13.2 Moment Distribution Method and Slope Deflection

13.2 Moment Distribution Method and Slope Deflection

While the Force Method treats redundant forces as primary unknowns, displacement methods formulate unknown joint rotations ($\theta$) and translations ($\Delta$). Because the Degree of Kinematic Indeterminacy ($DKI$) is often smaller than the static indeterminacy in multi-bay continuous systems, displacement methods are exceptionally well-suited for rigid frames. This section explores the Slope Deflection Method (the analytical precursor to modern stiffness matrix analysis) and the Moment Distribution Method (the renowned iterative relaxation algorithm developed by Hardy Cross in 1930).

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1. The Slope Deflection Method

Formulated by George A. Maney in 1915, the Slope Deflection Method expresses the internal bending moments at the ends of each structural member as linear functions of member end rotations, chord displacement, and applied span loads.

The Fundamental Slope Deflection Equations

For a prismatic member $ij$ of length $L$ and flexural rigidity $EI$:

$M_{ij} = \text{FEM}_{ij} + \frac{2EI}{L} \left( 2\theta_i + \theta_j - \frac{3\Delta}{L} \right)$ $M_{ji} = \text{FEM}_{ji} + \frac{2EI}{L} \left( 2\theta_j + \theta_i - \frac{3\Delta}{L} \right)$

Defining the chord rotation as $\psi = \frac{\Delta}{L}$:

$M_{ij} = \text{FEM}_{ij} + \frac{2EI}{L} \left( 2\theta_i + \theta_j - 3\psi \right)$

Where:

• $M_{ij}$ = internal bending moment acting on the end of member $ij$ at joint $i$ (clockwise positive).
• $\text{FEM}_{ij}$ = Fixed-End Moment at end $i$ resulting from transverse span loads (clockwise positive).
• $\theta_i, \theta_j$ = joint rotations at ends $i$ and $j$ in radians (clockwise positive).
• $\Delta$ = relative lateral displacement (drift) between ends $i$ and $j$ perpendicular to the member axis.
• $\psi = \Delta / L$ = rigid-body chord rotation angle (clockwise positive).

Modified Equation for a Pinned Far End

If far end $j$ is pinned or on a roller support, the end moment $M_{ji} = 0$. Substituting $M_{ji} = 0$ eliminates $\theta_j$ from the formulation, yielding the modified slope deflection equation for end $i$:

$M_{ij} = \text{FEM}_{ij} - \frac{\text{FEM}_{ji}}{2} + \frac{3EI}{L} (\theta_i - \psi)$

This modification reduces the number of simultaneous linear equations by one for every external pinned support.

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2. Table of Fixed-End Moments (FEM)

Fixed-end moments are internal reactions developed at fully clamped boundaries. In slope deflection analysis, clockwise end moments are positive and counterclockwise moments are negative.

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3. Analysis of Frames: Non-Sway vs. Sidesway

Non-Sway Frames

A frame is prevented from sidesway if:

1. It is symmetrically geometry-loaded and symmetrically supported, or
2. It is laterally braced against displacement by shear walls, bracing trusses, or external rigid supports. In non-sway analysis, chord rotation $\psi = 0$. The only unknowns are the joint rotations $\theta$. For each free joint $i$, formulate joint equilibrium:

$\sum M_i = 0 \implies \sum_{j} M_{ij} = 0$

Sway Frames

When a frame is unbraced, unsymmetrical, or subjected to lateral loads, lateral drift $\Delta$ occurs. Each independent lateral translation introduces an unknown chord rotation $\psi = \Delta / h$. To solve for $\psi$, an additional story shear equation is required from horizontal equilibrium:

$\sum H = 0 \implies \sum V_{\text{columns}} + P_{\text{lateral}} = 0$

For any column $AB$ of height $h$ with end moments $M_{AB}$ and $M_{BA}$:

$V_{AB} = \frac{M_{AB} + M_{BA}}{h}$

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4. The Moment Distribution Method (Hardy Cross)

The Moment Distribution Method solves continuous beams and rigid frames through successive numerical approximations without setting up simultaneous equations. Joints are initially locked against rotation, then systematically unlocked, balanced, and re-locked until unbalanced moments diminish to zero.

Member Stiffness Factor ($K$)

The flexural stiffness factor $K$ represents the moment required to produce a unit rotation ($\theta = 1\text{ rad}$) at one end while the far end boundary is maintained:

• Far End Fixed: $K = \frac{4EI}{L}$ (or relative stiffness $k = \frac{I}{L}$).
• Far End Pinned or Roller: $K = \frac{3EI}{L}$ (or relative stiffness $k = \frac{3}{4} \frac{I}{L} = 0.75 \frac{I}{L}$).

Distribution Factor ($DF$)

The Distribution Factor at a joint defines the proportion of an unbalanced moment absorbed by each framing member:

$DF_i = \frac{K_i}{\sum K}$

• At an ideal fixed wall support: $K = \infty \implies DF = 0$ (the fixed support absorbs all moment without rotating).
• At an exterior pinned/roller end: $K_{\text{support}} = 0 \implies DF = 1.0$.
• At an interior rigid joint connecting $m$ members: $\sum_{i=1}^m DF_i = 1.0$.

Carry-Over Factor ($COF$)

When a moment $M$ is applied to balance a joint, an induced moment travels to the opposite end of the member:

• To a rigidly fixed far end: $COF = +0.50$.
• To a pinned or roller far end (when modified stiffness $3EI/L$ is used): $COF = 0$.

