import math
from math import sin, cos, tan, pi, radians
#Initilization of Variables
L=5000 #mm #Length of strut
dell=10 #mm #Deflection
W=10 #N #Load
#Calculations
#Central Deflection of a simply supported beam with central concentrated load is
#dell=W*L**3*(48*E*I)**-1
#Let E*I=X
X=W*L**3*(48*dell)**-1 #mm
#Euler's Load
#Let Euler's Load be P
P=pi**2*X*(L**2)**-1
#Result
print"Critical Load of Bar is",round(P,2),"N"
import math
from math import sin, cos, tan, pi, radians
#Initilization of Variables
L=2000 #mm #Length of square column
E=12*10**3 #N/mm**2 #Modulus of Elasticity
sigma=12 #N/mm*2 #stress
W1=95*10**3 #N #Load1
W2=200*10**3 #N #Load2
FOS=3
#Calculations
#From Euler's Formula
#P=pi**2*E*I*(L**2)**-1 .........(1)
#Working Load
#W=P*(FOS)**-1
#Part-1
#At W1=95*10**3 #N
#W1=P*(3*L**2)**-1
#Let 'a' be the side of the square
#I=1*12**-1*a**4
#sub value of I in Equation 1 and further rearranging we get
a=(W1*3*12*L**2*(pi**2*E)**-1)**0.25 #mm
#From Consideration of direct crushing
#sigma*a**2=W1
#After Reaaranging the above equation we get
a2=(W1*(sigma)**-1)**0.5 #mm
#required size is 103.67*103.67 i.e a*a
#Part-2
#At W2=200*10**3 #N
#W2=P*(3*L**2)**-1
#After substituting values and further Rearranging the above equation we get
a3=(W2*3*12*L**2*(pi**2*E)**-1)**0.25 #mm
#From consideration of direct compression,size required is
a4=(W2*sigma**-1)**0.5
#required size is 129.10*129.10 i.e a4*a4
#Result
print"For W1 Load Required size is",round(a*a,2),"mm**2"
print"For W2 Load Required size is",round(a4*a4,2),"mm**2"
import math
from math import sin, cos, tan, pi, radians
#Initilization of Variables
#Flange
b=100 #mm #Width
D=80 #mm #Overall Depth
t=10 #mm #Thickness of web and flanges
L=3000 #mm #Length of strut
E=200*10**3 #N/mm**2 #Modulus of Elasticity
#Calculations
#Let centroid be at depth y_bar from top fibre
y_bar=(b*t*t*2**-1+(D-t)*t*((D-t)*2**-1+t))*(b*t+(D-t)*t)**-1 #mm
#M.I at x-x axis
I_x=1*12**-1*b*t**3+b*t*(y_bar-t*2**-1)**2+1*12**-1*t*((D-t))**3+t*((D-t))*((((D-t)*2**-1)+t)-y_bar)**2
#M.I at y-y axis
I_y=1*12**-1*t*b**3+1*12**-1*(D-t)*t**3 #mm**3
#Least M.I
I=I_y
#Since both ends are hinged
#Feective Length=Actual Length
L=l=3000 #mm
#Buckling Load
P=pi**2*E*I*(l**2)**-1*10**-3 #KN
#Result
print"The Buckling Load for strut of tee section",round(P,2),"KN"
import math
from math import sin, cos, tan, pi, radians
#Initilization of Variables
D=400 #mm #Overall Depth
#Flanges
b=300 #mm #Width
t=50 #mm #Thickness
t2=30 #mm #Web Thickness
dell=10 #mm #Deflection
w=40 #N/mm #Load
FOS=1.75 #Factor of safety
E=2*10**5 #N/mm**2
#Calculations
#M.I at x-x axis
I_x=1*12**-1*(b*D**3-(b-t2)*b**3) #mm**4
#Central Deflection
#dell=5*w*L**4*(384*E*I)**-1
#After sub values in above equation and further simplifying we get
L=(dell*384*E*I_x*(5*w)**-1)**0.25
#M.I aty-y axis
I=I_y=1*12**-1*t*b**3+1*12**-1*b*t2**3+1*12**-1*t*b**3 #mm**4
#Both the Ends of column are hinged
#Crippling Load
P=pi**2*E*I*(L**2)**-1 #N
#Safe Load
S=P*(FOS)**-1*10**-3 #N
#Result
