Slope-deflection method
The slope-deflection method was one of the first systematic ways to analyse continuous beams and rigid frames by hand. Instead of choosing redundant forces, it takes the unknown joint rotations (and sway displacements) as the main unknowns. Each member's end moments are written in terms of those rotations plus the fixed-end moments that the loads would cause if both ends were clamped. Joint equilibrium then gives as many equations as unknowns.
For a member AB of length L and stiffness E I, the end moment M_AB = M_F,AB + (2EI/L)(2θA + θB − 3ψ), where ψ is the chord rotation caused by relative sideways movement of the ends. A clockwise moment on the member end is usually taken as positive. At every rigid joint the sum of member end moments must be zero (or equal any external moment applied there). For a frame that sways, an extra shear equation for each storey closes the set.
Its strengths are clarity and a direct link to the stiffness method: the coefficients 4EI/L and 2EI/L are exactly the terms of a beam element stiffness matrix. It becomes tedious beyond a few joints, which is why moment distribution and then computers took over, but it remains the cleanest way to understand how stiffness, carry-over and sway interact.
Write all fixed-end moments first with signs; most errors come from a wrong sign on one of them.
M_AB = M_F,AB + (2EI/L)(2θA + θB − 3Δ/L)Fixed-end moments, UDL: ± wL²/12; central point load: ± PL/8Joint equilibrium: Σ M_ends = applied momentFar end pinned (modified): M_AB = M_F,AB − M_F,BA/2 + (3EI/L)(θA − Δ/L)
Forgetting the chord-rotation term 3Δ/L when a frame sways, which makes the joint equations look balanced while horizontal equilibrium is violated.
- Assumes prismatic, linear elastic members and neglects axial and shear deformation.
- Clockwise end moments positive; keep the same convention throughout.
Problem. A two-span continuous beam ABC, each span 6 m, same EI, is fixed at A and simply supported at B and C. Span AB carries 20 kN/m; BC is unloaded. Find the moment at B.
- Fixed-end moments AB: −60 and +60 kNm (wL²/12 = 20 × 36/12).
- θA = 0. M_BA = 60 + (2EI/6)(2θB); M_BC (C pinned, modified) = (3EI/6)θB.
- Equilibrium at B: 60 + (4EI/6)θB + (3EI/6)θB = 0, so EIθB = −51.43.
- M_BA = 60 − 34.29 = 25.7 kNm; M_BC = −25.7 kNm.
Answer. Hogging moment at B ≈ 25.7 kNm
Quick quiz
1What are the primary unknowns in slope-deflection?
Joint rotations and sway displacements.
2Fixed-end moment for a UDL?
wL²/12 at each end.
3Which modern method shares its coefficients?
The matrix stiffness method.
References and further reading
Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.
- R. C. Hibbeler, Structural Analysis, Pearson, latest edition.
- C. S. Reddy, Basic Structural Analysis, McGraw Hill Education (India), latest edition.
- Devdas Menon, Structural Analysis, Narosa, latest edition.
- A. K. Chopra, Dynamics of Structures, Pearson, latest edition.
Free and official sources
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