A compound bar consists of, a steel length of bar 1.72 metres and a diameter of 42mm, to which is brazed a copper tube of length 1.61 metres and has an outer diameter of 42mm and a inner diameter of 25mm, if the member carries a tensile load of 26kN, determine the stress in each material and the overall change of length.For steel E200GNm-2For copper E110GNm-2
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E stress/strain (F/A) / (dl/L) for steel stress 26 / (0.021)(0.021)(pi) 18.8 kn/m2 200000 18.8 / (dl/1.72) change in length dl 1.72 (18.8)/200000 0.00016 meters
Firstly I am assuming that the brazed copper tube is attached at the end of the steel bar and the force acts axially. So stress is equal to F/A. Once you know that this question is simple. S1 Stress in steel bar 26 000 / (pi * (.021^2)) 18 766 569.3 Pa 18.8 MPa S2 Stress in copper tube 26 000 / (pi * ((0.021^2) - (0.0125^2))) 29 064 291.6 Pa 29.06 MPa Then using Hooke's Law, find the strain. Hooke's Law: S eE; e is strain, E is elastic modulus. e1 18 766 569.3 / 200 000 000 000 9.38328465 × 10-5 e2 29 064 291.6 / 110 000 000 000 0.000264220833 Then find the change in length using e delta l / l original l1 9.38328465 × 10-5 * 1.72 0.000161392496 m l2 0.000264220833 * 1.61 0.000425395541 m lT l1 + l2 0.000586788037 m 0.586788037 mm Cheers.
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