Question

a steel rod of 20mm diameter passes centrally through a coper tube of 50mm external dia and and 40mm internal dia .The tube is enclosed at each end by rigid plates of negligible distance.The nuts are tighten lightly on the projecting parts of the load.if the temp. of assembly raised by 50°C . calculate the stresses developed in the copper and steel.E for steel and copper is 200GN/m2 and 100GN/m2.alpha steel=12x10^-6/°C and alpha copper= 18x
10^-6/°C​

Answer 1

Answer:

Here is the answer

Explanation:

Area of steel rod = Asr = π/4 (0.022) = 0.0003142 m2

Area of steel tube = Ast = π/4 (0.042 – 0.0252) = 0.000766 m2  

The load is equally applied on both tube and rod.

i.e.; σst1.Ast = σsr1.Asr = 2000  

Since compression in steel tube and tension in steel rod  

0003142 σsr1 = 0.000766 σst1 = 2000

σsr1 = 63.6 MN/m2 (T)  

σst1 = 26.1 MN/m2 (C)

Distance traveled by nut =1/4(1/0.4) = 0.625mm = 0.000625m = δLTotal  

δLTotal = δLRod + δLtub

= (σsr2 /E).Lsr + (σst2 /E).Lst ; Lst = Lsr  

0.000625 = L/E(σsr2 + σst2 ) ...(i)  

Again load are equal  

σst2.Ast = σsr2 .Asr  

σsr2 = σst2.Ast/Asr  

= σst2 .0.000766/0.0003142 = 2.44σst2 ......(ii)  

Now put in equation (i)

we get 0.000625 = (0.75/200 × 109)(2.44 σst2 + σst2)  

σst2 = 48.48MN/m2  

σsr2 = 118.19MN/m2