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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)
x
1
(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 +
3
(2000) - Cy = 0
5
+ ©F = 0;
:
x
-Cx +
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;
By (a) - P(a) = 0
+ c ©Fy = 0;
Cy - P - P = 0
+ ©F = 0;
:
x
Bx = 0
By = P
a
a
a
a
Cy = 2P
From the FBD of segment AB
a + ©MA = 0;
P(2a) - P(a) - MA = 0
+ c ©Fy = 0;
P - P = 0 (equilibrium is statisfied!)
MA = Pa
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 + 2xa 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;
1 4
a xb(x) - V = 0
2 3
2
V = e - x2 - 4 f kN
3
1 4
x
a xb(x)a b + 4x = 0
2 3
3
2
M = e - x3 - 4x f kN # m (2)
9
-4 -
a + ©M = 0; M +
(1)
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 (+) + wxa 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:
+ c ©Fy = 0;
V = 0
a + ©MNA = 0;
M + mx - mL = 0
M = m(L - x)
Shear and Moment Diagram:
6–27. Draw the shear and moment diagrams for the beam.
+ c ©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;
Cy - 150.0 - 225 = 0
+ ©F = 0;
:
x
Cx = 0
Cy = 375.0 lb
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
Ans.
w = 40.0 lb>ft
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
B
2m
a + ©MA = 0;
By (2) - 10.0(1) = 0
By = 5.00 kN
+ c ©Fy = 0;
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.
FR =
dA =
L
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
sina
w
2w0 L
p
xb 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
= 49.4 MPa
8(103)(0.01)
17.8133(10 - 6)
Ans.
= 4.49 MPa
361
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6–58. If the beam is subjected to an internal moment
of M = 100 kip # ft, determine the maximum tensile and
compressive bending stress in the beam.
3 in.
3 in.
6 in.
M
2 in.
1.5 in.
Section Properties: The neutral axis passes through centroid C of the cross section
as shown in Fig. a. The location of C is
©yA
y =
=
©A
4(8)(6) - 2 cp 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)a83 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 cp 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 # ina
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 # ina
1 ft
b = 101 kip # ft (controls)
12 in.
363
Ans.
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*6–60. The beam is constructed from four boards as
shown. If it is subjected to a moment of Mz = 16 kip # ft,
determine the stress at points A and B. Sketch a
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 = 2c (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 = 2c (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
6M
4M
=
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°)
sB =
=
= 25.9 ksi
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 =
4(103)(12)(6)
My
=
= 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
4 (0.045 )
= 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)
= 74.72 MPa = 74.7 MPa
Ans.
369
30 mm
(b)
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6–70. The simply supported truss is subjected to the central
distributed load. Neglect the effect of the diagonal lacing and
determine the absolute maximum bending stress in the truss.
The top member is a pipe having an outer diameter of 1 in.
3
and thickness of 16
in., and the bottom member is a solid rod
having a diameter of 12 in.
y =
6 ft
100 lb/ft
5.75 in.
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:
2.25 A 10 B (0.04)
Mmaxc
=
= 52.8 MPa
I
1.7038 A 10 - 6 B
3
sallow =
Ans.
*6–76. Determine the moment M that must be applied to
the beam in order to create a maximum stress of 80 MPa.Also
sketch the stress distribution acting over the cross section.
300 mm
20 mm
The moment of inertia of the 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
Mc
smax =
;
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 =
w
1
1
(8)(10.6)3 (7.7)(103) = 152.344 in4
12
12
8 ft
w
8 ft
8 in.
Mc
smax =
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 -
4
(4000) - 1200 = 0;
5
3
(4000) = 0;
5
FB = 4000 lb
Ay = 2000 lb
Ax = 2400 lb
6000(12) A 2 B
Mc
= 24(10)3
= 1
3
I
12 (2.5)(h )
h
smax =
5 ft
B
4 ft
4
F (3) - 1200(8) = 0;
5 B
3 ft
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:
t
7.5(12)a b
2
1.5 =
1
(12)t3
12
Mc
;
smax =
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
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
i
do
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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
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
i
do
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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
3
12 (b)(1.5b)
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
Ans.
= 19.1 ksi
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 4
12 a
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
Mmax c
smax =
=
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 =
©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)
I =
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 4
32 d
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 =
B
2m
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
Mmax c
I
100(103)(d)
p 4
32 d
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
0.375 4
4
4
p C A 1.25
2 B - A 2 B D = 0.11887 in
4
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
h
A Et
h2Ec
c =
=
Ec
2Et + 2Ec
1 +
A Et
[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 =
h2Ec
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 = -
(T)
Ans.
-14.14(12)(-8)
-14.14(12)(5)
+
1584
736
= -2.01 ksi = 2.01 ksi (C)
Ans.
Here, u = 180° + 45° = 225°
tan a =
tan a =
Iz
Iy
D
10 in.
M
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 =
= 0.057368 m = 57.4 mm
Iz =
Ans.
1
(0.4) A 0.023 B + (0.4)(0.02)(0.057368 - 0.01)2
12
+
1
(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
+
Iz
Iy
200(-0.2)
-480(-0.142632)
sA = -
+
-6
57.6014(10 )
0.366827(10 - 3)
= -1.298 MPa = 1.30 MPa (C)
200(0.2)
-480(0.057368)
sB = -
Ans.
+
-6
57.6014(10 )
0.366827(10 - 3)
= 0.587 MPa (T)
Ans.
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 =
= 0.057368 m = 57.4 mm
Iz =
Ans.
1
(0.4) A 0.023 B + (0.4)(0.02)(0.057368 - 0.01)2
12
1
(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
+
Iz
Iy
200(-0.2)
-480(-0.142632)
sA = -
+
57.6014(10 - 6)
0.366827(10 - 3)
= -1.298 MPa = 1.30 MPa (C) (Max)
200(0.2)
-480(0.057368)
sB = -
-6
57.6014(10 )
Ans.
+
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
My =
LA
z sx dA
=
LA
z(a + by + cz) dA
= a
dA + b
LA
y dA + c
LA
z dA + b
Mz =
LA
-y sx dA
=
-y(a + by + cz) dA
LA
= -a
LA
ydA - b
LA
LA
z dA
yz dA + c
y2 dA - c
LA
z2 dA
[2]
yz dA
[3]
LA
LA
[1]
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
x
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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) = 400 lb # ft = 4.80(103)lb # in.
Iy =
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.
10 mm
Internal Moment Components: Using method of section
y
200 mm
150 mm
10 mm
©Mz = 0;
Mz + P cos 30°(2) = 0
Mz = -1.732P
©My = 0;
My + P sin 30°(2) = 0
My = -1.00P
10 mm
A
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
(-1.732P)(0.085)
28.44583(10 - 6)
-1.00P(-0.1)
+
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.
10 mm
Internal Moment Components: Using method of sections
y
200 mm
150 mm
10 mm
©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
A
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
Iz
+
Iy
-600.0(-0.1)
-1039.32(0.085)
sA = -
-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
= -131 MPa = 131 MPa (C)(Max.)
Ans.
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
4 (d>2)
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)(-2.828)
(-2.121)(12)(1.155)
+
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 = 0.68218c
M(0.05)
7.7851(10 - 6)
d
M = 29215 N # m = 29.2 kN # m
415
50 mm
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*6–128. The composite beam is made of 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
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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
= 0.3655c
d = 13.3 ksi
I
30.8991
Ans.
At the seam,
saly = 0.6970 in. = n
Mmaxy
25.3125(12)(0.6970)
= 0.3655c
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 = 0.3655c
(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.
(Controls!)
Ans.
421
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6–134. The member has a brass core bonded to a steel
casing. If a couple moment of 8 kN # m is applied at its end,
determine the maximum bending stress in the member.
Ebr = 100 GPa, Est = 200 GPa.
8 kNm
3m
20 mm
100 mm
20 mm
n =
Ebr
100
=
= 0.5
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
423
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•6–137. If the beam is subjected to an internal moment of
M = 45 kN # m, determine the maximum bending stress
developed in the 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) + 0.2cp 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,
45 A 10 B (0.06185)
Mcst
=
= 154 MPa
I
18.08 A 10 - 6 B
3
(smax)st =
Ans.
For the aluminum alloy,
45 A 10 B (0.1882)
Mcal
= 0.3655 C
S = 171 MPa
I
18.08 A 10 - 6 B
3
(smax)al = n
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
=
= 8. It is required that both concrete and steel
in Fig. a. Here, n =
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 =
©yA
(1)(3)(2) + 3(0.6)(2) + 4.5(1.6875)(1)
=
= 1.9346 in.
©A
3(2) + 0.6(2) + 1.6875(1)
I =
1
1
(3)(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 =
©yA = 0;
29.0(103)
4.20(103)
8(h¿)a
(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
Ans.
My
29.0(103) 40(12)(13 - 5.517)
ba
b = a
b = 18.3 ksi
I
1358.781
4.20(103)
Ans.
(sst)max = na
10 kip
428
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
dA
1A 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
+ 1b 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 y ≤ dA
r LA 1 + r
r LA 1 + r
1A y dA = 0,
But,
y
y
Mr
dA
C A - ¢A dA ≤ - y
S
AI
r
+
y
r
LA
LA + y
y
Mr
dA
C
dA - y
S
AI LA r + y
r
LA + y
430
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6–143. Continued
y
=
y
As
r
y
Mr
dA
r
C
¢
¢
y ≤ dA y≤ S
AI LA 1 + r
r LA 1 + r
=
: 0
LA
¢
y
r
≤ dA = 0
1 + yr
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
=
dA
1A 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)
Ans.
= 224 kPa (C)
Ans.
No, because of localized stress concentration at the wall.
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
1.5 in.
8 in.
9(0.5)(2) + 8.6667 A 2 B (1)(2)
©rA
=
= 8.83333 in.
©A
2
1
r =
M
M
1(10)
dA
10
10
= 0.5 ln
+ c
cln
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
=
dA
1A 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)
Ans.
(Max)
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
=
dA
1A 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)
(Max)
Ans.
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
=
dA
1A 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
dA>r = (0.03)ln
0.625
0.630
0.605
+ (0.005)ln
+ (0.03)ln
= 0.650625(10 - 3) in.
0.6
0.605
0.625
A = 2(0.005)(0.03) + (0.02)(0.005) = 0.4(10 - 3) 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
(M)(6.25)
18(103) = 2.60c 1
d
3
12 (1)(12.5)
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 1
40(0.01)
3
12 (0.007)(0.02 )
R
K = 1.45
Stress Concentration Factor: From the graph in the text
w
80
r
with
=
= 4 and K = 1.45, then = 0.25.
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 1
17.5(0.01)
3
12 (0.007)(0.02 )
R
= 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.
b =
20 in.
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
20P(0.625)
36 = 1.92 c 1
d
3
12 (0.5)(1.25)
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
= 1.92c 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 1
M(0.005)
3
12 (0.015)(0.01 )
R
M = 41.7 N # m (Controls !)
Ans.
For the larger section
smax = sallow = K
Mc
;
I
200 A 106 B = 1.75 B 1
M(0.015)
3
12 (0.015)(0.03 )
R
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
3h 4bt(h - t) + t(h - 2t)2
c
d
2 bh3 - (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 =
MP
2.75a3sY
=
= 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.
a
1
1
(2) A 63 B +
(4) A 23 B = 38.667 in4
INA =
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 =
74.25 sY
MP
=
= 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
Mc
;
smax =
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 c2 a
4ro
4ri
b d - T2 c2 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 bsY =
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 48 B
sYI
b h2
=
s
=
h
c
24 Y
2
bh3
MY =
Shape factor:
k =
bh2
12 sY
Mp
MY
=
bh2
24 sY
= 2
Ans.
*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 hb (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
sYI
;
Mmax =
c
9 ft
8 in.
25(1866.67)
27w0(12) =
8
Ans.
w0 = 18.0 kip>ft
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
y C 20 tan 1 (0.0225y) D (2dy)
-
L0
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
20 mm
1
INA =
(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
Ans.
= 3.43 MPa (T)
= 1.62 MPa (C)
Ans.
464
250 mm
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*6–192. Determine the bending stress distribution in
the beam at section a–a. Sketch the distribution in three
dimensions acting over the cross section.
80 N
80 N
a
a
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
•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 =
75 mm
Ans.
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 :
20 mm
Mc
(sw)max =
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 yc 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 - 2xa b - M = 0
2
M = -x2 + 20x - 166
Ans.
6–199. Draw the shear and moment diagrams for the shaft
if it is subjected to the vertical loadings of the belt, gear, and
flywheel. The bearings at A and B exert only vertical
reactions on the shaft.
300 N
450 N
A
B
200 mm
400 mm
300 mm
200 mm
150 N
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 4
12 a
-Msin u ( - a2)
+
1 4
12 a
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 =
VQ
It
tmax =
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.
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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
12.8(10 ) C 4(36 - y ) D
VQ
=
It
1152 (8)
3
t =
2
= 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:
60 mm
1
1
INA =
(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
2y2c2 - 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
( c )(2 2c - y )
4
The maximum shear stress occur when y = 0
tmax =
4V
3 p c2
tavg =
V
V
=
A
p c2
4V
The faector =
tmax
4
3 pc2
= V =
tavg
3
2
pc
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 = 2c a
2h
x
a
1
2
2
b (x)(y) a h - yb d
2
3
3
4h2
2x
b (x2)a 1 b
a
3a
t = 2x
V(4h2>3a)(x2)(1 - a )
VQ
=
It
((1>36)(a)(h3))(2x)
2x
t =
t =
24V(x - a2 x2)
a2h
24V
4
dt
= 2 2 a 1 - 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
p
I = r4 = (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 y
3
=
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.
488
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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.
27.5(10 ) C 0.216(10 ) D
Vmax Qmax
=
It
27.0(10 - 6)(0.03)
3
tmax =
-3
= 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
6 ft
6 ft
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
B
6 ft
6 in.
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
6 ft
6 ft
The FBD is shown in Fig. a.
The shear diagram is shown in Fig. b, Vmax = 18.0 kip.
B
0.75 in.
6 ft
6 in.
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
VA 2 B
VQmax
3P
tmax =
=
=
2
3
It
4by¿
A by¿ B (b)
y¿ 3b
3
However,
A¿ = 2by¿
tmax =
3P
‚
2A¿
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:
-
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
INA =
(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
75 mm
50 mm 75 mm
25 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
Vmax QA
qallow =
;
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
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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
3
-3
1 300(10 ) C 0.3705(10 ) D
1 VQA
s
b = c
qA = 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)
= 13033 N>m = 13.0 kN>m
qB =
=
Ans.
1 VQB
c
d
2
I
1 18(103)(0.13125)(10 - 3)
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
+2 c
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
1 18(103)(0.5375)(10 - 3)
d
c
2
125.17(10 - 4)
= 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
+ 2c (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]a2.5 -
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)
b = 641 lb>in.
qmax = a
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 + 2c(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:
P A ht2 x1 B
VQ1
6P
q1 =
= 2
=
x1
th
I
C h + 6(b1 + b2) D h C h + 6(b1 + b2) D
12
q2 =
P A ht2 x2 B
VQ2
6P
= 2
=
x2
th
I
h
h
+
6(b
h
+
6(b
+
b
)
C
C
1 + b2) D
1
2 D
12
b1
(Ff)1 =
L0
q1 dx1 =
6P
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 Pb 2a
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)
q2 = q1 +
F2 =
=
1
3
4ta
V
a
V(a>2)(t)(a>4)
1
3
4ta
= 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)ar + 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 b t 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 =
h
q1 dy =
1
Pt
Pt hh1 2
2
(hy - 2h1 y + y2)dy =
c
- h31 d
2I L0
2I
2
3
Q2 = ©y¿A¿ =
1
1
h
(h - h1)h1 t + (x)(t) = t[h1 (h - h1) + hx]
2
2
2
V =
L
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 =
u
sin u du = r2 t (-cos u)|
Lp-a
= 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)
(2a - sin 2a)
2
L
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
VQ1
35(103)(104.25 - 4y21)
=
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
'
Q = y œ A¿ = 0.03048(0.115)(0.015) = 52.57705 A 10 - 6 B m3
B
Ans.
1
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.
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8–1. A spherical gas tank has an inner radius of r = 1.5 m.
If it is subjected to an internal pressure of p = 300 kPa,
determine its required thickness if the maximum normal
stress is not to exceed 12 MPa.
pr
;
2t
sallow =
12(106) =
300(103)(1.5)
2t
t = 0.0188 m = 18.8 mm
Ans.
8–2. A pressurized spherical tank is to be made of
0.5-in.-thick steel. If it is subjected to an internal pressure
of p = 200 psi, determine its outer radius if the maximum
normal stress is not to exceed 15 ksi.
sallow =
pr
;
2t
15(103) =
200 ri
2(0.5)
ri = 75 in.
ro = 75 in. + 0.5 in. = 75.5 in.
Ans.
8–3. The thin-walled cylinder can be supported in one of
two ways as shown. Determine the state of stress in the wall
of the cylinder for both cases if the piston P causes the
internal pressure to be 65 psi. The wall has a thickness of
0.25 in. and the inner diameter of the cylinder is 8 in.
P
Case (a):
s1 =
pr
;
t
s1 =
65(4)
= 1.04 ksi
0.25
Ans.
s2 = 0
Ans.
Case (b):
s1 =
pr
;
t
s1 =
65(4)
= 1.04 ksi
0.25
Ans.
s2 =
pr
;
2t
s2 =
65(4)
= 520 psi
2(0.25)
Ans.
532
P
8 in.
8 in.
(a)
(b)
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*8–4. The tank of the air compressor is subjected to an
internal pressure of 90 psi. If the internal diameter of
the tank is 22 in., and the wall thickness is 0.25 in.,
determine the stress components acting at point A. Draw a
volume element of the material at this point, and show the
results on the element.
Hoop Stress for Cylindrical Vessels: Since
A
11
r
=
= 44 7 10, then thin wall
t
0.25
analysis can be used. Applying Eq. 8–1
s1 =
pr
90(11)
=
= 3960 psi = 3.96 ksi
t
0.25
Ans.
Longitudinal Stress for Cylindrical Vessels: Applying Eq. 8–2
s2 =
pr
90(11)
=
= 1980 psi = 1.98 ksi
2t
2(0.25)
Ans.
•8–5.
The spherical gas tank is fabricated by bolting together
two hemispherical thin shells of thickness 30 mm. If the gas
contained in the tank is under a gauge pressure of 2 MPa,
determine the normal stress developed in the wall of the tank
and in each of the bolts.The tank has an inner diameter of 8 m
and is sealed with 900 bolts each 25 mm in diameter.
Normal Stress: Since
4
r
=
= 133.33 7 10, thin-wall analysis is valid. For the
t
0.03
spherical tank’s wall,
s =
Referring
pr
2(4)
=
= 133 MPa
2t
2(0.03)
to
the
free-body
diagram
p 2
6
P = pA = 2 A 10 B c A 8 B d = 32p A 10 B N. Thus,
4
Ans.
shown
in
Fig.
a,
6
+ c ©Fy = 0;
32p A 106 B - 450Pb - 450Pb = 0
Pb = 35.56 A 103 B p N
The normal stress developed in each bolt is then
35.56 A 10 B p
Pb
=
= 228 MPa
p
Ab
A 0.0252 B
4
3
sb =
Ans.
533
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8–6. The spherical gas tank is fabricated by bolting
together two hemispherical thin shells. If the 8-m inner
diameter tank is to be designed to withstand a gauge pressure
of 2 MPa, determine the minimum wall thickness of the
tank and the minimum number of 25-mm diameter bolts
that must be used to seal it. The tank and the bolts are made
from material having an allowable normal stress of 150 MPa
and 250 MPa, respectively.
Normal Stress: For the spherical tank’s wall,
sallow =
pr
2t
150 A 106 B =
2 A 106 B (4)
2t
t = 0.02667 m = 26.7 mm
Since
Ans.
r
4
=
= 150 7 10, thin-wall analysis is valid.
t
0.02667
Referring
the
free-body
diagram
p
P = pA = 2 A 106 B c A 82 B d = 32p A 106 B N. Thus,
4
+ c ©Fy = 0;
to
32p A 106 B n =
shown
in
Fig.
a,
n
n
(P )
- (Pb)allow = 0
2 b allow
2
32p A 106 B
(1)
(Pb)allow
The allowable tensile force for each bolt is
(Pb)allow = sallowAb = 250 A 106 B c
p
A 0.0252 B d = 39.0625 A 103 B pN
4
Substituting this result into Eq. (1),
n =
32p A 106 B
39.0625p A 103 B
= 819.2 = 820
Ans.
534
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8–7. A boiler is constructed of 8-mm thick steel plates that
are fastened together at their ends using a butt joint
consisting of two 8-mm cover plates and rivets having a
diameter of 10 mm and spaced 50 mm apart as shown. If the
steam pressure in the boiler is 1.35 MPa, determine (a) the
circumferential stress in the boiler’s plate apart from
the seam, (b) the circumferential stress in the outer cover plate
along the rivet line a–a, and (c) the shear stress in the rivets.
a
8 mm
50 mm
a)
s1 =
pr
1.35(106)(0.75)
=
= 126.56(106) = 127 MPa
t
0.008
Ans.
126.56 (106)(0.05)(0.008) = s1 ¿(2)(0.04)(0.008)
b)
s1 ¿ = 79.1 MPa
Ans.
c) From FBD(a)
+ c ©Fy = 0;
Fb - 79.1(106)[(0.008)(0.04)] = 0
Fb = 25.3 kN
(tavg)b =
Fb
25312.5
- p
= 322 MPa
2
A
4 (0.01)
Ans.
535
0.75 m
a
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*8–8. The gas storage tank is fabricated by bolting together
two half cylindrical thin shells and two hemispherical shells
as shown. If the tank is designed to withstand a pressure
of 3 MPa, determine the required minimum thickness of
the cylindrical and hemispherical shells and the minimum
required number of longitudinal bolts per meter length at
each side of the cylindrical shell. The tank and the 25 mm
diameter bolts are made from material having an allowable
normal stress of 150 MPa and 250 MPa, respectively. The
tank has an inner diameter of 4 m.
Normal Stress: For the cylindrical portion of the tank, the hoop stress is twice as
large as the longitudinal stress.
sallow =
pr
;
t
150 A 106 B =
3 A 106 B (2)
tc
tc = 0.04 m = 40 mm
For the hemispherical cap,
pr
;
sallow =
t
150 A 10 B =
6
Ans.
3 A 106 B (2)
2ts
ts = 0.02 m = 20 mm
Since
Ans.
r
6 10, thin-wall analysis is valid.
t
Referring to the free-body diagram of the per meter length of the cylindrical
portion, Fig. a, where P = pA = 3 A 106 B [4(1)] = 12 A 106 B N, we have
+ c ©Fy = 0;
12 A 106 B - nc(Pb)allow - nc(Pb)allow = 0
nc =
6 A 106 B
(1)
(Pb)allow
The allowable tensile force for each bolt is
(Pb)allow = sallowAb = 250 A 106 B c
p
A 0.0252 B d = 122.72 A 103 B N
4
Substituting this result into Eq. (1),
nc = 48.89 = 49 bolts>meter
Ans.
536
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•8–9.
The gas storage tank is fabricated by bolting together
two half cylindrical thin shells and two hemispherical shells
as shown. If the tank is designed to withstand a pressure of
3 MPa, determine the required minimum thickness of the
cylindrical and hemispherical shells and the minimum
required number of bolts for each hemispherical cap. The
tank and the 25 mm diameter bolts are made from material
having an allowable normal stress of 150 MPa and 250 MPa,
respectively. The tank has an inner diameter of 4 m.
Normal Stress: For the cylindrical portion of the tank, the hoop stress is twice as
large as the longitudinal stress.
sallow =
pr
;
t
150 A 106 B =
3 A 106 B (2)
tc
tc = 0.04 m = 40 mm
For the hemispherical cap,
sallow =
pr
;
t
150 A 106 B =
Ans.
3 A 106 B (2)
2ts
ts = 0.02 m = 20 mm
Since
Ans.
r
6 10, thin-wall analysis is valid.
t
The allowable tensile force for each bolt is
(Pb)allow = sallowAb = 250 A 106 B c
p
A 0.0252 B d = 122.72 A 103 B N
4
Referring to the free-body diagram of the hemispherical cap, Fig. b, where
p
P = pA = 3 A 106 B c A 42 B d = 12p A 106 B N,
4
+ ©F = 0;
:
x
12p A 106 B ns =
ns
ns
(Pb)allow (Pb)allow = 0
2
2
12p A 106 B
(1)
(Pb)allow
Substituting this result into Eq. (1),
ns = 307.2 = 308 bolts
Ans.
537
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8–10. A wood pipe having an inner diameter of 3 ft is
bound together using steel hoops each having a crosssectional area of 0.2 in2. If the allowable stress for the hoops
is sallow = 12 ksi, determine their maximum spacing s along
the section of pipe so that the pipe can resist an internal
gauge pressure of 4 psi. Assume each hoop supports the
pressure loading acting along the length s of the pipe.
s
4 psi
4 psi
s
s
Equilibrium for the steel Hoop: From the FBD
+ ©F = 0;
:
x
P = 72.0s
2P - 4(36s) = 0
Hoop Stress for the Steel Hoop:
s1 = sallow =
12(103) =
P
A
72.0s
0.2
s = 33.3 in.
Ans.
8–11. The staves or vertical members of the wooden tank
are held together using semicircular hoops having a
thickness of 0.5 in. and a width of 2 in. Determine the normal
stress in hoop AB if the tank is subjected to an internal
gauge pressure of 2 psi and this loading is transmitted
directly to the hoops. Also, if 0.25-in.-diameter bolts are used
to connect each hoop together, determine the tensile stress
in each bolt at A and B. Assume hoop AB supports the
pressure loading within a 12-in. length of the tank as shown.
18 in.
6 in.
6 in.
FR = 2(36)(12) = 864 lb
©F = 0; 864 - 2F = 0; F = 432 lb
sh =
sb =
F
432
=
= 432 psi
Ah
0.5(2)
Ans.
F
432
=
= 8801 psi = 8.80 ksi
p
Ab
(0.25)2
4
Ans.
538
12 in.
A
B
12 in.
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*8–12. Two hemispheres having an inner radius of 2 ft and
wall thickness of 0.25 in. are fitted together, and the inside
gauge pressure is reduced to -10 psi. If the coefficient
of static friction is ms = 0.5 between the hemispheres,
determine (a) the torque T needed to initiate the rotation
of the top hemisphere relative to the bottom one, (b) the
vertical force needed to pull the top hemisphere off
the bottom one, and (c) the horizontal force needed to slide
the top hemisphere off the bottom one.
0.25 in.
2 ft
Normal Pressure: Vertical force equilibrium for FBD(a).
+ c ©Fy = 0;
10 C p(242) D - N = 0
N = 5760p lb
The Friction Force: Applying friction formula
Ff = ms N = 0.5(5760p) = 2880p lb
a) The Required Torque: In order to initiate rotation of the two hemispheres
relative to each other, the torque must overcome the moment produced by the
friction force about the center of the sphere.
T = Ffr = 2880p(2 + 0.125>12) = 18190 lb # ft = 18.2 kip # ft
Ans.
b) The Required Vertical Force: In order to just pull the two hemispheres apart, the
vertical force P must overcome the normal force.
P = N = 5760p = 18096 lb = 18.1 kip
Ans.
c) The Required Horizontal Force: In order to just cause the two hemispheres to
slide relative to each other, the horizontal force F must overcome the friction force.
F = Ff = 2880p = 9048 lb = 9.05 kip
Ans.
•8–13. The 304 stainless steel band initially fits snugly around
the smooth rigid cylinder. If the band is then subjected to a
nonlinear temperature drop of ¢T = 20 sin2 u °F, where u is
in radians, determine the circumferential stress in the band.
1
in.
64
10 in.
Compatibility: Since the band is fixed to a rigid cylinder (it does not deform under
load), then
dF - dT = 0
2p
P(2pr)
a¢Trdu = 0
AE
L0
2pr P
a b = 20ar
E A
L0
2p
s = 10a
E c
L0
2p
sin2 udu
however,
P
= sc
A
2p
(1 - cos 2u)du
sc = 10aE
= 10(9.60) A 10 - 6 B 28.0 A 103 B = 2.69 ksi
Ans.
539
u
1 in.
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8–14. The ring, having the dimensions shown, is placed
over a flexible membrane which is pumped up with a
pressure p. Determine the change in the internal radius of
the ring after this pressure is applied. The modulus of
elasticity for the ring is E.
ro
ri
w
p
Equilibrium for the Ring: Form the FBD
+ ©F = 0;
:
x
2P - 2pri w = 0
P = pri w
Hoop Stress and Strain for the Ring:
s1 =
pri w
pri
P
=
=
rs - ri
A
(rs - ri)w
Using Hooke’s Law
e1 =
However,
e1 =
pri
s1
=
E
E(rs - ri)
[1]
2p(ri)1 - 2pri
(ri)1 - ri
dri
=
=
.
ri
ri
2pr
Then, from Eq. [1]
pri
dri
=
ri
E(rs - ri)
dri =
pr2i
E(rs - ri)
Ans.
540
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8–15. The inner ring A has an inner radius r1 and outer
radius r2. Before heating, the outer ring B has an inner
radius r3 and an outer radius r4, and r2 7 r3. If the outer ring
is heated and then fitted over the inner ring, determine the
pressure between the two rings when ring B reaches the
temperature of the inner ring. The material has a modulus of
elasticity of E and a coefficient of thermal expansion of a.
r1
A
Equilibrium for the Ring: From the FBD
+ ©F = 0;
:
x
P = priw
2P - 2priw = 0
Hoop Stress and Strain for the Ring:
s1 =
priw
pri
P
=
=
ro - ri
A
(ro - ri)w
Using Hooke’s law
e1 =
However,
e1 =
pri
s1
=
E
E(ro - ri)
[1]
2p(ri)1 - 2pri
(ri)1 - ri
dri
=
=
.
ri
ri
2pr
Then, from Eq. [1]
pri
dri
=
ri
E(ro - ri)
dri =
pr2i
E(ro - ri)
Compatibility: The pressure between the rings requires
dr2 + dr3 = r2 - r3
[2]
From the result obtained above
dr2 =
pr22
E(r2 - r1)
dr3 =
pr23
E(r4 - r3)
Substitute into Eq. [2]
pr22
pr23
+
= r2 - r3
E(r2 - r1)
E(r4 - r3)
p =
r4
r2
E(r2 - r3)
Ans.
r22
r23
+
r2 - r1
r4 - r3
541
r3
B
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*8–16. The cylindrical tank is fabricated by welding a
strip of thin plate helically, making an angle u with the
longitudinal axis of the tank. If the strip has a width w and
thickness t, and the gas within the tank of diameter d is
pressured to p, show that the normal stress developed along
the strip is given by su = (pd>8t)(3 - cos 2u).
w
u
Normal Stress:
sh = s1 =
pr
p(d>2)
pd
=
=
t
t
2t
sl = s2 =
p(d>2)
pd
pr
=
=
2t
2t
4t
Equilibrium: We will consider the triangular element cut from the strip
shown in Fig. a. Here,
Ah = (w sin u)t
and
Thus,
Al = (w cos u)t.
pd
pwd
and
(w sin u)t =
sin u
Fh = shAh =
2t
2
pwd
pd
(w cos u)t =
cos u.
4t
4
Fl = slAl =
Writing the force equation of equilibrium along the x¿ axis,
©Fx¿ = 0;
c
pwd
pwd
sin u d sin u + c
cos u d cos u - Nu = u
2
4
Nu =
pwd
A 2 sin2 u + cos2 u B
4
However, sin2 u + cos2 u = 1. This equation becomes
Nu =
Also, sin2 u =
pwd
A sin2 u + 1 B
4
1
(1 - cos 2u), so that
2
pwd
Nu =
(3 - cos 2u)
8
Since Au = wt, then
Nu
=
su =
Au
su =
pwd
(3 - cos 2u)
8
wt
pd
(3 - cos 2u)
8t
(Q.E.D.)
542
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8–17. In order to increase the strength of the pressure vessel,
filament winding of the same material is wrapped around the
circumference of the vessel as shown. If the pretension in the
filament is T and the vessel is subjected to an internal pressure
p, determine the hoop stresses in the filament and in the wall
of the vessel. Use the free-body diagram shown, and assume
the filament winding has a thickness t and width w for a
corresponding length of the vessel.
L
w
s1
t¿
T
p
t
s1
T
Normal Stress in the Wall and Filament Before the Internal Pressure is Applied:
The entire length w of wall is subjected to pretension filament force T. Hence, from
equilibrium, the normal stress in the wall at this state is
2T - (sl ¿)w (2wt) = 0
(sl ¿)w =
T
wt
and for the filament the normal stress is
(sl ¿)fil =
T
wt¿
Normal Stress in the Wall and Filament After the Internal Pressure is Applied: The
stress in the filament becomes
sfil = sl + (sl ¿)fil =
pr
T
+
(t + t¿)
wt¿
Ans.
sw = sl - (sl ¿)w =
pr
T
(t + t¿)
wt
Ans.
And for the wall,
8–18. The vertical force P acts on the bottom of the plate
having a negligible weight. Determine the shortest distance
d to the edge of the plate at which it can be applied so that
it produces no compressive stresses on the plate at section
a–a. The plate has a thickness of 10 mm and P acts along the
center line of this thickness.
300 mm
a
a
200 mm
500 mm
sA = 0 = sa - sb
0 =
0 =
P
Mc
A
I
P
(0.2)(0.01)
d
P(0.1 - d)(0.1)
1
3
12 (0.01)(0.2 )
P
P(-1000 + 15000 d) = 0
d = 0.0667 m = 66.7 mm
Ans.
543
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8–19. Determine the maximum and minimum normal
stress in the bracket at section a–a when the load is applied
at x = 0.
100 kN
15 mm
x
15 mm
200 mm
150 mm
a
Consider the equilibrium of the FBD of the top cut segment in Fig. a,
+ c ©Fy = 0;
N - 100 = 0
a + ©MC = 0;
100(0.1) - M = 0
A = 0.2(0.03) = 0.006 m2
I =
N = 100 kN
M = 10 kN # m
1
(0.03)(0.23) = 20.0(10 - 6) m4
12
The normal stress developed is the combination of axial and bending stress. Thus,
s =
My
N
;
A
I
For the left edge fiber, y = C = 0.1 m. Then
sL = -
100(103)
10(103)(0.1)
0.006
20.0(10 - 6)
= -66.67(106) Pa = 66.7 MPa (C) (Max)
Ans.
For the right edge fiber, y = 0.1 m. Then
sR = -
100 (103)
10(103)(0.1)
= 33.3 MPa (T)
+
0.006
20.0(10 - 6)
Ans.
544
a
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*8–20. Determine the maximum and minimum normal
stress in the bracket at section a–a when the load is applied
at x = 300 mm.
100 kN
15 mm
x
15 mm
200 mm
150 mm
a
Consider the equilibrium of the FBD of the top cut segment in Fig. a,
+ c ©Fy = 0;
N - 100 = 0
a + ©MC = 0;
M - 100(0.2) = 0
A = 0.2 (0.03) = 0.006 m2
I =
N = 100 kN
M = 20 kN # m
1
(0.03)(0.23) = 20.0(10 - 6) m4
12
The normal stress developed is the combination of axial and bending stress. Thus,
s =
My
N
;
A
I
For the left edge fiber, y = C = 0.1 m. Then
sC = -
100(103)
20.0(103)(0.1)
+
0.006
20.0(10 - 6)
= 83.33(106) Pa = 83.3 MPa (T)(Min)
Ans.
For the right edge fiber, y = C = 0.1 m. Thus
sR = -
100(103)
20.0(103)(0.1)
0.006
20.0(10 - 6)
= 117 MPa
Ans.
545
a
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•8–21.
The coping saw has an adjustable blade that is
tightened with a tension of 40 N. Determine the state of
stress in the frame at points A and B.
8 mm
75 mm
A
3 mm
8 mm
3 mm
B
100 mm
50 mm
sA = -
sB =
4(0.004)
P
Mc
40
= 123 MPa
+
= + 1
3
A
I
(0.008)(0.003)
(0.003)(0.008)
12
2(0.004)
Mc
= 62.5 MPa
= 1
3
I
12 (0.003)(0.008)
Ans.
Ans.
8–22. The clamp is made from members AB and AC,
which are pin connected at A. If it exerts a compressive
force at C and B of 180 N, determine the maximum
compressive stress in the clamp at section a–a. The screw EF
is subjected only to a tensile force along its axis.
30 mm
40 mm
F
C
180 N
15 mm
15 mm
Section a – a
a
a
B
A
E
There is no moment in this problem. Therefore, the compressive stress is produced
by axial force only.
smax =
P
240
=
= 1.07 MPa
A
(0.015)(0.015)
Ans.
546
180 N
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8–23. The clamp is made from members AB and AC,
which are pin connected at A. If it exerts a compressive
force at C and B of 180 N, sketch the stress distribution
acting over section a–a. The screw EF is subjected only to
a tensile force along its axis.
30 mm
40 mm
F
C
180 N
15 mm
15 mm
Section a – a
a
a
180 N
B
A
E
There is moment in this problem. Therefore, the compressive stress is produced by
axial force only.
smax =
240
P
=
= 1.07 MPa
A
(0.015)(0.015)
*8–24. The bearing pin supports the load of 700 lb.
Determine the stress components in the support member
at point A. The support is 0.5 in. thick.
0.75 in.
A
2 in.
30⬚
A
B
3 in.
©Fx = 0;
N - 700 cos 30° = 0;
N = 606.218 lb
©Fy = 0;
V - 700 sin 30° = 0;
V = 350 lb
a + ©M = 0;
M - 700(1.25 - 2 sin 30°) = 0;
sA =
1.25 in.
700 lb
M = 175 lb # in.
(175)(0.375)
N
Mc
606.218
=
- 1
3
A
I
(0.75)(0.5)
12 (0.5)(0.75)
sA = -2.12 ksi
Ans.
tA = 0
Ans.
(since QA = 0)
547
B
0.5 in.
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•8–25.
The bearing pin supports the load of 700 lb.
Determine the stress components in the support member
at point B. The support is 0.5 in. thick.
0.75 in.
A
2 in.
30⬚
A
B
B
0.5 in.
3 in.
©Fx = 0;
N - 700 cos 30° = 0;
N = 606.218 lb
©Fy = 0;
V - 700 sin 30° = 0;
V = 350 lb
a + ©M = 0;
M - 700(1.25 - 2 sin 30°) = 0;
sB =
1.25 in.
700 lb
M = 175 lb # in.
175(0.375)
N
Mc
606.218
+
=
+ 1
3
A
I
(0.75)(0.5)
12 (0.5)(0.75)
sB = 5.35 ksi
Ans.
tB = 0
Ans.
(since QB = 0)
8–26. The offset link supports the loading of P = 30 kN.
Determine its required width w if the allowable normal
stress is sallow = 73 MPa. The link has a thickness of 40 mm.
P
s due to axial force:
sa =
30(103)
750(103)
P
=
=
A
(w)(0.04)
w
w
3
sb =
w
50 mm
s due to bending:
w
30(10 )(0.05 + 2 )( 2 )
Mc
=
1
3
I
12 (0.04)(w)
4500 (103)(0.05 + w2)
=
w2
P
smax = sallow = sa + sb
w
73(106) =
4500(103)(0.05 + 2 )
750(103)
+
w
w2
73 w2 = 0.75 w + 0.225 + 2.25 w
73 w2 - 3 w - 0.225 = 0
w = 0.0797 m = 79.7 mm
Ans.
548
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8–27. The offset link has a width of w = 200 mm and a
thickness of 40 mm. If the allowable normal stress is
sallow = 75 MPa, determine the maximum load P that can
be applied to the cables.
P
A = 0.2(0.04) = 0.008 m2
I =
s =
1
(0.04)(0.2)3 = 26.6667(10 - 6) m4
12
w
50 mm
P
Mc
+
A
I
75(106) =
0.150 P(0.1)
P
+
0.008
26.6667(10 - 6)
P = 109 kN
Ans.
P
*8–28. The joint is subjected to a force of P 80 lb and
F 0. Sketch the normal-stress distribution acting over
section a–a if the member has a rectangular cross-sectional
area of width 2 in. and thickness 0.5 in.
a
B
s due to axial force:
s =
0.5 in.
P
80
=
= 80 psi
A
(0.5)(2)
2 in.
F
s due to bending:
s =
A a
100(0.25)
Mc
= 1
= 1200 psi
3
I
12 (2)(0.5)
1.25 in.
P
(smax)t = 80 + 1200 = 1280 psi = 1.28 ksi
Ans.
(smax)c = 1200 - 80 = 1120 psi = 1.12 ksi
Ans.
y
(0.5 - y)
=
1.28
1.12
y = 0.267 in.
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The joint is subjected to a force of P = 200 lb and
F = 150 lb. Determine the state of stress at points A and B
and sketch the results on differential elements located at
these points. The member has a rectangular cross-sectional
area of width 0.75 in. and thickness 0.5 in.
•8–29.
a
B
A a
0.5 in.
2 in.
F
1.25 in.
P
A = 0.5(0.75) = 0.375 in2
œ
QA = yA
A¿ = 0.125(0.75)(0.25) = 0.0234375 in3 ;
I =
QB = 0
1
(0.75)(0.53) = 0.0078125 in4
12
Normal Stress:
My
N
;
A
I
s =
sA =
200
+ 0 = 533 psi (T)
0.375
Ans.
sB =
50(0.25)
200
= -1067 psi = 1067 psi (C)
0.375
0.0078125
Ans.
Shear stress:
t =
VQ
It
tA =
150(0.0234375)
= 600 psi
(0.0078125)(0.75)
Ans.
tB = 0
Ans.
550
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8–30. If the 75-kg man stands in the position shown,
determine the state of stress at point A on the cross section
of the plank at section a–a. The center of gravity of the man
is at G. Assume that the contact point at C is smooth.
C
G
600 mm
A
a
50 mm
1.5 m
B
12.5 mm
30 a
600 mm
300 mm
Support Reactions: Referring to the free-body diagram of the entire plank, Fig. a,
a + ©MB = 0;
FC sin 30°(2.4) - 75(9.81) cos 30°(0.9) = 0
FC = 477.88 N
©Fx¿ = 0; Bx¿ - 75(9.81) sin 30° - 477.88 cos 30° = 0
Bx¿ = 781.73 N
©Fy¿ = 0; By¿ + 477.88 sin 30° - 75(9.81) cos 30° = 0
By¿ = 398.24 N
Internal Loadings: Consider the equilibrium of the free-body diagram of the plank’s
lower segment, Fig. b,
©Fx¿ = 0; 781.73 - N = 0
N = 781.73 N
©Fy¿ = 0; 398.24 - V = 0
V = 398.24 N
a + ©MO = 0;
M - 398.24(0.6) = 0
M = 238.94 N # m
Section Properties: The cross-sectional area and the moment of inertia about the
centroidal axis of the plank’s cross section are
A = 0.6(0.05) = 0.03 m2
I =
1
(0.6) A 0.053 B = 6.25 A 10 - 6 B m4
12
Referring to Fig. c, QA is
QA = y¿A¿ = 0.01875(0.0125)(0.6) = 0.140625 A 10 - 3 B m3
Normal Stress: The normal stress is the combination of axial and bending stress.
Thus,
s =
My
N
;
A
I
For point A, y = 0.0125 m. Then
sA =
238.94(0.0125)
-781.73
0.03
6.25 A 10 - 6 B
= -503.94 kPa = 504 kPa (C)
Ans.
551
Section a – a and b – b
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8–30.
Continued
Shear Stress: The shear stress is contributed by transverse shear stress. Thus,
VQA
tA =
=
It
398.24 c0.140625 A 10 - 3 B d
6.25 A 10 - 6 B (0.6)
Ans.
= 14.9 kPa
The state of stress at point A is represented on the element shown in Fig. d.
552
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8–31. Determine the smallest distance d to the edge of the
plate at which the force P can be applied so that it produces
no compressive stresses in the plate at section a–a. The
plate has a thickness of 20 mm and P acts along the
centerline of this thickness.
a
P
200 mm
d
300 mm
a
Consider the equilibrium of the FBD of the left cut segment in Fig. a,
+
: ©Fx = 0;
N - P = 0
a + ©MC = 0;
M - P(0.1 - d) = 0
A = 0.2 (0.02) = 0.004 m4
I =
N = P
M = P(0.1 - d)
1
(0.02)(0.23) = 13.3333(10 - 6) m4
12
The normal stress developed is the combination of axial and bending stress. Thus
s =
My
N
;
A
I
Since no compressive stress is desired, the normal stress at the top edge fiber must
be equal to zero. Thus,
0 =
P(0.1 - d)(0.1)
P
;
0.004
13.3333 (10 - 6)
0 = 250 P - 7500 P (0.1 - d)
d = 0.06667 m = 66.7 mm
Ans.
553
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*8–32. The horizontal force of P = 80 kN acts at the end
of the plate. The plate has a thickness of 10 mm and P acts
along the centerline of this thickness such that d = 50 mm.
Plot the distribution of normal stress acting along
section a–a.
a
P
200 mm
d
300 mm
Consider the equilibrium of the FBD of the left cut segment in Fig. a,
+
: ©Fx = 0;
N - 80 = 0
a + ©MC = 0;
M - 80(0.05) = 0
A = 0.01(0.2) = 0.002 m2
I =
N = 80 kN
M = 4.00 kN # m
1
(0.01)(0.23) = 6.667(10 - 6) m4
12
The normal stress developed is the combination of axial and bending stress. Thus,
s =
My
N
;
A
I
At point A, y = 0.1 m. Then
sA =
80(103)
4.00(103)(0.1)
0.002
6.667(10 - 6)
= -20.0(106) Pa = 20.0 Mpa (C)
At point B, y = 0.1 m. Then
sB =
80(103)
4.00(103)(0.1)
+
0.002
6.667(10 - 6)
= 100 (106) Pa = 100 MPa (T)
The location of neutral axis can be determined using the similar triangles.
554
a
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•8–33.
The pliers are made from two steel parts pinned
together at A. If a smooth bolt is held in the jaws and a
gripping force of 10 lb is applied at the handles, determine
the state of stress developed in the pliers at points B and C.
Here the cross section is rectangular, having the dimensions
shown in the figure.
0.18 in.
10 lb
D
0.2 in.
0.1 in.
D
3 in.
30
E
A B
0.2 in.
0.2 in.
B
E
C
C
0.2 in.
1.75 in.
2.5 in.
Q ©Fx = 0;
N - 10 sin 30° = 0;
N = 5.0 lb
a+ ©Fy = 0;
V - 10 cos 30° = 0;
V = 8.660 lb
a + ©MC = 0;
M - 10(3) = 0
+
10 lb
M = 30 lb # in.
A = 0.2(0.4) = 0.08 in2
I =
1
(0.2)(0.43) = 1.0667(10 - 3) in4
12
QB = 0
QC = y¿A¿ = 0.1(0.2)(0.2) = 4(10 - 3) in3
Point B:
sB =
My
30(0.2)
N
-5.0
= 5.56 ksi(T)
+
=
+
A
I
0.08
1.0667(10 - 3)
Ans.
VQ
= 0
It
Ans.
My
N
-5.0
+
=
+ 0 = -62.5 psi = 62.5 psi(C)
A
I
0.08
Ans.
VQ
8.660(4)(10 - 3)
= 162 psi
=
It
1.0667(10 - 3)(0.2)
Ans.
tB =
Point C:
sC =
Shear Stress :
tC =
4 in.
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8–34.
Solve Prob. 8–33 for points D and E.
0.18 in.
10 lb
D
0.2 in.
0.1 in.
D
3 in.
30
E
A B
0.2 in.
0.2 in.
B
E
C
C
0.2 in.
1.75 in.
2.5 in.
a + ©MA = 0;
-F(2.5) + 4(10) = 0;
F = 16 lb
10 lb
Point D:
sD = 0
tD =
Ans.
16(0.05)(0.1)(0.18)
VQ
= 667 psi
= 1
It
[12 (0.18)(0.2)3](0.18)
Ans.
My
28(0.1)
= 1
= 23.3 ksi (T)
3
I
12 (0.18)(0.2)
Ans.
Point E:
sE =
4 in.
tE = 0
Ans.
556
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8–35. The wide-flange beam is subjected to the loading
shown. Determine the stress components at points A and B
and show the results on a volume element at each of these
points. Use the shear formula to compute the shear stress.
500 lb
3000 lb
2500 lb
A
B
2 ft
2 ft
2 ft
4 ft
6 ft
A
0.5 in.
1
1
I =
(4)(73) (3.5)(63) = 51.33 in4
12
12
B
4 in.
2 in.
4 in.
0.5 in.
A = 2(0.5)(4) + 6(0.5) = 7 in2
QB = ©y¿A¿ = 3.25(4)(0.5) + 2(2)(0.5) = 8.5 in3
QA = 0
sA =
0.5 in.
-11500 (12)(3.5)
-Mc
=
= -9.41 ksi
I
51.33
Ans.
tA = 0
Ans.
sB =
My
11500(12)(1)
=
= 2.69 ksi
I
51.33
Ans.
tB =
VQB
2625(8.5)
=
= 0.869 ksi
It
51.33(0.5)
Ans.
557
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*8–36. The drill is jammed in the wall and is subjected to
the torque and force shown. Determine the state of stress at
point A on the cross section of drill bit at section a–a.
y
400 mm
a 20 N ·m
x
a
125 mm
y
A
z
5 mm
B
Section a – a
Internal Loadings: Consider the equilibrium of the free-body diagram of the drill’s
right cut segment, Fig. a,
4
©Fx = 0; N - 150 a b = 0
5
N = 120 N
3
©Fy = 0; 150 a b - Vy = 0
5
Vy = 90 N
T = 20 N # m
©Mx = 0; 20 - T = 0
4
3
©Mz = 0; -150 a b (0.4) + 150 a b(0.125) + Mz = 0
5
5
Mz = 21 N # m
Section Properties: The cross-sectional area, the moment of inertia about the z axis,
and the polar moment of inertia of the drill’s cross section are
A = p A 0.0052 B = 25p A 10 - 6 B m2
Iz =
p
A 0.0054 B = 0.15625p A 10 - 9 B m4
4
J =
p
A 0.0054 B = 0.3125p A 10 - 9 B m4
2
Referring to Fig. b, QA is
QA = 0
Normal Stress: The normal stress is a combination of axial and bending stress. Thus,
s =
Mzy
N
A
Iz
For point A, y = 0.005 m. Then
sA =
-120
25p A 10
-6
B
21(0.005)
-
0.15625p A 10 - 9 B
= -215.43 MPa = 215 MPa (C)
558
Ans.
3
5
4
150 N
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8–36. Continued
Shear Stress: The transverse shear stress developed at point A is
c A txy B V d
VyQA
=
A
Izt
Ans.
= 0
The torsional shear stress developed at point A is
C (txz)T D A =
20(0.005)
Tc
= 101.86 MPa
=
J
0.3125p A 10 - 9 B
Thus,
A txy B A = 0
Ans.
A txz B A = c A txz B T d = 102 MPa
Ans.
A
The state of stress at point A is represented on the element shown in Fig c.
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•8–37.
The drill is jammed in the wall and is subjected to
the torque and force shown. Determine the state of stress at
point B on the cross section of drill bit at section a–a.
y
400 mm
a 20 N ·m
x
a
125 mm
y
Internal Loadings: Consider the equilibrium of the free-body diagram of the drill’s
right cut segment, Fig. a,
4
©Fx = 0; N - 150 a b = 0
5
N = 120 N
3
©Fy = 0; 150 a b - Vy = 0
5
Vy = 90 N
z
Section a – a
4
3
©Mz = 0; -150 a b (0.4) + 150 a b (0.125) + Mz = 0
5
5
Mz = 21 N # m
Section Properties: The cross-sectional area, the moment of inertia about the z axis,
and the polar moment of inertia of the drill’s cross section are
A = p A 0.0052 B = 25p A 10 - 6 B m2
Iz =
p
A 0.0054 B = 0.15625p A 10 - 9 B m4
4
J =
p
A 0.0054 B = 0.3125p A 10 - 9 B m4
2
Referring to Fig. b, QB is
QB = y¿A¿ =
4(0.005) p
c A 0.0052 B d = 83.333 A 10 - 9 B m3
3p
2
Normal Stress: The normal stress is a combination of axial and bending stress. Thus,
Mzy
N
A
Iz
For point B, y = 0. Then
sB =
-120
25p A 10 - 6 B
- 0 = -1.528 MPa = 1.53 MPa(C)
560
5 mm
B
T = 20 N # m
©Mx = 0; 20 - T = 0
s =
A
Ans.
3
5
4
150 N
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8–37. Continued
Shear Stress: The transverse shear stress developed at point B is
c A txy B V d
90 c83.333 A 10 - 9 B d
VyQB
=
B
Izt
=
0.15625p A 10 - 9 B (0.01)
= 1.528 MPa
Ans.
The torsional shear stress developed at point B is
c A txy B T d
=
B
20(0.005)
TC
= 101.86 MPa
=
J
0.3125p A 10 - 9 B
Thus,
A tC B B = 0
Ans.
A txy B B = c A txy B T d - c A txy B V d
B
B
Ans.
= 101.86 - 1.528 = 100.33 MPa = 100 MPa
The state of stress at point B is represented on the element shown in Fig. d.
561
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8–38. Since concrete can support little or no tension, this
problem can be avoided by using wires or rods to prestress
the concrete once it is formed. Consider the simply
supported beam shown, which has a rectangular cross
section of 18 in. by 12 in. If concrete has a specific weight of
150 lb>ft3, determine the required tension in rod AB, which
runs through the beam so that no tensile stress is developed
in the concrete at its center section a–a. Neglect the size of
the rod and any deflection of the beam.
a
16 in.
B 2 in.
A
a
4 ft
Support Reactions: As shown on FBD.
Internal Force and Moment:
+
: ©Fx = 0;
T - N = 0
a + ©Mo = 0;
M + T(7) - 900(24) = 0
N = T
M = 21600 - 7T
Section Properties:
A = 18(12) = 216 in2
I =
1
(12) A 183 B = 5832 in4
12
Normal Stress: Requires sA = 0
sA = 0 =
0 =
N
Mc
+
A
I
(21600 - 7T)(9)
-T
+
216
5832
T = 2160 lb = 2.16 kip
Ans.
562
4 ft
18 in.
6 in. 6 in.
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8–39. Solve Prob. 8–38 if the rod has a diameter of 0.5 in.
Use the transformed area method discussed in Sec. 6.6.
Est = 2911032 ksi, Ec = 3.6011032 ksi.
a
16 in.
B 2 in.
A
a
4 ft
Support Reactions: As shown on FBD.
Section Properties:
n =
29(103)
Est
= 8.0556
=
Econ
3.6(103)
p
Acon = (n - 1)Aat = (8.0556 - 1) a b A 0.52 B = 1.3854 in2
4
A = 18(12) + 1.3854 = 217.3854 in2
y =
©yA
9(18)(12) + 16(1.3854)
=
= 9.04461 in.
©A
217.3854
I =
1
(12) A 183 B + 12(18)(9.04461 - 9)2 + 1.3854(16 - 9.04461)2
12
= 5899.45 in4
Internal Force and Moment:
+
: ©Fx = 0;
T - N = 0
a + ©Mo = 0;
M + T(6.9554) - 900(24) = 0
N = T
M = 21600 - 6.9554T
Normal Stress: Requires sA = 0
sA = 0 =
0 =
N
Mc
+
A
I
(21600 - 6.9554T)(8.9554)
-T
+
217.3854
5899.45
T = 2163.08 lb = 2.16 kip
Ans.
563
4 ft
18 in.
6 in. 6 in.
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*8–40. Determine the state of stress at point A when
the beam is subjected to the cable force of 4 kN. Indicate
the result as a differential volume element.
4 kN
250 mm
G
375 mm
D
2m
0.75 m
100 mm
B
1m
A
Support Reactions:
B
a + ©MD = 0;
150 mm
Cy = 0.6667 kN
+
: ©Fx = 0;
Cx - 4 = 0
Cx = 4.00 kN
Internal Forces and Moment:
+
: ©Fx = 0;
4.00 - N = 0
+ c ©Fy = 0;
V - 0.6667 = 0
a + ©Mo = 0;
M - 0.6667(1) = 0
N = 4.00 kN
V = 0.6667 kN
M = 0.6667 kN # m
Section Properties:
A = 0.24(0.15) - 0.2(0.135) = 9.00 A 10 - 3 B m2
I =
1
1
(0.15) A 0.243 B (0.135) A 0.23 B = 82.8 A 10 - 6 B m4
12
12
QA = ©y¿A¿ = 0.11(0.15)(0.02) + 0.05(0.1)(0.015)
= 0.405 A 10 - 3 B m3
Normal Stress:
s =
sA =
My
N
;
A
I
4.00(103)
-3
0.6667(103)(0)
+
9.00(10 )
82.8(10 - 6)
= 0.444 MPa (T)
Ans.
Shear Stress: Applying shear formula.
tA =
=
VQA
It
0.6667(103) C 0.405(10 - 3) D
82.8(10 - 6)(0.015)
= 0.217 MPa
Ans.
564
C
20 mm
15 mm
4(0.625) - Cy (3.75) = 0
A
200 mm
20 mm
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•8–41.
Determine the state of stress at point B when
the beam is subjected to the cable force of 4 kN. Indicate
the result as a differential volume element.
4 kN
250 mm
G
375 mm
D
2m
0.75 m
100 mm
B
1m
A
Support Reactions:
B
a + ©MD = 0;
150 mm
Cy = 0.6667 kN
+
: ©Fx = 0;
Cx - 4 = 0
Cx = 4.00 kN
Internal Forces and Moment:
+
: ©Fx = 0;
4.00 - N = 0
+ c ©Fy = 0;
V - 0.6667 = 0
a + ©Mo = 0;
M - 0.6667(1) = 0
N = 4.00 kN
V = 0.6667 kN
M = 0.6667 kN # m
Section Properties:
A = 0.24(0.15) - 0.2(0.135) = 9.00 A 10 - 3 B m2
I =
1
1
(0.15) A 0.243 B (0.135) A 0.23 B = 82.8 A 10 - 6 B m
12
12
QB = 0
Normal Stress:
s =
sB =
My
N
;
A
I
4.00(103)
9.00(10 - 3)
0.6667(103)(0.12)
-
82.8(10 - 6)
= -0.522 MPa = 0.522 MPa (C)
Ans.
Shear Stress: Since QB = 0, then
tB = 0
Ans.
565
C
20 mm
15 mm
4(0.625) - Cy (3.75) = 0
A
200 mm
20 mm
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8–42. The bar has a diameter of 80 mm. Determine the
stress components that act at point A and show the results
on a volume element located at this point.
200 mm
300 mm
B
A
©Fy = 0;
3
Vy - 5a b = 0
5
Vy = 3 kN
©Fz = 0;
4
Vz + 5 a b = 0
5
Vz = -4 kN
©My = 0;
4
My + 5a b(0.3) = 0
5
My = -1.2 kN # m
©Mz = 0;
3
Mz + 5 a b(0.3) = 0
5
Mz = -0.9 kN # m
Iy = It =
4
5 kN
p
(0.044) = 0.64(10 - 6)p m4
4
Referring to Fig. b,
(Qy)A = 0
(Qz)A = z¿A¿ =
4(0.04) p
c (0.042) d = 42.67(10 - 6) m3
3p
2
The normal stress is contributed by bending stress only. Thus,
s = -
Myz
Mzy
Iz
+
Iy
For point A, y = -0.04 m and z = 0. Then
s = -
-0.9(103)(-0.04)
0.64(10 - 6)p
+ 0 = -17.90(106)Pa = 17.9 MPa (C)
Ans.
The transverse shear stress developed at point A is
A txy B v =
Vy(Qy)A
A txz B v =
Vz(Qz)A
Iz t
Iy t
= 0
=
5
3
Consider the equilibrium of the FBD of bar’s left cut segment shown in Fig. a,
Ans.
4(103) C 42.67(10 - 6) D
0.64(10 - 6)p (0.08)
= 1.061(106) Pa = 1.06 MPa
Ans.
The state of stress for point A can be represented by the volume element shown in
Fig. c,
566
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8–42. Continued
567
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8–43. The bar has a diameter of 80 mm. Determine the
stress components that act at point B and show the results
on a volume element located at this point.
200 mm
300 mm
B
A
Consider the equilibrium of the FBD of bar’s left cut segment shown in Fig. a,
5
3
4
©Fy = 0;
3
Vy - 5a b = 0
5
Vy = 3 kN
©Fz = 0;
4
Vz + 5 a b = 0
5
Vz = -4 kN
©My = 0;
4
My + 5a b(0.3) = 0
5
My = -1.2 kN # m
©Mz = 0;
3
Mz + 5 a b (0.3) = 0
5
Mz = -0.9 kN # m
Iy = Iz =
p
(0.044) = 0.64(10 - 6)p m4
4
5 kN
Referring to Fig. b,
A Qy B B = y¿A¿ = c
4(0.04) p
d c (0.042) d = 42.67(10 - 6) m3
3p
2
A Qz B B = 0
The normal stress is contributed by bending stress only. Thus,
s = -
Myz
Mzy
Iz
+
Iy
For point B, y = 0 and z = 0.04 m. Then
s = -0 +
-1.2(103)(0.04)
0.64(10 - 6)p
= -23.87(106) Pa = 23.9 MPa (C)
Ans.
The transverse shear stress developed at point B is
A txy B v =
Vy(Qy)B
Iz t
=
3(103) C 42.67(10 - 6) D
0.64(10 - 6)p (0.08)
= 0.7958(106) MPa = 0.796 MPa
568
Ans.
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8–43. Continued
A txz B v =
Vz (Qz)B
Iy t
= 0
Ans.
The state of stress for point B can be represented by the volume element shown in
Fig. c
569
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*8–44. Determine the normal stress developed at points
A and B. Neglect the weight of the block.
6 kip
3 in.
Referring to Fig. a,
12 kip
6 in.
©Fx = (FR)x;
-6 - 12 = F
©My = (MR)y;
6(1.5) - 12(1.5) = My
©Mz = (MR)z;
12(3) - 6(3) = Mz
a
F = -18.0 kip
A
My = -9.00 kip # in
Mz = 18.0 kip # in
The cross-sectional area and moment of inertia about the y and z axes of the crosssection are
A = 6(3) = 18 in2
Iy =
1
(6)(3)3 = 13.5 in4
12
Iz =
1
(3)(63) = 54.0 in4
12
The normal stress developed is the combination of axial and bending stress. Thus,
s =
My z
Mz y
F
+
A
Iz
Iy
For point A, y = 3 in. and z = -1.5 in.
sA =
18.0(3)
-9.00(-1.5)
-18.0
+
18.0
54.0
13.5
Ans.
= -1.00 ksi = 1.00 ksi (C)
For point B, y = 3 in and z = 1.5 in.
sB =
18.0(3)
-9.00(1.5)
-18.0
+
18.0
54
13.5
= -3.00 ksi = 3.00 ksi (C)
Ans.
570
B
a
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•8–45.
Sketch the normal stress distribution acting over the
cross section at section a–a. Neglect the weight of the block.
6 kip
3 in.
12 kip
6 in.
a
A
B
a
Referring to Fig. a,
©Fx = (FR)x;
-6 - 12 = F
F = -18.0 kip
©My = (MR)y;
6(1.5) - 12(1.5) = My
©Mz = (MR)z;
12(3) - 6(3) = Mz
My = -9.00 kip # in
Mz = 18.0 kip # in
The cross-sectional area and the moment of inertia about the y and z axes of the
cross-section are
A = 3 (6) = 18.0 in2
Iy =
1
(6)(33) = 13.5 in4
12
Iz =
1
(3)(63) = 54.0 in4
12
The normal stress developed is the combination of axial and bending stress. Thus,
s =
Myz
Mzy
F
+
A
Iz
Iy
For point A, y = 3 in. and z = -1.5 in.
sA =
18.0(3)
-9.00(-1.5)
-18.0
+
18.0
54.0
13.5
= -1.00 ksi = 1.00 ksi (C)
For point B, y = 3 in. and z = 1.5 in.
sB =
18.0(3)
-9.00(1.5)
-18.0
+
18.0
54.0
13.5
= -3.00 ksi = 3.00 ksi (C)
571
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8–45. Continued
For point C, y = -3 in. and z = 1.5 in.
sC =
18.0(-3)
-9.00(1.5)
-18.0
+
18.0
54.0
13.5
= -1.00 ksi = 1.00 ksi (C)
For point D, y = -3 in. and z = -1.5 in.
sD =
18.0(-3)
-9.00(-1.5)
-18.0
+
18.0
54.0
13.5
= 1.00 ksi (T)
The normal stress distribution over the cross-section is shown in Fig. b
572
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8–46. The support is subjected to the compressive load P.
Determine the absolute maximum and minimum normal
stress acting in the material.
a
—
a 2
—
2
a a
—
2 —
2
Section Properties:
w = a +x
A = a(a + x)
I =
a
1
(a) (a + x)3 =
(a + x)3
12
12
Internal Forces and Moment: As shown on FBD.
Normal Stress:
s =
=
N
Mc
;
A
I
0.5Px C 12 (a + x) D
-P
;
a
3
a(a + x)
12 (a + x)
=
P
3x
-1
;
B
R
a a+x
(a + x)2
sA = -
P
1
3x
+
B
R
a a+x
(a + x)2
= sB =
P 4x + a
B
R
a (a + x)2
[1]
P
-1
3x
+
B
R
a a+x
(a + x)2
=
P 2x - a
B
R
a (a + x)2
In order to have maximum normal stress,
[2]
dsA
= 0.
dx
dsA
P (a + x)2(4) - (4x + a)(2)(a + x)(1)
= - B
R = 0
a
dx
(a + x)4
-
Since
P
(2a - 4x) = 0
a(a + x)3
P
Z 0, then
a(a + x)3
2a - 4x = 0
x = 0.500a
573
P
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8–46. Continued
Substituting the result into Eq. [1] yields
smax = -
P 4(0.500a) + a
B
R
a (a + 0.5a)2
= -
1.33P
1.33P
=
(C)
2
a
a2
In order to have minimum normal stress,
Ans.
dsB
= 0.
dx
dsB
P (a + x)2 (2) - (2x - a)(2)(a + x)(1)
=
B
R = 0
a
dx
(a + x)4
P
(4a - 2x) = 0
a(a + x)3
Since
P
Z 0, then
a(a + x)3
4a - 2x = 0
x = 2a
Substituting the result into Eq. [2] yields
smin =
P 2(2a) - a
P
B
R = 2 (T)
a (a + 2a)2
3a
Ans.
574
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8–47. The support is subjected to the compressive load P.
Determine the maximum and minimum normal stress acting
in the material. All horizontal cross sections are circular.
P
r
Section Properties:
d¿ = 2r + x
A = p(r + 0.5x)2
p
(r + 0.5x)4
4
I =
Internal Force and Moment: As shown on FBD.
Normal Stress:
s =
Mc
N
;
A
I
=
0.5Px(r + 0.5x)
–P
;
p
2
4
p(r + 0.5x)
4 (r + 0.5)
=
–1
P
2x
;
B
R
p (r + 0.5x)2
(r + 0.5x)3
sA = -
= -
sB =
=
P
1
2x
+
B
R
p (r + 0.5x)2
(r + 0.5x)3
P
r + 2.5x
B
R
p (r + 0.5x)3
[1]
P
–1
2x
+
B
R
p (r + 0.5x)2
(r + 0.5x)3
P 1.5x - r
B
R
p (r + 0.5x)3
In order to have maximum normal stress,
[2]
dsA
= 0.
dx
dsA
P (r + 0.5x)3 (2.5) - (r + 2.5x)(3)(r + 0.5x)2 (0.5)
= B
R = 0
p
dx
(r + 0.5x)6
-
Since
P
(r - 2.5x) = 0
p(r + 0.5x)4
P
Z 0, then
p(r + 0.5x)4
r - 2.5x = 0
x = 0.400r
Substituting the result into Eq. [1] yields
smax = -
P r + 2.5(0.400r)
B
R
p [r + 0.5(0.400r)]3
= -
0.368P
0.368P
=
(C)
r2
r2
Ans.
575
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8–47. Continued
In order to have minimum normal stress,
dsB
= 0.
dx
dsB
P (r + 0.5x)3 (1.5) - (1.5x - r)(3)(r + 0.5x)2 (0.5)
=
B
R = 0
p
dx
(r + 0.5x)6
P
(3r - 1.5x) = 0
p(r + 0.5x)4
Since
P
Z 0, then
p(r + 0.5x)4
x = 2.00r
3r - 1.5x = 0
Substituting the result into Eq. [2] yields
smin =
P 1.5(2.00r) - r
0.0796P
(T)
B
R =
p [r + 0.5(2.00r)]3
r2
Ans.
*8–48. The post has a circular cross section of radius c.
Determine the maximum radius e at which the load can be
applied so that no part of the post experiences a tensile
stress. Neglect the weight of the post.
P
c
e
Require sA = 0
sA = 0 =
P
Mc
+
;
A
I
e =
0 =
(Pe)c
-P
+ p 4
2
pc
4c
c
4
Ans.
576
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•8–49.
If the baby has a mass of 5 kg and his center of
mass is at G, determine the normal stress at points A and B
on the cross section of the rod at section a–a. There are two
rods, one on each side of the cradle.
500 mm
15⬚
G
a
75 mm
a
6 mm
A
B
Section a–a
Section Properties: The location of the neutral surface from the center of
curvature of the rod, Fig. a, can be determined from
A
R =
©
dA
LA r
where A = p A 0.0062 B = 36p A 10 - 6 B m2
©
dA
= 2p ¢ r - 2r2 - c2 ≤ = 2p ¢ 0.081 - 20.0812 - 0.0062 ≤ = 1.398184 A 10 - 3 B m
LA r
Thus,
R =
36p A 10 - 6 B
1.398184 A 10 - 3 B
= 0.080889 m
Then
e = r - R = 0.081 - 0.080889 = 0.111264 A 10 - 3 B m
Internal Loadings: Consider the equilibrium of the free-body diagram of the
cradle’s upper segment, Fig. b,
+ c ©Fy = 0;
-5(9.81) - 2N = 0
N = -24.525 N
a + ©MO = 0;
5(9.81)(0.5+ 0.080889) - 2M = 0
M = 14.2463 N # m
Normal Stress: The normal stress is the combination of axial and bending stress.
Thus,
s =
M(R - r)
N
+
A
Aer
Here, M = -14.1747 (negative) since it tends to increase the curvature of the
rod. For point A, r = rA = 0.075 m. Then,
sA =
-24.525
36p A 10
-6
B
-14.2463(0.080889 - 0.075)
+
36p A 10 - 6 B (0.111264) A 10 - 3 B (0.075)
= -89.1 MPa = 89.1 MPa (C)
Ans.
For point B, r = rB = 0.087 m. Then,
sB =
-24.525
36p A 10
-6
B
-14.2463(0.080889 - 0.087)
+
36p A 10 - 6 B (0.111264) A 10 - 3 B (0.087)
Ans.
= 79.3 kPa (T)
dA
5
= 0.25 ln = 0.055786
4
LA r
577
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8–50. The C-clamp applies a compressive stress on the
cylindrical block of 80 psi. Determine the maximum normal
stress developed in the clamp.
1 in.
4 in.
4.5 in.
0.75 in.
R =
A
dA
1 r
=
1(0.25)
= 4.48142
0.055786
P = sbA = 80p (0.375)2 = 35.3429 lb
M = 35.3429(8.98142) = 317.4205 lb # in.
s =
M(R - r)
P
+
Ar(r - R)
A
(st)max =
317.4205(4.48142 - 4)
35.3429
+
= 8.37 ksi
(1)(0.25)(4)(4.5 - 4.48142)
(1)(0.25)
(sc)max =
317.4205(4.48142 - 5)
35.3429
+
= -6.95 ksi
1(0.25)(5)(4.5 - 4.48142)
(1)(0.25)
578
Ans.
0.25 in.
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8–51. A post having the dimensions shown is subjected to
the bearing load P. Specify the region to which this load can
be applied without causing tensile stress to be developed at
points A, B, C, and D.
x
z
a
a
A
B
Equivalent Force System: As shown on FBD.
Section Properties:
1
A = 2a(2a) + 2 B (2a)a R = 6a2
2
Iz =
1
1
1
a 2
(2a)(2a)3 + 2 B (2a) a3 + (2a) aa a + b R
12
36
2
3
= 5a4
Iy =
=
1
1
1
a 2
(2a)(2a)3 + 2 B (2a) a3 + (2a) aa b R
12
36
2
3
5 4
a
3
Normal Stress:
s =
My z
Mzy
N
+
A
Iz
Iy
=
Peyy
Pez z
-P
+ 5
2
4
4
6a
5a
3a
=
P
A -5a2 - 6eyy + 18ez z B
30a4
At point A where y = -a and z = a, we require sA 6 0.
0 7
P
C -5a2 - 6(-a) ey + 18(a) ez D
30a4
0 7 -5a + 6ey + 18ez
6ey + 18ez 6 5a
When
ez = 0,
When
ey = 0,
Ans.
5
a
6
5
ez 6
a
18
ey 6
Repeat the same procedures for point B, C and D. The region where P can be
applied without creating tensile stress at points A, B, C and D is shown shaded in the
diagram.
579
a
P
a
D
ez
ey
C
a
a
y
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*8–52. The hook is used to lift the force of 600 lb.
Determine the maximum tensile and compressive stresses
at section a–a. The cross section is circular and has a
diameter of 1 in. Use the curved-beam formula to compute
the bending stress.
300 lb
a
300 lb
2.5 in.
a
1.5 in.
Section Properties:
r = 1.5 + 0.5 = 2.00 in.
dA
= 2p A r - 2r2 - c2 B
LA r
600 lb
= 2p A 2.00 - 22.002 - 0.52 B
= 0.399035 in.
A = p A 0.52 B = 0.25p in2
A
R =
dA
1A r
=
0.25p
= 1.968246 in.
0.399035
r - R = 2 - 1.968246 = 0.031754 in.
Internal Force and Moment: As shown on FBD. The internal moment must be
computed about the neutral axis. M = 1180.95 lb # in. is positive since it tends to
increase the beam’s radius of curvature.
Normal Stress: Applying the curved-beam formula.
For tensile stress
(st)max =
=
M(R - r1)
N
+
A
Ar1 (r - R)
1180.95(1.968246 - 1.5)
600
+
0.25p
0.25p(1.5)(0.031754)
= 15546 psi = 15.5 ksi (T)
Ans.
For compressive stress,
(sc)max =
=
M(R - r2)
N
+
A
Ar2 (r - R)
1180.95(1.968246 - 2.5)
600
+
0.25p
0.25p(2.5)(0.031754)
= -9308 psi = 9.31 ksi (C)
Ans.
580
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•8–53.
The masonry pier is subjected to the 800-kN load.
Determine the equation of the line y = f1x2 along which
the load can be placed without causing a tensile stress in the
pier. Neglect the weight of the pier.
800 kN
1.5 m y
1.5 m
2.25 m
y
2.25 m
x
x
C
A
B
A = 3(4.5) = 13.5 m2
Ix =
1
(3)(4.53) = 22.78125 m4
12
Iy =
1
(4.5)(33) = 10.125 m4
12
Normal Stress: Require sA = 0
sA =
0 =
Myx
Mxy
P
+
+
A
Ix
Iy
800(103)y(2.25)
800(103)x(1.5)
-800(103)
+
+
13.5
22.78125
10.125
0 = 0.148x + 0.0988y - 0.0741
y = 0.75 - 1.5 x
Ans.
581
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8–54. The masonry pier is subjected to the 800-kN load. If
x = 0.25 m and y = 0.5 m, determine the normal stress at
each corner A, B, C, D (not shown) and plot the stress
distribution over the cross section. Neglect the weight of
the pier.
800 kN
1.5 m y
1.5 m
2.25 m
y
2.25 m
x
x
A = 3(4.5) = 13.5 m2
Ix =
1
(3)(4.53) = 22.78125 m4
12
C
1
Iy =
(4.5)(33) = 10.125 m4
12
s =
sA =
A
B
Myx
Mxy
P
+
+
A
Ix
Iy
-800(103)
400(103)(2.25)
200(103)(1.5)
+
+
13.5
22.78125
10.125
Ans.
= 9.88 kPa (T)
3
sB =
3
3
400(10 )(2.25)
200(10 )(1.5)
-800(10 )
+
13.5
22.78125
10.125
Ans.
= -49.4 kPa = 49.4 kPa (C)
sC =
-800(103)
400(103)(2.25)
200(103)(1.5)
+
13.5
22.78125
10.125
Ans.
= -128 kPa = 128 kPa (C)
sD =
-800(103)
400(103)(2.25)
200(103)(1.5)
+
13.5
22.78125
10.125
= -69.1 kPa = 69.1 kPa (C)
Ans.
582
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8–55. The bar has a diameter of 40 mm. If it is subjected to
the two force components at its end as shown, determine
the state of stress at point A and show the results on a
differential volume element located at this point.
x
100 mm
150 mm
z
A
B
500 N
y
300 N
Internal Forces and Moment:
©Fx = 0;
Nx = 0
©Fy = 0;
Vy + 300 = 0
Vy = -300 N
©Fz = 0;
Vz - 500 = 0
Vz = 500 N
©Mx = 0;
Tx = 0
©My = 0;
My - 500(0.15) = 0
My = 75.0 N # m
©Mz = 0;
Mz - 300(0.15) = 0
Mz = 45.0 N # m
Section Properties:
A = p A 0.022 B = 0.400 A 10 - 3 B p m2
p
A 0.024 B = 40.0 A 10 - 9 B p m4
4
Ix = Iy =
J =
p
A 0.024 B = 80.0 A 10 - 9 B p m4
2
(QA)z = 0
4(0.02) 1
c p A 0.022 B d = 5.333 A 10 - 6 B m3
3p
2
(QA)y =
Normal Stress:
s =
Myz
Mzy
N
+
A
Iz
Iy
sA = 0 -
75.0(0.02)
45.0(0)
-9
+
40.0(10 )p
40.0(10 - 9)p
= 11.9 MPa (T)
Ans.
Shear Stress: The tranverse shear stress in the z and y directions can be obtained
VQ
using the shear formula, tV =
.
It
(txy)A = -tVy = -
300 C 5.333(10 - 6) D
40.0(10 - 9)p (0.04)
Ans.
= -0.318 MPa
(txz)A = tVz = 0
Ans.
583
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*8–56.
Solve Prob. 8–55 for point B.
x
100 mm
150 mm
z
A
B
500 N
y
300 N
Internal Forces and Moment:
©Fx = 0;
Nx = 0
©Fy = 0;
Vy + 300 = 0
Vy = -300 N
©Fz = 0;
Vz - 500 = 0
Vz = 500 N
©Mx = 0;
Tx = 0
©My = 0;
My - 500(0.15) = 0
My = 75.0 N # m
©Mz = 0;
Mz - 300(0.15) = 0
Mz = 45.0 N # m
Section Properties:
A = p A 0.022 B = 0.400 A 10 - 3 B p m2
Ix = Iy =
J =
p
A 0.024 B = 40.0 A 10 - 9 B p m4
4
p
A 0.024 B = 80.0 A 10 - 9 B p m4
2
(QB)y = 0
(QB)z =
4(0.02) 1
c p A 0.022 B d = 5.333 A 10 - 6 B m3
3p 2
Normal Strees:
s =
Myz
Mzy
N
+
A
Iz
Iy
sB = 0 -
75.0(0)
45.0(0.02)
-9
40.0(10 ) p
+
40.0(10 - 9) p
= -7.16 MPa = 7.16 MPa (C)
Ans.
Shear Stress: The tranverse shear stress in the z and y directions can be obtained
VQ
using the shear formula, tV =
.
It
(txz)B = tVz =
500 C 5.333(10 - 6) D
40.0(10 - 9) p (0.04)
Ans.
= 0.531 MPa
(txy)B = tVy = 0
Ans.
584
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•8–57. The 2-in.-diameter rod is subjected to the loads
shown. Determine the state of stress at point A, and show the
results on a differential element located at this point.
z
B
x
A
8 in.
y
600 lb
12 in.
Consider the equilibrium of the FBD of the right cut segment, Fig. a,
©Fy = 0 ;
Ny + 800 = 0
Ny = -800 lb
©Fz = 0 ;
Vz + 600 = 0
Vz = -600 lb
©Fx = 0 ;
Vx - 500 = 0
Vx = 500 lb
©My = 0 ;
Ty - 600(12) = 0
©Mz = 0 ;
Mz + 800(12) + 500(8) = 0
©Mx = 0 ;
Mx + 600(8) = 0
Mx = -4800 lb # in
p 4
p
(1 ) = in4
4
4
A = p(12) = p in2
Ix = Iz =
J =
500 lb
800 lb
Ty = 7200 lb # in
Mz = -13600 lb # in
p 4
p
(1 ) = in4
2
2
Referring to Fig. b,
(Qz)A = y¿A¿ =
(Qx)A = 0
4(1) p 2
c (1 ) d = 0.6667 in3
3p 2
The normal stress is contributed by axial and bending stress. Thus,
s =
Mzx
Mxz
N
+
A
Ix
Iz
For point A, z = 0 and x = 1 in.
4800(0)
-13600(1)
800
p
p>4
p>4
s =
= 17.57(103) psi = 17.6 ksi (T)
Ans.
The torsional shear stress developed at point A is
(tyz)T =
TyC
J
=
7200(1)
= 4.584(103) psi = 4.584 ksi T
p>2
The transverse shear stress developed at point A is.
(tyz)g =
(txy)g =
Vz(Qz)A
Ixt
=
600(0.6667)
= 254.64 psi = 0.2546 ksi T
p
(2)
4
Vx(Qx)A
500(0)
=
= 0
p
Izt
(2)
4
585
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8–57. Continued
Combining these two shear stress components,
tyz = A tyz B T + A tyz B g
= 4.584 + 0.2546
Ans.
= 4.838 ksi = 4.84 ksi
txy = 0
Ans.
The state of stress of point A can be represented by the volume element shown in
Fig. c.
586
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8–58. The 2-in.-diameter rod is subjected to the loads
shown. Determine the state of stress at point B, and show
the results on a differential element located at this point.
z
B
x
A
8 in.
y
600 lb
12 in.
Consider the equilibrium of the FBD of the right cut segment, Fig. a,
©Fy = 0;
Ny + 800 = 0
Ny = -800 lb
©Fz = 0;
Vz + 600 = 0
Vz = -600 lb
©Fx = 0;
Vx - 500 = 0
Vx = 500 lb
©My = 0;
Ty - 600(12) = 0
©Mz = 0;
Mz + 800(12) + 500(8) = 0 Mz = -13600 lb # in
©Mx = 0;
Mx + 600(8) = 0
800 lb
Ty = 7200 lb # in
Mx = -4800 lb # in.
The cross-sectional area the moment of inertia about x and Z axes and polar
moment of inertia of the rod are
¿
A = p(12) = p in2
Ix = Iz =
p 4
p
(1 ) = in4
4
4
J =
p 4
p
(1 ) = in4
2
2
Referring to Fig. b,
(Qz)B = 0 (Qx)B = z¿A¿ =
4(1) p 2
c (1 ) d = 0.6667 in4
3p 2
The normal stress is contributed by axial and bending stress. Thus,
s =
Mzx
Mxz
N
+
A
Ix
Iz
For point B, x = 0 and z = 1 in.
s =
4800 (1)
13600 (0)
800
+
p
p>4
p>4
= 5.86 ksi (C)
The torsional shear stress developed at point B is
(txy)T =
TyC
J
=
7200(1)
= 4.584(103) psi = 4.584 ksi :
p>2
The transverse shear stress developed at point B is.
(txy)v =
(tyz)v =
Vx(Qx)B
500 (0.6667)
=
= 212.21 psi = 0.2122 ksi :
p
Izt
(2)
4
Vz(Qz)B
Ixt
=
500 lb
600 (0)
= 0
p
(2)
4
587
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8–58.
Continued
Combining these two shear stress components,
txy = (txy)T + (txy)v
= 4.584 + 0.2122
= 4.796 ksi = 4.80 ksi
Ans.
tyz = 0
Ans.
The state of stress of point B can be represented by the volume element shown in
Fig. c.
588
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8–59. If P = 60 kN, determine the maximum normal
stress developed on the cross section of the column.
2P
150 mm
150 mm
Equivalent Force System: Referring to Fig. a,
+ c ©Fx = A FR B x;
F = 180 kN
-60 - 120 = -F
15 mm
©My = (MR)y;
-60(0.075) = -My
My = 4.5kN # m
©Mz = (MR)z;
-120(0.25) = -Mz
Mz = 30kN # m
100 mm
100 mm
Section Properties: The cross-sectional area and the moment of inertia about the y
and z axes of the cross section are
A = 0.2(0.3) - 0.185(0.27) = 0.01005 m2
Iz =
1
1
(0.2) A 0.33 B (0.185) A 0.273 B = 0.14655 A 10 - 3 B m4
12
12
Iy = 2c
1
1
(0.015) A 0.23 B d +
(0.27) A 0.0153 B = 20.0759 A 10 - 6 B m4
12
12
Normal Stress: The normal stress is the combination of axial and bending stress.
Here, F is negative since it is a compressive force. Also, My and Mz are negative
since they are directed towards the negative sense of their respective axes. By
inspection, point A is subjected to a maximum normal stress. Thus,
s =
Myz
Mzy
N
+
A
Iz
Iy
smax = sA =
-180 A 103 B
0.01005
-
C -30 A 103 B D ( -0.15)
0.14655 A 10 - 3 B
15 mm
15 mm
P
+
C -4.5 A 103 B D (0.1)
20.0759 A 10 - 6 B
= -71.0 MPa = 71.0 MPa(C)
Ans.
589
75 mm
100 mm
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*8–60. Determine the maximum allowable force P, if the
column is made from material having an allowable normal
stress of sallow = 100 MPa .
2P
150 mm
150 mm
Equivalent Force System: Referring to Fig. a,
+ c ©Fx = (FR)x;
15 mm
-P - 2P = -F
100 mm
100 mm
F = 3P
©My = (MR)y;
-P(0.075) = -My
My = 0.075 P
©Mz = (MR)z;
-2P(0.25) = -Mz
Mz = 0.5P
Section Properties: The cross-sectional area and the moment of inertia about the y
and z axes of the cross section are
A = 0.2(0.3) - 0.185(0.27) = 0.01005 m2
Iz =
1
1
(0.2) A 0.33 B (0.185) A 0.273 B = 0.14655 A 10 - 3 B m4
12
12
Iy = 2 c
1
1
(0.15) A 0.23 B d +
(0.27) A 0.0153 B = 20.0759 A 10 - 6 B m4
12
12
Normal Stress: The normal stress is the combination of axial and bending stress.
Here, F is negative since it is a compressive force. Also, My and Mz are negative
since they are directed towards the negative sense of their respective axes. By
inspection, point A is subjected to a maximum normal stress, which is in
compression. Thus,
s =
Myz
Mzy
N
+
A
Iz
Iy
-100 A 106 B = -
15 mm
15 mm
P
(-0.5P)(-0.15)
-0.075P(0.1)
3P
+
0.01005
0.14655 A 10 - 3 B
20.0759 A 10 - 6 B
P = 84470.40 N = 84.5 kN
Ans.
590
75 mm
100 mm
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•8–61.
The beveled gear is subjected to the loads shown.
Determine the stress components acting on the shaft at
point A, and show the results on a volume element located
at this point. The shaft has a diameter of 1 in. and is fixed to
the wall at C.
z
y
A
200 lb
C
B
x
8 in.
3 in.
©Fx = 0;
Vx - 125 = 0;
Vx = 125 lb
©Fy = 0;
75 - Ny = 0;
Ny = 75 lb
©Fz = 0;
Vz - 200 = 0;
Vz = 200 lb
©Mx = 0;
200(8) - Mx = 0;
Mx = 1600 lb # in.
©My = 0;
200(3) - Ty = 0;
Ty = 600 lb # in.
©Mz = 0;
Mz + 75(3) - 125(8) = 0;
75 lb
Mz = 775 lb # in.
A = p(0.52) = 0.7854 in2
J =
p
(0.54) = 0.098175 in4
2
I =
p
(0.54) = 0.049087 in4
4
(QA)x = 0
(QA)z =
4(0.5) 1
a b (p)(0.52) = 0.08333 in3
3p
2
(sA)y = -
= -
Ny
A
+
Mx c
I
1600(0.5)
75
+
0.7854
0.049087
= 16202 psi = 16.2 ksi (T)
Ans.
(tA)yx = (tA)V - (tA)twist
Tyc
Vx(QA)z
=
=
It
-
J
600(0.5)
125(0.08333)
0.049087 (1)
0.098175
= -2843 psi = -2.84 ksi
(tA)yz =
Vz(QA)x
It
Ans.
= 0
Ans.
591
125 lb
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8–62. The beveled gear is subjected to the loads shown.
Determine the stress components acting on the shaft at
point B, and show the results on a volume element located
at this point. The shaft has a diameter of 1 in. and is fixed to
the wall at C.
z
y
A
200 lb
C
B
x
8 in.
3 in.
75 lb
©Fx = 0;
Vx - 125 = 0;
Vx = 125 lb
©Fy = 0;
75 - Ny = 0;
Ny = 75 lb
©Fz = 0;
Vz - 200 = 0;
Vz = 200 lb
©Mx = 0;
200(8) - Mx = 0;
Mx = 1600 lb # in.
©My = 0;
200(3) - Ty = 0;
Ty = 600 lb # in.
©Mz = 0;
Mz + 75(3) - 125(8) = 0;
Mz = 775 lb # in.
A = p(0.52) = 0.7854 in2
J =
p
(0.54) = 0.098175 in4
2
I =
p
(0.54) = 0.049087 in4
4
(QB)z = 0
(QB)x =
4(0.5) 1
a b(p)(0.52) = 0.08333 in3
3p
2
(sB)y = -
= -
Ny
A
Mz c
+
I
775(0.5)
75
+
0.7854
0.049087
= 7.80 ksi (T)
Ans.
(tB)yz = (tB)V + (tB)twist
Ty c
Vz(QB)x
=
=
It
+
J
600(0.5)
200(0.08333)
+
0.049087 (1)
0.098175
Ans.
= 3395 psi = 3.40 ksi
(tB)yx =
Vx (QB)z
It
Ans.
= 0
592
125 lb
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8–63. The uniform sign has a weight of 1500 lb and
is supported by the pipe AB, which has an inner radius of
2.75 in. and an outer radius of 3.00 in. If the face of the sign
is subjected to a uniform wind pressure of p = 150 lb>ft2,
determine the state of stress at points C and D. Show the
results on a differential volume element located at each of
these points. Neglect the thickness of the sign, and assume
that it is supported along the outside edge of the pipe.
12 ft
B
150 lb/ft2
6 ft
F
E
3 ft
D
C
A
z
y
x
Section Properties:
A = p A 32 - 2.752 B = 1.4375p in2
Iy = Iz =
p 4
A 3 - 2.754 B = 18.6992 in4
4
(QC)z = (QD)y = 0
4(3) 1
4(2.75) 1
c (p) A 32 B d c (p) A 2.752 B d
3p 2
3p
2
(QC)y = (QD)z =
= 4.13542 in3
J =
p 4
A 3 - 2.754 B = 37.3984 in4
2
Normal Stress:
s =
My z
Mz y
N
+
A
Iz
Iy
sC =
(-64.8)(12)(0)
9.00(12)(2.75)
-1.50
+
1.4375p
18.6992
18.6992
Ans.
= 15.6 ksi (T)
sD =
(-64.8)(12)(3)
9.00(12)(0)
-1.50
+
1.4375p
18.6992
18.6992
Ans.
= 124 ksi (T)
Shear Stress: The tranverse shear stress in the z and y directions and the torsional
shear stress can be obtained using the shear formula and the torsion formula,
VQ
Tr
and ttwist =
, respectively.
tV =
It
J
(txz)D = ttwist =
64.8(12)(3)
= 62.4 ksi
37.3984
Ans.
(txy)D = tVy = 0
Ans.
593
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8–63.
Continued
(txy)C = tVy - ttwist
=
64.8(12)(2.75)
10.8(4.13542)
18.6992(2)(0.25)
37.3984
Ans.
= -52.4 ksi
(txz)C = tVz = 0
Ans.
Internal Forces and Moments: As shown on FBD.
©Fx = 0;
1.50 + Nx = 0
Nx = -15.0 kip
©Fy = 0;
Vy - 10.8 = 0
Vy = 10.8 kip
©Fz = 0;
Vz = 0
©Mx = 0;
Tx - 10.8(6) = 0
Tx = 64.8 kip # ft
©My = 0;
My - 1.50(6) = 0
My = 9.00 kip # ft
©Mz = 0;
10.8(6) + Mz = 0
Mz = -64.8 kip # ft
594
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*8–64.
Solve Prob. 8–63 for points E and F.
12 ft
B
150 lb/ft2
6 ft
F
E
3 ft
D
C
A
z
y
x
Internal Forces and Moments: As shown on FBD.
©Fx = 0;
1.50 + Nx = 0
Nx = -1.50 kip
©Fy = 0;
Vy - 10.8 = 0
Vy = 10.8 kip
©Fz = 0;
Vz = 0
©Mx = 0;
Tx - 10.8(6) = 0
Tx = 64.8 kip # ft
©My = 0;
My - 1.50(6) = 0
My = 9.00 kip # ft
©Mz = 0;
10.8(6) + Mz = 0
Mz = -64.8 kip # ft
Section Properties:
A = p A 32 - 2.752 B = 1.4375p in2
Iy = Iz =
p 4
A 3 - 2.754 B = 18.6992 in4
4
(QC)z = (QD)y = 0
(QC)y = (QD)z =
4(3) 1
4(2.75) 1
c (p) A 32 B d c (p) A 2.752 B d
3p 2
3p
2
= 4.13542 in3
J =
p 4
A 3 - 2.754 B = 37.3984 in4
2
Normal Stress:
s =
My z
Mzy
N
+
A
Iz
Iy
sF =
(-64.8)(12)(0)
9.00(12)(-3)
-1.50
+
1.4375p
18.6992
18.6992
Ans.
= -17.7 ksi = 17.7 ksi (C)
sE =
(-64.8)(12)(-3)
9.00(12)(0)
-1.50
+
1.4375p
18.6992
18.6992
= -125 ksi = 125 ksi (C)
Ans.
595
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8–64.
Continued
Shear Stress: The tranverse shear stress in the z and y directions and the torsional
shear stress can be obtained using the shear formula and the torsion formula,
VQ
Tr
and ttwist =
, respectively.
tV =
It
J
(txz)E = -ttwist = -
64.8(12)(3)
= -62.4 ksi
37.3984
Ans.
(txy)E = tVy = 0
Ans.
(txy)F = tVy + ttwist
=
64.8(12)(3)
10.8(4.13542)
+
18.6992(2)(0.25)
37.3984
= 67.2 ksi
Ans.
(txy)F = tVy = 0
Ans.
596
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•8–65.
Determine the state of stress at point A on the
cross section of the pipe at section a–a.
A
0.75 in.
B
y
50 lb
1 in.
Section a–a
x
a
60°
z
a
10 in.
Internal Loadings: Referring to the free - body diagram of the pipe’s right segment,
Fig. a,
©Fy = 0; Vy - 50 sin 60° = 0
Vy = 43.30 lb
©Fz = 0; Vz - 50 cos 60° = 0
Vz = 25 lb
©Mx = 0; T + 50 sin 60°(12) = 0
T = -519.62 lb # in
©My = 0; My - 50 cos 60°(10) = 0
My = 250 lb # in
©Mz = 0; Mz + 50 sin 60° (10) = 0
Mz = -433.01 lb # in
Section Properties: The moment of inertia about the y and z axes and the polar
moment of inertia of the pipe are
Iy = Iz =
J =
p 4
A 1 - 0.754 B = 0.53689 in4
4
p 4
A 1 - 0.754 B = 1.07379 in4
2
Referring to Fig. b,
A Qy B A = 0
A Qz B A = y1œ A1œ - y2œ A2œ =
4(1) p 2
4(0.75) p
c A1 B d c A 0.752 B d = 0.38542 in3
3p 2
3p
2
Normal Stress: The normal stress is contributed by bending stress only. Thus,
s = -
Myz
Mzy
Iz
+
Iy
For point A, y = 0.75 in and z = 0. Then
sA =
-433.01(0.75)
+ 0 = 604.89 psi = 605 psi (T)
0.53689
Shear Stress: The torsional shear stress developed at point A is
c A txz B T d
=
A
TrA
519.62(0.75)
=
= 362.93 psi
J
1.07379
597
Ans.
12 in.
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8–65.
Continued
The transverse shear stress developed at point A is
c A txy B V d
c A txz B V d
= 0
A
=
A
Vz A Qz B A
Iy t
=
25(0.38542)
= 35.89 psi
0.53689(2 - 1.5)
Combining these two shear stress components,
A txy B A = 0
Ans.
A txz B A = c A txz B T d - c A txz B V d
A
A
= 362.93 - 35.89 = 327 psi
Ans.
598
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8–66. Determine the state of stress at point B on the cross
section of the pipe at section a–a.
A
0.75 in.
B
y
50 lb
1 in.
Section a–a
x
a
60°
z
a
10 in.
Internal Loadings: Referring to the free - body diagram of the pipe’s right segment,
Fig. a,
©Fy = 0; Vy - 50 sin 60° = 0
Vy = 43.30 lb
©Fz = 0; Vz - 50 cos 60° = 0
Vz = 25 lb
©Mx = 0; T + 50 sin 60°(12) = 0
T = -519.62 lb # in
©My = 0; My - 50 cos 60°(10) = 0
My = 250 lb # in
©Mz = 0; Mz + 50 sin 60°(10) = 0
Mz = -433.01 lb # in
Section Properties: The moment of inertia about the y and z axes and the polar
moment of inertia of the pipe are
p 4
A 1 - 0.754 B = 0.53689 in4
4
Iy = Iz =
J =
p 4
A 1 - 0.754 B = 1.07379 in4
2
Referring to Fig. b,
A Qz B B = 0
A Qy B B = y1œ A1œ - y2œ A2œ =
4(1) p 2
4(0.75) p
c A1 B d c A 0.752 B d = 0.38542 in3
3p 2
3p
2
Normal Stress: The normal stress is contributed by bending stress only. Thus,
s = -
Myz
Mzy
Iz
+
Iy
For point B, y = 0 and z = -1. Then
sB = -0 +
250(1)
= -465.64 psi = 466 psi (C)
0.53689
Ans.
Shear Stress: The torsional shear stress developed at point B is
c A txy B T d
=
B
519.62(1)
TrC
=
= 483.91 psi
J
1.07379
599
12 in.
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8–66.
Continued
The transverse shear stress developed at point B is
c A txz B V d
c A txy B V d
= 0
B
=
B
Vy A Qy B B
Izt
=
43.30(0.38542)
= 62.17 psi
0.53689(2 - 1.5)
Combining these two shear stress components,
A txy B B = c A txy B T d - c A txy B V d
B
B
Ans.
= 483.91 - 62.17 = 422 psi
A txz B B = 0
Ans.
600
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•8–67. The eccentric force P is applied at a distance ey from
the centroid on the concrete support shown. Determine the
range along the y axis where P can be applied on the cross
section so that no tensile stress is developed in the material.
x
z
P
b
2
ey
b
2
2h
3
h
3
Internal Loadings: As shown on the free - body diagram, Fig. a.
Section Properties: The cross-sectional area and moment of inertia about the z axis
of the triangular concrete support are
A =
1
bh
2
Iz =
1
bh3
36
Normal Stress: The normal stress is the combination of axial and bending stress.
Thus,
s =
Mzy
N
A
Iz
A Pey B y
-P
1
1
bh
bh3
2
36
s =
2P 2
A h + 18eyy B
bh3
s = -
Here, it is required that sA … 0 and sB … 0. For point A, y =
(1)
h
, Then. Eq. (1) gives
3
2P 2
h
ch + 18ey a b d
3
3
bh
0 Ú -
0 … h2 + 6hey
ey Ú -
For Point B, y = -
h
6
2
h. Then. Eq. (1) gives
3
0 Ú -
2P 2
2
ch + 18ey a - hb d
3
3
bh
0 … h2 - 12hey
ey …
h
12
Thus, in order that no tensile stress be developed in the concrete support, ey must be
in the range of
-
h
h
… ey …
6
12
Ans.
601
y
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*8–68. The bar has a diameter of 40 mm. If it is subjected
to a force of 800 N as shown, determine the stress components
that act at point A and show the results on a volume element
located at this point.
150 mm
200 mm
1
1
I = p r4 = (p)(0.024) = 0.1256637 (10 - 6) m4
4
4
A
A = p r2 = p(0.022) = 1.256637 (10 - 3) m2
QA = y¿A¿ = a
z
B
4 (0.02) p (0.02)2
ba
b = 5.3333 (10 - 6) m3
3p
2
y
x
30⬚
800 N
P
Mz
sA =
+
A
I
400
+ 0 = 0.318 MPa
1.256637 (10 - 3)
Ans.
692.82 (5.3333) (10 - 6)
VQA
= 0.735 MPa
=
It
0.1256637 (10 - 6)(0.04)
Ans.
=
tA =
•8–69.
Solve Prob. 8–68 for point B.
150 mm
200 mm
A
z
B
y
x
30⬚
800 N
1
1
I = p r4 = (p)(0.024) = 0.1256637 (10 - 6) m4
4
4
A = p r2 = p(0.022) = 1.256637 (10 - 3) m2
QB = 0
sB =
138.56 (0.02)
P
Mc
400
= -21.7 MPa
=
-3
A
I
1.256637 (10 )
0.1256637 (10 - 6)
tB = 0
Ans.
Ans.
602
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8–70. The 43-in.-diameter shaft is subjected to the loading
shown. Determine the stress components at point A. Sketch
the results on a volume element located at this point. The
journal bearing at C can exert only force components Cy
and Cz on the shaft, and the thrust bearing at D can exert
force components Dx , Dy , and Dz on the shaft.
D
z
125 lb
2 in.
8 in.
125 lb
2 in.
A
20 in.
8 in.
B
C
10 in.
y
20 in.
x
A =
p
(0.752) = 0.44179 in2
4
I =
p
(0.3754) = 0.015531 in4
4
QA = 0
tA = 0
sA =
Ans.
My c
I
=
-1250(0.375)
= -30.2 ksi = 30.2 ksi (C)
0.015531
Ans.
8–71. Solve Prob. 8–70 for the stress components at point B.
D
z
125 lb
2 in.
8 in.
125 lb
2 in.
A
8 in.
B
C
10 in.
x
p
A = (0.752) = 0.44179 in2
4
I =
p
(0.3754) = 0.015531 in4
4
QB = y¿A¿ =
4(0.375) 1
a b(p)(0.3752) = 0.035156 in3
3p
2
sB = 0
tB =
Ans.
VzQB
It
=
125(0.035156)
= 0.377 ksi
0.015531(0.75)
Ans.
603
20 in.
20 in.
y
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*8–72. The hook is subjected to the force of 80 lb.
Determine the state of stress at point A at section a–a. The
cross section is circular and has a diameter of 0.5 in. Use the
curved-beam formula to compute the bending stress.
80 lb
1.5 in.
45⬚
The location of the neutral surface from the center of curvature of the hook, Fig. a,
can be determined from
where A = p(0.252) = 0.0625p in2
dA
LA r
= 2p A r - 2r2 - c2 B = 2p A 1.75 - 21.752 - 0.252 B = 0.11278 in.
Thus,
R =
0.0625p
= 1.74103 in.
0.11278
Then
e = r - R = 1.75 - 1.74103 = 0.0089746 in.
Referring to Fig. b, I and QA are
I =
p
(0.254) = 0.9765625(10 - 3)p in4
4
QA = 0
Consider the equilibrium of the FBD of the hook’s cut segment, Fig. c,
+ ©F = 0;
;
x
N - 80 cos 45° = 0
N = 56.57 lb
+ c ©Fy = 0;
80 sin 45° - V = 0
V = 56.57 lb
a + ©Mo = 0;
M - 80 cos 45°(1.74103) = 0
M = 98.49 lb # in
The normal stress developed is the combination of axial and bending stress. Thus,
s =
M(R - r)
N
+
A
Ae r
Here, M = 98.49 lb # in since it tends to reduce the curvature of the hook. For point
A, r = 1.5 in. Then
s =
(98.49)(1.74103 - 1.5)
56.57
+
0.0625p
0.0625p(0.0089746)(1.5)
= 9.269(103) psi = 9.27 ksi (T)
Ans.
The shear stress in contributed by the transverse shear stress only. Thus
t =
VQA
= 0
It
Ans.
The state of strees of point A can be represented by the element shown in Fig. d.
604
A
B
B a
A
R =
dA
©
LA r
©
a
A
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•8–73.
The hook is subjected to the force of 80 lb.
Determine the state of stress at point B at section a–a. The
cross section has a diameter of 0.5 in. Use the curved-beam
formula to compute the bending stress.
80 lb
1.5 in.
45⬚
The location of the neutral surface from the center of curvature of the the hook,
Fig. a, can be determined from
R =
dA
©
LA r
dA
= 2p A r - 2r2 - c2 B = 2p A 1.75 - 21.752 - 0.252 B = 0.11278 in.
