November 2008
Solution to Problem 209 Axial Deformation
Problem 209
An aluminum bar having a cross-sectional area of 0.5 in2 carries the axial loads applied at the positions shown in Fig. P-209. Compute the total change in length of the bar if E = 10 × 106 psi. Assume the bar is suitably braced to prevent lateral buckling.
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Solution to Problem 208 Axial Deformation
Problem 208
A steel tire, 10 mm thick, 80 mm wide, and 1500.0 mm inside diameter, is heated and shrunk onto a steel wheel 1500.5 mm in diameter. If the coefficient of static friction is 0.30, what torque is required to twist the tire relative to the wheel? Neglect the deformation of the wheel. Use E = 200 GPa.
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Solution to Problem 207 Axial Deformation
Problem 207
A steel wire 30 ft long, hanging vertically, supports a load of 500 lb. Neglecting the weight of the wire, determine the required diameter if the stress is not to exceed 20 ksi and the total elongation is not to exceed 0.20 in. Assume E = 29 × 106 psi.
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Solution to Problem 206 Axial Deformation
Problem 206
A steel rod having a cross-sectional area of 300 mm2 and a length of 150 m is suspended vertically from one end. It supports a tensile load of 20 kN at the lower end. If the unit mass of steel is 7850 kg/m3 and E = 200 × 103 MN/m2, find the total elongation of the rod.
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Solution to Problem 205 Axial Deformation
Problem 205
A uniform bar of length L, cross-sectional area A, and unit mass ρ is suspended vertically from one end. Show that its total elongation is δ = ρgL2/2E. If the total mass of the bar is M, show also that δ = MgL/2AE.
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Solution to Problem 204 Stress-strain Diagram
Problem 204
The following data were obtained during a tension test of an aluminum alloy. The initial diameter of the test specimen was 0.505 in. and the gage length was 2.0 in.
| Load (lb) | Elongation (in.) | Load (lb) | Elongation (in.) |
| 0 | 0 | 14 000 | 0.020 |
| 2 310 | 0.00220 | 14 400 | 0.025 |
| 4 640 | 0.00440 | 14 500 | 0.060 |
| 6 950 | 0.00660 | 14 600 | 0.080 |
| 9 290 | 0.00880 | 14 800 | 0.100 |
| 11 600 | 0.0110 | 14 600 | 0.120 |
| 12 600 | 0.0150 | 13 600 | Fracture |
Plot the stress-strain diagram and determine the following mechanical properties: (a) proportional limit; (b) modulus of elasticity; (c) yield point; (d) yield strength at 0.2% offset; (e) ultimate strength; and (f) rupture strength.
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