From Eq. (10.17), we can say

U_i=\int_0^{L_i} \frac{P^2}{2( AE )_i} d x ; \quad 1 \leq i \leq 3            (10.17)

U = Total strain – Energy of the stepped rod

=U_1+U_2=\frac{P^2(L / 2)}{2 A E}+\frac{P^2(L / 2)}{2(A / 2) E}

as the same force is acting on both rods. So,

U=\frac{P^2 L}{4 A E}+\frac{P^2 L}{2 A E}=\frac{3}{4} \frac{P^2 L}{A E}=\frac{3}{2} \frac{P^2 L}{2 A E}

\text { Thus, } U=1.5 U_0 \text { where } U_0=P^2 L / 2 A E is the strain energy of a prismatic rod of length L and cross-sectional area, A. It is observed that strain energy stored within the rod is 1.5 times that of an equivalent prismatic rod.

U=\frac{3}{4} \frac{P^2 L}{A E}