# Chapter10-Radial pressure-cylindrical and spherical shells¶

## Ex1-pg384¶

In [1]:
import math
#calculate The necessary thickness of metal for seamless pipe
d = 2.;## feet
p = 250.;## lb/in^2
f = 12000.;## lb/in^2
t_limit = p*d*12/(2*f) ;## inches
print'%s %.2f %s'%('The necessary thickness of metal for seamless pipe is',t_limit,'inches');

The necessary thickness of metal for seamless pipe is 0.25 inches


## Ex2-pg384¶

In [2]:
import math
#calculate Maximum intensity of shear stress induced
l = 8.;##feet
d = 3.;## feet
t = 1/2.;## inches
p = 200.;## lb/in**2
E = 30.*10**6.;## lb/in**2
PR = 0.3;## poisson's ratio
f1 = p*d*12./(2.*t);## lb/in**2
f2 = p*d*12./(4.*t);## lb/in**2
f_s = 0.5*(f1-f2);## lb/in**2
e1 = (f1/E)-(PR*f2/E);## lb/in**2
e2 = (f2/E)-(PR*f1/E);## lb/in**2
del_d = e1*d*12.;## inches
del_l = e2*l*12.;## inches
del_V = (e2+2.*e1)*0.25*math.pi*(12*d)**2. * l*12.;## cub. inches
print'%s %.d %s'%('Maximum intensity of shear stress induced =',f_s,'lb/in**2')

print'%s %.6f %s'%('del_d =',del_d,'inches')
print'%s %.6f %s'%('del_l = ',del_l,'inches')
print'%s %.1f %s'%('del_V =',del_V,'cub. inches')

Maximum intensity of shear stress induced = 1800 lb/in**2
del_d = 0.007344 inches
del_l =  0.004608 inches
del_V = 44.6 cub. inches


## Ex3-pg385¶

In [3]:
import math
#calculate Thickness of metal required
d = 30.;## inches
H = 300.;## feet
w = 62.5;
f = 2800.;
##intensity of water pressur
p = w*H/144.;## lb/in**2
t_limit = p*d/(2*f);## inches
print'%s %.4f %s'%('Thickness of metal required is',t_limit, 'inches')

##the answer is correct only, but it is approximated in the text book.

Thickness of metal required is 0.6975 inches


## Ex4-pg387¶

In [4]:
import math
#calcualte The permissible steam pressure and circumferential stress and longitudinal sretss
d = 78.;## inches
t = 3/4.;## inches
n1 = 70/100.;## efficiency of the longitudinal riveted joint
f = 6.;## tons/in^2
n2 = 60/100.;## efficiency of the circumferential riveted joint
p = f*2240./(d/(2.*t*n1));##lb/in^2
p = round(p-1);
f1 = p*d/(2.*t);## lb/in^2
f2 = p*d/(4.*t*n2);## lb/in^2
print'%s %.d %s'%('The permissible steam pressure, p = ',p,'lb/in^2');
print'%s %.d %s'%('The circumferential stress, f1 =',f1,'lb/in^2')
print'%s %.2f %s'%('The circumferential stress, f1= ',f1/2240,'tons/in^2');
print'%s %.d %s'%('The longitudinal stress, f2 =',f2,'lb/in^2')
print'%s %.2f %s'%('The longitudinal sretss, f2 =',f2/2240,'lb/in^2')

The permissible steam pressure, p =  180 lb/in^2
The circumferential stress, f1 = 9360 lb/in^2
The circumferential stress, f1=  4.18 tons/in^2
The longitudinal stress, f2 = 7800 lb/in^2
The longitudinal stress, f2 = 3.48 lb/in^2


## Ex5-pg389¶

In [5]:
import math
#calculate The thickness of the plate required
d = 4.;## feet
p = 200.;## lb/in^2
f = 15000.;## lb/in^2
n = 0.7;## efficiency
t_limit = p*d*12/(4*f*n);## inches
print'%s %.2f %s'%('The thickness of the plate required =',t_limit,'inches');

##the answer is correct only, but it is approximated in the text book.

