BEAM ANALYTICAL SOLUTIONS FOR THE WINKLER-SUBSOIL MODEL (the Winkler model, the Bleich method)
to Lecture no.3
1. Denotations 1.1. Internal forces M [MNm] and Q [MN]
M(x) & Q(x) along the beam must be found and a soil reaction r(x) [MN/m] first of all.
y(x) [m] is a beam settlement, B=const(x), EI=const(x).
Sign convention is presented.
The beam, is prismatic,
so the cross-section is constant.
1.2. Statical relations
∗ q(x) = r(x) – qo(x), dQ/dx = q, dM/dx = Q,
∗ the Euler-Bernoulli condition:
M = -EI/ρ ≅ -EI·d2y/dx2
∗
the Winkler assumption: r(x) = C·B·y(x), r[MN/m], where C – elastic subsoil coefficient [MN/m3].Bilateral bonds are assumed at the contact of the beam with the subsoil, i.e. the equality of the beam settlement y(x) and the settlement of the subsoil w(x).
1.3. Governing equation for the Winkler subsoil model
) ( )
) ( (
4 4
x y B C x q dx
x y
EI d = o −
(1) where EI, B, C do not depend on x; call this a prismatic beam.
Unknown function: y(x).
Thus the solving yielding from y(x) is complete because: r(x) = C⋅B⋅y(x), M = -EI⋅d2y/dx2, Q = -EI⋅d3y/dx3. 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 4 dξ
) ξ ( y
d + = = = (2)
The parametr Lw [m] is called the stiffness property.
If y(ξ) is found, so r(ξ) = C⋅B⋅y(ξ), M = -EI⋅d2y/dξ2⋅ , Q = -EI⋅d3y/dx3⋅ .
2. Solutions 2.1. General solution of the homogeneous equation
In the unloaded (open) interval where qo = 0 we have:
y(ξ)=e−ξ(C1cosξ +C2sinξ)+eξ(C3cosξ +C4sinξ)
(3)
This is not a foolish question,why we solve a beam that is unloaded?
0 Beam load qo(x) > 0
M(x) > 0
Q(x) > 0
Soil reaction r(x) > 0y > 0
It is not completely unloaded, only in open-internal cross-sections, where we can talk about derivatives at all;
some end-values can be pre-described of course. Then it turns out that in an indirect way we can come to solutions for beams loaded in any internal cross-sections as well, i.e. in this indirect way we can solve the general heterogeneous equation for qo(x)≠0.
There are 4 unknown constants Ci , i=1,2,3,4, for which 4 end-conditions must be introduced: for y(ξ) and its derivatives (internal forces etc.).
2.2. Fundamental solution of the first kind on a half-line ξ > 0.
A perpendicular concentrated force P at ξ = 0 on the infinite beam (−∞; +∞) is considered.
The solution is as follows.
Indeed, it suffices to adopt (3) focusing on ξ > 0 because the settlements y(ξ) must by a symmetrical function (even). The following condition can be formulated:
1) the Boussinesq condition: ξ → ∞ ⇒ y(ξ) → 0 ... thus C4 = C3 = 0 in (3),
2) symmetry: ξ = 0 ⇒ dy/dξ = 0 ... thus C2 = C1,
3) antisymmetry, discontinuity of the shearing force chart: ξ → 0+ (i.e. ξ → 0, ξ > 0) ⇒ Q(ξ) → −P/2 ... thus C1 =
BCL
wP
2
.The derivation is completed.
Remark 1: the following relations are true for the dimensionless functions introduced above Yo’ = -2e-ξ sinξ, Yo’’ = Mo , Yo’’’ = Qo, YoIV = -4Yo .
dy(ξ)/dx = dy(ξ)/dξ dξ/dx = y’/LW ,
d2y(ξ)/dx2 = d[y’(ξ)/LW]/dξ dξ/dx = y’’/(LW)2, itd.
so:
M = -EI d2y(x)/dx2 = = -EI d2y(ξ)/dξ2 /(LW)2 Q = -EI d3y(x)/dx3 = = -EI d3y(ξ)/dξ3 /(LW)3
Attention, Students often forget about this term LW in derivatives!
