---
title: "Slope deflection: B sinks, C rotates"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:slope-deflection-beam-4t-2tm-b"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/slope-deflection-beam-4t-2tm-b
site: BCT Engineering
---

# Slope deflection: B sinks, C rotates

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2066 Bhadra, 10 marks
- Video: [Rotation of Support A , Slope Deflection Method for Beam Analysis - Problem No 17](https://www.youtube.com/watch?v=0e-OyME20xo) — 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-4t-2tm-b
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Determine the member end moments using the slope deflection method and draw BMD and SFD for the beam loaded as shown in figure given below. Support B settles down by 5 mm and support C rotates clockwise by 0.02 radian and $EI = 20\ \text{t/mm}^2$. [Figure: beam A-B-C, A fixed, B support, C fixed; 4 t point load at 1 m from A; AB = 1 m + 3 m (I); BC = 6 m (2I) with 2 t/m UDL.]

## Short answer

Using slope-deflection with given settlement and rotation, equilibrium gives θB = -0.0055125 rad. For EI=20000 t m², end moments: AB=-94.88, BA=-147.00, BC=147.00, CB=499.17 tm. General: M=Mload+EI×coefficient.

## Answer

**Data and assumptions:** A is fixed; AB = 1 m + 3 m = 4 m (I) with 4 t at 1 m from A; BC = 6 m ($2I$) with 2 t/m; C is fixed. Support B settles 5 mm and support C rotates 0.02 rad clockwise. The value 'EI = 20 t/mm²' gives only E (20 t/mm² $=2\times10^7$ t/m²); I is not given. The result is therefore written as: moment = (load part, independent of EI) + EI × (settlement/rotation part), and a numerical example with $I=1\times10^{-3}$ m⁴ ($EI=20\,000$ t m²) is given. Units: tonne, metre.

#### Step 1: Chord rotations and support rotation
- $\psi_{AB}=\dfrac{0.005}{4}=0.001250$ (B moves down, chord turns clockwise)
- $\psi_{BC}=-\dfrac{0.005}{6}=-0.000833$ (C is higher than B, chord turns anticlockwise)
- $\theta_A=0$, $\theta_C=+0.02$ rad (clockwise)

#### Step 2: Fixed-end moments (tm, clockwise +)
- AB ($a=1$, $b=3$): $FEM_{AB}=-\dfrac{4\times1\times3^2}{4^2}=-2.250$, $FEM_{BA}=+\dfrac{4\times1^2\times3}{4^2}=0.750$
- BC: $\mp\dfrac{2\times6^2}{12}=\mp6.000$

#### Step 3: Slope-deflection equations (numerical example, $EI=20\,000$ t m²)
- $M_{AB}=10000.00\left(\theta_{B}\right) - 39.75$
- $M_{BA}=10000.00\left(2\theta_{B}\right) - 36.75$
- $M_{BC}=13333.33\left(2\theta_{B} + 0.02000\right) + 27.33$
- $M_{CB}=13333.33\left(0.04000 + \theta_{B}\right) + 39.33$

#### Step 4: Equilibrium at B
$M_{BA}+M_{BC}=0$ gives one equation in $\theta_B$:
$$
\begin{bmatrix}46666.6667\end{bmatrix}\begin{Bmatrix}\theta_{B}\end{Bmatrix}=\begin{Bmatrix}-257.250\end{Bmatrix}
$$
$$
\theta_{B}=-0.0055125
$$

#### Step 5: End moments
| End | Loads only | Movements, per unit EI | Total (EI = 20 000 t m²) |
|---|---|---|---|
| AB | -1.125 | -0.004688 | -94.875 |
| BA | 3.000 | -0.007500 | -147.000 |
| BC | -3.000 | 0.007500 | 147.000 |
| CB | 7.500 | 0.024583 | 499.167 |

(For any EI: $M=M_{load}+EI\times(\text{column 3})$.)

Check: $M_{BA}+M_{BC}=0.000$ ✓.

#### SFD and BMD (EI = 20 000 t m²)
| Member | BM at first end | BM at second end | Max within member |
|---|---|---|---|
| AB | -94.88 | 147.00 | - |
| BC | 147.00 | -499.17 | - |
Shears (t): AB: 63.47 at A, 59.47 after the 4 t load; BC: -101.69 at B falling at 2 t/m to -113.69 at C. Reactions: A: $H=0.00$, $V=63.47$, $M=94.88$; B: $H=0.00$, $V=-161.16$, $M=0.00$; C: $H=0.00$, $V=113.69$, $M=-499.17$.

**Answer: for $EI=20\,000$ t m²: $M_{AB}=-94.875$, $M_{BA}=-147.000$, $M_{BC}=147.000$, $M_{CB}=499.167$ tm; in general $M=M_{load}+EI\times(\text{movement coefficient})$.**

## Check yourself

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

1. B settles 5 mm and C stays level. What is psi_BC for BC (6 m)?
   A. +0.000833
   B. -0.000833 (C is higher than B, chord turns anticlockwise)
   C. -0.00125
   D. 0
   Answer: B. C is 5 mm above B, so the chord rotates anticlockwise: psi = -0.005/6. (Key point: Chord rotation of BC)

2. EI = 20 t/mm^2 gives only E, not I. How is the answer presented?
   A. Assume I = 1
   B. Ignore settlement effects
   C. As moment = load part (independent of EI) + EI x (settlement/rotation part)
   D. Use only the UDL part
   Answer: C. Load FEMs do not depend on EI while settlement and rotation moments scale with EI, so the result is split accordingly. (Key point: Handling missing I)

3. FEM_AB for the 4 t load at 1 m from A (L = 4 m) is...
   A. -PL/8 = -2
   B. -Pab^2/L^2 = -2.25
   C. -Pa^2b/L^2 = -0.75
   D. -wL^2/12
   Answer: B. At A the FEM is -Pab^2/L^2 = -4x1x9/16 = -2.25 tm. (Key point: FEM of point load)

4. What gives the single unknown theta_B?
   A. M_AB = 0
   B. M_CB = 0
   C. M_BA + M_BC = 0 with theta_A = 0 and theta_C = 0.02 rad known
   D. Shear equilibrium of the beam
   Answer: C. A is fixed and C's rotation is given, so only theta_B is unknown, found from moment equilibrium at B. (Key point: Unknowns and equilibrium)

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