FM_TOC 46060
6/22/10
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Page iii
CONTENTS
To the Instructor
iv
1 Stress
1
2 Strain
73
3 Mechanical Properties of Materials
92
4 Axial Load
122
5 Torsion
214
6 Bending
329
7 Transverse Shear
472
8 Combined Loadings
532
9 Stress Transformation
619
10 Strain Transformation
738
11 Design of Beams and Shafts
830
12 Deflection of Beams and Shafts
883
13 Buckling of Columns
1038
14 Energy Methods
1159
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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.
a + ©MA = 0;
4
F (3) - 1200(8) = 0;
5 A
+ c ©Fy = 0;
- Ay +
+ ©F = 0;
;
x
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 free-body 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;
C
V - 2(2 - x) = 0
V = {4 - 2x} kN‚
(1)
1
a + ©M = 0; - M - 2(2 - x) c (2 - x) d - 6 = 0 M = {-x2 + 4x - 10}kN # m‚(2)
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).
Vx = 0 = 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
C
A
B
2m
4m
6–7. Draw the shear and moment diagrams for the
compound beam which is pin connected at B.
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 free-body 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 = 150 a
x
b = 12.5x
12
Referring to Fig. b,
+ c ©Fy = 0;
a + ©M = 0; M +
275 -
1
(12.5x)(x) - V = 0
2
V = {275 - 6.25x2}lb‚ (1)
1
x
(12.5x)(x)a b - 275x = 0 M = {275x - 2.083x3}lb # ft‚(2)
2
3
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).
M x = 6.633 ft = 275(6.633) - 2.083(6.633)3 = 1216 lb # ft
332
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6–9. Draw the shear and moment diagrams for the beam.
Hint: The 20-kip load must be replaced by equivalent
loadings at point C on the axis of the beam.
15 kip
1 ft
A
C
4 ft
20 kip
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 free-body diagram of the frame shown
in Fig. a,
+ c ©Fy = 0;
P 150 lb
Ay - 150 = 0
C
A
Ay = 150 lb
a + ©MA = 0;
B
1.5 ft
1.5 ft
ND(1.5) - 150(3) = 0
D
ND = 300 lb
Shear and Moment Diagram: The couple moment acting on B due to ND 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
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) -
3
4
FDE(4) - FDE(2) = 0
5
5
6 ft
4 ft
FDE = 2000 lb
+ c ©Fy = 0;
- 800 +
+ ©F = 0;
:
x
- Cx +
3
(2000) - Cy = 0
5
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;
+ ©F = 0;
:
x
By (a) - P(a) = 0
Cy - P - P = 0
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
(two-force 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
M = (220x) lb ft‚
Ans.
For 3 ft 6 x 6 5 ft
+ c ©Fy = 0;
220 - 800 - V = 0
V = - 580 lb
a + ©MNA = 0;
Ans.
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
M = (500x - 3000) lb ft
Ans.
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 free-body 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 = 200 a b = 33.33x
6
Referring to Fig. b,
+ c ©Fy = 0;
- 300 -
a + ©M = 0; M +
1
(33.33x)(x) - V = 0
2
V = {- 300 - 16.67x2} lb (1)
1
x
(33.33x)(x)a b + 300x = 0 M = {-300x - 5.556x3} lb # ft (2)
2
3
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 kip⭈ft
Support Reactions: As shown on FBD.
Shear and Moment Function:
x
6 ft
For 0 … x 6 6 ft:
+ c ©Fy = 0;
4 ft
30.0 - 2x - V = 0
V = {30.0 - 2x} kip
Ans.
x
a + ©MNA = 0; 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 kip⭈ft
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 free-body diagram shown in Fig. b,
a + ©M = 0; Mx= 3 m +
1
(10)(3)(1) - 20(3) = 0
2
Mx= 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 free-body diagram of the beam shown
in Fig. a,
a + ©MA = 0;
3
FBC a b (2) - 2(3)(1.5) = 0
5
B
A
1.5 m
FBC = 7.5 kN
+ c ©Fy = 0;
C
3
Ay + 7.5 a b - 2(3) = 0
5
Ay = 1.5 kN
3
Shear and Moment Diagram: The vertical component of FBC 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 free-body
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
+ c ©Fy = 0;
-4 -
a + ©M = 0; M +
2
V = e - x2 - 4 f kN
3
1 4
a x b(x) - V = 0
2 3
1 4
x
a x b (x)a b + 4x = 0
2 3
3
(1)
2
M = e - x3 - 4x f kN # m (2)
9
Region 3 m 6 x … 6 m, Fig. c
+ c ©Fy = 0;
V - 4(6 - x) = 0
1
a + ©M = 0; - M - 4(6 - x) c (6 - x) d = 0
2
V = {24 - 4x} kN
(3)
M = { -2(6 - x)2}kN # m
(4)
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 - = - (32) - 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
Mx= 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
B
a
wL2
wL - wx = 0
2a
+ c ©Fy = 0;
x = L -
L
L2
2a
x
wL2
Mmax (+) + wx a b - a wL bx = 0
2
2a
a + ©M = 0;
Substitute x = L -
L2
;
2a
Mmax (+) = a wL =
wL2
L2
w
L2 2
b aL b aL b
2a
2a
2
2a
w
L2 2
aL b
2
2a
Mmax (-) - w(L - a)
©M = 0;
Mmax (-) =
(L - a)
= 0
2
w(L - a)2
2
To get absolute minimum moment,
Mmax (+) = Mmax (-)
L2 2
w
w
(L ) =
(L - a)2
2
2a
2
L a =
L2
= L - a
2a
L
22
‚
Ans.
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:
V = 0
+ c ©Fy = 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 ©Fy = 0;
w0
w0L
1 w0x
- a
b(x) = 0
4
2
L
B
x = 0.7071 L
a + ©MNA = 0;
M +
w0L
1 w0x
x
L
a
b (x)a b ax - b = 0
2
L
3
4
3
Substitute x = 0.7071L,
M = 0.0345 w0L2
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;
w0 L
w0 L
- V = 0
3
6
a + ©MNA = 0;
M +
V =
w0 L
6
w0 L L
w0 L L
a b a b = 0
6
9
3
3
M =
5w0 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 + ©MNA = 0;
9.375 - 0.5556x2 = 0
x = 4.108 m
M + (0.5556) A 4.1082 B a
4.108
b - 9.375(4.108) = 0
3
M = 25.67 kN # m
From FBD(b)
a + ©MNA = 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
6 ft
Support Reactions:
From the FBD of segment AB
a + ©MB = 0;
450(4) - Ay (6) = 0
Ay = 300.0 lb
+ c ©Fy = 0;
By - 450 + 300.0 = 0
By = 150.0 lb
+ ©F = 0;
:
x
Bx = 0
From the FBD of segment BC
a + ©MC = 0;
225(1) + 150.0(3) - MC = 0
MC = 675.0 lb # ft
+ c ©Fy = 0;
+ ©F = 0;
:
x
Cy - 150.0 - 225 = 0
Cy = 375.0 lb
Cx = 0
Shear and Moment Diagram: The maximum positive moment occurs when V = 0.
+ c ©Fy = 0;
a + ©MNA = 0;
150.0 - 12.5x2 = 0
x = 3.464 ft
150(3.464) - 12.5 A 3.4642 B a
3.464
b - Mmax = 0
3
Mmax = 346.4 lb # ft
346
C
B
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
V =
a + ©MNA = 0;
w0
(3L - 4x)
4
Ans.
7w0 L2
3w0 L
x
x + w0 xa b + M = 0
24
4
2
M =
w0
A - 12x2 + 18Lx - 7L2)
24
Ans.
For L>2 6 x … L
+ c ©Fy = 0;
V -
1 2w0
c
(L - x) d(L - x) = 0
2 L
V =
a + ©MNA = 0;
-M -
w0
(L - x)2
L
Ans.
1 2w0
L - x
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 w0 of
the leaves on the pin and draw the shear and moment diagram
for the pin.
0.4 kN/m
C
A
+ c ©Fy = 0;
B
w0
1
2(w0)(20)a b - 60(0.4) = 0
2
20 mm 60 mm 20 mm
w0 = 1.2 kN>m
Ans.
•6–33.
The ski supports the 180-lb 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-
1 200
c
(x - 3) d(x - 3) - V = 0
2 3
100 2
x + 500 f N
3
Ans.
Set V = 0, x = 3.873 m
a + ©MNA = 0;
M +
1 200
x - 3
c
(x - 3) d(x - 3) a
b
2 3
3
+ 200(x - 3)a
M = e-
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
100 3
x + 500x - 600 f N # m
9
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 dead-weight 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.
3000 lb
400 lb/ft
250 lb/ft
A
8 ft
2 ft
Support Reactions:
3 ft
15 000 lb
- 1.00 - 3 + 15 - 1.25 - 0.375 - Ay = 0
+ c ©Fy = 0;
Ay = 9.375 kip
Ans.
a + ©MA = 0;
1.00(7.667) + 3(5) - 15(3)
+ 1.25(2.5) + 0.375(1.667) + MA = 0
MA = 18.583 kip # ft = 18.6 kip # ft
Ans.
+ ©F = 0;
:
x
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 ©Fy = 0;
2wL
1w 2
x = 0
27
2L
x =
a + ©M = 0;
w
B
2/3 L
4
L = 0.385 L
A 27
M +
C
A
1w
1
2wL
(0.385L)2 a b(0.385L) (0.385L) = 0
2L
3
27
M = 0.0190 wL2
351
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
F
A
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 + ©MC = 0;
5.00(1) + 10.0(0) - Dy (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
FR =
x =
1
2
x dx = 21.33 kip
8 L0
1 8 3
8 10 x dx
21.33
8 kip/ft
1
w ⫽ x2
8
= 6.0 ft
x
B
A
8 ft
•6–45.
Draw the shear and moment diagrams for the beam.
L
FR =
dA =
LA
L0
w0
wdx =
w0
L
L L0
2
x2 dx =
w
w0 L
3
w
LA
2
x
A
w0L
w0x
= 0
12
3L2
1 1>3
x = a b L = 0.630 L
4
w0L
w0x3 1
a + ©M = 0;
(x) a xb - M = 0
12
3L2 4
M =
B
L
3
+ c ©Fy = 0;
w0
L
x3dx
L L0
3L
x =
=
=
w0 L
4
dA
3
LA
xdA
w0 2
x
L2
w0Lx
w0x4
12
12L2
Substitute x = 0.630L
M = 0.0394 w0L2
354
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6–46. Draw the shear and moment diagrams for the beam.
L
FR =
dA = w0
LA
L0
sin a
w
2w0 L
p
x b dx =
p
L
w0
A
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 cross-section about z and y axes are
1
(0.2)(0.153) = 56.25(10 - 6) m4
12
Iy =
1
(0.15)(0.23) = 0.1(10 - 3) m4
12
M
z
x
For the bending about z axis, c = 0.075 m.
smax =
90(103) (0.075)
Mc
= 120(106)Pa = 120 MPa
=
Iz
56.25 (10 - 6)
Ans.
For the bending about y axis, C = 0.1 m.
smax =
x
B
L
–
2
Iz =
p
w w0 sin – x
L
90(103) (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
3 in.
0.5 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 103 B + 0.5(10)(5.5 - 3.40)2
12
= 91.73 in4
Maximum Bending Stress: Applying the flexure formula
smax =
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)
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 103 B + 0.5(10)(5.5 - 3.40)2
12
= 91.73 in4
Maximum Bending Stress: Applying the flexure formula smax =
Mc
I
(st)max =
4(103)(12)(10.5 - 3.40)
= 3715.12 psi = 3.72 ksi
91.73
Ans.
(sc)max =
4(103)(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.
50 mm
C
5 mm
5 mm
y =
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)(53) + 50(5)(13.24 - 2.5)2 d
12
+ c
5 mm
5 mm
5 mm 7 mm 10 mm 7 mm
1
(34)(53) + 34(5)(13.24 - 7.5)2 d
12
+ 2c
1
1
(5)(203) + 5(20)(20 - 13.24)2 d + 2c (12)(53) + 12(5)(32.5 - 13.24)2 d
12
12
= 0.095883(10 - 6) m4
30(35 - 13.24)(10 - 3)
sA =
0.095883(10 - 6)
30(13.24 - 10)(10 - 3)
sB =
0.095883(10 - 6)
= 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
sallow = 175 MPa, determine the maximum bending moment
the strut will resist.
50 mm
C
5 mm
5 mm
B
-3
30(13.24)(10 )
sC =
-6
0.095883(10 )
= 4.14 MPa
©y2A
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)(53) + 50(5)(13.24 - 2.5)2 d + c (34)(53) + 34(5)(13.24 - 7.5)2 d
12
12
+ 2c
1
1
(5)(203) + 5(20)(20 - 13.24)2 d + 2c (12)(53) + 12(5)(32.5 - 13.24)2 d
12
12
= 0.095883(10 - 6) m4
s =
Mc
;
I
175(106) =
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 mm 7 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
M
D
Section Property:
I =
1
1
(0.2) A 0.23 B (0.15) A 0.153 B = 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)
91.14583(10 - 6)
= 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
sc a
M¿
b = 0.8457(100%) = 84.6 %
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 sD = 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
I =
(0.2) A 0.23 B (0.15) A 0.153 B = 91.14583 A 10 - 6 B m4
12
12
25 mm
25 mm
Bending Stress: Applying the flexure formula
s =
30 A 106 B =
My
I
M(0.075)
91.14583(10 - 6)
M = 36458 N # m = 36.5 kN # m
smax =
M
D
Ans.
36458(0.1)
Mc
= 40.0 MPa
=
I
91.14583(10 - 6)
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 three-dimensional 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)
200 mm M 600 Nm
1
(0.24)(0.0253) + (0.24)(0.025)(0.043752)
12
I =
+ 2a
20 mm
1
b (0.02)(0.153) + 2(0.15)(0.02)(0.043752)
12
= 34.53125 (10 - 6) m4
smax = sB =
Mc
I
600 (0.175 - 0.05625)
=
34.53125 (10 - 6)
= 2.06 MPa
sC =
Ans.
My
600 (0.05625)
= 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.
25 mm
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 =
20 mm
200 mm M 600 Nm
1
(0.24)(0.0253) + (0.24)(0.025)(0.043752)
12
I =
+ 2a
20 mm
1
b (0.02)(0.153) + 2(0.15)(0.02)(0.043752)
12
= 34.53125 (10 - 6) m4
s1 =
My
600(0.05625)
= 0.9774 MPa
=
I
34.53125(10 - 6)
sb =
My
600(0.05625 - 0.025)
= 0.5430 MPa
=
I
34.53125(10 - 6)
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 cross-sectional 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 0.223 B +
(0.1) A 0.023 B = 17.8133 A 10 - 6 B m4
12
12
Bending Stress: Applying the flexure formula s =
sA =
sB =
8(103)(0.11)
17.8133(10 - 6)
8(103)(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 cross-sectional 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 three-dimensional view of the stress distribution
acting over the entire cross-sectional 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 0.223 B +
(0.1) A 0.023 B = 17.8133 A 10 - 6 B m4
12
12
Bending Stress: Applying the flexure formula smax =
smax =
8(103)(0.11)
17.8133(10 - 6)
sy = 0.01m =
My
Mc
and s =
,
I
I
Ans.
= 49.4 MPa
8(103)(0.01)
17.8133(10 - 6)
= 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) - 2 c p A 1.52 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 + Ad2
=
1
1
(6)a 83 b + 6(8) A 4.3454 - 4 B 2 - B pa 1.54 b + pa 1.52 b 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.
A smax B T =
100(12)(4.3454)
Mc
=
= 23.8 ksi (T)
I
218.87
Ans.
A smax B C =
My
100(12)(8 - 4.3454)
=
= 20.0 ksi (C)
I
218.87
Ans.
362
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6–59. If the beam is made of material having an
allowable tensile and compressive stress of (sallow)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) - 2 c p A 1.52 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 + Ad2
=
1
1
(6) A 83 B + 6(8) A 4.3454 - 4 B 2 - B p A 1.54 B + p A 1.52 B A 4.3454 - 2 B 2 R
12
4
= 218.87 in4
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 # in a
1 ft
b = 109.79 kip # ft
12 in.
For the bottom edge,
A smax B t =
Mc
;
I
24 =
M(4.3454)
218.87
M = 1208.82 kip # in a
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
three-dimensional view of the stress distribution.
y
A
C
1 in. 10 in.
1 in.
10 in.
2[5(10)(1)] + 10.5(16)(1) + 16(10)(1)
y =
2(10)(1) + 16(1) + 10(1)
Mz 16 kipft
z
= 9.3043 in.
14 in.
1
1
I = 2 c (1)(103) + 1(10)(9.3043 - 5)2 d +
(16)(13) + 16(1)(10.5 - 9.3043)2
12
12
+
B
1 in.
x
1 in.
1
(1)(103) + 1(10)(16 - 9.3043)2 = 1093.07 in4
12
sA =
16(12)(21 - 9.3043)
Mc
=
= 2.05 ksi
I
1093.07
Ans.
sB =
My
16(12)(9.3043)
=
= 1.63 ksi
I
1093.07
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 kipft
z
14 in.
1
1
I = 2 c (1)(103) + (10)(9.3043 - 5)2 d +
(16)(13) + 16(1)(10.5 - 9.3043)2
12
12
+
1
(1)(103) + 1(10)(16 - 9.3043)2 = 1093.07 in4
12
sA =
16(12)(21 - 9.3043)
Mc
=
= 2.0544 ksi
I
1093.07
sD =
My
16(12)(11 - 9.3043)
=
= 0.2978 ksi
I
1093.07
(FR)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 cross-section 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, yA = C = 0.15 m.
sA =
10(103) (0.15)
MyA
= 6.207(106)Pa = 6.21 MPa (C)
=
I
0.2417(10 - 3)
Ans.
For point B, yB = 0.125 m.
sB =
MyB
10(103)(0.125)
= 5.172(106)Pa = 5.17 MPa (T)
=
I
0.2417(10 - 3)
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 smax B S =
M(a>2)
6M
Mc
= 3
= 4
IS
a >12
a
For the circular cross section, c = r.
A smax B c =
Mc
Mr
4M
=
Ic
1 4
pr3
pr
4
It is required that
A smax B S = A smax B C
4M
6M
=
a3
pr3
a = 1.677r
Ans.
*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 three-dimensional
view of the stress distribution acting over the cross section.
I =
A
B
p 4
p
r = (0.54) = 0.0490874 in4
4
4
sA =
M ⫽ 300 lb⭈ft
45⬚
300(12)(0.5)
Mc
=
= 36.7 ksi
I
0.0490874
Ans.
0.5 in.
My
300(12)(0.5 sin 45°)
=
= 25.9 ksi
sB =
I
0.0490874
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 three-dimensional view of the stress
distribution acting over the cross section.
A
1.5 in.
12 in.
The moment of inertia of the cross-section about the neutral axis is
12 in.
1
1
(12)(153) (10.5)(123) = 1863 in4
I =
12
12
M
1.5 in.
1.5 in.
Along the top edge of the flange y = c = 7.5 in. Thus
smax =
4(103)(12)(7.5)
Mc
=
= 193 psi
I
1863
Ans.
Along the bottom edge to the flange, y = 6 in. Thus
s =
My
4(103)(12)(6)
=
= 155 psi
I
1863
6–66. If M = 4 kip # ft, determine the resultant force the
bending stress produces on the top board A of the beam.
A
1.5 in.
The moment of inertia of the cross-section about the neutral axis is
12 in.
1
1
(12)(153) (10.5)(123) = 1863 in4
12
12
I =
12 in.
M
Along the top edge of the flange y = c = 7.5 in. Thus
1.5 in.
smax =
4(103)(12)(7.5)
Mc
=
= 193.24 psi
I
1863
Along the bottom edge of the flange, y = 6 in. Thus
s =
4(103)(12)(6)
My
=
= 154.59 psi
I
1863
The resultant force acting on board A is equal to the volume of the trapezoidal
stress block shown in Fig. a.
FR =
1
(193.24 + 154.59)(1.5)(12)
2
= 3130.43 lb
= 3.13 kip
Ans.
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 Mmax = 11.34 kN # m
as indicated on the moment diagram. Applying the flexure formula
smax =
Mmax c
I
11.34(103)(0.045)
=
p
4
(0.0454)
= 158 MPa
Ans.
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 sallow = 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
Mmax c
I
smax = sallow =
11.34(103) A d2 B
180 A 106 B =
p
4
A d2 B 4
d = 0.08626 m = 86.3 mm
Ans.
•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) m4
12
12
15 mm
(a)
For section (b)
I =
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)
smax =
150(103)(0.165)
0.21645(10 - 3)
= 114.3 MPa
For section (b)
smax =
150(103)(0.18)
0.36135(10 - 3)
Ans.
= 74.72 MPa = 74.7 MPa
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
having a diameter of 12 in.
y =
100 lb/ft
5.75 in.
6 ft
6 ft
6 ft
©yA
0 + (6.50)(0.4786)
=
= 4.6091 in.
©A
0.4786 + 0.19635
I = c
1
1
1
p(0.5)4 - p(0.3125)4 d + 0.4786(6.50 - 4.6091)2 + p(0.25)4
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
=
I
5.9271
= 22.1 ksi
Ans.
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
B
60 in.
10 in.
20 kip
smax =
200(2.75)
Mc
= 1
= 12.2 ksi
4
I
4 p(2.75)
Ans.
370
D
10 in.
20 kip
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*6–72. The steel beam has the cross-sectional 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
(8) A 10.63 B (7.7) A 103 B = 152.344 in4
12
12
Absolute Maximum Bending Stress: The maximum moment is Mmax = 48.0w0 as
indicated on the FBD. Applying the flexure formula
smax =
22 =
Mmax c
I
48.0w0 (12)(5.30)
152.344
w0 = 1.10 kip>ft
Ans.
•6–73.
The steel beam has the cross-sectional 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 103 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
smax =
=
Mmax c
I
24.0(12)(5.30)
152.344
= 10.0 ksi
Ans.
