Calculation Based on Yield strength and Ultimate strength

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SUMMARY

The discussion focuses on the calculation of shear capacity in mechanical design using two methods: shear yield strength and ultimate shear strength. Shear yield strength is calculated as 1/√3 times the yield strength, while ultimate shear strength is 0.75 times the ultimate strength. The significance of these methods lies in their different implications for safety and working load behavior. The shear yield strength indicates the limit of elastic behavior, whereas ultimate shear strength provides insights into safety but does not address working load performance.

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  • Understanding of shear strength calculations
  • Familiarity with yield strength and ultimate strength concepts
  • Knowledge of mechanical design principles
  • Basic grasp of plastic deformation in materials
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Mechanical engineers, design engineers, and students in mechanical engineering who are involved in structural analysis and material strength 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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