Mechanics of Materials Solutions Chapter11 Probs1 17

7/21/2019 Mechanics of Materials Solutions Chapter11 Probs1 17

http://slidepdf.com/reader/full/mechanics-of-materials-solutions-chapter11-probs1-17 1/28

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes onlyto 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.

11.1 A beam is loaded and supported asshown in Fig. P11.1. Use the double-

integration method to determine the

magnitude of the moment M 0 required to

make the slope at the left end of the beamzero.

Fig. P11.1

Solution

Moment equation:

0

2

0

( ) 02

( )2

a a

x M M x wx M

wx M x M

−

⎛ ⎞Σ = + − =⎜ ⎟

⎝ ⎠

∴ = −

Integration:2 2

02( )

2

d v wx EI M x M

dx= = −

3

0 16

dv wx EI M x C

dx= − +

2 4

01 2

2 24

M x wx EI v C x C = − + +

Boundary conditions and evaluate constants:3

0 1

3

1 0

( )at , 0 ( ) 0

6

6

dv w L x L M L C

dxwL

C M L

= = − + =

∴ = −

Beam slope equation:3 3

0 06 6

dv wx wL EI M x M L

dx= − + −

Constraint:

At x = 0, the slope of the beam is to be zero; therefore,

3 3

0 0

2

0

(0)(0) 06 6

6

A

dv w wL EI M M Ldx

wL M

= − + − =

∴ = Ans.





11.2 When moment M 0 is applied to the leftend 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 M 0.

Fig. P11.2

Solution

Moment equation:

0

0

( ) 0

( )

a a M M x Px M

M x Px M

−Σ = − + =

∴ = −

Integration:2

02( )

d v EI M x Px M

dx= = −

2

0 12

dv Px EI M x C

dx= − +

3 2

01 2

6 2

Px M x EI v C x C = − + +


0 1

2

1 0

( )at , 0 ( ) 02

2

dv P L x L M L C dx

PLC M L

= = − + =

∴ = − +

Beam slope equation:2 2

0 02 2

dv Px PL EI M x M L

dx= − − +

Constraint:

At x = 0, the slope of the beam is to be zero; therefore,2 2

0 0

0

(0)(0) 0

2 2

2

A

dv P PL EI M M L

dx

PL M

= − − + =

∴ = Ans.





11.3 When the load P is applied to the rightend 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 theload P .

Fig. P11.3

Solution

Moment equation:

( ) ( ) ( ) 02

a a

L x M w L x P L x M x−

−⎛ ⎞Σ = − − + − − =⎜ ⎟

⎝ ⎠

2( ) ( ) ( )2

w x L x P L x∴ = − − + −

Integration:2

2

2( ) ( ) ( )

2

d v w EI M x L x P L x

dx= = − − + −

3 2

1( ) ( )6 2

dv w P EI L x L x C

dx= − − − +

4 3

1 2( ) ( )24 6

w P EI v L x L x C x C = − − + − + +

Boundary conditions and evaluate constants:

3 2

1

3 2

1

at 0, 0 ( 0) ( 0) 06 2

6 2

dv w P x L L C dx

wL PLC

= = − − − + =

∴ = − +

4 3

1 2

4 3

2

at 0, 0 ( 0) ( 0) (0) 024 6

24 6

w P x v L L C C

wL PLC

= = − − + − + + =

∴ = −

Beam elastic curve equation:3 2 4 3

4 3

3 4 2 34 3

( ) ( )24 6 6 2 24 6

( ) ( )24 6 24 6 2 6

w P wLx PL x wL PL EI v L x L x

w wLx wL P PL x PL L x L x

= − − + − − + + −

= − − − + + − + −





Constraint:

At x = L, the deflection of the beam is to be zero; therefore,3 4 2 3

4 3( ) ( )( ) ( ) 0

24 6 24 6 2 6 B

w wL L wL P PL L PL EI v L L L L= − − − + + − + − =

which simplifies to4 4 3 3 4 3

06 24 2 6 8 3

B

wL wL PL PL wL PL EI v = − + + − = − + =

Therefore, the magnitude of P is3

8

wL P = Ans.







reactions at supports A and B.

