Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 1 (Q.No. 21)
21.
A cold drawn seamless steel tubing subject to internal pressure, has a diameter of 6 cm and wall thickness of 0.2 cm. The ultimate strength of steel is 3600 kg/cm2. The bursting pressure (in kg/cm2) is
Discussion:
7 comments Page 1 of 1.
Dofi said:
5 years ago
240 kg/cm^2.
Rahul khan said:
8 years ago
Then it will be thick cylindrical shell and problem of thick cylindrical shell can be solved by Lame's theory.
Deepa said:
8 years ago
@Amar, if the thickness is greater than 1/20*dia, which formula has to be used?
Amar said:
8 years ago
Since the thickness(0.2 cm) is less than 1/20of Dia ie; 1/20*0.6.Hence this problem comes under thin cylinders which can be solved by the formula.
Design of thin cylinders Pd/2t=permissible tensile strength.
Put the values and p=bursting pressure=240kg/cm2.
Design of thin cylinders Pd/2t=permissible tensile strength.
Put the values and p=bursting pressure=240kg/cm2.
(1)
Ashish panwar said:
8 years ago
According to Barlow's formula
Bursting pressure=2 * ultimate tensile strength * wall thickness/outside dia.
P = 2 * s * t/d.
P = 2 * 3600 * 0.2/30,
P = 240.
Bursting pressure=2 * ultimate tensile strength * wall thickness/outside dia.
P = 2 * s * t/d.
P = 2 * 3600 * 0.2/30,
P = 240.
RAHUL said:
9 years ago
Please explain the answer in detail.
(1)
Arunkumar said:
1 decade ago
P = 2*s*t / d.
= 2*3600*0.2/6 = 240.
= 2*3600*0.2/6 = 240.
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