Answer: Factor of Safety for Bolt Shear & Tensile Stress

[maali5]

Homework Statement

The material for the bolt shown in the angled joint above has an ultimate tensile strength of 500 MPa and a shear strength of 300 MPa. The diameter of the bolt is 8 mm. Determine the factor of safety in operation.

Homework Equations

The Attempt at a Solution

Shear stress= 5,5 . cos 60 (pi . 8 . 2/4)

= 0.0547kn

= 54.70 mPA



Shear Fs= 300/ 54.7 = 5.48



b) Tensile Stress = 5,5 . sin 60 (pi . 8. 2.4) = 94. 7 Mpa


Tensile Fs = 500/94.7 5.27



Is this correct?

 

[pongo38]

You have considered tension and shear independently, but actually they are acting together. To solve, you need to know the shear/tension strength characteristic. The graph of shear stress/ ult shear stress versus tensile stress/ ult tensile stress is sometimes assumed linear, but that is a 'safe' assumption. The experimental results were used to determine at least the relationship for the eurocode and is approximately parabolic or elliptical (I can't remember which without looking it up in ec3)