Direct Integration of A Beam - Sample Calculations

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,

Integrating,

where 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

Integrating the internal shear,

where, because there is no applied moment, .

Assuming a EI value of 1 kNmm (for simplicity, real EI values for structural members such as steel are normally greater by powers of ten)

* and

Because of the vertical supports at each end of the beam, the displacement at x = 0 and x = 15m is zero. Substituting (x = 0, v(0) = 0) and (x = 15m, v(15m) = 0), we can solve for constants =-1406.25 and =0, yielding

and

For the given EI value, the maximum displacement, at x=7.5m, is approximately 500 times the length of the beam. For a more realistic situation, such as a uniform load of 1kN and an EI value of 5,000 kN per square meter, the displacement would be approximately 1 cm.

• Note that for the rotation 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.