Mechanics of Materials Solutions Chapter11 Probs18 25

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11.18a For the beams and loadings shown below, assume that EI = 3.0 × 10

4 kN-m

2is

constant for each beam.

(a) For the beam in Fig. P11.18a, determine

the concentrated upward force P required tomake the total beam deflection at B equal to

zero (i.e., v B = 0).

Fig. P11.18a

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:3

2 2(4 7 3 )24

B

wav L aL a

LEI = − − +

Values: w = 15 kN/m, L = 7 m, a = 3.5 m,

EI = 3.0 × 104

kN-m2

Computation:3

2 2

3 32 2

(4 7 3 )24

(15 kN/m)(3.5 m) 234.472656 kN-m4(7 m) 7(3.5 m)(7 m) 3(3.5 m)

24(7 m)

B

wav L aL a

LEI

EI EI

= − − +

⎡ ⎤= − − + = −⎣ ⎦

Upward deflection at B due to concentrated load P .

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


48 B

PLv

EI =

Values: L = 7 m, EI = 3.0 × 10

4 kN-m

2

Computation:3 3 3(7 m) (7.145833 m )

48 48 B

PL P P v

EI EI EI = = =

Compatibility equation at B:3 3

3

3

234.472656 kN-m (7.145833 m )0

234.472656 kN-m32.8125 kN 32.8 kN

7.145833 m

P

EI EI

P

− + =

∴ = = = Ans.





11.18b For the beams and loadings shown below, assume that EI = 3.0 × 10

4 kN-m

2is


(b) For the beam in Fig. P11.18b, determinethe concentrated moment M required to

make the total beam slope at A equal to zero

(i.e., θ A = 0).

Fig. P11.18b

Solution

Slope at A due to 32 kN concentrated load.

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


2 A

PL

EI θ = (slope magnitude)

Values: P = 32 kN, L = 4 m, EI = 3.0 × 10

4 kN-m

2

Computation:2 2 2(32 kN)(4 m) 256 kN-m

(negative slope by inspection)2 2

A

PL

EI EI EI θ = = = −

Slope at A due to concentrated moment M .

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

Relevant equation from Appendix C:

A L EI

θ = (slope magnitude)

Values: L = 4 m, EI = 3.0 × 10

4 kN-m

2

Computation:

(4 m) (4 m)(positive slope by inspection) A

ML M M

EI EI EI θ = = =

Compatibility equation at A:2

2

256 kN-m (4 m)0

256 kN-m64.0 kN-m

4 m

M

EI EI

M

− + =

∴ = = Ans.






6 kip-in.

2is


(a) For the beam in Fig. P11.19a, determinethe concentrated upward force P required to

make the total beam deflection at B equal to

zero (i.e., v B = 0).

Fig. P11.19a

Solution

Downward deflection at B due to 4 kips/ft uniformly distributed load.

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


8 B

wLv

EI = −

Values: w = 4 kips/ft, L = 13 ft, EI = 5.0 × 10

6 kip-in.

2

Computation:4 4 3(4 kips/ft)(13 ft) 14,280.5 kip-ft

8 8 B

wLv

EI EI EI = − = − = −

Upward deflection at B due to concentrated load P .



3 B

PLv

EI

=

Values: L = 13 ft, EI = 5.0 × 10

6 kip-in.

2

Computation:3 3 3(13 ft) (732.333333 ft )

3 3 B

PL P P v

EI EI EI = = =


3

3

14,280.5 kip-ft (732.333333 ft )0

14,280.5 kip-ft19.50 kips

732.333333 ft

P

EI EI

P

− + =

∴ = = Ans.






6 kip-in.

2is



make the total beam slope at C equal to zero

(i.e., θ C = 0).

Fig. P11.19b

Solution

Slope at C due to 40-kip concentrated load.



16C

PL


Values: P = 40 kips, L = 18 ft, EI = 5.0 × 10

6 kip-in.

2

Computation:2 2 2(40 kips)(18 ft) 810 kip-ft

(negative slope by inspection)16 16

C

PL

EI EI EI θ = = =

Slope at C due to concentrated moment M .

[Appendix C, SS beam with concentrated moment at one end.]


3

C

L

EI

θ = (slope magnitude)

Values: L = 18 ft, EI = 5.0 × 10

6 kip-in.

