State and prove moment area theorems

The moment-area method uses the area and the moment of the $M/EI$ diagram to find slopes and deflections of beams.

Basis

For small deflections the curvature is $\dfrac{d^2y}{dx^2} = \dfrac{M}{EI}$. Consider two points A and B on the deflected beam and an elemental length $dx$ at a distance $x$ from A.

 A *
   |'.
   |  '.       elastic curve
 t |    '-._
   |        ''-.._____|_|________* B
 A1*-------------------------------- tangent at B (extended)
   |<-------- x ------>|dx|

The tangents at the two ends of $dx$ meet at an angle $d\theta$:

$$d\theta = \frac{dx}{\rho} = \frac{M}{EI}\,dx$$

Theorem I (slope)

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

Proof: integrate $d\theta$ from A to B:

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

Theorem II (deviation)

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

Proof: the tangents at the ends of the element $dx$ intercept a small vertical distance on the vertical through A:

$$d\Delta = x\,d\theta = x\,\frac{M}{EI}\,dx$$

where $x$ is the distance from A to the element. Adding up these intercepts from A to B gives the total deviation $t = AA_1$:

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

where $\bar x_A$ is the distance from A to the centroid of the $M/EI$ area between A and B.

Notes

• A positive $M/EI$ area means a counter-clockwise (positive) change of slope from A to B, and A lies above the tangent drawn at B.
• The theorems are valid for linear elastic beams with small slopes. The deviation is measured perpendicular to the original beam axis.