Remark 2: ξ > 0 is the dimensionless coordinate to the right or the distance from the point xo =0 loaded by the concentrated force; if xo ≠ 0 then ξ in the solution (4) must be replaced by ξ - ξo > 0, where ξo = xo /LW,
Remark 3: the superposition rule is true, so the action of several forces is just a sum of effects (solutions) from each force considered separately; due to the superposition law and the limit passage, the sum will be replaced by an integral if the loading is continuous like qo(x); this way, the solution of an inhomogeneous equation is reached – for doubly–infinite beam from -∞ to +∞, so far.
-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 M oQo Yo - even
Mo - even Qo- odd
) 8 ( cos
8 4 ) (
) 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
w o w
o w
o w w
P Q 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 (antisymmetrical for Q) or even (symmetrical for y, r, M).
(4)
Remark 4: fundamental solution of the second kind (also on this WWW) refers to a doubly-infinite beam but loaded by a concentrated moment.
Example:
Lack of detachment means that it is allowed to assume just the bilateral bonds which in this context do not differ from the unilateral (no-tension) ones.
Constant load qo causes a constant settlement equal to yo = qo /(C⋅B)> 0 along the entire beam as seen directly from equation (1). The solution (4) may, however, locally take negative values, yP <0, which indicates tension.
The point of this job is that the sum y =yo + yP has to be non-negative everywhere.
First, we need to figure out how much the beam moves up under the action of the force P.
We are looking for the extreme: dyP(ξ)/dξ = 0 based on the solution (4).
After differentiation, the equation sin(ξ)=0 can be obtained, i.e. the extremes are reached at points ξe = 0, ±π,
±2π, ... It is enough to take only "+", because the job is symmetrical;
ξe =0 means the maximum displacement, but downwards (under the force P); the only solution of interest is ξe = +π, because all subsequent extremes are either downward displacements or, if upward, they are smaller than that for ξe =+π (due to rapidly vanishing exponent term e-ξ).
The maximum upward displacement is therefore yP(ξe) = yP(π) = -P⋅e-π /(2BCLW).
From the condition yo +yP ≥ 0 we get qo /(C⋅B) - P⋅e-π/(2BCLW) ≥ 0, i.e.
qo ≥ P⋅e-π /(2LW) for the given P or P ≤ 2qo⋅e+π⋅LW for the given qo.
One could also look for the minimal LW, if both P and qo are given, i.e. search for stiffness EI or width B.
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 (fundamental solutions of the second kind); this way the exact mathematical solution will be reached.
Details are presented below.
3.1. Boundary conditions at the beam ends
Let 0 ≤ x ≤ L, 0 < ξ < L/LW
∗ at the left free end A: ΣMi = 0 , ΣQi = 0 (5)
∗ at the right free end B: ΣMi = 0 , ΣQi = 0
There are 4 boundary conditions to be set at these ends, this results from the 4th order of equation (2) and has an obvious relationship with 4 constants to be determined in the general solution (3).
At each of both ends of the beam, two conditions are set, i.e. values of:
- either y (settlement, linear vertical displacement) or Q (vertical force),
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.
B A
L Pi
xi
A double-infinite Winkler beam is loaded with a vertical force P [kN] concentrated at the cross-section ξ = 0.
There is also a constant load qo [kN/m] along the entire beam (dead weight and backfill on the foundation).
Calculate a maximal value of the force P, such that there is no tension under the beam (the lack of detachment, no gap).
