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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.
6–1. The load binder is used to support a load. If the force
applied to the handle is 225 N, determine the tensions T1 and
T2 in each end of the chain and then draw the shear and
moment diagrams for the arm ABC.
T1
A
C
B
225 N
300 mm
75 mm
a + ©MC = 0;
225(375) - T1 (75) = 0
Ans.
T1 = 1125 N
+ T ©Fy = 0;
T2
225 - 1125 + T2 = 0
Ans.
T2 = 900 N
-
Ans:
T1 = 1125 N, T2 = 900 N
V (N )
900
x
⫺225
M (N⭈m )
x
⫺67 . 5
463
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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.
6–2. Draw the shear and moment diagrams for the shaft.
The bearings at A and D exert only vertical reaction on the
shaft. The loading is applied to the pulleys at B and C and E.
500 mm
350 mm
375 mm 300 mm
A
E
B
C
D
160 N
360 N
500 N
Ans:
V (NN
N )
7
370
160
10
x
⫺225
⫺490
M (N⭈ m )
130
136
x
⫺48
464
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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.
6–3. The engine crane is used to support the engine, which
Drawthe
theshear
shearand
andmoment
moment diagrams
diagrams
has a weight
weightof
of1200
6 kN.lb.Draw
of the boom ABC when it is in the horizontal position shown.
a + ©MA = 0;
4
6(2.4) = =
0 0;
FB(0.9)
(3) -– 1200(8)
5 A
+ c ©Fy = 0;
-Ay +
+ ©F = 0;
;
x
Ax -
4
(4000)
= 0;
(20)
– 6-= 1200
0;
5
3
(4000)
(20)
= 0;= 0;
5
A
A
3 ftm
0.9
5 ftm
1.5
B
B
F
kN lb
FBA==204000
C
C
ft
1.24 m
Ayy==102000
A
kN lb
Ax = 2400
Ax =lb12 kN
0.9 m
1.5 m
V kN
6
FB
6 kN
–10
14 kN-m
–9
Ans:
x 0.9, V 10, M 9
465
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*6–4. Draw the shear and moment diagrams for the beam.
9 kN
1.2 m
466
9 kN
1.2 m
1.2 m
9 kN
9 kN
1.2 m
1.2 m
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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.
6–5. Draw the shear and moment diagrams for the beam.
10 kN
8 kN
15 kN⭈m
2m
3m
Ans:
V (kN)
18
8
x
M (kN⭈m)
x
⫺15
⫺39
⫺75
467
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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.
6–6. Express the internal shear and moment in terms of x
and then draw the shear and moment diagrams.
w0
Support Reactions: Referring to the free-body diagram of the entire beam
shown in Fig. a,
a + ©MA = 0;
+ c ©Fy = 0;
A
B
x
L
2
1
L 5
By(L) - w0 a b a L b = 0
2
2
6
By =
5
wL
24 0
Ay +
5
1
L
w L - w0 a b = 0
24 0
2
2
Ay =
w0L
24
Shear and Moment Function: For 0 … x 6
L
2
L
, we refer to the free-body
2
diagram of the beam segment shown in Fig. b.
w0L
- V = 0
24
+ c ©Fy = 0;
V =
a + ©M = 0; M -
w0L
24
Ans.
w0L
x = 0
24
M =
w0L
x
24
Ans.
L
6 x … L, we refer to the free-body diagram of the beam segment
2
shown in Fig. c.
For
w0L
1 w0
1
- c (2x - L) d c (2x - L) d - V = 0
24
2 L
2
+ c ©Fy = 0;
V =
a + ©M = 0; M +
Ans:
For 0 … x 6
w0
c L2 - 6(2x - L)2 d
24L
Ans.
M =
w0
L
6 x … L: V =
[L2 -6(2x - L)2],
2
24L
For
w0
1 w0
1
1
c (2x - L) d c (2x - L) d c (2x - L) d x = 0
2 L
2
6
24L
w0
c L2x - (2x - L)3 d
24L
w0L
w0L
L
:V =
,M =
x,
2
24
24
M =
Ans.
w0
[L2x - (2x - L)3]
24L
V
w0L
0.704 L
24
0
When V = 0, the shear function gives
0 = L2 - 6(2x - L)2
L
x
0.5 L
x = 0.7041L
Substituting this result into the moment equation,
⫺
5 wL
0
24
M|x = 0.7041L = 0.0265w0L2
Shear and Moment Diagrams: As shown in Figs. d and e.
M
0.0208 w0L2
0.0265 w0L2
x
0.5 L 0.704 L
468
L
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6–7. Draw the shear and moment diagrams for the
compound beam which is pin connected at B. (This structure
is not fully stable. But with the given loading, it is balanced
and will remain as shown if not disturbed.)
6 kip
30
kN
8 kip
40
kN
A
A
C
C
B
B
30 kN
40 kN
1m
1m
ft
14 m
6 ftm
1.5
ft
14 m
ft
14 m
1m
1.5 m
20 kN
20 kN
20 kN
50 kN
20
–30
–20
20
1.5 m
1m
–30
Ans:
x 1 , V 30, M 30
469
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*6–8. Express the internal shear and moment in terms of x
and then draw the shear and moment diagrams for the beam.
4000 N
6 kN/m
A
B
x
2m
Support Reactions: Referring to the free-body diagram of the entire beam shown
in Fig. a,
a + ©MA = 0;
By(3) - 6000(2)(1) - 4000(2) = 0
By = 6667 N
+ c ©Fy = 0;
Ay + 6667 - 6000(2) - 4000 = 0
Ay = 9333 N
Shear and Moment Function: For 0 … x 6 2 m, we refer to the free -body diagram
of the beam segment shown in Fig. b.
+ c ©Fy = 0;
9333 - 6000x - V = 0
Ans.
V = {9333 - 6000x} lb
a + ©M = 0;
x
M + 6000x a b - 9333x = 0
2
Ans.
M = {9333x - 3000x2 } N # m
For 2 m 6 x … 3 m, we refer to the free-body diagram of the beam segment shown
in Fig. c.
+ c ©Fy = 0;
V + 6667 = 0
Ans.
V = - 6667 N
a + ©M = 0;
6667(3 - x) - M = 0
Ans.
M = {6667(3 - x)} N# m
Shear and Moment Diagrams: As shown in Figs. d and e.
470
1m
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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.
6–9. Express the internal shear and moment in terms of x
and then draw the shear and moment diagrams for the
overhanging beam.
6 kN/m
A
B
x
2m
4m
Support Reactions: Referring to the free-body diagram of the entire beam shown
in Fig. a,
a + ©MA = 0;
By(4) - 6(6)(3) = 0
By = 27 kN
+ c ©Fy = 0;
Ay + 27 - 6(6) = 0
Ay = 9 kN
Shear and Moment Function: For 0 … x 6 4 m, we refer to the free-body diagram
of the beam segment shown in Fig. b.
+ c ©Fy = 0;
9 - 6x - V = 0
Ans.
V = {9 - 6x} kN
a + ©M = 0;
x
M + 6xa b - 9x = 0
2
Ans.
M = {9x - 3x2} kN # m
For 4 m 6 x … 6 m, we refer to the free-body diagram of the beam segment shown
in Fig. c.
+ c ©Fy = 0;
V - 6(6 - x) = 0
Ans.
V = {6(6 - x)} kN
a + ©M = 0;
- M - 6(6 - x)a
6 - x
b = 0
2
Ans.
M = - {3(6 - x)2} kN # m
Ans:
For 0 … x 6 4 m: V = { 9 - 6x} kN,
M = {9x - 3x2} kN # m,
Shear and Moment Diagrams: As shown in Figs. d and e.
For 4 m 6 x … 6 m: V = { 6(6 - x)} kN # m,
M = -{3(6 - x)2} kN # m
V (kN)
12
9
0
x (m)
1.5
4
6
⫺15
M (kN⭈m)
6.75
0
4
1.5
⫺12
471
6
x (m)
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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.
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;
 150
750 lb
N
P�
Ay - 750
150 = 0
C
150 N
lb
Ay = 750
a + ©MA = 0;
0.45
m
1.5 ft
B
A
0.45
m
1.5 ft
D
= (0.45)
– 750(0.9)
ND
- 150(3)
= 0= 0
D(1.5)
N
300 N
lb
NDD ==1500
0.45
m
1.5 ft
Shear and Moment Diagram: The couple moment acting on B due to ND is
300(1.5) == 450
MB ==1500(0.45)
. The
675 lb
N #· ftm.
Theloading
loadingacting
acting on
on member
member ABC
ABC is
is shown in Fig. b
and the shear and moment diagrams are shown in Figs. c and d.
750 N
0.45 m
0.45 m
0.45 m
0.45 m
0.45 m
V (N)
M(N · m)
337.5
750
x (m)
0.45
0.45
0.9
750 N
0.9
–337.5
1500 N
MB = 675 N· m
Ans:
x 0.45, V 750, M 337.5
472
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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.
6–11. Draw the shear and moment diagrams for the pipe.
The end screw is subjected to a horizontal force of 5 kN.
Hint: The reactions at the pin C must be replaced by an
equivalent loading at point B on the axis of the pipe.
C
A
80 mm
5 kN
B
400 mm
Ans:
V (kN)
x
⫺1
M (kN⭈m)
x
⫺0.4
473
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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.
*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
474
B
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6–13. Draw the shear and moment diagrams for the rod. It
is supported by a pin at A and a smooth plate at B. The plate
slides within the groove and so it cannot support a vertical
force, although it can support a moment.
15 kN
A
B
4m
2m
Ans:
V (kN)
15
x
M (kN⭈m)
60
x
⫺30
475
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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.
6–14. The industrial robot is held in the stationary position
6–14. The industrial robot is held in the stationary position
shown. Draw the shear and moment diagrams of the arm
shown. Draw the shear and moment diagrams of the arm ABC
ABC if it is pin connected at A and connected to a hydraulic
if it is pin connected at A and connected to a hydraulic cylinder
cylinder (two-force member) BD. Assume the arm and grip
(two-force member) BD. Assume the arm and grip have a
have a uniform weight of 0.3 N/mm and support the load of
uniform weight of 1.5 lb�in. and support the load of 40 lb at C.
200 N at C.
100 mm4 in.
250 mm
10 in.
A
B
50 in.
1250
mm
120�
120
200 N
250 mm
C
0.3 N/mm
D
1469 N
1250 mm
1469 N
100 mm
1864 N
2544 N
V (N)
575
200
–30
–1894
–1969
M (N · m)
–1.5
–484.4
Ans.
x 350 mm , V 575, M 484.4
476
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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.
*6–16. 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.
P
L
–
2
A
B
a
Support Reactions: As shown on FBD.
Absolute Minimum Moment: In order to get the absolute minimum moment, the
maximum positive and maximum negative moment must be equal that is
Mmax( + ) = Mmax( - ).
For the positive moment:
a + ©MNA = 0;
Mmax( + ) - a2P -
3PL L
ba b = 0
2a
2
Mmax( + ) = PL -
3PL2
4a
For the negative moment:
a + ©MNA = 0;
Mmax( - ) - P(L - a) = 0
Mmax( - ) = P(L - a)
Mmax( + ) = Mmax( - )
PL -
3PL2
4a
= P(L - a)
4a L - 3L2 = 4a L - 4a2
a =
P
L
–
2
23
L = 0.866L
2
Ans.
Shear and Moment Diagram:
477
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6–17. Express the internal shear and moment in the
cantilevered beam as a function of x and then draw the
shear and moment diagrams.
1.5
300kN
lb
3 kN/m
200
lb/ft
A
A
26m
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 x
3 a ab =b1.5x
= 33.33x
w = 200
2 6
1 (3)(2) kN
2
1.5 kN
4.5 kN
5 kN · m
4m
3
2m
3
Referring to Fig. b,
11
2
-300
(33.33x)(x)
V = V0 = {–1.5
V =
{- 300
- 16.67x2} lb (1)
–1.5 – - (1.5x)(x)
– V =-0 – 0.75x
} kN
22
x x
1
3
a b +b1.5
0 = M
0.25x} 5.556x
kN · m3} lb # ft (2)
a + ©M = 0; M + (1.5x)(x)
(33.33x)(x)a
+=
300x
0 =M{–1.5
= { -– 300x
3 3
2
+ c ©Fy = 0;
1.5 kN
1 (1.5x) x
2
The shear and moment diagrams shown in Figs. c and d are plotted using Eqs. (1)
and (2), respectively.
V (kN)
x (m)
–1.5
–4.5
M (kN · m)
x (m)
–5
Ans.
x 20, V 4.5, M 5,
V {1.5 0.75x 2} kN,
M {1.5x 0.25x 3} kN · m
478
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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.
6–18. Draw the shear and moment diagrams for the beam,
and determine the shear and moment throughout the beam
as functions of x.
10
50 kip
kN
2 kip/ft
30
kN/m
8 kip
40
kN
40
200kip�ft
kNm
Support Reactions: As shown on FBD.
Shear and Moment Function:
xx
6 ftm
1.8
ft:m
For 0 … x 6 61.8
+ c ©Fy = 0;
4 ftm
1.2
144 –-30x
= 0= 0
30.0
2x–-V V
30(1.8) = 54 kN
xx
a + ©MNA = 0; M ++458.6
216 ++ 30x
2xaa bb –-144x
30.0x= 0= 0
22
200 kN · m
458.6 kN · m
Ans.
V =={144
kN
{30.0– 30x}
- 2x}
kip
0.9 m
144 kN
+ 144x
30.0x– 216}kN
kip· # m
ft
M =={–15x
{ - x22 +
458.6}
a + ©MNA = 0;
0.9 m
1.2 m
30x
Ans.
458.6 kN · m
6 ft m6<xx …
ft:
For 1.8
# 10
3 m:
+ c ©Fy = 0;
40 kN
50 kN
144 kN
Ans.
V –-408 ==0 kNkip
0
V == 40
8.00
40 kN
200 kN · m
–M
M–-40(3
8(10– x)
- –x)200
- =
400 = 0
– 320}
kN ·kip
m # ft
M =={40x
{8.00x
- 120}
Ans.
3–x
V (kN)
144
M (kN · m)
90
40
1.8
3
x (m)
1.8
x (m)
3
–248
–200
–458.6
Ans.
V {144 30x} kN,
M {1.5x 2 144x 458.6} kN · m,
V 40 kN,
M {40x 320} kN · m
479
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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.
6–19. Draw the shear and moment diagrams for the beam.
30
kN/m
2 kip/ft
45 kip�ft
kNm
30
30 kN/m
B
45 kN · m
A
V (kN)
1.5 m
1.5 m
41.25 kN
1.5
5 ftm
1.5 m
3.75 kN
1.5
5 ftm
1.5
5 ftm
–3.75
–45
M (kN · m)
5.625
–33.75
–39.375
Ans.
x 1.5, V 45, M 33.75
480
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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.
*6–20. Draw the shear and moment diagrams for the
overhanging beam.
45 kN/m
A
Support Reactions: Referring to the free-body diagram of the beam shown in Fig. a,
a + ©MA = 0;
+ c ©Fy = 0;
1
(45)(4)(2.67) = 0
By(6) 2
By = 40.05 kN
1
(45)(4) = 0
2
Ay = 49.95 kN
Ay + 40.05 -
Shear and Moment Functions: For 0 … x 6 4 m, we refer to the free-body diagram
of the beam segment shown in Fig. b.
+ c ©Fy = 0;
49.95 -
1
a 11.25 1x b(x) - V = 0
2
V = e 49.95 - 5.625 x2 f kN
a + ©M = 0; M +
Ans.
x
1
a 11.25 1x b(x) a b - 49.95x = 0
3
2
M = e 49.95 x - 1.875 x3 f kN # m
Ans.
When V = 0, from the shear function,
0 = 49.95 - 5.625x2
x = 2.949 m
Substituting this result into the moment function,
M|x = 2.949 m = 99.21kN # m
For 4 m 6 x … 6 m, we refer to the free-body diagram of the beam segment
shown in Fig. c.
+ c ©Fy = 0;
V + 40.05 kN = 0
Ans.
V = - 40.05 kN
a + ©M = 0; 40.05(6 - x) - M = 0
Ans.
M = {40.05(6 - x)} kN # m
Shear and Moment Diagrams: As shown in Figs. d and e.
481
B
4m
2m
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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.
6–21. The 700-N man sits in the center of the boat, which
has a uniform width and a weight per linear foot of 45 N/m.
Determine the maximum bending moment exerted on the
boat. Assume that the water exerts a uniform distributed
load upward on the bottom of the boat.
2.25 m
2.25 m
Ans.
Mmax = 39 3.75 N # m
Ans:
Mmax = 39 3.75 N # m
V (N )
3 50
x
⫺3 50
#
M ( N m)
3 9 3 . 75
0
482
x
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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.
6–22. Draw the shear and moment diagrams for the
overhang beam.
4 kN/ m
A
B
3m
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
V冷x= 3 m - = - (32) - 4 = -10 kN
3
V冷x=3 m + = 24 - 4(3) = 12 kN
The moment diagram shown in Fig. e is plotted using Eqs. (2) and (4). The value of
the moment at support B is evaluated using either Eq. (2) or Eq. (4).
2
M冷x= 3 m = - (33) - 4(3) = -18 kN # m
9
or
M冷x= 3 m = -2(6 - 3)2 = -18 kN # m
Ans:
V (kN)
12
3
⫺4
x (m)
6
⫺10
M (kN⭈m)
3
⫺18
483
6
x (m)
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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.
6–23. The footing supports the load transmitted by the two
columns. Draw the shear and moment diagrams for the
footing if the reaction of soil pressure on the footing is
assumed to be uniform.
60 kN
60 kN
2m
4m
2m
Ans:
V (kN)
30
30
x
⫺30
⫺30
M (kzN⭈m)
90
90
x
484
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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.
*6–24. Express the shear and moment in terms of x and
then draw the shear and moment diagrams for the simply
supported beam.
300 N/m
A
B
3m
Support Reactions: Referring to the free-body diagram of the entire beam shown
in Fig. a,
a + ©MA = 0;
By(4.5) -
1
1
(300)(3)(2) - (300)(1.5)(3.5) = 0
2
2
By = 375 N
+ c ©Fy = 0;
1
1
(300)(3) - (300)(1.5) = 0
2
2
Ay + 375 Ay = 300 N
Shear and Moment Function: For 0 … x 6 3 m, we refer to the free-body diagram
of the beam segment shown in Fig. b.
+ c ©Fy = 0;
1
(100x)x - V = 0
2
V = {300 - 50x2} N
300 -
a + ©M = 0; M +
Ans.
1
x
(100x)xa b - 300x = 0
2
3
M = e 300x -
50 3
x fN#m
3
Ans.
When V = 0, from the shear function,
0 = 300 - 50x2
x = 26 m
Substituting this result into the moment equation,
M|x = 26 m = 489.90 N # m
For 3 m 6 x … 4.5 m, we refer to the free-body diagram of the beam segment
shown in Fig. c.
+ c ©Fy = 0;
V + 375 -
1
3200(4.5 - x)4(4.5 - x) = 0
2
V = e 100(4.5 - x)2 - 375 f N
Ans.
1
4.5 - x
[200(4.5 - x)](4.5 - x)a
b - M = 0
2
3
100
M = e 375(4.5 - x) Ans.
(4.5 - x)3 f N # m
3
a + ©M = 0; 375(4.5 - x) -
Shear and Moment Diagrams: As shown in Figs. d and e.
485
1.5 m
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6–25. Draw the shear and moment diagrams for the beam
and determine the shear and moment in the beam as
functions of x, where 1.2 m < x < 3 m.
2.25 kN/m
270 N.m
A
270 N.m
B
x
1.2 m
+ c ©Fy = 0;
1.2 m
- 2250(x - 1.2) - V + 2250 = 0
Ans.
V = 49 50 - 2250x
a + ©M = 0;
2m
- 270 - 2250(x - 1.2)
(x - 1.2)
- M + 2250(x - . 1 2) = 0
2
Ans.
M = - 11.2 5x2 + 49 50x - 49 50
Ans:
V = 49 50 - 2250x
M = - 1125x2 + 4 9 50x - 459 0
V (N )
2250
x
⫺2250
M ( N ⭈m )
8 55
x
⫺270
486
⫺270
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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.
6–27. Draw the shear and moment diagrams for the beam.
w0
B
L
3
+ c ©Fy = 0;
A
2L
3
w0L
1 w0x
- a
b(x) = 0
4
2
L
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
Ans:
V
7w0L
36
x
⫺w0L
18
⫺w0L
4
0.707 L
M
0.0345w0L2
x
⫺0.00617w0L2
487
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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.
*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
488
L
3
L
3
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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.
6–29. Draw the shear and moment diagrams for the double
overhanging beam.
8 kN/m
B
A
1.5 m
1.5 m
3m
Equations of Equilibrium: Referring to the free-body diagram shown in Fig. a,
a + ©MA = 0;
By(3) - 8(6)(1.5) = 0
By = 24 kN
+ c ©Fy = 0;
Ay + 24 - 8(6) = 0
Ay = 24 kN
Shear and Moment Diagram: As shown in Figs. b and c.
Ans:
V (kN)
12
0
1.5
12
4.5
3
⫺12
x (m)
6
⫺12
M (kN⭈m)
0
1.5
⫺9
489
3
4.5
⫺9
6
x (m)
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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.
6–30. The beam is bolted or pinned at A and rests on a bearing
pad at B that exerts a uniform distributed loading on the beam
over its 0.6-m length. Draw the shear and moment diagrams
for the beam if it supports a uniform loading of 30 kN/m.
30 kN/m
B
A
2.4 m
0.6 m
0.3 m
Ans:
V (kN )
3 6
x
⫺3 6
M (kN M ⫺m )
1 0 .8
5 4
1 0 .8
x
490
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6–31. The support at A allows the beam to slide freely along
the vertical guide so that it cannot support a vertical force.
Draw the shear and moment diagrams for the beam.
w
B
A
L
Equations of Equilibrium: Referring to the free-body diagram of the beam shown
in Fig. a,
a + ©MB = 0;
wLa
L
b - MA = 0
2
MA =
+ c ©Fy = 0;
wL2
2
By - wL = 0
By = wL
Shear and Moment Diagram: As shown in Figs. b and c.
Ans:
V
L
⫺wL
M
wL2
2
L
491
x
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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 kN>m 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
B
w0
w0
20 mm 60 mm 20 mm
+ c ©Fy = 0;
1
2(w0)(20)a b - 60(0.4) = 0
2
Ans.
w0 = 1.2 kN>m
492
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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.
6–33. The shaft is supported by a smooth thrust bearing at
A and smooth journal bearing at B. Draw the shear and
moment diagrams for the shaft.
400 N⭈m
B
A
1m
1m
1m
900 N
Equations of Equilibrium: Referring to the free-body diagram of the shaft shown
in Fig. a,
a + ©MA = 0;
By(2) + 400 - 900(1) = 0
By = 250 N
+ c ©Fy = 0;
Ay + 250 - 900 = 0
Ay = 650 N
Shear and Moment Diagram: As shown in Figs. b and c.
Ans:
V (N)
650
1
0
2
3
x (m)
⫺250
M (N⭈m)
650
400
0
493
x (m)
1
2
3
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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.
6–34. Draw the shear and moment diagrams for the
cantilever beam.
2 kN
A
3 kN⭈m
1.5 m
1.5 m
Equations of Equilibrium: Referring to the free-body diagram of the beam shown in
Fig. a,
+ c ©Fy = 0;
Ay - 2 = 0
Ay = 2 kN
a + ©MA = 0;
MA - 3 - 2(3) = 0
MA = 9 kN # m
Shear and Moment Diagram: As shown in Figs. b and c.
Ans:
V (kN)
2
0
x (m)
1.5
3
1.5
3
M (kN⭈m)
0
⫺3
⫺6
⫺9
494
x (m)
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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.
6–35. Draw the shear and moment diagrams for the beam
and determine the shear and moment as functions of x.
400 N/m
200 N/ m
A
B
x
3m
3m
Support Reactions: As shown on FBD.
