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10.1 For the loading shown, use the double-integration method to determine (a) theequation of the elastic curve for thecantilever beam, (b) the deflection at the freeend, and (c) the slope at the free end.Assume that EI is constant for each beam.

Fig. P10.1

SolutionIntegration of moment equation:

2

02 ( )d v

EI M x M dx

= = −

0 1

dv EI M x C

dx= − + (a)

20

1 22 M x

EI v C x C = − + + (b)

Boundary conditions:

0 at 0

0 at 0

dv x

dxv x

= =

= =

Evaluate constants:From Eq. (a), C 1 = 0. From Eq. (b), C 2 = 0

(a) Elastic curve equation:2 2

0 0

2 2 x M x

EI v v EI = − = − Ans.

(b) Deflection at the free end:2 2

0 0( )2 2 B

L M Lv

EI EI = − = − Ans.

(c) Slope at the free end:

0 0( ) B

B

dv M L M Ldx EI EI

θ = = − = − Ans.





10.2 For the loading shown, use the double-integration method to determine (a) theequation of the elastic curve for thecantilever beam, (b) the deflection at the freeend, and (c) the slope at the free end.Assume that EI is constant for each beam.

Fig. P10.2


2 2

2 ( )2

d v wx EI M x

dx= = −

3

16dv wx

EI C dx

= − + (a)

4

1 224wx

EI v C x C = − + + (b)


0 at

0 at

dv L

dxv x L

= =

= =

Evaluate constants:Substitute x = L and dv/dx = 0 into Eq. (a) to determine C 1:

3 3

1 1

( )(0)

6 6w L wL

EI C C = − + =

Substitute x = L and v = 0 into Eq. (b) to determine C 2:4 4 4 4

1 2 2 2

( )(0) ( )

24 24 6 8w L wL wL wL

EI C L C C C = − + + = − + + = −

(a) Elastic curve equation:4 3 4

4 3 44 324 6 8 24

wx wL x wL w EI v v x L x L

EI = − + − = − − + Ans.

(b) Deflection at the free end:4 4

4 3 4 3(0) 4 (0) 3

24 24 8 A

w wL wLv L L

EI EI EI = − + − = − = − Ans.

(c) Slope at the free end:3 3 3(0)

6 6 6 A A

dv w wL wLdx EI EI EI

θ = = − + = Ans.





10.3 For the loading shown, use the double-integration method to determine (a) theequation 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 isconstant for each beam.

Fig. P10.3


2 30

2 ( )6

d v w x EI M x

dx L= = −

40

124dv w x

EI C dx L

= − + (a)

50

1 2120w x

EI v C x C L

= − + + (b)


0 at

0 at

dv L

dxv x L

= =

= =


4 30 0

1 1

( )(0)

24 24w L w L

EI C C L

= − + =

Substitute x = L and v = 0 into Eq. (b) to determine C 2:5 5 3

0 0 01 2 2

4 4 40 0 0

2

( )(0) ( ) ( )

120 120 24

120 24 30

w L w L w L EI C L C L C

L Lw L w L w L

C

= − + + = − + +

= − = −


5 4 50 0 0 0 5 4120 24 30 120w x w L w L w

EI v x v x L x L L L EI

= − + − = − − + Ans.

(b) Deflection at the free end: 45 4 50 0(0) 5 (0) 4

120 30 A

w w Lv L L

L EI EI = − − + = − Ans.

(c) Slope at the free end:4 3 3

0 0 0(0)24 24 24 A

A

dv w w L w Ldx L EI EI EI

θ = = − + = Ans.





10.4 For the beam and loading shown inFig. P10.4, use the double-integrationmethod to determine (a) the equation othe elastic curve for segment AB of the

beam, (b) the deflection at B, and (c) theslope at A. Assume that EI is constant for the beam.

Fig. P10.4


2

2 ( )2

d v P EI M x x

dx= =

2

14dv Px

EI C dx

= + (a)

3

1 2

12

Px EI v C x C = + + (b)

Boundary conditions:0 at 0

0 at2

v x

dv L x

dx

= =

= =

Evaluate constants:Substitute x = L/2 and dv/dx = 0 into Eq. (a) to determine C 1:

2 2

1 1

( / 2)(0)

4 16

P L PL EI C C = + = −

Substitute x = 0 and v = 0 into Eq. (b) to determine C 2:3 2

2 2

(0) (0)(0) 0

12 16 P PL

EI C C = − + =

(a) Elastic curve equation:3 2

2 23 4 (0 )12 16 48 2

Px PL x P x L EI v v L x x

EI = − = − − ≤ ≤ Ans.

(b) Deflection at B :2

32( / 2) 3 448 2 48 B

P L L PLv L EI EI

= − − = − Ans.

(c) Slope at A :2 2 2(0)

4 16 16 A A

dv P PL PLdx EI EI EI

θ = = − = − Ans.





10.5 For the beam and loading shown inFig. P10.5, use the double-integrationmethod to determine (a) the equation othe elastic curve for the beam, (b) theslope at A, (c) the slope at B, and (d) thedeflection at midspan. Assume that EI isconstant for the beam.

