Solutions for the Winkler subsoil model
1. Denotations 1.1. Internal forces M and Q
M(x) & Q(x) along the beam must be found and a soil reaction r(x) first of all.
y(x) is a beam settlement.
1.2. Statical relations
• q(x) = r(x) – qo(x), dQ/dx = q, dM/dx = Q
• the Eulera-Bernoulliego condition:
M = -EI/ρ ≅ -EI·d2y/dx2
• the Winkler assumption: r(x) = C·B·y(x) , r [MN/m]
where C – elastic coefficient [MN/m3].
1.3. Governing equation for the Winkler subsoil model
) ( )
) ( (
4 4
x y B C x q dx
x y
EId = o −
(1) where EI, B, C do not depend on x. Unknown function: y(x).
After a simple reformulation, in dimensionless coordinates ξ instead of x:
W 4 W o
4 4
BC EI L 4
L , ξ x where BC ,
) ξ ( q ) 4 ξ ( y ξ 4
d ) ξ ( y
d + = = = (2)
The parametr Lw [m] is called a stiffness property.
If y(ξ) is found, all other functions r, M and Q will be known from statical relations (see 1.2).
2. Solutions 2.1. General solution
On the unloaded interval where qo = 0 we have:
) ξ sin C ξ cos C ξ( e ) ξ sin C ξ cos C ξ( e ) ξ (
y = − 1 + 2 + 3 + 4 (3)
There are 4 unknown constants Ci , i=1,2,3,4, for wich 4 end-conditions must be introduced.
2.2. Fundamental solution on a half-line ξ > 0.
A concentrated force P at ξ = 0 on the infinite beam (−∞; +∞) is considered. The solution is as follows.
2.3. The Bleich Method for finite beams
it's a simple and extremely useful application of the solutions (4) above, by making use of the superposition law.
Instead of a finite beam with two ends A and A', consider infinite beam and additionally apply two virtual forces T1,T2 on the left of A, and another two virtual forces T3,T4 on the right of A' (of course T1=T3, T2=T4 if you have symmetry);
find all the outside T forces from 4 boundary conditions at A, A', i.e. M=0 and Q=0 for all actions1.
In other words, in place of a real finite beam AA' loaded by n real forces P, consider the infinite beam loaded by n+4 forces and this way the exact mathematical solution will be reached.
1 Usually, dimensionless distances π/4,π/2 of the forces from the beam-end are in use;
note that other pair of boundary conditions can be used as well, like a perfect support for which yA = 0 and so on.
0 Beam load qo(x) > 0
M(x) > 0
Q(x) > 0 Soil reaction r(x) > 0
y > 0
-2,5 -2 -1,5 -1 -0,5 0 0,5 1 1,5 2 2,5 3 3,5 4 4,5
0 1 2 3 4
Yo Mo Qo Yo - even
Mo - even Qo- odd
) 8 ( cos 84
) (
) 8 ( ) cos (sin 8 2
) (
) 2 ( ) ( ) (
) 2 (
) sin 2 (cos
) (
ξ
−
= ξ
−
= ξ
ξ
−
= ξ
− ξ ξ
− −
= ξ
ξ
= ξ
= ξ
ξ
= ξ + ξ ξ
= − ξ
ξ
−
o
o w w
o w
o w w
PQ P e
Q
PL M PL e
M
L Y y P C B r
BCL Y e P
BCL y P
where ξ > 0;
for ξ < 0 the solutions are either odd (antysymmetrical) or even (symmetrical).
(4)
