Mechanics of Materials Solutions Chapter10 Probs47 58

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10.47 The simply supported beam shown inFig. P10.47 consists of a W410 × 60

structural steel wide-flange shape [ E = 200

GPa; I = 216 × 106 mm

4]. For the loading

shown, determine:

(a) the beam deflection at point B.

(b) the beam deflection at point C .

(c) the beam deflection at point D.

Fig. P10.47

Solution

(a) Beam deflection at point B

Consider concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C:

2 2(2 3 )6

B

M xv L Lx x

LEI = − − + (elastic curve)

Values: M = −180 kN-m, L = 6 m, x = 1.5 m,

EI = 4.32 × 10

4

kN-m

2

Computation:

2 2

2 2

4 2

(2 3 )6

( 180 kN-m)(1.5 m)2(6 m) 3(6 m)(1.5 m) (1.5 m) 0.008203 m

6(6 m)(4.32 10 kN-m )

B

M xv L Lx x

LEI = − − +

−⎡ ⎤= − − + =⎣ ⎦×

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

Relevant equation from Appendix C:

2 2 2( )6

B

Pabv L a b

LEI = − − −

Values: P = 70 kN, L = 6 m, a = 1.5 m, b = 4.5 m,

EI = 4.32 × 104 kN-m

2

Computation:

2 2 2

2 2 2

4 2

( )6

(70 kN)(1.5 m)(4.5 m)(6 m) (1.5 m) (4.5 m) 0.004102 m

6(6 m)(4.32 10 kN-m )

B

Pabv L a b

LEI = − − −

⎡ ⎤= − − − = −⎣ ⎦×

Consider uniformly distributed load.

[Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C:

23 2 2 2 2(2 6 4 )

24 B

wav x Lx a x L x a L

LEI = − − + + −

Values: w = 80 kN/m, L = 6 m, a = 3 m, x = 4.5 m,

EI = 4.32 × 104 kN-m

2





Computation:2

3 2 2 2 2

23 2 2 2 2

4 2

(2 6 4 )24

(80 kN/m)(3 m)2(4.5) 6(6)(4.5) (3) (4.5) 4(6) (4.5) (3) (6)

24(6.0 m)(4.32 10 kN-m )

0.010156 m

B


LEI = − − + + −

⎡ ⎤= − − + + −⎣ ⎦×

= −

Beam deflection at B

0.008203 m 0.004102 m 0.010156 m 0.006055 m 6.06 mm Bv = − − = − = ↓ Ans.

(b) Beam deflection at point C Consider concentrated moment. [Appendix C, SS beam with concentrated moment at one end.]


2 2(2 3 )6

C

M xv L Lx x


Values:

M =−

180 kN-m, L = 6 m, x = 3.0 m, EI = 4.32 × 104 kN-m

2

Computation:

2 2

2 2

4 2

(2 3 )6

( 180 kN-m)(3.0 m)2(6 m) 3(6 m)(3.0 m) (3.0 m) 0.009375 m

6(6 m)(4.32 10 kN-m )

C

M xv L Lx x

LEI = − − +

−⎡ ⎤= − − + =⎣ ⎦×


Relevant equation from Appendix C:2 2 2( )

6C

Pbxv L b x

LEI = − − − (elastic curve)

Values: P = 70 kN, L = 6 m, x = 3.0 m, b = 1.5 m,

EI = 4.32 × 104 kN-m

2

Computation:

2 2 2

2 2 24 2

( )6

(70 kN)(1.5 m)(3.0 m) (6 m) (1.5 m) (3.0 m) 0.005013 m6(6 m)(4.32 10 kN-m )

C

Pbxv L b x

LEI = − − −

⎡ ⎤= − − − = −⎣ ⎦×







32 2(4 7 3 )

24C

wav L aL a

LEI = − − +

Values: w = 80 kN/m, L = 6 m, a = 3 m,

EI = 4.32 × 10

4

kN-m

2

Computation:3

2 2

32 2

4 2

(4 7 3 )24

(80 kN/m)(3 m)4(6 m) 7(3 m)(6 m) 3(3 m) 0.015625 m

24(6.0 m)(4.32 10 kN-m )

C

wav L aL a

LEI = − − +

⎡ ⎤= − − + = −⎣ ⎦×

Beam deflection at C

0.009375 m 0.005013 m 0.015625 m 0.011263 m 11.26 mmC v = − − = − = ↓ Ans.

(c) Beam deflection at point D Consider concentrated moment. [Appendix C, SS beam with concentrated moment at one end.]


2 2(2 3 )6

D

M xv L Lx x


Values: M = −180 kN-m, L = 6 m, x = 4.5 m,

EI = 4.32 × 104 kN-m

2

Computation:2 2

2 2

4 2

(2 3 )6

( 180 kN-m)(4.5 m)2(6 m) 3(6 m)(4.5 m) (4.5 m) 0.005859 m

6(6 m)(4.32 10 kN-m )

D

M xv L Lx x

LEI = − − +

−⎡ ⎤= − − + =⎣ ⎦×



2 2 2( )6

D

Pbxv L b x


Values: P = 70 kN, L = 6 m, x = 1.5 m, b = 1.5 m,

EI = 4.32 × 104 kN-m

2





Computation:

2 2 2

2 2 2

4 2

( )6

(70 kN)(1.5 m)(1.5 m)(6 m) (1.5 m) (1.5 m) 0.003190 m

6(6 m)(4.32 10 kN-m )

D

Pbxv L b x

LEI = − − −

⎡ ⎤= − − − = −⎣ ⎦×



3 2 2 2 2 2 3 4( 4 2 4 4 )24

D

wxv Lx aLx a x a L a L a

LEI = − − + + − +

Values: w = 80 kN/m, L = 6 m, a = 3 m, x = 1.5 m,

EI = 4.32 × 104 kN-m

2

Computation:

3 2 2 2 2 2 3 4

3 2 2 2 2 2 3 4

4 2

( 4 2 4 4 )

24(80 kN/m)(1.5 m)

(6)(1.5) 4(3)(6)(1.5) 2(3) (1.5) 4(3) (6) 4(3) (6) (3)24(6.0 m)(4.32 10 kN-m )

0.012109 m

D

wxv Lx aLx a x a L a L a

LEI

= − − + + − +

⎡ ⎤= − − + + − +⎣ ⎦×

= −

Beam deflection at D

0.005859 m 0.003190 m 0.012109 m 0.009440 m 9.44 mm Dv = − − = − = ↓ Ans.





10.48 The simply supported beam shown inFig. P10.48 consists of a W530 × 66 structural

steel wide-flange shape [ E = 200 GPa; I = 351

× 106 mm

4]. For the loading shown,

determine:

(a) the beam deflection at point A.


(c) the beam deflection at point E .

Fig. P10.48

Solution

(a) Beam deflection at point A

Determine cantilever deflection due to concentrated load on overhang AB. [Appendix C, Cantilever beam with concentrated load.] Relevant equation from Appendix C:

3

3 A

PLv

EI = − (assuming fixed support at B)

Values:

P = 35 kN, L = 4 m, EI = 7.02 × 10

4

kN-m

2

Computation:

3 3

4 2

(35 kN)(4 m)0.0106363 m

3 3(7.02 10 kN-m ) A

PLv

EI = − = − = −

×

Consider deflection at A resulting from rotation at B caused by concentrated load on overhang

AB. [Appendix C, SS beam with concentrated moment.]


3 B

L

EI θ = (slope magnitude)

Values: M = (35 kN)(4 m) = 140 kN-m, L = 8 m,

EI = 7.02 × 104 kN-m

2

Computation:

4 2

(140 kN-m)(8 m)0.0053181 rad

3 3(7.02 10 kN-m )

(4 m)(0.0053181 rad) 0.0212726 m

B

A

ML

EI

v

θ = = =×

= − = −

Consider uniformly distributed loads between C and D. [Appendix C, SS beam with uniformly distributed load over portion of span.]

