A steel rod of cross-sectional area `1m^(2)` is acted upon by forces as shown in the Fig. Determine the total elongation of the bar. Take `Y = 2.0xx10^(11) N//m^(2)`

Explanation

The action of forces on each part of rod is shown in figure

We know that the extension due to external force `F` is given by
`e = (F l)/(AY) therefore e_(AB) = ((60xx10^(3))xx1.5)/(1xx2xx10^(11)) = 4.5xx10^(-7)m`
`e_(BC) = ((70xx10^(3))xx1)/(1xx2xx10^(11)) = 3.5xx10^(-7)m` and `e_(CD) = ((50xx10^(3))xx2)/(1xx2xx10^(11)) = 5.0xx10^(-7)m`
The total extension `e = e_(AB) + e_(BC) + e_(CD)`
`= 4.5xx10^(-7) + 3.5xx10^(-7) + 5.0xx10^(-7)`
`= 13xx10^(-7)m = 1.3mu m`