Mechanics of Materials Solutions Chapter11 Probs35 46

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11.35 The beam shown in Fig. P11.35consists of a W360 × 79 structural steel

wide-flange shape [ E = 200 GPa; I = 225 ×

106 mm

4]. 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. P11.35

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:3

2 2(4 7 3 )

24

B

wav L aL a

LEI

= − − +

Values: w = 80 kN/m, L = 9 m, a = 6 m

Calculation:3

2 2

3 32 2

(4 7 3 )24

(80 kN/m)(6 m) 4,320 kN-m4(9 m) 7(6 m)(9 m) 3(6 m)

24(9 m)

B

wav L aL a

LEI

EI 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:

2 2(2 3 ) (elastic curve)6

B

M xv L Lx x

LEI = − − +

Values: M = 240 kN-m, L = 9 m, x = 3 m

Calculation:

2 2

32 2

(2 3 )6

(240 kN-m)(3 m) 1,200 kN-m2(9 m) 3(9 m)(3 m) (3 m)

6(9 m)

B

M x

v L Lx x LEI

EI EI

= − − +

⎡ ⎤= − − + = −⎣ ⎦





Consider upward deflection of simply supported beam at B due to concentrated load R B.

[Appendix C, SS beam with concentrated load not at midspan.]


2 2 2( )6

B

Pabv L a b

LEI = − − −

Values: P = − R B, L = 9 m, a = 3 m, b = 6 m

Calculation:

2 2 2

32 2 2

( )6

( )(3 m)(6 m) (12 m )(9 m) (3 m) (6 m)

6(9 m)

B

B B

Pabv L a b

LEI

R R

EI EI

= − − −

−⎡ ⎤= − − − =⎣ ⎦

Compatibility equation for deflection at B:3 3 3

3

3

4,320 kN-m 1, 200 kN-m (12 m )

0

5,520 kN-m460 kN 460 kN

12 m

B

B

R

EI EI EI

R

− − + =

∴ = = = ↑ Ans.

Equilibrium equations for entire beam:

(3 m) (9 m) 240 kN-m (80 kN/m)(6 m)(6 m) 0 A B C M R RΣ = + − − =

240 kN-m (80 kN/m)(6 m)(6 m) (460 kN)(3 m)

9 m

193.3333 kN 193.3 kN

C R + −

∴ =

= = ↑ Ans.

(80 kN/m)(6 m) 0 y A B C F R R RΣ = + + − =

(80 kN/m)(6 m) 460 kN 193.3333 kN 173.3333 kN 173.3 kN A R∴ = − − = − = ↓ Ans.





Shear-force and bending-moment diagrams

(b) Magnitude of maximum bending stress:

Section properties (from Appendix B):6 4

3 3

225 10 mm 353 mm

1, 270 10 mm

I d

S

= × =

= ×

Maximum bending moment magnitude

M max = 280 kN-m

Bending stresses at maximum moment2

6 4

(280 kN-m)(353 mm/2)(1,000)

225 10 mm

220 MPa

xσ =×

= Ans.

or using the tabulated section modulus value:2

3 3(280 kN-m)(1,000)

1, 270 10 mm

220 MPa

xσ =×

= Ans.





11.36 The beam shown in Fig. P11.36consists of a W610 × 140 structural steel

wide-flange shape [ E = 200 GPa; I = 1,120

× 106 mm

4]. 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. P11.36

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.]


3 2 3( 2 ) (elastic curve)24

B

wxv L Lx x

EI = − − +

Values: w = 90 kN/m, L = 7.5 m, x = 1.5 m

Calculation:

3 2 3

33 2 3

( 2 )24

(90 kN/m)(1.5 m) 2,202.1875 kN-m(7.5 m) 2(7.5 m)(1.5 m) (1.5 m)

24

B

wxv L Lx x

EI

EI EI

= − − +

⎡ ⎤= − − + = −⎣ ⎦

Consider downward deflection of simply supported beam at B due to concentrated load.[Appendix C, SS beam with concentrated load not at midspan.]


2 2 2( ) (elastic curve)6

B

Pbxv L b x

LEI = − − −

Values: P = 160 kN, L = 7.5 m, b = 2.5 m, x = 1.5 m

Calculation:

2 2 2

32 2 2

( )6

(160 kN)(2.5 m)(1.5 m) 636.6667 kN-m(7.5 m) (2.5 m) (1.5 m)

6(7.5 m)

B

Pbxv L b x

LEI

EI EI

= − − −

⎡ ⎤= − − − = −⎣ ⎦





Consider upward deflection of simply supported beam at B due to concentrated load R B.



2 2 2( )6

B

Pabv L a b

LEI = − − −

Values: P = − R B, L = 7.5 m, a = 1.5 m, b = 6 m

Calculation:

2 2 2

32 2 2

( )6

( )(1.5 m)(6 m) (3.6 m )(7.5 m) (1.5 m) (6 m)

6(7.5 m)

B

B B

Pabv L a b

LEI

R R

EI EI

= − − −

−⎡ ⎤= − − − =⎣ ⎦


3

3

2,202.1875 kN-m 636.6667 kN-m (3.6 m )

0

2,838.8542 kN-m788.5706 kN 789 kN

3.6 m

B

B

R

EI EI EI

R

− − + =

∴ = = = ↑ Ans.


