Ch : 3 Mechanics of solids

exa 3-1 - Page 72

In [8]:
from math import pi, sqrt
d=10 
l=1500 
m=12 
h=50 
E=210*10**3 
sigut=450 
A=pi*d**2/4 
W=m*9.81 
sigi=W/A*(1+sqrt(1+(2*E*A*h)/(W*l))) 
deli=sigi*l/E 
siggradual=W/A 
sigsudden=2*siggradual 
print " sigi is %0.2f N/mm**2     "%(sigi) 
print "\n deli is %0.2f mm     "%(deli) 
print "\n siggradual is %0.2f N/mm**2     "%(siggradual) 

# The difference in the answer of sigi and siggradual is due to round-off errors.

 sigi is 146.37 N/mm**2     

 deli is 1.05 mm     

 siggradual is 1.50 N/mm**2

exa 3-2 - Page 73

In [9]:
from math import pi, sqrt
d=5 
A=pi*d**2/4 
l=100*10**3 
W=600 
E=210*10**3 
w=0.0784*10**-3 
del1=W*l/(A*E) 
del2=w*l**2/(2*E) 
Del=del1+del2 
print "del is %0.2f mm     "%(Del) 

del is 16.42 mm

exa 3-3 - Page 73

In [10]:
from math import pi, sqrt
m=25 
v=3 
E=210*10**3 
KE=0.5*m*v**2 
d=30 
L=2000 
A=pi*d**2/4 
U=A*L/(2*E) 
Del=4*10**-5*A 
W=A*Del 
sigi=sqrt(KE*10**3/(W+U)) 
print "del is %0.2f N/mm**2     "%(sigi) 

del is 69.41 N/mm**2

exa 3-4 - Page 74

In [11]:
P=40*10**3 
A=60*18 
sig=P/A 
r1=12 
b1=60 
SCF1=1.7 
sigmax1=sig*SCF1 
r2=24 
b2=60 
SCF2=1.5 
sigmax2=sig*SCF2 
print "sigmax1 is %0.2f N/mm**2     "%(sigmax1) 
print "\nsigmax2 is %0.2f N/mm**2     "%(sigmax2) 

sigmax1 is 62.96 N/mm**2     

sigmax2 is 55.56 N/mm**2

exa 3-5 - Page 75

In [12]:
from math import pi, sqrt
p=2.4 
#Let axial movement of nut be La
La=p*45/360 
d=20 
D=30 
L=500 
d1=18 
As=pi*d1**2/4 
Ac=pi*(D**2-d**2)/4 
sigt=120/(3.543) 
sigb=1.543*sigt 
print "sigt is %0.2f N/mm**2     "%(sigt) 
print "\nsigb is %0.2f N/mm**2     "%(sigb) 

sigt is 33.87 N/mm**2     

sigb is 52.26 N/mm**2

exa 3-6 - Page 76

In [13]:
from __future__ import division
delT=100 
ab=18*10**-6 
aa=23*10**-6 
delta=(360*ab*delT)+(450*aa*delT) 
lc=delta-0.6 
Ea=70*10**3 
Eb=105*10**3 
Aa=1600 
Ab=1300 
P=lc/((360/(Ab*Eb))+(450/(Aa*Ea))) 
P=P*10**-3 
#Let the change in length be delL
delL=(aa*450*delT)-(P*10**3*450/(Aa*Ea)) 
print " P is %0.2f kN     "%(P) 
print "\n delL is %0.2f mm     "%(delL) 
  
# The difference in the answer of delL is due to round-off errors.

 P is 162.73 kN     

 delL is 0.38 mm

exa 3-7 - Page 77

In [14]:
a=23*10**-6 
E=70*10**3 
l=750 
sig=35 
delT=((sig*l/E)+0.8)/(l*a) 
print "delT is %0.2f degC     "%(delT) 

