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Hoop Tension Core Concept

Hoop tension is the circumferential tensile force developed in a thin pipe or cylindrical vessel due to internal fluid pressure. It is normally analyzed by cutting the cylinder along a diametral plane and balancing pressure force with tensile resistance.

$$pDL = 2TL$$ $$T = \frac{pD}{2}$$ $$\sigma_h = \frac{T}{t} = \frac{pD}{2t}$$

Here $p$ is internal pressure, $D$ is pipe diameter, $L$ is length considered, $t$ is wall thickness, $T$ is hoop tension per unit length, and $\sigma_h$ is hoop stress.

Thin Cylinder Design Relations

For thin-walled vessels, the wall thickness is small compared with the diameter. Board-style questions often ask for required thickness or maximum allowable pressure.

$$t = \frac{pD}{2\sigma_{allow}}$$ $$p = \frac{2t\sigma_{allow}}{D}$$ $$D = \frac{2t\sigma_{allow}}{p}$$
Use consistent units. If $p$ is in kPa, convert it to $\text{kN/m}^2$ when using meters, or to $\text{N/mm}^2$ when using millimeters.

Riveted or Jointed Pipes

If a pipe has a longitudinal joint, the effective tensile strength is reduced by joint efficiency. Use the effective resisting stress in the hoop stress equation.

$$\sigma_{eff} = \eta\sigma_{allow}$$ $$t = \frac{pD}{2\eta\sigma_{allow}}$$

For a safe design, the developed hoop stress must not exceed the allowable stress of the shell or joint.

$$\sigma_h \leq \sigma_{allow}$$ $$\frac{pD}{2t} \leq \sigma_{allow}$$

Sample: Required Wall Thickness

A thin pipe has diameter $D = 1.2 \text{ m}$ and carries water pressure $p = 600 \text{ kPa}$. If allowable tensile stress is $120 \text{ MPa}$, compute the required wall thickness.

$$p = 600 \text{ kPa} = 0.6 \text{ MPa}$$ $$D = 1.2 \text{ m} = 1200 \text{ mm}$$ $$t = \frac{pD}{2\sigma_{allow}}$$ $$t = \frac{0.6(1200)}{2(120)} = 3.0 \text{ mm}$$

Answer: The required wall thickness is 3.0 mm, before adding corrosion allowance or code requirements.

Problem: Hoop Stress in a Water Main

A steel water main has an inside diameter of 600 mm and wall thickness of 8 mm. It operates at a working pressure of 800 kPa. Determine: (a) the hoop tension per unit length of pipe (kN/m), and (b) the hoop stress in the steel (MPa). Is this within an allowable stress of 120 MPa?

$$T = \frac{pD}{2} = \frac{800(0.600)}{2} = 240 \text{ kN/m}$$ $$\sigma_h = \frac{T}{t} = \frac{pD}{2t} = \frac{800(600)}{2(8)} = 30{,}000 \text{ kPa} = 30 \text{ MPa}$$

Answer: Hoop tension is 240 kN/m. Hoop stress is 30 MPa, well within the 120 MPa allowable. The pipe has considerable reserve capacity and could theoretically sustain up to 3200 kPa before reaching the allowable.

Problem: Riveted Pipe with Joint Efficiency

A riveted steel pipe has an inside diameter of 900 mm and carries water under 1.2 MPa internal pressure. The allowable tensile stress of the steel is 110 MPa and the longitudinal rivet joint has an efficiency of 80%. Calculate the required minimum wall thickness of the pipe.

$$t = \frac{pD}{2\eta\sigma_{allow}}$$ $$p = 1.2 \text{ MPa}, \quad D = 900 \text{ mm}, \quad \eta = 0.80, \quad \sigma_{allow} = 110 \text{ MPa}$$ $$t = \frac{1.2(900)}{2(0.80)(110)} = \frac{1080}{176} = 6.14 \text{ mm}$$

Answer: The required wall thickness is 6.14 mm (round up to 7 mm with standard plate sizes). The joint efficiency reduces the effective strength, requiring a thicker wall than a seamless pipe would need.

Problem: Burst Pressure of a Thin-Walled Cylinder

A thin-walled pressure vessel has an inside diameter of 500 mm and wall thickness of 5 mm. The ultimate tensile strength of the material is 400 MPa and the factor of safety is 4. Determine the maximum allowable working pressure in kPa and verify whether a working pressure of 1.5 MPa is safe.

$$\sigma_{allow} = \frac{\sigma_{ult}}{FS} = \frac{400}{4} = 100 \text{ MPa}$$ $$p_{allow} = \frac{2t\sigma_{allow}}{D} = \frac{2(5)(100)}{500} = 2.0 \text{ MPa} = 2000 \text{ kPa}$$

Since $p_{work} = 1.5 \text{ MPa} < p_{allow} = 2.0 \text{ MPa}$, the design is safe.

$$FS_{actual} = \frac{\sigma_{ult}}{\sigma_{work}} = \frac{400}{\frac{pD}{2t}} = \frac{400}{\frac{1.5(500)}{2(5)}} = \frac{400}{75} = 5.33$$

Answer: Maximum allowable working pressure is 2000 kPa. The 1.5 MPa operating pressure is safe (actual FS = 5.33 > required 4).

Problem: Longitudinal vs. Hoop Stress in a Closed Pressure Vessel

A closed cylindrical pressure vessel has inside diameter 400 mm, wall thickness 6 mm, and internal pressure 2.5 MPa. Compute both the hoop (circumferential) stress and the longitudinal (axial) stress. Explain which controls the design and why the vessel would fail first along a longitudinal seam.

$$\sigma_h = \frac{pD}{2t} = \frac{2.5(400)}{2(6)} = 83.33 \text{ MPa (hoop stress)}$$ $$\sigma_L = \frac{pD}{4t} = \frac{2.5(400)}{4(6)} = 41.67 \text{ MPa (longitudinal stress)}$$ $$\sigma_h = 2\sigma_L$$

Answer: Hoop stress is 83.33 MPa; longitudinal stress is 41.67 MPa. Hoop stress is exactly twice the longitudinal stress. The vessel would burst along a longitudinal line (splitting the cylinder lengthwise) because hoop stress governs — this is why pressure pipes are tested by bursting along the length, not the end caps.

Problem: Pipeline at Depth — Hoop Stress from Hydrostatic Head

A buried water main has an internal diameter of 750 mm and a wall thickness of 10 mm. The pipe is located 25 m below a water tower reservoir (this head drives the pressure). Neglecting velocity head and elevation corrections, determine the hoop stress in the pipe wall in MPa. What is the factor of safety if the steel yield stress is 250 MPa?

$$p = \gamma h = 9.81(25) = 245.25 \text{ kPa} = 0.2453 \text{ MPa}$$ $$\sigma_h = \frac{pD}{2t} = \frac{0.2453(750)}{2(10)} = 9.20 \text{ MPa}$$ $$FS = \frac{\sigma_y}{\sigma_h} = \frac{250}{9.20} = 27.2$$

Answer: The hoop stress is only 9.20 MPa — very low. The factor of safety is 27.2, which is excessively high. In practice, pipes at this depth are sized not by static head but by water hammer transients and external soil loads.