Shear and Moment Functions:
For 0 … x 6 3 m:
200 - V = 0
+ c ©Fy = 0;
a + ©MNA = 0;
Ans.
V = 200 N
M - 200 x = 0
M = 5200 x6 N # m
Ans.
For 3 m 6 x … 6 m:
+ c ©Fy = 0;
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-
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
Ans:
For 0 … x 6 3 m: V = 200 N, M = (200x) N # m,
100 2
For 3 m 6 x … 6 m: V = e x + 500 f N,
3
100 3
x + 500x - 600 f N # m
M = e9
V (N)
200
3.87
0
6
x (m)
3
⫺700
M (N⭈m)
600 691
x (m)
0
3 3.87
495
6
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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.
*6–36. The shaft is supported by a smooth thrust bearing at
A and a smooth journal bearing at B. Draw the shear and
moment diagrams for the shaft.
600 N⭈m
B
A
0.8 m
0.8 m
0.8 m
900 N
Equations of Equilibrium: Referring to the free-body diagram of the shaft shown in
Fig. a,
a + ©MA = 0;
By(1.6) - 600 - 900(2.4) = 0
By = 1725 N
+ c ©Fy = 0;
1725 - 900 - Ay = 0
Ay = 825 N
Shear and Moment Diagram: As shown in Figs. b and c.
496
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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.
6–37. Draw the shear and moment diagrams for the beam.
50 kN/m
50 kN/m
B
A
4.5 m
4.5 m
Ans:
V (kN)
112.5
x
⫺112.5
M (kN⭈m)
169
x
497
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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.
6–38. The beam is used to support a uniform load along CD
due to the 6-kN weight of the crate. If the reaction at
bearing support B can be assumed uniformly distributed
along its width, draw the shear and moment diagrams for
the beam.
0.5 m
0.75 m
2.75 m
2m
C
D
A
B
Equations of Equilibrium: Referring to the free-body diagram of the beam
shown in Fig. a,
a + ©MA = 0;
FB(3) - 6(5) = 0
FB = 10 kN
+ c ©Fy = 0;
10 - 6 - Ay = 0
Ay = 4 kN
Shear and Moment Diagram: The intensity of the distributed load at
FB
10
=
=
support B and portion CD of the beam are wB =
0.5
0.5
6
20 kN>m and wCD =
= 3 kN>m, Fig. b. The shear
2
and moment diagrams are shown in Figs. c and d.
Ans:
V (kN)
2.95
2.75
6
x (m)
0
⫺4
3.25 4
6
3.25
4
6
M (kN⭈m)
2.95
2.75
0
⫺6
⫺11
498
⫺10.5
⫺11.4
x (m)
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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.
6–39. Draw the shear and moment diagrams for the double
overhanging beam.
1800 N
1800 N
3000 N/m
A
B
1m
2m
1m
Equations of Equilibrium: Referring to the free-body diagram of the beam shown in
Fig. a,
a + ©MA = 0;
By(2) + 1800(1) - 3000(2)(1) - 1800(3) = 0
By = 4800 N
+ c ©Fy = 0;
Ay + 4800 - 1800 - 3000(2) - 1800 = 0
Ay = 4800 N
Shear and Moment Diagram: As shown in Figs. b and c.
Ans:
V (N)
3000
0
1
⫺1800
1800
3
2
4
x (m)
⫺3000
M (N ⭈m)
0
1
2
3
⫺300
⫺1800
499
⫺1800
4
x (m)
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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.
*6–40. Draw the shear and moment diagrams for the
simply supported beam.
10 kN
10 kN
15 kN⭈m
A
B
2m
500
2m
2m
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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.
6–41. The compound beam is fixed at A, pin connected at
B, and supported by a roller at C. Draw the shear and
moment diagrams for the beam.
600 N
400 N/m
A
C
B
2m
2m
2m
Support Reactions: Referring to the free-body diagram of segment BC shown
in Fig. a,
a +©MB = 0;
Cy(2) - 400(2)(1) = 0
Cy = 400 N
+ c ©Fy = 0;
By + 400 - 400(2) = 0
By = 400 N
Using the result of By and referring to the free-body diagram of segment AB, Fig. b,
+ c ©Fy = 0;
Ay - 600 - 400 = 0
Ay = 1000 N
a +©MA = 0;
MA - 600(2) - 400(4) = 0
MA = 2800 N
Shear and Moment Diagrams: As shown in Figs. c and d.
Ans:
V (N)
1000
400
0
5
2
6
4
x (m)
⫺400
M (N⭈m)
0
4
⫺800
⫺2800
501
200
2
5
6
x (m)
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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.
6–42. Draw the shear and moment diagrams for the
compound beam.
5 kN/m
A
B
2m
D
C
1m
1m
Support Reactions:
From the FBD of segment AB
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
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
+
: ©Fx = 0;
Bx = 0
From the FBD of segment AB
+
: ©Fx = 0;
Ax = 0
Shear and Moment Diagram:
Ans:
V (kN)
5.00
0
10.0
2
1
⫺5.00
3
5.00
x (m)
4
⫺10.0
M (kN⭈m)
2.50
0
3
1
4
2
⫺7.50
502
x (m)
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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.
6–43. The compound beam is fixed at A, pin connected at
B, and supported by a roller at C. Draw the shear and
moment diagrams for the beam.
2 kN
3 kN/m
C
B
A
3m
3m
Support Reactions: Referring to the free-body diagram of segment BC shown
in Fig. a,
a +©MB = 0;
Cy(3) - 3(3)(1.5) = 0
Cy = 4.5 kN
c ©Fy = 0;
By + 4.5 - 3(3) = 0
By = 4.5 kN
Using the result of By and referring to the free-body diagram of segment AB, Fig. b,
c ©Fy = 0;
Ay - 2 - 4.5 = 0
Ay = 6.5 kN
a +©MA = 0;
MA - 2(3) - 4.5(3) = 0
MA = 19.5 kN # m
Shear and Moment Diagrams: As shown in Figs. c and d.
Ans:
V (kN)
6.5
0
4.5
6
3
x (m)
4.5
⫺4.5
M (kN⭈m)
3.375
0
⫺19.5
503
x (m)
3
4.5
6
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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.
*6–44. Draw the shear and moment diagrams for the beam.
w
120 kN/m
1
w = – x2
8
B
A
2.4 m
2.4
FR =
1
2
x dx = 106.65 kN
8 L0
106.65 kN
192 kN # m
2.4
1
8
x =
3
x dx
L0
106.65
kN
= 1.8 m
106.65
kN # m
- 192
504
x
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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.
6–45. A short link at B is used to connect beams AB and
BC to form the compound beam shown. Draw the shear and
moment diagrams for the beam if the supports at A and B
are considered fixed and pinned, respectively.
15 kN
3 kN/m
B
A
4.5 m
C
1.5 m
1.5 m
Support Reactions: Referring to the free-body diagram of segment BC shown
in Fig. a,
a +©MC = 0;
15(1.5) - FB(3) = 0
FB = 7.5 kN
+ c ©Fy = 0;
Cy + 7.5 - 15 = 0
Cy = 7.5 kN
Using the result of FB and referring to the free-body diagram of segment AB, Fig. b,
+ c ©Fy = 0;
a +©MA = 0;
1
(3)(4.5) - 7.5 = 0
2
Ay = 14.25 kN
Ay -
1
(3)(4.5)(3) - 7.5(4.5) = 0
2
MA = 54 kN # m
MA -
Shear and Moment Diagrams: As shown in Figs. c and d.
Ans:
V (kN)
14.25
0
7.5
6
7.5
x (m)
4.5
⫺7.5
M (kN⭈m)
11.25
0
⫺54
505
x (m)
4.5
6
7.5
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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.
6–46. Determine the placement b of the hooks to minimize
the largest moment when the concrete member is being
hoisted. Draw the shear and moment diagrams. The
member has a square cross section of dimension a on each
side. The specific weight of concrete is g.
60⬚
60⬚
Support Reactions: The intensity of the uniform distributed load caused by its own
weight is w = ga2. Due to symmetry,
b
2Fy - ga2L = 0
+ c ©Fy = 0;
Fy =
ga2L
2
Absolute Minimum Moment: To obtain the absolute minimum moment, the
maximum positive moment must be equal to the maximum negative moment. The
maximum negative moment occurs at the supports. Referring to the free-body
diagram of the beam segment shown in Fig. b,
b
ga2ba b - Mmax( - ) = 0
2
a +©M = 0;
Mmax( - ) =
ga2b2
2
The maximum positive moment occurs between the supports. Referring to the
free-body diagram of the beam segment shown in Fig. c,
ga2L
- ga2x = 0
2
+ c ©Fy = 0;
x =
L
2
Using this result,
a+ ©M = 0; Mmax( + ) + ga2 a
ga2L L
L L
ba b a - bb = 0
2
4
2
2
Mmax( + ) =
b
L
ga2L
(L - 4b)
8
It is required that
Mmax( + ) = Mmax( - )
ga2L
ga2b2
(L - 4b) =
8
2
4b2 + 4Lb - L2 = 0
Solving for the positive result,
Ans.
b = 0.2071L = 0.207L
Shear and Moment Diagrams: Using the result for b, Fig. d, the shear and moment
diagrams are shown in Figs. e and f.
506
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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.
6–46. Continued
Ans:
b = 0.207L
V
0.293ga2L
0.793L
0.5L
0
0.207L
⫺0.207ga2L
M
0
x
L
⫺0.293ga2L
0.0214ga2L2
0.207L
0.793L
0.5L
⫺0.0214ga2L2
507
0.207ga2L
x
L
⫺0.0214ga2L2
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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.
6–47. If the A-36 steel sheet roll is supported as shown and
the allowable bending stress is 165 MPa, determine the
smallest radius r of the spool if the steel sheet has a width of
1 m and a thickness of 1.5 mm. Also, find the corresponding
maximum internal moment developed in the sheet.
r
Bending Stress-Curvature Relation:
sallow =
Ec
;
r
165(106) =
200(109)30.75(10 - 3)4
r
Ans.
r = 0.9091 m = 909 mm
Moment Curvature Relation:
1
M
=
;
r
EI
1
=
0.9091
M
200(109)c
1
(1)(0.00153) d
12
Ans.
M = 61.875 N # m = 61.9 N # m
Ans:
r = 909 mm, M = 61.9 N # m
508
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*6–48. Determine the moment M that will produce a
maximum stress of 70 MPa on the cross section.
12 mm
A
12 mm
75 mm
12 mm
B
C
75 mm
M
250 mm
D
12 mm
Section Properties:
y =
=
5
©yA
©A
0.25(4)(0.5)
2[2(3)(0.5)] + +5.5(10)(0.5)
(6)(99)(12) ++2[49.5(75)(12)]
137(250)(12)
=5
3.40
in.mm
84.7
4(0.5)
99(12)++2[(3)(0.5)]
2[(75)(12)]++10(0.5)
250(12)
1 1
3 3
INA
5 = (99)(12
) +B 99(12)(84.7
– 6)2
INA
(4) A 0.5
+ 4(0.5)(3.40
- 0.25)2
12 12
+ 2 c
+
1
1
3
+ 2 c 3(0.5)(3
) + 0.5(3)(3.40
(12)(75
) + (12)(75)(84.7
– 49.5)-2 d 2)2 d
12
12
1
1
(12)(250
+ (12)(250)(137
– 84.7)2 - 3.40)2
+ 3)(0.5)
A 103 B + 0.5(10)(5.5
12
12
= 91.73 in6) mm4
5 34.277(10
4
Maximum Bending Stress: Applying the flexure formula
smax =
70 =
Mc
I
M (262 - 84.7)
34.277 × 106
Ans.
M = 13.53 kN # m
509
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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.
•6–49.
Determine the maximum tensile and compressive
bending stress in the beam if it is subjected to a moment of
# ft.
M ==64kN
kip· m.
0.5mm
in.
12
A
12
0.5mm
in.
753mm
in.
12
0.5mm
in.
B
C
753 mm
in.
M
25010
mm
in.
D
12
0.5mm
in.
Section Properties:
y =
=
5
©yA
©A
0.25(4)(0.5)
2[2(3)(0.5)] + +5.5(10)(0.5)
(6)(99)(12) ++2[49.5(75)(12)]
137(250)(12)
=5
3.40
in.mm
84.7
4(0.5)
99(12)++2[(3)(0.5)]
2[(75)(12)]++10(0.5)
250(12)
1 1
3 3
INA
5 = (99)(12
) +B 99(12)(84.7
– 6)2
INA
(4) A 0.5
+ 4(0.5)(3.40
- 0.25)2
12 12
+ 2 c
+
1
1
3
+ 2 c 3(0.5)(3
) + 0.5(3)(3.40
(12)(75
) + (12)(75)(84.7
– 49.5)-2 d 2)2 d
12
12
1
1
(12)(250
+ (12)(250)(137
– 84.7)2 - 3.40)2
+ 3)(0.5)
A 103 B + 0.5(10)(5.5
12
12
= 91.73 in6) mm4
5 34.277(10
4
Maximum Bending Stress: Applying the flexure formula smax =
Mc
I
(st)max =
)(12)(10.5
- 3.40)
4(1063)(262
6(10
– 84.7)
3715.12 psi = 3.72 ksi
5 31.0=MPa
6
91.73)
34.277(10
Ans.
(sc)max =
)(12)(3.40)
4(1063)(4)(84.7)
6(10
1779.07
5 =14.8
MPa psi = 1.78 ksi
91.73 6)
34.277(10
Ans.
Ans:
y
= 84.7
3.40 mm,
in.,
I NA 34.277(10 6) mm 4,
(st) max 31.0 MPa, ( sc) max 14.8 MPa
510
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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.
6–50. 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 = 154 MPa and
(sallow)c = 105 MPa, respectively.
100 mm
100 mm
M
50 mm
50 mm
y (From base) =
I =
1
2100 2 - 50 2 = 28.867 mm.
3
1
(100) ( 2100 2 - 50 2 )3 = 1.8(106) mm 4
36
Assume failure due to tensile stress:
smax =
My
;
I
154 =
M(28.867)
1.8(106) mm 4
M = 9.6 kN # m. = 7.33 kN # m
Assume failure due to compressive stress:
smax =
Mc
;
I
105 =
M(86.6 - 28.867)
1.8(106) mm 4
M = 3.27 kN # m. = 2.50 kN # m
Ans.
(controls)
Ans:
M = 3.27 kN # m
511
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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.
6–51. A member has the triangular cross section shown. If a
moment of M = 1000 N · m is applied to the cross section,
determine the maximum tensile and compressive bending
stresses in the member. Also, sketch a three-dimensional view
of the stress distribution action over the cross section.
100 mm
100 mm
M
50 mm
50 mm
h = 2100 2 - 50 2 = 86.6 mm.
Ix =
1
(100)(86.6)3 = 1.8(106) mm 4
36
c =
2
(86.6) = 57.73 mm.
3
y =
1
(86.6) = 28.867 mm.
3
(smax)t =
My
106 (12)(28.867)
=
= 16 MPa
I
1.8(106)
Ans.
(smax)c =
106 (12)(57.73)
Mc
=
= 32 MPa
I
1.8(106)
Ans.
Ans:
(smax)t = 16 MPa, (smax)t = 32 MPa
512
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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.
*6–52. If the beam is subjected to an internal moment of
M = 30 kN # m, determine the maximum bending stress in
the beam. The beam is made from A992 steel. Sketch the
bending stress distribution on the cross section.
50 mm
50 mm
15 mm
A
10 mm
M
Section Properties: The moment of inertia of the cross section about the neutral
axis is
I =
1
1
(0.1)(0.153) (0.09)(0.123) = 15.165(10 - 6) m4
12
12
Maximum Bending Stress: The maximum bending stress occurs at the top and
bottom surfaces of the beam since they are located at the furthest distance from the
neutral axis. Thus, c = 75 mm = 0.075 m.
smax =
30(103)(0.075)
Mc
= 148 MPa
=
I
15.165(10 - 6)
Ans.
At y = 60 mm = 0.06 m,
s|y = 0.06 m =
My
30(103)(0.06)
= 119 MPa
=
I
15.165(10 - 6)
The bending stress distribution across the cross section is shown in Fig. a.
513
150 mm
15 mm
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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.
6–53. If the beam is subjected to an internal moment of
M = 30 kN # m, determine the resultant force caused by the
bending stress distribution acting on the top flange A.
50 mm
50 mm
15 mm
A
10 mm
M
150 mm
15 mm
Section Properties: The moment of inertia of the cross section about the neutral
axis is
1
1
(0.1)(0.153) (0.09)(0.123) = 15.165(10 - 6) m4
12
12
I =
Bending Stress: The distance from the neutral axis to the top and bottom surfaces of
flange A is yt = 75 mm = 0.075 m and yb = 60 mm = 0.06 m.
st =
Myt
30(103)(0.075)
= 148.37 = 148 MPa
=
I
15.165(10 - 6)
sb =
Myb
30(103)(0.06)
= 118.69 = 119 MPa
=
I
15.165(10 - 6)
Resultant Force: The resultant force acting on flange A is equal to the volume of the
trapezoidal stress block shown in Fig. a. Thus,
FR =
1
(148.37 + 118.69)(106)(0.1)(0.015)
2
Ans.
= 200 296.74 N = 200 kN
Ans:
FR = 200 kN
514
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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.
6–54. If the built-up beam is subjected to an internal
moment of M = 75 kN # m, determine the maximum tensile
and compressive stress acting in the beam.
150 mm
20 mm
150 mm
10 mm
150 mm
M
10 mm
300 mm
A
Section Properties: The neutral axis passes through centroid C of the cross section
as shown in Fig. a. The location of C is
y =
0.15(0.3)(0.02) + 230.225(0.15)(0.01)] + 230.295(0.01)(0.14)4
©y~A
=
= 0.2035 m
©A
0.3(0.02) + 2(0.15)(0.01) + 2(0.01)(0.14)
Thus, the moment of inertia of the cross section about the neutral axis is
I = ©I + Ad2
=
1
(0.02)(0.33) + 0.02(0.3)(0.2035 - 0.15)2
12
1
+ 2 c (0.01)(0.153) + 0.01(0.15)(0.225 - 0.2035)2 d
12
+ 2c
1
(0.14)(0.013) + 0.14(0.01)(0.295 - 0.2035)2 d
12
= 92.6509(10 - 6) m4
Maximum Bending Stress: The maximum compressive and tensile stress occurs at
the top and bottom-most fiber of the cross section.
(smax)c =
My
75(103)(0.3 - 0.2035)
= 78.1 MPa
=
I
92.6509(10 - 6)
Ans.
(smax)t =
75(103)(0.2035)
Mc
= 165 MPa
=
I
92.6509(10 - 6)
Ans.
Ans:
(smax)c = 78.1 MPa, (smax)t = 165 MPa
515
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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.
6–55. If the built-up beam is subjected to an internal moment of
M = 75 kN # m, determine the amount of this internal moment
resisted by plate A.
150 mm
20 mm
150 mm
10 mm
Section Properties: The neutral axis passes through centroid C of the cross section
as shown in Fig. a. The location of C is
y =
~A
0.15(0.3)(0.02) + 230.225(0.15)(0.01)] + 230.295(0.01)(0.14)4
©y
=
= 0.2035 m
©A
0.3(0.02) + 2(0.15)(0.01) + 2(0.01)(0.14)
150 mm
M
10 mm
300 mm
A
Thus, the moment of inertia of the cross section about the neutral axis is
I = I + Ad2
1
=
(0.02)(0.33) + 0.02(0.3)(0.2035 - 0.15)2
12
1
+ 2 c (0.01)(0.153) + 0.01(0.15)(0.225 - 0.2035)2 d
12
+ 2c
1
(0.14)(0.013) + 0.14(0.01)(0.295 - 0.2035)2 d
12
= 92.6509(10 - 6) m4
Bending Stress: The distance from the neutral axis to the top and
bottom of plate A is yt = 0.3 - 0.2035 = 0.0965 m and
yb = 0.2035 m.
st =
Myt
75(103)(0.0965)
= 78.14 MPa (C)
=
I
92.6509(10 - 6)
sb =
Myb
75(103)(0.2035)
= 164.71 MPa (T)
=
I
92.6509(10 - 6)
The bending stress distribution across the cross section of
plate A is shown in Fig. b. The resultant forces of the tensile
and compressive triangular stress blocks are
1
(164.71)(106)(0.2035)(0.02) = 335 144.46 N
2
1
(FR)c = (78.14)(106)(0.0965)(0.02) = 75 421.50 N
2
(FR)t =
Thus, the amount of internal moment resisted by plate A is
2
2
M = 335144.46c (0.2035) d + 75421.50 c (0.0965) d
3
3
= 50315.65 N # m = 50.3 kN # m
Ans.
Ans:
M = 50.3 kN # m
516
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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.
*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
Section Property:
B
20 mm
1
1
I =
(0.02)(0.223) +
(0.1)(0.023) = 17.8133(10 - 6) 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)
50 mm
50 mm
My
I
= 49.4 MPa (C)
Ans.
= 4.49 MPa (T)
Ans.
517
M ⫽ 8 kN⭈m
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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.
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
Section Property:
50 mm
1
1
I =
(0.02)(0.223) +
(0.1)(0.023) = 17.8133(10 - 6) m4
12
12
Bending Stress: Applying the flexure formula smax =
smax =
M ⫽ 8 kN⭈m
8(103)(0.11)
17.8133(10 - 6)
sy = 0.01m =
My
Mc
and s =
,
I
I
Ans.
= 49.4 MPa
8(103)(0.01)
17.8133(10 - 6)
50 mm
= 4.49 MPa
Ans:
smax = 49.4 MPa
518
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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.
6–58. The aluminum machine part is subjected to a moment
of M = 75 kN # m. Determine the bending stress created at
points B and C on the cross section. Sketch the results on a
volume element located at each of these points.
20 mm
20 mm
10 mm
10 mm
10 mm
10 mm
A
B
10 mm
N
C
M ⫽ 75 N⭈m
40 mm
y =
I =
0.005(0.08)(0.01) + 230.03(0.04)(0.01)4
0.08(0.01) + 2(0.04)(0.01)
= 0.0175 m
1
(0.08)(0.013) + 0.08(0.01)(0.01252)
12
+ 2c
1
(0.01)(0.043) + 0.01(0.04)(0.01252)d = 0.3633(10 - 5) m4
12
sB =
75(0.0175)
Mc
= 3.61 MPa
=
I
0.3633(10 - 6)
Ans.
sC =
My
75(0.0175 - 0.01)
= 1.55 MPa
=
I
0.3633(10 - 6)
Ans.
Ans:
sB = 3.61 MPa, sC = 1.55 MPa
519
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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.
6–59. The aluminum machine part is subjected to a moment
of M = 75 kN # m . Determine the maximum tensile and
compressive bending stresses in the part.
20 mm
10 mm
10 mm
A
B
y =
0.005(0.08)(0.01) + 230.03(0.04)(0.01)4
20 mm
10 mm
10 mm
10 mm
= 0.0175 m
0.08(0.01) + 2(0.04)(0.01)
1
I =
(0.08)(0.013) + 0.08(0.01)(0.01252)
12
1
+ 2 c (0.01)(0.043) + 0.01(0.04)(0.01252)d = 0.3633(10 - 5) m4
12
N
C
M ⫽ 75 N⭈m
40 mm
(smax)t =
75(0.050 - 0.0175)
Mc
= 6.71 MPa
=
I
0.3633(10 - 6)
Ans.
(smax)c =
My
75(0.0175)
= 3.61 MPa
=
I
0.3633(10 - 6)
Ans.
Ans:
(smax)t = 6.71 MPa, (smax)c = 3.61 MPa
520
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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.
*6–60. The beam is subjected to a moment of 20 kN · m.
Determine the resultant force the bending stress produces
on the top flange A and bottom flange B. Also compute the
maximum bending stress developed in the beam.
25 mm
125 mm
200 mm
A
M 20 kN.m
25 mm
25 mm
©y~A
12(25)(125) + 125(200)(25) + 237(75)(25)
y =
=
= 111 mm.
