---
title: "Beam reactions and mid-span deflection (2EI)"
subject: "Computational Techniques in Civil Engineering"
chapter: 4
key: "semester_8/computational_techniques_in_civil_engineering:beam-support-reactions-midspan-k8"
source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/beam-support-reactions-midspan-k8
site: BCT Engineering
---

# Beam reactions and mid-span deflection (2EI)

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

- Asked in: 2078 Kartik, 10 marks
- Video: [Conjugate Beam Method Slope and Deflection - Introduction and Step by Step Example](https://www.youtube.com/watch?v=QsNvwRQVkBE) — AF Math & Engineering
- 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-support-reactions-midspan-k8
- Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-4

## Question

Determine support reactions and deflections at mid span for the given structure. $E = 2\times10^5$ MPa, $I = 5\times10^6$ mm$^4$.

[Figure: beam fixed at the left end; 25 kN point load 3 m from the fixed end; $EI$ span of 7 m (from the fixed end) to an intermediate support carrying an applied moment of 25 kNm; $EI$ span of 5 m with 15 kN/m UDL to a roller support at the right end]

## Short answer

Use two beam elements with EI=1000 kN m^2, fixed and two roller supports, including equivalent nodal loads for the 25 kN point load and 15 kN/m UDL. Solving gives reactions R1=9.57 kN up, M1=11.44 kN m counter-clockwise, R2=56.83 kN up, R3=33.61 kN up. Mid-span deflections are 2.22 mm downward in span 1 and 91.65 mm downward in span 2.

## Answer

**Data:** $EI=2\times10^5\ \text{MPa}\times5\times10^6\ \text{mm}^4=10^{12}$ N mm$^2=1000$ kN m$^2$ (same for both spans). Nodes: 1 = fixed end (x = 0); 2 = intermediate support (x = 7 m) with an applied moment of 25 kN m (taken clockwise); 3 = right roller (x = 12 m). Span 1: $L=7$ m with 25 kN at 3 m from the fixed end; span 2: $L=5$ m with 15 kN/m UDL. Units: kN, m.

#### Element stiffness
$$
[k]=\frac{EI}{L^3}\begin{bmatrix}12&6L&-12&6L\\6L&4L^2&-6L&2L^2\\-12&-6L&12&-6L\\6L&2L^2&-6L&4L^2\end{bmatrix},\quad \{d\}=\{v_i,\theta_i,v_j,\theta_j\}^T
$$
(Upward forces and displacements, and counter-clockwise moments and rotations, are positive.)

#### Equivalent nodal loads
Span 1 ($a=3$, $b=4$, $L=7$): $\dfrac{Pb^2(3a+b)}{L^3}=\dfrac{25\times16\times13}{343}=15.16$ kN, $\dfrac{Pab^2}{L^2}=24.49$ kN m, $\dfrac{Pa^2(a+3b)}{L^3}=9.84$ kN, $\dfrac{Pa^2b}{L^2}=18.37$ kN m. Span 2: $wL/2=37.5$ kN, $wL^2/12=31.25$ kN m. Equivalent loads are the negatives of the fixed-end forces. The applied clockwise moment gives $-25$ kN m at $\theta_2$ (counter-clockwise positive).

#### Assembled equation
Boundary conditions: $v_1=\theta_1=0$, $v_2=0$, $v_3=0$. Free DOFs: $\theta_2,\theta_3$.
$$
\begin{bmatrix}1371.429 & 400.000 \\ 400.000 & 800.000\end{bmatrix}\begin{Bmatrix}\theta_2\\\theta_3\end{Bmatrix}=\begin{Bmatrix}-37.883 \\ 31.250\end{Bmatrix}
$$
$$
\theta_2=-0.04568\ \text{rad},\qquad \theta_3=0.06190\ \text{rad}
$$

#### Support reactions $\{f\}=[k]\{d\}-\{F_{eq}\}$
| Element | $V_i$ | $M_i$ | $V_j$ | $M_j$ |
|---|---|---|---|---|
| 1-2 | 9.567 | 11.439 | 15.433 | -44.469 |
| 2-3 | 41.394 | 19.469 | 33.606 | -0.000 |

- Fixed end: $R_1=9.567$ kN (up), $M_1=11.439$ kN m (counter-clockwise)
- Intermediate support: $R_2=56.827$ kN (up)
- Right roller: $R_3=33.606$ kN (up)
- Check: $\Sigma R=100.000$ kN $=25+15\times5=100$ kN.

