Common Beam Formulas
(http://structsource.com/analysis/types/beam.htm)
Beam formulas may be used to determine the deflection, shear and bending moment
in a beam based on the applied loading and boundary conditions.
PINNED-PINNED BEAM WITH UNIFORM LOAD
FIXED-FIXED BEAM WITH UNIFORM LOAD
PINNED-FIXED BEAM WITH UNIFORM LOAD
FREE-FIXED BEAM WITH UNIFORM LOAD
PINNED-PINNED BEAM WITH POINT LOAD
See definitions of < > step functions below.
FIXED-FIXED BEAM WITH POINT LOAD
PINNED-FIXED BEAM WITH POINT LOAD
See definitions of < > step functions below.
See definitions of < > step functions below.
See definitions of < > step functions below.
FREE-FIXED BEAM WITH POINT LOAD
Definitions of step functions:
If x < a then < x - a >0 = 0
If x < a then < x - a >n = 0
If x > a then < x - a >0 = 1
If x > a then < x - a >n = ( x - a )n
Boundary Conditions
(http://www.geom.uiuc.edu/education/calc-init/static-beam/boundary.html)
It is a general mathematical principle that the number of boundary conditions necessary to
determine a solution to a differential equation matches the order of the differential equation.
The static beam equation is fourth-order (it has a fourth derivative), so each mechanism for
supporting the beam should give rise to four boundary conditions.
Cantilevered Beams
Figure: A cantilevered beam.
For a cantilevered beam, the boundary conditions are as follows:
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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.
If a concentrated force is applied to the free end of the beam (for example, a weight of mass m
is hung on the free end), then this induces a shear on the end of the beam. Consequently, the
the fourth boundary condition is no longer valid, and is typically replaced by the condition
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w'''(L)= -mg
where g is the acceleration due to gravity (approximately 9.8 m/s2). We note that we could
actually use this boundary condition all the time, since if m=0 , it reduces to the previous
case.
Simply-Supported Beams
Figure: A simply-supported beam.
A simply-supported beam (or a simple beam , for short), has the following boundary
conditions:
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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.
Question
A simply-supported beam of length L is deflected by a uniform load of intensity q . We
assume that we know E, I, L and q . Let's use this fact to solve for the deflection of the beam
under the load.
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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.
Other Beam Supports
There are many other mechanisms for supporting beams. For example, both ends of the beam
may be clamped to a wall. Or one end may be bolted and the other end is free to rotate. Or the
beam may be clamped at one end but "overhang" a support placed at some point along its
length.
Figure 7: Other mechanisms for supporting beams.
Question
Each support mechanisms has an associated set of boundary conditions. In order to gain some
intuition for boundary conditions, sketch idealized beams whose support mechanism gives
rise to the following boundary conditions. The beams should be shown in a "deflected"
position, as shown in the figures on this page. In all cases, the beam is supported only at the
ends.
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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.