(1.5 m) (7.5 m) (90 kN/m)(7.5 m)(3.75 m) (160 kN)(5 m) 0 A B D M R RΣ = + − − =

(90 kN/m)(7.5 m)(3.75 m) (160 kN)(5 m) (788.5706 kN)(1.5 m)

7.5 m

286.4525 kN 286 kN

D R + −

∴ =

= = ↑ Ans.

(90 kN/m)(7.5 m) 160 kN 0 y A B D F R R RΣ = + + − − =

(90 kN/m)(7.5 m) 160 kN 788.5706 kN 286.4525 kN

240.0231 kN 240 kN

A R∴ = + − −

= − = ↓ Ans.








3 3

1,120 10 mm 617 mm

3, 640 10 mm

I d

S

= × =

= ×

Maximum bending moment magnitude M max = 461.28 kN-m


6 4

(461.28 kN-m)(617 mm/2)(1,000)

1,120 10 mm

127.1 MPa

xσ =×

= Ans.

or using the tabulated section modulus value:2

3 3

(461.28 kN-m)(1,000)

3, 640 10 mm

126.8 MPa

x

σ =

×

= Ans.





11.37 A propped cantilever beam is loadedas shown in Fig. P11.37. Determine the

reactions at A and D for the beam. Assume

EI = 12.8 × 106 lb-in.

2.

Fig. P11.37

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:4 3

and8 6

C C

wL wLv

EI EI θ = − = −

Values:

w = 20 lb/in., L = 72 in.

Calculation:4 4 3

3 3 2

3 2 3

(20 lb/in.)(48 in.) 13,271,040 lb-in.

8 8

(20 lb/in.)(48 in.) 368,640 lb-in.

6 6

13,271,040 lb-in. 368,640 lb-in. 22,118,400 lb-in.(24 in.)

C

C

D

wLv

EI EI EI

wL

EI EI EI

v EI EI EI

θ

= − = − = −

= − = − = −

⎛ ⎞= − − = −⎜ ⎟

⎝ ⎠

Consider downward deflection of cantilever beam at D due to the 320-lb concentrated load.

[Appendix C, Cantilever beam with concentrated load at tip.]


and3 2

B B

PL PLv

EI EI θ = − = −

Values: P = 320 lb, L = 24 in.

Calculation:3 3 3

2 2 2

3 2 3

(320 lb)(24 in.) 1,474,560 lb-in.3 3

(320 lb)(24 in.) 92,160 lb-in.

2 2

1,474,560 lb-in. 92,160 lb-in. 5,898,240 lb-in.(48 in.)

B

B

D

PLv EI EI EI

PL

EI EI EI

v EI EI EI

θ

= − = − = −

= − = − = −

= − − = −





Consider upward deflection of cantilever beam at D due to concentrated load R D.



3 D

PLv

EI = −

Values:

P = − R D, L = 72 in.

Calculation:3 3 3( )(72 in.) (124,416 in. )

3 3

D D D

PL R Rv

EI EI EI

−= − = − =

Compatibility equation for deflection at C :3 3 3

3

3

22,118,400 lb-in. 5,898,240 lb-in. (124,416 in. )0

28,016,640 lb-in.225.1852 lb 225 lb

124,416 in.

D

D

R

EI EI EI

R

− − + =

∴ = = = ↑

Ans.



(20 lb/in.)(48 in.) 320 lb 0

(20 lb/in.)(48 in.) 320 lb 225.1852 lb 1,054.8148 lb 1,055 lb

y A D

A

F R R

R

Σ = + − − =

∴ = + − = = ↑ Ans.

(20 lb/in.)(48 in.)(24 in.) (320 lb)(24 in.) (72 in.) 0 A A D M M RΣ = − − − + =

(225.1852 lb)(72 in.) (20 lb/in.)(48 in.)(24 in.) (320 lb)(24 in.)

14,506.6667 lb-in. 14,510 lb-in. (ccw)

A M ∴ = − −

= − = Ans.





11.38 A propped cantilever beam is loaded asshown in Fig. P11.38. Assume EI = 24 × 10

6

kip-in.2. Determine:

(a) the reactions at B and C for the beam.(b) the beam deflection at A.

Fig. P11.38

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.



8 B

wLv

EI = −

Values:

w = 6 kips/ft, L = 24 ft

Calculation:4 4 3(6 kips/ft)(24 ft) 248,832 kip-ft

8 8 B

wLv

EI EI EI = − = − = − (a)

Consider downward deflection of cantilever beam at B due to the 30-kip concentrated load.

[Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C:

2

(3 ) (elastic curve)6 B

Pxv L x EI = − −

Values: P = 30 kips, L = 36 ft, x = 24 ft

Calculation:

[ ]2 2 3(30 kips)(24 ft) 241,920 kip-ft

(3 ) 3(36 ft) (24 ft)6 6

B

Pxv L x

EI EI EI = − − = − − = −





Consider upward deflection of cantilever beam at B due to concentrated load R B.