LA r
Thus,
R =
0.0625p
= 1.74103 in
0.11278
Then
e = r - R = 1.75 - 1.74103 = 0.0089746 in
Referring to Fig. b, I and QB are computed as
p
(0.254) = 0.9765625(10 - 3)p in4
4
I =
QB = y¿A¿ =
4(0.25) p
c (0.252) d = 0.0104167 in3
3p
2
Consider the equilibrium of the FBD of the hook’s cut segment, Fig. c,
+ ©F = 0;
;
x
N - 80 cos 45° = 0
N = 56.57 lb
+ c ©Fy = 0;
80 sin 45° - V = 0
V = 56.57 lb
a + ©Mo = 0;
M - 80 cos 45° (1.74103) = 0
M = 98.49 lb # in
The normal stress developed is the combination of axial and bending stress. Thus,
s =
M(R - r)
N
+
A
Ae r
Here, M = 98.49 lb # in since it tends to reduce. the curvature of the hook. For point
B, r = 1.75 in. Then
s =
(98.49)(1.74103 - 1.75)
56.57
+
0.0625p
0.0625 p (0.0089746)(1.75)
= 1.62 psi (T)
Ans.
The shear stress is contributed by the transverse shear stress only. Thus,
t =
56.57 (0.0104167)
VQB
= 3.84 psi
=
It
0.9765625(10 - 3)p (0.5)
Ans.
The state of stress of point B can be represented by the element shown in Fig. d.
605
A
B
B a
A
Where A = p(0.252) = 0.0625p in2
©
a
A
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8–74. The block is subjected to the three axial loads
shown. Determine the normal stress developed at points A
and B. Neglect the weight of the block.
100 lb
250 lb
50 lb
2 in.4 in.
5 in.
2 in.
3 in.
5 in.
A
B
Mx = -250(1.5) - 100(1.5) + 50(6.5) = -200 lb # in.
My = 250(4) + 50(2) - 100(4) = 700 lb # in.
Ix =
1
1
(4)(133) + 2 a b (2)(33) = 741.33 in4
12
12
Iy =
1
1
(3)(83) + 2 a b (5)(43) = 181.33 in4
12
12
A = 4(13) + 2(2)(3) = 64 in2
s =
My x
Mx y
P
+
A
Iy
Ix
sA = -
700(4)
-200 (-1.5)
400
+
64
181.33
741.33
= -21.3 psi
sB = -
Ans.
700(2)
-200 (-6.5)
400
+
64
181.33
741.33
= -12.2 psi
Ans.
606
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8–75. The 20-kg drum is suspended from the hook
mounted on the wooden frame. Determine the state of
stress at point E on the cross section of the frame at
section a–a. Indicate the results on an element.
50 mm
25 mm
E
75 mm
Section a – a
0.5 m 0.5 m
1m
a
B
C
a
1m
30⬚
Support Reactions: Referring to the free-body diagram of member BC
shown in Fig. a,
a + ©MB = 0;
F sin 45°(1) - 20(9.81)(2) = 0
F = 554.94 N
+ ©F = 0;
:
x
554.94 cos 45° - Bx = 0
Bx = 392.4 N
+ c ©Fy = 0;
554.94 sin 45° - 20(9.81) - By = 0
By = 196.2 N
1m
b
b
75 mm
1m
D
F
A
25 mm
Internal Loadings: Consider the equilibrium of the free - body diagram of
the right segment shown in Fig. b.
Section b – b
+ ©F = 0;
:
x
N - 392.4 = 0
N = 392.4 N
+ c ©Fy = 0;
V - 196.2 = 0
V = 196.2 N
a + ©MC = 0;
196.2(0.5) - M = 0
M = 98.1 N # m
Section Properties: The cross -sectional area and the moment of inertia of the cross
section are
A = 0.05(0.075) = 3.75 A 10 - 3 B m2
I =
1
(0.05) A 0.0753 B = 1.7578 A 10 - 6 B m4
12
Referring to Fig. c, QE is
QE = y¿A¿ = 0.025(0.025)(0.05) = 3.125 A 10 - 6 B m3
Normal Stress: The normal stress is the combination of axial and bending stress.
Thus,
s =
My
N
;
A
I
For point A, y = 0.0375 - 0.025 = 0.0125 m. Then
sE =
392.4
3.75 A 10
-3
B
98.1(0.0125)
+
1.7578 A 10 - 6 B
= 802 kPa
Ans.
Shear Stress: The shear stress is contributed by transverse shear stress only. Thus,
tE =
196.2 C 31.25 A 10 - 6 B D
VQA
=
= 69.8 kPa
It
1.7578 A 10 - 6 B (0.05)
Ans.
The state of stress at point E is represented on the element shown in Fig. d.
607
75 mm
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8–75.
Continued
608
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*8–76. The 20-kg drum is suspended from the hook
mounted on the wooden frame. Determine the state of
stress at point F on the cross section of the frame at
section b–b. Indicate the results on an element.
50 mm
25 mm
E
75 mm
Section a – a
0.5 m 0.5 m
1m
a
B
C
a
1m
30⬚
1m
b
FBD sin 30°(3) - 20(9.81)(2) = 0
FBD = 261.6 N
+ c ©Fy = 0;
Ay - 261.6 cos 30° - 20(9.81) = 0
Ay = 422.75 N
+ ©F = 0;
:
x
Ax - 261.6 sin 30° = 0
Ax = 130.8 N
75 mm
1m
Support Reactions: Referring to the free-body diagram of the entire frame
shown in Fig. a,
a + ©MA = 0;
b
D
F
A
25 mm
Section b – b
Internal Loadings: Consider the equilibrium of the free - body diagram of the lower
cut segment, Fig. b,
+ ©F = 0;
:
x
130.8 - V = 0
V = 130.8 N
+ c ©Fy = 0;
422.75 - N = 0
N = 422.75 N
a + ©MC = 0;
130.8(1) - M = 0
M = 130.8 N # m
Section Properties: The cross -sectional area and the moment of inertia about the
centroidal axis of the cross section are
A = 0.075(0.075) = 5.625 A 10 - 3 B m2
I =
1
(0.075) A 0.0753 B = 2.6367 A 10 - 6 B m4
12
Referring to Fig. c, QE is
QF = y¿A¿ = 0.025(0.025)(0.075) = 46.875 A 10 - 6 B m3
Normal Stress: The normal stress is the combination of axial and bending stress.
Thus,
s =
My
N
;
A
I
For point F, y = 0.0375 - 0.025 = 0.0125 m. Then
sF =
-422.75
5.625 A 10
-3
B
130.8(0.0125)
-
2.6367 A 10 - 6 B
= -695.24 kPa = 695 kPa (C)
Ans.
609
75 mm
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8–76.
Continued
Shear Stress: The shear stress is contributed by transverse shear stress only. Thus,
130.8 c46.875 A 10 - 6 B d
VQA
=
= 31.0 kPa
tA =
It
2.6367 A 10 - 6 B (0.075)
Ans.
The state of stress at point A is represented on the element shown in Fig. d.
610
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•8–77. The eye is subjected to the force of 50 lb. Determine
the maximum tensile and compressive stresses at section a-a.
The cross section is circular and has a diameter of 0.25 in. Use
the curved-beam formula to compute the bending stress.
50 lb
0.25 in.
1.25 in.
a
Section Properties:
r = 1.25 +
0.25
= 1.375 in.
2
dA
= 2p A r - 2r2 - c2 B
LA r
= 2p A 1.375 - 21.3752 - 0.1252 B
= 0.035774 in.
A = p A 0.1252 B = 0.049087 in2
R =
A
dA
1A r
=
0.049087
= 1.372153 in.
0.035774
r - R = 1.375 - 1.372153 = 0.002847 in.
Internal Force and Moment: As shown on FBD. The internal moment must be
computed about the neutral axis. M = 68.608 lb # in is positive since it tends to
increase the beam’s radius of curvature.
Normal Stress: Applying the curved - beam formula, For tensile stress
(st)max =
=
M(Rr1)
N
+
A
Ar1(r - R)
68.608(1.372153 - 1.25)
50.0
+
0.049087
0.049087(1.25)(0.002847)
= 48996 psi = 49.0 ksi (T)
Ans.
For compressive stress
(sc)max =
=
M(R - r2)
N
+
A
Ar2(r - R)
68.608(1.372153 - 1.50)
50.0
+
0.049087
0.049087(1.50)(0.002847)
= -40826 psi = 40.8 ksi (C)
Ans.
611
a
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8–78. Solve Prob. 8–77 if the cross section is square, having
dimensions of 0.25 in. by 0.25 in.
50 lb
0.25 in.
1.25 in.
a
Section Properties:
r = 1.25 +
0.25
= 1.375 in.
2
r2
dA
1.5
= bln
= 0.25 ln
= 0.45580 in.
r1
1.25
LA r
A = 0.25(0.25) = 0.0625 in2
R =
0.0625
A
=
= 1.371204 in.
dA
0.045580
1A
r
r - R = 1.375 - 1.371204 = 0.003796 in.
Internal Force and Moment: As shown on FBD. The internal moment must be
computed about the neutral axis. M = 68.560 lb # in. is positive since it tends to
increase the beam’s radius of curvature.
Normal Stress: Applying the curved -beam formula, For tensile stress
(st)max =
=
M(R - r1)
N
+
A
Ar1(r - R)
68.560(1.371204 - 1.25)
50.0
+
0.0625
0.0625(1.25)(0.003796)
= 28818 psi = 28.8 ksi (T)
Ans.
For Compressive stress
(sc)max =
=
M(R - r2)
N
+
A
Ar2 (r - R)
68.560(1.371204 - 1.5)
50.0
+
0.0625
0.0625(1.5)(0.003796)
= -24011 psi = 24.0 ksi (C)
Ans.
612
a
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8–79. If the cross section of the femur at section a–a can
be approximated as a circular tube as shown, determine the
maximum normal stress developed on the cross section at
section a–a due to the load of 75 lb.
2 in.
75 lb
a
a
0.5 in.
1 in.
Section a – a
M
F
Internal Loadings: Considering the equilibrium for the free-body diagram of the
femur’s upper segment, Fig. a,
+ c ©Fy = 0;
N - 75 = 0
N = 75 lb
a + ©MO = 0;
M - 75(2) = 0
M = 150 lb # in
Section Properties: The cross-sectional area, the moment of inertia about the
centroidal axis of the femur’s cross section are
A = p A 12 - 0.52 B = 0.75p in2
I =
p 4
A 1 - 0.54 B = 0.234375p in4
4
Normal Stress: The normal stress is a combination of axial and bending stress. Thus,
s =
My
N
+
A
I
By inspection, the maximum normal stress is in compression.
smax =
150(1)
-75
= -236 psi = 236 psi (C)
0.75p
0.234375p
613
Ans.
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*8–80. The hydraulic cylinder is required to support a
force of P = 100 kN. If the cylinder has an inner diameter
of 100 mm and is made from a material having an allowable
normal stress of sallow = 150 MPa, determine the required
minimum thickness t of the wall of the cylinder.
P
t
100 mm
Equation of Equilibrium: The absolute pressure developed in the hydraulic
cylinder can be determined by considering the equilibrium of the free-body
diagram of the piston shown in Fig. a. The resultant force of the pressure on the
p
piston is F = pA = pc A 0.12 B d = 0.0025pp. Thus,
4
©Fx¿ = 0; 0.0025pp - 100 A 103 B = 0
p = 12.732 A 106 B Pa
Normal Stress: For the cylinder, the hoop stress is twice as large as the
longitudinal stress,
sallow =
pr
;
t
150 A 106 B =
12.732 A 106 B (50)
t
t = 4.24 mm
Since
Ans.
r
50
=
= 11.78 7 10, thin -wall analysis is valid.
t
4.24
614
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•8–81. The hydraulic cylinder has an inner diameter of
100 mm and wall thickness of t = 4 mm. If it is made from
a material having an allowable normal stress of sallow =
150 MPa, determine the maximum allowable force P.
P
t
100 mm
Normal Stress: For the hydraulic cylinder, the hoop stress is twice as large as the
longitudinal stress.
Since
50
r
=
= 12.5 7 10, thin-wall analysis can be used.
t
4
sallow =
pr
;
t
150 A 106 B =
p(50)
4
p = 12 A 106 B MPa
Ans.
Equation of Equilibrium: The resultant force on the piston is
F = pA = 12 A 106 B c
p
A 0.12 B d = 30 A 103 B p. Referring to the free-body diagram of
4
the piston shown in Fig. a,
©Fx¿ = 0; 30 A 103 B p - P = 0
P = 94.247 A 103 B N = 94.2 kN
Ans.
615
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8–82. The screw of the clamp exerts a compressive force
of 500 lb on the wood blocks. Determine the maximum
normal stress developed along section a-a. The cross
section there is rectangular, 0.75 in. by 0.50 in.
4 in.
Internal Force and Moment: As shown on FBD.
a
Section Properties:
A = 0.5(0.75) = 0.375 in2
I =
a
1
(0.5) A 0.753 B = 0.017578 in4
12
0.75 in.
Maximum Normal Stress: Maximum normal stress occurs at point A.
smax = sA =
=
Mc
N
+
A
I
2000(0.375)
500
+
0.375
0.017578
= 44000 psi = 44.0 ksi (T)
Ans.
8–83. Air pressure in the cylinder is increased by exerting
forces P = 2 kN on the two pistons, each having a radius
of 45 mm. If the cylinder has a wall thickness of 2 mm,
determine the state of stress in the wall of the cylinder.
p =
2(103)
P
= 314 380.13 Pa
=
A
p(0.0452)
s1 =
pr
314 380.13(0.045)
=
= 7.07 MPa
t
0.002
P
47 mm
Ans.
s2 = 0
P
Ans.
The pressure P is supported by the surface of the pistons in the longitudinal
direction.
*8–84. Determine the maximum force P that can be
exerted on each of the two pistons so that the circumferential
stress component in the cylinder does not exceed 3 MPa.
Each piston has a radius of 45 mm and the cylinder has a
wall thickness of 2 mm.
s =
pr
;
t
3(106) =
P
47 mm
p(0.045)
0.002
P = 133.3 kPa
Ans.
P = pA = 133.3 A 103 B (p)(0.045)2 = 848 N
Ans.
616
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•8–85.
The cap on the cylindrical tank is bolted to the tank
along the flanges. The tank has an inner diameter of 1.5 m
and a wall thickness of 18 mm. If the largest normal stress is
not to exceed 150 MPa, determine the maximum pressure
the tank can sustain. Also, compute the number of bolts
required to attach the cap to the tank if each bolt has a
diameter of 20 mm. The allowable stress for the bolts is
1sallow2b = 180 MPa.
Hoop Stress for Cylindrical Tank: Since
750
r
=
= 41.7 7 10, then thin wall
t
18
analysis can be used. Applying Eq. 8–1
s1 = sallow =
150 A 106 B =
pr
t
p(750)
18
p = 3.60 MPa
Ans.
Force Equilibrium for the Cap:
+ c ©Fy = 0;
3.60 A 106 B C p A 0.752 B D - Fb = 0
Fb = 6.3617 A 106 B N
Allowable Normal Stress for Bolts:
(sallow)b =
180 A 106 B =
P
A
6.3617(106)
n C p4 (0.022) D
n = 112.5
Use n = 113 bolts
Ans.
617
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8–86. The cap on the cylindrical tank is bolted to the tank
along the flanges. The tank has an inner diameter of 1.5 m
and a wall thickness of 18 mm. If the pressure in the tank is
p = 1.20 MPa, determine the force in each of the 16 bolts
that are used to attach the cap to the tank. Also, specify the
state of stress in the wall of the tank.
Hoop Stress for Cylindrical Tank: Since
750
r
=
= 41.7 7 10, then thin wall
t
18
analysis can be used. Applying Eq. 8–1
s1 =
pr
1.20(106)(750)
=
= 50.0 MPa
t
18
Ans.
Longitudinal Stress for Cylindrical Tank:
s2 =
pr
1.20(106)(750)
=
= 25.0 MPa
2t
2(18)
Ans.
Force Equilibrium for the Cap:
+ c ©Fy = 0;
1.20 A 106 B C p A 0.752 B D - 16Fb = 0
Fb = 132536 N = 133 kN
Ans.
618
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9–1. Prove that the sum of the normal stresses
sx + sy = sx¿ + sy¿ is constant. See Figs. 9–2a and 9–2b.
Stress Transformation Equations: Applying Eqs. 9-1 and 9-3 of the text.
sx¿ + sy¿ =
sx + sy
2
sx - sy
+
2
cos 2u + txy sin 2u
sx + sy
+
2
sx - sy
-
2
cos 2u - txy sin 2u
sx¿ + sy¿ = sx + sy
(Q.E.D.)
619
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9–2. The state of stress at a point in a member is shown on
the element. Determine the stress components acting on
the inclined plane AB. Solve the problem using the method
of equilibrium described in Sec. 9.1.
A
8 ksi
2 ksi
5 ksi
60⬚
B
Referring to Fig a, if we assume that the areas of the inclined plane AB is ¢A, then
the area of the horizontal and vertical of the triangular element are ¢A cos 60° and
¢A sin 60° respectively. The forces act acting on these two faces indicated on the
FBD of the triangular element, Fig. b.
¢Fx¿ + 2¢A sin 60° cos 60° + 5¢ A sin 60° sin 60°
+Q©Fx¿ = 0;
+ 2¢A cos 60° sin 60° - 8¢A cos 60° cos 60° = 0
¢Fx¿ = -3.482 ¢A
¢Fy¿ + 2¢A sin 60° sin 60° - 5¢ A sin 60° cos 60°
+a©Fy¿ = 0;
- 8¢A cos 60° sin 60° - 2¢A cos 60° cos 60° = 0
¢Fy¿ = 4.629 ¢A
From the definition,
sx¿ = lim¢A:0
¢Fx¿
= -3.48 ksi
¢A
tx¿y¿ = lim¢A:0
¢Fy¿
Ans.
Ans.
= 4.63 ksi
¢A
The negative sign indicates that sx¿, is a compressive stress.
620
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9–3. The state of stress at a point in a member is shown on
the element. Determine the stress components acting on
the inclined plane AB. Solve the problem using the method
of equilibrium described in Sec. 9.1.
500 psi
B
60⬚
A
Referring to Fig. a, if we assume that the area of the inclined plane AB is ¢A, then
the areas of the horizontal and vertical surfaces of the triangular element are
¢A sin 60° and ¢A cos 60° respectively. The force acting on these two faces are
indicated on the FBD of the triangular element, Fig. b
+R©Fx¿ = 0;
¢Fx¿ + 500 ¢A sin 60° sin 60° + 350¢A sin 60° cos 60°
+350¢A cos 60° sin 60° = 0
¢Fx¿ = -678.11 ¢A
+Q©Fy¿ = 0;
¢Fy¿ + 350¢A sin 60° sin 60° - 500¢A sin 60° cos 60°
-350¢A cos 60° cos 60° = 0
¢Fy¿ = 41.51 ¢A
From the definition
sx¿ = lim¢A:0
tx¿y¿ = lim¢A:0
¢Fx¿
= -6.78 psi
¢A
¢Fy¿
Ans.
Ans.
= 41.5 psi
¢A
The negative sign indicates that sx¿, is a compressive stress.
621
350 psi
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*9–4. The state of stress at a point in a member is shown
on the element. Determine the stress components acting on
the inclined plane AB. Solve the problem using the method
of equilibrium described in Sec. 9.1.
A
650 psi
¢Fx¿ - 400(¢Acos 60°)cos 60° + 650(¢ A sin 60°)cos 30° = 0
Q+ ©Fx¿ = 0
400 psi
60⬚
¢Fx¿ = -387.5¢A
¢Fy¿ - 650(¢Asin 60°)sin 30° - 400(¢ A cos 60°)sin 60° = 0
a+ ©Fy¿ = 0
B
¢Fy¿ = 455 ¢A
sx¿ = lim¢A:0
sx¿y¿ = lim¢A:0
¢Fx¿
= -388 psi
¢A
¢Fy¿
Ans.
Ans.
= 455 psi
¢A
The negative sign indicates that the sense of sx¿, is opposite to that shown on FBD.
•9–5.
Solve Prob. 9–4 using the stress-transformation
equations developed in Sec. 9.2.
sy = 400 psi
sx = -650 psi
sx¿ =
=
sx + sy
sx - sy
+
2
2
txy = 0
A
400 psi
u = 30°
650 psi
cos 2u + txy sin 2u
60⬚
-650 + 400
-650 - 400
+
cos 60° + 0 = -388 psi
2
2
Ans.
B
The negative sign indicates sx¿, is a compressive stress.
tx¿y¿ = = -a
sx - sy
2
sin 2u + txy cos 2u
-650 - 400
bsin 60° = 455 psi
2
Ans.
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9–6. The state of stress at a point in a member is shown on
the element. Determine the stress components acting on
the inclined plane AB. Solve the problem using the method
of equilibrium described in Sec. 9.1.
90 MPa
A
35 MPa
60⬚
30⬚
R+ ©Fy¿ = 0
B
50 MPa
¢Fy¿ - 50¢A sin 30° cos 30° - 35¢A sin 30° cos 60° +
90¢A cos 30° sin 30° + 35¢A cos 30° sin 60° = 0
¢Fy¿ = -34.82¢A
b+ ©Fx¿ = 0
¢Fx¿ - 50¢A sin 30° sin 30° + 35¢A sin 30° sin 60°
-90¢A cos 30° cos 30° + 35¢A cos 30° cos 60° = 0
¢Fx¿ = 49.69 ¢A
sx¿ = lim¢A:0
¢Fx¿
= 49.7 MPa
¢A
tx¿y¿ = lim¢A:0
¢Fy¿
Ans.
Ans.
= -34.8 MPa
¢A
The negative signs indicate that the sense of sx¿, and tx¿y¿ are opposite to the shown
on FBD.
9–7. Solve Prob. 9–6 using the stress-transformation
equations developed in Sec. 9.2. Show the result on a sketch.
90 MPa
A
35 MPa
60⬚
30⬚
sy = 50 MPa
sx = 90 MPa
sx¿ =
=
sx + sy
sx - sy
+
2
2
txy = -35 MPa
u = -150°
cos 2u + txy sin 2u
90 - 50
90 + 50
+
cos(-300°) + (-35) sin ( -300°)
2
2
= 49.7 MPa
tx¿y¿ = -
sx - sy
= -a
2
Ans.
sin 2u + txy cos 2u
90 - 50
bsin(-300°) + ( -35) cos ( -300°) = -34.8 MPa
2
The negative sign indicates tx¿y¿ acts in -y¿ direction.
623
Ans.
B
50 MPa
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*9–8. Determine the normal stress and shear stress acting
on the inclined plane AB. Solve the problem using the
method of equilibrium described in Sec. 9.1.
45 MPa
B
80 MPa
45⬚
A
Force Equllibrium: Referring to Fig. a, if we assume that the area of the inclined
plane AB is ¢A, then the area of the vertical and horizontal faces of the triangular
sectioned element are ¢A sin 45° and ¢A cos 45°, respectively. The forces acting on
the free-body diagram of the triangular sectioned element, Fig. b, are
©Fx¿ = 0;
¢Fx¿ + c45 A 106 B ¢A sin 45° dcos 45° + c45 A 106 B ¢A cos 45° dsin 45°
- c80 A 106 B ¢A sin 45° dcos 45° = 0
¢Fx¿ = -5 A 106 B ¢A
©Fy¿ = 0;
¢Fy¿ + c45 A 106 B ¢A cos 45° dcos 45° - c45 A 106 B ¢A sin 45° dsin 45°
- c80 A 106 B ¢ A sin 45° dsin 45° = 0
¢Fy¿ = 40 A 106 B ¢A
Normal and Shear Stress: From the definition of normal and shear stress,
sx¿ = lim¢A:0
¢Fx¿
= -5 MPa
¢A
tx¿y¿ = lim¢A:0
¢Fy¿
Ans.
Ans.
= 40 MPa
¢A
The negative sign indicates that sx¿ is a compressive stress.
624
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•9–9.
Determine the normal stress and shear stress acting
on the inclined plane AB. Solve the problem using the
stress transformation equations. Show the result on the
sectioned element.
45 MPa
80 MPa
45⬚
Stress Transformation Equations:
u = +135° (Fig. a)
sx = 80 MPa
sy = 0
txy = 45 MPa
we obtain,
sx¿ =
=
sx + sy
sx - sy
+
2
2
cos u + txysin 2u
80 - 0
80 + 0
+
cos 270 + 45 sin 270°
2
2
Ans.
= -5 MPa
tx¿y¿ = -
= -
sx - sy
2
B
sinu + txy cos 2u
80 - 0
sin 270° + 45 cos 270°
2
= 40 MPa
Ans.
The negative sign indicates that sx¿ is a compressive stress. These results are
indicated on the triangular element shown in Fig. b.
625
A
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9–10. The state of stress at a point in a member is shown
on the element. Determine the stress components acting on
the inclined plane AB. Solve the problem using the method
of equilibrium described in Sec. 9.1.
2 ksi
A
3 ksi
30⬚
4 ksi
B
Force Equllibrium: For the sectioned element,
¢Fy¿ - 3(¢A sin 30°) sin 60° + 4(¢ A sin 30°)sin 30°
a+ ©Fy¿ = 0;
-2(¢A cos 30°) sin 30° - 4(¢A cos 30°) sin 60° = 0
¢Fy¿ = 4.165 ¢A
¢Fx¿ + 3(¢A sin 30°) cos 60° + 4(¢ A sin 30°)cos 30°
Q+ ©Fx¿ = 0;
-2(¢A cos 30°) cos 30° + 4(¢A cos 30°) cos 60° = 0
¢Fx¿ = -2.714 ¢A
Normal and Shear Stress: For the inclined plane.
sx = lim¢A:0
tx¿y¿ = lim¢A:0
¢Fx¿
= -2.71 ksi
¢A
¢Fy¿
Ans.
Ans.
= 4.17 ksi
¢A
Negative sign indicates that the sense of sx¿, is opposite to that shown on FBD.
9–11. Solve Prob. 9–10 using the stress-transformation
equations developed in Sec. 9.2. Show the result on a sketch.
2 ksi
Normal and Shear Stress: In accordance with the established sign convention,
u = +60°
sx = -3 ksi
sy = 2 ksi
A
txy = -4 ksi
3 ksi
30⬚
Stress Transformation Equations: Applying Eqs. 9-1 and 9-2.
sx¿ =
=
sx + sy
sx - sy
+
2
2
B
cos 2u + txy sin 2u
-3 - 2
-3 + 2
+
cos 120° + (-4 sin 120°)
2
2
Ans.
= -2.71 ksi
tx¿y¿ = -
= -
sx - sy
2
4 ksi
sin 2u + txy cos 2u
-3 - 2
sin 120° + (-4 cos 120°)
2
= 4.17 ksi
Ans.
Negative sign indicates sx¿, is a compressive stress
626
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*9–12. Determine the equivalent state of stress on an
element if it is oriented 50° counterclockwise from the
element shown. Use the stress-transformation equations.
10 ksi
16 ksi
sy = 0
sx = -10 ksi
txy = -16 ksi
u = +50°
sx¿ =
=
sx + sy
= -a
=
2
cos 2u + txy sin 2u
-10 - 0
-10 + 0
+
cos 100° + ( -16)sin 100° = -19.9 ksi
2
2
tx¿y¿ = - a
sy¿ =
sx - sy
+
2
sx - sy
2
b sin 2u + txy cos 2u
-10 - 0
b sin 100° + (-16)cos 100° = 7.70 ksi
2
sx + sy
2
sx - sy
-
Ans.
2
Ans.
cos 2u - txy sin 2u
-10 + 0
-10 - 0
- a
bcos 100° - (-16)sin 100° = 9.89 ksi
2
2
627
Ans.
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•9–13.
Determine the equivalent state of stress on an
element if the element is oriented 60° clockwise from the
element shown. Show the result on a sketch.
350 psi
75 psi
200 psi
In accordance to the established sign covention,
u = -60° (Fig. a)
sx = 200 psi
sy = -350 psi
txy = 75 psi
Applying Eqs 9-1, 9-2 and 9-3,
sx¿ =
=
sx + sy
sx - sy
+
2
2
cos 2u + txy sin 2u
200 - ( -350)
200 + (-350)
+
cos ( -120°) + 75 sin (-120°)
2
2
= -277.45 psi = -277 psi
sy¿ =
=
sx + sy
sx - sy
-
2
2
Ans.
cos 2u - txy sin 2u
200 - ( -350)
200 + (-350)
cos ( -120°) - 75 sin ( -120°)
2
2
= 127.45 psi = 127 psi
tx¿y¿ = -
= -
sx - sy
2
Ans.
sin 2u + txy cos 2u
200 - (-350)
sin (-120°) + 75 cos (-120°)
2
= 200.66 psi = 201 psi
Ans.
Negative sign indicates that sx¿ is a compressive stress. These result, can be
represented by the element shown in Fig. b.
628
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9–14. The state of stress at a point is shown on the element.
Determine (a) the principal stress and (b) the maximum
in-plane shear stress and average normal stress at the point.
Specify the orientation of the element in each case. Show
the results on each element.
30 ksi
12 ksi
sx = -30 ksi
sy = 0
txy = -12 ksi
a)
sx + sy
s1, 2 =
;
2
C
a
sx - sy
2
2
b + txy 2 =
-30 + 0
-30 - 0 2
;
a
b + (-12)2
2
C
2
s1 = 4.21 ksi
Ans.
s2 = -34.2 ksi
Ans.
Orientation of principal stress:
txy
tan 2uP =
(sx - sy)>2
uP = 19.33° and
-12
= 0.8
(-30 -0)>2
=
-70.67°
Use Eq. 9-1 to determine the principal plane of s1 and s2.
sx + sy
sx¿ =
sx - sy
+
2
2
cos 2u + txy sin 2u
u = 19.33°
sx¿ =
-30 + 0
-30 - 0
+
cos 2(19.33°) + (-12)sin 2(19.33°) = -34.2 ksi
2
2
Therefore uP2 = 19.3°
Ans.
and uP1 = -70.7°
Ans.
b)
tmaxin-plane =
savg =
C
a
sx - sy
2
sx + sy
2
=
2
b + txy 2 =
-30 - 0 2
b + (-12)2 = 19.2 ksi
C
2
a
-30 + 0
= -15 ksi
2
Ans.
Ans.
Orientation of max, in - plane shear stress:
tan 2uP =
-(sx - sy)>2
=
txy
uP = -25.2°
and
-(-30 - 0)>2
= -1.25
-12
64.3°
Ans.
By observation, in order to preserve equllibrium along AB, tmax has to act in the
direction shown in the figure.
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9–15. The state of stress at a point is shown on the element.
Determine (a) the principal stress and (b) the maximum
in-plane shear stress and average normal stress at the point.
Specify the orientation of the element in each case. Show
the results on each element.
80 MPa
50 MPa
60 MPa
In accordance to the established sign convention,
sx = -60 MPa
s1, 2 =
=
sy = -80 MPa
sx + sy
;
2
C
a
sx - sy
2
txy = 50 MPa
2
b + txy 2
-60 + (-80)
-60 - (-80) 2
;
c
d + 502
2
C
2
= -70 ; 22600
s2 = -121 MPa
s1 = -19.0 MPa
txy
tan 2uP =
=
(sx - sy)>2
uP = 39.34°
Ans.
50
= 5
[-60 - (-80)]>2
and
-50.65°
Substitute u = 39.34° into Eq. 9-1,
sx¿ =
=
sx + sy
sx - sy
+
2
2
cos 2u + txy sin 2u
-60 + (-80)
-60 - ( -80)
+
cos 78.69° + 50 sin 78.69°
2
2
= -19.0 MPa = s1
Thus,
(uP)1 = 39.3°
Ans.
(uP)2 = -50.7°
The element that represents the state of principal stress is shown in Fig. a.
t max
in-plane
=
C
a
sx - sy
2
tan 2uS =
2
b + txy 2 =
-(sx - sy)>2
=
txy
-60 - (-80) 2
d + 502 = 51.0 MPa
C
2
c
-[-60 - (-80)]>2
= -0.2
50
uS = -5.65° and 84.3°
By Inspection, t max
Ans.
Ans.
has to act in the sense shown in Fig. b to maintain
in-plane
equilibrium.
savg =
sx + sy
2
=
-60 + (-80)
= -70 MPa
2
The element that represents the state of maximum in - plane shear stress is shown in
Fig. c.
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9–15. Continued
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*9–16. The state of stress at a point is shown on the
element. Determine (a) the principal stress and (b) the
maximum in-plane shear stress and average normal stress at
the point. Specify the orientation of the element in each case.
Sketch the results on each element.
60 MPa
30 MPa
45 MPa
sx = 45 MPa
sy = -60 MPa
txy = 30 MPa
a)
s1, 2 =
=
sx + sy
;
2
C
a
sx - sy
2
2
b + txy 2
45 - (-60) 2
45 - 60
a
;
b + (30)2
2
C
2
s1 = 53.0 MPa
Ans.
s2 = -68.0 MPa
Ans.
Orientation of principal stress:
tan 2uP =
txy
(sx - sy)>2
uP = 14.87,
=
30
= 0.5714
(45 - (-60))>2
-75.13
Use Eq. 9-1 to determine the principal plane of s1 and s2:
sx¿ =
=
sx + sy
sx - sy
+
2
2
cos 2u + txy sin 2u,
where u = 14.87°
45 + (-60)
45 - (-60)
+
cos 29.74° + 30 sin 29.74° = 53.0 MPa
2
2
Therefore uP1 = 14.9°
Ans.
and uP2 = -75.1°
Ans.
b)
tmaxin-plane =
savg =
C
a
sx - sy
2
sx - sy
2
=
2
b + txy 2 =
45 - (-60) 2
b + 302 = 60.5 MPa
C
2
a
45 + (-60)
= -7.50 MPa
2
Ans.
Ans.
Orientation of maximum in - plane shear stress:
tan 2uS =
-(sx - sy)>2
txy
=
-(45 - ( -60))>2
= -1.75
30
uS = -30.1°
Ans.
uS = 59.9°
Ans.
and
By observation, in order to preserve equilibrium along AB, tmax has to act in the
direction shown.
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•9–17.
Determine the equivalent state of stress on an
element at the same point which represents (a) the principal
stress, and (b) the maximum in-plane shear stress and the
associated average normal stress. Also, for each case,
determine the corresponding orientation of the element
with respect to the element shown. Sketch the results on
each element.
75 MPa
125 MPa
50 MPa
Normal and Shear Stress:
sx = 125 MPa
sy = -75 MPa
txy = -50 MPa
In - Plane Principal Stresses:
s1,2 =
=
sx - sy
;
2
B
a
sx - sy
2
2
b + txy 2
125 + (-75)
125 - (-75) 2
a
;
b + (-50)2
2
2
B
= 25; 212500
s2 = -86.8 MPa
s1 = 137 MPa
Ans.
Orientation of Principal Plane:
tan 2uP =
txy
A sx - sy B >2
-50
=
A 125 -(-75) B >2
= -0.5
up = -13.28° and 76.72°
Substitute u = -13.28° into
sx¿ =
=
sx + sy
sx - sy
+
2
2
cos 2u + txy sin 2u
125 + (-75)
125 - (-75)
+
cos(-26.57°)+(-50) sin(-26.57°)
2
2
= 137 MPa = s1
Thus,
A up B 1 = -13.3° and A up B 2 = 76.7°
Ans.
125 - (-75)>(-50)
The element that represents the state of principal stress is shown in Fig. a.
Maximum In - Plane Shear Stress:
t max
in-plane
=
C
¢
sx - sy
2
2
≤ + txy 2 =
-100 - 0 2
b + 252 = 112 MPa
2
B
a
Orientation of the Plane of Maximum In - Plane Shear Stress:
tan 2us = -
A sx - sy B >2
txy
= -
A 125 - (-75) B >2
= 2
-50
us = 31.7° and 122°
633
Ans.
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9–17.
Continued
By inspection, t max
has to act in the same sense shown in Fig. b to maintain
in-plane
equilibrium.
Average Normal Stress:
savg =
sx + sy
2
=
125 + (-75)
= 25 MPa
2
Ans.
The element that represents the state of maximum in - plane shear stress is shown in
Fig. c.
634
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sy
9–18. A point on a thin plate is subjected to the two
successive states of stress shown. Determine the resultant
state of stress represented on the element oriented as
shown on the right.
=
sx¿ + sy¿
sx¿ - sy¿
+
2
2
ⴙ
60⬚
Stress Transformation Equations: Applying Eqs. 9-1, 9-2, and 9-3
u = -30°, sx¿ = -200 MPa,
to
element
(a)
with
sy¿ = -350 MPa and tx¿y¿ = 0.
(sx)a =
350 MPa
cos 2u + tx¿y¿ sin 2u
-200 - (-350)
-200 + (-350)
+
cos (-60°) + 0
2
2
= -237.5 MPa
A sy B a =
=
sx¿ + sy¿
sx¿ - sy¿
-
2
2
cos 2u - tx¿y¿ sin 2u
-200 - (-350)
-200 + (-350)
cos (-60°) - 0
2
2
= -312.5 MPa
A txy B a = = -
sx¿ - sy¿
2
sin 2u + tx¿y¿ cos 2u
-200 - (-350)
sin (-60°) + 0
2
= 64.95 MPa
For element (b), u = 25°, sx¿ = sy¿ = 0 and sx¿y¿ = 58 MPa.
(sx)b =
sx¿ + sy¿
sx¿ - sy¿
+
2
2
cos 2u + tx¿y¿ sin 2u
= 0 + 0 + 58 sin 50°
= 44.43 MPa
A sy B b =
sx¿ + sy¿
sx¿ - sy¿
-
2
2
cos 2u - tx¿y¿ sin 2u
= 0 - 0 - 58 sin 50°
= -44.43 MPa
A txy B b = -
sx¿ - sy¿
2
58 MPa
200 MPa
sin 2u + tx¿y¿ cos 2u
= -0 + 58 cos 50°
= 37.28 MPa
Combining the stress components of two elements yields
ss = (sx)a + (sx)b = -237.5 + 44.43 = -193 MPa
Ans.
sy = A sy B a + A sy B b = -312.5 - 44.43 = -357 MPa
Ans.
txy = A txy B a + A txy B b = 64.95 + 37.28 = 102 MPa
Ans.
635
25⬚
ⴝ
txy
sx
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9–19. The state of stress at a point is shown on the element.
Determine (a) the principal stress and (b) the maximum
in-plane shear stress and average normal stress at the point.
Specify the orientation of the element in each case. Sketch
the results on each element.
160 MPa
120 MPa
In accordance to the established sign Convention,
sx = 0
sy = 160 MPa
s1, 2 =
=
sx + sy
;
2
B
a
txy = -120 MPa
sx - sy
2
2
b + t2xy
0 + 160
0 - 160 2
;
a
b + (-120)2
2
2
B
= 80 ; 220800
s2 = -64.2 MPa
s1 = 224 MPa
tan 2up =
txy
(sx - sy)>2
up = 28.15°
=
Ans.
-120
= 1.5
(0 - 160)>2
and -61.85°
Substitute u = 28.15° into Eq. 9-1,
sx¿ =
=
sx + sy
sx - sy
+
2
2
cos 2u + txy sin 2u
0 + 160
0 - 160
+
cos 56.31° + (-120) sin 56.31°
2
2
= -64.22 = s2
Thus,
(up)1 = -61.8°
Ans.
(up)2 = 28.2°
The element that represents the state of principal stress is shown in Fig. a
tmax
in-plane
=
B
a
sx - sy
2
tan 2us =
0 - 160 2
b + (-120)2 = 144 MPa
2
B
a
-(sx - sy)>2
us = -16.8°
By inspection,
equilibrium.
2
b + t2xy =
tmax
in-plane
savg =
txy
Ans.
-(0 - 160)>2
= -0.6667
-120
=
Ans.
and 73.2°
has to act in the sense shown in Fig. b to maintain
sx + sy
2
=
0 + 160
= 80 MPa
2
Ans.
The element that represents the state of Maximum in - plane shear stress is shown in
Fig. (c)
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9–19. Continued
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*9–20. The stress acting on two planes at a point is
indicated. Determine the normal stress sb and the principal
stresses at the point.
a
4 ksi
60⬚
45⬚
b
2 ksi
sb
a
Stress Transformation Equations: Applying Eqs. 9-2 and 9-1 with u = -135°,
sy = 3.464 ksi, txy = 2.00 ksi, tx¿y¿ = -2 ksi, and sx¿ = sb¿.,
tx¿y¿ = -
-2 = -
sx - sy
2
sin 2u + txy cos 2u
sx - 3.464
sin (-270°) + 2cos ( -270°)
2
sx = 7.464 ksi
sx¿ =
sy =
sx - sy
sx - sy
+
2
2
cos 2u + txy sin 2u
7.464 - 3.464
7.464 + 3.464
+
cos (-270°) + 2sin ( -270°)
2
2
Ans.
= 7.46 ksi
In - Plane Principal Stress: Applying Eq. 9-5.
s1, 2 =
=
sx + sy
2
;
B
a
sx - sy
2
2
b + t2xy
7.464 - 3.464 2
7.464 + 3.464
;
a
b + 22
2
2
B
= 5.464 ; 2.828
s1 = 8.29 ksi
s2 = 2.64 ksi
Ans.
638
b
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•9–21. The stress acting on two planes at a point is
indicated. Determine the shear stress on plane a–a and the
principal stresses at the point.
b
a
ta
45⬚
60 ksi
60⬚
txy = 60 cos 60° = 30 ksi
sa =
80 =
sx + sy
sx - sy
+
2
2
51.962 - sy
51.962 + sy
+
2
cos 2u + txy sin 2u
2
cos (90°) + 30 sin (90°)
sy = 48.038 ksi
ta = - a
= -a
sx - sy
2
b sin 2u + txy cos u
51.962 - 48.038
bsin (90°) + 30 cos (90°)
2
ta = -1.96 ksi
s1, 2 =
=
Ans.
sx + sy
2
;
C
a
sx - sy
2
2
b + t2xy
51.962 - 48.038 2
51.962 + 48.038
;
a
b + (30)2
2
C
2
s1 = 80.1 ksi
Ans.
s2 = 19.9 ksi
Ans.
639
90⬚
a
b
sx = 60 sin 60° = 51.962 ksi
80 ksi
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9–22. The T-beam is subjected to the distributed loading
that is applied along its centerline. Determine the principal
stress at point A and show the results on an element located
at this point.
100 kN/m
A
1m
0.5 m
200 mm
75 mm
'
©yA
0.1(0.2)(0.02) + 0.21(0.02)(0.2)
=
= 0.155 m
©A
0.2(0.02) + 0.02(0.2)
1
(0.02)(0.2 3) + 0.02(0.2)(0.155 - 0.1)2
12
I =
+
1
(0.2)(0.023) + 0.2(0.02)(0.21 - 0.155)2
12
= 37.6667(10 - 6) m4
Referring to Fig. b,
QA = y¿A¿ = 0.1175(0.075)(0.02) = 0.17625(10 - 3) m3
Using the method of sections and considering the FBD of the left cut segment of the
beam, Fig. c,
+ c ©Fy = 0;
V - 100(1) = 0
a + ©MC = 0;
100(1)(0.5) - M = 0 M = 50 kN # m
V = 100 kN
The normal stress developed is contributed by bending stress only. For point A,
y = 0.155 - 0.075 = 0.08 m. Thus
s =
My
50(103) (0.08)
= 106 MPa
=
I
37.6667(10 - 6)
The shear stress is contributed by the transverse shear stress only. Thus,
t =
100(103)[0.17625(10 - 3)]
VQA
= 23.40(106)Pa = 23.40 MPa
=
It
37.6667(10 - 6) (0.02)
The state of stress of point A can be represented by the element shown in Fig. c.
Here, sx = -106.19 MPa, sy = 0 and txy = 23.40 MPa.
s1, 2 =
=
sx + sy
2
;
B
a
sx - sy
2
2
b + txy 2
-106.19 - 0 2
-106.19 + 0
;
b + 23.402
a
2
2
B
= -53.10 ; 58.02
s1 = 4.93 MPa
20 mm
200 mm
20 mm
The location of the centroid c of the T cross-section, Fig. a, is
y =
A
s2 = -111 MPa
Ans.
640
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9–22.
Continued
tan 2up =
txy
(sx - sy)>2
up = -11.89°
=
ans
23.40
= -0.4406
( -106.19 - 0)>2
78.11°
Substitute u = -11.89°,
sx¿ =
=
sx + sy
2
sx - sy
+
2
cos 2u + txy sin 2u
-106.19 + 0
-106.19 - 0
+
cos (-23.78°) + 23.40 5m (-23.78°)
2
2
= -111.12 MPa = s2
Thus,
(up)1 = 78.1°
Ans.
(up)2 = -11.9°
The state of principal stress can be represented by the element shown in Fig. e.
641
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•9–23.
The wood beam is subjected to a load of 12 kN. If a
grain of wood in the beam at point A makes an angle of 25°
with the horizontal as shown, determine the normal and
shear stress that act perpendicular and parallel to the grain
due to the loading.
I =
12 kN
1m
2m
A
25⬚
300 mm
75 mm
1
(0.2)(0.3)3 = 0.45(10 - 3) m4
12
QA = yA¿ = 0.1125(0.2)(0.075) = 1.6875(10 - 3) m3
sA =
MyA
13.714(103)(0.075)
= 2.2857 MPa (T)
=
I
0.45(10 - 3)
tA =
6.875(103)(1.6875)(10 - 3)
VQA
= 0.1286 MPa
=
It
0.45(10 - 3)(0.2)
sx = 2.2857 MPa
sx¿ =
sx¿ =
sx + sy
sx - sy
+
2
sy = 0
2
txy = -0.1286 MPa
u = 115°
cos 2u + txy sin 2u
2.2857 - 0
2.2857 + 0
+
cos 230° + (-0.1286)sin 230°
2
2
Ans.
= 0.507 MPa
tx¿y¿ = -
sx - sy
= -a
2
sin 2u + txy cos 2u
2.2857 - 0
b sin 230° + (-0.1286)cos 230°
2
= 0.958 MPa
Ans.
642
4m
200 mm
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*9–24. The wood beam is subjected to a load of 12 kN.
Determine the principal stress at point A and specify the
orientation of the element.
12 kN
1m
2m
A
25⬚
I =
300 mm
75 mm
1
(0.2)(0.3)3 = 0.45(10 - 3) m4
12
QA = yA¿ = 0.1125(0.2)(0.075) = 1.6875(10 - 3) m3
sA =
13.714(103)(0.075)
MyA
= 2.2857 MPa (T)
=
I
0.45(10 - 3)
tA =
VQA
6.875(103)(1.6875)(10 - 3)
= 0.1286 MPa
=
It
0.45(10 - 3)(0.2)
sx = 2.2857 MPa
s1, 2 =
=
sx + sy
;
2
C
sy = 0
txy = -0.1286 MPa
sx - sy
b + t2xy
a
2
2
2.2857 - 0 2
2.2857 + 0
;
a
b + (-0.1286)2
2
C
2
s1 = 2.29 MPa
Ans.
s2 = -7.20 kPa
Ans.
tan 2up =
txy
(sx - sy)>2
=
-0.1286
(2.2857 - 0)>2
up = -3.21°
Check direction of principal stress:
sx¿ =
=
sx + sy
2
sx - sy
+
2
cos 2u + txy sin 2u
2.2857 + 0
2.2857 - 0
+
cos (-6.42°) - 0.1285 sin (-6.42)
2
2
= 2.29 MPa
643
4m
200 mm
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•9–25.
The bent rod has a diameter of 20 mm and is
subjected to the force of 400 N. Determine the principal
stress and the maximum in-plane shear stress that is
developed at point A. Show the results on a properly
oriented element located at this point.
100 mm
150 mm
400 N
400 N
250 mm
A
Using the method of sections and consider the FBD of the rod’s left cut segment,
Fig. a.
+ ©F = 0;
:
x
N - 400 = 0 N = 400 N
a + ©MC = 0;
400(0.25) - M = 0 M = 100 N # m
A = p(0.012) = 0.1(10 - 3) p m2
p
(0.014) = 2.5(10 - 9)p m4
4
I =
The normal stress developed is the combination of axial and bending stress. Thus,
My
N
;
A
I
s =
For point A, y = C = 0.01 m.
s =
100(0.01)
400
-3
0.1(10 )p
2.5(10 - 9)p
= -126.05 (106)Pa = 126.05 MPa (C)
Since no torque and transverse shear acting on the cross - section,
t = 0
The state of stress at point A can be represented by the element shown in Fig. b
Here, sx = -126.05 MPa, sy = 0 and txy = 0. Since no shear stress acting on the
element
s1 = sy = 0
s2 = sx = -126 MPa
Ans.
Thus, the state of principal stress can also be represented by the element shown in Fig. b.
tmax
in-plane
=
B
a
sx - sy
2
tan 2us = -
2
b + t2xy =
(sx - sy)>2
= -
txy
us = 45°
tx¿y¿ = -
-126.05 - 0 2
b + 02 = 63.0 MPa
2
B
a
(-126.05 - 0)>2
= q
0
and -45°
sx - sy
2
= -
= 63.0 =
sin 2u + txy cos 2u
-126.05 - 0
sin 90° + 0 cos 90°
2
tmax
in-plane
644
Ans.
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9–25. Continued
tmax
This indicates that in-plane
acts toward the positive sense of y¿ axis at the face of
element defined by us = 45°
savg =
sx + sy
2
=
-126.05 + 0
= -63.0 MPa
2
Ans.
The state of maximum In - plane shear stress can be represented by the element
shown in Fig. c
645
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9–26. The bracket is subjected to the force of 3 kip.
Determine the principal stress and maximum in-plane
shear stress at point A on the cross section at section a–a.
Specify the orientation of this state of stress and show the
results on elements.
3 kip
3 kip
a 3 in.
a
A
0.25 in.
2 in.
0.25 in.
Internal Loadings: Consider the equilibrium of the free - body diagram from the
bracket’s left cut segment, Fig. a.
B
1 in.
+ ©F = 0;
:
x
N - 3 = 0
N = 3 kip
Section a – a
M = 12 kip # in
©MO = 0; 3(4) - M = 0
Normal and Shear Stresses: The normal stress is the combination of axial and
bending stress. Thus,
s =
My
N
A
I
The cross - sectional area and the moment of inertia about the z axis of the bracket’s
cross section is
A = 1(2) - 0.75(1.5) = 0.875 in2
I =
1
1
(1) A 23 B (0.75) A 1.53 B = 0.45573 in4
12
12
For point A, y = 1 in. Then
sA =
(-12)(1)
3
= 29.76 ksi
0.875
0.45573
Since no shear force is acting on the section,
tA = 0
The state of stress at point A can be represented on the element shown in Fig. b.
In - Plane Principal Stress: sx = 29.76 ksi, sy = 0, and txy = 0. Since no shear
stress acts on the element,
s1 = sx = 29.8 ksi s2 = sy = 0
Ans.
The state of principal stresses can also be represented by the elements shown in Fig. b
Maximum In - Plane Shear Stress:
t max
in-plane
=
C
¢
sx - sy
2
2
≤ + txy 2 =
29.76 - 0 2
b + 02 = 14.9 ksi
2
B
a
Ans.
Orientation of the Plane of Maximum In - Plane Shear Stress:
tan 2us = -
A sx - sy B >2
txy
0.25 in.
= -
(29.76 - 0)>2
= -q
0
us = -45° and 45°
Ans.
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9–26.
Continued
Substituting u = -45° into
tx¿y¿ = -
= -
sx - sy
2
sin 2u + txy cos 2u
29.76 - 0
sin(-90°) + 0
2
= 14.9 ksi = t max
in-plane
This indicates that t max
is directed in the positive sense of the y¿ axes on the ace
in-plane
of the element defined by us = -45°.
Average Normal Stress:
savg =
sx + sy
2
=
29.76 + 0
= 14.9 ksi
2
Ans.
The state of maximum in - plane shear stress is represented by the element shown in
Fig. c.
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9–27. The bracket is subjected to the force of 3 kip.
Determine the principal stress and maximum in-plane
shear stress at point B on the cross section at section a–a.
Specify the orientation of this state of stress and show the
results on elements.
3 kip
3 kip
a 3 in.
a
A
0.25 in.
2 in.
0.25 in.
Internal Loadings: Consider the equilibrium of the free - body diagram of the
bracket’s left cut segment, Fig. a.
B
1 in.
+ ©F = 0;
:
x
N - 3 = 0
N = 3 kip
Section a – a
M = 12 kip # in
©MO = 0; 3(4) - M = 0
Normal and Shear Stresses: The normal stress is the combination of axial and
bending stress. Thus,
s =
My
N
A
I
The cross - sectional area and the moment of inertia about the z axis of the bracket’s
cross section is
A = 1(2) - 0.75(1.5) = 0.875 in2
I =
1
1
(1) A 23 B (0.75) A 1.53 B = 0.45573 in4
12
12
For point B, y = -1 in. Then
sB =
(-12)(-1)
3
= -22.90 ksi
0.875
0.45573
Since no shear force is acting on the section,
tB = 0
The state of stress at point A can be represented on the element shown in Fig. b.
In - Plane Principal Stress: sx = -22.90 ksi, sy = 0, and txy = 0. Since no shear
stress acts on the element,
s1 = sy = 0
s2 = sx = -22.90 ksi
Ans.
The state of principal stresses can also be represented by the elements shown in Fig. b.
Maximum In - Plane Shear Stress:
t max
in-plane
=
C
¢
sx - sy
2
2
≤ + txy 2 =
-22.90 - 0 2
b + 02 = 11.5 ksi
2
B
a
Ans.
Orientation of the Plane of Maximum In - Plane Shear Stress:
tan 2us = -
A sx - sy B >2
txy
0.25 in.
= -
(-22.9 - 0)>2
= -q
0
us = 45° and 135°
Ans.
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9–27. Continued
Substituting u = 45° into
tx¿y¿ = -
= -
sx - sy
2
sin 2u + txy cos 2u
-22.9 - 0
sin 90° + 0
2
= 11.5 ksi = t max
in-plane
is directed in the positive sense of the y¿ axes on the
This indicates that t max
in-plane
element defined by us = 45°.
Average Normal Stress:
savg =
sx + sy
2
=
-22.9 + 0
= -11.5 ksi
2
Ans.
The state of maximum in - plane shear stress is represented by the element shown in
Fig. c.
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*9–28. The wide-flange beam is subjected to the loading
shown. Determine the principal stress in the beam at point A
and at point B. These points are located at the top and
bottom of the web, respectively. Although it is not very
accurate, use the shear formula to determine the shear stress.
8 kN/m
A
B
1m
3m
B
Internal Forces and Moment: As shown on FBD(a).
200 mm
Section Properties:
A = 0.2(0.22) - 0.19(0.2) = 6.00 A 10 - 3 B m2
1
1
(0.2) A 0.223 B (0.19) A 0.22 B = 50.8 A 10 - 6 B m4
12
12
I =
QA = QB = y¿A¿ = 0.105(0.01)(0.2) = 0.210 A 10 - 3 B m3
Normal Stress:
s =
My
N
;
A
I
21.65(103)
=
73.5(103)(0.1)
;
6.00(10 - 3)
50.8(10 - 6)
sA = 3.608 + 144.685 = 148.3 MPa
sB = 3.608 - 144.685 = -141.1 MPa
VQ
.
It
Shear Stress: Applying the shear formula t =
tA = tB =
36.5(103) C 0.210(10 - 3) D
50.8(10 - 6)(0.01)
= 15.09 MPa
In - Plane Principal Stress: sx = 148.3 MPa, sy = 0, and txy = -15.09 MPa for
point A. Applying Eq. 9-5.
s1, 2 =
=
sx + sy
;
2
C
sx - sy
a
2
2
b + t2xy
148.3 - 0 2
148.3 + 0
;
a
b + (-15.09)2
2
C
2
= 74.147 ; 75.666
s1 = 150 MPa
s2 = -1.52 MPa
Ans.
sx = -141.1 MPa, sy = 0, and txy = -15.09 MPa for point B. Applying Eq. 9-5.
s1, 2 =
=
sx + sy
2
;
C
a
sx - sy
2
2
b + t2xy
( -141.1) - 0 2
-141.1 + 0
;
a
b + (-15.09)2
2
C
2
= -70.538 ; 72.134
s1 = 1.60 MPa
s2 = -143 MPa
Ans.
650
30⬚
25 kN
A
10 mm
110 mm
10 mm
200 mm
10 mm
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•9–29.
The wide-flange beam is subjected to the loading
shown. Determine the principal stress in the beam at
point A, which is located at the top of the web. Although
it is not very accurate, use the shear formula to determine
the shear stress. Show the result on an element located at
this point.
120 kN/m
30 kN
A
0.3 m
0.9 m
Using the method of sections and consider the FBD of the left cut segment of the
bean, Fig. a
V - 12 (90)(0.9) - 30 = 0
+ c ©Fy = 0;
M = 39.15 kN # m
150 mm
The moment of inertia of the cross - section about the bending axis is
I =
1
1
(0.15)(0.193) (0.13)(0.153) = 49.175(10 - 6) m4
12
12
Referring to Fig. b,
QA = y¿A¿ = 0.085 (0.02)(0.15) = 0.255 (10 - 3) m3
The normal stress developed is contributed by bending stress only. For point A,
y = 0.075 m. Thus,
s =
My
39.15(103)(0.075)
= 59.71(106)Pa = 59.71 MPa (T)
=
I
49.175(10 - 6)
The shear stress is contributed by the transverse shear stress only. Thus
70.5(10 ) C 0.255(10 ) D
VQA
= 18.28(106)Pa = 18.28 MPa
=
It
49.175(10 - 6) (0.02)
3
t =
-3
Here, sx = 59.71 MPa, sy = 0 and txy = 18.28 MPa.
s1, 2 =
=
sx + sy
;
2
C
a
sx - sy
2
2
b + txy
59.71 - 0 2
59.71 + 0
a
;
b + 18.282
2
C
2
= 29.86 ; 35.01
s2 = -5.15 MPa
s1 = 64.9 MPa
tan 2uP =
txy
(sx - sy)>2
uP = 15.74°
=
and
Ans.
18.28
= 0.6122
(59.71 - 0)>2
-74.26°
Substitute u = 15.74°,
sx¿ =
=
sx + sy
2
sx - sy
+
2
20 mm
150 mm
20 mm
V = 70.5 kN
1
2 (90)(0.9)(0.3) + 30(0.9) - M = 0
a + ©MC = 0;
A
20 mm
cos 2u + txy sin 2u
59.71 - 0
59.71 + 0
+
cos 31.48° + 18.28 sin 31.48°
2
2
= 64.9 MPa = s1
651
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9–29.
Continued
Thus,
A uP B 1 = 15.7°
A uP B 2 = -74.3°
Ans.
The state of principal stress can be represented by the element shown in Fig. d
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9–30. The cantilevered rectangular bar is subjected to the
force of 5 kip. Determine the principal stress at points A
and B.
1
I =
(3)(63) = 54 in4
12
1.5 in.
A
1.5 in.
A = (6)(3) = 18 in2
QA = 2.25(1.5)(3) = 10.125 in3
1.5 in.
1.5 in.
B 1 in.
3 in.
QB = 2(2)(3) = 12 in3
1 in.
Point A:
15 in.
3 in.
3 5
4
5 kip
45(1.5)
Mxz
4
P
+
=
+
= 1.472 ksi
sA =
A
I
18
54
tA =
Vz QA
=
It
3(10.125)
= 0.1875 ksi
54(3)
sx = 1.472 ksi
s1, 2 =
=
sy = 0
sx + sy
;
2
C
a
sx - sy
2
txy = 0.1875 ksi
2
b + txy 2
1.472 - 0 2
1.472 + 0
;
a
b + 0.18752
2
C
2
s1 = 1.50 ksi
Ans.
s2 = -0.0235 ksi
Ans.
Point B:
sB =
tB =
45(1)
Mxz
4
P
=
= -0.6111 ksi
A
I
18
54
Vz QB
=
It
3(12)
= 0.2222 ksi
54(3)
sx = -0.6111 ksi
s1, 2 =
=
sy = 0
sx + sy
2
;
C
a
sx - sy
2
txy = 0.2222 ksi
2
b + txy 2
-0.6111 - 0 2
-0.611 + 0
;
a
b + 0.2222
2
C
2
s1 = 0.0723 ksi
Ans.
s2 = -0.683 ksi
Ans.
653
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9–31. Determine the principal stress at point A on the
cross section of the arm at section a–a. Specify the
orientation of this state of stress and indicate the results on
an element at the point.
7.5 mm
A
50 mm
7.5 mm
Support Reactions: Referring to the free - body diagram of the entire arm shown
in Fig. a,
©MB = 0; FCD sin 30°(0.3) - 500(0.65) = 0
FCD = 2166.67 N
+ ©F = 0;
:
x
Bx - 2166.67 cos 30° = 0
Bx = 1876.39 N
+ c ©Fy = 0;
2166.67 sin 30° - 500 - By = 0
By = 583.33 N
20 mm
Section a – a
D
Internal Loadings: Consider the equilibrium of the free - body diagram of the
arm’s left segment, Fig. b.
+ ©F = 0;
:
x
1876.39 - N = 0
N = 1876.39 N
+ c ©Fy = 0;
V - 583.33 = 0
V = 583.33 N
583.33(0.15) - M = 0
M = 87.5N # m
+ ©MO = 0;
0.15 m
Referring to Fig. b,
QA = y¿A¿ = 0.02125(0.0075)(0.02) = 3.1875 A 10 - 6 B m3
Normal and Shear Stress: The normal stress is a combination of axial and bending
stress. Thus,
-1876.39
0.5625 A 10
-3
B
MyA
N
+
A
I
87.5(0.0175)
+
0.16367 A 10 - 6 B
= 6.020 MPa
The shear stress is caused by transverse shear stress.
tA =
583.33 C 3.1875 A 10 - 6 B D
VQA
=
= 1.515 MPa
It
0.16367 A 10 - 6 B (0.0075)
The share of stress at point A can be represented on the element shown in Fig. d.
In - Plane Principal Stress: sx = 6.020 MPa, sy = 0, and txy = 1.515 MPa. We have
s1,2 =
=
sx + sy
2
;
C
¢
sx - sy
2
2
≤ + txy 2
6.020 - 0 2
6.020 + 0
;
a
b + 1.5152
2
C
2
s1 = 6.38 MPa
s2 = -0.360 MPa
Ans.
654
C
0.15 m
0.35 m
500 N
1
1
(0.02) A 0.053 B (0.0125) A 0.0353 B = 0.16367 A 10 - 6 B m4
12
12
=
a
a
A = 0.02(0.05) - 0.0125(0.035) = 0.5625 A 10 - 3 B m2
sA =
60⬚
B
Section Properties: The cross - sectional area and the moment of inertia about the z
axis of the arm’s cross section are
I =
7.5 mm
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9–31. Continued
Orientation of the Principal Plane:
tan 2uP =
txy
A sx - sy B >2
=
1.515
= 0.5032
(6.020 - 0)>2
up = 13.36° and 26.71°
Substituting u = 13.36° into
sx¿ =
=
sx + sy
2
sx - sy
+
2
cos 2u + txy sin 2u
6.020 + 0
6.020 - 0
+
cos 26.71° + 1.515 sin 26.71°
2
2
= 6.38 MPa = s1
Thus, A uP B 1 = 13.4 and A uP B 2 = 26.71°
Ans.
The state of principal stresses is represented by the element shown in Fig. e.
655
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*9–32. Determine the maximum in-plane shear stress
developed at point A on the cross section of the arm at
section a–a. Specify the orientation of this state of stress and
indicate the results on an element at the point.
7.5 mm
A
50 mm
7.5 mm
20 mm
7.5 mm
Section a – a
Support Reactions: Referring to the free - body diagram of the entire arm shown
in Fig. a,
©MB = 0; FCD sin 30°(0.3) - 500(0.65) = 0
FCD = 2166.67 N
+ ©F = 0;
:
x
Bx - 2166.67 cos 30° = 0
Bx = 1876.39 N
+ c ©Fy = 0;
2166.67 sin 30° - 500 - By = 0
By = 583.33 N
D
60⬚
a
B
C
a
0.15 m
0.15 m
0.35 m
500 N
Internal Loadings: Considering the equilibrium of the free - body diagram of the
arm’s left cut segment, Fig. b,
+ ©F = 0;
:
x
1876.39 - N = 0
N = 1876.39 N
+ c ©Fy = 0;
V - 583.33 = 0
V = 583.33 N
+ ©MO = 0;
583.33(0.15) - M = 0
M = 87.5 N # m
Section Properties: The cross - sectional area and the moment of inertia about the z
axis of the arm’s cross section are
A = 0.02(0.05) - 0.0125(0.035) = 0.5625 A 10 - 3 B m2
1
1
(0.02) A 0.053 B (0.0125) A 0.0353 B = 0.16367 A 10 - 6 B m4
12
12
I =
Referring to Fig. b,
QA = y¿A¿ = 0.02125(0.0075)(0.02) = 3.1875 A 10 - 6 B m3
Normal and Shear Stress: The normal stress is a combination of axial and bending
stress. Thus,
sA =
MyA
N
+
A
I
-1876.39
=
0.5625 A 10
-3
B
87.5(0.0175)
+
0.16367 A 10 - 6 B
= 6.020 MPa
The shear stress is contributed only by transverse shear stress.
tA =
583.33 C 3.1875 A 10 - 6 B D
VQA
=
= 1.515 MPa
It
0.16367 A 10 - 6 B (0.0075)
Maximum In - Plane Shear Stress: sx = 6.020 MPa, sy = 0, and txy = 1.515 MPa.
tmax
in-plane
=
C
¢
sx - sy
2
2
≤ + txy 2 =
6.020 - 0 2
b + 1.5152 = 3.37 MPa
B
2
a
656
Ans.
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9–32.
Continued
Orientation of the Plane of Maximum In - Plane Shear Stress:
tan 2us = -
A sx - sy B >2
txy
= -
(6.020 - 0)>2
= -1.9871
1.515
us = -31.6° and 58.4°
Ans.
Substituting u = -31.6° into
tx¿y¿ = -
= -
sx - sy
2
sin 2u + txy cos 2u
6.020 - 0
sin(-63.29°) + 1.515 cos(-63.29°)
2
= 3.37 MPa = t max
in-plane
is directed in the positive sense of the y¿ axis on the face
This indicates that t max
in-plane
of the element defined by us = -31.6°.
Average Normal Stress:
savg =
sx + sy
2
=
6.020 + 0
= 3.01 MPa
2
Ans.
The state of maximum in - plane shear stress is represented on the element shown in
Fig. e.
657
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•9–33. The clamp bears down on the smooth surface at E
by tightening the bolt. If the tensile force in the bolt is 40
kN, determine the principal stress at points A and B and
show the results on elements located at each of these
points. The cross-sectional area at A and B is shown in the
adjacent figure.
300 mm
50 mm
30 mm
100 mm
B
A
Support Reactions: As shown on FBD(a).
E
Internal Forces and Moment: As shown on FBD(b).
Section Properties:
I =
1
(0.03) A 0.053 B = 0.3125 A 10 - 6 B m4
12
QA = 0
QB = y¿A¿ = 0.0125(0.025)(0.03) = 9.375 A 10 - 6 B m3
Normal Stress: Applying the flexure formula s = -
sA = -
sB = -
2.40(103)(0.025)
= -192 MPa
0.3125(10 - 6)
2.40(103)(0)
0.3125(10 - 6)
= 0
Shear Stress: Applying the shear formula t =
tA =
tB =
My
.
I
24.0(103)(0)
0.3125(10 - 6)(0.03)
VQ
It
= 0
24.0(103) C 9.375(10 - 6) D
0.3125(10 - 6)(0.03)
= 24.0 MPa
In - Plane Principal Stresses: sx = 0, sy = -192 MPa, and txy = 0 for point A.
Since no shear stress acts on the element.
s1 = sx = 0
Ans.
s2 = sy = -192 MPa
Ans.
sx = sy = 0 and txy = -24.0 MPa for point B. Applying Eq. 9-5
s1,2 =
sx + sy
2
;
C
a
sx - sy
2
2
b + t2xy
= 0 ; 20 + (-24.0)2
= 0 ; 24.0
s1 = 24.0
s2 = -24.0 MPa
Ans.
658
B
A
25 mm
100 mm
50 mm
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9–33. Continued
Orientation of Principal Plane: Applying Eq. 9-4 for point B.
tan 2up =
txy
=
-24.0
= -q
0
and
45.0°
A sx - sy B >2
up = -45.0°
Subsututing the results into Eq. 9-1 with u = -45.0° yields
sx¿ =
sx + sy
2
sx - sy
+
2
cos 2u + txy sin 2u
= 0 + 0 + [-24.0 sin (-90.0°)]
= 24.0 MPa = s1
Hence,
up1 = -45.0°
up2 = 45.0°
Ans.
659
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9–34. Determine the principal stress and the maximum inplane shear stress that are developed at point A in the
2-in.-diameter shaft. Show the results on an element located
at this point. The bearings only support vertical reactions.
300 lb
Using the method of sections and consider the FBD of shaft’s left cut segment, Fig. a,
+ ©F = 0;
:
x
N - 3000 = 0
N = 3000 lb
+ c ©Fy = 0;
75 - V = 0
V = 75 lb
a + ©MC = 0;
M - 75(24) = 0
M = 1800 lb # in
A = p(12) = p in2
I =
p 4
p
(1 ) = in4
4
4
Also,
QA = 0
The normal stress developed is the combination of axial and bending stress. Thus
My
N
;
A
I
s =
For point A, y = C = 1 in. Then
s =
1800(1)
3000
p
p>4
= -1.337 (103) psi = 1.337 ksi (c)
The shear stress developed is due to transverse shear force. Thus,
t =
VQA
= 0
It
The state of stress at point A, can be represented by the element shown in Fig. b.
Here, sx = -1.337 ksi, sy = 0 is txy = 0. Since no shear stress acting on the
element,
s1 = sy = 0
s2 = sx = -1.34 ksi
Ans.
Thus, the state of principal stress can also be represented by the element shown in Fig. b.
t
max
in-plane =
C
a
sx - sy
2
tan 2us = us = 45°
2
b + t2xy =
-1.337 - 0 2
b + 02 = 0.668 ksi - 668 psi Ans.
C
2
a
(sx - sy)>2
= -
txy
and
(-1.337 - 0)>2
= q
0
Ans.
-45°
Substitute u = 45°,
tx¿y¿ = -
= -
sx - sy
2
sin 2u + txy cos 2u
-1.337 - 0
sin 90° + 0
2
= 0.668 ksi = 668 psi =
A
3000 lb
tmax
in-plane
660
24 in.
3000 lb
12 in.
12 in.
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9–34. Continued
tmax
This indicates that in-plane
acts toward the positive sense of y¿ axis at the face of the
element defined by us = 45°.
Average Normal Stress.
The state of maximum in - plane shear stress can be represented by the element
shown in Fig. c.
661
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9–35. The square steel plate has a thickness of 10 mm and
is subjected to the edge loading shown. Determine the
maximum in-plane shear stress and the average normal
stress developed in the steel.
sx = 5 kPa
sy = -5 kPa
tmax
sx - sy
in-plane
savg =
a
b
txy = 0
200 mm
+ t2xy
C
=
5 + 5 2
b + 0 = 5 kPa
C
2
a
sx + sy
3
50 N/m
2
=
2
50 N/m
Ans.
200 mm
5 - 5
=
= 0
2
Ans.
Note:
tan 2us =
tan 2us =
-(sx - sy)>2
txy
-(5 + 5)>2
= q
0
us = 45°
*9–36. The square steel plate has a thickness of 0.5 in. and
is subjected to the edge loading shown. Determine the
principal stresses developed in the steel.
sx = 0
s1,2 =
sy = 0
txy = 32 psi
sx + sy
2
;
16 lb/in.
C
a
sx - sy
2
2
b + t2xy
= 0 ; 20 + 322
4 in.
s1 = 32 psi
Ans.
s2 = -32 psi
Ans.
Note:
tan 2up =
16 lb/in.
4 in.
txy
(sx - sy)>2
=
32
= q
0
up = 45°
662
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•9–37.
The shaft has a diameter d and is subjected to the
loadings shown. Determine the principal stress and the
maximum in-plane shear stress that is developed at point A.
The bearings only support vertical reactions.
P
F
F
A
L
2
Support Reactions: As shown on FBD(a).
Internal Forces and Moment: As shown on FBD(b).
Section Properties:
A =
p 2
d
4
p d 4
p 4
a b =
d
4 2
64
I =
QA = 0
Normal Stress:
N
Mc
;
A
I
s =
-F
= p 2 ;
4 d
sA =
A B
pL d
4
2
p 4
d
64
4
2PL
- Fb
a
2
d
pd
Shear Stress: Since QA = 0, tA = 0
In - Plane Principal Stress: sx =
4 2PL
a
- Fb.
pd2 d
sy = 0 and txy = 0 for point A. Since no shear stress acts on the element,
s1 = sx =
4
2PL
a
- Fb
d
pd2
Ans.
s2 = sy = 0
Ans.
Maximum In - Plane Shear Stress: Applying Eq. 9-7 for point A,
t max
in-plane
=
=
=
Q
£
B
a
4
2
pd
sx - sy
2
2
b + t2xy
A 2PL
d - FB - 0
2
2
≥ + 0
2PL
2
a
- Fb
d
pd2
Ans.
663
L
2
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9–38. A paper tube is formed by rolling a paper strip in
a spiral and then gluing the edges together as shown.
Determine the shear stress acting along the seam, which is
at 30° from the vertical, when the tube is subjected to an
axial force of 10 N. The paper is 1 mm thick and the tube has
an outer diameter of 30 mm.
30⬚
10 N
10 N
30 mm
10
P
= 109.76 kPa
= p
2
A
(0.03
- 0.0282)
4
s =
sx = 109.76 kPa
tx¿y¿ = -
= -
sx - sy
2
sy = 0
txy = 0
u = 30°
sin 2u + txy cos 2u
106.76 - 0
sin 60° + 0 = -47.5 kPa
2
Ans.
9–39. Solve Prob. 9–38 for the normal stress acting
perpendicular to the seam.
30⬚
10 N
10 N
30 mm
s =
sn =
=
10
P
= 109.76 kPa
= p
2
A
(0.03
- 0.0282)
4
sx + sy
2
sx - sy
+
2
cos 2u + txy sin 2u
109.76 - 0
109.76 + 0
+
cos (60°) + 0 = 82.3 kPa
2
2
Ans.
664
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*9–40. Determine the principal stresses acting at point A
of the supporting frame. Show the results on a properly
oriented element located at this point.
800 mm
B
A
300 mm
150 mm
12 mm
5
B
15 mm
130 mm
A
y =
0.065(0.13)(0.015) + 0.136(0.15)(0.012)
©yA
=
= 0.0991 m
©A
0.13(0.015) + 0.15(0.012)
I =
1
(0.015)(0.133) + 0.015(0.13)(0.0991 - 0.065)2
12
1
(0.15)(0.012 3) + 0.15(0.012)(0.136 - 0.0991)2 = 7.4862(10 - 6) m4
12
+
QA = 0
A = 0.13(0.015) + 0.15(0.012) = 3.75(10 - 3) m2
Normal stress:
s =
P
Mc
+
A
I
sA =
-3.6(103)
-3
3.75(10 )
5.2767(103)(0.0991)
-
7.4862(10 - 6)
= -70.80 MPa
Shear stress:
tA = 0
Principal stress:
s1 = 0
Ans.
s2 = -70.8 MPa
Ans.
665
4
3
6 kN
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•9–41. Determine the principal stress acting at point B,
which is located just on the web, below the horizontal
segment on the cross section. Show the results on a properly
oriented element located at this point. Although it is not very
accurate, use the shear formula to calculate the shear stress.
800 mm
B
A
300 mm
150 mm
12 mm
y =
©yA
0.065(0.13)(0.015) + 0.136(0.15)(0.012)
=
= 0.0991 m
©A
0.13(0.015) + 0.15(0.012)
I =
1
(0.015)(0.133) + 0.015(0.13)(0.0991 - 0.065)2
12
+
130 mm
A
1
(0.15)(0.0123) + 0.15(0.012)(0.136 - 0.0991)2 = 7.4862(10 - 6) m4
12
A = 0.13(0.015) + 0.15(0.012) = 3.75(10 - 3) m2
Normal stress:
s =
Mc
P
+
A
I
sB = -
3.6(103)
3.75(10 - 3)
5.2767(103)(0.130 - 0.0991)
+
7.4862(10 - 6)
= 20.834 MPa
Shear stress:
tB =
VQ
-4.8(103)(0.0369)(0.15)(0.012)
= -2.84 MPa
=
It
7.4862(10 - 6)(0.015)
Principal stress:
s1,2 = a
20.834 + 0
20.834 - 0 2
b ;
a
b + (-2.84)2
2
C
2
s1 = 21.2 MPa
Ans.
s2 = -0.380 MPa
Ans.
tan 2up =
A
-2.84
20.834 - 0
2
B
up = -7.63°
Ans.
666
5
B
15 mm
4
3
6 kN
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9–42. The drill pipe has an outer diameter of 3 in., a wall
thickness of 0.25 in., and a weight of 50 lb>ft. If it is
subjected to a torque and axial load as shown, determine
(a) the principal stress and (b) the maximum in-plane shear
stress at a point on its surface at section a.
1500 lb
800 lb⭈ft
20 ft
a
20 ft
Internal Forces and Torque: As shown on FBD(a).
Section Properties:
A =
p 2
A 3 - 2.52 B = 0.6875p in2
4
J =
p
A 1.54 - 1.254 B = 4.1172 in4
2
s =
N
-2500
=
= -1157.5 psi
A
0.6875p
Normal Stress:
Shear Stress: Applying the torsion formula.
t =
800(12)(1.5)
Tc
=
= 3497.5 psi
J
4.1172
a) In - Plane Principal Stresses: sx = 0, sy = -1157.5 psi and txy = 3497.5 psi for
any point on the shaft’s surface. Applying Eq. 9-5.
s1,2 =
=
sx + sy
2
;
C
a
sx - sy
2
2
b + t2xy
0 - (-1157.5) 2
0 + (-1157.5)
;
a
b + (3497.5)2
2
C
2
= -578.75 ; 3545.08
s1 = 2966 psi = 2.97 ksi
Ans.
s2 = -4124 psi = -4.12 ksi
Ans.
b) Maximum In - Plane Shear Stress: Applying Eq. 9-7
t max
in-plane
a
sx - sy
2
b + t2xy
=
C
=
0 - (-1157.5) 2
≤ + (3497.5)2
C
2
2
¢
= 3545 psi = 3.55 ksi
Ans.
667
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9–43. Determine the principal stress in the beam at
point A.
60 kN
50 mm
150 kN
A
A
60 mm
0.5 m
Using the method of sections and consider the FBD of the beam’s left cut segment,
Fig. a,
+ ©F = 0;
:
x
150 - N = 0
N = 150 kN
+ c ©Fy = 0;
V - 60 = 0
V = 60 kN
a + ©MC = 0;
60(0.5) - M = 0
M = 30 kN # m
A = 0.06(0.15) = 0.009 m2
1
(0.06)(0.153) = 16.875(10 - 6) m4
12
I =
Referring to Fig. b,
QA = y¿A¿ = 0.05 (0.05)(0.06) = 0.15(10 - 3) m3
The normal stress developed is the combination of axial and bending stress. Thus
My
N
;
A
I
s =
For point A, y = 0.075 - 0.05 = 0.025 m. Then
s =
30(103)(0.025)
-150(103)
0.009
16.875(10 - 6)
= -61.11(106) Pa = 61.11 MPa (c)
The shear stress developed is due to the transverse shear, Thus,
t =
60(103) C 0.15(10 - 3) D
VQA
= 8.889 MPa
=
It
16.875(10 - 6) (0.06)
Here, sx = -61.11 MPa, sy = 0 and txy = 8.889 MPa,
s1, 2 =
=
sx + sy
;
2
C
a
sx - sy
2
2
b + t2xy
-61.11 - 0 2
-61.11 + 0
;
a
b + 8.8892
2
C
2
= -30.56 ; 31.82
s2 = -62.4 MPa
s1 = 1.27 MPa
tan 2uP =
txy
(sx - sy)>2
uP = -8.11°
=
and
Ans.
8.889
= -0.2909
(-61.11 - 0)>2
81.89°
668
0.25 m
150 mm
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9–43.
Continued
Substitute u = -8.11°,
sx¿ =
=
sx + sy
2
sx - sy
+
2
cos 2u + txy sin 2u
-61.11 + 0
-61.11 - 0
+
cos (-16.22°) + 8.889 sin (-16.22°)
2
2
= -62.4 MPa = s2
Thus,
(uP)1 = 81.9°
(uP)2 = -8.11°
The state of principal stresses can be represented by the elements shown in Fig. (c)
669
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*9–44. Determine the principal stress at point A which is
located at the bottom of the web. Show the results on an
element located at this point.
150 kN/m
Using the method of sections, consider the FBD of the bean’s left cut segment, Fig. a,
+ c ©Fy = 0;
V -
1
(100)(0.6) = 0
2
V = 30 kN
a + ©MC = 0;
1
(100)(0.6)(0.2) - M = 0
2
M = 6 kN # m
I =
10 mm
A
1
1
(0.15)(0.223) (0.14)(0.23) = 39.7667(10 - 6) m4
12
12
150 mm
Referring to Fig. b
QA = y¿A¿ = 0.105 (0.01)(0.15) = 0.1575(10 - 3) m3
The normal stress developed is due to bending only. For point A, y = 0.1 m. Then
s =
My
6(103)(0.1)
=
I
= 15.09(106)Pa = 15.09 MPa (c)
39.7667(10 - 6)
The shear stress developed is due to the transverse shear. Thus,
t =
30(103) C 0.1575(10 - 3) D
VQA
= 11.88(106)Pa = 11.88 MPa
=
It
39.7667(10 - 6)(0.01)
Here, sx = -15.09 MPa, sy = 0 And txy = 11.88 MPa.
s1, 2 =
=
sx + sy
;
2
C
a
sx - sy
2
2
b + t2xy
-15.09 - 0 2
-15.09 + 0
;
a
b + 11.882
2
C
2
= -7.544 ; 14.074
s2 = -21.6 MPa
s1 = 6.53 MPa
tan 2uP =
txy
(sx - sy)>2
uP = -28.79°
=
Ans.
11.88
= -1.575
(-15.09 - 0)>2
and
61.21°
Substitute u = 61.21°,
sx¿ =
=
sx + sy
2
sx - sy
+
2
cos 2u + txy sin 2u
-15.09 + 0
-15.09 - 0
+
cos 122.42° + 11.88 sin 122.42°
2
2
= 6.53 MPa = s1
Thus,
(uP)1 = 61.2°
A
0.6 m
Ans.
(uP)2 = -28.8°
The state of principal stresses can be represented by the element shown in Fig. d.
670
0.3 m
10 mm
200 mm
10 mm
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9–44. Continued
671
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•9–45.
Determine the maximum in-plane shear stress in
the box beam at point A. Show the results on an element
located at this point.
10 kip
4 kip
A
B
Using the method of section, consider the FBD, of bean’s left cut segment, Fig. a,
8 - 10 + V = 0
+ c ©Fy = 0;
a + ©MC = 0;
M + 10(1.5) - 8(3.5) = 0
4 in.
A
M = 13 kip # ft
4 in.
The moment of inertia of the cross - section about the neutral axis is
Referring to Fig. b,
QA = 0
The normal stress developed is contributed by the bending stress only. For point A,
y = C = 3 in.
My
=
I
13(12)(3)
= 5.40 ksi (c)
86.6667
The shear stress is contributed by the transverse shear stress only. Thus
t =
VQA
= 0
It
The state of stress at point A can be represented by the element shown in Fig. c
Here, sx = -5.40 ksi, sy = 0 and txy = 0.
tmax
in-plane
=
C
a
sx - sy
2
tan 2us = -
2
b + txy 2 =
-5.40 - 0 2
b + 02 = 2.70 ksi
C
2
(sx - sy)>2
a
txy
and
-45°
us = 45°
Ans.
(-5.40 - 0)>2
= q
2
= -
Substitute u = 45°,
tx¿y¿ = -
sx - sy
= -
2
sin 2u + txy cos 2u
-5.40 - 0
sin 90° + 0
2
= 2.70 ksi =
tmax
in-plane
tmax
This indicates that in-plane
acts toward the positive sense of y¿ axis at the face of
element defined by us = 45°
savg =
sx + sy
2
=
-5.40 + 0
= -2.70 ksi
2
The state of maximum In - plane shear stress can be represented by the element
shown in Fig. d.
672
B
6 in.
1
1
I =
(6)(63) (4)(43) = 86.6667 in4
12
12
s =
1.5 ft
2 ft
V = 2 kip
2 ft
0.5 ft
3 in.
3 in.
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9–45. Continued
673
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9–46. Determine the principal stress in the box beam at
point B. Show the results on an element located at this point.
10 kip
4 kip
A
Using the method of sections, consider the FBD of bean’s left cut segment, Fig. a,
+ c ©Fy = 0;
8 - 10 + V = 0
a + ©MC = 0;
M + 10(1.5) - 8(3.5) = 0
I =
B
V = 2 kip
M = 13 kip # ft
4 in.
A
1
1
(6)(63) (4)(43) = 86.6667 in4
12
12
4 in.
Referring to Fig. b,
C
QB = 2y1œ A1œ + y2œ A2œ = 2 1(2)(1)
My
D + 2.5(1)(6) = 19 in
= 0
I
The shear stress is contributed by the transverse shear stress only. Thus
2(103)(19)
VQB
=
= 219.23 psi
It
86.6667(2)
t =
The state of stress at point B can be represented by the element shown in Fig. c
Here, sx = sy = 0 and txy = 219.23 psi.
s1, 2 =
sx + sy
;
2
C
a
sx - sy
2
2
b + txy 2
= 0 ; 20 + 219.232
s2 = -219 psi
s1 = 219 psi
tan 2uP =
txy
(sx - sy)>2
uP = 45°
and
sx + sy
sx - sy
=
Ans.
219.23
= q
0
-45°
Substitute u = 45°,
sx¿ =
2
+
2
cos 2u + txy sin 2u
= 0 + 0 + 219.23 sin 90°
= 219 psi = s1
Thus,
(uP)1 = 45°
Ans.
(uP)2 = -45°
The state of principal stress can be represented by the element shown in Fig. d.
674
B
6 in.
3
The normal stress developed is contributed by the bending stress only. For point B,
y = 0.
s =
1.5 ft
2 ft
2 ft
0.5 ft
3 in.
3 in.
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9–46. Continued
675
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9–47. The solid shaft is subjected to a torque, bending
moment, and shear force as shown. Determine the principal
stresses acting at point A.
Ix = Iy =
J =
p
(0.025)4 = 0.306796(10 - 6) m4
4
450 mm
300 N⭈m
p
(0.025)4 = 0.613592(10 - 6) m4
2
25 mm
45 N⭈m
QA = 0
800 N
60(0.025)
Mx c
= 4.889 MPa
sA =
=
I
0.306796(10 - 6)
tA =
Ty c
45(0.025)
=
J
0.613592(10 - 6)
sx = 4.889 MPa
s1, 2 =
=
*9–48.
sy = 0
sx + sy
;
2
= 1.833 MPa
C
a
txy = -1.833 MPa
sx - sy
2
2
b + txy 2
4.889 - 0 2
4.889 + 0
;
b + (-1.833)2
a
2
C
2
s1 = 5.50 MPa
Ans.
s2 = -0.611 MPa
Ans.
Solve Prob. 9–47 for point B.
Ix = Iy =
p
(0.025)4 = 0.306796(10 - 6) m4
4
450 mm
p
J = (0.025)4 = 0.613592(10 - 6) m4
2
QB = yA¿ =
300 N⭈m
4(0.025) 1
a b p (0.0252) = 10.4167(10 - 6) m3
3p
2
800 N
Ty c
VzQB
It
-
-6
800(10.4167)(10 )
=
J
0.306796(10 )(0.05)
sx = 0
s1, 2 =
sy = 0
sx + sy
2
;
C
a
sx - sy
2
45(0.025)
-
-6
0.61359(10 - 6)
= -1.290 MPa
txy = -1.290 MPa
2
b + txy 2
= 0 ; 2(0)2 + (-1.290)2
s1 = 1.29 MPa
Ans.
s2 = -1.29 MPa
Ans.
676
A
B
25 mm
45 N⭈m
sB = 0
tB =
A
B
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•9–49.
The internal loadings at a section of the beam are
shown. Determine the principal stress at point A. Also
compute the maximum in-plane shear stress at this point.
50 mm
A
200 mm
50 mm
50 mm
y
200 mm
Section Properties:
z
40 kN⭈m
30 kN⭈m
A = 0.2(0.3) - 0.15(0.2) = 0.030 m4
800 kN
1
1
Iz =
(0.2) A 0.33 B (0.15) A 0.23 B = 0.350 A 10 - 3 B m4
12
12
Iy =
1
1
(0.1) A 0.23 B +
(0.2) A 0.053 B = 68.75 A 10 - 6 B m4
12
12
(QA)y = 0
Normal Stress:
s =
Myz
Mzy
N
+
A
Iz
Iy
sA =
-30(103)(0.1)
-500(103)
40(103)(0.15)
+
3
0.030
0.350(10 )
68.75(10 - 6)
= -77.45 MPa
tA = 0.
Shear Stress: Since (QA)y = 0,
In - Plane Principal Stresses: sx = -77.45 MPa. sy = 0. and txy = 0 for point A.
Since no shear stress acts on the element.
s1 = sy = 0
Ans.
s2 = sz = -77.4 MPa
Ans.
Maximum In-Plane Shear Stress: Applying Eq. 9–7.
t max
in-plane
a
sx - sy
2
b + t2xy
=
C
=
-77.45 - 0 2
b + 0
C
2
2
a
= 38.7 MPa
Ans.
677
500 kN
x
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9–50. The internal loadings at a section of the beam
consist of an axial force of 500 N, a shear force of 800 N,
and two moment components of 30 N # m and 40 N # m.
Determine the principal stress at point A. Also calculate the
maximum in-plane shear stress at this point.
Ix =
A
1
(0.1)(0.2)3 = 66.67(10 - 6) in4
12
200 mm
40 N⭈m
B
C
50 mm
QA = 0
50 mm
100 mm
30(0.1)
Mz
P
500
= -20 kPa
=
sA =
A
Ix
(0.1)(0.2)
66.67(10 - 6)
30 N⭈m
500 N
800 N
tA = 0
Here, the principal stresses are
s1 = sy = 0
Ans.
s2 = sx = -20 kPa
Ans.
t max
in-plane=
=
C
a
sx - sy
2
2
b + txy 2
-20 - 0 2
b + 0 = 10 kPa
C
2
a
Ans.
9–51. Solve Prob. 9–4 using Mohr’s circle.
A
400 psi
650 psi
60⬚
sx + sy
2
-650 + 400
=
= -125
2
A(-650, 0)
B(400, 0)
C( -125, 0)
B
R = CA = = 650 - 125 = 525
sx¿ = -125 - 525 cos 60° = -388 psi
Ans.
tx¿y¿ = 525 sin 60° = 455 psi
Ans.
678
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*9–52.
Solve Prob. 9–6 using Mohr’s circle.
90 MPa
A
35 MPa
60⬚
30⬚
sx = 90 MPa
sx + sy
2
=
sy = 50 MPa
txy = -35 MPa
A(90, -35)
90 + 50
= 70
2
R = 2(90 - 70)2 + (35)2 = 40.311
Coordinates of point B:
f = tan - 1 a
35
b = 60.255°
20
c = 300° - 180° - 60.255° = 59.745°
sx¿ = 70 - 40.311 cos 59.745° = 49.7 MPa
Ans.
tx¿ = -40.311 sin 59.745° = -34.8 MPa
Ans.
679
B
50 MPa
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•9–53.
Solve Prob. 9–14 using Mohr’s circle.
30 ksi
12 ksi
sx + sy
2
=
-30 + 0
= -15
2
R = 2(30 - 15)2 + (12)2 = 19.21 ksi
s1 = 19.21 - 15 = 4.21 ksi
Ans.
s2 = -19.21 - 15 = -34.2 ksi
Ans.
2uP2 = tan - 1
tmax
in-plane
12
;
(30 - 15)
uP2 = 19.3°
Ans.
= R = 19.2 ksi
Ans.
savg = -15 ksi
2uP2 = tan - 1
12
+ 90°;
(30 - 15)
Ans.
us = 64.3°
Ans.
680
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9–54.
Solve Prob. 9–16 using Mohr’s circle.
350 psi
75 psi
200 psi
sx + sy
2
=
45 - 60
= -7.5 MPa
2
R = 2(45 + 7.5)2 + (30)2 = 60.467 MPa
s1 = 60.467 - 7.5 = 53.0 MPa
Ans.
s2 = -60.467 - 7.5 = -68.0 MPa
Ans.
2uP1 = tan - 1
uP1 = 14.9°
tmax
30
(45 + 7.5)
Ans.
counterclockwise
= 60.5 MPa
Ans.
savg = -7.50 MPa
Ans.
in-plane
2uP1 = 90° - tan - 1
us1 = 30.1°
30
(45 + 7.5)
Ans.
clockwise
681
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9–55.
Solve Prob. 9–12 using Mohr’s circle.
sx + sy
2
=
-10 + 0
= -5 ksi
2
10 ksi
16 ksi
R = 2(10 - 5)2 + (16)2 = 16.763 ksi
f = tan - 1
16
= 72.646°
(10 - 5)
a = 100 - 72.646 = 27.354°
sx¿ = -5 - 16.763 cos 27.354° = -19.9 ksi
Ans.
tx¿y¿ = 16.763 sin 27.354° = 7.70 ksi
Ans.
sy¿ = 16.763 cos 27.354° - 5 = 9.89 ksi
*9–56. Solve Prob. 9–11 using Mohr’s circle.
2 ksi
Construction of the Circle: In accordance with the sign convention, sx = -3 ksi,
sy = 2 ksi, and txy = -4 ksi. Hence,
savg =
sx + sy
2
=
-3 + 2
= -0.500 ksi
2
3 ksi
30⬚
The coordinates for reference point A and C are
A(-3, -4)
A
4 ksi
B
C(-0.500, 0)
The radius of circle is
R = 2(3 - 0.5)2 + 42 = 4.717 ksi
Stress on the Inclined Plane: The normal and shear stress components
A sx¿ and tx¿y¿ B are represented by the coordinates of point P on the circle.
sx¿ = -0.500 - 4.717 cos 62.01° = -2.71 ksi
Ans.
tx¿y¿ = 4.717 sin 62.01° = 4.17 ksi
Ans.
682
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9–57. Mohr’s circle for the state of stress in Fig. 9–15a is
shown in Fig. 9–15b. Show that finding the coordinates of
point P1sx¿ , tx¿y¿2 on the circle gives the same value as the
stress-transformation Eqs. 9–1 and 9–2.
A(sx, txy)
R =
sxœ =
C
Ca a
B(sy, -txy)
csx - a
sx + sy
2
sx + sy
+
2
C
a
sx + sy
2
2
b d + t2xy =
sx - sy
2
C
a
b, 0b
sx - sy
2
2
b + t2xy
2
b + t2xy cos u¿
(1)
u¿ = 2uP - 2u
(2)
cos (2uP - 2u) = cos 2uP cos 2u + sin 2up sin 2u
From the circle:
sx -
cos 2uP =
sin 2uP =
4A
4A
sx + sy
sx - sy
2
txy
sx - sy
2
B
2
2
(3)
+ t2xy
(4)
B 2 + t2xy
Substitute Eq. (2), (3) and into Eq. (1)
sx¿ =
tx¿y¿ =
sx + sy
C
sx - sy
2
+
a
sx - sy
2
2
cos 2u + txy sin 2u
QED
2
b + t2xy sin u¿
(5)
sin u¿ = sin (2uP - 2u)
(6)
= sin 2uP cos 2u - sin 2u cos 2uP
Substitute Eq. (3), (4), (6) into Eq. (5),
tx¿y¿ = -
sx - sy
2
sin 2u + txy cos 2u
QED
683
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9–58. Determine the equivalent state of stress if an element
is oriented 25° counterclockwise from the element shown.
550 MPa
A(0, -550)
B(0, 550)
C(0, 0)
R = CA = CB = 550
sx¿ = -550 sin 50° = -421 MPa
Ans.
tx¿y¿ = -550 cos 50° = -354 MPa
Ans.
sy¿ = 550 sin 50° = 421 MPa
Ans.
9–59. Determine the equivalent state of stress if an
element is oriented 20° clockwise from the element shown.
2 ksi
Construction of the Circle: In accordance with the sign convention, sx = 3 ksi,
sy = -2 ksi, and tx¿y¿ = -4 ksi. Hence,
savg =
sx + sy
2
=
3 + (-2)
= 0.500 ksi
2
4 ksi
The coordinates for reference points A and C are
A(3, -4)
3 ksi
C(0.500, 0)
The radius of the circle is
R = 2(3 - 0.500)2 + 42 = 4.717 ksi
Stress on the Rotated Element: The normal and shear stress components
A sx¿ and tx¿y¿ B are represented by the coordinate of point P on the circle, sy¿, can be
determined by calculating the coordinates of point Q on the circle.
sx¿ = 0.500 + 4.717 cos 17.99° = 4.99 ksi
Ans.
tx¿y¿ = -4.717 sin 17.99° = -1.46 ksi
Ans.
sy¿ = 0.500 - 4.717 cos 17.99° = -3.99 ksi
Ans.
684
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*9–60. Determine the equivalent state of stress if an
element is oriented 30° clockwise from the element shown.
Show the result on the element.
9 ksi
4 ksi
In accordance to the established sign convention, sx = -6 ksi, sy = 9 ksi and
txy = 4 ksi. Thus,
savg =
sx + sy
2
=
-6 + 9
= 1.50 ksi
2
Then, the coordinates of reference point A and C are
A(-6, 4)
C(1.5, 0)
The radius of the circle is
R = CA = 2(-6 - 1.5)2 + 42 = 8.50 ksi
Using these results, the circle shown in Fig. a can be constructed.
Referring to the geometry of the circle, Fig. a,
a = tan - 1 a
4
b = 28.07°
6 + 1.5
b = 60° - 28.07° = 31.93°
Then,
sx¿ = 1.5 - 8.50 cos 31.93° = -5.71 ksi
Ans.
tx¿y¿ = -8.5 sin 31.95° = -4.50 ksi
sy¿ = 8.71 ksi
Ans.
The results are shown in Fig. b.
685
6 ksi
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•9–61.
Determine the equivalent state of stress for an
element oriented 60° counterclockwise from the element
shown. Show the result on the element.
250 MPa
400 MPa
In accordance to the established sign convention, sx = -560 MPa, sy = 250 MPa
and txy = -400 MPa. Thus,
savg =
sx + sy
2
=
-560 + 250
= -155 MPa
2
Then, the coordinate of reference points A and C are
A(-560, -400)
C(-155, 0)
The radius of the circle is
R = CA = 3 C -560 - (-155) D 2 + (-400)2 = 569.23 MPa
Using these results, the circle shown in Fig. a can be constructed.
Referring to the geometry of the circle, Fig. a
a = tan - 1 a
400
b = 44.64°
560 - 155
b = 120° - 44.64° = 75.36°
Then,
sx¿ = -155 - 569.23 cos 75.36° = -299 MPa
Ans.
tx¿y¿ = 569.23 sin 75.36° = 551 MPa
Ans.
sy¿ = -155 + 569.23 cos 75.36° = -11.1 MPa
Ans.
The results are shown in Fig. b.
686
560 MPa
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9–62. Determine the equivalent state of stress for an
element oriented 30° clockwise from the element shown.
Show the result on the element.
5 ksi
In accordance to the established sign convention, sx = 2 ksi, sy = -5 ksi and
txy = 0. Thus,
savg =
sx + sy
2
=
2 + (-5)
= -1.50 ksi
2
Then, the coordinate of reference points A and C are
A(2, 0)
C(-1.5, 0)
The radius of the circle is
R = CA = 3 C 2 - (-1.5) D 2 + 02 = 3.50 ksi
Using these results, the circle shown in Fig. a can be constructed.
Referring to the geometry of the circle, Fig. a,
b = 60°
Then,
sx¿ = -1.50 + 3.50 cos 60° - 0.250 ksi
Ans.
tx¿y¿ = 3.50 sin 60° = 3.03 ksi
Ans.
sy¿ = -3.25 ksi
Ans.
The results are shown in Fig b.
687
2 ksi
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9–63. Determine the principal stress, the maximum in-plane
shear stress, and average normal stress. Specify the orientation
of the element in each case.
15 ksi
5 ksi
Construction of the Circle: In accordance with the sign convention, sx = 15 ksi,
sy = 0 and txy = -5 ksi. Hence,
sx + sy
savg =
=
2
15 + 0
= 7.50 ksi
2
Ans.
The coordinates for reference point A and C are
A(15, -5)
C(7.50, 0)
The radius of the circle is
R = 2(15 - 7.50)2 + 52 = 9.014 ksi
a)
In - Plane Principal Stress: The coordinates of points B and D represent s1 and
s2, respectively.
s1 = 7.50 + 9.014 = 16.5 ksi
Ans.
s2 = 7.50 - 9.014 = -1.51 ksi
Ans.
Orientation of Principal Plane: From the circle
tan 2uP1 =
5
= 0.6667
15 - 7.50
uP1 = 16.8° (Clockwise)
Ans.
b)
Maximum In - Plane Shear Stress: Represented by the coordinates of point E on
the circle.
tmax
in-plane
= -R = -9.01 ksi
Ans.
Orientation of the Plane for Maximum In - Plane Shear Stress: From the circle
tan 2us =
15 - 7.50
= 1.500
5
us = 28.2° (Counterclockwise)
Ans.
688
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*9–64. Determine the principal stress, the maximum
in-plane shear stress, and average normal stress. Specify the
orientation of the element in each case.
20 MPa
80 MPa
30 MPa
In accordance to the established sign convention, sx = 30 MPa, sy = -20 MPa and
txy = 80 MPa. Thus,
savg =
sx + sy
2
30 + ( -20)
= 5 MPa
2
=
Then, the coordinates of reference point A and the center C of the circle is
A(30, 80)
C(5, 0)
Thus, the radius of circle is given by
R = CA = 2(30 - 5)2 + (80 - 0)2 = 83.815 MPa
Using these results, the circle shown in Fig. a, can be constructed.
The coordinates of points B and D represent s1 and s2 respectively. Thus
s1 = 5 + 83.815 = 88.8 MPa
Ans.
s2 = 5 - 83.815 = -78.8 MPa
Ans.
Referring to the geometry of the circle, Fig. a
tan 2(uP)1 =
80
= 3.20
30 - 5
uP = 36.3° (Counterclockwise)
Ans.
The state of maximum in - plane shear stress is represented by the coordinate of
point E. Thus
tmax
in-plane = R = 83.8 MPa
Ans.
From the geometry of the circle, Fig. a,
tan 2us =
30 - 5
= 0.3125
80
us = 8.68° (Clockwise)
Ans.
The state of maximum in - plane shear stress is represented by the element in Fig. c
689
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9–64.
Continued
690
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•9–65. Determine the principal stress, the maximum inplane shear stress, and average normal stress. Specify the
orientation of the element in each case.
120 psi
300 psi
A(300, 120)
B(0, -120)
C(150, 0)
R = 2(300 - 150)2 + 1202 = 192.094
s1 = 150 + 192.094 = 342 psi
Ans.
s2 = 150 - 192.094 = -42.1 psi
Ans.
tan 2uP =
120
= 0.8
300 - 150
uP1 = 19.3° Counterclockwise
Ans.
savg = 150 psi
Ans.
tmax
Ans.
in-plane
= 192 psi
tan 2us =
300 - 150
= 1.25
120
us = -25.7°
Ans.
9–66. Determine the principal stress, the maximum in-plane
shear stress, and average normal stress. Specify the orientation
of the element in each case.
A(45, -50)
B(30, 50)
30 MPa
C(37.5, 0)
45 MPa
R = CA = CB = 27.52 + 502 = 50.56
50 MPa
a)
tan 2uP =
50
7.5
s1 = 37.5 + 50.56 = 88.1 MPa
Ans.
s2 = 37.5 - 50.56 = -13.1 MPa
Ans.
uP = -40.7°
2uP = 81.47°
b)
t max
in-plane
= R = 50.6 MPa
Ans.
savg = 37.5 MPa
Ans.
2us = 90 - 2uP
us = 4.27°
Ans.
691
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9–67. Determine the principal stress, the maximum in-plane
shear stress, and average normal stress. Specify the orientation
of the element in each case.
200 MPa
500 MPa
350 MPa
Construction of the Circle: In accordance with the sign convention, sx = 350 MPa,
sy = -200 MPa, and txy = 500 MPa. Hence,
savg =
sx + sy
2
=
350 + (-200)
= 75.0 MPa
2
Ans.
The coordinates for reference point A and C are
A(350, 500)
C(75.0, 0)
The radius of the circle is
R = 2(350 - 75.0)2 + 5002 = 570.64 MPa
a)
In - Plane Principal Stresses: The coordinate of points B and D represent s1 and s2
respectively.
s1 = 75.0 + 570.64 = 646 MPa
Ans.
s2 = 75.0 - 570.64 = -496 MPa
Ans.
Orientaion of Principal Plane: From the circle
tan 2uP1 =
500
= 1.82
350 - 75.0
uP1 = 30.6° (Counterclockwise)
Ans.
b)
Maximum In - Plane Shear Stress: Represented by the coordinates of point E on
the circle.
t max
in-plane
= R = 571 MPa
Ans.
Orientation of the Plane for Maximum In - Plane Shear Stress: From the circle
tan 2us =
350 - 75.0
= 0.55
500
us = 14.4° (Clockwise)
Ans.
692
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*9–68. Draw Mohr’s circle that describes each of the
following states of stress.
