Let's solve the **skirt thickness** for the vertical vessel step by step with explanations and full calculations. **Given Data:** - **Vessel Diameter, \( D \) = 1.6 m = 1600 mm** - **Vessel Height = 14 m** - **Skirt Height, \( h_{skirt} \) = 2.3 m = 2300 mm** - **Weight (including attachments), \( W \) = 80,000 kg** - **Diameter of skirt = Diameter of vessel = 1.6 m** - **Wind Pressure, \( P_{wind} \) = 100 kg/m²** - **Seismic Coefficient, \( C_{seismic} \) = 0.08** - **Permissible tensile stress of skirt, \( \sigma_{allow} = 960 \) kg/cm²** - **Yield stress = 2400 kg/cm²** - **K = 0.7 (usually "K" is the joint efficiency or a factor, here not used further as not relevant for skirt thickness if not welded joint)** --- ## **Step 1: Calculate Loads on Skirt** ### **1.1 Dead Load** - \( W_{dead} = 80,000 \) kg (This is the total weight supported by the skirt.) ### **1.2 Wind Load** #### **Projected area for wind:** - **Height exposed:** \( H = 14 \) m - **Diameter = 1.6 m** - **Projected area, \( A_{wind} = D \times H = 1.6 \times 14 = 22.4 \) m²** #### **Total wind force:** - \( F_{wind} = P_{wind} \times A_{wind} = 100 \ \text{kg/m}^2 \times 22.4 \ \text{m}^2 = 2240 \) kg #### **Bending Moment due to Wind (at base of skirt):** - **Wind acts at centroid of projected area, which is at height \( H/2 \) from base of vessel.** - \( M_{wind} = F_{wind} \times (H/2) = 2240 \ \text{kg} \times 7 \ \text{m} = 15,680 \ \text{kg}\cdot\text{m} \) --- ### **1.3 Seismic Load** #### **Seismic force:** - \( F_{seismic} = C_{seismic} \times W_{dead} = 0.08 \times 80,000 = 6,400 \ \text{kg} \) #### **Seismic moment (assuming mass acts at 2/3 vessel height):** - **Centroid at \( 2/3 \) of height:** \( h_{cg} = 14 \times (2/3) = 9.33 \) m - \( M_{seismic} = F_{seismic} \times h_{cg} = 6,400 \ \text{kg} \times 9.33 \ \text{m} = 59,712 \ \text{kg}\cdot\text{m} \) --- ### **1.4 Total Overturning Moment at base (Skirt Bottom)** - \( M_{total} = M_{wind} + M_{seismic} = 15,680 + 59,712 = 75,392 \ \text{kg}\cdot\text{m} \) --- ## **Step 2: Calculate Maximum Stress at Skirt Base** ### **2.1** The skirt is a vertical cylinder subjected to: - Axial compressive load (dead load) - Bending moment (wind + seismic) The maximum compressive stress at the skirt base: \[ \sigma_{max} = \frac{W_{dead}}{A} + \frac{M_{total} \times r}{I} \] Where: - \( A = \) cross-sectional area of skirt = \( \pi \times D \times t \) (since \( t \ll D \)) - \( I = \) moment of inertia = \( \frac{\pi D^3 t}{8} \) (thin-walled cylinder approximation) - \( r = D/2 \) ### **2.2** Let’s denote: - \( t \) = skirt thickness (unknown, in cm) --- ## **Step 3: Express All Quantities in Consistent Units** Convert all to **kg, cm, s** (since stress is in kg/cm²): - \( D = 160 \) cm - \( r = 80 \) cm - \( h_{skirt} = 230 \) cm - \( W_{dead} = 80,000 \) kg - \( M_{total} = 75,392 \ \text{kg}\cdot\text{m} = 7,539,200 \ \text{kg}\cdot\text{cm} \) --- ## **Step 4: Area and Moment of Inertia** - \( A = \pi D t \) - \( I = \frac{\pi D^3 t}{8} \) --- ## **Step 5: Write the Stress Equation** \[ \sigma_{max} = \frac{W_{dead}}{\pi D t} + \frac{M_{total} \times (D/2)}{\frac{\pi D^3 t}{8}} \] Let’s simplify the bending term: \[ \frac{M_{total} \times (D/2)}{\frac{\pi D^3 t}{8}} = \frac{M_{total} \times D/2 \times 8}{\pi D^3 t} = \frac{4 M_{total}}{\pi D^2 t} \] So, \[ \sigma_{max} = \frac{W_{dead}}{\pi D t} + \frac{4 M_{total}}{\pi D^2 t} \] --- ## **Step 6: Substitute Values** - \( W_{dead} = 80,000 \) kg - \( D = 160 \) cm - \( M_{total} = 7,539,200 \) kg·cm \[ \sigma_{max} = \frac{80,000}{\pi \times 160 \times t} + \frac{4 \times 7,539,200}{\pi \times (160)^2 \times t} \] Calculate denominators: - \( \pi \times 160 = 502.654 \) - \( 160^2 = 25,600 \) - \( \pi \times 25,600 = 80,424.8 \) So, - First term: \( 80,000 / (502.654 \times t) = 159.17 / t \) - Second term: \( (4 \times 7,539,200) / (80,424.8 \times t) = 30,156,800 / (80,424.8 \times t) = 374.98 / t \) \[ \sigma_{max} = \frac{159.17}{t} + \frac{374.98}{t} = \frac{534.15}{t} \] --- ## **Step 7: Equate to Allowable Stress and Solve for t** Allowable stress, \( \sigma_{allow} = 960 \) kg/cm² \[ \frac{534.15}{t} \leq 960 \] \[ t \geq \frac{534.15}{960} = 0.556 \ \text{cm} \] --- ## **Step 8: Check Against Minimum Fabrication Thickness** - For fabrication, a minimum thickness is often required (say, 6 mm = 0.6 cm or 8 mm = 0.8 cm). - Our calculated value is 0.56 cm, **which is less than typical minimums for skirt fabrication**. --- ## **Step 9: Recheck All Calculations** ### 9.1 **Dead load area calculation:** \[ A = \pi D t = 3.1416 \times 160 \times t = 502.654 \times t \] \[ 80,000 / (502.654 \times t) = 159.17 / t \] ### 9.2 **Bending stress calculation:** \[ 4 \times M_{total} = 4 \times 7,539,200 = 30,156,800 \] \[ \pi D^2 = 3.1416 \times (160)^2 = 3.1416 \times 25,600 = 80,424.8 \] \[ 30,156,800 / (80,424.8 \times t) = 374.98 / t \] ### 9.3 **Sum:** \[ (159.17 + 374.98) / t = 534.15 / t \] \[ t = 534.15 / 960 = 0.556 \ \text{cm} \] --- ## **Step 10: Final Answer and Recommendation** **Final Answer:** > The minimum calculated thickness of the skirt support is **0.56 cm (5.6 mm)** based on combined dead load, wind, and seismic loads. However, for practical fabrication and corrosion allowance, skirt thickness is typically not less than **6–8 mm**; therefore, you should specify **at least 8 mm** (0.8 cm) for manufacturing. > > **So, provide a skirt thickness of 0.8 cm (8 mm) minimum.** This ensures both structural safety and manufacturability. --- **Summary:** **The calculated minimum skirt thickness is 0.56 cm. For practical and fabrication reasons, specify the skirt thickness as 0.8 cm (8 mm) minimum.** This will safely handle the vessel's loads and meet construction standards.