# An Improved Winkler Foundation Modulus for a Beam in a Full Space
## 2 The Formula of the Winkler Foundation Modulus
### Beam Bending Response of the Beam on a Winkler Foundation  
As shown in Figure 1, an arbitrary displacement of the free field can be discretized into the standard displacement \(u_{g}\). The response of the structure under an arbitrary displacement of the free field can be transformed into the superposition of a series of standard displacements, which has the form  
\[u_{g}(x)=\left\{ \begin{array}{ll}0&x<-l\\ g_{0}&-l\leq x\leq l\\ 0&x\geq l\end{array} \right. \tag{1}\]  
where \(g_{0}\) and \(l\) are the height and half the width of the standard displacement, respectively. As illustrated in Figure 2, a long, linear underground structure embedded in a full-space foundation under the standard displacement can be simplified using a model of a Euler beam on a Winkler foundation. The [PERSON] elastic foundation model assumes that the foundation responds to the load on the beam in a linear elastic manner. For each position \(x\) along the beam axis, the beam experiences an elastic reaction force \(F_{g}(x)\) provided  
Figure 1: Displacement of the free field. by the foundation. According to the [PERSON] model theory, the relationship between the elastic reaction force and the deflection is given by:  
\[F_{s}(x)=K_{h}\big{[}u_{g}(x)-u_{b}(x)\big{]} \tag{2}\]  
where \(K_{h}\) is the [PERSON] foundation modulus, and \(u_{b}\) is the beam displacement. Based on [PERSON]'s beam theory, the relationship between the bending moment of the beam and the deflection has the form  
\[M(x)=-E_{b}l_{b}\frac{d^{2}u_{b}(x)}{dx^{2}} \tag{3}\]  
where \(M(x)\) is the bending moment, \(E_{b}\) is the elastic modulus of the beam, and \(l_{b}\) is the moment of inertia of the beam's cross-section. The equilibrium equation of the beam element can be written as follows:  
\[\frac{dQ(x)}{dx}=F_{s} \tag{4}\]  
By using the relationship \(Q=dM/dx\) and combining Equations (2)-(4), the governing equation of the Euler beam on the Winkler foundation can be derived as follows:  
\[\frac{d^{4}u_{b}}{dx^{4}}+4\xi^{4}u_{b}=4\xi^{4}u_{g}(x) \tag{5}\]  
where  
\[\xi=\sqrt[4]{K_{h}\Big{/}(4E_{b}l_{b})} \tag{6}\]  
Owing to the displacement of the free-field displacement from Equation (1) being divided into three segments, the response of the beam should be also represented by a three-segment function. However, this is not beneficial for the subsequent fitness to obtain the Winkler foundation modulus. Considering the displacement \(u_{g}\) with a finite interval, it can be represented by the Fourier integral as follows:  
\[u_{g}(x)=\frac{g_{0}}{\pi}\int_{0}^{\infty}\frac{2\sin(\lambda l)\cos(\lambda x )}{\lambda}d\lambda \tag{7}\]  
Owing to assumptions that the beam and foundation deform within the linear range, the beam response can be solved using the linear superposition method. From Equation (7), it is clear that the response of the beam under the standard displacement \(u_{g}(x)\) can be obtained by integrating the response of the beam under a cosine load \(\frac{g_{0}}{\pi}\frac{2\sin(\lambda l)\cos(\lambda x)}{\lambda}\). Thus, the free-field displacement can be assumed to be:  
\[u_{g1}(x)=a\cos(\lambda x) \tag{8}\]  
Figure 2: Euler beam on the Winkler foundation. where \(a=\frac{g_{0}}{\pi}\frac{2\sin(\lambda l)}{\lambda}\).