solutions manual_chapter 11_final.pdf

7/23/2019 Solutions Manual_Chapter 11_Final.pdf

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a

Ans.

Ans.T2 = 00

- + T2 = 0+ T ©Fy = 0;

T1 =

- T1 (75) = 0+ ©MC = 0; 225(375)

1125 N

225 1125

9 N

-

11–1. The load binder is used to support a load. If the force

applied to the handle is 225 N, determine the tensions T 1 and

T 2 in each end of the chain and then draw the shear and

moment diagrams for the arm ABC .

300 mm

225 N

T2

T1

75 mm

A

B

C

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$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1047



11–2. Draw the shear and moment diagrams for the shaft.The bearings at A and D exert only vertical reaction on theshaft.The loading is applied to the pulleys at B and C and E.

A

B

350 mm 500 mm 375 mm 300 mm

360 N

500 N

160 N

C D

E

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1048



! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

a

0;

3

5 (4000) 0;

2400 lb

0;

4

5 (4000) 1200 0;

2000 lb

4000 lb 0; 4

5 (3) 1200(8) 0;

The engine crane is used to support the engine, whichhas a weight of 1200 lb. Draw the shear and moment diagramsof the boom ABC when it is in the horizontal position shown.

5 ft3 ft

C B

4 ft

Ahas a weight of 6 kN. Draw the shear and moment diagrams

F B(0.9) – 6(2.4) = 0 F B = 20 kN

(20) – 6 = 0; A y = 10 kN

(20) = 0; A x = 12 kN

1.5 m0.9 m

C B

1.2 m

A

0.9 m 1.5 m

F B 6 kN

V kN

6

–10

–9

14 kN-m

11–3.

1049



*11–4. Draw the shear and moment diagrams for the beam.

1.2 m 1.2 m 1.2 m 1.2 m 1.2 m

9 kN 9 kN 9 kN 9 kN

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1050



11–5. Draw the shear and moment diagrams for the beam.

2 m 3 m

10 kN 8 kN

15 kNm

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1051



11–6. Express the internal shear and moment in terms of x

and then draw the shear and moment diagrams.

A B

x

L

2

L

2

w0

Support Reactions: Referring to the free-body diagram of the entire beamshown in Fig. a,

a

Shear and Moment Function: For we refer to the free-body

diagram of the beam segment shown in Fig. b.

Ans.

a

Ans.

For we refer to the free-body diagram of the beam segment

shown in Fig. c.

Ans.

a

Ans.

When V = 0, the shear function gives

Substituting this result into the moment equation,

Shear and Moment Diagrams:As shown in Figs. d and e.

M|x= 0.7041L = 0.0265w0L2

x = 0.7041L0 = L2 - 6(2x - L)2

M = w0

24LcL2x - (2x - L)3 d

M +1

2cw0

L(2x - L) d c1

2(2x - L) d c1

6(2x - L) d -

w0

24Lx = 0+ ©M = 0;

V = w0

24LcL2 - 6(2x - L)2 d

w0L

24 - 1

2 cw0

L(2x - L) d c12 (2x - L) d - V = 0+ c ©Fy = 0;

L

2 6 x … L,

M = w0L

24 x

M - w0L

24 x = 0+ ©M = 0;

V = w0L

24

w0L

24 - V = 0+ c ©Fy = 0;

0 … x 6 L

2,

Ay = w0L

24

Ay +5

24 w0L -

1

2 w0aL2 b = 0+ c ©Fy = 0;

By =5

24 w0L

By(L) -1

2 w0

aL

2 b a5

6L

b = 0+ ©MA = 0;

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1052



! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

4 ft

6 kip 8 kip

A

C

B

6 ft 4 ft 4 ft30 kN

40 kN

20 kN20 kN

20 kN50 kN

1.5 m1 m

1 m 1 m

1.5 m

1 m

20

20

–30

–30

–20

1 m

30 kN 40 kN

A

C

B

1.5 m 1 m 1 m

11–7. Draw the shear and moment diagrams for thecompound beam which is pin connected at B. (This structureis not fully stable. But with the given loading, it is balancedand will remain as shown if not disturbed.)

1053



Support Reactions: Referring to the free-body diagram of the entire beam shownin Fig. a,

a

Shear and Moment Function: For , we refer to the free -body diagramof the beam segment shown in Fig. b.

lb Ans.

a

Ans.

For we refer to the free-body diagram of the beam segment shownin Fig. c.

Ans.

a

Ans.


M = { ( - x)} N#( - x) - M = 0+ ©M = 0;

V = -

V + = 0+ c ©Fy = 0;

6 x … 3 m,

M = { x - 00x2} #

M + xax2b - x = 0+ ©M = 0;

V = {9333 - x}

- x - V = 0+ c ©Fy = 0;

0 … x 6

Ay =

Ay + - 00( ) - = 0+ c ©Fy = 0;

By =

By(3) - 00 ( )( ) - 00( ) = 0+ ©MA = 0;

*11–8. Express the internal shear and moment in terms of xand then draw the shear and moment diagrams for the beam.

40 26 0 2 1

6 67 N6

60 26667 4000

9333 N

2 m

6000

60009333

93336000

N m039333

2 m

6667

6667 N

6667 3

m6667 3

A

B

4000 N

6 kN/m

2 m 1 m

x

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1054



11–9. Express the internal shear and moment in terms of x

and then draw the shear and moment diagrams for theoverhanging beam.


a

Shear and Moment Function: For we refer to the free-body diagramof the beam segment shown in Fig.b.

Ans.

a

Ans.

For we refer to the free-body diagram of the beam segment shownin Fig. c.

Ans.

a

Ans.


M = -{3(6 - x)2} kN # m

-M - 6(6 - x)a6 - x2 b = 0+ ©M = 0;

V = {6(6 - x)} kN

V - 6(6 - x) = 0+ c ©Fy = 0;

4 m 6 x … 6 m,

M = {9x - 3x2} kN # m

M + 6xax2b - 9x = 0+ ©M = 0;

V = {9 - 6x} kN

9 - 6x - V = 0+ c ©Fy = 0;

0 … x 6 4 m,

Ay = 9 kN

Ay + 27 - 6(6) = 0+ c ©Fy = 0;

By = 27 kN

By(4) - 6(6)(3) = 0+ ©MA = 0;

A

B

6 kN/ m

x

4 m 2 m

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1055



! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

Equations of Equilibrium: Referring to the free-body diagram of the frame shownin Fig. a,

a

Shear and Moment Diagram: The couple moment acting on B due to N D

is

. The loading acting on member ABC is shown in Fig. band the shear and moment diagrams are shown in Figs.c and d. 300(1.5) 450 lb ft

300 lb

0;

(1.5) 150(3) 0

150 lb

0;

150 0

11–10. Members ABC and BD of the counter chair arerigidly connected at B and the smooth collar at D is allowedto move freely along the vertical slot. Draw the shear andmoment diagrams for member ABC.

A

D

B

C

P 150 lb

1.5 ft1.5 ft

1.5 ft

A

D

B

C

P 750 N

0.45 m0.45 m

0.45 m

750

750 N

N D = (0.45) – 750(0.9) = 0

N D = 1500 N

M B = 1500(0.45) = 675 N · m. The loading acting on member ABC is shown in Fig. b0.45 m 0.45 m

750 N

0.45 m

0.45 m 0.45 m750 N

M B = 675 N· m1500 N

V (N)

750

0.45 0.9

x (m)0.45

–337.5

0.9

337.5

M (N · m)

1056



11–11. Draw the shear and moment diagrams for the pipe.The end screw is subjected to a horizontal force of 5 kN.Hint: The reactions at the pin C must be replaced by anequivalent loading at point B on the axis of the pipe. 80 mm

400 mm

5 kN

A

B

C

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1057



*11–12. A reinforced concrete pier is used to support thestringers for a bridge deck. Draw the shear and momentdiagrams for the pier when it is subjected to the stringerloads shown. Assume the columns at A and B exert onlyvertical reactions on the pier.

1 m 1 m 1 m 1 m1.5 m

60 kN 60 kN35 kN 35 kN 35 kN1.5 m

A B

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1058



11–13. Draw the shear and moment diagrams for the rod. Itis supported by a pin at A and a smooth plate at B.The plateslides within the groove and so it cannot support a verticalforce,although it can support a moment.

4 m

A

B

2 m

15 kN

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1059



! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

10 in.4 in.

50 in. A B

C

D

120

6–14. The industrial robot is held in the stationary positionshown.Draw the shear and moment diagrams of the arm ABC

if it is pin connected at A and connected to a hydraulic cylinder(two-force member) BD. Assume the arm and grip have auniform weight of 1.5 lbin. and support the load of 40 lb atC .

250 mm100 mm

1250 mm A B

C

D

120

11–14. The industrial robot is held in the stationary positionshown. Draw the shear and moment diagrams of the arm

ABC if it is pin connected at A and connected to a hydrauliccylinder (two-force member) BD. Assume the arm and griphave a uniform weight of 0.3 N/mm and support the load of200 N at C .

200 N250 mm

100 mm1469 N

0.3 N/mm

1250 mm

1469 N

1864 N

2544 NV (N)

575

200

–30

–1969–1894

–484.4

–1.5

M (N · m)

1060



*11–16. Determine the placement distance a of the rollersupport so that the largest absolute value of the moment isa minimum. Draw the shear and moment diagrams for thiscondition.

Support Reactions: As shown on FBD.

Absolute Minimum Moment: In order to get the absolute minimum moment, themaximum positive and maximum negative moment must be equal that is

For the positive moment:

a

For the negative moment:

a

Ans.

Shear and Moment Diagram:

a = 2 3

2 L = 0.866L

4a L - 3L2 = 4a L - 4a

2

PL -3PL

2

4a= P(L - a)

Mmax( + ) = Mmax( - )

Mmax( - ) = P(L - a)

Mmax( - ) - P(L - a) = 0+ ©MNA = 0;

M max( + ) = PL -3PL

2

4a

Mmax( + ) - a2P -3PL

2a b aL

2 b = 0+ ©MNA = 0;

Mmax( + ) = Mmax( - ).

A

P

a

P

B

L–2

L–2

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1062



! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

(1)

a (2) 0;

1

2

(33.33)()

3 300 0 {300 5.5563} lb ft

0; 300 1

2 (33.33)() 0 {300 16.672} lb

300 lb 200 lb/ ft

A

6 ft

The free-body diagram of the beam’s left segment sectioned through an arbitrarypoint shown in Fig. b will be used to write the shear and moment equations. Theintensity of the triangular distributed load at the point of sectioning is

Referring to Fig.b,

w 2006 33.33

The shear and moment diagrams shown in Figs. c and d are plotted using Eqs. (1)and (2), respectively.

V (kN)

M (kN · m)

–1.5

–4.5

x (m)

x (m)

–5

1.5 kN

4.5 kN

5 kN · m

12

(3)(2) kN

43

m 23

m

12

(1.5 x) x

1.5 kN 3 kN/m

A

2 m

3 x2 = 1.5 x

–1.5 –1

2 (1.5 x)( x) – V = 0 V = {–1.5 – 0.75 x2} kN (1)

(1.5 x)( x) x3

+ 1.5 = 0 M = {–1.5 – 0.25 x3} kN · m (2)

1.5 kN

11–17. Express the internal shear and moment in thecantilevered beam as a function of x and then draw theshear and moment diagrams.

1063



! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8


Shear and Moment Function:

For :

Ans.

a

Ans.

For :

Ans.

a

Ans. {8.00 120} kip ft

0;

8(10 ) 40 0

0; 8 0 8.00 kip

6 ft 10 ft

{2 30.0 216} kip ft

0; 216 22 30.0 0

{30.0 2} kip

0;

30.0 2 0

0 6 ft

11–18. Draw the shear and moment diagrams for the beam,and determine the shear and moment throughout the beamas functions of x.

6 ft 4 ft

2 kip/ ft 8 kip

x

10 kip

40 kipft

1.8 m 1.2 m

30 kN/m40 kN

x

50 kN

200 kNm

458.6 kN · m

30(1.8) = 54 kN50 kN 40 kN

144 kN

200 kN · m

0.9 m 0.9 m 1.2 m

30 x

144 kN

458.6 kN · m

40 kN

200 kN · m

3 – x

V (kN)

14490

1.8 3

40 1.8 3 x (m) x (m)

–248–200

–458.6

M (kN · m)

1.8 m

144 – 30 x – V = 0

V = {144 – 30 x} kN

M + 458.6 + 30 x x2 – 144 x = 0

M = {–15 x2 + 144 x – 458.6} kN · m

For 1.8 m < x 3 m:

V – 40 = 0 V = 40 kN

–M – 40(3 – x) – 200 = 0

M = {40 x – 320} kN · m

1064



! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

A

30 kipft

B

5 ft 5 ft

2 kip/ ft

5 ft


30 kN/m 45 kN · m

3.75 kN

–3.75

–45

V (kN)

M (kN · m)

1.5 m 1.5 m 1.5 m

5.625

–39.375–33.75

A

45 kNm

B

1.5 m 1.5 m

30 kN/m

1.5 m41.25 kN

1065



*11–20. Draw the shear and moment diagrams for theoverhanging beam.

Support Reactions: Referring to the free-body diagram of the beam shown in Fig. a,

a

Shear and Moment Functions: For we refer to the free-body diagramof the beam segment shown in Fig. b.

Ans.

a

Ans.

When V = 0, from the shear function,

Substituting this result into the moment function,

For we refer to the free-body diagram of the beam segmentshown in Fig. c.

Ans.

a

Ans.

Shear and Moment Diagrams: As shown in Figs. d and e.

M = {40.05(6 - x)}

( - x) - M = 0+ ©M = 0;

V = - k

V + = 0+ c ©Fy = 0;

6 x … 6 m,

M|x= .94 = 9. kN #

x = .940 = - x2

M = e x - x3 f kN #

M + ax3b - x = 0+ ©M = 0;

V =

e - x2

f kN

-1

2a b(x) - V = 0+ c ©Fy = 0;

0 … x 6 ,

Ay = k

Ay + -1

2 ( )( ) = 0+ c ©Fy = 0;

By = k

By(6) -1

2 ( )( )( ) = 0+ ©MA = 0; 45 4 2.67

40.05 N

40.05 45 4

49.95 N

4 m

49.95 1x

49.95

49.95 m

49.95 2 9 m

2 9 m 9 21 m

4 m

40.05kN

40.05 N

40.05 6

11.25

49.95 5.625

1

2a b(x)1x11.25

1.875

5.625

kN # m

B A

4 m

45 kN/m

2 m

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1066



Ans.Mmax = #393.75 N m

11–21. The 700-N man sits in the center of the boat, which

has a uniform width and a weight per linear foot of 45 N/m.

Determine the maximum bending moment exerted on the

boat. Assume that the water exerts a uniform distributed

load upward on the bottom of the boat.

2.25 m 2.25 m

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1067



11–22. Draw the shear and moment diagrams for theoverhang beam.

Since the loading is discontinuous at support B, the shear and moment equations mustbe written for regions and of the beam.The free-bodydiagram of the beam’s segment sectioned through an arbitrary point within these tworegions is shown in Figs. b and c.

