The Three-Moment Equation

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The three-moment equation gives us the relation between the moments between any three points in a beam and their relative vertical distances or deviations. This method is widely used in finding the reactions in a continuous beam.
 

Consider three points on the beam loaded as shown.
 

000-three-moment-equation.gif

 

From proportions between similar triangles:
$ \dfrac{h_1 - t_{1/2}}{L_1} = \dfrac{t_{3/2} - h_3}{L_2} $
 

000-elastic-curve-three-moment-equation.gif

 

$ \dfrac{h_1}{L_1} - \dfrac{t_{1/2}}{L_1} = \dfrac{t_{3/2}}{L_2} - \dfrac{h_3}{L_2} $

$ \dfrac{t_{1/2}}{L_1} + \dfrac{t_{3/2}}{L_2} = \dfrac{h_1}{L_1} + \dfrac{h_3}{L_2} $   →   equation (1)
 

Values of t1/2 and t3/2
$ t_{1/2} = \dfrac{1}{E_1I_1}(\text{Area}_{1-2})\cdot\bar{X}_1 $

$ t_{1/2} = \dfrac{1}{E_1I_1}\left[ A_1\bar{a}_1 + (\frac{1}{2}M_1L_1)(\frac{1}{3}L_1) + (\frac{1}{2}M_2L_1)(\frac{2}{3}L_1) \right] $

$ t_{1/2} = \dfrac{1}{6E_1I_1}(6A_1\bar{a}_1 + M_1{L_1}^2 + 2M_2{L_1}^2) $
 

$ t_{3/2} = \dfrac{1}{E_2I_2}(\text{Area}_{2-3})\cdot\bar{X}_3 $

$ t_{3/2} = \dfrac{1}{E_2I_2}\left[ A_2\bar{b}_2 + (\frac{1}{2}M_2L_2)(\frac{2}{3}L_2) + (\frac{1}{2}M_3L_2)(\frac{1}{3}L_2) \right] $

$ t_{3/2} = \dfrac{1}{6E_2I_2}(6A_2\bar{b}_2 + 2M_2{L_2}^2 + M_3{L_2}^2) $

 

Substitute t1/2 and t3/2 to equation (1)
$ \dfrac{1}{6E_1I_1}\left(\dfrac{6A_1\bar{a}_1}{L_1} + M_1L_1 + 2M_2L_1 \right) + \dfrac{1}{6E_2I_2}\left( \dfrac{6A_2\bar{b}_2}{L_2} + 2M_2L_2 + M_3L_2 \right) = \dfrac{h_1}{L_1} + \dfrac{h_3}{L_2} $
 

Multiply both sides by 6
$ \dfrac{1}{E_1I_1}\left(\dfrac{6A_1\bar{a}_1}{L_1} + M_1L_1 + 2M_2L_1 \right) + \dfrac{1}{E_2I_2}\left( \dfrac{6A_2\bar{b}_2}{L_2} + 2M_2L_2 + M_3L_2 \right) = 6\left( \dfrac{h_1}{L_1} + \dfrac{h_3}{L_2} \right) $
 

Distribute 1/EI
$ \dfrac{6A_1\bar{a}_1}{E_1I_1L_1} + \dfrac{M_1L_1}{E_1I_1} + \dfrac{2M_2L_1}{E_1I_1} + \dfrac{6A_2\bar{b}_2}{E_2I_2L_2} + \dfrac{2M_2L_2}{E_2I_2} + \dfrac{M_3L_2}{E_2I_2} = 6\left( \dfrac{h_1}{L_1} + \dfrac{h_3}{L_2} \right) $
 

Combine similar terms and rearrange

$ \dfrac{M_1L_1}{E_1I_1} + 2M_2\left( \dfrac{L_1}{E_1I_1} + \dfrac{L_2}{E_2I_2} \right) + \dfrac{M_3L_2}{E_2I_2} + \dfrac{6A_1\bar{a}_1}{E_1I_1L_1} + \dfrac{6A_2\bar{b}_2}{E_2I_2L_2} = 6 \left( \dfrac{h_1}{L_1} + \dfrac{h_3}{L_2} \right) $

 

If E is constant this equation becomes,

$ \dfrac{M_1L_1}{I_1} + 2M_2\left( \dfrac{L_1}{I_1} + \dfrac{L_2}{I_2} \right) + \dfrac{M_3L_2}{I_2} + \dfrac{6A_1\bar{a}_1}{I_1L_1} + \dfrac{6A_2\bar{b}_2}{I_2L_2} = 6E \left( \dfrac{h_1}{L_1} + \dfrac{h_3}{L_2} \right) $

 

If E and I are constant then,

$ M_1L_1 + 2M_2(L_1 + L_2) + M_3L_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 6EI \left( \dfrac{h_1}{L_1} + \dfrac{h_3}{L_2} \right) $

 

For the application of three-moment equation to continuous beam, points 1, 2, and 3 are usually unsettling supports, thus h1 and h3 are zero. With E and I constants, the equation will reduce to

$ M_1L_1 + 2M_2(L_1 + L_2) + M_3L_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 0 $

 

Factors for the three-moment equation
The table below list the value of $ 6A\bar{a}/L $ and $ 6A\bar{b}/L $ for different types of loading.

Type of Loading $ \dfrac{6A\bar{a}}{L} $ $ \dfrac{6A\bar{b}}{L} $
Concentrated load anywhere on the span.
 
001-concentrated-load-anywhere.gif

 

$ \dfrac{Pa}{L}(L^2 - a^2) $ $ \dfrac{Pb}{L}(L^2 - b^2) $
Concentrated load at the midspan.
 
002-concentrated-load-midspan.gif

 

$ \dfrac{3PL^2}{8} $ $ \dfrac{3PL^2}{8} $
Uniform load over the entire span.
 
003-uniform-load.gif

 

$ \dfrac{w_oL^3}{4} $ $ \dfrac{w_oL^3}{4} $
Increasing triangular load on the whole span.
 
004-triangular-load-increasing.gif

 

$ \dfrac{8w_oL^3}{60} $ $ \dfrac{7w_oL^3}{60} $
Decreasing triangular load on the whole span.
 
005-triangular-load-decreasing.gif

 

$ \dfrac{7w_oL^3}{60} $ $ \dfrac{8w_oL^3}{60} $
Isosceles triangular load over the entire span.
 
006-triangular-load-symmetrical.gif

 

$ \dfrac{5w_oL^3}{32} $ $ \dfrac{5w_oL^3}{32} $
Moment load at any point on the span.
 
007-moment-load.gif

 

$ -\dfrac{M}{L}(3a^2 - L^2) $ $ +\dfrac{M}{L}(3b^2 - L^2) $
General uniform loading.
 
008-uniform-load-partial.gif

 

$ \dfrac{6A\bar{a}}{L} = \dfrac{w_o}{4L}[ \, b^2(2L^2 - b^2) - a^2(2L^2 - a^2) \, ] $
$ \dfrac{6A\bar{b}}{L} = \dfrac{w_o}{4L}[ \, d^2(2L^2 - d^2) - c^2(2L^2 - c^2) \, ] $

 

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