Application of Quintic Displacement Function in Static Analysis of Deep Beams on Elastic Foundation

Abstract

Beams supported by elastic foundations are one of the complex soil-structure interaction problems that have been studied using the "beam on elastic foundation" concept. To support the loads, the structural foundation and the soil continuum must work together. The development of more accurate foundation model and simpler procedure are critical for the safety and cost-effective construction of such type of structure. Three nodded beams based on Timoshenko beam theory and workable approaches for analysis of beams on Winkler foundation are attempted in this study. For the present formulation, a Matlab code has been developed. The results are then compared to similar studies done by other researchers, which demonstrate a high level of agreement. Parametric studies are followed to determine the response to various loading conditions, boundary conditions, and foundation parameters. The present formulations, regardless of boundary conditions, slenderness ratio and modulus of sub-grade reaction, have a higher convergence rate. It functions smoothly and efficiently for thin to moderately thick beams. The influence of the soil coefficient on the response of beams on elastic foundations is typically greater than the influence of beam physical and material properties.

Appendices

Appendix A

Development of governing differential equation

$$\therefore \begin{array}{c}\begin{array}{c}{\varepsilon }_{xx}=\frac{\partial u}{\partial x}=-z\frac{\partial {\theta }_{x}}{\partial x}\end{array}\end{array} ;{\gamma }_{xz}=\frac{\partial u}{\partial z}+\frac{\partial w}{\partial x}={-\theta }_{x}+\frac{\partial w}{\partial x};$$

The strain energy deformation in Timoshenko Beam corresponding to bending and shear deformation is

$$\mathrm{Strain\; energy},{ U }=\frac{1}{2}{{\upsigma }_{{ij}}\upvarepsilon }_{{ij}}{dV}=\frac{1}{2}{{\upsigma }_{{ij}}\upvarepsilon }_{{ij}}{dxdydz}=\frac{1}{2}\int \int \int \left({{\upsigma }_{{x}}\upvarepsilon }_{{x}}+{\uptau }_{{zx}}{\upgamma }_{{xz}}\right){dxdydz}$$

$${\mathrm{U}}_{\mathrm{b}} =\frac{1}{2}{\int }_{0}^{\mathrm{L}}{\int }_{-\frac{\mathrm{h}}{2}}^{\frac{\mathrm{h}}{2}}\left[{\mathrm{Ez}}^{2}{\left(\frac{\partial {\uptheta }_{\mathrm{x}}}{\partial \mathrm{x}}\right)}^{2}+{\mathrm{\alpha G}\left(-{\uptheta }_{\mathrm{x}}+\frac{\partial \mathrm{w}}{\partial \mathrm{x}}\right)}^{2}\right]bdxdz$$

$$\therefore {\mathrm{U}}_{\mathrm{b}} =\frac{\mathrm{E}}{2}{\int }_{0}^{\mathrm{L}}{\int }_{-\frac{\mathrm{h}}{2}}^{\frac{\mathrm{h}}{2}}{\mathrm{z}}^{2}{\left(\frac{\partial {\uptheta }_{\mathrm{x}}}{\partial \mathrm{x}}\right)}^{2}bdzdx+{\int }_{0}^{\mathrm{L}}{\int }_{-\frac{\mathrm{h}}{2}}^{\frac{\mathrm{h}}{2}}{\frac{\mathrm{\alpha {\rm A}}\mathrm{G}}{2}\left(-{\uptheta }_{\mathrm{x}}+\frac{\partial \mathrm{w}}{\partial \mathrm{x}}\right)}^{2}bdzdx$$

where E – is the Young’s modulus of beam material, I- is the second moment of inertia of the beam section G—is the modulus rigidity of beam material, b – is the beam width and α – is the shear correction factor.

