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hw1fsol.sp12

hw1fsol.sp12 - Name of Student ME 452 Machine Design II...

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1 Name of Student:_______________________ ME 452: Machine Design II Spring Semester 2012 Homework Set 1 Attempt by Wednesday, January 18th Solve the following problems from Chapter 3, Shigley’s Mechanical Engineering Design, Ninth Edition, R.G. Budynas, and J.K. Nisbett. 1. Problem 3-15, page 128. Parts (a) and (b) only. 2. Problem 3-20, page 129. 3. Problem 3-84, page 140. 4. Problem 3-90, page 141. 5. Problem 3-137, page 146. 6. Problem 3-139, page 146.

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2 Solution to Homework Set 1. 1. Problem 3-15, page 128. Part (a). Mohrs circle is shown in Figure 1a. The geometry is as follows: 20 10 5 2 C == kpsi and 20 10 15 2 CD + = = kpsi (1) The radius of Mohrs circle is 22 15 8 17 R = += kpsi (2) Therefore, the principal in-plane normal stresses are 1 5 17 22 kpsi σ =+ = and 2 5 17 12 kpsi = −= (3) The orientation is 1 18 tan 14.04 clockwise 21 5 p φ ⎛⎞ ⎜⎟ ⎝⎠ D (4) The principal in-plane shear stress is 1 17 R τ = = kpsi (5) The orientation is 45 14.04 30.96 counterclockwise s =− = DD D (6) Figure 1a. Mohrs circle. The principal in-plane normal stresses are shown on the stress element is shown in Figure 1b.
3 Figure 1b. The principal in-plane normal stresses. The principal in-plane shear stresses are shown on the stress element is shown in Figure 1c. Figure 1c. The principal in-plane shear stresses. Part (b). Mohrs circle is shown in Figure 1d. The geometry is as follows: 91 6 12.5 2 C + == kpsi and 16 9 3.5 2 CD kpsi (7) The radius of Mohrs circle is 22 5 3.5 6.10 R =+ = kpsi (8) Therefore, the principal in-plane normal stresses are 1 12.5 6.1 18.6 σ = kpsi and 2 12.5 6.1 6.4 = −= kpsi (9) The orientation is 1 15 tan 27.5 23 . 5 p φ ⎛⎞ ⎜⎟ ⎝⎠ D ccw (10) The principal in-plane shear stress is 1 6.10 R τ = = kpsi (11) The orientation is

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4 45 27.5 17.5 s φ =− = DDD cw (12) Figure 1d. Mohrs circle.
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hw1fsol.sp12 - Name of Student ME 452 Machine Design II...

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