Question

6- A 500 kg weight is supported by 2 aluminum hangers. Diameter of the wires is 5mm and they are 1000mm long each. What will be the final length of the bars after 500°C temperature increase in the room. Aa = 20x10-6/C, Ea = 70 GPa. Axial strain due to load is given by € = = 1000mal * im im &

Answer 1

Given - Weight W = 500 Kg, Diameter of aluminum wire d = 5 mm = 0.005 m , length of both hanger wire L = 1000 mm= 1 m

thermal expansion coefficient $\alpha = 20 \times 10{^{-6}}/C$ and modulus of elasticity = E_a = 70 GPa = $= 70 \times 10{^{9}} Pa$

Load is supported by both wire at equal distance from load of each wire is subjected to 250 Kg load and deflection in each wire will be same.

the total deflection in wire bars will be due to strain caused by weight 250 Kg and thermal expansion due to temperature change 500 C

Total deflection = Deflection due to weight 250 Kg + thermal expansion

= (250 X L)/(Area of wire X E_a) + ( Alpha X L X change in temperature)

= (250 X 1)/(0.0000196349 X 70 X 10⁹ ) + (20 X 10^-6 X 1 X 500 )

= 0.00018189136 + 0.01

= 0.0101819 m or = 10.1819 mm

Final length of wire (bar) = Initial + total deflection = 1.0101819 m = 1010.1819 mm