Math 2.0

Mechanics of deformable solids Thin bodies and structures Theorem

Three-moment equation

Builds on Euler–Bernoulli beam equationBending momentShear and moment diagramCentroidStatically indeterminate structureConjugate beam method

Statement

Let a continuous beam with constant bending stiffness EIEI rest on supports 0,1,…,N0, 1, \dots, N at the same level, which do not settle. Let span ii run from support i−1i - 1 to support ii and have length LiL_i, and assume the Euler–Bernoulli beam equation. Let MiM_i be the bending moment at support ii, with sagging taken positive. For each span ii, let AiA_i be the area of the moment diagram of its loads on the span taken as simply supported. Let aia_i be the distance of the centroid of that area from support i−1i - 1, and bi=Li−aib_i = L_i - a_i its distance from support ii. Then for each interior support nn, with 1≤n≤N−11 \le n \le N - 1, Mn−1Ln+2Mn(Ln+Ln+1)+Mn+1Ln+1=−6AnanLn−6An+1bn+1Ln+1.M_{n-1} L_n + 2 M_n (L_n + L_{n+1}) + M_{n+1} L_{n+1} = -\frac{6 A_n a_n}{L_n} - \frac{6 A_{n+1} b_{n+1}}{L_{n+1}}.

Remarks

The proof writes the moment in each span as the free moment plus the linear moment from the support moments. The slope of the beam at support nn is then computed from the left span and from the right span, for example by the moment-area theorems. Setting the two slopes equal, since the beam is continuous over the support, gives the equation.

Example: for a uniform load wiw_i on span ii, the free diagram is a parabola with Ai=wiLi3/12A_i = w_i L_i^3/12 and ai=bi=Li/2a_i = b_i = L_i/2, so the right side is −(wnLn3+wn+1Ln+13)/4-(w_n L_n^3 + w_{n+1} L_{n+1}^3)/4. For two equal spans LL with load ww and simply supported ends, M0=M2=0M_0 = M_2 = 0 gives 4M1L=−wL3/24 M_1 L = -wL^3/2, so M1=−wL2/8M_1 = -wL^2/8, a hogging moment over the middle support.

If the supports settle or the stiffness differs between spans, extra terms appear, and the equation above is false as written.

Going deeper

For a student

A continuous beam is a statically indeterminate structure, and the support moments are a natural choice of redundant unknowns. The equations form a tridiagonal linear system, which is easy to solve even for many spans. Once the MiM_i are known, each span is a simply supported beam with known end moments, and its shear and moment diagram follows by superposition. The same equation is often derived with the conjugate beam method.

Reading path

To understand this page from the ground up, read these 7 pages in order. The foundations come first.

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Each page uses the pages above it. The colour of the bar on a box shows how strong the evidence is, and the box also gives it as text. The list has the same pages.
  • no evidence level
  1. Bending moment concept
  2. Centroid concept
  3. Conjugate beam method concept
  4. Euler–Bernoulli beam equation concept
  5. Shear and moment diagram concept
  6. Statically indeterminate structure concept
  7. Three-moment equation statement

Further reading

  1. Web https://en.wikipedia.org/wiki/Theorem_of_three_moments
About this page
TagZGCR9KA1PT
URIm2://ZGCR9KA1PT
Kindstatement, theorem
WritingW0 stub
MSC74K10, 74G05
StatusThis copy of the site has no build data. Run m2 status ZGCR9KA1PT.
AboutBending moment
UsesEuler–Bernoulli beam equation, Bending moment, Shear and moment diagram, Centroid, Statically indeterminate structure, Conjugate beam method
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Changed3 November 2026

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