Question

The difference between two principal stresses is \(120 \text{ MPa}\), when a circular shaft is subjected to an axial force and shear force. What is the factor of safety based on maximum shear stress theory if elastic limit of the bar is \(300 \text{ MPa}\)?

• A. 1.03
• B. 2.5
• C. 3.25
• D. 5.0

Hint:

Under Tresca's maximum shear stress theory, for any two-dimensional stress case where the principal stresses have opposite signs or are standard planar outputs, the design relation simplifies directly to: \[ \text{F.O.S.} = \frac{\sigma_{\text{yield}}}{\sigma_1 - \sigma_2} \] This avoids calculating intermediate shear values.

Answer 1

The Correct Option is B

Concept: The Maximum Shear Stress Theory (also known as Tresca's Criteria) states that a ductile component will undergo plastic yield failure when the maximum absolute shear stress developed within the material under a multi-axial state of stress reaches a critical threshold value. This threshold equals the maximum shear stress experienced by a tensile test specimen of the same material at its elastic yielding limit. Mathematically, if the two principal stresses are \(\sigma_1\) and \(\sigma_2\) (with \(\sigma_1 \gt \sigma_2\)), the maximum structural shear stress (\(\tau_{\max}\)) in that plane is given by: \[ \tau_{\max} = \frac{\sigma_1 - \sigma_2}{2} \] According to Tresca's yielding condition, the allowable design shear stress (\(\tau_{\text{allowable}}\)) incorporates the safety factor: \[ \tau_{\text{allowable}} = \frac{\tau_y}{\text{F.O.S.}} = \frac{\sigma_y}{2 \cdot \text{F.O.S.}} \] Where:

• \(\sigma_y\) = Yield stress limit / elastic limit of the material under simple tension.

• \(\tau_y = \frac{\sigma_y}{2}\) = Yield stress threshold in pure shear.

• F.O.S. = Factor of safety.
Setting the maximum occurring shear stress equal to the design allowable limit gives: \[ \tau_{\max} = \frac{\sigma_1 - \sigma_2}{2} = \frac{\sigma_y}{2 \cdot \text{F.O.S.}} \] Canceling the common factor of 2 from both sides simplifies the expression to: \[ \sigma_1 - \sigma_2 = \frac{\sigma_y}{\text{F.O.S.}} \]

Step 1: Identifying values from the problem description.

• The difference between the principal stresses is: \(\sigma_1 - \sigma_2 = 120 \text{ MPa}\)

• The tensile elastic limit of the bar material is: \(\sigma_y = 300 \text{ MPa}\)

Step 2: Rearranging the simplified formula to solve for the factor of safety.
\[ \text{F.O.S.} = \frac{\sigma_y}{\sigma_1 - \sigma_2} \]

Step 3: Calculating the numerical value.
Substitute the parameters into the equation: \[ \text{F.O.S.} = \frac{300 \text{ MPa}}{120 \text{ MPa}} \] Simplifying the fraction by dividing numerator and denominator by 10, then by 6: \[ \text{F.O.S.} = \frac{30}{12} = \frac{5}{2} = 2.5 \] The calculated factor of safety is exactly 2.5, which matches Option (2).