Fig. P10.5

SolutionBeam FBD:

0

0 0

0

0

and

y y y

y y

A y

y y

F A B

A B

M B L M

M B A

L L

Σ = + =

= −

Σ = − =

= = −

Moment equation:

00 0

00

( ) ( ) 0

( )

a a y

M M M x A x M M x x M

L M x

M x M L

−Σ = − − = + − =

= −

Integration of moment equation:2

002 ( )d v M x EI M x M dx L

= = − 2

00 12

dv M x EI M x C

dx L= − + (a)

2 30 0

1 22 6 M x M x

EI v C x C L

= − + + (b)


0 at

v x

v x L

= =

= =

Evaluate constants:Substitute x = 0 and v = 0 into Eq. (b) to determine C 2:

2 30 0

1 2 2

(0) (0)(0) (0) 0

2 6 M M

EI C C C L

= − + + =

Substitute x = L and v = 0 into Eq. (b) to determine C 1:





10.6 For the beam and loading shown inFig. P10.6, use the double-integrationmethod to determine (a) the equation othe elastic curve for the beam, (b) themaximum deflection, and (c) the slope at

A. Assume that EI is constant for the beam.

Fig. P10.6

SolutionMoment equation:

2

2

( ) 02 2

( )2 2

a a

wLx wx M M x

wx wLx M x

−Σ = − + =

= − +

Integration of moment equation:2 2

2 ( )2 2

d v wx wLx EI M x

dx= = − +

3 2

16 4dv wx wLx

EI C dx

= − + + (a)

4 3

1 224 12wx wLx

EI v C x C = − + + + (b)


0 at

v x

v x L

= =

= =


4 3

1 2 2

(0) (0)(0) (0) 0

24 12w wL

EI C C C = − + + + =

Substitute x = L and v = 0 into Eq. (b) to determine C 1:4 3 4 4 3

1 1

( ) ( ) ( ) ( )(0) ( )

24 12 24 12 24w L wL L w L w L wL

EI C L C L L

= − + + = − = −


3 2 3224 12 24 24

wx wLx wL x wx EI v v x Lx L

EI = − + − = − − + Ans.

(b) Maximum deflection: At x = L/2:3 2 3 3 4

2 2max

( / 2) 52

24 2 2 48 8 2 384w L L L wL L L wL

v L L L EI EI EI

= − − + = − − + = −

Ans.

(c) Slope at A :3 2 3 3(0) (0)

6 4 24 24 A A

dv w wL wL wLdx

θ = = − + − = − Ans.






beam, (b) the deflection midway betweenthe two supports, (c) the slope at A, and(d) the slope at B. Assume that EI isconstant for the beam.

Fig. P10.7

SolutionBeam FBD:

0

30

23

and2 2

y y y

A y

y y

F A B P

L M B L P

P P B A

Σ = + − =

Σ = − =

= = −

Moment equation:

( ) 0 ( )2 2a a

P Px M M x x M x−

Σ = + = = −


2 ( )2

d v Px EI M x

dx= = −

2

14dv Px

EI C dx= − +

(a)3

1 212 Px

EI v C x C = − + + (b)


0 at

v x

v x L

= =

= =


3

1 2 2(0)(0) (0) 012

P EI C C C = − + + =

Substitute x = L and v = 0 into Eq. (b) to determine C 1:3 2

1 1

( )(0) ( )

12 12 P L PL

EI C L C = − + =





(a) Elastic curve equation for segment BC of the beam:

[ ]2

22 42 2

PLx PLx EI v PL x v L x

EI = − + = − Ans.

(b) Deflection at midspan:

[ ]3

/ 2

(2 ) 24 (2 )

2 x L

PL L PLv L L

EI EI =

= − = Ans.

(c) Slope at C :2 2(4 ) 2 2

C C

dv PL L PL PLdx EI EI EI

θ = = − + = − Ans.






beam, (b) the deflection midway between A and B, and (c) the slope at B. Assumethat EI is constant for the beam.

Fig. P10.9

SolutionBeam FBD:

2 50

2 45

2 40

2 4

A y

y

y y y

y

wL L M P B L

wL P B

F A B wL P

wL P A

Σ = − − + =

= +

Σ = + − − =

= −

Moment equation:2 2

2

( ) ( ) 02 2 2 4

( )2 2 4

a a y

wx wx wLx Px M M x A x M x

wx wLx Px M x

−Σ = + − = + − + =

= − + −


2( )

2 2 4

d v wx wLx Px EI M x

dx= = − + −

3 2 2

16 4 8dv wx wLx Px

EI C dx

= − + − + (a)

4 3 3

1 224 12 24wx wLx Px

EI v C x C = − + − + + (b)


0 at

v x

v x L

= =

= =


4 3 3

1 2 2

(0) (0) (0)(0) (0) 0

24 12 24w wL P

EI C C C = − + − + + =

Substitute x = L and v = 0 into Eq. (b) to determine C 1:4 3 3 3 2

1 1

( ) ( ) ( )(0) ( )

24 12 24 24 24w L wL L P L wL PL

EI C L C = − + − + = − +





(a) Elastic curve equation for segment AB of the beam:4 3 3 3 2

3 2 3 2 2

24 12 24 24 24

224 24

wx wLx Px wL x PL x EI v

wx Pxv x Lx L x L

EI EI

= − + − − +

= − − + − − Ans.

