solucionario hibbeler 8va edicion "resistencia de materiales " , cap 6 y 7 ,

To the Instructor iv

1 Stress 1

2 Strain 73

3 Mechanical Properties of Materials 92

4 Axial Load 122

5 Torsion 214

6 Bending 329

7 Transverse Shear 472

8 Combined Loadings 532

9 Stress Transformation 619

10 Strain Transformation 738

11 Design of Beams and Shafts 830

12 Deflection of Beams and Shafts 883

13 Buckling of Columns 1038

14 Energy Methods 1159

CONTENTS

FM_TOC 46060 6/22/10 11:26 AM Page iii

329

6–1. Draw the shear and moment diagrams for the shaft. Thebearings at A and B exert only vertical reactions on the shaft.

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

A B

250 mm800 mm

24 kN

6–2. Draw the shear and moment diagrams for the simplysupported beam.

A B

M � 2 kN�m

4 kN

2 m 2 m 2 m

06 Solutions 46060_Part1 5/27/10 3:51 PM Page 329

330


a

©Fx = 0; Ax -

35

(4000) = 0; Ax = 2400 lb;+

+ c ©Fy = 0; -Ay +

45

(4000) - 1200 = 0; Ay = 2000 lb

FA = 4000 lb+ ©MA = 0; 45

FA(3) - 1200(8) = 0;

6–3. 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

CB

4 ft

A

The free-body diagram of the beam’s right segment sectioned through an arbitrarypoint shown in Fig. a will be used to write the shear and moment equations of the beam.

*6–4. Draw the shear and moment diagrams for the canti-lever beam.

2 kN/m

6 kN�m2 m

A

‚ (1)

a ‚(2)+ ©M = 0; -M - 2(2 - x) c12

(2 - x) d - 6 = 0 M = {-x2 + 4x - 10}kN # m

+ c ©Fy = 0; V - 2(2 - x) = 0 V = {4 - 2x} kN

The shear and moment diagrams shown in Figs. b and c are plotted using Eqs. (1)and (2), respectively.The value of the shear and moment at is evaluated usingEqs. (1) and (2).

M�x= 0 = C -0 + 4(0) - 10 D = -10kN # m

V�x = 0= 4 - 2(0) = 4 kN

x = 0

06 Solutions 46060_Part1 5/27/10 3:51 PM 30

331


6–5. Draw the shear and moment diagrams for the beam.

2 m 3 m

10 kN 8 kN

15 kN�m

6–6. Draw the shear and moment diagrams for theoverhang beam.

A

B

C

4 m 2 m

8 kN/m

6–7. Draw the shear and moment diagrams for thecompound beam which is pin connected at B.

4 ft

6 kip 8 kip

A

CB

6 ft 4 ft 4 ft


332

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 = 150a x

12b = 12.5x

*6–8. Draw the shear and moment diagrams for the simplysupported beam.


A B

150 lb/ft

12 ft

300 lb�ft

‚ (1)

a ‚(2)+ ©M = 0; M +

12

(12.5x)(x)ax

3b - 275x = 0 M = {275x - 2.083x3}lb # ft

+ c ©Fy = 0; 275 -

12

(12.5x)(x) - V = 0 V = {275 - 6.25x2}lb

The shear and moment diagrams shown in Figs. c and d are plotted using Eqs. (1)and (2), respectively. The location where the shear is equal to zero can be obtainedby setting in Eq. (1).

The value of the moment at is evaluated using Eq. (2).

M�x= 6.633 ft = 275(6.633) - 2.083(6.633)3= 1216 lb # ft

x = 6.633 ft (V = 0)

0 = 275 - 6.25x2 x = 6.633 ft

V = 0


333

6–9. Draw the shear and moment diagrams for the beam.Hint: The 20-kip load must be replaced by equivalentloadings at point C on the axis of the beam.


B

4 ft

A

4 ft 4 ft

15 kip

20 kip

C

1 ft

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 ND is

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

ND = 300 lb

+ ©MA = 0; ND(1.5) - 150(3) = 0

Ay = 150 lb

+ c ©Fy = 0; Ay - 150 = 0

6–10. 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

BC

P � 150 lb

1.5 ft1.5 ft1.5 ft


Support Reactions:

a

Shear and Moment Diagram:

©Fx = 0; -Cx +

45

(2000) = 0 Cx = 1600 lb:+

+ c ©Fy = 0; -800 +

35

(2000) - Cy = 0 Cy = 400 lb

FDE = 2000 lb

+ ©MC = 0; 800(10) -

35

FDE(4) -

45

FDE(2) = 0

6–11. The overhanging beam has been fabricated with a projected arm BD on it. Draw the shear and momentdiagrams for the beam ABC if it supports a load of 800 lb.Hint: The loading in the supporting strut DE must be replacedby equivalent loads at point B on the axis of the beam.

334


800 lbD

BA

E

C

6 ft 4 ft

5 ft

2 ft

*6–12. 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 m60 kN 60 kN35 kN 35 kN 35 kN

1.5 m

A B


335


Support Reactions:

From the FBD of segment BD

a

From the FBD of segment AB

a

+ c ©Fy = 0; P - P = 0 (equilibrium is statisfied!)

+ ©MA = 0; P(2a) - P(a) - MA = 0 MA = Pa

©Fx = 0; Bx = 0:+

+ c ©Fy = 0; Cy - P - P = 0 Cy = 2P

+ ©MC = 0; By (a) - P(a) = 0 By = P

6–13. Draw the shear and moment diagrams for thecompound beam. It is supported by a smooth plate at A whichslides within the groove and so it cannot support a verticalforce,although it can support a moment and axial load.

a

A B

a a a

P P

CD

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 ABCif 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 lb�in. and support the load of 40 lb at C.


336


For

‚ Ans.

a

‚ Ans.

For

Ans.

a

‚ Ans.

For

‚ Ans.

a

Ans. M = (500x - 3000) lb ft

+ ©MNA = 0; -M - 500(5.5 - x) - 250 = 0

+ c ©Fy = 0; V - 500 = 0 V = 500 lb

5 ft 6 x … 6 ft

M = {-580x + 2400} lb ft

+ ©MNA = 0; M + 800(x - 3) - 220x = 0

V = -580 lb

+ c ©Fy = 0; 220 - 800 - V = 0

3 ft 6 x 6 5 ft

M = (220x) lb ft

+ ©MNA = 0. M - 220x = 0

+ c ©Fy = 0. 220 - V = 0 V = 220 lb

0 6 x 6 3 ft

*6–16. Draw the shear and moment diagrams for the shaftand determine the shear and moment throughout the shaft asa function of x. The bearings at A and B exert only verticalreactions on the shaft.

x

BA

800 lb500 lb

3 ft 2 ft0.5 ft

0.5 ft


337


(1)

a (2)+ ©M = 0; M + 12

(33.33x)(x)ax

3b + 300x = 0 M = {-300x - 5.556x3} lb # ft

+ c ©Fy = 0; -300 -

12

(33.33x)(x) - V = 0 V = {-300 - 16.67x2} lb

•6–17. Draw the shear and moment diagrams for thecantilevered beam.

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 = 200ax

6b = 33.33x

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


338


Support Reactions: As shown on FBD.

Shear and Moment Function:

For :

Ans.

a

Ans.

For :

Ans.

a

Ans. M = {8.00x - 120} kip # ft

+ ©MNA = 0; -M - 8(10 - x) - 40 = 0

+ c ©Fy = 0; V - 8 = 0 V = 8.00 kip

6 ft 6 x … 10 ft

M = {-x2+ 30.0x - 216} kip # ft

+ ©MNA = 0; M + 216 + 2xax

2b - 30.0x = 0

V = {30.0 - 2x} kip

+ c ©Fy = 0; 30.0 - 2x - V = 0

0 … x 6 6 ft

6–18. 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 kip�ft

A

30 kip�ft

B

5 ft 5 ft

2 kip/ ft

5 ft



339


Since the area under the curved shear diagram can not be computed directly, thevalue of the moment at will be computed using the method of sections. Byreferring to the free-body diagram shown in Fig. b,

a Ans.+ ©M = 0; M�x= 3 m +

12

(10)(3)(1) - 20(3) = 0 M�x= 3m = 45 kN # m

x = 3 m

*6–20. Draw the shear and moment diagrams for the simplysupported beam.

10 kN

10 kN/m

3 m

A B

3 m


•6–21. The beam is subjected to the uniform distributed loadshown. Draw the shear and moment diagrams for the beam.

340


BA

C2 m

1.5 m

1 m

2 kN/m

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

a

Shear and Moment Diagram: The vertical component of FBC is

. The shear and moment diagrams are shown in Figs. c and d.= 4.5 kN

AFBC By = 7.5a35b

Ay = 1.5 kN

+ c ©Fy = 0; Ay + 7.5a35b - 2(3) = 0

FBC = 7.5 kN

+ ©MA = 0; FBCa35b(2) - 2(3)(1.5) = 0


341


(1)

a (2)

Region , Fig. c

(3)

a (4)+ ©M = 0; -M - 4(6 - x) c12

(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 +

12a4

3xb(x)ax

3b + 4x = 0 M = e -

29

x3- 4x f kN # m

+ c ©Fy = 0; -4 -

12a4

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

23

x2- 4 f kN

6–22. Draw the shear and moment diagrams for theoverhang beam.

4 kN/m

3 m 3 m

AB

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

The shear diagram shown in Fig. d is plotted using Eqs. (1) and (3). The value ofshear 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 ofthe moment at support B is evaluated using either Eq. (2) or Eq. (4).

or

M�x= 3 m = -2(6 - 3)2= -18 kN # m

M�x=3 m = -

29

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

V�x=3 m += 24 - 4(3) = 12 kN

V�x= 3 m -= -

23

(32) - 4 = -10 kN


6–23. Draw the shear and moment diagrams for the beam.It is supported by a smooth plate at A which slides within thegroove and so it cannot support a vertical force, although itcan support a moment and axial load.

342


L

AB

w

a

Substitute ;

To get absolute minimum moment,

‚ Ans.a =

L

22

L -

L2

2a= L - a

w2

(L -

L2

2a)2

=

w2

(L - a)2

Mmax (+) = Mmax (-)

Mmax (-) =

w(L - a)2

2

©M = 0; Mmax (-) - w(L - a) (L - a)

2= 0

=

w2

aL -

L2

2ab2

Mmax (+) = awL -

wL2

2ab aL -

L2

2ab -

w2

aL -

L2

2ab2

x = L -

L2

2a

+ ©M = 0; Mmax (+) + wxax

2b - awL -

wL2

2abx = 0

x = L -

L2

2a

+ c ©Fy = 0; wL -

wL2

2a- wx = 0

*6–24. Determine the placement distance a of the rollersupport so that the largest absolute value of the momentis a minimum. Draw the shear and moment diagrams forthis condition.

a

w

L

AB


343



Shear and Moment Function:

a


+ ©MNA = 0; M + mx - mL = 0 M = m(L - x)

+ c ©Fy = 0; V = 0

6–25. The beam is subjected to the uniformly distributedmoment m ( ). Draw the shear and momentdiagrams for the beam.

moment>length

LA

m

a

Substitute ,

M = 0.0345 w0L2

x = 0.7071L

+ ©MNA = 0; M +

12

aw0x

Lb(x)ax

3b -

w0L

4 ax -

L

3b = 0

x = 0.7071 L

+ c ©Fy = 0; w0L

4-

12

aw0x

Lb(x) = 0


B

w0

A2L3

L3



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

9b -

w0 L

3 aL

3b = 0

+ c ©Fy = 0; w0 L

3-

w0 L

6- V = 0 V =

w0 L

6

x = L>3

344


*6–28. Draw the shear and moment diagrams for the beam.

–3L –

3L –

3L

w0

A B


345


From FBD(a)

a

From FBD(b)

a

M = 25.31 kN # m

+ ©MNA = 0; M + 11.25(1.5) - 9.375(4.5) = 0

M = 25.67 kN # m

+ ©MNA = 0; M + (0.5556) A4.1082 B a4.1083b - 9.375(4.108) = 0

+ c ©Fy = 0; 9.375 - 0.5556x2= 0 x = 4.108 m

•6–29. Draw the shear and moment diagrams for the beam.

BA4.5 m 4.5 m

5 kN/m5 kN/m


346


Support Reactions:


a

From the FBD of segment BC

a

Shear and Moment Diagram: The maximum positive moment occurs when .

a

Mmax = 346.4 lb # ft

+ ©MNA = 0; 150(3.464) - 12.5 A3.4642 B a3.4643b - Mmax = 0

+ c ©Fy = 0; 150.0 - 12.5x2= 0 x = 3.464 ft

V = 0

©Fx = 0; Cx = 0:+

+ c ©Fy = 0; Cy - 150.0 - 225 = 0 Cy = 375.0 lb

MC = 675.0 lb # ft

+ ©MC = 0; 225(1) + 150.0(3) - MC = 0

©Fx = 0; Bx = 0:+

+ c ©Fy = 0; By - 450 + 300.0 = 0 By = 150.0 lb

+ ©MB = 0; 450(4) - Ay (6) = 0 Ay = 300.0 lb

6–30. Draw the shear and moment diagrams for the compound beam.

BA

6 ft

150 lb/ft 150 lb/ft

3 ft

C


347



Shear and Moment Functions:

For

Ans.

a

Ans.

For

Ans.

a

Ans. M = -

w0

3L (L - x)3

+ ©MNA = 0; -M -

12

c2w0

L (L - x) d(L - x)aL - x

3b = 0

V =

w0

L (L - x)2

+ c ©Fy = 0; V -

12

c2w0

L (L - x) d(L - x) = 0

L>2 6 x … L

M =

w0

24 A -12x2

+ 18Lx - 7L2)

+ ©MNA = 0; 7w0 L2

24-

3w0 L

4 x + w0 xax

2b + M = 0

V =

w0

4 (3L - 4x)

+ c ©Fy = 0; 3w0 L

4 -w0x - V = 0

0 … x 6 L>2

6–31. Draw the shear and moment diagrams for the beam anddetermine the shear and moment in the beam as functions of x.

x

BA

w0

L–2

L–2


348

Ans. w0 = 1.2 kN>m + c ©Fy = 0; 2(w0)(20)a1

2b - 60(0.4) = 0

*6–32. The smooth pin is supported by two leaves A and Band subjected to a compressive load of 0.4 kN�m caused bybar C. Determine the intensity of the distributed load w0 ofthe leaves on the pin and draw the shear and moment diagramfor the pin.


20 mm

0.4 kN/m

w0

20 mm 60 mm

w0A B

C

Ski:

Ans.

Segment:

a+ ©M = 0; M - 30(0.5) = 0; M = 15.0 lb # ft

+ c ©Fy = 0; 30 - V = 0; V = 30.0 lb

w = 40.0 lb>ft + c ©Fy = 0; 1

2 w(1.5) + 3w +

12

w(1.5) - 180 = 0

•6–33. The ski supports the 180-lb weight of the man. Ifthe snow loading on its bottom surface is trapezoidal asshown, determine the intensity w, and then draw the shearand moment diagrams for the ski.

180 lb

w w1.5 ft 3 ft 1.5 ft

3 ft


349


6–34. Draw the shear and moment diagrams for thecompound beam.

3 m 3 m1.5 m 1.5 m

5 kN3 kN/m

AB C D

6–35. 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


Shear and Moment Functions:

For :

Ans.

a

Ans.

For :

Ans.

Set ,

a

Ans.

Substitute , M = 691 N # mx = 3.87 m

M = e - 1009

x3+ 500x - 600 f N # m

+ 200(x - 3)ax - 32b - 200x = 0

+ ©MNA = 0; M +

12

c2003

(x - 3) d(x - 3)ax - 33b

x = 3.873 mV = 0

V = e - 1003

x2+ 500 f N

+ c ©Fy = 0; 200 - 200(x - 3) -

12

c2003

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

3 m 6 x … 6 m

M = (200 x) N # m

+ ©MNA = 0; M - 200 x = 0

+ c ©Fy = 0; 200 - V = 0 V = 200 N

0 … x 6 3 m


350

*6–36. Draw the shear and moment diagrams for theoverhang beam.


AB

M � 10 kN�m2 m 2 m 2 m

6 kN18 kN


B

4.5 m 4.5 m

50 kN/m

A

50 kN/m

A


351


6–38. The dead-weight loading along the centerline of theairplane wing is shown. If the wing is fixed to the fuselage atA, determine the reactions at A, and then draw the shear andmoment diagram for the wing.

Support Reactions:

Ans.

a

Ans.

Ans.


©Fx = 0; Ax = 0:+

MA = 18.583 kip # ft = 18.6 kip # ft

+ 1.25(2.5) + 0.375(1.667) + MA = 0

+ ©MA = 0; 1.00(7.667) + 3(5) - 15(3)

Ay = 9.375 kip

+ c ©Fy = 0; -1.00 - 3 + 15 - 1.25 - 0.375 - Ay = 0

3 ft

400 lb/ft250 lb/ft

3000 lb

15 000 lb

2 ft8 ft

A

6–39. 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 CB

1/3 L

w

a

M = 0.0190 wL2

+ ©M = 0; M +

12

wL

(0.385L)2a13b(0.385L) -

2wL

27 (0.385L) = 0

x =

A

427

L = 0.385 L

+ c ©Fy = 0; 2wL

27-

12

wL

x2= 0


352


A

B

2 m

0.8 kN/m

1 m2 m

1 m2 m

1 m1 m

C D

E

F

3 kN 3 kN

AB

2 m 2 m

10 kN 10 kN

15 kN�m

2 m

*6–40. Draw the shear and moment diagrams for thesimply supported beam.

6–41. Draw the shear and moment diagrams for thecompound beam. The three segments are connected bypins at B and E.


353


Support Reactions:


a

From the FBD of segment BD

a



©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

6–42. Draw the shear and moment diagrams for thecompound beam.

BA CD

2 m 1 m 1 m

5 kN/m

A C

3 ft 8 ft

3 kip/ft

5 ft

B

8 kip6–43. Draw the shear and moment diagrams for the beam.The two segments are joined together at B.


354

*6–44. Draw the shear and moment diagrams for the beam.


•6–45. Draw the shear and moment diagrams for the beam.

a

Substitute

M = 0.0394 w0L2

x = 0.630L

M =

w0Lx

12-

w0x4

12L2

+ ©M = 0; w0L

12 (x) -

w0x3

3L2 a14

xb - M = 0

x = a14b1>3

L = 0.630 L

+ c ©Fy = 0; w0L

12-

w0x3

3L2 = 0

3L

4

w0

L2L

L

0x3dx

w0 L

3

=x =

LAxdA

LAdA

=

FR =

LAdA =

L

L

0wdx =

w0

L2L

L

0x2

dx =

w0 L

3

L

A Bx

w

w0w �

w0

L2 x2

x =

181

80 x3

dx

21.33= 6.0 ft

FR =

18L

8

0x2 dx = 21.33 kip

8 ft

A Bx

w

w � x2

8 kip/ ft18


355



FR =

LAdA = w0

L

L

0sinap

L xbdx =

2w0 L

p

w0 w � w0 sin x

BA

w

x

L–2

p–L

L–2

The moment of inertia of the cross-section about z and y axes are

For the bending about z axis, .

Ans.

For the bending about y axis, .

Ans.

The bending stress distribution for bending about z and y axes are shown in Fig. aand b respectively.

smax =

Mc

Iy=

90(103) (0.1)

0.1 (10- 3)= 90 (106)Pa = 90 MPa

C = 0.1 m

smax =

Mc

Iz=

90(103) (0.075)

56.25 (10- 6)= 120(106)Pa = 120 MPa

c = 0.075 m

Iy =

112

(0.15)(0.23) = 0.1(10- 3) m4

Iz =

112

(0.2)(0.153) = 56.25(10- 6) m4

6–47. A member having the dimensions shown is used toresist an internal bending moment of Determine the maximum stress in the member if the momentis applied (a) about the z axis (as shown) (b) about the y axis.Sketch the stress distribution for each case.

M = 90 kN # m.

200 mm

150 mm

z

y

x

M


356

Section Properties:

Maximum Bending Stress: Applying the flexure formula

Ans.M = 129.2 kip # in = 10.8 kip # ft

10 =

M (10.5 - 3.4)

91.73

smax =

Mc

I

= 91.73 in4

+

112

(0.5) A103 B + 0.5(10)(5.5 - 3.40)2

+ 2 c 112

(0.5)(33) + 0.5(3)(3.40 - 2)2 d

INA =

112

(4) A0.53 B + 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.

y =

© y A

©A

*6–48. Determine the moment M that will produce amaximum stress of 10 ksi on the cross section.


3 in.

D

A B

0.5 in.

M

0.5 in.

3 in.

C

10 in.

0.5 in.0.5 in.


357

Section Properties:


Ans.

Ans.(sc)max =

4(103)(12)(3.40)

91.73= 1779.07 psi = 1.78 ksi

(st)max =

4(103)(12)(10.5 - 3.40)

91.73= 3715.12 psi = 3.72 ksi

smax =

Mc

I

= 91.73 in4

+

112

(0.5) A103 B + 0.5(10)(5.5 - 3.40)2

+ 2 c 112

(0.5)(33) + 0.5(3)(3.40 - 2)2 d

INA =

112

(4) A0.53 B + 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.

y =

© y A

©A

•6–49. Determine the maximum tensile and compressivebending stress in the beam if it is subjected to a moment ofM = 4 kip # ft.


3 in.

D

A B

0.5 in.

M

0.5 in.

3 in.

C

10 in.

0.5 in.0.5 in.


358

Ans.

Ans.sB =

30(13.24 - 10)(10- 3)

0.095883(10- 6)= 1.01 MPa

sA =

30(35 - 13.24)(10- 3)

0.095883(10- 6)= 6.81 MPa

= 0.095883(10- 6) m4

+ 2 c 112

(5)(203) + 5(20)(20 - 13.24)2 d + 2 c 112

(12)(53) + 12(5)(32.5 - 13.24)2 d

+ c 112

(34)(53) + 34(5)(13.24 - 7.5)2 d

I = c 112

(50)(53) + 50(5)(13.24 - 2.5)2 d

= 13.24 mm

y =

2.5(50)(5) + 7.5(34)(5) + 2[20(5)(20)] + 2[(32.5)(12)(5)]

50(5) + 34(5) + 2[5(20)] + 2[(12)(5)]

Ans.

Ans.M = 771 N # m

s =

Mc

I ; 175(106) =

M(35 - 13.24)(10- 3)

0.095883(10- 6)

= 0.095883(10- 6) m4

+ 2 c 112

(5)(203) + 5(20)(20 - 13.24)2 d + 2 c 112

(12)(53) + 12(5)(32.5 - 13.24)2 d

I = c 112

(50)(53) + 50(5)(13.24 - 2.5)2 d + c 112

(34)(53) + 34(5)(13.24 - 7.5)2 d

y =

©y2A

©A=

2.5(50)(5) + 7.5(34)(5) + 2[20(5)(20)] + 2[(32.5)(12)(5)]

50(5) + 34(5) + 2[5(20)] + 2[(12)(5)]= 13.24 mm

sC =

30(13.24)(10- 3)

0.095883(10- 6)= 4.14 MPa

6–50. The channel strut is used as a guide rail for a trolley.If the maximum moment in the strut is determine the bending stress at points A, B, and C.

M = 30 N # m,

6–51. The channel strut is used as a guide rail for atrolley. If the allowable bending stress for the material is

determine the maximum bending momentthe strut will resist.sallow = 175 MPa,


50 mm

30 mm

A

B

C5 mm

5 mm

5 mm

5 mm5 mm

7 mm 7 mm10 mm

50 mm

30 mm

A

B

C5 mm

5 mm

5 mm

5 mm5 mm

7 mm 7 mm10 mm


359


Section Property:

Bending Stress: Applying the flexure formula

Resultant Force and Moment: For board A or B

Ans.scaM¿

Mb = 0.8457(100%) = 84.6 %

M¿ = F(0.17619) = 4.80M(0.17619) = 0.8457 M

= 4.800 M

F = 822.857M(0.025)(0.2) +

12

(1097.143M - 822.857M)(0.025)(0.2)

sD =

M(0.075)

91.14583(10- 6)= 822.857 M

sE =

M(0.1)

91.14583(10- 6)= 1097.143 M

s =

My

I

I =

112

(0.2) A0.23 B -

112

(0.15) A0.153 B = 91.14583 A10- 6 B m4

*6–52. The beam is subjected to a moment M. Determinethe percentage of this moment that is resisted by thestresses acting on both the top and bottom boards, A andB, of the beam.

150 mm

25 mm

25 mm

150 mm

M

25 mm

25 mm

B

A

D

Section Property:


Ans.

Ans.smax =

Mc

I=

36458(0.1)

91.14583(10- 6)= 40.0 MPa

M = 36458 N # m = 36.5 kN # m

30 A106 B =

M(0.075)

91.14583(10- 6)

s =

My

I

I =

112

(0.2) A0.23 B -

112

(0.15) A0.153 B = 91.14583 A10- 6 B m4

•6–53. Determine the moment M that should be appliedto the beam in order to create a compressive stress at pointD of Also sketch the stress distributionacting over the cross section and compute the maximumstress developed in the beam.

sD = 30 MPa.

150 mm

25 mm

25 mm

150 mm

M

25 mm

25 mm

B

A

D


360

Ans.

sC =

My

I=

600 (0.05625)

34.53125 (10- 6)= 0.977 MPa

= 2.06 MPa

=

600 (0.175 - 0.05625)

34.53125 (10- 6)

smax = sB =

Mc

I

= 34.53125 (10- 6) m4

+ 2 a 112b(0.02)(0.153) + 2(0.15)(0.02)(0.043752)

I =

112

(0.24)(0.0253) + (0.24)(0.025)(0.043752)

y =

(0.0125)(0.24)(0.025) + 2 (0.1)(0.15)(0.2)

0.24 (0.025) + 2 (0.15)(0.02)= 0.05625 m

Ans.F =

12

(0.025)(0.9774 + 0.5430)(106)(0.240) = 4.56 kN

sb =

My

I=

600(0.05625 - 0.025)

34.53125(10- 6)= 0.5430 MPa

s1 =

My

I=

600(0.05625)

34.53125(10- 6)= 0.9774 MPa

= 34.53125 (10- 6) m4

+ 2 a 112b(0.02)(0.153) + 2(0.15)(0.02)(0.043752)

I =

112

(0.24)(0.0253) + (0.24)(0.025)(0.043752)

y =

(0.0125)(0.24)(0.025) + 2 (0.15)(0.1)(0.02)

0.24 (0.025) + 2 (0.15)(0.02)= 0.05625 m

6–54. The beam is made from three boards nailed togetheras shown. If the moment acting on the cross section is

determine the maximum bending stress inthe beam. Sketch a three-dimensional view of the stressdistribution acting over the cross section.

M = 600 N # m,


determine the resultant force the bendingstress produces on the top board.M = 600 N # m,


25 mm

200 mm

150 mm

20 mm

20 mm

M � 600 N�m

25 mm

200 mm

150 mm

20 mm

20 mm

M � 600 N�m


361


Section Property:


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 =

112

(0.02) A0.223 B +

112

(0.1) A0.023 B = 17.8133 A10- 6 B m4

*6–56. 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 showthe results acting on volume elements located at these points.

M = 8 kN # m, A

20 mm

B

20 mm

100 mm

50 mm50 mm

100 mm

M � 8 kN�m

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 =

112

(0.02) A0.223 B +

112

(0.1) A0.023 B = 17.8133 A10- 6 B m4

•6–57. 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 in the beam, andsketch a three-dimensional view of the stress distributionacting over the entire cross-sectional area.

M = 8 kN # m, A

20 mm

B

20 mm

100 mm

50 mm50 mm

100 mm

M � 8 kN�m


362

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 bending stressoccurs at the top and bottom edges of the cross section.

Ans.

Ans.Asmax BC =

My

I=

100(12)(8 - 4.3454)

218.87= 20.0 ksi (C)

Asmax BT =

Mc

I=

100(12)(4.3454)

218.87= 23.8 ksi (T)

= 218.87 in4

=

112

(6)a83b + 6(8) A4.3454 - 4 B2 - B14pa1.54b + pa1.52b A4.3454 - 2 B2R

I = ©I + Ad2

y =

©yA

©A=

4(8)(6) - 2 cp A1.52 B d8(6) - p A1.52 B = 4.3454 in.