Region , Fig. b0 … x 6 3 m

3 m 6 x … 6 m0 … x 6 3 m

4 kN/ m

3 m 3 m

A

B

(1)

a (2)

Region , Fig.c

(3)

a (4)+ ©M = 0; -M - 4(6 - x) c 1

2 (6 - x) d = 0

M = {-2(6 - x)2}kN # m

+ c ©Fy = 0; V - 4(6 - x) = 0 V = {24 - 4x} kN

3 m 6 x … 6 m

+ ©M = 0; M +1

2a4

3xb(x)ax

3b + 4x = 0 M = e -

2

9 x3 - 4xf kN # m

+ c ©Fy = 0; -4 -1

2a 4

3 xb(x) - V = 0 V = e -

2

3 x2 - 4 f kN

The shear diagram shown in Fig. d is plotted using Eqs. (1) and (3). The value of shear just to the left and just to the right of the support is evaluated using Eqs. (1)and (3), respectively.

The moment diagram shown in Fig. e is plotted using Eqs. (2) and (4).The value of

the moment at support B is evaluated using either Eq. (2) or Eq. (4).

or

Mx= 3 m = -2(6 - 3)2 = -18 kN # m

Mx= 3 m = -2

9(33) - 4(3) = -18 kN # m

Vx=3 m + = 24 - 4(3) = 12 kN

Vx= 3 m - = -2

3 (32) - 4 = -10 kN

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1068



11–23. The footing supports the load transmitted by the twocolumns. Draw the shear and moment diagrams for thefooting if the reaction of soil pressure on the footing isassumed to be uniform.

2 m 4 m 2 m

60 kN60 kN

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1069



*11–24. Express the shear and moment in terms of x andthen draw the shear and moment diagrams for the simplysupported beam.

A B

3 m 1.5 m

300 N/ m


a

Shear and Moment Function: For we refer to the free-body diagram

of the beam segment shown in Fig. b.

Ans.

a

Ans.

When V = 0, from the shear function,

Substituting this result into the moment equation,

For we refer to the free-body diagram of the beam segmentshown in Fig. c.

Ans.

a

Ans.

Shear and Moment Diagrams: As shown in Figs. d and e.

M = e 375(4.5 - x) -100

3(4.5 - x)3f N # m

375(4.5 - x) -1

2 [200(4.5 - x)](4.5 - x)a4.5 - x

3 b - M = 0+ ©M = 0;

V =

e100(4.5 - x)2 - 375

f N

V + 375 -1

23200(4.5 - x)4(4.5 - x) = 0+ c ©Fy = 0;

3 m 6 x … 4.5 m,

M| x = 2 6 m = 489.90 N # m

x = 2 6 m0 = 300 - 50x2

M = e300x -50

3 x3f

N # m

M +1

2 (100x)xax

3b - 300x = 0+ ©M = 0;

V = {300 - 50x2} N

300 -1

2(100x)x - V = 0+ c ©Fy = 0;

0 … x 6 3 m,

Ay = 300 N

Ay + 375 -1

2 (300)(3) -

1

2 (300)(1.5) = 0+ c ©Fy = 0;

By = 375 N

By(4.5) -1

2 (300)(3)(2) -

1

2 (300)(1.5)(3.5) = 0+ ©MA = 0;

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1070



Ans.

a

Ans.M = - 5x2 + x -

- 0 - 50(x - (x -

2 - M + 50(x - ) = 0+ ©M = 0;

V = - 50x

+ c ©Fy = 0;

- 50(x - ) - V + 50 = 01.222 22

4950 22

22 1. 21.2)

22 1.2)

11.2 4950 4950

27

11–25. Draw the shear and moment diagrams for the beam

and determine the shear and moment in the beam as

functions of x, where 1.2 m < x < 3 m.

270 N.m

B

x

1.2 m 1.2 m

2.25 kN/m

2 m

270 N.m

A

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1071




a

Substitute ,

M = 0.0345 w0L2

x = 0.7071L

+ ©MNA = 0; M +1

2 aw0x

L b (x)ax

3b -

w0L

4 ax -

L

3 b = 0

x = 0.7071 L

+ c ©Fy = 0;

w0L

4 -

1

2 aw0x

L b (x) = 0

B

w0

A2L

3

L

3

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1072




Shear and Moment Diagram: Shear and moment at can be determinedusing the method of sections.

a

M =5w0 L2

54

+ ©MNA = 0; M + w0 L

6 aL

9 b -

w0 L

3 aL

3 b = 0

+ c ©Fy = 0; w0 L

3 -

w0 L

6 - V = 0 V =

w0 L

6

x = L>3

*11–28. Draw the shear and moment diagrams for the beam. w0

A B

L

3

L

3

L

3

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1073



Equations of Equilibrium: Referring to the free-body diagram shown in Fig. a,

a

Shear and Moment Diagram: As shown in Figs. b and c.

Ay = 24 kN

Ay + 24 - 8(6) = 0+ c ©Fy = 0;

By = 24 kN

By(3) - 8(6)(1.5) = 0+ ©MA = 0;

11–29. Draw the shear and moment diagrams for the doubleoverhanging beam.

B A

1.5 m 1.5 m3 m

8 kN/ m

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1074



11–30. The beam is bolted or pinned at A and rests on a bearing

pad at B that exerts a uniform distributed loading on the beam

over its 0.6-m length. Draw the shear and moment diagrams

for the beam if it supports a uniform loading of 30 kN/m.

2.4 m

0.3 m 0.6 m

A

B

30 kN/m

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1075



11–31. The support at A allows the beam to slide freely alongthe vertical guide so that it cannot support a vertical force.Draw the shear and moment diagrams for the beam.

B A

L

w

Equations of Equilibrium: Referring to the free-body diagram of the beam shownin Fig. a,

a


By = wL

By - wL = 0+ c ©Fy = 0;

MA = wL2

2

wLaL

2 b - MA = 0+ ©MB = 0;

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1076



Ans.w0 = 1.2 kN>m

+ c ©Fy = 0;

2(w0)(20)a 1

2b - 60(0.4) = 0

*11–32. The smooth pin is supported by two leaves A and Band subjected to a compressive load of caused bybar C . Determine the intensity of the distributed load of the leaves on the pin and draw the shear and momentdiagram for the pin.

w0

0.4 kN>m

20 mm

0.4 kN/ m

w0

20 mm 60 mm

w0 A B

C

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1077



Equations of Equilibrium: Referring to the free-body diagram of the shaft shownin Fig. a,

a


Ay = 650 N

Ay + 250 - 900 = 0+ c ©Fy = 0;

By = 250 N

By(2) + 400 - 900(1) = 0+ ©MA = 0;

11–33. The shaft is supported by a smooth thrust bearing at A and smooth journal bearing at B. Draw the shear andmoment diagrams for the shaft.

A B

1 m 1 m 1 m

900 N

400 Nm

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1078



Equations of Equilibrium: Referring to the free-body diagram of the beam shown inFig. a,

a


MA = 9 kN # m

MA - 3 - 2(3) = 0+ ©MA = 0;

Ay = 2 kN

Ay - 2 = 0+ c ©Fy = 0;

11–34. Draw the shear and moment diagrams for thecantilever beam.

A

2 kN

3 kNm

1.5 m 1.5 m

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1079




Shear and Moment Functions:

For :

Ans.

a

Ans.

For :

Ans.

Set ,

a

Ans.

Substitute , M = 691 N # mx = 3.87 m

M =

e -

100

9

x3 + 500x - 600

f N # m

+ 200(x - 3)ax - 3

2 b - 200x = 0

+ ©MNA = 0; M +1

2 c 200

3 (x - 3) d(x - 3)ax - 3

3 b

x = 3.873 mV = 0

V = e -

100

3 x2 + 500f N

+ c ©Fy = 0;

200 - 200(x - 3) -1

2 c200

3 (x - 3) d(x - 3) - V = 0

3 m 6 x … 6 m

M = 5200 x6 N # m

+ ©MNA = 0; M - 200 x = 0

+ c ©Fy = 0; 200 - V = 0 V = 200 N

0 … x 6 3 m

11–35. Draw the shear and moment diagrams for the beamand determine the shear and moment as functions of x.

3 m 3 m

x

A B

200 N/ m

400 N/ m

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1080



*11–36. The shaft is supported by a smooth thrust bearing at A and a smooth journal bearing at B. Draw the shear andmoment diagrams for the shaft.

Equations of Equilibrium: Referring to the free-body diagram of the shaft shown inFig. a,

a

Shear and Moment Diagram:As shown in Figs. b and c.

Ay = 825 N

1725 - 900 - Ay = 0+ c ©Fy = 0;

By = 1725 N

By(1.6) - 600 - 900(2.4) = 0+ ©MA = 0;

A B

900 N

600 Nm

0.8 m 0.8 m 0.8 m

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1081




B

4.5 m 4.5 m

50 kN/ m

A

50 kN/ m

A

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1082



11–38. The beam is used to support a uniform load along CDdue to the 6-kN weight of the crate. If the reaction atbearing support B can be assumed uniformly distributedalong its width, draw the shear and moment diagrams forthe beam.

2.75 m 2 m

0.75 m0.5 m

C

B

A

D

Equations of Equilibrium: Referring to the free-body diagram of the beamshown in Fig. a,

a

Shear and Moment Diagram: The intensity of the distributed load at

support B and portion CD of the beam are

Fig. b. The shear

and moment diagrams are shown in Figs.c and d.

= 3 kN>m,20 kN>m and wCD =6

2

100.5

=wB = FB0.5

=

Ay = 4 kN

10 - 6 - Ay = 0+ c ©Fy = 0;

FB = 10 kN

FB(3) - 6(5) = 0+ ©MA = 0;

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1083



Equations of Equilibrium: Referring to the free-body diagram of the beam shown inFig. a,

a


Ay =

Ay + 00 - 0 - 00( ) - 00 = 0+ c ©Fy = 0;

By = 00

By( ) + 00( ) - 00( )( ) - 00( ) = 0+ ©MA = 0;

11–39. Draw the shear and moment diagrams for the doubleoverhanging beam.

2 18 1 30 2 1 18 3

48 N

1830 248 180

4800 N

1 m 1 m

3000 N/m

1800 N

2 m

1800 N

B A

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1084



*11–40. Draw the shear and moment diagrams for thesimply supported beam.

AB

2 m 2 m

10 kN 10 kN

15 kNm

2 m

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1085



Support Reactions: Referring to the free-body diagram of segment BC shownin Fig. a,

a

Using the result of B y and referring to the free-body diagram of segment AB, Fig. b,

a

Shear and Moment Diagrams: As shown in Figs. c and d.

MA = 2800 N

MA - 600(2) - 400(4) = 0+ ©MA = 0;

Ay = 1000 N

Ay - 600 - 400 = 0+ c ©F y = 0;

By = 400 N

By + 400 - 400(2) = 0+ c ©F y = 0;

Cy = 400 N

Cy(2) - 400(2)(1) = 0+ ©MB = 0;

11–41. The compound beam is fixed at A, pin connected atB, and supported by a roller at C . Draw the shear andmoment diagrams for the beam.

A BC

2 m2 m2 m

400 N/ m

600 N

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1086



Support Reactions:

From the FBD of segment AB

a

From the FBD of segment BD

a

From the FBD of segment AB

Shear and Moment Diagram:

©Fx = 0; Ax = 0:+

©Fx = 0;

Bx = 0:+

Cy = 20.0 kN

+ c ©Fy = 0; Cy - 5.00 - 5.00 - 10.0 = 0

Dy = 5.00 kN

+ ©MC = 0; 5.00(1) + 10.0(0) - Dy (1) = 0

+ c ©Fy = 0; Ay - 10.0 + 5.00 = 0 Ay = 5.00 kN

+ ©MA = 0;

By (2) - 10.0(1) = 0

By = 5.00 kN

11–42. Draw the shear and moment diagrams for thecompound beam.

B A C D

2 m 1 m 1 m

5 kN/ m

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1087



11–43. The compound beam is fixed at A, pin connected atB, and supported by a roller at C . Draw the shear andmoment diagrams for the beam.


a

Using the result of B y and referring to the free-body diagram of segment AB, Fig.b,

a


MA = 19.5 kN # m

MA - 2(3) - 4.5(3) = 0+ ©MA = 0;

Ay = 6.5 kN

Ay - 2 - 4.5 = 0c ©Fy = 0;

By = 4.5 kN

By + 4.5 - 3(3) = 0c ©Fy = 0;

Cy = 4.5 kN

Cy(3) - 3(3)(1.5) = 0+ ©MB = 0;

A BC

3 kN/ m

2 kN

3 m 3 m

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1088



x =

18L

2.4

0

x3dx

106.65 = 1.8 m

FR =1

8L2.4

0

x dx2

= 106.65 kN

*11–44. Draw the shear and moment diagrams for the beam.

2.4 m

A B

x

w

w=1 x

2

120 kN/ m

1–8

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

106.65 kN

kN # m192

kN

106.65

kN #m

- 192

1089




a

Using the result of FB and referring to the free-body diagram of segment AB, Fig.b,

a


MA = 54 kN # m

MA -1

2(3)(4.5)(3) - 7.5(4.5) = 0+ ©MA = 0;

Ay = 14.25 kN

Ay -1

2(3)(4.5) - 7.5 = 0+ c ©F y = 0;

Cy = 7.5 kN

Cy + 7.5 - 15 = 0+ c ©F y = 0;

FB = 7.5 kN

15(1.5) - FB(3) = 0+ ©MC = 0;

11–45. A short link at B is used to connect beams AB andBC to form the compound beam shown. Draw the shear andmoment diagrams for the beam if the supports at A and Bare considered fixed and pinned, respectively.

A C B

4.5 m 1.5 m 1.5 m

15 kN

3 kN/ m

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1090



Support Reactions: The intensity of the uniform distributed load caused by its ownweight is Due to symmetry,

Absolute Minimum Moment: To obtain the absolute minimum moment, themaximum positive moment must be equal to the maximum negative moment. The

maximum negative moment occurs at the supports. Referring to the free-bodydiagram of the beam segment shown in Fig. b,

a

The maximum positive moment occurs between the supports. Referring to thefree-body diagram of the beam segment shown in Fig. c,

Using this result,

a

It is required that

Solving for the positive result,

Ans.

Shear and Moment Diagrams: Using the result for b, Fig. d, the shear and momentdiagrams are shown in Figs. e and f .

b = 0.2071L = 0.207L

4b2 + 4Lb - L2 = 0

ga2L

8 (L - 4b) =

ga2b2

2

Mmax( + ) = Mmax( - )

Mmax( + ) = ga2L

8(L - 4b)

Mmax( + ) + ga2aL

2 b aL

4 b -

ga2L

2 aL

2 - bb = 0+ ©M = 0;

x = L

2

ga2L

2 - ga2 x = 0+ c ©F y = 0;

Mmax( - ) = ga2b2

2

ga2bab

2b - Mmax( - ) = 0+ ©M = 0;

Fy = ga2L

2

2Fy - ga2L = 0+ c ©F y = 0;

w = ga2.

11–46. Determine the placement b of the hooks to minimizethe largest moment when the concrete member is beinghoisted. Draw the shear and moment diagrams. Themember has a square cross section of dimension a on eachside.The specific weight of concrete is .g

L

b b

60 60

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1091



11–46. Continued

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1092



Bending Stress-Curvature Relation:

Ans.