©A
25(125) + 200(25) + 75(25)
I =
75 mm
1
1
(125)(253) + 125(25)(111- 12)2 +
(25)(200)3 + 200(25)(125 - 111)2
12
12
1
+
(75)(253) + 75(25)(237 - 111)2
12
= 78.3(10 6) mm 4
Using flexure formula s =
My
I
sA =
20(10 6 )(111 - 25)
= 22 MPa
78.3(10 6 )
sB =
20(10 6 )(111)
78.3(10 6 )
sC =
= 28.35 MPa
20(10 6 )( 25 - 111)
78.3(10 6 )
smax =
= 29.12 MPa
20(10 6 ) (250 - 111)
78.3(10 6 )
= 4.9995 MPa = 35.5 MPa (Max)
Ans.
FA =
1
(22 + 28.35)(25)(125) = 78.67 kN
2
Ans.
FB =
1
(35.5 + 29.12)(25)(75) = 60.58 kN
2
Ans.
521
D
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.
6–61. The beam is subjected to a moment of 20 kN · m.
Determine the percentage of this moment that is resisted by
the web D of the beam.
25 mm
125 mm
200 mm
A
M 20 kN.m
25 mm
25 mm
©y~A
12(25)(125) + 125(200)(25) + 237(75)(25)
=
= 111 mm
©A
25(125) + 200(25) + 75(25)
1
1
I =
(125)(25 3) + 125(25)(111 - 12)2 +
(25)(2003 ) + 200(25)(125 - 111) 2
12
12
1
+
(75)(253) + 75(25)(237 - 111) 2
12
y =
D
75 mm
B
= 78.3(10 6 )
Using flexure formula s =
sA =
sB =
My
I
20(10 6 ) (111 - 25)
= 22 MPa
78.3(10 6 )
20(10 6 ) (225 - 111)
78.3(10 6 )
= 29.12 MPa
FC =
1
(22)(86)(25) = 23.65 kN
2
FT =
1
(29.12)(114)(25) = 41.5 kN
2
M = 23.65(58) + 41.5(76)
= 40.623 kip # in. = 4.52 kN # m
% of moment carried by web =
4.52
* 100 = 22.6 %
20
Ans.
Ans:
% of moment carried by web = 22.6 %
522
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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.
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
20 mm
25 mm
A
250 mm
25 mm
The moment of inertia of the cross-section about the neutral axis is
B
M ⫽ 10 kN⭈m
1
1
(0.2)(0.33) (0.16)(0.253) = 0.2417(10 - 3) m4.
I =
12
12
For point A, yA = C = 0.15 m.
sA =
MyA
10(103) (0.15)
= 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.
Ans:
sA = 6.21 MPa (C), sB = 5.17 MPa (T)
523
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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.
6–63. The beam is subjected to a moment of M = 40 N · m.
Determine the bending stress acting at point A and B. Also,
stetch a three-dimensional view of the stress distribution
acting over the entire cross-sectional area.
75 mm
A
B
25 mm
M 40 Nm
25 mm
y =
I =
50(100)(50) - 75(12)(50)(75)
100(50) - 12(50)(75)
= 35 mm
1
1
1
(50)(100 4 ) + (100)(50)(50 - 35) 2 - a (50)(75)3 + (50)(75)(75 - 35) 2 b = 1.7(10 6 ) mm 4
12
36
2
sA =
My
(40000)(12)(100 - 35)
=
= 1.53 MPa (C)
I
1.7(10 6 )
Ans.
sB =
My
40000 (12)(35 - 25)
=
= 0.235 MPa (T)
1.7(10 6 )
I
Ans.
sC =
My
40000 (12)(35)
=
= 0.823 MPa (T)
I
1.7(10 6 )
Ans:
sA = 1.53 MPa (C), sB = 0.253 MPa (T),
sC = 0.823 MPa (T)
524
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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.
*6–64. The axle of the freight car is subjected to wheel
loading of 90 kN. 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 140 mm.
C
A
250 mm
90 kN
smax =
22500(68.75)
Mc
= 88 MPa
= 1
4
I
4 p(68.75)
Ans.
525
B
D
1500 mm
250 mm
90 kN
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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.
6–65. A shaft is made of a polymer having an elliptical
cross-section. If it resists an internal moment of
M = 50 N # m, determine the maximum bending stress
developed in the material (a) using the flexure formula,
where lz = 14p(0.08 m)(0.04 m)3, (b) using integration.
Sketch a three-dimensional view of the stress distribution
acting over the cross-sectional area.
y
z2
y2
⫹ ———2 ⫽ 1
———
2
(80)
(40)
80 mm
M ⫽ 50 N⭈m
z
160 mm
x
(a)
1
1
p ab3 = p(0.08)(0.04)3 = 4.021238(10 - 6) m4
4
4
50(0.04)
Mc
smax =
= 497 kPa
=
I
4.021238(10 - 6)
I =
Ans.
(b)
M =
=
smax
y2dA
c LA
smax
y22zdy
c L
z = 20.0064 - 4y2 = 2 2(0.04)2 - y2
0.04
2
0.04
2
L- 0.04
2
y 2(0.04)2 - y2 dy
L- 0.04
0.04
(0.04)4
y
1
= 4c
sin - 1 a
b - y 2(0.04)2 - y2(0.042 - 2y2) d `
8
0.04
8
- 0.0 4
y zdy = 4
=
0.04
(0.04)4
y
sin - 1 a
b`
2
0.04 - 0.04
= 4.021238(10 - 6) m4
smax =
50(0.04)
4.021238(10 - 6)
Ans.
= 497 kPa
Ans:
(a) smax = 497 kPa,
(b) smax = 497 kPa
526
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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.
6–66. Solve Prob. 6–65 if the moment M = 50 kN # m is
applied about the y axis instead of the x axis. Here
Iy = 14 p(0.04 m)(0.08 m)3.
y
z2
y2
⫹ ———2 ⫽ 1
———
2
(80)
(40)
80 mm
M ⫽ 50 N⭈m
z
160 mm
x
(a)
1
1
p ab3 = p(0.04)(0.08)3 = 16.085(10 - 6) m4
4
4
50(0.08)
Mc
smax =
= 249 kPa
=
I
16.085(10 - 6)
I =
Ans.
(b)
M =
z(s dA) =
LA
50 = 2a
za
LA
smax
b (z)(2y)dz
0.08
0.08
1>2
smax
z2
b
z2 a 1 b
(0.04)dz
0.04 L0
(0.08)2
50 = 201.06(10 - 6)smax
Ans.
smax = 249 kPa
Ans:
(a) smax = 249 kPa,
(b) smax = 249 kPa
527
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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.
6–67. The shaft 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
1.5 m
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 )
Ans.
= 158 MPa
Ans:
smax = 158 MPa
528
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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.
*6–68. The shaft is supported by smooth journal bearings at
A and B that only exert vertical reactions on the shaft.
Determine its smallest diameter d if the allowable bending
stress is sallow = 180 MPa.
12 kN/m
d
A
B
3m
Allowable Bending Stress: The maximum moment is Mmax = 11.34 kN # m as
indicated on the moment diagram. Applying the flexure formula
smax = sallow =
180 A 106 B =
Mmax c
I
11.34(103) A d2 B
p
4
A d2 B 4
Ans.
d = 0.08626 m = 86.3 mm
529
1.5 m
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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.
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
300 mm
15 mm
15 mm
(a)
Section Property:
30 mm
(b)
For section (a)
I =
1
1
(0.2) A 0.333 B (0.17)(0.3)3 = 0.21645(10 - 3) m4
12
12
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)
smin =
150(103)(0.18)
0.36135(10 - 3)
Ans.
= 74.72 MPa = 74.7 MPa
Ans:
smin = 74.7 MPa
530
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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.
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.
top member
memberisisa pipe
a pipe
having
an outer
diameter
of
The top
having
an outer
diameter
of 1 in.
3
25 mm
and thickness
5 mm,
and themember
bottom is
member
is a
and
thickness
of 16
the bottom
a solid rod
in.,ofand
1
solid rod
having a of
diameter
of 12 mm.
having
a diameter
2 in.
y =
1.5
100kN/m
lb/ft
1.8
6 ftm
1.8
6 ftm
141.5
5.75mm
in.
1.8
6 ftm
©yA
(6.50)(0.4786)
0 ++ (160)(314.16)
=
4.6091
in.
5 =117.65
mm
©A
0.4786
+ 113.10
0.19635
314.16 +
11
1 1
1
1
4
4
II5=c c p(12.5)
p(0.5)4 4– - p(7.5)
p(0.3125)
d + 0.4786(6.50
d + 314.16(160
– 117.65)-2 +4.6091)
p(6)24 + p(0.25)4
4
44
4 4
4
+ +
113.10(117.65)22 == 2.1466(10
5.9271 in46) mm4
0.19635(4.6091)
– 0.5(1.5)(0.9
= 3.0375
Mmax = 1.35(2.7)
300(9 - 1.5)(12)
= 272)000
lb # in. kN · m
smax =
6
27
000(4.6091
+ 0.25)
Mc
3.0375(10
)(117.65
+ 6)
=
I
5.9271 6)
2.1466(10
Ans.
MPa
= 175.0
22.1 ksi
1.35 kN
0.45
2.7 m
1.35 kN
Ans:
smax 175.0 MPa
531
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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.
6–71. The boat has a weight of 10000 N 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.
G
B
0.3 m
C
A
D
0.9 m
1.5 m
0.3 m
45 mm
1.2 m
45 mm
75 mm
40 mm
Boat:
+
: ©Fx = 0;
Bx = 0
a +©MB = 0;
-NA(2.7) + 10000(1.5) = 0
NA = 5555.55 N
+ c ©Fy = 0;
5555.55 - 10000 + B y = 0
By = 4444.45 N
Assembly:
a +©MC = 0;
-ND(3) + 10000(2.7) = 0
ND = 9000 N
+ c ©Fy = 0;
Cy + 9000 - 10000 = 0
Cy = 1000 N
I =
1
1
(43.75)(75)3 (37.5)(43.75) 3 = 1.28(10 6 ) mm 4
12
12
smax =
5(10 6 ) (12)(37.5)
Mc
=
= 146.5 MPa
I
1.28(10 6 )
Ans.
Ans:
smax = 146.5 MPa
532
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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.
*6–72. Determine the absolute maximum bending stress
in the 40-mm-diameter shaft which is subjected to the
concentrated forces. The sleeve bearings at A and B support
only vertical forces.
1800 N
A
B
300 mm
450 mm
375 mm
Mmax = 506.25 N-m.
s =
506250(18.75)
Mc
= 98 MPa
=
I
1
p(18.75) 4
4
Ans.
533
1350 N
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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.
6–73. Determine the smallest allowable diameter of the
shaft which is subjected to the concentrated forces. The
sleeve bearings at A and B support only vertical forces, and
the allowable bending stress is sallow = 154 MPa.
1800 N
A
1350 N
B
300 mm
450 mm
375 mm
Mmax = 506.25 N-m.
s =
Mc
;
I
154(103) =
506250c
1 4
pc
4
c = 16 mm.
Ans.
d = 32 mm.
Ans:
d = 32 mm
534
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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.
6–74. The pin is used to connect the three links together.
Due to wear, the load is distributed over the top and bottom
of the pin as shown on the free-body diagram. If the
diameter of the pin is 10 mm, determine the maximum
bending stress on the cross-sectional area at the center
section a–a. For the solution it is first necessary to determine
the load intensities w1 and w2.
w2
3600 N
25 mm
a
w2
w1
25 mm
a
10 mm
40 mm
1800 N
1800 N
1
w (25) = 1800; w2 = 144 N> mm.
2 2
w1(40) = 3600;
w1 = 90 N> mm.
M = 1800(17.7) = 31860 N # mm
1
p(54) = 491 mm 4
4
31860 (5)
Mc
smax =
=
I
491
= 324.45 MPa
I =
Ans.
Ans:
smax = 324.45 MPa
535
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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.
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
C
B
0.75 m
1.5 m
3 kN
D
25 mm
0.75 m
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
smax =
Ans.
Ans:
smax = 52.8 MPa
536
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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.
*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
M
260 mm
20 mm 30 mm
The moment of inertia of the cross section about the neutral axis is
I =
1
1
(0.3)(0.33) (0.21)(0.263) = 0.36742(10 - 3) m4
12
12
Thus,
smax =
Mc
;
I
80(106) =
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.
537
Ans.
30 mm
30 mm
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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.
6–77. If the beam is subjected to an internal moment of
M = 3 kN # m, determine the maximum tensile and
compressive stress in the beam. Also, sketch the bending
stress distribution on the cross section.
A
Section Properties: The neutral axis passes through centroid C of the cross section
as shown in Fig. a. The location of C is
100 mm
25 mm
25 mm
M
75 mm
~A
©y
2[37.5(75)(25)] + 87.5(200)(25) + 150(100)(25)
= 85 mm.
=
y =
2(75)(25) + (200)(25) + (100)(25)
©A
25 mm
75 mm
75 mm
25 mm
Thus, the moment of inertia of the cross section about the neutral axis is
I = ©I + Ad2
1
1
= 2 c (25)(753) + 25(75)(85 - 37.5) 2d +
(200)(253) + 200(25)(87.5 + 85)2
12
12
+
1
(25)(1003 ) + (25)(100)(150 - 85)2
12
= 23.155(10 6) mm 4
Maximum Bending Stress: The maximum compressive and tensile stress occurs at
the top and bottom-most fibers of the cross section.
6
(smax)c =
3(10 )(12) (200 - 85)
Mc
=
= 13 MPa
I
23.155 (10 6)
Ans.
(smax)t =
My
3(10 6) (12)(85)
=
= 9.5 MPa
23.155 (10 6)
I
Ans.
The bending stresses at y = 15.3 mm and y = - 9.7 mm are
s|y = 0.6111 in. =
My
3(10 6) (12) (15.3)
= 1.7 MPa (C)
=
I
23.155 (10 6)
s|y = - 0.3889 in. =
3(10 6)(12) (9.7)
My
=
= 1.1 MPa (T)
I
23.155 (10 6)
The bending stress distribution across the cross section is shown in
Fig. b.
Ans:
(smax)c = 13 MPa,
(smax)t = 9.5 MPa
538
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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.
6–78. If the allowable tensile and compressive stress for
the beam are (sallow)t = 14 MPa and (sallow)c = 21 MPa,
respectively, determine the maximum allowable internal
moment M that can be applied on the cross section.
A
100 mm
25 mm
25 mm
M
Section Properties: The neutral axis passes through the centroid C of the cross
section as shown in Fig. a. The location of C is given by
y=
~A
©y
2[37.5(75) ( 25) + 87.5(200)(25) + 150(100)(25)]
=
= 85 mm.
©A
2(75)(25) + 200(25) + 100(25)
Thus, the moment of inertia of the cross section about the neutral axis is
75 mm
25 mm
75 mm
75 mm
25 mm
I = ©I + Ad2
= 2c
1
1
(25)(753) + 25(75)(85 - 37.5)2 d +
(200)(253 ) + 200(25)(87.5 + 85)2
12
12
+
1
(25)(1003 ) + (25)(100)(150 - 85) 2
12
= 23.155 (10 6) mm 4
Allowable Bending Stress: The maximum compressive and tensile stress occurs at
the top and bottom-most fibers of the cross section.
(sallow)c =
Mc
;
I
21 =
M(200 - 85)
23.155 (10 6)
M = 38.57 kip # ina
1 ft
b = 4.3 kN-m
12 in.
For the bottom-most fiber,
(sallow)t =
My
;
I
14 =
M(85)
23.155 (10 6)
M = 34.98 kip # ina
1 ft
b = 4 kN-m (controls)
12 in.
Ans.
Ans:
M = 4 kN # m
539
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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.
6–79. If the beam is subjected to an internal moment of
M = 3 kN # m, determine the resultant force of the bending
stress distribution acting on the top vertical board A.
A
100 mm
25 mm
25 mm
M
Section Properties: The neutral axis passes through the centroid C of the cross
section as shown in Fig. a. The location of C is given by
~
© yA
2[37.5(75)(25) + 87.5(200)(25) + 150(100)(25)]
=
= 85 mm
y =
©A
2(75)(25) + 200(25) + 100(25)
75 mm
25 mm
75 mm
75 mm
25 mm
Thus, the moment of inertia of the cross section about the neutral axis is
I = ©I + Ad2
= 2c
1
1
(25)(753) + 25(75)(85 - 37.5) 2 d +
(200)(253) + 200(25)(87.5 + 85) 2
12
12
+
1
(25)(1003) + (25)(100)(150 - 85)2
12
= 23.155 (10 6) mm 4
Bending Stress: The distance from the neutral axis to the top and bottom of board A
is yt = 200 - 85 = 115 mm and yb = 100 - 85 = 15 mm. We have
st =
3 (10 6)(12) (115)
Myt
=
= 13 MPa
I
23.155 (10 6)
sb =
3 (10 6)(12) (15)
Myb
=
= 1.7 MPa
I
23.155 (10 6)
Resultant Force: The resultant force acting on board A is equal to the volume of the
trapezoidal stress block shown in Fig. b. Thus,
FR =
1
(1.3 + 1.7)(25)(100) = 19 kN
2
Ans.
Ans:
FR = 19 kN
540
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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.
*6–80. If the beam is subjected to an internal moment of
M = 100 kN # m, determine the bending stress developed
at points A, B, and C. Sketch the bending stress distribution
on the cross section.
A
300 mm
M
30 mm
30 mm
C
150 mm
Section Properties: The neutral axis passes through centroid C of the cross section
as shown in Fig. a. The location of C is
y =
~
© yA
0.015(0.03)(0.3) + 0.18(0.3)(0.03)
=
= 0.0975 m
©A
0.03(0.3) + 0.3(0.03)
Thus, the moment of inertia of the cross section about the neutral axis is
I =
1
1
(0.3)(0.033) + 0.3(0.03)(0.0975 - 0.015)2 +
(0.03)(0.33)
12
12
+ 0.03(0.3)(0.18 - 0.0975)2
= 0.1907(10 - 3) m4
Bending Stress: The distance from the neutral axis to points A, B, and C is yA =
0.33 - 0.0975 = 0.2325 m, yB = 0.0975 m, and yC = 0.0975 - 0.03 = 0.0675 m.
sA =
MyA
100(103)(0.2325)
= 122 MPa (C)
=
I
0.1907(10 - 3)
Ans.
sB =
100(103)(0.0975)
MyB
= 51.1 MPa (T)
=
I
0.1907(10 - 3)
Ans.
sC =
MyC
100(103)(0.0675)
= 35.4 MPa (T)
=
I
0.1907(10 - 3)
Ans.
Using these results, the bending stress distribution across the cross section is shown
in Fig. b.
541
B
150 mm
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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.
6–81. If the beam is made of 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.
A
300 mm
M
30 mm
30 mm
B
150 mm
C
150 mm
Section Properties: The neutral axis passes through centroid C of the cross section
as shown in Fig. a. The location of C is
y =
~
© yA
0.015(0.03)(0.3) + 0.18(0.3)(0.03)
= 0.0975 m
=
©A
0.03(0.3) + 0.3(0.03)
Thus, the moment of inertia of the cross section about the neutral axis is
I =
1
1
(0.3)(0.033) + 0.3(0.03)(0.0975 - 0.015)2 +
(0.03)(0.33)
12
12
+ 0.03(0.3)(0.18 - 0.0975)2
= 0.1907(10 - 3) m4
Allowable Bending Stress: The maximum compressive and tensile stress occurs at
the top and bottom-most fibers of the cross section. For the top-most fiber,
(sallow)c =
Mc
I
150(106) =
M(0.33 - 0.0975)
0.1907(10 - 3)
M = 123024.19 N # m = 123 kN # m (controls)
Ans.
For the bottom-most fiber,
(sallow)t =
My
I
125(106) =
M(0.0975)
0.1907(10 - 3)
M = 244 471.15 N # m = 244 kN # m
Ans:
M = 123 kN # m
542
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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.
6–82. The shaft is supported by a smooth thrust bearing at
A and smooth journal bearing at C. If d = 75mm, determine
the absolute maximum bending stress in the shaft.
A
1m
d
C
B
1m
D
1m
8000 N
16000 N
Support Reactions: Shown on the free-body diagram of the shaft, Fig. a,
Maximum Moment: The shear and moment diagrams are shown in Figs. b and c. As
indicated on the moment diagram, the maximum moment is |Mmax| = 8000 N-m.
Section Properties: The moment of inertia of the cross section about the neutral
axis is
1
I = p(37 4) = 1.472 (10 6) mm 4
4
Absolute Maximum Bending Moment: Here, c =
smax =
3
= 37 mm.
2
8(10 6 )(12)(37)
Mmaxc
=
= 200 MPa
1.472 (10 6)
I
Ans:
smax = 200 MPa
543
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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.
6–83. The shaft is supported by a smooth thrust bearing
at A and smooth journal bearing at C. If the material has
an allowable bending stress of sallow = 168 MPa, determine
the required minimum diameter d of the shaft to the
nearest mm.
A
1m
d
C
B
1m
D
1m
8000 N
16000 N
Support Reactions: Shown on the free-body diagram of the shaft, Fig. a.
Maximum Moment: As indicated on the moment diagram, Figs. b and c, the
maximum moment is |Mmax| = 8000 N-m.
Section Properties: The moment of inertia of the cross section about the neutral
axis is
I =
1 d 4
p 4
pa b =
d
4
2
64
Absolute Maximum Bending Moment:
Mc
sallow =
;
I
Use
d
8000000(12) a b
2
168(103) =
p 4
d
64
d = 75 mm
Ans.
d = 75 mm
Ans:
Use d = 75 mm
544
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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.
*6–84. If the intensity of the load w = 15 kN> m, determine
the absolute maximum tensile and compressive stress in the
beam.
w
A
B
6m
300 mm
150 mm
Support Reactions: Shown on the free-body diagram of the beam, Fig. a.
Maximum Moment: The maximum moment occurs when V = 0. Referring to the
free-body diagram of the beam segment shown in Fig. b,
+ c g Fy = 0;
45 - 15x = 0
x = 3m
a + gM = 0;
3
Mmax + 15(3)a b - 45(3) = 0
2
Mmax = 67.5 kN # m
Section Properties: The moment of inertia of the cross section about the neutral
axis is
I =
1
(0.15)(0.33) = 0.1125(10 - 3) m4
36
Absolute Maximum Bending Stress: The maximum compressive and tensile stresses
occur at the top and bottom-most fibers of the cross section.
(smax)c =
67.5(103)(0.2)
Mc
= 120 MPa (C)
=
I
0.1125(10 - 3)
Ans.
(smax)t =
My
67.5(103)(0.1)
= 60 MPa (T)
=
I
0.1125(10 - 3)
Ans.
545
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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.
6–85. If the material of the beam has an allowable bending
stress of sallow = 150 MPa, determine the maximum
allowable intensity w of the uniform distributed load.
w
A
B
6m
300 mm
150 mm
Support Reactions: As shown on the free-body diagram of the beam, Fig. a,
Maximum Moment: The maximum moment occurs when V = 0. Referring to the
free-body diagram of the beam segment shown in Fig. b,
+ c gFy = 0;
3w - wx = 0
x = 3m
a + gM = 0;
3
Mmax + w(3) a b - 3w(3) = 0
2
Mmax =
9
w
2
Section Properties: The moment of inertia of the cross section about the neutral
axis is
I =
1
(0.15)(0.33) = 0.1125(10 - 3) m4
36
Absolute Maximum Bending Stress: Here, c =
sallow =
Mc
;
I
2
(0.3) = 0.2 m.
3
9
w(0.2)
2
150(106) =
0.1125(10 - 32
Ans.
w = 18750 N>m = 18.75 kN>m
Ans:
w = 18.75 kN>m
546
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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.
6–86. Determine the absolute maximum bending stress
-diametersh aft
shaftw hich
whichisis subjected
subjected to
to the
in the
the 50-mm
2-in.-di
ameter
concentrated forces. The journal bearings at A and B only
support vertical forces.
4 kN
800
lb
3 kN
600
lb
A
375
mm
15 in.
B
15 in.
375
mm
750
mm
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
lb # ·inm.
indicated on the moment diagram, Mmax = 15000
.