Symmetrical Loading Simplifications

When a structure and its loading possess geometric symmetry, computational effort is substantially reduced by analyzing half the structure using modified stiffness:

• Symmetric Deformation (Equal and opposite rotations, $\theta_j = -\theta_i$): Axis of symmetry behaves as a fixed guided support. Modified stiffness is: $K_{\text{sym}} = \frac{2EI}{L} = 0.5 K_{\text{standard}}$
• Antisymmetric Deformation (Identical rotations, $\theta_j = \theta_i$): Axis of symmetry behaves as a simple pin ($M = 0$). Modified stiffness is: $K_{\text{antisym}} = \frac{6EI}{L} = 1.5 K_{\text{standard}}$

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5. Comprehensive Worked Example

Worked Example: Two-Span Continuous Beam with Pinned Exterior End

Problem: A continuous beam $ABC$ rests on simple supports at $B$ and $C$ and is rigidly clamped at support $A$. Span $AB = 6.0\text{ m}$ with flexural rigidity $EI$; Span $BC = 4.0\text{ m}$ with flexural rigidity $EI$. Span $AB$ carries a uniform load $w = 24.0\text{ kN/m}$. Span $BC$ carries a central concentrated load $P = 40.0\text{ kN}$ at mid-span. Determine the final support moments using the Moment Distribution Method with modified stiffness for pinned end $C$.

[Fixed A]======== 6.0 m ========▲ B ======== 4.0 m ========▲ C (Pin)
           w = 24 kN/m                       P = 40 kN

Solution:

• Step 1: Member Stiffness and Distribution Factors:

  • Member $BA$ (far end $A$ is fixed): $K_{BA} = \frac{4EI}{L} = \frac{4EI}{6.0} = 0.667 EI$.
  • Member $BC$ (far end $C$ is pinned, using modified stiffness): $K_{BC} = \frac{3EI}{L} = \frac{3EI}{4.0} = 0.750 EI$.
  • Total stiffness at Joint $B$: $\sum K_B = 0.667 EI + 0.750 EI = 1.417 EI$.
  • Distribution Factors at $B$: $DF_{BA} = \frac{0.667}{1.417} = 0.471$ $DF_{BC} = \frac{0.750}{1.417} = 0.529$ Check: $0.471 + 0.529 = 1.000$.
  • At support $A$ (fixed): $DF_{AB} = 0$.
  • At support $C$ (pinned): Using $K = 3EI/L$, $C$ is released initially; $DF_{CB} = 1.0$, $COF_{B \to C} = 0$.

• Step 2: Fixed-End Moments (FEM):

  • Span $AB$ ($w = 24\text{ kN/m}$, $L = 6\text{ m}$): $\text{FEM}_{AB} = -\frac{wL^2}{12} = -\frac{24(6^2)}{12} = -72.00\text{ kN}\cdot\text{m}$ $\text{FEM}_{BA} = +\frac{wL^2}{12} = +72.00\text{ kN}\cdot\text{m}$
  • Span $BC$ ($P = 40\text{ kN}$, $L = 4\text{ m}$, pin at $C$): Standard fixed-end moment is $\text{FEM}_{BC} = -\frac{PL}{8} = -\frac{40(4)}{8} = -20.00\text{ kN}\cdot\text{m}$, and $\text{FEM}_{CB} = +20.00\text{ kN}\cdot\text{m}$. Releasing pinned joint $C$ upfront, the modified fixed-end moment at $B$ is: $\text{FEM}'_{BC} = \text{FEM}_{BC} - \frac{1}{2} \text{FEM}_{CB} = -20.00 - \frac{1}{2}(+20.00) = -30.00\text{ kN}\cdot\text{m}$ $\text{FEM}'_{CB} = 0.00\text{ kN}\cdot\text{m}$

• Step 3: Moment Distribution Table:

Computation Detail for Joint B Balance:

• Unbalanced moment at $B = +72.00 + (-30.00) = +42.00\text{ kN}\cdot\text{m}$.

• Balancing moments: $\Delta M_{BA} = -0.471 \times (+42.00) = -19.78\text{ kN}\cdot\text{m}$; $\Delta M_{BC} = -0.529 \times (+42.00) = -22.22\text{ kN}\cdot\text{m}$.

• Carry-over from $BA$ to fixed end $AB$: $-19.78 \times 0.50 = -9.89\text{ kN}\cdot\text{m}$.

• Carry-over from $BC$ to pinned end $CB$: $0$ (since modified stiffness $3EI/L$ was used).

• Because joint $B$ is balanced and no carry-overs return to $B$, the solution converges in a single distribution step!

• Step 4: Final Results:

  • $M_{AB} = -81.89\text{ kN}\cdot\text{m}$ (Counterclockwise on member, acts clockwise on support $A$).
  • $M_{BA} = +52.22\text{ kN}\cdot\text{m}$ (Clockwise).
  • $M_{BC} = -52.22\text{ kN}\cdot\text{m}$ (Counterclockwise).
  • $M_{CB} = 0.00\text{ kN}\cdot\text{m}$ (Pinned end).
  • Joint $B$ equilibrium: $M_{BA} + M_{BC} = +52.22 - 52.22 = 0$. (Exact equilibrium verified).

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6. Licensure Exam Pitfalls & Review Notes