Finding Maximum Shear Stress in a Circular Section

The question asks for the relationship between the maximum shear stress at the neutral axis of a circular section and its average shear stress. The maximum shear stress in a circular beam subjected to a shear force occurs at the neutral axis.

Understanding Average Shear Stress

The average shear stress (\(\tau_{avg}\)) over the cross-section of any beam is simply the total shear force (\(V\)) divided by the cross-sectional area (\(A\)). For a circular section with radius \(R\), the area is \(A = \pi R^2\). Thus, the average shear stress is:

\[\tau_{avg} = \frac{V}{A} = \frac{V}{\pi R^2}\]

Calculating Shear Stress Distribution

The shear stress (\(\tau\)) at any point in a beam cross-section is given by the formula:

\[\tau = \frac{VQ}{It}\]

Where:

• \(V\) is the shear force acting on the section.
• \(Q\) is the first moment of area of the section above or below the point where shear stress is calculated, taken about the neutral axis.
• \(I\) is the moment of inertia of the entire cross-section about the neutral axis.
• \(t\) is the thickness (width) of the section at the point where shear stress is calculated.

Finding Maximum Shear Stress at the Neutral Axis

For a circular section with radius \(R\) (and diameter \(D = 2R\)):

• The moment of inertia about the neutral axis is \(I = \frac{\pi R^4}{4}\).
• At the neutral axis, the thickness \(t\) is the full diameter, so \(t = 2R\).
• To find \(Q\) at the neutral axis, we consider the area of the semi-circle above (or below) the neutral axis. The area of this semi-circle is \(A_{semi} = \frac{1}{2}\pi R^2\). The centroid of a semi-circle is located at a distance of \(\frac{4R}{3\pi}\) from the diameter (neutral axis).
• So, \(Q\) at the neutral axis is the product of the semi-circle area and the distance of its centroid from the neutral axis:

\[Q = A_{semi} \times \bar{y} = \left(\frac{1}{2}\pi R^2\right) \times \left(\frac{4R}{3\pi}\right) = \frac{4\pi R^3}{6\pi} = \frac{2R^3}{3}\]

Now, substitute \(V\), \(Q\), \(I\), and \(t\) into the shear stress formula at the neutral axis (where shear stress is maximum for this shape):

\[\tau_{max} = \frac{V Q}{I t} = \frac{V \left(\frac{2R^3}{3}\right)}{\left(\frac{\pi R^4}{4}\right) (2R)}\]

Simplify the expression:

\[\tau_{max} = \frac{\frac{2VR^3}{3}}{\frac{2\pi R^5}{4}} = \frac{2VR^3}{3} \times \frac{4}{2\pi R^5} = \frac{8VR^3}{6\pi R^5} = \frac{4V}{3\pi R^2}\]

Relating Maximum Shear Stress to Average Shear Stress

We found that \(\tau_{max} = \frac{4V}{3\pi R^2}\) and we know that \(\tau_{avg} = \frac{V}{\pi R^2}\). We can rewrite the expression for \(\tau_{max}\) by factoring out \(\frac{V}{\pi R^2}\):

\[\tau_{max} = \frac{4}{3} \left(\frac{V}{\pi R^2}\right)\]

Since \(\frac{V}{\pi R^2} = \tau_{avg}\), we get:

\[\tau_{max} = \frac{4}{3} \tau_{avg}\]

Therefore, the maximum shear stress at the neutral axis for a circular section is 4/3 times the average shear stress.