Ch:8 Springs

exa 8-1 - Page 224

In [1]:
d=5#
D=30#
G=84*(10**3)#
Na=15#
#Axial Load W
W=300#
#Spring index C
C=30/5#
#Shear stress Augmentation factor Ks
Ks=((2*C)+1)/(2*C)#
#Wahl's factor Kw
Kw=(((4*C)-1)/((4*C)-4))+(0.615/C)#
#Curvature correction factor Kc
Kc=Kw/Ks#
#Spring stiffness k
k=(G*(d**4))/(8*(D**3)*Na)#
#Axial deflection delta
delta=W/k#
print " Ks is %0.4f "%(Ks)#
print "\n Kw is %0.4f "%(Kw)#
print "\n Kc is %0.3f "%(Kc)#
print "\n The Spring Stiffness is %0.1f N/mm"%(k)#
print "\n The Axial deflection is %0.3f mm"%(delta)#

 Ks is 1.0000 

 Kw is 1.1025 

 Kc is 1.103 

 The Spring Stiffness is 16.0 N/mm

 The Axial deflection is 18.000 mm

exa 8-2 - Page 224

In [2]:
from math import pi
W=196.2#
lenthofscale=50#
k=196.2/50#
C=8#
Ks=(1+(0.5/C))#

# Let us choose oil tempered wire 0.6-0.7 %C. Refer to Table 8-4 for constants A and m, relating strength wire 
#diameter.
G=77.2*(10**3)#
A=1855#
m=0.187#
# equating Tmax=0.5*sig(ut).
# Ks*(8*W*D/(pi*(d**3)))=0.5*A/(d**2)
d1=(Ks*(8*W*C/(pi*A*0.5)))#
d=d1**(1/1.813)#
D=C*d#
Na=G*(d**4)/(8*(D**3)*k)#
#Solid length = SL
SL=(Na-1)*d

print " wire diameter is %0.3f mm "%(d)#
print "\n mean diameter is %0.3f mm "%(D)#
print "\n Number of acting coils are %0.3f "%(Na)#

#The difference in the values of d,D and Na is due to rounding-off the digits.

 wire diameter is 2.314 mm 

 mean diameter is 18.516 mm 

 Number of acting coils are 11.117

exa 8-3 - Page 225

In [3]:
from math import pi
d=1.626#
A=2211#
m=0.145#
rm=3#
ri=(rm-(d/2))#
sigma=A/(d**m)#
W=(sigma*pi*(d**3)*ri)/(32*(rm**2))#
print " Ultimate tensile Strength is %0.1f MPa "%(sigma)#
print "\n Force at which the spring hook fails is %0.1f N "%(W)#

#The difference in the values of sigma and W is due to rounding-off the digits.

 Ultimate tensile Strength is 2060.5 MPa 

 Force at which the spring hook fails is 211.3 N

exa 8-4 - Page 226

In [4]:
from math import pi
Do=25#
# mean coil diameter D=25-d
W=150#
T=800#
G=81000#
# Substituting values in equation T=8*W*D/(pi*(d**3))
# therefore, the equation becomes d**3 + 0.477*d = 11.936
#consider d=2.2mm, (d can be taken between 2.2-2.3mm)
d=2.337# #(nearest available wire gauge)
C=9.5#
D=22.2# 
Do=D+d#
Ks=1+(0.5/C)#
Tmax=Ks*8*W*D/(pi*(d**3))#
# check for safety- Tmax<T#
Lo=100#
Ls=40#
#Lo=Ls+delta+0.15*delta
delta=(Lo-Ls)/1.15#
delta=50#
k=150/50#
Na=(G*d**4)/(8*(D**3)*k)#

