Vigas En Fundaciones Elasticas

Table of contents
Lecture notes: Structural Analysis II

Beams on elastic foundation
I. Basic concepts.
The beam lies on elastic foundation when under the applied external loads, the reaction forces of the foundation are proportional at every point to the deflection of the beam at this point. This assumption was introduced first by Winkler in 1867. Consider a straight beam supported alongits entire length by an elastic medium and subjected to vertical forces acting in the plane of symmetry of the cross section (Fig. 1) M A F q B x b y R(x) Figure 1 Beam on elastic foundation Because of the external loadings the beam will deflect producing continuously distributed reaction forces in the supporting medium. The intensity of these reaction forces at any point is proportional to thedeflection of the beam y(x) at this point via the constant k: R(x)=k·y(x). The reactions act vertically and opposing the deflection of the beam. Hence, where the deflection is acting downward there will be a compression in the supporting medium. Where the deflection happens to be upward in the supporting medium tension will be produced which is not possible. In spite of all it is assumed that thesupporting medium is elastic and is able to take up such tensile forces. In other words the foundation is made of material which follows Hooke’s law. Its elasticity is characterized by the force, which distributed over a unit area, will cause a unit deflection. This force is a constant of the supporting medium called the modulus of the foundation k0 [kN/m2/m]. Assume that the beam under considerationhas a constant cross section with constant width b which is supported by the foundation. A unit deflection of this beam will cause reaction equal to k0·b in the foundation, therefore the intensity of distributed reaction (per unit length of the beam) will be: R(x)= b·k0·y(x)= k·y(x), where k= k0·b is the constant of the foundation, known as Winkler’s constant, which includes the effect of thewidth of the beam, and has dimension kN/m/m.
plane of symmetry cross section

II. Differential equation of equilibrium of a beam on elastic foundation
Consider an infinitely small element enclosed between two vertical cross sections at distance dx apart on the beam into consideration (Fig. 2). Assume that this element was taken from a portion

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Lecture notes: Structural Analysis II

where the beam was acted upon by a distributed loading q(x). The internal forces that arise in section cuts are depicted in Fig. 2.

M Q dx

q(x)

M+dM Q+dQ R(x)

Figure 2 Differential element of length dx Considering the equilibrium of the differential element, the sum of the vertical forces gives: ΣV = 0Q − ( Q + dQ ) + R( x) ⋅ dx − q ( x) ⋅ dx = 0 ;
k⋅ y( x)

dQ = k⋅y−q . dx
Considering the equilibrium of moments along the left section of the element we get:
ΣM = 0 dM − ( Q + dQ ) ⋅ dx − q dx 2 dx 2 +r =0; 2 2 0 0

dM =q. dx
d2y
2

0

Using now the well known differential equation of a beam in bending:
M ’ EI dx it can be written: =− dQ d 2 M d4y = = − EI 4 . dx dx 2 dxFinally, from the summation of the vertical forces ΣV = 0 :
− EI d4y dx 4 = k⋅y−q;

y IV + 4

k q ( x) ⋅y= . 4 EI EI
α4

In the above equation the parameter α includes the flexural rigidity of the beam as well as the elasticity of the foundation. This factor is called the characteristic of the system with dimension length-1. In that respect 1/ α is referred to as the so called characteristiclength. Therefore, α ⋅ x will be an absolute number. The differential equation of equilibrium of an infinitely small element becomes:
2011 S. Parvanova, University of Architecture, Civil Engineering and Geodesy - Sofia

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Lecture notes: Structural Analysis II

q ( x) k , α =4 . EI 4 EI The solution of this differential equation could be expressed as: y ( x) = y0 ( x) + v( x) , y IV + 4α 4...