Relevant equations from Appendix C:2

2 2(2 )24

B

wa L a

LEI θ = − (slope magnitude)

Values: w = 80 kN/m, L = 8 m, a = 4 m,

EI = 7.02 × 104 kN-m

2





Computation:2 2

2 2 2 2

4 2

(80 kN/m)(4 m)(2 ) 2(8 m) (4 m) 0.0106363 rad

24 24(8 m)(7.02 10 kN-m )

(4 m)(0.0106363 rad) 0.0425451 m

B

A

wa L a

LEI

v

θ ⎡ ⎤= − = − =⎣ ⎦×

= =

Consider deflection at A resulting from rotation at B caused by uniform load on overhang DE .

[Appendix C, SS beam with concentrated moment.]


6 B

L


Values: M = (80 kN/m)(2 m)(1 m) = 160 kN-m,

L = 8 m, EI = 7.02 × 104 kN-m

2

Computation:

4 2

(160 kN-m)(8 m)0.0030389 rad

6 6(7.02 10 kN-m )

(4 m)(0.0030389 rad) 0.0121557 m

B

A

ML

EI

v

θ = = =×

= − = −

Beam deflection at A

0.0106363 m 0.0212726 m 0.0425451 m 0.0121557 m

0.0015195 m 1.520 mm

Av = − − + −

= − = ↓ Ans.

(b) Beam deflection at point C

Consider concentrated moment from overhang AB.



2 2(2 3 )6

C M xv L Lx x LEI

= − − + (elastic curve)

Values: M = (35 kN)(4 m) = −140 kN-m, L = 8 m,

x = 4 m, EI = 7.02 × 104 kN-m

2

Computation:

2 2

2 2

4 2

(2 3 )6

( 140 kN-m)(4 m)

2(8 m) 3(8 m)(4 m) (4 m) 0.0079772 m6(8 m)(7.02 10 kN-m )

C

M xv L Lx x

LEI = − − +

−⎡ ⎤= − − + =⎣ ⎦×







2 2(4 7 3 )24

C

wav L aL a

LEI = − − +

Values: w = 80 kN/m, L = 8 m, a = 4 m,

EI = 7.02 × 10

4

kN-m

2

Computation:3

2 2

32 2

4 2

(4 7 3 )24

(80 kN/m)(4 m)4(8 m) 7(4 m)(8 m) 3(4 m) 0.0303894 m

24(8 m)(7.02 10 kN-m )

C

wav L aL a

LEI = − − +

⎡ ⎤= − − + = −⎣ ⎦×

Consider concentrated moment from overhang DE .


Relevant equation from Appendix C:2 2(2 3 )

6C

M xv L Lx x



L = 8 m, x = 4 m, EI = 7.02 × 104 kN-m

2

Computation:

2 2

2 2

4 2

(2 3 )6

( 160 kN-m)(4 m) 2(8 m) 3(8 m)(4 m) (4 m) 0.0091168 m6(8 m)(7.02 10 kN-m )

C

M xv L Lx x

LEI = − − +

− ⎡ ⎤= − − + =⎣ ⎦×


0.0079772 m 0.0303894 m 0.0091168 m 0.0132954 m 13.30 mmC v = − + = − = ↓ Ans.

(c) Beam deflection at point E

Consider deflection at E resulting from rotation at D caused by concentrated load on overhang

AB. [Appendix C, SS beam with concentrated moment.]


6 D

L EI

θ = (slope magnitude)

Values: M = (35 kN)(4 m) = 140 kN-m, L = 8 m,

EI = 7.02 × 104 kN-m

2

Computation:





4 2

(140 kN-m)(8 m)0.0026591 rad

6 6(7.02 10 kN-m )

(2 m)(0.0026591 rad) 0.0053181 m

D

E

ML

EI

v

θ = = =×

= − = −


Relevant equations from Appendix C:

2 2(2 )24

D

wa L a LEI

θ = − (slope magnitude)

Values: w = 80 kN/m, L = 8 m, a = 4 m,

EI = 7.02 × 104 kN-m

2

Computation:

[ ]2 2

22

4 2

(80 kN/m)(4 m)(2 ) 2(8 m) (4 m) 0.0136752 rad

24 24(8 m)(7.02 10 kN-m )

(2 m)(0.0136752 rad) 0.0273504 m

D

E

wa L a

LEI

v

θ = − = − =×

= =

Consider deflection at E resulting from rotation at D caused by uniform load on overhang DE .



3 D

L



L = 8 m, EI = 7.02 × 104 kN-m

2

Computation:

4 2

(160 kN-m)(8 m)0.0060779 rad

3 3(7.02 10 kN-m )

(2 m)(0.0060779 rad) 0.0121557 m

D

E

ML

EI

v

θ = = =

×

= − = −

Determine cantilever deflection due to uniformly distributed load on overhang DE . [Appendix C, Cantilever beam with distributed load.]

Relevant equation from Appendix C:4

8 E

wLv

EI = − (assuming fixed support at D)

Values: w = 80 kN/m, L = 2 m, EI = 7.02 × 10

4 kN-m

2

Computation:4 4

4 2

(80 kN/m)(2 m)0.0022792 m

8 8(7.02 10 kN-m ) E

wLv

EI = − = − = −

×

Beam deflection at E

0.0053181 m 0.0273504 m 0.0121557 m 0.0022792 m

0.0075974 m 7.60 mm

E v = − + − −

= = ↑ Ans.





10.49 The simply supported beam shown inFig. P10.49 consists of a W16 × 40

structural steel wide-flange shape [ E =

29,000 ksi; I = 518 in.4]. For the loading

shown, determine:



(c) the beam deflection at point F .

Fig. P10.49

Solution


Consider 40-kip concentrated load at B. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C:

2 2 2( )6

B

Pabv L a b

LEI = − − −

Values:

P = 40 kips, L = 18 ft, a = 4 ft, b = 14 ft, EI = 1.5022 × 107 kip-in.

2

Computation:

2 2 2

32 2 2

7 2

( )6

(40 kips)(4 ft)(14 ft)(12 in./ft)(18 ft) (4 ft) (14 ft) 0.267213 in.

6(18 ft)(1.5022 10 kip-in. )

B

Pabv L a b

LEI = − − −

⎡ ⎤= − − − = −⎣ ⎦×

Consider 30-kip concentrated load at D.

[Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C:

2 2 2( )6

B

Pbxv L b x


Values: P = 30 kips, L = 18 ft, x = 4 ft, b = 6 ft,

EI = 1.5022 × 107 kip-in.

2

Computation:

2 2 2

32 2 2

7 2

( )6


6(18 ft)(1.5022 10 kip-in. )

B

Pbxv L b x

LEI = − − −

⎡ ⎤= − − − = −⎣ ⎦×





Consider 20-kip concentrated load at F . [Appendix C, SS beam with concentrated moment.]


2 2(2 3 )6

B

M xv L Lx x


Values: M = −(20 kips)(6 ft) = −120 kip-ft, L = 18 ft,

x = 14 ft, EI = 1.5022 × 107 kip-in.

2

Computation:

2 2

32 2

7 2

(2 3 )6

( 120 kip-ft)(14 ft)(12 in./ft)2(18 ft) 3(18 ft)(14 ft) (14 ft) 0.157465 in.

6(18 ft)(1.5022 10 kip-in. )

B

M xv L Lx x

LEI = − − +

−⎡ ⎤= − − + =⎣ ⎦×


0.267213 in. 0.208590 in. 0.157465 in. 0.318338 in. 0.318 in. Bv = − − + = − = ↓ Ans.

(b) Beam deflection at point C Consider 40-kip concentrated load at B. [Appendix C, SS beam with concentrated load not at midspan.]


2 2 2( )6

C

Pbxv L b x


Values: P = 40 kips, L = 18 ft, b = 4 ft, x = 10 ft,

EI = 1.5022 × 107 kip-in.

2

Computation:

2 2 2

32 2 2

7 2

( )6


6(18 ft)(1.5022 10 kip-in. )

C

Pbxv L b x

LEI = − − −

⎡ ⎤= − − − = −⎣ ⎦×

Consider 30-kip concentrated load at D. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C:

2 2 2( )

6

C

Pbxv L b x

LEI

= − − − (elastic curve)


EI = 1.5022 × 107 kip-in.

