2 16 12

(a) (b)

Boundary conditions: v 0 at x 0 dv 0 at x 0 dx Evaluate constants: Substitute x = 0 and v = 0 into Eq. (b) to determine C2 = 0. Next, substitute x = 0 and dv/dx = 0 into Eq. (b) to determine C1 = 0.


Elastic curve equation for beam segment AB: 3wL2 x 2 wLx 3 EI v 16 12 v

wLx 2 9L 4 x 48 EI

Slope at B: Let x = L/2 dv 3wL2 ( L / 2) B dx B 8EI

(0

wL( L / 2) 2 4EI

Deflection at B: Let x = L/2 wL( L / 2)2 L vB 9L 4 48EI 2

x

L / 2)

wL3 8EI

7wL4 192EI

Consider beam segment BC (L/2 ≤ x ≤ L) Moment equation:

Mb

b

M ( x) M A M ( x)

M ( x)

Ay x

3wL2 8 2

w x 2

L 2

wx 2 2

wLx

w x 2

L 2

wL x 2

w x 2

wL x 2

3wL2 8

2

L 2

x x

0

wL2 2

Integration of moment equation: d 2v wx 2 wL2 EI 2 M ( x) wLx dx 2 2 3 2 2 dv wx wLx wL x EI C3 dx 6 2 2 wx 4 wLx3 wL2 x 2 EI v C3 x C4 24 6 4 Continuity conditions: 7 wL4 v at 192 EI dv wL3 at dx 8EI

2

(c) (d)

L 2 L 2


Evaluate constants: Substitute the slope continuity condition into Eq. (c) for x = L/2 and solve for C3: dv w( L / 2)3 wL( L / 2) 2 wL2 ( L / 2) wL3 EI C3 dx 6 2 2 8 3 wL C3 48 Next, substitute the deflection continuity condition into Eq. (d) for x = L/2 and solve for C4 w( L / 2)4 wL( L / 2)3 wL2 ( L / 2)2 wL3 7 wL4 EI v ( L / 2) C4 24 6 4 48 192 4 wL C4 384 Elastic curve equation for beam segment BC: wx 4 wLx3 wL2 x 2 wL3 x wL4 EI v 24 6 4 48 384 w v 16 x 4 64 Lx3 96 L2 x 2 8L3 x 384 EI

L4

(L / 2

x

L)

(a) Elastic curve equations for entire beam: v

wLx 2 9L 4 x 48EI

v

w 16 x 4 384EI

(0

64Lx3

96L2 x 2

x

Ans.

L / 2)

8L3 x

L4

(L / 2

x

L)

Ans.

(b) Deflection at B: vB

7 wL4 192 EI

Ans.

(c) Deflection at free end of cantilever: vC

w 16( L)4 384 EI

64 L( L)3

96 L2 ( L) 2

(d) Slope at free end of cantilever: dv 8w( L)3 24wL( L) 2 24 wL2 ( L) EI dx 48 48 48 dv dx C

C

7 wL3 48 EI

8L3 ( L)

wL3 48

L4

41wL4 384 EI

Ans.

7 wL3 48

Ans.


10.18 For the beam and loading shown in Fig. P10.18, use the double-integration method to determine (a) the equation of the elastic curve for the beam, and (b) the deflection at B. Assume that EI is constant for the beam. Fig. P10.18

Solution Beam FBD:

MA

wL L 0 2 4 wL 8 wL Cy 0 2 3wL 8

C y ( L) Cy

Fy

Ay Ay

Consider beam segment AB (0 ≤ x ≤ L/2) Moment equation:

Ma

a

M ( x) M ( x)

wx 2 Ay x M ( x) 2 wx 2 3wLx 2 8

wx 2 2

3wL x 8

0

Integration of moment equation: d 2v wx 2 3wLx EI 2 M ( x) dx 2 8 3 2 dv wx 3wLx EI C1 dx 6 16 wx 4 wLx 3 EI v C1 x C2 24 16

(a) (b)

Boundary conditions: v 0 at x 0 Evaluate constants: Substitute x = 0 and v = 0 into Eq. (b) to determine C2 = 0. Slope at B: Let x = L/2 in Eq. (a). dv w( L / 2)3 3wL( L / 2) 2 EI EI B dx B 6 16

C1

wL3 48

3wL3 64

C1

5wL3 192

C1

(c)


Deflection at B: Let x = L/2 in Eq. (b).

w( L / 2)4 24

EI vB

wL( L / 2)3 16

C1 ( L / 2)

wL4 384

wL4 128

C1L 2

wL4 192

C1L 2

(d)

Consider beam segment BC (L/2 ≤ x ≤ L) Moment equation: Mb

M ( x) C y ( L

b

x)

wL (L 8 wLx 8

M ( x)

wL wL2 ( L x) 8 8 Integration of moment equation: d 2v wLx wL2 EI 2 M ( x) dx 8 8 2 2 dv wLx wL x EI C3 dx 16 8 wLx3 wL2 x 2 EI v C3 x C4 48 16 M ( x)

x)

0

(e) (f)

Boundary conditions: v 0 at x L Evaluate constants: Substitute x = L and v = 0 into Eq. (f) to find wL( L)3 wL2 ( L) 2 EI (0) C3 ( L) C4 48 16 C3 L C4

wL4 24

(g)

Slope at B: Let x = L/2 in Eq. (e).

dv EI dx

EI

B

B

wL( L / 2)2 16

wL2 ( L / 2) C3 8

wL3 64

wL3 16

C3

3wL3 64

C3

(h)

Deflection at B: Let x = L/2 in Eq. (f).

EI vB

wL( L / 2)3 48

wL2 ( L / 2) 2 16

C3 ( L / 2) C4

5wL4 384

C3 L 2

C4

Continuity conditions: Since the slope at B must be the same for both beam segments, equate Eqs. (c) and (h): 5wL3 3wL3 C1 C3 192 64

(i)

(j)


Further, the deflection at B must be the same for both segments; therefore, equate Eqs. (d) and (i): wL4 C1L 5wL4 C3 L C4 192 2 384 2

(k)

Evaluate constants: Solve Eqs. (g), (j), and (k) simultaneously to determine the values of constants C1, C3, and C4: 9wL3 17 wL3 wL4 C1 C3 C4 384 384 384 (a) Elastic curve equation for beam segment AB: wx 4 wLx 3 9wL3 x EI v 24 16 384 wx v 16 x3 24 Lx 2 9 L3 384 EI (a) Elastic curve equation for beam segment BC: wLx3 wL2 x 2 17 wL3 x wL4 EI v 48 16 384 384 wL v 8 x3 24 Lx 2 17 L2 x L3 384 EI

(0

x

(L / 2

L / 2)

Ans.

x

Ans.

L)

(b) Deflection at B: EI vB

wL4 192

9wL4 768

5wL4 768

vB

5wL4 768EI

Ans.


10.19 For the beam and loading shown in Fig. P10.19, use the double-integration method to determine (a) the equation of the elastic curve for the entire beam, (b) the deflection at C, and (c) the slope at B. Assume that EI is constant for the beam. Fig. P10.19


By (3L) wL 3L 7 wL 6 By wL

By Fy

Ay

L 2

0

0

wL 6

Ay

Consider beam segment AB (0 ≤ x ≤ 3L) Moment equation:

Ma

M ( x)

a

Ay x

wL x 6

M ( x)

0

wLx 6 Integration of moment equation: d 2v wLx EI 2 M ( x) dx 6 2 dv wLx EI C1 dx 12 wLx3 EI v C1 x C2 36 M ( x)

Boundary conditions: v 0 at x 0

(a) (b)

and

v 0 at

x 3L

Evaluate constants: Substitute x = 0 and v = 0 into Eq. (b) to determine C2 = 0. Next, substitute x = 3L and v = 0 into Eq. (b) and solve for C1: wL(3L)3 9wL3 wL3 EI (0) C1 (3L) C1 36 36 4 Slope at B: Let x = 3L in Eq. (a).

EI

dv dx

EI B

B

wL(3L)2 12

wL3 4

wL3 2

(c)


Consider beam segment BC (3L ≤ x ≤ 4L) Moment equation: Mb

w (4 L x) 2 2 w(4 L x) 2 2

M ( x)

b

M ( x)

0

Integration of moment equation: d 2v w(4 L x) 2 EI 2 M ( x) dx 2 3 dv w(4 L x) EI C3 dx 6 w(4 L x) 4 EI v C3 x C4 24

(e) (f)

Boundary conditions: v 0 at x 3L Substitute x = 3L and v = 0 into Eq. (f) to find w(4 L 3L)4 wL4 EI (0) C3 (3L) C4 24 24 4 wL C4 (3L)C3 24

C3 (3L) C4 (g)

Slope at B: Let x = 3L in Eq. (e).

EI

dv dx

EI B

B

w(4L 3L)3 6

C3

wL3 6

C3

Continuity conditions: Since the slope at B must be the same for both beam segments, equate Eqs. (c) and (h): wL3 wL3 2wL3 C3 C3 2 6 3 Backsubstitute this result into Eq. (g) to determine C4: wL4 wL4 2wL3 49wL4 C4 (3L)C3 (3L) 24 24 3 24

(h)

(i)


(a) Elastic curve equation for beam segment AB: wL x3 9wL3 x EI v 36 36 wL x 2 v x 9 L2 (0 x 3L) 36 EI Elastic curve equation for beam segment BC: w(4 L x) 4 2wL3 x 49wL4 EI v 24 3 24 w v (4 L x) 4 16 L3 x 49 L4 24 EI (b) Deflection at C: w vC (4 L 4 L) 4 16 L3 (4 L) 49 L4 24 EI vC

(3L

Ans.

x

w 64 L4 24 EI

5wL4 8 EI

Ans.

4 L)

49 L4

15wL4 24 EI

Ans.

(c) Slope at B: Let x = 3L in Eq. (a). EI

dv dx

EI B

B

wL3 2

dv dx

B B

wL3 2 EI

Ans.


10.20 For the beam and loading shown in Fig. P10.20, use the double-integration method to determine (a) the equation of the elastic curve for the beam, (b) the location of the maximum deflection, and (c) the maximum beam deflection. Assume that EI is constant for the beam. Fig. P10.20

Solution Beam FBD:

MA

By L By

Fy

Ay Ay

w0 L 2 L 0 2 3 w0 L 3 w0 L By 0 2 w0 L 6

Moment equation: Ma

a

M ( x)

w0 x 2 x 2L 3

Ay x

M ( x)

w0 x 2 x 2L 3

w0 Lx 6

M ( x)

w0 x 3 6L

0

w0 Lx 6

Integration of moment equation: d 2v w0 x3 w0 Lx EI 2 M ( x) dx 6L 6 4 2 dv w0 x w0 Lx EI C1 dx 24 L 12 w0 x5 w0 Lx3 EI v C1 x C2 120 L 36

(a) (b)

Boundary conditions: v 0 at x 0 v 0 at x L


Evaluate constants: Substitute x = 0 and v = 0 into Eq. (b) to determine C2 = 0. Next, substitute x = L and v = 0 into Eq. (b) and solve for C1: w0 ( L)5 w0 L( L)3 7 w0 L3 EI (0) C1 ( L) C1 120 L 36 360 (a) Elastic curve equation: w0 x5 w0 Lx3 7 w0 L3 x EI v 120 L 36 360

v

w0 x 3x 4 10 L2 x 2 360 L EI

7 L4

Ans.

(b) Location of maximum deflection: The maximum deflection occurs where the beam slope is zero. Therefore, set the beam slope equation [Eq. (a)] equal to zero: dv w0 x 4 w0 Lx 2 7 w0 L3 EI 0 dx 24 L 12 360 Multiply by −360L/w0 to obtain: 15 x 4 30 L2 x 2 7 L4 0 Solve this equation numerically to obtain: x = 0.51932962236L 0.51933L Ans. (c) Maximum beam deflection: w0 (0.51933L) vmax 3(0.51933L) 4 10 L2 (0.51933L) 2 360 L EI w0 (0.51933) 4.52118 L4 360 EI

(0.0065222) w0 L4 EI

7 L4 0.00652

w0 L4 EI

Ans.


10.21 For the beam and loading shown in Fig. P10.21, integrate the load distribution to determine (a) the equation of the elastic curve for the beam, and (b) the maximum deflection for the beam. Assume that EI is constant for the beam. Fig. P10.21

Solution Integrate the load distribution: d 4v w0 x EI 4 dx L 3 d v w0 x 2 EI 3 C1 dx 2L d 2v w0 x3 EI 2 C1 x C2 dx 6L dv w0 x 4 C1 x 2 EI C2 x C3 dx 24 L 2 w0 x5 C1 x 3 C2 x 2 EI v C3 x C4 120 L 6 2 Boundary conditions and evaluate constants: d 3v at x 0, V EI 3 0 dx d 2v at x 0, M EI 2 0 dx dv w0 ( L) 4 at x L, 0 C3 0 dx 24 L w0 ( L)5 w0 L3 ( L) at x L, v 0 C4 120 L 24 (a) Elastic curve equation: w0 x5 w0 L3 x w0 L4 EI v 120 L 24 30

v

C1

0

C2

0

C3 0

w0 x5 120 LEI

C4

w0 L3 24 w0 L4 30

5L4 x 4 L5

Ans.

(b) Maximum deflection: vmax

w0 (0)5 120 LEI

4

5

5L (0) 4 L

w0 L4 30 EI

Ans.


10.22 For the beam and loading shown in Fig. P10.22, integrate the load distribution to determine (a) the equation of the elastic curve for the beam, and (b) the deflection midway between the supports. Assume that EI is constant for the beam. Fig. P10.22

Solution Integrate the load distribution: d 4v w0 x EI 4 dx L 3 d v w0 x 2 EI 3 C1 dx 2L d 2v w0 x3 EI 2 C1 x C2 dx 6L dv w0 x 4 C1 x 2 EI C2 x C3 dx 24 L 2 w0 x5 C1 x 3 C2 x 2 EI v C3 x C4 120 L 6 2 Boundary conditions and evaluate constants: d 2v at x 0, M EI 2 0 dx d 2v w0 ( L)3 at x L, M EI 2 0 dx 6L at x 0, v 0 at x

L, v

0

w0 ( L)5 120 L

C2 C1 ( L)

C1 C4

w0 Lx 3 36

(a) Elastic curve equation: w0 x5 w0 Lx3 7 w0 L3 x EI v 120 L 36 360

0

C3 x

v

0

0 w0 L 6 0

C3

w0 3x5 10 L2 x3 360 L EI

7 w0 L3 360

7 L4 x

Ans.

(b) Deflection midway between the supports:

vx

L /2

w0 3( L / 2)5 10 L2 ( L / 2)3 7 L4 ( L / 2) 360 LEI

5w0 L4 768EI

Ans.


10.23 For the beam and loading shown in Fig. P10.23, integrate the load distribution to determine (a) the equation of the elastic curve, (b) the deflection at the left end of the beam, and (c) the support reactions By and MB. Assume that EI is constant for the beam. Fig. P10.23

Solution Integrate the load distribution: d 4v w0 x3 EI 4 dx L3 d 3v w0 x 4 EI 3 C1 dx 4 L3 d 2v w0 x5 EI 2 C1 x C2 dx 20 L3 dv w0 x 6 C1 x 2 EI C2 x C3 dx 120 L3 2 w0 x 7 C1 x3 C2 x 2 EI v C3 x C4 840 L3 6 2 Boundary conditions and evaluate constants: d 3v at x 0, V EI 3 0 dx d 2v at x 0, M EI 2 0 dx dv w0 ( L)6 at x L, 0 C3 0 dx 120 L3 w0 ( L)7 w0 L3 ( L) at x L, v 0 840 L3 120 (a) Elastic curve equation: w0 x 7 7 w0 L3 x 6w0 L4 EI v v 840 L3 840 840

C1

0

C2

0

C3 C4

0

C4

w0 x7 840 L3 EI

w0 L3 120 6w0 L4 840

7 L6 x 6 L7

Ans.

(b) Deflection at left end of beam:

vmax

w0 (0)7 7 L6 (0) 6 L7 3 840 L EI

(c) Support reactions By and MB: d 3v w0 ( L)4 w0 L VB EI 3 3 dx x L 4L 4 MB

d 2v EI 2 dx

x L

w0 ( L)5 20 L3

w0 L2 20

6w0 L7 840 L3 EI

By

w0 L4 140 EI

w0 L 4 MB

Ans.

Ans. w0 L2 (cw) 20

Ans.


10.24 For the beam and loading shown in Fig. P10.24, integrate the load distribution to determine (a) the equation of the elastic curve, (b) the deflection midway between the supports, and (c) the support reactions Ay and By. Assume that EI is constant for the beam. Fig. P10.24

Solution Integrate the load distribution: d 4v w0 x3 EI 4 dx L3 d 3v w0 x 4 EI 3 C1 dx 4 L3 d 2v w0 x5 EI 2 C1 x C2 dx 20 L3 dv w0 x 6 C1 x 2 EI C2 x C3 dx 120 L3 2 w0 x 7 C1 x3 C2 x 2 EI v C3 x C4 840 L3 6 2 Boundary conditions and evaluate constants: d 2v at x 0, M EI 2 0 dx d 2v w0 ( L)5 at x L, M EI 2 0 C1 ( L) dx 20 L3 at x 0, v 0 at x

L, v

w0 ( L)7 840 L3

0

w0 L( L)3 120

C2 0

C4

C3 ( L)

(a) Elastic curve equation: w0 x 7 w0 Lx3 6w0 L3 x EI v 840 L3 120 840

C1

v

0

C3

0 w0 L 20 0 6w0 L3 840

w0 x 7 7 L4 x3 6 L6 x 840 L3 EI

Ans.

13w0 L4 5120 EI

Ans.

(b) Deflection midway between the supports: vx

w0 840 L3 EI

L /2

L 2

7 4

7L

L 2

(c) Support reactions Ay and By: d 3v w0 (0)4 w0 L VA EI 3 dx x 0 4L3 20

VB

EI

d 3v dx3

x L

w0 ( L)4 4 L3

w0 L 20

3

6 L6

w0 L 20 4w0 L 20

L 2

Ay

w0 L 20

Ans.

By

w0 L 5

Ans.



Solution Integrate the load distribution: d 4v x EI 4 w0 cos dx 2L 3 d v 2w0 L x EI 3 sin C1 dx 2L d 2v 4w0 L2 x EI 2 cos C1 x C2 2 dx 2L dv 8w0 L3 x C1 x 2 EI sin C2 x C3 3 dx 2L 2 16w0 L4 x C1 x3 C2 x 2 EI v cos C3 x C4 4 2L 6 2 Boundary conditions and evaluate constants: d 3v at x 0, V EI 3 0 C1 0 dx d 2v 4w0 L2 (0) 4w0 L2 at x 0, M EI 2 0 cos C 0 C 2 2 2 2 dx 2L dv 8w0 L3 ( L) 4w0 L2 ( L) 4w0 L3 at x L, 0 sin C 0 C (2 3 3 3 2 3 dx 2L 16w0 L4 ( L) 4w0 L2 ( L) 2 4w0 L3 ( L) at x L, v 0 cos (2 ) C4 0 4 3 2L 2 2 2w0 L4 C4 (4 ) 3 (a) Elastic curve equation: 16w0 L4 x 4w0 L2 x 2 EI v cos 4 2L 2 2

v

w0 x 32 L4 cos 4 2 EI 2L

4w0 L3 3

(2

4 2 L2 x 2

)

2w0 L4

8 L3 x(2

3

(4

)

)

) 4 L4 (4

)

Ans.


(b) Deflection at left end of beam: w0 (0) vA 32 L4 cos 4 4 2 EI 2L

w0 32 L4 4 2 EI

4

4 L (4

2 2

L (0) 2

)

8 L3 (0)(2

) 4 L4 (4

w0 L4 32 4 (4 2 4 EI

)

)

w0 L4 0.1089 EI

Ans.

(c) Support reactions By and MB: d 3v 2w0 L ( L) VB EI 3 sin dx x L 2L MB

d 2v EI 2 dx

4w0 L2 2 x L

( L) cos 2L

2w0 L 4w0 L2 2

By 4w0 L2 2

MB

2w0 L 4w0 L2 2

Ans. (cw)

Ans.


10.26 For the beam and loading shown in Fig. P10.26, integrate the load distribution to determine (a) the equation of the elastic curve, (b) the deflection midway between the supports, (c) the slope at the left end of the beam, and (d) the support reactions Ay and By. Assume that EI is constant for the beam. Fig. P10.26

Solution Integrate the load distribution: d 4v x EI 4 w0 sin dx L 3 d v w0 L x EI 3 cos C1 dx L d 2v w0 L2 x EI 2 sin C1 x C2 2 dx L dv w0 L3 x C1 x 2 EI cos C2 x C3 3 dx L 2 w0 L4 x C1 x 3 C2 x 2 EI v sin C3 x C4 4 L 6 2 Boundary conditions and evaluate constants: d 2v at x 0, M EI 2 0 dx d 2v w0 L2 ( L) at x L, M EI 2 0 sin C1 ( L) 2 dx L at x 0, v 0 at x

L, v

w0 L4

0

4

( L) L

sin

C3 ( L)

0

0

C2

0

C1

0

C4

0

C3

0

(a) Elastic curve equation: w0 L4

EI v

4

x sin L

v

w0 L4 x sin 4 EI L

Ans.

(b) Deflection midway between the supports: vx

w0 L4 ( L / 2) sin 4 EI L

L/2

w0 L4 4 EI

Ans.

(c) Slope at the left end of the beam: EI

dv dx

EI A

w0 L3 A

3

cos

(0) L

w0 L3 3

A

w0 L3 3 EI

Ans.


(d) Support reactions Ay and By: d 3v w0 L (0) VA EI 3 cos dx x 0 L

VB

EI

d 3v dx3

w0 L x L

cos

( L) L

w0 L w0 L

Ay By

w0 L

Ans.

w0 L

Ans.


10.27 For the beam and loading shown in Fig. P10.27, integrate the load distribution to determine (a) the equation of the elastic curve, (b) the deflection midway between the supports, (c) the slope at the left end of the beam, and (d) the support reactions Ay and By. Assume that EI is constant for the beam. Fig. P10.27

Solution Integrate the load distribution: d 4v x EI 4 w0 sin dx 2L 3 d v 2w0 L x EI 3 cos C1 dx 2L d 2v 4w0 L2 x EI 2 sin C1 x C2 2 dx 2L dv 8w0 L3 x C1 x 2 EI cos C2 x C3 3 dx 2L 2 16w0 L4 x C1 x3 C2 x 2 EI v sin C3 x C4 4 2L 6 2 Boundary conditions and evaluate constants: d 2v at x 0, M EI 2 0 dx d 2v 4w0 L2 ( L) at x L, M EI 2 0 sin 2 dx 2L at x 0, v 0 at x

L, v

0

16w0 L4 4

sin

v

2w0 x 24 L4 sin 4 3 EI 2L

C1 ( L)

4w0 L( L)3 6 2

( L) 2L

(a) Elastic curve equation: 16w0 L4 x 4w0 Lx3 EI v sin 4 2L 6 2

C2

2w0 L3 x (24 3 4 2

Lx3

C3 ( L)

2

(24

0

0 4w0 L

C1

0

2

C4

0

C3

2w0 L3 (24 3 4

2

)

) 2

) L3 x

Ans.


(b) Deflection midway between the supports:

vx

2w0 ( L / 2) 24 L4 sin 4 3 EI 2L

L/2

2w0 L4 24sin 3 4 EI 4

2

2

2

(24

8

L 2

L

3 2

(24

) L3

L 2

)

2

4

2w0 L (1.2694611) 3 4 EI 0.0086882

w0 L4 EI

(c) Slope at the left end of the beam: dv 8w0 L3 (0) EI EI A cos 3 dx A 2L 8w0 L3

VB

EI By

2w0 L 2

d 3v dx3

(

4w0 L 2

2

2

w0 L3

)

8 3

4w0 L 2

2

)

16 4

2 3

2

0.026209 w0 L3

Ans.

2w0 L 2

(

2) Ans.

2) 2w0 L

x L

2 w0 L3 (24 3 4

w0 L3 0.0262 EI

(d) Support reactions Ay and By: d 3v 2w0 L (0) VA EI 3 cos dx x 0 2L

Ay

Ans.

2 w0 L(0) 2

2 w0 L3 (24 3 4

3

A

w0 L4 EI

0.00869

cos

( L) 2L

4w0 L 2

4w0 L 2

Ans.



Solution Integrate the load distribution: d 4v x EI 4 w0 sin dx 2L 3 d v 2w0 L x EI 3 cos C1 dx 2L d 2v 4w0 L2 x EI 2 sin C1 x C2 2 dx 2L dv 8w0 L3 x C1 x 2 EI cos C2 x C3 3 dx 2L 2 16w0 L4 x C1 x3 C2 x 2 EI v sin C3 x C4 4 2L 6 2 Boundary conditions and evaluate constants: d 3v 2w0 L (0) at x 0, V EI 3 0 cos C1 0 dx 2L d 2v 4w0 L2 (0) 2w0 L(0) at x 0, M EI 2 0 sin C2 0 2 dx 2L 8w0 L3 dv ( L) 2w0 L( L) 2 at x L, 0 cos C3 0 3 dx 2L 2 16 w0 L4 ( L) 2 w0 L( L)3 w0 L3 ( L) at x L, v 0 sin C4 4 2L 6 2 w0 L4 3 C4 (24 ) 3 4 (a) Elastic curve equation: 16w0 L4 x 2w0 Lx3 EI v sin 4 2L 6

v

w0 x 48L4 sin 4 3 EI 2L

w0 L3 x 3

2w0 L4 (24 3 4

Lx3 3 3 L3 x 2(24

3

2w0 L

C1 C2 C3

0 w0 L3

0

) 3

) L4

Ans.


(b) Deflection at left end of beam: w0 (0) 3 vA 48L4 sin L(0)3 4 3 EI 2L

w0 3 4 EI

3

2(24

3 3 L3 (0) 2(24

3

) L4

w0 L4 0.0479509 EI

4

)L

w0 L4 0.04795 EI

Ans.

(c) Support reactions By and MB: d 3v 2w0 L ( L) VB EI 3 cos dx x L 2L

MB

EI

d 2v dx 2 MB

4w0 L2 2 x L

2w0 L2 2

(

sin

( L) 2L 2)

2w0 L

2w0 L

By

4w0 L2

2 w0 L( L)

2w0 L

Ans.

2w0 L2

2

2w0 L2 2

(

2)

(cw)

Ans.


10.29 For the beam and loading shown, use discontinuity functions to compute the deflection of the beam at D. Assume a constant value of EI = 1,750 kip-ft2 for the beam.

Fig. P10.29

Solution Support reactions: A FBD of the beam is shown to the right. MA (5 kips)(4 ft) (3 kips)(13 ft) C y (10 ft) 0

Cy Fy

Ay

Cy Ay

5.90 kips 5 kips 3 kips

0

2.10 kips


1

1

5 kips x 4 ft

5.90 kips x 10 ft

Shear-force function V(x) and bending-moment function M(x): 0 0 V ( x) 2.10 kips x 0 ft 5 kips x 4 ft 5.90 kips x 10 ft

M ( x)

2.10 kips x 0 ft

1

5 kips x 4 ft

1

5.90 kips x 10 ft

0 1

1

3 kips x 13 ft

3 kips x 13 ft 3 kips x 13 ft

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 1 1 EI 2 M ( x) 2.10 kips x 0 ft 5 kips x 4 ft 5.90 kips x 10 ft dx Integrate the moment function to obtain an expression for the beam slope: dv 2.10 kips 5 kips 2 2 EI x 0 ft x 4 ft dx 2 2 5.90 kips 3 kips 2 2 x 10 ft x 13 ft C1 2 2 Integrate again to obtain the beam deflection function: 2.10 kips 5 kips 3 3 EI v x 0 ft x 4 ft 6 6 5.90 kips 3 kips 3 3 x 10 ft x 13 ft C1 x C2 6 6

1

1

0 1

3 kips x 13 ft

1

(a)

(b)

Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, the deflection v is known at the pin support (x = 0 ft) and at the roller support (x = 10 ft). Substitute the boundary condition v = 0 at x = 0 ft into Eq. (b) to obtain: C2 0


Next, substitute the boundary condition v = 0 at x = 10 ft into Eq. (b) to obtain: 2.10 kips 5 kips EI v (10 ft)3 (6 ft)3 C1 (10 ft) 0 6 6 C1 17 kip-ft 2 The beam slope and elastic curve equations are now complete: dv 2.10 kips 5 kips 2 2 EI x 0 ft x 4 ft dx 2 2 5.90 kips 3 kips 2 2 x 10 ft x 13 ft 17 kip-ft 2 2 2 2.10 kips 5 kips 3 3 EI v x 0 ft x 4 ft 6 6 5.90 kips 3 kips 3 3 x 10 ft x 13 ft (17 kip-ft 2 ) x 6 6

(c)

(d)

Beam deflection at D: At the tip of the overhang where x = 13 ft, the beam deflection is: 2.10 kips 5 kips 5.90 kips EI vD (13 ft)3 (9 ft)3 (3 ft)3 (17 kip-ft 2 )(13 ft) 6 6 6 3 33.000 kip-ft vD

33.000 kip-ft 3 1,750 kip-ft 2

0.018857 ft

0.226 in.

Ans.


10.30 The solid 30-mm-diameter steel [E = 200 GPa] shaft shown in Fig. P10.30 supports two pulleys. For the loading shown, use discontinuity functions to compute: (a) the shaft deflection at pulley B. (b) the shaft deflection at pulley C.

Fig. P10.30

Solution Support reactions: A FBD of the beam is shown to the right. Fy Ay 800 N 500 N 0

Ay MA

1,300 N

(800 N)(250 mm) (500 N)(600 mm) M A

MA 500,000 N-mm Load function w(x): w( x) 500,000 N-mm x 0 mm 1

800 N x 250 mm

2

0

1

1,300 N x 0 mm 1

500 N x 600 mm

Shear-force function V(x) and bending-moment function M(x): 1 0 V ( x) 500,000 N-mm x 0 mm 1,300 N x 0 mm

800 N x 250 mm

M ( x)

0

500 N x 600 mm

500,000 N-mm x 0 mm 800 N x 250 mm

1

0

0

1,300 N x 0 mm

500 N x 600 mm

1

1

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 0 EI 2 M ( x) 500,000 N-mm x 0 mm 1,300 N x 0 mm dx 1

1

1

800 N x 250 mm 500 N x 600 mm Integrate the moment function to obtain an expression for the beam slope: dv 1,300 N 1 2 EI 500,000 N-mm x 0 mm x 0 mm dx 2 800 N 500 N 2 2 x 250 mm x 600 mm C1 2 2 Integrate again to obtain the beam deflection function: 500,000 N-mm 1,300 N 2 3 EI v x 0 mm x 0 mm 2 6 800 N 500 N 3 3 x 250 mm x 600 mm C1 x C2 6 6

(a)

(b)


Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, both the slope dv/dx and the deflection v are known at the fixed support (x = 0 mm). Substitute the boundary condition dv/dx = 0 at x = 0 mm into Eq. (a) to obtain: C1 0 Next, substitute the boundary condition v = 0 at x = 0 mm into Eq. (b) to obtain: C2 0 The beam slope and elastic curve equations are now complete: dv 1,300 N 1 2 EI 500,000 N-mm x 0 mm x 0 mm dx 2 800 N 500 N 2 2 x 250 mm x 600 mm 2 2 500,000 N-mm 1,300 N 2 3 EI v x 0 mm x 0 mm 2 6 800 N 500 N 3 3 x 250 mm x 600 mm 6 6 Section properties: I

(30 mm) 4

39,750.782 mm 4

64 EI 7.9522 109 N-mm 2

E

200 GPa

200,000 N/mm 2

(a) Beam deflection at B: The beam deflection at B where x = 250 mm is: 500,000 N-mm 1,300 N EI vB (250 mm) 2 (250 mm)3 2 6 9 3 12.2396 10 N-mm vB 1.5392 mm 1.539 mm 7.9522 109 N-mm 2 (b) Beam deflection at C: The beam deflection at C where x = 600 mm is: 500,000 N-mm 1,300 N 800 N EI vC (600 mm) 2 (600 mm)3 (350 mm)3 2 6 6 48.9167 109 N-mm 3 vC 6.1514 mm 6.15 mm 7.9522 109 N-mm 2

Ans.

Ans.


10.31 For the beam and loading shown, use discontinuity functions to compute (a) the slope of the beam at C and (b) the deflection of the beam at C. Assume a constant value of EI = 560×106 Nmm2 for the beam. Fig. P10.31

Solution Support reactions: A FBD of the beam is shown to the right. M A (210 N-m)(1,000 mm/m) (1, 400 N)(450 mm) E y (700 mm) Ey Fy

Ay

600 N

E y 1, 400 N Ay

0

0

800 N

Load function w(x): w( x) 800 N x 0 mm

1

2

210,000 N-mm x 200 mm 1

1,400 N x 450 mm

1

600 N x 700 mm

Shear-force function V(x) and bending-moment function M(x): 0 1 V ( x) 800 N x 0 mm 210,000 N-mm x 200 mm

1,400 N x 450 mm M ( x) 800 N x 0 mm

1

0

600 N x 700 mm

210,000 N-mm x 200 mm

1,400 N x 450 mm

1

0

0

600 N x 700 mm

1

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 1 EI 2 M ( x) 800 N x 0 mm 210,000 N-mm x 200 mm dx 1

0

1

1, 400 N x 450 mm 600 N x 700 mm Integrate the moment function to obtain an expression for the beam slope: dv 800 N 2 1 EI x 0 mm 210,000 N-mm x 200 mm dx 2 1, 400 N 600 N 2 2 x 450 mm x 700 mm C1 2 2 Integrate again to obtain the beam deflection function: 800 N 210,000 N-mm 3 2 EI v x 0 mm x 200 mm 6 2 1, 400 N 600 N 3 3 x 450 mm x 700 mm C1 x C2 6 6

(a)

(b)


Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, the deflection v is known at the pin support (x = 0 mm) and at the roller support (x = 700 mm). Substitute the boundary condition v = 0 at x = 0 mm into Eq. (b) to obtain: C2 0 Next, substitute the boundary condition v = 0 at x = 700 mm into Eq. (b) to obtain: 800 N 210,000 N-mm 1,400 N 0 (700 mm)3 (500 mm)2 (250 mm)3 C1 (700 mm) 6 2 6 2 C1 22,625,000 N-mm The beam slope and elastic curve equations are now complete: dv 800 N 2 1 EI x 0 mm 210,000 N-mm x 200 mm dx 2 1, 400 N 600 N 2 x 450 mm x 700 mm 2 2 800 N 210,000 N-mm 3 2 EI v x 0 mm x 200 mm 6 2 1, 400 N 600 N 3 3 x 450 mm x 700 mm 6 6

2

22,625,000 N-mm 2

(22,625,000 N-mm 2 ) x

(a) Beam slope at C: The beam slope at C is: dv 800 N EI (350 mm) 2 (210,000 N-mm)(150 mm) 22,625,000 N-mm 2 dx C 2 dv dx C

5.125 106 N-mm 2 560 106 N-mm 2

0.009152 rad

(b) Beam deflection at C: The beam deflection at C is: 800 N 210,000 N-mm EI vC (350 mm)3 (150 mm)2 6 2 9 3 4.564583 10 N-mm vC

4.564583 109 N-mm3 560 106 N-mm 2

8.1510 mm

0.00915 rad

Ans.

(22,625,000 N-mm 2 )(350 mm)

8.15 mm

Ans.


10.32 The solid 30-mm-diameter steel [E = 200 GPa] shaft shown in Fig. P10.32 supports two belt pulleys. Assume that the bearing at A can be idealized as a pin support and that the bearing at E can be idealized as a roller support. For the loading shown, use discontinuity functions to compute: (a) the shaft deflection at pulley B. (b) the shaft deflection at point C. Fig. P10.32

Solution Support reactions: A FBD of the beam is shown to the right. MA (600 N)(300 mm) (800 N)(800 mm) E y (1,000 mm) Ey Fy

Ay

Ey Ay

0

820 N 600 N 800 N

0

580 N

Load function w(x): w( x) 580 N x 0 mm

1

1

600 N x 300 mm

Shear-force function V(x) and bending-moment function M(x): 0 0 V ( x) 580 N x 0 mm 600 N x 300 mm 800 N x 800 mm

M ( x) 580 N x 0 mm

1

600 N x 300 mm

1

800 N x 800 mm

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 1 EI 2 M ( x) 580 N x 0 mm 600 N x 300 mm dx 1

1

800 N x 800 mm

0 1

820 N x 1,000 mm

820 N x 1,000 mm 820 N x 1,000 mm

1

0 1

1

1

800 N x 800 mm 820 N x 1,000 mm Integrate the moment function to obtain an expression for the beam slope: dv 580 N 600 N 800 N 2 2 EI x 0 mm x 300 mm x 800 mm dx 2 2 2 820 N 2 x 1,000 mm C1 2 Integrate again to obtain the beam deflection function: 580 N 600 N 800 N 3 3 3 EI v x 0 mm x 300 mm x 800 mm 6 6 6 820 N 3 x 1,000 mm C1 x C2 6

2

(a)

(b)

Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, the Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

deflection v is known at the pin support (x = 0 mm) and at the roller support (x = 1,000 mm). Substitute the boundary condition v = 0 at x = 0 mm into Eq. (b) to obtain: C2 0 Next, substitute the boundary condition v = 0 at x = 1,000 mm into Eq. (b) to obtain: 580 N 600 N 800 N 0 (1,000 mm)3 (700 mm)3 (200 mm)3 C1 (1,000 mm) 6 6 6 6 2 C1 61.3 10 N-mm The beam slope and elastic curve equations are now complete: dv 580 N 600 N 800 N 2 2 EI x 0 mm x 300 mm x 800 mm dx 2 2 2 820 N 2 x 1,000 mm 61.3 106 N-mm 2 2 580 N 600 N 800 N 3 3 3 EI v x 0 mm x 300 mm x 800 mm 6 6 6 820 N 3 x 1,000 mm (61.3 106 N-mm 2 ) x 6

2

Section properties: I

(30 mm) 4 39,750.782 mm 4 64 EI 7.9522 109 N-mm 2

E

200 GPa

200,000 N/mm 2

(a) Beam deflection at B: The beam deflection at B where x = 300 mm is: 580 N EI vB (300 mm)3 (61.3 106 N-mm 2 )(300 mm) 6 15.7800 109 N-mm 3 vB 1.9844 mm 1.984 mm 7.9522 109 N-mm 2 (b) Beam deflection at C: The beam deflection at C where x = 500 mm is: 580 N 600 N EI vC (500 mm)3 (200 mm)3 (61.3 106 N-mm 2 )(500 mm) 6 6 9 3 19.3667 10 N-mm vC 2.4354 mm 2.44 mm 7.9522 109 N-mm 2

Ans.

Ans.


10.33 The cantilever beam shown in Fig. P10.33 consists of a W530 × 74 structural steel wideflange shape [E = 200 GPa; I = 410 × 106 mm4]. Use discontinuity functions to compute the deflection of the beam at C for the loading shown. Fig. P10.33

Solution Support reactions: A FBD of the beam is shown to the right. Fy Ay (30 kN/m)(3 m) 40 kN 0

Ay MA

130 kN

(30 kN/m)(3 m)(1.5 m) (40 kN)(5 m) M A

MA 335 kN-m Load function w(x): w( x) 130 kN x 0 m

1

2

335 kN-m x 0 m 0

30 kN/m x 0 m

30 kN/m x 3 m

0

0

1

40 kN x 5 m

Shear-force function V(x) and bending-moment function M(x): 0 1 V ( x) 130 kN x 0 m 335 kN-m x 0 m 1

30 kN/m x 0 m M ( x) 130 kN x 0 m

1

30 kN/m x 0m 2

30 kN/m x 3 m 335 kN-m x 0 m 2

1

40 kN x 5 m

0

0

30 kN/m x 3m 2

2

40 kN x 5 m

1

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 1 0 EI 2 M ( x) 130 kN x 0 m 335 kN-m x 0 m dx 30 kN/m 30 kN/m 2 2 x 0m x 3m 40 kN x 5 m 2 2 Integrate the moment function to obtain an expression for the beam slope: dv 130 kN 2 1 EI x 0m 335 kN-m x 0 m dx 2 30 kN/m 30 kN/m 40 kN 3 3 2 x 0m x 3m x 5m C1 6 6 2 Integrate again to obtain the beam deflection function: 130 kN 335 kN-m 3 2 EI v x 0m x 0m 6 2 30 kN/m 30 kN/m 40 kN 4 4 3 x 0m x 3m x 5m C1 x C2 24 24 6

1

(a)

(b)


Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, both the slope dv/dx and the deflection v are known at the fixed support (x = 0 m). Substitute the boundary condition dv/dx = 0 at x = 0 m into Eq. (a) to obtain: C1 0 Next, substitute the boundary condition v = 0 at x = 0 m into Eq. (b) to obtain: C2 0 The beam slope and elastic curve equations are now complete: dv 130 kN 2 1 EI x 0m 335 kN-m x 0 m dx 2 30 kN/m 30 kN/m 40 kN 3 3 x 0m x 3m x 5m 6 6 2 130 kN 335 kN-m 3 2 EI v x 0m x 0m 6 2 30 kN/m 30 kN/m 40 kN 4 4 3 x 0m x 3m x 5m 24 24 6

2

Beam deflection at C: For the W530 × 74 structural steel wide-flange shape, EI = 82,000 kN-m2. At the tip of the overhang where x = 5 m, the beam deflection is: 130 kN 335 kN-m 30 kN/m 30 kN/m EI vC (5 m)3 (5 m) 2 (5 m) 4 (2 m) 4 6 2 24 24 2, 240.416667 kN-m3 vC

2, 240.416667 kN-m3 82,000 kN-m 2

0.027322 m

27.3 mm

Ans.


10.34 The cantilever beam shown in Fig. P10.34 consists of a W21 × 50 structural steel wide-flange shape [E = 29,000 ksi; I = 984 in.4]. Use discontinuity functions to compute the deflection of the beam at D for the loading shown. Fig. P10.34

Solution Support reactions: A FBD of the beam is shown to the right.

MA

(9 kips)(4 ft) 1 2

4 kips/ft 9 ft (13 ft) M A

MA Fy

0

270.00 kip-ft 1 2

Ay

(9 kips)

Ay

27.00 kips

4 kips/ft 9 ft

0


270 kip-ft x 0 ft 4 kips/ft x 16 ft 9 ft

2

1

1

27 kips x 0 ft 4 kips/ft x 16 ft

1

9 kips x 4 ft


1

0


V ( x)

270 kip-ft x 0 ft 4 kips/ft x 16 ft 2(9 ft)

M ( x)

270 kip-ft x 0 ft 4 kips/ft x 16 ft 6(9 ft)

1

2

0

3

27 kips x 0 ft

0

4 kips/ft x 16 ft 27 kips x 0 ft

1

4 kips/ft x 16 ft 2

9 kips x 4 ft

0


2

1

9 kips x 4 ft

1


3

2

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 0 1 EI 2 M ( x) 270 kip-ft x 0 ft 27 kips x 0 ft 9 kips x 4 ft dx 4 kips/ft 4 kips/ft 4 kips/ft 3 3 2 x 7 ft x 16 ft x 16 ft 6(9 ft) 6(9 ft) 2 Integrate the moment function to obtain an expression for the beam slope:

1


27 kips 9 kips 2 x 0 ft x 2 2 4 kips/ft 4 kips/ft 4 kips/ft 4 4 x 7 ft x 16 ft 24(9 ft) 24(9 ft) 6 Integrate again to obtain the beam deflection function: 270 kip-ft 27 kips 9 kips 2 3 EI v x 0 ft x 0 ft x 2 6 6 4 kips/ft 4 kips/ft 4 kips/ft 5 5 x 7 ft x 16 ft 120(9 ft) 120(9 ft) 24 EI

dv dx

270 kip-ft x 0 ft

1

2

4 ft

(a) x 16 ft

4 ft

3

C1

3

x 16 ft

4

C1 x C2

(b)

Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, both the slope dv/dx and the deflection v are known at the fixed support (x = 0 ft). Substitute the boundary condition dv/dx = 0 at x = 0 ft into Eq. (a) to obtain: C1 0 Next, substitute the boundary condition v = 0 at x = 0 ft into Eq. (b) to obtain: C2 0 The beam slope and elastic curve equations are now complete: dv 27 kips 9 kips 4 kips/ft 1 2 2 4 EI 270 kip-ft x 0 ft x 0 ft x 4 ft x 7 ft dx 2 2 24(9 ft) 4 kips/ft 4 kips/ft 4 3 x 16 ft x 16 ft 24(9 ft) 6 270 kip-ft 27 kips 9 kips 4 kips/ft 2 3 3 5 EI v x 0 ft x 0 ft x 4 ft x 7 ft 2 6 6 120(9 ft) 4 kips/ft 4 kips/ft 5 4 x 16 ft x 16 ft 120(9 ft) 24 Beam deflection at D: For the W21 × 50 structural steel wide-flange shape, EI = 198,166.658 kip-ft2. At the tip of the overhang where x = 16 ft, the beam deflection is: 270 kip-ft 27 kips 9 kips 4 kips/ft EI vD (16 ft) 2 (16 ft)3 (12 ft)3 (9 ft)5 2 6 6 120(9 ft) 18,938.7 kip-ft 3 vD

18,938.7 kip-ft 3 198,166.658 kip-ft 2

0.095570 ft

1.147 in.

Ans.


