Circular Beam Shear Stress Ratio

When analyzing beams, understanding shear stress distribution is crucial, especially for different cross-sectional shapes. For beams with a circular cross-section, the distribution of shear stress is not uniform; it varies from zero at the extreme fibers to a maximum value at the neutral axis. The question asks for the ratio of the maximum shear stress to the average shear stress in such beams.

Understanding Shear Stress in Beams

Shear stress arises in a beam due to the applied transverse shear force. It acts parallel to the cross-section of the beam. The formula for shear stress (also known as the flexural shear formula) at any point in a beam's cross-section is given by:

$$\tau = \frac{V Q}{I b}$$

Where:

• \(\tau\) is the shear stress at the point.
• \(V\) is the shear force acting on the cross-section.
• \(Q\) (also written as \(A\bar{y}\)) is the first moment of area of the cross-section above or below the point where shear stress is being calculated, with respect to the neutral axis.
• \(I\) is the moment of inertia of the entire cross-section about the neutral axis.
• \(b\) is the width of the cross-section at the point where shear stress is being calculated.

Calculating Average Shear Stress for a Circular Cross-Section

The average shear stress in any beam cross-section is simply defined as the total shear force divided by the cross-sectional area.

For a circular cross-section, let \(R\) be the radius and \(D\) be the diameter.

Area of circular cross-section, \(A = \pi R^2 = \frac{\pi D^2}{4}\).

Therefore, the average shear stress \(\tau_{avg}\) is:

$$\tau_{avg} = \frac{V}{A} = \frac{V}{\frac{\pi D^2}{4}} = \frac{4V}{\pi D^2}$$

Determining Maximum Shear Stress for a Circular Cross-Section

For a circular cross-section, the maximum shear stress occurs at the neutral axis. To calculate this, we need the following properties:

• Moment of inertia \(I\) for a circular section about its neutral axis: $$I = \frac{\pi R^4}{4} = \frac{\pi D^4}{64}$$
• Width \(b\) at the neutral axis: For a circular section, the width at the neutral axis is the diameter, so \(b = D = 2R\).
• First moment of area \(Q\) (\(A\bar{y}\)) of the area above (or below) the neutral axis: This is the moment of the semicircle's area about the neutral axis. The area of a semicircle is \(\frac{1}{2}\pi R^2\), and its centroid is located at a distance of \(\frac{4R}{3\pi}\) from the diameter. $$Q = A\bar{y} = \left(\frac{1}{2}\pi R^2\right) \times \left(\frac{4R}{3\pi}\right) = \frac{2R^3}{3}$$ Substituting \(R = D/2\): $$Q = \frac{2(D/2)^3}{3} = \frac{2D^3/8}{3} = \frac{D^3}{12}$$

Now, substitute these values into the shear stress formula to find the maximum shear stress \(\tau_{max}\):

$$\tau_{max} = \frac{V Q}{I b} = \frac{V \left(\frac{D^3}{12}\right)}{\left(\frac{\pi D^4}{64}\right) (D)}$$

$$\tau_{max} = \frac{V D^3}{12} \times \frac{64}{\pi D^4 \cdot D} = \frac{V D^3}{12} \times \frac{64}{\pi D^5} = \frac{64V}{12\pi D^2} = \frac{16V}{3\pi D^2}$$

Ratio of Maximum to Average Shear Stress

Finally, we calculate the ratio of the maximum shear stress to the average shear stress:

$$\frac{\tau_{max}}{\tau_{avg}} = \frac{\frac{16V}{3\pi D^2}}{\frac{4V}{\pi D^2}}$$

$$\frac{\tau_{max}}{\tau_{avg}} = \frac{16V}{3\pi D^2} \times \frac{\pi D^2}{4V}$$

Cancel out common terms (\(V\) and \(\pi D^2\)):

$$\frac{\tau_{max}}{\tau_{avg}} = \frac{16}{3 \times 4} = \frac{16}{12} = \frac{4}{3}$$

Thus, for a beam with a circular cross-section, the ratio of the maximum shear stress to the average shear stress is 4:3.

This ratio is a standard result in the mechanics of materials for circular sections.