Chapter 6 hw edit
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Date
Apr 3, 2024
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Madelaine Tew
4/3/2023
HW 5
6.3 A specimen of aluminum having a rectangular cross section 10 mm * 12.7 mm (0.4
in.0.5 in.) is pulled in tension with 35,500 N (8000 lbf) force, producing only elastic
deformation. Calculate the resulting strain
Stress = 35500 / (10*12.7) = 279.53
Strain = 279.53 / 70E3 = 3.99 * 10^-3
6.7 For a bronze alloy, the stress at which plastic deformation begins is 275 MPa (40,000
psi), and the modulus of elasticity is 115 GPa (16.7 106 psi).
(a) What is the maximum load that may be applied to a specimen with a cross-sectional
area of 325 mm2 (0.5 in.2) without plastic deformation?
-
275 * 325 = 89,375
(b) If the original specimen length is 115 mm (4.5 in.), what is the maximum length to
which it may be stretched without causing plastic deformation?
-
115 * (1 + 275/115E3) = 115.28
6.15 A cylindrical specimen of aluminum having a diameter of 19 mm (0.75 in.) and length
of 200 mm (8.0 in.) is deformed elastically in tension with a force of 48,800 N (11,000 lbf).
Using the data contained in Table 6.1, determine the following:
(a) The amount by which this specimen will elongate in the direction of the applied
stress.
-
Area = pi r^2 = Pi (9.5)^2 = 283.53
-
E = youngs = 69E3
-
(48,800 * 200) / (285.53 * 69E3) = 0.50
(b) The change in diameter of the specimen. Will the diameter increase or decrease?
-
-(0.33 * 19 * 0.498) / 200 = -0.0156 which means it will be decreasing
6.18 A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its
original and final diameters are 20.000 and 20.025 mm, respectively, and its final length is
74.96 mm,
compute its original length if the deformation is totally elastic
. The elastic and
shear moduli for this alloy are 105 GPa and 39.7 GPa, respectively
-
6.21 A cylindrical metal specimen 12.7 mm (0.5 in.) in diameter and 250 mm (10 in.) long
is to be subjected to a tensile stress of 28 MPa (4000 psi); at this stress level, the
resulting deformation will be totally elastic.
(a) If the elongation must be less than 0.080 mm (3.2 10-3 in.), which of the metals in
Table 6.1 are suitable candidates? Why?
-
0.080 / 250 = 3.2E-4
-
28 / 3.2E-4 = 87.5 meaning anything except for aluminum or magnesium. This is
determined when we look at table 6.1
(b) If, in addition, the maximum permissible diameter decrease is 1.2 10-3 mm (4.7 10-5
in.) when the tensile stress of 28 MPa is applied, which of the metals that satisfy the
criterion in part (a) are suitable candidates? Why?
-
Because we are given the max permissible diameter we have to use it to find poisson's
ratio.
-
-1.2E-3 / 12.7 = -9.45E-5
-
Aluminum = (0.33 * 28E6) / (69 * 10E9) = 1.34E-4
-
Magnesium = (0.29 * 28E6) / (45 * 10E9) = 1.80E-4
-
Brass = (0.34 * 28E6) / (97 * 10E9) = 9.81E-5
Copper = (0.34 * 28E6) / (110 * 10E9) = 8.65E-5
Nickel = (0.31 * 28E6) / (207 * 10E9) = 4.19E-5
Steel = (0.30 * 28E6) / (207 * 10E9) = 4.06E-5
Titanium = (0.34 * 28E6) / (107 * 10E9) = 8.90E-5
Tungsten = (0.28 * 28E6) / (407 * 10E9) = 1.93E-5
6.27 A load of 85,000 N (19,100 lbf) is applied to a cylindrical specimen of a steel alloy
(displaying the stress–strain behavior shown in Figure 6.22) that has a cross-sectional
diameter of 15 mm (0.59 in.).
(a) Will the specimen experience elastic and/or plastic deformation? Why?