3 B

PLv

EI = −

Values:

P = − R B, L = 24 ft

Calculation:3 3 3( )(24 ft) (4,608 ft )

3 3

B B B

PL R Rv

EI EI EI

−= − = − = (b)


3

3

248,832 kip-ft 241,920 kip-ft (4,608 ft )0

490,752 kip-ft106.500 kips 106.5 kips

4,608 ft

B

B

R

EI EI EI

R

− − + =

∴ = = = ↑ Ans.


30 kips (6 kips/ft)(24 ft) 0

30 kips (6 kips/ft)(24 ft) 106.500 kips 67.5 kips

y B C

C

F R R

R

Σ = + − − =

∴ = + − = ↑ Ans.

(30 kips)(36 ft) (6 kips/ft)(24 ft)(12 ft) (24 ft) 0C C B M M RΣ = + + − =

(106.500 kips)(24 ft) (30 kips)(36 ft) (6 kips/ft)(24 ft)(12 ft)

252.0 kip-ft 252 kip-ft (cw)

C M ∴ = − −

= − = Ans.

(b) Beam deflection at A:

Consider downward deflection of cantilever beam at A due to uniformly distributed load.



and (slope magnitude)8 6

B B

wL wLv

EI EI θ = − =

Values: w = 6 kips/ft, L = 24 ft, EI = 24 × 10

6 kip-in.

2





Calculation:3

3 3 2

3 2 3

248,832 kip-ftcalculated previously in Eq. (a)

(6 kips/ft)(24 ft) 13,824 kip-ft

6 6

248,832 kip-ft 13,824 kip-ft 414,720 kip-ft(12 ft)

B

B

A

v EI

wL

EI EI EI

v

EI EI EI

θ

= −

= = =

⎛ ⎞= − − = −⎜ ⎟

⎝ ⎠

Consider downward deflection of cantilever beam at A due to the 30-kip concentrated load.


3

3 A

PLv

EI = −

Values: P = 30 kips, L = 36 ft, EI = 24 × 10

6 kip-in.

2

Calculation:3 3 3(30 kips)(36 ft) 466,560 kip-ft

3 3 A

PLv

EI EI EI = − = − = −

Consider upward deflection of cantilever beam at B due to concentrated load R B.

[Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C:

3 2


B B

PL PLv

EI EI θ = − =

Values:

P = − R B = 106.5 kips, L = 24 ft, EI = 24 × 106

kip-in.2

Calculation:3 3

2 2 2

3 2 3

(4,608 ft )(106.5 kips) 490,752 kip-ftusing the results from Eq. (b)

(106.5 kips)(24 ft) 30,672 kip-ft

2 2

490,752 kip-ft 30,672 kip-ft 858,816 kip-ft(12 ft)

B

B

A

v EI EI

PL

EI EI EI

v

EI EI EI

θ

= =

= = =

⎛ ⎞= + =⎜ ⎟

⎝ ⎠

Beam deflection at A.3 3 3

3 3 3

6 2

414,720 kip-ft 466,560 kip-ft 858,816 kip-ft

22,464 kip-ft (22,464 kip-ft )(12 in./ft)1.617408 in. 1.617 in.

24 10 kip-in.

Av EI EI EI

EI

= − − +

= − = − = − = ↓×

Ans.





11.39 A propped cantilever beam is loadedas shown in Fig. P11.39. Assume EI = 86.4

× 106 N-mm

2. Determine:

(a) the reactions at A and C for the beam.(b) the beam deflection at B.

Fig. P11.39

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.




C

wxv L Lx x

EI = − − +

Values:

w = 25 N/mm, L = 400 mm, x = 300 mm, EI = 86.4 × 10

6 N-mm

2

Calculation:2

2 2

22 2

9 3

(6 4 )24

(25 N/mm)(300 mm)6(400 mm) 4(400 mm)(300 mm) (300 mm)

24

53.43750 10 N-mm

C

wxv L Lx x

EI

EI

EI

= − − +

⎡ ⎤= − − +⎣ ⎦

×= −

Consider downward deflection of cantilever beam at C due to the 4,000-N concentrated load.




B B

PL PLv

EI EI θ = − =

Values: P = 4,000 N, L = 120 mm, EI = 86.4 × 10

6 N-mm

2

Calculation: 3 3 9 3

2 2 7 2

9 3 7 2 9 3

(4, 000 N)(120 mm) 2.304 10 N-mm

3 3

(4,000 N)(120 mm) 2.88 10 N-mm

2 2

2.304 10 N-mm 2.88 10 N-mm 7.488 10 N-mm(180 mm)

B

B

C

PLv

EI EI EI

PL

EI EI EI

v EI EI EI

θ

×= − = − = −

×= = =

⎛ ⎞× × ×= − − = −⎜ ⎟

⎝ ⎠





Consider upward deflection of cantilever beam at C due to concentrated load RC .


