## 2 Inconsistent Load-Deformation Problem of FLSPs  
### Action Mechanism and Load-Deformation Characteristics of FLSPs  
#### 2.1.1 The Action Mechanism and Simplified Model of a FLSP  
The force of FLSPs comprises two parts: on the one hand, the primary lining load (the axial force, the shear force, and the bending moment), and, on the other hand, the interaction between the FLSP and the surrounding rock (the friction along the direction of  
Figure 1: Feet-lock steel pipe.  
the pipe wall and the elastic stiffness perpendicular to the direction of the pipe wall). Under the primary lining axial load, FLSP exhibits displacement trends relative to the soil. The friction resistance between the soil and FLSP prevents the axial displacement of the FLSP. The FLSP bends under the shear and bending moment, and the soil resists deformation thanks to the elastic stiffness. Engineering practice shows that FLSPs mainly bear shear force, and the action between the FLSP and the primary lining is complicated. To simplify the calculation, the force between the FLSP and the primary lining is simplified as a pair of shear forces.  
The mechanical analysis models of the interaction between the FLSP and the soil under shear force includes the Winkler, elastic continuum, and two-parameter foundation models. The Winkler model is shown in Figure 2. The model assumes that the foundation reaction force received at any point of the FLSP is proportional to the deflection:  
\[\mathrm{P}(x)\ =KDy(x) \tag{1}\]  
where \(\mathrm{P}(x)\) is the foundation reaction force per unit length of the steel pipe, \(K\) is the elastic coefficient of the soil (MPa/m), \(D\) is the outer diameter of the FLSP (m), and \(y(x)\) is the deflection of the FLSP (m).  
When the FLSP is subjected to shear load at the end, the differential equation of its deflection curve is:  
\[\frac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}}+4\alpha^{4}y=0 \tag{2}\]  
where \(\alpha\ =\ \sqrt[4]{\frac{KD}{4E_{z}}}\) is the deformation coefficient of the FLSP, \(1/\alpha\) is the characteristic length (m), E is the elastic modulus of the FLSP (N/m\({}^{2}\)), and \(I_{z}\) is the cross-sectional moment of inertia of the FLSP (m\({}^{4}\)).  
The general solution of the above formula is:  
\[y\ =\ \mathrm{e}^{ax}(\mathrm{A}_{1}\mathrm{cos}(\alpha x)\ +\ \mathrm{B}_{1} \mathrm{sin}(\alpha x))\ +\ \mathrm{e}^{-\alpha x}(\mathrm{C}_{1}\mathrm{cos}(\alpha x)\ +\ \mathrm{D}_{1} \mathrm{sin}(\alpha x)) \tag{3}\]  
where \(\mathrm{A}_{1}\), \(\mathrm{B}_{1}\), \(\mathrm{C}_{1}\), and \(\mathrm{D}_{1}\) are constants determined by boundary conditions.  
In the Winkler model, an infinite beam when \(\alpha L\geq\pi\) can be considered. According to the force and displacement boundary conditions, the deflection curve equation of the FLSP under the action of concentrated force can be calculated by:  
\[y=\frac{2\alpha}{KD}\varphi_{1}F_{0} \tag{4}\]  
where \(\varphi_{1}=\mathrm{e}^{-\alpha x}\cos(\alpha x)\).  
At the end of the FLSP (\(x=0\)):  
\[y_{0}=\frac{2\alpha}{KD}F_{0} \tag{5}\]  
Figure 2: Winkler model of FLSP.  
To more intuitively express the influencing factors of the bearing characteristics of the FLSP, \(\alpha=\sqrt[4]{\frac{KD}{4E_{z}}}\) and \(I_{Z}=\frac{\pi(D^{4}-d^{4})}{64}\) can be substituted into Formula (5):  
\[y_{0}=F_{0}\frac{2\alpha}{KD}=\sqrt[4]{\frac{256}{E\pi(D^{4}-d^{4})(KD)^{3}}}F_{0} \tag{6}\]  
According to Formula (6), the lateral stiffness of the FLSP is:  
\[k=\frac{KD}{2\alpha}=\sqrt[4]{\frac{E\pi(D^{4}-d^{4})(KD)^{3}}{256}} \tag{7}\]