print"Safe Load if I-section is used as column with both Ends hhinged",round(S,2),"KN"
import math
from math import sin, cos, tan, pi, radians
#Initilization of Variables
D=200 #mm #External Diameter
t=20 #mm #hickness
d=200-2*t #mm #Internal Diameter
E=1*10**5 #N/mm**2
a=1*(1600)**-1 #Rankine's Constant
L=4.5 #m #Length
sigma=550 #N/mm**2 #Stress
FOS=2.5
#Calculations
#Moment of Inertia
I=pi*D**4*64**-1-pi*d**4*64**-1
#Both Ends are fixed
#Effective Length
l=1*2**-1*L*10**3 #mm
#Euler's Critical Load
P_E=pi**2*E*I*(l**2)**-1
A=pi*4**-1*(D**2-d**2) #mm*2
k=(I*A**-1)**0.5
#Rankine's Critical Load
P_R=sigma*A*(1+a*(l*k**-1)**2)**-1
X=P_E*P_R**-1
#Safe Load using Rankine's Formula
S=P_R*(FOS)**-1*10**-3 #KN
#Result
print"Safe Load by Rankine's Formula is",round(S,2),"KN"
import math
from math import sin, cos, tan, pi, radians
#Initilization of Variables
L=3000 #mm #Length of column
W=800*10**3 #N #Load
a=1*1600**-1 #Rankine's constant
FOS=4 #Factor of safety
sigma=550 #N/mm**2 #stress
#Calculations
#Effective Length
l=L*2**-1 #mm
#Let d1=outer diameter & d2=inner diameter
#d1=5*8**-1*d2
#M.I
#I=pi*64**-1*(d1**4-d2**4) #mm**4
#Area of section
#A=pi4**-1*(d1**2-d2**2) #mm**2
#k=(I*A**-1)
#substituting values in above equation
#k=1*16**-1*(d1**2-d2**2)
#after simplifying further we get
#k=0.2948119.d1
#X=l*k**-1
#substituting values in above equation and after simplifying further we get
#X=5087.9898*d1**-1
#Crtitcal Load
P=W*FOS #N
#From Rankine's Load
#P2=sigma*A*(1+a*(X)**2)**-1
#substituting values in above equation and after simplifying further we get
#d1**4-12156618*d1**4-1.96691*10**8=0
#Solving Quadratic Equation we get
#d1**2-12156618*d1-196691000=0
a=1
b=-12156.618
c=-196691000
Y=b**2-4*a*c
d1_1=((-b+Y**0.5)*(2*a)**-1)**0.5 #mm
d1_2=((-b-Y**0.5)*(2*a)**-1) #mm
d2=5*8**-1*d1_1
#Result
print"Section of cast iron hollow cylindrical column is:d1_1",round(d1_1,2),"mm"
print" :d2 ",round(d2,2),"mm"
import math
from math import sin, cos, tan, pi, radians
#Initilization of Variables
#Let X=(P*A**-1) #Average Stress at Failure
Lamda_1=70 #Slenderness Ratio
Lamda_2=170 #Slenderness Ratio
X1=200 #N/mm**2
X2=69 #N/mm**2
#Rectangular section
b=60 #mm #width
t=20 #mm #Thickness
L=1250 #mm #Length of strut
FOS=4 #Factor of safety
#Calculations
#Slenderness ratio
#Lamda=L*k**-1
#The Rankine's Formula for strut
#P=sigma*A*(1+a*(L*k**-1)**-1
#From test result 1,
#After sub values in above equation we get and further simplifying we get
#sigma_1=200+980000*a ...................(1)
#From test result 2,
#After sub values in above equation we get and further simplifying we get
#sigma_2=69+1994100*a ...................(2)
#Substituting it in equation (1) we get
a=131*1014100**-1
#Substituting a in equation 1
sigma_1=200+980000*a #N/mm**2
#Effective Length
l=1*2**-1*L #mm
#Least of M.I
I=1*12**-1*b*t**3 #mm**4
#Area
A=b*t #mm**2
k=(I*A**-1)**0.5
#Slenderness ratio
Lamda=l*k**-1
#From Rankine's Ratio
P=sigma_1*A*(1+a*(Lamda)**2)**-1
#Safe Load
S=P*(FOS)**-1*10**-3 #N
#Result
print"Constant in the Formula is:a ",round(a,6)