Moment Curvature Relation:

Ans.M = 61.875 N # m = 61.9 N # m

1

0.9091 =

M

200(109) c 1

12 (1)(0.00153) d

1

r =

M

EI;

r = 0.9091 m = 909 mm

165(106) =200(109)30.75(10- 3)4

rsallow =

Ec

r ;

11–47. If the A-36 steel sheet roll is supported as shown andthe allowable bending stress is 165 MPa, determine thesmallest radius r of the spool if the steel sheet has a width of 1 m and a thickness of 1.5 mm.Also, find the correspondingmaximum internal moment developed in the sheet.

r

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1093



Applying the flexure formula

Ans.M = 13.53 # m

70 = M (262 - 84.7)

smax = Mc

I

*11–48. Determine the moment M that will produce amaximum stress of 70 MPa on the cross section.

75 mm

D

AB

12 mm

M

12 mm

75 mm

C

250 mm

12 mm12 mm

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

Section Properties:

91.73 in4

1

12 (0.5) 103 0.5(10)(5.5 3.40)2

2 1

12 (0.5)(33) 0.5(3)(3.40 2)2

1

12 (4) 0.53 4(0.5)(3.40 0.25)2

0.25(4)(0.5) 2[2(3)(0.5)] 5.5(10)(0.5)

4(0.5) 2[(3)(0.5)] 10(0.5) 3.40 in.

(6)(99)(12) + 2[49.5(75)(12)] + 137(250)(12)

99(12) + 2[(75)(12)] + 250(12) 84.7 mm

I NA 1

12 (99)(123) + 99(12)(84.7 – 6)2

+ 2 1

12 (12)(753) + (12)(75)(84.7 – 49.5)2

+1

12 (12)(2503) + (12)(250)(137 – 84.7)2

34.277(106) mm4

Maximum Bending Stress:

34.277 106×

kN

1094



! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

Section Properties:

Maximum Bending Stress: Applying the flexure formula

Ans.

Ans.(s)max 4(103)(12)(3.40)

91.73 1779.07 psi 1.78 ksi

(s)max 4(103)(12)(10.5 3.40)

91.73 3715.12 psi 3.72 ksi

smax

91.73 in4

1

12 (0.5) 103 0.5(10)(5.5 3.40)2

2 1

12 (0.5)(33) 0.5(3)(3.40 2)2

1

12

(4) 0.53 4(0.5)(3.40 0.25)2

0.25(4)(0.5) 2[2(3)(0.5)] 5.5(10)(0.5)

4(0.5) 2[(3)(0.5)] 10(0.5) 3.40 in.

Determine the maximum tensile and compressivebending stress in the beam if it is subjected to a moment of 4 kip ft.

3 in.

D

AB

0.5 in.

M

0.5 in.

3 in.

C

10 in.

0.5 in.0.5 in. 75 mm

D

AB

12 mm

M

12 mm

75 mm

C

250 mm

12 mm12 mm

M = 6 kN · m.

(6)(99)(12) + 2[49.5(75)(12)] + 137(250)(12)

99(12) + 2[(75)(12)] + 250(12) 84.7 mm

I NA 1

12

(99)(123) + 99(12)(84.7 – 6)2

+ 2 1

12 (12)(753) + (12)(75)(84.7 – 49.5)2

+1

12 (12)(2503) + (12)(250)(137 – 84.7)2

34.277(106) mm4

6(106)(262 – 84.7)

34.277(106) 31.0 MPa

6(106)(4)(84.7)

34.277(106) 14.8 MPa

11–49.

1095



Assume failure due to tensile stress:

Assume failure due to compressive stress:

Ans.M = 3.27 kN # . = 2.50 kN # (controls)

1 5 = M( .6 - . 7)

smax = Mc

I ;

M = # . = 7.33 kN #

= M( . 7)

smax = My

I ;

I =

1

36(100) ( )3

= .8(10 )mm4

y(From base ) =

1

32 2

-2 = . 7 .28 86 mm50100

1

2886154

9.6 kN m

68 28 860

m

m

m

62 2 - 250100

.8(10 )mm41 6

.8(10 )mm41 6

11–50. A member has the triangular cross section shown.Determine the largest internal moment M that can beapplied to the cross section without exceeding allowabletensile and compressive stresses of (sallow)t = 154 MPa and(sallow)c = 105 MPa, respectively.

50 mm

50 mm

100 mm

M100 mm

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1096



Ans.

Ans.(smax)c = Mc

I =

(12)( . 3)=

(smax)t =

My

I =

. 7)=

y =

1

3( .6) = . 7 .

c =

2

3( .6) = . 3 .

Ix =

1

36(100)(86.6)3

=

h = = .6 .86 mm

57 7

86 28 86 mm

86

28 86(12)(16 MPa

57 732 MPa

mm

2 2 - 250100

.8(10 )mm41 6

106

.8(10 )1 6

106

.8(10 )1 6

11–51. A member has the triangular cross section shown. If amoment of M = 1000 N · m is applied to the cross section,determine the maximum tensile and compressive bendingstresses in the member. Also, sketch a three-dimensional viewof the stress distribution action over the cross section.

50 mm

50 mm

100 mm

M100 mm

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1097



*11–52. If the beam is subjected to an internal moment ofdetermine the maximum bending stress in

the beam. The beam is made from A992 steel. Sketch thebending stress distribution on the cross section.

M = 30 kN # m,

Section Properties: The moment of inertia of the cross section about the neutralaxis is

Maximum Bending Stress: The maximum bending stress occurs at the top andbottom surfaces of the beam since they are located at the furthest distance from theneutral axis.Thus, c = 75 mm = 0.075 m.

Ans.

At

The bending stress distribution across the cross section is shown in Fig.a.

s|y = 0.06 m = My

I =

30(103)(0.06)

15.165(10- 6)= 119 MPa

y = 60 mm = 0.06 m,

smax = Mc

I =

30(103)(0.075)

15.165(10 - 6)= 148 MPa

I =1

12 (0.1)(0.153) -

1

12 (0.09)(0.123) = 15.165(10 - 6) m4

50 mm

150 mm

15 mm

10 mm

15 mm

A

50 mm

M

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1098



11–53. If the beam is subjected to an internal moment ofdetermine the resultant force caused by the

bending stress distribution acting on the top flange A.M = 30 kN # m,


Bending Stress: The distance from the neutral axis to the top and bottom surfaces of flange A is yt = 75 mm = 0.075 m and yb = 60 mm = 0.06 m.

Resultant Force: The resultant force acting on flange A is equal to the volume of thetrapezoidal stress block shown in Fig.a. Thus,

Ans.= 200 296.74 N = 200 kN

FR =1

2 (148.37 + 118.69)(106)(0.1)(0.015)

sb = Myb

I =

30(103)(0.06)

15.165(10 - 6) = 118.69 = 119 MPa

st = Myt

I =

30(103)(0.075)

15.165(10 - 6) = 148.37 = 148 MPa

I =1

12 (0.1)(0.153) -

1

12 (0.09)(0.123) = 15.165(10- 6) m4

50 mm

150 mm

15 mm

10 mm

15 mm

A

50 mm

M

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1099



11–54. If the built-up beam is subjected to an internalmoment of determine the maximum tensileand compressive stress acting in the beam.

M = 75 kN # m,

300 mm

A

M

20 mm

10 mm

10 mm

150 mm

150 mm

150 mm

Section Properties: The neutral axis passes through centroid C of the cross sectionas shown in Fig. a. The location of C is

Thus, the moment of inertia of the cross section about the neutral axis is

Maximum Bending Stress: The maximum compressive and tensile stress occurs atthe top and bottom-most fiber of the cross section.

Ans.

Ans.(smax)t = Mc

I =

75(103)(0.2035)

92.6509(10 - 6)= 165 MPa

(smax)c = My

I =

75(103)(0.3 - 0.2035)

92.6509(10 - 6)= 78.1 MPa

= 92.6509(10 - 6) m 4

+ 2 c 1

12 (0.14)(0.013) + 0.14(0.01)(0.295 - 0.2035)2 d

+ 2 c 1

12 (0.01)(0.153) + 0.01(0.15)(0.225 - 0.2035)2 d

=1

12 (0.02)(0.33) + 0.02(0.3)(0.2035 - 0.15)2

I = ©I + Ad2

y = © y~A

©A

=0.15(0.3)(0.02) + 230.225(0.15)(0.01)] + 230.295(0.01)(0.14)4

0.3(0.02) + 2(0.15)(0.01) + 2(0.01)(0.14)

= 0.2035 m

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1100



11–55. If the built-up beam is subjected to an internal moment ofdetermine the amount of this internal moment

resisted by plate A.M = 75 kN # m,

300 mm

A

M

20 mm

10 mm

10 mm

150 mm

150 mm

150 mm



= 92.6509(10 - 6) m4

+ 2 c1

12(0.14)(0.01

3

) + 0.14(0.01)(0.295 - 0.2035)

2

d

+ 2 c 1

12(0.01)(0.153) + 0.01(0.15)(0.225 - 0.2035)2 d

=1

12 (0.02)(0.33) + 0.02(0.3)(0.2035 - 0.15)2

I = I + Ad2

y = © y~A

©A =

0.15(0.3)(0.02) + 230.225(0.15)(0.01)] + 230.295(0.01)(0.14)40.3(0.02) + 2(0.15)(0.01) + 2(0.01)(0.14)

= 0.2035 m

Bending Stress: The distance from the neutral axis to the top andbottom of plate A is and

The bending stress distribution across the cross section of plate A is shown in Fig. b. The resultant forces of the tensileand compressive triangular stress blocks are

Thus, the amount of internal moment resisted by plate A is

Ans.= 50315.65 N # m = 50.3 kN # mM = 335144.46 c

2

3 (0.2035) d + 75421.50 c2

3 (0.0965) d

(FR)c =1

2(78.14)(106)(0.0965)(0.02) = 75 421.50 N

(FR)t =1

2(164.71)(106)(0.2035)(0.02) = 335 144.46 N

sb = Myb

I =

75(103)(0.2035)

92.6509(10 - 6)= 164.71 MPa (T)

st = Myt

I =

75(103)(0.0965)

92.6509(10 - 6)= 78.14 MPa (C)

yb = 0.2035 m.yt = 0.3 - 0.2035 = 0.0965 m

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1101



Section Property:

Bending Stress: Applying the flexure formula

Ans.

Ans.sB =8(103)(0.01)

17.8133(10 - 6)= 4.49 MPa (T)

sA =8(103)(0.11)

17.8133(10- 6)= 49.4 MPa (C)

s = My

I

I =1

12 (0.02)(0.223) +

1

12 (0.1)(0.023) = 17.8133(10 - 6) m4

*11–56. The aluminum strut has a cross-sectional area in theform of a cross. If it is subjected to the moment

determine the bending stress acting at points A and B, and show the results acting on volume elementslocated at these points.

M = 8 kN # m,

A

20 mm

B

20 mm

100 mm

50 mm50 mm

100 mm

M 8 kNm

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1102



Section Property:

Bending Stress: Applying the flexure formula and ,

Ans.

sy= 0.01m =8(103)(0.01)

17.8133(10 - 6)= 4.49 MPa

smax =8(103)(0.11)

17.8133(10 - 6)= 49.4 MPa

s = My

Ismax =

Mc

I

I =1

12 (0.02)(0.223) +

1

12 (0.1)(0.023) = 17.8133(10 - 6) m4

11–57. The aluminum strut has a cross-sectional area in theform of a cross. If it is subjected to the moment

determine the maximum bending stress inthe beam, and sketch a three-dimensional view of the stressdistribution acting over the entire cross-sectional area.

M = 8 kN # m,

A

20 mm

B20 mm

100 mm

50 mm50 mm

100 mm

M 8 kNm

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1103



Ans.

Ans.sC = My

I =

75(0.0175 - 0.01)

0.3633(10 - 6)= 1.55 MPa

sB = Mc

I =

75(0.0175)

0.3633(10 - 6)= 3.61 MPa

+ 2 c 1

12 (0.01)(0.043) + 0.01(0.04)(0.01252)d = 0.3633(10 - 5) m4

I =

1

12

(0.08)(0.01

3

) +

0.08(0.01)(0.0125

2

)

y =0.005(0.08)(0.01) + 230.03(0.04)(0.01)4

0.08(0.01) + 2(0.04)(0.01) = 0.0175 m

11–58. The aluminum machine part is subjected to a momentof Determine the bending stress created atpoints B and C on the cross section. Sketch the results on avolume element located at each of these points.

M = 75 kN # m.

M 75 Nm40 mm

10 mm

10 mm10 mm

20 mm

20 mm10 mm

10 mm

B A

N

C

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1104



Ans.

Ans.(smax)c = My

I =

75(0.0175)

0.3633(10 - 6)= 3.61 MPa

(smax)t = Mc

I =

75(0.050 - 0.0175)

0.3633(10 - 6)= 6.71 MPa

+ 2 c 1

12 (0.01)(0.043) + 0.01(0.04)(0.01252)d = 0.3633(10 - 5) m4

I =1

12 (0.08)(0.013) + 0.08(0.01)(0.01252)

y =0.005(0.08)(0.01) + 230.03(0.04)(0.01)4

0.08(0.01) + 2(0.04)(0.01) = 0.0175 m

11–59. The aluminum machine part is subjected to a momentof . Determine the maximum tensile andcompressive bending stresses in the part.

M = 75 kN # m

M 75 Nm40 mm

10 mm

10 mm10 mm

20 mm

20 mm10 mm

10 mm

B A

N C

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1105



Using flexure formula

Ans.

Ans.

Ans.FB =1

2( .5 + )( )( ) = . k

FA =1

2(22 + = 7 . 7 k

smax =

(250 -

= 4.9995 = 5. (Max)

sC =( -

= .1

sB = = .

sA = - )

=

s = My

I

= 4

+1

12 (75)( 3 + - 2

I =1

12 (125)( 3 + - 2 +

1

12 (25)( 3 + - 2

y = © y~A

©A =

+ +

( 5) + ( ) + ( ) = .

125(200)(25)12(25)(125) 237(75)(25)111 mm

75 25200 2525 12

200(25)(125 111)200)25 ) 125(25)(111 12)

25 ) 75(25)(237 111)

78.3(10 ) mm

111 25

20(10 )(111)

22 MPa78.3(10 )

20(10 )(

28 35 MPa

111)29 2 MPa

111)

3 5 MPa

8 6 N28.35)(25)(125)

29.1235 60 58 N25 75

MPa

6

6

6

6

78.3(10 )6

20(10 )6 25

78.3(10 )6

78.3(10 )6

20(10 )6

*11–60. The beam is subjected to a moment of 20 kN · m.

Determine the resultant force the bending stress produces

on the top flange A and bottom flange B. Also compute the

maximum bending stress developed in the beam.

75 mm

125 mm

25 mm

25 mm

200 mm

M 20 kN.m

25 mm

A

B

D

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1106



Using flexure formula

% of moment carried by web Ans.= * 100 = 22.6 %

= 40.623 kip # in. = #M = . ( ) + .5(7 )

FT =1

2 ( .1 )( 4)( ) = 5 k

FC =1

2 ( )( )( ) = . k

sB =(225 -

= .1

sA =(111 -

=

s =

My

I

=

+1

12 (75)( 3

+ -2

I =1

12 (125)( 3

+ -2

+1

12 (25)( 3

+ -2

y =© y~A

©A =

+ +

+ + = 111 mm

75(25)200(25)25(125)

237(75)(25)125(200)(25)12(25)(125)

200(25)(125 111)200 )125(25)(111 12)25 )

25 ) 75(25)(237 111)

25)22 MPa

111)29 2 MPa

23 65 N86 2522

41. N251129 2

65823 65 414.52 kN m

4.52

20

78.3(10 )6

20(10 )6

78.3(10 )6

78.3(10 )6

20(10 )6

11–61. The beam is subjected to a moment of 20 kN · m.

