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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.
3–1. A tension test was performed on a steel specimen
having an original diameter of 12.5 mm and gauge length of
50 mm. The data is listed in the table. Plot the
stress–strain diagram and determine approximately the
modulus of elasticity, the yield stress, the ultimate stress, and
the rupture stress. Use a scale of 25 mm = 140 MPa and
25 mm= 0.05 mm >mm. Redraw the elastic region, using the
same stress scale but a strain scale of 25 mm = 0.001 mm >mm.
A =
1
p(12.5)2
4
L = 50 mm.
s(M Pa)
0
Load (kN) Elongation (mm)
0
7.0
21.0
36.0
50.0
53.0
53.0
54.0
75.0
90.0
97.0
87.8
83.3
= 122.65 mm2
e(mm> mm)
0
0.0125
0.0375
0.0625
0.0875
0.125
0.2
0.5
1.0
2.5
7.0
10.0
11.5
0
57.07
0.00025
171.21
0.00075
293.51
0.00125
407.66
0.00175
432.12
0.0025
432.12
0.0040
440.27
0.010
611.49
0.020
733.79
0.050
790.86
0.140
715.85
0.200
679.16
Eapprox
0.230
336
=
= 224(103) M Pa
0.0015
Ans.
Ans:
(sult)approx = 770 MPa, (sR) approx = 651.7 MPa,
(sY)approx = 385 MPa, Eapprox = 224(10 3) M Pa
139
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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.
3–2. Data taken from a stress–strain test for a ceramic are
given in the table. The curve is linear between the origin and
the first point. Plot the diagram, and determine the modulus
of elasticity and the modulus of resilience.
Modulus of Elasticity: From the stress–strain diagram
=
E5
33.2 –-00
232.4
55.3 A 103)3 BMPa
ksi = 387.3 GPa
5 =387.3(10
0.0006
0.0006 –-0 0
Ans.
Modulus of Resilience: The modulus of resilience is equal to the area under the
linear portion of the stress–strain diagram (shown shaded).
#
1
N1
mmlb
in.
3 in lb
ut = (232.4)a
b a0.0006
MJ/m3
ut = 2(33.2)
103 B ¢ 2b≤=¢0.0697
0.0006 N · ≤mm/mm
= 9.96 = 0.0697
A
2
mm2
mm
in.
in
in3
Ans.
S(MPa)
(ksi)
eP (mm/mm)
(in./in.)
00
232.4
33.2
45.5
318.5
49.4
345.8
51.5
360.5
53.4
373.8
0 0
0.0006
0.0006
0.0010
0.0010
0.0014
0.0014
0.0018
0.0018
0.0022
0.0022
420
350
280
232.4
210
140
70
e (mm/mm)
Ans:
E = 387.3 GPa , ur = 0.0697 MJ/m 3
140
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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.
3–3. Data taken from a stress–strain test for a ceramic are
given in the table. The curve is linear between the origin
and the first point. Plot the diagram, and determine
approximately the modulus of toughness. The rupture stress
is sr = 53.4
373.8ksi.
MPa
(MPa)
S (ksi)
00
232.4
33.2
45.5
318.5
49.4
345.8
51.5
360.5
53.4
373.8
Modulus of Toughness: The modulus of toughness is equal to the area under the
stress–strain diagram (shown shaded).
11
N lb
mm in.
(ut)approx == (232.4)a
+ 0.0010)a
(33.2) A 103 B 2¢b(0.0004
+ 0.0010) ¢b ≤
≤ (0.0004
22
mm in2
mm in.
e P(mm/mm)
(in./in.)
0 0
0.0006
0.0006
0.0010
0.0010
0.0014
0.0014
0.0018
0.0018
0.0022
0.0022
s (MPa)
420
N lb
mmin.
+
+ 318.5
45.5 A a103 B ¢2 b(0.0012)a
≤ (0.0012) ¢ b ≤
mm in2
mmin.
373.8
350
11
N lb
mm in.
+ (55.3)a
+
(7.90) A 103 B2¢b(0.0012)a
≤ (0.0012) ¢b ≤
22
mm in2
mm in.
318.5
280
232.4
11
N lb
mm in.
+
+ (86.1)a
(12.3) A 103 B2¢b(0.0004)a
≤ (0.0004) ¢b ≤
22
mm in2
mm in.
= 0.595in
N #·lb
mm/mm3
= 85.0
3
in
= 0.595 MJ/m3
210
Ans.
140
70
e (mm/mm)
Ans:
(u r) approx 0.595 MJ/m 3
141
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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.
*3–4. A tension test was performed on a specimen having
an original diameter of 12.5 mm and a gauge length of
50 mm. The data are listed in the table. Plot the stress–strain
diagram, and determine approximately the modulus of
elasticity, the ultimate stress, and the fracture stress. Use a
scale of 20 mm = 50 MPa and 20 mm = 0.05 mm>mm.
Redraw the linear-elastic region, using the same stress scale
but a strain scale of 20 mm = 0.001 mm>mm.
Stress and Strain:
s =
dL
P
(MPa) e =
(mm/mm)
A
L
0
0
90.45
0.00035
259.9
0.00120
308.0
0.00204
333.3
0.00330
355.3
0.00498
435.1
0.02032
507.7
0.06096
525.6
0.12700
507.7
0.17780
479.1
0.23876
Modulus of Elasticity: From the stress–strain diagram
(E)approx =
228.75(106) - 0
= 229 GPa
0.001 - 0
Ans.
Ultimate and Fracture Stress: From the stress–strain diagram
(sm)approx = 528 MPa
Ans.
(sf)approx = 479 MPa
Ans.
142
Load (kN)
Elongation (mm)
0
11.1
31.9
37.8
40.9
43.6
53.4
62.3
64.5
62.3
58.8
0
0.0175
0.0600
0.1020
0.1650
0.2490
1.0160
3.0480
6.3500
8.8900
11.9380
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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.
3–5. A tension test was performed on a steel specimen
having an original diameter of 12.5 mm and gauge length
of 50 mm. Using the data listed in the table, plot the
stress–strain diagram, and determine approximately the
modulus of toughness. Use a scale of 20 mm = 50 MPa and
20 mm = 0.05 mm>mm.
Stress and Strain:
s =
P
dL
(MPa) e =
(mm/mm)
A
L
0
0
90.45
0.00035
259.9
0.00120
308.0
0.00204
333.3
0.00330
355.3
0.00498
435.1
0.02032
507.7
0.06096
525.6
0.12700
507.7
0.17780
479.1
0.23876
Load (kN)
Elongation (mm)
0
11.1
31.9
37.8
40.9
43.6
53.4
62.3
64.5
62.3
58.8
0
0.0175
0.0600
0.1020
0.1650
0.2490
1.0160
3.0480
6.3500
8.8900
11.9380
Modulus of Toughness: The modulus of toughness is equal to the
total area under the stress–strain diagram and can be
approximated by counting the number of squares. The total
number of squares is 187.
