# Numerical - Factor of safety, given axial force & shear force

Q.  A circular bar is subjected to an axial force and shear force, the difference between two principle stresses is 120 Mpa. Based on maximum shear stress theory what is the factor of safety, if elastic limit of the bar is 300 Mpa?
- Published on 23 Sep 15

a. .5
b. 2
c. 2.5
d. 3

#### Discussion

• Dipanjan chakraborty   -Posted on 24 Dec 19
It;s a very help full.
• Vasugi   -Posted on 27 Sep 17
Thank you for explaining this.keep it up.
• Sravanthi   -Posted on 25 Nov 15
- According to Coulomb's theory (maximum shear stress theory), material subjected to complex stresses fails, if maximum shear stress induced in the material exceeds maximum shear stress at the yield point.

Given: Difference between two principle stresses (σ1 – σ2) = 120 Mpa, stress at elastic limit (σlimit) = 300 Mpa

Formula: Maximum shear stress (τ) = (σ1 – σ2) / 2

2) FOS = (σlimit) / (2 x τ)

Solution:

1) Maximum shear stress (τ) = (σ1 – σ2) / 2

= 120 / 2 = 60 Mpa

2) FOS = (σlimit) / (2 x τ)

= 300 / (2 x 60)

= 2.5

Factor of safety = 2.5

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