Three-Moment Equation Calculator

Solve a two-span continuous beam with the classical three-moment equation to find the interior support moment and midspan moments.

📏 Three-Moment Equation Calculator
Span 1 length (L1)6 m
m
115
Span 2 length (L2)6 m
m
115
UDL on span 1 (w1)10 kN/m
kN/m
150
UDL on span 2 (w2)10 kN/m
kN/m
150
Interior support moment (MB)
—
Midspan moment, span 1
—
Midspan moment, span 2
—
Step-by-step working

📏 What is the Three-Moment Equation?

The three-moment equation, also known as Clapeyron's theorem of three moments, is a classical structural analysis method that finds the unknown bending moments at the interior supports of a continuous beam. Developed by Benoit Paul Emile Clapeyron in the 19th century, it relates the moments at any three consecutive supports to the loads carried by the two spans between them, giving a direct algebraic equation instead of requiring integration of the full deflection curve.

Civil and structural engineers use this method when analyzing continuous beams and bridge girders that span across more than one support, a very common real-world configuration. A two-span floor beam resting on two end walls and one intermediate column, a bridge deck carried on three piers, and a continuous roof purlin over several rafters are all classic candidates for the three-moment equation. Before computer-based structural analysis software became standard, this was one of the primary hand-calculation methods taught for indeterminate beams.

A common point of confusion is assuming the three-moment equation and the moment distribution method give different answers because they use different setups. They do not, both are exact classical methods for the identical statically indeterminate beam problem and always converge on the same support and midspan moments. The three-moment equation solves the compatibility condition directly in closed form, while moment distribution reaches the same answer iteratively through distribution factors and moment balancing.

This calculator solves the most common textbook case directly: a two-span continuous beam with simple (pinned) outer supports and one continuous interior support, each span carrying its own uniformly distributed load. It returns the interior support moment M_B, the midspan moment in each span, and plots the full bending moment diagram across both spans.

📐 Formula

MAL1 + 2MB(L1+L2) + MCL2 = −(w1L1³)/4 − (w2L2³)/4
The general three-moment (Clapeyron) equation for consecutive supports A, B, C.
Since A and C are simple supports, MA = MC = 0, which reduces the equation to a single unknown, MB.
MB = −(w1L1³ + w2L2³) / (8(L1+L2))
MB = bending moment at the interior support (kN·m), negative means hogging
L1, L2 = span 1 and span 2 lengths (m)
w1, w2 = uniformly distributed loads on span 1 and span 2 (kN/m)
Example: L1 = L2 = 6 m, w1 = w2 = 10 kN/m → MB = −45 kN·m, matching the textbook symmetric result MB = −wL²/8.
Mmid = wL²/8 + MB/2
Midspan moment for a span carrying UDL w, with a simple far end (moment 0) and near-end moment MB.
Applied once for span 1 (using w1, L1) and once for span 2 (using w2, L2), per IS 456:2000 continuous-beam analysis conventions.

📖 How to Use This Calculator

Steps

1
Enter both span lengths. Type L1 and L2 in meters for the two spans of the continuous beam.
2
Enter both distributed loads. Type w1 and w2 in kilonewtons per meter for the uniformly distributed load on each span.
3
Read the interior support moment and midspan moments. See M_B at the interior support and the midspan moment for each span, with the full bending moment diagram plotted.

💡 Example Calculations

Example 1 — Symmetric Two-Span Beam

Equal spans: L1 = L2 = 6 m, equal UDL: w1 = w2 = 10 kN/m

1
MB = −(10×6³ + 10×6³) / (8×(6+6)) = −(2160+2160)/96
2
MB = −45.00 kN·m (matches textbook −wL²/8 = −45)
3
Midspan (each span, by symmetry) = 10×6²/8 + (−45)/2 = 45 − 22.5 = 22.50 kN·m
MB = −45.00 kN·m, midspan moment = 22.50 kN·m (both spans)
Try this example →