Determine the percentage of this moment that is resisted by

the web D of the beam.

75 mm

125 mm

25 mm

25 mm

200 mm

M 20 kN.m

25 mm

A

B

D

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1107



The moment of inertia of the cross-section about the neutral axis is

.

For point A, .

Ans.

For point B, .

Ans.

The state of stress at point A and B are represented by the volume element shownin Figs. a and b, respectively.

sB = MyB

I =

10(103)(0.125)

0.2417(10 - 3) = 5.172(106) Pa = 5.17 MPa (T)

yB = 0.125 m

sA = MyA

I =

10(103) (0.15)

0.2417(10 - 3) = 6.207(106) Pa = 6.21 MPa (C)

yA = C = 0.15 m

I =1

12 (0.2)(0.33) -

1

12 (0.16)(0.253) = 0.2417(10 - 3) m4

11–62. A box beam is constructed from four pieces of wood,glued together as shown. If the moment acting on the crosssection is , determine the stress at points A and Band show the results acting on volume elements located atthese points.

10 kN # m

20 mm 20 mm

250 mm

M 10 kNm

160 mm

25 mm

25 mm B

A

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1108



Ans.

Ans.

sC = My

I = = (T)

sB = My

I =

-= (T)

sA =My

I =

(40000)(12)( -= (C)

I =1

12 (50)(100 ) + (100)(50)(50 - 35) 2

- a 1

36 (50)(75)3

+ 1

2(50)(75)(75 - 35)2b =

4

y =-

12)(50)(75)

-12(50)(75)

=50(100)(50) 75(

35 mm100(50)

1.7(10 ) mm4

100 35)

40000(12)(35

1.53 MPa

25)0.235 MPa

40000(12)( )0.823 MPa

35

6

1.7(10 )6

1.7(10 )6

1.7(10 )6

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

11–63. The beam is subjected to a moment ofM = 40 N· m.Determine the bending stress acting at point A and B. Also,stetch a three-dimensional view of the stress distributionacting over the entire cross-sectional area. 75 mm

25 mm

25 mm

A

B

M 40 Nm

1109



Ans.smax = Mc

I =

14 p(68.75)4

=22500(68.75)

88 MPa

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

*11–64. The axle of the freight car is subjected to wheelloading of 90 kN. If it is supported by two journal bearings atC and D, determine the maximum bending stress developedat the center of the axle, where the diameter is 140 mm.

C D A B

90 kN 90 kN250 mm 250 mm

1500 mm

1110



(a)

Ans.

(b)

Ans.smax =

50(0.04)

4.021238(10-6)= 497 kPa

= 4.021238(10-6) m4

=

(0.04)4

2 sin-1

a y

0.04b ̀0.04

-0.04

= 4 c (0.04)4

8 sin-1a y

0.04b -

1

8y2 (0.04)2

- y2(0.042- 2y2) d ̀

-0.0

0.04

4

2L0.04

-0.04

y2zdy = 4L

0.04

-0.04

y22 (0.04)

2- y

2 dy

z = 2 0.0064 - 4y2= 22 (0.04)2

- y2

=

smax

c Ly22zdy

M =

s

maxc LA

y2dA

smax =

Mc

I =

50(0.04)

4.021238(10-6)= 497 kPa

I =

1

4 p ab3

=

1

4 p(0.08)(0.04)3

= 4.021238(10-6) m4

A shaft is made of a polymer having an ellipticalcross-section. If it resists an internal moment of

50 N m, determine the maximum bending stressdeveloped in the material (a) using the flexure formula,

where (b) using integration.

Sketch a three-dimensional view of the stress distributionacting over the cross-sectional area.

lz =14p(0.08 m)(0.04 m)3,

#M =

y

z x

M 50 Nm

80 mm

160 mm

y———(40)

2

2

z———(80)

2

2 1

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

11–65.

1111



(a)

Ans.

(b)

Ans.smax = 249 kPa

50 = 201.06(10-6)smax

50 = 2asmax

0.04 bL

0.08

0

z2a1 -

z2

(0.08)2b1>2

(0.04)dz

M =

LA

z(s dA) =

LA

zasmax

0.08

b(z)(2y)dz

smax =

Mc

I =

50(0.08)

16.085(10-6)= 249 kPa

I =

1

4 p ab3

=

1

4 p(0.04)(0.08)3

= 16.085(10-6) m4

Solve Prob. 6–65 if the moment isapplied about the y axis instead of the x axis. HereIy =

14 p(0.04 m)(0.08 m)3.

M = 50 kN m# y

z x

M 50 Nm

80 mm

160 mm

y———(40)

2

2

z———(80)

2

2 1

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

11–66.

1112



Absolute Maximum Bending Stress: The maximum moment isas indicated on the moment diagram.Applying the flexure formula

Ans.= 158 MPa

=11.34(103)(0.045)

p

4 (0.0454)

smax = Mmax c

I

Mmax = 11.34 kN # m

11–67. The shaft is supported by smooth journal bearings at A and B that only exert vertical reactions on the shaft. Ifd = 90 mm, determine the absolute maximum bending stressin the beam, and sketch the stress distribution acting overthe cross section. B

d

A

3 m 1.5 m

12 kN/ m

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1113



Allowable Bending Stress: The maximum moment is asindicated on the moment diagram.Applying the flexure formula

Ans.d = 0.08626 m = 86.3 mm

180A106 B =11.34(103) Ad2 B

p

4 Ad2 B4

smax = sallow = Mmax c

I

Mmax = 11.34 kN # m

*11–68. The shaft is supported by smooth journal bearings at A and B that only exert vertical reactions on the shaft.Determine its smallest diameter d if the allowable bendingstress is .sallow = 180 MPa

B

d

A

3 m 1.5 m

12 kN/ m

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1114



Section Property:

For section (a)

For section (b)

Maximum Bending Stress: Applying the flexure formula

For section (a)

For section (b)

Ans.smin =150(103)(0.18)

0.36135(10 - 3) = 74.72 MPa = 74.7 MPa

smax =150(103)(0.165)

0.21645(10 - 3) = 114.3 MPa

smax = Mc

I

I =1

12 (0.2) A0.363 B -

1

12 (0.185) A0.33 B = 0.36135(10 - 3) m4

I =1

12 (0.2) A0.333 B -

1

12 (0.17)(0.3)3 = 0.21645(10 - 3) m4

11–69. Two designs for a beam are to be considered.Determine which one will support a moment of

with the least amount of bending stress.What is that stress?M = 150 kN # m

200 mm

300 mm

(a) (b)

15 mm

30 mm

15 mm

200 mm

300 mm

30 mm

15 mm

30 mm

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1115



! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

Ans. 22.1 ksi

smax

27 000(4.6091 0.25)

5.9271

max 300(9 1.5)(12) 27 000 lb in.

0.19635(4.6091)2 5.9271 in4

14

p(0.5)4 1

4 p(0.3125)4 0.4786(6.50 4.6091)2

1

4p(0.25)4

0 (6.50)(0.4786)

0.4786 0.19635 4.6091 in.

11–70. The simply supported truss is subjected to the centraldistributed load. Neglect the effect of the diagonal lacing anddetermine the absolute maximum bending stress in the truss.The top member is a pipe having an outer diameter of 1 in.and thickness of and the bottom member is a solid rodhaving a diameter of 1

2 in.

316 in.,

6 ft

5.75 in.

6 ft 6 ft

100 lb ft

The top member is a pipe having an outer diameter of25 mm and thickness of 5 mm, and the bottom member is asolid rod having a diameter of 12 mm.

1.35 kN

2.7 m

1.35 kN

0.45

0 + (160)(314.16)

314.16 + 113.10 117.65 mm

I 1

4 (12.5)4 –

1

4 (7.5)4 + 314.16(160 – 117.65)2 +

1

4 (6)4

+ 113.10(117.65)2 = 2.1466(106) mm4

1.35(2.7) – 0.5(1.5)(0.92) = 3.0375 kN · m

3.0375(106)(117.65 + 6)

2.1466(106)

175.0 MPa

1.8 m

141.5 mm

1.8 m 1.8 m

1.5 kN/m

1116



Boat:

a

Assembly:

a

Ans.smax = Mc

I =

(12)(37.5)=

I =1

12 (43.75)(75)3 -

1

12 (37.5)(43.75) 3 = 4

Cy = 0

Cy + 0 0 - 00 = 0+ c ©F y = 0;

ND = 0 0

-ND( ) + 00( ) = 0+ ©MC = 0;

By = .

. - + By = 0+ c ©F y = 0;

NA = .

-NA(2.7) + = 0+ ©MB = 0;

Bx = 0:+ ©Fx = 0;

10000(1.5)

5555 55 N

5555 55 10000

4444 45 N

3 100 2.7

9 0 N

1009 0

100 N

1.28(10 ) mm

146.5 MPa

6

(10 )6

1.28(10 )6

5

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

11–71. The boat has a weight of 10000 N and a center ofgravity atG. If it rests on the trailer at the smooth contact A and can be considered pinned at B, determine the absolutemaximum bending stress developed in the main strut of thetrailer. Consider the strut to be a box-beam having thedimensions shown and pinned at C .

0.3 m

A

B

C

0.9 m0.3 m

1.5 m1.2 m

D

G

45 mm

75 mm 45 mm

40 mm

1117



Ans.s = Mc

I =

1

4 p(18.75)4

=

Mmax = 506.25 N-m.

506250(18.75)98 MPa

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

*11–72. Determine the absolute maximum bending stressin the 40-mm-diameter shaft which is subjected to theconcentrated forces. The sleeve bearings at A and B supportonly vertical forces.

300 mm

450 mm

B

A

1800 N

375 mm

1350 N

1118



Ans.d = .

c = 6 .

s = Mc

I; 154(10 )3 =

c

1

4 pc4

Mmax =

506250

1 mm

32 mm

506.25 N-m.

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

11–73. Determine the smallest allowable diameter of theshaft which is subjected to the concentrated forces. Thesleeve bearings at A and B support only vertical forces, andthe allowable bending stress is sallow = 154 MPa.

300 mm

450 mm

B

A

1800 N

375 mm

1350 N

1119



Ans.= 5

smax = Mc

I =

)

I = 14

p(5 )4 = 4

M = = #

w1 = > .w1( ) = 00;

w2 = > .1

2 w2( ) = 1800; 144 N mm25

3640 90 N mm

31860 N mm1800(17.7)

491 mm

531860 (

491324.4 MPa

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

11–74. The pin is used to connect the three links together.Due to wear, the load is distributed over the top and bottomof the pin as shown on the free-body diagram. If thediameter of the pin is 10 mm, determine the maximumbending stress on the cross-sectional area at the centersection a–a. For the solution it is first necessary to determinethe load intensitiesw1 and w2.

3600 N

1800 N 1800 N

25 mm

10 mm

40 mm

25 mm

a

a

w2 w2

w1

1120



Shear and Moment Diagrams:As shown in Fig. a.

Maximum Moment: Due to symmetry, the maximum moment occurs in region BC of the shaft. Referring to the free-body diagram of the segment shown in Fig.b.


Absolute Maximum Bending Stress:

Ans.smax = Mmaxc

I =

2.25 A103 B(0.04)

1.7038 A10- 6 B = 52.8 MPa

I = p

4 A0.044 - 0.0254 B = 1.7038 A10- 6 B m4

11–75. The shaft is supported by a smooth thrust bearing at A and smooth journal bearing atD. If the shaft has the crosssection shown, determine the absolute maximum bendingstress in the shaft.

AC

DB

3 kN 3 kN

0.75 m 0.75 m1.5 m

40 mm25 mm

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1121



The moment of inertia of the cross section about the neutral axis is

Thus,

Ans.

The bending stress distribution over the cross section is shown in Fig.a.

M = 195.96 (103) N # m = 196 kN # m

smax = Mc

I; 80(106) =

M(0.15)

0.36742(10 - 3)

I =1

12 (0.3)(0.33) -

1

12 (0.21)(0.263) = 0.36742(10 - 3) m4

*11–76. Determine the moment M that must be applied tothe beam in order to create a maximum stress of 80 MPa.Also sketch the stress distribution acting over the crosssection.

260 mm

20 mm30 mm

300 mm

M

30 mm

30 mm

20 mm

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1122



Section Properties: The neutral axis passes through centroid C of the cross section

as shown in Fig. a. The location ofC is


y = ©y~A

©A =

2[37.5(75)(25)] + +

2( )( ) + ( ) + ( ) = .

=

+ 1

12

(25)( 3+ (25)(100)(150 - 85)2

= 2 c 1

12(25)( 3

+ ( )( - .5) 2d + 1

12(200)( 3

+ 200(25)(87.5 + 85)

I = ©I + Ad2

Maximum Bending Stress: The maximum compressive and tensile stress occurs atthe top and bottom-most fibers of the cross section.

Ans.

Ans.

The bending stresses at y y

The bending stress distribution across the cross section is shown inFig. b.

s|y= - 0.3889 in. =My

I =

(9.7)= .1 (T)

s|y= 0.6111 in. =My

I = = .7 (C)

= -=

(smax)t =My

I =

)= .

(smax)c = Mc

I =

8 )= 13 MPa

150(100)(25)87.5(200)(25)85 mm

200 25 100 2575 25

25 ) 275 ) 25 75 85 37

100 )

23.155(10 ) mm6

200 53(10 )(12) (

23.155

59 MPa

15.3 mm and 9.7 are

5

mm

1 MPa

1 MPa

( ) ( )

(12)(8

(12)(15.3)

4

6-

(10 )6

3(10 )6

23.155(10 )6

3(10 )6

3(10 )(12)6

23.155(10 )6

23.155(10 )6

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

11–77. If the beam is subjected to an internal moment ofM = 3 kN # m, determine the maximum tensile andcompressive stress in the beam. Also, sketch the bendingstress distribution on the cross section. 100 mm

25 mm

75 mm

75 mm

75 mm

25 mm

25 mm

25 mm

A

M

1123



Section Properties: The neutral axis passes through the centroid C of the crosssection as shown in Fig. a. The location ofC is given by


y = ©y~A

©A =

2[37.5(75) 25) + +

2(75)(25) + + = 8 .

=

+1

12

(25)(

3+

(25)(100)(150 -

85)

2

+1

12 (200)(253) + 85)= 2c 1

12 (25)( 3

+ - .5)2 dI = ©I + Ad2

Allowable Bending Stress: The maximum compressive and tensile stress occurs atthe top and bottom-most fibers of the cross section.

For the bottom-most fiber,

Ans.M = (controls)

=M(85)

(sallow)t =My

I;

M =

=M(200 -

(sallow)c = Mc

I;

150(100)(25)]87.5(200)(25)5 mm

200(25) 100(25)

200(25)(87.53725(75)(8575 )

100 )

2

85)21

4.3 kN-m

14

4 kN-m

(

23.155(10 )6 mm4

23.155(10 )6

23.155(10 )6

+

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

11–78. If the allowable tensile and compressive stress forthe beam are (sallow)t = 14 MPa and (sallow)c = 21 MPa,respectively, determine the maximum allowable internalmoment M that can be applied on the cross section.