2





Computation:

2 2 2

32 2 2

7 2

( )6


6(18 ft)(1.5022 10 kip-in. )

C

Pbxv L b x

LEI = − − −

⎡ ⎤= − − − = −⎣ ⎦×

Consider 20-kip concentrated load at F . [Appendix C, SS beam with concentrated moment.]


6C

M xv L Lx x


Values: M = −(20 kips)(6 ft) = −120 kip-ft, L = 18 ft,

x = 10 ft, EI = 1.5022 × 107 kip-in.

2

Computation:

2 2

32 2

7 2

(2 3 )6


6(18 ft)(1.5022 10 kip-in. )

C

M xv L Lx x

LEI = − − +

− ⎡ ⎤= − − + =⎣ ⎦×


0.354467 in. 0.343560 in. 0.265850 in. 0.432177 in. 0.432 in.C v = − − + = − = ↓ Ans.

(c) Beam deflection at point F

Consider 40-kip concentrated load at B. [Appendix C, SS beam with concentrated load not at midspan.]

Relevant equation from Appendix C:2 2( )

6 E

Pa L a

LEI θ

−= (slope magnitude)

Values: P = 40 kips, L = 18 ft, a = 4 ft,

EI = 1.5022 × 107 kip-in.

2

Computation:2 2 2

2 2

7 2

( ) (40 kips)(4 ft)(12 in./ft)(18 ft) (4 ft) 0.0043740 rad

6 6(18 ft)(1.5022 10 kip-in. )

(6 ft)(12 in./ft)(0.0043740 rad) 0.314930 in.

E

F

Pa L a

LEI

v

θ −

⎡ ⎤= = − =⎣ ⎦×

= =





Consider 30-kip concentrated load at D. [Appendix C, SS beam with concentrated load not at midspan.]


6 E

Pa L a

LEI θ


Values: P = 30 kips, L = 18 ft, x = 8 ft, a = 12 ft,

EI = 1.5022 × 10

7

kip-in.

2

Computation:2 2 2

2 2

7 2


6 6(18 ft)(1.5022 10 kip-in. )

(6 ft)(12 in./ft)(0.0057516 rad) 0.414113 in.

E

F

Pa L a

LEI

v

θ −

⎡ ⎤= = − =⎣ ⎦×

= =

Consider deflection at F resulting from rotation at E caused by 20-kip load on overhang EF .



3 E L


Values: M = (20 kips)(6 ft) = 120 kip-ft, L = 18 ft,

EI = 1.5022 × 107 kip-in.

2

Computation:2

7 2

(120 kip-ft)(18 ft)(12 in./ft)0.0069019 rad

3 3(1.5022 10 kip-in. )

(6 ft)(12 in./ft)(0.0069019 rad) 0.496935 in.

E

F

ML

EI

v

θ = = =×

= − = −

Determine cantilever deflection due to concentrated load on overhang EF . [Appendix C, Cantilever beam with concentrated load.]


3 F

PLv

EI = − (assuming fixed support at E )

Values: P = 20 kips, L = 6 ft, EI = 1.5022 × 10

7 kip-in.

2

Computation:3 3 3

7 2

(20 kips)(6 ft) (12 in./ft)0.165645 in.

3 3(1.5022 10 kip-in. ) F

PLv

EI = − = − = −

×

Beam deflection at F

0.314930 in. 0.414113 in. 0.496935 in. 0.165645 in.

0.066463 in. 0.0665 in.

F v = + − −

= = ↑ Ans.





10.50 The cantilever beam shown in Fig.P10.50 consists of a rectangular structural

steel tube shape [ E = 200 GPa; I = 170 ×

106 mm


determine:


(b) the beam deflection at point B.

Fig. P10.50

Solution


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


8 A

wLv

EI = −

Values: w = −65 kN/m, L = 6 m, EI = 3.4 × 10

4 kN-m

2

Computation:4 4

4 2

( 65 kN/m)(6 m)0.309706 m

8 8(3.4 10 kN-m ) A

wLv

EI

−= − = − =

×

Consider 90-kN concentrated load at A. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C:

3

3 A

PLv

EI = −

Values: P = 90 kN, L = 6 m, EI = 3.4 × 10

4 kN-m

2

Computation:3 3

4 2

(90 kN)(6 m)0.190588 m

3 3(3.4 10 kN-m ) A

PLv

EI = − = − = −

×

Consider 30-kN concentrated load at B. [Appendix C, Cantilever beam with concentrated load at tip.]

Relevant equations from Appendix C:

3 2

and3 2

B B

PL PLv

EI EI θ = − = (magnitude)

Values: P = 30 kN, L = 3.5 m, EI = 3.4 × 10

4 kN-m

2





Computation:3 3

4 2

(30 kN)(3.5 m)0.012610 m

3 3(3.4 10 kN-m ) B

PLv

EI = − = − = −

× (a)

2 2

4 2

(30 kN)(3.5 m)0.0054044 rad

2 2(3.4 10 kN-m )

0.012610 m (2.5 m)(0.0054044 rad) 0.026121 m

B

A

PL

EI

v

θ = = =×

= − − = −

Consider 225 kN-m concentrated moment at B. [Appendix C, Cantilever beam with concentrated moment at tip.] Relevant equations from Appendix C:

2

and2

B B

L MLv

EI EI θ = − = (slope magnitude)

Values: M = 225 kN-m, L = 3.5 m, EI = 3.4 × 10

4 kN-m

2

Computation:2 2

4 2

(225 kN-m)(3.5 m)0.040533 m

2 2(3.4 10 kN-m ) B

MLv

EI = − = − = −

× (b)

4 2

(225 kN-m)(3.5 m)0.0231618 rad

(3.4 10 kN-m )

0.040533 m (2.5 m)(0.0231618 rad) 0.098438 m

B

A

ML

EI

v

θ = = =×

= − − = −


0.309706 m 0.190588 m 0.026121 m 0.098438 m 0.005441 m 5.44 mm Av = − − − = − = ↓ Ans.

(b) Beam deflection at point B

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


2 2(6 4 )24

B

wxv L Lx x

EI = − − + (elastic curve)

Values: w = −65 kN/m, L = 6 m, x = 3.5 m,

EI = 3.4 × 104 kN-m

2

Computation:2

2 2

22 2

4 2

(6 4 )24

( 65 kN/m)(3.5 m)6(6 m) 4(6 m)(3.5 m) (3.5 m) 0.140759 m

24(3.4 10 kN-m )

B

wxv L Lx x

EI = − − +

−⎡ ⎤= − − + =⎣ ⎦×





Consider 90-kN concentrated load at A. [Appendix C, Cantilever beam with concentrated load at tip.]


(3 )6

B

Pxv L x

EI = − − (elastic curve)

Values: P = 90 kN, L = 6 m, x = 3.5 m,

EI = 3.4 × 104 kN-m

2

Computation:

[ ]2 2

4 2

(90 kN)(3.5 m)(3 ) 3(6 m) (3.5 m) 0.078364 m

6 6(3.4 10 kN-m ) B

Pxv L x

EI = − − = − − = −

×

Consider 30-kN concentrated load at B. Previously calculated in Eq. (a).

Consider 225 kN-m concentrated moment at B. Previously calculated in Eq. (b).


0.140759 m 0.078364 m 0.012610 m 0.040533 m 0.009252 m 9.25 mm Bv = − − − = = ↑ Ans.





10.51 The simply supported beam shown inFig. P10.51 consists of a rectangular

structural steel tube shape [ E = 200 GPa; I =

350 × 106 mm


determine:


(b) the beam deflection at point C .(c) the beam deflection at point E .

Fig. P10.51

Solution


Consider 315 kN-m concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C:

2 2(2 3 )6

B

M xv L Lx x


Values:

M =−

315 kN-m, L = 9 m, x = 3 m, EI = 7.0 × 10

4 kN-m

2

Computation:

2 2

2 2

4 2

(2 3 )6

( 315 kN-m)(3 m)2(9 m) 3(9 m)(3 m) (3 m) 0.022500 m

6(9 m)(7.0 10 kN-m )

B

M xv L Lx x

LEI = − − +

−⎡ ⎤= − − + =⎣ ⎦×

Consider 120 kN/m uniformly distributed load.


