Slope-Deflection Method

Indeterminate Beams & Frames — Slope-Deflection Equations, Fixed End Moments, Sway Analysis & Solved Examples

Last Updated: March 2026

📌 Key Takeaways

  • The Slope-Deflection Method is a displacement-based method for analysing statically indeterminate beams and frames. The primary unknowns are joint rotations (slopes) and sway displacements — not forces.
  • The core tool is the slope-deflection equation: MAB = (2EI/L)(2θA + θB − 3ψ) + FEMAB, which expresses the end moment of any member in terms of the slopes at its ends and the chord rotation.
  • Fixed End Moments (FEM) are the moments that would develop at the ends of a fully fixed member under the applied loading — they are the “starting point” before any joint rotation is allowed.
  • The equilibrium condition at each joint — ΣM = 0 at every joint — gives the equations needed to solve for the unknown slopes and sway displacements.
  • For frames that can sway (sidesway), an additional shear equation (horizontal equilibrium of the storey) must be written to solve for the unknown sway displacement.
  • The Slope-Deflection Method is the conceptual foundation for the Moment Distribution Method — understanding SDM makes MDM far easier to grasp.
  • In GATE CE, SDM numerical problems appear as NAT (Numerical Answer Type) questions — typically asking for the end moment of a specific member or a support reaction.

1. Introduction — Why a Displacement Method?

Methods for solving statically indeterminate structures fall into two broad categories: force methods and displacement methods.

In a force method (such as the compatibility method or the three-moment theorem), the redundant forces are treated as unknowns. The number of unknowns equals the degree of static indeterminacy (DSI) — so for a highly indeterminate structure, there are many unknowns. Force methods become cumbersome as the DSI increases.

In a displacement method (such as the Slope-Deflection Method or the Stiffness Matrix Method), the joint displacements — rotations and translations — are the primary unknowns. The number of unknowns equals the degree of kinematic indeterminacy (the number of independent joint displacements), which is often much smaller than the DSI for typical building frames. This makes displacement methods more efficient for frames.

The Slope-Deflection Method (SDM), developed by G.A. Maney in 1915, was the first systematic displacement method for frame analysis. Its central idea is elegant: express the end moments of every member as a function of the unknown joint rotations and any sway displacement, then use equilibrium at each joint to set up a system of equations. Solving these equations gives the joint rotations, from which all end moments — and therefore the complete bending moment diagram — follow directly.

Understanding SDM is not just about solving exam problems. The Moment Distribution Method (MDM), which is faster for hand calculations and appears more frequently in GATE CE, is best understood as an iterative way of reaching the same SDM solution. The stiffness concepts of SDM (member stiffness = 4EI/L, carry-over factor = 0.5) are directly used in MDM. Invest in understanding SDM fully — it pays dividends in every method that follows.

2. The Slope-Deflection Equation

The slope-deflection equation expresses the moment at either end of a beam member in terms of:

  • The rotations at the near end (θA) and far end (θB) of the member
  • The chord rotation ψ (due to relative transverse displacement of the two ends)
  • The fixed end moment (FEM) due to any loads applied along the member
  • The flexural rigidity EI and length L of the member

Slope-Deflection Equations — General Form

MAB = (2EI/L)(2θA + θB − 3ψ) + FEMAB

MBA = (2EI/L)(2θB + θA − 3ψ) + FEMBA

Where:

MAB = moment at the near end A of member AB (positive = anticlockwise on the member end, by the standard SDM sign convention)

MBA = moment at the far end B of member AB

θA = rotation of joint A (positive = clockwise)

θB = rotation of joint B (positive = clockwise)

ψ = chord rotation = δ/L, where δ = relative transverse displacement between the two ends (positive when the far end moves to the right relative to the near end for a vertical member, or upward for a horizontal member)

FEMAB = fixed end moment at end A due to loads on the member (positive = anticlockwise)

EI = flexural rigidity of the member

L = length of the member

Physical Interpretation of Each Term

(2EI/L) × 2θA: Moment at A due to rotation of joint A alone — if only A rotates by θA, the moment at A is 4EIθA/L (the stiffness term).

(2EI/L) × θB: Moment at A due to rotation of far joint B alone — if only B rotates by θB, it induces a moment of 2EIθB/L at A (the carry-over term).

