Chapter 19:Designs by Limit State Method

Ex19.1:pg-1012

In [15]:
import math
fck=15 #in MPa
fy=250.0 #in MPa
l=4 #span, in m
MF=1.6
a=MF*20
D=l*10.0**3/a #in mm
W1=(D/10.0**3)*25 #self-weight, in kN/m
W2=1 #floor finish, in kN/m
W3=2 #live load, in kN/m
W=W1+W2+W3 #in kN/m
Wu=1.5*W #in kN/m
lef=4.125 #in m
Mu=Wu*lef**2/8 #in kN-m
d=math.sqrt(Mu*10**6/0.149/fck/10**3) #in mm
dia=12 #assume 12 mm dia bars
D=d+dia/2+15 #<125 mm (assumed value), hence OK
D=125 #in mm
d=D-dia/2-15 #in mm
#steel
#Xu=0.87*fy*Ast/0.36/fck/b = a*Ast
a=0.87*fy/0.36/fck/10**3
#using Mu=0.87 fy Ast (d-0.416 Xu), we get a quadratic equation
p=0.87*fy*0.416*a
q=-0.87*fy*d
r=Mu*10**6
Ast=(-q-math.sqrt(q**2-4*p*r))/2.0/p #in sq mm
s1=1000*0.785*dia**2/Ast #in mm
s1=105 #in mm
pt=1000*0.785*dia**2/s1/10**3/d*100.0 #in %
Ads=0.15/100*10**3*D #in sq mm
 #provide 8 mm dia bars
s2=1000*0.785*8**2/Ads #in mm
s2=265 #in mm
Vu=Wu*lef/2 #in kN
Tv=Vu*10**3/10**3/d #in MPa
 #for M15 and pt=1
Tc=0.6 #in MPa
 #for solid slabs
Tc=1.3*Tc #in MPa
 #as Tc>Tv, no shear reinforcement required
print "Summary of design:\nSlab thickness= ",round(D,2)," mm\nCover = 15 mm\nMain steel = 12 mm dia @ ",round(s1,2)," mm c/c\nDistribution steel = 8 mm dia @ ",round(s2,2)," mm c/c"

Summary of design:
Slab thickness=  125.0  mm
Cover = 15 mm
Main steel = 12 mm dia @  105.0  mm c/c
Distribution steel = 8 mm dia @  265.0  mm c/c

Ex19.2:pg-1013

In [16]:
import math
fck=15 #in MPa
fy=415.0 #in MPa
l=4.5 #span, in m
MF=1.4
a=MF*20.0
D=l*10**3.0/a #in mm
D=160.0 #in mm
W1=(D/10.0**3)*25 #self-weight, in kN/m
W2=1 #floor finish, in kN/m
W3=1 #partitions, in kN/m
W4=4 #live load, in kN/m
W=W1+W2+W3+W4 #in kN/m
Wu=1.5*W #in kN/m
lef=l+0.16 #in m
Mu=Wu*lef**2.0/8 #in kN-m
d=math.sqrt(Mu*10.0**6/0.138/fck/10.0**3) #in mm
dia=12 #assume 12 mm dia bars
D=d+dia/2.0+15 #=160 mm(assumed value), approximately
D=160 #in mm
d=140 #in mm
#steel
#Xu=0.87*fy*Ast/0.36/fck/b = a*Ast
a=0.87*fy/0.36/fck/10**3
#using Mu=0.87 fy Ast (d-0.416 Xu), we get a quadratic equation
p=0.87*fy*0.416*a
q=-0.87*fy*d
r=Mu*10**6
Ast=(-q-math.sqrt(q**2-4*p*r))/2/p #in sq mm
s1=1000*0.785*dia**2/Ast #in mm
s1=112 #in mm
pt=Ast/10**3/d*100 #in %
Ads=0.12/100*10**3*D #in sq mm
#provide 8 mm dia bars
s2=1000*0.785*8**2/Ads #in mm
s2=260 #in mm
Vu=Wu*lef/2 #in kN
Tv=Vu*10**3.0/10**3/d #in MPa
#for M15 and pt=0.718
Tc=0.53 #in MPa
#for solid slabs
Tc=1.25*Tc #in MPa
#as Tc>Tv, no shear reinforcement required
print "Summary of design:\nSlab thickness= ",(D)," mm\nCover = 15 mm\nMain steel = 12 mm dia @ ",(s1)," mm c/c\nDistribution steel = 8 mm dia @ ",round(s2,2)," mm c/c"