1.875 kN
The moment of inertia of the cross-section about the neutral axis is
I =
p 44
4 4
(1 ) )== 0.25
p inmm
(25
306796
4
Here, c = 1 in. Thus
smax =
Mmax c 6)(25)
1.875(10
I306796
15000(1)3
) psi
152.8 MPa
== 19.10(10
0.25 p
= 19.1 ksi
4 kN
375 mm
Ans.
3 kN
375 mm
750 mm
Ay = 4.5 kN
By = 2.5 kN
M (kN · m)
V (kN)
1.6875
1.875
4.5
0.5
1500
x (mm)
x (mm)
375
375
750
750
1500
–2.5
Ans:
smax 152.8 MPa
547
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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.
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.
MPa.
The allowable bending stress is sallow == 150
22 ksi.
800
lb
4 kN
600
lb
3 kN
A
A
15 mm
in.
375
B
B
15 mm
in.
375
30 mm
in.
750
The FBD of the shaft is shown in Fig. a
The shear and moment diagrams are shown in Fig. b and c respectively. As
1.875 kN
m.
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
6 d
15000(
1.875(10
) > 2)
150
53) = 4 4
22(10
pd pd
/64 >64
Ans.
d 5d50.3
mm in = 2 in.
= 1.908
4 kN
375 mm
3 kN
375 mm
750 mm
Ay = 4.5 kN
By = 2.5 kN
M (kN · m)
V (kN)
1.6875
1.875
4.5
0.5
1500
x (mm)
x (mm)
375
375
750
750
1500
–2.5
Ans:
d 50.3 mm
548
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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.
cross
section
of 225
*6–88. If the
the beam
beamhas
hasa asquare
square
cross
section
of 9mm
in. on
each side, determine the absolute maximum bending stress
in the beam.
6 kNlb
1200
15
800kN/m
lb/ft
B
A
2.5
8 ftm
2.5
8 ftm
Absolute Maximum Bending Stress: The maximum moment is Mmax = 76 875 kN # m
as indicated on moment diagram. Applying the flexure formula
6
smax =
)(112.5)
44.8(12)(4.5)
Mmax c 76.875(10
= 40.49
=
ksi MPa
3= 4.42
1
1
)
I
(9)(9)3
12 (225)(225
12
Ans.
V (kN)
M (kN · m)
43.5
6
8
x (m)
8
16
549
x (m)
–15
–76.875
16
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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.
6–89. If the compound beam in Prob. 6–42 has a square
cross section of side length a, determine the minimum value
of a if the allowable bending stress is sallow = 150 MPa.
Allowable Bending Stress: The maximum moment 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
Ans.
a = 0.06694 m = 66.9 mm
Ans:
a = 66.9 mm
550
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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.
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 moment is Mmax =
23w0 L2
as
216
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
Ans:
smax =
551
23w0 L2
36 bh2
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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, 冷 Mmax 冷 = 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
Ans.
= 119 MPa
Ans:
smax = 119 MPa
552
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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.
*6–92. Determine, to the nearest millimeter, 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, 冷 Mmax 冷 = 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
553
Ans.
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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.
6–93. The wing spar ABD of a light plane is made from
2014–T6 aluminum and has a cross-sectional area of
800 mm 2 , a depth of 75 mm, and a moment of inertia about
its neutral axis of 1.11(10-6) m4. Determine the absolute
maximum bending stress in the spar if the anticipated loading
is to be as shown. Assume A, B, and C are pins. Connection is
made along the central longitudinal axis of the spar.
smax =
Mc
;
I
smax =
5184000(37.5)
1.11(10 6 )
100 N/m
0.6 m
A
D
B
C
0.9 m
1.8 m
= 175.135 MPa Ans.
+
+
Note that 175.135 MPa 6 sY = 420 MPa
OK
Ans:
smax = 175.135 MPa
554
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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.
6–94. The beam has a rectangular cross section as shown.
Determine the largest load P that can be supported on its
overhanging ends so that the bending stress does not
exceed smax = 10 MPa.
P
P
1.5 m
1.5 m
1.5 m
250 mm
150 mm
I =
1
(0.15)(0.253) = 1.953125(10 - 4) m4
12
smax =
10(106) =
Mc
I
1.5P(0.125)
1.953125(10 - 4)
Ans.
P = 10.4 kN
Ans:
P = 10.4 kN
555
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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.
6–95. The beam has the rectangular cross section shown. If
P = 12 kN, determine the absolute maximum bending stress
in the beam. Sketch the stress distribution acting over the
cross section.
P
P
1.5 m
1.5 m
1.5 m
250 mm
150 mm
M = 1.5P = 1.5(12)(103) = 18000 N # m
I =
smax =
1
(0.15)(0.253) = 1.953125(10 - 4) m4
12
18000(0.125)
Mc
= 11.5 MPa
=
I
1.953125(10 - 4)
Ans.
Ans:
smax = 11.5 MPa
556
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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.
*6–96. A log that is 0.6 m in diameter is to be cut into a
rectangular section for use as a simply supported beam. If
the allowable bending stress for the wood is sallow = 56 MPa,
determine the required width b and height h of the beam
that will support the largest load possible. What is this load?
h
b
0.6 m
P
(600)2 = b2 + h2
2.4 m
P
Mmax = 2400 = 1200 P
2
sallow =
sallow =
bh2 =
Mmax(h2 )
Mc
= 1
3
I
12 (b)(h)
6 Mmax
bh2
6
(1200 P)
56
b(600) 2 - b3 = 128.57 P
(600) 2 - 3b2 = 128.57
dP
= 0
db
b = 346.41 mm.
Thus, from the above equations,
b = 346.41 mm
Ans.
h = 490 mm
Ans.
P = 647 kN
Ans.
557
2.4 m
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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.
6–97. A log that is 0.6 m in diameter is to be cut into a
rectangular section for use as a simply supported beam. If
the allowable bending stress for the wood is sallow = 56 MPa,
determine the largest load P that can be supported if the
width of the beam is b = 200 mm.
h
b
0.6 m
P
600 2 = h2 + 200 2
2.4 m
h = 566 mm.
Mmax =
P
(2400) = 1200 P
2
sallow =
Mmax c
I
56 =
2.4 m
1200P ( 566
2 )
1
3
12 (200)(566)
Ans.
P = 500 kN
Ans:
P = 500 kN
558
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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.
6–98. If the beam in Prob. 6–18 has a rectangular cross
section with a width of 200 mm and a height of 400 mm,
determine the absolute maximum bending stress in the beam.
400 mm
200 mm
Absolute Maximum Bending Stress: The maximum moment is Mmax = 294 kN # m
as indicated on moment diagram. Applying the flexure formula
6
smax =
294(10 )(200)
Mmax c
= 55 .1 MPa
= 1
3
I
12 (200)(400 )
Ans.
Ans:
smax = 55.1 MPa
559
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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.
section
of 150
6–99. If the
the beam
beamhas
hasa asquare
squarecross
cross
section
of 6mm
in. on
each side, determine the absolute maximum bending stress
in the beam.
400
lb/ft
6 kN/m
B
B
A
A
6 ftm
1.8
The maximum moment occurs at the fixed support A. Referring to the FBD shown
in Fig. a,
6 ftm
1.8
6(1.8) kN
1
2 (6)(1.8) kN
0.9 m
a + ©MA = 0;
Mmax - 400(6)(3)
6(1.8)(0.9)-–
1
(400)(6)(8)
(6)(1.8)(2.4)== 00
2
Mmax = 16800
lb # ·ftm
22.60 kN
The moment of inertia of the about the neutral axis is I =
smax =
1
3
(150)(150
) = 42.1875(10
(6)(6
) = 3108
in4. Thus, 6) mm4
12
2.4 m
16800(12)(3)
Mc
22.68(106)(75)
=
6
I
108
42.1875(10
)
Ans.
40.3 MPa
= 5600
psi = 5.60 ksi
Ans:
smax 40.3 MPa
560
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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.
*6–100. If d = 450 mm, determine the absolute maximum
bending stress in the overhanging beam.
12 kN
8 kN/m
125 mm
25 mm 25 mm
75 mm
d
A
B
4m
Support Reactions: Shown on the free-body diagram of the beam, Fig. a.
Maximum Moment: The shear and moment diagrams are shown in Figs. b and c.
As indicated on the moment diagram, Mmax = 24 kN # m.
Section Properties: The moment of inertia of the cross section about the neutral
axis is
I =
1
1
(0.175)(0.453) (0.125)(0.33)
12
12
= 1.0477(10 - 3) m4
Absolute Maximum Bending Stress: Here, c =
smax =
0.45
= 0.225 m.
2
Ans.
24(103)(0.225)
Mmax c
= 5.15 MPa
=
I
1.0477(10 - 3)
561
75 mm
2m
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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.
6–101. If wood used for the beam has an allowable
bending stress of sallow = 6 MPa, determine the minimum
dimension d of the beam’s cross-sectional area to the
nearest mm.
12 kN
8 kN/m
125 mm
25 mm 25 mm
75 mm
d
A
B
4m
75 mm
2m
Support Reactions: Shown on the free-body diagram of the beam, Fig. a.
Maximum Moment: The shear and moment diagrams are shown in Figs. b and c.
As indicated on the moment diagram, Mmax = 24 kN # m.
Section Properties: The moment of inertia of the cross section about the neutral
axis is
I =
1
1
(0.175)d3 (0.125)(d - 0.15)3
12
12
= 4.1667(10 - 3)d3 + 4.6875(10 - 3)d2 - 0.703125(10 - 3)d + 35.15625(10 - 6)
Absolute Maximum Bending Stress: Here, c =
sallow =
d
.
2
Mc
;
I
24(103)
6(106) =
d
2
4.1667(10 - 3)d3 + 4.6875(10 - 3)d2 - 0.703125(10 - 3)d + 35.15625(10 - 6)
4.1667(10 - 3)d3 + 4.6875(10 - 3)d2 - 2.703125(10 - 3)d + 35.15625(10 - 6) = 0
Solving,
Ans.
d = 0.4094 m = 410 mm
Ans:
d = 410 mm
562
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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.
6–102. If the concentrated force P = 2 kN is applied at the
free end of the overhanging beam, determine the absolute
maximum tensile and compressive stress developed in
the beam.
P
150 mm
25 mm
200 mm
A
B
2m
1m
25 mm 25 mm
Support Reactions: Shown on the free-body diagram of the beam, Fig. a.
Maximum Moment: The shear and moment diagrams are shown in Figs. b and c.
As indicated on the moment diagram, the maximum moment is ƒ Mmax ƒ = 2 kN # m.
Section Properties: The neutral axis passes through the centroid C of the cross
section as shown in Fig. d.
The location of C is given by
y =
gy~A
2[0.1(0.2)(0.025)] + 0.2125(0.025)(0.15)
=
= 0.13068 m
gA
2(0.2)(0.025) + 0.025(0.15)
Thus, the moment of inertia of the cross section about the neutral axis is
I = gI + Ad 2
= 2c
1
1
(0.025)(0.23) + 0.025(0.2)(0.13068 - 0.1)2 d +
(0.15)(0.0253) + 0.15(0.025)(0.2125 - 0.13068)2
12
12
= 68.0457(10 - 6) m4
Absolute Maximum Bending Stress: The maximum tensile and compressive stress
occurs at the top and bottom-most fibers of the cross section.
(smax)t =
Mmax y
2(103)(0.225 - 0.13068)
= 2.77 MPa
=
I
68.0457(10 - 6)
Ans.
(smax)c =
2(103)(0.13068)
Mmax c
= 3.84 MPa
=
I
68.0457(10 - 6)
Ans.
Ans:
(smax)t = 2.77 MPa, (smax)c = 3.84 MPa
563
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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.
6–103. If the overhanging beam is made of wood having
the allowable tensile and compressive stresses of
(sallow )t = 4 MPa and (sallow )c = 5 MPa, determine the
maximum concentrated force P that can applied at the
free end.
P
150 mm
25 mm
200 mm
A
B
2m
1m
25 mm 25 mm
Support Reactions: Shown on the free-body diagram of the beam, Fig. a.
Maximum Moment: The shear and moment diagrams are shown in Figs. b and c.
As indicated on the moment diagram, the maximum moment is ƒ Mmax ƒ = P.
Section Properties: The neutral axis passes through the centroid C of the cross
section as shown in Fig. d. The location of C is given by
~A
gy
2[0.1(0.2)(0.025)] + 0.2125(0.025)(0.15)
y=
=
= 0.13068 m
gA
2(0.2)(0.025) + 0.025(0.15)
The moment of inertia of the cross section about the neutral axis is
I = gI + Ad 2
= 2c
1
1
(0.025)(0.23) + 0.025(0.2)(0.13068 - 0.1)2 d +
(0.15)(0.0253) + 0.15(0.025)(0.2125 - 0.13068)2
12
12
= 68.0457(10 - 6) m4
Absolute Maximum Bending Stress: The maximum tensile and compressive stresses
occur at the top and bottom-most fibers of the cross section. For the top fiber,
(sallow )t =
Mmax y
;
I
4(106) =
P(0.225 - 0.13068)
68.0457(10 - 6)
P = 2885.79 N = 2.89 kN
For the top fiber,
(sallow )c =
P(0.13068)
Mmaxc
5(106) =
I
68.0457(10 - 6)
Ans.
P = 2603.49 N = 2.60 kN (controls)
Ans:
P = 2.60 kN
564
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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.
*6–104. The member has a square cross section and is
subjected to a resultant internal bending moment of
M = 850 N # m as shown. Determine the stress at each
corner and sketch the stress distribution produced by M.
Set u = 45°.
z
B
250 mm
125 mm
125 mm
E
A
C
M 850 Nm
u
D
y
My = 850 cos 45° = 601.04 N # m
Mz = 850 sin 45° = 601.04 N # m
Iz = Iy =
s = -
1
(0.25)(0.25)3 = 0.3255208(10 - 3) m4
12
Myz
Mzy
sA = -
sB = -
sD = -
sE = -
Iz
+
Iy
601.04(-0.125)
601.04(-0.125)
-3
+
0.3255208(10 )
601.04(0.125)
601.04(-0.125)
-3
+
0.3255208(10 - 3)
+
0.3255208(10 - 3)
0.3255208(10 )
601.04(-0.125)
601.04(0.125)
0.3255208(10 - 3)
601.04(0.125)
601.04(0.125)
-3
0.3255208(10 )
0.3255208(10 - 3)
+
0.3255208(10 - 3)
= 0
Ans.
= 462 kPa
Ans.
= -462 kPa
Ans.
= 0
Ans.
The negative sign indicates compressive stress.
565
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6–105. The member has a square cross section and is
subjected to a resultant internal bending moment of
M = 850 N # m as shown. Determine the stress at each
corner and sketch the stress distribution produced by M.
Set u = 30°.
z
B
250 mm
125 mm
125 mm
E
A
C
M 850 Nm
u
D
y
My = 850 cos 30° = 736.12 N # m
Mz = 850 sin 30° = 425 N # m
Iz = Iy =
s = -
1
(0.25)(0.25)3 = 0.3255208(10 - 3) m4
12
Myz
Mzy
sA = -
sB = -
sD = -
sE = -
Iz
+
Iy
736.12(- 0.125)
425(- 0.125)
-3
+
0.3255208(10 )
736.12(0.125)
425(- 0.125)
-3
0.3255208(10 - 3)
+
0.3255208(10 )
0.3255208(10 - 3)
736.12(- 0.125)
425(0.125)
-3
+
0.3255208(10 - 3)
0.3255208(10 - 3)
+
0.3255208(10 - 3)
0.3255208(10 )
736.12(0.125)
425(0.125)
= - 119 kPa
Ans.
= 446 kPa
Ans.
= - 446 kPa
Ans.
= 119 kPa
Ans.
The negative signs indicate compressive stress.
Ans:
sA = - 119 kPa, sB = 446 kPa, sD = -446 kPa,
sE = 119 kPa
566
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y
6–106. 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
z
My
dA
sC
y
Mz
s dA
My =
z s dA,
-y s dA,
Mz =
LA
LA
LA
determine the constants a, b, and c, and show that the
normal stress can be determined from the equation
s = [-(MzIy + MyIyz)y + (MyIz + MzIyz)z]> (IyIz - Iyz2),
where the moments and products of inertia are defined in
Appendix A.
0 =
x
z
Equilibrium Condition: sx = a + by + cz
0 =
LA
sx dA
0 =
LA
(a + by + cz) dA
dA + b
y dA + c
0 = a
LA
My =
LA
z sx dA
=
LA
z(a + by + cz) dA
= a
LA
LA
z dA + b
Mz =
LA
-y sx dA
=
-y(a + by + cz) dA
LA
= -a
LA
ydA - b
LA
LA
yz dA + c
y2 dA - c
LA
(1)
z dA
z2 dA
(2)
yz dA
(3)
LA
LA
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
Ans:
≤z
(Q.E.D.)
567
a = 0; b = - ¢
MzIy + MyIyz
IyIz - I2yz
≤; c =
MyIz + MzIyz
IyIz - I2yz
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6–107. If the beam is subjected to the internal moment of
M = 2 kN # m, determine the maximum bending stress
developed in the beam and the orientation of the neutral axis.
y
100 mm
50 mm
A
100 mm
M
200 mm
60°
y
Internal Moment Components: The y and z components of M are positive since they
are directed towards the positive sense of their respective axes, Fig. a. Thus,
x
z
My = 2 sin 60° = 1.732 kN # m
B
50 mm
Mz = 2 cos 60° = 1 kN # m
Section Properties: The location of the centroid of the cross-section is
~A
gy
0.025(0.05)(0.2) + 0.15(0.2)(0.05)
=
= 0.0875 m
gA
0.05(0.2) + 0.2(0.05)
y =
The moments of inertia of the cross section about the principal centroidal y and z axes are
Iy =
1
1
(0.05)(0.23) +
(0.2)(0.053) = 35.4167(10 - 6) m4
12
12
1
1
(0.2)(0.053) + 0.2(0.05)(0.0875 - 0.025)2 +
(0.05)(0.23) + 0.05(0.2)(0.15 - 0.0875)2
12
12
= 0.1135(10 - 3) m4
Iz =
Bending Stress: By inspection, the maximum bending stress occurs at either
corner A or B.
s = -
sA = -
Myz
Mzy
Iz
+
Iy
1(103)(0.0875)
0.1135(10 - 3)
1.732(103)(-0.1)
+
35.4167(10 - 6)
Ans.
= -5.66 MPa = 5.66 MPa (C) (Max.)
3
sB = -
3
1.732(10 )(0.025)
1(10 )(- 0.1625)
0.1135(10 - 3)
+
35.4167(10 - 6)
= 2.65 MPa (T)
Orientation of Neutral Axis: Here, u = 60°.
tan a =
tan a =
Iz
Iy
tan u
0.1135(10 - 3)
35.4167(10 - 6)
tan 60°
Ans.
a = 79.8°
The orientation of the neutral axis is shown in Fig. b.
Ans:
smax = 5.66 MPa (C), a = 79.8°
568
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y
*6–108. If the wood used for the T-beam has an allowable
tensile and compressive stress of (sallow)t = 4 MPa and
(sallow)c = 6 MPa, respectively, determine the maximum
allowable internal moment M that can be applied to the
beam.
100 mm
50 mm
A
100 mm
M
200 mm
60°
y
x
z
Internal Moment Components: The y and z components of M are positive since they
are directed towards the positive sense of their respective axes, Fig. a. Thus,
My = M sin 60° = 0.8660M
Mz = M cos 60° = 0.5M
Section Properties: The location of the centroid of the cross section is
©y~A
0.025(0.05)(0.2) + 0.15(0.2)(0.05)
y =
=
= 0.0875 m
©A
0.05(0.2) + 0.2(0.05)
The moments of inertia of the cross section about the principal centroidal y and
z axes are
Iy =
1
1
(0.05)(0.23) +
(0.2)(0.053) = 35.417(10 - 6) m4
12
12
Iz =
1
1
(0.2)(0.053) + 0.2(0.05)(0.0875 - 0.025)2 +
(0.05)(0.23) + 0.05(0.2)(0.15 - 0.0875)2
12
12
= 0.1135(10 - 3) m4
Bending Stress: By inspection, the maximum bending stress can occur at either
corner A or B. For corner A, which is in compression,
sA = (sallow)c = - 6(106) = -
MyzA
MzyA
+
Iz
0.5M(0.0875)
0.1135(10 - 3)
Iy
0.8660M(- 0.1)
+
35.417(10 - 6)
Ans.
M = 2119.71 N # m = 2.12 kN # m (controls)
For corner B which is in tension,
sB = (sallow)t = -
4(106) = -
MyzB
MzyB
Iz
+
Iy
0.8660M(0.025)
0.5M(-0.1625)
0.1135(10 - 3)
+
35.417(10 - 6)
M = 3014.53 N # m = 3.01 kN # m
569
B
50 mm
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6–109. The box beam is subjected to the internal moment
of M = 4 kN # m, which is directed as shown. Determine the
maximum bending stress developed in the beam and the
orientation of the neutral axis.
y
50 mm
25 mm
50 mm
25 mm
150 mm
50 mm
150 mm
45⬚
Internal Moment Components: The y component of M is negative since it is directed
towards the negative sense of the y axis, whereas the z component of M which is
directed towards the positive sense of the z axis is positive, Fig. a. Thus,
x
M
z
50 mm
My = - 4 sin 45° = -2.828 kN # m
Mz = 4 cos 45° = 2.828 kN # m
Section Properties: The moments of inertia of the cross section about the principal
centroidal y and z axes are
Iy =
1
1
(0.3)(0.153) (0.2)(0.13) = 67.7083(10 - 6) m4
12
12
Iz =
1
1
(0.15)(0.33) (0.1)(0.23) = 0.2708(10 - 3) m4
12
12
Bending Stress: By inspection, the maximum bending stress occurs at corners A and D.
s = -
smax = sA = -
Myz
Mzy
Iz
+
Iy
2.828(103)(0.15)
-3
(- 2.828)(103)(0.075)
+
67.7083(10 - 6)
0.2708(10 )
Ans.
= -4.70 MPa = 4.70 MPa (C)
3
smax = sD = -
3
2.828(10 )(-0.15)
-3
0.2708(10 )
(-2.828)(10 )(-0.075)
+
67.7083(10 - 6)
Ans.
= 4.70 MPa (T)
Orientation of Neutral Axis: Here, u = -45°.
tan a =
tan a =
Iz
Iy
tan u
0.2708(10 - 3)
67.7083(10 - 6)
tan (-45°)
Ans.
a = -76.0°
The orientation of the neutral axis is shown in Fig. b.
Ans:
smax = 4.70 MPa, a = -76.0°
570
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6–110. If the wood used for the box beam has an
allowable bending stress of (sallow) = 6 MPa , determine
the maximum allowable internal moment M that can be
applied to the beam.
y
50 mm
25 mm
50 mm
25 mm
150 mm
50 mm
150 mm
45⬚
Internal Moment Components: The y component of M is negative since it is directed
towards the negative sense of the y axis, whereas the z component of M, which is
directed towards the positive sense of the z axis, is positive, Fig. a. Thus,
x
M
z
50 mm
My = - M sin 45° = - 0.7071M
Mz = M cos 45° = 0.7071M
Section Properties: The moments of inertia of the cross section about the principal
centroidal y and z axes are
Iy =
1
1
(0.3)(0.153) (0.2)(0.13) = 67.708(10 - 6) m4
12
12
Iz =
1
1
(0.15)(0.33) (0.1)(0.23) = 0.2708(10 - 3) m4
12
12
Bending Stress: By inspection, the maximum bending stress occurs at corners A and D.
Here, we will consider corner D.
sD = sallow = 6(106) = -
My zD
Mz yD
Iz
+
Iy
(- 0.7071M)(- 0.075)
0.7071M(- 0.15)
0.2708(10 - 3)
+
67.708(10 - 6)
Ans.
M = 5106.88 N # m = 5.11 kN # m
Ans:
M = 5.11 kN # m
571
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6–111. 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
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
My = 1200 sin 30° = 600 kN # m
x
150 mm
z
Mz = -1200 cos 30° = -1039.23 kN # m
Section Properties: The location of the centroid of the cross-section is given by
y =
©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)
150 mm
The moments of inertia of the cross section about the principal centroidal y and z
axes are
1
1
Iy =
(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.