(2EI/L) × (−3ψ): Moment at A due to chord rotation (sway) ψ — if the member sways by δ = ψL, moments of −6EIψ/L develop at both ends (equal and opposite, causing a double-curvature S-shape).

FEMAB: The moment at A if the member were fully fixed at both ends under the applied loads — the “locked” state before any joint rotation is permitted.

Sign convention in SDM: End moments are positive when anticlockwise (acting on the member end). Joint rotations θ are positive when clockwise. This convention is used consistently throughout the examples below — changing it mid-problem is the most common source of sign errors.

3. Fixed End Moments — Complete Table

Fixed End Moments (FEM) are the moments that would develop at the ends of a beam member if both ends were fully fixed against rotation and translation, under the applied loading. They are the starting values in the slope-deflection equation — the moments in the “locked” state before joint equilibrium is enforced. The sign convention: FEM is positive (anticlockwise) when it acts in the direction that would resist the sagging of the beam (hogging at ends of a simply supported beam under downward load).

Loading CaseFEMAB (near end A)FEMBA (far end B)
Central point load W+WL/8−WL/8
UDL w over full span L+wL²/12−wL²/12
Point load W at distance a from A (b = L−a from B)+Wab²/L²−Wa²b/L²
UDL w over left half (0 to L/2)+5wL²/96−5wL²/96 × (11/10) ≈ −11wL²/192
Triangular load — zero at A, wmax at B+wmaxL²/20−wmaxL²/30
Triangular load — wmax at A, zero at B+wmaxL²/30−wmaxL²/20
Applied couple M₀ at distance a from A+M₀b(2a−b)/L²+M₀a(2b−a)/L²
Support settlement δ at B (far end sinks)+6EIδ/L²+6EIδ/L²

Memory rule for the most important FEM:

  • UDL: FEM = ±wL²/12 (near end positive, far end negative — hogging at both ends)
  • Central point load: FEM = ±WL/8
  • Eccentric point load at distance a from near end, b from far end: near end = +Wab²/L², far end = −Wa²b/L²

Note on signs: In the standard SDM sign convention, the near-end FEM is positive (anticlockwise) and the far-end FEM is negative (clockwise) for downward loading. This reflects the hogging nature of fixed-end moments — the beam is concave downward at its ends when fixed, meaning the top is in tension and the bottom is in compression at the supports.

4. Chord Rotation (Sway) — ψ

The chord rotation ψ accounts for the effect of relative transverse displacement between the two ends of a member. When the joints at both ends of a member remain at the same level (no sway), ψ = 0. When one end displaces transversely relative to the other — as happens in a frame that sways sideways under horizontal loading — ψ ≠ 0.

Chord Rotation — Definition

ψ = δ / L

Where δ = relative transverse displacement between the two ends of the member (far end relative to near end), L = member length.

For a column of height h in a frame that sways by Δ at the top:

ψcolumn = Δ / h

For a horizontal beam in a frame where both ends move by the same horizontal amount (no relative sway between floors):

ψbeam = 0 (both ends translate equally — no relative transverse displacement)

Sway vs no-sway frames: A frame is a no-sway frame (symmetric frame under symmetric loading, or frames with sway prevented by bracing) if there is no horizontal displacement of any joint. In a no-sway frame, ψ = 0 for all members and the SDM equations simplify considerably. A frame is a sway frame if horizontal joint displacements are possible — ψ becomes an additional unknown for each storey, and an additional shear equilibrium equation must be written per storey.

5. Step-by-Step Procedure

For no-sway beams and frames:

  1. Step 1 — Identify unknowns: List all unknown joint rotations (θ values). For a fixed end, θ = 0. For a pin or roller, θ is unknown. For a free end, θ is unknown.
  2. Step 2 — Write slope-deflection equations: Write MAB and MBA for every member using the slope-deflection equation. Express each in terms of the unknown θ values and the known FEMs. For no-sway frames, ψ = 0.
  3. Step 3 — Apply joint equilibrium (ΣM = 0 at each free joint): At every joint where rotation is unknown, the sum of all end moments from members meeting at that joint must equal any external applied moment at that joint (usually zero). This gives one equation per unknown rotation.
  4. Step 4 — Solve the system of equations: The equilibrium equations form a linear system in the unknown θ values. Solve simultaneously (substitution or matrix method).
  5. Step 5 — Back-substitute to find end moments: Substitute the solved θ values back into the slope-deflection equations to find all member end moments.
  6. Step 6 — Draw SFD and BMD: With all end moments known, treat each member as a simply supported beam with the end moments applied, and superimpose the simply supported free BMD to get the final BMD.
  7. Step 7 — Find support reactions: Use equilibrium of each member (taking the known end moments as applied loads) to find shear forces, then sum shear forces at each support joint to get support reactions.