-
Area = pi (7.5)^2 = 176.71
-
85,000 / 176.71 = 481 When looking at the graph we know that elastic deformation is
before the curve begins while elastic + plastic is during the curve. In this case it would be
both elastic and plastic.
(b) If the original specimen length is 250 mm (10 in.), how much will it increase in length
when this load is applied?
-
250 * 0.0140 = 3.5mm
6.30 A specimen of ductile cast iron having a rectangular cross section of dimensions 4.8
mm 15.9 mm (3/16 in. 5/8 in.) is deformed in tension. Using the load-elongation data
shown in the following table, complete problems (a) through (f)
Area: 0.0048 * 0.0159 = 7.63E-5 m
(a) Plot the data as engineering stress versus engineering strain.
-
Engineering stress = load / area
-
Engineering strain = (Deformed - original length) / original length
Load
Length
Length change
Area
Stress
Strain
0
7.50E-02
0.00E+00
7.63E+05
0
0.00E+00
4740
7.50E-02
2.50E-05
7.63E+05
6.21E-03 3.33E-04
9140
7.51E-02
5.00E-05
7.63E+05
1.20E-02 6.66E-04
12,920
7.51E-02
7.50E-05
7.63E+05
1.69E-02 9.99E-04
16,540
7.51E-02
1.13E-04
7.63E+05
2.17E-02 1.50E-03
18,300
7.52E-02
1.50E-04
7.63E+05
2.40E-02 2.00E-03
20,170
7.52E-02
2.25E-04
7.63E+05
2.64E-02 2.99E-03
22,900
7.54E-02
3.75E-04
7.63E+05
3.00E-02 4.98E-03
25,070
7.55E-02
5.25E-04
7.63E+05
3.29E-02 6.95E-03
26,800
7.58E-02
7.50E-04
7.63E+05
3.51E-02 9.90E-03
28,640
7.65E-02
1.50E-03
7.63E+05
3.75E-02 1.96E-02
30,240
7.80E-02
3.00E-03
7.63E+05
3.96E-02 3.85E-02
31,100
7.95E-02
4.50E-03
7.63E+05
4.08E-02 5.66E-02
31,280
8.10E-02
6.00E-03
7.63E+05
4.10E-02 7.41E-02
30,820
8.25E-02
7.50E-03
7.63E+05
4.04E-02 9.09E-02
29,180
8.40E-02
9.00E-03
7.63E+05
3.82E-02 1.07E-01
27,190
8.55E-02
1.05E-02
7.63E+05
3.56E-02 1.23E-01
24,140
8.70E-02
1.20E-02
7.63E+05
3.16E-02 1.38E-01
18,970
8.87E-02
1.37E-02
7.63E+05
2.49E-02 1.55E-01
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Related Questions
Compound axial member ABC has a uniform diameter of d = 1.45 in. Segment (1) is an aluminum [E₁ = 10,000 ksi] alloy rod with a
length of L₁ = 85 in. Segment (2) is a copper [E₂ = 17,000 ksi] alloy rod with a length of L₂ = 121 in. When axial force P is
applied, a strain gage attached to copper segment (2) measures a normal strain of 2 = 1800 μin./in. in the longitudinal direction.
What is the total elongation of member ABC?
SABC =
i
eTextbook and Media
GO Tutorial
Save for Later
L
Aluminum
in.
B
L₂2
Copper
CP
Attempts: 0 of 5 used Submit Answer
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●
Example 3.2
A 60mm diameter solid shaft has a strain gauge mounted at
65° to the axis of the shaft. In service a torque is applied to
the shaft and the strain gauge reads 200 x 10-6. Calculate
the value of the torque if the shaft is made from steel with E
= 207 GPa and v = 0.3.
(BC&A question 11.18)
Strain Transformations
Mechanical Engineering
Mechanics of Materials 2
Solution 3.2
Applied torque alone will produce only shear stress/strain
in the directions parallel and perpendicular to the torque.
The strain gauge is mounted at 65° to the axis of the shaft.
