---
title: "Slope deflection: beam with settlement of A and C"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:slope-deflection-beam-60kn-2tm-sinking"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/slope-deflection-beam-60kn-2tm-sinking
site: BCT Engineering
---

# Slope deflection: beam with settlement of A and C

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2068 Baishakh, 10 marks
- Video: [Slope Deflection Method for Beam Analysis - Problem No 13 ( Sinking / Settlement of Support B )](https://www.youtube.com/watch?v=HHJ7qNxRuw4) — 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-60kn-2tm-sinking
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Using the slope and deflection method, find support moments and draw bending moment diagram for the given beam. Support A sinks by 1 cm; support C sinks by 1.5 cm. Take $E = 2\times10^{5}$ MPa, $I = 10{,}000\ \text{cm}^4$. [Figure: beam ABCD, A fixed; 60 kN at 2 m from A; AB = 5 m (EI); BC = 6 m (2EI) with 2 t/m UDL; CD = 2 m (EI).]

## Short answer

Use slope-deflection with support settlements: psi_AB=-0.002, psi_BC=0.0025, EI=20000 kNm^2. Joint equilibrium yields theta_B=0.001708, theta_C=0.000689 rad. Final moments: M_AB=18.46, M_BA=104.13, M_BC=-104.13, M_CB=0 kNm; maximum sagging in BC is 43.90 kNm.

## Answer

**Data and assumptions:** A is fixed; AB = 5 m (EI) with 60 kN at 2 m from A; BC = 6 m (2EI) with 2 t/m $=2\times9.81=19.62$ kN/m; CD = 2 m (EI) is an unloaded overhang (it carries no moment). B and C are supports. Support A sinks 1 cm (without rotating) and C sinks 1.5 cm. $E=2\times10^5$ MPa $=2\times10^8$ kN/m², $I=10\,000$ cm⁴ $=10^{-4}$ m⁴, so $EI=20000$ kNm².

#### Chord rotations (clockwise +)
- $\psi_{AB}=\dfrac{\delta_B-\delta_A}{L}=\dfrac{0-(-0.010)}{5}\ \text{(B is higher than A by 10 mm)}\Rightarrow\psi_{AB}=-0.00200$ (anticlockwise)
- $\psi_{BC}=\dfrac{0.015}{6}=0.00250$ (C sinks relative to B, clockwise)
- Overhang CD: no moment.

#### Fixed-end moments (kNm, clockwise +)
- AB ($a=2$, $b=3$): $FEM_{AB}=-\dfrac{60\times2\times3^2}{5^2}=-43.20$, $FEM_{BA}=+\dfrac{60\times2^2\times3}{5^2}=28.80$
- BC: $\mp\dfrac{19.62\times6^2}{12}=\mp58.86$

#### Slope-deflection equations ($\theta_A=0$)
- $M_{AB}=8000.00\left(\theta_{B}\right) + 4.80$
- $M_{BA}=8000.00\left(2\theta_{B}\right) + 76.80$
- $M_{BC}=13333.33\left(2\theta_{B} + \theta_{C}\right) - 158.86$
- $M_{CB}=13333.33\left(2\theta_{C} + \theta_{B}\right) - 41.14$

#### Equilibrium
Joint B: $M_{BA}+M_{BC}=0$; joint C: $M_{CB}=0$ (the overhang CD is unloaded, so there is no moment from it).
$$
\begin{bmatrix}42666.6667 & 13333.3333 \\ 13333.3333 & 26666.6667\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C}\end{Bmatrix}=\begin{Bmatrix}82.060 \\ 41.140\end{Bmatrix}
$$
$$
\theta_{B}=0.00170806,\quad \theta_{C}=0.000688722
$$

#### Final end moments (kNm) and BMD
| End | $M$ (kNm) | End | $M$ (kNm) |
|---|---|---|---|
| AB | 18.46 | BA | 104.13 |
| BC | -104.13 | CB | 0.00 |
| CD | 0.00 | DC | 0.00 |

Checks: Joint B: $104.13+(-104.13)=0.00$ ✓; joint C: $0.00=0.00$ ✓.

| Member | BM at first end | BM at second end | Max within member |
|---|---|---|---|
| AB | 18.46 | -104.13 | 41.43 at 2.00 m |
| BC | -104.13 | 0.00 | 43.90 at 3.88 m |
| CD | 0.00 | 0.00 | - |

Reactions (kN, kNm): A: $H=0.00$, $V=11.48$, $M=-18.46$; B: $H=0.00$, $V=124.73$, $M=0.00$; C: $H=0.00$, $V=41.51$, $M=0.00$.

**Answer: $M_{AB}=18.46$, $M_{BA}=104.13$, $M_{BC}=-104.13$ kNm and $M_{CB}=0$ (support moment at C is zero since CD carries no load); BC sagging maximum 43.90 kNm.**

## Check yourself

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

1. Support A sinks 1 cm and B does not. What is psi_AB?
   A. -0.002 (anticlockwise), since B is higher than A by 10 mm over 5 m
   B. +0.002 (clockwise)
   C. +0.001
   D. Zero since A is fixed
   Answer: A. B is above A, so the chord rotates anticlockwise: psi = -0.010/5 = -0.002. (Key point: Chord rotation sign)

2. Why is the moment at C from CD zero and joint C condition M_CB = 0?
   A. CD is unloaded overhang so it carries no moment
   B. C is a fixed support
   C. CD is very stiff
   D. Because C settles
   Answer: A. An unloaded free overhang has no moment, so the end moment M_CB of span BC must be zero. (Key point: Overhang CD)

3. Which equation expresses equilibrium at joint B?
   A. M_BA + M_BC = 0
   B. M_AB + M_BA = 0
   C. M_BC = M_CB
   D. M_BA = M_BC
   Answer: A. No external moment acts at B, so the member end moments there sum to zero. (Key point: Equilibrium at B)

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