If a rod of length  , very small area of cross-section A, Young’s modulus of elasticity Y is acted upon by two parallel forces 3F and F respectively (as shown) and placed on a smooth horizontal plane. If the elastic limit is not crossed and then to study the change in length of rod $\left(\Delta l\right)$ and it’s elastic potential energy (U) the rod is segmented into four equal parts where magnitude of change in lengths are $\Delta l_1,\Delta l_2,\Delta l_3,\Delta l_4$ and elastic potential energy stored in each segment are $U_1,U_2,U_3,U_4$ respectively as shown then which is/are correct?

In steady state, the F.B.D. of the rod can be drawn by dividing it into four  equal parts as shown. We have  

$Y\frac{\Delta l}{l}={\left(stress\right)}_{average}$

$\Rightarrow \Delta l=\frac{l}{Y}.\frac{F_{avg}}{A}=\frac{l}{Y}.\frac{\left(F_1+F_2\right)}{2A}$  

For any rod . If  
 

Now,  $\Delta l_{1\left(compression\right)}=\frac{F_{avg}\left(length1\right)}{AY}=\frac{\left({\displaystyle \frac{F+0}{2}}\right)\frac{l}{4}}{AY}=\frac{Fl}{8AY}$

$;\Delta l_{2\left(elongation\right)}=\frac{Fl}{8AY}$

 $acceleration of the system=a=\frac{3F+F}{4m}=\frac{F}{m}$

From FBD of all blocks we get $N_1,N_2 and N_3 are the normal reactions betweeen 1^{st} and 2^{nd }blocks, 2^{nd } and 3^{rd }blocks, and 3^{rd } and 4^{th }blocks$

$on solving we get N_1=0, N_2=F and N_3=2F$

$\begin{array}{l}{\mathrm{\Delta l}}_{3\left(\text{ elongation }\right)}=\frac{\left({\displaystyle \frac{2F+F}{2}}\right){\displaystyle \frac{l}{4}}}{AY}=\frac{3Fl}{8AY},\end{array}$

${\mathrm{\Delta l}}_{4\left(\text{ elongation }\right)}=\frac{\left({\displaystyle \frac{2F+3F}{2}}\right){\displaystyle \frac{l}{4}}}{AY}=\frac{5Fl}{8AY}$

$\begin{array}{l}\mathrm{\Delta l}=\frac{8FI}{8AY}=\frac{Fl}{AY}\\ \end{array}$

$\mathrm{\Delta }U\propto {\text{Strain }}^2\Rightarrow \mathrm{\Delta }U\propto (\mathrm{\Delta l}{)}^2$