(b) Deflection at midspan:3 2 2

3 2/ 2

4 3

( / 2) ( / 2)2

24 2 2 24 2

5384 64

x L

w L L L P L Lv L L L

EI EI

wL PL EI EI

=

= − − + − −

= − + Ans.

(c) Slope at B :3 2 2 3 2 3 2( ) ( ) ( )

6 4 8 24 24 24 12 B B

dv w L wL L P L wL PL wL PLdx EI EI EI EI EI EI EI

θ = = − + − − + = − Ans.





10.10 For the beam and loading shown inFig. P10.10, use the double-integrationmethod to determine (a) the equation othe elastic curve for segment AC of the

beam, (b) the deflection at B, and (c) theslope at A. Assume that EI is constant forthe beam.

Fig. P10.10

SolutionBeam FBD:

3(3 ) (2 ) 0

29

4(3 ) 0

9 33

4 4

A y

y

y y y

y

L M w L C L

wLC

F A C w L

wL wL A wL

Σ = − + =

=

Σ = + − =

= − =

Moment equation:

2

3( ) ( ) 0

2 4 2

3( )

2 4

a a y

x wL x M M x A x wx M x x wx

wx wLx M x

−

Σ = − + = − + =

= − +

Integration of moment equation:

2 2

2

3( )

2 4

d v wx wLx EI M x

dx

= = +

3 2

1

36 8

dv wx wLx EI C

dx= − + + (a)

4 3

1 2

324 24

wx wLx EI v C x C = − + + + (b)


0 at 2

v x

v x L

= =

= =

Evaluate constants:Substitute x = 0 and v = 0 into Eq. (b) to determine C 2:4 3

1 2 2

(0) 3 (0)(0) (0) 0

24 24w wL

EI C C C = − + + + =

Substitute x = 2 L and v = 0 into Eq. (b) to determine C 1:





4 3

1

3 3 3

1

(2 ) 3 (2 )(0) (2 )

24 248 12

24 24 6

w L wL L EI C L

wL wL wLC

= − + +

= − = −

(a) Elastic curve equation for segment AC of the beam:4 3 3

3 2 3

3 2 3

33 424 24 6 24

3 424

wx wLx wL x wx EI v x Lx L

wxv x Lx L

EI

= − + − = − − +

= − − + Ans.


3 2 3( )( ) 3 ( ) 4

24 12 B

w L wLv L L L L

EI EI = − − + = − Ans.

(c) Slope at A :3 2 3 3

(0) 3 (0)6 8 6 6 A

A

dv w wL wL wLdx EI EI EI EI

θ = = − + − = − Ans.





10.11 For the simply supported steel beam[ E = 200 GPa; I = 129 × 10 6 mm 4] shownin Fig. P10.11, use the double-integrationmethod to determine the deflection at B.Assume L = 4 m, P = 60 kN, and w = 40kN/m.

Fig. P10.11

SolutionBeam FBD:

( ) 02 2

2 2( ) 0

2 2

A y

y

y y y

y

L L M wL P C L

wL P C

F A C w L P

wL P A

Σ = − − + =

= +

Σ = + − − =

= +

Moment equation:2 2

2

( ) ( ) 02 2 2 2

( )2 2 2

a a y

wx wx wLx Px M M x A x M x

wx wLx Px M x

−Σ = + − = + − − =

= − + +


2 ( )2 2 2

d v wx wLx Px EI M xdx

= = − + +

3 2 2

16 4 4dv wx wLx Px

EI C dx

= − + + + (a)

4 3 3

1 224 12 12wx wLx Px

EI v C x C = − + + + + (b)


0 at 2

v x

dv L

xdx

= =

= =

Evaluate constants:Substitute x = 0 and v = 0 into Eq. (b) to determine C 2 = 0. Next, substitute x = L/ 2 and dv/dx = 0 intoEq. (b) to determine C 1:





3 2 2

1

3 3 2 3 2

1

( / 2) ( / 2) ( / 2)(0)

6 4 4

48 16 16 24 16

w L wL L P L EI C

wL wL PL wL PLC

= − + + +

= − − = − −

Elastic curve equation:4 3 3 3 2

3 2 3 2 2

24 12 12 24 16

2 3 424 48

wx wLx Px wL x PL x EI v

wx Pxv x Lx L L x

EI EI

= − + + − −

= − − + − −

Deflection at B : At x = L/2:4 35

384 48 B

wL PLv

EI EI = − −

Let E = 200 GPa, I = 129 × 10 6 mm 4, w = 40 kN/m, P = 60 kN, and L = 4 m.4 3

2 6 4 2 6 4

5(40 N/mm)(4,000 mm) (60,000 N)(4,000 mm)

384(200,000 N/mm )(129 10 mm ) 48(200,000 N/mm )(129 10 mm )5.1680 mm 3.1008 mm

8.27 mm

Bv = − −

× ×= − −

= − Ans.





