Question A horizontal shaft AD supported in bearings at A and B and carrying pulleys at C and D is to transmit 75 kW at 500 r.p.m. from drive pulley D to off-take pulley C, as shown in figure below. Calculate the diameter of shaft. The data given is: P1 = 2 P2 (both horizontal), Q1 = 2 Q2 (both vertical), radius of pulley C = 220 mm, radius of pulley D = 160 mm, allowable shear stress = 45 MPа BP -- - -- 600 +-600— 300 | All dimensions in mm.

OCJO8I The Asker · Mechanical Engineering

Transcribed Image Text: A horizontal shaft AD supported in bearings at A and B and carrying pulleys at C and D is to transmit 75 kW at 500 r.p.m. from drive pulley D to off-take pulley C, as shown in figure below. Calculate the diameter of shaft. The data given is: P1 = 2 P2 (both horizontal), Q1 = 2 Q2 (both vertical), radius of pulley C = 220 mm, radius of pulley D = 160 mm, allowable shear stress = 45 MPа BP -- - -- 600 +-600— 300 | All dimensions in mm.
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Transcribed Image Text: A horizontal shaft AD supported in bearings at A and B and carrying pulleys at C and D is to transmit 75 kW at 500 r.p.m. from drive pulley D to off-take pulley C, as shown in figure below. Calculate the diameter of shaft. The data given is: P1 = 2 P2 (both horizontal), Q1 = 2 Q2 (both vertical), radius of pulley C = 220 mm, radius of pulley D = 160 mm, allowable shear stress = 45 MPа BP -- - -- 600 +-600— 300 | All dimensions in mm.
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The given problem is an application of design of shaft based oncombined bending moment and twisting moment. Equivalent twistingmoment is calculated and based on that diameter of shaft iscalculated. Please look into the images for solution in deQ) Given data" power "P=75kW=75 xx10^(3)Wspeed in R.P.M (N)=500 r.p.mAt pulley C,Q_(1)=2Q_(2) (both vertical)At pulley D,P_(I)=2P_(2) (both horizontal)Redius of pulley C,r_(c)=220mm=0.22mRadices of pulley D,gamma_(D)=160mm=0.16mAllowable shear stress (tau_(max))=45Mpa A and B are bearing.Step -(Figure - 1)we have the-formula power P=T xx((2pi N)/(60))=>" Torque "T=(60 xx P)/(2xx pi xx N)=(60 xx75 xx10^(3))/(2xx pi xx500)=>quad T=1432.4Nm". "T.At pulley D,T=(P_(1)-P_(2))xxgamma_(D)=(2P_(2)-P_(2))xxgamma_(D)=>T=P ... See the full answer