Thermal stress in a stepped bar

Thermal load in FEM

If an element is free to expand, there is no stress. If the expansion is prevented, the initial strain $\varepsilon_0 = \alpha\Delta T$ produces an equivalent nodal force vector

$$\{f_T\} = \int_V [B]^T[D]\{\varepsilon_0\}\,dV$$

For a bar element this gives $\{f_T\} = EA\alpha\Delta T\begin{Bmatrix} -1 \\ +1 \end{Bmatrix}$. Stress is found from the mechanical strain only: $\sigma = E(\varepsilon - \alpha\Delta T)$.

Element data

 wall|==[A1=500]==o==[A2=250]==|wall
      L1=300      2    L2=200
      1                       3

$$k_1 = \frac{500(200\times10^3)}{300} = 3.333\times10^5\ \text{N/mm},\qquad k_2 = \frac{250(200\times10^3)}{200} = 2.5\times10^5\ \text{N/mm}$$ $$E A_1\alpha\Delta T = 200\times10^3(500)(12\times10^{-6})(60) = 72\,000\ \text{N},\qquad EA_2\alpha\Delta T = 36\,000\ \text{N}$$

Equations

$u_1 = u_3 = 0$; only $u_2$ is unknown. The thermal force at node 2 is $+72\,000$ from element 1 and $-36\,000$ from element 2:

$$(k_1 + k_2)\,u_2 = 72\,000 - 36\,000 = 36\,000$$ $$u_2 = \frac{36\,000}{5.833\times10^5} = 0.0617\ \text{mm (to the right)}$$

Element forces and stresses

$$\begin{aligned} N_1 &= k_1(u_2 - 0) - EA_1\alpha\Delta T = 20\,571 - 72\,000 = -51\,429\ \text{N} \\ N_2 &= k_2(0 - u_2) - EA_2\alpha\Delta T = -15\,429 - 36\,000 = -51\,429\ \text{N} \end{aligned}$$

The two forces are equal, as equilibrium of the junction requires.

$$\sigma_1 = \frac{-51\,429}{500} = -102.9\ \text{MPa},\qquad \sigma_2 = \frac{-51\,429}{250} = -205.7\ \text{MPa}$$

Check

Free expansion would be $\alpha\Delta T(L_1 + L_2) = 0.36$ mm. It is cancelled by the compression of the two bars: $P\left(\dfrac{300}{500E} + \dfrac{200}{250E}\right) = \dfrac{P(1.4)}{200\times10^3} = 0.36$ gives $P = 51\,429$ N, which agrees.

Answer: $u_2 = 0.0617$ mm; $\sigma_1 = -102.9$ MPa and $\sigma_2 = -205.7$ MPa (compressive); wall reaction $= 51.43$ kN.