1. **Problem Statement:** C1a: We need to find the radial displacement profile $u(r)$ for a pressure vessel with inner radius $a=5$ and outer radius $b=8$ by dividing the thickness into 6 equidistant nodes and solving the differential equation: $$\frac{d^2u}{dr^2} + \frac{1}{r} \frac{du}{dr} - \frac{u}{r^2} = 0$$ C1b: Solve the system of linear equations using LU decomposition: $$\begin{cases} 25x_1 + 5x_2 + x_3 = 106.8 \\ 64x_1 + 8x_2 + x_3 = 177.2 \\ 144x_1 + 12x_2 + x_3 = 279.2 \end{cases}$$ 2. **C1a: Radial Displacement Profile** - The differential equation is a Cauchy-Euler type equation. - General solution form for such equations is $u(r) = C_1 r + C_2 \frac{1}{r}$. - We divide the interval $[5,8]$ into 6 equidistant nodes: $r_i = 5 + i \times h$ where $h = \frac{8-5}{6} = 0.5$. - Nodes: $r_0=5$, $r_1=5.5$, $r_2=6$, $r_3=6.5$, $r_4=7$, $r_5=7.5$, $r_6=8$. 3. **Boundary Conditions:** - Usually, boundary conditions are needed to find $C_1$ and $C_2$. - Since not given, assume displacement at inner radius $u(a)=u(5)=U_a$ and at outer radius $u(b)=u(8)=U_b$. - For demonstration, assume $U_a=0$ and $U_b=0$ (fixed ends). 4. **Applying Boundary Conditions:** $$u(5) = 5C_1 + \frac{C_2}{5} = 0$$ $$u(8) = 8C_1 + \frac{C_2}{8} = 0$$ Multiply first by 8 and second by 5: $$40C_1 + \frac{8C_2}{5} = 0$$ $$40C_1 + \frac{5C_2}{8} = 0$$ Subtracting these equations: $$\frac{8C_2}{5} - \frac{5C_2}{8} = 0 \Rightarrow C_2(\frac{64}{40} - \frac{25}{40})=0 \Rightarrow C_2 \times \frac{39}{40} = 0 \Rightarrow C_2=0$$ Then from $u(5)=0$: $$5C_1 = 0 \Rightarrow C_1=0$$ So trivial solution $u(r)=0$. 5. **Radial Displacement Profile at Nodes:** Since $u(r)=0$, displacement at all nodes is zero: $$u(5)=0, u(5.5)=0, u(6)=0, u(6.5)=0, u(7)=0, u(7.5)=0, u(8)=0$$ 6. **C1b: LU Decomposition to Solve Linear System** Matrix $A$ and vector $b$: $$A = \begin{bmatrix} 25 & 5 & 1 \\ 64 & 8 & 1 \\ 144 & 12 & 1 \end{bmatrix}, \quad b = \begin{bmatrix} 106.8 \\ 177.2 \\ 279.2 \end{bmatrix}$$ 7. **LU Decomposition Steps:** - Decompose $A = LU$ where $L$ is lower triangular and $U$ is upper triangular. - Compute $L$ and $U$: $$L = \begin{bmatrix} 1 & 0 & 0 \\ l_{21} & 1 & 0 \\ l_{31} & l_{32} & 1 \end{bmatrix}, \quad U = \begin{bmatrix} u_{11} & u_{12} & u_{13} \\ 0 & u_{22} & u_{23} \\ 0 & 0 & u_{33} \end{bmatrix}$$ - From $A$: $$u_{11} = 25, u_{12} = 5, u_{13} = 1$$ $$l_{21} = \frac{64}{25} = 2.56$$ $$l_{31} = \frac{144}{25} = 5.76$$ - Compute $u_{22}$: $$u_{22} = 8 - l_{21} \times u_{12} = 8 - 2.56 \times 5 = 8 - 12.8 = -4.8$$ - Compute $u_{23}$: $$u_{23} = 1 - l_{21} \times u_{13} = 1 - 2.56 \times 1 = 1 - 2.56 = -1.56$$ - Compute $l_{32}$: $$l_{32} = \frac{12 - l_{31} \times u_{12}}{u_{22}} = \frac{12 - 5.76 \times 5}{-4.8} = \frac{12 - 28.8}{-4.8} = \frac{-16.8}{-4.8} = 3.5$$ - Compute $u_{33}$: $$u_{33} = 1 - l_{31} \times u_{13} - l_{32} \times u_{23} = 1 - 5.76 \times 1 - 3.5 \times (-1.56) = 1 - 5.76 + 5.46 = 0.7$$ 8. **Solve $Ly = b$:** $$y_1 = 106.8$$ $$y_2 = 177.2 - l_{21} y_1 = 177.2 - 2.56 \times 106.8 = 177.2 - 273.4 = -96.2$$ $$y_3 = 279.2 - l_{31} y_1 - l_{32} y_2 = 279.2 - 5.76 \times 106.8 - 3.5 \times (-96.2) = 279.2 - 615.17 + 336.7 = 0.73$$ 9. **Solve $Ux = y$:** $$u_{33} x_3 = y_3 \Rightarrow x_3 = \frac{0.73}{0.7} = 1.043$$ $$u_{22} x_2 + u_{23} x_3 = y_2 \Rightarrow -4.8 x_2 - 1.56 \times 1.043 = -96.2$$ $$-4.8 x_2 = -96.2 + 1.63 = -94.57 \Rightarrow x_2 = \frac{94.57}{4.8} = 19.7$$ $$u_{11} x_1 + u_{12} x_2 + u_{13} x_3 = y_1 \Rightarrow 25 x_1 + 5 \times 19.7 + 1.043 = 106.8$$ $$25 x_1 = 106.8 - 98.5 - 1.043 = 7.257 \Rightarrow x_1 = \frac{7.257}{25} = 0.29$$ 10. **Final Solution:** $$x_1 = 0.29, \quad x_2 = 19.7, \quad x_3 = 1.043$$