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6–1. Draw the shear and moment diagrams for the shaft. The bearings at A and B exert only vertical reactions on the shaft.
B
A
800 mm
250 mm
24 kN
6–2. Draw the shear and moment diagrams for the simply supported beam.
4 kN M 2 kNm A
B 2m
329
2m
2m
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6–3. The engine crane is used to support the engine, which has a weight of 1200 lb. Draw the shear and moment diagrams of the boom ABC when it is in the horizontal position shown. 4 FA(3)  1200(8) = 0; 5
a + ©MA = 0; + c ©Fy = 0; + ©F = 0; ; x
Ay + Ax 
4 (4000)  1200 = 0; 5
3 (4000) = 0; 5
A
3 ft
5 ft B
FA = 4000 lb
4 ft
Ay = 2000 lb
Ax = 2400 lb
*6–4. Draw the shear and moment diagrams for the cantilever beam.
2 kN/m
A
6 kNm 2m
The freebody diagram of the beam’s right segment sectioned through an arbitrary point shown in Fig. a will be used to write the shear and moment equations of the beam. + c ©Fy = 0;
V = {4  2x} kN‚
V  2(2  x) = 0
1 a + ©M = 0;  M  2(2  x) c (2  x) d  6 = 0 2
(1)
M = {  x2 + 4x  10}kN # m‚(2)
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 x = 0 is evaluated using Eqs. (1) and (2). V
x= 0
C
= 4  2(0) = 4 kN
Mx= 0 = C  0 + 4(0)  10 D = 10kN # m
330
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6–5. Draw the shear and moment diagrams for the beam.
10 kN
8 kN
15 kNm 2m
3m
6–6. Draw the shear and moment diagrams for the overhang beam.
8 kN/m
A
C B 4m
6–7. Draw the shear and moment diagrams for the compound beam which is pin connected at B.
2m
6 kip
8 kip
A C B 4 ft
331
6 ft
4 ft
4 ft
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*6–8. Draw the shear and moment diagrams for the simply supported beam.
150 lb/ft 300 lbft A
B 12 ft
The freebody diagram of the beam’s left segment sectioned through an arbitrary point shown in Fig. b will be used to write the shear and moment equations. The intensity of the triangular distributed load at the point of sectioning is w = 150a Referring to Fig. b,
x b = 12.5x 12
1 (12.5x)(x)  V = 0 V = {275  6.25x2}lb‚ (1) 2 1 x a + ©M = 0; M + (12.5x)(x) a b  275x = 0 M = {275x  2.083x 3}lb # ft ‚(2) 2 3 + c ©Fy = 0;
275 
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 obtained by setting V = 0 in Eq. (1). 0 = 275  6.25x2
x = 6.633 ft
The value of the moment at x = 6.633 ft (V = 0) is evaluated using Eq. (2). Mx = 6.633 ft = 275(6.633)  2.083(6.633)3 = 1216 lb # ft
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6–9. Draw the shear and moment diagrams for the beam. Hint: The 20kip load must be replaced by equivalent loadings at point C on the axis of the beam.
15 kip 1 ft A
20 kip
C 4 ft
B
4 ft
4 ft
6–10. Members ABC and BD of the counter chair are rigidly connected at B and the smooth collar at D is allowed to move freely along the vertical slot. Draw the shear and moment diagrams for member ABC. Equations of Equilibrium: Referring to the freebody diagram of the frame shown in Fig. a, +c ©F y = 0;
P 150 lb
Ay  150 = 0 C
Ay = 150 lb a + ©M A = 0;
1.5 ft
N D(1.5)  150(3) = 0
A 1.5 ft D
ND = 300 lb Shear and Moment Diagram: The couple moment acting on B due to N D is MB = 300(1.5) = 450 lb # ft . The loading acting on member ABC is shown in Fig. b and the shear and moment diagrams are shown in Figs. c and d.
333
B 1.5 ft
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6–11. The overhanging beam has been fabricated with a projected arm BD on it. Draw the shear and moment diagrams for the beam ABC if it supports a load of 800 lb. Hint: The loading in the supporting strut DE must be replaced by equivalent loads at point B on the axis of the beam.
E
800 lb B
Support Reactions: a + ©MC = 0;
5 ft
D 2 ft
C
A
800(10) 
6 ft
3 4 F DE (4)  F DE(2) = 0 5 5
4 ft
FDE = 2000 lb + c ©Fy = 0;
 800 +
3 (2000)  C y = 0 5
+ ©F = 0; : x
C x +
4 (2000) = 0 5
Cy = 400 lb
Cx = 1600 lb
Shear and Moment Diagram:
*6–12. A reinforced concrete pier is used to support the stringers for a bridge deck. Draw the shear and moment diagrams for the pier when it is subjected to the stringer loads shown. Assume the columns at A and B exert only vertical reactions on the pier.
60 kN
60 kN 35 kN 35 kN 35 kN 1 m 1 m 1.5 m 1.5 m 1 m 1 m
A
334
B
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6–13. Draw the shear and moment diagrams for the compound beam. It is supported by a smooth plate at A which slides within the groove and so it cannot support a vertical force, although it can support a moment and axial load.
P
Support Reactions:
P
A
D
B C
From the FBD of segment BD a + ©MC = 0; + c ©Fy = 0;
By (a)  P(a) = 0 Cy  P  P = 0
+ ©F = 0; : x
By = P
a
a
a
a
Cy = 2P
Bx = 0
From the FBD of segment AB a + ©MA = 0; + c ©Fy = 0;
P(2a)  P(a)  MA = 0
MA = Pa
P  P = 0 (equilibrium is statisfied!)
6–14. The industrial robot is held in the stationary position shown. Draw the shear and moment diagrams of the arm ABC if it is pin connected at A and connected to a hydraulic cylinder (twoforce member) BD. Assume the arm and grip have a uniform weight of 1.5 lbin. and support the load of 40 lb at C.
4 in. A
10 in. B
50 in.
120 D
335
C
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*6–16. Draw the shear and moment diagrams for the shaft and determine the shear and moment throughout the shaft as a function of x. The bearings at A and B exert only vertical reactions on the shaft.
500 lb 800 lb A
B x 3 ft
For 0 6 x 6 3 ft + c ©Fy = 0.
220  V = 0
a + ©MNA = 0.
V = 220 lb ‚
Ans.
M  220x = 0 Ans.
M = (220x) lb ft ‚ For 3 ft 6 x 6 5 ft + c ©Fy = 0;
220  800  V = 0 Ans.
V = 580 lb a + ©MNA = 0;
M + 800(x  3)  220x = 0 M = {  580x + 2400} lb ft ‚
Ans.
For 5 ft 6 x … 6 ft + c ©Fy = 0; a + ©MNA = 0;
V  500 = 0
V = 500 lb ‚
Ans.
 M  500(5.5  x)  250 = 0 Ans.
M = (500x  3000) lb ft
336
2 ft
0.5 ft
0.5 ft
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•6–17.
Draw the shear and moment diagrams for the cantilevered beam.
300 lb
200 lb/ft
A 6 ft
The freebody diagram of the beam’s left segment sectioned through an arbitrary point shown in Fig. b will be used to write the shear and moment equations. The intensity of the triangular distributed load at the point of sectioning is x w = 200a b = 33.33x 6
Referring to Fig. b,
1 (33.33x)(x)  V = 0 V = {  300  16.67x2 } lb (1) 2 1 x a + ©M = 0; M + (33.33x)(x) a b + 300x = 0 M = {  300x  5.556x 3 } lb # ft (2) 2 3 + c©F y = 0;
 300 
The shear and moment diagrams shown in Figs. c and d are plotted using Eqs. (1) and (2), respectively.
337
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6–18. Draw the shear and moment diagrams for the beam, and determine the shear and moment throughout the beam as functions of x.
2 kip/ft
10 kip
8 kip 40 kipft
Support Reactions: As shown on FBD. Shear and Moment Function:
x 6 ft
For 0 … x 6 6 ft: + c ©Fy = 0;
30.0  2x  V = 0 V = {30.0  2x} kip
a + ©MNA = 0;
4 ft
Ans.
x M + 216 + 2x a b  30.0x = 0 2
M = {  x2 + 30.0x  216} kip # ft
Ans.
For 6 ft 6 x … 10 ft: + c ©Fy = 0; a + ©MNA = 0;
V  8 = 0
V = 8.00 kip
Ans.
 M  8(10  x)  40 = 0 M = {8.00x  120} kip # ft
Ans.
6–19. Draw the shear and moment diagrams for the beam.
2 kip/ ft 30 kipft B A 5 ft
338
5 ft
5 ft
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*6–20. Draw the shear and moment diagrams for the simply supported beam.
10 kN 10 kN/m
A
B
3m
Since the area under the curved shear diagram can not be computed directly, the value of the moment at x = 3 m will be computed using the method of sections. By referring to the freebody diagram shown in Fig. b, a + ©M = 0; Mx= 3 m +
1 (10)(3)(1)  20(3) = 0 2
M x= 3m = 45 kN # m
339
Ans.
3m
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•6–21.
The beam is subjected to the uniform distributed load shown. Draw the shear and moment diagrams for the beam.
2 kN/m
Equations of Equilibrium: Referring to the freebody diagram of the beam shown in Fig. a, a + ©M A = 0;
3 FBC a b(2)  2(3)(1.5) = 0 5
A
B
1.5 m
FBC = 7.5 kN + c ©Fy = 0;
C
3 A y + 7.5a b  2(3) = 0 5 A y = 1.5 kN
3 Shear and Moment Diagram: The vertical component of F BC is A FBC B y = 7.5a b 5 = 4.5 kN. The shear and moment diagrams are shown in Figs. c and d.
340
2m
1m
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6–22. Draw the shear and moment diagrams for the overhang beam.
4 kN/m
A B 3m
Since the loading is discontinuous at support B, the shear and moment equations must be written for regions 0 … x 6 3 m and 3 m 6 x … 6 m of the beam. The freebody diagram of the beam’s segment sectioned through an arbitrary point within these two regions is shown in Figs. b and c. Region 0 … x 6 3 m, Fig. b 1 4 a xb (x)  V = 0 2 3 1 4 x a + ©M = 0; M + a x b (x)a b + 4x = 0 2 3 3 + c©F y = 0;
Region 3 m 6 x … 6 m , Fig. c + c©F y = 0;
V = e
4 
(1)
2 M = e  x3  4x f kN # m (2) 9 V = {24  4x} kN
(3)
M = {  2(6  x)2 }kN # m
(4)
V  4(6  x) = 0
1 a + ©M = 0;  M  4(6  x)c (6  x) d = 0 2
2 2 x  4 f kN 3
The shear diagram shown in Fig. d is plotted using Eqs. (1) and (3). The value of shear just to the left and just to the right of the support is evaluated using Eqs. (1) and (3), respectively. 2 Vx= 3 m  =  (3 2 )  4 = 10 kN 3 Vx=3 m + = 24  4(3) = 12 kN The moment diagram shown in Fig. e is plotted using Eqs. (2) and (4). The value of the moment at support B is evaluated using either Eq. (2) or Eq. (4). 2 M x=3 m =  (33)  4(3) = 18 kN # m 9 or Mx= 3 m = 2(6  3) 2 = 18 kN # m
341
3m
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6–23. Draw the shear and moment diagrams for the beam. It is supported by a smooth plate at A which slides within the groove and so it cannot support a vertical force, although it can support a moment and axial load.
w
B
A
L
*6–24. Determine the placement distance a of the roller support so that the largest absolute value of the moment is a minimum. Draw the shear and moment diagrams for this condition.
w
A a
wL 2 wL  wx = 0 2a
+ c ©Fy = 0;
L
L2 2a 2 x wL + wxa b  a wL bx = 0 2 2a
x = L a + ©M = 0;
M max (+)
Substitute x = L 
L2 ; 2a
Mmax (+) = awL =
wL2 2a
b aL 
L2 2 w aL b 2 2a
© M = 0;
w L2 L2 2 b aL  b 2a 2 2a
M max ()  w(L  a) Mmax () =
(L  a) = 0 2
w(L  a) 2 2
To get absolute minimum moment, Mmax (+) = Mmax () w L2 2 w 2 (L ) = (L  a) 2 2a 2 L a =
B
L2 = L  a 2a L
Ans.
‚ 22 342
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6–25. The beam is subjected to the uniformly distributed moment m (moment> length ). Draw the shear and moment diagrams for the beam.
m A L
Support Reactions: As shown on FBD. Shear and Moment Function: + c ©Fy = 0;
V = 0
a + ©MNA = 0;
M + mx  mL = 0
M = m(L  x)
Shear and Moment Diagram:
6–27. Draw the shear and moment diagrams for the beam. + c©F y = 0;
w0 L 4

w0
1 w0 x a b(x) = 0 L 2
B
x = 0.7071 L a + ©MNA = 0;
M +
Substitute x = 0.7071L ,
w0L 1 w 0x x L a b (x) a b ax  b = 0 L 2 3 4 3
M = 0.0345 w0L 2
343
L 3
A
2L 3
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*6–28. Draw the shear and moment diagrams for the beam.
w0
B
A L – 3
Support Reactions: As shown on FBD. Shear and Moment Diagram: Shear and moment at x = L>3 can be determined using the method of sections. + c ©Fy = 0;
w 0L w0 L  V = 0 3 6
a + ©M NA = 0;
M +
V =
w 0L 6
w0 L L w0 L L a b a b = 0 6 9 3 3 M =
5w 0 L2 54
344
L – 3
L – 3
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•6–29.
Draw the shear and moment diagrams for the beam.
5 kN/m
5 kN/m
B
A 4.5 m
From FBD(a) + c ©Fy = 0; a + ©M NA = 0;
9.375  0.5556x 2 = 0
x = 4.108 m
M + (0.5556) A 4.1082B a M = 25.67 kN # m
4.108 b  9.375(4.108) = 0 3
From FBD(b) a + ©M NA = 0;
M + 11.25(1.5)  9.375(4.5) = 0 M = 25.31 kN # m
345
4.5 m
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6–30. Draw the shear and moment diagrams for the compound beam.
150 lb/ft
150 lb/ft
A
B 6 ft
Support Reactions: From the FBD of segment AB a + ©M B = 0;
450(4)  Ay (6) = 0
Ay = 300.0 lb
+ c ©F y = 0;
B y  450 + 300.0 = 0
B y = 150.0 lb
+ ©F = 0; : x
Bx = 0
From the FBD of segment BC a + ©MC = 0;
225(1) + 150.0(3)  M C = 0 MC = 675.0 lb # ft
+ c ©F y = 0; + ©F = 0; : x
C y  150.0  225 = 0
C y = 375.0 lb
Cx = 0
Shear and Moment Diagram: The maximum positive moment occurs when V = 0. + c ©F y = 0; a + ©MNA = 0;
150.0  12.5x2 = 0
x = 3.464 ft
150(3.464)  12.5 A 3.464 2B a
3.464 b  Mmax = 0 3
Mmax = 346.4 lb # ft
346
C 3 ft
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6–31. Draw the shear and moment diagrams for the beam and determine the shear and moment in the beam as functions of x.
w0
Support Reactions: As shown on FBD.
A
B x
Shear and Moment Functions:
L – 2
For 0 … x 6 L> 2 + c ©Fy = 0;
3w0 L  w0x  V = 0 4 w0
V = a + ©M NA = 0;
+ c ©Fy = 0;
(3L  4x)
V 
w0
24
A12x 2 + 18Lx  7L 2)
Ans.
1 2w0 c (L  x) d (L  x) = 0 2 L V =
a + ©M NA = 0;
Ans.
7w0 L 2 3w 0 L x x + w 0 xa b + M = 0 24 4 2 M =
For L>2 6 x … L
4
M 
w0 (L  x)2 L
Ans.
L  x 1 2w0 c (L  x) d (L  x) a b = 0 2 L 3
M = 
w0 (L  x) 3 3L
Ans.
347
L – 2
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*6–32. The smooth pin is supported by two leaves A and B and subjected to a compressive load of 0.4 kNm caused by bar C. Determine the intensity of the distributed load w 0 of the leaves on the pin and draw the shear and moment diagram for the pin. A
+ c ©Fy = 0;
0.4 kN/m
C
1 2(w 0)(20)a b  60(0.4) = 0 2
B
w0
20 mm 60 mm 20 mm
w 0 = 1.2 kN> m
Ans.
•6–33.
The ski supports the 180lb weight of the man. If the snow loading on its bottom surface is trapezoidal as shown, determine the intensity w, and then draw the shear and moment diagrams for the ski.
180 lb 3 ft
w 1.5 ft
Ski: + c ©Fy = 0;
1 1 w(1.5) + 3w + w(1.5)  180 = 0 2 2 w = 40.0 lb>ft
Ans.
Segment: + c ©Fy = 0;
30  V = 0;
a + ©M = 0;
M  30(0.5) = 0;
w0
V = 30.0 lb M = 15.0 lb # ft
348
w 3 ft
1.5 ft
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6–34. Draw the shear and moment diagrams for the compound beam.
5 kN 3 kN/m
A B 3m
6–35. Draw the shear and moment diagrams for the beam and determine the shear and moment as functions of x.
A x 3m
200  V = 0
V = 200 N
Ans.
M  200 x = 0 M = (200 x) N # m
Ans.
For 3 m 6 x … 6 m: 200  200(x  3) V = e
Set V = 0, x = 3.873 m a + ©M NA = 0;
M +
1 200 c (x  3) d (x  3)  V = 0 2 3
100 2 x + 500 f N 3
Ans.
x  3 1 200 c (x  3) d (x  3) a b 2 3 3 + 200(x  3)a
M = e
100 3 x + 500x  600 f N # m 9
1.5 m
B
For 0 … x 6 3 m:
+ c ©Fy = 0;
1.5 m
200 N/ m
Shear and Moment Functions:
a + ©MNA = 0;
3m
400 N/m
Support Reactions: As shown on FBD.
+ c ©Fy = 0;
D
C
x  3 b  200x = 0 2 Ans.
Substitute x = 3.87 m, M = 691 N # m
349
3m
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*6–36. Draw the shear and moment diagrams for the overhang beam.
18 kN
6 kN
A B 2m
6–37. Draw the shear and moment diagrams for the beam.
2m
M 10 kNm
2m
50 kN/m
50 kN/m
B A 4.5 m
350
4.5 m
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6–38. The deadweight loading along the centerline of the airplane wing is shown. If the wing is fixed to the fuselage at A, determine the reactions at A, and then draw the shear and moment diagram for the wing.
250 lb/ft
3000 lb
400 lb/ft
A
8 ft
Support Reactions: + c ©F y = 0;
2 ft
3 ft
15 000 lb
 1.00  3 + 15  1.25  0.375  Ay = 0
A y = 9.375 kip a + ©MA = 0;
Ans. 1.00(7.667) + 3(5)  15(3)
+ 1.25(2.5) + 0.375(1.667) + MA = 0 M A = 18.583 kip # ft = 18.6 kip # ft
Ans.
+ ©Fx = 0; :
Ans.
Ax = 0
Shear and Moment Diagram:
6–39. The compound beam consists of two segments that are pinned together at B. Draw the shear and moment diagrams if it supports the distributed loading shown. + c©F y = 0;
w
2wL 1w 2 x = 0 27 2L 4 L = 0.385 L A27 1w 1 2wL M + (0.385L) = 0 (0.385L)2 a b(0.385L) 2L 3 27 x =
a + ©M = 0;
C
A
M = 0.0190 wL2
351
B 2/ 3 L
1/ 3 L
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*6–40. Draw the shear and moment diagrams for the simply supported beam.
10 kN
10 kN 15 kNm
A
B 2m
6–41. Draw the shear and moment diagrams for the compound beam. The three segments are connected by pins at B and E.
3 kN
2m
2m
3 kN
0.8 kN/m B
E
A
F C 2m
352
1m
1m
D 2m
1m
1m
2m
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6–42. Draw the shear and moment diagrams for the compound beam.
5 kN/m
Support Reactions:
A
From the FBD of segment AB a + ©MA = 0; + c ©Fy = 0;
B 2m
By (2)  10.0(1) = 0
By = 5.00 kN
Ay  10.0 + 5.00 = 0
Ay = 5.00 kN
C 1m
D
1m
From the FBD of segment BD a + ©M C = 0;
5.00(1) + 10.0(0)  D y (1) = 0 Dy = 5.00 kN
+ c ©Fy = 0;
Cy  5.00  5.00  10.0 = 0 Cy = 20.0 kN
+ ©F = 0; : x
Bx = 0
From the FBD of segment AB + ©F = 0; : x
Ax = 0
Shear and Moment Diagram:
6–43. Draw the shear and moment diagrams for the beam. The two segments are joined together at B.
8 kip
3 kip/ft
A
C
B 3 ft
353
5 ft
8 ft
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*6–44. Draw the shear and moment diagrams for the beam.
w
8
1 2 x dx = 21.33 kip 8L 0
FR =
x =
1 8 3 8 10 x dx
21.33
8 kip/ ft 1 2 w x 8
= 6.0 ft
x B
A 8 ft
•6–45.
FR =
Draw the shear and moment diagrams for the beam. LA
dA =
xdA A L
L 0
L
wdx =
L
L L 0 w0 2
x2 dx =
w
w0 L
w
3
w0
L
w0
x 3dx 3L L2 L0 x = = = w0L 4 dA 3 LA + c ©Fy = 0;
w0 2 x L2
w 0L 12

w0 x
A L
3
3L2
= 0
1 1 >3 x = a b L = 0.630 L 4 w0 L w0x 3 1 a + ©M = 0; (x) 2 a xb  M = 0 12 3L 4 M =
w0 Lx 12

B
w 0x 4 12L 2
Substitute x = 0.630L M = 0.0394 w0L2
354
x
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6–46. Draw the shear and moment diagrams for the beam.
w
w0
p w w0 sin L –x
L
p 2w0 L sina x b dx = dA = w 0 L p LA L0
FR =
A L – 2
L – 2
6–47. A member having the dimensions shown is used to resist an internal bending moment of M = 90 kN # m . Determine the maximum stress in the member if the moment is applied (a) about the z axis (as shown) (b) about the y axis. Sketch the stress distribution for each case.
200 mm y 150 mm
The moment of inertia of the crosssection about z and y axes are Iz =
1 (0.2)(0.153) = 56.25(10  6 ) m4 12
Iy =
1 3 (0.15)(0.23) = 0.1(10 ) m4 12
M z x
For the bending about z axis, c = 0.075 m . 3
smax =
90(10 ) (0.075) Mc = = 120(10 6)Pa = 120 MPa 6 Iz 56.25 (10 )
Ans.
For the bending about y axis, C = 0.1 m . 3
smax =
x
B
90(10 ) (0.1) Mc = = 90 (106)Pa = 90 MPa Iy 0.1 (10 3 )
Ans.
The bending stress distribution for bending about z and y axes are shown in Fig. a and b respectively.
355
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*6–48. Determine the moment M that will produce a maximum stress of 10 ksi on the cross section.
0.5 in. A 0.5 in.
3 in.
0.5 in. B
C
3 in. M
10 in.
D 0.5 in.
Section Properties: y = =
© yA ©A 0.25(4)(0.5) + 2[2(3)(0.5)] + 5.5(10)(0.5) = 3.40 in. 4(0.5) + 2[(3)(0.5)] + 10(0.5)
INA =
1 (4) A 0.53 B + 4(0.5)(3.40  0.25)2 12 + 2c
+
1 (0.5)(33 ) + 0.5(3)(3.40  2)2 d 12
1 (0.5) A 10 3B + 0.5(10)(5.5  3.40)2 12
= 91.73 in4
Maximum Bending Stress: Applying the flexure formula s max = 10 =
Mc I M (10.5  3.4) 91.73
M = 129.2 kip # in = 10.8 kip # ft
Ans.
356
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•6–49.
Determine the maximum tensile and compressive bending stress in the beam if it is subjected to a moment of M = 4 kip # ft.
0.5 in. A 0.5 in.
3 in.
0.5 in. B
C
3 in. M 10 in.
D 0.5 in.
Section Properties: y = =
© yA ©A 0.25(4)(0.5) + 2[2(3)(0.5)] + 5.5(10)(0.5) = 3.40 in. 4(0.5) + 2[(3)(0.5)] + 10(0.5)
I NA =
1 (4) A 0.5 3B + 4(0.5)(3.40  0.25) 2 12 + 2c
1 3 2 (0.5)(3 ) + 0.5(3)(3.40  2) d 12
+ = 91.73 in4
1 (0.5) A 103 B + 0.5(10)(5.5  3.40)2 12
Maximum Bending Stress: Applying the flexure formula smax =
Mc I
(s t)max =
4(10 3)(12)(10.5  3.40) = 3715.12 psi = 3.72 ksi 91.73
Ans.
(s c)max =
4(10 3)(12)(3.40) = 1779.07 psi = 1.78 ksi 91.73
Ans.
357
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6–50. The channel strut is used as a guide rail for a trolley. If the maximum moment in the strut is M = 30 N # m, determine the bending stress at points A, B, and C. y=
50 mm C 5 mm 5 mm
B
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)]
30 mm
= 13.24 mm A
I = c
1 (50)(5 3) + 50(5)(13.24  2.5)2 d 12
+ c
5 mm
5 mm 5 mm 7 mm 10 mm7 mm
1 (34)(53) + 34(5)(13.24  7.5)2 d 12
+ 2c
1 1 3 2 3 2 (5)(20 ) + 5(20)(20  13.24) d + 2 c (12)(5 ) + 12(5)(32.5  13.24) d 12 12
= 0.095883(10 6 ) m4 sA =
sB =
30(35  13.24)(10  3 ) 0.095883(10
6
)
30(13.24  10)(10  3 ) 6
0.095883(10 )
= 6.81 MPa
Ans.
= 1.01 MPa
Ans.
6–51. The channel strut is used as a guide rail for a trolley. If the allowable bending stress for the material is s allow = 175 MPa, determine the maximum bending moment the strut will resist.
50 mm C 5 mm
5 mm
B
sC =
30(13.24)(10  3 ) 0.095883(10
6
)
= 4.14 MPa
©y2 A 2.5(50)(5) + 7.5(34)(5) + 2[20(5)(20)] + 2[(32.5)(12)(5)] = = 13.24 mm y= ©A 50(5) + 34(5) + 2[5(20)] + 2[(12)(5)] I = c
1 1 (50)(5 3) + 50(5)(13.24  2.5)2 d + c (34)(5 3 ) + 34(5)(13.24  7.5)2 d 12 12 + 2c
1 1 (5)(203 ) + 5(20)(20  13.24) 2 d + 2 c (12)(53 ) + 12(5)(32.5  13.24)2 d 12 12
= 0.095883(10 6 ) m4 s =
Mc ; I
175(10 6) =
30 mm
Ans.
M(35  13.24)(10  3 ) 0.095883(10  6 )
M = 771 N # m
Ans.
358
A
5 mm
5 mm 5 mm 7 mm 10 mm7 mm
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*6–52. The beam is subjected to a moment M. Determine the percentage of this moment that is resisted by the stresses acting on both the top and bottom boards, A and B, of the beam.
A 25 mm
I =
M
D
Section Property: 1 1 (0.2) A 0.23 B (0.15)A 0.15 3B = 91.14583 A 10  6 B m4 12 12
150 mm 25 mm 25 mm
Bending Stress: Applying the flexure formula
B 150 mm
25 mm
My I
s = sE =
sD =
M(0.1) 91.14583(10  6 ) M(0.075) 6
91.14583(10 )
= 1097.143 M
= 822.857 M
Resultant Force and Moment: For board A or B F = 822.857M(0.025)(0.2) +
1 (1097.143M  822.857M)(0.025)(0.2) 2
= 4.800 M M¿ = F(0.17619) = 4.80M(0.17619) = 0.8457 M M¿ b = 0.8457(100%) = 84.6 % sc a M
Ans.
•6–53.
Determine the moment M that should be applied to the beam in order to create a compressive stress at point D of s D = 30 MPa. Also sketch the stress distribution acting over the cross section and compute the maximum stress developed in the beam.
A 25 mm
Section Property:
150 mm
1 1 (0.2) A 0.2 3B (0.15) A 0.153 B = 91.14583 A 10 6 B m4 I = 12 12
25 mm 25 mm
Bending Stress: Applying the flexure formula s = 30A 10 6 B =
My I M(0.075) 91.14583(10  6 )
M = 36458 N # m = 36.5 kN # m
s max =
M
D
Ans.
36458(0.1) Mc =  6 = 40.0 MPa I 91.14583(10 )
Ans.
359
B 150 mm
25 mm
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6–54. The beam is made from three boards nailed together as shown. If the moment acting on the cross section is M = 600 N # m, determine the maximum bending stress in the beam. Sketch a threedimensional view of the stress distribution acting over the cross section.
25 mm
150 mm 20 mm
(0.0125)(0.24)(0.025) + 2 (0.1)(0.15)(0.2) = 0.05625 m y= 0.24 (0.025) + 2 (0.15)(0.02) I =
200 mm M 600 Nm
1 (0.24)(0.025 3) + (0.24)(0.025)(0.04375 2) 12 + 2a
20 mm
1 b(0.02)(0.153 ) + 2(0.15)(0.02)(0.04375 2) 12
= 34.53125 (10  6 ) m4 s max = s B = =
Mc I 600 (0.175  0.05625) 6
34.53125 (10 )
= 2.06 MPa sC =
Ans.
600 (0.05625) My = = 0.977 MPa I 34.53125 (10  6)
6–55. The beam is made from three boards nailed together as shown. If the moment acting on the cross section is M = 600 N # m, determine the resultant force the bending stress produces on the top board.
150 mm
(0.0125)(0.24)(0.025) + 2 (0.15)(0.1)(0.02) = 0.05625 m 0.24 (0.025) + 2 (0.15)(0.02)
y= I =
25 mm
20 mm 200 mm M 600 Nm
1 3 2 (0.24)(0.025 ) + (0.24)(0.025)(0.04375 ) 12 + 2a
20 mm
1 b(0.02)(0.15 3) + 2(0.15)(0.02)(0.04375 2) 12
= 34.53125 (10  6 ) m4 s1 =
600(0.05625) My = = 0.9774 MPa I 34.53125(10  6)
sb =
My 600(0.05625  0.025) = = 0.5430 MPa 6 I 34.53125(10 )
F =
1 (0.025)(0.9774 + 0.5430)(106)(0.240) = 4.56 kN 2
Ans.
360
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*6–56. The aluminum strut has a crosssectional area in the form of a cross. If it is subjected to the moment M = 8 kN # m, determine the bending stress acting at points A and B,and show the results acting on volume elements located at these points.
A 100 mm 20 mm 100 mm
B
M ⫽ 8 kN⭈m 20 mm 50 mm 50 mm
Section Property: I =
1 1 (0.02) A 0.22 3B + (0.1) A 0.023B = 17.8133 A 10  6 B m4 12 12
Bending Stress: Applying the flexure formula s = 8(10 3)(0.11)
sA =
sB =
6
17.8133(10 ) 8(10 3)(0.01) 17.8133(10
6
)
My I
= 49.4 MPa (C)
Ans.
= 4.49 MPa (T)
Ans.
•6–57.
The aluminum strut has a crosssectional area in the form of a cross. If it is subjected to the moment M = 8 kN # m, determine the maximum bending stress in the beam, and sketch a threedimensional view of the stress distribution acting over the entire crosssectional area.
A 100 mm 20 mm 100 mm
B 20 mm
M ⫽ 8 kN⭈m 50 mm
50 mm
Section Property: I =
1 1 (0.02) A 0.22 3B + (0.1) A 0.023B = 17.8133 A 10  6 B m4 12 12
Bending Stress: Applying the flexure formula s max = s max =
8(10 3 )(0.11) 6
17.8133(10 )
s y = 0.01m =
= 49.4 MPa
8(10 3)(0.01) 17.8133(10
Mc My and s = , I I
6
)
Ans.
= 4.49 MPa
361
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6–58. If the beam is subjected to an internal moment of M = 100 kip # ft, determine the maximum tensile and compressive bending stress in the beam.
3 in. 3 in. 6 in. M 2 in.
1.5 in.
Section Properties: The neutral axis passes through centroid C of the cross section as shown in Fig. a. The location of C is ©yA y = = ©A
4(8)(6)  2c p A1.5 2B d 8(6)  p A 1.5 2 B
= 4.3454 in.
Thus, the moment of inertia of the cross section about the neutral axis is I = ©I + Ad2 =
1 1 (6) a 83 b + 6(8) A 4.3454  4 B 2  B pa 1.54 b + p a 1.5 2b A 4.3454  2 B 2 R 12 4
= 218.87 in4
Maximum Bending Stress: The maximum compressive and tensile bending stress occurs at the top and bottom edges of the cross section. Mc
A smax BT = I = A smax BC =
100(12)(4.3454) = 23.8 ksi (T) 218.87
Ans.
My 100(12)(8  4.3454) = = 20.0 ksi (C) 218.87 I
362
Ans.
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6–59. If the beam is made of material having an allowable tensile and compressive stress of (s allow) t = 24 ksi and (sallow)c = 22 ksi, respectively, determine the maximum allowable internal moment M that can be applied to the beam.
3 in. 3 in. 6 in. M 2 in.
1.5 in.
Section Properties: The neutral axis passes through centroid C of the cross section as shown in Fig. a. The location of C is ©yA y = = ©A
4(8)(6)  2c pA 1.5 2 B d 8(6)  p A 1.52 B
= 4.3454 in.
Thus, the moment of inertia of the cross section about the neutral axis is I = ©I + Ad 2 =
1 1 (6) A 8 3 B + 6(8) A 4.3454  4B 2  B p A 1.5 4 B + p A1.5 2B A 4.3454  2 B 2 R 12 4
= 218.87 in 4
Allowable Bending Stress: The maximum compressive and tensile bending stress occurs at the top and bottom edges of the cross section. For the top edge, (sallow) c =
My I
;
22 =
M(8  4.3454) 218.87 M = 1317.53 kip # ina
For the bottom edge, Mc
A smax B t = I ;
24 =
1 ft b = 109.79 kip # ft 12 in.
M(4.3454) 218.87
M = 1208.82 kip # ina
1 ft b = 101 kip # ft (controls) 12 in.
363
Ans.
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*6–60. The beam is constructed from four boards as shown. If it is subjected to a moment of Mz = 16 kip # ft, determine the stress at points A and B. Sketch a threedimensional view of the stress distribution.
y
A
C
1 in. 10 in. 1 in. 10 in.
y=
2[5(10)(1)] + 10.5(16)(1) + 16(10)(1) 2(10)(1) + 16(1) + 10(1)
Mz 16 kip ft
z
= 9.3043 in.
14 in.
1 1 (1)(103 ) + 1(10)(9.3043  5)2 d + (16)(13) + 16(1)(10.5  9.3043) 2 I = 2c 12 12 +
B
1 in.
x
1 in.
1 (1)(103) + 1(10)(16  9.3043) 2 = 1093.07 in 4 12
sA =
16(12)(21  9.3043) Mc = = 2.05 ksi 1093.07 I
Ans.
sB =
16(12)(9.3043) My = = 1.63 ksi 1093.07 I
Ans.
•6–61.
The beam is constructed from four boards as shown. If it is subjected to a moment of Mz = 16 kip # ft, determine the resultant force the stress produces on the top board C.
y
A
C
1 in. 10 in. 1 in. 10 in.
y =
2[5(10)(1)] + 10.5(16)(1) + 16(10)(1) = 9.3043 in. 2(10)(1) + 16(1) + 10(1)
Mz 16 kip ft
z
14 in.
1 1 3 2 3 2 (1)(10 ) + (10)(9.3043  5) d + (16)(1 ) + 16(1)(10.5  9.3043) I = 2c 12 12 +
1 (1)(103 ) + 1(10)(16  9.3043) 2 = 1093.07 in 4 12
sA =
Mc 16(12)(21  9.3043) = = 2.0544 ksi 1093.07 I
sD =
My 16(12)(11  9.3043) = = 0.2978 ksi 1093.07 I
(F R)C =
1 (2.0544 + 0.2978)(10)(1) = 11.8 kip 2
Ans.
364
1 in.
B
1 in.
x
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6–62. A box beam is constructed from four pieces of wood, glued together as shown. If the moment acting on the cross section is 10 kN # m, determine the stress at points A and B and show the results acting on volume elements located at these points.
20 mm
160 mm
25 mm A 250 mm
25 mm
B
M 10 kNm
The moment of inertia of the crosssection about the neutral axis is I =
1 1 (0.2)(0.33 ) (0.16)(0.253) = 0.2417(10  3) m4 . 12 12
For point A, y A = C = 0.15 m. sA =
10(103) (0.15) My A = = 6.207(10 6)Pa = 6.21 MPa (C) 0.2417(10 3 ) I
Ans.
For point B, yB = 0.125 m. sB =
My B 10(10 3)(0.125) = = 5.172(106)Pa = 5.17 MPa (T) 0.2417(10 3 ) I
Ans.
The state of stress at point A and B are represented by the volume element shown in Figs. a and b respectively.
365
20 mm
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6–63. Determine the dimension a of a beam having a square cross section in terms of the radius r of a beam with a circular cross section if both beams are subjected to the same internal moment which results in the same maximum bending stress.
a a r
Section Properties: The moments of inertia of the square and circular cross sections about the neutral axis are 1 a4 a A a3 B = 12 12
IS =
IC =
1 4 pr 4
Maximum Bending Stress: For the square cross section, c = a>2.
A s maxB S =
Mc M(a>2) 6M = 4 = 3 IS a >12 a
A s maxB c =
4M Mc Mr =  3 1 Ic pr pr4 4
For the circular cross section, c = r .
It is required that
A smax B S = A smax BC
6M 4M = 3 a pr 3
Ans.
a = 1.677r
*6–64. The steel rod having a diameter of 1 in. is subjected to an internal moment of M = 300 lb # ft. Determine the stress created at points A and B. Also, sketch a threedimensional view of the stress distribution acting over the cross section. I =
B
p 4 p r = (0.5 4) = 0.0490874 in 4 4 4
sA = sB
A
M 300 lbft 45
300(12)(0.5) Mc = = 36.7 ksi 0.0490874 I
Ans. 0.5 in.
300(12)(0.5 sin 45°) My = = = 25.9 ksi 0.0490874 I
Ans.
366
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•6–65.
If the moment acting on the cross section of the beam is M = 4 kip # ft, determine the maximum bending stress in the beam. Sketch a threedimensional view of the stress distribution acting over the cross section.
A
1.5 in. 12 in.
The moment of inertia of the crosssection about the neutral axis is
12 in.
1 1 3 3 4 I = (12)(15 ) (10.5)(12 ) = 1863 in 12 12
M 1.5 in.
1.5 in.
Along the top edge of the flange y = c = 7.5 in . Thus 3
4(10 )(12)(7.5) Mc = = = 193 psi I 1863
smax
Ans.
Along the bottom edge to the flange, y = 6 in . Thus s =
4(10 3)(12)(6) My = = 155 psi 1863 I
6–66. If M = 4 kip # ft, determine the resultant force the bending stress produces on the top board A of the beam. The moment of inertia of the crosssection about the neutral axis is
12 in.
1 1 3 3 4 (12)(15 ) (10.5)(12 ) = 1863 in 12 12
I =
12 in. M
Along the top edge of the flange y = c = 7.5 in . Thus
1.5 in.
3
smax =
4(10 )(12)(7.5) Mc = = 193.24 psi I 1863
Along the bottom edge of the flange, y = 6 in . Thus s =
4(10 3)(12)(6) My = = 154.59 psi 1863 I
The resultant force acting on board A is equal to the volume of the trapezoidal stress block shown in Fig. a. FR =
A
1.5 in.
1 (193.24 + 154.59)(1.5)(12) 2
= 3130.43 lb Ans.
= 3.13 kip
367
1.5 in.
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6–67. The rod is supported by smooth journal bearings at A and B that only exert vertical reactions on the shaft. If d = 90 mm, determine the absolute maximum bending stress in the beam, and sketch the stress distribution acting over the cross section.
12 kN/m d A
B 3m
Absolute Maximum Bending Stress: The maximum moment is M max = 11.34 kN # m as indicated on the moment diagram. Applying the flexure formula smax =
Mmax c I 3
=
11.34(10 )(0.045) p (0.045 4) 4 Ans.
= 158 MPa
368
1.5 m
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*6–68. The rod is supported by smooth journal bearings at A and B that only exert vertical reactions on the shaft. Determine its smallest diameter d if the allowable bending stress is s allow = 180 MPa.
12 kN/m d A
B 3m
1.5 m
Allowable Bending Stress: The maximum moment is Mmax = 11.34 kN # m as indicated on the moment diagram. Applying the flexure formula smax = sallow = 180 A 10 6B =
Mmax c I 11.34(103) A d2 B p 4
A d2 B4
Ans.
d = 0.08626 m = 86.3 mm
•6–69.
Two designs for a beam are to be considered. Determine which one will support a moment of M = 150 kN # m with the least amount of bending stress. What is that stress?
200 mm
200 mm
30 mm
15 mm
300 mm 30 mm
Section Property:
300 mm 15 mm
For section (a) I =
1 1 (0.2) A 0.333 B  (0.17)(0.3) 3 = 0.21645(10  3) m 4 12 12
For section (b) I =
15 mm (a)
1 1 (0.2) A 0.363 B  (0.185) A 0.33 B = 0.36135(10  3 ) m4 12 12
Maximum Bending Stress: Applying the flexure formula smax =
Mc I
For section (a) s max =
150(103 )(0.165) = 114.3 MPa 0.21645(10  3)
For section (b) s max =
150(103 )(0.18) = 74.72 MPa = 74.7 MPa 0.36135(10  3 )
Ans.
369
30 mm (b)
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6–70. The simply supported truss is subjected to the central distributed load. Neglect the effect of the diagonal lacing and determine the absolute maximum bending stress in the truss. The top member is a pipe having an outer diameter of 1 in. 3 and thickness of 16 in., and the bottom member is a solid rod 1 having a diameter of 2 in. y =
100 lb/ft
6 ft
6 ft
5.75 in.
