For a cantilevered beam, the boundary conditions are as follows:

*w(0)=0*. This boundary condition says that the base of the beam (at the wall) does not experience any deflection.*w'(0)=0*. We also assume that the beam at the wall is horizontal, so that the derivative of the deflection function is zero at that point.*w''(L)=0*. This boundary condition models the assumption that there is no bending moment at the free end of the cantilever.*w'''(L)=0*. This boundary condition models the assumption that there is no shearing force acting at the free end of the beam.

*w'''(L)= -mg*

A simply-supported beam (or a

*w(0)=0*. Because the beam is pinned to its support, the beam cannot experience deflection at the left-hand support.*w(L)=0*. The beam is also pinned at the right-hand support.*w''(0)=0*. As for the cantilevered beam, this boundary condition says that the beam is free to rotate and does not experience any torque. In real life, there is usually a small torque due to friction between the beam and its pin, but if the pin is well-greased, this torque may be ignored.*w''(L)=0*. In the same way, the beam does not experience and bending moments on its right-hand attachment.

- Integrate the static beam equation
**twice**. (And please, please, please, remember the constants of integration!) - You now have an equation for
*w''*that depends on two arbitrary constants. Use two of the boundary conditions to solve for the two constants in terms of properties of the beam and load. (Cross off the boundary conditions that you use.) - The constants are now expressed in terms of known quantities, so
substitute back into the equation for
*w''*and integrate two more times to get an equation for*w*. - Use the remaining boundary conditions to solve for the constants of integration in terms of known quantities.
- Graph the deflection function (or
*-w*if you want your beam to sag down) over the interval*[0,L]*to see if your equation makes sense. - For what position does the beam experience its maximum deflection? Where does the beam experience the most torque (the largest bending moment)? Where does the beam experience the greatest shearing force? Interpret your answers in terms of the physical meanings of these quantities.

*w(0)=0, w(L)=0, w'(0)=0, w'(L)=0*; (This is called a*doubly-clamped*beam. Explain why.)*w(0)=0, w(L)=0, w'(0)= 0.2, w'(L)= -0.2*;*w(0)=0, w(L)=0, w'(L)= 0, w''(0)= 0*;*w(0)=0, w'(0)= 0, w''(L)=0, w'''(L)= -0.5*. (Hint: assume a cable is connected to the end of the beam at*x=L*.- Choose one of the above boundary conditions and find the deflection function
for a uniformly distributed load of intensity
*q*. Analyze the deflection function to determine the location of maximum deflection and maximum bending moment.

The Geometry Center Calculus Development Team

Copyright © 1996 by The Geometry Center. Last modified: Fri Apr 12 15:38:15 1996