(ut)approx = 187(25) A 106 B ¢
N
m
a 0.025 b = 117 MJ>m3
2≤
m
m
Ans.
Ans:
(ut) approx = 117 MJ >m 3
143
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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.
3–6. A specimen
specimen isisoriginally
originally300
1 ftmm
long,
hashas
a diameter
of
3–6.
long,
a diameter
0.5 12
in.,mm,
and isand
subjected
to a force
500 lb.
theWhen
force
of
is subjected
to aofforce
ofWhen
2.5 kN.
is increased
from 500 lb
to 1800
lb, the
the
force is increased
from
2.5 kN
to 9specimen
kN,, the elongates
specimen
0.009 in. 0.225
Determine
the modulus
of elasticity
for the
elongates
mm. Determine
the modulus
of elasticity
for
material
if it if
remains
linear
elastic.
the
material
it remains
linear
elastic.
Normal Stress and Strain: Applying s =
s1 5
=
dL
P
and e =
.
A
L
0.5003)
2.5(10
= 2.546 ksi
pp
2 5 22.10 MPa
(12)2)
44(0.5
3
1.80
9(10
)
9.167MPa
ksi
= pp
s2 5
5=79.58
22)
(0.5
44(12 )
0.009
¢e
= 0.225 5
= 0.000750
0.000750 mm/mm
in.>in.
∆e 5
12
300
Modulus of Elasticity:
E =
¢s
9.167 –-22.10
2.546
79.58
=
8.83 A 1033) BMPa
ksi = 76.64 GPa
5=76.64(10
¢e
0.000750
0.000750
Ans.
Ans:
E 76.64 GPa
144
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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.
3–7. A structural member in a nuclear reactor is made of a
zirconium alloy. If an
kN is to be supported
an axial
axial load
load of
of20
4 kip
by the member, determine its required cross-sectional area.
Use a factor of safety of 3 relative to yielding. What is the
load on the
the member
member ifif itit is
is 13 m
ft long and its elongation is
3
0.02mm?
in.? E
ksi.The
The material
material has
) ksi,sYsY= =400
57.5
0.5
Ezrzr =
= 14(10
100 GPa,
MPa.
elastic behavior.
Allowable Normal Stress:
F.S. =
3 =
sy
sallow
57.5
400
sallow
sallow = 19.17
133.33ksi
MPa
sallow =
19.17 5
=
133.33
P
A
3
4
20(10
)
AA
2 2
A == 150
0.2087
= 0.209 in2
mmin
Ans.
Stress–Strain Relationship: Applying Hooke’s law with
e =
0.02
d
0.5
== 0.0005
mm/mm
=
0.000555
in.>in.
3
L
3 (12)
1(10
)
33
)(0.0005)
= 50 MPa
s == Ee
Ee==100(10
14 A 10
= 7.778 ksi
B (0.000555)
Normal Force: Applying equation s =
P
.
A
P ==sA
sA= =50(150)
7.778 (0.2087)
kip
P
= 7500 N== 1.62
7.5 kN
Ans.
Ans:
A 150 mm 2, P 7.5 kN
145
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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.
*3–8. The strut is supported by a pin at C and an A-36
AB. If the
the wire
wire has
has aa diameter
diameter of
of 0.2
5 mm,
steel guy wire AB.
in.,
determine how much it stretches when the distributed load
acts on the strut.
A
60�
60
200
lb/ft
3.4 kN/m
C
Here, we are only interested in determining the force in wire AB.
11
(3.4)(2.7)(0.9)
FAB
cos 60°(2.7)
60°(9) -– (200)(9)(3)
= 0= 0 F
FAB
600kN
lb
AB cos
AB ==3.06
22
a + ©MC = 0;
The normal stress the wire is
sAB =
FAB
600 3)
3.06(10
3
= p p 22 ==19.10(10
) psi = 19.10 ksi
155.84 MPa
AAB
(0.2
)
(5
)
4
4
250ksi
MPa,
Hooke’s
Lawcan
canbe
beapplied
applied to determine
Since sAB 6 sy = 36
, Hooke’s
Law
determine the
thestrain
strain
in wire.
sAB = EPAB;
155.84== 29.0(10
200(1033)e
19.10
)PAB
AB
= 0.6586(10
0.7792(10–3
- 3) mm/mm
AB =
PeAB
) in>in
9(12)3)
2.7(10
== 124.71
inThus,
The unstretched length of the wire is LAB =
. Thus,
wire
3117.69.
thethe
wire
sin
sin 60°
60°
stretches
-3
0.7792(10–3
)(3117.69)
)(124.71)
dAB = PAB LAB = 0.6586(10
2.429 mm
= 0.0821
in.
Ans.
1
(3.4)(2.7) kN
2
0.9 m
1.8 m
146
B
9 ftm
2.7
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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.
3–9. The s-P diagram for elastic fibers that make up
human skin and muscle is shown. Determine the modulus of
elasticity of the fibers and estimate their modulus of
toughness and modulus of resilience.
s (MPa)
0.385
0.077
1
2 2.25
P (mm/mm)
s (MPa)
0.385
E =
0.077
= 0.0385 MPa
2
Ans.
ut =
1
1
(2)(0.077) + (0.385 + 0.77)(2.25 - 2) = 0.13475 MPa
2
2
Ans.
1
(0.077)(11) = 0.077 MPa
2
Ans.
ur =
0.077
1
2 2.25
P (mm/mm)
Ans:
E = 0.0385 MPa ut = 0.13475 MPa, ur = 0.077 M P a
147
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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.
3–10 The
alloy having
having an
an
3–10.
The stress-strain
stress–strain diagram
diagram for
for aa metal
metal alloy
original
diameter
of
12
mm
and
a
gauge
length
of
50
mm
original diameter of 0.5 in. and a gauge length of 2 in. is givenis
given
the figure.
Determine
approximately
modulus
in
theinfigure.
Determine
approximately
the the
modulus
of
of
elasticity
for
the
material,
the
load
on
the
specimen
that
elasticity for the material, the load on the specimen that causes
causes yielding,
and theload
ultimate
load the
yielding,
and the ultimate
the specimen
willspecimen
support. will
support.