delT is 68.12 degC

exa 3-8 - Page 78

In [15]:
OA=60 
AB=30 
OC=-20 
CD=-30 
theta=30 
angBEK=2*theta 
OM=14 
KM=49.5 
p1=70 
p2=-30 
angBEH=-37 
angBEI=143 
theta1=angBEH/2 
theta2=angBEI/2 
Tmax=50 
angBEL=53 
angBEN=233 
theta3=angBEL/2 
theta4=angBEN/2 
print " Stress on plane AB is %0.2f MPa     "%(OM) 
print "\n Stress on plane AB is %0.2f MPa     "%(KM) 
print "\n Principal stress p1 is %0.2f MPa     "%(p1) 
print "\n Principal stress p2 is %0.2f MPa     "%(p2) 
print "\n Principal angle theta1 is %0.2f deg     "%(theta1) 
print "\n Principal angle theta2 is %0.2f deg     "%(theta2) 
print "\n Maximum shear stress is %0.2f MPa     "%(Tmax) 
print "\n Direction of plane theta3 is %0.2f deg     "%(theta3) 
print "\n Direction of plane theta4 is %0.2f deg     "%(theta4) 

#The answers in the book are written in form of degrees and minutes.

 Stress on plane AB is 14.00 MPa     

 Stress on plane AB is 49.50 MPa     

 Principal stress p1 is 70.00 MPa     

 Principal stress p2 is -30.00 MPa     

 Principal angle theta1 is -18.50 deg     

 Principal angle theta2 is 71.50 deg     

 Maximum shear stress is 50.00 MPa     

 Direction of plane theta3 is 26.50 deg     

 Direction of plane theta4 is 116.50 deg

exa 3-9 - Page 78

In [16]:
E=200*10**3 
v=0.29 
E1=720*10**-6 
E2=560*10**-6 
p1=121.76 
p2=-76.69 
print "p1 is %0.2f MN/mm**2     "%(p1) 
print "\n p2 is %0.2f MN/mm**2     "%(p2) 

p1 is 121.76 MN/mm**2     

 p2 is -76.69 MN/mm**2

exa 3-10 - Page 79

In [17]:
from math import pi, sqrt
G=38*10**3 
d=10 
P=5*10**3 
A=pi*d**2/4 
sig=P/A 
deld=0.0002 
#Let the lateral strain be E1
E1=deld/d 
v=2*deld*G/(sig-(2*deld*G)) 
E=2*G*(1+v)*10**-3 
print "v is %0.4f     "%(v) 
print "\nE is %0.3f kN/mm**2     "%(E) 

v is 0.3136     

E is 99.837 kN/mm**2

exa 3-11 - Page 79

In [18]:
D=1500 
p=1.2 
sigt=100 
sigc=p*D/2 
siga=p*D/4 
P=sigc*2*10**3 
n=0.75 
t=sigc/(n*sigt) 
print "t is %0.1f mm    "%(t) 

t is 12.0 mm

exa 3-12 - Page 80

In [19]:
D=50 
t=1.25 
d=0.5 
n=1/d 
p=1.5 
siga=p*D/(4*t) 
sigc=20.27 
sigw=sigc/0.31416 
print "sigw is %0.2f N/mm**2    "%(sigw) 

sigw is 64.52 N/mm**2

exa 3-13 - Page 81

In [20]:
from math import pi, sqrt
R1=50 
p=75 
pmax=125 
R2=sqrt((pmax+p)*R1**2/(pmax-p)) 
t=R2-R1 
print "t is %0.1f mm    "%(t) 

t is 50.0 mm

exa 3-14 - Page 82

In [21]:
from math import pi, sqrt
R1=40 
R2=60 
B=50 
E=210*10**3 
e=41*10**-6 
sig=2*R1**2/(R2**2-R1**2) 
p=E*e/sig 
Fr=p*2*pi*R1*B 
u=0.2 
Fa=u*Fr*10**-3 
print "Fa is %0.2f kN    "%(Fa) 

Fa is 13.52 kN

exa 3-15 - Page 83

In [22]:
a1=10*1.5 
x1=15-0.75 
a2=1.5*(15-1.5) 
x2=(15-1.5)/2 
y1=((a1*x1)+(a2*x2))/(a1+a2) 
y2=a1-y1 
Ixx=(10*1.5**3)/12+(10*1.5*(5.06-1.5/2)**2)+(1.5*13.5**3/12)+(1.5*13.5*(9.94-6.75)**2) 
Z1=Ixx/y1 
Z2=Ixx/y2 
L=3 
sigc=50 
W=sigc*Z1/L*10**-3 
print "W is %0.3f kN    "%(W) 

