A rod AD consisting of three segments AB,BC and CD joined together is hanging vertically from a fixed support at A. The lengths of the segments are restectively 0.2 m , 0.3m and 0.15 m . The cross-section of the rod is uniformly `10^(-4) m^(2)` . A weight of 10 kg is hung from D.Calculate the displacement of points B,C and D if `Y_(AB)=3.5xx10^(10)Nm^(-2) , Y_(BC)=5xx 10^(10), Y_(CD)=2xx10^(10)Nm^(-2)`.
(Neglect the weight of the rod )

Explanation

Given, area , `A=10^(-14)m^(2)`
`Y_(AB)=3.5xx10^(10)Nm^(-2)`
`Y_(BC)=5xx10^(10)Nm^(-2)`
`Y_(CD)=2xx10^(10)Nm^(-2)`

Increase in length,
`DeltaL=(FL)/(AY)=(MgL)/(AY)=(10xx10L)/(10^(-4)Y)=10^(6)(L)/(Y)`
Now, increase in length of AB segment
`(DeltaL_(AB))=10^(6)xx(L_(AB))/(Y_(AB))=10^(6)xx(2)/(3.5xx10^(10))=5.7xx10^(-6)m`
Increase in length of BC segment
`(DeltaL)_(BC)=10^(6)xx(L_(BC))/(Y_(BC))`
`=10^(6)xx(0.3)/(5xx10^(10))=6xx10^(-6)m`
Increase in length of CD segment
`(DeltaL)_(CD)=10^(6)(L_(CD))/(Y_(CD))=10^(6)xx(0.15)xx(2xx10^(10))=7.5xx10^(-6)m`
So, Displacement of `B=L_(AB)=5.7xx10^(-6)m`
Displacement of `C=DeltaL_(AB)+DeltaL_(BC)=11.7xx10^(-6)m`
Displacement of `D=DeltaL_(AB)+DeltaL_(BC)+DeltaL_(CD)=26.7xx10^(-6)m`