统计代写|主成分分析代写Principal Component Analysis代考|Formulation of Finite Element for Elastic Bar

In summary, the formulation of a finite element for an elastic bar is developed under several key assumptions:

The bar is geometrically straight and only undergoes axial deformation. The material follows Hooke’s Law, where stress is proportional to strain. Forces are applied only at the endpoints of the bar. The bar only transmits axial loads, not bending, torsion, or shear forces. To model the behavior of this bar element, the displacement field u(x)u(x)u(x) along its length is approximated using nodal displacements u1u_1u 1 ​ and u2u_2u 2 ​ at the ends, with interpolation functions N1(x)N_1(x)N 1 ​ (x) and N2(x)N_2(x)N 2 ​ (x) such that u(x)=N1(x)u1+N2(x)u2u(x) = N_1(x)u_1 + N_2(x)u_2u(x)=N 1 ​ (x)u 1 ​ +N 2 ​ (x)u 2 ​ . These interpolation functions are chosen to be linear polynomials that satisfy the boundary conditions exactly, resulting in N1(x)=1−xLN_1(x) = 1 – \frac{x}{L}N 1 ​ (x)=1− L x ​ and N2(x)=xLN_2(x) = \frac{x}{L}N 2 ​ (x)= L x ​ .

From strength of materials principles, the relationship between the applied axial force PPP and the resulting deformation Δ\DeltaΔ is given by Δ=PLAE\Delta = \frac{PL}{AE}Δ= AE PL ​ , where AAA is the cross-sectional area and EEE is the modulus of elasticity. The spring constant kkk of the bar is then computed as k=AELk = \frac{AE}{L}k= L AE ​ .

However, instead of directly inferring the stiffness matrix from the spring constant, we derive it systematically through strain-displacement relationships. The normal strain εx\varepsilon_xε x ​ is found to be constant along the bar and related to nodal displacements by εx=u2−u1L\varepsilon_x = \frac{u_2 – u_1}{L}ε x ​ = L u 2 ​ −u 1 ​

​ . Using Hooke’s Law, the axial stress σx\sigma_xσ x ​ and the internal force PPP are calculated in terms of the nodal displacements.

The equilibrium conditions are established by relating the nodal forces f1f_1f 1 ​ and f2f_2f 2 ​ to the nodal displacements u1u_1u 1 ​ and u2u_2u 2 ​ . This leads to the following nodal force-displacement equations in matrix form:

1 & -1 -1 & 1 \end{bmatrix} \begin{bmatrix} u_1 u_2 \end{bmatrix} = \begin{bmatrix} f_1 f_2 \end{bmatrix} \] Thus, the element stiffness matrix for the bar element is: $[ke] = \frac{AE}{L} \begin{bmatrix} 1 & -1 -1 & 1 \end{bmatrix}$ This stiffness matrix is symmetric and singular, reflecting the 2 degrees of freedom associated with the two nodal displacements. Despite being a one-dimensional element, it can be effectively used in the analysis of higher-dimensional structures like frames and trusses by connecting multiple bar elements appropriately, usually through pin or ball-and-socket joints to ensure axial loading only. The transformation of this local stiffness matrix into the global coordinate system and its application to larger structures is discussed further in subsequent chapters.

MATLAB代写

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