100 mm

25 mm

75 mm

75 mm

75 mm

25 mm

25 mm

25 mm

A

M

1124



Section Properties: The neutral axis passes through the centroid C of the crosssection as shown in Fig. a. The location ofC is given by


y = ©~yA

©A =

2[37.5(75)(25) + +

2(75)(25) + + =

=

+1

12(25)( 3

+ (25)(100)(150 -2

= 2 c 1

12(25)( 3

+ -2d +

1

12(200)( 3

+

I = © I + Ad 2

Bending Stress: The distance from the neutral axis to the top and bottom of board Ais yt = 200 85 = 115 mm and yb = 100 85 = .We have

Resultant Force: The resultant force acting on board A is equal to the volume of thetrapezoidal stress block shown in Fig.b. Thus,

Ans.FR =1

2(1.3 + =

sb =

Myb

I =

(15)

= .7

st =Myt

I =

(115)= 1

--

150(100)(25)]87.5(200)(25)85 mm

100(25)200(25)

75 ) 25(75)(85 37.5) 25 ) 200(25)(87.5 85)

100 ) 85)

+2

15 mm

3 MPa

1 MPa

19 kN1.7)(25)(100)

23.155(10 )6 mm4

23.155(10 )6

23.155(10 )6

(10 )(12)63

(10 )(12)63

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

11–79. If the beam is subjected to an internal moment ofM = 3 kN # m, determine the resultant force of the bendingstress distribution acting on the top vertical board A. 100 mm

25 mm

75 mm

75 mm

75 mm

25 mm

25 mm

25 mm

A

M

1125



*11–80. If the beam is subjected to an internal moment ofdetermine the bending stress developed

at points A,B, and C . Sketch the bending stress distributionon the cross section.

M = 100 kN # m,

M

300 mm

150 mm

30 mm

150 mm

C

30 mm

B

A



y = © ~yA

©A =

0.015(0.03)(0.3) + 0.18(0.3)(0.03)

0.03(0.3) + 0.3(0.03) = 0.0975 m

= 0.1907(10 - 3) m4

+ 0.03(0.3)(0.18 - 0.0975)2

I =1

12 (0.3)(0.033) + 0.3(0.03)(0.0975 - 0.015)2 +

1

12(0.03)(0.33)

Bending Stress: The distance from the neutral axis to points A, B, and C is

Ans.

Ans.

Ans.

Using these results, the bending stress distribution across the cross section is shownin Fig. b.

sC = MyC

I =

100(103)(0.0675)

0.1907(10 - 3) = 35.4 MPa (T)

sB = MyB

I =

100(103)(0.0975)

0.1907(10 - 3) = 51.1 MPa (T)

sA = MyA

I =

100(103)(0.2325)

0.1907(10 - 3) = 122 MPa (C)

0.0675 m.yC = 0.0975 - 0.03 =yB = 0.0975 m, and0.33 - 0.0975 = 0.2325 m,yA =

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1126



11–81. If the beam is made of material having an allowabletensile and compressive stress of and

, respectively, determine the maximumallowable internal moment M that can be applied to thebeam.

(sallow)c = 150 MPa(sallow)t = 125 MPa

M

300 mm

150 mm

30 mm

150 mm

C

30 mm

B

A

Section Properties: The neutral axis passes through centroid C of the cross sectionas shown in Fig. a.The location of C is


=0.015(0.03)(0.3) + 0.18(0.3)(0.03)

0.03(0.3) + 0.3(0.03) = 0.0975 m y =

© ~yA

©A

= 0.1907(10 - 3) m4

+ 0.03(0.3)(0.18 - 0.0975)2

I =1

12 (0.3)(0.033) + 0.3(0.03)(0.0975 - 0.015)2 +

1

12 (0.03)(0.33)

Allowable Bending Stress: The maximum compressive and tensile stress occurs atthe top and bottom-most fibers of the cross section. For the top-most fiber,

Ans.

For the bottom-most fiber,

M = 244 471.15 N # m = 244 kN # m

125(106) = M(0.0975)

0.1907(10 - 3)(sallow)t =

My

I

M = 123024.19 N # m = 123 kN # m (controls)

150(106) = M(0.33 - 0.0975)

0.1907(10 - 3)(sallow)c =

Mc

I

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1127



Support Reactions: Shown on the free-body diagram of the shaft, Fig.a,

Maximum Moment: The shear and moment diagrams are shown in Figs. b and c. Asindicated on the moment diagram, the maximum moment is .


Absolute Maximum Bending Moment: Here,

smax = Mmaxc

I = = 2

c =3

2

= .

I =1

4 p(37 )4 = 4

|Mmax| = 8000 N-m

1.472 mm

37 mm

8(10 )(12)(37)00 MPa

(10 )6

6

1.472 (10 )6

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

11–82. The shaft is supported by a smooth thrust bearing at A and smooth journal bearing atC . If d = 75mm, determinethe absolute maximum bending stress in the shaft. A C D

d

B

1 m 1 m

16000 N

8000 N1 m

1128



Support Reactions: Shown on the free-body diagram of the shaft, Fig.a.

Maximum Moment: As indicated on the moment diagram, Figs. b and c, themaximum moment is


Absolute Maximum Bending Moment:

Use Ans.d = mm

d =

3 =

ad2b

p

64 d4

sallow = Mc

I;

I =1

4pad

2b 4

= p

64 d4

|Mmax| = 8000 N-m.

8000000(12)

168(10 )

75 mm

75

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

11–83. The shaft is supported by a smooth thrust bearingat A and smooth journal bearing at C . If the material hasan allowable bending stress of sallow = 168 MPa, determinethe required minimum diameter d of the shaft to thenearest mm.

A C D

d

B

1 m 1 m

16000 N

8000 N1 m

1129



*11–84. If the intensity of the load w = 15 kN m, determinethe absolute maximum tensile and compressive stress in thebeam.

>

6 m

150 mm

300 mm

AB

w

Support Reactions: Shown on the free-body diagram of the beam, Fig.a.

Maximum Moment: The maximum moment occurs when . Referring to thefree-body diagram of the beam segment shown in Fig. b,

a


Absolute Maximum Bending Stress: The maximum compressive and tensile stressesoccur at the top and bottom-most fibers of the cross section.

Ans.

Ans.(smax)t = My

I =

67.5(103)(0.1)

0.1125(10 - 3) = 60 MPa (T)

(smax)c = Mc

I =

67.5(103)(0.2)

0.1125(10 - 3) = 120 MPa (C)

I =1

36 (0.15)(0.33) = 0.1125(10 - 3) m4

Mmax = 67.5 kN # mMmax + 15(3)a 3

2b - 45(3) = 0+gM = 0;

x = 3 m45 - 15x = 0+ cg Fy = 0;

V = 0

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1130



11–85. If the material of the beam has an allowable bendingstress of , determine the maximumallowable intensity w of the uniform distributed load.

sallow = 150 MPa

Support Reactions: As shown on the free-body diagram of the beam, Fig.a,

Maximum Moment: The maximum moment occurs when . Referring to thefree-body diagram of the beam segment shown in Fig.b,

a


Absolute Maximum Bending Stress: Here,

Ans.w = 18750 N>m = 18.75 kN>m

150(106) =

9

2 w(0.2)

0.1125(10 - 32 sallow = Mc

I;

c =2

3 (0.3) = 0.2 m.

I =1

36(0.15)(0.33) = 0.1125(10 - 3) m4

Mmax =9

2 wMmax + w(3)a 3

2b - 3w(3) = 0+gM = 0;

x = 3 m3w - wx = 0+ cgFy = 0;

V = 0

6 m

150 mm

300 mm

AB

w

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1131



! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

11–86. Determine the absolute maximum bending stressin the 2-in.-diameter sh aft w hich i s subjected to theconcentrated forces. The journal bearings at A and B onlysupport vertical forces.

15 in.

15 in. B

A

800 lb

30 in.

600 lb

The FBD of the shaft is shown in Fig.a.

The shear and moment diagrams are shown in Fig.b and c, respectively. Asindicated on the moment diagram, .


Here, . Thus

Ans. 19.1 ksi

19.10(103) psi15000(1)

0.25 p

smax max

1 in

p

4 (14) 0.25 p 4

max 15000 lb in

4 kN 3 kN

375 mm 375 mm 750 mm

A y = 4.5 kN B y = 2.5 kN

M (kN · m)

V (kN)

4.5

0.5 1500

–2.5

375

x (mm) x (mm)

750375 1500

1.68751.875

375 mm

375 mm B

A

4 kN

750 mm

3 kNin the 50-mm-diameter shaft which is subjected to the

1.875 kN · m.

(254) = 306796 mm4

1.875(106)(25)

306796

152.8 MPa

750

1132



! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

Determine the smallest allowable diameter of theshaft which is subjected to the concentrated forces. The

journal bearings at A and B only support vertical forces.The allowable bending stress is sallow 22 ksi.

15 in.

15 in. B

A

800 lb

30 in.

600 lb

The FBD of the shaft is shown in Fig.a

The shear and moment diagrams are shown in Fig.b and c respectively. Asindicated on the moment diagram,


Here, . Thus

Ans. 1.908 in 2 in.

sallow max

;

22(103) 15000(2)

p464

2

p

4

24

p

64 4

max 15,000 lb in

The allowable bending stress isallow = 150 MPa.

1.875 kN · m.

150 1.875(106)

d4/64

d 50.3 mm

375 mm

375 mm B

A

4 kN

750 mm

3 kN

4 kN 3 kN

375 mm 375 mm 750 mm

A y = 4.5 kN B y = 2.5 kN

M (kN · m)

V (kN)

4.5

0.5 1500

–2.5

375

x (mm) x (mm)

750375 1500

1.68751.875

750

11–87 .

1133



! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

Absolute Maximum Bending Stress: The maximum moment isas indicated on moment diagram. Applying the flexure formula

Ans.smax max

44.8(12)(4.5)1

12 (9)(9)3 4.42 ksi

max

*6–88. If the beam has a square cross section of 9 in. oneach side, determine the absolute maximum bending stressin the beam.

A

B

8 ft 8 ft

800 lb/ ft1200 lb

43.5 M (kN · m)

V (kN)

6

8 16

–76.875

–15

8 16 x (m) x (m)

A

B

2.5 m 2.5 m

15 kN/m6 kN*11–88. If the beam has a square cross section of 225 mm on

76.875(106)(112.5)112(225)(2253)

= 40.49 MPa

76 875 # mkN

1134



Allowable Bending Stress: The maximum moment is asindicated on moment diagram.Applying the flexure formula

Ans.a = 0.06694 m = 66.9 mm

150A106 B =7.50(103) Aa2 B

112 a4

smax = sallow = Mmax c

I

Mmax = 7.50 kN # m

11–89. If the compound beam in Prob. 11–42 has a squarecross section of side length a, determine the minimum valueof a if the allowable bending stress is .sallow = 150 MPa

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1135



Absolute Maximum Bending Stress: The maximum moment is as

indicated on the moment diagram.Applying the flexure formula

Ans.smax = Mmax c

I =

23w0 L2

216 Ah2 B1

12 bh3 =

23w0 L2

36bh2

Mmax =23w0 L2

216

11–90. If the beam in Prob. 11–28 has a rectangular crosssection with a width b and a height h, determine theabsolute maximum bending stress in the beam.

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1136



The FBD of the shaft is shown in Fig. a

The shear and moment diagrams are shown in Fig. b and c, respectively. As

indicated on the moment diagram, .


Here, . Thus

Ans.= 119 MPa = 119.37(106) Pa

smax = Mmax c

I =

6(103)(0.04)

0.64(10 - 6)p

c = 0.04 m

I = p

4 (0.044) = 0.64(10 - 6)p m4

Mmax = 6 kN # m

11–91. Determine the absolute maximum bending stress inthe 80-mm-diameter shaft which is subjected to theconcentrated forces. The journal bearings at A and B onlysupport vertical forces.

0.5 m 0.6 m0.4 m

20 kN

A B

12 kN

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1137



The FBD of the shaft is shown in Fig. a.

The shear and moment diagrams are shown in Fig. b and c, respectively.As indicated

on the moment diagram, .


Here, . Thus

Ans.d = 0.07413 m = 74.13 mm = 75 mm

sallow =

Mmax c

I ;

150(106

) =

6(103)(d

>2)

pd4>64

c = d>2

I = p

4 ad

2b4

= pd4

64

Mmax = 6 kN # m

*11–92. Determine, to the nearest millimeter, the smallestallowable diameter of the shaft which is subjected to theconcentrated forces. The journal bearings at A and B onlysupport vertical forces. The allowable bending stress is

.sallow = 150 MPa0.5 m 0.6 m0.4 m

20 kN

A B

12 kN

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1138



Ans.

Note that 5. OK6 sY = 0

smax = = 5.smax = Mc

I;

5184000(37.5)17 135 MPa

1.11(10 )

17 135 MPa 42 MPa

6

++

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

11–93. The wing spar ABD of a light plane is made from2014–T6 aluminum and has a cross-sectional area of800 mm , a depth of 75 mm, and a moment of inertia about2

its neutral axis of 1.11(10-6) m4. Determine the absolutemaximum bending stress in the spar if the anticipated loadingis to be as shown. Assume A,B, andC are pins. Connection ismade along the central longitudinal axis of the spar.

0.6 m DB A

C

0.9 m 1.8 m

100 N/m

1139



Ans.P = 10.4 kN

10(106) =1.5P(0.125)

1.953125(10 - 4)

smax = Mc

I

I =1

12 (0.15)(0.253) = 1.953125(10 - 4) m4

11–94. The beam has a rectangular cross section as shown.Determine the largest load P that can be supported on itsoverhanging ends so that the bending stress does notexceed .smax = 10 MPa

P P

1.5 m 1.5 m 1.5 m

150 mm

250 mm

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1140



Ans.smax = Mc

I =

18000(0.125)

1.953125(10 - 4) = 11.5 MPa

I =1

12 (0.15)(0.253) = 1.953125(10 - 4) m4

M = 1.5P = 1.5(12)(103) = 18000 N # m

11–95. The beam has the rectangular cross section shown. If P = 12 kN, determine the absolute maximum bending stressin the beam. Sketch the stress distribution acting over thecross section.

P P

1.5 m 1.5 m 1.5 m

150 mm

250 mm

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1141



Thus, from the above equations,

Ans.

Ans.

Ans.P = 4 k

h =

b = .

b = .

(600)2 - 3b2 = dP

db = 0

b(600)2 - b3 = P

bh2 =6

(1200 P)

sallow =6 M max

bh2

sallow = Mc

I =

Mmax(h2)

112(b)(h)3

Mmax = P

2 =

(600)2 = b2 + h2

1200 P2400

56

128.57

128.57

346.41

346 41 mm

490 mm

6 7 N

mm

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

*11–96. A log that is 0.6 m in diameter is to be cut into arectangular section for use as a simply supported beam. Ifthe allowable bending stress for the wood issallow = 56 MPa,determine the required width b and height h of the beamthat will support the largest load possible. What is this load?

