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 \rightdx

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:
: $\theta (x) = \frac \int M(x)\, dx$

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,

: $w(x)= 10~\textrm/\textrm$

Integrating,

:$V(x)= -\int w(x)\, dx=-10x+C_1 (\textrm)$

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

:$V(x)=-10x+75 (\textrm)$

Integrating the internal shear,

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

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

:$\theta (x)= \int \frac\, dx= -\frac x^3 + \frac x^2 + C_3(\textrm/\textrm)$* and

:$\nu (x) = \int \theta (x)\, dx = -\frac x^4 + \frac x^3 + C_3 x + C_4 (\textrm)$

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

:$\theta (x)= \int \frac\, dx= -\frac x^3 + \frac x^2 -1406.25(\textrm/\textrm)$ and
:$\nu (x) = \int \theta (x)\, dx = -\frac x^4 + \frac x^3 -1406.25x (\textrm)$

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.

See also

* Bending
* Beam theory
* Euler–Bernoulli static beam equation
* Solid Mechanics
* Virtual Work

References

*Hibbeler, R.C., Mechanics Materials, sixth edition; Pearson Prentice Hall, 2005. {{ISBN, 0-13-191345-X.

External links

Beam Deflection by Double Integration Method

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