# Chapter 2: Axially Loaded Members¶

## Example 2.1, page no. 72¶

In [1]:
import math

#initialisation

W = 2.0                       #lb
b = 10.5                   #inch
c = 6.4                    #inch
k = 4.2                    #inch
p = 1.0/16.0                   #inch

#calculation

n = (W*b)/(c*k*p)          #inch

#result

print " No. of revolution required = ", n, "revolutions"

 No. of revolution required =  12.5 revolutions


## Example 2.2, page no. 74¶

In [3]:
import math
import numpy

#initialisation

Fce_ = 2.0                    #dummy variable
Fbd_ = 3.0                    #dummy variable
Lbd = 480.0                   #mm
Lce = 600.0                  #mm
E = 205e6                   #205Gpa
Abd = 1020.0                  #mm
Ace = 520.0                   #mm

#calculation
Dbd_ = (Fbd_*Lbd)/(E*Abd)   #dummy variable
Dce_ = (Fce_*Lce)/(E*Ace)   #dummy variable
Da = 1 #limiting value
P = ((((450+225)/225)*(Dbd_ + Dce_) - Dce_ )**(-1)) * Da
Fce = 2*P # Real value in newton
Fbd = 3*P #real value in newton
Dbd = (Fbd*Lbd)/(E*Abd) #print lacement in mm
Dce = (Fce*Lce)/(E*Ace) # print lacement in mm
a = numpy.degrees(numpy.arctan(((Da+Dce)/675)))  #alpha in degree

#result
print "alpha = ", round(a,2), "degree"

alpha =  0.11 degree


## Example 2.3, page no. 80¶

In [4]:
import math

#initialisation
P1 = 2100.0                       #lb
P2 = 5600.0                       #lb
b = 25.0                          #inch
a = 28.0                          #inch
A1 = 0.25                       #inch^2
A2 = 0.15                       #inch^2
L1 = 20.0                         #inch
L2 = 34.8                       #inch
E = 29e6                        #29Gpa

#Calculations
P3 = (P2*b)/a
Ra = P3-P1
N1 = -Ra
N2 = P1
D = ((N1*L1)/(E*A1)) + ((N2*L2)/(E*A2))     #print lacement

#Result
print  "Downward print lacement is = ", D, "inch"

Downward print lacement is =  0.0088 inch


## Example 2.6, page no. 90¶

In [5]:
import math

#Numerical calculation of allowable load

d1 = 4.0                      #mm
d2 = 3.0                      #mm
A1 = (math.pi*(d1**2))/4    #area
A2 = (math.pi*(d2**2))/4    #area
L1 = 0.4                    #meter
L2 = 0.3                    #meter
E1 = 72e9                   #Gpa
E2 = 45e9                   #Gpa
f1 = L1/(E1*A1) * 1e6       # To cpmpensate for the mm**2
f2 = L2/(E2*A2) * 1e6
s1 = 200e6                  #stress
s2 = 175e6                  #stress

#Calculations
P1 = ( (s1*A1*(4*f1 + f2))/(3*f2) ) * 1e-6      # To cpmpensate for the mm**2
P2 = ( (s2*A2*(4*f1 + f2))/(6*f1) ) * 1e-6

#Result
print "Newton Minimum allowable stress aomong the two P1 and P2 is smaller one, therefore MAS = ", P2

Newton Minimum allowable stress aomong the two P1 and P2 is smaller one, therefore MAS =  1264.49104307


## Example 2.10, page no. 113¶

In [4]:
import math

#initialisation
P = 90000.0                   #newton
A = 1200e-6                   # meter^2
s_x = -P/A                    #stress
t_1 = 25.0                    #for the stresses on ab and cd plane

#Calculations
s_1 = s_x*(math.cos(math.radians(t_1))**2)
T_1 = -s_x*math.cos(math.radians(t_1))*math.sin(math.radians(t_1))
t_2 = -65.0                   #for the stresses on ad and bc plane
s_2 = s_x*(math.cos(math.radians(t_2))**2)
T_2 = -s_x*math.cos(math.radians(t_2))*math.sin(math.radians(t_2))

#Result
print "The normal and shear stresses on the plane ab and cd are", round((T_1/1E+6),2), round((s_1/1E+6),2), "MPa respecively"
print "respecively The normal and shear stresses on the plane ad and bc are", round((T_2/1E+6),2), round((s_2/1E+6),2), "MPa respecively"

The normal and shear stresses on the plane ab and cd are 28.73 -61.6 MPa respecively
respecively The normal and shear stresses on the plane ad and bc are -28.73 -13.4 MPa respecively