700 psi
4 ksi
40 MPa
600 psi
(a)
a) Here, sx = 600 psi, sy = 700 psi and txy = 0. Thus,
savg =
sx + sy
=
2
600 + 700
= 650 psi
2
Thus, the coordinate of reference point A and center of circle are
A(600, 0)
C(650, 0)
Then the radius of the circle is
R = CA = 650 - 600 = 50 psi
The Mohr’s circle represents this state of stress is shown in Fig. a.
b) Here, sx = 0, sy = 4 ksi and txy = 0. Thus,
savg =
sx + sy
=
2
0 + 4
= 2 ksi
2
Thus, the coordinate of reference point A and center of circle are
A(0, 0)
C(2, 0)
Then the radius of the circle is
R = CA = 2 - 0 = 2 psi
c) Here, sx = sy = 0 and txy = -40 MPa. Thus,
savg =
sx + sy
2
= 0
Thus, the coordinate of reference point A and the center of circle are
A(0, -40)
C(0, 0)
Then, the radius of the circle is
R = CA = 40 MPa
The Mohr’s circle represents this state of stress shown in Fig. c
693
(b)
(c)
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9–68.
Continued
694
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9–69. The frame supports the distributed loading of
200 N兾m. Determine the normal and shear stresses at point
D that act perpendicular and parallel, respectively, to the
grain. The grain at this point makes an angle of 30° with the
horizontal as shown.
200 N/ m
B
30⬚
1m
200 mm
75 mm
D
1.5 m
C
100 mm
4m
60⬚
E
Support Reactions: As shown on FBD(a).
50 mm
30 mm
Internal Forces and Moment: As shown on FBD(b).
1.5 m
100 mm
Section Properties:
I =
A
1
(0.1) A 0.23 B = 66.667 A 10 - 6 B m4
12
QD = y¿A¿ = 0.0625(0.075)(0.1) = 0.46875 A 10 - 3 B m3
Normal Stress: Applying the flexure formula.
sD = -
My
150(-0.025)
= 56.25 kPa
= I
66.667(10 - 6)
Shear Stress: Applying the shear formula.
tD =
50.0 C 0.46875(10 - 3) D
VQD
= 3.516 kPa
=
It
66.667(10 - 6)(0.1)
Construction of the Circle: In accordance to the established sign convention,
sx = 56.25 kPa, sy = 0 and txy = -3.516 kPa. Hence.
savg =
sx + sy
2
=
56.25 + 0
= 28.125 kPa
2
The coordinates for reference point A and C are
A(56.25, -3.516)
C(28.125, 0)
The radius of the circle is
R = 2(56.25 - 28.125)2 + 3.5162 = 28.3439 kPa
Stresses on The Rotated Element: The normal and shear stress components
A sx¿ and tx¿y¿ B are represented by the coordinates of point P on the circle. Here,
u = 60°.
sx¿ = 28.125 - 28.3439 cos 52.875° = 11.0 kPa
Ans.
tx¿y¿ = -28.3439 sin 52.875° = -22.6 kPa
Ans.
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9–70. The frame supports the distributed loading of
200 N兾m. Determine the normal and shear stresses at point
E that act perpendicular and parallel, respectively, to the
grain. The grain at this point makes an angle of 60° with the
horizontal as shown.
200 N/ m
B
30⬚
1m
200 mm
75 mm
D
1.5 m
C
100 mm
4m
60⬚
E
Support Reactions: As shown on FBD(a).
50 mm
30 mm
Internal Forces and Moment: As shown on FBD(b).
1.5 m
100 mm
Section Properties:
A
A = 0.1(0.05) = 5.00 A 10 - 3 B m2
Normal Stress:
sE =
N
-250
= -50.0 kPa
=
A
5.00(10 - 3)
Construction of the Circle: In accordance with the sign convention. sx = 0,
sy = -50.0 kPa, and txy = 0. Hence.
savg =
sx + sy
2
=
0 + (-50.0)
= -25.0 kPa
2
The coordinates for reference points A and C are
A(0, 0)
C(-25.0, 0)
The radius of circle is R = 25.0 - 0 = 25.0 kPa
Stress on the Rotated Element: The normal and shear stress components
A sx¿ and tx¿y¿ B are represented by coordinates of point P on the circle. Here,
u = 150°.
sx = -25.0 + 25.0 cos 60° = -12.5 kPa
Ans.
tx¿y¿ = 25.0 sin 60° = 21.7 kPa
Ans.
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9–71. The stair tread of the escalator is supported on two
of its sides by the moving pin at A and the roller at B. If a
man having a weight of 300 lb stands in the center of the
tread, determine the principal stresses developed in the
supporting truck on the cross section at point C. The stairs
move at constant velocity.
1.25 ft
30⬚
A
C 1.5 ft
30⬚
0.5 ft B
0.5 ft
2 in.
Support Reactions: As shown on FBD (a).
Internal Forces and Moment: As shown on FBD (b).
Section Properties:
A = 2(0.5) = 1.00 in2
I =
1
(0.5) A 23 B = 0.3333 in4
12
QB = y¿A¿ = 0.5(1)(0.5) = 0.250 in3
Normal Stress:
s =
My
N
;
A
I
sC =
475.48(0)
-137.26
+
= -137.26 psi
1.00
0.3333
Shear Stress: Applying the shear formula t =
tC =
VQ
.
It
79.25(0.250)
= 118.87 psi
0.3333(0.5)
Construction of the Circle: In accordance with the sign convention, sx = 0,
sy = -137.26 psi, and txy = 118.87 psi. Hence,
savg =
sx + sy
2
=
0 + (-137.26)
= -68.63 psi
2
The coordinates for reference points A and C are
A(0, 118.87)
1 in.
C
C(-68.63, 0)
The radius of the circle is
R = 2(68.63 - 0)2 + 118.872 = 137.26 psi
In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2,
respectively.
s1 = -68.63 + 137.26 = 68.6 psi
Ans.
s2 = -68.63 - 137.26 = -206 psi
Ans.
697
0.5 in.
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*9–72. The thin-walled pipe has an inner diameter of
0.5 in. and a thickness of 0.025 in. If it is subjected to an
internal pressure of 500 psi and the axial tension and
torsional loadings shown, determine the principal stress at a
point on the surface of the pipe.
200 lb
200 lb
20 lb⭈ft
Section Properties:
A = p A 0.2752 - 0.252 B = 0.013125p in2
J =
p
A 0.2754 - 0.254 B = 2.84768 A 10 - 3 B in4
2
Normal Stress: Since
0.25
r
=
= 10, thin wall analysis is valid.
t
0.025
slong =
pr
500(0.25)
200
N
+
=
+
= 7.350 ksi
A
2t
0.013125p
2(0.025)
shoop =
pr
500(0.25)
=
= 5.00 ksi
t
0.025
Shear Stress: Applying the torsion formula,
t =
20(12)(0.275)
Tc
= 23.18 ksi
=
J
2.84768(10 - 3)
Construction of the Circle: In accordance with the sign convention sx = 7.350 ksi,
sy = 5.00 ksi, and txy = -23.18 ksi. Hence,
savg =
sx + sy
2
=
7.350 + 5.00
= 6.175 ksi
2
The coordinates for reference points A and C are
A(7.350, -23.18)
C(6.175, 0)
The radius of the circle is
R = 2(7.350 - 6.175)2 + 23.182 = 23.2065 ksi
In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2,
respectively.
s1 = 6.175 + 23.2065 = 29.4 ksi
Ans.
s2 = 6.175 - 23.2065 = -17.0 ksi
Ans.
698
20 lb⭈ft
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•9–73. The cantilevered rectangular bar is subjected to the
force of 5 kip. Determine the principal stress at point A.
1.5 in. A
1.5 in.
1 in.
B
1.5 in.
1.5 in.
3 in.
1 in.
15 in.
3 in.
3 5
4
5 kip
Internal Forces and Moment: As shown on FBD.
Section Properties:
A = 3(6) = 18.0 in2
I =
1
(3) A 63 B = 54.0 in4
12
QA = y¿A¿ = 2.25(1.5)(3) = 10.125 in3
Normal Stress:
s =
My
N
;
A
I
sA =
45.0(1.5)
4.00
+
= 1.4722 ksi
18.0
54.0
Shear Stress: Applying the shear formula t =
tA =
VQ
.
It
3.00(10.125)
= 0.1875 ksi
54.0(3)
Construction of the Circle: In accordance with the sign convention, sx = 1.4722 ksi,
sy = 0, and txy = -0.1875 ksi. Hence,
savg =
sx + sy
2
=
1.472 + 0
= 0.7361 ksi
2
The coordinates for reference points A and C are
A(1.4722, -0.1875)
C(0.7361, 0)
The radius of the circle is
R = 2(1.4722 - 0.7361)2 + 0.18752 = 0.7596 ksi
In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2,
respectively.
s1 = 0.7361 + 0.7596 = 1.50 ksi
Ans.
s2 = 0.7361 - 0.7596 = -0.0235 ksi
Ans.
699
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9–74. Solve Prob. 9–73 for the principal stress at point B.
1.5 in. A
1.5 in.
1 in.
B
1.5 in.
1.5 in.
3 in.
1 in.
15 in.
3 in.
3 5
4
5 kip
Internal Forces and Moment: As shown on FBD.
Section Properties:
A = 3(6) = 18.0 in2
1
(3) A 63 B = 54.0 in4
12
I =
QB = y¿A¿ = 2(2)(3) = 12.0 in3
Normal Stress:
s =
My
N
;
A
I
sB =
45.0(1)
4.00
= -0.6111 ksi
18.0
54.0
Shear Stress: Applying the shear formula t =
tB =
VQ
.
It
3.00(12.0)
= 0.2222 ksi
54.0(3)
Construction of the Circle: In accordance with the sign convention,
sx = -0.6111 ksi, sy = 0, and txy = -0.2222 ksi. Hence.
savg =
sx + sy
2
=
-0.6111 + 0
= -0.3055 ksi
2
The coordinates for reference points A and C are
A(-0.6111, -0.2222)
C(-0.3055, 0)
The radius of the circle is
R = 2(0.6111 - 0.3055)2 + 0.22222 = 0.3778 ksi
In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2,
respectively.
s1 = -0.3055 + 0.3778 = 0.0723 ksi
Ans.
s2 = -0.3055 - 0.3778 = -0.683 ksi
Ans.
700
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9–75. The 2-in.-diameter drive shaft AB on the helicopter
is subjected to an axial tension of 10 000 lb and a torque
of 300 lb # ft. Determine the principal stress and the
maximum in-plane shear stress that act at a point on the
surface of the shaft.
s =
10 000
P
=
= 3.183 ksi
A
p(1)2
t =
300(12)(1)
Tc
=
= 2.292 ksi
p
4
J
2 (1)
s1, 2 =
=
t max
sx + sy
;
2
A
(
sx - sy
2
B
A
)2 + t2xy
3.183 - 0 2
3.183 + 0
; (
) + (2.292)2
2
A
2
s1 = 4.38 ksi
Ans.
s2 = -1.20 ksi
Ans.
in-plane
=
A
(
=
A
(
sx - sy
2
)2 + t2xy
3.183 - 0 2
) + (2.292)2
2
= 2.79 ksi
Ans.
701
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*9–76. The pedal crank for a bicycle has the cross section
shown. If it is fixed to the gear at B and does not rotate
while subjected to a force of 75 lb, determine the principal
stress in the material on the cross section at point C.
75 lb
B
3 in.
A
4 in.
C
0.4 in.
0.4 in.
0.2 in.
0.3 in.
Internal Forces and Moment: As shown on FBD
Section Properties:
I =
1
(0.3) A 0.83 B = 0.0128 in3
12
QC = y¿A¿ = 0.3(0.2)(0.3) = 0.0180 in3
Normal Stress: Applying the flexure formula.
sC = -
My
-300(0.2)
= = 4687.5 psi = 4.6875 ksi
I
0.0128
Shear Stress: Applying the shear formula.
tC =
VQC
75.0(0.0180)
=
= 351.6 psi = 0.3516 ksi
It
0.0128(0.3)
Construction of the Circle: In accordance with the sign convention, sx = 4.6875 ksi,
sy = 0, and txy = 0.3516 ksi. Hence,
savg =
sx + sy
2
=
4.6875 + 0
= 2.34375 ksi
2
The coordinates for reference points A and C are
A(4.6875, 0.3516)
C(2.34375, 0)
The radius of the circle is
R = 2(4.6875 - 2.34375)2 + 0.35162 = 2.3670 ksi
In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2,
respectively.
s1 = 2.34375 + 2.3670 = 4.71 ksi
Ans.
s2 = 2.34375 - 2.3670 = -0.0262 ksi
Ans.
702
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•9–77.
A spherical pressure vessel has an inner radius of
5 ft and a wall thickness of 0.5 in. Draw Mohr’s circle for the
state of stress at a point on the vessel and explain the
significance of the result. The vessel is subjected to an
internal pressure of 80 psi.
Normal Stress:
s1 = s2 =
pr
80(5)(12)
=
= 4.80 ksi
2t
2(0.5)
Mohr’s circle:
A(4.80, 0)
B(4.80, 0)
C(4.80, 0)
Regardless of the orientation of the element, the shear stress is zero and the state of
stress is represented by the same two normal stress components.
9–78. The cylindrical pressure vessel has an inner radius
of 1.25 m and a wall thickness of 15 mm. It is made from
steel plates that are welded along the 45° seam. Determine
the normal and shear stress components along this seam if
the vessel is subjected to an internal pressure of 8 MPa.
sx =
45⬚
1.25 m
pr
8(1.25)
=
= 333.33 MPa
2t
2(0.015)
sy = 2sx = 666.67 MPa
A(333.33, 0)
sx¿ =
B(666.67, 0)
C(500, 0)
333.33 + 666.67
= 500 MPa
2
Ans.
tx¿y¿ = R = 666.67 - 500 = 167 MPa
Ans.
703
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•9–79.
Determine the normal and shear stresses at point
D that act perpendicular and parallel, respectively, to the
grains. The grains at this point make an angle of 30° with
the horizontal as shown. Point D is located just to the left of
the 10-kN force.
10 kN
A 100 mm
D
B
30⬚
1m
100 mm
D
100 mm
Using the method of section and consider the FBD of the left cut segment, Fig. a
+ c ©Fy = 0;
5 - V = 0
V = 5 kN
a + ©MC = 0;
M - 5(1) = 0
M = 5 kN # m
The moment of inertia of the rectangular cross - section about the neutral axis is
I =
1
(0.1)(0.33) = 0.225(10 - 3) m4
12
Referring to Fig. b,
QD = y¿A¿ = 0.1(0.1)(0.1) = 0.001 m3
The normal stress developed is contributed by bending stress only. For point D,
y = 0.05 m. Then
s =
My
5(103)(0.05)
= 1.111 MPa (T)
=
I
0.225(10 - 3)
The shear stress is contributed by the transverse shear stress only. Thus,
t =
5(103)(0.001)
VQD
= 0.2222 MPa
=
It
0.225(10 - 3)(0.1)
The state of stress at point D can be represented by the element shown in Fig. c
In accordance to the established sign convention, sx = 1.111 MPa, sy = 0 and
txy = -0.2222 MPa, Thus.
savg =
sx + sy
2
=
1.111 + 0
= 0.5556 MPa
2
Then, the coordinate of reference point A and the center C of the circle are
A(1.111, -0.2222)
C(0.5556, 0)
Thus, the radius of the circle is given by
R = 2(1.111 - 0.5556)2 + (-0.2222)2 = 0.5984 MPa
Using these results, the circle shown in Fig. d can be constructed.
Referring to the geometry of the circle, Fig. d,
a = tan - 1 a
0.2222
b = 21.80°
1.111 - 0.5556
b = 180° - (120° - 21.80°) = 81.80°
704
1m
300 mm
2m
C
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9–79. Continued
Then
sx¿ = 0.5556 - 0.5984 cos 81.80° = 0.4702 MPa = 470 kPa
Ans.
tx¿y¿ = 0.5984 sin 81.80° = 0.5922 MPa = 592 kPa
Ans.
705
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*9–80. Determine the principal stress at point D, which is
located just to the left of the 10-kN force.
10 kN
A 100 mm
D
B
30⬚
1m
100 mm
D
100 mm
Using the method of section and consider the FBD of the left cut segment, Fig. a,
+ c ©Fy = 0;
5 - V = 0
V = 5 kN
a + ©MC = 0;
M - 5(1) = 0
M = 5 kN # m
I =
1
(0.1)(0.33) = 0.225(10 - 3) m4
12
Referring to Fig. b,
QD = y¿A¿ = 0.1(0.1)(0.1) = 0.001 m3
The normal stress developed is contributed by bending stress only. For point D,
y = 0.05 m
s =
My
5(103)(0.05)
= 1.111 MPa (T)
=
I
0.225(10 - 3)
The shear stress is contributed by the transverse shear stress only. Thus,
t =
5(103)(0.001)
VQD
= 0.2222 MPa
=
It
0.225(10 - 3)(0.1)
The state of stress at point D can be represented by the element shown in Fig. c.
In accordance to the established sign convention, sx = 1.111 MPa, sy = 0, and
txy = -0.2222 MPa. Thus,
savg =
sx + sy
2
=
1.111 + 0
= 0.5556 MPa
2
Then, the coordinate of reference point A and center C of the circle are
A(1.111, -0.2222)
C(0.5556, 0)
Thus, the radius of the circle is
R = CA = 2(1.111 - 0.5556)2 + (-0.2222)2 = 0.5984 MPa
Using these results, the circle shown in Fig. d.
In-Plane Principal Stresses. The coordinates of points B and D represent s1 and s2,
respectively. Thus,
s1 = 0.5556 + 0.5984 = 1.15 MPa
Ans.
s2 = 0.5556 - 0.5984 = -0.0428 MPa
Ans.
706
1m
300 mm
2m
C
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9–80. Continued
Referring to the geometry of the circle, Fig. d,
tan (2uP)1 =
0.2222
= 0.4
1.111 - 0.5556
(uP)1 = 10.9° (Clockwise)
Ans.
The state of principal stresses is represented by the element show in Fig. e.
707
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•9–81.
Determine the principal stress at point A on the
cross section of the hanger at section a–a. Specify the
orientation of this state of stress and indicate the result on
an element at the point.
0.75 m
0.75 m
a
250 mm a
900 N
Internal Loadings: Considering the equilibrium of the free - body diagram of the
hanger’s left cut segment, Fig. a,
+ ©F = 0;
:
x
900 - N = 0
N = 900 N
+ c ©Fy = 0;
V - 900 = 0
V = 900 N
a + ©MO = 0;
900(1) - 900(0.25) - M = 0
M = 675 N # m
b
250 mm
Section Properties: The cross - sectional area and the moment of inertia about the
centroidal axis of the hanger’s cross section are
A = 0.05(0.1) - 0.04(0.09) = 1.4 A 10 - 3 B m2
1
1
(0.05) A 0.13 B (0.04) A 0.093 B = 1.7367 A 10 - 6 B m4
12
12
Referring to Fig. b,
QA = 2y1œ A1œ + y2œ A2œ = 2[0.0375(0.025)(0.005)] + 0.0475(0.005)(0.04)
= 18.875 A 10 - 6 B m3
Normal and Shear Stress: The normal stress is a combination of axial and bending
stresses. Thus,
sA =
675(0.025)
MyA
N
900
+
= +
= 9.074 MPa
-3
A
I
1.4 A 10 B
1.7367 A 10 - 6 B
The shear stress is caused by the transverse shear stress.
tA =
900 C 18.875 A 10 - 6 B D
VQA
=
= 0.9782 MPa
It
1.7367 A 10 - 6 B (0.01)
The state of stress at point A is represented by the element shown in Fig. c.
Construction of the Circle: sx = 9.074 MPa, sy = 0, and txy = 0.9782 MPa. Thus,
savg =
sx + sy
2
=
9.074 + 0
= 4.537 MPa
2
The coordinates of reference points A and the center C of the circle are
A(9.074, 0.9782)
C(4.537, 0)
Thus, the radius of the circle is
R = CA = 2(9.074 - 4.537)2 + 0.97822 = 4.641 MPa
Using these results, the circle is shown in Fig. d.
708
b
900 N
5 mm
25 mm
A
100 mm
5 mm
50 mm
I =
0.5 m
5 mm
Sections a – a
and b – b
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9–81. Continued
In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2,
respectively.
s1 = 4.537 + 4.641 = 9.18 MPa
Ans.
s2 = 4.537 - 4.641 = -0.104 MPa
Ans.
Orientaion of Principal Plane: Referring to the geometry of the circle, Fig. d,
tan 2 A uP B 1 =
0.9782
= 0.2156
9.074 - 4.537
A uP B 1 = 6.08° (counterclockwise)
Ans.
The state of principal stresses is represented on the element shown in Fig. e.
709
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9–82. Determine the principal stress at point A on the
cross section of the hanger at section b–b. Specify the
orientation of the state of stress and indicate the results on
an element at the point.
0.75 m
0.75 m
a
250 mm a
900 N
Internal Loadings: Considering the equilibrium of the free - body diagram of the
hanger’s left cut segment, Fig. a,
+ c ©Fy = 0;
V - 900 - 900 = 0
V = 1800 N
900(2.25) + 900(0.25) - M = 0
M = 2250 N # m
a + ©MO = 0;
Section Properties: The cross - sectional area and the moment of inertia about the
centroidal axis of the hanger’s cross section are
A = 0.05(0.1) - 0.04(0.09) = 1.4 A 10 - 3 B m2
I =
Referring to Fig. b.
QA = 2y1œ A1œ + y2œ A2œ = 2[0.0375(0.025)(0.005)] + 0.0475(0.005)(0.04)
= 18.875 A 10 - 6 B m3
Normal and Shear Stress: The normal stress is contributed by the bending stress
only.
MyA
2250(0.025)
=
= 32.39 MPa
I
1.7367 A 10 - 6 B
The shear stress is contributed by the transverse shear stress only.
tA =
1800 C 18.875 A 10 - 6 B D
VQA
=
= 1.956 MPa
It
1.7367 A 10 - 6 B (0.01)
The state stress at point A is represented by the element shown in Fig. c.
Construction of the Circle: sx = 32.39 MPa, sy = 0, and txy = 1.956 MPa. Thus,
savg =
sx + sy
2
=
32.39 + 0
= 16.19 MPa
2
The coordinates of reference point A and the center C of the circle are
A(32.39, 1.956)
b
250 mm
C(16.19, 0)
Thus, the radius of the circle is
R = CA = 2(32.39 - 16.19)2 + 1.9562 = 16.313 MPa
Using these results, the cricle is shown in Fig. d.
710
b
900 N
5 mm
25 mm
A
100 mm
5 mm
50 mm
5 mm
Sections a – a
and b – b
1
1
(0.05) A 0.13 B (0.04) A 0.093 B = 1.7367 A 10 - 6 B m4
12
12
sA =
0.5 m
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9–82. Continued
In - Plane Principal Stresses: The coordinates of reference point B and D represent
s1 and s2, respectively.
s1 = 16.19 + 16.313 = 32.5 MPa
Ans.
s2 = 16.19 - 16.313 = -0.118 MPa
Ans.
Orientaion of Principal Plane: Referring to the geometry of the circle, Fig. d,
tan 2 A uP B 1 =
1.956
= 0.1208
32.39 - 16.19
A uP B 1 = 3.44°
(counterclockwise)
Ans.
The state of principal stresses is represented on the element shown in Fig. e.
711
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9–83. Determine the principal stresses and the maximum
in-plane shear stress that are developed at point A. Show
the results on an element located at this point. The rod has a
diameter of 40 mm.
450 N
150 mm
Using the method of sections and consider the FBD of the member’s upper cut
segment, Fig. a,
+ c ©Fy = 0;
450 - N = 0
N = 450 N
a + ©MC = 0;
450(0.1) - M = 0
M = 45 N # m
B
100 mm
A
150 mm
A = p(0.022) = 0.4(10 - 3)p m2
I =
450 N
p
(0.024) = 40(10 - 9)p m4
4
The normal stress is the combination of axial and bending stress. Thus,
s =
My
N
+
A
I
For point A, y = C = 0.02 m.
s =
45 (0.02)
450
+
= 7.520 MPa
0.4(10 - 3)p
40(10 - 9)p
Since no transverse shear and torque is acting on the cross - section
t = 0
The state of stress at point A can be represented by the element shown in Fig. b.
In accordance to the established sign convention sx = 0, sy = 7.520 MPa and
txy = 0. Thus
savg =
sx + sy
2
=
0 + 7.520
= 3.760 MPa
2
Then, the coordinates of reference point A and the center C of the circle are
A(0, 0)
C(3.760, 0)
Thus, the radius of the circle is
R = CA = 3.760 MPa
Using this results, the circle shown in Fig. c can be constructed. Since no shear stress
acts on the element,
s1 = sy = 7.52 MPa
s2 = sx = 0
Ans.
The state of principal stresses can also be represented by the element shown in Fig. b.
The state of maximum in - plane shear stress is represented by point B on the circle,
Fig. c. Thus.
tmax
in-plane
= R = 3.76 MPa
Ans.
712
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9–83. Continued
From the circle,
2us = 90°
us = 45° (counter clockwise)
Ans.
The state of maximum In - Plane shear stress can be represented by the element
shown in Fig. d.
713
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*9–84. Draw the three Mohr’s circles that describe each of
the following states of stress.
5 ksi
(a) Here, smin = 0, sint = 3 ksi and smax = 5 ksi. The three Mohr’s circle
of this state of stress are shown in Fig. a
3 ksi
(b) Here, smin = 0, sint = 140 MPa and smax = 180 MPa. The three
Mohr’s circle of this state of stress are shown in Fig. b
(a)
•9–85.
Draw the three Mohr’s circles that describe the
following state of stress.
180 MPa
140 MPa
(b)
300 psi
Here, smin = -300 psi, sint = 0 and smax = 400 psi. The three Mohr’s circle for this
state of stress is shown in Fig. a.
400 psi
714
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z
9–86. The stress at a point is shown on the element.
Determine the principal stress and the absolute maximum
shear stress.
y
x
80 MPa
For y – z plane:
A(0, -80)
B(90, 80)
C(45, 0)
R = 2452 + 802 = 91.79
s1 = 45 + 91.79 = 136.79 MPa
s2 = 45 - 91.79 = -46.79 MPa
Thus,
tabs
max
=
s1 = 0
Ans.
s2 = 137 MPa
Ans.
s3 = -46.8 MPa
Ans.
136.79 - (-46.79)
smax - smin
=
= 91.8 MPa
2
2
Ans.
715
90 MPa
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z
9–87. The stress at a point is shown on the element.
Determine the principal stress and the absolute maximum
shear stress.
Mohr’s circle for the element in y - 7 plane, Fig. a, will be drawn first. In accordance
to the established sign convention, sy = 30 psi, sz = 120 psi and tyz = 70 psi. Thus
savg =
sy + sz
2
=
x
y
120 psi
70 psi
30 + 120
= 75 psi
2
30 psi
Thus the coordinates of reference point A and the center C of the circle are
A(30, 70)
C(75, 0)
Thus, the radius of the circle is
R = CA = 2(75 - 30)2 + 702 = 83.217 psi
Using these results, the circle shown in Fig. b.
The coordinates of point B and D represent the principal stresses
From the results,
smax = 158 psi
smin = -8.22 psi
sint = 0 psi
Ans.
Using these results, the three Mohr’s circle are shown in Fig. c,
From the geometry of the three circles,
tabs
max
=
158.22 - ( -8.22)
smax - smin
=
= 83.22 psi
2
2
716
Ans.
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z
*9–88. The stress at a point is shown on the element.
Determine the principal stress and the absolute maximum
shear stress.
Mohr’s circle for the element in x - z plane, Fig. a, will be drawn first. In accordance
to the established sign convention, sx = -2 ksi, sz = 0 and txz = 8 ksi. Thus
savg =
sx + sz
2
=
-2 + 0
= -1 ksi
2
2 ksi
8 ksi
Thus, the coordinates of reference point A and the center C of the circle are
A( -2, 8)
C(-1, 0)
Thus, the radius of the circle is
R = CA = 2[-2 - (-1)]2 + 82 = 265 ksi
Using these results, the circle in shown in Fig. b,
The coordinates of points B and D represent s1 and s2, respectively.
s = -1 + 265 = 7.062 ksi
smax = 7.06 ksi
sint = 0
smin = -9.06 ksi
From the results obtained,
sint = 0 ksi
smax = 7.06 ksi
smin = -9.06 ksi
Ans.
Using these results, the three Mohr’s circles are shown in Fig, c.
From the geometry of the cricle,
tabs
max
=
y
x
7.06 - (-9.06)
smax - smin
=
= 8.06 ksi
2
2
Ans.
717
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•9–89.
The stress at a point is shown on the element.
Determine the principal stress and the absolute maximum
shear stress.
z
y
x
For x – y plane:
150 MPa
120 MPa
R = CA = 2(120 - 60)2 + 1502 = 161.55
s1 = 60 + 161.55 = 221.55 MPa
s2 = 60 - 161.55 = -101.55 MPa
s1 = 222 MPa
tabs
max
=
s2 = 0 MPa
s3 = -102 MPa
Ans.
221.55 - (-101.55)
smax - smin
=
= 162 MPa
2
2
Ans.
9–90. The state of stress at a point is shown on the
element. Determine the principal stress and the absolute
maximum shear stress.
z
x
For y - z plane:
A(5, -4)
B(-2.5, 4)
4 ksi
s1 = 1.25 + 5.483 = 6.733 ksi
5 ksi
s2 = 1.25 - 5.483 = -4.233 ksi
Thus,
s1 = 6.73 ksi
Ans.
s2 = 0
Ans.
s3 = -4.23 ksi
Ans.
savg =
6.73 + (-4.23)
= 1.25 ksi
2
tabs
6.73 - (-4.23)
smax - smin
=
= 5.48 ksi
2
2
=
2.5 ksi
C(1.25, 0)
R = 23.752 + 42 = 5.483
max
y
Ans.
718
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*9–92. The solid shaft is subjected to a torque, bending
moment, and shear force as shown. Determine the principal
stress acting at points A and B and the absolute maximum
shear stress.
450 mm A
B
300 N⭈m
25 mm
45 N⭈m
800 N
Internal Forces and Moment: As shown on FBD.
Section Properties:
Iz =
p
A 0.0254 B = 0.306796 A 10 - 6 B m4
4
J =
p
A 0.0254 B = 0.613592 A 10 - 6 B m4
2
(QA)x = 0
(QB)y = y¿A¿
=
4(0.025) 1
c (p) A 0.0252 B d = 10.417 A 10 - 6 B m3
3p
2
Normal stress: Applying the flexure formula.
s = -
Mzy
Iz
-60.0(0.025)
sA = -
0.306796(10 - 6)
= 4.889 MPa
-60.0(0)
sB = -
0.306796(10 - 6)
= 0
Shear Stress: Applying the torsion formula for point A,
tA =
45.0(0.025)
Tc
= 1.833 MPa
=
J
0.613592(10 - 6)
The transverse shear stress in the y direction and the torsional shear stress can be
VQ
Tr
obtained using shear formula and torsion formula. tv =
and ttwist =
,
It
J
respectively.
tB = (tv)y - ttwist
=
800 C 10.417(10 - 6) D
-6
0.306796(10 )(0.05)
45.0(0.025)
-
0.613592(10 - 6)
= -1.290 MPa
Construction of the Circle: sx = 4.889 MPa, sz = 0, and txz = -1.833 MPa for
point A. Hence,
savg =
sx + sz
2
=
4.889 + 0
= 2.445 MPa
2
The coordinates for reference points A and C are A (4.889, –1.833) and C(2.445, 0).
719
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9–92.
Continued
The radius of the circle is
R = 2(4.889 - 2.445)2 + 1.8332 = 3.056 MPa
sx = sy = 0 and txy = -1.290 MPa for point B. Hence,
savg =
sx + sz
= 0
2
The coordinates for reference points A and C are A(0. ‚–1.290) and C(0,0).
The radius of the circle is R = 1.290 MPa
In - Plane Principal Stresses: The coordinates of point B and D represent s1 and s2,
respectively. For point A
s1 = 2.445 + 3.056 = 5.50 MPa
s2 = 2.445 - 3.506 = -0.611 MPa
For point B
s1 = 0 + 1.290 = 1.29 MPa
s2 = 0 - 1.290 = -1.290 MPa
Three Mohr’s Circles: From the results obtaired above, the principal stresses for
point A are
smax = 5.50 MPa
sint = 0
smin = -0.611 MPa
Ans.
sint = 0
smin = -1.29 MPa
Ans.
And for point B
smax = 1.29 MPa
Absolute Maximum Shear Stress: For point A,
tabs
max
=
5.50 - (-0.611)
smax - smin
=
= 3.06 MPa
2
2
Ans.
=
1.29 - (-1.29)
smax - smin
=
= 1.29 MPa
2
2
Ans.
For point B,
tabs
max
720
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•9–93.
The propane gas tank has an inner diameter of
1500 mm and wall thickness of 15 mm. If the tank is
pressurized to 2 MPa, determine the absolute maximum
shear stress in the wall of the tank.
Normal Stress: Since
750
r
=
= 50 7 10, thin - wall analysis can be used. We have
t
15
s1 =
2(750)
pr
=
= 100 MPa
t
15
s2 =
2(750)
pr
=
= 50 MPa
2t
2(15)
The state of stress of any point on the wall of the tank can be represented on the
element shown in Fig. a
Construction of Three Mohr’s Circles: Referring to the element,
smax = 100 MPa
sint = 50 MPa
smin = 0
Using these results, the three Mohr’s circles are shown in Fig. b.
Absolute Maximum Shear Stress: From the geometry of three circles,
tabs
max
=
smax - smin
100 - 0
=
= 50 MPa
2
2
Ans.
721
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9–94. Determine the principal stress and absolute
maximum shear stress developed at point A on the cross
section of the bracket at section a–a.
12 in.
6 in.
5
3
a
4
a
0.5 in.
B
0.25 in.
A
0.25 in.
0.25 in.
1.5 in.1.5 in.
Section a – a
Internal Loadings: Considering the equilibrium of the free - body diagram of the
bracket’s upper cut segment, Fig. a,
+ c ©Fy = 0;
3
N - 500 a b = 0
5
N = 300 lb
+ ©F = 0;
;
x
4
V - 500 a b = 0
5
V = 400 lb
3
4
©MO = 0; M - 500 a b(12) - 500 a b(6) = 0
5
5
M = 6000 lb # in
Section Properties: The cross - sectional area and the moment of inertia of the
bracket’s cross section are
A = 0.5(3) - 0.25(2.5) = 0.875 in2
I =
1
1
(0.5) A 33 B (0.25) A 2.53 B = 0.79948 in4
12
12
Referring to Fig. b.
QA = x1œ A1œ + x2œ A2œ = 0.625(1.25)(0.25) + 1.375(0.25)(0.5) = 0.3672 in3
Normal and Shear Stress: The normal stress is
sA =
N
300
= = -342.86 psi
A
0.875
The shear stress is contributed by the transverse shear stress.
tA =
400(0.3672)
VQA
=
= 734.85 psi
It
0.79948(0.25)
The state of stress at point A is represented by the element shown in Fig. c.
Construction of the Circle: sx = 0, sy = -342.86 psi, and txy = 734.85. Thus,
savg =
sx + sy
2
=
0 + (-342.86)
= -171.43 psi
2
The coordinates of reference point A and the center C of the circle are
A(0, 734.85)
C(-171.43, 0)
Thus, the radius of the circle is
R = CA = 2[0 - (-171.43)]2 + 734.852 = 754.58 psi
722
500 lb
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9–94. Continued
Using these results, the cricle is shown in Fig. d.
In - Plane Principal Stresses: The coordinates of reference point B and D represent
s1 and s2, respectively.
s1 = -171.43 + 754.58 = 583.2 psi
s2 = -171.43 - 754.58 = -926.0 psi
Three Mohr’s Circles: Using these results,
smax = 583 psi
sint = 0 smin = -926 psi
Ans.
Absolute Maximum Shear Stress:
tabs
max
=
583.2 - (-926.0)
smax - smin
=
- 755 psi
2
2
Ans.
723
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9–95. Determine the principal stress and absolute
maximum shear stress developed at point B on the cross
section of the bracket at section a–a.
12 in.
Internal Loadings: Considering the equilibrium of the free - body diagram of the 6 in.
bracket’s upper cut segment, Fig. a,
a
+ c ©Fy = 0;
+ ©F = 0;
;
x
3
N - 500 a b = 0
5
N = 300 lb
4
V - 500 a b = 0
5
V = 400 lb
1
1
(0.5) A 33 B (0.25) A 2.53 B = 0.79948 in4
12
12
Referring to Fig. b,
QB = 0
Normal and Shear Stress: The normal stress is a combination of axial and bending
stress.
6000(1.5)
MxB
N
300
+
= +
= 10.9 ksi
A
I
0.875
0.79948
Since QB = 0, tB = 0. The state of stress at point B is represented on the element
shown in Fig. c.
In - Plane Principal Stresses: Since no shear stress acts on the element,
s2 = 0
Three Mohr’s Circles: Using these results,
smax = 10.91 ksi
sint = smin = 0
Ans.
Absolute Maximum Shear Stress:
tabs
max
=
0.25 in.
1.5 in.1.5 in.
Section a – a
M = 6000 lb # in
smax - smin
10.91 - 0
=
= 5.46 ksi
2
2
Ans.
724
500 lb
0.25 in.
A
0.25 in.
A = 0.5(3) - 0.25(2.5) = 0.875 in2
s1 = 10.91 ksi
a
0.5 in.
Section Properties: The cross - sectional area and the moment of inertia about the
centroidal axis of the bracket’s cross section are
sB =
4
B
4
3
©MO = 0; M - 500 a b(12) - 500 a b(6) = 0
5
5
I =
5
3
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*9–96. The solid propeller shaft on a ship extends outward
from the hull. During operation it turns at v = 15 rad>s
when the engine develops 900 kW of power. This causes a
thrust of F = 1.23 MN on the shaft. If the shaft has an outer
diameter of 250 mm, determine the principal stresses at any
point located on the surface of the shaft.
0.75 m
A
T
Power Transmission: Using the formula developed in Chapter 5,
P = 900 kW = 0.900 A 106 B N # m>s
0.900(106)
P
=
= 60.0 A 103 B N # m
v
15
T0 =
Internal Torque and Force: As shown on FBD.
Section Properties:
A =
p
A 0.252 B = 0.015625p m2
4
J =
p
A 0.1254 B = 0.3835 A 10 - 3 B m4
2
s =
-1.23(106)
N
=
= -25.06 MPa
A
0.015625p
Normal Stress:
Shear Stress: Applying the torsion formula,
t =
60.0(103) (0.125)
Tc
= 19.56 MPa
=
J
0.3835(10 - 3)
In - Plane Principal Stresses: sx = -25.06 MPa, sy = 0 and txy = 19.56 MPa for
any point on the shaft’s surface. Applying Eq. 9-5,
s1,2 =
=
sx + sy
2
;
C
a
sx - sy
2
2
b + t2xy
-25.06 - 0 2
-25.06 + 0
;
a
b + (19.56)2
2
C
2
= -12.53 ; 23.23
s1 = 10.7 MPa
s2 = -35.8 MPa
Ans.
725
F
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•9–97. The solid propeller shaft on a ship extends outward
from the hull. During operation it turns at v = 15 rad>s
when the engine develops 900 kW of power. This causes a
thrust of F = 1.23 MN on the shaft. If the shaft has a
diameter of 250 mm, determine the maximum in-plane shear
stress at any point located on the surface of the shaft.
0.75 m
A
T
Power Transmission: Using the formula developed in Chapter 5,
P = 900 kW = 0.900 A 106 B N # m>s
T0 =
0.900(106)
P
=
= 60.0 A 103 B N # m
v
15
Internal Torque and Force: As shown on FBD.
Section Properties:
A =
p
A 0.252 B = 0.015625p m2
4
J =
p
A 0.1254 B = 0.3835 A 10 - 3 B m4
2
s =
-1.23(106)
N
=
= -25.06 MPa
A
0.015625p
Normal Stress:
Shear Stress: Applying the torsion formula.
t =
60.0(103) (0.125)
Tc
= 19.56 MPa
=
J
0.3835 (10 - 3)
Maximum In - Plane Principal Shear Stress: sx = -25.06 MPa, sy = 0, and
txy = 19.56 MPa for any point on the shaft’s surface. Applying Eq. 9-7,
t max
in-plane
a
sx - sy
2
b + t2xy
=
C
=
-25.06 - 0 2
b + (19.56)2
C
2
2
a
= 23.2 MPa
Ans.
726
F
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9–98. The steel pipe has an inner diameter of 2.75 in. and
an outer diameter of 3 in. If it is fixed at C and subjected to
the horizontal 20-lb force acting on the handle of the pipe
wrench at its end, determine the principal stresses in the
pipe at point A, which is located on the surface of the pipe.
20 lb
12 in.
10 in.
Internal Forces, Torque and Moment: As shown on FBD.
A
Section Properties:
B
I =
p
A 1.54 - 1.3754 B = 1.1687 in4
4
J =
p
A 1.54 - 1.3754 B = 2.3374 in4
2
C
y
z
(QA)z = ©y¿A¿
x
4(1.5) 1
4(1.375) 1
=
c p A 1.52 B d c p A 1.3752 B d
3p 2
3p
2
= 0.51693 in3
Normal Stress: Applying the flexure formula s =
sA =
My z
Iy
,
200(0)
= 0
1.1687
Shear Stress: The transverse shear stress in the z direction and the torsional shear
VQ
stress can be obtained using shear formula and torsion formula, tv =
and
It
Tr
ttwist =
, respectively.
J
tA = (tv)z - ttwist
=
20.0(0.51693)
240(1.5)
1.1687(2)(0.125)
2.3374
= -118.6 psi
In - Plane Principal Stress: sx = 0, sz = 0 and txz = -118.6 psi for point A.
Applying Eq. 9-5
s1,2 =
sx + sz
2
;
C
a
sx - sz
2
2
b + t2xz
= 0 ; 20 + (-118.6)2
s1 = 119 psi
s2 = -119 psi
Ans.
727
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9–99. Solve Prob. 9–98 for point B, which is located on the
surface of the pipe.
20 lb
12 in.
10 in.
A
B
Internal Forces, Torque and Moment: As shown on FBD.
Section Properties:
C
I =
p
A 1.54 - 1.3754 B = 1.1687 in4
4
y
z
x
p
J =
A 1.54 - 1.3754 B = 2.3374 in4
2
(QB)z = 0
Normal Stress: Applying the flexure formula s =
sB =
My z
Iv
,
200(1.5)
= 256.7 psi
1.1687
Shear Stress: Torsional shear stress can be obtained using torsion formula,
Tr
.
ttwist =
J
tB = ttwist =
240(1.5)
= 154.0 psi
2.3374
In - Plane Prinicipal Stress: sx = 256.7 psi, sy = 0, and txy = -154.0 psi for point B.
Applying Eq. 9-5
s1,2 =
=
sx + sy
2
;
C
sx - sy
a
2
2
b + t2xy
256.7 - 0 2
256.7 + 0
;
a
b + ( -154.0)2
2
C
2
= 128.35 ; 200.49
s1 = 329 psi
s2 = -72.1 psi
Ans.
728
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*9–100. The clamp exerts a force of 150 lb on the boards
at G. Determine the axial force in each screw, AB and CD,
and then compute the principal stresses at points E and F.
Show the results on properly oriented elements located at
these points. The section through EF is rectangular and is
1 in. wide.
A
150 lb
C
G
0.5 in.
E
Support Reactions: FBD(a).
F
a + ©MB = 0;
FCD(3) - 150(7) = 0
FCD = 350 lb
Ans.
+ c ©Fy = 0;
350 - 150 - FAB = 0
FAB = 200 lb
Ans.
B
1.5 in. 1.5 in.
Internal Forces and Moment: As shown on FBD(b).
Section Properties:
I =
1
(1) A 1.53 B = 0.28125 in4
12
QE = 0
QF = y¿A¿ = 0.5(0.5)(1) = 0.250 in3
Normal Stress: Applying the flexure formula s = -
My
,
I
sE = -
-300(0.75)
= 800 psi
0.28125
sF = -
-300(0.25)
= 266.67 psi
0.28125
VQ
,
It
Shear Stress: Applying the shear formula t =
tE =
200(0)
= 0
0.28125(1)
tF =
200(0.250)
= 177.78 psi
0.28125(1)
In - Plane Principal Stress: sx = 800 psi, sy = 0 and txy = 0 for point E. Since no
shear stress acts upon the element.
s1 = sx = 800 psi
Ans.
s2 = sy = 0
Ans.
sx = 266.67 psi, sy = 0, and txy = 177.78 psi for point F. Applying Eq. 9-5
s1,2 =
=
sx + sy
2
;
C
sx - sy
a
2
2
b + t2xy
266.67 - 0 2
266.67 + 0
;
a
b + 177.782
2
C
2
= 133.33 ; 222.22
s1 = 356 psi
s2 = -88.9 psi
Ans.
729
150 lb
D
4 in.
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9–100.
Continued
Orientation of Principal Plane: Applying Eq. 9-4 for point F,
tan 2up =
txy
A sx - sy B >2
up = 26.57°
=
and
177.78
= 1.3333
(266.67 - 0)>2
-63.43°
Substituting the results into Eq. 9-1 with u = 26.57° yields
sx¿ =
=
sx + sy
2
sx - sy
+
2
cos 2u + txy sin 2u
266.67 - 0
266.67 + 0
+
cos 53.13° + 177.78 sin 53.13°
2
2
= 356 psi = s1
Hence,
up1 = 26.6°
up2 = -63.4°
Ans.
730
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9–101. The shaft has a diameter d and is subjected to the
loadings shown. Determine the principal stress and
the maximum in-plane shear stress that is developed
anywhere on the surface of the shaft.
F
T0
F
T0
Internal Forces and Torque: As shown on FBD(b).
Section Properties:
A =
p 2
d
4
J =
p d 4
p 4
a b =
d
2 2
32
Normal Stress:
N
-F
4F
= p 2 = - 2
A
pd
4 d
s =
Shear Stress: Applying the shear torsion formula,
T0 A 2 B
16T0
Tc
= p 4 =
J
d
pd3
32
d
t =
16T0
4F
, sy = 0, and txy = for any point on
pd2
pd3
the shaft’s surface. Applying Eq. 9-5,
In - Plane Principal Stress: sx = -
s1,2 =
sx + sy
;
2
C
a
- 4F2 + 0
pd
=
;
2
D
sx - sy
2
¢
2
b + t2xy
- 4F2 - 0
pd
2
2
≤ + a-
16T0
3
pd
b
2
=
64T20
2
-F ;
F2 +
≤
2 ¢
C
pd
d2
s1 =
64T20
2
-F +
F2 +
≤
2 ¢
C
pd
d2
Ans.
64T20
2
F +
F2 +
≤
2 ¢
C
pd
d2
Ans.
s2 = -
Maximum In - Plane Shear Stress: Applying Eq. 9-7,
t max
in-plane
=
=
=
C
a
D
¢
sx - sy
2
- 4F2 - 0
pd
2
2
b + t2xy
2
≤ + a-
b
pd3
16T0
2
64T20
2
2
F
+
pd2 C
d2
Ans.
731
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9–102. The state of stress at a point in a member is shown
on the element. Determine the stress components acting on
the plane AB.
A
50 MPa
30⬚
28 MPa
100 MPa
B
Construction of the Circle: In accordance with the sign convention, sx = -50 MPa,
sy = -100 MPa, and txy = -28 MPa. Hence,
savg =
sx + sy
2
=
-50 + (-100)
= -75.0 MPa
2
The coordinates for reference points A and C are A(–50, –28) and C(–75.0, 0).
The radius of the circle is R = 2(75.0 - 50)2 + 282 = 37.54 MPa.
Stress on the Rotated Element: The normal and shear stress components
A sx¿ and tx¿y¿ B are represented by the coordinates of point P on the circle
sx¿ = -75.0 + 37.54 cos 71.76° = -63.3 MPa
Ans.
tx¿y¿ = 37.54 sin 71.76° = 35.7 MPa
Ans.
732
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–103. The propeller shaft of the tugboat is subjected to
the compressive force and torque shown. If the shaft has an
inner diameter of 100 mm and an outer diameter of 150 mm,
determine the principal stress at a point A located on the
outer surface.
10 kN
A
2 kN·m
Internal Loadings: Considering the equilibrium of the free - body diagram of the
propeller shaft’s right segment, Fig. a,
©Fx = 0; 10 - N = 0
N = 10 kN
©Mx = 0; T - 2 = 0
T = 2 kN # m
Section Properties: The cross - sectional area and the polar moment of inertia of the
propeller shaft’s cross section are
A = p A 0.0752 - 0.052 B = 3.125p A 10 - 3 B m2
J =
p
A 0.0754 - 0.054 B = 12.6953125p A 10 - 6 B m4
2
Normal and Shear Stress: The normal stress is a contributed by axial stress only.
sA =
10 A 103 B
N
= = -1.019 MPa
A
3.125p A 10 - 3 B
The shear stress is contributed by the torsional shear stress only.
tA =
2 A 103 B (0.075)
Tc
=
= 3.761 MPa
J
12.6953125p A 10 - 6 B
The state of stress at point A is represented by the element shown in Fig. b.
Construction of the Circle: sx = -1.019 MPa, sy = 0, and txy = -3.761 MPa.
Thus,
savg =
sx + sy
2
=
-1.019 + 0
= -0.5093 MPa
2
The coordinates of reference point A and the center C of the circle are
A(-1.019, -3.761)
C(-0.5093, 0)
Thus, the radius of the circle is
R = CA = 2[-1.019 - ( -0.5093)]2 + (-3.761)2 = 3.795 MPa
Using these results, the circle is shown is Fig. c.
In - Plane Principal Stress: The coordinates of reference points B and D represent
s1 and s2, respectively.
s1 = -0.5093 + 3.795 = 3.29 MPa
Ans.
s2 = -0.5093 - 3.795 = -4.30 MPa
Ans.
733
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9–103.
Continued
Orientation of the Principal Plane: Referring to the geometry of the circle, Fig. d,
tan 2 A up B 2 =
3.761
= 7.3846
1.019 - 0.5093
A up B 2 = 41.1° (clockwise)
Ans.
The state of principal stresses is represented on the element shown in Fig. d.
734
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*9–104. The box beam is subjected to the loading shown.
Determine the principal stress in the beam at points A and B.
6 in.
A
6 in. B 8 in.
8 in.
Support Reactions: As shown on FBD(a).
Internal Forces and Moment: As shown on FBD(b).
Section Properties:
I =
1
1
(8) A 83 B (6) A 63 B = 233.33 in4
12
12
QA = QB = 0
Normal Stress: Applying the flexure formula.
s = -
My
I
sA = -
-300(12)(4)
= 61.71 psi
233.33
sB = -
-300(12)(-3)
= -46.29 psi
233.33
1200 lb
800 lb
Shear Stress: Since QA = QB = 0, then tA = tB = 0.
In - Plane Principal Stress: sx = 61.71 psi, sy = 0, and txy = 0 for point A. Since
no shear stress acts on the element,
s1 = sx = 61.7 psi
Ans.
s2 = sy = 0
Ans.
sx = -46.29 psi, sy = 0, and txy = 0 for point B. Since no shear stress acts on the
element,
s1 = sy = 0
Ans.
s2 = sx = -46.3 psi
Ans.
735
A
B
3 ft
2.5 ft
2.5 ft
5 ft
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
•9–105.
The wooden strut is subjected to the loading
shown. Determine the principal stresses that act at point C
and specify the orientation of the element at this point.
The strut is supported by a bolt (pin) at B and smooth
support at A.
50 N
50 N
60⬚
C
100 mm
40 N
40 N
B
A
25 mm
50 mm
200 mm
200 mm 200 mm 200 mm
100 mm 100 mm
QC = y¿A¿ = 0.025(0.05)(0.025) = 31.25(10 - 6) m3
I =
1
(0.025)(0.13) = 2.0833(10 - 6) m4
12
Normal stress: sC = 0
Shear stress:
t =
VQC
44(31.25)(10 - 6)
= 26.4 kPa
=
It
2.0833(10 - 6)(0.025)
Principal stress:
sx = sy = 0;
s1,2 =
txy = -26.4 kPa
sx + sy
;
2
C
a
sx - sy
2
2
b + t2 xy
= 0 ; 20 + (26.4)2
s1 = 26.4 kPa
s2 = -26.4 kPa
;
Ans.
Orientation of principal stress:
tan 2up =
txy
(sx - sy)
= - q
2
up = +45° and -45°
Use
Eq. 9-1 to determine the principal
sx + sy
sx - sy
+
cos 2u + txy sin 2u
sx¿ =
2
2
plane
of
s1
and
s2
u = up = -45°
sx¿ = 0 + 0 + (-26.4) sin( -90°) = 26.4 kPa
Therefore, up1 = -45°;
up2 = 45°
Ans.
736
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–106. The wooden strut is subjected to the loading shown.
If grains of wood in the strut at point C make an angle of 60°
with the horizontal as shown, determine the normal and
shear stresses that act perpendicular and parallel to the
grains, respectively, due to the loading. The strut is supported
by a bolt (pin) at B and smooth support at A.
50 N
50 N
60⬚
C
100 mm
40 N
40 N
B
A
25 mm
50 mm
200 mm
200 mm 200 mm 200 mm
100 mm 100 mm
QC = y¿A¿ = 0.025(0.05)(0.025) = 31.25(10 - 6) m3
I =
1
(0.025)(0.13) = 2.0833(10 - 6) m4
12
Normal stress: sC = 0
Shear stress:
t =
VQC
44(31.25)(10 - 6)
= 26.4 kPa
=
It
2.0833(10 - 6)(0.025)
Stress transformation: sx = sy = 0;
sx¿ =
sx + sy
sx - sy
+
2
2
txy = -26.4 kPa;
u = 30°
cos 2u + txy sin 2u
= 0 + 0 + (-26.4) sin 60° = -22.9 kPa
tx¿y¿ = -
sx - sy
2
Ans.
sin 2u + txy cos 2u
= -0 + (-26.4) cos 60° = -13.2 kPa
Ans.
737
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10–1. Prove that the sum of the normal strains in
perpendicular directions is constant.
ex¿ =
ey¿ =
ex + ey
2
ex - ey
+
ex + ey
2
2
ex - ey
-
2
cos 2u +
cos 2u -
gxy
2
gxy
2
sin 2u
(1)
sin 2u
(2)
Adding Eq. (1) and Eq. (2) yields:
ex¿ + ey¿ = ex + ey = constant
QED
738
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10–2. The state of strain at the point has components
of Px = 200 110-62, Py = -300 110-62, and gxy = 400(10-62.
Use the strain-transformation equations to determine the
equivalent in-plane strains on an element oriented at an
angle of 30° counterclockwise from the original position.
Sketch the deformed element due to these strains within the
x–y plane.
y
x
In accordance to the established sign convention,
ex = 200(10 - 6),
ex¿ =
ex + ey
ex - ey
+
2
= c
ey = -300(10 - 6)
2
cos 2u +
gxy
2
gxy = 400(10 - 6)
u = 30°
sin 2u
200 - (-300)
200 + (-300)
400
+
cos 60° +
sin 60° d(10 - 6)
2
2
2
= 248 (10 - 6)
gx¿y¿
2
= -a
Ans.
ex - ey
2
b sin 2u +
gxy
2
cos 2u
gx¿y¿ = e - C 200 - ( -300) D sin 60° + 400 cos 60° f(10 - 6)
= -233(10 - 6)
ey¿ =
ex + ey
= c
2
Ans.
ex - ey
-
2
cos 2u -
gxy
2
sin 2u
200 - ( -300)
200 + (-300)
400
cos 60° sin 60° d(10 - 6)
2
2
2
= -348(10 - 6)
Ans.
The deformed element of this equivalent state of strain is shown in Fig. a
739
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10–3. A strain gauge is mounted on the 1-in.-diameter
A-36 steel shaft in the manner shown. When the shaft is
rotating with an angular velocity of v = 1760 rev>min, the
reading on the strain gauge is P = 800110-62. Determine
the power output of the motor. Assume the shaft is only
subjected to a torque.
v = (1760 rev>min)a
60⬚
2p rad
1 min
ba
b = 184.307 rad>s
60 sec
1 rev
ex = ey = 0
ex¿ =
ex + ey
2
ex - ey
+
2
800(10 - 6) = 0 + 0 +
cos 2u +
gxy
2
gxy
2
sin 2u
sin 120°
gxy = 1.848(10 - 3) rad
t = G gxy = 11(103)(1.848)(10 - 3) = 20.323 ksi
t =
Tc
;
J
T(0.5)
20.323 = p
;
4
2 (0.5)
T = 3.99 kip # in = 332.5 lb # ft
P = Tv = 0.332.5 (184.307) = 61.3 kips # ft>s = 111 hp
Ans.
740
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*10–4. The state of strain at a point on a wrench
has components Px = 120110-62, Py = -180110-62, gxy =
150110-62. Use the strain-transformation equations to
determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and average normal strain.
In each case specify the orientation of the element and show
how the strains deform the element within the x–y plane.
ex = 120(10 - 6)
e1, 2 =
a)
ey = -180(10 - 6)
gxy = 150(10 - 6)
Ex - Ey 2
ex + ey
gxy 2
;
a
b + a
b
2
A
2
2
120 + (-180)
120 - ( -180) 2
150 2
-6
;
a
b + a
b d 10
2
A
2
2
e1 = 138(10 - 6);
e2 = -198(10 - 6)
= c
Ans.
Orientation of e1 and e2
gxy
150
=
= 0.5
tan 2up =
ex - ey
[120 - (-180)]
up = 13.28° and -76.72°
Use Eq. 10.5 to determine the direction of e1 and e2
ex¿ =
ex + ey
ex - ey
+
2
2
cos 2u +
gxy
2
sin 2u
u = up = 13.28°
ex¿ = c
120 + ( -180)
120 - ( -180)
150
+
cos (26.56°) +
sin 26.56° d 10 - 6
2
2
2
= 138 (10 - 6) = e1
Therefore up1 = 13.3° ;
gmax
b)
a
in-plane
=
2
A
up2 = -76.7°
ex - ey
b + a
2
gxy
b
Ans.
2
2
2
150 2
120 - ( -180) 2
-6
-6
= 2c a
b + a
b d10 = 335 (10 )
in-plane
2
2
A
gmax
eavg =
ex + ey
2
= c
120 + (-180)
d 10 - 6 = -30.0(10 - 6)
2
Ans.
Ans.
Orientation of gmax
tan 2us =
-(ex - ey)
gxy
=
-[120 - ( -180)]
= -2.0
150
us = -31.7° and 58.3°
Ans.
gmax
Use Eq. 10–6 to determine the sign of in-plane
gx¿y¿
ex - ey
gxy
= sin 2u +
cos 2u
2
2
2
u = us = -31.7°
gx¿y¿ = 2 c -
120 - (-180)
150
sin (-63.4°) +
cos (-63.4°) d10 - 6 = 335(10 - 6)
2
2
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Page 742
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10–5. The state of strain at the point on the arm
has components Px = 250110-62, Py = -450110-62, gxy =
-825110-62. Use the strain-transformation equations to
determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and average normal strain.
In each case specify the orientation of the element and show
how the strains deform the element within the x–y plane.
ex = 250(10 - 6)
ey = -450(10 - 6)
y
gxy = -825(10 - 6)
x
a)
ex + ey
e1, 2 =
;
2
= c
A
ex - ey
a
2
2
b + a
gxy
2
b
2
250 - 450
250 - ( -450) 2
-825 2
-6
;
a
b + a
b d(10 )
2
A
2
2
e1 = 441(10 - 6)
Ans.
e2 = -641(10 - 6)
Ans.
Orientation of e1 and e2 :
gxy
tan 2up =
ex - ey
up = -24.84°
-825
250 - ( -450)
=
up = 65.16°
and
Use Eq. 10–5 to determine the direction of e1 and e2:
ex¿ =
ex + ey
ex - ey
+
2
2
cos 2u +
gxy
2
sin 2u
u = up = -24.84°
ex¿ = c
250 - (-450)
250 - 450
-825
+
cos (-49.69°) +
sin (-49.69°) d(10 - 6) = 441(10 - 6)
2
2
2
Therefore, up1 = -24.8°
Ans.
up2 = 65.2°
Ans.
b)
g
max
in-plane
2
=
A
a
ex - ey
max = 2c
a
in-plane
A
g
eavg =
ex + ey
2
2
b + a
2
gxy
2
b
2
250 - (-450) 2
-825 2
-6
-3
b + a
b d(10 ) = 1.08(10 )
2
2
Ans.
= a
Ans.
250 - 450
b (10 - 6) = -100(10 - 6)
2
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–6. The state of strain at the point has components of
Px = -100110-62, Py = 400110-62, and gxy = -300110-62.
Use the strain-transformation equations to determine the
equivalent in-plane strains on an element oriented at an
angle of 60° counterclockwise from the original position.
Sketch the deformed element due to these strains within
the x–y plane.
y
x
In accordance to the established sign convention,
ex = -100(10 - 6)
ex¿ =
ex + ey
= c
ex - ey
+
2
ey = 400(10 - 6)
2
gxy
cos 2u +
2
gxy = -300(10 - 6)
u = 60°
sin 2u
-100 - 400
-300
-100 + 400
+
cos 120° +
sin 120° d(10 - 6)
2
2
2
= 145(10 - 6)
gx¿y¿
2
= -a
Ans.
ex - ey
2
b sin 2u +
gxy
2
cos 2u
gx¿y¿ = c -(-100 - 400) sin 120° + (-300) cos 120° d(10 - 6)
= 583(10 - 6)
ey¿ =
ex + ey
= c
2
Ans.
ex - ey
-
2
cos 2u -
gxy
2
sin 2u
-100 - 400
-300
-100 + 400
cos 120° sin 120° d(10 - 6)
2
2
2
= 155 (10 - 6)
Ans.
The deformed element of this equivalent state of strain is shown in Fig. a
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Page 744
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–7. The state of strain at the point has components of
Px = 100110-62, Py = 300110-62, and gxy = -150110-62.
Use the strain-transformation equations to determine the
equivalent in-plane strains on an element oriented u = 30°
clockwise. Sketch the deformed element due to these
strains within the x–y plane.
y
x
In accordance to the established sign convention,
ex = 100(10 - 6)
ex¿ =
ex + ey
ex - ey
+
2
= c
ey = 300(10 - 6)
2
cos 2u +
gxy = -150(10 - 6)
gxy
2
u = -30°
sin 2u
100 - 300
-150
100 + 300
+
cos (-60°) +
sin ( -60°) d (10 - 6)
2
2
2
= 215(10 - 6)
gx¿y¿
2
= -a
Ans.
ex - ey
2
b sin 2u +
gxy
2
cos 2u
gx¿y¿ = c -(100 - 300) sin ( -60°) + ( -150) cos ( -60°) d(10 - 6)
= -248 (10 - 6)
ey¿ =
ex + ey
= c
2
Ans.
ex - ey
-
2
cos 2u -
gxy
2
sin 2u
100 - 300
-150
100 + 300
cos ( -60°) sin (-60°) d (10 - 6)
2
2
2
= 185(10 - 6)
Ans.
The deformed element of this equivalent state of strain is shown in Fig. a
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Page 745
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*10–8. The state of strain at the point on the bracket
has components Px = -200110-62, Py = -650110-62, gxy ⫽
-175110-62. Use the strain-transformation equations to
determine the equivalent in-plane strains on an element
oriented at an angle of u = 20° counterclockwise from the
original position. Sketch the deformed element due to these
strains within the x–y plane.
ex = -200(10 - 6)
ex¿ =
ex + ey
ex - ey
+
2
= c
ey = -650(10 - 6)
2
cos 2u +
gxy
2
y
x
gxy = -175(10 - 6)
u = 20°
sin 2u
( -200) - (-650)
(-175)
-200 + (-650)
+
cos (40°) +
sin (40°) d(10 - 6)
2
2
2
= -309(10 - 6)
ey¿ =
ex + ey
ex - ey
-
2
= c
Ans.
2
cos 2u -
gxy
2
sin 2u
-200 - ( -650)
( -175)
-200 + (-650)
cos (40°) sin (40°) d(10 - 6)
2
2
2
= -541(10 - 6)
gx¿y¿
2
ex - ey
= -
2
Ans.
sin 2u +
gxy
2
cos 2u
gx¿y¿ = [-(-200 - (-650)) sin (40°) + (-175) cos (40°)](10 - 6)
= -423(10 - 6)
Ans.
745
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Page 746
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–9. The state of strain at the point has components of
Px = 180110-62, Py = -120110-62, and gxy = -100110-62.
Use the strain-transformation equations to determine (a)
the in-plane principal strains and (b) the maximum in-plane
shear strain and average normal strain. In each case specify
the orientation of the element and show how the strains
deform the element within the x–y plane.
y
x
a)
In
accordance
to
the
established
sign
convention,
ex = 180(10 - 6),
ey = -120(10 - 6) and gxy = -100(10 - 6).
ex + ey
e1, 2 =
;
2
= b
a
A
ex - ey
2
2
b + a
gxy
2
b
2
180 + (-120)
180 - ( -120) 2
-100 2
-6
;
c
d + a
b r (10 )
2
A
2
2
= A 30 ; 158.11 B (10 - 6)
e1 = 188(10 - 6)
tan 2uP =
e2 = -128(10 - 6)
gxy
Ans.
-100(10 - 6)
ex - ey
C 180 - (-120) D (10 - 6)
=
uP = -9.217°
and
= -0.3333
80.78°
Substitute u = -9.217°,
ex + ey
ex¿ =
2
= c
ex - ey
+
2
cos 2u +
gxy
2
sin 2u
180 + ( -120)
180 - ( -120)
-100
+
cos (-18.43°) +
sin (-18.43) d(10 - 6)
2
2
2
= 188(10 - 6) = e1
Thus,
(uP)1 = -9.22°
(uP)2 = 80.8°
Ans.
The deformed element is shown in Fig (a).
gmax
ex - ey 2
gxy 2
in-plane
=
b)
a
b + a
b
2
A
2
2
180 - (-120) 2
-100 2
-6
-6
d + a
b r (10 ) = 316 A 10 B
A
2
2
gmax
= b2
tan 2us = - a
ex - ey
in-plane
gxy
c
b = -c
C 180 - (-120) D (10 - 6)
us = 35.78° = 35.8° and
-100(10 - 6)
Ans.
s = 3
Ans.
-54.22° = -54.2°
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Page 747
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–9.
Continued
gmax
The algebraic sign for in-plane
when u = 35.78°.
ex - ey
gxy
gx¿y¿
= -a
b sin 2u +
cos 2u
2
2
2
gx¿y¿ = e - C 180 - ( -120) D sin 71.56° + ( -100) cos 71.56° f(10 - 6)
= -316(10 - 6)
ex + ey
180 + (-120)
eavg =
= c
d(10 - 6) = 30(10 - 6)
2
2
Ans.
The deformed element for the state of maximum In-plane shear strain is shown is
shown in Fig. b
747
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Page 748
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–10. The state of strain at the point on the bracket
has components Px = 400110-62, Py = -250110-62, gxy ⫽
310110-62. Use the strain-transformation equations to
determine the equivalent in-plane strains on an element
oriented at an angle of u = 30° clockwise from the original
position. Sketch the deformed element due to these strains
within the x–y plane.
ex = 400(10 - 6)
ex¿ =
ex + ey
= c
ex - ey
+
2
ey = -250(10 - 6)
2
cos 2u +
gxy
2
gxy = 310(10 - 6)
y
x
u = -30°
sin 2u
400 - ( -250)
400 + ( -250)
310
+
cos (-60°) + a
b sin (-60°) d(10 - 6)
2
2
2
= 103(10 - 6)
ey¿ =
ex + ey
= c
ex - ey
-
2
Ans.
2
cos 2u -
gxy
2
sin 2u
400 - (-250)
400 + (-250)
310
cos (60°) sin (-60°) d(10 - 6)
2
2
2
= 46.7(10 - 6)
gx¿y¿
2
ex - ey
= -
2
Ans.
sin 2u +
gxy
2
cos 2u
gx¿y¿ = [-(400 - (-250)) sin (-60°) + 310 cos ( -60°)](10 - 6) = 718(10 - 6)
748
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Page 749
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–11. The state of strain at the point has components of
Px = -100110-62, Py = -200110-62, and gxy = 100110-62.
Use the strain-transformation equations to determine (a)
the in-plane principal strains and (b) the maximum in-plane
shear strain and average normal strain. In each case specify
the orientation of the element and show how the strains
deform the element within the x–y plane.
In accordance to the established
ey = -200(10 - 6) and gxy = 100(10 - 6).
ex + ey
e1, 2 =
;
2
= b
A
a
ex - ey
2
b + a
2
gxy
2
b
sign
y
x
convention,
ex = -100(10 - 6),
2
-100 + (-200)
100 2
-100 - (-200) 2
-6
;
c
d + a
b r (10 )
2
A
2
2
= A -150 ; 70.71 B (10 - 6)
e1 = -79.3(10 - 6)
tan 2uP =
e2 = -221(10 - 6)
gxy
100(10 - 6)
C -100 - (-200) D (10 - 6)
=
ex - ey
uP = 22.5°
and
Ans.
= 1
-67.5°
Substitute u = 22.5,
ex + ey
ex¿ =
ex - ey
cos 2u +
gxy
sin 2u
2
2
2
-100 + (-200)
-100 - (-200)
100
+
cos 45° +
sin 45° d(10 - 6)
= c
2
2
2
+
= -79.3(10 - 6) = e1
Thus,
(uP)1 = 22.5°
(uP)2 = -67.5°
Ans.
The deformed element of the state of principal strain is shown in Fig. a
gmax
ex - ey 2
gxy 2
in-plane
=
a
b + a
b
2
A
2
2
gmax
in-plane
= b2
tan 2us = - a
c
-100 - (-200) 2
100 2
-6
-6
d + a
b r (10 ) = 141(10 )
A
2
2
ex - ey
gxy
b = -c
us = -22.5°
and
The algebraic sign for
gx¿y¿
2
= -a
ex - ey
2
gmax
in-plane
b sin 2u +
C -100 - ( -200) D (10 - 6)
100(10 - 6)
Ans.
s = -1
Ans.
67.5°
when u = -22.5°.
gxy
2
cos 2u
gx¿y¿ = - C -100 - (-200) D sin ( -45°) + 100 cos (-45°)
= 141(10 - 6)
ex + ey
-100 + ( -200)
eavg =
= c
d(10 - 6) = -150(10 - 6)
2
2
Ans.
The deformed element for the state of maximum In-plane shear strain is shown in
Fig. b.
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Page 750
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–11.
Continued
*10–12. The state of plane strain on an element is given by
Px = 500110-62, Py = 300110-62, and gxy = -200110-62.
Determine the equivalent state of strain on an element at
the same point oriented 45° clockwise with respect to the
original element.
y
Pydy
dy gxy
2
Strain Transformation Equations:
ex = 500 A 10 - 6 B
ey = 300 A 10 - 6 B
gxy = -200 A 10 - 6 B
u = -45°
We obtain
ex¿ =
ex + ey
+
2
= c
ex - ey
2
cos 2u +
gxy
2
sin 2u
500 - 300
-200
500 + 300
+
cos (-90°) + a
b sin (-90°) d A 10 - 6 B
2
2
2
= 500 A 10 - 6 B
gx¿y¿
2
= -a
Ans.
ex - ey
2
b sin 2u +
gxy
2
cos 2u
gx¿y¿ = [-(500 - 300) sin ( -90°) + (-200) cos ( -90°)] A 10 - 6 B
= 200 A 10 - 6 B
ey¿ =
ex + ey
= c
2
Ans.
ex - ey
-
2
cos 2u -
gxy
2
sin 2u
500 + 300
500 - 300
-200
cos ( -90°) - a
b sin (-90°) d A 10 - 6 B
2
2
2
= 300 A 10 - 6 B
Ans.
The deformed element for this state of strain is shown in Fig. a.
750
gxy
2
dx
x
Pxdx
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Page 751
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–13. The state of plane strain on an element is
Px = -300110-62, Py = 0, and gxy = 150110-62. Determine
the equivalent state of strain which represents (a) the
principal strains, and (b) the maximum in-plane shear strain
and the associated average normal strain. Specify the
orientation of the corresponding elements for these states
of strain with respect to the original element.
y
gxy
dy 2
x
In-Plane Principal Strains: ex = -300 A 10 - 6 B , ey = 0, and gxy = 150 A 10 - 6 B . We
obtain
ex + ey
e1, 2 =
;
2
= C
C
¢
ex - ey
2
2
≤ + ¢
gxy
2
≤
2
-300 + 0
-300 - 0 2
150 2
;
¢
≤ + ¢
≤ S A 10 - 6 B
2
C
2
2
= ( -150 ; 167.71) A 10 - 6 B
e1 = 17.7 A 10 - 6 B
e2 = -318 A 10 - 6 B
Ans.
Orientation of Principal Strain:
tan 2up =
gxy
ex - ey
=
150 A 10 - 6 B
(-300 - 0) A 10 - 6 B
= -0.5
uP = -13.28° and 76.72°
Substituting u = -13.28° into Eq. 9-1,
ex¿ =
ex + ey
= c
ex - ey
+
2
2
cos 2u +
gxy
2
sin 2u
-300 + 0
-300 - 0
150
+
cos (-26.57°) +
sin (-26.57°) d A 10 - 6 B
2
2
2
= -318 A 10 - 6 B = e2
Thus,
A uP B 1 = 76.7° and A uP B 2 = -13.3°
Ans.
The deformed element of this state of strain is shown in Fig. a.
Maximum In-Plane Shear Strain:
gmax
ex - ey 2
gxy 2
in-plane
=
¢
≤ + ¢ ≤
2
C
2
2
gmax
in-plane
-300 - 0 2
150 2
-6
-6
b + a
b R A 10 B = 335 A 10 B
A
2
2
= B2
a
Ans.
Orientation of the Maximum In-Plane Shear Strain:
tan 2us = - ¢
ex - ey
gxy
≤ = -C
(-300 - 0) A 10 - 6 B
150 A 10 - 6 B
S = 2
us = 31.7° and 122°
Ans.
751
gxy
2
dx
Pxdx
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–13.
Continued
The algebraic sign for
gx¿y¿
2
= -¢
ex - ey
2
gmax
in-plane
≤ sin 2u +
when u = us = 31.7° can be obtained using
gxy
2
cos 2u
gx¿y¿ = [-(-300 - 0) sin 63.43° + 150 cos 63.43°] A 10 - 6 B
= 335 A 10 - 6 B
Average Normal Strain:
eavg =
ex + ey
2
= a
-300 + 0
b A 10 - 6 B = -150 A 10 - 6 B
2
Ans.
The deformed element for this state of strain is shown in Fig. b.
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–14. The state of strain at the point on a boom of an
hydraulic engine crane has components of Px = 250110-62,
Py = 300110-62, and gxy = -180110-62. Use the straintransformation equations to determine (a) the in-plane
principal strains and (b) the maximum in-plane shear strain
and average normal strain. In each case, specify the
orientation of the element and show how the strains deform
the element within the x–y plane.
y
a)
In-Plane Principal Strain: Applying Eq. 10–9,
ex + ey
e1, 2 =
;
2
= B
a
A
ex - ey
2
b + a
2
gxy
2
b
2
250 - 300 2
250 + 300
-180 2
-6
;
a
b + a
b R A 10 B
2
A
2
2
= 275 ; 93.41
e1 = 368 A 10 - 6 B
e2 = 182 A 10 - 6 B
Ans.
Orientation of Principal Strain: Applying Eq. 10–8,
gxy
tan 2uP =
-180(10 - 6)
ex - ey
=
(250 - 300)(10 - 6)
uP = 37.24°
and
= 3.600
-52.76°
Use Eq. 10–5 to determine which principal strain deforms the element in the x¿
direction with u = 37.24°.
ex¿ =
ex + ey
= c
2
ex - ey
+
2
cos 2u +
gxy
2
sin 2u
250 + 300
250 - 300
-180
+
cos 74.48° +
sin 74.48° d A 10 - 6 B
2
2
2
= 182 A 10 - 6 B = e2
Hence,
uP1 = -52.8°
and
uP2 = 37.2°
Ans.
b)
Maximum In-Plane Shear Strain: Applying Eq. 10–11,
g max
ex - ey 2
gxy 2
in-plane
=
a
b + a
b
2
A
2
2
g
max = 2 B
a
in-plane
A
-180 2
250 - 300 2
-6
b + a
b R A 10 B
2
2
= 187 A 10 - 6 B
Ans.
753
x
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10–14.
Continued
Orientation of the Maximum In-Plane Shear Strain: Applying Eq. 10–10,
tan 2us = -
ex - ey
us = -7.76°
2
ex - ey
= -
2
250 - 300
= -0.2778
-180
and
82.2°
g max
can be determined by substituting u = -7.76° into Eq. 10–6.
The proper sign of
gx¿y¿
= -
gxy
in-plane
sin 2u +
gxy
2
Ans.
cos 2u
gx¿y¿ = {-[250 - 300] sin (-15.52°) + (-180) cos (-15.52°)} A 10 - 6 B
= -187 A 10 - 6 B
Normal Strain and Shear strain: In accordance with the sign convention,
ex = 250 A 10 - 6 B
ey = 300 A 10 - 6 B
gxy = -180 A 10 - 6 B
Average Normal Strain: Applying Eq. 10–12,
eavg =
ex + ey
2
= c
250 + 300
d A 10 - 6 B = 275 A 10 - 6 B
2
Ans.
754
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*10–16. The state of strain at a point on a support
has components of Px = 350110-62, Py = 400110-62,
gxy = -675110-62. Use the strain-transformation equations
to determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and average normal strain.
In each case specify the orientation of the element and show
how the strains deform the element within the x–y plane.
a)
e1, 2 =
=
ex + ey
;
2
B
a
ex -ey
2
b + a
2
gxy
2
b
2
350 - 400 2
-675 2
350 + 400
;
a
b + a
b
2
A
2
2
e1 = 713(10 - 6)
Ans.
e2 = 36.6(10 - 6)
Ans.
tan 2uP =
gxy
ex - ey
=
-675
(350 - 400)
uP = 42.9°
Ans.
b)
(gx¿y¿)max
=
2
(gx¿y¿)max
=
2
A
a
ex - ey
2
b + a
2
gxy
2
b
2
a
350 - 400 2
-675 2
b + a
b
A
2
2
(gx¿y¿)max = 677(10 - 6)
eavg =
ex + ey
tan 2us =
2
=
Ans.
350 + 400
= 375(10 - 6)
2
-(ex - ey)
gxy
=
Ans.
350 - 400
675
us = -2.12°
Ans.
755
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•10–17.
Solve part (a) of Prob. 10–4 using Mohr’s circle.
ex = 120(10 - 6)
ey = -180(10 - 6)
gxy = 150(10 - 6)
A (120, 75)(10 - 6) C (-30, 0)(10 - 6)
R = C 2[120 - (-30)]2 + (75)2 D (10 - 6)
= 167.71 (10 - 6)
e1 = (-30 + 167.71)(10 - 6) = 138(10 - 6)
Ans.
e2 = (-30 - 167.71)(10 - 6) = -198(10 - 6)
Ans.
75
tan 2uP = a
b , uP = 13.3°
30 + 120
Ans.
756
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10–18.
Solve part (b) of Prob. 10–4 using Mohr’s circle.
ex = 120(10 - 6)
ey = -180(10 - 6)
gxy = 150(10 - 6)
A (120, 75)(10 - 6) C (-30, 0)(10 - 6)
R = C 2[120 - (-30)]2 + (75)2 D (10 - 6)
= 167.71 (10 - 6)
gxy
max
2 in-plane
gxy
= R = 167.7(10 - 6)
max
in-plane
= 335(10 - 6)
Ans.
eavg = -30 (10 - 6)
tan 2us =
120 + 30
75
Ans.
us = -31.7°
Ans.
757
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10–19.
Solve Prob. 10–8 using Mohr’s circle.
ex = -200(10 - 6)
ey = -650(10 - 6)
gxy = -175(10 - 6)
gxy
2
= -87.5(10 - 6)
u = 20°, 2u = 40°
A(-200, -87.5)(10 - 6)
C(-425, 0)(10 - 6)
R = [2(-200 - (-425))2 + 87.52 ](10 - 6) = 241.41(10 - 6)
tan a =
87.5
;
-200 - (-425)
a = 21.25°
f = 40 + 21.25 = 61.25°
ex¿ = (-425 + 241.41 cos 61.25°)(10 - 6) = -309(10 - 6)
Ans.
ey¿ = (-425 - 241.41 cos 61.25°)(10 - 6) = -541(10 - 6)
Ans.
-gx¿y¿
2
= 241.41(10 - 6) sin 61.25°
gx¿y¿ = -423(10 - 6)
Ans.
758
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*10–20.
Solve Prob. 10–10 using Mohr’s circle.
ex = 400(10 - 6)
ey = -250(10 - 6)
A(400, 155)(10 - 6)
C(75, 0)(10 - 6)
gxy = 310(10 - 6)
gxy
2
= 155(10 - 6)
R = [2(400 - 75)2 + 1552 ](10 - 6) = 360.1(10 - 6)
tan a =
155
;
400 - 75
a = 25.50°
f = 60 + 25.50 = 85.5°
ex¿ = (75 + 360.1 cos 85.5°)(10 - 6) = 103(10 - 6)
Ans.
ey¿ = (75 - 360.1 cos 85.5°)(10 - 6) = 46.7(10 - 6)
Ans.
gx¿y¿
2
= (360.1 sin 85.5°)(10 - 6)
gx¿y¿ = 718(10 - 6)
Ans.
759
u = 30°
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•10–21.
Solve Prob. 10–14 using Mohr’s circle.
Construction of the Circle: In accordance with the sign convention, ex = 250 A 10 - 6 B ,
gxy
ey = 300 A 10 - 6 B , and
= -90 A 10 - 6 B . Hence,
2
eavg =
ex + ey
2
= a
250 + 300
b A 10 - 6 B = 275 A 10 - 6 B
2
Ans.
The coordinates for reference points A and C are
A(250, -90) A 10 - 6 B
C(275, 0) A 10 - 6 B
The radius of the circle is
R = a 2(275 - 250)2 + 902 b A 10 - 6 B = 93.408
In-Plane Principal Strain: The coordinates of points B and D represent e1 and e2,
respectively.
e1 = (275 + 93.408) A 10 - 6 B = 368 A 10 - 6 B
Ans.
e2 = (275 - 93.408) A 10 - 6 B = 182 A 10 - 6 B
Ans.
Orientation of Principal Strain: From the circle,
tan 2uP2 =
90
= 3.600
275 - 250
2uP2 = 74.48°
2uP1 = 180° - 2uP2
uP1 =
180° - 74.78°
= 52.8° (Clockwise)
2
Ans.
Maximum In-Plane Shear Strain: Represented by the coordinates of point E on
the circle.
g max
in-plane
2
= -R = -93.408 A 10 - 6 B
g max
in-plane =
-187 A 10 - 6 B
Ans.
Orientation of the Maximum In-Plane Shear Strain: From the circle,
tan 2us =
275 - 250
= 0.2778
90
us = 7.76° (Clockwise)
Ans.
760
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10–22. The strain at point A on the bracket has
components Px = 300110-62, Py = 550110-62, gxy =
-650110-62. Determine (a) the principal strains at A in the
x– y plane, (b) the maximum shear strain in the x–y plane,
and (c) the absolute maximum shear strain.
ex = 300(10 - 6)
ey = 550(10 - 6)
A(300, -325)10 - 6
gxy = -650(10 - 6)
y
gxy
2
= -325(10 - 6)
A
C(425, 0)10 - 6
R = C 2(425 - 300)2 + (-325)2 D 10 - 6 = 348.2(10 - 6)
a)
e1 = (425 + 348.2)(10 - 6) = 773(10 - 6)
Ans.
e2 = (425 - 348.2)(10 - 6) = 76.8(10 - 6)
Ans.
b)
g
-6
-6
max
in-plane = 2R = 2(348.2)(10 ) = 696(10 )
Ans.
c)
gabs
max
=
2
773(10 - 6)
;
2
gabs
max
= 773(10 - 6)
Ans.
10–23. The strain at point A on the leg of the angle has
components Px = -140110-62, Py = 180110-62, gxy =
-125110-62. Determine (a) the principal strains at A in the
x–y plane, (b) the maximum shear strain in the x–y plane,
and (c) the absolute maximum shear strain.
ex = -140(10 - 6)
A( -140, -62.5)10 - 6
ey = 180(10 - 6)
gxy = -125(10 - 6)
A
gxy
2
= -62.5(10 - 6)
C(20, 0)10 - 6
R = A 2(20 - ( -140))2 + (-62.5)2 B 10 - 6 = 171.77(10 - 6)
a)
e1 = (20 + 171.77)(10 - 6) = 192(10 - 6)
Ans.
e2 = (20 - 171.77)(10 - 6) = -152(10 - 6)
Ans.
(b, c)
gabs
max
gmax
= in-plane
= 2R = 2(171.77)(10 - 6) = 344(10 - 6)
Ans.
761
x
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*10–24. The strain at point A on the pressure-vessel wall
has components Px = 480110-62, Py = 720110-62, gxy =
650110-62. Determine (a) the principal strains at A, in the
x– y plane, (b) the maximum shear strain in the x–y plane,
and (c) the absolute maximum shear strain.
ex = 480(10 - 6)
ey = 720(10 - 6)
A(480, 325)10 - 6
C(600, 0)10 - 6
gxy = 650(10 - 6)
y
A
gxy
2
= 325(10 - 6)
R = (2(600 - 480)2 + 3252 )10 - 6 = 346.44(10 - 6)
a)
e1 = (600 + 346.44)10 - 6 = 946(10 - 6)
Ans.
e2 = (600 - 346.44)10 - 6 = 254(10 - 6)
Ans.
b)
g
-6
max
= 693(10 - 6)
in-plane = 2R = 2(346.44)10
Ans.
c)
gabs
max
2
=
946(10 - 6)
;
2
gabs
max
= 946(10 - 6)
Ans.
762
x
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•10–25.
The 60° strain rosette is mounted on the bracket.
The following readings are obtained for each gauge:
Pa = -100110-62, Pb = 250110-62, and Pc = 150110-62.
Determine (a) the principal strains and (b) the maximum inplane shear strain and associated average normal strain. In
each case show the deformed element due to these strains.
b
c
60⬚
60⬚
This is a 60° strain rosette Thus,
ex = ea = -100(10 - 6)
1
A 2eb + 2ec - ea B
3
ey =
1
C 2(250) + 2(150) - (-100) D (10 - 6)
3
=
= 300(10 - 6)
gxy =
2
23
(eb - ec) =
2
23
(250 - 150)(10 - 6) = 115.47(10 - 6)
In accordance to the established sign convention, ex = -100(10 - 6), ey = 300(10 - 6)
gxy
and
= 57.74(10 - 6).
2
Thus,
eavg =
ex + ey
2
= a
-100 + 300
b(10 - 6) = 100(10 - 6)
2
Ans.
Then, the coordinates of reference point A and Center C of the circle are
A( -100, 57.74)(10 - 6)
C(100, 0)(10 - 6)
Thus, the radius of the circle is
R = CA = a 2(-100 - 100)2 + 208.16b(10 - 6) = 208.17(10 - 6)
Using these result, the circle is shown in Fig. a.
The coordinates of points B and D represent e1 and e2 respectively.
e1 = (100 + 208.17)(10 - 6) = 308(10 - 6)
Ans.
e2 = (100 - 208.17)(10 - 6) = -108(10 - 6)
Ans.
Referring to the geometry of the circle,
tan 2(uP)2 =
57.74(10 - 6)
(100 + 100)(10 - 6)
= 0.2887
A uP B 2 = 8.05° (Clockwise)
Ans.
The deformed element for the state of principal strain is shown in Fig. b.
763
a
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10–25.
Continued
gmax
The coordinates for point E represent eavg and
in-plane
2
. Thus,
gmax
in-plane
2
= R = 208.17(10 - 6)
gmax
in-plane
= 416(10 - 6)
Ans.
Referring to the geometry of the circle,
tan 2us =
100 + 100
57.74
us = 36.9° (Counter Clockwise)
Ans.
The deformed element for the state of maximum In-plane shear strain is shown in
Fig. c.
764
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10–26. The 60° strain rosette is mounted on a beam.
The following readings are obtained for each gauge:
Pa = 200110-62, Pb = - 450110-62, and Pc = 250110-62.
Determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and average normal strain.
In each case show the deformed element due to these strains.
b
a
30⬚ 30⬚
c
With ua = 60°, ub = 120° and uc = 180°,
ea = ex cos2 ua + ey sin2 ua + gxy sin ua cos ua
200(10 - 6) = ex cos2 60° + ey sin2 60° + gxy sin 60° cos 60°
0.25ex + 0.75ey + 0.4330 gxy = 200(10 - 6)
(1)
eb = ex cos2 ub + ey sin2 ub + gxy sin ub cos ub
-450(10 - 6) = ex cos2 120° + ey sin2 120° + gxy sin 120° cos 120°
0.25ex + 0.75ey - 0.4330 gxy = -450(10 - 6)
(2)
ec = ex cos2 uc + ey sin2 uc + gxy sin uc cos uc
250(10 - 6) = ex cos2 180° + ey sin2 180° + gxy sin 180° cos 180°
ex = 250(10 - 6)
Substitute this result into Eqs. (1) and (2) and solve them,
ey = -250 (10 - 6)
gxy = 750.56 (10 - 6)
In accordance to the established sign convention, ex = 250(10 - 6), ey = -250(10 - 6),
gxy
and
= 375.28(10 - 6), Thus,
2
eavg =
ex + ey
2
= c
250 + ( -250)
d(10 - 6) = 0
2
Ans.
Then, the coordinates of the reference point A and center C of the circle are
A(250, 375.28)(10 - 6)
C(0, 0)
Thus, the radius of the circle is
R = CA = A 2(250 - 0)2 + 375.282 B (10 - 6) = 450.92(10 - 6)
Using these results, the circle is shown in Fig. a.
The coordinates for points B and D represent e1 and e2, respectively. Thus,
e1 = 451(10 - 6)
e2 = -451(10 - 6)
Ans.
Referring to the geometry of the circle,
tan 2(uP)1 =
375.28
= 1.5011
250
(uP)1 = 28.2° (Counter Clockwise)
Ans.
The deformed element for the state of principal strains is shown in Fig. b.
765
60⬚
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10–26.
Continued
gmax
in-plane
The coordinates of point E represent eavg and
. Thus,
2
gmax
in-plane
gmax
= 902(10 - 6)
= R = 450.92(10 - 6)
in-plane
2
Ans.
Referring to the geometry of the circle,
tan 2us =
250
= 0.6662
375.28
us = 16.8° (Clockwise)
Ans.
766
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10–27. The 45° strain rosette is mounted on a steel shaft.
The following readings are obtained from each gauge:
Pa = 300110-62, Pb = -250110-62, and Pc = -450110-62.
Determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and average normal strain.
In each case show the deformed element due to these strains.
b
c
45⬚
With ua = 45°, ub = 90° and uc = 135°,
ea = ex cos2 ua + ey sin2 ua + gxy sin ua cos ua
300(10 - 6) = ex cos2 45° + ey sin2 45° + gxy sin 45° cos 45°
ex + ey + gxy = 600(10 - 6)
(1)
eb = ex cos2 ub + ey sin2 ub + gxy sin ub cos ub
-250(10 - 6) = ex cos2 90° + ey sin2 90° + gxy sin 90° cos 90°
ey = -250(10 - 6)
ec = ex cos2 uc + ey sin2 uc + gxy sin uc cos uc
-450(10 - 6) = ex cos2 135° + ey sin2 135° + gxy sin 135° cos 135°
ex + ey - gxy = -900(10 - 6)
(2)
Substitute the result of ey into Eq. (1) and (2) and solve them
ex = 100(10 - 6)
gxy = 750(10 - 6)
In accordance to the established sign convention, ex = 100(10 - 6), ey = -250(10 - 6)
gxy
and
= 375(10 - 6). Thus,
2
eavg =
ex + ey
2
= c
100 + (-250)
d(10 - 6) = -75(10 - 6)
2
Ans.
Then, the coordinates of the reference point A and the center C of the circle are
A(100, 375)(10 - 6)
C(-75, 0)(10 - 6)
Thus, the radius of the circle is
R = CA = a 2 C 100 - (-75) D 2 + 3752 b(10 - 6) = 413.82(10 - 6)
Using these results, the circle is shown in Fig. a.
The Coordinates of points B and D represent e1 and e2, respectively. Thus,
e1 = A -75 + 413.82 B (10 - 6) = 339(10 - 6)
Ans.
e2 = A -75 - 413.82 B (10 - 6) = -489(10 - 6)
Ans.
Referring to the geometry of the circle
tan 2(uP)1 =
375
= 2.1429
100 + 75
(uP)1 = 32.5° (Counter Clockwise)
Ans.
767
45⬚
a
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10–27.
Continued
The deformed element for the state of principal strains is shown in Fig. b.
gmax
The coordinates of point E represent eavg and
gmax
in-plane
2
= R = 413.82(106)
in-plane
gmax
2
in-plane
. Thus
= 828(10 - 6)
Ans.
Referring to the geometry of the circle
tan 2us =
-100 + 75
= 0.4667
375
us = 12.5° (Clockwise)
Ans.
768
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*10–28. The 45° strain rosette is mounted on the link of
the backhoe. The following readings are obtained from
each gauge: Pa = 650110-62, Pb = -300110-62, Pc = 480110-62.
Determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and associated average
normal strain.
a
45⬚
b
ea = 650(10 - 6);
ua = 180°;
eb = -300(10 - 6);
ub = 225°
c
uc = 270°
Applying Eq. 10–16, e = ex cos2 u + ey sin2 u + gxy sin u cos u
650(10 - 6) = ex cos2 (180°) + ey sin2 (180°) + gxy sin (180°) cos (180°)
ex = 650 (10 - 6)
480 (10 - 6) = ex cos2 (270°) + ey sin2 (270°) + gxy sin (270°) cos (270°)
ey = 480 (10 - 6)
-300 (10 - 6) = 650 (10 - 6) cos2 (225°) + 480 (10 - 6) sin2 (225°) + gxy sin (225°) cos (225°)
gxy = -1730 (10 - 6)
Therefore, ex = 650 (10 - 6)
gxy
2
ey = 480 (10 - 6)
gxy = -1730 (10 - 6)
= -865 (10 - 6)
Mohr’s circle:
A(650, -865) 10 - 6
C(565, 0) 10 - 6
R = CA = C 2(650 - 565)2 + 8652 D 10 - 6 = 869.17 (10 - 6)
(a)
(b)
e1 = [565 + 869.17]10 - 6 = 1434 (10 - 6)
Ans.
e2 = [565 - 869.17]10 - 6 = -304 (10 - 6)
Ans.
gmax
Ans.
in-plane
45⬚
ec = 480(10 - 6)
= 2 R = 2(869.17) (10 - 6) = 1738 (10 - 6)
eavg = 565(10 - 6)
Ans.
769
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10–30. For the case of plane stress, show that Hooke’s law
can be written as
sx =
E
E
1Px + nPy2, sy =
1Py + nPx2
11 - n22
11 - n22
Generalized Hooke’s Law: For plane stress, sz = 0. Applying Eq. 10–18,
ex =
1
A s - v sy B
E x
vEex = A sx - v sy B v
vEex = v sx - v2 sy
ey =
[1]
1
(s - v sx)
E y
E ey = -v sx + sy
[2]
Adding Eq [1] and Eq.[2] yields.
vE ex - E ey = sy - v2 sy
sy =
E
A vex + ey B
1 - v2
(Q.E.D.)
Substituting sy into Eq. [2]
E ey = -vsx +
sx =
E
A v ex + ey B
1 - v2
E A v ex + ey B
v (1 - v2)
Eey
-
v
E v ex + E ey - E ey + Eey v2
=
=
v(1 - v2)
E
(ex + v ey)
1 - v2
(Q.E.D.)
770
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10–31. Use Hooke’s law, Eq. 10–18, to develop the straintransformation equations, Eqs. 10–5 and 10–6, from the
stress-transformation equations, Eqs. 9–1 and 9–2.
Stress transformation equations:
sx + sy
sx¿ =
2
tx¿y¿ =
sy¿ =
sx - sy
+
sx - sy
2
(1)
sin 2u + txy cos 2u
2
sx + sy
sx - sy
-
2
cos 2u + txy sin 2u
2
(2)
cos 2u - txy sin 2u
(3)
Hooke’s Law:
ex =
v sy
sx
E
E
(4)
ey =
sy
-v sx
+
E
E
(5)
txy = G gxy
G =
(6)
E
2 (1 + v)
(7)
From Eqs. (4) and (5)
ex + ey =
ex - ey =
(1 - v)(sx + sy)
(8)
E
(1 + v)(sx - sy)
(9)
E
From Eqs. (6) and (7)
txy =
E
g
2 (1 + v) xy
(10)
From Eq. (4)
ex¿ =
v sy¿
sx¿
E
E
(11)
Substitute Eqs. (1) and (3) into Eq. (11)
ex¿ =
(1 - v)(sx - sy)
(1 + v)(sx - sy)
+
2E
2E
cos 2u +
(1 + v)txy sin 2u
E
(12)
By using Eqs. (8), (9) and (10) and substitute into Eq. (12),
ex¿ =
ex + ey
2
ex - ey
+
2
cos 2u +
gxy
2
QED
sin 2u
771
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10–31.
Continued
From Eq. (6).
gx¿y¿ = G gx¿y¿ =
E
g
2 (1 + v) x¿y¿
(13)
Substitute Eqs. (13), (6) and (9) into Eq. (2),
E (ex - ey)
E
E
gx¿y¿ = sin 2u +
g cos 2u
2 (1 + v)
2 (1 + v)
2 (1 + v) xy
gx¿y¿
(ex - ey)
= -
2
2
sin 2u +
gxy
2
QED
cos 2u
*10–32. A bar of copper alloy is loaded in a tension
machine and it is determined that Px = 940110-62 and
sx = 14 ksi, sy = 0, sz = 0. Determine the modulus of
elasticity, Ecu, and the dilatation, ecu, of the copper.
ncu = 0.35.
ex =
1
[s - v(sy + sz)]
E x
940(10 - 6) =
ecu =
1
[14(103) - 0.35(0 + 0)]
Ecu
Ecu = 14.9(103) ksi
Ans.
1 - 2(0.35)
1 - 2v
(14 + 0 + 0) = 0.282(10 - 3)
(sx + sy + sz) =
E
14.9(103)
Ans.
•10–33.
The principal strains at a point on the aluminum
fuselage of a jet aircraft are P1 = 780110-62 and P2 =
400110-62. Determine the associated principal stresses at
the point in the same plane. Eal = 1011032 ksi, nal = 0.33.
Hint: See Prob. 10–30.
Plane stress, s3 = 0
See Prob 10-30,
s1 =
E
(e1 + ve2)
1 - v2
10(103)
=
s2 =
1 - 0.332
Ans.
E
(e2 + ve1)
1 - v2
10(103)
=
(780(10 - 6) + 0.33(400)(10 - 6)) = 10.2 ksi
1 - 0.332
(400(10 - 6) + 0.33(780)(10 - 6)) = 7.38 ksi
Ans.
772
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10–34. The rod is made of aluminum 2014-T6. If it is
subjected to the tensile load of 700 N and has a diameter of
20 mm, determine the absolute maximum shear strain in the
rod at a point on its surface.
700 N
Normal Stress: For uniaxial loading, sy = sz = 0.
sx =
700
P
= 2.228 MPa
= p
2
A
4 (0.02 )
Normal Strain: Applying the generalized Hooke’s Law.
ex =
=
1
C s - v A sy + sz B D
E x
1
C 2.228 A 106 B - 0 D
73.1(109)
= 30.48 A 10 - 6 B
ey =
=
1
C s - v(sx + sz) D
E y
1
C 0 - 0.35 A 2.228 A 106 B + 0 B D
73.1(109)
= -10.67 A 10 - 6 B
ez =
=
1
C s - v A sx + sy B D
E z
1
C 0 - 0.35 A 2.228 A 106 B + 0 B D
73.1(109)
= -10.67 A 10 - 6 B
Therefore.
emax = 30.48 A 10 - 6 B
emin = -10.67 A 10 - 6 B
Absolute Maximum Shear Strain:
gabs
max = emax - emin
= [30.48 - (-10.67)] A 10 - 6 B = 41.1 A 10 - 6 B
Ans.
773
700 N
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10–35. The rod is made of aluminum 2014-T6. If it is
subjected to the tensile load of 700 N and has a diameter of
20 mm, determine the principal strains at a point on the
surface of the rod.
700 N
Normal Stress: For uniaxial loading, sy = sz = 0.
sx =
P
700
= 2.228 MPa
= p
2
A
4 (0.02 )
Normal Strains: Applying the generalized Hooke’s Law.
ex =
=
1
C s - v A sy + sz B D
E x
1
C 2.228 A 106 B - 0 D
73.1(109)
= 30.48 A 10 - 6 B
ey =
=
1
C s - v(sx + sz) D
E y
1
C 0 - 0.35 A 2.228 A 106 B + 0 B D
73.1(109)
= -10.67 A 10 - 6 B
ez =
=
1
C s - v A sx + sy B D
E z
1
C 0 - 0.35 A 2.228 A 106 B + 0 B D
73.1(109)
= -10.67 A 10 - 6 B
Principal Strains: From the results obtained above,
emax = 30.5 A 10 - 6 B
eint = emin = -10.7 A 10 - 6 B
Ans.
774
700 N
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*10–36. The steel shaft has a radius of 15 mm. Determine
the torque T in the shaft if the two strain gauges, attached to
the surface of the shaft, report strains of Px¿ = -80110-62
and Py¿ = 80110-62. Also, compute the strains acting in the x
and y directions. Est = 200 GPa, nst = 0.3.
ex¿ = -80(10 - 6)
y
T
ex = ey = 0
Ans.
ex¿ = ex cos2 u + ey sin2 u + gxy sin u cos u
u = 45°
-80(10 - 6) = 0 + 0 + gxy sin 45° cos 45°
gxy = -160(10 - 6)
Ans.
Also, u = 135°
80(10 - 6) = 0 + 0 + g sin 135° cos 135°
gxy = -160(10 - 6)
200(109)
E
=
= 76.923(109)
2(1 + V)
2(1 + 0.3)
t = Gg = 76.923(109)(160)(10 - 6) = 12.308(106) Pa
12.308(106) A
p
B (0.015)4
2
= 65.2 N # m
0.015
Ans.
10–37. Determine the bulk modulus for each of the
following materials: (a) rubber, Er = 0.4 ksi, nr = 0.48, and
(b) glass, Eg = 811032 ksi, ng = 0.24.
a) For rubber:
Kr =
Er
0.4
=
= 3.33 ksi
3 (1 - 2 vr)
3[1 - 2(0.48)]
Ans.
b) For glass:
Kg =
Eg
3 (1 - 2 vg)
=
x
T
Pure shear
tJ
T =
=
c
x¿
45⬚
ey¿ = 80(10 - 6)
G =
y¿
8(103)
= 5.13 (103) ksi
3[1 - 2(0.24)]
Ans.
775
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10–38. The principal stresses at a point are shown in the
figure. If the material is A-36 steel, determine the
principal strains.
12 ksi
e1 =
1
1
e 12 - 0.32 C 8 + (-20) D f = 546 (10-6)
C s - v(s2 + s3) D =
E 1
29.0(103)
e2 =
1
1
e 8 - 0.32 C 12 + (-20) D f = 364 (10-6)
C s - v(s1 + s3) D =
E 2
29.0(103)
e3 =
20 ksi
8 ksi
1
1
C s3 - v(s1 + s2) D =
C -20 - 0.32(12 + 8) D = -910 (10-6)
E
29.0(103)
emax = 546 (10 - 6)
eint = 346 (10 - 6)
emin = -910 (10 - 6)
Ans.
10–39. The spherical pressure vessel has an inner
diameter of 2 m and a thickness of 10 mm. A strain gauge
having a length of 20 mm is attached to it, and it is observed
to increase in length by 0.012 mm when the vessel
is pressurized. Determine the pressure causing this
deformation, and find the maximum in-plane shear stress,
and the absolute maximum shear stress at a point on the
outer surface of the vessel. The material is steel, for which
Est = 200 GPa and nst = 0.3.
20 mm
1000
r
=
= 100 7 10, the thin wall analysis is valid to
t
10
determine the normal stress in the wall of the spherical vessel. This is a plane stress
Normal Stresses: Since
problem where smin = 0 since there is no load acting on the outer surface of the wall.
smax = slat =
pr
p(1000)
=
= 50.0p
2t
2(10)
[1]
Normal Strains: Applying the generalized Hooke’s Law with
emax = elat =
0.012
= 0.600 A 10 - 3 B mm>mm
20
emax =
1
C s - V (slat + smin) D
E max
0.600 A 10 - 3 B =
1
[50.0p - 0.3 (50.0p + 0)]
200(104)
p = 3.4286 MPa = 3.43 MPa
Ans.
From Eq.[1] smax = slat = 50.0(3.4286) = 171.43 MPa
Maximum In-Plane Shear (Sphere’s Surface): Mohr’s circle is simply a dot. As the
result, the state of stress is the same consisting of two normal stresses with zero
shear stress regardless of the orientation of the element.
t max
Ans.
in-plane = 0
Absolute Maximum Shear Stress:
tabs
max
=
smax - smin
171.43 - 0
=
= 85.7MPa
2
2
Ans.
776
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*10–40. The strain in the x direction at point A on the
steel beam is measured and found to be Px = -100110-62.
Determine the applied load P. What is the shear strain gxy
at point A? Est = 2911032 ksi, nst = 0.3.
P
y
3 in.
A
3 ft
1
1
(5.5)(83) = 129.833 in4
(6)(9)3 12
12
Ix =
QA = (4.25)(0.5)(6) + (2.75)(0.5)(2.5) = 16.1875 in3
s = Eex = 29(103)(100)(10 - 6) = 2.90 ksi
My
,
I
s =
2.90 =
1.5P(12)(1.5)
129.833
P = 13.945 = 13.9 kip
tA =
0.5(13.945)(16.1875)
VQ
=
= 1.739 ksi
It
129.833(0.5)
G =
29(103)
E
=
= 11.154(103) ksi
2(1 + v)
2(1 + 0.3)
gxy =
txy
G
=
Ans.
