We come back to our original problem of a partially supported beam
as shown in Figure 6.1.
As we will see that gap in the middle of the foundation is designed such that
the stress in the middle of the beam is reduced.
In this case, the modulus **k** of the foundation is given in Mathematica as

4 EI beta^4 ( UnitStep[a + (l-b)/2 - x] + UnitStep[x - a - (l+b)/2 ] )and the differential equation becomes

4 b l 4 beta UnitStep[a - - + - - x] w[x] + 2 2 4 b l (4) 4 beta UnitStep[-a - - - - + x] w[x] + w [x] == 2 2 F (DiracDelta[-a + x] + DiracDelta[-a - l + x]) ----------------------------------------------- EI

Mathematica will not be able to solve all by itself the differential equation with this modulus of the foundation. But this does not mean that we have to give up. Actually, we can split the horizontal range into three regions for which we already know the solution:

{UnitStep[x-a-l]->0, d1->d11, d2->d12, d3->d13, d4->d14}in the general solution to obtain the following deflection in the left region:

d11 Cos[beta x] beta x --------------- + d13 E Cos[beta x] + beta x E d12 Sin[beta x] beta x --------------- + d14 E Sin[beta x] + beta x E 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)To obtain the deflection function in the right region we just apply the replacement rule

{UnitStep[x-a-l]->1, d1->d31, d2->d32, d3->d33, d4->d34}to the general solution of a beam on an elastic support.

The midsection is the special case of an unsupported beam. In this case the differential equation is

(4) F (DiracDelta[-a + x] + DiracDelta[-a - l + x]) w [x] == ----------------------------------------------- EIand can be easily solved by Mathematica; the result is

{{w[x] -> C[1] + x C[2] + x C[3] + x C[4] + 3 2 2 3 -(a F) a F x a F x F x (------- + ------ - ------ + ----) UnitStep[-a + x] + 6 EI 2 EI 2 EI 6 EI 3 2 2 3 -(F (a + l) ) F (a + l) x F (-a - l) x F x (------------- + ------------ + ------------- + ----) 6 EI 2 EI 2 EI 6 EIApplying the replacement rule

{UnitStep[x-a]->1, UnitStep[x-a-l]->0, C[1]->d21,C[2]->d22, C[3]->d23, C[4]->d24}we obtain the following deflection function for the middle region.

3 2 3 F (-a + x) d21 + d22 x + d23 x + d24 x + ----------- 6 EIBy differentiation we can also compute the slopes, bending moments, and shear forces in the threwe regions. We are left with twelve unknowns , where

`DiracDelta[a] -> 0`

, and `UnitStep[-a] -> 0`

to let Mathematica
know that we understand

eq[1] = Simplify[ m1[0]==0 /. simplificationrules ]; eq[2] = Simplify[ m3[L]==0 /. simplificationrules ]; eq[3] = Simplify[ q1[0] == 0 /. simplificationrules ]; eq[4] = Simplify[ q3[L] == 0 /. simplificationrules ]; eq[5] = Simplify[ w1[a+(l-b)/2] == w2[a+(l-b)/2] /. simplificationrules ]; eq[6] = Simplify[ h1[a+(l-b)/2] == h2[a+(l-b)/2] /. simplificationrules ] eq[7] = Simplify[ m1[a+(l-b)/2] == m2[a+(l-b)/2] /. simplificationrules ]; eq[8] = Simplify[ ( q1[a+(l-b)/2] == q2[a+(l-b)/2] ) /. simplificationrules ]; eq[9] = Simplify[ ( w2[a+l/2+b/2] == w3[a+l/2+b/2] ) /. simplificationrules ]; eq[10] = Simplify[ ( h2[a+l/2+b/2] == h3[a+l/2+b/2] ) /. simplificationrules ]; eq[11] = Simplify[ ( m2[a+l/2+b/2] == m3[a+l/2+b/2] ) /. simplificationrules ]; eq[12] = Simplify[ q2[a+l/2+b/2] == q3[a+l/2+b/2] /. simplificationrules ];If we take and the same data as in the previous section, Mathematica can solve the twelve equations in the 's. The deflection and the bending moment are shown below, together with the result without the gap. As you can clearly see from these figures, the stress is reduced in the middle section if the gap in the foundation is introduced.

In the design phase of course we don't know a ``good'' value for the gap
width **b**. But it is rather easy to graphically find such a value:
solve numerically the problem for several values of **b**, plot the function,
and in an animation step through the graphs to find the best choice for **b**.
This is what we have done in the following Mathematica code:

equations = Table[eq[i],{i,12}]; L=2600; l=1500; EI=6.33 10^(11); F=56.25 10^3; k=25; beta=(k/(4 EI))^(1/4); a=(L-l)/2; Do[ ( solutions = NSolve[ equations, {d11,d12,d13,d14,d21,d22,d23,d24,d31,d32,d33,d34} ][[1]]; p1 = a+(l-b)/2; p2 = a+(l+b)/2; W1 = w1[x] /. solutions; W2 = w2[x] /. solutions; W3 = w3[x] /. solutions; Clear[W]; Clear[M]; W[x_ /; Evaluate[x<=p1] ] = W1; W[x_ /; Evaluate[x>p1 && x<p2] ] = W2; W[x_ /; Evaluated[x>=p2] ] = W3; M1 = m1[x] /. solutions; M2 = m2[x] /. solutions; M3 = m3[x] /. solutions; M[x_ /; Evaluate[x<=p1] ] = M1; M[x_ /; Evaluate[x>p1 && x<p2] ] = M2; M[x_ /; Evaluate[x>=p2] ] = M3; plot[b] = myPlot[ W[x], {x,0,L}, PlotRange->{0,5.5}, DisplayFunction->Identity]; mplot[b] = myPlot[ M[x], {x,0,L}, PlotRange -> {-5 10^6,7 10^6}, DisplayFunction->Identity]), {b,0,1000,100} ];At the end of the article, the plots of the deflection function and bending moment can be easily drawn. They have been created with the following commands:

Show[ {plot[0],plot[600]}, DisplayFunction->$DisplayFunction ]; Show[ {mplot[0],mplot[600]}, DisplayFunction->$DisplayFunction ];

Sun Apr 23 10:32:10 MDT 1995