import math
#Initilization of Variables
P=40 #mm #Force applied to stretch a tape
L=30 #m #Length of steel tape
A=6*1 #mm #Cross section area
E=200*10**9*10**-6 #KN/m**2 #Modulus of Elasticity
#Calculations
sigma_L=(P*L*10**3)*(A*E)**-1 #mm
#Result
print"The Elongation of steel tape is",round(sigma_L,4),"mm"
import math
#Initilization of VAriables
#D=(D_0-2) #cm #Inside Diameter of cyclinder
#A=(pi*(D_0-1)) #cm**2 #Area of cross-section
#L=(pi*(D_0-1)*5400) #N #Crushing load for column
F=6 #Factor of safety
T=1 #cm #wall thickness of cyclinder
#S=L*F**-1
#After Simplifying,we get
S=600*10**3
#Calculations
D_0=(S*F)*(pi*54000)**-1+1 #cm #Outside diameter of cyclinder
#Result
print"The outside Diameter of cyclinder is",round(D_0,2),"cm"
import math
#Initilization of variables
P=800 #N #force applied to steel wire
L=150 #m #Length of steel wire
E=200 #GN/m**2 #Modulus of Elasticity
d=10 #mm #Diameter of steel wire
W=7.8*10**4 #N/m**3 #Weight Density of steel
#A=(pi*4**-1)*(d)**2 #m**2
#After simplifying Area,we get
A=7.85*10**-5 #m**2
#calculation (Part-1)
#Elongation Due to 800N Load
dell_L_1=(P*L*10**-3)*(A*E*10**9*10**-6)**-1 #mm
#calculation (Part-2)
#Elongation due to Weight of wire
dell_L_2=((pi*4**-1)*150*W*L*10**-3)*(2*A*E*10**7)**-1 #mm
#calculation (Part-3)
#Total Elongation of wire
dell_L_3=dell_L_1+dell_L_2
#Result
print"The Elongation due to 800N Load is",round(dell_L_1,2),"mm"
print"The Elongation due to Weight of wire is",round(dell_L_2,2),"mm"
print"Total Elongation of wire is",round(dell_L_3,2),"mm"
import math
#Intilization of variables
d=10 #mm #Diameter of Punching Hole
t=4 #mm #Thickness of Mild Steel Plate
tou=320 #N/mm**2 #Shear Strength of mild Steel
#Calculations
#Force Required for punching the hole
P=tou*pi*d*t #N
#Area of punch in contact with the plate surface
A=(pi*4**-1*d**2) #mm*2
#Compressive stress
sigma_c=P*A**-1 #N/mm*2
#Result
print"Force Required for punching the hole is",round(P,2),"N"
print"Compressive stress is",sigma_c,"N/mm*2"
import math
#Initilization of variables
P=200*10**3 #N
L_1=0.10 #mm #Length of of portin AB
L_2=0.16 #mm #Length of of portin BC
L_3=0.12 #mm #Length of of portin CD
E=200*10**9 #N
d_1=0.1 #cm
d_2=0.08 #cm
d_3=0.06 #cm
A_1=(pi*4**-1)*(0.1)**2 #mm**2
A_2=(pi*4**-1)*(0.08)**2 #mm**2
A_3=(pi*4**-1)*(0.06)**2 #mm**2
#Calculations
dell_L_1=(P*L_1*10**3)*(A_1*E)**-1 #mm
dell_L_2=(P*L_2*10**3)*(A_2*E)**-1 #mm
dell_L_3=(P*L_3*10**3)*(A_3*E)**-1 #mm
dell_L=dell_L_1+dell_L_2+dell_L_3 #mm
#Result
print"Total Elongation of steel bar is",round(dell_L,3),"mm"
import math
#Initilization of variables
#from F.B.D,we get
P_1=50 #KN
P_2=20 #KN
P_3=40 #KN
d=0.02 #mm #Diameter of steel bar
L_1=0.4 #mm
L_2=0.3 #mm
L_3=0.2 #mm
E=210*10**9 #N
#After simplifying Area,we get
A=pi*10**-4 #m**2 #Area of cross section
#Calculations
sigma_AB=P_1*1000*A #N/m**2
sigma_BA=P_2*1000*A #N/m**2
sigma_CD=P_3*1000*A #N/m**2
dell_L=((P_1*L_1+P_2*L_2+P_3*L_3)*(A*E)**-1)*10**6 #mm
#Result
print"Total Elongation of Steel bar is",round(dell_L,3),"mm"
