# Chapter 8:Springs¶

## Problem 8.1,Page no.206¶

In :
import math

#Initilization of variables

k=1 #KN/m #stiffness of spring
sigma_s=126 #MPa #Max shear stress
L=4.5 #cm #Lenght of spring
G=42 #GPa #Modulus of rigidity

#Calculations

#sigma_s_max=16*P*R*(pi*d**3)**-1 #Max shear stress

#After substituting values in above equation and simolifying we get
#1000=42*10**9*d**4*(64*R**3*n)**-1   (#Equation 1)

#R=0.175*10**6*pi*d**3 #Radius of spring (Equation 2)

#L=n*d #solid length of spring
#Thus simplifying above equation, n=L*d**-1

#substituting value of n and R in (equation 1) we get,

d=(42*10**9*(1000*64*4.5*10**-2*(0.175*pi)**3*(10**6)**3)**-1)**0.25*10**2 #cm #diameter of helical spring

#substituting value d in (equation 2) we get,
D=2*R #cm #Mean diameter of coil
n=0.045*0.00306**-1 #Number of turns

#Result
print"The Diameter of wire is",round(d,3),"cm"
print"The Mean Diameter of coil is",round(D,2),"cm"

The Diameter of coil is 0.306 cm
The Mean Diameter of coil is 3.15 cm


## Problem 8.2,Page no.207¶

In :
import math
import numpy as np

#Initilization of variables

L=15 #cm #Length of close coiled helical spring
U=50 #N*m #Strain energy
sigma_s=140 #MPa #Shear stress
D=10 #cm #Mean coil diameter
G=80 #GPa #Modulus of rigidity

#Calculations

#Let dell be the deflection of the spring when fully compressed
# 0.15-dell=n*d (Equation 1)

#U=(sigma_s)**2*V*(4*G)**-1 #Strain energy

#After substituting values in above equation and simolifying we get
V=50*4*80*10**9*((140*10**6)**2)**-1 #m**3 #Volume of spring

#But V=pi*4**-1*d**2*2*pi*R*n
#After substituting values in above equation and simolifying we get
#n=3.308*10**-3*(d**2)**-1 #Number of turns

#We know, T=P*R
#Now substituting values in T and simolifying we get

#U=P*dell*2**-1
#After substituting values in above equation and simolifying we get
#dell=0.18189*10**-6*d**3 #Deflection

#After substituting values in above equation and simolifying we get

#d**3-22.0533*10**-3*d**2-1.21261*10**-6=0

coeff=[1,-22.0533*10**-3,0,-1.21261*10**-6]
d=np.roots(coeff)  #Diameter of steel wire
n=3.308*10**-3*((d)**2)**-1 #no.of coils

#Result
print"Diameter of steel wire is",d,"m"
print"number of coils",n

Diameter of steel wire is (0.0241350343273+0j) m
number of coils (5.67897110767+0j)


## Problem 8.3,Page no.208¶

In :
import math

#Initilization of variables

k=10 #KN/m #stiffness
L=40 #cm #Length of coil when adjascent coil touch each other
G=80 #GPa #Modulus of rigidity
#dell=0.002*n #Max compression

#Calculation

#k=G*d**4*(8*D**3*n)**-1 #Stiffness
#After substituting values in above equation and simolifying we get
#d**4=D**3*n*10**-6     (Equation 1)

#L=n*d, #After substituting values we get
#n=0.4*d**-1            (Equation 2)

#Again, d*D**-1=1*10**-1
#After solving above ratios we get,D=10*d

#After substituting values in Equation 1 And Equation 2 we get
d=(10**3*0.4*10**-6)**0.5*100 #cm
D=10*d #cm #Mean Diameter
n=0.4*(d*10**-2)**-1 #Number of turns
dell=0.002*n*100 #Deflection

#k=P*dell**-1
#after solving above equation we get

sigma_s_max=16*P*R*10**-2*(pi*(d*10**-2)**3)**-1 #N/m**2 #Max shear stress

#Result
print"The wire diameter is",round(d,2),"cm"
print"The Mean diameter is",round(D,2),"cm"
print"Max shear stress is",round(sigma_s_max,0),"N/m**2"

The wire diameter is 2.0 cm
The Mean diameter is 20.0 cm
Max Load applied is 400.0 N
Max shear stress is 25464791.0 N/m**2


