Q2c) The steel alloy shaft having a diameter of 38 mm is used to transmit torque, T as shownin Figure Q2(c ). If the allowable shear stress of the shaft is 80kPa,Figure Q2(c )O2 c) (i)The torsion formula :7 =JТ (0.019)T (0.019)^4; Therefore, the max torque T =Nm80 x10 :2Q2 c) (ii) For point lying on the outside suurface of the shaft:T' (0.019)* (0.019 *-0.0125 4); Therefore the max torque T'Nm80x102Q2 c) (iii) The shear stress at point located at inner surface of shaft,O (0.0125)7 (0.019*-0.0125)T'cT'ckPa7 (e," -c4J22

Q2c) The steel alloy shaft having a diameter of 38 mm is used to transmit torque, T as shown in Figure Q2(c ). If the allowable shear stress of the shaft is 80kPa, Figure Q2(c) O2 c) (i) The torsion formula :7 = J

Q2c) The steel alloy shaft having a diameter of 38 mm is used to transmit torque, T as shown in Figure Q2(c ). If the allowable shear stress of the shaft is 80kPa, Figure Q2(c) O2 c) (i) The torsion formula :7 = J

Mechanics of Materials (MindTap Course List)

9th Edition

ISBN:9781337093347

Author:Barry J. Goodno, James M. Gere

Publisher:Barry J. Goodno, James M. Gere

Chapter3: Torsion

Section: Chapter Questions

Problem 3.3.19P: A solid brass bar of diameter d = 1.25 in. is subjected to torques T1as shown in part a of the...

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A stepped shaft ABC consisting of two solid, circular segments is subjected to uniformly distributed torque t1acting aver segment 1 and concentrated torque t2applied at C, as shown in the figure. Segment 1 of the shaft has a diameter of d1= 57 mm and length of L1= 0.75 m; segment 2 has a diameter d2— 44 mm and length L2= 0.5 m. Torque intensity /,"= 3100 N . m/m and T2= 1100 N. m. (a) Find reaction torque TAat support A. (b) Find the internal torque T(x) at two locations: .x = L1/2 and at .x = L1+ L2/2. Show these internal torques on properly drawn free-body diagrams.
A stepped shaft ABC consisting of two solid, circular segments is subjected to torques T}and T2acting in opposite directions, as shown in the figure. The larger segment of the shaft has a diameter of dv- 2.25 in. and a length Lt= 30 in.; the smaller segment has a diameter d2— 1.75 in. and a length L, = 20 in. The torques are T, = 21,000 lb-in. and fz=10.000 lb-in. (a) Find reaction torque TAat support A. (b) Find the internal torque T(x) at two locations: x = L1/2 and x = L1+ L2/2. Show these internal torques on properly drawn free-body diagrams (FBDs).
A nonprismatic bar ABC with a solid circular cross section is loaded by distributed torques (sec figure). The intensity of the torques, that is, the torque per unit distance, is denoted t(x) and varies linearly from zero at A to a maximum value T0/L at B. Segment BC has linearly distributed torque of intensity r(x) = T0/3L of opposite sign to that applied along AB. Also, the polar moment of inertia of AB is twice that of BC and the shear modulus of elasticity of the material is G. Find the reaction torque RA. Find internal torsional moments T(x) in segments AB and BC. Find the rotation t0 Find the maximum shear stress tmaxand its location along the bar, Draw the torsional moment diagram (TMD:T(x),0 < x < L).
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