Strain Transformations
Mechanical Engineering
Mechanics of Materials 2
Now
Strain
gauge
From the strain circle
65⁰
Strain Transformations
Note that for circular shaft
4
D
J = ² (2) ²*
33
Dr PJ Martin
Lecture Notes
34
Dr PJ Martin
Lecture Notes
35
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Problem F2
stress (ksi)
130
120
110
100
90
80
70
60
50
40
30
20
10
0
0.000 0.025 0.050
0.075 0.100 0.125 0.150 0.175
strain (in/in)
Above you will find the experimental
stress-strain diagram of 1045 steel.
Calculate the permanent set if a
cylindrical specimen with a diameter of
2 in is loaded to the ultimate stress and
then unloaded. Provide your answer in
units of in/in with 3 significant figures
after the decimal. The elastic modulus
of 1045 steel is 29,000 ksi.
arrow_forward
A rectangular steel block is 300mm in the x direction, 200mm in the y direction and 150mm in the z direction. The block is subjected to a triaxial loading consisting of three uniformly distributed forces as follows: 250kN tension in the x direction, 320kN compression in the y direction and 180kN tension in the z direction. For steel, v=0.30 and E=200,000MPa. Determine the total strain in the x direction due to the forces applied in the three directions.
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Example 1: (i) Find the compatibility condition for the strain tensor e; if e1 ,
e22 , e12 are independent of x3 and e31 = e32 = e33 = 0.
(ii) Find the condition under which the following are possible strain
components.
eii
= k(x,? – x2), e12 = k' x1 x2 , e22 = k x1 X2 ,
%3D
e31 = e32 = e33 = 0, k & k' are constants
(iii) when eij given above are possible strain components
corresponding displacements , given that u; = 0.
find the
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uniform diameter d= 1.75 in.
Segment (1) is an aluminum
Ę₁ = 10000 ksi
L₁ = 90 in.
Segment (2) is a copper
Ę₂ = 17000 ksi L₂ = 126 in.
When Pis applied, a strain gage attached to copper segment (2) measures a normal
strain of &2 = 1700 uin./in. in the longitudinal direction.
What is the elongation of segment (1)? Answer in inch rounded-off to 3 decimal places
L₁
A
(1)
Aluminum
B
L2
(2)
Copper
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On the free surface of a component, a
strain rosette was used to obtain the following normal strain
data: €a = 300µɛ, ɛp = 400µs, and Ec = 200µe. Calculate
the normal and shear strains in the x-y plane.
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Tensile test specimens are extracted from the "X" and "y" directions of a rolled sheet of metal. "x" is the rolling direction, "y" is
transverse to the rolling direction, and "z" is in the thickness direction. Both specimens were pulled to a longitudinal strain =
0.15 strain. For the sample in the x-direction, the width strain was measured to be ew= -0.0923 at that instant. For the sample in
the y-direction, the width strain was measured to be gw=-0.1000 at that instant.
The yield strength of the x-direction specimen was 50 kpsi and the yield strength of the y-direction specimen was 52 kpsi.
Determine the strain ratio for the x direction tensile test specimen. Determine the strain ratio for the y-direction tensile test
specimen. Determine the expected yield strength in the z-direction. Give your answer in units of kpsi (just the number). If the
sheet is plastically deformed in equal biaxial tension (a, = 0, to the point where & = 0.15, calculate the strain, 6, that would be
expected.
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Problem 8. Re-write the strain compatibility condition
€22,33 + €33,22
in the engineering notation.
= 2 e23,23
arrow_forward
Compound axial member ABC has a uniform diameter of d=1.45 in. Segment (1) is an aluminum [E1=10,000 ksi] alloy rod with a length of L1=88 in. Segment (2) is a copper [E2=17,000 ksi] alloy rod with a length of L2=128 in. When axial force P is applied, a strain gage attached to copper segment (2) measures a normal strain of ε2=2200 μin./in. in the longitudinal direction. What is the total elongation of member ABC?
answer is in inches
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400
350
300
250
200
150
100
50
0.002
0.004
0.006
0.008
0.01
Strain
Stress (MPa)
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124,000s
o(2)
0 s85 0.03 (o in MPa)
1+ 300ɛ
(a) Find the axial normal strain in the cable and its
elongation due to the load W = 6.8 kN.