Additional steps for sway frames:

  1. Step 8 — Introduce sway unknown: Let the sway displacement of the storey = Δ. Express ψ = Δ/h for all columns in that storey. Beams in the same storey have ψ = 0.
  2. Step 9 — Write shear equation: Take the horizontal equilibrium of the entire storey (sum of shear forces in all columns of that storey = applied horizontal load at that storey level). This gives one additional equation per unknown sway displacement.
  3. Step 10 — Solve and back-substitute as before.

6. Worked Example 1 — Propped Cantilever with Point Load

Problem: A propped cantilever AB is fixed at A and simply supported (roller) at B. Span = 6 m. A point load of 30 kN acts at midspan C (3 m from A). Find the end moments MAB and MBA and the support reactions. EI is constant.

Step 1 — Identify Unknowns

Fixed end A: θA = 0 (no rotation at fixed support).

Roller end B: θB = unknown (roller allows rotation).

No sway: ψ = 0.

One unknown: θB

Step 2 — Fixed End Moments

Point load W = 30 kN at a = 3 m from A, b = 3 m from B (central load):

FEMAB = +Wab²/L² = +30 × 3 × 9/36 = +22.5 kN·m

FEMBA = −Wa²b/L² = −30 × 9 × 3/36 = −22.5 kN·m

(Equal magnitude since load is central — which simplifies to ±WL/8 = ±30×6/8 = ±22.5 kN·m ✓)

Step 3 — Slope-Deflection Equations

Let k = 2EI/L = 2EI/6 = EI/3

MAB = k(2θA + θB − 3ψ) + FEMAB

= (EI/3)(2×0 + θB − 0) + 22.5

= (EI/3)θB + 22.5   … (i)

MBA = k(2θB + θA − 3ψ) + FEMBA

= (EI/3)(2θB + 0 − 0) + (−22.5)

= (2EI/3)θB − 22.5   … (ii)

Step 4 — Joint Equilibrium at B

At B (roller support): MBA = 0 (a roller cannot resist moment — the end moment must be zero).

From equation (ii): (2EI/3)θB − 22.5 = 0

θB = 22.5 × 3/(2EI) = 33.75/EI

Step 5 — Back-Substitute to Find End Moments

From equation (i): MAB = (EI/3) × (33.75/EI) + 22.5 = 11.25 + 22.5 = +33.75 kN·m

MBA = 0 (as enforced by boundary condition at B) ✓

Fixing moment at A = 33.75 kN·m (anticlockwise on member = hogging at A)

Step 6 — Support Reactions

Taking moments about A for member AB:

RB × 6 = 30 × 3 − MAB + MBA = 90 − 33.75 + 0 = 56.25

Wait — moment equation for a member with end moments: ΣMA = 0:

RB × 6 − 30 × 3 + MAB − MBA = 0

Careful with signs — using standard free body approach:

RB × 6 = 30 × 3 − MAB = 90 − 33.75 = 56.25

RB = 9.375 kN (upward)

ΣFy = 0: RA = 30 − 9.375 = 20.625 kN (upward)

Check using standard propped cantilever formula: RB = 3WL²×a/(2L³)… for central load: RB = 3W/16 = 3×30/16 = 5.625 kN

Hmm — rechecking the moment equation. For member AB with end moments MAB at A (acting on member, clockwise = positive in beam convention) and MBA at B:

ΣM about A (for the whole member): RB×6 − 30×3 + MAB = 0 (MBA = 0)

MAB here in beam moment convention (hogging = negative): MAB = −33.75 kN·m

RB × 6 = 90 − 33.75 = 56.25 → RB = 9.375 kN

Standard formula for propped cantilever, central load W: RB = 5W/16 = 150/16 = 9.375 kN ✓

RA = 30 − 9.375 = 20.625 kN ✓

7. Worked Example 2 — Two-Span Continuous Beam

Problem: A continuous beam ABC is fixed at A, simply supported at B (intermediate roller) and simply supported at C (end roller). Span AB = 4 m, span BC = 6 m. Span AB carries a UDL of 12 kN/m. Span BC carries a central point load of 24 kN. Find all end moments. EI is constant throughout.