From the stress–strain diagram, Fig. a,
MPa- –00
60 ksi
E 290
3
==
290 GPa
; ; EE== 30.0(10
) ksi
0.002
0.001 – 00
1
Ans.
sy ==290
MPa su>t
su/t==100
550ksi
GPa
60 ksi
Thus,
p 2 2
PYY ==ss
==
290[
)] )=D 32.80(10
) N == 32.80
kN
604C(12
= 11.783kip
11.8 kip
YA
YA
4 (0.5
p
Ans.
3
Pu/t
AA
= 500[
)] =262.20(10
)N
= 62.20
= 100
) D = 19.63
kip
= 19.6kN
kip
C p4 2(0.5
4 (12
u/tu>t
u>t ==ss
p
Ans.
 (MPa)
500
A
400
y = 290
300
200
E
100
0
1
0
p
B
0.05/ 0.08/ 0.15/ 0.20/ 0.25/ 0.30/ 0.35/
0.001 0.002 0.003 0.004 0.005 0.006 0.007
 (mm/m)
Elastic Recovery
(a)
Ans:
E 290 GPa, P Y 32.80 kN, P ult 62.20 kN
148
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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.
s (ksi)
3–11. The stress–strain diagram for a steel alloy having an
and aa gauge
gauge length of 50
mm
original diameter of 12
0.5mm
in. and
2 in.
is
is given
thefigure.
figure.If Ifthe
thespecimen
specimenisisloaded
loaded until
until itit is
given
in inthe
MPa,determine
determinethe
theapproximate
approximate amount
amount of
stressed to
to 500
90 ksi,
elastic recovery and the increase in the gauge length after it
is unloaded.
105
90
75
60
45
30
15
0
0
0
0.05 0.10 0.15 0.20 0.25 0.30 0.35
0.001 0.002 0.003 0.004 0.005 0.006 0.007
From the stress–strain diagram Fig. a, the modulus of elasticity for the steel alloy is
MPa- –00
E 290
60 ksi
==
= 290
GPa3) ksi
; EE =
30.0(10
0.002
0.001 – 00
1
when the specimen is unloaded, its normal strain recovered along line AB, Fig. a,
which has a gradient of E. Thus
ksi
50090 500 90
MPa
Elastic Recovery
= 0.001724
mm/mm
= 0.003 in>in
Elastic
Recovery = = =
3
3
E
30.0(10
) ksi
E
290(10 ) MPa
Amount ofset
Elastic
Recovery = 0.001724(50 mm) = 0.0862 mm
Thus, the permanent
is
Ans.
Ans.
PP = 0.05 = 0.047 in>in
Thus, the permanent
set0.003
is
Then, the increase in gauge length
is – 0.001724 = 0.078276 mm/mm
eP = 0.08
= 0.047(2)
= PPL
Then, the ¢L
increase
in gauge
length= is0.094 in
Ans.
∆L = ePL = 0.078276(50 mm) = 3.91379 mm
Ans.
 (MPa)
500
A
400
300
200
E
100
0
1
0
p
B
0.05/ 0.08/ 0.15/ 0.20/ 0.25/ 0.30/ 0.35/
0.001 0.002 0.003 0.004 0.005 0.006 0.007
 (mm/mm)
Elastic Recovery
(a)
Ans:
Elastic Recovery 0.08621 mm
욼L 3.91379 mm
149
P (in./in.)
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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.
*3–12. The stress–strain diagram for a steel alloy having an
original diameter
diameter of
of 12
0.5mm
in. and
and aagauge
gaugelength
lengthofof502mm
in.
is given in the figure. Determine approximately the modulus
of resilience and the modulus of toughness for the material.
The Modulus of resilience is equal to the area under the stress–strain diagram up to
the proportional limit.
MPa PPL
ePL= =0.002
0.001in>in.
mm/mm
sPL == 290
60 ksi
Thus,
(ui)r =
1
1
in # lb
= 0.145
MPa 3
sPLPPL = [(290)](0.001)
= 60.0
C 60(103) D (0.002)
2
2
in
Ans.
The modulus of toughness is equal to the area under the entire stress–strain
diagram. This area can be approximated by counting the number of squares. The
total number is 33.
38. Thus.
Thus,
lb
in
in
in
mm
mm
in # lb
in
3
3
= 33[100
b = 132
= i)38
c 15(10
) 2 d MPa]a0.04
a 0.05 b = 28.5(10
) MPa3
C (ui)t D approx[(u
t]approx
s (ksi)  (MPa)
105
500
90
A
400
75
60
PL = 290
45
200
30
100
15
0
300
0
0
E
1
B
P (in./in.)
0.050 00.100.05/
0.20 0.250.20/
0.30 0.350.30/ 0.35/  (mm/m)
0.150.08/
0.001 0.002 0.003 0.0040.15/
0.005 0.0060.25/
0.007
0.001 0.002 0.003 0.004 0.005 0.006 0.007
(a)
150
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.
3–13. A bar having a length of 125 mm and cross-sectional
area of 437.5 mm 2 is subjected to an axial force of 40 kN. If the
bar stretches 0.05 mm, determine the modulus of elasticity
of the material. The material has linear-elastic behavior.
40 kN
40 kN
125 mm
Normal Stress and Strain:
s =
P
40(103)
5 95.81 MPa
=
A
437.5
P =
dL
L
0.05
125
0.000400 mm/mm
Modulus of Elasticity:
E =
s
95.81
5 239.525 (103) MPa = 239.5 GPa
=
P 0.000400
Ans.
Ans:
s 95.81 MPa, 0.000400 mm/mm
E 239.5 GPa
151
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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.
3–14. The rigid pipe is supported by a pin at A and an
A-36 steel guy wire BD. If the wire has a diameter of
60.25
mm,
in.,determine
determinehow
howmuch
muchitit stretches
stretches when
when aa load
load of
P ==3600
kN acts
onon
thethe
pipe.
pipe.
lb acts
B
Here, we are only interested in determining the force in wire BD. Referring 1.2
4 ftm
to the FBD in Fig. a
a + ©MA = 0;
FBD A 45 B (0.9)
(3) -–600(6)
0
3(1.8) == 0 FF
7.51500
kN lb
BD
BD= =
A
D
C
The normal stress developed in the wire is
sBD
P
P
0.9
3 ftm
0.9
3 ftm
FBD
7.5(10
15003)
30.56(10
=
= p p 2 2 5=265.3
MPa3) psi = 30.56 ksi
ABD
(6 ) )
4 4(0.25
Since sBD 6 sy = 36
, Hooke’s
Law
cancan
bebe
applied
totodetermine
250ksi
MPa,
Hooke’s
Law
applied
determinethe
thestrain
straininin
the wire.
sBD = EPBD;
3
265.3 =
= 200(10
30.56
29.0(103)e
)PAB
BD
- 3–3
eBD==1.054(10
1.3265(10
mm/mm
PBD
) )in.>in.
