Structural Analysis

Displacement Method of Structural Analysis

A thorough guide to the displacement (stiffness) method for indeterminate beams and frames. Covers kinematic indeterminacy, slope-deflection equations, fixed-end moments, stiffness matrix, sway analysis, and fully worked numerical examples with BMD and SFD.

Stiffness Method Slope-Deflection Worked Examples

By Bimal Ghimire • Published July 18, 2025 • 22 min read

Introduction

What Is the Displacement Method?

The displacement method (also called the stiffness method or equilibrium method) is the second fundamental approach to analysing statically indeterminate structures. Instead of treating unknown forces as primary unknowns, the displacement method takes joint displacements and rotations as unknowns and enforces equilibrium at every free joint.

It was formalised by G. A. Maney (1915) as the slope-deflection method, then generalised into the direct stiffness method by Turner, Clough, Martin and Topp (1956) - the foundation of all modern FEM software including SAP2000, ETABS, and STAAD.Pro.

1915

Maney's slope-deflection

DKI

Kinematic indeterminacy

[K]

Global stiffness matrix

EI/L

Key stiffness parameter

Primary unknowns

Joint rotations $\theta$ and sway displacements $\Delta$ - one per kinematic DOF.

Governing equations

Joint equilibrium: $[K]\{\delta\} = \{F\}$. As many equations as DOF.

Fixed-end moments

Moments at locked member ends due to span loads - the starting point for each member.

Stiffness coefficients

Moment at DOF $i$ due to unit rotation at DOF $j$ - entries of the stiffness matrix.

Degree of kinematic indeterminacy (DKI)

$$\text{DKI} = 3j - r - C_i$$ 2D frames: $j$ = free joints, $r$ = restrained DOF from supports, $C_i$ = internal condition equations

$$\text{DKI (non-sway beam)} = \text{number of interior free joints with unknown } \theta$$ For continuous beams neglecting axial deformation

Contrast with force method: Force method needs DSI equations; displacement method needs DKI equations. For highly redundant structures DKI << DSI, making displacement method far more efficient. This is why all structural software uses it.

Theory

Key Theoretical Concepts

Member stiffness - rotational DOF

$$\begin{bmatrix} M_{AB} \\ M_{BA} \end{bmatrix} = \frac{2EI}{L}\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} \theta_A \\ \theta_B \end{bmatrix} + \begin{bmatrix} M_{AB}^F \\ M_{BA}^F \end{bmatrix}$$ Diagonal = $4EI/L$; off-diagonal = $2EI/L$ (carry-over stiffness)

Full 4×4 member stiffness matrix

$$[k] = \frac{EI}{L^3}\begin{bmatrix} 12 & 6L & -12 & 6L \\ 6L & 4L^2 & -6L & 2L^2 \\ -12 & -6L & 12 & -6L \\ 6L & 2L^2 & -6L & 4L^2 \end{bmatrix}$$ DOF order: $\{\delta_A,\theta_A,\delta_B,\theta_B\}$ - Euler-Bernoulli beam, no axial deformation

Key stiffness parameters

Parameter	Far end fixed	Far end pinned	Meaning
Near-end rotational stiffness $k$	$4EI/L$	$3EI/L$	Moment at near end per unit rotation
Carry-over factor $C$	$1/2$	$0$	Fraction of moment reaching far end
Sway stiffness	$12EI/L^3$	$3EI/L^3$	Shear per unit lateral displacement
End moment per unit chord rotation $\psi$	$-6EI/L$	$-3EI/L$	Moment induced by unit sway

Chord rotation and sway

$$\psi = \frac{\Delta}{L}$$ $\Delta$ = relative lateral displacement between member ends, $L$ = member length

Distribution factor (DF) for moment distribution

$$\text{DF}_i = \frac{k_i}{\displaystyle\sum_j k_j} \qquad k = \frac{4EI}{L}\text{ (far fixed)},\quad \frac{3EI}{L}\text{ (far pinned)}$$ $\sum\text{DF} = 1$ at every free joint; DF = 0 at fixed ends (infinite stiffness absorbs all moment)

Reference Data

Fixed-End Moments - Visual Reference Table

Fixed-end moments (FEM) are the moments at both ends of a member when all joint displacements are fully restrained and the member carries span loading. Sign convention: clockwise = positive at near end A.