N=Na+2#
Ls=N*d#
Lo=Ls+(1.15*delta)#
print  " d is %0.3fmm "%(d)#
print  "\n D is %0.2f mm"%(D)#
print  "\n Ls is %0.2f mm"%(Ls)#
print  "\n Lo is %0.2f mm"%(Lo)#
if (Do<=25):
    print '\nThe diameter is within space constraints'

 d is 2.337mm 

 D is 22.20 mm

 Ls is 26.18 mm

 Lo is 83.68 mm

The diameter is within space constraints

exa 8-5A - Page 227

In [5]:
from math import pi
Di=15#
Do=20#
d=2.3#
D=17.5#
C=D/d#
Ks=1+(0.5/C)#
Wmax=100#
Tmax=Ks*8*Wmax*D/(pi*(d**3))#
G=81000#
delmax=67.7/2.366#
k=100/28#
Na=G*(d**4)/(8*k*(D**3))#
Ls=Na+1# #(for plain ends)
delmax=28#
#TL= total working length
TL=Ls+delmax+(0.15*delmax)#
print  " d is %0.1fmm "%(d)#
print  "\n C is %0.1f "%(C)#
print  "\n Na is %0.1f "%(Na)#

 d is 2.3mm 

 C is 7.6 

 Na is 17.6

exa 8-6 - Page- 228

In [6]:
from math import pi
# 18 SWG=1.219MM in dia
d=1.219#
E=198.6*10**3#
G=80.7*10**3#
m=0.19#
A=1783#
sig=A/(d**m)#
Tys=(0.4*sig)#
Do=12.5#
D=Do-d#
C=D/d#
Ks=((2*C)+1)/(2*C)#
W=(Tys*pi*(d**3))/(8*D*Ks)#
Nt=13.5#
Na=Nt-2#
Del=(8*W*(D**3)*Na)/(G*(d**4))#
Ls=(Nt-1)*d#
Lo=Ls+Del+(0.15*Del)#
print  " Tys is %0.1f MPa "%(Tys)#
print  "\n W is %0.1f N "%(W)#
print  "\n del is %0.3f mm "%(Del)#
print  "\n Ls is %0.4f mm "%(Ls)#
print  "\n Lo is %0.2f mm "%(Lo)#
  
#Answers in the book for Torsional yeild strength have been rounded-off to the nearest whole number.

 Tys is 686.9 MPa 

 W is 41.1 N 

 del is 30.457 mm 

 Ls is 15.2375 mm 

 Lo is 50.26 mm

exa 8-7 - Page 228

In [7]:
from math import pi
d=1.016#
A=2211#
m=0.145#
G=81000#
Nt=16#
Na=16-2#
sig=A/(d**m)#
Tys=0.45*sig#
Do=12.6#
D=Do-d#
C=D/d#
Ks=1+(0.5/C)#
W=(Tys*pi*(d**3))/(8*D*Ks)#
k=(G*(d**4))/(8*(D**3)*Na)#
Del=W/k#
Ls=(Nt-1)*d#
Lo=Ls+(1.15*Del)#

print  "Tys is %0.1f MPa "%(Tys)#
print  "\n Do is %0.1f N "%(Do)#
print  "\n W is %0.1f N "%(W)#
print  "\n k is %0.3f N "%(k)#
print  "\n del is %0.2f mm "%(Del)#
print  "\n Ls is %0.2f mm "%(Ls)#
print  "\n Lo is %0.3f mm "%(Lo)#
  
if ((Lo/D)>=5.26):
    print 'The spring will fail under buckling'


#Values after the decimal point has not been considered for answer of Torsional yeild strength in the book, whereas answers for deflection and free-length is different as entire value of variables is taken for calculation in the code.

Tys is 992.7 MPa 

 Do is 12.6 N 

 W is 33.8 N 

 k is 0.496 N 

 del is 68.20 mm 

 Ls is 15.24 mm 

 Lo is 93.669 mm 
The spring will fail under buckling

exa 8-8 - Page 229

In [8]:
from math import pi
d=2#
Do=20#
D=Do-d#
C=D/d#
Na=9#
#Material hard drawn spring steel
A=1783#
m=0.19#
G=81000#
sig=A/(d**m)#
Tys=0.45*sig
Kf=1.5#
Ta=Tys/Kf#
Ks=1+(0.5/C)#
W=(Ta*pi*(d**3))/(8*D*Ks)#
k=(G*(d**4))/(8*(D**3)*Na)#
Del=W/k#
Lo=((Na+1)*d)+(1.15*Del)#
p=(Lo-d)/Na#
print  "k is %0.3f N/mm "%(k)#
print  "\n W is %0.1f N "%(W)#
print  "\n Lo is %0.3f mm "%(Lo)#
print  "\n p is %0.3f mm "%(p)
  
  
if ((Lo)>=47.34):
    print 'The spring will fail under buckling'

