# Chapter 5:Columns¶

## Ex5.1:pg-171¶

In [1]:
import math
sigma_cc=4 #in MPa
sigma_sc=130 #in MPa
Asc=4*0.785*25**2 #four 25 mm dia bars, in sq mm
b=300 #width, in mm
D=300 #depth, in mm
Ag=b*D #in sq mm
Ac=Ag-Asc #in sq mm
P=sigma_cc*Ac+sigma_sc*Asc #in N
print "Permissible load on the column = ",P/10**3,"kN"

Permissible load on the column =  607.275 kN


## Ex5.2:pg-172¶

In [2]:
import math
sigma_cc=5 #in MPa
sigma_sc=190 #in MPa
Asc=6*0.785*20**2 #six 20 mm dia bars, in sq mm
b=250 #width, in mm
D=400 #depth, in mm
Ag=b*D #in sq mm
Ac=Ag-Asc #in sq mm
P=sigma_cc*Ac+sigma_sc*Asc #in N
print "Permissible load on the column =",P/10**3," kN\n"
dia=20.0/4 #in mm
#as this is less than 6
dia=6 #in mm
s1=b #in mm
s2=16*20 #in mm
s3=48*dia #in mm
s=min(s1,s2,s3)
print "Provide ",(dia)," mm dia links at spacing equal to least of (i)Least lateral dimension = ",(b)," mm, (ii) 16 times longitudinal bar dia = ",(16*20)," mm, (iii) 48 times link bar dia = ",(48*dia)," mm, i.e., 250 mm\nHence, spacing or pitch = ",(s)," mm\n"
Pc=Asc*100/(b*D) #percentage steel
print "Percentage of steel is in between 0.8 to 4 as prescribed by IS code"

Permissible load on the column = 848.54  kN

Provide  6  mm dia links at spacing equal to least of (i)Least lateral dimension =  250  mm, (ii) 16 times longitudinal bar dia =  320  mm, (iii) 48 times link bar dia =  288  mm, i.e., 250 mm
Hence, spacing or pitch =  250  mm

Percentage of steel is in between 0.8 to 4 as prescribed by IS code


## Ex5.3:pg-173¶

In [3]:
import math
sigma_cc=5 #in MPa
sigma_sc=130 #in MPa
b=300 #width, in mm
D=400.0 #depth, in mm
Ag=b*D #in sq mm
Asc=(P*10.0**3-sigma_cc*Ag)/(sigma_sc-sigma_cc) #in sq mm
#provide 25 mm dia bars
n=round(Asc/(0.785*25**2))
print "Provide ",(n)," no. 25 mm dia bars\n"
dia=20.0/4 #in mm
#provide 8 mm dia links (available as per market size)
dia=8 #in mm
s1=b #in mm
s2=16*25 #in mm
s3=48*dia #in mm
s=min(s1,s2,s3)
print "Provide ",(dia)," mm dia links at spacing equal to least of (i)Least lateral dimension = ",(b)," mm, (ii) 16 times longitudinal bar dia = ",(16*25)," mm, (iii) 48 times link bar dia = ",(48*dia)," mm, i.e., 300 mm\nHence, spacing or pitch = ",(s)," mm\n"

Provide  7.0  no. 25 mm dia bars

Provide  8  mm dia links at spacing equal to least of (i)Least lateral dimension =  300  mm, (ii) 16 times longitudinal bar dia =  400  mm, (iii) 48 times link bar dia =  384  mm, i.e., 300 mm
Hence, spacing or pitch =  300  mm



## Ex5.4:pg-173¶

In [5]:
import math
sigma_cc=4.0 #in MPa
sigma_sc=130.0 #in MPa
Asc=6*0.785*12**2 #six 12 mm dia bars, in sq mm
D=200 #dia of column, in mm
Ag=0.785*D**2 #in sq mm
Ac=Ag-Asc #in sq mm
P=sigma_cc*Ac+sigma_sc*Asc #in N
dia=6 #dia of links used, in mm
s1=D #in mm
s2=16*12.0 #in mm
s3=48*dia #in mm
s=min(s1,s2,s3)
print "Permissible load on the column = ",P/10**3," kN\nProvide ",(dia)," mm dia links at spacing equal to least of (i)Least lateral dimension = ",(D)," mm, (ii) 16 times longitudinal bar dia = ",(16*12)," mm, (iii) 48 times link bar dia = ",(48*dia)," mm, i.e., ",(s)," mm\nHence, spacing or pitch = ",(s)," mm"

Permissible load on the column =  211.05824  kN
Provide  6  mm dia links at spacing equal to least of (i)Least lateral dimension =  200  mm, (ii) 16 times longitudinal bar dia =  192  mm, (iii) 48 times link bar dia =  288  mm, i.e.,  192.0  mm
Hence, spacing or pitch =  192.0  mm