1.739
= 0.156(10 - 3) rad
11.154(103)
Ans.
777
x
4 ft
7 ft
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•10–41.
The cross section of the rectangular beam is
subjected to the bending moment M. Determine an
expression for the increase in length of lines AB and CD.
The material has a modulus of elasticity E and Poisson’s
ratio is n.
C
D
B
h
For line AB,
sz = -
ey = -
A
M
12My
My
My
= 1
= 3
I
b h3
12 b h
v sz
=
E
E b h3
h
2
¢LAB =
=
b
12 v My
L0
h
ey dy =
2
12 v M
y dy
3
E b h L0
3vM
2Ebh
Ans.
For line CD,
sz = -
ex = -
M h2
Mc
6M
= - 1
= - 2
3
I
bh
b
h
12
v sz
E
=
6vM
E b h2
¢LCD = ex LCD =
=
6vM
(b)
E b h2
6vM
E h2
Ans.
10–42. The principal stresses at a point are shown in
the figure. If the material is aluminum for which
Eal = 1011032 ksi and nal = 0.33, determine the principal
strains.
26 ksi
ex =
1
1
(s - v(sy + sz)) =
(10 - 0.33(-15 - 26)) = 2.35(10 - 3)
E x
10(103)
ey =
1
1
(s - v(sx + sz)) =
(-15 - 0.33)(10 - 26)) = -0.972(10 - 3)Ans.
E y
10(103)
ez =
Ans.
1
1
(s - v(sx + sy)) =
(-26 - 0.33(10 - 15)) = -2.44(10 - 3) Ans.
E z
10(103)
778
15 ksi
10 ksi
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10–43. A single strain gauge, placed on the outer surface
and at an angle of 30° to the axis of the pipe, gives a reading
at point A of Pa = -200(10-6). Determine the horizontal
force P if the pipe has an outer diameter of 2 in. and an
inner diameter of 1 in. The pipe is made of A-36 steel.
1.5 ft
Using the method of section and consider the equilibrium of the FBD of the pipe’s
upper segment, Fig. a,
©Fz = 0;
Vz - p = 0
©Mx = 0;
Tx - p(1.5) = 0 Tx = 1.5p
©My = 0;
My - p(2.5) = 0 My = 2.5p
Vz = p
30⬚
A
The normal strees is due to bending only. For point A, z = 0. Thus
sx =
My z
Iy
= 0
The shear stress is the combination of torsional shear stress and transverse shear
stress. Here, J = p2 (14 - 0.54) = 0.46875 p in4. Thus, for point A
tt =
1.5p(12)(1)
38.4 p
Txc
=
=
p
J
0.46875p
Referring to Fig. b,
(QA)z = y1œ A1œ - y2œ A2œ =
4 (1) p 2
4(0.5) p
c (1 ) d c (0.52) d
3p 2
3p 2
= 0.5833 in3
Iy = p4 (14 - 0.54) = 0.234375 p in4
Combine these two shear stress components,
t = tt + tv =
P
2.5 ft
38.4P
2.4889P
40.8889P
+
=
p
p
p
Since no normal stress acting on point A, it is subjected to pure shear which can be
represented by the element shown in Fig. c.
For pure shear, ex = ez = 0,
ea = ex cos3 ua + ez sin2 ua + gxz sin ua cos ua
-200(10 - 6) = 0 + 0 + gxz sin 150° cos 150°
gxz = 461.88(10 - 6)
Applying the Hooke’s Law for shear,
txz = G gxz
40.8889P
= 11.0(103) C 461.88(10 - 6) D
p
P = 0.3904 kip = 390 lb
Ans.
779
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*10–44. A single strain gauge, placed in the vertical plane
on the outer surface and at an angle of 30° to the axis of the
pipe, gives a reading at point A of Pa = -200(10-6).
Determine the principal strains in the pipe at point A. The
pipe has an outer diameter of 2 in. and an inner diameter of
1 in. and is made of A-36 steel.
1.5 ft
P
2.5 ft
Using the method of sections and consider the equilibrium of the FBD of the pipe’s
upper segment, Fig. a,
©Fz = 0;
Vz - P = 0
Vz = P
©Mx = 0;
Tx - P(1.5) = 0 Tx = 1.5P
©My = 0;
My - P(2.5) = 0 My = 2.5P
By observation, no normal stress acting on point A. Thus, this is a case of pure shear.
For the case of pure shear,
ex = ez = ey = 0
ea = ex cos2 ua + ez sin2 ua + gxz sin ua cos ua
-200(10 - 6) = 0 + 0 + gxz sin 150° cos 150°
gxz = 461.88(10 - 6)
e1, 2 =
ex + ez
= B
2
+
A
a
ex - ez
2
b + a
2
gxz
2
b
2
0 - 0 2
461.88 2
0 + 0
-6
;
a
b + a
b R (10 )
2
A
2
2
e1 = 231(10 - 6)
e2 = -231(10 - 6)
Ans.
780
30⬚
A
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10–45. The cylindrical pressure vessel is fabricated using
hemispherical end caps in order to reduce the bending stress
that would occur if flat ends were used. The bending stresses
at the seam where the caps are attached can be eliminated
by proper choice of the thickness th and tc of the caps and
cylinder, respectively. This requires the radial expansion to
be the same for both the hemispheres and cylinder. Show
that this ratio is tc>th = 12 - n2>11 - n2.Assume that the
vessel is made of the same material and both the cylinder
and hemispheres have the same inner radius. If the cylinder
is to have a thickness of 0.5 in., what is the required thickness
of the hemispheres? Take n = 0.3.
tc
th
r
For cylindrical vessel:
s1 =
pr
;
tc
e1 =
1
[s - v (s2 + s3)]
E 1
=
s2 =
pr
2 tc
s3 = 0
vpr
pr
1
1 pr
a
b =
a1 - v b
E tc
2 tc
E tc
2
d r = e1 r =
p r2
1
a1 - v b
E tc
2
(1)
For hemispherical end caps:
s1 = s2 =
e1 =
=
pr
2 th
1
[s - v (s2 + s3)] ;
E 1
s3 = 0
vpr
pr
1 pr
a
b =
(1 - v)
E 2 th
2 th
2 E th
d r = e1 r =
p r2
(1 - v)
2 E th
(2)
Equate Eqs. (1) and (2):
p r2
p r2
1
a1 - vb =
(1 - v)
E tc
2
2 E th
1
2 (1 - 2 v)
tc
2 - v
=
=
th
1 - v
1 - v
th =
QED
(1 - v) tc
(1 - 0.3) (0.5)
=
= 0.206 in.
2 - v
2 - 0.3
Ans.
781
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10–46. The principal strains in a plane, measured
experimentally at a point on the aluminum fuselage of a jet
aircraft, are P1 = 630(10-6) and P2 = 350(10-6). If this is
a case of plane stress, determine the associated principal
stresses at the point in the same plane. Eal = 10(103) ksi
and nal = 0.33.
Normal Stresses: For plane stress, s3 = 0.
Normal Strains: Applying the generalized Hooke’s Law.
e1 =
1
C s - v (s2 + s3) D
E 1
630 A 10 - 6 B =
1
[s1 - 0.33(s2 + 0)]
10(103)
6.30 = s1 - 0.33s2
e2 =
[1]
1
C s - v (s1 + s3) D
E 2
350 A 10 - 6 B =
1
C s2 - 0.33(s1 + 0) D
10(103)
3.50 = s2 - 0.33s1
[2]
Solving Eqs.[1] and [2] yields:
s1 = 8.37 ksi
s2 = 6.26 ksi
Ans.
10–47. The principal stresses at a point are shown in
the figure. If the material is aluminum for which
Eal = 1011032 ksi and nal = 0.33, determine the principal
strains.
3 ksi
e1 =
1
1
e 8 - 0.33 C 3 + (-4) D f = 833 (10 - 6)
C s - v(s2 + s3) D =
E 1
10(103)
e2 =
1
1
e 3 - 0.33 C 8 + (-4) D f = 168 (10 - 6)
C s - v(s1 + s3) D =
E 2
10(103)
e3 =
1
1
C s3 - v(s1 + s2) D =
C -4 - 0.33(8 + 3) D = -763 (10 - 6)
E
10(103)
Using these results,
e1 = 833(10 - 6)
e2 = 168(10 - 6)
e3 = -763(10 - 6)
782
8 ksi
4 ksi
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*10–48. The 6061-T6 aluminum alloy plate fits snugly into
the rigid constraint. Determine the normal stresses sx and
sy developed in the plate if the temperature is increased by
¢T = 50°C. To solve, add the thermal strain a¢T to the
equations for Hooke’s Law.
y
400 mm
300 mm
x
Generalized Hooke’s Law: Since the sides of the aluminum plate are confined in the
rigid constraint along the x and y directions, ex = ey = 0. However, the plate is
allowed to have free expansion along the z direction. Thus, sz = 0. With the
additional thermal strain term, we have
ex =
0 =
1
cs - v A sy + sz B d + a¢T
E x
1
68.9 A 109 B
csx - 0.35 A sy + 0 B d + 24a 10 - 6 b(50)
sx - 0.35sy = -82.68 A 106 B
ey =
0 =
(1)
1
C s - v A sx + sz B D + a¢T
E y
1
68.9a 10 b
9
C sy - 0.35(sx + 0) D + 24 A 10 - 6 B (50)
sy - 0.35sx = -82.68 A 106 B
(2)
Solving Eqs. (1) and (2),
sx = sy = -127.2 MPa = 127.2 MPa (C)
Ans.
Since sx = sy and sy 6 sY, the above results are valid.
783
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•10–49.
Initially, gaps between the A-36 steel plate and
the rigid constraint are as shown. Determine the normal
stresses sx and sy developed in the plate if the temperature
is increased by ¢T = 100°F. To solve, add the thermal
strain a¢T to the equations for Hooke’s Law.
y
0.0015 in.
6 in.
8 in.
0.0025 in.
x
Generalized Hooke’s Law: Since there are gaps between the sides of the plate and the rigid
constraint, the plate is allowed to expand before it comes in contact with the constraint.
dy
dx
0.0025
0.0015
ey =
=
= 0.3125 A 10 - 3 B and
=
= 0.25 A 10 - 3 B .
Thus, ex =
Lx
8
Ly
6
However, the plate is allowed to have free expansion along the z direction. Thus, sz = 0.
With the additional thermal strain term, we have
ex =
1
csx - v A sy + sz B d + a¢T
E
0.3125 a 10 - 3 b =
1
29.0 a103 b
C sx - 0.32 A sy + 0 B D + 6.60 A 10 - 6 B (100)
sx - 0.32sy = -10.0775
ey =
(1)
1
C s - v A sx + sz B D + a¢T
E y
0.25 A 10 - 3 B =
1
29.0 A 103 B
C sy - 0.32(sx + 0) D + 6.60 A 10 - 6 B (100)
sy - 0.32sx = -11.89
(2)
Solving Eqs. (1) and (2),
sx = -15.5 ksi = 15.5 ksi (C)
Ans.
sy = -16.8 ksi = 16.8 ksi (C)
Ans.
Since sx 6 sY and sy 6 sY, the above results are valid.
784
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10–50. Two strain gauges a and b are attached to a plate
made from a material having a modulus of elasticity of
E = 70 GPa and Poisson’s ratio n = 0.35. If the gauges give
a reading of Pa = 450110-62 and Pb = 100110-62, determine
the intensities of the uniform distributed load wx and wy
acting on the plate. The thickness of the plate is 25 mm.
wy
b
45⬚
y
a
Normal Strain: Since no shear force acts on the plane along the x and y axes, gxy = 0.
With ua = 0 and ub = 45°, we have
2
2
ea = ex cos ua + ey sin ua + gxy sin ua cos ua
450 A 10 - 6 B = ex cos2 0° + ey sin2 0°+0
ex = 450 A 10 - 6 B
eb = ex cos2 ub + ey sin2 ub + gxy sin ub cos ub
100 A 10 - 6 B = 450 A 10 - 6 B cos2 45° + ey sin2 45° + 0
ey = -250 A 10 - 6 B
Generalized Hooke’s Law: This is a case of plane stress. Thus, sz = 0.
ex =
1
C s - v A sy + sz B D
E x
450 A 10 - 6 B =
1
70 A 109 B
C sy - 0.35 A sy + 0 B D
sx - 0.35sy = 31.5 A 106 B
ey =
(1)
1
C s - v A sx + sz B D
E y
-250 A 10-6 B =
1
70 A 109 B
C sy - 0.35 A sy + 0 B D
sy - 0.35sx = -17.5 A 106 B
(2)
Solving Eqs. (1) and (2),
sy = -7.379 A 106 B N>m2
sx = 28.917 A 106 B N>m2
Then,
wy = syt = -7.379 A 106 B (0.025) = -184 N>m
Ans.
wx = sxt = 28.917 A 106 B (0.025) = 723 N>m
Ans.
785
z
x
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10–51. Two strain gauges a and b are attached to the
surface of the plate which is subjected to the uniform
distributed load wx = 700 kN>m and wy = -175 kN>m.
If the gauges give a reading of Pa = 450110-62 and
Pb = 100110-62, determine the modulus of elasticity E,
shear modulus G, and Poisson’s ratio n for the material.
wy
b
45⬚
y
Normal Stress and Strain: The normal stresses along the x, y, and z axes are
sx =
700 A 103 B
0.025
sy = -
a
= 28 A 10 B N>m
6
175 A 103 B
0.025
2
= -7 A 106 B N>m2
z
sz = 0 (plane stress)
Since no shear force acts on the plane along the x and y axes, gxy = 0. With ua = 0°
and ub = 45°, we have
ea = ex cos2 ua + ey sin2 ua + gxy sin ua cos ua
450 A 10 - 6 B = ex cos2 0° + ey sin2 0° + 0
ex = 450 A 10 - 6 B
eb = ex cos2 ub + ey sin2 ub + gxy sin ub cos ub
100 A 10 - 6 B = 450 A 10 - 6 B cos2 45°+ ey sin2 45° + 0
ey = -250 A 10 - 6 B
Generalized Hooke’s Law:
ex =
1
C s - v A sy + sz B D
E x
450 A 10 - 6 B =
1
B 28 A 106 B - v C -7 A 106 B + 0 D R
E
450 A 10 - 6 B E - 7 A 106 B v = 28 A 106 B
ey =
(1)
1
[s - v(sx + sz)]
E y
-250 A 10 - 6 B =
1
b -7 A 106 B - v C 28 A 106 B + 0 D r
E
250 A 10 - 6 B E - 28 A 106 B v = 7 A 106 B
(2)
Solving Eqs. (1) and (2),
E = 67.74 A 109 B N>m2 = 67.7 GPa
Ans.
v = 0.3548 = 0.355
Ans.
Using the above results,
G =
67.74 A 109 B
E
=
2(1 + v)
2(1 + 0.3548)
= 25.0 A 109 B N>m2 = 25.0 GPa
Ans.
786
x
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*10–52. The block is fitted between the fixed supports. If the
glued joint can resist a maximum shear stress of tallow = 2 ksi,
determine the temperature rise that will cause the joint to fail.
Take E = 10 (103) ksi, n = 0.2, and Hint: Use Eq. 10–18 with
an additional strain term of a¢T (Eq. 4–4).
40⬚
Normal Strain: Since the aluminum is confined along the y direction by the rigid
frame, then ey = 0 and sx = sz = 0. Applying the generalized Hooke’s Law with
the additional thermal strain,
ey =
0 =
1
C s - v(sx + sz) D + a¢T
E y
1
C sy - 0.2(0 + 0) D + 6.0 A 10 - 6 B (¢T)
10.0(103)
sy = -0.06¢T
Construction of the Circle: In accordance with the sign convention. sx = 0,
sy = -0.06¢T and txy = 0. Hence.
savg =
sx + sy
2
=
0 + ( -0.06¢T)
= -0.03¢T
2
The coordinates for reference points A and C are A (0, 0) and C( -0.03¢T, 0).
The radius of the circle is R = 2(0 - 0.03¢T)2 + 0 = 0.03¢T
Stress on The inclined plane: The shear stress components tx¿y¿, are represented by
the coordinates of point P on the circle.
tx¿y¿ = 0.03¢T sin 80° = 0.02954¢T
Allowable Shear Stress:
tallow = tx¿y¿
2 = 0.02954¢T
¢T = 67.7 °F
Ans.
787
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z
•10–53.
The smooth rigid-body cavity is filled with liquid
6061-T6 aluminum. When cooled it is 0.012 in. from the top
of the cavity. If the top of the cavity is covered and the
temperature is increased by 200°F, determine the stress
components sx , sy , and sz in the aluminum. Hint: Use
Eqs. 10–18 with an additional strain term of a¢T (Eq. 4–4).
0.012 in.
4 in.
4 in.
6 in.
y
Normal Strains: Since the aluminum is confined at its sides by a rigid container and
0.012
allowed to expand in the z direction, ex = ey = 0; whereas ez =
= 0.002.
6
Applying the generalized Hooke’s Law with the additional thermal strain,
ex =
1
C s - v(sy + sz) D + a¢T
E x
0 =
1
C sx - 0.35 A sy + sz B D + 13.1 A 10 - 6 B (200)
10.0(103)
0 = sx - 0.35sy - 0.35sz + 26.2
[1]
ey =
1
C s - v(sx + sz) + a¢T
E y
0 =
1
C sy - 0.35(sx + sz) D + 13.1 A 10 - 6 B (200)
10.0(103)
0 = sy - 0.35sx - 0.35sz + 26.2
ez =
0.002 =
[2]
1
C s - v A sx + sy B D + a¢T
E z
1
C sz - 0.35 A sx + sy B D + 13.1 A 10 - 6 B (200)
10.0(103)
0 = sz - 0.35sx - 0.35sy + 6.20
[3]
Solving Eqs.[1], [2] and [3] yields:
sx = sy = -70.0 ksi
sz = -55.2 ksi
Ans.
788
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z
10–54. The smooth rigid-body cavity is filled with liquid
6061-T6 aluminum. When cooled it is 0.012 in. from the top
of the cavity. If the top of the cavity is not covered and the
temperature is increased by 200°F, determine the strain
components Px , Py , and Pz in the aluminum. Hint: Use
Eqs. 10–18 with an additional strain term of a¢T (Eq. 4–4).
0.012 in.
4 in.
4 in.
6 in.
y
Normal Strains: Since the aluminum is confined at its sides by a rigid container,
then
ex = ey = 0
Ans.
and since it is not restrained in z direction, sz = 0. Applying the generalized
Hooke’s Law with the additional thermal strain,
ex =
1
C s - v A sy + sz B D + a¢T
E x
0 =
1
C sx - 0.35 A sy + 0 B D + 13.1 A 10 - 6 B (200)
10.0(103)
0 = sx - 0.35sy + 26.2
[1]
ey =
1
C s - v(sx + sz) D + a¢T
E y
0 =
1
C sy - 0.35(sx + 0) D + 13.1 A 10 - 6 B (200)
10.0(103)
0 = sy - 0.35sx + 26.2
[2]
Solving Eqs. [1] and [2] yields:
sx = sy = -40.31 ksi
ez =
=
1
C s - v A sx + sy B D + a¢T
E z
1
{0 - 0.35[-40.31 + (-40.31)]} + 13.1 A 10 - 6 B (200)
10.0(103)
= 5.44 A 10 - 3 B
Ans.
789
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10–55. A thin-walled spherical pressure vessel having an
inner radius r and thickness t is subjected to an internal
pressure p. Show that the increase in the volume within
the vessel is ¢V = 12ppr4>Et211 - n2. Use a small-strain
analysis.
pr
2t
s1 = s2 =
s3 = 0
e1 = e2 =
1
(s - vs2)
E 1
e1 = e2 =
pr
(1 - v)
2t E
e3 =
1
(-v(s1 + s2))
E
e3 = -
V =
v pr
tE
4pr3
3
V + ¢V =
4p
4pr3
¢r 3
(r + ¢r)3 =
(1 +
)
r
3
3
where ¢V V V, ¢r V r
V + ¢V -
eVol =
¢r
4p r3
a1 + 3
b
r
3
¢V
¢r
= 3a b
V
r
Since e1 = e2 =
eVol = 3e1 =
2p(r + ¢r) - 2p r
¢r
=
r
2p r
3pr
(1 - v)
2t E
¢V = VeVol =
2pp r4
(1 - v)
Et
QED
790
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*10–56. A thin-walled cylindrical pressure vessel has an
inner radius r, thickness t, and length L. If it is subjected
to an internal pressure p, show that the increase in
its inner radius is dr = rP1 = pr211 - 12 n2>Et and the
increase in its length is ¢L = pLr112 - n2>Et. Using these
results, show that the change in internal volume becomes
dV = pr211 + P12211 + P22L - pr2L. Since P1 and P2 are
small quantities, show further that the change in volume
per unit volume, called volumetric strain, can be written as
dV>V = pr12.5 - 2n2>Et.
Normal stress:
pr
;
t
s1 =
s2 =
pr
2t
Normal strain: Applying Hooke’s law
e1 =
=
1
[s - v (s2 + s3)],
E 1
vpr
pr
1
1 pr
a
b =
a1 - vb
E
t
2t
Et
2
d r = et r =
e2 =
=
s3 = 0
p r2
1
a1 - v b
Et
2
1
[s - v (s1 + s3)],
E 2
QED
s3 = 0
vpr
pr 1
1 pr
a
b =
a - vb
E 2t
t
Et 2
¢L = e2 L =
pLr 1
a - vb
Et
2
V¿ = p(r + e1 r)2 (L + e2L) ;
QED
V = p r2 L
dV = V¿ - V = pr2 (1 + e1)2 (1 + e2)L - pr2 L
QED
(1 + e1)2 = 1 + 2 e1 neglect e21 term
(1 + e1)2 (1 + e2) = (1 + 2 e1)(1 + e2) = 1 + e2 + 2 e1 neglect e1 e2 term
dV
= 1 + e2 + 2 e1 - 1 = e2 + 2 e1
V
=
2pr
pr 1
1
a - vb +
a1 - v b
Et 2
Et
2
=
pr
(2.5 - 2 v)
Et
QED
791
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10–57. The rubber block is confined in the U-shape
smooth rigid block. If the rubber has a modulus of elasticity
E and Poisson’s ratio n, determine the effective modulus of
elasticity of the rubber under the confined condition.
P
Generalized Hooke’s Law: Under this confined condition, ex = 0 and sy = 0. We
have
ex =
1
C s - v A sy + sz B D
E x
0 =
1
(s - vsz)
E x
sx = vsz
(1)
ez =
1
C s - v A sx + sy B D
E z
ez =
1
[s - v(sx + 0)]
E z
ez =
1
(s - vsx)
E z
(2)
Substituting Eq. (1) into Eq. (2),
ez =
sz
E
A 1 - v2 B
The effective modulus of elasticity of the rubber block under the confined condition
can be determined by considering the rubber block as unconfined but rather
undergoing the same normal strain of ez when it is subjected to the same normal
stress sz, Thus,
sz = Eeff ez
Eeff =
sz
ez
sz
=
sz
E
A 1 - v2 B
=
E
1 - v2
Ans.
792
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z
10–58. A soft material is placed within the confines of a
rigid cylinder which rests on a rigid support. Assuming that
Px = 0 and Py = 0, determine the factor by which the
modulus of elasticity will be increased when a load is
applied if n = 0.3 for the material.
P
x
Normal Strain: Since the material is confined in a rigid cylinder. ex = ey = 0.
Applying the generalized Hooke’s Law,
ex =
1
C s - v(sy + sx) D
E z
0 = sx - v(sy + sz)
ey =
[1]
1
C s - v(sx + sz) D
E y
0 = sy - v(sx + sz)
[2]
Solving Eqs.[1] and [2] yields:
sx = sy =
v
s
1 - v z
Thus,
ez =
=
1
C s - v(sx + sy) D
E z
v
v
1
csz - v a
sz +
s bd
E
1 - v
1 - v z
sz
=
E
c1 -
2v2
d
1 - v
=
sz 1 - v - 2v2
c
d
E
1 - v
=
sz (1 + v)(1 - 2v
c
d
E
1 - v
Thus, when the material is not being confined and undergoes the same normal strain
of ez, then the requtred modulus of elasticity is
E¿ =
sz
ez
=
The increased factor is k =
1 - v
E
(1 - 2v)(1 + v)
E¿
1 - v
=
E
(1 - 2v)(1 + v)
=
1 - 0.3
[1 - 2(0.3)](1 + 0.3)
= 1.35
Ans.
793
y
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10–59. A material is subjected to plane stress. Express
the distortion-energy theory of failure in terms of sx , sy ,
and txy .
Maximum distortion energy theory:
(s21 - s1 s2 + s22) = s2Y
s1,2 =
sx + sy
;
2
Let a =
sx + sy
2
s1 = a + b;
A
a
(1)
sx - sy
2
and b =
A
a
2
2
b + txy
sx - sy
2
2
2
b + txy
s2 = a - b
s21 = a2 + b2 + 2 a b;
s22 = a2 + b2 - 2 a b
s1 s2 = a2 - b2
From Eq. (1)
(a2 + b2 + 2 a b - a2 + b2 + a2 + b2 - 2 a b) = s2y
(a2 + 3 b2) = s2Y
(sx + sy)2
4
+ 3
(sx - sy)2
4
+ 3 t2xy = s2Y
s2x + s2y - sxsy + 3 t2xy = s2Y
Ans.
*10–60. A material is subjected to plane stress. Express
the maximum-shear-stress theory of failure in terms of sx ,
sy , and txy . Assume that the principal stresses are of
different algebraic signs.
Maximum shear stress theory:
|s1 - s2| = sY
s1,2 =
(1)
sx + sy
2
` s1 - s2 ` = 2
a
sx - sy
;
A
a
sx - sy
A
2
2
2
2
b + txy
2
b + txy
2
From Eq. (1)
4 ca
sx - sy
2
2
b + t2xy d = s2Y
2
(sx - sy) + 4 t2xy = s2Y
Ans.
794
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•10–61.
An aluminum alloy 6061-T6 is to be used for
a solid drive shaft such that it transmits 40 hp at 2400
rev>min. Using a factor of safety of 2 with respect to
yielding, determine the smallest-diameter shaft that can be
selected based on the maximum-shear-stress theory.
v = a2400
T =
2p rad 1 min
rev
ba
ba
b = 80 p rad>s
min
rev
60s
40 (550) (12)
P
3300 #
=
=
lb in.
p
v
80 p
Tc
J
Applying t =
t =
A 3300
p B c
=
p 4
2 c
6600
p3 c3
The principal stresses:
s1 = t =
6600
;
p2 c3
s2 = -t =
6600
p2 c3
Maximum shear stress theory: Both principal stresses have opposite sign, hence,
` s1 - s2 ` =
2a
sY
;
F.S.
37 (103)
6600
b = `
`
2 3
2
pc
c = 0.4166 in.
d = 0.833 in.
Ans.
10–62. Solve Prob. 10–61 using the maximum-distortionenergy theory.
v = a2400
T =
2p rad 1 min
rev
ba
ba
b = 80 p rad>s
min
rev
60s
40 (550) (12)
P
3300
=
=
lb.in.
p
v
80 p
Applying t =
t =
A 3300
p B c
p 4
2 c
=
Tc
J
6600
p2 c3
The principal stresses:
s1 = t =
6600
;
p2 c3
s2 = - t = -
6600
p2 c3
The maximum distortion-energy theory:
s21 - s1 s2 + s22 = a
3B
sY 2
b
F.S.
37(103) 2
6600 2
=
a
b
R
2
p2 c3
c = 0.3971 in.
d = 0.794 in.
Ans.
795
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10–63. An aluminum alloy is to be used for a drive shaft
such that it transmits 25 hp at 1500 rev>min. Using a factor
of safety of 2.5 with respect to yielding, determine the
smallest-diameter shaft that can be selected based on the
maximum-distortion-energy theory. sY = 3.5 ksi.
1500(2p)
= 50p
60
T =
P
v
T =
25(550)(12)
3300
=
p
50p
t =
Tc
,
J
t =
3300
p c
p 4
2c
s1 =
v =
J =
=
p 4
c
2
6600
p2c3
6600
p2c3
s2 =
s21 - s1 s2 + s22 = a
3a
-6600
p2c3
sY 2
b
F.S.
3.5(103) 2
6600 2
b
b
=
a
2.5
p2c3
c = 0.9388 in.
d = 1.88 in.
Ans.
*10–64. A bar with a square cross-sectional area is made
of a material having a yield stress of sY = 120 ksi. If the bar
is subjected to a bending moment of 75 kip # in., determine
the required size of the bar according to the maximumdistortion-energy theory. Use a factor of safety of 1.5 with
respect to yielding.
Normal and Shear Stress: Applying the flexure formula,
75 A 2 B
Mc
450
= 1 4 = 3
I
a
12 a
a
s =
In-Plane Principal Stress: Since no shear stress acts on the element
s1 = sx =
450
a3
s2 = sy = 0
Maximum Distortion Energy Theory:
s21 - s1 s2 + s22 = s2allow
a
120 2
450 2
b - 0 + 0 = a
b
3
1.5
a
a = 1.78 in.
Ans.
796
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•10–65.
Solve Prob. 10–64 using the maximum-shearstress theory.
Normal and Shear Stress: Applying the flexure formula,
75 A 2 B
Mc
450
= 1 4 = 3
I
a
a
12
a
s =
In-Plane Principal Stress: Since no shear stress acts on the element.
s1 = sx =
450
a3
s2 = sx = 0
Maximum Shear Stress Theory:
|s2| = 0 6 sallow =
120
= 80.0 ksi
1.5
(O.K!)
|s1| = sallow
120
450
=
1.5
a3
a = 1.78 in.
Ans.
10–66. Derive an expression for an equivalent torque Te
that, if applied alone to a solid bar with a circular cross
section, would cause the same energy of distortion as
the combination of an applied bending moment M and
torque T.
t =
Te c
J
Principal stress:
s1 = tx ¿
ud =
s2 = -t
1 + v 2
(s1 - s1 s2 + s22)
3E
(ud)1 =
1 + v
1 + v 3 T2x c2
( 3 t2) =
a
b
3E
3E
J2
Bending moment and torsion:
s =
Mc
;
I
t =
Tc
J
Principal stress:
s1, 2 =
s1 =
s + 0
s - 0 2
2
;
a
b + t
2
A
2
s
s2
+
+ t2 ;
2
A4
s2 =
s
s2
+ t2
2
A4
797
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10–66.
Continued
Let a =
s
2
b =
s2
+ t2
A4
s21 = a2 + b2 + 2 a b
s1 s2 = a2 - b2
s22 = a2 + b2 - 2 a b
s21 - s1 s2 + s22 = 3 b2 + a2
ud =
1 + v 2
(s1 - s1 s2 + s22)
3E
(ud)2 =
=
1 + v
1 + v 3 s2
s2
(3 b2 + a2) =
a
+ 3t2 +
b
3E
3E
4
4
c2(1 + v) M2
1 + v 2
3 T2
b
(s + 3 t2) =
a 2 +
3E
3E
I
J2
(ud)1 = (ud)2
c3(1 + v) 3 Tx 2
c2(1 + v) M2
3 T2
=
a
+
b
3E
3E
J2
I2
J2
For circular shaft
p 4
J 3c
=p 4=2
I 4c
Te =
J2 M2
+ T2
A I2 3
Te =
4 2
M + T2
A3
Ans.
10–67. Derive an expression for an equivalent bending
moment Me that, if applied alone to a solid bar with a
circular cross section, would cause the same energy of
distortion as the combination of an applied bending
moment M and torque T.
Principal stresses:
s1 =
Me c
;
I
ud =
1 + v 2
(s1 - s1 s2 + s22)
3E
(ud)1 =
s2 = 0
1 + v M2e c2
a 2 b
3E
I
(1)
798
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10–67.
Continued
Principal stress:
s + 0
s - 0 2
3
;
a
b + t
2
A
2
s1, 2 =
s1 =
s
s2
+
+ t2;
2
A4
s2 =
s
s2
+ t2
2
A4
Distortion Energy:
s
s2
,b =
+ t2
2
A4
Let a =
s21 = a2 + b2 + 2 a b
s1 s2 = a2 - b2
s22 = a2 + b2 - 2 a b
s22 - s1 s2 + s22 = 3 b2 + a2
Apply s =
Mc
;
I
t =
Tc
J
(ud)2 =
3s2
1 + v
1 + v s2
(3 b2 + a2) =
a
+
+ 3 t2 b
3E
3E
4
4
=
1 + v 2
1 + v M2 c2
3 T2 c2
(s + 3 t2) =
a 2 +
b
3E
3E
I
J2
(2)
Equating Eq. (1) and (2) yields:
(1 + v) Me c2
1 + v M2 c2
3T2 c2
a 2 b =
a 2 +
b
3E
3E
I
I
J2
M2e
2
=
I
M1
3 T2
+
2
I
J2
M2e = M1 + 3 T2 a
I 2
b
J
For circular shaft
p 4
I
1
4 c
= p 4 =
J
2
c
2
1 2
Hence, M2e = M2 + 3 T2 a b
2
Me =
A
M2 +
3 2
T
4
Ans.
799
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*10–68. The short concrete cylinder having a diameter of
50 mm is subjected to a torque of 500 N # m and an axial
compressive force of 2 kN. Determine if it fails according to
the maximum-normal-stress theory. The ultimate stress of
the concrete is sult = 28 MPa.
A =
p
(0.05)2 = 1.9635(10 - 3) m2
4
J =
p
(0.025)4 = 0.61359(10 - 4) m4
2
2 kN
500 N⭈m
500 N⭈m
2 kN
3
s =
2(10 )
P
= 1.019 MPa
=
A
1.9635(10 - 3)
t =
500(0.025)
Tc
= 20.372 MPa
=
J
0.61359(10 - 6)
sx = 0
sy = -1.019 MPa
sx + sy
s1, 2 =
s1,2 =
2
;
A
a
sx - sy
2
txy = 20.372 MPa
2
b + txy
2
0 - 1.018
0 - (-1.019) 2
2
;
a
b + 20.372
2
A
2
s1 = 19.87 MPa
s2 = -20.89 MPa
Failure criteria:
|s1| 6 salt = 28 MPa
OK
|s2| 6 salt = 28 MPa
OK
No.
Ans.
•10–69.
Cast iron when tested in tension and compression
has an ultimate strength of 1sult2t = 280 MPa and
1sult2c = 420 MPa, respectively. Also, when subjected to
pure torsion it can sustain an ultimate shear stress of
tult = 168 MPa. Plot the Mohr’s circles for each case and
establish the failure envelope. If a part made of this
material is subjected to the state of plane stress shown,
determine if it fails according to Mohr’s failure criterion.
120 MPa
100 MPa
220 MPa
s1 = 50 + 197.23 = 247 MPa
s2 = 50 - 197.23 = -147 MPa
The principal stress coordinate is located at point A which is outside the shaded
region. Therefore the material fails according to Mohr’s failure criterion.
Yes.
Ans.
800
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10–69.
Continued
10–70. Derive an expression for an equivalent bending
moment Me that, if applied alone to a solid bar with a
circular cross section, would cause the same maximum
shear stress as the combination of an applied moment M
and torque T. Assume that the principal stresses are of
opposite algebraic signs.
Bending and Torsion:
Mc
Mc
4M
= p 4 =
;
I
c
p c3
4
s =
t =
Tc
Tc
2T
= p 4 =
J
c
p
c3
2
The principal stresses:
s1, 2 =
=
tabs
max
sx + sy
2
;
A
a
sx - sy
2
2
2
b + txy =
4M
4M
pc3 + 0
2
3
;
Q
¢pc
- 0
2
2
≤ + a
2T
3
pc
b
2
2
2M
;
2M2 + T2
p c3
p c3
= s1 - s2 = 2 c
2
2M2 + T2 d
p c3
(1)
Pure bending:
s1 =
tabs
max
Me c
4 Me
Mc
= p 4 =
;
I
c
p c3
4
= s1 - s2 =
s2 = 0
4 Me
(2)
p c3
Equating Eq. (1) and (2) yields:
4 Me
4
2M2 + T2 =
p c3
p c3
Me = 2M2 + T2
Ans.
801
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10–71. The components of plane stress at a critical point
on an A-36 steel shell are shown. Determine if failure
(yielding) has occurred on the basis of the maximum-shearstress theory.
60 MPa
40 MPa
In accordance to the established sign convention, sx = 70 MPa, sy = -60 MPa and
txy = 40 MPa.
s1, 2 =
=
sx + sy
2
;
A
a
sx - sy
2
70 MPa
2
b + txy
2
70 + (-60)
70 - (-60) 2
2
;
c
d + 40
2
A
2
= 5 ; 25825
s1 = 81.32 MPa
s2 = -71.32 MPa
In this case, s1 and s2 have opposite sign. Thus,
|s1 - s2| = |81.32 - (-71.32)| = 152.64 MPa 6 sy = 250 MPa
Based on this result, the steel shell does not yield according to the maximum shear
stress theory.
*10–72. The components of plane stress at a critical point
on an A-36 steel shell are shown. Determine if failure
(yielding) has occurred on the basis of the maximumdistortion-energy theory.
60 MPa
40 MPa
In accordance to the established sign convention, sx = 70 MPa, sy = -60 MPa and
txy = 40 MPa.
s1, 2 =
=
sx + sy
2
;
A
a
sx - sy
2
2
b + txy
2
70 + (-60)
70 - ( -60) 2
2
;
c
d + 40
2
A
2
= 5 ; 25825
s1 = 81.32 MPa
s2 = -71.32 MPa
s1 2 - s1 s2 + s2 2 = 81.322 - 81.32(-71.32) + (-71.32)2 = 17,500 6 sy 2 = 62500
Based on this result, the steel shell does not yield according to the maximum
distortion energy theory.
802
70 MPa
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•10–73.
If the 2-in. diameter shaft is made from brittle
material having an ultimate strength of sult = 50 ksi for
both tension and compression, determine if the shaft fails
according to the maximum-normal-stress theory. Use a
factor of safety of 1.5 against rupture.
30 kip
4 kip · ft
Normal Stress and Shear Stresses. The cross-sectional area and polar moment of
inertia of the shaft’s cross-section are
A = p A 12 B = pin2
J =
p 4
p
A 1 B = in4
2
2
The normal stress is caused by axial stress.
s =
N
30
= = -9.549 ksi
p
A
The shear stress is contributed by torsional shear stress.
t =
4(12)(1)
Tc
=
= 30.56 ksi
p
J
2
The state of stress at the points on the surface of the shaft is represented on the
element shown in Fig. a.
In-Plane Principal Stress. sx = -9.549 ksi, sy = 0 and txy = -30.56 ksi. We have
s1, 2 =
=
sx + sy
2
;
A
a
sx - sy
2
2
b + txy
2
-9.549 - 0 2
-9.549 + 0
2
;
a
b + ( -30.56)
2
A
2
= (-4.775 ; 30.929) ksi
s1 = 26.15 ksi
s2 = -35.70 ksi
Maximum Normal-Stress Theory.
sallow =
sult
50
=
= 33.33 ksi
F.S.
1.5
|s1| = 26.15 ksi 6 sallow = 33.33 ksi
(O.K.)
|s2| = 35.70 ksi 7 sallow = 33.33 ksi
(N.G.)
Based on these results, the material fails according to the maximum normal-stress
theory.
803
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10–74. If the 2-in. diameter shaft is made from cast
iron having tensile and compressive ultimate strengths
of 1sult2t = 50 ksi and 1sult2c = 75 ksi, respectively,
determine if the shaft fails in accordance with Mohr’s
failure criterion.
Normal Stress and Shear Stresses. The cross-sectional area and polar moment of
inertia of the shaft’s cross-section are
A = p A 12 B = p in2
J =
p 4
p
A 1 B = in4
2
2
The normal stress is contributed by axial stress.
s =
N
30
= = -9.549 ksi
p
A
The shear stress is contributed by torsional shear stress.
t =
4(12)(1)
Tc
=
= 30.56 ksi
p
J
2
The state of stress at the points on the surface of the shaft is represented on the
element shown in Fig. a.
In-Plane Principal Stress. sx = -9.549 ksi, sy = 0, and txy = -30.56 ksi. We have
s1, 2 =
=
sx + sy
2
;
A
a
sx - sy
2
2
b + txy
2
-9.549 - 0 2
-9.549 + 0
2
;
a
b + ( -30.56)
2
A
2
= ( -4.775 ; 30.929) ksi
s1 = 26.15 ksi
s2 = -35.70 ksi
Mohr’s Failure Criteria. As shown in Fig. b, the coordinates of point A, which
represent the principal stresses, are located inside the shaded region. Therefore, the
material does not fail according to Mohr’s failure criteria.
804
30 kip
4 kip · ft
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10–75. If the A-36 steel pipe has outer and inner
diameters of 30 mm and 20 mm, respectively, determine the
factor of safety against yielding of the material at point A
according to the maximum-shear-stress theory.
900 N
150 mm
A
100 mm
200 mm
Internal Loadings. Considering the equilibrium of the free - body diagram of the
post’s right cut segment Fig. a,
©Fy = 0; Vy + 900 - 900 = 0
Vy = 0
©Mx = 0; T + 900(0.4) = 0
T = -360 N # m
©Mz = 0; Mz + 900(0.15) - 900(0.25) = 0 Mz = 90 N # m
Section Properties. The moment of inertia about the z axis and the polar moment of
inertia of the pipe’s cross section are
Iz =
p
A 0.0154 - 0.014 B = 10.15625p A 10 - 9 B m4
4
J =
p
A 0.0154 - 0.014 B = 20.3125p A 10 - 9 B m4
2
Normal Stress and Shear Stress. The normal stress is contributed by bending stress.
Thus,
sY = -
MyA
90(0.015)
= = -42.31MPa
Iz
10.15625p A 10 - 9 B
The shear stress is contributed by torsional shear stress.
t =
360(0.015)
Tc
=
= 84.62 MPa
J
20.3125p A 10 - 9 B
The state of stress at point A is represented by the two - dimensional element shown
in Fig. b.
In - Plane Principal Stress. sx = -42.31 MPa, sz = 0 and txz = 84.62 MPa. We have
s1, 2 =
=
sx + sz
2
;
A
a
sx - sz
2
2
b + txz
2
-42.31 - 0 2
-42.31 + 0
2
;
a
b + 84.62
2
A
2
= (-21.16 ; 87.23) MPa
s1 = 66.07 MPa
s2 = -108.38 MPa
805
200 mm
900 N
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10–75.
Continued
Maximum Shear Stress Theory. s1 and s2 have opposite signs. This requires
|s1 - s2| = sallow
66.07 - (-108.38) = sallow
sallow = 174.45 MPa
The factor of safety is
F.S. =
sY
250
=
= 1.43
sallow
174.45
Ans.
806
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*10–76. If the A-36 steel pipe has an outer and inner
diameter of 30 mm and 20 mm, respectively, determine the
factor of safety against yielding of the material at point A
according to the maximum-distortion-energy theory.
900 N
150 mm
A
100 mm
200 mm
Internal Loadings: Considering the equilibrium of the free - body diagram of the
pipe’s right cut segment Fig. a,
©Fy = 0; Vy + 900 - 900 = 0
Vy = 0
©Mx = 0; T + 900(0.4) = 0
T = -360 N # m
©Mz = 0; Mz + 900(0.15) - 900(0.25) = 0 Mz = 90 N # m
Section Properties. The moment of inertia about the z axis and the polar moment of
inertia of the pipe’s cross section are
Iz =
p
A 0.0154 - 0.014 B = 10.15625p A 10 - 9 B m4
4
J =
p
A 0.0154 - 0.014 B = 20.3125p A 10 - 9 B m4
2
Normal Stress and Shear Stress. The normal stress is caused by bending stress. Thus,
sY = -
MyA
90(0.015)
= = -42.31MPa
Iz
10.15625p A 10 - 9 B
The shear stress is caused by torsional stress.
t =
360(0.015)
Tc
=
= 84.62 MPa
J
20.3125p A 10 - 9 B
The state of stress at point A is represented by the two -dimensional element shown
in Fig. b.
In - Plane Principal Stress. sx = -42.31 MPa, sz = 0 and txz = 84.62 MPa. We have
s1, 2 =
=
sx + sz
2
;
A
a
sx - sz
2
2
b + txz
2
-42.31 - 0 2
-42.31 + 0
2
;
a
b + 84.62
2
A
2
= (-21.16 ; 87.23) MPa
s1 = 66.07 MPa
s2 = -108.38 MPa
807
200 mm
900 N
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10–76.
Continued
Maximum Distortion Energy Theory.
s1 2 - s1s2 + s2 2 = sallow 2
66.072 - 66.07(-108.38) + (-108.38)2 = sallow 2
sallow = 152.55 MPa
Thus, the factor of safety is
F.S. =
sY
250
=
= 1.64
sallow
152.55
Ans.
808
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•10–77.
The element is subjected to the stresses shown. If
sY = 36 ksi, determine the factor of safety for the loading
based on the maximum-shear-stress theory.
sx = 4 ksi
s1, 2 =
=
sy = -12 ksi
sx + sy
;
2
A
a
sx - sy
2
12 ksi
txy = -8 ksi
4 ksi
2
b + txy
2
8 ksi
4 - (-12) 2
4 - 12
2
;
a
b + (-8)
2
A
2
s2 = -15.314 ksi
s1 = 7.314 ksi
tabsmax =
7.314 - (-15.314)
s1 - s2
=
= 11.314 ksi
2
2
tallow =
sY
36
=
= 18 ksi
2
2
F.S. =
tallow
18
=
= 1.59
abs
tmax
11.314
Ans.
10–78. Solve Prob. 10–77 using the maximum-distortionenergy theory.
sx = 4 ksi
s1, 2 =
=
sy = -12 ksi
sx + sy
2
;
A
a
sx - sy
txy = -8 ksi
4 ksi
2
b + txy
2
8 ksi
4 - (-12) 2
4 - 12
2
;
a
b + (-8)
2
A
2
s1 = 7.314 ksi
s2 = -15.314 ksi
s1 2 - s1 s2 + s2 2 = a
F.S. =
2
12 ksi
sY 2
b
F.S.
362
= 1.80
A (7.314)2 - (7.314)(-15.314) + (-15.314)2
Ans.
809
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10–79. The yield stress for heat-treated beryllium copper
is sY = 130 ksi. If this material is subjected to plane stress
and elastic failure occurs when one principal stress is 145 ksi,
what is the smallest magnitude of the other principal stress?
Use the maximum-distortion-energy theory.
Maximum Distortion Energy Theory : With s1 = 145 ksi,
s21 - s1 s2 + s22 = s2Y
1452 - 145s2 + s22 = 1302
s22 - 145s2 + 4125 = 0
s2 =
-(-145) ; 2( -145)2 - 4(1)(4125)
2(1)
= 72.5 ; 33.634
Choose the smaller root, s2 = 38.9 ksi
Ans.
sy ⫽ 0.5sx
*10–80. The plate is made of hard copper, which yields at
sY = 105 ksi. Using the maximum-shear-stress theory,
determine the tensile stress sx that can be applied to the
plate if a tensile stress sy = 0.5sx is also applied.
s1 = sx
sx
1
s2 = sx
2
|s1| = sY
sx = 105 ksi
Ans.
sy ⫽ 0.5sx
•10–81.
Solve Prob. 10–80 using the maximum-distortionenergy theory.
s1 = sx
s2 =
sx
2
sx
s21 - s1 s2 + s22 = s2Y
s2x -
s2x
s2x
+
= (105)2
2
4
sx = 121 ksi
Ans.
810
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10–82. The state of stress acting at a critical point on the
seat frame of an automobile during a crash is shown in the
figure. Determine the smallest yield stress for a steel that
can be selected for the member, based on the maximumshear-stress theory.
Normal and Shear Stress: In accordance with the sign convention.
sx = 80 ksi
sy = 0
25 ksi
txy = 25 ksi
80 ksi
In - Plane Principal Stress: Applying Eq. 9-5.
s1,2 =
=
sx + sy
;
2
A
a
sx - sy
2
2
b + txy
2
80 - 0 2
80 + 0
2
;
a
b + 25
2
A
2
= 40 ; 47.170
s1 = 87.170 ksi
s2 = -7.170 ksi
Maximum Shear Stress Theory: s1 and s2 have opposite signs so
|s1 - s2| = sY
|87.170 - (-7.170)| = sY
sY = 94.3 ksi
Ans.
10–83. Solve Prob. 10–82 using the maximum-distortionenergy theory.
Normal and Shear Stress: In accordance with the sign convention.
sx = 80 ksi
sy = 0
txy = 25 ksi
In - Plane Principal Stress: Applying Eq. 9-5.
s1,2 =
=
sx + sy
;
2
A
a
25 ksi
sx - s
2
b + txy
2
2
80 ksi
80 - 0 2
80 + 0
2
;
a
b + 25
2
A
2
= 40 ; 47.170
s1 = 87.170 ksi
s2 = -7.170 ksi
Maximum Distortion Energy Theory:
s21 - s1s2 + s22 = s2Y
87.1702 - 87.170(-7.170) + (-7.170)2 = s2Y
sY = 91.0 ksi
Ans.
811
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*10–84. A bar with a circular cross-sectional area is made
of SAE 1045 carbon steel having a yield stress of
sY = 150 ksi. If the bar is subjected to a torque of
30 kip # in. and a bending moment of 56 kip # in., determine
the required diameter of the bar according to the
maximum-distortion-energy theory. Use a factor of safety
of 2 with respect to yielding.
Normal and Shear Stresses: Applying the flexure and torsion formulas.
56 A d2 B
Mc
1792
=
=
p d 4
I
pd3
A B
s =
4
t =
2
30 A d2 B
Tc
480
=
=
3
p d 4
J
pd
A B
2
2
The critical state of stress is shown in Fig. (a) or (b), where
sx =
1792
pd3
sy = 0
txy =
480
pd3
In - Plane Principal Stresses : Applying Eq. 9-5,
s1,2 =
sx + sy
2
;
A
1792
3 + 0
pd
=
=
s1 =
2
;
D
a
¢
sx - sy
2
1792
3 - 0
pd
2
2
b + txy
2
2
≤ + a
480 2
b
pd3
896
1016.47
;
pd3
pd3
1912.47
pd3
s2 = -
120.47
pd3
Maximum Distortion Energy Theory :
s21 - s1s2 + s22 = s2allow
a
1912.47 2
1912.47
120.47
120.47 2
150 2
b - a
bab + ab = a
b
3
3
3
3
2
pd
pd
pd
pd
d = 2.30 in.
Ans.
812
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•10–85.
The state of stress acting at a critical point on a
machine element is shown in the figure. Determine the
smallest yield stress for a steel that might be selected for the
part, based on the maximum-shear-stress theory.
10 ksi
The Principal stresses:
s1,2 =
=
sx + sy
;
2
A
4 ksi
a
sx - sy
2
2
b + txy
2
8 ksi
8 - (-10) 2
8 - 10
2
;
a
b + 4
2
A
2
s1 = 8.8489 ksi
s2 = -10.8489 ksi
Maximum shear stress theory: Both principal stresses have opposite sign. hence,
|s1 - s2| = sY
8.8489 - (-10.8489) = sY
sY = 19.7 ksi
Ans.
10–86. The principal stresses acting at a point on a thinwalled cylindrical pressure vessel are s1 = pr>t, s2 = pr>2t,
and s3 = 0. If the yield stress is sY, determine the maximum
value of p based on (a) the maximum-shear-stress theory and
(b) the maximum-distortion-energy theory.
a) Maximum Shear Stress Theory: s1 and s2 have the same signs, then
|s2| = sg
2
pr
2 = sg
2t
p =
2t
s
r g
|s1| = sg
2
pr
2 = sg
t
p =
t
s (Controls!)
r g
Ans.
b) Maximum Distortion Energy Theory :
s21 - s1s2 + s22 = s2g
a
pr 2
pr pr
pr 2
b - a b a b + a b = s2g
t
t
2t
2t
p =
2t
23r
sg
Ans.
813
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10–87. If a solid shaft having a diameter d is subjected to
a torque T and moment M, show that by the maximumshear-stress theory the maximum allowable shear stress is
tallow = 116>pd322M2 + T2. Assume the principal stresses
to be of opposite algebraic signs.
T
T
M
M
Section properties :
I =
p d 4
pd4
a b =
;
4 2
64
J =
p d 4
pd4
a b =
2 2
32
Thus,
d
s =
M(2 )
Mc
32 M
=
=
p d4
I
pd3
64
d
T (2 )
Tc
16 T
t =
=
=
p d4
J
pd3
32
The principal stresses :
s1,2 =
=
sx + sy
;
2
A
a
sx - sy
2
2
b + txy
2
16 M
16 M 2
16 T 2
16 M
16
;
;
2M2 + T2
a
b + a
b =
3
3
A pd
pd
p d3
pd3
p d3
Assume s1 and s2 have opposite sign, hence,
2 C 163 2M2 + T2 D
s1 - s2
16
pd
tallow =
=
=
2M2 + T2
2
2
pd3
QED
*10–88. If a solid shaft having a diameter d is subjected to a
torque T and moment M, show that by the maximum-normalstress theory the maximum allowable principal stress is
sallow = 116>pd321M + 2M2 + T22.
T
M
M
Section properties :
I =
p d4
;
64
p d4
32
J =
Stress components :
d
s =
M (2 )
Mc
32 M
= p 4 =
;
I
d
p d3
64
t =
T(d2 )
Tc
16 T
= p 4 =
J
d
p d3
32
The principal stresses :
s1,2 =
=
sx + sy
2
;
A
a
sx - sy
2
2
b + txy =
2
32 M
3 + 0
pd
2
32 M
3
;
D
¢ pd
- 0
2
2
≤ + a
16 T 2
b
p d3
16 M
16
;
2M2 + T2
p d3
p d3
Maximum normal stress theory. Assume s1 7 s2
sallow = s1 =
=
16 M
16
+
2M2 + T2
p d3
p d3
16
[M + 2M2 + T2]
p d3
QED
814
T
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•10–89.
The shaft consists of a solid segment AB and a
hollow segment BC, which are rigidly joined by the
coupling at B. If the shaft is made from A-36 steel,
determine the maximum torque T that can be applied
according to the maximum-shear-stress theory. Use a factor
of safety of 1.5 against yielding.
A
B
T
C
Shear Stress: This is a case of pure shear, and the shear stress is contributed by
p
torsion. For the hollow segment, Jh =
A 0.054 - 0.044 B = 1.845p A 10 - 6 B m4. Thus,
2
(tmax)h =
T(0.05)
Tch
=
= 8626.28T
Jh
1.845p A 10 - 6 B
For the solid segment, Js =
(tmax)s =
p
A 0.044 B = 1.28p A 10 - 6 B m4. Thus,
2
T(0.04)
Tcs
=
= 9947.18T
Js
1.28p A 10 - 6 B
By comparision, the points on the surface of the solid segment are critical and their
state of stress is represented on the element shown in Fig. a.
In - Plane Principal Stress. sx = sy = 0 and txy = 9947.18T. We have
s1,2 =
=
sx + sy
2
;
C
¢
sx - sy
2
2
≤ + t2xy
0 - 0 2
0 + 0
;
¢
≤ + (9947.18T)2
2
C
2
s1 = 9947.18T
s2 = -9947.18T
Maximum Shear Stress Theory.
sallow =
80 mm
sY
250
=
= 166.67 MPa
F.S.
1.5
Since s1 and s2 have opposite sings,
|s1 - s2| = sallow
9947.18T - (-9947.18T) = 166.67 A 106 B
T = 8377.58 N # m = 8.38 kN # m
Ans.
815
80 mm
100 mm
T
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10–90. The shaft consists of a solid segment AB and a
hollow segment BC, which are rigidly joined by the
coupling at B. If the shaft is made from A-36 steel,
determine the maximum torque T that can be applied
according to the maximum-distortion-energy theory. Use a
factor of safety of 1.5 against yielding.
A
B
T
C
Shear Stress. This is a case of pure shear, and the shear stress is contributed by
p
torsion. For the hollow segment, Jh =
A 0.054 - 0.044 B = 1.845p A 10 - 6 B m4. Thus,
2
(tmax)h =
T(0.05)
Tch
=
= 8626.28T
Jh
1.845p A 10 - 6 B
For the solid segment, Js =
(tmax)s =
p
A 0.044 B = 1.28p A 10 - 6 B m4. Thus,
2
T(0.04)
Tcs
=
= 9947.18T
Js
1.28p A 10 - 6 B
By comparision, the points on the surface of the solid segment are critical and their
state of stress is represented on the element shown in Fig. a.
In - Plane Principal Stress. sx = sy = 0 and txy = 9947.18T. We have
s1,2 =
=
sx + sy
2
;
C
¢
sx - sy
2
2
≤ + t2xy
0 - 0 2
0 + 0
;
¢
≤ + (9947.18T)2
2
C
2
s1 = 9947.18T
s2 = -9947.18T
Maximum Distortion Energy Theory.
sallow =
80 mm
sY
250
=
= 166.67 MPa
F.S.
1.5
Then,
s1 2 - s1s2 + s2 2 = sallow 2
(9947.18T)2 - (9947.18T)( -9947.18T) + (-9947.18T)2 = C 166.67 A 106 B D 2
T = 9673.60 N # m = 9.67 kN # m
Ans.
816
80 mm
100 mm
T
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10–91. The internal loadings at a critical section along the
steel drive shaft of a ship are calculated to be a torque of
2300 lb # ft, a bending moment of 1500 lb # ft, and an axial
thrust of 2500 lb. If the yield points for tension and shear are
sY = 100 ksi and tY = 50 ksi, respectively, determine the
required diameter of the shaft using the maximum-shearstress theory.
2300 lb⭈ft
2500 lb
p
I = c4
4
A = p c2
sA =
p
J = c4
2
1500(12)(c)
P
Mc
2500
2500
72 000
+
b = -a
+
b
+
= -a
pc4
A
I
p c2
p c2
p c3
4
tA =
2300(12)(c)
Tc
55 200
=
=
p c4
J
p c3
2
s1,2 =
sx + sy
= -a
2
;
A
a
sx - sy
2
2
b + txy
2
2500 c + 72 000
2500c + 72 000 2
55200 2
b ;
a
b + a
b
3
3
A
2p c
2p c
p c3
(1)
Assume s1 and s2 have opposite signs:
|s1 - s2| = sg
2500c + 72 000 2
55 200 2
3
b + a
b = 100(10 )
3
A
2p c
p c3
2
a
(2500c + 72000)2 + 1104002 = 10 000(106)p2 c6
6.25c2 + 360c + 17372.16 - 10 000p2 c6 = 0
By trial and error:
c = 0.750 57 in.
Substitute c into Eq. (1):
s1 = 22 193 psi
s2 = -77 807 psi
s1 and s2 are of opposite signs
OK
Therefore,
d = 1.50 in.
Ans.
817
1500 lb⭈ft
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*10–92. The gas tank has an inner diameter of 1.50 m and
a wall thickness of 25 mm. If it is made from A-36 steel and
the tank is pressured to 5 MPa, determine the factor of
safety against yielding using (a) the maximum-shear-stress
theory, and (b) the maximum-distortion-energy theory.
(a) Normal Stress. Since
0.75
r
=
= 30 7 10, thin - wall analysis can be used.We have
t
0.025
s1 = sh =
5(0.75)
pr
=
= 150 MPa
t
0.025
s2 = slong =
pr
5(0.75)
=
= 75 MPa
2t
2(0.025)
Maximum Shear Stress Theory. s1 and s2 have the sign. Thus,
|s1| = sallow
sallow = 150 MPa
The factor of safety is
F.S. =
sY
250
=
= 1.67
sallow
150
Ans.
(b) Maximum Distortion Energy Theory.
s1 2 - s1s2 + s2 2 = sallow 2
1502 - 150(75) + 752 = sallow 2
sallow = 129.90 MPa
The factor of safety is
F.S. =
sY
250
=
= 1.92
sallow
129.90
Ans.
818
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•10–93.
The gas tank is made from A-36 steel and
has an inner diameter of 1.50 m. If the tank is designed
to withstand a pressure of 5 MPa, determine the required
minimum wall thickness to the nearest millimeter using
(a) the maximum-shear-stress theory, and (b) maximumdistortion-energy theory. Apply a factor of safety of 1.5
against yielding.
(a) Normal Stress. Assuming that thin - wall analysis is valid, we have
5 A 10 B (0.75)
3.75 A 10 B
pr
=
=
t
t
t
6
s1 = sh =
6
5 A 10 B (0.75)
1.875 A 10 B
pr
=
=
2t
2t
t
6
s2 = slong =
6
Maximum Shear Stress Theory.
250 A 10 B
sY
=
= 166.67 A 106 B Pa
FS.
1.5
6
sallow =
s1 and s2 have the same sign. Thus,
|s1| = sallow
3.75 A 106 B
= 166.67 A 106 B
t
t = 0.0225 m = 22.5 mm
Since
Ans.
0.75
r
=
= 33.3 7 10, thin - wall analysis is valid.
t
0.0225
(b) Maximum Distortion Energy Theory.
250 A 10 B
sY
=
= 166.67 A 106 B Pa
F.S.
1.5
6
sallow =
Thus,
s1 2 - s1s2 + s2 2 = sallow 2
C
3.75 A 106 B
t
3.2476 A 106 B
t
2
S - C
3.75 A 106 B
t
SC
1.875 A 106 B
t
S + C
1.875 A 106 B
t
2
S = c166.67 A 106 B d
= 166.67 A 106 B
t = 0.01949 m = 19.5 mm
Since
2
Ans.
0.75
r
=
= 38.5 7 10, thin - wall analysis is valid.
t
0.01949
819
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10–94. A thin-walled spherical pressure vessel has an
inner radius r, thickness t, and is subjected to an internal
pressure p. If the material constants are E and n, determine
the strain in the circumferential direction in terms of the
stated parameters.
s1 = s2 =
pr
2t
e1 = e2 = e =
e =
1
(s - vs)
E
pr
1 - v pr
1 - v
s =
a b =
(1 - v)
E
E
2t
2Et
Ans.
10–95. The strain at point A on the shell has components
Px = 250(10 - 6), Py = 400(10 - 6), gxy = 275(10 - 6), Pz = 0.
Determine (a) the principal strains at A, (b) the maximum
shear strain in the x–y plane, and (c) the absolute maximum
shear strain.
ex = 250(10 - 6)
A(250, 137.5)10 - 6
ey = 400(10 - 6)
gxy = 275(10 - 6)
y
A
gxy
2
= 137.5(10 - 6)
C(325, 0)10 - 6
R = a 2(325 - 250)2 + (137.5)2 b10 - 6 = 156.62(10 - 6)
a)
e1 = (325 + 156.62)10 - 6 = 482(10 - 6)
Ans.
e2 = (325 - 156.62)10 - 6 = 168(10 - 6)
Ans.
b)
g max
in-plane
= 2R = 2(156.62)(10 - 6) = 313(10 - 6)
Ans.
c)
gabs
max
2
gabs
max
=
482(10 - 6)
2
= 482(10 - 6)
Ans.
820
x
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*10–96. The principal plane stresses acting at a point are
shown in the figure. If the material is machine steel having a
yield stress of sY = 500 MPa, determine the factor of
safety with respect to yielding if the maximum-shear-stress
theory is considered.
100 MPa
150 MPa
Have, the in plane principal stresses are
s1 = sy = 100 MPa
s2 = sx = -150 MPa
Since s1 and s2 have same sign,
F.S =
sy
=
|s1 - s2|
500
= 2
|100 - (-150)|
Ans.
•10–97.
The components of plane stress at a critical point
on a thin steel shell are shown. Determine if failure
(yielding) has occurred on the basis of the maximumdistortion-energy theory. The yield stress for the steel is
sY = 650 MPa.
340 MPa
65 MPa
55 MPa
sx = -55 MPa
s1, 2 =
=
sy = 340 MPa
sx + sy
2
;
A
a
sx - sy
2
txy = 65 MPa
2
b + txy
2
-55 - 340 2
-55 + 340
2
;
a
b + 65
2
A
2
s1 = 350.42 MPa
s2 = -65.42 MPa
(s1 2 - s1s2 + s2 ) = [350.422 - 350.42(-65.42) + ( -65.42)2]
= 150 000 6 s2Y = 422 500
OK
No.
Ans.
821
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10–98. The 60° strain rosette is mounted on a beam.
The following readings are obtained for each gauge:
Pa = 600110-62, Pb = -700110-62, and Pc = 350110-62.
Determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and average normal strain. In
each case show the deformed element due to these strains.
a
60⬚
60⬚
b
Strain Rosettes (60º): Applying Eq. 10-15 with ex = 600 A 10 - 6 B ,
eb = -700 A 10
-6
c
B , ec = 350 A 10 B , ua = 150°, ub = -150° and uc = -90°,
-6
350 A 10 - 6 B = ex cos2 (-90°) + ey sin2( -90°) + gxy sin (-90°) cos ( -90°)
ey = 350 A 10 - 6 B
600 A 10 - 6 B = ex cos2 150° + 350 A 10 - 6 B sin2 150° + gxy sin 150° cos 150°
512.5 A 10 - 6 B = 0.75 ex - 0.4330 gxy
[1]
-787.5 A 10 - 6 B = 0.75ex + 0.4330 gxy
[2]
-700 A 10 - 6 B = ex cos2 ( -150°) + 350 A 10 - 6 B sin2(-150°) + gxy sin (-150°) cos (-150°)
Solving Eq. [1] and [2] yields ex = -183.33 A 10 - 6 B
gxy = -1501.11 A 10 - 6 B
Construction of she Circle: With ex = -183.33 A 10 - 6 B , ey = 350 A 10 - 6 B , and
gxy
= -750.56 A 10 - 6 B .
2
eavg =
ex + ey
2
= a
-183.33 + 350
b A 10 - 6 B = 83.3 A 10 - 6 B
2
Ans.
The coordinates for reference points A and C are
A( -183.33, -750.56) A 10 - 6 B
C(83.33, 0) A 10 - 6 B
The radius of the circle is
R = a 2(183.33 + 83.33)2 + 750.562 b A 10 - 6 B = 796.52 A 10 - 6 B
a)
In-plane Principal Strain: The coordinates of points B and D represent e1 and e2,
respectively.
e1 = (83.33 + 796.52) A 10 - 6 B = 880 A 10 - 6 B
Ans.
e2 = (83.33 - 796.52) A 10 - 6 B = -713 A 10 - 6 B
Ans.
Orientation of Principal Strain: From the circle,
tan 2uP1 =
750.56
= 2.8145
183.33 + 83.33
2uP2 = 70.44°
2uP1 = 180° - 2uP2
uP =
180° - 70.44°
= 54.8° (Clockwise)
2
Ans.
822
60⬚
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10–98.
Continued
b)
Maximum In - Plane Shear Strain: Represented by the coordinates of point E on
the circle.
g max
in-plane
2
g max
in-plane
= -R = -796.52 A 10 - 6 B
= -1593 A 10 - 6 B
Ans.
Orientation of Maximum In-Plane Shear Strain: From the circle.
tan 2uP =
183.33 + 83.33
= 0.3553
750.56
uP = 9.78° (Clockwise)
Ans.
10–99. A strain gauge forms an angle of 45° with the axis
of the 50-mm diameter shaft. If it gives a reading of
P = -200110-62 when the torque T is applied to the shaft,
determine the magnitude of T. The shaft is made from
A-36 steel.
T
45⬚
Shear Stress. This is a case of pure shear, and the shear stress developed is
p
contributed by torsional shear stress. Here, J =
A 0.0254 B = 0.1953125p A 10 - 6 B m4.
2
Then
0.128 A 10 B T
T(0.025)
Tc
=
=
t =
-6
p
J
0.1953125p A 10 B
T
6
The state of stress at points on the surface of the shaft can be represented by the
element shown in Fig. a.
Shear Strain: For pure shear ex = ey = 0. We obtain,
ea = ex cos2ua + ey sin2ua + gxysin ua cos ua
-200 A 10 - 6 B = 0 + 0 + gxy sin 45° cos 45°
gxy = -400 A 10 - 6 B
Shear Stress and Strain Relation: Applying Hooke’s Law for shear,
txy = Ggxy
-
0.128 A 106 B T
p
= 75 A 109 B C -400 A 10 - 6 B D
T = 736 N # m
Ans.
823
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*10–100. The A-36 steel post is subjected to the forces
shown. If the strain gauges a and b at point A give readings
of Pa = 300110-62 and Pb = 175110-62, determine the
magnitudes of P1 and P2.
P1
P2
a
A
Internal Loadings: Considering the equilibrium of the free - body diagram of the 1 in.
post’s segment, Fig. a,
P2 - V = 0
V = P2
+ c ©Fy = 0;
N - P1 = 0
N = P1
a + ©MO = 0;
M + P2(2) = 0
M = 2P2
Section Properties: The cross - sectional area and the moment of inertia about the
bending axis of the post’s cross - section are
A = 4(2) = 8 in2
I =
1
(2) A 43 B = 10.667 in4
12
Referring to Fig. b,
A Qy B A = x¿A¿ = 1.5(1)(2) = 3 in3
Normal and Shear Stress: The normal stress is a combination of axial and bending
stress.
sA =
2P2(12)(1)
MxA
P1
N
+
= +
= 2.25P2 - 0.125P1
A
I
8
10.667
The shear stress is caused by transverse shear stress.
tA =
VQA
P2(3)
=
= 0.140625P2
It
10.667(2)
Thus, the state of stress at point A is represented on the element shown in Fig. c.
Normal and Shear Strain: With ua = 90° and ub = 45°, we have
ea = ex cos2ua + ey sin2ua + gxysin ua cos ua
300 A 10 - 6 B = ex cos2 90° + ey sin2 90° + gxysin 90° cos 90°
ey = 300 A 10 - 6 B eb = ex cos2ub + ey sin2 ub + gxysin ub cos ub
175 A 10 - 6 B = ex cos2 45° + 300 A 10 - 6 B sin2 45° + gxy sin 45°cos 45°
ex + gxy = 50 A 10 - 6 B
(1)
824
A
1 in.
b 45⬚
c
+ ©F = 0;
:
x
2 in.
2 ft
A
4 in.
c
Section c– c
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10–100.
Continued
Since sy = sz = 0, ex = -vey = -0.32(300) A 10 - 6 B = -96 A 10 - 6 B
Then Eq. (1) gives
gxy = 146 A 10 - 6 B
Stress and Strain Relation: Hooke’s Law for shear gives
tx = Ggxy
0.140625P2 = 11.0 A 103 B C 146 A 10 - 6 B D
P2 = 11.42 kip = 11.4 kip
Ans.
Since sy = sz = 0, Hooke’s Law gives
sy = Eey
2.25(11.42) - 0.125P1 = 29.0 A 103 B C 300 A 10 - 6 B D
P1 = 136 kip
Ans.
825
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Page 826
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10–101. A differential element is subjected to plane strain
that has the following components: Px = 950110-62, Py =
420110-62, gxy = -325110-62. Use the strain-transformation
equations and determine (a) the principal strains and (b) the
maximum in-plane shear strain and the associated average
strain. In each case specify the orientation of the element
and show how the strains deform the element.
e1, 2 =
ex + ey
;
2
= c
A
a
ex - ey
2
2
b + gxy
2
950 - 420 2
-325 2
950 + 420
-6
;
a
b + a
b d(10 )
2
A
2
2
e1 = 996(10 - 6)
Ans.
e2 = 374(10 - 6)
Ans.
Orientation of e1 and e2 :
gxy
tan 2uP =
ex - ey
-325
950 - 420
=
uP = -15.76°, 74.24°
Use Eq. 10.5 to determine the direction of e1 and e2.
ex¿ =
ex + ey
ex - ey
+
2
2
cos 2u +
gxy
2
sin 2u
u = uP = -15.76°
ex¿ = b
( -325)
950 - 420
950 + 420
+
cos (-31.52°) +
sin (-31.52°) r (10 - 6) = 996(10 - 6)
2
2
2
uP1 = -15.8°
Ans.
uP2 = 74.2°
Ans.
b)
gmax
in-plane
2
gmax
in-plane
eavg =
=
A
= 2c
a
ex - ey
A
2
a
ex + ey
2
b + a
2
gxy
2
b
2
950 - 420 2
-325 2
-6
-6
b + a
b d(10 ) = 622(10 )
2
2
Ans.
= a
Ans.
950 + 420
b (10 - 6) = 685(10 - 6)
2
826
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10–101.
Continued
Orientation of gmax :
-(ex - ey)
tan 2uP =
-(950 - 420)
-325
=
gxy
uP = 29.2° and uP = 119°
Ans.
Use Eq. 10.6 to determine the sign of
ex - ey
gx¿y¿
= -
2
sin 2u +
2
gxy
2
g max
in-plane
:
cos 2u
u = uP = 29.2°
gx¿y¿ = 2 c
-(950 - 420)
-325
sin (58.4°) +
cos (58.4°) d(10 - 6)
2
2
gxy = -622(10 - 6)
10–102. The state of plane strain on an element is
Px = 400110-62, Py = 200110-62, and gxy = -300110-62.
Determine the equivalent state of strain on an element at
the same point oriented 30° clockwise with respect to the
original element. Sketch the results on the element.
y
Pydy
dy
Stress Transformation Equations:
ex = 400 A 10 - 6 B
ey = 200 A 10 - 6 B
gxy = -300 A 10 - 6 B
u = -30°
gxy
2
dx
We obtain,
ex¿ =
ex + ey
+
2
= B
ex - ey
2
cos 2u +
gxy
2
sin 2u
400 + 200
400 - 200
-300
+
cos (-60°) + a
b sin (-60°) R A 10 - 6 B
2
2
2
= 480 A 10 - 6 B
gx¿y¿
2
= -¢
Ans.
ex - ey
2
≤ sin 2u +
gxy
2
cos 2u
gx¿y¿ = [-(400 - 200) sin ( -60°) + (-300) cos ( -60°)] A 10 - 6 B
= 23.2 A 10 - 6 B
ey¿ =
ex + ey
= B
2
Ans.
ex - ey
-
2
cos 2u -
gxy
2
sin 2u
400 + 200
400 - 200
-300
cos ( -60°) - a
b sin (-60°) R A 10 - 6 B
2
2
2
= 120 A 10 - 6 B
Ans.
827
gxy
2
x
Pxdx
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Page 828
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10–103. The state of plane strain on an element is
Px = 400110-62, Py = 200110-62, and gxy = -300110-62.
Determine the equivalent state of strain, which represents
(a) the principal strains, and (b) the maximum in-plane
shear strain and the associated average normal strain.
Specify the orientation of the corresponding element at the
point with respect to the original element. Sketch the results
on the element.
y
Pydy
dy
Construction of the Circle: ex = 400 A 10 - 6 B , ey = 200 A 10 - 6 B , and
gxy
2
= -150 A 10 - 6 B .
Thus,
eavg =
ex + ey
2
= a
400 + 200
b A 10 - 6 B = 300 A 10 - 6 B
2
Ans.
The coordinates for reference points A and the center C of the circle are
A(400, -150) A 10 - 6 B
C(300, 0) A 10 - 6 B
The radius of the circle is
R = CA = 2(400 - 300)2 + (-150)2 = 180.28 A 10 - 6 B
Using these results, the circle is shown in Fig. a.
In - Plane Principal Stresses: The coordinates of points B and D represent e1 and e2,
respectively. Thus,
e1 = (300 + 180.28) A 10 - 6 B = 480 A 10 - 6 B
Ans.
e2 = (300 - 180.28) A 10 - 6 B = 120 A 10 - 6 B
Ans.
Orientation of Principal Plane: Referring to the geometry of the circle,
tan 2 A up B 1 =
150
= 1.5
400 - 300
A up B 1 = 28.2° (clockwise)
Ans.
The deformed element for the state of principal strains is shown in Fig. b.
Maximum In - Plane Shear Stress: The coordinates of point E represent eavg and
gmax
. Thus
in-plane
gmax
in-plane
2
gmax
in-plane =
= -R = -180.28 A 10 - 6 B
-361 A 10 - 6 B
Ans.
Orientation of the Plane of Maximum In - Plane Shear Strain: Referring to the
geometry of the circle,
tan 2us =
400 - 300
= 0.6667
150
uS = 16.8° (counterclockwise)
Ans.
The deformed element for the state of maximum in - plane shear strain is shown in
Fig. c.
828
gxy
2
gxy
2
dx
x
Pxdx
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10–103.
Continued
829
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11–1. The simply supported beam is made of timber that
has an allowable bending stress of sallow = 6.5 MPa and an
allowable shear stress of tallow = 500 kPa. Determine its
dimensions if it is to be rectangular and have a height-towidth ratio of 1.25.
8 kN/m
2m
Ix =
1
(b)(1.25b)3 = 0.16276b4
12
Qmax = y¿A¿ = (0.3125b)(0.625b)(b) = 0.1953125b3
Assume bending moment controls:
Mmax = 16 kN # m
sallow =
Mmax c
I
6.5(106) =
16(103)(0.625b)
0.16276b4
b = 0.21143 m = 211 mm
Ans.
h = 1.25b = 264 mm
Ans.
Check shear:
Qmax = 1.846159(10 - 3) m3
I = 0.325248(10 - 3) m4
tmax =
VQmax
16(103)(1.846159)(10 - 3)
= 429 kPa 6 500 kPa‚ OK
=
It
0.325248(10 - 3)(0.21143)
830
4m
2m
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Page 831
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11–2. The brick wall exerts a uniform distributed load
of 1.20 kip>ft on the beam. If the allowable bending stress
is sallow = 22 ksi and the allowable shear stress is
tallow = 12 ksi, select the lightest wide-flange section with
the shortest depth from Appendix B that will safely support
the load.
1.20 kip/
4 ft
10 ft
ft
6 ft
b
Bending Stress: From the moment diagram, Mmax = 44.55 kip # ft. Assuming
bending controls the design and applying the flexure formula.
Sreq d =
=
44.55 (12)
= 24.3 in3
22
W12 * 22
A Sx = 25.4 in3, d = 12.31 in., tw = 0.260 in. B
V
for the W12 * 22 wide tw d
flange section. From the shear diagram, Vmax = 6.60 kip.
Shear Stress: Provide a shear stress check using t =
tmax =
=
Vmax
tw d
6.60
0.260(12.31)
= 2.06 ksi 6 tallow = 12 ksi (O.K!)
Hence,
Use
9 in.
0.5 in.
Mmax
sallow
Two choices of wide flange section having the weight 22 lb>ft can be made. They
are W12 * 22 and W14 * 22. However, W12 * 22 is the shortest.
Select
0.5 in.
0.5 in.
Ans.
W12 * 22
831
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–3. The brick wall exerts a uniform distributed load
of 1.20 kip>ft on the beam. If the allowable bending stress
is sallow = 22 ksi, determine the required width b of the
flange to the nearest 14 in.
1.20 kip/
4 ft
10 ft
ft
6 ft
b
0.5 in.
0.5 in.
9 in.
0.5 in.
Section Property:
I =
1
1
(b) A 103 B (b - 0.5) A 93 B = 22.583b + 30.375
12
12
Bending Stress: From the moment diagram, Mmax = 44.55 kip # ft.
sallow =
22 =
Mmax c
I
44.55(12)(5)
22.583b + 30.375
b = 4.04 in.
Use
b = 4.25 in.
Ans.
832
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Page 833
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*11–4. Draw the shear and moment diagrams for the
shaft, and determine its required diameter to the nearest
1
4 in. if sallow = 7 ksi and tallow = 3 ksi. The bearings at A
and D exert only vertical reactions on the shaft. The loading
is applied to the pulleys at B, C, and E.
14 in.
20 in.
15 in.
12 in.
E
A
C
B
D
35 lb
80 lb
110 lb
sallow =
Mmax c
I
7(103) =
1196 c
p 4 ;
4 c
c = 0.601 in.
d = 2c = 1.20 in.
Use d = 1.25 in.
Ans.
Check shear:
2
tmax =
0.625
108(4(0.625)
Vmax Q
3p )(p)( 2 )
= 117 psi 6 3 ksi OK
=
p
4
It
4 (0.625) (1.25)
•11–5.
Select the lightest-weight steel wide-flange beam
from Appendix B that will safely support the machine loading
shown. The allowable bending stress is sallow = 24 ksi and
the allowable shear stress is tallow = 14 ksi.
2 ft
Bending Stress: From the moment diagram, Mmax = 30.0 kip # ft.
Assume bending controls the design. Applying the flexure formula.
Sreq¿d =
=
Select
W12 * 16
Mmax
sallow
30.0(12)
= 15.0 in3
24
A Sx = 17.1 in3, d = 11.99 in., tw = 0.220 in. B
V
for the W12 * 16 wide tw d
flange section. From the shear diagram, Vmax = 10.0 kip
Shear Stress: Provide a shear stress check using t =
tmax =
=
Vmax
tw d
10.0
0.220(11.99)
= 3.79 ksi 6 tallow = 14 ksi (O.K!)
Hence,
Use
5 kip
5 kip
Ans.
W12 * 16
833
2 ft
5 kip
2 ft
5 kip
2 ft
2 ft
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Page 834
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11–6. The compound beam is made from two sections,
which are pinned together at B. Use Appendix B and select
the lightest-weight wide-flange beam that would be safe for
each section if the allowable bending stress is sallow = 24 ksi
and the allowable shear stress is tallow = 14 ksi. The beam
supports a pipe loading of 1200 lb and 1800 lb as shown.
C
A
B
6 ft
Bending Stress: From the moment diagram, Mmax = 19.2 kip # ft for member AB.
Assuming bending controls the design, applying the flexure formula.
Sreq¿d =
=
Select
Mmax
sallow
19.2(12)
= 9.60 in3
24
A Sx = 10.9 in3, d = 9.87 in., tw = 0.19 in. B
W10 * 12
For member BC, Mmax = 8.00 kip # ft.
Sreq¿d =
=
Select
Mmax
sallow
8.00(12)
= 4.00 in3
24
A Sx = 5.56 in3, d = 5.90 in., tw = 0.17 in. B
W6 * 9
V
for the W10 * 12 widetw d
flange section for member AB. From the shear diagram, Vmax = 2.20 kip.
Shear Stress: Provide a shear stress check using t =
tmax =
=
Vmax
tw d
2.20
0.19(9.87)
= 1.17 ksi 6 tallow = 14 ksi (O.K!)
Use
Ans.
W10 * 12
For member BC (W6 * 9), Vmax = 1.00 kip.
tmax =
=
Vmax
tw d
1.00
0.17(5.90)
= 0.997 ksi 6 tallow = 14 ksi (O.K!)
Hence,
Use
1800 lb
1200 lb
W6 * 9
Ans.
834
6 ft
8 ft
10 ft
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Page 835
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11–7. If the bearing pads at A and B support only vertical
forces, determine the greatest magnitude of the uniform
distributed loading w that can be applied to the beam.
sallow = 15 MPa, tallow = 1.5 MPa.
w
A
B
1m
1m
150 mm
25 mm
150 mm
25 mm
The location of c, Fig. b, is
y =
0.1625(0.025)(0.15) + 0.075(0.15)(0.025)
©yA
=
©A
0.025(0.15) + 0.15(0.025)
= 0.11875 m
I =
1
(0.025)(0.153) + (0.025)(0.15)(0.04375)2
12
+
1
(0.15)(0.0253) + 0.15(0.025)(0.04375)2
12
= 21.58203125(10 - 6) m4
Referring to Fig. b,
Qmax = y¿A¿ = 0.059375 (0.11875)(0.025)
= 0.176295313(10 - 4) m3
Referring to the moment diagram, Mmax = 0.28125 w. Applying the Flexure
formula with C = y = 0.11875 m,
sallow =
Mmax c
;
I
15(106) =
0.28125w(0.11875)
21.582(10 - 6)
W = 9.693(103) N>m
Referring to shear diagram, Fig. a, Vmax = 0.75 w.
tallow =
Vallow Qmax
;
It
1.5(106) =
0.75w C 0.17627(10 - 3) D
21.582(10 - 6)(0.025)
W = 6.122(103) N>m
= 6.12 kN>m (Control!)
Ans.
835
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Page 836
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*11–8. The simply supported beam is made of timber that
has an allowable bending stress of sallow = 1.20 ksi and an
allowable shear stress of tallow = 100 psi. Determine its
smallest dimensions to the nearest 18 in. if it is rectangular
and has a height-to-width ratio of 1.5.
12 kip/ft
B
A
3 ft
3 ft
1.5 b
b
The moment of inertia of the beam’s cross-section about the neutral axis is
1
(b)(1.5b)3 = 0.28125b4. Referring to the moment diagram,
I =
12
Mmax = 45.375 kip # ft.
sallow =
Mmax c
;
I
1.2 =
45.375(12)(0.75b)
0.28125b4
b = 10.66 in
Referring to Fig. b, Qmax = y¿A¿ = 0.375b (0.75b)(b) = 0.28125b3. Referring to the
shear diagram, Fig. a, Vmax = 33 kip.
tmax =
Vmax Qmax
;
It
100 =
33(103)(0.28125b3)
0.28125b4(b)
b = 18.17 in (Control!)
Thus, use
b = 18
1
in
4
Ans.
836
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
•11–9.
Select the lightest-weight W12 steel wide-flange
beam from Appendix B that will safely support the loading
shown, where P = 6 kip. The allowable bending stress
is sallow = 22 ksi and the allowable shear stress is
tallow = 12 ksi.
P
P
9 ft
From the Moment Diagram, Fig. a, Mmax = 54 kip # ft.
Sreq¿d =
Mmax
sallow
=
54(12)
22
= 29.45 in3
Select W12 * 26
C Sx = 33.4 in3, d = 12.22 in and tw = 0.230 in. D
From the shear diagram, Fig. a, Vmax = 7.5 kip. Provide the shear-stress check
for W 12 * 26,
tmax =
=
Vmax
tw d
7.5
0.230(12.22)
= 2.67 ksi 6 tallow = 12 ksi (O.K!)
Hence
Use
Ans.
W12 * 26
837
6 ft
6 ft
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11–10. Select the lightest-weight W14 steel wide-flange
beam having the shortest height from Appendix B that
will safely support the loading shown, where P = 12 kip.
The allowable bending stress is sallow = 22 ksi and the
allowable shear stress is tallow = 12 ksi.
P
P
9 ft
From the moment diagram, Fig. a, Mmax = 108 kip # ft.
Mmax
sallow
Sreq¿d =
108(12)
22
=
= 58.91 in3
Select W14 * 43
C Sx = 62.7 in3, d = 13.66 in and tw = 0.305 in. D
From the shear diagram, Fig. a, Vmax = 15 kip . Provide the shear-stress check
for W14 * 43 ,
tmax =
=
Vmax
tw d
15
0.305(13.66)
= 3.60 ksi 6 tallow = 12 ksi‚ (O.K!)
Hence,
Use
Ans.
W14 * 43
838
6 ft
6 ft
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11–11. The timber beam is to be loaded as shown. If the ends
support only vertical forces, determine the greatest magnitude
of P that can be applied. sallow = 25 MPa, tallow = 700 kPa.
150 mm
30 mm
120 mm
40 mm
P
4m
A
y =
(0.015)(0.150)(0.03) + (0.09)(0.04)(0.120)
= 0.05371 m
(0.150)(0.03) + (0.04)(0.120)
I =
1
1
(0.150)(0.03)3 + (0.15)(0.03)(0.05371 - 0.015)2 +
(0.04)(0.120)3 +
12
12
B
(0.04)(0.120)(0.09 - 0.05371)2 = 19.162(10 - 6) m4
Maximum moment at center of beam:
Mmax =
P
(4) = 2P
2
Mc
;
I
s =
25(106) =
(2P)(0.15 - 0.05371)
19.162(10 - 6)
P = 2.49 kN
Maximum shear at end of beam:
Vmax =
P
2
VQ
;
t =
It
700(103) =
P 1
C (0.15 - 0.05371)(0.04)(0.15 - 0.05371) D
2 2
19.162(10 - 6)(0.04)
P = 5.79 kN
Thus,
P = 2.49 kN
Ans.
839
4m
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Page 840
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*11–12. Determine the minimum width of the beam to
the nearest 14 in. that will safely support the loading of
P = 8 kip. The allowable bending stress is sallow = 24 ksi
and the allowable shear stress is tallow = 15 ksi.
P
6 ft
6 ft
6 in.
B
A
Beam design: Assume moment controls.
sallow =
Mc
;
I
24 =
48.0(12)(3)
1
3
12 (b)(6 )
b = 4 in.
Ans.
Check shear:
8(1.5)(3)(4)
VQ
= 0.5 ksi 6 15 ksi OK
= 1
3
It
12 (4)(6 )(4)
tmax =
•11–13.
Select the shortest and lightest-weight steel wideflange beam from Appendix B that will safely support the
loading shown.The allowable bending stress is sallow = 22 ksi
and the allowable shear stress is tallow = 12 ksi.
10 kip
6 kip
4 kip
A
B
4 ft
Beam design: Assume bending moment controls.
Sreq¿d =
60.0(12)
Mmax
=
= 32.73 in3
sallow
22
Select a W 12 * 26
Sx = 33.4 in3, d = 12.22 in., tw = 0.230 in.
Check shear:
tavg =
V
10.5
=
= 3.74 ksi 6 12 ksi
Aweb
(12.22)(0.230)
Use W 12 * 26
Ans.
840
4 ft
4 ft
4 ft
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Page 841
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11–14. The beam is used in a railroad yard for loading and
unloading cars. If the maximum anticipated hoist load is
12 kip, select the lightest-weight steel wide-flange section
from Appendix B that will safely support the loading. The
hoist travels along the bottom flange of the beam,
1 ft … x … 25 ft, and has negligible size. Assume the beam
is pinned to the column at B and roller supported at A.
The allowable bending stress is sallow = 24 ksi and
the allowable shear stress is tallow = 12 ksi.
x
27 ft
A
B
12 kip
15 ft
C
Maximum moment occurs when load is in the center of beam.
Mmax = (6 kip)(13.5 ft) = 81 lb # ft
sallow =
M
;
S
24 =
81(12)
Sreq¿d
Sreq¿d = 40.5 in3
Select a W 14 * 30, Sx = 42.0 in3, d = 13.84 in, tw = 0.270 in.
At x = 1 ft, V = 11.56 kip
t =
11.36
V
=
= 3.09 ksi 6 12 ksi
Aweb
(13.84)(0.270)
Use W14 * 30
Ans.
841
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11–15. The simply supported beam is made of timber that
has an allowable bending stress of sallow = 960 psi and an
allowable shear stress of tallow = 75 psi. Determine its
dimensions if it is to be rectangular and have a heightto-width ratio of 1.25.
5 kip/ft
6 ft
1
I =
(b)(1.25b)3 = 0.16276b4
12
b
Assume bending moment controls:
Mmax = 60 kip # ft
960 =
Mmax
Sreq¿d
60(103)(12)
0.26042 b3
b = 14.2 in.
Check shear:
tmax =
1.5(15)(103)
1.5V
=
= 88.9 psi 7 75 psi NO
A
(14.2)(1.25)(14.2)
Shear controls:
tallow =
6 ft
1.25 b
I
0.16276b4
Sreq¿d =
=
= 0.26042b3
c
0.625b
sallow =
B
A
1.5(15)(103)
1.5V
=
A
(b)(1.25b)
b = 15.5 in.
Ans.
842
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Page 843
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*11–16. The simply supported beam is composed of two
W12 * 22 sections built up as shown. Determine the
maximum uniform loading w the beam will support if
the allowable bending stress is sallow = 22 ksi and the
allowable shear stress is tallow = 14 ksi.
w
Section properties:
24 ft
For W12 * 22 (d = 12.31 in. Ix = 156 in4 tw = 0.260 in. A = 6.48 in2)
I = 2c 156 + 6.48a
S =
12.31 2
b d = 802.98 in4
2
I
802.98
=
= 65.23 in3
c
12.31
Maximum Loading: Assume moment controls.
M = sallowS(72 w)(12) = 22(65.23)
w = 1.66 kip>ft
Check Shear:
tmax =
Ans.
(Neglect area of flanges.)
12(1.66)
Vmax
= 3.11 ksi 6 tallow = 14 ksi OK
=
Aw
2(12.31)(0.26)
•11–17.
The simply supported beam is composed of two
W12 * 22 sections built up as shown. Determine if the beam
will safely support a loading of w = 2 kip>ft. The allowable
bending stress is sallow = 22 ksi and the allowable shear
stress is tallow = 14 ksi.
w
24 ft
Section properties:
For W 12 * 22 (d = 12.31 in.
Ix = 156 in4
tw = 0.260 in.
A = 6.48 in2)
I = 2[156 + 6.48(6.1552)] = 802.98 in4
S =
802.98
I
=
= 65.23 in3
c
12.31
Bending stress:
smax =
144 (12)
Mallow
=
= 26.5 ksi 7 sallow = 22 ksi
S
65.23
No, the beam falls due to bending stress criteria.
Check shear:
tmax =
Ans.
(Neglect area of flanges.)
Vmax
24
=
= 3.75 ksi 6 tallow = 14 ksi OK
Aw
2(12.31)(0.26)
843
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11–18. Determine the smallest diameter rod that will
safely support the loading shown. The allowable bending
stress is sallow = 167 MPa and the allowable shear stress
is tallow = 97 MPa.
25 N/m
15 N/m
15 N/m
1.5 m
Bending Stress: From the moment diagram, Mmax = 24.375 N # m. Assume bending
controls the design. Applying the flexure formula.
sallow =
167 A 10 B =
6
Mmax c
I
24.375 A d2 B
p
4
A d2 B 4
d = 0.01141 m = 11.4 mm
Ans.
Shear Stress: Provide a shear stress check using the shear formula with
I =
p
A 0.0057074 B = 0.8329 A 10 - 9 B m4
4
Qmax =
4(0.005707) 1
c (p) A 0.0057062 B d = 0.1239 A 10 - 6 B m3
3p
2
From the shear diagram, Vmax = 30.0 N.
tmax =
=
Vmax Qmax
It
30.0 C 0.1239(10 - 6) D
0.8329 (10 - 9)(0.01141)
= 0.391 MPa 6 tallow = 97 MPa (O.K!)
844
1.5 m
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Page 845
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–19. The pipe has an outer diameter of 15 mm.
Determine the smallest inner diameter so that it will safely
support the loading shown. The allowable bending stress
is sallow = 167 MPa and the allowable shear stress is
tallow = 97 MPa.
25 N/m
15 N/m
15 N/m
1.5 m
Bending Stress: From the moment diagram, Mmax = 24.375 N # m. Q. Assume
bending controls the design. Applying the flexure formula.
sallow =
167 A 106 B =
Mmax c
I
24.375(0.0075)
p
4
C 0.00754 - A 2i B 4 D
d
di = 0.01297 m = 13.0 mm
Ans.
Shear Stress: Provide a shear stress check using the shear formula with
I =
p
A 0.00754 - 0.0064864 B = 1.0947 A 10 - 9 B m4
4
Qmax =
4(0.0075) 1
4(0.006486) 1
c (p) A 0.00752 B d c (p) A 0.0064862 B d
3p
2
3p
2
= 99.306 A 10 - 9 B m3
From the shear diagram, Vmax = 30.0 N. Q
tmax =
=
Vmax Qmax
It
30.0 C 99.306(10 - 9) D
1.0947(10 - 9)(0.015 - 0.01297)
= 1.34 MPa 6 tallow = 97 MPa (O.K!)
845
1.5 m
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Page 846
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*11–20. Determine the maximum uniform loading w
the W12 * 14 beam will support if the allowable bending
stress is sallow = 22 ksi and the allowable shear stress is
tallow = 12 ksi.
w
10 ft
10 ft
From the moment diagram, Fig. a, Mmax = 28.125 w. For W12 * 14, Sx = 14.9 in3,
d = 11.91 in and tw = 0.200 in.
sallow =
22 =
Mmax
S
28.125 w (12)
14.9
Ans.
w = 0.9712 kip>ft = 971 lb>ft
From the shear diagram, Fig. a, Vmax = 7.5(0.9712) = 7.284 kip. Provide a shear
stress check on W12 * 14,
tmax =
=
Vmax
tw d
7.284
0.200(11.91)
= 3.06 ksi 6 tallow = 12 ksi (O.K)
846
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
•11–21.
Determine if the W14 * 22 beam will safely
support a loading of w = 1.5 kip>ft. The allowable bending
stress is sallow = 22 ksi and the allowable shear stress
is tallow = 12 ksi.
w
10 ft
10 ft
For W14 * 22, Sx = 29.0 in3, d = 13.74 in and tw = 0.23 in. From the moment
diagram, Fig. a, Mmax = 42.1875 kip # ft.
smax =
=
Mmax
S
42.1875(12)
29.0
= 17.46 ksi 6 sallow = 22 ksi (O.K!)
From the shear diagram, Fig. a, Vmax = 11.25 kip.
tmax =
=
Vmax
tw d
11.25
0.23(13.74)
= 3.56 ksi 6 tallow = 12 ksi (O.K!)
Based on the investigated results, we conclude that W14 * 22 can safely support
the loading.
847
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Page 848
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11–22. Determine the minimum depth h of the beam to
the nearest 18 in. that will safely support the loading shown.
The allowable bending stress is sallow = 21 ksi and the
allowable shear stress is tallow = 10 ksi. The beam has a
uniform thickness of 3 in.
4 kip/ft
h
A
B
12 ft
The section modulus of the rectangular cross-section is
1
S =
3
I
12 (3)(h )
=
= 0.5 h2
C
h>2
From the moment diagram, Mmax = 72 kip # ft.
Sreq¿d =
Mmax
sallow
0.5h2 =
72(12)
21
h = 9.07 in
Use
h = 9 18 in
Ans.
From the shear diagram, Fig. a, Vmax = 24 kip . Referring to Fig. b,
9.125 9.125
ba
b (3) = 31.22 in3 and
Qmax = y¿A¿ = a
4
2
1
I =
(3) A 9.1253 B = 189.95 in4 . Provide the shear stress check by applying
12
shear formula,
tmax =
Vmax Qmax
It
=
24(31.22)
189.95(3)
= 1.315 ksi 6 tallow = 10 ksi (O.K!)
848
6 ft
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Page 849
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11–23. The box beam has an allowable bending stress
of sallow = 10 MPa and an allowable shear stress of
tallow = 775 kPa. Determine the maximum intensity w of the
distributed loading that it can safely support. Also, determine
the maximum safe nail spacing for each third of the length of
the beam. Each nail can resist a shear force of 200 N.
w
30 mm
250 mm
30 mm
150 mm
30 mm
Section Properties:
I =
1
1
(0.21) A 0.253 B (0.15) A 0.193 B = 0.1877 A 10 - 3 B m4
12
12
QA = y1 ¿A¿ = 0.11(0.03)(0.15) = 0.495 A 10 - 3 B m3
Qmax = ©y¿A¿ = 0.11(0.03)(0.15) + 0.0625(0.125)(0.06)
= 0.96375 A 10 - 3 B m3
Bending Stress: From the moment diagram, Mmax = 4.50w. Assume bending
controls the design. Applying the flexure formula.
sallow =
10 A 106 B =
Mmax c
I
4.50w (0.125)
0.1877 (10 - 3)
w = 3336.9 N>m
Shear Stress: Provide a shear stress check using the shear formula. From the shear
diagram, Vmax = 3.00w = 10.01 kN.
tmax =
=
Vmax Qmax
It
10.01(103) C 0.96375(10 - 3) D
0.1877(10 - 3)(0.06)
= 857 kPa 7 tallow = 775 kPa (No Good!)
Hence, shear stress controls.
tallow =
775 A 103 B =
Vmax Qmax
It
3.00w C 0.96375(10 - 3) D
0.1877(10 - 3)(0.06)
w = 3018.8 N>m = 3.02 kN>m
Ans.
Shear Flow: Since there are two rows of nails, the allowable shear flow is
2(200)
400
=
q =
.
s
s
849
6m
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11–23.
Continued
For 0 … x 6 2 m and 4 m 6 x … 6 m, the design shear force is
V = 3.00w = 9056.3 N.
q =
VQA
I
9056.3 C 0.495(10 - 3) D
400
=
s
0.1877(10 - 3)
s = 0.01675 m = 16.7 mm
Ans.
For 2 m 6 x 6 4 m, the design shear force is V = w = 3018.8 N.
q =
VQA
I
3018.8 C 0.495(10 - 3) D
400
=
s
0.1877(10 - 3)
s = 0.05024 m = 50.2 mm
Ans.
850
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*11–24. The simply supported joist is used in the
construction of a floor for a building. In order to keep the
floor low with respect to the sill beams C and D, the ends of
the joists are notched as shown. If the allowable shear stress
for the wood is tallow = 350 psi and the allowable bending
stress is sallow = 1500 psi, determine the height h that will
cause the beam to reach both allowable stresses at the same
time. Also, what load P causes this to happen? Neglect the
stress concentration at the notch.
P
2 in.
15 ft
B
h
15 ft
D
A
10 in.
C
Bending Stress: From the moment diagram, Mmax = 7.50P. Applying the flexure
formula.
Mmax c
I
salllow =
7.50P(12)(5)
1500 =
1
3
12 (2)(10 )
P = 555.56 lb = 556 lb
Ans.
Shear Stress: From the shear diagram, Vmax = 0.500P = 277.78 lb. The notch is the
critical section. Using the shear formula for a rectangular section.
tallow =
3Vmax
2A
350 =
3(277.78)
2(2) h
h = 0.595 in.
Ans.
11–25. The simply supported joist is used in the
construction of a floor for a building. In order to keep the
floor low with respect to the sill beams C and D, the ends of
the joists are notched as shown. If the allowable shear stress
for the wood is tallow = 350 psi and the allowable bending
stress is sallow = 1700 psi, determine the smallest height h
so that the beam will support a load of P = 600 lb. Also,
will the entire joist safely support the load? Neglect the
stress concentration at the notch.
P
B
tallow =
1.5V
;
A
350 =
D
A
10 in.
600
= 300 lb
2
1.5(300)
(2)(h)
h = 0.643 in.
smax =
Ans.
4500(12)(5)
Mmax c
= 1620 psi 6 1700 psi OK
= 1
3
I
12 (2)(10)
Yes, the joist will safely support the load.
Ans.
851
h
15 ft
C
The reaction at the support is
2 in.
15 ft
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–26. Select the lightest-weight steel wide-flange beam
from Appendix B that will safely support the loading
shown. The allowable bending stress is sallow = 22 ksi and
the allowable shear stress is tallow = 12 ksi.
5 kip
18 kip ft
B
A
6 ft
From the moment diagram, Fig. a, Mmax = 48 kip # ft.
Sreq¿d =
Mmax
sallow
=
48(12)
22
= 26.18 in3
Select W 14 * 22 C Sx = 29.0 in3, d = 13.74 in. and tw = 0.230 in. D
From the shear diagram, Fig. a, Vmax = 5 kip. Provide the shear stress check for
W 14 * 22,
tmax =
=
Vmax
twd
5
0.230(13.74)
= 1.58 ksi 6 tallow = 12 ksi‚ (O.K!)
Use
Ans.
W14 * 22
W12 * 22 would work also.
852
12 ft
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–27. The T-beam is made from two plates welded
together as shown. Determine the maximum uniform
distributed load w that can be safely supported on the beam
if the allowable bending stress is sallow = 150 MPa and the
allowable shear stress is tallow = 70 MPa.
w
A
1.5 m
1.5 m
200 mm
20 mm
200 mm
20 mm
The neutral axis passes through centroid c of the beam’s cross-section. The location
of c, Fig. b, is
y =
0.21(0.02)(0.2) + 0.1(0.2)(0.02)
©yA
=
©A
0.02(0.2) + 0.2(0.02)
= 0.155 m
I =
1
(0.2)(0.023) + 0.2(0.02)(0.055)2
12
+
1
(0.02)(0.23) + 0.02(0.2)(0.055)2
12
= 37.667 (10 - 6) m4
Referring to Fig. b,
Qmax = y¿A¿ = 0.0775(0.155)(0.02)
= 0.24025(10 - 3) m3
Referring to the moment diagram, Mmax = -3.375 w. Applying the flexure
formula with C = y = 0.155 m,
sallow =
Mmax c
;
I
150(106) =
3.375 w (0.155)
37.667(10 - 6)
w = 10.80(103) N>m
= 10.8 kN>m (Control!)
Ans.
Referring to the shear diagram, Vmax = 1.5w.
tallow =
Vmax Qmax
;
It
70(106) =
1.5 w C 0.24025(10 - 3) D
37.667(10 - 6)(0.02)
w = 146.33(103) N>m
= 146 kN>m
853
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*11–28. The beam is made of a ceramic material having
an allowable bending stress of sallow = 735 psi and an
allowable shear stress of tallow = 400 psi. Determine the
width b of the beam if the height h = 2b.
15 lb
10 lb
6 lb/in.
2 in.
6 in.
2 in.
h
b
Bending Stress: From the moment diagram, Mmax = 30.0 lb # in. Assume bending
controls the design. Applying the flexure formula.
sallow =
Mmax c
I
30.0 A 2b
2 B
735 = 1
3
12 (b) (2b)
b = 0.3941 in. = 0.394 in.
Ans.
Shear Stress: Provide a shear stress check using the shear formula for a rectangular
section. From the shear diagram, Vmax = 19.67 lb.
tmax =
=
3Vmax
2A
3(19.67)
2(0.3941)(2)(0.3941)
= 94.95 psi 6 tallow = 400 psi (O.K!)
854
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•11–29.
The wood beam has a rectangular cross section.
Determine its height h so that it simultaneously reaches
its allowable bending stress of sallow = 1.50 ksi and an
allowable shear stress of tallow = 150 psi. Also, what is the
maximum load P that the beam can then support?
P
P
B
A
1.5 ft
3 ft
1.5 ft
h
6 in.