10.12 For the cantilever steel beam [ E = 200GPa; I = 129 × 10 6 mm 4] shown in Fig.P10.12, use the double-integration method todetermine the deflection at A. Assume L = 2.5m, P = 50 kN, and w = 30 kN/m.

Fig. P10.12


2 2

( ) 0 ( )2 2a a

wx wx M x Px M x Px−

Σ = + + = = − −


2 ( )2

d v wx EI M x Px

dx= = − −

3 2

16 2dv wx Px EI C dx= − − + (a)

4 3

1 224 6wx Px

EI v C x C = − − + + (b)

Boundary conditions:0 at

0 at

v x L

dv L

dx

= =

= =

Evaluate constants:Substitute x = L and dv/dx = 0 into Eq. (a) to determine C 1:3 2 3 2

1 1

( ) ( )(0)

6 2 6 2w L P L wL PL

EI C C = − − + = +

Substitute x = L and v = 0 into Eq. (b) to determine C 2:4 3 3 2 4 4 3 3

2 2

4 3

2

( ) ( )(0) ( ) ( )

24 6 6 2 24 6 6 2

8 3

w L P L wL PL wL wL PL PL EI L L C C

wL PLC

= − − + + + = − + − + +

= − −

Elastic curve equation:4 3 4 3 2 3

4 3 4 3 2 3

24 6 8 6 2 3

4 3 3 224 6

wx wL x wL Px PL x PL EI v

w P v x L x L x L x L

EI EI

= − + − − + −

= − − + − − +





Deflection at A :4 3

4 3 4 3 2 3 3(0) 4 (0) 3 (0) 3 (0) 2

24 6 24 3 A

w P wL PLv L L L L

EI EI EI EI = − + − − − + = − −

Let E = 200 GPa, I = 129 × 10 6 mm 4, w = 30 kN/m, P = 50 kN, and L = 2.5 m.4 3

2 6 4 2 6 4

3(30 N/mm)(2,500 mm) (50,000 N)(2,500 mm)24(200,000 N/mm )(129 10 mm ) 3(200,000 N/mm )(129 10 mm )

5.6777 mm 10.0937 mm

= 15.77 mm

Av = − −× ×

= − −

− Ans.





10.13 For the cantilever steel beam [ E = 200GPa; I = 129 × 10 6 mm 4] shown in Fig.P10.13, use the double-integration method todetermine the deflection at B. Assume L = 3m, M 0 = 70 kN-m, and w = 15 kN/m.

Fig. P10.13


2

0

2

0

( )( ) 0

2( )

( )2

a a

w L x M M x M

w L x M x M

−

−Σ = − − − =

− = − −

Integration of moment equation:2 2 2 2

2 20 0 02

( )( ) 22 2 2 2

d v w L x w wL wx EI M x M L Lx x M wLx M dx

−

= = − − = − − + − = − + − − 2 2 3

0 12 2 6dv wL x wLx wx

EI M x C dx

= − + − − + (a)

2 2 3 4 20

1 24 6 24 2wL x wLx wx M x

EI v C x C = − + − − + + (b)


0 at 0

v x

dv x

dx

= =

= =

Evaluate constants:Substitute x = 0 and dv/dx = 0 into Eq. (a) to determine C 1 = 0. Next, substitute x = 0 and v = 0 into Eq.(b) to determine C 2 = 0.

Elastic curve equation:2 2 3 4 2

0

24 3 2 2 0

4 6 24 2

4 6

24 2

wL x wLx wx M x EI v

w M xv x Lx L x

EI EI

= − + − −

= − − + −

Deflection at B :2 4 2

4 3 2 2 0 0( )( ) 4 ( ) 6 ( )

24 2 8 2 B

w M L wL M Lv L L L L L

EI EI EI EI = − − + − = − −

Let E = 200 GPa, I = 129 × 10 6 mm 4, w = 15 kN/m, M 0 = 70 kN-m, and L = 3 m.





4 2

2 6 4 2 6 4

(15 N/mm)(3,000 mm) (70 kN-m)(1,000 N/kN)(1,000 mm/m)(3,000 mm)8(200,000 N/mm )(129 10 mm ) 2(200,000 N/mm )(129 10 mm )

5.8866 mm 12.2093 mm

= 18.10 mm

Bv = − −× ×

= − −

− Ans.





10.14 For the cantilever steel beam [ E = 200GPa; I = 129 × 10 6 mm 4] shown in Fig.P10.14, use the double-integration method todetermine the deflection at A. Assume L = 2.5m, P = 50 kN-m, and w0 = 90 kN/m.