0
yP(ξ)
yP
(ξ
e)
and
- either ϕ (angle of rotation, angular displacement) or M (moment),
because in mechanical systems displacement and the corresponding force cannot be pre-defined
simultaneously. These conditions set on y, ϕ, M, Q are de facto conditions set on the value of the function and its first 3 derivatives; there is no use to focus on higher derivatives (there is no need anyway), because as follows from (2), the fourth derivative is expressed by the value of the function y itself, the fifth derivative by the value of the first derivative of y, the sixth derivative by the value of the second derivative of y, etc.
Also some mixed boundary conditions are possible like a spring support QA = Cs
⋅
yA , Cs=const, linking the value of the function with the value of its third derivative – both at A.3.2. The Method of H.Bleich
• The end A is assumed at ξ = 0 whereas the end B occurs at ξ = λ= L/LW < ∞.
• Instead of the finite beam AB, consider a doubly-infinite beam being fictitiously extended to ±∞; thus the solutions (4) have the application in local coordinates.
• Now, A, B are not the ends of the beam (because it is doubly-infinite) but are just some fixed cross-sections;
at these cross-section the conditions (5) are generally not fulfilled – they need some corrections.
• Apply 4 virtual (vertical) forces Ti – all applied outside the AB interval corresponding to the real beam.
• Hence the doubly-infinite beam is loaded by real forces P1, P2 , ... , Pn on AB and T1, T2, T3, T4 outside AB.
Values of these 4 forces Ti should be found in such a way that the conditions (5) are satisfied at A,B; more generally, (5) describe here the free ends of the beam – some known non-zero values should be substituted instead of zeros for loaded ends; alternatively, we can pre-define some values of y (like zero for a support) or of ϕ (like zero for a fixed end).
• Superposition (sum) of actions of all n+4 forces determines the exact and unique mathematical solution of the problem on 0 ≤ x ≤ L.
• Strictly speaking, there are 8 unknowns in the Bleich method, not 4, because not only the values of 4 forces Ti are to be found but also their application points; so, the latter ones are usually taken arbitrary; potentially, a combined approach is possible as well – like only 2 forces but applied to specially selected points on the virtual extension of AB to the doubly-infinite beam (outside AB) – this seems to be more complicated, however.
• Traditionally, for free ends, the virtual forces are applied as follows:
ξA1 = ξB3 = π/4 , ξA2 = ξB4 = π/2 ,
because some terms M,Q yielding from (4) are reduced (more zeros in the 4x4 matrix below);
of course, this is not obligatory.
The set of governing equations for the unknown forces Ti which corresponds to (5) is as follows:
[ ( ) ( ) ( ) ( ) ] 0
8
1⋅ ξ 1 + 2⋅ ξ 2 + 3⋅ λ+ξ 3 + 4⋅ λ+ξ 4 = =−
∑
∞P W o A o A o B o B i
A
L T M T M T M T M M
M
[ ( ) ( ) ( ) ( ) ] 0
8 1
4 4
3 3
2 2
1
1⋅ ξ + ⋅ ξ − ⋅ λ+ξ − ⋅ λ+ξ = =
−
∑
∞P o A o A o B o B i
A
T Q T Q T Q T Q Q
Q
[ ( ) ( ) ( ) ( ) ] 0
8
1⋅ λ+ξ 1 + 2⋅ λ+ξ 2 + 3⋅ ξ 3 + 4⋅ ξ 4 = =−
∑
∞P W o A o A o B o B i
B
L T M T M T M T M M
M
[ ( ) ( ) ( ) ( ) ] 0
8 1
4 4
3 3
2 2
1
1⋅ λ+ξ + ⋅ λ+ξ − ⋅ ξ − ⋅ ξ = =
−
∑
∞P o A o A o B o B i
B
T Q T Q T Q T Q Q
Q
for λ = L/LW and the dimensionless functions Mo, Qo presented in (4) as well as:
P
M
∞A ,M
B∞P - resultant moments at the cross-sections A(x=0), B(x=L) due to the real forces Pi , i =1,2,...,n applied to the infinite beam,P
Q
A∞ ,Q
B∞P - resultant shearing forces at the cross-sections A(x=0), B(x=L) due to the real forces Pi , i =1,2,...,n applied to the infinite beam.B A
λ=L/LW
Pi
T2 T1 T3
T4
ξA1
ξ ξA2
ξB3
ξB4
Note that the solutions (4) and the corresponding chart refer only to the right half of the beam, so to cross- sections situated to the right of the applied force. For y, M the solutions are even and the same charts are on the left;
in contrast, for ϕ, Q and cross-sections on the left, the sign must be changed because the latter are odd
functions; that’s why there appear 4 minuses at T3, T4 in the above set of equations – cross-sections A,B (and all others between them) are situated on the left only from T3,T4, therefore the signs are changed.