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 box-beam 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
1.75 in.
3 in.
1.5 in.
Boat:
+ ©F = 0;
:
x
a + ©MB = 0;
Bx = 0
- NA(9) + 2300(5) = 0
NA = 1277.78 lb
+ c ©Fy = 0;
1277.78 - 2300 + By = 0
By = 1022.22 lb
Assembly:
a + ©MC = 0;
- ND(10) + 2300(9) = 0
ND = 2070 lb
+ c ©Fy = 0;
Cy + 2070 - 2300 = 0
Cy = 230 lb
I =
1
1
(1.75)(3)3 (1.5)(1.75)3 = 3.2676 in4
12
12
smax =
3833.3(12)(1.5)
Mc
=
= 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
D
C
1.5 m
25 mm
0.75 m
3 kN
3 kN
Shear and Moment Diagrams: As shown in Fig. a.
Maximum Moment: Due to symmetry, the maximum moment occurs in region BC
of the shaft. Referring to the free-body diagram of the segment shown in Fig. b.
Section Properties: The moment of inertia of the cross section about the neutral
axis is
I =
p
A 0.044 - 0.0254 B = 1.7038 A 10 - 6 B m4
4
Absolute Maximum Bending Stress:
sallow =
2.25 A 103 B (0.04)
Mmaxc
=
= 52.8 MPa
I
1.7038 A 10 - 6 B
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 cross-section about the neutral axis is
I =
M
1
1
(0.3)(0.33) (0.21)(0.263) = 0.36742(10 - 3) m4
12
12
260 mm
Thus,
20 mm 30 mm
smax
Mc
=
;
I
6
80(10 ) =
M(0.15)
0.36742(10 - 3)
M = 195.96 (103) N # m = 196 kN # m
The bending stress distribution over the cross-section is shown in Fig. a.
373
Ans.
30 mm
30 mm
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•6–77.
The steel beam has the cross-sectional area shown.
Determine the largest intensity of distributed load w that it
can support so that the bending stress does not exceed
smax = 22 ksi.
I =
smax
w
1
1
(8)(10.6)3 (7.7)(103) = 152.344 in4
12
12
8 ft
w
8 ft
8 in.
Mc
=
I
22 =
8 ft
0.30 in.
10 in.
0.3 in.
0.30 in.
32w(12)(5.3)
152.344
w = 1.65 kip>ft
Ans.
6–78. The steel beam has the cross-sectional 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. 6-78:
M = 32w = 32(5)(12) = 1920 kip # in.
I = 152.344 in4
smax =
1920(5.3)
Mc
=
= 66.8 ksi
I
152.344
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
smax =
46.7(103)(12)(3)
Mc
= 15.6 ksi
=
1
3
I
12 (6)(6 )
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
required height h to the nearest 14 in. if the allowable bending
stress is sallow = 24 ksi.
A
a + ©MA = 0;
+ c ©Fy = 0;
- Ay +
+ ©F = 0;
;
x
Ax -
smax =
4
(4000) - 1200 = 0;
5
3
(4000) = 0;
5
5 ft
B
4 ft
4
F (3) - 1200(8) = 0;
5 B
3 ft
FB = 4000 lb
Ay = 2000 lb
Ax = 2400 lb
6000(12) A h2 B
Mc
= 24(10)3
= 1
3
I
12 (2.5)(h )
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 Mmax = 7.5 kip # ft.
Absolute Maximum Bending Stress:
smax =
12 in.
7.5(12)(3)
Mmaxc
= 1.25 ksi
=
I
1
(12)(63)
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 sallow =
1.5 ksi, determine the required minimum thickness t of the
rectangular cross sectional area of the tie to the nearest 18 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 Mmax = 7.5 kip # ft.
Absolute Maximum Bending Stress:
smax
t
7.5(12)a b
2
1.5 =
1
(12)t3
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 di = 160 mm and do = 200 mm.
15 kN/m
60 kN m d
i do
A
B
3m
Section Property:
I =
p
A 0.14 - 0.084 B = 46.370 A 10 - 6 B m4
4
Absolute Maximum Bending Stress: The maximum moment is Mmax = 60.0 kN # m
as indicated on the moment diagram. Applying the flexure formula
smax =
Mmaxc
I
60.0(103)(0.1)
=
46.370(10 - 6)
= 129 MPa
Ans.
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 = 0.8do. Determine these required dimensions if the
allowable bending stress is sallow = 155 MPa.
15 kN/m
60 kN m d
i do
A
B
3m
Section Property:
I =
0.8do 4
do 4
dl 4
p do 4
p
- a
b R = 0.009225pd4o
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
smax = sallow =
155 A 106 B =
Thus,
Mmax c
I
60.0(103) A 2o B
d
0.009225pd4o
do = 0.1883 m = 188 mm
Ans.
dl = 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 sallow = 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 = sallow =
10 A 106 B =
Mmax c
I
562.5(0.75b)
1
12
(b)(1.5b)3
b = 0.05313 m = 53.1 mm
Ans.
380
2m
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6–86. Determine the absolute maximum bending stress
in the 2-in.-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 cross-section about the neutral axis is
I =
p 4
(1 ) = 0.25 p in4
4
Here, c = 1 in. Thus
smax =
=
Mmax c
I
15000(1)
0.25 p
= 19.10(103) 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 cross-section about the neutral axis is
I =
p 4
p d 4
a b =
d
4 2
64
Here, c = d>2. Thus
sallow =
Mmax c
;
I
22(103) =
15000(d> 2)
pd4>64
d = 1.908 in = 2 in.
Ans.
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 sallow = 150 MPa.
Allowable Bending Stress: The maximum moments is Mmax = 7.50 kN # m as
indicated on moment diagram. Applying the flexure formula
smax = sallow =
150 A 106 B =
Mmax c
I
7.50(103) A a2 B
1
12
a4
a = 0.06694 m = 66.9 mm
Ans.
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 =
23w0 L2
216
as indicated on the moment diagram. Applying the flexure formula
smax
Mmax c
=
=
I
A B
23w0 L2 h
2
216
1
3
12 bh
23w0 L2
=
Ans.
36bh2
384
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6–91. Determine the absolute maximum bending stress
in the 80-mm-diameter 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 cross-section about the neutral axis is
I =
p
(0.044) = 0.64(10 - 6)p m4
4
Here, c = 0.04 m. Thus
smax =
6(103)(0.04)
Mmax c
=
I
0.64(10 - 6)p
= 119.37(106) Pa
= 119 MPa
Ans.
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 cross-section about the neutral axis is
I =
pd4
p d 4
a b =
4 2
64
Here, c = d>2. Thus
sallow =
Mmax c
;
I
150(106) =
6(103)(d> 2)
pd4>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.
350 mm
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 m4
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
emax =
88.92(106)
smax
= 0.711 A 10 - 3 B mm>mm
=
E
125(109)
Ans.
387
B
2.5 m
C
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 sallow = 130 MPa.
20 kN/m
80 kN
A
B
2m
Section Property:
I = 2B
2m
p d 4
p
d 2
5p 4
a b + d2 a b R =
d
4 2
4
2
32
Allowable Bending Stress: The maximum moment is Mmax = 100 kN # m as
indicated on moment diagram. Applying the flexure formula
smax = sallow =
130 A 106 B =
Mmax c
I
100(103)(d)
5p
32
d4
d = 0.1162 m = 116 mm
Ans.
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 = sallow =
2m
Mmax c
I
100(103)(d)
p
32
B
2m
Allowable Bending Stress: The maximum moment is Mmax = 100 kN # m as
indicated on the moment diagram. Applying the flexure formula
130 A 106 B =
80 kN
d4
d = 0.1986 m = 199 mm
Ans.
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
a
0.5 in.
8 in.
c + ©M = 0;
M - 180(8) = 0
M = 1440 lb # in.
Ix =
1
1
(1)(33) (0.5)(2.53) = 1.59896 in4
12
12
smax =
1440 (1.5)
Mc
=
= 1.35 ksi
I
1.59896
Ans.
s (ksi)
•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
2.30
1.25
4 in.
0.02
I =
1
p
4
0.375 4
4
4
C A 1.25
2 B - A 2 B D = 0.11887 in
Mmax =
P
(4) = 2P
2
Require smax = 1.25 ksi
smax =
Mc
I
1.25 =
2P(1.25>2)
0.11887
P = 0.119 kip = 119 lb
Ans.
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.
16 in.
Absolute Maximum Bending Stress: The maximum moment is Mmax = 216 kip # ft
as indicated on moment diagram. Applying the flexure formula
smax =
216(12)(8)
Mmax c
= 7.59 ksi
= 1
3
I
12 (8)(16 )
8 in.
Ans.
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 + ©MA = 0;
Mmax - 400(6)(3) -
1
(400)(6)(8) = 0
2
Mmax = 16800 lb # ft
The moment of inertia of the about the neutral axis is I =
smax =
1
(6)(63) = 108 in4. Thus,
12
16800(12)(3)
Mc
=
I
108
= 5600 psi = 5.60 ksi
Ans.
390
6 ft
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*6–100. The steel beam has the cross-sectional 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.
12 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 = 27wo.
The moment of inertia of the cross-section about the neutral axis is
I =
1
1
(9)(12.53) (8.75)(123)
12
12
= 204.84375 in4
Here, ¢ = 6.25 in. Thus,
sallow =
Mmax c
;
I
22(103) =
(27wo)(12)(6.25)
204.84375
wo = 2 225.46 lb>ft
= 2.23 kip>ft
Ans.
391
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•6–101.
The steel beam has the cross-sectional 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.
12 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 cross-section about the bending axis is
I =
1
1
(9) A 12.53 B (8.75) A 123 B
12
12
= 204.84375 in4
Here, c = 6.25 in. Thus
smax =
=
Mmax c
I
54 (12)(6.25)
204.84375
= 19.77 ksi = 19.8 ksi
Ans.
392
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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.
+ ©MNA = 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 123 B = 438.1875 in4
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 cross-section is,
I =
1
(0.075) A 0.153 B = 21.09375 A 10 - 6 B m4
12
Here, c = 0.075 w. Thus,
sallow =
5 A 106 B =
Mmax c
;
I
0.125w(0.075)
21.09375 A 10 - 6 B
w = 11250 N>m = 11.25 kN>m
Ans.
394
1m
75 mm
0.5 m
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w
*6–104. If w = 10 kN>m, determine the maximum
bending stress in the beam. Sketch the stress distribution
acting over the cross section.
Support Reactions. The FBD of the beam is shown in Fig. a
0.5 m
75 mm
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 cross-section is
I =
1
(0.075) A 0.153 B = 21.09375 A 10 - 6 B m4
12
Here, c = 0.075 m. Thus
smax =
=
Mmax c
I
1.25 A 103 B (0.075)
21.09375 A 10 - 6 B
= 4.444 A 106 B Pa
= 4.44 MPa
Ans.
The bending stress distribution over the cross section is shown in Fig. d
395
1m
0.5 m
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400 lb/ft
•6–105.
If the allowable bending stress for the wood beam
is sallow = 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.
B
A
3 ft
The FBD of the beam is shown in Fig. a
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 =
2
1
(b)(2b)3 = b4
12
3
Here, c = 2b> 2 = b. Thus,
sallow =
150 =
Mmax c
;
I
3450(12)(b)
> 3 b4
2
b = 7.453 in = 7
1
in.
2
Ans.
396
3 ft
3 ft
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400 lb/ft
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.
B
A
The FBD of the beam is shown in Fig. a.
3 ft
The shear and moment diagrams are shown in Fig. 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
(7.5) A 153 B = 2109.375 in4
12
Here, c =
15
= 7.5 in. Thus
2
smax =
3450(12)(7.5)
Mmax c
=
= 147 psi
I
2109.375
Ans.
397
3 ft
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
s
P
Ec(emax)t (h - c)
c
Ec
Location of neutral axis:
+ ©F = 0;
:
1
1
- (h - c)(smax)c (b) + (c)(smax)t (b) = 0
2
2
(h - c)(smax)c = c(smax)t
(h - c)Ec (emax)t
[1]
(h - c)
= cEt (emax)t ;
c
Ec (h - c)2 = Etc2
Taking positive root:
Ec
c
=
h - c
A Et
Ec
A Et
h 2Ec
c =
=
Ec
2Et + 2Ec
1 +
A Et
h
[2] Ans.
©MNA = 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)(smax)c + c2b(smax)t
3
3
From Eq. [1]. (smax)c =
c
(s )
h - c max t
M =
c
1
1
(h - c)2 (b)a
b (smax)t + c2b(smax)t
3
h - c
3
M =
1
bc(smax)t (h - c + c) ;
3
(smax)t =
3M
bhc
From Eq. [2]
(smax)t =
b
Et
(emax)t (h - c)
(emax)c =
c
(smax)c = Ec(emax)c =
c
3M 2Et + 2Ec
£
≥
b h2
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
s
c
b
Et
P
Ec
See the solution to Prob. 6–107
c =
h 2Ec
Ans.
2Et + 2Ec
Since
(smax)c =
(smax)c =
c
(s ) =
h - c max t
2Ec
2Et
h2Ec
( 2Et + 2Ec) ch - a
h 1Ec
1Et + 1Ec
bd
(smax)t
(smax)t
(smax)c =
2Et + 2Ec
2Ec 3M
¢ 2≤¢
≤
bh
2Et
2Ec
(smax)c =
3M 2Et + 2Ec
¢
≤
bh2
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.
z
My = - 20 sin 45° = - 14.14 kip # ft
45
16 in.
Mz = - 20 cos 45° = - 14.14 kip # ft.
The moments of inertia of the cross-section about the principal centroidal y and z
axes are
Iy =
1
1
(16) A 103 B (14) A 83 B = 736 in4
12
12
Iz =
1
1
(10) A 163 B (8) A 143 B = 1584 in4
12
12
My z
Mz y
Iz
+
smax = sC = -
Iy
- 14.14(12)(8)
- 14.14(12)( - 5)
+
1584
736
= 2.01 ksi
smax = sA = -
Ans.
(T)
- 14.14(12)(- 8)
- 14.14(12)(5)
+
1584
736
Ans.
= - 2.01 ksi = 2.01 ksi (C)
Here, u = 180° + 45° = 225°
tan a =
tan a =
Iz
Iy
D
10 in.
M
By inspection, the bending stress occurs at corners A and C are
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,
14 in.
My = - M sin 45° = - 0.7071 M
z
45
16 in.
Mz = - M cos 45° = - 0.7071 M
The moments of inertia of the cross-section about principal centroidal y and z
axes are
Iy =
1
1
(16) A 103 B (14) A 83 B = 736 in4
12
12
Iz =
1
1
(10) A 163 B (8) A 143 B = 1584 in4
12
12
12 = -
Myzc
Mz yc
Iz
+
D
10 in.
M
By inspection, the maximum bending stress occurs at corners A and C. Here, we
will consider corner C.
sC = sallow = -
A
Iy
- 0.7071 M(12)( -5)
-0.7071 M (12)(8)
+
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 cross-sectional area must be
determined. Also, specify the orientation of the neutral axis.
y
M 520 Nm
12
20 mm
z
–y
5
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:
0.01(0.4)(0.02) + 2[(0.110)(0.18)(0.02)]
©yA
=
©A
0.4(0.02) + 2(0.18)(0.02)
y =
Ans.
= 0.057368 m = 57.4 mm
Iz =
1
(0.4) A 0.023 B + (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 m4
Iy =
1
1
(0.2) A 0.43 B (0.18) A 0.363 B = 0.366827 A 10 - 3 B m4
12
12
Maximum Bending Stress: Applying the flexure formula for biaxial at points A
and B
s = -
Myz
Mzy
sA = -
+
Iz
Iy
200( - 0.2)
- 480(- 0.142632)
+
-6
57.6014(10 )
0.366827(10 - 3)
Ans.
= - 1.298 MPa = 1.30 MPa (C)
200(0.2)
-480(0.057368)
sB = -
+
-6
57.6014(10 )
0.366827(10 - 3)
Ans.
= 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°)
a = - 3.74°
Ans.
402
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 cross-sectional area must be
determined. Also, specify the orientation of the neutral axis.
y
M 520 Nm
12
20 mm
z
–y
5
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:
0.01(0.4)(0.02) + 2[(0.110)(0.18)(0.02)]
©yA
=
©A
0.4(0.02) + 2(0.18)(0.02)
y =
Ans.
= 0.057368 m = 57.4 mm
Iz =
1
(0.4) A 0.023 B + (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 m4
Iy =
1
1
(0.2) A 0.43 B (0.18) A 0.363 B = 0.366827 A 10 - 3 B m4
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 = -
My z
Mz y
sA = -
+
Iz
Iy
200(- 0.2)
- 480(- 0.142632)
+
57.6014(10 - 6)
0.366827(10 - 3)
Ans.
= - 1.298 MPa = 1.30 MPa (C) (Max)
sB = -
200(0.2)
- 480(0.057368)
-6
57.6014(10 )
+
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°)
a = - 3.74°
Ans.
403
200 mm
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6–113. Consider the general case of a prismatic beam
subjected to bending-moment components My and Mz,
as shown, when the x, y, z axes pass through the centroid
of the cross section. If the material is linear-elastic, the
normal stress in the beam is a linear function of position
such that s = a + by + cz. Using the equilibrium conditions 0 = 1A s dA, My = 1A zs 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 = [-1MzIy + MyIyz2y + 1MyIz + MzIyz2z]>1IyIz - Iyz22,
where the moments and products of inertia are defined in
Appendix A.
y
z
My
dA
sC
y
Mz
z
Equilibrium Condition: sx = a + by + cz
0 =
LA
sx dA
0 =
LA
(a + by + cz) dA
0 = a
LA
dA + b
LA
y dA + c
My =
LA
z sx dA
=
LA
z(a + by + cz) dA
= a
Mz =
=
LA
= -a
LA
LA
z dA + b
LA
LA
z dA
yz dA + c
LA
[1]
z2 dA
[2]
- y sx dA
- y(a + by + cz) dA
LA
ydA - b
y2 dA - c
LA
LA
yz dA
[3]
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 + cIy
From Eq. [3]
Mz = - bIz - cIyz
Solving for a, b, c:
a = 0 (Since A Z 0)
b = -¢
Thus,
MzIy + My Iyz
sx = - ¢
Iy Iz -
I2yz
≤
Mz Iy + My Iyz
Iy Iz -
I2yz
c =
≤y + ¢
My Iz + Mz Iyz
Iy Iz - I2yz
My Iy + MzIyz
Iy Iz - I2yz
≤z
(Q.E.D.)
404
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6–114. The cantilevered beam is made from the Z-section
having the cross-section 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)
Iy =
= 400 lb # ft =
3
4.80(10 )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 in4
12
12
A
B
2.25 in.
1
1
Iz =
(0.25)(3.25)3 + 2 c (2)(0.25)3 + (0.25)(2)(1.5)2 d = 2.970378 in4
12
12
0.25 in.
3 in.
0.25 in.
Iyz = 2[1.5(1.125)(2)(0.25)] = 1.6875 in4
Using the equation developed in Prob. 6-113.
s = -a
sA =
Mz Iy + My Iyz
Iy Iz -
I2yz
by + a
My Iz + Mz Iyz
Iy 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]
= 8.95 ksi
Ans.
6–115. The cantilevered beam is made from the Z-section
having the cross-section 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.
50 lb
50 lb
3 ft
3
(My)max = 50(3) + 50(5) = 400 lb # ft = 4.80(10 )lb # in.
Iy =
1
1
(3.25)(0.25)3 + 2c (0.25)(2)3 + (0.25)(2)(1.125)2 d = 1.60319 in4
12
12
1
1
Iz =
(0.25)(3.25)3 + 2 c (2)(0.25)3 + (0.25)(2)(1.5)2 d = 2.970378 in4
12
12
2.25 in.
sB =
Iy Iz -
I2yz
by + a
My Iz + Mz Iyz
Iy Iz - I2yz
0.25 in.
3 in.
0.25 in.
Using the equation developed in Prob. 6-113.
Mz Iy + My Iyz
A
B
Iyz = 2[1.5(1.125)(2)(0.25)] = 1.6875 in4
s = -a
2 ft
0.25 in.
2 in.
bz
- [0 + (4.80)(103)(1.6875)](- 1.625) + [(4.80)(103)(2.976378) + 0](0.125)
[(1.60319)(2.970378) - (1.6875)2]
= 7.81 ksi
Ans.
405
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*6–116. The cantilevered wide-flange 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 sallow = 180 MPa.
200 mm
10 mm
150 mm
10 mm
Internal Moment Components: Using method of section
10 mm
A
y
©Mz = 0;
Mz + P cos 30°(2) = 0
Mz = - 1.732P
©My = 0;
My + P sin 30°(2) = 0
My = - 1.00P
z
Section Properties:
x
2m
30
1
1
Iz =
(0.2) A 0.173 B (0.19) A 0.153 B = 28.44583(10 - 6) m4
12
12
Iy = 2 c
P
1
1
(0.01) A 0.23 B d +
(0.15) A 0.013 B = 13.34583(10 - 6) 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.
sA = sallow = 180 A 106 B = -
Myz
Mzy
Iz
+
Iy
-1.00P( - 0.1)
( - 1.732P)(0.085)
28.44583(10 - 6)
+
13.34583(10 - 6)
P = 14208 N = 14.2 kN
Ans.
•6–117.
The cantilevered wide-flange 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
A
y
©Mz = 0;
Mz + 600 cos 30°(2) = 0
Mz = - 1039.23 N # m
©My = 0;
My + 600 sin 30°(2) = 0;
My = - 600.0 N # m
z
Section Properties:
x
1
1
Iz =
(0.2) A 0.173 B (0.19) A 0.153 B = 28.44583(10 - 6) m4
12
12
Iy = 2 c
Maximum Bending Stress: By inspection, maximum bending stress occurs at A and
B. Applying the flexure formula for biaxial bending at point A
s = -
Myz
Mzy
sA = -
Iz
+
Iy
- 600.0( - 0.1)
- 1039.32(0.085)
-6
28.44583(10 )
= 7.60 MPa (T)
+
13.34583(10 - 6)
(Max)
Ans.