10.35 The simply supported beam shown in Fig. P10.35 consists of a W410 × 85 structural steel wide-flange shape [E = 200 GPa; I = 316 × 106 mm4]. For the loading shown, use discontinuity functions to compute (a) the slope of the beam at A and (b) the deflection of the beam at midspan. Fig. P10.35

Solution Support reactions: A FBD of the beam is shown to the right. MA (75 kN/m)(2.5 m)(1.25 m) (75 kN/m)(2.5 m)(6.75 m) Dy Fy

Ay

Dy Ay

Dy (8 m)

0

187.5 kN (75 kN/m)(2.5 m) (75 kN/m)(2.5 m)

0

187.5 kN

Load function w(x): w( x) 187.5 kN x 0 m

75 kN/m x 5.5 m

1 0

75 kN/m x 0 m 75 kN/m x 8 m

0 0

75 kN/m x 2.5 m

75 kN/m x 5.5 m

1

1

1

187.5 kN x 8 m

Shear-force function V(x) and bending-moment function M(x): 0 1 V ( x) 187.5 kN x 0 m 75 kN/m x 0 m 75 kN/m x 2.5 m

0

1 0

75 kN/m x 8 m 187.5 kN x 8 m 75 kN/m 75 kN/m 1 2 2 M ( x) 187.5 kN x 0 m x 0m x 2.5 m 2 2 75 kN/m 75 kN/m 2 2 1 x 5.5 m x 8m 187.5 kN x 8 m 2 2 Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 75 kN/m 75 kN/m 1 2 2 EI 2 M ( x) 187.5 kN x 0 m x 0m x 2.5 m dx 2 2 75 kN/m 75 kN/m 2 2 x 5.5 m x 8m 187.5 kN x 8 m 2 2 Integrate the moment function to obtain an expression for the beam slope: dv 187.5 kN 75 kN/m 75 kN/m 2 3 3 EI x 0m x 0m x 2.5 m dx 2 6 6 75 kN/m 75 kN/m 187.5 kN 3 3 2 x 5.5 m x 8m x 8m C1 6 6 2 Integrate again to obtain the beam deflection function: 187.5 kN 75 kN/m 75 kN/m 3 4 4 EI v x 0m x 0m x 2.5 m 6 24 24 75 kN/m 75 kN/m 187.5 kN 4 4 3 x 5.5 m x 8m x 8m C1 x C2 24 24 6

1

(a)

(b)


Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, the deflection v is known at the pin support (x = 0 m) and at the roller support (x = 8 m). Substitute the boundary condition v = 0 at x = 0 m into Eq. (b) to obtain: C2 0 Next, substitute the boundary condition v = 0 at x = 8 m into Eq. (b) to obtain: 187.5 kN 75 kN/m 75 kN/m 75 kN/m 0 (8 m)3 (8 m) 4 (5.5 m) 4 (2.5 m) 4 C1 (8 m) 6 24 24 24 2 C1 742.1875 kN-m The beam slope and elastic curve equations are now complete: dv 187.5 kN 75 kN/m 75 kN/m 2 3 EI x 0m x 0m x 2.5 m dx 2 6 6 75 kN/m 75 kN/m 187.5 kN 3 3 x 5.5 m x 8m x 8m 6 6 2 187.5 kN 75 kN/m 75 kN/m 3 4 4 EI v x 0m x 0m x 2.5 m 6 24 24 75 kN/m 75 kN/m 187.5 kN 4 4 3 x 5.5 m x 8m x 8m 24 24 6

3

2

742.1875 kN-m 2

(742.1875 kN-m 2 ) x

(a) Beam slope at A: For the W410 × 85 structural steel wide-flange shape, EI = 63,200 kN-m2. The beam slope at A is: dv EI 742.1875 kN-m 2 dx A dv dx

A

742.1875 kN-m 2 63, 200 kN-m 2

0.011743 rad

0.01174 rad

Ans.

(b) Beam deflection at midspan: At midspan where x = 4 m, the beam deflection is: 187.5 kN 75 kN/m 75 kN/m EI vmidspan (4 m)3 (4 m) 4 (1.5 m) 4 (742.1875 kN-m 2 )(4 m) 6 24 24 3 1,752.929687 kN-m vmidspan

1,752.929687 kN-m3 63, 200 kN-m 2

0.027736 m

27.7 mm

Ans.


10.36 The simply supported beam shown in Fig. P10.36 consists of a W14 × 30 structural steel wide-flange shape [E = 29,000 ksi; I = 291 in.4]. For the loading shown, use discontinuity functions to compute (a) the slope of the beam at A and (b) the deflection of the beam at midspan. Fig. P10.36

Solution Support reactions: A FBD of the beam is shown to the right. MA (2.5 kips/ft)(12 ft)(12 ft) Dy (24 ft) 0

Dy Fy

Ay

Dy Ay

15 kips (2.5 kips/ft)(12 ft)

0

15 kips

Load function w(x): 1

w( x) 15 kips x 0 ft

0

2.5 kips/ft x 6 ft

2.5 kips/ft x 18 ft

0

15 kips x 24 ft

1


V ( x) 15 kips x 0 ft M ( x) 15 kips x 0 ft

0

1

1

2.5 kips/ft x 6 ft 2.5 kips/ft x 6 ft 2

1

2

0

2.5 kips/ft x 18 ft 15 kips x 24 ft 2.5 kips/ft 2 x 18 ft 15 kips x 24 ft 2

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 2.5 kips/ft 1 2 EI 2 M ( x) 15 kips x 0 ft x 6 ft dx 2 2.5 kips/ft 2 1 x 18 ft 15 kips x 24 ft 2 Integrate the moment function to obtain an expression for the beam slope: dv 15 kips 2.5 kips/ft 2 3 EI x 0 ft x 6 ft dx 2 6 2.5 kips/ft 15 kips 3 2 x 18 ft x 24 ft C1 6 2 Integrate again to obtain the beam deflection function: 15 kips 2.5 kips/ft 3 4 EI v x 0 ft x 6 ft 6 24 2.5 kips/ft 15 kips 4 3 x 18 ft x 24 ft C1 x C2 24 6

1

(a)

(b)

Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, the deflection v is known at the pin support (x = 0 ft) and at the roller support (x = 24 ft). Substitute the boundary condition v = 0 at x = 0 ft into Eq. (b) to obtain: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

C2 0 Next, substitute the boundary condition v = 0 at x = 24 ft into Eq. (b) to obtain: 15 kips 2.5 kips/ft 2.5 kips/ft 0 (24 ft)3 (18 ft)4 (6 ft)4 C1 (24 ft) 6 24 24 2 C1 990 kip-ft

The beam slope and elastic curve equations are now complete: dv 15 kips 2.5 kips/ft 2 3 EI x 0 ft x 6 ft dx 2 6 2.5 kips/ft 15 kips 3 2 x 18 ft x 24 ft 990 kip-ft 2 6 2 15 kips 2.5 kips/ft 3 4 EI v x 0 ft x 6 ft 6 24 2.5 kips/ft 15 kips 4 3 x 18 ft x 24 ft (990 kip-ft 2 ) x 24 6 (a) Beam slope at A: For the W14 × 30 structural steel wide-flange shape, EI = 58,604.164 kip-ft2. The beam slope at A is: dv EI 990 kip-ft 2 dx A dv dx

A

990 kip-ft 2 58,604.164 kip-ft 2

0.016893 rad

0.01689 rad

(b) Beam deflection at midspan: At midspan where x = 12 ft, the beam deflection is: 15 kips 2.5 kips/ft EI vmidspan (12 ft)3 (6 ft) 4 (990 kip-ft 2 )(12 ft) 7,695 kip-ft 3 6 24 7,695 kip-ft 3 vmidspan 0.131305 ft 1.576 in. 58,604.164 kip-ft 2

Ans.

Ans.


10.37 The simply supported beam shown in Fig. P10.37 consists of a W21 × 50 structural steel wide-flange shape [E = 29,000 ksi; I = 984 in.4]. For the loading shown, use discontinuity functions to compute (a) the slope of the beam at A and (b) the deflection of the beam at B. Fig. P10.37

Solution Support reactions: A FBD of the beam is shown to the right. MA (7 kips/ft)(11 ft)(5.5 ft) (4 kips/ft)(9 ft)(15.5 ft) C y (20 ft) Cy Fy

Ay

Cy Ay

0

49.075 kips (7 kips/ft)(11 ft) (4 kips/ft)(9 ft)

0

63.925 kips


4 kips/ft x 11 ft

0

1

7 kips/ft x 0 ft

4 kips/ft x 20 ft

0

0

M ( x)

1

1

1

49.075 kips x 20 ft

Shear-force function V(x) and bending-moment function M(x): 0 1 V ( x) 63.925 kips x 0 ft 7 kips/ft x 0 ft 7 kips/ft x 11 ft

4 kips/ft x 11 ft

0

7 kips/ft x 11 ft

1 0

4 kips/ft x 20 ft 49.075 kips x 20 ft 7 kips/ft 7 kips/ft 1 2 2 63.925 kips x 0 ft x 0 ft x 11 ft 2 2 4 kips/ft 4 kips/ft 2 2 x 11 ft x 20 ft 49.075 kips x 20 ft 2 2

1

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 7 kips/ft 7 kips/ft 1 2 2 EI 2 M ( x) 63.925 kips x 0 ft x 0 ft x 11 ft dx 2 2 4 kips/ft 4 kips/ft 2 2 x 11 ft x 20 ft 49.075 kips x 20 ft 2 2 Integrate the moment function to obtain an expression for the beam slope: dv 63.925 kips 7 kips/ft 7 kips/ft 2 3 3 EI x 0 ft x 0 ft x 11 ft dx 2 6 6 4 kips/ft 4 kips/ft 49.075 kips 3 3 2 x 11 ft x 20 ft x 20 ft C1 6 6 2 Integrate again to obtain the beam deflection function: 63.925 kips 7 kips/ft 7 kips/ft 3 4 4 EI v x 0 ft x 0 ft x 11 ft 6 24 24 4 kips/ft 4 kips/ft 49.075 kips 4 4 3 x 11 ft x 20 ft x 20 ft C1 x C2 24 24 6

1

(a)

(b)


Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, the deflection v is known at the pin support (x = 0 ft) and at the roller support (x = 20 ft). Substitute the boundary condition v = 0 at x = 0 ft into Eq. (b) to obtain: C2 0 Next, substitute the boundary condition v = 0 at x = 20 ft into Eq. (b) to obtain: 63.925 kips 7 kips/ft 7 kips/ft 4 kips/ft 0 (20 ft)3 (20 ft)4 (9 ft)4 (9 ft)4 C1 (20 ft) 6 24 24 24 2 C1 1,969.3396 kip-ft The beam slope and elastic curve equations are now complete: dv 63.925 kips 7 kips/ft 7 kips/ft 4 kips/ft 2 3 3 EI x 0 ft x 0 ft x 11 ft x 11 ft dx 2 6 6 6 4 kips/ft 49.075 kips 3 2 x 20 ft x 20 ft 1,969.3396 kip-ft 2 6 2 63.925 kips 7 kips/ft 7 kips/ft 4 kips/ft 3 4 4 4 EI v x 0 ft x 0 ft x 11 ft x 11 ft 6 24 24 24 4 kips/ft 49.075 kips 4 3 x 20 ft x 20 ft (1,969.3396 kip-ft 2 ) x 24 6

3

(a) Beam slope at A: For the W21 × 50 structural steel wide-flange shape, EI = 198,166.658 kip-ft2. The beam slope at A is: dv EI 1,969.3396 kip-ft 2 dx A dv dx

A

1,969.3396 kip-ft 2 198,166.658 kip-ft 2

0.009938 rad

0.00994 rad

(b) Beam deflection at B: At midspan where x = 11 ft, the beam deflection is: 63.925 kips 7 kips/ft EI vB (11 ft)3 (11 ft) 4 (1,969.3396 kip-ft 2 )(11 ft) 6 24 11,752.33123 kip-ft 3 vB 0.059305 ft 0.712 in. 198,166.658 kip-ft 2

Ans.

Ans.


10.38 The simply supported beam shown in Fig. P10.38 consists of a W200 × 59 structural steel wide-flange shape [E = 200 GPa; I = 60.8 × 106 mm4]. For the loading shown, use discontinuity functions to compute (a) the deflection of the beam at C and (b) the deflection of the beam at F. Fig. P10.38

Solution Support reactions: A FBD of the beam is shown to the right. MA (20 kN)(2 m) (8 kN/m)(6 m)(7 m) (10 kN)(12 m) Dy Fy

Ay

Dy Ay

Dy (8 m)

0

62 kN 20 kN (8 kN/m)(6 m) 10 kN

0

16 kN


1

8 kN/m x 10 m

1

20 kN x 2 m 0

8 kN/m x 4 m

0

1

10 kN x 12 m

Shear-force function V(x) and bending-moment function M(x): 0 0 1 V ( x) 16 kN x 0 m 20 kN x 2 m 8 kN/m x 4 m 62 kN x 8 m

8 kN/m x 10 m M ( x) 16 kN x 0 m

1

1

8 kN/m x 10 m 2

10 kN x 12 m 20 kN x 2 m 2

1

62 kN x 8 m

0

0

1

10 kN x 12 m

8 kN/m x 4m 2

2

62 kN x 8 m

1

1

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 8 kN/m 1 1 2 EI 2 M ( x) 16 kN x 0 m 20 kN x 2 m x 4m 62 kN x 8 m dx 2 8 kN/m 2 1 x 10 m 10 kN x 12 m 2 Integrate the moment function to obtain an expression for the beam slope: dv 16 kN 20 kN 8 kN/m 62 kN 2 2 3 2 EI x 0m x 2m x 4m x 8m dx 2 2 6 2 8 kN/m 10 kN 3 2 x 10 m x 12 m C1 6 2 Integrate again to obtain the beam deflection function:

1

(a)


EI v

16 kN 20 kN 8 kN/m 3 3 x 0m x 2m x 4m 6 6 24 8 kN/m 10 kN 4 3 x 10 m x 12 m C1 x C2 24 6

62 kN x 8m 6

4

3

(b)

Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, the deflection v is known at the pin support (x = 0 m) and at the roller support (x = 8 m). Substitute the boundary condition v = 0 at x = 0 m into Eq. (b) to obtain: C2 0 Next, substitute the boundary condition v = 0 at x = 8 m into Eq. (b) to obtain: 16 kN 20 kN 8 kN/m 0 (8 m)3 (6 m)3 (4 m) 4 C1 (8 m) 6 6 24 2 C1 70 kN-m The beam slope and elastic curve equations are now complete: dv 16 kN 20 kN 8 kN/m 2 2 EI x 0m x 2m x 4m dx 2 2 6 8 kN/m 10 kN 3 2 x 10 m x 12 m 70 kN-m 2 6 2 16 kN 20 kN 8 kN/m 3 3 4 EI v x 0m x 2m x 4m 6 6 24 8 kN/m 10 kN 4 3 x 10 m x 12 m (70 kN-m 2 ) x 24 6

3

62 kN x 8m 2

62 kN x 8m 6

2

3

(a) Beam deflection at C: For the W200 × 59 structural steel wide-flange shape, EI = 12,160 kN-m2. At C where x = 4 m, the beam deflection is: 16 kN 20 kN EI vC (4 m)3 (2 m)3 (70 kN-m 2 )(4 m) 6 6 3 136 kN-m vC

136 kN-m3 12,160 kN-m 2

0.011184 m

11.18 mm

Ans.

(b) Beam deflection at F: At F where x = 12 m, the beam deflection is: 16 kN 20 kN 8 kN/m 62 kN EI vF (12 m)3 (10 m)3 (8 m) 4 (4 m)3 6 6 24 6 8 kN/m (2 m) 4 (70 kN-m 2 )(12 m) 24 264 kN-m 3 vF

264 kN-m3 12,160 kN-m 2

0.021711 m

21.7 mm

Ans.


10.39 The solid 0.50-in.-diameter steel [E = 30,000 ksi] shaft shown in Fig. P10.39 supports two belt pulleys. Assume that the bearing at B can be idealized as a pin support and that the bearing at D can be idealized as a roller support. For the loading shown, use discontinuity functions to compute: (a) the shaft deflection at pulley A. (b) the shaft deflection at pulley C. Fig. P10.39

Solution Support reactions: A FBD of the beam is shown to the right. M B (90 lb)(5 in.) (120 lb)(10 in.) Dy (20 in.) 0

Dy Fy

By

Dy By

37.5 lb 90 lb 120 lb

0

172.5 lb

Load function w(x):

w( x)

1

90 lb x 0 in.

1

172.5 lb x 5 in.

120 lb x 15 in.

1

37.5 lb x 25 in.

1


V ( x) M ( x)

90 lb x 0 in.

0

90 lb x 0 in.

1

172.5 lb x 5 in. 172.5 lb x 5 in.

0 1

120 lb x 15 in. 120 lb x 15 in.

0 1

37.5 lb x 25 in. 37.5 lb x 25 in.

0 1

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 1 1 1 1 EI 2 M ( x) 90 lb x 0 in. 172.5 lb x 5 in. 120 lb x 15 in. 37.5 lb x 25 in. dx Integrate the moment function to obtain an expression for the beam slope: dv 90 lb 172.5 lb 120 lb 37.5 lb 2 2 2 2 EI x 0 in. x 5 in. x 15 in. x 25 in. C1 (a) dx 2 2 2 2 Integrate again to obtain the beam deflection function: 90 lb 172.5 lb 120 lb 3 3 3 EI v x 0 in. x 5 in. x 15 in. 6 6 6 37.5 lb 3 x 25 in. C1 x C2 (b) 6 Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, the deflection v is known at the pin support (x = 5 in.) and at the roller support (x = 25 in.). Substitute the boundary condition v = 0 at x = 5 in. into Eq. (b) to obtain:


90 lb (5 in.)3 C1 (5 in.) C2 6 C1 (5 in.) C2 1,875.0 lb-in.3 Next, substitute the boundary condition v = 0 at x = 25 in. into Eq. (b) to obtain: 90 lb 172.5 lb 120 lb 0 (25 in.)3 (20 in.)3 (10 in.)3 C1 (25 in.) C2 6 6 6 C1 (25 in.) C2 24,375.0 lb-in.3 0

(c)

(d)

Solve Eqs. (c) and (d) simultaneously for the two constants of integration C1 and C2: C1 1,125 lb-in.2 and C2 3,750 lb-in.3

The beam slope and elastic curve equations are now complete: dv 90 lb 172.5 lb 120 lb 2 2 2 EI x 0 in. x 5 in. x 15 in. dx 2 2 2 37.5 lb 2 x 25 in. 1,125 lb-in.2 2 90 lb 172.5 lb 120 lb 3 3 3 EI v x 0 in. x 5 in. x 15 in. 6 6 6 37.5 lb 3 x 25 in. (1,125 lb-in.2 ) x 3,750 lb-in.3 6 Section properties: I

(0.5 in.) 4

3.06796 10

64 EI 92.0388 103 lb-in.2

3

in.4

E

30,000 ksi

30 106 psi

(a) Beam deflection at A: The beam deflection at A where x = 0 in. is: EI vA 3,750 lb-in.3

vA

3,750 lb-in.3 92.0388 103 lb-in.2

0.040744 in.

0.0407 in.

(b) Beam deflection at C: The beam deflection at C where x = 15 in. is: 90 lb 172.5 lb EI vC (15 in.)3 (10 in.)3 (1,125 lb-in.2 )(15 in.) 3,750 lb-in.3 6 6 8,750 lb-in.3 vC 0.095069 in. 0.0951 in. 92.0388 103 lb-in.2

Ans.

Ans.


10.40 The cantilever beam shown in Fig. P10.40 consists of a W8 × 31 structural steel wide-flange shape [E = 29,000 ksi; I = 110 in.4]. For the loading shown, use discontinuity functions to compute (a) the slope of the beam at A and (b) the deflection of the beam at A. Fig. P10.40

Solution Support reactions: A FBD of the beam is shown to the right. Fy C y (3.5 kips/ft)(10 ft) 0

Cy MC

35 kips

75 kip-ft (3.5 kips/ft)(10 ft)(5 ft) M C

MC 100 kip-ft Load function w(x): w( x) 75 kip-ft x 0 ft

2 1

35 kips x 15 ft

0

3.5 kips/ft x 5 ft

0 2

100 kip-ft x 15 ft

Shear-force function V(x) and bending-moment function M(x): 1 1 V ( x) 75 kip-ft x 0 ft 3.5 kips/ft x 5 ft 3.5 kips/ft x 15 ft

35 kips x 15 ft M ( x)

0

75 kip-ft x 0 ft 35 kips x 15 ft

0

1

100 kip-ft x 15 ft 3.5 kips/ft x 5 ft 2 100 kip-ft x 15 ft

1

1

2

3.5 kips/ft x 15 ft 2

2

0

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 3.5 kips/ft 0 EI 2 M ( x) 75 kip-ft x 0 ft x 5 ft dx 2 1

0

3.5 kips/ft x 15 ft

2

3.5 kips/ft x 15 ft 2

2

0

35 kips x 15 ft 100 kip-ft x 15 ft Integrate the moment function to obtain an expression for the beam slope: dv 3.5 kips/ft 3.5 kips/ft 1 3 3 EI 75 kip-ft x 0 ft x 5 ft x 15 ft dx 6 6 35 kips 2 1 x 15 ft 100 kip-ft x 15 ft C1 2 Integrate again to obtain the beam deflection function: 75 kip-ft 3.5 kips/ft 3.5 kips/ft 2 4 4 EI v x 0 ft x 5 ft x 15 ft 2 24 24 35 kips 100 kip-ft 3 2 x 15 ft x 15 ft C1 x C2 6 2

(a)

(b)


Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, both the slope dv/dx and the deflection v are known at the fixed support (x = 15 ft). Substitute the boundary condition dv/dx = 0 at x = 15 ft into Eq. (a) to obtain: 3.5 kips/ft 0 (75 kip-ft)(15 ft)1 (10 ft)3 C1 6 C1 541.666667 kip-ft 2 Next, substitute the boundary condition v = 0 at x = 15 ft into Eq. (b) to obtain: 75 kip-ft 3.5 kips/ft 0 (15 ft)2 (10 ft)4 ( 541.666667 kip-ft 2 )(15 ft) C2 2 24 C2 1,145.833334 kip-ft 3 The beam slope and elastic curve equations are now complete: dv 3.5 kips/ft 3.5 kips/ft 1 3 3 EI 75 kip-ft x 0 ft x 5 ft x 15 ft dx 6 6 35 kips 2 1 x 15 ft 100 kip-ft x 15 ft 541.666667 kip-ft 2 2 75 kip-ft 3.5 kips/ft 3.5 kips/ft 2 4 4 EI v x 0 ft x 5 ft x 15 ft 2 24 24 35 kips 100 kip-ft 3 2 x 15 ft x 15 ft (541.666667 kip-ft 2 ) x 1,145.833334 kip-ft 3 6 2 (a) Beam slope at A: For the W8 × 31 structural steel wide-flange shape, EI = 22,152.777 kip-ft2. At the tip of the overhang where x = 0 ft, the beam slope is: dv EI 541.666667 kip-ft 2 dx A dv dx

A

541.666667 kip-ft 2 22,152.777 kip-ft 2

0.024451 rad

0.0245 rad

Ans.

(b) Beam deflection at A: At the tip of the overhang where x = 0 ft, the beam deflection is: EI vA 1,145.833334 kip-ft 3 vA

1,145.833334 kip-ft 3 22,152.777 kip-ft 2

0.051724 ft

0.621 in.

Ans.


10.41 The simply supported beam shown in Fig. P10.41 consists of a W14 × 34 structural steel wide-flange shape [E = 29,000 ksi; I = 340 in.4]. For the loading shown, use discontinuity functions to compute (a) the slope of the beam at E and (b) the deflection of the beam at C. Fig. P10.41

Solution Support reactions: A FBD of the beam is shown to the right. 8 ft M B 12 (6 kips/ft)(8 ft) (4 kips/ft)(10 ft)(13 ft) 3 E y (22 ft) 0 Ey Fy

By

Ey By

20.727 kips 1 2

(6 kips/ft)(8 ft) (4 kips/ft)(10 ft)

0

43.273 kips

Load function w(x): 6 kips/ft w( x) x 0 ft 8 ft

4 kips/ft x 16 ft


1

0

1

4 kips/ft x 26 ft

6 kips/ft x 8 ft 0

M ( x)

1

1

1

43.273 kips x 8 ft 1

20.727 kips x 30 ft

Shear-force function V(x) and bending-moment function M(x): 6 kips/ft 6 kips/ft 2 2 V ( x) x 0 ft x 8 ft 6 kips/ft x 8 ft 2(8 ft) 2(8 ft) 1

0

43.273 kips x 8 ft

0

0

4 kips/ft x 16 ft 4 kips/ft x 26 ft 20.727 kips x 30 ft 6 kips/ft 6 kips/ft 6 kips/ft 3 3 2 x 0 ft x 8 ft x 8 ft 43.273 kips x 8 ft 6(8 ft) 6(8 ft) 2 4 kips/ft 4 kips/ft 2 2 1 x 16 ft x 26 ft 20.727 kips x 30 ft 2 2

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 6 kips/ft 6 kips/ft 6 kips/ft 3 3 2 EI 2 M ( x) x 0 ft x 8 ft x 8 ft dx 6(8 ft) 6(8 ft) 2 4 kips/ft 4 kips/ft 1 2 43.273 kips x 8 ft x 16 ft x 26 ft 2 2

20.727 kips x 30 ft

1

2

1

Integrate the moment function to obtain an expression for the beam slope:


6 kips/ft 6 kips/ft 6 kips/ft 43.273 kips 4 4 3 x 0 ft x 8 ft x 8 ft x 8 ft 24(8 ft) 24(8 ft) 6 2 4 kips/ft 4 kips/ft 20.727 kips 3 3 2 x 16 ft x 26 ft x 30 ft C1 6 6 2 Integrate again to obtain the beam deflection function: 6 kips/ft 6 kips/ft 6 kips/ft 43.273 kips 5 5 4 3 EI v x 0 ft x 8 ft x 8 ft x 8 ft 120(8 ft) 120(8 ft) 24 6 4 kips/ft 4 kips/ft 20.727 kips 4 4 3 x 16 ft x 26 ft x 30 ft C1 x C2 24 24 6 EI

dv dx

2

(a)

(b)

Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, the deflection v is known at the pin support (x = 8 ft) and at the roller support (x = 30 ft). Substitute the boundary condition v = 0 at x = 8 ft into Eq. (b) to obtain: 6 kips/ft 0 (8 ft)5 C1 (8 ft) C2 120(8 ft)

C1 (8 ft) C2

204.80 kips-ft 3

Next, substitute the boundary condition v = 0 at x = 30 ft into Eq. (b) to obtain: 6 kips/ft 6 kips/ft 6 kips/ft 43.273 kips 0 (30 ft)5 (22 ft)5 (22 ft) 4 (22 ft) 3 120(8 ft) 120(8 ft) 24 6 4 kips/ft 4 kips/ft (14 ft) 4 (4 ft) 4 C1 (30 ft) C2 24 24 C1 (30 ft) C2 9,334.351 kip-ft 3

(c)

(d)

Solve Eqs. (c) and (d) simultaneously for the two constants of integration C1 and C2: C1 433.598 kip-ft 2 and C2 3,673.582 kip-ft 3 The beam slope and elastic curve equations are now complete: dv 6 kips/ft 6 kips/ft 6 kips/ft 43.273 kips 4 4 3 2 EI x 0 ft x 8 ft x 8 ft x 8 ft dx 24(8 ft) 24(8 ft) 6 2 4 kips/ft 4 kips/ft 20.727 kips 3 3 2 x 16 ft x 26 ft x 30 ft 433.598 kip-ft 2 6 6 2 6 kips/ft 6 kips/ft 6 kips/ft 43.273 kips 5 5 4 3 EI v x 0 ft x 8 ft x 8 ft x 8 ft 120(8 ft) 120(8 ft) 24 6 4 kips/ft 4 kips/ft 20.727 kips 4 4 3 x 16 ft x 26 ft x 30 ft 24 24 6 2 3 (433.598 kip-ft ) x 3,673.582 kip-ft


(a) Beam slope at E: For the W14 × 34 structural steel wide-flange shape, EI = 68,472.219 kip-ft2. The beam slope at E is: dv 6 kips/ft 6 kips/ft 6 kips/ft 43.273 kips EI (30 ft)4 (22 ft) 4 (22 ft)3 (22 ft) 2 dx E 24(8 ft) 24(8 ft) 6 2

dv dx

E

4 kips/ft 4 kips/ft (14 ft)3 (4 ft)3 433.598 kip-ft 2 6 6 2 907.801 kip-ft 0.013258 rad 0.01326 rad 68,472.219 kip-ft 2

Ans.

(b) Beam deflection at C: At C where x = 16 ft, the beam deflection is: 6 kips/ft 6 kips/ft 6 kips/ft 43.273 kips EI vC (16 ft)5 (8 ft)5 (8 ft) 4 (8 ft) 3 120(8 ft) 120(8 ft) 24 6 (433.598 kip-ft 2 )(16 ft) 3,673.582 kip-ft 3 vC

4,896.157 kip-ft 3 68, 472.219 kip-ft 2

0.071506 ft

0.858 in.

Ans.


10.42 For the beam and loading shown, use discontinuity functions to compute (a) the deflection of the beam at A and (b) the deflection of the beam at midspan (i.e., x = 2.5 m). Assume a constant value of EI = 1,500 kN-m2 for the beam. Fig. P10.42

Solution Support reactions: A FBD of the beam is shown to the right. M B 9 kN-m 12 (18 kN/m)(3 m)(1 m) C y (3 m) 0 Cy Fy

By

Cy By

6 kN 1 2

(18 kN/m)(3 m)

0

21 kN

Load function w(x): w( x) 9 kN-m x 0 m

18 kN/m x 1m 3m

2

1

21 kN x 1 m

18 kN/m x 4m 3m

1

1

1

6 kN x 4 m

Shear-force function V(x) and bending-moment function M(x): 1 0 V ( x) 9 kN-m x 0 m 21 kN x 1 m 18 kN/m x 1 m 18 kN/m x 1m 2(3 m) M ( x)

9 kN-m x 0 m 18 kN/m x 1m 6(3 m)

18 kN/m x 4m 2(3 m)

2

0

3

21 kN x 1 m

0

18 kN/m x 1 m

1

18 kN/m x 4m 6(3 m)

2

6 kN x 4 m

1

0

18 kN/m x 1m 2 3

6 kN x 4 m

2

1

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 18 kN/m 0 1 EI 2 M ( x) 9 kN-m x 0 m 21 kN x 1 m x 1m dx 2 18 kN/m 18 kN/m 3 3 1 x 1m x 4m 6 kN x 4 m 6(3 m) 6(3 m) Integrate the moment function to obtain an expression for the beam slope: dv 21 kN 18 kN/m 1 2 3 EI 9 kN-m x 0 m x 1m x 1m dx 2 6 18 kN/m 18 kN/m 6 kN 4 4 2 x 1m x 4m x 4m C1 24(3 m) 24(3 m) 2

2

(a)


Integrate again to obtain the beam deflection function: 9 kN-m 21 kN 18 kN/m 2 3 4 EI v x 0m x 1m x 1m 2 6 24 18 kN/m 18 kN/m 6 kN 5 5 3 x 1m x 4m x 4m C1 x C2 120(3 m) 120(3 m) 6

(b)

Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, the deflection v is known at the pin support (x = 1 m) and at the roller support (x = 4 m). Substitute the boundary condition v = 0 at x = 1 m into Eq. (b) to obtain: 9 kN-m 0 (1 m)2 C1 (1 m) C2 2 (c) C1 (1 m) C2 4.5 kN-m3 Next, substitute the boundary condition v = 0 at x = 4 m into Eq. (b) to obtain: 9 kN-m 21 kN 18 kN/m 18 kN/m 0 (4 m) 2 (3 m)3 (3 m) 4 (3 m)5 C1 (4 m) C2 2 6 24 120(3 m)

C1 (4 m) C2 26.10 kN-m3 Solve Eqs. (c) and (d) simultaneously for the two constants of integration C1 and C2: C1 7.2 kN-m 2 and C2 2.7 kN-m3

(d)

The beam slope and elastic curve equations are now complete: dv 21 kN 18 kN/m 1 2 3 EI 9 kN-m x 0 m x 1m x 1m dx 2 6 18 kN/m 18 kN/m 6 kN 4 4 2 x 1m x 4m x 4m 7.2 kN-m 2 24(3 m) 24(3 m) 2 9 kN-m 21 kN 18 kN/m 2 3 4 EI v x 0m x 1m x 1m 2 6 24 18 kN/m 18 kN/m 6 kN 5 5 3 x 1m x 4m x 4m (7.2 kN-m 2 ) x 2.7 kN-m 3 120(3 m) 120(3 m) 6 (a) Beam deflection at A: The beam deflection at A is: EI vA 2.7 kN-m3 vA

2.7 kN-m3 1,500 kN-m 2

0.001800 m

Ans.

1.800 mm

(b) Beam deflection at midspan: At x = 2.5 m, the beam deflection is: 9 kN-m 21 kN 18 kN/m EI vmidspan (2.5 m) 2 (1.5 m)3 (1.5 m) 4 2 6 24 18 kN/m (1.5 m)5 (7.2 kN-m 2 )(2.5 m) 2.7 kN-m3 120(3 m)

4.429688 kN-m3 vmidspan

4.429688 kN-m3 1,500 kN-m 2

0.002953 m

2.95 mm

Ans.


10.43 For the beam and loading shown, use discontinuity functions to compute (a) the slope of the beam at B and (b) the deflection of the beam at A. Assume a constant value of EI = 133,000 kipft2 for the beam. Fig. P10.43

Solution Support reactions: A FBD of the beam is shown to the right.

Fy

Cy

1 2

(4 kips/ft)(9 ft) 0

Cy MC

1 2

18 kips

(4 kips/ft)(9 ft)(11 ft) M C MC

0

198 kip-ft

Load function w(x):

w( x)

4 kips/ft x 0 ft


0

1

18 kips x 14 ft


1

2

198 kip-ft x 14 ft

Shear-force function V(x) and bending-moment function M(x): 4 kips/ft 4 kips/ft 1 2 V ( x) 4 kips/ft x 0 ft x 0 ft x 9 ft 2(9 ft) 2(9 ft) 0

M ( x)

18 kips x 14 ft 4 kips/ft x 0 ft 2 18 kips x 14 ft

2

1

2

1

198 kip-ft x 14 ft 4 kips/ft 3 x 0 ft 6(9 ft) 198 kip-ft x 14 ft

1


0

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 4 kips/ft 4 kips/ft 2 EI 2 M ( x) x 0 ft x 0 ft dx 2 6(9 ft) 1

3


2

198 kip-ft x 14 ft

3

0

18 kips x 14 ft 198 kip-ft x 14 ft Integrate the moment function to obtain an expression for the beam slope: dv 4 kips/ft 4 kips/ft 4 kips/ft 3 4 EI x 0 ft x 0 ft x 9 ft dx 6 24(9 ft) 24(9 ft)

18 kips x 14 ft 2

3

1

C1

4

(a)


Integrate again to obtain the beam deflection function: 4 kips/ft 4 kips/ft 4 kips/ft 4 5 EI v x 0 ft x 0 ft x 9 ft 24 120(9 ft) 120(9 ft) 18 kips 198 kip-ft 3 2 x 14 ft x 14 ft C1 x C2 6 2

5

(b)

Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, both the slope dv/dx and the deflection v are known at the fixed support (x = 14 ft). Substitute the boundary condition dv/dx = 0 at x = 14 ft into Eq. (a) to obtain: 4 kips/ft 4 kips/ft 4 kips/ft 0 (14 ft)3 (14 ft) 4 (5 ft) 4 C1 6 24(9 ft) 24(9 ft)

C1 1,129.5 kip-ft 2 Next, substitute the boundary condition v = 0 at x = 4 m into Eq. (b) to obtain: 4 kips/ft 4 kips/ft 4 kips/ft 0 (14 ft) 4 (14 ft)5 (5 ft)5 (1,129.5 kip-ft 2 )(14 ft) C2 24 120(9 ft) 120(9 ft) C2

11,390.7 kip-ft 3

The beam slope and elastic curve equations are now complete: dv 4 kips/ft 4 kips/ft 4 kips/ft 3 4 4 EI x 0 ft x 0 ft x 9 ft dx 6 24(9 ft) 24(9 ft) 18 kips 2 1 x 14 ft 198 kip-ft x 14 ft 1,129.5 kip-ft 2 2 4 kips/ft 4 kips/ft 4 kips/ft 4 5 5 EI v x 0 ft x 0 ft x 9 ft 24 120(9 ft) 120(9 ft) 18 kips 198 kip-ft 3 2 x 14 ft x 14 ft (1,129.5 kip-ft 2 ) x 11,390.7 kip-ft 3 6 2 (a) Beam slope at B: The beam slope at B (i.e., x = 9 ft) is: dv 4 kips/ft 4 kips/ft EI (9 ft)3 (9 ft) 4 1,129.5 kip-ft 2 dx B 6 24(9 ft) dv dx

B

765 kip-ft 2 133,000 kip-ft 2

0.005752 rad

0.00575 rad

765 kip-ft 2

Ans.

(b) Beam deflection at A: The beam deflection at A is: EI vA 11,390.7 kip-ft 3 vA

11,390.7 kip-ft 3 133,000 kip-ft 2

0.085644 ft

1.028 in.

Ans.


10.44 For the beam and loading shown, use discontinuity functions to compute (a) the slope of the beam at B and (b) the deflection of the beam at C. Assume a constant value of EI = 34×106 lb-ft2 for the beam. Fig. P10.44

Solution Support reactions: A FBD of the beam is shown to the right. 1 MB (7,000 lb/ft)(9 ft)(3 ft) Dy (14 ft) 0 2 Dy Fy

By

6,750 lbs 1 2

Dy By

(7,000 lb/ft)(9 ft)

0

24,750 lbs


24,750 lbs x 4 ft

1

7,000 lb/ft x 13 ft 9 ft

7,000 lb/ft x 4 ft 1

0

6,750 lbs x 18 ft

7,000 lb/ft x 4 ft 9 ft

1

1

Shear-force function V(x) and bending-moment function M(x): 7,000 lb/ft 0 1 2 V ( x) 24,750 lbs x 4 ft 7,000 lb/ft x 4 ft x 4 ft 2(9 ft) 7,000 lb/ft 2 0 x 13 ft 6,750 lbs x 18 ft 2(9 ft) 7,000 lb/ft 7,000 lb/ft 1 2 3 M ( x) 24,750 lbs x 4 ft x 4 ft x 4 ft 2 6(9 ft) 7,000 lb/ft 3 1 x 13 ft 6,750 lbs x 18 ft 6(9 ft)

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 7,000 lb/ft 7,000 lb/ft 1 2 EI 2 M ( x) 24,750 lbs x 4 ft x 4 ft x 4 ft dx 2 6(9 ft) 7,000 lb/ft 3 1 x 13 ft 6,750 lbs x 18 ft 6(9 ft) Integrate the moment function to obtain an expression for the beam slope: dv 24,750 lbs 7,000 lb/ft 7,000 lb/ft 2 3 4 EI x 4 ft x 4 ft x 4 ft dx 2 6 24(9 ft) 7,000 lb/ft 6,750 lbs 4 2 x 13 ft x 18 ft C1 24(9 ft) 2

3

(a)


Integrate again to obtain the beam deflection function: 24,750 lbs 7,000 lb/ft 7,000 lb/ft 3 4 EI v x 4 ft x 4 ft x 4 ft 6 24 120(9 ft) 7,000 lb/ft 6,750 lbs 5 3 x 13 ft x 18 ft C1 x C2 120(9 ft) 6

5

(b)

Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, the deflection v is known at the pin support (x = 4 ft) and at the roller support (x = 18 ft). Substitute the boundary condition v = 0 at x = 4 ft into Eq. (b) to obtain: C1 (4 ft) C2 0 (c) Next, substitute the boundary condition v = 0 at x = 18 ft into Eq. (b) to obtain: 24,750 lbs 7,000 lb/ft 7,000 lb/ft 7,000 lb/ft 0 (14 ft)3 (14 ft) 4 (14 ft)5 (5 ft)5 6 24 120(9 ft) 120(9 ft) C1 (18 ft) C2 C1 (18 ft) C2

3,579,975 lb-ft 3

(d)

Solve Eqs. (c) and (d) simultaneously for the two constants of integration C1 and C2: C1 255,712.5 lb-ft 2 and C2 1,022,850 lb-ft 3 The beam slope and elastic curve equations are now complete: dv 24,750 lbs 7,000 lb/ft 7,000 lb/ft 2 3 4 EI x 4 ft x 4 ft x 4 ft dx 2 6 24(9 ft) 7,000 lb/ft 6,750 lbs 4 2 x 13 ft x 18 ft 255,712.5 lb-ft 2 24(9 ft) 2 24,750 lbs 7,000 lb/ft 7,000 lb/ft 3 4 5 EI v x 4 ft x 4 ft x 4 ft 6 24 120(9 ft) 7,000 lb/ft 6,750 lbs 5 3 x 13 ft x 18 ft (255,712.5 lb-ft 2 ) x 1,022,850 lb-ft 3 120(9 ft) 6 (a) Beam slope at B: The beam slope at B is: dv EI 255,712.5 lb-ft 2 dx B dv dx

B

255,712.5 lb-ft 2 34 106 lb-ft 2

0.0075210 rad

0.00752 rad

Ans.

(b) Beam deflection at C: At C where x = 13 ft, the beam deflection is: 24,750 lbs 7,000 lb/ft 7,000 lb/ft EI vC (9 ft)3 (9 ft) 4 (9 ft)5 6 24 120(9 ft) (255,712.5 lb-ft 2 )(13 ft) 1,022,850 lb-ft 3 vC

825,187.5 kip-ft 3 34 106 lb-ft 2

0.0242702 ft

0.291 in.

Ans.


10.45 For the beam and loading shown, use discontinuity functions to compute (a) the slope of the beam at A and (b) the deflection of the beam at B. Assume a constant value of EI = 370,000 kipft2 for the beam. Fig. P10.45

Solution Support reactions: A FBD of the beam is shown to the right. 1 MA (8 kips/ft)(12 ft)(8 ft) 2 1 2

(8 kips/ft)(12 ft)(16 ft) C y (24 ft)

Cy Fy

Ay

Cy Ay

0

48 kips 1 2

(8 kips/ft)(24 ft)

0

48 kips

Load function w(x):

w( x)

48 kips x 0 ft

1


8 kips/ft x 12 ft

0

8 kips/ft 1 x 12 ft 12 ft 8 kips/ft 0 8 kips/ft x 12 ft x 12 ft 12 ft 1

8 kips/ft 1 1 x 24 ft 48 kips x 24 ft 12 ft 8 kips/ft 2(8 kips/ft) 1 1 48 kips x 0 ft x 0 ft x 12 ft 12 ft 12 ft 8 kips/ft 1 1 x 24 ft 48 kips x 24 ft 12 ft

1

1

Shear-force function V(x) and bending-moment function M(x): 8 kips/ft 2(8 kips/ft) 0 2 2 V ( x) 48 kips x 0 ft x 0 ft x 12 ft 2(12 ft) 2(12 ft) 8 kips/ft 2 0 x 24 ft 48 kips x 24 ft 2(12 ft) 8 kips/ft 2(8 kips/ft) 1 3 3 M ( x) 48 kips x 0 ft x 0 ft x 12 ft 6(12 ft) 6(12 ft) 8 kips/ft 3 1 x 24 ft 48 kips x 24 ft 6(12 ft)


Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 8 kips/ft 2(8 kips/ft) 1 3 EI 2 M ( x) 48 kips x 0 ft x 0 ft x 12 ft dx 6(12 ft) 6(12 ft) 8 kips/ft 3 1 x 24 ft 48 kips x 24 ft 6(12 ft) Integrate the moment function to obtain an expression for the beam slope: dv 48 kips 8 kips/ft 2(8 kips/ft) 2 4 4 EI x 0 ft x 0 ft x 12 ft dx 2 24(12 ft) 24(12 ft) 8 kips/ft 48 kips 4 2 x 24 ft x 24 ft C1 24(12 ft) 2 Integrate again to obtain the beam deflection function: 48 kips 8 kips/ft 2(8 kips/ft) 3 5 5 EI v x 0 ft x 0 ft x 12 ft 6 120(12 ft) 120(12 ft) 8 kips/ft 48 kips 5 3 x 24 ft x 24 ft C1 x C2 120(12 ft) 6

3

(a)

(b)

Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, the deflection v is known at the pin support (x = 0 ft) and at the roller support (x = 24 ft). Substitute the boundary condition v = 0 at x = 0 ft into Eq. (b) to obtain: C2 0 Next, substitute the boundary condition v = 0 at x = 24 ft into Eq. (b) to obtain: 48 kips 8 kips/ft 2(8 kips/ft) 0 (24 ft)3 (24 ft)5 (12 ft)5 C1 (24 ft) 6 120(12 ft) 120(12 ft)

C1

2,880 kip-ft 2

The beam slope and elastic curve equations are now complete: dv 48 kips 8 kips/ft 2(8 kips/ft) 2 4 4 EI x 0 ft x 0 ft x 12 ft dx 2 24(12 ft) 24(12 ft) 8 kips/ft 48 kips 4 2 x 24 ft x 24 ft 2,880 kip-ft 2 24(12 ft) 2 48 kips 8 kips/ft 2(8 kips/ft) 3 5 5 EI v x 0 ft x 0 ft x 12 ft 6 120(12 ft) 120(12 ft) 8 kips/ft 48 kips 5 3 x 24 ft x 24 ft (2,880 kip-ft 2 ) x 120(12 ft) 6 (a) Beam slope at A: The beam slope at A is: dv EI 2,880 kip-ft 2 dx A dv dx

A

2,880 kip-ft 2 370,000 kip-ft 2

0.0077838 rad

0.00778 rad

Ans.


(b) Beam deflection at B: At B where x = 12 ft, the beam deflection is: 48 kips 8 kips/ft EI vB (12 ft)3 (12 ft)5 (2,880 kip-ft 2 )(12 ft) 22,118.4 kip-ft 3 6 120(12 ft) vB

22,118.4 kip-ft 3 kip-ft 3 370,000 kip-ft 2

0.0597795 ft

0.717 in.

Ans.