Step 1 — Unknowns

θA = 0 (fixed end)  |  θB = unknown  |  θC = unknown (roller at C, free to rotate)

No sway: ψ = 0 throughout.

Two unknowns: θB and θC

Step 2 — Fixed End Moments

Span AB (UDL 12 kN/m, L = 4 m):

FEMAB = +wL²/12 = +12×16/12 = +16 kN·m

FEMBA = −wL²/12 = −12×16/12 = −16 kN·m

Span BC (Central point load 24 kN, L = 6 m):

FEMBC = +WL/8 = +24×6/8 = +18 kN·m

FEMCB = −WL/8 = −24×6/8 = −18 kN·m

Step 3 — Slope-Deflection Equations

Let kAB = 2EI/LAB = 2EI/4 = EI/2

Let kBC = 2EI/LBC = 2EI/6 = EI/3

Member AB:

MAB = (EI/2)(2×0 + θB) + 16 = (EI/2)θB + 16  …(i)

MBA = (EI/2)(2θB + 0) − 16 = EI·θB − 16  …(ii)

Member BC:

MBC = (EI/3)(2θB + θC) + 18 = (2EI/3)θB + (EI/3)θC + 18  …(iii)

MCB = (EI/3)(2θC + θB) − 18 = (2EI/3)θC + (EI/3)θB − 18  …(iv)

Step 4 — Joint Equilibrium Equations

At joint B (internal roller — free to rotate, but ΣMB = 0):

MBA + MBC = 0

(EI·θB − 16) + ((2EI/3)θB + (EI/3)θC + 18) = 0

EI·θB + (2EI/3)θB + (EI/3)θC + 2 = 0

(5EI/3)θB + (EI/3)θC = −2  …(v)

At joint C (end roller — MCB = 0):

MCB = 0 → (2EI/3)θC + (EI/3)θB − 18 = 0

(EI/3)θB + (2EI/3)θC = 18  …(vi)

Step 5 — Solve the System

From (vi): θB + 2θC = 54/EI → θB = (54 − 2EI·θC×…

Multiply (v) by 3/EI: 5θB + θC = −6/EI  …(v’)

Multiply (vi) by 3/EI: θB + 2θC = 54/EI  …(vi’)

From (vi’): θB = 54/EI − 2θC

Substitute into (v’): 5(54/EI − 2θC) + θC = −6/EI

270/EI − 10θC + θC = −6/EI

−9θC = −276/EI

θC = 276/(9EI) = 30.67/EI

θB = 54/EI − 2(30.67/EI) = 54/EI − 61.33/EI = −7.33/EI

Step 6 — Final End Moments

MAB = (EI/2)(−7.33/EI) + 16 = −3.67 + 16 = +12.33 kN·m

MBA = EI × (−7.33/EI) − 16 = −7.33 − 16 = −23.33 kN·m

MBC = (2EI/3)(−7.33/EI) + (EI/3)(30.67/EI) + 18 = −4.89 + 10.22 + 18 = +23.33 kN·m

MCB = 0 ✓

Check at B: MBA + MBC = −23.33 + 23.33 = 0 ✓

8. Worked Example 3 — Non-Sway Portal Frame

Problem: A portal frame ABCD has columns AB and CD (each 4 m tall) and beam BC (6 m span). All joints B and C are rigid. A = pin support, D = pin support. A UDL of 10 kN/m acts on beam BC. The frame is symmetric — no sway occurs. EI is constant throughout. Find all end moments.

Setup

Since A and D are pin supports: θA and θD are unknown BUT MAB = 0 and MDC = 0 (pins cannot resist moment). This means A and D are free to rotate — but the end moments must be zero there.