Summary of design:
Slab thickness=  160  mm
Cover = 15 mm
Main steel = 12 mm dia @  112  mm c/c
Distribution steel = 8 mm dia @  260.0  mm c/c

Ex19.3:pg-1013

In [17]:
import math
fck=15 #in MPa
fy=415 #in MPa
MF=1.4 #modification factor
 #let a be span to depth ratio
l=1 #span, in m
a=MF*7
D=l*1000/a #in mm
D=105 #assume, in mm
 #to calculate loading
W1=25*(D/10**3)*1.5 #self-weight, in kN/m
W2=0.5*1.5 #finish, in kN/m
W3=0.75*1.5 #live load, in kN/m
W=W1+W2+W3 #in kN/m
Wu=1.5*W #in kN/m
lef=l+0.23/2 #effective span, in m
Mu=Wu*lef/2 #in kN-m
 #check for depth
d=math.sqrt(Mu*10**6/(0.138*fck*1500)) #in mm
dia=12 #assume 12 mm dia bars
D=d+12/2+15 #<105, hence OK
D=100 #assume, in mm
d=D-dia/2-15 #in mm
#steel
#Xu=0.87*fy*Ast/0.36/fck/b = a*Ast
a=0.87*fy/0.36/fck/1.5/10**3
 #using Mu=0.87 fy Ast (d-0.416 Xu), we get a quadratic equation
p=0.87*fy*0.416*a
q=-0.87*fy*d
r=Mu*10**6
Ast=(-q-math.sqrt(q**2-4*p*r))/2/p #in sq mm
 #provide 8 mm dia bars
dia=8 #in mm
s1=1500*0.785*dia**2/Ast #>3d=3x79=237 mm
s1=235 #in mm
Ads=0.12/100*1000*D #distribution steel, in sq mm
 #assume 6 mm dia bars
s2=1000*0.785*6**2/Ads #in mm
s2=235 #round-off, in mm
Tbd=1.6 #in MPa
Ld=dia*0.87*fy/4/Tbd #in mm
Ld=452 #in mm
Tv=Wu*10**3/1500/d #in MPa
Ast=1500*0.785*8**2/235 #in sq mm
pt=Ast/1500/d*100 #in %
 #for M15 and pt=0.26
Tc=0.35 #in MPa
 #as Tc>Tv, no shear reinforcement required
print "Summary of design\nThickness of slab = ",(D)," mm\nCover = 15 mm\nMain steel = 8 mm dia @ ",(s1)," mm c/c\nDevelopment length = ",(Ld),"mm\nDistribution steel = 6 mm dia @ ",(s2)," mm c/c"

Summary of design
Thickness of slab =  100  mm
Cover = 15 mm
Main steel = 8 mm dia @  235  mm c/c
Development length =  452 mm
Distribution steel = 6 mm dia @  235  mm c/c

Ex19.4:pg-1015

In [18]:
import math
lx=3.5 #in m
ly=4 #in m
fck=15 #in MPa
fy=250 #in MPa
D=lx*10**3.0/35 #in mm
W1=(D/10**3)*25 #self-weight, in kN/m
W2=1.5 #live load, in kN/m
W=W1+W2 #in kN/m
Wu=1.5*W #in kN/m
a=ly/lx
Ax=0.078
Ay=0.0602
Mx=Ax*Wu*lx**2 #in kN-m
My=Ay*Wu*lx**2 #in kN-m
d=math.sqrt(Mx*10**6/0.149/fck/10**3) #in mm
d=51 #round-off, in mm
 #assume 10 mm dia bars
dia=10 #in mm
D=d+dia/2+15 #<100 mm assumed value
D=100 #in mm
d=D-dia/2-15 #in mm
#steel - short span
#Xu=0.87*fy*Ast/0.36/fck/b = a*Ast
a=0.87*fy/0.36/fck/10**3
 #using Mu=0.87 fy Ast (d-0.416 Xu), we get a quadratic equation
p=0.87*fy*0.416*a
q=-0.87*fy*d
r=Mx*10**6
Ast=(-q-math.sqrt(q**2-4*p*r))/2/p #in sq mm
s1=1000*0.785*dia**2/Ast #in mm
s1=220 #round-off, in mm
 #long span
d=d-dia/2-dia/2 #in mm
 #Xu=0.87*fy*Ast/0.36/fck/b = a*Ast
a=0.87*fy/0.36/fck/10**3
 #using Mu=0.87 fy Ast (d-0.416 Xu), we get a quadratic equation
p=0.87*fy*0.416*a
q=-0.87*fy*d
r=My*10**6
Ast=(-q-math.sqrt(q**2-4*p*r))/2/p #in sq mm
s2=1000*0.785*dia**2/Ast #in mm
s2=250 #round-off, in mm
print "Summary of design\nSlab thickness=",(D)," mm\nCover=15 mm\nSteel-\n(i)Short span = 10 mm dia @ ",(s1)," mm c/c\n(ii)Long span = 10 mm dia @ ",(s2)," mm c/c"