Fig. P10.14


0

30

( ) ( ) 02 3

( )6

a a

w x x M M x x Px

L

w x M x Px

L

−

Σ = + + =

= − −


02 ( )

6d v w x EI M x Pxdx L

= = − −

4 20

124 2dv w x Px

EI C dx L

= − − + (a)

5 30

1 2120 6w x Px

EI v C x C L

= − − + + (b)

Boundary conditions:0 at

0 at

v x L

dv Ldx

= =

= =


4 2 3 20 0

1 1

( ) ( )(0)

24 2 24 2w L P L w L PL

EI C C L

= − − + = +

Substitute x = L and v = 0 into Eq. (b) to determine C 2:5 3 3 2 4 4 3 3

0 0 0 02 2

4 3

02

( ) ( )(0) ( ) ( )

120 6 24 2 120 24 6 2

30 3

w L P L w L PL w L w L PL PL EI L L C C

L

w L PLC

= − − + + + = − + − + +

= − −

Elastic curve equation:5 3 3 2 4 3

0 0 0

5 4 5 3 2 30

120 6 24 2 30 3

5 4 3 2120 6

w x Px w L PL w L PL EI v x x

Lw P

v x L x L x L x L L EI EI

= − − + + − −

= − − + − − +





Deflection at A : Let E = 200 GPa, I = 129 × 10 6 mm 4, w0 = 90 kN/m, P = 50 kN, and L = 2.5 m.

5 4 5 3 2 30

4 30

4 3

2 6 4 2 6 4

(0) 5 (0) 4 (0) 3 (0) 2120 6

30 3(90 N/mm)(2,500 mm) (50,000 N)(2,500 mm)

30(200,000 N/mm )(129 10 mm ) 3(200,000 N/mm )(129 10 mm )

4.5422 mm 10.0937 mm

= 1

A

w P v L L L L

L EI EI

w L PL EI EI

= − − + − − +

= − −

= − −× ×

= − −

− 4.64 mm Ans.





10.15 For the beam and loading shown inFig. P10.15, use the double-integrationmethod to determine (a) the equation of theelastic curve for the cantilever beam AB, (b)the deflection at the free end, and (c) theslope at the free end. Assume that EI isconstant for each beam.

Fig. P10.15

SolutionBeam FBD:

0

20

0

0

20

2 3

3

02

2

A A

A

y y

y

w L L M M

w L M

w L F A

w L A

Σ = − − =

= −

Σ = − =

=

Moment equation:

0

20 0 0

3 20 0 0

( ) ( )2 3

( ) ( ) ( ) 03 2 3 2

( )

6 2 3

a a A y

w x x M M x M x A x

L

w L w x x w L M x x x

L

w x w Lx w L M x

L

−

Σ = − + −

= + + − =

= − + −

Integration of moment equation:2 3 2

0 0 02 ( )

6 2 3d v w x w Lx w L

EI M xdx L

= = − + −

4 2 20 0 0

124 4 3dv w x w Lx w L x

EI C dx L

= − + − + (a)

5 3 2 20 0 0

1 2120 12 6w x w Lx w L x

EI v C x C L

= − + − + + (b)


0 at 0

v x

dv x

dx

= =

= =

Evaluate constants:Substitute x = 0 and v = 0 into Eq. (b) to determine C 2 = 0. Next, substitute x = 0 and dv/dx = 0 into Eq.(b) to determine C 1 = 0.





(a) Elastic curve equation:5 3 2 2

0 0 0

5 2 3 3 20

120 12 6

10 20120

w x w Lx w L x EI v

L

wv x L x L x

L EI

= − + −

= − − + Ans.

(b) Deflection at the free end: 5

5 2 3 3 20 011( ) 10 ( ) 20 ( )

120 120 B

w w Lv L L L L L

L EI L EI = − − + = − Ans.

(c) Slope at the free end:4 2 2 3 3 3 3

0 0 0 0 0 0 0( ) ( ) ( ) 6 824 4 3 24 24 24 8 B

B

dv w L w L L w L L w L w L w L w Ldx L EI

θ = = − + − = − + − = − Ans.





10.16 For the beam and loading shown inFig. P10.16, use the double-integrationmethod to determine (a) the equation of theelastic curve for the cantilever beam AB, (b)the deflection at the free end, and (c) theslope at the free end. Assume that EI isconstant for each beam.

Fig. P10.16

SolutionBeam FBD:

0

20

0

0

02 3

6

02

2

A A

A

y y

y

w L L M M

w L M

w L F A

w L A

Σ = − − =

= −

Σ = − =

=

Moment equation:3

0 ( )( ) 0

2 3a a

w L x M M x

L−

−Σ = − − =

30

3 2 2 30

2 2 30 0 0 0

( ) ( )6

( 3 3 )6

6 2 2 6

w M x L x

Lw

L L x Lx x L

w L w Lx w x w x L

= − −

= − − + −

= − + − +

Integration of moment equation:2 3 2 2

0 0 0 02 ( )

6 2 2 6d v w x w x w Lx w L

EI M xdx L

= = − + −

4 3 2 20 0 0 0

124 6 4 6dv w x w x w Lx w L x

EI C dx L

= − + − + (a)

5 4 3 2 20 0 0 0

1 2

120 24 12 12

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

L

= − + − + + (b)


0 at 0

v x

dv x

dx

= =

= =






(a) Elastic curve equation:5 4 3 2 2

0 0 0 0

5 4 2 3 3 20

120 24 12 12

5 10 10120

w x w x w Lx w L x EI v

L

wv x Lx L x L x L EI

= − + −

= − + − Ans.

