Compatibility Conditions:

As the beam is continuous over the support the compatibility conditions stipulate that the slopes on the two sides are the same, i.e. ∠T₁ J ′ H = ∠T₂ J ′ H ′ = θ where HH ′ is a horizontal line through J, by geometry

tan T₁J ′ H = T₁ H/L_i and tan T₂ J ′ H = T2 H ′/L_j and both are equal to tan θ.

It can be also seen from the diagram that T₁ H = Δ _j - Δ_i - C_i and

 T₂ H ′ = Δ_k - Δ _j  + C_k

∴ We have Δ _j  - Δ_i- C_i/L_i = Δ_k - Δ _j+ C_k/L _j  = tan θ .

Let us assume that the E and I on the two spans are different (E_i, L_i and E_j L_j respectively). The free beam BM diagrams (M-diagram) can be easily drawn knowing the loads on it. Dividing the M-diagram of span L_i by E_i I_i, and that of span L_j by E_j I_j we get the two free M/EI diagram for the two adjacent spans as in Figure 2.18(b), having areas A_i/E_i L_i and A _j/E _j L _j. Let the CG of the area A_i/E_i L_i be  I from IH and of area A _j/E _j L _j be  j   from KH′.

Similarly, let as assume that M_i, M_j and M_k are the BM's at the supports I, J, K, the support BM diagrams are trapeziums with ordinates M_i, M_j and M_k. If these ordinates are divided by E_i, I_i in span L_i and E_j I_j in span L_j, we get the corresponding fixed M/EI   diagrams which are two trapezium with areas (M _i  + M _j/2 E_i I_i) L_i and (M _j  + M _k/ 2 E _j I _j) L_k. The CG of the former trapezium is (2M_j +M_i/M_i+M_j) (L_i/3) from vertical IH and CG of the second trapezium is  (2M_j+M_k /M_j+M_k) L_k/3 KH ′ from vertical  (Figure 18(c)).

Now, we know that the intercept Ci between the tangents at I and J is equal to the moment of the total (free + fixed) M/EI   diagram about vertical IH. Similarly, the intercept C_k is equal to the moment of these areas about vertical KH ′. This can be expressed by the following expression:

C_i =A_{i i}/E_iI_i +(M_i +M_j/2E_i I_j) L_i.(2M_j+M_i/M_i+M_j) L_i/3= (6A_{i I} + (2M_j+M_i) L²_i/6E_iI_i)

Similarly

C_k =A_{j j}/E _j I _j + (M_j+M_k/2E_jI_j)L_j . (2M_j+M_k/M_j+M_k)L_j/3 = 6Aj j +(2M_j+M_k)L²_j/6E_jI_j

Substituting these values in Eq. 2 we get

 Δ _j  - Δ _i/L_I - A_{i  i}/E_i I_i L_i -(2M_j +M_i)L_i/6E_i I_i =Δ_k-Δ_j/L_j +[(A_{j j} /E_jI_jL_j )+ (2M_j+M_k) L_j/6E_jI_j]

On rearranging M_i (L_i/E_i I_i)+2M(L_i/E_i I_i + L_j/E_j I_j) +M_k(L_j / E_j I_j)

+6 (Ai i/Ei Ii Li) +6(Aj j/Ej Ij Lj) = 6(Δ_j - Δ_i/L_i + Δ_j- Δ_k/L_j)

This is the Theorem of Three Moments.

On the right hand side of the equation Δi - Δ j is the relative settlement of the left support with respect to the central support and Δi - Δ j is negative if Δi < Δ j , i.e. left hand support is above the central support (since downward deflections are positive). Similarly, Δk - Δ j is the movement of the right hand support with reference to central support which is shown here as positive (downwards) since Δk  > Δ j in the diagram. The sign of the movements have to be properly taken when calculating the right hand side of the equation.

If there are no support settlements, which is the scope of this course, the RHS of Eq. is zero and we get

M_i(L_i/E_i I_i)+2M_j(L_i/E_i I_i+L_i/E_j I_j)+M_k(L_i/E_j I_j) =-6(A_{i i}/E_i I_i L_i )- 6(A_{j j}/E_j I_j L_j)