3

3C

PLv

EI = −

Values: P = − RC , L = 300 mm, EI = 86.4 × 10

6 N-mm

2

Calculation:3 3 6 3( )(300 mm) (9 10 mm )

3 3

C C C

PL R Rv

EI EI EI

− ×= − = − =

Compatibility equation for deflection at C :9 3 9 3 6 3

9 3

6 3

53.43750 10 N-mm 7.488 10 N-mm (9 10 mm )0

60.9255 10 N-mm

6,769.5 N 6,770 N9 10 mm

C

C

R

EI EI EI

R

× × ×− − + =

×

∴ = = = ↑× Ans.


4,000 N (25 N/mm)(400 mm) 0

4,000 N (25 N/mm)(400 mm) 6,769.5 N 7,230.5 N 7,230 N

y A C

A

F R R

R

Σ = + − − =

∴ = + − = = ↑ Ans.

(4,000 N)(120 mm) (25 N/mm)(400 mm)(200 mm) (300 mm) 0 A A C M M RΣ = − − − + =

(6,769.5 N)(300 mm) (4,000 N)(120 mm) (25 N/mm)(400 mm)(200 mm)

449,150 N-mm 449,000 N-mm (ccw)

A M ∴ = − −

= − = Ans.

(b) Beam deflection at B:

Consider downward deflection of cantilever beam at B due to uniformly distributed load.




B

wxv L Lx x

EI = − − +

Values: w = 25 N/mm, L = 400 mm, x = 120 mm,

EI = 86.4 × 106 N-mm

2





Calculation:2

2 2

22 2

9 3

(6 4 )24

(25 N/mm)(120 mm)6(400 mm) 4(400 mm)(120 mm) (120 mm)

24

11.736 10 N-mm

B

wxv L Lx x

EI

EI

EI

= − − +

⎡ ⎤= − − +⎣ ⎦

×= −

Consider downward deflection of cantilever beam at B due to the 4,000-N concentrated load.


3

3 B

PLv

EI = −

Values: P = 4,000 N, L = 120 mm, EI = 86.4 × 10

6 N-mm

2

Calculation:

3 3 9 3(4, 000 N)(120 mm) 2.304 10 N-mm3 3

B PLv EI EI EI

×= − = − = −

Consider upward deflection of cantilever beam at B due to concentrated load RC .



(3 )6

B

Pxv L x

EI = − −

Values:

P = − RC = −6,769.5 N, L = 300 mm, x = 120 mm,

EI = 86.4 × 106

N-mm2

Calculation:

[ ]2 2

9 3

( 6,769.5 N)(120 mm)(3 ) 3(300 mm) (120 mm)

6 6

12.6725 10 N-mm

B

Pxv L x

EI EI

EI

−= − − = − −

×=

Beam deflection at B.9 3 9 3 9 3

9 3 9 3

6 2

11.736 10 N-mm 2.304 10 N-mm 12.6725 10 N-mm

1.367496 10 N-mm 1.367496 10 N-mm15.8275 mm 15.83 mm

86.4 10 N-mm

Bv EI EI EI

EI

× × ×= − − +

× ×= − = − = − = ↓

× Ans.





11.40 The beam shown in Fig. P11.40 consistsof a W610 × 82 structural steel wide-flange

shape [ E = 200 GPa; I = 562 × 106 mm

4]. For

the loading shown, determine:

(a) the reaction force at C .(b) the beam deflection at A.

Fig. P11.40

Solution

(a) Reaction force at C . Choose the reaction force at C as the redundant; therefore, the released beamis 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.]



C

M xv L Lx x

LEI = − − +

Values: M = −(80 kN/m)(3 m)(1.5 m) = −360 kN-m,

L = 14 m, x = 7 m

Calculation:

2 2

32 2

(2 3 )6

( 360 kN-m)(7 m) 4,410 kN-m2(14 m) 3(14 m)(7 m) (7 m)

6(14 m)

C

M xv L Lx x

LEI

EI 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:

45

384C

wLv

EI = −

Values:

w = 80 kN/m, L = 14 m

Calculation:4 4 35 5(80 kN/m)(14 m) 40,016.6667 kN-m

384 384C

wLv

EI EI EI = − = − = −





Consider upward deflection of simply supported beam at C due to concentrated load RC .

[Appendix C, SS beam with concentrated load at midspan.]


48C

PLv

EI = −

Values:

P = − RC , L = 14 m

Calculation:3 3 3( )(14 m) (57.1667 m )

48 48

C C C

PL R Rv

EI EI EI

−= − = − =


3

3

4,410 kN-m 40,016.6667 kN-m (57.1667 m )0

35,606.6667 kN-m622.8571 kN 623 kN

57.1667 m

C

C

R

EI EI EI

R

− + =

∴ = = = ↑ 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.]


8 A

wLv

EI = −

Values: w = 80 kN/m, L = 3 m

Calculation: 4 4 3(80 kN/m)(3 m) 810 kN-m

8 8 A

wLv

EI EI 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.]


