---
title: "Slope deflection: continuous beam, SFD and BMD"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:slope-deflection-beam-50kn-10knm-3ei"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/slope-deflection-beam-50kn-10knm-3ei
site: BCT Engineering
---

# Slope deflection: continuous beam, SFD and BMD

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2069 Asar, 10 marks
- Video: [Slope Deflection Method for Beam Analysis - Problem No 8](https://www.youtube.com/watch?v=tuiMskZlEXA) — Stan Academy
- 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-10knm-3ei
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Analyse the continuous beam shown in figure below by the slope deflection method. Also draw shear force and bending moment diagram. [Figure: beam ABCD, A fixed; AB = 5 m + 3 m (2.5EI) with 50 kN at 5 m from A; BC = 6 m (3EI) with 10 kN/m UDL; overhang CD = 2 m (EI) with 20 kN at D. Printed as the alternative (OR) to the stiffness matrix question above.]

## Short answer

Overhang gives M_C=40 kNm. Slope-deflection equations give EI theta_B=-12.216, EI theta_C=11.108; final moments M_AB=-42.79, M_BA=43.32, M_BC=-43.32, M_CB=40.00, M_CD=-40.00 kNm. Reactions: R_B=61.87 kN, R_C=49.45 kN, V_A=18.68 kN.

## Answer

**Data:** A is fixed; AB = 5 m + 3 m = 8 m (2.5EI) with 50 kN at 5 m from A; BC = 6 m (3EI) with 10 kN/m; the overhang CD = 2 m (EI) carries 20 kN at D; B and C are simple supports. No settlement.

#### Step 1: Overhang CD
$M_C=20\times2=40$ kNm (hogging), a known moment at C.

#### Step 2: Fixed-end moments (clockwise +)
- AB ($a=5$, $b=3$): $FEM_{AB}=-\dfrac{50\times5\times3^2}{8^2}=-35.16$, $FEM_{BA}=+\dfrac{50\times5^2\times3}{8^2}=58.59$ kNm
- BC: $\mp\dfrac{10\times6^2}{12}=\mp30$ kNm

#### Step 3: Slope-deflection equations ($\theta_A=0$)
- $M_{AB}=\frac{2(2.5EI)}{8}\left(\theta_{B}\right) - 35.16$
- $M_{BA}=\frac{2(2.5EI)}{8}\left(2\theta_{B}\right) + 58.59$
- $M_{BC}=\frac{2(3EI)}{6}\left(2\theta_{B} + \theta_{C}\right) - 30.00$
- $M_{CB}=\frac{2(3EI)}{6}\left(2\theta_{C} + \theta_{B}\right) + 30.00$

#### Step 4: Equilibrium
Joint B: $M_{BA}+M_{BC}=0$. Joint C: $M_{CB}+M_{CD}=0$ with $M_{CD}=-40$ kNm.
$$
EI\begin{bmatrix}3.2500 & 1.0000 \\ 1.0000 & 2.0000\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C}\end{Bmatrix}=\begin{Bmatrix}-28.594 \\ 10.000\end{Bmatrix}
$$
$$
EI\theta_{B}=-12.2159,\quad EI\theta_{C}=11.1080
$$

#### Step 5: Final end moments (kNm)
| End | $M$ (kNm) | End | $M$ (kNm) |
|---|---|---|---|
| AB | -42.79 | BA | 43.32 |
| BC | -43.32 | CB | 40.00 |
| CD | -40.00 | DC | 0.00 |

#### Step 6: Shear force and bending moment
Shear forces follow from the equilibrium of each span with its end moments (positive when the left-hand part is pushed upward):

| Span | Shear at left end (kN) | Shear at right end (kN) |
|---|---|---|
| AB | 18.68 | -31.32 |
| BC | 30.55 | -29.45 |
| CD | 20.00 | 0.00 |

SFD: AB: 18.68 kN from A to the 50 kN load, then -31.32 kN to B; BC: 30.55 kN at B falling uniformly ($-10$ kN/m) to -29.45 kN at C; CD: $+20$ kN constant (dropping to zero at the free end after the 20 kN load). Reactions: A: $H=0.00$, $V=18.68$, $M=42.79$; B: $H=0.00$, $V=61.87$, $M=0.00$; C: $H=0.00$, $V=49.45$, $M=0.00$.

BMD (sagging +, kNm):
| Member | BM at first end | BM at second end | Max within member |
|---|---|---|---|
| AB | -42.79 | -43.32 | 50.63 at 5.00 m |
| BC | -43.32 | -40.00 | 3.35 at 3.06 m |
| CD | -40.00 | 0.00 | - |
BM under the 50 kN load: 50.63 kNm. The shear is zero in BC at 3.06 m from B, where the sagging moment is maximum.

**Answer: $M_{AB}=-42.79$, $M_{BA}=43.32$, $M_{BC}=-43.32$, $M_{CB}=40.00$ kNm, $M_{CD}=-40$ kNm; $R_B=61.87$ kN, $R_C=49.45$ kN, $V_A=18.68$ kN.**

## Check yourself

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

1. How is the overhang CD (20 kN at D, 2 m) handled?
   A. As a known hogging moment of 40 kNm at C, so M_CD = -40
   B. As an extra unknown rotation at D
   C. By ignoring it
   D. As a UDL on BC
   Answer: A. The overhang is statically determinate: the moment at C is 20 x 2 = 40 kNm, giving M_CD = -40 kNm. (Key point: Overhang moment)

2. Which equation is used at joint C?
   A. M_CB + M_CD = 0 with M_CD = -40
   B. M_CB = 0
   C. M_CB = M_BC
   D. M_CB - M_CD = 0
   Answer: A. Moments at C must sum to zero, so M_CB must equal +40 kNm to balance the overhang moment. (Key point: Equilibrium at C)

3. What are the FEMs for BC (6 m, 10 kN/m)?
   A. -30 and +30 kNm
   B. -60 and +60 kNm
   C. -15 and +15 kNm
   D. -45 and +45 kNm
   Answer: A. wL^2/12 = 10 x 36 / 12 = 30 kNm, with sign - at B and + at C in the clockwise-positive convention. (Key point: Fixed-end moments for BC)

4. What is M_BA with theta_A = 0, no settlement and EI of 2.5EI over 8 m?
   A. (2(2.5EI)/8)(2 theta_B) + FEM_BA
   B. (2(2.5EI)/8)(theta_B) + FEM_BA
   C. (4(2.5EI)/8)(theta_B) - FEM_BA
   D. (2(2.5EI)/8)(2 theta_B + theta_A) - FEM_AB
   Answer: A. M_BA = (2EI/L)(2 theta_B + theta_A - 3 psi) + FEM_BA; with theta_A = 0 and psi = 0 it reduces to (2EI/L)(2 theta_B) + 58.59. (Key point: Slope-deflection form)

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