Example 2 — Asymmetric Spans and Loads

Unequal spans: L1 = 5 m, L2 = 7 m, unequal UDL: w1 = 8 kN/m, w2 = 12 kN/m

1
MB = −(8×5³ + 12×7³) / (8×(5+7)) = −(1000+4116)/96
2
MB = −53.29 kN·m
3
Midspan span 1 = 8×5²/8 + (−53.29)/2 = 25 − 26.65 = −1.65 kN·m (still hogging, since span 1 is short relative to the heavier span 2). Midspan span 2 = 12×7²/8 + (−53.29)/2 = 73.5 − 26.65 = 46.85 kN·m
MB = −53.29 kN·m, midspan span 1 = −1.65 kN·m, midspan span 2 = 46.85 kN·m
Try this example →

Example 3 — Short Equal Spans, Heavy Load

Equal spans: L1 = L2 = 4.5 m (rounded example uses L1=4, L2=5), heavy UDL: w1 = w2 = 15 kN/m

1
MB = −(15×4³ + 15×5³) / (8×(4+5)) = −(960+1875)/72
2
MB = −39.38 kN·m
3
Midspan span 1 = 15×4²/8 + (−39.38)/2 = 30 − 19.69 = 10.31 kN·m. Midspan span 2 = 15×5²/8 + (−39.38)/2 = 46.88 − 19.69 = 27.19 kN·m
MB = −39.38 kN·m, midspan span 1 = 10.31 kN·m, midspan span 2 = 27.19 kN·m
Try this example →

❓ Frequently Asked Questions

What is the three-moment equation?+
The three-moment equation (Clapeyron's theorem) relates the bending moments at three consecutive supports of a continuous beam to the loads on the two spans between them: M_A*L1 + 2*M_B*(L1+L2) + M_C*L2 = -(w1*L1^3)/4 - (w2*L2^3)/4 for uniformly distributed loads.
What is the formula for M_B in a two-span beam with simple end supports?+
When the outer supports A and C are simple (M_A = M_C = 0), the equation reduces to M_B = -(w1*L1^3 + w2*L2^3) / (8*(L1+L2)), where w1 and w2 are the uniformly distributed loads on span 1 and span 2 and L1, L2 are the span lengths.
Why is M_B negative?+
A negative interior support moment means hogging, tension on the top fiber of the beam at that support, which is the expected behavior for a continuous beam carrying downward gravity loads. The magnitude, not the sign, is what matters for design checks.
What does the calculator assume about the end supports?+
This calculator assumes both outer supports (A and C) are simple supports, pins or rollers that carry no moment, so M_A = M_C = 0. This is the standard textbook two-span continuous beam case that the three-moment equation was originally developed to solve directly.
How do I find the midspan bending moment once I know M_B?+
For a span carrying UDL w with a simple far end (moment 0) and a near-end moment M_B, the midspan moment is M_mid = w*L^2/8 + M_B/2, the simple-beam midspan moment plus half the restraining end moment.
What is the sanity check for equal spans and equal loads?+
For two equal spans of length L each carrying the same UDL w, M_B = -(w*L^3 + w*L^3)/(8*(L+L)) = -w*L^2/8, the well-known textbook result for a symmetric two-span continuous beam, useful for verifying any three-moment equation calculation by hand.
Is the three-moment equation the same as the moment distribution method?+
No. Both solve the identical class of continuous-beam problem and give the same final moments, but the three-moment equation is a direct closed-form algebraic solution while moment distribution is an iterative method based on distribution factors and balancing fixed-end moments. They are two different classical techniques converging on the same answer.
What units does this calculator use?+
Span lengths L1 and L2 are entered in meters (m), and the uniformly distributed loads w1 and w2 are entered in kilonewtons per meter (kN/m). Results, the support moment M_B and both midspan moments, are shown in kilonewton-meters (kN·m).
Can the three-moment equation handle more than two spans?+
Yes, but each additional interior support adds another unknown moment and another three-moment equation, written between each successive triplet of supports, requiring a system of simultaneous equations. This calculator solves the simplest and most common case, a single interior support between two spans.
What causes the midspan moment in one span to become negative?+
If the interior support moment M_B is large relative to a span's own simple-beam moment (w*L^2/8), typically when that span is short or lightly loaded compared to the other span, the negative M_B/2 term can outweigh the positive simple-span term, leaving the whole span in hogging even at midspan.
Do I need to know the beam's material or cross-section to use this calculator?+
No. The three-moment equation for a uniformly loaded beam with simple end supports depends only on span lengths and applied UDLs, not on modulus of elasticity or moment of inertia, since EI cancels out of the compatibility equation when both spans share the same EI, the standard assumption.
Tags: Three Moment Equation Calculator Clapeyron Theorem Calculator Continuous Beam Moment Calculator Two Span Beam Support Moment Three Moment Theorem Calculator Formula Free Online

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What is the three-moment equation?

