Question

According to the maximum normal stress theory, the diameter of circular shaft subjected to bending moment M and torque T is (where σy is the yield stress in the uniaxial tensile test and N is the factor of safety)

• A. $$[\frac{N}{\pi \sigma_y}(16M+16\sqrt{M^2+T^2})]^{\frac 12}$$
• B. $$[\frac{1}{\pi N \sigma_y}(16M+16\sqrt{M^2+T^2})]^{\frac 12}$$
• C. $$[\frac{1}{\pi N \sigma_y}(16M+16\sqrt{M^2+T^2})]^{\frac 13}$$
• D. $$[\frac{N}{\pi \sigma_y}(16M+16\sqrt{M^2+T^2})]^{\frac 13}$$

Answer 1

Answer:

$$[\frac{N}{\pi \sigma_y}(16M+16\sqrt{M^2+T^2})]^{\frac 13}$$

The maximum normal stress theory states that for no failure, the maximum principal stress should be less than the yield stress under uniaxial loading. The diameter of a circular shaft subjected to bending moment M and torque T is given by $d = [\frac{N}{\pi \sigma_y}(16M+16\sqrt{M^2+T^2})]^{\frac 13}$ , where N is the factor 1 of safety, ay is the yield stress in uniaxial tensile test. This is derived using the formula for combined effect of bending and torsion, and the expression for maximum principal stress.