1.5=m.60
Thus,
the the
in. Thus,
The unstretched length of the wire is LBD = 23
0.922 ++ 14.222==5ft
wire stretches
3
dBD = PBD LBD = 1.3265(10
1.054(10 -–3
)(60)
)(1.5)(103)
= 1.98975
0.0632 inmm
Ans.
3 kN
0.9 m
0.9 m
Ans:
dBD 1.98975 mm
152
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3–15. The rigid pipe is supported by a pin at A and an
A-36 guy wire BD.
BD. If
If the
the wire
wire has
hasaadiameter
diameterofof0.25
6 mm,
in.,
determine the load
load PP ifif the
the end
end CCisisdisplaced
displaced1.875
0.075mm
in.
downward.
B
4 ftm
1.2
P
P
A
D
C
3 ftm
0.9
3 ftm
0.9
Here, we are only interested in determining the force in wire BD. Referring to the
FBD in Fig. a
FBD A 45 B (0.9)
(3) -–P(6)
P(1.8)= =0 0
a + ©MA = 0;
FBD = 2.50 P
The unstretched length for wire BD is LBD = 23
ft m.
= 60
in. From
From
the the
0.922 ++ 14.222==51.5
geometry shown in Fig. b, the stretched length of wire BD is
LBD¿ = 260
15002 2++ 0.075
1.87522 -− 22(60)(0.075)
(1500)(1.875)cos
cos143.13°
143.13°==60.060017
1501.500 mm
Thus, the normal strain is
PBD =
LBD¿ - LBD
1501.500 – 1500
60.060017
60
5
=
= 1.0000(10
1.0003(10–3- )3)mm/mm
in.>in.
1500
LBD
60
Then, the normal stress can be obtain by applying Hooke’s Law.
3
MPa
sBD = EPBD = 200(10
29(103)3C)[1.0000(10
1.0003(10 -–3
))]D =
= 200
29.01
ksi
MPa,
result
is valid.
36 ksi
Since sBD 6 sy = 250
, thethe
result
is valid.
sBD =
FBD
;
ABD
2.50P 2.50 P
200 5 p 3) 2 = p
29.01(10
2
4 (6 ) 4 (0.25 )
P ==2261.9
569.57Nlb= =2.26
570kN
lb
Ans.
LBD = 1.5 m
0.9 m
0.9 m
1.875 mm
Ans:
P 2.26 kN
153
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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.
*3–16. The wire has a diameter of 5 mm and is made from
A-36 steel. If a 80-kg man is sitting on seat C, determine the
elongation of wire DE.
E
W
600 mm
D
A
B
800 mm
Equations of Equilibrium: The force developed in wire DE can be determined by
writing the moment equation of equilibrium about A with reference to the freebody diagram shown in Fig. a,
a + ©MA = 0;
3
FDE a b(0.8) - 80(9.81)(1.4) = 0
5
FDE = 2289 N
Normal Stress and Strain:
sDE =
FDE
2289
=
= 116.58 MPa
p
ADE
(0.0052)
4
Since sDE < sY , Hooke’s Law can be applied
sDE = EPDE
116.58(106) = 200(109)PDE
PDE = 0.5829(10-3) mm>mm
The unstretched length of wire DE is LDE = 26002 + 8002 = 1000 mm. Thus, the
elongation of this wire is given by
dDE = PDELDE = 0.5829(10-3)(1000) = 0.583 mm
Ans.
154
C
600 mm
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3–17. A tension test was performed on a magnesium alloy
specimen having a diameter 12 mm and gauge length 50 mm.
The resulting stress–strain diagram is shown in the figure.
Determine the approximate modulus of elasticity and the
yield strength of the alloy using the 0.2% strain offset
method.
(MPa)
280
245
210
175
140
105
70
35
0
0.002
0.004
0.006
0.008
0.010
(mm/mm)
Modulus of Elasticity: From the stress–strain diagram, when P = 0.002 mm>mm, its
corresponding stress is s = 91 MPa. Thus,
Eapprox =
91 - 0
= 45.5 (103) M Pa
0.002 - 0
Ans.
Yield Strength: The intersection point between the stress–strain diagram and the
straight line drawn parallel to the initial straight portion of the stress–strain diagram
from the offset strain of P = 0.002 mmmm> mm i s the yield strength of the alloy. From the
stress–strain diagram,
sYS = 181.3 M Pa~~
Ans.
Ans:
Eapprox = 45.5(103) M Pa, sYS = 181.3 M Pa
155
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3–18. A tension test was performed on a magnesium alloy
specimen having a diameter 12 mm and gauge length of 50 mm.
The resulting stress–strain diagram is shown in the figure. If
the specimen is stressed to 210 MPa and unloaded, determine
the permanent elongation of the specimen.
(MPa)
280
245
210
175
140
105
70
35
0
0.002
0.004
0.006
0.008
0.010
(mm/mm)
Permanent Elongation: From the stress–strain diagram, the strain recovered is
along the straight line BC which is parallel to the straight line OA. Since
91 - 0
= 45.5(103) M Pa, then the permanent set for the specimen is
Eapprox =
0.002 - 0
210(103)
PP = 0.0078 = 0.00318 mm> mm.
45.5(106)
Thus,
dP = PPL = 0.00318(50) = 0.159 mm.
Ans.
Ans:
dP = 0.159 mm.
156
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P
3–19. The stress–strain diagram for a bone is shown, and
can be described by the equation P = 0.45110-62 s ⫹
0.36110-122 s3, where s is in kPa. Determine the yield
strength assuming a 0.3% offset.
s
P ⫽ 0.45(10⫺6)s + 0.36(10⫺12)s3
P
P
P = 0.45(10-6)s + 0.36(10-12)s3,
dP = A 0.45(10-6) + 1.08(10-12) s2 B ds
E =
ds
1
2 =
= 2.22(106) kPa = 2.22 GPa
dP
0.45(10 - 6)
s=0
The equation for the recovery line is s = 2.22(106)(P - 0.003).
This line intersects the stress–strain curve at sYS = 2027 kPa = 2.03 MPa
Ans.
Ans:
sYS = 2.03 MPa
157
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P
*3–20. The stress–strain diagram for a bone is shown
and can be described by the equation P = 0.45110-62 s ⫹
0.36110-122 s3, where s is in kPa. Determine the modulus of
toughness and the amount of elongation of a 200-mm-long region
just before it fractures if failure occurs at P = 0.12 mm>mm.
s
P ⫽ 0.45(10⫺6)s + 0.36(10⫺12)s3
P
When P = 0.12
120(10-3) = 0.45 s + 0.36(10-6)s3
Solving for the real root:
s = 6873.52 kPa
6873.52
ut =
LA
dA =
L0
(0.12 - P)ds
6873.52
ut =
L0
(0.12 - 0.45(10-6)s - 0.36(10-12)s3)ds
6873.52
= 0.12 s - 0.225(10-6)s2 - 0.09(10-12)s4|0
= 613 kJ>m3
Ans.
d = PL = 0.12(200) = 24 mm
Ans.
