Understanding Stress in a Solid Circular Shaft

This problem asks us to find the maximum principal stress in a solid circular shaft that is subjected to two types of loads simultaneously: an axial load causing direct axial stress and a torque causing shear stress.

To find the maximum principal stress, we first need to determine the stress state at a critical point on the shaft. For a circular shaft under axial load and torque, the critical point for maximum stress is usually on the outer surface, where the shear stress due to torque is maximum.

Calculating Stresses

We are given the axial stress directly:

• Axial stress, \(\sigma_x = 50\) MPa

The torque \(T\) causes shear stress \(\tau\) in the shaft. The maximum shear stress occurs at the outer radius \(r\). The formula for shear stress due to torque in a solid circular shaft is given by:

$$ \tau = \frac{Tr}{J} $$

Where:

• \(T\) is the applied torque.
• \(r\) is the radius of the shaft.
• \(J\) is the polar moment of inertia of the shaft's cross-section.

Given data:

• Diameter \(D = 100\) mm \( = 0.1\) m
• Radius \(r = D/2 = 100/2 = 50\) mm \( = 0.05\) m
• Torque \(T = 10\) kNm \( = 10 \times 10^3\) Nm

The polar moment of inertia \(J\) for a solid circular shaft is:

$$ J = \frac{\pi D^4}{32} $$

Let's calculate \(J\):

$$ J = \frac{\pi (0.1)^4}{32} = \frac{\pi \times 10^{-4}}{32} \text{ m}^4 $$

Now, let's calculate the maximum shear stress \(\tau\) at the outer surface:

$$ \tau = \frac{(10 \times 10^3 \text{ Nm}) \times (0.05 \text{ m})}{\frac{\pi (0.1)^4}{32} \text{ m}^4} $$

$$ \tau = \frac{500 \text{ Nm}^2}{\frac{\pi \times 10^{-4}}{32} \text{ m}^4} = \frac{500 \times 32}{\pi \times 10^{-4}} \text{ N/m}^2 $$

$$ \tau = \frac{16000}{\pi \times 10^{-4}} \text{ Pa} = \frac{16 \times 10^6}{\pi} \text{ Pa} $$

Converting to MPa:

$$ \tau = \frac{16}{\pi} \text{ MPa} \approx \frac{160}{3.14159} \text{ MPa} \approx 50.93 \text{ MPa} $$

So, the maximum shear stress \(\tau_{xy} = 50.93\) MPa.

Determining the Stress State

At a point on the surface of the shaft, subjected to axial load and torque, the stress state can be represented in 2D:

• Normal stress in the axial direction (\(\sigma_x\)) = 50 MPa
• Normal stress perpendicular to the axial direction (\(\sigma_y\)) = 0 MPa
• Shear stress (\(\tau_{xy}\)) = 50.93 MPa

Calculating Principal Stresses

The principal stresses (\(\sigma_1\) and \(\sigma_2\)) are the maximum and minimum normal stresses acting on planes where the shear stress is zero. They can be calculated using the formula for a 2D stress state:

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

Substitute the values \(\sigma_x = 50\) MPa, \(\sigma_y = 0\) MPa, and \(\tau_{xy} = 50.93\) MPa:

$$ \sigma_{1,2} = \frac{50 + 0}{2} \pm \sqrt{\left(\frac{50 - 0}{2}\right)^2 + (50.93)^2} $$

$$ \sigma_{1,2} = 25 \pm \sqrt{(25)^2 + (50.93)^2} $$

$$ \sigma_{1,2} = 25 \pm \sqrt{625 + 2593.8} $$

$$ \sigma_{1,2} = 25 \pm \sqrt{3218.8} $$

Calculating the square root:

$$ \sqrt{3218.8} \approx 56.73 $$

Now, find the two principal stresses:

$$ \sigma_1 = 25 + 56.73 = 81.73 \text{ MPa} $$

$$ \sigma_2 = 25 - 56.73 = -31.73 \text{ MPa} $$

The maximum principal stress is the larger of the two values, which is \(\sigma_1\).

Maximum principal stress \(\approx 81.73\) MPa.

Comparing with Options

We compare the calculated maximum principal stress (81.73 MPa) with the given options:

• 41 MPa
• 82 MPa
• 164 MPa
• 204 MPa

The value 81.73 MPa is closest to 82 MPa.

Therefore, the maximum principal stress experienced on the shaft is closest to 82 MPa.