A rod having cross-sectional area 100 × 10^–6 m² is subjected to a tensile load. Based on the Tresca failure criterion, if the uniaxial yield stress of the material is 200 MPa, the failure load is (a) 10 kN (b) 20 kN (c) 100 kN (d) 200 kN.

Question

A rod having cross-sectional area [latex]100 × 10^{-6} m^2[/latex] is subjected to a tensile load. Based on the Tresca failure criterion, if the uniaxial yield stress of the material is 200 MPa, the failure load is
(a) 10 kN (b) 20 kN (c) 100 kN (d) 200 kN.

Accepted Answer

(b)

EXPLANATION

[latex] A=100 imes 10^{-6} m ^2 [/latex]

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

Tensile stress due to load P, [latex] \sigma_1=\frac{P}{A} [/latex]

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

Max. shear stress due to stress system [latex] \left(\sigma_1, 0,0 ight)=\frac{1}{2}\left(\sigma_1-0 ight)=\frac{\sigma_1}{2} [/latex]

Uniaxial yield stress, [latex] \sigma_t^*=200 MPa =200 imes 10^6 N / m ^2 [/latex].

∴ For uniaxial yield stress, we have stress system as [latex] \left(\sigma_t^*, 0,0 ight) [/latex]

Max. shear stress due to uniaxial yield stress [latex] =\frac{1}{2}\left(\sigma_t^*-0 ight)=\frac{\sigma_t^*}{2} [/latex]

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

i.e., [latex] \frac{\sigma_1}{2}=\frac{\sigma_t^*}{2} \quad ext { or } \quad \sigma_1=\sigma_t^* [/latex]

or [latex] \frac{P}{A}=200 imes 10^6 N / m ^2 \quad\left(\because \sigma_t{ }^*=200 imes 10^6 N / m ^2 \quad ext { and } \quad \sigma_1=\frac{P}{A} ight) [/latex]

or [latex] P=200 imes 10^6 imes A \ =\left(200 imes 10^6 ight) imes\left(100 imes 10^{-6} ight)=20000=\pmb{20 kN.}[/latex]

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

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