Understanding Stress in Axially Loaded Bars

When a rectangular bar is subjected to an axial load, both normal stress and shear stress are developed within the material. The magnitude of these stresses depends on the orientation of the plane we are considering within the bar.

Normal Stress and Shear Stress

• Normal Stress ($\sigma$): This stress acts perpendicular to the surface of a plane. In an axially loaded bar, the normal stress is highest on planes perpendicular to the axis of loading.
• Shear Stress ($\tau$): This stress acts parallel to the surface of a plane. In an axially loaded bar, shear stress exists on planes that are inclined to the axis of loading.

Maximum Normal Stress

Consider a rectangular bar with cross-sectional area $A$ subjected to an axial tensile load $P$. The normal stress on the cross-section (a plane perpendicular to the axis of loading) is given by:

\(\sigma = \frac{P}{A}\)

This stress acts perpendicular to the cross-sectional area. For an axially loaded bar, the maximum normal stress ($\sigma_m$) occurs on this cross-sectional plane, where the entire load $P$ acts perpendicularly to the area $A$. So, we can say:

\(\sigma_m = \frac{P}{A}\)

Maximum Shear Stress

Now, let's consider a plane inclined at an angle \(\theta\) with respect to the cross-sectional plane (the plane perpendicular to the axis). The area of this inclined plane is \(A / \cos\theta\). The axial force $P$ can be resolved into components normal and parallel to this inclined plane.

The component of $P$ normal to the inclined plane is \(P\cos\theta\).

The component of $P$ tangential (parallel) to the inclined plane is \(P\sin\theta\).

The normal stress on the inclined plane is:

\(\sigma_\theta = \frac{P\cos\theta}{A/\cos\theta} = \frac{P}{A}\cos^2\theta\)

The shear stress on the inclined plane is:

\(\tau_\theta = \frac{P\sin\theta}{A/\cos\theta} = \frac{P}{A}\sin\theta\cos\theta\)

We know that \(\sigma_m = P/A\). Substituting this into the expressions for $\sigma_\theta$ and $\tau_\theta$:

\(\sigma_\theta = \sigma_m \cos^2\theta\)

\(\tau_\theta = \sigma_m \sin\theta\cos\theta\)

To find the maximum shear stress ($\tau_m$), we need to find the angle \(\theta\) at which \(\tau_\theta\) is maximum. We can rewrite \(\tau_\theta\) using the identity \(\sin(2\theta) = 2\sin\theta\cos\theta\):

\(\tau_\theta = \sigma_m \frac{1}{2}(2\sin\theta\cos\theta) = \frac{\sigma_m}{2}\sin(2\theta)\)

The term \(\sin(2\theta)\) has a maximum value of 1, which occurs when \(2\theta = 90^\circ\), or \(\theta = 45^\circ\). This means the maximum shear stress occurs on planes oriented at 45 degrees to the axis of the bar.

The maximum shear stress ($\tau_m$) is therefore:

\(\tau_m = \frac{\sigma_m}{2} \times 1 = \frac{\sigma_m}{2}\)

Relationship Between Maximum Stresses

From the derivation, we found that the maximum normal stress ($\sigma_m$) occurs on planes perpendicular to the axis and is equal to $P/A$. The maximum shear stress ($\tau_m$) occurs on planes inclined at 45 degrees to the axis and is equal to $\sigma_m / 2$.

Thus, the relation between maximum shear stress ($\tau_m$) and maximum normal stress ($\sigma_m$) in an axially loaded rectangular bar is:

\(\tau_m = \frac{\sigma_m}{2}\)

This relationship shows that the maximum shear stress is half of the maximum normal stress in this specific loading condition.