3 B

L

EI θ = (slope magnitude)

Values:

M = (80 kN/m)(3 m)(1.5 m) = 360 kN-m, L = 14 m

Computation:2

2 3

(360 kN-m)(14 m) 1,680 kN-m

3 3

1,680 kN-m 5,040 kN-m(3 m)

B

A

ML

EI EI EI

v EI EI

θ = = =

⎛ ⎞= − = −⎜ ⎟

⎝ ⎠





Consider upward deflection at A due to uniformly distributed load between B and D.

[Appendix C, SS beam with uniformly distributed load.]


(slope magnitude)24

B

wL

EI θ =

Values: w = 80 kN/m, L = 14 m

Calculation:3 3 2

2 3

(80 kN/m)(14 m) 9,146.6667 kN-m

24 24

9,146.6667 kN-m 27,440 kN-m(3 m)

B

A

wL

EI EI EI

v EI EI

θ = = =

⎛ ⎞= =⎜ ⎟

⎝ ⎠

Consider downward deflection at A due to concentrated load RC .

[Appendix C, SS beam with concentrated load at midspan.]


(slope magnitude)16

B

PL

EI θ =

Values: P = − RC = −622.8571 kN, L = 14 m

Calculation:2 2 2

2 3

(622.8571 kN)(14 m) 7,630 kN-m

16 16

7,630 kN-m 22,890 kN-m

(3 m)

B

A

PL

EI EI EI

v EI EI

θ = = =

⎛ ⎞= − = −

⎜ ⎟⎝ ⎠

Beam deflection at A.3 3 3 3

3 3

2

810 kN-m 5, 040 kN-m 27, 440 kN-m 22,890 kN-m

1,300 kN-m 1,300 kN-m0.011566 m 11.57 mm

112,400 kN-m

Av EI EI EI EI

EI

= − − + −

− −= = = − = ↓ Ans.





11.41 The beam shown in Fig. P11.41consists of a W8 × 15 structural steel wide-

flange shape [ E = 29,000 ksi; I = 48 in.4].

For the loading shown, determine:(a) the reactions at A and B.


stress in the beam.

( Reminder: The roller symbol implies that

both upward and downward displacement isrestrained.)

Fig. P11.41

Solution


Consider deflection of cantilever beam at B due to uniformly distributed load over entire beam

span. [Appendix C, Cantilever beam with uniformly distributed load.]


2 2

(6 4 ) (elastic curve)24 B

wx

v L Lx x EI = − − +

Values: w = −80 lb/in., L = 150 in., x = 100 in.

Calculation:2

2 2

22 2

9 3

(6 4 )24

( 80 lb/in.)(100 in.)6(150 in.) 4(150 in.)(100 in.) (100 in.)

24

2.8333 10 lb-in.

B

wxv L Lx x

EI

EI

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:

3

3 B

PLv

EI = −

Values: P = −½(50 in.)(60 lb/in.) = −1,500 lb, L = 100 in.

Calculation:3 3 6 3( 1,500 lb)(100 in.) 500 10 lb-in.

3 3 B

PLv

EI EI 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.]


2 B

Lv

EI = −

Values:

M = −½(50 in.)(60 lb/in.)[⅔(50 in.)]= −50,000 lb-in., L = 100 in.

Calculation:2 2 6 3( 50,000 lb-in.)(100 in.) 250 10 lb-in.

2 2 B

MLv

EI EI EI

− ×= − = − =

Consider deflection of cantilever beam at B due to concentrated load R B.



3 B

PLv

EI = −

Values: P = − R B, L = 100 in.

Calculation:3 3 3( )(100 in.) (333.3333 in. )

3 3

B B B

PL R Rv

EI EI EI

−= − = − =

Compatibility equation for deflection at B:9 3 6 3 6 3 3 3

9 3

3 3

2.8333 10 lb-in. 500 10 lb-in. 250 10 lb-in. (333.3333 10 in. ) 0

3.5833 10 lb-in.10,750 lb 10,750 lb

333.3333 10 in.

B

B

R EI EI EI EI

R

× × × ×+ + + =

×∴ = − = − = ↓

× Ans.


1(80 lb/in.)(150 in.) (140 lb/in. 80 lb/in.)(50 in.) 0

2

(60 lb/in.)(50 in.)(80 lb/in.)(150 in.) ( 10,750 lb)

2

2,750 lb 2,750 lb

y A B

A

F R R

R

Σ = + + + − =

∴ = − − − −

= − = ↓ Ans.





(80 lb/in.)(150 in.)(75 in.)

(140 lb/in. 80 lb/in.)(50 in.) 2100 in. (50 in.) (100 in.) 0

2 3

A A

B

M M

R

Σ = − +

−⎡ ⎤ ⎡ ⎤+ + + =⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦

(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)

A M ∴ = + + −

= = Ans.


Section properties (from Appendix B):4 348 in. 8.11 in. 11.8 in. I d S = = =


M max = 150,000 lb-in. (at B)

Bending stresses at maximum moment

4

(150,000 lb-in.)(8.11 in./2)12,571.875 psi 12,570 psi

48 in. x

σ = = = Ans.

or, using the tabulated value for the section modulus:

3

150,000 lb-in.12,711.864 psi 12,710 psi

11.8 in. xσ = = = Ans.