The thickness of the plate required = 0.23 inches


## Ex6-pg390¶

In [1]:
import math
#calculate The pressure exerted by fluid on the shell
d = 3.;## feet
t = 1/4.;## inches
del_V = 9.;## cub. inches
E = 30*10**6;## lb/in^2
PR = 0.3;## poisson's ratio
V = (math.pi/6.)*(12.*d)**3;## in^3
k = del_V/V;
f = k*E/(3.*(1.-PR));## lb/in^2
p = 4.*f*t/(12.*d);## lb/in^2
print'%s %.d %s'%('The pressure exerted by fluid on the shell, p =',p,'lb/in^2');

##there is a minute calculation error in the answer given in text book

The pressure exerted by fluid on the shell, p = 146 lb/in^2


## Ex7-pg390¶

In [7]:
import math
#calcualte the Pipe and Steel-wire of intiually due to p and finally due to p
d = 12.;## feet
t = 1/2.;## inches
d1 = 1/4.;## inches
p = 500.;## lb/in^2
E_c = 6000.;## tons/in^2
PR = 0.3;## Poisson's ratio
E_s = 13000.;## tons/in^2
f_t = 8000.;## lb/in^2
l = 8.;
P_c = l*(math.pi/64.)*f_t;## lb-wt
f_c = P_c/(2.*t);## lb/in^2
##bursting force per inch unit length
f_b = p*d;## lb-wt
f_p = (f_b + (l*0.049*PR*p*d/(4.*t))*(E_s/E_c))/(1 + (l*0.049)*E_s/E_c);## lb/in^2
f_w = (f_p - PR*p*d/(4.*t))*E_s/E_c;## lb/in^2

print('                  Pipe                         Steel-wire');
print'%s %.d %s'%(' Initually,',f_c,'lb/in^2.,compr.')
print'%s %.d %s'%('Intiually',f_t,'lb/in^2., tensile')
print'%s %.d %s'%( 'Due to p,  ',f_p,'lb/in^2.,tensile.')
print'%s %.d %s'%('Due to p',f_t,'lb/in^2., tensile')
print'%s %.d %s'%(' Finally, ',f_p-f_c,'lb/in^2.,tensile.')
print'%s %.d %s'%('Finally',f_w+f_t,'lb/in^2., tensile')

##there is a calculation error in the answer given in text book

                  Pipe                         Steel-wire
Initually, 3141 lb/in^2.,compr.
Intiually 8000 lb/in^2., tensile
Due to p,   3657 lb/in^2.,tensile.
Due to p 8000 lb/in^2., tensile
Finally,  516 lb/in^2.,tensile.
Finally 13975 lb/in^2., tensile


## Ex8-pg395¶

In [2]:
import math
#calculate 'The maximum  intensities of circumferential stresses are
d = 12.;## inches
t = 3.;## inches
p_x1 = 900.;## lb/in^2
x1 = 0.5*d;## inches
p_x2 = 0.;
x2 = 0.5*d+t;## inches
##from Lame's formulae
b = (p_x1-p_x2)/((1/x1**2)-(1/x2**2));
a = (b/x1**2)- p_x1;
f_x1 = (b/x1**2)+a;## lb/in^2
f_x2 = (b/x2**2)+a;## lb/in^2
print'%s %.d %s'%('The maximum  intensities of circumferential stresses are: f_6 =',f_x1,'lb/in^2.,tensile')
print'%s %.d %s'%('The minimum intensities of circumferential stresses are: f_6f_9 =',f_x2,'lb/in^2., tensile')

The maximum  intensities of circumferential stresses are: f_6 = 2340 lb/in^2.,tensile
The minimum intensities of circumferential stresses are: f_6f_9 = 1440 lb/in^2., tensile


## Ex9-pg396¶

In [9]:
import math
# calculate 'The maximum  intensities of circumferential stresses are and the outer diameter
d = 5.;## inches
p = 3.;## tons/in^2
f = 8.;## tons/in^2
x = 0.5*d;## inches
b = (p+f)/(2/x**2);
a = f-(b/x**2);
t = r-0.5*d;##thickness
D = 2.*t+d;##outer diameter
print'%s %.3f %s'%('The thickness of metal necessary, t =',t,'inches');
print'%s %.1f %s'%('the outer diameter will be, D =',D,'inches');

##the answer is correct, but it is approximated in the text book.

