Question
Download Solution PDFA machine element develops principal stresses of magnitudes 2P and P. What is the maximum magnitude of P before the material reaches the yield stress fy as per Distortion Shear Energy Theory?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Maximum shear strain energy / Distortion energy theory / Mises – Henky theory.
It states that inelastic action at any point in body, under any combination of stress begging, when the strain energy of distortion per unit volume absorbed at the point is equal to the strain energy of distortion absorbed per unit volume at any point in a bar stressed to the elastic limit under the state of uniaxial stress as occurs in a simple tension/compression test.
\(\frac{1}{2}\left[ {{{\left( {{{\rm{\sigma }}_1} - {{\rm{\sigma }}_2}} \right)}^2} + {{\left( {{{\rm{\sigma }}_2} - {{\rm{\sigma }}_3}} \right)}^2} + {{\left( {{{\rm{\sigma }}_3} - {{\rm{\sigma }}_1}} \right)}^2}} \right] \le {\rm{\sigma }}_{\rm{y}}^2\) for no failure
\(\frac{1}{2}\left[ {{{\left( {{{\rm{\sigma }}_1} - {{\rm{\sigma }}_2}} \right)}^2} + {{\left( {{{\rm{\sigma }}_2} - {{\rm{\sigma }}_3}} \right)}^2} + {{\left( {{{\rm{\sigma }}_3} - {{\rm{\sigma }}_1}} \right)}^2}} \right] \le {\left( {\frac{{{{\rm{\sigma }}_{\rm{y}}}}}{{{\rm{FOS}}}}} \right)^2}\) For design
It cannot be applied for material under hydrostatic pressure.
All theories will give the same results if loading is uniaxial.
Calculations:
Given, \({{\rm{\sigma }}_1} = 2{\rm{P}};{\rm{\;}}{{\rm{\sigma }}_2} = {\rm{P}};{\rm{\;}}{{\rm{\sigma }}_3} = 0{\rm{\;}}\)
\(\therefore \frac{1}{2}\left[ {{{\left( {2{\rm{P}} - {\rm{P}}} \right)}^2} + {{\left( {{\rm{P}} - 0} \right)}^2} + {{\left( {0 - 2{\rm{P}}} \right)}^2}} \right] \le {\rm{f}}_{\rm{y}}^2 \)
\(\frac{1}{2}\left[ {{{\rm{P}}^2} + {{\rm{P}}^2} + 4{{\rm{P}}^2}} \right] \le {\rm{f}}_{\rm{y}}^2 \)
\(3{{\rm{P}}^2} \le {\rm{f}}_{\rm{y}}^2\)
\({\rm{P}} \le \frac{{{{\rm{f}}_{\rm{y}}}}}{{\surd 3}}\)
∴ The maximum magnitude of P before the material reaches the yield stress is \(\frac{{{{\rm{f}}_{\rm{y}}}}}{{\surd 3}}\).
Last updated on Jul 2, 2025
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