6–58. If the beam is subjected to an internal moment of determine the maximum tensile andcompressive bending stress in the beam.

M = 100 kip # ft,


6 in.

3 in.

2 in.

3 in.

M

1.5 in.


363

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

Allowable Bending Stress: The maximum compressive and tensile bending stressoccurs at the top and bottom edges of the cross section. For the top edge,

For the bottom edge,

Ans. M = 1208.82 kip # ina 1 ft12 in.

b = 101 kip # ft (controls)

Asmax B t =

Mc

I ; 24 =

M(4.3454)

218.87

M = 1317.53 kip # ina 1 ft12 in.

b = 109.79 kip # ft

(sallow)c =

My

I ; 22 =

M(8 - 4.3454)

218.87

= 218.87 in4

=

112

(6) A83 B + 6(8) A4.3454 - 4 B2 - B14p A1.54 B + p A1.52 B A4.3454 - 2 B2R

I = ©I + Ad2

y =

©yA

©A=

4(8)(6) - 2 cp A1.52 B d8(6) - p A1.52 B = 4.3454 in.

6–59. If the beam is made of material having anallowable tensile and compressive stress of and respectively, determine the maximumallowable internal moment M that can be applied to the beam.

(sallow)c = 22 ksi,(sallow)t = 24 ksi


6 in.

3 in.

2 in.

3 in.

M

1.5 in.

06 Solutions 46060_Part1 5/27/10 3:51 PM 3

•6–61. The beam is constructed from four boards as shown.If it is subjected to a moment of determinethe resultant force the stress produces on the top board C.

Mz = 16 kip # ft,

364

*6–60. The beam is constructed from four boards asshown. If it is subjected to a moment of determine the stress at points A and B. Sketch a three-dimensional view of the stress distribution.

Mz = 16 kip # ft,


10 in.

10 in.

1 in.

14 in.

1 in.

1 in.

Mz � 16 kip�ft

y

z

x

1 in.

AC

B

10 in.

10 in.

1 in.

14 in.

1 in.

1 in.

Mz � 16 kip�ft

y

z

x

1 in.

AC

B

Ans.

Ans.sB =

My

I=

16(12)(9.3043)

1093.07= 1.63 ksi

sA =

Mc

I=

16(12)(21 - 9.3043)

1093.07= 2.05 ksi

+

112

(1)(103) + 1(10)(16 - 9.3043)2= 1093.07 in4

I = 2 c 112

(1)(103) + 1(10)(9.3043 - 5)2 d +

112

(16)(13) + 16(1)(10.5 - 9.3043)2

= 9.3043 in.

y =

2[5(10)(1)] + 10.5(16)(1) + 16(10)(1)

2(10)(1) + 16(1) + 10(1)

Ans.(FR)C =

12

(2.0544 + 0.2978)(10)(1) = 11.8 kip

sD =

My

I=

16(12)(11 - 9.3043)

1093.07= 0.2978 ksi

sA =

Mc

I=

16(12)(21 - 9.3043)

1093.07= 2.0544 ksi

+

112

(1)(103) + 1(10)(16 - 9.3043)2= 1093.07 in4

I = 2 c 112

(1)(103) + (10)(9.3043 - 5)2 d +

112

(16)(13) + 16(1)(10.5 - 9.3043)2

y =

2[5(10)(1)] + 10.5(16)(1) + 16(10)(1)

2(10)(1) + 16(1) + 10(1)= 9.3043 in.


365


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 =

112

(0.2)(0.33) -

112

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

6–62. A box beam is constructed from four pieces ofwood, glued together as shown. If the moment acting on thecross section is 10 kN m, determine the stress at points Aand B and show the results acting on volume elementslocated at these points.

#

20 mm 20 mm

250 mm

M � 10 kN�m

160 mm

25 mm

25 mm B

A


366

Section Properties: The moments of inertia of the square and circular cross sectionsabout the neutral axis are

Maximum Bending Stress: For the square cross section, .

For the circular cross section, .

It is required that

Ans.a = 1.677r

6M

a3 =

4M

pr3

Asmax BS = Asmax BC

Asmax B c =

Mc

Ic=

Mr

14

pr4-

4M

pr3

c = r

Asmax BS =

Mc

IS=

M(a>2)

a4>12=

6M

a3

c = a>2IS =

112

a Aa3 B =

a4

12 IC =

14

pr4

6–63. Determine the dimension a of a beam having asquare cross section in terms of the radius r of a beam witha circular cross section if both beams are subjected to thesame internal moment which results in the same maximumbending stress.


a

ra

Ans.

Ans.sB =

My

I=

300(12)(0.5 sin 45°)

0.0490874= 25.9 ksi

sA =

Mc

I=

300(12)(0.5)

0.0490874= 36.7 ksi

I =

p

4 r4

=

p

4 (0.54) = 0.0490874 in4

*6–64. The steel rod having a diameter of 1 in. is subjected toan internal moment of Determine the stresscreated at points A and B. Also, sketch a three-dimensionalview of the stress distribution acting over the cross section.

M = 300 lb # ft .

M � 300 lb�ft

A

BB

45�

0.5 in.


367



Along the top edge of the flange . Thus

Ans.

Along the bottom edge to the flange, . Thus

s =

My

I=

4(103)(12)(6)

1863= 155 psi

y = 6 in

smax =

Mc

I=

4(103)(12)(7.5)

1863= 193 psi

y = c = 7.5 in

I =

112

(12)(153) -

112

(10.5)(123) = 1863 in4

•6–65. If the moment acting on the cross section of the beamis determine the maximum bending stress inthe beam. Sketch a three-dimensional view of the stressdistribution acting over the cross section.

M = 4 kip # ft,

12 in.

12 in.

1.5 in.

1.5 in.

1.5 in.M

A


Along the top edge of the flange . Thus

Along the bottom edge of the flange, . Thus

The resultant force acting on board A is equal to the volume of the trapezoidalstress block shown in Fig. a.

Ans. = 3.13 kip

= 3130.43 lb

FR =

12

(193.24 + 154.59)(1.5)(12)

s =

My

I=

4(103)(12)(6)

1863= 154.59 psi

y = 6 in

smax =

Mc

I=

4(103)(12)(7.5)

1863= 193.24 psi

y = c = 7.5 in

I =

112

(12)(153) -

112

(10.5)(123) = 1863 in4

6–66. If determine the resultant force thebending stress produces on the top board A of the beam.

M = 4 kip # ft,

12 in.

12 in.

1.5 in.

1.5 in.

1.5 in.M

A


368


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

Ans. = 158 MPa

=

11.34(103)(0.045)p4 (0.0454)

smax =

Mmax c

I

Mmax = 11.34 kN # m

6–67. The rod is supported by smooth journal bearings at A and B that only exert vertical reactions on the shaft. If

determine the absolute maximum bendingstress in the beam, and sketch the stress distribution actingover the cross section.

d = 90 mm,

B

d

A

3 m 1.5 m

12 kN/m


369


Section Property:

For section (a)

For section (b)


For section (a)

For section (b)

Ans.smax =

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 =

112

(0.2) A0.363 B -

112

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

I =

112

(0.2) A0.333 B -

112

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

•6–69. 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 isthat stress? 150 kN # m

M =200 mm

300 mm

(a) (b)

15 mm

30 mm

15 mm

200 mm

300 mm

30 mm

15 mm

30 mm

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

Ans. d = 0.08626 m = 86.3 mm

180 A106 B =

11.34(103) Ad2 Bp4 Ad2 B4

smax = sallow =

Mmax c

I

Mmax = 11.34 kN # m

*6–68. The rod is supported by smooth journal bearings atA 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


370


Ans. = 22.1 ksi

smax =

Mc

I=

27 000(4.6091 + 0.25)

5.9271

Mmax = 300(9 - 1.5)(12) = 27 000 lb # in.

+ 0.19635(4.6091)2= 5.9271 in4

I = c 14

p(0.5)4-

14

p(0.3125)4 d + 0.4786(6.50 - 4.6091)2+

14p(0.25)4

y =

© y A

©A=

0 + (6.50)(0.4786)

0.4786 + 0.19635= 4.6091 in.

6–70. 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 12 in.

316 in.,

6 ft

5.75 in.

6 ft 6 ft

100 lb/ft

Ans.smax =

Mc

I=

200(2.75)14 p(2.75)4

= 12.2 ksi

6–71. The axle of the freight car is subjected to wheelloadings of 20 kip. 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 5.5 in.

C DA B

20 kip 20 kip

10 in. 10 in.60 in.


371



Internal Moment: The maximum moment occurs at mid span. The maximummoment is determined using the method of sections.

Section Property:

Absolute Maximum Bending Stress: The maximum moment is as indicated on the FBD. Applying the flexure formula

Ans. = 10.0 ksi

=

24.0(12)(5.30)

152.344

smax =

Mmax c

I

Mmax = 24.0 kip # ft

I =

112

(8) A10.63 B -

112

(7.7) A103 B = 152.344 in4

*6–72. The steel beam has the cross-sectional area shown.Determine the largest intensity of distributed load that itcan support so that the maximum bending stress in the beamdoes not exceed smax = 22 ksi.

w0

•6–73. The steel beam has the cross-sectional area shown. Ifdetermine the maximum bending stress in

the beam.w0 = 0.5 kip>ft,

10 in.

8 in.

0.30 in.

12 ft 12 ft

0.30 in.

0.3 in.

w0

10 in.

8 in.

0.30 in.

12 ft 12 ft

0.30 in.

0.3 in.

w0


Internal Moment: The maximum moment occurs at mid span. The maximummoment is determined using the method of sections.

Section Property:

Absolute Maximum Bending Stress: The maximum moment is asindicated on the FBD. Applying the flexure formula

Ans. w0 = 1.10 kip>ft 22 =

48.0w0 (12)(5.30)

152.344

smax =

Mmax c

I

Mmax = 48.0w0

I =

112

(8) A10.63 B -

112

(7.7) A103 B = 152.344 in4


372

Boat:

a

Assembly:

a

Ans.smax =

Mc

I=

3833.3(12)(1.5)

3.2676= 21.1 ksi

I =

112

(1.75)(3)3-

112

(1.5)(1.75)3= 3.2676 in4

Cy = 230 lb

+ c ©Fy = 0; Cy + 2070 - 2300 = 0

ND = 2070 lb

+ ©MC = 0; -ND(10) + 2300(9) = 0

By = 1022.22 lb

+ c ©Fy = 0; 1277.78 - 2300 + By = 0

NA = 1277.78 lb

+ ©MB = 0; -NA(9) + 2300(5) = 0

:+ ©Fx = 0; Bx = 0

6–74. The boat has a weight of 2300 lb and a center ofgravity at G. If it rests on the trailer at the smooth contact Aand can be considered pinned at B, determine the absolutemaximum bending stress developed in the main strut ofthe trailer. Consider the strut to be a box-beam having thedimensions shown and pinned at C.


1 ft

3 ftD

A

BC

1 ft5 ft 4 ft

G

1.75 in.

3 in. 1.75 in.

1.5 in.


373

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

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

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

Absolute Maximum Bending Stress:

Ans.sallow =

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 Bm4

6–75. The shaft is supported by a smooth thrust bearing atA and smooth journal bearing at D. If the shaft has the crosssection shown, determine the absolute maximum bendingstress in the shaft.


A C DB

3 kN 3 kN

0.75 m 0.75 m1.5 m

40 mm 25 mm


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 =

112

(0.3)(0.33) -

112

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

*6–76. Determine the moment M that must be applied tothe beam in order to create a maximum stress of 80 MPa.Alsosketch the stress distribution acting over the cross section.

260 mm

20 mm30 mm

300 mm

M

30 mm

30 mm

20 mm


374


Ans. w = 1.65 kip>ft 22 =

32w(12)(5.3)

152.344

smax =

Mc

I

I =

112

(8)(10.6)3-

112

(7.7)(103) = 152.344 in4

•6–77. The steel beam has the cross-sectional area shown.Determine the largest intensity of distributed load w that itcan support so that the bending stress does not exceedsmax = 22 ksi.

10 in.

8 in.

0.30 in.

8 ft 8 ft 8 ft

0.30 in.

0.3 in.

w w

From Prob. 6-78:

Ans.smax =

Mc

I=

1920(5.3)

152.344= 66.8 ksi

I = 152.344 in4

M = 32w = 32(5)(12) = 1920 kip # in.

6–78. The steel beam has the cross-sectional area shown.If determine the absolute maximum bendingstress in the beam.

w = 5 kip>ft,

10 in.

8 in.

0.30 in.

8 ft 8 ft 8 ft

0.30 in.

0.3 in.

w w


375


Ans.smax =

Mc

I=

46.7(103)(12)(3)112 (6)(63)

= 15.6 ksi


6–79. If the beam ACB in Prob. 6–9 has a square crosssection, 6 in. by 6 in., determine the absolute maximumbending stress in the beam.

B

4 ft

A

4 ft 4 ft

15 kip

20 kip

C

1 ft

a

Ans.

Ans.Use h = 2.75 in.

h = 2.68 in.

smax =

Mc

I=

6000(12) Ah2 B112(2.5)(h3)

= 24(10)3

;+ ©Fx = 0; Ax -

35

(4000) = 0; Ax = 2400 lb

+ c ©Fy = 0; -Ay +

45

(4000) - 1200 = 0; Ay = 2000 lb

+ ©MA = 0; 45

FB(3) - 1200(8) = 0; FB = 4000 lb

*6–80. If the crane boom ABC in Prob. 6–3 has arectangular cross section with a base of 2.5 in., determine itsrequired height h to the nearest if the allowable bendingstress is sallow = 24 ksi.

14 in.

5 ft3 ft

CB

4 ft

A


376


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

Maximum Moment: The shear and moment diagrams are shown in Figs. b and c. As

indicated on the moment diagram, the maximum moment is .


Ans.smax =

Mmaxc

I=

7.5(12)(3)

112

(12)(63)= 1.25 ksi

� Mmax � = 7.5 kip # ft

w = 3.75 kip>ft + c ©Fy = 0; w(8) - 2(15) = 0

•6–81. If the reaction of the ballast on the railway tie canbe assumed uniformly distributed over its length asshown, determine the maximum bending stress developedin the tie. The tie has the rectangular cross section withthickness t = 6 in.

5 ft1.5 ft 1.5 ft

15 kip 15 kip

12 in.

t

w


377


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

Maximum Moment: The shear and moment diagrams are shown in Figs. b and c. As

indicated on the moment diagram, the maximum moment is .


Ans.Use t = 5 12

in.

t = 5.48 in.

smax =

Mc

I ; 1.5 =

7.5(12)a t

2b

112

(12)t3

� Mmax � = 7.5 kip # ft

w = 3.75 kip>ft + c ©Fy = 0; w(8) - 2(15) = 0

6–82. The reaction of the ballast on the railway tie can beassumed uniformly distributed over its length as shown.If the wood has an allowable bending stress of 1.5 ksi, determine the required minimum thickness t of therectangular cross sectional area of the tie to the nearest in.1

8

sallow = 5 ft1.5 ft 1.5 ft

15 kip 15 kip

12 in.

t

w


378

Section Property:

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

Ans. = 129 MPa

=

60.0(103)(0.1)

46.370(10- 6)

smax =

Mmaxc

I

Mmax = 60.0 kN # m

I =

p

4 A0.14

- 0.084 B = 46.370 A10- 6 B m4

6–83. Determine the absolute maximum bending stressin the tubular shaft if and do = 200 mm.di = 160 mm


A B

di do

3 m 1 m

15 kN/m

60 kN � m


379


Section Property:


Ans.

Thus, Ans.dl = 0.8do = 151 mm

do = 0.1883 m = 188 mm

155 A106 B =

60.0(103) Ado2 B

0.009225pdo4

smax = sallow =

Mmax c

I

Mmax = 60.0 kN # m

I =

p

4 B ado

2b4

- adl

2b4R =

p

4Bdo

4

16- a0.8do

2b4R = 0.009225pdo

4

*6–84. The tubular shaft is to have a cross section suchthat its inner diameter and outer diameter are related by

Determine these required dimensions if theallowable bending stress is sallow = 155 MPa.di = 0.8do.

A B

di do

3 m 1 m

15 kN/m

60 kN � m


380



Ans. b = 0.05313 m = 53.1 mm

10 A106 B =

562.5(0.75b)112 (b)(1.5b)3

smax = sallow =

Mmax c

I

Mmax = 562.5 N # m

6–85. The wood beam has a rectangular cross section inthe proportion shown. Determine its required dimension bif the allowable bending stress is sallow = 10 MPa.

500 N/m

2 m 2 m

1.5b

bA B


381


6–86. Determine the absolute maximum bending stressin the 2-in.-diameter shaft which is 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) psi

=

15000(1)

0.25 p

smax =

Mmax c

I

c = 1 in

I =

p

4 (14) = 0.25 p in4

Mmax = 15000 lb # in


382


6–87. Determine the smallest allowable diameter of theshaft which is subjected to the concentrated forces. Thejournal 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. d = 1.908 in = 2 in.

sallow =

Mmax c

I; 22(103) =

15000(d>2)pd4>64

c = d>2I =

p

4 ad

2b4

=

p

64 d4

Mmax = 15,000 lb # in


383


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

Ans.smax =

Mmax c

I=

44.8(12)(4.5)112 (9)(9)3

= 4.42 ksi


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

AB

8 ft 8 ft

800 lb/ft1200 lb

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

Ans. a = 0.06694 m = 66.9 mm

150 A106 B =

7.50(103) Aa2 B112 a4

smax = sallow =

Mmax c

I

Mmax = 7.50 kN # m

•6–89. If the compound beam in Prob. 6–42 has a squarecross section, determine its dimension a if the allowablebending stress is sallow = 150 MPa.


384

6–90. If the beam in Prob. 6–28 has a rectangular cross sectionwith a width b and a height h, determine the absolute maximumbending stress in the beam.


Absolute Maximum Bending Stress: The maximum moments is

as indicated on the moment diagram. Applying the flexure formula

Ans.smax =

Mmax c

I=

23w0 L2

216 Ah2 B112 bh3

=

23w0 L2

36bh2

Mmax =

23w0 L2

216


385

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

6–91. Determine the absolute maximum bending stressin the 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


386


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>2I =

p

4 ad

2b4

=

pd4

64

� Mmax � = 6 kN # m

*6–92. Determine the smallest allowable diameter of theshaft which is subjected to the concentrated forces. Thejournal bearings at A and B only support vertical forces.The allowable bending stress is sallow = 150 MPa.


0.5 m 0.6 m0.4 m

20 kN

A B

12 kN


387

Internal Moment: The maximum moment occurs at support B. The maximummoment is determined using the method of sections.

Section Property:

Absolute Maximum Bending Stress: The maximum moment is as indicated on the FBD.Applying the flexure formula

Absolute Maximum Normal Strain: Applying Hooke’s law, we have

Ans.emax =

smax

E=

88.92(106)

125(109)= 0.711 A10- 3 B mm>mm

= 88.92 MPa

=

1912.95(0.05 - 0.012848)

0.79925(10- 6)

smax =

Mmax c

I

Mmax = 1912.95 N # m

= 0.79925 A10- 6 B m4

+

112

(0.03) A0.033 B + 0.03(0.03)(0.035 - 0.012848)2

I =

112

(0.35) A0.023 B + 0.35(0.02)(0.012848 - 0.01)2

=

0.01(0.35)(0.02) + 0.035(0.03)(0.03)

0.35(0.02) + 0.03(0.03)= 0.012848 m

y =

©yA

©A

•6–93. The man has a mass of 78 kg and stands motionless atthe end of the diving board. If the board has the cross sectionshown, determine the maximum normal strain developed inthe board. The modulus of elasticity for the material is

Assume A is a pin and B is a roller.E = 125 GPa.


B CA1.5 m 2.5 m

350 mm

20 mm30 mm

10 mm 10 mm 10 mm


388

Section Property:

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

Ans. d = 0.1162 m = 116 mm

130 A106 B =

100(103)(d)5p32 d4

smax = sallow =

Mmax c

I

Mmax = 100 kN # m

I = 2Bp4

ad

2b4

+

p

4 d2 ad

2b2R =

5p32

d4

Section Property:


Ans. d = 0.1986 m = 199 mm

130 A106 B =

100(103)(d)p32 d4

smax = sallow =

Mmax c

I

Mmax = 100 kN # m

I = 2Bp4

ad

2b4R =

p

32 d4

6–94. The two solid steel rods are bolted together alongtheir length and support the loading shown. Assume thesupport at A is a pin and B is a roller. Determine the requireddiameter d of each of the rods if the allowable bendingstress is sallow = 130 MPa.

6–95. Solve Prob. 6–94 if the rods are rotated so thatboth rods rest on the supports at A (pin) and B (roller).

90°


B

A

2 m

80 kN20 kN/m

2 m

B

A

2 m

80 kN20 kN/m

2 m


389


c

Ans.smax =

Mc

I=

1440 (1.5)

1.59896= 1.35 ksi

Ix =

112

(1)(33) -

112

(0.5)(2.53) = 1.59896 in4

M = 1440 lb # in.

+ ©M = 0; M - 180(8) = 0

*6–96. The chair is supported by an arm that is hinged soit rotates about the vertical axis at A. If the load on the chairis 180 lb and the arm is a hollow tube section having thedimensions shown, determine the maximum bending stressat section a–a.

1 in.

3 in.

a

a

A

180 lb

2.5 in.

0.5 in.8 in.

Require

Ans.P = 0.119 kip = 119 lb

1.25 =

2P(1.25>2)

0.11887

smax =

Mc

I

smax = 1.25 ksi

Mmax =

P

2 (4) = 2P

I =

14

p C A1.252 B4 - A0.375

2 B4 D = 0.11887 in4

•6–97. A portion of the femur can be modeled as a tubehaving an inner diameter of 0.375 in. and an outer diameterof 1.25 in. Determine the maximum elastic static force Pthat can be applied to its center. Assume the bone to beroller supported at its ends. The diagram for the bonemass is shown and is the same in tension as in compression.

s–P

4 in. 4 in.

2.30

1.25

0.02 0.05

Ps (ksi)

P (in./ in.)


390

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

Ans.smax =

Mmax c

I=

216(12)(8)112 (8)(163)

= 7.59 ksi

Mmax = 216 kip # ft

6–98. If the beam in Prob. 6–18 has a rectangular crosssection with a width of 8 in. and a height of 16 in., determinethe absolute maximum bending stress in the beam.


8 in.

16 in.

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 =

Mc

I=

16800(12)(3)

108

I =

112

(6)(63) = 108 in4

Mmax = 16800 lb # ft

+ ©MA = 0; Mmax - 400(6)(3) -

12

(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.

AB

6 ft 6 ft

400 lb/ft


391


Support Reactions. The FBD of the beam is shown in Fig. a.

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


Here, . Thus,

Ans. = 2.23 kip>ft wo = 2 225.46 lb>ft

22(103) =

(27wo)(12)(6.25)

204.84375

sallow =

Mmax c

I ;

¢ = 6.25 in

= 204.84375 in4

I =

112

(9)(12.53) -

112

(8.75)(123)

Mmax = 27wo

*6–100. The steel beam has the cross-sectional areashown. Determine the largest intensity of the distributedload that it can support so that the maximum bendingstress in the beam does not exceed sallow = 22 ksi.

w0

12 in.

9 in.

0.25 in.

9 ft 9 ft

0.25 in.

0.25 in.

w0


392


The FBD of the beam is shown in Fig. a


The moment of inertia of the I cross-section about the bending axis is

Here, . Thus

Ans. = 19.77 ksi = 19.8 ksi

=

54 (12)(6.25)

204.84375

smax =

Mmax c

I

c = 6.25 in

= 204.84375 in4

I =

112

(9) A12.53 B -

112

(8.75) A123 B

Mmax = 54 kip # ft

•6–101. The steel beam has the cross-sectional areashown. If determine the maximum bendingstress in the beam.

w0 = 2 kip>ft,

12 in.

9 in.

0.25 in.

9 ft 9 ft

0.25 in.

0.25 in.

w0


393



Internal Moment: Using the method of sections.

Section Property:


Ans.

Ans.sA =

72.0(12)(6)

438.1875= 11.8 ksi

sB =

72.0(12)(6.75)

438.1875= 13.3 ksi

s =

My

I

I =

112

(6) A13.53 B -

112

(5.5) A123 B = 438.1875 in4

M = 72.0 kip # ft

+ ©MNA = 0; M + 12.0(4) - 15.0(8) = 0

6–102. The bolster or main supporting girder of a truckbody is subjected to the uniform distributed load. Determinethe bending stress at points A and B.

1.5 kip/ft

12 in.

0.75 in.

0.75 in.

6 in.

0.5 in.

B

8 ft 12 ftF2

A

A

B

F1


394




The moment of inertia of the cross-section is,

Here, . Thus,

Ans. w = 11250 N>m = 11.25 kN>m

5 A106 B =

0.125w(0.075)

21.09375 A10- 6 B

sallow =

Mmax c

I ;

c = 0.075 w

I =

112

(0.075) A0.153 B = 21.09375 A10- 6 B m4

|Mmax| = 0.125 w

6–103. Determine the largest uniform distributed load wthat can be supported so that the bending stress in the beamdoes not exceed s allow = 5 MPa.

0.5 m 0.5 m1 m

75 mm

150 mm

w


395


Support Reactions. The FBD of the beam is shown in Fig. a

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

The moment of inertia of the cross-section is

Here, . Thus

Ans.

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

= 4.44 MPa

= 4.444 A106 B Pa

=

1.25 A103 B(0.075)

21.09375 A10- 6 B

smax =

Mmax c

I

c = 0.075 m

I =

112

(0.075) A0.153 B = 21.09375 A10- 6 B m4

|Mmax| = 1.25 kN # m

*6–104. If determine the maximumbending stress in the beam. Sketch the stress distributionacting over the cross section.

w = 10 kN>m,

0.5 m 0.5 m1 m

75 mm

150 mm

w


396



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

The moment of inertia of the cross section is

Here, . Thus,

Ans. b = 7.453 in = 7 12

in.

150 =

3450(12)(b)2>3 b4

sallow =

Mmax c

I ;

c = 2b>2 = b

I =

112

(b)(2b)3=

23

b4

Mmax = 3450 lb # ft

•6–105. If the allowable bending stress for the wood beamis determine the required dimension b tothe nearest in. of its cross section.Assume the support at Ais a pin and B is a roller.

14

sallow = 150 psi,

BA

3 ft3 ft3 ft

2b

b

400 lb/ft


397


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


The moment of inertia of the cross-section is

Here, . Thus

Ans.smax =

Mmax c

I=

3450(12)(7.5)

2109.375= 147 psi

c =

152

= 7.5 in

I =

112

(7.5) A153 B = 2109.375 in4

Mmax = 3450 lb # ft

6–106. The wood beam has a rectangular cross section inthe proportion shown. If b � 7.5 in., determine the absolutemaximum bending stress in the beam.

BA

3 ft3 ft3 ft

2b

b

400 lb/ft


398

Location of neutral axis:

[1]

Taking positive root:

[2] Ans.

From Eq. [1].

From Eq. [2]

Ans.(smax)t =

3M

b h2 £2Et + 2Ec

2Ec

≥

M =

13

bc(smax)t (h - c + c) ; (smax)t =

3M

b h c

M =

13

(h - c)2 (b)a c

h - cb(smax)t +

13

c2b(smax)t

(smax)c =

c

h - c (smax)t

M =

13

(h - c)2 (b)(smax)c +

13

c2b(smax)t

M = c12

(h - c)(smax)c (b) d a23b(h - c) + c1

2 (c)(smax)t(b) d a2

3b(c)

©MNA = 0;

c =

hA

Ec

Et

1 +

A

Ec

Et

=

h2Ec

2Et + 2Ec

c

h - c=

A

Ec

Et

(h - c)Ec (emax)t (h - c)

c= cEt (emax)t ; Ec (h - c)2

= Etc2

(h - c)(smax)c = c(smax)t

©F = 0; -

12

(h - c)(smax)c (b) +

12

(c)(smax)t (b) = 0:+

(smax)c = Ec(emax)c =

Ec(emax)t (h - c)

c

(emax)c =

(emax)t (h - c)

c

6–107. A beam is made of a material that has a modulus ofelasticity in compression different from that given fortension. Determine the location c of the neutral axis, andderive an expression for the maximum tensile stress in thebeam having the dimensions shown if it is subjected to thebending moment M.


hc

bEt

Ec

M

s

P


399


*6–108. The beam has a rectangular cross section and issubjected to a bending moment M. If the material fromwhich it is made has a different modulus of elasticity fortension and compression as shown, determine the location cof the neutral axis and the maximum compressive stress inthe beam.

hc

bEt

Ec

M

s

P

See the solution to Prob. 6–107

Ans.