## Problem 8.4,Page no.209¶

In :
import math

#Initilization of variables

G=80 #GPa #Modulus of rigidity
dell=10 #cm #Deflection
sigma_s=350 #MPa #Max shear stress
rho=78000 #N/m**3 #Density of materials

#Calculations

U=P*1000*dell*10**-2*2**-1 #N*m #Energy stored

#Again U=sigma_s**2*V*(4*G)**-1
#After substituting values in above equation and further simplifying we get
V=50*4*80*10**9*((350*10**6)**2)**-1 #m**3 #Volume

W=V*rho #N #Weight

#Now T=P*R=pi*d**3*sigma_s*16**-1
#After substituting values in above equation and further simplifying
D=(10**6*16*(2*pi*350*10**6)**-1)**0.5*10**2 #cm #Mean diameter of coil

k=P*10**3*(dell*10**-2)**-1 #stiffness

#Also k=D*n**-1*10**6
#After substituting values in above equation and further simplifying
n=10**6*D*10**-2*k**-1 #number of turns

#Result
print"The Value of weight is",round(W,3),"N"
print"Mean coil diameter is",round(D,2),"cm"
print"The number of Turns is",round(n,4)

The Value of weight is 10.188 N
Mean coil diameter is 8.53 cm
The number of Turns is 8.5297


## Problem 8.5,Page no.210¶

In :
import math

#Initilization of variables

d=6 #mm #Diameter of steel wire
n=50 #number of turns
D=5 #cm #Mean Diameter
G=80 #GPa #Modulus of Rigidity

#Calculation

#Dell=64*P*R**3*n*(G*d**4)**-1 #Deflection
#After substituting values in above equation and simplifying we get

#loss of potential Energy of the weight=Gain of strain Energy of the spring
#150*(0.05+dell)=P*dell*2**-1
#After substituting values in above equation we get

#dell**2-0.1446*dell-0.00723=0
#Above Equation is in the form of ax^2+bx+c=0

a=1
b=-0.1446
c=-0.00723

#First computing value of  b^2-4ac and store it in a variable say X
X=b**2-(4*a*c)
#now roots are given as x=(-b+X**0.5)/(2*a) and second root is negative sign before X

dell_1=(-b+X**0.5)*(2*a)**-1*10**2
dell_2=(-b-X**0.5)*(2*a)**-1*10**2

P=2073.6*dell_1*10**-2 #N

sigma=16*P*R*10**-2*(pi*(d*10**-3)**3)**-1 #N/m**2 #Max stress

#Result
print"The Max Extension of the Spring is",round(dell_1,2),"cm"
print"The Max stress is",sigma,"N/m**2"

The Max Extension of the Spring is 18.39 cm
The Max stress is 224797751.663 N/m**2


## Problem 8.6,Page no.209¶

In :
import math

#Initilization of variables

W=200 #N #weight
v=4 #m/s #velocity of spring
sigma=600 #MPa #max allowable stress in spring
G=80 #GPa #Modulus of rigidity
rho=78000 #N/m**3 #density
d=8 #mm #diameter of spring
D=5 #cm #Mean Diameter of coil

#Calculation

E=W*v**2*(2*9.81)**-1 #N*m #Kinetic Energy #Notification has been changed

#U=sigma_s**2*V*(4*G)**-1 #Strain Energy stored inthe spring

#After substituting values in above equation and simplifying we get
V=163.1*4*80*10**9*((600*10**6)**2)**-1  #Volume

W=rho*V #N #Weight of spring

#Now V=pi*4**-1*d**2*pi*D*n
#After substituting values in above equation and simplifying we get
n=0.000145*4*(pi**2*0.008**2*0.05)**-1 #number of turns of spring

#T=P*R=pi*16**-1*d**3*sigma_s #Torsion
#After substituting values in above equation and simplifying we get
P=pi*0.008**3*600*10**6*(0.025*16)**-1 #N

#Now U=P*dell*2**-1
#Again,After substituting values in above equation and simplifying we get
dell=163.1*2*(2412.743)**-1*10**2 #cm

#Result
print"The Max Deflection Produced is",round(dell,2),"cm"
print"Number of coil are",round(n,2),

The Max Deflection Produced is 13.52 cm
Number of coil are 18.36


## Problem 8.7,Page no.211¶

In :
import math

#Initilization of variables

n=12 #number of coils
d=3 #cm #mean diameter
k=720 #N/m #stiffness of spring
sigma_s=190 #MPa #Max shear stress
G=80 #GPa #Modulus of rigidity
D=3 #mm #Diameter of outer spring

