Question

A thick cylindrical pipe of outside diameter 300mm and internal diameter 200mm is subjected to an internal fluid pressure of 14 N/mm2. Determine the maximum hoop stress developed in the cross section. What is the percentage error if the maximum hoop stress is calculated by the equations for thin cylinder?

Answer 1

Answer:36.4 MPa

Explanation:

External diameter($D_0$)=300mm

Internal diameter($D_i$)=200mm

Internal Pressure($P_i$)=14N/$mm^2$

Now Hoop stress for Thick cylinders is given by

$\sigma _h$=$\frac{P_ir_i^2}{r_0^2-r_i^2}\left (\frac{r_0^2}{r^2}+1\right )$

Maximum hoop stress will be develop at $r=r_i$

$\sigma _h=36.4 MPa$

Now if we consider it as thin cylinder then Hoop stress is given by

$\sigma _h$=$\frac{P_{i}D}{2t}$

Where thickness =50 mm

$\sigma _h$=$\frac{14\times 300}{2\times 50}$

$\sigma _h=42MPa$

% error=$\frac{42-36.4}{36.4}$$\times 100$=15.38%