Fig. P11.4

Solution

Beam FBD:

0

0

0

y A B

A A B

F R R

M M R L M

Σ = + =

Σ = − + − =

Moment equation:

0 0( ) ( ) 0 ( ) ( )a a B B M x M R L x M x R L x M −Σ = − − + − = ∴ = − −

Integration:2

02( ) ( ) B

d v EI M x R L x M

dx= = − −

2

0 1( )2

Bdv R EI L x M x C

dx= − − − +

23 0

1 2( )6 2

B R M x EI v L x C x C = − − + +


2

0 1 1at 0, 0 ( 0) (0) 02 2

B Bdv R R L

x L M C C dx= = − − − + = ∴ =

2 33 0

1 2 2

(0)at 0, 0 ( 0) (0) 0

6 2 6

B B R M R L x v L C C C = = − − + + = ∴ = −

2 2 33 0

3 2

0 0 0

( )at , 0 ( ) ( ) 0

6 2 2 6

3 3

3 2 2 2

B B B

B B

R M L R L R L x L v L L L

R L M L M M R

L L

= = − − + − =

= ∴ = = ↑

Backsubstitute into equilibrium equations:

0 03 3

02 2

y A B A B A

M M F R R R R R

L LΣ = + = = − = − ∴ = ↓ Ans.

00 0 0

0 0

30 0

2

(cw)2 2

A A B A B

A

M M M R L M M R L M L M

L

M M M

⎛ ⎞Σ = − + − = = − = − =⎜ ⎟

⎝ ⎠

∴ = = Ans.





11.5 A beam is loaded and supported asshown 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 bending-moment diagrams for the beam.

Fig. P11.5

Solution

Beam FBD:

0

02

y A B

A B B

F R R wL

L M M R L wL

Σ = + − =

⎛ ⎞Σ = + − =⎜ ⎟

⎝ ⎠

Moment equation:2

( ) 0 ( )

2 2

a a A A

x wx M x wx R x M x R x−

⎛ ⎞Σ = + − = ∴ = − +⎜ ⎟

⎝ ⎠Integration:2 2

2( )

2 A

d v wx EI M x R x

dx= = − +

3 2

16 2

Adv wx R x EI C

dx= − + +

4 3

1 224 6

Awx R x EI v C x C = − + + +

Boundary conditions and evaluate constants:4 3

1 2 2

(0) (0)

at 0, 0 (0) 0 024 6

Aw R

x v C C C = = − + + + = ∴ =

3 2 3 2

1 1

( ) ( )at , 0 0

6 2 6 2

A Adv w L R L wL R L x L C C

dx= = − + + = ∴ = − (a)

4 3 3 2

1 1

( ) ( )at , 0 ( ) 0

24 6 24 6

A Aw L R L wL R L x L v C L C = = − + + = ∴ = − (b)

Solve Eqs. (a) and (b) simultaneously to find:3

1

3 3and

48 8 8 A

wL wL wLC R= − = = ↑ Ans.


3 5 50

8 8 8 y A B B A B

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

2 2 2 2

2 2

50

2 2 2 8 8

(cw)8 8

A B B B B

B

L wL wL wL wL M M R L wL M R L

wL wL M

⎛ ⎞Σ = + − = = − = − = −⎜ ⎟

⎝ ⎠

∴ = − = Ans.








Fig. P11.6

Solution

Beam FBD:

0

0

02

20

2 3

y A B

A B B

w L F R R

w L L M M R L

Σ = + − =

⎛ ⎞Σ = + − =⎜ ⎟

⎝ ⎠

Moment equation:2

0

3

0

( ) 02 3

( )6

a a A

A

w x x M M x R x

L

w x M x R x

L

−

⎛ ⎞Σ = + − =⎜ ⎟

⎝ ⎠

∴ = − +

Integration:2 3

0

2( )

6 A

d v w x EI M x R x

dx L= = − +

4 2

01

24 2

Adv w x R x EI C

dx L= − + +

5 3

01 2

120 6

Aw x R x EI v C x C

L= − + + +


01 2 2

(0) (0)at 0, 0 (0) 0 0

120 6

Aw R x v C C C

L

= = − + + + = ∴ =

4 2 3 2

0 01 1

( ) ( )at , 0 0

24 2 24 2

A Adv w L R L w L R L x L C C

dx L= = − + + = ∴ = − (a)

5 3 3 2

0 01 1

( ) ( )at , 0 ( ) 0

120 6 120 6

A Aw L R L w L R L x L v C L C

L= = − + + = ∴ = − (b)





Solve Eqs. (a) and (b) simultaneously to find:3

0 0 01 and

120 10 10 A

w L w L w LC R= − = = ↑ Ans.


0 0 0 0 0 04 20

2 2 2 10 10 5 y A B B A B

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

2 2 2 2

0 0 0 0 0

2 2

0 0

2 20

2 3 3 3 5 15

(cw)15 15

A B B B B

B

w L L w L w L w L w L M M R L M R L

w L w L M

⎛ ⎞Σ = + − = = − = − = −⎜ ⎟

⎝ ⎠

∴ = − = Ans.





