---
title: "Bar: nodal displacements, stresses, reactions"
subject: "Computational Techniques in Civil Engineering"
chapter: 4
key: "semester_8/computational_techniques_in_civil_engineering:axial-bar-stepped-p300"
source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/axial-bar-stepped-p300
site: BCT Engineering
---

# Bar: nodal displacements, stresses, reactions

Computational Techniques in Civil Engineering · Chapter 4: Finite element method

- Asked in: 2075 Bhadra, 10 marks
- Video: [FEM lecture How to derive Element stiffness matrix and stress matrix for a linear bar element](https://www.youtube.com/watch?v=fieYcv_aebU) — Iresh Bhavi (Dr. Iresh Bhavi)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/axial-bar-stepped-p300
- Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-4

## Question

Determine the nodal displacements, element stresses and support reactions for the bar as shown in the figure below. Take $E = 200$ GPa.

[Figure: bar fixed at both ends along the x-axis; left part of cross-sectional area 250 mm$^2$ and right part of area 300 mm$^2$; dimensions shown 150 mm, 150 mm and 300 mm; an axial load $P = 300$ kN acts in the +x direction at the junction of the two parts]

## Short answer

The two-element bar model gives nodal displacement u2 = 0.4091 mm (nodes 1 and 3 fixed). Element stresses are σ1 = 545.5 MPa (tension) in the 250 mm² segment and σ2 = -545.5 MPa (compression) in the 300 mm² segment. Support reactions: R1 = 136.36 kN and R3 = 163.64 kN, both acting opposite to the load.

## Answer

**Data (from the figure):** bar fixed at both ends; element 1: $A_1=250$ mm$^2$, $L_1=150$ mm; element 2: $A_2=300$ mm$^2$, $L_2=150$ mm (total 300 mm); $P=300$ kN at the junction node 2; $E=200$ GPa $=200\times10^3$ N/mm$^2$. Nodes: 1 (left wall), 2 (load), 3 (right wall). Units: N, mm.

#### Element stiffness
$$
[k]=\frac{AE}{L}\begin{bmatrix}1&-1\\-1&1\end{bmatrix}
$$
$$
\frac{A_1E}{L_1}=\frac{250\times200\times10^3}{150}=333333.3\ \text{N/mm},\qquad \frac{A_2E}{L_2}=\frac{300\times200\times10^3}{150}=400000\ \text{N/mm}
$$

#### Assembly
$$
\begin{bmatrix}k_1&-k_1&0\\-k_1&k_1+k_2&-k_2\\0&-k_2&k_2\end{bmatrix}\begin{Bmatrix}u_1\\u_2\\u_3\end{Bmatrix}=\begin{Bmatrix}R_1\\P\\R_3\end{Bmatrix}
$$
With $u_1=u_3=0$, the second equation gives $(k_1+k_2)u_2=P$:
$$
u_2=\frac{300\,000}{333333.3+400000}=0.40909\ \text{mm}
$$

#### Element stresses
$$
\sigma_1=E\,\frac{u_2-u_1}{L_1}=200\,000\times\frac{0.40909}{150}=545.45\ \text{MPa (tension)}
$$
$$
\sigma_2=E\,\frac{u_3-u_2}{L_2}=-545.45\ \text{MPa (compression)}
$$

#### Support reactions
$$
R_1=-k_1u_2=-136.36\ \text{kN},\qquad R_3=-k_2u_2=-163.64\ \text{kN}
$$
Check: $R_1+R_3+P=-136.36-163.64+300=0$.

**Answer:** $u_1=0$, $u_2=0.4091$ mm, $u_3=0$; $\sigma_1=+545.5$ MPa, $\sigma_2=-545.5$ MPa; $R_1=136.36$ kN and $R_3=163.64$ kN, both acting opposite to $P$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the axial stiffness of element 1 (A = 250 mm^2, L = 150 mm, E = 200 GPa)?
   A. 333 333 N/mm
   B. 200 000 N/mm
   C. 400 000 N/mm
   D. 1 666 667 N/mm
   Answer: A. k1 = AE/L = 250 x 200 000 / 150 = 333 333 N/mm. (Key point: Element stiffness)

2. With both ends fixed, how is the displacement u2 of the loaded junction found?
   A. u2 = P/k1
   B. u2 = P/(k1 + k2), because the two elements act in parallel against the load
   C. u2 = P/k2
   D. u2 = P(k1 + k2)
   Answer: B. Only the second equation survives: (k1 + k2) u2 = P, giving u2 = 300 000/733 333 = 0.409 mm. (Key point: Reduced equation)

3. Why is element 1 in tension and element 2 in compression?
   A. Element 1 has the smaller area
   B. Node 2 moves toward node 3, lengthening element 1 and shortening element 2
   C. P acts to the left
   D. Both are in tension because the load is tensile
   Answer: B. u2 is positive, so element 1 elongates (+545.45 MPa) while element 2 shortens (-545.45 MPa). (Key point: Stress signs)

4. Which is correct for the support reactions?
   A. R1 = R3 = 150 kN
   B. R1 = -k1 u2 = -136.36 kN and R3 = -k2 u2 = -163.64 kN, which with P sum to zero
   C. R1 = -163.64 kN and R3 = -136.36 kN
   D. R1 = P x A1/(A1 + A2) in the same direction as P
   Answer: B. Reactions oppose P and are shared by stiffness; -136.36 - 163.64 + 300 = 0. (Key point: Reactions and check)

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