## Superposition of Shear Stress

## Superposition of Shear Stress

(OP)

Hi all,

I'm having trouble solving this problem. I have a pipe, on a certain section this pipe has a shear load and torque applied, they both generate shear stress on the same plane.

A is the point at the top of the pipe section (12 o'clock), B at 3 o'clock etc, C at 6 o'clock and D at 9 o'clock.

We know that:

I need to find a tau xy to put in Von Mises equation.

Thank you

I'm having trouble solving this problem. I have a pipe, on a certain section this pipe has a shear load and torque applied, they both generate shear stress on the same plane.

A is the point at the top of the pipe section (12 o'clock), B at 3 o'clock etc, C at 6 o'clock and D at 9 o'clock.

We know that:

- shear stress
__from shear load__is max at B and D and 0 at A and C - shear stress
__from torque__is max at any point of the outer circumference so at A, B, C and D

**how do I sum them**?I need to find a tau xy to put in Von Mises equation.

Thank you

## RE: Superposition of Shear Stress

-It is true that ,shear stress from shear load is max at B and D and 0 at A and C.

-It is also true that ,shear stress from torque is constant at any point having the same radius and tangential to tge circumference so at A, B, C and D ..

Another truth is the shear stress must be tangential for circular sections at the extreme edges for equilibrium .

Assuming the shear load vertical and downward, and torsion clockwise,

-Betwen A to C; total shear stress = shear stress from shear load + shear stress from torque

-Between C to A ; total shear stress = shear stress from shear load - shear stress from torque

But, if you have shear load P, the same section SHALL experience bending stresses..

## RE: Superposition of Shear Stress

If I am understanding your problem (like a homework out of a mechanics of materials textbook), the shear stress from pure shear in a single plane and torsional shear are simply additive (not SRSS). The peak happens at a single point.

For steel, you can compare you demand (summation of shear stresses) with the capacity = tau_max = 0.577 * F_yield (also considering safety factors as req'd).

## RE: Superposition of Shear Stress

another day in paradise, or is paradise one day closer ?

## RE: Superposition of Shear Stress

## RE: Superposition of Shear Stress

EDIT: Unless I made a mistake, that doesn't seem to agree with (4) by r13.

If r1 = 1, r2 = 2 then the bracketed expression is (4 + 2 + 1)/(4 + 1) = 7/5 = 1.4

and Tau

_{max}= 4*1.4/3 *V/A = 1.867V/A (93% of r13's 2V/A)can anyone see the problem?

r13, what is your source and what is equation (1)?

BA

## RE: Superposition of Shear Stress

BA

## RE: Superposition of Shear Stress

Here is the link. Link Do you have anything on shear stress over solid rod? It looks like that's the form you got.

## RE: Superposition of Shear Stress

Yes, for solid rod we agree on 4V/3A for maximum shear stress.

The difference in the hollow cylinder involves an approximation. Your source used radius r to the middle of the ring whereas my source considered separate values for r1 and r2. In most cases, the value of t (thickness) is small compared to r, so it makes very little difference which way you do it.

For example, if r1 = 1 and r2 = 1.1, then the bracketed expression is (1.21 + 1.1 + 1)/(1.21 + 1) = 3.31/2.21 = 1.49774

and Taumax = 4*1.49774/3 *V/A = 1.997V/A (close enough to r13's 2V/A)

BA

## RE: Superposition of Shear Stress

Good work. Seems you still have lots of math in you, and having fun :)