11.7 A beam is loaded and supported asshown 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:4 2

0

4 2

d v w x EI

dx L= −

3 3

013 23

d v w x EI C

dx L= − +

2 4

01 22 212

d v w x EI C x C

dx L= − + +

5 20 1

2 3260 2

dv w x C x EI C x C

dx L= − + + +

6 3 2

0 1 23 42360 6 2

w x C x C x EI v C x C

L= − + + + +

Boundary conditions and evaluate constants:6 3 2

0 1 23 4 42

(0) (0) (0)at 0, 0 (0) 0 0

360 6 2

w C C x v C C C

L= = − + + + + = ∴ =

5 2

0 12 3 32

(0) (0)at 0, 0 (0) 0 0

60 2

dv w C x C C C

dx L= = − + + + = ∴ =

6 3 2 2

0 1 2 01 22

( ) ( ) ( )at , 0 0 3360 6 2 60w L C L C L w L x L v C L C

L= = − + + = ∴ + = (a)

2 4 2

0 01 2 1 22 2

( )at , 0 ( ) 0

12 12

d v w L w L x L M EI C L C C L C

dx L= = = − + + = ∴ + = (b)

Solve Eqs. (a) and (b) simultaneously to obtain:2 2 2 2

0 0 0 02 2

42

60 12 60 30

w L w L w L w LC C = − = − ∴ = −

2 2 2

0 0 0 01 1

7 7

12 30 60 60

w L w L w L w LC L C = + = ∴ =

Roller reaction at B:3 3

0 0 0 0 0 0

3 2

( ) 7 20 7 13 13

3 60 60 60 60 60 B B

x L

d v w L w L w L w L w L w LV EI R

dx L=

= = − + = − + = − ∴ = ↑ Ans.






integration method to determine the reaction

at roller support A.

Fig. P11.8

Solution

Integrate the load distribution:4 2

0

4 2

d v w x EI

dx L= −

3 3

013 23

d v w x EI C

dx L= − +

2 4

01 22 212

d v w x EI C x C

dx L= − + +

5 20 1

2 3260 2

dv w x C x EI C x C

dx L= − + + +

6 3 2

0 1 23 42360 6 2

w x C x C x EI v C x C

L= − + + + +


0 1 23 4 42

(0) (0) (0)at 0, 0 (0) 0 0

360 6 2

w C C x v C C C

L= = − + + + + = ∴ =

2 4

01 2 22 2

(0)at 0, 0 (0) 0 0

12

d v w x M EI C C C

dx L= = = − + + = ∴ =

5 2 3

20 1 03 1 32

( ) ( )at , 0 0 260 2 30

dv w L C L w L x L C C L C dx L

= = − + + = ∴ + = (a)

6 3 320 1 0

3 1 32

( ) ( )at , 0 ( ) 0 6

360 6 60

w L C L w L x L v C L C L C

L= = − + + = ∴ + = (b)

Solve Eqs. (a) and (b) simultaneously to obtain:3 3 3 3

0 0 0 03 34

30 60 60 240

w L w L w L w LC C − = − = ∴ = −

3 3 32 0 0 0 0

1 1

5 5

30 120 120 120

w L w L w L w LC L C = + = ∴ =

Roller reaction at A:3 3

0 0 0 0

3 2

0

(0) 5 5 5

3 120 120 120 A A

x

d v w w L w L w LV EI R

dx L=

= = − + = ∴ = ↑ Ans.







reaction at roller support A.

Fig. P11.9

Solution

Integrate the load distribution:4

04sin

2

d v x EI w

dx L

π = −

3

013

2cos

2

d v w L x EI C

dx L

π

π

= +

2 2

01 22 2

4sin

2

d v w L x EI C x C

dx L

π

π

= + +

3 20 1

2 33

8cos

2 2

dv w L x C x EI C x C

dx L

π

π

= − + + +

4 3 2

0 1 23 44

16sin

2 6 2

w L x C x C x EI v C x C

L

π

π

= − + + + +


0 1 23 4 44

16 (0) (0) (0)at 0, 0 sin (0) 0 0

2 6 2

w L C C x v C C C

L

π

π

= = − + + + + = ∴ =

2 2

01 2 22 2

4 (0)at 0, 0 sin (0) 0 0

2

d v w L x M EI C C C

dx L

π

π

= = = + + = ∴ =

3 220 1

3 1 33

8 ( ) ( )at , 0 cos 0 2 0

2 2

dv w L L C L x L C C L C

dx L

π

π

= = − + + = ∴ + = (a)

