--- title: "Stepped bar under axial loads by FEM" subject: "Finite Element Method" chapter: 3 key: "semester_8/finite_element_method:stepped-bar-displacement-stress" source: https://www.bctengineer.com/q/semester_8/finite_element_method/stepped-bar-displacement-stress site: BCT Engineering --- # Stepped bar under axial loads by FEM Finite Element Method · Chapter 3: Direct Stiffness Method: Discrete Finite Elements - Asked in: year not given, 8 marks - Video: [FEM #finite element method bar hindi #Nodal displacement,stress and reaction in bar in hindi](https://www.youtube.com/watch?v=ZBNqRsfU9Lw) — Candoanything - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/finite_element_method/stepped-bar-displacement-stress - Whole chapter: https://www.bctengineer.com/md/semester_8/finite_element_method/chapter-3 ## Question A stepped steel bar is fixed at its left end. The left portion has area $A_1 = 400\ \text{mm}^2$ and length 250 mm; the right portion has area $A_2 = 200\ \text{mm}^2$ and length 250 mm. $E = 200$ 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. ## 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. ## Answer #### Data and element stiffness $E = 200\times10^3$ N/mm$^2$; $L = 250$ mm for each element. ```text |==[ A1=400 ]==o==[ A2=200 ]==o--> 20 kN 1 L=250 2 L=250 3 --> 10 kN (axial, at node 2) ``` $$ \begin{aligned} k_1 &= \frac{A_1E}{L} = \frac{400(200\times10^3)}{250} = 3.2\times10^5\ \text{N/mm} \\ k_2 &= \frac{A_2E}{L} = \frac{200(200\times10^3)}{250} = 1.6\times10^5\ \text{N/mm} \end{aligned} $$ #### Global equations with $u_1 = 0$ $$ 10^5\begin{bmatrix} 3.2 + 1.6 & -1.6 \\ -1.6 & 1.6 \end{bmatrix}\begin{Bmatrix} u_2 \\ u_3 \end{Bmatrix} = \begin{Bmatrix} 10\,000 \\ 20\,000 \end{Bmatrix} $$ Adding the two equations: $3.2\times10^5\,u_2 = 30\,000$, so $u_2 = 0.09375$ mm. From the second equation: $1.6\times10^5(u_3 - u_2) = 20\,000$, so $u_3 - u_2 = 0.125$ and $u_3 = 0.21875$ mm. #### Stresses $$ \begin{aligned} \sigma_1 &= E\,\frac{u_2 - u_1}{L} = 200\times10^3\times\frac{0.09375}{250} = 75\ \text{MPa} \\ \sigma_2 &= E\,\frac{u_3 - u_2}{L} = 200\times10^3\times\frac{0.125}{250} = 100\ \text{MPa} \end{aligned} $$ #### Reaction $R_1 = -k_1(u_2 - u_1) = -3.2\times10^5(0.09375) = -30\,000$ N. Check: $\sigma_1A_1 = 75\times400 = 30$ kN and $\sigma_2A_2 = 100\times200 = 20$ kN, consistent with equilibrium ($10 + 20 = 30$ kN). **Answer:** $u_2 = 0.0938$ mm, $u_3 = 0.2188$ mm; $\sigma_1 = 75$ MPa, $\sigma_2 = 100$ MPa (both tensile); reaction = 30 kN, directed opposite to the loads. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the stiffness of element 1 (A = 400 mm^2, L = 250 mm, E = 200000 N/mm^2)? A. 3.2 x 10^5 N/mm B. 1.6 x 10^5 N/mm C. 6.4 x 10^5 N/mm D. 8 x 10^4 N/mm Answer: A. k1 = A1 E / L = 400 x 200000 / 250 = 3.2 x 10^5 N/mm. (Key point: Element stiffness) 2. Why does adding the two equations give u2 directly? A. u3 is zero B. The off-diagonal terms cancel, leaving k1 u2 = total load 30 000 N C. k2 is zero D. The loads are equal Answer: B. Adding rows eliminates the k2 terms, so 3.2 x 10^5 u2 = 10 000 + 20 000, giving u2 = 0.09375 mm. (Key point: Solving for u2) 3. How is the stress in an element found? A. sigma = k u B. sigma = A / F C. sigma = E (u_j - u_i)/L D. sigma = E u / A Answer: C. Stress is E times the strain, and the strain is the change in nodal displacement divided by L. (Key point: Element stress) 4. What are the stresses in elements 1 and 2? A. 100 MPa and 75 MPa B. 75 MPa and 50 MPa C. 30 MPa and 20 MPa D. 75 MPa and 100 MPa Answer: D. sigma1 = 200000(0.09375)/250 = 75 MPa and sigma2 = 200000(0.125)/250 = 100 MPa. 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