# System of bars subjected to temperature load

## Description

A system of bars, consisting of two bars in series are fixed between two walls. The bars are modeled as linearly-thermoelastic and at temperature $T_0$, the system is stress-free.

1. Determine the maximum stress in the system when the temperature is raised to $T_0 + \Delta T$.
2. Determine the movement of point B during the temperature rise.

Given data:

• Coefficient of thermal expansion $\alpha = 1.25 \cdot 10^{-5} 1/{}^{\circ}\rm{C}$
• Young’s modulus $E = 209.6\GPa$
• $L_1 = 300 \mm$
• $L_2 = 200 \mm$
• Cross-sectional area for bar 1: $A_1 = 200 \mm^2$
• Cross-sectional area for bar 2: $A_2 = 100 \mm^2$
• $\Delta T = 50{}^{\circ}\rm{C}$

1. Maximum stress in the system is: $\sigma = -187.1 \MPa$.
2. Point B moves $0.054 \mm$ to the right.

## Solution

### Our approach

The system is statically indeterminate since there are two unknown reaction forces but only one horizontal equilibrium equation. In other words, it isn’t enough to only consider equilibrium, instead, we must combine equilibrium with compatibility equations and material relations. Only then can stresses and deformations be determined.

### Equilibrium

Consider a free-body diagram around point B and establish equilibrium to determine a relationship between the normal forces in the bars:

$$\leftarrow: \quad N_1 – N_2 = 0 \Rightarrow N_1 = N_2$$

As we see, the normal force is the same in both bars.

### Compatibilty

The total deformation $n$ of the system is zero and is is calculated as the sum of deformation of each bar:

$$n = n_1 + n_2 = 0 \Rightarrow$$
$$\epsilon_1 L_1 + \epsilon_2 L_2 = 0 \gives$$
$$\epsilon_2 = – \epsilon_1 \frac{L_1}{L_2}$$

### Material relations

The material is modeled using Hooke’s law with thermal strains:
$$\sigma = E \paren{\epsilon – \epsilon_\term } = E \paren{\epsilon – \alpha \Delta T }$$

With this, the normal forces are written

$$N_1 = \sigma_1 A_1 = EA_1 \paren{ \epsilon_1 – \alpha \Delta T }$$
$$N_2 = \sigma_2 A_2= EA_2 \paren{\epsilon_2 – \alpha \Delta T }$$

Combine compatability, equilibrium and material

Equilibrium gives

$$N_1=N_2 \equivalent \sigma_1 A_1 = \sigma_2 A_2$$

Combine with the material realationship $\gives$

$$EA_1 \paren{ \epsilon_1 – \alpha \Delta T } = EA_2 \paren{\epsilon_2 – \alpha \Delta T}$$

Express (for example) $\epsilon_2$ in terms of $\epsilon_1$ and cancel out $E \gives$

$$A_1 \paren{ \epsilon_1 – \alpha \Delta T } = A_2 \paren{- \epsilon_1 \frac{L_1}{L_2} – \alpha \Delta T}$$

Solve for $\epsilon_1 \gives$

$$\epsilon_1 = \frac{ \alpha \Delta T \paren{ A_1-A_2 }}{ \paren{ A_1 + A_2\frac{L_1}{L_2} } } = \frac{1.25 \cdot 10^{-5} \cdot 50 \cdot \paren{ 200-100 }} { \paren{ 200 + 100\frac{300}{200} } } = 1.7857\cdot 10^{-4}$$

The strain in bar 2 then becomes $\epsilon_2 = -1.7857\cdot 10^{-4} \cdot\frac{300}{200} = -2.6786\cdot 10^{-4}$

### Calculating stresses

$$\sigma_1 = E(\epsilon_1 – \alpha \Delta T) = 209.6\cdot 10^9 \paren{ 1.7857\cdot 10^{-4} – 1.25 \cdot 10^{-5} \cdot 50} = -93.6 \MPa$$

$$\sigma_2 = E(\epsilon_2 – \alpha \Delta T) = 209.6\cdot 10^9 \paren{ -2.6786\cdot 10^{-4} – 1.25 \cdot 10^{-5} \cdot 50} = -187.1 \MPa$$

### Movement of point B

This can, for example, be determined from the deformation of bar 1:

$$n_1 = \epsilon_1 L_1 = 1.7857\cdot 10^{-4} \cdot 300 = 0.054 \mm$$

which corresponds to the displacement of B, since $n_1 = u_{\rm B} – u_{\rm A} = u_{\rm B} – 0 = u_{\rm B}$.

Note
From the expression of $\epsilon_1$, we can see that if $A_1=A_2$ the strain becomes zero and the problem reduces to the case of a bar fixed between two walls.