Summary of design
Slab thickness= 100  mm
Cover=15 mm
Steel-
(i)Short span = 10 mm dia @  220  mm c/c
(ii)Long span = 10 mm dia @  250  mm c/c

Ex19.5:pg-1016

In [19]:
import math
b=225 #width in mm
D=300 #depth in mm
fck=15 #in MPa
fy=415 #in MPa
l=4.2 #span, in m
W1=(b/10**3)*(D/10.0**3)*25 #self-weight, in kN/m
W2=6 #live load, in kN/m
W=W1+W2 #in kN/m
Wu=1.5*W #in kN/m
Mu=Wu*l**2.0/8 #in kN-m
d=270 #assume, in mm
Mulim=0.138*fck*b*d**2.0/10**6 #in kN-m
#as Mulim > Mu, it will be a singly reinforced beam
#Xu=0.87*fy*Ast/0.36/fck/b = a*Ast
a=0.87*fy/0.36/fck/b
#using Mu=0.87 fy Ast (d-0.416 Xu), we get a quadratic equation
p=0.87*fy*0.416*a
q=-0.87*fy*d
r=Mu*10**6
Ast=(-q-math.sqrt(q**2-4*p*r))/2/p #in sq mm
#provide 12 mm dia bars
n=Ast/0.785/12**2
n=3 #assume
Ast=n*0.785*12**2 #in sq mm
Vu=Wu*l/2 #in kN
Tv=Vu*10**3/b/d #in MPa
pt=Ast/b/d*100 #pt=0.56
#for M15 and pt=0.56
Tc=0.46 #in MPa
#as Tc>Tv, no shear reinforcement required
#provide nominal stirrups and provide 6 mm stirrups
Asv=2*0.785*6**2 #in sq mm
Sv=Asv*fy/0.4/b #in mm
Sv=260 #assume, in mm
Svmax=0.75*d #in mm
Svmax=200 #round-off, in mm
Sv=min(Sv,Svmax) #in mm
print "Summary of design:\nBeam size - ",(b)," x ",(D)," mm\nCover - 25 mm\nSteel - ",(n),"-12 mm dia bars\nStirrups - 6 mm dia @ ",(Sv)," mm c/c"
#deflection check
Ec=5700*math.sqrt(fck) #in MPa
Es=2*10.0**5 #in MPa
m=Es/Ec
fcr=0.7*math.sqrt(fck) #in MPa
#using b x x/2 = m Ast (d-x), we get a quadratic equation
#solving the quadratic equation
p=b/2
q=m*Ast
r=-m*Ast*d
x=(-q+math.sqrt(q**2-4*p*r))/2/p #in mm
z=d-x/3 #in mm
Ir=b*x**3.0/12+b*x*(x/2)**2+m*Ast*(d-x)**2 #in mm**4
Igr=b*D**3.0/12 #in mm**4
yt=D/2 #in mm
Mr=fcr*Igr/yt #in N-mm
M=W*l**2.0/8*10**6 #in N-mm
Ieff=Ir/(1.2-Mr/M*z/d*(1-x/d)*b/b) #in mm**4
#Ir<Ieff<Igr, hence OK
W1=W*l #in kN
u1=5/384.0*(W1*10**3)*(l*10**3)**3/Ec/Ieff #short-term deflection, in mm
#long-term deflection
#(i) deflection due to shrinkage
k3=0.125 #for simply supported beam
pt=0.56 #in %
pc=0 #in %
k4=0.72*(pt-pc)/math.sqrt(pt)
phi=k4*0.0003/D
u2=k3*phi*(l*10**3)**2 #in mm
#(ii) deflection due to creep
Ecc=Ec/(1+1.6) #in MPa
#assuming a permanent load of 60%
W2=0.6*W*l #in kN
u3=5/384.0*(W2*10**3)*(l*10**3)**3/Ecc/Ieff #in mm
u4=5/384.0*(W2*10**3)*(l*10**3)**3/Ec/Ieff #in mm
u5=u3-u4 #in mm
u=u1+u2+u5 #total deflection, in mm
v1=l*10**3.0/250 #permissible deflection, in mm
v2=l*10**3.0/350 #in mm
#assuming half the shrinkage strain occurs within the first 28 days, the deflection occurring after this time
v3=u2/2+u5 #< permissible value, hence OK

