---
title: "Slope deflection: continuous beam, 90 kN and 30 kN"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:slope-deflection-beam-15knm-90kn-30kn"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/slope-deflection-beam-15knm-90kn-30kn
site: BCT Engineering
---

# Slope deflection: continuous beam, 90 kN and 30 kN

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2073 Shrawan, 10 marks
- Video: [Slope Deflection Method | Example 1 | Structural Analysis](https://www.youtube.com/watch?v=j2eFkdNtFNs) — Being Civil
- 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-15knm-90kn-30kn
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Analyse the continuous beam shown in figure below using the slope deflection method. [Figure: beam A fixed; AB = 15 m (2EI) with 15 kN/m UDL; BC = 10 m (EI) with 90 kN at 4 m from B; overhang CD = 2 m with 30 kN at the end; supports B and C.]

## Short answer

The overhang moment at C is 60 kNm hogging. Slope-deflection equilibrium gives end moments: M_AB = -325.55, M_BA = 192.64, M_BC = -192.64, M_CB = 60.00, M_CD = -60.00 kNm.

## Answer

**Data:** A is fixed; AB = 15 m (2EI) carries 15 kN/m; B and C are simple supports; BC = 10 m (EI) carries 90 kN at 4 m from B; CD is a 2 m overhang with 30 kN at its free end D.

#### Step 1: Overhang CD
$M_C=30\times2=60$ kNm (hogging), a known moment at joint C.

#### Step 2: Fixed-end moments (clockwise +)
- AB: $\mp\dfrac{15\times15^2}{12}=\mp281.25$ kNm
- BC ($a=4$, $b=6$): $FEM_{BC}=-\dfrac{90\times4\times6^2}{10^2}=-129.60$, $FEM_{CB}=+\dfrac{90\times4^2\times6}{10^2}=86.40$ kNm

#### Step 3: Slope-deflection equations ($\theta_A=0$, $\psi=0$)
- $M_{AB}=\frac{2(2EI)}{15}\left(\theta_{B}\right) - 281.25$
- $M_{BA}=\frac{2(2EI)}{15}\left(2\theta_{B}\right) + 281.25$
- $M_{BC}=\frac{2EI}{10}\left(2\theta_{B} + \theta_{C}\right) - 129.60$
- $M_{CB}=\frac{2EI}{10}\left(2\theta_{C} + \theta_{B}\right) + 86.40$

#### Step 4: Joint equilibrium
- Joint B: $M_{BA}+M_{BC}=0$
- Joint C: $M_{CB}+M_{CD}=0$ with $M_{CD}=-60$ kNm, i.e. $M_{CB}=60$ kNm

$$
EI\begin{bmatrix}0.9333 & 0.2000 \\ 0.2000 & 0.4000\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C}\end{Bmatrix}=\begin{Bmatrix}-151.650 \\ -26.400\end{Bmatrix}
$$
$$
EI\theta_{B}=-166.1400,\quad EI\theta_{C}=17.0700
$$

#### Step 5: Final end moments (kNm)
| End | $M$ (kNm) | End | $M$ (kNm) |
|---|---|---|---|
| AB | -325.55 | BA | 192.64 |
| BC | -192.64 | CB | 60.00 |
| CD | -60.00 | DC | 0.00 |

Checks:
- Joint B: (192.64) + (-192.64) = 0.00 kNm (applied clockwise moment 0.00) ✓
- Joint C: (60.00) + (-60.00) = 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 | -325.55 | -192.64 | 165.39 at 8.10 m |
| BC | -192.64 | -60.00 | 76.41 at 4.00 m |
| CD | -60.00 | 0.00 | - |

Reactions: A: $H=0.00$, $V=121.36$, $M=325.55$; B: $H=0.00$, $V=170.90$, $M=0.00$; C: $H=0.00$, $V=52.74$, $M=0.00$.

**Answer: $M_{AB}=-325.55$, $M_{BA}=192.64$, $M_{BC}=-192.64$, $M_{CB}=60.00$ kNm and $M_{CD}=-60.00$ kNm (overhang moment, hogging 60 kNm).**

## Check yourself

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

1. How is the overhang CD treated?
   A. Ignored
   B. As a fixed beam
   C. As an extra unknown rotation
   D. Its moment 30 x 2 = 60 kNm (hogging) is a known moment at C
   Answer: D. The overhang is statically determinate, so it applies a known moment of 60 kNm at joint C. (Key point: Overhang CD)

2. FEM_BC for the 90 kN load at a = 4 m, b = 6 m, L = 10 m?
   A. -Pab^2/L^2 = -129.6 kNm
   B. -PL/8 = -112.5 kNm
   C. -Pa^2b/L^2 = -86.4 kNm
   D. -90 kNm
   Answer: A. FEM_BC = Pab^2/L^2 = 90x4x36/100 = 129.6, anticlockwise; FEM_CB = 86.4. (Key point: FEM for BC)

3. What is the condition at joint C?
   A. M_CB = -60
   B. M_CB + M_CD = 0 with M_CD = -60, so M_CB = 60
   C. M_CB = 0
   D. M_CB = M_BC
   Answer: B. The joint is in equilibrium with the overhang moment, so M_CB must equal +60 kNm. (Key point: Joint C equation)

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