2.4 m

0.6 m

h

b

2.4 m

P

1142



Ans.P = k

=P ( 2 )

112(200)(566)3

sallow = Mmax c

I

Mmax = P

2 (2400) = P

h = 6 .

2 = h2 + 2

600 200

56 mm

1200

5661200

56

500 N

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

11–97. A log that is 0.6 m in diameter is to be cut into arectangular section for use as a simply supported beam. Ifthe allowable bending stress for the wood is sallow = 56 MPa,determine the largest load P that can be supported if thewidth of the beam is b = 200 mm.

2.4 m

0.6 m

h

b

2.4 m

P

1143



Absolute Maximum Bending Stress: The maximum moment isas indicated on moment diagram.Applying the flexure formula

Ans.smax = Mmax c

I =

294(10 )(200)112 (200)(400 )3

= .

Mmax = 294 kN #

200 mm

400 mm

m

55 1 MPa

6

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

11–98. If the beam in Prob. 11–18 has a rectangular crosssection with a width of 200 mm and a height of 400 mm,determine the absolute maximum bending stress in the beam.

1144



! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

The maximum moment occurs at the fixed support A. Referring to the FBD shownin Fig. a,

a

The moment of inertia of the about the neutral axis is .Thus,

Ans. 5600 psi 5.60 ksi

smax

16800(12)(3)

108

1

12 (6)(63) 108 in4

max 16800 lb ft

0;

max 400(6)(3) 1

2 (400)(6)(8) 0

6–99. If the beam has a square cross section of 6 in. oneach side, determine the absolute maximum bending stressin the beam.

A

B

6 ft 6 ft

400 lb/ ft11–99. If the beam has a square cross section of 150 mm on

6(1.8)(0.9) –1

2 (6)(1.8)(2.4) = 0

22.60 kN · m

22.68(106)(75)

42.1875(106)

40.3 MPa

A

B

1.8 m 1.8 m

6 kN/m

0.9 m

6(1.8) kN 12 (6)(1.8) kN

2.4 m(150)(1503) = 42.1875(106) mm4

1145




Maximum Moment: The shear and moment diagrams are shown in Figs. b and c.As indicated on the moment diagram,


Absolute Maximum Bending Stress: Here,

Ans.smax =

M max c

I =

24(103)(0.225)

1.0477(10- 3)= 5.15 MPa

c =0.45

2 = 0.225 m.

= 1.0477(10- 3) m4

I =1

12 (0.175)(0.453) -

1

12 (0.125)(0.33)

Mmax = 24kN # m.

*11–100. If d = 450 mm, determine the absolute maximumbending stress in the overhanging beam.

4 m

8 kN/ m

2 m

12 kN

d

75 mm

25 mm

125 mm

25 mm

75 mm

A

B

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1146



11–101. If wood used for the beam has an allowablebending stress of determine the minimumdimension d of the beam’s cross-sectional area to thenearest mm.

sallow = 6 MPa,


Maximum Moment: The shear and moment diagrams are shown in Figs. b and c.As indicated on the moment diagram,


Mmax = 24kN # m.

Absolute Maximum Bending Stress: Here, c = d

2.

= 4.1667(10- 3)d3+ 4.6875(10 - 3)d2

- 0.703125(10 - 3)d + 35.15625(10 - 6)

I =1

12 (0.175)d3

-1

12 (0.125)(d - 0.15)3

Solving,

Ans.d = 0.4094 m= 410 mm

4.1667(10- 3

)d3

+ 4.6875(10- 3

)d2

- 2.703125(10- 3

)d + 35.15625(10- 6

) = 0

6(106) =

24(103)d

2

4.1667(10 - 3)d3+ 4.6875(10- 3)d2

- 0.703125(10 - 3)d + 35.15625(10- 6)

sallow = Mc

I ;

4 m

8 kN/ m

2 m

12 kN

d

75 mm

25 mm

125 mm

25 mm

75 mm

A

B

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1147



11–102. If the concentrated force P = 2 kN is applied at thefree end of the overhanging beam, determine the absolutemaximum tensile and compressive stress developed inthe beam.

= 68.0457(10- 6) m 4

+ 0.15(0.025)(0.2125 - 0.13068)2+

1

12 (0.15)(0.0253)= 2 c 1

12 (0.025)(0.23) + 0.025(0.2)(0.13068 - 0.1)2 d

I = gI + Ad 2

Absolute Maximum Bending Stress: The maximum tensile and compressive stressoccurs at the top and bottom-most fibers of the cross section.

Ans.

Ans.(smax)c =M max c

I =

2(103)(0.13068)

68.0457(10 - 6)= 3.84 MPa

(smax )t =M max y

I =

2(103)(0.225 - 0.13068)

68.0457(10- 6)= 2.77 MPa

2 m 1 m

150 mm

200 mm

25 mm

25 mm 25 mm

B

A

P


Maximum Moment: The shear and moment diagrams are shown in Figs. b and c.As indicated on the moment diagram, the maximum moment is

Section Properties: The neutral axis passes through the centroid C of the crosssection as shown in Fig. d.

The location of C is given by


y = gy~A

gA =2[0.1(0.2)(0.025)] + 0.2125(0.025)(0.15)

2(0.2)(0.025) + 0.025(0.15) = 0.13068 m

ƒMmax ƒ = 2 kN # m.

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1148



11–103. If the overhanging beam is made of wood havingthe allowable tensile and compressive stresses of

and determine themaximum concentrated force P that can applied at thefree end.

(sallow )c = 5 MPa,(sallow )t = 4 MPa


Maximum Moment: The shear and moment diagrams are shown in Figs. b and c.As indicated on the moment diagram, the maximum moment is

Section Properties: The neutral axis passes through the centroid C of the crosssection as shown in Fig.d.The location of C is given by

ƒMmax ƒ = P .


= 68.0457(10- 6) m4

+ 0.15(0.025)(0.2125 - 0.13068)2+

1

12 (0.15)(0.0253)= 2 c 1

12 (0.025)(0.23) + 0.025(0.2)(0.13068 - 0.1)2 d

I = gI + Ad2

y =

g y~AgA =

2[0.1(0.2)(0.025)] + 0.2125(0.025)(0.15)

2(0.2)(0.025) + 0.025(0.15) = 0.13068 m

2 m 1 m

150 mm

200 mm

25 mm

25 mm 25 mm

B

A

P

Absolute Maximum Bending Stress: The maximum tensile and compressive stressesoccur at the top and bottom-most fibers of the cross section. For the top fiber,

For the top fiber,

Ans.P = 2603.49 N = 2.60 kN (controls)

(sallow )c =M maxc

I 5(106) =

P (0.13068)

68.0457(10- 6)

P = 2885.79 N = 2.89 kN

4(106

) =

P(0.225 - 0.13068)

68.0457(10 - 6)(sallow )t =

M max y

I ;

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1149



*11–104. The member has a square cross section and issubjected to a resultant internal bending moment of

as shown. Determine the stress at eachcorner and sketch the stress distribution produced by M!

Set u = 45°.

M = 850 N # m

Ans.

Ans.

Ans.

Ans.

The negative sign indicates compressive stress.

sE = -601.04(0.125)

0.3255208(10 - 3)+

601.04(0.125)

0.3255208(10 - 3)= 0

sD = -601.04(0.125)

0.3255208(10 - 3)+

601.04( -0.125)

0.3255208(10 - 3)= -462kPa

sB = -601.04( -0.125)

0.3255208(10 - 3)+

601.04(0.125)

0.3255208(10 - 3)= 462kPa

sA = -601.04( -0.125)

0.3255208(10 - 3)+

601.04( -0.125)

0.3255208(10 - 3)= 0

s = -Mzy

Iz+

Myz

Iy

Iz = Iy =1

12(0.25)(0.25)3

= 0.3255208(10 - 3) m4

Mz = 850 sin 45° = 601.04 N # m

My = 850 cos 45° = 601.04 N # m

250 mm

125 mmB

A

z

y

E

M 850 NmC

125 mm

D

u

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1150



11–105. The member has a square cross section and issubjected to a resultant internal bending moment of

as shown. Determine the stress at eachcorner and sketch the stress distribution produced by M.Set u = 30°.

M = 850 N # m

Ans.

Ans.

Ans.

Ans.

The negative signs indicate compressive stress.

sE = -425(0.125)

0.3255208(10 - 3)+

736.12(0.125)

0.3255208(10 - 3)= 119 kPa

sD = -425(0.125)

0.3255208(10 - 3)+

736.12(- 0.125)

0.3255208(10 - 3)= - 446 kPa

sB = -425(-0.125)

0.3255208(10 - 3)+

736.12(0.125)

0.3255208(10 - 3)= 446 kPa

sA = -425(- 0.125)

0.3255208(10 - 3)+

736.12(- 0.125)

0.3255208(10 - 3)= - 119 kPa

s = -Mzy

Iz+

Myz

Iy

Iz = Iy =1

12 (0.25)(0.25)3

= 0.3255208(10 - 3) m4

Mz = 850 sin 30° = 425 N # m

My = 850 cos 30° = 736.12 N # m

250 mm

125 mmB

A

z

y

E

M 850 NmC

125 mm

D

u

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1151



11–106. Consider the general case of a prismatic beam

subjected to bending-moment components as

shown, when the x, y, z axes pass through the centroid of the

cross section. If the material is linear-elastic, the normal

stress in the beam is a linear function of position such

that . Using the equilibrium conditions

, ,

determine the constants a, b, and c, and show that the

normal stress can be determined from the equation

where the moments and products of inertia are defined in

Appendix A.

(IyIz - Iyz2),(MyIz + MzIyz)z]>MyIyz)y +[ -(MzIy +s =

Mz = LA

-y s dAMy = LA

z s dA0 = LA

s dA

s = a + by + cz

My and Mz,

Equilibrium Condition:

(1)

(2)

(3)

Section Properties: The integrals are defined in Appendix A. Note that

. Thus,

From Eq. (1)

From Eq. (2)

From Eq. (3)

Solving for a,b,c:

Thus, (Q.E.D.)sx = - ¢Mz Iy + My Iyz

Iy Iz - Iyz2 ≤y + ¢My Iy + MzIyz

Iy Iz - Iyz2 ≤z

b = - ¢MzIy + My Iyz

Iy Iz - Iyz2 ≤ c =

My Iz + Mz Iyz

Iy Iz - Iyz2

a = 0 (Since A Z 0)

Mz = -bIz - cIyz

My = bIyz + cIy

Aa = 0

LA y dA = LA z dA = 0

= -aLA ydA - bLA y2 dA - cLA yz dA

= LA -y(a + by + cz) dA

Mz = LA -y sx dA

= aLA z dA + bLA yz dA + cLA z2 dA

= LA z(a + by + cz) dA

My = LA z sx dA

0 = aLA dA + bLA y dA + cLA z dA

0 = LA

(a + by + cz) dA

0 = LA sx dA

sx = a + by + cz

y

y

z x

z

dA

M y

C

Mz

s

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1152



11–107. If the beam is subjected to the internal moment ofM = 2 kN m, determine the maximum bending stressdeveloped in the beam and the orientation of the neutral axis.

#

Internal Moment Components: The y and z components of M are positive since theyare directed towards the positive sense of their respective axes, Fig.a. Thus,

Section Properties: The location of the centroid of the cross-section is

The moments of inertia of the cross section about the principal centroidal y and z axes are

y =gy~ AgA =

0.025(0.05)(0.2) + 0.15(0.2)(0.05)

0.05(0.2) + 0.2(0.05) = 0.0875 m

Mz = 2 cos 60° = 1 kN # m

M y = 2 sin 60° = 1.732 kN # m

Bending Stress: By inspection, the maximum bending stress occurs at eithercorner A or B.

Ans.

Orientation of Neutral Axis: Here, .

Ans.

The orientation of the neutral axis is shown in Fig.b.

a = 79.8°

tan a =0.1135(10- 3)

35.4167(10 - 6) tan 60°

tan a =

I z

I y tan u

u = 60°

= 2.65 MPa (T)

sB = -1(103)(- 0.1625)

0.1135(10- 3)+

1.732(103)(0.025)

35.4167(10- 6)

= -5.66 MPa = 5.66 MPa (C) (Max.)

sA = -1(103)(0.0875)

0.1135(10 - 3)+

1.732(103)(-0.1)

35.4167(10- 6)

s = -Mzy

Iz

+

Myz

Iy

200 mm

50 mm

50 mm

y

z

A

B

x

y

100 mm

M

100 mm

60°

= 0.1135(10- 3) m4

+ 0.05(0.2)(0.15 - 0.0875)2+

1

12(0.05)(0.23)Iz =

1

12 (0.2)(0.053) + 0.2(0.05)(0.0875 - 0.025)2

Iy =1

12(0.05)(0.23) +

1

12 (0.2)(0.053) = 35.4167(10 - 6) m4

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1153



*11–108. If the wood used for the T-beam has an allowabletensile and compressive stress of and

respectively, determine the maximumallowable internal moment M that can be applied to thebeam.

(sallow)c = 6 MPa,(sallow)t = 4 MPa

Internal Moment Components: The y and z components of M are positive since theyare directed towards the positive sense of their respective axes, Fig.a.Thus,

Section Properties: The location of the centroid of the cross section is

The moments of inertia of the cross section about the principal centroidal y andz axes are

y = ©y~A©A

= 0.025(0.05)(0.2) + 0.15(0.2)(0.05)0.05(0.2) + 0.2(0.05)

= 0.0875 m

Mz = M cos 60° = 0.5M

My = M sin 60° = 0.8660M

= 0.1135(10- 3) m4

+ 0.05(0.2)(0.15 - 0.0875)2+

1

12(0.05)(0.23)Iz =

1

12(0.2)(0.053) + 0.2(0.05)(0.0875 - 0.025)2

Iy =1

12 (0.05)(0.23) +

1

12 (0.2)(0.053) = 35.417(10 - 6) m4

Bending Stress: By inspection, the maximum bending stress can occur at either

corner A or B. For corner A, which is in compression,

Ans.

For corner B which is in tension,

M = 3014.53 N # m = 3.01 kN # m

4(106) = - 0.5M(-

0.1625)0.1135(10- 3)

+ 0.8660M(0.025)35.417(10- 6)

sB = (sallow)t = -M z yB

I z+

M yzB

I y

M = 2119.71 N # m = 2.12 kN # m (controls)

- 6(106) = -0.5M(0.0875)

0.1135(10- 3)+

0.8660M(- 0.1)

35.417(10 - 6)

sA = (sallow)c = -M z y A

I z+

M yz A

I y

200 mm

50 mm

50 mm

y

z

A

B

x

y

100 mm

M

100 mm

60°

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1154



11–109. The box beam is subjected to the internal momentof M = 4 kN m, which is directed as shown. Determine themaximum bending stress developed in the beam and theorientation of the neutral axis.

#

Internal Moment Components: The y component of M is negative since it is directedtowards the negative sense of the y axis, whereas the z component of M which isdirected towards the positive sense of the z axis is positive, Fig.a. Thus,

Section Properties: The moments of inertia of the cross section about the principalcentroidal y and z axes are

Bending Stress: By inspection, the maximum bending stress occurs at corners A and D.

Ans.

Ans.


