Chapter5-Applications Of Energy Methods

Ex1-pg149

In [1]:
##calculate the Values of P and Q and small displacement component

## initialization of variables
import math
## part (b)
K1=2. ##N/mm (K1=E1A1/L1)
K2=3. ##N/mm (K2=E2A2/L2)
b1=400. ## mm (b1=h)
h=400. ## mm
b2=300. ##mm
u=30. ##mm
v=40. ##mm
## calculations
## Units convertion
K1=K1*10**3
K2=K2*10**3
b1=b1*10**-3
b2=b2*10**-3
h=h*10**-3
u=u*10**-3
v=v*10**-3
L1=math.sqrt(b1**2+h**2)
L2=math.sqrt(b2**2+h**2)
N1=math.sqrt((b1+u)**2+(h+v)**2)-L1
N2=math.sqrt((b1+u)**2+(h+v)**2)
N3=math.sqrt((b2-u)**2+(h+v)**2)-L2
N4=math.sqrt((b2-u)**2+(h+v)**2)
P=K1*(b1+u)*N1/N2-K2*(b2-u)*N3/N4
Q=K1*(h+v)*N1/N2+K2*(h+v)*N3/N4
## results
print('part (b)')
print"%s %.2f %s"%('\n P = ',P,' N')
print"%s %.2f %s"%('\n Q =',Q,' N')
part (b)

 P =  43.77  N

 Q = 112.37  N

Ex2-pg154

In [2]:
##calculate the small displacement components
## initialization of variables
import math
## Loads and stresses and dimensions
P=10. ##kN
Q=30. ##kN
S0=70. ##MPa
eps0=0.001
b1=400. ##mm
h=400. ##mm
b2=300. ##mm
A1=300. ##mm^2
A2=300. ##mm^2
## calculations
## Units convertion
P=P*10**3
Q=Q*10**3
S0=S0*10**6
b1=b1*10**-3
b2=b2*10**-3
h=h*10**-3
A1=A1*10**-6
A2=A2*10**-6
L1=math.sqrt(b1**2+h**2)
L2=math.sqrt(b2**2+h**2)
a=L1*(Q*b2+P*h)/(A1*S0*h*(b1+b2))
b=L2*(Q*b1-P*h)/(A2*S0*h*(b1+b2))
c=L1**2*eps0/(b1+b2)
d=L2**2*eps0/(b1+b2)
u=c*math.sinh(a)-d*math.sinh(b)
v=b2/h*c*math.sinh(a)+b1/h*d*math.sinh(b)
## results
print"%s %.2f %s"%('u = ',u*10**3,' mm')
print"%s %.2f %s"%('\n v =',v*10**3,' mm')
u =  0.47  mm

 v = 0.81  mm

Ex6-pg166

In [3]:
##calculate the deflection of the free end of the beam
## initialization of variables
import math
## Material constants
E=200. ##GPa
G=77.5 ## GPa
Lh=5. ## Lh = L/h
## part (b)
rhs1=1.8*Lh*E/G
rhs2=7.*12.*Lh**3./16.
LHS=1.8*Lh*E/G+7.*12.*Lh**3./16.
e=rhs1/LHS*100.
print"%s %.2f %s"%('The error in neglecting small terms is ',e,' per cent')
The error in neglecting small terms is  3.42  per cent

Ex7-pg167

In [4]:
##calculate the rotation of section B
## initialization of variables
import math
## Specifications
T=2. ##kN.m
E=72. ## G Pa
G=27. ## GPa
b=30. ##mm
h=40. ##mm
d=60. ##mm
l1=400. ##mm
l2=800. ##mm
## calculations
E=E*10**9
G=G*10**9
b=b*10**-3
h=h*10**-3
d=d*10**-3
l1=l1*10**-3
l2=l2*10**-3
T=T*10**3 ##N.m
Ix=b*h**3./12.
J=math.pi*d**4/32.
thB= 2.*l1**3./3.*0.001**2*T/(E*Ix)+T*l2/(G*J)
print"%s %.2f %s"%('The rotation of shaft B is th = ',thB,' rad')
## Wrong answer to an extent in the textbook
The rotation of shaft B is th =  0.05  rad