10.46 For the beam and loading shown, use discontinuity functions to compute (a) the slope of the beam at B and (b) the deflection of the beam at B. Assume a constant value of EI = 110,000 kNm2 for the beam. Fig. P10.46

Solution Support reactions: A FBD of the beam is shown to the right. Fy Ay (15 kN/m)(4 m) 12 (25 kN/m)(4 m) 0 Cy MA

110 kN

(15 kN/m)(4 m)(2 m) 1 2

2(4 m) 3 253.33 kN-m

(25 kN/m)(4 m)

MA

MA

0


2

253.33 kN-m x 0 m

1

110 kN x 0 m

25 kN/m 1 x 4m 25 kN/m x 4 m 4m Shear-force function V(x) and bending-moment function M(x): 15 kN/m x 4 m

V ( x)

15 kN/m x 4 m M ( x)

0

1

253.33 kN-m x 0 m

25 kN/m x 4m 2(4 m)

1

253.33 kN-m x 0 m 15 kN/m x 4m 2

110 kN x 0 m

2

0

2

3

1

1

25 kN/m x 0m 2(4 m)

2

1

15 kN/m x 0m 2 25 kN/m 2 x 4m 2

1

25 kN/m x 0m 4m

0

15 kN/m x 0 m

25 kN/m x 4 m

110 kN x 0 m

25 kN/m x 4m 6(4 m)

0

0

15 kN/m x 0 m

2

25 kN/m x 0m 6(4 m)

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 15 kN/m 0 1 2 EI 2 M ( x) 253.33 kN-m x 0 m 110 kN x 0 m x 0m dx 2 25 kN/m 15 kN/m 25 kN/m 25 kN/m 3 2 3 x 0m x 4m x 4m x 4m 6(4 m) 2 6(4 m) 2 Integrate the moment function to obtain an expression for the beam slope:

3

2


EI

dv dx

110 kN 15 kN/m 25 kN/m 2 3 4 x 0m x 0m x 0m 2 6 24(4 m) 25 kN/m 25 kN/m 4 3 (a) x 4m x 4m C1 24(4 m) 6

253.33 kN-m x 0 m 15 kN/m x 4m 6

3

1

Integrate again to obtain the beam deflection function: 253.33 kN-m 110 kN 15 kN/m 25 kN/m 2 3 4 5 EI v x 0m x 0m x 0m x 0m 2 6 24 120(4 m) 15 kN/m 25 kN/m 25 kN/m 4 5 4 (b) x 4m x 4m x 4m C1 x C2 24 120(4 m) 24 Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, both the slope dv/dx and the deflection v are known at the fixed support (x = 0 m). Substitute the boundary condition dv/dx = 0 at x = 0 m into Eq. (a) to obtain: C1 0 Next, substitute the boundary condition v = 0 at x = 0 m into Eq. (b) to obtain: C2 0 The beam slope and elastic curve equations are now complete: dv 110 kN 15 kN/m 25 kN/m 1 2 3 4 EI 253.33 kN-m x 0 m x 0m x 0m x 0m dx 2 6 24(4 m) 15 kN/m 25 kN/m 25 kN/m 3 4 3 x 4m x 4m x 4m 6 24(4 m) 6 253.33 kN-m 110 kN 15 kN/m 25 kN/m 2 3 4 5 EI v x 0m x 0m x 0m x 0m 2 6 24 120(4 m) 15 kN/m 25 kN/m 25 kN/m 4 5 4 x 4m x 4m x 4m 24 120(4 m) 24 (a) Beam slope at B: The beam slope at B is: dv 110 kN EI ( 253.33 kN-m)(4 m)1 (4 m)2 dx B 2

15 kN/m (4 m)3 6

25 kN/m (4 m)4 24(4 m)

360 kN-m 2 dv dx

B

2,120 kN-m 2 110,000 kN-m 2

0.003273 rad

0.00327 rad

(b) Beam deflection at B: The beam deflection at B is: 253.33 kN-m 110 kN 15 kN/m EI vB (4 m) 2 (4 m)3 (4 m) 4 2 6 24

Ans.

25 kN/m (4 m) 5 120(4 m)

1,066.67 kN-m3 vB

1,066.67 kN-m3 110,000 kN-m 2

0.009697 m

9.70 mm

Ans.


10.47 For the beam and loading shown, use discontinuity functions to compute (a) the deflection of the beam at A and (b) the deflection of the beam at C. Assume a constant value of EI = 24,000 kN-m2 for the beam. Fig. P10.47

Solution Support reactions: A FBD of the beam is shown to the right. M B (35 kN)(2.5 m) (25 kN/m)(4.0 m)(2.0 m) 1 2

(45 kN/m)(4.0 m)

Dy Fy

By

Dy 1 2

2(4.0 m) 3

0

64.09 kN 35 kN (25 kN/m)(4.0 m)

(45 kN/m)(4.0 m)

By

Dy (5.5 m)

0

160.91 kN

Load function w(x):

w( x)

35 kN x 0 m

1

1

160.91 kN x 2.5 m

25 kN/m x 6.5 m 45 kN/m x 6.5 m

0

0

25 kN/m x 2.5 m

45 kN/m x 2.5 m 4.0 m

1

45 kN/m x 6.5 m 4.0 m

45 kN/m x 6.5 m

M ( x)

35 kN x 0 m

1

1

1

45 kN/m x 2.5 m 2(4.0 m) 64.09 kN x 8 m

35 kN x 0 m

1

2

2

2

45 kN/m x 6.5 m 2(4.0 m)

2

0

1

64.09 kN x 8 m

160.91 kN x 2.5 m

70 kN/m x 6.5 m 2

2

1

25 kN/m 2 x 2.5 m 2 45 kN/m 45 kN/m 3 x 2.5 m x 6.5 m 6(4.0 m) 6(4.0 m)

160.91 kN x 2.5 m

25 kN/m x 6.5 m 2 45 kN/m x 6.5 m 2

1

1

64.09 kN x 8 m

Shear-force function V(x) and bending-moment function M(x): 0 0 V ( x) 35 kN x 0 m 160.91 kN x 2.5 m 25 kN/m x 2.5 m

25 kN/m x 6.5 m

0

1

3

1

25 kN/m x 2.5 m 2

45 kN/m x 6.5 m 6(4.0 m)

3

2

45 kN/m x 2.5 m 6(4.0 m)

64.09 kN x 8 m

3

1


Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 25 kN/m 1 1 EI 2 M ( x) 35 kN x 0 m 160.91 kN x 2.5 m x 2.5 m dx 2 45 kN/m 70 kN/m 45 kN/m 3 2 3 x 2.5 m x 6.5 m x 6.5 m 6(4.0 m) 2 6(4.0 m)

2

1

64.09 kN x 8 m Integrate the moment function to obtain an expression for the beam slope: dv 35 kN 160.91 kN 25 kN/m 2 2 3 EI x 0m x 2.5 m x 2.5 m dx 2 2 6 45 kN/m 70 kN/m 45 kN/m 4 3 x 2.5 m x 6.5 m x 6.5 m 24(4.0 m) 6 24(4.0 m) 64.09 kN 2 x 8m C1 2

4

Integrate again to obtain the beam deflection function: 35 kN 160.91 kN 25 kN/m 3 3 4 EI v x 0m x 2.5 m x 2.5 m 6 6 24 45 kN/m 70 kN/m 45 kN/m 5 4 x 2.5 m x 6.5 m x 6.5 m 120(4.0 m) 24 120(4.0 m) 64.09 kN 3 x 8m C1 x C2 6

(a)

5

(b)

Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, the deflection v is known at the pin support (x = 2.5 m) and at the roller support (x = 8 m). Substitute the boundary condition v = 0 at x = 2.5 m into Eq. (b) to obtain: 35 kN 0 (2.5 m)3 C1 (2.5 m) C2 6 (c) C1 (2.5 m) C2 91.145833 kN-m3 Next, substitute the boundary condition v = 0 at x = 8 m into Eq. (b) to obtain: 35 kN 160.91 kN 25 kN/m 45 kN/m 0 (8 m)3 (5.5 m)3 (5.5 m) 4 (5.5 m)5 6 6 24 120(4.0 m) 70 kN/m 45 kN/m (1.5 m) 4 (1.5 m)5 C1 (8 m) C2 24 120(4.0 m) C1 (8 m) C2

65.666667 kN-m3

(d)

Solve Eqs. (c) and (d) simultaneously for the two constants of integration C1 and C2: C1 28.511 kN-m 2 and C2 162.424 kN-m3


The beam slope and elastic curve equations are now complete: dv 35 kN 160.91 kN 25 kN/m 2 2 3 EI x 0m x 2.5 m x 2.5 m dx 2 2 6 45 kN/m 70 kN/m 45 kN/m 4 3 4 x 2.5 m x 6.5 m x 6.5 m 24(4.0 m) 6 24(4.0 m) 64.09 kN 2 x 8m 28.511 kN-m 2 2 35 kN 160.91 kN 25 kN/m 3 3 4 EI v x 0m x 2.5 m x 2.5 m 6 6 24 45 kN/m 70 kN/m 45 kN/m 5 4 x 2.5 m x 6.5 m x 6.5 m 120(4.0 m) 24 120(4.0 m) 64.09 kN 3 x 8m (28.511 kN-m 2 ) x 162.424 kN-m3 6

5

(a) Beam deflection at A: The beam deflection at A is: EI v A 162.424 kN-m3 vA

162.424 kN-m3 24,000 kN-m 2

0.006768 m

6.77 mm

Ans.

(b) Beam deflection at C: At x = 6.5 m, the beam deflection is: 35 kN 160.91 kN 25 kN/m EI vC (6.5 m)3 (4.0 m)3 (4.0 m) 4 6 6 24 45 kN/m (4.0 m)5 (28.511 kN-m 2 )(6.5 m) 162.424 kN-m3 120(4.0 m)

271.1797 kN-m3 vC

271.1797 kN-m3 24,000 kN-m 2

0.0011299 m

11.30 mm

Ans.


10.48 For the beam and loading shown, use discontinuity functions to compute (a) the slope of the beam at B and (b) the deflection of the beam at A. Assume a constant value of EI = 54,000 kN-m2 for the beam. Fig. P10.48

Solution Support reactions: A FBD of the beam is shown to the right. Fy C y (20 kN/m)(3 m) 12 (30 kN/m)(3 m) 0 Cy MC

105 kN

(20 kN/m)(3 m)(2.5 m) 1 2

(30 kN/m)(3 m)(2 m)

MC

MC

0

240 kN-m

Load function w(x):

w( x)

20 kN/m x 0 m

0

30 kN/m x 0m 3m

0

1

20 kN/m x 3 m

1

30 kN/m x 3 m 105 kN x 4 m 240 kN-m x 4 m Shear-force function V(x) and bending-moment function M(x): 30 kN/m 1 2 V ( x) 20 kN/m x 0 m x 0m 20 kN/m x 3 m 2(3 m) 30 kN/m x 3 m

M ( x)

1

105 kN x 4 m

0

240 kN-m x 4 m

30 kN/m x 3m 3m

0

1

2

30 kN/m x 3m 2(3 m)

1

2

1

20 kN/m 30 kN/m 20 kN/m 2 3 x 0m x 0m x 3m 2 6(3 m) 2 30 kN/m 2 1 0 x 3m 105 kN x 4 m 240 kN-m x 4 m 2

2

30 kN/m x 3m 6(3 m)

3

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 20 kN/m 30 kN/m 20 kN/m 2 3 2 EI 2 M ( x) x 0m x 0m x 3m dx 2 6(3 m) 2 30 kN/m 30 kN/m 3 2 1 0 x 3m x 3m 105 kN x 4 m 240 kN-m x 4 m 6(3 m) 2 Integrate the moment function to obtain an expression for the beam slope: dv 20 kN/m 30 kN/m 20 kN/m 30 kN/m 3 4 3 4 EI x 0m x 0m x 3m x 3m dx 6 24(3 m) 6 24(3 m) 30 kN/m 105 kN 3 2 1 (a) x 3m x 4m 240 kN-m x 4 m C1 6 2


Integrate again to obtain the beam deflection function: 20 kN/m 30 kN/m 20 kN/m 4 5 EI v x 0m x 0m x 3m 24 120(3 m) 24 30 kN/m 105 kN 240 kN-m 4 3 2 x 3m x 4m x 4m 24 6 2

4

30 kN/m x 3m 120(3 m)

5

(b)

C1 x C2

Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, both the slope dv/dx and the deflection v are known at the fixed support (x = 4 m). Substitute the boundary condition dv/dx = 0 at x = 4 m into Eq. (a) to obtain: 20 kN/m 30 kN/m 20 kN/m 30 kN/m 30 kN/m 0 (4 m)3 (4 m) 4 (1 m)3 (1 m) 4 (1 m)3 C1 6 24(3 m) 6 24(3 m) 6

C1 311.25 kN-m2 Next, substitute the boundary condition v = 0 at x = 4 m into Eq. (b) to obtain: 20 kN/m 30 kN/m 20 kN/m 30 kN/m 0 (4 m) 4 (4 m)5 (1 m) 4 (1 m)5 24 120(3 m) 24 120(3 m) 30 kN/m (1 m) 4 (311.25 kN-m 2 )(4 m) C2 24 C2 948.50 kN-m3 The beam slope and elastic curve equations are now complete: dv 20 kN/m 30 kN/m 20 kN/m 30 kN/m 3 4 3 4 EI x 0m x 0m x 3m x 3m dx 6 24(3 m) 6 24(3 m) 30 kN/m 105 kN 3 2 1 x 3m x 4m 240 kN-m x 4 m 311.25 kN-m 2 6 2 20 kN/m 30 kN/m 20 kN/m 30 kN/m 4 5 4 5 EI v x 0m x 0m x 3m x 3m 24 120(3 m) 24 120(3 m) 30 kN/m 105 kN 240 kN-m 4 3 2 x 3m x 4m x 4m 24 6 2 3 (311.25 kN-m)x 948.50 kN-m (a) Beam slope at B: The beam slope at B is: dv 20 kN/m 30 kN/m EI (3 m)3 (3 m) 4 dx B 6 24(3 m)

311.25 kN-m 2

187.5 kN-m 2 dv dx

B

187.5 kN-m 2 54,000 kN-m 2

0.003472 rad

0.00347 rad

Ans.

(b) Beam deflection at A: The beam deflection at A is: EI vA 948.50 kN-m3 vA

948.50 kN-m3 54,000 kN-m 2

0.017565 m

17.56 mm

Ans.


10.49a For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 8 × 104 kN-m2 is constant for each beam.

Fig. P10.49a

Solution Determine beam slope at A. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML (slope magnitude) A  6 EI Values: M = 150 kN-m, L = 8 m, EI = 8 × 104 kN-m2 Computation: ML (150 kN-m)(8 m) A    0.00250 rad 6 EI 6(8  104 kN-m 2 ) Determine beam deflection at H. [Skill 1]

vH  (3 m)(0.00250 rad)  0.00750 m  7.50 mm 

Ans.


10.49b For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 8 × 104 kN-m2 is constant for each beam.

Fig. P10.49b

Solution Determine beam deflection at A. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vA   8 EI Values: w = 6 kN/m, L = 4 m, EI = 8 × 104 kN-m2 Computation: wL4 (6 kN/m)(4 m)4 vA     0.00240 m 8EI 8(8  104 kN-m2 ) Determine beam slope at A. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL3 A  (slope magnitude) 6 EI Values: w = 6 kN/m, L = 4 m, EI = 8 × 104 kN-m2 Computation: wL3 (6 kN/m)(4 m)3 A    0.00080 rad 6EI 6(8  104 kN-m2 ) Determine beam deflection at H. [Skill 2]

vH  0.00240 m  (2 m)(0.00080 rad)  0.00400 m  4.00 mm 

Ans.


10.49c For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 8 × 104 kN-m2 is constant for each beam.

Fig. P10.49c

Solution Determine beam deflection at H. [Skill 3] [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vH   ( L  b 2  x 2 ) (elastic curve) 6 LEI Values: P = 30 kN-m, L = 12 m, b = 4 m, x = 4 m, EI = 8 × 104 kN-m2 Computation: Pbx 2 vH   ( L  b2  x2 ) 6 LEI



(30 kN)(4 m)(4 m) (12 m) 2  (4 m) 2  (4 m) 2  4 2  6(12 m)(8  10 kN-m )

 0.00933 m  9.33 mm 

Ans.


10.49d For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 8 × 104 kN-m2 is constant for each beam.

Fig. P10.49d

Solution Determine deflection of cantilever overhang. [Appendix C, Cantilever beam with concentrated load.] Relevant equation from Appendix C: PL3 vH ,cant   (assuming fixed support at B) 3EI Values: P = 15 kN, L = 4 m, EI = 8 × 104 kN-m2 Computation:

vH ,cant

PL3 (15 kN)(4 m)3    0.004000 m 3EI 3(8  104 kN-m2 )

Determine beam slope at B. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML (slope magnitude) B  3EI Values: M = (15 kN)(4 m) = 60 kN-m, L = 8 m, EI = 8 × 104 kN-m2 Computation: ML (60 kN-m)(8 m) B    0.002000 rad 3EI 3(8  104 kN-m 2 ) Determine beam deflection at H. [Skill 4]

vH  0.00400 m  (4 m)(0.00200 rad)  0.01200 m  12.00 mm 

Ans.


10.50a For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 1.2 × 107 kip-in.2 is constant for each beam.

Fig. P10.50a

Solution Determine beam deflection at B. [Appendix C, Cantilever beam with concentrated moment.] Relevant equation from Appendix C: ML2 vB   2 EI Values: M = 40 kip-ft, L = 9 ft, EI = 1.2 × 107 kip-in.2 Computation: ML2 (40 kip-ft)(9 ft)2 (12 in./ft)3 vB     0.23328 in. 2EI 2(1.2  107 kip-in.2 ) Determine beam slope at B. [Appendix C, Cantilever beam with concentrated moment.] Relevant equation from Appendix C: ML (slope magnitude) B  EI Values: M = 40 kip-ft, L = 9 ft, EI = 1.2 × 107 kip-in.2 Computation: ML (40 kip-ft)(9 ft)(12 in./ft)2 B    0.004320 rad EI (1.2  107 kip-in.2 ) Determine beam deflection at H. [Skill 2]

vH  0.23328 in.  (6 ft)(12 in./ft)(0.004320 rad)  0.54432 in.  0.544 in. 

Ans.


10.50b For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 1.2 × 107 kipin.2 is constant for each beam.

Fig. P10.50b

Solution Determine beam slope at C. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pa( L2  a 2 ) (slope magnitude) C  6 LEI Values: P = 25 kips, L = 18 ft, a = 12 ft, EI = 1.2 × 107 kip-in.2 Computation: Pa( L2  a 2 ) C  6 LEI (25 kips)(12 ft) (18 ft) 2  (12 ft) 2  (12 in./ft) 2  0.00600 rad  7 2  6(18 ft)(1.2  10 kip-in. ) Determine beam deflection at H. [Skill 1]

vH  (7 ft)(12 in./ft)(0.00600 rad)  0.5040 in.  0.504 in. 

Ans.


10.50c For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 1.2 × 107 kip-in.2 is constant for each beam.

Fig. P10.50c

Solution Determine beam deflection at H. [Skill 3] [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wx 2 vH   (6 L2  4 Lx  x 2 ) (elastic curve) 24 EI Values: w = 2.5 kips/ft, L = 15 ft, x = 9 ft, EI = 1.2 × 107 kip-in.2 Computation: wx 2 vH   (6 L2  4 Lx  x 2 ) 24 EI (2.5 kips/ft)(9 ft)2 6(15 ft)2  4(15 ft)(9 ft)  (9 ft)2  (12 in./ft)3  7 2  24(1.2  10 kip-in. )

 1.082565 in.  1.083 in. 

Ans.


10.50d For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 1.2 × 107 kip-in.2 is constant for each beam.

Fig. P10.50d

Solution Determine deflection of cantilever overhang. [Appendix C, Cantilever beam with uniform load.] Relevant equation from Appendix C: wL4 vH ,cant   (assuming fixed support at A) 8EI Values: w = 5 kips/ft, L = 8 ft, EI = 1.2 × 107 kip-in.2 Computation:

vH ,cant

wL4 (5 kips/ft)(8 ft)4 (12 in./ft)3    0.36864 in. 8EI 8(1.2  107 kip-in.2 )

Determine beam slope at A. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML (slope magnitude) A  3EI Values: M = (5 kips/ft)(8 ft)(4 ft) = 160 kip-ft, L = 22 ft, EI = 1.2 × 107 kip-in.2 Computation: ML (160 kip-ft)(22 ft)(12 in./ft)2 A    0.014080 rad 3EI 3(1.2  107 kip-in.2 ) Determine beam deflection at H. [Skill 4]


Ans.


10.51a For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 6 × 104 kN-m2 is constant for each beam.

Fig. P10.51a

Solution Determine beam deflection at H. [Skill 3] [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: Mx vH   (2 L2  3Lx  x 2 ) (elastic curve) 6 LEI Values: M = −60 kN-m, L = 12 m, x = 6 m, EI = 6 × 104 kN-m2 Computation: Mx vH   (2 L2  3Lx  x 2 ) 6 LEI



( 60 kN-m)(6 m)  2(12 m) 2  3(12 m)(6 m)  (6 m) 2  6(12 m)(6  104 kN-m 2 ) 

 0.009000 m  9.00 mm 

Ans.


10.51b For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 6 × 104 kN-m2 is constant for each beam.

Fig. P10.51b

Solution Determine deflection of cantilever overhang. [Appendix C, Cantilever beam with uniform load.] Relevant equation from Appendix C: wL4 vH ,cant   (assuming fixed support at A) 8EI Values: w = 7.5 kN/m, L = 3 m, EI = 6 × 104 kN-m2 Computation:

vH ,cant  

wL4 (7.5 kN/m)(3 m)4   0.00126563 m 8EI 8(6  104 kN-m2 )

Determine beam slope at A. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML (slope magnitude) A  3EI Values: M = (7.5 kN/m)(3 m)(1.5 m) = 33.75 kN-m, L = 6 m, EI = 6 × 104 kN-m2 Computation: ML (33.75 kN-m)(6 m) A    0.001125 rad 3EI 3(6  104 kN-m 2 ) Determine beam deflection at H. [Skill 4]

vH  0.00126563 m  (3 m)(0.001125 rad)  0.00464063 m  4.64 mm 

Ans.


10.51c For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 6 × 104 kN-m2 is constant for each beam.

Fig. P10.51c

Solution Determine beam deflection at B. [Appendix C, Cantilever beam with concentrated load.] Relevant equation from Appendix C: PL3 vB   3EI Values: P = 30 kN, L = 3 m, EI = 6 × 104 kN-m2 Computation: PL3 (30 kN)(3 m)3 vB     0.004500 m 3EI 3(6  104 kN-m2 ) Determine beam slope at B. [Appendix C, Cantilever beam with concentrated load.] Relevant equation from Appendix C: PL2 B  (slope magnitude) 2 EI Values: P = 30 kN, L = 3 m, EI = 6 × 104 kN-m2 Computation: PL2 (30 kN)(3 m)2 B    0.002250 rad 2EI 2(6  104 kN-m2 ) Determine beam deflection at H. [Skill 2]

vH  0.004500 m  (3 m)(0.002250 rad)  0.01125 m  11.25 mm 

Ans.

Alternative solution for beam deflection at B. [Appendix C, Cantilever beam with concentrated load at midspan.] 5 PL3 Relevant equation from Appendix C: vH   48 EI 4 Values: P = 30 kN, L = 6 m, EI = 6 × 10 kN-m2 5PL3 5(30 kN)(6 m)3 Computation: vH     0.011250 m  11.25 mm  48EI 48(6  104 kN-m2 )


10.51d For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 6 × 104 kN-m2 is constant for each beam.

Fig. P10.51d

Solution Determine beam slope at C. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 C  (2 L  a)2 (slope magnitude) 24 LEI Values: w = 5 kN/m, L = 9 m, a = 6 m, EI = 6 × 104 kN-m2 Computation: wa 2 (5 kN/m)(6 m)2 2 C  (2L  a)2  2(9 m)  (6 m)  0.00200 rad 4 2  24LEI 24(9 m)(6 10 kN-m ) Determine beam deflection at H. [Skill 1]

vH  (3 m)(0.00200 rad)  0.00600 m  6.00 mm 

Ans.


10.52a For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 3.0 × 106 kipin.2 is constant for each beam.

Fig. P10.52a

Solution Determine deflection of cantilever overhang. [Appendix C, Cantilever beam with concentrated moment.] Relevant equation from Appendix C: ML2 vH ,cant   (assuming fixed support at A) 2 EI Values: M = 50 kip-ft, L = 6 ft, EI = 3.0 × 106 kip-in.2 Computation:

vH ,cant  

ML2 (50 kip-ft)(6 ft)2 (12 in./ft)3   0.51840 in. 2EI 2(3.0  106 kip-in.2 )

Determine beam slope at A. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML (slope magnitude) A  3EI Values: M = 50 kip-ft, L = 18 ft, EI = 3.0 × 106 kip-in.2 Computation: ML (50 kip-ft)(18 ft)(12 in./ft)2 A    0.01440 rad 3EI 3(3.0  106 kip-in.2 ) Determine beam deflection at H. [Skill 4]


Ans.


10.52b For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 3.0 × 106 kip-in.2 is constant for each beam.

Fig. P10.52b

Solution Determine beam deflection at H. [Skill 3] [Appendix C, Cantilever beam with concentrated load.] Relevant equation from Appendix C: Px 2 vH   (3L  x) (elastic curve) 6 EI Values: P = 10 kips, L = 10 ft, x = 7 ft, EI = 3.0 × 106 kip-in.2 Computation: Px 2 vH   (3L  x) 6 EI (10 kips)(7 ft) 2 (12 in./ft) 3  [3(10 ft)  (7 ft)]  1.081920 in.  1.082 in.  6(3.0  106 kip-in.2 )

Ans.


10.52c For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 3.0 × 106 kip-in.2 is constant for each beam.

Fig. P10.52c

Solution Determine beam slope at A. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML (slope magnitude) A  6 EI Values: M = (2 kips/ft)(8 ft)(4 ft) = 64 kip-ft, L = 18 ft, EI = 3.0 × 106 kip-in.2 Computation: ML (64 kip-ft)(18 ft)(12 in./ft) 2 A    0.009216 rad 6EI 6(3.0  106 kip-in.2 ) Determine beam deflection at H. [Skill 1]

vH  (6 ft)(12 in./ft)(0.009216 rad)  0.663552 in.  0.664 in. 

Ans.


10.52d For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 3.0 × 106 kip-in.2 is constant for each beam.

Fig. P10.52d

Solution Determine beam deflection at B. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vB   8 EI Values: w = 1.5 kips/ft, L = 10 ft, EI = 3.0 × 106 kip-in.2 Computation: wL4 (1.5 kips/ft)(10 ft)4 (12 in./ft)3 vB     1.0800 in. 8EI 8(3.0  106 kip-in.2 ) Determine beam slope at B. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL3 B  (slope magnitude) 6 EI Values: w = 1.5 kips/ft, L = 10 ft, EI = 3.0 × 106 kip-in.2 Computation: wL3 (1.5 kips/ft)(10 ft)3 (12 in./ft)2 B    0.01200 rad 6EI 6(3.0  106 kip-in.2 ) Determine beam deflection at H. [Skill 2]


Ans.


10.53 The simply supported beam shown in Fig. P10.53 consists of a W24 × 94 structural steel wide-flange shape [E = 29,000 ksi; I = 2,700 in.4]. For the loading shown, determine the beam deflection at point C. Fig. P10.53

Solution Consider distributed load. [Appendix C, SS beam with uniformly distributed load over portion of span.] Relevant equation from Appendix C: wa3 vC   (4L2  7aL  3a 2 ) 24 LEI Values: w = 3.2 kips/ft, L = 28 ft, a = 21 ft, EI = 7.830 × 107 kip-in.2 Computation: wa 3 vC   (4 L2  7aL  3a 2 ) 24 LEI 

(3.2 kips/ft)(21 ft)3 (12 in./ft)3  4(28 ft) 2  7(21 ft)(28 ft)  3(21 ft)2   0.333822 in. 24(28 ft)(7.830  107 kip-in.2 ) 

Consider concentrated load. [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: Px vC   (3L2  4 x 2 ) (elastic curve) 48EI Values: P = 36 kips, L = 28 ft, x = 7 ft, EI = 7.830 × 107 kip-in.2 Computation: Px vC   (3L2  4 x 2 ) 48EI (36 kips)(7 ft)(12 in./ft)3 3(28 ft)2  4(7 ft) 2   0.249799 in.  7 2  48(7.830  10 kip-in. ) Beam deflection at C

vC  0.333822 in.  0.249799 in.  0.583620 in.  0.584 in. 

Ans.


10.54 The simply supported beam shown in Fig. P10.54 consists of a W460 × 82 structural steel wide-flange shape [E = 200 GPa; I = 370 × 106 mm4]. For the loading shown, determine the beam deflection at point C. Fig. P10.54

Solution Consider distributed load. [Appendix C, SS beam with uniformly distributed load over portion of span.] Relevant equation from Appendix C: wa3 vC   (4L2  7aL  3a 2 ) 24 LEI Values: w = 26 kN/m, L = 8 m, a = 6 m, EI = 7.4 × 104 kN-m2 Computation: wa 3 vC   (4 L2  7aL  3a 2 ) 24 LEI 

(26 kN/m)(6 m)3  4(8 m) 2  7(6 m)(8 m)  3(6 m) 2   0.011068 m 24(8 m)(7.40  104 kN-m 2 ) 

Consider concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vC   ( L  b 2  x 2 ) (elastic curve) 6 LEI Values: P = 60 kN, L = 8 m, b = 3 m, x = 2 m, EI = 7.4 × 104 kN-m2 Computation: Pbx 2 vC   ( L  b2  x 2 ) 6 LEI (60 kN)(3 m)(2 m) (8 m) 2  (3 m) 2  (2 m) 2   0.005169 m  4 2  6(8 m)(7.40  10 kN-m ) Beam deflection at C

vC  0.011068 m  0.005169 m  0.016237 m  16.24 mm 

Ans.


10.55 The simply supported beam shown in Fig. P10.55 consists of a W410 × 60 structural steel wide-flange shape [E = 200 GPa; I = 216 × 106 mm4]. For the loading shown, determine the beam deflection at point B. Fig. P10.55

Solution Consider concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 vB   ( L  a 2  b2 ) 6 LEI Values: P = 60 kN, L = 9 m, a = 3 m, b = 6 m, EI = 4.32 × 104 kN-m2 Computation: Pab 2 vB   ( L  a 2  b2 ) 6 LEI (60 kN)(3 m)(6 m) (9 m) 2  (3 m)2  (6 m) 2   0.016667 m  4 2  6(9 m)(4.32  10 kN-m ) Consider concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vB   (2 L2  3Lx  x 2 ) (elastic curve) 6 LEI Values: M = −45 kN-m, L = 9 m, x = 6 m, EI = 4.32 × 104 kN-m2 Computation: Mx vB   (2 L2  3Lx  x 2 ) 6 LEI (45 kN-m)(6 m)  2(9 m) 2  3(9 m)(6 m)  (6 m) 2   0.004167 m  4 2  6(9 m)(4.32  10 kN-m ) Beam deflection at B

vB  0.016667 m  0.004167 m  0.012500 m  12.50 mm 

Ans.


10.56 The simply supported beam shown in Fig. P10.56 consists of a W21 × 44 structural steel wide-flange shape [E = 29,000 ksi; I = 843 in.4]. For the loading shown, determine the beam deflection at point B. Fig. P10.56

Solution Consider uniformly distributed load. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa3 vB   (4L2  7aL  3a 2 ) 24 LEI Values: w = 5 kips/ft, L = 24 ft, a = 16 ft, EI = 2.4447 × 107 kip-in.2 Computation: wa 3 vB   (4 L2  7 aL  3a 2 ) 24 LEI 

(5 kips/ft)(16 ft)3 (12 in./ft)3  4(24 ft) 2  7(16 ft)(24 ft)  3(16 ft) 2   0.965066 in. 24(24 ft)(2.4447  107 kip-in.2 ) 

Consider concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vB   (2 L2  3Lx  x 2 ) (elastic curve) 6 LEI Values: M = −200 kip-ft, L = 24 ft, x = 8 ft, EI = 2.4447 × 107 kip-in.2 Computation: Mx vB   (2 L2  3Lx  x 2 ) 6 LEI (200 kip-ft)(8 ft)(12 in./ft)3  2(24 ft) 2  3(24 ft)(8 ft)  (8 ft) 2   0.502638 in.  7 2  6(24 ft)(2.4447  10 kip-in. )

Beam deflection at B

vB  0.965066 in.  0.502638 in.  0.462428 in.  0.462 in. 

Ans.


10.57 The cantilever beam shown in Fig. P10.57 consists of a rectangular structural steel tube shape [E = 29,000 ksi; I = 476 in.4]. For the loading shown, determine: (a) the beam deflection at point B. (b) the beam deflection at point C. Fig. P10.57

Solution (a) Beam deflection at point B Consider uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vB   8 EI Values: w = 2 kips/ft, L = 6 ft, EI = 1.3804 × 107 kip-in.2 Computation: wL4 (2 kips/ft)(6 ft)4 (12 in./ft)3 vB     0.040559 in. 8EI 8(1.3804  107 kip-in.2 ) Consider concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: Px 2 vB   (3L  x) (elastic curve) 6 EI Values: P = 12 kips, L = 10 ft, x = 6 ft, EI = 1.3804 × 107 kip-in.2 Computation: Px 2 (12 kips)(6 ft)2 (12 in./ft)3 vB   (3L  x)   3(10 ft)  (6 ft)  0.216313 in. 6EI 6(1.3804  107 kip-in.2 ) Beam deflection at B

vB  0.040559 in.  0.216313 in.  0.256872 in.  0.257 in. 

Ans.


(b) Beam deflection at point C Consider uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL3 B   6 EI Values: w = 2 kips/ft, L = 6 ft, EI = 1.3804 × 107 kip-in.2 Computation: wL3 (2 kips/ft)(6 ft)3 (12 in./ft) 2 B     751.0866  106 rad 7 2 6 EI 6(1.3804  10 kip-in. ) vC  0.040559 in.  (4 ft)(12 in./ft)(751.0866  106 rad)  0.076611 in.

Consider concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vC   3EI Values: P = 12 kips, L = 10 ft, EI = 1.3804 × 107 kip-in.2 Computation: PL3 (12 kips)(10 ft)3 (12 in./ft)3 vC     0.500724 in. 3EI 3(1.3804  107 kip-in.2 ) Beam deflection at C

vC  0.076611 in.  0.500724 in.  0.577336 in.  0.577 in. 

Ans.


10.58 The cantilever beam shown in Fig. P10.58 consists of a rectangular structural steel tube shape [E = 200 GPa; I = 400 × 106 mm4]. For the loading shown, determine: (a) the beam deflection at point A. (b) the beam deflection at point B. Fig. P10.58

Solution (a) Beam deflection at point A Consider uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vA   8 EI Values: w = 25 kN/m, L = 4 m, EI = 8.0 × 104 kN-m2 Computation: wL4 (25 kN/m)(4 m)4 vA     0.010000 m 8EI 8(8.0  104 kN-m2 ) Consider concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 PL2 vB   and  B   3EI 2 EI Values: P = 55 kN, L = 2.5 m, EI = 8.0 × 104 kN-m2 Computation: PL3 (55 kN)(2.5 m)3 vB     0.003581 m 3EI 3(8.0  104 kN-m 2 )

B 

PL2 (55 kN)(2.5 m) 2   0.002148 rad 2 EI 2(8.0  104 kN-m 2 )

vA  0.003581 m  (1.5 m)(0.002148 rad)  0.006803 m Beam deflection at A

vA  0.010000 m  0.006803 m  0.016803 m  16.80 mm 

Ans.


(b) Beam deflection at point B Consider uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wx 2 vB   (6 L2  4 Lx  x 2 ) (elastic curve) 24 EI Values: w = 25 kN/m, L = 4 m, x = 2.5 m, EI = 8.0 × 104 kN-m2 Computation: wx 2 vB   (6 L2  4 Lx  x 2 ) 24 EI (25 kN/m)(2.5 m) 2 6(4.0 m) 2  4(4.0 m)(2.5 m)  (2.5 m) 2   0.005066 m  24(8.0  104 kN-m 2 )  Consider concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 vB   3EI Values: P = 55 kN, L = 2.5 m, EI = 8.0 × 104 kN-m2 Computation: PL3 (55 kN)(2.5 m)3 vB     0.003581 m 3EI 3(8.0  104 kN-m2 ) Beam deflection at B

vB  0.005066 m  0.003581 m  0.008647 m  8.65 mm 

Ans.


10.59 The solid 1.25-in.-diameter steel [E = 29,000 ksi] shaft shown in Fig. P10.59 supports two pulleys. For the loading shown, determine: (a) the shaft deflection at point B. (b) the shaft deflection at point C.

Fig. P10.59




64

(1.25 in.)4  0.119842 in.4

(a) Shaft deflection at point B Consider concentrated load at pulley B. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vB   3EI Values: P = 200 lb, L = 10 in., EI = 3.47543 × 106 lb-in.2 Computation: PL3 (200 lb)(10 in.)3 vB     0.019182 in. 3EI 3(3.47543  106 lb-in.2 ) Consider concentrated load at pulley C. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: Px 2 vB   (3L  x) (elastic curve) 6 EI Values: P = 120 lb, L = 25 in., x = 10 in., EI = 3.47543 × 106 lb-in.2 Computation: Px 2 (120 lb)(10 in.)2 vB   (3L  x)   3(25 in.)  (10 in.)  0.037405 in. 6EI 6(3.47543  106 lb-in.2 ) Shaft deflection at B

vB  0.019182 in.  0.037405 in.  0.056588 in.  0.0566 in. 

Ans.


(b) Shaft deflection at point C Consider concentrated load at pulley B. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 PL2 vB   and  B  (magnitude) 3EI 2 EI Values: P = 200 lb, L = 10 in., EI = 3.47543 × 106 lb-in.2 Computation: PL3 (200 lb)(10 in.)3 vB     0.019182 in. 3EI 3(3.47543  106 lb-in.2 )

B 

PL2 (200 lb)(10 in.) 2   0.0028773 rad 2 EI 2(3.47543  106 lb-in.2 )

vC  0.019182 in.  (15 in.)(0.0028773 rad)  0.062342 in.

Consider concentrated load at pulley C. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vC   3EI Values: P = 120 lb, L = 25 in., EI = 3.47543 × 106 lb-in.2 Computation: PL3 (120 lb)(25 in.)3 vC     0.179834 in. 3EI 3(3.47543  106 lb-in.2 ) Shaft deflection at C

vC  0.062342 in.  0.179834 in.  0.242176 in.  0.242 in. 

Ans.


10.60 The cantilever beam shown in Fig. P10.60 consists of a rectangular structural steel tube shape [E = 29,000 ksi; I = 1,710 in.4]. For the loading shown, determine: (a) the beam deflection at point A. (b) the beam deflection at point B. Fig. P10.60

Solution (a) Beam deflection at point A Consider concentrated moment. [Appendix C, Cantilever beam with concentrated moment.] Relevant equation from Appendix C: ML2 vA   2 EI Values: M = −200 kip-ft, L = 15 ft, EI = 4.959 × 107 kip-in.2 Computation: ML2 (200 kip-ft)(15 ft) 2 (12 in./ft)3 vA     0.784029 in. 2EI 2(4.959  107 kip-in.2 )

Consider concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 PL2 vB   and  B  (slope magnitude) 3EI 2 EI Values: P = −18 kips, L = 9 ft, EI = 4.959 × 107 kip-in.2 Computation: PL3 (18 kips)(9 ft)3 (12 in./ft)3 vB     0.152415 in. 3EI 3(4.959  107 kip-in.2 )

PL2 (18 kips)(9 ft)2 (12 in./ft)2 B    0.0021169 rad 2 EI 2(4.959  107 kip-in.2 ) v A  0.152415 in.  (6 ft)(12 in./ft)(0.0021169 rad)  0.304830 in. Beam deflection at A

vA  0.784029 in.  0.304830 in.  1.088860 in.  1.089 in. 

Ans.


(b) Beam deflection at point B Consider concentrated moment. [Appendix C, Cantilever beam with concentrated moment.] Relevant equation from Appendix C: Mx 2 vB   (elastic curve) 2 EI Values: M = −200 kip-ft, L = 15 ft, x = 9 ft, EI = 4.959 × 107 kip-in.2 Computation: Mx 2 (200 kip-ft)(9 ft)2 (12 in./ft)3 vB     0.282250 in. 2EI 2(4.959  107 kip-in.2 )

Consider concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vB   3EI Values: P = −18 kips, L = 9 ft, EI = 4.959 × 107 kip-in.2 Computation: PL3 (18 kips)(9 ft)3 (12 in./ft)3 vB     0.152415 in. 3EI 3(4.959  107 kip-in.2 ) Beam deflection at B

vB  0.282250 in.  0.152415 in.  0.434665 in.  0.435 in. 

Ans.


10.61 The simply supported beam shown in Fig. P10.61 consists of a W21 × 44 structural steel wide-flange shape [E = 29,000 ksi; I = 843 in.4]. For the loading shown, determine: (a) the beam deflection at point A. (b) the beam deflection at point C. Fig. P10.61

Solution (a) Beam deflection at point A Determine cantilever deflection due to uniformly distributed load on overhang. [Appendix C, Cantilever beam with uniform load.] Relevant equation from Appendix C: wL4 vA   (assuming fixed support at B) 8 EI Values: w = 4 kips/ft, L = 8 ft, EI = 2.4447 × 107 kip-in.2 Computation: wL4 (4 kips/ft)(8 ft)4 (12 in./ft)3 vA     0.144760 in. 8EI 8(2.4447  107 kip-in.2 ) Consider deflection at A resulting from rotation at B caused by distributed load on overhang. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML (slope magnitude) B  3EI Values: M = (4 kips/ft)(8 ft)(4 ft) = 128 kip-ft, L = 22 ft, EI = 2.4447 × 107 kip-in.2 Computation: ML (128 kip-ft)(22 ft)(12 in./ft) 2 B    0.0055290 rad 3EI 3(2.4447  107 kip-in.2 ) vA  (8 ft)(12 in./ft)(0.0055290 rad)  0.530786 in.


Consider concentrated load. [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL2 B  (slope magnitude) 16 EI Values: P = 45 kips, L = 22 ft, EI = 2.4447 × 107 kip-in.2 Computation: PL2 (45 kips)(22 ft) 2 (12 in./ft) 2 B    0.0080182 rad 16 EI 16(2.4447  107 kip-in.2 ) v A  (8 ft)(12 in./ft)(0.0080182 rad)  0.769744 in.

Beam deflection at A

vA  0.144760 in.  0.530786 in.  0.769744 in.  0.094198 in.  0.0942 in. 

Ans.

(b) Beam deflection at point C Consider distributed load on overhang. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: Mx vC   (2 L2  3Lx  x 2 ) (elastic curve) 6 LEI Values: M = (4 kips/ft)(8 ft)(4 ft) = −128 kip-ft, L = 22 ft, x = 11 ft, EI = 2.4447 × 107 kip-in.2 Computation: Mx vC   (2 L2  3Lx  x 2 ) 6 LEI (128 kip-ft)(11 ft)(12 in./ft)3  2(22 ft) 2  3(22 ft)(11 ft)  (11 ft) 2   0.273687 in.  7 2  6(22 ft)(2.4447  10 kip-in. )

Consider concentrated load. [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL3 vC   48 EI Values: P = 45 kips, L = 22 ft, EI = 2.4447 × 107 kip-in.2 Computation: PL3 (45 kips)(22 ft)3 (12 in./ft)3 vC     0.705598 in. 48EI 48(2.4447 107 kip-in.2 ) Beam deflection at C

vC  0.273687 in.  0.705598 in.  0.431912 in.  0.432 in. 

Ans.


10.62 The simply supported beam shown in Fig. P10.62 consists of a W530 × 66 structural steel wide-flange shape [E = 200 GPa; I = 351 × 106 mm4]. For the loading shown, determine: (a) the beam deflection at point B. (b) the beam deflection at point D. Fig. P10.62

Solution (a) Beam deflection at point B Consider distributed load between supports. [Appendix C, SS beam with uniformly distributed load.] Relevant equation from Appendix C: 5wL4 vB   384 EI Values: w = 55 kN/m, L = 7.2 m, EI = 7.02 × 104 kN-m2 Computation: 5wL4 5(55 kN/m)(7.2 m)4 vB     0.027415 m 384EI 384(7.02  104 kN-m2 ) Consider distributed load on overhang. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: Mx vB   (2 L2  3Lx  x 2 ) (elastic curve) 6 LEI Values: M = (55 kN/m)(2.8 m)(1.4 m) = −215.6 kN-m, L = 7.2 m, x = 3.6 m, EI = 7.02 × 104 kN-m2 Computation: Mx vB   (2 L2  3Lx  x 2 ) 6 LEI (215.6 kN-m)(3.6 m)  2(7.2 m) 2  3(7.2 m)(3.6 m)  (3.6 m) 2   0.009951 m  6(7.2 m)(7.02  107 kN-m 2 )  Beam deflection at B

vB  0.027415 m  0.009951 m  0.017464 m  17.46 mm 

Ans.


(b) Beam deflection at point D Consider distributed load between supports. [Appendix C, SS beam with uniformly distributed load.] Relevant equation from Appendix C: wL3 C  (slope magnitude) 24 EI Values: w = 55 kN/m, L = 7.2 m, EI = 7.02 × 104 kN-m2 Computation: wL3 (55 kN/m)(7.2 m)3 C    0.0121846 rad 24 EI 24(7.02  104 kN-m 2 ) vD  (2.8 m)(0.0121846 rad)  0.034117 m

Consider deflection at D resulting from rotation at C caused by distributed load on overhang. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML (slope magnitude) C  3EI Values: M = (55 kN/m)(2.8 m)(1.4 m) = 215.6 kN-m, L = 7.2 m, EI = 7.02 × 104 kN-m2 Computation: ML (215.6 kN-m)(7.2 m) C    0.0073709 rad 3EI 3(7.02  107 kN-m 2 )

vD  (2.8 m)(0.0073709 rad)  0.020639 m Determine cantilever deflection due to uniformly distributed load on overhang. [Appendix C, Cantilever beam with uniform load.] Relevant equation from Appendix C: wL4 vD   (assuming fixed support at C) 8EI Values: w = 55 kN/m, L = 2.8 m, EI = 7.02 × 104 kN-m2 Computation: wL4 (55 kN/m)(2.8 m) 4 vD     0.006020 m 8EI 8(7.02  104 kN-m2 ) Beam deflection at D

vD  0.034117 m  0.020639 m  0.006020 m  0.007459 m  7.46 mm 

Ans.


10.63 The simply supported beam shown in Fig. P10.63 consists of a W21 × 44 structural steel wide-flange shape [E = 29,000 ksi; I = 843 in.4]. For a loading of w = 6 kips/ft, determine: (a) the beam deflection at point A. (b) the beam deflection at point C. Fig. P10.63

Solution (a) Beam deflection at point A Determine cantilever deflection due to linearly distributed load on overhang. [Appendix C, Cantilever beam with linear load.] Relevant equation from Appendix C: w0 L4 vA (assuming fixed support at B) 30 EI Values: w0 = −6 kips/ft, L = 12 ft, EI = 2.4447 × 107 kip-in.2 Computation: w0 L4 vA 30EI

( 6 kips/ft)(12 ft)4 (12 in./ft)3 30(2.4447 107 kip-in.2 )

0.293139 in.

Consider deflection at A resulting from rotation at B caused by linear load on overhang. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML (slope magnitude) B 3EI Values: M = ½(6 kips/ft)(12 ft)(4 ft) = 144 kip-ft, L = 18 ft, EI = 2.4447 × 107 kip-in.2 Computation: ML (144 kip-ft)(18 ft)(12 in./ft) 2 B 3EI 3(2.4447 107 kip-in.2 ) vA

(12 ft)(12 in./ft)(0.0050892 rad)

0.0050892 rad

0.732847 in.