Myz
Mzy
s = +
Iz
Iy
sA = -
c -1039.23 A 103 B d(0.2893)
5.2132 A 10 - 3 B
+
600 A 103 B (0.15)
1.3078 A 10 - 3 B
= 126 MPa (T)
sB = -
c -1039.23 A 103 B d(-0.3107)
5.2132 A 10 - 3 B
+
600 A 103 B (-0.15)
1.3078 A 10 - 3 B
Ans.
= -131 MPa = 131 MPa (C)(Max.)
Orientation of Neutral Axis: Here, u = -30°.
Iz
tan u
tan a =
Iy
tan a =
5.2132 A 10 - 3 B
1.3078 A 10 - 3 B
tan (-30°)
Ans.
a = -66.5°
The orientation of the neutral axis is shown in Fig. b.
Ans:
smax = 131 MPa (C), a = -66.5°
572
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y
*6–112. 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.
150 mm
150 mm
M
300 mm
30⬚
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,
x
150 mm
z
My = M sin 30° = 0.5M
150 mm
Mz = -M cos 30° = -0.8660M
Section Properties: The location of the centroid of the cross section is
y =
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
Ans.
M = 1185 906.82 N # m = 1186 kN # m (controls)
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
573
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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.
6–113. The board is used as a simply supported floor joist.
If a bending moment of M = 1000 N · m is applied 3° from
the z axis, determine the stress developed in the board at the
corner A. Compare this stress with that developed by the same
moment applied along the z axis (u = 0°). What is the angle a
for the neutral axis when u = 3°? Comment: Normally, floor
boards would be nailed to the top of the beams so that u ≈ 0°
and the high stress due to misalignment would not occur.
50 mm
A
M 1000 Nm
150 mm
z
u 3
x
y
Mz = 1000 cos 3° = 998.63 N # m
My = - 1000 sin 3° = - 52.33 N-m
Iz =
4
6
1
(50 )(150 3) = 14(10 ) mm ;
12
s = -
sA = -
tan a =
+
Iy
998630(- 75)
14.(10 6 )
Iz
Iy
1
(150)(503 ) = 1.5625(10 6 ) mm 4
12
My z
Mzy
Iz
Iy =
tan u;
- 52330(12)( - 25)
+
1.5625(10 6 )
tan a =
Ans.
= 6.187 MPa
14.(10 6 )
1.5625(10 6 )
tan (- 3°)
Ans.
a = - 25.3°
When u = 0°
sA =
1000000(12)(75)
Mc
=
= 5.36 MPa
I
14 (10 6 )
Ans.
Ans:
When u = 3°: sA = 6.187 MPa, a = -25.3°,
When u = 0°: sA = 5.36 MPa
574
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y
6–114. The T-beam is subjected to a bending moment of
M = 15 kN · m directed as shown. Determine the maximum
bending stress in the beam and the orientation of the neutral
axis. The location y of the centroid, C, must be determined.
M 15 kNm
150 mm 150 mm
50 mm
y–
z
200 mm C
50 mm
60
My = 15(10 6 ) sin 60° = 13(10 6 ) N # mm
Mz = - 15(10 6 ) cos 60° = - 7.5(10 6 ) N-mm
y =
(25)(300)(50) + (150)(200)(50)
= 75 mm
300(50) + 200(50)
Iy =
1
1
(50)(3003) +
(200)(503) = 114.58(10 6) mm 4
12
12
Iz =
1
1
(300)(503) + 300(50)(50 2) +
(50)(2003) + 50(200)(752) = 130.2(106) mm4
12
12
s = -
sA =
sD =
Myz
Mzy
+
Iz
Iy
-(-7.5)(10 6 ) (75)
13(10 6 ) (150)
+
6
114.58(10 6)
130.2 (10 )
-(-7.5)(10 6 ) (- 175)
+
6
13(10 6 ) ( - 150)
-(-7.5)(10 6 ) (75)
sB =
tan a =
6
130.2 (10 )
Iz
Iy
13(10 6 ) (- 25)
114.58(10 6)
130.2 (10 )
+
114.58(10 6)
Ans.
= 21.34 MPa
= - 12.92 MPa
= -12.7 MPa
6
tan u =
130.2 (10 )
tan (-60°)
114.58(10 6)
Ans.
a = -63.1°
Ans:
smax = 21.34 MPa (T), a = -63.1°
575
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6–115. The beam has a rectangular cross section. If it is
subjected to a bending moment of M = 3500 N # m directed
as shown, determine the maximum bending stress in the
beam and the orientation of the neutral axis.
y
z 150 mm
150 mm
M 3500 Nm
x
30
150 mm
My = 3500 sin 30° = 1750 N # m
Mz = -3500 cos 30° = -3031.09 N # m
Iy =
1
(0.3)(0.153) = 84.375(10 - 6) m4
12
Iz =
1
(0.15)(0.33) = 0.3375(10 - 3) m4
12
s = -
Myz
Mzy
Iz
+
Iy
1750(0.075)
-3031.09(0.15)
sA = -
0.3375(10 - 3)
+
84.375(10 - 6)
1750(-0.075)
-3031.09(-0.15)
sB = -
+
-3
0.3375(10 )
-3
0.3375(10 )
84.375(10 - 6)
1750(-0.075)
-3031.09(0.15)
sC = -
+
84.375(10 - 6)
Ans.
= 2.90 MPa (max)
= -2.90 MPa (max)
Ans.
= -0.2084 MPa
sD = 0.2084 MPa
tan a4 =
Iz
Iy
tan u =
3.375 (10 - 4)
8.4375(10 - 5)
tan (-30°)
Ans.
a = -66.6°
Ans:
smax = 2.90 MPa, a = -66.6°
576
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*6–116. For the section, Iy¿ = 31.7(10 - 6) m4, Iz¿ = 114(10 - 6) m4,
Iy¿z¿ = 15.1(10 - 6) m4. Using the techniques outlined in
Appendix A, the member’s cross-sectional area has principal
moments of inertia of Iy = 29.0(10 - 6) m4 and Iz =
117(10 - 6) m4, computed about the principal axes of inertia
y and z, respectively. If the section is subjected to a moment of
M = 2500 N # m directed as shown, determine the stress
produced at point A, using Eq. 6–17.
y
60 mm
y¿
60 mm
60 mm
M 2500 Nm
z¿
10.10
z
80 mm
C
140 mm
60 mm
A
Iz = 117(10 - 6) m4
Iy = 29.0(10 - 6) m4
My = 2500 sin 10.1° = 438.42 N # m
Mz = 2500 cos 10.1° = 2461.26 N # m
y = -0.06 sin 10.1° - 0.14 cos 10.1° = -0.14835 m
z = 0.14 sin 10.1° - 0.06 cos 10.1° = -0.034519 m
sA =
My z
-Mzy
Iz
+
Iy
438.42(-0.034519)
-2461.26(-0.14835)
=
-6
117(10 )
+
29.0(10 - 6)
= 2.60 MPa (T)
577
Ans.
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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.
6–117. Solve Prob. 6–116 using the equation developed in
Prob. 6–106.
y
60 mm
y¿
60 mm
60 mm
M 2500 Nm
z¿
10.10
z
80 mm
C
140 mm
60 mm
sA =
=
-(Mz¿Iy¿ + My¿Iy¿z¿)y¿ + (My¿Iz¿ + Mz¿Iy¿z¿)z¿
A
Iy¿Iz¿ - Iy¿z¿ 2
-32500(31.7)(10 - 6) + 04(-0.14) + 30 + 2500(15.1)(10 - 6)4(-0.06)
31.7(10 - 6)(114)(10 - 6) - 3(15.1)(10 - 6)42
= 2.60 MPa (T)
Ans.
Ans:
sA = 2.60 MPa
578
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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.
6–118. If the applied distributed loading of w = 4 kN>m can
be assumed to pass through the centroid of the beam’s cross
sectional area, determine the absolute maximum bending
stress in the joist and the orientation of the neutral axis. The
beam can be considered simply supported at A and B.
15⬚
Internal Moment Components: The uniform distributed load w can be resolved into
its y and z components as shown in Fig. a.
6m
wz = 4 sin 15° = 1.035 kN>m
wy and wz produce internal moments in the beam about the z and y axes,
respectively. For the simply supported beam subjected to the uniform distributed
wL2
. Thus,
load, the maximum moment in the beam is Mmax =
8
(Mz)max =
(My)max =
8
=
3.864(62)
= 17.387 kN # m
8
=
1.035(62)
= 4.659 kN # m
8
wzL2
8
w(6 m)
B
15 mm
100 mm
100 mm
15⬚
15⬚
wy = 4 cos 15° = 3.864 kN>m
wyL2
w
A
10 mm
15 mm
15⬚
100 mm
As shown in Fig. b, (Mz)max and (My)max are positive since they are directed towards
the positive sense of their respective axes.
Section Properties: The moment of inertia of the cross section about the principal
centroidal y and z axes are
Iy = 2 c
Iz =
1
1
(0.015)(0.1 3) d +
(0.17)(0.01 3) = 2.5142(10 - 6) m4
12
12
1
1
(0.1)(0.2 3) (0.09)(0.17 3) = 29.8192(10 - 6) m4
12
12
Bending Stress: By inspection, the maximum bending stress occurs at points A and B.
s = -
(My)max z
(Mz)max y
+
Iz
Iy
17.387(103)(- 0.1)
smax = sA = -
4.659(103)(0.05)
+
-6
29.8192 (10 )
2.5142(10 - 6)
Ans.
= 150.96 MPa = 151 MPa (T)
smax = sB = -
17.387(103)(0.1)
4.659(103)(- 0.05)
+
29.8192 (10 - 6)
2.5142(10 - 6)
= -150.96 MPa = 151 MPa (C)
Orientation of Neutral Axis: Here, u = tan - 1 c
tan a =
tan a =
Iz
Iy
(My)max
(Mz)max
Ans.
d = tan - 1 a
4.659
b = 15°.
17.387
tan u
29.8192(10 - 6)
2.5142(10 - 6)
tan 15°
Ans.
a = 72.5°
The orientation of the neutral axis is shown in Fig. c.
579
Ans:
smax = 151 MPa, a = 72.5°
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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.
6–119. Determine the maximum allowable intensity w of
the uniform distributed load that can be applied to the
beam. Assume w passes through the centroid of the beam’s
cross sectional area and the beam is simply supported at A
and B. The beam is made of material having an allowable
bending stress of sallow = 165 MPa .
w
A
15⬚
6m
w(6 m)
B
15 mm
100 mm
100 mm
15⬚
15⬚
10 mm
15 mm
15⬚
Internal Moment Components: The uniform distributed load w can be resolved into
its y and z components as shown in Fig. a.
100 mm
wy = w cos 15° = 0.9659w
wz = w sin 15° = 0.2588w
wy and wz produce internal moments in the beam about the z and y axes,
respectively. For the simply supported beam subjected to the a uniform distributed
wL2
load, the maximum moment in the beam is Mmax =
. Thus,
8
(Mz)max =
(My)max =
wyL2
8
=
0.9659w(62)
= 4.3476w
8
=
0.2588w(62)
= 1.1647w
8
wzL2
8
As shown in Fig. b, (Mz)max and (My)max are positive since they are directed
towards the positive sense of their respective axes.
Section Properties: The moment of inertia of the cross section about the principal
centroidal y and z axes are
Iy = 2 c
Iz =
1
1
(0.015)(0.13) d +
(0.17)(0.013) = 2.5142(10 - 6) m4
12
12
1
1
(0.1)(0.23) (0.09)(0.173) = 29.8192(10 - 6) m4
12
12
Bending Stress: By inspection, the maximum bending stress occurs at points A and B.
We will consider point A.
sA = sallow = -
165(106) = -
(My)maxzA
(Mz)maxyA
+
Iz
1.1647w(0.05)
4.3467w(-0.1)
-6
Iy
+
29.8192(10 )
2.5142(10 - 6)
w = 4372.11 N>m = 4.37 kN>m
Ans.
Ans:
w = 4.37 kN>m
580
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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.
*6–120. The composite beam is made of steel (A) bonded
to brass (B) and has the cross section shown. If it is
subjected to a moment of M = 6.5 kN # m, determine the
maximum bending stress in the brass and steel. Also, what is
the stress in each material at the seam where they are
bonded together? Ebr = 100 GPa. Est = 200 GPa.
A
y
50 mm
200 mm
M
B
z
x
175 mm
n =
200(109)
Est
= 2
=
Ebr
100(109)
y =
(350)(50)(25) + (175)(200)(150)
= 108.33 mm
350(50) + 175(200)
I =
1
1
(0.35)(0.053) + (0.35)(0.05)(0.083332) +
(0.175)(0.23) +
12
12
(0.175)(0.2)(0.041672) = 0.3026042(10 - 3) m4
Maximum stress in brass:
(sbr)max =
6.5(103)(0.14167)
Mc1
= 3.04 MPa
=
I
0.3026042(10 - 3)
Ans.
Maximum stress in steel:
(sst)max =
(2)(6.5)(103)(0.10833)
nMc2
= 4.65 MPa
=
I
0.3026042(10 - 3)
Ans.
Stress at the junction:
sbr =
6.5(103)(0.05833)
Mr
= 1.25 MPa
=
I
0.3026042(10 - 3)
Ans.
Ans.
sst = nsbr = 2(1.25) = 2.51 MPa
581
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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.
6–121. The composite beam is made of steel (A) bonded
to brass (B) and has the cross section shown. If the
allowable bending stress for the steel is (sallow)st = 180 MPa,
and for the brass (sallow)br = 60 MPa , determine the
maximum moment M that can be applied to the beam.
Ebr = 100 GPa, Est = 200 GPa.
A
y
50 mm
200 mm
M
B
z
x
175 mm
n =
200(109)
Est
= 2
=
Ebr
100(109)
y =
(350)(50)(25) + (175)(200)(150)
= 108.33 mm
350(50) + 175(200)
I =
1
1
(0.35)(0.053) + (0.35)(0.05)(0.083332) +
(0.175)(0.23) +
12
12
(0.175)(0.2)(0.041672) = 0.3026042(10 - 3) m4
(sst)allow =
nMc2
;
I
180(106) =
(2) M(0.10833)
0.3026042(10 - 3)
M = 251 kN # m
(sbr)allow =
Mc1
;
I
60(106) =
M(0.14167)
0.3026042(10 - 3)
Ans.
M = 128 kN # m (controls)
Ans:
M = 128 kN # m
582
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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.
•6–122. Segment A of the composite beam is made from
2014-T6 aluminum alloy and segment B is A-36 steel. If
w =
= 15
0.9 kN/m,
determine the
the absolute
absolute maximum bending
kip>ft, determine
stress developed in the aluminum and steel. Sketch the
stress distribution on the cross section.
w
w
15
4.5ftm
A
A
375in.
mm
B
B
375in.
mm
in.
753 mm
Maximum Moment: For the simply-supported beam subjected to the uniform
0.9 A 1522)B
wL2 15(4.5
=
distributed load, the maximum moment in the beam is Mmax =
8
88
= 25.3125
37.97 kNkip
· m.# ft.
Section Properties: The cross section will be transformed into that of steel as
Eal
73.1
10.6
== 0.3655
=
0.3655.
shown in Fig. a. Here, n =
Est
29
200
75 mm
(75)== 1.0965
27.4125inmm.
location
of the
centroid
of the
Then bst = nbal = 0.3655(3)
. TheThe
location
of the
centroid
of the
transformed section is
75 mm
y =
©yA
1.5(3)(3) + 4.5(3)(1.0965)
37.5(75)(75)
+ 112.5(75)(27.4125)
=
= 2.3030
in.
= 57.575
mm
©A
3(3)
+ 3(1.0965)
75(75)
+ 75(27.4125)
75 mm
The moment of inertia of the transformed section about the neutral axis is
I = ©I + Ad2 =
1
2
2
(75)(75
) +3(3)(2.3030
75(75)(57.575
– 37.5)
(3) A 33 B 3+
- 1.5)
12
+
1
2
2
(1.0965)
- 2.3030)
(27.4125)(75
(27.4125)(75)(112.5
– 57.575)
A 33 B 3+) +1.0965(3)(4.5
12
106.3 MPa
= 12.0696(10
30.8991 in46) mm4
Maximum Bending Stress: For the steel,
20.0 MPa
6
25.3125(12)(2.3030)
Mmaxcst
37.97(10
)(57.575)
= 181.1
(smax)st =
=
= 22.6MPa
ksi
12.0696(10
I
30.89916)
Ans.
54.8 MPa
At the seam,
sst�y = 17.425
0.6970 mm
in. =
181.1 MPa
Mmaxy
25.3125(12)(0.6970)
37.97(106)(17.425)
==
54.8
MPa
=
6.85
ksi
I
30.89916)
12.0696(10
For the aluminium,
(smax)al = n
At the seam,
25.3125(12)(6
2.3030)
Mmaxcal
37.97(106)(150 -– 57.575)
dd ==106.3
= 0.3655 cc
13.3 MPa
ksi
6
12.0696(10
I
30.8991 )
sal�y = 17.425
0.6970 mm
in. = n
Mmaxy
25.3125(12)(0.6970)
37.97(106)(17.425)
d =
MPa
= 0.3655 cc
d 20.0
= 2.50
ksi
I
30.8991 6)
12.0696(10
Ans.
The bending stress across the cross section of the composite beam is shown in Fig. b.
Ans:
M max 37.97 kN · m, y = 57.575
2.3030 mm,
in.,
I 12.0696(10 6) mm 4, (smax ) st 181.1 MPa,
(smax ) al 106.3 MPa
583
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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.
6–12 3. 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
)al
sallow
are (s
(sallow
)al==100
15MPa
ksi and ( (s
22 MPa,
ksi, determine
allow
allow))stst==150
the maximum allowable intensity w of the uniform
distributed load.
w
w
15
4.5ftm
A
A
3 in.
75
mm
B
B
3 in.
75
mm
in.
753mm
Maximum Moment: For the simply-supported beam subjected to the uniform
distributed load, the maximum moment in the beam is
w
A 1522B)
wL2
w(4.5
2.53125w.
=
== 28.125w
Mmax =
.
8
88
Section Properties: The cross section will be transformed into that of steel as
Eal
10.6
73.1
== 0.3655.
=
0.3655.
shown in Fig. a. Here, n =
Est
29
200
75 mm
Then bst = nbal = 0.3655(3)
. TheThe
location
of the
centroid
of the
(75)== 1.0965
27.4125inmm.
location
of the
centroid
of the
transformed section is
75 mm
y =
©yA
1.5(3)(3)
+ 4.5(3)(1.0965)
37.5(75)(75)
+ 112.5(75)(27.4125)
=
= 2.3030
in.
= 57.575
mm
©A
3(3)
+ 3(1.0965)
75(75)
+ 75(27.4125)
75 mm
The moment of inertia of the transformed section about the neutral axis is
I = ©I + Ad2 =
1
1 1
2
(75)(75
) +3(3)(2.3030
75(75)(57.575
– 37.5)
(27.4125)(75
(3) A 33 B 3+
- 1.5)
+ 2 + (1.0965)
A 33 B 3)
12
12 12
27.4125(75)(112.5
– 57.575)2- 2.3030)2
+ 1.0965
A 33 B + 1.0965(3)(4.5
12.0696(10
= 30.8991
in46) mm4
Bending Stress: Assuming failure of steel,
(sallow)st =
Mmax cst
;
I
22 5
=
150
6
(28.125w)(12)(2.3030)
(2.53125w)(10
)(57.575)
30.8991 6)
12.0696(10
w5
= 12.42
0.875 kN/m
kip>ft (controls)
Ans.
Assuming failure of aluminium alloy,
(sallow)al = n
Mmax cal
;
I
(28.125w)(12)(6
- 2.3030)
(2.53125w)(106)(150
– 57.575)
100
0.3655 c
15 =50.3655c
d d
30.8991 6)
12.0696(10
14.12
kN/m
ww=51.02
kip>ft
Ans:
w 12.42 kN/m
584
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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.
*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 = 0.060(10 - 3) m4 and Iz = 0.471(10 - 3) 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 N⭈m
200 mm
50 mm
z
z¿
300 mm
Internal Moment Components:
My¿ = 250 cos 32.9° = 209.9 N # m
Mz¿ = 250 sin 32.9° = 135.8 N # m
Section Property:
y¿ = 0.15 cos 32.9° + 0.175 sin 32.9° = 0.2210 m
z¿ = 0.15 sin 32.9° - 0.175 cos 32.9° = -0.06546 m
Bending Stress: Applying the flexure formula for biaxial bending
s =
sB =
My¿z¿
Mz¿y¿
Iz¿
+
Iy¿
209.9(-0.06546)
135.8(0.2210)
0.471(10 - 3)
-
0.060(10 - 3)
Ans.
= 293 kPa = 293 kPa (T)
585
B
50 mm
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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.
6–125. The wooden section of the beam is reinforced with
two steel plates as shown. Determine the maximum internal
moment M that the beam can support if the allowable
stresses for the wood and steel are (sallow)w = 6 MPa , and
(sallow)st = 150 MPa , respectively. Take Ew = 10 GPa and
Est = 200 GPa.
15 mm
150 mm
M
15 mm
100 mm
Section Properties: The cross section will be transformed into that of steel as shown
Ew
10
= 0.05. Thus, bst = nbw = 0.05(0.1) = 0.005 m . The
=
in Fig. a. Here, n =
Est
200
moment of inertia of the transformed section about the neutral axis is
I =
1
1
(0.1)(0.183) (0.095)(0.153) = 21.88125(10 - 6) m4
12
12
Bending Stress: Assuming failure of the steel,
(sallow)st =
Mcst
;
I
150(106) =
M(0.09)
21.88125(10 - 6)
M = 36 468.75 N # m = 36.5 kN # m
Assuming failure of wood,
(sallow)w = n
Mcw
;
I
6(106) = 0.05c
M(0.075)
21.88125(10 - 6)
d
M = 35 010 N # m = 35.0 kN # m (controls)
Ans.
Ans:
M = 35.0 kN # m
586
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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.
6–126. The wooden section of the beam is reinforced with
two steel plates as shown. If the beam is subjected to an
internal moment of M = 30 kN # m, determine the maximum
bending stresses developed in the steel and wood. Sketch
the stress distribution over the cross section. Take
Ew = 10 GPa and Est = 200 GPa.
15 mm
150 mm
M
15 mm
100 mm
Section Properties: The cross section will be transformed into that of steel as shown
Ew
10
in Fig. a. Here, n =
=
= 0.05. Thus, bst = nbw = 0.05(0.1) = 0.005 m. The
Est
200
moment of inertia of the transformed section about the neutral axis is
I =
1
1
(0.1)(0.183) (0.095)(0.153) = 21.88125(10 - 6) m4
12
12
Maximum Bending Stress: For the steel,
(smax)st =
30(103)(0.09)
Mcst
= 123 MPa
=
I
21.88125(10 - 6)
sst|y = 0.075 m =
Ans.
My
30(103)(0.075)
= 103 MPa
=
I
21.88125(10 - 6)
For the wood,
(smax)w = n
30(103)(0.075)
Mcw
d = 5.14 MPa
= 0.05 c
I
21.88125(10 - 6)
Ans.
The bending stress distribution across the cross section is shown in Fig. b.
Ans:
(smax)st = 123 MPa, (smax)w = 5.14 MPa
587
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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.
6–127. 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
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
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)
Ans:
smax = 20.1 MPa
588
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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.
6–128 . 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
3
M ==1.2
850kN
lb· #m
ft.EEstst==200
29(10
)E
ksi,
= 1600 ksi.
GPa,
12
w =E
w GPa.
y =
100
4 in.mm
12
0.5mm
in.
(0.5)(16)(0.25)
+ 2(3.5)(0.5)(2.25)
+ (0.8276)(3.5)(2.25)
12(399)(6) + 2(88)(12)(56)
+ (22.5)(88)(56)
= 29.04 mm = 1.1386 in.