406
2m
30
P
1
1
(0.01) A 0.23 B d +
(0.15) A 0.013 B = 13.34583(10 - 6) m4
12
12
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
30
My = 1200 sin 30° = 600 kN # m
150 mm
Mz = - 1200 cos 30° = - 1039.23 kN # m
z
x
150 mm
Section Properties: The location of the centroid of the cross-section 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 =
1
1
(0.6) A 0.33 B (0.15) A 0.153 B = 1.3078 A 10 - 3 B m4
12
12
Iz =
1
(0.3) A 0.63 B + 0.3(0.6)(0.3 - 0.2893)2
12
- c
1
(0.15) A 0.153 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 occurs at either corner
A or B.
s = -
Myz
Mzy
sA = -
+
Iz
Iy
c - 1039.23 A 103 B d (0.2893)
5.2132 A 10 - 3 B
+
600 A 103 B (0.15)
1.3078 A 10 - 3 B
= 126 MPa (T)
sB = -
c - 1039.23 A 103 B d ( -0.3107)
5.2132 A 10 - 3 B
+
600 A 103 B ( -0.15)
1.3078 A 10 - 3 B
Ans.
= - 131 MPa = 131 MPa (C)(Max.)
Orientation of Neutral Axis: Here, u = - 30°.
tan a =
tan a =
Iz
Iy
tan u
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.
407
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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 (sallow)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,
30
150 mm
z
My = M sin 30° = 0.5M
x
150 mm
Mz = - M cos 30° = - 0.8660M
Section Properties: The location of the centroid of the cross section is
y =
150 mm
0.3(0.6)(0.3) - 0.375(0.15)(0.15)
©yA
=
= 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 =
1
1
(0.6) A 0.33 B (0.15) A 0.153 B = 1.3078 A 10 - 3 B m4
12
12
Iz =
1
(0.3) A 0.63 B + 0.3(0.6)(0.3 - 0.2893)2
12
- c
1
(0.15) A 0.153 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 = (sallow)t = 125 A 106 B = -
My zA
Mz yA
Iz
+
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 = (sallow)c = - 150 A 106 B = -
My zB
Mz yB
Iz
+
Iy
(- 0.8660M)( -0.3107)
5.2132 A 10 - 3 B
0.5M( -0.15)
+
1.3078 A 10 - 3 B
M = 1376 597.12 N # m = 1377 kN # m
408
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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
y
0.5 m
0.5 m
C
0.5 m
200 N
The FBD of the shaft is shown in Fig. a.
A
200 N 300 N
The shaft is subjected to two bending moment components Mz and My, Figs. b and c,
respectively.
Since all the axes through the centroid of the circular cross-section of the shaft are
principal axes, then the resultant moment M = 2My 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,
sallow =
Mmax C
;
I
150(106) =
230.49(d>2)
p
4
(d>2)4
d = 0.02501 m = 25 mm
Ans.
409
300 N
0.5 m
D
B
E
x
150 N 150 N
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•6–121.
The 30-mm-diameter 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.
1m
1m
1m
1m
A
D
150 N
150 N
Support Reactions: As shown on FBD.
Internal Moment Components: The shaft is subjected to two bending moment
components My and Mz. The moment diagram for each component is drawn.
Maximum Bending Stress: Since all the axes through the circle’s center for circular
shaft are principal axis, then the resultant moment M = 2My 2 + Mz 2 can be used
to determine the maximum bending stress. The maximum resultant moment occurs
at E Mmax = 24002 + 1502 = 427.2 N # m.
Applying the flexure formula
Mmax c
I
427.2(0.015)
=
p
4
A 0.0154 B
= 161 MPa
Ans.
410
E
C
B
400 N
100 mm
400 N
60 mm
x
smax =
y
z
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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 = 0.060110-32 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
z
My = 250 cos 32.9° = 209.9 N # m
z¿
300 mm
Mz = 250 sin 32.9° = 135.8 N # m
200 mm
50 mm
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 = -
Myz
Mzy
+
Iz
Iy
209.9(- 0.06546)
-135.8(0.2210)
=
0.471(10 - 3)
+
60.0(10 - 6)
= - 293 kPa = 293 kPa (C)
Ans.
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 0.053 B + 2c (0.05) A 0.153 B + 0.05(0.15) A 0.12 B d
12
12
= 0.18125 A 10
Iz =
-3
z
z¿
300 mm
Bm
4
1
1
(0.05) A 0.33 B + 2c (0.15) A 0.053 B + 0.15(0.05) A 0.1252 B d
12
12
= 0.350(10 - 3) m4
Iyz = 0.15(0.05)(0.125)(- 0.1) + 0.15(0.05)( - 0.125)(0.1)
= - 0.1875 A 10 - 3 B m4
Bending Stress: Using formula developed in Prob. 6-113
s =
sA =
- (Mz Iy + My Iyz)y + (My Iz + MzIyz)z
IyIz - I2yz
-[0 + 250( - 0.1875)(10 - 3)](0.15) + [250(0.350)(10 - 3) + 0]( -0.175)
0.18125(10 - 3)(0.350)(10 - 3) - [0.1875(10 - 3)]2
= - 293 kPa = 293 kPa (C)
Ans.
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 Iy = 0.060110-32 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 B. Solve the
problem using Eq. 6–17.
50 mm
y
A
200 mm
32.9
y¿
250 Nm
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 =
My¿z¿
Mz¿y¿
Iz¿
+
Iy¿
209.9(- 0.06546)
135.8(0.2210)
0.471(10 - 3)
-
0.060(10 - 3)
= 293 kPa = 293 kPa (T)
Ans.
412
200 mm
50 mm
B
50 mm
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z
•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 in4 and I¿y = 2.295 in4,
where z¿ and y¿ are the principal axes. Solve the problem
using Eq. 6–17.
1.183 in.
0.5 in.
z¿
A
4 in.
45
C
y
1.183 in.
0.5 in.
M 3 kip ft
y′
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
= 2.817 cos 45° - 1.183 sin 45° = 1.155 in.
zA
œ
yA
= - (2.817 sin 45° + 1.183 cos 45°) = - 2.828 in.
Bending Stress:
sA = -
= -
œ
My¿zA
œ
Mz¿yA
Iz¿
+
Iy¿
(- 2.121)(12)(1.155)
( - 2.121)(12)( -2.828)
+
8.828
2.295
= - 20.97 ksi = 21.0 ksi (C)
Ans.
413
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z
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
are Iz = Iy = 5.561 in4 and the product of inertia of the
cross sectional area with respect to the z and y axes is
Iyz 3.267 in4. (See Appendix A)
1.183 in.
0.5 in.
z¿
A
4 in.
45
C
y
1.183 in.
0.5 in.
M 3 kip ft
y′
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 =
=
- A MzIy + MyIyz B yA + A MyIz + MzIyz B zA
IyIz - Iyz 2
- C 0(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
= - 20.97 ksi = 21.0 ksi
Ans.
414
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6–127. The composite beam is made of 6061-T6 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 1sallow2al = 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)
7.7851(10 - 6)
M = 6598 N # m = 6.60 kN # m (controls!)
Ans.
Assume failure of aluminium
(sallow)al = n
Mc
INA
128 A 106 B = 0.68218 c
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 6061-T6 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 1sallow2br = 35 MPa.
h
B
A
Section Properties: For transformed section.
150 mm
68.9(109)
Eal
= 0.68218
=
n =
Ebr
101.0(109)
bbr = nbal = 0.68218(0.15) = 0.10233 m
y =
=
©yA
©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)
7.45799(10 - 6)
M = 6412 N # m = 6.41 kN # m (controls!)
Ans.
Assume failure of aluminium
(sallow)al = n
Mc
INA
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
2014-T6 aluminum alloy and segment B is A-36 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 simply-supported beam subjected to the uniform
0.9 A 152 B
wL2
=
distributed load, the maximum moment in the beam is Mmax =
8
8
= 25.3125 kip # ft.
Section Properties: The cross section will be transformed into that of steel as
Eal
10.6
=
= 0.3655.
shown in Fig. a. Here, n =
Est
29
Then bst = 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
(1.0965) A 33 B + 1.0965(3)(4.5 - 2.3030)2
12
= 30.8991 in4
Maximum Bending Stress: For the steel,
(smax)st =
25.3125(12)(2.3030)
Mmaxcst
=
= 22.6 ksi
I
30.8991
Ans.
At the seam,
ssty = 0.6970 in. =
Mmaxy
25.3125(12)(0.6970)
=
= 6.85 ksi
I
30.8991
For the aluminium,
(smax)al = n
25.3125(12)(6 - 2.3030)
Mmaxcal
= 0.3655 c
d = 13.3 ksi
I
30.8991
Ans.
At the seam,
saly = 0.6970 in. = n
Mmaxy
25.3125(12)(0.6970)
= 0.3655 c
d = 2.50 ksi
I
30.8991
The bending stress across the cross section of the composite beam is shown in Fig. b.
417
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6–130. Segment A of the composite beam is made from
2014-T6 aluminum alloy and segment B is A-36 steel. If
the allowable bending stress for the aluminum and steel
are (sallow)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 simply-supported beam subjected to the uniform
distributed load, the maximum moment in the beam is
w A 152 B
wL2
=
= 28.125w.
Mmax =
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 =
Est
29
Then bst = 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
1
(3) A 33 B + 3(3)(2.3030 - 1.5)2 +
(1.0965) A 33 B
12
12
+ 1.0965 A 33 B + 1.0965(3)(4.5 - 2.3030)2
= 30.8991 in4
Bending Stress: Assuming failure of steel,
(sallow)st =
Mmax cst
;
I
22 =
(28.125w)(12)(2.3030)
30.8991
w = 0.875 kip>ft (controls)
Ans.
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
418
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6–131. The Douglas fir beam is reinforced with A-36
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
=
Est
29.0(103)
bst = nbw = 0.065517(4) = 0.26207 in.
INA =
1
(1.5 + 0.26207) A 63 B = 31.7172 in4
12
Maximum Bending Stress: Applying the flexure formula
(smax)st =
7.5(12)(3)
Mc
=
= 8.51 ksi
I
31.7172
(smax)w = n
Ans.
7.5(12)(3)
Mc
= 0.065517c
d = 0.558 ksi
I
31.7172
Ans.
419
2 in.
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*6–132. The top plate is made of 2014-T6 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
= 0.55789
=
Ek
19.0(103)
bk = n bal = 0.55789(12) = 6.6947 in.
y =
0.25(13)(0.5) + 2[(3.25)(5.5)(0.5)] + 5.75(6.6947)(0.5)
©yA
=
©A
13(0.5) + 2(5.5)(0.5) + 6.6947(0.5)
= 2.5247 in.
INA =
1
(13) A 0.53 B + 13(0.5)(2.5247 - 0.25)2
12
+
1
(1) A 5.53 B + 1(5.5)(3.25 - 2.5247)2
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
900(12)(6 - 2.5247)
Mc
=
= 439 psi
I
85.4168
Ans.
Ans.
420
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•6–133.
The top plate made of 2014-T6 aluminum is used
to reinforce a Kevlar 49 plastic beam. If the allowable
bending stress for the aluminum is (sallow)al = 40 ksi and
for the Kevlar (sallow)k = 8 ksi, determine the maximum
moment M that can be applied to the beam.
6 in.
0.5 in.
0.5 in.
Section Properties:
n =
10.6(103)
Eal
= 0.55789
=
Ek
19.0(103)
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
(13) A 0.53 B + 13(0.5)(2.5247 - 0.25)2
12
+
1
(1) A 5.53 B + 1(5.5)(3.25 - 2.5247)2
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
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!)
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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
Est
200
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(103)(0.07)
Mc1
= 20.1 MPa
=
I
27.84667(10 - 6)
Ans.
(max)
Maximum stress in brass:
(sbr)max =
0.5(8)(103)(0.05)
nMc2
= 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.
y =
4 in.
0.5 in.
(0.5)(16)(0.25) + 2(3.5)(0.5)(2.25) + (0.8276)(3.5)(2.25)
= 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
I =
(16)(0.53) + (16)(0.5)(0.88862) + 2 a b(0.5)(3.53) + 2(0.5)(3.5)(1.11142)
12
12
+
1
(0.8276)(3.53) + (0.8276)(3.5)(1.11142) = 20.914 in4
12
Maximum stress in steel:
(sst) =
850(12)(4 - 1.1386)
Mc
=
= 1395 psi = 1.40 ksi
I
20.914
Ans.
Maximum stress in wood:
(sw) = n(sst)max
= 0.05517(1395) = 77.0 psi
Ans.
422
0.5 in.
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*6–136. A white spruce beam is reinforced with A-36 steel
straps at its top and bottom as shown. Determine the
bending moment M it can support if (sallow)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(103)
Ew
= 0.048276
=
Est
29.0(103)
bst = nbw = 0.048276(3) = 0.14483 in.
INA =
1
1
(3) A 53 B (3 - 0.14483) A 43 B = 16.0224 in4
12
12
Allowable Bending Stress: Applying the flexure formula
Assume failure of steel
(sallow)st =
22 =
Mc
I
M(2.5)
16.0224
M = 141.0 kip # in
= 11.7 kip # ft (Controls !)
Ans.
Assume failure of wood
(sallow)w = n
My
I
2.0 = 0.048276 c
M(2)
d
16.0224
M = 331.9 kip # in = 27.7 kip # ft
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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 A-36 steel section A and the 2014-T6
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.
73.1 A 109 B
Eal
=
= 0.3655. Thus, bst = nbal = 0.3655(0.015) = 0.0054825 m. The
Here, n =
Est
200 A 109 B
location of the transformed section is
©yA
y =
=
©A
0.075(0.15)(0.0054825) + 0.2 cp A 0.052 B d
0.15(0.0054825) + p A 0.052 B
= 0.1882 m
The moment of inertia of the transformed section about the neutral axis is
I = ©I + Ad2 =
1
(0.0054825) A 0.153 B + 0.0054825(0.15)(0.1882 - 0.075)2
12
+
1
p A 0.054 B + p A 0.052 B (0.2 - 0.1882)2
4
= 18.08 A 10 - 6 B m4
Maximum Bending Stress: For the steel,
(smax)st =
45 A 103 B (0.06185)
Mcst
=
= 154 MPa
I
18.08 A 10 - 6 B
Ans.
For the aluminum alloy,
(smax)al = n
45 A 103 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 20-mm
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 (sallow)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 Est = 200 GPa, respectively.
200 mm
M
Bending Stress: The cross section will be transformed into that of concrete as shown
Est
200
in Fig. a. Here, n =
=
= 8. It is required that both concrete and steel
Econ
25
achieve their allowable stress simultaneously. Thus,
(sallow)con =
12.5 A 106 B =
Mccon
;
I
Mccon
I
M = 12.5 A 106 B ¢
(sallow)st = n
I
≤
ccon
220 A 106 B = 8 B
Mcst
;
I
(1)
M(d - ccon)
R
I
M = 27.5 A 106 B ¢
I
≤
d - ccon
(2)
Equating Eqs. (1) and (2),
12.5 A 106 B ¢
I
I
≤ = 27.5 A 106 B ¢
≤
ccon
d - ccon
ccon = 0.3125d (3)
Section Properties: The area of the steel bars is Ast = 3c
(3)
p
A 0.022 B d = 0.3 A 10 - 3 B p m2.
4
Thus, the transformed area of concrete from steel is (Acon)t = nAs = 8 C 0.3 A 10 - 3 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,
0.2(ccon)(ccon>2) = 2.4 A 10 - 3 B p (d - ccon)
0.1ccon 2 = 2.4 A 10 - 3 B pd - 2.4 A 10 - 3 B pccon
ccon 2 = 0.024pd - 0.024pccon
(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
(0.2) A 0.16593 B + 2.4 A 10 - 3 B p(0.5308 - 0.1659)2
3
425
d
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6–138. Continued
= 1.3084 A 10 - 3 B m4
Substituting this result into Eq. (1),
M = 12.5 A 106 B C
1.3084 A 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 = n1 bEs =
160
(3) = 0.6 in.
800
(bbk)2 = n2 bpvc =
450
(3) = 1.6875 in.
800
500 lb
PVC EPVC 450 ksi
Escon EE 160 ksi
Bakelite EB 800 ksi
3 ft
y =
(1)(3)(2) + 3(0.6)(2) + 4.5(1.6875)(1)
©yA
=
= 1.9346 in.
©A
3(2) + 0.6(2) + 1.6875(1)
I =
1
1
(3)(23) + 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)(2.56542) = 20.2495 in4
12
(smax)pvc = n2
500 lb
450 1500(12)(3.0654)
Mc
= a
b
I
800
20.2495
= 1.53 ksi
Ans.
426
3 ft
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*6–140. The low strength concrete floor slab is integrated
with a wide-flange A-36 steel beam using shear studs (not
shown) to form the composite beam. If the allowable
bending stress for the concrete is (sallow)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
Econ
22.1
that
of
steel.
Here,
Thus,
=
= 0.1105.
n =
Est
200
bst = 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 0.0153 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
= 647.93 A 10 - 6 B m4
Bending Stress: Assuming failure of steel,
(sallow)st =
M(0.3222)
Mcst
; 165 A 106 B =
I
647.93 A 10 - 6 B
M = 331 770.52 N # m = 332 kN # m
Assuming failure of concrete,
(sallow)con = n
Mccon
;
I
10 A 106 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 A-36 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
Ast = 3(p)(0.5)2 = 2.3562 in2
Est = 29.0(103) ksi
Econ = 4.20(103) ksi
A¿ = nAst =
29.0(103)
4.20(103)
8(h¿) a
©yA = 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
(8)(5.517)3 + 8(5.517)(5.517>2)2 d + 16.2690(13 - 5.517)2
12
= 1358.781 in4
(scon)max =
My
40(12)(5.517)
=
= 1.95 ksi
I
1358.781
(sst)max = n a
10 kip
Ans.
My
29.0(103) 40(12)(13 - 5.517)
ba
b = a
b = 18.3 ksi
I
1358.781
4.20(103)
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 (sst)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.
1-in. diameter rods
Ast = 2(p)(0.5)2 = 1.5708 in2
A¿ = nAst =
©yA = 0;
29(103)
3.8(103)
(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:
(scon)allow =
My
;
I
3 =
M(4.15731)
3535.69
M = 2551 kip # in.
Assume steel fails:
(sst)allow = na
My
b;
I
40 = ¢
29(103)
3
3.8(10 )
≤¢
M(16 - 0.15731)
≤
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 curved-beam
formula, Eq. 6–24, reduces to the flexure formula, Eq. 6–13.
Normal Stress: Curved-beam formula
M(R - r)
s =
where A¿ =
Ar(r - R)
dA
LA r
and R =
A
1A
dA
r
=
A
A¿
M(A - rA¿)
s =
[1]
Ar(rA¿ - A)
r = r + y
rA¿ = r
[2]
dA
r
=
a
- 1 + 1 b dA
LA r + y
LA r
=
LA
a
= A -
r - r - y
r + y
y
+ 1 b dA
dA
LA r + y
[3]
Denominator of Eq. [1] becomes,
y
Ar(rA¿ - A) = Ar ¢ A -
LA r + y
dA - A ≤ = - Ar
y
LA r + y
dA
Using Eq. [2],
Ar(rA¿ - A) = - A
= A
=
¢
ry
LA r + y
y2
LA r + y
+ y - y ≤ dA - Ay
LA r + y
dA - A 1A y dA - Ay
y
LA r + y
as
y
r
: 0
A
I
r
Then,
Ar(rA¿ - A) :
Eq. [1] becomes
s =
Mr
(A - rA¿)
AI
Using Eq. [2],
s =
Mr
(A - rA¿ - yA¿)
AI
Using Eq. [3],
s =
=
dA
dA
y2
y
Ay
A
¢
¢
y ≤ dA - A 1A y dA r LA 1 + r
r LA 1 +
1A y dA = 0,
But,
y
y
Mr
dA
C A - ¢A dA ≤ - y
S
AI
r
+
y
r
LA
LA + y
y
dA
Mr
C
dA - y
S
AI LA r + y
r
LA + y
430
y≤
r
dA
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6–143. Continued
y
=
y
As
r
Mr
r
C
¢
AI LA 1 +
y ≤ dA
r
y
-
r LA
¢
dA
≤S
1 + yr
=
: 0
¢
y
r
LA 1 +
y≤
r
dA = 0
s =
Therefore,
and
y
r LA
¢
y
yA
dA
A dA =
y≤ = 1
1 + r
r
r
yA
My
Mr
b = aAI
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
LA r
100 mm
2p(0.0375)
=
(0.175 - 20.1752 - 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
=
B
2.8125(10 - 3)p
= 0.166556941
0.053049301
r - R = 0.175 - 0.166556941 = 0.0084430586
sA =
sB =
M(R - rA)
50(0.166556941 - 0.1)
=
2.8125(10 - 3)p (0.1)(0.0084430586)
=
2.8125(10 - 3)p (0.25)(0.0084430586)
ArA (r - R)
M(R - rB)
ArB (r - R)
50(0.166556941 - 0.25)
= 446k Pa (T)
= 224 kPa (C)
No, because of localized stress concentration at the wall.
Ans.
Ans.
Ans.
431
M
Mr
AI
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•6–145.
The member has an elliptical cross section. If the
allowable bending stress is sallow = 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
b = 0.0375 m
a = 0.075 m;
A = p(0.075)(0.0375) = 0.0028125 p
2p(0.0375)
dA
2pb
(0.175 - 20.1752 - 0.0752)
=
(r - 2r2 - a2) =
r
a
0.075
LA
= 0.053049301 m
R =
A
dA
1A r
=
0.0028125p
= 0.166556941 m
0.053049301
r - R = 0.175 - 0.166556941 = 8.4430586(10 - 3) 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 (controls)
Ans.
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
©
dA
0.26
0.41
0.42
= 0.15 ln
+ 0.01 ln
+ 0.075 ln
0.25
0.26
0.41
LA 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 curved-beam formula
Assume tension failure
(sallow)t =
120 A 106 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 106 B =
M(R - r)
Ar(r - R)
0.16P(0.306243 - 0.42)
0.00375(0.42)(0.012757)
P = 55195 N = 55.2 kN (Controls !)
Ans.
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
10 mm
10 mm
160 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
©
dA
0.41
0.42
0.26
= 0.15 ln
+ 0.01 ln
+ 0.075 ln
0.25
0.26
0.41
LA 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 curved-beam 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)
= - 5.44 MPa
Ans.
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
30
20 mm
15 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
r =
©
2.18875 (10 - 3)
©rA
=
= 0.5150 m
©A
0.00425
dA
0.57
0.55
= 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 curved-beam formula
sA =
M(R - rA)
-900(0.509067 - 0.57)
=
ArA (r - R)
0.00425(0.57)(5.933479)(10 - 3)
Ans.
= 3.82 MPa (T)
sB =
M(R - rB)
- 900(0.509067 - 0.4)
=
ArB (r - R)
0.00425(0.4)(5.933479)(10 - 3)
= - 9.73 MPa = 9.73 MPa (C)
Ans.
435
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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
30
20 mm
15 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 m2
©rA = 0.475(0.15)(0.015) + 0.56(0.1)(0.02) = 2.18875(10 - 3) m
r =
©
2.18875 (10 - 3)
©rA
=
= 0.5150 m
©A
0.00425
dA
0.57
0.55
= 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 curved-beam formula
sC =
M(R - rC)
-900(0.509067 - 0.55)
=
ArC(r - R)
0.00425(0.55)(5.933479)(10 - 3)
= 2.66 MPa (T)
Ans.
436
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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
M 25 lbin.
1 in.
a
dA
= ©2p (r - 2r2 - c2)
LA r
= 2p(1.75 - 21.752 - 0.752) - 2p (1.75 - 21.752 - 0.632)
0.63 in.
0.75 in.
= 0.32375809 in.
A = p(0.752) - p(0.632) = 0.1656 p
R =
A
dA
1A r
=
M = 25 lbin.
0.1656 p
= 1.606902679 in.
0.32375809
r - R = 1.75 - 1.606902679 = 0.14309732 in.
(smax)t =
M(R - rA)
=
ArA(r - R)
(smax)c = =
25(1.606902679 - 1)
= 204 psi (T)
0.1656 p(1)(0.14309732)
M(R - rB)
=
ArB(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.
0.5 in.
B
2 in.
A
1
A = 0.5(2) + (1)(2) = 2 in2
2
r =
1.5 in.
8 in.
9(0.5)(2) + 8.6667 A 12 B (1)(2)
©rA
=
= 8.83333 in.
©A
2
M
M
1(10)
dA
10
10
= 0.5 ln
+ c
c ln
d - 1 d = 0.22729 in.
r
8
(10 - 8)
8
LA
R =
A
dA
1A r
=
2
= 8.7993 in.
0.22729
r - R = 8.83333 - 8.7993 = 0.03398 in.
s =
M(R - r)
Ar(r - R)
sA =
600(12)(8.7993 - 8)
= 10.6 ksi (T)
2(8)(0.03398)
Ans.
sB =
600(12)(8.7993 - 10)
= - 12.7 ksi = 12.7 ksi (C)
2(10)(0.03398)
Ans.
437
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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 + ©MO = 0;
M - 250 cos 60° (0.075) - 250 sin 60° (0.15) = 0
M = 41.851 N # m
r2
dA
0.2375
= b ln
= 0.05 ln
= 0.018974481 m
r
r
0.1625
1
LA
A = (0.075)(0.05) = 3.75(10 - 3) m2
R =
A
1A
dA
r
=
3.75(10 - 3)
= 0.197633863 m
0.018974481
r - R = 0.2 - 0.197633863 = 0.002366137
sA =
M(R - rA)
41.851(0.197633863 - 0.2375)
=
ArA(r - R)
3.75(10 - 3)(0.2375)(0.002366137)
= - 791.72 kPa
Ans.
= 792 kPa (C)
sB =
M(R - rB)
41.851 (0.197633863 - 0.1625)
=
ArB(r - R)
3.75(10 - 3)(0.1625)(0.002366137)
= 1.02 MPa (T)
438
Ans.
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•6–153.
The ceiling-suspended C-arm is used to support
the X-ray 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
100 mm
20 mm
40 mm
Section Properties:
r =
©
1.22(0.1)(0.04) + 1.25(0.2)(0.02)
©rA
=
= 1.235 m
©A
0.1(0.04) + 0.2(0.02)
dA
1.26
1.24
= 0.1 ln
+ 0.2 ln
= 6.479051 A 10 - 3 B m
r
1.20
1.24
LA
A = 0.1(0.04) + 0.2(0.02) = 0.008 m2
R =
A
dA
1A r
=
0.008
= 1.234749 m
6.479051 (10 - 3)
r - R = 1.235 - 1.234749 = 0.251183 A 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 curved-beam 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)
ArB (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
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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
0.200 + 0.210
r =
= 0.205 m
2
r2
dA
0.21
= 0.02 ln
= b ln
= 0.97580328 A 10 - 3 B m
r
r
0.20
1
LA
A = (0.01)(0.02) = 0.200 A 10 - 3 B m2
R =
0.200(10 - 3)
A
1A
dA
r
=
0.97580328(10 - 3)
= 0.204959343 m
r - R = 0.205 - 0.204959343 = 0.040657 A 10 - 3 B m
Maximum Normal Stress: Applying the curved-beam formula
sC =
M(R - r2)
Ar2(r - R)
0.660(0.204959343 - 0.21)
=
0.200(10 - 3)(0.21)(0.040657)(10 - 3)
= - 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)
440
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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
r2
dA
0.21
= b ln
= 0.02 ln
= 0.97580328 A 10 - 3 B m
r1
0.20
LA r
A = (0.01)(0.02) = 0.200 A 10 - 3 B m2
R =
0.200(10 - 3)
A
1A
dA
r
=
0.97580328(10 - 3)
= 0.204959 m
r - R = 0.205 - 0.204959343 = 0.040657 A 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 curved-beam formula
Assume compression failure
sc = sallow =
- 4 A 106 B =
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 = sallow =
4 A 106 B =
M(R - r1)
Ar1 (r - R)
0.424959P(0.204959 - 0.2)
0.200(10 - 3)(0.2)(0.040657)(10 - 3)
P = 3.09 N (Controls !)
Ans.
441
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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 two-dimensional view of the
stress distribution.
16 Nm
5 mm
20 mm
5 mm
0.6 m
5 mm
30 mm
LA
0.625
0.630
0.605
+ (0.005)ln
+ (0.03)ln
= 0.650625(10 - 3) in.
0.6
0.605
0.625
dA>r = (0.03)ln
A = 2(0.005)(0.03) + (0.02)(0.005) = 0.4(10 - 3) in2
R =
0.4(10 - 3)
A
1A dA>r
=
0.650625(10 - 3)
= 0.6147933
(sc)max =
M(R - rc)
16(0.6147933 - 0.630)
= - 4.67 MPa
=
ArA(r - R)
0.4(10 3)(0.630)(0.615 - 0.6147933)
(ss)max =
M(R - rs)
16(0.6147933 - 0.6)
= 4.77 MPa
=
ArA(r - R)
0.4(10 - 3)(0.6)(0.615 - 0.6147933)
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 sallow = 18 ksi.
•6–157.
14.5 in.
M
b =
14.5 - 12.5
= 1.0 in.
2
r
0.5
=
= 0.04
h
12.5
1
b
=
= 2.0
r
0.5
From Fig. 6-44:
K = 2.60
smax = K
Mc
I
18(103) = 2.60 c
(M)(6.25)
1
3
12 (1)(12.5)
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
M
12.5 in.
r
0.5
=
= 0.04
h
12.5
1
b
= 2.0
=
r
0.5
1 in.
From Fig. 6-44:
K = 2.60
smax = K
(10)(12)(6.25)
Mc
= 2.60 c 1
d = 12.0 ksi
3
I
12 (1)(12.5)
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 sallow = 124 MPa is not
exceeded.
80 mm
7 mm
20 mm
r
M
M
r
Allowable Bending Stress:
sallow = K
Mc
I
124 A 106 B = K B
40(0.01)
R
1
3
12 (0.007)(0.02 )
K = 1.45
Stress Concentration Factor: From the graph in the text
w
80
r
=
= 4 and K = 1.45, then = 0.25.
with
h
20
h
r
= 0.25
20
r = 5.00 mm
Ans.
*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
7 mm
20 mm
r
M
M
Stress Concentration Factor: From the graph in the text with
r
w
80
5
=
= 4 and =
= 0.25, then K = 1.45.
h
20
h
20
r
Maximum Bending Stress:
smax = K
Mc
I
= 1.45 B
17.5(0.01)
R
1
3
12 (0.007)(0.02 )
= 54.4 MPa
Ans.
443
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•6–161. The simply supported notched bar is subjected to
two forces P. Determine the largest magnitude of P that can
be applied without causing the material to yield.The material
is A-36 steel. Each notch has a radius of r = 0.125 in.
P
P
0.5 in.
1.75 in.
1.25 in.
20 in.
20 in.
b =
20 in.
20 in.
1.75 - 1.25
= 0.25
2
0.25
b
=
= 2;
r
0.125
r
0.125
=
= 0.1
h
1.25
From Fig. 6-44. K = 1.92
sY = K
Mc
;
I
36 = 1.92 c
20P(0.625)
1
3
12 (0.5)(1.25)
d
P = 122 lb
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 bending-stress 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
b
0.25
=
= 2;
r
0.125
r
0.125
=
= 0.1
h
1.25
From Fig. 6-44, K = 1.92
smax = K
1.75 in.
1.25 in.
20 in.
b =
P
2000(0.625)
Mc
= 1.92 c 1
d = 29.5 ksi
3
I
12 (0.5)(1.25)
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.
7 mm
350 N
60 mm
A
r
7
=
= 0.175
h
40
60
w
=
= 1.5
h
40
200 mm
40 mm
7 mm
C
L
2
B
L
2
200 mm
From Fig. 6-43, K = 1.5
(sA)max = K
(35)(0.02)
MAc
d = 19.6875 MPa
= 1.5c 1
3
I
12 (0.01)(0.04 )
(sB)max = (sA)max =
19.6875(106) =
MB c
I
175(0.2 + L2 )(0.03)
1
3
12 (0.01)(0.06 )
L = 0.95 m = 950 mm
Ans.
*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
h
10
h
10
obtained from the graph in the text.
For the smaller section with
w
45
3
r
=
= 1.5 and =
= 0.1, we have K = 1.75
h
30
h
30
obtained from the graph in the text.
For the larger section with
Allowable Bending Stress:
For the smaller section
smax = sallow = K
Mc
;
I
200 A 106 B = 1.2 B
M(0.005)
R
1
3
12 (0.015)(0.01 )
M = 41.7 N # m (Controls !)
Ans.
For the larger section
smax = sallow = K
Mc
;
I
200 A 106 B = 1.75 B
M(0.015)
R
1
3
12 (0.015)(0.03 )
M = 257 N # m
445
10 mm
6 mm
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•6–165.
The beam is made of an elastic plastic material for
which sY = 250 MPa. Determine the residual stress in the
beam at its top and bottom after the plastic moment Mp is
applied and then released.
15 mm
1
1
(0.2)(0.23)3 (0.18)(0.2)3 = 82.78333(10 - 6)m4
12
12
20 mm
200 mm
Ix =
Mp
C1 = T1 = sY (0.2)(0.015) = 0.003sY
15 mm
C2 = T2 = sY (0.1)(0.02) = 0.002sY
200 mm
Mp = 0.003sY (0.215) + 0.002sY (0.1) = 0.000845 sY
= 0.000845(250)(106) = 211.25 kN # m
s =
Mp c
211.25(103)(0.115)
=
I
82.78333(10 - 6)
y
0.115
=
;
250
293.5
= 293.5 MPa
y = 0.09796 m = 98.0 mm
stop = sbottom = 293.5 - 250 = 43.5 MPa
Ans.
6–166. The wide-flange member is made from an elasticplastic material. Determine the shape factor.
t
Plastic analysis:
T1 = C1 = sY bt;
h
T2 = C2 = sY a
MP = sY bt(h - t) + sY a
h - 2t
bt
2
t
t
h - 2t
h - 2t
b (t)a
b
2
2
b
t
= sY c bt(h - t) + (h - 2t)2 d
4
Elastic analysis:
I =
=
1
1
bh3 (b - t)(h - 2t)3
12
12
1
[bh3 - (b - t)(h - 2 t)3]
12
MY =
sy I
c
=
=
1
sY A 12
B [bh3 - (b - t)(h - 2t)3]
h
2
bh3 - (b - t)(h - 2t)3
sY
6h
Shape factor:
k =
[bt(h - t) + 4t (h - 2t)2]sY
MP
=
bh3 - (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
a
a
sY (2.41667a4)
sYI
=
= 1.6111a3sY
c
1.5a
Plastic Moment:
MP = sY (a)(a)(2a) + sY (0.5a)(3a)(0.5a)
= 2.75a3sY
Shape Factor:
k =
2.75a3sY
MP
=
= 1.71
MY
1.6111a3sY
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 sY = 36 ksi.
a
a
Maximum Elastic Moment: The moment of inertia about neutral axis must be
determined first.
INA
a
1
1
(2) A 63 B +
(4) A 23 B = 38.667 in4
=
12
12
Applying the flexure formula with s = sY, we have
sY = =
MY =
a
MY c
I
36(38.667)
sY I
=
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 sY = 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:
MP = 250 A 106 B (0.2)(0.025)(0.175) + 250 A 106 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 m4
sr =
289062.5 (0.1)
MP c
= 317.41 MPa
=
I
91.14583 A 10 - 6 B
Residual Bending Stress: As shown on the diagram.
œ
œ
= sbot
= sr - sY
stop
= 317.14 - 250 = 67.1 MPa
Ans.
6–170. Determine the shape factor for the wideflange beam.
15 mm
1
1
(0.2)(0.23)3 (0.18)(0.2)3 = 82.78333 A 10 - 6 B m4
12
12
Ix =
20 mm
200 mm
C1 = T1 = sY(0.2)(0.015) = 0.003sY
Mp
C2 = T2 = sY(0.1)(0.02) = 0.002sY
Mp = 0.003sY(0.215) + 0.002sY(0.1) = 0.000845 sY
15 mm
200 mm
sY =
MY =
k =
MY c
I
sY A 82.78333)10 - 6 B
0.115
Mp
MY
=
= 0.000719855 sY
0.000845sY
= 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 cross-section is
y =
7.5(3)(6) + 3(6)(3)
©yA
=
= 5.25 in.
©A
3(6) + 6(3)
6 in.
The moment of inertia of the cross-section about the neutral axis is
I =
1
1
(3) A 63 B + 3(6)(5.25 - 3)2 +
(6) A 33 B + 6(3)(7.5 - 5.25)2
12
12
1.5 in. 3 in.
1.5 in.
4
= 249.75 in
Here smax = sY and c = y = 5.25 in. Thus
smax =
Mc
;
I
sY =
MY (5.25)
249.75
MY = 47.571sY
Referring to the stress block shown in Fig. b,
sdA = 0;
LA
T - C1 - C2 = 0
d(3)sY - (6 - d)(3)sY - 3(6)sY = 0
d = 6 in.
Since d = 6 in., c1 = 0, Fig. c. Here
T = C = 3(6) sY = 18 sY
Thus,
MP = T(4.5) = 18 sY (4.5) = 81 sY
Thus,
k =
MP
81 sY
=
= 1.70
MY
47.571 sY
Ans.
449
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*6–172. 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 sY = 36 ksi.
3 in.
Referring to Fig. a, the location of centroid of the cross-section is
6 in.
7.5(3)(6) + 3(6)(3)
©yA
y =
=
= 5.25 in.
©A
3(6) + 6(3)
The moment of inertia of the cross-section about the neutral axis is
1.5 in. 3 in.
1.5 in.
I =
1
1
(3)(63) + 3(6)(5.25 - 3)2 +
(6)(33) + 6(3)(7.5 - 5.25)2
12
12
= 249.75 in4
Here, smax = sY = 36 ksi and ¢ = y = 5.25 in. Then
smax =
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,
sdA = 0;
LA
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 H-beam.
Ix =
1
1
(0.2)(0.023) + 2 a b (0.02)(0.23) = 26.8(10 - 6)m4
12
12
200 mm
C1 = T1 = sY(2)(0.09)(0.02) = 0.0036sy
20 mm
C2 = T2 = sY(0.01)(0.24) = 0.0024sy
Mp
20 mm
200 mm
Mp = 0.0036sY(0.11) + 0.0024sY(0.01) = 0.00042sY
20 mm
MYc
sY =
I
MY =
k =
sY(26.8)(10 - 6)
= 0.000268sY
0.1
Mp
MY
=
0.00042sY
= 1.57
0.000268sY
Ans.
6–174. The H-beam is made of an elastic-plastic material
for which sY = 250 MPa. Determine the residual stress in
the top and bottom of the beam after the plastic moment
Mp is applied and then released.
200 mm
Ix =
1
1
(0.2)(0.023) + 2 a b (0.02)(0.23) = 26.8(10 - 6)m4
12
12
20 mm
C1 = T1 = sY(2)(0.09)(0.02) = 0.0036sy
200 mm
C2 = T2 = sY(0.01)(0.24) = 0.0024sy
20 mm
Mp = 0.0036sY(0.11) + 0.0024sY(0.01) = 0.00042sY
Mp = 0.00042(250) A 106 B = 105 kN # m
s¿ =
Mp c
I
y
0.1
=
;
250
392
105(103)(0.1)
=
26.8(10 - 6)
Mp
= 392 MPa
y = 0.0638 = 63.8 mm
sT = sB = 392 - 250 = 142 MPa
Ans.
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 cross-section about the neutral axis is
I =
3 in.
1
1
(3)(93) +
(6) (33) = 195.75 in4
12
12
3 in.
Here, smax = sY and c = 4.5 in. Then
smax =
Mc
;
I
sY =
MY(4.5)
195.75
3 in.
MY = 43.5 sY
Referring to the stress block shown in Fig. a,
T1 = C1 = 3(3)sY = 9 sY
T2 = C2 = 1.5(9)sY = 13.5 sY
Thus,
MP = T1(6) + T2(1.5)
= 9sY(6) + 13.5sY(1.5) = 74.25 sY
k =
MP
74.25 sY
=
= 1.71
MY
43.5 sY
Ans.
452
3 in.
3 in.
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*6–176. 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 sY = 36 ksi.
3 in.
3 in.
The moment of inertia of the cross-section about the neutral axis is
I =
3 in.
1
1
(3)(93) +
(6)(33) = 195.75 in4
12
12
Here, smax = sY = 36 ksi and c = 4.5 in. Then
smax
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,
T1 = C1 = 3(3)(36) = 324 kip
T2 = C2 = 1.5(9)(36) = 486 kip
Thus,
MP = T1(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 cross-section about the neutral axis is
I =
5 in.
p 4
p
A r - r4i B = A 64 - 54 B = 167.75 p in4
4 o
4
6 in.
Here, smax = sY and C = ro = 6 in,
smax =
Mc
;
I
sY =
MY (6)
167.75 p
MY = 87.83 sY
The plastic Moment of the table’s cross-section can be determined by super posing
the moment of the stress block of the solid circular cross-section with radius
ro = 6 in and ri = 5 in. as shown in Figure a, Here,
T1 = C1 =
1
p(62)sY = 18psY
2
T2 = C2 =
1
p(52)sY = 12.5p sY
2
Thus,
MP = T1 b 2 c
4(6)
4(5)
d r - T2 b 2 c
dr
3p
3p
= (18psY) a
16
40
b - 12.5psY a b
p
3p
= 121.33 sY
k =
121.33 sY
MP
=
= 1.38
MY
87.83 sY
Ans.
454
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6–178. The beam is made from elastic-perfectly plastic
material. Determine the shape factor for the thick-walled tube.
ro
Maximum Elastic Moment. The moment of inertia of the cross-section about the
neutral axis is
I =
p
A r 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
A r 4 - ri 4 B sY
4ro 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 r0 and ri
as shown in Fig. a, Referring to the stress block shown in Fig. a,
T1 = c1 =
p 2
r s
2 o Y
T2 = c2 =
p 2
r s
2 i Y
MP = T1 c 2 a
4ro
4ri
b d - T2 c 2 a b d
3p
3p
=
8ro
8ri
p 2
p
r s a
b - ri 2sY a b
2 o Y 3p
2
3p
=
4
A r 3 - ri 3 B sY
3 o
Shape Factor.
4
A r 3 - ri 3 B sY
16ro A ro 3 - ri 3 B
MP
3 o
=
=
k =
p
MY
3p A ro 4 - ri 4 B
A ro 4 - ri 4 B sY
4ro
Ans.
455
ri
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6–179.
Determine the shape factor for the member.
Plastic analysis:
T = C =
–h
2
h
1
bh
(b) a b sY =
s
2
2
4 Y
–h
2
b h2
bh
h
MP =
sY a b =
s
4
3
12 Y
Elastic analysis:
I = 2c
1
h 3
b h3
(b)a b d =
12
2
48
b
sY A bh
sYI
48 B
b h2
=
s
=
h
c
24 Y
2
3
MY =
Shape factor:
k =
Mp
MY
=
bh2
12
sY
bh2
24
sY
Ans.
= 2
*6–180. The member is made from an elastic-plastic
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., sY = 36 ksi.
–h
2
Elastic analysis:
I = 2c
1
(4)(3)3 d = 18 in4
12
MY =
36(18)
sYI
=
= 216 kip # in. = 18 kip # ft
c
3
–h
2
Ans.
b
Plastic analysis:
T = C =
1
(4)(3)(36) = 216 kip
2
6
Mp = 2160 a 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
11
11a h2
2
1
sY a h b (a)a hb =
sY
2
3
18
54
Ans.
6–182. The box beam is made from an elastic-plastic
material for which sY = 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 in4
12
12
Mmax
sYI
=
;
c
9 ft
8 in.
25(1866.67)
27w0(12) =
8
w0 = 18.0 kip>ft
Ans.
Plastic analysis:
16 in.
12 in.
6 in.
C1 = T1 = 25(8)(2) = 400 kip
C2 = T2 = 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 elastic-plastic
material for which sY = 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.
P
8 ft
6 ft
6 ft
The moment of inertia of the beam’s cross-section about the neutral axis is
6 in.
1
1
(6)(123) (5)(103) = 447.33 in4
I =
12
12
12 in.
10 in.
Here, smax = sY = 36 ksi and c = 6 in.
smax =
Mc
;
I
36 =
5 in.
MY (6)
447.33
MY = 2684 kip # in = 223.67 kip # ft
It is required that
Mmax = MY
6P = 223.67
P = 37.28 kip = 37.3 kip
Ans.
Referring to the stress block shown in Fig. b,
T1 = C1 = 6(1)(36) = 216 kip
T2 = C2 = 5(1)(36) = 180 kip
Thus,
MP = T1(11) + T2(5)
= 216(11) + 180(5)
= 3276 kip # in = 273 kip # ft
It is required that
Mmax = MP
6P = 273
P = 45.5 kip
Ans.
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 tan-1115P2] ksi, where tan-1115P2
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
s(ksi)
8 ft
s 20 tan1(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 smax = 0.8994 ksi
Resultant Internal Moment: The resultant internal moment M can be evaluated
from the integal
M = 2
LA
ysdA.
ysdA
2in
= 2
LA
L0
y C 20 tan
-1
(0.0225y) D (2dy)
2in
= 80
L0
= 80 B
y tan - 1 (0.0225y) dy
1 + (0.0225)2y2
2(0.0225)2
tan - 1 (0.0225y) -
2in.
y
R2
2(0.0225) 0
= 4.798 kip # in
Equating
M = 4.00P(12) = 4.798
P = 0.100 kip = 100 lb
Ans.
459
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•6–185.
The plexiglass bar has a stress–strain curve that
can be approximated by the straight-line 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 - T1 = 0
1
1
d
1 d
sc (0.02 - d)(0.02) d - 40 A 106 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,
0.04
e
=
0.02 - 0.010334
0.010334
e = 0.037417 mm>mm
From the stress–strain diagram,
80
s
=
0.037417
0.04
s = 74.833 MPa (OK! Close to assumed value)
Therefore,
1
C = 74.833 A 106 B c (0.02 - 0.010334)(0.02) d = 7233.59 N
2
T1 =
1
0.010334
(60 + 40) A 106 B c (0.02) a
b d = 5166.85 N
2
2
1
0.010334
b d = 2066.74 N
T2 = 40 A 106 B c (0.02) a
2
2
y1 =
2
(0.02 - 0.010334) = 0.0064442 m
3
y2 =
2 0.010334
a
b = 0.0034445 m
3
2
y3 =
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 =
=
sA I
c
1
140 C 12
(2)(33) D
1.5
= 420 kip # in = 35.0 kip # ft
b)
Ans.
The Ultimate Moment :
C1 = T1 =
1
(140 + 180)(1.125)(2) = 360 kip
2
C2 = 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)
s 10(106)P1/ 4
emax = 0.02
smax = 10 A 106 B (0.02)1>4 = 3.761 MPa
M
100 mm
30 mm
P (mm/ mm)
e
0.02
=
y
0.05
e = 0.4 y
s = 10 A 106 B (0.4)1>4y1>4
y(7.9527) A 106 B y1>4(0.03)dy
0.05
M =
y s dA = 2
LA
M = 0.47716 A 106 B
L0
4
y5>4dy = 0.47716 A 106 B a b(0.05)9>4
5
0.05
L0
M = 251 N # m
Ans.
*6–188. The beam has a rectangular cross section and is
made of an elastic-plastic 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 = T1 = 200 A 106 B (0.1)(0.2) = 4000 kN
C2 = T2 =
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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s (ksi)
90
80
•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.
90 - 80
s - 80
=
;
0.03 - 0.025
0.05 - 0.025
60
4 in. M
s = 82 ksi
C1 = T1 =
1
(0.3333)(80 + 82)(3) = 81 kip
2
C2 = T2 =
1
(1.2666)(60 + 80)(3) = 266 kip
2
C3 = T3 =
1
(0.4)(60)(3) = 36 kip
2
0.006
0.025
0.05
P (in./ in.)
3 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:
y =
0.0075(0.29)(0.015) + 2[0.0775(0.125)(0.02)]
0.29(0.015) + 2(0.125)(0.02)
M 650 Nm
20 mm
125 mm
= 0.044933 m
INA
20 mm
1
=
(0.29) A 0.0153 B + 0.29(0.015) (0.044933 - 0.0075)2
12
+
1
(0.04) A 0.1253 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 - 6)
650(0.044933)
17.99037(10 - 6)
My
I
= 1.0815 MPa
= 1.6234 MPa
Resultant Force:
FR =
1
(1.0815 + 1.6234) A 106 B (0.015)(0.29)
2
= 5883 N = 5.88 kN
Ans.
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 0.1253 B + 0.04(0.125)(0.0775 - 0.044933)2
12
= 17.99037 A 10 - 6 B m4
Maximum Bending Stress: Applying the flexure formula s =
(smax)t =
(smax)c =
650(0.14 - 0.044933)
17.99037(10 - 6)
650(0.044933)
17.99037(10 - 6)
My
I
= 3.43 MPa (T)
Ans.
= 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
300 mm
400 mm
a + ©M = 0;
300 mm
400 mm
80 N
M - 80(0.4) = 0
80 N
15 mm
M = 32 N # m
100 mm
1
1
Iz =
(0.075)(0.0153) + 2 a b (0.015)(0.13) = 2.52109(10 - 6)m4
12
12
smax =
32(0.05)
Mc
= 635 kPa
=
I
2.52109(10 - 6)
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(109)
Est
= 18.182
=
Ew
11(109)
M
1
(0.80227)(0.1253) = 0.130578(10 - 3)m4
I =
12
x
75 mm
Failure of wood :
(sw)max
20 mm
Mc
=
I
20(106) =
M(0.0625)
0.130578(10 - 3)
;
M = 41.8 kN # m
Failure of steel :
(sst)max =
20 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(104)
= 0.055
1
1
(0.125)(0.1153) (0.118125)(0.0753) = 11.689616(10 - 6)
12
12
Failure of wood :
(sw)max =
nMc2
I
20(106) =
0.055(M)(0.0375)
11.689616(10 - 6)
;
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 three-dimensional view of
the stress distribution acting over the cross-sectional area.
Hint: The moment of inertia is determined using Eq. A–3 of
Appendix A.
y
100 mm
y 100 – z 2/ 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 =
LA
50 mm
50 mm
y2 dA
100mm
= 2
L0
y2 (2z) dy
100mm
= 20
L0
y2 2100 - y dy
100 mm
3
5
7
3
8
16
y (100 - y)2 (100 - y)2 R 2
= 20 B - y2 (100 - y)2 2
15
105
0
= 30.4762 A 10 - 6 B mm4 = 30.4762 A 10 - 6 B m4
Thus,
smax =
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 y c a
M =
smax
by d(2z dy) r
100
smax 100mm 2
y 2100 - y dy
5 L0
125 A 103 B =
100 mm
smax
3
5
7
3
8
16
y(100 - y)2 (100 - y)2 R 2
B - y2(100 - y)2 5
2
15
105
0
125 A 103 B =
smax
(1.5238) A 106 B
5
smax = 0.410 N>mm2 = 0.410 MPa
Ans.
467
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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 cross-sectional area is shown in
the figure.
20
45 lb
a
5 in.
4 in.
3 in.
0.75 in.
A
a
0.50 in.
45 lb
a + ©M = 0;
M - 45(5 + 4 cos 20°) = 0
M = 394.14 lb # in.
394.14(0.375)
Mc
= 8.41 ksi
= 1
3
I
12 (0.5)(0.75 )
smax =
Ans.
M 85 Nm
•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.
100 mm
A
r2
0.57
0.59
dA
0.42
+ 0.015 ln
+ 0.1 ln
= b ln
= 0.1 ln
r1
0.40
0.42
0.57
LA r
400 mm
= 0.012908358 m
=
LA
dA
r
6.25(10 - 3)
= 0.484182418 m
0.012908358
r - R = 0.495 - 0.484182418 = 0.010817581 m
sA =
M(R - rA)
85(0.484182418 - 0.59)
=
ArA(r - R)
6.25(10 - 3)(0.59)(0.010817581)
= - 225.48 kPa
sA = 225 kPa (C)
sB =
Ans.
M(R - rB)
85(0.484182418 - 0.40)
=
ArB(r - R)
150 mm
6.25(10 - 3)(0.40)(0.010817581)
20 mm
B
2
A = 2(0.1)(0.02) + (0.15)(0.015) = 6.25(10 ) m
A
20 mm
30
-3
R =
A
15 mm
B
= 265 kPa (T)
468
Ans.
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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 ©Fy = 0;
20 - 2x - V = 0
V = 20 - 2x
c + ©MNA = 0;
4 ft
Ans.
x
20x - 166 - 2x a 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
469
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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 (sallow)t = 22 ksi and
(sallow)c = 15 ksi, respectively.
4 in.
4 in.
M
2 in.
2 in.
y (From base) =
I =
1
242 - 22 = 1.1547 in.
3
1
(4)(242 - 22)3 = 4.6188 in4
36
Assume failure due to tensile stress :
smax =
My
;
I
22 =
M(1.1547)
4.6188
M = 88.0 kip # in. = 7.33 kip # ft
Assume failure due to compressive stress:
smax =
Mc
;
I
15 =
M(3.4641 - 1.1547)
4.6188
M = 30.0 kip # in. = 2.50 kip # ft
(controls)
Ans.
470
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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:
Mz = - M cos u
My = - 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 = -
My z
Mzy
+
Iz
Iy
- M cos u (a2)
= -
=
1
12
a4
- Msin u ( - a2)
+
1
12
a4
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
Iy
tan u
tan a = (1) tan(45°)
a = 45°
Ans.
471
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•7–1.
If the wide-flange beam is subjected to a shear of
V = 20 kN, determine the shear stress on the web at A.
Indicate the shear-stress components on a volume element
located at this point.
200 mm
A
20 mm
20 mm
B
V
300 mm
200 mm
The moment of inertia of the cross-section about the neutral axis is
I =
1
1
(0.2)(0.343) (0.18)(0.33) = 0.2501(10 - 3) m4
12
12
From Fig. a,
QA = y¿A¿ = 0.16 (0.02)(0.2) = 0.64(10 - 3) m3
Applying the shear formula,
VQA
20(103)[0.64(10 - 3)]
=
tA =
It
0.2501(10 - 3)(0.02)
= 2.559(106) Pa = 2.56 MPa
Ans.
The shear stress component at A is represented by the volume element shown in
Fig. b.
472
20 mm
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7–2. If the wide-flange beam is subjected to a shear of
V = 20 kN, determine the maximum shear stress in the beam.
200 mm
A
20 mm
20 mm
B
V
300 mm
200 mm
The moment of inertia of the cross-section about the neutral axis is
I =
1
1
(0.2)(0.343) (0.18)(0.33) = 0.2501(10 - 3) m4
12
12
From Fig. a.
Qmax = ©y¿A¿ = 0.16 (0.02)(0.2) + 0.075 (0.15)(0.02) = 0.865(10 - 3) m3
The maximum shear stress occurs at the points along neutral axis since Q is
maximum and thicknest t is the smallest.
tmax =
VQmax
20(103) [0.865(10 - 3)]
=
It
0.2501(10 - 3) (0.02)
= 3.459(106) Pa = 3.46 MPa
Ans.
473
20 mm
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7–3. If the wide-flange beam is subjected to a shear of
V = 20 kN, determine the shear force resisted by the web
of the beam.
200 mm
A
20 mm
20 mm
B
V
300 mm
200 mm
The moment of inertia of the cross-section about the neutral axis is
I =
1
1
(0.2)(0.343) (0.18)(0.33) = 0.2501(10 - 3) m4
12
12
For 0 … y 6 0.15 m, Fig. a, Q as a function of y is
Q = ©y¿A¿ = 0.16 (0.02)(0.2) +
1
(y + 0.15)(0.15 - y)(0.02)
2
= 0.865(10 - 3) - 0.01y2
For 0 … y 6 0.15 m, t = 0.02 m. Thus.
t =
20(103) C 0.865(10 - 3) - 0.01y2 D
VQ
=
It
0.2501(10 - 3) (0.02)
=
E 3.459(106) - 39.99(106) y2 F Pa.
The sheer force resisted by the web is,
0.15 m
Vw = 2
L0
0.15 m
tdA = 2
L0
C 3.459(106) - 39.99(106) y2 D (0.02 dy)
= 18.95 (103) N = 19.0 kN
Ans.
474
20 mm
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*7–4. If the T-beam is subjected to a vertical shear of
V = 12 kip, determine the maximum shear stress in the
beam. Also, compute the shear-stress jump at the flangeweb junction AB. Sketch the variation of the shear-stress
intensity over the entire cross section.
4 in.
4 in.
3 in.
4 in.
B
6 in.
A
V ⫽ 12 kip
Section Properties:
y =
INA =
1.5(12)(3) + 6(4)(6)
©yA
=
= 3.30 in.
©A
12(3) + 4(6)
1
1
(12) A 33 B + 12(3)(3.30 - 1.5)2 +
(4) A 63 B + 4(6)(6 - 3.30)2
12
12
= 390.60 in4
Qmax = y1œ A¿ = 2.85(5.7)(4) = 64.98 in3
QAB = y2œ A¿ = 1.8(3)(12) = 64.8 in3
Shear Stress: Applying the shear formula t =
tmax =
VQ
It
VQmax
12(64.98)
=
= 0.499 ksi
It
390.60(4)
Ans.
(tAB)f =
VQAB
12(64.8)
=
= 0.166 ksi
Itf
390.60(12)
Ans.
(tAB)W =
VQAB
12(64.8)
=
= 0.498 ksi
I tW
390.60(4)
Ans.
475
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•7–5.
If the T-beam is subjected to a vertical shear of
V = 12 kip, determine the vertical shear force resisted by
the flange.
4 in.
4 in.
3 in.
4 in.
B
6 in.
A
V ⫽ 12 kip
Section Properties:
y =
©yA
1.5(12)(3) + 6(4)(6)
=
= 3.30 in.
©A
12(3) + 4(6)
INA =
1
1
(12) A 33 B + 12(3)(3.30 - 1.5)2 +
(4) A 63 B + 6(4)(6 - 3.30)2
12
12
= 390.60 in4
Q = y¿A¿ = (1.65 + 0.5y)(3.3 - y)(12) = 65.34 - 6y2
Shear Stress: Applying the shear formula
t =
VQ
12(65.34 - 6y2)
=
It
390.60(12)
= 0.16728 - 0.01536y2
Resultant Shear Force: For the flange
Vf =
tdA
LA
3.3 in
=
L0.3 in
A 0.16728 - 0.01536y2 B (12dy)
= 3.82 kip
Ans.
476
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7–6. If the beam is subjected to a shear of V = 15 kN,
determine the web’s shear stress at A and B. Indicate the
shear-stress components on a volume element located
at these points. Show that the neutral axis is located at
y = 0.1747 m from the bottom and INA = 0.2182110-32 m4.
200 mm
A
30 mm
25 mm
V
(0.015)(0.125)(0.03) + (0.155)(0.025)(0.25) + (0.295)(0.2)(0.03)
y =
= 0.1747 m
0.125(0.03) + (0.025)(0.25) + (0.2)(0.03)
I =
1
(0.125)(0.033) + 0.125(0.03)(0.1747 - 0.015)2
12
+
1
(0.025)(0.253) + 0.25(0.025)(0.1747 - 0.155)2
12
+
1
(0.2)(0.033) + 0.2(0.03)(0.295 - 0.1747)2 = 0.218182 (10 - 3) m4
12
B
250 mm
30 mm
125 mm
œ
QA = yAA
= (0.310 - 0.015 - 0.1747)(0.2)(0.03) = 0.7219 (10 - 3) m3
QB = yABœ = (0.1747 - 0.015)(0.125)(0.03) = 0.59883 (10 - 3) m3
tA =
15(103)(0.7219)(10 - 3)
VQA
= 1.99 MPa
=
It
0.218182(10 - 3)(0.025)
Ans.
tB =
VQB
15(103)(0.59883)(10 - 3)
= 1.65 MPa
=
It
0.218182(10 - 3)0.025)
Ans.
7–7. If the wide-flange beam is subjected to a shear of
V = 30 kN, determine the maximum shear stress in the beam.
200 mm
A
30 mm
25 mm
V
B
250 mm
30 mm
Section Properties:
I =
1
1
(0.2)(0.310)3 (0.175)(0.250)3 = 268.652(10) - 6 m4
12
12
Qmax = © y¿A = 0.0625(0.125)(0.025) + 0.140(0.2)(0.030) = 1.0353(10) - 3 m3
tmax =
VQ
30(10)3(1.0353)(10) - 3
= 4.62 MPa
=
It
268.652(10) - 6 (0.025)
Ans.
477
200 mm
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*7–8. If the wide-flange beam is subjected to a shear of
V = 30 kN, determine the shear force resisted by the web
of the beam.
200 mm
A
30 mm
1
1
(0.2)(0.310)3 (0.175)(0.250)3 = 268.652(10) - 6 m4
12
12
I =
Q = a
25 mm
V
B
0.155 + y
b (0.155 - y)(0.2) = 0.1(0.024025 - y2)
2
250 mm
30(10)3(0.1)(0.024025 - y2)
tf =
268.652(10)
-6
30 mm
200 mm
(0.2)
0.155
Vf =
L
tf dA = 55.8343(10)6
L0.125
= 11.1669(10)6[ 0.024025y -
(0.024025 - y2)(0.2 dy)
1 3 0.155
y ]
2 0.125
Vf = 1.457 kN
Vw = 30 - 2(1.457) = 27.1 kN
Ans.
•7–9. Determine the largest shear force V that the member
can sustain if the allowable shear stress is tallow = 8 ksi.
3 in.
1 in.
V
3 in. 1 in.
1 in.
y =
(0.5)(1)(5) + 2 [(2)(1)(2)]
= 1.1667 in.
1 (5) + 2 (1)(2)
I =
1
(5)(13) + 5 (1)(1.1667 - 0.5)2
12
+ 2a
1
b (1)(23) + 2 (1)(2)(2 - 1.1667)2 = 6.75 in4
12
Qmax = ©y¿A¿ = 2 (0.91665)(1.8333)(1) = 3.3611 in3
tmax = tallow =
8 (103) = -
VQmax
It
V (3.3611)
6.75 (2)(1)
V = 32132 lb = 32.1 kip
Ans.
478
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7–10. If the applied shear force V = 18 kip, determine the
maximum shear stress in the member.
3 in.
1 in.
V
3 in. 1 in.
1 in.
y =
(0.5)(1)(5) + 2 [(2)(1)(2)]
= 1.1667 in.
1 (5) + 2 (1)(2)
I =
1
(5)(13) + 5 (1)(1.1667 - 0.5)2
12
+ 2a
1
b (1)(23) + 2 (1)(2)(2 - 1.1667) = 6.75 in4
12
Qmax = ©y¿A¿ = 2 (0.91665)(1.8333)(1) = 3.3611 in3
tmax =
18(3.3611)
VQmax
=
= 4.48 ksi
It
6.75 (2)(1)
Ans.
7–11. The wood beam has an allowable shear stress of
tallow = 7 MPa. Determine the maximum shear force V that
can be applied to the cross section.
50 mm
50 mm
100 mm
50 mm
200 mm
V
50 mm
I =
1
1
(0.2)(0.2)3 (0.1)(0.1)3 = 125(10 - 6) m4
12
12
tallow =
7(106) =
VQmax
It
V[(0.075)(0.1)(0.05) + 2(0.05)(0.1)(0.05)]
125(10 - 6)(0.1)
V = 100 kN
Ans.
479
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*7–12. The beam has a rectangular cross section and is
made of wood having an allowable shear stress of tallow =
200 psi. Determine the maximum shear force V that can be
developed in the cross section of the beam. Also, plot the
shear-stress variation over the cross section.
V
12 in.
8 in.
Section Properties The moment of inertia of the cross-section about the neutral axis is
I =
1
(8) (123) = 1152 in4
12
Q as the function of y, Fig. a,
Q =
1
(y + 6)(6 - y)(8) = 4 (36 - y2)
2
Qmax occurs when y = 0. Thus,
Qmax = 4(36 - 02) = 144 in3
The maximum shear stress occurs of points along the neutral axis since Q is
maximum and the thickness t = 8 in. is constant.
tallow =
VQmax
;
It
200 =
V(144)
1152(8)
V = 12800 16 = 12.8 kip
Ans.
Thus, the shear stress distribution as a function of y is
t =
12.8(103) C 4(36 - y2) D
VQ
=
It
1152 (8)
=
E 5.56 (36 - y2) F psi
480
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7–13. Determine the maximum shear stress in the strut if
it is subjected to a shear force of V = 20 kN.
12 mm
Section Properties:
INA
60 mm
1
1
=
(0.12) A 0.0843 B (0.04) A 0.063 B
12
12
V
= 5.20704 A 10 - 6 B m4
12 mm
80 mm
Qmax = ©y¿A¿
20 mm
20 mm
= 0.015(0.08)(0.03) + 0.036(0.012)(0.12)
= 87.84 A 10 - 6 B m3
Maximum Shear Stress: Maximum shear stress occurs at the point where the
neutral axis passes through the section.
Applying the shear formula
tmax =
VQmax
It
20(103)(87.84)(10 - 6)
=
5.20704(10 - 6)(0.08)
= 4 22 MPa
Ans.
7–14. Determine the maximum shear force V that the
strut can support if the allowable shear stress for the
material is tallow = 40 MPa.