Consider uniformly distributed loads between supports. [Appendix C, SS beam with uniformly distributed load.] Relevant equation from Appendix C: wL3 (slope magnitude) B 24 EI Values: w = 6 kips/ft, L = 18 ft, EI = 2.4447 × 107 kip-in.2


Computation: wL3 B 24 EI vA

(6 kips/ft)(18 ft)3 (12 in./ft) 2 24(2.4447 107 kip-in.2 )

(12 ft)(12 in./ft)(0.0085880 rad)

0.0085880 rad 1.236679 in.

Consider deflection at A resulting from rotation at B caused by uniform load on overhang DE. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML (slope magnitude) B 6 EI Values: M = (6 kips/ft)(6 ft)(3 ft) = 108 kip-ft, L = 18 ft, EI = 2.4447 × 107 kip-in.2 Computation: ML (108 kip-ft)(18 ft)(12 in./ft) 2 B 6 EI 6(2.4447 107 kip-in.2 ) vA

(12 ft)(12 in./ft)(0.0019085 rad)

0.0019085 rad

0.274818 in.


vA

0.293139 in. 0.732847 in. 1.236679 in. 0.274818 in. 0.064124 in.

0.0641 in.

Ans.


(b) Beam deflection at point C Consider deflection at C from moment caused by linear load on overhang. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: Mx vC (2 L2 3Lx x 2 ) (elastic curve) 6 LEI Values: M = ½(6 kips/ft)(12 ft)(4 ft) = 144 kip-ft, L = 18 ft, x = 9 ft, EI = 2.4447 × 107 kip-in.2 Computation: Mx vC (2 L2 3Lx x 2 ) 6 LEI (144 kip-ft)(9 ft)(12 in./ft) 3 2(18 ft) 2 7 2 6(18 ft)(2.4447 10 kip-in. )

3(18 ft)(9 ft) (9 ft) 2

0.206112 in.

Consider uniformly distributed loads between supports. [Appendix C, SS beam with uniformly distributed load.] Relevant equation from Appendix C: 5wL4 vC 384 EI Values: w = −6 kips/ft, L = 18 ft, EI = 2.4447 × 107 kip-in.2 Computation: 5wL4 5( 6 kips/ft)(18 ft)4 (12 in./ft)2 vC 0.579693 in. 384EI 384(2.4447 107 kip-in.2 ) Consider deflection at C resulting from moment caused by uniform load on overhang DE. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: Mx vC (2 L2 3Lx x 2 ) (elastic curve) 6 LEI Values: M = (6 kips/ft)(6 ft)(3 ft) = 108 kip-ft, L = 18 ft, x = 9 ft, EI = 2.4447 × 107 kip-in.2 Computation: Mx vC (2 L2 3Lx x 2 ) 6 LEI (108 kip-ft)(9 ft)(12 in./ft) 3 2(18 ft) 2 7 2 6(18 ft)(2.4447 10 kip-in. )

3(18 ft)(9 ft) (9 ft) 2

0.154585 in.

Beam deflection at C

vC

0.206112 in. 0.579693 in. 0.154585 in. 0.218995 in.

0.219 in.

Ans.


10.64 The simply supported beam shown in Fig. P10.64 consists of a W21 × 44 structural steel wide-flange shape [E = 29,000 ksi; I = 843 in.4]. For a loading of w = 8 kips/ft, determine: (a) the beam deflection at point C. (b) the beam deflection at point E. Fig. P10.64

Solution (a) Beam deflection at point C Consider deflection at C from moment caused by linear load on overhang. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: Mx vC (2 L2 3Lx x 2 ) (elastic curve) 6 LEI Values: M = ½(8 kips/ft)(12 ft)(4 ft) = 192 kip-ft, L = 18 ft, x = 9 ft, EI = 2.4447 × 107 kip-in.2 Computation: Mx vC (2 L2 3Lx x 2 ) 6 LEI (192 kip-ft)(9 ft)(12 in./ft)3 2(18 ft) 2 7 2 6(18 ft)(2.4447 10 kip-in. )

3(18 ft)(9 ft) (9 ft) 2

0.274816 in.

Consider uniformly distributed loads between supports. [Appendix C, SS beam with uniformly distributed load.] Relevant equation from Appendix C: 5wL4 vC 384 EI Values: w = −8 kips/ft, L = 18 ft, EI = 2.4447 × 107 kip-in.2 Computation: 5wL4 5( 8 kips/ft)(18 ft) 4 (12 in./ft) 2 vC 0.772924 in. 384 EI 384(2.4447 107 kip-in.2 )

Consider deflection at C resulting from moment caused by uniform load on overhang DE. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: Mx vC (2 L2 3Lx x 2 ) (elastic curve) 6 LEI Values: M = (8 kips/ft)(6 ft)(3 ft) = 144 kip-ft, L = 18 ft, x = 9 ft, EI = 2.4447 × 107 kip-in.2 Computation:


vC

Mx (2 L2 6 LEI

3Lx

x2 )

(144 kip-ft)(9 ft)(12 in./ft)3 2(18 ft) 2 7 2 6(18 ft)(2.4447 10 kip-in. )

3(18 ft)(9 ft) (9 ft) 2

0.206113 in.


vC

0.274816 in. 0.772924 in. 0.206113 in. 0.291995 in.

0.292 in.

Ans.

(b) Beam deflection at point E Consider deflection at E resulting from rotation at D caused by linear load on overhang. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML (slope magnitude) D 6 EI Values: M = ½(8 kips/ft)(12 ft)(4 ft) = 192 kip-ft, L = 18 ft, EI = 2.4447 × 107 kip-in.2 Computation: ML D 6 EI vE

(192 kip-ft)(18 ft)(12 in./ft) 2 6(2.4447 107 kip-in.2 )

(6 ft)(12 in./ft)(0.0033928 rad)

0.0033928 rad

0.244282 in.

Consider uniformly distributed loads between supports. [Appendix C, SS beam with uniformly distributed load.] Relevant equation from Appendix C: wL3 (slope magnitude) D 24 EI Values: w = 8 kips/ft, L = 18 ft, EI = 2.4447 × 107 kip-in.2 Computation: wL3 D 24 EI vE


(6 ft)(12 in./ft)(0.0114507 rad)

0.0114507 rad 0.824448 in.

Consider deflection at E resulting from rotation at D caused by uniform load on overhang DE. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML (slope magnitude) D 3EI Values: M = (8 kips/ft)(6 ft)(3 ft) = 144 kip-ft, L = 18 ft, EI = 2.4447 × 107 kip-in.2 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Computation: ML (144 kip-ft)(18 ft)(12 in./ft) 2 D 3EI 3(2.4447 107 kip-in.2 ) vE

(6 ft)(12 in./ft)(0.0050892 rad)

0.0050892 rad

0.366422 in.

Determine cantilever deflection due to uniformly distributed load on overhang DE. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vE (assuming fixed support at D) 8EI Values: w = −8 kips/ft, L = 6 ft, EI = 2.4447 × 107 kip-in.2 Computation: wL4 vE 8 EI

( 8 kips/ft)(6 ft) 4 (12 in./ft)3 8(2.4447 107 kip-in.2 )

0.091605 in.

Beam deflection at E vE 0.244282 in. 0.824448 in. 0.366422 in. 0.091605 in. 0.122139 in.

0.1221 in.

Ans.


10.65 The solid 30-mm-diameter steel [E = 200 GPa] shaft shown in Fig. P10.65 supports two belt pulleys. Assume that the bearing at B can be idealized as a roller support and that the bearing at D can be idealized as a pin support. For the loading shown, determine: (a) the shaft deflection at pulley A. (b) the shaft deflection at pulley C. Fig. P10.65

Solution Section properties: I

64

(30 mm)4

39,760.78 mm4

(a) Shaft deflection at pulley A Determine cantilever deflection due to pulley A load. [Appendix C, Cantilever beam with concentrated load.] Relevant equation from Appendix C: PL3 vA (assuming fixed support at B) 3EI Values: P = 700 N, L = 500 mm, EI = 7.95216 × 109 N-mm2 Computation: PL3 vA 3EI

(700 N)(500 mm)3 3(7.95216 109 N-mm2 )

3.6678 mm

Consider deflection at A resulting from rotation at B caused by pulley A load. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML (slope magnitude) B 3EI Values: M = (700 N)(500 mm) = 350,000 N-mm, L = 1,800 mm, EI = 7.95216 × 109 N-mm2 Computation: ML (350,000 N-mm)(1,800 mm) B 3EI 3(7.95216 109 N-mm 2 ) vA

(500 mm)(0.0264079 rad)

0.0264079 rad

13.2040 mm

Consider deflection at A resulting from rotation at B caused by pulley C load. [Appendix C, SS beam with concentrated load.]


Relevant equation from Appendix C: PL2 (slope magnitude) B 16 EI Values: P = 1,000 N, L = 1,800 mm, EI = 7.95216 × 109 N-mm2 Computation: PL2 (1,000 N)(1,800 mm) 2 B 16 EI 16(7.95216 109 N-mm 2 ) vA

(500 mm)(0.0254648 rad)

0.0254648 rad

12.7324 mm

Shaft deflection at A

vA

3.6678 mm 13.2040 mm 12.7324 mm

4.1393 mm

4.14 mm

Ans.

(b) Shaft deflection at pulley C Consider pulley A load. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: Mx vC (2 L2 3Lx x 2 ) (elastic curve) 6 LEI Values: M = (700 N)(500 mm) = −350,000 N-mm, L = 1,800 mm, x = 900 mm, EI = 7.95216 × 109 N-mm2 Computation: Mx vC (2 L2 3Lx x 2 ) 6 LEI ( 350,000 N-mm)(900 mm) 2(1,800 mm) 2 3(1,800 mm)(900 mm) (900 mm) 2 9 2 6(1,800 mm)(7.95216 10 N-mm ) 8.9127 mm

Consider pulley C load. [Appendix C, SS beam with concentrated load.] Relevant equation from Appendix C: PL3 vC 48 EI Values: P = 1,000 N, L = 1,800 mm, EI = 7.95216 × 109 N-mm2 Computation: PL3 vC 48EI

(1,000 N)(1,800 mm)3 48(7.95216 109 N-mm2 )

15.2789 mm

Shaft deflection at C

vC

8.9127 mm 15.2789 mm

6.3662 mm

6.37 mm

Ans.


10.66 The cantilever beam shown in Fig. P10.66 consists of a W530 × 92 structural steel wideflange shape [E = 200 GPa; I = 552 × 106 mm4]. For the loading shown, determine: (a) the beam deflection at point A. (b) the beam deflection at point B. Fig. P10.66

Solution (a) Beam deflection at point A Consider an upward 85 kN/m uniformly distributed load acting over entire 4-m span. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vA 8 EI Values: w = −85 kN/m, L = 4 m, EI = 1.104 × 105 kN-m2 Computation: wL4 vA 8EI

( 85 kN/m)(4 m)4 8(1.104 105 kN-m2 )

0.024638 m

Consider a downward 85 kN/m uniformly distributed load acting over span BC. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 wL3 vB and (magnitude) B 8EI 6 EI Values: w = 85 kN/m, L = 2.5 m, EI = 1.104 × 105 kN-m2 Computation: wL4 vB 8EI B

vA

wL3 6 EI

(85 kN)(2.5 m) 4 8(1.104 105 kN-m 2 ) (85 kN)(2.5 m)3 6(1.104 105 kN-m 2 )

0.003759 m

0.0020050 rad

0.003759 m (1.5 m)(0.0020050 rad)

0.006767 m


vA

0.024638 m 0.006767 m

0.017871 m

17.87 mm

Ans.


(b) Beam deflection at point B Consider an upward 85 kN/m uniformly distributed load acting over entire 4-m span. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wx 2 vB (6 L2 4 Lx x 2 ) (elastic curve) 24 EI Values: w = −85 kN/m, L = 4 m, x = 2.5 m, EI = 1.104 × 105 kN-m2 Computation: wx 2 vB (6 L2 4 Lx x 2 ) 24 EI ( 85 kN/m)(2.5 m) 2 6(4 m) 2 5 2 24(1.104 10 kN-m )

4(4 m)(2.5 m) (2.5 m) 2

0.012481 m

Consider a downward 85 kN/m uniformly distributed load acting over span BC. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vB 8 EI Values: w = 85 kN/m, L = 2.5 m, EI = 1.104 × 105 kN-m2 Computation: wL4 vB 8EI

(85 kN)(2.5 m)4 8(1.104 105 kN-m2 )

0.003759 m


vB

0.012481 m 0.003759 m

0.008722 m

8.72 mm

Ans.


10.67 The solid 30-mm-diameter steel [E = 200 GPa] shaft shown in Fig. P10.67 supports two belt pulleys. Assume that the bearing at A can be idealized as a pin support and that the bearing at E can be idealized as a roller support. For the loading shown, determine the shaft deflection at pulley B. Fig. P10.67


(30 mm)4

39,760.78 mm4

64 Shaft deflection at pulley B Consider pulley B load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 vB ( L a 2 b2 ) 6 LEI Values: P = 750 N, L = 1,000 mm, a = 300 mm, b = 700 mm, EI = 7.95216 × 109 N-mm2 Computation: Pab 2 vB ( L a 2 b2 ) 6 LEI (750 N)(300 mm)(700 mm) (1,000 mm) 2 (300 mm) 2 (700 mm) 2 9 2 6(1,000 mm)(7.95216 10 N-mm ) 1.38642 mm

Consider pulley D load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vB ( L b 2 x 2 ) (elastic curve) 6 LEI Values: P = 500 N, L = 1,000 mm, x = 300 mm, b = 200 mm, EI = 7.95216 × 109 N-mm2 Computation: Pbx 2 vB ( L b2 x2 ) 6 LEI (500 N)(200 mm)(300 mm) (1,000 mm) 2 (200 mm) 2 (300 mm) 2 9 2 6(1,000 mm)(7.95216 10 N-mm ) 0.54702 mm

Shaft deflection at B

vB

1.38642 mm 0.54702 mm

1.93344 mm

1.933 mm

Ans.


10.68 The solid 30-mm-diameter steel [E = 200 GPa] shaft shown in Fig. P10.68 supports two belt pulleys. Assume that the bearing at A can be idealized as a pin support and that the bearing at E can be idealized as a roller support. For the loading shown, determine the shaft deflection at pulley D. Fig. P10.68


64

(30 mm)4

39,760.78 mm4

Shaft deflection at pulley D Consider pulley B load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vD ( L b 2 x 2 ) (elastic curve) 6 LEI Values: P = 750 N, L = 1,000 mm, x = 200 mm, b = 300 mm, EI = 7.95216 × 109 N-mm2 Computation: Pbx 2 vD ( L b2 x2 ) 6 LEI (750 N)(300 mm)(200 mm) (1,000 mm) 2 (300 mm) 2 (200 mm) 2 9 2 6(1,000 mm)(7.95216 10 N-mm ) 0.82053 mm

Consider pulley D load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 vD ( L a 2 b2 ) 6 LEI Values: P = 500 N, L = 1,000 mm, a = 800 mm, b = 200 mm, EI = 7.95216 × 109 N-mm2 Computation: Pab 2 vD ( L a 2 b2 ) 6 LEI (500 N)(800 mm)(200 mm) (1,000 mm) 2 (800 mm) 2 (200 mm) 2 9 2 6(1,000 mm)(7.95216 10 N-mm ) 0.53654 mm

Shaft deflection at D

vD

0.82053 mm 0.53654 mm

1.35707 mm

1.357 mm

Ans.


10.69 The simply supported beam shown in Fig. P10.69 consists of a W410 × 60 structural steel wide-flange shape [E = 200 GPa; I = 216 × 106 mm4]. For the loading shown, determine the beam deflection at point B.

Fig. P10.69

Solution Beam deflection at point B Consider concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vB (2 L2 3Lx x 2 ) (elastic curve) 6 LEI Values: M = −180 kN-m, L = 6 m, x = 1.5 m, EI = 4.32 × 104 kN-m2 Computation: Mx vB (2 L2 3Lx x 2 ) 6 LEI ( 180 kN-m)(1.5 m) 2(6 m) 2 3(6 m)(1.5 m) (1.5 m) 2 0.008203 m 4 2 6(6 m)(4.32 10 kN-m ) Consider concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 vB ( L a 2 b2 ) 6 LEI Values: P = 70 kN, L = 6 m, a = 1.5 m, b = 4.5 m, EI = 4.32 × 104 kN-m2 Computation: Pab 2 vB ( L a 2 b2 ) 6 LEI (70 kN)(1.5 m)(4.5 m) (6 m) 2 (1.5 m) 2 (4.5 m) 2 0.004102 m 6(6 m)(4.32 104 kN-m 2 ) Consider uniformly distributed load. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 vB (2 x3 6Lx 2 a 2 x 4L2 x a 2 L) 24LEI Values: w = 80 kN/m, L = 6 m, a = 3 m, x = 4.5 m, EI = 4.32 × 104 kN-m2

Computation: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

vB

wa 2 (2 x3 6 Lx 2 24 LEI

a 2 x 4 L2 x a 2 L)

(80 kN/m)(3 m) 2 2(4.5)3 6(6)(4.5) 2 4 2 24(6.0 m)(4.32 10 kN-m )

(3) 2 (4.5) 4(6) 2 (4.5) (3) 2 (6)

0.010156 m Beam deflection at B

vB

0.008203 m 0.004102 m 0.010156 m

0.006055 m

6.06 mm

Ans.


10.70 The simply supported beam shown in Fig. P10.70 consists of a W410 × 60 structural steel wide-flange shape [E = 200 GPa; I = 216 × 106 mm4]. For the loading shown, determine the beam deflection at point C.

Fig. P10.70

Solution Beam deflection at point C Consider concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vC (2 L2 3Lx x 2 ) (elastic curve) 6 LEI Values: M = −180 kN-m, L = 6 m, x = 3.0 m, EI = 4.32 × 104 kN-m2 Computation: Mx vC (2 L2 3Lx x 2 ) 6 LEI ( 180 kN-m)(3.0 m) 2(6 m) 2 4 2 6(6 m)(4.32 10 kN-m )

3(6 m)(3.0 m) (3.0 m) 2

0.009375 m

Consider concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vC ( L b 2 x 2 ) (elastic curve) 6 LEI Values: P = 70 kN, L = 6 m, x = 3.0 m, b = 1.5 m, EI = 4.32 × 104 kN-m2 Computation: Pbx 2 vC ( L b2 x2 ) 6 LEI (70 kN)(1.5 m)(3.0 m) (6 m)2 4 2 6(6 m)(4.32 10 kN-m )

(1.5 m) 2

(3.0 m) 2

0.005013 m


Consider uniformly distributed load. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa3 vC (4L2 7aL 3a 2 ) 24 LEI Values: w = 80 kN/m, L = 6 m, a = 3 m, EI = 4.32 × 104 kN-m2 Computation: wa 3 vC (4 L2 24 LEI

7aL 3a 2 )

(80 kN/m)(3 m)3 4(6 m) 2 24(6.0 m)(4.32 10 4 kN-m 2 )

7(3 m)(6 m) 3(3 m) 2

0.015625 m


vC

0.009375 m 0.005013 m 0.015625 m

0.011263 m

11.26 mm

Ans.


10.71 The simply supported beam shown in Fig. P10.71 consists of a W530 × 66 structural steel wide-flange shape [E = 200 GPa; I = 351 × 106 mm4]. If w = 80 kN/m, determine (a) the beam deflection at point A. (b) the beam deflection at point C. Fig. P10.71

Solution (a) Beam deflection at point A Determine cantilever deflection due to concentrated load on overhang AB. [Appendix C, Cantilever beam with concentrated load.] Relevant equation from Appendix C: PL3 vA (assuming fixed support at B) 3EI Values: P = 35 kN, L = 4 m, EI = 7.02 × 104 kN-m2 Computation: PL3 (35 kN)(4 m)3 vA 0.0106363 m 3EI 3(7.02 104 kN-m2 ) Consider deflection at A resulting from rotation at B caused by concentrated load on overhang AB. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML (slope magnitude) B 3EI Values: M = (35 kN)(4 m) = 140 kN-m, L = 8 m, EI = 7.02 × 104 kN-m2 Computation: ML (140 kN-m)(8 m) 0.0053181 rad B 3EI 3(7.02 104 kN-m 2 )

vA

(4 m)(0.0053181 rad)

0.0212726 m

Consider uniformly distributed loads between C and D. [Appendix C, SS beam with uniformly distributed load over portion of span.] Relevant equations from Appendix C: wa 2 (2L2 a 2 ) (slope magnitude) B 24LEI Values: w = 80 kN/m, L = 8 m, a = 4 m, EI = 7.02 × 104 kN-m2


Computation: wa 2 (2 L2 B 24 LEI vA

a2 )

(80 kN/m)(4 m) 2 2(8 m) 2 4 2 24(8 m)(7.02 10 kN-m )

(4 m)(0.0106363 rad)

(4 m) 2

0.0106363 rad

0.0425451 m

Consider deflection at A resulting from rotation at B caused by uniform load on overhang DE. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML (slope magnitude) B 6 EI Values: M = (80 kN/m)(2 m)(1 m) = 160 kN-m, L = 8 m, EI = 7.02 × 104 kN-m2 Computation: ML (160 kN-m)(8 m) B 6 EI 6(7.02 104 kN-m 2 )

vA

(4 m)(0.0030389 rad)

0.0030389 rad 0.0121557 m

Beam deflection at A vA 0.0106363 m 0.0212726 m 0.0425451 m 0.0121557 m 0.0015195 m

Ans.

1.520 mm

(b) Beam deflection at point C Consider concentrated moment from overhang AB. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: Mx vC (2 L2 3Lx x 2 ) (elastic curve) 6 LEI Values: M = (35 kN)(4 m) = −140 kN-m, L = 8 m, x = 4 m, EI = 7.02 × 104 kN-m2 Computation: Mx vC (2 L2 3Lx x 2 ) 6 LEI ( 140 kN-m)(4 m) 2(8 m) 2 6(8 m)(7.02 104 kN-m 2 )

3(8 m)(4 m) (4 m)2

0.0079772 m


Consider uniformly distributed loads between C and D. [Appendix C, SS beam with uniformly distributed load over portion of span.] Relevant equations from Appendix C: wa3 vC (4L2 7aL 3a 2 ) 24 LEI Values: w = 80 kN/m, L = 8 m, a = 4 m, EI = 7.02 × 104 kN-m2 Computation: wa 3 vC (4 L2 24 LEI

7aL 3a 2 )

(80 kN/m)(4 m)3 4(8 m) 2 4 2 24(8 m)(7.02 10 kN-m )

7(4 m)(8 m) 3(4 m) 2

0.0303894 m

Consider concentrated moment from overhang DE. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: Mx vC (2 L2 3Lx x 2 ) (elastic curve) 6 LEI Values: M = (80 kN/m)(2 m)(1 m) = 160 kN-m, L = 8 m, x = 4 m, EI = 7.02 × 104 kN-m2 Computation: Mx vC (2 L2 3Lx x 2 ) 6 LEI ( 160 kN-m)(4 m) 2(8 m) 2 4 2 6(8 m)(7.02 10 kN-m )

3(8 m)(4 m) (4 m)2

0.0091168 m


vC

0.0079772 m 0.0303894 m 0.0091168 m

0.0132954 m

13.30 mm

Ans.


10.72 The simply supported beam shown in Fig. P10.72 consists of a W530 × 66 structural steel wide-flange shape [E = 200 GPa; I = 351 × 106 mm4]. If w = 90 kN/m, determine: (a) the beam deflection at point C. (b) the beam deflection at point E. Fig. P10.72

Solution (a) Beam deflection at point C Consider concentrated moment from overhang AB. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: Mx vC (2 L2 3Lx x 2 ) (elastic curve) 6 LEI Values: M = (35 kN)(4 m) = −140 kN-m, L = 8 m, x = 4 m, EI = 7.02 × 104 kN-m2 Computation: Mx vC (2 L2 3Lx x 2 ) 6 LEI ( 140 kN-m)(4 m) 2(8 m) 2 4 2 6(8 m)(7.02 10 kN-m )

3(8 m)(4 m) (4 m)2

0.0079772 m

Consider uniformly distributed loads between C and D. [Appendix C, SS beam with uniformly distributed load over portion of span.] Relevant equations from Appendix C: wa3 vC (4L2 7aL 3a 2 ) 24 LEI Values: w = 90 kN/m, L = 8 m, a = 4 m, EI = 7.02 × 104 kN-m2 Computation: wa 3 vC (4 L2 24 LEI

7aL 3a 2 )

(90 kN/m)(4 m)3 4(8 m) 2 4 2 24(8 m)(7.02 10 kN-m )

7(4 m)(8 m) 3(4 m) 2

0.0341881 m


Consider concentrated moment from overhang DE. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: Mx vC (2 L2 3Lx x 2 ) (elastic curve) 6 LEI Values: M = (90 kN/m)(2 m)(1 m) = 180 kN-m, L = 8 m, x = 4 m, EI = 7.02 × 104 kN-m2 Computation: Mx vC (2 L2 6 LEI

3Lx

x2 )

( 180 kN-m)(4 m) 2(8 m)2 4 2 6(8 m)(7.02 10 kN-m )

3(8 m)(4 m) (4 m) 2

0.0102564 m


vC

0.0079772 m 0.0341881 m 0.0102564 m

0.0159545 m

15.95 mm

Ans.

(b) Beam deflection at point E Consider deflection at E resulting from rotation at D caused by concentrated load on overhang AB. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML (slope magnitude) D 6 EI Values: M = (35 kN)(4 m) = 140 kN-m, L = 8 m, EI = 7.02 × 104 kN-m2 Computation: ML (140 kN-m)(8 m) D 6 EI 6(7.02 104 kN-m 2 )

vE

(2 m)(0.0026591 rad)

0.0026591 rad

0.0053181 m

Consider uniformly distributed loads between C and D. [Appendix C, SS beam with uniformly distributed load over portion of span.] Relevant equations from Appendix C: wa 2 (2 L a)2 (slope magnitude) D 24 LEI Values: w = 90 kN/m, L = 8 m, a = 4 m, EI = 7.02 × 104 kN-m2 Computation: wa 2 (90 kN/m)(4 m)2 2 2 (2 L a ) 2(8 m) (4 m) 0.0153846 rad D 4 2 24 LEI 24(8 m)(7.02 10 kN-m ) vE

(2 m)(0.0153846 rad)

0.0307692 m


Consider deflection at E resulting from rotation at D caused by uniform load on overhang DE. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML (slope magnitude) D 3EI Values: M = (90 kN/m)(2 m)(1 m) = 180 kN-m, L = 8 m, EI = 7.02 × 104 kN-m2 Computation: ML (180 kN-m)(8 m) 0.0068376 rad D 3EI 3(7.02 104 kN-m 2 ) vE

(2 m)(0.0068376 rad)

0.0136753 m

Determine cantilever deflection due to uniformly distributed load on overhang DE. [Appendix C, Cantilever beam with distributed load.] Relevant equation from Appendix C: wL4 vE (assuming fixed support at D) 8EI Values: w = 90 kN/m, L = 2 m, EI = 7.02 × 104 kN-m2 wL4 (90 kN/m)(2 m) 4 0.0025641 m Computation: vE 8 EI 8(7.02 104 kN-m 2 ) Beam deflection at E vE 0.0053181 m 0.0307692 m 0.0136753 m 0.0025641 m 0.0092117 m

9.21 mm

Ans.


10.73 The simply supported beam shown in Fig. P10.73 consists of a W16 × 40 structural steel wide-flange shape [E = 29,000 ksi; I = 518 in.4]. For the loading shown, determine: (a) the beam deflection at point C. (b) the beam deflection at point F. Fig. P10.73

Solution (a) Beam deflection at point C Consider 40-kip concentrated load at B. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vC ( L b 2 x 2 ) (elastic curve) 6 LEI Values: P = 40 kips, L = 18 ft, b = 4 ft, x = 10 ft, EI = 1.5022 × 107 kip-in.2 Computation: Pbx 2 vC (L 6 LEI

b2

x2 )

(40 kips)(4 ft)(10 ft)(12 in./ft)3 (18 ft) 2 6(18 ft)(1.5022 107 kip-in.2 )

(4 ft) 2

(10 ft) 2

0.354467 in.

Consider 30-kip concentrated load at D. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vC ( L b 2 x 2 ) (elastic curve) 6 LEI Values: P = 30 kips, L = 18 ft, b = 6 ft, x = 8 ft, EI = 1.5022 × 107 kip-in.2

Computation: Pbx 2 vC (L 6 LEI

b2

x2 )

(30 kips)(6 ft)(8 ft)(12 in./ft)3 (18 ft)2 7 2 6(18 ft)(1.5022 10 kip-in. )

(6 ft) 2

(8 ft) 2

0.343560 in.


Consider 20-kip concentrated load at F. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: Mx vC (2 L2 3Lx x 2 ) (elastic curve) 6 LEI Values: M = −(20 kips)(6 ft) = −120 kip-ft, L = 18 ft, x = 10 ft, EI = 1.5022 × 107 kip-in.2 Computation: Mx vC (2 L2 6 LEI

3Lx

x2 )

( 120 kip-ft)(10 ft)(12 in./ft)3 2(18 ft) 2 6(18 ft)(1.5022 107 kip-in.2 )

3(18 ft)(10 ft) (10 ft) 2

0.265850 in.


vC

0.354467 in. 0.343560 in. 0.265850 in.

0.432177 in.

0.432 in.

Ans.

(b) Beam deflection at point F Consider 40-kip concentrated load at B. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pa( L2 a 2 ) (slope magnitude) E 6 LEI Values: P = 40 kips, L = 18 ft, a = 4 ft, EI = 1.5022 × 107 kip-in.2 Computation: Pa( L2 a 2 ) E 6 LEI vF

(40 kips)(4 ft)(12 in./ft) 2 (18 ft) 2 7 2 6(18 ft)(1.5022 10 kip-in. )

(6 ft)(12 in./ft)(0.0043740 rad)

(4 ft) 2

0.0043740 rad

0.314930 in.

Consider 30-kip concentrated load at D. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pa( L2 a 2 ) (slope magnitude) E 6 LEI Values: P = 30 kips, L = 18 ft, x = 8 ft, a = 12 ft, EI = 1.5022 × 107 kip-in.2 Computation: Pa( L2 a 2 ) E 6 LEI vF


(6 ft)(12 in./ft)(0.0057516 rad)

(12 ft) 2

0.0057516 rad

0.414113 in.


Consider deflection at F resulting from rotation at E caused by 20-kip load on overhang EF. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML (slope magnitude) E 3EI Values: M = (20 kips)(6 ft) = 120 kip-ft, L = 18 ft, EI = 1.5022 × 107 kip-in.2 Computation: ML (120 kip-ft)(18 ft)(12 in./ft) 2 E 3EI 3(1.5022 107 kip-in.2 ) vF

(6 ft)(12 in./ft)(0.0069019 rad)

0.0069019 rad 0.496935 in.

Determine cantilever deflection due to concentrated load on overhang EF. [Appendix C, Cantilever beam with concentrated load.] Relevant equation from Appendix C: PL3 vF (assuming fixed support at E) 3EI Values: P = 20 kips, L = 6 ft, EI = 1.5022 × 107 kip-in.2

Computation: vF

PL3 3EI

(20 kips)(6 ft)3 (12 in./ft)3 3(1.5022 107 kip-in.2 )

0.165645 in.

Beam deflection at F vF 0.314930 in. 0.414113 in. 0.496935 in. 0.165645 in.

0.066463 in.

0.0665 in.

Ans.


10.74 The cantilever beam shown in Fig. P10.74 consists of a rectangular structural steel tube shape [E = 200 GPa; I = 170 × 106 mm4]. For the loading shown, determine: (a) the beam deflection at point A. (b) the beam deflection at point B. Fig. P10.74

Solution (a) Beam deflection at point A Consider uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vA 8 EI Values: w = −65 kN/m, L = 6 m, EI = 3.4 × 104 kN-m2 Computation: wL4 vA 8EI

( 65 kN/m)(6 m)4 8(3.4 104 kN-m2 )

0.309706 m

Consider 90-kN concentrated load at A. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 vA 3EI Values: P = 90 kN, L = 6 m, EI = 3.4 × 104 kN-m2

Computation: PL3 vA 3EI

(90 kN)(6 m)3 3(3.4 104 kN-m2 )

0.190588 m

Consider 30-kN concentrated load at B. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 PL2 vB and (magnitude) B 3EI 2 EI Values: P = 30 kN, L = 3.5 m, EI = 3.4 × 104 kN-m2

Computation: PL3 vB 3EI

(30 kN)(3.5 m)3 3(3.4 104 kN-m2 )

0.012610 m

(a)


B

vA

PL2 2 EI

(30 kN)(3.5 m) 2 2(3.4 104 kN-m 2 )

0.0054044 rad

0.012610 m (2.5 m)(0.0054044 rad)

0.026121 m

Consider 225 kN-m concentrated moment at B. [Appendix C, Cantilever beam with concentrated moment at tip.] Relevant equations from Appendix C: ML2 ML vB and (slope magnitude) B 2 EI EI Values: M = 225 kN-m, L = 3.5 m, EI = 3.4 × 104 kN-m2

Computation: ML2 (225 kN-m)(3.5 m)2 vB 0.040533 m 2EI 2(3.4 104 kN-m2 ) ML (225 kN-m)(3.5 m) 0.0231618 rad B EI (3.4 104 kN-m 2 ) vA

0.040533 m (2.5 m)(0.0231618 rad)

(b)

0.098438 m


vA

0.309706 m 0.190588 m 0.026121 m 0.098438 m

0.005441 m

5.44 mm

Ans.

(b) Beam deflection at point B Consider uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wx 2 vB (6 L2 4 Lx x 2 ) (elastic curve) 24 EI Values: w = −65 kN/m, L = 6 m, x = 3.5 m, EI = 3.4 × 104 kN-m2 Computation: wx 2 vB (6 L2 4 Lx x 2 ) 24 EI ( 65 kN/m)(3.5 m) 2 6(6 m) 2 4 2 24(3.4 10 kN-m )

4(6 m)(3.5 m) (3.5 m) 2

0.140759 m


Consider 90-kN concentrated load at A. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: Px 2 vB (3L x) (elastic curve) 6 EI Values: P = 90 kN, L = 6 m, x = 3.5 m, EI = 3.4 × 104 kN-m2 Computation: Px 2 vB (3L x) 6EI

(90 kN)(3.5 m)2 3(6 m) (3.5 m) 6(3.4 104 kN-m2 )

0.078364 m

Consider 30-kN concentrated load at B. Previously calculated in Eq. (a). Consider 225 kN-m concentrated moment at B. Previously calculated in Eq. (b). Beam deflection at B

vB

0.140759 m 0.078364 m 0.012610 m 0.040533 m 0.009252 m

9.25 mm

Ans.


10.75 The simply supported beam shown in Fig. P10.75 consists of a rectangular structural steel tube shape [E = 200 GPa; I = 350 × 106 mm4]. For the loading shown, determine: (a) the beam deflection at point C. (b) the beam deflection at point E. Fig. P10.75

Solution (a) Beam deflection at point C Consider 315 kN-m concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vC (2 L2 3Lx x 2 ) (elastic curve) 6 LEI Values: M = −315 kN-m, L = 9 m, x = 6 m, EI = 7.0 × 104 kN-m2 Computation: Mx vC (2 L2 3Lx x 2 ) 6 LEI ( 315 kN-m)(6 m) 2(9 m)2 4 2 6(9 m)(7.0 10 kN-m )

3(9 m)(6 m) (6 m) 2

0.018000 m

Consider 120 kN/m uniformly distributed load. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 vC (2 x3 6 Lx 2 a 2 x 4L2 x a 2 L) 24LEI Values: w = 120 kN/m, L = 9 m, a = 3 m, x = 6 m, EI = 7.0 × 104 kN-m2 Computation: wa 2 vC (2 x3 6 Lx 2 a 2 x 4 L2 x a 2 L) 24 LEI (120 kN/m)(3 m)2 2(6 m)3 24(9 m)(7.0 104 kN-m 2 )

6(9 m)(6 m) 2

(3 m) 2 (6 m) 4(9 m) 2 (6 m) (3 m) 2 (9 m)

0.028929 m


Consider 100-kN concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 vC (L a 2 b2 ) 6 LEI Values: P = 100 kN, L = 9 m, a = 6 m, b = 3 m, EI = 7.0 × 104 kN-m2 Computation: Pab 2 vC ( L a 2 b2 ) 6 LEI (100 kN)(6 m)(3 m) (9 m) 2 (6 m) 2 (3 m) 2 0.017143 m 4 2 6(9 m)(7.0 10 kN-m ) Consider 60 kN/m uniformly distributed load on overhang DE. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: Mx vC (2 L2 3Lx x 2 ) (elastic curve) 6 LEI Values: M = −(60 kN/m)(3 m)(1.5 m) = −270 kN-m, L = 9 m, x = 3 m, EI = 7.0 × 104 kN-m2 Computation: Mx vC (2 L2 3Lx x 2 ) 6 LEI ( 270 kN-m)(3 m) 2(9 m) 2 4 2 6(9 m)(7.0 10 kN-m )

3(9 m)(3 m) (3 m) 2

0.019286 m


vC

0.018000 m 0.028929 m 0.017143 m 0.019286 m

0.008786 m

8.79 mm

Ans.

(b) Beam deflection at point E Consider 315 kN-m concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: ML (slope magnitude) D 6 EI Values: M = −315 kN-m, L = 9 m, EI = 7.0 × 104 kN-m2 Computation: ML ( 315 kN-m)(9 m) D 6 EI 6(7.0 104 kN-m 2 )

vE

(3 m)( 0.0067500 rad)

0.0067500 rad 0.020250 m


Consider 120 kN/m uniformly distributed load. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 (2L2 a 2 ) (slope magnitude) D 24LEI Values: w = 120 kN/m, L = 9 m, a = 3 m, EI = 7.0 × 104 kN-m2 Computation: wa 2 (2 L2 D 24 LEI vE

a2 )

(120 kN/m)(3 m)2 2(9 m) 2 4 2 24(9 m)(7.0 10 kN-m )

(3 m)(0.0109286 rad)

(3 m) 2

0.0109286 rad

0.032786 m

Consider 100-kN concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pa( L2 a 2 ) (slope magnitude) D 6 LEI Values: P = 100 kN, L = 9 m, a = 6 m, EI = 7.0 × 104 kN-m2 Computation: Pa( L2 a 2 ) D 6 LEI vE

(100 kN)(6 m) (9 m) 2 6(9 m)(7.0 104 kN-m 2 )

(3 m)(0.0071429 rad)

(6 m) 2

0.0071429 rad

0.021429 m

Consider 60 kN/m uniformly distributed load on overhang DE. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML (slope magnitude) D 3EI Values: M = −(60 kN/m)(3 m)(1.5 m) = −270 kN-m, L = 9 m, EI = 7.0 × 104 kN-m2 Computation: ML D 3EI

vE

(270 kN-m)(9 m) 3(7.0 104 kN-m 2 )

(3 m)(0.0115714 rad)

0.0115714 rad 0.034714 m


Determine cantilever deflection due to 60 kN/m uniformly distributed load on overhang DE. [Appendix C, Cantilever beam with concentrated load.] Relevant equation from Appendix C: wL4 vE (assuming fixed support at D) 8EI Values: w = 60 kN/m, L = 3 m, EI = 7.0 × 104 kN-m2

Computation: vE

wL4 8EI

(60 kN-m)(3 m)4 8(7.0 104 kN-m2 )

0.008679 m

Beam deflection at E vE 0.020250 m 0.032786 m 0.021429 m 0.034714 m 0.008679 m 0.009429 m

9.43 mm

Ans.


10.76 The cantilever beam shown in Fig. P10.76 consists of a rectangular structural steel tube shape [E = 200 GPa; I = 95 × 106 mm4]. For the loading shown, determine the beam deflection at point B. Fig. P10.76

Solution Consider the downward 50 kN/m uniformly distributed load acting over span AB. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vB 8 EI Values: w = 50 kN/m, L = 2 m, EI = 1.9 × 104 kN-m2 Computation: wL4 (50 kN/m)(2 m)4 vB 0.0052632 m 8EI 8(1.9 104 kN-m2 ) Consider an upward 25 kN/m uniformly distributed load acting over entire 5-m span. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wx 2 vB (6 L2 4 Lx x 2 ) (elastic curve) 24 EI Values: w = −25 kN/m, L = 5 m, x = 2 m, EI = 1.9 × 104 kN-m2 Computation: wx 2 vB (6 L2 4 Lx x 2 ) 24 EI ( 25 kN/m)(2 m) 2 6(5 m) 2 4(5 m)(2 m) (2 m) 2 0.0250000 m 24(1.9 104 kN-m 2 ) Consider a downward 25 kN/m uniformly distributed load acting over span AB. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vB 8 EI Values: w = 25 kN/m, L = 2 m, EI = 1.9 × 104 kN-m2 Computation: wL4 vB 8EI

(25 kN/m)(2 m)4 8(1.9 104 kN-m2 )

0.0026316 m


Consider 20-kN concentrated load at B. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 vB 3EI Values: P = −20 kN, L = 2 m, EI = 1.9 × 104 kN-m2 Computation: PL3 ( 20 kN)(2 m)3 vB 0.0028070 m 3EI 3(1.9 104 kN-m2 ) Consider 50-kN concentrated load at C. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: Px 2 vB (3L x) (elastic curve) 6 EI Values: P = 50 kN, L = 5 m, x = 2 m, EI = 1.9 × 104 kN-m2 Computation: Px2 (50 kN)(2 m)2 vB (3L x) 3(5 m) (2 m) 0.0228070 m 6EI 6(1.9 104 kN-m2 ) Beam deflection at B vB 0.0052632 m 0.0250000 m 0.0026316 m 0.0028070 m 0.0228070 m 0.0028947 m

2.89 mm

Ans.


10.77 The cantilever beam shown in Fig. P10.77 consists of a rectangular structural steel tube shape [E = 200 GPa; I = 95 × 106 mm4]. For the loading shown, determine the beam deflection at point C. Fig. P10.77

Solution Consider the downward 50 kN/m uniformly distributed load acting over span AB. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 wL3 vB and B (slope magnitude) 8 EI 6 EI Values: w = 50 kN/m, L = 2 m, EI = 1.9 × 104 kN-m2 Computation: wL4 (50 kN/m)(2 m) 4 vB 0.0052632 m 8 EI 8(1.9 10 4 kN-m 2 ) wL3 (50 kN/m)(2 m)3 0.0035088 rad B 6 EI 6(1.9 104 kN-m 2 ) vC

0.0052632 m (3 m)(0.0035088 rad)

0.0157895 m

Consider an upward 25 kN/m uniformly distributed load acting over entire 5-m span. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vC 8EI Values: w = −25 kN/m, L = 5 m, EI = 1.9 × 104 kN-m2 Computation: wL4 vC 8EI

( 25 kN/m)(5 m)4 8(1.9 104 kN-m2 )

0.1027961 m

Consider a downward 25 kN/m uniformly distributed load acting over span AB. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 wL3 vB and B (slope magnitude) 8 EI 6 EI Values: w = 25 kN/m, L = 2 m, EI = 1.9 × 104 kN-m2

Computation: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

vB B

vC

wL4 8 EI

wL3 6 EI

(25 kN/m)(2 m) 4 8(1.9 10 4 kN-m 2 )

(25 kN)(2 m)3 6(1.9 104 kN-m 2 )

0.0026316 m

0.0017544 rad

0.0026316 m (3 m)(0.0017544 rad)

0.0078948 m

Consider 20-kN concentrated load at B. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 PL2 vB and B (slope magnitude) 3EI 2 EI Values: P = −20 kN, L = 2 m, EI = 1.9 × 104 kN-m2 Computation: PL3 ( 20 kN)(2 m)3 vB 0.0028070 m 3EI 3(1.9 10 4 kN-m 2 ) PL2 (20 kN)(2 m) 2 0.0021053 rad B 2 EI 2(1.9 104 kN-m 2 ) vC

0.0028070 m (3 m)(0.0021053 rad)

0.0091228 m

Consider 50-kN concentrated load at C. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 vC 3EI Values: P = 50 kN, L = 5 m, EI = 1.9 × 104 kN-m2 Computation: PL3 vC 3EI

(50 kN)(5 m)3 3(1.9 104 kN-m2 )

0.1096491 m

Beam deflection at C vC 0.0157895 m 0.1027961 m 0.0078948 m 0.0091228 m 0.1096491 m 0.0214145 m

21.4 mm

Ans.


10.78 The simply supported beam shown in Fig. P10.78 consists of a W10 × 30 structural steel wide-flange shape [E = 29,000 ksi; I = 170 in.4]. If w = 5 kips/ft, determine: (a) the beam deflection at point A. (b) the beam deflection at point C. Fig. P10.78

Solution (a) Beam deflection at point A Consider cantilever beam deflection of 85 kip-ft concentrated moment. [Appendix C, Cantilever beam with concentrated moment at one end.] Relevant equation from Appendix C: ML2 vA 2 EI Values: M = 85 kip-ft, L = 3 ft, EI = 4.93 × 106 kip-in.2 Computation: ML2 vA 2EI

(85 kip-ft)(3 ft)2 (12 in./ft)3 2(4.93 106 kip-in.2 )

0.134069 in.

Consider rotation at B caused by 85 kip-ft concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: ML (slope magnitude) B 3EI Values: M = 85 kip-ft, L = 15 ft, EI = 4.93 × 106 kip-in.2 Computation: ML (85 kip-ft)(15 ft)(12 in./ft) 2 B 3EI 3(4.93 106 kip-in.2 ) vA

(3 ft)(12 in./ft)(0.0124138 rad)

0.0124138 rad 0.446897 in.

Consider cantilever beam deflection of 5 kips/ft uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vA 8 EI Values: w = 5 kips/ft, L = 3 ft, EI = 4.93 × 106 kip-in.2


Computation: wL4 vA 8EI

(5 kips/ft)(3 ft)4 (12 in./ft)3 8(4.93 106 kip-in.2 )

0.017744 in.

Consider rotation at B caused by 5 kips/ft uniformly distributed load. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: ML (slope magnitude) B 3EI Values: M = (5 kips/ft)(3 ft)(1.5 ft) = 22.5 kip-ft, L = 15 ft, EI = 4.93 × 106 kip-in.2 Computation: ML (22.5 kip-ft)(15 ft)(12 in./ft) 2 0.0032860 rad B 3EI 3(4.93 106 kip-in.2 ) vA

(3 ft)(12 in./ft)(0.0032860 rad)

0.118296 in.