Summary of design:
Beam size -  225  x  300  mm
Cover - 25 mm
Steel -  3 -12 mm dia bars
Stirrups - 6 mm dia @  200  mm c/c

Ex19.6:pg-1018

In [20]:
import math
l=7 #span, in m
fck=15 #in MPa
fy=250 #in MPa
b=300 #assume, in mm
W=35 #live load, in kN/m
Wu=1.5*W #in kN/m
Mu=Wu*l**2/8.0 #in kN-m
d=(Mu*10**6/0.149/fck/b)**0.5 #in mm
d=1.1*d #increase depth by 10% for self-weight
d=750 #assume, in mm
c=50 #cover, in mm
D=d+c #in mm
W1=(b/10.0**3)*(D/10**3)*25 #self-weight, in kN/m
W2=35 #live load, in kN/m
W=W1+W2 #in kN/m
Wu=1.5*W #in kN/m
Mu=Wu*l**2/8 #in kN-m
d=(Mu*10**6/0.149/fck/b)**0.5 #<750 mm, hence OK
d=750 #in mm
#steel
#Xu=0.87*fy*Ast/0.36/fck/b = a*Ast
a=0.87*fy/0.36/fck/b
#using Mu=0.87 fy Ast (d-0.416 Xu), we get a quadratic equation
p=0.87*fy*0.416*a
q=-0.87*fy*d
r=Mu*10**6
Ast=(-q-math.sqrt(q**2-4*p*r))/2/p #in sq mm
#provide 20 mm dia bars
n=Ast/0.785/20**2
#provide 8-20 mm + 2-18 mm dia bars
Ast=8*0.785*20**2+2*0.785*18**2 #in sq mm
pt=Ast/b/d*100 #pt=1.34
Vu=Wu*l/2 #in kN
Tv=Vu*10**3/b/d #in MPa
#for M15 and pt=1.34
Tc=0.65 #in MPa
#as Tv>Tc, shear reinforcement required
#provide 6 mm stirrups
Vus=Vu-Tc*b*d/10.0**3 #in kN
Asv=2*0.785*6**2 #in sq mm
Sv=Asv*0.87*fy*d/Vus/10.0**3 #in mm
Sv=130 #assume, in mm
Svmin=Asv*fy/0.4/b #in mm
Svmin=115 #assume, in mm
Sv=min(Sv,Svmin) #in mm
print "Summary of design:\nBeam size - ",(b)," x ",(D)," mm\nCover - 50 mm\nSteel - 8-20 mm + 2-18 mm dia bars\nStirrups - 6 mm dia @ ",(Sv)," mm c/c"
#deflection check
Ec=5700*math.sqrt(fck) #in MPa
Es=2*10**5 #in MPa
m=Es/Ec
fcr=0.7*math.sqrt(fck) #in MPa
#using b x x/2 = m Ast (d-x), we get a quadratic equation
#solving the quadratic equation
p=b/2
q=m*Ast
r=-m*Ast*d
x=(-q+math.sqrt(q**2-4*p*r))/2/p #in mm
x=290 #assume, in mm
z=d-x/3 #in mm
Ir=b*x**3/12+b*x*(x/2)**2+m*Ast*(d-x)**2 #in mm**4
Igr=b*D**3/12 #in mm**4
yt=D/2 #in mm
Mr=fcr*Igr/yt #in N-mm
M=W*l**2/8*10**6 #in N-mm
Ieff=Ir/(1.2-Mr/M*z/d*(1-x/d)*b/b) #in mm**4
 #Ir>Ieff
Ieff=Ir #in mm**4
W1=W*l #in kN
u1=5.0/384*(W1*10**3)*(l*10**3)**3/Ec/Ieff #short-term deflection, in mm
#long-term deflection
#(i) deflection due to shrinkage
k3=0.125 #for simply supported beam
pt=1.34 #in %
pc=0 #in %
k4=0.65*(pt-pc)/math.sqrt(pt)
phi=k4*0.0003/D
u2=k3*phi*(l*10**3)**2 #in mm
#(ii) deflection due to creep
Ecc=Ec/(1+1.6) #in MPa
#assuming a permanent load of 60%
W2=0.6*W*l #in kN
u3=5/384*(W2*10**3)*(l*10**3)**3/Ecc/Ieff #in mm
u4=5/384*(W2*10**3)*(l*10**3)**3/Ec/Ieff #in mm
u5=u3-u4 #in mm
u=u1+u2+u5 #total deflection, in mm
v1=l*10**3/250 #permissible deflection, in mm
v2=l*10**3/350 #in mm
 #assuming half the shrinkage strain occurs within the first 28 days, the deflection occurring after this time
v3=u2/2+u5 #< permissible value, hence OK