12 mm
60 mm
Section Properties:
INA =
V
1
1
(0.12) A 0.0843 B (0.04) A 0.063 B
12
12
12 mm
= 5.20704 A 10 - 6 B m4
80 mm
Qmax = ©y¿A¿
20 mm
= 0.015(0.08)(0.03) + 0.036(0.012)(0.12)
= 87.84 A 10 - 6 B m3
Allowable shear stress: Maximum shear stress occurs at the point where the neutral
axis passes through the section.
Applying the shear formula
tmax = tallow =
40 A 106 B =
VQmax
It
V(87.84)(10 - 6)
5.20704(10 - 6)(0.08)
V = 189 692 N = 190 kN
Ans.
481
20 mm
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7–15. Plot the shear-stress distribution over the cross
section of a rod that has a radius c. By what factor is the
maximum shear stress greater than the average shear stress
acting over the cross section?
c
y
V
x = 2c2 - y2 ;
p 4
c
4
I =
t = 2 x = 2 2c2 - y2
dA = 2 x dy = 22c2 - y2 dy
dQ = ydA = 2y 2c2 - y2 dy
x
Q =
Ly
2y 2c2 - y2 dy = -
3 x
2
2 2
2
(c - y2)2 | y = (c2 - y2)3
3
3
3
V[23 (c2 - y2)2]
VQ
4V 2
t =
=
=
[c - y2)
p 4
2
2
It
3pc4
( 4 c )(2 2c - y )
The maximum shear stress occur when y = 0
tmax =
4V
3 p c2
tavg =
V
V
=
A
p c2
The faector =
tmax
=
tavg
4V
3 pc2
V
pc2
=
4
3
Ans.
482
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*7–16. A member has a cross section in the form of an
equilateral triangle. If it is subjected to a shear force V,
determine the maximum average shear stress in the member
using the shear formula. Should the shear formula actually be
used to predict this value? Explain.
I =
V
1
(a)(h)3
36
y
h
;
=
x
a>2
Q =
a
LA¿
Q = a
y =
y dA = 2 c a
2h
x
a
1
2
2
b (x)(y) a h - y b d
2
3
3
4h2
2x
b (x2) a 1 b
a
3a
t = 2x
t =
t =
V(4h2>3a)(x2)(1 - 2x
VQ
a)
=
It
((1>36)(a)(h3))(2x)
24V(x - a2 x2)
a2h
24V
4
dt
= 2 2 a1 - xb = 0
a
dx
ah
At x =
y =
a
4
h
2h a
a b =
a 4
2
tmax =
24V a
2 a
a b a1 - a b b
a 4
a2h 4
tmax =
3V
ah
Ans.
No, because the shear stress is not perpendicular to the boundary. See Sec. 7-3.
483
h
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•7–17.
Determine the maximum shear stress in the strut if
it is subjected to a shear force of V = 600 kN.
30 mm
150 mm
V
100 mm
100 mm
100 mm
The moment of inertia of the cross-section about the neutral axis is
I =
1
1
(0.3)(0.213) (0.2)(0.153) = 0.175275(10 - 3) m4
12
12
From Fig. a,
Qmax = ©y¿A¿ = 0.09(0.03)(0.3) + 0.0375(0.075)(0.1)
= 1.09125(10 - 3) m3
The maximum shear stress occurs at the points along the neutral axis since Q is
maximum and thickness t = 0.1 m is the smallest.
tmax =
VQmax
600(103)[1.09125(10 - 3)]
=
It
0.175275(10 - 3) (0.1)
= 37.36(106) Pa = 37.4 MPa
Ans.
484
30 mm
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7–18. Determine the maximum shear force V that the strut
can support if the allowable shear stress for the material is
tallow = 45 MPa.
30 mm
150 mm
V
100 mm
100 mm
100 mm
The moment of inertia of the cross-section about the neutral axis is
I =
1
1
(0.3)(0.213) (0.2)(0.153) = 0.175275 (10 - 3) m4
12
12
From Fig. a
Qmax = ©y¿A¿ = 0.09(0.03)(0.3) + 0.0375 (0.075)(0.1)
= 1.09125 (10 - 3) m3
The maximum shear stress occeurs at the points along the neutral axis since Q is
maximum and thickness t = 0.1 m is the smallest.
tallow =
VQmax
;
It
45(106) =
V C 1.09125(10 - 3) D
0.175275(10 - 3)(0.1)
V = 722.78(103) N = 723 kN
Ans.
485
30 mm
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7–19. Plot the intensity of the shear stress distributed over
the cross section of the strut if it is subjected to a shear force
of V = 600 kN.
30 mm
The moment of inertia of the cross-section about the neutral axis is
I =
1
1
(0.3)(0.213) (0.2)(0.153) = 0.175275 (10 - 3) m4
12
12
For 0.075 m 6 y … 0.105 m, Fig. a, Q as a function of y is
Q = y¿A¿ =
1
(0.105 + y) (0.105 - y)(0.3) = 1.65375(10 - 3) - 0.15y2
2
For 0 … y 6 0.075 m, Fig. b, Q as a function of y is
Q = ©y¿A¿ = 0.09 (0.03)(0.3) +
1
(0.075 + y)(0.075 - y)(0.1) = 1.09125(10 - 3) - 0.05 y2
2
For 0.075 m 6 y … 0.105 m, t = 0.3 m. Thus,
t =
600 (103) C 1.65375(10 - 3) - 0.15y2 D
VQ
= (18.8703 - 1711.60y2) MPa
=
It
0.175275(10 - 3) (0.3)
At y = 0.075 m and y = 0.105 m,
t|y = 0.015 m = 9.24 MPa
ty = 0.105 m = 0
For 0 … y 6 0.075 m, t = 0.1 m. Thus,
t =
VQ
600 (103) [1.09125(10 - 3) - 0.05 y2]
= (37.3556 - 1711.60 y2) MPa
=
It
0.175275(10 - 3) (0.1)
At y = 0 and y = 0.075 m,
t|y = 0 = 37.4 MPa
ty = 0.075 m = 27.7 MPa
The plot shear stress distribution over the cross-section is shown in Fig. c.
486
150 mm
V
100 mm
100 mm
100 mm
30 mm
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*7–20. The steel rod is subjected to a shear of 30 kip.
Determine the maximum shear stress in the rod.
The moment of inertia of the ciralor cross-section about the neutral axis (x axis) is
p 4
p
I =
r = (24) = 4 p in4
4
4
30 kip
dQ = ydA = y (2xdy) = 2xy dy
1
However, from the equation of the circle, x = (4 - y2)2 , Then
1
dQ = 2y(4 - y2)2 dy
Thus, Q for the area above y is
2 in
1
2y (4 - y2)2 dy
Ly
3 2 in
2
= - (4 - y2)2 3
y
=
3
2
(4 - y2)2
3
1
Here, t = 2x = 2 (4 - y2)2 . Thus
30 C 23 (4 - y2)2 D
VQ
=
t =
1
It
4p C 2(4 - y2)2 D
3
t =
5
(4 - y2) ksi
2p
By inspecting this equation, t = tmax at y = 0. Thus
¿=
tmax
A
2 in.
Q for the differential area shown shaded in Fig. a is
Q =
1 in.
20
10
= 3.18 ksi
=
p
2p
Ans.
487
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•7–21.
The steel rod is subjected to a shear of 30 kip.
Determine the shear stress at point A. Show the result on a
volume element at this point.
1 in.
A
The moment of inertia of the circular cross-section about the neutral axis (x axis) is
I =
2 in.
p 4
p
r = (24) = 4p in4
4
4
30 kip
Q for the differential area shown in Fig. a is
dQ = ydA = y (2xdy) = 2xy dy
1
However, from the equation of the circle, x = (4 - y2)2 , Then
1
dQ = 2y (4 - y2)2 dy
Thus, Q for the area above y is
2 in.
1
Q =
Ly
= -
2y (4 - y2)2 dy
2 in.
3
3
2
2
(4 - y2)2 `
= (4 - y2)2
3
3
y
1
Here t = 2x = 2 (4 - y2)2 . Thus,
30 C 23 (4 - y2)2 D
VQ
=
t =
1
It
4p C 2(4 - y2)2 D
3
t =
5
(4 - y2) ksi
2p
For point A, y = 1 in. Thus
tA =
5
(4 - 12) = 2.39 ksi
2p
Ans.
The state of shear stress at point A can be represented by the volume element
shown in Fig. b.
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7–22. Determine the shear stress at point B on the web of
the cantilevered strut at section a–a.
2 kN
250 mm
a
250 mm
4 kN
300 mm
a
20 mm
70 mm
(0.01)(0.05)(0.02) + (0.055)(0.07)(0.02)
y =
= 0.03625 m
(0.05)(0.02) + (0.07)(0.02)
I =
+
B
20 mm
50 mm
1
(0.05)(0.023) + (0.05)(0.02)(0.03625 - 0.01)2
12
1
(0.02)(0.073) + (0.02)(0.07)(0.055 - 0.03625)2 = 1.78625(10 - 6) m4
12
yBœ = 0.03625 - 0.01 = 0.02625 m
QB = (0.02)(0.05)(0.02625) = 26.25(10 - 6) m3
tB =
6(103)(26.25)(10 - 6)
VQB
=
It
1.78622(10 - 6)(0.02)
= 4.41 MPa
Ans.
7–23. Determine the maximum shear stress acting at
section a–a of the cantilevered strut.
2 kN
250 mm
a
250 mm
4 kN
300 mm
a
20 mm
70 mm
y =
(0.01)(0.05)(0.02) + (0.055)(0.07)(0.02)
= 0.03625 m
(0.05)(0.02) + (0.07)(0.02)
I =
1
(0.05)(0.023) + (0.05)(0.02)(0.03625 - 0.01)2
12
+
20 mm
50 mm
1
(0.02)(0.073) + (0.02)(0.07)(0.055 - 0.03625)2 = 1.78625(10 - 6) m4
12
Qmax = y¿A¿ = (0.026875)(0.05375)(0.02) = 28.8906(10 - 6) m3
tmax =
B
VQmax
6(103)(28.8906)(10 - 6)
=
It
1.78625(10 - 6)(0.02)
= 4.85 MPa
Ans.
489
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*7–24. Determine the maximum shear stress in the T-beam
at the critical section where the internal shear force is
maximum.
10 kN/m
A
1.5 m
3m
The shear diagram is shown in Fig. b. As indicated, Vmax = 27.5 kN
150 mm
The neutral axis passes through centroid c of the cross-section, Fig. c.
'
0.075(0.15)(0.03) + 0.165(0.03)(0.15)
© y A
=
y =
©A
0.15(0.03) + 0.03(0.15)
150 mm
1
(0.03)(0.153) + 0.03(0.15)(0.12 - 0.075)2
12
+
1
(0.15)(0.033) + 0.15(0.03)(0.165 - 0.12)2
12
= 27.0 (10 - 6) m4
From Fig. d,
Qmax = y¿A¿ = 0.06(0.12)(0.03)
= 0.216 (10 - 3) m3
The maximum shear stress occurs at points on the neutral axis since Q is maximum
and thickness t = 0.03 m is the smallest.
tmax =
27.5(103) C 0.216(10 - 3) D
Vmax Qmax
=
It
27.0(10 - 6)(0.03)
= 7.333(106) Pa
= 7.33 MPa
Ans.
490
30 mm
30 mm
= 0.12 m
I =
B
C
The FBD of the beam is shown in Fig. a,
1.5 m
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•7–25.
Determine the maximum shear stress in the
T-beam at point C. Show the result on a volume element
at this point.
10 kN/m
A
B
C
1.5 m
3m
150 mm
150 mm
30 mm
using the method of sections,
+ c ©Fy = 0;
VC + 17.5 -
1
(5)(1.5) = 0
2
VC = - 13.75 kN
The neutral axis passes through centroid C of the cross-section,
0.075 (0.15)(0.03) + 0.165(0.03)(0.15)
©yA
=
©A
0.15(0.03) + 0.03(0.15)
y =
= 0.12 m
I =
1
(0.03)(0.15) + 0.03(0.15)(0.12 - 0.075)2
12
+
1
(0.15)(0.033) + 0.15(0.03)(0.165 - 0.12)2
12
= 27.0 (10 - 6) m4
Qmax = y¿A¿ = 0.06 (0.12)(0.03)
= 0.216 (10 - 3) m3 490
The maximum shear stress occurs at points on the neutral axis since Q is maximum
and thickness t = 0.03 m is the smallest.
tmax =
30 mm
13.75(103) C 0.216(10 - 3) D
VC Qmax
=
It
27.0(10 - 6) (0.03)
= 3.667(106) Pa = 3.67 MPa
Ans.
491
1.5 m
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7–26. Determine the maximum shear stress acting in the
fiberglass beam at the section where the internal shear
force is maximum.
200 lb/ft
150 lb/ft
D
A
6 ft
6 ft
2 ft
4 in.
6 in.
0.5 in.
4 in.
Support Reactions: As shown on FBD.
Internal Shear Force: As shown on shear diagram, Vmax = 878.57 lb.
Section Properties:
INA =
1
1
(4) A 7.53 B (3.5) A 63 B = 77.625 in4
12
12
Qmax = ©y¿A¿
= 3.375(4)(0.75) + 1.5(3)(0.5) = 12.375 in3
Maximum Shear Stress: Maximum shear stress occurs at the point where the
neutral axis passes through the section.
Applying the shear formula
tmax =
=
VQmax
It
878.57(12.375)
= 280 psi
77.625(0.5)
Ans.
492
0.75 in.
0.75 in.
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7–27. Determine the shear stress at points C and D
located on the web of the beam.
3 kip/ft
D
A
C
B
6 ft
6 ft
6 in.
0.75 in.
The FBD is shown in Fig. a.
Using the method of sections, Fig. b,
+ c ©Fy = 0;
18 -
1
(3)(6) - V = 0
2
V = 9.00 kip.
The moment of inertia of the beam’s cross section about the neutral axis is
I =
1
1
(6)(103) (5.25)(83) = 276 in4
12
12
QC and QD can be computed by refering to Fig. c.
QC = ©y¿A¿ = 4.5 (1)(6) + 2 (4)(0.75)
= 33 in3
QD = y3œ A¿ = 4.5 (1)(6) = 27 in3
Shear Stress. since points C and D are on the web, t = 0.75 in.
tC =
VQC
9.00 (33)
=
= 1.43 ksi
It
276 (0.75)
Ans.
tD =
VQD
9.00 (27)
=
= 1.17 ksi
It
276 (0.75)
Ans.
493
6 ft
1 in.
C
D
4 in.
4 in.
6 in.
1 in.
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*7–28. Determine the maximum shear stress acting in the
beam at the critical section where the internal shear force
is maximum.
3 kip/ft
D
A
C
B
6 ft
6 ft
6 in.
The FBD is shown in Fig. a.
The shear diagram is shown in Fig. b, Vmax = 18.0 kip.
0.75 in.
6 ft
1 in.
C
D
4 in.
4 in.
6 in.
1 in.
The moment of inertia of the beam’s cross-section about the neutral axis is
I =
1
1
(6)(103) (5.25)(83)
12
12
= 276 in4
From Fig. c
Qmax = ©y¿A¿ = 4.5 (1)(6) + 2(4)(0.75)
= 33 in3
The maximum shear stress occurs at points on the neutral axis since Q is the
maximum and thickness t = 0.75 in is the smallest
tmax =
Vmax Qmax
18.0 (33)
=
= 2.87 ksi
It
276 (0.75)
Ans.
494
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7–30. The beam has a rectangular cross section and is
subjected to a load P that is just large enough to develop a
fully plastic moment Mp = PL at the fixed support. If the
material is elastic-plastic, then at a distance x 6 L the
moment M = Px creates a region of plastic yielding with
an associated elastic core having a height 2y¿. This situation
has been described by Eq. 6–30 and the moment M is
distributed over the cross section as shown in Fig. 6–48e.
Prove that the maximum shear stress developed in the beam
is given by tmax = 321P>A¿2, where A¿ = 2y¿b, the crosssectional area of the elastic core.
P
x
Plastic region
2y¿
h
b
Elastic region
Force Equilibrium: The shaded area indicares the plastic zone. Isolate an element in
the plastic zone and write the equation of equilibrium.
; ©Fx = 0;
tlong A2 + sg A1 - sg A1 = 0
tlong = 0
This proves that the longitudinal shear stress. tlong, is equal to zero. Hence the
corresponding transverse stress, tmax, is also equal to zero in the plastic zone.
Therefore, the shear force V = P is carried by the malerial only in the elastic zone.
Section Properties:
INA =
1
2
(b)(2y¿)3 = b y¿ 3
12
3
Qmax = y¿ A¿ =
y¿
y¿ 2b
(y¿)(b) =
2
2
Maximum Shear Stress: Applying the shear formula
V A y¿2 b B
3
tmax
However,
VQmax
=
=
It
A¿ = 2by¿
tmax =
3P
‚
2A¿
A by¿ B (b)
2
3
3
=
3P
4by¿
hence
(Q.E.D.)
495
L
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7–31. The beam in Fig. 6–48f is subjected to a fully plastic
moment Mp . Prove that the longitudinal and transverse
shear stresses in the beam are zero. Hint: Consider an element
of the beam as shown in Fig. 7–4c.
P
x
Plastic region
2y¿
h
b
Elastic region
L
Force Equilibrium: If a fully plastic moment acts on the cross section, then an
element of the material taken from the top or bottom of the cross section is
subjected to the loading shown. For equilibrium
; ©Fx = 0;
sg A1 + tlong A2 - sg A1 = 0
tlong = 0
Thus no shear stress is developed on the longitudinal or transverse plane of the
element. (Q. E. D.)
*7–32. The beam is constructed from two boards fastened
together at the top and bottom with two rows of nails
spaced every 6 in. If each nail can support a 500-lb shear
force, determine the maximum shear force V that can be
applied to the beam.
6 in.
6 in.
2 in.
2 in.
V
6 in.
Section Properties:
I =
1
(6) A 43 B = 32.0 in4
12
Q = y¿A¿ = 1(6)(2) = 12.0 in4
Shear Flow: There are two rows of nails. Hence, the allowable shear flow
2(500)
= 166.67 lb>in.
q =
6
q =
166.67 =
VQ
I
V(12.0)
32.0
V = 444 lb
Ans.
496
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•7–33.
The beam is constructed from two boards
fastened together at the top and bottom with two rows of
nails spaced every 6 in. If an internal shear force of
V = 600 lb is applied to the boards, determine the shear
force resisted by each nail.
6 in.
6 in.
2 in.
2 in.
Section Properties:
I =
1
(6) A 43 B = 32.0 in4
12
V
6 in.
Q = y¿A¿ = 1(6)(2) = 12.0 in4
Shear Flow:
q =
VQ
600(12.0)
=
= 225 lb>in.
I
32.0
There are two rows of nails. Hence, the shear force resisted by each nail is
q
225 lb>in.
F = a bs = a
b(6 in.) = 675 lb
2
2
Ans.
7–34. The beam is constructed from two boards fastened
together with three rows of nails spaced s = 2 in. apart. If
each nail can support a 450-lb shear force, determine the
maximum shear force V that can be applied to the beam. The
allowable shear stress for the wood is tallow = 300 psi.
s
s
1.5 in.
The moment of inertia of the cross-section about the neutral axis is
I =
V
1
(6)(33) = 13.5 in4
12
6 in.
Refering to Fig. a,
QA = Qmax = y¿A¿ = 0.75(1.5)(6) = 6.75 in3
The maximum shear stress occurs at the points on the neutral axis where Q is
maximum and t = 6 in.
tallow =
VQmax
;
It
300 =
V(6.75)
13.5(6)
V = 3600 lb = 3.60 kips
Shear Flow: Since there are three rows of nails,
F
450
b = 675 lb>in.
qallow = 3 a b = 3 a
s
2
VQA
V(6.75)
;
675 =
qallow =
I
13.5
V = 1350 lb = 1.35 kip
497
Ans.
1.5 in.
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7–35. The beam is constructed from two boards fastened
together with three rows of nails. If the allowable shear
stress for the wood is tallow = 150 psi, determine the
maximum shear force V that can be applied to the beam.
Also, find the maximum spacing s of the nails if each nail
can resist 650 lb in shear.
s
s
1.5 in.
V
6 in.
The moment of inertia of the cross-section about the neutral axis is
I =
1
(6)(33) = 13.5 in4
12
Refering to Fig. a,
QA = Qmax = y¿A¿ = 0.75(1.5)(6) = 6.75 in3
The maximum shear stress occurs at the points on the neutral axis where Q is
maximum and t = 6 in.
tallow =
VQmax
;
It
150 =
V(6.75)
13.5(6)
V = 1800 lb = 1.80 kip
Since there are three rows of nails, qallow = 3 a
qallow =
VQA
;
I
Ans.
F
650
1950 lb
b = 3¢
b
≤ = a
s
s
s
in.
1800(6.75)
1950
=
s
13.5
s = 2.167 in = 2
1
in
8
Ans.
498
1.5 in.
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*7–36. The beam is fabricated from two equivalent
structural tees and two plates. Each plate has a height of
6 in. and a thickness of 0.5 in. If a shear of V = 50 kip is
applied to the cross section, determine the maximum spacing
of the bolts. Each bolt can resist a shear force of 15 kip.
0.5 in.
s
3 in.
1 in.
A
Section Properties:
INA =
V
6 in.
1
1
(3) A 93 B (2.5) A 83 B
12
12
0.5 in.
N
1
1
(0.5) A 23 B +
(1) A 63 B
12
12
3 in.
= 93.25 in4
Q = ©y¿A¿ = 2.5(3)(0.5) + 4.25(3)(0.5) = 10.125 in3
Shear Flow: Since there are two shear planes on the bolt, the allowable shear flow is
2(15)
30
.
=
q =
s
s
VQ
q =
I
50(10.125)
30
=
s
93.25
s = 5.53 in.
Ans.
•7–37.
The beam is fabricated from two equivalent
structural tees and two plates. Each plate has a height of
6 in. and a thickness of 0.5 in. If the bolts are spaced at
s = 8 in., determine the maximum shear force V that can
be applied to the cross section. Each bolt can resist a
shear force of 15 kip.
