Direct integration of a beam

Sample calculations

Take the beam shown at right supported by a fixed pin at the left and a roller at the right. There are no applied moments, the weight is a constant 10 kN, and - due to symmetry - each support applies a 75 kN vertical force to the beam. Taking x as the distance from the pin,

${\displaystyle w(x)=10\text{kN}/\text{m}}$

Integrating,

${\displaystyle V(x)=-\int w(x)dx=-10x+C_1(\text{kN})}$

where ${\displaystyle C_1}$ represents the applied loads. For these calculations, the only load having an effect on the beam is the 75 kN load applied by the pin, applied at x=0, giving

${\displaystyle V(x)=-10x+75(\text{kN})}$

Integrating the internal shear,

${\displaystyle M(x)=\int V(x)dx=-5x^2+75x(\text{kN}\cdot \text{m})}$ where, because there is no applied moment, ${\displaystyle C_2=0}$.

Assuming an EI value of 1 kN${\displaystyle \cdot }$m${\displaystyle \cdot }$m (for simplicity, real EI values for structural members such as steel are normally greater by powers of ten)

${\displaystyle \theta (x)=\int \frac{M(x)}{EI}dx=-\frac{5}{3}{x}^3+\frac{75}{2}{x}^2+C_3(\text{m}/\text{m})}$* and

${\displaystyle \nu (x)=\int \theta (x)dx=-\frac{5}{12}{x}^4+\frac{75}{6}{x}^3+C_{3}x+C_4(\text{m})}$

Because of the vertical supports at each end of the beam, the displacement (${\displaystyle \nu }$) at x = 0 and x = 15m is zero. Substituting (x = 0, ν(0) = 0) and (x = 15m, ν(15m) = 0), we can solve for constants ${\displaystyle C_3}$=-1406.25 and ${\displaystyle C_4}$=0, yielding

${\displaystyle \theta (x)=\int \frac{M(x)}{EI}dx=-\frac{5}{3}{x}^3+\frac{75}{2}{x}^2-1406.25(\text{m}/\text{m})}$ and

${\displaystyle \nu (x)=\int \theta (x)dx=-\frac{5}{12}{x}^4+\frac{75}{6}{x}^3-1406.25x(\text{m})}$

For the given EI value, the maximum displacement, at x=7.5m, is approximately 440 times the length of the beam. For a more realistic situation, such as a uniform load of 1 kN and an EI value of 5,000 kN·m², the displacement would be approximately 13 cm.

• Note that for the rotation ${\displaystyle \theta }$ the units are meters divided by meters (or any other units of length which reduce to unity). This is because rotation is given as a slope, the vertical displacement divided by the horizontal change.