Moment Distribution Method Calculator

Solve a two-span continuous beam under UDLs for the support moment and midspan moments using the moment distribution method.

📊 Moment Distribution Method Calculator
m
m
kN/m
kN/m
Support moment (MB)
—
Midspan moment, span 1
—
Midspan moment, span 2
—
Step-by-step working

📊 What is the Moment Distribution Method?

The moment distribution method is a classical hand-calculation technique, developed by Hardy Cross in 1930, for finding bending moments in continuous beams and rigid frames without solving a full system of simultaneous equations. It works by imagining every joint temporarily locked against rotation, computing the fixed-end moment each locked span would carry, then releasing each joint one at a time and distributing its unbalanced moment to the connected members in proportion to their relative stiffness, repeating until the structure settles into equilibrium.

This calculator applies the method to the simplest genuinely continuous case: a two-span beam with a pin or roller support at each outer end and one interior support that ties the two spans together, each span carrying its own uniformly distributed load, with constant flexural stiffness (EI) throughout. This is the standard first worked example in nearly every structural analysis course, because it needs only a single balancing step at the one interior joint, no iteration back and forth is required.

A common misconception is that fixed-end moments are always wL squared over 12. That value only applies when both ends of a span are genuinely fixed against rotation. When one end is a real pin (as with the outer supports here), the correct fixed-end moment at the other end, treating that span in isolation as a propped cantilever, is wL squared over 8, a modified-stiffness shortcut that lets a two-span problem like this one be solved in one pass instead of iterating.

This calculator carries out the full moment distribution narrative, fixed-end moments, modified stiffness, distribution factors, and the final balanced support moment, then cross-checks that result against the exact closed-form three-moment equation solution, and reports the midspan moment in each span along with the complete bending moment diagram.

📐 Formula

MB = −(w1L1³ + w2L2³) ÷ (8(L1+L2))
MB = balanced bending moment at the interior support (kN·m, negative = hogging)
w1, w2 = uniformly distributed load on span 1 and span 2 (kN/m)
L1, L2 = length of span 1 and span 2 (m)
Example: L1 = L2 = 6 m, w1 = w2 = 10 kN/m → MB = −(10×216 + 10×216)÷(8×12) = −45.00 kN·m, matching the classic wL²÷8 result for two equal spans.
Mmid = wL²÷8 + MB÷2
Applies separately to each span, using that span's own w and L, superposing the simple-beam midspan moment with the linear contribution of the support moment
Example: w = 10 kN/m, L = 6 m, MB = −45.00 kN·m → Mmid = 10×36÷8 + (−45.00)÷2 = 45.00 − 22.50 = 22.50 kN·m.

📖 How to Use This Calculator

Steps

1
Enter the span lengths. Type the length of span 1 (L1) and span 2 (L2), both in meters.
2
Enter the distributed loads. Type the uniformly distributed load on span 1 (w1) and span 2 (w2), both in kilonewtons per meter.
3
Read the balanced moments. Click Calculate to see the balanced support moment at the interior support, the midspan moments for both spans, and the full bending moment diagram.

💡 Example Calculations

Example 1 — Two Equal Spans, Equal Loads

L1 = L2 = 6 m, w1 = w2 = 10 kN/m

1
Modified FEM: FEM'BA = 10×6²÷8 = 45.00 kN·m; FEM'BC = 10×6²÷8 = 45.00 kN·m
2
KBA = KBC = 3÷6 = 0.5000, so DFBA = DFBC = 0.5000 (symmetric)
3
Unbalanced moment = 45.00 − 45.00 = 0.00, so no correction is needed by symmetry; MB = −45.00 kN·m
MB = −45.00 kN·m, midspan (both spans) = 22.50 kN·m
Try this example →

Example 2 — Unequal Spans and Loads

L1 = 6 m, L2 = 4 m, w1 = 10 kN/m, w2 = 8 kN/m

1
Modified FEM: FEM'BA = 10×6²÷8 = 45.00 kN·m; FEM'BC = 8×4²÷8 = 16.00 kN·m
2
KBA = 3÷6 = 0.5000; KBC = 3÷4 = 0.7500; DFBA = 0.4000; DFBC = 0.6000
3
Unbalanced = 45.00 − 16.00 = 29.00; balance = −29.00; ΔMBA = 0.4×(−29.00) = −11.60
4
MB = −(45.00 + (−11.60)) = −33.40 kN·m (matches the closed-form check exactly)
MB = −33.40 kN·m, midspan span 1 = 28.30 kN·m, midspan span 2 = −0.70 kN·m
Try this example →