$${{U}}_{{b}} =\frac{{EI}}{2}{\int }_{0}^{{L}}{\left(\frac{\partial {\uptheta }_{{x}}}{\partial {x}}\right)}^{2}dx+\frac{{\alpha AG}}{2}{\int }_{0}^{{L}}{\left(-{\uptheta }_{{x}}+\frac{\partial {w}}{\partial {x}}\right)}^{2}dx;A-is\; the\; cross\; sectional\; area\; of\; the\; beam.$$

$$\Pi =\frac{{EI}}{2}{\int }_{0}^{{L}}{\left(\frac{\partial {\uptheta }_{{x}}}{\partial {x}}\right)}^{2}dx+\frac{{\alpha AG}}{2}{\int }_{0}^{{L}}{\left(-{\uptheta }_{{x}}+\frac{\partial {w}}{\partial {x}}\right)}^{2}dx+\frac{1}{2}{\int }_{0}^{{L}}{{kw}}^{2}\mathrm{dx}-{\int }_{0}^{{L}}{qwdx}$$

Now, to obtain the equilibrium equations, we can use the calculus of variations. The first variation of the potential energy must be zero (δΠ = 0) in order for a stable equilibrium state to be achieved. From 0 to L we consider the first beam soil system and take variations on w. We get the following by making variations on w.

$$\mathrm{F}=\frac{\mathrm{EI}}{2}{\left(\frac{\partial {\uptheta }_{\mathrm{x}}}{\partial \mathrm{x}}\right)}^{2}+\frac{\mathrm{\alpha AG}}{2}{\left(-{\uptheta }_{\mathrm{x}}+\frac{\partial \mathrm{w}}{\partial \mathrm{x}}\right)}^{2}+\frac{1}{2}{\mathrm{kw}}^{2}-\mathrm{qw}$$

$$\mathrm{F}=\frac{\mathrm{EI}}{2}{\left(\frac{\partial {\uptheta }_{\mathrm{x}}}{\partial \mathrm{x}}\right)}^{2}+\frac{\mathrm{\alpha AG}{{\uptheta }_{\mathrm{x}}}^{2}}{2}-\mathrm{KAG}{\uptheta }_{\mathrm{x}}\frac{\partial \mathrm{w}}{\partial \mathrm{x}}+\frac{\mathrm{\alpha AG}}{2}{\left(\frac{\mathrm{dw}}{\mathrm{dx}}\right)}^{2}+\frac{1}{2}{\mathrm{kw}}^{2}-\mathrm{qw};$$

$$\frac{\partial \mathrm{F}}{\partial \mathrm{w}}-\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\partial \mathrm{F}}{\partial \left(\frac{\mathrm{dw}}{\mathrm{dx}}\right)}\right)=0$$

(1)

$$\frac{\partial \mathrm{F}}{\partial {\uptheta }_{\mathrm{x}}}-\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\partial \mathrm{F}}{\partial \left(\frac{\partial {\uptheta }_{\mathrm{x}}}{\partial \mathrm{x}}\right)}\right)=0$$

(2)

is known as Euler – Lagrange equation and the governing static equations of equilibrium state of elastic body.

$$\frac{\partial {F}}{\partial \mathrm{w}}={kw}-{q}; \frac{\partial {F}}{\partial \left(\frac{{dw}}{{dx}}\right)}={\alpha AG}\left(-{\uptheta }_{{x}}+\frac{\partial {w}}{\partial {x}}\right);\frac{\partial {F}}{\partial {\uptheta }_{{x}}}={\alpha AG}{\uptheta }_{{x}}-{\alpha AG}\frac{\partial {w}}{\partial {x}}{ and }\frac{\partial {F}}{\partial \left(\frac{\partial {\uptheta }_{{x}}}{\partial {x}}\right)}={EI}\frac{\partial {\uptheta }_{{x}}}{\partial {x}}$$