#Calculations
R=D*2**-1 #mm #Radius of outer spring

#Dell_1=64*P*(R*10**-3**3*n*(G*10**9*(d*10**-3)**4)**-1 #m #Extension of first spring
#After substituting values and further simplifying we get
#Dell_1=0.0004*P #m

#Dell_2=64*P*(R*10**-3**3*n*(G*10**9*(d_1)**4)**-1 #m #Extension of first spring
#After substituting values and further simplifying we get
#Dell_2=3.24*10**-14*P*(d_1**4)**-1 #m #where d_1 is diameter of inner spring

#Dell=Dell_1+Dell_2
#After substituting values and further simplifying we get
#dell=0.0004*P+3.24*10**-14*P*((d)**4)**-1

#But dell=P*k**-1=P*720**-1

#Now substituting value of dell in above equation we get
d_1=(3.24*10**-14*(1*720**-1-0.0004)**-1)**0.25 #cm #diameter of inner spring

#Now T=P*R=pi*d_1**3*dell_s*sigma_s*16**-1
#simplifying above equation further
#P=pi*d**3*sigma_s*(16*R)**-1
#Now substituting values and further simplifying we get

dell=P*k**-1*10**2 #cm #Total Elongation

#Result
print"Greatest Load that can be carried by composite spring is",round(P,0),"N"
print"Extension in spring is",round(dell,2),"cm"

Greatest Load that can be carried by composite spring is 34.0 N
Extension in spring is 4.73 cm


## Problem 8.8,Page no.212¶

In :
import math

#Initilization of variables

#Outer spring
n_1=10 #number of coils
D_1=3 #cm #Diameter of coil
d_1=3 #mm #diameter of wire
dell_1=2 #cm #deflection of spring

#Inner spring
n_2=8 #number of coils

G=80 #GPa #Modulus of rigidity

#Calculation

R_1=D_1*2**-1
P_1=G*10**9*dell_1*10**-2*(d_1*10**-3)**4*(64*(R_1*10**-2)**3*n_1)**-1 #Load carried outer spring for compression of 2 cm

P_2=100-P_1 #N #Load carried by inner spring
k_2=P_2*0.01**-1 #N/m #stiffness of inner spring

#D_2=D_1*10**-2-d_1*10**-3-2*dell_1*10**-2-d_2 #Diameter of inner spring
#Further simplifying above equation we get
#D_2=0.023-d_2

#Now from stiffness equation of inner spring
#k=G*d_2**4*(8*D_2**3*n_2)**-1
#Now substituting values and further simplifying we get
#d**4=(0.023-d)**3*312500**-1

#As d is small compared with 0.023,as a first appromixation
d_2_1=(0.023**3*312500**-1)**0.25 #m

#Second Approximation
d_2_2=((0.023-d_2_1)**3*312500**-1)**0.25 #m

#Final approximation
d_2_3=((0.023-d_2_2)**3*312500**-1)**0.25*100 #cm

#Result
print P_1
print"Stiffness of inner spring is",round(k_2,2),"N/m"
print"Wire Diameter of inner spring is",round(d_2_3,3),"cm"

60.0
Stiffness of inner spring is 4000.0 N/m
Wire Diameter of inner spring is 0.231 cm


## Problem 8.9,Page no.212¶

In :
import math

#Initilization of variables

L= 3 #m #Length of rod
d_1=25*10**-3 #m #Diameter of rod
n= 5 #no. of coils
sigma=70*10**6 #MPa  #instantaneous stress
E=70*10**9 #Pa
G=80*10**9 #Pa
D=24*10**-2 #Spring diameter
d=4*10**-2 #diameter of steel

#Calculations

dell_1=sigma*L*(E)**-1

#Let P be the equivalent applied Load which will produce same stress of 70 MPa
P=pi*4**-1*(d_1)**2*E*10**-3 #KN

#deflection of the spring is given by
dell_2=P*64*R**3*n*(G*d**4)**-1

#Now Loss of Potential Energy of the weight=strain energy stored in the rod and the spring
#Height measured from top of uncompressed  spring
h=((P*dell_1*2**-1+P*dell_2*2**-1)*(5.5*10**3)**-1-(dell_1+dell_2))*10**2

#Shear stress in the spring is given by
sigma_s=16*P*R*(pi*d**3)**-1*10**-6 #MPa

#Result
print"Height measured from top of uncompressed  spring",round(h,2),"cm"
print"max shearing stress is",round(sigma_s,2),"MPa"