6 ft
©yA 0 + (6.50)(0.4786) = = 4.6091 in. ©A 0.4786 + 0.19635
I = c
1 1 1 4 4 2 4 p(0.5)  p(0.3125) d + 0.4786(6.50  4.6091) + p(0.25) 4 4 4
+ 0.19635(4.6091)2 = 5.9271 in4
Mmax = 300(9  1.5)(12) = 27 000 lb # in. smax =
27 000(4.6091 + 0.25) Mc = 5.9271 I Ans.
= 22.1 ksi
6–71. The axle of the freight car is subjected to wheel loadings of 20 kip. If it is supported by two journal bearings at C and D, determine the maximum bending stress developed at the center of the axle, where the diameter is 5.5 in. A
C
10 in. 20 kip
s max =
200(2.75) Mc =1 = 12.2 ksi I p(2.75) 4 4
Ans.
370
B
60 in.
D
10 in. 20 kip
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*6–72. The steel beam has the crosssectional area shown. Determine the largest intensity of distributed load w0 that it can support so that the maximum bending stress in the beam does not exceed smax = 22 ksi.
w0
12 ft
12 ft 8 in. 0.30 in. 10 in.
0.3 in.
Support Reactions: As shown on FBD.
0.30 in.
Internal Moment: The maximum moment occurs at mid span. The maximum moment is determined using the method of sections. Section Property: I =
1 1 3 3 4 (8) A 10.6 B (7.7)A 10 B = 152.344 in 12 12
Absolute Maximum Bending Stress: The maximum moment is Mmax = 48.0w 0 as indicated on the FBD. Applying the flexure formula smax = 22 =
Mmax c I 48.0w 0 (12)(5.30) 152.344
w 0 = 1.10 kip>ft
Ans.
•6–73.
The steel beam has the crosssectional area shown. If w0 = 0.5 kip>ft, determine the maximum bending stress in the beam.
w0
12 ft
12 ft 8 in.
Support Reactions: As shown on FBD.
0.3 in.
0.30 in.
Internal Moment: The maximum moment occurs at mid span. The maximum moment is determined using the method of sections. Section Property: I =
1 1 (8) A 10.63 B (7.7)A 10 3 B = 152.344 in4 12 12
Absolute Maximum Bending Stress: The maximum moment is Mmax = 24.0 kip # ft as indicated on the FBD. Applying the flexure formula s max = =
Mmax c I
24.0(12)(5.30) 152.344 Ans.
= 10.0 ksi 371
0.30 in. 10 in.
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6–74. The boat has a weight of 2300 lb and a center of gravity at G. If it rests on the trailer at the smooth contact A and can be considered pinned at B, determine the absolute maximum bending stress developed in the main strut of the trailer. Consider the strut to be a boxbeam having the dimensions shown and pinned at C.
B 1 ft
G
C
A 3 ft
D
5 ft
4 ft
1.75 in.
1 ft 3 in.
1.75 in. 1.5 in.
Boat: + ©F = 0; : x a + ©M B = 0;
Bx = 0 NA(9) + 2300(5) = 0 NA = 1277.78 lb
+ c ©Fy = 0;
1277.78  2300 + B y = 0 By = 1022.22 lb
Assembly: a + ©M C = 0;
ND(10) + 2300(9) = 0 ND = 2070 lb
+ c ©Fy = 0;
C y + 2070  2300 = 0 Cy = 230 lb
I =
1 1 (1.75)(3)3 (1.5)(1.75) 3 = 3.2676 in4 12 12
s max =
Mc 3833.3(12)(1.5) = = 21.1 ksi I 3.2676
Ans.
372
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6–75. The shaft is supported by a smooth thrust bearing at A and smooth journal bearing at D. If the shaft has the cross section shown, determine the absolute maximum bending stress in the shaft.
40 mm A
B 0.75 m
C 1.5 m
0.75 m
3 kN
Shear and Moment Diagrams: As shown in Fig. a.
25 mm
D
3 kN
Maximum Moment: Due to symmetry, the maximum moment occurs in region BC of the shaft. Referring to the freebody diagram of the segment shown in Fig. b. Section Properties: The moment of inertia of the cross section about the neutral axis is I =
p A0.044  0.0254 B = 1.7038 A 10  6 Bm 4 4
Absolute Maximum Bending Stress:
2.25A 10 B (0.04) M maxc = = = 52.8 MPa 6 I 1.7038A 10 B 3
sallow
Ans.
*6–76. Determine the moment M that must be applied to the beam in order to create a maximum stress of 80 MPa.Also sketch the stress distribution acting over the cross section.
300 mm
20 mm
The moment of inertia of the crosssection about the neutral axis is M
1 1 (0.3)(0.33 ) (0.21)(0.263) = 0.36742(10 3 ) m4 I = 12 12
260 mm 30 mm
Thus, s max
Mc = ; I
80(10 6) =
20 mm 30 mm
M(0.15) 3
0.36742(10 )
M = 195.96 (10 3) N # m = 196 kN # m The bending stress distribution over the crosssection is shown in Fig. a.
373
Ans.
30 mm
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•6–77.
The steel beam has the crosssectional area shown. Determine the largest intensity of distributed load w that it can support so that the bending stress does not exceed s max = 22 ksi. I =
w
1 1 (8)(10.6)3 (7.7)(103 ) = 152.344 in4 12 12
s max = 22 =
8 ft
w
8 ft
8 ft 8 in.
Mc I
0.30 in. 0.3 in.
10 in. 0.30 in.
32w(12)(5.3) 152.344 Ans.
w = 1.65 kip>ft
6–78. The steel beam has the crosssectional area shown. If w = 5 kip>ft, determine the absolute maximum bending stress in the beam.
w
8 ft
w
8 ft
8 ft 8 in. 0.3 in.
0.30 in. 10 in. 0.30 in.
From Prob. 678: M = 32w = 32(5)(12) = 1920 kip # in. I = 152.344 in 4 s max =
Mc 1920(5.3) = = 66.8 ksi 152.344 I
Ans.
374
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6–79. If the beam ACB in Prob. 6–9 has a square cross section, 6 in. by 6 in., determine the absolute maximum bending stress in the beam.
15 kip 1 ft A
20 kip
C 4 ft
B
4 ft
4 ft
Mmax = 46.7 kip # ft s max =
46.7(103 )(12)(3) Mc = = 15.6 ksi 1 I (6)(63 ) 12
Ans.
*6–80. If the crane boom ABC in Prob. 6–3 has a rectangular cross section with a base of 2.5 in., determine its 1 required height h to the nearest 4 in. if the allowable bending stress is s allow = 24 ksi.
A
a + ©MA = 0; + c©F y = 0;
Ay +
+ ©F = 0; ; x
Ax 
smax =
4 (4000)  1200 = 0; 5
3 (4000) = 0; 5
5 ft B
4 ft
4 FB(3)  1200(8) = 0; 5
3 ft
FB = 4000 lb A y = 2000 lb
Ax = 2400 lb
6000(12) A h2 B Mc = 1 = 24(10) 3 3 I (2.5)(h ) 12
h = 2.68 in.
Ans.
Use h = 2.75 in.
Ans.
375
C
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•6–81.
If the reaction of the ballast on the railway tie can be assumed uniformly distributed over its length as shown, determine the maximum bending stress developed in the tie. The tie has the rectangular cross section with thickness t = 6 in.
15 kip 1.5 ft
15 kip 5 ft
1.5 ft
t
w
Support Reactions: Referring to the free  body diagram of the tie shown in Fig. a, we have + c ©Fy = 0;
w(8)  2(15) = 0 w = 3.75 kip>ft
Maximum Moment: The shear and moment diagrams are shown in Figs. b and c. As indicated on the moment diagram, the maximum moment is M max = 7.5 kip # ft. Absolute Maximum Bending Stress: s max =
12 in.
7.5(12)(3) Mmax c = = 1.25 ksi 1 I 3 (12)(6 ) 12
Ans.
376
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6–82. The reaction of the ballast on the railway tie can be assumed uniformly distributed over its length as shown. If the wood has an allowable bending stress of s allow = 1.5 ksi, determine the required minimum thickness t of the 1 rectangular cross sectional area of the tie to the nearest 8 in.
15 kip 1.5 ft
15 kip 5 ft
1.5 ft
t
w
Support Reactions: Referring to the freebody diagram of the tie shown in Fig. a, we have + c ©Fy = 0;
w(8)  2(15) = 0 w = 3.75 kip>ft
Maximum Moment: The shear and moment diagrams are shown in Figs. b and c. As indicated on the moment diagram, the maximum moment is Mmax
= 7.5 kip # ft.
Absolute Maximum Bending Stress:
smax =
t 7.5(12)a b 2 1.5 = 1 3 (12)t 12
Mc ; I
t = 5.48 in. Use
t = 5
12 in.
1 in. 2
Ans.
377
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6–83. Determine the absolute maximum bending stress in the tubular shaft if d i = 160 mm and d o = 200 mm.
15 kN/m 60 kN m d i do A
B 3m
Section Property: I =
p A 0.14  0.084B = 46.370 A10  6 B m 4 4
Absolute Maximum Bending Stress: The maximum moment is Mmax = 60.0 kN # m as indicated on the moment diagram. Applying the flexure formula s max = =
Mmaxc I
60.0(10 3)(0.1) 6
46.370(10 ) Ans.
= 129 MPa
378
1m
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*6–84. The tubular shaft is to have a cross section such that its inner diameter and outer diameter are related by di = 0.8d o . Determine these required dimensions if the allowable bending stress is s allow = 155 MPa.
15 kN/m 60 kN m d i do A
B 3m
Section Property: I =
4 0.8d o 4 do 4 dl 4 p p do 4 =  a b R = 0.009225pdo Ba b  a b R B 4 2 2 4 16 2
Allowable Bending Stress: The maximum moment is Mmax = 60.0 kN # m as indicated on the moment diagram. Applying the flexure formula s max = s allow =
Mmax c I d
155 A 10 B = 6
Thus,
60.0(10 3) A 2o B
4
0.009225pd o
do = 0.1883 m = 188 mm
Ans.
d l = 0.8do = 151 mm
Ans.
379
1m
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6–85. The wood beam has a rectangular cross section in the proportion shown. Determine its required dimension b if the allowable bending stress is s allow = 10 MPa.
500 N/m
1.5b A
B b 2m
Allowable Bending Stress: The maximum moment is Mmax = 562.5 N # m as indicated on the moment diagram. Applying the flexure formula smax = s allow = 10 A 10 6 B =
Mmax c I 562.5(0.75b) 1 (b)(1.5b) 3 12
Ans.
b = 0.05313 m = 53.1 mm
380
2m
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6–86. Determine the absolute maximum bending stress in the 2in.diameter shaft which is subjected to the concentrated forces. The journal bearings at A and B only support vertical forces.
800 lb 600 lb
A
15 in. B
15 in. 30 in.
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, Mmax = 15000 lb # in. The moment of inertia of the crosssection about the neutral axis is I =
p 4 4 (1 ) = 0.25 p in 4
Here, c = 1 in. Thus s max = =
Mmax c I 15000(1) 0.25 p
= 19.10(10 3) psi = 19.1 ksi
Ans.
381
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6–87. Determine the smallest allowable diameter of the shaft which is subjected to the concentrated forces. The journal bearings at A and B only support vertical forces. The allowable bending stress is sallow = 22 ksi.
800 lb 600 lb
A
15 in. B
15 in. 30 in.
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, Mmax = 15,000 lb #in The moment of inertia of the crosssection about the neutral axis is 4
I =
p d p 4 d a b = 4 2 64
Here, c = d>2 . Thus s allow =
Mmax c ; I
22(103) =
15000( d >2 ) pd 4>64
Ans.
d = 1.908 in = 2 in.
382
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*6–88. If the beam has a square cross section of 9 in. on each side, determine the absolute maximum bending stress in the beam.
1200 lb
800 lb/ft
B A 8 ft
Absolute Maximum Bending Stress: The maximum moment is Mmax = 44.8 kip # ft as indicated on moment diagram. Applying the flexure formula smax =
44.8(12)(4.5) Mmax c = = 4.42 ksi 1 3 I 12 (9)(9)
Ans.
•6–89.
If the compound beam in Prob. 6–42 has a square cross section, determine its dimension a if the allowable bending stress is s allow = 150 MPa. Allowable Bending Stress: The maximum moments is Mmax = 7.50 kN # m as indicated on moment diagram. Applying the flexure formula smax = s allow = 150 A 10 B = 6
M max c I 7.50(10 3) A a2 B 1 12
a4
Ans.
a = 0.06694 m = 66.9 mm
383
8 ft
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6–90. If the beam in Prob. 6–28 has a rectangular cross section with a width b and a height h, determine the absolute maximum bending stress in the beam.
Absolute Maximum Bending Stress: The maximum moments is Mmax = as indicated on the moment diagram. Applying the flexure formula
smax
Mmax c = = I
23w L 2 h 0 216 2 1 3 12 bh
A B
=
23w 0 L 2 36bh2
23w0 L 2 216
Ans.
384
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6–91. Determine the absolute maximum bending stress in the 80mmdiameter shaft which is subjected to the concentrated forces. The journal bearings at A and B only support vertical forces.
A
0.5 m
B
0.4 m
0.6 m
12 kN 20 kN
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, Mmax = 6 kN # m.
The moment of inertia of the crosssection about the neutral axis is I =
p 4
(0.04 4) = 0.64(10  6 )p m 4
Here, c = 0.04 m. Thus s max =
6(10 3)(0.04) Mmax c = I 0.64(10  6)p = 119.37(106 ) Pa Ans.
= 119 MPa
385
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*6–92. Determine the smallest allowable diameter of the shaft which is subjected to the concentrated forces. The journal bearings at A and B only support vertical forces. The allowable bending stress is sallow = 150 MPa.
A
0.5 m
B
0.4 m
0.6 m
12 kN 20 kN
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, Mmax = 6 kN # m .
The moment of inertia of the crosssection about the neutral axis is I =
p d 4 pd 4 a b = 4 2 64
Here, c = d>2 . Thus s allow =
Mmax c ; I
150(106 ) =
6(10 3)( d> 2 ) pd 4>64
d = 0.07413 m = 74.13 mm = 75 mm
386
Ans.
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•6–93.
The man has a mass of 78 kg and stands motionless at the end of the diving board. If the board has the cross section shown, determine the maximum normal strain developed in the board. The modulus of elasticity for the material is E = 125 GPa. Assume A is a pin and B is a roller.
30 mm A
1.5 m
Internal Moment: The maximum moment occurs at support B. The maximum moment is determined using the method of sections. Section Property: y = = I =
©yA ©A 0.01(0.35)(0.02) + 0.035(0.03)(0.03) = 0.012848 m 0.35(0.02) + 0.03(0.03) 1 (0.35) A 0.023 B + 0.35(0.02)(0.012848  0.01)2 12 +
1 (0.03)A 0.033 B + 0.03(0.03)(0.035  0.012848)2 12
= 0.79925 A 10  6 B m 4
Absolute Maximum Bending Stress: The maximum moment is Mmax = 1912.95 N # m as indicated on the FBD. Applying the flexure formula smax = =
Mmax c I 1912.95(0.05  0.012848) 0.79925(10  6)
= 88.92 MPa Absolute Maximum Normal Strain: Applying Hooke’s law, we have 6
e max =
s max 88.92(10 ) = = 0.711A 10  3B mm>mm E 125(109 )
Ans.
387
B
2.5 m
C
350 mm 20 mm
10 mm 10 mm 10 mm
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6–94. The two solid steel rods are bolted together along their length and support the loading shown. Assume the support at A is a pin and B is a roller. Determine the required diameter d of each of the rods if the allowable bending stress is s allow = 130 MPa.
20 kN/m
A B
2m
Section Property: I = 2B
80 kN
2m
p 5p 4 d 4 d 2 d a b + d2 a b R = 4 2 4 2 32
p
Allowable Bending Stress: The maximum moment is Mmax = 100 kN # m as indicated on moment diagram. Applying the flexure formula s max = s allow = 130 A10 6 B =
M max c I 100(10 3)(d) 5p 32
d4 Ans.
d = 0.1162 m = 116 mm
6–95. Solve Prob. 6–94 if the rods are rotated 90° so that both rods rest on the supports at A (pin) and B (roller).
20 kN/m
Section Property: I = 2B
A
p d 4 p 4 a b R = d 4 2 32
smax = s allow =
B
2m 2m
Allowable Bending Stress: The maximum moment is Mmax = 100 kN # m as indicated on the moment diagram. Applying the flexure formula
130A 10 6 B =
80 kN
Mmax c I
100(103)(d) p 4 d 32 Ans.
d = 0.1986 m = 199 mm
388
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*6–96. The chair is supported by an arm that is hinged so it rotates about the vertical axis at A. If the load on the chair is 180 lb and the arm is a hollow tube section having the dimensions shown, determine the maximum bending stress at section a–a.
180 lb
1 in. a 3 in.
A 8 in.
c + © M = 0;
Ix =
0.5 in.
M  180(8) = 0
M = 1440 lb # in.
1 1 (1)(3 3) (0.5)(2.53) = 1.59896 in4 12 12
smax =
1440 (1.5) Mc = = 1.35 ksi 1.59896 I
Ans.
•6–97.
A portion of the femur can be modeled as a tube having an inner diameter of 0.375 in. and an outer diameter of 1.25 in. Determine the maximum elastic static force P that can be applied to its center. Assume the bone to be roller supported at its ends. The s– P diagram for the bone mass is shown and is the same in tension as in compression.
P
s (ksi) 2.30 1.25 4 in.
0.02
I =
a
1 4 0.375 4 4 p C A1.25 2 B  A 2 B D = 0.11887 in 4
Mmax =
P (4) = 2P 2
Require smax = 1.25 ksi s max =
Mc I
1.25 =
2P(1.25> 2) 0.11887 Ans.
P = 0.119 kip = 119 lb
389
2.5 in.
0.05
P (in./ in.)
4 in.
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6–98. If the beam in Prob. 6–18 has a rectangular cross section with a width of 8 in. and a height of 16 in., determine the absolute maximum bending stress in the beam. Absolute Maximum Bending Stress: The maximum moment is Mmax as indicated on moment diagram. Applying the flexure formula smax =
216(12)(8) M max c = 1 = 7.59 ksi 3 I 12 (8)(16 )
16 in.
= 216 kip # ft
Ans.
8 in.
6–99. If the beam has a square cross section of 6 in. on each side, determine the absolute maximum bending stress in the beam.
400 lb/ft
B A 6 ft
The maximum moment occurs at the fixed support A. Referring to the FBD shown in Fig. a, a + ©M A = 0;
Mmax  400(6)(3) 
M max = 16800 lb # ft
1 (400)(6)(8) = 0 2
The moment of inertia of the about the neutral axis is I = s max =
1 (6)(63) = 108 in 4. Thus, 12
Mc 16800(12)(3) = 108 I Ans.
= 5600 psi = 5.60 ksi
390
6 ft
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*6–100. The steel beam has the crosssectional area shown. Determine the largest intensity of the distributed load w0 that it can support so that the maximum bending stress in the beam does not exceed sallow = 22 ksi.
w0
9 ft
9 ft
9 in. 0.25 in. 0.25 in.
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. As indicated on the moment diagram, Mmax = 27w o. The moment of inertia of the crosssection about the neutral axis is I =
1 1 (9)(12.5 3 ) (8.75)(123) 12 12
= 204.84375 in4 Here, ¢ = 6.25 in. Thus, s allow =
Mmax c ; I
22(10 3) =
(27wo )(12)(6.25) 204.84375
wo = 2 225.46 lb>ft Ans.
= 2.23 kip>ft
391
12 in. 0.25 in.
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•6–101.
The steel beam has the crosssectional area shown. If w0 = 2 kip>ft, determine the maximum bending stress in the beam.
w0
9 ft
9 ft
9 in. 0.25 in. 0.25 in.
The FBD of the beam is shown in Fig. a The shear and moment diagrams are shown in Fig. b and c, respectively. As indicated on the moment diagram, Mmax = 54 kip # ft . The moment of inertia of the I crosssection about the bending axis is I =
1 1 (9)A 12.53 B (8.75) A 12 3 B 12 12
= 204.84375 in 4
Here, c = 6.25 in . Thus s max = =
Mmax c I 54 (12)(6.25) 204.84375 Ans.
= 19.77 ksi = 19.8 ksi
392
12 in. 0.25 in.
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6–102. The bolster or main supporting girder of a truck body is subjected to the uniform distributed load. Determine the bending stress at points A and B.
1.5 kip/ft
A 8 ft
B 12 ft F2
F1 0.75 in. 6 in.
12 in.
0.5 in. A B
0.75 in.
Support Reactions: As shown on FBD. Internal Moment: Using the method of sections. + ©M NA = 0;
M + 12.0(4)  15.0(8) = 0 M = 72.0 kip # ft
Section Property: I =
1 1 (6) A 13.53 B (5.5) A 12 3B = 438.1875 in 4 12 12
Bending Stress: Applying the flexure formula s =
My I
sB =
72.0(12)(6.75) = 13.3 ksi 438.1875
Ans.
sA =
72.0(12)(6) = 11.8 ksi 438.1875
Ans.