Summary of design:
Beam size -  300  x  800  mm
Cover - 50 mm
Steel - 8-20 mm + 2-18 mm dia bars
Stirrups - 6 mm dia @  115  mm c/c

Ex19.7:pg-1019

In [21]:
import math
l=10 #span, in m
fck=15 #in MPa
fy=250 #in MPa
Df=100 #slab thickness, in mm 
D=l*10**3.0/15 #depth of beam, in mm
D=600 #assume, in mm
d=D-50 #cover=50 mm
bw=300 #beam width, in mm
bf=l*10**3/6+bw+6*Df #>2500 mm c/c distance of beams
bf=2500 #in mm
W1=(bw/10**3)*(D-Df)/10**3*25 #web, in kN/m
W2=(Df/10**3)*(bf/10**3)*25 #slab, in kN/m
W3=(bf/10**3)*5 #imposed load, in kN/m
W=W1+W2+W3 #in kN/m
Wu=1.5*W #in kN/m
Mu=Wu*l**2.0/8 #in kN-m
Vu=Wu*l/2 #in kN
Asf=0.36*fck*bf*Df/0.87/fy #steel required only for flange, in sq mm
Asf=6210 #round-off, in sq mm
#verification of trial section
xu=100 #assume, in mm
Ast=Asf #in sq mm
Mulim=0.87*fy*Ast*(d-0.416*xu)/10**6 #in kN-m
#Mulim > Mu, hence OK
#keeping the assumed trial section, work out the steel required
#Xu=0.87*fy*Ast/0.36/fck/b = a*Ast
a=0.87*fy/0.36/fck/bf
 #using Mu=0.87 fy Ast (d-0.416 Xu), we get a quadratic equation
p=0.87*fy*0.416*a
q=-0.87*fy*d
r=Mu*10**6
Ast=(-q-math.sqrt(q**2-4*p*r))/2/p #in sq mm
 #provide 5-25 mm dia + 3-22 mm dia bars
pt=Ast*100/(bw*d+(bf-bw)*Df) #pt=1%, approximately
 #check for shear
Tv=Vu*10**3/bw/d #in MPa
 #for M15 grade concrete and pt=1%
Tc=0.6 #in MPa
 #as Tv > Tc, shear reinforcement required
Vus=Vu-Tc*bw*d/10**3 #in kN
 #provide 6 mm dia stirrups
Asv=2*0.785*6**2 #in sq mm
Sv=Asv*0.87*fy*d/Vus/10**3 #in mm
Sv=90 #round-off, in mm
print "T beam:bf=",bf," mm\nDf=",Df," mm\nd=",d," mm\nD=",D," mm\nCover = 50 mm\nSteel= 5-25 mm dia + 3-22 mm dia bars\nStirrups = 6 mm dia @ ",Sv," mm c/c throughout"
#answer in textbook for spacing of stirrups is incorrect
#deflection check
Ec=5700*math.sqrt(fck) #in MPa
Es=2*10**5 #in MPa
m=Es/Ec #modular ratio
fcr=0.7*math.sqrt(fck) #in MPa
#using bf Df (x-Df/2) = m Ast (d-x), we get a quadratic equation
x=(m*Ast*d+bf*Df**2/2)/(bf*Df+m*Ast) #in mm
z=0.87*d #assume, in mm
#refer Fig. 19.5 of textbook
Ir=bf*x**3.0/12+bf*Df*(x/2)**2+m*Ast*(d-x)**2 #in mm**4
y=(bf*Df*Df/2+(D-Df)*bw*((D-Df)/2+Df))/(bf*Df+(D-Df)*bw) #c.g. from top, in mm (neglecting steel)
Igr=bf*Df**3.0/12+bf*Df*(Df/2-y)**2+bw*(D-Df)**3/12+bw*(D-Df)*((D-Df)/2+Df-y)**2 #in mm**4
yt=d/2 #in mm
Mr=fcr*Igr/yt #in N-mm
M=W*l**2/8*10**6 #in N-mm
Ieff=Ir/(1.2-Mr/M*z/d*(1-x/d)*bw/bf) #in mm**4
 #Ir > Ieff
Ieff=Ir #in mm**4
W1=W*l #in kN
u1=5.0/384*(W1*10**3)*(l*10**3)**3/Ec/Ieff #short term deflection, in mm
 #deflection due to shrinkage
k3=0.125 #for simply supported beam
pt=1 #in %
pc=0 #in %
k4=0.65*(pt-pc)/math.sqrt(pt)
phi=k4*0.0003/D
u2=k3*phi*(l*10**3)**2 #in mm
 #deflection due to creep
Ecc=Ec/(1+1.6) #in MPa
 #assuming a permanent load of 60%
W2=0.6*W*l #in kN
u3=5/384*(W2*10**3)*(l*10**3)**3/Ecc/Ieff #in mm
u4=5/384*(W2*10**3)*(l*10**3)**3/Ec/Ieff #in mm
u5=u3-u4 #in mm
u=u1+u2+u5 #total deflection, in mm
v1=l*10**3.0/250 #permissible deflection, in mm
v2=l*10**3.0/350 #>20 mm
v2=20 #in mm
 #assuming half the shrinkage strain occurs within the first 28 days, the deflection occurring after this time
v3=u2/2+u5 #< permissible value, hence OK
print "V3=",v3,"\nThus the value of V3 is permissible"