Since

Ans.(smax)c =

3M

bh2 ¢2Et + 2Ec

2Et

≤

(smax)c =

2Ec

2Et

¢3M

bh2 ≤ ¢2Et + 2Ec

2Ec

≤

(smax)c =

2Ec

2Et

(smax)t

(smax)c =

c

h - c (smax)t =

h2Ec

(2Et + 2Ec) ch - a h1Ec

1Et + 1Ec

b d (smax)t

c =

h2Ec

2Et + 2Ec


400


The y and z components of M are negative, Fig. a. Thus,

The moments of inertia of the cross-section about the principal centroidal y and zaxes are

By inspection, the bending stress occurs at corners A and C are

Ans.

Ans.

Here,

Ans.

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

a = 65.1°

tan a =

1584736

tan 225°

tan a =

Iz

Iy tan u

u = 180° + 45° = 225°

= -2.01 ksi = 2.01 ksi (C)

smax = sA = -

-14.14(12)(-8)

1584+

-14.14(12)(5)

736

= 2.01 ksi (T)

smax = sC = -

-14.14(12)(8)

1584+

-14.14(12)( - 5)

736

s = -

Mz y

Iz+

My z

Iy

Iz =

112

(10) A163 B -

112

(8) A143 B = 1584 in4

Iy =

112

(16) A103 B -

112

(14) A83 B = 736 in4

Mz = -20 cos 45° = -14.14 kip # ft.

My = -20 sin 45° = -14.14 kip # ft

•6–109. The beam is subjected to a bending moment ofdirected as shown. Determine the maximum

bending stress in the beam and the orientation of theneutral axis.

M = 20 kip # ft

16 in.

10 in.

8 in.

14 in.

y

z

M

B C

A D

45�


401


The y and z components of M are negative, Fig. a. Thus,

The moments of inertia of the cross-section about principal centroidal y and zaxes are

By inspection, the maximum bending stress occurs at corners A and C. Here, wewill consider corner C.

Ans.M = 119.40 kip # ft = 119 kip # ft

12 = -

-0.7071 M (12)(8)

1584+

-0.7071 M(12)(-5)

736

sC = sallow = -

Mz yc

Iz+

Myzc

Iy

Iz =

112

(10) A163 B -

112

(8) A143 B = 1584 in4

Iy =

112

(16) A103 B -

112

(14) A83 B = 736 in4

Mz = -M cos 45° = -0.7071 M

My = -M sin 45° = -0.7071 M

6–110. Determine the maximum magnitude of thebending moment M that can be applied to the beam so thatthe bending stress in the member does not exceed 12 ksi.

16 in.

10 in.

8 in.

14 in.

y

z

M

B C

A D

45�


402

Internal Moment Components:

Section Properties:

Ans.

Maximum Bending Stress: Applying the flexure formula for biaxial at points Aand B

Ans.

Ans.

Orientation of Neutral Axis:

Ans. a = -3.74°

tan a =

57.6014(10- 6)

0.366827(10- 3) tan (-22.62°)

tan a =

Iz

Iy tan u

= 0.587 MPa (T)

sB = -

-480(0.057368)

57.6014(10- 6)+

200(0.2)

0.366827(10- 3)

= -1.298 MPa = 1.30 MPa (C)

sA = -

-480(-0.142632)

57.6014(10- 6)+

200(-0.2)

0.366827(10- 3)

s = -

Mzy

Iz+

Myz

Iy

Iy =

112

(0.2) A0.43 B -

112

(0.18) A0.363 B = 0.366827 A10- 3 B m4

= 57.6014 A10- 6 B m4

+ 1

12(0.04) A0.183 B + 0.04(0.18)(0.110 - 0.057368)2

Iz =

112

(0.4) A0.023 B + (0.4)(0.02)(0.057368 - 0.01)2

= 0.057368 m = 57.4 mm

y =

©yA

©A=

0.01(0.4)(0.02) + 2[(0.110)(0.18)(0.02)]

0.4(0.02) + 2(0.18)(0.02)

Mz = -

1213

(520) = -480 N # m My =

513

(520) = 200 N # m

6–111. If the resultant internal moment acting on thecross section of the aluminum strut has a magnitude of

and is directed as shown, determine thebending stress at points A and B. The location of thecentroid C of the strut’s cross-sectional area must bedetermined.Also, specify the orientation of the neutral axis.

yM = 520 N # m


20 mm20 mm

zB

C

–y

200 mm

y

M � 520 N�m

125 13

200 mm 200 mm

A

20 mm


403



Section Properties:

Ans.

Maximum Bending Stress: By inspection, the maximum bending stress can occur ateither point A or B. Applying the flexure formula for biaxial bending at points Aand B

Ans.


Ans. a = -3.74°

tan a =

57.6014(10- 6)

0.366827(10- 3) tan (-22.62°)

tan a =

Iz

Iy tan u

= 0.587 MPa (T)

sB = -

-480(0.057368)

57.6014(10- 6)+

200(0.2)

0.366827(10- 3)

= -1.298 MPa = 1.30 MPa (C) (Max)

sA = -

-480(-0.142632)

57.6014(10- 6)+

200(-0.2)

0.366827(10- 3)

s = -

Mz y

Iz+

My z

Iy

Iy =

112

(0.2) A0.43 B -

112

(0.18) A0.363 B = 0.366827 A10- 3 B m4

= 57.6014 A10- 6 B m4

+

112

(0.04) A0.183 B + 0.04(0.18)(0.110 - 0.057368)2

Iz =

112

(0.4) A0.023 B + (0.4)(0.02)(0.057368 - 0.01)2

= 0.057368 m = 57.4 mm

y =

© y A

©A=

0.01(0.4)(0.02) + 2[(0.110)(0.18)(0.02)]

0.4(0.02) + 2(0.18)(0.02)

Mz = -

1213

(520) = -480 N # m My =

513

(520) = 200 N # m

*6–112. The resultant internal moment acting on thecross section of the aluminum strut has a magnitude of

and is directed as shown. Determinemaximum bending stress in the strut. The location of thecentroid C of the strut’s cross-sectional area must bedetermined.Also, specify the orientation of the neutral axis.

yM = 520 N # m

20 mm20 mm

zB

C

–y

200 mm

y

M � 520 N�m

125 13

200 mm 200 mm

A

20 mm


404


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

6–113. Consider the general case of a prismatic beamsubjected to bending-moment components and as shown, when the x, y, z axes pass through the centroidof the cross section. If the material is linear-elastic, thenormal stress in the beam is a linear function of positionsuch that Using the equilibrium con-ditions determine the constants a, b, and c, and show that thenormal stress can be determined from the equation

where the moments and products of inertia are defined inAppendix A.

s = [-1MzIy + MyIyz2y + 1MyIz + MzIyz2z]>1IyIz - Iyz 22,

1A - ys dA,Mz =My = 1A zs dA,0 = 1As dA,s = a + by + cz.

Mz,My

y

y

z x

z

dAMy

C

Mz

s


405


Using the equation developed in Prob. 6-113.

Ans. = 8.95 ksi

sA =

{- [0 + (4.80)(103)(1.6875)](1.625) + [(4.80)(103)(2.970378) + 0](2.125)}

[1.60319(2.970378) - (1.6875)2]

s = - aMz Iy + My Iyz

Iy Iz - Iyz2 by + aMy Iz + Mz Iyz

Iy Iz - Iyz2 bz

Iyz = 2[1.5(1.125)(2)(0.25)] = 1.6875 in4

Iz =

112

(0.25)(3.25)3+ 2 c 1

12 (2)(0.25)3

+ (0.25)(2)(1.5)2 d = 2.970378 in4

Iy =

112

(3.25)(0.25)3+ 2 c 1

12 (0.25)(2)3

+ (0.25)(2)(1.125)2 d = 1.60319 in4

(My)max = 50(3) + 50(5) = 400 lb # ft = 4.80(103)lb # in.

6–114. The cantilevered beam is made from the Z-sectionhaving the cross-section shown. If it supports the twoloadings, determine the bending stress at the wall in thebeam at point A. Use the result of Prob. 6–113.

2 ft

50 lb

50 lb

3 ft

3 in.

0.25 in.

0.25 in.

0.25 in.2.25 in.

2 in.A

B

Using the equation developed in Prob. 6-113.

Ans. = 7.81 ksi

sB =

- [0 + (4.80)(103)(1.6875)](-1.625) + [(4.80)(103)(2.976378) + 0](0.125)

[(1.60319)(2.970378) - (1.6875)2]

s = - aMz Iy + My Iyz

Iy Iz - Iyz2 by + aMy Iz + Mz Iyz

Iy Iz - Iyz2 bz

Iyz = 2[1.5(1.125)(2)(0.25)] = 1.6875 in4

Iz =

112

(0.25)(3.25)3+ 2 c 1

12 (2)(0.25)3

+ (0.25)(2)(1.5)2 d = 2.970378 in4

Iy =

112

(3.25)(0.25)3+ 2 c 1

12 (0.25)(2)3

+ (0.25)(2)(1.125)2 d = 1.60319 in4

(My)max = 50(3) + 50(5) = 400 lb # ft = 4.80(103)lb # in.

6–115. The cantilevered beam is made from the Z-sectionhaving the cross-section shown. If it supports the twoloadings, determine the bending stress at the wall in thebeam at point B. Use the result of Prob. 6–113.

2 ft

50 lb

50 lb

3 ft

3 in.

0.25 in.

0.25 in.

0.25 in.2.25 in.

2 in.A

B


406


Internal Moment Components: Using method of section

Section Properties:

Allowable Bending Stress: By inspection, maximum bending stress occurs at pointsA and B. Applying the flexure formula for biaxial bending at point A.

Ans. P = 14208 N = 14.2 kN

180 A106 B = -

(-1.732P)(0.085)

28.44583(10- 6)+

-1.00P(-0.1)

13.34583(10- 6)

sA = sallow = -

Mzy

Iz+

Myz

Iy

Iy = 2 c 112

(0.01) A0.23 B d +

112

(0.15) A0.013 B = 13.34583(10- 6) m4

Iz =

112

(0.2) A0.173 B -

112

(0.19) A0.153 B = 28.44583(10- 6) m4

©My = 0; My + P sin 30°(2) = 0 My = -1.00P

©Mz = 0; Mz + P cos 30°(2) = 0 Mz = -1.732P

Internal Moment Components: Using method of sections

Section Properties:

Maximum Bending Stress: By inspection, maximum bending stress occurs at A andB. Applying the flexure formula for biaxial bending at point A

Ans. = 7.60 MPa (T) (Max)

sA = -

-1039.32(0.085)

28.44583(10- 6)+

-600.0(-0.1)

13.34583(10- 6)

s = -

Mzy

Iz+

Myz

Iy

Iy = 2 c 112

(0.01) A0.23 B d +

112

(0.15) A0.013 B = 13.34583(10- 6) m4

Iz =

112

(0.2) A0.173 B -

112

(0.19) A0.153 B = 28.44583(10- 6) m4

©My = 0; My + 600 sin 30°(2) = 0; My = -600.0 N # m

©Mz = 0; Mz + 600 cos 30°(2) = 0 Mz = -1039.23 N # m

*6–116. The cantilevered wide-flange steel beam issubjected to the concentrated force P at its end. Determinethe largest magnitude of this force so that the bending stressdeveloped at A does not exceed sallow = 180 MPa.

•6–117. The cantilevered wide-flange steel beam issubjected to the concentrated force of at its end.Determine the maximum bending stress developed in thebeam at section A.

P = 600 N

150 mm

10 mm

10 mm

10 mm

200 mm

30�

z

y

x

A

P

2 m

150 mm

10 mm

10 mm

10 mm

200 mm

30�

z

y

x

A

P

2 m


407


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 zaxes are

Bending Stress: By inspection, the maximum bending stress occurs at either cornerA or B.

Ans.

Orientation of Neutral Axis: Here, .

Ans.

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

a = -66.5°

tan a =

5.2132 A10- 3 B1.3078 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 A103 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 112

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

Iz =

112

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

Iy =

112

(0.6) A0.33 B -

112

(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

6–118. If the beam is subjected to the internal moment ofdetermine the maximum bending stress

acting on the beam and the orientation of the neutral axis.M = 1200 kN # m,

150 mm

150 mm

150 mm

150 mm

300 mm

150 mm

y

xz

M

30�


408


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 zaxes 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 112

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

Iz =

112

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

Iy =

112

(0.6) A0.33 B -

112

(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

6–119. If the beam is made from a material having an allowable tensile and compressive stress of

and respectively,determine the maximum allowable internal moment M 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�


409



The shaft is subjected to two bending moment components Mz and My, Figs. b and c,

respectively.

Since all the axes through the centroid of the circular cross-section of the shaft are

principal axes, then the resultant moment can be used for

design. The maximum moment occurs at . Then,

Then,

Ans. d = 0.02501 m = 25 mm

sallow =

Mmax C

I; 150(106) =

230.49(d>2)p4 (d>2)4

Mmax = 21502+ 1752

= 230.49 N # m

D (x = 1m)

M = 2My 2

+ Mz 2

*6–120. The shaft is supported on two journal bearingsat A and B which offer no resistance to axial loading.Determine the required diameter d of the shaft if the allowable bending stress for the material issallow = 150 MPa.

300 N

300 N

150 N 150 N

200 N

200 N 0.5 m

0.5 m

0.5 m

0.5 m

A

D

E

C

z

x

B

y


410



Internal Moment Components: The shaft is subjected to two bending momentcomponents My and Mz. The moment diagram for each component is drawn.

Maximum Bending Stress: Since all the axes through the circle’s center for circular

shaft are principal axis, then the resultant moment can be usedto determine the maximum bending stress. The maximum resultant moment occurs

at .

Applying the flexure formula

Ans. = 161 MPa

=

427.2(0.015)p4 A0.0154 B

smax =

Mmax c

I

E Mmax = 24002+ 1502

= 427.2 N # m

M = 2My 2

+ Mz 2

•6–121. The 30-mm-diameter shaft is subjected to thevertical and horizontal loadings of two pulleys as shown. Itis supported on two journal bearings at A and B which offerno resistance to axial loading. Furthermore, the coupling tothe motor at C can be assumed not to offer any supportto the shaft. Determine the maximum bending stressdeveloped in the shaft.

1 m

150 N

400 N

400 N

60 mm

100 mm

150 N

1 mA

D

E

BC

yz

x

1 m

1 m


411


Ans. = -293 kPa = 293 kPa (C)

sA = -

Mzy

Iz+

Myz

Iy=

-135.8(0.2210)

0.471(10- 3)+

209.9(-0.06546)

60.0(10- 6)

z = -(0.175 cos 32.9° - 0.15 sin 32.9°) = -0.06546 m

y = 0.15 cos 32.9° + 0.175 sin 32.9° = 0.2210 m

Mz = 250 sin 32.9° = 135.8 N # m

My = 250 cos 32.9° = 209.9 N # m


Section Properties:

Bending Stress: Using formula developed in Prob. 6-113

Ans. = -293 kPa = 293 kPa (C)

sA =

- [0 + 250(-0.1875)(10- 3)](0.15) + [250(0.350)(10- 3) + 0](-0.175)

0.18125(10- 3)(0.350)(10- 3) - [0.1875(10- 3)]2

s =

-(Mz Iy + My Iyz)y + (My Iz + MzIyz)z

IyIz - Iyz2

= -0.1875 A10- 3 B m4

Iyz = 0.15(0.05)(0.125)(-0.1) + 0.15(0.05)(-0.125)(0.1)

= 0.350(10- 3) m4

Iz =

112

(0.05) A0.33 B + 2 c 112

(0.15) A0.053 B + 0.15(0.05) A0.1252 B d = 0.18125 A10- 3 B m4

Iy =

112

(0.3) A0.053 B + 2 c 112

(0.05) A0.153 B + 0.05(0.15) A0.12 B d

My = 250 N # m Mz = 0

6–122. Using the techniques outlined in Appendix A,Example A.5 or A.6, the Z section has principal moments ofinertia of and computed about the principal axes of inertia y and z,respectively. If the section is subjected to an internalmoment of directed horizontally as shown,determine the stress produced at point A. Solve theproblem using Eq. 6–17.

M = 250 N # m

Iz = 0.471110-32 m4,Iy = 0.060110-32 m4

6–123. Solve Prob. 6–122 using the equation developed inProb. 6–113.

200 mm

50 mm

50 mm

300 mm

z¿z

250 N�m

y

y¿

A

32.9�

50 mm200 mm

B

200 mm

50 mm

50 mm

300 mm

z¿z

250 N�m

y

y¿

A

32.9�

50 mm200 mm

B


412



Section Property:

Bending Stress: Applying the flexure formula for biaxial bending

Ans. = 293 kPa = 293 kPa (T)

sB =

135.8(0.2210)

0.471(10- 3)-

209.9(-0.06546)

0.060(10- 3)

s =

Mz¿y¿

Iz¿

+

My¿z¿

Iy¿

z¿ = 0.15 sin 32.9° - 0.175 cos 32.9° = -0.06546 m

y¿ = 0.15 cos 32.9° + 0.175 sin 32.9° = 0.2210 m

Mz¿= 250 sin 32.9° = 135.8 N # m

My¿= 250 cos 32.9° = 209.9 N # m

*6–124. Using the techniques outlined in Appendix A,Example A.5 or A.6, the Z section has principal moments ofinertia of and computed about the principal axes of inertia y and z,respectively. If the section is subjected to an internalmoment of directed horizontally as shown,determine the stress produced at point B. Solve theproblem using Eq. 6–17.

M = 250 N # m

Iz = 0.471110-32 m4,Iy = 0.060110-32 m4

200 mm

50 mm

50 mm

300 mm

z¿z

250 N�m

y

y¿

A

32.9�

50 mm200 mm

B


413


•6–125. Determine the bending stress at point A of thebeam, and the orientation of the neutral axis. Using themethod in Appendix A, the principal moments of inertia ofthe cross section are and where and are the principal axes. Solve the problemusing Eq. 6–17.

y¿z¿

I¿y = 2.295 in4,I¿z = 8.828 in4

4 in.

4 in.

0.5 in.

0.5 in.1.183 in.

1.183 in.

A

C

z

z¿

y

y′

45�

M � 3 kip � ft

Internal Moment Components: Referring to Fig. a, the and components of Mare negative since they are directed towards the negative sense of their respectiveaxes. Thus,

Section Properties: Referring to the geometry shown in Fig. b,

Bending Stress:

Ans. = -20.97 ksi = 21.0 ksi (C)

= -

(-2.121)(12)(-2.828)

8.828+

(-2.121)(12)(1.155)

2.295

sA = -

Mz¿yA

œ

Iz¿

+

My¿zA

œ

Iy¿

yAœ

= -(2.817 sin 45° + 1.183 cos 45°) = -2.828 in.

zAœ

= 2.817 cos 45° - 1.183 sin 45° = 1.155 in.

z¿y¿


414


Internal Moment Components: Since M is directed towards the negative sense of the y axis, itsy component is negative and it has no z component. Thus,

Bending Stress:

Ans. = -20.97 ksi = 21.0 ksi

=

- C0(5.561) + (-3)(12)(-3.267) D(-1.183) + C -3(12)(5.561) + 0(-3.267) D(2.817)

5.561(5.561) - (-3.267)2

sA =

- AMzIy + MyIyz ByA + AMyIz + MzIyz BzA

IyIz - Iyz 2

My = -3 kip # ft Mz = 0

6–126. Determine the bending stress at point A of thebeam using the result obtained in Prob. 6–113. The momentsof inertia of the cross sectional area about the z and y axesare and the product of inertia of the cross sectional area with respect to the z and y axes is Iyz � �3.267 in4. (See Appendix A)

Iz = Iy = 5.561 in4

4 in.

4 in.

0.5 in.

0.5 in.1.183 in.

1.183 in.

A

C

z

z¿

y

y′

45�

M � 3 kip � ft


415


Section Properties:

Ans.

Allowable Bending Stress: Applying the flexure formula

Assume failure of red brass

Ans.

Assume failure of aluminium

M = 29215 N # m = 29.2 kN # m

128 A106 B = 0.68218 c M(0.05)

7.7851(10- 6)d

(sallow)al = n Mc

INA

M = 6598 N # m = 6.60 kN # m (controls!)

35 A106 B =

M(0.04130)

7.7851(10- 6)

(sallow)br =

Mc

INA

= 7.7851 A10- 6 B m4

+ 1

12(0.15) A0.041303 B + 0.15(0.04130)(0.070649 - 0.05)2

INA =

112

(0.10233) A0.053 B + 0.10233(0.05)(0.05 - 0.025)2

h = 0.04130 m = 41.3 mm

0.05 =

0.025(0.10233)(0.05) + (0.05 + 0.5h)(0.15)h

0.10233(0.05) + (0.15)h

y =

©yA

©A

bbr = nbal = 0.68218(0.15) = 0.10233 m

n =

Eal

Ebr=

68.9(109)

101(109)= 0.68218

6–127. The composite beam is made of 6061-T6 aluminum(A) and C83400 red brass (B). Determine the dimension h ofthe brass strip so that the neutral axis of the beam is locatedat the seam of the two metals. What maximum moment willthis beam support if the allowable bending stress for thealuminum is and for the brass1sallow2br = 35 MPa?

1sallow2al = 128 MPa

B

A50 mm

150 mm

h


416


Section Properties: For transformed section.


Assume failure of red brass

Ans.


M = 28391 N # m = 28.4 kN # m

128 A106 B = 0.68218 c M(0.049289)

7.45799(10- 6)d

(sallow)al = n Mc

INA

M = 6412 N # m = 6.41 kN # m (controls!)

35 A106 B =

M(0.09 - 0.049289)

7.45799(10- 6)

(sallow)br =

Mc

INA

= 7.45799 A10- 6 B m4

+ 1

12(0.15) A0.043 B + 0.15(0.04)(0.07 - 0.049289)2

INA =

112

(0.10233) A0.053 B + 0.10233(0.05)(0.049289 - 0.025)2

= 0.049289 m

=

0.025(0.10233)(0.05) + (0.07)(0.15)(0.04)

0.10233(0.05) + 0.15(0.04)

y =

©yA

©A

bbr = nbal = 0.68218(0.15) = 0.10233 m

n =

Eal

Ebr=

68.9(109)

101.0(109)= 0.68218

*6–128. The composite beam is made of 6061-T6 aluminum(A) and C83400 red brass (B). If the height determine the maximum moment that can be applied to thebeam if the allowable bending stress for the aluminum is

and for the brass 1sallow2br = 35 MPa.1sallow2al = 128 MPa

h = 40 mm,

B

A50 mm

150 mm

h


417


Maximum Moment: For the simply-supported beam subjected to the uniform

distributed load, the maximum moment in the beam is

.

Section Properties: The cross section will be transformed into that of steel as

shown in Fig. a. Here, .

Then . The location of the centroid of thetransformed section is

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

Maximum Bending Stress: For the steel,

Ans.

At the seam,

For the aluminium,

Ans.

At the seam,

The bending stress across the cross section of the composite beam is shown in Fig. b.

sal� y = 0.6970 in. = nMmaxy

I= 0.3655 c25.3125(12)(0.6970)

30.8991d = 2.50 ksi

(smax)al = n Mmaxcal

I= 0.3655 c25.3125(12)(6 - 2.3030)

30.8991d = 13.3 ksi

sst� y = 0.6970 in. =

Mmaxy

I=

25.3125(12)(0.6970)

30.8991= 6.85 ksi

(smax)st =

Mmaxcst

I=

25.3125(12)(2.3030)

30.8991= 22.6 ksi

= 30.8991 in4

+

112

(1.0965) A33 B + 1.0965(3)(4.5 - 2.3030)2

I = ©I + Ad2=

112

(3) A33 B + 3(3)(2.3030 - 1.5)2

y =

©yA

©A=

1.5(3)(3) + 4.5(3)(1.0965)

3(3) + 3(1.0965)= 2.3030 in.

bst = nbal = 0.3655(3) = 1.0965 in

n =

Eal

Est=

10.629

= 0.3655

= 25.3125 kip # ft

Mmax =

wL2

8=

0.9 A152 B8

•6–129. Segment A of the composite beam is made from2014-T6 aluminum alloy and segment B is A-36 steel. If

, determine the absolute maximum bendingstress developed in the aluminum and steel. Sketch thestress distribution on the cross section.

w = 0.9 kip>ftw

A

B

15 ft

3 in.

3 in.

3 in.


418


Maximum Moment: For the simply-supported beam subjected to the uniform

distributed load, the maximum moment in the beam is

.

Section Properties: The cross section will be transformed into that of steel as

shown in Fig. a. Here, .

Then . The location of the centroid of thetransformed section is


Bending Stress: Assuming failure of steel,

Ans.

Assuming failure of aluminium alloy,

w = 1.02 kip>ft (sallow)al = n

Mmax cal

I; 15 = 0.3655 c (28.125w)(12)(6 - 2.3030)

30.8991d

w = 0.875 kip>ft (controls)

(sallow)st =

Mmax cst

I; 22 =

(28.125w)(12)(2.3030)

30.8991

= 30.8991 in4

+ 1.0965 A33 B + 1.0965(3)(4.5 - 2.3030)2

I = ©I + Ad2=

112

(3) A33 B + 3(3)(2.3030 - 1.5)2+

112

(1.0965) A33 B

y =

©yA

©A=

1.5(3)(3) + 4.5(3)(1.0965)

3(3) + 3(1.0965)= 2.3030 in.

bst = nbal = 0.3655(3) = 1.0965 in

n =

Eal

Est=

10.629

= 0.3655

Mmax =

wL2

8=

w A152 B8

= 28.125w

6–130. Segment A of the composite beam is made from2014-T6 aluminum alloy and segment B is A-36 steel. Ifthe allowable bending stress for the aluminum and steelare and , determinethe maximum allowable intensity w of the uniformdistributed load.

(sallow)st = 22 ksi(sallow)al = 15 ksi

w

A

B

15 ft

3 in.

3 in.

3 in.


419


Section Properties: For the transformed section.


Ans.

Ans.(smax)w = n Mc

I= 0.065517 c7.5(12)(3)

31.7172d = 0.558 ksi

(smax)st =

Mc

I=

7.5(12)(3)

31.7172= 8.51 ksi

INA =

112

(1.5 + 0.26207) A63 B = 31.7172 in4

bst = nbw = 0.065517(4) = 0.26207 in.

n =

Ew

Est=

1.90(103)

29.0(103)= 0.065517

6–131. The Douglas fir beam is reinforced with A-36straps at its center and sides. Determine the maximumstress developed in the wood and steel if the beam is subjected to a bending moment of .Sketch the stress distribution acting over the cross section.

Mz = 7.50 kip # ft

y

z6 in.

0.5 in. 0.5 in.

2 in. 2 in.

0.5 in.


420


Section Properties:


Ans.

