Let's solve each part step by step: --- **5.14 Consider a six-node element in \(\mathbb{R}^2\) shown in Figure (a). Its map in the natural coordinate space \(\xi \eta\) is shown in Figure (b).** --- ### (a) **Construct interpolation functions for the six-node element in the natural coordinate space \(\xi\eta\).** #### **Step 1: Identify Nodes and Their Natural Coordinates** From the diagram (b): - Node 1: (0, 0) - Node 2: (0.5, 0) - Node 3: (1, 0) - Node 4: (1, 1) - Node 5: (0.5, 1) - Node 6: (0, 1) #### **Step 2: Quadratic Lagrange Interpolation Functions** For a 6-node quadrilateral element, the standard shape functions in the natural coordinates \((\xi, \eta)\) are: Let’s denote the nodes as follows: - Corner nodes: 1, 3, 4, 6 - Midside nodes: 2, 5 The quadratic interpolation shape functions are: \[ \begin{align*} N_1(\xi, \eta) &= (1-\xi)(1-\eta)(-1+\xi+\eta)/4 \\ N_2(\xi, \eta) &= (1-\xi^2)(1-\eta)/2 \\ N_3(\xi, \eta) &= (1+\xi)(1-\eta)(-1-\xi+\eta)/4 \\ N_4(\xi, \eta) &= (1+\xi)(1+\eta)(-1-\xi-\eta)/4 \\ N_5(\xi, \eta) &= (1-\xi^2)(1+\eta)/2 \\ N_6(\xi, \eta) &= (1-\xi)(1+\eta)(-1+\xi-\eta)/4 \\ \end{align*} \] Alternatively, using the nodes at: - \((\xi, \eta)\) = (0,0), (0.5,0), (1,0), (1,1), (0.5,1), (0,1) The shape functions can be written using Lagrange polynomials: - For \(\xi\): at 0, 0.5, 1 - For \(\eta\): at 0, 1 So: - \(L_1(\xi) = (1-\xi)(1-2\xi)\) at \(\xi=0\) - \(L_2(\xi) = 4\xi(1-\xi)\) at \(\xi=0.5\) - \(L_3(\xi) = \xi(2\xi-1)\) at \(\xi=1\) - \(M_1(\eta) = 1-\eta\) at \(\eta=0\) - \(M_2(\eta) = \eta\) at \(\eta=1\) Thus, the shape/interpolation functions are: \[ \begin{align*} N_1(\xi,\eta) &= L_1(\xi) M_1(\eta) = (1-\xi)(1-2\xi)(1-\eta) \\ N_2(\xi,\eta) &= L_2(\xi) M_1(\eta) = 4\xi(1-\xi)(1-\eta) \\ N_3(\xi,\eta) &= L_3(\xi) M_1(\eta) = \xi(2\xi-1)(1-\eta) \\ N_4(\xi,\eta) &= L_3(\xi) M_2(\eta) = \xi(2\xi-1)\eta \\ N_5(\xi,\eta) &= L_2(\xi) M_2(\eta) = 4\xi(1-\xi)\eta \\ N_6(\xi,\eta) &= L_1(\xi) M_2(\eta) = (1-\xi)(1-2\xi)\eta \\ \end{align*} \] --- ### (b) **Is \(f(\xi, \eta)\) a complete polynomial in \(\xi\) and \(\eta\)?** #### **Step 1: What is a "Complete Polynomial"?** A complete polynomial of degree \(p\) in 2 variables contains all monomials up to degree \(p\): - Degree 1: \(1, \xi, \eta\) - Degree 2: \(1, \xi, \eta, \xi^2, \xi\eta, \eta^2\) #### **Step 2: What does the 6-node interpolation produce?** Look at the shape functions above: They include terms like \(\xi, \eta, \xi^2, \xi\eta\). However, the interpolation is **quadratic in \(\xi\)** (since there are 3 points in \(\xi\): 0, 0.5, 1), but only **linear in \(\eta\)** (since there are only 2 distinct values: 0, 1). Thus, **the interpolation is not a complete quadratic polynomial in both variables**—it is quadratic in \(\xi\) and linear in \(\eta\). #### **Answer:** > **No, \(f(\xi, \eta)\) interpolated using these functions is not a complete polynomial in \(\xi\) and \(\eta\).