393
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6–103. Determine the largest uniform distributed load w that can be supported so that the bending stress in the beam does not exceed s allow = 5 MPa.
w
The FBD of the beam is shown in Fig. a
0.5 m
The shear and moment diagrams are shown in Fig. b and c, respectively. As indicated on the moment diagram, Mmax  = 0.125 w.
150 mm
The moment of inertia of the crosssection is, I =
1 3 4 6 (0.075) A0.15 B = 21.09375 A 10 B m 12
Here, c = 0.075 w . Thus, s allow = 5 A 10 6 B =
Mmax c ; I 0.125w(0.075) 21.09375 A 10  6 B
Ans.
w = 11250 N> m = 11.25 kN> m
394
1m 75 mm
0.5 m
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*6–104. If w = 10 kN>m, determine the maximum bending stress in the beam. Sketch the stress distribution acting over the cross section.
w
Support Reactions. The FBD of the beam is shown in Fig. a
0.5 m
The shear and moment diagrams are shown in Figs. b and c, respectively. As indicated on the moment diagram, Mmax  = 1.25 kN # m.
150 mm
The moment of inertia of the crosssection is I =
1 (0.075) A 0.153 B = 21.09375A 10 6 B m4 12
Here, c = 0.075 m. Thus s max =
=
Mmax c I 3 1.25 A 10 B (0.075)
21.09375A 10
6
= 4.444 A 10 6 B Pa
B
Ans.
= 4.44 MPa
The bending stress distribution over the cross section is shown in Fig. d
395
1m 75 mm
0.5 m
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•6–105.
If the allowable bending stress for the wood beam is s allow = 150 psi, determine the required dimension b to the nearest 14 in. of its cross section. Assume the support at A is a pin and B is a roller.
400 lb/ft
B
A 3 ft
The FBD of the beam is shown in Fig. a
3 ft
The shear and moment diagrams are shown in Figs. b and c, respectively. As indicated on the moment diagram, Mmax = 3450 lb # ft.
2b b
The moment of inertia of the cross section is I =
1 2 (b)(2b)3 = b4 12 3
Here, c = 2b> 2 = b. Thus, s allow =
150 =
Mmax c ; I
3450(12)(b) > 3 b4
2
b = 7.453 in = 7
1 in. 2
Ans.
396
3 ft
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6–106. The wood beam has a rectangular cross section in the proportion shown. If b 7.5 in., determine the absolute maximum bending stress in the beam.
400 lb/ ft
B
A
The FBD of the beam is shown in Fig. a.
3 ft
3 ft
The shear and moment diagrams are shown in Fig. b and c, respectively. As indicated on the moment diagram, M max = 3450 lb # ft.
2b b
The moment of inertia of the crosssection is I =
1 (7.5) A 153 B = 2109.375 in 4 12
Here, c =
15 = 7.5 in. Thus 2
s max =
3450(12)(7.5) Mmax c = = 147 psi 2109.375 I
Ans.
397
3 ft
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6–107. A beam is made of a material that has a modulus of elasticity in compression different from that given for tension. Determine the location c of the neutral axis, and derive an expression for the maximum tensile stress in the beam having the dimensions shown if it is subjected to the bending moment M.
M h
P
Ec(emax )t (h  c) c
Ec
Location of neutral axis: + © F = 0; :
1 1  (h  c)(smax) c (b) + (c)(s max )t (b) = 0 2 2
(h  c)(s max)c = c(smax )t (h  c)E c (e max) t
[1]
(h  c) = cEt (e max) t ; c
E c (h  c) 2 = Etc2
Taking positive root: Ec c = A Et h  c
Ec A Et h 2Ec c = = Ec 2E t + 2Ec 1 + A Et h
[2] Ans.
©M NA = 0;
1 2 1 2 M = c (h  c)(smax) c (b) d a b (h  c) + c (c)(smax )t (b) d a b(c) 2 3 2 3 M =
1 1 (h  c)2 (b)(s max) c + c 2b(smax )t 3 3
From Eq. [1]. (smax )c = M = M =
c (s ) h  c max t
1 1 c (h  c)2 (b) a b(smax )t + c 2b(s max)t 3 3 h  c 1 bc(smax ) t (h  c + c) ; 3
(smax )t =
3M bhc
From Eq. [2]
(s max )t =
b
Et
(emax) t (h  c) (emax) c = c (smax )c = Ec(emax) c =
c
s
3M 2Et + 2Ec ≥ 2£ bh 2Ec
Ans.
398
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*6–108. The beam has a rectangular cross section and is subjected to a bending moment M. If the material from which it is made has a different modulus of elasticity for tension and compression as shown, determine the location c of the neutral axis and the maximum compressive stress in the beam.
M h c
s
b
Et
P Ec
See the solution to Prob. 6–107 c =
h 2E c
Ans.
2E t + 2Ec
Since
(smax ) c =
(smax ) c =
(smax ) c =
(s max ) c =
c h  c
2E c 2Et
(smax) t =
h 2Ec
(2Et + 2Ec)c h  a
h 1Ec
1Et + 1Ec
(smax )t bd
(smax ) t
2E c 3M 2Et + 2Ec ¢ 2≤¢ ≤ 2Et bh 2Ec 3M 2E t + 2Ec ¢ ≤ bh 2 2Et
Ans.
399
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•6–109.
The beam is subjected to a bending moment of M = 20 kip # ft directed as shown. Determine the maximum bending stress in the beam and the orientation of the neutral axis.
y 8 in. C
B
The y and z components of M are negative, Fig. a. Thus,
14 in.
My = 20 sin 45° = 14.14 kip # ft
z
45 16 in.
Mz = 20 cos 45° = 14.14 kip # ft. The moments of inertia of the crosssection about the principal centroidal y and z axes are Iy = Iz =
1 1 (10) A 16 3B (8) A 14 3 B = 1584 in4 12 12
By inspection, the bending stress occurs at corners A and C are Mz y My z + Iz Iy
s max = sC = 
14.14(12)(8)  14.14(12)(  5) + 1584 736
= 2.01 ksi s max = sA = 
Ans.
(T)
 14.14(12)(  8) 14.14(12)(5) + 1584 736
= 2.01 ksi = 2.01 ksi (C)
Ans.
Here, u = 180° + 45° = 225° tan a =
tan a =
Iz Iy
D 10 in. M
1 1 3 3 4 (16)A 10 B (14) A 8 B = 736 in 12 12
s = 
A
tan u
1584 tan 225° 736
a = 65.1°
Ans.
The orientation of neutral axis is shown in Fig. b.
400
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6–110. Determine the maximum magnitude of the bending moment M that can be applied to the beam so that the bending stress in the member does not exceed 12 ksi.
y 8 in. C
B
The y and z components of M are negative, Fig. a. Thus, My = M sin 45° = 0.7071 M
14 in. z
45
Mz = M cos 45° = 0.7071 M
16 in.
The moments of inertia of the crosssection about principal centroidal y and z axes are Iy = Iz =
1 1 (16) A 10 3B (14) A 83 B = 736 in 4 12 12
By inspection, the maximum bending stress occurs at corners A and C. Here, we will consider corner C.
12 = 
D 10 in. M
1 1 (10) A 16 3B (8) A 14 3 B = 1584 in4 12 12
s C = s allow = 
A
My zc Mz yc + Iz Iy 0.7071 M (12)(8)  0.7071 M(12)(  5) + 1584 736
M = 119.40 kip # ft = 119 kip # ft
Ans.
401
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6–111. If the resultant internal moment acting on the cross section of the aluminum strut has a magnitude of M = 520 N # m and is directed as shown, determine the bending stress at points A and B. The location y of the centroid C of the strut’s crosssectional area must be determined. Also, specify the orientation of the neutral axis.
y M 520 Nm 20 mm z
– y
5
12 13
B C
200 mm 20 mm 200 mm
Internal Moment Components: Mz = 
12 (520) = 480 N # m 13
5 (520) = 200 N # m 13
My =
Section Properties: ©y A
y =
©A
=
0.01(0.4)(0.02) + 2[(0.110)(0.18)(0.02)] 0.4(0.02) + 2(0.18)(0.02)
= 0.057368 m = 57.4 mm Iz =
Ans.
1 (0.4) A 0.023 B + (0.4)(0.02)(0.057368  0.01)2 12
+
1 3 2 (0.04) A 0.18 B + 0.04(0.18)(0.110  0.057368) 12
= 57.6014A 10  6 B m4
Iy =
1 1 (0.2)A 0.43 B (0.18) A 0.363 B = 0.366827 A 10  3 B m 4 12 12
Maximum Bending Stress: Applying the flexure formula for biaxial at points A and B s = 
Myz Mz y + Iz Iy
sA = 
 480(  0.142632) 6
57.6014(10 )
+
200(  0.2) 3
0.366827(10 ) Ans.
= 1.298 MPa = 1.30 MPa (C) sB = 
480(0.057368) 6
57.6014(10 )
+
200(0.2) 3
0.366827(10 ) Ans.
= 0.587 MPa (T) Orientation of Neutral Axis: tan a =
tan a =
Iz Iy
tan u
57.6014(10  6 ) tan ( 22.62°) 0.366827(10  3 ) Ans.
a = 3.74°
402
20 mm A 200 mm
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*6–112. The resultant internal moment acting on the cross section of the aluminum strut has a magnitude of M = 520 N # m and is directed as shown. Determine maximum bending stress in the strut. The location y of the centroid C of the strut’s crosssectional area must be determined. Also, specify the orientation of the neutral axis.
y M 520 Nm 20 mm z
– y
5
12 13
B C
200 mm 20 mm
20 mm A
200 mm
Internal Moment Components: Mz = 
12 (520) = 480 N # m 13
My =
5 (520) = 200 N # m 13
Section Properties: ©yA 0.01(0.4)(0.02) + 2[(0.110)(0.18)(0.02)] = ©A 0.4(0.02) + 2(0.18)(0.02)
y =
= 0.057368 m = 57.4 mm Iz =
Ans.
1 (0.4) A0.02 3B + (0.4)(0.02)(0.057368  0.01)2 12
1 (0.04)A 0.183 B + 0.04(0.18)(0.110  0.057368) 2 12
+
= 57.6014 A 10  6 B m 4
Iy =
1 1 3 3 4 3 (0.2) A 0.4 B (0.18) A 0.36 B = 0.366827 A 10 B m 12 12
Maximum Bending Stress: By inspection, the maximum bending stress can occur at either point A or B. Applying the flexure formula for biaxial bending at points A and B s = 
Mz y
sA = 
+
Iz
My z Iy
 480(  0.142632) 57.6014(10
6
)
+
200(  0.2) 0.366827(10  3)
= 1.298 MPa = 1.30 MPa (C) (Max) sB = 
480(0.057368) 6
57.6014(10 )
+
Ans.
200(0.2) 0.366827(10  3 )
= 0.587 MPa (T) Orientation of Neutral Axis: tan a =
tan a =
Iz Iy
tan u
57.6014(10  6) 0.366827(10
3
)
tan ( 22.62°) Ans.
a = 3.74°
403
200 mm
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6–113. Consider the general case of a prismatic beam subjected to bendingmoment components M y and M z, as shown, when the x, y, z axes pass through the centroid of the cross section. If the material is linearelastic, the normal stress in the beam is a linear function of position such that s = a + by + cz. Using the equilibrium conditions 0 = 1As dA, My = 1Azs dA, Mz = 1A  ys dA, determine the constants a, b, and c, and show that the normal stress can be determined from the equation s = [  1M zI y + M y Iyz2y + 1M yIz + M zIyz2z]>1Iy I z  Iyz2 2, where the moments and products of inertia are defined in Appendix A.
y z My dA s C y Mz z
Equilibrium Condition: sx = a + by + cz 0 =
L A
sx dA
0 =
A L
(a + by + cz) dA
0 = a
LA
dA + b
LA
y dA + c
My =
LA
z s x dA
=
A L
z(a + by + cz) dA
= a Mz = =
LA
= a
LA
LA
z dA + b
LA
LA
[1]
z dA
yz dA + c
LA
2
[2]
yz dA
[3]
z dA
 y sx dA
 y(a + by + cz) dA LA
ydA  b
LA
LA
2 y dA  c
Section Properties: The integrals are defined in Appendix A. Note that LA
y dA =
LA
z dA = 0.Thus,
From Eq. [1]
Aa = 0
From Eq. [2]
My = bIyz + cI y
From Eq. [3]
Mz = bIz  cI yz
Solving for a, b, c: a = 0 (Since A Z 0) b = ¢
Thus,
MzI y + My Iyz
sx =  ¢
I y Iz  I2yz
≤
Mz Iy + My Iyz Iy Iz  I2yz
c =
≤y + ¢
M y Iz + M z I yz I y Iz  I2yz
My I y + MzIyz Iy Iz  I 2yz
(Q.E.D.)
≤z
404
x
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6–114. The cantilevered beam is made from the Zsection having the crosssection shown. If it supports the two loadings, determine the bending stress at the wall in the beam at point A. Use the result of Prob. 6–113.
50 lb 50 lb 3 ft
(My) max = 50(3) + 50(5) = 400 lb # ft = 4.80(10 3)lb # in.
2 ft
0.25 in. 2 in.
1 1 (3.25)(0.25)3 + 2c (0.25)(2)3 + (0.25)(2)(1.125)2 d = 1.60319 in 4 Iy = 12 12
A
B
0.25 in.
2.25 in.
1 1 (0.25)(3.25)3 + 2c (2)(0.25) 3 + (0.25)(2)(1.5) 2d = 2.970378 in4 Iz = 12 12
3 in.
0.25 in.
Iyz = 2[1.5(1.125)(2)(0.25)] = 1.6875 in4
Using the equation developed in Prob. 6113. s = a sA =
Mz I y + My Iyz I y Iz 
I2yz
by + a
My Iz + Mz Iyz I y Iz  I2yz
bz
{  [0 + (4.80)(103)(1.6875)](1.625) + [(4.80)(103 )(2.970378) + 0](2.125)} [1.60319(2.970378)  (1.6875) 2] Ans.
= 8.95 ksi
6–115. The cantilevered beam is made from the Zsection having the crosssection shown. If it supports the two loadings, determine the bending stress at the wall in the beam at point B. Use the result of Prob. 6–113. (My) max = 50(3) + 50(5) = 400 lb # ft = Iy = Iz =
50 lb 50 lb 3 ft
)lb # in.
4.80(10 3
1 1 3 3 2 4 (3.25)(0.25) + 2c (0.25)(2) + (0.25)(2)(1.125) d = 1.60319 in 12 12
1 1 (0.25)(3.25)3 + 2c (2)(0.25) 3 + (0.25)(2)(1.5) 2d = 2.970378 in4 12 12
2 in.
2.25 in.
sB =
I y I z  I 2yz
by + a
My I z + Mz I yz Iy Iz  I 2yz
bz
 [0 + (4.80)(10 3 )(1.6875)](  1.625) + [(4.80)(103 )(2.976378) + 0](0.125) [(1.60319)(2.970378)  (1.6875)2 ] Ans.
= 7.81 ksi
405
0.25 in. 3 in.
0.25 in.
Using the equation developed in Prob. 6113. Mz I y + My I yz
A
B
Iyz = 2[1.5(1.125)(2)(0.25)] = 1.6875 in4
s = a
2 ft
0.25 in.
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*6–116. The cantilevered wideflange steel beam is subjected to the concentrated force P at its end. Determine the largest magnitude of this force so that the bending stress developed at A does not exceed s allow = 180 MPa.
200 mm 10 mm 150 mm 10 mm
Internal Moment Components: Using method of section
A
y
©M z = 0;
Mz + P cos 30°(2) = 0
Mz = 1.732P
©M y = 0;
My + P sin 30°(2) = 0
My = 1.00P
Section Properties:
z
x
2m
30
1 1 (0.2) A 0.173 B (0.19) A 0.153 B = 28.44583(10 6 ) m4 Iz = 12 12 Iy = 2c
10 mm
P
1 1 6 (0.01) A 0.2 3 B d + (0.15) A 0.013 B = 13.34583(10 ) m4 12 12
Allowable Bending Stress: By inspection, maximum bending stress occurs at points A and B. Applying the flexure formula for biaxial bending at point A. s A = sallow = 180 A 106 B = 
Mzy Iz
My z
+
Iy
(  1.732P)(0.085) 6
28.44583(10 )
+
 1.00P(  0.1) 13.34583(10  6) Ans.
P = 14208 N = 14.2 kN
•6–117.
The cantilevered wideflange steel beam is subjected to the concentrated force of P = 600 N at its end. Determine the maximum bending stress developed in the beam at section A.
200 mm 10 mm 150 mm 10 mm
Internal Moment Components: Using method of sections ©M z = 0;
Mz + 600 cos 30°(2) = 0
©M y = 0;
My + 600 sin 30°(2) = 0;
z
600.0 N # m
Section Properties:
x
1 1 6 (0.2) A 0.173 B (0.19) A 0.153 B = 28.44583(10 ) m4 Iz = 12 12 Iy = 2c
1 1 3 3 4 6 (0.01) A 0.2 B d + (0.15) A 0.01 B = 13.34583(10 ) m 12 12
Mz y
sA = 
Iz
+
Myz Iy
1039.32(0.085) 6
28.44583(10 )
= 7.60 MPa (T)
+
 600.0(  0.1) 13.34583(10  6 )
(Max)
Ans. 406
2m
30 P
Maximum Bending Stress: By inspection, maximum bending stress occurs at A and B. Applying the flexure formula for biaxial bending at point A s = 
A
y
Mz = 1039.23 N # m My =
10 mm
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6–118. If the beam is subjected to the internal moment of M = 1200 kN # m, determine the maximum bending stress acting on the beam and the orientation of the neutral axis.
y 150 mm 150 mm
Internal Moment Components: The y component of M is positive since it is directed towards the positive sense of the y axis, whereas the z component of M, which is directed towards the negative sense of the z axis, is negative, Fig. a. Thus,
M 300 mm
My = 1200 sin 30° = 600 kN # m
30 150 mm
Mz = 1200 cos 30° = 1039.23 kN # m
x 150 mm
z
Section Properties: The location of the centroid of the crosssection is given by ©yA 0.3(0.6)(0.3)  0.375(0.15)(0.15) = = 0.2893 m ©A 0.6(0.3)  0.15(0.15)
y =
150 mm
The moments of inertia of the cross section about the principal centroidal y and z axes are Iy = Iz =
1 1 (0.6) A 0.3 3B (0.15) A 0.153 B = 1.3078 A 10  3 B m 4 12 12 1 (0.3) A 0.63 B + 0.3(0.6)(0.3  0.2893)2 12
1  c (0.15) A 0.153 B + 0.15(0.15)(0.375  0.2893)2 d 12
= 5.2132 A 10  3 B m 4
Bending Stress: By inspection, the maximum bending stress occurs at either corner A or B. s = 
Mz y
sA = 
Iz
+
Myz Iy
c 1039.23 A 103 B d(0.2893) 5.2132 A 10
= 126 MPa (T)
sB = 
3
B
+
3 c 1039.23 A 10 B d(  0.3107)
5.2132 A 10  3 B
3 600 A 10 B (0.15)
1.3078 A 10
+
3
B
600 A 103 B (  0.15)
= 131 MPa = 131 MPa (C)(Max.)
1.3078 A 10  3 B
Ans.
Orientation of Neutral Axis: Here, u = 30°. tan a =
Iz tan u Iy
tan a =
5.2132 A 10  3 B 1.3078 A 10  3 B
tan( 30°)
a = 66.5°
Ans.
The orientation of the neutral axis is shown in Fig. b.
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6–119. If the beam is made from a material having an allowable tensile and compressive stress of (sallow) t = 125 MPa and (s allow) c = 150 MPa, respectively, determine the maximum allowable internal moment M that can be applied to the beam.
y 150 mm 150 mm M 300 mm
Internal Moment Components: The y component of M is positive since it is directed towards the positive sense of the y axis, whereas the z component of M, which is directed towards the negative sense of the z axis, is negative, Fig. a. Thus, My = M sin 30° = 0.5M
30 150 mm x 150 mm
z
Mz = M cos 30° = 0.8660M Section Properties: The location of the centroid of the cross section is 150 mm
©yA 0.3(0.6)(0.3)  0.375(0.15)(0.15) y = = = 0.2893 m ©A 0.6(0.3)  0.15(0.15) The moments of inertia of the cross section about the principal centroidal y and z axes are Iy = Iz =
1 1 3 (0.6) A 0.33 B (0.15) A 0.153 B = 1.3078 A 10 B m 4 12 12 1 (0.3) A 0.63 B + 0.3(0.6)(0.3  0.2893)2 12  c
1 (0.15) A 0.15 3 B + 0.15(0.15)(0.375  0.2893)2 d 12
= 5.2132 A 10  3 B m4
Bending Stress: By inspection, the maximum bending stress can occur at either corner A or B. For corner A which is in tension, sA = (s allow) t = 125 A 106 B = 
Mz yA Iz
+
My zA Iy
(  0.8660M)(0.2893) 5.2132 A 10  3 B
+
0.5M(0.15) 1.3078 A 10 3 B
M = 1185 906.82 N # m = 1186 kN # m (controls)
Ans.
