SolutionsHW4

# SolutionsHW4 - HW#4 Problem 6.8 a To Find The diameter of...

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HW #4 Problem 6.8 a. To Find: The diameter of the test cylinder,d 0 . b. Given: Load, F = 6660 N (1500 lb) Modulus of elasticity, E = 110 GPa (16 *10 6 psi) Yield strength, y = 240 MPa (35,000 psi) Initial length of rod,l 0 = 380 mm (15.0 in) Elongation, l = 0.50 mm (0.020 in) c. Assumptions: (i) The applied stress is in the elastic regime, i.e, the deformation is elastic. If this assumption is valid, F/( * d 0 2 /4) < y (ii) The applied force is uniaxial and tensile/compressive and in a direction parallel to the length of the rod. (iii) The material is isotropic and hence, the value of E is independent of the direction in which elongation occurs. (iv) The experiment is conducted at room temperature. This is relevant because we do not wish to consider time-dependent plastic deformation (creep) that occurs at   y and comes into the picture at higher temperatures. d. Solution: A 0 = * d 0 2 /4 4 = = 2 0 0 d F A F …(1)

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0 = l l E …(2) From (1) and (2), 4 2 0 d F = 0 l l E l E F l d 0 0 4 = d 0 = 7.65 10 -3 m = 7.65 mm (0.30 in.)
Problem 6.22 a. To Find: (a) Specimen elongation (b) Reduction in specimen diameter b. Given: (a) Stress-strain behavior of the alloy is as shown in Fig. 6.12 (b) Diameter of cylinder, d 0 = 6 mm (0.24 in.) (c) Length of specimen,l 0 = 50 mm (2 in.) (d) Tensile force, F=5000N (1125 lb) (e) Poisson’s ratio = 0.3 c. Assumptions: (i) Fig.6.12 is an accurate representation of the stress vs. strain behavior for the given alloy. (ii) The applied force is uniaxial and in a direction parallel to the length of the cylinder. (iii) The material is isotropic. (iv) The experiment is conducted at room temperature. This is relevant because we do not wish to consider time-dependent plastic deformation that comes into the picture at higher temperatures. d. Solution: (a) Applied stress, = F A 0 = F d 0 2 2 = 5000 N 6 10 3 m 2 2 = 177 10 6 N/m 2 =177 MPa (25,000 psi) From the stress-strain curve in Figure 6.12, at this stress, strain ≈ 2.0 10 -3 .

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= l/l in.) 10 (4 mm 0.10 = mm) (50 10 2.0 = = -3 -3 0 ) ( l l (b) Fig.1. The cylinder before and after (red color) elongation. In part (a), the elongation/strain is in a direction parallel to the direction of the applied force/stress. In a lateral direction, from equation 6.8, x    z …(1) Also, by definition, x = ∆d/d 0 …(2) From (1) and (2),
d = d 0 x = d 0  z = (6 mm)(0.30) ( 2.0 10 -3 ) ∆d = – 3.6 10 -3 mm ( 1.4 10 -4 in.) a   l = 0.1 mm (4 x 10 -3 in.) b   d = 3.6 10 -3 mm ( 1.4 10 -4 in.)

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Problem 6.29
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