Steel Tank Calculation

Steel Tank Calculation: Precision Engineering for Structural Integrity

Steel tank calculation is the process of determining the structural requirements for safe, efficient storage. It involves complex formulas and standards to ensure durability and compliance.

This article covers detailed formulas, variable explanations, common values, and real-world examples for expert-level steel tank design and analysis.

Calculadora con inteligencia artificial (IA) para Steel Tank Calculation

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  • Calculate required shell thickness for a 10-meter diameter steel tank storing water at atmospheric pressure.
  • Determine the maximum allowable working pressure for a cylindrical steel tank with 5 mm shell thickness.
  • Compute the wind load effect on a 15-meter tall steel storage tank located in a high-wind zone.
  • Estimate the settlement stress on the tank foundation for a 20,000-liter steel tank filled with oil.

Common Parameters and Values in Steel Tank Calculation

ParameterSymbolTypical UnitsCommon Values / RangeDescription
Tank DiameterDm (meters)2 – 50Diameter of the cylindrical tank shell
Shell Thicknesstmm (millimeters)3 – 25Thickness of the steel shell plate
Tank HeightHm (meters)3 – 30Height of the cylindrical shell
Design PressurePkPa (kilopascals)0 – 500Internal pressure the tank is designed to withstand
Allowable StressĻƒaMPa (megapascals)140 – 250Maximum permissible stress for the steel material
Material Yield StrengthĻƒyMPa250 – 450Yield strength of the steel used
Corrosion Allowancecmm1 – 3Extra thickness added to compensate for corrosion
Weld EfficiencyEDimensionless (0 – 1)0.7 – 1.0Factor representing weld quality
Density of Stored FluidĻkg/mĀ³800 – 1200Density of the liquid stored inside the tank
Wind PressurePwkPa0.5 – 2.0Pressure exerted by wind on the tank surface
Seismic Load FactorSDimensionless0 – 0.4Factor accounting for seismic forces

Fundamental Formulas for Steel Tank Calculation

1. Shell Thickness for Internal Pressure

The shell thickness required to withstand internal pressure is calculated by the formula:

t = (P Ɨ D) / (2 Ɨ Ļƒa Ɨ E – 1.2 Ɨ P) + c
  • t: Required shell thickness (mm)
  • P: Design internal pressure (kPa)
  • D: Tank diameter (mm)
  • Ļƒa: Allowable stress of steel (MPa)
  • E: Weld efficiency (dimensionless)
  • c: Corrosion allowance (mm)

Explanation: This formula is derived from the thin shell theory for cylindrical pressure vessels, accounting for weld efficiency and corrosion allowance. The factor 1.2 Ɨ P in the denominator accounts for additional safety margin.

2. Hoop Stress in the Shell

The hoop stress generated in the tank shell due to internal pressure is:

Ļƒh = (P Ɨ D) / (2 Ɨ t Ɨ E)
  • Ļƒh: Hoop stress (MPa)
  • P: Internal pressure (kPa)
  • D: Diameter (mm)
  • t: Shell thickness (mm)
  • E: Weld efficiency

This stress must be less than the allowable stress to ensure safety.

3. Longitudinal Stress

The longitudinal stress in the shell is given by:

Ļƒl = (P Ɨ D) / (4 Ɨ t Ɨ E)
  • Ļƒl: Longitudinal stress (MPa)
  • Other variables as defined above

Longitudinal stress is typically half the hoop stress in cylindrical tanks.

4. Base Plate Thickness

For flat base plates supporting the tank, thickness is calculated by:

tbase = āˆš( (P Ɨ DĀ²) / (4 Ɨ Ļƒa) )
  • tbase: Base plate thickness (mm)
  • P: Pressure or load (kPa)
  • D: Diameter (mm)
  • Ļƒa: Allowable stress (MPa)

5. Wind Load Calculation

Wind pressure on the tank surface is calculated by:

Pw = 0.613 Ɨ VĀ² Ɨ Cd
  • Pw: Wind pressure (kPa)
  • V: Wind velocity (m/s)
  • Cd: Drag coefficient (typically 0.6 – 1.2)

This pressure is used to calculate stresses induced by wind loads on the tank shell and foundation.