Condition	$M_{AB}^F$ (near A)	$M_{BA}^F$ (far B)
UDL $w$, full span $L$	$+\dfrac{wL^2}{12}$	$-\dfrac{wL^2}{12}$
Central point load $P$ at $L/2$	$+\dfrac{PL}{8}$	$-\dfrac{PL}{8}$
Point load $P$ at $a$ from A ($b=L-a$)	$+\dfrac{Pab^2}{L^2}$	$-\dfrac{Pa^2b}{L^2}$
Triangular load: 0 at A, $w_0$ at B	$+\dfrac{w_0L^2}{20}$	$-\dfrac{w_0L^2}{30}$
Support settlement $\delta$ at B, no span load	$+\dfrac{6EI\delta}{L^2}$	$+\dfrac{6EI\delta}{L^2}$ (same sign)
Temperature gradient; $T_{bot} - T_{top} = \Delta T$, depth $d$	$-\dfrac{EI\alpha\Delta T}{d}$	$+\dfrac{EI\alpha\Delta T}{d}$

CW positive at near end A. Settlement and temperature FEM have the same sign at both ends. For symmetric loadings $|M_{AB}^F| = |M_{BA}^F|$ but opposite signs.

Core Equations

Slope-Deflection Equations

The slope-deflection equations express every member-end moment in terms of joint rotations, chord rotation, and fixed-end moments. For member AB:

$$M_{AB} = \frac{2EI}{L}(2\theta_A + \theta_B - 3\psi) + M_{AB}^F$$ $$M_{BA} = \frac{2EI}{L}(2\theta_B + \theta_A - 3\psi) + M_{BA}^F$$ Slope-deflection equations (Maney, 1915) - $\psi = \Delta/L$ chord rotation, CW positive

Modified SDE - far end pinned

When far end B is a pin or roller ($M_{BA}=0$), eliminate $\theta_B$ to get:

$$M_{AB} = \frac{3EI}{L}(\theta_A - \psi) + M_{AB}^{F*}$$ $M_{AB}^{F*} = M_{AB}^F - \tfrac{1}{2}M_{BA}^F$; stiffness reduces from $4EI/L$ to $3EI/L$

Special cases at a glance

Condition	$M_{AB}$	$M_{BA}$
$\theta_A=1$, far fixed, $\theta_B=\psi=0$, no load	$4EI/L$	$2EI/L$ (carry-over = ½)
$\theta_A=1$, far pinned, $\psi=0$, no load	$3EI/L$	$0$
Unit sway: $\psi=1$, $\theta_A=\theta_B=0$, far fixed	$-6EI/L$	$-6EI/L$
Unit sway: $\psi=1$, $\theta_A=0$, far pinned	$-3EI/L$	$0$
Settlement $\delta$ at B, $\psi=\delta/L$, $\theta=0$	$+6EI\delta/L^2$	$+6EI\delta/L^2$

Why the factor 3 in $3\psi$? A chord rotation $\psi$ is geometrically equivalent to applying $-\psi$ to both end rotations simultaneously (rigid body rotation). Substituting: $M_{AB} = \frac{2EI}{L}(2(\theta_A-\psi)+(\theta_B-\psi)) = \frac{2EI}{L}(2\theta_A+\theta_B-3\psi)$. The 3 comes from the $2+1$ pattern of the SDE coefficients.

Matrix Formulation

Stiffness Matrix Assembly

The stiffness matrix approach systematises the slope-deflection method for any structure. Members contribute local stiffness matrices that are assembled by DOF mapping into the global system.

