Answer:
1.228 x [tex]10^{-6}[/tex] mm
Explanation:
diameter of aluminium bar D = 40 mm
radius of aluminium bar R = D/2 = 40/2 = 20 mm
diameter of hole d = 30 mm
radius of the hole r = d/2 = 30/2 = 15 mm
compressive Load F = 180 kN = 180 x [tex]10^{3}[/tex] N
modulus of elasticity E = 85 GN/m^2 = 85 x [tex]10^{9}[/tex] Pa
length of bar L = 600 mm
length of hole = 100 mm
true length of bar = 600 - 100 = 500 mm
area of the bar A = [tex]\pi r^{2}[/tex] = 3.142 x [tex]20^{2}[/tex] = 1256.8 mm^2
area of hole a = [tex]\pi (R^{2} - r^{2})[/tex] = [tex]3.142* (20^{2} - 15^{2})[/tex] = 549.85 mm^2
Total contraction of the bar = [tex]\frac{F*L}{AE} + \frac{Fl}{aE}[/tex]
total contraction = [tex]\frac{F}{E} * (\frac{L}{A} +\frac{l}{a})[/tex]
==> [tex]\frac{180*10^{3}}{85*10^{9}} *( \frac{500}{1256.8} + \frac{100}{549.85})[/tex] = 1.228 x [tex]10^{-6}[/tex] mm