Ans.


a = -76.0°

tana =

0.2708(10- 3)

67.7083(10 - 6) tan (-

45°)

tan a =Iz

Iy tan u

u = -45°

= 4.70 MPa (T)

smax = sD = -2.828(103)(-0.15)

0.2708(10- 3)+

(-2.828)(103)(-0.075)

67.7083(10 - 6)

= -4.70 MPa = 4.70 MPa (C)

smax = s A = -2.828(103)(0.15)

0.2708(10- 3

)

+(- 2.828)(103)(0.075)

67.7083(10- 6

)

s = -Mzy

Iz+

Myz

Iy

Iz =1

12(0.15)(0.33) -

1

12(0.1)(0.23) = 0.2708(10- 3) m4

Iy =1

12(0.3)(0.153) -

1

12(0.2)(0.13) = 67.7083(10 - 6) m4

Mz = 4 cos 45° = 2.828 kN # m

My = - 4 sin 45° = -2.828 kN # m

z

y

x

150 mm

150 mm

25 mm50 mm

50 mm

50 mm

50 mm

45

25 mm

M

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1155



11–110. If the wood used for the box beam has anallowable bending stress of , determinethe maximum allowable internal moment M that can beapplied to the beam.

(sallow) = 6 MPa

Internal Moment Components: The y component of M is negative since it is directedtowards the negative sense of the y axis, whereas the z component of M, which isdirected towards the positive sense of the z axis, is positive, Fig.a. Thus,

Section Properties: The moments of inertia of the cross section about the principalcentroidal y and z axes are

Bending Stress: By inspection, the maximum bending stress occurs at corners A and D.Here,we will consider cornerD.

Ans.M = 5106.88 N # m = 5.11 kN # m

6(106) = -0.7071M(- 0.15)

0.2708(10 - 3)+

(- 0.7071M)(- 0.075)

67.708(10- 6)

sD = s allow = -M z yD

I z+

M yzD

I y

Iz =1

12(0.15)(0.33) -

1

12(0.1)(0.23) = 0.2708(10 - 3) m4

Iy =1

12(0.3)(0.153) -

1

12(0.2)(0.13) = 67.708(10- 6) m4

Mz = M cos 45° = 0.7071M

My = -M sin 45° = - 0.7071M

z

y

x

150 mm

150 mm

25 mm50 mm

50 mm

50 mm

50 mm

45

25 mm

M

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1156



Internal Moment Components: The y component of M is positive since it is directedtowards the positive sense of the y axis, whereas the z component of M, which isdirected towards the negative sense of the z axis, is negative, Fig.a. Thus,

Section Properties: The location of the centroid of the cross-section is given by

The moments of inertia of the cross section about the principal centroidal y and z

axes are

Bending Stress: By inspection, the maximum bending stress occurs at eithercorner A or B.

Ans.


Ans.


a = -66.5°

tan a =

5.2132 A10- 3 B

1.3078 A10- 3 B tan ( -30°)

tan a =

Iz

Iy tan u

u = -30°

= -131 MPa = 131 MPa (C)(Max.)

sB = -

c -1039.23 A103 B d(-0.3107)

5.2132 A10- 3 B +

600 A103 B( -0.15)

1.3078 A10- 3 B

= 126 MPa (T)

sA = - c-

1039.23 A10

3

B d(0.2893)

5.2132 A10- 3 B +

600 A103 B(0.15)

1.3078 A10- 3 B

s = -Mzy

Iz+

Myz

Iy

= 5.2132 A10 - 3 Bm4

- c 1

12(0.15) A0.153 B + 0.15(0.15)(0.375 - 0.2893)2 d

Iz =1

12 (0.3) A0.63 B + 0.3(0.6)(0.3 - 0.2893)2

Iy =1

12 (0.6) A0.33 B -

1

12 (0.15) A0.153 B = 1.3078 A10- 3 B m4

y =©yA

©A =

0.3(0.6)(0.3) - 0.375(0.15)(0.15)

0.6(0.3) - 0.15(0.15) = 0.2893 m

Mz = -1200 cos 30° = -1039.23 kN # m

My = 1200 sin 30° = 600 kN # m

11–111. If the beam is subjected to the internal moment of M = 1200 kN m, determine the maximum bending stressacting on the beam and the orientation of the neutral axis.

#

150 mm

150 mm

150 mm

150 mm

300 mm

150 mm

y

xz

M

30

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1157



Internal Moment Components: The y component of M is positive since it is directedtowards the positive sense of the y axis, whereas the z component of M, which isdirected towards the negative sense of the z axis, is negative, Fig.a. Thus,

Section Properties: The location of the centroid of the cross section is

The moments of inertia of the cross section about the principal centroidal y and z

axes are

Bending Stress: By inspection, the maximum bending stress can occur at eithercorner A or B. For corner A which is in tension,

Ans.

For corner B which is in compression,

M = 1376 597.12 N # m = 1377 kN # m

-150 A106 B = -(-0.8660M)(-0.3107)

5.2132 A10- 3 B +

0.5M(-0.15)

1.3078 A10 - 3 B

sB

= (sallow

)c

= -Mz yB

Iz+

My zB

Iy

M = 1185 906.82 N # m = 1186 kN # m (controls)

125 A106 B = -(-0.8660M)(0.2893)

5.2132 A10 - 3 B +

0.5M(0.15)

1.3078 A10- 3 B

sA = (sallow)t = -Mz yA

Iz+

My zA

Iy

= 5.2132 A10 - 3 Bm4

- c 1

12 (0.15) A0.153 B + 0.15(0.15)(0.375 - 0.2893)2 d

Iz =1

12 (0.3) A0.63 B + 0.3(0.6)(0.3 - 0.2893)2

Iy =1

12 (0.6) A0.33 B -

1

12 (0.15) A0.153 B = 1.3078 A10- 3 B m4

y =©yA

©A =

0.3(0.6)(0.3) - 0.375(0.15)(0.15)

0.6(0.3) - 0.15(0.15) = 0.2893 m

Mz = -M cos 30° = -0.8660M

My = M sin 30° = 0.5M

*11–112. If the beam is made from a material havingan allowable tensile and compressive stress of

and , respectively,determine the maximum allowable internal momentM thatcan be applied to the beam.

(sallow)c = 150 MPa(sallow)t = 125 MPa

150 mm

150 mm

150 mm

150 mm

300 mm

150 mm

y

xz

M

30

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1158



Ans.

Ans.

Ans.sA = Mc

I = =

When u = 0°

a = - 25.3°

tan a = tan (- 3°)tan a =

I z

I y tan u;

sA = -

998630(-

+

- -

=

s = -

M z y

I z+

Myz

Iy

Iy =

1

12

(150)(50

3

)=

Iz =

1

12

(50)(150

3

) =

;

My = - 00 sin 3° = -

Mz = 00 cos 3° = #998.63 N m10

52.33 N-m10

1.5625(10 ) mm

6

75) 25)52330(12)(6.187 MPa

14.(10 )

1000000(12)(75)5.36 MPa

46 414(10 ) mm

1.5625(10 )6

1.5625(10 )6

6

14.(10 )6

14 (10 )6

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

11–113. The board is used as a simply supported floor joist.If a bending moment of M = 1000 N · m is applied 3° fromthe z axis, determine the stress developed in the board at thecorner A. Compare this stress with that developed by the samemoment applied along the z axis (u = 0°). What is the angle a for the neutral axis when u = 3°? Comment: Normally, floorboards would be nailed to the top of the beams so that u ≈ 0°and the high stress due to misalignment would not occur.

M 1000 Nm

150 mm

y

x

z

50 mm

A

u 3

1159



Ans.

Ans.a = -63.1°

tan a = I z

I y tan u = tan ( -60°)

sB = + = -1 .

sD =-(-7.5) ( -

+(-

= - 2. 2

sA =-( -7.5) (75)

+

(150)= .3

s = -

Mzy

Iz+

Myz

Iy

Iz =1

12(300)( 3

+2

+1

12(50)( 3

+2

=

Iy =1

12 (50)( 3

+1

12 (200)( 3

=

y =

(25)(300)(50) + (150)(200)(50)

+=

Mz = - cos 60° = -

My = sin 60° = N # mm

N-mm

75 mm200(50)300(50)

50 )300 )

50 ) 300(50)(50 ) 200 ) 50(200)(75 )

21 4 MPa130.2

175) 25)9 MPa

2 7 MPa

1

15(10 )6 13(10 )6

67.5(10 )15(10 )6

6 41 10 ) mm14.58(

6 41 10 ) mm30.2(

(10 )6

(10 )6 61 10 )14.58(

610 )13(

130.2(10 )6 61 10 )14.58(

610 )13((10 )6

( 150)

61 10 )14.58(

610 )13(-(-7.5) (75)

130.2

(10 )6

(10 )6

130.2(10 )6

61 10 )14.58(

-

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

11–114. The T-beam is subjected to a bending moment ofM = 15 kN · m directed as shown. Determine the maximumbending stress in the beam and the orientation of the neutralaxis. The location y of the centroid,C , must be determined.

200 mm

50 mm

50 mm

– y

y

z

M 15 kNm

C

60

150 mm150 mm

1160



11–115. The beam has a rectangular cross section. If it issubjected to a bending moment of M = 3500 N m directedas shown, determine the maximum bending stress in thebeam and the orientation of the neutral axis.

#

Ans.

Ans.

Ans.a = -66.6°

tan a4

= I z

I y tan u =

3.375 (10- 4)

8.4375(10 - 5) tan ( -30°)

sD = 0.2084 MPa

sC = --3031.09(0.15)

0.3375(10- 3)+

1750( -0.075)

84.375(10- 6)= -0.2084 MPa

sB = --3031.09( -0.15)

0.3375(10 - 3)+

1750( -0.075)

84.375(10- 6)= -2.90 MPa (max)

sA = --3031.09(0.15)

0.3375(10- 3)+

1750(0.075)

84.375(10 - 6)= 2.90 MPa (max)

s = - Mzy

Iz+Myz

Iy

Iz =1

12(0.15)(0.33) = 0.3375(10- 3) m4

Iy =1

12(0.3)(0.153) = 84.375(10 - 6) m4

Mz = -3500 cos 30° = -3031.09 N # m

My = 3500 sin 30° = 1750 N # m

150 mm

150 mmz

y

x

M 3500 Nm

30

150 mm

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1161



*11–116. For the section, 114(10- 6) m ,4Iz¿ =Iy¿ = 31.7(10 ) m ,- 6 4

Ans.=-2461.26( -0.14835)

117(10 - 6)+

438.42( -0.034519)

29.0(10- 6)= 2.60 MPa (T)

sA =-Mzy

Iz+

Myz

I y

z = 0.14 sin 10.1° - 0.06 cos 10.1° = -0.034519 m

y = -0.06 sin 10.1° - 0.14 cos 10.1° = -0.14835 m

Mz = 2500 cos 10.1° = 2461.26 N # m

My = 2500 sin 10.1° = 438.42 N # m

Iy = 29.0(10- 6) m4Iz = 117(10- 6) m4

60 mm

60 mm

60 mm 60 mm

140 mm

80 mm

z¿

y¿

10.10

M 2500 Nm C

A

z

y

Using the techniques outlined inAppendix A, the member’s cross-sectional area has principalmoments of inertia of and Iz =Iy = 29.0(10- 6) m4

Iy¿z¿ = 15.1(10- 6) m4.

computed about the principal axes of inertia y and z, respectively. If the section is subjected to a moment of

directed as shown, determine the stressproduced at point A, using Eq.11–17.M = 2500 N # m

117(10 - 6) m4,

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1162



11–117. Solve Prob. 11–116 using the equation developed inProb. 6–106.

60 mm

60 mm

60 mm 60 mm

140 mm

80 mm

z¿

y¿

10.10

M 2500 Nm C

A

z

y

Ans.=-32500(31.7)(10- 6) + 04(-0.14) + 30 + 2500(15.1)(10- 6)4( -0.06)

31.7(10- 6)(114)(10- 6) - 3(15.1)(10- 6)42 = 2.60 MPa (T)

sA =

-(Mz¿Iy¿ + My¿Iy¿z¿)y¿ + (My¿Iz¿ + Mz¿Iy¿z¿)z¿

Iy¿Iz¿ - Iy¿z¿2

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1163



11–118. If the applied distributed loading ofbe assumed to pass through the centroid of the beam’s crosssectional area, determine the absolute maximum bendingstress in the joist and the orientation of the neutral axis.Thebeam can be considered simply supported at A and B.

w = 4 kN>m can

Internal Moment Components: The uniform distributed load w can be resolved intoits y and z components as shown in Fig. a.

w y and wz produce internal moments in the beam about the z and y axes,

respectively. For the simply supported beam subjected to the uniform distributed

load, the maximum moment in the beam is Thus,

As shown in Fig.b, and are positive since they are directed towardsthe positive sense of their respective axes.

Section Properties: The moment of inertia of the cross section about the principalcentroidal y and z axes are

Bending Stress: By inspection,the maximum bending stress occurs at points A and B.

Ans.

Ans.

Orientation of Neutral Axis: Here,

Ans.

The orientation of the neutral axis is shown in Fig.c.

a = 72.5°

tan a =29.8192(10 - 6)

2.5142(10- 6) tan 15°

tan a = I z

I y tan u

u = tan - 1 c (M y)max

(M z)max

d = tan - 1 a 4.659

17.387b = 15°.

= -150.96 MPa = 151 MPa (C)

smax = s B = -

17.387(103)(0.1)

29.8192 (10- 6)+

4.659(103)(- 0.05)

2.5142(10- 6)

= 150.96 MPa = 151 MPa (T)

smax = s A = -

17.387(103)(- 0.1)

29.8192 (10- 6)+

4.659(103)(0.05)

2.5142(10 - 6)

s = -(Mz)max y

I z+

(M y )maxz

I y

Iz =1

12 (0.1)(0.23) -

1

12 (0.09)(0.173) = 29.8192(10 - 6) m4

Iy = 2 c 1

12 (0.015)(0.13) d +

1

12 (0.17)(0.013) = 2.5142(10 - 6) m4

(My)max(Mz)max

(My)max =

wzL2

8 =

1.035(62)

8 = 4.659 kN # m

(Mz)max =

w yL2

8 =

3.864(62)

8 = 17.387 kN # m

M max = wL2

8.

wz = 4 sin 15° = 1.035 kN>m

wy = 4 cos 15° = 3.864 kN>m

A

B

6 m

(6 m)

15

15

15

15

w

w

100 mm

100 mm

100 mm

15 mm

15 mm10 mm

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1164



Internal Moment Components: The uniform distributed load w can be resolved intoits y and z components as shown in Fig. a.

w y and wz produce internal moments in the beam about the z and y axes,respectively. For the simply supported beam subjected to the a uniform distributed

load, the maximum moment in the beam is . Thus,

As shown in Fig. b, and are positive since they are directedtowards the positive sense of their respective axes.

Section Properties: The moment of inertia of the cross section about the principalcentroidal y and z axes are

Bending Stress: By inspection, the maximum bending stress occurs at points A and B.We will consider point A.

