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 hollow segment of length L/3 had outside and inside diameters of 60 mm and 35 mm, respectively. The solid segment of length 2L/3 has diameter of 60 mm. A strain gauge is placed on the outside of the hollow segment of the bar to measure normal strains in the longitudinal direction. a.) if the measured strain in the hollow segment is Eh= 470 * 10 ^-6, what is the strain Es 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 the hollow to that of the solid segment.) b.) what is the overall shortening of S of the bar? c.) If the compressive stress in the bar cannot exceed 48 MPa, what is the maximum permissible value of load P?

Explanation / Answer

a) ratio of areas is 0.66 Es = Eh x ratio= 310.07 X 10^-6 b) overall shortening is Eh X L/3 + Es x 2L/3 = 65800 + 86819.6 = 152619.6*10^-9 c) 48*10^6 = P*(1/A1 + 1/A2) where A1 = 3.14/4*(60*10^-3)^2 A2 = 3.14/4*[(60*10^-3)^2 - (35*10^-3)^2] P = 48*10^6/(353.86 + 536.37) = 53.92 kN

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