Find shear strength from tensile strength

[diblazing]

I would like to know if there is a rule of thumb for determining shear strength from tensile strength. For some reason I thought I could multiply tensile strength by 0.6 to get a shear strength value but may just be making that up. The material I am using is a 303 stainless steel with yield strength of about 35 ksi. I will be using it as a pin and need to determine the appropriate diameter of the pin. Thanks for you time.

 

[corus]

I believe that shear stress is assessed against half the yield stress. Check with an appropriate design code to be sure.

 

[CoryPad]

A rule of thumb for engineering alloys is ultimate shear strength = 0.6 * ultimate tensile strength, although austenitic stainless steels can range up to ~ 0.8. If your application is critical, perform a test, since it is a simple test to perform.


Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[diblazing]

This is the first time I have used this forum. Thank you for your quick reply's and your time, you we're both very helpful.
diblazing

 

[rastogisk]

The maximum-shear-stress theory of yielding predict the yield strength in shear to be:
Ssy = 0.5 Sy

The distortion-energy theory of yielding predicts the yield strength in shear to be:
Ssy = 0.577 Sy (this rounds off to be 0.6 Sy)

Ssy = 0.5 Sy is generally used for design purposes since it is conservative and provides some margin to failure. However, if you are investigating a failure, you should use Ssy = 0.577 Sy since it is more accurate and matches well with the test results.

I hope it helps

 

[CoryPad]

corus and rastogisk,

Your information relates to yielding in ductile materials. diblazing's application, a pin, routinely is sized to provide a known shear resistance at fracture, hence the relationship to ultimate tensile strength.


Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[diblazing]

I appreciate the help. I used 0.6 * Su as a starting point but the pin is going to be safety critical so I am going to run a simple test to verify, and then probably use a fair factor of safety. Thank you for your time.
diblazing

 

[strokersix]

We routinely use cotter pins for overload protection on drive systems. Torque level at shear varies widely, affected by hole alignment, hole size, clearance between moving elements, pin hardness, multiple shears wallowing the holes, and wear of the pin over time. You may want to investigate the effect of the above mentioned factors in your application.

 

[PeterCharles]

Its a sad fact of life that many problems require a knowledge of the ultimate shear stress of a material, yet only the stresses in tension are quoted.

The real answer is to test a material sample from the batch you are using as quoted figures are only an approximation anyway.

But what sort of factor of safety will be used? In many cases the higher the factor the less accurate the ultimate figure need be.