Two vertical rods of 20 mm diameter, one of steel and the other of copper, are rigidly fixed at the top 500 mm apart. A cross bar fixed to the rods at the lower ends carries a load of 5 KN such that the cross bar remains horizontal even after loading. Find the stress in each rod and the position of load on the bars, if the bars are 4 m long, Young's modulus of steel 200 GPa and of copper 100 GPa.I've found the stresses but don't know how to find the positionplease help!
Let: T1 tension on the steel rod T2 Tension on the copper rod d1 deformation of the steel rod d2 deformation of the copper rod A area of each rod pi(10)^2 100pi mm^2 Since the bottom bar remains horizontal, the deformation of both bars are equal. Hence; d1 T1/(A E1) T1/(200A) d2 T2/(A E2) T2/(100A) But d1 d2, Hence; T1/(200A) T2/(100A) T1 2T2 T1 + T2 500,000 2T2 + T2 500.000 T2 166,667 N S2 166667/(100pi) 530.52 N/mm^2 S1 2x166667/(100pi) 1061.035N/mm^2 Let x distance of the load from the steel bar 500,000x 166667(500) x 166.67 mm
It sounds like the idea is to position the load on the horizontal bar so that the amount of stress on each bar amounts to an identical change in the length of the copper and steel bars. Considering the difference in Young's modulus I would off-hand say the load should be located 1/3 the bars length from the steel side. To show this use stress/strain equation to identify the change in length of each bar with stress applied to each an independent variable. Set change in length to be equal for both bars. This gives 3 eq's w/ 3 unknowns. You should be able to find a ratio of stress with this. Use this ratio w/ your mechanics to identify the position of the 5 kN load on the horizontal bar.
