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

[joe465]

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

 

[pongo38]

I suggest you check the quantities and units in the question. Also I think you have some typos in your relevant equations and solutions.

 

[joe465]

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 .