Three-Moment Equation

An analytical method, derived from the Force Method, specifically formulated for the rapid analysis of continuous beams.

Introduction

Definition

The Three-Moment Equation, introduced by Clapeyron, is a powerful tool for analyzing statically indeterminate continuous beams. It directly relates the internal bending moments at three consecutive supports (or joints) to the loads applied on the two spans connecting those supports.

Purpose and Scope

  • It simplifies the Force Method for continuous beams by automatically satisfying rotation compatibility over interior supports.
  • It sets up a system of linear equations where the unknown variables are the bending moments over the supports.
  • Once the support moments are found, the continuous beam is broken down into simple spans, and the reactions and internal forces are calculated using basic statics.
  • It can account for variable member lengths, variable cross-sections (EIEI), and support settlements.

Clapeyron's Theorem of Three Moments

Historical Context & Principle

Originally formulated by French engineer Émile Clapeyron in 1857, the Theorem of Three Moments (or Clapeyron's Theorem) expresses the relationship between the bending moments at three consecutive supports of a continuous beam. It is fundamentally derived by enforcing the geometric compatibility condition that the slope of the elastic curve must be continuous (the same on just the left and just the right) over any interior support.

Treatment of Fixed Ends

How to apply the Three-Moment Equation when a continuous beam terminates at a fixed support.

The Imaginary Span Method

The standard Three-Moment Equation requires three supports to formulate one equation. If a beam ends at a fixed support (e.g., Support AA), there is no adjacent span to the left to form a triplet (Left-Middle-Right).
To solve this, we introduce an imaginary span (or dummy span) extending outward from the fixed support.
  • The length of this imaginary span is zero (L0=0L_0 = 0).
  • The moment of inertia of this imaginary span is infinite (I0=∞I_0 = \infty), making it perfectly rigid so no rotation occurs at the fixed end.
  • No loads are placed on this imaginary span, so its area term (A0A_0) is zero.
By applying the Three-Moment Equation to the newly created triplet (Imaginary Support - Actual Fixed Support - First Interior Support), we generate the necessary extra equation to solve for the unknown bending moment acting at the fixed support.

The General Equation

Consider three consecutive supports AA, BB, and CC of a continuous beam, separating two spans L1L_1 (between AA and BB) and L2L_2 (between BB and CC). The moments over the supports are MAM_A, MBM_B, and MCM_C.
MAL1I1+2MB(L1I1+L2I2)+MCL2I2=−6A1xˉ1L1I1−6A2xˉ2L2I2−6E(ΔA−ΔBL1+ΔC−ΔBL2) M_A \frac{L_1}{I_1} + 2M_B \left( \frac{L_1}{I_1} + \frac{L_2}{I_2} \right) + M_C \frac{L_2}{I_2} = - \frac{6A_1 \bar{x}_1}{L_1 I_1} - \frac{6A_2 \bar{x}_2}{L_2 I_2} - 6E \left( \frac{\Delta_A - \Delta_B}{L_1} + \frac{\Delta_C - \Delta_B}{L_2} \right)
Where:
  • MA,MB,MCM_A, M_B, M_C: Internal bending moments at supports A, B, and C.
  • L1,L2L_1, L_2: Lengths of the spans AB and BC.
  • I1,I2I_1, I_2: Moments of inertia for spans AB and BC.
  • EE: Modulus of elasticity (assumed constant for both spans).
  • A1,A2A_1, A_2: Areas of the simple-beam bending moment diagrams due to applied loads on spans AB and BC, assuming they are simply supported.
  • xˉ1\bar{x}_1: Distance from support A to the centroid of area A1A_1.
  • xˉ2\bar{x}_2: Distance from support C to the centroid of area A2A_2.
  • ΔA,ΔB,ΔC\Delta_A, \Delta_B, \Delta_C: Vertical displacements (settlements) of supports A, B, and C (positive downwards).