print" :sigma_1",round(sigma_1,2)
print"Safe Load is",round(S,2),"KN"
import math
from math import sin, cos, tan, pi, radians
#Initilization of Variables
D=200 #mm #Depth
b=140 #mm #width
#Plate
b2=160 #mm #Width
t2=10 #mm #Thickness
L=l=4000 #mm #Length
FOS=4 #Factor of safety
sigma=315 #N/mm**2 #stress
a2=1*7500**-1
I_xx=26.245*10**6 #mm**4 #M.I at x-x
I_yy=3.288*10**6 #mm**4 #M.I at y-y
a=3671 #mm**2 #Area
k_x=84.6#mm
k_y=29.9 #mm
#Calculations
#Total Area
A=a+2*t2*b2 #mm**2
#M.I
I=I_yy+2*12**-1*t2*b2**3 #mm**4
k=(I*A**-1)**0.5 #mm
#Let X=L*k**-1
X=L*k**-1
#Appliying Rankine's Formula
P=sigma*A*(1+a2*(X)**2)**-1 #N
#Safe Load
S=P*(FOS)**-1*10**-3 #KN
#Result
print"Safe axial Load is",round(S,2),"KN"
import math
from math import sin, cos, tan, pi, radians
#Initilization of Variables
E=200*10**3 #N/mm**2 #Modulus of elasticity
sigma=330 #N/mm**2 #Stress
a=1*7500**-1 #Rankine's constant
A=5205 #mm**2 #area of column
I_xx=59.431*10**6 #mm**4 #M.I at x-x axis
I_yy=8.575*10**6 #mm**24#M.I at y-y axis
#Calculations
#Total M.I
I=I_xx+I_yy #mm**4
#Area of compound Section
A2=2*A #mm**2
k=(I*A2**-1)**0.5 #mm
#Equating Euler's Load to Rankine's Load we get
#pi**2*E*I*(L**2)**-1=sigma*A*(1+a*(L*k)**2)**-1
#After Substitt=uting values and further simplifying we get
L=(39076198*(1-0.7975432)**-1)**0.5*10**-3 #m
#Result
print"Length of column for which Rankine's formula and Euler's Formula give the same result is",round(L,2),"m"
import math
from math import sin, cos, tan, pi, radians
#Initilization of Variables
sigma=326 #N/mm**2 #stress
E=2*10**5 #N/mm**2 #Modulus of Elasticity
FOS=2 #Factor of safety
a=1*7500**-1 #Rankine's constant
D=350 #mm #Overall Depth
#Cover plates
b1=500 #mm #width
t1=10 #mm #Thickness
d=220 #mm #Distance between two channels
L=6000 #mm #Length of column
A=5366 #mm**2 #Area of Column section
I_xx=100.08*10**6 #mm**4 #M.I of x-x axis
I_yy=4.306*10**6 #mm**4 #M.I of y-y axis
C_yy=23.6 #mm #Centroid at y-y axis
#Calculations
#Symmetric axes are the centroidal axes is
#M.I of Channel at x-x axis
I_xx_1=2*I_xx+2*(1*12**-1*b1*t1**3+b1*t1*(D*2**-1+t1*2**-1)**2)
#M.I of Channel at y-y axis
I_yy_1=2*(I_yy+A*(d*2**-1+C_yy)**2)+2*12**-1*t1*b1**3
#As I_yy<I_xx
#So
I=I_yy_1 #mm**4
A2=2*A+2*t1*b1 #Area of channel
k=(I*A2**-1)**0.5 #mm
#Critical Load
P=sigma*A2*(1+a*(L*k**-1)**2)**-1
#Safe Load
S=P*2**-1*10**-3 #KN
#Result
print"Safe Load carrying Capacity is",round(S,2),"KN"
import math
from math import sin, cos, tan, pi, radians
#Initilization of Variables
I=4.085*10**8 #mm**4 #M.I
A=20732.0 #mm**2 #area of column
f_y=250 #N/mm**2
L=6000 #mm #Length of column
#Calculations
k=(I*A**-1)**0.5 #mm
lamda=L*k**-1 #Slenderness ratro
#From Indian standard table
lamda_1=40
sigma_a_c_1=139 #N/mm**2
lamda_2=50
sigma_a_c_2=132 #N/mm**2
#Linearly interpolating between these values for lambda=42.744
sigma_a_c_3=sigma_a_c_1-2.744*10**-1*(sigma_a_c_1-sigma_a_c_2)
#Safe Load carrying capacity of column
P=sigma_a_c_3*A*10**-3
#Result
print"Safe Load carrying capacity is",round(P,2),"KN"