Solid Steel Loading

what would be the toque strength of a shaft?

what would be the toque strength of a shaft,the shaft between the drive and load is only 1 inch long and is made of 314L stainless steel it has a 3/8 diameter and is of solid rod,need the answer in horse power,

There’s a pretty good example of the calculation you need at http://ocw.mit.edu/NR/rdonlyres/Materials-Science-and-Engineering/3-11Mechanics-of-MaterialsFall1999/Modules/torsion.pdf. Search the text for Example 4.

The only difference is, you’re working backwards.

First of all, you need material properties. Unfortunately, 314 L has a range of strengths, depending on how much it’s been work hardened. I’ll assume it’s annealed. In that condition, it has an ultimate tensile strength of 100 ksi, and a yield tensile strength of 50 ksi. (ksi means 1000 psi.) But since you’re putting torque on the shaft and no axial load, it’s more convenient to use shear allowables.

The tensile yield strength and shear yield strength are always related for any ductile material by a factor of 1/sqrt(3), or 0.577. So, the shear yield strength for our material is 0.577 * 50 = 28.9 ksi.

The tensile ultimate and tensile yield strength don’t have a constant relationship, but it’s generally conservative to assume a 0.5 factor for steels. So, 0.5*100 = 50 ksi shear ultimate strength.

Now, since you didn’t include any safety factor in your problem, we pick the lower of these two, which is yield. (You might have specified a safety factor of 2 on yield failure and 5 on ultimate failure, for example, in which case we would have used the lower of 28.9/2 = 14.5 ksi and 50/5 = 10 ksi.)

Now, we know that when the stress anywhere in the shaft reaches τ = 28.9 ksi, it will experience permanent deformation, which we will consider to be failure. We can find the torque, T, on the shaft that will produce this stress, using

τ = T*r / J

where r is the radius of the shaft, and J is the polar moment inertia of the cross section, equal to π*r^4 / 2. (Note that this formula gives you the shear stress at any radius in the cross section. However, the *maximum* shear stress occurs at the maximum value of r, on the outside surface of the bar.) So, we can solve that for T.

T = τ*J / r
T = τ*(π*r^4 / 2) / r
T = τ*(π*r^3) / 2

We know everything on the right hand side, so we can solve for T.

T = (28.9 * 1000 lb/in^2) * [π * (3/8 in / 2)^3] / 2
T = 300 lb-in

So that’s the torque the shaft can withstand.

Now, you want an answer in horsepower, but horsepower is torque times rotational velocity. So if you want an answer in horsepower, you need to specify the RPM. Let’s say it’s 500 RPM.

Then,

T = W/w

where W is the work (horsepower), and w is the rotational velocity. So,

W = T*w
W = 300 lb-in * 500 rev/min

Now, it’s essentially a unit conversion problem. We know that 1 hp = 33,000 ft-lb/min, by definition. So, we need to change inches to feet. That’s easy, 12 in = 1 ft. We also need to do something about those revolutions. That’s a little trickier. 1 revolution = 2 pi radians, and a radian is a unitless dimension, so it just drops out. Essentially, we could write 1 revolution = 2 pi (with no units). Now, we can finish the problem.

W = 300 lb-in * (1 ft / 12 in) * 500 rev/min * (2 pi / 1 rev)
W = 78540 lb-ft/min * (1 hp / 33000 ft-lb/min)
W = 2.4 hp (at 500 rpm)

There’s your answer. Of course, you have to match the shaft to both the power and the rotational velocity of your motor – it could also be a 1 hp motor at 1200 rpm, or a 5 hp motor at 240 rpm, or a 400 hp motor at 3 rpm. (Good luck finding that motor, though!)

Note that the length of the shaft doesn’t come into play in this calculation. It’s only important if you care about the “springiness” of the shaft – the amount of twist, or the amount of energy it stores.

I hope that helps!

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