Three Moment Equation - Civil Engineering (CE) PDF Download

 Three Moment Equation 

The continuous beams are very common in the structural design and it is necessary to develop simplified force method known as three moment equation for their analysis. This equation is a relationship that exists between the moments at three points in continuous beam. The points are considered as three supports of the indeterminate beams. Consider three points on the beam marked as 1, 2 and 3 as shown in Figure 5.25(a). Let the bending moment at these points is M₁,M₂, and M₃ and the corresponding vertical displacement of these points are Δ₁,Δ₂ and Δ₃ respectively. Let L₁ and L₂ be the distance between points 1 – 2 and 2 – 3, respectively.

The continuity of deflected shape of the beam at point 2 gives

θ₂₁ = θ₂₃         (5.4)

From the Figure 5.25(d)

θ₂₁ = θ1 - β21 and θ₂₃ = θ₃ - β₂₃  (5.5)

where

  and 

Using the bending moment diagrams shown in Figure 5.25(c) and the second moment area theorem,

  (5.7)

  (5.8)

where A₁ and A₂ are the areas of the bending moment diagram of span 1-2 and 2-3, respectively considering the applied loading acting as simply supported beams.

Substituting from Eqs. (5.7) and Eqs. (5.8) in Eqs. (5.4) and Eqs. (5.5).

The above is known as three moment equation .

Sign Conventions

The M₁, M₂ and M₃ are positive for sagging moment and negative for hogging moment. Similarly, areas A₁,A₂ and A₃  are positive if it is sagging moment and negative for hogging moment. The displacements Δ₁,Δ₂ and Δ₃ are positive if measured downward from the reference axis.

Example 5.22 Analyze the continuous beam shown in Figure 5.26(a) by the three moment equation. Draw the shear force and bending moment diagram.

Solution: The simply supported bending moment diagram on AB and AC are shown in Fig 5.26 (b). Since supports A and C are simply supported

M_A  = M_C =0

Applying the three moment equation to span AB and BC (Δ₁= Δ₂ = Δ₃) 

or   M_B  =-56.25 kN.m

The reactions at support A , B and C are given as

V_B= 120 + 40 3 – 41.25 – 41.25 = 157.5 kN

The bending moment and shear force diagram are shown in Figures 5.26(c) and (d), respectively

Example 5.23 Analyze the continuous beam shown in Figure 5.27(a) by the three moment equation. Draw the shear force and bending moment diagram.

Solution: The effect of a fixed support is reproduced by adding an imaginary span A₀ A as shown in Figure 5.27 (b). The moment of inertia, I₀ of the imaginary span is infinity so that it will never deform and the compatibility condition at the end A , that slope should be is zero, is satisfied.

Applying three moment equation to the span A₀A and AB :

or 2M_A + M_B =-135      (i)  

Span AB and BC :

or  M_A + 4M_B = -225    (iii)

Solving Eqs. (i) and (ii), MA = -45 kNm and M_B = – 45 kNm

The shear force and bending moment diagram are shown in Figures 5.27(d) and (e), respectively.

Example 5.24 Analyze the continuous beam shown in Figure 5.28(a) by the three moment equation. Draw the shear force and bending moment diagram.

Solution: The simply supported moment diagram on AB , BC and CD are shown in Figure 5.28(b). Since the support A is simply supported, MA = 0 The moment at D is M_D = -20 x 2 = - 40 kNm.

Applying three moment equation to the span AB and BC:

or  6M_B + M_C = -456

Span BC and CD : (M_D = - 20kNM)

or M_B + 5M_c = -556

Solving Eqs. (i) and (ii) will give M_B = -59.448 kNm and M_c = - 99310 kNm.

The bending moment and shear force diagram are shown in Figures 5.28(d) and (c), respectively.

Example 5.25 Analyze the continuous beam show in Fig. 5.29(a) by the three moment equation method if support B sinks by an amount of 10 mm. Draw the shear force and bending moment diagram. Take flexural rigidity EI=48000kNm.

Solution: Since support A and D are simply supported, M_A = M_D = 0

Applying the three moment equation for span AB and BC : (M_A = 0)

or 6M_B + M_C = 600    (i)

Span BC and CD :

or

M_B + 5M_C =-240    (ii)

Solving Eqs. (i) and (ii), M_B = 111.72 kNm and M_C = -70.344 kNm

The bending moment diagram is shown in Figure 5.29(b).