As the hollow steel shaft shown rotates at 180 rpm, astroboscopic measurement in

Question

As the hollow steel shaft shown rotates at 180 rpm, astroboscopic measurement indicates that the angle of twist of theshaft is 3 degree. Knowing that G = 77.2 GPa, determine (a) thepower being transmitted, (b) the maximum shearing stress in theshaft. From the Fig.3.68 we know L=5 m Outer diameter is 60 mm. As the hollow steel shaft shown rotates at 180 rpm, astroboscopic measurement indicates that the angle of twist of theshaft is 3 degree. Knowing that G = 77.2 GPa, determine (a) thepower being transmitted, (b) the maximum shearing stress in theshaft. From the Fig.3.68 we know L=5 m Outer diameter is 60 mm. Inner diameter is 25 mm.

Explanation / Answer

N = 180 rpm = 2N/60 = 2*180/60 = 18.85 rad/sec = 3 degrees = 3 * /180 = 0.05236 rad L = 5m J = (0.064-0.0254)/2 J = 1.9744 *10-5 m4 we know that T/J = G/L T = 77.2*109*0.05236*1.9744*10-5/5 T = 15.962 kN.m (a) Power,P= T P = 15962*18.85 P = 300 kW (b) max =T*r/J max = 15962*0.03/1.9744*10-5 max = 24.2 MPa

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