Question 12.3: Determine the pressure required to burst a standard DN200 Sc...
Determine the pressure required to burst a standard DN200 Schedule 40 steel pipe if the ultimate tensile strength of the steel is 276 MPa.
Objective Compute the pressure required to burst the steel pipe.
Given Ultimate tensile strength of steel = s_{u} = 276 MPa
Pipe is a standard DN200 Schedule 40 steel pipe.
The dimensions of the pipe are found in Appendix A–9(b) to be
Outside diameter = 219.1 mm = D_{o}
Inside diameter = 202.7 mm = D_{i}
Wall thickness = 8.18 mm = t
Analysis We should first check to determine if the pipe should be called a thin-walled cylinder by computing the ratio of the mean diameter to the wall thickness.
D_{m} = Mean diameter = \frac{D_{0} + D_{i}}{2} = 210.9 mm
\frac{D_{m}}{t} = \frac{210.9 mm}{8.18 mm} = 25.8
Since this ratio is greater than 20, the thin-wall equations can be used. The hoop stress is the maximum stress and should be used to compute the bursting pressure.
Results Use Equation (12–20).
\sigma = \frac{pD_{m}}{2t} (12-20)
Letting σ = 276 MPa and using the mean diameter gives the bursting pressure to be
p = \frac{2t \sigma}{D_{m}} = \frac{(2)(8.18 mm)(276 MPa)}{210.9 mm} = 12.4 MPa
Comment A design factor of 6 or greater is usually applied to the bursting pressure to get an allowable operating pressure. This cylinder would be limited to approximately 3500 kPa of internal pressure.