Height measured from top of uncompressed  spring 20.34 cm
max shearing stress is 328.13 MPa


## Problem 8.10,Page no.213¶

In :
import math

#Initilization of variables

L=75 #cm #Legth of Leaf spring
y_c=20 #mm #Deflection
sigma=200 #MPa #Bending stress
E=200 #GPa #modulus of Elasticity
#b=12*t

#Calculation

#y_c=sigma*L**2*(4*E*t)**-1
#After substituting values and further simplifying we get
t=200*10**6*(75*10**-2)**2*(4*200*10**9*0.02)**-1*10**2 #Thickness of plate

b=12*t #width of plate

#Now using relation we get
#sigma=3*P*L*(2*n*b*t**2)**-1
#After substituting values and further simplifying we get
n=3*8*10**3*0.75*(2*200*10**6*0.084*0.007**2)**-1

#Y_c=L**2*(8*R)**-1

#Result
print"The thickness of plate is",round(t,2),"cm"
print"The width of plate is",round(b,2),"cm"
print"The number of plate is",round(n,2)

The thickness of plate is 0.7 cm
The width of plate is 8.44 cm
The number of plate is 10.93
The Radius of plate is 3.52


## Problem 8.11,Page no.214¶

In :
import math

#Initilization of variables

L=75 #cm #span of laminated steel spring
y_c=5 #cm #Central Deflection
sigma=400 #MPa #Bending stress
E=200 #GPa #Modulus of Elasticity
#b=12*t

#Calculations

#y_c=3*P*L**3*(8*E*n*b*t**3)**-1 #Deflection
#After substituting values and further simplifying we get
#nt**4=9.887*10**-9   (Equation 1)

#We Know sigma=3*P*L*(2*n*b*t**3)**-1 #bending stress
#Again after substituting values and further simplifying we get
#nt**3=1.757*10**-6   (Equation 2)

#After Divviding (Equation 1) by (Equation 2) we have
t=9.887*10**-9*(1.757*10**-6)**-1*10**2 #cm

#substituting value of t in Equation 2) we get
n=1.757*10**-6*((t*10**-2)**3)**-1 #Number of plates

#Result
print"The thickness of Plates is",round(t,2),"cm"
print"The Number of Plates is",round(n,2)
print"The Radius of Curvature of Plates is",round(R,2),"m"

The thickness of Plates is 0.56 cm
The Number of Plates is 9.86
The Radius of Curvature of Plates is 1.41 m


## Problem 8.12,Page no.214¶

In :
import math

#Initilization of variables

L=1.3 #m #Length of carriage spring
b=10 #cm #width of spring
t=12 #mm #thickness of spring
sigma=150 #MPa #Bending stresses
E=200 #GPa #Modulus of Elasticity
U=120 #N*m #Strain Energy

#Calculation

#V=n*b*t*L #Volume of carriage spring
#U=sigma**2*(6*E)**-1*V
#After substituting values in above equation and further simplifying we get
n=120*6*200*10**9*2*((150*10**6)**2*10*10**-2*12*10**-3*1.3)**-1

sigma_1=(120*6*200*10**9*2*(9*0.1*0.012*1.3)**-1)**0.5*10**-6 #MPa #Actual Bending stress

R=E*t*(2*sigma_1)**-1 #m

#Result
print"The number of plates is",round(n,2)

The number of plates is 8.21


## Problem 8.13,Page no.215¶

In :
import math

#Initilization of variables

h=10 #cm #Height of Load dropped
n=10 #Number of turns
b_1=5 #cm #width of plates
t=6 #mm #thickness of plates
L=75 #cm #Length of plates
E=200 #GPa #Modulus of Elasticity

#Calculaion

#Let P be the equivalent gradually applied load whuch would cause the same stress as is caused by the impact Load
#200(0.1+dell)=P*dell*2**-1 (Equation 1)

#dell=3*P*L**3*(8*E*n*b*t**3)**-1 #Deflection
#After substituting values in above equation and further simplifying we get
#P=136533.33*dell

#After substituting values of P in (equation 1) and further simplifying we get
#200(0.1+dell)=136533.33*dell**2*2**-1

#simplifying above equation we get
#dell**2-2.93*10**-3*dell-2.93*10**-4=0
#The Above Equation is in the form of ax**2+bx+c=0
a=1
b=-2.93*10**-3
c=-2.93*10**-4

