mechanics of materials solutions chapter10 probs29 46

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10.29a For the beams and loadings shown below, determine the beam deflection at

point H . Assume that EI = 8 × 104 kN-m

2is

constant for each beam.

Fig. P10.29a

Solution

Determine beam slope at A.

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

6 A

L

EI θ = (slope magnitude)

Values: M = 150 kN-m, L = 8 m, EI = 8 × 10

4 kN-m

2

Computation:

4 2

(150 kN-m)(8 m) 0.00250 rad6 6(8 10 kN-m )

A

ML

EI θ = = =

×

Determine beam deflection at H . [Skill 1]

(3 m)(0.00250 rad) 0.00750 m 7.50 mm H v = = = ↑ Ans.





10.29b For the beams and loadings shown below, determine the beam deflection at point

H . Assume that EI = 8 × 104 kN-m

2 is constant

for each beam.

Fig. P10.29b

Solution

Determine beam deflection at A. [Appendix C, Cantilever beam with uniformly distributed load.]

Relevant equation from Appendix C:4

8 A

wLv

EI = −

Values: w = 6 kN/m, L = 4 m, EI = 8 × 10

4 kN-m

2

Computation:4 4

4 2

(6 kN/m)(4 m)0.00240 m

8 8(8 10 kN-m ) A

wLv

EI = − = − = −

×

Determine beam slope at A. [Appendix C, Cantilever beam with uniformly distributed load.]


6 A

wL


Values: w = 6 kN/m, L = 4 m, EI = 8 × 10

4 kN-m

2

Computation:3 3

4 2

(6 kN/m)(4 m)0.00080 rad

6 6(8 10 kN-m ) A

wL

EI θ = = =

×


0.00240 m (2 m)(0.00080 rad) 0.00400 m 4.00 mm H v = − − = − = ↓ Ans.





10.29c For the beams and loadings shown below, determine the beam deflection at


2is


Fig. P10.29c

Solution

Determine beam deflection at H . [Skill 3] [Appendix C, SS beam with concentrated load not at midspan.]

Relevant equation from Appendix C:

2 2 2( )6

H

Pbxv L b x

LEI = − − − (elastic curve)

Values: P = 30 kN-m, L = 12 m, b = 4 m, x = 4 m, EI = 8 × 10

4 kN-m

2

Computation:

2 2 2

2 2 2

4 2

( )6

(30 kN)(4 m)(4 m)(12 m) (4 m) (4 m)

6(12 m)(8 10 kN-m )

0.00933 m 9.33 mm

H

Pbxv L b x

LEI = − − −

⎡ ⎤= − − −⎣ ⎦×

= = ↓ Ans.





10.29c For the beams and loadings shown below, determine the beam deflection at


2is


Fig. P10.29d

Solution

Determine deflection of cantilever overhang. [Appendix C, Cantilever beam with concentrated load.]


,cant3

H

PLv

EI = − (assuming fixed support at B)

Values: P = 15 kN, L = 4 m, EI = 8 × 10

4 kN-m

2

Computation: 3 3

,cant 4 2

(15 kN)(4 m)0.004000 m

3 3(8 10 kN-m ) H

PLv

EI = − = − = −

×

Determine beam slope at B. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C:

3 B

L


Values: M = (15 kN)(4 m) = 60 kN-m, L = 8 m,

EI = 8 × 104 kN-m2

Computation:

4 2

(60 kN-m)(8 m)0.002000 rad

3 3(8 10 kN-m ) B

ML

EI θ = = =

×


0.00400 m (4 m)(0.00200 rad) 0.01200 m 12.00 mm H v = − − = − = ↓ Ans.






point H . Assume that EI = 1.2 × 107 kip-in.

2

is constant for each beam.

Fig. P10.30a

Solution

Determine beam deflection at B. [Appendix C, Cantilever beam with concentrated moment.]


2 B

Lv

EI = −

Values: M = 40 kip-ft, L = 9 ft, EI = 1.2 × 10

7 kip-in.

2

Computation:2 2 3

7 2

(40 kip-ft)(9 ft) (12 in./ft)0.23328 in.

2 2(1.2 10 kip-in. ) B

MLv

EI = − = − = −

×

Determine beam slope at B. [Appendix C, Cantilever beam with concentrated moment.]


B

L



7 kip-in.

2

Computation:2

7 2

(40 kip-ft)(9 ft)(12 in./ft)0.004320 rad

(1.2 10 kip-in. ) B

ML

EI θ = = =

×


0.23328 in. (6 ft)(12 in./ft)(0.004320 rad) 0.54432 in. 0.544 in. H v = − − = − = ↓ Ans.





10.30b For the beams and loadings shown below, determine the beam deflection at

point H . Assume that EI = 1.2 × 107 kip-

in.2 is constant for each beam.

Fig. P10.30b

Solution

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

2 2( )

6C

Pa L a

LEI θ

−= (slope magnitude)

Values: P = 25 kips, L = 18 ft, a = 12 ft, EI = 1.2 × 10

7 kip-in.

2

Computation:2 2

2 2 2

7 2

( )

6

(25 kips)(12 ft)(18 ft) (12 ft) (12 in./ft) 0.00600 rad

6(18 ft)(1.2 10 kip-in. )

C

Pa L a

LEI θ

−=

⎡ ⎤= − =⎣ ⎦×


(7 ft)(12 in./ft)(0.00600 rad) 0.5040 in. 0.504 in. H v = = = ↑ Ans.





10.30c For the beams and loadings shown below, determine the beam deflection at point

H . Assume that EI = 1.2 × 107 kip-in.

2is


Fig. P10.30c

Solution


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


2 2(6 4 )24

H

wxv L Lx x

EI = − − + (elastic curve)

Values: w = 2.5 kips/ft, L = 15 ft, x = 9 ft, EI = 1.2 × 10

7 kip-in.

