blob: 769bbd3c54d551644c9a9a8c71127714b976a038 (
plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
|
clear
//
//
//Initilization of Variables
d=250 //mm //Diameter iron pipe
t=10 //mm //Thickness
d2=6 //mm //Diameter of steel
p=80 //N/mm**2 //stress
P=3 //N/mm**2 //Pressure
E_c=1*10**5 //N/mm**2
mu=0.3 //Poissons ratio
E_s=2*10**5 //N/mm**2
n=1 //No.of wires
//Calculations
L=6 //mm //Length of cyclinder
//Force Exerted by steel wire at diameterical section
F=p*2*%pi*d2**2*1*4**-1 //N
//Initial stress in cyclinder
f_c=F*(2*t*d2)**-1 //N/mm**2
//LEt due to fluid pressure alone stresses developed in steel wire be F_w and in cyclinder f1 and f2
f2=P*d*(4*t)**-1 //N/mm**2
//Considering the equilibrium of half the cyclinder, 6mm long we get
//F_w*2*%pi*4**-1*d2**2*n+f1*2*t*d2=P*d*d2
//After further simplifying we get
//F_w+2.122*f1=79.58 . ......................................(1)
//Equating strain in wire to circumferential strain in cyclinder
//F_w=(f1-mu*f2)*E_s*E_c**-1 //N/mm**2
//After further simplifying we get
//F_w=2*f1-11.25 ....................................(2)
//Sub in equation in1 we get
f1=(79.58+11.25)*(4.122)**-1 //N/mm**2
F_w=2*f1-11.25 //N/mm**2
//Final stresses
//1) In steel Wir
sigma=F_w+p //N/mm**2
//2) In Cyclinde
sigma2=f1-f_c
//Result
printf("\n Final Stresses developed in:cyclinder is %0.2f N/mm**2",sigma)
printf("\n :Steel is %0.2f N/mm**2",sigma2)
|