Ex9-pg168

In [5]:
##calculate the deflection of the free end of curved beam in he direction and error deflection
## initialization of variables
import math
## specification
R=65. ##mm
E=200. ##GPa
G=77.5 ##GPa
v=0.29
P=6.##kN
##calculations
R=R*10**-3
E=E*10**9
G=G*10**9
P=P*10**3
A=30**2*10**-6
I=30**4./12.*10**-12
q_p1=3*math.pi*P*R/(4.*E*A)+1.2*3.*math.pi*P*R/(4.*G*A)+(9.*math.pi/4.+2.)*P*R**3./(E*I)
print('part (a)')
print"%s %.2f %s"%('\n qp = ',q_p1*10**3,' mm')
##part (b)
## if Un and Us are neglected
q_p2=(9.*math.pi/4.+2.)*P*R**3./(E*I)
e=(q_p1-q_p2)/q_p1*100.
print('\n part (b)')
print"%s %.2f %s"%('\n error = ',e,' per cent')
part (a)

 qp =  1.13  mm

 part (b)

 error =  1.85  per cent

Ex10-pg169

In [6]:
##calculate the  effect of neglecting the strain energy Us due to shear and vertical deflection
## initialization of variables
import math
## part (b)
## Specifications
P=150. ##N
R=200. ##mm
d=20. ##mm
E=200. ##GPa
G=77.5 ##GPa
##calculations
R=R*10**-3.
d=d*10**-3.
E=E*10**9.
G=G*10**9.
r1=R+d/2.
r2=R-d/2.
A=314.*10**-6 
I=7850.*10**-12 ##m**4
Ax=3.*math.pi/4.*P*R/(E*A)
Sh=3.*math.pi/4.*1.33*P*R/(G*A)
M=(7.*math.pi/4.+1.)*P*R**3./(E*I)
##qc=3*%pi/4*P*R/(E*A)+3*%pi/4*1.33*P*R/(G*A)+(7*%pi/4+1)*P*R**3/(E*I)
qc=Ax+Sh+M
print"%s %.2f %s %.2f %s %.2f %s %.2f %s"%('qc = ',qc*10**3,' mm 'and 'among which due to Axial is ',Ax*10**3,' mm'and '',Sh*10**3, 'mm'and ' is due to shear, and ',M*10**3,' mm is due to moment')

print('\n which means The concentrations of axial loads and shear are negligible')
qc =  4.97 among which due to Axial is  0.00  0.00  is due to shear, and  4.97  mm is due to moment

 which means The concentrations of axial loads and shear are negligible

Ex12-pg174

In [7]:
##calculate the rotation of member BE caused by the loads P and Q
## initialization of variables
import math
## Material properties and dimensions
E=72. ##G Pa
P=10. ##kN
Q=5. ##kN
Aab=150. ##mm**2
Abc=900. ##mm**2
Acd=900. ##mm**2
Ade=900. ##mm**2
Abd=150. ##mm**2
Abe=150. ##mm**2
Lab=2. ##m
Lbc=2.5 ##m
Lbd=1.5 ##m
Lbe=2.5 ##m
Lcd=2. ##m
Lde=2. ##m
##calculations
E=E*10**9.
P=P*10**3.
Q=Q*10**3.
Aab=150.
Aab=Aab*10**-6.
Abc=Abc*10**-6.
Acd=Acd*10**-6.
Ade=Ade*10**-6.
Abd=Abd*10**-6.
Abe=Abe*10**-6.
M=0.
Nab=4./3.*(Q+2.*P)-5.*M/(3.*Lbe)
dNab=-5./(3.*Lbe)
Nbc=-5./3.*(Q+P)
dNbc=0.
Nbd=Q
dNbd=0.
Nbe=5.*P/3.-4./3.*M/Lbe
dNbe=-4./(3.*Lbe)
Ncd=-4.*P/3.+5./3.*M/Lbe
dNcd=5./(3.*Lbe)
Nde=Ncd
thBE=Nab*Lab*dNab/(E*Aab)+Nbc*Lbc*dNbc/(E*Abc)+Nbd*Lbd*dNbd/(E*Abd)+Nbe*Lbe*dNbe/(E*Abe)+2.*Ncd*Lcd*dNcd/(E*Lcd)
print"%s %.2f %s"%('The rotation of member BE is ',thBE,' rad')
## Wrong answer in the text
The rotation of member BE is  -0.01  rad