For corner B which is in compression, sB = (s allow) c = 150 A 10 6 B = 
Mz yB Iz
+
My zB Iy
(  0.8660M)( 0.3107) 5.2132 A 10 B 3
M = 1376 597.12 N # m = 1377 kN # m
+
0.5M(  0.15) 1.3078 A 10  3 B
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*6–120. The shaft is supported on two journal bearings at 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 is sallow = 150 MPa.
z
0.5 m 200 N
Since all the axes through the centroid of the circular crosssection of the shaft are principal axes, then the resultant moment M = 2M y 2 + Mz 2 can be used for design. The maximum moment occurs at D (x = 1m) . Then, Mmax = 21502 + 1752 = 230.49 N # m
Then,
6 150(10 ) =
A
200 N 300 N
The shaft is subjected to two bending moment components M z and My , Figs. b and c, respectively.
Mmax C ; I
0.5 m
C
The FBD of the shaft is shown in Fig. a.
s allow =
y
0.5 m
230.49(d> 2) p 4 4 (d> 2)
Ans.
d = 0.02501 m = 25 mm
409
300 N
0.5 m D B E x 150 N
150 N
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•6–121.
The 30mmdiameter shaft is subjected to the vertical and horizontal loadings of two pulleys as shown. It is supported on two journal bearings at A and B which offer no resistance to axial loading. Furthermore, the coupling to the motor at C can be assumed not to offer any support to the shaft. Determine the maximum bending stress developed in the shaft.
y
z 1m 1m 1m E
1m A
60 mm x 150 N 150 N
Internal Moment Components: The shaft is subjected to two bending moment components M y 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 M = 2M y2 + M z 2 can be used to determine the maximum bending stress. The maximum resultant moment occurs at E M max = 2400 2 + 1502 = 427.2 N # m . Applying the flexure formula s max = =
Mmax c I 427.2(0.015) p 4
A 0.0154 B
Ans.
= 161 MPa
410
400 N
100 mm
D
Support Reactions: As shown on FBD.
C B
400 N
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6–122. Using the techniques outlined in Appendix A, Example A.5 or A.6, the Z section has principal moments of inertia of Iy = 0.060110 3 2 m4 and Iz = 0.471110 32 m4, computed about the principal axes of inertia y and z, respectively. If the section is subjected to an internal moment of M = 250 N # m directed horizontally as shown, determine the stress produced at point A. Solve the problem using Eq. 6–17.
50 mm y
A
200 mm
32.9
y¿ 250 Nm
My = 250 cos 32.9° = 209.9 N # m
200 mm
50 mm
z z¿ 300 mm
Mz = 250 sin 32.9° = 135.8 N # m
B 50 mm
y = 0.15 cos 32.9° + 0.175 sin 32.9° = 0.2210 m z = (0.175 cos 32.9°  0.15 sin 32.9°) = 0.06546 m sA = 
Mzy
+
Iz
My z Iy
=
135.8(0.2210) 0.471(10
3
)
+
209.9(  0.06546) 60.0(10  6) Ans.
= 293 kPa = 293 kPa (C)
6–123. Solve Prob. 6–122 using the equation developed in Prob. 6–113.
50 mm y
A
Internal Moment Components: My = 250 N # m
200 mm
Mz = 0
32.9
y¿
Section Properties: Iy =
250 Nm
1 1 (0.3) A 0.053B + 2c (0.05) A 0.153 B + 0.05(0.15) A 0.1 2 B d 12 12
z z¿ 300 mm
= 0.18125A 10  3 B m 4
Iz =
1 1 (0.05) A 0.3 3B + 2c (0.15) A 0.053 B + 0.15(0.05) A 0.125 2B d 12 12
= 0.350(10  3 ) m4
I yz = 0.15(0.05)(0.125)(  0.1) + 0.15(0.05)(  0.125)(0.1) = 0.1875A 10  3 B m4
Bending Stress: Using formula developed in Prob. 6113 s =
(M z Iy + My Iyz)y + (My Iz + MzIyz)z Iy I z  I 2yz 3
sA =
3
 [0 + 250(  0.1875)(10 )](0.15) + [250(0.350)(10 ) + 0]( 0.175) 0.18125(10  3 )(0.350)(10  3)  [0.1875(10 3 )]2 Ans.
= 293 kPa = 293 kPa (C)
411
200 mm
50 mm B
50 mm
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*6–124. Using the techniques outlined in Appendix A, Example A.5 or A.6, the Z section has principal moments of inertia of I y = 0.0601103 2 m 4 and Iz = 0.47111032 m 4, computed about the principal axes of inertia y and z, respectively. If the section is subjected to an internal moment of M = 250 N # m directed horizontally as shown, determine the stress produced at point B. Solve the problem using Eq. 6–17.
50 mm y
A
200 mm
32.9
y¿ 250 Nm
200 mm
50 mm
z z¿ 300 mm
Internal Moment Components: My¿ = 250 cos 32.9° = 209.9 N # m Mz¿ = 250 sin 32.9° = 135.8 N # m Section Property: y¿ = 0.15 cos 32.9° + 0.175 sin 32.9° = 0.2210 m z¿ = 0.15 sin 32.9°  0.175 cos 32.9° = 0.06546 m Bending Stress: Applying the flexure formula for biaxial bending s =
sB =
M y¿z¿ Mz¿ y¿ + Iz¿ Iy¿ 135.8(0.2210) 3
0.471(10 )

209.9(  0.06546) 0.060(10
3
) Ans.
= 293 kPa = 293 kPa (T)
412
B 50 mm
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•6–125.
Determine the bending stress at point A of the beam, and the orientation of the neutral axis. Using the method in Appendix A, the principal moments of inertia of the cross section are I¿z = 8.828 in 4 and I¿y = 2.295 in4, where z¿ and y¿ are the principal axes. Solve the problem using Eq. 6–17.
z 1.183 in. 0.5 in. z¿
A
4 in.
45 C y 1.183 in. 0.5 in.
M 3 kip ft 4 in.
Internal Moment Components: Referring to Fig. a, the y¿ and z¿ components of M are negative since they are directed towards the negative sense of their respective axes. Thus, Section Properties: Referring to the geometry shown in Fig. b, œ zA = 2.817 cos 45°  1.183 sin 45° = 1.155 in.
yœA = (2.817 sin 45° + 1.183 cos 45°) = 2.828 in. Bending Stress: œ
sA = 
= 
œ
M y¿ zA Mz¿y A + I z¿ I y¿ (  2.121)(12)(  2.828) (  2.121)(12)(1.155) + 8.828 2.295 Ans.
= 20.97 ksi = 21.0 ksi (C)
413
y′
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6–126. Determine the bending stress at point A of the beam using the result obtained in Prob. 6–113. The moments of inertia of the cross sectional area about the z and y axes 4 are I z = Iy = 5.561 in and the product of inertia of the cross sectional area with respect to the z and y axes is I yz 3.267 in 4. (See Appendix A)
z 1.183 in. 0.5 in. z¿
A
45
4 in. C
y 1.183 in. 0.5 in. M 3 kip ft 4 in.
Internal Moment Components: Since M is directed towards the negative sense of the y axis, its y component is negative and it has no z component. Thus, My = 3 kip # ft
Mz = 0
Bending Stress: sA =
=
 AM zIy + My Iyz B y A + A MyIz + Mz Iyz B zA I y I z  Iyz 2
 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
Ans.
= 20.97 ksi = 21.0 ksi
414
y′
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6–127. The composite beam is made of 6061T6 aluminum (A) and C83400 red brass (B). Determine the dimension h of the brass strip so that the neutral axis of the beam is located at the seam of the two metals. What maximum moment will this beam support if the allowable bending stress for the aluminum is 1s allow2 al = 128 MPa and for the brass 1sallow2br = 35 MPa?
h B A 150 mm
Section Properties: n =
68.9(109) Eal = = 0.68218 Ebr 101(109)
bbr = nbal = 0.68218(0.15) = 0.10233 m y = 0.05 =
©yA ©A 0.025(0.10233)(0.05) + (0.05 + 0.5h)(0.15)h 0.10233(0.05) + (0.15)h
h = 0.04130 m = 41.3 mm INA =
Ans.
1 (0.10233) A 0.053 B + 0.10233(0.05)(0.05  0.025)2 12
+
1 (0.15)A 0.041303 B + 0.15(0.04130)(0.070649  0.05)2 12
= 7.7851 A 10  6 B m4
Allowable Bending Stress: Applying the flexure formula Assume failure of red brass (sallow) br = 35 A 106 B =
Mc INA M(0.04130) 6
7.7851(10 )
M = 6598 N # m = 6.60 kN # m (controls!)
Ans.
Assume failure of aluminium (sallow) al = n
Mc I NA
6 128 A 10 B = 0.68218c
M(0.05) 7.7851(10 6 )
d
M = 29215 N # m = 29.2 kN # m
415
50 mm
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*6–128. The composite beam is made of 6061T6 aluminum (A) and C83400 red brass (B). If the height h = 40 mm, determine the maximum moment that can be applied to the beam if the allowable bending stress for the aluminum is 1sallow2al = 128 MPa and for the brass 1s allow2 br = 35 MPa.
h B A
Section Properties: For transformed section.
150 mm
68.9(109) Eal n = = = 0.68218 Ebr 101.0(109) b br = nb al = 0.68218(0.15) = 0.10233 m y = =
©y A ©A 0.025(0.10233)(0.05) + (0.07)(0.15)(0.04) 0.10233(0.05) + 0.15(0.04)
= 0.049289 m INA =
1 (0.10233) A 0.053 B + 0.10233(0.05)(0.049289  0.025) 2 12 +
1 (0.15)A 0.043 B + 0.15(0.04)(0.07  0.049289)2 12
= 7.45799 A 10 6 B m4
Allowable Bending Stress: Applying the flexure formula Assume failure of red brass (sallow )br = 35 A 106 B =
Mc INA M(0.09  0.049289) 6
7.45799(10 )
M = 6412 N # m = 6.41 kN # m (controls!)
Ans.
Assume failure of aluminium (s allow) al = n
Mc I NA
128 A 106 B = 0.68218c
M(0.049289) 7.45799(10  6 )
d
M = 28391 N # m = 28.4 kN # m
416
50 mm
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•6–129.
Segment A of the composite beam is made from 2014T6 aluminum alloy and segment B is A36 steel. If w = 0.9 kip> ft, determine the absolute maximum bending stress developed in the aluminum and steel. Sketch the stress distribution on the cross section.
w
15 ft A
3 in.
B
3 in.
3 in.
Maximum Moment: For the simplysupported beam subjected to the uniform 0.9 A 15 2 B wL 2 M = = distributed load, the maximum moment in the beam is max 8 8 # = 25.3125 kip ft. Section Properties: The cross section will be transformed into that of steel as Eal 10.6 shown in Fig. a. Here, n = = = 0.3655 . 29 Est Then b st = nbal = 0.3655(3) = 1.0965 in. The location of the centroid of the transformed section is y =
©yA 1.5(3)(3) + 4.5(3)(1.0965) = = 2.3030 in. ©A 3(3) + 3(1.0965)
The moment of inertia of the transformed section about the neutral axis is I = ©I + Ad2 =
1 (3) A 33 B + 3(3)(2.3030  1.5)2 12 +
1 3 2 (1.0965) A 3 B + 1.0965(3)(4.5  2.3030) 12
= 30.8991 in4
Maximum Bending Stress: For the steel, (s max )st =
25.3125(12)(2.3030) Mmax cst = = 22.6 ksi 30.8991 I
Ans.
At the seam, s st y = 0.6970 in. =
25.3125(12)(0.6970) M maxy = = 6.85 ksi 30.8991 I
For the aluminium, (s max )al = n At the seam,
25.3125(12)(6  2.3030) Mmax cal = 0.3655c d = 13.3 ksi 30.8991 I
25.3125(12)(0.6970) Mmax y s aly = 0.6970 in. = n = 0.3655c d = 2.50 ksi 30.8991 I
Ans.
The bending stress across the cross section of the composite beam is shown in Fig. b.
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6–130. Segment A of the composite beam is made from 2014T6 aluminum alloy and segment B is A36 steel. If the allowable bending stress for the aluminum and steel are (s allow) al = 15 ksi and (sallow)st = 22 ksi , determine the maximum allowable intensity w of the uniform distributed load.
w
15 ft A
3 in.
B
3 in.
3 in.
Maximum Moment: For the simplysupported beam subjected to the uniform distributed load, the maximum moment in the beam is wA 152 B wL 2 Mmax = = = 28.125w . 8 8 Section Properties: The cross section will be transformed into that of steel as Eal 10.6 = = 0.3655. shown in Fig. a. Here, n = 29 E st Then b st = nbal = 0.3655(3) = 1.0965 in . The location of the centroid of the transformed section is y =
©y A 1.5(3)(3) + 4.5(3)(1.0965) = = 2.3030 in. ©A 3(3) + 3(1.0965)
The moment of inertia of the transformed section about the neutral axis is I = ©I + Ad2 =
1 1 (3)A 33 B + 3(3)(2.3030  1.5)2 + (1.0965) A 3 3B 12 12 + 1.0965 A 33B + 1.0965(3)(4.5  2.3030)2
= 30.8991 in 4
Bending Stress: Assuming failure of steel, (sallow )st =
Mmax c st ; I
22 =
(28.125w)(12)(2.3030) 30.8991 Ans.
w = 0.875 kip>ft (controls) Assuming failure of aluminium alloy, (sallow )al = n
Mmax cal ; I
15 = 0.3655 c
(28.125w)(12)(6  2.3030) d 30.8991
w = 1.02 kip>ft
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6–131. The Douglas fir beam is reinforced with A36 straps at its center and sides. Determine the maximum stress developed in the wood and steel if the beam is subjected to a bending moment of Mz = 7.50 kip # ft. Sketch the stress distribution acting over the cross section.
y
0.5 in.
0.5 in.
0.5 in.
z
6 in.
2 in.
Section Properties: For the transformed section. n =
1.90(103 ) Ew = = 0.065517 29.0(103 ) Est
bst = nbw = 0.065517(4) = 0.26207 in. INA =
1 (1.5 + 0.26207)A 63 B = 31.7172 in 4 12
Maximum Bending Stress: Applying the flexure formula (smax )st =
Mc 7.5(12)(3) = = 8.51 ksi 31.7172 I
(smax )w = n
Ans.
7.5(12)(3) Mc = 0.065517c d = 0.558 ksi 31.7172 I
Ans.
419
2 in.
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*6–132. The top plate is made of 2014T6 aluminum and is used to reinforce a Kevlar 49 plastic beam. Determine the maximum stress in the aluminum and in the Kevlar if the beam is subjected to a moment of M = 900 lb # ft.
6 in. 0.5 in. 0.5 in. 12 in. M
0.5 in. 0.5 in.
Section Properties: n =
10.6(103 ) Eal = 3 = 0.55789 Ek 19.0(10 )
b k = n bal = 0.55789(12) = 6.6947 in. y =
©y A 0.25(13)(0.5) + 2[(3.25)(5.5)(0.5)] + 5.75(6.6947)(0.5) = ©A 13(0.5) + 2(5.5)(0.5) + 6.6947(0.5) = 2.5247 in.
INA =
1 (13) A 0.5 3 B + 13(0.5)(2.5247  0.25)2 12 +
1 3 2 (1) A 5.5 B + 1(5.5)(3.25  2.5247) 12 +
1 (6.6947) A 0.53 B + 6.6947(0.5)(5.75  2.5247)2 12
= 85.4170 in4
Maximum Bending Stress: Applying the flexure formula (smax )al = n (smax )k =
900(12)(6  2.5247) Mc = 0.55789 c d = 245 psi I 85.4170
Mc 900(12)(6  2.5247) = = 439 psi 85.4168 I
Ans.
Ans.
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•6–133.
The top plate made of 2014T6 aluminum is used to reinforce a Kevlar 49 plastic beam. If the allowable bending stress for the aluminum is (s allow) al = 40 ksi and for the Kevlar (s allow) k = 8 ksi, determine the maximum moment M that can be applied to the beam.
6 in. 0.5 in.
Section Properties:
0.5 in.
Eal 10.6(10 3) = = 0.55789 19.0(10 3) Ek
n =
12 in.
bk = n bal = 0.55789(12) = 6.6947 in. y =
0.5 in.
© yA 0.25(13)(0.5) + 2[(3.25)(5.5)(0.5)] + 5.75(6.6947(0.5) = ©A 13(0.5) + 2(5.5)(0.5) + 6.6947(0.5) = 2.5247 in.
INA =
1 3 2 (13) A 0.5 B + 13(0.5)(2.5247  0.25) 12 +
1 (1)A 5.5 3B + 1(5.5)(3.25  2.5247)2 12 +
= 85.4170 in4
1 (6.6947) A 0.5 3B + 6.6947(0.5)(5.75  2.5247)2 12
Maximum Bending Stress: Applying the flexure formula Assume failure of aluminium (sallow) al = n
Mc I
40 = 0.55789 c
M(6  2.5247) d 85.4170
M = 1762 kip # in = 146.9 kip # ft Assume failure of Kevlar 49 (sallow) k = 8 =
Mc I M(6  2.5247) 85.4170
M = 196.62 kip # in = 16.4 kip # ft
M
0.5 in.
Ans.
(Controls!)
421
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6–134. The member has a brass core bonded to a steel casing. If a couple moment of 8 kN # m is applied at its end, determine the maximum bending stress in the member. Ebr = 100 GPa, Est = 200 GPa.
8 kNm
3m 20 mm 100 mm 20 mm
n =
Ebr 100 = = 0.5 200 E st
I =
1 1 (0.14)(0.14)3 (0.05)(0.1)3 = 27.84667(10  6 )m4 12 12
20 mm
100 mm
20 mm
Maximum stress in steel: (sst ) max =
8(10 3)(0.07) Mc 1 = = 20.1 MPa I 27.84667(10 6 )
(max)
Ans.
Maximum stress in brass: (sbr ) max =
0.5(8)(103 )(0.05) nMc 2 = = 7.18 MPa I 27.84667(10  6 )
6–135. The steel channel is used to reinforce the wood beam. Determine the maximum stress in the steel and in the wood if the beam is subjected to a moment of M = 850 lb # ft. Est = 29(103 ) ksi, Ew = 1600 ksi.
4 in.
0.5 in.
(0.5)(16)(0.25) + 2(3.5)(0.5)(2.25) + (0.8276)(3.5)(2.25) y= = 1.1386 in. 0.5(16) + 2(3.5)(0.5) + (0.8276)(3.5)
15 in. M 850 lbft
0.5 in.
1 1 3 2 3 2 I = (16)(0.5 ) + (16)(0.5)(0.8886 ) + 2 a b(0.5)(3.5 ) + 2(0.5)(3.5)(1.1114 ) 12 12 +
1 (0.8276)(3.53) + (0.8276)(3.5)(1.1114 2) = 20.914 in 4 12
Maximum stress in steel: (sst ) =
Mc 850(12)(4  1.1386) = = 1395 psi = 1.40 ksi 20.914 I
Ans.
Maximum stress in wood: (sw ) = n(s st) max Ans.
= 0.05517(1395) = 77.0 psi
422
0.5 in.
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*6–136. A white spruce beam is reinforced with A36 steel straps at its top and bottom as shown. Determine the bending moment M it can support if (s allow)st = 22 ksi and (sallow)w = 2.0 ksi .
y 0.5 in.
4 in.
M
0.5 in. x
z 3 in.
Section Properties: For the transformed section. n =
1.40(10 3) Ew = = 0.048276 Est 29.0(10 3)
b st = nbw = 0.048276(3) = 0.14483 in. INA =
1 1 (3) A 5 3 B (3  0.14483) A 43 B = 16.0224 in 4 12 12
Allowable Bending Stress: Applying the flexure formula Assume failure of steel (s allow)st = 22 =
Mc I M(2.5) 16.0224
M = 141.0 kip # in = 11.7 kip # ft (Controls !)
Ans.
Assume failure of wood (s allow)w = n
My I
2.0 = 0.048276c
M(2) d 16.0224
M = 331.9 kip # in = 27.7 kip # ft
423
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•6–137. If the beam is subjected to an internal moment of
M = 45 kN # m, determine the maximum bending stress developed in the A36 steel section A and the 2014T6 aluminum alloy section B.
A 50 mm
M 15 mm 150 mm
Section Properties: The cross section will be transformed into that of steel as shown in Fig. a. 9 E al 73.1 A 10 B = = 0.3655 . Thus, b st = nb al = 0.3655(0.015) = 0.0054825 m . The Here, n = E st 200A 109 B location of the transformed section is ©yA y = = ©A
0.075(0.15)(0.0054825) + 0.2 c p A 0.052 B d
= 0.1882 m
0.15(0.0054825) + p A0.052 B
The moment of inertia of the transformed section about the neutral axis is I = ©I + Ad2 =
1 3 2 (0.0054825) A 0.15 B + 0.0054825(0.15)(0.1882  0.075) 12 +
1 p A 0.054 B + p A 0.052 B (0.2  0.1882) 2 4
= 18.08 A10  6 B m4
Maximum Bending Stress: For the steel, 45 A 10 B (0.06185) Mc st = = 154 MPa 6 I 18.08 A 10 B 3
(s max) st =
Ans.
