# NCEES Practice Exam Problem



## EnvPERose (Sep 15, 2018)

The Pressure gauge in an air cylinder reads 1.680 kPA.   The cylinder is constructed of a 12-mm rolled steel plate with an internal diameter of 700 mm.  The tangential stress (MPa) inside the tank is most nearly:

A.  25

B. 50

C. 77

D. 100

The answer provided in the solutions is B.  I think it is wrong, should the answer be .05 MPa?

Thank you!


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## EnvPERose (Sep 15, 2018)

The Solution is as follows:

Refer to Cylindrical Pressure Vessel in the Mechanics of Materials section of the FE Reference Manual (top left of page 81). 

The cylinder can be considered thin-walled if t &lt; do/2.  In this case, t = 12 mm and ro = do/2 = 362 mm.   Thus,

Tangential stress = Pi * r / t 

where r = (ro + ri) /2     

do = 700mm (di)  + 12 mm + 12 mm = 724 mm   =&gt; ro =  724 mm/ 2  = 362 mm

di = 700 mm = &gt;  ri =  700 mm/ 2 = 350 mm

r =( 362 mm  + 350 mm ) / 2  =  712 mm / 2  = 356 mm

Tangential stress = Pi * r / t  =  1.680 kPa * 356 mm / 12 mm = 50 kPa  

Therefore,  answer is .050 MPa.  (not 50 MPa?)

Am I missing something here?


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## cvanwy02 (Jan 25, 2019)

I think you're correct.  The solution could have been a little simpler by using (PD)/(2T) to obtain 49kPa.

49kPA equals .049 MPa.


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## EnvPERose (May 26, 2019)

Sorry, slight typo on my part:

The problem provided pressure as 1,680 kPA (not 1.680kPa)  This would be 1.680MPa.   So the answer is correct, 50 MPa.


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## Fawaz (Dec 26, 2022)

cvanwy02 said:


> I think you're correct. The solution could have been a little simpler by using (PD)/(2T) to obtain 49kPa.
> 
> 49kPA equals .049 MPa.


I think, Hoop stress = PD/2t
P=1.68MPa
D= 0.35
t=0.012
Then PD/2t = 1.68 MPa *0.35 m / (2 * 0.012m) = 24.5 MPa nearly to 25 MPa


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