Two bars absorbing the same energy

Assume both bars are linearly elastic, have the same modulus $E$, carry the same axial force along their length, and absorb the same strain energy $U$.

Bar 1 (uniform area $A$, length $L$)

Stress $\sigma_1 = P_1/A$ throughout.

$$U_1 = \frac{\sigma_1^2}{2E}\times AL = \frac{\sigma_1^2 AL}{2E}$$

Bar 2 (upper half $2A$, lower half $A$, each $0.5L$)

The same force $P_2$ acts in both halves. Maximum stress occurs in the smaller area (lower half):

$$\sigma_2 = \frac{P_2}{A}\quad(\text{lower half}),\qquad \frac{\sigma_2}{2} = \frac{P_2}{2A}\quad(\text{upper half})$$ $$\begin{aligned} U_2 &= \frac{\sigma_2^2}{2E}\,(A)(0.5L) + \frac{(\sigma_2/2)^2}{2E}\,(2A)(0.5L)\\ &= \frac{\sigma_2^2 AL}{4E} + \frac{\sigma_2^2 AL}{8E} = \frac{3\sigma_2^2AL}{8E} \end{aligned}$$

Equal energy

$$U_1 = U_2 \;\Rightarrow\; \frac{\sigma_1^2}{2} = \frac{3\sigma_2^2}{8} \;\Rightarrow\; \sigma_2^2 = \frac{4}{3}\sigma_1^2$$ $$\sigma_2 = \frac{2}{\sqrt3}\,\sigma_1 = 1.155\,\sigma_1$$

Answer: The maximum stress in bar 2 (in its lower, smaller-area half) is $1.155\,\sigma_1$, i.e. 15.5% higher than in bar 1; the upper half of bar 2 carries only $0.577\,\sigma_1$. For the same energy absorbed, the bar of uniform section (bar 1) is stressed less, so it is the better energy absorber. Reducing the area over part of the length increases stress and wastes material in the larger portion.