A cantilever bar of length of 05m length and diameter 002m c

A cantilever bar of length of 0.5m length and diameter 0.02m carries a point load of 250N at the free end. The bar is made of AISI 1040 CD steel. If a torque of 250N.m is applied on the bar, calculate the factor of safety using a relatively realistic approach.

Solution

Initially the Maximum Bending stress and the Maximum torsional shear stress on the bar is to be determined.

For a point load of 250N at the free end of the cantilever beam, the maximum bending moment occurs at the fixed end with a magnitude of 250*0.5 = 125N-m.

The bending stress = Bending Moment / Sectional modulus of the bar.

Sectional modulud of the circular bar = (Pi* diameter^3)/32.

By calculating, the sectional modulus of the bar is 7.853*10^-7 m^3. And the Maximum bending stress is 159.15MPa.

The maximum Twisting Moment occurs at the top surface of the bar. The maxmimum torsional shear stress is calculated by t = (16*T)/(Pi*diameter^3). and by calculating the maximum torsional shear stress is obtained as 159.15MPa.

By taking tha normal stress and shear stress at the critical cross-sectional , the Principal Stress are to be determined.

The Principal stress obtained are 257.51MPa & -98.36Mpa.

For the bar AISI 1040 CD steel, considering the Yield strength as 415MPa.

Therefore, The factor of safety = (Yield strength/ Max induced stress)

= (415/257.51) = 1.611.