0.5 in.
s
3 in.
1 in.
A
Section Properties:
INA
-
1
1
(0.5) A 23 B +
(1) A 63 B
12
12
3 in.
Q = ©y¿A¿ = 2.5(3)(0.5) + 4.25(3)(0.5) = 10.125 in3
Shear Flow: Since there are two shear planes on the bolt, the allowable shear flow is
2(15)
= 3.75 kip>in.
q =
8
3.75 =
0.5 in.
N
= 93.25 in4
q =
V
6 in.
1
1
=
(3) A 93 B (2.5) A 83 B
12
12
VQ
I
V(10.125)
93.25
y = 34.5 kip
Ans.
499
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7–38. The beam is subjected to a shear of V = 2 kN.
Determine the average shear stress developed in each nail
if the nails are spaced 75 mm apart on each side of the
beam. Each nail has a diameter of 4 mm.
The neutral axis passes through centroid C of the cross-section as shown in Fig. a.
'
0.175(0.05)(0.2) + 0.1(0.2)(0.05)
© y A
y =
=
= 0.1375 m
©A
0.05(0.2) + 0.2(0.05)
200 mm
25 mm
75 mm
50 mm 75 mm
V
200 mm
Thus,
I =
1
(0.2)(0.053) + 0.2 (0.05)(0.175 - 0.1375)2
12
+
25 mm
1
(0.05)(0.23) + 0.05(0.2)(0.1375 - 0.1)2
12
= 63.5417(10 - 6) m4
Q for the shaded area shown in Fig. b is
Q = y¿A¿ = 0.0375 (0.05)(0.2) = 0.375(10 - 3) m3
Since there are two rows of nails q = 2a
q =
VQ
;
I
26.67 F =
F
2F
b =
= (26.67 F) N>m.
s
0.075
2000 C 0.375 (10 - 3) D
63.5417 (10 - 6)
F = 442.62 N
Thus, the shear stress developed in the nail is
tn =
F
442.62
=
= 35.22(106)Pa = 35.2 MPa
p
A
2
(0.004 )
4
Ans.
500
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7–39. A beam is constructed from three boards bolted
together as shown. Determine the shear force developed
in each bolt if the bolts are spaced s = 250 mm apart and the
applied shear is V = 35 kN.
25 mm
25 mm
100 mm 250 mm
2 (0.125)(0.25)(0.025) + 0.275 (0.35)(0.025)
y =
= 0.18676 m
2 (0.25)(0.025) + 0.35 (0.025)
I = (2) a
+
1
b (0.025)(0.253) + 2 (0.025)(0.25)(0.18676 - 0.125)2
12
V
1
(0.025)(0.35)3 + (0.025)(0.35)(0.275 - 0.18676)2
12
350 mm
s = 250 mm
= 0.270236 (10 - 3) m4
25 mm
-3
3
Q = y¿A¿ = 0.06176(0.025)(0.25) = 0.386(10 ) m
q =
35 (0.386)(10 - 3)
VQ
= 49.997 kN>m
=
I
0.270236 (10 - 3)
F = q(s) = 49.997 (0.25) = 12.5 kN
Ans.
*7–40. The double-web girder is constructed from two
plywood sheets that are secured to wood members at its top
and bottom. If each fastener can support 600 lb in single
shear, determine the required spacing s of the fasteners
needed to support the loading P = 3000 lb. Assume A is
pinned and B is a roller.
2 in.
2 in.
s
10 in.
A
4 ft
2 in.
2 in.
6 in.
0.5 in.
0.5 in.
Support Reactions: As shown on FBD.
Internal Shear Force: As shown on shear diagram, Vmax = 1500 lb.
Section Properties:
INA =
P
1
1
(7) A 183 B (6) A 103 B = 2902 in4
12
12
Q = y¿A¿ = 7(4)(6) = 168 in3
Shear Flow: Since there are two shear planes on the bolt, the allowable shear flow is
2(600)
1200
=
q =
.
s
s
VQ
q =
I
1500(168)
1200
=
s
2902
s = 13.8 in.
Ans.
501
4 ft
B
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•7–41. The double-web girder is constructed from two
plywood sheets that are secured to wood members at its top
and bottom. The allowable bending stress for the wood is
sallow = 8 ksi and the allowable shear stress is tallow = 3 ksi.
If the fasteners are spaced s = 6 in. and each fastener can
support 600 lb in single shear, determine the maximum load
P that can be applied to the beam.
2 in.
2 in.
s
10 in.
A
4 ft
2 in.
2 in.
6 in.
0.5 in.
0.5 in.
Support Reactions: As shown on FBD.
Internal Shear Force and Moment: As shown on shear and moment diagram,
Vmax = 0.500P and Mmax = 2.00P.
Section Properties:
INA =
P
1
1
(7) A 183 B (6) A 103 B = 2902 in4
12
12
Q = y2œ A¿ = 7(4)(6) = 168 in3
Qmax = ©y¿A¿ = 7(4)(6) + 4.5(9)(1) = 208.5 in3
Shear Flow: Assume bolt failure. Since there are two shear planes on the bolt, the
2(600)
= 200 lb>in.
allowable shear flow is q =
6
VQ
q =
I
0.500P(168)
200 =
2902
P = 6910 lb = 6.91 kip (Controls !)
Ans.
Shear Stress: Assume failure due to shear stress.
VQmax
It
0.500P(208.5)
3000 =
2902(1)
tmax = tallow =
P = 22270 lb = 83.5 kip
Bending Stress: Assume failure due to bending stress.
Mc
I
2.00P(12)(9)
8(103) =
2902
smax = sallow =
P = 107 ksi
502
4 ft
B
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7–42. The T-beam is nailed together as shown. If the nails
can each support a shear force of 950 lb, determine the
maximum shear force V that the beam can support and the
corresponding maximum nail spacing s to the nearest 18 in.
The allowable shear stress for the wood is tallow = 450 psi.
2 in.
s
The neutral axis passes through the centroid c of the cross-section as shown in Fig. a.
'
13(2)(12) + 6(12)(2)
© y A
y =
=
= 9.5 in.
©A
2(12) + 12(2)
I =
1
(2)(123) + 2(12)(9.5 - 6)2
12
+
= 884 in4
Refering to Fig. a, Qmax and QA are
Qmax = y1œ A1œ = 4.75(9.5)(2) = 90.25 in3
QA = y2œ A2œ = 3.5 (2)(12) = 84 in3
The maximum shear stress occurs at the points on the neutral axis where Q is
maximum and t = 2 in.
VQmax
;
It
450 =
V (90.25)
884 (2)
V = 8815.51 lb = 8.82 kip
Here, qallow =
F
950
=
lb>in. Then
s
s
VQA
;
qallow =
I
Ans.
8815.51(84)
950
=
s
884
s = 1.134 in = 1
12 in.
V
2 in.
1
(12)(23) + 12(2)(13 - 9.5)2
12
tallow =
s
12 in.
1
in
8
Ans.
503
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7–43. Determine the average shear stress developed in the
nails within region AB of the beam. The nails are located on
each side of the beam and are spaced 100 mm apart. Each
nail has a diameter of 4 mm. Take P = 2 kN.
P
2 kN/m
A
B
C
1.5 m
The FBD is shown in Fig. a.
As indicated in Fig. b, the internal shear force on the cross-section within region AB
is constant that is VAB = 5 kN.
1.5 m
100 mm
The neutral axis passes through centroid C of the cross section as shown in Fig. c.
'
0.18(0.04)(0.2) + 0.1(0.2)(0.04)
© y A
=
y =
©A
0.04(0.2) + 0.2(0.04)
40 mm
= 0.14 m
200 mm
1
I =
(0.04)(0.23) + 0.04(0.2)(0.14 - 0.1)2
12
1
+
(0.2)(0.043) + 0.2(0.04)(0.18 - 0.14)2
12
200 mm 20 mm
20 mm
= 53.333(10 - 6) m4
Q for the shaded area shown in Fig. d is
Q = y¿A¿ = 0.04(0.04)(0.2) = 0.32(10 - 3) m3
Since there are two rows of nail, q = 2 a
q =
VAB Q
;
I
20F =
F
F
b = 2a
b = 20F N>m.
s
0.1
5(103) C 0.32(10 - 3) D
53.333(10 - 6)
F = 1500 N
Thus, the average shear stress developed in each nail is
A tnail B avg =
F
1500
=
= 119.37(106)Pa = 119 MPa
p
Anail
2
(0.004 )
4
504
Ans.
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*7–44. The nails are on both sides of the beam and each
can resist a shear of 2 kN. In addition to the distributed
loading, determine the maximum load P that can be applied
to the end of the beam. The nails are spaced 100 mm apart
and the allowable shear stress for the wood is tallow = 3 MPa.
P
2 kN/m
A
B
C
1.5 m
1.5 m
100 mm
The FBD is shown in Fig. a.
40 mm
As indicated the shear diagram, Fig. b, the maximum shear occurs in region AB of
Constant value, Vmax = (P + 3) kN.
The neutral axis passes through Centroid C of the cross-section as shown in Fig. c.
'
0.18(0.04)(0.2) + 0.1(0.2)(0.04)
© y A
=
y =
©A
0.04(0.2) + 0.2(0.04)
= 0.14 m
I =
1
(0.04)(0.23) + 0.04(0.2)(0.14 - 0.1)2
12
1
+
(0.2)(0.043) + 0.2(0.04)(0.18 - 0.142)
12
Refering to Fig. d,
Qmax = y1œ A1œ = 0.07(0.14)(0.04) = 0.392(10 - 3) m3
QA = y2œ A2œ = 0.04(0.04)(0.2) = 0.32(10 - 3) m3
The maximum shear stress occurs at the points on Neutral axis where Q is maximum
and t = 0.04 m.
Vmax Qmax
;
It
3(106) =
(P + 3)(103) C 0.392(10 - 3) D
53.333(10 - 6)(0.04)
P = 13.33 kN
Since there are two rows of nails qallow = 2 a
qallow
Vmax QA
=
;
I
40 000 =
200 mm 20 mm
20 mm
= 53.333(10 - 6) m4
tallow =
200 mm
2(103)
F
d = 40 000 N>m.
b = 2c
s
0.1
(P + 3)(103) C 0.32(10 - 3) D
53.333(10 - 6)
P = 3.67 kN (Controls!)
Ans.
505
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7–44. Continued
506
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•7–45.
The beam is constructed from four boards which
are nailed together. If the nails are on both sides of the beam
and each can resist a shear of 3 kN, determine the maximum
load P that can be applied to the end of the beam.
3 kN
A
P
B
C
2m
2m
100 mm
30 mm
150 mm
30 mm
250 mm 30 mm
30 mm
Support Reactions: As shown on FBD.
Internal Shear Force: As shown on shear diagram, VAB = (P + 3) kN.
Section Properties:
INA =
1
1
(0.31) A 0.153 B (0.25) A 0.093 B
12
12
= 72.0 A 10 - 6 B m4
Q = y¿A¿ = 0.06(0.25)(0.03) = 0.450 A 10 - 3 B m3
Shear Flow: There are two rows of nails. Hence the allowable shear flow is
3(2)
= 60.0 kN>m.
q =
0.1
VQ
q =
I
(P
+ 3)(103)0.450(10 - 3)
60.0 A 103 B =
72.0(10 - 6)
P = 6.60 kN
Ans.
507
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7–47. The beam is made from four boards nailed together
as shown. If the nails can each support a shear force of
100 lb., determine their required spacing s and s if the beam
is subjected to a shear of V = 700 lb.
D
1 in.
1 in.
2 in.
s¿
s¿
s
A
C
s
10 in.
1 in.
10 in.
V
B
1.5 in.
Section Properties:
y =
©yA
0.5(10)(1) + 1.5(2)(3) + 6(1.5)(10)
=
©A
10(1) + 2(3) + 1.5(10)
= 3.3548 in
INA =
1
(10) A 13 B + 10(1)(3.3548 - 0.5)2
12
1
+
(2) A 33 B + 2(3)(3.3548 - 1.5)2
12
= 337.43 in4
QC = y1 ¿A¿ = 1.8548(3)(1) = 5.5645 in3
QD = y2 ¿A¿ = (3.3548 - 0.5)(10)(1) + 2 C (3.3548 - 1.5)(3)(1) D = 39.6774 in3
Shear Flow: The allowable shear flow at points C and D is qC =
100
, respectively.
qB =
s¿
VQC
qC =
I
700(5.5645)
100
=
s
337.43
s = 8.66 in.
VQD
qD =
I
700(39.6774)
100
=
s¿
337.43
100
and
s
Ans.
s¿ = 1.21 in.
Ans.
508
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*7–48. The box beam is constructed from four boards that
are fastened together using nails spaced along the beam
every 2 in. If each nail can resist a shear of 50 lb, determine
the greatest shear V that can be applied to the beam without
causing failure of the nails.
1 in.
12 in.
5 in.
V
2 in.
1 in.
6 in.
1 in.
y =
©yA
0.5 (12)(1) + 2 (4)(6)(1) + (6.5)(6)(1)
=
= 3.1 in.
©A
12(1) + 2(6)(1) + (6)(1)
I =
1
(12)(13) + 12(1)(3.1 - 0.5)2
12
+ 2a
+
1
b (1)(63) + 2(1)(6)(4 - 3.1)2
12
1
(6)(13) + 6(1)(6.5 - 3.1)2 = 197.7 in4
12
QB = y1œ A¿ = 2.6(12)(1) = 31.2 in3
qB =
V(31.2)
1 VQB
a
b =
= 0.0789 V
2
I
2(197.7)
qB s = 0.0789V(2) = 50
V = 317 lb (controls)
Ans.
QA = y2œ A¿ = 3.4(6)(1) = 20.4 in3
qA =
V(20.4)
1 VQA
a
b =
= 0.0516 V
2
I
2(197.7)
qA s = 0.0516V(2) = 50
V = 485 lb
509
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7–50. A shear force of V = 300 kN is applied to the box
girder. Determine the shear flow at points A and B.
90 mm
90 mm
C
A
D
200 mm
B
190 mm
V
200 mm
10 mm
180 mm
10 mm
The moment of inertia of the cross-section about the neutral axis is
I =
1
1
(0.2)(0.43) (0.18)(0.383) = 0.24359(10 - 3) m4
12
12
Refering to Fig. a Fig. b,
QA = y1œ A1œ = 0.195 (0.01)(0.19) = 0.3705 (10 - 3) m3
QB = 2yzœ A2œ + y3œ A3œ = 2 [0.1(0.2)(0.01)] + 0.195(0.01)(0.18) = 0.751(10 - 3) m3
Due to symmety, the shear flow at points A and A¿ , Fig. a, and at points B and B¿ ,
Fig. b, are the same. Thus
qA
3
-3
1 300(10 ) C 0.3705(10 ) D
1 VQA
s
b = c
= a
2
I
2
0.24359(10 - 3)
= 228.15(103) N>m = 228 kN>m
qB =
Ans.
3
-3
1 VQB
1 300(10 ) C 0.751(10 ) D
s
a
b = c
2
I
2
0.24359(10 - 3)
= 462.46(103) N>m = 462 kN>m
Ans.
510
100 mm
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7–51. A shear force of V = 450 kN is applied to the box
girder. Determine the shear flow at points C and D.
90 mm
90 mm
C
A
D
200 mm
B
190 mm
V
200 mm
10 mm
180 mm
10 mm
The moment of inertia of the cross-section about the neutral axis is
I =
1
1
(0.2)(0.43) (0.18)(0.383) = 0.24359(10 - 3) m4
12
12
Refering to Fig. a, due to symmetry ACœ = 0. Thus
QC = 0
Then refering to Fig. b,
QD = y1œ A1œ + y2œ A2œ = 0.195 (0.01)(0.09) + 0.15(0.1)(0.01)
= 0.3255(10 - 3) m3
Thus,
qC =
qD =
VQC
= 0
I
Ans.
450(103) C 0.3255(10 - 3) D
VQD
=
I
0.24359(10 - 3)
= 601.33(103) N>m = 601 kN>m
Ans.
100 mm
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*7–52. A shear force of V = 18 kN is applied to the
symmetric box girder. Determine the shear flow at A and B.
10 mm
30 mm
10 mm
A
100 mm
C
B
100 mm
150 mm
10 mm
10 mm
V
150 mm
10 mm 125 mm
10 mm
Section Properties:
INA =
1
1
(0.145) A 0.33 B (0.125) A 0.283 B
12
12
+ 2c
1
(0.125) A 0.013 B + 0.125(0.01) A 0.1052 B d
12
= 125.17 A 10 - 6 B m4
QA = y2œ A¿ = 0.145(0.125)(0.01) = 0.18125 A 10 - 3 B m3
QB = y1œ A¿ = 0.105(0.125)(0.01) = 0.13125 A 10 - 3 B m3
Shear Flow:
qA =
=
1 VQA
c
d
2
I
1 18(103)(0.18125)(10 - 3)
d
c
2
125.17(10 - 6)
Ans.
= 13033 N>m = 13.0 kN>m
qB =
=
1 VQB
c
d
2
I
3
-3
1 18(10 )(0.13125)(10 )
d
c
2
125.17(10 - 6)
= 9437 N>m = 9.44 kN>m
Ans.
512
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A shear force of V = 18 kN is applied to the box
girder. Determine the shear flow at C.
•7–53.
10 mm
30 mm
10 mm
A
100 mm
C
B
100 mm
150 mm
10 mm
10 mm
V
150 mm
10 mm 125 mm
10 mm
Section Properties:
INA =
1
1
(0.145) A 0.33 B (0.125) A 0.283 B
12
12
+2c
1
(0.125) A 0.013 B + 0.125(0.01) A 0.1052 B d
12
= 125.17 A 10 - 6 B m4
QC = ©y¿A¿
= 0.145(0.125)(0.01) + 0.105(0.125)(0.01) + 0.075(0.15)(0.02)
= 0.5375 A 10 - 3 B m3
Shear Flow:
qC =
=
1 VQC
c
d
2
I
3
-3
1 18(10 )(0.5375)(10 )
d
c
4
2
125.17(10 )
= 38648 N>m = 38.6 kN>m
Ans.
513
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7–54. The aluminum strut is 10 mm thick and has the cross
section shown. If it is subjected to a shear of, V = 150 N,
determine the shear flow at points A and B.
10 mm
2[0.005(0.03)(0.01)] + 2[0.03(0.06)(0.01)] + 0.055(0.04)(0.01)
= 0.027727 m
y =
2(0.03)(0.01) + 2(0.06)(0.01) + 0.04(0.01)
I = 2c
1
(0.03)(0.01)3 + 0.03(0.01)(0.027727 - 0.005)2 d
12
+ 2c
+
40 mm
10 mm
30 mm
1
(0.01)(0.06)3 + 0.01(0.06)(0.03 - 0.027727)2 d
12
B
A
V
40 mm
10 mm
30 mm
10 mm
1
(0.04)(0.01)3 + 0.04(0.01)(0.055 - 0.027727)2 = 0.98197(10 - 6) m4
12
yB ¿ = 0.055 - 0.027727 = 0.027272 m
yA ¿ = 0.027727 - 0.005 = 0.022727 m
QA = yA ¿A¿ = 0.022727(0.04)(0.01) = 9.0909(10 - 6) m3
QB = yB ¿A¿ = 0.027272(0.03)(0.01) = 8.1818(10 - 6) m3
qA =
VQA
150(9.0909)(10 - 6)
= 1.39 kN>m
=
I
0.98197(10 - 6)
Ans.
qB =
VQB
150(8.1818)(10 - 6)
= 1.25 kN>m
=
I
0.98197(10 - 6)
Ans.
7–55. The aluminum strut is 10 mm thick and has the cross
section shown. If it is subjected to a shear of V = 150 N,
determine the maximum shear flow in the strut.
y =
2[0.005(0.03)(0.01)] + 2[0.03(0.06)(0.01)] + 0.055(0.04)(0.01)
2(0.03)(0.01) + 2(0.06)(0.01) + 0.04(0.01)
10 mm
40 mm
B
A
= 0.027727 m
I = 2c
10 mm
1
(0.03)(0.01)3 + 0.03(0.01)(0.027727 - 0.005)2 d
12
30 mm
1
+ 2 c (0.01)(0.06)3 + 0.01(0.06)(0.03 - 0.027727)2 d
12
+
10 mm
1
(0.04)(0.01)3 + 0.04(0.01)(0.055 - 0.027727)2
12
= 0.98197(10 - 6) m4
Qmax = (0.055 - 0.027727)(0.04)(0.01) + 2[(0.06 - 0.027727)(0.01)]a
0.06 - 0.0277
b
2
= 21.3(10 - 6) m3
qmax =
V
40 mm
1 150(21.3(10 - 6))
1 VQmax
b = 1.63 kN>m
a
b = a
2
I
2 0.98197(10 - 6)
Ans.
514
30 mm
10 mm
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*7–56. The beam is subjected to a shear force of
V = 5 kip. Determine the shear flow at points A and B.
0.5 in.
C
5 in.
5 in. 0.5 in.
0.5 in.
2 in.
0.25(11)(0.5) + 2[4.5(8)(0.5)] + 6.25(10)(0.5)
©yA
y =
=
= 3.70946 in.
©A
11(0.5) + 2(8)(0.5) + 10(0.5)
A
D
8 in.
1
1
(11)(0.53) + 11(0.5)(3.70946 - 0.25)2 + 2c (0.5)(83) + 0.5(8)(4.5 - 3.70946)2 d
12
12
I =
+
0.5 in.
V
B
1
(10)(0.53) + 10(0.5)(6.25 - 3.70946)2
12
= 145.98 in4
œ
= 3.70946 - 0.25 = 3.45946 in.
yA
yBœ = 6.25 - 3.70946 = 2.54054 in.
œ
QA = yA
A¿ = 3.45946(11)(0.5) = 19.02703 in3
QB = yBœ A¿ = 2.54054(10)(0.5) = 12.7027 in3
qA =
1 VQA
1 5(103)(19.02703)
a
b = a
b = 326 lb>in.