0.5(16)
+ 2(3.5)(0.5)
+ (0.8276)(3.5)
12(399)
+ 2(88)(12)
+ (22.5)(88)
12
0.5mm
in.
11
1
2
3
2
I =
(16)(0.5
(16)(0.5)(0.88862)2)++2 a2 a b (12)(88
b(0.5)(3.5
+ 2(0.5)(3.5)(1.1114
)
(399)(1233)) +
+ (399)(12)(23.04
) +3)2(12)(88)(26.96
)
12
12
12
+
1
2
2
(22.5)(883) +3)(22.5)(88)(26.96
) = 8.214(10
) mm4 in4
(0.8276)(3.5
+ (0.8276)(3.5)(1.1114
) = 620.914
12
12
0.5mm
in.
R(375) = 22.5 mm
12 mm
Maximum stress in steel:
(sst) =
375
15 mm
in.
M
M
� 1.2
850kNm
lb�ft
100 mm
399 mm
6
850(12)(4
- –1.1386)
1.2(10
)(100
29.04)
Mc
=
1395 MPa
psi = 1.40 ksi
== 10.37
I
20.914 6)
8.214(10
Ans.
Maximum stress in wood:
(sw) = n(sst)max
12
= 0.05517(1395)
77.0MPa
psi
a
b (10.37) ==0.62
200
Ans.
589
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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.
6–129. A wood beam is reinforced with steel straps at its
top and bottom as shown. Determine the maximum
bending stress developed in the wood and steel if the beam
is subjected to a bending moment of M = 5 kN # m. Sketch
the stress distribution acting over the cross section. Take
Ew = 11 GPa, Est = 200 GPa.
y
20 mm
300 mm
M 5 kNm
20 mm
x
z
n =
200
= 18.182
11
I =
1
1
(3.63636)(0.34)3 (3.43636)(0.3)3 = 4.17848(10 - 3) m4
12
12
200 mm
Maximum stress in steel:
(sst)max =
18.182(5)(103)(0.17)
nMc1
= 3.70 MPa
=
I
4.17848(10 - 3)
Ans.
Maximum stress in wood:
(sw)max =
5(103)(0.15)
Mc2
= 0.179 MPa
=
I
4.17848(10 - 3)
Ans.
(sst) = n(sw)max = 18.182(0.179) = 3.26 MPa
Ans:
(sst)max = 3.70 MPa, (sw)max = 0.179 MPa
590
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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.
6–130 . 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 =
1.12
160
(75)= =0.6
15 in.
mm
(3)
5.6
800
(bbk)2 = n2 bpvc =
3.15
450
(75)== 1.6875
42.1875in.mm
(3)
5.6
800
2.5
500 kN
lb
PVC
�
450
ksiGPa
PVCEE
3.15
PVC
PVC
Escon
�
160
ksiGPa
EsconEE
1.12
EE
800
Bakelite
BakeliteEBE�
5.6ksi
GPa
B
3 ftm
0.9
y =
©yA
(1)(3)(2) + 3(0.6)(2)
+ 4.5(1.6875)(1)
(25)(75)(50)
+ 75(15)(50)
+ 112.5(42.1875)(25)
=
= 1.9346
= in.
48.365 mm
©A
3(2)
+ 0.6(2)
+ 1.6875(1)
75(50)
+ 15(50)
+ 42.1875(25)
I =
1
1 1
3 3
2
3 3
2 2
(75)(50
75(50)(23.365
(15)(50
+ 15(50)(26.635
(3)(2
) +) +3(2)(0.9346
) +2) + (0.6)(2
) +) 0.6(2)(1.0654
) )
12
12 12
+
4 ftm
1.2
3 ft m
0.9
in.
251mm
in.
502mm
2
in.
50 mm
3 in.
75 mm
1
2
(42.1875)(25
(42.1875(25)(64.135
) = 7.910(10
(1.6875)(13) 3+) +1.6875(1)(2.5654
) = 220.2495
in4 6) mm4
12
(smax)pvc = n2
2.5 lb
kN
500
2.5 kN
0.9 m
2.5 kN
1.2 m
0.9 m
6
1500(12)(3.0654)
3.15
Mc
450 2.25(10 )(76.635)
= a
b
I
800
20.24956)
5.6
7.910(10
2.5 kN
Ans.
= 1.53
12.26ksi
MPa
2.5 kN
0.9 m
2.5 kN
Mmax = 2.25 kN · m
2.5 kN
25 mm
50 mm
25 mm
75 mm
Ans:
(smax ) pvc 12.26 MPa
591
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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.
6–131. 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)con = 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. Here,
Est
200
n =
=
= 8. It is required that both concrete and steel achieve their
Econ
25
allowable stress simultaneously. Thus,
(sallow)con =
Mccon
;
I
12.5 A 106 B =
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
(3)
ccon = 0.3125d
Section Properties: The area of the steel bars is Ast = 3c
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
(4)
ccon 2 = 0.024pd - 0.024pccon
Solving Eqs. (3) and (4),
Ans.
d = 0.5308 m = 531 mm
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
592
d
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6–131.
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
Ans.
= 98 594.98 N # m = 98.6 kN # m
Ans:
d = 531 mm, M = 98.6 kN # m
593
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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.
*6–132. The wide-flange section is reinforced with two
wooden boards as shown. If this composite beam is
subjected to an internal moment of M = 100 kN # m,
determine the maximum bending stress developed in the
steel and the wood. Take Ew = 10 GPa and Est = 200 GPa.
100 mm
10 mm
100 mm
300 mm
M
10 mm
Section Properties: The cross section will be transformed into that of steel
Ew
10
=
= 0.05. Thus, bst = 0.01 + 0.05bw =
as shown in Fig. a. Here, n =
Est
200
0.01 + 0.05(0.19) = 0.0195 m. The moment of inertia of the transformed section
about the neutral axis is
I =
1
1
(0.2)(0.33) (0.1805)(0.283) = 119.81(10 - 6) m4
12
12
Maximum Bending Stress: For the steel,
(smax)st =
100(103)(0.15)
Mcst
= 125 MPa
=
I
119.81(10 - 6)
Ans.
For the wood,
(smax)w = na
100(103)(0.14)
Mcw
d = 5.84 MPa
b = 0.05 c
I
119.81(10 - 6)
Ans.
594
10 mm
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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.
6–133. The wide-flange section is reinforced with two
wooden boards as shown. If the steel and wood have an
allowable bending stress of (sallow)st = 150 MPa and
(sallow)w = 6 MPa, determine the maximum allowable
internal moment M that can be applied to the beam. Take
Ew = 10 GPa and Est = 200 GPa.
100 mm
10 mm
100 mm
300 mm
M
10 mm
Section Properties: The cross section will be transformed into that of steel
Ew
10
as shown in Fig. a. Here, n =
=
= 0.5. Thus, bst = 0.01 + nbw =
Est
200
0.01 + 0.05(0.19) = 0.0195 m. The moment of inertia of the transformed section
10 mm
about the neutral axis is
I =
1
1
(0.2)(0.33) (0.1805)(0.283) = 119.81(10 - 6) m4
12
12
Bending Stress: Assuming failure of steel,
(sallow)st =
Mcst
;
I
150(106) =
M (0.15)
119.81(10 - 6)
M = 119 805.33 N # m = 120 kN # m
Assuming failure of wood,
(sallow)w = n
Mcw
;
I
6(106) = 0.05 c
M (0.14)
119.81(10 - 6)
d
M = 102 690.29 N # m = 103 kN # m (controls)
Ans.
Ans:
M = 103 kN # m
595
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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.
6–134. 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
Section Properties: The cross section will be transformed into that of steel as shown
73.1 A 109 B
Eal
in Fig. a. Here, n =
=
= 0.3655. Thus, bst = nbal = 0.3655(0.015) =
Est
200 A 109 B
0.0054825 m. The location of the transformed section is
©yA
y =
=
©A
150 mm
B
0.075(0.15)(0.0054825) + 0.2 cp A 0.052 B d
0.15(0.0054825) + p A 0.052 B
= 0.1882 m
The moment of inertia of the transformed section about the neutral axis is
I = ©I + Ad2 =
1
(0.0054825) A 0.153 B + 0.0054825(0.15)(0.1882 - 0.075)2
12
+
1
p A 0.054 B + p A 0.052 B (0.2 - 0.1882)2
4
= 18.08 A 10 - 6 B m4
Maximum Bending Stress: For the steel,
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
S = 171 MPa
= 0.3655 C
I
18.08 A 10 - 6 B
3
(smax)al = n
Ans.
Ans:
(smax)st = 154 MPa, (smax)al = 171 MPa
596
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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.
6–135 .\ 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 toto aa bending
bending moment
moment of
of M
Mzz == 7.50
10 kN
m.
kip·# ft
Sketch the stress distribution acting over the cross section.
y
12
0.5mm
in.
12
0.5mm
in.
12
0.5mm
in.
z
150
mm
6 in.
2 in.
50 mm
2 in.
50 mm
Section Properties: For the transformed section.
n =
1.90(103)
Ew
13.1
= 0.0655
= 0.065517
=
3
Est
29.0(10
)
200
bst = nbw = 0.065517(4)
0.26207
0.0655(100) == 6.55
mm in.
1
6
(36 ++6.55)(150
) mm
(1.5
0.26207)3A)6=3 B11.967(10
= 31.7172
in4 4
12
INA =
Maximum Bending Stress: Applying the flexure formula
(smax)st =
6
7.5(12)(3)
Mc
10(10
)(75)
62.7 ksi
MPa
=
== 8.51
I
31.7172 6)
11.967(10
(smax)w = n
Ans.
6
7.5(12)(3)
(10)(10
)(75)
Mc
0.0655 c c
5 0.558
4.1 MPa
= 0.065517
dd =
ksi
11.967(10
I
31.71726)
Ans.
75 mm
75 mm
bst = 6.55 mm
36 mm
4.1 MPa
62.7 MPa
Ans:
(smax ) st 62.7 MPa, ( smax ) w 4.1 MPa
597
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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.
*6–136. 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)
[2]
r = r + y
rA¿ = r
dA
r
=
a
- 1 + 1b dA
r
LA r + y
LA
=
LA
a
= A -
r - r - y
r + y
y
LA r + y
+ 1 b dA
[3]
dA
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
=
LA
¢
ry
r + y
y2
LA r + y
+ y - y ≤ dA - Ay
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
¢
¢
A y dA y ≤ dA - A 1
y ≤ dA
r LA 1 + r
r LA 1 + r
1A y dA = 0,
But,
y
LA r + 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
598
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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.
6–136. Continued
y
=
As
y
r
y
dA
Mr
r
C
¢
≤ dA ¢
y≤ S
AI LA 1 + yr
r LA 1 + r
: 0
LA
Therefore,
¢
y
r
y ≤ dA = 0
1 + r
s =
and
y
y
yA
dA
¢
1A dA =
y≤ =
r LA 1 + r
r
r
yA
My
Mr
ab = AI
I
r
(Q.E.D.)
599
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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.
6–137. The curved member is subjected to the internal
moment of M = 50 kN # m. Determine the percentage
error introduced in the computation of maximum bending
stress using the flexure formula for straight members.
M
100 mm
M
200 mm
200 mm
Straight Member: The maximum bending stress developed in the straight member
smax =
50(103)(0.1)
Mc
=
= 75 MPa
I
1
(0.1)(0.23)
12
Curved Member: When r = 0.2 m, r = 0.3 m, rA = 0.2 m and rB = 0.4 m, Fig. a.
The location of the neutral surface from the center of curvature of the curve
member is
R =
0.1(0.2)
A
=
= 0.288539 m
dA
0.4
0.1ln
0.2
LA r
Then
e = r - R = 0.011461 m
Here, M = 50 kN # m. Since it tends to decrease the curvature of the curved
member.
sB =
M(R-rB)
50(103)(0.288539 - 0.4)
=
AerB
0.1(0.2)(0.011461)(0.4)
= -60.78 MPa = 60.78 MPa (C)
sA =
M(R - rA)
50(103)(0.288539 - 0.2)
=
AerA
0.1(0.2)(0.011461)(0.2)
= 96.57 MPa (T) (Max.)
Thus,
% of error = a
96.57 - 75
b100 = 22.3%
96.57
Ans.
Ans:
% of error = 22.3%
600
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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.
6–138. The curved member is made from material
having an allowable bending stress of sallow = 100 MPa.
Determine the maximum allowable internal moment M
that can be applied to the member.
M
100 mm
M
200 mm
200 mm
Internal Moment: M is negative since it tends to decrease the curvature of the
curved member.
Section Properties: Referring to Fig. a, the location of the neutral surface from the
center of curvature of the curve beam is
R =
0.1(0.2)
A
=
= 0.288539 m
dA
0.4
0.1ln
0.2
LA r
Then
e = r - R = 0.3 - 0.288539 = 0.011461 m
Allowable Bending Stress: The maximum stress occurs at either point A or B. For
point A which is in tension,
sallow =
M(R - rA)
;
AerA
100(106) =
M(0.288539 - 0.2)
0.1(0.2)(0.011461)(0.2)
M = 51 778.27 N # m = 51.8 kN # m (controls)
Ans.
For point B which is in compression,
sallow =
M(R - rB)
;
AerB
-100(106) =
M(0.288539 - 0.4)
0.1(0.2)(0.011461)(0.4)
M = 82 260.10 N # m = 82.3 kN # m
Ans:
M = 51.8 kN # m
601
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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.
6–139. The curved beam is subjected to a bending moment
of M = 50 N # m. Determine the maximum bending stress in
the beam. Also, sketch a two-dimensional view of the stress
distribution acting on section a–a.
50 mm
a
12 mm
50 mm
125 mm
a
75 mm
12 mm
140 mm
M
M
Section properties:
r =
100(50)(12) + 131.25(50)(12)
= 115.625 mm
50(12) + 50(12)
125
dA
137
= 12.5 ln
+ 50 ln
= 11.1 mm
©
r
75
125
LA
1250
A
= 112.1 mm
=
dA
8.1
LA r
R =
r - R = 115.625 - 112.1 = 3.525 mm
s =
M(R - r)
Ar(r - R)
sA =
500000(12)(112.1 - 75)
= 5.613 MPa (T) (Max)
1250(75)(3.525)
sB =
5000(12)(112.1 - 137.5)
= - 2.1 MPa = 2.1 MPa (C)
1250(137.5)(3.525)
Ans.
Ans:
smax = 5.613 MPa (T)
602
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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.
50 mm
*6–140. The curved beam is made from material having
an allowable bending stress of sallow = 168 MPa. Determine
the maximum moment M that can be applied to the beam.
a
12 mm
50 mm
125 mm
a
75 mm
12 mm
140 mm
M
r =
100(50)(12) + 131.25(50)(12)
= 115.625 mm
50(12) + 50(12)
dA
137
125
= 12.5 ln
+ 50 ln
= 11.1 mm
©
75
125
LA r
R =
1250
A
=
= 112.1 mm
dA
11.1
LA r
r - R = 115.625 - 112.1 = 3.525 mm
s =
M(R - r)
Ar(r - R)
Assume tension failure:
168 =
M(112.1 - 75)
1250(75)(3.525)
Ans.
M = 1.5 kN # m. = 1.5 kN # m (controls)
Assume compression failure:
- 168 =
M(112.1 - 137.5)
;
1250(137.5)(3.525)
M = 4 kN # m
603
M
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6–141. If P = 3 kN, determine the bending stress developed
at points A, B, and C of the cross section at section a-a.
Using these results, sketch the stress distribution on
section a-a.
D
600 mm
E
a 300 mm
A
B
C
a
P
20 mm
50 mm
25 mm
25 mm
25 mm
Section a – a
Internal Moment: The internal moment developed at section a–a can be determined by
writing the moment equation of equilibrium about the neutral axis of the cross section at
a–a. Using the free-body diagram shown in Fig. a,
a+ ©MNA = 0;
3(0.6) - Ma - a = 0
Ma - a = 1.8 kN # m
Here, M a - a is considered negative since it tends to reduce the curvature of the
curved segment of the beam.
Section Properties: Referring to Fig. b, the location of the centroid of the cross
section from the center of the beam’s curvature is
r =
0.31(0.02)(0.075) + 0.345(0.05)(0.025)
©rA
=
= 0.325909 m
©A
0.02(0.075) + 0.05(0.025)
The location of the neutral surface from the center of the beam’s curvature can be
determined from
R =
A
dA
©LA
r
where A = 0.02(0.075) + 0.05(0.025) = 2.75(10 - 3) m2
dA
0.32
0.37
©
+ 0.025 ln
= 8.46994(10 - 3) m
= 0.075 ln
r
0.3
0.32
LA
Thus,
R =
2.75(10 - 3)
8.46994(10 - 3)
= 0.324678 m
then
e = r - R = 0.325909 - 0.324678 = 1.23144(10 - 3) m
Normal Stress:
sA =
M(R - rA)
1.8(103)(0.324678 - 0.3)
= 43.7 MPa (T)
=
AerA
2.75(10 - 3)(1.23144)(10 - 3)(0.3)
Ans.
sB =
M(R - rB)
1.8(103)(0.324678 - 0.32)
= 7.77 MPa (T)
=
AerB
2.75(10 - 3)(1.23144)(10 - 3)(0.32)
Ans.
sC =
M(R - rC)
1.8(103)(0.324678 - 0.37)
= -65.1 MPa (C)
=
AerC
2.75(10 - 3)(1.23144)(10 - 3)(0.37)
Ans.
604
Ans:
sA = 43.7 MPa (T), sB = 7.77 MPa (T),
sC = -65.1 MPa (C)
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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.
6–142. If the maximum bending stress at section a-a is
not allowed to exceed sallow = 150 MPa, determine the
maximum allowable force P that can be applied to the
end E.
D
600 mm
E
a 300 mm
A
B
C
a
P
20 mm
50 mm
25 mm
25 mm
25 mm
Section a – a
Internal Moment: The internal Moment developed at section a–a can be
determined by writing the moment equation of equilibrium about the neutral axis of
the cross section at a–a.
a + ©MNA = 0;
P(0.6) - Ma - a = 0
Ma - a = 0.6P
Here, Ma–a is considered positive since it tends to decrease the curvature of the
curved segment of the beam.
Section Properties: Referring to Fig. b, the location of the centroid of the cross
section from the center of the beam’s curvature is
r =
0.31(0.02)(0.075) + 0.345(0.05)(0.025)
©r~A
=
= 0.325909 m
©A
0.02(0.075) + 0.05(0.025)
The location of the neutral surface from the center of the beam’s curvature can be
determined from
A
R =
©
dA
L
A r
where A = 0.02(0.075) + 0.05(0.025) = 2.75(10 - 3) m2
©L
dA
A r
= 0.075 ln
0.32
0.37
+ 0.025 ln
= 8.46994(10 - 3) m
0.3
0.32
Thus,
R =
2.75(10 - 3)
8.46994(10 - 3)
= 0.324678 m
then
e = r - R = 0.325909 - 0.324678 = 1.23144(10 - 3) m
Allowable Normal Stress: The maximum normal stress occurs at either points A or C.
For point A which is in tension,
sallow =
M(R - rA)
;
AerA
150(106) =
0.6P(0.324678 - 0.3)
2.75(10 - 3)(1.23144)(10 - 3)(0.3)
P = 10 292.09 N = 10.3 kN
For point C which is in compression,
sallow =
M(R - rC)
;
AerC
-150(106) =
0.6P(0.324678 - 0.37)
2.75(10 - 3)(1.23144)(10 - 3)(0.37)
P = 6911.55 N = 6.91 kN (controls)
605
Ans.
Ans:
P = 6.91 kN
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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.
6–143. The elbow of the pipe has an outer radius of
20 mm
0.75
in. and an inner radius of 17
0.63mm.
in. If the assembly is
to the
themoments
momentsofofMM= =253 lb
N #· in.,
m, determine
subjected to
determine the
maximum stress developed at section a- a.
a
30
30�
mm
125in.
M�
 25
3 Nm
lb�in.
a
dA
= ©2p (r - 2r2 - c2)
LA r
2
2
2
2
− 2022 )-– 0.75
5
– -4521.75
2p(45
) 21.752 - 0.632)
45(1.75
− 17= 2p(45
2p(1.75
) -– 2p
17 mm
0.63
in.
20 mm
0.75
in.
= 8.507713
0.32375809
in.
5
mm
2
2 2
2
) –) p(17
) 5 111p
A5
= p(20
p(0.75
- p(0.63
) = 0.1656 p
R =
A
1A r
=
dA
3 Nm
M = 25
lb�in.
111p p
0.1656
= 1.606902679
5 40.98831
mm in.
0.32375809
8.507713
– 40.98831
= 4.01169
mm
r - R = 45
1.75
- 1.606902679
= 0.14309732
in.
(smax)t =
M(R - rA)
ArA(r - R)
(smax)c = =
=
M(R - rB)
ArB(r - R)
3
25(1.606902679
- 1)
3(10
)(40.98831 – 25)
= 204
psi (T)
5 1.37
MPa
(T)
0.1656
p(1)(0.14309732)
111p(25)(4.01169)
=
Ans. Ans.
3
25(1.606902679
3(10
)(40.98831 – -65)2.5)
= 120
psi=(C)
5 –0.79
MPa
0.79 MPa Ans.
(C) Ans.
0.1656p(2.5)(0.14309732)
111p(65)(4.01169)
Ans:
(smax ) t 1.37 MPa (T), ( smax ) c 0.79 MPa (C)
606
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6–144. The curved member is symmetric and is subjected
M == 900
600 Nlb ·# ft.
to a moment
moment of
of M
m. Determine
Determine the bending
stress in the member at points A and B. Show the stress
acting on volume elements located at these points.
0.5mm
in.
12
B
2 in.
50
mm
A
11
(1)(2) == 1225
2 in2 mm2
A == 12(50)
0.5(2) ++ (25)(50)
22
1.5mm
in.
37
8 in.mm
200
1
9(0.5)(2) + +
8.6667
A 2 B (1)(2)
©rA
225(12)(50)
216.667(
2 )(25)(50)
5 220.75
=
= 8.83333
in. mm
©A
21225
1
r =
M
M
1(10)
250
25(250)
dA
10
10250
12 ln
+ cc
c ln d -d –1 25
5.5707 in.
mm
= 0.5
ln
+
c ln
d =d =0.22729
200
(250
–
200)
r
8
(10
8)
8200
LA
R =
A
dA
1A r
=
12 mm
1225
2
8.7993mm
in.
5=219.90
0.22729
5.5707
50 mm
250 mm
220.75 – 219.90
= 0.85
mm in.
r - R = 8.83333
8.7993
= 0.03398
s =
37 mm
200 mm
M(R - r)
Ar(r - R)
sA =
3
600(12)(8.7993
8)
900(10
)(219.90 -– 200)
= 510.6
(T) (T)
86.0ksi
MPa
1225(200)(0.85)
2(8)(0.03398)
sB =
3
900(10
)(219.90 -– 250)
600(12)(8.7993
10)
=5 -12.7
= =
12.7
ksi MPa
(C)
–104.1ksi
MPa
104.1
1225(250)(0.85)
2(10)(0.03398)
86.0 MPa
Ans.
Ans.
(C)
TA = 86.0 MPa
104.1 MPa
TB = 104.1 MPa
607
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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.
6–145. 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)
Ans.
Ans:
(smax)c = 792 kPa, (smax)t = 1.02 MPa
608
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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.
6–146. The fork is used as part of a nosewheel assembly for
an airplane. If the maximum wheel reaction at the end of the
fork is 3750 N, determine the maximum bending stress in the
curved portion of the fork at section a–a. There the crosssectional area is circular, having a diameter of 50 mm.
a
250 mm
a
30
a +©MC = 0;
M - 3750 N(150 - 250 sin 30°) = 0
M = 93750 N # mm
150 mm
3750 N
dA
= 2p(r - 2 r- 2 - c2)
LA r
= 2p(250 - 2(2502- (25)2)
= 7.873 mm.
A = p c2 = p (25)2 = 1963.5 mm2
A
R =
=
dA
L
Ar
1
1963.5
= 249.3966 mm.