2

Computation:2

2 2

22 2 3

7 2

(6 4 )24

(2.5 kips/ft)(9 ft)6(15 ft) 4(15 ft)(9 ft) (9 ft) (12 in./ft)

24(1.2 10 kip-in. )

1.082565 in. 1.083 in.

H

wxv L Lx x

EI = − − +

⎡ ⎤= − − +⎣ ⎦×

= − = ↓ Ans.





10.30d For the beams and loadings shown below, determine the beam deflection at point


2 is


Fig. P10.30d

Solution

Determine deflection of cantilever overhang. [Appendix C, Cantilever beam with uniform load.]


,cant8

H

wLv

EI = − (assuming fixed support at A)

Values: w = 5 kips/ft, L = 8 ft, EI = 1.2 × 10

7 kip-in.

2

Computation: 4 4 3

,cant 7 2

(5 kips/ft)(8 ft) (12 in./ft)0.36864 in.

8 8(1.2 10 kip-in. ) H

wLv

EI = − = − = −

×

Determine beam slope at A. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C:

3 A

L


Values: M = (5 kips/ft)(8 ft)(4 ft) = 160 kip-ft, L = 22 ft, EI = 1.2 × 10

7 kip-in.

2

Computation:2

7 2

(160 kip-ft)(22 ft)(12 in./ft)0.014080 rad

3 3(1.2 10 kip-in. ) A

ML

EI θ = = =

×









2

is constant for each beam.

Fig. P10.31a

Solution


[Appendix C, SS beam with concentrated moment.]


2 2(2 3 )6

H

M xv L Lx x

LEI = − − + (elastic curve)

Values: M = −60 kN-m, L = 12 m, x = 6 m, EI = 6 × 10

4 kN-m

2

Computation:

2 2

2 2

4 2

(2 3 )6

( 60 kN-m)(6 m)2(12 m) 3(12 m)(6 m) (6 m)

6(12 m)(6 10 kN-m )

0.009000 m 9.00 mm

H

M xv L Lx x

LEI = − − +

−⎡ ⎤= − − +⎣ ⎦×

= = ↑ Ans.







2 is constant

for each beam.

Fig. P10.31b

Solution

Determine deflection of cantilever overhang. [Appendix C, Cantilever beam with uniform load.]


,cant8

H

wLv


Values: w = 7.5 kN/m, L = 3 m, EI = 6 × 10

4 kN-m

2

Computation:4 4

,cant 4 2

(7.5 kN/m)(3 m)0.00126563 m

8 8(6 10 kN-m ) H

wLv

EI = − = − = −

×

Determine beam slope at A. [Appendix C, SS beam with concentrated moment.]


3 A

L


Values: M = (7.5 kN/m)(3 m)(1.5 m) = 33.75 kN-m,

L = 6 m, EI = 6 × 104

kN-m2

Computation:

4 2

(33.75 kN-m)(6 m)0.001125 rad

3 3(6 10 kN-m ) A

ML

EI θ = = =

×


0.00126563 m (3 m)(0.001125 rad) 0.00464063 m 4.64 mm H v = − − = − = ↓ Ans.







2is


Fig. P10.31c

Solution

Determine beam deflection at B. [Appendix C, Cantilever beam with concentrated load.]


3 B

PLv

EI = −

Values: P = 30 kN, L = 3 m, EI = 6 × 10

4 kN-m

2

Computation:3 3

4 2

(30 kN)(3 m)0.004500 m

3 3(6 10 kN-m ) B

PLv

EI = − = − = −

×

Determine beam slope at B. [Appendix C, Cantilever beam with concentrated load.]


2 B

PL


Values: P = 30 kN, L = 3 m, EI = 6 × 10

4 kN-m

2

Computation:2 2

4 2

(30 kN)(3 m)0.002250 rad

2 2(6 10 kN-m ) B

PL

EI θ = = =

×


0.004500 m (3 m)(0.002250 rad) 0.01125 m 11.25 mm H v = − − = − = ↓ Ans.

Alternative solution for beam deflection at B.

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


48 H

PLv

EI = −

Values: P = 30 kN, L = 6 m, EI = 6 × 104 kN-m

2

Computation:3 3

4 2

5 5(30 kN)(6 m)0.011250 m 11.25 mm

48 48(6 10 kN-m ) H

PLv

EI = − = − = − = ↓

×





10.31d For the beams and loadings shown below, determine the beam deflection at


2is


Fig. P10.31d

Solution

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


2(2 )24

C

wa L a

LEI θ = − (slope magnitude)

Values: w = 5 kN/m, L = 9 m, a = 6 m, EI = 6 × 10

4 kN-m

2

Computation:

[ ]2 2

22

4 2

(5 kN/m)(6 m)(2 ) 2(9 m) (6 m) 0.00200 rad

24 24(9 m)(6 10 kN-m )C

wa L a

LEI θ = − = − =

×


(3 m)(0.00200 rad) 0.00600 m 6.00 mm H v = = = ↑ Ans.






point H . Assume that EI = 3.0 × 106 kip-

in.2 is constant for each beam.

Fig. P10.32a

Solution

Determine deflection of cantilever overhang. [Appendix C, Cantilever beam with concentrated moment.]


,cant2

H

Lv



6 kip-in.

2

Computation:2 2 3

,cant 6 2

(50 kip-ft)(6 ft) (12 in./ft)0.51840 in.

2 2(3.0 10 kip-in. ) H

MLv

EI = − = − = −

×



3 A

L



6 kip-in.

2

Computation:2

6 2

(50 kip-ft)(18 ft)(12 in./ft)0.01440 rad

3 3(3.0 10 kip-in. ) A

ML

EI θ = = =

×









2is


Fig. P10.32b

Solution


[Appendix C, Cantilever beam with concentrated load.]


(3 )6

H

Pxv L x

EI = − − (elastic curve)

Values: P = 10 kips, L = 10 ft, x = 7 ft,

EI = 3.0 × 106 kip-in.