For the aluminum alloy,
(s max) al = n
45 A 10 3 B (0.1882) Mcal = 0.3655 C S = 171 MPa I 18.08 A 10  6 B
424
Ans.
B
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6–138. The concrete beam is reinforced with three 20mm diameter steel rods. Assume that the concrete cannot support tensile stress. If the allowable compressive stress for concrete is (sallow)con = 12.5 MPa and the allowable tensile stress for steel is (s allow) st = 220 MPa, determine the required dimension d so that both the concrete and steel achieve their allowable stress simultaneously. This condition is said to be ‘balanced’. Also, compute the corresponding maximum allowable internal moment M that can be applied to the beam. The moduli of elasticity for concrete and steel are Econ = 25 GPa and E st = 200 GPa, respectively.
200 mm
M
Bending Stress: The cross section will be transformed into that of concrete as shown Est 200 = = 8 . It is required that both concrete and steel in Fig. a. Here, n = Econ 25 achieve their allowable stress simultaneously. Thus, (sallow) con =
(sallow) st = n
Mccon ; I
Mc con I
M = 12.5 A 10 6 B ¢
Mcst ; I
I ≤ ccon
6 220 A 10 B = 8 B
Equating Eqs. (1) and (2), 12.5A 10 6 B ¢
6 12.5 A 10 B =
(1)
M(d  c con) R I
M = 27.5 A 106 B ¢
I ≤ d  ccon
(2)
I I ≤ = 27.5 A 10 6B ¢ ≤ ccon d  ccon
(3)
ccon = 0.3125d (3)
Section Properties: The area of the steel bars is A st = 3c
p A0.02 2 B d = 0.3 A 10  3 Bp m 2. 4
3 Thus, the transformed area of concrete from steel is (Acon )t = nAs = 8 C 0.3 A 10 B p D
= 2.4 A 10  3 B p m2 . Equating the first moment of the area of concrete above and below the neutral axis about the neutral axis, 3 0.2(ccon )(ccon >2) = 2.4 A10 B p (d  ccon)
0.1c con2 = 2.4 A 10  3 B pd  2.4 A 10  3B pc con ccon 2 = 0.024pd  0.024pc con
(4)
Solving Eqs. (3) and (4), d = 0.5308 m = 531 mm
Ans.
ccon = 0.1659 m Thus, the moment of inertia of the transformed section is I =
1 3 2 3 (0.2) A 0.1659 B + 2.4 A 10 B p(0.5308  0.1659) 3 425
d
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6–138. Continued
= 1.3084A 10  3 B m 4
Substituting this result into Eq. (1),
M = 12.5 A 10 B C 6
1.3084A 10  3 B 0.1659
S
= 98 594.98 N # m = 98.6 kN # m‚
Ans.
6–139. The beam is made from three types of plastic that are identified and have the moduli of elasticity shown in the figure. Determine the maximum bending stress in the PVC. (bbk )1 = n 1 b Es =
160 (3) = 0.6 in. 800
(bbk )2 = n 2 b pvc =
450 (3) = 1.6875 in. 800
500 lb
PVC EPVC 450 ksi Escon E E 160 ksi Bakelite E B 800 ksi 3 ft
y=
©yA (1)(3)(2) + 3(0.6)(2) + 4.5(1.6875)(1) = = 1.9346 in. ©A 3(2) + 0.6(2) + 1.6875(1)
I =
1 1 (3)(2 3) + 3(2)(0.93462) + (0.6)(23) + 0.6(2)(1.06542 ) 12 12 +
4 ft 1 in. 2 in. 2 in. 3 in.
1 (1.6875)(13) + 1.6875(1)(2.5654 2) = 20.2495 in 4 12
(smax )pvc = n2
500 lb
450 1500(12)(3.0654) Mc = a b 800 20.2495 I
Ans.
= 1.53 ksi
426
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*6–140. The low strength concrete floor slab is integrated with a wideflange A36 steel beam using shear studs (not shown) to form the composite beam. If the allowable bending stress for the concrete is (s allow)con = 10 MPa, and allowable bending stress for steel is (sallow) st = 165 MPa, determine the maximum allowable internal moment M that can be applied to the beam.
1m
100 mm
15 mm 400 mm M 15 mm 15 mm
Section Properties: The beam cross section will be transformed into E 22.1 n = con = = 0.1105 . that of steel. Here, Thus, Est 200 b st = nbcon = 0.1105(1) = 0.1105 m . The location of the transformed section is y = =
©yA ©A 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)
= 0.3222 m The moment of inertia of the transformed section about the neutral axis is I = ©I + Ad2 =
1 (0.2)A 0.0153 B 12
+ 0.2(0.015)(0.3222  0.0075)2 + + +
1 (0.015)A 0.373 B + 0.015(0.37)(0.3222  0.2)2 12
1 (0.2)A 0.015 3 B + 0.2(0.015)(0.3925  0.3222)2 12 1 (0.1105)A 0.13 B + 0.1105(0.1)(0.45  0.3222)2 12
= 647.93A 10  6 B m 4
Bending Stress: Assuming failure of steel, (sallow) st =
M(0.3222) Mcst ; 165A 106 B = 6 I 647.93 A 10 B
M = 331 770.52 N # m = 332 kN # m
Assuming failure of concrete,
(sallow) con = n
Mccon ; I
10 A 10 6 B = 0.1105C
M(0.5  0.3222) 647.93 A 10  6 B
S
M = 329 849.77 N # m = 330 kN # m (controls) Ans.
427
200 mm
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•6–141.
The reinforced concrete beam is used to support the loading shown. Determine the absolute maximum normal stress in each of the A36 steel reinforcing rods and the absolute maximum compressive stress in the concrete. Assume the concrete has a high strength in compression and yet neglect its strength in supporting tension.
10 kip
8 in. 15 in. 4 ft
8 ft
Mmax = (10 kip)(4 ft) = 40 kip # ft A st = 3(p)(0.5)2 = 2.3562 in2 Est = 29.0(103) ksi Econ = 4.20(10 3) ksi A¿ = nAst =
29.0(103) 4.20(103) 8(h¿ ) a
©y A = 0;
(2.3562) = 16.2690 in2
h¿ b  16.2690(13  h ¿ ) = 0 2
h¿ 2 + 4.06724h  52.8741 = 0
Solving for the positive root: h¿ = 5.517 in. I = c
1 3 2 2 (8)(5.517) + 8(5.517)(5.517> 2) d + 16.2690(13  5.517) 12
= 1358.781 in 4
(scon ) max =
40(12)(5.517) My = = 1.95 ksi 1358.781 I
(sst ) max = na
My I
b = a
10 kip
Ans.
29.0(103 ) 40(12)(13  5.517) ba b = 18.3 ksi 4.20(103 ) 1358.781
428
Ans.
4 ft
2 in. 1 in. diameter rods
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6–142. The reinforced concrete beam is made using two steel reinforcing rods. If the allowable tensile stress for the steel is (s st )allow = 40 ksi and the allowable compressive stress for the concrete is (sconc) allow = 3 ksi , determine the maximum moment M that can be applied to the section. Assume the concrete cannot support a tensile stress. Est = 29(103) ksi, Econc = 3.8(103) ksi.
8 in. 6 in. 4 in.
8 in.
M 18 in. 2 in. 1in. diameter rods
Ast = 2(p)(0.5)2 = 1.5708 in2 A¿ = nAst = ©yA = 0;
29(10 3 ) 3
3.8(10 )
(1.5708) = 11.9877 in2
22(4)(h¿ + 2) + h ¿ (6)(h¿> 2)  11.9877(16  h ¿ ) = 0 3h2 + 99.9877h¿  15.8032 = 0
Solving for the positive root: h¿ = 0.15731 in. I = c
1 1 (22)(4)3 + 22(4)(2.15731)2 d + c (6)(0.15731) 3 + 6(0.15731)(0.15731> 2)2 d 12 12 + 11.9877(16  0.15731) 2 = 3535.69 in4
Assume concrete fails: (s con )allow =
My ; I
3 =
M(4.15731) 3535.69
M = 2551 kip # in. Assume steel fails: (s st) allow = na
My b; I
29(103) M(16  0.15731) ≤¢ ≤ 40 = ¢ 3.8(103 ) 3535.69 M = 1169.7 kip # in. = 97.5 kip # ft (controls) Ans.
429
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6–143. For the curved beam in Fig. 6–40a, show that when the radius of curvature approaches infinity, the curvedbeam formula, Eq. 6–24, reduces to the flexure formula, Eq. 6–13. Normal Stress: Curvedbeam formula M(R  r) Ar(r  R)
s =
where A¿ =
dA A A and R = = dA A¿ A r L 1A r
M(A  rA¿)
s =
[1]
Ar(rA¿  A)
r = r + y rA¿ = r
[2]
dA r a  1 + 1b dA = r r + y LA LA =
L A
a
= A 
r r  y + 1b dA r + y y
LA r + y
[3]
dA
Denominator of Eq. [1] becomes, y
LA r + y
Ar(rA¿  A) = Ar ¢ A 
dA  A ≤ = Ar
y
LA r + y
dA
Using Eq. [2], Ar(rA¿  A) = A
= A
=
LA
¢
ry r + y
y2
A r + y L
+ y  y ≤ dA  Ay
LA r + y
dA  A 1A y dA  Ay
y
LA r + y
as
y : 0 r
Then,
Ar(r A¿  A) :
Eq. [1] becomes
s =
Mr (A  rA¿ ) AI
Using Eq. [2],
s =
Mr (A  rA ¿  yA¿) AI
A I r
Using Eq. [3],
s =
=
dA
dA
A y2 Ay y ¢ ¢ ≤ dA A y dA y ≤ dA  A 1 r LA 1 + r r LA 1 + yr 1A y dA = 0,
But,
y
Mr y dA C A  ¢A S dA≤  y AI LA r + y LA r + y
y Mr dA dA  y C S AI LA r + y LA r + y
430
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6–143. Continued
= y
As
r
y y Mr dA r C ¢ ¢ y≤ dA y≤ S AI L 1 + r r L A A 1 + r
=
: 0 y r
LA 1 +
¢
y≤ r
s =
Therefore,
and
dA = 0
y dA y yA ¢ y ≤ = 1A dA = r LA 1 + r r r
yA My Mr ab = AI I r
(Q.E.D.)
*6–144. The member has an elliptical cross section. If it is subjected to a moment of M = 50 N # m, determine the stress at points A and B. Is the stress at point A¿ , which is located on the member near the wall, the same as that at A? Explain.
75 mm
150 mm A¿ 250 mm A
dA 2p b = (r  2r2  a2 ) a A r L =
100 mm
2p(0.0375) (0.175  20.175 2  0.0752 ) = 0.053049301 m 0.075
A = p ab = p(0.075)(0.0375) = 2.8125(10 3 )p R =
A
1A
dA r
=
2.8125(10  3 )p 0.053049301
B
= 0.166556941
r  R = 0.175  0.166556941 = 0.0084430586 sA =
sB =
M(R  rA) ArA (r  R)
=
50(0.166556941  0.1) 2.8125(10  3 )p (0.1)(0.0084430586)
= 446k Pa (T)
M(R  r B) 50(0.166556941  0.25) = = 224 kPa (C) 2.8125(10  3 )p (0.25)(0.0084430586) ArB (r  R)
Ans.
Ans. Ans.
No, because of localized stress concentration at the wall.
431
M
Mr AI
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•6–145.
The member has an elliptical cross section. If the allowable bending stress is s allow = 125 MPa determine the maximum moment M that can be applied to the member.
75 mm
150 mm A¿ 250 mm A 100 mm
B
a = 0.075 m;
b = 0.0375 m
A = p(0.075)(0.0375) = 0.0028125 p 2p(0.0375) 2pb dA = (r  2r 2  a2 ) = (0.175  20.1752  0.075 2) 0.075 a LA r = 0.053049301 m R =
A
dA 1A r
=
0.0028125p = 0.166556941 m 0.053049301
3 r  R = 0.175  0.166556941 = 8.4430586(10 ) m
s =
M(R  r) Ar(r  R)
Assume tension failure. 125(106 ) =
M(0.166556941  0.1) 0.0028125p(0.1)(8.4430586)(10  3)
M = 14.0 kN # m
Ans.
(controls)
Assume compression failure: 125(106) =
M(0.166556941  0.25) 0.0028125p(0.25)(8.4430586)(10  3 )
M = 27.9 kN # m
432
M
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6–146. Determine the greatest magnitude of the applied forces P if the allowable bending stress is (sallow) c = 50 MPa in compression and (sallow )t = 120 MPa in tension.
75 mm P
10 mm
10 mm 160 mm
10 mm
P 150 mm 250 mm
Internal Moment: M = 0.160P is positive since it tends to increase the beam’s radius of curvature. Section Properties: r = =
©yA ©A 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)
= 0.3190 m A = 0.15(0.01) + 0.15(0.01) + 0.075(0.01) = 0.00375 m2 ©
0.26 0.41 0.42 dA = 0.15 ln + 0.01 ln + 0.075 ln 0.25 0.26 0.41 L A r = 0.012245 m
R =
A
© 1A dA r
=
0.00375 = 0.306243 m 0.012245
r  R = 0.319  0.306243 = 0.012757 m Allowable Normal Stress: Applying the curvedbeam formula Assume tension failure (sallow )t = 6 120 A 10 B =
M(R  r) Ar(r  R) 0.16P(0.306243  0.25) 0.00375(0.25)(0.012757)
P = 159482 N = 159.5 kN
Assume compression failure (sallow) t = 50 A 10 6B =
M(R  r) Ar(r  R) 0.16P(0.306243  0.42) 0.00375(0.42)(0.012757) Ans.
P = 55195 N = 55.2 kN (Controls !)
433
150 mm
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6–147. If P = 6 kN, determine the maximum tensile and compressive bending stresses in the beam.
75 mm P 160 mm
10 mm 10 mm
10 mm
P 150 mm 250 mm
Internal Moment: M = 0.160(6) = 0.960 kN # m is positive since it tends to increase the beam’s radius of curvature. Section Properties: r = =
©yA ©A 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)
= 0.3190 m A = 0.15(0.01) + 0.15(0.01) + 0.075(0.01) = 0.00375 m2 ©
0.26 0.41 0.42 dA = 0.15 ln + 0.01 ln + 0.075 ln 0.25 0.26 0.41 L A r = 0.012245 m
R =
A
©1A dA r
=
0.00375 = 0.306243 m 0.012245
r  R = 0.319  0.306243 = 0.012757 m Normal Stress: Applying the curvedbeam formula (smax )t =
=
M(R  r) Ar(r  R) 0.960(103 )(0.306243  0.25) 0.00375(0.25)(0.012757)
= 4.51 MPa (smax )c = =
Ans.
M(R  r) Ar(r  R) 0.960(103 )(0.306243  0.42) 0.00375(0.42)(0.012757) Ans.
= 5.44 MPa
434
150 mm
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*6–148. The curved beam is subjected to a bending moment of M = 900 N # m as shown. Determine the stress at points A and B, and show the stress on a volume element located at each of these points.
A C B
100 mm C
A 15 mm
30
20 mm 150 mm
400 mm B M
Internal Moment: M = 900 N # m is negative since it tends to decrease the beam’s radius curvature. Section Properties: © A = 0.15(0.015) + 0.1(0.02) = 0.00425 m2 ©rA = 0.475(0.15)(0.015) + 0.56(0.1)(0.02) = 2.18875(10 3 ) m3 3
2.18875 (10 ) ©rA r = = = 0.5150 m 0.00425 ©A ©
dA 0.55 0.57 = 0.015 ln + 0.1 ln = 8.348614(10  3 ) m 0.4 0.55 A r L
R =
A
©1A dA r
=
0.00425 = 0.509067 m 8.348614(10 3 )
3 r  R = 0.515  0.509067 = 5.933479(10 ) m
Normal Stress: Applying the curvedbeam formula sA =
M(R  r A)  900(0.509067  0.57) = 0.00425(0.57)(5.933479)(10 3) Ar A (r  R) = 3.82 MPa (T)
sB =
Ans.
M(R  r B)  900(0.509067  0.4) = 3 Ar B (r  R) 0.00425(0.4)(5.933479)(10 ) Ans.
= 9.73 MPa = 9.73 MPa (C)
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•6–149.
The curved beam is subjected to a bending moment of M = 900 N # m . Determine the stress at point C. A C B
100 mm C
A 15 mm
30
20 mm 150 mm
400 mm B M
Internal Moment: M = 900 N # m is negative since it tends to decrease the beam’s radius of curvature. Section Properties: © A = 0.15(0.015) + 0.1(0.02) = 0.00425 m 2 ©r A = 0.475(0.15)(0.015) + 0.56(0.1)(0.02) = 2.18875(10 3 ) m 2.18875 (10 ©rA r = = 0.00425 ©A ©
3
)
= 0.5150 m
0.55 0.57 dA = 0.015 ln + 0.1 ln = 8.348614(10  3) m 0.4 0.55 LA r
R =
A
©1A dA r
=
0.00425 = 0.509067 m 8.348614(10  3 )
r  R = 0.515  0.509067 = 5.933479(10  3 ) m Normal Stress: Applying the curvedbeam formula sC =
M(R  r C)  900(0.509067  0.55) = 0.00425(0.55)(5.933479)(10  3 ) ArC(r  R) Ans.
= 2.66 MPa (T)
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6–150. The elbow of the pipe has an outer radius of 0.75 in. and an inner radius of 0.63 in. If the assembly is subjected to the moments of M = 25 lb # in., determine the maximum stress developed at section a  a.
a
30
1 in.
M 25 lbin. a
dA = ©2p (r  2r2  c2 ) L A r = 2p(1.75  21.752  0.75 2)  2p (1.75  21.752  0.632)
0.63 in. 0.75 in.
= 0.32375809 in.
A = p(0.75 2)  p(0.632) = 0.1656 p R =
1A
A dA r
=
M = 25 lbin.
0.1656 p = 1.606902679 in. 0.32375809
r  R = 1.75  1.606902679 = 0.14309732 in. M(R  rA) 25(1.606902679  1) = = 204 psi (T) 0.1656 p(1)(0.14309732) ArA(r  R)
(smax ) t =
(s max ) c = =
M(R  r B ) Ar B(r  R)
=
Ans.
25(1.606902679  2.5) = 120 psi (C) 0.1656p(2.5)(0.14309732)
Ans.
6–151. The curved member is symmetric and is subjected to a moment of M = 600 lb # ft. Determine the bending stress in the member at points A and B. Show the stress acting on volume elements located at these points. A = 0.5(2) +
0.5 in. B 2 in. A
1 (1)(2) = 2 in 2 2
1.5 in. 8 in.
9(0.5)(2) + 8.6667 A 12 B (1)(2) ©rA r = = = 8.83333 in. ©A 2
M
M
10 1(10) 10 dA + c c ln d  1 d = 0.22729 in. = 0.5 ln r (10 8) 8 8 L A
R =
A
dA 1A r
=
2 = 8.7993 in. 0.22729
r  R = 8.83333  8.7993 = 0.03398 in. s =
sA = sB =
M(R  r) Ar(r  R) 600(12)(8.7993  8) 2(8)(0.03398)
= 10.6 ksi (T)
Ans.
600(12)(8.7993  10) = 12.7 ksi = 12.7 ksi (C) 2(10)(0.03398)
Ans.
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*6–152. The curved bar used on a machine has a rectangular cross section. If the bar is subjected to a couple as shown, determine the maximum tensile and compressive stress acting at section a  a. Sketch the stress distribution on the section in three dimensions.
a 75 mm a
50 mm
162.5 mm
250 N 60
150 mm
60 250 N 75 mm
a+ ©M O = 0;
M  250 cos 60° (0.075)  250 sin 60° (0.15) = 0 M = 41.851 N # m
0.2375 dA r2 = b ln = 0.05 ln = 0.018974481 m 0.1625 r r 1 LA A = (0.075)(0.05) = 3.75(10  3 ) m2 R =
A
1A
dA r
=
3.75(10  3 ) 0.018974481
= 0.197633863 m
r  R = 0.2  0.197633863 = 0.002366137 sA =
M(R  r A) Ar A(r  R)
=
41.851(0.197633863  0.2375) 3.75(10  3 )(0.2375)(0.002366137)
= 791.72 kPa
= 792 kPa (C) sB =
Ans.
M(R  rB ) 41.851 (0.197633863  0.1625) = = 1.02 MPa (T) 3.75(10  3)(0.1625)(0.002366137) Ar B(r  R)
438
Ans.
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•6–153.
The ceilingsuspended Carm is used to support the Xray camera used in medical diagnoses. If the camera has a mass of 150 kg, with center of mass at G, determine the maximum bending stress at section A.