Unknowns: θB and θC. By symmetry (symmetric frame, symmetric load): θB = −θC. One unknown: θB.

No sway (symmetric): ψ = 0 for all members.

Fixed End Moments

Columns AB and CD: No load on columns → FEM = 0 for both columns.

Beam BC (UDL 10 kN/m, L = 6 m):

FEMBC = +wL²/12 = +10×36/12 = +30 kN·m

FEMCB = −wL²/12 = −30 kN·m

Slope-Deflection Equations

kcol = 2EI/4 = EI/2  |  kbeam = 2EI/6 = EI/3

Column AB (near end A = pin, θA unknown but MAB = 0):

MAB = (EI/2)(2θA + θB) = 0 → θA = −θB/2

MBA = (EI/2)(2θB + θA) = (EI/2)(2θB − θB/2) = (EI/2)(3θB/2) = (3EI/4)θB

(Alternatively, for a member with a pin at the far end, use the modified stiffness: MBA = (3EI/L)θB = (3EI/4)θB ✓)

By symmetry: MDC = 0, MCD = (3EI/4)θC = −(3EI/4)θB

Beam BC:

MBC = (EI/3)(2θB + θC) + 30 = (EI/3)(2θB − θB) + 30 = (EI/3)θB + 30

MCB = (EI/3)(2θC + θB) − 30 = (EI/3)(−2θB + θB) − 30 = −(EI/3)θB − 30

Joint Equilibrium at B and Solving

ΣMB = 0: MBA + MBC = 0

(3EI/4)θB + (EI/3)θB + 30 = 0

EI·θB(3/4 + 1/3) = −30

EI·θB(9/12 + 4/12) = −30

EI·θB(13/12) = −30

EI·θB = −30 × 12/13 = −360/13 = −27.69 kN·m

Final end moments:

MBA = (3/4)(−27.69) = −20.77 kN·m

MBC = (1/3)(−27.69) + 30 = −9.23 + 30 = +20.77 kN·m ✓ (ΣMB = 0)

MCD = −(3/4)(−27.69) = +20.77 kN·m  |  MCB = −(1/3)(−27.69) − 30 = 9.23 − 30 = −20.77 kN·m

MAB = MDC = 0 (pin supports) ✓

9. Worked Example 4 — Sway Frame with Horizontal Load

Problem: A portal frame ABCD has columns AB and CD (each 3 m tall, fixed at A and D) and a rigid beam BC (4 m span, no load on beam). A horizontal load of 30 kN acts to the right at joint B. EI is constant. Find all end moments.

Setup — Sway Frame

Fixed ends A and D: θA = θD = 0. Rigid joints B and C: θB and θC unknown.

Horizontal load causes sway Δ to the right. Both columns sway by the same Δ (beam BC is rigid and inextensible — both B and C move right by Δ).

Chord rotation for columns: ψAB = ψCD = Δ/3 (same for both columns)

Chord rotation for beam BC: ψBC = 0 (both ends move by same horizontal amount)

No loads on any member → all FEM = 0.

Three unknowns: θB, θC, and Δ (or ψ = Δ/3)

Let ψ = Δ/3 for simplicity.

Slope-Deflection Equations (FEM = 0 throughout)

kcol = 2EI/3  |  kbeam = 2EI/4 = EI/2

Column AB (A fixed, B rigid):

MAB = (2EI/3)(2×0 + θB − 3ψ) = (2EI/3)(θB − 3ψ)  …(i)

MBA = (2EI/3)(2θB + 0 − 3ψ) = (2EI/3)(2θB − 3ψ)  …(ii)

Column CD (D fixed, C rigid):

MDC = (2EI/3)(2×0 + θC − 3ψ) = (2EI/3)(θC − 3ψ)  …(iii)

MCD = (2EI/3)(2θC + 0 − 3ψ) = (2EI/3)(2θC − 3ψ)  …(iv)

Beam BC (no load, no sway):

MBC = (EI/2)(2θB + θC)  …(v)

MCB = (EI/2)(2θC + θB)  …(vi)

Equilibrium Equations

ΣMB = 0: MBA + MBC = 0

(2EI/3)(2θB − 3ψ) + (EI/2)(2θB + θC) = 0

(4EI/3)θB − 2EIψ + EI·θB + (EI/2)θC = 0

(7EI/3)θB + (EI/2)θC − 2EIψ = 0  …(A)

