Civil Engineering - Steel Structure Design - Discussion
Discussion Forum : Steel Structure Design - Section 3 (Q.No. 20)
20.
The safe working pressure for a spherical vessel 1.5 m diameter and having 1.5 cm thick wall not to exceed tensile stress 50kg/cm2 , is
Discussion:
14 comments Page 1 of 2.
Muhammd usman said:
3 years ago
Pd/4t = T,
P=?
D=1.5m = 150cm.
t = 1.5cm.
t = 50.
By solving.
50 * 4 * 1.5/150 = 2kg/cm2.
The date is not correct,
If t = 500 or d=.15m,
Then answer will be equal to 20.
P=?
D=1.5m = 150cm.
t = 1.5cm.
t = 50.
By solving.
50 * 4 * 1.5/150 = 2kg/cm2.
The date is not correct,
If t = 500 or d=.15m,
Then answer will be equal to 20.
(2)
Pradeep Singh said:
5 years ago
Longitudinal stress = pd/4t , p=50,d=1.5m, t=1.5cm.
50*1.5/4*1.5=25 for safe partial safety factor 1.25 then 25/1.25=20.
50*1.5/4*1.5=25 for safe partial safety factor 1.25 then 25/1.25=20.
(1)
Chandu said:
7 years ago
Tensile stress will develop along the longitudinal Axis so from the formula of longitudinal stress pd/4t.
Pradeep Singh said:
5 years ago
Hoop stress = pd/2t.
50*1.5/2*1.5 = 25 partial safety factor 1.25.
25/1.25 = 20.
50*1.5/2*1.5 = 25 partial safety factor 1.25.
25/1.25 = 20.
Engr Syed Junaid Ali said:
6 years ago
I get just 2kg/cm2 : (kindly someone specifies some complete solution. Please).
Malathi said:
6 years ago
By using formula of PD/4t. I can not get a answer 20kg/cm2.
Anujit said:
9 years ago
I think PD/4T is the formula of the solution. So I get 22.
Chhaya said:
8 years ago
@ Anujit.
By using the formula PD/4t. We Get the answer.
By using the formula PD/4t. We Get the answer.
Azzu said:
7 years ago
How can you use that formula?
Anyone explanation.
Anyone explanation.
Eneka said:
7 years ago
20 kg/cm2 is the correct answer. I agree.
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