# Chapter 14:Energy Methods¶

## Example 14.1 Page No 721¶

In [29]:
#Given
sigma_y = 44        #stress ,ksi
db =0.731           #inch, diameter
import math
Ab = math.pi*(rb**2)
E = 29*10**3    #N/mm**2, stress
da1 = 0.875     #inch, diameter
ra1 = da1/2.0
La1 = 2           #inch
La2= 0.25            #inch
da2 =0.731          #inch
ra2 = da2/2.0
Lb = 2.25         #inch

#Calculation
#Bolt A
Aa2 = math.pi*(ra2**2)
Aa1 = math.pi*(ra1**2)
P_max = sigma_y*Ab
Uia = (P_max**2/(2*E))*(La1/Aa1 + La2/Aa2) #Ui = (N**2L)/(2AE)
#Bolt B
Uib = (P_max**2/(2*E))*(Lb/Ab)

#Display
print'The greatest amount of strain energy absorbed by bolt A    = ',round(Uia,4),"J"
print'The greatest amount of strain energy absorbed by bolt B    = ',round(Uib,4),"J"

0.0230568851609
The greatest amount of strain energy absorbed by bolt A    =  0.0231 J
The greatest amount of strain energy absorbed by bolt B    =  0.0315 J


## Example 14.5 Page No 728¶

In [2]:
#Given
G = 75*10**9    #N/m**2, stress
ro = 80/1000.0  #m, outside radius
t = 15/1000.0   #m, thickness
ri = ro - t     #inside radius
l1 = 750/1000.0   #m, length
l2 = 300/1000.0   #m
T1 = 40         #Nm. torque
T2 =15          #Nm

#Calculations
import math
J = (math.pi/2.0)*(ro**4 - ri**4)
#Eqn 14-22
U1 = (T1**2*l1)/(2*G*J)
U2 = (T2**2*l2)/(2*G*J)
Ui = U1 + U2
Ui = Ui*10**6    #in micro Joule

#Display
print'The strain energy stored in the shaft    = ',round(Ui,0),"micro J"

The strain energy stored in the shaft    =  233.0 micro J


## Example 14.6 Page No 735¶

In [3]:
#Given:
F=5.0       #kip, horizontal distance
A=0.20    #inch**2, area
E=29*10**3   #ksi, stress
Nab=2.89   #kip, normal stress
Nac=5.77   #kip
L1=2         #ft, length
L2=4       #ft
L3=3.46     #ft

#calculation
#Appling equation 14-24
dBh=(Nab**2*L1/(2*A*E)+(-Nac)**2*L2/(2*A*E)+F**2*L3/(2*A*E))*(2/F)

#result
print"The horizontal displacement is",round(dBh*12,4),"inch"

The horizontal displacement is 0.0978 inch


## Example 14.8 Page No 743¶

In [3]:
#Given
ro = 3         #inch, outside radius
ri = 2.5       #inch, inside radius
E = 10*10**3   #ksi, stress
W = 150        #kip, force
L = 12         #inch, length
h = 0

#Part a
import math
A = (math.pi)*(ro**2 - ri**2)
del_st= (W*L)/(A*E)
#Part b
del_max = del_st*(1 + math.sqrt(1 + 2*(h/del_st)))

#Display
print'The maximum displacement at the top of the pipe for gradually applied load   = ',round(del_st,4),"inch"
print'The maximum displacement at the top of the pipe for suddenly applied load    = ',round(del_max,4),"inch"

The maximum displacement at the top of the pipe for gradually applied load   =  0.0208 inch
The maximum displacement at the top of the pipe for suddenly applied load    =  0.0417 inch


## Example 14.9 Page No 744¶

In [22]:
#Given
W = 1.5           #kip, force
h = 2             #inch, height
E = 29*1000       #N/mm**2, stress
L = 16           #ft, length
I = 209          #inch**2, area

#Calculations
import math
del_st = (W*L**3*12**3)/(48*E*I)
del_max = del_st*(1 + math.sqrt(1 + 2*(h/del_st)))
c = 9.92/2.0
Pmax=48*E*I/(L**3*12**3)
Mmax=Pmax*L/4.0
sigma_max = (12*E*del_max*c)/(L**2*12**2)

#Display
print'The maximum bending stress in the steel beam   = ',round(sigma_max,1),"ksi"
print'The maximum deflection in the beam             = ',round(del_max,2),"inch"

The maximum bending stress in the steel beam   =  19.7 ksi
The maximum deflection in the beam             =  0.42 inch


## Example 14.10 Page No 745¶

In [16]:
#Given
m = 80*1000 #kg
v = 0.2           #m/s
l_ac = 1.5        #m
E = 200*10**9     #N/m**2
w = 0.2           #m
I = (1/12.0)*(w**4)
l_ab = 1000      #mm

#Calculations
import math
del_Amax = math.sqrt((m*v**2*l_ac**3)/(3*E*I))

P_max = (3*E*I*del_Amax)/(l_ac**3)
theta_A = (P_max*l_ac**2)/(2*E*I)
del_Amax = del_Amax*1000
del_Bmax = del_Amax + (theta_A*l_ab)

#Display
print'The maximum horizontal displacement of the post at B due to impact   =',round(del_Bmax,2),"mm"


The maximum horizontal displacement of the post at B due to impact   = 23.24 mm


## Example 14.11 Page No 758¶

In [6]:
#Given
A = 400*10**-6 #m**2
E = 200*10**6 #kN/m**2
P = 100 #kN

#Virtual Work Equation
n1 = 0
n2=0
n3=-1.414
n4=1
N1=-100
N2=-141.4
N3=-141.4
N4=200
L1=4
L2=2.828
L3=2.828
L4=2
Sum=n1*N1*L1+n2*N2*L2+n3*N3*L3+n4*N4*L4
del_cv = Sum/(A*E)

#Display
print'The vertical displacement of joint C of the steel truss  ',round(del_cv*1000,1),"mm"

The vertical displacement of joint C of the steel truss   12.1 mm


## Example 14.12 Page no 759¶

In [8]:
#given
A=250*10**(-6)    #m**2
E=200*10**6            #pa
alpha=12*10**(-6)
l=4               #m
l1=-1.155         #KN
l2=-12            #KN

#calculation
x=0              #x=n*N*L/A*E
y=0              #y=n*alpha*delta*T*L
z=(l1*l2*l)/(A*E)
deltabh=(x+y+z)

#result
print "The horizontal displacement is ",round(deltabh*1000,2),"m"

The horizontal displacement is  1.11 m


## Example 14.15 Page No 774¶

In [5]:
#Given
E = 200*10**6 #kN/m**2
P = 0         #N
A = 400*10**-6 #m**2

#Calculation
N_by_P1 = 0
N_by_P2=0
N_by_P3=-1.414
N_by_P4=1
L1=4
L2=2.828
L3=2.828
L4=2
N1=-100
N2=141.4
N3=-141.4
N4=200

Sum = N_by_P1*L1*N1+N_by_P2*L2*N2+N_by_P3*L3*N3+N_by_P4*L4*N4
del_ch = Sum/(E*A)

#Display
print'The vertical displacement of joint C of the steel truss   =',round(del_ch*1000,1),"mm"

The vertical displacement of joint C of the steel truss   = 12.1 mm