Load Terms (Right Side of Equation)

The terms 6AxˉL\frac{6A\bar{x}}{L} are standard load terms derived from the moment-area theorems. Common cases include:
  • Uniform distributed load ww over entire span LL: 6AxˉL=wL34\frac{6A\bar{x}}{L} = \frac{wL^3}{4}
  • Point load PP at distance aa from the left and bb from the right support:
    • Left side term (using distance from left support): Pab(L+a)L\frac{Pab(L+a)}{L}
    • Right side term (using distance from right support): Pab(L+b)L\frac{Pab(L+b)}{L}

Analysis Procedure

Procedure

  1. Identify all supports and the continuous spans between them.
  2. Apply the Three-Moment Equation to every set of three consecutive supports (e.g., spans 1-2, 2-3, 3-4, etc.).
  3. For fixed end supports, imagine an adjacent "dummy" span of zero length (L=0L=0, I=∞I=\infty) to create a three-support system. The moment at the actual fixed end becomes an unknown to solve.
  4. For pinned or roller exterior supports, the end moment is known to be zero (Mexterior=0M_{exterior} = 0).
  5. Calculate the load terms for each span based on the applied loading.
  6. Solve the resulting system of simultaneous equations to find the unknown support moments.
  7. Once support moments are known, break the continuous beam into individual simply supported spans. Apply the loads and the calculated support moments (as external couples) to find the final vertical support reactions using statics.

Interactive Tool: Three-Moment Equation Simulator

Visualize how the Three-Moment Equation determines internal moments over supports for continuous beams, demonstrating the relationship between span lengths, applied loads, and support moments.

Three-Moment Equation Interactive Lab

Adjust the span lengths and uniform loads for a two-span continuous beam. Exterior supports are simple pins/rollers ($M_A=0, M_C=0$). Observe how the internal moment at the middle support ($M_B$) changes.

Calculation

$M_A L_1 + 2M_B(L_1 + L_2) + M_C L_2 = dots$

$dots - rac{w_1 L_1^3}{4} - rac{w_2 L_2^3}{4}$

$0(5) + 2M_B(5 + 5) + 0(5) = dots$

$dots - rac{(10)(5)^3}{4} - rac{(10)(5)^3}{4}$

$20 M_B = -625.00$

Internal Moment $M_B$-31.25 kNm

Support Settlements

Incorporating Yielding Supports

The basic Clapeyron's equation assumes rigid, unyielding supports. If supports settle, the resulting relative displacements induce additional bending moments. The equation is modified by adding a settlement term to the right side:
M1L1+2M2(L1+L2)+M3L2=−6[A1xˉ1L1+A2xˉ2L2]+6EI[h1−h2L1+h3−h2L2]M_1 L_1 + 2M_2 (L_1 + L_2) + M_3 L_2 = -6 \left[ \frac{A_1 \bar{x}_1}{L_1} + \frac{A_2 \bar{x}_2}{L_2} \right] + 6EI \left[ \frac{h_1 - h_2}{L_1} + \frac{h_3 - h_2}{L_2} \right]
Where h1,h2,h3h_1, h_2, h_3 are the vertical downward settlements of supports 1, 2, and 3, respectively.
Key Takeaways
  • The Three-Moment Equation relates bending moments at three consecutive supports to the loads on the connecting spans.
  • It simplifies the analysis of continuous beams by enforcing slope compatibility at interior supports.
  • The equation requires calculating "load terms" which are derived from the simple-beam moment diagrams of each span.
  • Solving the simultaneous equations yields the support moments, from which all other internal forces and reactions can be found using statics.
  • It easily accommodates support settlements by including displacement terms in the general equation.
  • The Three-Moment Equation provides a relationship between internal bending moments at three consecutive supports of a continuous beam.
  • It is an application of the Force Method (consistency of deformation) specifically tailored for continuous beams.
  • Fixed ends are handled by introducing a fictitious span of zero length with an infinite moment of inertia.
  • Support settlements directly induce significant bending moments, which are integrated into the expanded equation.
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