#The answer for value of spring rate 'k' is misprinted in the book. Due to this all subsequent values of del,Lo,p is calucated incorrectly in the book.

k is 3.000 N/mm 

 W is 77.5 N 

 Lo is 49.720 mm 

 p is 5.302 mm 
The spring will fail under buckling

exa 8-9 - Page 230

In [9]:
from math import pi
# for music wire
d1=11.5#
A=2211#
d=1.5#
m=0.145#
sigut=A/(d**m)#
sigy=0.78*sigut#
Do=16#
E=2*(10**5)#
Nb=4.25#
D=Do-d#
C=D/d#
Ki=((4*(C**2))-C-1)/(4*C*(C-1))#
Mmax=(sigy*pi*(d**3))/(32*Ki)#
kc=((d**4)*E)/(10.8*D*Nb)#
theta3=Mmax/kc#
l1=20#
l2=20#
Ne=(l1+l2)/(3*pi*D)#
Na=Nb+Ne#
k=((d**4)*E)/(10.8*Na*D)#
thetat=Mmax/k#
ke=(3*pi*(d**4)*E)/(10.8*(l1+l2))#
# angdisp=theta1+theta2=Mmax/ke#
angdisp=Mmax/ke#
#D1 is final coil diameter
D1=(Nb*D)/(Nb+theta3)#
#IRC=Initial radial clearance
IRC=((D-d)-d1)/2#
#FRC=Final radial clearance
FRC=((D1-d)-d1)/2#



print " Maximum Torque is %0.2f Nmm "%(Mmax)#
print "\n theta3 is %0.3f turns "%(theta3)#
print "\n Ne is %0.3f turns "%(Ne)#
print "\n ke is %0.1f N/mm "%(ke)#
print "\n theta1+theta2 is %0.4f turns "%(angdisp)#
print "\n D1 is %0.2f mm "%(D1)#
print "\n IRC is %0.2f mm "%(IRC)#
print "\n FRC is %0.2f mm "%(FRC)#

 Maximum Torque is 497.25 Nmm 

 theta3 is 0.327 turns 

 Ne is 0.293 turns 

 ke is 22089.3 N/mm 

 theta1+theta2 is 0.0225 turns 

 D1 is 13.46 mm 

 IRC is 0.75 mm 

 FRC is 0.23 mm

exa 8-10 - Page 231

In [10]:
from math import pi
A=1783#
m=0.190#
d=1.5#
D=15#
M=300#
E=20800#
k=30#
#sigult= ultimate strength of the material
# sigy= yield strength of the material
sigult=A/(d**m)#
sigy=0.7*sigult#
#siga= allowable yield strength of the material
siga=sigy/2#
C=D/d#
Ki=(4*(C**2)-C-1)/(4*C*(C-1))#
Z=pi*(d**3)/32#
#sigb=bending strength of the material#
sigb=Ki*M/Z#
while (sigb>=siga) :
    d=d+0.15#
    D=15#
    C=D/d#
    sigult=A/(d**m)#
    sigy=0.7*sigult#
    siga=sigy/2#
    Ki=(4*(C**2)-C-1)/(4*C*(C-1))#
    Z=pi*(d**3)/32#
    sigb=Ki*M/Z#

d=2## rounding off the value of the diameter.
Na=(d**4)*E/(64*D*k)#
print  " d is %0.1f mm "%(d)#
print  "\n D is %0.1f mm "%(D)#
print  "\n Na is %0.2f mm "%(Na)#

