Question

A shaft is subjected to pure torsion. The state of stress at any point on the surface of the shaft, relative to the axis of the shaft, consists of

• A. pure normal stress
• B. pure shear stress
• C. equal normal and shear stresses
• D. zero stress

Hint:

While the stress state relative to the shaft axis is pure shear, if you rotate the element by $45^\circ$, the stress state becomes pure normal stress (tensile and compressive stresses of magnitude equal to $\tau$).
This explains why brittle materials fail at $45^\circ$ under torsion.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the state of stress at any point on the outer surface of a circular shaft when it is subjected to a pure twisting moment (pure torsion).

Step 2: Key Formula or Approach:
According to the torsion equation for a circular shaft:
\[ \frac{T}{J} = \frac{\tau}{r} = \frac{G \theta}{L} \]
where:
$T$ is the applied torque.
$J$ is the polar moment of inertia.
$\tau$ is the shear stress at radius $r$.

Step 3: Detailed Explanation:

• Under pure torsion, the only load applied to the shaft is a torque acting about its longitudinal axis.

• There are no axial loads, bending moments, or external pressures acting on the shaft.

• Consequently, the normal stress ($\sigma$) along the longitudinal, radial, and circumferential directions of the shaft is zero ($\sigma = 0$).

• The applied torque produces a shear stress ($\tau$) on planes parallel and perpendicular to the longitudinal axis of the shaft.

• This shear stress is maximum at the outer surface ($r = R$) and varies linearly to zero at the central axis.

• Thus, relative to the coordinate system aligned with the axis of the shaft, the stress state consists entirely of pure shear stress.

Step 4: Final Answer:
The state of stress at any point on the surface of the shaft consists of pure shear stress.