Moment-area theorems with proof

Basis

For small deflections $\dfrac{d^2y}{dx^2} = \dfrac{M}{EI}$, so $\dfrac{d\theta}{dx} = \dfrac{M}{EI}$, where $\theta$ is the slope. The quantity $M/EI$ plotted along the beam is the $M/EI$ diagram.

        tangent at A
   A ___.----__           
      \        ---.__ B     theta_AB : change in slope
 tangent at B      t_BA : deviation of B from the
                   tangent drawn at A

Theorem I (change of slope)

The change in slope between two points A and B on the elastic curve equals the area of the $M/EI$ diagram between them.

Proof: from $d\theta = \dfrac{M}{EI}dx$, integrate from A to B:

$$\theta_{B} - \theta_{A} = \int_A^B \frac{M}{EI}\,dx = \text{Area}_{AB}\ \text{of the }M/EI\text{ diagram}$$

Theorem II (tangential deviation)

The vertical deviation of point B on the elastic curve from the tangent drawn at A equals the moment of the area of the $M/EI$ diagram between A and B, taken about B.

Proof: consider a small element $dx$ at distance $x$ from B. The change in slope across the element is $d\theta = \dfrac{M}{EI}dx$. The tangents at the ends of the element intercept a vertical distance $dt = x\,d\theta$ on the vertical through B. Integrating,

$$t_{B/A} = \int_A^B x\,\frac{M}{EI}\,dx = \text{Area}_{AB}\times\bar x_B$$

where $\bar x_B$ is the distance of the centroid of the $M/EI$ area from B.

Use

1. Draw the BM diagram, divide by $EI$.
2. Choose a reference tangent where the slope is known: at the fixed end of a cantilever (zero slope) or at the centre of a symmetrically loaded beam.
3. Slope: use Theorem I from the reference point. Deflection: use Theorem II (deviation from the tangent). For a simply supported beam, find the deviation of the supports from a tangent at one support, get the slope there, then find the deflection at a section from geometry.

Sign: positive area (sagging) gives an anticlockwise rotation from A to B and B lies above the tangent at A.