 d is 2.0 mm 

 D is 15.0 mm 

 Na is 11.00 mm

exa 8-11 - Page 231

In [11]:
from math import pi
L=1180#
W=40*(10**3)#
Nf=2#
Ng=8#
E=207*(10**3)#
#sigut is ultimate strength
sigut=1400#
FOS=2#
#siga= allowable yield strength of the material
siga=1400/2#
#sigbf=bending strength in full length
sigbf=700#
b=75#
t=((4.5*W*L)/(((3*Nf)+(2*Ng))*sigbf))**(0.5)#
t=14#
I=(Nf*b*(t**3))/12#
Wf=(3*Nf*W)/((3*Nf)+(2*Ng))#
Del=(Wf*(L**3))/(48*E*I)#
print " t is %0.0f mm "%(t)#
print "\n Wf is %0.0f N "%(Wf)#
print "\n I is %0.0f mm**4 "%(I)#
print "\n del is %0.1f mm "%(Del)#

 t is 14 mm 

 Wf is 10909 N 

 I is 34300 mm**4 

 del is 52.0 mm

exa 8-12A - Page 232

In [12]:
W=80000#
sigbfr=500#
L=1100#
Nf=3#
Ng=10#
N=Nf+Ng#
t=((1.5*W*L)/(N*6*sigbfr))**(1/3)#
t=15#
b=6*t#
E=207*10**3#
deli=(W*(L**3))/(8*E*N*b*(t**3))#
Wi=(W*Nf*Ng)/(N*((3*Nf)+(2*Ng)))#
print " t is %0.1f mm "%(t)#
print "\n deli is %0.1f mm "%(deli)#
print "\n Wi is %0.0f N "%(Wi)#

 t is 15.0 mm 

 deli is 16.0 mm 

 Wi is 6366 N

exa 8-13 - Page 233

In [13]:
from math import pi
#ultimate strength=sigut
sigut=1500#
C=7#
d=3#
D=C*d#
Ks=1+(0.5/C)#
Kw=(((4*C)-1)/((4*C)-4))+(0.615/C)#
Pmax=120#
Pmin=40#
Pm=80#
Tm=(Ks*8*Pm*D)/(pi*(d**3))#
Ta=(Kw*8*Pmin*D)/(pi*(d**3))#
Tse=0.22*sigut#
Tys=0.45*sigut#
x=(Tys-(0.5*Tse))/(0.5*Tse)#
y=((x)*Ta)+Tm#
FOS=(Tys/y)#
print  " Tm is %0.2f MPa "%(Tm)#
print  "\n Ta is %0.1f MPa "%(Ta)#
print  "\n FOS is %0.3f "%(FOS)#

 Tm is 169.77 MPa 

 Ta is 86.2 MPa 

 FOS is 1.548

exa 8-14 - Page 234

In [14]:
from math import pi
Tse=360#
Tys=660#
d=25#
P=0.03#
m=40#
Pmin=((pi*(d**2)*P)/4)+(m*9.81/1000)#
k=6#
#Additional load= Padd=k*further compression in spring
Padd=k*10#
Pmax=Padd+Pmin#
Pm=(Pmax+Pmin)/2#
Pa=(Pmax-Pmin)/2#
d=2#
D=12#
C=6#
Ks=1+(0.5/C)#
Ks=1.083#
Kw=(((4*C)-1)/((4*C)-4))+(0.615/C)#
Ta=(Kw*8*Pa*D)/(pi*(d**3))#
Tm=(Ks*8*Pm*D)/(pi*(d**3))#
x=(Tys-(0.5*Tse))/(0.5*Tse)#
y=((x)*Ta)+Tm#
FOS=(Tys/y)#
print  " Tm is %0.2f MPa "%(Tm)#
print  "\n Ta is %0.3f MPa "%(Ta)#
print  "\n FOS is %0.2f "%(FOS)#

 Tm is 186.64 MPa 

 Ta is 126.337 MPa 

 FOS is 1.26