158
P
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–21. The two bars are made of polystyrene, which has the
stress–strain diagram shown. If the cross-sectional area of
bar AB is 975 mm2 and BC is 2600 mm2, determine the largest
force P that can be supported before any member ruptures.
Assume that buckling does not occur.
P
1.2 m
C
B
1m
A
(MPa)
175
+ c gFy = 0;
+
; ©Fx = 0;
1
FAB - P = 0;
1.56
FBC
FAB = 1.56 P
(1)
140
105
1.2
(1.56P) = 0;
1.56
FBC = 1.2 P
0
From the stress–strain diagram (sR)t = 35 M Pa
FBC
;
ABC
35 =
tension
35
Assuming failure of bar BC:
s =
compression
70
(2)
FBC
;
2600
0
0.20
0.40
0.60
0.80
(mm/mm)
FBC = 91 kN
From Eq. (2), P = 75.83 kN
Assuming failure of bar AB:
From stress–strain diagram (sR)c = 175 M Pa
s =
FAB
;
AAB
175 =
FAB
;
975
FAB = 170.625 kN
From Eq. (1), P ⫽ 109 .37 5 kN
Choose the smallest value
P = 75.83 kN
Ans.
Ans:
P = 75.83 kN
159
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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.
3–22. The two bars are made of polystyrene, which has the
stress–strain diagram shown. Determine the cross-sectional
area of each bar so that the bars rupture simultaneously
when the load P = 13.5 kN. Assume that buckling does
not occur.
P
1.2 m
C
B
1m
A
(MPa)
175
+ c ©Fy = 0;
+
: ©Fx = 0;
1
FBA a
b - 13.5 = 0;
1.56
.06 a
-FBC + 21.06
140
FBA = 21.06 kN
105
1.2
b = 0; FBC = 16.2 kN
1.56
compression
70
tension
35
For member BC:
0
(smax)t =
FBC
;
ABC
ABC =
(smax)c =
FBA
;
ABA
ABA =
16.2 kN
= 462.85 mm2
35 MPa
0
0.20
0.40
0.60
0.80
(mm/mm)
Ans.
For member BA:
21.06 kN
= 120.34 mm2
175 M Pa
Ans.
Ans:
ABC = 462.85 mm2 , ABA = 120.34 mm2
160
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*3–23. The stress–strain diagram for many metal alloys
can be described analytically using the Ramberg-Osgood
three parameter equation P = s>E + ksn, where E, k, and
n are determined from measurements taken from the
diagram. Using the stress–strain diagram shown in the
3
figure, take E
E == 30110
210 GPa
and determine
determine the
the other two
2 ksiand
parameters k and n and thereby obtain an analytical
expression for the curve.
ss(MPa)
(ksi)
560
80
420
60
280
40
140
20
0.1 0.2 0.3 0.4 0.5
–6)
P (10–6
Choose,
s == 280
MPa, e e= =0.1
0.1
40 ksi,
s == 420
MPa, e e= =0.3
0.3
60 ksi,
0.1==
0.1
40
280
nn
k(280)
++ k(40)
33
30(10
))
210(10
0.3==
0.3
60
420
nn
k(420)
++ k(60)
33
30(10
))
210(10
0.098667 == k(280)
k(40)nn
0.29800 == k(420)
k(60)nn
0.3310962 = (0.6667)n
ln (0.3310962) = n ln (0.6667)
n = 2.73
Ans.
k = 4.23(10-6)
Ans.
Ans:
n = 2.73, k = 4.23(10 - 6)
161
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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.
3–24. The wires AB and BC have original lengths of 0.6 m
and 0.9 m, and diameters of 3 mm and 5 mm, respectively. If
these wires are made of a material that has the approximate
stress–strain diagram shown, determine the elongations of
the wires after the 6750-N load is placed on the platform.
C
Equations of Equilibrium: The forces developed in wires AB and BC can be
determined by analyzing the equilibrium of joint B, Fig. a,
+
: ©Fx = 0;
+ c ©Fy = 0;
FBC sin 30° - FAB sin 45° = 0
(1)
FBC cos 30° + FAB cos 45° = 6750
(2)
A
0.9 m
45⬚
30⬚
0.6 m
B
Solving Eqs. (1) and (2),
FAB = 3494.07 N
FBC = 4941.36 N
Normal Stress and Strain:
sAB =
FAB
3494.07
=
= 494.56 MPa
p
AAB
(3)2
4
s (MPa)
sBC =
FBC
4941.36
=
= 251.78 MPa
p
ABC
2
(5)
4
560
406
The corresponding normal strain can be determined from the stress–strain diagram,
Fig. b.
251.78
406
=
;
PBC
0.002
PBC = 0.00124 mm0> mm.
494 .56 - 406
560 - 406
=
;
PAB - 0.002
0.01 - 0.002
PAB = 0.0066 mm> mm.
0.002
Thus, the elongations of wires AB and BC are
dAB = PABLAB = 0.0066(600) = 3.96
Ans.
dBC = PBCLBC = 0.00124(900) = 0.1116
Ans.
162
0.01
P (mm/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.
3–25. The acrylic plastic rod is 200 mm long and 15 mm in
diameter. If an axial load of 300 N is applied to it, determine
the change in its length and the change in its diameter.
Ep = 2.70 GPa, np = 0.4.
s =
P
=
A
Plong =
300
p
2
4 (0.015)
300 N
300 N
200 mm
= 1.678 MPa
1.678(106)
s
= 0.0006288
=
E
2.70(109)
d = Plong L = 0.0006288 (200) = 0.126 mm
Ans.
Plat = -nPlong = -0.4(0.0006288) = -0.0002515
¢d = Platd = -0.0002515 (15) = -0.00377 mm
Ans.
Ans:
d = 0.126 mm, ¢d = -0.00377 mm
163
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3–26. The thin-walled tube is subjected to an axial force of
40 kN. If the tube elongates 3 mm and its circumference
decreases 0.09 mm, determine the modulus of elasticity,
Poisson’s ratio, and the shear modulus of the tube’s
material. The material behaves elastically.