(b) If the forces are removed, what is the permanent
set of the cable?
Hint: Start with constructing the stress-strain dia-
gram and determine the modulus of elasticity, E, and
the 0.2% offset yield stress.
D
2 m
A 1.5 m
B
1.5 m
E
m--1 m-
0.75 m
0.25 m
W = 6.8 kN
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A bar BD is held in a horizontal position when there is no
load by the steel wire AB. When a weight Wis added at C,
also hanging from a steel wire CE, then point C is displace
down by 0.025in. The cross-sectional area of the wires is
0.002in². The wires are made of a steel alloy with
E = 29000ksi and σy = 70ksi. Calculate the strain in both
wires, and the weight W.
B
A
N
E
Parameter ft
L_1
4
L_2
1
L_3 2
L_4 4
The strain in AB is € AB =
The strain in CE is ECE =
The weight W = 30
W
cc 080
BY NO SA
2021 Cathy Zupke
Xlb
D
in./in.
in./in.
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An initially rectangular element of a material is deformed into the shape shown in
the figure.
Deformed
Undeformed
16.6
0.1992 mm
15.7
0.2 mm
Normal strain in the x- direction *
-4.0 x 10^-3
4.0 x 10^-3
10 x 10^-3
-10 x 10^-3
Normal strain in the y- direction
O -4.0 x 10^-3
4.0 x 10^-3
O 10 x 10^-3
O -10 x 10^-3
0.1515 mm
0.15 mm
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In uniaxial loading the poisson's ratio is 0.30 and the strain in the x- direction is
250 x 10^-6. What is the strain in the z- direction? *
A -8.5 x 10^-5
B -7.5 x 10^-5
C 7.5 x 10^-5
9.5 x 10^-5
D
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Brinell hardness number = 0 BHN
Rockwell hardness = .
(a) The naval brass for which the stress-strain behavior is shown in Animated Figure 6.12. You might need to use Animated
figure 6.19
Rockwell hardness
Tensile strength = 0 MPA
Tensile strength = 0 ksi
60 70 80 90
100 HRB
Brinell hardness
i
HB
20 30
40
50 HRC
Rockwell hardness
i
HRB
250
1500
200
Steels
150
1000
(b) The steel alloy for which the stress-strain behavior is shown in Animated Figure 6.22. You might need to use Animated
100
Figure 6.19.
500
Brass
Cast iron (nodular)
50
Brinell hardness
i
HB
Rockwell hardness
i
HRB
100
200
300
400
500
Brinell hardness number
Tensile strength (MPa)
Tensile strength (ksi)
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Q. // (A) - An element of bone in plane strain where the state of strain are
(Ex = 8 × 10¬4, €y = 5 × 10¬4, Yxy = 12 × 10-4). Find (a) the stresses at
orientation a counterclockwise angle (40 deg.) from the x-axis, b) the strain
energy and (c) the failure stress by Von Misses criteria. Known
Modulus of elasticity = 2 × 10° N /m² and Poisson's ratio = 0.3
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If an isotropic material has a shear modulus of 125 Gpa and a Poisson's ratio of
0.2, calculate its Young's modulus.
Select one:
O E= 80 Gpa
O E= 450 Gpa
O E= 60 Gpa
%3D
O E= 30 Gpa
%3D
O E= 300 Gpa
%3D
If a rubber material is deformed as shown in the following figure, determine the
normal strain along diagonal BD.
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Stress (MPa)
600
500-
400
300
200
100-
0
0.00
Stress (MPa)
0.04
500
400
300
200
100
0.000
1
0.08
0.002 0.004
Strain
Strain
0.12
0.006
1
0.16
0.20
Figure 6.22 Tensile stress-strain behavior for a steel alloy.
