Derive slope deflection equations

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 +)

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.