32 2(4 7 3 )

24 B

wav L aL a

LEI = − − +

Values: w = 120 kN/m, L = 9 m, a = 3 m,

EI = 7.0 × 104 kN-m

2

Computation:3

2 2

32 2

4 2

(4 7 3 )

24

(120 kN/m)(3 m)4(9 m) 7(3 m)(9 m) 3(3 m) 0.034714 m

24(9 m)(7.0 10 kN-m )

B

wav L aL a

LEI

= − − +

⎡ ⎤= − − + = −⎣ ⎦×





Consider 100-kN concentrated load. [Appendix C, SS beam with concentrated load not at midspan.]


2 2 2( )6

B

Pbxv L b x


Values: P = 100 kN, L = 9 m, b = 3 m, x = 3 m,

EI = 7.0 × 104 kN-m

2

Computation:

2 2 2

2 2 2

4 2

( )6

(100 kN)(3 m)(3 m)(9 m) (3 m) (3 m) 0.015000 m

6(9 m)(7.0 10 kN-m )

B

Pbxv L b x

LEI = − − −

⎡ ⎤= − − − = −⎣ ⎦×

Consider 60 kN/m uniformly distributed load on overhang DE . [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C:

2 2(2 3 )6

B M xv L Lx x LEI = − − + (elastic curve)

Values: M = −(60 kN/m)(3 m)(1.5 m) = −270 kN-m,

L = 9 m, x = 6 m, EI = 7.0 × 104 kN-m

2

Computation:

2 2

2 2

4 2

(2 3 )6

( 270 kN-m)(6 m)2(9 m) 3(9 m)(6 m) (6 m) 0.015429 m

6(9 m)(7.0 10 kN-m )

B

M xv L Lx x

LEI = − − +

−⎡ ⎤= − − + =

⎣ ⎦×


0.022500 m 0.034714 m 0.015000 m 0.015429 m 0.011785 m 11.79 mm Bv = − − + = − = ↓ Ans.


Consider 315 kN-m concentrated moment. [Appendix C, SS beam with concentrated moment at one end.]


2 2(2 3 )6

C

M xv L Lx x


Values: M = −315 kN-m, L = 9 m, x = 6 m,

EI = 7.0 × 104 kN-m

2





Computation:

2 2

2 2

4 2

(2 3 )6

( 315 kN-m)(6 m)2(9 m) 3(9 m)(6 m) (6 m) 0.018000 m

6(9 m)(7.0 10 kN-m )

C

M xv L Lx x

LEI = − − +

−⎡ ⎤= − − + =⎣ ⎦×



23 2 2 2 2(2 6 4 )

24C


LEI = − − + + −

Values: w = 120 kN/m, L = 9 m, a = 3 m, x = 6 m,

EI = 7.0 × 104 kN-m

2

Computation:2

3 2 2 2 2

23 2 2 2 2

4 2

(2 6 4 )24

(120 kN/m)(3 m)2(6 m) 6(9 m)(6 m) (3 m) (6 m) 4(9 m) (6 m) (3 m) (9 m)

24(9 m)(7.0 10 kN-m )

0.028929 m

C


LEI = − − + + −

⎡ ⎤= − − + + −⎣ ⎦×

= −

Consider 100-kN concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C:

2 2 2( )6

C

Pabv L a b

LEI = − − −

Values:

P = 100 kN, L = 9 m, a = 6 m, b = 3 m, EI = 7.0 × 104 kN-m

2

Computation:

2 2 2

2 2 2

4 2

( )6

(100 kN)(6 m)(3 m)(9 m) (6 m) (3 m) 0.017143 m

6(9 m)(7.0 10 kN-m )

C

Pabv L a b

LEI = − − −

⎡ ⎤= − − − = −⎣ ⎦×

Consider 60 kN/m uniformly distributed load on overhang DE . [Appendix C, SS beam with concentrated moment.]


6C

M xv L Lx x


Values: M = −(60 kN/m)(3 m)(1.5 m) = −270 kN-m, L = 9 m, x = 3 m, EI = 7.0 × 10

4 kN-m

2





Computation:

2 2

2 2

4 2

(2 3 )6

( 270 kN-m)(3 m)2(9 m) 3(9 m)(3 m) (3 m) 0.019286 m

6(9 m)(7.0 10 kN-m )

C

M xv L Lx x

LEI = − − +

−⎡ ⎤= − − + =⎣ ⎦×


0.018000 m 0.028929 m 0.017143 m 0.019286 m 0.008786 m 8.79 mmC v = − − + = − = ↓ Ans.

(c) Beam deflection at point E

Consider 315 kN-m concentrated moment. [Appendix C, SS beam with concentrated moment at one end.]


6 D

L


Values: M = −315 kN-m, L = 9 m,

EI = 7.0 × 104

kN-m2

Computation:

4 2

( 315 kN-m)(9 m)0.0067500 rad

6 6(7.0 10 kN-m )

(3 m)( 0.0067500 rad) 0.020250 m

D

E

ML

EI

v

θ −

= = = −×

= − = −


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


2 2(2 )24

D

wa L a


Values: w = 120 kN/m, L = 9 m, a = 3 m,

EI = 7.0 × 104 kN-m

2

Computation:2 2

2 2 2 2

4 2

(120 kN/m)(3 m)(2 ) 2(9 m) (3 m) 0.0109286 rad

24 24(9 m)(7.0 10 kN-m )

(3 m)(0.0109286 rad) 0.032786 m

D

E

wa L a

LEI

v

θ ⎡ ⎤= − = − =⎣ ⎦×

= =







6 D

Pa L a

LEI θ


Values: P = 100 kN, L = 9 m, a = 6 m,

EI = 7.0 × 104 kN-m

2

Computation:2 2

2 2

4 2

( ) (100 kN)(6 m)(9 m) (6 m) 0.0071429 rad

6 6(9 m)(7.0 10 kN-m )

(3 m)(0.0071429 rad) 0.021429 m

D

E

Pa L a

LEI

v

θ −

⎡ ⎤= = − =⎣ ⎦×

= =

Consider 60 kN/m uniformly distributed load on overhang DE . [Appendix C, SS beam with concentrated moment.]


3 D

L

EI


Values: M = −(60 kN/m)(3 m)(1.5 m) = −270 kN-m,

L = 9 m, EI = 7.0 × 104 kN-m

2

Computation:

4 2

(270 kN-m)(9 m)0.0115714 rad

3 3(7.0 10 kN-m )

(3 m)(0.0115714 rad) 0.034714 m

D

E

ML

EI

v

θ = = =×

= − = −

Determine cantilever deflection due to 60 kN/m uniformly distributed load on overhang DE . [Appendix C, Cantilever beam with concentrated load.] Relevant equation from Appendix C:

4

8 E

wLv

EI = − (assuming fixed support at D)

Values: w = 60 kN/m, L = 3 m, EI = 7.0 × 10

4 kN-m

2

Computation:4 4

4 2

(60 kN-m)(3 m)0.008679 m

8 8(7.0 10 kN-m ) E

wLv

EI = − = − = −

×

Beam deflection at E

0.020250 m 0.032786 m 0.021429 m 0.034714 m 0.008679 m

0.009429 m 9.43 mm

E v = − + + − −

= − = ↓ Ans.





10.52 The cantilever beam shown in Fig.P10.52 consists of a rectangular structural

steel tube shape [ E = 200 GPa; I = 95 × 106

mm4]. For the loading shown, determine:



Fig. P10.52

Solution


Consider the downward 50 kN/m uniformly distributed load acting over span AB.

[Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C:

4

8 B

wLv

EI = −

Values: w = 50 kN/m, L = 2 m, EI = 1.9 × 10

4 kN-m

2

Computation:4 4

4 2

(50 kN/m)(2 m)0.0052632 m

8 8(1.9 10 kN-m ) B

wLv

EI = − = − = −

× (a)

Consider an upward 25 kN/m uniformly distributed load acting over entire 5-m span.