11.42 The beam shown in Fig. P11.42consists of a W24 × 94 structural steel wide-

flange shape [ E = 29,000 ksi; I = 2,700 in.4].

For the loading shown, determine:(a) the reactions at A and D.


stress in the beam.

Fig. P11.42

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.




B B

PL PLv

EI EI θ = − =

Values:

P = 50 kips, L = 20 ft

Calculation:3 3 3

2 2 3

3 3 3

(50 kips)(20 ft) 133,333.333 kip-ft

3 3

(50 kips)(20 ft) 10,000 kip-ft

2 2

133,333.333 kip-ft 10,000 kip-ft 193,333.333 kip-ft(6 ft)

B

B

A

PLv

EI EI EI

PL

EI EI EI

v EI EI EI

θ

= − = − = −

= = =

⎛ ⎞= − − = −⎜ ⎟

⎝ ⎠

Consider downward deflection of cantilever beam at A due to the uniformly distributed load.




C C

wL wLv

EI EI θ = − =

Values:

w = 4 kips/ft, L = 14 ft

Calculation:

4 4 3

3 3 3

3 3 3

(4 kips/ft)(14 ft) 19,208 kip-ft

8 8

(4 kips/ft)(14 ft) 1,829.333 kip-ft

6 6

19,208 kip-ft 1,829.333 kip-ft 41,160 kip-ft(12 ft)

C

C

A

wLv

EI EI EI

wL

EI EI EI

v EI EI EI

θ

= − = − = −

= = =

⎛ ⎞= − − = −⎜ ⎟

⎝ ⎠





Consider upward deflection of cantilever beam at A due to concentrated load R A.



3 A

PLv

EI = −

Values:

P = − R A, L = 26 ft

Calculation:3 3 3( )(26 ft) (5,858.667 ft )

3 3

A A A

PL R Rv

EI EI EI

−= − = − = (b)

Compatibility equation for deflection at A:3 3 3

3

3

193,333.333 kip-ft 41,160 kip-ft (5,858.667 ft )0

234,493.333 kip-ft40.025 kips 40.0 kips

5,858.667 ft

A

A

R

EI EI EI

R

− − + =

∴ = = = ↑ Ans.


50 kips (4 kips/ft)(14 ft) 0

50 kips (4 kips/ft)(14 ft) 40.025 kips 65.975 kips 66.0 kips

y A D

D

F R R

R

Σ = + − − =

∴ = + − = = ↑ Ans.

(50 kips)(20 ft) (4 kips/ft)(14 ft)(7 ft) (26 ft) 0 D D A M M RΣ = + + − =

(40.025 kips)(26 ft) (50 kips)(20 ft) (4 kips/ft)(14 ft)(7 ft)

351.350 kip-ft 351 kip-ft (cw)

D M ∴ = − −

= − = Ans.







Section properties (from Appendix B):4

3

2,700 in.

24.3 in.

222 in.

I

d

S

=

=

=


M max = 351.350 kip-ft


4

(351.350 kip-ft)(24.3 in./2)(12 in./ft)

2,700 in.

18.97 ksi

xσ =

= Ans.

or, using the tabulated section modulus:

3(351.350 kip-ft)(12 in./ft)

222 in.

18.99 ksi

xσ =

= Ans.





11.43 The solid 20-mm-diameter steel [ E = 200GPa] shaft shown in Fig. P11.43 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. P11.43

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:

2 2 2

( ) (elastic curve)6C

Pbx

v L b x LEI = − − −

Values: P = 750 N, L = 2,000 mm, b = 600 mm,

x = 1,000 mm

Calculation:

2 2 2

2 2 2

9 3

( )6

(750 N)(600 mm)(1,000 mm)(2,000 mm) (600 mm) (1,000 mm)

6(2,000 mm)

99.0 10 N-mm

C

Pbxv L b x

LEI

EI

EI

= − − −

⎡ ⎤= − − −⎣ ⎦

×= −

Consider downward deflection of simply supported beam at C due to pulley D load.



C

Pbxv L b x

LEI = − − −

Values:

P = 500 N, L = 2,000 mm, b = 400 mm, x = 1,000 mm





Calculation:

2 2 2

2 2 2

9 3

( )6

(500 N)(400 mm)(1,000 mm)(2,000 mm) (400 mm) (1,000 mm)

6(2,000 mm)

47.3333 10 N-mm

C

Pbxv L b x

LEI

EI

EI

= − − −

⎡ ⎤= − − −⎣ ⎦

×= −


[Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C:

3

48C

PLv

EI = −

Values: P = − RC , L = 2,000 mm

Calculation:3 3 6 3( )(2,000 mm) (166.6667 10 mm )

48 48

C C C

PL R Rv

EI EI EI

− ×= − = − =

Compatibility equation for deflection at C :9 3 9 3 6 3

9 3

6 3

99.0 10 N-mm 47.3333 10 N-mm (166.6667 10 mm )0

146.3333 10 N-mm878 N 878 N

166.6667 10 mm

C

C

R

EI EI EI

R

× × ×− − + =

×∴ = = = ↑

× Ans.


