Problem Statement
Two rods, one of brass ($L_1$, $\alpha_1$) and one of steel ($L_2$, $\alpha_2$), are placed end to end between rigid walls with no initial stress. The composite system is heated by $\Delta T$. Find the stress in the rods if $L_1=20\,\text{cm}$, $L_2=30\,\text{cm}$, $\alpha_1=1.9\times10^{-5}$ /°C, $\alpha_2=1.2\times10^{-5}$ /°C, $Y_1=10^{11}\,\text{N/m}^2$, $Y_2=2\times10^{11}\,\text{N/m}^2$, $\Delta T=100\,\text{°C}$. (Same cross-sectional area.)
Given Information
- $L_1=0.20\,\text{m}$ (brass), $\alpha_1=1.9\times10^{-5}$ /°C, $Y_1=10^{11}\,\text{N/m}^2$
- $L_2=0.30\,\text{m}$ (steel), $\alpha_2=1.2\times10^{-5}$ /°C, $Y_2=2\times10^{11}\,\text{N/m}^2$
- $\Delta T=100\,\text{°C}$
- Same cross-sectional area $A$
Physical Concepts & Formulas
Total free thermal expansion = constraint. The rods are in series (same force $F$). The sum of mechanical compressions equals the total thermal expansion.
- Total free expansion $= (L_1\alpha_1+L_2\alpha_2)\Delta T$
- Compression in each rod: $\Delta L_i=FL_i/(Y_iA)$
- Constraint: $\sum\Delta L_{mech}=(L_1\alpha_1+L_2\alpha_2)\Delta T$
Step-by-Step Solution
Step 1 — Total free thermal expansion: Sum of individual expansions.
$\Delta L_{free}=(0.20\times1.9\times10^{-5}+0.30\times1.2\times10^{-5})\times100=7.4\times10^{-4}\,\text{m}$
Step 2 — Mechanical constraint equation: Total compression = total thermal expansion.
$\frac{FL_1}{Y_1 A}+\frac{FL_2}{Y_2 A}=7.4\times10^{-4}$
Step 3 — Solve for stress $\sigma=F/A$: Rearrange.
$\sigma\left(\frac{L_1}{Y_1}+\frac{L_2}{Y_2}\right)=7.4\times10^{-4}\Rightarrow\sigma=\frac{7.4\times10^{-4}}{0.20/10^{11}+0.30/(2\times10^{11})}=\frac{7.4\times10^{-4}}{2\times10^{-12}+1.5\times10^{-12}}=\frac{7.4\times10^{-4}}{3.5\times10^{-12}}\approx2.11\times10^8\,\text{Pa}$
Worked Calculation
$$\sigma=7.4\times10^{-4}/(L_1/Y_1+L_2/Y_2)\approx2.11\times10^8\,\text{Pa}=211\,\text{MPa}$$
Answer
$$\boxed{\sigma\approx211\,\text{MPa}}$$
Both rods are under approximately 211 MPa compressive stress.
Physical Interpretation
The stress is the same in both rods (same force, same area). Brass contributes more to the thermal expansion problem ($\alpha_1 > \alpha_2$) and is less stiff ($Y_1 < Y_2$), so it bears more mechanical compression than steel.
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