The section modulus of the rectangular cross-section about the neutral axis is
S =
1
3
I
12 (6) h
=
= h2
C
h>2
From the moment diagram, Fig. a, Mmax = 1.5P.
Mmax = sallow S
1.5P(12) = 1.50(103) h2
P = 83.33h2
(1)
h h
1
a b (6) = 0.75 h2 and I =
(6) h3 = 0.5h3.
4 2
12
From the shear diagram, Fig. a, Vmax = P.
Referring to Fig. b, Qmax = y¿A¿ =
tmax =
150 =
Vmax Qmax
It
P (0.75 h2)
0.5 h3 (6)
P = 600 h
(2)
Solving Eq (1) and (2)
h = 7.20 in
P = 4320 lb = 4.32 kip
Ans.
855
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11–30. The beam is constructed from three boards as
shown. If each nail can support a shear force of 300 lb,
determine the maximum allowable spacing of the nails, s,
s¿, s– , for regions AB, BC, and CD respectively. Also, if the
allowable bending stress is sallow = 1.5 ksi and the allowable
shear stress is tallow = 150 psi, determine if it can safely
support the load.
1500 lb
500 lb
s¿
s
A
s¿¿
C
B
6 ft
6 ft
6 ft
10 in.
4 in.
10 in.
The neutral axis passes through centroid c of the beam’s cross-section. The location
of c, Fig. b, is
y =
12(4)(10) + 2 C 5(10)(2) D
©yA
=
©A
4(10) + 2(10)(2)
= 8.50 in
The moment of inertia of the beam’s cross-section about the neutral axis is
I = 2c
1
(2)(103) + 2(10)(3.50)2 d
12
+
1
(10)(43) + 10(4)(3.50)2
12
= 1366.67 in4
Referring to Fig. b,
Qmax = 2y2œ A2œ = 2 C 4.25(8.50)(2) D = 144.5 in3
QA = y1œ A1œ = 3.50(4)(10) = 140 in3
Referring to the moment diagram, Fig. a, Mmax = 9000 lb # ft. Applying flexure
formula with C = y = 8.50 in,
smax =
=
Mmax c
I
9000(12)(8.50)
1366.67
= 671.70 psi 6 sallow = 1.50 ksi (O.K!)
Referring to shear diagram, Fig. a, Vmax = 1500 lb.
tmax =
=
Vmax Qmax
It
1500 (144.5)
= 39.65 psi 6 tallow = 150 psi (O.K!)
1366.67 (4)
1
S– = 11 in. Yes, it can support the load.
2
Ans.
856
2 in.
D
2 in.
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11–30.
Continued
Since there are two rows of nails, the allowable shear flow is
2(300)
600
2F
qallow =
=
=
. For region AB, V = 1500 lb. Thus
S
S
S
qallow =
VQA
;
I
1500 (140)
600
=
S
1366.67
Use
S = 3.904 in
S = 3 34 in
Ans.
For region BC, V = 1000 lb. Thus
qallow =
VQA
;
I
1000(140)
600
=
S¿
1366.67
Use
S¿ = 5.85 in
S¿ = 5 34 in
Ans.
For region CD, V = 500 lb. Thus
qallow =
VQA
;
I
500 (140)
600
=
S–
1366.67
Use
S– = 11.71 in
S– = 1112 in
Ans.
857
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Page 858
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11–31. The tapered beam supports a concentrated force P
at its center. If it is made from a plate that has a constant
width b, determine the absolute maximum bending stress in
the beam.
2h0
h0
L
2
L
2
P
Section Properties:
h - h0
h0
= L
x
2
I =
S =
h =
h0
(2x + L)
L
h30
1
(b) a 3 b(2x + L)3
12
L
A B
h30
1
3
12 (b) L3 (2x + L)
bh20
=
h3
2L (2x + L)
6L2
(2x + L)2
Bending Stress: Applying the flexure formula.
s =
M
= 2
bh0
S
2
6L
Px
2
3PL2x
=
(2x + L)2
[1]
bh20 (2x + L)2
In order to have the absolute maximum bending stress,
ds
= 0.
dx
3PL2 (2x + L)2(1) - x(2)(2x + L)(2)
ds
=
c
d = 0
dx
bh20
(2x + L)4
x =
Substituting x =
L
2
L
into Eq. [1] yields
2
smax =
3PL
8bh20
Ans.
858
h0
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*11–32. The beam is made from a plate that has a constant
thickness b. If it is simply supported and carries a uniform
load w, determine the variation of its depth as a function of
x so that it maintains a constant maximum bending stress
sallow throughout its length.
w
h0
y
x
L
––
2
Moment Function: As shown on FBD(b).
Section Properties:
I =
1 3
by
12
S =
1
3
I
1
12 by
= y = by2
c
6
2
Bending Stress: Applying the flexure formula.
w
2
2
M
8 (L - 4x )
=
1
2
S
6 by
sallow =
sallow =
3w (L2 - 4x2)
[1]
4by2
At x = 0, y = h0. From Eq. [1],
sallow =
3wL2
4bh20
[2]
Equating Eq. [1] and [2] yields
y2 =
h20
L2
A L2 - 4x2 B
y2
4x2
+
= 1
h20
L2
Ans.
The beam has a semi-elliptical shape.
859
L
––
2
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•11–33.
The beam is made from a plate having a constant
thickness t and a width that varies as shown. If it supports a
concentrated force P at its center, determine the absolute
maximum bending stress in the beam and specify its
location x, 0 6 x 6 L>2.
P
P
—
2
b0
L
—
2
x
b
L
—
2
t
P
—
2
Section properties:
b
x
= L;
b0
2
b =
2b0
x
L
I =
b0 t3
1 2b0
a
xb t3 =
x
12 L
6L
S =
b0 t2
I
6L x
x
= t =
c
3L
2
b0 t
Bending stress:
s =
P
M
3PL
2x
= 2 =
b0 t
S
2b0t2
3L x
Ans.
The bending stress is independent of x. Therefore, the stress is constant throughout
the span.
Ans.
860
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11–34. The beam is made from a plate that has a constant
thickness b. If it is simply supported and carries the
distributed loading shown, determine the variation of its
depth as a function of x so that it maintains a constant
maximum bending stress sallow throughout its length.
w0
A
L
––
2
Moment Function: The distributed load as a function of x is
w0
w
=
x
L>2
w =
2w0
x
L
The free-body diagram of the beam’s left cut segment is shown in Fig. b.
Considering the moment equilibrium of this free-body diagram,
d+ ©MO = 0;
M +
1 2w0
x
1
x R x ¢ ≤ - w0Lx = 0
B
2 L
3
4
M =
w0
A 3L2x - 4x3 B
12L
Section Properties: At position x, the height of the beam’s cross section is h. Thus
1
bh3
12
I =
Then
1
bh3
I
12
1
S =
=
= bh2
c
h>2
6
Bending Stress: The maximum bending stress smax as a function of x can be
obtained by applying the flexure formula.
w0
A 3L2x - 4x3 B
w0
M
12L
smax =
=
=
A 3L2x - 4x3 B ‚
S
1 2
2bh2L
bh
6
At x =
(1)
L
, h = h0. From Eq. (1),
2
smax =
w0L2
(2)
2bh0 2
Equating Eqs. (1) and (2),
w0
2
2bh L
h =
A 3L2x - 4x3 B =
h0
L3>2
w0L2
2bh0 2
A 3L2x - 4x3 B 1>2
Ans.
861
h0
B
x
Support Reactions: As shown on the free-body diagram of the entire beam, Fig. a.
C
h
L
––
2
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11–35. The beam is made from a plate that has a constant
thickness b. If it is simply supported and carries the
distributed loading shown, determine the maximum
bending stress in the beam.
w0
h0
h0
2h0
L
–
2
Support Reactions: As shown on the free - body diagram of the entire beam, Fig. a.
Moment Function: The distributed load as a function of x is
w0
w
=
;
x
L>2
w =
2w0
x
L
The free - body diagram of the beam’s left cut segment is shown in Fig. b.
Considering the moment equilibrium of this free - body diagram,
d+ ©MO = 0;
M +
w0L
x
1 2w0
a
xbxa b x = 0
2 L
3
4
M =
w0
A 3L2x - 4x3 B
12L
Section Properties: Referring to the geometry shown in Fig. c,
h - h0
h0
=
;
x
L>2
h =
h0
(2x + L)
L
At position x, the height of the beam’s cross section is h. Thus
I =
1
bh3
12
Then
1
bh3
bh0 2
12
I
1
=
S =
= bh2 =
(2x + L)2
c
h>2
6
6L2
Bending Stress: Applying the flexure formula,
w0
A 3L2x - 4x3 B
M
12L
smax =
=
S
bh0 2
(2x + L)2
6L2
smax =
w0L
2bh0
2 B
3L2x - 4x3
R
(2x + L)2
In order to have absolute maximum bending stress,
(1)
dsmax
= 0.
dx
2
2
2
2
3
dsmax
w0L (2x + L) A 3L - 12x B - A 3L x - 4x B (2)(2x + L)(2)
=
C
S = 0
dx
2bh0 2
(2x + L)4
w0L
2bh0
2 B
3L3 - 8x3 - 6L2x - 12Lx2
R = 0
(2x + L)3
862
L
–
2
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11–35.
Since
Continued
w0L
2bh0 2
Z 0, then
3L3 - 8x3 - 6L2x - 12Lx2 = 0
Solving by trial and error,
x = 0.2937L = 0.294L
Substituting this result into Eq. (1),
sabs =
max
0.155w0L2
Ans.
bh0 2
863
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*11–36. Determine the variation of the radius r of the
cantilevered beam that supports the uniform distributed
load so that it has a constant maximum bending stress smax
throughout its length.
w
r0
r
Moment Function: As shown on FBD.
Section Properties:
I =
p 4
r
4
I
=
c
S =
r
x
L
p 4
4 r
=
p 3
r
4
Bending Stress: Applying the flexure formula.
wx2
smax =
M
2
= p 3
S
4r
smax =
2wx2
pr3
[1]
At x = L, r = r0. From Eq. [1],
smax =
2wL2
pr30
[2]
Equating Eq. [1] and [2] yields
r3 =
r30
L2
x2
Ans.
•11–37.
Determine the variation in the depth d of a
cantilevered beam that supports a concentrated force P at
its end so that it has a constant maximum bending stress
sallow throughout its length. The beam has a constant
width b0 .
P
d0
d
L
Section properties:
I =
1
(b )(d3)
12 0
sallow =
1
S =
3
b0d2
I
12 (b0)(d )
=
=
c
d>2
6
M
Px
=
S
b0d2>6
(1)
PL
b0d0 2>6
(2)
At x = L
sallow =
Equate Eqs. (1) and (2):
PL
Px
=
b0d2>6
b0 d0 2>6
d2 = a
d0 2
bx ;
L
x
d = d0
AL
Ans.
864
x
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11–38. Determine the variation in the width b as a
function of x for the cantilevered beam that supports a
uniform distributed load along its centerline so that it has
the same maximum bending stress sallow throughout its
length. The beam has a constant depth t.
b
—0
2
b
—0
2
b
—
2
w
L
x
t
Section properties:
I =
1
b t3
12
S =
1
3
I
t2
12 b t
=
= b
t
c
6
2
Bending stress:
w x2
M
3wx2
2
sallow =
= 2 = 2
t
S
tb
b
(1)
6
At x = L, b = b0
sallow =
3wL2
t2b0
(2)
Equating Eqs. (1) and (2) yields:
3wL2
3wx2
= 2
2
t b
t b0
b =
b0
L2
x2
Ans.
865
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z
11–39. The shaft is supported on journal bearings that do
not offer resistance to axial load. If the allowable normal
stress for the shaft is sallow = 80 MPa, determine to the
nearest millimeter the smallest diameter of the shaft that
will support the loading. Use the maximum-distortionenergy theory of failure.
A
150 mm
D
30
250 mm 50 N
x
30
C
30 150 N
100 mm
500 mm
100 N
30
250 N
250 mm
Torque and Moment Diagrams: As shown.
In-Plane Principal Stresses: Applying Eq. 9–5 with sy = 0, sx =
txy =
Mc
4M
, and
=
I
pc3
2T
Tc
.
=
J
pc3
s1, 2 =
sx + sy
;
2
Aa
sx - sy
2
2
2
b + txy
=
2M
2M 2
2T 2
;
A a pc3 b + a pc3 b
pc3
=
2M
2
;
2M2 + T2
pc3
pc3
2
2M
2M2 + T2, then
and b =
pc3
pc3
s21 = a2 + b2 + 2ab, s1s2 = a2 - b2, s22 = a2 + b2 - 2ab, and
s21 - s1 s2 + s22 = 3b2 + a2.
Maximum Distortion Energy Theory: Let a =
s21 - s1 s2 + s22 = s2allow
3a
2
2
2M 2
2
2
2M
+
T
b
+
a
b = s2allow
pc3
pc3
6
4
A 4M2 + 3T2 B R
p2s2allow
1
c = B
Shaft Design: By observation, the critical section is located just to the left of gear
C, where M = 239.06252 + 46.012 = 60.354 N # m and T = 15.0 N # m. Using the
maximum distortion energy theory,
6
4
A 4M2 + 3T2 B R
p2s2allow
1
c = B
= b
4
C 4(60.354)2 + 3(15.0)2 D r
6 2
1
6
p2 [80(10 )]
= 0.009942 m
d = 2c = 2(0.009942) = 0.01988 m = 19.88 mm
Use
d = 20 mm
Ans.
866
B
y
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z
*11–40. The shaft is supported on journal bearings that do
not offer resistance to axial load. If the allowable shear
stress for the shaft is tallow = 35 MPa, determine to the
nearest millimeter the smallest diameter of the shaft that
will support the loading. Use the maximum-shear-stress
theory of failure.
A
x
150 mm
D
30
250 mm 50 N
30
C
30 150 N
100 mm
500 mm
100 N
30
250 N
250 mm
Shaft Design: By observation, the critical section is located just to the left of gear C,
where M = 239.06252 + 46.012 = 60.354 N # m and T = 15.0 N # m. Using the
maximum shear stress theory.
c = a
= B
1
3
2
2M2 + T2 b
ptallow
2
p(35)(106)
260.3542 + 15.02 R
1
3
= 0.01042 m
d = 2c = 2(0.01042) = 0.02084 m = 20.84 mm
Use
d = 21 mm
Ans.
867
B
y
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z
•11–41.
The end gear connected to the shaft is subjected
to the loading shown. If the bearings at A and B exert only
y and z components of force on the shaft, determine the
equilibrium torque T at gear C and then determine the
smallest diameter of the shaft to the nearest millimeter that
will support the loading. Use the maximum-shear-stress
theory of failure with tallow = 60 MPa.
100 mm
T
250 mm C
50 mm
150 mm
A
x
100 mm
Fz 1.5 kN
From the free - body diagrams:
T = 100 N # m
Ans.
Critical section is at support A.
1
1
3
3
2
2
c = c
22252 + 1502 d
2M2 + T2 d = c
p tallow
p(60)(106)
= 0.01421 m
d = 2c = 0.0284 m = 28.4 mm
Use d = 29 mm
Ans.
868
B
75 mm
y
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Page 869
z
11–42. The end gear connected to the shaft is subjected to
the loading shown. If the bearings at A and B exert only y
and z components of force on the shaft, determine the
equilibrium torque T at gear C and then determine the
smallest diameter of the shaft to the nearest millimeter
that will support the loading. Use the maximum-distortionenergy theory of failure with sallow = 80 MPa.
100 mm
T
250 mm C
50 mm
150 mm
A
x
T = 100 N # m
Ans.
Critical section is at support A.
s1, 2 =
sx
s2x
2
;
2
A 4 + txy
sx
s2x
2
,b =
A 4 + txy
2
Let a =
s1 = a + b, s2 = a - b
Require,
s21 - s1 s2 + s22 = s2allowa2 + 2ab + b2 - [a2 - b2] + a2 - 2ab + b2 = s2allow
a2 + 3b2 = s2allow
s2x
s2x
+ 3a
+ t2xy b = s2allow
4
4
s2x + 3t2xy = s2allow
Mt 2
Tc 2
a p 4 b + 3a p 4 b = s2allow
4 c
2 c
4M 2
2T 2
1
ca
b + 3a b d = s2allow
4
p
p
c
c4 =
16
s2allow p2
c = a
= c
M2 +
4
s2allow p2
12T2
s2allow p2
(4M + 3T ) b
2
4
(80(106))2(p)2
2
100 mm
Fz 1.5 kN
From the free-body diagrams:
1
2
(4(225) + 3(150) ) d
2
2
1
2
= 0.01605 m
d = 2c = 0.0321 m = 32.1 mm
Use d = 33 mm
Ans.
869
B
75 mm
y
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11–43. The shaft is supported by bearings at A and B that
exert force components only in the x and z directions on
the shaft. If the allowable normal stress for the shaft is
sallow = 15 ksi, determine to the nearest 18 in. the smallest
diameter of the shaft that will support the loading. Use the
maximum-distortion-energy theory of failure.
z
C
F¿x 100 lb
6 in.
A
x
8 in.
12 in.
Critical moment is just to the right of D.
T = 1200 lb # in.
Both states of stress will yield the same result.
Let
s
s 2
2
;
a
2
A 2b + t
2
s
= A and s + t2 = B
2
A4
s2a = (A + B)2, s2b = (A - B)2
sa sb = (A + B)(A - B) = A2 - B2
s2a - sa sb + s2b = A2 + B2 + 2AB - A2 + B2 + A2 + B2 - 2AB
= A2 + 3B2 =
s2
s2
+ 3a
+ t2 b = s2 + 3t2
4
4
s2a - sa sb + s2b = s2allow
s2 + 3t2 = s2allow‚
s =
Mc
Mc
4M
= p 4 =
I
c
pc3
4
t =
Tc
Tc
2T
= p 4 =
J
c
p
c3
2
(1)
From Eq. (1)
16M2
2
6
p c
c = a
12T2
+
p2 c6
= s2allow
16(2396)2 + 12(12002) 1>6
16M2 + 12T2 1>6
b
=
c
d
= 0.605 in.
p2s2allow
p2((15)(103))2
d = 2c = 1.210 in.
Use d = 1
2 in.
Fz 300 lb
10 in.
6 in.
M = 220572 + 12292 = 2396 lb # in.
sa, b =
F y 300 lb
D
1
in.
4
Ans.
870
4 in.
E
B
y
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*11–44. The shaft is supported by bearings at A and B
that exert force components only in the x and z directions
on the shaft. If the allowable normal stress for the shaft is
sallow = 15 ksi, determine to the nearest 18 in. the smallest
diameter of the shaft that will support the loading. Use the
maximum-shear-stress theory of failure. Take tallow = 6 ksi.
z
C
F¿x 100 lb
6 in.
A
x
8 in.
12 in.
Critical moment is just to the right of D.
T = 1200 lb # in.
Use Eq. 11-2,
1>3
2
2M2 + T2 b
p tallow
c = a
1>3
2
2(2396)2 + (1200)2 b
= 0.6576 in.
3
p(6)(10 )
2 in.
Fz 300 lb
10 in.
6 in.
M = 2(2057)2 + (1229)2 = 2396 lb # in.
c = a
F y 300 lb
D
dreq¿d = 2c = 1.315 in.
3
Use d = 1 in.
8
Ans.
871
4 in.
E
B
y
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z
•11–45. The bearings at A and D exert only y and z
components of force on the shaft. If tallow = 60 MPa,
determine to the nearest millimeter the smallest-diameter
shaft that will support the loading. Use the maximum-shearstress theory of failure.
350 mm
D
400 mm
200 mm
B
A
Critical moment is at point B:
M = 2(473.7)2 + (147.4)2 = 496.1 N # m
x
T = 150 N # m
c = a
1>3
1>3
2
2
2
2
2496.1
2M2 + T2 b
= a
+
150
b
= 0.0176 m
p tallow
p(60)(106)
c = 0.0176 m = 17.6 mm
d = 2c = 35.3 mm
Use d = 36 mm
Ans.
872
y
C
75 mm
Fy ⫽ 3 kN
50 mm
Fz ⫽ 2 kN
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z
11–46. The bearings at A and D exert only y and z
components of force on the shaft. If tallow = 60 MPa,
determine to the nearest millimeter the smallest-diameter
shaft that will support the loading. Use the maximumdistortion-energy theory of failure. sallow = 130 MPa.
350 mm
D
400 mm
200 mm
B
A
The critical moment is at B.
M = 2(473.7)2 + (147.4)2 = 496.1 N # m
x
T = 150 N # m
Since,
sa, b =
Let
s
s 2
2
;
a
2
A 2b + t
s
= A
2
and
s 2
2 = B
a
A 2b + t
s2a = (A + B)2
s2b = (A - B)2
sa sb = (A + B)(A - B)
s2a - sa sb + s2b = A2 + B2 + 2AB - A2 + B2 + A2 + B2 - 2AB
= A2 + 3B2
=
s2
s2
+ 3a
+ t2 b
4
4
= s2 + 3t2
s2a - sasb + s2b = s2allow
s2 + 3t2 = s2allow
(1)
s =
Mc
Mc
4M
= p 4 =
I
c
pc3
4
t =
Tc
Tc
2T
= p 4 =
J
pc3
2 c
From Eq (1)
12T2
16M2
+ 2 4 = s2allow
2 4
pc
pc
c = a
= a
16M2 + 12T2 1>6
b
p2s2allow
16(496.1)2 + 12(150)2
2
4
2
p ((130)(10 ))
b
1>4
= 0.01712 m
d = 2c = 34.3 mm
Ans.
873
y
C
75 mm
Fy ⫽ 3 kN
50 mm
Fz ⫽ 2 kN
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11–47. Draw the shear and moment diagrams for the shaft,
and then determine its required diameter to the nearest
millimeter if sallow = 140 MPa and tallow = 80 MPa. The
bearings at A and B exert only vertical reactions on the shaft.
1500 N
800 N
A
B
600 mm
125 mm
Bending Stress: From the moment diagram, Mmax = 111 N # m. Assume bending
controls the design. Applying the flexure formula.
sallow =
140 A 106 B =
Mmax c
I
111 A d2 B
p
4
A d2 B 4
d = 0.02008 m = 20.1 mm
d = 21 mm
Use
Ans.
Shear Stress: Provide a shear stress check using the shear formula with
I =
p
A 0.01054 B = 9.5466 A 10 - 9 B m4
4
Qmax =
4(0.0105) 1
c (p)(0.0105)2 d = 0.77175 A 10 - 6 B m3
3p
2
From the shear diagram, Vmax = 1484 N.
tmax =
=
Vmax Qmax
It
1484 C 0.77175(10 - 6) D
9.5466(10 - 9)(0.021)
= 5.71 MPa 6 tallow = 80 MPa (O.K!)
874
75 mm
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*11–48. The overhang beam is constructed using two 2-in.
by 4-in. pieces of wood braced as shown. If the allowable
bending stress is sallow = 600 psi, determine the largest
load P that can be applied. Also, determine the associated
maximum spacing of nails, s, along the beam section AC if
each nail can resist a shear force of 800 lb. Assume the beam
is pin-connected at A, B, and D. Neglect the axial force
developed in the beam along DA.
D
2 ft
3 ft
A
2 ft
Section properties:
I =
1
(4)(4)3 = 21.33 in4
12
S =
21.33
I
=
= 10.67 in3
c
2
Mmax = sallow S
3P(12) = 600(10.67)
P = 177.78 = 178 lb
Ans.
Nail Spacing:
V = P = 177.78 lb
Q = (4)(2)(1) = 8 in3
q =
177.78(8)
VQ
=
= 66.67 lb>in.
I
21.33
S =
800 lb
= 12.0 in.
66.67 lb>in.
Ans.
875
2 in.
2 in.
s
B
MA = Mmax = 3P
P
C
4 in.
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z
•11–49. The bearings at A and B exert only x and z
components of force on the steel shaft. Determine the
shaft’s diameter to the nearest millimeter so that it can
resist the loadings of the gears without exceeding an
allowable shear stress of tallow = 80 MPa. Use the
maximum-shear-stress theory of failure.
Fx 5 kN
A
75 mm
x
50 mm
150 mm
350 mm
B
Fz 7.5 kN
250 mm
Maximum resultant moment M = 212502 + 2502 = 1274.75 N # m
1
1
3
3
2
2
21274.752 + 3752 d = 0.0219 m
2M2 + T2 d = c
c = c
6
p tallow
p(80)(10 )
d = 2c = 0.0439 m = 43.9 mm
Use d = 44 mm
Ans.
876
y
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z
11–50. The bearings at A and B exert only x and z
components of force on the steel shaft. Determine the
shaft’s diameter to the nearest millimeter so that it can resist
the loadings of the gears without exceeding an allowable
shear stress of tallow = 80 MPa. Use the maximumdistortion-energy theory of failure with sallow = 200 MPa.
Fx 5 kN
A
75 mm
x
50 mm
150 mm
350 mm
Maximum resultant moment M = 212502 + 2502 = 1274.75 N # m
s1, 2 =
sx
s2x
2
;
2
A 4 + txy
sx
s2x
2
,b =
2
A 4 + txy
Let a =
s1 = a + b,
s2 = a - b
Require,
s21 - s1 s2 + s22 = s2allow
a2 + 2ab + b2 - [a2 - b2] + a2 - 2ab + b2 = s2allow
a2 + 3b2 = s2allow
s2x
s2x
+ 3a
+ t2xy b = s2allow
4
4
s2x + 3t2xy = s2allow
Mc 2
Tc 2
a p 4 b + 3a p 4 b = s2allow
4 c
2 c
1
6
c
Ba
c6 =
4M 2
2T 2
b + 3a b R = s2allow
p
p
12T2
2
M
+
s2allow p2
s2allow p2
c = B
= B
16
4
(4M2 + 3 T2) R
s2allow p2
4
1
4
(4(1274.75)2 + 3(375)2) R
(200(106))2(p)2
1
4
= 0.0203 m = 20.3 mm
d = 40.6 mm
Ans.
Use
d = 41 mm
Ans.
877
B
Fz 7.5 kN
250 mm
y
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11–51. Draw the shear and moment diagrams for the
beam. Then select the lightest-weight steel wide-flange
beam from Appendix B that will safely support the loading.
Take sallow = 22 ksi, and tallow = 12 ksi.
3 kip/ft
1.5 kip ft
A
B
12 ft
Bending Stress: From the moment diagram, Mmax = 18.0 kip # ft. Assume bending
controls the design. Applying the flexure formula.
Sreq¿d =
=
Select
Mmax
sallow
18.0(12)
= 9.82 in3
22
A Sx = 10.9 in3, d = 9.87 in., tw = 0.19 in. B
W10 * 12
V
for the W10 * 12 wide twd
flange section. From the shear diagram, Vmax = 9.00 kip
Shear Stress: Provide a shear stress check using t =
tmax =
=
Vmax
tw d
9.00
0.19(9.87)
= 4.80 ksi 6 tallow = 12 ksi (O.K!)
Hence,
Use
W10 * 12
Ans.
878
6 ft
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*11–52. The beam is made of cypress having an allowable
bending stress of sallow = 850 psi and an allowable shear
stress of tallow = 80 psi. Determine the width b of the beam
if the height h = 1.5b.
300 lb
75 lb/ft
B
A
5 ft
5 ft
h 1.5b
b
Ix =
1
(b)(1.5b)3 = 0.28125 b4
12
Qmax = y¿A¿ = (0.375b) (0.75b)(b) = 0.28125 b3
Assume bending controls.
Mmax = 527.34 lb # ft
sallow =
Mmax c
;
I
850 =
527.34(12)(0.75 b)
0.28125 b4
b = 2.71 in.
Ans.
Check shear:
I = 15.12 in4
tmax =
Qmax = 5.584 in3
VQmax
281.25(5.584)
=
It
15.12(2.71)
= 38.36 psi 6 80 psi
OK
879
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•11–53.
The tapered beam supports a uniform distributed
load w. If it is made from a plate and has a constant width b,
determine the absolute maximum bending stress in the beam.
w
h0
L
––
2
Support Reactions: As shown on FBD(a).
Moment Function: As shown on FBD(b).
Section Properties:
h - h0
h0
= L
x
2
I =
S =
h =
h0
(2x + L)
L
h30
1
(b) a 3 b(2x + L)3
12
L
A B
h30
1
3
12 (b) L3 (2x + L)
h0
2L (2x + L)
bh20
=
6L2
(2x + L)2
Bending Stress: Applying the flexure formula.
s =
w
2
3wL2 (Lx - x2)
M
2 (Lx - x )
= 2
=
bh
S
bh20 (2x + L)2
0
2
2 (2x + L)
[1]
6L
In order to have the absolute maximum bending stress,
ds
= 0.
dx
3wL2 (2x + L)2(L - 2x) - (Lx - x2)(2)(2x + L)(2)
ds
=
c
d = 0
dx
bh20
(2x + L)4
x =
Substituting x =
L
4
L
into Eq. [1] yields
4
smax =
wL2
4bh20
Ans.
880
h0
2 h0
L
––
2
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11–54. The tubular shaft has an inner diameter of 15 mm.
Determine to the nearest millimeter its outer diameter if it
is subjected to the gear loading. The bearings at A and B
exert force components only in the y and z directions on the
shaft. Use an allowable shear stress of tallow = 70 MPa, and
base the design on the maximum-shear-stress theory of
failure.
z
100 mm
B
500 N
150 mm
A
200 mm
150 mm
x
I =
p 4
p
(c - 0.00754) and J = (c4 - 0.00754)
4
2
tallow =
Aa
sx - sy
tallow =
Aa
Mc 2
Tc 2
b + a b
2I
J
t2allow =
M2 c2
T2 c2
+
2
4I
J2
¢
2
100 mm
2
b + t2xy
c4 - 0.00754 2
4M2
4T2
≤ = 2 + 2
c
p
p
c4 - 0.00754
2
=
2M2 + T2
c
p tallow
c4 - 0.00754
2
2752 + 502
=
c
p(70)(106)
c4 - 0.00754 = 0.8198(10 - 6)c
Solving, c = 0.0103976 m
d = 2c = 0.0207952 m = 20.8 mm
Use d = 21 mm
Ans.
881
500 N
y
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11–55. Determine to the nearest millimeter the diameter
of the solid shaft if it is subjected to the gear loading. The
bearings at A and B exert force components only in
the y and z directions on the shaft. Base the design on
the maximum-distortion-energy theory of failure with
sallow = 150 MPa.
z
100 mm
B
500 N
150 mm
A
200 mm
150 mm
x
s1, 2 =
100 mm
sx
2
;
2
A 4 + txy
s2x
sx
s2x
2
,b =
A 4 + txy
2
Let a =
s1 = a + b, s2 = a - b
Require,
s21 - s1 s2 + s21 = s2allow
a2 + 2ab + b2 - [a2 - b2] + a2 - 2ab + b2 = sallow
a2 + 3b2 = s2allow
s2x
s2x
+ 3a
+ t2xy b = s2allow
4
4
s2x + 3t2xy = s2allow
Mc 2
Tc 2
a p 4 b + 3a p 4 b = s2allow
4 c
2 c
1
6
c
ca
c6 =
4M 2
2T 2
b + 3a b d = s2allow
p
p
12T2
2
M
+
s2allow p2
s2allow p2
16
c = a
= c
(4M2 + 3T2) b
s2allow p2
4
4
(150(106))2(p)2
1
4
1
4
(4(75) + 3(50) ) d = 0.009025 m
2
2
d = 2c = 0.0181 m
Use d = 19 mm
Ans.
882
500 N
y
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•12–1. An A-36 steel strap having a thickness of 10 mm and
a width of 20 mm is bent into a circular arc of radius r = 10 m.
Determine the maximum bending stress in the strap.
Moment-Curvature Relationship:
M
1
=
r
EI
however,
M =
I
s
c
1
1
c s
=
r
EI
s =
0.005
c
E = a
b C 200 A 109 B D = 100 MPa
r
10
12–2. A picture is taken of a man performing a pole vault,
and the minimum radius of curvature of the pole is
estimated by measurement to be 4.5 m. If the pole is 40 mm
in diameter and it is made of a glass-reinforced plastic for
which Eg = 131 GPa, determine the maximum bending
stress in the pole.
r ⫽ 4.5 m
Moment-Curvature Relationship:
M
1
=
r
EI
however,
M =
I
s
c
I
1
c s
=
r
EI
s =
0.02
c
E = a
b C 131 A 109 B D = 582 MPa
r
4.5
Ans.
883
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12–3. When the diver stands at end C of the diving board,
it deflects downward 3.5 in. Determine the weight of the
diver. The board is made of material having a modulus of
elasticity of E = 1.5(103) ksi.
B
A
3.5 in.
2 in.
C
9 ft
3 ft
Support Reactions and Elastic Curve. As shown in Fig. a.
Moment Functions. Referring to the free-body diagrams of the diving board’s cut
segments, Fig. b, M A x1 B is
a + ©MO = 0;
and M A x2 B is
a + ©MO = 0;
M A x1 B + 3Wx1 = 0
M A x1 B = -3Wx1
-M A x2 B - Wx2 = 0
M A x2 B = -Wx2
Equations of Slope and Elastic Curve.
EI
d2v
= M(x)
dx2
For coordinate x1,
EI
d2v1
dx1 2
= -3Wx1
d2v1
3
= - Wx1 2 + C1
dx1
2
(1)
1
EIv1 = - Wx1 3 + C1x1 + C2
2
(2)
EI
For coordinate x2
EI
EI
d2v2
dx2 2
= -Wx2
dv2
1
= - Wx2 2 + C3
dx2
2
EIv2 = -
(3)
1
Wx2 3 + C3x2 + C4
6
(4)
Boundary Conditions. At x1 = 0, v1 = 0. Then, Eq. (2) gives
1
EI(0) = - W A 03 B + C1(0) + C2
2
C2 = 0
At x1 = 3 ft, v1 = 0. Then, Eq. (2) gives
1
EI(0) = - W A 33 B + C1(3) + 0
2
C1 = 4.5W
884
18 in.
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12-3.
Continued
At x2 = 9 ft, v2 = 0. Then, Eq. (4) gives
1
EI(0) = - W A 93 B + C3(9) + C4
6
9C3 + C4 = 121.5W
Continuity Conditions. At x1 = 3 ft and x2 = 9 ft,
(5)
dv2
dv1
. Thus, Eqs. (1) and
= dx1
dx2
(3) give
1
3
- W A 32 B + 4.5W = - c - W A 92 B + C3 d
2
2
C3 = 49.5W
Substituting the value of C3 into Eq. (5),
C4 = -324W
Substituting the values of C3 and C4 into Eq. (4),
v2 =
1
1
a - Wx2 3 + 49.5Wx2 - 324Wb
EI
6
At x2 = 0, v2 = -3.5 in. Then,
-324W(1728)
-3.5 =
1.5 A 106 B c
1
(18) A 2 3 B d
12
W = 112.53 lb = 113 lb
Ans.
885
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*12–4. Determine the equations of the elastic curve using
the x1 and x2 coordinates. EI is constant.
P
A
EI
d2v1
dx1 2
= M1 (x)
M1(x) = 0;
EI
d v1
dx1
2
L
= 0
x3
dv1
= C1
dx1
(1)
EI v1 = C1x1 + C2
(2)
EI
M2(x) = Px2 - P(L - a)
EI
EI
d2 v2
dx2 2
= Px2 - P(L - a)
dv2
P 2
=
x - P(L - a)x2 + C3
dx2
2 2
EI v2 =
(3)
P(L - a)x22
P 3
x2 + C3x2 + C4
6
2
(4)
Boundary conditions:
At x2 = 0,
dv2
= 0
dx2
From Eq. (3), 0 = C3
At x2 = 0, v2 = 0
0 = C4
Continuity condition:
At x1 = a, x2 = L - a;
dv1
dv2
= dx1
dx2
From Eqs. (1) and (3),
C1 = - c
P(L - a)2
- P(L - a)2 d ;
2
C1 =
P(L - a)2
2
At x1 = a, x2 = L - a, v1 = v2
From Eqs. (2) and (4),
a
P(L - a)3
P(L - a)3
P(L - a)2
b a + C2 =
2
6
2
C2 = -
Pa(L - a)2
P(L - a)3
2
3
From Eq. (2),
v1 =
P
[3(L - a)2x1 - 3a(L - a)2 - 2(L - a)3]
6EI
Ans.
For Eq. (4),
v2 =
B
x1
2
P
[x2 - 3(L - a)x33]
6EI 2
Ans.
886
L
2
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11:52 AM
Page 887
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•12–5.
Determine the equations of the elastic curve for
the beam using the x1 and x2 coordinates. EI is constant.
P
A
B
x1
L
Moment Functions. Referring to the FBDs of the beam’s cut segments shown in
Fig. b and c,
1
P(x1) = 0
2
a + ©MO = 0;
M(x1) +
a + ©MO = 0;
-Px2 - M(x2) = 0
M(x1) = -
P
x
2 1
And
EI
M(x2) = -Px2
d2v
= M(x)
dx2
For coordinate x1,
EI
EI
d2v1
dx1 2
= -
P
x
2 1
dv1
P
= - x1 2 + C1
dx1
4
EI v1 = -
(1)
P 3
x + C1x + C2
12 1
(2)
For coordinate x2,
EI
EI
d2v2
dx2 2
= -Px2
dv2
P
= - x2 2 + C3
dx2
2
EI v2 = -
(3)
P 3
x + C3x2 + C4
6 2
(4)
At x1 = 0, v1 = 0. Then, Eq (2) gives
EI(0) = -
P
(0) + C1(0) + C2
12
C2 = 0
At x1 = L, v1 = 0. Then, Eq (2) gives
EI(0) = At x2 =
P
(L3) + C1L + 0
12
C1 =
PL2
12
L
, v2 = 0. Then Eq (4) gives
2
EI(0) = -
P L 3
L
a b + C3 a b + C4
6 2
2
C3L + 2C4 =
PL3
24
(5)
887
x2
L
2
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11:52 AM
Page 888
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•12–5.
Continued
At x1 = L and x2 =
-
dv2
L dv1
,
= . Thus, Eqs. (1) and (3) gives
2 dx1
dx2
P 2
P L 2
PL2
= - c - a b + C3 d
AL B +
4
12
2 2
C3 =
7PL2
24
Substitute the result of C3 into Eq. (5)
C4 = -
PL3
8
Substitute the values of C1 and C2 into Eq. (2) and C3 and C4 into Eq (4),
v1 =
P
A -x1 3 + L2x1 B
12EI
Ans.
v2 =
P
A -4x2 3 + 7L2x2 - 3L3 B
24EI
Ans.
888
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11:52 AM
Page 889
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–6. Determine the equations of the elastic curve for the
beam using the x1 and x3 coordinates. Specify the beam’s
maximum deflection. EI is constant.
P
A
Support Reactions and Elastic Curve: As shown on FBD(a).
L
Slope and Elastic Curve:
For M(x1) = -
x3
d2v
= M(x)
dx2
EI
P
x.
2 1
EI
d2y1
dx21
EI y1 = For M(x3) = Px3 -
= -
P
x
2 1
dy1
P
= - x21 + C1
dx1
4
EI
[1]
P 3
x + C1x1 + C2
12 1
[2]
3PL
.
2
EI
d2y3
dx23
= Px3 -
3PL
2
dy3
P 2
3PL
=
x3 x3 + C3
dx3
2
2
EI
EI y3 =
[3]
P 3
3PL 3
x x3 + C3x3 + C4
6 3
4
[4]
Boundary Conditions:
y1 = 0 at x1 = 0. From Eq. [2], C2 = 0
y1 = 0 at x1 = L. From Eq. [2].
0 = -
PL3
+ C1L
12
C1 =
PL2
12
y3 = 0 at x3 = L. From Eq. [4].
0 =
PL3
3PL3
+ C3L + C4
6
4
0 = -
7PL3
+ C3L + C4
12
[5]
Continuity Condition:
At x1 = x3 = L,
-
dy1
dy3
. From Eqs. [1] and [3],
=
dx1
dx3
PL2
PL2
PL2
3PL2
+
=
+ C3
4
12
2
2
From Eq. [5], C4 = -
B
x1
Moment Function: As shown on FBD(b) and (c).
C3 =
5PL2
6
PL3
4
889
L
2
12 Solutions 46060
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11:52 AM
Page 890
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12–6.
Continued
The Slope: Substitute the value of C1 into Eq. [1],
dy1
P
=
A L2 - 3x21 B
dx1
12EI
dy1
P
= 0 =
A L2 - 3x21 B
dx1
12EI
x1 =
L
23
The Elastic Curve: Substitute the values of C1, C2, C3, and C4 into Eqs. [2] and [4],
respectively.
y1 =
Px1
A -x21 + L2 B
12EI
yO = y1 |x1 =
y3 =
L
23
=
PA L B
23
12EI
Ans.
a-
0.0321PL3
L3
+ L2 b =
3
EI
P
A 2x33 - 9Lx23 + 10L2x3 - 3L3 B
12EI
Ans.
yC = y3 |x3 = 32 L
=
2
P
3 3
3
3
c2 a L b - 9La Lb + 10L2 a L b - 3L3 d
12EI
2
2
2
= -
PL3
8EI
ymax =
PL3
8EI
Hence,
Ans.
890
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11:52 AM
Page 891
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12–7. The beam is made of two rods and is subjected to
the concentrated load P. Determine the maximum
deflection of the beam if the moments of inertia of the rods
are IAB and IBC , and the modulus of elasticity is E.
EI
P
B
A
C
l
d2y
= M(x)
dx2
L
M1(x) = - Px1
EIBC
EIBC
d2y1
dx1 2
= - Px1
dy1
Px21
= + C1
dx1
2
EIBC y1 = -
(1)
Px31
+ C1x1 + C2
6
(2)
M2(x) = - Px2
EIAB
EIAB
d2y2
dx2 2
= - Px2
dy2
P
= - x2 2 + C3
dx2
2
EIAB y2 = -
(3)
P 3
x + C3x2 + C4
2 2
(4)
Boundary conditions:
At x2 = L,
0 = -
dy2
= 0
dx2
PL2
+ C3;
2
C3 =
PL2
2
At x2 = L, y = 0
0 = -
PL3
PL3
+
+ C4;
6
2
C4 = -
PL3
3
Continuity Conditions:
At x1 = x2 = l,
dy1
dy2
=
dx1
dx2
891
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11:52 AM
Page 892
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–7.
Continued
From Eqs. (1) and (3),
1
PI 2
1
PI 2
PL2
cc+ C1 d =
+
d
EIBC
2
EIAB
2
2
C1 =
IBC
PL2
Pl2
Pl2
c+
d +
IAB
2
2
2
At x1 = x2 = l, y1 = y2
From Eqs. (2) and (4),
IBC
PL2
Pl2
1
Pl3
Pl2
e+ c
a+
b +
dl + C2 f
EIBC
6
IAB
2
2
2
=
1
PL2l
PL3
Pl3
c+
d
EIAB
6
2
3
C2 =
IBC PL3
IBC Pl3
Pl3
IAB 3
IAB 3
3
Therefore,
y1 =
Px1 3
IBC
1
Pl2
PL2
Pl2
e+ c
a+
b +
dx1
EIBC
6
IAB
2
2
2
+
IBC PL3
IBC Pl3
Pl3
f
IAB 3
IAB 3
3
At x1 = 0, y1 |x = 0 = ymax
ymax =
=
IBC Pl3
IBC PL3
IAB 3
I
Pl3
P
e
f =
e l3 - L3 - a
bl f
EIBC IAB 3
IAB 3
3
3EIAB
IBC
IAB 3
P
e a1 b l - L3 f
3EIAB
IBC
Ans.
892
12 Solutions 46060
6/11/10
11:52 AM
Page 893
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*12–8. Determine the equations of the elastic curve for
the beam using the x1 and x2 coordinates. EI is constant.
P
Referring to the FBDs of the beam’s cut segments shown in Fig. b and c,
a + ©MO = 0;
M(x1) +
PL
- Px1 = 0
2
M(x1) = Px1 -
PL
2
x1
x2
And
a + ©MO = 0;
EI
L
2
M(x2) = 0
d2v
= M(x)
dx2
For coordinate x1,
EI
EI
d2v1
dx21
= Px1 -
PL
2
dv1
P 2
PL
=
x x + C1
dx1
2 1
2 1
EI v1 =
(1)
P 3
PL 2
x x + C1x1 + C2
6 1
4 1
(2)
For coordinate x2,
EI
dx22
= 0
dv2
= C3
dx2
(3)
EI v2 = C3x2 = C4
(4)
EI
At x1 = 0,
d2v2
dv1
= 0. Then, Eq.(1) gives
dx1
EI(0) =
PL
P 2
(0 ) (0) + C1
2
2
C1 = 0
At x1 = 0, v1 = 0. Then, Eq(2) gives
EI(0) =
At x1 = x2 =
PL 2
P 3
(0 ) (0 ) + 0 + C2
6
4
C2 = 0
dv2
L dv1
=
,
. Thus, Eqs.(1) and (3) gives
2 dx1
dx2
P L 2
PL L
a b a b = C3
2 2
2
2
Also, at x1 = x2 =
C3 = -
PL2
8
L
, v = v2. Thus, Eqs, (2) and (4) gives
2 1
PL L 2
PL2 L
P L 3
a b a b = ab a b + C4
6 2
4 2
8
2
C4 =
PL3
48
Substitute the values of C1 and C2 into Eq. (2) and C3 and C4 into Eq (4),
v1 =
P
A 2x31 - 3Lx21 B
12EI
Ans.
v2 =
PL2
(-6x2 + L)
48EI
Ans.
893
L
2
12 Solutions 46060
6/11/10
11:52 AM
Page 894
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•12–9.
Determine the equations of the elastic curve using
the x1 and x2 coordinates. EI is constant.
EI
d2y
= M(x)
dx2
M1 =
EI
EI
P
A
x1
Pb
x
L 1
d2y1
dx21
=
a
EI y3 =
L
(1)
Pb 3
x + C3x1 + C2
6L 1
(2)
Pb
x - P(x2 - a)
L 2
But b = L - a. Thus
M2 = Paa1 -
EI
EI
d2y2
dx2 2
x2
b
L
= Pa a1 -
x2
b
L
dy2
x22
= Paax2 b + C3
dx2
2L
EI y2 = Pa a
(3)
x22
x22
b + C3x2 + C4
2
6L
(4)
Applying the boundary conditions:
y1 = 0 at x1 = 0
Therefore,C2 = 0,
y2 = 0 at x2 = L
0 =
b
x2
Pb
x
L 1
dy1
Pb 2
=
x + C1
dx1
2L 1
M2 =
B
Pa L2
+ C3L + C4
3
(5)
Applying the continuity conditions:
y1 |x1 = a = y2 |x2 = a
Pb 3
a2
a3
a + C1a = Pa a
b + C3a + C4
6L
2
6L
(6)
dy1
dy2
2
2
=
dx1 x1 = a
dx2 x2 = a
a2
Pb 2
a + C1 = Pa a a b + C3
2L
2L
(7)
894
12 Solutions 46060
6/11/10
11:52 AM
Page 895
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•12–9.
Continued
Solving Eqs. (5), (6) and (7) simultaneously yields,
C1 = C4 =
Pb 2
A L - b2 B ;
6L
C3 = -
Pa
A 2L2 + a2 B
6L
Pa3
6
Thus,
EIy1 =
Pb 3
Pb 2
x A L - b 2 B x1
6L 1
6L
or
v1 =
Pb
A x3 - A L2 - b2 B x1 B
6EIL 1
Ans.
and
EIy2 = Pa a
y2 =
x22
x32
Pa
Pa3
b A 2L2 + a2 B x2 +
2
6L
6L
6
Pa
C 3x22 L - x32 - A 2L2 + a2 B x2 + a2L D
6EIL
Ans.
895
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11:52 AM
Page 896
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12–10. Determine the maximum slope and maximum
deflection of the simply supported beam which is subjected
to the couple moment M0 . EI is constant.
M0
A
B
L
Support Reactions and Elastic Curve: As shown on FBD(a).
Moment Function: As shown on FBD(b).
Slope and Elastic Curve:
EI
d2y
= M(x)
dx2
EI
M0
d2y
=
x
2
L
dx
EI
M0 2
dy
=
x + C1
dx
2L
EI y =
[1]
M0 3
x + C1x + C2
6L
[2]
Boundary Conditions:
y = 0 at x = 0. From Eq. [2].
0 = 0 + 0 + C2
C2 = 0
y = 0 at x = L. From Eq. [2].
0 =
M0 3
A L B + C1 (L)
6L
C1 = -
M0L
6
The Slope: Substitute the value of C1 into Eq. [1],
M0
dy
=
A 3x2 - L2 B
dx
6LEI
M0
dy
= 0 =
A 3x2 - L2 B
dx
6LEI
uB =
x =
23
L
3
M0L
dy
2
= dx x = 0
6EI
umax = uA =
M0L
dy 2
=
dx x = L
3EI
Ans.
The Elastic Curve: Substituting the values of C1 and C2 into Eq. [2],
y =
ymax occurs at x =
M0
A x3 - L2x B
6LEI
23
L,
3
ymax = -
23M0L2
Ans
27EI
The negative sign indicates downward displacement.
896
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6/11/10
11:52 AM
Page 897
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12–11. Determine the equations of the elastic curve for the
beam using the x1 and x2 coordinates. Specify the beam’s
maximum deflection. EI is constant.
P
Referring to the FBDs of the beam’s cut segments shown in Fig. b and c,
2
M(x1) - Px1 = 0
3
a + ©M0 = 0;
A
B
x1
2P
x
M(x1) =
3 1
a
2a
x2
And
1
P
P(3a - x2) - M(x2) = 0 M(x2) = Pa x
3
3 2
a + ©M0 = 0;
EI
d2y
= M(x)
dx2
For coordinate x1,
EI
EI
d2y1
dx21
=
2P
x
3 1
dy1
P 2
=
x + C1
dx1
3 1
EI y1 =
(1)
P 3
x = C1x1 + C2
9 1
(2)
For coordinate x2,
EI
EI
d2y2
dx2
2
= Pa -
P
x
3 2
dy2
P 2
= Pax2 x + C3
dx2
6 2
EI y2 =
(3)
Pa 2
P 3
x x + C3x2 + C4
2 2
18 2
(4)
At x1 = 0, y1 = 0. Then, Eq (2) gives
EI(0) =
P 3
A 0 B + C1(0) + C2
9
C2 = 0
At x2 = 3a, y2 = 0. Then Eq (4) gives
EI(0) =
Pa
P
(3a)2 (3a)3 + C3(3a) + C4
2
18
C3(3a) + C4 = -3Pa3
At x1 = x2 = a,
(5)
dy1
dy2
=
. Thus, Eq. (1) and (3) gives
dx1
dx2
P 2
P 2
a + C1 = Pa(a) a + C3
3
6
C1 - C3 =
Pa2
2
(6)
897
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11:52 AM
Page 898
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12–11.
Continued
Also, At x1 = x2 = a, y1 = y2. Thus, Eqs, (2) and (4) gives.
P 3
Pa 2
P 3
a + C1a =
(a ) a + C3a + C4
9
2
18
C1a - C3a - C4 =
Pa3
3
(7)
Solving Eqs. (5), (6) and (7),
C4 =
Pa3
6
C3 = -
19 Pa2
18
C1 = -
5Pa2
9
Substitute the values of C1 into Eq. (1) and C3 into Eq. (3),
dy1
P
=
A 3x1 2 - 5a2 B
dx1
9EI
dy1
P
= 0 =
A 3x1 2 - 5a2 B
dx1
9EI
x1 =
5
a 7 a (Not Valid)
A3
And
dy2
P
=
A 18ax2 - 3x2 2 - 19a2 B
dx2
18EI
dy2
P
= 0 =
A 18ax2 - 3x2 2 - 19a2)
dx2
18EI
x2 = 4.633a 7 3a (Not Valid)
x2 = 1.367a
Substitute the values of C1 and C2 into Eq. (2) and C3 and C4 into Eq.(4),
y1 =
P
A x 3 - 5a2x1 B
9EI 1
Ans.
y2 =
P
A -x2 3 + 9ax2 2 - 19a2x2 + 3a3 B
18EI
Ans.
Vmax occurs at x2 = 1.367a. Thus.
ymax = -
0.484 Pa3
0.484 Pa3
=
T
EI
EI
Ans.
898
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11:52 AM
Page 899
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*12–12. Determine the equations of the elastic curve for
the beam using the x1 and x2 coordinates. Specify the slope at
A and the maximum displacement of the shaft. EI is constant.
P
P
a
a
A
B
Referring to the FBDs of the beam’s cut segments shown in Fig. b and c,
a + ©M0 = 0;
M(x1) - Px1 = 0
x1
M(x1) = Px1
x2
And
L
a + ©M0 = 0;
M(x2) - Pa = 0
EI
M(x2) = Pa
d2y
= M(x)
dx2
For coordinate x1,
EI
EI
d2y1
= Px1
dx21
dy1
P 2
=
x + C1
dx1
2 1
(1)
P 3
x + C1x1 + C2
6 1
EI y1 =
(2)
For coordinate x2,
EI
EI
d2y2
dx2 2
= Pa
dy2
= Pax2 + C3
dx2
(3)
Pa 2
x + C3x2 + C4
2 2
EI y2 =
(4)
At x1 = 0, y1 = 0. Then, Eq. (2) gives
EI (0) =
Due to symmetry, at x2 =
P 3
(0 ) + C1(0) + C2
6
L dv2
= 0. Then, Eq. (3) gives
,
2 dx2
EI (0) = Pa a
At x1 = x2 = a,
C2 = 0
L
b + C3
2
C3 = -
PaL
2
dy1
dy2
=
. Thus, Eqs(1) and (3) give
dx1
dx2
P 2
PaL
a + C1 = Pa (a) + a b
2
2
C1 =
Pa2
PaL
2
2
899
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Page 900
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*12–12.
Continued
Also, at x1 = x2 = a, y1 = y2. Thus, Eq. (2) and (4) give
P 3
Pa2
PaL
Pa 2
PaL
a + a
ba =
(a ) + a ba + C4
6
2
2
2
2
C4 =
Pa3
6
Substituting the value of C1 and C2 into Eq. (2) and C3 and C4 into Eq.(4),
y1 =
P
C x 3 + a(3a - 3L)x1 D
6EI 1
Ans.
y2 =
Pa
A 3x2 2 - 3Lx2 + a2 B
6EI
Ans.
Due to symmetry, ymax occurs at x2 =
ymax =
L
. Thus
2
Pa
Pa
A 4a2 - 3L2 B =
A 3L2 - 4a2 B T
24EI
24EI
Ans.
Substitute the value C1 into Eq (1),
dy1
P
=
A x 2 + a2 - aL B
dx1
2EI 1
At point A, x1 = 0. Then
uA =
dy1
Pa
Pa
2
=
(a - L) =
(L - a) T
dx1 x1 = 0
2EI
2EI
900
Ans.
12 Solutions 46060
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11:52 AM
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12–13. The bar is supported by a roller constraint at B, which
allows vertical displacement but resists axial load and moment.
If the bar is subjected to the loading shown, determine the
slope at A and the deflection at C. EI is constant
d2y1
EI
EI
dx21
P
C
A
L
2
= M1 = Px1
dy1
Px21
=
+ C1
dx1
2
EI y1 =
Px31
+ C1x1 + C2
6
EI
d2y2
PL
= M2 =
dx2
2
EI
dy2
PL
=
x + C3
dx2
2 2
EI y2 =
PL 2
x + C3x2 + C4
4 2
Boundary conditions:
At x1 = 0, y1 = 0
0 = 0 + 0 + C2 ;
At x2 = 0,
At x1 =
P(L2 )3
C3 = 0
P A L2 B 2
2
dy1
dy2
L
L
= , x = , y1 = y2,
2 2
2
dx1
dx2
+ C1 a
6
L 2
PL( 2 )
L
b =
+ C4
2
4
+ C1 = -
C4 = -
C2 = 0
dy2
= 0
dx2
0 + C3 = 0 ;
PL A L2 B
2
;
3
C1 = - PL2
8
11
PL3
48
At x1 = 0
dy1
3 PL2
= uA = dx1
8 EI
At x1 =
yC =
yC =
Ans.
L
2
P A L2 B 3
6EI
- a
B
3 PL2 L
ba b + 0
8 EI
2
-PL2
6EI
Ans.
901
L
2
12 Solutions 46060
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12–14. The simply supported shaft has a moment of inertia
of 2I for region BC and a moment of inertia I for regions
AB and CD. Determine the maximum deflection of the
beam due to the load P.
M1 (x) =
P
x
2 1
M2(x) =
P
x
2 2
P
B
A
L
–
4
Elastic curve and slope:
EI
EI
EI
d2v
= M(x)
dx2
d2v1
dx1
=
2
P
x
2 1
dv1
Px21
=
+ C1
dx1
4
EIv1 =
2EI
2EI
(1)
Px31
+ C1x1 + C2
12
d2v2
=
dx2 2
(2)
P
x
2 2
dv2
Px22
=
+ C3
dx1
4
2EIv2 =
(3)
Px32
+ C3x2 + C4
12
(4)
Boundary Conditions:
v1 = 0 at x1 = 0
From Eq. (2), C2 = 0
dv2
L
= 0 at x2 =
dx2
2
From Eq. (3),
0 =
PL2
+ C3
16
C3 =
PL2
16
Continuity conditions:
dv1
dv2
L
=
at x1 = x2 =
dx1
dx2
4
902
C
L
–
4
L
–
4
D
L
–
4
12 Solutions 46060
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12–14.
Continued
From Eqs. (1) and (3),
PL2
1 PL2
PL2
+ C1 =
- a
b
64
128
2 16
C1 =
-5PL2
128
v1 = v2 at x1 = x2 =
L
4
From Eqs. (2) and (4)
PL3
5PL2 L
PL3
1 PL2 L
1
a b =
- a
b a b + C4
768
128
4
1536
2 16
4
2
C4 =
-PL3
384
v2 =
P
A 32x32 - 24L2 x2 - L3 B
768EI
vmax = v2 2
=
x2 = L2
-3PL3
3PL3
=
T
256EI
256EI
Ans.
903
12 Solutions 46060
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11:52 AM
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12–15. Determine the equations of the elastic curve for the
shaft using the x1 and x3 coordinates. Specify the slope at A
and the deflection at the center of the shaft. EI is constant.
P
P
A
Support Reactions and Elastic Curve: As shown on FBD(a).
x1
Moment Function: As shown on FBD(b) and (c).
x3
a
Slope and Elastic Curve:
EI
d2 y
= M(x)
dx2
For M(x1) = -Px1,
EI
EI
d2y1
dx21
= -Px1
dy1
P
= - x21 + C1
dx1
2
EI y1 = -
[1]
P 3
x + C1x1 + C2
6 1
[2]
For M(x3) = -Pa,
EI
EI
d2y3
dx23
= -Pa
dy3
= -Pax3 + C3
dx3
EI y3 = -
[3]
Pa 2
x + C3x3 + C4
2 3
[4]
Boundary Conditions:
y1 = 0 at x1 = a. From Eq. [2],
0 = Due to symmetry,
Pa3
+ C1a + C2
6
[5]
dy3
b
= 0 at x3 = . From Eq. [3]
dx3
2
b
0 = -Paa b + C3
2
C3 =
Pab
2
y3 = 0 at x3 = 0 From Eq.[4]. C4 = 0
Continuity Condition:
At x1 = a and x3 = 0,
dy1
dy3
=
. From Eqs. [1] and [3],
dx1
dx3
-
Pa2
Pab
+ C1 =
2
2
C1 =
From Eq. [5]
C2 = -
Pa2
(2a + 3b)
6
B
Pa
(a + b)
2
904
b
a
12 Solutions 46060
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12–15. Continued
The Slope: Either Eq. [1] or [3] can be used. Substitute the value of C1 into Eq. [1],
dy1
P
=
C -x21 + a(a + b) D
dx1
2EI
uA =
dy1
P
Pab
2
=
C -a2 + a(a + b) D =
dx1 x1 = a
2EI
2EI
Ans.
The Elastic Curve: Substitute the values of C1, C2, C3, and C4 into Eqs. [2] and [4],
respectively,
y1 =
P
C -x31 + 3a(a + b)x1 - a2(2b + 3b) D
6EI
Ans.
y3 =
Pax3
( -x3 + b)
2EI
Ans.
yC = y3 |x3 = b2
=
=
Pa A b2 B
2EI
a-
b
+ bb
2
Pab2
8EI
Ans.
905
12 Solutions 46060
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11:52 AM
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*12–16. The fence board weaves between the three smooth
fixed posts. If the posts remain along the same line, determine
the maximum bending stress in the board. The board has a
width of 6 in. and a thickness of 0.5 in. E = 1.60(103) ksi.
Assume the displacement of each end of the board relative
to its center is 3 in.
4 ft
3 in.
A
Support Reactions and Elastic Curve: As shown on FBD(a).
Moment Function: As shown on FBD(b).
Slope and Elastic Curve:
EI
d2y
= M(x)
dx2
EI
EI
d2y
P
=
x
2
dx2
dy
P 2
=
x + C1
dx
4
EI y =
[1]
P 3
x + C1x + C2
12
Boundary Conditions: Due to symmetry,
[2]
L
dy
= 0 at x = .
dx
2
Also, y = 0 at x = 0.
From Eq. [1] 0 =
P L 2
a b + C1
4 2
From Eq. [2] 0 = 0 + 0 + C2
C1 = -
PL2
16
C2 = 0
The Elastic Curve: Substitute the values of C1 and C2 into Eq. [2],
y =
Px
A 4x2 - 3L2 B
48EI
[1]
Require at x = 48 in., y = -3 in. From Eq.[1],
P(48)
-3 =
1
48(1.60) A 106 B A 12
B (6) A 0.53 B
C 4 A 482 B - 3 A 962 B D
P = 16.28 lb
Maximum Bending Stress: From the moment diagram, the maximum moment is
Mmax = 390.625 lb # in. Applying the flexure formula,
smax =
4 ft
390.625(0.25)
Mc
=
= 1562.5 psi = 1.56 ksi
1
I
(6) A 0.53 B
Ans.
12
906
B
C
12 Solutions 46060
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•12–17.
Determine the equations of the elastic curve for
the shaft using the x1 and x2 coordinates. Specify the slope
at A and the deflection at C. EI is constant.
a + ©MO = 0;
a + ©MO = 0;
MO - M(x2) = 0
M(x2) = MO
d2y
= M(x)
dx2
For coordinate x1,
EI
EI
d2y1
dx1 2
=
MO
x
L 1
dy1
MO 2
=
x + C1
dx1
2L 1
EI y1 =
(1)
MO 3
x + C1x1 + C2
6L 1
(2)
For coordinate x2,
EI
EI
d2y2
dx2 2
= MO
dy2
= MOx2 + C3
dx2
EI y2 =
(3)
MO 2
x + C3x2 + C4
2 2
(4)
At x1 = 0, y1 = 0. Then, Eq. (2) gives
EI (0) =
MO 3
(0 ) + C1(0) + C2
6L
C2 = 0
At x1 = L, y1 = 0. Then, Eq. (2) gives
C1 =
Also, at x2 =
-ML
6
L
, y = 0. Then Eq. (4) gives.
2 2
EI(0) =
MO L 2
L
a b + C3 a b + C4
2
2
2
C3L + 2C4 = -
MOL2
4
(5)
907
C
x2
L
MO
M(x1) =
x
L 1
And
EI
B
x1
Referring to the FBDs of the shaft’s cut segments shown in Fig. b and c,
MO
M(x1) x = 0
L 1
M0
A
L
2
12 Solutions 46060
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11:52 AM
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•12–17.
Continued
At x1 = L and x2 =
dy2
L dy1
,
= . Then, Eq. (1) and (3) give
2 dx1
dx2
MO 2
MOL
L
= - cMO a b + C3 d
AL B 2L
6
2
C3 = -
5MOL
6
Substitute the result of C3 into Eq. (5),
C4 =
7M0L2
24
Substitute the value of C1 into Eq. (1),
dy1
MO
=
A 3x1 2 - L2 B
dx1
6LEI
At A, x1 = 0. Thus
uA =
MOL
dy1
MO
2
= =
dx1 x1 = 0
6EI
6EI
Ans.
Substitute the values of C1 and C2 into Eq (2) and C3 and C4 into Eq. (4),
y1 =
MO
A x 3 - L2x1 B
6EIL 1
Ans.
y2 =
MO
A 12x2 2 - 20 Lx2 + 7L2 B
24EI
Ans.
7MOL2
24EI
Ans.
At C, x2 = 0. Thus
yC = y2 2
=
x2 = 0
c
908
12 Solutions 46060
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11:52 AM
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12–18. Determine the equation of the elastic curve for the
beam using the x coordinate. Specify the slope at A and the
maximum deflection. EI is constant.
M0
A
M0
x
B
L
Referring to the FBD of the beam’s cut segment shown in Fig. b,
a + ©MO = 0;
M(x) +
2MO
x - MO = 0
L
EI
d2y
= M(x)
dx2
EI
2MO
d2y
= MO x
L
dx2
EI
MO 2
dy
= MOx x + C1
dx
L
M(x) = MO -
2MO
x
L
(1)
MO 3
MO 2
x x + C1x + C2
2
3L
EI y =
(2)
At x = 0, y = 0. Then Eq (2) gives
EI(0) =
MO 3
MO 2
A0 B A 0 B + C1(0) + C2
2
3L
C2 = 0
Also, at x = L, y = 0. Then Eq (2) gives
EI(0) =
MO 2
MO 3
AL B A L B + C1L + 0
2
3L
C1 = -
MOL
6
Substitute the value of C1 into Eq (1),
MO
dy
=
A 6Lx - 6x2 - L2 B
dx
6EIL
MO
dy
= 0 =
A 6Lx - 6x2 - L2 B
dx
6EIL
x = 0.2113 L
and
0.7887 L
At A, x = 0. Thus
uA = -
MOL
6EI
Ans.
Substitute the values of C1 and C2 into Eq (2)
y =
MO
A 3Lx2 - 2x3 - L2x B
6EIL
Ans.
909
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12–18.
Continued
vmax occurs at x = 0.21132 L
ymax =
or
0.7887 L. Thus,
MO
c3L(0.2113L)2 - 2(0.2113L)3 - L2(0.2113L) d
6EIL
= -
0.0160 MOL2
0.0160 MOL2
=
EI
EI
Ans.
T
and
ymax =
=
MO
c3L(0.7887L)2 - 2(0.7887L)3 - L2(0.7887L) d
6EIL
0.0160 MOL2
EI
Ans.
c
910
12 Solutions 46060
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11:52 AM
Page 911
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12–19. Determine the deflection at the center of the beam
and the slope at B. EI is constant.
M0
A
M0
x
B
L
Referring to FBD of the beam’s cut segment shown in Fig. b,
a + ¢Mo = 0;
M(x) +
2Mo
x - Mo = 0
L
EI
d2v
= M(x)
dx2
EI
2Mo
d2v
= Mo x
L
dx2
EI
Mo 2
dv
= Mox x + C1
dx
L
EI v =
M(x) = Mo -
2Mo
x
L
(1)
Mo 3
Mo 2
x x + C1x + C2
2
3L
(2)
At x = 0, v = 0. Then Eq. (2) gives
EI(0) =
Mo 3
Mo 2
(0 ) A 0 B + C1(0) + C2
2
3L
C2 = 0
Also, at x = L, v = 0. Then Eq. (2) gives
EI (0) =
Mo 2
Mo 3
AL B A L B + C1L + 0
2
3L
C1 = -
MoL
6
Substitute the value of C1 into Eq. (1),
Mo
dv
=
A 6Lx - 6x2 - L2 B
dx
6EIL
At B, x = L. Thus
uB =
MoL
MoL
dv 2
= =
dx x = L
6EI
6EI
Ans.
Substitute the values of C1 and C2 into Eq. (2),
v =
Mo
A 3Lx2 - 2x3 - L2x B
6EIL
At the center of the beam, x =
L
. Thus
2
v冷x = L2 = 0
Ans.
911
12 Solutions 46060
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11:52 AM
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*12–20. Determine the equations of the elastic curve
using the x1 and x2 coordinates, and specify the slope at A
and the deflection at C. EI is constant.
8 kip
A
C
B
x1
x2
20 ft
Referring to the FBDs of the beam’s cut segments shown in Fig. b, and c,
M(x1) = (-5x1) kip # ft
a + ©Mo = 0;
M(x1) + 5x1 = 0
a + ©Mo = 0;
-M(x2) - 8x2 - 20 = 0 M(x2) = ( -8x2 - 20) kip # ft
And
EI
d2v
= M(x)
dx2
For coordinate x1,
EI
EI
d2v1
dx21
= (-5x1) kip # ft
dv1
5
= a - x21 + C1 b kip # ft2
dx1
2
(1)
5
EI v1 = a - x1 3 + C1x1 + C2 b kip # ft3
6
(2)
For coordinate x2,
EI
EI
d2v2
dx2 2
= (-8x2 - 20) kip # ft
dv2
= A -4x2 2 - 20x2 + C3 B kip # ft2
dx2
(3)
4
EI v2 = a - x2 3 - 10x2 2 + C3x2 + C4 b kip # ft3
3
(4)
At x1 = 0, v1 = 0. Then, Eq (2) gives
EI(0) = -
5 3
A 0 B + C1(0) + C2
6
C2 = 0
Also, at x1 = 20 ft, v1 = 0. Then, Eq (2) gives
EI(0) = -
5
A 203 B + C1 (20) + 0
6
C1 = 333.33 kip # ft2
Also, at x2 = 10 ft, v2 = 0. Then, Eq. (4) gives
EI(0) = -
4
A 103 B - 10 A 102 B + C3(10) + C4
3
10C3 + C4 = 2333.33
(5)
912
10 ft
20 kip⭈ft
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*12–20.
Continued
At x1 = 20 ft and x2 = 10 ft,
-
dv1
dv2
= . Then Eq. (1) and (3) gives
dx1
dx2
5
A 202 B + 333.33 = - C -4 A 102 B - 20(10) + C3 D
2
C3 = 1266.67 kip # ft2
Substitute the value of C3 into Eq (5),
C4 = -10333.33 kip # ft3
Substitute the value of C1 into Eq. (1),
dv1
5
1
a - x1 2 + 333.33b kip # ft2
=
dx1
EI
2
At A, x1 = 0. Thus,
uA =
333 kip # ft2
dv1
2
=
dx1 x1 = 0
EI
uA
Ans.
Substitute the values of C1 and C2 into Eq. (2) and C3 and C4 into Eq (4),
v1 =
1
5
a - x1 3 + 333 x1 b kip # ft3
EI
6
Ans.
v2 =
4
1
a - x2 3 - 10x2 2 + 1267x2 - 10333b kip # ft3
EI
3
Ans.
At C, x2 = 0. Thus
vC = v2 冷x2 = 0 = -
10 333 kip # ft3
10 333 kip # ft3
=
T
EI
EI
913
Ans.
12 Solutions 46060
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11:52 AM
Page 914
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•12–21.
Determine the elastic curve in terms of the and
coordinates and the deflection of end C of the overhang
beam. EI is constant.
w
A
C
Support Reactions and Elastic Curve. As shown in Fig. a.
B
Moment Functions. Referring to the free-body diagrams of the beam’s cut
segments, Fig. b, M(x1) is
a + ©MO = 0;
M(x1) +
wL
x = 0
8 1
M(x1) = -
-M(x2) - wx2 a
x2
b = 0
2
M(x2) = -
w 2
x
2 2
Equations of Slope and Elastic Curve.
EI
d2v
= M(x)
dx2
For coordinate x1,
EI
EI
d2v1
dx1 2
= -
wL
x
8 1
dv1
wL 2
= x + C1
dx1
16 1
EIv1 = -
(1)
wL 3
x + C1x1 + C2
48 1
(2)
For coordinate x2,
EI
EI
d2v2
dx2 2
= -
w 2
x
2 2
dv2
w
= - x2 3 + C3
dx2
6
EIv2 = -
(3)
w 4
x + C3x2 + C4
24 2
(4)
Boundary Conditions. At x1 = 0, v1 = 0. Then, Eq. (2) gives
EI(0) = -
wL 3
A 0 B + C1(0) + C2
48
C2 = 0
At x1 = L, v1 = 0. Then, Eq. (2) gives
EI(0) = At x2 =
wL 3
A L B + C1L + 0
48
C1 =
wL3
48
L
, v = 0. Then, Eq. (4) gives
2 2
EI(0) = -
w L 4
L
a b + C3 a b + C4
24 2
2
L
wL4
C3 + C4 =
2
384
(5)
914
x2
L
wL
x
8 1
and M(x2) is
a + ©MO = 0;
x1
L
2
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Page 915
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•12–21.
Continued
Continuity Conditions. At x1 = L Land x2 =
dv2
L dv1
,
= . Thus, Eqs. (1) and
2 dx1
dx2
(3) give
-
w L 3
wL3
wL 2
= - C - a b + C3 S
AL B +
16
48
6 2
C3 =
wL3
16
Substituting the value of C3 into Eq. (5),
C4 = -
11wL4
384
Substituting the values of C3 and C4 into Eq. (4),
v2 =
w
A -16x2 4 + 24L3x2 - 11L4 B
384EI
At C, x2 = 0. Thus,
vC = v2冷x2 = 0 = -
11wL4
11wL4
=
T
384EI
384EI
Ans.
915
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Page 916
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12–22. Determine the elastic curve for the cantilevered
beam using the x coordinate. Specify the maximum slope and
maximum deflection. E = 29(103) ksi.
3 kip/ft
A
Referring to the FBD of the beam’s cut segment shown in Fig. b,
a + ©Mo = 0;
M(x) + 81 + 12
A
B
A B - 13.5x = 0
9 ft
M(x) = A 13.5x - 0.05556x3 - 81 B kip # ft.
EI
d2v
= M(x)
dx2
EI
d2v
= A 13.5x - 0.05556x3 - 81 B kip # ft
dx2
EI
dv
= A 6.75x2 - 0.01389x4 - 81x + C1 B kip # ft2
dx
(1)
EI v = A 2.25x3 - 0.002778x5 - 40.5x2 + C1x + C2 B kip # ft3 (2)
At x = 0,
dv
= 0. Then, Eq (1) gives
dx
EI(0) = 6.75 A 02 B - 0.01388 A 04 B - 81(0) + C1
C1 = 0
Also, at x = 0, v = 0. Then Eq. (2) gives
EI(0) = 2.25 A 03 B - 0.002778 A 05 B - 40.5 A 02 B + 0 + C2
C2 = 0
Substitute the value of C1 into Eq (1) gives.
1
dv
=
A 6.75x2 - 0.01389x4 - 81x B kip # ft2
dx
EI
The Maximum Slope occurs at x = 9 ft. Thus,
umax =
273.375 kip # ft2
dv 2
= dx x = 9ft
EI
=
273.375 kip # ft2
EI
umax
For W14 * 30, I = 291 in4. Thus
u = 273.375 A 12 2 B = 0.00466 rad
Ans.
Substitute the values of C1 and C2 into Eq (2),
v =
1
A 2.25x3 - 0.002778x5 - 40.5x2 B kip # ft3
EI
The maximum deflection occurs at x = 9 ft, Thus,
vmax = v 冷x = 9 ft = -
=
=
B
x
1
x
3 x (x) 3
1804.275 kip # ft3
EI
1804.275 kip # ft3
T
EI
1804.275 A 12 3 B
29.0 A 103 B (291)
= 0.369 in T
Ans.
916
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12–23. The beam is subjected to the linearly varying
distributed load. Determine the maximum slope of the
beam. EI is constant.
w0
A
B
x
L
EI
d2y
= M(x)
dx2
EI
w0
d2y
=
A L2x - x3 B
2
6L
dx
EI
w0 L2x2
x4
dy
=
a
b + C1
dx
6L
2
4
EI y =
(1)
w0 L2x3
x5
a
b + C1x + C2
6L
6
20
(2)
Boundary conditions:
At x = 0, y = 0.
From Eq. (2), C2 = 0
At x = L, y = 0
From Eq. (2),
0 =
w0 L5
L5
a
b + C1L ;
6L 6
20
C1 = -
7w0L3
360
The slope:
From Eq.(1),
w0
dy
L2x2
x4
7L4
=
a
b
dx
6EIL
2
4
60
umax =
w0L3
dy 2
=
dx x = L
45EI
Ans.
917
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Page 918
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*12–24. The beam is subjected to the linearly varying
distributed load. Determine the maximum deflection of the
beam. EI is constant.
w0
A
B
x
L
EI
d2y
= M(x)
dx2
EI
w0
d2y
=
A L2x - x3 B
2
6L
dx
EI
w0 L2x2
x4
dy
=
a
b + C1
dx
6L
2
4
EI y =
(1)
w0 L2x3
x5
a
b + C1x + C2
6L
6
20
(2)
Boundary conditions:
y = 0 at x = 0.
From Eq. (2), C2 = 0
y = 0 at x = L.
From Eq. (2),
0 =
w0 L2
L5
a
b + C1L;
6L 6
20
C1 = -
7w0L3
360
w0
dy
L2x2
x4
7L4
=
a
b
dx
6EIL
2
4
60
dy
L2x2
x4
7L4
= 0 = a
b
dx
2
4
60
15x4 - 30L2x2 + 7L4 = 0;
y =
x = 0.5193L
w0x
A 10L2x2 - 3x4 - 7L4 B
360EIL
Substitute x = 0.5193L into y,
ymax = -
0.00652w0L4
EI
Ans.
918
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Page 919
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•12–25.
Determine the equation of the elastic curve
for the simply supported beam using the x coordinate.
Determine the slope at A and the maximum deflection. EI
is constant.
12 kN/m
A
B
x
6m
Referring to the FBD of the beam’s cut segment shown in Fig. b,
a + ©Mo = 0; M(x) +
1
1
(2x)(x) A x3 B - 36x = 0 M(x) = a36x - x3 bkN # m
2
3
EI
d2v
= M(x)
dx2
EI
d2v
1
= a36x - x3 b kN # m
2
3
dx
dv
1 4
= a18x2 x + C1 b kN # m2
dx
12
EI
EI v = a6x3 Due to the Symmetry,
(1)
1 5
x + C1x + C2 b kN # m3
60
(2)
dv
= 0 at x = 6 m. Then, Eq (1) gives
dx
EI(0) = 18 A 62 B -
1
A 64 B + C1
12
C1 = -540 kN # m2
Also, at x = 0, v = 0. Then, Eq (2) gives
EI(0) = 6 A 03 B -
1
A 05 B + C1(0) + C2
60
C2 = 0
Substitute the value of C1 into Eq. (1),
dv
1
1 4
=
a18x2 x - 540b kN # m2
dx
EI
12
At A, x = 0. Then
uA =
dv
540 kN # m2
540kN # m2
2
= =
dx x = 0
EI
EI
Ans.
Substitute the values of C1 and C2 into Eq (2)
v =
1
1 5
a6x3 x - 540xb kN # m3
EI
60
Ans.
Due to Symmetry, vmax occurs at mind span x = 6 m. Thus,
vmax =
1
1
c6 A 63 B A 65 B - 540(6) d
EI
60
= -
2074 kN # m3
2073.6 kN # m3
=
EI
EI
Ans.
T
919
6m
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12–26. Determine the equations of the elastic curve using
the coordinates x1 and x2 , and specify the slope and deflection
at B. EI is constant.
w
C
A
B
x1
a
x2
L
EI
d2y
= M(x)
dx2
For M1(x) = -
EI
EI
d2y1
= 2
dx1
w 2
wa2
x1 + wax1 2
2
w 2
wa2
x1 + wax1 2
2
dy1
w
wa 2
wa2
= - x21 +
x1 x + C1
dx1
6
2
2 1
EI y1 = -
(1)
w 4
wa 3
wa2 2
x1 +
x1 x1 + C1x1 + C2
24
6
4
For M2(x) = 0 ;
EI
d2y2
dx2 2
(2)
= 0
dy2
= C3
dx2
(3)
EI y2 = C3x2 + C4
(4)
EI
Boundary conditions:
At x1 = 0.
dy1
= 0
dx1
From Eq. (1), C1 = 0
At x1 = 0. y1 = 0
From Eq. (2): C2 = 0
Continuity conditions:
At x1 = a,
dy1
dy2
=
dx1
dx2
x2 = a ;
From Eqs. (1) and (3),
-
wa3
wa3
wa3
+
= C3;
6
2
2
C3 = -
wa3
6
From Eqs. (2) and (4),
At x1 = a, x2 = a
-
y1 = y2
wa4
wa4
wa4
wa4
+
= + C4 ;
24
6
4
6
C4 =
wa4
24
920
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6/11/10
11:52 AM
Page 921
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12–26.
Continued
The slope, from Eq. (3).
uB =
dy2
wa3
= dx2
6EI
Ans.
The elastic curve:
y1 =
w
a -x41 + 4ax31 - 6a2 x21 b
24EI
Ans.
y2 =
wa3
a -4x2 + ab
24EI
Ans.
yB = y2 2
=
x3 = L
wa3
a -4L + ab
24EI
Ans.
921
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Page 922
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12–27. Wooden posts used for a retaining wall have a
diameter of 3 in. If the soil pressure along a post varies
uniformly from zero at the top A to a maximum of 300 lb>ft
at the bottom B, determine the slope and displacement at
the top of the post. Ew = 1.6(103) ksi.
A
6 ft
Moment Function: As shown on FBD.
Slope and Elastic Curve:
B
d2y
EI 2 = M(y)
dy
EI
EI
d2y
= -8.333y3
dy2
dy
= -2.0833y4 + C1
dy
[1]
EI y = -0.4167y5 + C1y + C2
Boundary Conditions:
[2]
dy
= 0 at y = 6 ft and y = 0 at y = 6 ft
dy
From Eq. [1], 0 = -2.0833 A 64 B + C1
C1 = 2700
From Eq. [2], 0 = -0.4167 A 65 B + 2700(6) + C2
C2 = -12960
The Slope: Substituting the value of C1 into Eq. [1],
1
dy
=
b A -2.0833y4 + 2700 B r lb # ft2
dy
EI
uA =
dy 2
2700 lb # ft2
=
dy y = 0
EI
2700(144)
=
1.6 A 106 B A p4 B A 1.54 B
= 0.0611 rad
Ans.
The Elastic Curve: Substituting the values of C1 C2 into Eq. [2],
y =
1
E A -0.4167y5 + 2700y - 12960 B F lb # ft3
EI
yA = y|y = 0 = -
12960 lb # ft3
EI
12960(1728)
= -
1.6 A 106 B A p4 B A 1.54 B
= -3.52 in.
Ans.
The negative sign indicates leftward displacement.
922
300 lb/ft
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Page 923
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*12–28. Determine the slope at end B and the maximum
deflection of the cantilevered triangular plate of constant
thickness t. The plate is made of material having a modulus
of elasticity E.
b
2
b
2
L
Section Properties. Referring to the geometry shown in Fig. a,
b(x)
b
=
;
x
L
b(x) =
A
b
x
L
t
Thus, the moment of the plate as a function of x is
I(x) =
1
bt3
x
C b(x) D t3 =
12
12L
x
Moment Functions. Referring to the free-body diagram of the plate’s cut
segments, Fig. b,
+ ©MO = 0;
x
-M(x) - w(x) a b = 0
2
M(x) = -
w 2
x
2
Equations of Slope and Elastic Curve.
E
M(x)
d 2v
=
I(x)
dx2
d 2v
E 2 =
dx
E
w 2
x
6wL
2
= - 3 x
3
bt
bt
x
12L
-
dv
3wL
= - 3 x2 + C1
dx
bt
Ev = -
(1)
wL 3
x + C1x + C2
bt3
Boundary Conditions. At x = L,
E(0) = -
(2)
dv
= 0. Then Eq. (1) gives
dx
3wL 2
A L B + C1
bt3
C1 =
3wL3
bt3
At x = L, v = 0. Then Eq. (2) gives
E(0) = -
wL 3
A L B + C1(L) + C2
bt3
C2 = -
2wL4
bt3
Substituting the value of C1 into Eq. (1),
dv
3wL
=
A -x2 + L2 B
dx
Ebt3
At B, x = 0. Thus,
uB =
dv 2
3wL3
=
dx x = 0
Ebt3
Substituting the values of C1 and C2 into Eq. (2),
v =
wL
A -x3 + 3L2x - 2L3 B
Ebt3
vmax occurs at x = 0. Thus,
vmax = v冷x = 0 = -
2wL4
2wL4
=
3
Ebt
Ebt3
w
Ans.
T
923
B
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Page 924
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•12–29.
The beam is made of a material having a specific
weight g. Determine the displacement and slope at its end
A due to its weight. The modulus of elasticity for the
material is E.
L
Section Properties:
h
bh 2
1 h
a xb(x)(b) =
x
2 L
2L
h(x) =
h
x
L
I(x) =
3
h
bh3 3
1
(b)a xb =
x
12
L
12L3
V(x) =
Moment Function: As shown on FBD.
Slope and Elastic Curve:
E
E
M(x)
d2y
=
I(x)
dx2
bhg 3
2gL2
d2y
6L x
= - 3
= - 2
2
bh
3
dx
h
3x
12L
E
2gL2
dy
= - 2 x + C1
dx
h
Ey = -
gL2
h2
x2 + C1 x + C2
From Eq. [2], 0 = -
[2]
dy
= 0 at x = L and y = 0 at x = L.
dx
Boundary Conditions:
From Eq. [1], 0 = -
[1]
2gL2
2
h
gL2
2
h
(L) + C1
A L2 B +
C1 =
2gL3
2
h
2gL3
(L) + C2
h3
C2 = -
gL4
h2
The Slope: Substituting the value of C1 into Eq. [1],
2gL2
dy
= 2 ( -x + L)
dx
hE
uA =
2gL3
dy 2
= 2
dx x = 0
hE
Ans.
The Elastic Curve: Substituting the values of C1 and C2 into Eq. [2],
y =
gL2
h2E
A -x2 + 2Lx - L2 B
gL4
yA |x = 0 = - 2
hE
Ans.
The negative sign indicates downward displacement.
924
A
b
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Page 925
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12–30. The beam is made of a material having a specific
weight of g. Determine the displacement and slope at its
end A due to its weight. The modulus of elasticity for the
material is E.
r
Section Properties:
r(x) =
r
x
L
4
I(x) =
A
2
p r
pr2 3
a xb x =
x
3 L
3L2
V(x) =
L
4
pr 4
p r
a xb =
x
4 L
4L4
Moment Function: As shown on FBD.
Slope and Elastic Curve:
M(x)
d2y
=
I(x)
dx2
E
2
pr g2 4
gL2
d 2y
12L x
E 2 = - 4
= - 2
pr 4 4
dx
3r
x
4L
E
gL2
dy
= - 2 x + C1
dx
3r
Ey = Boundary Conditions:
From Eq. [1], 0 = -
From Eq. [2], 0 = -
6r2
x2 + C1x + C2
[2]
dy
= 0 at x = L and y = 0 at x = L.
dx
gL2
3r
gL2
[1]
(L) + C1
C1 =
L2 B + ¢
2 A
L + C2
2 ≤
2
gL2
gL3
6r
3r
gL3
3r2
C2 = -
gL4
6r2
The Slope: Substituting the value of C1 into Eq. [1],
gL2
dy
= 2 (-x + L)
dx
3r E
uA =
gL3
dy 2
= 2
dx x = 0
3r E
Ans.
The Elastic Curve: Substituting the values of C1 and C2 into Eq. [2],
y =
gL2
6r2E
A -x2 + 2Lx - L2 B
yA |x = 0 = -
gL4
Ans.
6r2E
The negative sign indicates downward displacement.
925
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Page 926
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12–31. The tapered beam has a rectangular cross section.
Determine the deflection of its free end in terms of the load
P, length L, modulus of elasticity E, and the moment of
inertia I0 of its fixed end.
b
A
P
Moment function:
M(x) = -Px
L
Moment of inertia:
w =
b
x;
L
I =
l0
1 b
1
x
a xb t3 =
b t3 a b =
x
12 L
12
L
L
Slope and elastic curve:
EI(x)
d2y
= M(x)
dx2
Ea
l0
d2y
bx 2 = -Px ;
L
dx
El0
dy
= -PLx + C1
dx
El0 y =
El0
d2y
= -PL
dx2
(1)
-PL 2
x + C1x + C2
2
(2)
Boundary conditions:
dy
= 0, x = L
dx
From Eq. (1), 0 = -PL2 + C1 ;
C1 = PL2
y = 0, x = L
From Eq. (2),
0 = y =
PL3
+ PL3 + C2 ;
2
C2 = -
PL3
2
PL
( -x2 + Lx - L2)
2El0
x = 0,
ymax = y 2
= u=0
PL3
2El0
Ans.
The negative sign indicates downward displacement.
926
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*12–32. The beam is made from a plate that has a constant
thickness t and a width that varies linearly. The plate is cut
into strips to form a series of leaves that are stacked to
make a leaf spring consisting of n leaves. Determine the
deflection at its end when loaded. Neglect friction between
the leaves.
P
b
L
Use the triangular plate for the calculation.
M = Px
I =
1 b
a xb(t)3
12 L
d2v
M
Px
=
=
1
EI
dx2
E A B A b B x(t)3
12
L
d2v
12PL
=
2
dx
Ebt3
dv
12PL
=
x + C1
dx
Ebt3
v =
6PL 2
x + C1x + C2
Ebt3
dv
= 0 at x = L
dx
C1 =
-12PL2
Ebt3
v = 0 at x = L
C2 =
6PL3
Ebt3
When x = 0
vmax =
6PL3
Ebt3
Ans.
927
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Page 928
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•12–33.
The tapered beam has a rectangular cross section.
Determine the deflection of its center in terms of the load
P, length L, modulus of elasticity E, and the moment of
inertia Ic of its center.
P
b
L
—
2
Moment of inertia:
L
—
2
2b
w =
x
L
I =
2IC
1 2b
1
2x
a xb (t3) =
(b) A t3 B a b = a
bx
12 L
12
L
L
Elastic curve and slope:
EI(x)
Ea
d2v
= M(x)
dx2
2IC
d2v
P
b(x) 3 = x
L
2
dx
EIC
dv
PL
=
x + C1
dx
4
(1)
PL 2
x + C1x + C2
8
EICv1 =
(2)
Boundary condition:
Due to symmetry:
dv
= 0
dx
x =
at
L
2
From Eq. (1),
0 =
PL2
+ C1
8
C1 = -
PL2
8
v = 0 at x = 0
C2 = 0
v =
PLx
(x - L)
8EIC
vC = v `
= x = L2
PL3
32EIC
Ans.
The negative sign indicates downward displacement.
928
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Page 929
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12–34. The leaf spring assembly is designed so that it is
subjected to the same maximum stress throughout its length.
If the plates of each leaf have a thickness t and can slide
freely between each other, show that the spring must be in
the form of a circular arc in order that the entire spring
becomes flat when a large enough load P is applied. What
is the maximum normal stress in the spring? Consider
the spring to be made by cutting the n strips from the
diamond-shaped plate of thickness t and width b.The modulus
of elasticity for the material is E. Hint: Show that the radius
of curvature of the spring is constant.
nb
b
x
P
L
2
Section Properties: Since the plates can slide freely relative to each other, the plates
resist the moment individually. At an arbitrary distance x from the support, the
2nx
nx
numbers, of plates is L =
. Hence,
L
2
I(x) =
1 2nx
nbt3
a
b (b) A t3 B =
x
12 L
6L
Moment Function: As shown on FBD.
Bending Stress: Applying the flexure formula,
M(x) c
smax =
=
I(x)
Px
2
A 2t B
nbt3
6L x
=
3PL
2nbt2
Ans.
Moment - Curvature Relationship:
Px
M(x)
1
3PL
2
= Constant (Q.E.D.)
=
=
=
3
nbt3
r
EI(x)
nbt
E
EA xB
6L
929
x
L
2
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Page 930
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12–35. The shaft is made of steel and has a diameter of
15 mm. Determine its maximum deflection. The bearings at
A and B exert only vertical reactions on the shaft.
Est = 200 GPa.
15 mm
A
B
200 mm
300 mm
250 N
M = -(-201.43) 6 x - 0 7 -250 6 x - 0.2 7 - 80 6 x - 0.5 7
M = 201.43x - 250 6 x - 0.2 7 - 80 6 x - 0.5 7
Elastic curve and slope:
EI
d2v
= M = 201.43x - 250 6 x - 0.2 7 - 80 6 x - 0.5 7
dx3
EI
dv
= 100.71x2 - 125 6 x - 0.2 7 2 - 40 6 x - 0.5 7 2 + C1
dx
EIv = 33.57x3 - 41.67 6 x - 0.2 7 3 - 13.33 6 x - 0.5 7 3 + C1x + C2 (1)
Boundary conditions:
v = 0
at
x = 0
From Eq. (1)
C2 = 0
v = 0
at
x = 0.7 m
0 = 11.515 - 5.2083 - 0.1067 + 0.7C1
C1 = -8.857
dv
1
=
C 100.71x2 - 125 6 x - 0.2 7 2 - 40 6 x - 0.5 7 2 - 8.857 D
dx
EI
Assume vmax occurs at 0.2 m 6 x 6 0.5 m
dv
1
= 0 =
C 100.71x2 - 125(x - 0.2)2 - 8.857 D
dx
EI
24.28x2 - 50x + 13.857 = 0
x = 0.3300 m
v =
O.K.
1
C 33.57x3 - 41.67 6 x - 0.2 7 3 - 13.33 6 x - 0.5 7 3 - 8.857x D
EI
Substitute x = 0.3300 m into the elastic curve:
vmax = -
1.808N # m3
1.808
= = -0.00364 = -3.64 mm Ans.
9 p
EI
200 A 10 B (0.0075)4
4
The negative sign indicates downward displacement.
930
200 mm
80 N
12 Solutions 46060
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Page 931
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*12–36. The beam is subjected to the loads shown.
Determine the equation of the elastic curve. EI is constant.
4 kip
2 kip
4 kip⭈ft
A
B
x
8 ft
M = -(-2.5) 6 x - 0 7 - 2 6 x - 8 7 - 4 6 x - 16 7
M = 2.5x - 2 6 x - 8 7 - 4 6 x - 16 7
Elastic curve and slope:
EI
d2v
= M = 2.5x - 2 6 x - 8 7 - 4 6 x - 16 7
dx2
EI
dv
= 1.25x2 - 6 x - 8 7 2 - 2 6 x - 16 7 2 + C1
dx
EIv = 0.417x3 - 0.333 6 x - 8 7 3 - 0.667 6 x - 16 7 3 + C1x + C2
(1)
Boundary conditions:
v = 0
at
x = 0
From Eq. (1), C2 = 0
v = 0
at
x = 24 ft
0 = 5760 - 1365.33 - 341.33 + 24C1
C1 = -169
v =
1
C 0.417x3 - 0.333 6 x - 8 7 3 - 0.667 6 x - 16 7 3 - 169x D kip # ft3 Ans.
EI
931
8 ft
8 ft
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•12–37.
Determine the deflection at each of the pulleys
C, D, and E. The shaft is made of steel and has a diameter
of 30 mm. The bearings at A and B exert only vertical
reactions on the shaft. Est = 200 GPa.
C
E
D
A
B
250 mm
250 mm
250 mm
250 mm
M = -(-180) 6 x - 0 7 - 150 6 x - 0.25 7
150 N
- 60 6 x - 0.5 7 -150 6 x - 0.75 7
M = 180x - 150 6 x - 0.25 7 - 60 6 x - 0.5 7 -150 6 x - 0.75 7
Elastic curve and slope:
EI
d2v
= M = 180x - 150 6 x - 0.25 7 - 60 6 x - 0.5 7
dx2
-150 6 x - 0.75 7
EI
dv
= 90x2 - 75 6 x - 0.25 7 2 - 30 6 x - 0.50 7 2
dx
- 75 6 x - 0.75 7 2 + C1
(1)
EIv = 30x3 - 25 6 x - 0.25 7 3 - 10 6 x - 0.50 7 3
- 25 6 x - 0.75 7 3 + C1x + x2
(2)
Boundary conditions:
v = 0
at
x = 0
From Eq. (2)
C2 = 0
v = 0
x = 1.0 m
at
0 = 30 - 10.55 - 1.25 - 0.39 + C1
C1 = -17.8125
dv
1
=
C 90x2 - 75 6 x - 0.25 7 2 - 30 6 x - 0.5 7 2
dx
EI
- 75 6 x - 0.75 7 2 - 17.8125 D
v =
(3)
1
C 30x3 - 25 6 x - 0.25 7 3 - 10 6 x - 0.5 7 3
EI
- 25 6 x - 0.75 7 3 - 17.8125x D
vC = v `
=
x = 0.25m
-3.984
-3.984
=
= -0.000501 m
EI
200 A 109 B p (0.015)4
4
Ans.
= -0.501 mm
vD = v `
x = 0.5m
vE = v `
x = 0.75 m
-5.547
=
=
200 A 109 B p4 (0.015)4
= -0.000698 m = -0.698 mm
-3.984
= -0.501 mm
EI
Ans.
Ans. (symmetry check !)
The negative signs indicate downward displacement.
932
60 N
150 N
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Page 933
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12–38. The shaft supports the two pulley loads shown.
Determine the equation of the elastic curve.The bearings at A
and B exert only vertical reactions on the shaft. EI is constant.
A
B
x
20 in.
20 in.
40 lb
M = -10 6 x - 0 7 -40 6 x - 20 7 -(-110) 6 x - 40 7
M = -10x - 40 6 x - 20 7 + 110 6 x - 40 7
Elastic curve and slope:
EI
d2v
= M
dx2
EI
d2v
= -10x - 40 6 x - 20 7 + 110 6 x - 40 7
dx2
EI
dv
= -5x2 - 20 6 x - 20 7 2 + 55 6 x - 40 7 2 + C1
dx
EIv = -1.667x3 - 6.667 6 x - 20 7 3 + 18.33 6 x - 40 7 3 + C1x + C2 (1)
Boundary conditions:
v = 0 at x = 0
From Eq. (1):
C2 = 0
v = 0 at x = 40 in.
0 = -106,666.67 - 53,333.33 + 0 + 40C1.
C1 = 4000
v =
1
C -1.67x3 - 6.67 6 x - 20 7 3 + 18.3 6 x - 40 7 3 + 4000x D lb # in3
EI
933
Ans.
20 in.
60 lb
12 Solutions 46060
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Page 934
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12–39. Determine the maximum deflection of the simply
supported beam. E = 200 GPa and I = 65.0(106) mm4.
30 kN
15 kN
Support Reactions and Elastic Curve. As shown in Fig. a.
Moment Function. From Fig. a, we obtain
A
M = -(-25)(x - 0) - 30(x - 2) - 15(x - 4)
2m
= 25x - 30(x - 2) - 15(x - 4)
Equations of Slope and Elastic Curve.
EI
d2v
= M
dx2
EI
d2v
= 25x - 30(x - 2) - 15(x - 4)
dx2
EI
dv
= 12.5x2 - 15(x - 2)2 - 7.5(x - 4)2 + C1
dx
(1)
EIv = 4.1667x3 - 5(x - 2)3 - 2.5(x - 4)3 + C1x + C2
(2)
Boundary Conditions. At x = 0, v = 0. Then, Eq. (2) gives
0 = 0 - 0 - 0 + C1(0) + C2
C2 = 0
At x = 6 m, v = 0. Then Eq. (2) gives
0 = 4.1667 A 63 B - 5(6 - 2)3 - 2.5(6 - 4)3 + C1(6) + C2
C1 = -93.333 kN # m3
Substituting the value of C1 into Eq. (1),
dv
1
=
c12.5x2 - 15(x - 2)2 - 7.5(x - 4)2 - 93.333 d
dx
EI
Assuming that
B
dv
= 0 occurs in the region 2 m 6 x 6 4 m. Then
dx
dv
1
= 0 =
c12.5x2 - 15(x - 2)2 - 93.333 d
dx
EI
12.5x2 - 15(x - 2)2 - 93.333 = 0
2.5x2 - 60x + 153.333 = 0
Solving for the root 2 m 6 x 6 4 m,
x = 2.9079 ft O.K.
934
2m
2m
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Page 935
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12–39. Continued
Substituting the values of C1 and C2 into Eq. (2),
v =
1
c4.1667x3 - 5(x - 2)3 - 2.5(x - 4)3 - 93.333x d
EI
vmax occurs at x = 2.9079 m, where
Ans.
dv
= 0. Thus,
dx
vmax = v|x = 2.9079 ft
=
1
c4.1667 A 2.90793 B - 5(2.9079 - 2)3 - 0 - 93.333(2.9079) d
EI
= -
172.69 A 103 B
172.69kN # m3
= EI
200 A 109 B C 65.0 A 10 - 6 B D
= -0.01328 m = 13.3 mm T
Ans.
935
12 Solutions 46060
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11:52 AM
Page 936
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*12–40. Determine the eqution of the elastic curve, the
slope at A, and the deflection at B of the simply supported
beam. EI is constant.
M0
M0
A
B
L
3
Support Reactions and Elastic Curve. As shown in Fig. a.
Moment Function.
M = -(-MO)a x = MO a x -
L 0
2 0
b - MO ax - L b
3
3
L 0
2 0
b - MO a x - L b
3
3
Equations of Slope and Elastic Curve.
EI
d2v
= M
dx2
EI
0
d2v
L 0
2
=
M
a
x
b
M
a
x
Lb
O
O
3
3
dx2
EI
dv
L
2
= MO ax - b - MO a x - Lb + C1
dx
3
3
EIv
(1)
2
MO
MO
L 2
2
ax - b a x - Lb + C1x + C2
2
3
2
3
Boundary Conditions. Due to symmetry,
EI(0) = MO a
L
L
- b - 0 + C1
2
3
(2)
dv
L
= 0 at x = . Then Eq. (1) gives
dx
2
C1 = -
MOL
6
At x = 0, v = 0. Then, Eq. (2) gives
EI(0) = 0 - 0 + C1(0) + C2
C2 = 0
Substituting the value of C1 into Eq. (1),
2
MO
L
2
dv
=
B 6 ax - b - 6 ax - Lb - L R
dx
6EI
3
3
At A, x = 0. Thus,
uA =
MO
MO L
MOL
dv
=
=
C 6(0) - 6(0) - L D = `
dx x = 0
6EI
6EI
6EI
Ans.
Substituting the values of C1 and C2 into Eq. (2),
v =
MO
L 2
2 2
B 3 a x - b - 3 ax - L b - Lx R
6EI
3
3
At B, x =
Ans.
L
. Thus,
3
vB = v|x = L3 =
= -
MO
L
B 3(0) - 3(0) - L a b R
6EI
3
MOL2
MOL2
=
18EI
18EI
Ans.
T
936
D
C
L
3
L
3
12 Solutions 46060
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11:52 AM
Page 937
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
•12–41.
Determine the equation of the elastic curve and
the maximum deflection of the simply supported beam. EI
is constant.
M0
M0
A
B
Support Reactions and Elastic Curve. As shown in Fig. a.
L
3
Moment Function.
M = -(-MO)a x -
L 0
2
b - MO ax - L b
3
3
L 0
2
b - MO a x - L b
3
3
= MO a x -
Equations of Slope and Elastic Curve.
EI
d2v
= M
dx2
EI
d2v
L 0
2
=
M
a
x
b - MO a x - Lb
O
3
3
dx2
EI
dv
L
2
= MO ax - b - MO a x - Lb + C1
dx
3
3
EIv
(1)
2
MO
MO
L 2
2
ax - b a x - Lb + C1x + C2
2
3
2
3
Boundary Conditions. Due to symmetry,
EI(0) = MO a
(2)
dv
L
= 0 at x = . Then Eq. (1) gives
dx
2
L
L
- b - 0 + C1
2
3
C1 = -
MOL
6
At x = 0, v = 0. Then, Eq. (2) gives
EI(0) = 0 - 0 + C1(0) + C2
C2 = 0
Substituting the values of C1 and C2 into Eq. (2),
v =
2
MO
L 2
2
B 3 a x - b - 3 ax - Lb - Lx R
6EI
3
3
vmax occurs at x =
L
dv
= 0. Then,
, where
2
dx
vmax = v|x = L2 =
= -
Ans.
MO
L
L 2
L
B3a - b - 0 - La b R
6EI
2
3
2
5MOL2
5MOL2
=
T
72EI
72EI
Ans.
937
D
C
L
3
L
3
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11:52 AM
Page 938
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12–42. Determine the equation of the elastic curve, the
slope at A, and the maximum deflection of the simply
supported beam. EI is constant.
P
Moment Function.
= Px - Pax -
L
3
L
2
b - Pa x - Lb
3
3
L
2
b - Pax - Lb
3
3
Equations of Slope and Elastic Curve.
EI
d2v
= M
dx2
EI
2
d2v
L
= Px - Pax - b - Pax - Lb
3
3
dx2
EI
2
P
dv
P
L 2
P
2
= x2 ax - b a x - Lb + C1
dx
2
2
3
2
3
EIv =
(1)
3
P 3
P
L 3
P
2
x ax - b a x - Lb + C1x + C2
6
6
3
6
3
Boundary Conditions. Due to symmetry,
EI(0) =
(2)
dv
L
= 0 at x = . Then Eq. (1) gives
dx
2
P L 2
P L
L 2
a b a - b - 0 + C1
2 2
2 2
3
C1 = -
PL2
9
At x = 0, v = 0. Then, Eq. (2) gives
EI(0) = 0 - 0 - 0 + C1(0) + C2
C2 = 0
Substituting the value of C1 into Eq. (1),
2
P
L 2
2
dv
=
B 9x2 - 9a x - b - 9ax - Lb - 2L2 R
dx
18EI
3
3
At A, x = 0. Thus,
uA =
dv
P
PL2
PL2
=
=
C 0 - 0 - 0 - 2L2 D = `
dx x = 0
18EI
9EI
9EI
Ans.
SubStituting the values of C1 and C2 into Eq. (2),
v =
3
P
L 3
2
B 3x3 - 3ax - b - 3ax - Lb - 2L2x R
18EI
3
3
vmax occurs at x =
= -
Ans.