2

Computation:2

2 3

6 2

(3 )6

(10 kips)(7 ft) (12 in./ft)[3(10 ft) (7 ft)] 1.081920 in. 1.082 in.

6(3.0 10 kip-in. )

H

Pxv L x

EI = − −

= − − = − = ↓×

Ans.







2is


Fig. P10.32c

Solution



6 A

L


Values: M = (2 kips/ft)(8 ft)(4 ft) = 64 kip-ft, L = 18 ft, EI = 3.0 × 10

6 kip-in.

2

Computation:2

6 2

(64 kip-ft)(18 ft)(12 in./ft)0.009216 rad

6 6(3.0 10 kip-in. ) A

ML

EI θ = = =

×


(6 ft)(12 in./ft)(0.009216 rad) 0.663552 in. 0.664 in. H v = = = ↑ Ans.





10.32d For the beams and loadings shown below, determine the beam deflection at point


2is


Fig. P10.32d

Solution

Determine beam deflection at B. [Appendix C, Cantilever beam with uniformly distributed load.]


8 B

wLv

EI = −

Values: w = 1.5 kips/ft, L = 10 ft, EI = 3.0 × 10

6 kip-in.

2

Computation:4 4 3

6 2

(1.5 kips/ft)(10 ft) (12 in./ft)1.0800 in.

8 8(3.0 10 kip-in. ) B

wLv

EI = − = − = −

×

Determine beam slope at B. [Appendix C, Cantilever beam with uniformly distributed load.]


6 B

wL


Values: w = 1.5 kips/ft, L = 10 ft, EI = 3.0 × 10

6 kip-in.

2

Computation:3 3 2

6 2

(1.5 kips/ft)(10 ft) (12 in./ft)0.01200 rad

6 6(3.0 10 kip-in. ) B

wL

EI θ = = =

×







10.33 The simply supported beam shownin Fig. P10.33 consists of a W24 × 94

structural steel wide-flange shape [ E =

29,000 ksi; I = 2,700 in.4]. For the loading

shown, determine the beam deflection at point C .

Fig. P10.33

Solution

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


2 2(4 7 3 )24

C

wav L aL a

LEI = − − +

Values: w = 3.2 kips/ft, L = 28 ft, a = 21 ft,

EI = 7.830 × 107 kip-in.

2

Computation:3

2 2

3 32 2

7 2

(4 7 3 )24

(3.2 kips/ft)(21 ft) (12 in./ft)4(28 ft) 7(21 ft)(28 ft) 3(21 ft) 0.333822 in.

24(28 ft)(7.830 10 kip-in. )

C

wav L aL a

LEI = − − +

⎡ ⎤= − − + = −⎣ ⎦×

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


2 2

(3 4 )48C

Pxv L x EI = − − (elastic curve)

Values: P = 36 kips, L = 28 ft, x = 7 ft, EI = 7.830 × 10

7 kip-in.

2

Computation:

2 2

32 2

7 2

(3 4 )48

(36 kips)(7 ft)(12 in./ft)3(28 ft) 4(7 ft) 0.249799 in.

48(7.830 10 kip-in. )

C

Pxv L x

EI = − −

⎡ ⎤= − − = −⎣ ⎦×

Beam deflection at C

0.333822 in. 0.249799 in. 0.583620 in. 0.584 in.C v = − − = − = ↓ Ans.






structural steel wide-flange shape [ E = 200

GPa; I = 370 × 106 mm

4]. For the loading

shown, determine the beam deflection at point C .

Fig. P10.34

Solution

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


2 2(4 7 3 )24

C

wav L aL a

LEI = − − +

Values: w = 26 kN/m, L = 8 m, a = 6 m,

EI = 7.4 × 104 kN-m

2

Computation:3

2 2

32 2

4 2

(4 7 3 )24

(26 kN/m)(6 m)4(8 m) 7(6 m)(8 m) 3(6 m) 0.011068 m

24(8 m)(7.40 10 kN-m )

C

wav L aL a

LEI = − − +

⎡ ⎤= − − + = −⎣ ⎦×

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


2 2 2

( )6C

Pbxv L b x LEI = − − − (elastic curve)

Values: P = 60 kN, L = 8 m, b = 3 m, x = 2 m, EI = 7.4 × 10

4 kN-m

2

Computation:

2 2 2

2 2 2

4 2

( )6

(60 kN)(3 m)(2 m)(8 m) (3 m) (2 m) 0.005169 m

6(8 m)(7.40 10 kN-m )

C

Pbxv L b x

LEI = − − −

⎡ ⎤= − − − = −⎣ ⎦×


0.011068 m 0.005169 m 0.016237 m 16.24 mmC v = − − = − = ↓ Ans.






structural steel wide-flange shape [ E = 200

GPa; I = 216 × 106 mm

4]. For the loading

shown, determine the beam deflection at

point B.

Fig. P10.35

Solution

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


2 2 2( )6

B

Pabv L a b

LEI = − − −

Values: P = 60 kN, L = 9 m, a = 3 m, b = 6 m, EI = 4.32 × 10

4 kN-m

2

Computation:

2 2 2

2 2 2

4 2

( )6

(60 kN)(3 m)(6 m)(9 m) (3 m) (6 m) 0.016667 m

6(9 m)(4.32 10 kN-m )

B

Pabv L a b

LEI = − − −

⎡ ⎤= − − − = −⎣ ⎦×

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


2 2(2 3 )6

B

M xv L Lx x


Values: M = −45 kN-m, L = 9 m, x = 6 m,

EI = 4.32 × 104 kN-m

2

Computation:

2 2

2 2

4 2

(2 3 )6

( 45 kN-m)(6 m)2(9 m) 3(9 m)(6 m) (6 m) 0.004167 m

6(9 m)(4.32 10 kN-m )

B

M xv L Lx x

LEI = − − +

−⎡ ⎤= − − + =⎣ ⎦×

Beam deflection at B

0.016667 m 0.004167 m 0.012500 m 12.50 mm Bv = − + = − = ↓ Ans.