4 3 320 1 0

3 1 34 4

16 ( ) ( ) 96at , 0 sin ( ) 0 6

2 6

w L L C L w L x L v C L C L C

L

π

π π

= = − + + = ∴ + = (b)

Solve Eqs. (a) and (b) simultaneously to obtain:3 3

0 03 34 4

96 244

w L w LC C

π π

− = − ∴ =

3

2 0 01 14 424 482 w L w LC L C

π π

⎛ ⎞= − ∴ = −⎜ ⎟

⎝ ⎠

Roller reaction at A:3

0 0 0 0

3 4 4

0

2 (0) 48 2 48cos

2 A A

x

d v w L w L w L w LV EI R

dx L

π

π π π π =

= = − ∴ = − Ans.








Fig. P11.10

Solution


04cos

2

d v x EI w

dx L

π =

3

013

2sin

2

d v w L x EI C

dx L

π

π

= +

2 2

01 22 2

4cos

2

d v w L x EI C x C

dx L

π

π

= − + +

3 2

0 12 33

8sin

2 2


dx L

π

π

= − + + +

4 3 2

0 1 23 44

16cos

2 6 2


L

π

π

= + + + +


0 12 3 33

8 (0) (0)at 0, 0 sin (0) 0 0

2 2

dv w L C x C C C

dx L

π

π

= = − + + + = ∴ =

4 3 2 4

0 1 2 0

4 44 4

16 (0) (0) (0) 16at 0, 0 cos 0

2 6 2

w L C C w L x v C C

L

π

π π

= = + + + = ∴ = −

3 2 2

0 1 02 1 23 3

8 ( ) ( ) 16at , 0 sin ( ) 0 2

2 2

dv w L L C L w L x L C L C L C

dx L

π

π π

= = − + + = ∴ + = (a)

4 3 2 4 2

0 1 2 0 01 24 4 4

16 ( ) ( ) ( ) 16 96at , 0 cos 0 3

2 6 2

w L L C L C L w L w L x L v C L C

L

π

π π π

= = + + − = ∴ + = (b)

Solve Eqs. (a) and (b) simultaneously to obtain:

[ ]2 2 2

0 0 02 23 4 4

16 96 166

w L w L w LC C π

π π π

− = − ∴ = −

[ ]2 2

0 0 01 13 4 4

48 192 484

w L w L w LC L C π

π π π

− = − + ∴ = −





Reactions at supports A and B

[ ] [ ]

[ ]

3

0 0 0

3 4 4

0

0

4

2 (0) 48 48sin 4 4

2

484

A

x

A

d v w L w L w LV EI

dx L

w L R

π π π

π π π

π

π

=

= = + − = −

∴ = − ↓ Ans.

[ ]3

30 0 03 4 4

30

4

2 ( ) 48 2sin 4 24 962

296 24

B

x L

B

d v w L L w L w LV EI dx L

w L R

π π π π

π π π

π π

π

=

⎡ ⎤= = + − = + −⎣ ⎦

⎡ ⎤∴ = − + ↓⎣ ⎦ Ans.

[ ] [ ]

[ ]

2 2 2

0 0 0

2 2 4 4

0

2 2

0 0

2 4

220

4

4 (0) 48 (0) 16cos 4 6

2

4 166

424 4 (cw)

A

x

A

d v w L w L w L M EI

dx L

w L w L

w L M

π π π

π π π

π

π π

π π

π

=

= = − + − + −

= − + −

⎡ ⎤∴ = − −⎣ ⎦ Ans.

[ ] [ ]

[ ] [ ] [ ] [ ]

[ ]

2 2 2

0 0 0

2 2 4 4

2 2 2 2

0 0 0 0

4 4 4 4

2

0

4

4 ( ) 48 ( ) 16cos 4 6

2

48 16 16 164 6 3 12 6 2 6

323 (ccw)

B

x L

B

d v w L L w L L w L M EI

dx L

w L w L w L w L

w L M

π π π

π π π

π π π π π

π π π π

π

π

=

= = − + − + −

= − + − = − + − = −

∴ = − Ans.








Fig. P11.11

Solution


04sin

d v x EI w

dx L

π =

3

013

cosd v w L x

EI C dx L

π

π

= − +

2 2

01 22 2

sind v w L x

EI C x C

dx L

π

π

= − + +

3 2

0 12 33

cos2


dx L

π

π

= + + +

4 3 2

0 1 23 44

sin6 2


L

π

π

= − + + + +


0 1 02 3 33 3

(0) (0)at 0, 0 cos (0) 0

2

dv w L C w L x C C C

dx L

π

π π

= = + + + = ∴ = −

4 3 2

0 1 23 4 4

4

(0) (0) (0)at 0, 0 sin (0) 0 0

6 2

w L C C x v C C C

L

π

π

= = − + + + + = ∴ =

3 2 3 2

0 1 0 02 1 23 3 3

( ) ( ) 4at , 0 cos ( ) 0 2

2

dv w L L C L w L w L x L C L C L C

dx L

π

π π π

= = + + − = ∴ + = (a)

4 3 2 3 2

0 1 2 0 01 24 3 3

( ) ( ) ( ) 6at , 0 sin ( ) 0 3

6 2

w L L C L C L w L w L x L v L C L C

L

π

π π π

= = − + + − = ∴ + = (b)

Solve Eqs. (a) and (b) simultaneously to obtain:2 2 2

0 0 02 23 3 3

6 4 2w L w L w LC C

π π π

= − ∴ =

2 2

0 01 13 3

4 22 0

w L w LC L C

π π

⎛ ⎞= − ∴ =⎜ ⎟

⎝ ⎠





Reactions at supports A and B 3

0 0

3

0

0

(0)cos A

x

A

d v w L w LV EI

dx L

w L R

π

π π

π

=

= = − = −

∴ = ↓ Ans.

3

0 03

0

( )cos B

x L

B

d v w L L w LV EI dx L

w L R

π

π π

π

=

= = − =

∴ = ↓ Ans.

2 2 2 2

0 0 0

2 2 3 3

0

2

0

3

(0) 2 2sin

2 (cw)

A

x

A

d v w L w L w L M EI

dx L

w L M

π

π π π

π

=

= = − + =

∴ = Ans.

2 2 2 2

0 0 0

2 2 3 3

2

0

3

( ) 2 2sin

2 (ccw)

B

x L

B

d v w L L w L w L M EI

dx L

w L M

π

π π π

π

=

= = − + =

∴ = Ans.







determine the reactions at supports A and C .

(b) Draw the shear-force and bending-moment diagrams for the beam.

(c) Determine the deflection in the middle

of the span.

Fig. P11.12

Solution

Beam FBD:from symmetry,

2

and

A C

A C

P R R

M M

= =

=

Moment equation:

( ) 0 ( )2 2

a a A A

P Px M x M x M x M −Σ = − − = ∴ = +

Integration:2

2( )

2 A

d v Px EI M x M

dx= = +

2

14

A

dv Px EI M x C

dx= + +

3 2

1 212 2

A Px M x EI v C x C = + + +


2

1 1(0)at 0, 0 (0) 0 04

Adv P x M C C dx

= = + + = ∴ =

3 2

2 2

(0) (0)at 0, 0 0 0

12 2

A P M x v C C = = + + = ∴ =





(a) Beam reaction forces:

2 A C

P R R= = Ans.

(a) Beam reaction moments:2( / 2)

at , 0 02 4 2

(ccw) (cw)8 8 8

A

A C

L dv P L L x M

dx

PL PL PL M M

⎛ ⎞= = + =⎜ ⎟

⎝ ⎠

= − = = Ans.

Elastic curve equation:

[ ]

[ ]

3 2 3 2 2

2

3 412 2 12 16 48

3 448

A Px M x Px PLx Px EI v L x

Pxv L x

EI

= + = − = − −

∴ = − −

(c) Midspan deflection:

[ ]2 3( / 2)

3 4( / 2)48 192

B

P L PLv L L

EI EI = − − = − Ans.







determine the reactions at supports A and B.(b) Draw the shear-force and bending-

moment diagrams for the beam.


of the span.

Fig. P11.13

Solution

Beam FBD:from symmetry,

2

and

A B

A B

wL R R

M M

= =

=

Moment equation:

2

( ) 02 2

( )2 2

a a A

A

x wL M M x M wx x

wx wLx M x M

−

⎛ ⎞ ⎛ ⎞Σ = − + − =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

∴ = − + +

Integration:2 2

2( )

2 2 A

d v wx wLx EI M x M

dx= = − + +

3 2

16 4

A

dv wx wLx EI M x C

dx= − + + +

4 3 2

1 224 12 2

Awx wLx M x EI v C x C = − + + + +


3 2

1 1

(0) (0)at 0, 0 (0) 0 06 4

A

dv w wL x M C C dx

= = − + + + = ∴ =

4 3 2

2 2

(0) (0) (0)at 0, 0 0 0

24 12 2

Aw wL M x v C C = = − + + + = ∴ =






2 A B

wL R R= = Ans.