The thickness of metal necessary, t = 1.208 inches
the outer diameter will be, D = 7.4 inches


## Ex10-pg399¶

In [10]:
import math
# calculate The outer radius in all four cases
d = 9.;## inches
p = 5000/2240.;## lb/in**2
f = 8.;## tons/in**2
PR = 0.3;## Poisson's ratio
##(i) Maximum principal stress hypothesis:
k_limit1 = math.sqrt((f + p)/(f - p));##k_limit = r1/r2
r_limit1 = k_limit1*0.5*d;##inches
print'%s %.3f %s'%('The outer radius in case(i), r2 =',r_limit1,'inches');
##(ii) Maximum principal strain:
k_limit2 = math.sqrt(((f/p - PR)+1)/(f/p - PR -1));
r_limit2 = k_limit2*0.5*d;## inches
print'%s %.3f %s'%(' The outer radius in case(ii), r2 =',r_limit2,'inches');
##(iii) Maximum shear stress:
k_limit3 = math.sqrt(f/(2*p) /((f/(2*p)) - 1));
r_limit3 = k_limit3*0.5*d;## inches
print'%s %.3f %s'%('The outer radius in case(iii), r2 =',r_limit3,' inches');
##(iv) Maximum strain energy
K1 = (f**2 /p**2)/(2*((f**2 /(2*p**2)) - (1+PR)));
K2 = K1**2;
K3 = ((f**2 /(2*p**2)) - (1-PR))/((f**2 /(2*p**2)) - (1+PR));
k_limit4 =math. sqrt(K1+math.sqrt(K2-K3));
r_limit4 = k_limit4*0.5*d;## inches
print'%s %.3f %s'%(' The outer radius in case(iv), r2 =',r_limit4,' inches');

##there are calculation errors in the answer given in text book

The outer radius in case(i), r2 = 5.994 inches
The outer radius in case(ii), r2 = 6.163 inches
The outer radius in case(iii), r2 = 6.769  inches
The outer radius in case(iv), r2 = 6.249  inches


## Ex11-pg404¶

In [3]:
import math
# calculate hoop stress at x = 3,6 inches,x = 5 inches initially on inner tube and outer tube
r1 = 0.5*6.;## inches
r2 = 0.5*12.;## inches
r3 = 0.5*10.;## inches
p = 1500.;## lb/in**2
p_f = 12000.;## lb/in**2
##Initially, for the inner tube
b = -p/((1./r1**2.) - (1./r3**2.));
a = b/r1**2.;
f_3 = (b/r1**2.) +a;## lb/in**2
f_5 = (b/r3**2.) +a;## lb/in**2
##for the outer tube
b1 = p/((1/r3**2.)-(1/r2**2.));
a1 = b1/r2**2.;
f1_5 = (b1/r3**2.)+a1;## lb/in**2
f1_6 = (b1/r2**2.)+a1;## lb/in**2
##When the fluid pressure of 12000 lb/in**2, is admitted into the compound tube
B = p_f/((1./r1**2.)-(1/r2**2.));
A = B/(r2**2.);
f_3_ = (B/r1**2.)+A;## lb/in**2
f_5_ = (B/r3**2.)+A;## lb/in**2
f_6_ = (B/r2**2.)+A;## lb/in**2

print('The hoop stresse are');
print'%s %.1f %s %.1f %s '%(' at x = 3 inches,x = 5 inches initially on inner tube are ',-f_3,' lb/in^2.., compressive,  ',-f_5,' lb/in^2..,compressive respectively');
print'%s %.1f %s %.1f %s'%(' at x = 5 inches,x = 6 inches initially on outer tube are ',f1_5,' lb/in^2.., tensile,  ',f1_6,' lb/in^2..,tensile respectively');
print'%s %.d %s %.d %s %.d %s'%(' at x = 3 inches,x = 5 inches and x = 6 inches due to fluid pressure are ',f_3_,' lb/in^2.., tensile,  ',f_5_,' lb/in^2..,tensile, and',f_6_,' lb/in^2..,tensile respectively')
print'%s %.1f %s %.1f %s'%(' at x = 3 inches,x = 5 inches finally on inner tube are ',f_3_+f_3,' lb/in^2.., tensile,  ',f_5_+f_5,' lb/in^2..,tensile respectively');
print'%s %.d %s %.d %s'%(' at x = 5 inches,x = 6 inches finally on outer tube are',f1_5+f_5_,' lb/in^2.., tensile,  ',f1_6+f_6_,' lb/in^2.., tensile respectively');