import math
import numpy as np
#Initilization of variables
#R_a+R_c=25 #KN #R_a,R_b are reactions at supports A and C respectively
L_ab=2 #m
L_bc=3 #m
#Calculation
#From F.B.D,we get
#dell_L_AB=(R_a*L_AB)*(A*E)**-1 #Elongation of portion AB
#dell_L_BC=(R_c*L_BC)*(A*E)**-1 #Compression of portion BC
#After simplifying above equations we get,
#R_a=(1.5)*R_c #KN
#R_a+R_c=25 #KN
#Solving the above simultaneous equations using matrix method
A=np.array([[1,-1.5],[1,1]]) #Here the coefficients of the first equations of unknowns are setup
B=np.array([0,25]) #Here the RHS of both equations are setup
C=np.linalg.solve(A,B)
#print C[0] #Prints the first element in the vector C
#print C[1] #Prints the second element in the vector C
#Result
print"The reaction at support A is",round(C[0],2),"KN"
print"The reaction at support C is",round(C[1],2),"KN"
import math
#Initilization of variables
#P is the force acting on the bar BC compressive in nature and force on AB is (100-P) Tensile in nature
E=200*10*9 #N
A_1=3*10**-4 #cm**2 #Area of AB
A_2=4*10**-4 #cm**2 #Area of BC
L=1.5 #cm #Length of bar
#Calculations
#The total elongation of bar
#(((100-P)*10**3*1.5)*(3*10**-4*E)**-1)-((P*10**3*1.5)*(4*10**-4*E)**-1)=0
#The total elongation of bar is limited to 1
#(25-0.4375*P)*10**-4=1*10**-3
#After simplifying above equation we get,
P=-(10-25)*0.4375**-1 #KN #Total elongation of bar
F_AB=100-P #KN #force in AB
F_BC=P #KN #Force in BC
sigma_AB=(((F_AB)*(3*10**-4)**-1)*10**-3) #KN #Stress in AB
sigma_BC=((F_BC)*(4*10**-4)**-1*10**-3) #KN #Stress in Bc
#Result
print"F_AB",round(F_AB,2),"KN"
print"F_BC",round(F_BC,2),"KN"
print"sigma_AB",round(sigma_AB,2),"KN"
print"sigma_BC",round(sigma_BC,2),"KN"
import math
#Initilization of variables
P=500 #KN #Safe Load
d=20 #mm #steel rod diameter
n=4 #number of steel rod
sigma_c=4 #N/mm**2 #stress in concrete
#E_S*E_c**-1=15
#Calculations
A_s=4*pi*4**-1*d**2 #mm**2 #Area os steel rod
sigma_s=15*sigma_c #N/mm**2 #stress in steel
#P=sigma_s*A_s+sigma_c*A_c
#After substituting and simplifying above equation we get,
A_c=(P*10**3-sigma_s*1256)*(sigma_c)**-1 #mm**2 #Area of the concrete
X=(A_s+A_c)**0.5 #mm #Total cross sectional area
P_s=A_s*sigma_s*10**-3 #KN #Load carried by steel
#Result
print"Load carried by steel is",round(P_s,2),"KN"
print"stress induced in steel is",round(sigma_s,3),"KN"
print"cross sectional area of column is",round(X,2),"mm"
import math
#Initilization of variables
A_s=500 #mm**2
E_s=200000
E_al=80000
A_al=1000
#Calculations
#(P_al*L_al)*(A_al*E_al)**-1+(P_s*L_s)*(A_s*E_s)**-1=1*2**-1
P=1*1000**-1*((A_s*E_s*A_al*E_al)*(A_s*E_s+A_al*E_al)**-1) #N
P_s=P_al=P #N
sigma_t=P_s*A_s**-1 #N/mm**2 #Tensile stress in bolt
sigma_c=P_al*A_al**-1 #N/mm**2 #Compressive stress in Aluminium tube
#result
print"Tensile stress in bolt is",round(sigma_t,2),"N/mm**2"
print"Compressive stress in Aluminium tube is",round(sigma_c,2),"N/mm**2"
import math
#Initilization of variables
A=1600 #mm**2 #Area of the Bar
P=480*10**3 #N #Load