(750 N)(600 mm) (500 N)(1,600 mm) (1,000 mm) (2,000 mm) 0 A C E M R RΣ = − − + + =

(750 N)(600 mm) (500 N)(1,600 mm) (878 N)(1,000 mm)186 N

2,000 mm

E R + −

∴ = = ↑ Ans.

750 N 500 N 0

750 N 500 N 878 N 186 N 186 N

y A C E

A

F R R R

R

Σ = + + − − =

∴ = + − − = ↑ Ans.







Section properties:

4 4(20 mm) 7,853.9816 mm64

I π

= =


M max = 114,000 N-mm


4

(114,000 N-mm)(20 mm/2)

7,853.9816 mm

145.1 MPa

xσ =

= Ans.





11.44 The solid 1.00-in.-diameter steel [ E = 29,000 ksi] shaft shown in Fig. P11.44

supports three belt pulleys. Assume that

the bearing at A can be idealized as a pinsupport 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. P11.44

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.



2 2 2

( ) (elastic curve)6C

Pbx

v L b x LEI = − − −

Values: P = 200 lb, L = 60 in., b = 15 in., x = 30 in.

Calculation:

2 2 2

2 2 2

3

( )6

(200 lb)(15 in.)(30 in.)(60 in.) (15 in.) (30 in.)

6(60 in.)

618,750 lb-in.

C

Pbxv L b x

LEI

EI

EI

= − − −

⎡ ⎤= − − −⎣ ⎦

= −




C

Pbxv L b x

LEI = − − −


Calculation:2 2 2

2 2 2

3

( )6

(200 lb)(15 in.)(30 in.)(60 in.) (15 in.) (30 in.)

6(60 in.)

618,750 lb-in.

C

Pbxv L b x

LEI

EI

EI

= − − −

⎡ ⎤= − − −⎣ ⎦

= −





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:


C

M xv L Lx x

LEI = − − +

Values: M = −(120 lb)(10 in.) = −1,200 lb-in., L = 60 in., x = 30 in.

Calculation:

2 2

2 2

3

(2 3 )6

( 1,200 lb-in.)(30 in.)2(60 in.) 3(60 in.)(30 in.) (30 in.)

6(60 in.)

270,000 lb-in.

C

M xv L Lx x

LEI

EI

EI

= − − +

−⎡ ⎤= − − +⎣ ⎦

=


[Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C:

3

48C

PLv

EI = −

Values: P = − RC , L = 60 in.

Calculation:3 3 3( )(60 in.) (4,500 in. )

48 48

C C C

PL R Rv

EI EI EI

−= − = − =

Compatibility equation for deflection at C :3 3 3 3

3

3

618,750 lb-in. 618,750 lb-in. 270,000 lb-in. (4,500 in. )0

967,500 lb-in.215 lb 215 lb

4,500 in.

C

C

R

EI EI EI EI

R

− − + + =

∴ = = = ↑ Ans.






(200 lb)(15 in.) (200 lb)(45 in.) (120 lb)(70 in.) (30 in.) (60 in.) 0 A C E M R RΣ = − − − + + = (200 lb)(15 in.) (200 lb)(45 in.) (120 lb)(70 in.) (215 lb)(30 in.)

60 in.

232.50 lb 233 lb

E R + + −

∴ =

= = ↑ Ans.

200 lb 200 lb 120 lb 0

200 lb 200 lb 120 lb 215 lb 232.50 lb 72.5 lb

y A C E

A

F R R R

R

Σ = + + − − − =

∴ = + + − − = ↑ Ans.


(b) Magnitude of maximum bendingstress:

Section properties:

4 4(1.00 in.) 0.0490874 in.64

I π

= =


M max = 1,200 lb-in.


4

(1,200 lb-in.)(1.00 in./2)

0.0490874 in.

12,223.1 psi

12,220 psi

xσ =

=

= Ans.





11.45 The solid 1.00-in.-diameter steel [ E =29,000 ksi] shaft shown in Fig. P11.45

supports two belt pulleys. Assume that the

bearing at E can be idealized as a pinsupport 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. P11.45

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.]



C

M xv L Lx x

LEI = − − +

Values: M = −(110 lb)(7 in.) = −770 lb-in.,

L = 45 in., x = 15 in.

Calculation:

2 2

2 2

3

(2 3 )6

( 770 lb-in.)(15 in.)

2(45 in.) 3(45 in.)(15 in.) (15 in.)6(45 in.)

96,250 lb-in.

C

M xv L Lx x

LEI

EI

EI

= − − +

−

⎡ ⎤= − − +⎣ ⎦

=





C

Pbxv L b x

LEI = − − −






Calculation:

2 2 2

2 2 2

3

( )6

(260 lb)(15 in.)(15 in.)(45 in.) (15 in.) (15 in.)

6(45 in.)

341,250 lb-in.