Substitute in Eq. (1) one get, ${kw}-{q}-\frac{{d}}{{dx}}\left({\alpha AG}\left({-\uptheta }_{{x}}+\frac{\partial {w}}{{\partial }_{{x}}}\right)\right)=0\therefore -{\alpha AG}\left(-\frac{\partial {\uptheta }_{{x}}}{\partial {x}}+\frac{{{d}}^{2}{w}}{{{dx}}^{2}}\right)+{kw}-{q}=0$

$$\therefore \mathrm{\alpha AG}\frac{{\mathrm{d}}^{2}\mathrm{w}}{{\mathrm{dx}}^{2}}-\mathrm{\alpha AG}\frac{\partial {\uptheta }_{\mathrm{x}}}{\partial \mathrm{x}}-\mathrm{kw}+\mathrm{q}=0;$$

(3)

Substitute in Eq. (2) one get, ${\alpha AG}{\uptheta }_{{x}}-{\alpha AG}\frac{\partial {w}}{\partial {x}}-\frac{{d}}{{dx}}\left({EI}\frac{\partial {\uptheta }_{{x}}}{\partial {x}}\right)=0\therefore {\alpha AG}{\uptheta }_{{x}}-{\alpha AG}\frac{\partial {w}}{\partial {x}}-{EI}\frac{{\mathrm{d}}^{2}{\uptheta }_{{x}}}{{{dx}}^{2}}=0$

$$\therefore \mathrm{EI}\frac{{\mathrm{d}}^{2}{\uptheta }_{\mathrm{x}}}{{\mathrm{dx}}^{2}}+\mathrm{\alpha AG}\frac{\partial \mathrm{w}}{\partial \mathrm{x}}-\mathrm{\alpha AG}{\uptheta }_{\mathrm{x}}=0;$$

(4)

Equations (3) and (4) are couple governing differential equation of beams on Winkler foundation.

Decoupling the differential equations of equilibrium

$$\frac{{dM}}{{dx}}={Q\; and\; }\frac{{dQ}}{{dx}}=-{q};{M}=-{EI}\frac{\partial {\uptheta }_{{x}}}{\partial {x}}\therefore \frac{{dM}}{{dx}}=-{EI}\frac{{{d}}^{2}{\uptheta }_{{x}}}{{{dx}}^{2}}={Q};\therefore \frac{{dQ}}{{dx}}=-{EI}\frac{{{d}}^{3}{\uptheta }_{{x}}}{{{dx}}^{3}}=-{q}\therefore \frac{{{d}}^{3}{\uptheta }_{{x}}}{{{dx}}^{3}}=\frac{{q}}{{EI}}$$

From Eq. (3), ${\alpha AG}\frac{{{d}}^{2}{w}}{{{dx}}^{2}}-{\alpha AG}\frac{\partial {\uptheta }_{{x}}}{\partial {x}}-{kw}+{q}=0 \therefore {\alpha AG}\frac{{{d}}^{3}{\uptheta }_{{x}}}{{{dx}}^{3}}-{\alpha AG}\frac{{{d}}^{4}{w}}{{{dx}}^{4}}+{k}\frac{{{d}}^{2}{w}}{{{dx}}^{2}}=\frac{{{d}}^{2}{q}}{{{dx}}^{2}}$

$$\mathrm{Substituting }\frac{{\mathrm{d}}^{3}{\uptheta }_{\mathrm{x}}}{{\mathrm{dx}}^{3}}=\frac{\mathrm{q}}{\mathrm{EI}} \therefore \frac{\mathrm{\alpha AGq}}{\mathrm{EI}}-\mathrm{\alpha AG}\frac{{\mathrm{d}}^{4}\mathrm{w}}{{\mathrm{dx}}^{4}}+\mathrm{k}\frac{{\mathrm{d}}^{2}\mathrm{w}}{{\mathrm{dx}}^{2}}=\frac{{\mathrm{d}}^{2}\mathrm{q}}{{\mathrm{dx}}^{2}}$$

