Let's solve the problem step by step. --- ## **Given:** - **Internal radius, \( r_i \) = 100 mm** - **External radius, \( r_o \) = 125 mm** - **External pressure, \( p_o \) = 40 bar = 4,000 kN/m² (since 1 bar = 100 kN/m²)** - **Initial internal pressure, \( p_i = \) (since not mentioned initially)** - **Additional internal pressure applied, \( p_i = 200 bar = 20,000 kN/m² \)** - **Asked:** What will be the maximum stress set up in the cylinder in both cases? --- # **Step 1: Stresses in Thick Cylinder (Lame's Equations)** The radial and hoop (circumferential) stresses at radius \( r \) in a thick-walled cylinder: \[ \sigma_r = A - \frac{B}{r^2} \] \[ \sigma_\theta = A + \frac{B}{r^2} \] Where \( A \) and \( B \) are constants to be found using boundary conditions. --- ## **Case 1: Only External Pressure** ### **Boundary Conditions** - At \( r = r_i \), \( \sigma_r = p_i = \) - At \( r = r_o \), \( \sigma_r = -p_o \) (negative because it's compressive) ### **Apply boundary conditions:** \[ \begin{align*} \text{At } r_i = 100 \text{ mm}: & \quad = A - \frac{B}{(100)^2} \\ \text{At } r_o = 125 \text{ mm}: & \quad -400 = A - \frac{B}{(125)^2} \\ \end{align*} \] Let’s solve these equations: 1. \( = A - \frac{B}{100^2} \implies A = \frac{B}{100^2} \) 2. Substitute \( A \) into the second equation: \[ -400 = \frac{B}{100^2} - \frac{B}{125^2} \] \[ -400 = B \left( \frac{1}{100^2} - \frac{1}{125^2} \right ) \] Calculate values: - \( 100^2 = 10,000 \) - \( 125^2 = 15,625 \) \[ \frac{1}{10,000} - \frac{1}{15,625} = \frac{15,625 - 10,000}{10,000 \times 15,625} = \frac{5,625}{156,250,000} = .000036 \] \[ -400 = B \times .000036 \implies B = \frac{-400}{.000036} = -111,111,111 \] Now, \( A = \frac{B}{10,000} = \frac{-111,111,111}{10,000} = -11,111.11 \) --- ### **Maximum Stress Location** Maximum hoop (circumferential) stress occurs at the inner wall (\( r = r_i \)). \[ \sigma_\theta (r_i) = A + \frac{B}{r_i^2} \] \[ = -11,111.11 + \frac{-111,111,111}{10,000} = -11,111.11 - 11,111.11 = -22,222.22 \text{ kN/m}^2 \] **But this is negative, which means it's compressive.** --- ### **Summary for Case 1** - **Maximum (hoop) stress at inner wall:** \[ \boxed{-22,222.22 \text{ kN/m}^2 \text{ (compressive)}} \] --- ## **Case 2: Add Internal Pressure (200 bar)** Now, \( p_i = 20,000 \) kN/m² (tensile), \( p_o = 4,000 \) kN/m² (compressive). ### **Boundary Conditions** - At \( r = r_i \), \( \sigma_r = -p_i = -20,000 \) kN/m² (internal pressure acts outward, so negative sign for compressive radial stress) - At \( r = r_o \), \( \sigma_r = -p_o = -4,000 \) kN/m² ### **Apply boundary conditions:** 1. \( -20,000 = A - \frac{B}{(100)^2} \) 2. \( -400 = A - \frac{B}{(125)^2} \) Subtract equations: \[ (-20,000) - (-400) = \left( A - \frac{B}{10,000} \right ) - \left( A - \frac{B}{15,625} \right ) \] \[ -16,000 = -\frac{B}{10,000} + \frac{B}{15,625} \] \[ -16,000 = B \left( -\frac{1}{10,000} + \frac{1}{15,625} \right ) \] \[ -16,000 = B \left( \frac{15,625 - 10,000}{10,000 \times 15,625} \right ) = B \left( \frac{5,625}{156,250,000} \right ) = B \times .000036 \] \[ B = \frac{-16,000}{.000036} = -444,444,444 \] Now, substitute \( B \) in 1st equation: \[ -20,000 = A - \frac{-444,444,444}{10,000} \] \[ -20,000 = A + 44,444.44 \] \[ A = -20,000 - 44,444.44 = -64,444.44 \] --- ### **Maximum Hoop Stress** \[ \sigma_\theta (r_i) = A + \frac{B}{r_i^2} = -64,444.44 + \frac{-444,444,444}{10,000} = -64,444.44 - 44,444.44 = -108,888.88 \text{ kN/m}^2 \] **Again, it's compressive.** --- ## **Final Answers** ### **Case 1 (Only External Pressure):** \[ \boxed{-22,222 \text{ kN/m}^2 \text{ (compressive, hoop stress at inner wall)}} \] ### **Case 2 (Internal + External Pressure):** \[ \boxed{-108,889 \text{ kN/m}^2 \text{ (compressive, hoop stress at inner wall)}} \] --- ### **Note:** - All stresses are compressive due to the high external and internal pressures. - If you want the hoop stress at the outer wall or at another radius, substitute the respective \( r \) value in the Lame’s equation. --- Let me know if you need the stresses at other locations or a diagram!