---
title: "Slope deflection: beam with settlement of C"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:slope-deflection-settlement-c-4mm"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/slope-deflection-settlement-c-4mm
site: BCT Engineering
---

# Slope deflection: beam with settlement of C

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2075 Asoj, 10 marks
- Video: [Slope Deflection Method for Beam Analysis - Problem No 18 ( Support C Sinking )](https://www.youtube.com/watch?v=4d4nqQmX0ds) — 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-settlement-c-4mm
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Analyse the continuous beam loaded as shown in figure below using the slope deflection method considering settlement of support C by 4 mm downward. Take $EI = 1\times10^{9}\ \text{t mm}^2$. [Figure: continuous beam ABCD, D fixed; 2 t point load near A; 3 t/m UDL on BC (3I); 5 t point load on CD (I); dimensions 2 m, 1.5 m, 2 m, 4 m, 3 m and 1 m as marked.]

## Short answer

Using slope-deflection with chord rotations from 4 mm settlement, fixed-end moments are computed and joint equilibrium equations solved. Final end moments: M_AB=0, M_BA=3.21, M_BC=-3.21, M_CB=0.27, M_CD=-0.27, M_DC=3.90 tm.

## Answer

**Assumptions (dimensions are not fully legible):** A is a pin (free to rotate), AB = 3.5 m with the 2 t load 2 m from A; BC = 4 m (3I) with 3 t/m; CD = 4 m (I) with the 5 t load 3 m from C (1 m from D); D is fixed; B and C are simple supports. Units: tonne, metre. $EI=1\times10^9\ \text{t mm}^2=1000$ t m². Support C settles $\delta=4$ mm = 0.004 m.

#### Chord rotations (clockwise +, a member whose right end sinks rotates clockwise)
$\psi_{BC}=\dfrac{0.004}{4}=0.0010$ (C goes down relative to B), $\psi_{CD}=-\dfrac{0.004}{4}=-0.0010$ (D is higher than C). $\psi_{AB}=0$.

#### Fixed-end moments (tm, clockwise +)
- AB: $FEM_{AB}=-\dfrac{2\times2\times1.5^2}{3.5^2}=-0.735$, $FEM_{BA}=+\dfrac{2\times2^2\times1.5}{3.5^2}=0.980$
- BC: $\mp\dfrac{3\times4^2}{12}=\mp4.000$
- CD: $FEM_{CD}=-\dfrac{5\times3\times1^2}{4^2}=-0.938$, $FEM_{DC}=+\dfrac{5\times3^2\times1}{4^2}=2.812$

#### Slope-deflection equations ($M_{ij}=\tfrac{2EI}{L}(2\theta_i+\theta_j-3\psi)+FEM_{ij}$, numerical)
- $M_{AB}=571.43\left(2\theta_{A} + \theta_{B}\right) - 0.73$
- $M_{BA}=571.43\left(2\theta_{B} + \theta_{A}\right) + 0.98$
- $M_{BC}=1500.00\left(2\theta_{B} + \theta_{C}\right) - 8.50$
- $M_{CB}=1500.00\left(2\theta_{C} + \theta_{B}\right) - 0.50$
- $M_{CD}=500.00\left(2\theta_{C}\right) + 0.56$
- $M_{DC}=500.00\left(\theta_{C}\right) + 4.31$

#### Joint equilibrium
$M_{AB}=0$; $M_{BA}+M_{BC}=0$; $M_{CB}+M_{CD}=0$ (θ in radians)
$$
\begin{bmatrix}1142.8571 & 571.4286 & 0.0000 \\ 571.4286 & 4142.8571 & 1500.0000 \\ 0.0000 & 1500.0000 & 4000.0000\end{bmatrix}\begin{Bmatrix}\theta_{A} \\ \theta_{B} \\ \theta_{C}\end{Bmatrix}=\begin{Bmatrix}0.735 \\ 7.520 \\ -0.062\end{Bmatrix}
$$
$$
\theta_{A}=-0.000446259,\quad \theta_{B}=0.00217823,\quad \theta_{C}=-0.000832462
$$

#### Final end moments (tm, clockwise on the member end +)
| End | $M$ (tm) | End | $M$ (tm) |
|---|---|---|---|
| AB | 0.00 | BA | 3.21 |
| BC | -3.21 | CB | 0.27 |
| CD | -0.27 | DC | 3.90 |

Checks:
- Pin A: $M_{AB}=0.00$ tm ✓
- Joint B: (3.21) + (-3.21) = 0.00 tm ✓
- Joint C: (0.27) + (-0.27) = 0.00 tm ✓

#### Bending moments (sagging +, tm)
| Member | BM at first end | BM at second end | Max within member |
|---|---|---|---|
| AB | 0.00 | -3.21 | -0.12 at 2.00 m (under the 2 t load) |
| BC | -3.21 | -0.27 | 4.35 at 2.25 m |
| CD | -0.27 | -3.90 | 0.76 at 3.00 m |

Reactions (t, tm): A: $H=0.00$, $V=-0.06$, $M=0.00$; B: $H=0.00$, $V=8.80$, $M=0.00$; C: $H=0.00$, $V=5.61$, $M=0.00$; D: $H=0.00$, $V=4.66$, $M=-3.90$.

**Answer: $M_{BA}=3.214$, $M_{BC}=-3.214$, $M_{CB}=0.270$, $M_{CD}=-0.270$, $M_{DC}=3.896$ tm; $M_{AB}=0$.**

## Check yourself

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

1. The chord rotation of BC when C settles 4 mm over 4 m is:
   A. 0.004
   B. 0.001 clockwise
   C. 0.016
   D. 0.001 anticlockwise
   Answer: B. ψ = δ/L = 0.004/4 = 0.001; the right end sinking rotates the member clockwise. (Key point: Chord rotation)

2. Why is ψCD = -0.001?
   A. CD has smaller EI
   B. D is fixed
   C. D is higher than C relative to the chord, giving anticlockwise rotation
   D. The load is larger
   Answer: C. C sinks relative to D, so CD rotates opposite to BC. (Key point: CD rotation sign)

3. EI = 1 x 10^9 t mm^2 equals:
   A. 10^9 t m^2
   B. 10^6 t m^2
   C. 1 t m^2
   D. 1000 t m^2
   Answer: D. 1 mm^2 = 10^-6 m^2, so 10^9 t mm^2 = 1000 t m^2. (Key point: Units)

4. What does pin A add?
   A. MAB = 0, so θA is an extra unknown
   B. Sway
   C. A fixed-end moment
   D. Nothing
   Answer: A. A pin is free to rotate and carries no moment. (Key point: Pin A)

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