Shear and Tensile stress due to bending and torsion together

[ketanco]

I was solving an AM problem on PPI exam cafe online . (Question number 59)

Think of a simple mechanism similar to a door knob, fixed to a wall only and free hanging like cantilever, with the part where you hold with your hand in the perpendicular plane. Imagine you are pressing with your finger vertically at the tip of the doorknob handle downwards with say (2000 pounds). so you are both creating torsion and a vertical reaction and a bending moment at the fixed connection at the wall.

The question asks for the maximum shear stress and also tensile stress.

so in the solution, first they calculate a shear stress due to Torsion, from shear stress=Tr/J... I understand that one

Then it calculates a stress due to bending from sigma = My/I ... I understand that one too

Then here is the part I dont understand...

From the Mohrs circle formula, they calculate a maximum shear, with sigma x being the stress we just calculated, and the shear stress we just calculated above. But where did the vertical reaction go? why didnt we take into account the 2000 pound force in the mohrs circle? i mean okay we took that into account when calculating bending stress which is normal to the shaft plane, but this vertical 2000 reaction still exists, to keep the whole thing in equilibrium. so shouldnt we also divide 2000 with vertical cross section area and obtain another shear stress and take that also into account?

 

[MA_PE]

if you're pushing down verticaly on the tip in the center then I don't see where the torsion comes from. you have a two-dimensional problem of fixed cantilever with a concentrated load at the end.

bending stress is Mc/I

shear stress = V/Q

 

[ketanco]

no it is not two dimensional

the handle part is on the other plane as I wrote so it creates torsion in the shaft. i can not draw here unfortunately

 

[MA_PE]

Use superposition. Calculate the shear and bending stress assuming the 2dimensional case and then add the torsion from the eccentric load relative to the main longitudinal axis of the shaft

 

[civilized_naah]

The transverse and longitudinal shear produced by the shear force itself (the force pushing down) vanishes at the outer planes (where the bending stress is maximum). These shear stress components are maximum at the neutral axis, where the bending stress is zero.

On the other hand, the torsional shear stress is maximum at the outer edges, where it coexists with the maximum bending stresses (normal). This is therefore taken as the critical (worst) combination.