$$\mathrm{\alpha AGq}-\mathrm{EI\alpha AG}\frac{{\mathrm{d}}^{4}\mathrm{w}}{{\mathrm{dx}}^{4}}+\mathrm{EIk}\frac{{\mathrm{d}}^{2}\mathrm{w}}{{\mathrm{dx}}^{2}}=\mathrm{EI}\frac{{\mathrm{d}}^{2}\mathrm{q}}{{\mathrm{dx}}^{2}}\therefore \mathrm{EI\alpha AG}\frac{{\mathrm{d}}^{4}\mathrm{w}}{{\mathrm{dx}}^{4}}-\mathrm{EIk}\frac{{\mathrm{d}}^{2}\mathrm{w}}{{\mathrm{dx}}^{2}}+\mathrm{EI}\frac{{\mathrm{d}}^{2}\mathrm{q}}{{\mathrm{dx}}^{2}}=\mathrm{\alpha AGq}$$

Beam without foundation k = 0;

$$\therefore \mathrm{EI}\frac{{\mathrm{d}}^{4}\mathrm{w}}{{\mathrm{dx}}^{4}}+\frac{\mathrm{EI}}{\mathrm{\alpha AG}}\frac{{\mathrm{d}}^{2}\mathrm{q}}{{\mathrm{dx}}^{2}}=\mathrm{q}$$

Appendix B

$${{N}}_{{w}1}=\frac{1}{\left(1+5{\varphi }-20{{\varphi }}^{2}\right)\left(4{\varphi }+1\right)}\left[\frac{24\left(4{\varphi }+1\right){{x}}^{5}}{{{L}}^{5}}-\frac{4{{x}}^{4}\left(-40{{\varphi }}^{2}+70{\varphi }+17\right)}{{{L}}^{4}}+\frac{6{{x}}^{3}\left(-80{{\varphi }}^{2}+40{\varphi }+11\right)}{{{L}}^{3}}-\frac{{{x}}^{2}\left(160{{\varphi }}^{3}-440{{\varphi }}^{2}+32{\varphi }+23\right)}{{{L}}^{2}}-\frac{3{x}\left(-80{{\varphi }}^{3}+40{{\varphi }}^{2}+11{\varphi }\right)}{{L}}+\left(1+5{\varphi }-20{{\varphi }}^{2}\right)\left(4{\varphi }+1\right)\right]$$

$${{N}}_{{w}2}=\frac{1}{\left(1+5{\varphi }-20{{\varphi }}^{2}\right)}\left[-\frac{4\left(2{\varphi }-1\right){{x}}^{5}}{{{L}}^{4}}-\frac{6\left(-20{{\varphi }}^{2}+5{\varphi }+2\right){{x}}^{4}}{\left(4{\varphi }+1\right){{L}}^{3}}+\frac{\left(-160{{\varphi }}^{2}+36{\varphi }+13\right){{x}}^{3}}{\left(4{\varphi }+1\right){{L}}^{2}}-\frac{3\left(-60{{\varphi }}^{2}+11{\varphi }+4\right){{x}}^{2}}{2{L}\left(4{\varphi }+1\right)}+\frac{\left(-36{{\varphi }}^{2}+5{\varphi }+2\right){x}}{2\left(4{\varphi }+1\right)}\right]$$

$${\mathrm{N}}_{\mathrm{w}3}=\frac{1}{\left(4\mathrm{\varphi }+1\right)}\left[\frac{16{\mathrm{x}}^{4}}{{\mathrm{L}}^{4}}-\frac{32{\mathrm{x}}^{3}}{{\mathrm{L}}^{3}}-\frac{16{\mathrm{x}}^{2}\left(\mathrm{\varphi }-1\right)}{{\mathrm{L}}^{2}}+\frac{16\mathrm{\varphi x}}{\mathrm{L}}\right]$$