(a) Beam reaction moments:3 2

2 2 2 2

( / 2) ( / 2)at , 0 0

2 6 4 2

(ccw) (cw)12 12 12 12

A

A B

L dv w L wL L L x M

dx

wL wL wL wL M M

⎛ ⎞= = − + + =⎜ ⎟

⎝ ⎠

= − = = − = Ans.

Elastic curve equation:4 3 2 4 3 2 2 2 2

2 2 2

22

2 ( )24 12 2 24 12 24 24 24

( )24

Awx wLx M x wx wLx wL x wx wx EI v x Lx L x L

wxv x L

EI

⎡ ⎤= − + + = − + − = − − + = − −⎣ ⎦

∴ = − −

(c) Midspan deflection:22 4

/ 2

( / 2))

24 2 384 x L

w L L wLv L

EI EI =

⎡ ⎤⎛ ⎞= − − = −⎜ ⎟⎢ ⎥

⎝ ⎠⎣ ⎦ Ans.







determine the reactions at supports A and C .(b) Determine the deflection in the middle

of the span.

Fig. P11.14

Solution

Beam FBD:

0

from symmetry,

2

and

A C

A C

w L R R

M M

= =

=

Moment equation:2

0 0

3

0 0

( ) 02 3 2

( )6 2

a a A

A

w x x w L M M x M x

L

w x w Lx M x M

L

−

⎛ ⎞ ⎛ ⎞Σ = − + − =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

∴ = − + +

Integration:2 3

0 0

2( )

6 2 A

d v w x w Lx EI M x M

dx L= = − + +

4 2

0 01

24 4 A

dv w x w Lx EI M x C

dx L= − + + +

5 3 2

0 01 2

120 12 2

Aw x w Lx M x EI v C x C

L= − + + + +


0 01 1

(0) (0)at 0, 0 (0) 0 0

24 4

A

dv w w L x M C C

dx L

= = − + + + = ∴ =

5 3 2

0 02 2

(0) (0) (0)at 0, 0 0 0

120 12 2

Aw w L M x v C C

L= = − + + + = ∴ =


0

2 A C

w L R R= = Ans.





(a) Beam reaction moments:4 2

0 0

2 2 2 2

0 0 0 0

( ) ( )at , 0 ( ) 0

24 4

5 5 5 5 (ccw) (cw)

24 24 24 24

A

A C

dv w L w L L x L M L

dx L

w L w L w L w L M M

= = − + + =

= − = = − = Ans.

Elastic curve equation:5 3 2 5 3 2 2

0 0 0 0 0

5 2 3 3 2 23 2 30 0 0 0

23 2 30

5

120 12 2 120 12 48

2 20 252 20 25

240 240 240 240

2 20 25240

Aw x w Lx M x w x w Lx w L x EI v

L L

w x w L x w L x w x L x L

L L L L

w xv x L x L

L EI

= − + + = − + −

⎡ ⎤= − + − = − − +⎣ ⎦

⎡ ⎤∴ = − − +⎣ ⎦


3 2 30 0( ) 7

2( ) 20 ( ) 25240 240 B

w L w L

v L L L L L EI EI ⎡ ⎤= − − + = −⎣ ⎦ Ans.







determine the reactions at supports A and C .(b) Draw the shear-force and bending-



of the span.

Fig. P11.15

Solution

Beam FBD:

0

02

y A C

A C C

F R R P

L M M R L P

Σ = + − =

⎛ ⎞Σ = + − =⎜ ⎟

⎝ ⎠

Moment equation:

( ) 0

( ) 02

a a A

A

M M x R x

L M x R x x

−Σ = − =

⎛ ⎞∴ = ≤ ≤⎜ ⎟

⎝ ⎠

( ) 02

( )2 2

b b A

A

L M M x R x P x

PL L x R x Px x L

−

⎛ ⎞Σ = − + − =⎜ ⎟

⎝ ⎠

⎛ ⎞∴ = − + ≤ ≤⎜ ⎟

⎝ ⎠

Integration:

For beam segment AB:2

2( ) A

d v EI M x R x

dx= =

2

12

Adv R x EI C

dx= +

3

1 2

6

A R x EI v C x C = + +

For beam segment BC :2

2( )

