A stepped shaft ABC consisting of two solid circular segments is subjected to to

Question

A stepped shaft ABC consisting of two solid circular segments is subjected to torques T1 and T2 acting in opposite directions. as shown . The larger segment of the shaft has diameter d1 = 58 mm and length L1 = 760 mm; the smaller segment has diameter d2 = 45 mm and length L2 = 510 mm. The material is steel with shear modulus G = 76 GPa. and the torques are T1 = 2300 N -m and T2 = 900 N 'm. Calculate the following quantities: (a) the maximum shear stress tau max in the shaft, and (b) the angle of twist phi c (in degrees) at end C.

Explanation / Answer

Data: d1 = 58 x 10^-3 m d2 = 45 x 10^-3 m L1 = 760 mm      = 760 x 10^-3 m L2 = 510 mm      = 510 x 10^-3 m Shear modulus, G = 76 GPa                             = 76 x 10^9 Pa Torque, T1 = 2300 N.m Torque, T2 = 900 N.m Solution: Segment AB: Tab = T2 - T1 = 900 - 2300                        = - 1400 N.m ab = |16 Tab /d1^3|       = 16 * 1400 / [ * (58 x 10^-3)^3 ]       = 3.65 x 10^7 Pa ab = Tab L1 / G (Ip)ab        = Tab L1 / [ G * (/32) d1^4 ]        = - 1400 * 0.760 / [ 76 x 10^9 * (/32) * 0.058^4 ]        = - 0.0126013 rad         Segment BC: Tbc = T2 = 900 N.m bc = |16 Tbc /d2^3|       = 16 * 900 / [ * (45 x 10^-3)^3 ]       = 5.03 x 10^7 Pa bc = Tbc L2 / G (Ip)ab        = Tbc L2 / [ G * (/32) d2^4 ]        =  900 * 0.510 / [ 76 x 10^9 * (/32) * 0.045^4 ]        = 0.0150019 rad (a) Max shear stress = bc = 5.03 x 10^7 Pa (b) Angle of twist at the end of c, = ab + bc              = - 0.0126013 rad + 0.0150019 rad              = 0.0024 rad              = 0.0024 * (180/) deg              = 0.14 deg                 Segment BC: Tbc = T2 = 900 N.m bc = |16 Tbc /d2^3|       = 16 * 900 / [ * (45 x 10^-3)^3 ]       = 5.03 x 10^7 Pa bc = Tbc L2 / G (Ip)ab        = Tbc L2 / [ G * (/32) d2^4 ]        =  900 * 0.510 / [ 76 x 10^9 * (/32) * 0.045^4 ]        = 0.0150019 rad (a) Max shear stress = bc = 5.03 x 10^7 Pa (b) Angle of twist at the end of c, = ab + bc              = - 0.0126013 rad + 0.0150019 rad              = 0.0024 rad              = 0.0024 * (180/) deg              = 0.14 deg                 bc = Tbc L2 / G (Ip)ab        = Tbc L2 / [ G * (/32) d2^4 ]        =  900 * 0.510 / [ 76 x 10^9 * (/32) * 0.045^4 ]        = 0.0150019 rad (a) Max shear stress = bc = 5.03 x 10^7 Pa (b) Angle of twist at the end of c, = ab + bc              = - 0.0126013 rad + 0.0150019 rad              = 0.0024 rad              = 0.0024 * (180/) deg              = 0.14 deg                

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