The same happens also at A from Pi but this sign changes are “hidden” in
Q
A∞P. Remark 1:For „long” beams and „great” distances of T1, T2 from the right end B, say ξ > 4÷5, the forces T1, T2 can not significantly influence M,Q at B because exp{-ξ } is small, so they can be ignored; the same for T3, T4 at A;
if so, the above matrix 4x4 is diagonal and four equations are separated, each with one variable, but this is only approximate solution.
Remark 2:
The Bleich method is not a very complex tool but requires some practice and skills. Therefore, there are numerous examples and testing questions attached..
The idea of virtual loadings is very popular – in mechanics, heat conduction, diffusion and not only. This is correct and fully exact mathematical approach.
Many applications can be found elsewhere on this WWW.
4. Direct solving of finite Winkler beams
Consider a finite interval (a,b). Solving consists of 2 following stages.
Stage no.I. Homogeneous case
Solution of the homogeneous equation (qo = 0) for the finite beam is of course in the form of (3), but the terms with C3 and C4 do not disappear this time (at least one), because the vanishing of solution in infinity cannot be assumed.
The general (“mathematical”) method imposes 4 conditions on the solutions (3) and/or its derivatives by making use of statical relations given in the section 1.2. Denote such a solution as ya,b(x) and 4 boundary conditions as b1a, b2a and b3b, b4b.
Stage no. II. Inhomogeneous case
To take into account a loading of the beam qo(x) along the interval (a,b), not only at the ends, it suffices to focus only on one concentrated force P applied at a certain internal point xo in (a,b); indeed, the solution for q(xo) will be a sum (integral) of such solution due to the superposition law, replacing P by dP=qo(x)dx and so on.
Let the finite beam to be solved has the following boundary conditions B1a, B2a oraz B3b, B4b respectively on both ends of the beam a, b and consider the force P applied to a xo , ξo = xo /LW.
Symbols B1, B2, B3, B4 denote any rational boundary conditions, like for example M=0, Q=0 for free ends of the beam.
First, the fundamental solution of the first kind should be found for the force P applied to xo located between cross-sections a and b. The beam is generally the same but extended to ±∞. In other words, the solution (4) is used but with shifted central coordinate to xo, ξo = xo/LW. Denote this solution as yP(x). Such the solution will not satisfy (yet) the required boundary conditions B1a, B2a and B3b, B4b at the distinguished cross-sections a and b, because yP(x) generate there some other values called B1Pa, B2Pa oraz B3Pb, B4Pb; they should be corrected by making use of the homogeneous solution (I) taking at the ends:
b1a = B1a - B1Pa b2a = B2a – B2Pa
and
b3b = B3b – B3Pb b4b = B4b – B4Pb
Sum of these two solutions I and II satisfies the differential equation itself as well as the required boundary conditions B1a, B2a and B3b, B4b at a and b; this completes the construction because there is only one such solution of the differential equation. The method presented above seems to be “purely mathematical”, less clear than the Bleich method which has a very intuitive engineering sense. It should be emphasized that both
methods form pt.no.3 and pt.no.4 are exact, so their results must be the same.
As a summary, analyze the Fig. below (Winkler springs under the beams are omitted).
