---
title: "Slope deflection: beam with rotations and settlements"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:slope-direction-beam-fbd-bmd-sfd"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/slope-direction-beam-fbd-bmd-sfd
site: BCT Engineering
---

# Slope deflection: beam with rotations and settlements

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2065 Shrawan, 14 marks
- Video: [SA32: Slope-Deflection Method and Support Settlement](https://www.youtube.com/watch?v=dwXudm5IPDE) — Dr. Structure
- 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-direction-beam-fbd-bmd-sfd
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Use the slope-direction (slope deflection) method to analyze the continuous beam shown in the figure. Draw free body diagram, BMD and SFD. The support "a" rotates by 0.001 radian clockwise and support "b" rotates by 0.001 radian anticlockwise. Support "b" and "c" both settle down by 10 mm. [Figure: beam a-b-c-d, a and d fixed; 20 kN at 2 m from a (ab = 2 m + 2 m, 3EI); bc = 5 m (4EI) with 2 kN/m UDL; cd = 2 m + 2 m (3EI) with 20 kN at mid-span.]

## Short answer

Using slope-deflection with given support rotations and settlements, joint rotations are θ_b = 0.001993 rad, θ_c = -0.001993 rad. Final end moments: M_ab = -62.61, M_ba = -27.72, M_bc = 27.72, M_cb = -27.72, M_cd = 27.72, M_dc = 62.61 kN·m.

## Answer

#### Assumptions
- $EI$ is not given. Take $EI = 10{,}000$ kN·m² (the moments caused by the support rotation and settlement scale with $EI$; load moments do not).
- "Support b rotates" is read as a typing slip for the other fixed support d (b is a free interior joint), so the prescribed rotations are $\theta_a = 0.001$ rad clockwise and $\theta_d = 0.001$ rad anticlockwise. The joints $b$ and $c$ rotate freely and settle 10 mm.
- Spans: ab = 4 m ($3EI$), bc = 5 m ($4EI$), cd = 4 m ($3EI$). Clockwise end moments are positive.

#### Data
- Chord rotations (clockwise +): $\psi_{ab} = 0.01/4 = +0.0025$, $\psi_{bc} = 0$, $\psi_{cd} = -0.01/4 = -0.0025$.
- FEM: ab: $\pm PL/8 = \pm 10$; bc: $\pm wL^2/12 = \pm 2\times 25/12 = \pm 4.1667$; cd: $\pm 10$.
- $\theta_a = +0.001$, $\theta_d = -0.001$; unknowns $\theta_b$, $\theta_c$.

#### Slope-deflection equations
$M_{near} = FEM + \dfrac{2EI}{L}(2\theta_{near} + \theta_{far} - 3\psi)$

$$
\begin{aligned}
M_{ab} &= -10 + 15000(2\times0.001 + \theta_b - 0.0075) = -92.5 + 15000\,\theta_b \\
M_{ba} &= +10 + 15000(2\theta_b + 0.001 - 0.0075) = -87.5 + 30000\,\theta_b \\
M_{bc} &= -4.1667 + 16000(2\theta_b + \theta_c) \\
M_{cb} &= +4.1667 + 16000(2\theta_c + \theta_b) \\
M_{cd} &= -10 + 15000(2\theta_c - 0.001 + 0.0075) = 87.5 + 30000\,\theta_c \\
M_{dc} &= +10 + 15000(-0.002 + \theta_c + 0.0075) = 92.5 + 15000\,\theta_c
\end{aligned}
$$

#### Joint equilibrium
$$
\begin{aligned}
M_{ba} + M_{bc} = 0 &: \; 62000\,\theta_b + 16000\,\theta_c = 91.667 \\
M_{cb} + M_{cd} = 0 &: \; 16000\,\theta_b + 62000\,\theta_c = -91.667
\end{aligned}
$$

Solving: $\theta_b = +0.0019928$ rad (clockwise), $\theta_c = -0.0019928$ rad.

#### Final end moments (kN·m)
| End | Moment | Meaning |
|---|---|---|
| $M_{ab}$ | -62.61 | hogging (tension top) at a |
| $M_{ba}$ | -27.72 | sagging (tension bottom) at b |
| $M_{bc}$ | +27.72 | sagging at b |
| $M_{cb}$ | -27.72 | sagging at c |
| $M_{cd}$ | +27.72 | sagging at c |
| $M_{dc}$ | +62.61 | hogging at d |

The settlement of b and c turns the support moments into sagging moments. The result is symmetrical about the mid-point of bc, as the loading, rotations and settlements are.

#### Free body diagrams: shears and reactions
Span-wise (shear at left end, taking moments of each span):
- ab: $V_{a} = 32.58$ kN up; after the 20 kN load $V = 12.58$ kN; $V_b$(left) = 12.58 kN.
- bc: $V_b$(right) = +5.00 kN, falling at 2 kN/m to $-5.00$ kN at c.
- cd: $V_c$ = -12.58 kN; after the 20 kN load -32.58 kN at d.

Reactions: $R_a = 32.58$ kN up with $M_a = 62.61$ kN·m anticlockwise; $R_b = 7.58$ kN **down**; $R_c = 7.58$ kN **down**; $R_d = 32.58$ kN up with $M_d = 62.61$ kN·m clockwise.
Check: $32.58 + 32.58 - 7.58 - 7.58 = 50 = 20+20+2\times5$.

#### BMD and SFD values
| Section | BM (kN·m) | SF (kN) |
|---|---|---|
| a | -62.61 | +32.58 |
| 2 m from a (load) | +2.55 | +32.58 / +12.58 |
| b | +27.72 | +12.58 / +5.00 |
| mid of bc (2.5 m) | +33.97 | 0 |
| c | +27.72 | -5.00 / -12.58 |
| 2 m from d (load) | +2.55 | -12.58 / -32.58 |
| d | -62.61 | -32.58 |

BMD is linear in ab and cd between the loads, parabolic in bc ($M = 27.72 + 5x - x^2$, $x$ from b); SFD is stepped in ab and cd and a straight line through zero at mid-bc.

**Answer:** $\theta_b = 0.001993$ rad, $\theta_c = -0.001993$ rad; $M_{ab} = -62.61$, $M_{ba} = -27.72$, $M_{bc} = 27.72$, $M_{cb} = -27.72$, $M_{cd} = 27.72$, $M_{dc} = 62.61$ kN·m.

## Check yourself

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

1. Supports b and c settle 10 mm; ab = 4 m. What are psi_ab and psi_cd?
   A. +0.0025 and +0.0025
   B. -0.0025 and +0.0025
   C. +0.0025 and -0.0025
   D. 0 and 0
   Answer: C. b goes down relative to a (clockwise, +); c goes down relative to d (anticlockwise, -). psi_bc = 0 as b and c settle equally. (Key point: Chord rotations)

2. Which equations determine theta_b and theta_c?
   A. M_ab = 0 and M_dc = 0
   B. R_b = R_c
   C. M_ba + M_bc = 0 and M_cb + M_cd = 0
   D. Sum of vertical forces = 50 kN
   Answer: C. Joints b and c are free with no applied moment, so end moments sum to zero at each. (Key point: Joint equilibrium)

3. What does the settlement of b and c do to the support moments?
   A. Nothing, since EI is not given
   B. Makes all moments hogging
   C. Doubles the load FEMs
   D. Turns the moments at b and c to sagging, giving a symmetrical result; this part scales with EI
   Answer: D. Settlement and rotation moments scale with EI (assumed 10000) and here produce sagging +27.72 at b and c. (Key point: Effect of settlement)

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