Problem 81 - An Introduction to the Mechanics of Solids in SI Units

Question

The cross section of a solid circular shaft of radius \(r\) is acted on by a bending moment \(M_{b}\) and a twisting moment \(M_{t}\). Show that the maximum shear stress in the shaft is given by $$ \tau_{\max }=\frac{r}{2 I_{y v}} \sqrt{M_{b}^{2}+M_{t}^{2}} $$ Show also that this shear stress acts on planes whose normals make an angle $$ \psi=\frac{1}{2} \tan ^{-1} \frac{M_{b}}{M_{t}} $$ with the axial and tangential directions. Finally, verify that these results are valid for a hollow circular shaft (not necessarily thin-walled).

Step-by-Step Solution

Answer

The maximum shear stress \(\tau_{\max }\) in a solid or hollow circular shaft under bending and twisting moments is given by \(\tau_{\max }=\frac{r}{2 I_{yv}} \sqrt{M_{b}^{2}+M_{t}^{2}}\). It acts at an angle \(\psi = \frac{1}{2} \tan^{-1} \frac{M_{b}}{M_{t}}\) with the axial and tangential directions.

Step 1: Understanding Shear Stress

The shear stress caused by the bending and twisting moments can be expressed by two equations: \(\tau_{b} = \frac{M_{b}r}{I_{yv}}\) and \(\tau_{t} = \frac{M_{t}r}{J}\), where \(I_{yv}\) is the moment of inertia of the cross section about the transverse axis and \(J\) is the polar moment of inertia. Since the values of \(I_{yv}\) and \(J\) for a solid circular shaft are \(\frac{\pi}{4} r^4\) and \(\frac{\pi}{2} r^4\) respectively, these equations become \(\tau_{b} = \frac{4M_{b}}{\pi r^3}\) and \(\tau_{t} = \frac{2M_{t}}{\pi r^3}\).

Step 2: Finding Maximum Shear Stress

The total shear stress at any point is the vector sum of the bending and twisting shear stresses: \(\tau = \sqrt{\tau_{b}^2+\tau_{t}^2}\). Substituting the values of \(\tau_{b}\) and \(\tau_{t}\) from step 1, the equation becomes \(\tau = \sqrt{(\frac{4M_{b}}{\pi r^3})^2 + (\frac{2M_{t}}{\pi r^3})^2}\) = \(\frac{r}{2 I_{yv}} \sqrt{M_{b}^{2}+M_{t}^{2}}\). This is the required expression for the maximum shear stress.

Step 3: Determining Direction of Shear Stress

The direction of the total shear stress \(\tau\) is given by the angle \(\psi\) which satisfies the equation \(\tan \psi = \frac{\tau_{b}}{\tau_{t}}\). Substituting the values of \(\tau_{b}\) and \(\tau_{t}\) from step 1, the equation becomes \(\tan \psi = \frac{2M_{b}}{M_{t}}\). It simplifies to \(\psi = \frac{1}{2} \tan^{-1} \frac{M_{b}}{M_{t}}\), which is the required expression for the angle.

Step 4: Extending to a Hollow Shaft

For a hollow circular shaft, where the inner radius is \(a\) and outer radius is \(b\), the values of \(I_{yv}\) and \(J\) are \(\frac{\pi}{4} (b^4 - a^4)\) and \(\frac{\pi}{2} (b^4 - a^4)\) respectively. Following the same steps 1-3 with these values, we would again obtain \(\tau_{\max }=\frac{r}{2 I_{yv}} \sqrt{M_{b}^{2}+M_{t}^{2}}\) and \(\psi = \frac{1}{2} \tan ^{-1} \frac{M_{b}}{M_{t}}\), thus verifying that the results apply to hollow circular shafts as well.

Key Concepts

Bending Moment

When a shaft experiences a bending moment, it indicates that a force is causing the shaft to bend. Imagine trying to bend a plastic ruler; this bending force is what is scientifically referred to as the bending moment. It quantifies the tendency of a force to cause rotation about a point or axis. For circular shafts, this moment creates tensile and compressive stresses across the material. The formula to calculate the bending stress in the shaft due to a bending moment \(M_b\) is given by the equation \(\tau_b = \frac{M_b r}{I_{yv}}\), where \(r\) is the radius of the shaft and \(I_{yv}\) is the moment of inertia about the axis. This relationship highlights how a larger shaft (greater radius) or a design with more resistance to bending (larger moment of inertia) will affect the amount of bending stress generated.

Twisting Moment

A twisting moment, also known as a torque, occurs when a force causes an object to rotate around an axis. Picture using a wrench to turn a bolt; the force applied at a distance generates torque. In mechanical shafts, twisting moments lead to shear stresses as different parts of the shaft want to twist at varying rates.
These are calculated using the formula \(\tau_t = \frac{M_t r}{J}\), where \(M_t\) is the twisting moment, \(r\) is the radius of the shaft, and \(J\) is the polar moment of inertia. The polar moment of inertia is a measure of an object's ability to resist twisting. The greater this value, the more resistant the shaft is to twisting, which directly influences the resulting shear stress.

Moment of Inertia

The moment of inertia is akin to the measure of how difficult it is to change an object's state of rotation. If you've ever tried to spin a heavy wheel, you've felt the effects of a high moment of inertia. For shafts, the moment of inertia \(I_{yv}\) about a transverse axis determines resistance to bending, while the polar moment of inertia \(J\) affects resistance to twisting. Both play a crucial role in determining the shaft's ability to withstand loads without failing. A solid circular shaft has a moment of inertia \(I_{yv} = \frac{\pi}{4} r^4\) and a polar moment of inertia \(J = \frac{\pi}{2} r^4\). These values are calculated from the geometric properties and dimensions of the shaft. The distribution of material in a cross-section dramatically influences these moments and, consequently, the shaft's reaction to bending and twisting forces.