Question

An aluminum bar 600mm long with diameter 40mm, has a hole drilled in the center of the bsr.The hole is 30mm in diameter and is 100mm long. If the modulus of elasticity for the aluminum is 85GN/m2, calculate the total contraction on the bar due to a compressive load of 180KN

Answer 1

Answer:

1.228 x $10^{-6}$ 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 $10^{3}$ N

modulus of elasticity E = 85 GN/m^2  = 85 x $10^{9}$ 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 = $\pi r^{2}$ = 3.142 x $20^{2}$ = 1256.8 mm^2

area of hole a = $\pi (R^{2} - r^{2})$ = $3.142* (20^{2} - 15^{2})$ = 549.85 mm^2

Total contraction of the bar = $\frac{F*L}{AE} + \frac{Fl}{aE}$

total contraction = $\frac{F}{E} * (\frac{L}{A} +\frac{l}{a})$

==> $\frac{180*10^{3}}{85*10^{9}} *( \frac{500}{1256.8} + \frac{100}{549.85})$ = 1.228 x $10^{-6}$ mm