A curved-sided isoparametric LST uses natural coordinates (r,s), with L_1=1-r-s, L_2=r…

A curved-sided isoparametric LST uses natural coordinates \((r,s)\), with \(L_1=1-r-s\), \(L_2=r\), \(L_3=s\), \(N_i=L_i(2L_i-1)\) for \(i=1,2,3\), \(N_4=4L_1L_2\), \(N_5=4L_2L_3\), and \(N_6=4L_3L_1\). Its nodal coordinates in metres are \((0,0)\), \((2,0)\), \((0,2)\), \((1,-0.20)\), \((1.10,1.10)\), and \((-0.10,1.00)\), where nodes 4, 5, and 6 lie on sides 1-2, 2-3, and 3-1. The nodal displacements \((u_i,v_i)\), in mm, are \((0,0)\), \((1.2,0.2)\), \((-0.3,0.8)\), \((0.4,-0.1)\), \((0.7,0.6)\), and \((-0.2,0.3)\). At \((r,s)=(0.20,0.30)\), calculate the physical coordinates, the Jacobian \(\mathbf J=\begin{bmatrix}x_{,r}&y_{,r}\\x_{,s}&y_{,s}\end{bmatrix}\), its determinant, all six pairs of physical shape-function derivatives \((N_{i,x},N_{i,y})\), the interpolated displacement, and the engineering strain vector. Use \([N_{i,r},N_{i,s}]^T=\mathbf J[N_{i,x},N_{i,y}]^T\).

Use the quadratic shape functions for both coordinates and displacements.