$${{N}}_{{w}4}=\frac{32{{x}}^{3}}{-20{{L}}^{2}{{\varphi }}^{2}+5{{L}}^{2}{\varphi }+{{L}}^{2}}-\frac{40{{x}}^{4}\left({\varphi }+1\right)}{-20{{L}}^{3}{{\varphi }}^{2}+5{{L}}^{3}{\varphi }+{{L}}^{3}}+\frac{16{{x}}^{5}\left({\varphi }+1\right)}{-20{{L}}^{4}{{\varphi }}^{2}+5{{L}}^{4}{\varphi }+{{L}}^{4}}-\frac{16{\varphi x}}{-20{{\varphi }}^{2}+5{\varphi }+1}+\frac{8{{x}}^{2}\left(5{\varphi }-1\right)}{-20{L}{{\varphi }}^{2}+5{L\varphi }+{L}}$$

$${{N}}_{{w}5}=-\frac{24{{x}}^{5}}{-20{{L}}^{5}{{\varphi }}^{2}+5{{L}}^{5}{\varphi }+{{L}}^{5}}+\frac{4\left(40{{\varphi }}^{2}+50{\varphi }+13\right){{x}}^{4}}{\left(4{\varphi }+1\right)\left(-20{{L}}^{4}{{\varphi }}^{2}+5{{L}}^{4}{\varphi }+{{L}}^{4}\right)}-\frac{2\left(80{{\varphi }}^{2}+40{\varphi }+17\right){{x}}^{3}}{\left(4{\varphi }+1\right)\left(-20{{L}}^{3}{{\varphi }}^{2}+5{{L}}^{3}{\varphi }+{{L}}^{3}\right)}-\frac{\left(160{{\varphi }}^{3}+40{{\varphi }}^{2}+32{\varphi }-7\right){{x}}^{2}}{\left(4{\varphi }+1\right)\left(-20{{L}}^{2}{{\varphi }}^{2}+5{{L}}^{2}{\varphi }+{{L}}^{2}\right)}+\frac{\left(80{{\varphi }}^{3}+40{{\varphi }}^{2}+17{\varphi }\right){x}}{\left(4{\varphi }+1\right)\left(-20{L}{{\varphi }}^{2}+5{L\varphi }+{L}\right)}$$

$${{N}}_{{w}6}=-\frac{4{{x}}^{5}\left(2{\varphi }-1\right)}{-20{{L}}^{4}{{\varphi }}^{2}+5{{L}}^{4}{\varphi }+{{L}}^{4}}-\frac{{x}\left(-4{{\varphi }}^{2}+5{\varphi }\right)}{2\left(4{\varphi }+1\right)\left(-20{{\varphi }}^{2}+5{\varphi }+1\right)}-\frac{{{x}}^{3}\left(4{\varphi }-5\right)}{\left(4{\varphi }+1\right)\left(-20{{L}}^{2}{{\varphi }}^{2}+5{{L}}^{2}{\varphi }+{{L}}^{2}\right)}-\frac{{{x}}^{2}\left(20{{\varphi }}^{2}-17{\varphi }+2\right)}{2\left(4{\varphi }+1\right)\left(-20{L}{{\varphi }}^{2}+5{L\varphi }+{L}\right)}-\frac{2{{x}}^{4}\left(-20{{\varphi }}^{2}+5{\varphi }+4\right)}{\left(4{\varphi }+1\right)\left(-20{{L}}^{3}{{\varphi }}^{2}+5{{L}}^{3}{\varphi }+{{L}}^{3}\right)}$$

