# 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