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Stepped bar under axial loads by FEM

Short answer

Using two elements with k1=3.2×10^5 N/mm, k2=1.6×10^5 N/mm, solving yields u2=0.0938 mm, u3=0.2188 mm. Stresses: σ1 = 75 MPa, σ2 = 100 MPa (both tensile). Reaction at support: 30 kN.

A stepped steel bar is fixed at its left end. The left portion has area A1=400 mm2A_1 = 400\ \text{mm}^2 and length 250 mm; the right portion has area A2=200 mm2A_2 = 200\ \text{mm}^2 and length 250 mm. E=200E = 200 GPa. Axial loads of 10 kN and 20 kN act at the junction (node 2) and at the free end (node 3) respectively. Using two bar elements, find the nodal displacements, the stress in each element and the reaction at the support.

Data and element stiffness

E=200×103E = 200\times10^3 N/mm2^2; L=250L = 250 mm for each element.

 |==[ A1=400 ]==o==[ A2=200 ]==o--> 20 kN
 1      L=250   2     L=250    3
                --> 10 kN (axial, at node 2)
k1=A1EL=400(200×103)250=3.2×105 N/mmk2=A2EL=200(200×103)250=1.6×105 N/mm\begin{aligned} k_1 &= \frac{A_1E}{L} = \frac{400(200\times10^3)}{250} = 3.2\times10^5\ \text{N/mm} \\ k_2 &= \frac{A_2E}{L} = \frac{200(200\times10^3)}{250} = 1.6\times10^5\ \text{N/mm} \end{aligned}

Global equations with u1=0u_1 = 0

105[3.2+1.6−1.6−1.61.6]{u2u3}={10 00020 000}10^5\begin{bmatrix} 3.2 + 1.6 & -1.6 \\ -1.6 & 1.6 \end{bmatrix}\begin{Bmatrix} u_2 \\ u_3 \end{Bmatrix} = \begin{Bmatrix} 10\,000 \\ 20\,000 \end{Bmatrix}

Adding the two equations: 3.2×105 u2=30 0003.2\times10^5\,u_2 = 30\,000, so u2=0.09375u_2 = 0.09375 mm.

From the second equation: 1.6×105(u3−u2)=20 0001.6\times10^5(u_3 - u_2) = 20\,000, so u3−u2=0.125u_3 - u_2 = 0.125 and u3=0.21875u_3 = 0.21875 mm.

Stresses

σ1=E u2−u1L=200×103×0.09375250=75 MPaσ2=E u3−u2L=200×103×0.125250=100 MPa\begin{aligned} \sigma_1 &= E\,\frac{u_2 - u_1}{L} = 200\times10^3\times\frac{0.09375}{250} = 75\ \text{MPa} \\ \sigma_2 &= E\,\frac{u_3 - u_2}{L} = 200\times10^3\times\frac{0.125}{250} = 100\ \text{MPa} \end{aligned}

Reaction

R1=−k1(u2−u1)=−3.2×105(0.09375)=−30 000R_1 = -k_1(u_2 - u_1) = -3.2\times10^5(0.09375) = -30\,000 N.

Check: σ1A1=75×400=30\sigma_1A_1 = 75\times400 = 30 kN and σ2A2=100×200=20\sigma_2A_2 = 100\times200 = 20 kN, consistent with equilibrium (10+20=3010 + 20 = 30 kN).

Answer: u2=0.0938u_2 = 0.0938 mm, u3=0.2188u_3 = 0.2188 mm; σ1=75\sigma_1 = 75 MPa, σ2=100\sigma_2 = 100 MPa (both tensile); reaction = 30 kN, directed opposite to the loads.

✓ Answer checked by Claude Opus 5.5 (high effort)

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