Calculating Maximum Shear Stress in a Shaft under Combined Loads

When a shaft is subjected to both a bending load (represented by bending moment, M) and a torsional load (represented by torsional moment, T), stresses are induced in the shaft material. Bending load causes normal stress (bending stress), while torsional load causes shear stress (torsional shear stress). At any point on the outer surface of the shaft, these stresses act simultaneously.

Let's consider a point on the outer surface of the shaft. Due to the bending moment M, the maximum bending stress occurs at the outer fibers and is given by:

\(\sigma_b = \frac{M}{Z}\)

where Z is the section modulus. For a solid circular shaft of diameter d, the section modulus is \(Z = \frac{I}{y_{max}}\). The moment of inertia \(I = \frac{\pi d^4}{64}\) and \(y_{max} = \frac{d}{2}\).

So, the maximum bending stress is:

\(\sigma_b = \frac{M}{\frac{\pi d^4/64}{d/2}} = \frac{M}{\frac{\pi d^3}{32}} = \frac{32M}{\pi d^3}\)

This stress is a normal stress (tensile or compressive depending on the location relative to the neutral axis and the direction of bending).

Due to the torsional moment T, the maximum shear stress occurs at the outer surface and is given by:

\(\tau = \frac{T}{Z_p}\)

where \(Z_p\) is the polar section modulus. For a solid circular shaft of diameter d, the polar moment of inertia \(J = \frac{\pi d^4}{32}\) and \(r_{max} = \frac{d}{2}\). \(Z_p = \frac{J}{r_{max}}\).

So, the maximum torsional shear stress is:

\(\tau = \frac{T}{\frac{\pi d^4/32}{d/2}} = \frac{T}{\frac{\pi d^3}{16}} = \frac{16T}{\pi d^3}\)

This stress is a shear stress acting tangentially on the outer surface.

At a point on the outer surface, we have a state of plane stress with a normal stress (\(\sigma_b\)) and a shear stress (\(\tau\)). To find the maximum shear stress under this combined loading, we use the formula derived from Mohr's Circle or principal stress equations. The maximum shear stress (\(\tau_{max}\)) is given by:

\(\tau_{max} = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}\)

In this case, we can consider the state of stress at a point on the outer surface as having a normal stress \(\sigma_x = \sigma_b\), \(\sigma_y = 0\), and a shear stress \(\tau_{xy} = \tau\).

Substituting the values:

\(\tau_{max} = \sqrt{\left(\frac{\sigma_b - 0}{2}\right)^2 + \tau^2}\)

\(\tau_{max} = \sqrt{\left(\frac{\sigma_b}{2}\right)^2 + \tau^2}\)

Now, substitute the expressions for \(\sigma_b\) and \(\tau\):

\(\tau_{max} = \sqrt{\left(\frac{1}{2} \times \frac{32M}{\pi d^3}\right)^2 + \left(\frac{16T}{\pi d^3}\right)^2}\)

\(\tau_{max} = \sqrt{\left(\frac{16M}{\pi d^3}\right)^2 + \left(\frac{16T}{\pi d^3}\right)^2}\)

\(\tau_{max} = \sqrt{\frac{16^2 M^2}{(\pi d^3)^2} + \frac{16^2 T^2}{(\pi d^3)^2}}\)

\(\tau_{max} = \sqrt{\frac{16^2 (M^2 + T^2)}{(\pi d^3)^2}}\)

\(\tau_{max} = \frac{16\sqrt{M^2 + T^2}}{\pi d^3}\)

This formula represents the maximum shear stress on the outer surface of a solid circular shaft subjected to combined bending moment M and torsional moment T.

Reviewing the options, the expression \(\frac{16\sqrt{M^2~+~T^2}}}{{\pi d^3}}\) matches the derived maximum shear stress formula.