# Chapter 18:Columns¶

## Ex18.1:pg-977¶

In [13]:
import math
Pu=3000 #in kN
fck=20 #in MPa
fy=415 #in MPa
l=3 #unsupported length, in m
#assume 1% steel
Ag=Pu*10**3/(0.4*fck*0.99+0.67*fy*0.01) #in sq mm
L=math.sqrt(Ag) #assuming a square column
L=530 #in mm
Asc=0.01*L**2 #in sq mm
emin=l*10**3/500+L/30 #in mm
ep=0.05*L #>emin, hence OK
print "Column size - ",L," x ",L," mm"

Column size -  530  x  530  mm


## Ex18.2:pg-978¶

In [14]:
import math
Pu=1500 #in kN
fck=15 #in MPa
fy=250 #in MPa
l=2.75 #unsupported length, in m
#assume 1% steel
Ag=Pu*10**3/(0.4*fck*0.99+0.67*fy*0.01) #in sq mm
L1=225 #assuming a square column
L2=Ag/L1 #in mm
L2=880 #in mm
Asc=0.01*L1*L2 #in sq mm
e1=l*10**3/500+L1/30 #in mm
e2=l*10**3/500+L2/30 #in mm
ep1=0.05*L1 #<e1
ep2=0.05*L2 #>e2, hence Ok
print "The column is safe on long dimension side but not on short dimension side. As such, the column be checked for eccentricity in short direction."

The column is safe on long dimension side but not on short dimension side. As such, the column be checked for eccentricity in short direction.


## Ex18.3:pg-978¶

In [15]:
import math
b=225 #in mm
D=500 #in mm
c=45 #cover, in mm
Asc=2463 #in sq mm
Ast=Asc
fck=15 #in MPa
fy=250 #in MPa
fcc=0.446*fck #in MPa
#(i)
xu=1.1*D #in mm
m=0.43*D #in mm
esc1=0.002*(xu-c)/(xu-m)
esc2=0.002*(xu-D+c)/(xu-m)
#by interpolation
fsc1=217.5 #in MPa
fsc2=217.5*esc2/0.0010875 #in MPa
#stress block parameters for xu / D = 1.1
n=0.384
l=0.443
A=n*fck*D #area of stress block
r=l*D #distance of c.g., in mm
Pu=(A*b+Asc*(fsc1-fcc)+Ast*fsc2)/10**3
Mu=(A*b*(D/2-r)+Asc*(fsc1-fcc)*(D/2-c)-Ast*fsc2*(D/2-c))/10**6
print "(i) For xu = 1.1 D\nP=",Pu," kN\nMu=",Mu," kN-m\n"
#answer in textbook is incorrect
#(ii)
xu=330 #in mm
esc=0.0035*(xu-c)/xu
est=0.0035*(D-c-xu)/xu
#by interpolation
fsc=217.5 #in MPa
fst=217.5 #in MPa
Pu=(0.36*fck*b*xu+Asc*(fsc-fcc)-Ast*fst)/10**3 #in kN
Mu=(0.36*fck*b*xu*(D/2-0.416*xu)+Asc*(fsc-fcc)*(D/2-c)+Ast*fst*(D/2-c))/10**6 #in kN-m
print "(ii) For xu = 330 mm\nP=",Pu," kN\nMu=",Mu," kN-m"

