# Chapter 8: Applications of Plane Stress Pressure Vessels Beams and Combined Loadings¶

## Example 8.1, page no. 546¶

In [2]:
import math

#initialisation
d = 18                                  # inner idameter of the hemisphere in inch
t = 1.0/4.0                                 # thickness of the hemisphere in inch

#calculation
# Part (a)
sa = 14000                                # Allowable tensile stress in Psi
Pa = (2*t*sa)/(d/2.0)                     # Maximum permissible air pressure in Psi
print "Maximum permissible air pressure in the tank (Part(a)) is", round(Pa,1), "psi"

# Part (b)
sb = 6000                                  # Allowable shear stress in Psi
Pb = (4*t*sb)/(d/2.0)                      # Maximum permissible air pressure in Psi
print "Maximum permissible air pressure in the tank (Part(b)) is", round(Pb,1), "psi"

# Part (c)
e = 0.0003                                # Allowable Strain in Outer sufrface of the hemisphere
E = 29e06                                 # Modulus of epasticity of the steel in Psi
v = 0.28                                  # Poissions's ratio of the steel
Pc = (2*t*E*e)/((d/2.0)*(1-v))            # Maximum permissible air pressure in Psi
print "Maximum permissible air pressure in the tank (Part(c)) is", round(Pc,1), "psi"

# Part (d)
Tf = 8100                                 # failure tensile load in lb/in
n = 2.5                                   # Required factor of safetty against failure of the weld
Ta = Tf / n                               # Allowable load in ld/in
sd = (Ta*(1))/(t*(1))                     # Allowable tensile stress in Psi
Pd = (2*t*sd)/(d/2.0)                       # Maximum permissible air pressure in Psi
print "Maximum permissible air pressure in the tank (Part(d)) is", round(Pd,1), "psi"

# Part (e)
Pallow = Pb
print "Maximum permissible air pressure in the tank (Part(e)) is", round(Pb,1) ,"psi"

Maximum permissible air pressure in the tank (Part(a)) is 777.8 psi
Maximum permissible air pressure in the tank (Part(b)) is 666.7 psi
Maximum permissible air pressure in the tank (Part(c)) is 671.3 psi
Maximum permissible air pressure in the tank (Part(d)) is 720.0 psi
Maximum permissible air pressure in the tank (Part(e)) is 666.7 psi


## Example 8.2, page no. 552¶

In [13]:
import math

#initialisation
a = 55                              # Angle made by helix with longitudinal axis in degree
r = 1.8                             # Inner radius of vessel in m
t = 0.02                            # thickness of vessel in m
E = 200e09                          # Modulus of ealsticity of steel in Pa
v = 0.3                             # Poission's ratio of steel
P = 800e03                          # Pressure inside the tank in Pa

#calculation
# Part (a)
s1 = (P*r)/t                        # Circumferential stress in Pa
s2 = (P*r)/(2*t)                    # Longitudinal stress in Pa

print "Circumferential stress is ", s1, "Pa"
print "Longitudinal stress is ", s2, "Pa"

# Part (b)
t_max_z = (s1-s2)/2.0               # Maximum inplane shear stress in Pa
t_max = s1/2.0                      # Maximum out of plane shear stress in Pa

print "Maximum inplane shear stress is ", t_max_z, "Pa"
print "Maximum inplane shear stress is ", t_max, "Pa"

# Part (c)
e1 = (s1/(2*E))*(2-v)               # Strain in circumferential direction
e2 = (s2/E)*(1-(2*v))               # Strain in longitudinal direction

print "Strain in circumferential direction is %e"%(e1)
print "Strain in longitudinal direction is ", e2

# Part (d)
# x1 is the direction along the helix
theta = 90 - a
sx1 = ((P*r)/(4*t))*(3-math.cos(math.radians(2*theta))) # Stress along x1 direction
tx1y1 = ((P*r)/(4*t))*(math.sin(math.radians(2*theta))) # Shear stress in x1y1 plane
sy1 = s1+s2-sx1  # Stress along y1 direction

print "Stress along y1 direction is ", sy1

# Mohr Circle Method
savg = (s1+s2)/2.0                    # Average stress in Pa
R = (s1 - s2 )/2.0                    # Radius of Mohr's Circle in Pa
sx1_ = savg - R*math.cos(math.radians(2*theta))             # Stress along x1 direction
tx1y1_ = R*math.sin(math.radians(2*theta))                  # Shear stress in x1y1 plane
print "Stress along x1 direction is ", sx1_, "Pa"

Circumferential stress is  72000000.0 Pa
Longitudinal stress is  36000000.0 Pa
Maximum inplane shear stress is  18000000.0 Pa
Maximum inplane shear stress is  36000000.0 Pa
Strain in circumferential direction is 3.060000e-04
Strain in longitudinal direction is  7.2e-05
Stress along y1 direction is  60156362.5799
Stress along x1 direction is  47843637.4201 Pa


## Example 8.3, page no. 562¶

In [1]:
%matplotlib inline
from matplotlib import *
from pylab import *
import numpy