Consider 5 kips/ft uniformly distributed load on segment BC. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 (2 L a)2 (slope magnitude) B 24 LEI Values: w = 5 kips/ft, L = 15 ft, a = 5 ft, EI = 4.93 × 106 kip-in.2 Computation: B

vA

wa 2 (2 L a) 2 24 LEI

(5 kips/ft)(5 ft)2 (12 in./ft)2 2(15 ft) (5 ft) 24(15 ft)(4.93 106 kip-in.2 )

(3 ft)(12 in./ft)(0.0063387 rad)

2

0.0063387 rad

0.228195 in.

Consider 25-kip concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pb( L2 b2 ) (slope magnitude) B 6 LEI Values: P = 25 kips, L = 15 ft, b = 5 ft, EI = 4.93 × 106 kip-in.2 Computation: Pb( L2 b 2 ) B 6 LEI vA

(25 kips)(5 ft)(12 in./ft) 2 (15 ft) 2 6(15 ft)(4.93 106 kip-in.2 )

(3 ft)(12 in./ft)(0.0081136 rad)

(5 ft) 2

0.0081136 rad

0.292089 in.


Beam deflection at A vA 0.134069 in. 0.446897 in. 0.017744 in. 0.118296 in. 0.228195 in. 0.292089 in. 0.196722 in.

Ans.

0.1967 in.

(b) Beam deflection at point C Consider 85 kip-ft concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vC (2 L2 3Lx x 2 ) (elastic curve) 6 LEI Values: M = −85 kip-ft, L = 15 ft, x = 5 ft, EI = 4.93 × 106 kip-in.2 Computation: Mx vC (2 L2 6 LEI

3Lx

x2 )

( 85 kip-ft)(5 ft)(12 in./ft)3 2(15 ft) 2 6 2 6(15 ft)(4.93 10 kip-in. )

3(15 ft)(5 ft) (5 ft) 2

0.413793 in.

Consider moment at B caused by 5 kips/ft uniformly distributed load on overhang AB. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vC (2 L2 3Lx x 2 ) (elastic curve) 6 LEI Values: M = −(5 kips/ft)(3 ft)(1.5 ft) = −22.5 kip-ft, L = 15 ft, x = 5 ft, EI = 4.93 × 106 kip-in.2 Computation: Mx vC (2 L2 6 LEI

3Lx

x2 )

( 22.5 kip-ft)(5 ft)(12 in./ft) 3 2(15 ft) 2 6(15 ft)(4.93 106 kip-in.2 )

3(15 ft)(5 ft) (5 ft) 2

0.109533 in.

Consider 5 kips/ft uniformly distributed load on segment BC. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa3 vC (4L2 7aL 3a 2 ) 24 LEI Values: w = 5 kips/ft, L = 15 ft, a = 5 ft, EI = 4.93 × 106 kip-in.2


Computation: wa 3 vC (4 L2 24 LEI

7aL 3a 2 )

(5 kips/ft)(5 ft)3 (12 in./ft) 3 4(15 ft) 2 24(15 ft)(4.93 106 kip-in.2 )

7(5 ft)(15 ft) 3(5 ft) 2

0.273834 in.

Consider 25-kip concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vC ( L b 2 x 2 ) (elastic curve) 6 LEI Values: P = 25 kips, L = 15 ft, b = 5 ft, x = 5 ft, EI = 4.93 × 106 kip-in.2 Computation: Pbx 2 vC (L 6 LEI

b2

x2 )

(25 kips)(5 ft)(5 ft)(12 in./ft)3 (15 ft)2 6 2 6(15 ft)(4.93 10 kip-in. )

(5 ft) 2

(5 ft) 2

0.425963 in.

Beam deflection at C vC 0.413793 in. 0.109533 in. 0.273834 in. 0.425963 in. 0.176471 in.

0.1765 in.

Ans.


10.79 The simply supported beam shown in Fig. P10.79 consists of a W10 × 30 structural steel wide-flange shape [E = 29,000 ksi; I = 170 in.4]. If w = 9 kips/ft, determine: (a) the beam deflection at point A. (b) the beam deflection at point D. Fig. P10.79

Solution (a) Beam deflection at point A Consider cantilever beam deflection of 85 kip-ft concentrated moment. [Appendix C, Cantilever beam with concentrated moment at one end.] Relevant equation from Appendix C: ML2 vA 2 EI Values: M = 85 kip-ft, L = 3 ft, EI = 4.93 × 106 kip-in.2 Computation: ML2 vA 2EI

(85 kip-ft)(3 ft)2 (12 in./ft)3 2(4.93 106 kip-in.2 )

0.134069 in.

Consider rotation at B caused by 85 kip-ft concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: ML (slope magnitude) B 3EI Values: M = 85 kip-ft, L = 15 ft, EI = 4.93 × 106 kip-in.2 Computation: ML (85 kip-ft)(15 ft)(12 in./ft) 2 B 3EI 3(4.93 106 kip-in.2 ) vA

(3 ft)(12 in./ft)(0.0124138 rad)

0.0124138 rad 0.446897 in.

Consider cantilever beam deflection of 9 kips/ft uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vA 8 EI Values: w = 9 kips/ft, L = 3 ft, EI = 4.93 × 106 kip-in.2


Computation: wL4 vA 8 EI

(9 kips/ft)(3 ft) 4 (12 in./ft)3 8(4.93 106 kip-in.2 )

0.031939 in.

Consider rotation at B caused by 9 kips/ft uniformly distributed load. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: ML (slope magnitude) B 3EI Values: M = (9 kips/ft)(3 ft)(1.5 ft) = 40.5 kip-ft, L = 15 ft, EI = 4.93 × 106 kip-in.2 Computation: ML (40.5 kip-ft)(15 ft)(12 in./ft)2 0.0059148 rad B 3EI 3(4.93 106 kip-in.2 ) vA

(3 ft)(12 in./ft)(0.0059148 rad)

0.212933 in.

Consider 9 kips/ft uniformly distributed load on segment BC. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 (2 L a)2 (slope magnitude) B 24 LEI Values: w = 9 kips/ft, L = 15 ft, a = 5 ft, EI = 4.93 × 106 kip-in.2 Computation: B

vA

wa 2 (2 L a) 2 24 LEI

(9 kips/ft)(5 ft) 2 (12 in./ft) 2 2(15 ft) (5 ft) 24(15 ft)(4.93 106 kip-in.2 )

(3 ft)(12 in./ft)(0.0114097 rad)

2

0.0114097 rad

0.410748 in.



(3 ft)(12 in./ft)(0.0081136 rad)

(5 ft) 2

0.0081136 rad

0.292089 in.


Beam deflection at A vA 0.134069 in. 0.446897 in. 0.031939 in. 0.212933 in. 0.410748 in. 0.292089 in. 0.123001 in.

Ans.

0.1230 in.

(b) Beam deflection at point D Consider 85 kip-ft concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vD (2 L2 3Lx x 2 ) (elastic curve) 6 LEI Values: M = −85 kip-ft, L = 15 ft, x = 10 ft, EI = 4.93 × 106 kip-in.2 Computation: Mx vD (2 L2 6 LEI

3Lx

x2 )

( 85 kip-ft)(10 ft)(12 in./ft) 3 2(15 ft) 2 6 2 6(15 ft)(4.93 10 kip-in. )

3(15 ft)(10 ft) (10 ft) 2

0.331034 in.

Consider moment at B caused by 9 kips/ft uniformly distributed load on overhang AB. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vD (2 L2 3Lx x 2 ) (elastic curve) 6 LEI Values: M = −(9 kips/ft)(3 ft)(1.5 ft) = −40.5 kip-ft, L = 15 ft, x = 10 ft, EI = 4.93 × 106 kip-in.2 Computation: Mx vD (2 L2 6 LEI

3Lx

x2 )

( 40.5 kip-ft)(10 ft)(12 in./ft)3 2(15 ft) 2 6(15 ft)(4.93 106 kip-in.2 )

3(15 ft)(10 ft) (10 ft) 2

0.157729 in.

Consider 9 kips/ft uniformly distributed load on segment BC. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 vD (2 x3 6Lx 2 a 2 x 4L2 x a 2 L) 24 LEI Values: w = 9 kips/ft, L = 15 ft, a = 5 ft, x = 10 ft, EI = 4.93 × 106 kip-in.2


Computation: wa 2 vD (2 x3 24 LEI

6 Lx 2

a 2 x 4 L2 x a 2 L)

(9 kips/ft)(5 ft) 2 (12 in./ft)3 2(10 ft)3 24(15 ft)(4.93 106 kip-in.2 )

6(15 ft)(10 ft) 2

(5 ft) 2 (10 ft) 4(15 ft)2 (10 ft) (5 ft) 2 (15 ft)

0.410751 in.

Consider 25-kip concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 vD ( L a 2 b2 ) 6 LEI Values: P = 25 kips, L = 15 ft, a = 10 ft, b = 5 ft, EI = 4.93 × 106 kip-in.2 Computation: Pab 2 vD (L 6 LEI

a2

b2 )

(25 kips)(10 ft)(5 ft)(12 in./ft)3 (15 ft) 2 6 2 6(15 ft)(4.93 10 kip-in. )

(10 ft) 2

(5 ft) 2

0.486815 in.

Beam deflection at D vD 0.331034 in. 0.157729 in. 0.410751 in. 0.486815 in. 0.408803 in.

0.409 in.

Ans.


10.80 The simply supported beam shown in Fig. P10.80 consists of a W10 × 30 structural steel wide-flange shape [E = 29,000 ksi; I = 170 in.4]. For the loading shown, determine: (a) the beam deflection at point A. (b) the beam deflection at point C.

Fig. P10.80

Solution (a) Beam deflection at point A Consider cantilever beam deflection of linearly distributed load on overhang AB. [Appendix C, Cantilever beam with linearly distributed load.] Relevant equation from Appendix C: w0 L4 vA 30 EI Values: w0 = 8 kips/ft, L = 9 ft, EI = 4.93 × 106 kip-in.2 Computation: w0 L4 vA 30EI


0.613247 in.

Consider rotation at B caused by linearly distributed load on overhang AB. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: ML (slope magnitude) B 3EI Values: M = ½(8 kips/ft)(9 ft)(3 ft) = 108 kip-ft, L = 18 ft, EI = 4.93 × 106 kip-in.2 Computation: ML (108 kip-ft)(18 ft)(12 in./ft) 2 B 3EI 3(4.93 106 kip-in.2 ) vA

(9 ft)(12 in./ft)(0.0189274 rad)

0.0189274 rad 2.044157 in.


Consider linearly distributed load from 8 kips/ft to 0 kips/ft over span BD. [Appendix C, SS beam with linearly distributed load.] Relevant equation from Appendix C: w0 L3 (slope magnitude) B 45 EI Values: w0 = 8 kips/ft, L = 18 ft, EI = 4.93 × 106 kip-in.2 Computation: w0 L3 B 45EI vA


(9 ft)(12 in./ft)(0.0302838 rad)

0.0302838 rad

3.270652 in.

Consider 4 kips/ft uniformly distributed load on segment CD. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 (2L2 a 2 ) (slope magnitude) B 24LEI Values: w = 4 kips/ft, L = 18 ft, a = 9 ft, EI = 4.93 × 106 kip-in.2 Computation: wa 2 (2 L2 B 24 LEI

a2 )

(4 kips/ft)(9 ft) 2 (12 in./ft) 2 2(18 ft) 2 24(18 ft)(4.93 106 kip-in.2 ) vA

(9 ft)(12 in./ft)(0.0124211 rad)

(9 ft) 2

0.0124211 rad

1.341478 in.


vA

0.613247 in. 2.044157 in. 3.270652 in. 1.341478 in. 1.954726 in.

1.955 in.

Ans.

(b) Beam deflection at point C Consider moment at B caused by linearly distributed load on overhang AB. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vC (2 L2 3Lx x 2 ) 6 LEI Values: M = −½(8 kips/ft)(9 ft)(3 ft) = −108 kip-ft, L = 18 ft, x = 9 ft, EI = 4.93 × 106 kip-in.2


Computation: Mx vC (2 L2 6 LEI

3Lx

x2 )

( 108 kip-ft)(9 ft)(12 in./ft)3 2(18 ft) 2 6(18 ft)(4.93 106 kip-in.2 )

3(18 ft)(9 ft) (9 ft) 2

0.766559 in.

Consider linearly distributed load from 8 kips/ft to 0 kips/ft over span BD. [Appendix C, SS beam with linearly distributed load.] Relevant equation from Appendix C: w0 x vC (7 L4 10 L2 x 2 3 x 4 ) 360 LEI Values: w0 = 8 kips/ft, L = 18 ft, x = 9 ft, EI = 4.93 × 106 kip-in.2 Computation: w0 x vC (7 L4 10 L2 x 2 360 LEI

3x 4 )

(8 kips/ft)(9 ft)(12 in./ft)3 7(18 ft) 4 10(18 ft) 2 (9 ft) 2 6 2 360(18 ft)(4.93 10 kip-in. )

3(9 ft) 4

1.916398 in.

Consider 4 kips/ft uniformly distributed load on segment CD. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa3 vC (4L2 7aL 3a 2 ) 24 LEI Values: w = 4 kips/ft, L = 18 ft, a = 9 ft, EI = 4.93 × 106 kip-in.2 Computation: wa 3 vC (4 L2 24 LEI

7aL 3a 2 )

(4 kips/ft)(9 ft)3 (12 in./ft) 3 4(18 ft) 2 24(18 ft)(4.93 106 kip-in.2 )

7(9 ft)(18 ft) 3(9 ft) 2

0.958199 in.


vC

0.766559 in. 1.916398 in. 0.958199 in.

2.108037 in.

2.11 in.

Ans.


10.81 The simply supported beam shown in Fig. P10.81 consists of a W21 × 44 structural steel wide-flange shape [E = 29,000 ksi; I = 843 in.4]. For the loading shown, determine: (a) the beam deflection at point A. (b) the beam deflection at point C. Fig. P10.81

Solution (a) Beam deflection at point A Consider cantilever beam deflection of downward 4 kips/ft uniform load over AB. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vA 8 EI Values: w = 4 kips/ft, L = 12 ft, EI = 2.4447 × 107 kip-in.2 Computation: wL4 vA 8EI


0.732847 in.

Consider cantilever beam deflection of upward 4 kips/ft uniform load over 6-ft segment. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 wL3 v and (slope magnitude) 8 EI 6 EI Values: w = −4 kips/ft, L = 6 ft, EI = 2.4447 × 107 kip-in.2 Computation: wL4 v 8EI

wL3 6 EI vA

( 4 kips/ft)(6 ft) 4 (12 in./ft)3 8(2.4447 107 kip-in.2 ) (4 kips/ft)(6 ft)3 (12 in./ft) 2 6(2.4447 107 kip-in.2 )

0.045803 in.

0.0008482 rad

0.045803 in. (6 ft)(12 in./ft)(0.0008482 rad)

0.106873 in.


Consider rotation at B caused by downward 4 kips/ft uniform load. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: ML (slope magnitude) B 3EI Values: M = (4 kips/ft)(6 ft)(9 ft) = 216 kip-ft, L = 24 ft, EI = 2.4447 × 107 kip-in.2 Computation: ML (216 kip-ft)(24 ft)(12 in./ft) 2 B 3EI 3(2.4447 107 kip-in.2 ) vA

(12 ft)(12 in./ft)(0.0101784 rad)

0.0101784 rad 1.465693 in.



(12 ft)(12 in./ft)(0.0077929 rad)

(18 ft) 2

0.0077929 rad

1.122172 in.

Consider 4 kips/ft uniformly distributed load on 6-ft segment near D. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 (2L2 a 2 ) (slope magnitude) B 24LEI Values: w = 4 kips/ft, L = 24 ft, a = 6 ft, EI = 2.4447 × 107 kip-in.2 Computation: wa 2 (2 L2 B 24 LEI vA

2

a )

(4 kips/ft)(6 ft) 2 (12 in./ft) 2 2(24 ft) 2 7 2 24(24 ft)(2.4447 10 kip-in. )

(12 ft)(12 in./ft)(0.0016434 rad)

(6 ft) 2

0.0016434 rad

0.236648 in.

Beam deflection at A vA 0.732847 in. 0.106873 in. 1.465693 in. 1.122172 in. 0.236648 in. 0.732847 in.

0.733 in.

Ans.


(b) Beam deflection at point C Consider moment at B caused by downward 4 kips/ft uniform load. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vC (2 L2 3Lx x 2 ) (elastic curve) 6 LEI Values: M = −(4 kips/ft)(6 ft)(9 ft) = −216 kip-ft, L = 24 ft, x = 12 ft, EI = 2.4447 × 107 kip-in.2 Computation: Mx vC (2 L2 6 LEI

3Lx

x2 )

( 216 kip-ft)(12 ft)(12 in./ft) 3 2(24 ft) 2 6(24 ft)(2.4447 107 kip-in.2 )

3(24 ft)(12 ft) (12 ft) 2

0.549635 in.

Consider 42-kip concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vC ( L b 2 x 2 ) (elastic curve) 6 LEI Values: P = 42 kips, L = 24 ft, b = 6 ft, x = 12 ft, EI = 2.4447 × 107 kip-in.2 Computation: Pbx 2 vC (L 6 LEI

b2

x2 )

(42 kips)(6 ft)(12 ft)(12 in./ft)3 (24 ft) 2 7 2 6(24 ft)(2.4447 10 kip-in. )

(6 ft) 2

(12 ft) 2

0.587804 in.

Consider 4 kips/ft uniformly distributed load on 6-ft segment near D. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 vC (2 x3 6 Lx 2 a 2 x 4L2 x a 2 L) 24LEI (elastic curve) Values: w = 4 kips/ft, L = 24 ft, a = 6 ft, x = 12 ft, EI = 2.4447 × 107 kip-in.2


Computation: wa 2 vC (2 x 3 24 LEI

6 Lx 2

a 2 x 4 L2 x a 2 L)

2(12 ft)3 6(24 ft)(12 ft) 2 (6 ft) 2 (12 ft) (4 kips/ft)(6 ft) 2 (12 in./ft) 3 24(24 ft)(2.4447 107 kip-in.2 ) 4(24 ft) 2 (12 ft) (6 ft) 2 (24 ft) 0.175578 in.


vC

0.549635 in. 0.587804 in. 0.175578 in.

0.213747 in.

0.214 in.

Ans.


10.82 The simply supported beam shown in Fig. P10.82 consists of a W530 × 66 structural steel wide-flange shape [E = 200 GPa; I = 351 × 106 mm4]. If w = 85 kN/m, determine the beam deflection at point B.

Fig. P10.82

Solution Beam deflection at point B Consider 300 kN-m concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vB (2 L2 3Lx x 2 ) (elastic curve) 6 LEI Values: M = −300 kN-m, L = 9 m, x = 4 m, EI = 7.02 × 104 kN-m2 Computation: Mx vB (2 L2 3Lx x 2 ) 6 LEI ( 300 kN-m)(4 m) 2(9 m) 2 4 2 6(9 m)(7.02 10 kN-m )

3(9 m)(4 m) (4 m) 2

0.022159 m

Consider 85 kN/m uniformly distributed load on segment AB. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa3 vB (4L2 7aL 3a 2 ) 24 LEI Values: w = 85 kN/m, L = 9 m, a = 4 m, EI = 7.02 × 104 kN-m2 Computation: wa 3 vB (4 L2 24 LEI

7aL 3a 2 )

(85 kN/m)(4 m)3 4(9 m) 2 4 2 24(9 m)(7.02 10 kN-m )

7(4 m)(9 m) 3(4 m) 2

0.043052 m


Consider 140-kN concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vB ( L b 2 x 2 ) (elastic curve) 6 LEI Values: P = 140 kN, L = 9 m, b = 3 m, x = 4 m, EI = 7.02 × 104 kN-m2 Computation: Pbx 2 vB ( L b2 x 2 ) 6 LEI (140 kN)(3 m)(4 m) (9 m) 2 6(9 m)(7.02 104 kN-m 2 )

(3 m)2

(4 m)2

0.024818 m

Consider 175 kN-m concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vB (2 L2 3Lx x 2 ) (elastic curve) 6 LEI Values: M = −175 kN-m, L = 9 m, x = 5 m, EI = 7.02 × 104 kN-m2 Computation: Mx vB (2 L2 3Lx x 2 ) 6 LEI ( 175 kN-m)(5 m) 2(9 m) 2 6(9 m)(7.02 104 kN-m 2 )

3(9 m)(5 m) (5 m) 2

0.012003 m


vB

0.022159 m 0.043052 m 0.024818 m 0.012003 m

0.033708 m

33.7 mm

Ans.


10.83 The simply supported beam shown in Fig. P10.83 consists of a W530 × 66 structural steel wide-flange shape [E = 200 GPa; I = 351 × 106 mm4]. If w = 115 kN/m, determine the beam deflection at point C.

Fig. P10.83

Solution (b) Beam deflection at point C Consider 300 kN-m concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vC (2 L2 3Lx x 2 ) (elastic curve) 6 LEI Values: M = −300 kN-m, L = 9 m, x = 6 m, EI = 7.02 × 104 kN-m2 Computation: Mx vC (2 L2 3Lx x 2 ) 6 LEI ( 300 kN-m)(6 m) 2(9 m) 2 4 2 6(9 m)(7.02 10 kN-m )

3(9 m)(6 m) (6 m) 2

0.017094 m

Consider 115 kN/m uniformly distributed load on segment AB. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 vC (2 x3 6 Lx 2 a 2 x 4L2 x a 2 L) 24LEI Values: w = 115 kN/m, L = 9 m, a = 4 m, x = 6 m, EI = 7.02 × 104 kN-m2 Computation: wa 2 vC (2 x3 24 LEI

6 Lx 2

a 2 x 4 L2 x a 2 L)

(115 kN/m)(4 m) 2 2(6 m)3 4 2 24(9 m)(7.02 10 kN-m )

6(9 m)(6 m) 2

(4 m) 2 (6 m) 4(9 m) 2 (6 m) (4 m)2 (9 m)

0.046597 m


Consider 140-kN concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 vC (L a 2 b2 ) 6 LEI Values: P = 140 kN, L = 9 m, a = 6 m, b = 3 m, EI = 7.02 × 104 kN-m2 Computation: Pab 2 vC ( L a 2 b2 ) 6 LEI (140 kN)(6 m)(3 m) (9 m) 2 6(9 m)(7.02 104 kN-m 2 )

(6 m) 2

(3 m) 2

0.023932 m

Consider 175 kN-m concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vC (2 L2 3Lx x 2 ) (elastic curve) 6 LEI Values: M = −175 kN-m, L = 9 m, x = 3 m, EI = 7.02 × 104 kN-m2 Computation: Mx vC (2 L2 3Lx x 2 ) 6 LEI ( 175 kN-m)(3 m) 2(9 m) 2 6(9 m)(7.02 104 kN-m 2 )

3(9 m)(3 m) (3 m) 2

0.012464 m


vC

0.017094 m 0.046597 m 0.023932 m 0.012464 m

0.040971 m

41.0 mm

Ans.


10.84 A 25-ft-long soldier beam is used as a key component of an earth retention system at an excavation site. The soldier beam is subjected to a soil loading that is linearly distributed from 520 lb/ft to 260 lb/ft, as shown in Fig. P10.84. The soldier beam can be idealized as a cantilever with a fixed support at A. Added support is supplied by a tieback anchor at B, which exerts a force of 5,000 lb on the soldier beam. Determine the horizontal deflection of the soldier beam at point C. Assume EI = 5 × 108 lbin.2.

Fig. P10.84

Solution Consider 260 lb/ft uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vC 8EI Values: w = 260 lb/ft, L = 25 ft, EI = 5.0 × 108 lb-in.2 Computation: wL4 vC 8EI

(260 lb/ft)(25 ft)4 (12 in./ft)3 8(5.0 108 lb-in.2 )

43.875 in.

Consider a linearly distributed load that varies from 260 lb/ft at A to 0 lb/ft at C. [Appendix C, Cantilever beam with linearly distributed load.] Relevant equation from Appendix C: w0 L4 vC 30 EI Values: w0 = 260 lb/ft, L = 25 ft, EI = 5.0 × 108 lb-in.2 Computation: w0 L4 vC 30EI

(260 lb/ft)(25 ft)4 (12 in./ft)3 30(5.0 108 lb-in.2 )

11.700 in.


Consider 5,000-lb concentrated load. [Appendix C, Cantilever beam with concentrated load.] Relevant equations from Appendix C: PL3 PL2 vB and (slope magnitude) B 3EI 2 EI Values: P = 5,000 lb, L = 18 ft, EI = 5.0 × 108 lb-in.2 Computation: PL3 (5,000 lb)(18 ft)3 (12 in./ft)3 vB 3EI 3(5.0 108 lb-in.2 ) B

vC

PL2 2 EI

(5,000 lb)(18 ft) 2 (12 in./ft) 2 2(5.0 108 lb-in.2 )

33.592320 in. 0.2332800 rad

33.592320 in. (7 ft)(12 in./ft)(0.2332800 rad)

Beam deflection at C vC 43.875 in. 11.700 in. 53.187840 in.

53.187840 in.

2.387160 in.

2.39 in.

Ans.


10.85 A 25-ft-long soldier beam is used as a key component of an earth retention system at an excavation site. The soldier beam is subjected to a uniformly distributed soil loading of 260 lb/ft, as shown in Fig. P10.85. The soldier beam can be idealized as a cantilever with a fixed support at A. Added support is supplied by a tieback anchor at B, which exerts a force of 4,000 lb on the soldier beam. Determine the horizontal deflection of the soldier beam at point C. Assume EI = 5 × 108 lb-in.2. Fig. P10.85

Solution Consider 260 lb/ft uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vC 8EI Values: w = 260 lb/ft, L = 25 ft, EI = 5.0 × 108 lb-in.2 Computation: wL4 vC 8EI

(260 lb/ft)(25 ft)4 (12 in./ft)3 8(5.0 108 lb-in.2 )

43.875 in.

Consider 4,000-lb concentrated load. [Appendix C, Cantilever beam with concentrated load.] Relevant equations from Appendix C: PL3 PL2 vB and (slope magnitude) B 3EI 2 EI Values: P = 4,000 lb, L = 18 ft, EI = 5.0 × 108 lb-in.2 Computation: PL3 (4,000 lb)(18 ft)3 (12 in./ft)3 vB 26.873856 in. 3EI 3(5.0 108 lb-in.2 ) B

vC

PL2 2 EI

(4,000 lb)(18 ft) 2 (12 in./ft) 2 2(5.0 108 lb-in.2 )

0.1866240 rad

26.873856 in. (7 ft)(12 in./ft)(0.1866240 rad)

Beam deflection at C vC 43.875 in. 42.550272 in.

1.324728 in.

42.550272 in.

1.325 in.

Ans.


11.1 A beam is loaded and supported as shown in Fig. P11.1. Use the doubleintegration method to determine the magnitude of the moment M0 required to make the slope at the left end of the beam zero. Fig. P11.1


Ma

a

M ( x) wx M ( x)

M0

x 2

M0

0

wx 2 2

Integration: d 2v wx 2 EI 2 M ( x) M 0 dx 2 3 dv wx EI M0x C1 dx 6 M 0 x 2 wx 4 EI v C1 x C2 2 24 Boundary conditions and evaluate constants: dv w( L)3 at x L, 0 M 0 ( L) C1 0 dx 6 C1

wL3 6

Beam slope equation: dv wx 3 EI M0x dx 6

M 0L

wL3 6

M 0L

Constraint: At x = 0, the slope of the beam is to be zero; therefore, dv w(0)3 wL3 EI M 0 (0) M 0L 0 dx A 6 6

M0

wL2 6

Ans.


11.2 When moment M0 is applied to the left end of the cantilever beam shown in Fig. P11.2, the slope of the beam is zero. Use the double-integration method to determine the magnitude of the moment M0.

Fig. P11.2

Solution Moment equation: M a a M ( x) Px M 0

M ( x)

0

Px M 0

Integration: d 2v EI 2 M ( x) Px M 0 dx dv Px 2 EI M 0 x C1 dx 2 Px3 M 0 x 2 EI v C1 x C2 6 2 Boundary conditions and evaluate constants: dv P ( L) 2 at x L, 0 M 0 ( L) C1 0 dx 2 C1

PL2 2

Beam slope equation: dv Px 2 EI M0x dx 2

M 0L

PL2 2

M 0L

Constraint: At x = 0, the slope of the beam is to be zero; therefore, dv P (0) 2 PL2 EI M 0 (0) M 0L 0 dx A 2 2 M0

PL 2

Ans.


11.3 When the load P is applied to the right end of the cantilever beam shown in Fig. P11.3, the deflection at the right end of the beam is zero. Use the double-integration method to determine the magnitude of the load P.

Fig. P11.3

Solution Moment equation: Ma

a

w( L

L

x)

x 2

w ( L x) 2 2

M ( x)

P( L

x) M ( x)

0

P( L x)

Integration: d 2v w EI 2 M ( x) ( L x) 2 P( L x) dx 2 dv w P EI ( L x )3 ( L x) 2 C1 dx 6 2 w P EI v ( L x) 4 ( L x)3 C1x C2 24 6 Boundary conditions and evaluate constants: dv w P at x 0, 0 ( L 0)3 ( L 0) 2 dx 6 2 C1 at x

0, v

wL3 6

0 C2

wL4 24

PL2 2 w ( L 0) 4 24

C1

0

P ( L 0)3 C1 (0) C2 6

0

PL3 6

Beam elastic curve equation: w P wLx3 PL2 x wL4 EI v ( L x) 4 ( L x )3 24 6 6 2 24 3 4 w wLx wL P PL2 x 4 3 ( L x) ( L x) 24 6 24 6 2

PL3 6 PL3 6


Constraint: At x = L, the deflection of the beam is to be zero; therefore, w wL( L)3 wL4 P PL2 ( L) 4 3 EI vB ( L L) ( L L) 24 6 24 6 2 which simplifies to wL4 wL4 PL3 PL3 wL4 PL3 EI vB 0 6 24 2 6 8 3 Therefore, the magnitude of P is 3wL P 8

PL3 6

0

Ans.


11.4 A beam is loaded and supported as shown in Fig. P11.4. Use the doubleintegration method to determine the reactions at supports A and B.

Fig. P11.4


MA

By MA

0 By L M 0

Moment equation: Ma a M ( x) M 0

0

By ( L

x)

0

M ( x)

By ( L

x)

M0

Integration: d 2v EI 2 M ( x) By ( L x) M 0 dx By dv EI ( L x) 2 M 0 x C1 dx 2 By M 0 x2 3 EI v ( L x) C1 x C2 6 2 Boundary conditions and evaluate constants: at x at x

at x

0,

dv dx

0, v

L, v

By

0

2 By

0

6

By

0

6

( L 0) 2

( L 0)

3

(L

3

L)

M 0 (0) C1 M 0 (0) 2 2

M 0 L2 3M 0 By 3 2 2L Backsubstitute into equilibrium equations:

Fy MA

Ay

By

0

MA

By L

MA

M0 2

Ay M0

0

M0 (cw) 2

C1 (0) C2

M 0 ( L) 2 2

By L3

0

By L2 2

( L)

By L2

C1 0

2 By L3

C2

By L3 6

6

0

3M 0 2L

By

3M 0 2L

MA

By L

Ans.

Ay M0

3M 0 L 2L

3M 0 2L M0

Ans. 0

Ans.


11.5 A beam is loaded and supported as shown in Fig. P11.5. (a) Use the double-integration method to determine the reactions at supports A and B. (b) Draw the shear-force and bendingmoment diagrams for the beam. Fig. P11.5


MA

By

MB

wL

0

By L wL

L 2

0

Moment equation: Ma

M ( x)

a

wx

x 2

Ay x

0

wx 2 2

M ( x)

Integration: d 2v wx 2 EI 2 M ( x) Ay x dx 2 2 dv wx 3 Ay x EI C1 dx 6 2 3 wx 4 Ay x EI v C1 x C2 24 6 Boundary conditions and evaluate constants: 3 w(0) 4 Ay (0) at x 0, v 0 C1 (0) C2 24 6 at x

dv L, dx

Ay ( L) 2

w( L)3 6

0

2

Ay x

0

C1

0

C2

0

C1

wL3 6

Ay L2

C1

wL3 24

Ay L2

3

w( L) 4 Ay ( L) at x L, v 0 C1 ( L) 24 6 Solve Eqs. (a) and (b) simultaneously to find:

wL3 3wL and Ay 48 8 Backsubstitute into equilibrium equations: C1

Fy MA

Ay

By

MB MB

wL

0

By L wL wL2 8

By L 2

0

wL2 (cw) 8

wL

Ay

wL MB

0

2 6

3wL 8

3wL 8 wL2 2

(a) (b)

Ans.

5wL 8 By L

By wL2 2

5wL2 8

5wL 8

Ans.

wL2 8 Ans.


11.6 A beam is loaded and supported as shown in Fig. P11.6. Use the doubleintegration method to determine the reactions at supports A and B.

Fig. P11.6

Solution Beam FBD:

Fy

Ay

MA

w0 L 0 2 w0 L 2 L By L 2 3

By

MB

0

Moment equation:

Ma

a

M ( x)

w0 x 2 x 2L 3

M ( x)

w0 x3 6L

Ay x

0

Ay x

Integration: d 2v w0 x3 EI 2 M ( x) Ay x dx 6L 2 dv w0 x 4 Ay x EI C1 dx 24 L 2 3 w0 x5 Ay x EI v C1 x C2 120 L 6 Boundary conditions and evaluate constants: 3 w0 (0)5 Ay (0) at x 0, v 0 C1 (0) C2 120 L 6 at x at x

dv L, dx L, v

0 0

w0 ( L)4 24 L

Ay ( L)2

w0 ( L)5 120 L

Ay ( L)3

2 6

C1

0

C1 ( L)

0

0

C2

0

C1

w0 L3 24

Ay L2

C1

w0 L3 120

Ay L2

2 6

(a) (b)


Solve Eqs. (a) and (b) simultaneously to find: C1

w0 L3 120

and

Ay

w0 L 10

Backsubstitute into equilibrium equations: w0 L w0 L Fy Ay By 0 By Ay 2 2

MA

MB MB

By L

w0 L 2 L 2 3 w0 L2 15

0

w0 L2 (cw) 15

w0 L 10

w0 L 2 MB

w0 L 10 w0 L2 3

Ans.

4w0 L 10 By L

w0 L2 3

By 2w0 L2 5

2w0 L 5

Ans.

w0 L2 15 Ans.


11.7 A beam is loaded and supported as shown in Fig. P11.7. Use the fourth-order integration method to determine the reaction at roller support B.

Fig. P11.7

Solution Integrate the load distribution: d 4v w0 x 2 EI 4 dx L2 d 3v w0 x 3 EI 3 C1 dx 3L2 d 2v w0 x 4 EI 2 C1 x C2 dx 12 L2 dv w0 x5 C1 x 2 EI C2 x C3 dx 60 L2 2 w0 x 6 C1 x3 C2 x 2 EI v C3 x C4 360 L2 6 2 Boundary conditions and evaluate constants: w0 (0)6 C1 (0)3 C2 (0) 2 at x 0, v 0 C3 (0) C4 360 L2 6 2 dv w0 (0)5 C1 (0) 2 at x 0, 0 C2 (0) C3 0 dx 60 L2 2 w0 ( L)6 C1 ( L)3 C2 ( L) 2 at x L, v 0 0 360 L2 6 2

d 2v w0 ( L)4 0 C1 ( L) C2 dx 2 12 L2 Solve Eqs. (a) and (b) simultaneously to obtain: at x

L, M

EI

2C2

w0 L2 60

w0 L2 12

C1L

w0 L2 12

w0 L2 30

4w0 L2 60

Roller reaction at B: d 3v w0 ( L)3 VB EI 3 dx x L 3L2

C1

7 w0 L 60

C4

0

C3

0

C1L 3C2

0

C1L C2

w0 L2 60 w0 L2 12

(a) (b)

w0 L2 30

C2

7w0 L2 60

0

7w0 L 60

20w0 L 60

7w0 L 60

13w0 L 60

By

13w0 L 60

Ans.


11.8 A beam is loaded and supported as shown in Fig. P11.8. Use the fourth-order integration method to determine the reaction at roller support A.

Fig. P11.8

Solution Integrate the load distribution: d 4v w0 x 2 EI 4 dx L2 d 3v w0 x 3 EI 3 C1 dx 3L2 d 2v w0 x 4 EI 2 C1 x C2 dx 12 L2 dv w0 x5 C1 x 2 EI C2 x C3 dx 60 L2 2 w0 x 6 C1 x3 C2 x 2 EI v C3 x C4 360 L2 6 2 Boundary conditions and evaluate constants: w0 (0)6 C1 (0)3 C2 (0) 2 at x 0, v 0 C3 (0) C4 360 L2 6 2 d 2v w0 (0) 4 at x 0, M EI 2 0 C1 (0) C2 0 dx 12 L2 dv w0 ( L)5 C1 ( L)2 at x L, 0 C3 0 dx 60 L2 2

at x

L, v

w0 ( L)6 360 L2

0

C1 ( L)3 6

C3 ( L)

0

0

C4

0

C2

0

C1L2

2C3

w0 L3 30

(a)

C1L2

6C3

w0 L3 60

(b)

Solve Eqs. (a) and (b) simultaneously to obtain:

4C3 C1L2

w0 L3 30

w0 L3 60

w0 L3 60

C3

w0 L3 30

w0 L3 120

5w0 L3 120

C1

Roller reaction at A: d 3v w0 (0)3 VA EI 3 dx x 0 3L2

w0 L 24

w0 L 24

w0 L3 240 5w0 L 120

w0 L 24

Ay

w0 L 24

Ans.


11.9 A beam is loaded and supported as shown in Fig. P11.9. Use the fourth-order integration method to determine the reaction at roller support A.

Fig. P11.9

Solution Integrate the load distribution: d 4v x EI 4 w0 sin dx 2L 3 d v 2w0 L x EI 3 cos C1 dx 2L d 2v 4w0 L2 x EI 2 sin C1 x C2 2 dx 2L dv 8w0 L3 x C1 x 2 EI cos C2 x C3 3 dx 2L 2 16w0 L4 x C1 x3 C2 x 2 EI v sin C3 x C4 4 2L 6 2 Boundary conditions and evaluate constants: 16w0 L4 (0) C1 (0)3 C2 (0) 2 at x 0, v 0 sin C3 (0) C4 0 4 2L 6 2 d 2v 4w0 L2 (0) at x 0, M EI 2 0 sin C1 (0) C2 0 2 dx 2L dv 8w0 L3 ( L) C1 ( L) 2 at x L, 0 cos C3 0 C1L2 2C3 3 dx 2L 2

at x

L, v

16w0 L4

0

4

sin

C1 ( L)3 6

( L) 2L

C3 ( L)

0

C1L2

6C3

C4

0

C2

0

0

(a)

96w0 L3 4

(b)


4C3 2

C1L

96w0 L3 4

2

24w0 L3 4

Roller reaction at A: d 3v 2w0 L (0) VA EI 3 cos dx x 0 2L

C3

24w0 L3 4

48w0 L

C1

4

48w0 L 4

Ay

2w0 L

48w0 L 4

Ans.


11.10 A beam is loaded and supported as shown in Fig. P11.10. Use the fourth-order integration method to determine the reactions at supports A and B.

Fig. P11.10

Solution Integrate the load distribution: d 4v x EI 4 w0 cos dx 2L 3 d v 2w0 L x EI 3 sin C1 dx 2L d 2v 4w0 L2 x EI 2 cos C1 x C2 2 dx 2L dv 8w0 L3 x C1x 2 EI sin C2 x C3 3 dx 2L 2 16w0 L4 x C1 x3 C2 x 2 EI v cos C3 x C4 4 2L 6 2 Boundary conditions and evaluate constants: dv 8w0 L3 (0) C1 (0) 2 at x 0, 0 sin 3 dx 2L 2 4 16w0 L (0) C1 (0)3 at x 0, v 0 cos 4 2L 6

at x

dv L, dx

at x

L, v

8w0 L3

0

3

16w0 L4

0

4

C2 (0) C3 C2 (0) 2 2

C4

( L) sin 2L

C1 ( L) 2 2

C2 ( L)

( L) 2L

C1 ( L)3 6

C2 ( L)2 2

cos

0

C3 0

C4

0

0 16 w0 L4

C1L 2C2

16w0 L4 4

0

C1L 3C2

4

16w0 L2 3

96w0 L2 4

(a) (b)


C2 C1L

16w0 L2

96w0 L2

3

C2

4

48w0 L2

192w0 L2

3

C1

4

Reactions at supports A and B d 3v 2w0 L (0) VA EI 3 sin dx x 0 2L

Ay

48w0 L 4

48w0 L 4

4

16w0 L2 4

6

48w0 L

4

4

4

48w0 L 4

4 Ans.


VB

EI

d 3v dx3

2w0 L

sin

x L

2w0 L

By

MA

EI

d 2v dx 2

( L) 2L

2

cos

x 0

4 w0 L2

(0) 2L

MB

EI

d 2v dx 2

4

4 w0 L2 4

4 w0 L2 2 x L

48w0 L2 4

MB

4 32 w0 L2 4

3

24

4

96 Ans.

4

4

16 w0 L2 4

6

6 2

24 4

( L) 2L

cos

2w0 L

3

48w0 L(0)

16 w0 L2

2

MA

4

4

96 24

4

4 w0 L2

48w0 L

48w0 L( L ) 4

16 w0 L2 4

6

3 (ccw)

(cw)

Ans.

16w0 L2

4

4

16 w0 L2 4

3

6 12 6

16 w0 L2 4

2

6

Ans.


11.11 A beam is loaded and supported as shown in Fig. P11.11. Use the fourth-order integration method to determine the reactions at supports A and B.

Fig. P11.11

Solution Integrate the load distribution: d 4v x EI 4 w0 sin dx L 3 d v w0 L x EI 3 cos C1 dx L d 2v w0 L2 x EI 2 sin C1 x C2 2 dx L dv w0 L3 x C1 x 2 EI cos C2 x C3 3 dx L 2 w0 L4 x C1 x 3 C2 x 2 EI v sin C3 x C4 4 L 6 2 Boundary conditions and evaluate constants: dv w0 L3 (0) C1 (0) 2 at x 0, 0 cos C2 (0) C3 0 3 dx L 2 w0 L4 (0) C1 (0)3 C2 (0) 2 at x 0, v 0 sin C3 (0) C4 4 L 6 2

dv dx

at x

L,

at x

L, v

w0 L3

0

3

cos

w0 L4

0

4

sin

C1 ( L)2 2

( L) L ( L) L

C2 ( L)

C1 ( L)3 6

C2 ( L)2 2

w0 L3

0

0

3

w0 L3 3

( L)

w0 L3

C3 C4

3

0

C1L 2C2 0

C1L 3C2

4w0 L2 3

6w0 L2 3

(a) (b)


C2 C1L

6w0 L2

4w0 L2

3

C2

3

4w0 L2 3

2

2w0 L2

C1

3

Reactions at supports A and B d 3v w0 L (0) VA EI 3 cos dx x 0 L

Ay

w0 L

2w0 L2 3

0

w0 L

Ans.


VB

EI

d 3v dx3

w0 L x L

EI

d 2v dx 2

w0 L2 2

EI

d 2v dx 2

sin

x 0

Ans.

3

w0 L2 2 x L

MB

sin

2w0 L2

(0) L

2w0 L2

MA

MB

w0 L

w0 L

By

MA

( L) L

cos

3

3

3

Ans.

(cw)

( L) L

2w0 L2

2w0 L2

2w0 L2 3

(ccw)

2w0 L2 3

Ans.


11.12 A beam is loaded and supported as shown in Fig. P11.12. (a) Use the double-integration method to determine the reactions at supports A and C. (b) Draw the shear-force and bendingmoment diagrams for the beam. (c) Determine the deflection in the middle of the span. Fig. P11.12

Solution Beam FBD: from symmetry, Ay

P 2

Cy and

MA

MC

Moment equation: Ma

a

M ( x) M A

P x 2

0

M ( x)

Px 2

MA

Integration: d 2v Px EI 2 M ( x) MA dx 2 dv Px 2 EI M A x C1 dx 4 Px3 M A x 2 EI v C1 x C2 12 2 Boundary conditions and evaluate constants: dv P(0) 2 at x 0, 0 M A (0) C1 0 dx 4 P(0)3 M A (0) 2 at x 0, v 0 C2 0 12 2 (a) Beam reaction forces: P Ay C y 2

C1

0

C2

0

Ans.


(a) Beam reaction moments: L dv at x , 0 2 dx MA

PL 8

P ( L / 2) 2 4

MA

L 2

PL (ccw) 8

Elastic curve equation: Px3 M A x 2 Px3 PLx 2 EI v 12 2 12 16 2 Px v 3L 4 x 48EI

0 MC

PL (cw) 8

Ans.

Px 2 3L 4 x 48

(c) Midspan deflection: vB

P( L / 2) 2 3L 4( L / 2) 48EI

PL3 192 EI

Ans.


11.13 A beam is loaded and supported as shown in Fig. P11.13. (a) Use the double-integration method to determine the reactions at supports A and B. (b) Draw the shear-force and bendingmoment diagrams for the beam. (c) Determine the deflection in the middle of the span. Fig. P11.13


wL 2

By and

MA

MB

Moment equation:

Ma

a

M ( x) M A M ( x)

wx

wx 2 2

x 2 wLx 2

wL x 2

0

MA

Integration: d 2v wx 2 wLx EI 2 M ( x) MA dx 2 2 dv wx3 wLx 2 EI M A x C1 dx 6 4 wx 4 wLx3 M A x 2 EI v C1 x C2 24 12 2 Boundary conditions and evaluate constants: dv w(0)3 wL(0) 2 at x 0, 0 M A (0) C1 0 dx 6 4 w(0) 4 wL(0)3 M A (0) 2 at x 0, v 0 C2 0 24 12 2 (a) Beam reaction forces: wL Ay By 2

C1

0

C2

0

Ans.


(a) Beam reaction moments: L dv at x , 0 2 dx

MA

wL2 12

w( L / 2)3 6

wL( L / 2) 2 4

wL2 (ccw) 12

Elastic curve equation: wx 4 wLx 3 M A x 2 EI v 24 12 2 2 wx v ( x L) 2 24 EI

MB

wx 4 24

wLx 3 12

wL2 x 2 24

MA

L 2 wL2 12

wx 2 2 x 24

0 wL2 (cw) 12

2 Lx

L2

Ans.

wx 2 ( x L) 2 24

(c) Midspan deflection: vx

L/2

w( L / 2) 2 24 EI

L 2

2

L)

wL4 384 EI

Ans.