6. A steel alloy with a rectangular cross section 13.8 mm x 6.4 mm and length 560 mm has the stress-strain
behavior shown in Figure 6.22 above. This specimen is subjected to a tensile force of 42,000 N. (a) Determine
the elastic and plastic strain in the specimen in the loaded condition. (b) Determine the final length of the
specimen after the load is released.
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Help me please
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A bar of a uniform cross section is subjected to uniaxial tension and develops a strain in the direction of the force of 1/800. Calculate the change of volume per unit volume. assume v= 1/3
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Calculate the strain (e) for a change in length (ō) and a length (L), where 8 = 0.015
in and L = 5 ft.
0.0025
0.00025
0.00215
0.00125
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h.w:
A compressive force was applied in the longitudinal direction of a
cross-sectional bar ( A= 10 cm) and this resulted in a longitudinal
strain of (0.0007) Calculate the amount of force? If you know that the
elastic modulus (E= 206*10°
m2
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PROPAGATION OF ERRORS
Determine the range of values to which the strain will be calculated.F = 150000 +/- 150 lbfSide length = 1.2 +/- 0.025 inModulus of Elasticity = 25000 +/- 11.5 ksiDetermine the relative error. Is it above or below the accepted 5% relative error?
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Question 2
A torsion circular bar made of aluminum in Figure 1 is loaded by its weight and P kNm
torsion (P = 10). Determine the strain energy stored
in the bar. Given that the diameter of bar AB and BC is d₁= 0.5 m and d₂=0.3 m
respectively. Take the specific weight of the aluminum, y as 27 kN/m³, Poisson's ratio,
v= 0.3 and shear modulus, G as 25 GPa.
2.5 m
1m
B
C
A
Figure 1
d₁
d2
T=P kNm
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A solid bar of length L = 4 m and diameter 200 mm is heated from 20 to 320 degrees celsius and restrained between two solid immovable walls.
Young's modulus of the material is 90 GPa and the coefficient of thermal expansion is 22 x 10-6/°C.
→Calculate the thermal strain, e, in micro-strain correct to two decimal places.
micro-strain.
Calculate the thermal stress, o, in megapascals (MPa) correct to two decimal places.
E:
σ:
MPa.
Hence calculate the force exerted by the bar in meganewtons (MN) correct to two decimal places:
F
MN.
L
arrow_forward
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- Compound axial member ABC has a uniform diameter of d = 1.45 in. Segment (1) is an aluminum [E₁ = 10,000 ksi] alloy rod with a length of L₁ = 85 in. Segment (2) is a copper [E₂ = 17,000 ksi] alloy rod with a length of L₂ = 121 in. When axial force P is applied, a strain gage attached to copper segment (2) measures a normal strain of 2 = 1800 μin./in. in the longitudinal direction. What is the total elongation of member ABC? SABC = i eTextbook and Media GO Tutorial Save for Later L Aluminum in. B L₂2 Copper CP Attempts: 0 of 5 used Submit Answerarrow_forward● Example 3.2 A 60mm diameter solid shaft has a strain gauge mounted at 65° to the axis of the shaft. In service a torque is applied to the shaft and the strain gauge reads 200 x 10-6. Calculate the value of the torque if the shaft is made from steel with E = 207 GPa and v = 0.3. (BC&A question 11.18) Strain Transformations Mechanical Engineering Mechanics of Materials 2 Solution 3.2 Applied torque alone will produce only shear stress/strain in the directions parallel and perpendicular to the torque. The strain gauge is mounted at 65° to the axis of the shaft. Strain Transformations Mechanical Engineering Mechanics of Materials 2 Now Strain gauge From the strain circle 65⁰ Strain Transformations Note that for circular shaft 4 D J = ² (2) ²* 33 Dr PJ Martin Lecture Notes 34 Dr PJ Martin Lecture Notes 35arrow_forwardProblem F2 stress (ksi) 130 120 110 100 90 80 70 60 50 40 30 20 10 0 0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 strain (in/in) Above you will find the experimental stress-strain diagram of 1045 steel. Calculate the permanent set if a cylindrical specimen with a diameter of 2 in is loaded to the ultimate stress and then unloaded. Provide your answer in units of in/in with 3 significant figures after the decimal. The elastic modulus of 1045 steel is 29,000 ksi.arrow_forward