2

Computation:

(18 ft) (6 ft)(positive slope by inspection)

3 3C

ML M M

EI EI EI θ = = =

Compatibility equation at C :2

2

810 kip-ft (6 ft)0

810 kip-ft135.0 kip-ft

6 ft

M

EI EI

M

− + =

∴ = = Ans.






4 kN-m

2

is constant for each beam.

(a) For the beam in Fig. P11.20a,

determine the concentrated downwardforce P required to make the total beam

deflection at B equal to zero (i.e., v B = 0).

Fig. P11.20a

Solution

Upward deflection at B due to 105 kN-m concentrated moment.

[Appendix C, SS beam with concentrated moment at one end.]


2 2(2 3 )6

B

M xv L Lx x

LEI = − − + (elastic curve)

Values: M = −105 kN-m, L = 8 m, x = 4 m,

EI = 5.0 × 104 kN-m

2

Computation:

2 2

32 2

(2 3 )6

( 105 kN-m)(4 m) 420 kN-m2(8 m) 3(8 m)(4 m) (4 m)

6(8 m)

B

M xv L Lx x

LEI

EI EI

= − − +

−⎡ ⎤= − − + =⎣ ⎦

Downward deflection at B due to concentrated load P .



48 B

PLv

EI = −

Values: L = 8 m, EI = 5.0 × 10

4 kN-m

2

Computation:3 3 3(8 m) (10.666667 m )

48 48 B

PL P P v

EI EI EI = − = − = −


3

3

420 kN-m (10.666667 m )0

420 kN-m39.375 kN 39.4 kN

10.666667 m

P

EI EI

P

− =

∴ = = = Ans.






4 kN-m

2is



make the total beam slope at A equal to zero

(i.e., θ A = 0).

Fig. P11.20b

Solution

Slope at A due to 6 kN/m uniformly distributed load.



6 A

wL


Values: w = 6 kN/m, L = 5 m, EI = 5.0 × 10

4 kN-m

2

Computation:3 3 2(6 kN/m)(5 m) 125 kN-m

(positive slope by inspection)6 6

A

wL

EI EI EI θ = = =

Slope at A due to concentrated moment M .



A

L


Values: L = 5 m, EI = 5.0 × 10

4 kN-m

2

Computation:

(5 m)(negative slope by inspection) A

ML M

EI EI θ = =


2

125 kN-m (5 m)0

125 kN-m25.0 kN-m

5 m

M

EI EI

M

− =

∴ = = Ans.






6 kip-in.

2


(a) For the beam in Fig. P11.21a,determine the concentrated downward

force P required to make the total beam

deflection at B equal to zero (i.e., v B = 0).

Fig. P11.21a

Solution

Upward deflection at B due to 125 kip-ft concentrated moment.



2 B

Lv

EI = −

Values: M = −125 kip-ft, L = 15 ft, EI = 8.0 × 10

6 kip-in.

2

Computation:2 2 3( 125 kip-ft)(15 ft) 14,062.5 kip-ft

2 2 B

MLv

EI EI EI

−= − = − =

Downward deflection at B due to concentrated load P .



3 B

PLv

EI = −

Values: L = 15 ft, EI = 8.0 × 106 kip-in.

2

Computation:3 3 3(15 ft) (1,125 ft )

3 3 B

PL P P v

EI EI EI = − = − = −


3

3

14,062.5 kip-ft (1,125 ft )0

14,062.5 kip-ft12.50 kips

1,125 ft

P

EI EI

P

− =

∴ = = Ans.






6 kip-in.

2


(b) For the beam in Fig. P11.21b,determine the concentrated moment

required to make the total beam slope at A

equal to zero (i.e., θ A = 0).

Fig. P11.21b

Solution

Slope at A due to 7 kips/ft uniformly distributed load.

[Appendix C, SS beam with uniformly distributed load over portion of span.]


2 2(2 )24

A

wa L a

LEI θ = − (slope magnitude)

Values: w = 7 kips/ft, L = 23 ft, a = 15 ft,

EI = 8.0 × 106 kip-in.