29,000 ksi; I = 843 in.4]. For the loading

shown, determine the beam deflection at

point B.

Fig. P10.36

Solution

Consider uniformly distributed load.

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


2 2(4 7 3 )24

B

wav L aL a

LEI = − − +

Values: w = 5 kips/ft, L = 24 ft, a = 16 ft,

EI = 2.4447 × 107 kip-in.

2

Computation:3

2 2

3 32 2

7 2

(4 7 3 )24

(5 kips/ft)(16 ft) (12 in./ft)4(24 ft) 7(16 ft)(24 ft) 3(16 ft) 0.965066 in.

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

B

wav L aL a

LEI = − − +

⎡ ⎤= − − + = −⎣ ⎦×

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


2 2

(2 3 )6 B

M x

v L Lx x LEI = − − +

(elastic curve)

Values: M = −200 kip-ft, L = 24 ft, x = 8 ft,

EI = 2.4447 × 107 kip-in.

2

Computation:

2 2

32 2

7 2

(2 3 )6

( 200 kip-ft)(8 ft)(12 in./ft)2(24 ft) 3(24 ft)(8 ft) (8 ft) 0.502638 in.

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

B

M xv L Lx x

LEI = − − +

−⎡ ⎤= − − + =⎣ ⎦×


0.965066 in. 0.502638 in. 0.462428 in. 0.462 in. Bv = − + = − = ↓ Ans.





10.37 The cantilever beam shown in Fig.P10.37 consists of a rectangular structural

steel tube shape [ E = 29,000 ksi; I = 476

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

(a) the beam deflection at point B.

(b) the beam deflection at point C .

Fig. P10.37

Solution

(a) Beam deflection at point B




8 B

wLv

EI = −


7 kip-in.

2

Computation:4 4 3

7 2

(2 kips/ft)(6 ft) (12 in./ft)0.040559 in.

8 8(1.3804 10 kip-in. ) B

wLv

EI = − = − = −

×

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


(3 )6

B

Pxv L x


Values:

P = 12 kips, L = 10 ft, x = 6 ft, EI = 1.3804 × 107 kip-in.

2

Computation:

[ ]2 2 3

7 2

(12 kips)(6 ft) (12 in./ft)(3 ) 3(10 ft) (6 ft) 0.216313 in.

6 6(1.3804 10 kip-in. ) B

Pxv L x

EI = − − = − − = −

×


0.040559 in. 0.216313 in. 0.256872 in. 0.257 in. Bv = − − = − = ↓ Ans.





(b) Beam deflection at point C




6 B

wL

EI θ = −


7 kip-in.

2

Computation:3 3 3

7 2

(2 kips/ft)(6 ft) (12 in./ft)0.0090130 rad

6 6(1.3804 10 kip-in. )

0.040559 in. (4 ft)(0.0090130 rad) 0.076611 in.

B

C

wL

EI

v

θ = − = − = −×

= − − = −



3C

PLv

EI = −

Values: P = 12 kips, L = 10 ft, EI = 1.3804 × 10

7 kip-in.

2

Computation:3 3 3

7 2

(12 kips)(10 ft) (12 in./ft)0.500724 in.

3 3(1.3804 10 kip-in. )C

PLv

EI = − = − = −

×


0.076611 in. 0.500724 in. 0.577336 in. 0.577 in.C v = − − = − = ↓ 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


Values: w = 25 kN/m, L = 4 m, x = 2.5 m, EI = 8.0 × 10

4 kN-m

2

Computation:2

2 2

22 2

4 2

(6 4 )24

(25 kN/m)(2.5 m)6(4.0 m) 4(4.0 m)(2.5 m) (2.5 m) 0.005066 m

24(8.0 10 kN-m )

B

wxv L Lx x

EI = − − +

⎡ ⎤= − − + = −⎣ ⎦×


Relevant equations from Appendix C:3

3 B PLv

EI = −

Values: P = 55 kN, L = 2.5 m, EI = 8.0 × 10

4 kN-m

2

Computation:3 3

4 2

(55 kN)(2.5 m)0.003581 m

3 3(8.0 10 kN-m ) B

PLv

EI = − = − = −

×


0.005066 m 0.003581 m 0.008647 m 8.65 mm Bv = − − = − = ↓ Ans.





10.39 The solid 1.25-in.-diameter steel [ E =29,000 ksi] shaft shown in Fig. P10.39

supports two pulleys. For the loading

shown, determine:(a) the shaft deflection at point B.

(b) the shaft deflection at point C .

Fig. P10.39

Solution

Section properties:

4 4(1.25 in.) 0.119842 in.64

I π

= =

(a) Shaft deflection at point B

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

3

3 B

PLv

EI = −

Values: P = 200 lb, L = 10 in., EI = 3.47543 × 10

6 lb-in.

2

Computation:3 3

6 2

(200 lb)(10 in.)0.019182 in.

3 3(3.47543 10 lb-in. ) B

PLv

EI = − = − = −

×

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


(3 )6

B

Pxv L x


Values: P = 120 lb, L = 25 in., x = 10 in., EI = 3.47543 × 10

6 lb-in.

2

Computation:

[ ]2 2

6 2

(120 lb)(10 in.)(3 ) 3(25 in.) (10 in.) 0.037405 in.

6 6(3.47543 10 lb-in. ) B

Pxv L x

EI = − − = − − = −

×

Shaft deflection at B

0.019182 in. 0.037405 in. 0.056588 in. 0.0566 in. Bv = − − = − = ↓ Ans.





(b) Shaft deflection at point C

Consider concentrated load at pulley 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 = 200 lb, L = 10 in., EI = 3.47543 × 10

6 lb-in.

2

Computation:3 3

6 2

(200 lb)(10 in.)0.019182 in.

3 3(3.47543 10 lb-in. ) B

PLv

EI = − = − = −

×

2 2

6 2

(200 lb)(10 in.)0.0028773 rad

2 2(3.47543 10 lb-in. )

0.019182 in. (15 in.)(0.0028773 rad) 0.062342 in.

