03-01ChapGere.0010

# 03-01ChapGere.0010 - 5 50 MPa M IN DIAMETER BASED UPON...

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Solution 3.3-9 Three circular bars 190 CHAPTER 3 Torsion d 1 5 diameter of bars 5 0.5 in. d 2 5 diameter of disks 5 3.0 in. P 1 5 28 lb T 1 5 P 1 d 2 T 2 5 P 2 d 2 T 3 5 P 3 d 2 T HE THREE TORQUES MUST BE IN EQUILIBRIUM T 3 is the largest torque M AXIMUM SHEAR STRESS (Eq. 3-12) t max 5 16(28 lb)(3.0 in.) Ï 2 p (0.5 in.) 3 5 4840 psi t max 5 16 T p d 3 5 16 T 3 p d 1 3 5 16 P 1 d 2 Ï 2 p d 1 3 T 3 5 T 1 Ï 2 5 P 1 d 2 Ï 2 C A B 135 ° 135 ° 90 ° T 1 T 3 T 2 45 ° T 1 T 3 T 2 Problem 3.3-10 The steel axle of a large winch on an ocean liner is subjected to a torque of 1.5 kN ? m (see figure). What is the minimum required diameter d min if the allowable shear stress is 50 MPa and the allowable rate of twist is 0.8°/m? (Assume that the shear modulus of elasticity is 80 GPa.) Solution 3.3-10 Axle of a large winch T T d T 5 1.5 kN ? m G 5 80 GPa t allow
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Unformatted text preview: 5 50 MPa M IN . DIAMETER BASED UPON SHEAR STRESS d 5 0.05346 m Ê d min 5 53.5 mm d 3 5 16(1.5 kN ? m) p (50 MPa) 5 152.789 3 10 2 6 m 3 t 5 16 T p d 3 Ê d 3 5 16 T p t allow 5 0.013963 rad / m u allow 5 0.8 8 / m 5 (0.8 8 ) ¢ p 180 ≤ rad / m M IN . DIAMETER BASED UPON RATE OF TWIST R ATE OF TWIST GOVERNS d min 5 60.8 mm d 5 0.0608 m Ê d min 5 60.8 mm 5 0.00001368 m 4 d 4 5 32(1.5 kN ? m) p (80 GPa)(0.013963 rad / m) u 5 T GI p 5 32 T G p d 4 Ê d 4 5 32 T p G u allow T T d A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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## This note was uploaded on 09/20/2009 for the course COE 3001 taught by Professor Armanios during the Spring '08 term at Georgia Tech.

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