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please help me solve Q1(b) Fig.2 (b) which is a three-noded triangular element, related to the subject of 'Finite Element Analysis'?

To solve part (b) for the three-noded triangular elements shown in Fig. 2(b) in the context of finite element analysis, we need to compute the approximate values of the temperature at the points E \((\frac{1}{2}, \frac{1}{2})\), F \((\frac{3}{4}, \frac{1}{4})\), and G \((\frac{3}{4}, \frac{3}{4})\).

### Step-by-Step Solution:

1. **Define the Nodes and their Coordinates:**
   For the triangular elements in Fig. 2(b), we have the following nodes and coordinates:
   - \(A (0, 0)\)
   - \(B (1, 0)\)
   - \(C (1, 1)\)
   - \(D (0, 1)\)

   The domain is divided into two triangular elements:
   - Triangle 1: \(A (0,0)\), \(B (1,0)\), and \(D (0,1)\)
   - Triangle 2: \(B (1,0)\), \(C (1,1)\), and \(D (0,1)\)

2. **Shape Functions for Triangular Elements:**
   The shape functions \(N_i\) for a triangular element with nodes at \((x_i, y_i)\), \((x_j, y_j)\), and \((x_k, y_k)\) are given by:
   \[
   N_i = \frac{(a_i + b_i x + c_i y)}{2A}
   \]
   where \(a_i\), \(b_i\), and \(c_i\) are constants determined by the coordinates of the nodes, and \(A\) is the area of the triangle.

3. **Calculate Shape Functions for the Points:**

   - For Triangle 1 (\(A, B, D\)):
     \[
     \begin{aligned}
     A_1 &= \frac{1}{2} \left| x_A(y_B - y_D) + x_B(y_D - y_A) + x_D(y_A - y_B) \right| \\
        &= \frac{1}{2} \left| 0(0 - 1) + 1(1 - 0) + 0(0 - 0) \right| = \frac{1}{2}
     \end{aligned}
     \]

     Shape functions:
     \[
     \begin{aligned}
     N_A &= \frac{1 - x - y}{2A_1} = 1 - x - y \\
     N_B &= \frac{x}{2A_1} = x \\
     N_D &= \frac{y}{2A_1} = y
     \end{aligned}
     \]

   - For Triangle 2 (\(B, C, D\)):
     \[
     \begin{aligned}
     A_2 &= \frac{1}{2} \left| x_B(y_C - y_D) + x_C(y_D - y_B) + x_D(y_B - y_C) \right| \\
        &= \frac{1}{2} \left| 1(1 - 1) + 1(1 - 0) + 0(0 - 1) \right| = \frac{1}{2}
     \end{aligned}
     \]

     Shape functions:
     \[
     \begin{aligned}
     N_B &= \frac{(1 - x)}{2A_2} = 1 - x \\
     N_C &= \frac{(x - y)}{2A_2} = x - y \\
     N_D &= \frac{y}{2A_2} = y
     \end{aligned}
     \]

4. **Evaluate the Shape Functions at the Given Points:**

   - **Point \(E\left(\frac{1}{2}, \frac{1}{2}\right)\)**:
     \(E\) lies on the boundary between both triangles. We use both shape functions to get temperature \(T_E\):
     \[
     T_E = N_A T_A + N_B T_B + N_D T_D \quad \text{(for Triangle 1)}
     \]
     \[
     T_E = N_B T_B + N_C T_C + N_D T_D \quad \text{(for Triangle 2)}
     \]

     Using \(x = \frac{1}{2}\) and \(y = \frac{1}{2}\):
     \[
     T_E = (1 - \frac{1}{2} - \frac{1}{2})T_A + (\frac{1}{2})T_B + (\frac{1}{2})T_D = 0
     \]

     Since \(E\) lies on both elements, we average the results.

   - **Point \(F\left(\frac{3}{4}, \frac{1}{4}\right)\)**:
     \(F\) lies in Triangle 1 (\(A, B, D\)):
     \[
     T_F = (1 - \frac{3}{4} - \frac{1}{4})T_A + (\frac{3}{4})T_B + (\frac{1}{4})T_D
     \]
     \[
     T_F = 0T_A + \frac{3}{4}T_B + \frac{1}{4}T_D = \frac{3}{4}T_B + \frac{1}{4}T_D
     \]

   - **Point \(G\left(\frac{3}{4}, \frac{3}{4}\right)\)**:
     \(G\) lies in Triangle 2 (\(B, C, D\)):
     \[
     T_G = (1 - \frac{3}{4})T_B + (\frac{3}{4} - \frac{3}{4})T_C + (\frac{3}{4})T_D
     \]
     \[
     T_G = \frac{1}{4}T_B + 0T_C + \frac{3}{4}T_D = \frac{1}{4}T_B + \frac{3}{4}T_D
     \]

### Summary:
- \(T_E = 0\)
- \(T_F = \frac{3}{4}T_B + \frac{1}{4}T_D\)
- \(T_G = \frac{1}{4}T_B + \frac{3}{4}T_D\)

These are the approximate temperature values at the given points using the finite element method with two three-noded triangular elements. 

Do you have any questions or need more details on any part of the solution?

Here are some questions you might consider next:
1. How are shape functions derived for different elements in finite element analysis?
2. What are the common types of elements used in finite element analysis?
3. How does the choice of element type affect the accuracy of the finite element solution?
4. What are the methods to evaluate the accuracy of a finite element solution?
5. How can boundary conditions be incorporated into finite element analysis?

**Tip:** In finite element analysis, always check the compatibility and completeness of the shape functions to ensure accurate interpolation within the elements.

Explore the solution for Q1(b) Fig.2 (b) involving three-noded triangular elements in the context of finite element analysis. Learn how to compute temperatures at specific points using shape functions and triangular element division.

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