L
dv
, where
= 0. Then,
2
dx
vmax = v|x = L2 =
B
A
Support Reactions and Elastic Curve. As shown in Fig. a.
M = -(-P)(x - 0) - Pax -
P
P
L 3
L
L 3
L
B 3 a b - 3a - b - 0 - 2L2 a b R
18EI
2
2
3
2
23PL3
23PL3
=
T
648EI
648EI
Ans.
938
L
3
L
3
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Page 939
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12–43. Determine the maximum deflection of the
cantilevered beam. The beam is made of material having an
E = 200 GPa and I = 65.0(106) mm6.
15kN
30 kN/m
A
Support Reactions and Elastic Curve. As shown in Fig. a.
Moment Function. From Fig. b, we obtain
M = -(-37.5)(x - 0) - 67.5(x - 0)0 - a-
1.5 m
20
(x - 0)3
6
20
30
b (x - 1.5)3 - a - b(x - 1.5)2
6
2
= 37.5x - 67.5 -
10 3
10
x +
(x - 1.5)3 + 15(x - 1.5)2
3
3
Equations of Slope and Elastic Curve.
EI
d2v
= M
dx2
EI
d2v
10 3
10
= 37.5x - 67.5 x +
(x - 1.5)3 + 15(x - 1.5)2
3
3
dx2
EI
dv
5
5
= 18.75x2 - 67.5x - x4 + (x - 1.5)4 + 5(x - 1.5)3 + C1
dx
6
6
EIv = 6.25x3 - 33.75x2 -
(1)
1 5
1
5
x + (x - 1.5)5 + (x - 1.5)4 + C1x + C2 (2)
6
6
4
Boundary Conditions. At x = 0,
dv
= 0 Then Eq. (1) gives
dx
0 = 0 - 0 - 0 + 0 + 0 + C1
C1 = 0
At x = 0, v = 0. Then Eq. (2) gives
0 = 0 - 0 - 0 + 0 + 0 + 0 + C2
C2 = 0
Substituting the values of C1 and C2 into Eq. (2),
v =
1
1
1
5
c6.25x3 - 33.75x2 - x5 + (x - 1.5)5 + (x - 1.5)4 d
EI
6
6
4
Ans.
vmax occurs at x = 3 m Thus
vmax = v|x = 3 m
=
1
1
1
5
c6.25 A 33 B - 33.75 A 32 B - A 35 B + (3 - 1.5)5 + (3 - 1.5)4 d
EI
6
6
4
= -
167.91kN # m3
= EI
167.91 A 103 B
200 A 109 B c65.0 A 10 - 6 B d
= -0.01292 m = 12.9 mm T
Ans.
939
1.5 m
12 Solutions 46060
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Page 940
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*12–44. The beam is subjected to the load shown. Determine
the equation of the elastic curve. EI is constant.
50 kN
3 kN/m
Support Reactions and Elastic Curve: As shown on FBD.
Moment Function: Using discontinuity function,
B
A
x
M = 24.6 6 x - 0 7 - 1.5 6 x - 0 7 2 - (-1.5) 6 x - 4 7 2
4m
- 50 6 x - 7 7
= 24.6x - 1.5x2 + 1.5 6 x - 47 2 - 50 6 x - 7 7
Slope and Elastic Curve:
EI
EI
EI
d2 y
= M
dx2
d2 y
= 24.6x - 1.5x2 + 1.5 6 x - 47 2 - 50 6 x - 7 7
dx2
dy
= 12.3x2 - 0.5x3 + 0.5 6 x - 4 7 3 - 25 6 x - 7 7 2 + C1
dx
[1]
EI y = 4.10x3 - 0.125x4 + 0.125 6 x - 4 7 4 - 8.333 6 x - 7 7 3
+ C1x + C2
[2]
Boundary Conditions:
y = 0 at x = 0. From Eq.[2], C2 = 0
y = 0 at x = 10 m. From Eq.[2],
0 = 4.10 A 103 B - 0.125 A 104 B + 0.125(10 - 4)4 - 8.333(10 - 7)3 + C1 (10)
C1 = -278.7
The Elastic Curve: Substituting the values of C1 and C2 into Eq.[2],
y =
1
{4.10x3 - 0.125x4 + 0.125 6 x - 4 7 4
EI
- 8.33 6 x - 7 7 3 - 279x} kN # m3
Ans.
940
3m
3m
12 Solutions 46060
6/11/10
11:52 AM
Page 941
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•12–45.
The beam is subjected to the load shown.
Determine the displacement at x = 7 m and the slope at A.
EI is constant.
50 kN
3 kN/m
Support Reactions and Elastic Curve: As shown on FBD.
B
A
Moment Function: Using the discontinuity function.
x
2
M = 24.6 6 x - 0 7 - 1.5 6 x - 0 7 - (-1.5) 6 x - 4 7
4m
2
- 50 6 x - 7 7
= 24.6x - 1.5x2 + 1.5 6 x - 47 2 - 50 6 x - 7 7
Slope and Elastic Curve:
EI
EI
EI
d2 y
= M
dx2
d2 y
= 24.6x - 1.5x2 + 1.5 6 x - 47 2 - 50 6 x - 7 7
dx2
dy
= 12.3x2 - 0.5x3 + 0.5 6 x - 4 7 3 - 25 6 x - 7 7 2 + C1
dx
[1]
EI y = 4.10x3 - 0.125x4 + 0.125 6 x - 4 7 4 - 8.333 6 x - 7 7 3
+ C1x + C2
[2]
Boundary Conditions:
y = 0 at x = 0. From Eq.[2], C2 = 0
y = 0 at x = 10 m. From Eq.[2],
0 = 4.10 A 103 B - 0.125 A 104 B + 0.125(10 - 4)4 - 8.333(10 - 7)3 + C1 (10)
C1 = -278.7
The Slope: Substituting the value of C1 into Eq.[1],
1
dy
E 12.3x2 - 0.5x3 + 0.5 6 x - 4 7 3 - 25 6 x - 7 7 2 - 278.7 F kN # m2
=
dx
EI
uA =
279kN # m2
dy
1
{0 - 0 + 0 - 0 - 278.7} = =
`
dx x = 0
EI
EI
Ans.
The Elastic Curve: Substituting the values of C1 and C2 into Eq.[2],
y =
1
E 4.10x3 - 0.125x4 + 0.125 6 x - 4 7 4 - 8.33 6 x - 7 7 3
EI
- 278.7x F kN # m3
y |x = 7 m =
1
E 4.10 A 73 B - 0.125 A 74 B + 0.125(7 - 4)4 - 0 - 278.7(7) F kN # m3
EI
= -
835 kN # m3
EI
Ans.
941
3m
3m
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11:52 AM
Page 942
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12–46. Determine the maximum deflection of the simply
supported beam. E = 200 GPa and I = 65.0(106) mm4.
20 kN
15 kN/m
Support Reactions and Elastic Curve. As shown in Fig. a.
A
Moment Function. From Fig. b, we obtain
M = -(-22.5)(x - 0) - 20(x - 1.5) -
15
5
(x - 3)2 - a - b(x - 3)3
2
6
= 22.5x - 20(x - 1.5) - 7.5(x - 3)2 +
1.5 m
5
(x - 3)3
6
Equations of Slope and Elastic Curve.
EI
d2v
= M
dx2
EI
d2v
5
= 22.5x - 20(x - 1.5) - 7.5(x - 3)2 + (x - 3)3
2
6
dx
EI
dv
5
(x - 3)4 + C1
= 11.25x2 - 10(x - 1.5)2 - 2.5(x - 3)3 +
dx
24
EIv = 3.75x3 -
(1)
10
1
(x - 1.5)3 - 0.625(x - 3)4 +
(x - 3)5 + C1x + C2 (2)
3
24
Boundary Conditions. At x = 0, v = 0. Then, Eq. (2) gives
0 = 0 - 0 - 0 + 0 + C1(0) + C2
C2 = 0
At x = 6 m, v = 0. Then Eq. (2) gives
0 = 3.75 A 63 B -
10
1
(6 - 1.5)3 - 0.625(6 - 3)4 +
(6 - 3)5 + C1(6) + C2
3
24
C1 = -77.625 kN # m2
Substituting the value of C1 into Eq. (1),
dv
1
5
=
c11.25x2 - 10(x - 1.5)2 - 2.5(x - 3)3 +
(x - 3)4 - 77.625 d
dx
EI
24
Assuming that
dv
= 0 occurs in the region 1.5 m 6 x 6 3 m, then
dx
dv
1
= 0 =
c11.25x2 - 10(x - 1.5)2 - 0 + 0 - 77.625 d
dx
EI
Solving for the root 1.5 m 6 x 6 3 m,
x = 2.970 m O.K.
Substituting the values of C1 and C2 into Eq. (2),
v =
1
10
1
c3.75x3 (x - 1.5)3 - 0.625(x - 3)4 +
(x - 3)5 - 77.625x d Ans.
EI
3
24
vmax occurs at x = 2.970 m, where
dv
= 0. Thus,
dx
942
1.5 m
3m
B
12 Solutions 46060
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11:52 AM
Page 943
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12–46.
Continued
vmax = v ƒ x = 2.970 m
=
1
10
c3.75 A 2.9703 B (2.970 - 1.5)3 - 0 + 0 - 77.625(2.970) d
EI
3
= -
142.89kN # m3
= EI
142.89 A 103 B
200 A 109 B c65.0 A 10 - 6 B d
= -0.01099 m = 11.0 mm T
Ans.
943
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6/11/10
11:52 AM
Page 944
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12–47. The wooden beam is subjected to the load
shown. Determine the equation of the elastic curve. If
Ew = 12 GPa, determine the deflection and the slope at
end B.
M = -63 6 x - 0 7 0 - (-16) 6 x - 0 7 -
6 kN
4 kN
2 kN/m
A
B
x
2
6 x - 0 72
2
3m
1.5 m
1.5 m
2
- a - b 6 x - 3 7 2 - 4 6 x - 4.5 7
2
400 mm
M = -63 + 16x - x2 + 6 x - 3 7 2 - 4 6 x - 4.5 7
200 mm
Elastic curve and Slope:
EI
d2v
= M = -63 + 16x - x2 + 6 x - 3 7 2 - 4 6 x - 4.5 7
dx2
EI
1
dv
x3
= -63x + 8x2 + 6 x - 3 7 3 - 2 6 x - 4.5 7 2 + C1
dx
3
3
EIv = -31.5x2 +
(1)
1
8 3
x4
2
x +
6 x - 3 7 4 - 6 x - 4.5 7 3
3
12
12
3
+ C1x + C2
(2)
Boundary condition:
dv
= 0
dx
at
x = 0
From Eq. (1), C1 = 0
v = 0
x = 0
at
From Eq. (2), C2 = 0
1
x3
1
dv
=
c -63x + 8x2 + 6 x - 3 7 3 - 2 6 x - 4.5 7 2 d
dx
EI
3
3
v =
(3)
1
8
x4
1
c -31.5x2 + x3 +
6 x - 3 74
EI
3
12
12
-
2
6 x - 4.5 7 3 d kN # m3 (4)
3
Ans.
1
(0.20)(0.40)3 = 1.067 A 10 - 3 B m4
12
I =
At point B, x = 6m
uB =
-157.5 A 103 B
dv
-157.5
=
=
= -0.0123 rad = -0.705° Ans.
`
dx x = 6m
EI
12 A 103 B (1.067) A 10 - 3 B
The negative sign indicates clockwise rotation.
vB =
-661.5 A 103 B
-661.5
=
= -0.0517m = -51.7 mm
EI
12 A 103 B (1.067) A 10 - 3 B
944
Ans.
12 Solutions 46060
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11:52 AM
Page 945
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*12–48. The beam is subjected to the load shown.
Determine the slopes at A and B and the displacement at
C. EI is constant.
30 kN
A
The negative sign indicates downward displacement.
C
Moment Function: Using the discontinuity function,
3m
M = 66.75 6 x - 0 7 -6 6 x - 0 7 2 - 30 6 x - 3 7
= 66.75x - 6x2 - 30 6 x - 3 7
Slope and Elastic Curve:
EI
d2y
= M
dx2
d2y
= 66.75x - 6x2 - 30 6 x - 3 7
dx2
dy
= 33.375x2 - 2x3 - 15 6 x - 3 7 2 + C1
dx
[1]
EI y = 11.125x3 - 0.5x4 - 5 6 x - 3 7 3 + C1x + C2
[2]
EI
Boundary Conditions:
y = 0 at x = 0. From Eq.[2], C2 = 0
y = 0 at x = 8 m. From Eq.[2],
0 = 11.125 A 83 B - 0.5 A 84 B - 5(8 - 3)3 + C1 (8)
C1 = -377.875
The Slope: Substituting the value of C1 into Eq.[1],
1
dy
=
E 33.375x2 - 2x3 - 15 6 x - 3 7 2 - 377.875 F kN # m2
dx
EI
uA =
dy
1
378 kN # m2
=
{0 - 0 - 0 - 377.875} = `
dx x = 0
EI
EI
uB =
dy
`
dx x = 8 m
=
1
E 33.375 A 82 B - 2 A 83 B - 15(8 - 3)2 - 377.875 F
EI
=
359 kN # m2
EI
Ans.
Ans.
The Elastic Curve: Substituting the values of C1 and C2 into Eq.[2],
y =
1
E 11.125x3 - 0.5x4 - 5 6 x - 3 7 3 - 377.875 x F kN # m3
EI
yC = y|x = 3 m =
1
E 11.125 A 33 B - 0.5 A 34 B - 0 - 377.875(3) F
EI
= -
B
x
Support Reactions and Elastic Curve: As shown on FBD.
EI
12 kN/m
874 kN # m3
EI
Ans.
945
5m
12 Solutions 46060
6/11/10
11:52 AM
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•12–49.
Determine the equation of the elastic curve of the
simply supported beam and then find the maximum
deflection. The beam is made of wood having a modulus of
elasticity E = 1.5(103) ksi.
600 lb
500 lb/ft
3 in.
A
B
6 ft
Support Reactions and Elastic Curve. As shown in Fig. a.
Moment Function. From Fig. b, we obtain
M = -(-2400)(x - 0) - 600(x - 9) -
500
500
(x - 0)2 - ¢ ≤ (x - 6)
2
2
= 2400x - 600(x - 9) - 250x2 + 250(x - 6)2
Equations of Slope and Elastic Curve.
EI
d2v
= M
dx2
EI
d2v
= 2400x - 600(x - 9) - 250x2 + 250(x - 6)2
dx2
EI
dv
250 3
250
= 1200x2 - 300(x - 9)2 x +
(x - 6)3 + C1
dx
3
3
EIv = 400x3 - 100(x - 9)3 -
125 4
125
x +
(x - 6)4 + C1x + C2
6
6
Boundary Conditions. At x = 0, v = 0. Then Eq.(2) gives
C2 = 0
At x = 12ft, v = 0. Then Eq.(2) gives
0 = 400 A 12 3 B - 100(12 - 9)3 -
125
125
(12)4 +
(12 - 6)4 + C1(12)
6
6
C1 = -23625 lb # ft2
Substituting the value of C1 into Eq.(1),
dv
1
250 3
250
=
x +
(x - 6)3 - 23625 R
B 1200x2 - 300(x - 9)2 dx
EI
3
3
Assuming that
dv
= 0 occurs in the region 0 6 x 6 6 ft. Then
dx
dv
1
250 3
= 0 =
c1200x2 x - 23625 d
dx
EI
3
1200x2 -
250 3
x - 23625 = 0
3
Solving
x = 5.7126 ft O.K.
946
(1)
(2)
3 ft
3 ft
6 in.
12 Solutions 46060
6/11/10
11:52 AM
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•12–49.
Continued
Substituting the values of C1 and C2 into Eq.(2),
v =
1
125 4
125
c400x3 - 100(x - 9)3 x +
(x - 6)4 - 23625x d
EI
6
6
vmax occurs at x = 5.7126 ft, where
Ans.
dv
= 0. Thus,
dx
vmax = v|x = 5.7126 ft
=
1
125
c400 A 5.71263 B - 0 A 5.71264 B + 0 - 23625(5.7126) d
EI
6
= -
82.577.41lb # ft3
= EI
82577.41 A 12 3 B
1.5 A 106 B c
1
(3) A 63 B d
12
= -1.76 in = 1.76 in T
Ans.
947
12 Solutions 46060
6/11/10
11:52 AM
Page 948
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12–50. The beam is subjected to the load shown.
Determine the equations of the slope and elastic curve. EI
is constant.
2 kN/m
8 kN⭈m
A
B
Support Reactions and Elastic Curve: As shown on FBD.
x
Moment Function: Using the discontinuity function,
5m
1
1
M = 0.200 6 x - 0 7 - (2) 6 x - 0 7 2 - ( -2) 6 x - 5 7 2
2
2
- (-17.8) 6 x - 5 7
= 0.200x - x2 + 6 x - 5 7 2 + 17.8 6 x - 5 7
Slope and Elastic Curve:
EI
EI
EI
d2y
= M
dx2
d2y
= 0.200x - x2 + 6 x - 5 7 2 + 17.8 6 x - 5 7
dx2
dy
= 0.100x2 - 0.3333x3 + 0.3333 6 x - 5 7 3 + 8.90 6 x - 5 7 2 + C1 [1]
dx
EI y = 0.03333x3 - 0.08333x4 + 0.08333 6 x - 5 7 4
+ 2.9667 6 x - 5 7 3 + C1x + C2
[2]
Boundary Conditions:
y = 0 at x = 0. From Eq.[2], C2 = 0
y = 0 at x = 5 m. From Eq.[2],
0 = 0.03333 A 53 B - 0.08333 A 54 B + 0 + 0 + C1 (5)
C1 = 9.5833
The Slope: Substituting the value of C1 into Eq.[1],
dy
1
=
E 0.100x2 - 0.333x3 + 0.333 6 x - 5 7 3 + 8.90 6 x - 5 7 2
dx
EI
+ 9.58 F kN # m2
Ans.
948
3m
12 Solutions 46060
6/11/10
11:52 AM
Page 949
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12–51. The beam is subjected to the load shown. Determine
the equation of the elastic curve. EI is constant.
The Elastic Curve: Substituting the values of C1 and C2 into Eq.[2],
y =
+ 9.58x F kN # m3
1.5 m
Ans.
6
6
6 x - 0 7 2 -(-1.25) 6 x - 1.5 7 -(- ) 6 x - 1.5 7 2
2
2
- (-27.75) 6 x - 4.5 7
M = -3x2 + 1.25 6 x - 1.5 7 +3 6 x - 1.5 7 2 + 27.75 6 x - 4.5 7
Elastic curve and slope:
EI
d2v
= M
dx2
= -3x2 + 1.25 6 x - 1.5 7 + 3 6 x - 1.5 7 2 + 27.75 6 x - 4.5 7
EI
dv
= -x3 + 0.625 6 x - 1.5 7 2 + 6 x - 1.5 7 3
dx
+ 13.875 6 x - 4.5 7 2 + C1
EIv = -0.25x4 + 0.208 6 x - 1.5 7 3 + 0.25 6 x - 1.5 7 4
+ 4.625 6 x - 4.5 7 3 + C1x + C2
(1)
Boundary conditions:
v = 0
at
x = 1.5 m
From Eq.(1)
0 = -1.266 + 1.5C1 + C2
1.5C1 + C2 = 1.266
v = 0
at
(2)
x = 4.5 m
From Eq.(1)
0 = -102.516 + 5.625 + 20.25 + 4.5C1 + C2
4.5C1 + C2 = 76.641
(3)
Solving Eqs. (2) and (3) yields:
C1 = 25.12
C2 = -36.42
v =
Ans.
1
C -0.25x4 + 0.208 6 x - 1.5 7 3 + 0.25 6 x - 1.5 7 4
EI
+ 4.625 6 x - 4.5 7 3 + 25.1x - 36.4 D kN # m3
949
B
A
1
E 0.0333x3 - 0.0833x4 + 0.0833 6 x - 5 7 4 + 2.97 6 x - 5 7 3
EI
M = -
20 kN
6 kN/m
3m
1.5 m
12 Solutions 46060
6/11/10
11:52 AM
Page 950
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*12–52. The wooden beam is subjected to the load shown.
Determine the equation of the elastic curve. Specify the
deflection at the end C. Ew = 1.6(103) ksi.
0.8 kip/ft
1.5 kip
A
12 in.
C
B
x
9 ft
M = -0.3 6 x - 0 7 -a-
1 1.6
a
b 6 x - 0 7 3 -(-5.4) 6 x - 9 7
6 18
0.8
0.8
1
b 6 x - 9 72 - a b 6 x - 9 73
2
6
9
M = -0.3x - 0.0148x3 + 5.4 6 x - 9 7 +0.4 6 x - 9 7 2
+ 0.0148 6 x - 9 7 3
Elastic curve and slope:
EI
d2v
= M = -0.3x - 0.0148x3 + 5.4 6 x - 9 7 +0.4 6 x - 9 7 2
dx2
+0.0148 6 x - 9 7 3
EI
dv
= -0.15x2 - 0.003704x4 + 2.7 6 x - 9 7 2 + 0.1333 6 x - 9 7 3
dx
+ 0.003704 6 x - 9 7 4 + C1
EIv = -0.05x3 + 0.0007407x5 + 0.9 6 x - 9 7 3 + 0.03333 6 x - 9 7 4
+ 0.0007407 6 x - 9 7 5 + C1x + C2
(1)
Boundary conditions:
v = 0
at
x = 0
at
x = 9 ft
From Eq.(1)
C2 = 0
v = 0
From Eq.(1)
0 = -36.45 - 43.74 + 0 + 0 + 0 + 9C1
C1 = 8.91
v =
1
C -0.05x3 - 0.000741x5 + 0.9 6 x - 9 7 3 + 0.0333 6 x - 9 7 4
EI
+ 0.000741 6 x - 9 7 5 + 8.91x D kip # ft3
Ans.
At point C, x = 18 ft
vC =
-612.29 A 12 3 B
-612.29kip # ft3
=
= -0.765 in.
EI
1.6 A 103 B A 1 B (6) A 12 3 B
Ans.
12
The negative sign indicates downward displacement.
950
9 ft
6 in.
12 Solutions 46060
6/11/10
11:52 AM
Page 951
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12–53. Determine the displacement at C and the slope at
A of the beam.
8 kip/ft
Support Reactions and Elastic Curve: As shown on FBD.
C
Moment Function: Using the discontinuity function,
M = -
x
1
1
8
(8) 6 x - 0 7 2 - a - b 6 x - 6 7 3 - (-88) 6 x - 6 7
2
6
9
= -4x2 +
6 ft
4
6 x - 6 7 3 + 88 6 x - 6 7
27
Slope and Elastic Curve:
EI
EI
EI
d2y
= M
dx2
d 2y
4
= -4x2 +
6 x - 6 7 3 + 88 6 x - 6 7
27
dx2
dy
4
1
= - x3 +
6 x - 6 7 4 + 44 6 x - 6 7 2 + C1
dx
3
27
1
1
44
EI y = - x4 +
6 x - 6 75 +
6 x - 6 7 3 + C1x + C2
3
135
3
[1]
[2]
Boundary Conditions:
y = 0 at x = 6 ft. From Eq.[2],
0 = -
1 4
A 6 B + 0 + 0 + C1 (6) + C2
3
432 = 6C1 + C2
[3]
y = 0 at x = 15 ft. From Eq.[2],
0 = -
1
1
44
(15 - 6)3 +
(15 - 6)3 + C1 (15) + C2
A 154 B +
3
135
3
5745.6 = 15C1 + C2
[4]
Solving Eqs. [3] and [4] yields,
C1 = 590.4
C2 = -3110.4
The Slope: Substitute the value of C1 into Eq.[1],
dy
1
1
4
=
6 x - 6 7 4 + 44 6 x - 6 7 2 + 590.4 r kip # ft2
b - x3 +
dx
EI
3
27
uA =
302 kip # ft2
dy
1
4
=
b - A 63 B + 0 + 0 + 590.4 r =
`
dx x = 6 ft
EI
3
EI
Ans.
The Elastic Curve: Substitute the values of C1 and C2 into Eq. [2],
y =
1
1
1
44
6 x - 6 75 +
6 x - 6 73
b - x4 +
EI
3
135
3
+ 590.4x - 3110.4 r kip # ft3
yC = y |x = 0 =
B
A
3110kip # ft3
1
{-0 + 0 + 0 + 0 - 3110.4} kip # ft3 = EI
EI
951
Ans.
9 ft
12 Solutions 46060
6/11/10
11:52 AM
Page 952
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–54. The beam is subjected to the load shown. Determine
the equation of the elastic curve. EI is constant.
6 kip/ft
B
A
x
9 ft
M = -
1 16
1 10
a b 6 x - 0 7 3 -(-77.4) 6 x - 9 7 - a - b a b 6 x - 9 7 3
6 24
6 15
M = -0.1111x3 + 77.4 6 x - 9 7 +0.1111 6 x - 9 7 3
Elastic curve and slope:
EI
d2v
= M = -0.1111x3 + 77.4 6 x - 9 7 +0.1111 6 x - 9 7 3
dx2
EI
dv
= -0.02778x4 + 38.7 6 x - 9 7 2 + 0.02778 6 x - 9 7 4 + C1
dx
EIv = -0.005556x5 + 12.9 6 x - 9 7 3 + 0.005556 6 x - 9 7 5 + C1x + C2
(1)
Boundary conditions:
v = 0
at
x = 9 ft
From Eq.(1)
0 = -328.05 + 0 + 0 + 9C1 + C2
9C1 + C2 = 328.05
v = 0
at
(2)
x = 24 ft
0 = -44236.8 + 43537.5 + 4218.75 + 24C1 + C2
24C1 + C2 = -3519.45
(3)
Solving Eqs. (2) and (3) yields,
C1 = -256.5 C2 = 2637
v =
1
C -0.00556x5 + 12.9 6 x - 9 7 3 + 0.00556 6 x - 9 7 5
EI
- 265.5x + 2637 D kip # ft3
Ans.
952
15 ft
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12–55. Determine the slope and deflection at C. EI is
constant.
15 kip
A
C
B
30 ft
|tB>A|
uA =
30
tB>A =
-33 750
1 -225
a
b(30)(10) =
2
EI
EI
uA =
1125
EI
uC>A =
1 -225
-5062.5
5062.5
1 -225
a
b(30) + a
b(15) =
=
2 EI
2 EI
EI
EI
uC = uC>A + uA
uC =
5062.5
1125
3937.5
=
EI
EI
EI
¢ C = |tC>A| -
Ans.
45
|t |
30 B>A
tC>A =
1
-225
1
225
101 250
ab(30)(25) + a b(15)(10) = 2
EI
2
EI
EI
¢C =
45 33 750
50 625
101.250
b =
a
EI
30
EI
EI
Ans.
953
15 ft
12 Solutions 46060
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*12–56. Determine the slope and deflection at C. EI is
constant.
10 kN
A
C
B
6m
Referring to Fig. b,
|uC>A| =
1 30
135 kN # m2
a
b A9B =
2 EI
EI
|tB>A| =
6 1 30
180 kN # m3
c a
b A6B d =
3 2 EI
EI
|tC>A| = a
=
6
1 30
2
1 30
+ 3b c a
b A 6 B d + c (3) d c a
b A3B d
3
2 EI
3
2 EI
540 kN # m3
EI
From the geometry shown in Fig. b,
uA =
|tB>A|
=
6
180>EI
30 kN # m2
=
6
EI
Here,
+ buC = uA + uC>A
uC = uC =
135
30
+
EI
EI
105 kN # m2
EI
uC
Ans.
9
yC = 2 tC>A 2 - 2 tB>A 2 a b
6
=
540
180 9
a b
EI
EI 6
=
270 kN # m3
T
EI
Ans.
954
3m
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11:52 AM
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•12–57. Determine the deflection of end B of the
cantilever beam. E is constant.
P
P
B
A
L
2
Support Reactions and
M
Diagram. As shown in Fig. a.
EI
Moment Area Theorem. Since A is a fixed support, uA = 0. Referring to the
geometry of the elastic curve, Fig. b,
uB = |uB>A| =
=
1 3PL
PL
L
1 PL
L
+
B
R¢ ≤ + B
R¢ ≤
2 2EI
2EI
2
2 2EI
2
5PL2
8 EI
Ans.
¢ B = |tB>A| - ¢
=
7PL3
16EI
3L
PL
L
L
5L 1 PL L
L 1 PL
≤¢
≤¢ ≤ +
B ¢
≤¢ ≤R + B ¢
≤¢ ≤R
4
2EI
2
6 2 EI
2
3 2 2EI
2
Ans.
T
955
L
2
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12–58. Determine the slope at A and the maximum
deflection. EI is constant.
20 kip⭈ft
20 kip⭈ft
A
6 ft
Point D is located at the mid span of the beam. Due to symmetry, the slope at D is
zero. Referring to Fig. b,
|uD>A| = a
120 kip # ft2
20
b(6) =
EI
EI
|tD>A| = 3 a
360 kip # ft3
20
b A6B =
EI
EI
|tC>D| = 6a
1440 kip # ft3
20
b A 12 B =
EI
EI
From the geomtry shown in Fig. b
uA = |uD>A| =
120 kip # ft2
EI
uA
Ans.
y D = uA(6) - |tD>A|
=
360
120
(6) EI
EI
=
360 kip # ft3
EI
c
yC = |tC>D| - 4D
=
1440
360
EI
EI
=
1080 kip # ft
EI
T (max)
Ans.
956
C
B
12 ft
6 ft
12 Solutions 46060
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11:52 AM
Page 957
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12–59. Determine the slope and deflection at C. EI is
constant.
20 kip⭈ft
20 kip⭈ft
A
Referring to Fig. b,
|uC>A| = a
360 kip # ft2
20
b A 18 B =
EI
EI
|uB>A| = 6a
|tC>A| = 9 a
6 ft
b
1440 kip # ft3
20
b A 12 B = =
EI
EI
3240 kip # ft3
20
b A 18 B =
EI
EI
From the geometry shown in Fig. b
uA =
|tB>A|
12
=
1440>EI
12
=
120 kip # ft2
EI
uA
Here,
+ b uC = uA + uC>A
uC = -
uC =
360
120
+
EI
EI
240 kip # ft2
EI
yC = |tC>A| - |tB>A| a
uC
Ans.
18
b
12
=
1440 18
3240
a b
EI
EI 12
=
1080 kip # ft3
T
EI
Ans.
957
C
B
12 ft
6 ft
12 Solutions 46060
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*12–60. If the bearings at A and B exert only vertical
reactions on the shaft, determine the slope at A and the
maximum deflection of the shaft. EI is constant.
50 lb⭈ft
B
C
2 ft
Point E is located at the mid span of the shaft. Due to symmetry, the slope at E is
zero. Referring to Fig. b,
|uE>A| =
100 lb # ft2
50
(2) =
EI
EI
|tE>A| = (1)a
100 lb # ft3
50
b (2) =
EI
EI
Here,
uA = |uE>A| =
100 lb # ft2
EI
uA
Ans.
ymax = uA (4) - |tE>A|
=
100
100
(4) EI
EI
=
300 lb # ft3
EI
50 lb⭈ft
A
Ans.
c
958
D
4 ft
2 ft
12 Solutions 46060
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Page 959
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•12–61.
Determine the maximum slope and the maximum
deflection of the beam. EI is constant.
M0
M0
B
A
M0L
M0 L
uC>A =
a b =
EI 2
2EI
L
uC = uC>A + uA
0 =
M0 L
+ uA
2EI
umax = uA =
M0 L
-M0 L
=
2EI
2EI
¢ max = |tB>C| =
Ans.
M0 L2
M0 L L
a ba b =
EI 2
4
8EI
Ans.
12–62. Determine the deflection and slope at C. EI is
constant.
A
B
C
M0
M0L2
1 -M0
1
b(L)a b(L) = tB>A = a
2 EI
3
6EI
L
¢ C = |tC>A| - 2|tB>A|
tC>A =
-M0
7M0L2
L
L
1 -M0
a
b(L)aL + b + a
b (L) a b = 2 EI
3
EI
2
6EI
¢C =
7M0 L2
M0L2
5M0L2
- (2) a
b =
6EI
6EI
6EI
uA =
|tB>A|
L
uC>A =
=
Ans.
M0L
6EI
M0
M0
3M0L
3M0L
1
ab(L) + a b(L) = =
2
EI
EI
2EI
2EI
uC = uC>A + uA
uC =
3M0L
M0L
4M0L
=
2EI
6EI
3EI
Ans.
959
L
12 Solutions 46060
6/11/10
11:52 AM
Page 960
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12–63. Determine the slope at A of the overhang beam.
E = 200 GPa and I = 45.5(106) mm4.
30 kN
30 kN⭈m
M
Support Reactions and
Diagram. As shown in Fig. a.
EI
A
Moment Area Theorem. Referring to Fig. b,
4m
1 30
30
1
1
|tB>A| = c (4) d B ¢
(4) R
≤ (4) R + c (4) d B
3
2 EI
2
EI
=
320 kN # m3
EI
From the geometry of the elastic curve, Fig. b,
uA =
=
|tB>A|
LAB
=
320>EI
80 kN # m2
=
4
EI
80 A 103 B
200 A 109 B C 45.5 A 10 - 6 B D
C
B
Ans.
= 0.00879 rad
960
2m
12 Solutions 46060
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11:52 AM
Page 961
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*12–64. Determine the deflection at C of the overhang
beam. E = 200 GPa and I = 45.5(106) mm4.
30 kN
30 kN⭈m
1
1 30
1
30
|tB>A| = c (4) d B ¢
(4) R
≤ (4) R + c (4) d B
3
2 EI
2
EI
=
A
320 kN # m3
EI
4m
1 30
30
1
1
|tC>A| = c (4) + 2 d B ¢
(4) R
≤ (4) R + c (4) + 2 d B
3
2 EI
2
EI
1 60
2
+ c (2) d B ¢
≤ (2) R
3
2 EI
=
760 kN # m3
EI
¢ C = |tC>A| - |tB>A ¢
C
B
L
≤
LAB
=
760
320 6
¢ ≤
EI
EI 4
=
280 A 103 B
280 kN # m3
=
EI
200 A 109 B C 45.5 A 10 - 6 B D
= 0.03077 m = 30.8 mm T
Ans.
961
2m
12 Solutions 46060
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Page 962
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•12–65. Determine the position a of roller support B in
terms of L so that the deflection at end C is the same as the
maximum deflection of region AB of the overhang beam.
EI is constant.
P
L
C
A
B
M
Support Reactions and
Diagram. As shown in Fig. a.
EI
Moment Area Theorem. Referring to Fig. b,
|tB>A| =
Pa2(L - a)
a 1 P(L - a)
B ¢
≤ (a) R =
3 2
EI
6EI
|tC>A| = aL -
=
a
2(L - a) 1 P(L - a)
1 P(L - a)
2
ab B ¢
≤ (a) R +
B ¢
≤ (L - a) R
3
2
EI
3
2
EI
P(L - a) A 2L2 - aL B
6EI
From the geometry shown in Fig. b,
¢ C = |tC>A| -
=
=
uA =
|tB>A|
a
L
PL(L - a) A 2L - a B
6EI
-
Pa2 (L - a) L
¢ ≤
a
6EI
PL(L - a)2
3EI
|tB>A|
a
Pa2(L - a)
Pa(L - a)
6EI
=
=
a
6EI
The maximum deflection in region AB occurs at point D, where the slope of the
elastic curve is zero (uD = 0).
Thus,
|uD>A| = uA
Pa(L - a)
1 P(L - a)
x R (x) =
B
2
EIa
6EI
x =
23
a
3
Also,
¢ D = |t4>D| = a
23Pa2(L - a)
2 23
1 P(L - a) 23
23
ab B c
a
ab d R a
ab =
9
2
EIa
3
3
27EI
It is required that
¢C = ¢D
PL(L - a)2
23Pa2(L - a)
=
3EI
27EI
23 2
a + La - L2 = 0
9
Solving for the positive root,
a = 0.858L
Ans.
962
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Page 963
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12–66. Determine the slope at A of the simply supported
beam. EI is constant.
Support Reactions and
P
M
Diagram. As shown in Fig. a.
EI
A
Moment Area Theorem.
2L
3
1 2PL 2
2 1 2PL L
5
tB>A = a Lb c a
b a L b d + Lc a
ba bd
9
2 9EI
3
9 2 9EI
3
=
B
4PL3
81EI
Referring to the geometry of the elastic curve, Fig. b,
4PL3
|tB>A|
81EI
4PL2
uA =
=
=
L
L
81EI
Ans.
963
L
3
12 Solutions 46060
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11:52 AM
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12–67. The beam is subjected to the load P as shown.
Determine the magnitude of force F that must be applied
at the end of the overhang C so that the deflection at C is
zero. EI is constant.
F
P
B
A
C
3
3
tB>A =
1
Fa
2
2Fa
1 Pa
Pa
a
b(2a)(a) + a b(2a)a ab =
2 2EI
2
EI
3
2EI
3EI
tC>A =
1 -Fa
2a
1 -Fa
2a
1 Pa
a
b(2a)(2a) + a
b(2a)a a +
b + a
b(a) a b
2 2EI
2 EI
3
2 EI
3
=
a
a
a
2Fa3
Pa3
EI
EI
¢ C = tC>A -
3
t
= 0
2 B>A
Pa3
2Fa3
3 Pa3
2Fa3
- a
b = 0
EI
EI
2 2EI
3EI
F =
P
4
Ans.
M0 ⫽ Pa
*12–68. If the bearings at A and B exert only vertical
reactions on the shaft, determine the slope at A and the
maximum deflection.
A
C
a
a
Pa
17Pa3
1 Pa
tB>A = a
b(a)a3a + b + a
b (2a)(a + a) =
2 EI
3
EI
3EI
uA =
|tB>A|
4a
=
B
Ans.
Assume ¢ max is at point E located at 0 6 x 6 2a
uE>A =
1 Pa
Pa
Pa2
Pax
a
b(a) + a
b(x) =
+
2 EI
EI
2EI
EI
uE = 0 = uE>A + uA
0 =
Pax
-17Pa2
Pa2
+
+ a
b
2EI
EI
12EI
x =
11
a
12
¢ max = |tB>E| = a
2a
P
17Pa2
12EI
11
(2a - 12a)
Pa
11
481Pa3
b a 2a ab c
+ ad =
EI
12
2
288EI
964
Ans.
D
a
12 Solutions 46060
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•12–69.
The beam is subjected to the loading shown.
Determine the slope at A and the displacement at C. Assume
the support at A is a pin and B is a roller. EI is constant.
P
Support Reactions and Elastic Curve: As shown.
P
A
C
a
M/EI Diagram: As shown.
P
a
B
a
a
Moment - Area Theorems: Due to symmetry, the slope at midspan (point C) is zero.
Hence the slope at A is
uA = uA>C =
=
1 3Pa
3Pa
1 Pa
a
b(a) + a
b(a) + a
b(a)
2 2EI
2EI
2 2EI
5Pa2
2EI
Ans.
The displacement at C is
¢ C = tA>C =
=
2a
3Pa
a
1 Pa
2a
1 3Pa
a
b (a)a b + a
b aa + b + a
b(a)a a +
b
2 2EI
3
2EI
2
2 2EI
3
19Pa3
T
6EI
Ans.
12–70. The shaft supports the gear at its end C.
Determine the deflection at C and the slopes at the
bearings A and B. EI is constant.
A
L
––
2
1 -PL L L
-PL3
tB>A = a
ba ba b =
2 2EI
2
6
48EI
L
-PL3
1 -PL
a
b(L)a b =
2 2EI
2
8EI
tC>A =
L
¢ C = |tC>A| - a L b|tB>A|
2
=
uA =
PL3
PL3
PL3
- 2a
b =
8EI
48EI
12EI
|tB>A|
L
2
uB>A =
=
PL3
48 EI
L
2
=
Ans.
PL2
24EI
Ans.
1 -PL L
-PL2
PL2
a
ba b =
=
2 2EI
2
8EI
8EI
uB = uB>A + uA
uB =
B
PL2
PL2
PL2
=
8EI
24EI
12EI
Ans.
965
C
L
––
2
P
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Page 966
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12–71. The shaft supports the gear at its end C. Determine
its maximum deflection within region AB. EI is constant.
The bearings exert only vertical reactions on the shaft.
A
B
L
––
2
uD>A =
C
L
––
2
P
tB>A
A L2 B
1 Px
2 A 2 B A 2EI B A 3 B A 2 B
a
;
bx =
2 EI
ALB
1 L
PL
1
L
x = 0.288675 L
2
¢ max =
2
1 P(0.288675 L)
a
b (0.288675 L)a b(0.288675 L)
2
EI
3
¢ max =
0.00802PL3
EI
Ans.
*12–72. Determine the value of a so that the displacement
at C is equal to zero. EI is constant.
P
P
A
C
B
Moment-Area Theorems:
(¢ C)1 = (tA>C)1 =
(tB>A)2 =
a
1
Pa
2
PaL2
ab(L)a Lb = 2
EI
3
3EI
(tC>A)2 = a -
(¢ C)2 =
1 PL
L L
PL3
a
ba ba b =
2 4EI
2
3
48EI
Pa
Pa
L L
1
L L
5PaL2
b¢ ≤¢ ≤ + ab¢ ≤¢ ≤ = 2EI
2
4
2
2EI
2
3
48EI
PaL2
1
1 PaL2
5PaL2
|(tB>A)2| - |(tC>A)2| = ¢
=
≤ 2
2 3EI
48EI
16EI
Require,
¢ C = 0 = (¢ C)1 - (¢ C)2
0 =
PaL2
PL3
48EI
16EI
a =
L
3
Ans.
966
L
2
L
2
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•12–73.
The shaft is subjected to the loading shown. If the
bearings at A and B only exert vertical reactions on the
shaft, determine the slope at A and the displacement at C.
EI is constant.
A
B
C
a
M/EI Diagram: As shown.
Moment-Area Theorems:
M0
M0
1
a
1
a
ab(a)a b + a b(a)aa + b
2
EI
3
2
EI
3
tB>A =
= -
5M0 a2
6EI
M0
M0 a2
a
1
ab(a)a b = 2
EI
3
6EI
tC>A =
The slope at A is
uA =
5M0a 2
|tB>A|
L
=
6EI
2a
=
5M0 a
12EI
Ans.
The displacement at C is,
¢C = `
1
t
` - |tC>A|
2 B>A
=
M0 a2
1 5M0 a2
¢
≤ 2
6EI
6EI
=
M0 a2
4EI
M0
M0
Ans.
c
967
a
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11:52 AM
Page 968
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12–74. Determine the slope at A and the maximum
deflection in the beam. EI is constant.
12 kip
24 kip⭈ft
A
B
6 ft
Here,
tB>A = 20 c
=
1 12
36
1 24
1 60
a
b(6) d + 12 c
(12) d + 10 c a
b(12) d + 4c a
b(6) d
2 EI
EI
2 EI
2 EI
8064 kip # ft3
EI
From the geometry of the elastic curve diagram, Fig. b,
uA =
tB>A
=
L
8064>EI
336 kip # ft2
=
uA
24
EI
Ans.
Assuming that the zero slope of the elastic curve occurs in the region 6ft 6 x = 18ft
such as point C where the maximum deflection occurs, then
uC>A = uA
1 12
36
1 2x
336
a
b (6) + a
bx + a
b(x) =
2 EI
EI
2 EI
EI
x2 + 36x - 300 = 0
Solving for the root 0 6 x 6 12 ft,
x = 6.980 ft O.K.
Thus,
ymax = tA>C = 4 c
=
1 12
36
1
a
b(6) d + 9.490 c (6.980) d + 10.653 c (13.960)(6.980) d
2 EI
EI
2
3048 kip # ft3
T
EI
Ans.
968
12 ft
6 ft
12 Solutions 46060
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11:52 AM
Page 969
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12–75. The beam is made of a ceramic material. In order to
obtain its modulus of elasticity, it is subjected to the elastic
loading shown. If the moment of inertia is I and the beam has
a measured maximum deflection ¢ , determine E. The
supports at A and D exert only vertical reactions on the beam.
P
P
B
C
A
D
a
a
L
Moment-Area Theorems: Due to symmetry, the slope at midspan (point E) is zero.
Hence the maximum displacement is,
¢ max = tA>E = a
=
L - 2a
1 Pa
2
Pa L - 2a
ba
b aa +
b + a
b(a)a ab
EI
2
4
2 EI
3
Pa
A 3L2 - 4a2 B
24EI
Require, ¢ max = ¢ , then,
¢ =
Pa
A 3L2 - 4a2 B
24EI
E =
Pa
A 3L2 - 4a2 B
24¢I
Ans.
*12–76. The bar is supported by a roller constraint at B,
which allows vertical displacement but resists axial load and
moment. If the bar is subjected to the loading shown,
determine the slope at A and the deflection at C. EI is
constant.
P
C
A
B
L
—
2
uA>B =
1 PL
L
PL L
3PL2
a
ba b +
a b =
2 2EI
2
2EI 2
8EI
uA = uA>B
uA =
3PL2
8EI
Ans.
tA>B =
1 PL
L L
PL L L
L
11PL3
a
ba ba b +
a ba + b =
2 2EI
2
3
2EI 2
2
4
48EI
tC>B =
PL L L
PL3
a ba b =
2EI 2
4
16EI
¢ C = tA>B - tC>B =
11PL3
PL3
PL3
=
48EI
16EI
6EI
Ans.
969
L
—
2
12 Solutions 46060
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11:52 AM
Page 970
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•12–77. The bar is supported by the roller constraint at C,
which allows vertical displacement but resists axial load
and moment. If the bar is subjected to the loading shown,
determine the slope and displacement at A. EI is constant.
P
B
C
A
Support Reactions and Elastic Curve: As shown.
a
M/EI Diagram: As shown.
2a
Moment-Area Theorems:
uA>C = a -
Pa
1
Pa
5Pa2
b (2a) + a b(a) = EI
2
EI
2EI
tB>C = a -
Pa
2Pa3
b(2a)(a) =
EI
EI
tA>C = a -
1
2
Pa
Pa
13Pa3
b(2a)(2a) + a b (a)a ab = EI
2
EI
3
3EI
Due to the moment constraint, the slope at support C is zero. Hence, the slope at A is
uA = |uA>C| =
5Pa2
2EI
Ans.
and the displacement at A is
¢ A = |tA>C| - |tB>C|
2Pa3
7Pa3
13Pa3
=
3EI
EI
3EI
=
Ans.
T
12–78. The rod is constructed from two shafts for which
the moment of inertia of AB is I and of BC is 2I. Determine
the maximum slope and deflection of the rod due to the
loading. The modulus of elasticity is E.
P
L
2
uA>C =
1 -PL L
-PL L
-5PL2
5PL2
1 -PL L
a
ba b + a
ba b + a
ba b =
=
2 2EI
2
2 4EI
2
4EI
2
16EI
16EI
uA = uA>C + uC
umax = uA =
5PL2
5PL2
+ 0 =
16EI
16EI
Ans.
¢ max = ¢ A = |tA>C|
= `
1 -PL L L
1 -PL L L
L
a
ba ba b + a
ba ba + b
2 2 EI
2
3
2 4EI
2
2
3
+ a
=
-PL L L
L
ba ba + b `
4EI
2
2
4
3PL3
16EI
Ans.
970
C
B
A
L
2
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11:52 AM
Page 971
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12–79. Determine the slope at point D and the deflection
at point C of the simply supported beam. The beam is made
of material having a modulus of elasticity E. The moment of
inertia of segments AB and CD of the beam is I, while the
moment of inertia of segment BC of the beam is 2I.
A
L
4
Moment Area Theorem. Referring to Fig. b,
=
tC>D =
=
L
L PL L
5L 1 PL
L
L 1 PL
c a
ba bd +
c
a bd +
c a
ba bd
6 2 4EI
4
2 8EI 2
6 2 4EI
4
PL3
16EI
L 1 PL
L
c a
ba bd
12 2 4EI
4
PL3
384EI
From the geometry of Fig. b,
uD =
|tA>D|
L
PL3
PL2
18EI
=
=
L
16EI
¢ C + tC>D =
Ans.
tA>D
4
PL3
PL
16EI
¢C =
=
384EI
4
3
¢C =
D
C
B
M
Support Reactions and
Diagram. As shown in Fig. a.
EI
tA>D =
P
P
5PL3
384EI
Ans.
971
L
2
L
4
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11:52 AM
Page 972
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*12–80. Determine the slope at point A and the maximum
deflection of the simply supported beam. The beam is made
of material having a modulus of elasticity E. The moment of
inertia of segments AB and CD of the beam is I, while the
moment of inertia of segment BC is 2I.
A
L
4
Moment Area Theorem. Due to symmetry, the slope at the midspan of the beam,
i.e., point E, is zero (uE = 0). Thus the maximum deflection occurs here. Referring
to the geometry of the elastic curve, Fig. b,
=
1 PL
L
PL L
a
ba b +
a b
2 4EI
4
8EI 4
PL2
16EI
¢ max = ¢ E = |tA>E| =
=
D
C
B
M
Support Reactions and
Diagram. As shown in Fig. a.
EI
uA = |uA>E| =
P
P
Ans.
L 1 PL
L
3 PL L
Lc
a bd + c a
ba bd
8 8EI 4
6 2 4EI
4
13PL3
T
768EI
Ans.
972
L
2
L
4
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11:52 AM
Page 973
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•12–81. Determine the position a of roller support B in
terms of L so that deflection at end C is the same as the
maximum deflection of region AB of the simply supported
overhang beam. EI is constant.
Support Reactions and
A
B
a
L
M
Diagram. As shown in Fig. a.
EI
Moment Area Theorem. Referring to Fig. b,
|tB>A| =
MOa2
a 1 MO
c a
b (a) d =
3 2 EI
6EI
|tC>A| = aL =
2
1 MO
L - a MO
ab c a
b (a) d + a
bc
(La) d
3
2 EI
2
EI
MO 2
A a + 3L2 - 3La B
6EI
From the geometry shown in Fig. b,
¢ C = |tC>A| -
|tB>A|
a
L
=
MO 2
MOa2 L
a b
A a + 3L2 - 3La B 6EI
6EI a
=
MO 2
A a + 3L2 - 4La B
6EI
MOa2
|tB>A|
MO a
6EI
uA =
=
=
a
a
6EI
The maximum deflection in region AB occurs at point D, where the slope of the
elastic curve is zero (uD = 0).
Thus,
|uD>A| = uA
MOa
1 MO
a
b (x)2 =
2 EIa
6EI
x =
23
a
3
Also,
¢ D = |tA>D| =
23MOa2
2 23
1 MO
23
23
a≤ B a
b¢
a≤ R ¢
a≤ =
¢
3
3
2 EIa
3
3
27EI
It is required that
¢C = ¢D
MO 2
23MO a2
A a + 3L2 - 4La B =
6EI
27EI
0.6151a2 - 4La + 3L2 = 0
Solving for the root 6 L,
a = 0.865L
Ans.
973
C
M0
12 Solutions 46060
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2:06 PM
Page 974
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12–82. The W10 * 15 cantilevered beam is made of A-36
steel and is subjected to the loading shown. Determine the
slope and displacement at its end B.
3 kip/ft
B
A
Here,
uB = 冷uB>A冷 =
1 54
a
b (6)
3 EI
=
108 kip # ft2
EI
6 ft
uC
For W 10 * 15 I = 68.9 in4, and for A36 steel E = 29.0 A 103 B ksi. Thus
uB =
108 A 12 2 B
29 A 103 B (68.9)
= 0.00778 rad uB
Ans.
1 54
3
b (6) d
yB = 冷 tB>A冷 = c (6) + 6 d c a
4
3 EI
=
=
1134 kip # ft3
EI
1134 A 12 3 B
29 A 103 B (68.9)
= 0.981 in.
Ans.
T
974
6 ft
12 Solutions 46060
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11:52 AM
Page 975
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12–83. The cantilevered beam is subjected to the loading
shown. Determine the slope and displacement at C. Assume
the support at A is fixed. EI is constant.
P
w
Support Reactions and Elastic Curve: As shown.
A
C
B
a
M/EI Diagrams: The M/EI diagrams for the uniform distributed load and
concentrated load are drawn separately as shown.
a
Moment-Area Theorems: The slope at support A is zero. The slope at C is
2Pa
1
wa 2
1
ab (2a) + a b(a)
2
EI
3
2EI
uC = 冷uC>A冷 =
a2
(12P + wa)
6EI
=
Ans.
The displacement at C is
¢ C = 冷tC>A 冷 =
=
2Pa
4
1
wa 2
3
1
ab(2a) a ab + a b(a)a a + a b
2
EI
3
3
2EI
4
a3
(64P + 7wa)
24EI
Ans.
T
*12–84. Determine the slope at C and deflection at B. EI
is constant.
w
C
A
B
Support Reactions and Elastic Curve: As shown.
a
M/EI Diagram: As shown.
Moment-Area Theorems: The slope at support A is zero. The slope at C is
uC = 冷uC>A冷 =
=
1
wa2
wa2
ab(a) + a b (a)
2
EI
2EI
wa3
EI
Ans.
The displacement at B is
¢ B = 冷tB>A冷
=
1
wa2
2
wa2
a
1
wa2
3
ab(a)aa + ab + a b(a)a a + b + a b(a)a ab
2
EI
3
2EI
2
3
2EI
4
=
41wa4
24EI
Ans.
T
975
a
12 Solutions 46060
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11:52 AM
Page 976
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•12–85.
Determine the slope at B and the displacement
at C. The member is an A-36 steel structural tee for which
I = 76.8 in4.
5 kip
1.5 kip/ft
B
A
C
3 ft
3 ft
Support Reactions and Elastic Curve: As shown.
M/EI Diagrams: The M/EI diagrams for the uniform distributed load and
concentrated load are drawn separately as shown.
Moment-Area Theorems: Due to symmetry, the slope at midspan C is zero. Hence
the slope at B is
uB = 冷uB>C冷 =
=
2 6.75
1 7.50
a
b (3) + a
b (3)
2 EI
3 EI
24.75 kip # ft2
EI
24.75(144)
=
29.0 A 103 B (76.8)
= 0.00160 rad
Ans.
The dispacement at C is
¢ C = 冷tA>C冷 =
=
2
2 6.75
5
1 7.50
a
b (3)a b (3) + a
b (3) a b(3)
2 EI
3
3 EI
8
47.8125 kip # ft3
EI
47.8125(1728)
=
29.0 A 103 B (76.8)
= 0.0371 in.
Ans.
T
12–86. The A-36 steel shaft is used to support a rotor that
exerts a uniform load of 5 kN兾m within the region CD of the
shaft. Determine the slope of the shaft at the bearings A and
B. The bearings exert only vertical reactions on the shaft.
uE>A =
5 kN/m
A
1 75
4.6875
2 3.5156
4.805
a
b(0.1) + a
b(0.15) + a
b(0.15) =
2 EI
EI
3
EI
EI
uA = uE>A =
4.805
=
EI
4.805
= 0.00306rad = 0.175°
200 (109)(0.01)4
976
B
C
20 mm
100 mm
Ans.
40 mm
300 mm
D
20 mm
100 mm
12 Solutions 46060
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Page 977
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12–87. The W12 * 45 simply supported beam is made of
A-36 steel and is subjected to the loading shown. Determine
the deflection at its center C.
12 kip
50 kip⭈ft
B
A
C
12 ft
A ¢C B 1 =
¢ 2 (x) =
12 A 24 B
3456
PL3
=
=
T
48EI
48EI
EI
3
Mx
A L2 - x2 B
6LEI
At point C, x =
A ¢C B 2 =
L
2
MN A L2 B
6LEI
A L2 - A L2 B 2 B
50 A 24 B
1800
ML2
=
=
T
16EI
16EI
EI
2
=
¢C = A ¢C B 1 + A ¢C B 2 =
5256(1728)
=
29 A 103 B (350)
3456
1800
5256
+
=
EI
EI
EI
= 0.895 in. T
Ans.
977
12 ft
12 Solutions 46060
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11:52 AM
Page 978
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*12–88. The W10 * 15 cantilevered beam is made of
A-36 steel and is subjected to the loading shown. Determine
the displacement at B and the slope at B.
6 kip
4 kip
A
B
6 ft
Using the table in appendix, the required slopes and deflections for each load case
are computed as follow:
5(4) A 12 B
720 kip # in.3
5PL3
=
=
T
48EI
48EI
EI
3
(¢ B)1 =
4 A 12 B
72 kip # in.2
PL2
=
=
8EI
8EI
EI
2
(uB)1 =
(uB)1
6 A 12 B
3456 kip # in.3
PL3
=
=
T
3EI
3EI
EI
3
(¢ B)2 =
6 A 12 B
432 kip # in.2
PL2
=
=
2EI
2EI
EI
2
(uB)2 =
(uB)2
Then the slope and deflection at B are
uB = (uB)1 + (uB)2
=
432
72
+
EI
EI
=
504 kip # ft2
EI
¢ B = (¢ B)1 + (¢ B)2
=
3456
720
+
EI
EI
=
4176 kip # in.3
EI
For A36 steel W10 * 15, I = 68.9 in4 And E = 29.0 A 103 B ksi
uB =
504
29.0 A 103 B (68.9)
= 0.252(10-3) rad
Ans.
4176
¢B =
29.0 A 103 B (68.9)
= 0.00209 in
Ans.
978
6 ft
12 Solutions 46060
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11:52 AM
Page 979
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•12–89.
Determine the slope and deflection at end C of
the overhang beam. EI is constant.
w
A
C
D
a
Elastic Curves. The uniform distributed load on the beam is equivalent to the sum of
the seperate loadings shown in Fig.a. The elastic curve for each seperate loading is
shown Fig. a.
Method of Superposition. Using the table in the appendix, the required slopes and
deflections are
(uC)1 = (uB)1 =
w(2a)3
wa3
wL3
=
=
24EI
24EI
3EI
(¢ C)1 = (uB)1(a) =
wa3
wa4
(a) =
3EI
3EI
(uC)2 =
wa3
wL3
=
6EI
6EI
(¢ C)2 =
wL4
wa4
=
8EI
8EI
T
MOL
(uC)3 = (uB)3 =
=
3EI
(¢ C)3 = (uB)3 (a) =
c
¢
wa2
2
≤ (2a)
3EI
wa3
wa4
(a) =
3EI
3EI
=
wa3
3EI
T
Then the slope and deflection of C are
uC = (uC)1 + (uC)2 + (uC)3
= -
=
wa3
wa 3
wa3
+
+
3EI
6EI
3EI
wa3
6EI
Ans.
¢ C = (¢ C)1 + (¢ C)2 + (¢ C)3
= -
=
wa4
wa 4
wa4
+
+
3EI
8EI
3EI
wa4
T
8EI
Ans.
979
B
a
a
12 Solutions 46060
6/11/10
11:52 AM
Page 980
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12–90. Determine the slope at A and the deflection at
point D of the overhang beam. EI is constant.
w
A
C
D
a
Elastic Curves. The uniform distributed load on the deformation of span AB is
equivalent to the sum of the seperate loadings shown in Fig. a. The elastic curve for
each seperate loading is shown in Fig. a.
Method of Superposition. Using the table in the appendix, the required slope and
deflections are
(uA)1 =
w(2a)3
wL3
wa3
=
=
24EI
24EI
3EI
(¢ D)1 =
5w(2a)4
5wa4
5wL4
=
=
384EI
384EI
24EI
T
wa2
(2a)
MOL
2
wa3
(uA)2 =
=
=
6EI
6EI
6EI
¢
(¢ D)2 =
=
wa2
≤ (a)
2
MOx
A L2 - x2 B =
C (2a)2 - a2 D
6EIL
6EI(2a)
wa4
8EI
c
Then the slope and deflection of point D are
uA = (uA)1 + (uA)2
=
wa3
wa3
wa3
=
3EI
6EI
6EI
Ans.
¢ D = (¢ D)1 + (¢ D)2
=
5wa4
wa4
wa4
=
T
24EI
8EI
12EI
Ans.
980
B
a
a
12 Solutions 46060
6/11/10
11:52 AM
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–91. Determine the slope at B and the deflection at
point C of the simply supported beam. E = 200 GPa and
I = 45.5(106) mm4.
9 kN/m
10 kN
A
B
C
3m
Elastic Curves. The loading system on the beam is equivalent to the sum of the
seperate loadings shown in Fig. a. The elastic curves for each loading are shown in
Fig. a.
Method of Superposition. Using the table in the appendix, the required slope and
deflections are
(uB)1 =
(¢ C)1 =
9 A 63 B
wOL3
43.2kN # m2
=
=
45EI
45EI
EI
wOx
A 3x4 - 10L2x2 + 7L4 B
360EIL
=
9(3)
C 3 A 34 B - 10 A 62 B A 32 B + 7 A 64 B D
360EI(6)
=
75.9375kN # m3
EI
T
10 A 6 B
PL2
22.5kN # m2
=
=
16EI
16EI
EI
2
(uB)2 =
10 A 6 B
PL3
45kN # m3
=
=
48EI
48EI
EI
3
(¢ C)2 =
T
Then the slope at B and deflection at C are
uB = (uB)1 + (uB)2
=
65.7 A 103 B
43.2
22.5
65.7kN # m2
+
=
=
= 0.00722 rad
EI
EI
EI
200 A 109 B C 45.5 A 10 - 6 B D
Ans.
¢ C = (¢ C)1 + (¢ C)2
=
120.9375 A 103 B
75.9375
45
120.9375 kN # m3
+
=
=
EI
EI
EI
200 A 109 B C 45.5 A 10 - 6 B D
= 0.01329 m = 13.3 mm T
Ans.
981
3m
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*12–92. Determine the slope at A and the deflection
at point C of the simply supported beam. The modulus of
elasticity of the wood is E = 10 GPa .
3 kN
100 m
C
A
1.5 m
Method of Superposition. Using the table in the appendix, the required slopes and
deflections are
(uA)1 =
Pab(L + b)
3(1.5)(4.5)(6 + 4.5)
5.90625kN # m2
=
=
6EIL
6EI(6)
EI
(¢ C)1 =
3(4.5)(1.5) 2
Pbx
A L2 - b2 - x2 B =
A 6 - 4.52 - 1.52 B
6EIL
6EI(6)
(uA)2 =
(¢ C)2 =
7.594kN # m3
EI
T
3 A 62 B
PL2
6.75 kN # m2
=
=
16EL
16EI
EI
3(1.5)
Px
9.281
a3(6)2 - 4(1.5)2 b =
A 3L2 - 4x2 B =
48EI
48EI
EI
Then the slope and deflection at C are
uA = (uA)1 + (uA)2
=
6.75
5.90625
+
EI
EI
12.65625kN # m2
=
=
EI
12.6525 A 103 B
= 0.0190 rad
Ans.
16.88 A 103 B
= 0.0253 m = 25.3 mm
Ans.
10 A 109 B c
1
(0.1) A 0.2 3 B d
12
and
¢ C = (¢ C)1 + (¢ C)2
=
9.281
7.594
+
=
EI
EI
10 A 109 B c
1
(0.1) A 0.2 3 B d
12
982
B
1.5 m
Elastic Curves. The two concentrated forces P are applied seperately on the beam
and the resulting elastic curves are shown in Fig. a.
=
3 kN
3m
200 m
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•12–93.
The W8 * 24 simply supported beam is made
of A-36 steel and is subjected to the loading shown.
Determine the deflection at its center C.
6 kip/ft
5 kip⭈ft
A
B
I = 82.8 in4
C
8 ft
5(6) A 16 B
2560
5wL4
(¢ C)1 =
=
=
T
768EI
768EI
EI
8 ft
4
¢ 2 (x) =
Mx
A L2 - x2 B
6LEI
At point C, x =
(¢ C)2 =
M A L2 B
6LEI
L
2
A L2 - A L2 B 2 B
5 A 16 B
80
ML2
=
=
16EI
16EI
EI
2
=
¢ C = (¢ C)1 + (¢ C)2 =
2640(1728)
=
29 A 103 B (82.8)
T
80
2640
2560
+
=
EI
EI
EI
Ans.
= 1.90 in.
12–94. Determine the vertical deflection and slope at the
end A of the bracket. Assume that the bracket is fixed
supported at its base, and neglect the axial deformation of
segment AB. EI is constant.
3 in.
B
6 in.
A
3
3
¢A =
8(3)
72
PL
=
=
3EI
3EI
EI
Ans.
8A3 B
PL2
36
=
=
2EI
2EI
EI
2
uA =
Ans.
983
8 kip
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12–95. The simply supported beam is made of A-36 steel
and is subjected to the loading shown. Determine the
deflection at its center C. I = 0.1457(10-3) m4.
20 kN
4 kN/m
A
5m
Using the table in appendix, the required deflections for each load case are
computed as follow:
5(4) A 10 B
5wL4
=
768EI
768 EI
4
(yC)1 =
=
260.42 kN # m3
EI
T
20N A 10 B
PL3
416.67 kN # m3
=
=
T
48EI
48EI
EI
3
(yC)2 =
Then the deflection of point C is
yC = (yC)1 + (yC)2
=
260.42
416.67
+
EI
EI
=
677.08 kN # m3
EI
T
= 0.1457 A 10 - 3 B m4
and E = 200GPa
¢C =
677.08 A 103 B
200 A 109 B C 0.1457 A 10 - 3 B D
Ans.
= 0.0232 m = 23.2 m T
984
B
C
5m
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*12–96. Determine the deflection at end E of beam CDE.
The beams are made of wood having a modulus of elasticity
of E = 10 GPa .
2m
1.5 m
A
C
1m
3 kN
Method of Superposition. Referring to the table in the appendix, the deflection of
point D is
4.5 A 3 B
2.53125 kN # m3
PL3
=
=
48EI
48EI
EI
3
T
Subsequently,
3
2.53125 3
3.796875 kN # m3
(¢ E)1 = ¢ D a b =
a b =
T
2
EI
2
EI
Also,
3A1 B
1 kN # m3
PL3
=
=
T
3EI
3EI
EI
3
(¢ E)2 =
(uD)3 =
3(2)
MOL
2 kN # m2
=
=
3EI
3EI
EI
(¢ E)3 = (uD)3L =
2
2 kN # m3
(1) =
T
EI
EI
Thus, the deflection of end E is
¢ E = (¢ E)1 + (¢ E)2 + (¢ E)3
=
3.796875
1
2
6.796875kN # m3
+
+
=
=
EI
EI
EI
EI
6.796875 A 103 B
10 A 109 B c
1
(0.075) A 0.153 B d
12
= 0.03222 m = 32.2 mm T
Ans.
985
a
a
E
75 mm
D
a
¢D =
1.5 m
a
150 mm
Section a – a
B
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•12–97.
The pipe assembly consists of three equal-sized
pipes with flexibility stiffness EI and torsional stiffness GJ.
Determine the vertical deflection at point A.
L
–
2
C
¢B =
P A L2 B 3
3EI
(¢ A)1 =
=
P A L2 B 3
3EI
PL3
24EI
=
L
–
2
P
PL3
24EI
L
–
2
A
B
(PL>2) A 2 B
PL2
TL
u =
=
=
JG
JG
4JG
L
(¢ A)2 = u a
L
PL3
b =
2
8JG
¢ A = ¢ B + (¢ A)1 + (¢ A)2
=
PL3
PL3
PL3
+
+
24EI
24EI
8JG
= PL3 a
1
1
+
b
12EI
8JG
Ans.
12–98. Determine the vertical deflection at the end A of
the bracket. Assume that the bracket is fixed supported at
its base B and neglect axial deflection. EI is constant.
u =
a
A
ML
Pab
=
EI
EI
b
Pa2b
(¢ A)1 = u(a) =
EI
(¢ A)2 =
B
PL3
Pa3
=
3EI
3EI
¢ A = (¢ A)1 + (¢ A)2 =
P
Pa2 (3b + a)
Pa3
Pa2b
+
=
EI
3EI
3EI
Ans.
986
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12–99. Determine the vertical deflection and slope at
the end A of the bracket. Assume that the bracket is fixed
supported at its base, and neglect the axial deformation of
segment AB. EI is constant.
20 lb/in.
80 lb
B
4 in.
3 in.
C
Elastic Curve: The elastic curves for the concentrated load, uniform distibuted load,
and couple moment are drawn separately as shown.
Method of Superposition: Using the table in Appendix C, the required slopes and
displacements are
20 A 4 B
wL3AB
213.33 lb # in2
=
=
6EI
6EI
EI
3
(uA)1 =
(uA)2 = (uB)2 =
160(3)
M0 LBC
480 lb # in2
=
=
EI
EI
EI
80 A 3 B
PL2BC
360 lb # in2
=
=
2EI
2EI
EI
2
(uA)3 = (uB)3 =
20 A 4 B
wL4AB
640 lb # in3
=
=
8EI
8EI
EI
4
(¢ A)v1 =
T
(¢ A)v2 = (uB)2 (LAB) =
480
1920lb # in3
(4) =
EI
EI
T
(¢ A)v3 = (uB)3 (LAB) =
1440lb # in3
360
(4) =
EI
EI
T
The slope at A is
uA = (uA)1 + (uA)2 + (uA)3
=
213.33
480
360
+
+
EI
EI
EI
=
1053 lb # in2
EI
Ans.
The vertical displacement at A is
(¢ A)v = (¢ A)v1 + (¢ A)v2 (¢ A)v3
=
640
1920
1440
+
+
EI
EI
EI
=
4000 lb # in3
EI
Ans.
T
987
A
12 Solutions 46060
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*12–100. The framework consists of two A-36 steel
cantilevered beams CD and BA and a simply supported
beam CB. If each beam is made of steel and has a moment
of inertia about its principal axis of Ix = 118 in4, determine
the deflection at the center G of beam CB.
A
15 kip
B
D
C
G
8 ft
16 ft
8 ft
7.5 A 16 B
PL3
10.240
=
=
T
3EI
3EI
EI
3
¢C =
15 A 16 B
PL3
1.280
=
=
T
48EI
48EI
EI
3
¢ ¿G =
¢G = ¢C + ¢ ¿G
=
1,280
11,520
10,240
+
=
EI
EI
EI
11,520(1728)
=
29 A 103 B (118)
Ans.
= 5.82 in. T
•12–101.
The wide-flange beam acts as a cantilever. Due
to an error it is installed at an angle u with the vertical.
Determine the ratio of its deflection in the x direction to its
deflection in the y direction at A when a load P is applied at
this point. The moments of inertia are Ix and Iy. For the
solution, resolve P into components and use the method of
superposition. Note: The result indicates that large lateral
deflections (x direction) can occur in narrow beams,
Iy V Ix, when they are improperly installed in this
manner. To show this numerically, compute the deflections
in the x and y directions for an A-36 steel W10 * 15, with
P = 1.5 kip, u = 10°, and L = 12 ft.
ymax =
P cos L3
;
3EIx
xmax =
u
Vertical
P y
u
L
A
x
P sinu L3
3EIy
P sin u L3
3 EIy
xmax
Ix
=
=
tan u
Pcosu L3
ymax
Iy
Ans.
3 EIx
W 10 * 15
ymax =
xmax =
Ix = 68.9 in4
1.5( cos 10°)(144)3
3(29) A 103 B (68.9)
1.5( sin 10°)(144)3
3(29) A 103 B (2.89)
Iy = 2.89 in4
= 0.736 in.
Ans.
= 3.09 in.
Ans.
988
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12–102. The simply supported beam carries a uniform load
of 2 kip>ft. Code restrictions, due to a plaster ceiling, require
the maximum deflection not to exceed 1>360 of the span
length. Select the lightest-weight A-36 steel wide-flange
beam from Appendix B that will satisfy this requirement
and safely support the load. The allowable bending stress
is sallow = 24 ksi and the allowable shear stress is
tallow = 14 ksi. Assume A is a pin and B a roller support.
8 kip
A
Strength criterion:
24 =
M
Sreq’d
96(12)
Sreq’d
Sreq’d = 48 in3
Choose W14 * 34, S = 48.6 in3, tw = 0.285 in., d = 13.98 in., I = 340 in4.
tallow =
14 Ú
V
A web
24
= 6.02 ksi O.K.
(13.98)(0.285)
Deflection criterion:
Maximum is at center.
vmax =
P(4)(8)
5wL4
+ (2)
C (16)2 - (4)2 - (8)2) D (12)3
384EI
6EI(16)
= c
117.33(8)
5(2)(16)4
+
d(12)3
384EI
EI
4.571(106)
=
29(106)(340)
= 0.000464 in. 6
B
4 ft
Mmax = 96 kip # ft
sallow =
8 kip
2 kip/ft
1
(16)(12) = 0.533 in. O.K.
360
Use W14 * 34
Ans.
989
8 ft
4 ft
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12–103. Determine the reactions at the supports A and B,
then draw the moment diagram. EI is constant.
M0
A
B
L
Support Reactions: FBD(a).
+ ©F = 0;
:
x
Ax = 0
Ans.
+ c ©Fy = 0;
A y - By = 0
[1]
M0 - A y L + MB = 0
[2]
a + ©MB = 0;
Moment Function: FBD(b)
a + ©MNA = 0;
M(x) + M0 - A y x = 0
M(x) = A y x - M0
Slope and Elastic Curve:
EI
EI
EI
d2y
= M(x)
dx2
d 2y
= A y x - M0
dx2
Ay
dy
=
x2 - M0x + C1
dx
2
EI y =
Ay
6
x3 -
[3]
M0 2
x + C1x + C2
2
[4]
Boundary Conditions:
At x = 0, y = 0. From Eq.[4], C2 = 0
At x = L,
0 =
dy
= 0. From Eq. [3],
dx
A y L2
- M0 L + C1
2
[5]
At x = L, y = 0. From Eq. [4],
0 =
A y L3
-
6
M0 L2
+ C1 L
2
[6]
Solving Eqs. [5] and [6] yields,
Ay =
3M0
2L
C1 =
Ans.
M0 L
4
Substituting Ay, into Eqs. [1] and [2] yields:
By =
3M0
2L
MB =
M0
2
Ans.
990
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*12–104. Determine the value of a for which the
maximum positive moment has the same magnitude as the
maximum negative moment. EI is constant.
P
a
L
+ c ©Fy = 0;
A y + By - P = 0
[1]
a + ©MA = 0;
MA + By L - Pa = 0
[2]
Moment Functions: FBD(b) and (c).
M(x1) = Byx1
M(x2) = Byx2 - Px2 + PL - Pa
Slope and Elastic Curve:
EI
d2y
= M(x)
dx2
For M(x1) = Byx1,
EI
EI
d2y1
dx21
= Byx1
By
dy1
=
x2 + C1
dx1
2 1
EI y1 =
By
6
[3]
x31 + C1x1 + C2
[4]
For M(x2) = Byx2 - Px2 + PL - Pa,
EI
EI
d2y2
dx22
= Byx2 - Px2 + PL - Pa
By
dy2
P
=
x2 - x22 + PLx2 - Pax2 + C3
dx2
2 2
2
EI y2 =
By
6
x32 -
[5]
P 3
PL 2
Pa 2
x +
x x + C3x2 + C4
6 2
2 2
2 2
[6]
Boundary Conditions:
y1 = 0 at x1 = 0. From Eq.[4], C2 = 0
dy2
dx2
= 0 at x2 = L. From Eq.[5]
0 =
ByL2
2
-
C3 = -
PL2
+ PL2 - PaL + C3
2
By L2
2
-
PL2
+ PaL
2
991
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*12–104.
Continued
y2 = 0 at x2 = L. From Eq.[6],
0 =
By L3
6
2
By L
PL3
PL3
PaL2
PL2
+
+ a+ PaL bL + C4
6
2
2
2
2
C4 =
By L3
3
+
PL3
PaL2
6
2
Continuity Conditions:
At x1 = x2 = L - a,
By
2
dy1
dy2
. From Eqs.[3] and [5],
=
dx1
dx2
(L - a)2 + C1 =
By
2
(L - a)2 -
P
(L - a)2 + PL(L - a)
2
- Pa(L - a) + a -
By L2
2
-
PL2
+ PaL b
2
2
C1 =
By L
Pa2
2
2
At x1 = x2 = L - a, y1 = y2. From Eqs.[4] and [6],
2
By L
Pa2
(L - a) + a
b(L - a)
6
2
2
By
3
By
=
6
(L - a)3 -
+ a-
By L2
2
P
PL
Pa
(L - a)3 +
(L - a)2 (L - a)2
6
2
2
3
-
By L
PL2
PL3
PaL2
+ PaL b(L - a) +
+
2
3
6
2
3
By L
Pa3
Pa2L
+
= 0
6
2
3
By =
3Pa2
Pa3
Pa2
=
(3L - a)
2
3
2L
2L
2L3
Substituting By into Eqs.[1] and [2], we have
Ay =
P
A 2L3 - 3a2L + a3 B
2L3
MA =
Pa
A -3aL + a2 + 2L2 B
2L2
Require |Mmax( + )| = |Mmax( - )|. From the moment diagram,
Pa2
Pa
(3L - a)(L - a) =
A -3aL + a2 + 2L2 B
2L3
2L2
a2 - 4aL + 2L2 = 0
a = A 2 - 22 B L
Ans.
992
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•12–105. Determine the reactions at the supports A, B, and
C; then draw the shear and moment diagrams. EI is constant.
P
A
L
2
Support Reactions: FBD(a).
+ ©Fx = 0;
:
Ax = 0
Ans.
+ c ©Fy = 0;
A y + By + Cy - 2P = 0
[1]
a + ©MA = 0;
By L + Cy (2L) - Pa
3L
L
b - Pa
b = 0
2
2
[2]
Moment Function: FBD(b) and (c).
M(x1) = Cy x1
M(x2) = Cy x2 - Px2 +
PL
2
Slope and Elastic Curve:
d2y
= M(x)
dx2
EI
For M(x1) = Cy x1,
EI
d2y1
= Cyx1
dx21
Cy
dy1
=
x2 + C1
dx1
2 1
EI
EI y1 =
Cy
6
x31 + C1x1 + C2
For M(x2) = Cyx2 - Px2 +
EI
EI
[3]
d2y2
PL
,
2
= Cyx2 - Px2 +
dx22
[4]
PL
2
Cy
dy2
P
PL
=
x22 - x22 +
x + C3
dx2
2
2
2 2
EI y2 =
Cy
6
x32 -
[5]
P 3
PL 2
x +
x + C3x2 + C4
6 2
4 2
[6]
Boundary Conditions:
y1 = 0 at x1 = 0. From Eq.[4], C2 = 0
Due to symmetry,
0 =
Cy L2
2
-
dy2
= 0 at x2 = L. From Eq.[5],
dx2
PL2
PL2
+
+ C3
2
2
C3 = -
Cy L2
2
993
P
B
L
2
C
L
2
L
2
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•12–105.
Continued
y2 = 0 at x2 = L. From Eq. [6],
0 =
Cy L3
2
-
6
Cy L
PL3
PL3
+
+ abL + C4
6
4
2
Cy L3
C4 =
-
3
PL3
12
Continuity Conditions:
At x1 = x2 =
dy2
L dy1
,
. From Eqs.[3] and [5],
=
2 dx1
dx2
Cy
2
a
2
Cy L 2
Cy L
L 2
P L 2
PL L
b + C1 =
a b - a b +
a b 2
2 2
2 2
2
2
2
2
C1 =
At x1 = x2 =
L
, y = y2. From Eqs.[4] and [6],
2 1
2
a
Cy L
PL2
L 3
L
b + a
ba b
2
8
2
2
Cy
a
Cy
6
=
Cy L
PL2
8
2
6
2
3
Cy L
Cy L
L 3
P L 3
PL L 2
L
PL3
b a b +
a b + aba b +
2
6 2
4 2
2
2
3
12
Cy =
5
P
16
Ans.
Substituting Cy into Eqs.[1] and [2],
By =
11
P
8
Ay =
5
P
16
Ans.
994
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12–106. Determine the reactions at the supports, then
draw the shear and moment diagram. EI is constant.
P
A
B
L
Support Reactions: FBD(a).
+ ©F = 0;
:
x
Ax = 0
Ans.
+ c ©Fy = 0;
By - A y - P = 0
[1]
a + ©MB = 0;
A y L - MA - PL = 0
[2]
Moment Functions: FBD(b) and (c).
M(x1) = -Px1
M(x2) = MA - A yx2
Slope and Elastic Curve:
EI
d2y
= M(x)
dx2
For M(x1) = -Px1,
EI
d2y1
dx21
= -Px1
dy1
P
= - x21 + C1
dx1
2
[3]
P
EI y1 = - x31 + C1x1 + C2
6
[4]
EI
For M(x2) = MA - A yx2,
EI
EI
EI y2 =
d2y2
dx22
= MA - A yx2
Ay
dy2
= MAx2 x2 + C3
dx2
2 2
[5]
Ay
MA 2
x2 x3 + C3x2 + C4
2
6 2
[6]
995
L
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12–106.
Continued
Boundary Conditions:
y2 = 0 at x2 = 0. From Eq.[6], C4 = 0
dy2
= 0 at x2 = 0. From Eq.[5], C3 = 0
dx2
y2 = 0 at x2 = L. From Eq. [6],
3
0 =
Ay L
MA L2
2
6
[7]
Solving Eqs.[2] and [7] yields,
MA =
PL
2
Ay =
3P
2
Ans.
Substituting the value of Ay into Eq.[1],
By =
5P
2
Ans.
Note: The other boundary and continuity conditions can be used to determine the
constants C1 and C2 which are not needed here.
996
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12–107. Determine the moment reactions at the supports
A and B. EI is constant.
P
P
A
B
a
Support Reactions: FBD(a).
a
L
a + ©MB = 0;
Pa + P(L - a) + MA - A y L - MB = 0
PL + MA - A y L - MB = 0
[1]
Moment Functions: FBD(b) and (c).
M(x1) = A y x1 - MA
M(x2) = A y x2 - Px2 + Pa - MA
Slope and Elastic Curve:
d2y
= M(x)
dx2
EI
For M(x1) = A y x1 - MA,
EI
EI
d2y1
dx21
= A y x1 - MA
Ay
dy1
=
x2 - MA x1 + C1
dx1
2 1
EI y1 =
Ay
6
x31 -
[2]
MA 2
x + C1x1 + C2
2 1
[3]
For M(x2) = A y x2 - Px2 + Pa - Ma,
EI
EI
EI y2 =
d2y
= A y x2 - Px2 + Pa - MA
dx22
Ay
dy2
P
=
x2 - x22 + Pax2 - MA x2 + C3
dx2
2 2
2
Ay
6
x32 -
[4]
MA 2
P 3
Pa 2
x +
x x + C3 x2 + C4
6 2
2 2
2 2
[5]
Boundary Conditions:
dy1
= 0 at x1 = 0. From Eq.[2], C1 = 0
dx1
y1 = 0 at x1 = 0. From Eq.[3], C2 = 0
Due to symmetry,
0 =
Ay
2
a
dy2
dx2
= 0 at x2 =
L
. From Eq.[4],
2
L 2
P L 2
L
L
b a b + Paa b - MA a b + C3
2
2 2
2
2
C3 = -
A y L2
8
+
MA L
PL2
PaL
+
8
2
2
997
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12–107.
Continued
Due to symmetry,
A y a2
dy1
dy2
= at x1 = a and x2 = L - a. From Eqs.[2] and [4],
dx1
dx2
- MA a = -
2
Ay
2
(L - a)2 +
+ MA (L - a) +
-A y a2 -
3A y L2
8
+ A y aL +
P
(L - a)2 - Pa(L - a)
2
A y L2
8
-
MA L
PL2
PaL
+
8
2
2
MAL
3PaL
3Pa2
3PL2
+
+
= 0
8
2
2
2
[6]
Continuity Conditions:
At x1 = x2 = a,
A y a2
2
dy1
dy2
. From Eqs.[2] and [4],
=
dx1
dx2
- MA a
A y a2
=
2
2
Ay L
MA L
Pa2
PL2
PaL
+ Pa2 - MA a +
+
2
8
8
2
2
-
2
Ay L
MA L
Pa2
PL2
PaL
+
+
= 0
2
8
8
2
2
[7]
Solving Eqs.[6] and [7] yields,
MA =
Pa
(L - a)
L
Ans.
Ay = P
Substitute the value of MA and Ay obtained into Eqs.[1],
MB =
Pa
(L - a)
L
Ans.
998
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*12–108. Determine the reactions at roller support A and
fixed support B.
w
Equations of Equilibrium. Referring to the free-body diagram of the entire beam, Fig. a,
A
+ ©F = 0;
:
x
Bx = 0
+ c ©Fy = 0;
A y + By - wL = 0
a+ ©MB = 0;
wL a
(1)
L
2
b - A y a Lb - MB = 0
2
3
MB =
wL2
2
- A yL
2
3
(2)
Moment Functions. Referring to the free-body diagram of the beam’s segment, Fig. b,
x
L
L
M(x) + wxa b + wa b ¢ x + ≤ - A yx = 0
2
3
6
a+ ©MO = 0;
M(x) = A yx -
w 2
wL
wL2
x x 2
3
18
Equations of Slope and Elastic Curves.
EI
d2v
= M(x)
dx2
EI
wL2
d2v
w 2
wL
x
x
=
A
x
y
2
3
18
dx2
EI
Ay
dv
w 3
wL 2
wL2
=
x2 x x x + C1
dx
2
6
6
18
EIv =
Ay
6
x3 -
(3)
w 4
wL 3
wL2 2
x x x + C1x + C2
24
18
36
(4)
Boundary Conditions. At x = 0, v = 0. Then Eq. (4) gives
0 = 0 - 0 - 0 - 0 + 0 + C2
0 =
C2 = 0At x =
2 dv
L,
= 0. Then Eq. (3) gives
3 dx
Ay
2 2
w 2 3
wL 2 2
wL2 2
a Lb a Lb a Lb a L b + C1
2 3
6 3
6 3
18 3
2
C1 =
At x =
0 =
2A yL
13wL3
81
9
(5)
2
L, v = 0. Then Eq. (4) gives
3
Ay
2 3
w 2 4
wL 2 3
wL2 2 2
2
a Lb a Lb a Lb a Lb + C1 a Lb
6 3
24 3
18 3
36 3
3
2
C1 =
B
Ans.
2A yL
wL3
18
27
(6)
Solving Eqs. (5) and (6),
Ay =
17wL
wL
Ans.C1 =
Substituting the result of Ay into Eqs. (1) and (2),
24
324
By =
7wL
24
MB =
wL2
36
Ans.
The shear and moment diagrams are shown in Figs. c and d, respectively.
999
L
3
2L
3
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•12–109.
Use discontinuity functions and determine the
reactions at the supports, then draw the shear and moment
diagrams. EI is constant.
3 kip/ft
C
A
B
8 ft
+ ©F = 0
:
x
Cx = 0
+ c ©Fy = 0
A y + By + Cy - 24 = 0
(1)
a + ©MA = 0
18 Cy + 8By - 24(4) = 0
(2)
Ans.
Bending Moment M(x):
M(x) = -(-Cy) 6 x - 0 7 -( -By) 6 x - 10 7 = Cyx + By 6 x - 10 7 -
3
6 x - 10 7 2
2
3
6 x - 10 7 2
2
Elastic curve and slope:
EI
d2v
3
= M(x) = Cyx + By 6 x - 10 7 - 6 x - 10 7 2
2
dx2
EI
Cyx2
By
dv
1
=
+
6 x - 10 7 2 - 6 x - 10 7 3 + C1
dx
2
2
2
EIv =
Cyx3
6
By
+
6
6 x - 10 7 3 -
1
6 x - 10 7 4 + C1x + C2
8
(3)
(4)
Boundary conditions:
v = 0
at
x = 0
From Eq. (4)
C2 = 0
v = 0
at
x = 10 ft
From Eq. (4)
0 = 166.67 Cy + 10C1
v = 0
at
(5)
x = 18 ft
0 = 972Cy + 85.33By - 512 + 18C1
(6)
Solving Eqs. (2),(5) and (6) yields:
By = 14.4 kip
Ans.
Cy = -1.07 kip = 1.07 kip T
Ans.
C1 = 17.78
From Eq. (1):
A y = 10.7 kip
Ans.
1000
10 ft
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12–110. Determine the reactions at the supports, then
draw the shear and moment diagrams. EI is constant.
w0
A
C
B
L
Support Reaction: FBD(b).
+ ©F = 0;
:
x
Ax = 0
Ans.
+ c ©Fy = 0;
A y + By + Cy - w0L = 0
[1]
a + ©MA = 0;
By L + Cy (2L) - w0 L(L) = 0
[2]
Moment Function: FBD(b).
a + ©MNA = 0;
-M(x) -
x
1 w0
a xbx a b + Cyx = 0
2 L
3
M(x) = Cyx -
w0 3
x
6L
Slope and Elastic Curve:
EI
EI
EI
d 2y
= M(x)
dx2
w0 3
d 2y
= Cyx x
2
6L
dx
Cy
w0 4
dy
=
x2 x + C1
dx
2
24L
EI y =
Cy
6
x3 -
[3]
w0 5
x + C1x + C2
120L
[4]
Boundary Conditions:
At x = 0, y = 0. From Eq.[4], C2 = 0
Due to symmetry,
0 =
Cy L2
-
2
C1 = -
dy
= 0 at x = L. From Eq. [3],
dx
w0L3
+ C1
24
CyL2
2
+
w0L3
24
At x = L, y = 0. From Eq. [4],
0 =
Cy L3
6
2
-
Cy L
w0L4
w0L3
+ a+
bL
120
2
24
Cy =
w0L
10
Ans.
1001
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12–110.
Continued
Substituting Cy into Eqs. [1] and [2] yields:
By =
4w0L
5
Ay =
w0L
10
Ans.
Shear and Moment diagrams: The maximum span (positive) moment occurs when
the shear force V = 0. From FBD(c),
+ c ©Fy = 0;
w0L
1 w0
- a xbx = 0
10
2 L
x =
+ ©MNA = 0;
At x =
25
L,
5
At x = L,
M +
25
L
5
w0L
1 w0
x
a xb (x) a b (x) = 0
2 L
3
10
M =
w0L
w0 3
x x
10
6L
M =
25w0L2
75
M = -
w0L2
15
1002
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12–111. Determine the reactions at pin support A and
roller supports B and C. EI is constant.
w
Equations of Equilibrium. Referring to the free-body diagram of the entire beam,
Fig. a,
+ ©F = 0;
:
x
Ax = 0
+ c ©Fy = 0;
A y + By + Cy - wL = 0
a + ©MB = 0;
Cy (L) + wLa
A y - Cy =
Ans.
(1)
L
b - A y(L) = 0
2
wL
2
(2)
Moment Functions. Referring to the free-body diagram of the beam’s segment, Fig. b,
M(x1) is
M(x1) + wx1 ¢
a + ©MO = 0;
x1
≤ - A yx1 = 0
2
M(x1) = A yx1 -
w 2
x
2 1
and M(x2) is given by
a + ©MO = 0;
Cyx2 - M(x2) = 0
M(x2) = Cyx2
Equations of Slope and Elastic Curves.
EI
d2v
= M(x)
dx2
For coordinate x1,
EI
d2v
w 2
= A yx1 x
2 1
dx21
EI
Ay
dv
w 3
x 2 x + C1
=
dx1
2 1
6 1
EIv =
Ay
6
x1 3 -
(3)
w 4
x + C1x1 + C2
24 1
(4)
For coordinate x2,
EI
d2v
= Cyx2
dx22
EI
Cy
dv
=
x 2 + C3
dx2
2 2
EIv =
Cy
6
(5)
x2 3 + C3x2 + C4
(6)
1003
A
C
B
L
L
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12–111.
Continued
Boundary Conditions. At x1 = 0, v1 = 0. Then Eq.(4) gives
0 = 0 - 0 + 0 + C2
C2 = 0
At x1 = L, v1 = 0. Then Eq. (4) gives
0 =
Ay
6
A L3 B -
w
A L4 B + C1L
24
C1 =
A yL
wL3
24
6
2
At x2 = 0, v2 = 0. Then Eq. (6) gives
0 = 0 + 0 + C4
C4 = 0
At x2 = L, v2 = 0. Then Eq. (6) gives
0 =
Cy
6
A L B + C3L
3
Continuity Conditions. At x1 = x2 = L,
Ay
2
A L2 B -
A y + Cy =
C3 = -
CyL2
6
dv1
dv2
. Then Eqs.(3) and (5) give
= dx1
dx2
Cy
A yL2
CyL2
wL3
w 3
≤ = - B A L2 B R
AL B + ¢
6
24
6
2
6
3wL
8
(7)
Solving Eqs. (2) and (7),
Ay =
7wL
16
Cy = -
wL
16
Ans.
The negative sign indicates that Cy acts in the opposite sense to that shown on freebody diagram. Substituting these results into Eq. (1),
By =
5wL
8
Ans.
The shear and moment diagrams are shown in Figs. c and d, respectively.
1004
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*12–112. Determine the moment reactions at fixed
supports A and B. EI is constant.
w0
Equations of Equilibrium. Due to symmetry, A y = By = R and MA = MB = M.
Referring to the free-body diagram of the entire beam, Fig. a,
+ c ©Fy = 0;
2R R =
A
1
wL = 0
2 0
L
2
w0L
4
Moment Function. Referring to the free-body diagram of the beam’s segment, Fig. b,
M(x) + c
w0L
x
1 2w0
a
xb (x) d a b + M x = 0
2 L
3
4
M(x) =
w0L
w0 3
x x - M
4
3L
Equations of Slope and Elastic Curves.
EI
EI
EI
d2v
= M(x)
dx2
d2v
dx 2
=
w0L
w0 3
x x - M
4
3L
w0L 2
w0 4
dv
=
x x - Mx + C1
dx
8
12L
EIv =
Due to symmetry,
(2)
dv
= 0. Then Eq. (1) gives
dx
0 = 0 - 0 - 0 + C1
0 =
(1)
w0L 3
w0 5
M 2
x x x + C1x + C2
24
60L
2
Boundary Conditions. At x = 0,
C1 = 0
dv
L
= 0 at x = . Then Eq. (1) gives
dx
2
w0 L 4
w0L L 2
L
a b a b - Ma b
8
2
12L 2
2
MA = MB = M =
B
5w0L2
96
Note. The boundary condition v = 0 at x = 0 can be used to determine C2 using
Eq.(2).
1005
L
2
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The beam has a constant E1I1 and is supported
by the fixed wall at B and the rod AC. If the rod has a
cross-sectional area A2 and the material has a modulus of
elasticity E2 , determine the force in the rod.
•12–113.
C
w
L2
B
A
L1
+ c ©Fy = 0
TAC + By - wL1 = 0
c + ©MB = 0
TAC(L1) + MB MB =
wL1 2
= 0
2
(1)
wL1 2
- TACL1
2
(2)
Bending Moment M(x):
wx2
2
M(x) = TACx -
Elastic curve and slope:
EI
d2v
wx2
= M(x) = TACx 2
2
dx
EI
TACx2
dv
wx3
=
+ C1
dx
2
6
EIv =
(3)
TACx3
wx4
+ C1x + C2
6
24
(4)
Boundary conditions:
v =
TACL2
A 2E2
x = 0
From Eq. (4)
-E2I1 a
TACL2
b = 0 - 0 + 0 + C2
A 2E2
C2 = a
v = 0
-E1I1L2
b TAC
A 2E2
at
x = L1
From Eq. (4)
0 =
TACL1 3
wL1 4
E1I1L2
+ C1L1 T
6
24
A 2E2 AC
dv
= 0
dx
at
(5)
x = L1
From Eq. (3)
0 =
TACL1 2
wL1 3
+ C1
2
6
(6)
Solving Eqs. (5) and (6) yields:
TAC =
3A 2E2wL1 4
8 A A 2E2L1 3 + 3E1I1L2 B
Ans.
1006
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–114. The beam is supported by a pin at A, a roller at B,
and a post having a diameter of 50 mm at C. Determine the
support reactions at A, B, and C. The post and the beam are
made of the same material having a modulus of elasticity
E = 200 GPa, and the beam has a constant moment of
inertia I = 255(106) mm4.
15 kN/m
A
1m
6m
Equations of Equilibrium. Referring to the free-body diagram of the entire beam, Fig. a,
+ ©F = 0;
:
x
Ax = 0
+ c ©Fy = 0;
A y + By + FC - 15(12) = 0
a+ ©MB = 0;
15(12)(6) - FC(6) - A y(12) = 0
Ans.
(1)
2A y + FC = 180
(2)
Moment Functions. Referring to the free-body diagram of the beam’s segment, Fig. b,
x
M(x) + 15xa b - A yx = 0
2
a+ ©MO = 0;
M(x) = A yx - 7.5x2
Equations of Slope and Elastic Curves.
EI
d2v
= M(x)
dx2
EI
d2v
= A yx - 7.5x2
dx2
EI
Ay
dv
=
x2 - 2.5x3 + C1
dx
2
EIv =
Ay
6
(3)
x3 - 0.625x4 + C1x + C2
(4)
Boundary Conditions. At x = 0, v = 0. Then Eq. (4) gives
C2 = 0
0 = 0 - 0 + 0 + C2
At x = 6 m, v = - ¢ C = -
E C 255 A 10 - 6 B D a -
FC(1)
FCLC
1600FC
. Then Eq. (4) gives
= = p
A CE
pE
2
0.05
E
A
B
4
Ay
1600FC
b =
A 63 B - 0.625 A 64 B + C1(6)
pE
6
C1 = 135 - 6A y - 0.02165FC
Due to symmetry,
0 =
Ay
2
dv
= 0 at x = 6 m. Then Eq. (3) gives
dx
A 62 B - 2.5 A 63 B + 135 - 6A y - 0.02165FC
(5)
12A y - 0.02165FC = 405
Solving Eqs. (2) and (5),
FC = 112.096 kN = 112 kN
A y = 33.95 kN = 34.0 kN
Ans.
Substituting these results into Eq. (1),
By = 33.95 kN = 34.0 kN
Ans.
1007
B
C
6m
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12–115. Determine the moment reactions at the supports
A and B, then draw the shear and moment diagrams. EI is
constant.
A
B
L
Support Reaction: FBD(a).
+ ©F = 0;
:
x
Ax = 0
Ans.
+ c ©Fy = 0;
By - A y = 0
[1]
a + ©MA = 0;
ByL - MA - M0 = 0
[2]
Elastic Curve: As shown.
M/EI Diagrams: M/EI diagrams for By and M0 acting on a cantilever beam are shown.
Moment-Area Theorems: From the elastic curve, tB>A = 0.
tB>A = 0 =
M0
2
L
1 By L
a
b(L)a Lb + a b (L)a b
2 EI
3
EI
2
By =
3M0
2L
Ans.
Substituting the value of By into Eqs.[1] and [2] yields,
Ay =
3M0
2L
MA =
M0
2
Ans.
1008
M0
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*12–116. The rod is fixed at A, and the connection at B
consists of a roller constraint which allows vertical
displacement but resists axial load and moment. Determine
the moment reactions at these supports. EI is constant.
w
B
A
L
Support Reaction: FBD(a).
a + ©MA = 0;
MB + MA - wLa
L
b = 0
2
[1]
Elastic Curve: As shown.
M/EI Diagrams: M/EI diagrams for MB and the uniform distributed load acting on
a cantilever beam are shown.
Moment-Area Theorems: Since both tangents at A and B are horizontal
(parallel), uB>A = 0.
uB>A = 0 = a
MB
1
wL2
b(L) + a b(L)
EI
3
2EI
MB =
wL2
6
Ans.
Substituting MB into Eq.[1],
MA =
wL2
3
Ans.
1009
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•12–117. Determine the value of a for which the
maximum positive moment has the same magnitude as the
maximum negative moment. EI is constant.
P
a
L
(tA>B)1 =
2(L - a)
-P(L - a)2(2L + a)
1 -P(L - a)
a
b (L - a)a a +
b =
2
EI
3
6EI
(tA>B)2 =
A yL3
2L
1 A yL
a
b(L)a
b =
2 EI
3
3EI
tA>B = 0 = (tA>B)1 + (tA>B)2
3
0 =
A yL
-P(L - a)2(2L + a)
+
6EI
3EI
Ay =
P(L - a)2(2L + a)
2L3
Require:
|M1| = |M2|
Pa(L - a)2(2L + a)
3
2L
Pa(L - a)(L + a)
=
2L2
a2 + 2La - L2 = 0
a = 0.414L
Ans.
1010
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12–118. Determine the reactions at the supports, then
draw the shear and moment diagrams. EI is constant.
M0
M0
A
L
Require:
tA>B = 0 = a
M0
L
1 -A y L
2L
b(L)a b + a
b(L)a
b
EI
2
2
EI
3
3
0 =
Ay L
M0L2
;
2EI
3EI
Ay =
3M0
2L
Ans.
Equilibrium:
a+ ©MB = 0;
3M0
(L) - Cy (L) = 0
2L
Cy =
+ c ©Fy = 0;
By -
Ans.
3M0
3M0
= 0
2L
2L
By =
+ ©F = 0;
:
x
3M0
2L
3M0
L
Ans.
Cx = 0
Ans.
1011
C
B
L
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12–119. Determine the reactions at the supports, then
draw the shear and moment diagrams. EI is constant.
Support B is a thrust bearing.
P
A
B
L
Support Reactions: FBD(a).
+ ©F = 0;
:
x
Bx = 0
Ans.
+ c ©Fy = 0;
-A y + By + Cy - P = 0
a+ ©MA = 0;
By (L) + Cy (2L) - Pa
[1]
3L
b = 0
2
[2]
Elastic Curve: As shown.
M/EI Diagrams: M/EI diagrams for P and By acting on a simply supported beam
are drawn separately.
Moment-Area Theorems:
(tA>C)1 =
=
1 3PL L 3L
L
1 3PL 3L 2 3L
a
ba
ba ba
b + a
ba ba
+ b
2 8EI
2
3
2
2 8EI
2
2
6
7PL3
16EI
(tA>C)2 =
By L
By L3
1
ab (2L)(L) = 2
2EI
2EI
(tB>C)1 =
1 PL
L 2 L
PL
L L
a
ba ba ba b + a
ba ba b
2 8EI
2
3
2
4EI
2
4
+
=
(tB>C)2 =
1 3PL L L
L
a
ba ba + b
2 8EI
2
2
6
5PL3
48EI
By L
By L3
1
L
ab(L)a b = 2
2EI
3
12EI
3
tA>C = (tA>C)1 + (tA>C)2 =
By L
7PL3
16EI
2EI
tB>C = (tB>C)1 + (tB>C)2 =
By L3
5PL3
48EI
12EI
From the elastic curve,
tA>C = 2tB>C
By L3
By L3
5PL3
7PL3
= 2a
b
16EI
2EI
48EI
12EI
By =
11P
16
Ans.
Substituting By into Eqs. [1] and [2] yields,
Cy =
13P
32
Ay =
3P
32
Ans.
1012
C
L
2
L
2
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*12–120. Determine the moment reactions at the supports
A and B. EI is constant.
w
B
A
L
–
2
-MA
1 -wL2 L
1 Ay L
a
b(L) + a
b (L) + a
ba b
2
EI
EI
3 8EI
2
uB>A = 0 =
Ay L
0 =
tB>A = 0 =
0 =
2
- MA -
wL2
48
(1)
-MA
L
L
1 -wL2 L L
1 Ay L
a
b(L)a b + a
b(L)a b + a
ba ba b
2
EI
3
EI
2
3 8EI
2
8
Ay L
6
-
MA
wL2
2
384
(2)
Solving Eqs. (1) and (2) yields:
Ay =
3wL
32
MA =
5wL2
192
c + ©MB = 0;
Ans.
MB +
3wL
5wL2
wL L
(L) a b = 0
32
192
2
4
MB =
11wL2
192
Ans.
1013
L
–
2
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•12–121.
Determine the reactions at the bearing supports
A, B, and C of the shaft, then draw the shear and moment
diagrams. EI is constant. Each bearing exerts only vertical
reactions on the shaft.
A
1m
1m
400 N
Support Reactions: FBD(a).
+ c ©Fy = 0;
A y + By + Cy - 800 = 0
[1]
a + ©MA = 0;
By (2) + Cy (4) - 400(1) - 400(3) = 0
[2]
Method of superposition: Using the table in Appendix C, the required
displacements are
yB œ =
Pbx
A L2 - b2 - x2 B
6EIL
=
400(1)(2) 2
A 4 - 12 - 2 2 B
6EI(4)
=
366.67 N # m3
EI
T
By A 4 B
1.3333By m
PL3
=
=
=
48EI
48EI
EI
3
yB
fl
3
c
The compatibility condition requires
(+ T)
0 = 2yB ¿ + yB –
0 = 2a
1.3333By
366.67
b + ab
EI
EI
By = 550 N
Ans.
Substituting By into Eqs. [1] and [2] yields,
A y = 125 N
Cy = 125 N
Ans.
1014
C
B
1m
1m
400 N
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12–122. Determine the reactions at the supports A and B.
EI is constant.
P
A
B
L
Referring to the FBD of the beam, Fig. a
+ ©F = 0;
:
x
Ax = 0
Ans.
By - P - A y = 0
+ c ©Fy = 0;
A y = By - P
(1)
3
a+ ©MA = 0; -MA + By L - Pa Lb = 0
2
MA = By L -
3
PL
2
(2)
Referring to Fig. b and the table in appendix, the necessary deflections are computed
as follow:
yP =
Px2
(3LAC - x)
6EI
=
P(L2)
3
c 3a L b - L d
6EI
2
=
7PL3
12EI
T
3
yBy =
By L
PL3AB
c
=
3EI
3EI
The compatibility condition at support B requires that
(+ T)
0 = vP + vBy
3
0 =
-By L
7PL3
+ a
b
12EI
3EI
By =
7P
4
Ans.
Substitute this result into Eq (1) and (2)
Ay =
3P
4
MA =
PL
4
Ans.
1015
L
2
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12–123. Determine the reactions at the supports A, B, and
C, then draw the shear and moment diagrams. EI is constant.