B

C

PL

EI

v

θ = = =×

= − − = −

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


3C

PLv EI

= −

Values: P = 120 lb, L = 25 in.,

EI = 3.47543 × 106 lb-in.

2

Computation:3 3

6 2

(120 lb)(25 in.)0.179834 in.

3 3(3.47543 10 lb-in. )C

PLv

EI = − = − = −

×

Shaft deflection atC

0.062342 in. 0.179834 in. 0.242176 in. 0.242 in.C v = − − = − = ↓ Ans.





10.40 The cantilever beam shown in Fig.P10.40 consists of a rectangular structural

steel tube shape [ E = 29,000 ksi; I = 1,710

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

(a) the beam deflection at point A.

(b) the beam deflection at point B.

Fig. P10.40

Solution

(a) Beam deflection at point A

Consider concentrated moment. [Appendix C, Cantilever beam with concentrated moment.]


2 A

Lv

EI = −

Values: M = −200 kip-ft, L = 15 ft,

EI = 4.959 × 107 kip-in.

2

Computation:2 2 3

7 2

( 200 kip-ft)(15 ft) (12 in./ft)0.784029 in.

2 2(4.959 10 kip-in. ) A

MLv

EI

−= − = − =

×


Relevant equations from Appendix C:3 2

and3 2

B B

PL PLv

EI EI θ = − = (slope magnitude)

Values: P = −18 kips, L = 9 ft, EI = 4.959 × 10

7 kip-in.

2

Computation:3 3 3

7 2

2 2 2

7 2

( 18 kips)(9 ft) (12 in./ft)0.152415 in.

3 3(4.959 10 kip-in. )

(18 kips)(9 ft) (12 in./ft)0.0021169 rad

2 2(4.959 10 kip-in. )

0.152415 in. (6 ft)(12 in./ft)(0.0021169 rad) 0.3

B

B

A

PLv

EI

PL

EI

v

θ

−= − = − =

×

= = =×

= + = 04830 in.

Beam deflection at A

0.784029 in. 0.304830 in. 1.088860 in. 1.089 in. Av = + = = ↑ Ans.






Consider concentrated moment. [Appendix C, Cantilever beam with concentrated moment.]


2 B

xv

EI = − (elastic curve)

Values: M = −200 kip-ft, L = 15 ft, x = 9 ft, EI = 4.959 × 10

7 kip-in.

2

Computation:2 2 3

7 2

( 200 kip-ft)(9 ft) (12 in./ft)0.282250 in.

2 2(4.959 10 kip-in. ) B

Mxv

EI

−= − = − =

×



3 B

PLv

EI = −

Values: P = −18 kips, L = 9 ft,

EI = 4.959 × 107 kip-in.

2

Computation:3 3 3

7 2

( 18 kips)(9 ft) (12 in./ft)0.152415 in.

3 3(4.959 10 kip-in. ) B

PLv

EI

−= − = − =

×


0.282250 in. 0.152415 in. 0.434665 in. 0.435 in. Bv = + = = ↑ Ans.





10.41 The simply supported beam shown inFig. P10.41 consists of a W21 × 44


29,000 ksi; I = 843 in.4]. For the loading

shown, determine:



Fig. P10.41

Solution


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


8 A

wLv



7 kip-in.

2

Computation:4 4 3

7 2

(4 kips/ft)(8 ft) (12 in./ft)0.144760 in.

8 8(2.4447 10 kip-in. ) A

wLv

EI = − = − = −

×

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



3 B

L


Values: M = (4 kips/ft)(8 ft)(4 ft) = 128 kip-ft,

L = 22 ft, EI = 2.4447 × 107 kip-in.

2

Computation:2

7 2

(128 kip-ft)(22 ft)(12 in./ft)0.0055290 rad

3 3(2.4447 10 kip-in. )

(8 ft)(12 in./ft)(0.0055290 rad) 0.530786 in.

B

A

ML

EI

v

θ = = =×

= = −







16 B

PL



7 kip-in.

2

Computation:

2 2 2

7 2

(45 kips)(22 ft) (12 in./ft)0.0080182 rad

16 16(2.4447 10 kip-in. )

(8 ft)(12 in./ft)(0.0080182 rad) 0.769744 in.

B

A

PL

EI

v

θ = = =×

= =


0.144760 in. 0.530786 in. 0.769744 in. 0.094198 in. 0.0942 in. Av = − − + = = ↑ Ans.


Consider distributed load on overhang.


2 2(2 3 )6

C

M xv L Lx x


Values: M = (4 kips/ft)(8 ft)(4 ft) = −128 kip-ft,

L = 22 ft, x = 11 ft, EI = 2.4447 × 107 kip-in.

2

Computation:

2 2

32 2

7 2

(2 3 )6

( 128 kip-ft)(11 ft)(12 in./ft)2(22 ft) 3(22 ft)(11 ft) (11 ft) 0.273687 in.

6(22 ft)(2.4447 10 kip-in. )

C

M xv L Lx x

LEI = − − +

−⎡ ⎤= − − + =⎣ ⎦×



48C

PLv

EI = −


7 kip-in.

2

Computation:3 3 3

7 2

(45 kips)(22 ft) (12 in./ft)0.705598 in.

48 48(2.4447 10 kip-in. )C

PLv

EI = − = − = −

×


0.273687 in. 0.705598 in. 0.431912 in. 0.432 in.C v = − = − = ↓ Ans.







200 GPa; I = 351 × 106 mm

4]. For the

loading shown, determine:

(a) the beam deflection at point B.

(b) the beam deflection at point D.

Fig. P10.42

Solution

(a) Beam deflection at point B

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


384 B

wLv

EI = −

Values: w = 55 kN/m, L = 7.2 m, EI = 7.02 × 10

4 kN-m

2

Computation:4 4

4 2

5 5(55 kN/m)(7.2 m)0.027415 m

384 384(7.02 10 kN-m ) B

wLv

EI = − = − = −

×

Consider distributed load on overhang. [Appendix C, SS beam with concentrated moment.]