Ans.(smax)k =

Mc

I=

900(12)(6 - 2.5247)

85.4168= 439 psi

(smax)al = n Mc

I= 0.55789 c900(12)(6 - 2.5247)

85.4170d = 245 psi

= 85.4170 in4

+ 1

12 (6.6947) A0.53 B + 6.6947(0.5)(5.75 - 2.5247)2

+ 1

12 (1) A5.53 B + 1(5.5)(3.25 - 2.5247)2

INA =

112

(13) A0.53 B + 13(0.5)(2.5247 - 0.25)2

= 2.5247 in.

y =

©yA

©A=

0.25(13)(0.5) + 2[(3.25)(5.5)(0.5)] + 5.75(6.6947)(0.5)

13(0.5) + 2(5.5)(0.5) + 6.6947(0.5)

bk = n bal = 0.55789(12) = 6.6947 in.

n =

Eal

Ek=

10.6(103)

19.0(103)= 0.55789

*6–132. The top plate is made of 2014-T6 aluminum and isused to reinforce a Kevlar 49 plastic beam. Determine themaximum stress in the aluminum and in the Kevlar if thebeam is subjected to a moment of M = 900 lb # ft.

0.5 in.

6 in.

0.5 in.

0.5 in.

12 in.M

0.5 in.


421


Section Properties:



Assume failure of Kevlar 49

Ans. = 16.4 kip # ft (Controls!)

M = 196.62 kip # in

8 =

M(6 - 2.5247)

85.4170

(sallow)k =

Mc

I

M = 1762 kip # in = 146.9 kip # ft

40 = 0.55789 cM(6 - 2.5247)

85.4170d

(sallow)al = n Mc

I

= 85.4170 in4

+ 1

12 (6.6947) A0.53 B + 6.6947(0.5)(5.75 - 2.5247)2

+ 1

12 (1) A5.53 B + 1(5.5)(3.25 - 2.5247)2

INA =

112

(13) A0.53 B + 13(0.5)(2.5247 - 0.25)2

= 2.5247 in.

y =

© yA

©A=

0.25(13)(0.5) + 2[(3.25)(5.5)(0.5)] + 5.75(6.6947(0.5)

13(0.5) + 2(5.5)(0.5) + 6.6947(0.5)

bk = n bal = 0.55789(12) = 6.6947 in.

n =

Eal

Ek=

10.6(103)

19.0(103)= 0.55789

•6–133. The top plate made of 2014-T6 aluminum is usedto reinforce a Kevlar 49 plastic beam. If the allowablebending stress for the aluminum is andfor the Kevlar , determine the maximummoment M that can be applied to the beam.

(sallow)k = 8 ksi(sallow)al = 40 ksi

0.5 in.

6 in.

0.5 in.

0.5 in.

12 in.M

0.5 in.


422


Maximum stress in steel:

Ans.

Maximum stress in brass:

(sbr)max =

nMc2

I=

0.5(8)(103)(0.05)

27.84667(10- 6)= 7.18 MPa

(sst)max =

Mc1

I=

8(103)(0.07)

27.84667(10- 6)= 20.1 MPa (max)

I =

112

(0.14)(0.14)3-

112

(0.05)(0.1)3= 27.84667(10- 6)m4

n =

Ebr

Est=

100200

= 0.5

Maximum stress in steel:

Ans.

Maximum stress in wood:

Ans. = 0.05517(1395) = 77.0 psi

(sw) = n(sst)max

(sst) =

Mc

I=

850(12)(4 - 1.1386)

20.914= 1395 psi = 1.40 ksi

+

112

(0.8276)(3.53) + (0.8276)(3.5)(1.11142) = 20.914 in4

I =

112

(16)(0.53) + (16)(0.5)(0.88862) + 2a 112b(0.5)(3.53) + 2(0.5)(3.5)(1.11142)

y =

(0.5)(16)(0.25) + 2(3.5)(0.5)(2.25) + (0.8276)(3.5)(2.25)

0.5(16) + 2(3.5)(0.5) + (0.8276)(3.5)= 1.1386 in.

6–134. The member has a brass core bonded to a steelcasing. If a couple moment of is applied at its end,determine the maximum bending stress in the member.Ebr = 100 GPa, Est = 200 GPa.

8 kN # m

6–135. The steel channel is used to reinforce the woodbeam. Determine the maximum stress in the steel and inthe wood if the beam is subjected to a moment of

Est = 29(103) ksi, Ew = 1600 ksi.M = 850 lb # ft.

3 m

100 mm

20 mm100 mm20 mm

20 mm 20 mm

8 kN�m

0.5 in.

4 in.

0.5 in.

0.5 in.

15 in.M � 850 lb�ft


423


Section Properties: For the transformed section.


Assume failure of steel

Ans.

Assume failure of wood

M = 331.9 kip # in = 27.7 kip # ft

2.0 = 0.048276 c M(2)

16.0224d

(sallow)w = n My

I

= 11.7 kip # ft (Controls !)

M = 141.0 kip # in

22 =

M(2.5)

16.0224

(sallow)st =

Mc

I

INA =

112

(3) A53 B -

112

(3 - 0.14483) A43 B = 16.0224 in4

bst = nbw = 0.048276(3) = 0.14483 in.

n =

Ew

Est=

1.40(103)

29.0(103)= 0.048276

*6–136. A white spruce beam is reinforced with A-36 steelstraps at its top and bottom as shown. Determine thebending moment M it can support if

.and (sallow)w = 2.0 ksi(sallow)st = 22 ksi

3 in.

0.5 in.

0.5 in.

4 in.

x

y

z

M


424


Section Properties: The cross section will be transformed into that of steel as shown in Fig. a.

Here, . Thus, . The

location of the transformed section is


Maximum Bending Stress: For the steel,

Ans.

For the aluminum alloy,

Ans.(smax)al = n Mcal

I= 0.3655 C 45 A103 B(0.1882)

18.08 A10- 6 B S = 171 MPa

(smax)st =

Mcst

I=

45 A103 B(0.06185)

18.08 A10- 6 B = 154 MPa

= 18.08 A10- 6 B m4

+

14p A0.054 B + p A0.052 B(0.2 - 0.1882)2

I = ©I + Ad2=

112

(0.0054825) A0.153 B + 0.0054825(0.15)(0.1882 - 0.075)2

= 0.1882 m

y =

©yA

©A=

0.075(0.15)(0.0054825) + 0.2 cp A0.052 B d0.15(0.0054825) + p A0.052 B

bst = nbal = 0.3655(0.015) = 0.0054825 mn =

Eal

Est=

73.1 A109 B200 A109 B = 0.3655

•6–137. If the beam is subjected to an internal moment of, determine the maximum bending stress

developed in the A-36 steel section A and the 2014-T6aluminum alloy section B.

M = 45 kN # m

M

150 mm

15 mm

B

A

50 mm


425


Bending Stress: The cross section will be transformed into that of concrete as shown

in Fig. a. Here, . It is required that both concrete and steel

achieve their allowable stress simultaneously. Thus,

(1)

(2)

Equating Eqs. (1) and (2),

(3)

Section Properties: The area of the steel bars is .

Thus, the transformed area of concrete from steel is

. Equating the first moment of the area of concrete above and belowthe neutral axis about the neutral axis,

(4)

Solving Eqs. (3) and (4),

Ans.

Thus, the moment of inertia of the transformed section is

I =

13

(0.2) A0.16593 B + 2.4 A10- 3 Bp(0.5308 - 0.1659)2

ccon = 0.1659 m

d = 0.5308 m = 531 mm

ccon 2

= 0.024pd - 0.024pccon

0.1ccon 2

= 2.4 A10- 3 Bpd - 2.4 A10- 3 Bpccon

0.2(ccon)(ccon>2) = 2.4 A10- 3 Bp (d - ccon)

= 2.4 A10- 3 Bp m2

(Acon)t = nAs = 8 C0.3 A10- 3 Bp DAst = 3 cp

4 A0.022 B d = 0.3 A10- 3 Bp m2

ccon = 0.3125d (3)

12.5 A106 B ¢ Iccon≤ = 27.5 A106 B ¢ I

d - ccon≤

M = 27.5 A106 B ¢ I

d - ccon≤

(sallow)st = n Mcst

I ; 220 A106 B = 8BM(d - ccon)

IR

M = 12.5 A106 B ¢ Iccon≤

(sallow)con =

Mccon

I ; 12.5 A106 B =

Mccon

I

n =

Est

Econ=

20025

= 8

6–138. The concrete beam is reinforced with three 20-mmdiameter steel rods. Assume that the concrete cannotsupport tensile stress. If the allowable compressive stressfor concrete is and the allowabletensile stress for steel is , determinethe required dimension d so that both the concrete and steelachieve their allowable stress simultaneously. This conditionis said to be ‘balanced’. Also, compute the correspondingmaximum allowable internal moment M that can be appliedto the beam. The moduli of elasticity for concrete and steelare and , respectively.Est = 200 GPaEcon = 25 GPa

(sallow)st = 220 MPa(sallow)con = 12.5 MPa

d

200 mm

M


426


Substituting this result into Eq. (1),

‚ Ans. = 98 594.98 N # m = 98.6 kN # m

M = 12.5 A106 B C 1.3084 A10- 3 B0.1659

S

= 1.3084 A10- 3 B m4

6–138. Continued

Ans. = 1.53 ksi

(smax)pvc = n2 Mc

I= a450

800b1500(12)(3.0654)

20.2495

+

112

(1.6875)(13) + 1.6875(1)(2.56542) = 20.2495 in4

I =

112

(3)(23) + 3(2)(0.93462) +

112

(0.6)(23) + 0.6(2)(1.06542)

y =

©yA

©A=

(1)(3)(2) + 3(0.6)(2) + 4.5(1.6875)(1)

3(2) + 0.6(2) + 1.6875(1)= 1.9346 in.

(bbk)2 = n2 bpvc =

450800

(3) = 1.6875 in.

(bbk)1 = n1 bEs =

160800

(3) = 0.6 in.

6–139. The beam is made from three types of plastic thatare identified and have the moduli of elasticity shown in thefigure. Determine the maximum bending stress in the PVC.

PVC EPVC � 450 ksi

Escon EE � 160 ksi

Bakelite EB � 800 ksi

3 ft 4 ft

500 lb 500 lb

3 ft

3 in.

1 in.2 in.

2 in.


427


Section Properties: The beam cross section will be transformed into

that of steel. Here, . Thus,

. The location of the transformed section is


Bending Stress: Assuming failure of steel,

Assuming failure of concrete,

Ans. M = 329 849.77 N # m = 330 kN # m (controls)

(sallow)con = n Mccon

I ; 10 A106 B = 0.1105CM(0.5 - 0.3222)

647.93 A10- 6 B S

M = 331 770.52 N # m = 332 kN # m

(sallow)st =

Mcst

I ; 165 A106 B =

M(0.3222)

647.93 A10- 6 B

= 647.93 A10- 6 B m4

+ 1

12(0.1105) A0.13 B + 0.1105(0.1)(0.45 - 0.3222)2

+ 1

12 (0.2) A0.0153 B + 0.2(0.015)(0.3925 - 0.3222)2

+ 1

12(0.015) A0.373 B + 0.015(0.37)(0.3222 - 0.2)2

+ 0.2(0.015)(0.3222 - 0.0075)2

I = ©I + Ad2=

112

(0.2) A0.0153 B

= 0.3222 m

=

0.0075(0.015)(0.2) + 0.2(0.37)(0.015) + 0.3925(0.015)(0.2) + 0.45(0.1)(0.1105)

0.015(0.2) + 0.37(0.015) + 0.015(0.2) + 0.1(0.1105)

y =

©yA

©A

bst = nbcon = 0.1105(1) = 0.1105 m

n =

Econ

Est=

22.1200

= 0.1105

*6–140. The low strength concrete floor slab is integratedwith a wide-flange A-36 steel beam using shear studs (notshown) to form the composite beam. If the allowablebending stress for the concrete is andallowable bending stress for steel is determine the maximum allowable internal moment M thatcan be applied to the beam.

(sallow)st = 165 MPa,(sallow)con = 10 MPa,

M

100 mm

400 mm15 mm

15 mm

15 mm

200 mm

1 m


428

Solving for the positive root:

Ans.

Ans.(sst)max = naMy

Ib = a29.0(103)

4.20(103)b a 40(12)(13 - 5.517)

1358.781b = 18.3 ksi

(scon)max =

My

I=

40(12)(5.517)

1358.781= 1.95 ksi

= 1358.781 in4

I = c 112

(8)(5.517)3+ 8(5.517)(5.517>2)2 d + 16.2690(13 - 5.517)2

h¿ = 5.517 in.

h¿2

+ 4.06724h - 52.8741 = 0

©yA = 0; 8(h¿)ah¿

2b - 16.2690(13 - h¿) = 0

A¿ = nAst =

29.0(103)

4.20(103) (2.3562) = 16.2690 in2

Econ = 4.20(103) ksi

Est = 29.0(103) ksi

Ast = 3(p)(0.5)2= 2.3562 in2

Mmax = (10 kip)(4 ft) = 40 kip # ft

•6–141. The reinforced concrete beam is used to supportthe loading shown. Determine the absolute maximumnormal stress in each of the A-36 steel reinforcing rods andthe absolute maximum compressive stress in the concrete.Assume the concrete has a high strength in compressionand yet neglect its strength in supporting tension.


15 in.

8 in.

2 in.1 in. diameter rods

4 ft8 ft4 ft

10 kip10 kip


429


Solving for the positive root:

Assume concrete fails:

Assume steel fails:

Ans. M = 1169.7 kip # in. = 97.5 kip # ft (controls)

(sst)allow = naMy

Ib ; 40 = ¢ 29(103)

3.8(103)≤ ¢M(16 - 0.15731)

3535.69≤

M = 2551 kip # in.

(scon)allow =

My

I ; 3 =

M(4.15731)

3535.69

+ 11.9877(16 - 0.15731)2= 3535.69 in4

I = c 112

(22)(4)3+ 22(4)(2.15731)2 d + c 1

12 (6)(0.15731)3

+ 6(0.15731)(0.15731>2)2 d

h¿ = 0.15731 in.

3h2+ 99.9877h¿ - 15.8032 = 0

©yA = 0; 22(4)(h¿ + 2) + h¿(6)(h¿>2) - 11.9877(16 - h¿) = 0

A¿ = nAst =

29(103)

3.8(103) (1.5708) = 11.9877 in2

Ast = 2(p)(0.5)2= 1.5708 in2

6–142. The reinforced concrete beam is made using twosteel reinforcing rods. If the allowable tensile stress for the steel is and the allowablecompressive stress for the concrete is ,determine the maximum moment M that can be applied tothe section. Assume the concrete cannot support a tensilestress. Est = 29(103) ksi, Econc = 3.8(103) ksi.

(sconc)allow = 3 ksi(sst)allow = 40 ksi

4 in.

18 in.

8 in.8 in. 6 in.

2 in.

1-in. diameter rods

M


430

Normal Stress: Curved-beam formula

[1]

[2]

[3]

Denominator of Eq. [1] becomes,

Using Eq. [2],

But,

Then,

Eq. [1] becomes

Using Eq. [2],

Using Eq. [3],

=

Mr

AI CLA

y

r + y dA - y

LA

dA

r + yS

s =

Mr

AI CA - ¢A -

LA

y

r + y dA≤ - y

LA

dA

r + yS

s =

Mr

AI (A - rA¿ - yA¿)

s =

Mr

AI (A - rA¿)

Ar(rA¿ - A) : A

r I

1A y dA = 0, as y

r: 0

=

A

r LA

¢ y2

1 +yr

≤dA - A 1A y dA -

Ay

r LA ¢ y

1 +yr

≤ dA

= ALA

y2

r + y dA - A 1A y dA - Ay

LA

y

r + y dA

Ar(rA¿ - A) = -ALA

¢ ry

r + y+ y - y≤dA - Ay

LA

y

r + y dA

Ar(rA¿ - A) = Ar¢A -

LA

y

r + y dA - A≤ = -Ar

LA

y

r + y dA

= A -

LA

y

r + y dA

=

LA a r - r - y

r + y+ 1b dA

rA¿ = r LA

dAr

=

LAa r

r + y- 1 + 1bdA

r = r + y

s =

M(A - rA¿)

Ar(rA¿ - A)

s =

M(R - r)

Ar(r - R) where A¿ =

LA dAr and R =

A

1AdAr

=

A

A¿

6–143. For the curved beam in Fig. 6–40a, show that whenthe radius of curvature approaches infinity, the curved-beamformula, Eq. 6–24, reduces to the flexure formula, Eq. 6–13.



431


As

Therefore, (Q.E.D.)s =

Mr

AI a -

yA

rb = -

My

I

LA ¢ y

r

1 +yr

≤ dA = 0 and y

r LA

¢ dA

1 +yr

≤ =

y

r1A dA =

yA

r

y

r : 0

=

Mr

AI CLA

¢ yr

1 +yr

≤dA -

y

r LA

¢ dA

1 +yr

≤ S

Ans.

Ans.

No, because of localized stress concentration at the wall. Ans.

sB =

M(R - rB)

ArB (r - R)=

50(0.166556941 - 0.25)

2.8125(10- 3)p (0.25)(0.0084430586)= 224 kPa (C)

sA =

M(R - rA)

ArA (r - R)=

50(0.166556941 - 0.1)

2.8125(10- 3)p (0.1)(0.0084430586)= 446k Pa (T)

r - R = 0.175 - 0.166556941 = 0.0084430586

R =

A

1AdAr

=

2.8125(10- 3)p

0.053049301= 0.166556941

A = p ab = p(0.075)(0.0375) = 2.8125(10- 3)p

=

2p(0.0375)

0.075 (0.175 - 20.1752

- 0.0752 ) = 0.053049301 m

LA

dAr

=

2p ba

(r - 2r2- a2 )

6–143. Continued

*6–144. The member has an elliptical cross section. If it issubjected to a moment of , determine thestress at points A and B. Is the stress at point , which islocated on the member near the wall, the same as that at A?Explain.

A¿

M = 50 N # m

B

A¿

A

100 mm

250 mm

150 mm

M

75 mm

=

Mr

AI


432


B

A¿

A

100 mm

250 mm

150 mm

M

75 mm•6–145. The member has an elliptical cross section. If theallowable bending stress is determine themaximum moment M that can be applied to the member.

sallow = 125 MPa

Assume tension failure.

Ans.

Assume compression failure:

M = 27.9 kN # m

-125(106) =

M(0.166556941 - 0.25)

0.0028125p(0.25)(8.4430586)(10- 3)

M = 14.0 kN # m (controls)

125(106) =

M(0.166556941 - 0.1)

0.0028125p(0.1)(8.4430586)(10- 3)

s =

M(R - r)

Ar(r - R)

r - R = 0.175 - 0.166556941 = 8.4430586(10- 3) m

R =

A

1AdAr

=

0.0028125p0.053049301

= 0.166556941 m

= 0.053049301 m

LA

dAr

=

2pba

(r - 2r2- a2) =

2p(0.0375)

0.075 (0.175 - 20.1752

- 0.0752)

A = p(0.075)(0.0375) = 0.0028125 p

a = 0.075 m; b = 0.0375 m


433


Internal Moment: is positive since it tends to increase the beam’sradius of curvature.

Section Properties:

Allowable Normal Stress: Applying the curved-beam formula

Assume tension failure

Assume compression failure

Ans. P = 55195 N = 55.2 kN (Controls !)

-50 A106 B =

0.16P(0.306243 - 0.42)

0.00375(0.42)(0.012757)

(sallow)t =

M(R - r)

Ar(r - R)

P = 159482 N = 159.5 kN

120 A106 B =

0.16P(0.306243 - 0.25)

0.00375(0.25)(0.012757)

(sallow)t =

M(R - r)

Ar(r - R)

r - R = 0.319 - 0.306243 = 0.012757 m

R =

A

© 1AdAr

=

0.003750.012245

= 0.306243 m

= 0.012245 m

©LA

dAr

= 0.15 ln 0.260.25

+ 0.01 ln 0.410.26

+ 0.075 ln 0.420.41

A = 0.15(0.01) + 0.15(0.01) + 0.075(0.01) = 0.00375 m2

= 0.3190 m

=

0.255(0.15)(0.01) + 0.335(0.15)(0.01) + 0.415(0.075)(0.01)

0.15(0.01) + 0.15(0.01) + 0.075(0.01)

r =

©yA

©A

M = 0.160P

6–146. Determine the greatest magnitude of the appliedforces P if the allowable bending stress is in compression and in tension.(sallow)t = 120 MPa

(sallow)c = 50 MPa75 mm

250 mm

150 mm

10 mm

10 mm

10 mm

150 mm

160 mm

P

P


434


Internal Moment: is positive since it tends to increasethe beam’s radius of curvature.

Section Properties:

Normal Stress: Applying the curved-beam formula

Ans.

Ans. = -5.44 MPa

=

0.960(103)(0.306243 - 0.42)

0.00375(0.42)(0.012757)

(smax)c =

M(R - r)

Ar(r - R)

= 4.51 MPa

=

0.960(103)(0.306243 - 0.25)

0.00375(0.25)(0.012757)

(smax)t =

M(R - r)

Ar(r - R)

r - R = 0.319 - 0.306243 = 0.012757 m

R =

A

©1A dAr

=

0.003750.012245

= 0.306243 m

= 0.012245 m

© LA

dAr

= 0.15 ln 0.260.25

+ 0.01 ln 0.410.26

+ 0.075 ln 0.420.41

A = 0.15(0.01) + 0.15(0.01) + 0.075(0.01) = 0.00375 m2

= 0.3190 m

=

0.255(0.15)(0.01) + 0.335(0.15)(0.01) + 0.415(0.075)(0.01)

0.15(0.01) + 0.15(0.01) + 0.075(0.01)

r =

©yA

©A

M = 0.160(6) = 0.960 kN # m

6–147. If determine the maximum tensile andcompressive bending stresses in the beam.

P = 6 kN, 75 mm

250 mm

150 mm

10 mm

10 mm

10 mm

150 mm

160 mm

P

P


435


Internal Moment: is negative since it tends to decrease the beam’sradius curvature.

Section Properties:


Ans.

Ans. = -9.73 MPa = 9.73 MPa (C)

sB =

M(R - rB)

ArB (r - R)=

-900(0.509067 - 0.4)

0.00425(0.4)(5.933479)(10- 3)

= 3.82 MPa (T)

sA =

M(R - rA)

ArA (r - R)=

-900(0.509067 - 0.57)

0.00425(0.57)(5.933479)(10- 3)

r - R = 0.515 - 0.509067 = 5.933479(10- 3) m

R =

A

©1AdAr

=

0.004258.348614(10- 3)

= 0.509067 m

©

LA dAr

= 0.015 ln 0.550.4

+ 0.1 ln 0.570.55

= 8.348614(10- 3) m

r =

©rA

©A=

2.18875 (10- 3)

0.00425= 0.5150 m

©rA = 0.475(0.15)(0.015) + 0.56(0.1)(0.02) = 2.18875(10- 3) m3

©A = 0.15(0.015) + 0.1(0.02) = 0.00425 m2

M = -900 N # m

*6–148. The curved beam is subjected to a bendingmoment of as shown. Determine the stressat points A and B, and show the stress on a volume elementlocated at each of these points.

M = 900 N # m

30�

B

A

100 mm

150 mm

20 mm15 mm

400 mm

B

A

M

C

C


436

Internal Moment: is negative since it tends to decrease the beam’sradius of curvature.

Section Properties:


Ans. = 2.66 MPa (T)

sC =

M(R - rC)

ArC(r - R)=

-900(0.509067 - 0.55)

0.00425(0.55)(5.933479)(10- 3)

r - R = 0.515 - 0.509067 = 5.933479(10- 3) m

R =

A

©1AdAr

=

0.004258.348614(10- 3)

= 0.509067 m

©

LA dAr

= 0.015 ln 0.550.4

+ 0.1 ln 0.570.55

= 8.348614(10- 3) m

r =

©rA

©A=

2.18875 (10- 3)

0.00425= 0.5150 m

©rA = 0.475(0.15)(0.015) + 0.56(0.1)(0.02) = 2.18875(10- 3) m

©A = 0.15(0.015) + 0.1(0.02) = 0.00425 m2

M = -900 N # m

•6–149. The curved beam is subjected to a bendingmoment of . Determine the stress at point C.M = 900 N # m


30�

B

A

100 mm

150 mm

20 mm15 mm

400 mm

B

A

M

C

C


437


Ans.

Ans.(smax)c = =

M(R - rB)

ArB(r - R)=

25(1.606902679 - 2.5)

0.1656p(2.5)(0.14309732)= 120 psi (C)

(smax)t =

M(R - rA)

ArA(r - R)=

25(1.606902679 - 1)

0.1656 p(1)(0.14309732)= 204 psi (T)

r - R = 1.75 - 1.606902679 = 0.14309732 in.

R =

A

1AdAr

=

0.1656 p0.32375809

= 1.606902679 in.

A = p(0.752) - p(0.632) = 0.1656 p

= 0.32375809 in.

= 2p(1.75 - 21.752- 0.752) - 2p (1.75 - 21.752

- 0.632)

LA

dAr

= ©2p (r - 2r2- c2)

6–150. The elbow of the pipe has an outer radius of0.75 in. and an inner radius of 0.63 in. If the assembly issubjected to the moments of determine themaximum stress developed at section .a-a

M = 25 lb # in.,� 25 lb�in.M

= 25 lb�in.M

1 in.30�

a

a

0.75 in.

0.63 in.

Ans.

Ans.sB =

600(12)(8.7993 - 10)

2(10)(0.03398)= -12.7 ksi = 12.7 ksi (C)

sA =

600(12)(8.7993 - 8)

2(8)(0.03398)= 10.6 ksi (T)

s =

M(R - r)

Ar(r - R)

r - R = 8.83333 - 8.7993 = 0.03398 in.

R =

A

1AdAr

=

20.22729

= 8.7993 in.

LA dAr

= 0.5 ln 108

+ c 1(10)

(10 - 8) c ln

108

d - 1 d = 0.22729 in.

r =

©rA

©A=

9(0.5)(2) + 8.6667 A12 B(1)(2)

2= 8.83333 in.

A = 0.5(2) +

12

(1)(2) = 2 in2

6–151. The curved member is symmetric and is subjectedto a moment of Determine the bendingstress in the member at points A and B. Show the stressacting on volume elements located at these points.

M = 600 lb # ft.

8 in.

A

MM

B

2 in.

1.5 in.

0.5 in.


438


a

Ans.

Ans.sB =

M(R - rB)

ArB(r - R)=

41.851 (0.197633863 - 0.1625)

3.75(10- 3)(0.1625)(0.002366137)= 1.02 MPa (T)

= 792 kPa (C)

sA =

M(R - rA)

ArA(r - R)=

41.851(0.197633863 - 0.2375)

3.75(10- 3)(0.2375)(0.002366137)= -791.72 kPa

r - R = 0.2 - 0.197633863 = 0.002366137

R =

A

1AdAr

=

3.75(10- 3)

0.018974481= 0.197633863 m

A = (0.075)(0.05) = 3.75(10- 3) m2

LA dAr

= b ln r2

r1= 0.05 ln

0.23750.1625

= 0.018974481 m

M = 41.851 N # m

+ ©MO = 0; M - 250 cos 60° (0.075) - 250 sin 60° (0.15) = 0

*6–152. The curved bar used on a machine has arectangular cross section. If the bar is subjected to a coupleas shown, determine the maximum tensile and compressivestress acting at section . Sketch the stress distributionon the section in three dimensions.

a-a75 mm

50 mm

150 mm

162.5 mm

a

a

60� 60�

250 N

250 N

75 mm


439


Section Properties:

Internal Moment: The internal moment must be computed about the neutral axis asshown on FBD. is negative since it tends to decrease thebeam’s radius of curvature.

Maximum Normal Stress: Applying the curved-beam formula

Ans. = -26.2 MPa = 26.2 MPa (C) (Max)

=

-1816.93(1.234749 - 1.20)

0.008(1.20)(0.251183)(10- 3)

sB =

M(R - rB)

ArB (r - R)

= 18.1 MPa (T)

=

-1816.93(1.234749 - 1.26)

0.008(1.26)(0.251183)(10- 3)

sA =

M(R - rA)

ArA (r - R)

M = -1816.93 N # m

r - R = 1.235 - 1.234749 = 0.251183 A10- 3 B mR =

A

1AdAr

=

0.0086.479051 (10- 3)

= 1.234749 m

A = 0.1(0.04) + 0.2(0.02) = 0.008 m2

©

LA dAr

= 0.1 ln 1.241.20

+ 0.2 ln 1.261.24

= 6.479051 A10- 3 Bm

r =

©rA

©A=

1.22(0.1)(0.04) + 1.25(0.2)(0.02)

0.1(0.04) + 0.2(0.02)= 1.235 m

•6–153. The ceiling-suspended C-arm is used to supportthe X-ray camera used in medical diagnoses. If the camerahas a mass of 150 kg, with center of mass at G, determinethe maximum bending stress at section A.

