Find out maximum stress and elongation of the bar:
A rectangular bar having uniform cross-section of 4 cm × 2.5 cm and of length 2 m is hanging vertically from rigid support. lt is subjected to the axial tensile loading of 10KN. Take the density of steel as 7850 kg/m^{3}. And E = 200 GN/m^{2}. Find out maximum stress and elongation of the bar.
Sol.: Data, cross sectional area A = 4 cm. × 2.Scm.
length L = 2m
axial load
P = I0KN
Density of steel σ= 7850 Kg/m3
E = 200 GN/m2
As δL = PL/AE + L^{2}/2E
= (10 × 103 × 2)/(4 × 2.5 × 10^{-4} × 200 × 10^{9}) + (7850 × 9.81 × 22)/(2 × 200 × 10^{9})
10 kN
= 0.0001 m
= 0.1 mm stress ( σ _{max}) = E × strain
= 200 × 10^{9} × δ L/L =200 × 10^{9} × (0.0001/2) = 10.08 mpa .......ANS