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Question

According to maximum shear stress theory, the yield strength in shear $(\tau_s)$ and yield strength in tension $(\sigma_t)$ is related as :

The correct answer is
$\tau_s = \frac{\sigma_t}{2}$

Maximum Shear Stress Theory: Yield Relation

The Maximum Shear Stress Theory, also known as the Tresca criterion or Guest's rule, provides a condition for yielding under combined stresses.

It states that yielding begins when the maximum shear stress in the material reaches the value of the shear stress at the yield point in pure shear, or equivalently, half the tensile yield strength.

Derivation Basis

Consider the yielding condition under different stress states:

  • Uniaxial Tension: When a material yields under a tensile stress $(\sigma_t)$, the principal stresses are $\sigma_1 = \sigma_t$, $\sigma_2 = 0$, and $\sigma_3 = 0$. The maximum shear stress is calculated as: $ \tau_{max, \text{tension}} = \frac{\sigma_1 - \sigma_3}{2} = \frac{\sigma_t - 0}{2} = \frac{\sigma_t}{2} $ This value, $\frac{\sigma_t}{2}$, represents the maximum shear stress the material can withstand before yielding under simple tension.
  • Pure Shear: Let the yield strength in shear be denoted by $(\tau_s)$. In a state of pure shear, the principal stresses are $\sigma_1 = \tau_s$, $\sigma_2 = 0$, and $\sigma_3 = -\tau_s$. The maximum shear stress is: $ \tau_{max, \text{shear}} = \frac{\sigma_1 - \sigma_3}{2} = \frac{\tau_s - (-\tau_s)}{2} = \frac{2\tau_s}{2} = \tau_s $

According to the Maximum Shear Stress Theory, yielding occurs when the maximum shear stress in the component equals the maximum shear stress at yield determined from the uniaxial tension test. Therefore, we equate the two maximum shear stress values:

$ \tau_{max, \text{shear}} = \tau_{max, \text{tension}} $ $ \tau_s = \frac{\sigma_t}{2} $

This equation relates the yield strength in shear $(\tau_s)$ to the yield strength in tension $(\sigma_t)$ according to this theory.

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