## Ex5.5:pg-174¶

In [7]:
import math
dia=300 #in mm
Asc=8*0.785*20**2 #8-20 mm dia bars, in sq mm
helical_dia=8 #in mm
pitch=25 #in mm
cover=40 #in mm
sigma_cc=5 #in MPa
sigma_sc=130 #in MPa
fck=25 #in MPa
fy=250 #in MPa
Ag=0.785*dia**2 #in sq mm
Ac=Ag-Asc #in sq mm
P=sigma_cc*Ac + sigma_sc*Asc #in N
#to find volume of helical reinforcement
core_dia=dia-2*cover+2*helical_dia #in mm
l=math.pi*core_dia #length of helical steel for one revolution, in mm
Ab=l*0.785*helical_dia**2/pitch #volume of helical reinforcement per mm height of column, in mm**3
Ak=0.785*core_dia**2-Asc #in sq mm
Ac=0.785*core_dia**2 #in sq mm
m=Ab/Ak
n=0.36*(Ag/Ac-1)*fck/fy
#as m > n
P=1.05*P #in N

Safe load= 700.6125  kN


## Ex5.6:pg-175¶

In [8]:
import math
b=250 #width, in mm
D=350 #depth, in mm
Asc=4*0.785*22**2 #four 22 mm dia bars, in sq mm
Lef=5 #effective length of column, in m
sigma_cc=4 #in MPa
sigma_sc=130 #in MPa
a=Lef*10**3/b
#as Lef/b > 12, it is a long column
Cr=1.25-Lef*1000/(48*b) #reduction coefficient
sigma_cc=Cr*sigma_cc #in MPa
sigma_sc=Cr*sigma_sc #in MPa
Ag=b*D #in sq mm
Ac=Ag-Asc #in sq mm
P=sigma_cc*Ac+sigma_sc*Asc #in N
print "The safe load on the column=",P/10**3," kN"

The safe load on the column= 676.8622  kN


## Ex5.7:pg-176¶

In [1]:
import math

dia=500 #in mm
Asc=6*math.pi/4*25**2 #six 25 mm dia bars, in sq mm
Lef=8 #effective length of column, in m
sigma_cc=5 #in MPa
sigma_sc=190 #in MPa
a=Lef*10**3/dia
#as Lef/b >12, it is a long column
Cr=1.25-Lef*1000.0/(48*dia) #reduction coefficient
sigma_cc=Cr*sigma_cc #in MPa
sigma_sc=Cr*sigma_sc #in MPa
Ag=math.pi/4*dia**2 #in sq mm
Ac=Ag-Asc #in sq mm
P=sigma_cc*Ac+sigma_sc*Asc #in N
print "The safe load on the column=",P/10**3,"kN"
#the answer doesn't match with that given in textbook due to round-off error

The safe load on the column= 1399.3995401 kN


## Ex5.8:pg-177¶

In [10]:
import math
P=850 #in kN
sigma_cc=4 #in MPa
m=18.66 #modular ratio
sigma_sc=130 #in MPa
Lef=5*1.001 #effective length, in m
#assume 1% steel
Ag=P*10**3/(sigma_cc*0.99+sigma_sc*0.01) #in sq mm
l=math.sqrt(Ag) #in mm
l=400 #approximately, in mm
a=Lef*1000/l
#as a>12, it is a long column
#Method I-section to be changed
b=Lef*1000/12 #in mm
b=420 #approximately, in mm
Ag=b**2 #in sq mm
Asc=(P*1000-sigma_cc*Ag)/(sigma_sc-sigma_cc) #in sq mm
minimum_steel=0.8/100*b**2 #in sq mm
#as Asc < minimum steel
Asc=minimum_steel #in sq mm
#assume 20 mm dia bars
n=Asc/(math.pi/4*20**2) #no. of bars
n=5 #round-off
dia=1/4.0*20 #in mm
#as dia < 6 mm, provide 6 mm diameter links
dia=6 #in mm
spacing=min(b,16*20,48*dia,300) #in mm
print "Method I\nColumn size ",(b)," x ",(b)," mm\nMain steel =",(n),"-20 mm dia bars\nLinks=6 mm dia links @ ",(spacing)," mm c/c\n"
#Method II-same section
b=400 #in mm
Ag=b**2 #in sq mm
Cr=1.25-Lef*1000/(48*b) #reduction coefficient
sigma_cc=Cr*sigma_cc #in MPa
sigma_sc=Cr*sigma_sc #in MPa
Asc=(P*1000-sigma_cc*Ag)/(sigma_sc-sigma_cc) #in MPa
n=round(Asc/(math.pi/4*20.0**2)) #no. of bars
dia=1/4*20 #in mm
#as dia < 6 mm, provide 6 mm diameter links
dia=6 #in mm
spacing=min(b,16*20,48*dia,300) #in mm
print "Method II\nColumn size ",(b)," x ",(b)," mm\nMain steel =",(n),"-20 mm dia bars\nLinks=6 mm dia links @ ",(spacing)," mm c/c"

