Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 1 (Q.No. 17)
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
Discussion:
8 comments Page 1 of 1.
Walkfrom0to1 said:
8 years ago
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.
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.
(1)
Guru said:
5 years ago
@Tanu @Abinash Dash.
Here, the Assuming Factor of safety is four.
Here, the Assuming Factor of safety is four.
(1)
Praveen kumar said:
1 decade ago
Stress = (pd/2t)*f.o.s.
2000 = (10*100/2*t)*4.
= 1cm.
= 10mm.
2000 = (10*100/2*t)*4.
= 1cm.
= 10mm.
Sneha said:
9 years ago
Can you please explain clearly?
Snehal said:
9 years ago
In cylinder hoop, stress is pd/2t and longitudinal stress is pd/4t.
The maximum stress is (p*fos)d/2t.
The maximum stress is (p*fos)d/2t.
R.K. said:
9 years ago
Please explain the answer.
Abinash dash said:
6 years ago
Why taking 4 here? Please explain.
Tanu said:
6 years ago
Why we taking fos as 4? please explain it.
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