Ans.w = 4372.11 N>m = 4.37 kN>m

165(106) = -4.3467w(-0.1)

29.8192(10 - 6)+

1.1647w(0.05)

2.5142(10 - 6)

s A = sallow = -

(M z)max y A

I z +

(M y)maxz A

I y

Iz =1

12 (0.1)(0.23) -

1

12 (0.09)(0.173) = 29.8192(10- 6) m4

Iy = 2 c 1

12 (0.015)(0.13) d +

1

12 (0.17)(0.013) = 2.5142(10- 6) m4

(My)max(Mz)max

(My)max = wzL

2

8 =

0.2588w(62)

8 = 1.1647w

(Mz)max = w yL

2

8 =

0.9659w(62)

8 = 4.3476w

Mmax = wL2

8

wz = w sin 15° = 0.2588w

w y = w cos 15° = 0.9659w

11–119. Determine the maximum allowable intensityw of the uniform distributed load that can be applied to thebeam.Assume w passes through the centroid of the beam’scross sectional area and the beam is simply supported at Aand B. The beam is made of material having an allowablebending stress of .sallow = 165 MPa

A

B

6 m

(6 m)

15

15

15

15

w

w

100 mm

100 mm

100 mm

15 mm

15 mm10 mm

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1165



From Fig. 6-43,

Ans.r = 0.2(40) = 8.0 mm

r

h = 0.2

w

h = 60

40 = 1.5

K = 1.46

120(106) = K J (153)(0.02)

112(0.007)(0.04)3 K

smax = K Mc

I

60 mm40 mm 7 mm

M M

r

r

11–120. The bar is subjected to a moment ofDetermine the smallest radius r of the fillets so that anallowable bending stress of is not exceeded.sallow = 120 MPa

M = 153 N m.#

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1166



From Fig. 6–43,

Ans.smax = K Mc

I = 1.57

J

17.5(0.02)112(0.007)(0.04)3

K = 14.7 MPa

K = 1.57

r

h =

6

40 = 0.15

w

h =

60

40 = 1.5;

11–121. The bar is subjected to a moment of If r = 6 mm determine the maximum bending stress in thematerial.

M = 17.5 N m.# 60 mm40 mm 7 mm

M M

r

r

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1167



80 mm

20 mm7 mm

M M

r

r

Allowable Bending Stress:

Stress Concentration Factor: From the graph in the text

with and , then .

Ans.r = 5.00 mm

r

20 = 0.25

r

h = 0.25K = 1.45

w

h =

80

20 = 4

K = 1.45

124 A106 B = KB 40(0.01)1

12 (0.007)(0.023)R

sallow = KMc

I

*11–122. The bar is subjected to a moment ofDetermine the smallest radius r of the fillets so

that an allowable bending stress of is notexceeded.

sallow = 124 MPa40 N # m.

M = 80 mm

20 mm7 mm

M M

r

r

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1168



Stress Concentration Factor: From the graph in the text with

and , then .

Maximum Bending Stress:

Ans.= 54.4 MPa

= 1.45

B17.5(0.01)

112 (0.007)(0.023) R

smax = KMc

I

K = 1.45r

h =

5

20 = 0.25

w

h =

80

20 = 4

11–123. The bar is subjected to a moment of If determine the maximum bending stress in thematerial.r = 5 mm,

M = 17.5 N m.# 80 mm

20 mm7 mm

M M

r

r

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1169



! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

From Fig. 6-44.

Ans. 122 lb

s

; 36 1.92 20(0.625)

112 (0.5)(1.25)3

1.92

0.25

0.125 2;

0.125

1.25 0.1

1.75 1.25

2 0.25

11–1 24. The simply supported notched bar is subjected totwo forces P. Determine the largest magnitude ofP that canbe applied without causing the material to yield.The materialis A-36 steel. Each notch has a radius of 0.125 in.

20 in. 20 in.

1.75 in.

0.5 in.

P P

1.25 in.

20 in. 20 in.500 mm 500 mm

42 mm

12 mm

P P

30 mm

500 mm 500 mm

M = 0.5 P

0.5 m

500mm

500mm

1000 mm

r = 3 mm.

b 42 – 30

2 6 mm

6

3 2;

3

30 0.1

250 = 1.92 0.5P (106)(15)112(12)(30)3

P = 0.469 kN

1170



! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

From Fig. 6-44,

Ans.smax

1.92 2000(0.625)

112 (0.5)(1.25)3

29.5 ksi

1.92

0.25

0.125 2;

0.125

1.25 0.1

1.75 1.25

2 0.25

11–1 5 . The simply supported notched bar is subjected tothe two loads, each having a magnitude ofDetermine the maximum bending stress developed in thebar, and sketch the bending-stress distribution acting overthe cross section at the center of the bar. Each notch has aradius of 0.125 in.

100 lb.

20 in. 20 in.

1.75 in.

0.5 in.

P P

1.25 in.

20 in. 20 in.500 mm 500 mm

42 mm

12 mm

P P

30 mm

500 mm 500 mm

500 N

250 N · m

500 mm

500 N500 N

500mm

500mm

1000

mm

500 N500 N 500 N

the two loads, each having a magnitude of P = 500 N.

radiius of r = 3 mm.

6

3 2;

3

30 0.1

1.92 250(103)(15)1

12(12)(30)3 266.7 MPa

b 42 – 30

2 6 mm

2

1171



From Fig. 6-43,

Ans.L = 0.95 m = 950 mm

19.6875(106) =

175(0.2 + L

2)(0.03)112 (0.01)(0.063)

(sB)max = (sA)max =MB

c

I

(sA)max = KMAc

I = 1.5 c

(35)(0.02)1

12(0.01)(0.043) d = 19.6875 MPa

K = 1.5

w

h =

60

40 = 1.5

r

h =

7

40 = 0.175

*11–126. Determine the length L of the center portion of the bar so that the maximum bending stress at A,B,and C isthe same.The bar has a thickness of 10 mm.

200 mm 200 mm

7 mm40 mm60 mm

350 N

B A C

7 mm

L

2

L

2

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1172



Stress Concentration Factor:

For the smaller section with and , we have

obtained from the graph in the text.

For the larger section with and , we have

obtained from the graph in the text.

Allowable Bending Stress:

For the smaller section

Ans.

For the larger section

M = 254 N # m

200 A106 B = 1.77B M(0.015)1

12 (0.015)(0.033)R

smax = sallow = K Mc

I ;

M = 41.7 N # m (Controls ! )

200 A106 B = 1.2B M(0.005)1

12 (0.015)(0.013)R

smax = sallow = K Mc

I;

K = 1.77r

h =

3

30 = 0.1

w

h =

45

30 = 1.5

K = 1.2r

h =

6

10 = 0.6

w

h =

30

10 = 3

11–127. The stepped bar has a thickness of 15 mm.Determine the maximum moment that can be applied to itsends if it is made of a material having an allowable bendingstress of sallow = 200 MPa.

M

10 mm30 mm

45 mm

3 mm6 mm

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1173



a

M = 0.0190 wL2

+ ©M = 0; M +1

2 w

L (0.385L)2a 1

3b(0.385L) -

2wL

27 (0.385L) = 0

x = A 4

27 L = 0.385 L

+ c ©Fy = 0; 2wL

27 -

1

2 w

Lx2 = 0

11–128. The compound beam consists of two segments thatare pinned together at B. Draw the shear and momentdiagrams if it supports the distributed loading shown.

2/ 3 L

A C

B

1/ 3 L

w

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1174



! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

Problem 11-129

The beam is constructed from four pieces of wood, glued together as shown. If the internal bendingmoment is M = 120 kN·m, determine the maximum bending stress in the beam. Sketch a

three-dimensional view of the stress distribution acting over the cross section.

Given: bo 300 mm:= do 300 mm:=

bi 250 mm:= di 250 mm:=

M 120kN m⋅:=

Solution:

Section Property :

I1

12 bo do

3⋅ bi di

3⋅−⋅:=

Maximum Bending Stress: σM c⋅

I=

cmax 0.5do:=

σmax

M cmax⋅

I:=

σmax 51.51 MPa= Ans.

1175



! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

Problem 11-13

For the section, I y =31.7(10-6) m4, I z = 114(10-6) m4, I yz = 15.1(10-6) m4

in Appendix A, the member's cross-sectional area has principal moments of inertia of I y' = 29(10-6

z' -6 4

is subjected to a moment of M

(a) using Eq. 6-11 and (b) using the equation developed in Prob. 6-111.

Given: M 2000N m⋅:= 10.10deg:=

b1 80mm:= b2 140mm:=

h1 60mm:= h2 60mm:=

Iy 31.7 106− m

4:= Iz 114 10

6− m4

:=

Iyz 15.1 106− m

4:=

Iy' 29.0 106− m

4:= Iz' 117 10

6− m4

:=

Solution: ' −:=

Coordinates of Point A : yA b2:= zA h2:=

y'A

z'A

⎛

⎝

⎞

⎠

cos '

sin '

sin ' −

cos '

⎛

⎝

⎞

⎠

yA

zA

⎛

⎝

⎞

⎠⋅:=

y'A

z'A

⎛

⎝

⎞

⎠

148.35

34.52

⎛

⎝

⎞

⎠mm=

a) Using Eq. 6-11

Internal Moment Components :

My' M sin ⋅:= Mz' M cos ⋅:=

Bending Stress:

σA

Mz' y'A⋅

Iz'−

My' z'A⋅

Iy'

⎛

⎝

⎞

⎠:= σA 2.079− MPa= (C) Ans

b) Using the equation developed in Prob. 6-111

Internal Moment Components :

My 0:= Mz M:=

Bending Stress: Using formula developed in Prob. 6-111.

D Iy Iz⋅ Iyz2

−:=

σA1

DMz My

Iy

Iyz

Iyz

Iz

⎛

⎝

⎞

⎠⋅

yA−

zA

⎛

⎝

⎞

⎠⋅:=

σA 2.086 MPa= − (C) Ans.

and I = 117(10 ) m , computed about the principal axes of inertia y' and z'

. Using the techniques outline

) m

= 2 kN·m directed as shown, determine the stress produced at point A,

, respectively. If the section

0

1176



Maximum Bending Stress: The moment of inertia about y axis must be determinedfirst in order to use flexure formula

Thus,

Ans.

Maximum Bending Stress: Using integration

Ans.smax = 0.410 N>mm2= 0.410 MPa

125 A103 B =smax

5 (1.5238) A106 B

125 A103 B =smax

5 B -

3

2 y2(100 - y)

32 -

8

15 y(100 - y)

52 -

16

105 (100 - y)

72R 2 100 mm

0

M =

smax

5 L100 mm

0

y22 100 - y dy

dM = 2[y(s dA)] = 2by c asmax

100 by d(2z dy) r

smax = McI

= 125(0.1)30.4762(10-6)

= 0.410 MPa

= 30.4762 A106 B mm4= 30.4762 A10-6 B m4

= 20B -

3

2 y2

(100 - y)32 -

8

15 y (100 - y)

52 -

16

105(100 - y)

72R 2 100 mm

0

= 20L100 mm

0

y22 100 - y dy

= 2L

100 mm

0

y2 (2z) dy

I = LA y2 dA

*11–13 . A shaft is made of a polymer having a parabolicupper and lower cross section. If it resists an internal momentof , determine the maximum bending stressdeveloped in the material (a) using the flexure formula and(b) using integration. Sketch a three-dimensional view of thestress distribution acting over the cross-sectional area.Hint:Themoment of inertia is determined using Eq. A–3 of Appendix A.

M = 125 N # m y

z

x

M 125 N· m

50 mm

100 mm

50 mm

y 100 – z

2/ 25

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1

1177



! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

Problem 11-132

The beam is made from three boards nailed together as shown. If the moment acting on the crosssection is M = 650 N·m, determine the resultant force the bending stress produces on the top board.

Given: bf 290 mm:= tf 15 mm:=

tw 20 mm:= dw 125 mm:=

M 650 N m⋅:=

Solution: D dw tf :=

y yi Ai⋅ ⋅

Ai ⋅=

yc

bf tf ⋅ 0.5⋅ tf 2 dw tw⋅ 0.5dw tf ⋅

bf tf ⋅ 2dw tw⋅:=

y

c

44.933mm=

If 1

12 bf ⋅ tf

3⋅ bf tf ⋅ yc 0.5tf −

2⋅:=

Iw1

122tw ⋅ dw

3⋅ 2tw dw⋅ yc 0.5dw tf −

2⋅:=

I If Iw:= I 17990374.89mm4

=

Bending Stress: σ Mc

I⋅=

At B: cB yc tf −:= σB M

cB

I⋅:= σB 1.0815MPa=

At A: cA yc:= σA McA

I⋅:= σA 1.6235MPa=

Resultant Force: For the top board.

F 0.5 σA σB bf tf ⋅ ⋅:= F 5.883 kN= Ans.

1178



! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

Ans.

c

Ans. 2 20 166

0; 20 166 22 0

20 2

0;

20 2 0

11–13 . Draw the shear and moment diagrams for the

beam and determine the shear and moment in the beam as

functions of x, where 0 6 ft .

6 ft 4 ft

2 kip/ ft

50 kipft

8 kip

x

1. 8 m.

94 – 30 x – V = 0

V = 94 – 30 x

94 – 243.6 – 30 x x2 – M = 0

M = –15 x2 + 94 x – 243.6

243.6 kN · m 30 kN/m40 kN

75 kN · m

94 kN

1.8 m 1.2 m

M (kN · m)

V (kN)

40

–243.6

–48

–123

30 x

94 kN

1.8 m 1.2 m

30 kN/m

75 kNm

40 kN

x

3

1179



Case (a):

Case (b):

Case (a) provides higher strength since the resulting maximum stress is less for agiven M and a.

Case (a) Ans.

Ans.¢smax =8.4853 M

a3 -

6M

a3 = 2.49aM

a3 b

smax = Mc

I =

Ma 1

2 2 ab0.08333 a4

=8.4853 M

a3

I = 2 c1

36a2

2 2 ab a

1

2 2 ab

3

+1

2 a

2

2 2 ab a

1

2 2 ab c a

1

2 2 ab a

1

3bd2

d = 0.08333 a4

smax = Mc

I =

M (a>2)112(a)4

=6M

a3

*11–134. A wooden beam has a square cross section asshown. Determine which orientation of the beam providesthe greatest strength at resisting the moment M. What is thedifference in the resulting maximum stress in both cases?

a

a

a a

M M

(a) (b)

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1180



11–135. Draw the shear and moment diagrams for the shaftif it is subjected to the vertical loadings of the belt, gear, andflywheel. The bearings at A and B exert only verticalreactions on the shaft.

A B

200 mm

450 N

150 N

300 N

200 mm400 mm 300 mm

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

1181



Internal Moment Components:

Section Property:

Maximum Bending Stress: By Inspection, Maximum bending stress occurs at Aand B. Applying the flexure formula for biaxial bending at point A

Ans.

Ansu 45°

cos u - sin u = 0

ds

du=

6M

a3 (-sin u + cos u) = 0

=6M

a3 (cos u + sin u)

= --M cos u (a2)

112 a4

+-Msin u (-a

2)1

12 a4

s = -MzyIz

+ My zIy

Iy = Iz =1

12 a4

Mz = -M cos u

My = -M sin u

11–136. The strut has a square cross section a by a and issubjected to the bending moment M applied at an angleas shown. Determine the maximum bending stress in termsof a, M , and . What angle will give the largest bendingstress in the strut? Specify the orientation of the neutralaxis for this case.

uu

u

M

x

z

y

a

a

! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?

$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8

solutions manual_chapter 11_final.pdf

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