Calculating thickness of a cast iron cylinder carrying pressure of 30Nmm^2

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SUMMARY

The discussion centers on calculating the thickness of a cast iron cylinder subjected to a pressure of 30 N/mm² and a unit working stress of 25 N/mm². The formula used is σ₁ = pd/2t, leading to the conclusion that the required thickness (t) is 180 mm. Despite initial confusion regarding the calculations, the final assessment confirms that the thickness should indeed be 180 mm based on the provided parameters. The participants emphasized the importance of checking units and equations to avoid errors in calculations.

PREREQUISITES
  • Understanding of stress and pressure calculations in mechanical engineering
  • Familiarity with the formula σ₁ = pd/2t for thin-walled cylinders
  • Knowledge of unit conversions, particularly in N/mm² and mm
  • Basic principles of material strength, specifically for cast iron
NEXT STEPS
  • Review the derivation of the formula for thin-walled cylinder stress
  • Explore the impact of material properties on cylinder design
  • Learn about safety factors in pressure vessel design
  • Investigate common mistakes in engineering calculations and how to avoid them
USEFUL FOR

Mechanical engineers, engineering students, and professionals involved in pressure vessel design and analysis will benefit from this discussion.

joe465
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1. Homework Statement

A cast-iron cylinder of 300mm bore carries a pressure of 30 Newtons per square
millimetre; what should be its thickness for a unit working stress of 25 Newtons per
square millimetre?


2. Homework Equations

ó1 = pd/2t

t=pd/2tó1


3. The Attempt at a Solution

t=pd/2tó1

t= 30*300/2*25

=180mm

Its got to be wrong as its way too big but I am unsure on where I am going wrong, any help would be much appreciated.

Heres another attempt

ó1/pd=2t

25/(30*3000) = 2t

2t = 0.00277778

t= 0.0014mm (4dp)

Maybes this one, I am not sure.

Thanks, Joe
 
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I suggest you check the quantities and units in the question. Also I think you have some typos in your relevant equations and solutions.
 
Thankyou for your reply.

This is the best i can get.

Assuming that the cylinder is a thin cylinder

o1 = pd/2t

where o1=Stress on the cylinder

This stress is the circumferential stress on the cylinder

p= Pressure

d=Diameter of he cylinder bore

t= thickness

Now we have been given

o1=25 N/mm^2

P=30N/mm^2

d=300mm

SO

We know

o1 = pd/2t

So the design criteria for the thickness becomes

SO t=pd/2o1 = (30x300)/(2x25)= 9000/50 = 180

Its thickness should be 180 mm .
 
Last edited:

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