G 1.2 m A
200 mm 20 mm 40 mm
Section Properties: r = ©
©rA 1.22(0.1)(0.04) + 1.25(0.2)(0.02) = = 1.235 m ©A 0.1(0.04) + 0.2(0.02)
1.24 1.26 dA 3 = 0.1 ln + 0.2 ln = 6.479051 A 10 B m 1.20 1.24 LA r
A = 0.1(0.04) + 0.2(0.02) = 0.008 m2 R =
A
dA 1A r
=
0.008  3 = 1.234749 m 6.479051 (10 )
r  R = 1.235  1.234749 = 0.251183A 10  3 B m
Internal Moment: The internal moment must be computed about the neutral axis as shown on FBD. M = 1816.93 N # m is negative since it tends to decrease the beam’s radius of curvature. Maximum Normal Stress: Applying the curvedbeam formula sA =
=
M(R  rA) ArA (r  R)  1816.93(1.234749  1.26) 0.008(1.26)(0.251183)(10 3 )
= 18.1 MPa (T) sB =
=
M(R  rB) Ar B (r  R)  1816.93(1.234749  1.20) 0.008(1.20)(0.251183)(10  3 )
= 26.2 MPa = 26.2 MPa (C)
(Max)
Ans.
439
100 mm
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6–154. The circular spring clamp produces a compressive force of 3 N on the plates. Determine the maximum bending stress produced in the spring at A. The spring has a rectangular cross section as shown.
10 mm 20 mm
Internal Moment: As shown on FBD, M = 0.660 N # m is positive since it tends to increase the beam’s radius of curvature.
210 mm
200 mm A
Section Properties: 220 mm
r =
0.200 + 0.210 = 0.205 m 2
r2 0.21 dA = b ln = 0.02 ln = 0.97580328 A 10 3 B m 0.20 r r 1 LA
A = (0.01)(0.02) = 0.200 A10  3 B m2 R =
A
1A
dA r
=
0.200(10  3 )
3
0.97580328(10 )
= 0.204959343 m
r  R = 0.205  0.204959343 = 0.040657 A 10 3 B m
Maximum Normal Stress: Applying the curvedbeam formula sC = =
M(R  r2 ) Ar 2 (r  R) 0.660(0.204959343  0.21) 0.200(10
3
3
)(0.21)(0.040657)(10 )
= 1.95MPa = 1.95 MPa (C) st = =
M(R  r1) Ar1 (r  R) 0.660(0.204959343  0.2) 0.200(10  3 )(0.2)(0.040657)(10  3 )
= 2.01 MPa (T)
Ans.
(Max)
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6–155. Determine the maximum compressive force the spring clamp can exert on the plates if the allowable bending stress for the clamp is sallow = 4 MPa.
10 mm 20 mm
210 mm
200 mm A
220 mm
Section Properties: r =
0.200 + 0.210 = 0.205 m 2
0.21 dA r2 = b ln = 0.02 ln = 0.97580328A 10  3 B m r r 0.20 LA 1
A = (0.01)(0.02) = 0.200 A 10  3 B m2 R =
3
A
1A
dA r
=
0.200(10 ) = 0.204959 m 0.97580328(10  3 )
r  R = 0.205  0.204959343 = 0.040657A 10  3 B m
Internal Moment: The internal moment must be computed about the neutral axis as shown on FBD. Mmax = 0.424959P is positive since it tends to increase the beam’s radius of curvature. Allowable Normal Stress: Applying the curvedbeam formula Assume compression failure sc = s allow = 4 A 106B =
M(R  r2 ) Ar2 (r  R) 0.424959P(0.204959  0.21) 0.200(10  3 )(0.21)(0.040657)(10  3 )
P = 3.189 N
Assume tension failure st = s allow = 4 A 10 6 B =
M(R  r 1) Ar1 (r  R) 0.424959P(0.204959  0.2) 0.200(10  3 )(0.2)(0.040657)(10  3) Ans.
P = 3.09 N (Controls !)
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*6–156. While in flight, the curved rib on the jet plane is subjected to an anticipated moment of M = 16 N # m at the section. Determine the maximum bending stress in the rib at this section, and sketch a twodimensional view of the stress distribution. 16 Nm 5 mm 20 mm 5 mm
0.6 m 5 mm
30 mm
LA
dA>r = (0.03)ln
0.605 0.625 0.630 + (0.005)ln + (0.03)ln = 0.650625(10  3) in. 0.6 0.605 0.625
A = 2(0.005)(0.03) + (0.02)(0.005) = 0.4(10  3) in 2 R =
A
1A dA>r
=
0.4(10 3 ) = 0.6147933 0.650625(10  3 )
(sc) max =
M(R  r c) 16(0.6147933  0.630) = = 4.67 MPa 0.4(10 3 )(0.630)(0.615  0.6147933) Ar A(r  R)
(s s) max =
M(R  rs ) 16(0.6147933  0.6) = = 4.77 MPa 3 ( r R ) 0.4(10 )(0.6)(0.615  0.6147933) ArA
Ans.
If the radius of each notch on the plate is r = 0.5 in., determine the largest moment that can be applied. The allowable bending stress for the material is s allow = 18 ksi. •6–157.
14.5 in.
M
b =
14.5  12.5 = 1.0 in. 2
b 1 = = 2.0 0.5 r
r 0.5 = = 0.04 12.5 h
From Fig. 644: K = 2.60 s max = K
Mc I
3 18(10 ) = 2.60 c
(M)(6.25) 1 (1)(12.5) 3 12
d
M = 180 288 lb # in. = 15.0 kip # ft
Ans.
442
1 in.
12.5 in.
M
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6–158. The symmetric notched plate is subjected to bending. If the radius of each notch is r = 0.5 in. and the applied moment is M = 10 kip # ft, determine the maximum bending stress in the plate.
14.5 in.
M
b 1 = = 2.0 0.5 r
1 in.
M
12.5 in.
r 0.5 = = 0.04 12.5 h
From Fig. 644: K = 2.60 smax = K
(10)(12)(6.25) Mc = 2.60c 1 d = 12.0 ksi I (1)(12.5)3 12
Ans.
6–159. The bar is subjected to a moment of M = 40 N # m . Determine the smallest radius r of the fillets so that an allowable bending stress of s allow = 124 MPa is not exceeded.
80 mm 7 mm
20 mm r
M
M r
Allowable Bending Stress: Mc s allow = K I 124 A 10 6B = KB
40(0.01)
1 3 R 12 (0.007)(0.02 )
K = 1.45
Stress Concentration Factor: From the graph in the text w 80 r with = = 4 and K = 1.45 , then = 0.25 . 20 h h r = 0.25 20 Ans.
r = 5.00 mm
*6–160. The bar is subjected to a moment of M = 17.5 N # m . If r = 5 mm, determine the maximum bending stress in the material.
80 mm r M
M
Stress Concentration Factor: From the graph in the text with w 80 5 r = = 4 and = = 0.25, then K = 1.45 . h h 20 20
r
Maximum Bending Stress: s max = K
Mc I
= 1.45 B 1
12
17.5(0.01) (0.007)(0.023 )
7 mm
20 mm
R Ans.
= 54.4 MPa
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•6–161. The simply supported notched bar is subjected to
P
two forces P. Determine the largest magnitude of P that can be applied without causing the material to yield.The material is A36 steel. Each notch has a radius of r = 0.125 in.
P 0.5 in. 1.75 in.
1.25 in.
20 in.
b =
20 in.
20 in.
20 in.
1.75  1.25 = 0.25 2
0.25 b = = 2; 0.125 r
0.125 r = = 0.1 1.25 h
From Fig. 644. K = 1.92 sY = K
Mc ; I
36 = 1.92c
20P(0.625) 1 3 12 (0.5)(1.25)
P = 122 lb
d
Ans.
6–162. The simply supported notched bar is subjected to the two loads, each having a magnitude of P = 100 lb. Determine the maximum bending stress developed in the bar, and sketch the bendingstress distribution acting over the cross section at the center of the bar. Each notch has a radius of r = 0.125 in.
P
0.5 in.
1.75  1.25 = 0.25 2
0.25 b = = 2; r 0.125
0.125 r = = 0.1 h 1.25
From Fig. 644, K = 1.92 s max = K
1.75 in.
1.25 in.
20 in.
b =
P
2000(0.625) Mc = 1.92c 1 d = 29.5 ksi I (0.5)(1.25) 3 12
Ans.
444
20 in.
20 in.
20 in.
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6–163. Determine the length L of the center portion of the bar so that the maximum bending stress at A, B, and C is the same. The bar has a thickness of 10 mm. w 60 = = 1.5 h 40
350 N 7 mm
60 mm
A
7 r = = 0.175 h 40
200 mm
40 mm 7 mm C
L 2
B L 2
200 mm
From Fig. 643, K = 1.5 (sA) max = K
MAc = 1.5c I
(sB )max = (sA) max = 19.6875(106) =
(35)(0.02) 1 (0.01)(0.043 ) 12
MB c I
d = 19.6875 MPa
175(0.2 + L2)(0.03) 1 (0.01)(0.06 3) 12
Ans.
L = 0.95 m = 950 mm
*6–164. The stepped bar has a thickness of 15 mm. Determine the maximum moment that can be applied to its ends if it is made of a material having an allowable bending stress of sallow = 200 MPa .
45 mm 30 mm 3 mm
M
M
Stress Concentration Factor: w 30 6 r = = 3 and = = 0.6, we have K = 1.2 10 10 h h obtained from the graph in the text. For the smaller section with
r w 45 3 = = 1.5 and = = 0.1, we have K = 1.75 30 30 h h obtained from the graph in the text. For the larger section with
Allowable Bending Stress: For the smaller section s max = sallow = K
Mc ; I
200A 10 6 B = 1.2 B 1
12
M(0.005) (0.015)(0.01 3)
R
M = 41.7 N # m (Controls !)
Ans.
For the larger section s max = sallow = K
Mc ; I
6 200A 10 B = 1.75B
M(0.015)
R 1 (0.015)(0.03 3 ) 12
M = 257 N # m 445
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•6–165.
The beam is made of an elastic plastic material for which s Y = 250 MPa. Determine the residual stress in the beam at its top and bottom after the plastic moment M p is applied and then released. Ix =
15 mm
20 mm 200 mm
1 1 6 (0.2)(0.23) 3 (0.18)(0.2) 3 = 82.78333(10 )m4 12 12
Mp
C1 = T 1 = sY (0.2)(0.015) = 0.003sY C2 = T 2 = sY (0.1)(0.02) = 0.002sY
15 mm 200 mm
Mp = 0.003s Y (0.215) + 0.002sY (0.1) = 0.000845 sY = 0.000845(250)(106) = 211.25 kN # m s =
Mp c I
211.25(103 )(0.115)
=
82.78333(10  6 )
y 0.115 ; = 250 293.5
= 293.5 MPa
y = 0.09796 m = 98.0 mm
s top = s bottom = 293.5  250 = 43.5 MPa
Ans.
6–166. The wideflange member is made from an elasticplastic material. Determine the shape factor.
t
Plastic analysis: T 1 = C1 = sY bt;
h
T 2 = C2 = s Y a
MP = s Y bt(h  t) + sY a
h  2t bt 2
t t
h  2t h  2t b (t) a b 2 2
b
t = s Y cbt(h  t) + (h  2t)2 d 4
Elastic analysis: I = =
1 1 3 (b  t)(h  2t) 3 bh 12 12 1 [bh3  (b  t)(h  2 t)3 ] 12
MY =
sy I c
=
=
1 3 3 sY A 12 B [bh  (b  t)(h  2t) ] h 2
bh3  (b  t)(h  2t) 3 6h
sY
Shape factor: k =
[bt(h  t) + 4t (h  2t)2 ]s Y MP = bh 3  (b  t)(h  2t)3 MY s 6h
=
Y
2 3h 4bt(h  t) + t(h  2t) c 3 d 2 bh  (b  t)(h  2t) 3
Ans.
446
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6–167. Determine the shape factor for the cross section. Maximum Elastic Moment: The moment of inertia about neutral axis must be determined first.
a
1 1 (a)(3a)3 + (2a) A a3 B = 2.41667a4 12 12
INA =
a a
Applying the flexure formula with s = sY, we have sY =
MY c I
MY =
a
sY I c
=
s Y (2.41667a 4) 1.5a
a
a
= 1.6111a3 sY
Plastic Moment: M P = s Y (a)(a)(2a) + sY (0.5a)(3a)(0.5a) = 2.75a3 sY Shape Factor: k =
2.75a3s Y MP = = 1.71 MY 1.6111a3 sY
Ans.
*6–168. The beam is made of elastic perfectly plastic material. Determine the maximum elastic moment and the plastic moment that can be applied to the cross section. Take a = 2 in. and s Y = 36 ksi.
a a
Maximum Elastic Moment: The moment of inertia about neutral axis must be determined first.
a
INA
1 1 (2) A 6 3B + (4) A 23 B = 38.667 in4 = 12 12
Applying the flexure formula with s = sY, we have sY = = MY =
a
MY c I
s Y I 36(38.667) = c 3
= 464 kip # in = 38.7 kip # ft
Ans.
Plastic Moment: MP = 36(2)(2)(4) + 36(1)(6)(1) = 792 kip # in = 66.0 kip # ft
Ans.
447
a
a
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•6–169.
The box beam is made of an elastic perfectly plastic material for which s Y = 250 MPa. Determine the residual stress in the top and bottom of the beam after the plastic moment Mp is applied and then released.
Plastic Moment: 6 6 MP = 250 A 10 B (0.2)(0.025)(0.175) + 250 A 10 B (0.075)(0.05)(0.075)
25 mm
= 289062.5 N # m
150 mm
Modulus of Rupture: The modulus of rupture sr can be determined using the flexure formula with the application of reverse, plastic moment MP = 289062.5 N #m . I =
25 mm
25 mm 150 mm 25 mm
1 1 (0.2) A 0.23 B (0.15) A 0.153 B 12 12
= 91.14583 A 10  6 B m 4
s r=
289062.5 (0.1) M Pc = = 317.41 MPa 6 I 91.14583 A 10 B
Residual Bending Stress: As shown on the diagram. œ œ s top = sbot = sr  sY
Ans.
= 317.14  250 = 67.1 MPa
6–170. Determine the shape factor for the wideflange beam. Ix =
15 mm
1 1 3 3 4 6 (0.2)(0.23) (0.18)(0.2) = 82.78333A 10 B m 12 12
20 mm 200 mm
C1 = T 1 = sY(0.2)(0.015) = 0.003sY
Mp
C2 = T 2 = sY(0.1)(0.02) = 0.002s Y Mp = 0.003s Y(0.215) + 0.002sY(0.1) = 0.000845 s Y
15 mm 200 mm
MY c sY = I MY = k =
sY A 82.78333)10  6 B
Mp MY
0.115
=
= 0.000719855 s Y
0.000845s Y = 1.17 0.000719855sY
Ans.
448
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6–171. Determine the shape factor of the beam’s cross section. 3 in.
Referring to Fig. a, the location of centroid of the crosssection is y =
©yA 7.5(3)(6) + 3(6)(3) = = 5.25 in. ©A 3(6) + 6(3)
6 in.
The moment of inertia of the crosssection about the neutral axis is I =
1 1 (3) A 6 3B + 3(6)(5.25  3) 2 + (6) A 3 3B + 6(3)(7.5  5.25) 2 12 12
1.5 in. 3 in. 1.5 in.
= 249.75 in4
Here s max = s Y and c = y = 5.25 in . Thus s max =
Mc ; I
sY =
MY (5.25) 249.75
MY = 47.571sY Referring to the stress block shown in Fig. b, A L
sdA = 0;
T  C1  C2 = 0
d(3)s Y  (6  d)(3)s Y  3(6)s Y = 0 d = 6 in. Since d = 6 in. , c1 = 0 , Fig. c. Here T = C = 3(6) s Y = 18 sY Thus, M P = T(4.5) = 18 sY (4.5) = 81 s Y Thus, k =
81 sY MP = = 1.70 47.571 s Y MY
Ans.
449
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*6–172. The beam is made of elasticperfectly plastic material. Determine the maximum elastic moment and the plastic moment that can be applied to the cross section. Take sY = 36 ksi .
3 in.
Referring to Fig. a, the location of centroid of the crosssection is 6 in.
©y A 7.5(3)(6) + 3(6)(3) y = = = 5.25 in. ©A 3(6) + 6(3) The moment of inertia of the crosssection about the neutral axis is
1.5 in. 3 in. 1.5 in.
1 1 (3)(63 ) + 3(6)(5.25  3) 2 + (6)(33 ) + 6(3)(7.5  5.25)2 I = 12 12 = 249.75 in 4 Here, smax = sY = 36 ksi and ¢ = y = 5.25 in . Then s max =
Mc ; I
36 =
MY (5.25) 249.75
MY = 1712.57 kip # in = 143 kip # ft
Ans.
Referring to the stress block shown in Fig. b, A L
sdA = 0;
T  C1  C2 = 0
d(3) (36)  (6  d)(3)(36)  3(6) (36) = 0 d = 6 in. Since d = 6 in. , c1 = 0 , Here, T = C = 3(6)(36) = 648 kip Thus, MP = T(4.5) = 648(4.5) = 2916 kip # in = 243 kip # ft
Ans.
450
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•6–173.
Determine the shape factor for the cross section of the Hbeam. 1 1 3 3 4 6 (0.2)(0.02 ) + 2 a b(0.02)(0.2 ) = 26.8(10 )m 12 12
Ix =
200 mm
C1 = T 1 = s Y(2)(0.09)(0.02) = 0.0036sy
20 mm M p
C2 = T 2 = s Y(0.01)(0.24) = 0.0024s y
200 mm
Mp = 0.0036sY (0.11) + 0.0024sY (0.01) = 0.00042sY
20 mm
MYc I
sY = MY =
s Y(26.8)(10  6 ) 0.1
Mp
k =
20 mm
MY
=
= 0.000268s Y
0.00042sY = 1.57 0.000268sY
Ans.
6–174. The Hbeam is made of an elasticplastic material for which sY = 250 MPa. Determine the residual stress in the top and bottom of the beam after the plastic moment M p is applied and then released. 200 mm
1 1 (0.2)(0.023) + 2 a b(0.02)(0.23) = 26.8(10  6 )m4 12 12
Ix =
20 mm
C1 = T 1 = sY (2)(0.09)(0.02) = 0.0036sy
200 mm
C2 = T 2 = sY (0.01)(0.24) = 0.0024sy
20 mm
Mp = 0.0036sY (0.11) + 0.0024sY (0.01) = 0.00042s Y Mp = 0.00042(250) A 10 6B = 105 kN # m s¿ = y 250
=
Mp c I
0.1 ; 392
=
105(10 3)(0.1) 26.8(10  6)
Mp
= 392 MPa
y = 0.0638 = 63.8 mm Ans.
sT = s B = 392  250 = 142 MPa
451
20 mm
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6–175. Determine the shape factor of the cross section. 3 in.
The moment of inertia of the crosssection about the neutral axis is I =
1 1 (3)(9 3) + (6) (33) = 195.75 in 4 12 12
3 in. 3 in.
Here, s max = sY and c = 4.5 in. Then s max =
Mc ; I
sY =
MY (4.5) 195.75
3 in.
MY = 43.5 sY Referring to the stress block shown in Fig. a, T 1 = C1 = 3(3)sY = 9 sY T 2 = C2 = 1.5(9)s Y = 13.5 sY Thus, MP = T 1 (6) + T2(1.5) = 9sY(6) + 13.5sY (1.5) = 74.25 sY k =
74.25 s Y MP = = 1.71 MY 43.5 sY
Ans.
452
3 in.
3 in.
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*6–176. The beam is made of elasticperfectly plastic material. Determine the maximum elastic moment and the plastic moment that can be applied to the cross section. Take s Y = 36 ksi.
3 in. 3 in.
The moment of inertia of the crosssection about the neutral axis is
3 in.
1 1 (3)(93 ) + (6)(3 3) = 195.75 in4 12 12
I =
Here, smax = s Y = 36 ksi and c = 4.5 in. Then s max
Mc ; = I
3 in.
MY (4.5) 36 = 195.75
MY = 1566 kip # in = 130.5 kip # ft
Ans.
Referring to the stress block shown in Fig. a, T 1 = C1 = 3(3)(36) = 324 kip T 2 = C2 = 1.5(9)(36) = 486 kip Thus, M P = T 1(6) + T2 (1.5) = 324(6) + 486(1.5) = 2673 kip # in. = 222.75 kip # ft = 223 kip # ft
Ans.
453
3 in.
3 in.
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•6–177.
Determine the shape factor of the cross section for the tube. The moment of inertia of the tube’s crosssection about the neutral axis is I =
5 in.
p 4 p A r  r4i B = A 6 4  5 4B = 167.75 p in 4 4 o 4
6 in.
