306_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

# 306_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: Sec_3.5-3.7.qxd 300 9/27/08 1:10 PM Page 300 CHAPTER 3 Torsion Segment AB: 0.4018 fBC d4 32TAB LAB fAB pGd 4 32(17,330 lb in.)(6 ft)(12 in./ft) p(11.5 * 106 psi)d 4 ( AC)allow 1.1052 fAB d d4 1.5070 d4 and d 2.75 in. Angle of twist governs d 32 TBCLBC pGd 1.5070 0.02618 rad 0.02618 4 Segment BC: fBC fAB + fBC From A to C: fAC ; 2.75 in. 4 32(9450 lb-in.)(4 ft)(12 in./ft) p(11.5 * 106 psi)d 4 Problem 3.7-10 The shaft ABC shown in the figure is driven by a motor that delivers 300 kW at a rotational speed of 32 Hz. The gears at B and C take out 120 and 180 kW, respectively. The lengths of the two parts of the shaft are L1 1.5 m and L2 0.9 m. Determine the required diameter d of the shaft if the allowable shear stress is 50 MPa, the allowable angle of twist between points A and C is 4.0°, and G 75 GPa. Solution 3.7-10 Motor-driven shaft At point A: TA 300,000 W 2p(32 Hz) At point B: TB 120 T 300 A 596.8 N # m 180 T 300 A 895.3 N # m L1 1.5 m L2 0.9 m d diameter At point C: TC f 32 Hz FREE-BODY DIAGRAM allow G ( 50 MPa 75 GPa AC)allow 4° 0.06981 rad TORQUES ACTING ON THE SHAFT TA 1492 N m newton meters TB 596.8 N m P 2pf TC 895.3 N m P 2 fT P T T watts f Hz d diameter 1492 N # m ...
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## This note was uploaded on 12/22/2011 for the course MEEG 310 taught by Professor Staff during the Fall '11 term at University of Delaware.

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