---
title: "Slope deflection: beam with 50 kN and 30 kN"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:slope-deflection-beam-50kn-30kn-bmd"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/slope-deflection-beam-50kn-30kn-bmd
site: BCT Engineering
---

# Slope deflection: beam with 50 kN and 30 kN

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2072 Chaitra, 12 marks
- Video: [Slope deflection method explained](https://www.youtube.com/watch?v=ftjehUrCloM) — Engineer4Free
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/slope-deflection-beam-50kn-30kn-bmd
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Analyse the beam loaded as shown in the figure below by the slope deflection method. Also draw the bending moment diagram (BMD). [Figure: beam ABCD, A fixed; AB = 2.5 m + 1.5 m (3EI) with 50 kN at 2.5 m from A; BC = 4 m (2EI) with 10 kN/m UDL; CD = 2 m + 2 m (1.5EI) with 30 kN at mid-span; D roller.]

## Short answer

Slope-deflection method: solving for rotations, the end moments are M_AB = -23.63, M_BA = 17.18, M_BC = -17.18, M_CB = 17.75, M_CD = -17.75, M_DC = 0 kNm; sagging moments under concentrated loads are 27.27 kNm and 21.13 kNm respectively.

## Answer

**Data:** A fixed; AB = 2.5 m + 1.5 m = 4 m (3EI) with 50 kN at 2.5 m from A; BC = 4 m (2EI) with 10 kN/m; CD = 2 m + 2 m = 4 m (1.5EI) with 30 kN at mid-span; D is a roller; B and C simple supports. No settlement.

#### Fixed-end moments (clockwise +)
- AB ($a=2.5$, $b=1.5$): $FEM_{AB}=-\dfrac{50\times2.5\times1.5^2}{4^2}=-17.58$, $FEM_{BA}=+\dfrac{50\times2.5^2\times1.5}{4^2}=29.30$ kNm
- BC: $\mp\dfrac{10\times4^2}{12}=\mp13.33$ kNm
- CD: $\mp\dfrac{30\times4}{8}=\mp15$ kNm

#### Slope-deflection equations
- $M_{AB}=\frac{2(3EI)}{4}\left(\theta_{B}\right) - 17.58$
- $M_{BA}=\frac{2(3EI)}{4}\left(2\theta_{B}\right) + 29.30$
- $M_{BC}=\frac{2(2EI)}{4}\left(2\theta_{B} + \theta_{C}\right) - 13.33$
- $M_{CB}=\frac{2(2EI)}{4}\left(2\theta_{C} + \theta_{B}\right) + 13.33$
- $M_{CD}=\frac{2(1.5EI)}{4}\left(2\theta_{C} + \theta_{D}\right) - 15.00$
- $M_{DC}=\frac{2(1.5EI)}{4}\left(2\theta_{D} + \theta_{C}\right) + 15.00$

#### Equilibrium
- Joint B: $M_{BA}+M_{BC}=0$ Joint C: $M_{CB}+M_{CD}=0$ Roller D: $M_{DC}=0$

$$
EI\begin{bmatrix}5.0000 & 1.0000 & 0.0000 \\ 1.0000 & 3.5000 & 0.7500 \\ 0.0000 & 0.7500 & 1.5000\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C} \\ \theta_{D}\end{Bmatrix}=\begin{Bmatrix}-15.964 \\ 1.667 \\ -15.000\end{Bmatrix}
$$
$$
EI\theta_{B}=-4.0378,\quad EI\theta_{C}=4.2254,\quad EI\theta_{D}=-12.1127
$$

#### Final end moments (kNm)
| End | $M$ (kNm) | End | $M$ (kNm) |
|---|---|---|---|
| AB | -23.63 | BA | 17.18 |
| BC | -17.18 | CB | 17.75 |
| CD | -17.75 | DC | 0.00 |

Checks:
- Joint B: (17.18) + (-17.18) = 0.00 kNm (applied clockwise moment 0.00) ✓
- Joint C: (17.75) + (-17.75) = 0.00 kNm (applied clockwise moment 0.00) ✓
- Joint D: (0.00) = 0.00 kNm (applied clockwise moment 0.00) ✓

#### Bending moment diagram ordinates (sagging +, kNm)
| Member | BM at first end | BM at second end | Max within member |
|---|---|---|---|
| AB | -23.63 | -17.18 | 27.27 at 2.50 m |
| BC | -17.18 | -17.75 | 2.54 at 1.99 m |
| CD | -17.75 | 0.00 | 21.13 at 2.00 m |

Reactions: A: $H=0.00$, $V=20.36$, $M=23.63$; B: $H=0.00$, $V=49.50$, $M=0.00$; C: $H=0.00$, $V=39.58$, $M=0.00$; D: $H=0.00$, $V=10.56$, $M=0.00$.

**Answer: $M_{AB}=-23.63$, $M_{BA}=17.18$, $M_{BC}=-17.18$, $M_{CB}=17.75$, $M_{CD}=-17.75$, $M_{DC}=0$ kNm; BM at the load points: 27.27 kNm under 50 kN, 21.13 kNm under 30 kN.**

## Check yourself

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

1. What condition applies at roller D?
   A. theta_D = 0
   B. M_DC = 15
   C. Delta = 0
   D. M_DC = 0, theta_D is an unknown
   Answer: D. A roller has free rotation and no moment, so M_DC = 0 gives a third equation. (Key point: Roller end D)

2. FEM_AB for 50 kN at a = 2.5, b = 1.5, L = 4?
   A. -Pab^2/L^2 = -17.58 kNm
   B. -Pa^2b/L^2 = -29.30
   C. -PL/8 = -25
   D. -12.5
   Answer: A. Pab^2/L^2 = 50x2.5x2.25/16 = 17.58; the near-end FEM uses the distance squared of the far side. (Key point: FEM_AB)

3. How many unknown rotations?
   A. Four
   B. Three: theta_B, theta_C, theta_D
   C. Two
   D. One
   Answer: B. A is fixed, B, C and D rotate; three joint equations are written. (Key point: Unknowns)

4. Joint B check: M_BA + M_BC = 0 because?
   A. B is a roller
   B. Load acts at B
   C. No external moment at B
   D. B is fixed
   Answer: C. B has no applied moment, so end moments balance (17.18 - 17.18 = 0). (Key point: Check)

## Related in Theory of Structures II

- [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
- [Slope deflection: continuous beam, 90 kN and 30 kN](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aslope-deflection-beam-15knm-90kn-30kn) — Theory of Structures II, chapter 3
- [Slope deflection: continuous beam, B sinks 19 mm](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aslope-deflection-beam-sinks-19mm) — Theory of Structures II, chapter 3
- [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
- [Slope deflection: continuous beam, I = 4x10^7 mm^4](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aslope-deflection-beam-i4e7-60kn) — Theory of Structures II, chapter 3
- [Slope deflection: B sinks, C rises](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aslope-deflection-beam-settlement-b-c) — Theory of Structures II, chapter 3

## Related in other subjects

- [Moment area: slope and deflection under 50 kN](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Amoment-area-12-ss-defl-4m-d) — Theory of Structures I, chapter 4
- [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×
- [Conjugate beam: deflection at B, slope at D](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aconjugate-50kn-20knm-3-spans) — Theory of Structures I, 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