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


2 2(6 4 )24

B

wxv L Lx x

EI = − − + (elastic curve)

Values: w = −25 kN/m, L = 5 m, x = 2 m,

EI = 1.9 × 104 kN-m

2

Computation:2

2 2

22 2

4 2

(6 4 )24

( 25 kN/m)(2 m)6(5 m) 4(5 m)(2 m) (2 m) 0.0250000 m

24(1.9 10 kN-m )

B

wxv L Lx x

EI = − − +

−⎡ ⎤= − − + =⎣ ⎦×

Consider a downward 25 kN/m uniformly distributed load acting over span AB.

[Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C:4

8 B

wLv

EI = −

Values: w = 25 kN/m, L = 2 m, EI = 1.9 × 10

4 kN-m

2





Computation:4 4

4 2

(25 kN/m)(2 m)0.0026316 m

8 8(1.9 10 kN-m ) B

wLv

EI = − = − = −

× (b)

Consider 20-kN concentrated load at B. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C:

3

3 B

PLv

EI = −

Values: P = −20 kN, L = 2 m, EI = 1.9 × 10

4 kN-m

2

Computation:3 3

4 2

( 20 kN)(2 m)0.0028070 m

3 3(1.9 10 kN-m ) B

PLv

EI

−= − = − =

× (c)

Consider 50-kN concentrated load at C . [Appendix C, Cantilever beam with concentrated load at tip.]


(3 )6 B

Pxv L x EI = − − (elastic curve)

Values: P = 50 kN, L = 5 m, x = 2 m, EI = 1.9 × 10

4 kN-m

2

Computation:

[ ]2 2

4 2

(50 kN)(2 m)(3 ) 3(5 m) (2 m) 0.0228070 m

6 6(1.9 10 kN-m ) B

Pxv L x

EI = − − = − − = −

×


0.0052632 m 0.0250000 m 0.0026316 m 0.0028070 m 0.0228070 m

0.0028947 m 2.89 mm

Bv = − + − + −

= − = ↓ Ans.






Consider the downward 50 kN/m uniformly distributed load acting over span AB.


3

6 B

wL


Values: w = 50 kN/m, L = 2 m, EI = 1.9 × 10

4 kN-m

2

Computation: [v B previously calculated in Eq. (a)]3 3

4 2

(50 kN/m)(2 m)0.0035088 rad

6 6(1.9 10 kN-m )

0.0052632 m (3 m)(0.0035088 rad) 0.0157895 m

B

C

wL

EI

v

θ = = =×

= − − = −

Consider an upward 25 kN/m uniformly distributed load acting over entire 5-m span.

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


4

8C

wLv EI

= −

Values: w = −25 kN/m, L = 5 m, EI = 1.9 × 10

4 kN-m

2

Computation:4 4

4 2

( 25 kN/m)(5 m)0.1027961 m

8 8(1.9 10 kN-m )C

wLv

EI

−= − = − =

×

Consider a downward 25 kN/m uniformly distributed load acting over span AB.


3

6 B

wL


Values: w = 25 kN/m, L = 2 m, EI = 1.9 × 10

4 kN-m

2

Computation: [v B previously calculated in Eq. (b)]3 3

4 2

(25 kN)(2 m)0.0017544 rad

6 6(1.9 10 kN-m )0.0026316 m (3 m)(0.0017544 rad) 0.0078948 m

B

C

wL

EI v

θ = = =×

= − − = −





Consider 20-kN concentrated load at B. [Appendix C, Cantilever beam with concentrated load at tip.]


2 B

PL

EI θ =

Values: P = 20 kN, L = 2 m, EI = 1.9 × 10

4 kN-m

2

Computation: [v B previously calculated in Eq. (c)] 2 2

4 2

(20 kN)(2 m)0.0021053 rad

2 2(1.9 10 kN-m )

0.0028070 m (3 m)(0.0021053 rad) 0.0091228 m

B

C

PL

EI

v

θ = = =×

= + =

Consider 50-kN concentrated load at C . [Appendix C, Cantilever beam with concentrated load at tip.]


3C

PLv

EI = −

Values: P = 50 kN, L = 5 m, EI = 1.9 × 104 kN-m2

Computation:3 3

4 2

(50 kN)(5 m)0.1096491 m

3 3(1.9 10 kN-m )C

PLv

EI = − = − = −

×


0.0157895 m 0.1027961 m 0.0078948 m 0.0091228 m 0.1096491 m

0.0214145 m 21.4 mm

C v = − + − + −

= − = ↓ Ans.





10.53 The simply supported beam shownin Fig. P10.53 consists of a W10 × 30



shown, determine:


(b) the beam deflection at point C .(c) the beam deflection at point D.

Fig. P10.53

Solution


Consider cantilever beam deflection of 85 kip-ft concentrated moment. [Appendix C, Cantilever beam with concentrated moment at one end.] Relevant equation from Appendix C:

2

2 A

Lv

EI = −

Values:

M = 85 kip-ft, L = 3 ft, EI = 4.93 × 106

kip-in.2

Computation:2 2 3

6 2

(85 kip-ft)(3 ft) (12 in./ft)0.134069 in.

2 2(4.93 10 kip-in. ) A

MLv

EI = − = − = −

×

Consider rotation at B caused by 85 kip-ft concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C:

3 B

L

EI


Values: M = 85 kip-ft, L = 15 ft, EI = 4.93 × 10

6 kip-in.

2

Computation:2

6 2

(85 kip-ft)(15 ft)(12 in./ft)0.0124138 rad

3 3(4.93 10 kip-in. )

(3 ft)(12 in./ft)(0.0124138 rad) 0.446897 in.

B

A

ML

EI

v

θ = = =×

= − = −

Consider cantilever beam deflection of 5 kips/ft uniformly distributed load.


4

8 A

wLv

EI = −

Values: w = 5 kips/ft, L = 3 ft, EI = 4.93 × 10

6 kip-in.

2





Computation:4 4 3

6 2

(5 kips/ft)(3 ft) (12 in./ft)0.017744 in.

8 8(4.93 10 kip-in. ) A

wLv

EI = − = − = −

×

Consider rotation at B caused by 5 kips/ft uniformly distributed load. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C:

3 B L


Values: M = (5 kips/ft)(3 ft)(1.5 ft) = 22.5 kip-ft, L = 15 ft, EI = 4.93 × 10

6 kip-in.

2

Computation:2

6 2

(22.5 kip-ft)(15 ft)(12 in./ft)0.0032860 rad

3 3(4.93 10 kip-in. )

(3 ft)(12 in./ft)(0.0032860 rad) 0.118296 in.

B

A

ML

EI

v

θ = = =×

= − = −

Consider 5 kips/ft uniformly distributed load on segment BC .[Appendix C, SS beam with uniformly distributed load over a portion of the span.]


2(2 )24

B

wa L a


Values: w = 5 kips/ft, L = 15 ft, a = 5 ft,

EI = 4.93 × 106 kip-in.

2

Computation:

[ ]2 2 2

22

6 2(5 kips/ft)(5 ft) (12 in./ft)(2 ) 2(15 ft) (5 ft) 0.0063387 rad

24 24(15 ft)(4.93 10 kip-in. )

(3 ft)(12 in./ft)(0.0063387 rad) 0.228195 in.

B

A

wa L a LEI

v

θ = − = − =×

= =

Consider 25-kip concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C:

2 2( )

6 B

Pb L b

LEI θ


Values: P = 25 kips, L = 15 ft, b = 5 ft,

EI = 4.93 × 106 kip-in.2

Computation:2 2 2

2 2

6 2


6 6(15 ft)(4.93 10 kip-in. )

(3 ft)(12 in./ft)(0.0081136 rad) 0.292089 in.

B

A

Pb L b

LEI

v

θ −

⎡ ⎤= = − =⎣ ⎦×

= =






0.134069 in. 0.446897 in. 0.017744 in. 0.118296 in. 0.228195 in. 0.292089 in.

0.196722 in. 0.1967 in.

Av = − − − − + +

= − = ↓ Ans.


Consider 85 kip-ft concentrated moment. [Appendix C, SS beam with concentrated moment at one end.]


6C

M xv L Lx x


Values: M = −85 kip-ft, L = 15 ft, x = 5 ft, EI = 4.93 × 10

6 kip-in.