DKI = 1: single interior joint

For beam ABC (fixed A, roller B, pin C), using modified SDE for span BC:

$$\underbrace{\left(\frac{4EI}{L_1} + \frac{3EI}{L_2}\right)}_{K_{BB}}\,\theta_B = -\underbrace{(M_{BA}^F + M_{BC}^{F*})}_{\text{unbalanced FEM at B}}$$ Single equation: stiffness × unknown rotation = negative of unbalanced FEM

DKI = 2: two interior joints

$$\begin{bmatrix} \dfrac{4EI}{L_1}+\dfrac{4EI}{L_2} & \dfrac{2EI}{L_2} \\[8pt] \dfrac{2EI}{L_2} & \dfrac{4EI}{L_2}+\dfrac{4EI}{L_3} \end{bmatrix} \begin{Bmatrix} \theta_B \\ \theta_C \end{Bmatrix} = -\begin{Bmatrix} M_{BA}^F+M_{BC}^F \\ M_{CB}^F+M_{CD}^F \end{Bmatrix}$$ $2\times2$ system; off-diagonal entry $2EI/L_2$ = carry-over coupling between adjacent joints

Properties of the global stiffness matrix $[K]$

Property	Description
Symmetric	$K_{ij}=K_{ji}$ - Maxwell's reciprocal theorem
Banded	Non-zeros only near the diagonal for sequential DOF numbering
Positive definite	All eigenvalues positive if properly supported - unique solution guaranteed
Size	$n\times n$ where $n=\text{DKI}$

Step-by-Step

General Procedure for the Displacement Method

Sketch and label the structure. Identify all joints and members. Mark boundary conditions: $\theta=0$ at fixed ends, $M=0$ at pins/rollers, and known sway = 0 for symmetric non-sway frames.
Determine DKI. Count unknown rotations and sway displacements. Use modified SDE for any member with a pinned far end to reduce the count.
Compute fixed-end moments for every member. Apply the actual span loads with both ends locked. Use the FEM table. Sign: CW positive at near end A.
Write slope-deflection equations for all member ends. Express $M_{AB}$ and $M_{BA}$ using the full SDE (or modified form for pinned far ends). Include chord rotation $\psi$ for sway frames.
Apply equilibrium at every free joint. $\sum M = M_{ext}$ (usually zero - no external moment applied at that joint). One equation per unknown rotation.
For sway frames, add the storey shear condition. $\sum V_{col} = H_{storey}$, where $V_{col} = (M_{top}+M_{bottom})/h$. One additional equation per sway DOF $\Delta$.
Solve the simultaneous equations. DKI = 1: direct substitution. DKI ≥ 2: Gaussian elimination or matrix inversion $\{\theta\} = [K]^{-1}\{-M_{unbalanced}\}$.
Back-substitute into the SDEs. Compute all member-end moments $M_{AB}$, $M_{BA}$, etc.
Find reactions and draw BMD and SFD. Free body of each span gives vertical reactions. Verify: $\sum M_{joint}=0$ at each free joint and overall $\sum F=0$.

Example 1

Worked Example: Two-Span Continuous Beam

Problem: Beam ABC - fixed at A, roller at interior support B, pin at C. Span AB = $4\,\text{m}$ (UDL $w=30\,\text{kN/m}$), span BC = $6\,\text{m}$ (central load $P=60\,\text{kN}$). Constant $EI$. Find all end moments, reactions, BMD and SFD.

Step 1: Unknowns - DKI = 1

$\theta_A=0$ (fixed). $\theta_B$ = unknown. Far end C pinned: use modified SDE for BC (eliminates $\theta_C$). One unknown: $\theta_B$.

Step 2: Fixed-end moments

$M_{AB}^F = +wL_1^2/12 = +30\times16/12$ $= +40\,\text{kN·m}$

$M_{BA}^F = -wL_1^2/12$ $= -40\,\text{kN·m}$

$M_{BC}^F = +PL_2/8 = +60\times6/8$ $= +45\,\text{kN·m}$

$M_{CB}^F = -PL_2/8$ $= -45\,\text{kN·m}$

Modified FEM for BC: $M_{BC}^{F*} = M_{BC}^F - \tfrac{1}{2}M_{CB}^F = 45 - \tfrac{1}{2}(-45)$ $= +67.5\,\text{kN·m}$

Step 3: Slope-deflection equations ($\psi=0$)

$$M_{AB} = \frac{2EI}{4}(0 + \theta_B) + 40 = 0.5EI\,\theta_B + 40$$ $$M_{BA} = \frac{2EI}{4}(2\theta_B + 0) - 40 = EI\,\theta_B - 40$$ $$M_{BC} = \frac{3EI}{6}\,\theta_B + 67.5 = 0.5EI\,\theta_B + 67.5 \quad\text{(modified SDE)}$$ Three end-moment expressions in terms of the single unknown $\theta_B$

