Question

An alloy steel shaft of 40 mm diameter is subjected to a fluctuating bending moment from 1800 Nm to 3000 Nm. What will be the factor of safety for a steel shaft, if goodman criteria are used? Take the corrected endurance limit of the component as 270 MPa and the ultimate tensile strength of the material of the shaft as 650 MPa.

• A. 2.42
• B. 1.06
• C. 1.92
• D. 2.82

Answer 1

The correct option is B 1.06

$(M_2{)}_{mean}=\frac{(M_b{)}_{max}+(M_b{)}_{min}}{2}=\frac{3000+1800}{2}=2400N-m$

$(M_b{)}_{variable}=\frac{(M_b{)}_{max}-(M_b{)}_{min}}{2}=\frac{3000-1800}{2}=600N-m$

$({\sigma }_b{)}_{\infty }=\frac{32\times (M_b{)}_m}{\pi d^3}=\frac{32\times 2400\times {10}^3}{\pi \times (40{)}^3}=381.971N/m{m}^2$

$({\sigma }_b{)}_v=\frac{32\times (M_b{)}_n}{\pi d^3}=\frac{32\times 600\times {10}^5}{\pi \times (40{)}^3}=95.493N/m{m}^2$

Using Good man's crieteria,

$\frac{{\sigma }_m}{S_{ut}}+\frac{{\sigma }_v}{S_c}=\frac{1}{FOS}$

$\frac{381.971}{650}+\frac{95.492}{270}=\frac{1}{FOS}$

FOS = 1.062