2

Computation:2

2 2

22 2

2

(2 )24

(7 kips/ft)(15 ft)2(23 ft) (15 ft)

24(23 ft)

2,376.766304 kip-ft(negative slope by inspection)

A

wa L a

LEI

EI

EI

θ = −

⎡ ⎤= −⎣ ⎦

= −

Slope at A due to concentrated moment M .[Appendix C, SS beam with concentrated moment at one end.]


3 A

L


Values: L = 23 ft, EI = 8.0 × 10

6 kip-in.

2

Computation:

(23 ft) (7.666667 ft)(positive slope by inspection)

3 3 A

ML M M

EI EI EI θ = = =


2

2,376.766304 kip-ft (7.666667 ft)0

2,376.766304 kip-ft310 kip-ft

7.666667 ft

M

EI EI

M

− + =

∴ = = Ans.





11.22 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. P11.22

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 M 0.


Relevant equation from Appendix C:2 2

0

2 2 B

L M Lv

EI EI = − = −

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



3 3

B B

PL R Lv

EI EI = =

Compatibility equation for deflection at B:2 3

0 030

2 3 2

B B

M L R L M R

EI EI L− + = ∴ = ↑ Ans.

Equilibrium equations

for entire beam:

0 03 30

2 2 y A B A B

M M F R R R R

L LΣ = + = ∴ = − = − = ↓ Ans.

0 0 A A B M M M R LΣ = − − + =

0 0 00 0 0

3 3( ) (cw)

2 2 2 A B

M M M M R L M L M M

L∴ = − = − = − = Ans.





11.23 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. P11.23

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.]


0

30 B

w Lv

EI = −

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



3 3

B B

PL R Lv

EI EI = =

Compatibility equation for deflection at B:4 3

0 0030 3 10

B B

w L R L w L R

EI EI − + = ∴ = ↑ Ans.

Equilibrium equations for entire beam:

0 0 0 0 04 20

2 2 10 10 5

y A B A

w L w L w L w L w L F R R RΣ = + − = ∴ = − = = ↑ Ans.

0 02 3

A A B

w L L M M R L

⎛ ⎞Σ = − − + =⎜ ⎟

⎝ ⎠

2 2 2 2

0 0 0 0 0( ) (ccw)6 10 6 15 15

A B

w L w L w L w L w L M R L L∴ = − = − = − = Ans.





11.24 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. P11.24

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 .



(3 )6

A

Pxv L x

EI = − − (elastic curve)

2 3

3Let ,2

( ) 3 73

6 2 12 A

L x L L

P L L PLv L

EI EI

= =

⎡ ⎤⎛ ⎞∴ = − − = −⎜ ⎟⎢ ⎥

⎝ ⎠⎣ ⎦

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



3 3

A A

PL R Lv

EI EI = =

Compatibility equation for deflection at A:

3 37 70

12 3 4

A A

PL R L P R

EI EI − + = ∴ = ↑ Ans.


0

7 3 34 4 4

y A B

B

F R R P

P P P R P

Σ = + − =

∴ = − = − = ↓ Ans.

3

02

B B A

L M M R L P

⎛ ⎞Σ = − + =⎜ ⎟

⎝ ⎠

3 7 3( ) (ccw)

2 4 2 4 4 B A

L P PL PL PL M R L P L

⎛ ⎞∴ = − = − = =⎜ ⎟

⎝ ⎠ Ans.





11.25 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. P11.25

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.



2 2(6 4 )24

A

wxv L Lx x

EI = − − + (elastic curve)

22 42

3Let ,

2

( ) 3 3 176 4 ( ) ( )

24 2 2 48 A

L x L L

w L L L wLv L L

EI EI

= =

⎡ ⎤⎛ ⎞ ⎛ ⎞∴ = − − + = −⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

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



3 3

A A

PL R Lv

EI EI = =

Compatibility equation for deflection at A:4 317 17

048 3 16

A A

wL R L wL R

EI EI − + = ∴ = ↑ Ans.






3 3 17 7 702 2 16 16 16

y A B B

wL wL wL wL wL F R R RΣ = + − = ∴ = − = = ↑ Ans.

3 30

2 4 B B A

wL L M M R L

⎛ ⎞⎛ ⎞Σ = − + =⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

2 2 2 29 17 9( ) (cw)

8 16 8 16 16 B A

wL wL wL wL wL M R L L∴ = − = − = − = Ans.

Mechanics of Materials Solutions Chapter11 Probs18 25

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