---
title: "Stepped bar displacements and support forces"
subject: "Computational Techniques in Civil Engineering"
chapter: 4
key: "semester_8/computational_techniques_in_civil_engineering:stepped-bar-nodal-displacements"
source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/stepped-bar-nodal-displacements
site: BCT Engineering
---

# Stepped bar displacements and support forces

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

- Asked in: 2071 Bhadra, 8 marks
- Video: [Deflection of Stepped Bar : Problem 1 - Stress and Strain - Strength of Materials](https://www.youtube.com/watch?v=iUF_VTfKWLY) — Ekeeda
- 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/stepped-bar-nodal-displacements
- Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-4

## Question

For the given stepped bar obtain the nodal displacements at nodes 2, 3 and 4. Also obtain the forces developed at the supports. Take $E$ = constant and cross-sectional areas as indicated in the figure.

[Figure: bar fixed at both ends (nodes 1 and 5), four equal elements each of length $l/4$ between nodes 1-2-3-4-5; cross-sectional areas $A$ (element 1), $3A/2$ (element 2), $A/2$ (element 3), $3A/4$ (element 4); a point load $P$ acts at node 2 in the +x direction]

## Short answer

Use four bar elements of length l/4 with areas A, 1.5A, 0.5A, 0.75A and fixed ends. Nodal displacements are u2=Pl/(5AE), u3=Pl/(6AE), u4=Pl/(15AE). Support reactions are 0.8P and 0.2P, both opposite to the applied load.

## Answer

Take $k=AE/l$ as the unit of stiffness. Each element has length $l/4$, so $k_e = A_eE/(l/4)=4A_eE/l$.

#### Element stiffnesses
| Element | Nodes | Area | $k_e$ |
|---|---|---|---|
| 1 | 1-2 | $A$ | $4k$ |
| 2 | 2-3 | $3A/2$ | $6k$ |
| 3 | 3-4 | $A/2$ | $2k$ |
| 4 | 4-5 | $3A/4$ | $3k$ |

Each element matrix is $k_e\begin{bmatrix}1&-1\\-1&1\end{bmatrix}$.

#### Assembled global equations
$$
k\begin{bmatrix}4&-4&0&0&0\\-4&10&-6&0&0\\0&-6&8&-2&0\\0&0&-2&5&-3\\0&0&0&-3&3\end{bmatrix}\begin{bmatrix}u_1\\u_2\\u_3\\u_4\\u_5\end{bmatrix}=\begin{bmatrix}R_1\\P\\0\\0\\R_5\end{bmatrix}
$$

#### Boundary conditions: $u_1=u_5=0$
Deleting rows and columns 1 and 5:
$$
\frac{AE}{l}\begin{bmatrix}10&-6&0\\-6&8&-2\\0&-2&5\end{bmatrix}\begin{bmatrix}u_2\\u_3\\u_4\end{bmatrix}=\begin{bmatrix}P\\0\\0\end{bmatrix}
$$
From row 3: $u_4=0.4u_3$. Row 2: $-6u_2+8u_3-2(0.4u_3)=0\Rightarrow u_3=\dfrac{6}{7.2}u_2=\dfrac{5}{6}u_2$. Row 1: $10u_2-6\cdot\dfrac56u_2=5u_2=\dfrac{Pl}{AE}$.

$$
u_2=\frac{Pl}{5AE},\qquad u_3=\frac{Pl}{6AE},\qquad u_4=\frac{Pl}{15AE}
$$

#### Support reactions
$$
R_1=4k(u_1-u_2)=-\frac{4P}{5}=-0.8P,\qquad R_5=3k(u_5-u_4)=-\frac{P}{5}=-0.2P
$$
Both act in the $-x$ direction (towards the left). Check: $P-0.8P-0.2P=0$.

#### Element forces (check)
$F_1=k_1(u_2-u_1)=+0.8P$ (tension); $F_2=6k(u_3-u_2)=-0.2P$; $F_3=2k(u_4-u_3)=-0.2P$; $F_4=3k(u_5-u_4)=-0.2P$ (compression). Equilibrium at node 2: $P-0.8P-0.2P=0$.

**Answer:** $u_2=\dfrac{Pl}{5AE},\ u_3=\dfrac{Pl}{6AE},\ u_4=\dfrac{Pl}{15AE}$; $R_1=0.8P$ and $R_5=0.2P$, both acting opposite to $P$.

## Check yourself

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

1. Each element has length l/4. What is the stiffness k_e of an element of area A_e?
   A. A_e E / l
   B. A_e E / (4l)
   C. 4 A_e E / l
   D. A_e E l / 4
   Answer: C. k_e = A_e E / (element length) = A_e E / (l/4) = 4 A_e E / l. (Key point: Element stiffness)

2. What is done to the global equations because the bar is fixed at both ends?
   A. Rows and columns 1 and 5 are deleted since u_1 = u_5 = 0
   B. The load P is set to zero
   C. Rows 2 to 4 are deleted
   D. Rows 1 and 5 are doubled
   Answer: A. Fixed nodes have zero displacement, so their rows and columns are removed, leaving a 3 x 3 system for u_2, u_3 and u_4. (Key point: Boundary conditions)

3. From the reduced system, what is u_2?
   A. Pl/(10AE)
   B. Pl/(6AE)
   C. Pl/(4AE)
   D. Pl/(5AE)
   Answer: D. Back substitution gives u_4 = 0.4u_3 and u_3 = (5/6)u_2, so 5u_2 = Pl/AE, hence u_2 = Pl/(5AE). (Key point: Solving displacements)

4. What are the support reactions and their directions?
   A. R_1 = 0.5P and R_5 = 0.5P, along +x
   B. R_1 = 0.8P and R_5 = 0.2P, both opposite to P
   C. R_1 = 0.2P and R_5 = 0.8P, both opposite to P
   D. R_1 = P and R_5 = 0, opposite to P
   Answer: B. R_1 = 4k(u_1 - u_2) = -0.8P and R_5 = 3k(u_5 - u_4) = -0.2P, so P - 0.8P - 0.2P = 0 balances. (Key point: Support reactions)

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