2

Computation:

2 2

3 2 2

6 2

(2 3 )6

( 85 kip-ft)(5 ft)(12 in./ft) 2(15 ft) 3(15 ft)(5 ft) (5 ft) 0.413793 in.6(15 ft)(4.93 10 kip-in. )

C

M xv L Lx x

LEI = − − +

− ⎡ ⎤= − − + =⎣ ⎦×

Consider moment at B caused by 5 kips/ft uniformly distributed load on overhang AB. [Appendix C, SS beam with concentrated moment at one end.]


2 2(2 3 )6

C

M xv L Lx x


Values: M = −(5 kips/ft)(3 ft)(1.5 ft) = −22.5 kip-ft,

L = 15 ft, x = 5 ft, EI = 4.93 × 106

kip-in.2

Computation:

2 2

32 2

6 2

(2 3 )6

( 22.5 kip-ft)(5 ft)(12 in./ft)2(15 ft) 3(15 ft)(5 ft) (5 ft) 0.109533 in.

6(15 ft)(4.93 10 kip-in. )

C

M xv L Lx x

LEI = − − +

−⎡ ⎤= − − + =⎣ ⎦×

Consider 5 kips/ft uniformly distributed load on segment BC .



2 2(4 7 3 )24

C

wav L aL a

LEI = − − +


EI = 4.93 × 106 kip-in.

2





Computation:3

2 2

3 32 2

6 2

(4 7 3 )24

(5 kips/ft)(5 ft) (12 in./ft)4(15 ft) 7(5 ft)(15 ft) 3(5 ft) 0.273834 in.

24(15 ft)(4.93 10 kip-in. )

C

wav L aL a

LEI = − − +

⎡ ⎤= − − + = −⎣ ⎦×

Consider 25-kip concentrated load. [Appendix C, SS beam with concentrated load not at midspan.]


2 2 2( )6

C

Pbxv L b x



EI = 4.93 × 106 kip-in.2

Computation:

2 2 2

32 2 2

6 2

( )6


6(15 ft)(4.93 10 kip-in. )

C

Pbxv L b x

LEI = − − −

⎡ ⎤= − − − = −⎣ ⎦×


0.413793 in. 0.109533 in. 0.273834 in. 0.425963 in.

0.176471 in. 0.1765 in.

C v = + − −

= − = ↓ Ans.

(c) Beam deflection at point D

Consider 85 kip-ft concentrated moment.

[Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C:

2 2(2 3 )6

D

M xv L Lx x


Values: M = −85 kip-ft, L = 15 ft, x = 10 ft, EI = 4.93 × 10

6 kip-in.

2

Computation:

2 2

32 2

6 2

(2 3 )6


6(15 ft)(4.93 10 kip-in. )

D

M xv L Lx x

LEI = − − +

−⎡ ⎤= − − + =⎣ ⎦×





Consider moment at B caused by 5 kips/ft uniformly distributed load on overhang AB. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C:

2 2(2 3 )6

D

M xv L Lx x


Values: M = −(5 kips/ft)(3 ft)(1.5 ft) = −22.5 kip-ft,

L = 15 ft, x = 10 ft, EI = 4.93 × 106 kip-in.

2

Computation:

2 2

32 2

6 2

(2 3 )6

( 22.5 kip-ft)(10 ft)(12 in./ft)2(15 ft) 3(15 ft)(10 ft) (10 ft) 0.087627 in.

6(15 ft)(4.93 10 kip-in. )

D

M xv L Lx x

LEI = − − +

−⎡ ⎤= − − + =⎣ ⎦×

Consider 5 kips/ft uniformly distributed load on segment BC .


Relevant equation from Appendix C:23 2 2 2 2(2 6 4 )

24 D


LEI = − − + + −

Values: w = 5 kips/ft, L = 15 ft, a = 5 ft, x = 10 ft,

EI = 4.93 × 106 kip-in.

2

Computation:2

3 2 2 2 2

2 33 2 2 2 2

6 2

(2 6 4 )24

(5 kips/ft)(5 ft) (12 in./ft) 2(10 ft) 6(15 ft)(10 ft) (5 ft) (10 ft) 4(15 ft) (10 ft) (5 ft) (15 ft)24(15 ft)(4.93 10 kip-in. )

0.228195 in.

D


LEI = − − + + −

⎡ ⎤= − − + + −⎣ ⎦×

= −


2 2 2( )6

D

Pabv L a b

LEI = − − −

Values: P = 25 kips, L = 15 ft, a = 10 ft, b = 5 ft,

EI = 4.93 × 106 kip-in.2





Computation:

2 2 2

32 2 2

6 2

( )6


6(15 ft)(4.93 10 kip-in. )

D

Pabv L a b

LEI = − − −

⎡ ⎤= − − − = −⎣ ⎦×

Beam deflection at D

0.331034 in. 0.087627 in. 0.228195 in. 0.486815 in.

0.296349 in. 0.296 in.

Dv = + − −

= − = ↓ Ans.





10.54 The simply supported beam shown inFig. P10.54 consists of a W10 × 30 structural

steel wide-flange shape [ E = 29,000 ksi; I =

170 in.4]. For the loading shown, determine:



Fig. P10.54

Solution


Consider cantilever beam deflection of linearly distributed load on overhang AB. [Appendix C, Cantilever beam with linearly distributed load.]


0

30 A

w Lv

EI = −

Values: w0 = 8 kips/ft, L = 9 ft, EI = 4.93 × 10

6 kip-in.

2

Computation:4 4 3

0

6 2

(8 kips/ft)(9 ft) (12 in./ft)0.613247 in.

30 30(4.93 10 kip-in. ) A

w Lv

EI = − = − = −

×

Consider rotation at B caused by linearly distributed load on overhang AB. [Appendix C, SS beam with concentrated moment at one end.]


3 B

L


Values: M = ½(8 kips/ft)(9 ft)(3 ft) = 108 kip-ft, L = 18 ft, EI = 4.93 × 10

6 kip-in.

2

Computation:2

6 2

(108 kip-ft)(18 ft)(12 in./ft)0.0189274 rad

3 3(4.93 10 kip-in. )

(9 ft)(12 in./ft)(0.0189274 rad) 2.044157 in.

B

A

ML

EI

v

θ = = =×

= − = −





Consider linearly distributed load from 8 kips/ft to 0 kips/ft over span BD. [Appendix C, SS beam with linearly distributed load.] Relevant equation from Appendix C:

3

0

45 B

w L


Values: w0 = 8 kips/ft, L = 18 ft, EI = 4.93 × 10

6 kip-in.

2

Computation:3 3 2

0

6 2

(8 kips/ft)(18 ft) (12 in./ft)0.0302838 rad

45 45(4.93 10 kip-in. )

(9 ft)(12 in./ft)(0.0302838 rad) 3.270652 in.

B

A

w L

EI

v

θ = = =×

= =

Consider 4 kips/ft uniformly distributed load on segment CD.


22 2(2 )

24 B

wa L a



EI = 4.93 × 106 kip-in.

2

Computation:2

2 2

2 22 2

6 2

(2 )24

(4 kips/ft)(9 ft) (12 in./ft)2(18 ft) (9 ft) 0.0124211 rad

24(18 ft)(4.93 10 kip-in. )

(9 ft)(12 in./ft)(0.0124211 rad) 1.341478 in.

B

A

wa L a

LEI

v

θ = −

⎡ ⎤= − =⎣ ⎦×

= =


0.613247 in. 2.044157 in. 3.270652 in. 1.341478 in. 1.954726 in. 1.955 in. Av = − − + + = = ↑ Ans.


Consider moment at B caused by linearly distributed load on overhang AB. [Appendix C, SS beam with concentrated moment at one end.]


2 2(2 3 )6

C M xv L Lx x LEI = − − +

Values: M = −½(8 kips/ft)(9 ft)(3 ft) = −108 kip-ft,

L = 18 ft, x = 9 ft, EI = 4.93 × 106 kip-in.

2





Computation:

2 2

32 2

6 2

(2 3 )6


6(18 ft)(4.93 10 kip-in. )

C

M xv L Lx x

LEI = − − +

−⎡ ⎤= − − + =⎣ ⎦×

Consider linearly distributed load from 8 kips/ft to 0 kips/ft over span BD.