C

Pbxv L b x

LEI

EI

EI

= − − −

⎡ ⎤= − − −⎣ ⎦

= −



2 2 2( )6

C

Pabv L a b

LEI = − − −

Values:

P = − RC , L = 45 in., a = 15 in., b = 30 in.

Calculation:2 2 2

2 2 2

3

( )6

( )(15 in.)(30 in.)(45 in.) (15 in.) (30 in.)

6(45 in.)

(1,500 in. )

C

C

C

Pabv L a b

LEI

R

EI

R

EI

= − − −

−⎡ ⎤= − − −⎣ ⎦

=

Compatibility equation for deflection at C :

3 3 3

3

3

96,250 lb-in. 341,250 lb-in. (1,500 in. ) 0

245,000 lb-in.163.3333 lb 163.3 lb

1,500 in.

C

C

R EI EI EI

R

− + =

∴ = = = ↑ Ans.


(260 lb)(15 in.) (110 lb)(52 in.) (45 in.) (30 in.) 0 E B C M R RΣ = + − − =

(260 lb)(15 in.) (110 lb)(52 in.) (163.3333 lb)(30 in.)

45 in.

104.8889 lb 104.9 lb

B R + −

∴ =

= = ↑ Ans.





110 lb 260 lb 0

110 lb 260 lb 163.3333 lb 104.8889 lb 101.7778 lb 101.8 lb

y B C E

E

F R R R

R

Σ = + + − − =

∴ = + − − = = ↑ Ans.



Section properties:

4 4(1.00 in.) 0.0490874 in.64

I π

= =


M max = 1,526.7 lb-in.


4

(1,526.7 lb-in.)(1.00 in./2)

0.0490874 in.15,550.8 psi

15,550 psi

xσ =

=

= Ans.





11.46 The beam shown in Fig. P11.46 consistsof a W360 × 101 structural steel wide-flange

shape [ E = 200 GPa; I = 301 × 106 mm

4]. For

the loading shown, determine:(a) the reactions at A and B.


stress in the beam.

Fig. P11.46

Solution


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:2


B

wxv L Lx x

EI = − − +

Values: w = 30 kN/m, L = 8 m, x = 5.5 m

Calculation:2

2 2

22 2

3

(6 4 )24

(30 kN/m)(5.5 m)6(8 m) 4(8 m)(5.5 m) (5.5 m)

24

9,008.828125 kN-m

B

wxv L Lx x

EI

EI

EI

= − − +

⎡ ⎤= − − +⎣ ⎦

= −

Consider deflection of cantilever beam at B due a linearly distributed load.

[Appendix C, Cantilever beam with linearly distributed load.]


3 2 2 30 (10 10 5 )120

B

w xv L L x Lx x

LEI = − − + − (elastic curve)

Values:

w0 = 60 kN/m, L = 8 m, x = 5.5 m

Calculation: 23 2 2 30

23 2 2 3

3

(10 10 5 )120

(60 kN/m)(5.5 m)10(8 m) 10(8 m) (5.5 m) 5(8 m)(5.5 m) (5.5 m)

120(8 m)

4,998.103516 kN-m

B

w xv L L x Lx x

LEI

EI

EI

= − − + −

⎡ ⎤= − − + −⎣ ⎦

= −





Consider deflection of cantilever beam at B due to concentrated load R B.


3

3 B

PLv

EI = −

Values: P = − R B, L = 5.5 m

Calculation:3 3 3( )(5.5 m) (55.458333 m )

3 3

B B B

PL R Rv

EI EI EI

−= − = − =


3

3

9,008.828125 kN-m 4,998.103516 kN-m (55.458333 m )0

14,006.93164 kN-m

252.56676 kN 253 kN55.458333 m

B

B

R

EI EI EI

R

− − + =

∴ = = = ↑ Ans.


1

(30 kN/m)(8 m) (90 kN/m 30 kN/m)(8 m) 02

1(30 kN/m)(8 m) (90 kN/m 30 kN/m)(8 m) 252.56676 kN

2

227.43324 kN 227 kN

y A B

A

F R R

R

Σ = + − − − =

∴ = + − −

= = ↑ Ans.

(90 kN/m 30 kN/m)(8 m) 8 m(30 kN/m)(8 m)(4 m) (5.5 m) 0

2 3 A A B M M R

−⎛ ⎞⎛ ⎞Σ = − − − + =⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

960 kN-m 640 kN-m (252.56676 kN)(5.5 m)

210.88281 kN-m 211 kN-m (ccw)

A M ∴ = − − +

= − = Ans.






3 3

301 10 mm

356 mm

1, 690 10 mm

I

d

S

= ×

=

= ×


M max = 210.88281 kN-m


6 4

(210.88281 kN-m)(356 mm/2)(1,000)

301 10 mm

124.7 MPa

xσ =×

=

or, using the tabulated value for the section

modulus: 2

3 3

(210.88281 kN-m)(1,000)

1, 690 10 mm

124.8 MPa

xσ =×

= Ans.

Mechanics of Materials Solutions Chapter11 Probs35 46

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