The figure shows a shaft mounted in bearings at A and D and

The figure shows a shaft mounted in bearings at A and D and having pulleys at B and C. The forces shown action on the pulley surfaces represent the belt tensions. The shaft is to be made of AISI 1035 CD steel. Using a conservation failure theory with a design factor of 2, determine the minimum shaft diameter to avoid yielding.

Solution

a) first find the reactions at the bearings, then evaluate the moment in xy plane and x-z plane

Using the Bending moment diagram, find where the maximum moment occurs ( it is zero at the two bearings and rises up and down between.

The resultant moment at each B and C can be found from summing the squares and taking the square roots

Find which point has largest BM, You will find that C has the largest

This part deals with the Bending moment due to the uneven forces.

Now there is torsion between the two pulleys, the torque due to pulley at B is ((1200-200)*0.2 Nm = 200 Nm

and the reverse torque from pulley at C is 1480*0.15 Nm = 222 Nm

The net torque will be the sum of these two acting between B and C

The stress caused by this is 16*T/(pi d^3), where the stress casued by BM is

32 M/(pid^3)

Maximum stress due to bending and torsion is found as theresultant of these two.

Using this stress as the maximum as a simple guide, reduce it by a safety factor for the yield stress found from tables or ruleof thumb. 2 is suggested in the problem.

Doing this the value of d^3 is found.

Regret I cannot give you the numerical solution, this is a real life problem so use your brain.