A shaft is subjected to a maximum torque of 12 kN-m and a maximum bending moment 16 kN-m at a particular section. What will be the diameter of the shaft according to maximum shear stress theory (Guest's & Tresca's theory)? If the elastic limit in simple tension is 160 MPa.

\(\left[\left(\frac{4}{\pi}\right)^{1 / 3}=1.08\right]\)

Concept:

According to the Maximum Shear Stress Theory (Guest’s or Tresca’s theory), the maximum shear stress in a shaft subjected to combined bending and torsion is given by:

\( \tau_{max} = \frac{1}{2} \sqrt{\sigma_b^2 + 4\tau^2} \)

To ensure safety, this is set equal to half the yield strength in simple tension:

\( \tau_{max} = \frac{\sigma_y}{2} \)

Calculation:

Given:

Bending moment, \(M = 16~kN\cdot m = 16 \times 10^3~N\cdot m\)

Torque, T = \(12~kN\cdot m = 12 \times 10^3~N\cdot m\)

Elastic limit in tension, \(\sigma_y = 160~MPa = 160 \times 10^6~Pa\)

The combined equivalent moment is calculated using:

\( M_e = \sqrt{M^2 + T^2} = \sqrt{(16)^2 + (12)^2} \times 10^3 = \sqrt{256 + 144} \times 10^3 = 20 \times 10^3~N\cdot m \)

Using the formula:

\( \frac{32 M_e}{\pi d^3} = \sigma_y \Rightarrow d^3 = \frac{32 M_e}{\pi \sigma_y} \)

\( d^3 = \frac{32 \cdot 20 \times 10^3}{\pi \cdot 160 \times 10^6} = \frac{640 \times 10^3}{502.65 \times 10^6} = 1.273 \times 10^{-3}~m^3 \)

\( d = (1.273 \times 10^{-3})^{1/3} = 0.108~m = 108~mm \)