7.873
r - R = 250 - 249.3966 = 0.6034 mm.
sA =
M(R - rA)
93750(249.3966 - 225)
=
= 8.579 MPa (T) (max)
ArA(r - R)
1963.5(225)(0.6034)
sB =
M(R - rB)
93750(249.3966 - 275)
=
= - 7.367 MPa (C)
ArB(r - R)
1963.5(275)(0.6034)
Ans.
Ans:
(smax)t = 8.579 MPa, (s max)c = - 7.367 MPa
609
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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.
6–147. If the curved member is subjected to the internal
moment of M = 800 N # m, determine the bending stress
developed at points A, B and C. Using these results, sketch
the stress distribution on the cross section.
C
50 mm
C
B
A
B
A
20 mm
30 mm
25 mm
150 mm
M
M
Internal Moment: M = -600 ft is negative since it tends to increase the curvature of
the curved member.
Section Properties: The location of the centroid of the cross section from the center
of the beam’s curvature, Fig. a, is
~
r =
165.625(31.25)(25) + 190.625(18.75)(50)
©rA
=
= 179.26 mm
©A
31.25(25) + 18.75(50)
The location of the neutral surface from the center of the beam’s curvature can be
determined from
A
R =
©
L
A
dA
r
where
A = 31.25(25) + 18.75(50) = 1718.75 mm 2
©L
dA
A r
= (25) ln
181.25
200
= 9.653 mm
+ 50 ln
150
181.25
Thus,
R =
1718.75
= 178 mm
9.653
and
e = r - R = 1.26 mm
Normal Stress:
sA =
M(R - rA)
- 800000(12)(178 - 150)
= - 69 MPa = 69 MPa (C)
=
AerA
1718.75(1.26)(150)
Ans.
sB =
M(R - rB)
- 800000(12)(178 - 181.25)
=
= 6.6 MPa (T)
AerB
1718.75(1.26)(181.25)
Ans.
sC =
M(R - rC)
- 800000(12)(178 - 200)
=
= 40.63 MPa (T)
AerC
1718.75(1.26)(200)
Ans.
The normal stress distribution across the cross section is shown in Fig. b.
Ans:
sA = 69 MPa (C), sB = 6.6 MPa (T),
sC = 40.63 MPa (T)
610
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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.
*6–148. If the curved member is made from material
having an allowable bending stress of sallow = 105 MPa,
determine the maximum allowable internal moment M that
can be applied to the member.
C
50 mm
C
B
A
B
A
25 mm
150 mm
M
M
Internal Moment: M is negative since it tends to increase the curvature of the
curved member.
Section Properties: The location of the centroid of the cross section from the center
of the beam’s curvature, Fig. a, is
©r~A
r =
=
©A
165.625(31.25)(25) + 190.625(18.75)(50)
= 179.26 mm
31.25(25) + 18.75(50)
The location of the neutral surface from the center of the beam’s curvature can be
determined from
A
R =
©
dA
L
A r
where
A = 31.25(25) + 18.75(50) = 1718.75 mm 2
©L
dA
A r
= (25)ln
181.25
200
= 9.653 mm
+ 50 ln
150
181.25
Thus,
R =
1718.75
= 178 mm
9.653
and
e = r - R = 1.26 mm
Normal Stress: The maximum normal stress can occur at either point A or C. For
point A which is in compression,
sallow =
M(rA - R)
;
AerA
-105 =
M(178 - 150)
1718.75(1.26)(150)
M = -1.2 kN # m
Ans.
(controls)
For point C which is in tension,
sallow =
M(rC - R)
;
AerC
105 =
M(178 - 200)
1718.75(1.26)(200)
M = 2.1 N # m
611
20 mm
30 mm
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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.
6–149. A 100-mm-diameter circular rod is bent into an
S shape. If it is subjected to the applied moments
M = 125 N # m at its ends, determine the maximum tensile
and compressive stress developed in the rod.
M ⫽125 N⭈m
400 mm
400 mm
M ⫽125 N⭈m
dA
LA r
= 2p(r - 2 r- 2 - c2)
= 2p(0.45 - 2 0.452 - 0.052) = 0.01750707495 m
A = p c2 = p (0.052) = 2.5 (10 - 3)p m2
R =
2.5(10 - 3)p
A
=
= 0.448606818
dA
0.017507495
1A r
r - R = 0.45 - 0.448606818 = 1.39318138(10 - 3) m
On the upper edge of each curve:
sA =
M(R - rA)
-125(0.448606818 - 0.4)
= -1.39 MPa (max) Ans.
=
ArA(r - R)
2.5(10 - 3)p(0.4)(1.39318138)(10 - 3)
sD =
M(R - rD)
125(0.448606818 - 0.5)
= -1.17 MPa
=
ArD(r - R)
2.5(10 - 3)p(0.5)(1.39318138)(10 - 3)
On the lower edge of each curve:
sB =
M(R - rB)
-125(0.448606818 - 0.5)
= 1.17 MPa
=
ArB(r - R)
2.5(10 - 3)p(0.5)(1.39318138)(10 - 3)
sC =
M(R - rC)
125(0.448606818 - 0.4)
= 1.39 MPa
=
ArC(r - R)
2.5(10 - 3)p(0.4)(1.39318138)(10 - 3)
Ans.
Ans:
smax = ; 1.39 MPa
612
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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.
6–150. The bar is subjected to a moment of M = 153 N # m.
Determine the smallest radius r of the fillets so that an
allowable bending stress of sallow = 120 MPa is not exceeded.
60 mm
40 mm
r
7 mm
M
M
r
smax = K
Mc
I
120(106) = K
J 1 (0.007)(0.04)3 K
(153)(0.02)
12
K = 1.46
60
w
=
= 1.5
h
40
From Fig. 6-43,
r
= 0.2
h
Ans.
r = 0.2(40) = 8.0 mm
Ans:
r = 8.0 mm
613
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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.
6–151. The bar is subjected to a moment of M = 17.5 N # m.
If r = 6 mm determine the maximum bending stress in the
material.
60 mm
40 mm
r
7 mm
M
M
r
w
60
=
= 1.5;
h
40
r
6
=
= 0.15
h
40
From Fig. 6–43,
K = 1.57
smax = K
17.5(0.02)
Mc
= 1.57 1
= 14.7 MPa
I
J 12(0.007)(0.04)3 K
Ans.
Ans:
smax = 14.7 MPa
614
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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.
*6–152. 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
80 mm
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
with
r
r
r
r
M M
Allowable Bending Stress:
sallow = K
7 mm
7 mm
20 mm
20 mm
r
w
80
=
= 4 and K = 1.45, then = 0.25.
h
20
h
r
= 0.25
20
Ans.
r = 5.00 mm
615
M M
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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.
6–153. 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
r
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
Maximum Bending Stress:
smax = K
Mc
I
= 1.45 B 1
17.5(0.01)
3
12 (0.007)(0.02 )
R
Ans.
= 54.4 MPa
Ans:
smax = 54.4 MPa
616
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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.
•6–15 4. 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
mm. in.
is A-36 steel. Each notch has a radius of r == 30.125
P
0.5mm
in.
12
1.75
in.
42 mm
1.25
in.
30 mm
20 mm
in.
500
b5
=
P
20mm
in.
500
20 in.
500
mm
20 in.
500
mm
1.75
- 1.25
42
– 30
= 0.25
5 6 mm
2 2
b
60.25
= 5 2; = 2;
r
0.125
3
M = 0.5 P
r
0.125
3
5 0.1
=
= 0.1
h
1.25
30
0.5 m
500
mm
From Fig. 6-44. K = 1.92
sY = K
Mc
;
I
122 lb
P ==0.469
kN
6
20P(0.625)
0.5P(10
)(15)
36
250== 1.92
1.92cc 1 1
dd
(12)(30)3 3
(0.5)(1.25)
1212
1000 mm
500
mm
Ans.
Ans:
K = 1.92, P 0.469 kN
617
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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.
6–155 5 . The simply supported notched bar is subjected to
the two loads, each having a magnitude
magnitude of P
500 lb.
N.
P =
= 100
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
radiiusof
ofrr =
= 30.125
mm. in.
P
0.5mm
in.
12
1.75
- 1.25
42
– 30
= 0.25
5 6 mm
2 2
b
60.25
= 5 2; = 2;
r
0.125
3
1.75
in.
42 mm
1.25
in.
30 mm
20 mm
in.
500
=
b5
P
20mm
in.
500
500 N
20 in.
500
mm
500 N
20 in.
500
mm
500 N
250 N · m
r
0.125
3
5 0.1
=
= 0.1
h
1.25
30
500 mm
500 N 500
mm
1000
mm
500 500 N
mm
500 N
From Fig. 6-44, K = 1.92
smax = K
3
2000(0.625)
250(10
)(15)
Mc
= 1.92 cc 1 1
d =266.7
29.5MPa
ksi
3 3d 5
I
(12)(30)
1212(0.5)(1.25)
Ans.
Ans:
smax 266.7 MPa
618
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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.
*6–156. 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
200 mm
r
7
=
= 0.175
h
40
60
w
=
= 1.5
h
40
From Fig. 6-43, K = 1.5
(sA)max = K
(35)(0.02)
MAc
= 1.5 c 1
d = 19.6875 MPa
3
I
(0.01)(0.04
)
12
(sB)max = (sA)max =
19.6875(106) =
MB c
I
175(0.2 + L2 )(0.03)
1
3
12 (0.01)(0.06 )
Ans.
L = 0.95 m = 950 mm
619
40 mm
7 mm
C
L
2
B
L
2
200 mm
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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.
6–157. 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
10 mm
6 mm
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
r
w
45
3
=
= 1.5 and
=
= 0.1, we have K = 1.77
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
Ans.
M = 41.7 N # m (Controls !)
For the larger section
smax = sallow = K
Mc
;
I
200 A 106 B = 1.77 B 1
M(0.015)
3
12 (0.015)(0.03 )
R
M = 254 N # m
Ans:
M = 41.7 N # m
620
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6–158.
beam.
Determine the shape factor for the wide-flange
15 mm
20 mm
200 mm
Mp
15 mm
200 mm
1
1
(0.2)(0.23)3 (0.18)(0.2)3 = 82.78333(10 - 6) m4
12
12
Ix =
C1 = T1 = sY(0.2)(0.015) = 0.003 sY
C2 = T2 = sY(0.1)(0.02) = 0.002 sY
Mp = 0.003 sY(0.215) + 0.002 sY(0.1) = 0.000845 sY
sY =
MYc
I
MY =
sY(82.78333)(10 - 6)
= 0.000719855 sY
0.115
k =
Mp
MY
=
0.000845 sY
= 1.17
0.000719855 sY
Ans.
Ans:
k = 1.17
621
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6–159. 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
20 mm
200 mm
Mp
15 mm
200 mm
Ix =
1
1
(0.2)(0.23)3 (0.18)(0.2)3 = 82.78333(10 - 6) m4
12
12
C1 = T1 = sY(0.2)(0.015) = 0.003 sY
C2 = T2 = sY(0.1)(0.02) = 0.002 sY
Mp = 0.003 sY(0.215) + 0.002 sY(0.1) = 0.000845 sY
= 0.000845(250)(106) = 211.25 kN # m
s¿ =
Mpc
I
211.25(103)(0.115)
=
y
0.115
=
;
250
293.5
82.78333(10 - 6)
= 293.5 MPa
y = 0.09796 m = 98.0 mm
Ans.
stop = sbottom = 293.5 - 250 = 43.5 MPa
Ans:
stop = sbottom = 43.5 MPa
622
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*6–160. Determine the shape factor for the cross section
of the H-beam.
200 mm
20 mm
Mp
200 mm
20 mm
1
1
(0.2)(0.023) + 2 a b(0.02)(0.23) = 26.8(10 - 6) m4
12
12
lx =
C1 = T1 = sY(2)(0.09)(0.02) = 0.0036 sY
C2 = T2 = sY(0.01)(0.24) = 0.0024 sY
Mp = 0.0036 sY(0.11) + 0.0024 sY(0.01) = 0.00042 sY
sY =
MYc
I
MY =
sY(26.8)(10 - 6)
= 0.000268 sY
0.1
k =
Mp
MY
=
0.00042 sY
= 1.57
0.000268 sY
Ans.
623
20 mm
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6–161. 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
20 mm
Mp
20 mm
200 mm
20 mm
Ix =
1
1
(0.2)(0.023) + 2a b(0.02)(0.23) = 26.8(10 - 6) m4
12
12
C1 = T1 = sY(2)(0.09)(0.02) = 0.0036 sY
C2 = T2 = sY(0.01)(0.24) = 0.0024 sY
Mp = 0.0036 sY(0.11) + 0.0024 sY(0.01) = 0.00042 sY
Mp = 0.00042(250)(106) = 105 kN # m
s¿ =
Mpc
I
105(103)(0.1)
=
y
0.1
=
;
250
392
26.8(10 - 6)
= 392 MPa
y = 0.0638 = 63.8 mm
Ans.
sT = sB = 392 - 250 = 142 MPa
Ans:
stop = sbottom = 142 MPa
624
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6–162. 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.
25 mm
150 mm
25 mm
Plastic Moment:
25 mm
150 mm
25 mm
MP = 250 A 106 B (0.2)(0.025)(0.175) + 250 A 106 B (0.075)(0.05)(0.075)
= 289062.5 N # m
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 =
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
stop
= sbot
= sr - sY
Ans.
= 317.14 - 250 = 67.1 MPa
Ans:
stop = sbottom = 67.1 MPa
625
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6–163. Determine the plastic moment Mp that can be
supported by a beam having the cross section shown.
sY = 210 MPa.
50 mm
25 mm
Mp
250 mm
1 s dA = 0
25 mm
C1 + C2 - T1 = 0
p(50 2 - 252)(210) + (250 - d)(25)(210) - d(25)(210) = 0
d = 242.8 mm. 6 250 mm
Mp = p(502 - 252 )(210)(57.2)
OK
+ (7.2)(25)(210)(3.6)
+ (242.8)(25)(210)(121.4)
= 226 N # m .
Ans.
Ans:
Mp = 226 N # m .
626
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6–164. Determine the shape factor of the beam’s cross
section.
in.
503 mm
Referring to Fig. a, the location of centroid of the cross-section is
y =
©yA
7.5(3)(6)
+ 3(6)(3)
125(50)(100)
+ 50(100)(50)
=
= 5.25 in.
5 87.5 mm
©A
3(6)
+ 6(3)
50(100)
+ 100(50)
in.
1006mm
The moment of inertia of the cross-section about the neutral axis is
I =
1 3
1
1
3
(3) A 63 B +3) 3(6)(5.25
- 3)2–+50)2 +(6) A 3(100)(50
- 5.25)2 – 87.5)2
(50)(100
+ 50(100)(87.5
) + 100(50)(125
B + 6(3)(7.5
12
12 12
25 mm
1.5
in. 503 mm
in.
25 mm
1.5
in.
= 19.271(10
249.75 in46) mm4
87.5.in
Thus
Here smax = sY and c = y = 5.25
. Thus
smax =
Mc
;
I
sY =
M
MYY(5.25)
(87.5)
249.75 6)
19.271(10
6
0.2202(10
MY = 47.571s
Y )sY
Referring to the stress block shown in Fig. b,
sdA = 0;
LA
T - C1 - C2 = 0
d(50)sYY-– (100
d)(50)s
50(100)s
d(3)s
(6 - –d)(3)s
- –3(6)s
0= 0
Y Y
Y = Y
d ==100
mm
6 in.
6 in.mm,
0,0,
Since d == 100
, c1 c=1 =
Fig.
c. c.
Here
Fig.
Here
T ==CC= =50(100)s
5000s
3(6) sYY == 18
sY Y
Thus,
T(4.5)
= 18 sYY(75)
(4.5)
= 81 ss
MP ==T(75)
= 5000s
= 375000
YY
Thus,
k =
MP
81
sY Y
375000s
=
= 1.70
5 1.70
MY
47.571
sY6)sY
0.2202(10
Ans.
50 mm
50 mm
75 mm
100 – d
100 mm
627
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*6–165. 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 ==250
36 MPa.
ksi.
503 mm
in.
Referring to Fig. a, the location of centroid of the cross-section is
1006mm
in.
135(50)(100)
+ 50(100)(50)
©yA
7.5(3)(6)
+ 3(6)(3)
5 87.5 mm
y =
=
= 5.25 in.
50(100)
+
©A
3(6) + 6(3)100(50)
25 mm
1.5
in. 503 mm
in.
The moment of inertia of the cross-section about the neutral axis is
25 mm
1.5
in.
1 3
1
1
3
(50)(100
+ 50(100)(87.5
+ (100)(50
) + 100(50)(125
I =
(3)(63) +3)3(6)(5.25
- 3)2 –+50)2 (6)(3
) + 6(3)(7.5
- 5.25)2 – 87.5)2
12
12
12
6
4
= 19.271(10
249.75 in4 ) mm
MPa
87.5inmm.
Then
Here, smax = sY = 250
andand
. Then
36 ksi
¢ =¢ y= y= =5.25
smax =
Mc
;
I
250==
36
(87.5)
MMYY(5.25)
19.271(10
249.75 6)
6 #
55.06(10kip
) Nin
· mm
= 55.06
MM
= 143
kip # kN
ft · m
YY== 1712.57
Ans.
Referring to the stress block shown in Fig. b,
sdA = 0;
LA
T - C1 - C2 = 0
d(50)(250)
(100
– d)(50)(250)
– 50(100)(250)
d(3)
(36) -–(6
- d)(3)(36)
- 3(6)
(36) = 0 = 0
d == 100
6 in.mm
6 in.mm,
0,0,
Since d == 100
, c1 c=1 =
Here,
T
1250
T ==CC= =50(100)(250)
3(6)(36) = =648
kip000 N = 1250 kN
Thus,
# inkN
kN (0.75
m) =kip
93.75
· m kip # ft
T(4.5)= =1250
648(4.5)
= 2916
= 243
MP ==T(4.5)
Ans.
50 mm
50 mm
100 – d
75 mm
100 mm
Ans:
MY = 55.06 kN # m, MP = 93.75 kN # m
628
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6–166. Determine the shape factor for the cross section of
the beam.
10 mm
15 mm
200 mm
Mp
10 mm
I =
200 mm
1
1
(0.2)(0.22)3 - (0.185)(0.2)3 = 54.133(10 - 6) m4
12
12
C1 = sY(0.01)(0.2) = (0.002) sY
C2 = sY(0.1)(0.015) = (0.0015) sY
Mp = 0.002 sY(0.21) + 0.0015 sY(0.1) = 0.0005 sY
sY =
MYc
I
MY =
sY(54.133)(10 - 6)
= 0.000492 sY
0.11
k =
Mp
MY
=
0.00057 sY
= 1.16
0.000492 sY
Ans.
Ans:
k = 1.16
629
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6–167. The beam is made of an elastic-plastic material for
which sY = 200 MPa. If the largest moment in the beam
occurs within the center section a-a, determine the
magnitude of each force P that causes this moment to be
(a) the largest elastic moment and (b) the largest plastic
moment.
P
P
a
a
2m
2m
2m
2m
200 mm
100 mm
(1)
M = 2P
a) Elastic moment
I =
1
(0.1)(0.23) = 66.667(10 - 6) m4
12
sY =
MYc
I
MY =
200(106)(66.667)(10 - 6)
0.1
= 133.33 kN # m
From Eq. (1)
133.33 = 2 P
Ans.
P = 66.7 kN
b) Plastic moment
b h2
sY
4
Mp =
=
0.1(0.22)
(200)(106)
4
= 200 kN # m
From Eq. (1)
200 = 2 P
Ans.
P = 100 kN
Ans:
Elastic: P = 66.7 kN,
Plastic: P = 100 kN
630
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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.
6–168 .
Determine the shape factor of the cross section.
Let a = 50 mm.
The moment of inertia of the cross-section about the neutral axis is
503mm
in.
1
11
29
33 +
4
II == 1(a)(3a)
(2a)(a3)3 = a4
1212 (3)(9 ) +1212 (6) (3 ) =12195.75 in
Here, smax = sY and c = 1.5a. Then
Here, smax = sY and c = 4.5 in. Then
Mc
M (1.5a)
smax = Mc;
sY =M Y(4.5) 4
Y
I
smax =
sY = (29/12)a
;
I
195.75
29
a saY3b
Y =43.5
MYM=
18
503mm
in.
503mm
in.
50 3mm
in.
Referring to the stress block shown in Fig. a,
2
T11 ==CC1 1= =a(a)s
3(3)s
Y a=s9YsY
T
Y=
1.5(9)sY = =13.5
s2Y)s
T2 ==CC2= =(0.5a)(3a)s
T
(1.5a
2
2
Y
Y
Thus,
Thus,
MP = T1(6) + T2(1.5)
MP = T1(2a) + T2(0.5a)
= 9sY(6) + 13.5sY(1.5) = 74.25 sY
= (a2)sY(2a) + (1.5a2)sY(0.5a) = (2.75a3)sY
MP
74.25 s
3Y
(2.75a
)sY= 1.71
k k== MP =
=
MY
43.5 sY = 1.707
29
MY
a a3bsY
18
Ans.
a
0.5a
0.5a
2a
0.5a
a
631
in.
503mm
in.
503 mm
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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.
*6–169 . 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 ==250
36 MPa.
ksi .
50
mm
3 in.
50
mm
3 in.
The moment of inertia of the cross-section about the neutral axis is
50
mm
3 in.
1 1
1
3
3
3
6
(3)(9
) +3) + (6)(3
) = 195.75
in4
I =
(50)(150
(100)(50
) = 15.104(10
) mm4
12
12 12
Here, smax = sY = 36
andand
. Then
c =c 4.5
250ksi
MPa
= 75inmm.
Then
Mc
;
smax =
I
50
mm
3 in.
50
mm
3 in.
50
mm
3 in.
MM
Y (4.5)
Y(75)
36 ==
250
195.75 6)
15.104(10
MMYY == 1566
kip 6# )inN =· mm
130.5
kip # ftkN · m
50.35(10
= 50.35
Ans.
Referring to the stress block shown in Fig. a,
T1 ==CC1 1= =50(50)(250)
625kip
000 N = 625 kN
3(3)(36) ==324
1.5(9)(36) == 486
T2 ==CC2 2= =25(150)(250)
937 kip
500 N = 937.5 kN
Thus,
+ T2(1.5)
MPP ==TT1(100
M
+ T2(25 mm)
1(6) mm)
324(6) ++ 937.5(0.025)
486(1.5)
==625(0.1)
# in. = 222.75 kip # ft = 223 kip # ft
2673kN
kip· m
==85.9
Ans.
50 mm
25 mm
25 mm
100 mm
25 mm
50 mm
Ans:
MY = 50.35 kN # m, Mp = 85.9 kN # m
632
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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.
6–17 0 . The box beam is made from an elastic-plastic
Determine the
the magnitude
material for
for which
whichssYY==250
36MPa.
ksi. Determine
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
2.4
m
6 ft
1.8
m
The moment of inertia of the beam’s cross-section about the neutral axis is
1.86 m
ft
6 in.
150
mm
1
3 11
6
3
(150)(300
) – (5)(10
(125)(250
) = 174.74(10
) mm4
(6)(123) ) = 3447.33
in4
I =
1212
12
12 in.
300
mm
in.
25010mm
250ksi
MPa
150 mm
Here, smax = sY = 36
andand
c =c 6= in.
smax =
Mc
;
I
250==
36
5 in.
125
mm
MM
(150)
Y Y(6)
447.33 6)
174.74(10
291.2(10
= 291.2
MM
kip 6#)inN =· mm
223.67
kip # kN
ft · m
YY== 2684
It is required that
Mmax = MY
6P 5
= 291.2
223.67
1.8P
Ans.