2 2(2 3 )6

B

M xv L Lx x


Values: M = (55 kN/m)(2.8 m)(1.4 m) = −215.6 kN-m,

L = 7.2 m, x = 3.6 m, EI = 7.02 × 104

kN-m2

Computation:

2 2

2 2

7 2

(2 3 )6

( 215.6 kN-m)(3.6 m)2(7.2 m) 3(7.2 m)(3.6 m) (3.6 m) 0.009951 m

6(7.2 m)(7.02 10 kN-m )

B

M xv L Lx x

LEI = − − +

−⎡ ⎤= − − + =⎣ ⎦×


0.027415 m 0.009951 m 0.017464 m 17.46 mm Bv = − + = − = ↓ Ans.





(b) Beam deflection at point D

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


24C

wL


Values: w = 55 kN/m, L = 7.2 m, EI = 7.02 × 10

4 kN-m

2

Computation:3 3

4 2

(55 kN/m)(7.2 m)0.0121846 rad

24 24(7.02 10 kN-m )

(2.8 m)(0.0121846 rad) 0.034117 m

C

D

wL

EI

v

θ = = =×

= =

Consider deflection at D resulting from rotation at C caused by distributed load on overhang.



3C

L


Values: M = (55 kN/m)(2.8 m)(1.4 m) = 215.6 kN-m,

L = 7.2 m, EI = 7.02 × 104 kN-m

2

Computation:

7 2

(215.6 kN-m)(7.2 m)0.0073709 rad

3 3(7.02 10 kN-m )

(2.8 m)(0.0073709 rad) 0.020639 m

C

D

ML

EI

v

θ = = =×

= − = −

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


8 D

wLv

EI = − (assuming fixed support at C )

Values: w = 55 kN/m, L = 2.8 m, EI = 7.02 × 10

4 kN-m

2

Computation:4 4

4 2

(55 kN/m)(2.8 m)0.006020 m

8 8(7.02 10 kN-m ) D

wLv

EI

= − = − = −

×

Beam deflection at D

0.034117 m 0.020639 m 0.006020 m 0.007459 m 7.46 mm Dv = − − = = ↑ Ans.





10.43 The simply supported beam shown inFig. P10.43 consists of a W21 × 44 structural

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

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



(c) the beam deflection at point E .Fig. P10.43

Solution


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

4

0

30 A

w Lv


Values: w0 = −6 kips/ft, L = 12 ft,

EI = 2.4447 × 10

7

kip-in.

2

Computation:4 4 3

0

7 2

( 6 kips/ft)(12 ft) (12 in./ft)0.293139 in.

30 30(2.4447 10 kip-in. ) A

w Lv

EI

−= − = − =

×

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


3 B

L


Values: M = ½(6 kips/ft)(12 ft)(4 ft) = 144 kip-ft,

L = 18 ft, EI = 2.4447 × 107 kip-in.

2

Computation:2

7 2

(144 kip-ft)(18 ft)(12 in./ft)0.0050892 rad

3 3(2.4447 10 kip-in. )

(12 ft)(12 in./ft)(0.0050892 rad) 0.732847 in.

B

A

ML

EI

v

θ = = =×

= =





Consider uniformly distributed loads between supports. [Appendix C, SS beam with uniformly distributed load.]


24 B

wL



7 kip-in.

2

Computation:3 3 2

7 2


24 24(2.4447 10 kip-in. )

(12 ft)(12 in./ft)(0.0085880 rad) 1.236679 in.

B

A

wL

EI

v

θ = = =×

= − = −

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



6 B

L



L = 18 ft, EI = 2.4447 × 107 kip-in.

2

Computation:2

7 2

(108 kip-ft)(18 ft)(12 in./ft)0.0019085 rad

6 6(2.4447 10 kip-in. )

(12 ft)(12 in./ft)(0.0019085 rad) 0.274818 in.

B

A

ML

EI

v

θ = = =×

= =


0.293139 in. 0.732847 in. 1.236679 in. 0.274818 in. 0.064124 in. 0.0641 in. Av = + − + = = ↑ Ans.






Consider deflection at C from moment caused by linear load on overhang.


2 2(2 3 )6

C

M xv L Lx x



L = 18 ft, x = 9 ft, EI = 2.4447 × 107 kip-in.2

Computation:

2 2

32 2

7 2

(2 3 )6

(144 kip-ft)(9 ft)(12 in./ft)2(18 ft) 3(18 ft)(9 ft) (9 ft) 0.206112 in.

6(18 ft)(2.4447 10 kip-in. )

C

M xv L Lx x

LEI = − − +

⎡ ⎤= − − + = −⎣ ⎦×



384C

wLv

EI = −

Values: w = −6 kips/ft, L = 18 ft, EI = 2.4447 × 10

7 kip-in.

2

Computation:4 4 2

7 2

5 5( 6 kips/ft)(18 ft) (12 in./ft)0.579693 in.

384 384(2.4447 10 kip-in. )C

wLv

EI

−= − = − =

×

Consider deflection at C resulting from moment caused by uniform load on overhang DE .



2 2(2 3 )6

C

M xv L Lx x



L = 18 ft, x = 9 ft, EI = 2.4447 × 107 kip-in.

2

Computation:

2 2

32 2

7 2

(2 3 )6

(108 kip-ft)(9 ft)(12 in./ft)2(18 ft) 3(18 ft)(9 ft) (9 ft) 0.154585 in.

6(18 ft)(2.4447 10 kip-in. )

C

M xv L Lx x

LEI = − − +

⎡ ⎤= − − + = −⎣ ⎦×


0.206112 in. 0.579693 in. 0.154585 in. 0.218995 in. 0.219 in.C v = − + − = = ↑ Ans.