(i) For xu = 1.1 D
P= 1446.61010463  kN
Mu= 67.6351908515  kN-m

(ii) For xu = 330 mm
P= 384.47253  kN
Mu= 261.45522765  kN-m


## Ex18.4:pg-979¶

In [16]:
import math
b=300 #in mm
D=400 #in mm
c=30 #cover, in mm
Asc=452 #in sq mm
Ast=Asc
fck=15 #in MPa
fy=415 #in MPa
fcc=0.446*fck #in MPa
#(i)
xu=1.4*D #in mm
m=0.43*D #in mm
esc1=0.002*(xu-c)/(xu-m)
esc2=0.002*(xu-D+c)/(xu-m)
#by interpolation
fsc1=356.8 #in MPa
fsc2=238.68 #in MPa
#stress block parameters for xu / D = 1.4
n=0.417
l=0.475
A=n*fck*D #area of stress block
r=l*D #distance of c.g., in mm
Pu=(A*b+Asc*(fsc1-fcc)+Ast*fsc2)/10**3 #in kN
Mu=(A*b*(D/2-r)+Asc*(fsc1-fcc)*(D/2-c)-Ast*fsc2*(D/2-c))/10**6 #in kN-m
print "(i) For xu = 1.4 D\nP=",Pu," kN\nMu=",Mu," kN-m\n"
#(ii)
xu=370 #in mm
esc=0.0035*(xu-c)/xu
est=0.0035*(D-c-xu)/xu
#by interpolation
fsc=355.8 #in MPa
Pu=(0.36*fck*b*xu+Asc*(fsc-fcc))/10**3 #in kN
Mu=(0.36*fck*b*xu*(D/2-0.416*xu)+Asc*(fsc-fcc)*(D/2-c))/10**6 #in kN-m
print "(ii) For xu = 370 mm\nP=",Pu," kN\nMu=",Mu," kN-m"

(i) For xu = 1.4 D
P= 1016.73308  kN
Mu= 16.0682812  kN-m

(ii) For xu = 370 mm
P= 757.19772  kN
Mu= 54.4459644  kN-m


## Ex18.5:pg-980¶

In [17]:
import math
b=225 #in mm
D=500 #in mm
c=50 #cover, in mm
Asc=1520 #in sq mm
Ast=Asc
fck=20 #in MPa
fy=500 #in MPa
fcc=0.446*fck #in MPa
#(i)
xu=1.3*D #in mm
m=0.43*D #in mm
esc1=0.002*(xu-c)/(xu-m)
esc2=0.002*(xu-D+c)/(xu-m)
#by interpolation
fsc1=412.515 #in MPa
fsc2=183.794 #in MPa
#stress block parameters for xu / D = 1.3
n=0.409
l=0.468
A=n*fck*D #area of stress block
r=l*D #distance of c.g., in mm
Pu=(A*b+Asc*(fsc1-fcc)+Ast*fsc2)/10**3 #in kN
Mu=(A*b*(D/2-r)+Asc*(fsc1-fcc)*(D/2-c)-Ast*fsc2*(D/2-c))/10**6 #in kN-m
print "(i) For xu = 1.3 D\nP=",Pu," kN\nMu=",Mu," kN-m\n"
#(ii)
xu=400 #in mm
esc=0.0035*(xu-c)/xu
est=0.0035*(D-c-xu)/xu
#by interpolation
fsc=422.11 #in MPa
fst=87.45 #in MPa
Pu=(0.36*fck*b*xu+Asc*(fsc-fcc)-Ast*fst)/10**3 #in kN
Mu=(0.36*fck*b*xu*(D/2-0.416*xu)+Asc*(fsc-fcc)*(D/2-c)+Ast*fst*(D/2-c))/10**6 #in kN-m
print "(ii) For xu = 400 mm\nP=",Pu," kN\nMu=",Mu," kN-m"
#answer in textbook for Mu in (ii) is incorrect

(i) For xu = 1.3 D
P= 1813.08128  kN
Mu= 81.543504  kN-m

(ii) For xu = 400 mm
P= 1143.1248  kN
Mu= 206.36736  kN-m


## Ex18.6:pg-981¶

In [18]:
import math
b=250.0 #column width, in mm
D=500.0 #column depth, in mm
lex=4.0 #in m
ley=4.0 #in m
Pu=300.0 #in kN
Asc=1472.0 #in sq mm
Ast=1472.0 #in sq mm
fck=15.0 #in MPa
fy=250.0 #in MPa
c=50 #cover, in mm
Max=Pu*10**3*D/2000*(lex/(D/10**3))**2.0/10**6 #in kN-m
May=Pu*10.0**3*b/2000*(ley/(b/10**3))**2.0/10**6 #in kN-m
Puz=(0.45*fck*(b*D-(Asc+Ast))+0.75*fy*(Asc+Ast))/10**3 #in kN
#to find Pb
xu=(D-c)/(1+0.002/0.0035) #in mm
fsc=217.5 #in MPa
fst=217.5 #in MPa
Pb=(0.36*fck*b*xu+fsc*Asc-fst*Ast)/10**3 #in kN
k=(Puz-Pu)/(Puz-Pb) #>1
k=1
Max=k*Max #in kN-m
May=k*May #in kN-m
print "Additional Moments are:\nMax=",Max," kN/m\nMay=",May," kN-m"

Additional Moments are:
Max= 4.8  kN/m
May= 9.6  kN-m