Answer:
Total contraction on the Bar = 1.22786 mm
Explanation:
Given that:
Total Length for aluminum bar
= 600 mm
Diameter for aluminum bar = 40 mm
Hole diameter = 30 mm
Hole length = 100 mm
elasticity for the aluminum is 85GN/m² = 85 × 10³ N/mm²
compressive load P = 180 KN = 180 × 10³ N
Calculate the total contraction on the bar = ???
The relation used in calculating the contraction on the bar is:
![\delta L = \dfrac{P *L }{A*E}](https://tex.z-dn.net/?f=%5Cdelta%20L%20%3D%20%5Cdfrac%7BP%20%2AL%20%7D%7BA%2AE%7D)
The relation used in calculating the total contraction on the bar can be expressed as :
Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)
i.e
Total contraction on the Bar = ![\dfrac{P *L_1 }{A_1*E} + \dfrac{P *L_2 }{A_2 *E}](https://tex.z-dn.net/?f=%5Cdfrac%7BP%20%2AL_1%20%7D%7BA_1%2AE%7D%20%2B%20%20%5Cdfrac%7BP%20%2AL_2%20%7D%7BA_2%20%2AE%7D)
Let's find the area of cross section without the hole and with the hole
Area of cross section without the hole is :
Using A = πd²/4
A = π (40)²/4
A = 1256.64 mm²
Area of cross section with the hole is :
A = π (40²-30²)/4
A = 549.78 mm²
Total contraction on the Bar = ![\dfrac{P *L_1 }{A_1*E} + \dfrac{P *L_2 }{A_2 *E}](https://tex.z-dn.net/?f=%5Cdfrac%7BP%20%2AL_1%20%7D%7BA_1%2AE%7D%20%2B%20%20%5Cdfrac%7BP%20%2AL_2%20%7D%7BA_2%20%2AE%7D)
Total contraction on the Bar = ![\dfrac{180 *10^3 \N }{85*10^3 \ N/mm^2} [\dfrac{500}{1256.64}+ \dfrac{100}{549.78}]](https://tex.z-dn.net/?f=%5Cdfrac%7B180%20%2A10%5E3%20%5CN%20%20%7D%7B85%2A10%5E3%20%5C%20N%2Fmm%5E2%7D%20%5B%5Cdfrac%7B500%7D%7B1256.64%7D%2B%20%5Cdfrac%7B100%7D%7B549.78%7D%5D)
Total contraction on the Bar = 2.117( 0.398 + 0.182)
Total contraction on the Bar = 2.117*(0.58)
Total contraction on the Bar = 1.22786 mm