2 A

d v PL EI M x R x Px

dx= = − +

2 2

32 2 2

Adv R x Px PLx EI C

dx= − + +

3 3 2

3 4

6 6 4

A R x Px PLx EI v C x C = − + + +


1 2 2

(0)at 0, 0 (0) 0 0

6

A R x v C C C = = + + = ∴ =

2 2 2

3 3

( ) ( ) ( )at , 0 0

2 2 2 2

A Adv R L P L PL L R L x L C C

dx= = − + + = ∴ = −





3 3 2 2 3 3

4 4

( ) ( ) ( )at , 0 ( ) 0

6 6 4 2 3 12

A A A R L P L PL L R L R L PL x L v L C C = = − + − + = ∴ = −

Slope continuity condition at x = L/2:

2 2 2 2

1

2 2

1

at ,2

( / 2) ( / 2) ( / 2) ( / 2)

2 2 2 2 2

8 2

AB BC

A A A

A

L dv dv x

dx dx

R L R L P L PL L R LC

PL R LC

= =

+ = − + −

∴ = −

Deflection continuity condition at x = L/2:

3 2 2 3 3 2 2 3 3

at ,2

6 8 2 6 6 4 2 3 12

B B AB BC

A A A A A

L x v v

R x PL x R L x R x Px PLx R L x R L PL

= =

+ − = − + − + −

eliminate terms and rearrange:3 3 2 2 3

3 6 4 8 12

A R L Px PLx PL x PL= − + +

Substitute x = L/2 to obtain:3 3 2 2 3 3( / 2) ( / 2) ( / 2) 5

3 6 4 8 12 48

5

16

A

A

R L P L PL L PL L PL PL

P R

= − + + =

∴ =


5 1116 16

A C P P R R= = Ans.

(a) Beam reaction moment:

11 3 3 3 (cw)

2 2 16 16 16 16C C C

L PL PL PL PL PL M P R L M

⎛ ⎞= − = − = − = − =⎜ ⎟

⎝ ⎠ Ans.

Elastic curve equation for beam segment AB:3 2 2 3 2 2 3 2

2 2

5 5 5 3

6 8 2 96 8 32 96 96

5 396

A A R x PL x R L x Px PL x PL x Px PL x EI v

Pxv x L

EI

= + − = + − = −

⎡ ⎤∴ = −⎣ ⎦


2( / 2) 75 3

96 2 768 B

P L L PLv L

EI EI

⎡ ⎤⎛ ⎞= − = −⎢ ⎥⎜ ⎟

⎝ ⎠⎢ ⎥⎣ ⎦ Ans.









Fig. P11.16

Solution

Beam FBD:

0

0

0

y A C C A

A A C

F R R R R

M M R L M

Σ = + = ∴ = −

Σ = − + − =

Moment equation:

( ) 0

( ) 02

a a A A

A A

M M x R x M

L M x R x M x

−Σ = − − =

⎛ ⎞∴ = + ≤ ≤⎜ ⎟

⎝ ⎠

0

0

( ) 0

( )2

b b A A

A A

M M x R x M M

L x R x M M x L

−Σ = − − − =

⎛ ⎞∴ = + + ≤ ≤⎜ ⎟

⎝ ⎠

Integration:

For beam segment AB:2

2( ) A A

d v EI M x R x M

dx= = +

2

12

A A

dv R x EI M x C

dx= + +

3 2

1 26 2

A A R x M x EI v C x C = + + +


02( ) A A

d v EI M x R x M M

dx= = + +

2

0 32

A A

dv R x EI M x M x C

dx= + + +

3 2 20

3 46 2 2

A A R x M x M x EI v C x C = + + + +

Boundary conditions and evaluate constants for segment AB:3 2

1 2 2

(0) (0)at 0, 0 (0) 0 0

6 2

A A R M x v C C C = = + + + = ∴ =





2

1 1

(0)at 0, 0 (0) 0 0

2

A A

dv R x M C C

dx= = + + = ∴ =


2 2

0 3

03

at ,2

2 2

2

AB BC

A A

A A

L dv dv x

dx dx

R x R x

x M x M x C

M LC

= =

+ = + + +

∴ = −


3 2 3 2 2

0 04

20

4

at ,2

6 2 6 2 2 2

8

B B AB BC

A A A A

L x v v

R x M x R x M x M x M LxC

M LC

= =

+ = + + − +

∴ =

Boundary condition for segment BC :3 2 2 2

0 0 0 0( ) ( ) ( ) 3at , 0 ( ) 0 3

6 2 2 2 8 4

A A A A

R L M L M L M L M L M x L v L R L M = = + + − + = ∴ + = −

Also, the beam moment equilibrium equation can be written as:

0 A A R L M M + = −

(a) Beam Reactions: Solve these two equations simultaneously to obtain:0 0 0 0 09 9 9

(cw)8 8 8 8 8

A A C

M M M M M M R R

L L L= = = − = ↓ = ↑ Ans.