W is 1.333 kN

exa 3-16 - Page 83

In [23]:
from math import pi, sqrt
D=22 
d=20 
r=1 
K=2.2 
sigmax=130 
sigmax=sigmax/K 
Z=pi*d**3/32 
M=sigmax*Z*10**-3 
print "M is %0.3f Nm    "%(M) 

M is 46.410 Nm

exa 3-17 - Page 84

In [24]:
from math import pi, sqrt
A=(12*2)+(12*2)+(30-4) 
B=sqrt(A/2) 
D=2*B 
B1=12 
D1=30 
d=26 
b=1 
Z1=((B1*D1**3)-((B1-b)*d**3))/(B1*D1/2) 
Zr=B*D**2/6 
#Let the ratio of both the sections be x
x=Z1/Zr 
M=30*10**6 
sigmax=M/(Z1*10**3) 
print "Z1/Zr is %0.2f     "%(x) 
print "\nsigmax is %0.2f N/mm**2    "%(sigmax) 

Z1/Zr is 4.84     

sigmax is 41.33 N/mm**2

exa 3-19 - Page 85

In [25]:
from math import pi, sqrt
x1=((13*3*1.5)+(2*15*8))/(39+30) 
x2=13-x1 
A=30+39 
E=2*10**7 
Iyy=995.66 
e=54.32 
x=x2-3 
sigb=e*x/Iyy 
sigd=1/69 
sigr=sigd+sigb 
#Let the strain be E1
E1=800*10**-6 
P=E1*E/sigr 
P=P*10**-3 
print "P is %0.2f kN    "%(P) 

P is 49.38 kN

exa 3-20 - Page 86

In [26]:
from math import pi, sqrt
H=20 
D=5 
d=3 
rho=21 
sigd=rho*H 
p=2 
A=D*H 
P=p*A 
M=P*H/2 
Z=pi*(D**4-d**4)/(32*D) 
sigb=M/Z 
sigmax=420+sigb 
sigmin=420-sigb 
print "sigmax is %0.2f kN/m**2    "%(sigmax) 
print "\nsigmin is %0.2f kN/m**2    "%(sigmin) 

sigmax is 607.24 kN/m**2    

sigmin is 232.76 kN/m**2

exa 3-21 - Page 87

In [27]:
from math import pi, sqrt
D=30 
R=15 
T=0.56*10**6 
G=82*10**3 
J=pi*R**4/2 
T1=T*R/J 
l=1000 
theta=T*l/(G*J)*180/pi 
r=10 
Tr=T1*r/R 
print " T1 is %0.2f N/mm**2    "%(T1) 
print "\n theta is %0.2f deg    "%(theta) 
print "\n Tr is %0.2f N/mm**2    "%(Tr) 

 T1 is 105.63 N/mm**2    

 theta is 4.92 deg    

 Tr is 70.42 N/mm**2

exa 3-22 - Page 87

In [28]:
from math import pi, sqrt
T=8*10**3 
d=80 
D=110 
l=2000 
Gst=80*10**3 
Gcop=Gst/2 
Js=pi*d**4/32 
Jc=pi*(D**4-d**4)/32 
#Ts=0.777*Tc
Tc=T/1.777*10**3 
Ts=0.777*Tc 
Ts1=Ts/Js*d/2 
Tc1=Tc/Jc*D/2 
#Let tl be Angular twist per unit length
tl=Ts*10**3/(Js*Gst)*180/pi 
# Let the maximum stress developed when the Torque is acting in the centre of the shaft be Ts2 & Tc2 resp. for steel and copper
Ts2=Ts1/2 
Tc2=Tc1/2 
print " Ts1 is %0.3f N/mm**2    "%(Ts1) 
print "\n Tc1 is %0.1f N/mm**2    "%(Tc1) 
print "\n theta/length is %0.3f deg/m    "%(tl) 
print "\n Ts2 is %0.3f N/mm**2    "%(Ts2) 
print "\n Tc2 is %0.2f N/mm**2    "%(Tc2) 