Step 4: Joint equilibrium at B

$$M_{BA} + M_{BC} = 0 \;\Rightarrow\; (EI\,\theta_B - 40) + (0.5EI\,\theta_B + 67.5) = 0$$ $$1.5EI\,\theta_B = -27.5 \;\Rightarrow\; \boxed{\theta_B = \frac{-18.33}{EI}}\,\text{rad (anticlockwise)}$$

Step 5: Final end moments

$M_{AB}$ $0.5EI\times(-18.33/EI)+40 = \mathbf{+30.83\,\text{kN·m}}$

$M_{BA}$ $EI\times(-18.33/EI)-40 = \mathbf{-58.33\,\text{kN·m}}$

$M_{BC}$ $0.5EI\times(-18.33/EI)+67.5 = \mathbf{+58.33\,\text{kN·m}}$

$M_{CB}$ $\mathbf{0}$ (pinned end C)

Check at B: $M_{BA} + M_{BC} = -58.33 + 58.33 = 0$ ✓

Reactions

$V_A$ (moment about B on span AB: $V_A\times4 = wL_1\times2 - (M_{BA}-M_{AB})$) $V_A = (120 - 27.5)/4 = \mathbf{66.87\,\text{kN}}$ $\uparrow$

$V_B^{AB} = wL_1 - V_A$ $= 120 - 66.87 = \mathbf{53.13\,\text{kN}}$ $\uparrow$

$V_C$ (moment about B on span BC: $V_C\times6 = P\times3 - M_{BC}$) $V_C = (180-58.33)/6 = \mathbf{20.28\,\text{kN}}$ $\uparrow$

$V_B^{BC} = P - V_C$ $= 60 - 20.28 = \mathbf{39.72\,\text{kN}}$ $\uparrow$

$V_B$ total $53.13 + 39.72 = \mathbf{91.85\,\text{kN}}$ $\uparrow$

Diagrams

BMD and SFD - Two-Span Continuous Beam

Section	BM (kN·m)	Nature
A (fixed end)	+30.83 (CW moment = hogging at wall)	Hogging
Max sagging in AB: $x=V_A/w=66.87/30=2.23\,\text{m}$ from A	$66.87\times2.23 - 30\times2.23^2/2 - 30.83 = +43.5$	Sagging
B (interior support)	−58.33	Hogging
Mid-BC: $x=3\,\text{m}$ from B	$39.72\times3 - 58.33 = +60.83$	Sagging
C (pinned end)	0	-

Bending moment diagram

Shear force diagram

Example 2

Worked Example: Symmetric Portal Frame

Problem: Portal frame ABCD - pinned at A and D (bases). Columns AB and DC: height $h=3\,\text{m}$, stiffness $EI$. Beam BC: span $L=6\,\text{m}$, stiffness $2EI$. UDL $w=20\,\text{kN/m}$ on beam BC. No sway (symmetric). Find all end moments and draw the BMD.

Step 1: Unknowns - by symmetry DKI = 1

Pinned bases A and D: $M_{AB}=M_{DC}=0$. By symmetry: $\theta_B = \theta_C$ and $\theta_A = \theta_D$. Use modified SDE for columns (pinned base). One unknown: $\theta_B$.

Step 2: FEM for beam BC (UDL $w=20\,\text{kN/m}$, span 6 m, stiffness $2EI$)

$$M_{BC}^F = +\frac{wL^2}{12} = +\frac{20\times36}{12} = +60\,\text{kN·m} \qquad M_{CB}^F = -60\,\text{kN·m}$$ Columns carry no span load: $M^F=0$ for both columns

Step 3: Slope-deflection equations

Column AB (modified SDE, pinned at A, $\psi=0$):

$$M_{BA} = \frac{3EI}{h}\,\theta_B = \frac{3EI}{3}\,\theta_B = EI\,\theta_B$$ Pinned base: stiffness $3EI/h$ eliminates $\theta_A$ term entirely

