Answer to Question #156342 in Other for ali

Question

A tie bar of length 2.5 m and diameter 10 mm carries an axial load of 12 kN. The modulus of

elasticity of the bar material is 180 GPa. Determine the induced tensile stress, the tensile

strain and the change in length that occurs.

Answer 1

$A$ is the cross sectional area of tie bar

$A=\dfrac{\pi d^2}{4}=\dfrac{\pi(10^{-2}m)^2}{4}=7.854\times10^{-5}m^2$

$\sigma$ is the induced tensile stress

$\sigma=\dfrac{F}{A}=\dfrac{12\times10^3N}{7.854\times10^{-5}m^2}=1.528\times10^9\ Pa$

$E$  is the modulus of elasticity of the material. 

$\varepsilon$ is the tensile strain. 

$\varepsilon=\dfrac{\sigma}{E}=\dfrac{1.528\times10^9N/m^2}{180\times10^9Pa}=0.008488$

The change in length that occurs

$x=\varepsilon L=0.008488\times2.5m=0.02122m=21.22mm$

$\sigma=1.528\times10^9\ Pa=1.528\ GPa$

$\varepsilon=0.008488$

$x=21.22\ mm$