## Example 2.11, page no. 114¶

In [1]:
import math

# Value of s_x based on allowable stresses on glued joint

#initialisation
s_t = -750.0                                  #psi
t = -50.0                                     #degree
T_t = -500.0                                  #psi

sg_x_1 = s_t/(math.cos(math.radians(t))**2)
sg_x_2 = -T_t/(math.cos(math.radians(t))*math.sin(math.radians(t)))

# Value of s_x based on allowable stresses on plastic

sp_x_1 = -1100.0                              #psi
T_t_p = 600.0                                  #psi
t_p = 45.0                                    #degree
sp_x_2 = -T_t_p/(math.cos(math.radians(t_p))*math.sin(math.radians(t_p)))

# Minimum width of bar

P = 8000.0                                    #lb
A = P/sg_x_2
b_min = math.sqrt(abs(A))                        #inch
print "The minimum width of the bar is", round(b_min,2), "inch"

The minimum width of the bar is 2.81 inch


## Example 2.15, page no. 126¶

In [3]:
import math
#Bolt with reduced shank diameter

#initialisation
g = 1.50                     # inch
d = 0.5                      #inch
t = 0.25                     #inch
d_r = 0.406                  #inch
L = 13.5                     #inch

#calculation
ratio = ((g*(d**2))/(((g-t)*(d_r**2))+(t*(d**2))))              #U2/U1

print "The energy absorbing capacity of the bolts with reduced shank diameter", round(ratio,2)
ratio_1 = ( (((L-t)*(d_r**2))+(t*(d**2))) / ((2*(g-t)*(d_r**2))+2*(t*(d**2))) )         #U3/2U1
print "The energy absorbing capacity of the long bolts", round(ratio_1,2)

The energy absorbing capacity of the bolts with reduced shank diameter 1.4
The energy absorbing capacity of the long bolts 4.18


## Example 2.16, page no. 133¶

In [7]:
import math

#initialisation
# Maximum elongation
M = 20                      #kg
g = 9.81                    #m/s^2
L = 2                       #meter
E = 210e9                   #210Gpa
h = 0.15                    #meter
diameter = 0.015            #milimeter

#Calculations & Result
A = (math.pi/4)*(diameter**2)               #area
D_st = ((M*g*L)/(E*A))
D_max = D_st*(1+(1+(2*h/D_st))**0.5)
D_max_1 = math.sqrt(2*h*D_st)                # another approach to find D_max
i = D_max / D_st                             # Impact factor
print "Maximum elongation is",round((D_max/1E-3),2), "mm"  # Maximum tensile stress
s_max = (E*D_max)/L                          #Maximum tensile stress
s_st = (M*g)/A                               #static stress
i_1 = s_max / s_st                           #Impact factor
print "Maximum tensile stress is ", round((s_max/1E+6),2), "MPa"

Maximum elongation is 1.79 mm
Maximum tensile stress is  188.13 MPa


## Example 2.18, page no. 148¶

In [3]:
import math

#initialisation
P1 = 108000.0                 #Newton
P2 = 27000.0                  #Newton
L = 2.2                     #meter
A = 480.0                     #mm^2

#calculations

# Displacement due to load P1 acting alone
s = (P1/A)                      #stress in MPa
e = (s/70000) + (1/628.2)*((s/260)**10)         #strain
D_b = e*L*1e3                                   #elongation in mm
print "elongation when only P1 load acting is = ", round(D_b,2), " mm"

# Displacement due to load P2 acting alone
s_1 = (P2/A) #stress in MPa
e_1 = (s_1/70000) + (1/628.2)*((s_1/260)**10) #strain
D_b_1 = e_1*(L/2)*1e3 #elongation in mm (no elongation in lower half)
print "elongation when only P2 load acting is = ", round(D_b_1,2), " mm"

# Displacement due to both load acting simonmath.taneously
#upper half
s_2 = (P1/A) #stress in MPa
e_2 = (s_2/70000) + (1/628.2)*((s_2/260)**10) #strain

#lower half
s_3 = (P1+P2)/A #stress in MPa
e_3 = (s_3/70000) + (1/628.2)*((s_3/260)**10) #strain
D_b_2 = ((e_2*L)/2 + (e_3*L)/2) * 1e3 # elongation in mm
print "elongation when P1 and P2 both loads are acting is = ", round(D_b_2,2), " mm"

elongation when only P1 load acting is =  7.9  mm
elongation when only P2 load acting is =  0.88  mm
elongation when P1 and P2 both loads are acting is =  12.21  mm