12 kip
A
Support Reaction: FBD(b).
Cx = 0
Ans.
+ c ©Fy = 0;
A y + By + Cy - 12 - 36.0 = 0
[1]
a + ©MA = 0;
By (12) + Cy (24) - 12(6) - 36.0(18) = 0
[2]
Method of superposition: Using the table in Appendix C, the required
displacements are
5(3) A 24 B
6480 kip # ft3
5wL4
=
=
768EI
768EI
EI
4
yB ¿ =
yB – =
=
Pbx
A L2 - b2 - x2 B
6EIL
By A 24 B
288By ft
PL3
=
=
48EI
48EI
EI
3
yB –¿ =
T
2376 kip # ft3
12(6)(12)
A 24 2 - 62 - 12 2 B =
6EI(24)
EI
T
3
c
The compatibility condition requires
(+ T)
0 = yB ¿ + yB – + yB –¿
0 =
288By
2376
6480
+
+ ab
EI
EI
EI
By = 30.75 kip
Ans.
Substituting By into Eqs.[1] and [2] yields,
A y = 2.625 kip
Cy = 14.625 kip
Ans.
1016
C
B
6 ft
+ ©F = 0;
:
x
3 kip/ft
6 ft
12 ft
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*12–124. The assembly consists of a steel and an aluminum
bar, each of which is 1 in. thick, fixed at its ends A and B, and
pin connected to the rigid short link CD. If a horizontal
force of 80 lb is applied to the link as shown, determine
the moments created at A and B. Est = 2911032 ksi,
Eal = 1011032 ksi.
C
80 lb
D
1 in.
Steel
30 in.
Aluminum
0.5 in.
A
; ©Fx = 0
Pal + Pst - 80 = 0
(1)
Compatibility condition:
¢ st = ¢ al
PstL3
Pal L3
=
3EstIst
3EalIal
Pst = a
1
(29) A 103 B A 12
B (1) A 0.53 B
EstIst
b (Pal) =
Pal
EalIal
(10) A 103 B A 1 B (1) A 13 B (N)
12
Pst = 0.3625 Pal
(2)
Solving Eqs. (1) and (2) yields:
Pal = 58.72 lb
Pst = 21.28 lb
MA = Pst (30) = 639 lb # in. = 0.639 kip # in.
Ans.
MB = Pal (30) = 1761 lb # in. = 1.76 kip # in.
Ans.
1017
B
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•12–125. Determine the reactions at the supports A, B,
and C, then draw the shear and moment diagrams. EI is
constant.
A
3m
Cx = 0
Ans.
a+ ©MC = 0; A y(12) + By(16) - 10(3) - 10(9) = 0
2A y + By = 20
+ c ©Fy = 0;
(1)
A y + By + Cy - 10 - 10 = 0
A y + By + Cy = 20
(2)
Referring to Fig. b and table in appendix, the necessary deflections are:
(vP)1 = (vP)2 =
Pbx
A L2 - b2 - x 2 B
6EILAC AC
=
10(3)(6)
A 12 2 - 32 - 62 B
6EI(12)
=
247.5 kN # m3
EI
T
3
By(12 )
36 By
PL3AC
c
(vB)y =
=
=
48EI
48EI
EI
The compatibility condition at support B requires that
(+ T) 0 = (vP)1 + (vP)2 + (vB)y
0 =
36 By
247.5
247.5
+
+ ab
EI
EI
EI
By = 13.75 kN
Ans.
Substitute this result into Eq. (1) and (2) and solve,
A y = Cy = 3.125 kN
Ans.
The shear And moment diagrams are shown in Fig. b and c respectively.
1018
C
B
Referring to the FBD of the beam, Fig. a,
+ ©F = 0;
;
x
10 kN
10 kN
3m
3m
3m
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12–126. Determine the reactions at the supports A and B.
EI is constant.
M0
A
B
L
Referring to the FBD of the beam, Fig. a,
+ ©F = 0;
:
x
Ax = 0
Ans.
+ c ©Fy = 0;
By - A y = 0
(1)
a+ ©MA = 0;
By(L) - Mo - MA = 0
MA = ByL - Mo
(2)
Referring to Fig. b and the table in the appendix, the necessary deflections are:
vMo =
MoL2
2EI
T
ByL3
PL3
vBy =
=
3EI
3EI
c
Compatibility condition at roller support B requires
(+ T)
0 = vMo + (vB)y
0 =
ByL3
MoL2
+ ab
2EI
3EI
By =
3Mo
2L
Ans.
Substitute this result into Eq. (1) and (2)
Ay =
3Mo
2L
MA =
Mo
2
Ans.
1019
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12–127. Determine the reactions at support C. EI is
constant for both beams.
P
D
B
A
C
L
2
Support Reactions: FBD (a).
+ ©F = 0;
:
x
Cx = 0
a + ©MA = 0;
Cy(L) - By a
Ans.
L
b = 0
2
[1]
Method of superposition: Using the table in Appendix C, the required
displacements are
yB =
By L3
PL3
=
48EI
48EI
T
PA 2 B
PL3BD
PL3
=
=
3EI
3EI
24EI
L 3
yB ¿ =
yB – =
By L3
PL3BD
=
3EI
24EI
T
c
The compatibility condition requires
yB = yB ¿ + yB –
(+ T)
By L3
48EI
=
By =
By L3
PL3
+ ab
24EI
24EI
2P
3
Substituting By into Eq. [1] yields,
Cy =
P
3
Ans.
1020
L
2
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*12–128. The compound beam segments meet in the
center using a smooth contact (roller). Determine the
reactions at the fixed supports A and B when the load P is
applied. EI is constant.
P
A
C
B
L
L
(P - R)L3
RL3
=
3EI
3EI
¢C =
R =
P
2
Member AC:
©Fy = 0;
Ay =
P
2
Ans.
©Fx = 0;
Ax = 0
Ans.
© MA = 0;
MA =
PL
2
Ans.
Member BC:
©Fy = 0;
By =
P
2
Ans.
©Fx = 0;
Bx = 0
Ans.
©MB = 0;
MB =
PL
2
Ans.
The beam has a constant E1 I1 and is supported
by the fixed wall at B and the rod AC. If the rod has a crosssectional area A2 and the material has a modulus of
elasticity E2, determine the force in the rod.
•12–129.
(¢ A)¿ =
dA =
wL41
;
8E1I1
¢A =
C
L2
TACL2
A 2E2
L1
By superposition:
¢ A = (¢ A)¿ - dA
TACL2
wL41
TACL31
=
A 2E2
8E1I1
3E1I1
TAC a
L2
L31
wL41
+
b =
A 2E2
3E1I1
8E1I1
TAC =
B
A
TAC L31
3E1I1
(+ T)
w
3wA 2E2L41
8 C 3E1I1L2 + A 2E2L31 D
Ans.
1021
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12–130. Determine the reactions at A and B. Assume the
support at A only exerts a moment on the beam. EI is
constant.
P
A
2
(uA)1 =
PL
;
8EI
(uA)2 =
B
MAL
EI
L
–
2
By superposition:
L
–
2
0 = (uA)1 - (uA)2
0 =
MAL
PL2
8EI
EI
MA =
PL
8
Ans.
Equilibrium:
a + ©MB = 0;
MB =
-
PL
PL
+
- MB = 0
8
2
3PL
8
Ans.
+ ©F = 0 ;
:
x
Bx = 0
Ans.
+ c ©Fy = 0 ;
By = P
Ans.
12–131. The beam is supported by the bolted supports at its
ends. When loaded these supports do not provide an actual
fixed connection, but instead allow a slight rotation a before
becoming fixed. Determine the moment at the connections
and the maximum deflection of the beam.
P
u - u¿ = a
L
—
2
ML
ML
PL2
= a
16EI
3EI
6EI
ML = a
M = a
PL2
- a b(2EI)
16EI
2EI
PL
ab
8
L
¢ max = ¢ - ¢ ¿ =
Ans.
M(L2 )
PL3
- 2c
C L2 - (L>2)2 D d
48EI
6EIL
¢ max =
PL3
L2 PL
2EIa
a
b
48EI
8EI
8
L
¢ max =
aL
PL3
+
192EI
4
Ans.
1022
L
—
2
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*12–132. The beam is supported by a pin at A, a spring
having a stiffness k at B, and a roller at C. Determine the
force the spring exerts on the beam. EI is constant.
w
A
B
Method of Superposition: Using the table in appendix C, the required
displacements are
5w(2L)4
5wL4AC
5wL4
=
=
384EI
384EI
24EI
T
Fsp (2L)3
Fsp L3
PL3AC
=
=
yB – =
48EI
48EI
6EI
c
yB ¿ =
Using the spring formula, ysp =
Fsp
k
L
C
k
L
.
The compatibility condition requires
ysp = yB ¿ + yB –
(+ T)
Fsp
k
=
Fsp =
Fsp L3
5wL4
+ ab
24EI
6EI
5wkL4
4 A 6EI + kL3 B
Ans.
•12–133.
The beam is made from a soft linear elastic
material having a constant EI. If it is originally a distance
¢ from the surface of its end support, determine the
distance a at which it rests on this support when it is
subjected to the uniform load w0 , which is great enough to
cause this to happen.
w0
⌬
a
L
The curvature of the beam in region BC is zero, therefore there is no bending
moment in the region BC, The reaction F is at B where it touches the support. The
slope is zero at this point and the deflection is ¢ where
¢ =
R(L - a)3
w0(L - a)4
8EI
3EI
u1
=
w0(L - a)3
R(L - a)2
6EI
2EI
Thus,
1
8¢EI 4
R = a
b
9w30
Ans.
1
72¢EI 4
L - a = a
b
w0
1
72¢EI 4
a = L - a
b
w0
Ans.
1023
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12–134. Before the uniform distributed load is applied on
the beam, there is a small gap of 0.2 mm between the beam
and the post at B. Determine the support reactions at A, B,
and C. The post at B has a diameter of 40 mm, and the
moment of inertia of the beam is I = 875(106) mm4. The
post and the beam are made of material having a modulus
of elasticity of E = 200 GPa .
30 kN/m
A
6m
Equations of Equilibrium. Referring to the free-body diagram of the beam, Fig. a,
+ ©F = 0;
:
x
Ax = 0
+ c ©Fy = 0;
A y + FB + Cy - 30(12) = 0
(1)
a + ©MA = 0;
FB(6) + Cy(12) - 30(12)(6) = 0
(2)
Ans.
Method of superposition: Referring to Fig. b and the table in the Appendix, the
necessary deflections are
5(30) A 12 B
8100kN # m3
5wL4
=
=
T
384EI
384EI
EI
4
(vB)1 =
FB A 12 B
36FB
PL3
=
=
48EI
48EI
EI
3
(vB)2 =
c
The deflection of point B is
vB = 0.2 A 10 - 3 B +
FB(a)
FBLB
= 0.2 A 10 - 3 B +
AE
AE
T
The compatibility condition at support B requires
A+TB
vB = (vB)1 + (vB)2
0.2 A 10 - 3 B +
FB (1)
36FB
8100
=
+ ab
AE
EI
EI
0.2 A 10 - 3 B E +
FB
p
A 0.04 2 B
4
+
FB
36FB
8100
=
A
I
I
36FB
875 A 10 - 6 B
8100
=
875 A 10 - 6 B
-
C
1m
0.2 A 10 - 3 B C 200 A 109 B D
1000
FB = 219.78 kN = 220 kN
Ans.
Substituting the result of FB into Eqs. (1) and (2),
A y = Cy = 70.11 kN = 70.1 kN
Ans.
1024
B
0.2 mm
6m
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12–135. The 1-in.-diameter A-36 steel shaft is supported by
unyielding bearings at A and C. The bearing at B rests on a
simply supported steel wide-flange beam having a moment
of inertia of I = 500 in4. If the belt loads on the pulley are
400 lb each, determine the vertical reactions at A, B, and C.
3 ft
5 ft
A
2 ft
5 ft
B
For the shaft:
(¢ b)1 =
(¢ b)2 =
400
lb
800(3)(5)
13200
A -52 - 32 + 102 B =
6EIs(10)
EIs
By A 103 B
48EIs
C
5 ft
20.833By
=
EIs
For the beam:
¢b =
By A 103 B
48EIb
20.833By
=
EIb
Compatibility condition:
+ T ¢ b = (¢ b)1 - (¢ b)2
20.833By
EIb
Is =
=
20.833By
13200
EIs
EIs
p
(0.5)4 = 0.04909 in4
4
20.833By (0.04909)
500
400
lb
= 13200 - 20.833By
By = 634 lb
Ans.
Form the free-body digram,
A y = 243 lb
Ans.
Cy = 76.8 lb
Ans.
1025
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*12–136. If the temperature of the 75-mm-diameter post
CD is increased by 60°C, determine the force developed in
the post. The post and the beam are made of A-36 steel, and
the moment of inertia of the beam is I = 255(106) mm4.
3m
3m
A
B
C
3m
D
Method of Superposition. Referring to Fig. a and the table in the Appendix, the
necessary deflections are
(vC)1 =
FCD A 33 B
9FCD
PLBC 3
c
=
=
3EI
3EI
EI
(vC)2 = (uB)2LBC =
3FCD (3)
9FCD
MOLAB
c
(LBC) =
(3) =
3EI
3EI
EI
The compatibility condition at end C requires
A+cB
vC = (vC)1 + (vC)2
=
9FCD
9FCD
18FCD
c
+
=
EI
EI
EI
Referring to Fig. b, the compatibility condition of post CD requires that
dFCD + vC = dT
dFCD =
(1)
FCD (3)
FCD LCD
=
AE
AE
dT = a¢TL = 12 A 10 - 6 B (60)(3) = 2.16 A 10 - 3 B m
Thus, Eq. (1) becomes
3FCD
18FCD
+
= 2.16 A 10 - 3 B
AE
EI
3FCD
p
A 0.0752 B
4
+
18FCD
255 A 10 - 6 B
= 2.16 A 10 - 3 B C 200 A 109 B D
FCD = 6061.69N = 6.06 kN
Ans.
1026
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
•12–137.
The shaft supports the two pulley loads shown.
Using discontinuity functions, determine the equation of
the elastic curve. The bearings at A and B exert only vertical
reactions on the shaft. EI is constant.
x
A
12 in.
B
12 in.
70 lb
180 lb
M = -180 6 x - 0 7 - (-277.5) 6 x - 12 7 - 70 6 x - 24 7
M = -180x + 277.5 6 x - 12 7 - 70 6 x - 24 7
Elastic curve and slope:
EI
d2v
= M = -180x + 277.5 6 x - 12 7 - 70 6 x - 24 7
dx2
EI
dv
= -90x2 + 138.75 6 x - 12 7 2 - 35(x - 24 7 2 + C1
dx
EIv = -30x3 + 46.25 6 x - 12 7 3 - 11.67 6 x - 24 7 3 + C1x + C2 (1)
Boundary conditions:
v = 0
at
x = 12 in,
From Eq. (1)
0 = -51,840 + 12C1 + C2
12C1 + C2 = 51 840
v = 0
at
(2)
x = 60 in.
From Eq.(1)
0 = -6 480 000 + 5 114 880 - 544 320 + 60C1 + C2
60C1 + C2 = 1909440
(3)
Solving Eqs. (2) and (3) yields:
C1 = 38 700
v =
C2 = -412 560
1
[-30x3 + 46.25 6 x - 12 7 3 - 11.7 6 x - 24 7 3
EI
+ 38 700x - 412 560]
Ans.
1027
36 in.
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12–138. The shaft is supported by a journal bearing at A,
which exerts only vertical reactions on the shaft, and by a
thrust bearing at B, which exerts both horizontal and
vertical reactions on the shaft. Draw the bending-moment
diagram for the shaft and then, from this diagram, sketch
the deflection or elastic curve for the shaft’s centerline.
Determine the equations of the elastic curve using the
coordinates x1 and x2 . EI is constant.
80 lb
A
x1
EI
d2v1
dx21
= 26.67x1
dv1
= 13.33x21 + C1
dx1
(1)
EIv1 = 4.44x31 + C1x1 + C2
(2)
EI
For M2 (x) = -26.67x2
EI
d2v2
dx22
= -26.67x2
dv2
= -13.33x22 + C3
dx2
(3)
EIv2 = -4.44x32 + C3x2 + C4
(4)
EI
Boundary conditions:
v1 = 0
at
x1 = 0
at
x2 = 0
From Eq.(2)
C2 = 0
v2 = 0
C4 = 0
Continuity conditions:
dv1
dv2
= dx1
dx2
at
x1 = x2 = 12
From Eqs. (1) and (3)
1920 + C1 = -( -1920 + C3)
C1 = -C3
v1 = v2
(5)
x1 = x2 = 12
at
7680 + 12C1 = -7680 + 12C3
C3 - C1 = 1280
(6)
Solving Eqs. (5) and (6) yields:
C3 = 640
80 lb
12 in.
For M1 (x) = 26.67 x1
C1 = -640
v1 =
1
A 4.44x31 - 640x1 B lb # in3
EI
Ans
v2 =
1
A -4.44x32 + 640x2 B lb # in3
EI
Ans.
1028
B
4 in.
4 in.
x2
12 in.
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12–139. The W8 * 24 simply supported beam is subjected
to the loading shown. Using the method of superposition,
determine the deflection at its center C. The beam is made
of A-36 steel.
6 kip/ft
5 kip⭈ft
A
B
C
8 ft
Elastic Curves: The elastic curves for the uniform distributed load and couple
moment are drawn separately as shown.
Method of superposition: Using the table in Appendix C, the required
displacements are
-5(6) A 16 B
2560 kip # ft3
-5wL4
=
=
T
768EI
768EI
EI
4
(¢ C)1 =
(¢ C)2 = -
= -
=
M0x
A L2 - x2 B
6EIL
5(8)
C (16)2 - (8)2 D
6EI(16)
80 kip # ft3
EI
T
The displacement at C is
¢ C = (¢ C)1 + (¢ C)2
=
80
2560
+
EI
EI
=
2640 kip # ft3
EI
2640(1728)
=
29 A 103 B (82.8)
= 1.90 in.
Ans.
T
1029
8 ft
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*12–140. Using the moment-area method, determine the
slope and deflection at end C of the shaft. The 75-mmdiameter shaft is made of material having E = 200 GPa.
Support Reactions and
B
A
M
Diagram. As shown in Fig. a.
EI
1m
1m
15 kN
冷tB>A 冷 = (1) B 2 a EI b (2) R + c 3 (2) d B 2 a - EI b(2) R
=
7.5
1
1
3
5.5 kN # m3
EI
冷tC>A 冷 = (1 + 1) B
1 7.5
1
1
3
a
b(2) R + c (2) + 1 d B a b(2) R
2 EI
3
2
EI
2
3
1
+ c (1) d B a b(1) R
3
2
EI
=
9 kN # m3
EI
冷uC>A 冷 = 2 a EI b(2) + 2 a - EI b(3)
1
=
7.5
1
3
3 kN # m3
EI
Referring to the geometry of the elastic curve, Fig. b,
5.5
EI
2.75kN # m2
=
=
uA =
LAB
2
EI
冷tB>A 冷
uC = uC>A - uA =
=
3
2.75
EI
EI
0.25 kN # m2
=
EI
0.25 A 103 B
200 A 109 B c
p
A 0.03754 B d
4
= 0.805 A 10 - 3 B rad
Ans.
and
¢ C = 冷 tC>A冷 - 冷 tB>A ¢
=
9
5.5 3
a b
EI
EI 2
=
0.75 kN # m3
=
EI
LAC
≤
LAB
0.75 A 103 B
1m
3 kN
Moment Area Theorem. Referring to Fig. b,
1
C
p
200 A 10 B c A 0.03754 B d
4
= 0.002414 m = 2.41 mm c Ans.
9
1030
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•12–141. Determine the reactions at the supports. EI is
constant. Use the method of superposition.
w
A
wL
C L3 - 2(3L)L2 + (3L)3 D
24EI
¢B = ¢C =
D
B
L
4
11wL
12EI
=
Due to symmetry, By = Cy
By (L)(2L)
¢ BB = ¢ CC =
6EI(3L)
C (3L)2 - (2L)2 - L2 D
4By L3
=
9EI
By (L)(L)
¢ BC = ¢ CB =
6EI(3L)
C -L2 - L2 + (3L)2 D
7By L3
=
18EI
By superposition:
+T
0 = ¢ B - ¢ BB - ¢ BC
0 =
4By L3
7By L3
11wL4
12EI
9EI
18EI
By = Cy =
11wL
10
Ans.
Equilibrium:
a+ ©MD = 0;
Ay =
c + ©Fy = 0;
Dy =
+ ©F = 0;
;
x
3wLa
3L
11wL
11wL
b (L) (2L) - A y (3L) = 0
2
10
10
2wL
5
Ans.
2wL
11wL
11wL
+
+
+ Dy - 3wL = 0
5
10
10
2wL
5
Ans.
Dx = 0
Ans.
1031
C
L
L
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Page 1032
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12–142. Determine the moment reactions at the supports
A and B. Use the method of integration. EI is constant.
w0
A
B
L
Support Reactions: FBD(a).
w0L
= 0
2
+ c ©Fy = 0;
A y + By -
a + ©MA = 0;
ByL + MA - MB -
[1]
w0L L
a b = 0
2
3
[2]
Moment Function: FBD(b).
a+ ©MNA = 0;
-M(x) -
x
1 w0
a xb x a b - MB + Byx = 0
2 L
3
M(x) = Byx -
w0 3
x - MB
6L
Slope and Elastic Curve:
EI
EI
EI
EI y =
d2y
= M(x)
dx2
w0 3
d2y
= Byx x - MB
6L
dx2
By
w0 4
dy
=
x2 x - MBx + C1
dx
2
24L
By
6
x3 -
[3]
w0 5
MB 2
x x + C1x + C2
120L
2
[4]
Boundary Conditions:
dy
= 0
dx
From Eq.[3],
C1 = 0
At x = 0, y = 0.
From Eq.[4],
C2 = 0
At x = 0,
At x = L,
0 =
dy
= 0.
dx
By L2
2
-
From Eq. [3].
w0L3
- MBL
24
0 = 12By L - w0 L2 - 24MB
At x = L, y = 0.
0 =
By L3
6
-
[5]
From Eq. [4],
w0 L4
MB L2
120
2
0 = 20By L - w0 L2 - 60MB
[6]
1032
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12–142.
Continued
Solving Eqs. [5] and [6] yields,
MB =
w0 L2
30
By =
3w0L
20
Ans.
Substituting By and MB into Eqs. [1] and [2] yields,
MA =
w0L2
20
Ay =
7w0 L
20
Ans.
1033
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12–143. If the cantilever beam has a constant thickness t,
determine the deflection at end A. The beam is made of
material having a modulus of elasticity E.
L
w0
x
Section Properties: Referring to the geometry shown in Fig. a,
A
h(x)
h0
=
;
x
L
h0
x
h(x) =
L
h0
B
Thus, the moment of inertia of the tapered beam as a function of x is
I(x) =
3
h0
th0 3 3
1
1
t C h(x) D 3 =
t¢ x≤ =
x
12
12 L
12L3
Moment Function. Referring to the free-body diagram of the beam’s segment, Fig. b,
M(x) + B
a + ©MO = 0;
1 w0
x
a xbx R a b = 0
2 L
3
M(x) = -
w0 3
x
6L
Equations of slope and Elastic Curve.
E
M(x)
d2v
=
2
I(x)
dx
w0 3
x
2w0L2
dv
6L
E 2 =
=
dx
th0 3 3
th0 3
x
3
12L
-
2
E
2w0L2
dv
= x + C1
dx
th0 3
Ev = -
w0L2
th0 3
(1)
x2 + C1x + C2
Boundary conditions. At x = L,
0 = -
(2)
dv
= 0. Then Eq. (1) gives
dx
2w0L2
th0
(L) + C1
3
C1 =
2w0L3
th0 3
At x = L, v = 0. Then Eq. (2) gives
0 = -
w0L2
th0
3
A L2 B +
2w0L3
th0
3
(L) + C2
C2 = -
w0L4
th0 3
Substituting the results of C1 and C2 into Eq. (2),
v =
w0L2
Eth0 3
A -x2 + 2Lx - L2 B
At A, x = 0. Then
vA = v冷x = 0 = -
w0L4
w0L4
Eth0
Eth0 3
=
3
Ans.
T
1034
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*12–144. Beam ABC is supported by beam DBE and
fixed at C. Determine the reactions at B and C. The beams
are made of the same material having a modulus of
elasticity E = 200 GPa, and the moment of inertia of both
beams is I = 25.0(106) mm4.
100 lb/ft
a
A
B
C
D
E
a
4 ft
4 ft
6 ft
6 in.
Section a – a
+ ©F = 0;
:
x
Cx = 0
+ c ©Fy = 0;
By + Cy - 9(6) = 0
a+ ©MC = 0;
9(6)(3) - By(4) - MC = 0
Ans.
(1)
MC = 162 - 4By
(2)
Method of superposition: Referring to Fig. b and the table in the appendix, the
deflections are
By A 6 B
4.5By
PLDE 3
=
=
T
48EI
48EI
EI
3
vB =
(vB)1 =
=
9 A 42 B
wx2
A x2 - 4Lx + 6L2 B =
C 4 2 - 4(6)(4) + 6 A 62 B D
24EI
24EI
816 kN # m3
T
EI
By A 4 B
21.3333By
PLBC 3
c
=
=
3EI
3EI
EI
3
(vB)2 =
The compatibility condition at support B requires that
A+TB
vB = (vB)1 + (vB)2
4.5By
EI
=
21.3333By
816
+ ab
EI
EI
By = 31.59 kN = 31.6 kN
Ans.
Substituting the result of By into Eqs. (1) and (2),
MC = 35.65 kN # m = 35.7 kN # m
Ans.
Cy = 22.41 kN = 22.4 kN
Ans.
1035
a
6 ft
3 in.
Equation of Equilibrium. Referring to the free-body diagram of the beam, Fig. a,
a
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•12–145.
Using the method of superposition, determine
the deflection at C of beam AB. The beams are made of
wood having a modulus of elasticity of E = 1.5(103) ksi.
100 lb/ft
a
A
B
C
D
E
a
4 ft
4 ft
6 ft
6 in.
Section a – a
Method of superposition. Referring to Fig. b and the table in the appendix, the
deflection of point B is
600 A 8 B
PLDE 3
6400 lb # ft3
=
=
T
48EI
48EI
EI
3
¢B =
Subsequently, referring to Fig. c,
(¢ C)1 = ¢ B a
6
6400 6
3200 lb # ft3
b =
a b =
T
12
EI 12
EI
5(100) A 12 B
5wL4
27000 lb # ft3
=
=
T
384EI
384EI
EI
4
(¢ C)2 =
Thus, the deflection of point C is
A+TB
¢ C = (¢ C)1 + (¢ C)2
=
3200
27000
+
EI
EI
30200 lb # ft3
=
=
EI
30200 A 12 3 B
1.5 A 106 B c
1
(3) A 63 B d
12
= 0.644 in T
Ans.
1036
a
6 ft
3 in.
Support Reactions: The reaction at B is shown on the free-body diagram of beam
AB, Fig. a.
a
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12–146. The rim on the flywheel has a thickness t, width b,
and specific weight g. If the flywheel is rotating at a
constant rate of v, determine the maximum moment
developed in the rim. Assume that the spokes do not
deform. Hint: Due to symmetry of the loading, the slope of
the rim at each spoke is zero. Consider the radius to be
sufficiently large so that the segment AB can be considered
as a straight beam fixed at both ends and loaded with a
uniform centrifugal force per unit length. Show that this
force is w ⫽ btgv2r>g.
A
B
v
r
Centrifugal Force: The centrifugal force action on a unit length of the rim rotating
at a constant rate of v is
g
btgv2r
w = mv2 r = bta bv2r =
g
g
(Q.E.D.)
Elastic Curve: Member AB of the rim is modeled as a straight beam with both of
its ends fixed and subjected to a uniform centrifigal force w.
Method of Superposition: Using the table in Appendix C, the required
displacements are
uB ¿ =
wL3
6EI
yB ¿ =
wL4
c
8EI
uB – =
yB – =
MBL
EI
uB ¿– =
MBL2
c
2EI
yB –¿ =
ByL2
2EI
ByL3
3EI
T
Computibility requires,
0 = uB ¿ + uB – + uB ¿–
0 =
By L2
MBL
wL3
+
+ ab
6EI
EI
2EI
0 = wL2 + 6MB - 3By L
[1]
0 = yB ¿ + yB – + yB –¿
(+ c )
0 =
By L3
MB L2
wL4
+
+ ab
8EI
2EI
3EI
0 = 3wL2 + 12MB - 8By L
[2]
Solving Eqs. [1] and [2] yields,
By =
wL
2
MB =
Due to symmetry, A y =
wL
2
wL2
12
MA =
wL2
12
Maximum Moment: From the moment diagram, the maximum moment occurs at
btgv2r
pr
the two fixed end supports. With w =
and L = ru =
.
g
3
wL2
Mmax =
=
12
A B
btgv2r pr 2
g
3
12
=
t
p2btgv2r3
108g
Ans.
1037
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•13–1.
Determine the critical buckling load for the column.
The material can be assumed rigid.
P
L
2
k
Equilibrium: The disturbing force F can be determined by summing moments
about point A.
a + ©MA = 0;
P(Lu) - F a
L
b = 0
2
A
F = 2Pu
Spring Formula: The restoring spring force F1 can be determine using spring
formula Fs = kx.
Fs = ka
L
kLu
ub =
2
2
Critical Buckling Load: For the mechanism to be on the verge of buckling, the
disturbing force F must be equal to the restoring spring force F1.
2Pcr u =
kLu
2
Pcr =
kL
4
L
2
Ans.
1038
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13–2. Determine the critical load Pcr for the rigid bar and
spring system. Each spring has a stiffness k.
P
Equilibrium: The disturbing forces F1 and F2 can be related to P by writing the
moment equation of equlibrium about point A. Using small angle ananlysis, where
cos u ⬵ 1 and sin u = u,
+ ©MA = 0;
F2 a
L
3
k
L
2
b + F1 a Lb - PLu = 01
3
3
L
3
F2 + 2F1 = 3Pu
k
(1)
Spring Force. The restoring spring force A Fsp B 1 and A Fsp B 2 can be determined using
the spring formula,
2
1
Lu and x2 = Lu, Fig. b. Thus,
3
3
2
2
A Fsp B 1 = kx1 = ka Lu b = kLu
3
3
L
3
A
Fsp = kx, where x1 =
A Fsp B 2 = kx2 = ka Lu b =
1
3
Critical Buckling Load. When the mechanism is on the verge of buckling the
disturbing force F must be equal to the restoring force of the spring Fsp. Thus,
F1 = A Fsp B 1 =
2
kLu
3
F2 = A Fsp B 2 =
1
kLu
3
Substituting this result into Eq. (1),
2
1
kLu + 2 a kLu b = 3Pcr u
3
3
Pcr =
5
kL
9
Ans.
1039
1
kLu
3
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13–3. The leg in (a) acts as a column and can be modeled
(b) by the two pin-connected members that are attached to a
torsional spring having a stiffness k (torque兾rad). Determine
the critical buckling load. Assume the bone material is rigid.
a + ©MA = 0;
-P(u)a
P
L
b + 2ku = 0
2
L
—
2
Require:
k
Pcr =
4k
L
Ans.
L
—
2
(a)
*13–4. Rigid bars AB and BC are pin connected at B. If
the spring at D has a stiffness k, determine the critical load
Pcr for the system.
(b)
P
A
Equilibrium. The disturbing force F can be related P by considering the equilibrium
of joint A and then the equilibrium of member BC,
a
B
Joint A (Fig. b)
+ c ©Fy = 0;
FAB cos f - P = 0
FAB =
a
P
cos f
k
D
Member BC (Fig. c)
a
©MC = 0; F(a cos u) -
P
P
cos f (2a sin u) sin f(2a cos u) = 0
cos f
cos f
C
F = 2P(tan u + tan f)
Since u and f are small, tan u ⬵ u and tan f ⬵ f. Thus,
F = 2P(u + f)
(1)
Also, from the geometry shown in Fig. a,
2au = af
f = 2u
Thus Eq. (1) becomes
F = 2P(u + 2u) = 6Pu
Spring Force. The restoring spring force Fsp can be determined using the spring
formula, Fsp = kx, where x = au, Fig. a. Thus,
Fsp = kx = kau
1040
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13–4.
Continued
Critical Buckling Load. When the mechanism is on the verge of buckling the
disturbing force F must be equal to the restoring spring force Fsp.
F = Fsp
6Pcru = kau
Pcr =
ka
6
Ans.
1041
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•13–5.
An A-36 steel column has a length of 4 m and is
pinned at both ends. If the cross sectional area has the
dimensions shown, determine the critical load.
25 mm
Section Properties:
A = 0.01(0.06) + 0..05(0.01) = 1.10 A 10 - 3 B m2
Ix = Iy =
10 mm
1
1
(0.01) A 0.063 B +
(0.05) A 0.013 B = 0.184167 A 10 - 6 B m4
12
12
25 mm
Critical Buckling Load: K = 1 for pin supported ends column. Applying Euler’s
formula,
Pcr =
25 mm
25 mm
10 mm
p2EI
(KL)2
p2 (200)(109)(0.184167)(10 - 6)
=
[1(4)]2
Ans.
= 22720.65 N = 22.7 kN
Critical Stress: Euler’s formula is only valid if scr 6 sg.
scr =
Pcr
22720.65
= 20.66 MPa 6 sg = 250 MPa
=
A
1.10(10 - 3)
O.K.
13–6. Solve Prob. 13–5 if the column is fixed at its bottom
and pinned at its top.
25 mm
Section Properties:
A = 0.01(0.06) + 0.05(0.01) = 1.10 A 10 - 3 B m2
10 mm
1
1
Ix = Iy =
(0.01) A 0.063 B +
(0.05) A 0.013 B = 0.184167 A 10 - 6 B m4
12
12
25 mm
Critical Buckling Load: K = 0.7 for one end fixed and the other end pinned
column. Applying Euler’s formula,
Pcr =
p EI
(EL)2
p2 (200)(109)(0.184167)(10 - 6)
=
[0.7(4)]2
Ans.
= 46368.68 N = 46.4 kN
Critical Stress: Euler’s formula is only valid if scr 6 sg.
scr =
25 mm
25 mm
10 mm
2
Pcr
46368.68
= 42.15 MPa 6 sg = 250 MPa
=
A
1.10(10 - 3)
1042
O.K.
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13–7. A column is made of A-36 steel, has a length of 20 ft,
and is pinned at both ends. If the cross-sectional area has
the dimensions shown, determine the critical load.
6 in.
0.25 in.
The cross sectional area and moment of inertia of the square tube is
5.5 in.
A = 6(6) - 5.5(5.5) = 5.75 in2
I =
0.25 in.
1
1
(6)(63) (5.5)(5.53) = 31.74 in4
12
12
0.25 in.
0.25 in.
The column is pinned at both of its end, k = 1. For A36 steel, E = 29.0(103) ksi and
sg = 36 ksi (table in appendix). Applying Euler’s formula,
p C 29.0(10 ) D (31.74)
p2EI
=
2
(KL)
C 1(20)(12) D 2
2
Pcr =
3
Ans.
= 157.74 kip = 158
Critical Stress. Euler’s formula is valid only if scr 6 sg.
scr =
Pcr
157.74
=
= 27.4 ksi 6 sg = 36 ksi
A
5.75
O.K.
*13–8. A column is made of 2014-T6 aluminum, has a
length of 30 ft, and is fixed at its bottom and pinned at its
top. If the cross-sectional area has the dimensions shown,
determine the critical load.
6 in.
0.25 in.
5.5 in.
The cross-sectional area and moment of inertia of the square tube is
0.25 in.
A = 6(6) - 5.5(5.5) = 5.75 in2
0.25 in.
1
1
I =
(6)(63) (5.5)(5.53) = 31.74 in4
12
12
The column is fixed at one end, K = 0.7. For 2014–76 aluminium, E = 10.6(103) ksi
and sg = 60 ksi (table in appendix). Applying Euler’s formula,
p C 10.6(10 ) D (31.74)
p2EI
=
2
(KL)
C 0.7(30)(12) D 2
2
Pcr =
3
Ans.
= 52.29 kip = 52.3 kip
Critical Stress. Euler’s formula is valid only if scr 6 sg.
scr =
Pcr
52.3
=
= 9.10 ksi 6 sg = 60 ksi
A
5.75
O.K.
1043
0.25 in.
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•13–9.
The W14 * 38 column is made of A-36 steel and is
fixed supported at its base. If it is subjected to an axial load
of P = 15 kip, determine the factor of safety with respect to
buckling.
P
From the table in appendix, the cross-sectional area and moment of inertia about
weak axis (y-axis) for W14 * 38 are
20 ft
A = 11.2 in2
Iy = 26.7 in4
The column is fixed at its base and free at top, k = 2. Here, the column will buckle
about the weak axis (y axis). For A36 steel, E = 29.0(103) ksi and sy = 36 ksi.
Applying Euler’s formula,
p2 C 29.0(103) D (26.7)
p2EIy
Pcr =
C 2 (20)(12) D 2
=
(KL)2
= 33.17 kip
Thus, the factor of safety with respect to buckling is
F.S =
Pcr
33.17
=
= 2.21
P
15
Ans.
The Euler’s formula is valid only if scr 6 sg.
scr =
Pcr
33.17
=
= 2.96 ksi 6 sg = 36 ksi
A
11.2
O.K.
13–10. The W14 * 38 column is made of A-36 steel.
Determine the critical load if its bottom end is fixed
supported and its top is free to move about the strong axis
and is pinned about the weak axis.
P
From the table in appendix, the cross-sectional area and moment of inertia about
weak axis (y-axis) for W14 * 38 are
A = 11.2 in2
Ix = 385 in4
Iy = 26.7 in4
The column is fixed at its base and free at top about strong axis. Thus, kx = 2. For
A36 steel, E = 29.0(103) ksi and sg = 36 ksi.
Pcr =
p2EIx
(KxLx)
2
=
p2 C 29.0(103) D (385)
C 2 (20)(12) D 2
= 478.28 kip
The column is fixed at its base and pinned at top about weak axis. Thus, ky = 0.7.
Pcr =
p2EIy
2
(KyLy)
=
p2 C 29.0(103) D (26.7)
C 0.7(20)(12) D 2
Ans.
= 270.76 kip = 271 kip (Control)
The Euler’s formula is valid only if scr 6 sg.
scr =
Pcr
270.76
=
= 24.17 ksi 6 sg = 36 ksi
A
11.2
O.K.
1044
20 ft
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13–11. The A-36 steel angle has a cross-sectional area of
A = 2.48 in2 and a radius of gyration about the x axis of
rx = 1.26 in. and about the y axis of ry = 0.879 in. The
smallest radius of gyration occurs about the z axis and is
rz = 0.644 in. If the angle is to be used as a pin-connected
10-ft-long column, determine the largest axial load that can
be applied through its centroid C without causing it to buckle.
y
z
C
x
x
z
y
The least radius of gyration:
r2 = 0.644 in.
scr =
p2E
2
A KL
r B
controls.
K = 1.0
;
p2 (29)(103)
(120) 2
C 1.00.644
D
= 8.243 ksi 6 sg
O.K.
Pcr = scr A = 8.243 (2.48) = 20.4 kip
Ans.
=
*13–12. An A-36 steel column has a length of 15 ft and is
pinned at both ends. If the cross-sectional area has the
dimensions shown, determine the critical load.
8 in.
0.5 in.
0.5 in.
6 in.
0.5 in.
Ix =
1
1
(8)(73) (7.5)(63) = 93.67 in4
12
12
Iy = 2 a
Pcr =
1
1
b(0.5)(83) +
(6)(0.53) = 42.729 in4 (controls)
12
12
p2(29)(103)(42.729)
p2EI
=
2
(EL)
[(1.0)(15)(12)]2
= 377 kip
Ans.
Check:
A = (2)(8)(0.5) + 6(0.5) = 11 in2
scr =
Pcr
377
=
= 34.3 ksi 6 sg
A
11
Therefore, Euler’s formula is valid
1045
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•13–13.
An A-36 steel column has a length of 5 m and is
fixed at both ends. If the cross-sectional area has the
dimensions shown, determine the critical load.
I =
10 mm
10 mm
50 mm
1
1
(0.1)(0.053) (0.08)(0.033) = 0.86167 (10 - 6) m4
12
12
Pcr =
100 mm
p2(200)(109)(0.86167)(10 - 6)
p2EI
=
2
(KL)
[(0.5)(5)]2
= 272 138 N
= 272 kN
scr =
=
Pcr
;
A
Ans.
A = (0.1)(0.05) - (0.08)(0.03) = 2.6(10 - 3) m2
272 138
= 105 MPa 6 sg
2.6 (10 - 3)
Therefore, Euler’s formula is valid.
13–14. The two steel channels are to be laced together
to form a 30-ft-long bridge column assumed to be pin
connected at its ends. Each channel has a cross-sectional
area of A = 3.10 in2 and moments of inertia Ix = 55.4 in4,
Iy = 0.382 in4. The centroid C of its area is located in
the figure. Determine the proper distance d between the
centroids of the channels so that buckling occurs about the
x–x and y¿ – y¿ axes due to the same load. What is the value
of this critical load? Neglect the effect of the lacing.
Est = 2911032 ksi, sY = 50 ksi.
y
0.269 in.
C
d
y
In order for the column to buckle about x - x and y - y at the same time, Iy must
be equal to Ix
Iy = Ix
0.764 + 1.55 d2 = 110.8
d = 8.43 in.
Ans.
Check:
d 7 2(1.231) = 2.462 in.
O.K.
3
p (29)(10 )(110.8)
p2 EI
=
2
(KL)
[1.0(360)]2
= 245 kip
Ans.
Check stress:
scr =
x
C
d 2
Iy = 2(0.382) + 2 (3.10)a b = 0.764 + 1.55 d2
2
Pcr =
1.231 in.
x
Ix = 2(55.4) = 110.8 in.4
2
y¿
Pcr
245
=
= 39.5 ksi 6 sg
A
2(3.10)
Therefore, Euler’s formula is valid.
1046
y¿
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13–15. An A-36-steel W8 * 24 column is fixed at one end
and free at its other end. If it is subjected to an axial load
of 20 kip, determine the maximum allowable length of the
column if F.S. = 2 against buckling is desired.
Section Properties. From the table listed in the appendix, the cross-sectional area
and moment of inertia about the y axis for a W8 * 24 are
A = 7.08 in2
Iy = 18.3 in4
Critical Buckling Load. The critical buckling load is
Pcr = Pallow (F.S) = 20(2) = 40 kip
Applying Euler’s formula,
p2 EIy
Pcr =
40 =
(KL)2
p2 C 29 A 103 B D (18.3)
(2L)2
L = 180.93 in = 15.08 ft = 15.1 ft
Ans.
Critical Stress. Euler’s formula is valid only if scr 6 sY.
scr =
Pcr
40
=
= 5.65 ksi 6 sY = 36 ksi
A
7.08
O.K.
*13–16. An A-36-steel W8 * 24 column is fixed at one
end and pinned at the other end. If it is subjected to an axial
load of 60 kip, determine the maximum allowable length of
the column if F.S. = 2 against buckling is desired.
Section Properties. From the table listed in the appendix, the cross-sectional area
and moment of inertia about the y axis for a W8 * 24 are
A = 7.08 in2
Iy = 18.3 in4
Critical Buckling Load. The critical buckling load is
Pcr = Pallow (F.S.) = 60(2) = 120 kip
Applying Euler’s formula,
Pcr =
120 =
p2EIy
(KL)2
p2 C 24 A 103 B D (18.3)
(0.7L)2
L = 298.46 in = 24.87 ft = 24.9 ft
Ans.
Critical Stress. Euler’s formula is valid only if scr 6 sY.
scr =
Pcr
120
=
= 16.95 ksi 6 sY = 36 ksi
A
7.08
O.K.
1047
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•13–17.
The 10-ft wooden rectangular column has the
dimensions shown. Determine the critical load if the ends
are assumed to be pin connected. Ew = 1.611032 ksi,
sY = 5 ksi.
Section Properties:
10 ft
A = 4(2) = 8.00 in2
4 in.
Ix =
1
(2) A 43 B = 10.667 in4
12
Iy =
1
(4) A 23 B = 2.6667 in4 (Controls !)
12
2 in.
Critical Buckling Load: K = 1 for pin supported ends column. Applying Euler’s
formula,.
Pcr =
p2EI
(KL)2
p2(1.6)(103)(2.6667)
=
[1(10)(12)]2
Ans.
= 2.924 kip = 2.92 kip
Critical Stress: Euler’s formula is only valid if scr 6 sg.
scr =
Pcr
2.924
=
= 0.3655 ksi 6 sg = 5 ksi
A
8.00
O.K.
13–18. The 10-ft column has the dimensions shown.
Determine the critical load if the bottom is fixed and the
top is pinned. Ew = 1.611032 ksi, sY = 5 ksi.
Section Properties:
A = 4(2) = 8.00 in2
10 ft
1
(2) A 43 B = 10.667 in4
Ix =
12
4 in.
2 in.
1
Iy =
(4) A 23 B = 2.6667 in4 (Controls!)
12
Critical Buckling Load: K = 0.7 for column with one end fixed and the other end
pinned. Applying Euler’s formula.
Pcr =
p2EI
(KL)2
p2 (1.6)(103)(2.6667)
=
[0.7(10)(12)]2
Ans.
= 5.968 kip = 5.97 kip
Critical Stress: Euler’s formula is only valid if scr 6 sg.
scr =
Pcr
5.968
=
= 0.7460 ksi 6 sg = 5 ksi
A
8.00
O.K.
1048
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13–19. Determine the maximum force P that can be
applied to the handle so that the A-36 steel control rod BC
does not buckle. The rod has a diameter of 25 mm.
P
350 mm
A
250 mm
45⬚
Support Reactions:
a + ©MA = 0;
P(0.35) - FBC sin 45°(0.25) = 0
FBC = 1.9799P
Section Properties:
A =
p
A 0.0252 B = 0.15625 A 10 - 3 B p m2
4
I =
p
A 0.01254 B = 19.17476 A 10 - 9 B m4
4
Critical Buckling Load: K = 1 for a column with both ends pinned. Appyling
Euler’s formula,
Pcr = FBC =
1.9799P =
p2EI
(KLBC)2
p2(200)(109) C 19.17476(10 - 9) D
[1(0.8)]2
P = 29 870 N = 29.9 kN
Ans.
Critical Stress: Euler’s formula is only valid if scr 6 sg.
scr =
1.9799(29 870)
Pcr
= 120.5 MPa 6 sg = 250 MPa
=
A
0.15625(10 - 3)p
1049
O.K.
C
B
800 mm
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*13–20. The W10 * 45 is made of A-36 steel and is used
as a column that has a length of 15 ft. If its ends are assumed
pin supported, and it is subjected to an axial load of 100 kip,
determine the factor of safety with respect to buckling.
P
Critical Buckling Load: Iy = 53.4 in4 for a W10 * 45 wide flange section and
K = 1 for pin supported ends column. Applying Euler’s formula,
Pcr =
15 ft
p2EI
(KL)2
p2 (29)(103)(53.4)
=
[1(15)(12)]2
P
= 471.73 kip
Critical Stress: Euler’s formula is only valid if scr 6 sg. A = 13.3 in2 for the
W10 * 45 wide-flange section.
scr =
Pcr
471.73
=
= 35.47 ksi 6 sg = 36 ksi
A
13.3
O.K.
Pcr
471.73
=
= 4.72
P
100
Ans.
Factor of Safety:
F.S =
The W10 * 45 is made of A-36 steel and is used
as a column that has a length of 15 ft. If the ends of the
column are fixed supported, can the column support the
critical load without yielding?
•13–21.
P
Critical Buckling Load: Iy = 53.4 in4 for W10 * 45 wide flange section and
K = 0.5 for fixed ends support column. Applying Euler’s formula,
Pcr =
15 ft
p2EI
(KL)2
p2 (29)(103)(53.4)
=
[0.5(15)(12)]2
P
= 1886.92 kip
Critical Stress: Euler’s formula is only valid if scr 6 sg. A = 13.3 in2 for W10 * 45
wide flange section.
scr =
Pcr
1886.92
=
= 141.87 ksi 7 sg = 36 ksi (No!)
A
13.3
Ans.
The column will yield before the axial force achieves the critical load Pcr and so
Euler’s formula is not valid.
1050
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13–22. The W12 * 87 structural A-36 steel column has a
length of 12 ft. If its bottom end is fixed supported while
its top is free, and it is subjected to an axial load of
P = 380 kip, determine the factor of safety with respect to
buckling.
W 12 * 87
A = 25.6 in2
Ix = 740 in4
P
Iy = 241 in4 (controls)
12 ft
K = 2.0
Pcr =
p2(29)(103)(241)
p2EI
=
= 831.63 kip
2
(KL)
[(2.0)(12)(12)]2
F.S. =
Pcr
831.63
=
= 2.19
P
380
Ans.
Check:
scr =
=
Pcr
A
831.63
= 32.5 ksi 6 sg
25.6
O.K.
13–23. The W12 * 87 structural A-36 steel column has a
length of 12 ft. If its bottom end is fixed supported while its
top is free, determine the largest axial load it can support.
Use a factor of safety with respect to buckling of 1.75.
W 12 * 87
A = 25.6 in2
Ix = 740 in4
P
Iy = 241 in4
(controls)
K = 2.0
12 ft
p2(29)(103)(241)
p2EI
=
= 831.63 kip
Pcr =
2
(KL)
(2.0(12)(12))2
P =
Pcr
831.63
=
= 475 ksi
F.S
1.75
Ans.
Check:
scr =
P
831.63
=
= 32.5 ksi 6 sg
A
25.6
O.K.
1051
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*13–24. An L-2 tool steel link in a forging machine is pin
connected to the forks at its ends as shown. Determine the
maximum load P it can carry without buckling. Use a factor
of safety with respect to buckling of F.S. = 1.75. Note from
the figure on the left that the ends are pinned for buckling,
whereas from the figure on the right the ends are fixed.
P
P
1.5 in.
0.5 in.
24 in.
Section Properties:
A = 1.5(0.5) = 0.750 in2
Ix =
1
(0.5) A 1.53 B = 0.140625 in4
12
Iy =
1
(1.5) A 0.53 B = 0.015625 in4
12
P
Critical Buckling Load: With respect to the x - x axis, K = 1 (column with both
ends pinned). Applying Euler’s formula,
Pcr =
p2EI
(KL)2
p2(29.0)(103)(0.140625)
=
[1(24)]2
= 69.88 kip
With respect to the y - y axis, K = 0.5 (column with both ends fixed).
Pcr =
p2EI
(KL)2
p2(29.0)(103)(0.015625)
=
[0.5(24)]2
= 31.06 kip
(Controls!)
Critical Stress: Euler’s formula is only valid if scr 6 sg.
scr =
Pcr
31.06
=
= 41.41 ksi 6 sg = 102 ksi
A
0.75
O.K.
Factor of Safety:
F.S =
Pcr
P
1.75 =
31.06
P
P = 17.7 kip
Ans.
1052
P
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
The W14 * 30 is used as a structural A-36 steel
column that can be assumed pinned at both of its ends.
Determine the largest axial force P that can be applied
without causing it to buckle.
•13–25.
P
From the table in appendix, the cross-sectional area and the moment of inertia
about weak axis (y-axis) for W14 * 30 are
A = 8.85 in2
Iy = 19.6 in4
25 ft
Critical Buckling Load: Since the column is pinned at its base and top, K = 1. For
A36 steel, E = 29.0(103) ksi and sg = 36 ksi. Here, the buckling occurs about the
weak axis (y-axis).
P = Pcr =
p2EIy
=
(KL)2
p2 C 29.0(103) D (19.6)
C 1(25)(12) D 2
Ans.
= 62.33 kip = 62.3 kip
Euler’s formula is valid only if scr 6 sg.
scr =
Pcr
62.33
=
= 7.04 ksi 6 sg = 36 ksi
A
8.85
O.K.
13–26. The A-36 steel bar AB has a square cross section.
If it is pin connected at its ends, determine the maximum
allowable load P that can be applied to the frame. Use a
factor of safety with respect to buckling of 2.
a + ©MA = 0;
C
FBC sin 30°(10) - P(10) = 0
FBC = 2 P
+
: ©Fx = 0;
A
1.5 in.
30⬚
B
1.5 in.
FA - 2P cos 30° = 0
1.5 in.
10 ft
FA = 1.732 P
P
Buckling load:
Pcr = FA(F.S.) = 1.732 P(2) = 3.464 P
L = 10(12) = 120 in.
I =
1
(1.5)(1.5)3 = 0.421875 in4
12
Pcr =
p2 EI
(KL)2
3.464 P =
p2 (29)(103)(0.421875)
[(1.0)(120)]2
P = 2.42 kip
Ans.
Pcr = FA(F.S.) = 1.732(2.42)(2) = 8.38 kip
Check:
scr =
Pcr
8.38
=
= 3.72 ksi 6 sg
A
1.5 (1.5)
O.K.
1053
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13–27. Determine the maximum allowable intensity w of
the distributed load that can be applied to member BC
without causing member AB to buckle. Assume that AB is
made of steel and is pinned at its ends for x–x axis buckling
and fixed at its ends for y–y axis buckling. Use a factor
of safety with respect to buckling of 3. Est = 200 GPa,
sY = 360 MPa.
w
C
1.5 m
B
0.5 m
2m
30 mm
x
Ix =
1
(0.02)(0.033) = 45.0(10 - 9)m4
12
Iy =
1
(0.03)(0.023) = 20(10 - 9) m4
12
x
x-x axis:
Pcr = FAB (F.S.) = 1.333w(3) = 4.0 w
K = 1.0,
Pcr =
L = 2m
p2EI
(KL)2
4.0w =
p2(200)(109)(45.0)(10 - 9)
[(1.0)(2)]2
w = 5552 N>m = 5.55 kN>m
Ans.
(controls)
y-y axis
K = 0.5,
4.0w =
L = 2m
p2 (200)(109)(20)(10 - 9)
[(0.5)(2)]2
w = 9870 N>m = 9.87 kN>m
Check:
scr =
20 mm
y
y
Moment of inertia:
4(5552)
Pcr
=
= 37.0 MPa 6 sg
A
(0.02)(0.03)
O.K.
1054
30 mm
A
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*13–28. Determine if the frame can support a load of
w = 6 kN>m if the factor of safety with respect to buckling
of member AB is 3. Assume that AB is made of steel and is
pinned at its ends for x–x axis buckling and fixed at its ends
for y–y axis buckling. Est = 200 GPa, sY = 360 MPa.
w
C
B
1.5 m
0.5 m
Check x-x axis buckling:
Ix =
2m
1
(0.02)(0.03)3 = 45.0(10 - 9) m4
12
K = 1.0
30 mm
x
20 mm
y
y
L = 2m
p2(200)(109)(45.0)(10 - 9)
p2EI
=
Pcr =
2
(KL)
((1.0)(2))2
x
A
30 mm
Pcr = 22.2 kN
a + ©MC = 0;
FAB(1.5) - 6(2)(1) = 0
FAB = 8 kN
Preq’d = 8(3) = 24 kN 7 22.2 kN
No, AB will fail.
Ans.
The beam supports the load of P = 6 kip. As a
result, the A-36 steel member BC is subjected to a
compressive load. Due to the forked ends on the member,
consider the supports at B and C to act as pins for x–x axis
buckling and as fixed supports for y–y axis buckling.
Determine the factor of safety with respect to buckling
about each of these axes.
•13–29.
P
4 ft
A
3 ft
x-x axis buckling:
2
3
1
3
p (29)(10 )(12)(1)(3)
p2EI
=
= 178.9 kip
(KL)2
(1.0(5)(12))2
F.S. =
178.9
= 8.94
20
Ans.
y-y axis buckling:
2
3
1
3 in.
y
FBC = 20 kip
Pcr =
3
Pcr =
p (29)(10 )(12)(3)(1)
p2EI
=
= 79.51
(KL)2
(0.5(5)(12))2
F.S. =
79.51
= 3.98
20
B
C x
3
FBC a b(4) - 6000(8) = 0
5
a + ©MA = 0;
4 ft
Ans.
1055
1 in.
x
y
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13–30. Determine the greatest load P the frame will
support without causing the A-36 steel member BC to
buckle. Due to the forked ends on the member, consider the
supports at B and C to act as pins for x–x axis buckling and
as fixed supports for y–y axis buckling.
P
4 ft
A
3 ft
3
FBC a b(4) - P(8) = 0
5
a + ©MA = 0;
4 ft
B
y
3 in.
C x
1 in.
FBC = 3.33 P
y
x-x axis buckling:
Pcr =
x
1
)(1)(3)3
p2(29)(103)(12
p2EI
=
= 178.9 kip
(KL)2
(1.0(5)(12))2
y -y axis buckling:
Pcr =
1
)(3)(1)3
p2(29)(103)(12
p2EI
=
= 79.51 kip
(KL)2
(0.5(5)(12))2
Thus,
3.33 P = 79.51
P = 23.9 kip
Ans.
13–31. Determine the maximum distributed load that can
be applied to the bar so that the A-36 steel strut AB does
not buckle. The strut has a diameter of 2 in. It is pin
connected at its ends.
w
C
A
2 ft
The compressive force developed in member AB can be determined by writing the
moment equation of equilibrium about C.
a + ©MC = 0;
FAB(2) - w(2)(3) = 0
A = p(12) = p in2
I =
FAB = 3w
4 ft
p 4
p
(1 ) = in4
4
4
Since member AB is pinned at both ends, K = 1. For A36 steel, E = 29.0(103) ksi
and sg = 36 ksi.
Pcr =
p EI
;
(KL)2
p C 29.0(10 ) D (p>4)
2
2
3w =
3
C 1(4)(12) D 2
Ans.
w = 32.52 kip>ft = 32.5 kip>ft
The Euler’s formula is valid only if scr 6 sg.
scr =
3(32.52)
Pcr
=
= 31.06 ksi 6 sg = 36 ksi
p
A
O.K.
1056
B
2 ft
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*13–32. The members of the truss are assumed to be pin
connected. If member AC is an A-36 steel rod of 2 in.
diameter, determine the maximum load P that can be
supported by the truss without causing the member to buckle.
P
C
B
4 ft
D
A
3 ft
Section the truss through a-a, the FBD of the top cut segment is shown in Fig. a. The
compressive force developed in member AC can be determined directly by writing
the force equation of equilibrium along x axis.
+
: ©Fx = 0;
3
FAC a b - P = 0
5
A = p(12) = p in2
I =
FAC =
5
P (C)
3
p 4
p
(1 ) = in4
4
4
Since both ends of member AC are pinned, K = 1. For A-36 steel, E = 29.0(103) ksi
and sg = 36 ksi. The length of member AC is LAC = 232 + 42 = 5 ft.
Pcr =
p2EI
;
(KL)2
p C 29.0(10 ) D (p>4)
5
P =
3
C 1(5)(12) D 2
2
3
P = 37.47 kip = 37.5 kip
Ans.
Euler’s formula is valid only if scr 6 sg.
5
(37.47)
Pcr
3
scr =
=
= 19.88 ksi 6 sg = 36 ksi
p
A
O.K.
1057
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
•13–33.
The steel bar AB of the frame is assumed to be pin
connected at its ends for y–y axis buckling. If w = 3 kN>m,
determine the factor of safety with respect to buckling about
the y–y axis due to the applied loading. Est = 200 GPa,
sY = 360 MPa.
6m
w
B
C
40 mm
40 mm
3m
40 mm
y
x
A
4m
The force with reference to the FBD shown in Fig. a.
a + ©MC = 0;
3
3(6)(3) - FAB a b(6) = 0
5
A = 0.04(0.08) = 3.2(10 - 3) m2
Iy =
FAB = 15 kN
1
(0.08)(0.043) = 0.4267(10 - 6)m4
12
The length of member AB is L = 232 + 42 = 5m. Here, buckling will occur about
the weak axis, (y-axis). Since both ends of the member are pinned, Ky = 1.
Pcr =
p2EIy
=
(KyLy)2
p2 C 200(109) D C 0.4267(10 - 6) D
C 1.0(5) D 2
= 33.69 kN
Euler’s formula is valid only if scr 6 sg.
scr =
33.69(103)
Pcr
= 10.53(106)Pa = 10.53 MPa 6 sg = 360 MPa
=
A
3.2(10 - 3)
O.K.
Thus, the factor of safety against buckling is
F.S =
Pcr
33.69
=
= 2.25
FAB
15
Ans.
1058
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13–34. The members of the truss are assumed to be pin
connected. If member AB is an A-36 steel rod of 40 mm
diameter, determine the maximum force P that can be
supported by the truss without causing the member to buckle.
2m
C
E
D
1.5 m
B
A
2m
P
By inspecting the equilibrium of joint E, FAB = 0. Then, the compressive force
developed in member AB can be determined by analysing the equilibrium of joint
A, Fig. a.
+ c ©Fy = 0;
3
FAC a b - P = 0
5
+
: ©Fx = 0;
5
4
P a b - FAB = 0
3
5
A = p(0.022) = 0.4(10 - 3)p m2
I =
FAC =
5
P (T)
3
FAB =
4
P(c)
3
p
(0.024) = 40(10 - 9) p m4
4
Since both ends of member AB are pinned, K = 1. For A36 steel, E = 200 GPa and
sg = 250 MPa.
Pcr =
p2EI
;
(KL)2
p2 C 200(109) D C 40(10 - 9)p D
4
P =
3
C 1(2) D 2
P = 46.51(103) N = 46.5 kN
Ans.
The Euler’s formula is valid only if scr 6 sg.
4
(46.51)(103)
Pcr
3
= 49.35(106) Pa = 49.35 MPa 6 sg = 250 MPa O.K.
scr =
=
A
0.4(10 - 3)p
1059
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13–35. The members of the truss are assumed to be pin
connected. If member CB is an A-36 steel rod of 40 mm
diameter, determine the maximum load P that can be
supported by the truss without causing the member to buckle.
2m
C
E
D
1.5 m
B
A
2m
P
Section the truss through a–a, the FBD of the left cut segment is shown in Fig. a. The
compressive force developed in member CB can be obtained directly by writing the
force equation of equilibrium along y axis.
+ c ©Fy = 0;
FCB - P = 0
A = p(0.022) = 0.4(10 - 3)p m2
FCB = P (C)
I =
p
(0.024) = 40(10 - 9)p m4
4
Since both ends of member CB are pinned, K = 1. For A36 steel, E = 200 GPa and
sg = 250 MPa.
Pcr =
p2EI
;
(KL)2
P =
p2 C 200(109) D C 40(10 - 9)p D
C 1(1.5) D 2
= 110.24(103) N = 110 kN
Ans.
The Euler’s formula is valid only if scr 6 sg.
scr =
110.24(103)
Pcr
= 87.73(106) Pa = 87.73 MPa 6 sg = 250 MPa
=
A
0.4(10 - 3)p
1060
O.K.
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–36. If load C has a mass of 500 kg, determine the
required minimum diameter of the solid L2-steel rod AB
to the nearest mm so that it will not buckle. Use F.S. = 2
against buckling.
A
45°
4m
D
Equilibriun. The compressive force developed in rod AB can be determined by
analyzing the equilibrium of joint A, Fig. a.
©Fy¿ = 0; FAB sin 15° - 500(9.81) cos 45° = 0
FAB = 13 400.71 N
Section Properties. The cross-sectional area and moment of inertia of the solid
rod are
A =
p 2
d
4
I =
p d 4
p 4
a b =
d
4 2
64
Critical Buckling Load. Since the rod is pinned at both of its ends, K = 1. Here,
Pcr = FAB (F.S.) = 13400.71(2) = 26801.42 N. Applying Euler’s formula,
Pcr =
p2EIy
(KL)2
26801.42 =
p2 C 200 A 109 B D c
p 4
d d
64
[1(4)]2
d = 0.04587 m = 45.87 mm
Use d = 46 mm
Ans.
Critical Stress. Euler’s formula is valid only if scr 6 sY.
scr =
Pcr
26801.42
=
= 16.13 MPa 6 sY = 703 MPa
p
A
2
A 0.046 B
4
1061
O.K.
60°
B
C
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•13–37. If the diameter of the solid L2-steel rod AB is
50 mm, determine the maximum mass C that the rod can
support without buckling. Use F.S. = 2 against buckling.
A
45°
4m
D
Equilibrium. The compressive force developed in rod AB can be determined by
analyzing the equilibrium of joint A, Fig. a.
©Fy¿ = 0; FAB sin 15° - m(9.81) cos 45° = 0
FAB = 26.8014m
B
Section Properties. The cross-sectional area and moment of inertia of the rod are
A =
p
A 0.052 B = 0.625 A 10 - 3 B pm2
4
I =
p
A 0.0254 B = 97.65625 A 10 - 9 B pm4
4
Critical Buckling Load. Since the rod is pinned at both of its ends, K = 1. Here,
Pcr = FAB (F.S.) = 26.8014m(2) = 53.6028m. Applying Euler’s formula,
Pcr =
p2EIy
(KL)2
53.6028m =
p2 c200 A 109 B d c97.65625 A 10 - 9 B p d
[1(4)]2
m = 706.11 kg = 7.06 kg
Ans.
Critical Stress. Euler’s formula is valid only if scr 6 sY.
scr =
53.6028(706.11)
Pcr
=
= 19.28 MPa 6 sY = 703 MPa
A
p 0.625 A 10 - 3 B
1062
60°
O.K.
C
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13–38. The members of the truss are assumed to be pin
connected. If member GF is an A-36 steel rod having a
diameter of 2 in., determine the greatest magnitude of load
P that can be supported by the truss without causing this
member to buckle.
H
G
F
B
C
12 ft
A
16 ft
16 ft
P
Support Reactions: As shown on FBD(a).
Member Forces: Use the method of sections [FBD(b)].
+ ©MB = 0;
FGF (12) - P(16) = 0
FGF = 1.3333P (C)
Section Properties:
A =
p 2
A 2 B = p in2
4
I =
p 4
A 1 B = 0.250p in4
4
Critical Buckling Load: K = 1 for a column with both ends pinned. Applying
Euler’s formula,
Pcr = FGF =
1.3333P =
p2EI
(KLGF)2
p2 (29)(103)(0.250p)
[1(16)(12)]2
P = 4.573 kip = 4.57 kip
Ans.
Critical Stress: Euler’s formula is only valid if scr 6 sg.
scr =
E
1.3333(4.573)
Pcr
=
= 1.94 ksi 6 sg = 36 ksi
p
A
1063
O.K.
D
16 ft
P
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Page 1064
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13–39. The members of the truss are assumed to be pin
connected. If member AG is an A-36 steel rod having a
diameter of 2 in., determine the greatest magnitude of load
P that can be supported by the truss without causing this
member to buckle.
H
G
F
B
C
12 ft
A
16 ft
16 ft
P
Support Reactions: As shown on FBD(a).
Member Forces: Use the method of joints [FBD(b)].
P -
+ c ©Fy = 0;
3
= 0
F
5 AG
FAG = 1.6667P (C)
Section Properties:
LAG = 2162 + 122 = 20.0 ft
A =
p 2
A 2 B = p in2
4
I =
p 4
A 1 B = 0.250p in4
4
Critical Buckling Load: K = 1 for a column with both ends pinned. Applying
Euler’s formula,
Pcr = FGF =
1.6667P =
p2EI
(KLGF)2
p2 (29)(103)(0.250p)
[1(20)(12)]2
P = 2.342 kip = 2.34 kip
Ans.
Critical Stress: Euler’s formula is only valid if scr = sg.
scr =
E
1.6667(2.342)
Pcr
=
= 1.24 ksi 6 sg = 36 ksi
p
A
1064
O.K.
D
16 ft
P
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Page 1065
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*13–40. The column is supported at B by a support that
does not permit rotation but allows vertical deflection.
Determine the critical load Pcr . EI is constant.
L
B
Pcr
A
Elastic curve:
EI
d2y
= M = -P y
dx2
P
d2y
+
y = 0
EI
dx2
y = C1 sin c
P
P
x d + C2 cos c
xd
A EI
A EI
Boundry conditions:
At x = 0;
0 = 0 + C2;
At x = L;
y = 0
C2 = 0
dv
= 0
dx
P
P
dv
= C1
cos c
L] d = 0;
dx
A EI
A EI
cos c
P
L d = 0;
A EI
For n = 1 ;
Pcr =
C1
P
p
L = na b
A EI
2
P
Z 0
A EI
n = 1, 3, 5
p2
P
=
EI
4L2
p2EI
4L2
Ans.
1065
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Page 1066
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w
The ideal column has a weight w (force兾length)
and rests in the horizontal position when it is subjected to the
axial load P. Determine the maximum moment in the column
at midspan. EI is constant. Hint: Establish the differential
equation for deflection, Eq. 13–1, with the origin at the mid
span. The general solution is v = C1 sin kx + C2 cos kx +
1w>12P22x2 - 1wL>12P22x - 1wEI>P22 where k2 = P>EI.
•13–41.
P
L
Moment Functions: FBD(b).
a + ©Mo = 0;
M(x) =
wL
x
wx a b - M(x) - a
bx - Pv = 0
2
2
w 2
A x - Lx B - Pv
2
[1]
Differential Equation of The Elastic Curve:
EI
d2y
= M(x)
dx2
EI
d2y
w 2
=
A x - Lx B - Py
2
dx2
w
d2y
P
y =
+
A x2 - Lx B
EI
2EI
dx2
The solution of the above differential equation is of the form
v = C1 sin a
P
P
w 2
wL
wEI
xb + C2 cos ¢
xb +
x x A EI
A EI
2P
2P
P2
[2]
dv
P
P
P
P
w
wL
= C1
cos ¢
x ≤ - C2
sin ¢
x≤ +
xdx
A EI
A EI
A EI
A EI
P
2P
[3]
and
The integration constants can be determined from the boundary conditions.
Boundary Condition:
At x = 0, y = 0. From Eq. [2],
0 = C2 -
wEI
P2
C2 =
wEI
P2
At x =
L dy
= 0. From Eq.[3],
,
2 dx
0 = C1
P
wEI
P
w L
P L
P L
wL
cos ¢
sin ¢
≤ ≤ + a b A EI
A EI 2
A EI 2
P 2
2P
P2 A EI
C1 =
wEI
P L
tan ¢
≤
A EI 2
P2
1066
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Page 1067
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13–41.
Continued
Elastic Curve:
y =
w EI
P L
P
EI
P
x2
L
EI
x≤ +
x≤ +
tan ¢
cos ¢
- x B
≤ sin ¢
R
P P
A EI 2
A EI
P
A EI
2
2
P
However, y = ymax at x =
ymax =
=
L
. Then,
2
P L
P L
P L
EI
w EI
EI
L2
tan ¢
cos ¢
B
≤ sin ¢
≤ +
≤ R
P P
A EI 2
A EI 2
P
A EI 2
8
P
wEI
P L
PL2
sec
- 1R
B
¢
≤
A EI 2
8EI
P2
Maximum Moment: The maximum moment occurs at x =
Mmax =
L
. From, Eq.[1],
2
w L2
L
- L a b R - Pymax
B
2 4
2
= -
wL2
wEI
P L
PL2
- P b 2 B sec ¢
- 1R r
≤ 8
A EI 2
8EI
P
= -
PL
wEI
B sec ¢
≤ - 1R
P
A EI 2
Ans.
13–42. The ideal column is subjected to the force F at its
midpoint and the axial load P. Determine the maximum
moment in the column at midspan. EI is constant. Hint:
Establish the differential equation for deflection, Eq. 13–1.
The general solution is v = C1 sin kx + C2 cos kx - c2x>k2,
where c2 = F>2EI, k2 = P>EI.
F
P
L
2
Moment Functions: FBD(b).
a + ©Mo = 0;
M(x) +
F
x + P(v) = 0
2
M(x) = -
F
x - Pv
2
[1]
Differential Equation of The Elastic Curve:
EI
d2y
= M(x)
dx2
EI
F
d2y
= - x - Py
2
2
dx
d2y
F
P
y = x
+
EI
2EI
dx2
The solution of the above differential equation is of the form,
v = C1 sin a
P
P
F
xb + C2 cos ¢
xb x
A EI
A EI
2P
1067
[2]
L
2
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13–42.
Continued
and
dv
P
P
P
P
F
= C1
cos ¢
x ≤ - C2
sin ¢
x≤ dx
A EI
A EI
A EI
A EI
2P
[3]
The integration constants can be determined from the boundary conditions.
Boundary Conditions:
At x = 0, y = 0. From Eq.[2], C2 = 0
At x =
L dy
= 0. From Eq.[3],
,
2 dx
0 = C1
P L
P
F
cos ¢
≤ A EI
A EI 2
2P
C1 =
F
EI
P L
sec ¢
≤
2P A P
A EI 2
Elastic Curve:
y =
F
EI
P L
P
F
sec ¢
x≤ x
≤ sin ¢
2P A P
A EI 2
A EI
2P
=
F
EI
P L
P
sec ¢
x≤ - xR
B
≤ sin ¢
2P A P
A EI 2
A EI
However, y = ymax at x =
ymax =
=
L
. Then,
2
F
EI
P L
P L
L
sec ¢
B
≤ sin ¢
≤ - R
2P A P
A EI 2
A EI 2
2
F
EI
P L
L
tan ¢
B
≤ - R
2P A P
A EI 2
2
Maximum Moment: The maximum moment occurs at x =
Mmax = -
L
. From Eq.[1],
2
F L
a b - Pymax
2 2
= -
FL
F
EI
P L
L
- Pb
tan ¢
B
≤ - Rr
4
2P A P
A EI 2
2
= -
P L
F EI
tan ¢
≤
2 AP
A EI 2
Ans.
1068
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Page 1069
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13–43. The column with constant EI has the end
constraints shown. Determine the critical load for the
column.
P
L
Moment Function. Referring to the free-body diagram of the upper part of the
deflected column, Fig. a,
a + ©MO = 0;
M + Pv = 0
M = -Pv
Differential Equation of the Elastic Curve.
EI
d2v
= M
dx2
EI
d2v
= -Pv
dx2
d2v
P
+
v = 0
2
EI
dx
The solution is in the form of
v = C1 sin a
P
P
xb + C2 cos ¢
xb
A EI
A EI
(1)
dv
P
P
P
P
= C1
cos ¢
x ≤ - C2
sin ¢
x≤
dx
A EI
A EI
A EI
A EI
(2)
Boundary Conditions. At x = 0, v = 0. Then Eq. (1) gives
0 = 0 + C2
At x = L,
C2 = 0
dv
= 0. Then Eq. (2) gives
dx
0 = C1
P
P
cos ¢
L≤
A EI
A EI
C1 = 0 is the trivial solution, where v = 0. This means that the column will remain
straight and buckling will not occur regardless of the load P. Another possible
solution is
cos ¢
P
L≤ = 0
A EI
P
np
L =
A EI
2
n = 1, 3, 5
The smallest critical load occurs when n = 1, then
p
Pcr
L =
A EI
2
Pcr =
p2EI
4L2
Ans.
1069
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Page 1070
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*13–44. Consider an ideal column as in Fig. 13–10c, having
both ends fixed. Show that the critical load on the column
is given by Pcr = 4p2EI>L2. Hint: Due to the vertical
deflection of the top of the column, a constant moment
M¿ will be developed at the supports. Show that
d2v>dx2 + 1P>EI2v = M¿>EI. The solution is of the form
v = C1 sin11P>EIx2 + C2 cos11P>EIx2 + M¿>P.
Moment Functions:
M(x) = M¿ - Py
Differential Equation of The Elastic Curve:
EI
d2y
= M(x)
dx2
d2y
= M¿ - Py
dx2
EI
M¿
d2y
P
+
y =
2
EI
EI
dx
(Q.E.D.)
The solution of the above differential equation is of the form
v = C1 sin a
P
P
M¿
xb + C2 cos ¢
xb +
A EI
P
A EI
[1]
and
dv
P
P
P
P
= C1
cos ¢
x ≤ - C2
sin ¢
x≤
dx
A EI
A EI
A EI
A EI
[2]
The integration constants can be determined from the boundary conditions.
Boundary Conditions:
At x = 0, y = 0. From Eq.[1], C2 = At x = 0,
M¿
P
dy
= 0. From Eq.[2], C1 = 0
dx
Elastic Curve:
y =
M¿
P
x≤ R
B 1 - cos ¢
P
A EI
and
M¿
P
P
dy
=
sin ¢
x≤
dx
P A EI
A EI
However, due to symmetry
sin B
L
dy
= 0 at x = . Then,
dx
2
P L
a bR = 0
A EI 2
or
P L
a b = np
A EI 2
where n = 1, 2, 3,...
The smallest critical load occurs when n = 1.
Pce =
4p2EI
L2
(Q.E.D.)
1070
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Page 1071
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•13–45. Consider an ideal column as in Fig. 13–10d, having
one end fixed and the other pinned. Show that the critical load
on the column is given by Pcr = 20.19EI>L2. Hint: Due to the
vertical deflection at the top of the column,a constant moment
M¿ will be developed at the fixed support and horizontal
reactive forces R¿ will be developed at both supports. Show
that d2v>dx2 + 1P>EI2v = 1R¿>EI21L - x2. The solution
is of the form v = C1 sin 11P>EIx2 + C2 cos 11P>EIx2 +
1R¿>P21L - x2. After application of the boundary conditions
show that tan 11P>EIL2 = 1P>EI L. Solve by trial and
error for the smallest nonzero root.
Equilibrium. FBD(a).
Moment Functions: FBD(b).
M(x) = R¿(L - x) - Py
Differential Equation of The Elastic Curve:
EI
d2y
= M(x)
dx2
EI
d2y
= R¿(L - x) - Py
dx2
d2y
P
R¿
+
y =
(L - x)
2
EI
EI
dx
(Q.E.D.)
The solution of the above differential equation is of the form
v = C1 sin a
P
P
R¿
(L - x)
xb + C2 cos ¢
xb +
A EI
P
A EI
[1]
and
dv
P
P
P
P
R¿
= C1
cos ¢
x ≤ - C2
sin ¢
x≤ dx
A EI
A EI
A EI
A EI
P
The integration constants can be determined from the boundary conditions.
Boundary Conditions:
At x = 0, y = 0. From Eq.[1], C2 = -
At x = 0,
R¿L
P
dy
R¿ EI
= 0. From Eq.[2], C1 =
dx
P AP
Elastic Curve:
y =
=
R¿ EI
P
R¿L
P
R¿
sin ¢
x≤ cos ¢
x≤ +
(L - x)
P AP
A EI
P
A EI
P
EI
P
P
R¿
sin ¢
x ≤ - L cos ¢
x ≤ + (Lx) R
B
P AP
A EI
A EI
1071
[2]
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13–45.
Continued
However, y = 0 at x = L. Then,
0 =
P
P
EI
sin ¢
L ≤ - L cos ¢
L≤
A EI
A EI
AP
tan ¢
P
P
L≤ =
L
A EI
A EI
(Q.E.D.)
By trial and error and choosing the smallest root, we have
P
L = 4.49341
A EI
Then,
Pcr =
20.19EI
L2
(Q.E.D.)
13–46. Determine the load P required to cause the A-36
steel W8 * 15 column to fail either by buckling or by
yielding. The column is fixed at its base and free at its top.
1 in.
P
Section properties for W8 * 15:
A = 4.44 in2
Ix = 48.0 in4
rx = 3.29 in.
d = 8.11 in.
Iy = 3.41 in4
8 ft
Buckling about y-y axis:
L = 8(12) = 96 in.
K = 2.0
P = Pcr =
p2EIy
p2(29)(103)(3.41)
=
(KL)2
[(2.0)(96)]2
Check: scr =
= 26.5 kip
(controls)
Pcr
26.5
=
= 5.96 ksi 6 sg
A
4.44
Ans.
O.K.
Check yielding about x-x axis:
smax =
P
ec
KL P
c1 + 2 sec a
bd
A
2r A EA
r
26.5
P
=
= 5.963 ksi
A
4.44
(1) A 2 B
ec
=
= 0.37463
2
r
(3.29)2
8.11
2.0(96)
P
26.5
KL
=
= 0.4184
2r A EA
2(3.29) A 29(103)(4.44)
smax = 5.963[1 + 0.37463 sec (0.4184)] = 8.41 ksi 6 sg = 36 ksi
1072
O.K.
13 Solutions 46060
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Page 1073
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13–47. The hollow red brass C83400 copper alloy shaft is
fixed at one end but free at the other end. Determine the
maximum eccentric force P the shaft can support without
causing it to buckle or yield. Also, find the corresponding
maximum deflection of the shaft.
2m
a
a
P
150 mm
30 mm
Section Properties.
A = p A 0.032 - 0.022 B = 0.5 A 10 - 3 B p m2
I =
20 mm
Section a – a
p
A 0.034 - 0.024 B = 0.1625 A 10 - 6 B p m4
4
0.1625 A 10 B p
I
=
= 0.01803 m
C 0.5 A 10 - 3 B p
AA
-6
r =
e = 0.15 m
c = 0.03 m
For a column that is fixed at one end and free at the other, K = 2. Thus,
KL = 2(2) = 4 m
Yielding. In this case, yielding will occur before buckling. Applying the secant
formula,
smax =
P
ec
KL P
B 1 + 2 sec ¢
≤R
A
2rx A EA
rx
70.0 A 106 B =
70.0 A 106 B =
P
0.5 A 10
-3
Bp
P
0.5 A 10 - 3 B p
D1 +
0.15(0.03)
0.018032
secC
P
4
ST
2(0.01803)A 101 A 109 B C 0.5 A 10 - 3 B p D
a1 + 13.846 sec 8.8078 A 10 - 3 B 2Pb
Solving by trial and error,
P = 5.8697 kN = 5.87 kN
Ans.
Maximum Deflection.
vmax = e B sec ¢
P KL
≤ - 1R
A EI 2
= 0.15 D sec C
5.8697 A 103 B
4
a b S - 1T
C 101 A 109 B C 0.1625 A 10 - 6 B p D 2
= 0.04210 m = 42.1 mm
Ans.
1073
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Page 1074
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*13–48. The hollow red brass C83400 copper alloy shaft is
fixed at one end but free at the other end. If the eccentric
force P = 5 kN is applied to the shaft as shown, determine
the maximum normal stress and the maximum deflection.
2m
a
a
P
150 mm
30 mm
Section Properties.
A = p A 0.032 - 0.022 B = 0.5 A 10 - 3 B p m2
I =
20 mm
Section a – a
p
A 0.034 - 0.024 B = 0.1625 A 10 - 6 B p m4
4
0.1625 A 10 B p
I
= 0.01803 m
=
r =
C 0.5 A 10 - 3 B p
AA
-6
e = 0.15 m
c = 0.03 m
For a column that is fixed at one end and free at the other, K = 2. Thus,
KL = 2(2) = 4 m
Yielding. Applying the secant formula,
smax =
P
ec
KL P
B 1 + 2 sec ¢
≤R
A
2r A EA
r
5 A 103 B
5 A 103 B
4
ST
secC
D1 +
=
2(0.01803) C 101 A 109 B C 0.5 A 10 - 3 B p D
0.018032
0.5 A 10 - 3 B p
0.15(0.03)
Ans.
= 57.44 MPa = 57.4 MPa
Since smax 6 sY = 70 MPa, the shaft does not yield.
Maximum Deflection.
vmax = e B sec ¢
P KL
≤ - 1R
A EI 2
= 0.15 Dsec C
5 A 103 B
4
a b S - 1T
C 101 A 109 B C 0.1625 A 10 - 6 B p D 2
= 0.03467 m = 34.7 mm
Ans.
1074
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•13–49.
The tube is made of copper and has an outer
diameter of 35 mm and a wall thickness of 7 mm. Using a
factor of safety with respect to buckling and yielding of
F.S. = 2.5, determine the allowable eccentric load P. The
tube is pin supported at its ends. Ecu = 120 GPa, sY =
750 MPa.
2m
P
14 mm
Section Properties:
A =
p
(0.0352 - 0.0212) = 0.61575(10 - 3) m2
4
I =
p
(0.01754 - 0.01054) = 64.1152(10 - 9) m4
4
r =
I
64.1152(10 - 9)
=
= 0.010204 m
AA
A 0.61575(10 - 3)
For a column pinned at both ends, K = 1. Then KL = 1(2) = 2 m.
Buckling: Applying Euler’s formula,
Pmax = Pcr =
p2 (120)(109) C 64.1152(10 - 9) D
p2EI
=
= 18983.7 N = 18.98 kN
(KL)2
22
Critical Stress: Euler’s formula is only valid if scr 6 sg.
scr =
Pcr
18983.7
= 30.83 MPa 6 sg = 750 MPa
=
A
0.61575(10 - 3)
O.K.
Yielding: Applying the secant formula,
smax =
(KL) Pmax
Pmax
ec
B 1 + 2 sec ¢
≤R
A
2r A EA
r
750 A 106 B =
0.61575(10 - 3)
750 A 106 B =
0.61575(10 - 3)
Pmax
Pmax
B1 +
0.014(0.0175)
0.0102042
sec ¢
Pmax
2
≤R
2(0.010204)A 120(109)[0.61575(10 - 3)]
A 1 + 2.35294 sec 0.01140062Pmax B
Solving by trial and error,
Pmax = 16 885 N = 16.885 kN (Controls!)
Factor of Safety:
P =
Pmax
16.885
=
= 6.75 kN
F.S.
2.5
Ans.
1075
P
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Page 1076
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–50. The tube is made of copper and has an outer
diameter of 35 mm and a wall thickness of 7 mm. Using a
factor of safety with respect to buckling and yielding of
F.S. = 2.5, determine the allowable eccentric load P that it
can support without failure. The tube is fixed supported at
its ends. Ecu = 120 GPa, sY = 750 MPa.
2m
P
14 mm
Section Properties:
A =
p
A 0.0352 - 0.0212 B = 0.61575 A 10 - 3 B m2
4
I =
p
A 0.01754 - 0.01054 B = 64.1152 A 10 - 9 B m4
4
r =
I
64.1152(10 - 9)
=
= 0.010204 ms
AA
A 0.61575(10 - 3)
For a column fixed at both ends, K = 0.5. Then KL = 0.5(2) = 1 m.
Buckling: Applying Euler’s formula,
Pmax = Pcr =
p2(120)(109) C 64.1152(10 - 9) D
p2EI
=
= 75 935.0 N = 75.93 kN
(KL)2
12
Critical Stress: Euler’s formula is only valid if scr 6 sg.
scr =
Pcr
75 935.0
= 123.3 MPa 6 sg = 750 MPa
=
A
0.61575(10 - 3)
O. K.
Yielding: Applying the secant formula,
smax =
(KL) Pmax
Pmax
ec
B 1 + 2 sec ¢
≤R
A
2r A EA
r
750 A 106 B =
0.61575(10 - 3)
750 A 106 B =
0.61575(10 - 3)
Pmax
Pmax
B1 +
0.014(0.0175)
0.0102042
sec ¢
2
Pmax
≤R
2(0.010204)A 120(109)[0.61575(10 - 3)]
A 1 + 2.35294 sec 5.70032 A 10 - 3 B 2P B
Solving by trial and error,
Pmax = 50 325 N = 50.325 kN (Controls!)
Factor of Safety:
P =
Pmax
50.325
=
= 20.1 kN
F.S.
2.5
Ans.
1076
P
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13–51. The wood column is fixed at its base and can be
assumed pin connected at its top. Determine the maximum
eccentric load P that can be applied without causing the
column to buckle or yield. Ew = 1.811032 ksi, sY = 8 ksi.
P
y
4 in.
x
x
P
y 10 in.
10 ft
Section Properties:
A = 10(4) = 40 in2
ry =
Iy =
1
(4)(103) = 333.33 in4
12
Ix =
1
(10)(43) = 53.33 in4
12
Ix
333.33
=
= 2.8868 in.
AA
A 40
Buckling about x-x axis:
P = Pcr =
p2(1.8)(103)(53.33)
p2EI
=
= 134 kip
(KL)2
[(0.7)(10)(12)]2
Check: scr =
Pcr
134
=
= 3.36 ksi 6 sg
A
40
O.K.
Yielding about y -y axis:
smax =
P
ec
KL
P
a1 + 2 seca
b
b
A
2r A EA
r
5(5)
ec
=
= 3.0
r2
2.88682
a
0.7(10)(12)
P
P
KL
b
=
= 0.0542212P
2r A EA
2(2.8868) A 1.8(103)(40)
8(40) = P[1 + 3.0 sec (0.0542212P)]
By trial and error:
P = 73.5 kip
Ans.
(controls)
1077
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Page 1078
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–52. The wood column is fixed at its base and can
be assumed fixed connected at its top. Determine the
maximum eccentric load P that can be applied without
causing the column to buckle or yield. Ew = 1.811032 ksi,
sY = 8 ksi.
P
y
4 in.
x
x
P
y 10 in.
10 ft
Section Properties:
A = 10(4) = 40 in2
ry =
Iy =
1
(4)(103) = 333.33 in4
12
Ix =
1
(10)(43) = 53.33 in4
12
Iy
333.33
=
= 2.8868 in.
AA
A 40
Buckling about x-x axis:
P = Pcr =
p2(1.8)(103)(53.33)
p2EI
=
= 263 kip
2
(KL)
[(0.5)(10)(12)]2
Check: scr =
Pcr
263
=
= 6.58 ksi 6 sg
A
40
O.K.
Yielding about y-y axis:
smax =
P
ec
KL P
a1 + 2 seca
bb
A
2r A EA
r
5(5)
ec
=
= 3.0
r2
2.88682
a
0.5(10)(12)
P
P
KL
b
=
= 0.0387292P
2r A EA
2(2.8868) A 1.8(103)(40)
8(40) = P[1 + 3.0 sec (0.0387292P)]
By trial and error:
P = 76.5 kip
Ans.
(controls)
1078
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The W200 * 22 A-36-steel column is fixed at its
base. Its top is constrained to rotate about the y–y axis and
free to move along the y–y axis. Also, the column is braced
along the x–x axis at its mid-height. Determine the
allowable eccentric force P that can be applied without
causing the column either to buckle or yield. Use F.S. = 2
against buckling and F.S. = 1.5 against yielding.
•13–53.
100 mm
y
A = 2860 mm2 = 2.86 A 10 - 3 B m2
ry = 22.3 mm = 0.0223 m
Ix = 20.0 A 106 B mm4 = 20.0 A 10 - 6 B m4
c =
bf
2
=
102
= 51 mm = 0.051 m
2
e = 0.1m
Buckling About the Strong Axis. Since the column is fixed at the base and free at
the top, Kx = 2. Applying Euler’s formula,
Pcr =
p2EIx
(KL)x 2
=
p2 c200 A 109 B d c20.0 A 10 - 6 B d
[2(10)]2
= 98.70kN
Euler’s formula is valid if scr 6 sY.
scr =
98.70 A 103 B
Pcr
=
= 34.51 MPa 6 sY = 250MPa
A
2.86 A 10 - 3 B
O.K.
Then,
Pallow =
Pcr
98.70
=
= 49.35 kN
F.S.
2
Yielding About Weak Axis. Since the support provided by the bracing can be
considered a pin connection, the upper portion of the column is pinned at both of its
ends. Then Ky = 1 and L = 5 m. Applying the secant formula,
smax =
A KL B y Pmax
Pmax
ec
C 1 + 2 sec B
RS
A
2ry A EA
ry
250 A 106 B =
250 A 106 B =
Pmax
2.86 A 10
-3
Pmax
B
2.86 A 10 - 3 B
D1 +
0.1(0.051)
0.02232
secC
1(5)
Pmax
ST
2(0.0223)A 200 A 109 B C 2.86 A 10 - 3 B D
c1 + 10.2556 sec 4.6875 A 10 - 3 B 2Pmax d
Solving by trial and error,
Pmax = 39.376 kN
Then,
Pallow =
Pmax
39.376
=
= 26.3 kN (controls)
1.5
1.5
Ans.
1079
5m
y
x
5m
Section Properties. From the table listed in the appendix, the necessary section
properties for a W200 * 22 are
P
x
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Page 1080
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–54. The W200 * 22 A-36-steel column is fixed at its
base. Its top is constrained to rotate about the y–y axis and
free to move along the y–y axis. Also, the column is braced
along the x–x axis at its mid-height. If P = 25 kN,
determine the maximum normal stress developed in the
column.
100 mm
y
Section Properties. From the table listed in the appendix, necessary section
properties for a W200 * 22 are
A = 2860 mm2 = 2.86 A 10 - 3 B m2
ry = 22.3 mm = 0.0223 m
Ix = 20.0 A 106 B mm4 = 20.0 A 10 - 6 B m4
c =
bf
2
=
102
= 51 mm = 0.051 m
2
e = 0.1m
Buckling About the Strong Axis. Since the column is fixed at the base and free at
the top, Kx = 2. Applying Euler’s formula,
Pcr =
p2EIx
(KL)x 2
=
p2 c200 A 109 B d c20.0 A 10 - 6 B d
[2(10)]2
= 98.70kN
Euler’s formula is valid only if scr 6 sY.
scr =
98.70 A 103 B
Pcr
=
= 34.51 MPa 6 sY = 250 MPa
A
2.86 A 10 - 3 B
O.K.
Since P = 25 kN 6 Pcr, the column does not buckle.
Yielding About Weak Axis. Since the support provided by the bracing can be
considered a pin connection, the upper portion of the column is pinned at both of its
ends. Then Ky = 1 and L = 5 m. Applying the secant formula,
smax =
=
(KL) P
P
ec
C 1 + 2 sec B
RS
A
2ry A EA
ry
2.5 A 103 B
2.86 A 10 - 3 B
D1 +
0.1(0.051)
0.02232
secC
25 A 103 B
1(5)
ST
2(0.0223) C 200 A 109 B C 2.86 A 10 - 3 B D
= 130.26 MPa = 130 MPa
Ans.
Since smax 6 sY = 250 MPa, the column does not yield.
1080
y
x
5m
5m
P
x
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Page 1081
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13–55. The wood column is fixed at its base, and its top
can be considered pinned. If the eccentric force P = 10 kN
is applied to the column, investigate whether the column
is adequate to support this loading without buckling or
yielding. Take E = 10 GPa and sY = 15 MPa.
x
P
150 mm
25 mm
yx
25 mm
75 mm
5m
Section Properties.
A = 0.05(0.15) = 7.5 A 10 - 3 B m2
Ix =
rx =
1
(0.05) A 0.153 B = 14.0625 A 10 - 6 B m4
12
14.0625 A 10 - 6 B
Ix
= 0.04330 m
=
AA
C 7.5 A 10 - 3 B
1
(0.15) A 0.053 B = 1.5625 A 10 - 6 B m4
12
e = 0.15 m
c = 0.075 m
Iy =
For a column that is fixed at one end and pinned at the other K = 0.7. Then,
(KL)x = (KL)y = 0.7(5) = 3.5 m
Buckling About the Weak Axis. Applying Euler’s formula,
Pcr =
p2EIy
(KL)y
=
2
p2 C 10 A 109 B D C 1.5625 A 10 - 6 B D
3.52
= 12.59 kN
Euler’s formula is valid if scr 6 sY.
12.59 A 10 B
Pcr
=
= 1.68 MPa 6 sY = 15 MPa
A
7.5 A 10 - 3 B
3
scr =
O.K.
Since Pcr 7 P = 10 kN, the column will not buckle.
Yielding About Strong Axis. Applying the secant formula.
smax =
=
(KL)x P
ec
P
C 1 + 2 sec B
RS
A
2rx A EA
rx
10 A 103 B
7.5 A 10 - 3 B
D1 +
0.15(0.075)
2
0.04330
secC
10 A 103 B
3.5
ST
2(0.04330) C 10 A 109 B C 7.5 A 10 - 3 B D
= 10.29 MPa
Since smax 6 sY = 15 MPa , the column will not yield.
Ans.
1081
75 mm
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Page 1082
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*13–56. The wood column is fixed at its base, and its
top can be considered pinned. Determine the maximum
eccentric force P the column can support without
causing it to either buckle or yield. Take E = 10 GPa
and sY = 15 MPa .
x
P
150 mm
25 mm
yx
25 mm
75 mm
5m
Section Properties.
A = 0.05(0.15) = 7.5 A 10 - 3 B m2
Ix =
1
(0.05) A 0.153 B = 14.0625 A 10 - 6 B m4
12
14.0625 A 10
Ix
=
C 7.5 A 10 - 3 B
AA
-6
rx =
B
= 0.04330 m
1
(0.15) A 0.053 B = 1.5625 A 10 - 6 B m4
12
e = 0.15 m
c = 0.075 m
Iy =
For a column that is fixed at one end and pinned at the other K = 0.7. Then,
(KL)x = (KL)y = 0.7(5) = 3.5 m
Buckling About the Weak Axis. Applying Euler’s formula,
Pcr =
p2EIy
(KL)y
2
=
p2 C 10 A 109 B D C 1.5625 A 10 - 6 B D
3.52
= 12.59 kN = 12.6 kN
Ans.
Euler’s formula is valid if scr 6 sY.
12.59 A 10 B
Pcr
scr =
=
= 1.68 MPa 6 sY = 15 MPa
A
7.5 A 10 - 3 B
3
O.K.
Yielding About Strong Axis. Applying the secant formula with P = Pcr = 12.59 kN,
smax =
=
(KL)x P
P
ec
C B 1 + 2 sec B
RS
A
2rx A EA
rx
12.59 A 103 B
7.5 A 10 - 3 B
D1 +
0.15(0.075)
0.043302
secC
12.59 A 103 B B
3.5
ST
2(0.04330) C 10 A 109 B C 7.5 A 10 - 3 B D
= 13.31 MPa 6 sY = 15 MPa
O.K.
1082
75 mm
13 Solutions 46060
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Page 1083
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The W250 * 28 A-36-steel column is fixed at its
base. Its top is constrained to rotate about the y–y axis and
free to move along the y–y axis. If e = 350 mm, determine
the allowable eccentric force P that can be applied without
causing the column either to buckle or yield. Use F.S. = 2
against buckling and F.S. = 1.5 against yielding.
•13–57.
P e
x
x
y
6m
Section Properties. From the table listed in the appendix, necessary section
properties for a W250 * 28 are
A = 3620 mm2 = 3.62 A 10 - 3 B m2
rx = 105 mm = 0.105 m
Iy = 1.78 A 106 B mm4 = 1.78 A 10 - 6 B m4
c =
260
d
=
= 130 mm = 0.13 m
2
2
e = 0.35 m
Buckling About the Strong Axis. Since the column is fixed at the base and pinned at
the top, Kx = 0.7. Applying Euler’s formula,
Pcr =
p2EIy
(KL)y
=
2
p2 C 200 A 109 B D C 1.78 A 10 - 6 B D
[0.7(6)]2
= 199.18 kN
Euler’s formula is valid only if scr 6 sY.
199.18 A 10 B
Pcr
=
= 55.02 MPa 6 sY = 250 MPa
A
3.62 A 10 - 3 B
3
scr =
O.K.
Thus,
Pallow =
Pcr
199.18
=
= 99.59 kN
F.S.
2
Yielding About Strong Axis. Since the column is fixed at its base and free at its top,
Kx = 2. Applying the secant formula,
smax =
(KL)x Pmax
Pmax
ec
C 1 + 2 sec B
RS
A
2rx A EA
rx
250 A 106 B =
250 A 106 B =
Pmax
3.62 A 10
-3
Pmax
B
3.62 A 10 - 3 B
D1 +
0.35(0.13)
0.1052
secC
2(6)
Pmax
ST
2(0.105)A 200 A 109 B C 3.62 A 10 - 3 B D
A 1 + 4.1270 sec (0.0021237)2Pmax B
Solving by trial and error,
Pmax = 133.45 kN
Then,
Pallow =
y
Pmax
133.45
=
= 88.97 kN = 89.0 kN (controls)
1.5
1.5
Ans.
1083
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Page 1084
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–58. The W250 * 28 A-36-steel column is fixed at its
base. Its top is constrained to rotate about the y–y axis and
free to move along the y–y axis. Determine the force P and
its eccentricity e so that the column will yield and buckle
simultaneously.
P e
x
x
y
6m
Section Properties. From the table listed in the appendix, the necessary section
properties for a W250 * 28 are
A = 3620 mm2 = 3.62 A 10 - 3 B m2
rx = 105 mm = 0.105 m
Iy = 1.78 A 106 B mm4 = 1.78 A 10 - 6 B m4
c =
d
260
=
= 130 mm = 0.13 m
2
2
Buckling About the Weak Axis. Since the column is fixed at the base and pinned at
its top, Kx = 0.7. Applying Euler’s formula,
Pcr =
p2EIy
(KL)y
=
2
p2 C 200 A 109 B D C 1.78 A 10 - 6 B D
[0.7(6)]2
= 199.18 kN = 199 kN
Ans.
Euler’s formula is valid only if scr 6 sY.
199.18 A 10 B
Pcr
=
= 55.02 MPa 6 sY = 250 MPa
A
3.62 A 10 - 3 B
3
scr =
O.K.
Yielding About Strong Axis. Since the column is fixed at its base and free at its top,
Kx = 2. Applying the secant formula with P = Pcr = 199.18 kN,
smax =
(KL)x P
P
ec
C 1 + 2 sec B
RS
A
2rx A EA
rx
250 A 106 B =
199.18 A 103 B
3.62 A 10 - 3 B
D1 +
y
199.18 A 103 B
2(6)
secC
ST
2(0.105) C 200 A 109 B C 3.62 A 10 - 3 B D
0.1052
e(0.13)
e = 0.1753 m = 175 mm
Ans.
1084
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Page 1085
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13–59. The steel column supports the two eccentric
loadings. If it is assumed to be pinned at its top, fixed at the
bottom, and fully braced against buckling about the y–y
axis, determine the maximum deflection of the column
and the maximum stress in the column. Est = 200 GPa,
sY = 360 MPa.
130 kN 50 kN
80 mm
120 mm
100 mm
10 mm
6m
A = 0.12(0.1) - (0.1)(0.09) = 3.00 A 10 - 3 B m2
Ix =
1
1
(0.1) A 0.123 B (0.09) A 0.13 B = 6.90 A 10 - 6 B m4
12
12
rx =
Ix
6.90(10 - 6)
=
= 0.047958 m
AA
A 3.00(10 - 3)
For a column fixed at one end and pinned at the other end, K = 0.7.
(KL)x = 0.7(6) = 4.2 m
The eccentricity of the two applied loads is,
e =
130(0.12) - 50(0.08)
= 0.06444 m
180
Yielding About x–x Axis: Applying the secant formula,
(KL)x P
P
ec
B 1 + 2 sec ¢
≤R
A
2rx A EA
rx
180(103)
=
-3
3.00(10 )
y
B1 +
0.06444(0.06)
0.047958
2
sec ¢
180(103)
4.2
≤R
2(0.047958)A 200(109)(3.00)(10 - 3)
= 199 MPa
Ans.
Since smax 6 sg = 360 MPa, the column does not yield.
Maximum Displacement:
ymax = e B sec ¢
P KL
≤ - 1R
A EI 2
= 0.06444 B sec ¢
4.2
180(103)
a
b ≤ - 1R
A 200(109)[6.90(10 - 6)] 2
= 0.02433 m = 24.3 mm
Ans.
1085
10 mm
100 mm
x
Section Properties:
smax =
x
10 mm
y
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Page 1086
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–60. The steel column supports the two eccentric
loadings. If it is assumed to be fixed at its top and bottom,
and braced against buckling about the y–y axis, determine
the maximum deflection of the column and the maximum
stress in the column. Est = 200 GPa, sY = 360 MPa.
130 kN 50 kN
80 mm
120 mm
100 mm
10 mm
6m
A = 0.12(0.1) - (0.1)(0.09) = 3.00 A 10 - 3 B m2
Ix =
1
1
(0.1) A 0.123 B (0.09) A 0.013 B = 6.90 A 10 - 6 B m4
12
12
rx =
6.90(10 - 6)
Ix
=
= 0.047958 m
AA
A 3.00(10 - 3)
For a column fixed at both ends, K = 0.5.
(KL)x = 0.5(6) = 3.00 m
The eccentricity of the two applied loads is,
e =
130(0.12) - 50(0.08)
= 0.06444 m
180
Yielding About x–x Axis: Applying the secant formula,
(KL)x P
P
ec
B 1 + 2 sec ¢
≤R
A
2rx A EA
rx
180(103)
=
-3
3.00(10 )
y
B1 +
0.06444(0.06)
0.047958
2
sec ¢
180(103)
3.00
≤R
2(0.047958)A 200(109)(3.00)(10 - 3)
= 178 MPa
Ans.
Since smax 6 sg = 360 MPa, the column does not yield.
Maximum Displacement:
ymax = e B sec ¢
P KL
≤ - 1R
A EI 2
= 0.06444 B sec ¢
3
180(103)
a b ≤ - 1R
9
6
A 200(10 )[6.90(10 )] 2
= 0.01077 m = 10.8 mm
Ans.
1086
10 mm
100 mm
x
Section Properties:
smax =
x
10 mm
y
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13–61. The W250 * 45 A-36-steel column is pinned at its
top and fixed at its base. Also, the column is braced along
its weak axis at mid-height. If P = 250 kN, investigate
whether the column is adequate to support this loading.
Use F.S. = 2 against buckling and F.S. = 1.5 against
yielding.
P
4
250 mm
P
250 mm
4m
Section Properties. From the table listed in the appendix, the necessary section
properties for a W250 * 45 are
A = 5700 mm = 5.70 A 10
2
-3
Bm
2
rx = 112 mm = 0.112 m
Iy = 7.03 A 106 B mm4 = 7.03 A 10 - 6 B m4
c =
266
d
=
= 133 mm = 0.133 m
2
2
The eccentricity of the equivalent force P¿ = 250 +
250
= 312.5 kN is
4
250
(0.25)
4
= 0.15 m
250
250 +
4
250(0.25) e =
Buckling About the Weak Axis. The column is braced along the weak axis at
midheight and the support provided by the bracing can be considered as a pin. The
top portion of the column is critical is since the top is pinned so Ky = 1 and
L = 4 m Applying Euler’s formula,
Pcr =
p2EIy
(KL)y
=
2
p2 C 200 A 109 B D C 7.03 A 10 - 6 B D
[1(4)]2
= 867.29 kN
Euler’s equation is valid only if scr 6 sY.