(b) Deflection at the free end: 5 4

5 4 2 3 3 20 0 04( ) 5 ( ) 10 ( ) 10 ( )

120 120 30 B

w w L w Lv L L L L L L L

L EI L EI EI = − + − = − = − Ans.

(c) Slope at the free end:4 3 2 2 3

0 0 0 0 0( ) ( ) ( ) ( )

24 6 4 6 24 B

B

dv w L w L w L L w L L w L

dx L EI EI EI EI EI

θ = = − + − = − Ans.





10.17 For the beam and loading shown inFig. P10.17, use the double-integrationmethod to determine (a) the equation of theelastic curve for the cantilever beam, (b) thedeflection at B, (c) the deflection at the freeend, and (d) the slope at the free end. Assumethat EI is constant for the beam.

Fig. P10.17

SolutionBeam FBD:

2

02 2 4

38

02

2

A A

A

y y

y

wL L L M M

wL M

wL F A

wL A

Σ = − − + =

= −

Σ = − =

=

Consider beam segment AB (0 ≤ x ≤ L /2)Moment equation:

2

2

3( ) ( ) 0

8 23

( )8 2

a a A y

wL wL M M x M A x M x x

wL wLx M x

−Σ = − − = + − =

= − +


2

3( )

8 2d v wL wLx

EI M xdx

= = − +

2 2

1

38 4

dv wL x wLx EI C

dx= − + + (a)

2 2 3

1 2

316 12

wL x wLx EI v C x C = − + + + (b)


0 at 00 at 0

v xdv

xdx

= =

= =






Elastic curve equation for beam segment AB :

[ ]

2 2 3

2

316 12

9 4 (0 / 2)48

wL x wLx EI v

wL xv L x x L

EI

= − +

= − − ≤ ≤

Slope at B : Let x = L /2 2 2 33 ( / 2) ( / 2)

8 4 8 B B

dv wL L wL L wLdx EI EI EI

θ = = − + = −

Deflection at B : Let x = L /22 4( / 2) 7

9 448 2 192 B

wL L L wLv L

EI EI

= − − = −

Consider beam segment BC ( L /2 ≤ x ≤ L )

Moment equation: 2

22

( )2 2

3( ) 0

8 2 2 2

b b A y

w L M M x M A x x

wL wL w L M x x x

−

Σ = − − + −

= + − + − =

2 2

2 2

3( )

2 2 2 8

2 2

w L wL wL M x x x

wx wLwLx

= − − + −

= − + −

Integration of moment equation:2 2 2

2 ( )2 2

d v wx wL EI M x wLx

dx= = − + −

3 2 2

36 2 2dv wx wLx wL x

EI C dx

= − + − + (c)

4 3 2 2

3 424 6 4wx wLx wL x

EI v C x C = − + − + + (d)

Continuity conditions:4

3

7at

192 2

at8 2

wL Lv x

EI dv wL L

xdx EI

= − =

= − =

Evaluate constants:Substitute the slope continuity condition into Eq. (c) for x = L/ 2 and solve for C 3:





3 2 2 3

3

3

3

( / 2) ( / 2) ( / 2)6 2 2 8

48

dv w L wL L wL L wL EI C

dxwL

C

= − + − + = −

=

Next, substitute the deflection continuity condition into Eq. (d) for x = L/ 2 and solve for C 4 4 3 2 2 3 4

4

4

4

( / 2) ( / 2) ( / 2) 7( / 2)

24 6 4 48 192

384

w L wL L wL L wL wL EI v L C

wLC

= − + − + + = −

= −

Elastic curve equation for beam segment BC :4 3 2 2 3 4

4 3 2 2 3 4

24 6 4 48 384

16 64 96 8 ( / 2 )384

wx wLx wL x wL x wL EI v

wv x Lx L x L x L L x L

EI

= − + − − −

= − − + − + ≤ ≤

(a) Elastic curve equations for entire beam:

[ ]2

9 4 (0 / 2)48wL x

v L x x L EI

= − − ≤ ≤ Ans.

4 3 2 2 3 416 64 96 8 ( / 2 )384

wv x Lx L x L x L L x L

EI = − − + − + ≤ ≤ Ans.


192 B

wLv

EI = − Ans.

(c) Deflection at free end of cantilever:4

4 3 2 2 3 4 4116( ) 64 ( ) 96 ( ) 8 ( )

384 384C

w wLv L L L L L L L L

EI EI = − − + − + = − Ans.

