We can use Mathematica to sketch the situation:
<<Graphics`Arrow`
beam = { Thickness[0.02],
Line[{ {-4,-1}, {4,-1}, {4,1}, {-4,1}, {-4,-1} }] };
foundation = { GrayLevel[0.8], Rectangle[ {-5,-5}, {5,-1}] };
xArrow = Arrow[{-4,0}, {5, 0}];
wArrow = Arrow[{-4,0}, {-4,-4}];
leftPointLoad = Arrow[{-2.5,2.5}, {-2.5,1}];
rightPointLoad = Arrow[{ 2.5,2.5}, { 2.5,1}];
LArrow = { Arrow[{0,-2}, {4, -2}], Arrow[{0,-2}, {-4,-2}] };
lArrow = { Arrow[{0, 2}, {2.5,2}], Arrow[{0, 2},{-2.5,2}] };
aArrows = {
Arrow[{ 2.5,2}, { 4,2}], Arrow[{ 4,2}, { 2.5,2}],
Arrow[{-2.5,2}, {-4,2}], Arrow[{-4,2}, {-2.5,2}]
};
text = {
Text[ FontForm["x",{"Helvetica-Bold",14}], {5.5,0} ],
Text[ FontForm["w",{"Helvetica-Bold",14}], {-4,-4.5} ],
Text[ FontForm["L",{"Helvetica-Bold",14}], {0,-1.75} ],
Text[ FontForm["l",{"Helvetica-Bold",14}], {0,1.7} ],
Text[ FontForm["a",{"Helvetica-Bold",14}], {3.25,1.7} ],
Text[ FontForm["a",{"Helvetica-Bold",14}], {-3.25,1.7} ],
Text[ FontForm["F",{"Helvetica-Bold",14}], {-2.5,2.8} ],
Text[ FontForm["F",{"Helvetica-Bold",14}], { 2.5,2.8} ]
};
Show[Graphics[ { foundation, beam, xArrow, wArrow, leftPointLoad,
rightPointLoad, LArrow, lArrow, aArrows, text },
PlotRange -> All ]];
It may look overwhelming, but it is merely a
specification of the graphics objects and it will produce
Figure 6.3.
Figure 6.3: Beam on Elastic Foundation
By the way, this is also how Figure 6.1 was created.
The differential equation 6.6 is entered in Mathematica as
4 beta^4 w[x] + D[w[x],{x,4}] == load / EI
and displays as
4 (4) load
4 beta w[x] + w [x] == ----
EI
The load is in this case
<<Calculus`DiracDelta`
load = F DiracDelta[x-a] + F DiracDelta[x-a-l];
The Mathematica command DSolve is powerful enough to solve the
differential equation. the result is after some simplification
d1 Cos[beta x] beta x d2 Sin[beta x]
-------------- + d3 E Cos[beta x] + -------------- +
beta x beta x
E E
beta x
d4 E Sin[beta x] +
beta (a - x)
(E F (Cos[beta (-a + x)] -
2 beta (-a + x)
E Cos[beta (-a + x)] +
Sin[beta (-a + x)] +
2 beta (-a + x)
E Sin[beta (-a + x)]) UnitStep[-a + x]
3
) / (8 beta EI) + (F
(-Cos[beta (a + l - x)] +
2 beta (a + l - x)
E Cos[beta (a + l - x)] -
Sin[beta (a + l - x)] -
2 beta (a + l - x)
E Sin[beta (a + l - x)])
3 beta (a + l - x)
UnitStep[-a - l + x]) / (8 beta E EI)
We can then define the deflection, slope, bending moment, and shear
force functions.
W[x_] = %;
H[x_] = D[ W[x], x ];
M[x_] = -EI D[ H[x], x ];
Q[x_] = D[ M[x], x ];
The ends of the beam are free so that the bending moment and shear
force are at these points equal to zero. This gives us 4 equations in
the four unknowns d1, d2, d3, and d4,
which Mathematica can solve. Even after simplification the results are
large: typically formulae that fill one A4 page.
Instead of a we may also use in the formulae the indeterminates l and
L via the relation 
.
Furthermore, if we take concrete values for the unknowns,
says
,
,
,
, and
,
then the general solution specializes to
{d1 -> 0.537024, d2 -> 0.216867, d3 -> 0.970758,
d4 -> 0.216867}
We define a new plot routine to plot the graphs of the beam functions in our
downward oriented coordinate system. This shows a bit of
Mathematica's programming facilities.
myPlot[ function_, domain_, options___ ] :=
Module[ {picture, range, markers},
picture = Plot[ function, domain, options,
DisplayFunction -> Identity ];
range = FullOptions[ picture, PlotRange ];
range = { range[[1]], { - range[[2,2]], - range[[2,1]] } };
picture = Plot[ -function, domain, PlotRange -> range, options,
DisplayFunction -> Identity ];
markers = FullOptions[ picture, Ticks ];
markers = {
Partition[Flatten[
Map[ ( If[ NumberQ[#[[2]]], {#[[1]],#[[2]]},{} ] )&,
markers[[1]] ] ], {2} ],
Partition[Flatten[
Map[ ( If[ NumberQ[#[[2]]], {#[[1]],-#[[2]]},{} ] )&,
markers[[2]] ] ], {2} ]
};
Show[ picture, Ticks -> markers, PlotRange -> range, options,
DisplayFunction -> $DisplayFunction ]
];
So, this procedure is used to graph the deflection, slope, bending
moment, and shear force of the beam.
Figure 6.4: Deflection of Supported Beam
Figure 6.5: Slope of Supported Beam
Figure 6.6: Bending Moment of Supported Beam
Figure 6.7: Shear Force of Supported Beam
Notice that in the solution method we have nowhere used the symmetry of the problem. Therefore, the symmetry shown in the graphs supports the correctness of the solution.
Having solved the general problem with indeterminates instead of
concrete values allows us to analyze the dependency of the solution
on certain parameters. For example, we could look at the dependency of
the deflection on the stiffness of the foundation, i.e., on the parameter
. We could make a surface plot or have an animation showing this
dependency.
Below is a picture consisting of four graphs from an animation of
the bending moment in which 
ranges from 
to 
.
Going from left to right, and from top to bottom, you see that when the
foundation becomes more stiff, the bending moment reduces and gets flatter
in the middle.
Figure 6.8: Bending Moments for Different