A circular aluminum tube with a length of L = 420 mm is loaded in compression by

Question

A circular aluminum tube with a length of L = 420 mm is loaded in compression by forces P (see figure). The hallow segment of length L/3 had outside and inside diametrs 60 mm and 35 mm. respectively. The solid segment of length 2L/3 has a diameter of 60 mm. A strain gage is placed on the outside of the hollow segment of the bar to measure normal strains in the longitudinal direction. If the measured strain in the hollow segment is epsilon h = 470 times 10-6. what is the strain epsilon s in the solid part ? (Hint: The strain in the solid segment is equal to that in the hollow segment multiplied by the ratio of the area of hollow to that of the solid segment.) What is the overall shortening delta of the bat? If the compressive stress in the bar cannot exceed 48 MPa. what is the maximum permissible value of load P ?

Explanation / Answer

Area of hollow segment Ah = /4*(60^2 - 35^2) = 1864.375 mm^2

Area of solid segment As = /4*60^2 = 2826 mm^2

Ratio of area = 1864.375/2826 = 0.66

a) Strain in solid part s = h*Area ratio = (470*10^-6)*0.66 = 310.07*10^-6

b) Overall shortening of bar = Shortening of hollow+solid portion = hLh + sLs

Overall shortening of bar = (470*10^-6)*(420/3) + (310.07*10^-6)*(2*420/3) = 0.1526 mm

c) For hollow portion, stress = P/Ah

48*10^6 = P/(1864.375*10^-6)

This gives P = 89490 N

For solid portion, stress = P/As

48*10^6 = P/(2826*10^-6)

This gives P = 135648 N

Hence, max.a llowed P = 89490 N.

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