#First computing value of  b^2-4ac and store it in a variable say X
X=b**2-(4*a*c)
#now roots are given as x=(-b+X**0.5)/(2*a) and second root is negative sign before X

dell_1=(-b+X**0.5)*(2*a)**-1
dell_2=(-b-X**0.5)*(2*a)**-1

#Now deflection cannot be negative so consider value of dell_1

P=136533.33*dell_1
sigma=3*P*L*10**-2*(2*n*b_1*10**-2*(t*10**-3)**2)**-1*10**-6 #MPa #Max instantaneous stress

#Result
print"Max instantaneous stress in plates is",round(sigma,2),"MPa"
print"Radius of curvature of spring is",round(R,2),"m"

Max instantaneous stress in plates is 159.1 MPa
Radius of curvature of spring is 3.77 m


## Problem 8.14,Page no.215¶

In :
import math

#Initilization of variables

L=70 #cm #Length of Longest plate
y_c=1.8 #cm #central deflection
sigma=190 #MPa #allowable bending stress
#b=12*t
E=200 #GPa #Modulus of Elasticity

#Calculation

y_c=3*P*L**3*(8*n*E*b*t**3)**-1 #Deflection (#Equation 1)
sigma=3*P*L*(2*n*b*t**2)**-1 #stress
#Dividing Equation 1 by Equation 2 we get
#y_c*sigma**-1=L**2*(4*E*t)**-1
#After substituting values in above equation and further simplifying we get
t=190*10**6*0.7**2*(1.8*10**-2*4*200*10**9)**-1*10**3 #thickness of plate
b=12*t #Width of plates

#sigma=3*2**-1*P*L*(n*b*t**2)**-1 #stress
#After substituting values in above equation and further simplifying we get
n=3*3.5*10**3*0.7*(2*190*10**6*0.077583*(6.465*10**-3)**2)**-1

#Now sigma*y**-1=E*R**-1
#simplifying above equationwe get
a=L*10**-2*(2*n)**-1*10**3 #Overlap

#Result
print"size of the plate is:",round(b,2),"mm"
print"                    :",round(t,2),"mm"
print"Overlap of plates is",round(a,2),"mm"
print"Number of plates is",round(n,2)

size of the plate is: 77.58 mm
: 6.47 mm
Overlap of plates is 58.68 mm
Number of plates is 5.96
The Radius of curvature is 3.403 m


## Problem 8.15,Page no.216¶

In :
import math

#Initilization of variables

alpha=30 #degree #helix angle
dell=2.3*10**-2 #m #Vertical displacement
d=6*10**-3 #steel rod diameter
E=200*10**9 #Pa
G=80*10**9 #Pa

#Calculations

#from equation of deflection of the spring under the Load we get
#R**3*n=8.49*10**-4

#Let R**3*n=X
X=8.49*10**-4  #Equation 1

#from equation of angular rotation
#R**2*n=8.1*10**-3

#Let R**2*n=Y
Y=8.1*10**-3   #Equation 2

#After dividing equation 1 by equation 2 we get R
#Let Z=R

Z=X*Y**-1

#Result
print"Mean Radius of Open coiled spring of helix angle is",round(R,2),"cm"

Mean Radius of Open coiled spring of helix angle is 10.48 cm


## Problem 8.16,Page no.217¶

In :
import math

#Initilization of variables

n=10 #Number of coils
sigma=100 #MPa #Bending stress
sigma_s=110 #MPa #Twisting stress
#D=8*d
dell=1.8 #cm #Max extension of of wire
E=200 #GPa #Modulus of Elasticity
G=80 #GPa #Modulus od Rigidity

#Calculation

#M=W*R*sin_alpha=pi*d**3*sigma_1*32**-1 #(Equation 1) #Bending moment
#As D=8*d
#then R=D*2**-1
#Therefore, R=4*d

#Now substituting values in equation 1 we get
#W*sin_alpha=2454369.3*d**2  (Equation 2)

#T=W*R*cos_alpha=pi*d**3*sigma_s #Torque  (Equation 3)
#Now substituting values in equation 3 we get
#W*cos_alpha=5399612.4*d**2    (Equation 4)

#Dividing Equation 2 by Equation 4 we get,
#tan_alpha=0.4545
alpha=arctan(0.4545)*180*pi**-1

#From Equation 2 we get
#W=2454369.3*d**2*(sin24.443)**-1
#W=5931241.1*d**2   (Equation 5)

