Bolt / Pin Design

[Nabil Baig]

Hi there,

I calculated single shear stress in a pin as:

Diameter of Pin = 100mm
Material: IS:2062
Load / Force = 16000N
so
Shear Stress = Force/Area
=16000N/7854mm2
=2.04 N/mm2
Shear Stress =2.04 MPa

Now,
Shear Strength(Yield) of IS:2062 = Tensile Strength(Yield) x 0.58
= 410 MPa x 0.58
= 238 MPa

Now my question is, which value is to be compared with which value to ensure that the Pin will withstand the load?
Like what I see 2.04MPa is way less than 238MPa.

Please help clearing my concept.

Best regards,
Nabil Baig.

 

[desertfox]

Hi Nabil Baig

The calculated value should be compared with the 238Mpa I can’t see any other values in your post to compare the calculated value with.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[BrianE22]

Your shear stress of 2.04 MPa is well below the shear yield stress of the material (238MPa). I don't see a problem.

 

[Nabil Baig]

Hi, BriamE22,

Does that mean 238/2.04 = 116.66. Is the pin 116.66 times stronger than the applied load?

 

[MJCronin]

Its a really big pin, Nabil ....


MJCronin
Sr. Process Engineer

 

[dvd]

Is it a pivot pin, or simply a pinned connection? There may be a need to observe a much lower bearing stress due to pressure.

 

[Nabil Baig]

@dvd
Yes, it is supposed to be a pivot pin.
Like U-hook lifts the load fixing in it. It is noy always under liad.

 

[MintJulep]

For any real system with 3 dimensions the assumptions that come along with single-shear are usually not valid.

Real loads are usually eccentric - consider bending.

Real parts that pivot always have some clearance - consider contact stress at the limits of movement.

 

[dvd]

A description of the application along with a sketch would produce better suggestions.