(d) Slope at free end of cantilever: 3 2 2 3 3

3

8 ( ) 24 ( ) 24 ( ) 748 48 48 48 48

7

48C

C

dv w L wL L wL L wL wL EI

dx

dv wL

dx EI

θ

= − + − + = −

= = − Ans.





10.18 For the beam and loading shown inFig. P10.18, use the double-integrationmethod to determine (a) the equation othe elastic curve for the beam, and (b) thedeflection at B. Assume that EI is constantfor the beam.

Fig. P10.18

SolutionBeam FBD:

( ) 02 4

8

02

38

A y

y

y y y

y

wL L M C L

wLC

wL F A C

wL A

Σ = − =

=

Σ = + − =

=

Consider beam segment AB (0 ≤ x ≤ L /2)Moment equation:

2 2

2

3( ) ( ) 0

2 2 83

( )2 8

a a y

wx wx wL M M x A x M x x

wx wLx M x

−Σ = + − = + − =

= − +


2

3( )

2 8d v wx wLx

EI M xdx

= = − +

3 2

1

36 16

dv wx wLx EI C

dx= − + + (a)

4 3

1 224 16wx wLx

EI v C x C = − + + + (b)


0 at 0v x= =

Evaluate constants:Substitute x = 0 and v = 0 into Eq. (b) to determine C 2 = 0.

Slope at B : Let x = L /2 in Eq. (a). 3 2 3 3 3

1 1 1

( / 2) 3 ( / 2) 3 56 16 48 64 192 B

B

dv w L wL L wL wL wL EI EI C C C

dxθ = = − + + = − + + = + (c)





Deflection at B : Let x = L /2 in Eq. (b).4 3 4 4 4

1 11

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

24 16 384 128 2 192 2 B

w L wL L wL wL C L wL C L EI v C L= − + + = − + + = + (d)

Consider beam segment BC ( L /2 ≤ x ≤ L )

Moment equation:

2

( ) ( ) ( ) ( ) 08

( ) ( )8 8 8

b b y

wL M M x C L x M x L x

wL wL wLx M x L x

−Σ = − + − = − + − =

= − = −


2 ( )8 8

d v wLx wL EI M x

dx= = − +

2 2

3

16 8

dv wLx wL x EI C

dx= − + + (e)

3 2 2

3 448 16wLx wL x

EI v C x C = − + + + (f)

Boundary conditions:0 atv x L= =

Evaluate constants:Substitute x = L and v = 0 into Eq. (f) to find

3 2 2

3 4

4

3 4

( ) ( )(0) ( )

48 16

24

wL L wL L EI C L C

wLC L C

= − + + +

+ = − (g)

Slope at B : Let x = L /2 in Eq. (e). 2 2 3 3 3

3 3 3

( / 2) ( / 2) 316 8 64 16 64 B

B

dv wL L wL L wL wL wL EI EI C C C

dxθ = = − + + = − + + = + (h)

Deflection at B : Let x = L /2 in Eq. (f).3 2 2 4

33 4 4

( / 2) ( / 2) 5( / 2)48 16 384 2 B

wL L wL L wL C L EI v C L C C = − + + + = + + (i)

Continuity conditions:Since the slope at B must be the same for both beam segments, equate Eqs. (c) and (h):

3 3

1 3

5 3192 64wL wL

C C + = + (j)





Further, the deflection at B must be the same for both segments; therefore, equate Eqs. (d) and (i):4 4

1 34

5192 2 384 2wL C L wL C L

C + = + + (k)

Evaluate constants: Solve Eqs. (g), (j), and (k) simultaneously to determine the values of constants C 1,C 3, and C 4:

3 3 4

1 3 4

9 17

384 384 384

wL wL wLC C C = − = − =

(a) Elastic curve equation for beam segment AB :4 3 3

3 2 3

924 16 384

16 24 9 (0 / 2)384

wx wLx wL x EI v

wxv x Lx L x L

EI

= − + −

= − − + ≤ ≤ Ans.

(a) Elastic curve equation for beam segment BC :3 2 2 3 4

3 2 2 3

1748 16 384 384

8 24 17 ( / 2 )384

wLx wL x wL x wL EI v

wLv x Lx L x L L x L

EI

= − + − +

= − − + − ≤ ≤ Ans.

(b) Deflection at B :4 4 4 49 5 5

192 768 768 768 B B

wL wL wL wL EI v v

EI = − = − = − Ans.





