---
title: "Beam deflection and stresses at B (FEM)"
subject: "Computational Techniques in Civil Engineering"
chapter: 4
key: "semester_8/computational_techniques_in_civil_engineering:beam-fem-deflection-bmd-a"
source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/beam-fem-deflection-bmd-a
site: BCT Engineering
---

# Beam deflection and stresses at B (FEM)

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

- Asked in: 2079 Jestha
- 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/beam-fem-deflection-bmd-a
- Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-4

## Question

For the given beam determine the deflection at point B and the stresses at the same point.

[Figure: beam A-B-D-C of uniform $EI$; 10 kN/m uniformly distributed load over AB (4 m); B to D is 2 m with a 40 kN downward point load at D; D to C is 2 m; A fixed end, C supported at the right end]

## Short answer

Using three beam elements and solving the assembled stiffness equations gives deflection at B: v_B = -230/EI m (downward). Bending moment at B is 38.75 kN m, so bending stress σ = 38.75×10^6 y / I N/mm^2 (tension bottom, compression top).

## Answer

**Assumptions (figure data):** A is fixed, C is a roller (simple support), $EI$ is constant. Three beam elements: AB (4 m, 10 kN/m UDL), BD (2 m) and DC (2 m, no load); the 40 kN downward load acts at node D. Positive: $v$ upward, $\theta$ counter-clockwise. Units: kN, m.

#### Element stiffness
Each element uses
$$
[k]=\frac{EI}{L^3}\begin{bmatrix}12&6L&-12&6L\\6L&4L^2&-6L&2L^2\\-12&-6L&12&-6L\\6L&2L^2&-6L&4L^2\end{bmatrix},\quad \{d\}=\{v_i,\theta_i,v_j,\theta_j\}^T
$$
(Upward forces and displacements, and counter-clockwise moments and rotations, are positive.)
with $L=4$ m (AB), $L=2$ m (BD, DC). Nodes: 1 = A, 2 = B, 3 = D, 4 = C.

#### Equivalent nodal loads
Element AB: $wL/2=20$ kN, $wL^2/12=13.333$ kN m, so the equivalent loads are $-20$ kN and $-13.333$ kN m at A, and $-20$ kN and $+13.333$ kN m at B. The 40 kN load is a direct nodal load at D.

#### Assembled equation (free DOFs $v_B,\theta_B,v_D,\theta_D,\theta_C$)
With $v_A=\theta_A=0$ and $v_C=0$:
$$
EI\,\begin{bmatrix}1.688 & 1.125 & -1.500 & 1.500 & 0 \\ 1.125 & 3.000 & -1.500 & 1.000 & 0 \\ -1.500 & -1.500 & 3.000 & 0 & 1.500 \\ 1.500 & 1.000 & 0 & 4.000 & 1.000 \\ 0 & 0 & 1.500 & 1.000 & 2.000\end{bmatrix}\begin{Bmatrix}v_B\\\theta_B\\v_D\\\theta_D\\\theta_C\end{Bmatrix}=\begin{Bmatrix}-20.000 \\ 13.333 \\ -40.000 \\ 0 \\ 0\end{Bmatrix}
$$
Solving:
$$
\begin{aligned}
v_B&=-\frac{230.00}{EI}\ \text{m},&\theta_B&=-\frac{34.167}{EI}\ \text{rad}\\
v_D&=-\frac{207.083}{EI}\ \text{m},&\theta_D&=\frac{63.958}{EI}\ \text{rad},\quad \theta_C=\frac{123.333}{EI}\ \text{rad}
\end{aligned}
$$

**Deflection at B = 230.0/EI m downward** (with $EI$ in kN m$^2$; e.g. $EI=2\times10^4$ kN m$^2$ gives 11.5 mm).

#### Element end forces $\{f\}=[k]\{d\}-\{F_{eq}\}$
| Element | $V_i$ (kN) | $M_i$ (kN m) | $V_j$ (kN) | $M_j$ (kN m) |
|---|---|---|---|---|
| AB | 50.3125 | 82.500 | -10.3125 | 38.750 |
| BD | 10.3125 | -38.750 | -10.3125 | 59.375 |
| DC | -29.6875 | -59.375 | 29.6875 | 0.000 |

Reactions: $R_A=50.312$ kN (up), $M_A=82.50$ kN m (counter-clockwise), $R_C=29.687$ kN. Check: $50.3125+29.6875=80$ kN = $10\times4+40$.

#### Bending moment (sagging +) and shear force
- A: $-82.50$ (hogging); at 2 m from A: $-1.875$; B: $+38.75$; D: $+59.375$; C: $0$.
- Shear: $+50.31$ at A, $+10.31$ at B (just before and after), $-29.69$ right of D.

```text
 A(fixed)      B          D         C(roller)
 |-------------|----------|---------|
 BMD: -82.5   +38.75     +59.375     0   (kN m)
 SFD: +50.31 -> +10.31 (A to D) -> -29.69 (D to C)
```
#### Stresses at B
At B, $M_B=38.75$ kN m (sagging) and $V_B=10.31$ kN. The bending stress at a fibre at distance $y$ from the neutral axis is
$$
\sigma_B=\frac{M_B\,y}{I}=\frac{38.75\times10^6\ y}{I}\ \text{N/mm}^2
$$
tension at the bottom and compression at the top. The section size is not given, so for a rectangular section $b\times h$ use $\sigma_{max}=6M_B/(bh^2)$; the average shear stress is $\tau=V_B/A$ (maximum $1.5V_B/A$ for a rectangle).

**Answer:** $v_B=-230.00/EI$ m (downward), $\theta_B=-34.167/EI$, $M_B=38.75$ kN m; $\sigma_B=M_By/I$.

## Check yourself

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

1. For element AB with 10 kN/m over 4 m, what are the equivalent nodal loads at B?
   A. -20 kN and -13.333 kN m
   B. -20 kN and +13.333 kN m
   C. -40 kN and +13.333 kN m
   D. -10 kN and +20 kN m
   Answer: B. wL/2 = 20 kN downward, and wL^2/12 = 13.333 kN m, which is counter-clockwise (positive) at the right end B. (Key point: Equivalent nodal loads)

2. With A fixed and C a roller, which DOFs are unknown?
   A. v_B, theta_B, v_D, theta_D, theta_C
   B. v_B, v_D, v_C only
   C. theta_A, theta_B, theta_D
   D. All displacements and rotations at all four nodes
   Answer: A. v_A = theta_A = 0 (fixed) and v_C = 0 (roller); the rest are free. (Key point: Free degrees of freedom)

3. How are the element end forces obtained after solving for displacements?
   A. {f} = [k]{d} - {F_eq}
   B. {f} = [k]^-1 {d}
   C. {f} = [k]{d} + {d}
   D. {f} = {F_eq} only
   Answer: A. The stiffness force from the nodal displacements is corrected by subtracting the equivalent loads of the span. (Key point: Element end forces)

4. How is the bending stress at B found with M_B = 38.75 kN m sagging?
   A. sigma = V y / I with compression at the bottom
   B. sigma = M y / I, tension at the bottom and compression at the top
   C. sigma = M / A in every fibre equally
   D. sigma = EI v_B
   Answer: B. Flexure formula M y / I; sagging puts the bottom in tension and the top in compression. (Key point: Stress at B)

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