Question:
A steel plate having yield strength of \( 550 \) MPa is subjected to a biaxial state of stress as \( \sigma_x = \sigma \) and \( \sigma_y = -2\sigma \). Using the distortion energy theory (von Mises criterion) and taking the factor of safety as \( 2 \), find the permissible value of \( \sigma \) that can be applied to the plate. Give the answer in MPa, rounded off to 1 decimal place.
A steel plate having yield strength of \( 550 \) MPa is subjected to a biaxial state of stress as \( \sigma_x = \sigma \) and \( \sigma_y = -2\sigma \). Using the distortion energy theory (von Mises criterion) and taking the factor of safety as \( 2 \), find the permissible value of \( \sigma \) that can be applied to the plate. Give the answer in MPa, rounded off to 1 decimal place.
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Write the von Mises equivalent stress for this biaxial state and set it equal to yield strength divided by the factor of safety.
Updated On: Aug 14, 2026
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Correct Answer: 103.9
Solution and Explanation
Step 1: Write the distortion energy (von Mises) condition.
For a biaxial stress state with principal stresses \( \sigma_x \) and \( \sigma_y \) and no stress in the third direction, the von Mises equivalent stress is \( \sigma_{eq} = \sqrt{\sigma_x^2 - \sigma_x \sigma_y + \sigma_y^2} \).
Here \( \sigma_x = \sigma \) and \( \sigma_y = -2\sigma \), so both principal stresses depend on the single unknown \( \sigma \).
Step 2: Substitute the given stresses.
\( \sigma_{eq} = \sqrt{\sigma^2 - \sigma(-2\sigma) + (-2\sigma)^2} = \sqrt{\sigma^2 + 2\sigma^2 + 4\sigma^2} = \sqrt{7\sigma^2} = \sigma\sqrt{7} \).
This shows the equivalent stress grows as a fixed multiple of \( \sigma \), so one equation is enough to solve for \( \sigma \).
Step 3: Apply the factor of safety.
The design rule for yielding under distortion energy theory is \( \sigma_{eq} = \dfrac{S_{yt}}{FOS} \), where \( S_{yt} = 550 \) MPa and \( FOS = 2 \).
This gives the allowable equivalent stress as \( \sigma_{eq} = 550/2 = 275 \) MPa.
Step 4: Solve for \( \sigma \).
\( \sigma\sqrt{7} = 275 \), so \( \sigma = 275/\sqrt{7} \).
Since \( \sqrt{7} = 2.6458 \), we get \( \sigma = 275/2.6458 = 103.94 \) MPa.
Final Answer:
This is the largest \( \sigma \) the plate can carry without yielding at the given factor of safety. \[ \boxed{\sigma \approx 103.9 \text{ MPa}} \]
For a biaxial stress state with principal stresses \( \sigma_x \) and \( \sigma_y \) and no stress in the third direction, the von Mises equivalent stress is \( \sigma_{eq} = \sqrt{\sigma_x^2 - \sigma_x \sigma_y + \sigma_y^2} \).
Here \( \sigma_x = \sigma \) and \( \sigma_y = -2\sigma \), so both principal stresses depend on the single unknown \( \sigma \).
Step 2: Substitute the given stresses.
\( \sigma_{eq} = \sqrt{\sigma^2 - \sigma(-2\sigma) + (-2\sigma)^2} = \sqrt{\sigma^2 + 2\sigma^2 + 4\sigma^2} = \sqrt{7\sigma^2} = \sigma\sqrt{7} \).
This shows the equivalent stress grows as a fixed multiple of \( \sigma \), so one equation is enough to solve for \( \sigma \).
Step 3: Apply the factor of safety.
The design rule for yielding under distortion energy theory is \( \sigma_{eq} = \dfrac{S_{yt}}{FOS} \), where \( S_{yt} = 550 \) MPa and \( FOS = 2 \).
This gives the allowable equivalent stress as \( \sigma_{eq} = 550/2 = 275 \) MPa.
Step 4: Solve for \( \sigma \).
\( \sigma\sqrt{7} = 275 \), so \( \sigma = 275/\sqrt{7} \).
Since \( \sqrt{7} = 2.6458 \), we get \( \sigma = 275/2.6458 = 103.94 \) MPa.
Final Answer:
This is the largest \( \sigma \) the plate can carry without yielding at the given factor of safety. \[ \boxed{\sigma \approx 103.9 \text{ MPa}} \]
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