P
= 161.8
37.28 kN
kip = 37.3 kip
P5
1.8 m
2.4 m
1.8 m
Referring to the stress block shown in Fig. b,
6(1)(36) = =216
kip N = 937.5 kN
T11 ==CC1 1= =150(25)(250)
937500
T22 ==CC2 2= =125(25)(250)
781250
5(1)(36) = =180
kip N = 781.25 kN
1.8P
Thus,
1.8P
MP = T1(11) + T2(5)
x (m)
= 216(11)
+ 180(5)
937.5(275)
+ 781.25(125)
# ft kN · m
355468.75
kN=· mm
= 355.5
= 3276
kip # in
273 kip
It is required that
Mmax = MP
25 mm
1.8P
6P 5
= 355.5
273
Ans.
P
= 197.5
45.5 kip
P5
kN
125 mm
125
mm
275 mm
125 mm
25 mm
Ans:
(a) P 161.8 kN (b) P 197.5 kN
633
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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.
6–171. 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 =
ri
p
A r 4 - r4i B
4 o
With c = ro and smax = sY,
smax =
Mc
;
I
sY =
MY =
MY(ro)
p
A r 4 - ri 4 B
4 o
p
A r 4 - ri 4 B sY
4ro o
Plastic Moment. The plastic moment of the cross section can be determined by
superimposing the moment of the stress block of the solid beam with radius r0 and ri
as shown in Fig. a, Referring to the stress block shown in Fig. a,
T1 = C1 =
p 2
r s
2 o Y
T2 = C2 =
p 2
r s
2 i Y
MP = T1 c 2a
4ro
4ri
b d - T2 c 2 a
bd
3p
3p
=
8ro
8ri
p 2
p
r s a
b - ri 2sY a
b
2 o Y 3p
2
3p
=
4
A ro 3 - ri 3 B sY
3
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 r 4 - ri 4 B sY
4ro o
Ans.
Ans:
k =
634
16ro A ro3 - r3i B
3p A ro4 - r4i B
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*6–172.
Determine the shape factor for the member.
–h
2
–h
2
Plastic analysis:
T = C =
MP =
b
1
h
bh
(b)a bsY =
s
2
2
4 Y
bh
h
b h2
sY a b =
s
4
3
12 Y
Elastic analysis:
I = 2c
h 3
1
b h3
(b)a b d =
12
2
48
sY A 48 B
sYI
b h2
MY =
=
=
s
h
c
24 Y
2
bh3
Shape Factor:
k =
Mp
MY
=
bh2
12 sY
bh2
24 sY
Ans.
= 2
635
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*6–173 . 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.
mm,
h =6 150
250 MPa.
Take b == 100
sY =sY36= ksi.
4 in.,
h =
in., mm,
–h
2
Elastic analysis:
I = 2 cc
MY =
–h
2
1
(4)(3)3 d 3)=d =
187031250
in4
(100)(75
mm4
12
36(18)
sYI
250(7031250)
6
# in. = 18
# ft = 23.44 kN · m
=
= 216 5
kip23.44(10
kip
)N
· mm
c
3 75
Ans.
b
Plastic analysis:
T = C =
1
(100)(75)(250)
= 937500
N = 937.5 kN
(4)(3)(36) = 216
kip
2
150
6
Mp =
= 937.5
2160 a b b=46875
432 kip
= =3646.875
kip # ftkN · m
kN# ·in.
mm
33
Ans.
h
3
Ans:
MY = 23.44 kN # m,
Mp = 46.875 kN # m
636
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w
6–174. The beam is made of an elastic plastic material for
which sY 5 210 MPa. If the largest moment in the beam
occurs at the center section a–a, determine the intensity of
the distributed load w that causes this moment to be (a) the
largest elastic moment and (b) the largest plastic moment.
a
a
3m
3m
200 mm
200 mm
M = 4.5 w
(1)
(a) Elastic moment
I =
1
(200)(200 3) = 133.33(10 6) mm4
12
sY =
MYc
I
MY =
210 (133.33(10 6 ))
100
= 280000 N.m
From Eq. (1),
280000 = 4.5 w
w = 62.22 kN> m
Ans.
(b) Plastic moment
C = T = 210(200)(100) = 4200 kN
Mp = 4200 (0.1) = 420 kN.m
From Eq. (1)
420 = 4.5 w
w = 93.33 kN> m
Ans.
Ans:
Elastic: w = 62.22 kN> m,
Plastic: w = 93.33 kN> m
637
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6–175. The box beam is made from an elastic-plastic
Determinethe
the intensity of
material for which
which ssYY==175
25MPa.
ksi. Determine
the distributed load w0 that will cause the moment to be
(a) the largest elastic moment and (b) the largest plastic
moment.
w00
w
Elastic analysis:
I =
ft
39 m
1
3
3 11
3
6
(200)(400
) – (6)(12
(150)(300
) = 729.167(10
) mm4
(8)(163) ) = 1866.67
in4
1212
12
Mmax =
sYI
;
c
ft
39 m
8 in.mm
200
25(1866.67) 6)
175(729.167)(10
6
27w
3w0(10
) ==
0(12)
8200
16 in.
400
mm
12mm
in.
300
Ans.
18.0 kip>ft
w0 ==212.67
kN/m
Plastic analysis:
6 in.
150
mm
25(8)(2) = =
400
kip 6) N = 1750 kN
C1 == TT1 1==175(200)(50)
1.75(10
1 w (6) = 3w
0
2 0
6
C2 == TT2 2==175(150)(50)
1.3125(10
) N = 1312.5 kN
25(6)(2) = =
300
kip
+ 300(6)
= 7400 kip= #809.375
in.
MP = 400(14)
1750(0.350)
+ 1312.5(0.150)
kN · m
= 7400
27w
3w0 0=(12)
809.375
kN · m
3m
1.5w0
3m
1.5w0
Ans.
w0 == 269.79
22.8 kip>ft
kN/m
1.5w0
Mmax = 3w0
2m
1.5w0
150 350 mm
mm
Ans:
(a) w 0 212.67 kN/m (b) w 0 269.79 kN/m
638
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*6–176. The wide-flange member is made from an elasticplastic material. Determine the shape factor.
t
h
t
t
b
Plastic analysis:
T1 = C1 = sYb t ;
T2 = C2 = sY a
Mp = sY b t(h - t) + sY a
= sY[b t(h - t) +
h - 2t
bt
2
h - 2t
h - 2t
b(t) a
b
2
2
t
(h - 2t)2]
4
Elastic analysis:
I =
1
1
b h3 (b - t)(h - 2t)3
12
12
=
1
[b h3 - (b - t)(h - 2t)3]
12
MY =
1
) [bh3 - (b - t)(h - 2t)3]
sY(12
sYI
=
h
c
2
=
bh3 - (b - t)(h - 2t)3
sY
6h
Shape Factor:
k =
Mp
MY
=
=
[b t(h - t) + 14(h - 2t)2]sY
bh3 - (b - t)(h - 2t)3
sY
6h
3h 4b t(h - t) + t(h - 2t)2
c
d
2 b h3 - (b - t)(h - 2t)3
Ans.
639
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*6–177. The beam is made of a polyester that has the
stress–strain curve shown. If the curve can be represented
–1
–1
by the equation s == [140
tan-1
(15e) MPa,where
tan-1
(15e)
[20 tan
115P2]
ksi, where tan
115P2
is in radians, determine the magnitude of the force P that
can be applied to the beam without causing the maximum
strain in its fibers at the critical section to exceed
Pmax = 0.003 mm/mm.
in.>in.
P
in.
502 mm
4100
in.mm
8 ftm
2.4
�s(ksi)
s(MPa)
8 ftm
2.4
�1
�140
20 tan
(15P)
P)
ss
tan1(15
P(in./in.)
P(mm/mm)
occurs
at at
4.00P
Maximum Internal Moment: The maximum internal moment M ==1.2P
occurs
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
.
0.00006y.
Mmax = 1.2P
2.4 m
= 140
tan-–11 (15e)
s =
20 tan
= 140
tan-–11 [15(0.0015y)]
[15(0.00006y)]
=
20 tan
e = 0.00006y
= 140
tan-–11 (0.0225y)
(0.0009y)
=
20 tan
smax
When emax == 0.003
andand
0.003mm/mm,
in.>in., y y== 250in.mm
= =
0.8994
ksi
smax
6.30 MPa
–6.30
s (MPa)
e (mm/mm)
Resultant Internal Moment: The resultant internal moment M can be evaluated
from the integal
M = 2
LA
LA
00
L
50 mm
ysdA
6.30 MPa
50
2inmm
= 2
6.30
ysdA.
-
1 –1
(0.0009y)](50dy)
y Cy[140
20 tantan
(0.0225y)
D (2dy)
50 mm
2in 50 mm
–1
= 80
(0.0225y)
dy dy
14000 y tan -y1tan
(0.0009y)
L0 0
50 mm
6.30 MPa
2 22 2
50 mm
2in.
1 +1 (0.0225)
y y - 1 –1
+ (0.0009)
y
2
tan
(0.0009y)
–
d
p
tan
(0.0225y)
= 14000
80 B c
R
2 2
2(0.009) 0 0
2(0.009)
2(0.0225)
2(0.0225)
= 4.798
kip #3)inN · mm = 524.79 N · m
524.79(10
Equating
M ==1.2P
4.00P(12)
=N
4.798
M
= 524.79
·m
Ans.
P
0.100 kip
P=
= 437.32
N = 100 lb
Ans:
P = 437.32 N
640
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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.
6–178. 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
0.04
compression
Ultimate Moment:
80
100
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
Since
P
0.04
=
, then 0.02 - d = 25Pd.
d
0.02 -d
40(106)
s
s
.
=
= 2(109), then P =
P
0.02
2(109)
And since
So then 0.02 - d =
23sd
= 1.25(10 - 8)sd.
2(109)
Substituting for 0.02 - d, then solving for s, yields s = 74.833 MPa. Then
P = 0.037417 mm>mm and d = 0.010334 m.
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)
Ans.
= 94.7 N # m
Ans:
M = 94.7 N # m
641
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6–179. 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.
in.
75 3mm
M
in.
502 mm
�s(MPa)
(ksi)
s
B
B
180
ss
�1260
BB
� 980
140
sAA 
A
A
0.01
0.01
0.04
0.04
(in./in.)
P (mm/mm)
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
–1260
9.375 mm
sA I
M =
c
=
s (MPa)
980
11
140
(2)(33) D3
980[C 12
12 (50)(75 )]
1.5
37.5
1260
# ft kN · m
= 45.94(10
420 kip # 6in
kip
) N=· 35.0
mm =
45.94
b)
9.375 mm
e (mm/mm)–980
Ans.
28.125 mm
The Ultimate Moment :
9.375 mm
C1 = T1 =
1
(1260++ 180)(1.125)(2)
980)(28.125)(50)
1.575(10
(140
= =360
kip 6) N = 1575 kN
2
C2 = T2 =
1
(980)(9.375)(50)
229687.5
(140)(0.375)(2) == 52.5
kip N = 229.7 kN
2
12.5 mm
48.05 mm
28.125 mm
M == 1575(0.04805)
360(1.921875)++229.7(0.0125)
52.5(0.5)
718.125
== 78.54
kN kip
· m # in = 59.8 kip # ft
Ans.
Note: The centroid of a trapezodial area was used in calculation of moment.
Ans:
(a) M 45.94 kN · m, (b) M 78.54 kN · m
642
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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.
*6–180. 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
100 mm
M
30 mm
P (mm/ mm)
Pmax = 0.02
smax = 10 A 106 B (0.02)1>4 = 3.761 MPa
P
0.02
=
y
0.05
P = 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 =
LA
y s dA = 2
M = 0.47716 A 106 B
L0
0.05
L0
4
y5>4dy = 0.47716 A 106 B a b (0.05)9>4
9
Ans.
M = 251 N # m
643
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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.
•6–181.
�s
(ksi)
s(MPa)
90
630
80
560
in. M
1004mm
MPa
s == 574
82 ksi
C1 = T1 =
1
(0.3333)(80 ++ 574)(75)
82)(3) == 81
kip N = 354.4 kN
(8.333)(560
354361
2
C2 = T2 =
1
(31.667)(420
560)(75)
1163762
(1.2666)(60 ++ 80)(3)
= =266
kip N = 1163.8 kN
2
C3 = T3 =
1
(10)(420)(75)==36
157500
(0.4)(60)(3)
kip N = 157.5 kN
2
0.006
0.006
0.025
0.025
0.05
0.05
P (in./in.)
(mm/mm)
y1 = 10 mm
yz = 41.667 mm
in.
753 mm
53.17 mm
13.33 mm
- 80
80
ss –-560
63090– 560
5=
; ;
0.03 –-0.025
0.025 0.05
0.05– 0.025
- 0.025
60
420
91.70 mm
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.
420 MPa
560 MPa 574 MPa
M == 354.4(0.09170)
81(3.6680) + 266(2.1270)
+ 36(0.5333)
+ 1163.8(0.05317)
+ 157.5(0.01333)
882.09
== 96.45
kNkip
· m# in. = 73.5 kip # ft
Ans.
Note: The centroid of a trapezodial area was used in calculation of moment areas.
Ans:
s 574 MPa, M 96.48 kN · m
644
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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.
•6–182 .
�s
(ksi)
s(MPa)
90
630
80
560
in. M
1004mm
MPa
s == 574
82 ksi
0.006
0.006
C1 = T1 =
1
(0.3333)(80 ++ 574)(75)
82)(3) == 81
kip N = 354.4 kN
(8.333)(560
354361
2
C2 = T2 =
1
(31.667)(420
560)(75)
1163762
(1.2666)(60 ++ 80)(3)
= =266
kip N = 1163.8 kN
2
C3 = T3 =
1
(10)(420)(75)==36
157500
(0.4)(60)(3)
kip N = 157.5 kN
2
0.025
0.025
0.05
0.05
P (in./in.)
(mm/mm)
y1 = 10 mm
yz = 41.667 mm
in.
753 mm
53.17 mm
13.33 mm
- 80
80
ss –-560
63090– 560
5=
; ;
0.03 –-0.025
0.025 0.05
0.05– 0.025
- 0.025
60
420
91.70 mm
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 = 0.05.
420 MPa
560 MPa 574 MPa
M == 354.4(0.09170)
81(3.6680) + 266(2.1270)
+ 36(0.5333)
+ 1163.8(0.05317)
+ 157.5(0.01333)
882.09
== 96.45
kNkip
· m# in. = 73.5 kip # ft
Ans.
Note: The centroid of a trapezodial area was used in calculation of moment areas.
Ans:
s 574 MPa, M 96.48 kN · m
645
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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.
6–183. Determine the shape factor for the wide-flange beam.
20 mm
30 mm
180 mm
Mp
20 mm
180 mm
I =
1
1
(0.18)(0.223) (0.15)(0.183)
12
12
= 86.82(10 - 6) m4
Plastic moment:
Mp = sY(0.18)(0.02)(0.2) + sY(0.09)(0.03)(0.09)
= 0.963(10 - 3)sY
Shape Factor:
MY =
k =
sY(86.82)(10 - 6)
sYI
=
= 0.789273(10 - 3)sY
c
0.11
Mp
MY
0.963(10 - 3)sY
=
0.789273(10 - 3)sY
Ans.
= 1.22
Ans:
k = 1.22
646
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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.
*6–184. 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.
20 mm
30 mm
180 mm
Mp
20 mm
180 mm
I =
1
1
(0.18)(0.223) (0.15)(0.183)
12
12
= 86.82(10 - 6) m4
Plastic moment:
Mp = 250(106)(0.18)(0.02)(0.2)
+ 250(106)(0.09)(0.03)(0.09)
= 240750 N # m
Applying a reverse Mp = 240750 N # m
sp =
Mpc
I
240750(0.11)
=
86.82(10 - 6)
= 305.03 MPa
Ans.
¿
= 305 - 250 = 55.0 MPa
s¿top = sbottom
647
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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.
w
6–185. 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.
A
2/3 L
+ c ©Fy = 0;
a + ©M = 0;
2wL
1 w 2
x = 0
27
2 L
x =
4
L = 0.385 L
A 27
M +
1w
1
2wL
(0.385L) = 0
(0.385L)2 a b (0.385L) 2L
3
27
M = 0.0190 wL2
648
C
B
1/3 L
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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.
Problem 6-186
The beam is constructed from four pieces of wood, glued together as shown. If the internal bending
moment is M = 120 kN·m, determine the maximum bending stress in the beam. Sketch a
three-dimensional view of the stress distribution acting over the cross section.
Given:
bo := 300 mm
do := 300 mm
bi := 250 mm
di := 250 mm
M := 120kN⋅ m
Solution:
Section Property :
I :=
1 ⎛
3
3
⋅ ⎝ bo⋅ do − bi⋅ di ⎞⎠
12
Maximum Bending Stress:
σ=
M⋅ c
I
cmax := 0.5do
M⋅ cmax
σ max :=
I
σ max = 51.51 MPa
Ans.
Ans:
n = 18.182, I = 0.130578(10 - 3) m4,
M = 14.9 kN # m
649
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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.
Problem 6-18 7
For the section, Iy =31.7(10-6) m4, Iz = 114(10-6) m4, Iyz = 15.1(10-6) m4. Using the techniques outline
in Appendix A, the member's cross-sectional area has principal moments of inertia of Iy' = 29(10-6) m
and Iz' = 117(10-6) m4, computed about the principal axes of inertia y' and z', respectively. If the section
is subjected to a moment of M = 2 kN·m directed as shown, determine the stress produced at point A,
(a) using Eq. 6-11 and (b) using the equation developed in Prob. 6-111.
Given:
M := 2000N⋅ m
θ := 10.10deg
b1 := 80mm
b2 := 140mm
h1 := 60mm
h2 := 60mm
( − 6 ) m4
−6 4
Iyz := 15.1 ( 10 ) m
−6 4
Iy' := 29.0 ( 10 ) m
Iy := 31.7 10
Solution:
( − 6 ) m4
Iz := 114 10
( − 6 ) m4
Iz' := 117 10
θ' := −θ
yA := b2
Coordinates of Point A :
zA := h2
⎛⎜ y'A ⎞ ⎛ cos ( θ' ) −sin ( θ' ) ⎞ ⎛⎜ yA ⎞
:= ⎜
⋅
⎜ z'A
⎜ zA
(
)
(
)
sin
θ'
cos
θ'
⎝
⎠
⎝ ⎠
⎝ ⎠
⎛⎜ y'A ⎞ ⎛ 148.35 ⎞
=⎜
mm
⎜ z'A
34.52
⎝
⎠
⎝ ⎠
a) Using Eq. 6-11
Internal Moment Components :
My' := M⋅ sin ( θ )
Mz' := M⋅ cos ( θ )
Bending Stress:
⎛ Mz'⋅ y'A My'⋅ z'A ⎞
σ A := ⎜ −
+
Iz'
Iy'
⎝
⎠
σ A = −2.079 MPa (C)
Ans
b) Using the equation developed in Prob. 6-111
Internal Moment Components :
My := 0
Mz := M
Bending Stress: Using formula developed in Prob. 6-111.
2
D := Iy⋅ Iz − Iyz
⎛⎜ Iy Iyz ⎞ ⎛⎜ −yA ⎞
1
σ A :=
M My ⋅
⋅
⎜ Iyz Iz ⎜ zA
D z
⎝
⎠⎝
⎠
(
)
σ A= −2.086 MPa (C)
Ans.
Ans:
M = 26.4 kN # m
650
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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.
*6–188. A shaft is made of a polymer having a parabolic
upper and lower 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
50 mm
50 mm
Maximum Bending Stress: The moment of inertia about y axis must be determined
first in order to use flexure formula
I =
LA
y2 dA
100 mm
= 2
L0
y2 (2z) dy
100 mm
= 20
L0
y2 2100 - y dy
100 mm
3
5
7
3
8
16
= 20 B - y2 (100 - y)2 y (100 - y)2 (100 - y)2 R 2
2
15
105
0
= 30.4762 A 106 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
b y d (2z dy) r
100
smax 100 mm 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
Ans.
smax = 0.410 N>mm2 = 0.410 MPa
651
x
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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.
Problem 6-189
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.
Given:
bf := 290 mm
tf := 15 mm
tw := 20 mm
dw := 125 mm
M := 650 N⋅ m
Solution: D := dw + tf
⎯
⎯ Σ ⋅ yi⋅ Ai
y=
Σ ⋅ ( Ai)
(
yc :=
)
(bf⋅ tf)⋅ 0.5tf + 2(dw⋅ tw)⋅ (0.5dw + tf)
bf⋅ tf + 2dw⋅ tw
yc = 44.933 mm
(
)(
)
If :=
1
3
2
⋅ bf⋅ tf + bf⋅ tf ⋅ yc − 0.5tf
12
Iw :=
1
3
2
⋅ 2tw ⋅ dw + 2tw⋅ dw ⋅ ⎡⎣yc − 0.5dw + tf ⎤⎦
12
( )
(
(
)
4
I := If + Iw
Bending Stress:
)
I = 17990374.89 mm
σ = M⋅
c
I
At B:
cB := yc − tf
σ B := M⋅
cB
I
σ B = 1.0815 MPa
At A:
cA := yc
σ A := M⋅
cA
I
σ A = 1.6235 MPa
F = 5.883 kN
Ans.
Resultant Force: For the top board.
(
)(
F := 0.5 σ A + σ B ⋅ bf⋅ tf
)
Ans:
FR = 5.88 kN
652
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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.
6–190. The curved beam is subjected to a bending
moment of M = 85 N # m as shown. Determine the stress at
points A and B and show the stress on a volume element
located at these points.
M 85 Nm
100 mm
A
400 mm
B
A
20 mm
15 mm
150 mm
30
B
20 mm
r2
0.57
0.59
dA
0.42
= b ln
+ 0.015 ln
+ 0.1 ln
= 0.1 ln
r
r
0.40
0.42
0.57
1
LA
= 0.012908358 m
A = 2(0.1)(0.02) + (0.15)(0.015) = 6.25(10 - 3) m2
R =
A
=
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
Ans.
sA = 225 kPa (C)
sB =
M(R - rB )
85(0.484182418 - 0.40)
=
ArB (r - R)
6.25(10 - 3)(0.40)(0.010817581)
= 265 kPa (T)
Ans.
Ans:
sA = 225 kPa (C), sB = 265 kPa (T)
653
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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.
6–191. Draw the shear and moment diagrams for the
beam and determine the shear and moment in the beam as
8 .m.
functions of x, where 0 … x 6 1.
6 ft
40
kN
8 kip
30
kN/m
2 kip/ft
75
50 kNm
kip�ft
x
1.8
6 ftm
+ c ©Fy = 0;
20 –-30x
2x –-VV= =
94
0 0
40 kN
243.6 kN · m
30 kN/m
75 kN · m
1.8 m
1.2 m
Ans.
– 30x
V =V
20= -942x
c + ©MNA = 0;
1.2
4 ftm
94 kN
xx
20x– 243.6
- 166– -30x
2xaa b b– M
-=
M0 = 0
94
22
V (kN)
Ans.
20x –-243.6
166
M == –15x
- x22 +
+ 94x
40
M (kN · m)
–48
–123
30x
–243.6
94 kN
Ans:
V 94 30x, M 15x 2 94x 243.6
654
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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.
*6–192. A wooden beam has a square cross section as
shown. Determine which orientation of the beam provides
the greatest strength at resisting the moment M. What is the
difference in the resulting maximum stress in both cases?
a
M
a
M
a
(a)
Case (a):
smax =
M(a>2)
Mc
6M
= 1 4 = 3
I
a
12 (a)
Case (b):
I = 2c
3
1 2
1
1
1
2
1
1
2
ab a bd d = 0.08333 a4
ab c a
ab a
ab + a
ab a
a
3
2 22
36 22
22
22
22
1
ab
Ma
8.4853 M
Mc
22
=
smax =
=
I
0.08333 a4
a3
Case (a) provides higher strength since the resulting maximum stress is less for a
given M and a.
Ans.
Case (a)
¢smax =
8.4853 M
6M
M
- 3 = 2.49a 3 b
a3
a
a
Ans.
655
a
(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.
6–193. 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
656
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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.
6–194. 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
Ans.
u = 45°
Orientation of Neutral Axis:
tan a =
Iz
Iy
tan u
tan a = (1) tan (45°)
Ans.
a = 45°
Ans:
smax =
6M
(cos u + sin u),
a3
u = 45°, a = 45°
657