The three-moment equation, which was initially presented by Clapeyron in 1857, provides a convenient tool for analyzing continuous beams. The three-moment equation represents, in a general form, the compatibility condition that the slope of the elastic curve be continuous at an interior support of the continuous beam. Since the equation involves three moments—the bending moments at the support under consideration and at the two adjacent supports—it commonly is referred to as the three-moment equation. When using this method, the bending moments at the interior (and any fixed) supports of the continuous beam are treated as the redundants. The three-moment equation is then applied at the location of each redundant to obtain a set of compatibility equations which can be solved for the unknown redundant moments.

The following step-by-step procedure can be used for analyzing continuous beams by the three-moment equation. **1**. **Select the unknown bending moments at all interior supports of the beam as the redundants. 2. By treating each interior support successively as the intermediate support c, write a three-moment equation. When writing these equations, it should be realized that bending moments at the simple end supports are known. For such a support with a cantilever overhang, the bending moment equals that due to the external loads acting on the cantilever portion about the end support. The total number of three-moment equations thus obtained must be equal to the number of redundant support bending moments, which must be the only unknowns in these equations. 3. Solve the system of three-moment equations for the unknown support bending moments. 4. Compute the span end shears. For each span of the beam, (a) draw a free-body diagram showing the external loads and end moments and (b) apply the equations of equilibrium to calculate the shear forces at the ends of the span. 5. Determine support reactions by considering the equilibrium of the support joints of the beam. 6. If so desired, draw shear and bending moment diagrams of the beam by using the beam sign convention.**

**Continuous beams with fixed supports**

Sometimes, a continuous beam is fixed at its one or both ends. If the beam is fixed at the left end A, then an imaginary Zero span is taken to the left of A and the three moments theorem is applied as usual. Similarly, if the beam is fixed at the right end, then an imaginary Zero span is taken after the right end support and the three moments theorem is applied as usual.

**Evaluate the bending moment and shear force diagrams of the beam shown**

Redraw the beam by assuming an imaginary support to the left and right

0MA’ +2MA(0+6) + 6MB = 6a1x1/L1. **a1= Area of the bending moment diagram for the span AB x1=distance of centroid of the free bending moment diagram. a2= Area of the bending moment diagram for the span BC. x2= distance of centroid of the free bending moment diagram for the span BC** a1= 2/3(L)(WL ^2/8)=2/3*6*2*6^2/8=36KNM^2 x1 =L/2= 6/2=3m. a2= 1/2*Pab/L *L=1/2*12*3*3/6 *6=54KNM^2. x2= L + a/3= 6 + 3/3=3m. 12MA + 6MB= -6*36*3/6=-108…………..(1). 6MA +2MB(6+6)+6MC=-(6a1x1/L1 + 6a2x2/L2 ). 6MA + 24MB + 6MC = -(6*36*3/6 + 6*54*3/6)=-270………………(2). 6MB + 2MC(6+0) + MC(0)= -6a2x2/L2. 6MB +12MC =-162………….(3)

**Re-arranging the equation. 12MA+6MB+0MC=-108………..(1). 6MA+24MB+6MC=-270………..(2). 0MA+6MB+12MC=-162…………(3**)

**Solution of simultaneous equation equation in 3 variables using determinant method**

Below is a rough sketch of bending moment and shear force