The hoop stresse are
at x = 3 inches,x = 5 inches initially on inner tube are  4687.5  lb/in^2.., compressive,   3187.5  lb/in^2..,compressive respectively
at x = 5 inches,x = 6 inches initially on outer tube are  8318.2  lb/in^2.., tensile,   6818.2  lb/in^2..,tensile respectively
at x = 3 inches,x = 5 inches and x = 6 inches due to fluid pressure are  20000  lb/in^2.., tensile,   9760  lb/in^2..,tensile, and 8000  lb/in^2..,tensile respectively
at x = 3 inches,x = 5 inches finally on inner tube are  15312.5  lb/in^2.., tensile,   6572.5  lb/in^2..,tensile respectively
at x = 5 inches,x = 6 inches finally on outer tube are 18078  lb/in^2.., tensile,   14818  lb/in^2.., tensile respectively


## Ex12-pg407¶

In [4]:
import math
#calculate The minimum temperature to which outer tube should be heated before it can be slipped
p = 1500.;## lb/in^2
E = 30*10**6;## lb/in^2
f1_5 = 8318.;## lb/in^2
f2_5 = 3187.5;## lb/in^2
alpha = 0.0000062;## per F
r3 = 6;## inches
del_r3 = r3*(f1_5+f2_5)/E;## inches
t = ((f1_5+f2_5)/E)/(alpha);## inches
print'%s %.2f %s'%('The minimum temperature to which outer tube should be heated before it can be slipped on, t =',t,'F');

The minimum temperature to which outer tube should be heated before it can be slipped on, t = 61.86 F


## Ex13-pg408¶

In [5]:
import math
#calculate The hoop stresses are as under for inner tube at x=1/5 and 3 and 4.5 inches
r1 = 0.5*9.;## inches
r2 = 0.5*3.;## inches
r3 = 0.5*6.;## inches
del_r3 = 0.5*0.003;## inches
E = 13000.;## tons/in^2
k1 = r1/r3;
k2 = r2/r3;
a1 = (del_r3/r3)*E/((k1**2 +1)- (k2**2 +1)*(k1**2 -1)/(k2**2 -1));
a = a1*(k1**2 -1)/(k2**2 -1);
b1 = a1*r1**2;
b = a*r2**2;
p_ = (b/r3**2) -a;## tons/in^2
## for the inner tube
f_x1 = (b/r2**2) +a;## tons/in^2
f_x2 = (b/r3**2) +a;## tons/in^2
## for the outer tube
f_x3 = (b1/r3**2) +a1;## tons/in^2
f_x4 = (b1/r1**2) +a1;## tons/in^2
print ('The hoop stresses are as under:');
print'%s %.2f %s'%('For the inner tube, at x = 1/5 inches, f =',-f_x1,'tons/in^2., compressive')
print'%s %.2f %s'%('at x = 3 inches, f =',-f_x2,'tons/in^2.,compressive')
print'%s %.2f %s'%(' For the outer tube, at x = 3 inches, f =',f_x3,'tons/in^2., tensile')

print'%s %.2f %s'%('at x = 4.5 inches, f =',f_x4,' tons/in^2.,tensile');

The hoop stresses are as under:
For the inner tube, at x = 1/5 inches, f = 4.06 tons/in^2., compressive
at x = 3 inches, f = 2.54 tons/in^2.,compressive
For the outer tube, at x = 3 inches, f = 3.96 tons/in^2., tensile
at x = 4.5 inches, f = 2.44  tons/in^2.,tensile


## Ex14-pg414¶

In [6]:
import math
#calculate the thickness of the shell required
r1 = 0.5*5.;## inches
p = 5000.;## lb/in^2
f = 5.;## tons/in^2
b = (f + p/2240.)/((1/r1**3) + (2/r1**3));
a = f - (b/r1**3);
##external diameter
r = (2*b/a)**(1/3);## inches
t = r - r1;## inches
print'%s %.3f %s'%('The thickness of the shell required, t =',t,'inches');

##the answer is approximated in the text book

The thickness of the shell required, t = -1.500 inches