import math
#Initilization of variables
k=1 #KN/m #stiffness of spring
P=45 #N #Maximum Load
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,
R=0.175*10**6*pi*(d)**3*10**-6*100 #cm #Radius of coil
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"
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
R=D*2**-1 #cm #Mean coil Radius
#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
#P=549.7787*10**6*d**3 #Load
#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[0])**2)**-1 #no.of coils
#Result
print"Diameter of steel wire is",d[0],"m"
print"number of coils",n
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
R=D*2**-1 #cm #Mean Radius
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
P=k*10**3*dell*10**-2 #N #Load
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 Load applied is",round(P,2),"N"
print"Max shear stress is",round(sigma_s_max,0),"N/m**2"
import math
#Initilization of variables
G=80 #GPa #Modulus of rigidity
P=1 #KN #Load
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)
import math
#Initilization of variables
d=6 #mm #Diameter of steel wire
n=50 #number of turns
D=5 #cm #Mean Diameter
R=D*2**-1 #cm #Radius of coil
G=80 #GPa #Modulus of Rigidity
P=150 #KN #Load
#Calculation
#Dell=64*P*R**3*n*(G*d**4)**-1 #Deflection
#After substituting values in above equation and simplifying we get
#P=2073.6*dell #Gradually applied equivalent Load
#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"
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),
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
P=pi*d_1**3*sigma_s*10**6*(16*R*10**-2)**-1 #N #Limiting Load
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"
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"
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
R=d_2*2**-1 #spring radius
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"
import math
#Initilization of variables
L=75 #cm #Legth of Leaf spring
P=8 #KN #Load
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
R=(L*10**-2)**2*(8*y_c*10**-3)**-1 #m #Radius of spring
#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)
print"The Radius of plate is",round(R,2)
import math
#Initilization of variables
L=75 #cm #span of laminated steel spring
P=7.5 #KN #Load
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
R=(L*10**-2)**2*(8*y_c*10**-2)**-1 #Radius of curvature
#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"
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)
print"Radius of curvature is",round(R,3)
import math
#Initilization of variables
P=200 #N #Load
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
R=(L*10**-2)**2*(8*dell_1)**-1 #Radius of curvature
#Result
print"Max instantaneous stress in plates is",round(sigma,2),"MPa"
print"Radius of curvature of spring is",round(R,2),"m"
import math
#Initilization of variables
L=70 #cm #Length of Longest plate
P=3.5 #KN #central Load
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
R=200*10**9*6.465*10**-3*(2*190*10**6)**-1 #Radius of Curvature
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)
print"The Radius of curvature is",round(R,3),"m"
import math
#Initilization of variables
alpha=30 #degree #helix angle
dell=2.3*10**-2 #m #Vertical displacement
W=40 #N #Axial Load
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
R=Z*10**2 #cm #Mean Radius
#Result
print"Mean Radius of Open coiled spring of helix angle is",round(R,2),"cm"
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
W=33140.016*d #N #MAx Permissible Load
#Result
print"The Max Permissible Load is",round(W,2),"N"
print"Thw Wire Diameter is",round(d,6),"m"
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"The axial Load is",round(W,2),"N"
print"Necesscary torque is",round(T,2),"N*m"
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=150 #Load
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,
P=dell*10**-2*G*10**9*(d*10**-3)**4*(64*(R*10**-2)**3*n)**-1 #N #Axial Load
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"Axial Load is",round(P,2),"N"
print"Height from which 150 N load falls is",round(h,2),"cm"
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)