# Chapter 2:Axial strains and Deformations in bars¶

## Example 2.1 page number 77¶

In :
l_ob = 2000 #mm - length of rod ob
l_bc = 1000 #mm - length of rod bc
l_cd = 1500 #mm - length of rod cd
p_ob = 100  #kN - Force in rods
p_bc = -150 #KN
p_cd = 50   #KN
A_ob = 1000 #mm2 - Area of rod ob
A_bc = 2000 #mm2 - Area of rod bc
A_cd = 1000 #mm2 - Area of rod cd
E = 200.0 #GPA
# the total deflection is algebraic sums of deflection in each module
e_1 = p_ob*l_ob/(A_ob*E)
e_2 = p_bc*l_bc/(A_bc*E)
e_3 = p_cd*l_cd/(A_cd*E)
#All units are satisfied
e_total = e_1+ e_2 + e_3
print "The total deflection is :",round(e_total,3) ,"mm"

The total deflection is : 1.0 mm


## Example 2.4 page number 80¶

In :
p_app = 3 #Kips - applied force
P_A = 2.23 #Kips
p_B = -2.83 #kips - compressive force
l_ab = 6.71 #inch
l_bc = 8.29 #inch
s_ab = 17.8 #ksi - tensile stress
s_bc = -12.9 #ksi - compressive stress
E = 10.6 * pow(10,3) #ksi -youngs modulus
e_ab = s_ab*l_ab/E

e_bc = s_bc*l_bc/E
x = e_ab/e_bc #the Ratio of cosines of the deflected angles
# t_1 and t_2 be deflected angles
#t_2 = 180-45-26.6-t_1 the sum of angles is 360
#t_1 = 52.2 degress
import math
k  = p_app/e # kips/in vertical stiffness of the combination
print "The vertical stiffness of the combination is",round(k,3),"kips/inch"

0.0112677358491
The vertical stiffness of the combination is 113.14 kips/inch


## Example 2.6 page number 83¶

In :
dia = 50 #mm - diameter of aluminium
p = 100 # KN - instant force applid
dia_c = 0.1215 #mm- change in diameter
l_c = 0.219 #mm - change in length
l = 300 #mm - length
strain_dia = dia_c/dia # lateral strain
strain_l = l_c/l #longitudinal strain
po = strain_dia/strain_l # poission ratio
area = 3.14*dia*dia/4 #mm2 area
E = p*l/(area*l_c) #N/mm2 youngs modulus
print "The lateral strain is:",strain_dia,"no units"
print "The longitudinal strain is:",strain_l,"no units"
print "The poissions ratio is:",po,"no units"
print "Youngs modulus:",round(E,2),"N/mm2"

The lateral strain is: 0.00243 no units
The longitudinal strain is: 0.00073 no units
The poissions ratio is: 3.32876712329 no units
Youngs modulus: 69.8 N/mm2


## Example 2.7 page number 86¶

In :
T = 12.9*pow(10,-6) #/F
t = 100.00 # F
l_ab = 6.71 #inch
l_bc = 8.29 #inch
e_ab = T*t*l_ab #in-elongation
e_bc = T*t*l_bc #in-elongation
k = e_ab/e_bc # ratio of cosines of deflected angles
# t_1 and t_2 be deflected angles
#t_2 = 180-45-26.6-t_1 the sum of angles is 360
t_1 = 26.6
import math
print "The displacement in point B is :",e ,"in"

The displacement in point B is : 0.00795578950395 in


## Example 2.11 page number¶

In :
mass = 4 #kg
dist = 1 #mt freely falling distance
l = 1500 #mm length of rod
d = 15 #mm diameter
l_ab = 6.71 #inch
l_bc = 8.29 #inch
E =  200  #GPA youngs modulus
k = 4.5 # N/mm stiffness costant
F =  mass*9.81# The force applying
Area =  3.14*(d**2)/4
# Two cases
#youngs modulus
e_y = F*l/(Area*E*pow(10,3))
# stiffness
e_f = F/k
#total
e = e_y +e_f
k = 1+(2/(e*pow(10,-3)))
stress_max_1 = F*(1+pow(k,0.5))/Area
print "The maximum stress is:",stress_max_1,"Mpa"

The maximum stress is: 3.59377281766 Mpa


## Example 2.12 page number 103¶

In :
flex_a = 1#f
flex_b = 2#f
#removing lower support and solving FBD
e = -2 -(2+1)#fp
#e_1 = (2+1+1)*R
#e_1 = -e Making the elongations zero since the both ends are fixed
R = e/(2+1+1.0) #p
print "The reactions at bottom is",R,"p"

The reactions at bottom is -1.25 p


## Example 2.19 page number 113¶

In :
#Given
l = 30 #in - The length of the rod
p_1 = 80 #kips - The Force on the end
p_2  = 125 #kips - The force on the other end
A_s = 0.5 #in2 - The crossection of the steel rod
A_a = 0.5 #in2 - The crossection of the aluminium
E_a = 10*(10**6) #psi - The youngs modulus of the aluminium
E_s = 30*(10**6) #psi - The youngs modulus of the steel
#Internally stastically indeterminant
p_a = p_1/4 #From solving we get p_s = 3*P_a
#From material properties point of view
#stress_steel = stress_aluminium
e = p_a*l*(10**3)/(A_a*E_a) #The end deflection
print "The end deflection is",e,"in"

The end deflection is 0.12 in
`