A

G

20 mm100 mm

200 mm

40 mm

1.2 m


440


Internal Moment: As shown on FBD, is positive since it tends toincrease the beam’s radius of curvature.

Section Properties:

Maximum Normal Stress: Applying the curved-beam formula

Ans. = 2.01 MPa (T) (Max)

=

0.660(0.204959343 - 0.2)

0.200(10- 3)(0.2)(0.040657)(10- 3)

st =

M(R - r1)

Ar1 (r - R)

= -1.95MPa = 1.95 MPa (C)

=

0.660(0.204959343 - 0.21)

0.200(10- 3)(0.21)(0.040657)(10- 3)

sC =

M(R - r2)

Ar2(r - R)

r - R = 0.205 - 0.204959343 = 0.040657 A10- 3 B mR =

A

1AdAr

=

0.200(10- 3)

0.97580328(10- 3)= 0.204959343 m

A = (0.01)(0.02) = 0.200 A10- 3 B m2

LA dAr

= b ln r2

r1= 0.02 ln

0.210.20

= 0.97580328 A10- 3 B m

r =

0.200 + 0.2102

= 0.205 m

M = 0.660 N # m

6–154. The circular spring clamp produces a compressiveforce of 3 N on the plates. Determine the maximum bendingstress produced in the spring at A. The spring has arectangular cross section as shown.

200 mm210 mm

220 mm

10 mm

20 mm

A


441


Section Properties:

Internal Moment: The internal moment must be computed about the neutral axis asshown on FBD. is positive since it tends to increase the beam’sradius of curvature.

Allowable Normal Stress: Applying the curved-beam formula

Assume compression failure

Assume tension failure

Ans. P = 3.09 N (Controls !)

4 A106 B =

0.424959P(0.204959 - 0.2)

0.200(10- 3)(0.2)(0.040657)(10- 3)

st = sallow =

M(R - r1)

Ar1 (r - R)

P = 3.189 N

-4 A106 B =

0.424959P(0.204959 - 0.21)

0.200(10- 3)(0.21)(0.040657)(10- 3)

sc = sallow =

M(R - r2)

Ar2(r - R)

Mmax = 0.424959P

r - R = 0.205 - 0.204959343 = 0.040657 A10- 3 B mR =

A

1AdAr

=

0.200(10- 3)

0.97580328(10- 3)= 0.204959 m

A = (0.01)(0.02) = 0.200 A10- 3 B m2

LA dAr

= b ln r2

r1= 0.02 ln

0.210.20

= 0.97580328 A10- 3 B m

r =

0.200 + 0.2102

= 0.205 m

6–155. Determine the maximum compressive force thespring clamp can exert on the plates if the allowablebending stress for the clamp is sallow = 4 MPa.

200 mm210 mm

220 mm

10 mm

20 mm

A


442


From Fig. 6-44:

Ans.M = 180 288 lb # in. = 15.0 kip # ft

18(103) = 2.60 c (M)(6.25)112 (1)(12.5)3

d

smax = KMc

I

K = 2.60

br

=

10.5

= 2.0 r

h=

0.512.5

= 0.04

b =

14.5 - 12.52

= 1.0 in.

*6–156. While in flight, the curved rib on the jet plane issubjected to an anticipated moment of at thesection. Determine the maximum bending stress in the ribat this section, and sketch a two-dimensional view of thestress distribution.

M = 16 N # m

•6–157. If the radius of each notch on the plate is determine the largest moment that can be applied. Theallowable bending stress for the material is sallow = 18 ksi.

r = 0.5 in.,

0.6 m

5 mm20 mm

5 mm

30 mm

5 mm

16 N�m

12.5 in.

14.5 in.1 in.

MM

Ans.(ss)max =

M(R - rs)

ArA(r - R)=

16(0.6147933 - 0.6)

0.4(10- 3)(0.6)(0.615 - 0.6147933)= 4.77 MPa

(sc)max =

M(R - rc)

ArA(r - R)=

16(0.6147933 - 0.630)

0.4(10- 3)(0.630)(0.615 - 0.6147933)= -4.67 MPa

R =

A

1A dA>r =

0.4(10- 3)

0.650625(10- 3)= 0.6147933

A = 2(0.005)(0.03) + (0.02)(0.005) = 0.4(10- 3) in2

LA dA>r = (0.03)ln

0.6050.6

+ (0.005)ln 0.6250.605

+ (0.03)ln 0.6300.625

= 0.650625(10- 3) in.


443


From Fig. 6-44:

Ans.smax = KMc

I= 2.60 c (10)(12)(6.25)

112(1)(12.5)3

d = 12.0 ksi

K = 2.60

br

=

10.5

= 2.0 r

h=

0.512.5

= 0.04

6–158. The symmetric notched plate is subjected tobending. If the radius of each notch is and theapplied moment is determine the maximumbending stress in the plate.

M = 10 kip # ft,r = 0.5 in.

12.5 in.

14.5 in.1 in.

MM

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

wh

=

8020

= 4

K = 1.45

124 A106 B = KB 40(0.01)112 (0.007)(0.023)

R

sallow = KMc

I

6–159. The bar is subjected to a moment of Determine the smallest radius r of the fillets so that an allowable bending stress of is notexceeded.

sallow = 124 MPa

M = 40 N # m. 80 mm

20 mm 7 mm

M Mr

r

Stress Concentration Factor: From the graph in the text with

and , then .

Maximum Bending Stress:

Ans. = 54.4 MPa

= 1.45B 17.5(0.01)112 (0.007)(0.023)

R

smax = KMc

I

K = 1.45r

h=

520

= 0.25wh

=

8020

= 4

*6–160. The bar is subjected to a moment of If determine the maximum bending

stress in the material.r = 5 mm,17.5 N # m.

M = 80 mm

20 mm 7 mm

M Mr

r


444


From Fig. 6-44.

Ans.P = 122 lb

sY = K Mc

I ; 36 = 1.92 c 20P(0.625)

112 (0.5)(1.25)3

d

K = 1.92

br

=

0.250.125

= 2; r

h=

0.1251.25

= 0.1

b =

1.75 - 1.252

= 0.25

•6–161. The simply supported notched bar is subjected totwo forces P. Determine the largest magnitude of P that canbe applied without causing the material to yield.The materialis A-36 steel. Each notch has a radius of r = 0.125 in.

20 in. 20 in.

1.75 in.

0.5 in.

P P

1.25 in.

20 in. 20 in.

From Fig. 6-44,

Ans.smax = K Mc

I= 1.92 c 2000(0.625)

112 (0.5)(1.25)3

d = 29.5 ksi

K = 1.92

br

=

0.250.125

= 2; r

h=

0.1251.25

= 0.1

b =

1.75 - 1.252

= 0.25

6–162. The simply supported notched bar is subjected tothe two loads, each having a magnitude of Determine 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 r = 0.125 in.

P = 100 lb.

20 in. 20 in.

1.75 in.

0.5 in.

P P

1.25 in.

20 in. 20 in.


445


From Fig. 6-43,

Ans. L = 0.95 m = 950 mm

19.6875(106) =

175(0.2 +L2)(0.03)

112 (0.01)(0.063)

(sB)max = (sA)max =

MB c

I

(sA)max = KMAc

I= 1.5 c (35)(0.02)

112(0.01)(0.043)

d = 19.6875 MPa

K = 1.5

wh

=

6040

= 1.5 r

h=

740

= 0.175

6–163. Determine the length L of the center portion ofthe 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

BA C

7 mm

L2

L2

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 = 257 N # m

200 A106 B = 1.75B M(0.015)112 (0.015)(0.033)

R

smax = sallow = K Mc

I ;

M = 41.7 N # m (Controls !)

200 A106 B = 1.2B M(0.005)112 (0.015)(0.013)

R

smax = sallow = K Mc

I ;

K = 1.75r

h=

330

= 0.1wh

=

4530

= 1.5

K = 1.2r

h=

610

= 0.6wh

=

3010

= 3

*6–164. 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 mm

M

30 mm45 mm

3 mm6 mm


446


Ans.stop = sbottom = 293.5 - 250 = 43.5 MPa

y

250=

0.115293.5

; y = 0.09796 m = 98.0 mm

s =

Mp c

I=

211.25(103)(0.115)

82.78333(10- 6)= 293.5 MPa

= 0.000845(250)(106) = 211.25 kN # m

Mp = 0.003sY (0.215) + 0.002sY (0.1) = 0.000845 sY

C2 = T2 = sY (0.1)(0.02) = 0.002sY

C1 = T1 = sY (0.2)(0.015) = 0.003sY

Ix =

112

(0.2)(0.23)3-

112

(0.18)(0.2)3= 82.78333(10- 6)m4

•6–165. The beam is made of an elastic plastic material forwhich Determine the residual stress in thebeam at its top and bottom after the plastic moment isapplied and then released.

Mp

sY = 250 MPa.

200 mm

15 mm

15 mm

20 mm

200 mm

Mp

Plastic analysis:

Elastic analysis:

Shape factor:

Ans. =

3h

2 c4bt(h - t) + t(h - 2t)2

bh3- (b - t)(h - 2t)3 d

k =

MP

MY=

[bt(h - t) +t4 (h - 2t)2]sY

bh3- (b - t)(h - 2t)3

6h sY

=

bh3- (b - t)(h - 2t)3

6h sY

MY =

sy I

c=

sY A 112 B [bh3

- (b - t)(h - 2t)3]h2

=

112

[bh3- (b - t)(h - 2 t)3]

I =

112

bh3-

112

(b - t)(h - 2t)3

= sY cbt(h - t) +

t

4 (h - 2t)2 d

MP = sY bt(h - t) + sY ah - 2t

2b(t)ah - 2t

2b

T1 = C1 = sY bt; T2 = C2 = sY ah - 2t

2b t

6–166. The wide-flange member is made from an elastic-plastic material. Determine the shape factor.

t

b

h

t

t


447


Maximum Elastic Moment: The moment of inertia about neutral axis must bedetermined first.

Applying the flexure formula with , we have

Plastic Moment:

Shape Factor:

Ans.k =

MP

MY=

2.75a3sY

1.6111a3sY= 1.71

= 2.75a3sY

MP = sY (a)(a)(2a) + sY (0.5a)(3a)(0.5a)

MY =

sYI

c=

sY (2.41667a4)

1.5a= 1.6111a3sY

sY =

MY c

I

s = sY

INA =

112

(a)(3a)3+

112

(2a) Aa3 B = 2.41667a4

6–167. Determine the shape factor for the cross section.

aa

a

a

a

a

Maximum Elastic Moment: The moment of inertia about neutral axis must bedetermined first.

Applying the flexure formula with , we have

Ans.

Plastic Moment:

Ans. = 792 kip # in = 66.0 kip # ft

MP = 36(2)(2)(4) + 36(1)(6)(1)

= 464 kip # in = 38.7 kip # ft

MY =

sY Ic

=

36(38.667)

3

sY = =

MY c

I

s = sY

INA =

112

(2) A63 B +

112

(4) A23 B = 38.667 in4

*6–168. The beam is made of elastic perfectly plasticmaterial. Determine the maximum elastic moment and theplastic moment that can be applied to the cross section.Take and sY = 36 ksi.a = 2 in.

aa

a

a

a

a


448

Plastic Moment:

Modulus of Rupture: The modulus of rupture can be determined using the flexureformula with the application of reverse, plastic moment .

Residual Bending Stress: As shown on the diagram.

Ans. = 317.14 - 250 = 67.1 MPa

sœ

top = sœ

bot = sr - sY

sr =

MP c

I=

289062.5 (0.1)

91.14583 A10- 6 B = 317.41 MPa

= 91.14583 A10- 6 B m4

I =

112

(0.2) A0.23 B -

112

(0.15) A0.153 BMP = 289062.5 N # m

sr

= 289062.5 N # m

MP = 250 A106 B (0.2)(0.025)(0.175) + 250 A106 B (0.075)(0.05)(0.075)

•6–169. The box beam is made of an elastic perfectlyplastic material for which Determine theresidual stress in the top and bottom of the beam after theplastic moment is applied and then released.Mp

sY = 250 MPa.


25 mm

150 mm150 mm

25 mm 25 mm

25 mm

Ans.k =

Mp

MY=

0.000845sY

0.000719855sY= 1.17

MY =

sY A82.78333)10- 6 B0.115

= 0.000719855 sY

sY =

MY c

I

Mp = 0.003sY(0.215) + 0.002sY(0.1) = 0.000845 sY

C2 = T2 = sY(0.1)(0.02) = 0.002sY

C1 = T1 = sY(0.2)(0.015) = 0.003sY

Ix =

112

(0.2)(0.23)3-

112

(0.18)(0.2)3= 82.78333 A10- 6 B m4

6–170. Determine the shape factor for the wide-flange beam.

200 mm

15 mm

15 mm

20 mm

200 mm

Mp


449


Referring to Fig. a, the location of centroid of the cross-section is


Here and . Thus

Referring to the stress block shown in Fig. b,

Since , , Fig. c. Here

Thus,

Thus,

Ans.k =

MP

MY=

81 sY

47.571 sY= 1.70

MP = T(4.5) = 18 sY (4.5) = 81 sY

T = C = 3(6) sY = 18 sY

c1 = 0d = 6 in.

d = 6 in.

d(3)sY - (6 - d)(3)sY - 3(6)sY = 0

LAsdA = 0; T - C1 - C2 = 0

MY = 47.571sY

smax =

Mc

I; sY =

MY (5.25)

249.75

c = y = 5.25 insmax = sY

= 249.75 in4

I =

112

(3) A63 B + 3(6)(5.25 - 3)2+

112

(6) A33 B + 6(3)(7.5 - 5.25)2

y =

©yA

©A=

7.5(3)(6) + 3(6)(3)

3(6) + 6(3)= 5.25 in.

6–171. Determine the shape factor of the beam’s crosssection.

3 in.

3 in.1.5 in.

1.5 in.

6 in.


450

Referring to Fig. a, the location of centroid of the cross-section is


Here, and . Then

Ans.


Since , ,

Here,

Thus,

Ans.MP = T(4.5) = 648(4.5) = 2916 kip # in = 243 kip # ft

T = C = 3(6)(36) = 648 kip

c1 = 0d = 6 in.

d = 6 in.

d(3) (36) - (6 - d)(3)(36) - 3(6) (36) = 0

LAsdA = 0; T - C1 - C2 = 0

MY = 1712.57 kip # in = 143 kip # ft

smax =

Mc

I ; 36 =

MY (5.25)

249.75

¢ = y = 5.25 insmax = sY = 36 ksi

= 249.75 in4

I =

112

(3)(63) + 3(6)(5.25 - 3)2+

112

(6)(33) + 6(3)(7.5 - 5.25)2

y =

©yA

©A=

7.5(3)(6) + 3(6)(3)

3(6) + 6(3)= 5.25 in.

*6–172. The beam is made of elastic-perfectly plasticmaterial. Determine the maximum elastic moment and theplastic moment that can be applied to the cross section.Take .sY = 36 ksi


3 in.

3 in.1.5 in.

1.5 in.

6 in.


451

Ans.k =

Mp

MY=

0.00042sY

0.000268sY= 1.57

MY =

sY(26.8)(10- 6)

0.1= 0.000268sY

sY =

MYc

I

Mp = 0.0036sY(0.11) + 0.0024sY(0.01) = 0.00042sY

C2 = T2 = sY(0.01)(0.24) = 0.0024sy

C1 = T1 = sY(2)(0.09)(0.02) = 0.0036sy

Ix =

112

(0.2)(0.023) + 2a 112b(0.02)(0.23) = 26.8(10- 6)m4

•6–173. Determine the shape factor for the cross sectionof the H-beam.


200 mm

Mp 20 mm

20 mm

200 mm

20 mm

Ans.sT = sB = 392 - 250 = 142 MPa

y

250=

0.1392

; y = 0.0638 = 63.8 mm

s¿ =

Mp c

I=

105(103)(0.1)

26.8(10- 6)= 392 MPa

Mp = 0.00042(250) A106 B = 105 kN # m

Mp = 0.0036sY(0.11) + 0.0024sY(0.01) = 0.00042sY

C2 = T2 = sY(0.01)(0.24) = 0.0024sy

C1 = T1 = sY(2)(0.09)(0.02) = 0.0036sy

Ix =

112

(0.2)(0.023) + 2a 112b(0.02)(0.23) = 26.8(10- 6)m4

6–174. The H-beam is made of an elastic-plastic materialfor which . Determine the residual stress inthe top and bottom of the beam after the plastic moment

is applied and then released.Mp

sY = 250 MPa

200 mm

Mp 20 mm

20 mm

200 mm

20 mm


452


Here, and . Then

Referring to the stress block shown in Fig. a,

Thus,

Ans.k =

MP

MY=

74.25 sY

43.5 sY= 1.71

= 9sY(6) + 13.5sY(1.5) = 74.25 sY

MP = T1(6) + T2(1.5)

T2 = C2 = 1.5(9)sY = 13.5 sY

T1 = C1 = 3(3)sY = 9 sY

MY = 43.5 sY

smax =

Mc

I; sY =

MY(4.5)

195.75

c = 4.5 insmax = sY

I =

112

(3)(93) +

112

(6) (33) = 195.75 in4

6–175. Determine the shape factor of the cross section.


3 in.

3 in.

3 in.

3 in. 3 in.3 in.


453



Here, and . Then

Ans.

Referring to the stress block shown in Fig. a,

Thus,

Ans. = 2673 kip # in. = 222.75 kip # ft = 223 kip # ft

= 324(6) + 486(1.5)

MP = T1(6) + T2(1.5)

T2 = C2 = 1.5(9)(36) = 486 kip

T1 = C1 = 3(3)(36) = 324 kip

MY = 1566 kip # in = 130.5 kip # ft

smax =

Mc

I; 36 =

MY (4.5)

195.75

c = 4.5 insmax = sY = 36 ksi

I =

112

(3)(93) +

112

(6)(33) = 195.75 in4

*6–176. The beam is made of elastic-perfectly plasticmaterial. Determine the maximum elastic moment and theplastic moment that can be applied to the cross section.Take sY = 36 ksi .

3 in.

3 in.

3 in.

3 in. 3 in.3 in.


454

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

Here, and ,

The plastic Moment of the table’s cross-section can be determined by super posingthe moment of the stress block of the solid circular cross-section with radius

and as shown in Figure a, Here,

Thus,

Ans.k =

MP

MY=

121.33 sY

87.83 sY= 1.38

= 121.33 sY

= (18psY)a16pb - 12.5psYa 40

3pb

MP = T1b2 c4(6)

3pd r - T2b2 c4(5)

3pd r

T2 = C2 =

12

p(52)sY = 12.5p sY

T1 = C1 =

12

p(62)sY = 18psY

ri = 5 in.ro = 6 in

MY = 87.83 sY

smax =

Mc

I ; sY =

MY (6)

167.75 p

C = ro = 6 insmax = sY

I =

p

4 Aro

4- ri

4 B =

p

4 A64

- 54 B = 167.75 p in4

•6–177. Determine the shape factor of the cross sectionfor the tube.


6 in.

5 in.


455


Maximum Elastic Moment. The moment of inertia of the cross-section about theneutral axis is

With and ,

Plastic Moment. The plastic moment of the cross section can be determined bysuperimposing the moment of the stress block of the solid beam with radius r0 and ri

as shown in Fig. a, Referring to the stress block shown in Fig. a,

Shape Factor.

Ans.k =

MP

MY=

43

Aro 3

- ri 3 BsY

p

4ro Aro

4- ri

4 BsY

=

16ro Aro 3

- ri 3 B

3p Aro 4

- ri 4 B

=

43

Aro 3

- ri 3 BsY

=

p

2 ro

2sYa8ro

3pb -

p

2ri

2sYa 8ri

3pb

MP = T1 c2a4ro

3pb d - T2 c2a 4ri

3pb d

T2 = c2 =

p

2 ri

2sY

T1 = c1 =

p

2 ro

2sY

MY =

p

4ro Aro

4- ri

4 BsY

smax =

Mc

I ; sY =

MY(ro)

p

4 Aro

4- ri

4 B

smax = sYc = ro

I =

p

4 Aro

4- ri

4 B

6–178. The beam is made from elastic-perfectly plasticmaterial. Determine the shape factor for the thick-walled tube.

ri

ro


456

Plastic analysis:

Elastic analysis:

Shape factor:

Ans.k =

Mp

MY=

bh2

12 sY

bh2

24 sY

= 2

MY =

sYI

c=

sY Abh3

48 Bh2

=

b h2

24 sY

I = 2 c 112

(b)ah

2b3 d =

b h3

48

MP =

b h4

sYah

3b =

b h2

12 sY

T = C =

12

(b)ah

2bsY =

b h4

sY

6–179. Determine the shape factor for the member.


–2

–2

h

b

h

Elastic analysis:

Ans.

Plastic analysis:

Ans.Mp = 2160a63b = 432 kip # in. = 36 kip # ft

T = C =

12

(4)(3)(36) = 216 kip

MY =

sYI

c=

36(18)

3= 216 kip # in. = 18 kip # ft

I = 2 c 112

(4)(3)3 d = 18 in4

*6–180. The member is made from an elastic-plasticmaterial. Determine the maximum elastic moment and theplastic moment that can be applied to the cross section.Take sY = 36 ksi.h = 6 in.,b = 4 in.,

–2

–2

h

b

h


457


Ans.M =

12sYa2

3 hb(a)a11

18 hb =

11a h2

54 sY

d =

23

h

12

sY(d)(a) - sY(h - d)a = 0

LAsdA = 0; C - T = 0

•6–181. The beam is made of a material that can beassumed perfectly plastic in tension and elastic perfectlyplastic in compression. Determine the maximum bendingmoment M that can be supported by the beam so that thecompressive material at the outer edge starts to yield.

�

h

a

MsY

�sY

Elastic analysis:

Ans.

Plastic analysis:

Ans.w0 = 22.8 kip>ft27w0(12) = 7400

MP = 400(14) + 300(6) = 7400 kip # in.

C2 = T2 = 25(6)(2) = 300 kip

C1 = T1 = 25(8)(2) = 400 kip

w0 = 18.0 kip>ftMmax =

sYI

c ; 27w0(12) =

25(1866.67)

8

I =

112

(8)(163) -

112

(6)(123) = 1866.67 in4

6–182. The box beam is made from an elastic-plasticmaterial for which . Determine the intensity ofthe distributed load that will cause the moment to be (a) the largest elastic moment and (b) the largest plasticmoment.

w0

sY = 25 ksi

9 ft

16 in.12 in.

8 in.

w0

6 in.

9 ft


458

From the moment diagram shown in Fig. a, .

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

Here, and

It is required that

Ans.


Thus,

It is required that

Ans. P = 45.5 kip

6P = 273

Mmax = MP

= 3276 kip # in = 273 kip # ft

= 216(11) + 180(5)

MP = T1(11) + T2(5)

T2 = C2 = 5(1)(36) = 180 kip

T1 = C1 = 6(1)(36) = 216 kip

P = 37.28 kip = 37.3 kip

6P = 223.67

Mmax = MY

MY = 2684 kip # in = 223.67 kip # ft

smax =

Mc

I ; 36 =

MY (6)

447.33

c = 6 in.smax = sY = 36 ksi

I =

112

(6)(123) -

112

(5)(103) = 447.33 in4

Mmax = 6 P

6–183. The box beam is made from an elastic-plasticmaterial for which . Determine the magnitudeof each concentrated force P that will cause the moment tobe (a) the largest elastic moment and (b) the largest plasticmoment.

sY = 36 ksi


6 ft 8 ft

12 in.10 in.

6 in.

5 in.

6 ft

PP


459


Maximum Internal Moment: The maximum internal moment occurs atthe mid span as shown on FBD.

Stress–Strain Relationship: Using the stress–strain relationship. the bending stresscan be expressed in terms of y using .

When , and

Resultant Internal Moment: The resultant internal moment M can be evaluated

from the integal .

Equating

Ans.P = 0.100 kip = 100 lb

M = 4.00P(12) = 4.798

= 4.798 kip # in

= 80B1 + (0.0225)2y2

2(0.0225)2 tan- 1 (0.0225y) -

y

2(0.0225)R 2 2in.

0

= 80L

2in

0 y tan- 1 (0.0225y) dy

= 2L

2in

0 y C20 tan- 1 (0.0225y) D(2dy)

M = 2LA

ysdA

LA ysdA

smax = 0.8994 ksiy = 2 in.emax = 0.003 in.>in.

= 20 tan- 1 (0.0225y)

= 20 tan- 1 [15(0.0015y)]

s = 20 tan- 1 (15e)

e = 0.0015y

M = 4.00P

*6–184. The beam is made of a polyester that has thestress–strain curve shown. If the curve can be representedby the equation where is in radians, determine the magnitude of the force P thatcan be applied to the beam without causing the maximumstrain in its fibers at the critical section to exceedPmax = 0.003 in.>in.

tan-1115P2s = [20 tan-1115P2] ksi,

8 ft 8 ft

P

2 in.

4 in.

P(in./in.)

�s(ksi)s � 20 tan�1(15 P)


460

Ultimate Moment:

Assume. ;

From the strain diagram,

From the stress–strain diagram,

(OK! Close to assumed value)

Therefore,

Ans. = 94.7 N # m

M = 7233.59(0.0064442) + 2066.74(0.0034445) + 5166.85(0.0079255)

y3 =

0.0103342

+ c1 -

13

a2(40) + 60

40 + 60b d a0.010334

2b = 0.0079225m

y2 =

23

a0.0103342

b = 0.0034445 m

y1 =

23

(0.02 - 0.010334) = 0.0064442 m

T2 = 40 A106 B c12

(0.02)a0.0103342

b d = 2066.74 N

T1 =

12

(60 + 40) A106 B c(0.02)a0.0103342

b d = 5166.85 N

C = 74.833 A106 B c12

(0.02 - 0.010334)(0.02) d = 7233.59 N

s

0.037417=

800.04 s = 74.833 MPa

e

0.02 - 0.010334=

0.040.010334 e = 0.037417 mm>mm

d = 0.010334 ms = 74.833 MPa

s - 50s d - 3500(106)d = 0

s c12

(0.02 - d)(0.02) d - 40 A106 B c12

ad

2b(0.02) d -

12

(60 + 40) A106 B c(0.02) d

2d = 0

LA s dA = 0; C - T2 - T1 = 0

•6–185. The plexiglass bar has a stress–strain curve thatcan be approximated by the straight-line segments shown.Determine the largest moment M that can be applied to thebar before it fails.


20 mm

20 mm

M

�0.06 �0.04

0.02 0.04

60

�80

compression

tension

failure

s (MPa)

P (mm/mm)

�100

40


461


a) Maximum Elastic Moment : Since the stress is linearly related to strain up topoint A, the flexure formula can be applied.

Ans.

b) The Ultimate Moment :

Ans.

Note: The centroid of a trapezodial area was used in calculation of moment.

= 718.125 kip # in = 59.8 kip # ft

M = 360(1.921875) + 52.5(0.5)

C2 = T2 =

12

(140)(0.375)(2) = 52.5 kip

C1 = T1 =

12

(140 + 180)(1.125)(2) = 360 kip

= 420 kip # in = 35.0 kip # ft

=

140 C 112 (2)(33) D

1.5

M =

sA I

c

sA =

Mc

I

6–186. The stress–strain diagram for a titanium alloy canbe approximated by the two straight lines. If a strut made ofthis material is subjected to bending, determine the momentresisted by the strut if the maximum stress reaches a valueof (a) and (b) .sBsA 3 in.

M

2 in.

0.040.01

sB � 180

sA � 140

B

A

P (in./in.)

