| Practice Problem 3-4 (a) |
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Given: Under a certain applied torque, a thin walled titanium (G = 40 GPa) shaft of length L = 2 m, radius R = 80 mm, and thickness t = 5 mm, experiences an angle of twist of q = 5°. Req'd: |
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Solution: First convert 5° to radians: q = 5° (p radians / 180°) = 0.0873 radians The shear strain is related to the angle of twist by: g = Rq/L = (80 mm)(0.0873 radians) / 2000 mm = 0.00349 The shear stress is then: t = gG = (0.00349)(40 GPa) = 0.140 GPa = 140 MPa |
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