Conjugate beam theorems and derivation

The conjugate beam method converts the problem of finding the slope and deflection of a real beam into finding the shear and moment in a fictitious beam (the conjugate beam), which is loaded with the $M/EI$ diagram of the real beam.

Theorems

• Theorem 1: The slope at a point in the real beam is numerically equal to the shear in the conjugate beam at the corresponding point.
• Theorem 2: The deflection at a point in the real beam is numerically equal to the bending moment in the conjugate beam at the corresponding point.

Proof

For a real beam, $\dfrac{d^2y}{dx^2} = \dfrac{M}{EI}$, with $\dfrac{d^2M}{dx^2} = w$, $\dfrac{dM}{dx} = V$ (load, shear, moment).

For the conjugate beam, loaded by $w' = \dfrac{M}{EI}$:

$$\frac{dV'}{dx} = w' = \frac{M}{EI},\qquad \frac{d^2M'}{dx^2} = w' = \frac{M}{EI}$$

Comparing with the real beam relations:

$$\frac{d\theta}{dx} = \frac{M}{EI}\ \Rightarrow\ \theta = \int\frac{M}{EI}dx = V'$$ $$\frac{d^2y}{dx^2} = \frac{M}{EI}\ \Rightarrow\ y = M'$$

The mathematical relations are the same, so $\theta = V'$ and $y = M'$ (provided the boundary conditions also match).

Supports of the conjugate beam

Procedure and example

1. Find $M$ of the real beam and draw the $M/EI$ diagram.
2. Load the conjugate beam with $M/EI$ (sagging $M/EI$ acts downward).
3. Find the reactions, then the shear (slope) and moment (deflection).

Example: simply supported beam, span $L$, central load $P$. $M_{max} = PL/4$.

Conjugate beam: simply supported, triangular load with peak $\dfrac{PL}{4EI}$. Total load $= \tfrac12\cdot L\cdot\dfrac{PL}{4EI} = \dfrac{PL^2}{8EI}$, so each reaction is $\dfrac{PL^2}{16EI}$.

$$\theta_A = V'_A = \frac{PL^2}{16EI}$$

At mid-span, the conjugate moment is

$$y_C = M'_C = \frac{PL^2}{16EI}\times\frac{L}{2} - \frac{PL^2}{16EI}\times\frac{L}{6}\ \ \Rightarrow\ \ y_C = \frac{PL^3}{32EI} - \frac{PL^3}{96EI} = \frac{PL^3}{48EI}$$

which is the known mid-span deflection.