�s (ksi)


462

Ans.M = 251 N # m

M = 0.47716 A106 BL

0.05

0y5>4dy = 0.47716 A106 B a4

5b(0.05)9>4

M =

LAy s dA = 2

L

0.05

0y(7.9527) A106 By1>4(0.03)dy

s = 10 A106 B(0.4)1>4y1>4

e = 0.4 y

0.020.05

=

e

y

smax = 10 A106 B(0.02)1>4= 3.761 MPa

emax = 0.02

6–187. A beam is made from polypropylene plastic and hasa stress–strain diagram that can be approximated by the curveshown. If the beam is subjected to a maximum tensile andcompressive strain of determine themaximum moment M.

P = 0.02 mm>mm,


P (mm/mm)

M

M100 mm

30 mm

�s(Pa)

s� 10(106)P1/4

Ans.M = 4000(0.3) + 2000(0.1333) = 1467 kN # m = 1.47 MN # m

C2 = T2 =

12

(200) A106 B(0.1)(0.2) = 2000 kN

C1 = T1 = 200 A106 B(0.1)(0.2) = 4000 kN

*6–188. The beam has a rectangular cross section and ismade of an elastic-plastic material having a stress–straindiagram as shown. Determine the magnitude of themoment M that must be applied to the beam in order tocreate a maximum strain in its outer fibers of P max = 0.008.

400 mm

200 mm

M

0.004

200

P (mm/mm)

�s(MPa)


463


Ans.

Note: The centroid of a trapezodial area was used in calculation of moment areas.

= 882.09 kip # in. = 73.5 kip # ft

M = 81(3.6680) + 266(2.1270) + 36(0.5333)

C3 = T3 =

12

(0.4)(60)(3) = 36 kip

C2 = T2 =

12

(1.2666)(60 + 80)(3) = 266 kip

C1 = T1 =

12

(0.3333)(80 + 82)(3) = 81 kip

s - 800.03 - 0.025

=

90 - 800.05 - 0.025

; s = 82 ksi

•6–189. The bar is made of an aluminum alloy having astress–strain diagram that can be approximated by thestraight line segments shown. Assuming that this diagram isthe same for both tension and compression, determine themoment the bar will support if the maximum strain at thetop and bottom fibers of the beam is P max = 0.03.

90

0.050.006 0.025

80

604 in. M

3 in.P(in./ in.)

�s(ksi)

Section Properties:


Resultant Force:

Ans. = 5883 N = 5.88 kN

FR =

12

(1.0815 + 1.6234) A106 B (0.015)(0.29)

sA =

650(0.044933)

17.99037(10- 6)= 1.6234 MPa

sB =

650(0.044933 - 0.015)

17.99037(10- 6)= 1.0815 MPa

s =

My

I

= 17.99037 A10- 6 B m4

+

112

(0.04) A0.1253 B + 0.04(0.125)(0.0775 - 0.044933)2

INA =

112

(0.29) A0.0153 B + 0.29(0.015) (0.044933 - 0.0075)2

= 0.044933 m

y =

0.0075(0.29)(0.015) + 2[0.0775(0.125)(0.02)]

0.29(0.015) + 2(0.125)(0.02)


determine the resultant force the bendingstress produces on the top board.M = 650 N # m,

M � 650 N�m

250 mm

15 mm

125 mm 20 mm

20 mm


464


Section Properties:


Ans.

Ans.(smax)c =

650(0.044933)

17.99037(10- 6)= 1.62 MPa (C)

(smax)t =

650(0.14 - 0.044933)

17.99037(10- 6)= 3.43 MPa (T)

s =

My

I

= 17.99037 A10- 6 B m4

+

112

(0.04) A0.1253 B + 0.04(0.125)(0.0775 - 0.044933)2

INA =

112

(0.29) A0.0153 B + 0.29(0.015)(0.044933 - 0.0075)2

= 0.044933 m

y =

0.0075(0.29)(0.015) + 2[0.0775(0.125)(0.02)]

0.29(0.015) + 2(0.125)(0.02)

6–191. The beam is made from three boards nailed togetheras shown. Determine the maximum tensile and compressivestresses in the beam.

M � 650 N�m

250 mm

15 mm

125 mm 20 mm

20 mm


465


a

Ans.smax =

Mc

I=

32(0.05)

2.52109(10- 6)= 635 kPa

Iz =

112

(0.075)(0.0153) + 2a 112b(0.015)(0.13) = 2.52109(10- 6)m4

M = 32 N # m

+ ©M = 0; M - 80(0.4) = 0

*6–192. Determine the bending stress distribution inthe beam at section a–a. Sketch the distribution in threedimensions acting over the cross section.

15 mm

400 mm

80 N80 N

15 mm

100 mm

75 mm

80 N 80 N

400 mm300 mm300 mm

a

a

Failure of wood :

Failure of steel :

Ans. M = 14.9 kN # m (controls)

130(106) =

18.182(M)(0.0625)

0.130578(10- 3)

(sst)max =

nMc

I

20(106) =

M(0.0625)

0.130578(10- 3) ; M = 41.8 kN # m

(sw)max =

Mc

I

I =

112

(0.80227)(0.1253) = 0.130578(10- 3)m4

n =

Est

Ew=

200(109)

11(109)= 18.182

•6–193. The composite beam consists of a wood core andtwo plates of steel. If the allowable bending stress forthe wood is , and for the steel

, determine the maximum moment thatcan be applied to the beam. Ew = 11 GPa, Est = 200 GPa.(sallow)st = 130 MPa

(sallow)w = 20 MPa

125 mm

20 mm

20 mm75 mm

z

x

y

M


466


Failure of wood :

Failure of steel :

Ans. M = 26.4 kN # m (controls)

130(106) =

M(0.0575)

11.689616(10- 6)

(sst)max =

Mc1

I

20(106) =

0.055(M)(0.0375)

11.689616(10- 6); M = 113 kN # m

(sw)max =

nMc2

I

I =

112

(0.125)(0.1153) -

112

(0.118125)(0.0753) = 11.689616(10- 6)

n =

11(109)

200(104)= 0.055

6–194. Solve Prob. 6–193 if the moment is applied aboutthe y axis instead of the z axis as shown.

125 mm

20 mm

20 mm75 mm

z

x

y

M


467


Maximum Bending Stress: The moment of inertia about y axis must bedetermined first 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 -

32

y2(100 - y)32 -

815

y(100 - y)52 -

16105

(100 - y)72R 2 100 mm

0

M =

smax

5 L

100mm

0y22100 - y dy

dM = 2[y(s dA)] = 2by c asmax

100by d(2z dy) r

smax =

Mc

I=

125(0.1)

30.4762(10- 6)= 0.410 MPa

= 30.4762 A10- 6 B mm4= 30.4762 A10- 6 B m4

= 20B -

32

y2 (100 - y)

32 -

815

y (100 - y)52 -

16105

(100 - y)72R 2 100 mm

0

= 20L

100mm

0y22100 - y dy

= 2L

100mm

0y2 (2z) dy

I =

LA y

2 dA

6–195. A shaft is made of a polymer having a paraboliccross section. If it resists an internal moment of

, determine the maximum bending stressdeveloped in the material (a) using the flexure formula and(b) using integration. Sketch a three-dimensional view ofthe stress distribution acting over the cross-sectional area.Hint: The moment of inertia is determined using Eq. A–3 ofAppendix A.

M = 125 N # my

z

x

M � 125 N· m

50 mm

100 mm

50 mm

y � 100 – z

2/ 25


468

a

Ans.smax =

Mc

I=

394.14(0.375)112 (0.5)(0.753)

= 8.41 ksi

M = 394.14 lb # in.

+ ©M = 0; M - 45(5 + 4 cos 20°) = 0

*6–196. Determine the maximum bending stress in thehandle of the cable cutter at section a–a. A force of 45 lb isapplied to the handles. The cross-sectional area is shown inthe figure.


4 in.

45 lb20�

a

a

3 in.

5 in.

A

45 lb

0.75 in.

0.50 in.

Ans.

Ans.sB =

M(R - rB)

ArB(r - R)=

85(0.484182418 - 0.40)

6.25(10- 3)(0.40)(0.010817581)= 265 kPa (T)

sA = 225 kPa (C)

sA =

M(R - rA)

ArA(r - R)=

85(0.484182418 - 0.59)

6.25(10- 3)(0.59)(0.010817581)= -225.48 kPa

r - R = 0.495 - 0.484182418 = 0.010817581 m

R =

A

L

A

dAr

=

6.25(10- 3)

0.012908358= 0.484182418 m

A = 2(0.1)(0.02) + (0.15)(0.015) = 6.25(10- 3) m2

= 0.012908358 m

LA

dAr

= b ln r2

r1= 0.1 ln

0.420.40

+ 0.015 ln 0.570.42

+ 0.1 ln 0.590.57

•6–197. The curved beam is subjected to a bendingmoment of as shown. Determine the stress atpoints A and B and show the stress on a volume elementlocated at these points.

M = 85 N # m

30�

M � 85 N�m

B

A

100 mm

150 mm

20 mm

20 mm

15 mm

400 mm

B

A


469


Ans.

c

Ans. M = -x2+ 20x - 166

+ ©MNA = 0; 20x - 166 - 2xax

2b - M = 0

V = 20 - 2x

+ c ©Fy = 0; 20 - 2x - V = 0

6–198. Draw the shear and moment diagrams for thebeam and determine the shear and moment in the beam asfunctions of x, where 0 … x 6 6 ft .

6 ft 4 ft

2 kip/ ft

50 kip�ft

8 kip

x

6–199. 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


470

Assume failure due to tensile stress :

Assume failure due to compressive stress:

Ans.M = 30.0 kip # in. = 2.50 kip # ft (controls)

smax =

Mc

I ; 15 =

M(3.4641 - 1.1547)

4.6188

M = 88.0 kip # in. = 7.33 kip # ft

smax =

My

I; 22 =

M(1.1547)

4.6188

I =

136

(4)(242- 22)3

= 4.6188 in4

y (From base) =

13242

- 22= 1.1547 in.

*6–200. A member has the triangular cross sectionshown. Determine the largest internal moment M that canbe applied to the cross section without exceeding allowabletensile and compressive stresses of and

, respectively.(sallow)c = 15 ksi(sallow)t = 22 ksi


2 in.2 in.

4 in.

M4 in.


471


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.

Ans.


Ans. a = 45°

tan a = (1) tan(45°)

tan a =

Iz

Iy tan u

u = 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 (-a2)

112 a4

s = -

Mzy

Iz+

My z

Iy

Iy = Iz =

112

a4

Mz = -M cos u My = -M sin u

•6–201. The strut has a square cross section a by a and issubjected to the bending moment M applied at an angle as 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


Mx

z

y

a

a

�


472



From Fig. a,

Applying the shear formula,

Ans.

The shear stress component at A is represented by the volume element shown inFig. b.

= 2.559(106) Pa = 2.56 MPa

tA =

VQA

It=

20(103)[0.64(10- 3)]

0.2501(10- 3)(0.02)

QA = y¿A¿ = 0.16 (0.02)(0.2) = 0.64(10- 3) m3

I =

112

(0.2)(0.343) -

112

(0.18)(0.33) = 0.2501(10- 3) m4

•7–1. If the wide-flange beam is subjected to a shear ofdetermine the shear stress on the web at A.

Indicate the shear-stress components on a volume elementlocated at this point.

V = 20 kN,

A

BV

20 mm

20 mm

20 mm

300 mm

200 mm

200 mm

07 Solutions 46060 5/26/10 2:04 PM Page 472

473



From Fig. a.

The maximum shear stress occurs at the points along neutral axis since Q ismaximum and thicknest t is the smallest.

Ans. = 3.459(106) Pa = 3.46 MPa

tmax =

VQmax

It=

20(103) [0.865(10- 3)]

0.2501(10- 3) (0.02)

Qmax = ©y¿A¿ = 0.16 (0.02)(0.2) + 0.075 (0.15)(0.02) = 0.865(10- 3) m3

I =

112

(0.2)(0.343) -

112

(0.18)(0.33) = 0.2501(10- 3) m4

7–2. If the wide-flange beam is subjected to a shear ofdetermine the maximum shear stress in the beam.V = 20 kN,

A

BV

20 mm

20 mm

20 mm

300 mm

200 mm

200 mm



For , Fig. a, Q as a function of y is

For , . Thus.

The sheer force resisted by the web is,

Ans. = 18.95 (103) N = 19.0 kN

Vw = 2L

0.15 m

0tdA = 2

L

0.15 m

0C3.459(106) - 39.99(106) y2 D (0.02 dy)

= E3.459(106) - 39.99(106) y2F Pa.

t =

VQ

It=

20(103) C0.865(10- 3) - 0.01y2 D0.2501(10- 3) (0.02)

t = 0.02 m0 … y 6 0.15 m

= 0.865(10- 3) - 0.01y2

Q = ©y¿A¿ = 0.16 (0.02)(0.2) +

12

(y + 0.15)(0.15 - y)(0.02)

0 … y 6 0.15 m

I =

112

(0.2)(0.343) -

112

(0.18)(0.33) = 0.2501(10- 3) m4

7–3. If the wide-flange beam is subjected to a shear ofdetermine the shear force resisted by the web

of the beam.V = 20 kN,

474


A

BV

20 mm

20 mm

20 mm

300 mm

200 mm

200 mm


475


Section Properties:

Shear Stress: Applying the shear formula

Ans.

Ans.

Ans.(tAB)W =

VQAB

I tW=

12(64.8)

390.60(4)= 0.498 ksi

(tAB)f =

VQAB

Itf=

12(64.8)

390.60(12)= 0.166 ksi

tmax =

VQmax

It=

12(64.98)

390.60(4)= 0.499 ksi

t =

VQ

It

QAB = yœ

2 A¿ = 1.8(3)(12) = 64.8 in3

Qmax = yœ

1 A¿ = 2.85(5.7)(4) = 64.98 in3

= 390.60 in4

INA =

112

(12) A33 B + 12(3)(3.30 - 1.5)2+

112

(4) A63 B + 4(6)(6 - 3.30)2

y =

©yA

©A=

1.5(12)(3) + 6(4)(6)

12(3) + 4(6)= 3.30 in.

*7–4. If the T-beam is subjected to a vertical shear ofdetermine the maximum shear stress in the

beam. Also, compute the shear-stress jump at the flange-web junction AB. Sketch the variation of the shear-stressintensity over the entire cross section.

V = 12 kip,

BB

V � 12 kip

6 in.

3 in.

4 in.

4 in.4 in.

A


Section Properties:

Shear Stress: Applying the shear formula

Resultant Shear Force: For the flange

Ans. = 3.82 kip

=

L

3.3 in

0.3 inA0.16728 - 0.01536y2 B(12dy)

Vf =

LAtdA

= 0.16728 - 0.01536y2

t =

VQ

It=

12(65.34 - 6y2)

390.60(12)

Q = y¿A¿ = (1.65 + 0.5y)(3.3 - y)(12) = 65.34 - 6y2

= 390.60 in4

INA =

112

(12) A33 B + 12(3)(3.30 - 1.5)2+

112

(4) A63 B + 6(4)(6 - 3.30)2

y =

©yA

©A=

1.5(12)(3) + 6(4)(6)

12(3) + 4(6)= 3.30 in.

•7–5. If the T-beam is subjected to a vertical shear ofdetermine the vertical shear force resisted by

the flange.V = 12 kip,

476


BB

V � 12 kip

6 in.

3 in.

4 in.

4 in.4 in.

A


477


Ans.

Ans.tB =

VQB

I t=

15(103)(0.59883)(10- 3)

0.218182(10- 3)0.025)= 1.65 MPa

tA =

VQA

I t=

15(103)(0.7219)(10- 3)

0.218182(10- 3)(0.025)= 1.99 MPa

QB = yAœ

B = (0.1747 - 0.015)(0.125)(0.03) = 0.59883 (10- 3) m3

QA = yAœ

A = (0.310 - 0.015 - 0.1747)(0.2)(0.03) = 0.7219 (10- 3) m3

+

112

(0.2)(0.033) + 0.2(0.03)(0.295 - 0.1747)2= 0.218182 (10- 3) m4

+

112

(0.025)(0.253) + 0.25(0.025)(0.1747 - 0.155)2

I =

112

(0.125)(0.033) + 0.125(0.03)(0.1747 - 0.015)2

y =

(0.015)(0.125)(0.03) + (0.155)(0.025)(0.25) + (0.295)(0.2)(0.03)

0.125(0.03) + (0.025)(0.25) + (0.2)(0.03)= 0.1747 m

7–6. If the beam is subjected to a shear of determine the web’s shear stress at A and B. Indicate theshear-stress components on a volume element locatedat these points. Show that the neutral axis is located at

from the bottom and INA = 0.2182110-32 m4.y = 0.1747 m

V = 15 kN,

A

B

V

30 mm25 mm

30 mm

250 mm

200 mm

125 mm

A

BV

30 mm25 mm

30 mm

250 mm

200 mm

200 mm

Section Properties:

Ans.tmax =

VQ

I t=

30(10)3(1.0353)(10)- 3

268.652(10)- 6 (0.025)= 4.62 MPa

Qmax = © y¿A = 0.0625(0.125)(0.025) + 0.140(0.2)(0.030) = 1.0353(10)- 3 m3

I =

112

(0.2)(0.310)3-

112

(0.175)(0.250)3= 268.652(10)- 6 m4

7–7. If the wide-flange beam is subjected to a shear ofdetermine the maximum shear stress in the beam.V = 30 kN,


478


Ans.Vw = 30 - 2(1.457) = 27.1 kN

Vf = 1.457 kN

= 11.1669(10)6[ 0.024025y -

12

y3 ]0.155

0.125

Vf =

Ltf dA = 55.8343(10)6

L

0.155

0.125 (0.024025 - y2)(0.2 dy)

tf =

30(10)3(0.1)(0.024025 - y2)

268.652(10)- 6 (0.2)

Q = a

0.155 + y

2 b(0.155 - y)(0.2) = 0.1(0.024025 - y2)

I =

112

(0.2)(0.310)3-

112

(0.175)(0.250)3= 268.652(10)- 6 m4

*7–8. If the wide-flange beam is subjected to a shear ofdetermine the shear force resisted by the web

of the beam.V = 30 kN,

A

BV

30 mm25 mm

30 mm

250 mm

200 mm

200 mm

Ans.V = 32132 lb = 32.1 kip

8 (103) = -

V (3.3611)

6.75 (2)(1)

tmax = tallow =

VQmax

I t

Qmax = ©y¿A¿ = 2 (0.91665)(1.8333)(1) = 3.3611 in3

+ 2 a 112b(1)(23) + 2 (1)(2)(2 - 1.1667)2

= 6.75 in4

I =

112

(5)(13) + 5 (1)(1.1667 - 0.5)2

y =

(0.5)(1)(5) + 2 [(2)(1)(2)]

1 (5) + 2 (1)(2)= 1.1667 in.

•7–9. Determine the largest shear force V that the membercan sustain if the allowable shear stress is tallow = 8 ksi.

V3 in. 1 in.

1 in.

1 in.

3 in.


479


Ans.tmax =

VQmax

I t=

18(3.3611)

6.75 (2)(1)= 4.48 ksi

Qmax = ©y¿A¿ = 2 (0.91665)(1.8333)(1) = 3.3611 in3

+ 2 a 112b(1)(23) + 2 (1)(2)(2 - 1.1667) = 6.75 in4

I =

112

(5)(13) + 5 (1)(1.1667 - 0.5)2

y =

(0.5)(1)(5) + 2 [(2)(1)(2)]

1 (5) + 2 (1)(2)= 1.1667 in.

7–10. If the applied shear force determine themaximum shear stress in the member.

V = 18 kip,

V3 in. 1 in.

1 in.

1 in.

3 in.

Ans.V = 100 kN

7(106) =

V[(0.075)(0.1)(0.05) + 2(0.05)(0.1)(0.05)]

125(10- 6)(0.1)

tallow =

VQmax

It

I =

112

(0.2)(0.2)3-

112

(0.1)(0.1)3= 125(10- 6) m4

7–11. The wood beam has an allowable shear stress ofDetermine the maximum shear force V that

can be applied to the cross section.tallow = 7 MPa.

50 mm

50 mm

200 mm

100 mm50 mm

V

50 mm


480


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

Q as the function of y, Fig. a,

Qmax occurs when . Thus,

The maximum shear stress occurs of points along the neutral axis since Q ismaximum and the thickness is constant.

Ans.

Thus, the shear stress distribution as a function of y is

= E5.56 (36 - y2)F psi

t =

VQ

It=

12.8(103) C4(36 - y2) D1152 (8)

V = 12800 16 = 12.8 kip

tallow =

VQmax

It; 200 =

V(144)

1152(8)

t = 8 in.

Qmax = 4(36 - 02) = 144 in3

y = 0

Q =

12

(y + 6)(6 - y)(8) = 4 (36 - y2)

I =

112

(8) (123) = 1152 in4

*7–12. The beam has a rectangular cross section and ismade of wood having an allowable shear stress of 200 psi. Determine the maximum shear force V that can bedeveloped in the cross section of the beam. Also, plot theshear-stress variation over the cross section.

tallow =

V

12 in.

8 in.


481


Section Properties:

Maximum Shear Stress: Maximum shear stress occurs at the point where theneutral axis passes through the section.

Applying the shear formula

Ans. = 4 22 MPa

=

20(103)(87.84)(10- 6)

5.20704(10- 6)(0.08)

tmax =

VQmax

It

= 87.84 A10- 6 B m3

= 0.015(0.08)(0.03) + 0.036(0.012)(0.12)

Qmax = ©y¿A¿

= 5.20704 A10- 6 B m4

INA =

112

(0.12) A0.0843 B -

112

(0.04) A0.063 B

7–13. Determine the maximum shear stress in the strut ifit is subjected to a shear force of V = 20 kN.

V60 mm

12 mm

20 mm

20 mm

80 mm

12 mm

Section Properties:

Allowable shear stress: Maximum shear stress occurs at the point where the neutralaxis passes through the section.


Ans. V = 189 692 N = 190 kN

40 A 106 B =

V(87.84)(10- 6)

5.20704(10- 6)(0.08)

tmax = tallow =

VQmax

It

= 87.84 A 10- 6 B m3

= 0.015(0.08)(0.03) + 0.036(0.012)(0.12)

Qmax = ©y¿A¿

= 5.20704 A 10- 6 B m4

INA =

112

(0.12) A0.0843 B -

112

(0.04) A0.063 B

7–14. Determine the maximum shear force V that thestrut can support if the allowable shear stress for thematerial is tallow = 40 MPa.

V60 mm

12 mm

20 mm

20 mm

80 mm

12 mm


482


The maximum shear stress occur when

Ans.The faector =

tmax

tavg=

4V3 pc2

Vpc2

=

43

tavg =

V

A=

V

p c2

tmax =

4V

3 p c2

y = 0

t =

VQ

I t=

V[23 (c2

- y2)32]

(p4 c4)(22c2- y2)

=

4V

3pc4 [c2- y2)

Q =

L

x

y2y2c2

- y2 dy = - 23

(c2- y2)

32 | x

y =

23

(c2- y2)

23

dQ = ydA = 2y2c2- y2 dy

dA = 2 x dy = 22c2- y2 dy

t = 2 x = 22c2- y2

x = 2c2- y2 ; I =

p

4 c4

7–15. Plot the shear-stress distribution over the crosssection of a rod that has a radius c. By what factor is themaximum shear stress greater than the average shear stressacting over the cross section?

c

V

y


483


Ans.

No, because the shear stress is not perpendicular to the boundary. See Sec. 7-3.

tmax =

3V

ah

tmax =

24V

a2h aa

4b a1 -

2a

aa

4b b

y =

2ha

aa

4b =

h

2

At x =

a

4

dt

dx=

24V

a2h2 a1 -

4a

xb = 0

t =

24V(x -2a x2)

a2h

t =

VQ

It=

V(4h2>3a)(x2)(1 -2xa )

((1>36)(a)(h3))(2x)

t = 2x

Q = a 4h2

3a b(x2)a1 -

2xab

Q =

LA¿

y dA = 2 c a 12

b(x)(y)a 23

h -

23

yb d

y

x=

h

a>2 ; y =

2ha

x

I =

136

(a)(h)3

*7–16. A member has a cross section in the form of anequilateral triangle. If it is subjected to a shear force V,determine the maximum average shear stress in the memberusing the shear formula. Should the shear formula actually beused to predict this value? Explain. V

a

h


484



From Fig. a,

The maximum shear stress occurs at the points along the neutral axis since Q ismaximum and thickness is the smallest.

Ans. = 37.36(106) Pa = 37.4 MPa

tmax =

VQmax

It=

600(103)[1.09125(10- 3)]

0.175275(10- 3) (0.1)

t = 0.1 m

= 1.09125(10- 3) m3

Qmax = ©y¿A¿ = 0.09(0.03)(0.3) + 0.0375(0.075)(0.1)

I =

112

(0.3)(0.213) -

112

(0.2)(0.153) = 0.175275(10- 3) m4

•7–17. Determine the maximum shear stress in the strut ifit is subjected to a shear force of V = 600 kN.

V

150 mm

30 mm

100 mm100 mm

100 mm

30 mm


485



From Fig. a

The maximum shear stress occeurs at the points along the neutral axis since Q ismaximum and thickness is the smallest.

Ans. V = 722.78(103) N = 723 kN

tallow =

VQmax

It ; 45(106) =

V C1.09125(10- 3) D0.175275(10- 3)(0.1)

t = 0.1 m

= 1.09125 (10- 3) m3

Qmax = ©y¿A¿ = 0.09(0.03)(0.3) + 0.0375 (0.075)(0.1)

I =

112

(0.3)(0.213) -

112

(0.2)(0.153) = 0.175275 (10- 3) m4

7–18. Determine the maximum shear force V that the strutcan support if the allowable shear stress for the material istallow = 45 MPa.

V

150 mm

30 mm

100 mm100 mm

100 mm

30 mm


486



For , Fig. a, Q as a function of y is

For , Fig. b, Q as a function of y is0 … y 6 0.075 m

Q = y¿A¿ =

12

(0.105 + y) (0.105 - y)(0.3) = 1.65375(10- 3) - 0.15y2

0.075 m 6 y … 0.105 m

I =

112

(0.3)(0.213) -

112

(0.2)(0.153) = 0.175275 (10- 3) m4

7–19. Plot the intensity of the shear stress distributed overthe cross section of the strut if it is subjected to a shear forceof V = 600 kN.

V

150 mm

30 mm

100 mm100 mm

100 mm

30 mm

Q = ©y¿A¿ = 0.09 (0.03)(0.3) +

12

(0.075 + y)(0.075 - y)(0.1) = 1.09125(10- 3) - 0.05 y2

For , . Thus,

At and ,

For , . Thus,

At and ,

The plot shear stress distribution over the cross-section is shown in Fig. c.

t|y = 0 = 37.4 MPa ty = 0.075 m = 27.7 MPa

y = 0.075 my = 0

t =

VQ

It=

600 (103) [1.09125(10- 3) - 0.05 y2]

0.175275(10- 3) (0.1)= (37.3556 - 1711.60 y2) MPa

t = 0.1 m0 … y 6 0.075 m

t|y = 0.015 m = 9.24 MPa ty = 0.105 m = 0

y = 0.105 my = 0.075 m

t =

VQ

It=

600 (103) C1.65375(10- 3) - 0.15y2 D0.175275(10- 3) (0.3)

= (18.8703 - 1711.60y2) MPa

t = 0.3 m0.075 m 6 y … 0.105 m


487


The moment of inertia of the ciralor cross-section about the neutral axis (x axis) is

Q for the differential area shown shaded in Fig. a is

However, from the equation of the circle, , Then

Thus, Q for the area above y is

Here, . Thus

By inspecting this equation, at . Thus

Ans.tmax¿ =

202p

=

10p

= 3.18 ksi

y = 0t = tmax

t =

52p

(4 - y2) ksi

t =

VQ

It=

30 C23 (4 - y2)32 D

4p C2(4 - y2)12 D

t = 2x = 2 (4 - y2)12

=

23

(4 - y2)32

= - 23

(4 - y2)32 �

2 in

y

Q =

L

2 in

y2y (4 - y2)

12 dy

dQ = 2y(4 - y2)12 dy

x = (4 - y2)12

dQ = ydA = y (2xdy) = 2xy dy

I =

p

4 r4

=

p

4 (24) = 4 p in4

*7–20. The steel rod is subjected to a shear of 30 kip.Determine the maximum shear stress in the rod.

