Question

A brittle material of 4 sm cross section carries an axial tensile load of 20 tonnes. what will be the maximum shear stress in the block?

Answer 1

shear stress=force/area
shear stress=2,00,000/4
50,000N/sq.m

Answer 2

Given:

A brittle material of 4 m² cross section carries an axial tensile load of 20 tonnes.

To Find:

What will be the maximum shear stress in the block?

Solution:

Area of cross-section (A) = 4 m²

Axial tensile force (F) = 20 ×1000 kg-f = 20000×10 N = 200000 N

Maximum value of tensile normal stress is :

$\mathbf{Normal\ Stress(\sigma _{max})=\dfrac{Force}{Area}}$

$\mathbf{Normal\ Stress(\sigma _{max})=\dfrac{200000}{4}}$

$\mathbf{Normal\ Stress(\sigma _{max})=50000 \ \dfrac{N}{m^{2}}}$

We know relation between maximum shear stress and normal stress is:

$\mathbf{Maximum\ Shear\ Stress(\tau _{max})= \dfrac{\sigma _{max}}{2}}$

$\mathbf{Maximum\ Shear\ Stress(\tau _{max})= \dfrac{50000}{2} =25000 \ \dfrac{N}{m^{2}}= 25000 \ Pa}$

25000 Pa  will be the maximum shear stress in the block.