1. An aluminium rod is to withstand an applied force of 200 kN. To ...

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1. An aluminium rod is to withstand an applied force of 200 kN. To ensure a sufficient factor of safety, the maximum allowable stress on the rod is limited to 170 MPa (170MN/m2 ) with a strain of 0.0025mm.mm-1. The rod must be at least 3.8 m long but must deform elastically no more than 6 mm when the force is applied. Design an appropriate rod. 2. From stress-strain diagram below for Aluminium alloy, calculate the E, YS (σY) and UTS (σU), and percent elongation at failure. 3. A 10-mm diameter rod of carbon steel is subjected to a tensile load of 50,000 N, taking beyond its yield point (YS = 600 MPa and UTS = 750 MPa). Calculate the elastic recovery that would occur upon removal of the tensile load. E = 200 x 103 MPa. 4. Table 1 below shows the results of a tensile test of a 12.5 mm diameter aluminium alloy test bar with original cross sectional area Ao = 122.7 mm2 ) Measured Calculated Load (kN) Gauge Length (mm) Eng. Stress (MPa) Eng. Strain (mm.mm-1) 0 50 0 0 5 50.03 40.7 0.0006 10 50.06 81.5 0.0012 15 50.09 122.2 0.0018 20 50.12 163 0.0024 25 50.15 203.7 0.0030 30 50.185 244.5 0.0037 35.3 52 287.7 0.04 35.6 (max) 53 290.1 0.06 33.8 (fracture) 55.3 275.4 0.104 a. Convert the load-gauge length data in Table 1 to engineering stress and strain and plot a stress-strain curve b. From the data in Table 1, calculate the modulus of elasticity of the aluminium alloy. Use the modulus to determine the length of a 1.25 m bar to which a stress of 210 MPa is applied. c. Compare engineering stress and strain with true stress and strain for the aluminium alloy (on a – above), and determine the (i) maximum load and (ii) fracture. The diameter at maximum load is 12.3 mm and fracture is 9.85 mm.

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