Fig. P11.17

Solution

Beam FBD:

02

02 4

y A C

A C C

wL F R R

wL L M M R L

Σ = + − =

⎛ ⎞Σ = + − =⎜ ⎟

⎝ ⎠

Moment equation:

2

( ) 02

( ) 02 2

a a A

A

x M M x wx R x

wx L M x R x x

− ⎛ ⎞Σ = + − =⎜ ⎟⎝ ⎠

⎛ ⎞∴ = − + ≤ ≤⎜ ⎟

⎝ ⎠

( ) 02 4

( )2 4 2

b b A

A

wL L M M x x R x

wL L L x x R x x L

−

⎛ ⎞Σ = + − − =⎜ ⎟

⎝ ⎠

⎛ ⎞ ⎛ ⎞∴ = − − + ≤ ≤⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

Integration:

For beam segment AB:2 2

2( )

2 A

d v wx EI M x R x

dx= = − +

3 2

16 2

Adv wx R x EI C

dx= − + +

4 3

1 2

24 6

Awx R x EI v C x C = − + + +


2( )

2 4 A

d v wL L EI M x x R x

dx

⎛ ⎞= = − − +⎜ ⎟

⎝ ⎠

2 2

34 4 2

Adv wL L R x EI x C

dx

⎛ ⎞= − − + +⎜ ⎟

⎝ ⎠

3 3

3 412 4 6

AwL L R x

EI v x C x C

⎛ ⎞

= − − + + +⎜ ⎟⎝ ⎠

Boundary conditions and evaluate constants for segment AB:4 3

1 2 2

(0) (0)at 0, 0 (0) 0 0

24 6

Aw R x v C C C = = − + + + = ∴ =






Equate the slope expressions for the two beam segments:23 2 2

1 36 2 4 4 2

A Awx R x wL L R xC x C

⎛ ⎞− + + = − − + +⎜ ⎟

⎝ ⎠

Set x = L/2 and solve for the constant C 1:2 23 3 3 3

1 3 3 3

3

1 3

( / 2)

6 4 4 6 4 2 4 48 64

192

wx wL L w L wL L L wL wLC C x C C

wLC C

⎛ ⎞ ⎛ ⎞= + − − = + − − = + −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

∴ = +


Equate the deflection expressions for the two beam segments:34 3 3

1 3 424 6 12 4 6

A Awx R x wL L R xC x x C x C

⎛ ⎞− + + = − − + + +⎜ ⎟

⎝ ⎠

Set x = L/2 and solve for the constant C 4:34 3

3 3 4

4 4 4

3 3 4

4

4

( / 2)

24 192 2 12 2 4 2

384 2 384 768 2

768

w L wL L wL L L L

C C C

wL L wL wL LC C C

wLC

⎡ ⎤ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + + = − − + +

⎢ ⎥ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

⎛ ⎞ ⎛ ⎞− + + = − + +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

∴ =

Boundary conditions and evaluate constants for segment BC :2 2 3 2

3 3

( ) 9at , 0 0

4 4 2 64 2

A Adv wL L R L wL R L x L L C C

dx

⎛ ⎞= = − − + + = ∴ = −⎜ ⎟

⎝ ⎠

at x = L, v = 03 3 4

3

4 3 3 2 4

4 3 4 3 4

3 3 4 4

3 4

( )( ) 0

12 4 6 768

27 9( ) 0

768 6 64 2 768

27 90

768 6 64 2 768

3 26 1086 6 768 768

82 41

3 768 128

A

A A

A A

A A

A A

wL L R L wL L C L

wL R L wL R L wL L

wL R L wL R L wL

R L R L wL wL

R L wL wL R

⎛ ⎞− − + + + =⎜ ⎟

⎝ ⎠

⎡ ⎤− + + − + =⎢ ⎥

⎣ ⎦

− + + − + =

− = −

= ∴ =



Solve for C 3:3 3 3

3

9 41 5

64 256 256

wL wL wLC = − = −

and for C 1:3 3 3

1

5 11

256 192 768

wL wL wLC = − + = −

(a) Beam force reactions:

41

128 A

wL R = Ans.

41 23 23

2 2 128 128 128C A C

wL wL wL wL wL R R R= − = − = ∴ = ↑ Ans.

Beam moment reaction:2 2 2 2 2 223 7 7 7

(cw)8 8 128 128 128 128

C C C

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

Mechanics of Materials Solutions Chapter11 Probs1 17

Documents