---
title: "Two bars absorbing the same energy"
subject: "Theory of Structures I"
chapter: 2
key: "semester_4/theory_of_structures_i:two-plastic-bars-same-energy"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/two-plastic-bars-same-energy
site: BCT Engineering
---

# Two bars absorbing the same energy

Theory of Structures I · Chapter 2: Analysis by the Strain Energy Method

- Asked in: 2070 Magh, 10 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/two-plastic-bars-same-energy
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-2

## Question

Two plastic bars as shown in the figure are to absorb the same amount of energy delivered by the axial forces. Neglecting stress concentrations, compare the stresses in the two bars. [Figure: bar 1 of uniform cross-section area A and length L, hanging from a support with stress $\sigma_1$; bar 2 of length L fixed at the top, upper half (length 0.5L) of cross-section area 2A and lower half (length 0.5L) of area A.]

## Short answer

Equating strain energies gives maximum stress in bar 2 (lower half) as σ2 = (2/√3) σ1 ≈ 1.155σ1, 15.5% higher than in bar 1; upper half stress is 0.577σ1. The uniform bar 1 is stressed less, making it the better energy absorber.

## Answer

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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For bar 1 (uniform area A, length L, stress sigma1) the strain energy is:
   A. sigma1^2 A L / E
   B. sigma1^2 A L / (2E)
   C. sigma1 A L / (2E)
   D. sigma1^2 L / (2 A E)
   Answer: B. U = (sigma^2 / 2E) x volume = sigma1^2 A L / (2E). (Key point: Strain energy of uniform bar)

2. In bar 2 the same force acts in both halves. If the lower half (area A) has stress sigma2, the upper half (area 2A) has stress:
   A. 2 sigma2
   B. sigma2
   C. sigma2 / 2
   D. sigma2 / 4
   Answer: C. Same force over double the area gives half the stress, sigma2/2. (Key point: Stress in the larger-area half)

3. What is the total strain energy of bar 2?
   A. sigma2^2 A L / (2E)
   B. 3 sigma2^2 A L / (8E)
   C. sigma2^2 A L / (4E)
   D. 3 sigma2^2 A L / (4E)
   Answer: B. Lower half gives sigma2^2 AL/(4E), upper half gives sigma2^2 AL/(8E); the sum is 3 sigma2^2 AL/(8E). (Key point: Strain energy of stepped bar)

4. Equating the energies of the two bars, the maximum stress in bar 2 compared with bar 1 is:
   A. 0.866 sigma1
   B. 1.155 sigma1
   C. 1.5 sigma1
   D. Equal to sigma1
   Answer: B. sigma1^2/2 = 3 sigma2^2/8 gives sigma2 = 2/sqrt(3) sigma1 = 1.155 sigma1, so the stepped bar is stressed more and is a poorer energy absorber. (Key point: Resulting stress comparison)

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