40 kN
900 mm
10 mm
40 kN
12.5 mm
Normal Stress and Strain:
s =
40(103)
P
= 226.35 MPa
=
A
p(0.01252 - 0.012)
Pa =
d
3
=
= 3.3333 (10-3) mm>mm
L
900
Applying Hooke’s law,
s = EPa;
226.35(106) = E [3.3333(10-3)]
E = 67.91(106) Pa = 67.9 GPa
Ans.
Poisson’s Ratio: The circumference of the loaded tube is 2p(12.5) - 0.09 =
78.4498 mm. Thus, the outer radius of the tube is
r =
78.4498
= 12.4857 mm
2p
The lateral strain is
Plat =
r - r0
12.4857 - 12.5
=
= -1.1459(10-3) mm>mm
r0
12.5
n = -
-1.1459(10-3)
Plat
d = 0.3438 = 0.344
= -c
Pa
3.3333(10-3)
G =
Ans.
67.91(109)
E
=
= 25.27(109) Pa = 25.3 GPa
2(1 + n)
2(1 + 0.3438)
Ans.
Ans:
E = 67.9 GPa, v = 0.344, G = 25.3 GPa
164
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3–27. When the two forces are placed on the beam, the
diameter of the A-36 steel rod BC decreases from 40 mm to
39.99 mm. Determine the magnitude of each force P.
C
P
1m
A
P
1m
1m
1m
B
0.75 m
Equations of Equilibrium: The force developed in rod BC can be determined by
writing the moment equation of equilibrium about A with reference to the
free-body diagram of the beam shown in Fig. a.
4
FBC a b (3) - P(2) - P(1) = 0
5
a + ©MA = 0;
FBC = 1.25P
Normal Stress and Strain: The lateral strain of rod BC is
Plat =
d - d0
39.99 - 40
=
= -0.25(10 - 3) mm>mm
d0
40
Plat = -nPa;
-0.25(10-3) = -(0.32)Pa
Pa = 0.78125(10-3) mm>mm
Assuming that Hooke’s Law applies,
sBC = EPa;
sBC = 200(109)(0.78125)(10-3) = 156.25 MPa
Since s 6 sY, the assumption is correct.
sBC =
FBC
;
ABC
156.25(106) =
1.25P
p
A 0.042 B
4
P = 157.08(103)N = 157 kN
Ans.
Ans:
P = 157 kN
165
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*3–28. If P = 150 kN, determine the elastic elongation of
rod BC and the decrease in its diameter. Rod BC is made of
A-36 steel and has a diameter of 40 mm.
C
P
1m
A
P
1m
1m
1m
B
0.75 m
Equations of Equilibrium: The force developed in rod BC can be determined by
writing the moment equation of equilibrium about A with reference to the freebody diagram of the beam shown in Fig. a.
a + ©MA = 0;
4
FBC a b(3) - 150(2) - 150(1) = 0
5
FBC = 187.5 kN
Normal Stress and Strain: The lateral strain of rod BC is
sBC =
187.5(103)
FBC
=
= 149.21 MPa
p
ABC
A 0.042 B
4
Since s 6 sY, Hooke’s Law can be applied. Thus,
sBC = EPBC;
149.21(106) = 200(109)PBC
PBC = 0.7460(10-3) mm>mm
The unstretched length of rod BC is LBC = 27502 + 10002 = 1250 mm. Thus the
elongation of this rod is given by
dBC = PBCLBC = 0.7460(10-3)(1250) = 0.933 mm
Ans.
We obtain,
Plat = -nPa ;
Plat = -(0.32)(0.7460)(10-3)
= -0.2387(10-3) mm>mm
Thus,
dd = Plat dBC = -0.2387(10-3)(40) = -9.55(10-3) mm
166
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.
3–29. The friction pad A is used to support the member,
which is subjected to an axial force of P = 2 kN. The pad is
made from a material having a modulus of elasticity of
E = 4 MPa and Poisson’s ratio n = 0.4. If slipping does not
occur, determine the normal and shear strains in the pad.
The width is 50 mm. Assume that the material is linearly
elastic. Also, neglect the effect of the moment acting on
the pad.
P
60⬚
25 mm
A
100 mm
Internal Loading: The normal force and shear force acting on the friction pad can be
determined by considering the equilibrium of the pin shown in Fig. a.
+
: ©Fx = 0;
V - 2 cos 60° = 0
V = 1 kN
+ c ©Fy = 0;
N - 2 sin 60° = 0
N = 1.732 kN
Normal and Shear Stress:
t =
1(103)
V
=
= 200 kPa
A
0.1(0.05)
s =
1.732(103)
N
=
= 346.41 kPa
A
0.1(0.05)
Normal and Shear Strain: The shear modulus of the friction pad is
G =
4
E
=
= 1.429 MPa
2(1 + n)
2(1 + 0.4)
Applying Hooke’s Law,
s = EP;
346.41(103) = 4(106)P
P = 0.08660 mm>mm
Ans.
t = Gg;
200(103) = 1.429(106)g
g = 0.140 rad
Ans.
Ans:
P = 0.08660 mm>mm, g = 0.140 rad
167
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3–30. The lap joint is connected together using a 30 mm
diameter bolt. If the bolt is made from a material having a
shear stress–strain diagram that is approximated as shown,
determine the shear strain developed in the shear plane of
the bolt when P = 340 kN.
P
2
P
P
2
t (MPa)
525
350
Internal Loadings: The shear force developed in the shear planes of the bolt can be
determined by considering the equilibrium of the free-body diagram shown in Fig. a.
+
: ©Fx = 0;
340 - 2V = 0
0.005
0.05
g (rad)
V = 170 kN
Shear Stress and Strain:
t =
V
170
=
p
A
A 302 B
4
= 240.62 MPa
Using this result, the corresponding shear strain can be obtained from the shear
stress–strain diagram, Fig. b.
240.62 350
=
;
g
0.005
g = 3.43(10-3) rad
Ans.
Ans:
g = 3.43(10-3) rad
168
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3–31. The lap joint is connected together using a 30 mm
diameter bolt. If the bolt is made from a material having a
shear stress–strain diagram that is approximated as shown,
determine the permanent shear strain in the shear plane of
the bolt when the applied force P = 680 kN is removed.
P
2
P
P
2
t (MPa)
525
350
Internal Loadings: The shear force developed in the shear planes of the bolt can be
determined by considering the equilibrium of the free-body diagram shown in Fig. a.
+
: ©Fx = 0;
680 - 2V = 0
V = 340 N
0.005
Shear Stress and Strain:
t =
V
340
=
p
A
A 30 2 B
4
0.05
g (rad)
= 481.24 MPa
Using this result, the corresponding shear strain can be obtained from the shear
stress–strain diagram, Fig. b.