dell_L=0.4 #mm #Contraction of metal bar
L=200 #mm #Length of metal bar
sigma_t=0.04 #mm #Guage Length
t=40
#Calculations
sigma_L=dell_L*L**-1
E=((P*L)*(A*dell_L)**-1*10**-3) #N/mm**2 #Young's Modulus
m=t*sigma_t**-1*sigma_L
#Result
print"The value of Young's Modulus is",round(E,2),"N/mm*2"
print"The value of Poissoin's ratio is",round(m,2)
import math
#Initilization of variables
A_s=0.003848 #m**2 #Area of steel bar
A_al=0.003436 #m**2 #Area of Aluminium tube
E=220*10*9 #N #Young's modulus of steel
E=70*10*9 #N #Young's modulus of aluminium
P=600*10**3 #N #Load applied to the bar
#dell_L_al-dell_L_s=0.00015 #mm #difference between strain in aluminium bar and steel bar
#Calculations
#Let the aluminium tube be compressed by dell_L_al and steel bar by by dellL_s
#dell_L_al=sigma_al*E_al**-1*L_al
#dell_L_s=sigma_s*E_s**-1*L_s
#After substituting and simplifying above equation we get,
#((sigma_al*70**-1)-(sigma_s*220**-1))=300000 #(equation 1)
#After simplifying above equation we get,
#sigma_al=17462.165*10**4-1.1199*sigma_s #(equation 2)
#Now substituting sigma_al in equation(1)
#((17462.165*10**4-1.1199*sigma_s)*(70)**-1)-(sigma_s*220**-1)=300000
#After simplifying above equation we get,
sigma_s=-((300000-249.4594*10**4)*0.0205444**-1)*10**-6 #MN/m**2 #stress developed in steel bar
#sigma_al=17462.165*10**4-1.1199*sigma_s
sigma_al=(17462.165*10**4-1.1199*106822005.02)*10**-6
#Result
print"stress developed in steel bar is",round(sigma_s,2),"MN/m**2"
print"stress developed in aluminium bar is",round(sigma_al,2),"MN/m**2"
import math
#Initilization of variables
E=200 #GN/m**2 #Modulus of elasticity
alpha=11*10**-6 #per degree celsius #coeffecient o flinear expansion of steel bar
L=6 #m #Length of rod
#Calculations
#(Part-1) #IF the walls do not yield
t=58 #degree celsius #Fall in temperature #(t=80-22)
dell=alpha*t #strain
sigma=E*10**9*dell*10**-6 #MN/m**2 #Stress
A=pi*4**-1*6.25*10**-4 #mm**2 #Area of wall and rod
P=sigma*10**6*A*10**-3 #KN #Pull Exerted
#(Part-2) #IF the walls yield together at the two ends is 1.15 mm
L_2=L*(1-alpha*t) #m #Length of rod at 22 degree celsius
L_3=L-L_2 #m #Decrease in Length
#As the walls yield by 1.5 mm, actual decrease in length is
L_4=L_3-0.0015 #m
dell_2=L_4*L**-1 #strain
P_2=E*10**9*dell_2*A*10**-3 #KN
#Result
print"Pull Exerted by the rod:when walls do not yield",round(P,2),"KN"
print" :when total yield together at two ends is 1.5 mm",round(P_2,2),"KN"
import math
#Initilization of variables
D=4.5 #cm #External Diameter of tube
d=3 #cm #Internal diameter of tube
t=3 #mm #thickness of tube
t_1=30 #degree celsius
t_2=180 #degree celsius #when metal heated
L=30 #cm #Original LEngth
alpha_s=1.08*10**-5 #Per degree celsius #coefficient of Linear expansion of steel tube
alpha_c=1.7*10**-5 #Per degree celsius #coefficient of Linear expansion of copper tube
E_s=210 #GPa #Modulus of Elasticity of steel
E_c=110 #GPA #Modulus of Elasticity of copper
#Calculation
#For Equilibrium of the system, Total tension in steel=Total tension in copper
