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# Superposition of Shear Stress4

## 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:
• 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
When the shear stress ARE NOT directed in the same direction, but they are on the same plane (for example between A and B at 45 degrees) 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

#### Quote (flyforever85 ...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)

-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

I'm with HT (I think).
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

I'd've drawn the shear flows along the thin wall CL. I don't think there's vector addition, only scalar addition.

### RE: Superposition of Shear Stress

The derivation below profs me wrong. Since the shear flow/distribution manners are the same, scalar addition is correct. Apologize for my ignorance. (Previous erroneous comments have been deleted)

### RE: Superposition of Shear Stress

Shear stress due to bending of a hollow circular tube is shown in the first sketch below. An expression for the maximum value is shown in the second sketch. Other values can be determined using the relationship s = VQ/Ib where s is shear stress due to bending. Shear stress is always tangential to the circle, so the combined stress due to torsion and bending are found by scalar addition.

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 Taumax = 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

I see the problem now. r13 is using a line element for the circle whereas I was using the exact area.

BA

### RE: Superposition of Shear Stress

BA,

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

r13,

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

BA,

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

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