11.14 A beam is loaded and supported as shown in Fig. P11.14. (a) Use the double-integration method to determine the reactions at supports A and C. (b) Determine the deflection in the middle of the span. Fig. P11.14


w0 L 2

Cy and MA

MC

Moment equation: Ma

a

M ( x) M A M ( x)

w0 x 2 x 2L 3

w0 x3 6L

w0 Lx 2

w0 L x 2

0

MA

Integration: d 2v w0 x3 w0 Lx EI 2 M ( x) MA dx 6L 2 dv w0 x 4 w0 Lx 2 EI M A x C1 dx 24 L 4 w0 x5 w0 Lx3 M A x 2 EI v C1 x C2 120 L 12 2 Boundary conditions and evaluate constants: dv w0 (0) 4 w0 L(0) 2 at x 0, 0 dx 24 L 4 5 w0 (0) w0 L(0)3 at x 0, v 0 120 L 12 (a) Beam reaction forces: w0 L Ay C y 2

M A (0) C1 M A (0) 2 2

C2

0 0

C1

0

C2

0

Ans.


(a) Beam reaction moments: dv at x L, 0 dx MA

5w0 L2 24

w0 ( L) 4 24 L

w0 L( L) 2 4

5w0 L2 (ccw) 24

M A ( L) MC

Elastic curve equation: w0 x 5 w0 Lx 3 M A x 2 w0 x 5 EI v 120 L 12 2 120 L 5 2 3 2 w0 x 20w0 L x 25w0 L3 x 2 240 L 240 L 240 L 2 w0 x v 2 x 3 20 L2 x 25L3 240 L EI

0 5w0 L2 24

5w0 L2 (cw) 24

Ans.

w0 Lx 3 5w0 L2 x 2 12 48 2 w0 x 2 x 3 20 L2 x 25 L3 240 L


w0 ( L)2 2( L)3 240 L EI

2

3

20 L ( L) 25L

7 w0 L4 240 EI

Ans.


11.15 A beam is loaded and supported as shown in Fig. P11.15. (a) Use the double-integration method to determine the reactions at supports A and C. (b) Draw the shear-force and bendingmoment diagrams for the beam. (c) Determine the deflection in the middle of the span. Fig. P11.15


MA

Cy

MC

P

0

Cy L

P

Moment equation: M a a M ( x) Ay x

M ( x)

Mb

b

M ( x) M ( x)

L 2

0

0

Ay x

Ay x Ay x

0

P x

L 2

Px

PL 2

x

L 2

0

Integration: For beam segment AB: d 2v EI 2 M ( x) Ay x dx 2 dv Ay x EI C1 dx 2 Ay x3 EI v C1 x C2 6

L 2

x

L

For beam segment BC: d 2v PL EI 2 M ( x) Ay x Px dx 2 2 2 dv Ay x Px PLx EI C3 dx 2 2 2 Ay x 3 Px3 PLx 2 EI v C3 x C4 6 6 4

Boundary conditions and evaluate constants: Ay (0)3 at x 0, v 0 C1 (0) C2 0 6 at x at x

L,

dv dx

L, v

0 0

Ay ( L)2 2 Ay ( L) 6

3

P ( L) 2 2 P ( L )3 6

PL( L) 2 PL( L) 2 4

C2

C3

0

C3

2

Ay L 2

( L) C4

0

C4

0

Ay L2 2 Ay L3 3

PL3 12


Slope continuity condition at x = L/2: L dv dv at x , 2 dx AB dx BC Ay ( L / 2) 2

C1

2

Ay ( L / 2) 2

P( L / 2) 2 2

2 C1

PL2 8

Ay L 2

6 8 2 6 eliminate terms and rearrange: Ay L3 Px 3 PLx 2 PL2 x

6

3 6 4 8 Substitute x = L/2 to obtain: Ay L3 P( L / 2)3 PL( L / 2) 2

6 Ay

2

2

Deflection continuity condition at x = L/2: L at x , vB AB vB BC 2 Ay x3 PL2 x Ay L2 x Ay x3 Px3 PLx 2

3

Ay L2

PL( L / 2) 2

4

Ay L2 x

Ay L3

2

3

PL3 12

PL3 12

4

PL2 ( L / 2) 8

PL3 12

5 PL3 48

5P 16

(a) Beam reaction forces: 5P Ay Cy 16

11P 16

(a) Beam reaction moment: L PL 11PL MC P Cy L 2 2 16

Ans.

3PL 16

Elastic curve equation for beam segment AB: Ay x3 PL2 x Ay L2 x 5Px3 PL2 x 5PL2 x EI v 6 8 2 96 8 32 Px v 5 x 2 3L2 96 EI

MC

5Px 3 96

3PL 16

3PL (cw) 16

Ans.

3PL2 x 96


P( L / 2) L 5 96 EI 2

2

3L2

7 PL3 768EI

Ans.


11.16 A beam is loaded and supported as shown in Fig. P11.16. (a) Use the double-integration method to determine the reactions at supports A and C. (b) Draw the shear-force and bendingmoment diagrams for the beam. Fig. P11.16


MA

Cy MA

0

Cy L M 0

Moment equation: M a a M ( x) Ay x

M ( x)

Mb

b

M ( x) M ( x)

Cy

Ay x

MA

Ay 0

0

MA

Ay x M A Ay x M A

M0 M0

Integration: For beam segment AB: d 2v EI 2 M ( x) Ay x M A dx 2 dv Ay x EI M A x C1 dx 2 Ay x 3 M A x 2 EI v C1 x C2 6 2

0

x

L 2

L 2

x

L

0

For beam segment BC: d 2v EI 2 M ( x) Ay x M A M 0 dx 2 dv Ay x EI M A x M 0 x C3 dx 2 Ay x3 M A x 2 M 0 x 2 EI v C3 x C4 6 2 2

Boundary conditions and evaluate constants for segment AB: Ay (0)3 M A (0) 2 at x 0, v 0 C1 (0) C2 0 6 2 Ay (0) 2 dv at x 0, 0 M A (0) C1 0 dx 2

C2

0

C1

0


Slope continuity condition at x = L/2: L dv dv at x , 2 dx AB dx BC Ay x 2 2

M Ax

Ay x 2 2 C3

M Ax

M 0 x C3

M 0L 2

Deflection continuity condition at x = L/2: L at x , vB AB vB BC 2 Ay x3 M A x 2 Ay x 3 M A x 2 M 0 x 2 M 0 Lx 6 2 6 2 2 2

C4

C4

M 0 L2 8

Boundary condition for segment BC: Ay ( L)3 M A ( L)2 M 0 ( L)2 M 0 L M 0 L2 at x L, v 0 ( L) 6 2 2 2 8 Also, the beam moment equilibrium equation can be written as: Ay L M A M0

0

(a) Beam Reactions: Solve these two equations simultaneously to obtain: M0 M0 9M 0 9M 0 MA (cw) Ay 8 8 8L 8L

Ay L 3M A

Cy

9M 0 8L

3M 0 4

Ans.


11.17 A beam is loaded and supported as shown in Fig. P11.17. (a) Use the double-integration method to determine the reactions at supports A and C. (b) Draw the shear-force and bending-moment diagrams for the beam. Fig. P11.17

Solution Beam FBD:

Fy

Ay

MA

Cy

MC

wL 2

Cy L

0 wL L 2 4

0

Moment equation: Ma

a

M ( x) wx

M ( x)

Mb

b

M ( x)

M ( x)

x 2

wx 2 2

Ay x

Ay x

wL x 2 wL x 2

0

L 4 L 4

0

Ay x Ay x

Integration: For beam segment AB: d 2v wx 2 EI 2 M ( x) Ay x dx 2 2 dv wx 3 Ay x EI C1 dx 6 2 3 wx 4 Ay x EI v C1 x C2 24 6

L 2

x

0 L 2

x

L

For beam segment BC: d 2v wL EI 2 M ( x) x dx 2

dv EI dx EI v

wL x 4 wL x 12

Boundary conditions and evaluate constants for segment AB: 3 w(0) 4 Ay (0) at x 0, v 0 C1 (0) C2 0 24 6

L 4 L 4

3

2

L 4

Ay x 2 2 Ay x3 6

Ay x

C3 C3 x C4

C2

0


Slope continuity condition at x = L/2: Equate the slope expressions for the two beam segments: 2 2 Ay x 2 wx3 Ay x wL L C1 x C3 6 2 4 4 2 Set x = L/2 and solve for the constant C1:

C1

wx3 6

C3 C1

wL x 4

2

L 4

w( L / 2)3 6

C3

wL L 4 2

L 4

2

C3

wL3 48

wL3 64

C3

9wL3 64

wL3 192

C3

Deflection continuity condition at x = L/2: Equate the deflection expressions for the two beam segments: 3 3 Ay x3 wx 4 Ay x wL L C1 x x C3 x C4 24 6 12 4 6 Set x = L/2 and solve for the constant C4: 3 w( L / 2) 4 wL3 L wL L L L C3 C3 C4 24 192 2 12 2 4 2

wL4 384

wL4 384

L 2

C3

wL4 768

L 2

C3

C4

wL4 768

C4

Boundary conditions and evaluate constants for segment BC: at x

dv L, dx

wL L 4

0

L 4

2

Ay ( L)2 2

C3

0

Ay L2 2

at x = L, v = 0 wL L 12 27 wL4 768

L 4

Ay L3 6

27 wL4 768

3

Ay ( L)3

C3 ( L)

6 9wL3 64 Ay L3 6

Ay L2

9wL4 64

( L)

2

Ay L3 2 Ay L3 6

wL4 768

0

wL4 768

0

wL4 768 3 Ay L3 6 Ay L3 3

0 26 wL4 768 82wL4 768

108wL4 768 Ay

41wL 128


Solve for C3:

9wL3 C3 64 and for C1: C1

5wL3 256

41wL3 256

5wL3 256

wL3 192

11wL3 768

(a) Beam force reactions: 41wL Ay 128

Ans.

wL wL 41wL Ay 2 2 128 Beam moment reaction: Cy

MC

wL2 8

Cy L

wL2 8

23wL 128

23wL2 128

Cy 7 wL2 128

23wL 128

MC

Ans.

7 wL2 128

7 wL2 (cw) 128

Ans.


11.18 A propped cantilever beam is loaded as shown in Fig. P11.18. Assume EI = 200,000 kN-m2. Use discontinuity functions to determine: (a) the reactions at A and C. (b) the beam deflection at B. Fig. P11.18

Solution Moment equation: (a) Support reactions: Fy Ay C y 150 kN

Ay MA

0

C y 150 kN (150 kN)(7 m) C y (12 m) M A

MA

C y (12 m)

(a)

0 (b)

1,050 kN-m

Discontinuity expressions:

w x

2

MA x 0 m

V ( x)

w x dx

MA x 0 m

M x

V ( x)dx

MA x 0 m

d 2v M ( x) M A dx 2 dv 1 EI MA x 0 m dx MA 2 EI v x 0m 2 EI

x 0m Ay 2 Ay 6

1

0

0

0

Ay x 0 m Ay x 0 m Ay x 0 m

x 0m x 0m

1

Ay x 0 m

3

2

1

1

1

150 kN x 7 m 150 kN x 7 m 150 kN x 7 m 150 kN x 7 m

1

1

0

C y x 12 m C y x 12 m

C y x 12 m C y x 12 m

1

0

1

1

Cy 150 kN 2 2 x 7m x 12 m C1 2 2 Cy 150 kN 3 3 x 7m x 12 m C1 x C2 6 6

(c) (d)

Boundary conditions and evaluate constants: dv at x 0 m, 0 C1 0 dx at x 0 m, v 0 C2 0

at x 12 m, v

0

M A (72 m 2 )

Ay MA (12 m) 2 (12 m)3 2 6 Ay (288 m3 ) 3,125 kN-m3 0

150 kN (5 m)3 6 (e)

(a) Solve for Ay, Cy, and MA: Solve equations (a), (b), and (c) simultaneously to obtain the results:

Ay MA

88.3247 kN

88.3 kN

309.8958 kN-m

310 kN-m (ccw)

Cy

61.6753 kN

61.7 kN Ans.


(b) Beam deflection at B: From Eq. (d), the beam deflection at B (x = 7 m) is computed as follows: 309.8958 kN-m 88.3247 kN EI vB (7 m) 2 (7 m)3 2 6 3 2,543.2219 kN-m vB

2,543.2219 kN-m3 200, 000 kN-m 2

0.0127161 m

12.72 mm

Ans.


11.19 A propped cantilever beam is loaded as shown in Fig. P11.19. Assume EI = 200,000 kNm2. Use discontinuity functions to determine: (a) the reactions at A and B. (b) the beam deflection at C. Fig. P11.19

Solution Moment equation: (a) Support reactions: Fy Ay By 0

Ay

MA

(a)

By

750 kN-m By (5 m) M A MA

By (5 m)

0 (b)

750 kN-m


w x

2

MA x 0 m

V ( x)

w x dx

MA x 0 m

M x

V ( x)dx

MA x 0 m

d 2v EI 2 M ( x) M A dx dv 1 EI MA x 0 m dx MA 2 EI v x 0m 2

x 0m Ay 2 Ay 6

1

0

0

Ay x 0 m Ay x 0 m 2

3

By

6

By x 5 m

1

By x 5 m

x 5m

2 By

By x 5 m

1

x 5m

1

By x 5 m

0

Ay x 0 m

x 0m x 0m

1

Ay x 0 m

3

2

1

1

0

2

750 kN-m x 7.5 m 750 kN-m x 7.5 m

750 kN-m x 7.5 m 750 kN-m x 7.5 m 1

750 kN-m x 7.5 m 750 kN-m x 7.5 m 2

2

1

0

0

C1

(c)

C1 x C2

(d)

Boundary conditions and evaluate constants: dv at x 0 m, 0 C1 0 dx at x 0 m, v 0 C2 0

at x

5 m, v

0

Ay

MA (5 m) 2 2 Ay (20.83333 m3 ) 0 0

2

M A (12.5 m )

6

(5 m)3 (e)

(a) Solve for Ay, By, and MA: Solve equations (a), (b), and (c) simultaneously to obtain the results:

Ay

225.000 kN

MA

375.000 kN-m

225 kN 375 kN-m (cw)

By

225.000 kN

225 kN Ans.


(b) Beam deflection at C: From Eq. (d), the beam deflection at C (x = 7.5 m) is computed as follows: Ay By MA EI vC (7.5 m) 2 (7.5 m)3 (2.5 m) 3 2 6 6 4,687.500 kN-m3 vC

4,687.500 kN-m3 200,000 kN-m 2

0.023438 m

23.4 mm

Ans.


11.20 A propped cantilever beam is loaded as shown in Fig. P11.20. Assume EI = 100,000 kip-ft2. Use discontinuity functions to determine: (a) the reactions at A and E. (b) the beam deflection at C.

Fig. P11.20

Solution Moment equation: (a) Support reactions: Fy Ay E y 20 kips 30 kips 20 kips

Ay ME

Ey

0 (a)

70 kips

Ay (28 ft) 20 kips(21 ft) 30 kips(14 ft) 20 kips(7 ft) Ay (28 ft)

ME

ME

0

980 kip-ft

(b)

Discontinuity expressions: w x

1

Ay x 0 ft

2

M E x 28 ft V ( x)

w x dx

Ay x 0 ft

0

1

V ( x)dx

Ay x 0 ft

1

M E x 28 ft d 2v EI 2 dx

M ( x)

Ay x 0 ft

1

M E x 28 ft

EI

dv dx

EI v

Ay

0

E y x 28 ft

20 kips x 7 ft 0

0

E y x 28 ft

30 kips x 14 ft

1

30 kips x 14 ft

2

20 kips x 21 ft

1

0

0

20 kips x 21 ft

1

20 kips x 21 ft

1

1

1

30 kips x 14 ft

1

20 kips x 21 ft

1

1

20 kips 30 kips 2 x 7 ft x 14 ft 2 2 2 Ey 1 2 M E x 28 ft x 28 ft C1 2 Ay 20 kips 30 kips 3 3 3 x 0 ft x 7 ft x 14 ft 6 6 6 Ey ME 2 3 x 28 ft x 28 ft C1 x C2 2 6 x 0 ft

1

0

E y x 28 ft

20 kips x 7 ft

30 kips x 14 ft

1

E y x 28 ft

20 kips x 7 ft

M E x 28 ft M x

1

20 kips x 7 ft

2

20 kips x 21 ft 2

2

(c)

20 kips x 21 ft 6

3

(d)

Boundary conditions and evaluate constants: at x 0 ft, v 0 C2 0


at x 28 ft, v 0 Ay 20 kips (28 ft)3 (21 ft)3 6 6 dv at x 28 ft, 0 dx Ay 20 kips (28 ft) 2 (21 ft) 2 2 2

30 kips (14 ft)3 6

20 kips (7 ft)3 6

C1 (28 ft)

30 kips (14 ft) 2 2

20 kips (7 ft) 2 2

C1

0

(e)

0

(f)

(a) Solve for Ay, Ey, and ME: Solve equations (e) and (f) simultaneously to obtain: C1 1,470.000 kip-ft 2

Ay

23.7500 kips

Ans.

23.8 kips

With the value of Ay, calculate Ey and ME from equations (a) and (b), respectively.

Ey ME

46.2500 kips

46.3 kips

315.000 kip-ft

Ans.

315 kip-ft (cw)

(b) Beam deflection at C: From Eq. (d), the beam deflection at C (x = 14 ft) is computed as follows: 23.75 kips 20 kips EI vC (14 ft)3 (7 ft)3 (1, 470 kip-ft 2 )(14 ft) 6 6 3 10,861.6667 kip-ft vC

10,861.6667 kip-ft 3 100,000 kip-ft 2

0.108617 ft

1.3034 in.

1.303 in.

Ans.


11.21 A propped cantilever beam is loaded as shown in Fig. P11.21. Assume EI = 100,000 kipft2. Use discontinuity functions to determine: (a) the reactions at A and B. (b) the beam deflection at x = 7 ft. Fig. P11.21

Solution Moment equation: (a) Support reactions: 1 Fy Ay By (12 kips/ft)(16 ft) 2 Ay By 96 kips MB

Ay (16 ft)

0 (a)

1 2(16 ft) (12 kips/ft)(16 ft) 2 3 MB 0

Ay (16 ft) M B

1,024 kip-ft

(b)


w x

1

Ay x 0 ft

2

M B x 16 ft V ( x)

w x dx

Ay x 0 ft

0

1

V ( x)dx

Ay x 0 ft

1

M B x 16 ft d 2v EI 2 dx

M ( x)

Ay x 0 ft

1

M B x 16 ft EI

dv dx

EI v

Ay

By x 16 ft

By x 16 ft

1


2


2


2

3

3

1

12 kips/ft 12 kips/ft 3 x 0 ft x 0 ft 2 6 24(16 ft) By 1 2 M B x 16 ft x 16 ft C1 2 Ay 12 kips/ft 12 kips/ft 3 4 5 x 0 ft x 0 ft x 0 ft 6 24 120(16 ft) By MB 2 3 x 16 ft x 16 ft C1 x C2 2 6 x 0 ft

2

1

12 kips/ft x 0 ft 2 0

1

0

By x 16 ft

12 kips/ft x 0 ft 2 0


1

By x 16 ft

12 kips/ft x 0 ft

M B x 16 ft M x

0

12 kips/ft x 0 ft

4

(c)

(d)


Boundary conditions and evaluate constants: at x 0 ft, v 0 C2 0

at x 16 ft, v 0 Ay 12 kips/ft (16 ft)3 (16 ft)4 6 24 dv at x 16 ft, 0 dx Ay 12 kips/ft (16 ft)2 (16 ft)3 2 6

12 kips/ft (16 ft)5 120(16 ft)

C1 (16 ft)

12 kips/ft (16 ft) 4 24(16 ft)

C1

(e)

0

(f)

0

(a) Solve for Ay, By, and MB: Solve equations (e) and (f) simultaneously to obtain: C1 614.4000 kip-ft 2

Ay

52.8000 kips

Ans.

52.8 kips

With the value of Ay, calculate By and MB from equations (a) and (b), respectively. By MB

43.200 kips

43.2 kips

179.200 kip-ft

Ans.

179.2 kip-ft (cw)

(b) Beam deflection at x = 7 ft: From Eq. (d), the beam deflection at x = 7 ft is computed as follows: 52.8 kips 12 kips/ft 12 kips/ft EI v (7 ft)3 (7 ft) 4 (7 ft)5 (614.400 kip-ft 2 )(7 ft) 6 24 120(16 ft) 2,377.85625 kip-ft 3 v

2,377.85625 kip-ft 3 100,000 kip-ft 2

0.023779 ft

0.2853 in.

0.285 in.

Ans.


11.22 A propped cantilever beam is loaded as shown in Fig. P11.22. Assume EI = 200,000 kNm2. Use discontinuity functions to determine: (a) the reactions at A and B. (b) the beam deflection at C. Fig. P11.22

Solution Moment equation: (a) Support reactions: 1 Fy Ay By (120 kN/m)(8 m) 2 Ay By 480 kN

0 (a)

1 2(8 m) (120 kN/m)(8 m) 2 3 MA 0

MA

By (6 m) M A

By (6 m)

2,560 kN-m

(b)


w x V ( x)

w x dx

MA x 0 m

M x

V ( x)dx

MA x 0 m

M ( x)

MA x 0 m

d 2v EI 2 dx EI

dv dx

EI v

MA x 0 m MA x 0m 2

2

MA x 0 m

2

1

Ay 2 Ay 6

1

0

0

0

Ay x 0 m Ay x 0 m

3

2

1

120 kN/m 1 1 x 0m By x 6 m 8m 120 kN/m 2 0 x 0m By x 6 m 2(8 m) 120 kN/m 3 1 x 0m By x 6 m 6(8 m)

120 kN/m 3 1 x 0m By x 6 m 6(8 m) By 120 kN/m 4 2 x 0m x 6m C1 24(8 m) 2 By 120 kN/m 5 3 x 0m x 6m C1 x C2 120(8 m) 6

Ay x 0 m

x 0m x 0m

1

Ay x 0 m

1

(c) (d)

Boundary conditions and evaluate constants: dv at x 0 m, 0 C1 0 dx at x 0 m, v 0 C2 0 at x

6 m, v

0

MA (6 m) 2 2

Ay 6

(6 m)3

120 kN/m (6 m)5 120(8 m)

0

(e)


(a) Solve for Ay, By, and MA: Solve equations (a), (b), and (e) simultaneously to obtain:

Ay MA

66.5000 kN

66.5 kN

79.0000 kN-m

By

413.5000 kN

414 kN Ans.

79.0 kN-m (ccw)

(b) Beam deflection at C: From Eq. (d), the beam deflection at C (x = 8 m) is computed as follows: 79.0 kN-m 66.5 kN 120 kN/m 413.5 kN EI v (8 m) 2 (8 m)3 (8 m)5 (2 m)3 2 6 120(8 m) 6 398.0000 kN-m3 v

398.0000 kN-m3 200, 000 kN-m 2

0.001990 m

1.990 mm

1.990 mm

Ans.


11.23 For the beam shown in Fig. P11.23, assume EI = 200,000 kN-m2 and use discontinuity functions to determine: (a) the reactions at A, C, and D. (b) the beam deflection at B. Fig. P11.23

Solution Moment equation: (a) Support reactions: Fy Ay C y Dy

Ay MA

Cy

(120 kN/m)(6 m)

Dy

0 (a)

720 kN

(120 kN/m)(6 m)(3 m) C y (6 m) Dy (10 m) C y (6 m)

0

Dy (10 m)

2,160 kN-m

(b)


1

Ay x 0 m

1

Cy x 6 m V ( x)

w x dx

0

Ay x 0 m Cy x 6 m

M x

V ( x)dx

Ay x 0 m

EI

d 2v dx 2

M ( x)

Ay x 0 m

dv dx

EI v

Ay

Dy x 10 m

1

Dy x 10 m

Dy x 10 m

1

120 kN/m x 6 m

2

120 kN/m x 6m 2

2

120 kN/m x 6m 2

2

2

1

120 kN/m 120 kN/m 3 x 0m x 6m 2 6 6 Cy Dy 2 2 x 6m x 10 m C1 2 2 Ay 120 kN/m 120 kN/m 3 4 4 x 0m x 0m x 6m 6 24 24 Cy Dy 3 3 x 6m x 10 m C1 x C2 6 6 x 0m

1

1

120 kN/m x 0m 2 1

0

0

120 kN/m x 0m 2

1

Cy x 6 m

EI

0

120 kN/m x 6 m

1

Dy x 10 m

120 kN/m x 0 m

1

Cy x 6 m

0

120 kN/m x 0 m

2

3

(c)

(d)

Boundary conditions and evaluate constants: at x 0 m, v 0 C2 0 at x

6 m, v

0

Ay 6

(6 m)3

120 kN/m (6 m) 4 24

C1 (6 m)

0

(e)


at x 10 m, v 0 Ay (10 m)3 6

120 kN/m (10 m) 4 24

120 kN/m (4 m) 4 24

Cy 6

(4 m)3

C1 (10 m)

0

(f)

(a) Solve for Ay, Cy, and Dy: Solve equations (a), (b), (e), and (f) simultaneously to obtain: C1 756.000 kN-m 2

Ay

306.0000 kN

306 kN

Cy

495.0000 kN

495 kN

Dy

81.0000 kN

Ans.

81.0 kN

(b) Beam deflection at B: From Eq. (d), the beam deflection at B (x = 3 m) is computed as follows: 306.00 kN 120 kN/m EI vB (3 m)3 (3 m) 4 (756.000 kN-m 2 )(3 m) 6 24 3 1, 296.0000 kN-m vB

1, 296.0000 kN-m3 200,000 kN-m 2

0.006480 m

6.48 mm

6.48 mm

Ans.


11.24 For the beam shown in Fig. P11.24, assume EI = 100,000 kip-ft2 and use discontinuity functions to determine: (a) the reactions at A, C, and D. (b) the beam deflection at B. Fig. P11.24

Solution Moment equation: (a) Support reactions: Fy Ay C y Dy

Ay MA

Cy

75 kips (7 kips/ft)(16 ft)

Dy

(a)

187 kips

(75 kips)(8 ft) (7 kips/ft)(16 ft)(24 ft) C y (16 ft) C y (16 ft)

Dy (32 ft)

Dy (32 ft)

1

Ay x 0 ft

0

V ( x)dx

Ay x 0 ft

1

M ( x)

Ay x 0 ft

1

1

1

C y x 16 ft

7 kips/ft x 32 ft 0

2

1

0

1

75 kips x 8 ft

1

1

0

Dy x 32 ft

C y x 16 ft

7 kips/ft x 32 ft 2

1

Dy x 32 ft

C y x 16 ft

7 kips/ft x 32 ft

75 kips x 8 ft

7 kips/ft x 16 ft 2 d 2v EI 2 dx

0

75 kips x 8 ft

7 kips/ft x 16 ft M x

(b)

75 kips x 8 ft

7 kips/ft x 16 ft w x dx

0

3, 288 kip-ft

Discontinuity expressions: w x Ay x 0 ft

V ( x)

0

2

0

1

Dy x 32 ft

C y x 16 ft

1

1

7 kips/ft 7 kips/ft 2 2 x 16 ft x 32 ft Dy x 32 ft 2 2 Cy dv Ay 75 kips 2 2 2 EI x 0 ft x 8 ft x 16 ft dx 2 2 2 Dy 7 kips/ft 7 kips/ft 3 3 2 x 16 ft x 32 ft x 32 ft C1 6 6 2 Ay Cy 75 kips 3 3 3 EI v x 0 ft x 8 ft x 16 ft 6 6 6 Dy 7 kips/ft 7 kips/ft 4 4 3 x 16 ft x 32 ft x 32 ft C1 x C2 24 24 6

1

(c)

(d)

Boundary conditions and evaluate constants: at x 0 ft, v 0 C2 0 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

at x 16 ft, v

0

at x

0

32 ft, v Ay 6

(32 ft)3

Ay 6

(16 ft)3

75 kips (24 ft)3 6

Cy 6

75 kips (8 ft)3 6 (16 ft) 3

C1 (16 ft)

7 kips/ft (16 ft) 4 24

0

C1 (32 ft)

(e)

0

(f)

(a) Solve for Ay, Cy, and Dy: Solve equations (a), (b), (e), and (f) simultaneously to obtain: C1 601.3333 kip-ft 2

Ay

23.4688 kips

23.5 kips

Cy

121.5625 kips

121.6 kips

Dy

41.9688 kips

42.0 kips

Ans.

(b) Beam deflection at B: From Eq. (d), the beam deflection at B (x = 8 ft) is computed as follows: 23.4688 kips EI vB (8 ft)3 (601.3333 kip-ft 2 )(8 ft) 6 2,808.0000 kip-ft 3 vB

2,808.0000 kip-ft 3 100,000 kip-ft 2

0.028080 ft

0.3370 in.

0.337 in.

Ans.


11.25 For the propped cantilever beam shown in Fig. P11.25, assume EI = 100,000 kip-ft2 and use discontinuity functions to determine: (a) the reactions at B and D. (b) the beam deflection at C. Fig. P11.25

Solution Moment equation: (a) Support reactions: 1 Fy By Dy (5 kips/ft)(10 ft) 2 (5 kips/ft)(12 ft) 0

By

Dy

(a)

85 kips

1 10 ft (5 kips/ft)(10 ft) 20 ft 2 3 (5 kips/ft)(12 ft)(14 ft) By (20 ft) M D 0

MD

By (20 ft) M D

1,423.3333 kip-ft

Discontinuity expressions: 5 kips/ft w x x 0 ft 10 ft By x 10 ft M D x 30 ft

1

V ( x)

w x dx

2

2

5 kips/ft x 0 ft 2(10 ft) 5 kips/ft x 22 ft

M x

EI

V ( x)dx

d 2v dx 2

M ( x)

5 kips/ft 6(10 ft) 5 kips/ft 2 5 kips/ft 6(10 ft) 5 kips/ft 2


5 kips/ft x 10 ft

5 kips/ft x 0 ft 10 ft M D x 30 ft

1

(b)

x 0 ft x 22 ft x 0 ft x 22 ft

Dy x 30 ft 1

1

2

3

2

0

0

1

1

5 kips/ft x 22 ft

0

1

5 kips/ft x 10 ft 2(10 ft) M D x 30 ft

3

5 kips/ft x 10 ft

5 kips/ft x 22 ft


Dy x 30 ft 2

0

1

1


0


0

2

By x 10 ft

Dy x 30 ft 3

3

0

By x 10 ft

Dy x 30 ft

1

1

By x 10 ft

Dy x 30 ft

0

1

1


EI

dv dx

EI v

5 kips/ft 5 kips/ft 4 4 x 0 ft x 10 ft 24(10 ft) 24(10 ft) Dy 1 2 M D x 30 ft x 30 ft C1 2 5 kips/ft 5 kips/ft 5 5 x 0 ft x 10 ft 120(10 ft) 120(10 ft) Dy MD 2 3 x 30 ft x 30 ft C1 x C2 2 6

By 2

x 10 ft

2

5 kips/ft x 22 ft 6

3

(c)

By 6

x 10 ft

3

5 kips/ft x 22 ft 24

4

(d)

Boundary conditions and evaluate constants: at x 10 ft, v 0

at x

at x

5 kips/ft (10 ft)5 120(10 ft) 30 ft, v 0

C1 (10 ft) C2

(e)

0

By 5 kips/ft 5 kips/ft (30 ft)5 (20 ft)5 (20 ft)3 120(10 ft) 120(10 ft) 6 5 kips/ft (8 ft) 4 C1 (30 ft) C2 0 24 dv 30 ft, 0 dx By 5 kips/ft 5 kips/ft 5 kips/ft (30 ft) 4 (20 ft) 4 (20 ft) 2 (8 ft)3 24(10 ft) 24(10 ft) 2 6

(f)

C1

(g)

0

(a) Solve for By, Dy, and MD: Solve equations (a), (b), (e), (f), and (g) simultaneously to obtain: C1 59.0000 kip-ft 2 C2

1,006.6667 kip-ft 3

By

65.8700 kips

65.9 kips

Dy

19.1300 kips

19.13 kips

MD

105.9333 kip-ft

105.9 kip-ft (cw)

Ans.

(b) Beam deflection at C: From Eq. (d), the beam deflection at C (x = 22 ft) is computed as follows: 5 kips/ft 5 kips/ft 65.8700 kips EI vC (22 ft)5 (12 ft)5 (12 ft)3 120(10 ft) 120(10 ft) 6 (59.0000 kip-ft 2 )(22 ft) 1,006.6667 kip-ft 3 1,757.4400 kip-ft 3 vC

1,757.4400 kip-ft 3 100,000 kip-ft 2

0.017574 ft

0.2109 in.

0.211 in.

Ans.


11.26 For the beam shown in Fig. P11.26, assume EI = 200,000 kN-m2 and use discontinuity functions to determine: (a) the reactions at B, C, and D. (b) the beam deflection at A. Fig. P11.26

Solution Moment equation: (a) Support reactions: Fy By C y Dy 120 kN (60 kN/m)(12 m)

By MB

Cy

Dy

0 (a)

840 kN

(120 kN)(3 m) (60 kN/m)(12 m)(6 m) C y (6 m) C y (6 m)

Dy (12 m)

Dy (12 m)

0

3,960 kN-m

(b)


1

Cy x 9 m V ( x)

w x dx

V ( x)dx

d 2v dx 2

M ( x)

dv dx

EI v

1

1

0

60 kN/m x 3 m

1

60 kN/m x 3m 2

1

60 kN/m x 3m 2

2

2

1

By 120 kN 60 kN/m 2 2 x 0m x 3m x 3m 2 2 6 Cy Dy 2 2 x 9m x 15 m C1 2 2 By 120 kN 60 kN/m 3 3 4 x 0m x 3m x 3m 6 6 24 Cy Dy 3 3 x 9m x 15 m C1 x C2 6 6

Boundary conditions and evaluate constants: at x 3 m, v 0 120 kN (3 m)3 C1 (3 m) C2 0 6

1

1

By x 3 m

Dy x 15 m

0

0

By x 3 m

Dy x 15 m

120 kN x 0 m Cy x 9 m

EI

1

1

60 kN/m x 3 m

1

By x 3 m

Dy x 15 m

120 kN x 0 m Cy x 9 m

EI

0

0

1

By x 3 m

Dy x 15 m

120 kN x 0 m Cy x 9 m

M x

1

120 kN x 0 m

3

(c)

(d)

(e)


at x

9 m, v

0

120 kN (9 m)3 6 at x 15 m, v 0

By

120 kN (15 m)3 6

6

(6 m)3

By 6

(12 m)3

60 kN/m (6 m) 4 24

C1 (9 m) C2

60 kN/m (12 m) 4 24

Cy 6

(6 m)3

0

C1 (15 m) C2

(f)

0

(g)

(a) Solve for By, Cy, and Dy: Solve equations (a), (b), (e), (f), and (g) simultaneously to obtain: C1 900.0000 kN-m 2 C2

2,160.0000 kN-m3

By

330.0000 kN

330 kN

Cy

360.0000 kN

360 kN

Dy

150.0000 kN

150.0 kN

Ans.

(b) Beam deflection at A: From Eq. (d), the beam deflection at A (x = 0 m) is computed as follows: EI v A 2,160.0000 kN-m3 vA

2,160.0000 kN-m3 200,000 kN-m 2

0.010800 m

10.80 mm

10.80 mm

Ans.


11.27 For the beam shown in Fig. P11.27, assume EI = 200,000 kN-m2 and use discontinuity functions to determine: (a) the reactions at B, C, and D. (b) the beam deflection at A. Fig. P11.27

Solution Moment equation: (a) Support reactions: Fy By C y Dy

By MB

Cy

Dy

(60 kN/m)(6 m)

0 (a)

360 kN

420 kN-m (60 kN/m)(6 m)(3 m) C y (6 m) C y (6 m)

Dy (12 m)

Dy (12 m)

0

660 kN-m

Discontinuity expressions: w x 420 kN-m x 0 m 1

Cy x 9 m V ( x)

w x dx

V ( x)dx

d 2v dx 2

M ( x)

0

1

0

0

0

1

By x 3 m

60 kN/m x 9m 2

420 kN-m x 0 m

0

2

By x 3 m

1

60 kN/m x 3 m

Dy x 15 m 1

2

1

60 kN/m x 3m 2

60 kN/m 2 1 x 9m Dy x 15 m 2 By dv 60 kN/m 1 2 3 EI 420 kN-m x 0 m x 3m x 3m dx 2 6 Cy D 60 kN/m 2 3 2 y x 9m x 9m x 15 m C1 2 6 2 By 420 kN-m 60 kN/m 2 3 4 EI v x 0m x 3m x 3m 2 6 24 Cy Dy 60 kN/m 3 4 3 x 9m x 9m x 15 m C1 x C2 6 24 6 Cy x 9 m

1

0

60 kN/m x 3m 2 Dy x 15 m

1

0

60 kN/m x 3 m

Dy x 15 m

By x 3 m

60 kN/m x 9 m

420 kN-m x 0 m Cy x 9 m

EI

1

1

By x 3 m

60 kN/m x 9 m

420 kN-m x 0 m Cy x 9 m

M x

2

(b)

2

1

(c)

(d)


Boundary conditions and evaluate constants: at x 3 m, v 0 420 kN-m (3 m) 2 C1 (3 m) C2 0 2 at x 9 m, v 0 By 420 kN-m 60 kN/m (9 m) 2 (6 m)3 (6 m) 4 2 6 24 at x 15 m, v 0

(e)

C1 (9 m) C2

0

(f)

By 420 kN-m 60 kN/m (15 m) 2 (12 m)3 (12 m) 4 2 6 24 Cy 60 kN/m (6 m)3 (6 m) 4 C1 (15 m) C2 0 6 24

(g)

(a) Solve for By, Cy, and Dy: Solve equations (a), (b), (e), (f), and (g) simultaneously to obtain: C1 1,590.0000 kN-m 2 C2

2,880.0000 kN-m3

By

245.0000 kN

245 kN

Cy

120.0000 kN

120 kN

Dy

5.0000 kN

5.00 kN

Ans.

(b) Beam deflection at A: From Eq. (d), the beam deflection at A (x = 0 m) is computed as follows: EI vA 2,880.0000 kN-m3 vA

2,880.0000 kN-m3 200,000 kN-m2

0.014400 m

14.40 mm

14.40 mm

Ans.


11.28a For the beams and loadings shown below, assume that EI = 3.0 × 104 kN-m2 is constant for each beam. (a) For the beam in Fig. P11.28a, determine the concentrated upward force P required to make the total beam deflection at B equal to zero (i.e., vB = 0). Fig. P11.28a

Solution Downward deflection at B due to 15 kN/m uniformly distributed load. [Appendix C, SS beam with uniformly distributed load over portion of span.] Relevant equation from Appendix C: wa3 vB (4L2 7aL 3a 2 ) 24 LEI Values: w = 15 kN/m, L = 7 m, a = 3.5 m, EI = 3.0 × 104 kN-m2 Computation: wa3 vB (4 L2 24 LEI

7aL 3a 2 )

(15 kN/m)(3.5 m)3 4(7 m) 2 24(7 m)EI

7(3.5 m)(7 m) 3(3.5 m) 2

234.472656 kN-m3 EI

Upward deflection at B due to concentrated load P. [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL3 vB 48 EI Values: L = 7 m, EI = 3.0 × 104 kN-m2

Computation:

vB

PL3 48EI

P(7 m)3 48EI

P(7.145833 m3 ) EI

Compatibility equation at B: 234.472656 kN-m3 P(7.145833 m3 ) EI EI 3 234.472656 kN-m P 32.8125 kN 7.145833 m3

0 32.8 kN

Ans.


11.28b For the beams and loadings shown below, assume that EI = 3.0 × 104 kN-m2 is constant for each beam. (b) For the beam in Fig. P11.28b, determine the concentrated moment M required to make the total beam slope at A equal to zero (i.e., A = 0). Fig. P11.28b

Solution Slope at A due to 32 kN concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL2 (slope magnitude) A 2 EI Values: P = 32 kN, L = 4 m, EI = 3.0 × 104 kN-m2

Computation: A

PL2 2 EI

(32 kN)(4 m)2 2 EI

256 kN-m2 EI

(negative slope by inspection)

Slope at A due to concentrated moment M. [Appendix C, Cantilever beam with concentrated moment at tip.] Relevant equation from Appendix C: ML (slope magnitude) A EI Values: L = 4 m, EI = 3.0 × 104 kN-m2

Computation: ML M (4 m) A EI EI

M (4 m) EI

Compatibility equation at A: 256 kN-m 2 M (4 m) 0 EI EI 256 kN-m 2 M 64.0 kN-m 4m

(positive slope by inspection)

Ans.


11.29a For the beams and loadings shown below, assume that EI = 5.0 × 106 kip-in.2 is constant for each beam. (a) For the beam in Fig. P11.29a, determine the concentrated upward force P required to make the total beam deflection at B equal to zero (i.e., vB = 0). Fig. P11.29a

Solution Downward deflection at B due to 4 kips/ft uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vB 8 EI Values: w = 4 kips/ft, L = 13 ft, EI = 5.0 × 106 kip-in.2 Computation:

vB

wL4 8EI

(4 kips/ft)(13 ft) 4 8EI

14,280.5 kip-ft 3 EI

Upward deflection at B due to concentrated load P. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vB 3EI Values: L = 13 ft, EI = 5.0 × 106 kip-in.2

Computation:

vB

PL3 3EI

P(13 ft)3 3EI

P(732.333333 ft 3 ) EI

Compatibility equation at B: 14,280.5 kip-ft 3 P(732.333333 ft 3 ) EI EI 3 14,280.5 kip-ft P 19.50 kips 732.333333 ft 3

0 Ans.


11.29b For the beams and loadings shown below, assume that EI = 5.0 × 106 kip-in.2 is constant for each beam. (b) For the beam in Fig. P11.29b, determine the concentrated moment M required to make the total beam slope at C equal to zero (i.e., C = 0). Fig. P11.29b

Solution Slope at C due to 40-kip concentrated load. [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL2 (slope magnitude) C 16 EI Values: P = 40 kips, L = 18 ft, EI = 5.0 × 106 kip-in.2

Computation: C

PL2 16 EI

(40 kips)(18 ft)2 16 EI

810 kip-ft 2 EI


Slope at C due to concentrated moment M. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: ML (slope magnitude) C 3EI Values: L = 18 ft, EI = 5.0 × 106 kip-in.2 Computation: ML M (18 ft) C 3EI 3EI

M (6 ft) EI

Compatibility equation at C: 810 kip-ft 2 M (6 ft) 0 EI EI 810 kip-ft 2 M 135.0 kip-ft 6 ft


Ans.


11.30a For the beams and loadings shown below, assume that EI = 5.0 × 104 kN-m2 is constant for each beam. (a) For the beam in Fig. P11.30a, determine the concentrated downward force P required to make the total beam deflection at B equal to zero (i.e., vB = 0). Fig. P11.30a

Solution Upward deflection at B due to 105 kN-m concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vB (2 L2 3Lx x 2 ) (elastic curve) 6 LEI Values: M = −105 kN-m, L = 8 m, x = 4 m, EI = 5.0 × 104 kN-m2 Computation: Mx vB (2 L2 6 LEI

3Lx

x2 )

( 105 kN-m)(4 m) 2(8 m)2 3(8 m)(4 m) (4 m)2 6(8 m)EI

420 kN-m3 EI

Downward deflection at B due to concentrated load P. [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL3 vB 48 EI Values: L = 8 m, EI = 5.0 × 104 kN-m2 Computation:

vB

PL3 48EI

P(8 m)3 48EI

P(10.666667 m3 ) EI

Compatibility equation at B: 420 kN-m3 P(10.666667 m3 ) 0 EI EI 420 kN-m3 P 39.375 kN 39.4 kN 10.666667 m3

Ans.


11.30b For the beams and loadings shown below, assume that EI = 5.0 × 104 kN-m2 is constant for each beam. (b) For the beam in Fig. P11.30b, determine the concentrated moment M required to make the total beam slope at A equal to zero (i.e., A = 0). Fig. P11.30b

Solution Slope at A due to 6 kN/m uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL3 (slope magnitude) A 6 EI Values: w = 6 kN/m, L = 5 m, EI = 5.0 × 104 kN-m2

Computation: A

wL3 6 EI

(6 kN/m)(5 m)3 6 EI

125 kN-m2 EI


Slope at A due to concentrated moment M. [Appendix C, Cantilever beam with concentrated moment at tip.] Relevant equation from Appendix C: ML (slope magnitude) A EI Values: L = 5 m, EI = 5.0 × 104 kN-m2 Computation: ML M (5 m) A EI EI


Compatibility equation at A: 125 kN-m2 M (5 m) 0 EI EI 125 kN-m 2 M 25.0 kN-m 5m

Ans.


11.31a For the beams and loadings shown below, assume that EI = 8.0 × 106 kip-in.2 is constant for each beam. (a) For the beam in Fig. P11.31a, determine the concentrated downward force P required to make the total beam deflection at B equal to zero (i.e., vB = 0). Fig. P11.31a

Solution Upward deflection at B due to 125 kip-ft concentrated moment. [Appendix C, Cantilever beam with concentrated moment at tip.] Relevant equation from Appendix C: ML2 vB 2 EI Values: M = −125 kip-ft, L = 15 ft, EI = 8.0 × 106 kip-in.2 Computation:

vB

ML2 2 EI

( 125 kip-ft)(15 ft) 2 2 EI

14,062.5 kip-ft 3 EI

Downward deflection at B due to concentrated load P. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vB 3EI Values: L = 15 ft, EI = 8.0 × 106 kip-in.2 Computation:

vB

PL3 3EI

P(15 ft)3 3EI

P(1,125 ft 3 ) EI

Compatibility equation at B: 14,062.5 kip-ft 3 P(1,125 ft 3 ) 0 EI EI 14,062.5 kip-ft 3 P 12.50 kips 1,125 ft 3

Ans.


