# How Much Torque Can a Stepped Shaft Withstand Before Shearing?

• acc0untnam3
In summary, the torque that can be applied to the larger side before the material starts to shear is constant throughout the shaft, so failure occurs in the smaller shaft, probably at the abrupt diameter change.
acc0untnam3
Hi there,

I'm trying to find a solution to a problem that I thought was straight forward, but am now struggling with.

I've got a shaft with a fixed end set up, where there is a small diameter fixed and then it steps up to a larger diameter which is having a torque applied to it.

I'm trying to find out the maximum torque that can be applied to the larger side before the material starts to shear.

The only things that i know about the shaft are the diameters and material type.

So far I've been using:

T=(τ*J)/r
and
τ=0.557*σ

Where:
T = maximum torque
τ = allowable shear stress
J = polar moment of inertia
σ = ultimate tensile strength.

Can anybody help me out? I'm not sure if I've got the equations right or whether I'm going down the correct route.

Cheers

acc0untnam3 said:
Hi there,

I'm trying to find a solution to a problem that I thought was straight forward, but am now struggling with.

I've got a shaft with a fixed end set up, where there is a small diameter fixed and then it steps up to a larger diameter which is having a torque applied to it.

I'm trying to find out the maximum torque that can be applied to the larger side before the material starts to shear.
Do you mean starts to yield in shear?
The only things that i know about the shaft are the diameters and material type.

So far I've been using:

T=(τ*J)/r
OK, within elastic range
and
τ=0.557*σ
this is a good approximation for the shear yield stress for steel...not always true for other materials...best to look it up...but note that σ is the tensile yield stress, not the ultimate stress.
Where:
T = maximum torqueat yield
τ = allowable shear stressat yield
J = polar moment of inertiaof smaller diameter shaft
r = radiusof smaller diameter shaft
σ = ultimate tensile strength.yield stress, not ultimate

Can anybody help me out? I'm not sure if I've got the equations right or whether I'm going down the correct route.

Cheers
The torque is constant throughout the shaft, so failure occurs in the smaller shaft, probably at the abrupt diameter change.

1 person
acc0untnam3: I agree with PhanthomJay, except I probably would use 0.577, instead of 0.557. Also, I probably would define Stu = tensile ultimate strength, Sty = tensile yield strength, FSu = ultimate factor of safety, and FSy = yield factor of safety. Then define Ssa = allowable shear stress = min(0.60*Stu/FSu, 0.577*Sty/FSy). You want to ensure shear stress, tau2, does not exceed Ssa.

Nominal shear stress, tau1 = T*r/J, where r = radius of smaller-diameter shaft, as pointed out by PhanthomJay. And, tau2 = Kn*tau1, where Kn = stress concentration factor, from here. However, because you are using FSu and FSy, you might be able to use Kn = 1.00 (?).

Aside: The main problem in post 1 is, you said allowable shear stress (Ssa) is equal to shear ultimate strength (Ssu). That is incorrect. Allowable shear stress (Ssa) should be less than (not equal to) shear yield strength (Ssy), where Ssy = 0.577*Sty, and Ssu = 0.60*Stu.

You claimed in post 1 you want to "find the maximum torque that can be applied before the material starts to [rupture]." If this is truly what you want (and you do not want the allowable shear stress, for design), then you would set tau2 = Ssu, where tau2 = Kn*tau1, and Ssu = 0.60*Stu. This would tell you when the material starts to rupture. You, of course, would never use this for design.

Or, in post 1, did you mean you want to know when the material starts to yield in shear? If this is truly what you want (and you do not want the allowable shear stress, for design), then you would set tau2 = Ssy, where tau2 = Kn*tau1, and Ssy = 0.577*Sty. This would tell you when the material starts to yield. You, of course, would never use this for design.​

Last edited:
Thanks for both of your answers - I was after the Torque that would lead to shearing of the material. Problem solved :)

## 1. What is torsional stress on a shaft?

Torsional stress on a shaft is a type of stress that occurs when a twisting force, also known as torque, is applied to a shaft. This can cause the shaft to deform or break if the stress exceeds its strength.

## 2. What are the causes of torsional stress on a shaft?

Torsional stress on a shaft can be caused by various factors such as uneven distribution of weight or pressure on the shaft, sudden changes in direction or speed of rotation, and external forces acting on the shaft.

## 3. How is torsional stress calculated?

Torsional stress is calculated by dividing the applied torque by the polar moment of inertia of the shaft. The polar moment of inertia is a measure of the shaft's resistance to torsion and is dependent on the geometry of the shaft.

## 4. What are the effects of torsional stress on a shaft?

The effects of torsional stress on a shaft can range from minor deformations to complete failure of the shaft. It can also cause vibrations, noise, and instability in the rotating system.

## 5. How can torsional stress on a shaft be prevented?

Torsional stress on a shaft can be prevented by using materials with high strength and stiffness, ensuring proper alignment and balance of the shaft, and avoiding sudden changes in torque or speed. Regular maintenance and inspection of the shaft can also help prevent torsional stress.

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