Understanding Stress in a Shaft Subjected to Torsion

When a shaft is subjected to torsion, the material on the surface experiences a state of pure shear stress. This shear stress, denoted by $\tau$, is maximum on the surface and acts on planes perpendicular to the shaft's axis and on planes containing the shaft's axis.

Pure Shear Stress State

A pure shear state exists when the only non-zero stress components relative to a set of axes are shear stresses. For a shaft in torsion, if we consider an element on the surface, the stress state can be represented as pure shear $\tau$ relative to axes aligned with the shaft's axis and perpendicular to it (on the surface). Visualizing this, imagine the shear stress acting on faces perpendicular to the shaft length and on faces parallel to the shaft length.

Finding Principal Stresses from Pure Shear

Principal stresses are the maximum and minimum normal stresses that exist at a point, and they occur on planes where the shear stress is zero. For a state of pure shear $\tau$, the principal planes are always oriented at 45° to the planes on which the shear stress acts. The magnitudes of the principal stresses are equal to the magnitude of the shear stress $\tau$, but with opposite signs.

Let's consider the stress transformation equations or Mohr's Circle for a pure shear state where $\sigma_x = 0$, $\sigma_y = 0$, and $\tau_{xy} = \tau$.

The principal stresses ($\sigma_{1,2}$) are given by the formula:

$$\sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}$$

Substituting $\sigma_x = 0$, $\sigma_y = 0$, and $\tau_{xy} = \tau$ for pure shear:

$$\sigma_{1,2} = \frac{0 + 0}{2} \pm \sqrt{\left(\frac{0 - 0}{2}\right)^2 + \tau^2}$$

$$\sigma_{1,2} = 0 \pm \sqrt{0 + \tau^2}$$

$$\sigma_{1,2} = \pm \tau$$

So, the two principal stresses are $\sigma_1 = +\tau$ and $\sigma_2 = -\tau$.

The maximum principal stress is the positive value, which is $\sigma_1 = +\tau$. These principal stresses occur on planes oriented at 45° and 135° to the original shear planes.

Relating to the Given Expression

The question asks for the value of the maximum principal stress on the surface, which is at 45° to the axis. As established, the principal stress planes are indeed at 45° to the original shear planes (which are aligned with or perpendicular to the shaft axis). The maximum principal stress has a value of $\tau$.

Let's evaluate the provided expression: $2\tau \sin 45\degree \cos 45\degree$.

We know that $\sin 45\degree = \frac{1}{\sqrt{2}}$ and $\cos 45\degree = \frac{1}{\sqrt{2}}$.

Substituting these values into the expression:

$$2\tau \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{\sqrt{2}}\right)$$

$$2\tau \left(\frac{1}{\sqrt{2} \cdot \sqrt{2}}\right)$$

$$2\tau \left(\frac{1}{2}\right)$$

$$\tau$$

The expression $2\tau \sin 45\degree \cos 45\degree$ simplifies to $\tau$, which is the magnitude of the maximum principal stress in a state of pure shear $\tau$. The plane where this maximum principal stress acts is at 45° to the original shear planes, which aligns with the description "at 45° to the axis" for a shaft in torsion.

Therefore, the maximum principal stress on the surface at 45° to the axis has a value equal to $\tau$.