11.31b For the beams and loadings shown below, assume that EI = 8.0 × 106 kip-in.2 is constant for each beam. (b) For the beam in Fig. P11.31b, determine the concentrated moment M required to make the total beam slope at A equal to zero (i.e., A = 0). Fig. P11.31b

Solution Slope at A due to 7 kips/ft uniformly distributed load. [Appendix C, SS beam with uniformly distributed load over portion of span.] Relevant equation from Appendix C: wa 2 (2L2 a 2 ) (slope magnitude) A 24LEI Values: w = 7 kips/ft, L = 23 ft, a = 15 ft, EI = 8.0 × 106 kip-in.2 Computation: wa 2 (2 L2 A 24 LEI

a2 )

(7 kips/ft)(15 ft) 2 2(23 ft) 2 24(23 ft)EI

(15 ft) 2

2,376.766304 kip-ft 2 EI


Slope at A due to concentrated moment M. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: ML (slope magnitude) A 3EI Values: L = 23 ft, EI = 8.0 × 106 kip-in.2 Computation: ML M (23 ft) A 3EI 3EI

M (7.666667 ft) EI

Compatibility equation at A: 2,376.766304 kip-ft 2 M (7.666667 ft) EI EI 2 2,376.766304 kip-ft M 310 kip-ft 7.666667 ft


0 Ans.


11.32 For the beam and loading shown below, derive an expression for the reactions at supports A and B. Assume that EI is constant for the beam.

Fig. P11.32

Solution Choose the reaction force at B as the redundant; therefore, the released beam is a cantilever. Consider downward deflection of cantilever beam at B due to concentrated moment M0. [Appendix C, Cantilever beam with concentrated moment at tip.] Relevant equation from Appendix C: ML2 M 0 L2 vB 2 EI 2 EI

Consider upward deflection of cantilever beam at B due to concentrated load By. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: 3 PL3 By L vB 3EI 3EI

Compatibility equation for deflection at B: 3 M 0 L2 By L 3M 0 0 By 2EI 3EI 2L

Ans.

Equilibrium equations for entire beam:

Fy MA

MA

Ay

By MA

0 M0

By L M 0

Ay

By

3M 0 ( L) M 0 2L

3M 0 2

By L

3M 0 2L

3M 0 2L

Ans.

0

M0

M0 2

(cw)

Ans.



Fig. P11.33

Solution Choose the reaction force at B as the redundant; therefore, the released beam is a cantilever. Consider downward deflection of cantilever beam at B due to linearly distributed load. [Appendix C, Cantilever beam with linearly distributed load.] Relevant equation from Appendix C: w0 L4 vB 30 EI

Consider upward deflection of cantilever beam at B due to concentrated load By. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: 3 PL3 By L vB 3EI 3EI

Compatibility equation for deflection at B: 3 w0 L4 By L w0 L 0 By 30 EI 3EI 10

Equilibrium equations for entire beam: w0 L Fy Ay By 0 Ay 2 w0 L L MA MA By L 0 2 3 MA

By L

w0 L2 6

w0 L ( L) 10

w0 L2 6

Ans.

w0 L 2

w0 L2 15

w0 L 10

w0 L2 15

4w0 L 10

(ccw)

2w0 L 5

Ans.

Ans.



Fig. P11.34

Solution Choose the reaction force at A as the redundant; therefore, the released beam is a cantilever. Consider downward deflection of cantilever beam at A due to concentrated load P. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: Px 2 vA (3L x) (elastic curve) 6 EI 3L Let x L, L 2 P ( L) 2 3L 7 PL3 vA 3 L 6 EI 2 12 EI Consider upward deflection of cantilever beam at A due to concentrated load Ay. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: 3 PL3 Ay L vA 3EI 3EI

Compatibility equation for deflection at A:

Ay L3

7 PL3 12 EI

0

3EI

Ay

7P 4

Ans.

Equilibrium equations for entire beam: Fy Ay By P 0 By MB

MB

MB

P

7P 4 Ay L

Ay L P

3P 4 3L P 2

3L 2

3P 4

Ans.

0

7P 3PL ( L) 4 2

PL 4

PL (ccw) 4

Ans.



Fig. P11.35

Solution Choose the reaction force at A as the redundant; therefore, the released beam is a cantilever. Consider downward deflection of cantilever beam at A due to uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wx 2 vA (6 L2 4 Lx x 2 ) (elastic curve) 24 EI

Let x

L, L vA

3L 2 w( L)2 3L 6 24 EI 2

2

3L 4 ( L) ( L) 2 2

17 wL4 48EI

Consider upward deflection of cantilever beam at A due to concentrated load Ay. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: 3 PL3 Ay L vA 3EI 3EI

Compatibility equation for deflection at A: 3 17wL4 Ay L 17 wL 0 Ay 48EI 3EI 16

Ans.


Equilibrium equations for entire beam: 3wL Fy Ay By 0 By 2 3wL 3L M B M B Ay L 0 2 4 MB

Ay L

9wL2 8

17 wL ( L) 16

9wL2 8

3wL 2

17wL 16

wL2 16

wL2 16

7wL 16

(cw)

7 wL 16

Ans.

Ans.



Fig. P11.36

Solution Choose the reaction force at A as the redundant; therefore, the released beam is a cantilever. Consider downward deflection of cantilever beam at A due to concentrated load P. [Appendix C, Cantilever beam with concentrated load at midspan.] Relevant equation from Appendix C: 5 PL3 vA 48 EI Let L 2 L 5 P (2 L)3 48 EI

vA

5 PL3 6 EI

Consider upward deflection of cantilever beam at A due to concentrated load Ay. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vA 3EI Let L 2 L, P Ay

vA

( Ay )(2 L)3

8 Ay L3

3EI

3EI

Compatibility equation for deflection at A: 3 5PL3 8 Ay L 5P 0 Ay 6 EI 3EI 16

Ans.

Equilibrium equations for entire beam: Fy Ay By P 0 By MB

P

MB

MB

5 P 11P 11P 16 16 16 Ay (2 L) P( L) 0

Ay (2L) P( L)

5PL 8

Ans.

PL

3PL 8

3PL (cw) 8

Ans.


11.37 For the beam and loading shown below, derive an expression for the reactions at supports A and C. Assume that EI is constant for the beam.

Fig. P11.37

Solution Choose the reaction force at C as the redundant; therefore, the released beam is a cantilever. Consider downward deflection of cantilever beam at C due to concentrated moment M0. [Appendix C, Cantilever beam with concentrated moment.] Relevant equations from Appendix C: ML2 ML vB and B 2 EI EI 2 M 0L M 0L 3M 0 L2 vC ( L) 2 EI EI 2 EI Consider upward deflection of cantilever beam at C due to concentrated load Cy. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vC 3EI Let L 2 L, P Cy

vC

( C y )(2 L)3

8C y L3

3EI

3EI

Compatibility equation for deflection at C: 3 3M 0 L2 8C y L 9M 0 0 Cy 2 EI 3EI 16 L

Ans.

Equilibrium equations for entire beam: Fy Ay C y 0 Ay MA

MA

9M 0 16 L M A M0

9M 0 16 L C y (2 L)

C y (2L) M 0

Ans. 0

9M 0 (2 L) M 0 16 L

M0 8

M0 8

(cw)

Ans.


11.38 For the beam and loading shown below, derive an expression for the reactions at supports A and C. Assume that EI is constant for the beam.

Fig. P11.38

Solution Choose the reaction force at C as the redundant; therefore, the released beam is a cantilever. Consider downward deflection of cantilever beam at C due to uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equations from Appendix C: wL4 wL3 vB and B 8EI 6 EI 4 3 wL wL 7 wL4 vC ( L) 8EI 6 EI 24 EI Consider upward deflection of cantilever beam at C due to concentrated load Cy. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vC 3EI Let L 2 L, P Cy

vC

( C y )(2 L)3

8C y L3

3EI

3EI

Compatibility equation for deflection at C: 3 7 wL4 8C y L 7 wL 0 Cy 24 EI 3EI 64

Ans.

Equilibrium equations for entire beam: Fy Ay C y wL 0

MA

Ay

wL

7 wL 64

MA

wL

L 2

MA

C y (2 L)

57 wL 64 C y (2 L)

wL2 2

Ans. 0

7 wL (2 L) 64

wL2 2

18wL2 64

9wL2 32

9wL2 32

(ccw)

Ans.


11.39 For the beam and loading shown below, derive an expression for the reaction forces at A, C, and D. Assume that EI is constant for the beam. (Reminder: The roller symbol implies that both upward and downward displacement is restrained.) Fig. P11.39

Solution Choose the reaction force at C as the redundant; therefore, the released beam is simply supported. Consider downward deflection of simply supported beam at C due to P. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vC ( L b2 x2 ) (elastic curve) 6 LEI Let L 3L, b L, x L P ( L)( L) vC (3L) 2 ( L) 2 ( L) 2 6(3L) EI 7 PL3 18 EI

PL 7 L2 18 EI

Consider upward deflection of simply supported beam at C due to concentrated load Cy. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 vC (L a 2 b2 ) 6 LEI Let L 3L, a 2 L, b L, P Cy vC

( C y )(2 L)( L) 6(3L) EI 2C y L 18 EI

4 L2

(3L) 2

(2 L) 2

( L) 2

8C y L3 18 EI

Compatibility equation for deflection at C: 3 7 PL3 8C y L 7P 0 Cy 18EI 18EI 8

Ans.

Equilibrium equations for entire beam: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

MA

PL C y (2 L)

Dy (3L)

PL C y (2 L) Dy

Fy

Ay

Dy (3L)

Cy Ay

P 4

P 4

Dy

P

P Cy

PL

0

7P (2 L) 8

3PL 4 Ans.

0 Dy

P

7P 8

P 4

P

5P 8

3P 8

3P 8

Ans.


11.40 For the beam and loading shown below, derive an expression for the reaction force at B. Assume that EI is constant for the beam. (Reminder: The roller symbol implies that both upward and downward displacement is restrained.) Fig. P11.40

Solution Choose the reaction force at B as the redundant; therefore, the released beam is simply supported. Consider upward deflection of simply supported beam at B due to M0. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vB (2 L2 3Lx x 2 ) (elastic curve) 6 LEI

Let L

2 L, x vB

L, M

M0

( M 0 )( L) 2(2 L) 2 3(2 L)( L) ( L) 2 6(2 L) EI

M0 3L2 12 EI

M 0 L2 4 EI

Consider upward deflection of simply supported beam at B due to concentrated load By. [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL3 vB 48 EI Let L 2 L, P By

vB

( By )(2 L)3

By L3

48EI

6 EI

Compatibility equation for deflection at B: 3 M 0 L2 By L 3M 0 0 By 4 EI 6EI 2L

3M 0 2L

Ans.


11.41 For the beam and loading shown below, derive an expression for the reaction force at B. Assume that EI is constant for the beam.

Fig. P11.41

Solution Choose the reaction force at B as the redundant; therefore, the released beam is simply supported. Consider downward deflection of simply supported beam at B due to uniformly distributed load. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa3 vB (4L2 7aL 3a 2 ) 24 LEI

Let L 3L, a vB

2L w(2 L)3 4(3L)2 24(3L) EI

wL2 6 L2 9 EI

7(2 L)(3L) 3(2 L) 2

2wL4 3EI

Consider upward deflection of simply supported beam at B due to concentrated load By. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 vB ( L a 2 b2 ) 6 LEI

Let L

3 L, a vB

L, b

2 L, P

( By )( L)(2 L) 6(3L) EI

By (3L) 2

( L) 2

Compatibility equation for deflection at B: 3 2wL4 4By L 3wL 0 By 3EI 9EI 2

(2 L) 2

3wL 2

By L 9 EI

4 L2

4 By L3 9 EI

Ans.


11.42 For the beam and loading shown below, derive an expression for the reaction force at B. Assume that EI is constant for the beam. (Reminder: The roller symbol implies that both upward and downward displacement is restrained.) Fig. P11.42

Solution Choose the reaction force at B as the redundant; therefore, the released beam is simply supported. Consider upward deflection of simply supported beam at B due to uniformly distributed load. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vB (2 L2 3Lx x 2 ) (elastic curve) 6 LEI

Let L

2 L, x

vB

L, M

w

L 2

L 4

wL2 8

wL2 ( L) 8 2(2 L)2 3(2 L)( L) ( L) 2 6(2 L) EI

wL2 3L2 96 EI

wL4 32 EI

Consider upward deflection of simply supported beam at B due to concentrated load By. [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL3 vB 48 EI Let L 2 L, P By

vB

( By )(2 L)3

By L3

48EI

6 EI

Compatibility equation for deflection at B: 3 wL4 By L 3wL 0 By 32EI 6EI 16

3wL 16

Ans.



Fig. P11.43

Solution Choose the reaction force at B as the redundant; therefore, the released beam is simply supported. Consider downward deflection of simply supported beam at B due to one concentrated load P. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vB ( L b2 x2 ) (elastic curve) 6 LEI

Let L

4 L, b vB

L, x

2L

P( L)(2 L) (4 L)2 6(4 L) EI

( L)

2

(2 L)

2

PL 11L2 12 EI

11PL3 12 EI

The second concentrated load will cause an additional deflection at B of the same magnitude. Consider upward deflection of simply supported beam at B due to concentrated load By. [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL3 vB 48 EI Let L 4 L, P By

vB

( By )(4 L)3

64 By L3

16 By L3

48EI

48EI

12 EI

Compatibility equation for deflection at B: 3 11PL3 11PL3 16By L 0 12 EI 12 EI 12 EI

By

11P 8

Ans.



Fig. P11.44

Solution Choose the reaction force at B as the redundant; therefore, the released beam is simply supported. Consider downward deflection of simply supported beam at B due to uniformly distributed load. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa3 vB (4L2 7aL 3a 2 ) 24 LEI

Let L 5L, a vB

3L w(3L)3 4(5L) 2 24(5L) EI

7(3L)(5L) 3(3L) 2

27 wL2 22 L2 120 EI

99wL4 20 EI

Consider downward deflection of simply supported beam at B due to concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vB ( L b2 x2 ) (elastic curve) 6 LEI

Let L

5L, b

vB

L, x

3L, P

wL 3

wL ( L)(3L) 3 (5L)2 6(5L) EI

( L) 2

(3L)2

wL2 15L2 30 EI

wL4 2 EI


Consider upward deflection of simply supported beam at B due to concentrated load By. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 vB ( L a 2 b2 ) 6 LEI

Let L

5L, a vB

3L, b

2 L, P

( By )(3L)(2 L) 6(5L) EI

By (5L)

2

(3L)

2

Compatibility equation for deflection at B: 3 99wL4 wL4 12 By L 0 By 20 EI 2 EI 5EI

(2 L)

2

109wL 48

By L 5EI

2

12 L

2.270833wL

12 By L3 5EI

Ans.


11.45 The beam shown in Fig. P11.45 consists of a W360 × 79 structural steel wide-flange shape [E = 200 GPa; I = 225 × 106 mm4]. For the loading shown, determine: (a) the reactions at A, B, and C. (b) the magnitude of the maximum bending stress in the beam. Fig. P11.45

Solution (a) Reactions at A, B, and C. Choose the reaction force at B as the redundant; therefore, the released beam is simply supported between A and C. Consider downward deflection of simply supported beam at B due to uniformly distributed load. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa3 vB (4L2 7aL 3a 2 ) 24 LEI Values: w = 90 kN/m, L = 9 m, a = 6 m Calculation: wa3 vB (4 L2 24 LEI

7aL 3a 2 )

(90 kN/m)(6 m)3 4(9 m) 2 24(9 m) EI

7(6 m)(9 m) 3(6 m) 2

4,860 kN-m3 EI

Consider downward deflection of simply supported beam at B due to concentrated moment. [Appendix C, SS beam with concentrated moment at one end of span.] Relevant equation from Appendix C: Mx vB (2 L2 3Lx x 2 ) (elastic curve) 6 LEI Values: M = 180 kN-m, L = 9 m, x = 3 m Calculation: Mx vB (2 L2 6 LEI

3Lx

x2 )

(180 kN-m)(3 m) 2(9 m) 2 6(9 m) EI

3(9 m)(3 m) (3 m) 2

900 kN-m 3 EI


Consider upward deflection of simply supported beam at B due to concentrated load By. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 vB ( L a 2 b2 ) 6 LEI Values: P = −By, L = 9 m, a = 3 m, b = 6 m

Calculation: Pab 2 vB (L 6 LEI

a2

b2 )

( By )(3 m)(6 m) 6(9 m) EI

(9 m)2

(3 m) 2

(6 m) 2

(12 m3 ) By EI

Compatibility equation for deflection at B: 3 4,860 kN-m3 900 kN-m3 (12 m ) By 0 EI EI EI 5,760 kN-m3 By 480 kN 480 kN 12 m3

Ans.

Equilibrium equations for entire beam: M A By (3 m) C y (9 m) 180 kN-m (90 kN/m)(6 m)(6 m)

Cy

180 kN-m (90 kN/m)(6 m)(6 m) (480 kN)(3 m) 9m 220.0 kN

Fy

0

Ay

By

Ay

Cy

Ans.

220 kN (90 kN/m)(6 m)

0

(90 kN/m)(6 m) 480 kN 220 kN

160.0 kN

160.0 kN

Ans.


Shear-force and bending-moment diagrams (b) Magnitude of maximum bending stress: Section properties (from Appendix B): I 225 106 mm4 d 353 mm

S 1,270 103 mm3 Maximum bending moment magnitude Mmax = 300 kN-m Bending stresses at maximum moment (300 kN-m)(353 mm/2)(1,000)2 x 225 106 mm 4

235 MPa

Ans.

or using the tabulated section modulus value: (300 kN-m)(1,000) 2 x 1, 270 103 mm3 236 MPa

Ans.


11.46 The beam shown in Fig. P11.46 consists of a W610 × 140 structural steel wide-flange shape [E = 200 GPa; I = 1,120 × 106 mm4]. For the loading shown, determine: (a) the reactions at A, B, and D. (b) the magnitude of the maximum bending stress in the beam. Fig. P11.46

Solution (a) Reactions at A, B, and D. Choose the reaction force at B as the redundant; therefore, the released beam is simply supported between A and D. Consider downward deflection of simply supported beam at B due to uniformly distributed load. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wx vB ( L3 2 Lx 2 x3 ) (elastic curve) 24 EI Values: w = 60 kN/m, L = 7.5 m, x = 1.5 m Calculation: wx vB ( L3 24 EI

2 Lx 2

x3 )

(60 kN/m)(1.5 m) (7.5 m)3 24 EI

2(7.5 m)(1.5 m) 2

(1.5 m)3

1, 468.125 kN-m3 EI

Consider downward deflection of simply supported beam at B due to concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vB ( L b2 x2 ) (elastic curve) 6 LEI Values: P = 125 kN, L = 7.5 m, b = 2.5 m, x = 1.5 m Calculation: Pbx 2 vB (L 6 LEI

b2

x2 )

(125 kN)(2.5 m)(1.5 m) (7.5 m) 2 6(7.5 m) EI

(2.5 m) 2

(1.5 m) 2

497.396 kN-m 3 EI


Consider upward deflection of simply supported beam at B due to concentrated load By. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 vB ( L a 2 b2 ) 6 LEI Values: P = −By, L = 7.5 m, a = 1.5 m, b = 6 m

Calculation: Pab 2 vB (L 6 LEI

a2

b2 )

( By )(1.5 m)(6 m) 6(7.5 m) EI

(7.5 m)

2

(1.5 m)

2

(6 m)

2

(3.6 m3 ) By EI

Compatibility equation for deflection at B: 3 1, 468.125 kN-m3 497.396 kN-m3 (3.6 m ) By 0 EI EI EI 1,965.521 kN-m3 By 545.978 kN 546 kN 3.6 m3

Equilibrium equations for entire beam: M A By (1.5 m) Dy (7.5 m) (60 kN/m)(7.5 m)(3.75 m) (125 kN)(5 m)

Dy

Ay

By

Ay

0

(60 kN/m)(7.5 m)(3.75 m) (125 kN)(5 m) (545.978 kN)(1.5 m) 7.5 m 199.138 kN

Fy

Ans.

Dy

Ans.

199.1 kN

(60 kN/m)(7.5 m) 125 kN

0

(60 kN/m)(7.5 m) 125 kN 545.978 kN 199.138 kN 170.116 kN

170.1 kN

Ans.


Shear-force and bending-moment diagrams (b) Magnitude of maximum bending stress: Section properties (from Appendix B): I 1,120 106 mm4 d 617 mm

S

3,640 103 mm3

Maximum bending moment magnitude Mmax = 322.67 kN-m Bending stresses at maximum moment (322.67 kN-m)(617 mm/2)(1,000)2 x 1,120 106 mm 4 88.9 MPa

Ans.

or using the tabulated section modulus value: (322.67 kN-m)(1,000) 2 x 3,640 103 mm3 88.6 MPa

Ans.


11.47 A propped cantilever beam is loaded as shown in Fig. P11.47. Determine the reactions at A and D for the beam. Assume EI = 12.8 × 106 lb-in.2.

Fig. P11.47

Solution Choose the reaction force at D as the redundant; therefore, the released beam is a cantilever. Consider downward deflection of cantilever beam at D due to uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equations from Appendix C: wL4 wL3 vC and C 8EI 6 EI Values: w = 25 lb/in., L = 72 in. Calculation: wL4 vC 8EI wL3 C 6 EI

vD

(25 lb/in.)(48 in.) 4 8EI (25 lb/in.)(48 in.)3 6 EI

16,588,800 lb-in.3 EI

(24 in.)

16,588,800 lb-in.3 EI 460,800 lb-in.2 EI 460,800 lb-in.2 EI

27,648,000 lb-in.3 EI

Consider downward deflection of cantilever beam at D due to the 360-lb concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 PL2 vB and B 3EI 2 EI Values: P = 360 lb, L = 24 in. Calculation: PL3 vB 3EI PL2 B 2 EI

vD

(360 lb)(24 in.)3 3EI (360 lb)(24 in.) 2 2 EI

1,658,880 lb-in.3 EI

1,658,880 lb-in.3 EI 103,680 lb-in.2 EI

103,680 lb-in.2 (48 in.) EI

6,635,520 lb-in.3 EI


Consider upward deflection of cantilever beam at D due to concentrated load Dy. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vD 3EI Values: P = −Dy, L = 72 in. Calculation: vD

PL3 3EI

( Dy )(72 in.)3

(124,416 in.3 ) Dy

3EI

EI

Compatibility equation for deflection at D: 3 27,648,000 lb-in.3 6,635,520 lb-in.3 (124,416 in. ) Dy EI EI EI 3 34,283,520 lb-in. Dy 275.556 lb 276 lb 124,416 in.3

0 Ans.


Equilibrium equations for entire beam:

Fy

MA

Ay

Dy

(25 lb/in.)(48 in.) 360 lb

0

Ay

(25 lb/in.)(48 in.) 360 lb 275.556 lb 1,284.444 lb

1, 284 lb

MA

(25 lb/in.)(48 in.)(24 in.) (360 lb)(24 in.)

0

MA

(275.556 lb)(72 in.) (25 lb/in.)(48 in.)(24 in.) (360 lb)(24 in.) 17,600 lb-in.

17,600 lb-in. (ccw)

Dy (72 in.)

Ans.

Ans.


11.48 A propped cantilever beam is loaded as shown in Fig. P11.48. Assume EI = 24 × 106 kip-in.2. Determine: (a) the reactions at B and C for the beam. (b) the beam deflection at A. Fig. P11.48

Solution Choose the reaction force at B as the redundant; therefore, the released beam is a cantilever. Consider downward deflection of cantilever beam at B due to uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vB 8 EI Values: w = 8 kips/ft, L = 24 ft Calculation:

vB

wL4 8EI

(8 kips/ft)(24 ft)4 8EI

331,776 kip-ft 3 EI

(a)

Consider downward deflection of cantilever beam at B due to the 40-kip concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: Px 2 vB (3L x) (elastic curve) 6 EI Values: P = 40 kips, L = 36 ft, x = 24 ft Calculation:

vB

Px 2 (3L 6 EI

x)

(40 kips)(24 ft) 2 3(36 ft) (24 ft) 6 EI

322,560 kip-ft 3 EI


Consider upward deflection of cantilever beam at B due to concentrated load By. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vB 3EI Values: P = −By, L = 24 ft Calculation: vB

PL3 3EI

( By )(24 ft)3

(4,608 ft 3 ) By

3EI

EI

(b)

Compatibility equation for deflection at B: 3 331,776 kip-ft 3 322,560 kip-ft 3 (4,608 ft ) By 0 EI EI EI 654,336 kip-ft 3 By 142.0 kips 142.0 kips 4,608 ft 3

Equilibrium equations for entire beam: Fy By C y 40 kips (8 kips/ft)(24 ft)

Cy MC

MC

0

40 kips (8 kips/ft)(24 ft) 142.0 kips

90.0 kips

(40 kips)(36 ft) (8 kips/ft)(24 ft)(12 ft)

By (24 ft)

MC

Ans.

Ans. 0

(142.0 kips)(24 ft) (40 kips)(36 ft) (8 kips/ft)(24 ft)(12 ft) 336.0 kip-ft

336 kip-ft (cw)

Ans.

(b) Beam deflection at A: Consider downward deflection of cantilever beam at A due to uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equations from Appendix C: wL4 wL3 vB and (slope magnitude) B 8EI 6 EI Values: w = 8 kips/ft, L = 24 ft, EI = 24 × 106 kip-in.2


Calculation: 331,776 kip-ft 3 vB calculated previously in Eq. (a) EI wL3 (8 kips/ft)(24 ft)3 18, 432 kip-ft 2 B 6 EI 6 EI EI

vA

331,776 kip-ft 3 EI

(12 ft)

18, 432 kip-ft 2 EI

552,960 kip-ft 3 EI

Consider downward deflection of cantilever beam at A due to the 40-kip concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vA 3EI Values: P = 40 kips, L = 36 ft, EI = 24 × 106 kip-in.2 Calculation:

vA

PL3 3EI

(40 kips)(36 ft)3 3EI

622,080 kip-ft 3 EI

Consider upward deflection of cantilever beam at B due to concentrated load By. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 PL2 vB and (slope magnitude) B 3EI 2 EI Values: P = −By = 142 kips, L = 24 ft, EI = 24 × 106 kip-in.2

Calculation: (4,608 ft 3 )(142.0 kips) 654,336 kip-ft 3 vB EI EI 2 2 PL (142.0 kips)(24 ft) 40,896 kip-ft 2 B 2 EI 2 EI EI

vA

654,336 kip-ft 3 EI

Beam deflection at A. 552,960 kip-ft 3 vA EI 29,952 kip-ft 3 EI

40,896 kip-ft 2 (12 ft) EI

using the results from Eq. (b)

1,145,088 kip-ft 3 EI

622,080 kip-ft 3 1,145,088 kip-ft 3 EI EI 3 (29,952 kip-ft )(12 in./ft)3 2.156544 in. 24 106 kip-in.2

2.16 in.

Ans.


11.49 A propped cantilever beam is loaded as shown in Fig. P11.49. Assume EI = 86.4 × 106 N-mm2. Determine: (a) the reactions at A and C for the beam. (b) the beam deflection at B.

Fig. P11.49

Solution Choose the reaction force at C as the redundant; therefore, the released beam is a cantilever. Consider downward deflection of cantilever beam at C due to uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wx 2 vC (6 L2 4 Lx x 2 ) (elastic curve) 24 EI Values: w = 25 N/mm, L = 400 mm, x = 300 mm, EI = 86.4 × 106 N-mm2 Calculation: wx 2 vC (6 L2 4 Lx x 2 ) 24 EI (25 N/mm)(300 mm)2 6(400 mm)2 24 EI

4(400 mm)(300 mm) (300 mm) 2

53.43750 109 N-mm3 EI Consider downward deflection of cantilever beam at C due to the 4,000-N concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 PL2 vB and (slope magnitude) B 3EI 2 EI Values: P = 4,000 N, L = 120 mm, EI = 86.4 × 106 N-mm2 Calculation: PL3 (4,000 N)(120 mm)3 2.304 109 N-mm3 vB 3EI 3EI EI 2 2 7 PL (4,000 N)(120 mm) 2.88 10 N-mm 2 B 2 EI 2 EI EI

vC

2.304 109 N-mm3 EI

(180 mm)

2.88 107 N-mm2 EI

7.488 109 N-mm3 EI


Consider upward deflection of cantilever beam at C due to concentrated load Cy. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vC 3EI Values: P = −Cy, L = 300 mm, EI = 86.4 × 106 N-mm2 Calculation: vC

PL3 3EI

( C y )(300 mm)3

(9 106 mm3 )C y

3EI

EI

Compatibility equation for deflection at C: 6 3 53.43750 109 N-mm3 7.488 109 N-mm3 (9 10 mm )C y EI EI EI 9 3 60.9255 10 N-mm Cy 6,769.5 N 6, 770 N 9 106 mm3

Equilibrium equations for entire beam: Fy Ay C y 4,000 N (25 N/mm)(400 mm)

MA

0 Ans.

0

4,000 N (25 N/mm)(400 mm) 6,769.5 N

MA

(4,000 N)(120 mm) (25 N/mm)(400 mm)(200 mm) C y (300 mm)

MA

(6,769.5 N)(300 mm) (4,000 N)(120 mm) (25 N/mm)(400 mm)(200 mm) 449,150 N-mm

7,230.5 N

Ans.

Ay

7,230 N 0

449,000 N-mm (ccw)

Ans.

(b) Beam deflection at B: Consider downward deflection of cantilever beam at B due to uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wx 2 vB (6 L2 4 Lx x 2 ) (elastic curve) 24 EI Values: w = 25 N/mm, L = 400 mm, x = 120 mm, EI = 86.4 × 106 N-mm2


Calculation: wx 2 vB (6 L2 4 Lx x 2 ) 24 EI (25 N/mm)(120 mm) 2 6(400 mm)2 24 EI

4(400 mm)(120 mm) (120 mm) 2

11.736 109 N-mm3 EI Consider downward deflection of cantilever beam at B due to the 4,000-N concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vB 3EI Values: P = 4,000 N, L = 120 mm, EI = 86.4 × 106 N-mm2 Calculation:

vB

PL3 3EI

(4,000 N)(120 mm)3 3EI

2.304 109 N-mm3 EI

Consider upward deflection of cantilever beam at B due to concentrated load Cy. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: Px 2 vB (3L x) 6 EI Values: P = −Cy = −6,769.5 N, L = 300 mm, x = 120 mm, EI = 86.4 × 106 N-mm2 Calculation: Px 2 vB (3L 6 EI

x)

( 6,769.5 N)(120 mm) 2 3(300 mm) (120 mm) 6 EI

12.6725 109 N-mm3 EI

Beam deflection at B. 11.736 109 N-mm3 2.304 109 N-mm3 12.6725 109 N-mm3 vB EI EI EI 9 3 9 3 1.367496 10 N-mm 1.367496 10 N-mm 15.8275 mm EI 86.4 106 N-mm2

15.83 mm

Ans.


11.50 The beam shown in Fig. P11.50 consists of a W610 × 82 structural steel wide-flange shape [E = 200 GPa; I = 562 × 106 mm4]. For the loading shown, determine: (a) the reaction force at C. (b) the beam deflection at A. Fig. P11.50

Solution (a) Reaction force at C. Choose the reaction force at C as the redundant; therefore, the released beam is simply supported between B and D. Consider upward deflection of simply supported beam at C due to uniformly distributed load on overhang AB. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vC (2 L2 3Lx x 2 ) (elastic curve) 6 LEI Values: M = −(105 kN/m)(3 m)(1.5 m) = −472.5 kN-m, L = 14 m, x = 7 m Calculation: Mx vC (2 L2 6 LEI

3Lx

x2 )

( 472.5 kN-m)(7 m) 2(14 m) 2 6(14 m)EI

3(14 m)(7 m) (7 m)

2

5,788.125 kN-m 3 EI

Consider downward deflection of simply supported beam at C due to uniformly distributed load. [Appendix C, SS beam with uniformly distributed load.] Relevant equation from Appendix C: 5wL4 vC 384 EI Values: w = 105 kN/m, L = 14 m Calculation:

vC

5wL4 384 EI

5(105 kN/m)(14 m) 4 384 EI

52,521.875 kN-m3 EI


Consider upward deflection of simply supported beam at C due to concentrated load Cy. [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL3 vC 48 EI Values: P = −Cy, L = 14 m Calculation: vC

PL3 48EI

( C y )(14 m)3

(57.1667 m3 )C y

48EI

EI

Compatibility equation for deflection at C: 3 5,788.125 kN-m3 52,521.875 kN-m 3 (57.1667 m )C y EI EI EI 3 46,733.750 kN-m Cy 817.5 kN 818 kN 57.1667 m3

0 Ans.

(b) Beam deflection at A. Consider downward cantilever beam deflection caused by uniformly distributed load on overhang AB. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vA 8 EI Values: w = 105 kN/m, L = 3 m Calculation:

vA

wL4 8EI

(105 kN/m)(3 m)4 8EI

1,063.125 kN-m3 EI

Consider downward deflection at A resulting from rotation at B caused by concentrated load on overhang AB. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: ML (slope magnitude) B 3EI Values: M = (105 kN/m)(3 m)(1.5 m) = 472.5 kN-m, L = 14 m Computation: ML (472.5 kN-m)(14 m) B 3EI 3EI vA

2, 205 kN-m 2 (3 m) EI

2,205 kN-m 2 EI 6,615 kN-m 3 EI


Consider upward deflection at A due to uniformly distributed load between B and D. [Appendix C, SS beam with uniformly distributed load.] Relevant equation from Appendix C: wL3 (slope magnitude) B 24 EI Values: w = 105 kN/m, L = 14 m Calculation: wL3 B 24 EI vA

(3 m)

(105 kN/m)(14 m)3 24 EI 12,005 kN-m 2 EI

12,005 kN-m 2 EI

36,015 kN-m 3 EI

Consider downward deflection at A due to concentrated load Cy. [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL2 (slope magnitude) B 16 EI Values: P = −Cy = −817.5 kN, L = 14 m Calculation: PL2 B 16 EI vA

(817.5 kN)(14 m) 2 16 EI

10,014.375 kN-m 2 (3 m) EI

Beam deflection at A. 1,063.125 kN-m3 vA EI 1,706.25 kN-m3 EI

10,014.375 kN-m 2 EI 30,043.125 kN-m 3 EI

6,615 kN-m3 36,015 kN-m3 30,043.125 kN-m3 EI EI EI 3 1,706.25 kN-m 0.015180 m 15.18 mm 112,400 kN-m 2

Ans.


11.51 The beam shown in Fig. P11.51 consists of a W8 × 15 structural steel wideflange shape [E = 29,000 ksi; I = 48 in.4]. For the loading shown, determine: (a) the reactions at A and B. (b) the magnitude of the maximum bending stress in the beam. (Reminder: The roller symbol implies that both upward and downward displacement is restrained.) Fig. P11.51

Solution Choose the reaction force at B as the redundant; therefore, the released beam is a cantilever. Consider deflection of cantilever beam at B due to uniformly distributed load over entire beam span. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wx 2 vB (6 L2 4 Lx x 2 ) (elastic curve) 24 EI Values: w = −80 lb/in., L = 150 in., x = 100 in. Calculation: wx 2 vB (6 L2 4 Lx x 2 ) 24 EI ( 80 lb/in.)(100 in.) 2 6(150 in.) 2 24 EI

4(150 in.)(100 in.) (100 in.) 2

2.8333 109 lb-in.3 EI Consider deflection of cantilever beam at B due to the force caused by the linear portion of the distributed load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vB 3EI Values: P = −½(50 in.)(60 lb/in.) = −1,500 lb, L = 100 in. Calculation:

vB

PL3 3EI

( 1,500 lb)(100 in.)3 3EI

500 106 lb-in.3 EI


Consider deflection of cantilever beam at B due to the moment caused by the linear portion of the distributed load. [Appendix C, Cantilever beam with concentrated moment at tip.] Relevant equation from Appendix C: ML2 vB 2 EI Values: M = −½(50 in.)(60 lb/in.)[⅔(50 in.)] = −50,000 lb-in., L = 100 in. Calculation:

vB

ML2 2 EI

( 50,000 lb-in.)(100 in.) 2 2 EI

250 106 lb-in.3 EI

Consider deflection of cantilever beam at B due to concentrated load By. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vB 3EI Values: P = −By, L = 100 in.

Calculation: vB

PL3 3EI

( By )(100 in.)3

(333.3333 in.3 ) By

3EI

EI

Compatibility equation for deflection at B: 3 3 2.8333 109 lb-in.3 500 106 lb-in.3 250 106 lb-in.3 (333.3333 10 in. ) By EI EI EI EI 9 3 3.5833 10 lb-in. By 10,750 lb 10,750 lb 333.3333 103 in.3

0 Ans.

Equilibrium equations for entire beam: 1 Fy Ay By (80 lb/in.)(150 in.) (140 lb/in. 80 lb/in.)(50 in.) 0 2 (60 lb/in.)(50 in.) Ay (80 lb/in.)(150 in.) ( 10,750 lb) 2 2,750 lb

2, 750 lb

Ans.


MA

MA

MA

(80 lb/in.)(150 in.)(75 in.)

(140 lb/in. 80 lb/in.)(50 in.) 2 100 in. (50 in.) By (100 in.) 0 2 3 (60 lb/in.)(50 in.) (80 lb/in.)(150 in.)(75 in.) (133.3333 in.) ( 10,750 in.)(100 in.) 2 25,000 lb-in. 25,000 lb-in. (cw)

Ans.

(b) Magnitude of maximum bending stress: Section properties (from Appendix B): I 48 in.4 d 8.11 in. S 11.8 in.3 Maximum bending moment magnitude Mmax = 150,000 lb-in. (at B) Bending stresses at maximum moment (150,000 lb-in.)(8.11 in./2) 12,671.875 psi 12,670 psi x 48 in.4 or, using the tabulated value for the section modulus: 150,000 lb-in. 12,711.864 psi 12,710 psi x 11.8 in.3

Ans.

Ans.


11.52 The beam shown in Fig. P11.52 consists of a W24 × 94 structural steel wideflange shape [E = 29,000 ksi; I = 2,700 in.4]. For the loading shown, determine: (a) the reactions at A and D. (b) the magnitude of the maximum bending stress in the beam. Fig. P11.52

Solution Choose the reaction force at A as the redundant; therefore, the released beam is a cantilever. Consider downward deflection of cantilever beam at A due to the 50-kip concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 PL2 vB and (slope magnitude) B 3EI 2 EI Values: P = 50 kips, L = 20 ft Calculation: PL3 (50 kips)(20 ft)3 133,333.333 kip-ft 3 vB 3EI 3EI EI 2 2 PL (50 kips)(20 ft) 10,000 kip-ft 3 B 2 EI 2 EI EI

vA

133,333.333 kip-ft 3 EI

(6 ft)

10,000 kip-ft 3 EI

193,333.333 kip-ft 3 EI

Consider downward deflection of cantilever beam at A due to the uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equations from Appendix C: wL4 wL3 vC and (slope magnitude) C 8EI 6 EI Values: w = 4 kips/ft, L = 14 ft Calculation: wL4 (4 kips/ft)(14 ft) 4 19, 208 kip-ft 3 vC 8EI 8EI EI 3 3 wL (4 kips/ft)(14 ft) 1,829.333 kip-ft 3 C 6 EI 6 EI EI vA

19, 208 kip-ft 3 EI

1,829.333 kip-ft 3 (12 ft) EI

41,160 kip-ft 3 EI


Consider upward deflection of cantilever beam at A due to concentrated load Ay. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vA 3EI Values: P = −Ay, L = 26 ft Calculation: vA

PL3 3EI

( Ay )(26 ft)3

(5,858.667 ft 3 ) Ay

3EI

EI

(b)

Compatibility equation for deflection at A: 3 193,333.333 kip-ft 3 41,160 kip-ft 3 (5,858.667 ft ) Ay 0 EI EI EI 234, 493.333 kip-ft 3 Ay 40.025 kips 40.0 kips 5,858.667 ft 3

Equilibrium equations for entire beam: Fy Ay Dy 50 kips (4 kips/ft)(14 ft)

Dy MD

MD

MD

Ans.

0

50 kips (4 kips/ft)(14 ft) 40.025 kips (50 kips)(20 ft) (4 kips/ft)(14 ft)(7 ft)

65.975 kips Ay (26 ft)

66.0 kips

Ans.

0

(40.025 kips)(26 ft) (50 kips)(20 ft) (4 kips/ft)(14 ft)(7 ft) 351.350 kip-ft

351 kip-ft (cw)

Ans.


Shear-force and bending-moment diagrams (b) Magnitude of maximum bending stress: Section properties (from Appendix B): I 2,700 in.4

d

24.3 in.

S

222 in.3

Maximum bending moment magnitude Mmax = 351.350 kip-ft Bending stresses at maximum moment (351.350 kip-ft)(24.3 in./2)(12 in./ft) x 2,700 in.4

18.97 ksi

Ans.

or, using the tabulated section modulus: (351.350 kip-ft)(12 in./ft) x 222 in.3

18.99 ksi

Ans.


11.53 The solid 20-mm-diameter steel [E = 200 GPa] shaft shown in Fig. P11.53 supports two belt pulleys. Assume that the bearing at A can be idealized as a pin support and that the bearings at C and E can be idealized as roller supports. For the loading shown, determine: (a) the reaction forces at bearings A, C, and E. (b) the magnitude of the maximum bending stress in the shaft. Fig. P11.53

Solution (a) Reaction forces at A, C, and E. Choose the reaction force at C as the redundant; therefore, the released beam is simply supported between A and E. Consider downward deflection of simply supported beam at C due to pulley B load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vC ( L b2 x2 ) (elastic curve) 6 LEI Values: P = 850 N, L = 2,000 mm, b = 600 mm, x = 1,000 mm Calculation: vC

Pbx 2 (L 6 LEI

b2

x2 )

(850 N)(600 mm)(1,000 mm) (2,000 mm) 2 6(2,000 mm)EI

(600 mm) 2

(1,000 mm) 2

112.2 109 N-mm3 EI

Consider downward deflection of simply supported beam at C due to pulley D load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vC ( L b2 x2 ) (elastic curve) 6 LEI Values: P = 600 N, L = 2,000 mm, b = 400 mm, x = 1,000 mm


Calculation: Pbx 2 vC (L 6 LEI

b2

x2 )

(600 N)(400 mm)(1,000 mm) (2,000 mm) 2 6(2,000 mm)EI

(400 mm) 2

(1,000 mm) 2

56.8 109 N-mm3 EI

Consider upward deflection of simply supported beam at C due to concentrated load Cy. [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL3 vC 48 EI Values: P = −Cy, L = 2,000 mm Calculation: vC

PL3 48EI

( C y )(2,000 mm)3

(166.6667 106 mm3 )C y

48EI

EI

Compatibility equation for deflection at C: 6 3 112.2 109 N-mm3 56.8 109 N-mm3 (166.6667 10 mm )C y EI EI EI 9 3 169.0 10 N-mm Cy 1,014 N 1,014 N 166.6667 106 mm3

Equilibrium equations for entire beam: MA (850 N)(600 mm) (600 N)(1,600 mm) C y (1,000 mm) Ey

Fy

0

E y (2,000 mm)

(850 N)(600 mm) (600 N)(1,600 mm) (1,014 N)(1,000 mm) 2,000 mm

Ay

Cy

Ey

850 N 600 N

Ay

850 N 600 N 1,014 N 228 N

Ans.

0

228 N

Ans.

0 208 N

Ans.


Shear-force and bending-moment diagrams (b) Magnitude of maximum bending stress: Section properties: I

64

(20 mm)4

7,853.9816 mm4

Maximum bending moment magnitude Mmax = 132,000 N-mm Bending stresses at maximum moment (132,000 N-mm)(20 mm/2) x 7,853.9816 mm 4

168.1 MPa

Ans.


11.54 The solid 1.00-in.-diameter steel [E = 29,000 ksi] shaft shown in Fig. P11.54 supports three belt pulleys. Assume that the bearing at A can be idealized as a pin support and that the bearings at C and E can be idealized as roller supports. For the loading shown, determine: (a) the reaction forces at bearings A, C, and E. (b) the magnitude of the maximum bending stress in the shaft.

Fig. P11.54

Solution (a) Reaction forces at A, C, and E. Choose the reaction force at C as the redundant; therefore, the released beam is simply supported between A and E. Consider downward deflection of simply supported beam at C due to pulley B load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vC ( L b2 x2 ) (elastic curve) 6 LEI Values: P = 200 lb, L = 60 in., b = 15 in., x = 30 in. Calculation: Pbx 2 vC ( L b2 x2 ) 6 LEI (200 lb)(15 in.)(30 in.) (60 in.) 2 6(60 in.)EI

(15 in.) 2

(30 in.) 2

618,750 lb-in.3 EI

Consider downward deflection of simply supported beam at C due to pulley D load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vC ( L b2 x2 ) (elastic curve) 6 LEI Values: P = 280 lb, L = 60 in., b = 15 in., x = 30 in. Calculation: Pbx 2 vC (L 6 LEI

b2

x2 )

(280 lb)(15 in.)(30 in.) (60 in.) 2 6(60 in.)EI

(15 in.) 2

(30 in.) 2

866, 250 lb-in.3 EI Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Consider upward deflection of simply supported beam at C due to pulley F load. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vC (2 L2 3Lx x 2 ) (elastic curve) 6 LEI Values: M = −(120 lb)(10 in.) = −1,200 lb-in., L = 60 in., x = 30 in. Calculation: Mx vC (2 L2 3Lx x 2 ) 6 LEI ( 1, 200 lb-in.)(30 in.) 2(60 in.) 2 3(60 in.)(30 in.) (30 in.) 2 6(60 in.)EI 270,000 lb-in.3 EI

Consider upward deflection of simply supported beam at C due to concentrated load Cy. [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL3 vC 48 EI Values: P = −Cy, L = 60 in. Calculation: vC

PL3 48EI

( C y )(60 in.)3

(4,500 in.3 )C y

48EI

EI

Compatibility equation for deflection at C: 618,750 lb-in.3 866, 250 lb-in.3 270,000 lb-in.3 EI EI EI 3 1, 215,000 lb-in. Cy 270 lb 270 lb 4,500 in.3

(4,500 in.3 )C y EI

Equilibrium equations for entire beam: MA (200 lb)(15 in.) (280 lb)(45 in.) (120 lb)(70 in.) C y (30 in.)

0 Ans.

E y (60 in.)

0


Ey

(200 lb)(15 in.) (280 lb)(45 in.) (120 lb)(70 in.) (215 lb)(30 in.) 60 in. 265.0 lb

Fy

Ey

Ans.

265 lb

Ay

Cy

200 lb 280 lb 120 lb

0

Ay

200 lb 280 lb 120 lb 270 lb 265 lb

Ans.

65.0 lb


64

(1.00 in.) 4

0.0490874 in.4

Maximum bending moment magnitude Mmax = 1,200 lb-in. Bending stresses at maximum moment (1, 200 lb-in.)(1.00 in./2) x 0.0490874 in.4 12, 223.1 psi 12, 220 psi

Ans.


