Civil Engineering - UPSC Civil Service Exam Questions - Discussion

17. 

A cylindrical shell made of mild steel plate of 100 cm diameter is to be subjected to an internal pressure of 10 kg/cm2. If the material yields at 2000 kg/cm2, assuming factor of safety as four and using maximum principal stress theory, thickness of the plate will be

[A]. 5 mm
[B]. 10 mm
[C]. 15 mm
[D]. 20 mm

Answer: Option B

Explanation:

No answer description available for this question.

Praveen Kumar said: (Jan 28, 2015)  
Stress = (pd/2t)*f.o.s.

2000 = (10*100/2*t)*4.
= 1cm.
= 10mm.

Sneha said: (Oct 21, 2016)  
Can you please explain clearly?

Snehal said: (Dec 26, 2016)  
In cylinder hoop, stress is pd/2t and longitudinal stress is pd/4t.

The maximum stress is (p*fos)d/2t.

R.K. said: (Jan 4, 2017)  
Please explain the answer.

Walkfrom0To1 said: (Apr 2, 2017)  
According to max principal stress theory.
Yield stress = applied stress * f.o.s.

In above problem, applied stress is a hoop stress type and it is equal to (pd/2t).
So,
2000=pd/2t * f.o.s.
2000=(10 * 100/2 * t) * 4.
t = 1cm = 10mm.

Abinash Dash said: (Jul 23, 2019)  
Why taking 4 here? Please explain.

Tanu said: (Jul 30, 2019)  
Why we taking fos as 4? please explain it.

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