A steel bar ABCD 40 cm long is made up of three parts AB, BC and CD, as shown in figure.The rod is subjected to a pull of 25 kN. The total extension of the rod is (Young's modulus for steel`= 2 xx 10^(11) N m^(-2)`).

Explanation

Method1: The axial force `25 kN` is transmitted to each of the three bars.
Stress in part `AB` is
`(25000N)/(pi/4(50)^(2)mm^(2))=40/pi=12.73N//mm^(2)`
Stress in part `BC =25000/(pi/4(25)^(2))=50.93N//mm^(2)`
Stress in part `CD =12.73N//mm^(2)`
Therefore, total extension of the rod `=` extension in the parts `AB+BC+CA`
`=((12.73N/mm^(2))/(2xx10^(5)N//mm^(2))xx10mm)xx2+(50.93N/mm^(2))/(2xx10^(5)N//mm^(2))xx200mm`
`=12732/(2xx10^(5))mm=0.0637mm`
Method 2: The steel bar `ABCD` can be replaced with series combination of three springs.
Equivalent spring contact can be written as `1/k_(eq)=1/(k_(1))+1/(k_(2))+1/(k_(1))`
Here `k_(1)=(YA_(1))/(l_(1))` and `k_(2)=(YA_(2))/(l_(2))`

which gives `k_(eq)=(Y(A_(1)A_(2)))/((2A_(2)l_(1)+A_(1)l_(2)))`
Extension of composite rod `x=F/k_(eq)`
This gives `x=(F(2A_(2)l_(2)+A_(1)l_(2)))/(YA_(1)A_(2))=1/(5pi)mm=0.0637mm`
In series combination of springs the force in each spring should be equal. hence, each rod will experiecne same force `F`.
Hence stress is rods `AB` and `CD` is
`sigma_(AB)=sigma_(CD)=F/(A_(1))=(25xx10^(3))/((pi/4xx50^(2)))=40/piN//mm^(2)`
Stress in rod `BC` is
`(25xx10^(3))/((pi/4xx25^(2)))=160/(pi) N//mm^(2)`