[Appendix C, SS beam with linearly distributed load.] Relevant equation from Appendix C:

4 2 2 40 (7 10 3 )360

C

w xv L L x x

LEI = − − +

Values: w0 = 8 kips/ft, L = 18 ft, x = 9 ft,

EI = 4.93 × 106 kip-in.

2

Computation:

4 2 2 40

34 2 2 4

6 2

(7 10 3 )360

(8 kips/ft)(9 ft)(12 in./ft)7(18 ft) 10(18 ft) (9 ft) 3(9 ft) 1.916398 in.

360(18 ft)(4.93 10 kip-in. )

C

w xv L L x x

LEI = − − +

⎡ ⎤= − − + = −⎣ ⎦×

Consider 4 kips/ft uniformly distributed load on segment CD.



2 2(4 7 3 )24

C

wav L aL a

LEI = − − +


EI = 4.93 × 106 kip-in.

2

Computation:3

2 2

3 32 2

6 2

(4 7 3 )24

(4 kips/ft)(9 ft) (12 in./ft)4(18 ft) 7(9 ft)(18 ft) 3(9 ft) 0.958199 in.

24(18 ft)(4.93 10 kip-in. )

C

wav L aL a

LEI = − − +

⎡ ⎤= − − + = −⎣ ⎦×

Beam deflection at C 0.766559 in. 1.916398 in. 0.958199 in. 2.108037 in. 2.11 in.C v = − − = − = ↓ Ans.








shown, determine:



Fig. P10.55

Solution


Consider cantilever beam deflection of downward 4 kips/ft uniform load over AB. [Appendix C, Cantilever beam with uniformly distributed load.]


8 A

wLv

EI = −

Values:

w = 4 kips/ft, L = 12 ft, EI = 2.4447 × 107

kip-in.2

Computation:4 4 3

7 2

(4 kips/ft)(12 ft) (12 in./ft)0.732847 in.

8 8(2.4447 10 kip-in. ) A

wLv

EI = − = − = −

×

Consider cantilever beam deflection of upward 4 kips/ft uniform load over 6-ft segment. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C:

4 3

and8 6

wL wL

v EI EI θ = − = (slope magnitude)

Values: w = −4 kips/ft, L = 6 ft, EI = 2.4447 × 10

7 kip-in.

2

Computation:4 4 3

7 2

3 3 2

7 2

( 4 kips/ft)(6 ft) (12 in./ft)0.045803 in.

8 8(2.4447 10 kip-in. )

(4 kips/ft)(6 ft) (12 in./ft)0.0008482 rad

6 6(2.4447 10 kip-in. )

0.045803 in. (6 ft)(12 in./ft)(0.0008482 rad) A

wLv

EI

wL

EI

v

θ

−= − = − =

×

= = =×

= + 0.106873 in.=





Consider rotation at B caused by downward 4 kips/ft uniform load. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C:

3 B

L


Values: M = (4 kips/ft)(6 ft)(9 ft) = 216 kip-ft,

L = 24 ft, EI = 2.4447 × 107 kip-in.

2

Computation:2

7 2

(216 kip-ft)(24 ft)(12 in./ft)0.0101784 rad

3 3(2.4447 10 kip-in. )

(12 ft)(12 in./ft)(0.0101784 rad) 1.465693 in.

B

A

ML

EI

v

θ = = =×

= − = −


2 2( )

6

B

Pb L b

LEI

θ −

= (slope magnitude)

Values: P = 42 kips, L = 24 ft, b = 18 ft,

EI = 2.4447 × 107 kip-in.

2

Computation:2 2 2

2 2

7 2


6 6(24 ft)(2.4447 10 kip-in. )

(12 ft)(12 in./ft)(0.0077929 rad) 1.122172 in.

B

A

Pb L b

LEI

v

θ −

⎡ ⎤= = − =⎣ ⎦×

= =

Consider 4 kips/ft uniformly distributed load on 6-ft segment near D.[Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C:

22 2(2 )

24 B

wa L a



EI = 2.4447 × 107 kip-in.

2

Computation:2 2 2

2 2 2 27 2(4 kips/ft)(6 ft) (12 in./ft)(2 ) 2(24 ft) (6 ft) 0.0016434 rad

24 24(24 ft)(2.4447 10 kip-in. )

(12 ft)(12 in./ft)(0.0016434 rad) 0.236648 in.

B

A

wa L a LEI

v

θ ⎡ ⎤= − = − =⎣ ⎦×

= =


0.732847 in. 0.106873 in. 1.465693 in. 1.122172 in. 0.236648 in.

0.732847 in. 0.733 in.

Av = − + − + +

= − = ↓ Ans.






Consider moment at B caused by downward 4 kips/ft uniform load. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C:

2 2(2 3 )6

C

M xv L Lx x


Values: M = −(4 kips/ft)(6 ft)(9 ft) = −216 kip-ft, L = 24 ft, x = 12 ft, EI = 2.4447 × 107 kip-in.2

Computation:

2 2

32 2

7 2

(2 3 )6


6(24 ft)(2.4447 10 kip-in. )

C

M xv L Lx x

LEI = − − +

−⎡ ⎤= − − + =⎣ ⎦×

Consider 42-kip concentrated load. [Appendix C, SS beam with concentrated load not at midspan.]

Relevant equation from Appendix C:2 2 2( )

6C

Pbxv L b x


Values: P = 42 kips, L = 24 ft, b = 6 ft, x = 12 ft, EI = 2.4447 × 10

7 kip-in.

2

Computation:

2 2 2

3

2 2 2

7 2

( )6

(42 kips)(6 ft)(12 ft)(12 in./ft) (24 ft) (6 ft) (12 ft) 0.587804 in.6(24 ft)(2.4447 10 kip-in. )

C

Pbxv L b x

LEI = − − −

⎡ ⎤= − − − = −⎣ ⎦×

Consider 4 kips/ft uniformly distributed load on 6-ft segment near D.



3 2 2 2 2(2 6 4 )24

C


LEI = − − + + −

(elastic curve)

Values:

w = 4 kips/ft, L = 24 ft, a = 6 ft, x = 12 ft, EI = 2.4447 × 10

7 kip-in.

2





Computation:2

3 2 2 2 2

2 33 2 2 2 2

7 2

(2 6 4 )24

(4 kips/ft)(6 ft) (12 in./ft)2(12 ft) 6(24 ft)(12 ft) (6 ft) (12 ft) 4(24 ft) (12 ft) (6 ft) (24

24(24 ft)(2.4447 10 kip-in. )

0.175578 in.

C


LEI = − − + + −

⎡ = − − + + −⎣ ×

= −

2 2 22 2 2 2

7 2

(4 kips/ft)(6 ft) (12 in./ft)(2 ) 2(24 ft) (6 ft) 0.0016434 rad

24 24(24 ft)(2.4447 10 kip-in. )

(12 ft)(12 in./ft)(0.0016434 rad) 0.236648 in.

B

A

wa L a

LEI

v

θ ⎡ ⎤= − = − =⎣ ⎦×

= =


0.549635 in. 0.587804 in. 0.175578 in. 0.213747 in. 0.214 in.C v = − − = − = ↓ Ans.






structural steel wide-flange shape [ E = 200

GPa; I = 351 × 106 mm

4]. For the loading

shown, determine:



Fig. P10.56

Solution



2 2(2 3 )6

B

M xv L Lx x


Values: M = −300 kN-m, L = 9 m, x = 4 m, EI = 7.02 × 104 kN-m2

Computation:

2 2

2 2

4 2

(2 3 )6

( 300 kN-m)(4 m)2(9 m) 3(9 m)(4 m) (4 m) 0.022159 m

6(9 m)(7.02 10 kN-m )

B

M xv L Lx x

LEI = − − +

−⎡ ⎤= − − + =⎣ ⎦×

Consider 85 kN/m uniformly distributed load on segment AB.


