---
title: "Slope deflection: D rotates, C settles 4 mm"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:slope-deflection-beam-25kn-d-rotation"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/slope-deflection-beam-25kn-d-rotation
site: BCT Engineering
---

# Slope deflection: D rotates, C settles 4 mm

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2068 Chaitra, 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-25kn-d-rotation
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Analyse the beam shown in figure below by the slope deflection method. Draw BM diagram considering given external loading and rotation of support D by (1/10) clockwise, support C settles down by 4 mm. [Figure: beam ABCD, D fixed; 25 kN at the free end A; AB = 3 m (EI); BC = 2 m + 2 m + 2 m (2.5EI) with 50 kN and 40 kN point loads and 10 kN/m UDL on the right part of BC; CD = 3 m + 3 m (3EI).]

## Short answer

Slope-deflection with support settlement and rotation gives M_BA=75, M_BC=-75, M_CB=-18.43, M_CD=18.43, M_DC=117.75 kNm for EI=30,000 kNm². General form includes load and movement terms.

## Answer

**Data and assumptions:** the figure is only described. A is the free end of the cantilever AB (3 m, 25 kN at A); BC = 6 m (2.5EI) carries 50 kN at 2 m and 40 kN at 4 m from B and 10 kN/m over the last 2 m; CD = 6 m (3EI) is unloaded; B and C are supports; D is fixed. Support C settles 4 mm; support D rotates $\tfrac{1}{10}$ degree $=0.001745$ rad clockwise (the unit is taken as degrees). EI is not given, so the answer is split into the load part (independent of EI) and the support-movement part (proportional to EI), and a numerical example with $EI=30\,000$ kNm² is worked out.

#### Step 1: Overhang AB
$M_B=25\times3=75$ kNm (hogging), known moment at B.

#### Step 2: Fixed-end moments (kNm, clockwise +)
BC: 50 kN at $a=2$ ($b=4$), 40 kN at $a=4$ ($b=2$) and UDL on the last 2 m:
$FEM_{BC}=-65.556$, $FEM_{CB}=70.000$ (sum of the point-load formulas $\mp Pab^2/L^2$, $\pm Pa^2b/L^2$ and the integrated UDL contribution). CD: 0.

#### Step 3: Support movement
- $\psi_{BC}=+\dfrac{0.004}{6}=0.000667$ (C moves down relative to B, so the chord of BC turns clockwise) and $\psi_{CD}=-\dfrac{0.004}{6}=-0.000667$ (D is higher than C, so the chord of CD turns anticlockwise); $\psi_{AB}$ does not enter (overhang).
- $\theta_D=+0.001745$ rad (clockwise) appears in $M_{CD}$ and $M_{DC}$.

#### Step 4: Slope-deflection equations ($EI=30\,000$ kNm², numerical)
- $M_{BC}=25000.00\left(2\theta_{B} + \theta_{C}\right) - 115.56$
- $M_{CB}=25000.00\left(2\theta_{C} + \theta_{B}\right) + 20.00$
- $M_{CD}=30000.00\left(2\theta_{C} + 0.00175\right) + 60.00$
- $M_{DC}=30000.00\left(0.00349 + \theta_{C}\right) + 60.00$

#### Step 5: Equilibrium and solution
Joint B: $M_{BC}=-M_{BA}$ with $M_{BA}=+75$, so $M_{BC}=-75$. Joint C: $M_{CB}+M_{CD}=0$.
$$
\begin{bmatrix}50000.0000 & 25000.0000 \\ 25000.0000 & 110000.0000\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C}\end{Bmatrix}=\begin{Bmatrix}40.556 \\ -132.360\end{Bmatrix}
$$
$$
\theta_{B}=0.00159387,\quad \theta_{C}=-0.00156551
$$

#### Step 6: Final end moments (kNm)
| End | Loads only (EI-independent) | Support movements, per unit EI | Total for EI = 30 000 |
|---|---|---|---|
| BA | 75.00 | -0.000000 | 75.00 |
| BC | -75.00 | -0.000000 | -75.00 |
| CB | 40.17 | -0.001953 | -18.43 |
| CD | -40.17 | 0.001953 | 18.43 |
| DC | -20.09 | 0.004595 | 117.75 |

(The middle column multiplied by EI gives the effect of the movements for any EI: $M=M_{load}+EI\times(\text{column 3})$.)

Check: joint B: $75.00+(-75.00)=0$ ✓; joint C: $-18.43+(18.43)=0$ ✓.

#### Bending moment diagram (EI = 30 000)
| Member | BM at first end | BM at second end | Max within member |
|---|---|---|---|
| AB | 0.00 | -75.00 | - |
| BC | -75.00 | 18.43 | 87.29 at 4.00 m |
| CD | 18.43 | -117.75 | - |

Reactions: B: $H=0.00$, $V=90.57$, $M=0.00$; C: $H=0.00$, $V=21.73$, $M=0.00$; D: $H=0.00$, $V=22.70$, $M=-117.75$.

**Answer: for $EI=30\,000$ kNm²: $M_{BA}=75.00$, $M_{BC}=-75.00$, $M_{CB}=-18.43$, $M_{CD}=18.43$, $M_{DC}=117.75$ kNm; 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. How is the overhang AB (25 kN at free end A, 3 m) treated?
   A. It is included in the slope-deflection equations with its own rotation
   B. It gives a known moment M_B = 25x3 = 75 kNm (hogging) applied at joint B
   C. It is ignored because A is free
   D. It gives M_B = 25/3 kNm
   Answer: B. The cantilever moment is statically determinate, so it is a known external moment at B. (Key point: Overhang moment)

2. Support C settles 4 mm. What chord rotations result for BC and CD (L = 6 m each)?
   A. psi_BC = +0.000667 (clockwise) and psi_CD = -0.000667 (anticlockwise)
   B. Both +0.000667
   C. psi_BC = -0.000667 and psi_CD = +0.000667
   D. Both zero
   Answer: A. C moves down relative to B (clockwise chord) and up relative to... D being higher than C turns CD anticlockwise. (Key point: Chord rotation from settlement)

3. EI is not given. How does the answer handle this?
   A. It assumes EI = 1
   B. It splits moments into a load part (independent of EI) and a movement part proportional to EI
   C. It ignores support movements
   D. It leaves the answer without numbers
   Answer: B. M = M_load + EI x (movement coefficient); a numerical case with EI = 30000 kNm^2 is worked. (Key point: Result structure with unknown EI)

4. Which equilibrium conditions are used to solve for theta_B and theta_C?
   A. M_BC = -75 at B (since M_BA = +75) and M_CB + M_CD = 0 at C
   B. M_BC = 0 and M_CB = 0
   C. M_BC + M_CB = 0
   D. M_CD = M_DC
   Answer: A. At B the span moment must balance the overhang moment; at C the two member moments sum to zero. (Key point: Joint equilibrium conditions)

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