An polyethylene bar (E = 1.4 GPa, v = 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:
A) 18 kN
B) 25 kN
C) 44 kN bar Steeltube
D) 60 kN Polyethylene d1 d2

Question

Quizplus · Accepted Answer

To determine when the gap between the steel tube and the polyethylene bar will close, we need to calculate the compressive load required to cause enough deformation in the polyethylene bar so that it fills the gap. The gap is 0.2 mm (80.2 mm - 80 mm). The deformation needed can be calculated using the formula for radial deformation under axial load, considering the material's Young's modulus (E) and Poisson's ratio (v).Given:- Diameter of polyethylene bar = 80 mm- Gap = 0.2 mm- E (Young's modulus for polyethylene) = 1.4 GPa = 1400 MPa- v (Poisson's ratio for polyethylene) = 0.4The radial deformation ($\Delta r$) can be related to the axial stress ($\sigma$) by the relation $\Delta r = \frac{\sigma \cdot r}{E} \cdot v$, where $r$ is the initial radius of the bar. To find the stress ($\sigma$), we rearrange the formula to $\sigma = \frac{\Delta r \cdot E}{r \cdot v}$.Substituting the given values and converting units appropriately:- $\Delta r = 0.1$ mm (since the gap is 0.2 mm, but the deformation is radial, so it's half the gap)- $r = 40$ mm (radius of the bar)- $E = 1400$ MPa- $v = 0.4$$\sigma = \frac{0.1 \cdot 1400}{40 \cdot 0.4} = 8.75$ MPaThe force ($P$) required to achieve this stress is found by multiplying the stress by the cross-sectional area of the bar ($A = \pi r^2$):$P = \sigma \cdot A = 8.75 \cdot \pi \cdot (40)^2 \cdot 10^{-6}$ kN$P \approx 44$ kNTherefore, a compressive load of approximately 44 kN is required to close the gap between the steel tube and the polyethylene bar.

An Polyethylene Bar (E = 1