#initialisation
L = 6.0                               # Span of the beam in ft
P = 10800                             # Pressure acting in lb
c = 2.0                               # in ft
b = 2.0                               # Width of cross section of the beam in inch
h = 6.0                               # Height of the cross section of the beam in inch
x = 9.0                               # in inch

#calculation
Ra = P/3.0                                          # Reaction at point at A
V = Ra                                              # Shear force at section mn
M = Ra*x                                            # Bending moment at the section mn
I = (b*h**3)/12.0                                   # Moment of inertia in in4
y = linspace(-3, 3, 61)
sx = -(M/I)*y                                       # Normal stress on  crossection  mn
Q = (b*(h/2-y))*(y+((((h/2.0)-y)/2.0)))               # First moment of recmath.tangular cross section
txy = (V*Q)/(I*b)                                   # Shear stress acting on x face of the stress element
s1 = (sx/2.0)+numpy.sqrt((sx/2.0)**2+(txy)**2)       # Principal Tesile stress on the cross section
s2 = (sx/2.0)-numpy.sqrt((sx/2.0)**2+(txy)**2)       # Principal Compressive stress on the cross section
tmax = numpy.sqrt((sx/2)**2+(txy)**2)                # Maximum shear stress on the cross section
plot(sx,y,'o',color='c')
plot(txy,y,'+',color='m')
plot(s1,y,'--',color='y')
plot(s2,y,'<',color='k')
plot(tmax,y,label="Maximum shear stress on cross section")
legend()
show()
#print "Principal Tesile stress on the cross section", s1, "psi"
#print "Principal Compressive stress on the cross section", s2, "psi"

# Conclusions
s1_max = 14400.0                            # Maximum tensile stress in Psi
txy_max = 900.0                             # Maximum shear stress in Psi
t_max = 14400.0/2.0                         # Largest shear stress at 45 degree plane

Populating the interactive namespace from numpy and matplotlib


## Example 8.4, page no. 570¶

In [5]:
import math

#initialisation
d = 0.05                                # Diameter of shaft in m
T = 2400                                # Torque transmitted by the shaft in N-m
P = 125000                              # Tensile force

#calculation
s0 = (4*P)/(math.pi*d**2)               # Tensile stress in
t0 = (16*T)/(math.pi*d**3)              # Shear force
# Stresses along x and y direction
sx = 0
sy = s0
txy = -t0
s1 = (sx+sy)/2.0 + math.sqrt(((sx-sy)/2.0)**2 + (txy)**2)   # Maximum tensile stress
s2 = (sx+sy)/2.0 - math.sqrt(((sx-sy)/2.0)**2 + (txy)**2)   # Maximum compressive stress
tmax =  math.sqrt(((sx-sy)/2)**2 + (txy)**2)            # Maximum in plane shear stress
print "Maximum tensile stress %e" %s1, "Pa"
print "Maximum compressive stress %e" %s2, "Pa"
print "Maximum in plane shear stress %e " %tmax, "Pa"

Maximum tensile stress 1.346662e+08 Pa
Maximum compressive stress -7.100421e+07 Pa
Maximum in plane shear stress 1.028352e+08  Pa


## Example 8.5, page no. 573¶

In [3]:
import math

#initialisation
P = 12                              # Axial load in K
r = 2.1                             # Inner radius of the cylinder in inch
t = 0.15                            # Thickness of the cylinder in inch
ta = 6500                           # Allowable shear stress in Psi

#calculation
p1 = (ta - 3032)/3.5                # allowable internal pressure
p2 = (ta + 3032)/3.5                # allowable internal pressure
p3 = 6500/7.0                       # allowable internal pressure

prs_allowable = min(p1,p2,p3)              # Minimum pressure would govern the design
print "Maximum allowable internal pressure ", round(prs_allowable), "psi"

Maximum allowable internal pressure  929.0 psi


## Example 8.6, page no. 574¶

In [8]:
import math

#initialisation
d1 = 0.18                                   # Inner diameter of circular pole in m
d2 = 0.22                                   # Outer diameter of circular pole in m
P = 2000                                    # Pressure of wind in Pa
b = 1.5                                     # Distance between centre line of pole and board in m
h = 6.6                                     # Distance between centre line of board and bottom of the ploe in m

#calculation
W = P*(2*1.2)                               # Force at the midpoint of sign
T = W*b                                     # Torque acting on the pole
M = W*h                                     # Moment at the bottom of the pole
I = (math.pi/64.0)*(d2**4-d1**4)            # Momet of inertia of cross section of the pole
sa = (M*d2)/(2*I)                           # Tensile stress at A
Ip = (math.pi/32.0)*(d2**4-d1**4)           # Polar momet of inertia of cross section of the pole
t1 = (T*d2)/(2*Ip)                          # Shear stress at A and B
r1 = d1/2.0                                 # Inner radius of circular pole in m
r2 = d2/2.0                                 # Outer radius of circular pole in m
A = math.pi*(r2**2-r1**2)                   # Area of the cross section
t2 = ((4*V)/(3*A))*((r2**2 + r1*r2 +r1**2)/(r2**2+r1**2))  # Shear stress at point B

