---
title: "Derive slope deflection equations"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:slope-deflection-equation-derive"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/slope-deflection-equation-derive
site: BCT Engineering
---

# Derive slope deflection equations

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2078 Kartik, 2 marks; 2076 Chaitra, 4 marks; 2076 Asoj, 6 marks; 2073 Shrawan, 5 marks (asked 4 times)
- Video: [Slope Deflection Method | Equation Derivation | Structural Analysis](https://www.youtube.com/watch?v=9Uwx4XWZlNQ) — Being Civil
- 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-equation-derive
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Derive the slope deflection equations for a beam of span L and flexural rigidity EI (continuous beams), mentioning all symbols used. Assume other data if required.

## Short answer

End moments are derived by superimposing fixed-end moments and effects of joint rotations θ and chord rotation ψ = Δ/L. Using moment-area theorems yields M_AB = (2EI/L)(2θ_A + θ_B – 3Δ/L) + FEM_AB and M_BA = (2EI/L)(2θ_B + θ_A – 3Δ/L) + FEM_BA, with clockwise positive sign convention.

## Answer

The **slope-deflection equations** express the end moments of a member in terms of the end rotations, the chord rotation and the loads on the member. They are derived here for a prismatic member AB of length $L$ and flexural rigidity $EI$ (the same derivation applies to every span of a continuous beam).

#### Symbols and sign convention
- $M_{AB}, M_{BA}$: end moments at A and B, **clockwise on the member end = positive**
- $\theta_A,\theta_B$: rotations of the tangents at A and B, clockwise positive
- $\Delta$: relative transverse displacement of B with respect to A (e.g. support settlement); $\psi=\Delta/L$ is the chord rotation, positive clockwise (B moves down relative to A)
- $FEM_{AB}, FEM_{BA}$: fixed-end moments due to the loads on the span (clockwise +)
- Assumptions: linear elastic material, small deformations, axial and shear deformations neglected

#### Derivation (conjugate beam / moment-area)
Treat the member as simply supported at A and B carrying the end moments $M_{AB}$, $M_{BA}$ and the span loads. The end moments are then the sum of the effect of the loads on a fixed-ended member and the effect of the end displacements.

1. **Rotation due to end moments only.** For end moments $M_{AB}$ at A and $M_{BA}$ at B (clockwise), the slopes of the tangent relative to the chord are, by the moment-area theorem,
$$
\theta_A-\psi=\frac{L}{6EI}\left(2M_{AB}-M_{BA}\right),\qquad \theta_B-\psi=\frac{L}{6EI}\left(2M_{BA}-M_{AB}\right)
$$
(the end moments act with opposite sense on the beam, so one is sagging and the other hogging; this is the origin of the minus signs).

2. **Solve for the moments.** Multiply the first equation by 2 and add the second:
$$
2(\theta_A-\psi)+(\theta_B-\psi)=\frac{L}{6EI}\left(3M_{AB}\right)\;\Rightarrow\;M_{AB}=\frac{2EI}{L}\left(2\theta_A+\theta_B-3\psi\right)
$$
In the same way
$$
M_{BA}=\frac{2EI}{L}\left(2\theta_B+\theta_A-3\psi\right)
$$

3. **Add the effect of the loads on the span.** With the span loaded, the member is first held fixed at both ends; this gives the fixed-end moments $FEM_{AB}$ and $FEM_{BA}$. The end rotations and the chord rotation are then applied. By superposition:

$$
\boxed{M_{AB}=\frac{2EI}{L}\left(2\theta_A+\theta_B-\frac{3\Delta}{L}\right)+FEM_{AB}}
$$
$$
\boxed{M_{BA}=\frac{2EI}{L}\left(2\theta_B+\theta_A-\frac{3\Delta}{L}\right)+FEM_{BA}}
$$

#### Standard fixed-end moments (clockwise +)
| Load | $FEM_{AB}$ | $FEM_{BA}$ |
|---|---|---|
| UDL $w$ over the span | $-wL^2/12$ | $+wL^2/12$ |
| Central point load $P$ | $-PL/8$ | $+PL/8$ |
| Point load $P$ at $a$ from A, $b=L-a$ | $-Pab^2/L^2$ | $+Pa^2b/L^2$ |

#### Special cases
- Far end hinged or roller ($M_{BA}=0$): $M_{AB}=\dfrac{3EI}{L}\left(\theta_A-\psi\right)+FEM_{AB}-\dfrac{FEM_{BA}}{2}$
- Fixed end: $\theta=0$ for that end. Rigid support without settlement: $\psi=0$.

#### Use in a continuous beam
Write the two equations for every span, express each end moment in terms of the unknown joint rotations, and apply the joint equilibrium conditions $\sum M=0$ at each support (including any overhang moment). Solve the simultaneous equations for the rotations, then back-substitute to get the final end moments. For frames with sway, one more equation (the shear / storey equilibrium) is added for each independent sway.

## Check yourself

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

1. In the derivation, what is the chord rotation psi, and what is its positive sense?
   A. psi = theta_A + theta_B, anticlockwise positive
   B. psi = Delta/L, clockwise positive when B moves down relative to A
   C. psi = Delta x L, clockwise positive when B moves up
   D. psi = FEM/EI, anticlockwise positive
   Answer: B. psi = Delta/L, the rotation of the chord due to the relative transverse displacement. End moments and rotations are clockwise positive. (Key point: Sign convention and chord rotation)

2. Which equation is the final slope-deflection equation for M_AB?
   A. M_AB = (2EI/L)(theta_A + 2 theta_B - 3Delta/L) + FEM_AB
   B. M_AB = (4EI/L)(theta_A + theta_B) + FEM_AB
   C. M_AB = (2EI/L)(2 theta_A + theta_B - 3Delta/L) + FEM_AB
   D. M_AB = (6EI/L)(2 theta_A + theta_B) - FEM_AB
   Answer: C. Solving the moment-area relations gives M_AB = (2EI/L)(2 theta_A + theta_B - 3 psi), and the fixed-end moment is added by superposition. (Key point: Slope-deflection equation)

3. How are the unknown joint rotations of a continuous beam found after writing the slope-deflection equations?
   A. By setting the shear at each support to zero
   B. By equating the fixed-end moments of adjacent spans
   C. By assuming the rotations equal to the chord rotations
   D. By applying the joint equilibrium equation, sum of moments equal to zero, at each support and solving simultaneously
   Answer: D. Each support gives a moment equilibrium equation in terms of the unknown rotations, which are solved and then back-substituted for the end moments. (Key point: Use in continuous beam)

## Related in Theory of Structures II

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- [Slope deflection: continuous beam, I = 4x10^7 mm^4](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aslope-deflection-beam-i4e7-60kn) — Theory of Structures II, chapter 3
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