867.29 A 10 B
Pcr
=
= 152.16 MPa 6 sY = 250 MPa
A
5.70 A 10 - 3 B
3
scr =
O.K.
Then,
œ
Pallow
=
Pcr
867.29
=
= 433.65 kN
F.S.
2
œ
Since Pallow
7 P¿ , the column does not buckle.
Yielding About Strong Axis. Since the column is fixed at its base and pinned at its
top, Kx = 0.7 and L = 8 m. Applying the secant formula with
œ
= P¿(F.S.) = 312.5(1.5) = 468.75 kN
Pmax
smax =
=
œ
œ
(KL)x Pmax
Pmax
ec
C 1 + 2 sec B
RS
A
2rx A EA
rx
468.75 A 103 B
5.70 A 10 - 3 B
C1 +
0.15(0.133)
0.112
2
sec B
468.75 A 103 B
0.7(8)
RS
2(0.112) C 200 A 109 B C 5.70 A 10 - 3 B D
= 231.84 MPa
Since smax 6 sY = 250 MPa, the column does not yield.
1087
4m
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The W250 * 45 A-36-steel column is pinned at its
top and fixed at its base. Also, the column is braced along
its weak axis at mid-height. Determine the allowable force
P that the column can support without causing it either
to buckle or yield. Use F.S. = 2 against buckling and
F.S. = 1.5 against yielding.
•13–62.
P
4
250 mm
P
250 mm
4m
Section Properties. From the table listed in the appendix, the necessary section
properties for a W250 * 45 are
A = 5700 mm2 = 5.70 A 10 - 3 B m2
rx = 112 mm = 0.112 m
Iy = 7.03 A 106 B mm4 = 7.03 A 10 - 6 B m4
c =
The eccentricity of the equivalent force P¿ = P +
d
266
=
= 133 mm = 0.133 m
2
2
P
= 1.25P is
4
P
(0.25)
4
= 0.15 m
P
P +
4
P(0.25) e =
Buckling About the Weak Axis. The column is braced along the weak axis at
midheight and the support provided by the bracing can be considered as a pin. The
top portion of the column is critical is since the top is pinned so Ky = 1 and
L = 4 m. Applying Euler’s formula,
Pcr =
p2EIy
(KL)y
2
=
p2 C 200 A 109 B D C 7.03 A 10 - 6 B D
[1(4)]2
= 867.29 kN
Euler’s equation is valid only if scr 6 sY.
867.29 A 10 B
Pcr
scr =
=
= 152.16 MPa 6 sY = 250 MPa
A
5.70 A 10 - 3 B
3
O.K.
Then,
œ
=
Pallow
Pcr
F.S.
867.29
2
Pallow = 346.92 kN
1.25Pallow =
Yielding About Strong Axis. Since the column is fixed at its base and pinned at its
top, Kx = 0.7 and L = 8 m. Applying the secant formula,
smax =
œ
œ
(KL)x Pmax
Pmax
ec
C 1 + 2 sec B
RS
A
2rx A EA
rx
250 A 106 B =
250 A 106 B =
1.25Pmax
5.70 A 10
-3
1.25Pmax
B
5.70 A 10 - 3 B
C1 +
0.15(0.133)
0.1122
sec B
0.7(8)
1.25Pmax
RS
2(0.112) A 200 A 109 B C 5.70 A 10 - 3 B D
A 1 + 1.5904 sec (0.00082783)2Pmax B
Solving by trial and error,
Pmax = 401.75 kN
Then,
Pallow =
401.75
= 267.83 kN = 268 kN (controls)
1.5
Ans.
1088
4m
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13–63. The W14 * 26 structural A-36 steel member is
used as a 20-ft-long column that is assumed to be fixed at
its top and fixed at its bottom. If the 15-kip load is applied
at an eccentric distance of 10 in., determine the maximum
stress in the column.
15 kip
Section Properties for W 14 * 26
A = 7.69 in2
d = 13.91 in.
10 in.
20 ft
rx = 5.65 in.
Yielding about x-x axis:
smax =
P
ec
KL
P
b d;
c1 + 2 sec a
A
2 r AE A
r
15
P
=
= 1.9506 ksi;
A
7.69
K = 0.5
10 A 2 B
ec
=
= 2.178714
2
r
(5.65)2
13.91
0.5 (20)(12)
15
KL P
=
= 0.087094
2 r A EA
2(5.65) A 29 (103)(7.69)
smax = 1.9506[1 + 2.178714 sec (0.087094)]
= 6.22 ksi 6 sg = 36 ksi
O.K.
Ans.
*13–64. The W14 * 26 structural A-36 steel member is
used as a column that is assumed to be fixed at its top and
pinned at its bottom. If the 15-kip load is applied at an
eccentric distance of 10 in., determine the maximum stress
in the column.
15 kip
Section Properties for W 14 * 26
A = 7.69 in2
d = 13.91 in.
20 ft
rx = 5.65 in.
Yielding about x-x axis:
smax =
P
ec
KL
P
c1 + 2 sec a
b d;
A
2 r AE A
r
15
P
=
= 1.9506 ksi ;
A
7.69
K = 0.7
10 A 2 B
ec
=
= 2.178714
2
r
(5.65)2
13.91
0.7 (20)(12)
15
KL P
=
= 0.121931
2 r A EA
2(5.65) A 29 (103)(7.69)
smax = 1.9506[1 + 2.178714 sec (0.121931)]
= 6.24 ksi 6 sg = 36 ksi
10 in.
O.K.
Ans.
1089
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•13–65. Determine the maximum eccentric load P the
2014-T6-aluminum-alloy strut can support without causing
it either to buckle or yield. The ends of the strut are
pin-connected.
P
150 mm
100 mm
a
a
3m
50 mm
100 mm
Section a – a
Section Properties. The necessary section properties are
A = 0.05(0.1) = 5 A 10 - 3 B m2
Iy =
1
(0.1) A 0.053 B = 1.04167 A 10 - 6 B m4
12
4.1667 A 10
Ix
=
C 5 A 10 - 3 B
AA
-6
rx =
B
= 0.02887 m
For a column that is pinned at both of its ends K = 1. Thus,
(KL)x = (KL)y = 1(3) = 3 m
Buckling About the Weak Axis. Applying Euler’s formula,
Pcr =
p2EIy
(KL)y
=
2
p2 C 73.1 A 109 B D C 1.04167 A 10 - 6 B D
32
= 83.50 kN = 83.5 kN
Ans.
Critical Stress: Euler’s formula is valid only if scr 6 sY.
83.50 A 10 B
Pcr
=
= 16.70 MPa 6 sY = 414 MPa
A
5 A 10 - 3 B
3
scr =
O.K.
Yielding About Strong Axis. Applying the secant formula,
smax =
=
(KL)x P
P
ec
C 1 + 2 sec B
RS
A
2rx A EA
rx
83.50 A 103 B
5 A 10 - 3 B
D1 +
83.50 A 103 B
3
secC
ST
2(0.02887) C 73.1 A 109 B C 5 A 10 - 3 B D
0.028872
0.15(0.05)
= 229.27 MPa 6 sY = 414 MPa
O.K.
1090
P
150 mm
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13–66. The W8 * 48 structural A-36 steel column is fixed
at its bottom and free at its top. If it is subjected to the
eccentric load of 75 kip, determine the factor of safety with
respect to either the initiation of buckling or yielding.
75 kip
y
12 ft
Section Properties: For a wide flange section W8 * 48,
A = 14.1 in2
rx = 3.61 in.
Iy = 60.9 in4
d = 8.50 in.
For a column fixed at one end and free at the other and, K = 2.
(KL)y = (KL)x = 2(12)(12) = 288 in.
Buckling About y–y Axis: Applying Euler’s formula,
P = Pcr =
p2EIy
(KL)2y
p2 (29.0)(103)(60.9)
=
2882
= 210.15 kip
Critical Stress: Euler’s formula is only valid if scr 6 sg.
scr =
Pcr
210.15
=
= 14.90 ksi 6 sg = 36 ksi
A
14.1
O. K.
Yielding About x–x Axis: Applying the secant formula,
smax =
(KL)x Pmax
Pmax
ec
B 1 + 2 sec ¢
≤R
A
2rx A EA
rx
8A 2 B
Pmax
Pmax
288
sec ¢
B1 +
≤R
14.1
2(3.61)A 29.0(103)(14.1)
3.612
1.50
36 =
36(14.1) = Pmax A 1 + 2.608943 sec 0.06238022Pmax B
Solving by trial and error,
Pmax = 117.0 kip (Controls!)
Factor of Safety:
F.S. =
Pmax
117.0
=
= 1.56
P
75
Ans.
1091
8 in.
y
x
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13–67. The W8 * 48 structural A-36 steel column is fixed
at its bottom and pinned at its top. If it is subjected to the
eccentric load of 75 kip, determine if the column fails by
yielding. The column is braced so that it does not buckle
about the y–y axis.
75 kip
y
12 ft
Section Properties: For a wide flange section W8 * 48,
A = 14.1 in2
rx = 3.61 in.
d = 8.50 in.
For a column fixed at one end and pinned at the other end, K = 0.7.
(KL)x = 0.7(12)(12) = 100.8 in.
Yielding About x–x Axis: Applying the secant formula,
smax =
(KL)x P
P
ec
B 1 + 2 sec ¢
≤R
A
2rx A EA
rx
8A 2 B
75
100.8
75
sec ¢
B1 +
≤R
2
14.1
2(3.61)A
3.61
29.0(103)(14.1)
1.50
=
= 19.45 ksi 6 sg = 36 ksi
O.K.
Hence, the column does not fail by yielding.
Ans.
1092
8 in.
y
x
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*13–68. Determine the load P required to cause the steel
W12 * 50 structural A-36 steel column to fail either by
buckling or by yielding. The column is fixed at its bottom
and the cables at its top act as a pin to hold it.
2 in.
P
25 ft
Section Properties: For a wide flange section W12 * 50,
A = 14.7 in2
rx = 5.18 in.
Iy = 56.3 in4
d = 12.19 in.
For a column fixed at one end and pinned at the other end, K = 0.7.
(KL)y = (KL)x = 0.7(25)(12) = 210 in.
Buckling About y–y Axis: Applying Euler’s formula,
P = Pcr =
p2EIy
(KL)2y
p2 (29.0)(103)(56.3)
=
2102
= 365.40 kip
Critical Stress: Euler’s formula is only valid if scr 6 sg.
scr =
Pcr
365.4
=
= 24.86 ksi 6 sg = 36 ksi
A
14.7
O. K.
Yielding About x–x Axis: Applying the secant formula,
smax =
(KL)x P
P
ec
B 1 + 2 sec ¢
≤R
A
2rx A EA
rx
2A 2 B
P
210
P
sec ¢
B1 +
≤R
14.7
2(5.18)A 29.0(103)(14.7)
5.182
12.19
36 =
36(14.7) = P A 1 + 0.454302 sec 0.03104572P B
Solving by trial and error,
Pmax = 343.3 kip = 343 kip (Controls!)
Ans.
1093
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•13–69.
Solve Prob. 13–68 if the column is an A-36 steel
W12 * 16 section.
2 in.
P
25 ft
Section Properties: For a wide flange section W12 * 16,
A = 4.71 in2
rx = 4.67 in.
Iy = 2.82 in4
d = 11.99 in.
For a column fixed at one end and pinned at the other end, K = 0.7.
(KL)y = (KL)x = 0.7(25)(12) = 210 in.
Buckling About y–y Axis: Applying Euler’s formula,
P = Pcr =
p 2EIy
(KL)2y
p2 (29.0)(103)(2.82)
=
2102
= 18.30 kip = 18.3 kip‚
Ans.
Critical Stress: Euler’s formula is only valid if scr 6 sg.
scr =
Pcr
18.30
=
= 3.89 ksi 6 sg = 36 ksi
A
4.71
Yielding About x–x Axis: Applying the secant formula,
smax =
(KL)x P
P
ec
B 1 + 2 sec ¢
≤R
A
2rx A EA
rx
2A 2 B
18.30
210
18.30
seca
bR
B1 +
4.71
2(4.67)A 29.0(103)(4.71)
4.672
11.99
=
= 6.10 ksi 6 sg = 36 ksi
O.K.
1094
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13–70. A column of intermediate length buckles when the
compressive stress is 40 ksi. If the slenderness ratio is 60,
determine the tangent modulus.
p2 Et
A KrL B 2
scr =
40 =
;
a
KL
b = 60
r
p2 Et
(60)2
Et = 14590 ksi = 14.6 (103) ksi ‚
Ans
s(ksi)
13–71. The 6-ft-long column has the cross section shown
and is made of material which has a stress-strain diagram
that can be approximated as shown. If the column is
pinned at both ends, determine the critical load Pcr for the
column.
0.5 in.
55
0.5 in.
5 in.
25
0.5 in.
3 in.
Section Properties: The neccessary section properties are
A = 2[0.5(3)] + 5(0.5) = 5.5 in2
P (in./in.)
0.001
I = 2B
ry =
1
1
(0.5) A 33 B R +
(5) A 0.53 B = 2.3021 in4
12
12
2.3021
Iy
=
= 0.6470 in.
AA
A 5.5
For the column pinned at both of its ends, K = 1. Thus,
1(6)(12)
KL
= 111.29
=
ry
0.6470
Critical Stress. Applying Engesser’s equation,
scr =
p2Et
a
KL
r
b
=
p2Et
111.292
= 0.7969 A 10 - 3 B Et
(1)
From the stress - strain diagram, the tangent moduli are
(Et)1 =
25 ksi
= 25 A 103 B ksi
0.001
(Et)2 =
(55 - 25) ksi
= 10 A 103 B (ksi)
0.004 - 0.001
0 … s 6 25 ksi
25ksi 6 s … 40 ksi
Substituting (Et)1 = 25 A 103 B into Eq. (1),
scr = 0.7969 A 10 - 3 B c25 A 103 B d = 19.92 ksi
Since scr 6 sY = 25 ksi, elastic buckling occurs. Thus,
Pcr = scr A = 19.92(5.5) = 109.57 kip = 110 kip‚
Ans.
1095
0.004
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s(ksi)
*13–72. The 6-ft-long column has the cross section shown
and is made of material which has a stress-strain diagram
that can be approximated as shown. If the column is fixed
at both ends, determine the critical load Pcr for the column.
0.5 in.
55
0.5 in.
5 in.
25
0.5 in.
3 in.
Section Properties. The neccessary section properties are
A = 2[0.5(3)] + 5(0.5) = 5.5 in2
P (in./in.)
0.001
I = 2B
ry =
1
1
(0.5) A 33 B R +
(5) A 0.53 B = 2.3021 in4
12
12
Iy
2.3021
=
= 0.6470 in.
AA
A 5.5
For the column fixed at its ends, K = 0.5. Thus,
0.5(6)(12)
KL
= 55.64
=
ry
0.6470
Critical Stress. Applying Engesser’s equation,
From the stress - strain diagram, the tangent moduli are
(Et)1 =
25 ksi
= 25 A 103 B ksi
0.001
0 … s 6 25 ksi
(Et)2 =
(55 - 25)ksi
= 10 A 103 B ksi
0.004 - 0.001
25 ksi 6 s … 40 ksi
Substituting (Et)1 = 25 A 103 B ksi into Eq. (1),
scr = 3.1875 A 10 - 3 B c25 A 103 B d = 79.69 ksi
Since scr 7 sY = 25 ksi, the inelastic buckling occurs. Substituting (Et)2 into Eq. (1),
scr = 3.1875 A 10 - 3 B c10 A 103 B d = 31.88 ksi
Since 25 ksi 6 scr 6 55 ksi, this result can be used to calculate the critical load.
Pcr = scr A = 31.88(5.5) = 175.31 kip = 175 kip
Ans.
1096
0.004
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s (MPa)
•13–73.
The stress-strain diagram of the material of a
column can be approximated as shown. Plot P兾A vs. KL兾r
for the column.
350
200
Tangent Moduli. From the stress - strain diagram,
(Et)1 =
(Et)2 =
200 A 106 B
0.001
0 … s 6 200MPa
= 200 GPa
(350 - 200) A 106 B
0.004 - 0.001
= 50 GPa
0
P (in./in.)
0.001
0.004
200MPa 6 s … 350 MPa
Critical Stress. Applying Engesser’s equation,
scr
P
=
A
p2Et
(1)
KL 2
a
b
r
If Et = (Et)1 = 200 GPa, Eq. (1) becomes
p C 200 A 10 B D
1.974 A 10 B
P
=
=
MPa
2
A
KL
KL 2
a
b
a
b
r
r
2
when scr =
9
6
P
= sY = 200 MPa, this equation becomes
A
200 A 106 B =
p2 C 200 A 109 B D
a
KL 2
b
r
KL
= 99.346 = 99.3
r
If Et = (Et)2 = 50 GPa, Eq. (1) becomes
P
=
A
p2 c50 A 109 B d
0.4935 A 106 B
MPa
KL 2
KL 2
¢
≤
¢
≤
r
r
P
when scr =
= sY = 200 MPa, this
A
equation gives
200 A 106 B =
=
p2 C 50 A 109 B D
a
KL 2
b
r
KL
= 49.67 = 49.7
r
Using these results, the graphs of
P
KL
vs.
is shown in Fig. a can be plotted.
r
A
1097
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s (MPa)
13–74. Construct the buckling curve, P兾A versus L兾r,
for a column that has a bilinear stress–strain curve in
compression as shown. The column is pinned at its ends.
260
140
0.001
Tangent modulus: From the stress–strain diagram,
(Et)1 =
140(106)
= 140 GPa
0.001
(Et)2 =
(260 - 140)(106)
= 40 GPa
0.004 - 0.001
Critical Stress: Applying Engesser’s equation,
scr
p2Et
P
=
A
L 2
a b
r
[1]
Substituting (Et)1 = 140 GPa into Eq. [1], we have
p C 140(10 ) D
P
=
A
ALB2
2
9
r
P
=
A
When
1.38(106)
A Lr B 2
MPa
L
P
= 140 MPa, = 99.3
r
A
Substitute (Et)2 = 40 GPa into Eq. [1], we have
p C 40(10 ) D
P
=
A
ALB2
2
9
r
P
=
A
When
0.395(106)
A Lr B 2
MPa
L
P
= 140 MPa, = 53.1
r
A
1098
0.004
P (mm/mm)
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13–75. The stress-strain diagram for a material can be
approximated by the two line segments shown. If a bar
having a diameter of 80 mm and a length of 1.5 m is made
from this material, determine the critical load provided the
ends are pinned. Assume that the load acts through the axis
of the bar. Use Engesser’s equation.
s (MPa)
1100
200
0.001
E1 =
200 (106)
= 200 GPa
0.001
E2 =
1100 (106) - 200 (106)
= 150 GPa
0.007 - 0.001
Section properties:
I =
p 4
c;
4
r =
p 4
I
0.04
c
4 c
=
= =
= 0.02 m
AA
C p c2 2
2
A = pc2
Engesser’s equation:
1.0(1.5)
KL
=
= 75
r
0.02
scr =
p2 Et
A
B
KL 2
r
=
p2 Et
(75)2
= 1.7546(10 - 3) Et
Assume Et = E1 = 200 GPa
scr = 1.7546 (10 - 3)(200)(109) = 351 MPa 7 200 MPa
Therefore, inelastic buckling occurs:
Assume Et = E2 = 150 GPa
scr = 1.7546 (10 - 3)(150)(109) = 263.2 MPa
200 MPa 6 scr 6 1100 MPa
O.K.
Critical load:
Pcr = scr A = 263.2 (106)(p)(0.042) = 1323 kN‚
Ans
1099
0.007
P (mm/mm)
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*13–76. The stress-strain diagram for a material can be
approximated by the two line segments shown. If a bar
having a diameter of 80 mm and a length of 1.5 m is made
from this material, determine the critical load provided the
ends are fixed. Assume that the load acts through the axis of
the bar. Use Engesser’s equation.
s (MPa)
1100
200
0.001
E1 =
200 (106)
= 200 GPa
0.001
E2 =
1100 (106) - 200 (106)
= 150 GPa
0.007 - 0.001
Section properties:
I =
p 4
c;
4
r =
p 4
I
0.04
c
4 c
=
= =
= 0.02 m
AA
C p c2 2
2
A = pc2
Engesser’s equation:
0.5 (1.5)
KL
=
= 37.5
r
0.02
scr =
p2 Et
A
B
KL 2
r
=
p2 Et
(37.5)2
= 7.018385(10 - 3) Et
Assume Et = E1 = 200 GPa
scr = 7.018385 (10 - 3)(200)(109) = 1403.7 MPa 7 200 MPa
NG
Assume Et = E2 = 150 GPa
scr = 7.018385 (10 - 3)(150)(109) = 1052.8 MPa
200 MPa 6 scr 6 1100 MPa
O.K.
Critical load:
Pcr = scr A = 1052.8 (106)(p)(0.042) = 5292 kN
Ans.
1100
0.007
P (mm/mm)
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•13–77.
The stress-strain diagram for a material can be
approximated by the two line segments shown. If a bar
having a diameter of 80 mm and length of 1.5 m is made
from this material, determine the critical load provided one
end is pinned and the other is fixed. Assume that the load
acts through the axis of the bar. Use Engesser’s equation.
s (MPa)
1100
200
0.001
E1 =
200 (106)
= 200 GPa
0.001
E2 =
1100 (106) - 200 (106)
= 150 GPa
0.007 - 0.001
Section properties:
I =
p 4
c ;
4
r =
p 4
I
0.04
c
4 c
=
= =
= 0.02 m
AA
C pc2
2
2
A = pc2
Engesser’s equation:
0.7 (1.5)
KL
=
= 52.5
r
0.02
scr =
p2 Et
A
B
KL 2
r
=
p2Et
(52.5)2
= 3.58081 (10 - 3) Et
Assume Et = E1 = 200 GPa
scr = 3.58081 (10 - 3)(200)(109) = 716.2 MPa 7 200 MPa
NG
Assume Et = E2 = 150 GPa
scr = 3.58081 (10 - 3)(150)(109) = 537.1 MPa
200 MPa 6 scr 6 1100 MPa
O.K.
Critical load:
Pcr = scr A = 537.1 (106)(p)(0.042) = 2700 kN
Ans.
1101
0.007
P (mm/mm)
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13–78. Determine the largest length of a structural A-36
steel rod if it is fixed supported and subjected to an axial
load of 100 kN. The rod has a diameter of 50 mm. Use the
AISC equations.
Section Properties:
A = p A 0.0252 B = 0.625 A 10 - 3 B p m2
I =
p
A 0.0254 B = 97.65625 A 10 - 9 B p m4
4
r =
97.65625(10 - 9)p
I
=
= 0.0125 m
AA
A 0.625(10 - 3)p
Slenderness Ratio: For a column fixed at both ends, K = 0.5. Thus,
0.5L
KL
=
= 40.0L
r
0.0125
AISC Column Formula: Assume a long column.
sallow =
100(103)
0.625(10 - 3)p
=
12p2E
2
23 A KL
r B
12p2 C 200(109) D
23(40.0L)3
L = 3.555 m
KL
KL
2p2E
= 40.0(3.555) = 142.2 and for A–36 steel, a
b =
r
r e A sg
KL
KL
2p2[200(109)]
=
= 125.7. Since a
b …
… 200, the assumption is correct.
r e
r
A 250(106)
Here,
Thus,
L = 3.56 m
Ans.
1102
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13–79. Determine the largest length of a W10 * 45
structural steel column if it is pin supported and subjected
to an axial load of 290 kip. Est = 29(103) ksi, sY = 50 ksi.
Use the AISC equations.
Section Properties: For a W10 * 45 wide flange section,
A = 13.3 in2
ry = 2.01 in
Slenderness Ratio: For a column pinned at both ends, K = 1. Thus,
a
1(L)
KL
b =
= 0.49751L
r y
2.01
AISC Column Formula: Assume a long column,
sallow =
12p2E
2
23 A KL
r B
12p2 C 29(103) D
290
=
13.3
23(0.49751L)2
L = 166.3 in.
KL
KL
2p2E
= 0.49751 (166.3) = 82.76 and for grade 50 steel, a
b =
r
r c A sg
KL
KL
2p2[29(103)]
6 a
b , the assumption is not correct.
=
= 107.0. Since
r
r c
50
A
Thus, the column is an intermediate column.
Here,
Applying Eq. 13–23,
B1 sallow =
(KL>r)2
2(KL>r)2c
R sg
3(KL>r)
(KL>r)3
5
+
3
8(KL>r)c
8(KL>r)3c
B1 -
290
=
13.3
5
3
+
(0.49751L)2
2(107.02)
R (50)
3(0.49751L)
(0.49751L)3
8(107.0)
8(107.03)
0 = 12.565658 A 10 - 9 B L3 - 24.788132 A 10 - 6 B L2 - 1.743638 A 10 - 3 B L + 0.626437
Solving by trial and error,
L = 131.12 in. = 10.9 ft
Ans.
1103
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*13–80. Determine the largest length of a W10 * 12
structural A-36 steel section if it is pin supported and is
subjected to an axial load of 28 kip. Use the AISC equations.
For a W 10 * 12,
A = 3.54 in2
ry = 0.785 in.
s =
28
P
=
= 7.91 ksi
A
3.54
Assume a long column:
sallow =
a
a
12p2E
23(KL>r)2
KL
12p2(29)(103)
12p2E
=
= 137.4
b =
r
A 23(7.91)
A 23sallow
KL
2p2(29)(103)
2p2E
=
= 126.1,
b =
r c A sg
A
36
KL
KL
7 a
b
r
r c
Long column.
KL
= 137.4
r
L = 137.4 ¢
r
0.785
≤ = 137.4 ¢
≤ = 107.86 in.
K
1
= 8.99 ft
Ans.
•13–81.
Using the AISC equations, select from Appendix B
the lightest-weight structural A-36 steel column that is 14 ft
long and supports an axial load of 40 kip.The ends are pinned.
Take sY = 50 ksi.
Try, W6 * 15 (A = 4.43 in2
ry = 1.46 in.)
a
KL
2p2(29)(103)
2p2E
b =
=
= 107
r c A sY
50
A
a
(1.0)(14)(12)
KL
b =
= 115.1,
ry
1.46
a
KL
KL
b 7 a
b
ry
r c
Long column
sallow =
12p2(29)(103)
12 p2E
=
= 11.28 ksi
2
23(KL>r)
23(115.1)2
Pallow = sallowA
= 11.28(4.43) = 50.0 kip 7 40 kip
O.K.
Use W6 * 15
Ans.
1104
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13–82. Using the AISC equations, select from Appendix B
the lightest-weight structural A-36 steel column that is 12 ft
long and supports an axial load of 40 kip. The ends are fixed.
Take sY = 50 ksi.
A = 2.68 in2
Try W6 * 9
a
ry = 0.905 in.
KL
2p2(29)(103)
2p2E
b =
=
= 107
r c A sY
A
50
0.5(12)(12)
KL
=
= 79.56
ry
0.905
KL
KL
6 a
b
ry
r c
Intermediate column
sallow =
KL>r 2
C 1 - 12 A (KL>r)c
B D sg
KL>r
KL>r 3
C 53 + 38 A (KL>r)c
B - 18 A (KL>r)c
B D
=
2
C 1 - 12 A 79.56
126.1 B D 36 ksi
1 79.56 3
C 53 + 38 A 79.56
126.1 B - 8 A 12.61 B D
= 15.40 ksi
Pallow = sallowA
= 15.40(2.68)
= 41.3 kip 7 40 kip
O.K.
Use W6 * 9
Ans.
1105
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13–83. Using the AISC equations, select from Appendix B
the lightest-weight structural A-36 steel column that is 24 ft
long and supports an axial load of 100 kip.The ends are fixed.
Section Properties: Try a W8 * 24 wide flange section,
A = 7.08 in2
ry = 1.61 in
Slenderness Ratio: For a column fixed at both ends, K = 0.5. Thus,
a
AISC
Column
0.5(24)(12)
KL
b =
= 89.44
r y
1.61
Formula:
For
A–36
steel,
a
KL
2p2E
b =
r c A sg
KL
KL
2p2[29(103)]
6 a
b , the column is an intermediate
= 126.1. Since
r
r c
A
36
column. Applying Eq. 13–23,
=
(KL>r)2
B1 sallow =
R sg
3(KL>r)
(KL>r)3
5
+
3
8(KL>r)c
8(KL>r)3c
B1 =
2(KL>r)2c
(89.442)
2(126.12)
R (36)
3(89.44)
(89.443)
5
+
3
8(126.1)
8(126.13)
= 14.271 ksi
The allowable load is
Pallow = sallowA
= 14.271(7.08)
= 101 kip 7 P = 100 kip
Thus, Use
O.K.
Ans.
W8 * 24
1106
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*13–84. Using the AISC equations, select from Appendix B
the lightest-weight structural A-36 steel column that is 30 ft
long and supports an axial load of 200 kip.The ends are fixed.
a
A = 14.1 in2
ry = 2.08 in.
Try W8 * 48
KL
2 p2 (29)(103)
2 p2E
b =
=
= 126.1
r c A sg
A
36
0.5 (30)(12)
KL
=
= 86.54
ry
2.08
a
KL
KL
b 6 a
b intermediate column.
ry
r c
KL
r
2
b 1 - 12 B KL R r sg
sallow =
A r Bc
KL
r
KL
r
b 53 + 38 B KL R - 18 B KL
=
A r Bc
A r Bc
D
3
- 18 86.54
f
126.1
2
e 1 - 12 C 86.54
126.1 D f36
e 53 + 38
C
86.54
126.1
C
D
3
R r
= 14.611 ksi
Pallow = sallow A = 14.611 (14.1) = 206 kip 7 P = 200 kip
Use
O.K.
W 8 * 48
Ans.
1107
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•13–85.
A W8 * 24 A-36-steel column of 30-ft length is
pinned at both ends and braced against its weak axis at midheight. Determine the allowable axial force P that can be
safely supported by the column. Use the AISC column
design formulas.
Section Properties. From the table listed in the appendix, the necessary section
properties for a W8 * 24 are
A = 7.08 in2
rx = 3.42 in.
ry = 1.61 in.
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Here,
Lx = 30(12) = 360 in. and Ly = 15(12) = 180 in. Thus,
¢
1(360)
KL
= 105.26
≤ =
r x
3.42
¢
1(180)
KL
= 111.80 (controls)
≤ =
r y
1.61
AISC
=
C
Column
2p2 C 29 A 103 B D
36
Formulas.
For
= 126.10. Since
¢
A-36
steel
¢
KL
KL
≤ 6 ¢
≤ , the
r y
r c
KL
2p2E
≤ =
r c A sY
column
is
an
intermediate column.
(KL>r)2
B1 sallow =
R sY
3(KL>r)
(KL>r)3
5
+
3
8(KL>r)c
8(KL>r)c 3
C1 -
=
2(KL>r)c 2
111.802
2 A 126.102 B
S(36)
3(111.80)
5
111.803
+
3
8(126.10)
8 A 126.103 B
= 11.428 ksi
Thus, the allowable force is
Pallow = sallowA = 11.428(7.08) = 80.91 kip = 80.9 kip
1108
Ans.
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13–86. Check if a W10 * 39 column can safely support an
axial force of P = 250 kip. The column is 20 ft long and is
pinned at both ends and braced against its weak axis at
mid-height. It is made of steel having E = 29(103) ksi and
sY = 50 ksi. Use the AISC column design formulas.
Section Properties. From the table listed in the appendix, the necessary section
properties for a W10 * 39 are
A = 11.5 in2
rx = 4.27 in.
ry = 1.98 in.
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Here,
Lx = 20(12) = 240 in. and Ly = 10(12) = 120 in. Thus,
¢
1(240)
KL
= 56.21
≤ =
r x
4.27
¢
1(120)
KL
= 60.606 (controls)
≤ =
r y
1.98
AISC
Column
=
2p2 c29 A 103 B d
S
50
Formulas.
For
A-36
= 107.00 . Since ¢
steel
¢
KL
2p2E
≤ =
r c A sY
KL
KL
≤ 6 ¢
≤ , the column is an
r y
r c
intermediate column.
B1 sallow =
2(KL>r)c 2
R sY
3(KL>r)
(KL>r)3
5
+
3
8(KL>r)c
8(KL>r)c 3
C1 -
=
(KL>r)2
60.6062
2 A 107.002 B
S(50)
3(60.606)
5
60.6063
+
3
8(107.00)
8 A 107.003 B
= 22.614 ksi
Thus, the allowable force is
Pallow = sallowA = 22.614(11.5) = 260.06 kip 7 P = 250 kip
Thus, a W10 * 39 column is adequate.
1109
O.K.
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13–87. A 5-ft-long rod is used in a machine to transmit
an axial compressive load of 3 kip. Determine its smallest
diameter if it is pin connected at its ends and is made of a
2014-T6 aluminum alloy.
Section properties:
p d 4
pd4
a b =
4 2
64
A =
p 2
d ;
4
r =
I
d
64
=
=
AA
C p4 d2
4
I =
pd4
sallow =
P
3
3.820
= p 2 =
A
d
d2
4
Assume long column:
1.0 (5)(12)
240
KL
=
=
d
r
d
4
sallow =
54 000
A
B
KL 2
r
;
3.820
54000
=
d2
C 240 D 2
d
d = 1.42 in.
Ans.
KL
240
=
= 169 7 55
r
1.42
O.K.
1110
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*13–88. Check if a W10 * 45 column can safely support
an axial force of P = 200 kip. The column is 15 ft long and
is pinned at both of its ends. It is made of steel having
E = 29(103) ksi and sY = 50 ksi. Use the AISC column
design formulas.
Section Properties. Try W10 * 45. From the table listed in the appendix, the
necessary section properties are
A = 13.3 in2
ry = 2.01 in.
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Thus,
a
1(15)(12)
KL
b =
= 89.552
r y
2.01
KL
2p2E
AISC Column Formulas. Here, a
b =
=
r c A sY
S
KL
KL
Since a
b 6 a
b , the
r y
r c
2p2 c29 A 103 B d
50
= 107.00.
column is an intermediate column.
(KL>r)2
B1 sallow =
R sY
(KL>r)3
3(KL>r)
5
+
3
8(KL>r)c
8(KL>r)c 3
C1 -
=
2(KL>r)c 2
89.5522
2 A 107.002 B
S(50)
3(89.552)
5
89.5523
+
3
8(107.00)
8 A 107.003 B
= 17.034 ksi
Thus, the allowable force is
Pallow = sallowA = 17.034(13.3) = 226.55 kip 7 P = 200 kip
O.K.
Thus,
A W10 * 45 can be used
Ans.
1111
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•13–89.
Using the AISC equations, check if a column
having the cross section shown can support an axial force of
1500 kN. The column has a length of 4 m, is made from A-36
steel, and its ends are pinned.
20 mm
350 mm
300 mm
10 mm
Section Properties:
A = 0.3(0.35) - 0.29(0.31) = 0.0151 m2
Iy =
1
1
(0.04) A 0.33 B +
(0.31) A 0.013 B = 90.025833 A 10 - 6 B m4
12
12
ry =
Iy
90.02583(10 - 6)
=
= 0.077214 m
AA
A
0.0151
Slenderness Ratio: For a column pinned at both ends, K = 1. Thus,
a
AISC
=
Column
2p2[200(109)]
1(4)
KL
b =
= 51.80
r y
0.077214
Formula:
= 125.7. Since
A 250(10 )
column. Applying Eq. 13–23,
6
For
a
KL
2p2E
b =
r c A sg
(KL>r)2
2(KL>r)2c
R sg
3(KL>r)
(KL>r)3
5
+
3
8(KL>r)c
8(KL>r)3c
B1 =
steel,
KL
KL
6 a
b , the column is an intermediate
r
r c
B1 sallow =
A–36
(51.802)
2(125.72)
20 mm
R (250)(106)
3(51.80)
(51.803)
5
+
3
8(125.7)
8(125.73)
= 126.2 MPa
The allowable load is
Pallow = sallowA
= 126.2 A 106 B (0.0151)
= 1906 kN 7 P = 1500 kN
O.K.
Thus, the column is adequate.
Ans.
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13–90. The A-36-steel tube is pinned at both ends. If it
is subjected to an axial force of 150 kN, determine the
maximum length that the tube can safely support using the
AISC column design formulas.
100 mm
80 mm
Section Properties.
A = p A 0.052 - 0.042 B = 0.9 A 10 - 3 B p m2
I =
p
A 0.054 - 0.044 B = 0.9225 A 10 - 6 B p m4
4
0.9225 A 10 B p
I
=
= 0.03202 m
AA
C 0.9 A 10 - 3 B p
-6
r =
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Thus,
1(L)
KL
=
= 31.23L
r
0.03202
AISC Column Formulas.
sallow =
12p2E
23(KL>r)2
150 A 103 B
.9 A 10 - 3 B p
=
12p2 C 200 A 109 B D
23(31.23L)2
L = 4.4607 m = 4.46 m
Here,
KL
= 31.23(4.4607) = 139.33.
r
2p2 C 200 A 109 B D
= 125.66. Since a
250 A 10 B
long column is correct.
=
C
6
Ans.
For
A-36
steel
a
KL
2p2E
b =
r c A sY
KL
KL
b 6
6 200, the assumption of a
r c
r
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13–91. The bar is made of a 2014-T6 aluminum alloy.
Determine its smallest thickness b if its width is 5b. Assume
that it is pin connected at its ends.
600 lb
b
5b
8 ft
Section Properties:
A = b(5b) = 5b2
Iy =
1
5 4
(5b) A b3 B =
b
12
12
ry =
5 4
Iy
23
12 b
=
=
b
AA
C 5b2
6
600 lb
Slenderness Ratio: For a column pinned at both ends, K = 1. Thus,
a
1(8)(12)
332.55
KL
=
b =
23
b
r y
b
6
Aluminum (2014 - T6 alloy) Column Formulas: Assume a long column and apply
Eq. 13–26.
sallow =
54 000
(KL>r)2
0.600
54 000
=
2
5b
A 332.55 B 2
b
b = 0.7041 in.
Here,
KL
KL
332.55
=
7 55, the assumption is correct. Thus,
= 472.3. Since
r
r
0.7041
b = 0.704 in.
Ans.
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*13–92. The bar is made of a 2014-T6 aluminum alloy.
Determine its smallest thickness b if its width is 5b. Assume
that it is fixed connected at its ends.
600 lb
b
5b
8 ft
Section Properties:
A = b(5b) = 5b2
Iy =
1
5 4
(5b) A b3 B =
b
12
12
ry =
5 4
Iy
23
12 b
=
=
b
AA
C 5b2
6
600 lb
Slenderness Ratio: For a column fixed at both ends, K = 0.5. Thus,
a
0.5(8)(12)
166.28
KL
=
b =
r y
b
23
6 b
Aluminum (2014 - T6 alloy) Column Formulas: Assume a long column and apply
Eq. 13–26.
sallow =
54 000
(KL>r)2
0.600
54 000
=
5b2
A 166.28 B 2
b
b = 0.4979 in.
Here,
KL
KL
166.28
= 334.0. Since
=
7 55, the assumption is correct.
r
r
0.4979
Thus,
b = 0.498 in.
Ans.
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•13–93.
The 2014-T6 aluminum column of 3-m length has
the cross section shown. If the column is pinned at both
ends and braced against the weak axis at its mid-height,
determine the allowable axial force P that can be safely
supported by the column.
15 mm
170 mm
15 mm
15 mm
100 mm
Section Properties.
A = 0.1(0.2) - 0.085(0.17) = 5.55 A 10 - 3 B m2
Ix =
1
1
(0.1) A 0.23 B (0.085) A 0.173 B = 31.86625 A 10 - 6 B m4
12
12
Iy = 2 c
1
1
(0.015) A 0.13 B d +
(0.17) A 0.0153 B = 2.5478 A 10 - 6 B m4
12
12
31.86625 A 10
Ix
=
AA
C 5.55 A 10 - 3 B
-6
rx =
ry =
B
= 0.07577
2.5478 A 10 - 6 B
Iy
=
= 0.02143 m
AA
C 5.55 A 10 - 3 B
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Here, Lx = 3 m
and Ly = 1.5 m. Thus,
a
(1)(3)
KL
b =
= 39.592
r x
0.07577
a
(1)(1.5)
KL
b =
= 70.009 (controls)
r y
0.02143
2014-T6 Alumimum Alloy Column Formulas. Since a
KL
b 7 55, the column can
r y
be classified a long column,
sallow = D
373 A 103 B
T Mpa
= C
373 A 103 B
S MPa
a
KL 2
b
r
70.0092
= 76.103 MPa
Thus, the allowed force is
Pallow = sallowA = 76.103 A 106 B C 5.55 A 10 - 3 B D = 422.37 kN = 422 kN
1116
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13–94. The 2014-T6 aluminum column has the cross
section shown. If the column is pinned at both ends and
subjected to an axial force P = 100 kN, determine the
maximum length the column can have to safely support the
loading.
15 mm
170 mm
15 mm
15 mm
100 mm
Section Properties.
A = 0.1(0.2) - 0.085(0.17) = 5.55 A 10 - 3 B m2
Iy = 2c
ry =
1
1
(0.015) A 0.13 B d +
(0.17) A 0.0153 B = 2.5478 A 10 - 6 B m4
12
12
2.5478 A 10 - 6 B
Iy
=
= 0.02143 m
AA
C 5.55 A 10 - 3 B
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Then,
a
1(L)
KL
b =
= 46.6727L
r y
0.02143
2014-T6 Alumimum Alloy Column Formulas. Assuming a long column,
sallow = D
373 A 103 B
100 A 103 B
= C
5.55 A 10 - 3 B
a
KL 2
b
r
T MPa
373 A 103 B
(46.672L)2
S A 106 B Pa
L = 3.083 m = 3.08 m
Since a
Ans.
KL
b = 46.6727(3.083) = 143.88 7 55, the assumption is correct.
r y
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13–95. The 2014-T6 aluminum hollow section has the
cross section shown. If the column is 10 ft long and is fixed
at both ends, determine the allowable axial force P that can
be safely supported by the column.
4 in.
3 in.
Section Properties.
A = p A 22 - 1.52 B = 1.75p in2
r =
I =
p 4
A 2 - 1.54 B = 2.734375p in4
4
I
2.734375p
=
= 1.25 in.
AA
A 1.75p
Slenderness Ratio. For a column fixed at both of its ends, K = 0.5. Thus,
0.5(10)(12)
KL
=
= 48
r
1.25
2014-T6 Aluminum Alloy Column Formulas. Since 12 6
KL
6 55, the column can
r
be classified as an intermediate column.
sallow = c30.7 - 0.23 a
KL
b d ksi
r
= [30.7 - 0.23(48)] ksi
= 19.66 ksi
Thus, the allowable load is
Pallow = sallowA = 19.66 A 106 B (1.75p) = 108.09 kip = 108 kip
1118
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*13–96. The 2014-T6 aluminum hollow section has the
cross section shown. If the column is fixed at its base and
pinned at its top, and is subjected to the axial force
P = 100 kip, determine the maximum length of the column
for it to safely support the load.
4 in.
3 in.
Section Properties.
A = p A 22 - 1.52 B = 1.75p in2
r =
I =
p 4
A 2 - 1.54 B = 2.734375p in4
4
I
2.734375p
=
= 1.25 in.
AA
A 1.75p
Slenderness Ratio. For a column fixed at its base and pinned at its top, K = 0.7.
Thus,
0.7(L)
KL
=
= 0.56L
r
1.25
2014-T6 Aluminum Alloy Column Formulas. Assuming an intermediate column,
sallow = c30.7 - 0.23a
KL
b d ksi
r
100
= 30.7 - 0.23(0.56L)
1.75p
L = 97.13 in. = 8.09 ft
Ans.
KL
= 0.56(97.13) = 54.39 6 55, the assumption of an intermediate column
r
is correct.
Since
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•13–97.
The tube is 0.25 in. thick, is made of a 2014-T6
aluminum alloy, and is fixed at its bottom and pinned at its
top. Determine the largest axial load that it can support.
P
x
y
6 in.
x
6 in.
y
10 ft
Section Properties:
P
A = 6(6) - 5.5(5.5) = 5.75 in2
I =
1
1
(6) A 63 B (5.5) A 5.53 B = 31.7448 in4
12
12
r =
I
31.7448
=
= 2.3496 in.
A 5.75
AA
Slenderness Ratio: For a column fixed at one end and pinned at the other end,
K = 0.7. Thus,
0.7(10)(12)
KL
=
= 35.75
r
2.3496
Aluminium (2014 –∑ T6 alloy) Column Formulas: Since 12 6
KL
6 55, the
r
column is classified as an intermediate column. Applying Eq. 13–25,
sallow = c30.7 - 0.23 a
KL
b d ksi
r
= [30.7 - 0.23(33.75)]
= 24.48 ksi
The allowable load is
Pallow = sallowA = 22.48(5.75) = 129 kip
Ans.
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13–98. The tube is 0.25 in. thick, is made of a 2014-T6
aluminum alloy, and is fixed connected at its ends.
Determine the largest axial load that it can support.
P
x
y
6 in.
x
6 in.
y
10 ft
Section Properties:
P
A = 6(6) - 5.5(5.5) = 5.75 in2
I =
1
1
(6) A 63 B (5.5) A 5.53 B = 31.7448 in4
12
12
r =
I
31.7448
=
= 2.3496 in.
AA
A 5.75
Slenderness Ratio: For column fixed at both ends, K = 0.5. Thus,
0.5(10)(12)
KL
=
= 25.54
r
2.3496
Aluminium (2014 – T6 alloy) Column Formulas: Since 12 6
KL
6 55, the
r
column is classified as an intermediate column. Applying Eq. 13–25,
sallow = c30.7 - 0.23 a
KL
b d ksi
r
= [30.7 - 0.23(25.54)]
= 24.83 ksi
The allowable load is
Pallow = sallowA = 24.83(5.75) = 143 kip
Ans.
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13–99. The tube is 0.25 in. thick, is made of 2014-T6
aluminum alloy and is pin connected at its ends. Determine
the largest axial load it can support.
P
x
y
6 in.
x
6 in.
y
Section Properties:
A = 6(6) - 5.5(5.5) = 5.75 in2
I =
1
1
(6) A 63 B (5.5) A 5.53 B = 31.7448 in4
12
12
r =
I
31.7448
=
= 2.3496 in.
AA
A 5.75
10 ft
P
Slenderness Ratio: For a column pinned as both ends, K = 1. Thus,
1(10)(12)
KL
=
= 51.07
r
2.3496
Aluminum (2014 – T6 alloy) Column Formulas: Since 12 6
KL
6 55, the
r
column is classified as an intermediate column. Applying Eq. 13–25,
sallow = c30.7 - 0.23 a
KL
b d ksi
r
= [30.7 - 0.23(51.07)]
= 18.95 ksi
The allowable load is
Pallow = sallowA = 18.95(5.75) = 109 kip
Ans.
*13–100. A rectangular wooden column has the cross
section shown. If the column is 6 ft long and subjected to an
axial force of P = 15 kip, determine the required minimum
1
dimension a of its cross-sectional area to the nearest 16
in.
so that the column can safely support the loading. The
column is pinned at both ends.
a
2a
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Then,
(1)(6)(12)
KL
72
=
=
a
a
d
NFPA Timber Column Formula. Assuming an intermediate column,
sallow = 1.20 c1 -
1 KL>d 2
a
b d ksi
3 26.0
15
1 72>a 2
= 1.20 c1 - a
b d
2a(a)
3 26.0
a = 2.968 in.
Use a = 3 in.
Ans.
KL
72
KL
=
= 24. Since 11 6
6 26, the assumption is correct.
d
3
d
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•13–101.
A rectangular wooden column has the cross
section shown. If a = 3 in. and the column is 12 ft long,
determine the allowable axial force P that can be safely
supported by the column if it is pinned at its top and fixed at
its base.
a
2a
Slenderness Ratio. For a column fixed at its base and pinned at its top K = 0.7.
Then,
0.7(12)(12)
KL
=
= 33.6
d
3
NFPA Timer Column Formula. Since 26 6
KL
6 50, the column can be classified
d
as a long column.
sallow =
540 ksi
540
=
= 0.4783 ksi
2
(KL>d)
33.62
The allowable force is
Pallow = sallowA = 0.4783(3)(6) = 8.61 kip
Ans.
13–102. A rectangular wooden column has the cross
section shown. If a = 3 in. and the column is subjected to
an axial force of P = 15 kip, determine the maximum
length the column can have to safely support the load. The
column is pinned at its top and fixed at its base.
a
2a
Slenderness Ratio. For a column fixed at its base and pinned at its top, K = 0.7.
Then,
KL
0.7L
=
= 0.2333L
d
3
NFPA Timber Column Formula. Assuming an intermediate column,
sallow = 1.20 c1 -
1 KL>d 2
a
b d ksi
3 26.0
15
1 0.2333L 2
= 1.20c1 - a
b d
3(6)
3
26.0
L = 106.68 in. = 8.89 ft
Ans.
KL
KL
= 0.2333(106.68) = 24.89. Since 11 6
6 26, the assumption is
d
d
correct.
Here,
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13–103. The timber column has a square cross section and
is assumed to be pin connected at its top and bottom. If it
supports an axial load of 50 kip, determine its smallest side
dimension a to the nearest 12 in. Use the NFPA formulas.
14 ft
a
Section properties:
A = a2
sallow = s =
50
P
= 2
A
a
Assume long column:
sallow =
540
2
A KL
d B
540
50
=
(1.0)(14)(12) 2
a2
C
D
a
a = 7.15 in.
(1.0)(14)(12)
KL
KL
=
= 23.5,
6 26
d
7.15
d
Assumption NG
Assume intermediate column:
sallow = 1.20 B 1 -
1 KL>d 2
a
b R
3 26.0
2
50
1
a
b R
= 1.20 B 1 - a
2
a
26.0
3
1.0(14)(12)
a = 7.46 in.
1.0(14)(12)
KL
KL
=
= 22.53, 11 6
6 26
d
7.46
d
Assumption O.K.
1
Use a = 7 in.
2
Ans.
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*13–104. The wooden column shown is formed by gluing
together the 6 in. * 0.5 in. boards. If the column is pinned
at both ends and is subjected to an axial load P = 20 kip,
determine the required number of boards needed to form
the column in order to safely support the loading.
P
6 in.
0.5 in.
9 ft
Slenderness Ratio. For a column pinned at both of its ends, K = 1. If the number of
the boards required is n and assuming that n(0.5) 6 6 in. Then, d = n(0.5). Thus,
(1)(9)(12)
KL
216
=
=
n
d
n(0.5)
P
NFPA Timber Column Formula. Assuming an intermediate column,
sallow = 1.20 B 1 -
1 KL>d 2
a
b R ksi
3 26.0
1 216>n 2
20
= 1.20 B 1 - a
b R
[n(0.5)](6)
3 26.0
n2 - 5.5556n - 23.01 = 0
Solving for the positive root,
n = 8.32
Use n = 9
Ans.
KL
KL
216
=
= 24. Since n(0.5) = 9(0.5) = 4.5 in. 6 6 in. and 11 6
6 26,
d
9
d
the assumptions made are correct.
Here,
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•13–105.
The column is made of wood. It is fixed at its
bottom and free at its top. Use the NFPA formulas to
determine its greatest allowable length if it supports an
axial load of P = 2 kip.
P
2 in.
y
y
x
x
4 in.
Slenderness Ratio: For a column fixed at one end and free at the other end, K = 2.
Thus,
L
2(L)
KL
=
= 1.00L
d
2
NFPA Timber Column Formulas: Assume a long column. Apply Eq. 13–29,
sallow =
540
ksi
(KL>d)2
2
540
=
2(4)
(1.00L)2
L = 46.48 in
Here,
KL
KL
= 1.00(46.48) = 46.48. Since 26 6
6 50, the assumption is correct.
d
d
Thus,
L = 46.48 in. = 3.87 ft
Ans.
13–106. The column is made of wood. It is fixed at its
bottom and free at its top. Use the NFPA formulas to
determine the largest allowable axial load P that it can
support if it has a length L = 4 ft.
P
2 in.
y
x
y
x
4 in.
Slenderness Ratio: For a column fixed at one end and free at the other end, K = 2.
Thus,
L
2(4)(12)
KL
=
= 48.0
d
2
NFPA Timber Column Formulas: Since 26 6
KL
6 50, it is a long column. Apply
d
Eq. 13–29,
sallow =
=
540
ksi
(KL>d)2
540
48.02
= 0.234375 ksi
The allowable axial force is
Pallow = sallowA = 0.234375[2(4)] = 1.875 kip
Ans.
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13–107. The W14 * 53 structural A-36 steel column
supports an axial load of 80 kip in addition to an eccentric
load P. Determine the maximum allowable value of P based
on the AISC equations of Sec. 13.6 and Eq. 13–30. Assume
the column is fixed at its base, and at its top it is free to sway
in the x–z plane while it is pinned in the y–z plane.
z
80 kip
x
y
12 ft
Section Properties: For a W14 * 53 wide flange section.
A = 15.6 in2
Ix = 541 in4
d = 13.92 in.
rx = 5.89 in.
ry = 1.92 in.
Slenderness Ratio: By observation, the largest slenderness ratio is about y-y axis.
For a column fixed at one end and free at the other end, K = 2. Thus,
a
2(12)(12)
KL
b =
= 150
r y
1.92
Allowable Stress: The allowable stress can be determined using AISC Column
2p2[29(103)]
2p2E
KL
b =
=
= 126.1. Since
Formulas. For A–36 steel, a
r c
B sY
B
36
KL
KL
b …
… 200, the column is a long column. Applying Eq. 13–21,
a
r c
r
sallow =
12p2E
23(KL>r)2
12p2(29.0)(103)
=
23(1502)
= 6.637 ksi
Maximum Stress: Bending is about x-x axis. Applying we have
smax = sallow =
Mc
P
+
A
I
P(10) A 2
P + 80
+
15.6
541
13.92
6.637 =
B
P = 7.83 kip
Ans.
1127
P
y
x 10 in.
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*13–108. The W12 * 45 structural A-36 steel column
supports an axial load of 80 kip in addition to an eccentric
load of P = 60 kip. Determine if the column fails based on
the AISC equations of Sec. 13.6 and Eq. 13–30. Assume that
the column is fixed at its base, and at its top it is free to sway
in the x–z plane while it is pinned in the y–z plane.
z
80 kip
x
y
12 ft
Section Properties: For a W12 * 45 wide flange section,
A = 13.2 in2
d2 = 12.06 in.
Ix = 350 in4
rx = 5.15 in.
ry = 1.94 in.
Slenderness Ratio: By observation, the largest slenderness ratio is about y -y axis.
For a column fixed at one end and free at the other end, K = 2. Thus,
a
2(12)(12)
KL
b =
= 148.45
r y
1.94
Allowable Stress: The allowable stress can be determined using AISC Column
2p2[29(103)]
2p2E
KL
b =
=
= 126.1. Since
Formulas. For A–36 steel, a
r c
B sY
B
36
KL
KL
b …
… 200, the column is a long column. Applying Eq. 13–21,
a
r c
r
sallow =
12p2E
23(KL>r)2
12p2(29.0)(103)
=
23(148.452)
= 6.776 ksi
Maximum Stress: Bending is about x-x axis. Applying Eq. 1 we have
smax =
Mc
P
+
A
I
60(10) A 2
140
+
13.2
350
12.06
=
B
= 20.94 ksi
Since smax 7 sallow, the column is not adequate.
1128
P
y
x 10 in.
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•13–109.
The W14 * 22 structural A-36 steel column is
fixed at its top and bottom. If a horizontal load (not shown)
causes it to support end moments of M = 10 kip # ft,
determine the maximum allowable axial force P that can be
applied. Bending is about the x–x axis. Use the AISC
equations of Sec. 13.6 and Eq. 13–30.
P
x
M
y
y
x
12 ft
Section properties for W14 * 22:
A = 6.49 in2
d = 13.74 in2
Ix = 199 in4
ry = 1.04 in.
M
Allowable stress method:
P
0.5(12)(12)
KL
=
= 69.231
ry
1.04
a
KL
KL
2p2E
KL
2p2(29)(103)
b =
=
= 126.1,
6 a
b
r c
r
r c
B sY
B
36
y
Hence,
A
B
A KL
r Bc
- 18
KL 2
r
1
B 1 - 2 ¢ KL 2 ≤ R sY
r c
(sa)allow =
B 53 + 38
smax = (sa)allow =
KL
r
A
B
3
A KL
r B
3
A KL
r Bc
=
R
2
c1 - 12 A 69.231
126.1 B d36
1 69.231 3
c 53 + 38 A 69.231
126.1 B - 8 A 126.1 B d
= 16.510 ksi
My c
P
+
A
Ix
13.74
16.510 =
10(12)( 2 )
P
+
6.49
199
P = 80.3 kip
Ans.
1129
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13–110. The W14 * 22 column is fixed at its top and
bottom. If a horizontal load (not shown) causes it to support
end moments of M = 15 kip # ft, determine the maximum
allowable axial force P that can be applied. Bending is
about the x–x axis. Use the interaction formula with
1sb2allow = 24 ksi.
P
x
M
y
y
x
12 ft
Section Properties for W 14 * 22:
A = 6.49 in2
d = 13.74 in2
Ix = 199 in4
ry = 1.04 in.
M
P
Interaction method:
0.5(12)(12)
KL
=
= 69.231
ry
1.04
a
KL
2p2E
KL
KL
2p2(29)(103)
b =
=
= 126.1,
6 a
b
r c
ry
r c
B sY
B
36
Hence,
2
A KL
r B
B 1 - 12 ¢ KL 2 ≤ R sY
(sa)allow =
B 53 + 38
sa =
KL
r
A r Bc
A KL
r Bc
- 18
P
P
=
= 0.15408 P
A
6.49
15(12) A 2
Mxc
=
sb =
Ix
199
13.74
B
3
A KL
r B
3
A KL
r Bc
=
R
2
c1 - 12 A 69.231
126.1 B d36
1 69.231 3
c 53 + 38 A 69.231
126.1 B - 8 A 126.1 B d
= 16.510 ksi
= 6.214 ksi
sb
sa
+
= 1.0
(sa)allow
(sb)allow
0.15408 P
6.2141
+
= 1.0
16.510
24
P = 79.4 kip
Ans.
1130
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13–111. The W14 * 43 structural A-36 steel column
is fixed at its bottom and free at its top. Determine the
greatest eccentric load P that can be applied using
Eq. 13–30 and the AISC equations of Sec. 13.6.
40 kip
16 in.
10 ft
Section properties for W14 * 43:
A = 12.6 in2
d = 13.66 in.
Iy = 45.2 in4
ry = 1.89 in.
b = 7.995
Allowable stress method:
2(10)(12)
KL
=
= 126.98
ry
1.89
a
2p2E
KL
KL
KL
2p2 (29)(103)
b =
=
= 126.1, 200 7
7 a
b
r c
ry
r c
B sY
B
36
(sa)allow =
12p2(29)(103)
12p2E
=
= 9.26 ksi
2
23(KL>r)
23(126.98)2
smax = (sa)allow =
P
My c
P
+
A
Iy
P(16) A A B
P + 40
9.26 =
+
12.6
45.2
7.995
P = 4.07 kip
Ans.
1131
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*13–112. The W10 * 45 structural A-36 steel column is
fixed at its bottom and free at its top. If it is subjected to a
load of P = 2 kip, determine if it is safe based on the AISC
equations of Sec. 13.6 and Eq. 13–30.
40 kip
16 in.
10 ft
Section Properties for W10 * 45:
A = 13.3 in2
d = 10.10 in.
Iy = 53.4 in4
ry = 2.01 in.
b = 8.020 in.
Allowable stress method:
2.0(10)(12)
KL
=
= 119.4
ry
2.01
a
KL
2p2E
2p2(29)(103)
b =
=
= 126.1
r c
B sY
B
36
KL
KL
6 a
b
r
r c
(sa)allow Ú
(sa)allow =
c1 - 12
A
(KL>r)2
3
(KL>r)c
5
3 KL>r
3 + 8 KL>rc
My c
P
+
A
Iy
(KL>r)
B - 18(KL>r
3
c)
3
2(16) A 2
42
+
13.3
53.4
8.020
10.37 Ú
dsY
=
2
C 1 - 12 A 119.4
126.1 B D 36
5
3
3 + 8
1 119.4 3
A 119.4
136.1 B - 8 A 126.1 B
P
= 10.37 ksi
B
O.K.
10.37 Ú 5.56
Column is safe.
Yes.
Ans.
1132
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•13–113.
The A-36-steel W10 * 45 column is fixed at its
base. Its top is constrained to move along the x–x axis but
free to rotate about and move along the y–y axis. Determine
the maximum eccentric force P that can be safely supported
by the column using the allowable stress method.
12 in.
y
Section Properties. From the table listed in the appendix, the section properties for
a W10 * 45 are
bf = 8.02 in.
rx = 4.32 in.
Iy = 53.4 in4
ry = 2.01 in.
Slenderness Ratio. Here, Lx = 24(12) = 288 in. and for a column fixed at its base
and free at its top, Kx = 2. Thus,
a
2(288)
KL
b =
= 133.33 (controls)
r x
4.32
Since the column is fixed at its base and pinned at its top, Ky = 0.7 and
Ly = 24(12) = 288 in. Then,
a
0.7(288)
KL
b =
= 100.30
r y
2.01
Allowable Stress. The allowable stress will be determined using the AISC column
formulas. For A-36 steel,
2p2 C 29 A 103 B D
KL
2p2E
KL
KL
b =
=
= 126.10. Since a
b 6 a
b 6 200,
r c
r c
r x
B sY
C
36
the column is classified as a long column.
a
sallow =
=
12p2E
23(KL>r)2
12p2 C 29 A 103 B D
23(133.332)
= 8.400 ksi
Maximum Stress. Bending is about the weak axis. Since M = P(12) and
bf
8.02
=
= 4.01 in,
c =
2
2
sallow =
Mc
P
+
A
I
8.400 =
[P(12)](4.01)
P
+
13.3
53.4
y
x
24 ft
A = 13.3 in2
P
P = 8.604 kip = 8.60 kip
Ans.
1133
x
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13–114. The A-36-steel W10 * 45 column is fixed at its
base. Its top is constrained to move along the x–x axis but
free to rotate about and move along the y–y axis. Determine
the maximum eccentric force P that can be safely supported
by the column using an interaction formula. The allowable
bending stress is (sb)allow = 15 ksi.
12 in.
bf = 8.02 in.
rx = 4.32 in.
y
24 ft
Iy = 53.4 in4
ry = 2.01 in.
Slenderness Ratio. Here, Lx = 24(12) = 288 in and for a column fixed at its base
and free at its top, Kx = 2. Thus,
a
2(288)
KL
b =
= 133.33 (controls)
r x
4.32
Since the column is fixed at its base and pinned at its top, Ky = 0.7 and
Ly = 24(12) = 288 in. Then,
a
0.7(288)
KL
b =
= 100.30
r y
2.01
Allowable Stress. The allowable stress will be determined using the AISC column
2p2 C 29 A 103 B D
2p2E
KL
= 126.10. Since
formulas. For A-36 steel, a
b =
=
C
r c
B sY
36
a
KL
KL
b 6 a
b 6 200, the column is classified as a long column.
r c
r x
sallow =
=
12p2E
23(KL>r)2
12p2 C 29 A 103 B D
23 A 133.332 B
= 8.400 ksi
Interaction Formula. Bending is about the weak axis. Here, M = P(12) and
bf
8.02
=
= 4.01 in.
c =
2
2
P>A
Mc>Ar2
+
= 1
(sa)allow
(sb)allow
P>13.3
+
8.400
P(12)(4.01) n C 13.3 A 2.012 B D
15
y
x
Section Properties. From the table listed in the appendix, the section properties for
a W10 * 45 are
A = 13.3 in2
P
= 1
P = 14.57 kip = 14.6 kip
Ans.
14.57>13.3
sa
=
= 0.1304 6 0.15
(sa)allow
8.400
O.K.
1134
x
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13–115. The A-36-steel W12 * 50 column is fixed at its
base. Its top is constrained to move along the x–x axis but
free to rotate about and move along the y–y axis. If the
eccentric force P = 15 kip is applied to the column,
investigate if the column is adequate to support the loading.
Use the allowable stress method.
12 in.
bf = 8.08 in.
rx = 5.18 in.
y
24 ft
Iy = 56.3 in4
ry = 1.96 in.
Slenderness Ratio. Here, Lx = 24(12) = 288 in. and for a column fixed at its base
and pinned at its top, Kx = 2. Thus,
a
2(288)
KL
b =
= 111.20 (controls)
r x
5.18
Since the column is fixed at its base and pinned at its top, Ky = 0.7 and
Ly = 24(12) = 288 in. Thus,
a
0.7(288)
KL
b =
= 102.86
r y
1.96
Allowable Stress. The allowable stress will be determined using the AISC column
formulas. For A-36 steel, a
a
2p C 29 A 10 B D
KL
2p2E
= 126.10. Since
b =
=
C
r c A sY
36
2
3
KL
KL
b 6 a
b , the column can be classified as an intermediate column.
r x
r c
B1 sallow =
2(KL>r)C 2
R sY
(KL>r)3
3(KL>r)
5
+
3
8(KL>r)C
8(KL>r)C 3
C1 -
=
(KL>r)2
111.202
2 A 126.102 B
S(36)
3(111.20)
5
111.203
+
3
8(126.10)
8 A 126.103 B
= 11.51 ksi
Maximum Stress. Bending is about the weak axis. Since, M = 15(12) = 180 kip # in.
bf
8.08
and c =
=
= 4.04 in.,
2
2
smax =
y
x
Section Properties. From the table listed in the appendix, the section properties for
a W12 * 50 are
A = 14.7 in2
P
180(4.04)
P
Mc
15
+
=
+
= 13.94 ksi
A
I
14.7
56.3
Since smax 7 sallow, the W12 * 50 column is inadequate according to the allowable
stress method.
1135
x
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*13–116. The A-36-steel W12 * 50 column is fixed at its
base. Its top is constrained to move along the x–x axis but
free to rotate about and move along the y–y axis. If the
eccentric force P = 15 kip is applied to the column,
investigate if the column is adequate to support the loading.
Use the interaction formula. The allowable bending stress is
(sb)allow = 15 ksi.
12 in.
bf = 8.08 in.
rx = 5.18 in.
y
24 ft
ry = 1.96 in.
Slenderness Ratio. Here, Lx = 24(12) = 288 in and for a column fixed at its base
and pinned at its top, Kx = 2. Thus,
a
2(288)
KL
b =
= 111.20 (controls)
r x
5.18
Since the column is fixed at its base and pinned at its top, Ky = 0.7 and
Ly = 24(12) = 288 in. Then,
a
Allowable
=
C
Axial
2p2 C 29 A 103 B D
36
0.7(288)
KL
b =
= 102.86
r y
1.96
Stress.
For
= 126.10. Since a
A-36
steel,
a
KL
2p2E
b =
r c A sY
KL
KL
b 6 a
b , the column can be
r x
r c
classified as an intermediate column.
C1 sallow =
2(KL>r)c 2
SsY
3(KL>r)
(KL>r)3
5
+
3
8(KL>r)c
8(KL>r)c 3
C1 -
=
(KL>r)2
111.202
2 A 126.102 B
S(36)
3(111.20)
5
111.203
+
3
8(126.10)
8 A 126.103 B
= 11.51 ksi
Interaction Formula. Bending is about the weak axis. Here, M = 15(12)
bf
8.08
= 180 kip # in. and c =
=
= 4.04 in.
2
2
P>A
Mc>Ar2
15>14.7
+
+
=
(sa)allow
(sb)allow
11.51
180(4.04) n C 14.7 A 1.962 B D
15
y
x
Section Properties. From the table listed in the appendix, the section properties for
a W12 * 50 are
A = 14.7 in2
P
= 0.9471 6 1
P = 14.57 kip = 14.6 kip
Ans.
15>14.7
sa
=
= 0.089 6 0.15
(sa)allow
11.51
O.K.
Thus, a W12 * 50 column is adequate according to the interaction formula.
1136
x
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•13–117.
A 16-ft-long column is made of aluminum alloy
2014-T6. If it is fixed at its top and bottom, and a
compressive load P is applied at point A, determine the
maximum allowable magnitude of P using the equations of
Sec. 13.6 and Eq. 13–30.
P
A
4.25 in.
x
0.5 in.
y
8 in.
Section properties:
A = 2(0.5)(8) + 8(0.5) = 12 in2
Ix =
1
1
(8)(93) (7.5)(83) = 166 in4
12
12
Iy = 2 a
ry =
1
1
b(0.5)(83) +
(8)(0.53) = 42.75 in4
12
12
Iy
42.75
=
= 1.8875 in.
AA
A 12
Allowable stress method:
0.5(16)(12)
KL
KL
= 50.86, 12 6
=
6 55
ry
ry
1.8875
sallow = c30.7 - 0.23a
KL
bd
r
= [30.7 - 0.23(50.86)] = 19.00 ksi
smax = sallow =
19.00 =
Mx c
P
+
A
Ix
P(4.25)(4.5)
P
+
12
166
P = 95.7 kip
Ans.
1137
y
x 8 in.
0.5 in.
0.5 in.
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13–118. A 16-ft-long column is made of aluminum alloy
2014-T6. If it is fixed at its top and bottom, and a
compressive load P is applied at point A, determine the
maximum allowable magnitude of P using the equations of
Sec. 13.6 and the interaction formula with 1sb2allow = 20 ksi.
P
A
4.25 in.
x
0.5 in.
y
8 in.
Section Properties:
A = 2(0.5)(8) + 8(0.5) = 12 in2
Ix =
1
1
(8)(93) (7.5)(83) = 166 in4
12
12
Iy = 2 a
ry =
1
1
b(0.5)(83) +
(8)(0.53) = 42.75 in4
12
12
Iy
42.75
=
= 1.8875 in.
AA
A 12
Interaction method:
0.5(16)(12)
KL
KL
= 50.86, 12 6
=
6 55
ry
ry
1.8875
sallow = c30.7 - 0.23a
KL
bd
r
= [30.7 - 0.23(50.86)]
= 19.00 ksi
sa =
P
P
=
= 0.08333P
A
12
sb =
P(4.25)(4.50)
Mc
=
= 0.1152P
Ix
166
sb
sa
+
= 1.0
(sa)allow
(sb)allow
0.08333P
0.1152P
+
= 1
19.00
20
P = 98.6 kip
Ans.
1138
y
x 8 in.
0.5 in.
0.5 in.
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13–119. The 2014-T6 hollow column is fixed at its base
and free at its top. Determine the maximum eccentric
force P that can be safely supported by the column. Use the
allowable stress method. The thickness of the wall for the
section is t = 0.5 in.
6 in.
P
3 in.
6 in.
8 ft
Section Properties.
A = 6(3) - 5(2) = 8 in2
Ix =
1
1
(3) A 63 B (2) A 53 B = 33.1667 in4
12
12
rx =
33.1667
Ix
=
= 2.036 in.
AA
A
8
Iy =
1
1
(6) A 33 B (5) A 23 B = 10.1667 in4
12
12
ry =
Iy
10.1667
=
= 1.127 in.
AA
A
8
Slenderness Ratio. For a column fixed at its base and free at its top, K = 2. Thus,
a
2(8)(12)
KL
b =
= 170.32
r y
1.127
Allowable Stress. Since a
KL
b 7 55, the column can be classified as a long
r y
column.
sallow =
54 000 ksi
54 000 ksi
=
= 1.862 ksi
(KL>r)2
170.312
Maximum Stress. Bending occurs about the strong axis so that M = P(6) and
6
c = = 3 in.
2
sallow =
1.862 =
Mc
P
+
A
I
C P(6) D (3)
P
+
8
33.1667
P = 2.788 kip = 2.79 kip
Ans.
1139
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*13–120. The 2014-T6 hollow column is fixed at its base
and free at its top. Determine the maximum eccentric force
P that can be safely supported by the column. Use the
interaction formula. The allowable bending stress is
(sb)allow = 30 ksi. The thickness of the wall for the section is
t = 0.5 in.
6 in.
P
3 in.
6 in.
8 ft
Section Properties.
A = 6(3) - 5(2) = 8 in2
Ix =
1
1
(3) A 63 B (2) A 53 B = 33.1667 in4
12
12
rx =
33.1667
Ix
=
= 2.036 in.
AA
A
8
Iy =
1
1
(6) A 33 B (5) A 23 B = 10.1667 in4
12
12
ry =
Iy
10.1667
=
= 1.127 in.
AA
A
8
Slenderness Ratio. For a column fixed at its base and pinned at its top, K = 2. Thus,
a
2(8)(12)
KL
b =
= 170.32
r y
1.127
KL
b 7 55, the column can be classified as the column is
r y
classified as a long column.
Allowable Stress. Since a
sallow =
54000 ksi
54000 ksi
=
= 1.862 ksi
(KL>r)2
170.312
Interaction Formula. Bending is about the strong axis. Since M = P(6) and
6
c = = 3 in,
2
P>A
Mc>Ar2
+
= 1
(sa)allow
(sb)allow
P>8
+
1.862
[P(6)](3) n C 8 A 2.0362 B D
30
= 1
P = 11.73 kip = 11.7 kip
Ans.
1140
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•13–121. The 10-ft-long bar is made of aluminum alloy
2014-T6. If it is fixed at its bottom and pinned at the top,
determine the maximum allowable eccentric load P that can
be applied using the formulas in Sec. 13.6 and Eq. 13–30.
P
x
1.5 in.
1.5 in.
x
Section Properties:
A = 6(4) = 24.0 in2
Ix =
1
(4) A 63 B = 72.0 in4
12
Iy =
1
(6) A 43 B = 32.0 in4
12
ry =
Iy
32.0
=
= 1.155 in.
AA
A 24
Slenderness Ratio: The largest slenderness ratio is about y-y axis. For a column
pinned at one end fixed at the other end, K = 0.7. Thus,
a
0.7(10)(12)
KL
b =
= 72.75
r y
1.155
Allowable Stress: The allowable stress can be determined using aluminum
KL
(2014 –T6 alloy) column formulas. Since
7 55, the column is classified as a long
r
column. Applying Eq. 13–26,
sallow = c
=
54 000
d ksi
(KL>r)2
54 000
72.752
= 10.204 ksi
Maximum Stress: Bending is about x-x axis. Applying Eq. 13–30, we have
smax = sallow =
10.204 =
P
Mc
+
A
I
P(1.5)(3)
P
+
24.0
72.0
P = 98.0 kip
Ans.
1141
y
3 in.
2 in.
y
2 in.
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13–122. The 10-ft-long bar is made of aluminum alloy
2014-T6. If it is fixed at its bottom and pinned at the top,
determine the maximum allowable eccentric load P that
can be applied using the equations of Sec. 13.6 and the
interaction formula with 1sb2allow = 18 ksi.
P
x
1.5 in.
1.5 in.
x
Section Properties:
A = 6(4) = 24.0 in2
Ix =
1
(4) A 63 B = 72.0 in4
12
Iy =
1
(6) A 43 B = 32.0 in4
12
rx =
Ix
72.0
=
= 1.732 in.
AA
A 24.0
ry =
Iy
32.0
=
= 1.155 in.
AA
A 24.0
Slenderness Ratio: The largest slenderness radio is about y-y axis. For a column
pinned at one end and fixed at the other end, K = 0.7. Thus
a
0.7(10)(12)
KL
b =
= 72.75
r y
1.155
Allowable Stress: The allowable stress can be determined using aluminum
KL
(2014 –T6 alloy) column formulas. Since
7 55, the column is classified as a long
r
column. Applying Eq. 13–26,
(sa)allow = c
=
54 000
d ksi
(KL>r)2
54 000
72.752
= 10.204 ksi
Interaction Formula: Bending is about x-x axis. Applying Eq. 13–31, we have
Mc>Ar2
P>A
+
= 1
(sa)allow
(sb)allow
P(1.5)(3)>24.0(1.7322)
P>24.0
+
= 1
10.204
18
P = 132 kip
Ans.
1142
y
3 in.
2 in.
y
2 in.
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13–123. The rectangular wooden column can be
considered fixed at its base and pinned at its top. Also, the
column is braced at its mid-height against the weak axis.
Determine the maximum eccentric force P that can be safely
supported by the column using the allowable stress method.
6 in.
6 in.
Section Properties.
Ix =
dx = 6 in.
5 ft
1
(3) A 63 B = 54 in4
12
dy = 3 in.
Slenderness Ratio. Here, Lx = 10(12) = 120 in. and for a column fixed at its base
and pinned at its top, K = 0.7. Thus,
a
0.7(120)
KL
b =
= 14
d x
6
Since the bracing provides support equivalent to a pin, Ky = 1 and
Ly = 5(12) = 60 in. Then
a
1(60)
KL
b =
= 20 (controls)
d y
3
KL
6 26, the column can be classified as the column
d
is classified as an intermediate column.
Allowable Stress. Since 11 6
sallow = 1.20c 1 = 1.20 c 1 -
1 KL>d 2
a
b d ksi
3 26.0
1 20 2
a
b d ksi = 0.9633 ksi
3 26.0
Maximum Stress. Bending occurs about the strong axis. Here, M = P(6) and
6
c = = 3 in.
2
sallow =
P
Mc
+
A
I
0.9633 =
[P(6)](3)
P
+
18
54
6 in.
3 in.
5 ft
A = 6(3) = 18 in2
P
P = 2.477 kip = 2.48 kip
Ans.
1143
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*13–124. The rectangular wooden column can be
considered fixed at its base and pinned at its top. Also, the
column is braced at its mid-height against the weak axis.
Determine the maximum eccentric force P that can be
safely supported by the column using the interaction
formula. The allowable bending stress is (sb)allow = 1.5 ksi.
6 in.
6 in.
5 ft
Section Properties.
rx =
Ix =
1
(3) A 63 B = 54 in4
12
Ix
54
=
= 1.732 in.
AA
A 18
dx = 6 in.
dy = 3 in.
Slenderness Ratio. Here, Lx = 10(12) = 120 in. and for a column fixed at its base
pinned at its top, K = 0.7. Thus,
a
0.7(120)
KL
b =
= 14
d x
6
Since the bracing provides support equivalent to a pin, Ky = 1 and
Ly = 5(12) = 60 in. Then
a
1(60)
KL
b =
= 20 (controls)
d y
3
KL
6 26, the column can be classified as the
d
column is classified as an intermediate column.
Allowable Axial Stress. Since 11 6
sallow = 1.20c 1 = 1.20 c 1 -
1 KL>d 2
a
b d ksi
3 26.0
1 20 2
a
b d ksi = 0.9633 ksi
3 26.0
Interaction Formula. Bending occurs about the strong axis. Since M = P(6) and
6
c = = 3 in.
2
P>A
Mc>Ar2
+
= 1
(sa)allow
(sb)allow
P>18
+
0.9633
[P(6)](3) n C 18 A 1.7322 B D
1.5
6 in.
3 in.
5 ft
A = 6(3) = 18 in2
P
= 1
P = 3.573 kip = 3.57 kip
Ans.
1144
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•13–125.
The 10-in.-diameter utility pole supports the
transformer that has a weight of 600 lb and center of gravity
at G. If the pole is fixed to the ground and free at its
top, determine if it is adequate according to the NFPA
equations of Sec. 13.6 and Eq. 13–30.
G
15 in.
18 ft
2(18)(12)
KL
=
= 43.2 in.
d
10
26 6 43.2 … 50
Use Eq. 13–29,
sallow =
540
540
=
= 0.2894 ksi
(KL>d)
(43.2)2
smax =
Mc
P
+
A
I
smax =
(600)(15)(5)
600
+
2
p (5)
A p B (5)4
4
smax = 99.31 psi 6 0.289 ksi
O.K.
Yes.
Ans.
1145
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13–126. Using the NFPA equations of Sec. 13.6 and
Eq. 13–30, determine the maximum allowable eccentric
load P that can be applied to the wood column. Assume that
the column is pinned at both its top and bottom.
P
0.75 in.
6 in.
3 in.
12 ft
Section Properties:
A = 6(3) = 18.0 in2
Iy =
1
(6) A 33 B = 13.5 in4
12
Slenderness Ratio: For a column pinned at both ends, K = 1.0. Thus,
a
1.0(12)(12)
KL
b =
= 48.0
d y
3
Allowable Stress: The allowable stress can be determined using NFPA timber
KL
column formulas. Since 26 6
6 50, it is a long column. Applying Eq. 13–29,
d
sallow =
=
540
ksi
(KL>d)2
540
= 0.234375 ksi
48.02
Maximum Stress: Bending is about y -y axis. Applying Eq. 13–30, we have
smax = sallow =
0.234375 =
P
Mc
+
A
I
P(0.75)(1.5)
P
+
18.0
13.5
P = 1.69 kip
Ans.
1146
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13–127. Using the NFPA equations of Sec. 13.6 and
Eq. 13–30, determine the maximum allowable eccentric
load P that can be applied to the wood column. Assume that
the column is pinned at the top and fixed at the bottom.
P
0.75 in.
6 in.
3 in.
12 ft
Section Properties:
A = 6(3) = 18.0 in2
Iy =
1
(6) A 33 B = 13.5 in4
12
Slenderness Ratio: For a column pinned at one end and fixed at the other end,
K = 0.7. Thus,
a
0.7(12)(12)
KL
b =
= 33.6
d y
3
Allowable Stress: The allowable stress can be determined using NFPA timber
KL
column formulas. Since 26 6
6 50, it is a long column. Applying Eq. 13–29,
d
sallow =
=
540
ksi
(KL>d)2
540
= 0.4783 ksi
33.62
Maximum Stress: Bending is about y -y axis. Applying Eq. 13–30, we have
smax = sallow =
0.4783 =
Mc
P
+
A
I
P(0.75)(1.5)
P
+
18.0
13.5
P = 3.44 kip
Ans.
1147
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*13–128. The wood column is 4 m long and is required to
support the axial load of 25 kN. If the cross section is square,
determine the dimension a of each of its sides using a factor
of safety against buckling of F.S. = 2.5. The column is
assumed to be pinned at its top and bottom. Use the Euler
equation. Ew = 11 GPa, and sY = 10 MPa.
25 kN
a
4m
a
1
a4
(a) A a3 B =
, P = (2.5)25 = 62.5 kN and K = 1
12
12 cr
for pin supported ends column. Applying Euler’s formula,
Critical Buckling Load: I =
Pcr =
62.5 A 10 B =
3
p2EI
(KL)2
a
p2(11)(109) A 12
B
4
[1(4)]2
a = 0.1025 m = 103 mm
Ans.
Critical Stress: Euler’s formula is only valid if scr 6 sY.
scr =
62.5(103)
Pcr
=
= 5.94 MPa 6 s Y = 10 MPa
A
0.1025(0.1025)
1148
O.K.
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•13–129. If the torsional springs attached to ends A and
C of the rigid members AB and BC have a stiffness k,
determine the critical load Pcr.
P
k
A
L
2
B
Equilibrium. When the system is given a slight lateral disturbance, the configuration
shown in Fig. a is formed. The couple moment M can be related to P by considering
the equilibrium of members AB and BC.
Member AB
+ c ©Fy = 0;
a + ©MA = 0;
By - P = 0
(1)
By a
(2)
L
L
sin u b + Bx a cos ub - M = 0
2
2
Member BC
a + ©MC = 0; -By a
L
L
sin u b + Bx a cos ub + M = 0
2
2
(3)
Solving Eqs. (1), (2), and (3), we obtain
Bx = 0
By =
2M
L sin u
M =
PL
sin u
2
Since u is very small, the small angle analysis gives sin u ⬵ u. Thus,
M =
PL
u
2
(4)
Torslonal Spring Moment. The restoring couple moment Msp can be determined
using the torsional spring formula, M = ku. Thus,
Msp = ku
Critical Buckling Load. When the mechanism is on the verge of bucklling M must
equal Msp.
M = Msp
Pcr L
u = ku
2
Pcr =
2k
L
Ans.
1149
L
2
k
C
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13–129.
Continued
1150
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13–130. Determine the maximum intensity w of the
uniform distributed load that can be applied on the beam
without causing the compressive members of the supporting
truss to buckle. The members of the truss are made from
A-36-steel rods having a 60-mm diameter. Use F.S. = 2
against buckling.
w
B
A
C
2m
Equilibrium. The force developed in member BC can be determined by considering
the equilibrium of the free-body diagram of the beam AB, Fig. a.
3
w(5.6)(2.8) - FBC a b(5.6) = 0 FBC = 4.6667w
5
a + ©MA = 0;
The Force developed in member CD can be obtained by analyzing the equilibrium
of joint C, Fig. b,
+ c ©Fy = 0;
FAC a
5
3
b - 4.6667wa b = 0
13
5
FAC = 7.28w (T)
+
: ©Fx = 0;
4
12
4.6667wa b + 7.28a b w-FCD = 0
5
13
FCD = 10.4533w (C)
Section Properties. The cross-sectional area and moment of inertia of the solid
circular rod CD are
A = p A 0.032 B = 0.9 A 10 - 3 B p m2
I =
p
A 0.034 B = 0.2025 A 10 - 6 B p m4
4
Critical Buckling Load. Since both ends of member CD are pinned, K = 1. The
critical buckling load is
Pcr = FCD (F.S.) = 10.4533w(2) = 20.9067w
Applying Euler’s formula,
Pcr =
p2EI
(KL)2
20.9067w =
p2 C 200 A 109 B D C 0.2025 A 10 - 6 B p D
[1(3.6)]2
Ans.
w = 4634.63 N>m = 4.63 kN>m
Critical Stress: Euler’s formula is valid only if scr 6 sY.
scr =
20.907(4634.63)
Pcr
=
= 34.27 MPa 6 sY = 250 MPa
A
0.9 A 10 - 3 B p
1151
O.K.
3.6 m
D
1.5 m
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13–131. The W10 * 45 steel column supports an axial load
of 60 kip in addition to an eccentric load P. Determine the
maximum allowable value of P based on the AISC equations
of Sec. 13.6 and Eq. 13–30. Assume that in the x–z plane
Kx = 1.0 and in the y–z plane Ky = 2.0. Est = 2911032 ksi,
sY = 50 ksi.
z
P
60 kip
x
y
10 ft
Section properties for W 10 * 45:
A = 13.3 in2
d = 10.10 in.
rx = 4.32 in.
ry = 2.01 in.
Ix = 248 in4
Allowable stress method:
a
1.0(10)(12)
KL
b =
= 27.8
r x
4.32
a
2.0(10)(12)
KL
b =
= 119.4
r y
2.01
a
2p2(29)(103)
KL
2p2E
b =
=
= 107
r c
B sg
B
50
(controls)
KL
KL
7 a
b
r
r c
(sa)allow =
12p2(29)(103)
12p2E
=
= 10.47 ksi
2
23(KL>r)
23(119.4)4
smax = (sa)allow =
Mc
P
+
A
I
P(8) A 2
P + 60
+
10.47 =
13.3
248
10.10
B
P = 25.0 kip
Ans.
1152
y
x 8 in.
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*13–132. The A-36-steel column can be considered pinned
at its top and fixed at its base. Also, it is braced at its
mid-height along the weak axis. Investigate whether a
W250 * 45 section can safely support the loading shown.
Use the allowable stress method.
600 mm
10 kN
4.5 m
Section Properties. From the table listed in the appendix, the necessary section
properties for a W250 * 45 are
A = 5700 mm2 = 5.70 A 10 - 3 B m2
d = 266 mm = 0.266 m
Ix = 71.1 A 106 B mm4 = 71.1 A 10 - 6 B m4
rx = 112 mm = 0.112 m
ry = 35.1 mm = 0.0351 mm
Slenderness Ratio. Here, Lx = 9 m and for a column fixed at its base and pinned at
its top, Kx = 0.7. Thus,
¢
0.7(9)
KL
= 56.25
≤ =
r x
0.112
Since the bracing provides support equivalent to a pin, Ky = 1 and Ly = 4.5 m.
Then,
¢
1(4.5)
KL
= 128.21 (controls)
≤ =
r y
0.0351
Allowable Stress. For A-36 steel, a
Since a
2p C 200 A 10 B D
KL
2p2E
b =
=
= 125.66.
r c
C 250 A 106 B
B sY
2
9
KL
KL
b 6 a
b 6 200, the column can be classified as a long column.
r c
r y
12p C 200 A 10 B D
12p2E
=
= 62.657 MPa
23(KL>r)2
23(128.21)2
2
sallow =
9
Maximum Stress. Bending occurs about the strong axis. Here, P = 10 + 40
0.266
d
= 50 kN, M = 40(0.6) = 24 kN # m and c =
=
= 0.133 m,
2
2
smax =
50 A 103 B
24 A 103 B (0.133)
P
Mc
+
=
+
= 53.67 MPa
A
I
5.70 A 10 - 3 B
71.1 A 10 - 6 B
Since smax 6 sallow, the column is adequate according to the allowable stress
method.
1153
4.5 m
40 kN
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•13–133.
The A-36-steel column can be considered pinned
at its top and fixed at its base. Also, it is braced at
its mid-height along the weak axis. Investigate whether a
W250 * 45 section can safely support the loading shown.
Use the interaction formula. The allowable bending stress is
(sb)allow = 100 MPa.
600 mm
10 kN
4.5 m
Section Properties. From the table listed in the appendix, the necessary section
properties for a W250 * 45 are
A = 5700 mm2 = 5.70 A 10 - 3 B m2
Ix = 71.1 A 10 B mm = 71.1 A 10
6
4
-6
d = 266 mm = 0.266 m
4.5 m
Bm
4
rx = 112 mm = 0.112 m
ry = 35.1 mm = 0.0351 mm
Slenderness Ratio. Here, Lx = 9 m and for a column fixed at its base and pinned at
its top, Kx = 0.7. Thus,
a
0.7(9)
KL
b =
= 56.25
r x
0.112
Since the bracing provides support equivalent to a pin, Ky = 1 and Ly = 4..5 m.
Then,
a
Allowable
=
C
1(4.5)
KL
b =
= 128.21 (controls)
r y
0.0351
Axial
2p2 C 200 A 109 B D
250 A 10 B
6
Stress.
For
= 125.66. Since a
A–36
steel,
a
KL
2p2E
b =
r c
B sY
KL
KL
b 6 a
b 6 200, the column can be
r c
r y
classified as a long column.
12p C 200 A 10 B D
12p2E
=
= 62.657 MPa
2
23(KL>r)
23(128.21)2
2
sallow =
9
Interaction Formula. Bending is about the strong axis. Here, P = 10 + 40 = 50 kN,
d
0.266
M = 40(0.6) = 24 kN # m and c =
=
= 0.133 m,
2
2
Mc>Ar2
P>A
+
=
(sa)allow
(sb)allow
50 A 103 B n 5.70 A 10 - 3 B
62.657 A 106 B
+
24 A 103 B (0.133) n C 5.70 A 10 - 3 B A 0.1122 B D
100 A 106 B
= 0.5864 6 1
sa
=
(sa)allow
50 A 103 B n 5.7 A 10 - 3 B
62.657 A 106 B
O.K.
= 0.140 6 0.15
O.K.
Thus, a W250 * 45 column is adequate according to the interaction formula.
1154
40 kN
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13–134. The member has a symmetric cross section. If it is
pin connected at its ends, determine the largest force it can
support. It is made of 2014-T6 aluminum alloy.
0.5 in.
2 in.
P
5 ft
Section properties:
A = 4.5(0.5) + 4(0.5) = 4.25 in2
P
1
1
(0.5)(4.53) +
(4)(0.5)3 = 3.839 in4
I =
12
12
r =
I
3.839
=
= 0.9504 in.
AA
A 4.25
Allowable stress:
1.0(5)(12)
KL
=
= 63.13
r
0.9504
KL
7 55
r
Long column
sallow =
54000
54000
=
= 13.55 ksi
(KL>r)2
63.132
Pallow = sallowA
= 13.55(4.25) = 57.6 kip
Ans.
1155
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13–135. The W200 * 46 A-36-steel column can be
considered pinned at its top and fixed at its base. Also, the
column is braced at its mid-height against the weak axis.
Determine the maximum axial load the column can support
without causing it to buckle.
6m
Section Properties. From the table listed in the appendix, the section properties for
a W200 * 46 are
A = 5890 mm2 = 5.89 A 10 - 3 B m2
Iy = 15.3 A 106 B mm4 = 15.3 A 10 - 6 B m4
Ix = 45.5 A 106 B mm4 = 45.5 A 10 - 6 B m4
Critical Buckling Load. For buckling about the strong axis, Kx = 0.7 and Lx = 12 m.
Since the column is fixed at its base and pinned at its top,
Pcr =
p2EIx
(KL)x 2
=
p2 c200 A 109 B d c45.5 A 10 - 6 B d
[0.7(12)]2
= 1.273 A 106 B N = 1.27 MN
For buckling about the weak axis, Ky = 1 and Ly = 6 m since the bracing provides
a support equivalent to a pin. Applying Euler’s formula,
Pcr =
p2EIy
(KL)y 2
=
p2 c200 A 109 B d c15.3 A 10 - 6 B d
[1(6)]2
= 838.92 kN = 839 kN (controls)Ans.
Critical Stress. Euler’s formula is valid only if scr 6 sY.
838.92 A 10 B
Pcr
=
= 142.43 MPa 6 sY = 250 MPa
A
5.89 A 10 - 3 B
3
scr =
1156
O.K.
6m
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*13–136. The structural A-36 steel column has the cross
section shown. If it is fixed at the bottom and free at the top,
determine the maximum force P that can be applied at A
without causing it to buckle or yield. Use a factor of safety
of 3 with respect to buckling and yielding.
P
20 mm
A
10 mm
Section properties:
100 mm
-3
2
4m
©A = 0.2(0.01) + 0.15 (0.01) + 0.1(0.01) = 4.5(10 ) m
150 mm
A
100 mm
10 mm
0.005 (0.2)(0.01) + 0.085 (0.15)(0.01) + 0.165 (0.1)(0.01)
©xA
x =
=
©A
4.5(10 - 3)
= 0.06722 m
1
(0.2)(0.013) + 0.2 (0.01)(0.06722 - 0.005)2
12
Iy =
+
1
(0.01)(0.153) + 0.01 (0.15)(0.085 - 0.06722)2
12
+
1
(0.1)(0.013) + 0.1 (0.01)(0.165 - 0.06722)2
12
= 20.615278 (10 - 6) m4
1
1
1
(0.01)(0.23) +
(0.15)(0.013) +
(0.01)(0.13)
12
12
12
Ix =
= 7.5125 (10 - 6) m4
ry =
Iy
20.615278(10 - 6)
=
= 0.0676844
AA
A
4.5 (10 - 3)
Buckling about x-x axis:
Pcr =
p2(200)(109)(7.5125)(10 - 6)
p2 EI
=
2
(KL)
[2.0(4)]2
= 231.70 kN
scr =
(controls)
231.7 (103)
Pcr
= 51.5 MPa 6 sg = 250 MPa
=
A
4.5 (10 - 3)
Yielding about y-y axis:
smax =
P
ec
KL P
c1 + 2 sec a
b d;
A
2r A EA
r
e = 0.06722 - 0.02 = 0.04722 m
0.04722 (0.06722)
ec
=
= 0.692919
0.0676844
r2
2.0 (4)
P
P
KL
=
= 1.96992 P (10 - 3) 2P
2r A EA
2(0.0676844) A 200 (109)(4.5)(10 - 3)
250(106)(4.5)(10 - 3) = P[1 + 0.692919 sec (1.96992P (10 - 3)2P)]
By trial and error:
P = 378.45 kN
Hence,
Pallow =
231.70
= 77.2 kN
3
Ans.
1157
10 mm
100 mm
13 Solutions 46060
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•13–137.
The structural A-36 steel column has the cross
section shown. If it is fixed at the bottom and free at the top,
determine if the column will buckle or yield when the load
P = 10 kN. Use a factor of safety of 3 with respect to
buckling and yielding.
P
20 mm
A
10 mm
100 mm
Section properties:
4m
©A = 0.2 (0.01) + 0.15 (0.01) + 0.1 (0.01) = 4.5 (10 - 3) m2
0.005 (0.2)(0.01) + 0.085 (0.15)(0.01) + 0.165 (0.1)(0.01)
©xA
= 0.06722 m
=
©A
4.5 (10 - 3)
Iy =
1
(0.2)(0.013) + 0.2 (0.01)(0.06722 - 0.005)2
12
ry =
+
1
(0.01)(0.153) + 0.01 (0.15)(0.085 - 0.06722)2
12
+
1
(0.1)(0.013) + 0.1 (0.01)(0.165 - 0.06722)2 = 20.615278 (10 - 6) m4
12
1
1
1
(0.01)(0.23) +
(0.15)(0.013) +
(0.01)(0.13) = 7.5125 (10 - 6) m4
12
12
12
Iy
BA
20.615278 (10 - 6)
=
B
4.5 (10 - 3)
= 0.067843648 m
Buckling about x-x axis:
Pcr =
p2(200)(109)(7.5125)(10 - 6)
p2 EI
=
= 231.70 kN
2
(KL)
[2.0(4)]2
scr =
231.7 (103)
Pcr
= 51.5 MPa 6 sg = 250 MPa
=
A
4.5 (10 - 3)
Pallow =
O.K.
Pcr
231.7
=
= 77.2 kN 7 P = 10 kN
FS
3
Hence the column does not buckle.
Yielding about y-y axis:
smax =
P =
P
KL
ec
P
bd
c1 + 2 sec a
A
2r A EA
r
A
100 mm
10 mm
x =
Ix =
150 mm
e = 0.06722 - 0.02 = 0.04722 m
10
= 3.333 kN
3
3.333 (103)
P
= 0.7407 MPa
=
A
4.5 (10 - 3)
0.04722 (0.06722)
ec
= 0.689815
=
(0.067844)
r2
2.0 (4)
P
KL
3.333 (103)
=
= 0.1134788
2 r AE A
2(0.06783648) A 200 (109)(4.5)(10 - 3)
smax = 0.7407 [1 + 0.692919 sec (0.1134788)] = 1.25 MPa 6 sg = 250 MPa
Hence the column does not yield!
No.
Ans.
1158
10 mm
100 mm
FM_TOC 46060
6/22/10
11:26 AM
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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sy
14–1. A material is subjected to a general state of plane
stress. Express the strain energy density in terms of the
elastic constants E, G, and n and the stress components sx ,
sy , and txy .
txy
sx
‘Strain Energy Due to Normal Stresses: We will consider the application of normal
stresses on the element in two successive stages. For the first stage, we apply only sx
on the element. Since sx is a constant, from Eq. 14-8, we have
s2x
s2x V
dV =
2E
Lv 2E
(Ui)1 =
When sy is applied in the second stage, the normal strain ex will be strained by
ex ¿ = - vey = -
vsy
E
. Therefore, the strain energy for the second stage is
(Ui)2 =
=
s2y
¢
Lv 2E
B
s2y
Lv 2E
+ sx ex ¿ ≤ dV
+ sx a -
vsy
E
b R dV
Since sx and sy are constants,
(Ui)2 =
V
(s2y - 2vsx sy)
2E
Strain Energy Due to Shear Stresses: The application of txy does not strain the
element in normal direction. Thus, from Eq. 14–11, we have
(Ui)3 =
t2xy
Lv 2G
dV =
t2xy V
2G
The total strain energy is
Ui = (Ui)1 + (Ui)2 + (Ui)3
2
=
txy V
s2x V
V
+
(s2y - 2vsx sy) +
2E
2E
2G
=
t2xy V
V
(s2x + s2y - 2vsx sy) +
2E
2G
and the strain energy density is
t2xy
Ui
1
=
(s2x + s2y - 2vsx sy) +
V
2E
2G
Ans.
1159
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14–2. The strain-energy density must be the same whether
the state of stress is represented by sx , sy , and txy , or
by the principal stresses s1 and s2 . This being the case,
equate the strain–energy expressions for each of these two
cases and show that G = E>[211 + n2].
U =
1
v
1 2
(s2x + s2y) - sxsy +
t R dV
E
2 G xy
Lv 2 E
U =
1
v
(s21 + s22) s s R dV
B
E 1 2
Lv 2 E
B
Equating the above two equations yields.
v
1 2
1
v
1
(s2x + s2y) sxsy +
txy =
(s21 + s22) s s
2E
E
2G
2E
E 1 2
However, s1, 2 =
sx + sy
2
;
A
a
sx - sy
2
(1)
2
b + txy
2
Thus, A s21 + s22 B = s2x + s2y + 2 t2xy
s1 s2 = sxsy - t2xy
Substitute into Eq. (1)
v
1 2
1
v
v 2
1
t =
(s2 + s2y + 2t2xy) ss +
t
A s2 + s2y B - sxsy +
2E x
E
2 G xy
2E x
E x y
E xy
2
txy
v 2
1 2
txy =
+
txy
2G
E
E
1
v
1
=
+
2G
E
E
1
1
=
(1 + v)
2G
E
G =
E
2(1 + v)
QED
1160
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14–3. Determine the strain energy in the stepped
rod assembly. Portion AB is steel and BC is
brass. Ebr = 101 GPa, Est = 200 GPa, (sY)br = 410 MPa,
(sY)st = 250 MPa.
100 mm
A
B
30 kN
30 kN
1.5 m
Referring to the FBDs of cut segments in Fig. a and b,
+ ©F = 0;
:
x
NBC - 20 = 0
+ ©F = 0;
:
x
NAB - 30 - 30 - 20 = 0
NBC = 20 kN
NAB = 80 kN
p
The cross-sectional area of segments AB and BC are AAB = (0.12) = 2.5(10 - 3)p m2 and
4
p
ABC = (0.0752) = 1.40625(10 - 3)p m2.
4
(Ui)a = ©
NAB 2LAB
NBC 2LBC
N2L
=
+
2AE
2AAB Est
2ABC Ebr
=
C 80(103) D 2 (1.5)
2 C 2.5(10 - 3)p D C 200(109) D
+
C 20(103) D 2(0.5)
2 C 1.40625(10 - 3) p D C 101(109) D
= 3.28 J
Ans.
This result is valid only if s 6 sy.
sAB =
80(103)
NAB
= 10.19(106)Pa = 10.19 MPa 6 (sy)st = 250 MPa
=
AAB
2.5(10 - 3)p
O.K.
sBC =
20 (103)
NBC
= 4.527(106)Pa = 4.527 MPa 6 (sy)br = 410 MPa
=
ABC
1.40625(10 - 3) p
O.K.
1161
0.5 m
75 mm
C 20 kN
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*14–4. Determine the torsional strain energy in the A-36
steel shaft. The shaft has a diameter of 40 mm.
900 N⭈m
200 N⭈m
0.5 m
Referring to the FBDs of the cut segments shown in Fig. a, b and c,
300 N⭈m
0.5 m
TAB = 300 N # m
©Mx = 0;
TAB - 300 = 0
©Mx = 0;
TBC - 200 - 300 = 0
©Mx = 0;
TCD - 200 - 300 + 900 = 0 TCD = - 400 N # m
0.5 m
TBC = 500 N # m
The shaft has a constant circular cross-section and its polar moment of inertia is
p
J = (0.024) = 80(10 - 9)p m4.
2
(Ui)t = ©
TAB 2 LAB
TBC 2LBC
TCD LCD
T2L
=
+
+
2GJ
2GJ
2GJ
2GJ
1
=
2 C 75(10 ) 80 (10 - 9)p D
9
c 3002 (0.5) + 5002 (0.5) + ( - 400)2 (0.5) d
= 6.63 J
Ans.
•14–5.
Determine the strain energy in the rod assembly.
Portion AB is steel, BC is brass, and CD is aluminum.
Est = 200 GPa, Ebr = 101 GPa, and Eal = 73.1 GPa.
15 mm
A
20 mm
2 kN B
25 mm
D
5 kN C
2 kN
5 kN
3 kN
300 mm
N2 L
Ui = ©
2AE
[3 (103) ]2 (0.3)
=
2 (p4 )(0.0152)(200)(109)
[7 (103) ]2 (0.4)
+
2(p4 )(0.022)(101)(109)
[-3 (103) ]2 (0.2)
+
2 (p4 )(0.0252)(73.1)(109)
= 0.372 N # m = 0.372 J
Ans.
1162
400 mm
200 mm
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14–6. If P = 60 kN, determine the total strain energy
stored in the truss. Each member has a cross-sectional area
of 2.511032 mm2 and is made of A-36 steel.
2m
B
C
Normal Forces. The normal force developed in each member of the truss can be
determined using the method of joints.
1.5 m
Joint A (Fig. a)
D
+ ©F = 0;
:
x
FAD = 0
+ c ©Fy = 0;
FAB - 60 = 0
FAB = 60 kN (T)
P
Joint B (Fig. b)
+ c ©Fy = 0;
3
FBD a b - 60 = 0
5
FBD = 100 kN (C)
+ ©F = 0;
:
x
4
100 a b - FBC = 0
5
FBC = 80 kN (T)
Axial
Strain
Energy.
LBD = 222 + 1.52 = 2.5 m
(Ui)a = ©
=
A = 2.5 A 103 B mm2 = 2.5 A 10 - 3 B m2
and
N2L
2AE
2 C 2.5 A 10
1
-3
B D C 200 A 109 B D
c C 60 A 103 B D 2 (1.5) + C 100 A 103 B D 2 (2.5)
+ C 80 A 103 B D 2 (2) d
= 43.2 J
Ans.
This result is only valid if s 6 sY. We only need to check member BD since it is
subjected to the greatest normal force
sBD =
A
100 A 103 B
FBD
=
= 40 MPa 6 sY = 250 MPa
A
2