Example 3 — Longer, Lighter Second Span

L1 = 5 m, L2 = 7 m, w1 = 12 kN/m, w2 = 6 kN/m

1
Modified FEM: FEM'BA = 12×5²÷8 = 37.50 kN·m; FEM'BC = 6×7²÷8 = 36.75 kN·m
2
KBA = 3÷5 = 0.6000; KBC = 3÷7 = 0.4286; DFBA = 0.5833; DFBC = 0.4167
3
Unbalanced = 37.50 − 36.75 = 0.75; balance = −0.75; ΔMBA = 0.5833×(−0.75) ≈ −0.4375
4
MB = −(37.50 + (−0.4375)) = −37.06 kN·m
MB = −37.06 kN·m, midspan span 1 = 18.97 kN·m, midspan span 2 = 18.22 kN·m
Try this example →

❓ Frequently Asked Questions

What is the moment distribution method?+
The moment distribution method is a hand-calculation technique for finding bending moments in continuous beams and frames. It starts by locking every joint against rotation and computing fixed-end moments, then releases each joint in turn, distributing the unbalanced moment to connected members in proportion to their relative stiffness, until the structure reaches equilibrium.
What is the formula for the support moment in a two-span continuous beam?+
For a two-span beam with pinned outer supports, span lengths L1 and L2, and uniformly distributed loads w1 and w2, the exact support moment is M_B = minus (w1 L1 cubed + w2 L2 cubed) divided by (8 times (L1+L2)), a hogging (negative) moment derived from the three-moment equation.
Why are the fixed-end moments here wL squared over 8 instead of wL squared over 12?+
This calculator uses the modified-stiffness shortcut for a member with a pinned far end, since supports A and C are simple (pinned/roller) supports, not other continuous joints. The correct fixed-end moment for a propped cantilever (fixed at one end, pinned at the other) under a UDL is wL squared over 8, not the wL squared over 12 value used for a beam fixed at both ends.
What are distribution factors and how are they calculated?+
A distribution factor tells you what share of an unbalanced joint moment each connected member absorbs, in proportion to its relative stiffness. For this two-span case, using modified stiffness K = 3 divided by span length (the 3/4 factor for a pinned far end already applied), DF_BA = K_BA divided by (K_BA + K_BC), and the two distribution factors always sum to exactly 1.0.
How do you find the midspan moment once the support moment is known?+
For a span with a simple support at one end and the balanced support moment M_B at the other, the moment at midspan is M_mid = wL squared over 8 (the ordinary simply supported midspan moment) plus M_B divided by 2 (half of the support moment, since the end-moment effect interpolates linearly from 0 at the simple end to M_B at the continuous end).
Does this calculator work for beams with more than two spans?+
No, this calculator is scoped to exactly two spans with one continuous interior support and simple (pinned/roller) supports at both outer ends. Beams with three or more spans have multiple interior joints that interact through carry-over moments, requiring several iterative balancing rounds instead of the single-step shortcut used here.
Do I need to know EI to find the support moment?+
No, as long as EI (modulus of elasticity times moment of inertia) is constant across both spans, it cancels out of the stiffness ratio used to compute the distribution factors, so the support moment depends only on the span lengths and the loads. EI is only needed if you go on to compute deflections, which this calculator does not cover.
What sign convention does this calculator use for bending moment?+
This calculator uses the sagging-positive convention: a positive moment means the beam sags (tension on the bottom fiber), and a negative moment means the beam hogs (tension on the top fiber), which is why the support moment over the continuous interior support is reported as a negative number.
What does it mean if the support moment matches wL squared over 8 exactly?+
That happens specifically when both spans have equal length and equal UDL, since the beam is then fully symmetric about the interior support and the classic two-equal-span continuous beam result, M_B = w L squared over 8 (hogging), applies exactly. It is a standard textbook check for this calculator's arithmetic.
Why is the interior support moment always hogging (negative)?+
A continuous beam over an interior support is pulled down on both sides by its own UDLs, which bends the beam upward at that support (tension on top, compression on bottom), the definition of hogging. This happens regardless of how unequal the two spans or loads are, as long as both spans carry a downward UDL.
Tags: Moment Distribution Method Calculator Two Span Continuous Beam Calculator Distribution Factor Stiffness Calculator Continuous Beam Support Moment Fixed End Moment Calculator Calculator Formula Free Online

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What is the moment distribution method?