481.24 - 350
525 - 350
=
;
g - 0.005
0.05 - 0.005
g = 0.0387 rad
When force P is removed, the shear strain recovers linearly along line BC, Fig. b,
with a slope that is the same as line OA. This slope represents the shear modulus.
G =
350
= 70(103) MPa
0.005
Thus, the elastic recovery of shear strain is
t = Ggr;
481.24 = (70)(103)gr
gr = 6.874(10-3) rad
And the permanent shear strain is
gP = g - gr = 0.0387 - 6.874(10 -3) = 0.031826 rad
Ans.
Ans:
gP = 0.031826 rad
169
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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.
*3–32. A shear spring is made by bonding the rubber
annulus to a rigid fixed ring and a plug. When an axial load
P is placed on the plug, show that the slope at point y in
the rubber is dy>dr = -tan g = -tan1P>12phGr22. For small
angles we can write dy>dr = -P>12phGr2. Integrate this
expression and evaluate the constant of integration using
the condition that y = 0 at r = ro. From the result compute
the deflection y = d of the plug.
P
h
ro
y
d
ri
r
y
Shear Stress–Strain Relationship: Applying Hooke’s law with tA
g =
P
=
.
2p r h
tA
P
=
G
2p h G r
dy
P
= -tan g = -tan a
b
dr
2p h G r
(Q.E.D)
If g is small, then tan g = g. Therefore,
dy
P
= dr
2p h G r
At r = ro,
y = -
dr
P
2p h G L r
y = -
P
ln r + C
2p h G
0 = -
P
ln ro + C
2p h G
y = 0
C =
Then, y =
ro
P
ln
r
2p h G
At r = ri,
y = d
d =
P
ln ro
2p h G
ro
P
ln
ri
2p h G
Ans.
170
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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.
3–33. The aluminum block has a rectangular cross section
and is subjected to an axial compressive force of 36 kN. If
the 37-mm side changed its length to 37.5033 mm, determine Poisson’s ratio and the new length of the 50-mm.
Eal = 70 GPa.
s =
50 mm
36 kN
36 kN
75 mm
P
36
=
= 19.45 MPa
A (50)(37)
Plong =
Plat =
n =
37 mm
s
- 19.45
= -0.0002778
=
E
70000
37.5033 - 37
37
= 0.00008918
-0.00008918
= 0.321
-0. 0002778
Ans.
h¿ = 50 + 0.00008918(2) = 50.000178 mm.
Ans.
Ans:
n = 0.321, h¿ = 50.000178 mm.
171
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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.
3–34. A shear spring is made from two blocks of rubber,
each having a height h, width b, and thickness a. The
blocks are bonded to three plates as shown. If the plates
are rigid and the shear modulus of the rubber is G,
determine the displacement of plate A if a vertical load P is
applied to this plate. Assume that the displacement is small
so that d = a tan g L ag.
P
d
A
h
Average Shear Stress: The rubber block is subjected to a shear force of V =
P
.
2
a
a
P
t =
V
P
2
=
=
A
bh
2bh
Shear Strain: Applying Hooke’s law for shear
P
g =
t
P
2bh
=
=
G
G
2bhG
Thus,
d = ag = =
Pa
2bhG
Ans.
Ans:
d =
172
Pa
2bhG
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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.
3–35. The s– Pdiagram for elastic fibers that make up
human skin and muscle is shown. Determine the modulus
of elasticity of the fibers and estimate their modulus of
toughness and modulus of resilience.
s(psi)
s(MPa)
55
385
11
77
11
77
11
Eal=5 =55.5
38.5
psiMPa
22
Ans.
11
11
utt 5
MJ/m
= (2)(77)
(2)(11)++ (385
(55++77)(2.25
11)(2.25– 2)
- 5
2) 134.75
= 19.25
psi 3
22
22
Ans.
11
utt 5
= (2)(77)
(2)(11)5= 77
11MJ/m
psi 3
22
Ans.
22 2.25
2.25
P(in./in.)
P(mm/mm)
Ans:
E 38.5 MPa, u t 134.75 MJ/m 3, u r 77 MJ/m 3
173
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s (ksi)
(MPa)
*3–36. The elastic portion of the tension stress–strain
diagram for an aluminum alloy is shown in the figure. The
502mm
specimen used for
for the
thetest
testhas
hasaagauge
gaugelength
lengthofof
in. and a
mm.
theapplied
appliedload
loadisis10
50kip,
kN, determine
diameter of 12.5
0.5 in.
If If
the
the new diameter of the specimen. The shear modulus is
GPa.32 ksi.
Gal == 28
3.8110
500
70
0.00614
0.00614
10 3)
PP 50(10
= 5= pp
= 407.44
50.9296MPa
ksi
s5
5
A 44(12.5)
(0.5)22
A
stress–strain diagram
diagram
From the stress-strain
=
E5
70
500
11400.653ksi
5= 81.433(10
) MPa
0.00614
0.00614
=
elong 5
=
G5
407.44
50.9296
ss
=50.0044673
==
0.0050033in.>in.
mm/mm
E 81.433(10
11400.653)
E
3
11400.65
81.433(10
)
EE
3
3.8(10
v = 0.500
; ; 28(10
)5 3) =
; v; 5 0.45416
2(12(1
+ v)+ v)
2(1 ++v)v)
= –ve
- ve
- 0.500(0.0044673) 5= –0.002272
- 0.002234 in.>in.
elat 5
5=–0.45416(0.0050033)
long
long
¢d5= eelatlatdd5=–0.002272(12.5)
- 0.002234(0.5)5=–0.0284
- 0.001117
∆d
mm in.
d¿ 5
= dd++∆d¢d
0.5– 0.001117
= 0.4989
d9
5=
12.5
0.0284
5 12.4716
mmin.
Ans.
174
P (mm/mm)
(in./in.)
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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.
3–37. The rigid beam rests in the horizontal position on
two 2014-T6 aluminum cylinders having the unloaded
lengths shown. If each cylinder has a diameter of 30 mm.
determine the placement x of the applied 80-kN load so
that the beam remains horizontal. What is the new diameter
of cylinder A after the load is applied? nal = 0.35.
80 kN
x
A
B
220 mm
210 mm
3m
a +©MA = 0;
FB(3) - 80(x) = 0;
a +©MB = 0;
-FA(3) + 80(3 - x) = 0;
FB =
80x
3
FA =
(1)
80(3 - x)
3
(2)
Since the beam is held horizontally, dA = dB
s =
P
P
;
A
s
A
=
E
E
P =
d = PL = a
P
A
E
bL =
80(3 - x)
3
dA = dB;
PL
AE
(220)
AE
=
80x
3
(210)
AE
80(3 - x)(220) = 80x(210)
x = 1.53 m
Ans.
