Chapter 6 hw edit
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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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(b) Compute the modulus of elasticity.
-
169.29 - 119.76 / 0.001 - 0.0005 = 99060
(c) Determine the yield strength at a strain offset of 0.002.
-
264.28
(d) Determine the tensile strength of this alloy.
-
409.85
(e) Compute the modulus of resilience.
-
264.28 * 0.003 / 2 = 0.396
(f) What is the ductility, in percent elongation?
-
88.725 - 75 / 75 * 100 = 18.30%
6.37 Determine the modulus of resilience for each of the following alloys:
-
Steel alloy: 550^2 / 400E3 = 0.76
-
Brass alloy: 350^2 / 200E3 = 0.61
-
Aluminum alloy: 250^2 / 140E3 = 0.45
-
Titanium alloy: 800^2 / 220E3 = 2.91
6.42 A tensile test is performed on a metal specimen, and it is found that a true plastic
strain of 0.20 is produced when a true stress of 575 MPa (83,500 psi) is applied; for the
same metal, the value of K in Equation 6.19 is 860 MPa (125,000 psi). Calculate the true
strain that results from the application of a true stress of 600 MPa (87,000 psi).
-
True strain = ln (1 + Engineering strain)
-
True stress = engineering stress ( 1 + engineering strain)
-
575 = 860 *0.20^n -> 0.668 = 0.20^n -> ln0.668 = n ln0.20 n = 0.25
-
600 = 860 * x^0.25 -> 0.698 = x^0.25 x = 0.24
6.50 A steel alloy specimen having a rectangular cross section of dimensions 12.7 mm ×
6.4 mm (0.5 in. × 0.25 in.) has the stress–strain behavior shown in Figure 6.22. If this
specimen is subjected to a tensile force of 38,000 N (8540 lbf) then
(a) Determine the elastic and plastic strain values.
-
Elastic: 38000 / (12.7 * 6.4) = 467.52 ->
467.52 / 207E3 = 0.0023
-
Plastic: 0.010 - 0.0023 = 0.0077
(b) If its original length is 460 mm (18.0 in.), what will be its final length after the load in
part (a) is applied and then released?
-
Final length: (460)(1 + 0.0077) = 463.54
6.51 (a) A 10-mm-diameter Brinell hardness indenter produced an indentation 1.62 mm in
diameter in a steel alloy when a load of 500 kg was used. Compute the HB of this
material.
-
HB = (2 * 500) / (pi * 10 (10 - (10^2 - 1.62^2)^(½)) ) = 241
(b) What will be the diameter of an indentation to yield a hardness of 450 HB when a 500
kg load is used?
-
450 = 500 / diameter of an indentation
-
(100 - (10 - (2*500 / 450*pi*10))^2)^½ = 1.20
6.54 Estimate the Brinell and Rockwell hardnesses for the following:
(a) The naval brass for which the stress–strain behavior is shown in Figure 6.12.
-
450mpa / 3.45 = 130.43
(b) The steel alloy for which the stress–strain behavior is shown in Figure 6.22.
-
500mpa / 3.45 = 144.93
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stress (ksi)
130
120
110
100
90
80
70
60
50
40
30
20
10
0
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●
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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
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Strain Transformations
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D
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Dr PJ Martin
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