Question 26.3.45: A rod having cross-sectional area 100 × 10^–6 m² is subjecte...

(b)

EXPLANATION

A=100 \times 10^{-6}  m ^2

Let P = tensile load at failure ∴ For one-dimensional stress system, we have stresses as \left(\frac{P}{A}, 0,0\right)

Tensile stress due to load P, \sigma_1=\frac{P}{A}

This stress is in one-direction only i.e., (σ, 0, 0)

Max. shear stress due to stress system \left(\sigma_1, 0,0\right)=\frac{1}{2}\left(\sigma_1-0\right)=\frac{\sigma_1}{2}

Uniaxial yield stress, \sigma_t^*=200  MPa =200 \times 10^6  N / m ^2 .

∴ For uniaxial yield stress, we have stress system as \left(\sigma_t^*, 0,0\right)

Max. shear stress due to uniaxial yield stress =\frac{1}{2}\left(\sigma_t^*-0\right)=\frac{\sigma_t^*}{2}

According to Tresca failure criterion,
Max. shear stress developed = Max. shear due to yield stress

i.e.,  \frac{\sigma_1}{2}=\frac{\sigma_t^*}{2} \quad \text { or } \quad \sigma_1=\sigma_t^*

or    \frac{P}{A}=200 \times 10^6  N / m ^2 \quad\left(\because \sigma_t{ }^*=200 \times 10^6  N / m ^2 \quad \text { and } \quad \sigma_1=\frac{P}{A}\right)

or    P=200 \times 10^6 \times A \\ =\left(200 \times 10^6\right) \times\left(100 \times 10^{-6}\right)=20000=\pmb{20  kN.}