- A rectangular steel block is 300mm in the x direction, 200mm in the y direction and 150mm in the z direction. The block is subjected to a triaxial loading consisting of three uniformly distributed forces as follows: 250kN tension in the x direction, 320kN compression in the y direction and 180kN tension in the z direction. For steel, v=0.30 and E=200,000MPa. Determine the total strain in the x direction due to the forces applied in the three directions.arrow_forwardExample 1: (i) Find the compatibility condition for the strain tensor e; if e1 , e22 , e12 are independent of x3 and e31 = e32 = e33 = 0. (ii) Find the condition under which the following are possible strain components. eii = k(x,? – x2), e12 = k' x1 x2 , e22 = k x1 X2 , %3D e31 = e32 = e33 = 0, k & k' are constants (iii) when eij given above are possible strain components corresponding displacements , given that u; = 0. find thearrow_forwarduniform diameter d= 1.75 in. Segment (1) is an aluminum Ę₁ = 10000 ksi L₁ = 90 in. Segment (2) is a copper Ę₂ = 17000 ksi L₂ = 126 in. When Pis applied, a strain gage attached to copper segment (2) measures a normal strain of &2 = 1700 uin./in. in the longitudinal direction. What is the elongation of segment (1)? Answer in inch rounded-off to 3 decimal places L₁ A (1) Aluminum B L2 (2) Copperarrow_forward
- On the free surface of a component, a strain rosette was used to obtain the following normal strain data: €a = 300µɛ, ɛp = 400µs, and Ec = 200µe. Calculate the normal and shear strains in the x-y plane.arrow_forwardTensile test specimens are extracted from the "X" and "y" directions of a rolled sheet of metal. "x" is the rolling direction, "y" is transverse to the rolling direction, and "z" is in the thickness direction. Both specimens were pulled to a longitudinal strain = 0.15 strain. For the sample in the x-direction, the width strain was measured to be ew= -0.0923 at that instant. For the sample in the y-direction, the width strain was measured to be gw=-0.1000 at that instant. The yield strength of the x-direction specimen was 50 kpsi and the yield strength of the y-direction specimen was 52 kpsi. Determine the strain ratio for the x direction tensile test specimen. Determine the strain ratio for the y-direction tensile test specimen. Determine the expected yield strength in the z-direction. Give your answer in units of kpsi (just the number). If the sheet is plastically deformed in equal biaxial tension (a, = 0, to the point where & = 0.15, calculate the strain, 6, that would be expected.arrow_forwardProblem 8. Re-write the strain compatibility condition €22,33 + €33,22 in the engineering notation. = 2 e23,23arrow_forward
- Compound axial member ABC has a uniform diameter of d=1.45 in. Segment (1) is an aluminum [E1=10,000 ksi] alloy rod with a length of L1=88 in. Segment (2) is a copper [E2=17,000 ksi] alloy rod with a length of L2=128 in. When axial force P is applied, a strain gage attached to copper segment (2) measures a normal strain of ε2=2200 μin./in. in the longitudinal direction. What is the total elongation of member ABC? answer is in inchesarrow_forward400 350 300 250 200 150 100 50 0.002 0.004 0.006 0.008 0.01 Strain Stress (MPa)arrow_forward124,000s o(2) 0 s85 0.03 (o in MPa) 1+ 300ɛ (a) Find the axial normal strain in the cable and its elongation due to the load W = 6.8 kN. (b) If the forces are removed, what is the permanent set of the cable? Hint: Start with constructing the stress-strain dia- gram and determine the modulus of elasticity, E, and the 0.2% offset yield stress. D 2 m A 1.5 m B 1.5 m E m--1 m- 0.75 m 0.25 m W = 6.8 kNarrow_forward
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