6. Settlement Stress on Foundation

Settlement stress due to tank weight and fluid load is:

Ļƒsettlement = (Wtank + Wfluid) / A
  • Ļƒsettlement: Stress on foundation (kPa)
  • Wtank: Weight of empty tank (N)
  • Wfluid: Weight of stored fluid (N)
  • A: Area of foundation (mĀ²)

Detailed Explanation of Variables and Typical Values

  • Design Pressure (P): Usually atmospheric or slightly above for storage tanks; typical values range from 0 to 500 kPa depending on contents.
  • Diameter (D): Common diameters vary widely; 2 to 50 meters is typical for industrial tanks.
  • Shell Thickness (t): Determined by pressure, material strength, and corrosion allowance; usually between 3 mm and 25 mm.
  • Allowable Stress (Ļƒa): Depends on steel grade and temperature; common values are 140 MPa for mild steel to 250 MPa for higher grades.
  • Weld Efficiency (E): Reflects quality of welds; full penetration welds approach 1.0, while partial welds may be 0.7.
  • Corrosion Allowance (c): Added thickness to compensate for corrosion over tank life; typically 1-3 mm.
  • Wind Velocity (V): Site-specific; design codes specify values based on geographic location.
  • Drag Coefficient (Cd): Depends on tank shape and surface roughness; usually between 0.6 and 1.2.

Real-World Application Examples

Example 1: Calculating Shell Thickness for a Water Storage Tank

A cylindrical steel tank with a diameter of 10 meters stores water at atmospheric pressure. The design pressure is 10 kPa (due to slight overpressure). The steel allowable stress is 160 MPa, weld efficiency is 0.85, and corrosion allowance is 2 mm. Calculate the minimum shell thickness.

Given:

  • D = 10,000 mm
  • P = 10 kPa
  • Ļƒa = 160 MPa
  • E = 0.85
  • c = 2 mm

Calculation:

t = (P Ɨ D) / (2 Ɨ Ļƒa Ɨ E – 1.2 Ɨ P) + c

Convert units for consistency: P in MPa = 10 kPa / 1000 = 0.01 MPa

t = (0.01 Ɨ 10,000) / (2 Ɨ 160 Ɨ 0.85 – 1.2 Ɨ 0.01) + 2

Calculate denominator:

2 Ɨ 160 Ɨ 0.85 = 272 MPa
1.2 Ɨ 0.01 = 0.012 MPa
Denominator = 272 – 0.012 = 271.988 MPa

Calculate numerator:

0.01 Ɨ 10,000 = 100

Calculate thickness:

t = 100 / 271.988 + 2 ā‰ˆ 0.368 + 2 = 2.368 mm

Minimum shell thickness is approximately 2.37 mm. Considering manufacturing standards, a 3 mm plate would be selected.

Example 2: Wind Load Effect on a Tall Steel Tank

A 15-meter tall cylindrical steel tank with a diameter of 8 meters is located in a region with a design wind speed of 40 m/s. The drag coefficient is 0.8. Calculate the wind pressure and the total force exerted on the tank.

Given:

  • V = 40 m/s
  • Cd = 0.8
  • D = 8 m
  • H = 15 m

Step 1: Calculate wind pressure

Pw = 0.613 Ɨ VĀ² Ɨ Cd

Calculate VĀ²:

40Ā² = 1600

Calculate wind pressure:

Pw = 0.613 Ɨ 1600 Ɨ 0.8 = 785.6 Pa = 0.7856 kPa

Step 2: Calculate projected area

Projected area A = Height Ɨ Diameter = 15 m Ɨ 8 m = 120 mĀ²

Step 3: Calculate total wind force

F = Pw Ɨ A = 0.7856 kPa Ɨ 120 mĀ² = 94.27 kN

The tank must be designed to resist a lateral wind force of approximately 94.3 kN.

Additional Considerations in Steel Tank Calculation

  • Seismic Loads: In seismic zones, additional lateral forces must be considered using seismic load factors and dynamic analysis.
  • Thermal Expansion: Temperature variations cause expansion and contraction, requiring flexible joints or expansion loops.
  • Fatigue Analysis: Repeated loading cycles, especially in pressure tanks, necessitate fatigue life assessment.
  • Foundation Design: Settlement and bearing capacity must be evaluated to prevent structural failure.
  • Corrosion Protection: Coatings, cathodic protection, and corrosion allowances extend tank life.
  • Code Compliance: Calculations must adhere to standards such as API 650, ASME Section VIII, and EN 14015.

References and Further Reading