---
title: "Slope deflection: B sinks, A rotates"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:slope-deflection-beam-b-sinks-7"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/slope-deflection-beam-b-sinks-7
site: BCT Engineering
---

# Slope deflection: B sinks, A rotates

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2070 Chaitra (old course), 16 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-b-sinks-7
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Analyse the continuous beam shown in figure below by the slope deflection method. Support B sinks by 7.5 mm. Support A rotates by 5° anticlockwise. $E = 5\times10^{5}$ MPa, $I = 3\times10^{7}\ \text{mm}^4$. [Figure: beam ABCD, A fixed; AB = 5 m (1.5I) with 5 kN/m UDL; BC = 2 m + 2 m (2I) with 15 kN at mid-span; CD = 4 m (I) with 5 kN/m UDL; D fixed.]

## Short answer

By slope-deflection method accounting for support settlement and rotation, the end moments are: M_AB = -1461.18 kNm, M_BA = -494.41 kNm, M_BC = 494.41 kNm, M_CB = 124.26 kNm, M_CD = -124.26 kNm, M_DC = -52.13 kNm.

## Answer

**Data:** $E=5\times10^5$ MPa $=5\times10^8$ kN/m², $I=3\times10^7$ mm⁴ $=3\times10^{-5}$ m⁴, so $EI=15000$ kNm². AB = 5 m (1.5I) with 5 kN/m; BC = 4 m (2I) with 15 kN at mid-span; CD = 4 m (I) with 5 kN/m. A and D are fixed; support A rotates by $5^\circ=0.08727$ rad anticlockwise and B sinks 7.5 mm.

#### Support movements
- $\theta_A=-0.08727$ rad (anticlockwise, so negative in the clockwise-positive convention)
- $\psi_{AB}=\dfrac{0.0075}{5}=0.00150$, $\psi_{BC}=-\dfrac{0.0075}{4}=-0.00187$ (C is higher than B), $\psi_{CD}=0$

#### Fixed-end moments (kNm, clockwise +)
- AB: $\mp\dfrac{5\times5^2}{12}=\mp10.42$
- BC: $\mp\dfrac{15\times4}{8}=\mp7.5$
- CD: $\mp\dfrac{5\times4^2}{12}=\mp6.67$

#### Slope-deflection equations
With $\dfrac{2EI}{L}$: AB = 9000, BC = 15000, CD = 7500 kNm/rad.
- $M_{AB}=9000.00\left(-0.17453 + \theta_{B}\right) - 50.92$
- $M_{BA}=9000.00\left(2\theta_{B} - 0.08727\right) - 30.08$
- $M_{BC}=15000.00\left(2\theta_{B} + \theta_{C}\right) + 76.88$
- $M_{CB}=15000.00\left(2\theta_{C} + \theta_{B}\right) + 91.88$
- $M_{CD}=7500.00\left(2\theta_{C}\right) - 6.67$
- $M_{DC}=7500.00\left(\theta_{C}\right) + 6.67$
(The constants include the FEM, the effect of the rotation of A and the chord rotation.)

#### Equilibrium
Joint B: $M_{BA}+M_{BC}=0$; joint C: $M_{CB}+M_{CD}=0$.
$$
\begin{bmatrix}48000.0000 & 15000.0000 \\ 15000.0000 & 45000.0000\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C}\end{Bmatrix}=\begin{Bmatrix}738.606 \\ -85.208\end{Bmatrix}
$$
$$
\theta_{B}=0.0178374,\quad \theta_{C}=-0.00783933
$$

#### Final end moments (kNm)
| End | $M$ (kNm) | End | $M$ (kNm) |
|---|---|---|---|
| AB | -1461.18 | BA | -494.41 |
| BC | 494.41 | CB | 124.26 |
| CD | -124.26 | DC | -52.13 |

Checks:
- Joint B: (-494.41) + (494.41) = 0.00 kNm (applied clockwise moment 0.00) ✓
- Joint C: (124.26) + (-124.26) = 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 | -1461.18 | 494.41 | - |
| BC | 494.41 | -124.26 | - |
| CD | -124.26 | 52.13 | - |

Reactions: A: $H=0.00$, $V=403.62$, $M=1461.18$; B: $H=0.00$, $V=-525.78$, $M=0.00$; C: $H=0.00$, $V=216.26$, $M=0.00$; D: $H=0.00$, $V=-34.10$, $M=52.13$.

The rotation of 5° of support A is very large for a 5 m span, so the end moments are dominated by the support movements.

**Answer: $M_{AB}=-1461.18$, $M_{BA}=-494.41$, $M_{BC}=494.41$, $M_{CB}=124.26$, $M_{CD}=-124.26$, $M_{DC}=-52.13$ kNm.**

## Check yourself

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

1. Support A rotates 5 degrees anticlockwise. With clockwise taken as positive, how is theta_A entered?
   A. +0.08727 rad
   B. -0.08727 rad
   C. 0 because A is fixed
   D. -5 rad
   Answer: B. Anticlockwise is negative in the clockwise-positive convention, and 5 degrees = 0.08727 rad, so theta_A = -0.08727 rad. (Key point: Sign of support rotation)

2. B sinks 7.5 mm. What are the chord rotations of AB (5 m) and BC (4 m, C higher than B)?
   A. psi_AB = -0.0015 and psi_BC = +0.00187
   B. psi_AB = +0.0015 and psi_BC = -0.00187
   C. Both +0.0015
   D. Both zero, since the supports only rotate
   Answer: B. Sinking of B turns AB clockwise (0.0075/5 = +0.0015); relative to B, C is higher so BC rotates the other way (-0.0075/4 = -0.00187). (Key point: Chord rotation from settlement)

3. Which conditions give the equations for theta_B and theta_C?
   A. M_AB = 0 and M_DC = 0
   B. M_BA + M_BC = 0 and M_CB + M_CD = 0
   C. M_AB + M_BA = 0 and M_CB + M_DC = 0
   D. Sum of all vertical forces = 0 only
   Answer: B. No external moment is applied at joints B and C, so the end moments of the members meeting there must sum to zero. (Key point: Equilibrium equations for unknowns)

4. Why are the end moments so large (for example M_AB = -1461 kNm)?
   A. The UDL on AB is very large
   B. The 5 degree rotation of A and the settlement dominate, because the rotation is very large for a 5 m span
   C. EI was taken in the wrong units
   D. Because A is fixed, carry-over doubles the moment
   Answer: B. A rotation of 0.087 rad on a 5 m span with EI = 15000 kNm^2 produces moments far larger than those from the loads. (Key point: Effect of large support movement)

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