11.55 The solid 1.00-in.-diameter steel [E = 29,000 ksi] shaft shown in Fig. P11.55 supports two belt pulleys. Assume that the bearing at E can be idealized as a pin support and that the bearings at B and C can be idealized as roller supports. For the loading shown, determine: (a) the reaction forces at bearings B, C, and E. (b) the magnitude of the maximum bending stress in the shaft.

Fig. P11.55

Solution (a) Reaction forces at B, C, and E. Choose the reaction force at C as the redundant; therefore, the released beam is simply supported between B and E. Consider upward deflection of simply supported beam at C due to pulley A load. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vC (2 L2 3Lx x 2 ) (elastic curve) 6 LEI Values: M = −(90 lb)(7 in.) = −630 lb-in., L = 45 in., x = 15 in. Calculation: Mx vC (2 L2 6 LEI

3Lx

x2 )

( 630 lb-in.)(15 in.) 2(45 in.) 2 6(45 in.)EI

3(45 in.)(15 in.) (15 in.) 2

78,750 lb-in.3 EI

Consider downward deflection of simply supported beam at C due to pulley D load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vC ( L b2 x2 ) (elastic curve) 6 LEI Values: P = 240 lb, L = 45 in., b = 15 in., x = 15 in.


Calculation: Pbx 2 vC (L 6 LEI

b2

x2 )

(240 lb)(15 in.)(15 in.) (45 in.) 2 6(45 in.)EI

(15 in.) 2

(15 in.) 2

315,000 lb-in.3 EI

Consider upward deflection of simply supported beam at C due to concentrated load Cy. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 vC (L a 2 b2 ) 6 LEI Values: P = −Cy, L = 45 in., a = 15 in., b = 30 in. Calculation: Pab 2 vC (L 6 LEI

a2

b2 )

( C y )(15 in.)(30 in.) 6(45 in.)EI

(45 in.)

2

(15 in.)

2

(30 in.)

2

(1,500 in.3 )C y EI

Compatibility equation for deflection at C: 3 78,750 lb-in.3 315,000 lb-in.3 (1,500 in. )C y 0 EI EI EI 236, 250 lb-in.3 Cy 157.50 lb 157.5 lb 1,500 in.3

Equilibrium equations for entire beam: M E (240 lb)(15 in.) (90 lb)(52 in.)

By

By (45 in.) C y (30 in.)

0

(240 lb)(15 in.) (90 lb)(52 in.) (157.5 lb)(30 in.) 45 in. 79.0 lb

Fy

Ans.

Ey

Ans.

79.0 lb

By

Cy

90 lb 240 lb

0

Ey

90 lb 240 lb 157.5 lb 79.0 lb

93.5 lb

93.5 lb

Ans.



64

(1.00 in.) 4

0.0490874 in.4

Maximum bending moment magnitude Mmax = 1,402.5 lb-in. Bending stresses at maximum moment (1, 402.5 lb-in.)(1.00 in./2) x 0.0490874 in.4 14, 285.74 psi 14, 290 psi

Ans.


11.56 The beam shown in Fig. P11.56 consists of a W360 × 101 structural steel wide-flange shape [E = 200 GPa; I = 301 × 106 mm4]. For the loading shown, determine: (a) the reactions at A and B. (b) the magnitude of the maximum bending stress in the beam. Fig. P11.56

Solution Choose the reaction force at B as the redundant; therefore, the released beam is a cantilever. Consider deflection of cantilever beam at B due to uniformly distributed load over entire beam span. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equations from Appendix C: wx 2 vB (6 L2 4 Lx x 2 ) (elastic curve) 24 EI Values: w = 30 kN/m, L = 8 m, x = 5.5 m Calculation: wx 2 vB (6 L2 4 Lx x 2 ) 24 EI (30 kN/m)(5.5 m)2 6(8 m) 2 24 EI

4(8 m)(5.5 m) (5.5 m) 2

9,008.828125 kN-m3 EI Consider deflection of cantilever beam at B due a linearly distributed load. [Appendix C, Cantilever beam with linearly distributed load.] Relevant equation from Appendix C: w0 x 2 vB (10L3 10L2 x 5Lx 2 x3 ) (elastic curve) 120 LEI Values: w0 = 60 kN/m, L = 8 m, x = 5.5 m Calculation: w0 x 2 vB (10 L3 10 L2 x 5 Lx 2 120 LEI

x3 )

(60 kN/m)(5.5 m) 2 10(8 m)3 10(8 m) 2 (5.5 m) 5(8 m)(5.5 m) 2 120(8 m)EI

(5.5 m)3

4,998.103516 kN-m 3 EI Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Consider deflection of cantilever beam at B due to concentrated load By. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vB 3EI Values: P = −By, L = 5.5 m Calculation: vB

PL3 3EI

( By )(5.5 m)3

(55.458333 m3 ) By

3EI

EI

Compatibility equation for deflection at B: 3 9,008.828125 kN-m3 4,998.103516 kN-m3 (55.458333 m ) By EI EI EI 3 14,006.93164 kN-m By 252.56676 kN 253 kN 55.458333 m3

0 Ans.

Equilibrium equations for entire beam: 1 Fy Ay By (30 kN/m)(8 m) (90 kN/m 30 kN/m)(8 m) 0 2 1 Ay (30 kN/m)(8 m) (90 kN/m 30 kN/m)(8 m) 252.56676 kN 2 227.43324 kN MA

MA

MA

227 kN

(90 kN/m 30 kN/m)(8 m) 2 960 kN-m 640 kN-m (252.56676 kN)(5.5 m)

(30 kN/m)(8 m)(4 m)

210.88281 kN-m

211 kN-m (ccw)

Ans. 8m 3

By (5.5 m)

0

Ans.


Shear-force and bending-moment diagrams (b) Magnitude of maximum bending stress: Section properties (from Appendix B): I 301 106 mm 4 d 356 mm S

1,690 103 mm3

Maximum bending moment magnitude Mmax = 210.88281 kN-m Bending stresses at maximum moment (210.88281 kN-m)(356 mm/2)(1,000) 2 x 301 106 mm4

124.7 MPa or, using the tabulated value for the section modulus: (210.88281 kN-m)(1,000) 2 x 1,690 103 mm3 124.8 MPa

Ans.


11.57 A W530 × 92 structural steel wideflange shape [E = 200 GPa; I = 554 × 106 mm4] is loaded and supported as shown in Fig. P11.57. Determine: (a) the force and moment reactions at supports A and C. (b) the maximum bending stress in the beam. (c) the deflection of the beam at B. Fig. P11.57

Solution (a) Reactions at A and C. Choose the moment reactions at A and C as the redundants. This will leave a simply supported beam between A and C as the released beam. Determine the slopes at A and C caused by the 150-kN concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equations from Appendix C: Pb( L2  b2 ) Pa( L2  a 2 ) A   and C  6LEI 6LEI Values: P = 150 kN, L = 10 m, a = 6 m, b = 4 m Calculation:

A   C 

Pb( L2  b 2 ) (150 kN)(4 m) 840 kN-m 2 (10 m) 2  (4 m) 2     6 LEI 6(10 m)EI EI

Pa( L2  a 2 ) (150 kN)(6 m) 960 kN-m 2 (10 m) 2  (6 m) 2    6 LEI 6(10 m)EI EI

Determine the slopes at A and C caused by moment reaction MA. [Appendix C, SS beam with concentrated moment at one end.] Relevant equations from Appendix C: ML ML A   and C  3EI 6 EI Values: M = MA, L = 10 m Calculation: ML M (10 m) (3.333333 m)M A A    A   3EI 3EI EI

C 

ML M A (10 m) (1.666667 m)M A   6 EI 6 EI EI


Determine the slopes at A and C caused by moment reaction MC. [Appendix C, SS beam with concentrated moment at one end.] Relevant equations from Appendix C: ML ML A   and C  6 EI 3EI Values: M = MC, L = 10 m Calculation: ML M (10 m) (1.666667 m)M C A    C   6 EI 6 EI EI

C 

ML M C (10 m) (3.333333 m)M C   3EI 3EI EI

Compatibility equation for slope at A: 840 kN-m 2 (3.333333 m)M A (1.666667 m)M C    0 EI EI EI Compatibility equation for slope at C: 960 kN-m 2 (1.666667 m)M A (3.333333 m)M C   0 EI EI EI

(a)

(b)

Solve Equations (a) and (b). Equations (a) and (b) can be rewritten as: (3.333333 m)M A  (1.666667 m)M C  840 kN-m2

(1.666667 m)M A  (3.333333 m)M C  960 kN-m2 and solved simultaneously for MA and MC: M A  144 kN-m  144 kN-m (ccw) M C  216 kN-m  216 kN-m (cw)

Ans. Ans.

Equilibrium equations for entire beam: M A   M A  M C  (150 kN)(6 m)  C y (10 m)  0

 Cy 

(150 kN)(6 m)  ( 144 kN-m)  ( 216 kN-m) 10 m

 97.2 kN  97.2 kN  Fy  Ay  C y  150 kN  0

 Ay  150 kN  97.2 kN  52.8 kN  52.8 kN 

Ans. Ans.


Shear-force and bending-moment diagrams (b) Magnitude of maximum bending stress: Section properties (from Appendix B): I  554  106 mm 4 d  533 mm S  2,080  103 mm3

Maximum bending moment magnitude Mmax = 216 kN-m (at C) Bending stresses at maximum moment (216 kN-m)(533 mm/2)(1,000)2 x  554  106 mm 4 Ans.  103.9 MPa or using the tabulated value for the section modulus: (216 kN-m)(1,000) 2 x  2,080  103 mm3  103.8 MPa

Ans.

(c) Beam deflection at B: Determine the deflection at B caused by the 150-kN concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 vB   ( L  a 2  b2 ) 6 LEI Values: P = 150 kN, L = 10 m, a = 6 m, b=4m Calculation:

Pab 2 (150 kN)(6 m)(4 m) 2,880 kN-m3 2 2 2 2 2   vB   (L  a  b )   (10 m)  (6 m)  (4 m)    6 LEI 6(10 m)EI EI Determine the deflection at B caused by MA. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vB   (2 L2  3Lx  x 2 ) 6 LEI Values: M = −144 kN-m, L = 10 m, x = 6 m


Calculation: Mx vB   (2 L2  3Lx  x 2 ) 6 LEI



(144 kN-m)(6 m) 806.4 kN-m3  2(10 m) 2  3(10 m)(6 m)  (6 m) 2   6(10 m)EI EI

Determine the deflection at B caused by MC. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vB   (2 L2  3Lx  x 2 ) 6 LEI Values: M = −216 kN-m, L = 10 m, x = 4 m Calculation: Mx vB   (2 L2  3Lx  x 2 ) 6 LEI



(216 kN-m)(4 m) 1,382.4 kN-m3  2(10 m) 2  3(10 m)(4 m)  (4 m) 2   6(10 m)EI EI

Beam deflection vB: 2,880 kN-m3 806.4 kN-m3 1,382.4 kN-m3 vB     EI EI EI 3 3 691.2 kN-m 691.2 kN-m    0.006238 m  6.24 mm  EI 110,800 kN-m 2

Ans.


11.58 A W530 × 92 structural steel wideflange shape [E = 200 GPa; I = 554 × 106 mm4] is loaded and supported as shown in Fig. P11.58. Determine: (a) the force and moment reactions at supports A and C. (b) the maximum bending stress in the beam. (c) the deflection of the beam at B. Fig. P11.58

Solution (a) Reactions at A and C. Choose the moment reactions at A and C as the redundants. This will leave a simply supported beam between A and C as the released beam. Determine the slopes at A and C caused by the 80 kN/m uniformly distributed load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equations from Appendix C: wa 2 A   (2 L  a) 2 and 24 LEI

C 

wa 2 (2 L2  a 2 ) 24 LEI

Values: w = 80 kN/m, L = 9 m, a = 4.5 m Calculation:

A   C 

wa 2 (80 kN/m)(4.5 m)2 1,366.875 kN-m 2 2 (2 L  a) 2   2(9 m)  (4.5 m)     24 LEI 24(9 m)EI EI

wa 2 (80 kN/m)(4.5 m)2 1,063.125 kN-m 2  2(9 m) 2  (4.5 m) 2   (2 L2  a 2 )  24 LEI 24(9 m)EI EI

Determine the slopes at A and C caused by moment reaction MA. [Appendix C, SS beam with concentrated moment at one end.] Relevant equations from Appendix C: ML ML A   and C  3EI 6 EI Values: M = MA, L = 9 m Calculation: ML M (9 m) (3 m)M A A    A   3EI 3EI EI

C 

ML M A (9 m) (1.5 m)M A   6 EI 6 EI EI

Determine the slopes at A and C caused by moment reaction MC. [Appendix C, SS beam with concentrated moment at one end.] Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Relevant equations from Appendix C: ML ML A   and C  6 EI 3EI Values: M = MC, L = 9 m Calculation: ML M (9 m) (1.5 m)M C A    C   6 EI 6 EI EI

C 

ML M C (9 m) (3 m)M C   3EI 3EI EI

Compatibility equation for slope at A: 1,366.875 kN-m 2 (3 m)M A (1.5 m)M C    0 EI EI EI Compatibility equation for slope at C: 1,063.125 kN-m 2 (1.5 m)M A (3 m)M C   0 EI EI EI

(a)

(b)

Solve Equations (a) and (b). Equations (a) and (b) can be rewritten as: (3 m)M A  (1.5 m)M C  1,366.875 kN-m 2

(1.5 m)M A  (3 m)M C  1,063.125 kN-m 2 and solved simultaneously for MA and MC: M A  371.25 kN-m  371.25 kN-m (ccw) M C  168.75 kN-m  168.8 kN-m (cw)

Ans. Ans.

Equilibrium equations for entire beam: M A   M A  M C  (80 kN/m)(4.5 m)(2.25 m)  C y (9 m)  0

 Cy 

(80 kN/m)(4.5 m)(2.25 m)  (371.25 kN-m)  (168.75 kN-m) 9m

 67.5 kN  67.5 kN 

Ans.

Fy  Ay  C y  (80 kN/m)(4.5 m)  0

 Ay  (80 kN/m)(4.5 m)  67.5 kN  292.5 kN  293 kN 

Ans.


Shear-force and bending-moment diagrams (b) Magnitude of maximum bending stress: Section properties (from Appendix B): I  554  106 mm 4 d  533 mm S  2,080  103 mm3

Maximum bending moment magnitude Mmax = 371.25 kN-m (at A) Bending stresses at maximum moment (371.25 kN-m)(533 mm/2)(1,000)2 x  554  106 mm 4 Ans.  178.6 MPa or using the tabulated value for the section modulus: (371.25 kN-m)(1,000) 2 x  2,080  103 mm3  178.5 MPa

Ans.

(c) Beam deflection at B: Determine the deflection at B caused by the 80 kN/m uniformly distributed load. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa3 vB   (4L2  7aL  3a 2 ) 24 LEI Values: w = 80 kN/m, L = 9 m, a = 4.5 m Calculation: wa3 vB   (4 L2  7aL  3a 2 ) 24 LEI



(80 kN/m)(4.5 m)3 3,417.1875 kN-m3  4(9 m)2  7(4.5 m)(9 m)  3(4.5 m) 2    24(9 m)EI EI


Determine the deflection at B caused by MA. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vB   (2 L2  3Lx  x 2 ) 6 LEI Values: M = −371.25 kN-m, L = 9 m, x = 4.5 m Calculation: Mx vB   (2 L2  3Lx  x 2 ) 6 LEI



(371.25 kN-m)(4.5 m) 1,879.4531 kN-m3  2(9 m)2  3(9 m)(4.5 m)  (4.5 m) 2   6(9 m)EI EI

Determine the deflection at B caused by MC. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vB   (2 L2  3Lx  x 2 ) 6 LEI Values: M = −168.75 kN-m, L = 9 m, x = 4.5 m Calculation: Mx vB   (2 L2  3Lx  x 2 ) 6 LEI



(168.75 kN-m)(4.5 m) 854.2969 kN-m3  2(9 m)2  3(9 m)(4.5 m)  (4.5 m) 2   6(9 m)EI EI

Beam deflection vB: 3, 417.1875 kN-m3 1,879.4531 kN-m3 854.2969 kN-m3 vB     EI EI EI 3 3 683.4375 kN-m 683.4375 kN-m    0.006168 m  6.17 mm  EI 110,800 kN-m 2

Ans.


11.59 A timber [E = 1,800 ksi] beam is loaded and supported as shown in Fig. P11.59. The cross section of the timber beam is 4-in. wide and 8-in. deep. The beam is supported at B by a ½-in.-diameter steel [E = 30,000 ksi] rod, which has no load before the distributed load is applied to the beam. After a distributed load of 900 lb/ft is applied to the beam, determine: (a) the force carried by the steel rod. (b) the maximum bending stress in the timber beam. (c) the deflection of the beam at B. Fig. P11.59

Solution Section properties: Beam:

I beam 

Rod (1):

A1 

 4

(4 in.)(8 in.)3  170.6667 in.4 12

Ebeam  1,800,000 psi

(0.50 in.) 2  0.1963495 in.2

E1  30,000,000 psi

(a) Force carried by the steel rod. The reaction force from rod (1) will be taken as the redundant, leaving a simply supported beam between A and C as the released beam. For this analysis, a tension force is assumed to exist in axial member (1). Beam free-body diagram Downward deflection of wood beam at B due to 900 lb/ft uniformly distributed load. [Appendix C, SS beam with uniformly distributed load.] Relevant equation from Appendix C: 5wL4 vB   384 EI Values: w = 900 lb/ft, L = 14 ft, EI = 307.2 ×106 lb-in.2 Calculation: 5wL4 5(900 lb/ft)(14 ft)4 (12 in./ft)3 vB     2.532305 in. 384EI 384(307.2  106 lb-in.2 )


Upward deflection of wood beam at B due to force F1 in rod (1). [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL3 vB   48 EI Values: P = −F1, L = 14 ft, EI = 307.2 ×106 lb-in.2 Calculation: PL3 ( F1 )(14 ft)3 (12 in./ft)3 vB     (321.5625  106 in./lb)F1 6 2 48EI 48(307.2  10 lb-in. ) Elongation of steel rod (1) due to force F1. FL (16 ft)(12 in./ft) 1  1 1  F1  (32.59493  10 6 in./lb)F1 2 6 A1E1 (0.1963495 in. )(30  10 psi) The deflection of the wood beam at B will not equal zero in this instance because the rod that supports the beam at B will elongate, thus permitting the wood beam to deflect downward. Therefore, vB  1  (32.59493  106 in./lb)F1

Compatibility equation at B: The sum of the downward deflection caused by the uniformly distributed load and the upward deflection caused by the force in rod (1) must equal the elongation of the steel rod. Elongation of the steel rod will produce a downward (i.e., negative) deflection of the wood beam at B.

2.532305 in.  (321.5625  106 in./lb)F1  (32.59493  106 in./lb)F1  F1 

2.532305 in.  7,150.22 lb  7,150 lb (T) 354.1574  106 in./lb

Ans.


(b) Determine maximum bending stress in wood beam: Maximum bending moment magnitude Mmax = 4,125 lb-ft = 49,500 lb-in. Bending stress at maximum moment (49,500 lb-in.)(8 in./2) x   1,160 psi 170.6667 in.4

(c) Beam deflection at B. The beam deflection at B is equal to the elongation of the steel rod: FL (7,150.22 lb)(16 ft)(12 in./ft) vB  1   1 1    0.233 in.  0.233 in.  A1E1 (0.1963495 in.2 )(30  106 psi)

Ans.

Ans.


11.60 A W360 × 72 structural steel [E = 200 GPa] wide-flange shape is loaded and supported as shown in Fig. P11.60. The beam is supported at B by 20-mm-diameter solid aluminum [E = 70 GPa] rod. After a concentrated load of 40 kN is applied to the tip of the cantilever, determine: (a) the force produced in the aluminum rod. (b) the maximum bending stress in the beam. (c) the deflection of the beam at B. Fig. P11.60

Solution Section properties: W360 × 72: I beam  201  106 mm 4

d  351 mm

Ebeam  200,000 MPa

Sbeam  1,150  103 mm3 Rod (1):

A1 

 4

(20 mm) 2  314.159266 mm 2

E1  70,000 MPa

(a) Force in the aluminum rod. The reaction force from rod (1) will be taken as the redundant, leaving a cantilever beam as the released beam. For this analysis, a tension force is assumed to exist in axial member (1).

Downward deflection of W360 × 72 beam at B due to 40-kN concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: Px 2 vB   (3L  x) (elastic curve) 6 EI Values: P = 40 kN, L = 5 m, x = 3.6 m, EI = 40,200 kN-m2 Calculation: Px 2 (40 kN)(3.6 m)2 vB   (3L  x)   3(5 m)  (3.6 m)  24.50149  103 m 2  6EI 6(40,200 kN-m )


Upward deflection of W360 × 72 beam at B due to force F1 in rod (1). [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vB   3EI Values: P = −F1, L = 3.6 m, EI = 40,200 kN-m2 Calculation: PL3 ( F1 )(3.6 m)3 vB     (386.8657  106 m/kN)F1 3EI 3(40,200 kN-m2 ) Elongation of aluminum rod (1) due to force F1. FL (3 m) 1  1 1  F1  (136.4185  10 6 m/kN)F1 2 2 A1E1 (314.159266 mm )(70,000 N/mm )(1 kN/1,000 N) The deflection of the W360 × 72 beam at B will not equal zero in this instance because the rod that supports the beam at B will elongate, thus permitting the W360 × 72 beam to deflect downward. Therefore, vB  1  (136.4185  106 m/kN)F1

Compatibility equation at B: The sum of the downward deflection caused by the 40-kN concentrated load and the upward deflection caused by the force in rod (1) must equal the elongation of the aluminum rod. Elongation of the aluminum rod will produce a downward (i.e., negative) deflection of the W360 × 72 beam at B.

24.50149  103 m  (386.8657  106 m/kN)F1  (136.4185  10 6 m/kN)F1 24.50149  103 m  F1   46.82254 kN  46.8 kN (T) 523.2842  106 m/kN

Ans.


(b) Determine maximum bending stress in the W360 × 72 beam: Maximum bending moment magnitude Mmax = 56 kN-m (at B) Bending stress at maximum moment (56 kN-m)(351 mm/2)(1,000) 2 x  201  106 mm4 Ans.  48.9 MPa or using the tabulated value for the section modulus of the W360 × 72 beam: (56 kN-m)(1,000) 2 x   48.7 MPa Ans. 1,150  103 mm3

(c) Beam deflection at B. The beam deflection at B is equal to the elongation of the aluminum rod: FL (46,822.54 N)(3,000 mm) vB  1   1 1    6.38746 mm  6.39 mm  A1E1 (314.159266 mm 2 )(70,000 N/mm 2 )

Ans.


11.61 A W18 × 55 structural steel [E = 29,000 ksi] wide-flange shape is loaded and supported as shown in Fig. P11.61. The beam is supported at C by a ¾-in.-diameter aluminum [E = 10,000 ksi] rod, which has no load before the distributed load is applied to the beam. After a distributed load of 4 kips/ft is applied to the beam, determine: (a) the force carried by the aluminum rod. (b) the maximum bending stress in the steel beam. (c) the deflection of the beam at C. Fig. P11.61

Solution Section properties: Beam: I beam  890 in.4

d  18.1 in.

Ebeam  29,000 ksi

Sbeam  98.3 in.

3

A1 



(0.75 in.) 2  0.4417865 in.2 4 (a) Force in the aluminum rod. The reaction force from rod (1) will be taken as the redundant, leaving a cantilever beam as the released beam. Rod (1):

E1  10,000 ksi

For this analysis, a tension force is assumed to exist in axial member (1). Downward deflection of W18 × 55 beam at C due to 4 kips/ft uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equations from Appendix C: wL4 wL3 vB   and  B   8EI 6 EI Values: w = 4 kips/ft, L = 11 ft, EI = 25.81×106 kip-in.2 Calculation: wL4 (4 kips/ft)(11 ft) 4 (12 in./ft)3 vB     0.490113 in. 8EI 8(25.81  106 kip-in.2 )

B  

wL3 (4 kips/ft)(11 ft)3 (12 in./ft) 2   4.950639  103 rad 6 2 6 EI 6(25.81  10 kip-in. )

vC  0.490113 in.  (5 ft)(12 in./ft)(  4.950639  103 rad)  0.787152 in.


Upward deflection of W18 × 55 beam at C due to force F1 in rod (1). [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vC   3EI Values: P = −F1, L = 16 ft, EI = 25.81×106 kip-in.2 Calculation: PL3 ( F1 )(16 ft)3 (12 in./ft)3 vC     (91.41015  103 in./kip)F1 6 2 3EI 3(25.81  10 kip-in. ) Elongation of aluminum rod (1) due to force F1. FL (14 ft)(12 in./ft) 1  1 1  F1  (38.02742  10 3 in./kip)F1 2 A1E1 (0.4417865 in. )(10,000 ksi) The deflection of the W18 × 55 beam at C will not equal zero in this instance because the rod that supports the beam at C elongates, thus permitting the W18 × 55 beam to deflect downward. Therefore, vC  1  (38.02742  103 in./kip)F1

Compatibility equation at C: The sum of the downward deflection caused by the uniformly distributed load and the upward deflection caused by the force in rod (1) must equal the elongation of the aluminum rod. Elongation of the aluminum rod will produce a downward (i.e., negative) deflection of the W18 × 55 beam at C.

0.787152 in.  (91.41015  103 in./kip)F1   (38.02742  103 in./kip)F1  F1 

0.787152 in.  6.081323 kips  6.08 kips (T) 129.4376  103 in./kip

Ans.

(b) Determine maximum bending stress in W18 × 55 beam: Maximum bending moment magnitude Mmax = 144.70 kip-ft (at A) Bending stress at maximum moment (144.70 kip-ft)(18.1 in./2)(12 in./ft) x  890 in.4 Ans.  17.66 ksi or using the tabulated value for the section modulus of the W18 × 55 beam: (144.70 kip-ft)(12 in./ft) x  98.3 in.3 Ans.  17.66 ksi


(c) Beam deflection at C. The beam deflection at C is equal to the elongation of the aluminum rod: FL (6.081323 kips)(14 ft)(12 in./ft) vC  1   1 1    0.231257 in.  0.231 in.  A1E1 (0.4417865 in.2 )(10,000 ksi)

Ans.


11.62 A W250 × 32.7 structural steel [E = 200 GPa] wide-flange shape is loaded and supported as shown in Fig. P11.62. A uniformly distributed load of 16 kN/m is applied to the beam, causing the roller support at B to settle downward (i.e., displace downward) by 15 mm. Determine: (a) the reactions at supports A, B, and C. (b) the maximum bending stress in the beam. Fig. P11.62

Solution Section properties: W 250  32.7 :

I  49.1 106 mm4

d  259 mm

S  380  103 mm3

(a) Reactions at A, B, and C. Choose the reaction force at B as the redundant; therefore, the released beam is simply supported between A and C. Consider downward deflection of simply supported beam at B due to uniformly distributed load. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wx vB   ( L3  2 Lx 2  x3 ) (elastic curve) 24 EI Values: w = 16 kN/m, L = 10 m, x = 4 m Calculation: wx vB   ( L3  2 Lx 2  x3 ) 24 EI 

(16 kN/m)(4 m) 1,984 kN-m3 (10 m)3  2(10 m)(4 m) 2  (4 m)3    24 EI EI

Consider upward deflection of simply supported beam at B due to concentrated load By. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 vB   ( L  a 2  b2 ) 6 LEI Values: P = −By, L = 10 m, a = 4 m, b = 6 m Calculation: Pab 2 vB   ( L  a 2  b2 ) 6 LEI



(  By )(4 m)(6 m) 6(10 m) EI

(10 m)  (4 m)  (6 m)   2

2

2

(19.2 m3 ) By EI


Compatibility equation for deflection at B: EI  (200,000 N/mm 2 )(49.1  106 mm 4 )  9.82  1012 N-mm 2  9.82  103 kN-m 2 

3 1,984 kN-m3 (19.2 m ) By   0.015 m EI EI ( 0.015 m)(9.82  103 kN-m 2 )  1,984 kN-m3  By   95.6615 kN  95.7 kN  19.2 m3

Ans.

Equilibrium equations for entire beam: M A  By (4 m)  C y (10 m)  (16 kN/m)(10 m)(5 m)  0

 Cy 

(16 kN/m)(10 m)(5 m)  (95.6615 kN)(4 m) 10 m

 41.7354 kN  41.7 kN 

Ans.

Fy  Ay  By  C y  (16 kN/m)(10 m)  0

 Ay  (16 kN/m)(10 m)  95.6615 kN  41.7354 kN  22.6031 kN  22.6 kN 

Ans.

Shear-force and bending-moment diagrams (b) Determine maximum bending stress in W250 × 32.7 beam: Maximum bending moment magnitude Mmax = 54.4327 kN-m Bending stress at maximum moment (54.4327 kN-m)(259 mm/2)(1,000)2 x  49.1  106 mm4 Ans.  143.6 MPa or using the tabulated value for the section modulus of the W250 × 32.7 beam: (54.4327 kN-m)(1,000)2 x  380  103 mm3 Ans.  143.2 MPa


11.63 A W10 × 22 structural steel [E = 29,000 ksi] wide-flange shape is loaded and supported as shown in Fig. P11.63. The beam is supported at C by a timber [E = 1,800 ksi] post having a crosssectional area of 16 in.2. After a concentrated load of 10 kips is applied to the beam, determine: (a) the reactions at supports A and C. (b) the maximum bending stress in the beam. (c) the deflection of the beam at C. Fig. P11.63

Solution Section properties: W10  22: I beam  118 in.4

d  10.2 in.

Ebeam  29,000 ksi

S beam  23.2 in.

3

Post (1):

A1  16 in.

E1  1,800 ksi

(a) Reactions at supports A and C. The reaction force from post (1) will be taken as the redundant, leaving a cantilever beam as the released beam. To be consistent with earlier sign conventions (e.g., Chapter 5), we will assume that the force in the axial member is tension (even though intuitively we recognize that the post must be in compression). Beam free-body diagram Downward deflection of W10 × 22 beam at C due to 10-kip concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 PL2 vB   and  B   3EI 2 EI Values: P = 10 kips, L = 14 ft, EI = 3,422,000 kip-in.2 Calculation: PL3 (10 kips)(14 ft)3 (12 in./ft)3 vB     4.618773 in. 3EI 3(3, 422,000 kip-in.2 )

B  

PL2 (10 kips)(14 ft) 2 (12 in./ft) 2   0.0412390 rad 2 EI 2(3, 422,000 kip-in.2 )

vC  4.618773 in.  (0.0412390 rad)(6 ft)(12 in./ft)  7.587984 in. Since we are assuming that a tension force exists in post (1), the tension force from the post will cause a downward deflection of W10 × 22 beam at C. [Appendix C, Cantilever beam with concentrated load at tip.] Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Relevant equation from Appendix C: PL3 vB   3EI Values: P = F1, L = 20 ft, EI = 3,422,000 kip-in.2 Calculation: PL3 ( F )(20 ft)3 (12 in./ft)3 vB    1  (1.346581 in./kip)F1 3EI 3(3,422,000 kip-in.2 ) Axial deformation of post (1) due to force F1. FL (12 ft)(12 in./ft) 1  1 1  F1  (0.005 in./kip)F1 A1E1 (16 in.2 )(1,800 ksi) The deflection of the steel beam will not equal zero at C in this instance because the post deforms. If we are consistent and assume that there is tension in the post, then the steel beam must deflect upward at C. Therefore, vC  1  (0.005 in./kip)F1 Compatibility equation at C: The sum of the downward deflection caused by the 10-kip concentrated load and the upward deflection caused by tension in post (1) must equal the elongation of the post.

7.587984 in.  (1.346581 in./kip)F1  (0.005 in./kip)F  F1 

7.587984 in.  5.614154 kips  5.61 kips (C)  C y 1.351582 in./kip

Ans.

Equilibrium equations for entire beam: M A   M A  F1 (20 ft)  (10 kips)(14 ft)  0

 M A  (5.614154 kips)(20 ft)  (10 kips)(14 ft)  27.7169 kip-ft  27.7 kip-ft (ccw)

Ans.

Fy  Ay  F1  (10 kips)  0

 Ay  10 kips  (5.614154 kips)  4.3858 kips  4.39 kips 

Ans.


(b) Determine maximum bending stress in the beam: Maximum bending moment magnitude Mmax = 33.685 kip-ft = 404.218 kip-in. (at B) Bending stress at maximum moment (404.218 kip-in.)(10.2 in./2) x  118 in.4 Ans.  17.47 ksi or using the tabulated value for the section modulus of the W10 × 22 beam: 404.218 kip-in. Ans. x   17.42 ksi 23.2 in.3

(c) Beam deflection at C. The beam deflection at C is equal to the deformation of the post: FL (  5.614154 kips)(12 ft)(12 in./ft) vC  1  1 1   0.0281 in.  0.0281 in.  A1E1 (16 in.2 )(1,800 ksi)

Ans.


11.64 A timber [E = 12 GPa] beam is loaded and supported as shown in Fig. P11.64. The cross section of the timber beam is 100-mm. wide and 300-mm deep. The beam is supported at B by a 12-mm-diameter steel [E = 200 GPa] rod, which has no load before the distributed load is applied to the beam. After a distributed load of 7 kN/m is applied to the beam, determine: (a) the force carried by the steel rod. (b) the maximum bending stress in the timber beam. (c) the deflection of the beam at B. Fig. P11.64


I beam 

Rod (1):

A1 

 4

(100 mm)(300 mm)3  225  106 mm 4 12


(12 mm) 2  113.097336 mm 2

E1  200,000 MPa

(a) Force carried by the steel rod. The reaction force from rod (1) will be taken as the redundant, leaving a simply supported beam between A and C as the released beam. For this analysis, a tension force is assumed to exist in axial member (1).

Downward deflection of wood beam at B due to 7 kN/m uniformly distributed load. [Appendix C, SS beam with uniformly distributed load.] Relevant equation from Appendix C: wx vB   ( L3  2 Lx 2  x3 ) (elastic curve) 24 EI Values: w = 7 kN/m, L = 6 m, x = 4 m, EI = 2,700 kN-m2 Calculation: wx vB   ( L3  2 Lx 2  x3 ) 24 EI (7 kN/m)(4 m) (6 m)3  2(6 m)(4 m)2  (4 m)3   38.02469  103 m  2  24(2,700 kN-m )


Upward deflection of wood beam at B due to force F1 in rod (1). [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 vB   ( L  a 2  b2 ) 6 LEI Values: P = −F1, L = 6 m, a = 4 m, b = 2 m, EI = 2,700 kN-m2 Calculation: Pab 2 vB   ( L  a 2  b2 ) 6 LEI ( F1 )(4 m)(2 m) (6 m) 2  (4 m) 2  (2 m) 2   (1.316872  10 3 m/kN)F1  2  6(6 m)(2,700 kN-m ) Elongation of steel rod (1) due to force F1. FL (5 m) 1  1 1  F1  (221.0485  10 6 m/kN)F1 2 A1E1 (113.097336 mm )(200,000 N/mm 2 )(1 kN/1,000 N) The deflection of the wood beam at B will not equal zero in this instance because the rod that supports the beam at B will elongate, thus permitting the wood beam to deflect downward. Therefore, vB  1  (221.0485  106 m/kN)F1

Compatibility equation at B: The sum of the downward deflection caused by the uniformly distributed load and the upward deflection caused by the force in rod (1) must equal the elongation of the steel rod. Elongation of the steel rod will produce a downward (i.e., negative) deflection of the wood beam at B.

38.02469  103 m  (1.316872  10 3 m/kN)F1  (221.0485  10 6 m/kN)F1 38.02469  103 m  F1   24.72474 kN  24.7 kN (T) 1.537921  103 m/kN

Ans.


(b) Determine maximum bending stress in wood beam: Maximum bending moment magnitude Mmax = 11.6270 kN-m Bending stress at maximum moment (11.6270 kN-m)(300 mm/2)(1,000) 2 x  225  106 mm4

 7.75 MPa

Ans.

(c) Beam deflection at B. The beam deflection at B is equal to the elongation of the steel rod: FL vB  1   1 1 A1E1



(24.72474 kN)(5 m)(1,000 N/kN)(1,000 mm/m) (113.097336 mm 2 )(200,000 N/mm 2 )

 5.465367 mm  5.47 mm 

Ans.


11.65 A W360 × 72 structural steel [E = 200 GPa] wide-flange shape is loaded and supported as shown in Fig. P11.65. The beam is supported at B by a timber [E = 12 GPa] post having a cross-sectional area of 20,000 mm2. After a uniformly distributed load of 50 kN/m is applied to the beam, determine: (a) the reactions at supports A, B, and C. (b) the maximum bending stress in the beam. (c) the deflection of the beam at B. Fig. P11.65

Solution Section properties: W360 × 72: I beam  201  106 mm 4 S beam  1,150  10 mm 3

Post (1):

d  351 mm


3

A1  20,000 mm 2

E1  12,000 MPa

(a) Reactions at supports A, B, and C. The reaction force from post (1) will be taken as the redundant, leaving a simply supported beam as the released beam. To be consistent with earlier sign conventions (e.g., Chapter 5), we will assume that the force in the axial member is tension (even though intuitively we recognize that the post must be in compression).

Downward deflection of W360 × 72 beam at B due to 50 kN/m uniformly distributed load. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa3 vB   (4L2  7aL  3a 2 ) 24 LEI Values: P = 50 kN/m, L = 13 m, a = 6 m, EI = 40,200 kN-m2 Calculation: wa 3 vB   (4 L2  7aL  3a 2 ) 24 LEI (50 kN/m)(6 m)3  4(13 m) 2  7(6 m)(13 m)  3(6 m) 2   204.937  10 3 m  2  24(13 m)(40, 200 kN-m )

Since we are assuming that a tension force exists in post (1), the tension force from the post will cause a downward deflection of W360 × 72 beam at B. [Appendix C, SS beam with concentrated load not at midspan.] Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Relevant equation from Appendix C: Pab 2 vB   ( L  a 2  b2 ) 6 LEI Values: P = F1, L = 13 m, a = 6 m, b = 7 m, EI = 40,200 kN-m2 Calculation: Pab 2 vB   ( L  a 2  b2 ) 6 LEI ( F1 )(6 m)(7 m) (13 m) 2  (6 m) 2  (7 m) 2   (1.12514  10 3 m/kN)F1  2  6(13 m)(40, 200 kN-m ) Deformation of wood post (1) due to force F1. FL (5 m) 1  1 1  F1  (20.8333  10 6 m/kN)F1 2 2 A1E1 (20,000 mm )(12,000 N/mm )(1 kN/1,000 N) The deflection of the W360 × 72 beam at B will not equal zero in this instance because the post deforms. If we are consistent and assume that there is tension in the post, then the steel beam must deflect upward at B. Therefore, vB  1  (20.8333  106 m/kN)F1

Compatibility equation at B: The sum of the downward deflection caused by the 50 kN/m uniformly distributed load and the downward deflection caused by the force in post (1) must equal the deformation of the wood post.

204.937  103 m  (1.12514  103 m/kN)F1  (20.8333  10 6 m/kN)F1  F1 

204.937  103 m  178.832 kN  178.8 kN (C)  By 1.14598  103 m/kN

Ans.

Equilibrium equations for entire beam: M A   F1 (6 m)  C y (13 m)  (50 kN/m)(6 m)(3 m)  0  Cy 

(50 kN/m)(6 m)(3 m)  ( 178.832 kN)(6 m)  13.3071 kN  13.31 kN  13 m

Ans.

Fy  Ay  F1  C y  (50 kN/m)(6 m)  0

 Ay  (50 kN/m)(6 m)  ( 178.832 kN)  ( 13.3071 kN)  134.475 kN  134.5 kN 

Ans.


(b) Determine maximum bending stress in the W360 × 72 beam: Maximum bending moment magnitude Mmax = 180.836 kN-m Bending stress at maximum moment (180.836 kN-m)(351 mm/2)(1,000) 2 x  201  106 mm4 Ans.  157.9 MPa or using the tabulated value for the section modulus of the W360 × 72 beam: (180.836 kN-m)(1,000) 2 x  1,150  103 mm3  157.2 MPa

Ans.

(c) Beam deflection at B. The beam deflection at B is equal to the deformation of the wood post: FL (  178.832 kN)(5 m)(1,000 N/kN)(1,000 mm/m) vB   1  1 1  A1E1 (20,000 mm 2 )(12,000 N/mm 2 )  3.7257 mm  3.73 mm 

Ans.


11.66 A timber [E = 1,800 ksi] beam is loaded and supported as shown in Fig. P11.66. The cross section of the timber beam is 4-in. wide and 8-in. deep. The beam is supported at B by a ¾-in.-diameter aluminum [E = 10,000 ksi] rod, which has no load before the distributed load is applied to the beam. After a distributed load of 800 lb/ft is applied to the beam, determine: (a) the force carried by the aluminum rod. (b) the maximum bending stress in the timber beam. (c) the deflection of the beam at B. Fig. P11.66


I beam 

Rod (1):

A1 

 4

(4 in.)(8 in.)3  170.6667 in.4 12

Ebeam  1,800,000 psi

(0.75 in.) 2  0.441786 in.2

E1  10,000,000 psi

(a) Force carried by the aluminum rod. The reaction force from rod (1) will be taken as the redundant, leaving a cantilever beam between A and C as the released beam. For this analysis, a tension force is assumed to exist in axial member (1). Downward deflection of wood beam at B due to 800 lb/ft uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wx 2 vB   (6 L2  4 Lx  x 2 ) (elastic curve) 24 EI Values: w = 800 lb/ft, L = 16 ft, x = 12 ft, EI = 307.2×106 lb-in.2 Calculation: wx 2 vB   (6 L2  4 Lx  x 2 ) 24 EI (800 lb/ft)(12 ft) 2 (12 in./ft)3 6(16 ft) 2  4(16 ft)(12 ft)  (12 ft) 2   24.624 in.  24(307.2  106 lb-in.2 ) 


Upward deflection of wood beam at B due to force F1 in rod (1). [Appendix C, SS beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vB   3EI Values: P = −F1, L = 12 ft, EI = 307.2×106 lb-in.2 Calculation: PL3 vB   3EI ( F1 )(12 ft)3 (12 in./ft)3   (3.24  103 in./lb)F1 6 2 3(307.2  10 lb-in. ) Elongation of aluminum rod (1) due to force F1. FL (14 ft)(12 in./ft) 1  1 1  F1  (38.02746  10 6 in./lb)F1 2 6 A1E1 (0.441786 in. )(10  10 psi) The deflection of the wood beam at B will not equal zero in this instance because the rod that supports the beam at B will elongate, thus permitting the wood beam to deflect downward. Therefore, vB  1  (38.02746  106 in./lb)F1

Compatibility equation at B: The sum of the downward deflection caused by the uniformly distributed load and the upward deflection caused by the force in rod (1) must equal the elongation of the aluminum rod. Elongation of the aluminum rod will produce a downward (i.e., negative) deflection of the wood beam at B.

24.624 in.  (3.24  103 in./lb)F1  (38.02746  106 in./lb)F1  F1 

24.624 in.  7,511.835 lb  7,510 lb (T) 3.278027  103 in./lb

Ans.


(b) Determine maximum bending stress in wood beam: Maximum bending moment magnitude Mmax = 12,258 lb-ft Bending stress at maximum moment (12, 258 lb-ft)(8 in./2)(12 in./ft) x  170.6667 in.4  3, 447.56 psi  3, 450 psi

Ans.

(c) Beam deflection at B. The beam deflection at B is equal to the elongation of the aluminum rod: FL vB  1   1 1 A1E1



(7,511.835 lb)(14 ft)(12 in./ft) (0.441786 in.2 )(10  106 psi)

 0.28566 in.  0.286 in. 

Ans.


11.67 A W530 × 66 structural steel [E = 200 GPa] wide-flange shape is loaded and supported as shown in Fig. P11.67. A uniformly distributed load of 70 kN/m is applied to the beam, causing the roller support at B to settle downward (i.e., displace downward) by 10 mm. Determine: (a) the reactions at supports A and B. (b) the maximum bending stress in the beam.

Fig. P11.67

Solution Section properties: W530  66: I  351  106 mm4

d  526 mm

S  1,340  103 mm3

(a) Reactions at supports A and B. The reaction force at B will be taken as the redundant, leaving a cantilever beam between A and C as the released beam. Downward deflection of beam at B due to 70 kN/m uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wx 2 vB   (6 L2  4 Lx  x 2 ) (elastic curve) 24 EI Values: w = 70 kN/m, L = 6 m, x = 4.5 m, EI = 70,200 kN-m2 Calculation: wx 2 vB   (6 L2  4 Lx  x 2 ) 24 EI (70 kN/m)(4.5 m) 2 6(6 m) 2  4(6 m)(4.5 m)  (4.5 m) 2   0.107903 m  24(70, 200 kN-m 2 )  Upward deflection of beam at B due to reaction force By. [Appendix C, SS beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vB   3EI Values: P = −By, L = 4.5 m, EI = 70,200 kN-m2 Calculation: PL3 vB   3EI (  By )(4.5 m)3   (432.6923  10 6 m/kN)By 3(70, 200 kN-m 2 ) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Compatibility equation at B: The sum of the downward deflection caused by the uniformly distributed load and the upward deflection caused by the reaction force By must equal the support settlement:

0.107903 m  (432.6923  106 m/kN)By  0.010 m  By 

97.90264  103 m  226.2639 kN  226 kN  432.6923  106 m/kN

Ans.

Equilibrium equations for entire beam: M A   M A  By (4.5 m)  (70 kN/m)(6 m)(3 m)  0  M A  (226.2639 kN)(4.5 m)  (70 kN/m)(6 m)(3 m)  241.8125 kN-m  242 kN-m (ccw)

Ans.

Fy  Ay  By  (70 kN/m)(6 m)  0

 Ay  (70 kN/m)(6 m)  226.2639 kN  193.7361 kN  193.7 kN 

Ans.

(b) Determine maximum bending stress in beam: Maximum bending moment magnitude Mmax = 241.8125 kN-m (at A) Bending stress at maximum moment (241.8125 kN-m)(526 mm/2)(1,000) 2 x  351  106 mm 4 Ans.  181.187 MPa  181.2 MPa or using the tabulated value for the section modulus: (241.8125 kN-m)(1,000)2 x  1,340  103 mm3  180.457 MPa  180.5 MPa

Ans.


Philpot MoM 2nd Ch07-11 ISM

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