$${{N}}_{\uptheta }=\left\{\begin{array}{c}\frac{120{{x}}^{4}}{5{{L}}^{5}{\varphi }+{{L}}^{5}}-\frac{2{x}\left(100{\varphi }+23\right)}{\left(5{\varphi }+1\right)\left(4{{L}}^{2}{\varphi }+{{L}}^{2}\right)}-\frac{16{{x}}^{3}\left(70{\varphi }+17\right)}{\left(5{\varphi }+1\right)\left(4{{L}}^{4}{\varphi }+{{L}}^{4}\right)}+\frac{6{{x}}^{2}\left(140{\varphi }+33\right)}{\left(5{\varphi }+1\right)\left(4{{L}}^{3}{\varphi }+{{L}}^{3}\right)}\\ -\frac{20\left(2{\varphi }-1\right){{x}}^{4}}{5{{L}}^{4}{\varphi }+{{L}}^{4}}-\frac{8\left(-40{{\varphi }}^{2}+15{\varphi }+6\right){{x}}^{3}}{\left(5{\varphi }+1\right)\left(4{{L}}^{3}{\varphi }+{{L}}^{3}\right)}+\frac{\left(-160{{\varphi }}^{2}+128{\varphi }+39\right){{x}}^{2}}{\left(5{\varphi }+1\right)\left(4{{L}}^{2}{\varphi }+{{L}}^{2}\right)}-\frac{\left(20{{\varphi }}^{2}+57{\varphi }+12\right){x}}{\left(5{\varphi }+1\right)\left({L}+4{L\varphi }\right)}+1\\ \frac{64{{x}}^{3}}{4{{L}}^{4}{\varphi }+{{L}}^{4}}-\frac{96{{x}}^{2}}{4{{L}}^{3}{\varphi }+{{L}}^{3}}+\frac{32{x}}{4{{L}}^{2}{\varphi }+{{L}}^{2}}\\ \frac{80{{x}}^{4}\left({\varphi }+1\right)}{{{L}}^{4}\left(5{\varphi }+1\right)}-\frac{160{{x}}^{3}\left({\varphi }+1\right)}{{{L}}^{3}\left(5{\varphi }+1\right)}-\frac{16{x}}{{L}\left(5{\varphi }+1\right)}+\frac{16{{x}}^{2}\left(5{\varphi }+6\right)}{{{L}}^{2}\left(5{\varphi }+1\right)}\\ -\frac{120{{x}}^{4}}{5{{L}}^{5}{\varphi }+{{L}}^{5}}+\frac{16\left(50{\varphi }+13\right){{x}}^{3}}{\left(5{\varphi }+1\right)\left(4{{L}}^{4}{\varphi }+{{L}}^{4}\right)}-\frac{6\left(60{\varphi }+17\right){{x}}^{2}}{\left(5{\varphi }+1\right)\left(4{{L}}^{3}{\varphi }+{{L}}^{3}\right)}+\frac{2\left(20{\varphi }+7\right){x}}{\left(5{\varphi }+1\right)\left(4{{L}}^{2}{\varphi }+{{L}}^{2}\right)}\\ -\frac{20\left(2{\varphi }-1\right){{x}}^{4}}{5{{L}}^{4}{\varphi }+{{L}}^{4}}-\frac{8\left(-40{{\varphi }}^{2}+5{\varphi }+4\right){{x}}^{3}}{\left(5{\varphi }+1\right)\left(4{{L}}^{3}{\varphi }+{{L}}^{3}\right)}+\frac{\left(-160{{\varphi }}^{2}+8{\varphi }+15\right){{x}}^{2}}{\left(5{\varphi }+1\right)\left(4{{L}}^{2}{\varphi }+{{L}}^{2}\right)}+\frac{\left(20{{\varphi }}^{2}+{\varphi }-2\right){x}}{\left(5{\varphi }+1\right)\left({L}+4{L\varphi }\right)}\end{array}\right.$$

where φ = 12D/(SL²) and D = EI, S = αAG.

