polyethylene bar ( $( E = 1.4 \mathrm { GPa } , v = 0.4 )$ 0.4) of diameter 80 mm is inserted in a steel tube of inside diameter 80.2 mm and then compressed by axial force P. The gap between steel tube and polyethylene bar will close when compressive load P is approximately:

Question

Quizplus · Accepted Answer

The deformation of the polyethylene bar can be calculated using the formula:
δ = (P*L)/(A*E*(1-v^2))
Where,
P = compressive load applied
L = length of the polyethylene bar
A = cross-sectional area of the polyethylene bar
E = Young's modulus of the polyethylene bar
v = Poisson's ratio of the polyethylene bar

The length of the polyethylene bar can be assumed to be equal to the length of the steel tube, which is not provided in the question. However, since the gap between the steel tube and the polyethylene bar will close when compressed, we can assume that the length of the polyethylene bar is equal to the length of the steel tube plus the initial gap between them. Let's assume the initial gap to be 0.1 mm.

Length of polyethylene bar = Length of steel tube + Initial gap
= π*(80.2/2)*10^-3 + 0.1*10^-3
= 0.12608 m

Cross-sectional area of polyethylene bar = (π/4)*(80/2)^2*10^-6
= 4.02*10^-4 m^2

Putting these values along with the given values of Young's modulus and Poisson's ratio of polyethylene bar, we get:
δ = (P*0.12608)/(4.02*10^-4*1.4*10^9*(1-0.4^2))

We want to find the compressive load P required to close the gap, that is, the value of δ should be equal to 0.1 mm. So, we can solve the above equation for P as:
P = (0.1*10^-3)*(4.02*10^-4*1.4*10^9*(1-0.4^2))/0.12608
≈ 44 kN

Therefore, the compressive load required to close the gap is approximately 44 kN, which corresponds to choice C.

Polyethylene Bar (E=1.4GPa,v=0.4)( E = 1.4 \mathrm { GPa } , v = 0.4 )(E=1.4GPa,v=0.4)

Polyethylene Bar $( E = 1.4 \mathrm { GPa } , v = 0.4 )$