ΣMC = 0: MCD + MCB = 0

(2EI/3)(2θC − 3ψ) + (EI/2)(2θC + θB) = 0

(EI/2)θB + (7EI/3)θC − 2EIψ = 0  …(B)

Shear equation — horizontal equilibrium of storey:

ΣFx = 0 for entire frame: 30 = HAB + HCD

Shear in column AB: HAB = (MAB + MBA)/h = (MAB + MBA)/3

Shear in column CD: HCD = (MDC + MCD)/3

(MAB + MBA + MDC + MCD)/3 = 30

MAB + MBA + MDC + MCD = 90  …(C)

Substituting expressions:

(2EI/3)(θB−3ψ) + (2EI/3)(2θB−3ψ) + (2EI/3)(θC−3ψ) + (2EI/3)(2θC−3ψ) = 90

(2EI/3)[3θB + 3θC − 12ψ] = 90

2EI(θB + θC − 4ψ) = 90  …(C)

By symmetry of frame geometry (but asymmetric loading at B): Note that equations (A) and (B) are symmetric in θB and θC exchange but not identical due to load at B only… For this symmetric frame with load at B, let us note θC can be solved.

Subtracting (B) from (A): (7EI/3 − EI/2)(θB − θC) = 0 → θB = θC? No — equations are symmetric so subtracting gives (7/3 − 1/2)EI·θB + (1/2 − 7/3)EI·θC = 0 → same coefficient → θB = θC.

With θB = θC = θ, from (A): (7EI/3 + EI/2)θ = 2EIψ → (17EI/6)θ = 2EIψ → ψ = 17θ/12

From (C): 2EI(2θ − 4 × 17θ/12) = 90 → 2EI(2θ − 17θ/3) = 90 → 2EI·θ(6/3 − 17/3) = 90

2EI·θ(−11/3) = 90 → EI·θ = −90×3/(2×11) = −270/22 = −12.27 kN·m

ψ = 17×(−12.27/EI)/12… EI·ψ = 17×(−12.27)/12 = −17.38 kN·m

Final moments:

MAB = (2EI/3)(θ − 3ψ)/EI × EI = (2/3)(−12.27 − 3×(−17.38)) = (2/3)(−12.27 + 52.14) = (2/3)(39.87) = +26.58 kN·m

MBA = (2/3)(2×(−12.27) − 3×(−17.38)) = (2/3)(−24.54 + 52.14) = (2/3)(27.6) = +18.4 kN·m

By symmetry: MDC = +26.58 kN·m, MCD = +18.4 kN·m

MBC = (EI/2)(2θ + θ)/EI × EI = (3/2)(−12.27) = −18.4 kN·m ✓ (ΣMB = 18.4 − 18.4 = 0)

10. Special Cases — Far End Conditions

The standard slope-deflection equation assumes both ends of the member are rigid joints (capable of transmitting moment). When the far end has a special condition — pin, roller, or free end — the equation can be simplified using a modified slope-deflection equation, which reduces the number of unknowns.

Modified Slope-Deflection Equation — Far End Pinned or Roller

When the far end B is a pin or roller (MBA = 0), substitute this condition into the standard pair of equations and eliminate θB. The result for the near-end moment is:

MAB = (3EI/L)(θA − ψ) + FEMAB − FEMBA/2

The effective stiffness of a member pinned at the far end is 3EI/L (instead of 4EI/L for both ends fixed). The carry-over to the pinned end is zero — this is exactly why the carry-over factor for a far-pinned member is 0 in the Moment Distribution Method.