 Ts1 is 34.796 N/mm**2    

 Tc1 is 23.9 N/mm**2    

 theta/length is 0.623 deg/m    

 Ts2 is 17.398 N/mm**2    

 Tc2 is 11.96 N/mm**2

exa 3-23 - Page 88

In [29]:
from math import pi, sqrt
D=100 
d=75 
r=6 
K=1.45 
P=20*746 
N=400 
w=2*pi*N/60 
T=P/w 
Ts=16*T*10**3/(pi*d**3) 
Tmax=K*Ts 
print "Tmax is %0.3f MPa    "%(Tmax) 

Tmax is 6.235 MPa

exa 3-24 - Page 88

In [30]:
from math import pi, sqrt
G=84*10**3 
T=28*10**3 
l=1000 
theta=pi/180 
J=T*l/(G*theta) 
d=(J*32/pi)**(1/4) 
print "d is %0.1f mm    "%(d) 

d is 21.0 mm

exa 3-25 - Page 89

In [31]:
from math import pi, sqrt
P=2*10**6 
N=200 
w=2*pi*N/60 
Tm=P/w 
W=5*10**3*9.81 
l=1800 
Mmax=W*l/4 
Tmax=1.8*Tm*10**3 
Me=(Mmax+sqrt(Mmax**2+Tmax**2))/2 
Te=sqrt(Mmax**2+Tmax**2) 
sig=60 
Ts=40 
d1=(32*Me/(pi*sig))**(1/3) 
d2=(16*Te/(pi*Ts))**(1/3) 
print "d is %0.1f mm    "%(d2) 

d is 280.5 mm

exa 3-26 - Page 90

In [32]:
from math import pi, sqrt
Q=4*10**3 
P=8*10**3 
sig=P 
T=Q 
p1=(sig/2+sqrt((sig/2)**2+T**2)) 
p2=(sig/2-sqrt((sig/2)**2+T**2)) 
sigyp=285 
FOS=3 
siga=sigyp/3 
A1=p1/siga 
d1=sqrt(4*A1/pi) 
A2=(p1-p2)*2/(siga*2) 
d2=sqrt(4*A2/pi) 
v=0.3 
A3=sqrt(p1**2+p2**2-(2*v*p1*p2))/siga 
d3=sqrt(4*A3/pi) 
print " d1 is %0.2f mm    "%(d1) 
print "\n d2 is %0.1f mm    "%(d2) 
print "\n d3 is %0.2f mm    "%(d3) 

 d1 is 11.38 mm    

 d2 is 12.3 mm    

 d3 is 11.74 mm

exa 3-27 - Page 91

In [33]:
from math import pi, sqrt
sigx=-105 
Txy=105 
sigy=270 
p1=(sigx/2+sqrt((sigx/2)**2+Txy**2)) 
p2=(sigx/2-sqrt((sigx/2)**2+Txy**2)) 
p3=0 
Tmax=(p1-p2)/2 
siga=sigy/2 
if (Tmax<=siga) :
    print "The component is safe"
print "\nTmax is %0.1f MPa    "%(Tmax) 

The component is safe

Tmax is 117.4 MPa

exa 3-28 - Page 91

In [34]:
from math import pi, sqrt
rho=0.0078*9.81*10**-6 
sigc=150 
g=9.81 
V=sqrt(sigc*g/rho)*10**-3 
R=1 
w=V/R 
N=w*60/(2*pi) 
print "N is %0.3f rpm    "%(N) 

N is 1324.249 rpm

exa 3-29 - Page 92

In [35]:
from math import pi, sqrt
%matplotlib inline
from matplotlib.pyplot import plot, xlabel, ylabel, show, grid

R1=50 
R2=200 
N=6*10**3 
w=2*pi*N/60 
v=0.28 
rho=7800*10**-9 
g=9810 
k1=(3+v)/8 
k2=(1+(3*v))/8 
W=rho*9.81 
x=k1*w**2*W*(R1**2+R2**2)/g 
y=k1*w**2*W*(R1*R2)**2/g 
y1=k1*w**2*W/g 
z=k2*w**2*W/g 
r=sqrt(R1*R2) 
sigrmax=x-(y/r**2)-(r**2*y1) 
r=range(50,201)
n=len(r) 
sigr = range(0,n)
for i in range(0,n):
    sigr[i]=x-(y/r[i]**2)-(r[i]**2*y1)

