---
title: "Beam BMD, SFD and mid-span deflection (FEM)"
subject: "Computational Techniques in Civil Engineering"
chapter: 4
key: "semester_8/computational_techniques_in_civil_engineering:beam-bmd-sfd-midspan-concrete"
source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/beam-bmd-sfd-midspan-concrete
site: BCT Engineering
---

# Beam BMD, SFD and mid-span deflection (FEM)

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

- Asked in: 2078 Chaitra, 10 marks
- Video: [Shear Force and Bending Moment Made EASY!](https://www.youtube.com/watch?v=dnIhP4ndtoE) — Math Physics Engage
- 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-bmd-sfd-midspan-concrete
- Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-4

## Question

Draw Bending Moment and Shear force diagrams of the concrete beam shown in the following figure using the finite element method. Take the beam section $500\ \text{mm} \times 1000\ \text{mm}$. Also determine the deflections at the mid span.

[Figure: beam of total length 10 m (5 m + 5 m), fixed at the left end and roller supported at the right end, with a 20 kN point load at mid-span]

## Short answer

Model as two 5 m beam elements with assumed M25 concrete giving EI=1.042e6 kN m^2; fixed and roller supports leave v2, theta2, theta3 free. Solving gives mid-span deflection 0.175 mm downward, matching 7PL^3/768EI. Support reactions are 13.75 kN at A, 37.5 kN m at A, and 6.25 kN at C; SFD is +13.75 then -6.25 kN and BMD has -37.5, +31.25, 0 kN m.

## Answer

**Data:** $L=10$ m, two elements of 5 m, fixed at A (left), roller at C (right), 20 kN at mid-span B. Section $b\times h=500\times1000$ mm: $I=\dfrac{bh^3}{12}=\dfrac{0.5\times1^3}{12}=0.041667$ m$^4$. $E$ of concrete is not given; assume M25 concrete, $E_c=5000\sqrt{f_{ck}}=25\,000$ MPa (IS 456, cl. 6.2.3.1), so
$$
EI=25\times10^6\times0.041667=1.042\times10^{6}\ \text{kN m}^2
$$
Nodes: 1 = A, 2 = B (mid-span), 3 = C. 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.)
with $L=5$ m for both elements.

#### Assembly and boundary conditions
Boundary conditions: $v_1=\theta_1=0$ (fixed), $v_3=0$ (roller). Free DOFs: $v_2,\theta_2,\theta_3$. Load vector: $F_{v2}=-20$ kN.
$$
EI\,\begin{bmatrix}0.1920 & 0 & 0.2400 \\ 0 & 1.6000 & 0.4000 \\ 0.2400 & 0.4000 & 0.8000\end{bmatrix}\begin{Bmatrix}v_2\\\theta_2\\\theta_3\end{Bmatrix}=\begin{Bmatrix}-20.0 \\ 0 \\ 0\end{Bmatrix}
$$
(the matrix shown is $[K]/EI$). Solving:
$$
v_2=-0.1750\ \text{mm},\quad \theta_2=-1.5\times10^{-5}\ \text{rad},\quad \theta_3=6\times10^{-5}\ \text{rad}
$$
**Mid-span deflection = 0.175 mm downward.** Check with the standard result for a propped cantilever with a central load: $\dfrac{7PL^3}{768EI}=\dfrac{7\times20\times1000}{768\times1.042\times10^{6}}=0.175$ mm. It agrees.

#### End forces
| Element | $V_i$ (kN) | $M_i$ (kN m) | $V_j$ (kN) | $M_j$ (kN m) |
|---|---|---|---|---|
| AB | 13.75 | 37.50 | -13.75 | 31.25 |
| BC | -6.25 | -31.25 | 6.25 | 0.00 |

Reactions: $R_A=13.75$ kN, $M_A=37.5$ kN m (counter-clockwise), $R_C=6.25$ kN. Check: $13.75+6.25=20$ kN.

#### Shear force diagram (kN)
- A to B: $+13.75$ (constant)
- B to C: $-6.25$ (constant)

#### Bending moment diagram (kN m, sagging +)
- A: $-37.50$ (hogging); B: $+31.25$ (13.75 x 5 - 37.5); C: $0$.
- Zero moment at $x=37.5/13.75=2.73$ m from A.

```text
 A(fixed)                    B(mid)                C(roller)
 |---------------------------|---------------------|
 BMD: -37.5                  +31.25                0   (kN m)
 SFD: +13.75 (constant)      -6.25 (constant)
```

## Check yourself

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

1. What is I for the 500 mm x 1000 mm section (1000 mm deep)?
   A. 0.041667 m^4
   B. 0.5 m^4
   C. 0.0104 m^4
   D. 0.1667 m^4
   Answer: A. I = b h^3/12 = 0.5 x 1^3/12 = 0.041667 m^4. (Key point: Section property)

2. E of concrete was not given. What was assumed?
   A. E = 25,000 MPa (M25, 5000 sqrt(fck))
   B. E = 200 GPa
   C. E = 2.1x10^5 MPa
   D. E = 5000 MPa
   Answer: A. IS 456 gives Ec = 5000 sqrt(fck) = 25,000 MPa for M25. (Key point: Assumed modulus)

3. How was the mid-span deflection of 0.175 mm checked?
   A. Using 7PL^3/(768EI) for a propped cantilever with central load
   B. Using PL^3/(48EI) for a simply supported beam
   C. Using PL^3/(3EI) for a cantilever
   D. Using 5wL^4/(384EI)
   Answer: A. A fixed-roller beam with central load deflects 7PL^3/768EI under the load, matching the FEM result. (Key point: Verification of deflection)

4. What are the end reactions and the moment at the fixed end?
   A. R_A = 10, R_C = 10, M_A = 0
   B. R_A = 13.75 kN, R_C = 6.25 kN, M_A = 37.5 kN m
   C. R_A = 6.25 kN, R_C = 13.75 kN, M_A = 31.25 kN m
   D. R_A = 20, R_C = 0, M_A = 50
   Answer: B. They sum to 20 kN; BM at B is 13.75x5 - 37.5 = 31.25 kN m. (Key point: Reactions and moment)

## Related in Computational Techniques in Civil Engineering

- [Beam reactions and mid-span deflection (2EI)](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Abeam-support-reactions-midspan-k8) — Computational Techniques in Civil Engineering, chapter 4, asked 2×
- [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×
- [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
- [Beam deflection and stresses at B (FEM)](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Abeam-fem-deflection-bmd-a) — 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
- [Basic steps in finite element analysis](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Afem-basic-steps) — Computational Techniques in Civil Engineering, chapter 1

## Related in other subjects

- [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
- [Slope deflection: BMD and SFD of beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aslope-deflection-beam-30knm-20knm-100kn) — Theory of Structures II, chapter 3
- [ILD: bending moment at fixed support, 10 m beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-bm-fixed-support-10m-1) — Theory of Structures II, chapter 4
- [Slope deflection: continuous beam, SFD and BMD](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aslope-deflection-beam-50kn-10knm-3ei) — Theory of Structures II, chapter 3