$${\mathrm{U}}_{\mathrm{Beam}} =\frac{\mathrm{EI}}{2}{\int }_{0}^{\mathrm{L}}{\left(\frac{\partial {\uptheta }_{\mathrm{x}}}{\partial \mathrm{x}}\right)}^{2}dx+\frac{\mathrm{\alpha AG}}{2}{\int }_{0}^{\mathrm{L}}{\left(-{\uptheta }_{\mathrm{z}}+\frac{\partial \mathrm{w}}{\partial \mathrm{x}}\right)}^{2}dz$$

$${\mathrm{U}}_{\mathrm{Beam}} =\frac{1}{2}{\int }_{0}^{\mathrm{L}}\mathrm{EI}{\left\{\mathrm{d}\right\}}^{\mathrm{T}}{\left[{N}_{\varphi }\mathrm{^{\prime}}\right]}^{\mathrm{T}}\left[{N}_{\varphi }\mathrm{^{\prime}}\right]\left\{\mathrm{d}\right\}dx+\frac{1}{2}{\int }_{0}^{\mathrm{L}}\mathrm{\alpha AG}{\left\{\mathrm{d}\right\}}^{\mathrm{T}}{\left(-\left[{N}_{\varphi }\right]+\left[{N}_{w}\mathrm{^{\prime}}\right]\right)}^{\mathrm{T}}\left(-\left[{N}_{\varphi }\right]+\left[{N}_{w}\mathrm{^{\prime}}\right]\right)\left\{\mathrm{d}\right\}dx$$

$${\mathrm{U}}_{\mathrm{Beam}} =\frac{1}{2}{\left\{\mathrm{d}\right\}}^{\mathrm{T}}\left[{\mathrm{K}}_{\mathrm{b}}\right]\left\{\mathrm{d}\right\}+\frac{1}{2}{\left\{\mathrm{d}\right\}}^{\mathrm{T}}\left[{\mathrm{K}}_{\mathrm{s}}\right]\left\{\mathrm{d}\right\}$$

$$\mathrm{Bending stiffness matrix}, \left[{\mathrm{K}}_{\mathrm{b}}\right]={\int }_{0}^{\mathrm{L}}\mathrm{EI}{\left[{N}_{\varphi }\mathrm{^{\prime}}\right]}^{\mathrm{T}}\left[{N}_{\varphi }\mathrm{^{\prime}}\right]dx$$

$$and\; {shear\; stiffness\; matrix}, \left[{{K}}_{{s}}\right]={\int }_{0}^{{L}}{EI}{\left(-\left[{N}_{\varphi }\right]+\left[{N}_{w}{^{\prime}}\right]\right)}^{{T}}\left(-\left[{N}_{\varphi }\right]+\left[{N}_{w}{^{\prime}}\right]\right)dx$$

$${\uptheta }_{{x}}=\left[{N}_{\varphi }\right]\left\{{d }\right\}{ and w}=\left[{N}_{w}\right]\left\{{d }\right\}{ where\; d\; is\; the\; displacement\; vector}.$$

According to the theory of elasticity, the strain energy U_r in a linear spring is given as 1/2Kw².

$${\mathrm{U}}_{\mathrm{sk}}=\frac{1}{2}{\int }_{0}^{\mathrm{L}}{\mathrm{Kw}}^{2}\mathrm{dA}=\frac{\mathrm{Kb}}{2}{\int }_{0}^{\mathrm{L}}{\mathrm{w}}^{2}\mathrm{dx}=\frac{\mathrm{k}}{2}{\int }_{0}^{\mathrm{L}}{\mathrm{w}}^{2}\mathrm{dx };\mathrm{ K is the soil parameter}.$$

$$\mathrm{Strain energy }{\mathrm{U}}_{\mathrm{sk}}=\frac{1}{2}{\int }_{0}^{\mathrm{L}}\mathrm{k}{\left\{\mathrm{d}\right\}}^{\mathrm{T}}{\left[{N}_{w}\right]}^{\mathrm{T}}\left[{N}_{w}\right]\left\{\mathrm{d}\right\}\mathrm{dx }= \frac{1}{2}{\left\{\mathrm{d}\right\}}^{\mathrm{T}}\left[{\mathrm{K}}_{\mathrm{f}}\right]\left\{\mathrm{d}\right\}$$