sigc = range(0,n)
for j in range(0,n):
    sigc[j]=y/r[j]**2-(r[j]**2*z)

plot(r,sigr) 
plot (r,sigc) 
xlabel('r in mm') 
ylabel('stress in N/mm**2') 
grid()
show()
print "sigrmax is %0.1f MPa  "%(sigrmax) 

sigrmax is 28.4 MPa

exa 3-30 - Page 93

In [36]:
from numpy import exp
r=500 
to=15 
N=3500 
w=2*pi*N/60 
sig=80 
w1=0.07644*10**-3 
g=9810 
a=w1*w**2*r**2/(2*sig*g) 
t=to*exp(-a) 
print "t is %0.3f mm  "%(t) 

t is 2.923 mm

exa 3-31 - Page 93

In [37]:
from numpy import log
M=60*10**3 
y1=((5*1*2.5)+(6*1*5.5))/(5+6) 
y2=6-y1 
R=12 
R1=R-y2 
R1=10.136
R2=11.136 
R3=R1+6 
B=6 
b=1 
A=(B*b)+((B-1)*b) 
#Let x= h**2/R**2
x=R/A*((B*log(R2/R1))+(b*log(R3/R2)))-1 
x=1/x 
#Let Maximum compressive stress at B be sigB
sigB=M/(A*R)*(1+(x*y1/(R+y1)))*10**-2 
#Let Maximum tensile stress at A be sigA
sigA=M/(A*R)*((y2*x/(R-y2))-1)*10**-2 
print "sigB is %0.1f MPa     "%(sigB) 
print "\nsigA is %0.0f MPa     "%(sigA) 
  
#The answer to R**2/h**2 is calculated incorrectly in the book.

sigB is 61.5 MPa     

sigA is 36 MPa

exa 3-32 - Page 94

In [38]:
from numpy import log
R1=24 
R2=30 
R3=50 
R4=54 
F=200 
y1=((16*4*2)+(2*20*14*4)+(24*6*27))/((16*4)+(2*20*4)+(24*6)) 
y2=30-y1 
R=24+y2 
A=(24*6)+(2*4*20)+(4*16) 
#Let x= h**2/R**2
x=R/A*((24*log(R2/R1))+(2*4*log(R3/R2))+(16*log(R4/R3)))-1 
x=1/x 
M=F*(60+R) 
sigd=F/A 
#Let bending stress at a be sigA
sigA=M/(A*R)*((y2*x/(R-y2))-1) 
#Let bending stress at b be sigB
sigB=M/(A*R)*(1+(x*y1/(R+y1))) 
#Let resultant at a be Ra
Ra=(sigA+sigd)*10 
#Let resultant at b be Rb
Rb=(sigB-sigd)*10 
print "Ra is %0.2f N/mm**2     "%(Ra) 
print "\nRb is %0.2f N/mm**2      "%(Rb) 
#The difference in the answers are due to rounding-off of values.

Ra is 96.70 N/mm**2     

Rb is 70.14 N/mm**2

exa 3-33 - Page 95

In [39]:
from __future__ import division
from numpy import log
F=50 
B1=4 
B2=8 
D=12 
y1=D/3*(B1+(2*B2))/(B1+B2) 
y2=12-y1 
R=6+y2 
A=(B1+B2)/2*D 
#Let x= h**2/R**2
a=(B1+((B2-B1)*(y1+R)/D))*log((R+y1)/(R-y2))
x=R/(A)*(a -(B2-B1)) 
x=x-1 
x=1/x 
KG=y2+8 
M=F*KG 
sigd=F/A 
#Let bending stress at a be sigA
sigA=M/(A*R)*(1+(x*y1/(R+y1))) 
#Let bending stress at b be sigB
sigB=M/(A*R)*((y2*x/(R-y2))-1) 
sigA=(sigA-sigd)*10 
sigB=(sigB+sigd)*10 
print "sigA is %0.2f MPa     "%(sigA) 
print "\nsigB is %0.2f MPa     "%(sigB) 
#The difference in the answers are due to rounding-off of values.

sigA is 31.59 MPa     

sigB is 71.64 MPa