Slope Deflection Method Notes

Last reviewed 16 Sept 2026 · 5 min read

The displacement approach

In the slope deflection method (G. A. Maney, 1915) the unknowns are joint displacements — rotations of rigid joints and, in sway frames, lateral translations. Member end moments are written in terms of these displacements, and joint equilibrium equations are solved for them. It is the basis of moment distribution and of the stiffness matrix method.

It is convenient when the kinematic indeterminacy $D_{k}$ is small, even if the static indeterminacy is large (a fixed beam with many spans, a multi-bay frame).

Assumptions: members are prismatic and linearly elastic; axial and shear deformations are neglected; joints are rigid (all members at a joint rotate by the same angle).

Sign convention

Moments at member ends: clockwise positive (acting on the member).
Rotations $θ$ of joints: clockwise positive.
Chord rotation $ψ = Δ/ L$ : clockwise positive (the far end moving so that the chord rotates clockwise).

Slope-deflection equations

For member AB of length $L$ and flexural rigidity $E I$ :

FormulaSlope-deflection equations

M_{A B} = M_{F A B} + \frac{2 E I}{L} (2 θ_{A} + θ_{B} - 3 ψ)

M_{B A} = M_{F B A} + \frac{2 E I}{L} (2 θ_{B} + θ_{A} - 3 ψ)

$M_{F A B}$ , $M_{F B A}$ = fixed-end moments due to loads on the span (clockwise positive: e.g. UDL gives $- w L^{2} /12$ at A and $+ w L^{2} /12$ at B); $ψ = Δ/ L$ with $Δ$ the relative transverse displacement of B with respect to A.

Where the equations come from

Superpose on a fixed-ended member: (1) the loads with ends fixed → fixed-end moments; (2) rotation $θ_{A}$ with B fixed → $4 E I θ_{A} / L$ at A and $2 E I θ_{A} / L$ at B (carry-over factor ½); (3) rotation $θ_{B}$ similarly; (4) relative translation $Δ$ with no rotation → $- 6 E I Δ/ L^{2}$ at both ends.

Modified equation for a pinned far end

If end B is a pin or roller (so $M_{B A} = 0$ ), eliminate $θ_{B}$ :

M_{A B} = (M_{F A B} - \frac{M _{F B A}}{2}) + \frac{3 E I}{L} (θ_{A} - ψ)

This saves one unknown per pinned end.

Procedure

Identify unknown joint rotations (and sway $Δ$ if the frame can sway). Fixed supports have $θ = 0$ .
Compute fixed-end moments for every loaded member.
Write slope-deflection equations for every member end.
Joint equilibrium: at each rigid joint with unknown rotation, $\sum M_{m e mb er e n d s} =$ external moment at the joint (often 0).
Shear (sway) equation for each sway degree: horizontal equilibrium of the storey, using column shears from their end moments.
Solve the simultaneous equations; substitute back for end moments.
Find reactions, shear and moment diagrams by statics.

Frames without and with sway

A frame does not sway when it is symmetric in geometry and loading, or when lateral movement is prevented (a brace or a support).

For a portal with columns AB and DC (feet A, D) and beam BC, with horizontal load $P$ at beam level, the shear equation is:

\frac{M _{A B} + M _{B A}}{h _{1}} + \frac{M _{D C} + M _{C D}}{h _{2}} + P = 0

(sign depends on the chosen sway direction; column shear $= - (M_{t o p} + M_{b o tt o m}) / h$ for clockwise-positive moments, plus any load on the column).

Worked examples

Worked ExampleExample 1 — two-span continuous beam

A beam ABC is fixed at A and simply supported at B and C. AB = 6 m carries a UDL of 20 kN/m; BC = 4 m carries a central load of 40 kN. $E I$ constant. Find the support moments.

Solution. Unknowns: $θ_{B}$ , $θ_{C}$ ( $θ_{A} = 0$ ).

Fixed-end moments: $M_{F A B} = - \frac{20 \times 36}{12} = - 60$ , $M_{F B A} = + 60$ ; $M_{F B C} = - \frac{40 \times 4}{8} = - 20$ , $M_{F C B} = + 20$ (kN·m).

Use the modified equation for BC (C is a simple support, $M_{C B} = 0$ ):

$M_{B C} = (- 20 - \frac{20}{2}) + \frac{3 E I}{4} θ_{B} = - 30 + 0.75 E I θ_{B}$

$M_{B A} = 60 + \frac{2 E I}{6} (2 θ_{B}) = 60 + 0.6667 E I θ_{B}$

$M_{A B} = - 60 + \frac{2 E I}{6} θ_{B} = - 60 + 0.3333 E I θ_{B}$

Joint B: $M_{B A} + M_{B C} = 0$ → $30 + 1.4167 E I θ_{B} = 0$ → $E I θ_{B} = - 21.18$

$M_{A B} = - 60 - 7.06 = - 67.06 kN \cdot m$ $M_{B A} = 60 - 14.12 = + 45.88 kN \cdot m$ , $M_{B C} = - 30 - 15.88 = - 45.88 kN \cdot m$ ✔ (equal and opposite)

So hogging moments: 67.06 kN·m at A and 45.88 kN·m at B.

Worked ExampleExample 2 — non-sway portal frame

A symmetric portal frame ABCD has fixed feet A and D, columns 4 m high and beam BC 6 m long carrying a UDL of 24 kN/m. All members have the same $E I$ . Find the moments.

Solution. Symmetry → no sway, and $θ_{C} = - θ_{B}$ .

FEM for BC: $M_{F B C} = - \frac{24 \times 36}{12} = - 72$ , $M_{F C B} = + 72$ .

$M_{A B} = \frac{2 E I}{4} θ_{B} = 0.5 E I θ_{B}$ ; $M_{B A} = E I θ_{B}$

$M_{B C} = - 72 + \frac{2 E I}{6} (2 θ_{B} + θ_{C}) = - 72 + \frac{E I}{3} θ_{B}$ (using $θ_{C} = - θ_{B}$ )

Joint B: $E I θ_{B} - 72 + \frac{E I θ _{B}}{3} = 0$ → $E I θ_{B} = 54$

$M_{B A} = 54 kN \cdot m$ , $M_{B C} = - 54 kN \cdot m$ , $M_{A B} = 27 kN \cdot m$

Mid-span moment of beam $= \frac{w L ^{2}}{8} - 54 = 108 - 54 = 54 kN \cdot m$ (sagging).

Worked ExampleExample 3 — effect of support settlement

A two-span continuous beam ABC (AB = BC = $L$ , constant $E I$ , simply supported at A and C and at B) has no load, but support B settles by $δ$ . Find the moment at B.

Solution. Chord rotations: AB: B moves down relative to A → $ψ_{A B} = + δ / L$ (clockwise); BC: B lower than C → $ψ_{B C} = - δ / L$ .

With pinned ends at A and C (modified equations), by symmetry $θ_{B} = 0$ :

$M_{B A} = \frac{3 E I}{L} (0 - ψ_{A B}) = - \frac{3 E I δ}{L ^{2}}$ ; $M_{B C} = \frac{3 E I}{L} (0 - ψ_{B C}) = + \frac{3 E I δ}{L ^{2}}$

Joint B balances ✔. Support moment at B $= \frac{3 E I δ}{L ^{2}}$ (sagging at B — the settling support "hangs" from the beam).

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Slope Deflection Method