This question was previously asked in

PGCIL DT Civil 2020 Official Paper

Option 2 : Trusses and frames

**Explanation:**

__Clapeyron’____s Theorem:__

Clapeyron’s Theorem or the Theorem of Three Moments describes the relationship among the** bending moments at three consecutive supports of a horizontal continuous beam.**

Let the bending moment at these points is MA, MB and MC and the corresponding vertical displacement of these points are ΔA, ΔB and ΔC, respectively. Let L1 and L2 be the distance between points AB and BC, respectively.

\({M_A}\left( {\frac{{{L_1}}}{{{I_1}}}} \right) + 2{M_B}\left( {\frac{{{L_1}}}{{{I_1}}} + \frac{{{L_2}}}{{{I_2}}}} \right) + {M_C}\left( {\frac{{{L_2}}}{{{I_2}}}} \right) = - \frac{{6{A_1}{X_1}}}{{{L_1}{I_1}}} - \frac{{6{A_2}{X_2}}}{{{L_2}{I_2}}} \)

\(+\; 6E\left( {\frac{{{{\rm{\Delta }}_B} - {{\rm{\Delta }}_A}}}{{{L_1}}} + \frac{{{{\rm{\Delta }}_B} - {{\rm{\Delta }}_C}}}{{{L_2}}}} \right)\)

where

A1 and A2 are the areas of BMD’s,

X̅1 and X̅2 are the centroidal distance of areas A1 and A2 measured from A and C and

I1 and I2 are the Moments of Inertia of the beam AB and BC respectively.

Three moment’s equation is derived using Mohr’s first and second-moment theorems.

This theorem can be comfortably used for simply supported and continuous beams (even when there is support settlements).

**Note:**

It is **not applicable to trusses and frames.**

It is **not applicable to fixed supports.**

But in case of a fixed end of a continuous beam, to apply the Clapeyron’s Theorem of Three Moments we assume the following things:

1) Imaginary length (Zero length)

2) Infinite stiffness

3) Replacing fixed end with simply supported end.