#sigma_s*A_s=sigma_c*A_c (equation 1)
A_c=pi*4**-1*d**2 #cm**2 #Area of copper
A_s=pi*4**-1*(D**2-d**2) #cm**2 #Area of steel
#simplifying equation 1
#sigma_s=1.785*sigma_c
T=t_2-t_1 #change in temperature
#Actual expansion of steel=Actual expansion of copper
#alpha_s*T*L+sigma_s*E_s**-1*L=alpha_c*T*L-sigma_c*E_c**-1*L
#After substituting values in above equation and simplifying we get
sigma_c=(930*10**5*1.7591**-1)*10**-6 #MN/m**2 #Stress in copper
sigma_s=1.785*sigma_c #MN/m**2 #Stress in steel
#Increase in Length of either component
L_2=(alpha_s*T+sigma_s*10**6*(E_s*10**9)**-1)*L
#Result
print"stress in copper bar is",round(sigma_c,2),"MN/m**2"
print"stress in steel bar is",round(sigma_s,2),"MN/m**2"
print"Increase in Length is",round(L_2,3),"cm"
import math
#Initilization of variables
t_1=15 #degree celsius #temperature of steel bar
t_2=315 #degree celsius #raised temperature
E_s=210 #GPa #Modulus of Elasticity of steel bar
E_c=100 #GPa #Modulus of Elasticity of copper bar
dell_L=0.15 #cm #Increase in Length of bar
#Calculation
#For Equilibrium of the system, Tension in steel bar = Tension in copper bar
#sigma_s*A_s = sigma_c*2*A_c
#sigma_S=2*sigma_c
#Actual expansion of steel = Actual expansion of copper
#L*alpha_s*T+sigma_s*E_s**-1*L = L*alpha_c*T-sigma_c*E_c**-1*L (Equation 1)
T=t_2-t_1 #per degree celsius #change in temperature
#After substituting values in above equation and simplifying we get
sigma_c=(1650*10**5*1.9524**-1)*10**-6 #MN/m**2 #Stress in copper
sigma_s=2*sigma_c #MN/m**2 #Stress in steel
#Actual Expansion of steel bar
#L*alpha_s*T+sigma_s*E_s**-1*L = L*alpha_c*T-sigma_c*E_c**-1*L
#After substituting values in above equation and simplifying we get
L=0.15*10**-2*0.0044048**-1 #m
#Result
print"Stress in copper bar is",round(sigma_c,2),"MN/m**2"
print"Stress in steel bar is",round(sigma_s,2),"MN/m**2"
print"Original Length of bar is",round(L,2),"m"
import math
#Initilization of variables
L=100 #m #Length of rod
A=2 #cm**2 #cross sectional area
rho=80 #KN/m**3
#Calculatiom
W=A*10**-4*L*rho #KN
sigma_s=10+1.6 #KN #Rod experiencing max tensile stress when it is at top performing upstroke
sigma_s_2=sigma_s*10**3*200**-1 #N/mm**2 #corresponding stress at this moment
sigma_c=1 #KN ##Rod experiencing max compressive stress at its lower end,free from its own weight
sigma_c_2=sigma_c*10**3*200**-1 #corresponding stress at this moment
#Result
print"Tensile stress in bar",round(sigma_s_2,2),"N/mm**2"
print"Compressive stress in bar",round(sigma_c_2,2),"N/mm**2"
import math
#Initilization of variables
sigma=0.012 #strain
P=150 #KN #Total Load on the Post
E=1.4*10**4 #N/mm**2 #modulus of elasticity
#b be the width of the post in mm
#2b is the longer dimension of the post in mm
#Calculations
#We know,
#sigma=(P*(A*E)**-1)
#After substituting values and simplifying, we get
b=((150*10**3)*(0.012*1.4*2*10**4)**-1)**0.5
q=2*b #mm #Longer dimension of post
#Result
print"Width of post is",round(b,2),"mm"
print"Longer dimension of post is",round(q,2),"mm"