Here, s max = sY and C = r o = 6 in , s max =
Mc ; I
sY =
MY (6) 167.75 p
M Y = 87.83 s Y The plastic Moment of the table’s crosssection can be determined by super posing the moment of the stress block of the solid circular crosssection with radius ro = 6 in and ri = 5 in. as shown in Figure a, Here, T 1 = C1 =
1 p(6 2)s Y = 18psY 2
T 2 = C2 =
1 2 p(5 )s Y = 12.5p s Y 2
Thus, MP = T 1b 2c
4(6) 4(5) d r  T2 b 2c dr 3p 3p
16 40 b = (18psY )a b  12.5psY a p 3p = 121.33 sY k =
121.33 s Y MP = = 1.38 87.83 s Y MY
Ans.
454
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6–178. The beam is made from elasticperfectly plastic material. Determine the shape factor for the thickwalled tube. Maximum Elastic Moment. The moment of inertia of the crosssection about the neutral axis is I =
p Ar 4  r4i B 4 o
With c = ro and smax = sY, smax =
Mc ; I
sY =
MY =
MY (ro ) p A r 4  ri 4 B 4 o
p Ar 4  ri 4 BsY 4r o o
Plastic Moment. The plastic moment of the cross section can be determined by superimposing the moment of the stress block of the solid beam with radius r 0 and ri as shown in Fig. a, Referring to the stress block shown in Fig. a, T1 = c 1 = T2 = c 2 =
p 2 p 2
MP = T 1 c 2 a = = Shape Factor.
ro 2sY ri 2sY 4r o 4ri b d  T2 c 2 a b d 3p 3p
p 2 p 8ro 8ri r s a b  ri 2sY a b 2 o Y 3p 2 3p 4 A r 3  ri 3 B sY 3 o
4 Aro 3  ri 3 BsY 16ro A ro 3  ri3 B MP 3 = = k = p MY 3p A ro 4  ri 4 B ro 4  ri 4 B s Y A 4ro
Ans.
455
ro ri
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6–179. Determine the shape factor for the member. Plastic analysis: T = C =
h – 2
1 bh h s (b) a bs Y = 2 2 4 Y
h – 2
bh b h2 h MP = sY a b = s 4 3 12 Y Elastic analysis: I = 2c
1 b h3 h 3 (b) a b d = 12 2 48
b
3
s YA bh 48 B sY I b h2 MY = = = s h 24 Y c 2 Shape factor: k =
Mp MY
=
bh2 12
sY
bh2 24
sY
= 2
Ans.
*6–180. The member is made from an elasticplastic material. Determine the maximum elastic moment and the plastic moment that can be applied to the cross section. Take b = 4 in., h = 6 in., s Y = 36 ksi.
h – 2
Elastic analysis: I = 2c MY =
h – 2
1 (4)(3)3 d = 18 in 4 12
sY I 36(18) = = 216 kip # in. = 18 kip # ft c 3
Ans. b
Plastic analysis: T = C =
1 (4)(3)(36) = 216 kip 2
6 M p = 2160a b = 432 kip # in. = 36 kip # ft 3
Ans.
456
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•6–181.
The beam is made of a material that can be assumed perfectly plastic in tension and elastic perfectly plastic in compression. Determine the maximum bending moment M that can be supported by the beam so that the compressive material at the outer edge starts to yield.
h sY M
sdA = 0; LA
C  T = 0
sY
a
1 s (d)(a)  sY (h  d)a = 0 2 Y d = M =
2 h 3
1 2 11 11a h2 sY a hb (a) a hb = sY 2 3 18 54
Ans.
6–182. The box beam is made from an elasticplastic material for which s Y = 25 ksi. Determine the intensity of the distributed load w0 that will cause the moment to be (a) the largest elastic moment and (b) the largest plastic moment.
w0
Elastic analysis: I =
9 ft
1 1 (8)(163) (6)(123) = 1866.67 in 4 12 12
Mmax
sYI ; = c
9 ft
8 in.
25(1866.67) 27w0 (12) = 8
w0 = 18.0 kip>ft
Ans.
Plastic analysis:
12 in.
16 in.
6 in.
C1 = T 1 = 25(8)(2) = 400 kip C2 = T 2 = 25(6)(2) = 300 kip MP = 400(14) + 300(6) = 7400 kip # in. 27w0(12) = 7400 w0 = 22.8 kip>ft
Ans.
457
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6–183. The box beam is made from an elasticplastic material for which s Y = 36 ksi. Determine the magnitude of each concentrated force P that will cause the moment to be (a) the largest elastic moment and (b) the largest plastic moment.
P
From the moment diagram shown in Fig. a, Mmax = 6 P.
6 ft
P
8 ft
6 ft
The moment of inertia of the beam’s crosssection about the neutral axis is 6 in.
1 1 3 3 4 I = (6)(12 ) (5)(10 ) = 447.33 in 12 12
10 in.
12 in.
Here, smax = sY = 36 ksi and c = 6 in. s max =
Mc ; I
36 =
MY (6) 447.33
5 in.
MY = 2684 kip # in = 223.67 kip # ft It is required that M max = MY 6P = 223.67 P = 37.28 kip = 37.3 kip
Ans.
Referring to the stress block shown in Fig. b, T 1 = C1 = 6(1)(36) = 216 kip T 2 = C2 = 5(1)(36) = 180 kip Thus, M P = T 1(11) + T 2(5) = 216(11) + 180(5) = 3276 kip # in = 273 kip # ft It is required that M max = MP 6P = 273 Ans.
P = 45.5 kip
458
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*6–184. The beam is made of a polyester that has the stress–strain curve shown. If the curve can be represented by the equation s = [20 tan1 115P2] ksi, where tan1 115P2 is in radians, determine the magnitude of the force P that can be applied to the beam without causing the maximum strain in its fibers at the critical section to exceed Pmax = 0.003 in.>in.
P 2 in. 4 in.
8 ft
8 ft
s(ksi) s 20 tan 1 (15 P)
P(in./in.)
Maximum Internal Moment: The maximum internal moment M = 4.00P occurs at the mid span as shown on FBD. Stress–Strain Relationship: Using the stress–strain relationship. the bending stress can be expressed in terms of y using e = 0.0015y. s = 20 tan  1 (15e) = 20 tan  1 [15(0.0015y)] = 20 tan
1
(0.0225y)
When emax = 0.003 in.>in. , y = 2 in. and s max = 0.8994 ksi Resultant Internal Moment: The resultant internal moment M can be evaluated from the integal M = 2
A L
A L
ysdA .
ysdA
2in
= 2
0 L
= 80
0 L
= 80B
yC 20 tan
1
2in
(0.0225y)D (2dy)
y tan  1 (0.0225y) dy
1 + (0.0225)2 y2 2(0.0225)2
1
tan
(0.0225y) 
y R2 2(0.0225)
2in. 0
= 4.798 kip # in Equating M = 4.00P(12) = 4.798 Ans.
P = 0.100 kip = 100 lb
459
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•6–185.
The plexiglass bar has a stress–strain curve that can be approximated by the straightline segments shown. Determine the largest moment M that can be applied to the bar before it fails.
s (MPa)
20 mm
M 20 mm
failure
60 40
tension
0.06 0.04
P (mm/ mm) 0.02
compression 80 100
Ultimate Moment: LA
s dA = 0;
C  T2  T 1 = 0
1 1 d 1 d sc (0.02  d)(0.02) d  40 A 10 6 B c a b (0.02) d  (60 + 40) A 106 B c(0.02) d = 0 2 2 2 2 2 s  50s d  3500(106 )d = 0
Assume.s = 74.833 MPa; d = 0.010334 m From the strain diagram, e 0.04 = 0.02  0.010334 0.010334
e = 0.037417 mm>mm
From the stress–strain diagram, s 80 = 0.037417 0.04
s = 74.833 MPa (OK! Close to assumed value)
Therefore, 1 C = 74.833 A 10 6 B c (0.02  0.010334)(0.02) d = 7233.59 N 2 T1 =
1 0.010334 b d = 5166.85 N (60 + 40) A 106 B c (0.02) a 2 2
T 2 = 40 A 10 6 B c y1 = y2 = y3 =
1 0.010334 (0.02)a b d = 2066.74 N 2 2
2 (0.02  0.010334) = 0.0064442 m 3 2 0.010334 a b = 0.0034445 m 3 2
0.010334 1 2(40) + 60 0.010334 + c1  a bda b = 0.0079225m 2 3 40 + 60 2
M = 7233.59(0.0064442) + 2066.74(0.0034445) + 5166.85(0.0079255) = 94.7 N # m
Ans.
460
0.04
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6–186. The stress–strain diagram for a titanium alloy can be approximated by the two straight lines. If a strut made of this material is subjected to bending, determine the moment resisted by the strut if the maximum stress reaches a value of (a) sA and (b) sB .
3 in. M
2 in. s (ksi) B
sB 180 sA 140
A
0.01
a) Maximum Elastic Moment : Since the stress is linearly related to strain up to point A, the flexure formula can be applied. sA =
Mc I M =
=
s AI c 1 140 C 12 (2)(3 3) D
1.5
= 420 kip # in = 35.0 kip # ft b)
Ans.
The Ultimate Moment : C 1 = T1 =
1 (140 + 180)(1.125)(2) = 360 kip 2
C 2 = T2 =
1 (140)(0.375)(2) = 52.5 kip 2
M = 360(1.921875) + 52.5(0.5) = 718.125 kip # in = 59.8 kip # ft
Ans.
Note: The centroid of a trapezodial area was used in calculation of moment.
461
0.04
P (in./in.)
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6–187. A beam is made from polypropylene plastic and has a stress–strain diagram that can be approximated by the curve shown. If the beam is subjected to a maximum tensile and compressive strain of P = 0.02 mm> mm, determine the maximum moment M.
M
s(Pa)
emax = 0.02
s 10(106 )P1/ 4
M
100 mm
30 mm
1 >4 6 s max = 10 A 10 B (0.02) = 3.761 MPa
P (mm/ mm)
e 0.02 = y 0.05
e = 0.4 y s = 10 A 10 6 B (0.4) 1>4 y 1>4 M =
LA
y s dA = 2
M = 0.47716 A 10 6B
M = 251 N # m
0 L
L0
0.05
0.05
y(7.9527) A 10 6 B y1>4 (0.03)dy
4 y5>4 dy = 0.47716 A 106 B a b(0.05) 9>4 5
Ans.
*6–188. The beam has a rectangular cross section and is made of an elasticplastic material having a stress–strain diagram as shown. Determine the magnitude of the moment M that must be applied to the beam in order to create a maximum strain in its outer fibers of P max = 0.008.
400 mm M
200 mm
s(MPa)
200
0.004
C1 = T 1 = 200A 106 B(0.1)(0.2) = 4000 kN C2 = T 2 =
1 (200) A 106 B (0.1)(0.2) = 2000 kN 2
M = 4000(0.3) + 2000(0.1333) = 1467 kN # m = 1.47 MN # m
Ans.
462
P (mm/mm)
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•6–189.
The bar is made of an aluminum alloy having a stress–strain diagram that can be approximated by the straight line segments shown. Assuming that this diagram is the same for both tension and compression, determine the moment the bar will support if the maximum strain at the top and bottom fibers of the beam is P max = 0.03. s  80 90  80 ; = 0.03  0.025 0.05  0.025
s(ksi) 90 80 60 4 in. M
s = 82 ksi
C1 = T 1 =
1 (0.3333)(80 + 82)(3) = 81 kip 2
C2 = T 2 =
1 (1.2666)(60 + 80)(3) = 266 kip 2
C3 = T 3 =
1 (0.4)(60)(3) = 36 kip 2
3 in. 0.006
0.025
0.05
P(in./ in.)
M = 81(3.6680) + 266(2.1270) + 36(0.5333) = 882.09 kip # in. = 73.5 kip # ft
Ans.
Note: The centroid of a trapezodial area was used in calculation of moment areas.
6–190. The beam is made from three boards nailed together as shown. If the moment acting on the cross section is M = 650 N # m, determine the resultant force the bending stress produces on the top board. 15 mm
Section Properties: 0.0075(0.29)(0.015) + 2[0.0775(0.125)(0.02)] 0.29(0.015) + 2(0.125)(0.02)
y =
M 650 Nm 20 mm
125 mm
= 0.044933 m 20 mm
1 (0.29) A 0.0153 B + 0.29(0.015) (0.044933  0.0075) 2 = 12
I NA
+
1 (0.04) A0.1253 B + 0.04(0.125)(0.0775  0.044933)2 12
= 17.99037 A 10
6
B m4
Bending Stress: Applying the flexure formula s = sB =
sA =
650(0.044933  0.015) 17.99037(10 650(0.044933) 17.99037(10  6)
6
)
My I
= 1.0815 MPa
= 1.6234 MPa
Resultant Force: FR =
1 (1.0815 + 1.6234) A 10 6 B (0.015)(0.29) 2
Ans.
= 5883 N = 5.88 kN
463
250 mm
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6–191. The beam is made from three boards nailed together as shown. Determine the maximum tensile and compressive stresses in the beam. 15 mm M 650 Nm 20 mm
125 mm 20 mm
Section Properties: y =
0.0075(0.29)(0.015) + 2[0.0775(0.125)(0.02)] 0.29(0.015) + 2(0.125)(0.02)
= 0.044933 m INA =
1 (0.29) A 0.0153 B + 0.29(0.015)(0.044933  0.0075)2 12 +
1 (0.04) A 0.1253B + 0.04(0.125)(0.0775  0.044933)2 12
= 17.99037 A10
6
B m4
Maximum Bending Stress: Applying the flexure formula s = (smax )t =
(smax )c =
650(0.14  0.044933) 6
17.99037(10 ) 650(0.044933) 17.99037(10  6 )
My I Ans.
= 3.43 MPa (T)
= 1.62 MPa (C)
Ans.
464
250 mm
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*6–192. Determine the bending stress distribution in the beam at section a– a. Sketch the distribution in three dimensions acting over the cross section.
80 N
80 N
a
a 400 mm
a + ©M = 0;
300 mm
80 N 15 mm 100 mm
1 1 3 3 4 6 Iz = (0.075)(0.015 ) + 2 a b(0.015)(0.1 ) = 2.52109(10 )m 12 12 32(0.05) Mc = = 635 kPa 6 I 2.52109(10 )
15 mm 75 mm
Ans.
•6–193.
The composite beam consists of a wood core and two plates of steel. If the allowable bending stress for the wood is (sallow) w = 20 MPa, and for the steel (sallow) st = 130 MPa , determine the maximum moment that can be applied to the beam. Ew = 11 GPa, Est = 200 GPa. n =
y
z 125 mm
200(10 9) Est = = 18.182 11(10 9) Ew
M
1 3 4 3 I = (0.80227)(0.125 ) = 0.130578(10 )m 12
x
Failure of wood : (sw) max
M(0.0625) ; 0.130578(10  3 )
M = 41.8 kN # m
Failure of steel : (sst )max =
20 mm 75 mm 20 mm
Mc = I
20(106 ) =
400 mm
80 N
M  80(0.4) = 0 M = 32 N # m
smax =
300 mm
nMc I 130(106 ) =
18.182(M)(0.0625) 0.130578(10  3)
M = 14.9 kN # m (controls)
Ans.
465
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6–194. Solve Prob. 6–193 if the moment is applied about the y axis instead of the z axis as shown.
y
z
125 mm
M x
20 mm 75 mm 20 mm
n = I =
11(109 ) 200(10 4)
= 0.055
1 1 (0.125)(0.115 3) (0.118125)(0.0753 ) = 11.689616(10  6 ) 12 12
Failure of wood : (sw ) max =
nMc2 I 0.055(M)(0.0375) ; 6 11.689616(10 )
20(106) =
M = 113 kN # m
Failure of steel : (sst ) max =
Mc1 I 130(106) =
M(0.0575) 11.689616(10  6 )
M = 26.4 kN # m (controls)
Ans.
466
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6–195. A shaft is made of a polymer having a parabolic cross section. If it resists an internal moment of M = 125 N # m , determine the maximum bending stress developed in the material (a) using the flexure formula and (b) using integration. Sketch a threedimensional view of the stress distribution acting over the crosssectional area. Hint: The moment of inertia is determined using Eq. A–3 of Appendix A.
y
100 mm y 100 – z2 /25 M 125 N·m z
Maximum Bending Stress: The moment of inertia about y axis must be determined first in order to use Flexure Formula I =
A L
= 2
50 mm 50 mm
y 2 dA
L0
100mm
y2 (2z) dy 100mm
= 20
0 L
y 22100  y dy
100 mm 3 8 16 3 5 7 = 20B  y2 (100  y)2 y (100  y)2 (100  y)2 R 2 2 15 105 0
Thus,
= 30.4762 A 10  6 B mm4 = 30.4762 A 10  6 B m4
s max =
125(0.1) Mc = = 0.410 MPa I 30.4762(10  6)
Ans.
Maximum Bending Stress: Using integration dM = 2[y(s dA)] = 2 b yc a M = 125 A 10 3 B = 125 A 10 3 B =
5 L0
smax
smax 5 smax 5
100mm
s max 100
b y d (2z dy) r
y2 2100  y dy
3 2
3
B  y2 (100  y)2 
100 mm 8 16 5 7 y(100  y) 2 (100  y)2 R 2 15 105 0
(1.5238) A10 6 B
s max = 0.410 N>mm2 = 0.410 MPa
Ans.
467
x
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*6–196. Determine the maximum bending stress in the handle of the cable cutter at section a–a. A force of 45 lb is applied to the handles. The crosssectional area is shown in the figure.
20
45 lb
a
5 in.
4 in. 3 in. A
a
0.75 in. 0.50 in.
45 lb
a+ ©M = 0;
M  45(5 + 4 cos 20°) = 0
M = 394.14 lb # in.
Mc 394.14(0.375) = 1 = 8.41 ksi 3 I 12 (0.5)(0.75 )
s max =
Ans.
•6–197.
The curved beam is subjected to a bending moment of M = 85 N # m as shown. Determine the stress at points A and B and show the stress on a volume element located at these points.
M 85 Nm
100 mm
A
r 0.42 0.57 0.59 dA = b ln 2 = 0.1 ln + 0.015 ln + 0.1 ln r r 0.40 0.42 0.57 1 LA
400 mm
R =
LA
dA r
B
3
6.25(10 ) = = 0.484182418 m 0.012908358
r  R = 0.495  0.484182418 = 0.010817581 m sA =
M(R  r A) Ar A(r  R)
=
85(0.484182418  0.59) 6.25(10  3 )(0.59)(0.010817581)
= 225.48 kPa
s A = 225 kPa (C) sB =
150 mm
30
A = 2(0.1)(0.02) + (0.15)(0.015) = 6.25(10  3 ) m2 A
20 mm
15 mm
B
= 0.012908358 m
A
Ans.
M(R  rB) 85(0.484182418  0.40) = = 265 kPa (T) ArB (r  R) 6.25(10  3 )(0.40)(0.010817581)
468
Ans.
20 mm
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6–198. Draw the shear and moment diagrams for the beam and determine the shear and moment in the beam as functions of x, where 0 … x 6 6 ft.
8 kip
2 kip/ ft 50 kipft
x 6 ft
+ c ©F y = 0;
20  2x  V = 0 V = 20  2x
c+ ©M NA = 0;
4 ft
Ans.
x 20x  166  2xa b  M = 0 2
M = x2 + 20x  166
Ans.
6–199. Draw the shear and moment diagrams for the shaft if it is subjected to the vertical loadings of the belt, gear, and flywheel. The bearings at A and B exert only vertical reactions on the shaft.
300 N 450 N A
B
200 mm
400 mm
300 mm
200 mm 150 N
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*6–200. A member has the triangular cross section shown. Determine the largest internal moment M that can be applied to the cross section without exceeding allowable tensile and compressive stresses of (s allow) t = 22 ksi and (s allow) c = 15 ksi, respectively.
4 in. 4 in.
2 in.
y (From base) = I =
1 242  2 2 = 1.1547 in. 3
1 (4)( 242  2 2)3 = 4.6188 in4 36
Assume failure due to tensile stress : s max =
My ; I
22 =
M(1.1547) 4.6188
M = 88.0 kip # in. = 7.33 kip # ft Assume failure due to compressive stress: s max =
Mc ; I
15 =
M(3.4641  1.1547) 4.6188
M = 30.0 kip # in. = 2.50 kip # ft
Ans.
(controls)
470
M 2 in.
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•6–201.
The strut has a square cross section a by a and is subjected to the bending moment M applied at an angle u as shown. Determine the maximum bending stress in terms of a, M, and u . What angle u will give the largest bending stress in the strut? Specify the orientation of the neutral axis for this case.
y
a
z
x
a M
Internal Moment Components: M z = M cos u
M y = M sin u
Section Property: Iy = Iz =
1 4 a 12
Maximum Bending Stress: By Inspection, Maximum bending stress occurs at A and B. Applying the flexure formula for biaxial bending at point A s = 
= =
Mzy
+
Iz
My z Iy
M cos u ( a2) 1 4 a 12
+
 Msin u (  a2 ) 1 4 a 12
6M (cos u + sin u) a3
Ans.
6M ds = 3 (  sin u + cos u) = 0 du a cos u  sin u = 0 u = 45°
Ans.
Orientation of Neutral Axis: tan a =
Iz tan u Iy
tan a = (1) tan(45°) Ans.
a = 45°
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