30 kip

2 in.

1 in.A


The moment of inertia of the circular cross-section about the neutral axis (x axis) is

Q for the differential area shown in Fig. a is

However, from the equation of the circle, , Then

Thus, Q for the area above y is

Here . Thus,

For point A, . Thus

Ans.

The state of shear stress at point A can be represented by the volume elementshown in Fig. b.

tA =

52p

(4 - 12) = 2.39 ksi

y = 1 in

t =

52p

(4 - y2) ksi

t =

VQ

It=

30 C23 (4 - y2)32 D

4p C2(4 - y2)12 D

t = 2x = 2 (4 - y2)12

= - 23 (4 - y2)

32 ` 2 in.

y=

23

(4 - y2)32

Q =

L

2 in.

y2y (4 - y2)

12 dy

dQ = 2y (4 - y2)12 dy

x = (4 - y2)12

dQ = ydA = y (2xdy) = 2xy dy

I =

p

4 r4

=

p

4 (24) = 4p in4

•7–21. The steel rod is subjected to a shear of 30 kip.Determine the shear stress at point A. Show the result on avolume element at this point.

488


30 kip

2 in.

1 in.A


489


y =

(0.01)(0.05)(0.02) + (0.055)(0.07)(0.02)

(0.05)(0.02) + (0.07)(0.02)= 0.03625 m

7–22. Determine the shear stress at point B on the web ofthe cantilevered strut at section a–a.

a

a

2 kN 4 kN

250 mm 250 mm 300 mm

20 mm50 mm

70 mm

20 mm

B

Ans. = 4.41 MPa

tB =

VQB

I t=

6(103)(26.25)(10- 6)

1.78622(10- 6)(0.02)

QB = (0.02)(0.05)(0.02625) = 26.25(10- 6) m3

yœ

B = 0.03625 - 0.01 = 0.02625 m

+

112

(0.02)(0.073) + (0.02)(0.07)(0.055 - 0.03625)2= 1.78625(10- 6) m4

I =

112

(0.05)(0.023) + (0.05)(0.02)(0.03625 - 0.01)2

Ans. = 4.85 MPa

tmax =

VQmax

I t=

6(103)(28.8906)(10- 6)

1.78625(10- 6)(0.02)

Qmax = y¿A¿ = (0.026875)(0.05375)(0.02) = 28.8906(10- 6) m3

+

112

(0.02)(0.073) + (0.02)(0.07)(0.055 - 0.03625)2= 1.78625(10- 6) m4

I =

112

(0.05)(0.023) + (0.05)(0.02)(0.03625 - 0.01)2

y =

(0.01)(0.05)(0.02) + (0.055)(0.07)(0.02)

(0.05)(0.02) + (0.07)(0.02)= 0.03625 m

7–23. Determine the maximum shear stress acting atsection a–a of the cantilevered strut.

a

a

2 kN 4 kN

250 mm 250 mm 300 mm

20 mm50 mm

70 mm

20 mm

B


490


*7–24. Determine the maximum shear stress in the T-beamat the critical section where the internal shear force ismaximum.

3 m 1.5 m1.5 m

10 kN/m

A

150 mm

150 mm 30 mm

30 mm

BCThe FBD of the beam is shown in Fig. a,

The shear diagram is shown in Fig. b. As indicated,

The neutral axis passes through centroid c of the cross-section, Fig. c.

From Fig. d,

The maximum shear stress occurs at points on the neutral axis since Q is maximumand thickness is the smallest.

Ans. = 7.33 MPa

= 7.333(106) Pa

tmax =

Vmax Qmax

I t=

27.5(103) C0.216(10- 3) D27.0(10- 6)(0.03)

t = 0.03 m

= 0.216 (10- 3) m3

Qmax = y¿A¿ = 0.06(0.12)(0.03)

= 27.0 (10- 6) m4

+

112

(0.15)(0.033) + 0.15(0.03)(0.165 - 0.12)2

I =

112

(0.03)(0.153) + 0.03(0.15)(0.12 - 0.075)2

= 0.12 m

y =

©

'

y A

©A=

0.075(0.15)(0.03) + 0.165(0.03)(0.15)

0.15(0.03) + 0.03(0.15)

Vmax = 27.5 kN


491


using the method of sections,

The neutral axis passes through centroid C of the cross-section,

490

The maximum shear stress occurs at points on the neutral axis since Q is maximumand thickness t = 0.03 m is the smallest.

Ans. = 3.667(106) Pa = 3.67 MPa

tmax =

VC Qmax

It=

13.75(103) C0.216(10- 3) D27.0(10- 6) (0.03)

= 0.216 (10- 3) m3

Qmax = y¿A¿ = 0.06 (0.12)(0.03)

= 27.0 (10- 6) m4

+ 1

12 (0.15)(0.033) + 0.15(0.03)(0.165 - 0.12)2

I =

112

(0.03)(0.15) + 0.03(0.15)(0.12 - 0.075)2

= 0.12 m

y =

©yA

©A=

0.075 (0.15)(0.03) + 0.165(0.03)(0.15)

0.15(0.03) + 0.03(0.15)

VC = -13.75 kN

+ c ©Fy = 0; VC + 17.5 -

12

(5)(1.5) = 0

•7–25. Determine the maximum shear stress in the T-beam at point C. Show the result on a volume elementat this point.

3 m 1.5 m1.5 m

10 kN/m

A

150 mm

150 mm 30 mm

30 mm

BC


492



Internal Shear Force: As shown on shear diagram, .

Section Properties:

Maximum Shear Stress: Maximum shear stress occurs at the point where theneutral axis passes through the section.


Ans. =

878.57(12.375)

77.625(0.5)= 280 psi

tmax =

VQmax

It

= 3.375(4)(0.75) + 1.5(3)(0.5) = 12.375 in3

Qmax = ©y¿A¿

INA =

112

(4) A7.53 B -

112

(3.5) A63 B = 77.625 in4

Vmax = 878.57 lb

7–26. Determine the maximum shear stress acting in thefiberglass beam at the section where the internal shearforce is maximum.

A

150 lb/ft

D

0.75 in.

0.75 in.4 in.

6 in.

2 ft6 ft6 ft

0.5 in.

200 lb/ft

4 in.


493


The FBD is shown in Fig. a.

Using the method of sections, Fig. b,

The moment of inertia of the beam’s cross section about the neutral axis is

QC and QD can be computed by refering to Fig. c.

Shear Stress. since points C and D are on the web, .

Ans.

Ans.tD =

VQD

It=

9.00 (27)

276 (0.75)= 1.17 ksi

tC =

VQC

It=

9.00 (33)

276 (0.75)= 1.43 ksi

t = 0.75 in

QD = yœ

3A¿ = 4.5 (1)(6) = 27 in3

= 33 in3

QC = ©y¿A¿ = 4.5 (1)(6) + 2 (4)(0.75)

I =

112

(6)(103) -

112

(5.25)(83) = 276 in4

V = 9.00 kip.

+ c ©Fy = 0; 18 -

12

(3)(6) - V = 0

7–27. Determine the shear stress at points C and Dlocated on the web of the beam.

A

3 kip/ft

D

D

C

C

B

1 in.

1 in.6 in.

4 in.

4 in.

6 ft 6 ft6 ft

0.75 in.

6 in.


494



The shear diagram is shown in Fig. b, .

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

From Fig. c

The maximum shear stress occurs at points on the neutral axis since Q is themaximum and thickness is the smallest

Ans.tmax =

Vmax Qmax

It=

18.0 (33)

276 (0.75)= 2.87 ksi

t = 0.75 in

= 33 in3

Qmax = ©y¿A¿ = 4.5 (1)(6) + 2(4)(0.75)

= 276 in4

I =

112

(6)(103) -

112

(5.25)(83)

Vmax = 18.0 kip

*7–28. Determine the maximum shear stress acting in thebeam at the critical section where the internal shear forceis maximum.

A

3 kip/ft

D

D

C

C

B

1 in.

1 in.6 in.

4 in.

4 in.

6 ft 6 ft6 ft

0.75 in.

6 in.


495


Force Equilibrium: The shaded area indicares the plastic zone. Isolate an element inthe plastic zone and write the equation of equilibrium.

This proves that the longitudinal shear stress. , is equal to zero. Hence thecorresponding transverse stress, , is also equal to zero in the plastic zone.Therefore, the shear force is carried by the malerial only in the elastic zone.

Section Properties:

Maximum Shear Stress: Applying the shear formula

However, hence

‚ (Q.E.D.)tmax =

3P

2A¿

A¿ = 2by¿

tmax =

VQmax

It=

V Ay ¿3b

2 BA23 by¿

3 B(b)=

3P

4by¿

Qmax = y¿ A¿ =

y¿

2 (y¿)(b) =

y¿2b

2

INA =

112

(b)(2y¿)3=

23

b y¿3

V = Ptmax

tlong

tlong = 0

; ©Fx = 0; tlong A2 + sg A1 - sg A1 = 0

7–30. The beam has a rectangular cross section and issubjected to a load P that is just large enough to develop afully plastic moment at the fixed support. If thematerial is elastic-plastic, then at a distance themoment creates a region of plastic yielding withan associated elastic core having a height This situationhas been described by Eq. 6–30 and the moment M isdistributed over the cross section as shown in Fig. 6–48e.Prove that the maximum shear stress developed in the beamis given by where the cross-sectional area of the elastic core.

A¿ = 2y¿b,tmax =321P>A¿2,

2y¿.M = Px

x 6 LMp = PL

h

b L

Px

Plastic region

2y¿

Elastic region


496


Force Equilibrium: If a fully plastic moment acts on the cross section, then anelement of the material taken from the top or bottom of the cross section issubjected to the loading shown. For equilibrium

Thus no shear stress is developed on the longitudinal or transverse plane of theelement. (Q. E. D.)

tlong = 0

; ©Fx = 0; sg A1 + tlong A2 - sg A1 = 0

7–31. The beam in Fig. 6–48f is subjected to a fully plasticmoment Prove that the longitudinal and transverseshear stresses in the beam are zero. Hint: Consider an elementof the beam as shown in Fig. 7–4c.

Mp .

Section Properties:

Shear Flow: There are two rows of nails. Hence, the allowable shear flow

.

Ans. V = 444 lb

166.67 =

V(12.0)

32.0

q =

VQ

I

q =

2(500)

6= 166.67 lb>in

Q = y¿A¿ = 1(6)(2) = 12.0 in4

I =

112

(6) A43 B = 32.0 in4

*7–32. The beam is constructed from two boards fastenedtogether at the top and bottom with two rows of nailsspaced every 6 in. If each nail can support a 500-lb shearforce, determine the maximum shear force V that can beapplied to the beam.

V2 in.

6 in.

6 in.

6 in.

2 in.

h

b L

Px

Plastic region

2y¿

Elastic region


497


Section Properties:

Shear Flow:

There are two rows of nails. Hence, the shear force resisted by each nail is

Ans.F = aq

2bs = a225 lb>in.

2b(6 in.) = 675 lb

q =

VQ

I=

600(12.0)

32.0= 225 lb>in.

Q = y¿A¿ = 1(6)(2) = 12.0 in4

I =

112

(6) A43 B = 32.0 in4

•7–33. The beam is constructed from two boardsfastened together at the top and bottom with two rows ofnails spaced every 6 in. If an internal shear force of

is applied to the boards, determine the shearforce resisted by each nail.V = 600 lb

V2 in.

6 in.

6 in.

6 in.

2 in.

7–34. The beam is constructed from two boards fastenedtogether with three rows of nails spaced Ifeach nail can support a 450-lb shear force, determine themaximum shear force V that can be applied to the beam. Theallowable shear stress for the wood is tallow = 300 psi.

s = 2 in. apart.

V

1.5 in.

s

s

6 in.

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

Refering to Fig. a,

The maximum shear stress occurs at the points on the neutral axis where Q ismaximum and .

Shear Flow: Since there are three rows of nails,

Ans. V = 1350 lb = 1.35 kip

qallow =

VQA

I ; 675 =

V(6.75)

13.5

qallow = 3aFsb = 3a450

2b = 675 lb>in.

V = 3600 lb = 3.60 kips

tallow =

VQmax

It ; 300 =

V(6.75)

13.5(6)

t = 6 in

QA = Qmax = y¿A¿ = 0.75(1.5)(6) = 6.75 in3

I =

112

(6)(33) = 13.5 in4


498



Refering to Fig. a,


Ans.

Since there are three rows of nails,

Ans. s = 2.167 in = 2 18

in

qallow =

VQA

I ; 1950

s=

1800(6.75)

13.5

qallow = 3aFsb = 3¢650

s≤ = a1950

sb

lbin.

V = 1800 lb = 1.80 kip

tallow =

VQmax

It ; 150 =

V(6.75)

13.5(6)

t = 6 in

QA = Qmax = y¿A¿ = 0.75(1.5)(6) = 6.75 in3

I =

112

(6)(33) = 13.5 in4

7–35. The beam is constructed from two boards fastenedtogether with three rows of nails. If the allowable shearstress for the wood is determine themaximum shear force V that can be applied to the beam.Also, find the maximum spacing s of the nails if each nailcan resist 650 lb in shear.

tallow = 150 psi,

V

1.5 in.

s

s

6 in.

1.5 in.


499


Section Properties:

Shear Flow: Since there are two shear planes on the bolt, the allowable shear flow is

.

Ans. s = 5.53 in.

30s

=

50(10.125)

93.25

q =

VQ

I

q =

2(15)

s=

30s

Q = ©y¿A¿ = 2.5(3)(0.5) + 4.25(3)(0.5) = 10.125 in3

= 93.25 in4

- 1

12 (0.5) A23 B +

112

(1) A63 B INA =

112

(3) A93 B -

112

(2.5) A83 B

*7–36. The beam is fabricated from two equivalentstructural tees and two plates. Each plate has a height of 6 in. and a thickness of 0.5 in. If a shear of isapplied to the cross section, determine the maximum spacingof the bolts. Each bolt can resist a shear force of 15 kip.

V = 50 kip

3 in.

3 in.

A

V

0.5 in.

1 in.

0.5 in.

6 in.

ss

NN

Section Properties:


.

Ans. y = 34.5 kip

3.75 =

V(10.125)

93.25

q =

VQ

I

q =

2(15)

8= 3.75 kip>in

Q = ©y¿A¿ = 2.5(3)(0.5) + 4.25(3)(0.5) = 10.125 in3

= 93.25 in4

- 1

12 (0.5) A23 B +

112

(1) A63 B INA =

112

(3) A93 B -

112

(2.5) A83 B

•7–37. The beam is fabricated from two equivalentstructural tees and two plates. Each plate has a height of 6 in. and a thickness of 0.5 in. If the bolts are spaced at

determine the maximum shear force V that canbe applied to the cross section. Each bolt can resist ashear force of 15 kip.

s = 8 in.,

3 in.

3 in.

A

V

0.5 in.

1 in.

0.5 in.

6 in.

ss

NN


500


The neutral axis passes through centroid C of the cross-section as shown in Fig. a.

Thus,

Q for the shaded area shown in Fig. b is

Since there are two rows of nails .

Thus, the shear stress developed in the nail is

Ans.tn =

F

A=

442.62p

4 (0.0042)

= 35.22(106)Pa = 35.2 MPa

F = 442.62 N

q =

VQ

I ; 26.67 F =

2000 C0.375 (10- 3) D63.5417 (10- 6)

q = 2aFsb =

2F

0.075= (26.67 F) N>m

Q = y¿A¿ = 0.0375 (0.05)(0.2) = 0.375(10- 3) m3

= 63.5417(10- 6) m4

+ 1

12 (0.05)(0.23) + 0.05(0.2)(0.1375 - 0.1)2

I =

112

(0.2)(0.053) + 0.2 (0.05)(0.175 - 0.1375)2

y =

©

'

y A

©A=

0.175(0.05)(0.2) + 0.1(0.2)(0.05)

0.05(0.2) + 0.2(0.05)= 0.1375 m

7–38. The beam is subjected to a shear of Determine the average shear stress developed in each nailif the nails are spaced 75 mm apart on each side of thebeam. Each nail has a diameter of 4 mm.

V = 2 kN.

75 mm75 mm

50 mm

25 mm

200 mm

200 mm

25 mm V


501


Ans.F = q(s) = 49.997 (0.25) = 12.5 kN

q =

VQ

I=

35 (0.386)(10- 3)

0.270236 (10- 3)= 49.997 kN>m

Q = y¿A¿ = 0.06176(0.025)(0.25) = 0.386(10- 3) m3

= 0.270236 (10- 3) m4

+

112

(0.025)(0.35)3+ (0.025)(0.35)(0.275 - 0.18676)2

I = (2)a 112b(0.025)(0.253) + 2 (0.025)(0.25)(0.18676 - 0.125)2

y =

2 (0.125)(0.25)(0.025) + 0.275 (0.35)(0.025)

2 (0.25)(0.025) + 0.35 (0.025)= 0.18676 m

7–39. A beam is constructed from three boards boltedtogether as shown. Determine the shear force developedin each bolt if the bolts are spaced apart and theapplied shear is V = 35 kN.

s = 250 mm

s = 250 mm

250 mm100 mm

25 mm

25 mm

25 mm

350 mmV



Section Properties:


.

Ans. s = 13.8 in.

1200

s=

1500(168)

2902

q =

VQ

I

q =

2(600)

s=

1200s

Q = y¿A¿ = 7(4)(6) = 168 in3

INA =

112

(7) A183 B -

112

(6) A103 B = 2902 in4

Vmax = 1500 lb

*7–40. The double-web girder is constructed from twoplywood sheets that are secured to wood members at its topand bottom. If each fastener can support 600 lb in singleshear, determine the required spacing s of the fastenersneeded to support the loading Assume A ispinned and B is a roller.

P = 3000 lb.

P

B

s

A

2 in.2 in.

10 in.

6 in.0.5 in. 0.5 in.

2 in.2 in.

4 ft 4 ft


502



Internal Shear Force and Moment: As shown on shear and moment diagram,and .

Section Properties:

Shear Flow: Assume bolt failure. Since there are two shear planes on the bolt, the

allowable shear flow is .

kip (Controls !) Ans.

Shear Stress: Assume failure due to shear stress.

Bending Stress: Assume failure due to bending stress.

P = 107 ksi

8(103) =

2.00P(12)(9)

2902

smax = sallow =

Mc

I

P = 22270 lb = 83.5 kip

3000 =

0.500P(208.5)

2902(1)

tmax = tallow =

VQmax

It

P = 6910 lb = 6.91

200 =

0.500P(168)

2902

q =

VQ

I

q =

2(600)

6= 200 lb>in

Qmax = ©y¿A¿ = 7(4)(6) + 4.5(9)(1) = 208.5 in3

Q = yœ

2A¿ = 7(4)(6) = 168 in3

INA =

112

(7) A183 B -

112

(6) A103 B = 2902 in4

Mmax = 2.00PVmax = 0.500P

•7–41. The double-web girder is constructed from twoplywood sheets that are secured to wood members at its topand bottom. The allowable bending stress for the wood is

and the allowable shear stress is If the fasteners are spaced and each fastener cansupport 600 lb in single shear, determine the maximum loadP that can be applied to the beam.

s = 6 in.tallow = 3 ksi.sallow = 8 ksi

P

B

s

A

2 in.2 in.

10 in.

6 in.0.5 in. 0.5 in.

2 in.2 in.

4 ft 4 ft


503


The neutral axis passes through the centroid c of the cross-section as shown in Fig. a.

Refering to Fig. a, Qmax and QA are


Ans.

Here, . Then

Ans. s = 1.134 in = 1 18

in

qallow =

VQA

I ; 950

s=

8815.51(84)

884

qallow =

Fs

=

950s

lb>in

V = 8815.51 lb = 8.82 kip

tallow =

VQmax

It ; 450 =

V (90.25)

884 (2)

t = 2 in

QA = y2œ A2

œ

= 3.5 (2)(12) = 84 in3

Qmax = y1œ A1

œ

= 4.75(9.5)(2) = 90.25 in3

= 884 in4

+

112

(12)(23) + 12(2)(13 - 9.5)2

I =

112

(2)(123) + 2(12)(9.5 - 6)2

y =

©

'

y A

©A=

13(2)(12) + 6(12)(2)

2(12) + 12(2)= 9.5 in.

7–42. The T-beam is nailed together as shown. If the nailscan each support a shear force of 950 lb, determine themaximum shear force V that the beam can support and thecorresponding maximum nail spacing s to the nearest in.The allowable shear stress for the wood is .tallow = 450 psi

18

12 in.

12 in.2 in.

2 in.

V

s

s


504


7–43. Determine the average shear stress developed in thenails within region AB of the beam. The nails are located oneach side of the beam and are spaced 100 mm apart. Eachnail has a diameter of 4 mm. Take P = 2 kN.


As indicated in Fig. b, the internal shear force on the cross-section within region ABis constant that is .

The neutral axis passes through centroid C of the cross section as shown in Fig. c.

Q for the shaded area shown in Fig. d is

Since there are two rows of nail, .

Thus, the average shear stress developed in each nail is

Ans.Atnail Bavg =

F

Anail=

1500p

4 (0.0042)

= 119.37(106)Pa = 119 MPa

F = 1500 N

q =

VAB Q

I ; 20F =

5(103) C0.32(10- 3) D53.333(10- 6)

q = 2 aFsb = 2 a F

0.1b = 20F N>m

Q = y¿A¿ = 0.04(0.04)(0.2) = 0.32(10- 3) m3

= 53.333(10- 6) m4

+

112

(0.2)(0.043) + 0.2(0.04)(0.18 - 0.14)2

I =

112

(0.04)(0.23) + 0.04(0.2)(0.14 - 0.1)2

= 0.14 m

y =

©

'

y A

©A=

0.18(0.04)(0.2) + 0.1(0.2)(0.04)

0.04(0.2) + 0.2(0.04)

VAB = 5 kN

P

1.5 m 1.5 m

B CA

40 mm

20 mm20 mm

100 mm

200 mm

200 mm

2 kN/m


505



As indicated the shear diagram, Fig. b, the maximum shear occurs in region AB ofConstant value, .

The neutral axis passes through Centroid C of the cross-section as shown in Fig. c.

Refering to Fig. d,

The maximum shear stress occurs at the points on Neutral axis where Q is maximumand .

Since there are two rows of nails .

Ans. P = 3.67 kN (Controls!)

qallow =

Vmax QA

I ; 40 000 =

(P + 3)(103) C0.32(10- 3) D53.333(10- 6)

qallow = 2 aFsb = 2 c2(103)

0.1d = 40 000 N>m

P = 13.33 kN

tallow =

Vmax Qmax

It ; 3(106) =

(P + 3)(103) C0.392(10- 3) D53.333(10- 6)(0.04)

t = 0.04 m

QA = y2œ A2

œ

= 0.04(0.04)(0.2) = 0.32(10- 3) m3

Qmax = y1œ A1

œ

= 0.07(0.14)(0.04) = 0.392(10- 3) m3

= 53.333(10- 6) m4

+

112

(0.2)(0.043) + 0.2(0.04)(0.18 - 0.142)

I =

112

(0.04)(0.23) + 0.04(0.2)(0.14 - 0.1)2

= 0.14 m

y =

©

'

y A

©A=

0.18(0.04)(0.2) + 0.1(0.2)(0.04)

0.04(0.2) + 0.2(0.04)

Vmax = (P + 3) kN

*7–44. The nails are on both sides of the beam and eachcan resist a shear of 2 kN. In addition to the distributedloading, determine the maximum load P that can be appliedto the end of the beam. The nails are spaced 100 mm apartand the allowable shear stress for the wood is .tallow = 3 MPa

P

1.5 m 1.5 m

B CA

40 mm

20 mm20 mm

100 mm

200 mm

200 mm

2 kN/m


506


7–44. Continued


507




Section Properties:

Shear Flow: There are two rows of nails. Hence the allowable shear flow is

.

Ans. P = 6.60 kN

60.0 A103 B =

(P + 3)(103)0.450(10- 3)

72.0(10- 6)

q =

VQ

I

q =

3(2)

0.1= 60.0 kN>m

Q = y¿A¿ = 0.06(0.25)(0.03) = 0.450 A10- 3 B m3

= 72.0 A10- 6 B m4

INA =

112

(0.31) A0.153 B -

112

(0.25) A0.093 B

VAB = (P + 3) kN

•7–45. The beam is constructed from four boards whichare nailed together. If the nails are on both sides of the beamand each can resist a shear of 3 kN, determine the maximumload P that can be applied to the end of the beam.

P

2 m 2 m

3 kN

B CA

30 mm

30 mm30 mm

100 mm

250 mm30 mm

150 mm


508


Section Properties:

Shear Flow: The allowable shear flow at points C and D is and

, respectively.

Ans.

Ans. s¿ = 1.21 in.

100s¿

=

700(39.6774)

337.43

qD =

VQD

I

s = 8.66 in.

100

s=

700(5.5645)

337.43

qC =

VQC

I

qB =

100s¿

qC =

100s

QD = y2¿A¿ = (3.3548 - 0.5)(10)(1) + 2 C(3.3548 - 1.5)(3)(1) D = 39.6774 in3

QC = y1¿A¿ = 1.8548(3)(1) = 5.5645 in3

= 337.43 in4

+

112

(2) A33 B + 2(3)(3.3548 - 1.5)2

INA =

112

(10) A13 B + 10(1)(3.3548 - 0.5)2

= 3.3548 in

y =

©yA

©A=

0.5(10)(1) + 1.5(2)(3) + 6(1.5)(10)

10(1) + 2(3) + 1.5(10)

7–47. The beam is made from four boards nailed togetheras shown. If the nails can each support a shear force of 100 lb., determine their required spacing s� and s if the beamis subjected to a shear of .V = 700 lb

1.5 in.

10 in.

2 in.

B

V

1 in.

10 in.

1 in.

A

1 in.

C

ss

D

s¿

s¿


509


Ans.

V = 485 lb

qA s = 0.0516V(2) = 50

qA =

12

aVQA

Ib =

V(20.4)

2(197.7)= 0.0516 V

QA = y2œ A¿ = 3.4(6)(1) = 20.4 in3

V = 317 lb (controls)

qB s = 0.0789V(2) = 50

qB =

12

aVQB

Ib =

V(31.2)

2(197.7)= 0.0789 V

QB = y1œ A¿ = 2.6(12)(1) = 31.2 in3

+

112

(6)(13) + 6(1)(6.5 - 3.1)2= 197.7 in4

+ 2a 112b(1)(63) + 2(1)(6)(4 - 3.1)2

I =

112

(12)(13) + 12(1)(3.1 - 0.5)2

y =

©yA

©A=

0.5 (12)(1) + 2 (4)(6)(1) + (6.5)(6)(1)

12(1) + 2(6)(1) + (6)(1)= 3.1 in.

*7–48. The box beam is constructed from four boards thatare fastened together using nails spaced along the beamevery 2 in. If each nail can resist a shear of 50 lb, determinethe greatest shear V that can be applied to the beam withoutcausing failure of the nails.

6 in.

1 in.

1 in.

1 in.

5 in.V

12 in.

2 in.


510510



Refering to Fig. a Fig. b,

Due to symmety, the shear flow at points A and , Fig. a, and at points B and ,Fig. b, are the same. Thus

Ans.

Ans. = 462.46(103) N>m = 462 kN>m

qB =

12

aVQB

Ib =

12

c 300(103) C0.751(10- 3) D0.24359(10- 3)

s

= 228.15(103) N>m = 228 kN>m

qA =

12

aVQA

Ib =

12

c 300(103) C0.3705(10- 3) D0.24359(10- 3)

s

B¿A¿

QB = 2yœ

zA2œ

+ y3œ A3

œ

= 2 [0.1(0.2)(0.01)] + 0.195(0.01)(0.18) = 0.751(10- 3) m3

QA = y1œ A1

œ

= 0.195 (0.01)(0.19) = 0.3705 (10- 3) m3

I =

112

(0.2)(0.43) -

112

(0.18)(0.383) = 0.24359(10- 3) m4

7–50. A shear force of is applied to the boxgirder. Determine the shear flow at points A and B.

