Transcribed Text
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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