Far End ConditionEffective Stiffness (near end)Carry-Over FactorM at Far End
Fixed (built-in)4EI/L0.5MBA = (1/2) × M applied at A (plus FEM)
Pin or roller3EI/L0MBA = 0 (always zero)
Free end3EI/L (same as pinned — no moment at free end)0MBA = 0 (no moment at free end)
Guided (prevents rotation, allows translation)EI/L−0.5MBA = −(1/2) × M applied at A

11. Common Mistakes Students Make

  • Setting θ = 0 at a pin or roller support: A fixed support has θ = 0 (no rotation). A pin or roller support has zero moment (M = 0 there) but the rotation θ is non-zero and unknown — the joint is free to rotate. Setting θ = 0 at a simply supported end is one of the most common and consequential errors in SDM problems. The correct condition at a pin or roller end is M = 0, not θ = 0.
  • Wrong sign for Fixed End Moments: The standard convention is FEMAB positive (anticlockwise on the near end) and FEMBA negative (clockwise on the far end) for downward loading on a horizontal member. Reversing these signs leads to wrong equations that give completely incorrect end moments. Always draw the fixed beam with the loads and verify the direction of the FEMs before substituting.
  • Forgetting the sway term (−3ψ) for columns in sway frames: In a sway frame, the −3ψ term in the slope-deflection equation is critical — omitting it makes the equations represent a no-sway frame and gives wrong results for the sway case. Always check at the start: can the frame sway? If yes, introduce ψ as an unknown for each storey.
  • Incorrect shear equation for sway frames: The shear equation is the horizontal equilibrium of the entire storey, not of a single column. The storey shear equals the applied horizontal load at that storey level, and it must equal the sum of the shear forces in all columns of that storey. Writing the shear equation for only one column is wrong unless the frame has only one column (which is unusual).
  • Not checking joint equilibrium after solving: After finding all end moments, verify that ΣM = 0 at every free joint. If the sum of end moments at any joint is not zero (or not equal to the applied moment at that joint), an error was made in setting up or solving the equations. This check takes thirty seconds and catches almost every mistake.

12. Frequently Asked Questions

What is the difference between the Slope-Deflection Method and the Moment Distribution Method?

Both methods are displacement-based and produce the same final end moments. The Slope-Deflection Method is a direct (exact) method — it sets up a system of simultaneous linear equations in the unknown joint rotations and solves them exactly. The Moment Distribution Method is an iterative (approximate-to-exact) method — it starts by locking all joints (computing FEMs), then unlocks them one at a time and distributes the out-of-balance moments iteratively until convergence. MDM avoids solving simultaneous equations, making it faster for hand calculations on beams and simple frames. However, SDM is conceptually cleaner — understanding SDM makes the logic behind MDM (distribution factors, carry-over factors, stiffness) immediately clear. For GATE CE, MDM is more commonly tested as a direct numerical question, but SDM questions also appear regularly.

When is a frame considered a no-sway frame?

A frame is a no-sway frame when horizontal displacement of the joints is prevented — either physically (by a sway brace or rigid wall) or by symmetry. A symmetric frame under symmetric loading has no tendency to sway (the horizontal forces from the two columns cancel), so ψ = 0 even without bracing. An unsymmetric frame, or a symmetric frame under asymmetric loading (such as a horizontal load at one joint), will sway unless braced. In the Slope-Deflection Method, the distinction matters because sway introduces an additional unknown (ψ per storey) and requires an additional equation (the storey shear equation).

What is the physical meaning of the carry-over factor of 0.5?

The carry-over factor of 0.5 means that when a moment M is applied at the near end of a beam with the far end fixed, a moment of M/2 is induced at the far end — in the same direction (both cause hogging or both sagging). This comes directly from the slope-deflection equations: if you apply a moment at end A of a fixed-far-end member while keeping end B fixed (θB = 0), the ratio MBA/MAB = 1/2. Physically, the applied moment at A causes the beam to bend, and this bending induces a reaction moment at the fixed end B. Since B is fixed, it cannot rotate — it must apply a moment to the beam to prevent rotation, and this moment is exactly half the applied moment at A. This 0.5 ratio is a fundamental property of the prismatic elastic beam and is the basis for the carry-over step in the Moment Distribution Method.

How do you handle a beam with varying EI (non-prismatic beam) in the Slope-Deflection Method?

For a non-prismatic beam (where EI varies along the length), the standard slope-deflection equations cannot be used directly — they assume constant EI throughout the member. The correct approach is to use the stiffness coefficients for the non-prismatic member, which must be derived from integration of the beam equation with the variable EI. The resulting slope-deflection equations have the same form but with modified stiffness coefficients and carry-over factors (which are no longer equal to 0.5). For most GATE CE problems, EI is assumed constant unless explicitly stated otherwise.

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