Beam BC (by symmetry $\theta_C = \theta_B$, $\psi=0$, stiffness $2EI$):

$$M_{BC} = \frac{2(2EI)}{6}(2\theta_B + \theta_C) + 60 = \frac{4EI}{6}\times3\theta_B + 60 = 2EI\,\theta_B + 60$$ $\theta_C = \theta_B$ used by symmetry; $2EI/L_2 = 2EI/6 = EI/3$, so $2\times(EI/3)\times3 = 2EI$

Step 4: Joint equilibrium at B

$$M_{BA} + M_{BC} = 0 \;\Rightarrow\; EI\,\theta_B + 2EI\,\theta_B + 60 = 0$$ $$3EI\,\theta_B = -60 \;\Rightarrow\; \boxed{\theta_B = \frac{-20}{EI}}$$

Step 5: Final end moments

$M_{AB} = M_{DC} = 0$ (pinned bases)$\mathbf{0}$

$M_{BA}$ (top of left column)$EI\times(-20/EI) = \mathbf{-20\,\text{kN·m}}$

$M_{BC}$ (left end of beam)$2EI\times(-20/EI)+60 = -40+60 = \mathbf{+20\,\text{kN·m}}$

$M_{CB}$, $M_{CD}$ (by symmetry)$\mathbf{-20\,\text{kN·m}}$, $\mathbf{+20\,\text{kN·m}}$

Check at B: $M_{BA}+M_{BC}=-20+20=0$ ✓. The beam end moment is reduced from 60 kN·m (FEM for fixed-fixed) to 20 kN·m due to column flexibility.

Midspan beam moment: $M_{mid} = wL^2/8 - M_{BC} = 20\times36/8 - 20 = 90-20 = \mathbf{70\,\text{kN·m}}$ (sagging).

Diagram

Portal Frame BMD

BMD plotted on tension side. Columns: linear from 0 at base to 20 kN·m at top. Beam: parabolic from 20 kN·m (hogging at ends) to 70 kN·m (sagging at midspan).

Advanced Topic

Sway Frames: Adding the Shear Condition

When a frame has asymmetric loading or geometry, or carries a direct horizontal force, it sways laterally. Each storey sway displacement $\Delta$ becomes an additional unknown, requiring an extra equation from horizontal force equilibrium.

Chord rotation due to sway

$$\psi = \frac{\Delta}{h}$$ For a column of height $h$ with relative sway $\Delta$ at the top

Column shear from end moments

$$V_{col} = \frac{M_{top} + M_{bottom}}{h}$$ Horizontal shear in column - from equilibrium of the column free body

Storey shear equilibrium (the sway equation)

$$\sum V_{col} = H_{applied,\,storey}$$ Sum of all column shears in a storey = applied horizontal force at that storey level

SDEs for a fixed-base portal frame with sway

Columns of height $h$, beam span $L$, all $EI$, horizontal load $H$ at B, fixed bases. Unknowns: $\theta_B$, $\theta_C$, $\Delta$.

$$M_{AB} = \frac{2EI}{h}(\theta_B - 3\psi) \qquad M_{BA} = \frac{2EI}{h}(2\theta_B - 3\psi)$$ $$M_{DC} = \frac{2EI}{h}(\theta_C - 3\psi) \qquad M_{CD} = \frac{2EI}{h}(2\theta_C - 3\psi)$$ $\theta_A=\theta_D=0$ (fixed bases); $\psi=\Delta/h$ (same for both columns)

Three equations: $\sum M_B=0$, $\sum M_C=0$, and the sway condition $V_{AB}+V_{DC}=H$. Solved simultaneously for $\theta_B$, $\theta_C$, $\Delta$.

Sway vs non-sway identification

Condition	Sways?	Reason
Symmetric frame, symmetric vertical loading	No	Column horizontal forces cancel by symmetry
Symmetric frame, asymmetric loading	Yes	Net horizontal force at beam level
Any frame with horizontal load	Yes	Direct storey shear from applied force
One column stiffer than the other	Yes (even under symmetric load)	Unequal stiffness creates net horizontal force
Multi-storey frame	Each storey has own $\Delta_i$	One extra DOF and one shear equation per storey

Common error: Forgetting $-3\psi$ in the SDE for sway frames. This violates horizontal equilibrium and gives wrong column moments. Always verify by summing column shears and comparing to applied horizontal force.