# Principle stresses
sxa = 0
sya = sa
txya = t1
sxb = 0
syb = 0
txyb = t1+t2

# Stresses at A
s1a = (sxa+sya)/2.0 + math.sqrt(((sxa-sya)/2)**2 + (txya)**2)                 # Maximum tensile stress
s2a = (sxa+sya)/2.0 - math.sqrt(((sxa-sya)/2)**2 + (txya)**2)                 # Maximum compressive stress
tmaxa =  math.sqrt(((sxa-sya)/2)**2 + (txya)**2)                            # Maximum in plane shear stress

print "Maximum tensile stress at point A is", s1a, "Pa"
print "Maximum compressive stress at point A is", s2a, "Pa"
print "Maximum in plane shear stress at point A is", tmaxa, "Pa"

# Stress at B
s1b = (sxb+syb)/2.0 + math.sqrt(((sxb-syb)/2)**2 + (txyb)**2)                 # Maximum tensile stress
s2b = (sxb+syb)/2.0 - math.sqrt(((sxb-syb)/2)**2 + (txyb)**2)                 # Maximum compressive stress
tmaxb =  math.sqrt(((sxb-syb)/2.0)**2 + (txyb)**2)                            # Maximum in plane shear stress
print "Maximum tensile stress at point B is", s1b, "Pa"
print "Maximum compressive stress at point B is", s2b, "Pa"
print "Maximum in plane shear stress at point B is", tmaxb, "Pa"

Maximum tensile stress at point A is 55613361.197 Pa
Maximum compressive stress at point A is -700178.455718 Pa
Maximum in plane shear stress at point A is 28156769.8263 Pa
Maximum tensile stress at point B is 6999035.59641 Pa
Maximum compressive stress at point B is -6999035.59641 Pa
Maximum in plane shear stress at point B is 6999035.59641 Pa


## Example 8.7, page no. 578¶

In [5]:
import math

#initialisation
b = 6                           # Outer dimension of the pole in inch
t = 0.5                         # thickness of the pole
P1 = 20*(6.75*24)               # Load acting at the midpoint of the platform
d = 9                           # Distance between longitudinal axis of the post and midpoint of platform
P2 = 800                        # Load in lb
h = 52                          # Distance between base and point of action of P2

#calculation
M1 = P1*d                       # Moment due to P1
M2 = P2*h                       # Moment due to P2
A = b**2 - (b-2*t)**2           # Area of the cross section
sp1 = P1/A                      # Comoressive stress due to P1 at A and B
I = (1.0/12.0)*(b**4 - (b-2*t)**4)  # Moment of inertia of the cross section
sm1 = (M1*b)/(2*I)              # Comoressive stress due to M1 at A and B
Aweb = (2*t)*(b-(2*t))          # Area of the web
tp2 = P2/Aweb                   # Shear stress at point B by lpad P2
sm2 = (M2*b)/(2*I)              # Comoressive stress due to M2 at A
sa = sp1+sm1+sm2                # Total Compressive stress at point A
sb = sp1+sm1                    # Total compressive at point B
tb = tp2                        # Shear stress at point B

# Principle stresses
sxa = 0
sya = -sa
txya = 0
sxb = 0
syb = -sb
txyb = tp2

# Stresses at A
s1a = (sxa+sya)/2 + math.sqrt(((sxa-sya)/2)**2 + (txya)**2)             # Maximum tensile stress
s2a = (sxa+sya)/2 - math.sqrt(((sxa-sya)/2)**2 + (txya)**2)             # Maximum compressive stress
tmaxa =  math.sqrt(((sxa-sya)/2)**2 + (txya)**2)                        # Maximum in plane shear stress
print "Maximum tensile stress at point A is", s1a,"Psi"
print "Maximum compressive stress at point A is", round(s2a,2), "Psi"
print "Maximum in plane shear stress at point A is", round(tmaxa,2), "Psi"

# Stress at B
s1b = (sxb+syb)/2 + math.sqrt(((sxb-syb)/2)**2 + (txyb)**2)             # Maximum tensile stress
s2b = (sxb+syb)/2 - math.sqrt(((sxb-syb)/2)**2 + (txyb)**2)             # Maximum compressive stress
tmaxb =  math.sqrt(((sxb-syb)/2)**2 + (txyb)**2)                        # Maximum in plane shear stress
print "Maximum tensile stress at point B is", round(s1b,2), "Psi"
print "Maximum compressive stress at point B is", round(s2b,2), "Psi"
print "Maximum in plane shear stress at point B is", round(tmaxb,2), "Psi"

Maximum tensile stress at point A is 0.0 Psi
Maximum compressive stress at point A is -4090.91 Psi
Maximum in plane shear stress at point A is 2045.45 Psi
Maximum tensile stress at point B is 13.67 Psi
Maximum compressive stress at point B is -1872.69 Psi
Maximum in plane shear stress at point B is 943.18 Psi