(c) Beam deflection at point E

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


6 D

L



L = 18 ft, EI = 2.4447 × 107 kip-in.2

Computation:2

7 2

(144 kip-ft)(18 ft)(12 in./ft)0.0025446 rad

6 6(2.4447 10 kip-in. )

(6 ft)(12 in./ft)(0.0025446 rad) 0.183212 in.

D

E

ML

EI

v

θ = = =×

= =



24 D

wL



7 kip-in.

2

Computation:3 3 2

7 2


24 24(2.4447 10 kip-in. )

(6 ft)(12 in./ft)(0.0085880 rad) 0.618336 in.

D

E

wL

EI

v

θ = = =×

= − = −

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


3 D

L



L = 18 ft, EI = 2.4447 × 107 kip-in.

2

Computation:2

7 2

(108 kip-ft)(18 ft)(12 in./ft)0.0038169 rad

3 3(2.4447 10 kip-in. )

(6 ft)(12 in./ft)(0.0038169 rad) 0.274818 in.

D

E

ML

EI

v

θ = = =×

= =





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


8 E

wLv

EI = − (assuming fixed support at D)

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

7 kip-in.

2

Computation:4 4 3

7 2

( 6 kips/ft)(6 ft) (12 in./ft)0.068704 in.

8 8(2.4447 10 kip-in. ) E

wLv

EI

−= − = − =

×

Beam deflection at E

0.183212 in. 0.618336 in. 0.274818 in. 0.068704 in.

0.091602 in. 0.0916 in.

E v = − + +

= − = ↓ Ans.





10.44 The solid 30-mm-diameter steel [ E = 200GPa] shaft shown in Fig. P10.44 supports two

belt pulleys. Assume that the bearing at B can

be idealized as a roller support and that the bearing at D can be idealized as a pin support.

For the loading shown, determine:

(a) the shaft deflection at pulley A.(b) the shaft deflection at pulley C .

Fig. P10.44

Solution

Section properties:

4 4(30 mm) 39,760.78 mm64

I π

= =

(a) Shaft deflection at pulley A

Determine cantilever deflection due to pulley A load. [Appendix C, Cantilever beam with concentrated load.]


3 A

PLv


Values: P = 700 N, L = 500 mm, EI = 7.95216 × 10

9 N-mm

2

Computation:3 3

9 2

(700 N)(500 mm)3.6678 mm

3 3(7.95216 10 N-mm ) A

PLv

EI = − = − = −

×

Consider deflection at A resulting from rotation at B caused by pulley A load.


3 B

L


Values: M = (700 N)(500 mm) = 350,000 N-mm,

L = 1,800 mm, EI = 7.95216 × 109 N-mm

2

Computation:

9 2

(350,000 N-mm)(1,800 mm)0.0264079 rad

3 3(7.95216 10 N-mm )

(500 mm)(0.0264079 rad) 13.2040 mm

B

A

ML

EI

v

θ = = =×

= − = −

Consider deflection at A resulting from rotation at B caused by pulley C load.

[Appendix C, SS beam with concentrated load.]






16 B

PL


Values: P = 1,000 N, L = 1,800 mm,

EI = 7.95216 × 109 N-mm

2

Computation:2 2

9 2(1,000 N)(1,800 mm) 0.0254648 rad16 16(7.95216 10 N-mm )

(500 mm)(0.0254648 rad) 12.7324 mm

B

A

PL EI

v

θ = = =×

= =

Shaft deflection at A

3.6678 mm 13.2040 mm 12.7324 mm 4.1393 mm 4.14 mm Av = − − + = − = ↓ Ans.

(b) Shaft deflection at pulley C

Consider pulley A load. [Appendix C, SS beam with concentrated moment.]


2 2(2 3 )6

C M xv L Lx x LEI

= − − + (elastic curve)

Values: M = (700 N)(500 mm) = −350,000 N-mm,

L = 1,800 mm, x = 900 mm,

EI = 7.95216 × 109 N-mm

2

Computation:

2 2

2 2

9 2

(2 3 )6

( 350,000 N-mm)(900 mm)2(1,800 mm) 3(1,800 mm)(900 mm) (900 mm)

6(1,800 mm)(7.95216 10 N-mm )

8.9127 mm

C

M xv L Lx x

LEI = − − +

−⎡ ⎤= − − +⎣ ⎦×

=

Consider pulley C load. [Appendix C, SS beam with concentrated load.]


48C

PLv

EI = −

Values: P = 1,000 N, L = 1,800 mm,

EI = 7.95216 × 109 N-mm

2

Computation:3 3

9 2

(1,000 N)(1,800 mm)15.2789 mm

48 48(7.95216 10 N-mm )C

PLv

EI = − = − = −

×

Shaft deflection at C

8.9127 mm 15.2789 mm 6.3662 mm 6.37 mmC v = − = − = ↓ Ans.





10.45 The cantilever beam shown in Fig.P10.45 consists of a W530 × 92 structural steel

wide-flange shape [ E = 200 GPa; I = 552 × 106

mm4]. For the loading shown, determine:

(a) the beam deflection at point A.(b) the beam deflection at point B.

Fig. P10.45

Solution


Consider an upward 85 kN/m uniformly distributed load acting over entire 4-m span.



8 A

wLv

EI = −

Values: w = −85 kN/m, L = 4 m, EI = 1.104 × 10

5 kN-m

2

Computation:4 4

5 2

( 85 kN/m)(4 m)0.024638 m

8 8(1.104 10 kN-m ) A

wLv

EI

−= − = − =

×

Consider a downward 85 kN/m uniformly distributed load acting over span BC .