Method I
Column size  420  x  420  mm
Main steel = 5 -20 mm dia bars

Method II
Column size  400  x  400  mm
Main steel = 6.0 -20 mm dia bars


## Ex5.9:pg-179¶

In [11]:
import math
P=400 #in kN
b=200 #width, in mm
sigma_cc=4 #in MPa
sigma_sc=190 #in MPa
Lef=3.5 #effective length, in m
#assume 1% steel
Ag=P*10**3/(sigma_cc*0.99+sigma_sc*0.01) #in sq mm
D=Ag/b #in mm
D=340 #round-off, in mm
a=Lef*1000/b
#as a > 12, it is a long column
Cr=1.25-Lef*1000/(48*b) #reduction coefficient
sigma_cc=Cr*sigma_cc #in MPa
sigma_sc=Cr*sigma_sc #in MPa
Asc=(P*1000-sigma_cc*Ag)/(sigma_sc-sigma_cc) #in sq mm
#assume 18 mm dia bars
n=Asc/(math.pi/4*18**2) #no. of bars
n=4 #round-off
dia=1/4*20 #in mm
#as dia < 6 mm, provide 6 mm diameter links
dia=6 #in mm
spacing=min(b,16*20,48*dia,300) #in mm
print "Column size ",(b)," x ",(D)," mm\nMain steel =",(n),"-18 mm dia bars\nLinks=6 mm dia links @ ",(spacing)," mm c/c\n"

Column size  200  x  340  mm
Main steel = 4 -18 mm dia bars



## Ex5.10:pg-180¶

In [13]:
import math
P=280 #in kN
e=50 #eccentricity, in mm
b=300 #width, in mm
D=300 #depth, in mm
sigma_cc=4 #in MPa
sigma_cbc=5 #in MPa
m=18.66 #modular ratio
cover=50 #in mm
Asc=4*0.785*20**2 #four 20 mm dia bars, in sq mm
Ag=b*D #in sq mm
Ac=Ag-Asc #in sq mm
sigma_cc_cal=P*10**3/(Ac+1.5*m*Asc) #in MPa
I=b*D**3.0/12 + (m-1)*Asc*(D/2-cover)**2 #in mm**4
z=I/(D/2) #in mm**3
sigma_cbc_cal=P*10**3*e/z #in MPa
sigma_max=sigma_cc_cal + sigma_cbc_cal #in MPa
sigma_min=sigma_cc_cal - sigma_cbc_cal #in MPa
print "Maximum stress = ",sigma_max,"MPa (compressive)\nMinimum stress = ",sigma_min," MPa (tensile)"

Maximum stress =  4.60153138615 MPa (compressive)
Minimum stress =  -0.0817369463968  MPa (tensile)


## Ex5.11:pg-181¶

In [14]:
import math
P=200 #in kN
b=200 #width, in mm
D=350 #depth, in mm
sigma_cc=5 #in MPa
sigma_cbc=7 #in MPa
m=13.33 #modular ratio
Mxx=6 #in kN-m
Myy=4 #in kN-m
cover=40 #in mm
eff_cover=cover+25/2 #in mm
Asc=4*0.785*25**2 #four 25 mm dia bars, in sq mm
Ag=b*D #in sq mm
Ac=Ag-Asc #in sq mm
sigma_cc_cal=P*10**3/(Ac+1.5*m*Asc) #in MPa
#to find bending stress on XX axis
Ixx=b*D**3.0/12 + (m-1)*Asc*(D/2-eff_cover)**2 #in mm**4
Zxx=Ixx/(D/2) #in mm**3
sigma_cbc_xx=Mxx*10**6/Zxx #in MPa
#to find bending stress on YY axis
Iyy=D*b**3.0/12 + (m-1)*Asc*(b/2-eff_cover)**2 #in mm**4
Zyy=Iyy/(b/2) #in mm**3
sigma_cbc_yy=Myy*10**6/Zyy #in MPa
sigma_cbc_cal=sigma_cbc_xx + sigma_cbc_yy #in MPa
sigma_max=sigma_cc_cal + sigma_cbc_cal #in MPa
sigma_min=sigma_cc_cal - sigma_cbc_cal #in MPa
print "Maximum stress = ",sigma_max," MPa (compressive)\nMinimum stress = ",sigma_min," MPa (tensile)"

Maximum stress =  4.21961845282  MPa (compressive)
Minimum stress =  -0.490977303652  MPa (tensile)