State and explain moment area theorems

The moment-area method finds slopes and deflections of beams from the area and the moment of the $M/EI$ diagram.

Theorem I

The change of slope between two points A and B of the elastic curve equals the area of the $M/EI$ diagram between those points:

$$\theta_B - \theta_A = \int_A^B \frac{M}{EI}\,dx$$

Theorem II

The vertical deviation of a point A from the tangent drawn at another point B equals the moment of the $M/EI$ area between A and B about the vertical through A:

$$\Delta_{A/B} = \int_A^B x\,\frac{M}{EI}\,dx = \text{Area}\times\bar{x}_A$$

Here $x$ is measured from A, and $\bar x_A$ is the horizontal distance from A to the centroid of the area.

 fixed A *----------------------------* B1   tangent at A (horizontal)
         |'''--..__                   |
         |         ''--..__           | deviation of B from the
         |    elastic curve ''--..    | tangent at A = deflection at B
         |                        ''--* B (free end)

Simple example

A cantilever of length $L$ with a point load $P$ at the free end B, fixed at A. The tangent at A is horizontal.

$M = -Px$ ($x$ from B), so the $M/EI$ diagram is a triangle with base $L$ and peak $\dfrac{PL}{EI}$ at A.

$$\text{Area} = \frac12\times L\times\frac{PL}{EI} = \frac{PL^2}{2EI}$$

Slope at B (Theorem I, since $\theta_A = 0$):

$$\theta_B = \frac{PL^2}{2EI}$$

Deflection at B (Theorem II, deviation of B from the tangent at A): the centroid of the triangle is at $\dfrac{2L}{3}$ from B:

$$\Delta_B = \frac{PL^2}{2EI}\times\frac{2L}{3} = \frac{PL^3}{3EI}$$

These are the standard cantilever results.