2
I
2
145.98
Ans.
qB =
1 VQB
1 5(103)(12.7027)
a
b = a
b = 218 lb>in.
2
I
2
145.98
Ans.
•7–57.
The beam is constructed from four plates and is
subjected to a shear force of V = 5 kip. Determine the
maximum shear flow in the cross section.
0.5 in.
C
5 in.
5 in. 0.5 in.
y =
©yA
0.25(11)(0.5) + 2[4.5(8)(0.5)] + 6.25(10)(0.5)
=
= 3.70946 in.
©A
11(0.5) + 2(8)(0.5) + 10(0.5)
I =
1
1
(11)(0.53) + 11(0.5)(3.45952) + 2 c (0.5)(83) + 0.5(8)(0.79052) d
12
12
0.5 in.
2 in.
A
D
8 in.
V
+
1
(10)(0.53) + 10(0.5)(2.54052)
12
= 145.98 in4
Qmax = 3.4594 (11)(0.5) + 2[(1.6047)(0.5)(3.7094 - 0.5)]
= 24.177 in3
qmax =
1 VQmax
1 5(103)(24.177)
a
b = a
b
2
I
2
145.98
= 414 lb>in.
Ans.
515
0.5 in.
B
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7–58. The channel is subjected to a shear of V = 75 kN.
Determine the shear flow developed at point A.
30 mm
400 mm
A
200 mm
30 mm
V ⫽ 75 kN
30 mm
y =
©yA
0.015(0.4)(0.03) + 2[0.13(0.2)(0.03)]
=
= 0.0725 m
©A
0.4(0.03) + 2(0.2)(0.03)
I =
1
(0.4)(0.033) + 0.4(0.03)(0.0725 - 0.015)2
12
+ 2c
1
(0.03)(0.23) + 0.03(0.2)(0.13 - 0.0725)2 d = 0.12025(10 - 3) m4
12
œ
A¿ = 0.0575(0.2)(0.03) = 0.3450(10 - 3) m3
QA = yA
q =
qA =
VQ
I
75(103)(0.3450)(10 - 3)
0.12025(10 - 3)
= 215 kN>m
Ans.
7–59. The channel is subjected to a shear of V = 75 kN.
Determine the maximum shear flow in the channel.
30 mm
400 mm
A
V ⫽ 75 kN
30 mm
y =
©yA
0.015(0.4)(0.03) + 2[0.13(0.2)(0.03)]
=
©A
0.4(0.03) + 2(0.2)(0.03)
= 0.0725 m
1
I =
(0.4)(0.033) + 0.4(0.03)(0.0725 - 0.015)2
12
1
+ 2 c (0.03)(0.23) + 0.03(0.2)(0.13 - 0.0725)2 d
12
= 0.12025(10 - 3) m4
Qmax = y¿A¿ = 0.07875(0.1575)(0.03) = 0.37209(10 - 3) m3
qmax =
75(103)(0.37209)(10 - 3)
0.12025(10 - 3)
= 232 kN>m
Ans.
516
200 mm
30 mm
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*7–60. The angle is subjected to a shear of V = 2 kip.
Sketch the distribution of shear flow along the leg AB.
Indicate numerical values at all peaks.
A
5 in.
5 in.
45⬚ 45⬚
0.25 in.
Section Properties:
b =
0.25
= 0.35355 in.
sin 45°
h = 5 cos 45° = 3.53553 in.
INA = 2c
1
(0.35355) A 3.535533 B d = 2.604167 in4
12
Q = y¿A¿ = [0.25(3.53553) + 0.5y] a 2.5 -
y
b (0.25)
sin 45°
= 0.55243 - 0.17678y2
Shear Flow:
VQ
I
2(103)(0.55243 - 0.17678y2)
=
2.604167
q =
= {424 - 136y2} lb>in.
At y = 0,
Ans.
q = qmax = 424 lb>in.
Ans.
517
B
V
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•7–61.
The assembly is subjected to a vertical shear of
V = 7 kip. Determine the shear flow at points A and B and
the maximum shear flow in the cross section.
A
0.5 in.
B
V
2 in.
0.5 in.
0.5 in.
6 in.
6 in.
2 in.
0.5 in.
y =
©yA
(0.25)(11)(0.5) + 2(3.25)(5.5)(0.5) + 6.25(7)(0.5)
=
= 2.8362 in.
©A
0.5(11) + 2(0.5)(5.5) + 7(0.5)
I =
1
1
(11)(0.53) + 11(0.5)(2.8362 - 0.25)2 + 2a b(0.5)(5.53) + 2(0.5)(5.5)(3.25 - 2.8362)2
12
12
+
1
(7)(0.53) + (0.5)(7)(6.25 - 2.8362)2 = 92.569 in4
12
QA = y1 ¿A1 ¿ = (2.5862)(2)(0.5) = 2.5862 in3
QB = y2 ¿A2 ¿ = (3.4138)(7)(0.5) = 11.9483 in3
Qmax = ©y¿A¿ = (3.4138)(7)(0.5) + 2(1.5819)(3.1638)(0.5) = 16.9531 in3
q =
VQ
I
7(103)(2.5862)
= 196 lb>in.
92.569
1 7(103)(11.9483)
qB = a
b = 452 lb>in.
2
92.569
1 7(103)(16.9531)
qmax = a
b = 641 lb>in.
2
92.569
qA =
Ans.
Ans.
Ans.
518
0.5 in.
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7–62. Determine the shear-stress variation over the cross
section of the thin-walled tube as a function of elevation y and
show that t max = 2V>A, where A = 2prt. Hint: Choose a
differential area element dA = Rt du. Using dQ = y dA,
formulate Q for a circular section from u to (p - u) and show
that Q = 2R2t cos u, where cos u = 2R2 - y2>R.
ds
du
y
u
t
dA = R t du
dQ = y dA = yR t du
Here y = R sin u
Therefore dQ = R2 t sin u du
p-u
Q =
p-u
R2 t sin u du = R2 t(- cos u) |
Lu
u
2
= R t [- cos (p - u) - ( - cos u)] = 2R2 t cos u
dI = y2 dA = y2 R t du = R3 t sin2 u du
2p
I =
L0
2p
R3 t sin2 u du = R3 t
2p
=
t =
sin 2u
R3 t
[u ]
2
2 0
R3 t
[2p - 0] = pR3 t
2
VQ
V(2R2t cos u)
V cos u
=
=
3
It
pR t
pR t(2t)
Here cos u =
t =
=
L0
(1 - cos 2u)
du
2
2R2 - y2
R
V
2R2 - y2
pR2t
Ans.
tmax occurs at y = 0; therefore
tmax =
V
pR t
A = 2pRt; therefore
tmax =
2V
A
QED
519
R
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7–63. Determine the location e of the shear center,
point O, for the thin-walled member having the cross
section shown where b2 7 b1. The member segments have
the same thickness t.
t
h
e
b2
Section Properties:
I =
1
h 2
t h2
t h3 + 2 c (b1 + b2)ta b d =
C h + 6(b1 + b2) D
12
2
12
Q1 = y¿A¿ =
h
ht
(x )t =
x
2 1
2 1
Q2 = y¿A¿ =
h
ht
(x )t =
x
2 2
2 2
Shear Flow Resultant:
VQ1
q1 =
=
I
q2 =
VQ2
=
I
P A ht2 x1 B
P A ht2 x2 B
h C h + 6(b1 + b2) D
h C h + 6(b1 + b2) D
6P
t h2
12
C h + 6(b1 + b2) D
=
t h2
12
C h + 6(b1 + b2) D
=
6P
b1
(Ff)1 =
L0
q1 dx1 =
6P
x1
x2
b1
h C h + 6(b1 + b2) D L0
x1 dx1
3Pb21
=
b2
(Ff)2 =
L0
q2 dx2 =
h C h + 6(b1 + b2) D
6P
b2
h C h + 6(b1 + b2) D L0
x2 dx2
3Pb22
=
h C h + 6(b1 + b2) D
Shear Center: Summing moment about point A.
Pe = A Ff B 2 h - A Ff B 1 h
Pe =
e =
3Pb22
h C h + 6(b1 + b2) D
3(b22 - b21)
h + 6(b1 + b2)
(h) -
3Pb21
h C h + 6(b1 + b2) D
(h)
Ans.
Note that if b2 = b1, e = 0 (I shape).
520
b1
O
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*7–64. Determine the location e of the shear center,
point O, for the thin-walled member having the cross
section shown. The member segments have the same
thickness t.
b
d
45⬚
O
e
Section Properties:
I =
=
t
1
a
b (2d sin 45°)3 + 2 C bt(d sin 45°)2 D
12 sin 45°
td2
(d + 3b)
3
Q = y¿A¿ = d sin 45° (xt) = (td sin 45°)x
Shear Flow Resultant:
qf =
P(td sin 45°)x
VQ
3P sin 45°
=
=
x
td2
I
d(d + 3b)
(d
+
3b)
3
b
Ff =
L0
b
qfdx =
2
3P sin 45°
3b sin 45°
P
xdx =
d(d + 3b) L0
2d(d + 3b)
Shear Center: Summing moments about point A,
Pe = Ff(2d sin 45°)
Pe = c
e =
3b2 sin 45°
P d (2d sin 45°)
2d(d + 3b)
3b2
2(d + 3b)
Ans.
521
45⬚
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•7–65.
Determine the location e of the shear center,
point O, for the thin-walled member having a slit along its
side. Each element has a constant thickness t.
a
e
a
t
a
Section Properties:
I =
1
10 3
(2t)(2a)3 + 2 C at A a2 B D =
a t
12
3
Q1 = y1œ A¿ =
y
t
(yt) = y2
2
2
Q2 = ©y¿A¿ =
a
at
(at) + a(xt) =
(a + 2x)
2
2
Shear Flow Resultant:
q1 =
P A 12 y2 B
VQ1
3P 2
= 10 3 =
y
3
I
20a
a
t
3
P C at2 (a + 2x) D
VQ2
3P
=
=
(a + 2x)
q2 =
10 3
2
I
20a
a
t
3
a
(Fw)1 =
L0
a
q1 dy =
a
Ff =
L0
3P
P
y2 dy =
20
20a3 L0
a
q2 dx =
3P
3
(a + 2x)dx =
P
2
10
20a L0
Shear Center: Summing moments about point A.
Pe = 2(Fw)1 (a) + Ff(2a)
Pe = 2 a
e =
3
P
b a + a Pb2a
20
10
7
a
10
Ans.
522
O
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7–66. Determine the location e of the shear center,
point O, for the thin-walled member having the cross
section shown.
a
60⬚
O
a
60⬚
a
e
Summing moments about A.
Pe = F2 a
I =
13
ab
2
t
1
1
1
(t)(a)3 +
a
b (a)3 = t a3
12
12 sin 30°
4
q1 =
V(a)(t)(a>4)
1
4
q2 = q1 +
F2 =
=
3
ta
V
a
V(a>2)(t)(a>4)
1
4
ta
3
= q1 +
V
2a
V
4V
2 V
(a) + a b (a) =
a
3 2a
3
e =
223
a
3
Ans.
523
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7–67. Determine the location e of the shear center,
point O, for the thin-walled member having the cross
section shown. The member segments have the same
thickness t.
b
t
h
2
O
e
h
2
b
Shear Flow Resultant: The shear force flows through as Indicated by F1, F2, and F3
on FBD (b). Hence, The horizontal force equilibrium is not satisfied (©Fx Z 0). In
order to satisfy this equilibrium requirement. F1 and F2 must be equal to zero.
Shear Center: Summing moments about point A.
Pe = F2(0)
e = 0
Ans.
Also,
The shear flows through the section as indicated by F1, F2, F3.
+ ©F Z 0
However, :
x
To satisfy this equation, the section must tip so that the resultant of
:
:
:
:
F1 + F2 + F3 = P
Also, due to the geometry, for calculating F1 and F3, we require F1 = F3.
Hence, e = 0
Ans.
524
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*7–68. Determine the location e of the shear center,
point O, for the beam having the cross section shown. The
thickness is t.
1
—
r
2
e
r
O
I = (2) c
1
r 2
(t)(r>2)3 + (r>2)(t) a r + b d + Isemi-circle
12
4
= 1.583333t r3 + Isemi-circle
p>2
Isemi-circle =
p>2
2
L-p>2
(r sin u) t r du = t r3
L-p>2
sin2 u du
p
Isemi-circle = t r3 a b
2
Thus,
p
I = 1.583333t r3 + t r3 a b = 3.15413t r3
2
r
r
Q = a bt a + rb +
2
4
Lu
p>2
r sin u (t r du)
Q = 0.625 t r2 + t r2 cos u
q =
VQ
P(0.625 + cos u)t r2
=
I
3.15413 t r3
Summing moments about A:
p>2
Pe =
L-p>2
(q r du)r
p>2
Pe =
e =
Pr
(0.625 + cos u)du
3.15413 L-p>2
r (1.9634 + 2)
3.15413
e = 1.26 r
Ans.
525
1
—
r
2
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•7–69. Determine the location e of the shear center,
point O, for the thin-walled member having the cross
section shown. The member segments have the same
thickness t.
h1
h
O
e
h1
b
Summing moments about A.
Pe = F(h) + 2V(b)
h 2
1
1
(t)(h3) + 2b(t)a b +
(t)[h3 - (h - 2h1)3]
12
2
12
I =
=
(1)
t(h - 2h1)3
bth2
th3
+
6
2
12
Q1 = y¿A¿ =
t(hy - 2h1 y + y2)
1
(h - 2h1 + y)yt =
2
2
VQ
Pt(hy - 2h1 y + y2)
=
I
2I
q1 =
V =
L
h1
Pt
Pt hh1 2
2
(hy - 2h1 y + y2)dy =
c
- h31 d
2I L0
2I
2
3
q1 dy =
Q2 = ©y¿A¿ =
1
1
h
(h - h1)h1 t + (x)(t) = t[h1 (h - h1) + hx]
2
2
2
VQ2
Pt
=
(h (h - h1) + hx)
I
2I 1
q2 =
b
F =
L
q2 dx =
Pt
Pt
hb2
[h1 (h - h1) + hx]dx =
ah1 hb - h21 b +
b
2I L0
2I
2
From Eq, (1).
Pe =
h2b2
4
Pt
[h1 h2b - h21 hb +
+ hh21 b - h31 b]
2I
2
3
I =
t
(2h3 + 6bh2 - (h - 2h1)3)
12
e =
b(6h1 h2 + 3h2b - 8h31)
t
(6h1 h2b + 3h2b2 - 8h1 3b) =
12I
2h3 + 6bh2 - (h - 2h1)3
526
Ans.
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7–70. Determine the location e of the shear center, point O,
for the thin-walled member having the cross section shown.
t
r
a
O
a
e
Summing moments about A.
Pe = r
dF
L
dA = t ds = t r du
(1)
y = r sin u
dI = y2 dA = r2 sin2 u(t r du) = r3 t sin2 udu
p+a
I = r3 t
L
sin2 u du = r3 t
Lp - a
1 - cos 2u
du
2
=
sin 2u p + a
r3 t
(u )
2
2 p - a
=
sin 2(p + a)
sin 2(p - a)
r3 t
c ap + a b - ap - a bd
2
2
2
=
r3 t
r3 t
2 (2a - 2 sin a cos a) =
(2a - sin 2a)
2
2
dQ = y dA = r sin u(t r du) = r2 t sin u du
u
Q = r2 t
q =
L
u
Lp-a
sin u du = r2 t (- cos u)|
= r2 t( - cos u - cos a) = - r2 t(cos u + cos a)
p-a
P(- r2t)(cos u + cos a)
- 2P(cos u + cos a)
VQ
=
=
r3t
I
r(2a - sin 2a)
2 (2a - sin 2a)
dF =
L
q ds =
L
q r du
p+p
L
=
dF =
2P r
- 2P
(cos u + cos a) du =
(2a cos a - 2 sin a)
r(2a - sin 2a) Lp - a
2a - sin 2a
4P
(sin a - a cos a)
2a - sin 2a
4P
(sin a - a cos a) d
2a - sin 2a
4r (sin a - a cos a)
e =
2a - sin 2a
From Eq. (1); P e = r c
Ans.
527
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7–71. Sketch the intensity of the shear-stress distribution
acting over the beam’s cross-sectional area, and determine
the resultant shear force acting on the segment AB. The
shear acting at the section is V = 35 kip. Show that
INA = 872.49 in4.
C
V
8 in.
B
A
6 in.
Section Properties:
y =
4(8)(8) + 11(6)(2)
©yA
=
= 5.1053 in.
©A
8(8) + 6(2)
INA =
2 in.
1
(8) A 83 B + 8(8)(5.1053 - 4)2
12
+
1
(2) A 63 B + 2(6)(11 - 5.1053)2
12
= 872.49 in4 (Q.E.D)
Q1 = y1œ A¿ = (2.55265 + 0.5y1)(5.1053 - y1)(8)
= 104.25 - 4y21
Q2 = y2œ A¿ = (4.44735 + 0.5y2)(8.8947 - y2)(2)
= 79.12 - y22
Shear Stress: Applying the shear formula t =
tCB =
VQ
,
It
35(103)(104.25 - 4y21)
VQ1
=
It
872.49(8)
= {522.77 - 20.06y21} psi
At y1 = 0,
tCB = 523 psi
At y1 = - 2.8947 in.
tCB = 355 psi
tAB =
VQ2
35(103)(79.12 - y22)
=
It
872.49(2)
= {1586.88 - 20.06y22} psi
At y2 = 2.8947 in.
tAB = 1419 psi
Resultant Shear Force: For segment AB.
VAB =
L
tAB dA
0.8947 in
=
L2.8947 in
0.8947 in
=
L2.8947 in
3 in.
3 in.
A 1586.88 - 20.06y22 B (2dy)
A 3173.76 - 40.12y22 B dy
= 9957 lb = 9.96 kip
Ans.
528
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*7–72. The beam is fabricated from four boards nailed
together as shown. Determine the shear force each nail
along the sides C and the top D must resist if the nails are
uniformly spaced at s = 3 in. The beam is subjected to a
shear of V = 4.5 kip.
1 in.
1 in.
3 in.
10 in.
A
1 in.
12 in.
V
B
Section Properties:
y =
0.5(10)(1) + 2(4)(2) + 7(12)(1)
© yA
=
= 3.50 in.
©A
10(1) + 4(2) + 12(1)
INA =
1
(10) A 13 B + (10)(1)(3.50 - 0.5)2
12
+
1
(2) A 43 B + 2(4)(3.50 - 2)2
12
1
+
(1) A 123 B + 1(12)(7 - 3.50)2
12
= 410.5 in4
QC = y1œ A¿ = 1.5(4)(1) = 6.00 in2
QD = y2œ A¿ = 3.50(12)(1) = 42.0 in2
Shear Flow:
qC =
VQC
4.5(103)(6.00)
=
= 65.773 lb>in.
I
410.5
qD =
VQD
4.5(103)(42.0)
=
= 460.41 lb>in.
I
410.5
Hence, the shear force resisted by each nail is
FC = qC s = (65.773 lb>in.)(3 in.) = 197 lb
Ans.
FD = qD s = (460.41 lb>in.)(3 in.) = 1.38 kip
Ans.
529
1 in.
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•7–73.
The member is subjected to a shear force of
V = 2 kN. Determine the shear flow at points A, B, and C.
The thickness of each thin-walled segment is 15 mm.
200 mm
B
100 mm
A C
V ⫽ 2 kN
Section Properties:
y =
=
© yA
©A
0.0075(0.2)(0.015) + 0.0575(0.115)(0.03) + 0.165(0.3)(0.015)
0.2(0.015) + 0.115(0.03) + 0.3(0.015)
= 0.08798 m
1
(0.2) A 0.0153 B + 0.2(0.015)(0.08798 - 0.0075)2
12
1
+
(0.03) A 0.1153 B + 0.03(0.115)(0.08798 - 0.0575)2
12
1
+
(0.015) A 0.33 B + 0.015(0.3)(0.165 - 0.08798)2
12
INA =
= 86.93913 A 10 - 6 B m4
QA = 0
'
QB = y 1œ A¿ = 0.03048(0.115)(0.015) = 52.57705 A 10 - 6 B m3
Ans.
QC = ©y¿A¿
= 0.03048(0.115)(0.015) + 0.08048(0.0925)(0.015)
= 0.16424 A 10 - 3 B m3
Shear Flow:
qA =
VQA
= 0
I
Ans.
qB =
VQB
2(103)(52.57705)(10 - 6)
= 1.21 kN>m
=
I
86.93913(10 - 6)
Ans.
qC =
VQC
2(103)(0.16424)(10 - 3)
= 3.78 kN>m
=
I
86.93913(10 - 6)
Ans.
530
300 mm
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7–74. The beam is constructed from four boards glued
together at their seams. If the glue can withstand
75 lb>in., what is the maximum vertical shear V that the
beam can support?
3 in.
0.5 in.
Section Properties:
INA =
1
1
(1) A 103 B + 2c (4) A 0.53 B + 4(0.5) A 1.752 B d
12
12
3 in.
0.5 in.
= 95.667 in4
V
Q = y¿A¿ = 1.75(4)(0.5) = 3.50 in3
4 in.
Shear Flow: There are two glue joints in this case, hence the allowable shear flow is
2(75) = 150 lb>in.
q =
150 =
3 in.
0.5 in.
0.5 in.
VQ
I
V(3.50)
95.667
V = 4100 lb = 4.10 kip
Ans.
7–75. Solve Prob. 7–74 if the beam is rotated 90° from the
position shown.
3 in.
0.5 in.
3 in.
0.5 in.
V
3 in.
4 in.
0.5 in.
Section Properties:
INA =
1
1
(10) A 53 B (9) A 43 B = 56.167 in4
12
12
Q = y¿A¿ = 2.25(10)(0.5) = 11.25 in3
Shear Flow: There are two glue joints in this case, hence the allowable shear flow is
2(75) = 150 lb>in.
q =
150 =
VQ
I
V(11.25)
56.167
V = 749 lb
Ans.
531
0.5 in.