Direct integration of a beam

Direct integration is a structural analysis method for measuring internal shear, internal moment, rotation, and deflection of a beam.

For a beam with an applied weight $$w(x) $$, taking downward to be positive, the internal shear force is given by taking the negative integral of the weight:
 * $$V(x) = -\int w(x)\, dx$$

The internal moment $$M(x)$$ is the integral of the internal shear:
 * $$M(x) = \int V(x)\, dx$$ = $$ -\int \left[\int w(x)\, dx \right] dx $$

The angle of rotation from the horizontal, $$ \theta$$, is the integral of the internal moment divided by the product of the Young's modulus and the area moment of inertia:
 * $$\theta (x) = \frac{1}{EI} \int M(x)\, dx $$

Integrating the angle of rotation obtains the vertical displacement $$ \nu $$:
 * $$\nu (x) = \int \theta (x)\, dx $$

Integrating
Each time an integration is carried out, a constant of integration needs to be obtained. These constants are determined by using either the forces at supports, or at free ends.
 * For internal shear and moment, the constants can be found by analyzing the beam's free body diagram.


 * For rotation and displacement, the constants are found using conditions dependent on the type of supports. For a cantilever beam, the fixed support has zero rotation and zero displacement. For a beam supported by a pin and roller, both the supports have zero displacement.

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,


 * $$ w(x)= 10~\textrm{kN}/\textrm{m}$$

Integrating,


 * $$ V(x)= -\int w(x)\, dx=-10x+C_1 (\textrm{kN})$$

where $$C_1$$ 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


 * $$ V(x)=-10x+75 (\textrm{kN}) $$

Integrating the internal shear,


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

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


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


 * $$ \nu (x) = \int \theta (x)\, dx = -\frac{5}{12} x^4 + \frac{75}{6} x^3 + C_3 x + C_4 (\textrm{m})$$

Because of the vertical supports at each end of the beam, the displacement ($$\nu$$) 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 $$C_3$$=-1406.25 and $$C_4$$=0, yielding


 * $$ \theta (x)= \int \frac{M(x)}{EI}\, dx= -\frac{5}{3} x^3 + \frac{75}{2} x^2 -1406.25(\textrm{m}/\textrm{m})$$ and
 * $$ \nu (x) = \int \theta (x)\, dx = -\frac{5}{12} x^4 + \frac{75}{6} x^3 -1406.25x (\textrm{m})$$

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 $$\theta$$ 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.