From Eq. (2),
FA = 39.07 kN
sA =
39.07(103)
FA
=
= 55.27 MPa
p
A
(0.032)
4
Plong =
sA
E
55.27(106)
= -
73.1(109)
= -0.000756
Plat = -nPlong = -0.35(-0.000756) = 0.0002646
dA¿ = dA + d Plat = 30 + 30(0.0002646) = 30.008 mm
Ans.
Ans:
x = 1.53 m, dA
¿ = 30.008 mm
175
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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.
3–38. The wires each have a diameter of 12 mm, length of
0.6 m, and are made from 304 stainless steel. If P = 30 kN,
determine the angle of tilt of the rigid beam AB.
D
C
0.6 m
P
0.6 m
A
0.3 m
B
Equations of Equilibrium: Referring to the free-body diagram of beam AB shown
in Fig. a,
a +©MA = 0;
FBC (0.9) - 30(0.6) = 0 FBC = 20 kN
+ c ©MB = 0;
30(0.3) - FAD (0.9) = 0 FAD = 10 kN
Normal Stress and Strain:
sBC =
sAD =
20000
FBC
=
= 176.92 MPa
ABC
p (12)2
4
10000
FAD
=
= 88.46 MPa
AAD
2
p
(12)
4
Since sBC 6 sY and sA 6 sY, Hooke’s Law can be applied.
sBC = EPBC;
176.92 = 196(103)PBC
PBC = 0.9026(10-3) mm> mm.
sAD = EPAD;
88.46 = 196(10 3)PAD
PAD = 0.4513(10-3) mm> mm.
Thus, the elongation of cables BC and AD are given by
dBC = PBCLBC = 0.9026(10-3)(600) = 0.54156 mm.
dAD = PADLAD = 0.4513(10-3)(600) = 0.27078 mm.
Referring to the geometry shown in Fig. b and using small angle analysis,
u =
/ /
dBC - dAD (0.54156 - 0.27078)>900
180°
=
= 0.3(10-3) rada
b = 0.0172°
36
36
prad
Ans.
Ans:
u = 0.0172°
176
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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.
3–39. The wires each have a diameter of 12 mm, length of
0.6 m, and are made from 304 stainless steel. Determine the
magnitude of force P so that the rigid beam tilts 0.015°.
D
C
0.6 m
P
0.6 m
A
0.3 m
B
Equations of Equilibrium: Referring to the free-body diagram of beam AB shown
in Fig. a,
a +©MA = 0;
FBC(0.9) - P(0.6) = 0
FBC = 0.6667P
+ c ©MB = 0;
P(0.3) - FAD(0.9) = 0
FAD = 0.3333P
Normal Stress and Strain:
sBC =
sAD =
FBC
0.6667P
=
= 0.00589P
ABC
2
p
(12)
4
FAD
0.3333P
=
= 0.00294P
AAD
2
p
(12)
4
Assuming that sBC 6 sY and sAD 6 sY and applying Hooke’s Law,
sBC = EPBC;
0.0~0589P = 196(103)PBC
PBC = 0.0300510(10- 6)P
sAD = EPAD;
0.00294P = 196(10 3)PAD
PAD = 15(10 -9)P
Thus, the elongation of cables BC and AD are given by
dBC = PBCLBC = 0.0300510(10- 6)P(600) = 18.0306(10-6)P
dAD = PADLAD = 15(10 -9)P(600) = 9(10 -6 )P
Here, the angle of the tile is u = 0.015°a
prad
b = 0.2618(10-3) rad. Using small
180°
angle analysis,
u =
dBC - dAD
;
36
0.2618(10-3) =
(18.0306(10-6)P - 9(10 -6)P)>900
36
P = 26091.28 N = 26.091 kN
Ans.
Since sBC = 0.00589(26091.28/) = 153.67 MPa 6 sY and sAD = 0.00294 (26091. 28) = 76 .70 MP a
6 sY, the assumption is correct.
Ans:
P = 26.091 kN
177
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45.0°.
44.9°.
178
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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.
Ans:
254.175 mm
179
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3–42. The pipe with two rigid caps attached to its ends is
subjected to an axial force P. If the pipe is made from a
material having a modulus of elasticity E and Poisson’s
ratio n, determine the change in volume of the material.
ri
ro
L
P
a
Section a – a
a
P
Normal Stress: The rod is subjected to uniaxial loading. Thus, slong =
P
and slat = 0.
A
dV = AdL + 2prLdr
= APlong L + 2prLPlatr
Using Poisson’s ratio and noting that AL = pr2L = V,
dV = PlongV - 2nPlongV
= Plong (1 - 2n)V
slong
=
E
(1 - 2n)V
Since slong = P>A,
dV =
=
P
(1 - 2n)AL
AE
PL
(1 - 2n)
E
Ans.
Ans:
dV =
180
PL
(1 - 2n)
E
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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.
3–43. The 8-mm-diameter bolt is made of an aluminum
alloy. It fits through a magnesium sleeve that has an inner
diameter of 12 mm and an outer diameter of 20 mm. If the
original lengths of the bolt and sleeve are 80 mm and
50 mm, respectively, determine the strains in the sleeve and
the bolt if the nut on the bolt is tightened so that the tension
in the bolt is 8 kN. Assume the material at A is rigid.
Eal = 70 GPa, Emg = 45 GPa.
50 mm
A
30 mm
Normal Stress:
8(103)
sb =
P
=
Ab
p
2
4 (0.008 )
ss =
P
=
As
p
2
4 (0.02
= 159.15 MPa
8(103)
- 0.0122)
= 39.79 MPa
Normal Strain: Applying Hooke’s Law
Pb =
159.15(106)
sb
= 0.00227 mm>mm
=
Eal
70(109)
Ans.
Ps =
39.79(106)
ss
= 0.000884 mm>mm
=
Emg
45(109)
Ans.
Ans:
Pb = 0.00227 mm>mm, Ps = 0.000884 mm>mm
181
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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.
*3–44. An acetal polymer block is fixed to the rigid plates
at its top and bottom surfaces. If the top plate displaces
2 mm horizontally when it is subjected to a horizontal force
P = 2 kN, determine the shear modulus of the polymer.
The width of the block is 100 mm. Assume that the polymer
is linearly elastic and use small angle analysis.
400 mm
P ⫽ 2 kN
200 mm
Normal and Shear Stress:
t =
2(103)
V
=
= 50 kPa
A
0.4(0.1)
Referring to the geometry of the undeformed and deformed shape of the block
shown in Fig. a,
g =
2
= 0.01 rad
200
Applying Hooke’s Law,
t = Gg;
50(103) = G(0.01)
G = 5 MPa
Ans.
182