#dell=64*W*R**3*n*sec_alpha*(d**4)**-1*((cos_alpha)**2*G**-1+2*sin_alpha**2*E**-1)
#Now substituting values in above equation  we get
#W=33140.016*d  (Equation 6)

#From Equation 5 and Equation 6 we get
#5931241.1*d**2=33140.016*d
#After simplifying above equation we get
d=33140.016*5931241.1**-1 #m #Diameter of wire

#Result
print"Thw Wire Diameter is",round(d,6),"m"

The Max Permissible Load is 185.17 N
Thw Wire Diameter is 0.005587 m


## Problem 8.18,Page no.218¶

In :
import math

#Initilization of variables

#Calculation

n=10 #number of coils
d=2*10**-2 #m #Diameter of wire
D=12*10**-2 #m #Diameter of coiled spring
R=0.06 #m #Radius of coiled spring
dell=0.5*10**-2 #Deflection
E=200*10**9 #Pa
G=80*10**9 #Pa
alpha=30 #degree

#Calculations

#beta=64*W*R**2*n*sinalpha*(d**4)**-1*(1*G**-1-2*E**-1)+64*T*R*n*secalpha*(d**4)**-1*(sin**2alpha*G**-1+2*cos**2alpha*E**-1)=0
#From above equation anf simplifying we get

#T=-6.11*10**-3*W

#dell=64*W*R**3*n*sec(alpha)*(d**4)**-1*[(cos(alpha))**2*G**-1+2*(sin(alpha))**2*E**-1]+64*T*R**2*n*sin(alpha)*(d**4)**-1*[1*G**-1+2*E**-1]

#After substituting Values and further simplifying we get
#1.1847*10**-5*W+1.62*10**-4*T=0.005

#Now substituting value of T in above equation we get
#1.1847*10**-5*W-9.8982*10**-7*W=0.005
W=0.005*(1.1847*10**-5-9.8982*10**-7)**-1 #N
T=-6.11*10**-3*W #N*m

#Result
print"Necesscary torque is",round(T,2),"N*m"

The axial Load is 460.52 N
Necesscary torque is -2.81 N*m


## Problem 8.19,Page no.219¶

In :
import math

#Initilization of variables

d=6 #mm #Diameter of steel wire
n=1 #number of turns
D=6.5 #cm #Mean of diameter
G=80 #GPa #modulus of rigidity
p=1.5 #cm #Pitch of coil

#Calculation

R=D*2**-1
#For one turn deflection is
dell=p-d*10**-1 #cm

#dell=64*P*R**3*n*(G*d**4)**-1
#Now, after simplifying further we get,

dell_2=dell*8 #cm #Total Displacement #Notification has been changed
U=P*dell_2*10**-2*2**-1 #N-m #Strain Energy

#Potential Energy given by 150N Load is
#U=150*(h+0.072)

#After simplifying above equation we get
h=(U*P_1**-1-0.072)*10**2 #cm #Height from which 150 N load falls

#Result
print"Height from which 150 N load falls is",round(h,2),"cm"

Axial Load is 424.72 N
Height from which 150 N load falls is 2.99 cm


## Problem 8.20,Page no.219¶

In :
import math

#Initilization of variables

alpha=30 #degree
E=200*10**9 #Pa
G=80*10**9 #pa

#Calculations

#For alpha=30 #Degree
#dell=64*W*R**3*n*sec(alpha)*(d**4)**-1*(cos(alpha)**2*G**-1+2*sin(alpha)**2*E**-1)
#Now substituting values in above equation we get

#dell_1=64*W*R**3*n*(d**4)**-1*1330*(10**9)**-1  (equation 1)

#For alpha=0 #Degree
#dell=64*W*R**3*n*sec(alpha)*(d**4)**-1*(cos(alpha)**2*G**-1+2*sin(alpha)**2*E**-1)
#Now substituting values in above equation we get

#dell_2=64*W*R**3*n*(d**4)**-1*1250*(10**9)**-1  (equation 2)

#subtracting equation 1 and equation 2 we get
#Let dell_1-dell_2=X
#X=64*W*R**3*n*(d**4)**-1*80*(10**9)

#Let Y=X*dell_1**-1*100
Y=80*1330**-1*100 #% under estimation of axial extension

#Result
print"% under estimation of axial extension is",round(Y,2)

% under estimation of axial extension is 6.02