# 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

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.

 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.

 Guru said: (Apr 30, 2020) @Tanu @Abinash Dash. Here, the Assuming Factor of safety is four.