Calculation Based on Yield strength and Ultimate strength

AI Thread Summary
The discussion revolves around the methods for calculating shear capacity in mechanical design, specifically using yield strength and ultimate strength. Shear yield strength is derived as one-third of the yield strength, while ultimate shear strength is 75% of ultimate strength. The significance of these methods lies in their differing implications for safety and performance; yield strength relates to elastic behavior under working loads, whereas ultimate strength assesses safety during failure. Participants clarify that the factors of 0.577 and 0.75 are not purely safety factors, with the former indicating elastic limits and the latter providing a safety margin. Understanding these distinctions is crucial for effective mechanical design calculations.
shinoysivan
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Hello Gurus,
I am new to mechanical design calculations.

Since I have came across one interesting fact there are two methods using for finding the shear capacity of the components. Yield Strength and Ultimate strength.
Shear yeild strength=1/√3* Yield strength
Ultimate shear strength= .75 of of Ultimate strength
Up on these shear capcacity calculations are made.

What is the signficance of the using two methods for the design calculations? Either one is enough?
 
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The first is (the limit of) elastic behaviour and helps you decide what to do at working loads. But it says nothing about safety. The second tells you about safety but nothing about working load behaviour.
 
Pongo,
Thanks for the reply.

can you be little elaborate?

I assumed that mutliplying the yield and ultimate strength with 0.577 and 0.75 respectively are matters of saftey.

I also believed that considering yield strength is for safty purpose, as it never consider plastic region in the calculation part.

Regards
S~
 
shinoysivan said:
Pongo,
Thanks for the reply.

can you be little elaborate?

I assumed that mutliplying the yield and ultimate strength with 0.577 and 0.75 respectively are matters of saftey.

I also believed that considering yield strength is for safty purpose, as it never consider plastic region in the calculation part.

Regards
S~

See my post in this thread: https://www.physicsforums.com/showthread.php?t=521969
The .577 is not a safety factor.
The .75 on the other hand, is probably a safety factor.
 
Shear yield strength:

At the onset of yielding, the magnitude of the shear stress in pure shear is 3^0.5 times lower than the tensile stress in the case of simple tension.

When something fails in yielding, the result is plastic deformation of the part.
 
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