An aluminum bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 40mm in diameter and 100mm long. If modulus of elasticity is for the aluminum is 85GN/m^2, calculate the total contraction on the bar due to compressive load of 180kN?

1.228 x mm

Explanation:

diameter of aluminium bar D = 40 mm  

diameter of hole d = 30 mm

compressive Load F = 180 kN = 180 x N

modulus of elasticity E = 85 GN/m^2  = 85 x 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 = =   = 1256.8 mm^2

area of hole a = = = 549.85 mm^2

Total contraction of the bar =

total contraction =

==> = 1.228 x mm

since you're a real machine, your ama is always less than your ima, because you can never totally eliminate friction.

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