#### Deflection at mid span
The displacement within a span is the Hermite interpolation of the nodal values plus the fixed-end beam deflection due to the load. Evaluating (checked by refining the mesh with nodes at the mid-points):
- Mid of span 1 (x = 3.5 m): $v=-2.22$ mm (downward)
- Mid of span 2 (x = 9.5 m): $v=-91.65$ mm (downward)

The value is large because $EI=1000$ kN m$^2$ is small for these spans.

**Answer:** $R_1=9.57$ kN, $M_1=11.44$ kN m, $R_2=56.83$ kN, $R_3=33.61$ kN; mid-span deflections $2.22$ mm (span 1) and $91.65$ mm (span 2), both downward.

## Check yourself

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

1. EI for E = 2x10^5 MPa and I = 5x10^6 mm^4 is
   A. 1000 kN m^2
   B. 10 kN m^2
   C. 10^6 kN m^2
   D. 100 kN m^2
   Answer: A. 2x10^5 x 5x10^6 = 10^12 N mm^2 = 1000 kN m^2. (Key point: Flexural rigidity)

2. With v1 = theta1 = 0 and v2 = v3 = 0, the unknowns are
   A. theta2 and theta3
   B. v2 and v3
   C. theta1 and theta2
   D. v2, theta2, theta3
   Answer: A. Only the rotations at the intermediate and right supports are free. (Key point: Free DOFs)

3. How is the 25 kN m clockwise moment at node 2 entered in the load vector?
   A. -25 kN m at theta2 (counter-clockwise positive)
   B. +25 kN m at theta2
   C. -25 kN at v2
   D. It is added to the UDL load
   Answer: A. Clockwise is negative under the counter-clockwise-positive convention. (Key point: Applied nodal moment)

4. What check confirms the reactions?
   A. Sum = 25 + 15x5 = 100 kN
   B. Sum = 25 kN
   C. Sum = 15 kN
   D. Sum of moments is zero at the fixed end only
   Answer: A. 9.567 + 56.827 + 33.606 = 100 kN equals total load. (Key point: Reaction check)

## Close variants set in other exams

- [Beam reactions, deflection, BMD (30 kN, 20 kN/m)](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Abeam-reactions-midspan-bmd-a2) — Computational Techniques in Civil Engineering, chapter 4, asked 2×

## Related in Computational Techniques in Civil Engineering

- [Beam BMD, SFD and mid-span deflection (FEM)](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Abeam-bmd-sfd-midspan-concrete) — Computational Techniques in Civil Engineering, chapter 4
- [Rotation and deflection at free end, BMD](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Abeam-fem-rotation-deflection-free-end) — Computational Techniques in Civil Engineering, chapter 4
- [Beam: deflection, slope under load, BMD](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Abeam-fem-s79) — Computational Techniques in Civil Engineering, chapter 4

## Related in other subjects

- [Slope deflection: beam, 360 kN load, fixed end](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aslope-deflection-beam-10kn-20knm-360kn) — Theory of Structures II, chapter 3
- [Mid-span deflection of beam, 10 kN load](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ass-beam-deflection-mid-span-4m) — Strength of Materials, chapter 7
- [Beam mid-span deflection with temperature gradient](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Abeam-midspan-defl-temperature) — Theory of Structures I, chapter 3
- [Slope deflection: three-span beam with settlement](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aslope-deflection-three-span-fixed-ends) — Theory of Structures II, chapter 3