** > > It is quadratic in \(\xi\) and linear in \(\eta\), so not all terms up to degree 2 in both variables are present (e.g., there is no \(\eta^2\) term). --- ### (c) **Degrees of interpolation in \(\xi\) and \(\eta\), \(p_\xi\) and \(p_\eta\):** - **In \(\xi\):** There are three nodes (0, 0.5, 1), so the interpolation is quadratic (\(p_\xi = 2\)). - **In \(\eta\):** There are two nodes (0, 1), so the interpolation is linear (\(p_\eta = 1\)). #### **Reasoning:** The degree of interpolation is determined by the number of distinct nodes in each coordinate direction. Three points allow quadratic interpolation, two points allow linear interpolation. --- ## **Summary** ### **(a)** The interpolation (shape) functions are: \[ \begin{align*} N_1(\xi,\eta) &= (1-\xi)(1-2\xi)(1-\eta) \\ N_2(\xi,\eta) &= 4\xi(1-\xi)(1-\eta) \\ N_3(\xi,\eta) &= \xi(2\xi-1)(1-\eta) \\ N_4(\xi,\eta) &= \xi(2\xi-1)\eta \\ N_5(\xi,\eta) &= 4\xi(1-\xi)\eta \\ N_6(\xi,\eta) &= (1-\xi)(1-2\xi)\eta \\ \end{align*} \] ### **(b)** No, \(f(\xi,\eta)\) is not a complete polynomial in \(\xi\) and \(\eta\), because the interpolation is quadratic in \(\xi\) and linear in \(\eta\). ### **(c)** Degrees of interpolation: \(p_\xi = 2\), \(p_\eta = 1\). --- **If you need further clarification or algebraic steps for the shape functions, let me know!**

Let's proceed with a detailed, step-by-step calculation for **Part (a): Constructing the interpolation functions** for the six-node element in the natural coordinate space \((\xi, \eta)\). --- ### Step 1: Identify the Nodes and Their Natural Coordinates From the diagram (Figure b), the six nodes are located at: | Node | \(\xi\) | \(\eta\) | |---------|---------|---------| | 1 | 0 | 0 | | 2 | 0.5 | 0 | | 3 | 1 | 0 | | 4 | 1 | 1 | | 5 | 0.5 | 1 | | 6 | 0 | 1 | --- ### Step 2: Recognize the Type of Element and Shape Functions This is a **six-node (quadratic) quadrilateral element** with nodes at the corners and midpoints of sides. The shape functions are constructed such that: \[ f(\xi, \eta) = \sum_{i=1}^6 N_i(\xi, \eta) f_i \] where \(N_i(\xi, \eta)\) are the shape functions associated with nodes \(i\), satisfying: \[ N_i(\xi_j, \eta_j) = \delta_{ij} \] --- ### Step 3: Define the Interpolation Functions Since the element is quadratic, the shape functions \(N_i(\xi, \eta)\) are quadratic polynomials in \(\xi\) and \(\eta\). They can be written as products of **Lagrange polynomials** in each coordinate: \[ N_i(\xi, \eta) = L_{i,\xi}(\xi) \times L_{i,\eta}(\eta) \] --- ### Step 4: Construct Lagrange Polynomials in \(\xi\) and \(\eta\) For \(\xi\): - Nodes at \(\xi = 0, 0.5, 1\) Lagrange basis functions in \(\xi\): \[ \begin{aligned} l_0(\xi) &= \frac{(\xi - 0.5)(\xi - 1)}{(0 - 0.5)(0 - 1)} = \frac{(\xi - 0.5)(\xi - 1)}{(-0.5)(-1)} = \frac{(\xi - 0.5)(\xi - 1)}{0.5} \\ l_{0.5}(\xi) &= \frac{(\xi - 0)(\xi - 1)}{(0.5 - 0)(0.5 - 1)} = \frac{\xi (\xi - 1)}{(0.5)(-0.5)} = \frac{\xi (\xi - 1)}{-0.25} \\ l_1(\xi) &= \frac{(\xi - 0)(\xi - 0.5)}{(1 - 0)(1 - 0.5)} = \frac{\xi (\xi - 0.5)}{(1)(0.5)} = 2 \xi (\xi - 0.5) \\ \end{aligned} \] Similarly for \(\eta\): - Nodes at \(\eta = 0, 1\) Lagrange basis functions: \[ \begin{aligned} m_0(\eta) &= 1 - \eta \\ m_1(\eta) &= \eta \\ \end{aligned} \] --- ### Step 5: Write the Shape Functions \(N_i(\xi, \eta)\) Using the above basis functions, the shape functions are: | Node | \(\xi_i, \eta_i\) | Shape Function \(N_i(\xi,\eta)\) | |---------|-------------------|----------------------------------| | 1 | (0, 0) | \(N_1 = l_0(\xi) \times m_0(\eta)\) | | 2 | (0.5, 0) | \(N_2 = l_{0.5}(\xi) \times m_0(\eta)\) | | 3 | (1, 0) | \(N_3 = l_1(\xi) \times m_0(\eta)\) | | 4 | (1, 1) | \(N_4 = l_1(\xi) \times m_1(\eta)\) | | 5 | (0.5, 1) | \(N_5 = l_{0.5}(\xi) \times m_1(\eta)\) | | 6 | (0, 1) | \(N_6 = l_0(\xi) \times m_1(\eta)\) | --- ### **Explicit Forms of \(l_0(\xi), l_{0.5}(\xi), l_1(\xi)\):** \[ \boxed{ \begin{aligned} l_0(\xi) &= \frac{(\xi - 0.5)(\xi - 1)}{0.5} = 2 (\xi - 0.5)(\xi - 1) \\ l_{0.5}(\xi) &= \frac{\xi (\xi - 1)}{-0.25} = -4 \xi (\xi - 1) \\ l_1(\xi) &= 2 \xi (\xi - 0.5) \\ \end{aligned} } \] Similarly: \[ \boxed{ \begin{aligned} m_0(\eta) &= 1 - \eta \\ m_1(\eta) &= \eta \\ \end{aligned} } \] --- ### **Final Expressions for the Six Shape Functions:** \[ \boxed{ \begin{aligned} N_1(\xi, \eta) &= l_0(\xi) \times m_0(\eta) = 2 (\xi - 0.5)(\xi - 1) \times (1 - \eta) \\ N_2(\xi, \eta) &= l_{0.5}(\xi) \times m_0(\eta) = -4 \xi (\xi - 1) \times (1 - \eta) \\ N_3(\xi, \eta) &= l_1(\xi) \times m_0(\eta) = 2 \xi (\xi - 0.5) \times (1 - \eta) \\ N_4(\xi, \eta) &= l_1(\xi) \times m_1(\eta) = 2 \xi (\xi - 0.5) \times \eta \\ N_5(\xi, \eta) &= l_{0.5}(\xi) \times m_1(\eta) = -4 \xi (\xi - 1) \times \eta \\ N_6(\xi, \eta) &= l_0(\xi) \times m_1(\eta) = 2 (\xi - 0.5)(\xi - 1) \times \eta \\ \end{aligned} } \] --- ### **Summary of Part (a):** The **interpolation functions** (shape functions) for the six-node quadratic element in natural coordinates \((\xi, \eta)\) are: \[ \boxed{ \begin{aligned} \boxed{ N_1(\xi, \eta) = 2 (\xi - 0.5)(\xi - 1) (1 - \eta) } } \quad \boxed{ N_2(\xi, \eta) = -4 \xi (\xi - 1) (1 - \eta) } \quad \boxed{ N_3(\xi, \eta) = 2 \xi (\xi - 0.5) (1 - \eta) } \] \[ \boxed{ N_4(\xi, \eta) = 2 \xi (\xi - 0.5) \eta } \quad \boxed{ N_5(\xi, \eta) = -4 \xi (\xi - 1) \eta } \quad \boxed{ N_6(\xi, \eta) = 2 (\xi - 0.5)(\xi - 1) \eta } \] --- **This completes the detailed solution for Part (a).** If you'd like, I can help with further parts or clarify the derivation.