The three-moment equation (Clapeyron's theorem) relates the bending moments at three consecutive supports of a continuous beam to the loads on the two spans between them: M_A*L1 + 2*M_B*(L1+L2) + M_C*L2 = -(w1*L1^3)/4 - (w2*L2^3)/4 for uniformly distributed loads.

What is the formula for M_B in a two-span beam with simple end supports?

When the outer supports A and C are simple (M_A = M_C = 0), the equation reduces to M_B = -(w1*L1^3 + w2*L2^3) / (8*(L1+L2)), where w1 and w2 are the uniformly distributed loads on span 1 and span 2 and L1, L2 are the span lengths.

Why is M_B negative?

A negative interior support moment means hogging, tension on the top fiber of the beam at that support, which is the expected behavior for a continuous beam carrying downward gravity loads. The magnitude, not the sign, is what matters for design checks.

What does the calculator assume about the end supports?

This calculator assumes both outer supports (A and C) are simple supports, pins or rollers that carry no moment, so M_A = M_C = 0. This is the standard textbook two-span continuous beam case that the three-moment equation was originally developed to solve directly.

How do I find the midspan bending moment once I know M_B?

For a span carrying UDL w with a simple far end (moment 0) and a near-end moment M_B, the midspan moment is M_mid = w*L^2/8 + M_B/2, the simple-beam midspan moment plus half the restraining end moment.

What is the sanity check for equal spans and equal loads?

For two equal spans of length L each carrying the same UDL w, M_B = -(w*L^3 + w*L^3)/(8*(L+L)) = -w*L^2/8, the well-known textbook result for a symmetric two-span continuous beam, useful for verifying any three-moment equation calculation by hand.

Is the three-moment equation the same as the moment distribution method?

No. Both solve the identical class of continuous-beam problem and give the same final moments, but the three-moment equation is a direct closed-form algebraic solution while moment distribution is an iterative method based on distribution factors and balancing fixed-end moments. They are two different classical techniques converging on the same answer.

What units does this calculator use?

Span lengths L1 and L2 are entered in meters (m), and the uniformly distributed loads w1 and w2 are entered in kilonewtons per meter (kN/m). Results, the support moment M_B and both midspan moments, are shown in kilonewton-meters (kN·m).

Can the three-moment equation handle more than two spans?

Yes, but each additional interior support adds another unknown moment and another three-moment equation, written between each successive triplet of supports, requiring a system of simultaneous equations. This calculator solves the simplest and most common case, a single interior support between two spans.

What causes the midspan moment in one span to become negative?

If the interior support moment M_B is large relative to a span's own simple-beam moment (w*L^2/8), typically when that span is short or lightly loaded compared to the other span, the negative M_B/2 term can outweigh the positive simple-span term, leaving the whole span in hogging even at midspan.

Do I need to know the beam's material or cross-section to use this calculator?

No. The three-moment equation for a uniformly loaded beam with simple end supports depends only on span lengths and applied UDLs, not on modulus of elasticity or moment of inertia, since EI cancels out of the compatibility equation when both spans share the same EI, the standard assumption.

📌 Quick Tips

💡The three-moment equation directly relates the unknown support moments to the loads on the two spans it connects, one equation per interior support, so a beam with more spans needs one equation per additional interior support solved simultaneously.
💡When both outer supports are simple (pins or rollers), M_A and M_C drop out of the general equation entirely, leaving a single equation for the one unknown interior moment M_B, exactly the case this calculator solves.
💡Equal spans with equal loading reduce to the textbook result M_B = -wL^2/8, a useful mental check any time you plug numbers into the general formula.
💡A negative M_B means hogging (tension on top) at the interior support, the normal condition for a continuous beam under gravity loads.
💡Moment distribution method and slope-deflection method solve the exact same class of problem by a different, more iteration-friendly route, useful for beams with more than two spans.