-∞ +∞
= +
5. Testing questions
1. There is a loading qo(x) along the finite beam. What is the settlement of the Winkler subsoil out of the beam (unloaded surface)? Is the answer dependent on the stiffness of the beam EI?
2. Check that the uniform loading of the Winkler beam causes its uniform settlement and this independent of the beam stiffness 0 ≤ EI ≤ +∞ and the beam length L.
3. Can we place the virtual forces Ti at any distances?
4. Can we place – theoretically - all the virtual forces Ti on one side of a finite beam? And practically?
5. Is this possible the only 3 virtual forces are in use, not four? (the beam is finite) Is this true that 6 virtual forces will assure greater accuracy of results?
6. How about the Ti forces if the beam is half-infinite?
7. Can we use the Bleich method if a finite beam is loaded by concentrated moments on AB, not by the forces Pi
?
8. How should we proceed if the beam is loaded at the end by MA?
9. Develop an analytical method to solve a finite beam which is not prismatic, i.e. it is composed of several prismatic segments Eii ≠ const(i) or Bi ≠ const(i) or Ci ≠ const(i). The Bleich metod can be used or just modify (3).
10. Fundamental solution of the second kind can be expressed as:
) sin(
)
(ξ = 2 e−ξ ξ L
C B y M
W
Derive this formula by a simple modification of the general method presented in the section 2.2.
11. Can we propose „a new Bleich method” which uses virtual concentrated moments Mi (fundamental solution of the second kind as in question 10) instead of virtual concentrated forces Ti?
12. Is this true that the Bleich method for finite beams is only approximate?
Hints and answers to the questions:
1. There are zero settlements y=0 out of the beam in the Winkler model. This is independent of qo , EI and all shallow foundations – assumption that: y = q/C and loading q=0 outside the beam.
2. y = const and qo = const both satisfy (1).
3. Yes, we can but not overlapping ones because 2 overlapping forces mean de facto one force, etc.
4. Theoretically – always yes but in practice could be not rational.
If, e.g., virtual forces were very far from the beam ends (say 10 x LW or more), they would have to be very large to correct cross-sectional forces to zero in appropriate cross-sections A,B. A small inaccuracy in determining the value of such a virtual force (like rounding errors of the order of E-05) could significantly affect the accuracy of the solution between these sections. This is called a problem of bad numerical conditioning.
5. For the finite beam there are 4 equations and exactly 4 virtual forces to be found. The obtained values can be positive, can be negative, so sometimes one or more could be zero-valued. In this sense we can have less than 4 virtual forces. Never 6. You cannot find 6 unknowns from 4 linear equations.
6. There is only one end point, so 2 boundary conditions, so 2 virtual forces to be found from two equations (2x2).
7. Possible. Only the boundary values to be corrected at the ends like M∞APare used, no matter where they come from.
8. Put this MA instead of 0 in the right equation in section 3.2.
9. Each of the n prismatic sections can be solved separately, where Lwi = const. Basic solutions (3) then give you 4n constants to be determined. They are determined from 4n equations: 2 conditions at the left end of the beam, 2 conditions at the right end, 4⋅(n-1) continuity conditions at connections of the n-1 prismatic segments. These bonding conditions are: continuity of the y-deflection line (because the beam does not break), continuity of the derivative y (ϕ angle, rotation), continuity of moments (no concentrated moment in the contact cross-section) and continuity of transverse forces (no transverse concentrated force in the contact cross-section). These four conditions of continuity are quite obvious, maybe not entirely when it comes to rotation angle continuity. We simply assume that the deflection line is smooth (there are no undifferentiated "edges" in the places of bonding).
10. As for P: solve only for ξ > 0 finding four constants Ci, for y(+∞)=0, y(0)=0, M/2 as the moment at ξ > 0, ξ → 0.
Can be found in attached examples, somewhere.
11. Certainly, yes.
12. No, it is exact. It is approximate for foundation slabs, not beams.
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