32 2(4 7 3 )

24 B

wav L aL a

LEI = − − +

Values: w = 85 kN/m, L = 9 m, a = 4 m,

EI = 7.02 × 104 kN-m

2

Computation:3

2 2

32 2

4 2

(4 7 3 )24

(85 kN/m)(4 m)4(9 m) 7(4 m)(9 m) 3(4 m) 0.043052 m

24(9 m)(7.02 10 kN-m )

B

wav L aL a

LEI = − − +

⎡ ⎤= − − + = −⎣ ⎦×







2 2 2( )6

B

Pbxv L b x


Values: P = 140 kN, L = 9 m, b = 3 m, x = 4 m,

EI = 7.02 × 104 kN-m

2

Computation:

2 2 2

2 2 2

4 2

( )6

(140 kN)(3 m)(4 m)(9 m) (3 m) (4 m) 0.024818 m

6(9 m)(7.02 10 kN-m )

B

Pbxv L b x

LEI = − − −

⎡ ⎤= − − − = −⎣ ⎦×


2 2(2 3 )6

B M xv L Lx x LEI = − − + (elastic curve)

Values: M = −175 kN-m, L = 9 m, x = 5 m,

EI = 7.02 × 104 kN-m

2

Computation:

2 2

2 2

4 2

(2 3 )6

( 175 kN-m)(5 m)2(9 m) 3(9 m)(5 m) (5 m) 0.012003 m

6(9 m)(7.02 10 kN-m )

B

M xv L Lx x

LEI = − − +

−⎡ ⎤= − − + =⎣ ⎦

×


0.022159 m 0.043052 m 0.024818 m 0.012003 m 0.033708 m 33.7 mm Bv = − − + = − = ↓ Ans.







2 2(2 3 )6

C

M xv L Lx x


Values: M = −300 kN-m, L = 9 m, x = 6 m, EI = 7.02 × 104 kN-m2

Computation:

2 2

2 2

4 2

(2 3 )6

( 300 kN-m)(6 m)2(9 m) 3(9 m)(6 m) (6 m) 0.017094 m

6(9 m)(7.02 10 kN-m )

C

M xv L Lx x

LEI = − − +

−⎡ ⎤= − − + =⎣ ⎦×

Consider 85 kN/m uniformly distributed load on segment AB.



3 2 2 2 2(2 6 4 )24

C


LEI = − − + + −

Values: w = 85 kN/m, L = 9 m, a = 4 m, x = 6 m,

EI = 7.02 × 104 kN-m

2

Computation:2

3 2 2 2 2

23 2 2 2 2

4 2

(2 6 4 )24

(85 kN/m)(4 m)2(6 m) 6(9 m)(6 m) (4 m) (6 m) 4(9 m) (6 m) (4 m) (9 m)

24(9 m)(7.02 10 kN-m )

0.034441 m

C


LEI = − − + + −

⎡ ⎤= − − + + −⎣ ⎦×

= −

Consider 140-kN concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C:

2 2 2( )6

C

Pabv L a b

LEI = − − −

Values:

P = 140 kN, L = 9 m, a = 6 m, b = 3 m, EI = 7.02 × 10

4 kN-m

2





Computation:

2 2 2

2 2 2

4 2

( )6

(140 kN)(6 m)(3 m)(9 m) (6 m) (3 m) 0.023932 m

6(9 m)(7.02 10 kN-m )

C

Pabv L a b

LEI = − − −

⎡ ⎤= − − − = −⎣ ⎦×

Consider 175 kN-m concentrated moment.

[Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C:

2 2(2 3 )6

C

M xv L Lx x


Values: M = −175 kN-m, L = 9 m, x = 3 m,

EI = 7.02 × 104 kN-m

2

Computation:

2 2

2 2

4 2

(2 3 )

6( 175 kN-m)(3 m)

2(9 m) 3(9 m)(3 m) (3 m) 0.012464 m6(9 m)(7.02 10 kN-m )

C

M xv L Lx x

LEI

= − − +

−⎡ ⎤= − − + =⎣ ⎦×


0.017094 m 0.034441 m 0.023932 m 0.012464 m 0.028814 m 28.8 mmC

v = − − + = − = ↓ Ans.





10.57 A 25-ft-long soldier beam is used as a keycomponent of an earth retention system at an

excavation site. The soldier beam is subjected to a

uniformly distributed soil loading of 260 lb/ft, asshown in Fig. P10.57. The soldier beam can be

idealized as a cantilever with a fixed support at A.

Added support is supplied by a tieback anchor at B,which exerts a force of 4,000 lb on the soldier beam.

Determine the horizontal deflection of the soldier beam at point C . Assume EI = 5 × 10

8lb-in.

2.

Fig. P10.57

Solution

Consider 260 lb/ft uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.]


8C

wLv

EI = −

Values: w = 260 lb/ft, L = 25 ft, EI = 5.0 × 10

8 lb-in.

2

Computation:4 4 3

8 2

(260 lb/ft)(25 ft) (12 in./ft)43.875 in.

8 8(5.0 10 lb-in. )C

wLv

EI = − = − = −

×

Consider 4,000-lb concentrated load. [Appendix C, Cantilever beam with concentrated load.] Relevant equations from Appendix C:

3 2

and3 2

B B

PL PLv

EI EI θ = = (slope magnitude)

Values: P = 4,000 lb, L = 18 ft, EI = 5.0 × 10

8 lb-in.

2

Computation:3 3 3

8 2

2 2 2

8 2

(4,000 lb)(18 ft) (12 in./ft)26.873856 in.

3 3(5.0 10 lb-in. )(4,000 lb)(18 ft) (12 in./ft)

0.1866240 rad2 2(5.0 10 lb-in. )

26.873856 in. (7 ft)(12 in./ft)(0.1866240 rad) 42.550

B

B

C

PLv

EI PL

EI

v

θ

= = =

×

= = =×

= + = 272 in.


43.875 in. 42.550272 in. 1.324728 in. 1.325 in.C v = − + = − = → Ans.





10.58 A 25-ft-long soldier beam is used as a keycomponent of an earth retention system at an

excavation site. The soldier beam is subjected to

a soil loading that is linearly distributed from520 lb/ft to 260 lb/ft, as shown in Fig. P10.58.

The soldier beam can be idealized as a

cantilever with a fixed support at A. Addedsupport is supplied by a tieback anchor at B,

which exerts a force of 5,000 lb on the soldier beam. Determine the horizontal deflection of the

soldier beam at point C . Assume EI = 5 × 108lb-

in.2.

Fig. P10.58

Solution

Consider 260 lb/ft uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.]


8C

wLv

EI = −

Values: w = 260 lb/ft, L = 25 ft, EI = 5.0 × 10

8 lb-in.

2

Computation:4 4 3

8 2

(260 lb/ft)(25 ft) (12 in./ft)43.875 in.

8 8(5.0 10 lb-in. )C

wLv

EI = − = − = −

×

Consider a linearly distributed load that varies from 260 lb/ft at A to 0 lb/ft at B. [Appendix C, Cantilever beam with linearly distributed load.]


0

30C

w Lv

EI = −

Values: w0 = 260 lb/ft, L = 25 ft, EI = 5.0 × 108 lb-in.2

Computation:4 4 3

0

8 2

(260 lb/ft)(25 ft) (12 in./ft)11.700 in.

30 30(5.0 10 lb-in. )C

w Lv

EI = − = − = −

×



Consider 5,000-lb concentrated load. [Appendix C, Cantilever beam with concentrated load.] Relevant equations from Appendix C:

3 2

and3 2

B B

PL PLv

EI EI θ = = (slope magnitude)

Values: P = 5,000 lb, L = 18 ft, EI = 5.0 × 10

8 lb-in.

2

Computation:3 3 3

8 2

2 2 2

8 2

(5,000 lb)(18 ft) (12 in./ft)33.592320 in.

3 3(5.0 10 lb-in. )

(5,000 lb)(18 ft) (12 in./ft)0.2332800 rad

2 2(5.0 10 lb-in. )

33.592320 in. (7 ft)(12 in./ft)(0.2332800 rad) 53.187

B

B

C

PLv

EI

PL

EI

v

θ

= = =×

= = =×

= + = 840 in.


43.875 in. 11.700 in. 53.187840 in. 2.387160 in. 2.39 in.C v = − − + = − = → Ans.

Mechanics of Materials Solutions Chapter10 Probs47 58

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Transcript of Mechanics of Materials Solutions Chapter10 Probs47 58