10.19 For the beam and loading shown inFig. P10.19, use the double-integrationmethod to determine (a) the equation othe 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

SolutionBeam FBD:

(3 ) 3 02

76

0

6

A y

y

y y y

y

L M B L wL L

wL B

F A B wL

wL A

Σ = − + =

=

Σ = + − =

= −

Consider beam segment AB (0 ≤ x ≤ 3 L )Moment equation:

( ) ( ) 06

( )6

a a y

wL M M x A x M x x

wLx M x

−

Σ = − = + =

= −


2 ( ) 6d v wLx

EI M xdx = = − 2

112dv wLx

EI C dx

= − + (a)

3

1 236wLx

EI v C x C = − + + (b)

Boundary conditions:0 at 0 and 0 at 3v x v x L= = = =

Evaluate constants:Substitute x = 0 and v = 0 into Eq. (b) to determine C 2 = 0. Next, substitute x = 3 L and v = 0 into Eq. (b)and solve for C 1:

3 3 3

1 1

(3 ) 9(0) (3 )

36 36 4wL L wL wL

EI C L C = − + = =





Slope at B : Let x = 3 L in Eq. (a). 2 3 3(3 )

12 4 2 B B

dv wL L wL wL EI EI

dxθ = = − + = − (c)

Consider beam segment BC (3 L ≤ x ≤ 4 L )Moment equation:

2( ) (4 ) 02b bw M M x L x−

Σ = − − − =

2(4 )( )

2w L x

M x −

= −


2

(4 )( )

2d v w L x

EI M xdx

−= = −

3

3

(4 )

6

dv w L x EI C

dx

−= + (e)

4

3 4

(4 )24

w L x EI v C x C

−= − + + (f)

Boundary conditions:0 at 3v x L= =

Substitute x = 3 L and v = 0 into Eq. (f) to find4 4

3 4 3 4

4

4 3

(4 3 )(0) (3 ) (3 )

24 24

(3 )24

w L L wL EI C L C C L C

wLC L C

−= − + + = − + +

= − (g)

Slope at B : Let x = 3 L in Eq. (e). 3 3

3 3

(4 3 )6 6 B

B

dv w L L wL EI EI C C

dxθ

−= = + = + (h)

Continuity conditions:Since the slope at B must be the same for both beam segments, equate Eqs. (c) and (h):

3 3 3

3 3

22 6 3

wL wL wLC C − = + = − (i)

Backsubstitute this result into Eq. (g) to determine C 4:4 4 3 4

4 3

2 49(3 ) (3 )

24 24 3 24wL wL wL wL

C L C L −

= − = − =





(a) Elastic curve equation for beam segment AB :3 3

2 2

936 36

9 (0 3 )36

wLx wL x EI v

wLxv x L x L

EI

= − +

= − − ≤ ≤ Ans.

(a) Elastic curve equation for beam segment BC :4 3 4

4 3 4

(4 ) 2 4924 3 24

(4 ) 16 49 (3 4 )24

w L x wL x wL EI v

wv L x L x L L x L

EI

−= − − +

= − − + − ≤ ≤ Ans.

(b) Deflection at C :4

4 3 4 4 4

4

15(4 4 ) 16 (4 ) 49 64 49

24 24 24

58

C

C

w w wLv L L L L L L L

EI EI EI

wLv EI

= − − + − = − − = −

= − Ans.

(c) Slope at B : Let x = 3 L in Eq. (a). 3 3

2 2 B B B B

dv wL dv wL EI EI

dx dx EI θ θ = = − = = − Ans.





10.20 For the beam and loading shown inFig. P10.20, use the double-integrationmethod to determine (a) the equation othe elastic curve for the beam, (b) thelocation of the maximum deflection, and(c) the maximum beam deflection.Assume that EI is constant for the beam.

Fig. P10.20

SolutionBeam FBD:

0

0

0

0

20

2 3

3

02

6

A y

y

y y y

y

w L L M B L

w L B

w L F A B

w L A

Σ = − =

=

Σ = + − =

=

Moment equation:2

0

20 0

( )2 3

( ) 02 3 6

a a y

w x x M M x A x

L

w x x w Lx M x

L

−

Σ = + −

= + − =

30 0( )

6 6w x w Lx

M x L

= − +


0 02 ( )

6 6d v w x w Lx

EI M xdx L

= = − +

4 20 0

124 12dv w x w Lx

EI C dx L

= − + + (a)

5 30 0

1 2120 36w x w Lx

EI v C x C L

= − + + + (b)


0 at

v x

v x L

= =

= =



Evaluate constants:Substitute x = 0 and v = 0 into Eq. (b) to determine C 2 = 0. Next, substitute x = L and v = 0 into Eq. (b)and solve for C 1:

5 3 30 0 0

1 1

( ) ( ) 7(0) ( )

120 36 360w L w L L w L

EI C L C L

= − + + = −


4 2 2 40 0 0 07 3 10 7120 36 360 360w x w Lx w L x w x EI v v x L x L

L L EI = − + − = − − + 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:

4 2 30 0 07

024 12 360

dv w x w Lx w L EI

dx L= − + − =

Multiply by −360 L/w0 to obtain:4 2 2 415 30 7 0 x L x L− + =

Solve this equation numerically to obtain:

= 0.51932962236 0.51933 L L= Ans.

(c) Maximum beam deflection:4 2 2 40

max

4 440 0 0

(0.51933 )3(0.51933 ) 10 (0.51933 ) 7

360

(0.51933) (0.0065222)4.52118 0.00652

360

w Lv L L L L

L EI

w w L w L L

EI EI EI

= − − +

= − = − = − Ans.

mechanics of materials solutions chapter10 probs1 20

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