V = 300 kN

100 mm

90 mm90 mm

200 mm

200 mm

180 mm

190 mm

10 mm

10 mm

V

A DC

B




Refering to Fig. a, due to symmetry . Thus

Then refering to Fig. b,

Thus,

Ans.

Ans. = 601.33(103) N>m = 601 kN>m qD =

VQD

I=

450(103) C0.3255(10- 3) D0.24359(10- 3)

qC =

VQC

I= 0

= 0.3255(10- 3) m3

QD = y1œ A1

œ

+ y2œ A2

œ

= 0.195 (0.01)(0.09) + 0.15(0.1)(0.01)

QC = 0

ACœ

= 0

I =

112

(0.2)(0.43) -

112

(0.18)(0.383) = 0.24359(10- 3) m4

7–51. A shear force of is applied to the boxgirder. Determine the shear flow at points C and D.

V = 450 kN

100 mm

90 mm90 mm

200 mm

200 mm

180 mm

190 mm

10 mm

10 mm

V

A DC

B


512


Section Properties:

Shear Flow:

Ans.

Ans. = 9437 N>m = 9.44 kN>m =

12

c18(103)(0.13125)(10- 3)

125.17(10- 6)d

qB =

12

cVQB

Id

= 13033 N>m = 13.0 kN>m =

12

c18(103)(0.18125)(10- 3)

125.17(10- 6)d

qA =

12

cVQA

Id

QB = y1œ A¿ = 0.105(0.125)(0.01) = 0.13125 A10- 3 B m3

QA = y2œ A¿ = 0.145(0.125)(0.01) = 0.18125 A10- 3 B m3

= 125.17 A10- 6 B m4

+ 2 c 112

(0.125) A0.013 B + 0.125(0.01) A0.1052 B d

INA =

112

(0.145) A0.33 B -

112

(0.125) A0.283 B

*7–52. A shear force of is applied to thesymmetric box girder. Determine the shear flow at A and B.

V = 18 kN

C

A

150 mm

10 mm

10 mm

100 mm

100 mm

10 mm

10 mm

125 mm

150 mm

10 mm30 mm

B

V10 mm


513


Section Properties:

Shear Flow:

Ans. = 38648 N>m = 38.6 kN>m =

12

c18(103)(0.5375)(10- 3)

125.17(10- 4)d

qC =

12

cVQC

Id

= 0.5375 A10- 3 B m3

= 0.145(0.125)(0.01) + 0.105(0.125)(0.01) + 0.075(0.15)(0.02)

QC = ©y¿A¿

= 125.17 A10- 6 B m4

+2 c 112

(0.125) A0.013 B + 0.125(0.01) A0.1052 B d

INA =

112

(0.145) A0.33 B -

112

(0.125) A0.283 B

•7–53. A shear force of is applied to the boxgirder. Determine the shear flow at C.

V = 18 kN

C

A

150 mm

10 mm

10 mm

100 mm

100 mm

10 mm

10 mm

125 mm

150 mm

10 mm30 mm

B

V10 mm


514


Ans.

Ans.qB =

VQB

I=

150(8.1818)(10- 6)

0.98197(10- 6)= 1.25 kN>m

qA =

VQA

I=

150(9.0909)(10- 6)

0.98197(10- 6)= 1.39 kN>m

QB = yB¿A¿ = 0.027272(0.03)(0.01) = 8.1818(10- 6) m3

QA = yA¿A¿ = 0.022727(0.04)(0.01) = 9.0909(10- 6) m3

yA¿ = 0.027727 - 0.005 = 0.022727 m

yB¿ = 0.055 - 0.027727 = 0.027272 m

+

112

(0.04)(0.01)3+ 0.04(0.01)(0.055 - 0.027727)2

= 0.98197(10- 6) m4

+ 2 c 112

(0.01)(0.06)3+ 0.01(0.06)(0.03 - 0.027727)2 d

I = 2 c 112

(0.03)(0.01)3+ 0.03(0.01)(0.027727 - 0.005)2 d

y =

2[0.005(0.03)(0.01)] + 2[0.03(0.06)(0.01)] + 0.055(0.04)(0.01)

2(0.03)(0.01) + 2(0.06)(0.01) + 0.04(0.01)= 0.027727 m

7–54. The aluminum strut is 10 mm thick and has the crosssection shown. If it is subjected to a shear of, ,determine the shear flow at points A and B.

V = 150 N

30 mm40 mm

30 mm

V

AB40 mm

10 mm

10 mm

10 mm10 mm

= 0.98197(10- 6) m4

+

112

(0.04)(0.01)3+ 0.04(0.01)(0.055 - 0.027727)2

+ 2 c 112

(0.01)(0.06)3+ 0.01(0.06)(0.03 - 0.027727)2 d

I = 2 c 112

(0.03)(0.01)3+ 0.03(0.01)(0.027727 - 0.005)2 d

= 0.027727 m

y =

2[0.005(0.03)(0.01)] + 2[0.03(0.06)(0.01)] + 0.055(0.04)(0.01)

2(0.03)(0.01) + 2(0.06)(0.01) + 0.04(0.01)

7–55. The aluminum strut is 10 mm thick and has the crosssection shown. If it is subjected to a shear of ,determine the maximum shear flow in the strut.

V = 150 N

30 mm40 mm

30 mm

V

AB40 mm

10 mm

10 mm

10 mm10 mm

Qmax = (0.055 - 0.027727)(0.04)(0.01) + 2[(0.06 - 0.027727)(0.01)]a0.06 - 0.02772

b

Ans. qmax =

12

aVQmax

Ib =

12

a150(21.3(10- 6))

0.98197(10- 6)b = 1.63 kN>m

= 21.3(10- 6) m3


515


y =

©yA

©A=

0.25(11)(0.5) + 2[4.5(8)(0.5)] + 6.25(10)(0.5)

11(0.5) + 2(8)(0.5) + 10(0.5)= 3.70946 in.

*7–56. The beam is subjected to a shear force of. Determine the shear flow at points A and B.V = 5 kip

A

V

0.5 in.

0.5 in.

5 in.

5 in.

0.5 in.

2 in. 0.5 in.

8 in.

B

A

C

D

I =

112

(11)(0.53) + 11(0.5)(3.70946 - 0.25)2+ 2 c 1

12 (0.5)(83) + 0.5(8)(4.5 - 3.70946)2 d

Ans.

Ans.qB =

12

aVQB

Ib =

12

a5(103)(12.7027)

145.98b = 218 lb>in.

qA =

12

aVQA

Ib =

12

a5(103)(19.02703)

145.98b = 326 lb>in.

QB = yBœ A¿ = 2.54054(10)(0.5) = 12.7027 in3

QA = yAœ A¿ = 3.45946(11)(0.5) = 19.02703 in3

yBœ

= 6.25 - 3.70946 = 2.54054 in.

yAœ

= 3.70946 - 0.25 = 3.45946 in.

= 145.98 in4

+

112

(10)(0.53) + 10(0.5)(6.25 - 3.70946)2

Ans. = 414 lb>in.

qmax =

12

aVQmax

Ib =

12

a5(103)(24.177)

145.98b

= 24.177 in3

Qmax = 3.4594 (11)(0.5) + 2[(1.6047)(0.5)(3.7094 - 0.5)]

= 145.98 in4

+

112

(10)(0.53) + 10(0.5)(2.54052)

I =

112

(11)(0.53) + 11(0.5)(3.45952) + 2 c 112

(0.5)(83) + 0.5(8)(0.79052) d

y =

©yA

©A=

0.25(11)(0.5) + 2[4.5(8)(0.5)] + 6.25(10)(0.5)

11(0.5) + 2(8)(0.5) + 10(0.5)= 3.70946 in.

•7–57. The beam is constructed from four plates and issubjected to a shear force of . Determine themaximum shear flow in the cross section.

V = 5 kip

A

V

0.5 in.

0.5 in.

5 in.

5 in.

0.5 in.

2 in. 0.5 in.

8 in.

B

A

C

D


516


Ans.qA =

75(103)(0.3450)(10- 3)

0.12025(10- 3)= 215 kN>m

q =

VQ

I

QA = yAœ A¿ = 0.0575(0.2)(0.03) = 0.3450(10- 3) m3

+ 2 c 112

(0.03)(0.23) + 0.03(0.2)(0.13 - 0.0725)2 d = 0.12025(10- 3) m4

I =

112

(0.4)(0.033) + 0.4(0.03)(0.0725 - 0.015)2

y =

©yA

©A=

0.015(0.4)(0.03) + 2[0.13(0.2)(0.03)]

0.4(0.03) + 2(0.2)(0.03)= 0.0725 m

7–58. The channel is subjected to a shear of .Determine the shear flow developed at point A.

V = 75 kN

200 mm

30 mm

30 mm

V � 75 kN

30 mm

400 mm

A

Ans.qmax =

75(103)(0.37209)(10- 3)

0.12025(10- 3)= 232 kN>m

Qmax = y¿A¿ = 0.07875(0.1575)(0.03) = 0.37209(10- 3) m3

= 0.12025(10- 3) m4

+ 2 c 112

(0.03)(0.23) + 0.03(0.2)(0.13 - 0.0725)2 d I =

112

(0.4)(0.033) + 0.4(0.03)(0.0725 - 0.015)2

= 0.0725 m

y =

©yA

©A=

0.015(0.4)(0.03) + 2[0.13(0.2)(0.03)]

0.4(0.03) + 2(0.2)(0.03)

7–59. The channel is subjected to a shear of .Determine the maximum shear flow in the channel.

V = 75 kN

200 mm

30 mm

30 mm

V � 75 kN

30 mm

400 mm

A


517


Section Properties:

Shear Flow:

Ans.

At Ans.y = 0, q = qmax = 424 lb>in.

= {424 - 136y2} lb>in.

=

2(103)(0.55243 - 0.17678y2)

2.604167

q =

VQ

I

= 0.55243 - 0.17678y2

Q = y¿A¿ = [0.25(3.53553) + 0.5y]a2.5 -

y

sin 45°b(0.25)

INA = 2 c 112

(0.35355) A3.535533 B d = 2.604167 in4

h = 5 cos 45° = 3.53553 in.

b =

0.25sin 45°

= 0.35355 in.

*7–60. The angle is subjected to a shear of .Sketch the distribution of shear flow along the leg AB.Indicate numerical values at all peaks.

V = 2 kip

45� 45�

V

A

B

5 in.5 in.

0.25 in.


518


y =

©yA

©A=

(0.25)(11)(0.5) + 2(3.25)(5.5)(0.5) + 6.25(7)(0.5)

0.5(11) + 2(0.5)(5.5) + 7(0.5)= 2.8362 in.

•7–61. The assembly is subjected to a vertical shear of. Determine the shear flow at points A and B and

the maximum shear flow in the cross section.V = 7 kip

6 in.

0.5 in.

2 in.

2 in.6 in.0.5 in.

V

A

B

0.5 in.

0.5 in.0.5 in.

I =

112

(11)(0.53) + 11(0.5)(2.8362 - 0.25)2+ 2a 1

12b(0.5)(5.53) + 2(0.5)(5.5)(3.25 - 2.8362)2

Ans.

Ans.

Ans.qmax =

12

a7(103)(16.9531)

92.569b = 641 lb>in.

qB =

12

a7(103)(11.9483)

92.569b = 452 lb>in.

qA =

7(103)(2.5862)

92.569= 196 lb>in.

q =

VQ

I

Qmax = ©y¿A¿ = (3.4138)(7)(0.5) + 2(1.5819)(3.1638)(0.5) = 16.9531 in3

QB = y2¿A2¿ = (3.4138)(7)(0.5) = 11.9483 in3

QA = y1¿A1¿ = (2.5862)(2)(0.5) = 2.5862 in3

+ 1

12 (7)(0.53) + (0.5)(7)(6.25 - 2.8362)2

= 92.569 in4


519


Here

Therefore

Here

Ans.

occurs at ; therefore

; therefore

QEDtmax =

2V

A

A = 2pRt

tmax =

V

pR t

y = 0tmax

t =

V

pR2t 2R2

- y2

cos u =

2R2- y2

R

t =

VQ

I t=

V(2R2t cos u)

pR3 t(2t)=

V cos u

pR t

=

R3 t2

[u -

sin 2u2

] �2p

0=

R3 t2

[2p - 0] = pR3 t

I =

L

2p

0R3 t sin2

u du = R3 tL

2p

0 (1 - cos 2u)

2 du

dI = y2 dA = y2 R t du = R3 t sin2 u du

= R2 t [-cos (p - u) - (-cos u)] = 2R2

t cos u

Q =

L

p-u

u

R2 t sin u du = R2 t(-cos u) |p- u

u

dQ = R2 t sin u du

y = R sin u

dQ = y dA = yR t du

dA = R t du

7–62. Determine the shear-stress variation over the crosssection of the thin-walled tube as a function of elevation y andshow that , where Hint: Choose adifferential area element . Using formulate Q for a circular section from to and showthat where cos u = 2R2

- y2>R.Q = 2R2t cos u,(p - u)u

dQ = y dA,dA = Rt duA = 2prt.t max = 2V>A

t

ydu

ds

R

u


520


Section Properties:

Shear Flow Resultant:

Shear Center: Summing moment about point A.

Ans.

Note that if , (I shape).e = 0b2 = b1

e =

3(b22

- b12)

h + 6(b1 + b2)

Pe =

3Pb22

h Ch + 6(b1 + b2) D (h) -

3Pb12

h Ch + 6(b1 + b2) D (h)

Pe = AFf B2 h - AFf B1 h

=

3Pb22

h Ch + 6(b1 + b2) D

(Ff)2 =

L

b2

0q2 dx2 =

6P

h Ch + 6(b1 + b2) D Lb2

0x2 dx2

=

3Pb12

h Ch + 6(b1 + b2) D

(Ff)1 =

L

b1

0q1 dx1 =

6P

h Ch + 6(b1 + b2) D Lb1

0x1 dx1

q2 =

VQ2

I=

P Aht2 x2 B

t h2

12 Ch + 6(b1 + b2) D =

6P

h Ch + 6(b1 + b2) D x2

q1 =

VQ1

I=

P Aht2 x1 B

t h2

12 Ch + 6(b1 + b2) D =

6P

h Ch + 6(b1 + b2) D x1

Q2 = y¿A¿ =

h

2 (x2)t =

h t2

x2

Q1 = y¿A¿ =

h

2 (x1)t =

h t2

x1

I =

112

t h3+ 2 c(b1 + b2)tah

2b2 d =

t h2

12 Ch + 6(b1 + b2) D

7–63. Determine the location e of the shear center,point O, for the thin-walled member having the crosssection shown where . The member segments havethe same thickness t.

b2 7 b1

Oh

b2 b1

t

e


521


Section Properties:


Shear Center: Summing moments about point A,

Ans. e =

3b2

2(d + 3b)

Pe = c 3b2 sin 45°2d(d + 3b)

P d(2d sin 45°)

Pe = Ff(2d sin 45°)

Ff =

L

b

0qfdx =

3P sin 45°d(d + 3b)

L

b

0xdx =

3b2 sin 45°2d(d + 3b)

P

qf =

VQ

I=

P(td sin 45°)xtd2

3 (d + 3b)=

3P sin 45°d(d + 3b)

x

Q = y¿A¿ = d sin 45° (xt) = (td sin 45°)x

=

td2

3 (d + 3b)

I =

112

a t

sin 45°b(2d sin 45°)3

+ 2 Cbt(d sin 45°)2 D

*7–64. Determine the location e of the shear center,point O, for the thin-walled member having the crosssection shown. The member segments have the samethickness t.

45�

45�

d

O

b

e


522


Section Properties:


Shear Center: Summing moments about point A.

Ans. e =

710

a

Pe = 2a P

20ba + a 3

10 Pb2a

Pe = 2(Fw)1 (a) + Ff(2a)

Ff =

L

a

0q2 dx =

3P

20a2 L

a

0 (a + 2x)dx =

310

P

(Fw)1 =

L

a

0q1 dy =

3P

20a3 L

a

0 y2 dy =

P

20

q2 =

VQ2

I=

P Cat2 (a + 2x) D

103 a3 t

=

3P

20a2 (a + 2x)

q1 =

VQ1

I=

P A12 y2 B103 a3 t

=

3P

20a3 y2

Q2 = ©y¿A¿ =

a

2 (at) + a(xt) =

at

2 (a + 2x)

Q1 = y1œ A¿ =

y

2 (yt) =

t

2 y2

I =

112

(2t)(2a)3+ 2 Cat Aa2 B D =

103

a3 t

•7–65. Determine the location e of the shear center,point O, for the thin-walled member having a slit along itsside. Each element has a constant thickness t.

t

a

e

a

a

O


523


Summing moments about A.

Ans. e =

2233

a

F2 =

Va

(a) +

23

a V

2ab(a) =

4V

3

q2 = q1 +

V(a>2)(t)(a>4)14 t a3

= q1 +

V

2a

q1 =

V(a)(t)(a>4)14 t a3

=

Va

I =

112

(t)(a)3+

112

a t

sin 30°b(a)3

=

14

t a3

Pe = F2 a132

ab

7–66. Determine the location e of the shear center,point O, for the thin-walled member having the crosssection shown. a

a

a

60�

60�

O

e


524


Shear Flow Resultant: The shear force flows through as Indicated by F1, F2, and F3on FBD (b). Hence, The horizontal force equilibrium is not satisfied . Inorder to satisfy this equilibrium requirement. F1 and F2 must be equal to zero.

Shear Center: Summing moments about point A.

Ans.

Also,

The shear flows through the section as indicated by F1, F2, F3.

However,

To satisfy this equation, the section must tip so that the resultant of

Also, due to the geometry, for calculating F1 and F3, we require .

Hence, Ans.e = 0

F1 = F3

= P:

+ F3:

+ F2:

F1:

+:©Fx Z 0

Pe = F2(0) e = 0

(©Fx Z 0)

7–67. Determine the location e of the shear center,point O, for the thin-walled member having the crosssection shown. The member segments have the samethickness t.

b

b

O

t

e

h2

h2


525


Thus,

Summing moments about A:

Ans.e = 1.26 r

e =

r (1.9634 + 2)

3.15413

Pe =

Pr

3.15413 L

p>2

-p>2 (0.625 + cos u)du

Pe =

L

p>2

-p>2 (q r du)r

q =

VQ

I=

P(0.625 + cos u)t r2

3.15413 t r3

Q = 0.625 t r2+ t r2 cos u

Q = a r

2b t a r

4+ rb +

L

p>2

u

r sin u (t r du)

I = 1.583333t r3+ t r3ap

2b = 3.15413t r3

Isemi-circle = t r3 ap2b

Isemi-circle =

L

p>2

-p>2 (r sin u)2 t r du = t r3

L

p>2

-p>2 sin2

u du

= 1.583333t r3+ Isemi-circle

I = (2) c 112

(t)(r>2)3+ (r>2)(t)ar +

r

4b2 d + Isemi-circle

*7–68. Determine the location e of the shear center,point O, for the beam having the cross section shown. Thethickness is t.

e

O

r

1—2 r

1—2 r


526



(1)

From Eq, (1).

Ans.e =

t

12I (6h1 h

2b + 3h2b2- 8h1

3b) =

b(6h1 h2

+ 3h2b - 8h13)

2h3+ 6bh2

- (h - 2h1)3

I =

t

12 (2h3

+ 6bh2- (h - 2h1)

3)

Pe =

Pt

2I [h1 h

2b - h1 2 hb +

h2b2

2+ hh1

2 b -

43

h1 3 b]

F =

L q2 dx =

Pt

2I L

b

0 [h1 (h - h1) + hx]dx =

Pt

2I ah1 hb - h1

2 b +

hb2

2b

q2 =

VQ2

I=

Pt

2I (h1 (h - h1) + hx)

Q2 = ©y¿A¿ =

12

(h - h1)h1 t +

h

2 (x)(t) =

12

t[h1 (h - h1) + hx]

V =

Lq1 dy =

Pt

2IL

h1

0 (hy - 2h1 y + y2)dy =

Pt

2I chh1

2

2-

23

h13 d

q1 =

VQ

I=

Pt(hy - 2h1 y + y2)

2I

Q1 = y¿A¿ =

12

(h - 2h1 + y)yt =

t(hy - 2h1 y + y2)

2

=

th3

6+

bth2

2-

t(h - 2h1)3

12

I =

112

(t)(h3) + 2b(t)ah

2b2

+

112

(t)[h3- (h - 2h1)

3]

Pe = F(h) + 2V(b)

•7–69. Determine the location e of the shear center,point O, for the thin-walled member having the crosssection shown. The member segments have the samethickness t.

h

b

eO

h1

h1


527



(1)

dQ = y dA = r sin u(t r du) = r2 t sin u du

=

r3 t2

2 (2a - 2 sin a cos a) =

r3 t2

(2a - sin 2a)

=

r3 t2

c ap + a -

sin 2(p + a)

2b - ap - a -

sin 2(p - a)

2b d

=

r3 t2

(u -

sin 2u2

) �p+a

p-a

I = r3 tL

sin2 u du = r3 tL

p+a

p-a

1 - cos 2u

2 du

dI = y2 dA = r2 sin2 u(t r du) = r3 t sin2

udu

y = r sin u

dA = t ds = t r du

P e = rL

dF

7–70. Determine the location e of the shear center, point O,for the thin-walled member having the cross section shown.

e

r Oa

a

t

Q = r2 tL

u

p-a

sin u du = r2 t (-cos u)|u

p-a

= r2 t(-cos u - cos a) = -r2 t(cos u + cos a)

LdF =

Lq ds =

Lq r du

q =

VQ

I=

P(-r2t)(cos u + cos a)r3t2 (2a - sin 2a)

=

-2P(cos u + cos a)

r(2a - sin 2a)

L

dF =

2P rr(2a - sin 2a)

L

p+p

p-a

(cos u + cos a) du =

-2P

2a - sin 2a (2a cos a - 2 sin a)

From Eq. (1);

Ans.e =

4r (sin a - a cos a)

2a - sin 2a

P e = r c 4P

2a - sin 2a (sin a - a cos a) d

=

4P

2a - sin 2a (sin a - a cos a)


528


Section Properties:

(Q.E.D)

Shear Stress: Applying the shear formula ,

At

At

At

Resultant Shear Force: For segment AB.

Ans. = 9957 lb = 9.96 kip

=

L

0.8947 in

2.8947 in A3173.76 - 40.12y2

2 B dy

=

L

0.8947 in

2.8947 in A1586.88 - 20.06y2

2 B (2dy)

VAB =

L tAB dA

y2 = 2.8947 in. tAB = 1419 psi

= {1586.88 - 20.06y22} psi

tAB =

VQ2

It=

35(103)(79.12 - y22)

872.49(2)

y1 = -2.8947 in. tCB = 355 psi

y1 = 0, tCB = 523 psi

= {522.77 - 20.06y12} psi

tCB =

VQ1

It=

35(103)(104.25 - 4y12)

872.49(8)

t =

VQ

It

= 79.12 - y22

Q2 = y2œ A¿ = (4.44735 + 0.5y2)(8.8947 - y2)(2)

= 104.25 - 4y12

Q1 = y1œ A¿ = (2.55265 + 0.5y1)(5.1053 - y1)(8)

= 872.49 in4

+

112

(2) A63 B + 2(6)(11 - 5.1053)2

INA =

112

(8) A83 B + 8(8)(5.1053 - 4)2

y =

©yA

©A=

4(8)(8) + 11(6)(2)

8(8) + 6(2)= 5.1053 in.

7–71. Sketch the intensity of the shear-stress distributionacting over the beam’s cross-sectional area, and determinethe resultant shear force acting on the segment AB. Theshear acting at the section is . Show that INA = 872.49 in4.

V = 35 kip

2 in.

3 in.

3 in.

6 in.

8 in.

A

B

C

V


529


Section Properties:

Shear Flow:

Hence, the shear force resisted by each nail is

Ans.

Ans.FD = qD s = (460.41 lb>in.)(3 in.) = 1.38 kip

FC = qC s = (65.773 lb>in.)(3 in.) = 197 lb

qD =

VQD

I=

4.5(103)(42.0)

410.5= 460.41 lb>in.

qC =

VQC

I=

4.5(103)(6.00)

410.5= 65.773 lb>in.

QD = yœ

2A¿ = 3.50(12)(1) = 42.0 in2

QC = yœ

1A¿ = 1.5(4)(1) = 6.00 in2

= 410.5 in4

+

112

(1) A123 B + 1(12)(7 - 3.50)2

+

112

(2) A43 B + 2(4)(3.50 - 2)2

INA =

112

(10) A13 B + (10)(1)(3.50 - 0.5)2

y =

© y A

©A=

0.5(10)(1) + 2(4)(2) + 7(12)(1)

10(1) + 4(2) + 12(1)= 3.50 in.

*7–72. The beam is fabricated from four boards nailedtogether as shown. Determine the shear force each nailalong the sides C and the top D must resist if the nails areuniformly spaced at The beam is subjected to ashear of V = 4.5 kip.

s = 3 in.

1 in.

12 in.

3 in.

B

V

1 in.

10 in.

A

1 in.

1 in.


530


Section Properties:

Ans.

Shear Flow:

Ans.

Ans.

Ans.qC =

VQC

I=

2(103)(0.16424)(10- 3)

86.93913(10- 6)= 3.78 kN>m

qB =

VQB

I=

2(103)(52.57705)(10- 6)

86.93913(10- 6)= 1.21 kN>m

qA =

VQA

I= 0

= 0.16424 A10- 3 B m3

= 0.03048(0.115)(0.015) + 0.08048(0.0925)(0.015)

QC = ©y¿A¿

QB =

'

y œ

1A¿ = 0.03048(0.115)(0.015) = 52.57705 A10- 6 B m3

QA = 0

= 86.93913 A10- 6 B m4

+

112

(0.015) A0.33 B + 0.015(0.3)(0.165 - 0.08798)2

+

112

(0.03) A0.1153 B + 0.03(0.115)(0.08798 - 0.0575)2

INA =

112

(0.2) A0.0153 B + 0.2(0.015)(0.08798 - 0.0075)2

= 0.08798 m

=

0.0075(0.2)(0.015) + 0.0575(0.115)(0.03) + 0.165(0.3)(0.015)

0.2(0.015) + 0.115(0.03) + 0.3(0.015)

y =

© yA

©A

•7–73. The member is subjected to a shear force of. Determine the shear flow at points A, B, and C.

The thickness of each thin-walled segment is 15 mm.V = 2 kN

300 mmA

B

C

100 mm

200 mm

V � 2 kN


531


Section Properties:

Shear Flow: There are two glue joints in this case, hence the allowable shear flow is.

Ans. V = 4100 lb = 4.10 kip

150 =

V(3.50)

95.667

q =

VQ

I

2(75) = 150 lb>in

Q = y¿A¿ = 1.75(4)(0.5) = 3.50 in3

= 95.667 in4

INA =

112

(1) A103 B + 2 c 112

(4) A0.53 B + 4(0.5) A1.752 B d

7–74. The beam is constructed from four boards gluedtogether at their seams. If the glue can withstand

what is the maximum vertical shear V that thebeam can support?75 lb>in.,

3 in.

4 in.

3 in.

3 in.

0.5 in.

0.5 in.

0.5 in.0.5 in.

V

Section Properties:

Shear Flow: There are two glue joints in this case, hence the allowable shear flow is.

Ans. V = 749 lb

150 =

V(11.25)

56.167

q =

VQ

I

2(75) = 150 lb>in

Q = y¿A¿ = 2.25(10)(0.5) = 11.25 in3

INA =

112

(10) A53 B -

112

(9) A43 B = 56.167 in4

7–75. Solve Prob. 7–74 if the beam is rotated 90° from theposition shown.

3 in.

4 in.

3 in.

3 in.

0.5 in.

0.5 in.

0.5 in.0.5 in.

V


solucionario hibbeler 8va edicion "resistencia de materiales " , cap 6 y 7 ,

Engineering

Transcript of solucionario hibbeler 8va edicion "resistencia de materiales " , cap 6 y 7 ,