---
title: "Slope deflection: beam with overhang"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:slope-deflection-beam-40kn-30kn"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/slope-deflection-beam-40kn-30kn
site: BCT Engineering
---

# Slope deflection: beam with overhang

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2079 Baishakh, 12 marks
- Video: [Slope Deflection Method for Beam Analysis - Problem No 9 ( Overhanging on the Right Side)](https://www.youtube.com/watch?v=AHNDCWSfrCQ) — 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-40kn-30kn
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Draw the BMD for the given beam using the slope deflection method. [Figure: continuous beam ABCD; dimensions 2 m, 4 m, 10 m, 2 m; AB with 1.5EI and 40 kN point load, BC with 2EI carrying 10 kN/m UDL over 10 m, CD with EI and 30 kN at the end D.]

## Short answer

Solving slope-deflection equations for unknown rotations θ_B and θ_C yields M_AB = –11.42, M_BA = 66.04, M_BC = –66.04, M_CB = 60.00 kNm. Overhang CD causes a fixed moment of –60 kNm at C. Maximum sagging moment in BC is approximately 62.00 kNm.

## Answer

**Assumptions:** the figure shows AB = 2 m + 4 m = 6 m with the 40 kN load 2 m from A, BC = 10 m, CD = 2 m overhang with 30 kN at D. A is taken as a fixed end; B and C are simple supports. Stiffness: AB = 1.5EI, BC = 2EI, CD = EI.

#### Step 1: Overhang CD
Moment at C due to the overhang: $M_C=30\times2=60$ kNm (hogging). This is a known moment acting on joint C.

#### Step 2: Fixed-end moments (clockwise +)
- AB: $FEM_{AB}=-\dfrac{40\times2\times4^2}{6^2}=-35.56$, $FEM_{BA}=+\dfrac{40\times2^2\times4}{6^2}=17.78$ kNm
- BC: $FEM_{BC}=-\dfrac{10\times10^2}{12}=-83.33$, $FEM_{CB}=+83.33$ kNm

No support settles ($\psi=0$) and $\theta_A=0$.

#### Step 3: Slope-deflection equations
- $M_{AB}=\frac{2(1.5EI)}{6}\left(\theta_{B}\right) - 35.56$
- $M_{BA}=\frac{2(1.5EI)}{6}\left(2\theta_{B}\right) + 17.78$
- $M_{BC}=\frac{2(2EI)}{10}\left(2\theta_{B} + \theta_{C}\right) - 83.33$
- $M_{CB}=\frac{2(2EI)}{10}\left(2\theta_{C} + \theta_{B}\right) + 83.33$

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

$$
EI\begin{bmatrix}1.8000 & 0.4000 \\ 0.4000 & 0.8000\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C}\end{Bmatrix}=\begin{Bmatrix}65.556 \\ -23.333\end{Bmatrix}
$$
$$
EI\theta_{B}=48.2639,\quad EI\theta_{C}=-53.2986
$$

#### Step 5: Final end moments (kNm)
| End | $M$ (kNm) | End | $M$ (kNm) |
|---|---|---|---|
| AB | -11.42 | BA | 66.04 |
| BC | -66.04 | CB | 60.00 |
| CD | -60.00 | DC | 0.00 |
(The overhang moment $M_{CD}=-60.00$ kNm.)

#### Bending moment diagram
| Member | BM at first end | BM at second end | Max within member |
|---|---|---|---|
| AB | -11.42 | -66.04 | 23.70 at 2.00 m |
| BC | -66.04 | -60.00 | 62.00 at 5.06 m |
| CD | -60.00 | 0.00 | - |

Reactions: A: $H=0.00$, $V=17.56$, $M=11.42$; B: $H=0.00$, $V=73.04$, $M=0.00$; C: $H=0.00$, $V=79.40$, $M=0.00$.

**Answer: $M_{AB}=-11.42$, $M_{BA}=66.04$, $M_{BC}=-66.04$, $M_{CB}=60.00$ kNm (A fixed); $M_C=60$ kNm hogging; span BC sagging maximum about 62.00 kNm.**

## Check yourself

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

1. What is the moment at C due to the overhang CD with 30 kN at D, 2 m away, and how is it used?
   A. 30 kNm; it is ignored
   B. 60 kNm hogging; it is a known moment on joint C, so M_CD = -60 and M_CB = 60
   C. 60 kNm sagging; M_CB = -60
   D. 15 kNm hogging; shared between CB and CD
   Answer: B. Statics gives M_C = 30 x 2 = 60 kNm hogging. The joint C equation is M_CB + M_CD = 0 with M_CD = -60, so M_CB = 60 kNm. (Key point: Overhang moment)

2. For the 40 kN load 2 m from A on AB (L = 6 m), what is FEM_BA?
   A. -35.56 kNm
   B. +30 kNm
   C. +17.78 kNm
   D. -17.78 kNm
   Answer: C. FEM_BA = +P a^2 b/L^2 = 40 x 4 x 4/36 = +17.78 kNm. FEM_AB is -35.56. (Key point: Fixed-end moment for point load)

3. In the joint B equation, what is the coefficient of theta_B (in EI)?
   A. 1.0
   B. 2.8
   C. 1.8
   D. 0.4
   Answer: C. M_BA contributes (2 x 1.5EI/6) x 2 = 1.0 EI and M_BC contributes (2 x 2EI/10) x 2 = 0.8 EI, giving 1.8 EI. (Key point: Joint B equation)

4. Why does the slope-deflection equations for AB have no theta_A or psi terms?
   A. Because A is a fixed end (theta_A = 0) and the supports do not settle (psi = 0)
   B. Because the load at AB is small
   C. Because AB has the lowest stiffness
   D. Because theta_A is cancelled by the overhang
   Answer: A. A is taken as fixed so theta_A = 0, and with no settlement the chord rotations are zero, leaving only theta_B and the fixed-end moments as unknowns. (Key point: Boundary assumptions)

## Related in Theory of Structures II

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