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

4 3

and8 6

B B

wL wLv

EI EI θ = − = (magnitude)

Values: w = 85 kN/m, L = 2.5 m, EI = 1.104 × 10

5 kN-m

2

Computation:4 4

5 2

3 3

5 2

(85 kN)(2.5 m)0.003759 m

8 8(1.104 10 kN-m )

(85 kN)(2.5 m)0.0020050 rad

6 6(1.104 10 kN-m )

0.003759 m (1.5 m)(0.0020050 rad) 0.006767 m

B

B

A

wLv

EI

wL

EI

v

θ

= − = − = −×

= = =×

= − − = −


0.024638 m 0.006767 m 0.017871 m 17.87 mm Av = − = = ↑ Ans.






Consider an upward 85 kN/m uniformly distributed load acting over entire 4-m span.


22 2(6 4 )

24 B

wxv L Lx x


Values: w = −85 kN/m, L = 4 m, x = 2.5 m,

EI = 1.104 × 105 kN-m2

Computation:2

2 2

22 2

5 2

(6 4 )24

( 85 kN/m)(2.5 m)6(4 m) 4(4 m)(2.5 m) (2.5 m) 0.012481 m

24(1.104 10 kN-m )

B

wxv L Lx x

EI = − − +

−⎡ ⎤= − − + =⎣ ⎦×

Consider a downward 85 kN/m uniformly distributed load acting over span BC .


4

8 B

wLv

EI = −

Values: w = 85 kN/m, L = 2.5 m, EI = 1.104 × 10

5 kN-m

2

Computation:4 4

5 2

(85 kN)(2.5 m)0.003759 m

8 8(1.104 10 kN-m ) B

wLv

EI = − = − = −

×


0.012481 m 0.003759 m 0.008722 m 8.72 mm Bv = − = = ↑ Ans.





10.46 The solid 30-mm-diameter steel [ E = 200 GPa] shaft shown in Fig. P10.46

supports two belt pulleys. Assume that

the bearing at A can be idealized as a pin

support and that the bearing at E can beidealized as a roller support. For the

loading shown, determine:

(a) the shaft deflection at pulley B.

(b) the shaft deflection at point C .(c) the shaft deflection at pulley D.

Fig. P10.46

Solution

Section properties:

4 4(30 mm) 39,760.78 mm64

I π

= =

(a) Shaft deflection at pulley B

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


2 2 2( )6

B

Pabv L a b

LEI = − − −

Values: P = 750 N, L = 1,000 mm, a = 300 mm,

b = 700 mm, EI = 7.95216 × 109 N-mm

2

Computation:

2 2 2

2 2 2

9 2

( )6

(750 N)(300 mm)(700 mm)(1,000 mm) (300 mm) (700 mm)

6(1,000 mm)(7.95216 10 N-mm )1.38642 mm

B

Pabv L a b

LEI = − − −

⎡ ⎤= − − −⎣ ⎦×

= −

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


2 2 2( )6

B

Pbxv L b x


Values: P = 500 N, L = 1,000 mm, x = 300 mm,b = 200 mm, EI = 7.95216 × 10

9 N-mm

2

Computation:2 2 2

2 2 2

9 2

( )6

(500 N)(200 mm)(300 mm)(1,000 mm) (200 mm) (300 mm)

6(1,000 mm)(7.95216 10 N-mm )

0.54702 mm

B

Pbxv L b x

LEI = − − −

⎡ ⎤= − − −⎣ ⎦×

= −





Shaft deflection at B

1.38642 mm 0.54702 mm 1.93344 mm 1.933 mm Bv = − − = − = ↓ Ans.

(b) Shaft deflection at point C

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

2 2 2( )

6C

Pbxv L b x

LEI

= − − − (elastic curve)

Values: P = 750 N, L = 1,000 mm, x = 500 mm,

b = 300 mm, EI = 7.95216 × 109 N-mm

2

Computation:

2 2 2

2 2 2

9 2

( )6

(750 N)(300 mm)(500 mm)(1,000 mm) (300 mm) (500 mm)

6(1,000 mm)(7.95216 10 N-mm )

1.55618 mm

C

Pbxv L b x

LEI = − − −

⎡ ⎤= − − −⎣ ⎦×

= −



2 2 2( )6

C

Pbxv L b x


Values: P = 500 N, L = 1,000 mm, x = 500 mm,

b = 200 mm, EI = 7.95216 × 109 N-mm

2

Computation:2 2 2

2 2 2

9 2

( )6

(500 N)(200 mm)(500 mm)(1,000 mm) (200 mm) (500 mm)

6(1,000 mm)(7.95216 10 N-mm )

0.74403 mm

C

Pbxv L b x

LEI = − − −

⎡ ⎤= − − −⎣ ⎦×

= −

Shaft deflection at C

1.55618 mm 0.74403 mm 2.30021 mm 2.30 mmC v = − − = − = ↓ Ans.



(c) Shaft deflection at pulley D

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


2 2 2( )6

D

Pbxv L b x


Values: P = 750 N, L = 1,000 mm, x = 200 mm,

b = 300 mm, EI = 7.95216 × 109 N-mm

2

Computation:

2 2 2

2 2 2

9 2

( )6

(750 N)(300 mm)(200 mm)(1,000 mm) (300 mm) (200 mm)

6(1,000 mm)(7.95216 10 N-mm )

0.82053 mm

D

Pbxv L b x

LEI = − − −

⎡ ⎤= − − −⎣ ⎦×

= −


Relevant equation from Appendix C:2 2 2( )

6 D

Pabv L a b

LEI = − − −

Values: P = 500 N, L = 1,000 mm, a = 800 mm,b = 200 mm, EI = 7.95216 × 10

9 N-mm

2

Computation:

2 2 2

2 2 2

9 2

( )6

(500 N)(800 mm)(200 mm) (1,000 mm) (800 mm) (200 mm)6(1,000 mm)(7.95216 10 N-mm )

0.53654 mm

D

Pabv L a b

LEI = − − −

⎡ ⎤= − − −⎣ ⎦×

= −

Shaft deflection at D

0.82053 mm 0.53654 mm 1.35707 mm 1.357 mm Dv = − − = − = ↓ Ans.

mechanics of materials solutions chapter10 probs29 46

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