Methods Compared

Displacement Method vs Force Method

Both methods analyse the same structures and give identical results. The choice affects computational effort only.

Aspect	Displacement method (stiffness)	Force method (flexibility)
Primary unknowns	Joint rotations/displacements $\{\theta,\Delta\}$	Redundant forces/moments $\{X\}$
Governing equations	Equilibrium: $[K]\{\delta\}=\{F\}$	Compatibility: $[f]\{X\}=-\{\Delta_0\}$
Matrix properties	Symmetric, positive definite, banded	Symmetric (Maxwell's theorem)
Number of equations	DKI	DSI
Preferred when	DKI << DSI; computer use; high redundancy	DSI << DKI; hand calc; few redundants
Primary structure	Fully locked structure (all DOF = 0)	Released structure (redundants removed)
Support settlement	Known non-zero DOF in $\{\delta\}$ vector	Non-zero RHS in compatibility equation
Temperature effects	Equivalent nodal loads from thermal FEM	Additional free displacements $\Delta_{i,T}$
Computer implementation	Direct stiffness - basis of all FEM software	Rarely used in software
Iterative variant	Moment distribution (Hardy Cross, 1930)	No standard iterative form

Moment distribution method

The moment distribution method (Hardy Cross, 1930) is an iterative solution of the displacement method equations without forming the stiffness matrix explicitly. Starting from the locked structure (FEMs at all joints), it cycles through joints repeatedly: distribute the unbalanced moment to connected members using DFs, carry over half to far ends, repeat until convergence. Each cycle reduces the residual unbalanced moment, and the method converges to the exact displacement method answer.

$$\text{Distributed moment to member } i = \text{DF}_i \times (-M_{unbalanced,\,joint})$$ $$\text{Carry-over moment} = \frac{1}{2} \times \text{distributed moment} \quad \text{(far end fixed)}$$ Moment distribution cycle - repeated until unbalanced moment at each joint < 0.5% of initial FEM

When to use which method: For hand analysis of frames with DKI ≤ 3, the slope-deflection method (direct solution) is fastest. For multi-span beams, moment distribution is often quickest. For any computer analysis or DKI > 4, the direct stiffness method (matrix formulation) is used - it is the basis of all structural engineering software.

Help

Frequently Asked Questions

1. What is the displacement method of structural analysis?

The displacement method (stiffness method) takes joint displacements and rotations as primary unknowns and enforces equilibrium at each free joint to find them. The governing equation is [K]{delta} = {F}. It was formalised by G.A. Maney (1915) as the slope-deflection method and generalised into the direct stiffness method by Turner et al. (1956) - the basis of all modern FEM software.

2. What is the degree of kinematic indeterminacy (DKI)?

DKI is the number of independent joint displacements and rotations that are unknown. It equals the number of equilibrium equations to solve. For 2D frames: DKI = 3j - r - C_i where j = free joints, r = restrained DOF, C_i = internal condition equations. For beams neglecting axial deformation, DKI counts rotational DOF at interior supports plus sway DOF per storey.

3. What are fixed-end moments and why are they needed?

Fixed-end moments (FEM) are the moments at both ends of a member when all joint displacements are locked (zero) and the member carries span loading. They form the starting point for every displacement method analysis. Without FEM, the slope-deflection equations only capture joint rotation effects and miss the direct contribution of span loads. Common examples: wL²/12 for UDL, PL/8 for central point load.

4. What is the slope-deflection equation?

M_AB = (2EI/L)(2theta_A + theta_B - 3psi) + M_AB^F. It expresses the end moment of member AB in terms of both end rotations (theta_A, theta_B), chord rotation due to sway (psi = Delta/L), and the fixed-end moment. G.A. Maney derived it in 1915 as a systematic approach to displacement method analysis.

