Chapter 1: Tension Comprssion and Shear¶

Example 1.1, page no. 9¶

In [5]:
import math

#initialisation

d_1 = 4                 # inner diameter (inch)
d_2 = 4.5               #outer diameter (inch)
P = 26000               # pressure in pound
L = 16                  # Length of cylinder (inch)
my_del = 0.012             # shortening of post (inch)

#calculation
A = (math.pi/4)*((d_2**2)-(d_1**2))     #Area (inch^2)
s = P/A                                 # stress

print "compressive stress in the post is ", round(s), "psi"

e = my_del/L                            # strain

print "compressive strain in the post is %e" %e

compressive stress in the post is  7789.0 psi
compressive strain in the post is 7.500000e-04


Example 1.2, page no. 10¶

In [16]:
import math

#initialisation
W = 1500                    # weight (Newton)
d = 0.008                   #diameter(meter)
g = 77000                   # Weight density of steel
L = 40                      # Length of bar (m)

#calculation

A = (math.pi/4)*(d**2)      # Area
s_max = (1500/A) + (g*L)    # maximum stress

#result
print "Therefore the maximum stress in the rod is ", round(s_max,1), "Pa"

Therefore the maximum stress in the rod is  32921551.8 Pa


Example 1.3. page no. 26¶

In [7]:
import math

#initialisation
d1 = 4.5                            # diameter in inch
d2 = 6                              # diameter in inch
A = (math.pi/4)*((d2**2)-(d1**2))   # Area
P = 140                             # pressure in K
s = -P/A                            # stress (compression)
E = 30000                           # young's modulus in Ksi
e = s/E                             # strain

#calculation

# Part (a)
my_del = e*4*12                        # del = e*L
print "Change in length of the pipe is", round(my_del,3), "inch"

# Part (b)
v = 0.30                            # Poissio's ratio
e_ = -(v*e)
print "Lateral strain in the pipe is %e" %e_

# Part (c)
del_d2 = e_*d2
del_d1 = e_*d1
print "Increase in the inner diameter is ", round(del_d1,6), "inch"

# Part (d)
t = 0.75
del_t = e_*t
print "Increase in the wall thicness is %f" %del_t, "inch"
del_t1 = (del_d2-del_d1)/2
print "del_t1 = del_t"

Change in length of the pipe is -0.018 inch
Lateral strain in the pipe is 1.131768e-04
Increase in the inner diameter is  0.000509 inch
Increase in the wall thicness is 0.000085 inch
del_t1 = del_t


Example 1.4, page no. 35¶

In [37]:
import math

#initialisation
d = 0.02                            # diameter in m
t = 0.008                           # thickness in m
A = math.pi*d*t                     # shear area
P = 110000                          # prassure in Newton

#calculation
A1 = (math.pi/4)*(d**2)             # Punch area
t_aver = P/A                        # Average shear stress

print "Average shear stress in the plate is ", t_aver, "Pa"
s_c = P/A1  # compressive stress
print "Average compressive stress in the plate is ", s_c, "Pa"

Average shear stress in the plate is  218838046.751 Pa
Average compressive stress in the plate is  350140874.802 Pa


Eample 1.5, page no. 36¶

In [39]:
import math

#initialisation

P = 12.0                              # Pressure in K
t = 0.375                           # thickness of wall in inch
theta = 40.0                          # angle in degree
d_pin = 0.75                        # diameter of pin in inch
t_G = 0.625                         # thickness of gusset in inch
t_B = 0.375                         #thickness of base plate in inch
d_b = 0.50                          # diameter of bolt in inch

#calculation

#Part (a)
s_b1 = P/(2*t*d_pin)                  # bearing stress
print "Bearing stress between strut and pin", round(s_b1,1), "ksi"

#Part (b)
t_pin = (4*P)/(2*math.pi*(d_pin**2))  # average shear stress in the
print "Shear stress in pin is ", round(t_pin,1), "ksi"

# Part (c)
s_b2 = P/(2*t_G*d_pin)                # bearing stress between pin and gusset
print "Bearing stress between pin and gussets is", s_b2, "ksi"

# Part (d)
s_b3 = (P*math.cos(math.radians(40))/(4*t_B*d_b)) # bearing stress between anchor bolt and base plate
print "Bearing stress between anchor bolts & base plate", round(s_b3,1), "ksi"

# Part (e)
t_bolt = (4*math.cos(math.radians(40))*P)/(4*math.pi*(d_b**2))  # shear stress in anchor bolt
print "Shear stress in anchor bolts is", round(t_bolt,1), "ksi"

Bearing stress between strut and pin 21.3 ksi
Shear stress in pin is  13.6 ksi
Bearing stress between pin and gussets is 12.8 ksi
Bearing stress between anchor bolts & base plate 12.3 ksi
Shear stress in anchor bolts is 11.7 ksi


Example 1.7, page no. 42¶

In [42]:
import math

#initialisation
b1 = 1.5                            # width of recmath.tangular crosssection in inch
t = 0.5                             # thickness of recmath.tangular crosssection in inch
b2 = 3.0                            # width  of enlarged recmath.tangular crosssection in inch
d = 1.0                             # diameter in inch

#calculation

# Part (a)
s_1 = 16000 # maximum allowable tensile stress in Psi
P_1 = s_1*t*b1
print "The allowable load  P1 is", P_1, "lb"

# Part (b)
s_2 = 11000 # maximum allowable tensile stress in Psi
P_2 = s_2*t*(b2-d)
print "allowable load P2 at this section is", P_2, "lb"

#Part (c)
s_3 = 26000 # maximum allowable tensile stress in Psi
P_3 = s_3*t*d
print "The allowable load based upon bearing between the hanger and the bolt is", P_3, "lb"

# Part (d)
s_4 = 6500 # maximum allowable tensile stress in Psi
P_4 = (math.pi/4)*(d**2)*2*s_4
print "the allowable load  P4 based upon shear in the bolt is", round(P_4), "lb"

The allowable load  P1 is 12000.0 lb
allowable load P2 at this section is 11000.0 lb
The allowable load based upon bearing between the hanger and the bolt is 13000.0 lb
the allowable load  P4 based upon shear in the bolt is 10210.0 lb


Example 1.8, page no. 46¶

In [9]:
import math

#initialisation
R_ah = (2700*0.8 + 2700*2.6)/2                  # Horizontal component at A in N
R_ch = R_ah                                     # Horizontal component at C in N
R_cv = (2700*2.2 + 2700*0.4)/3                  # vertical component at C in N
R_av = 2700 + 2700 - R_cv                       # vertical component at A in N
R_a = math.sqrt((R_ah**2)+(R_av**2))
R_c = math.sqrt((R_ch**2)+(R_cv**2))
Fab = R_a                                       # Tensile force in bar AB
Vc = R_c                                        # Shear force acting on the pin at C
s_allow = 125000000                             # allowable stress in tension
t_allow = 45000000                              # allowable stress in shear

#calculation
Aab = Fab / s_allow                             # required area of bar
Apin = Vc / (2*t_allow)                         # required area of pin

print "Required area of bar is %f" %Apin, "m^2"
d = math.sqrt((4*Apin)/math.pi)                 # diameter in meter
print "Required diameter of pin is %f" %d, "m"

Required area of bar is 0.000057 m^2
Required diameter of pin is 0.008537 m

In [ ]: