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Structural Analysis Fundamentals / Statically indeterminate structures

The slope-deflection method

Write every end moment in terms of the joint rotations, then solve for them.

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Why this matters

The force method treats forces as the unknowns. Slope-deflection flips that round and takes displacements as the unknowns — the rotations of the joints. That change of viewpoint scales far better, and it is the direct ancestor of the stiffness method that every analysis program uses today.

By the end of this lesson you should be able to

  • Write the slope-deflection equation for a member
  • Identify the unknown rotations in a continuous beam
  • Solve a two-span continuous beam and check the result

What you should already know

  • Fixed-end moments (this module, lesson 2)
  • Moment–curvature and EI (Module 13)
The slope-deflection equation
MAB=2EIL(2θA+θB−3ψ)+FEMABM_{AB} = \frac{2EI}{L}\left(2\theta_A + \theta_B - 3\psi\right) + \text{FEM}_{AB}

What it calculates: the moment at end A of member AB

θA, θB
rotations of the two ends of the member (radians)
ψ
sway angle: relative transverse displacement of the ends, divided by L (radians)
FEMAB
fixed-end moment at A from the loads on the member (kNm)
EI/L
member stiffness (kNm)

This assumes

  • Linearly elastic, prismatic member
  • Small rotations, so ψ ≈ tan ψ
  • Axial deformation of the member is neglected

In plain terms: Read it as three contributions: what the loads would do if the ends were locked (the FEM), plus what the actual end rotations add, plus what any sideways movement adds. Lock everything (θ = ψ = 0) and only the fixed-end moment survives — which is exactly what a fixed-end moment means.

The method is a short recipe:

  1. 1.Identify the unknown joint rotations. A rigid joint between members shares one rotation; a fully fixed support has θ = 0.
  2. 2.Write a slope-deflection expression for every member end.
  3. 3.Impose equilibrium at each joint: the moments from all members meeting there must sum to zero. At a simple end support, the moment must be zero.
  4. 4.Solve for the rotations.
  5. 5.Substitute back to get every end moment, and then the reactions.

Step 3 is the crucial reversal. In the force method the extra equations were compatibility conditions; here compatibility is built in from the start — the members share a rotation because the joint is rigid — and it is equilibrium that supplies the equations to solve.

Worked example

Two-span continuous beam by slope-deflection

Given

  • Two equal spans AB and BC, each L = 6.00 m
  • Simple supports at the outer ends A and C, continuous over the central support B
  • UDL w = 20.0 kN/m over both spans
  • Uniform EI throughout

Find

The moment over the central support and all three reactions.

Assumptions

  • Symmetry about B, so the rotation at B is zero
  • No sway: all supports stay at the same level, so ψ = 0

    Practice

    A two-span continuous beam has equal spans of 6.00 m and carries a UDL of 20.0 kN/m over both spans. What is the magnitude of the hogging moment over the central support, in kNm?

    Practice

    For the same beam, what is the reaction at each outer support, in kN?

    Practice

    For the same beam, what is the reaction at the central support, in kN?

    Summary

    • Slope-deflection takes joint rotations as the unknowns, not forces
    • MAB = (2EI/L)(2θA + θB − 3ψ) + FEMAB
    • Lock everything and only the fixed-end moment remains
    • Compatibility is built in; equilibrium at the joints supplies the equations
    • Two equal spans under a full UDL: M over the support = wL²/8, central reaction 1.25wL