5. When should the modified slope-deflection equation be used?

The modified SDE is used when the far end of a member is a pin or roller (zero moment condition). M_AB = (3EI/L)(theta_A - psi) + M_AB^F*, where M_AB^F* = M_AB^F - (1/2)M_BA^F. It reduces the near-end stiffness from 4EI/L to 3EI/L, eliminates the far-end rotation as an unknown, and simplifies the analysis by reducing DKI.

6. What is the carry-over factor?

The carry-over factor (COF) is the fraction of an applied moment at one end that is transferred to the far end. For a prismatic beam with the far end fixed, COF = 1/2: applying moment M at the near end produces M/2 at the fixed far end. For a beam with the far end pinned, COF = 0. The COF appears as the off-diagonal terms (2EI/L) in the member stiffness matrix.

7. What is the distribution factor in moment distribution?

The distribution factor (DF) for member i at a joint is the fraction of unbalanced moment absorbed by that member: DF_i = k_i / sum(k_j), where k = 4EI/L (far end fixed) or 3EI/L (far end pinned). The sum of DFs at any free joint is always 1. At a fixed end, DF = 0 (fixed support absorbs all moment with zero rotation).

8. What is a sway frame and how is it handled?

A sway frame is one where joints displace laterally under loading. Each storey sway Delta introduces a chord rotation psi = Delta/h in all columns of that storey and adds one unknown. One additional equation is provided by horizontal force equilibrium: sum of column shears = applied horizontal force. This 'sway equation' is V_col = (M_top + M_bottom)/h for each column.

9. How does support settlement affect the displacement method?

Settlement at a support is a known non-zero displacement, not an unknown. In the slope-deflection equation, the chord rotation psi = delta/L is treated as a known value. This modifies the FEM: equivalent additional FEM due to settlement = 6EI*delta/L^2 at both ends (same sign). The equilibrium equations are then solved as normal with this modified FEM on the RHS.

10. How do you check displacement method results?

Three checks: (1) Joint equilibrium - sum of all member-end moments at each free joint must equal the applied external moment (usually zero). (2) Global equilibrium - sum of vertical reactions = total vertical load; sum of horizontal reactions = total horizontal load. (3) For sway frames: verify sum of column shears equals the applied storey force.

11. What is the relationship between slope-deflection and stiffness matrix methods?

They are the same method expressed differently. Assembling all slope-deflection equations and equilibrium conditions produces exactly the matrix equation [K]{theta} = {-M_unbalanced}. The direct stiffness method formalises this assembly process and extends it to include translational DOF, 3D structures, and axial deformation. Every FEM program uses the stiffness matrix formulation.

12. What is moment distribution and how does it relate to the displacement method?

Moment distribution (Hardy Cross, 1930) is an iterative solution of the displacement method equations without explicitly forming [K]. Starting from the locked structure (FEMs computed), it distributes unbalanced moments to connected members using distribution factors, carries over half to far ends, and repeats until convergence. The final result is identical to the slope-deflection solution. It is most efficient for non-sway beams and frames with few joints.

13. Why is the stiffness matrix always positive definite?

The stiffness matrix [K] is positive definite when the structure is properly supported (no rigid body modes). This guarantees that the strain energy is positive for any non-zero displacement: {delta}^T [K] {delta} > 0. In practice it means [K] is always invertible and the system [K]{delta} = {F} always has a unique solution. The flexibility matrix [f] is positive definite by a dual argument.

14. When does the force method have fewer equations than the displacement method?

The force method has DSI equations; the displacement method has DKI equations. The force method wins (fewer equations) when DSI < DKI, which occurs for lightly redundant structures with many free joints. Example: a propped cantilever has DSI = 1 but DKI = 1 (equal). A continuous beam with 5 spans has DSI = 4 and DKI = 4 (also equal). A multi-bay multi-storey frame typically has DKI << DSI, favouring the displacement method strongly.

15. Can the displacement method handle non-prismatic (variable section) members?

Yes. For non-prismatic members, the stiffness coefficients (4EI/L, 2EI/L) are replaced by general stiffness factors computed by integration of the varying EI(x) over the member length using the conjugate beam or virtual work method. The slope-deflection equations take the same form but with modified stiffness coefficients. FEM software handles this automatically through numerical integration of element stiffness matrices.

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