In which the foundation stiffness matrix for the element is $[{\mathrm{K}}_{\mathrm{f}}] ={\int }_{0}^{\mathrm{L}}\mathrm{k}{\left[{N}_{w}\right]}^{\mathrm{T}}\left[{N}_{w}\right]\mathrm{dx}$

Hence the element stiffness matrix is

$$\therefore \left[{\mathrm{k}}_{\mathrm{e}}\right]=\left[{\mathrm{K}}_{\mathrm{b}}\right]+\left[{\mathrm{K}}_{\mathrm{s}}\right]+\left[{\mathrm{K}}_{\mathrm{f}}\right]$$

By using standard finite element procedure the global stiffness matrix is obtained

$$\therefore \left[{\mathrm{K}}_{\mathrm{S}}\right]=\sum_{1}^{\mathrm{n}}\left(\left[{\mathrm{K}}_{\mathrm{b}}\right]+\left[{\mathrm{K}}_{\mathrm{s}}\right]+\left[{\mathrm{K}}_{\mathrm{f}}\right]\right)$$

(5)

The nodal load vector is determined by the virtual work principle using the Lagrangian interpolation. The force vector for uniformly varying load of intensity w₁, w₂ and w₃ at three nodes respectively is the quadratic polynomial that passes through the three data points. This element has nodes at x = (0, L/2, L).

$${fe}=\left[\begin{array}{ccc}\frac{L\left(-4480{\varphi}^3-680{\varphi}^2+374\varphi +57\right)}{420\left(-80{\varphi}^3+9\varphi +1\right)}& \frac{L\left(28\varphi +11\right)}{105\left(4\varphi +1\right)}& \frac{L\left(1120{\varphi}^3+440{\varphi}^2+64\varphi -3\right)}{420\left(-80{\varphi}^3+9\varphi +1\right)}\\ {}\frac{L^2\left(-140{\varphi}^2+7\varphi +6\right)}{840\left(-80{\varphi}^3+9\varphi +1\right)}& \frac{L^2}{105\left(4\varphi +1\right)}& \frac{L^2\varphi \left(20\varphi +23\right)}{840\left(-80{\varphi}^3+9\varphi +1\right)}\\ {}\frac{4L\left(7\varphi +1\right)}{105\left(4\varphi +1\right)}& \frac{16L\left(14\varphi +3\right)}{105\left(4\varphi +1\right)}& \frac{4L\left(7\varphi +1\right)}{105\left(4\varphi +1\right)}\\ {}-\frac{2{L}^2\left(15\varphi +1\right)}{105\left(-20{\varphi}^2+5\varphi +1\right)}& 0& \frac{2{L}^2\left(15\varphi +1\right)}{105\left(-20{\varphi}^2+5\varphi +1\right)}\\ {}\frac{L\left(1120{\varphi}^3+440{\varphi}^2+64\varphi -3\right)}{420\left(-80{\varphi}^3+9\varphi +1\right)}& \frac{L\left(28\varphi +11\right)}{105\left(4\varphi +1\right)}& \frac{L\left(-4480{\varphi}^3-680{\varphi}^2+374\varphi +57\right)}{420\left(-80{\varphi}^3+9\varphi +1\right)}\\ {}-\frac{L^2\varphi \left(20\varphi +23\right)}{840\left(-80{\varphi}^3+9\varphi +1\right)}& -\frac{L^2}{105\left(4\varphi +1\right)}& -\frac{L^2\left(-140{\varphi}^2+7\varphi +6\right)}{840\left(-80{\varphi}^3+9\varphi +1\right)}\end{array}\right]\left\lfloor \begin{array}{c}{{w}}_1\\ {}{{w}}_2\\ {}{{w}}_3\end{array}\right\rfloor$$