The moment distribution method is a hand-calculation technique for finding bending moments in continuous beams and frames. It starts by locking every joint against rotation and computing fixed-end moments, then releases each joint in turn, distributing the unbalanced moment to connected members in proportion to their relative stiffness, until the structure reaches equilibrium.

What is the formula for the support moment in a two-span continuous beam?

For a two-span beam with pinned outer supports, span lengths L1 and L2, and uniformly distributed loads w1 and w2, the exact support moment is M_B = minus (w1 L1 cubed + w2 L2 cubed) divided by (8 times (L1+L2)), a hogging (negative) moment derived from the three-moment equation.

Why are the fixed-end moments here wL squared over 8 instead of wL squared over 12?

This calculator uses the modified-stiffness shortcut for a member with a pinned far end, since supports A and C are simple (pinned/roller) supports, not other continuous joints. The correct fixed-end moment for a propped cantilever (fixed at one end, pinned at the other) under a UDL is wL squared over 8, not the wL squared over 12 value used for a beam fixed at both ends.

What are distribution factors and how are they calculated?

A distribution factor tells you what share of an unbalanced joint moment each connected member absorbs, in proportion to its relative stiffness. For this two-span case, using modified stiffness K = 3 divided by span length (the 3/4 factor for a pinned far end already applied), DF_BA = K_BA divided by (K_BA + K_BC), and the two distribution factors always sum to exactly 1.0.

How do you find the midspan moment once the support moment is known?

For a span with a simple support at one end and the balanced support moment M_B at the other, the moment at midspan is M_mid = wL squared over 8 (the ordinary simply supported midspan moment) plus M_B divided by 2 (half of the support moment, since the end-moment effect interpolates linearly from 0 at the simple end to M_B at the continuous end).

Does this calculator work for beams with more than two spans?

No, this calculator is scoped to exactly two spans with one continuous interior support and simple (pinned/roller) supports at both outer ends. Beams with three or more spans have multiple interior joints that interact through carry-over moments, requiring several iterative balancing rounds instead of the single-step shortcut used here.

Do I need to know EI to find the support moment?

No, as long as EI (modulus of elasticity times moment of inertia) is constant across both spans, it cancels out of the stiffness ratio used to compute the distribution factors, so the support moment depends only on the span lengths and the loads. EI is only needed if you go on to compute deflections, which this calculator does not cover.

What sign convention does this calculator use for bending moment?

This calculator uses the sagging-positive convention: a positive moment means the beam sags (tension on the bottom fiber), and a negative moment means the beam hogs (tension on the top fiber), which is why the support moment over the continuous interior support is reported as a negative number.

What does it mean if the support moment matches wL squared over 8 exactly?

That happens specifically when both spans have equal length and equal UDL, since the beam is then fully symmetric about the interior support and the classic two-equal-span continuous beam result, M_B = w L squared over 8 (hogging), applies exactly. It is a standard textbook check for this calculator's arithmetic.

Why is the interior support moment always hogging (negative)?

A continuous beam over an interior support is pulled down on both sides by its own UDLs, which bends the beam upward at that support (tension on top, compression on bottom), the definition of hogging. This happens regardless of how unequal the two spans or loads are, as long as both spans carry a downward UDL.

📌 Quick Tips

💡The moment distribution method is an iterative hand technique, but a two-span beam with only one interior joint needs just a single balancing step, since there is no second joint to carry moment back and forth with.
💡Using the pinned-far-end shortcut, modified fixed-end moments come out to wL squared over 8, not the usual wL squared over 12, because a propped cantilever (one end truly pinned) carries more moment at its fixed end than a beam fixed at both ends.
💡For two equal spans with equal UDL, the support moment reduces exactly to the classic wL squared over 8 result, a good way to sanity-check any two-span moment distribution by hand.
💡Distribution factors always sum to exactly 1.0 at a joint. If your two distribution factors do not add to 1.0, recheck the relative stiffness values before trusting the balanced moment.
💡The support moment is independent of EI as long as EI is the same in both spans, since EI cancels out of the stiffness ratio used to compute the distribution factors.