T beam:bf= 2500  mm
Df= 100  mm
d= 550  mm
D= 600  mm
Cover = 50 mm
Steel= 5-25 mm dia + 3-22 mm dia bars
Stirrups = 6 mm dia @  90  mm c/c throughout
V3= 2.03125 
Thus the value of V3 is permissible

Ex19.8:pg-1021

In [22]:
import math
l=2.7+1 #span, in m
R=0.15 #rise, in m
t=0.27 #tread, in m
fck=15 #in MPa
fy=415 #in MPa
D=200 #assume, in mm
W1=D/10**3*25*math.sqrt(R**2+t**2)/t #slab load on plan, in kN/m
W2=1/2*R*t*25/t #load of step per metre, in kN/m
W3=3 #live load, in kN/m
W=W1+W2+W3 #in kN/m
Wu=1.5*W #in kN/m
Mu=Wu*l**2/8 #in kN-m
d=math.sqrt(Mu*10**6/0.138/fck/10**3) #in mm
d=115 #round-off, in mm
 #assume 10 mm dia bars
dia=10 #in mm
D=d+dia/2+25 #< 200 mm, hence OK
D=l*10**3/24 #depth required for deflection, in mm
D=155 #round-off, in mm
d=D-dia/2-25 #in mm
#steel
#Xu=0.87*fy*Ast/0.36/fck/b = a*Ast
a=0.87*fy/0.36/fck/10**3
#using Mu=0.87 fy Ast (d-0.416 Xu), we get a quadratic equation
p=0.87*fy*0.416*a
q=-0.87*fy*d
r=Mu*10**6
Ast=(-q-math.sqrt(q**2-4*p*r))/2/p #in sq mm
s1=1000*0.785*dia**2/Ast #spacing of 10 mm dia bars
s1=110 #round-off, in mm
Ads=0.12/100*D*10**3 #distribution steel, in sq mm
#provide 8 mm dia bars
s2=1000*0.785*8**2/Ads #in mm
s2=270 #round-off, in mm
print "Summary of design\nSlab thickness=",D," mm\nCover = 25 mm\nMain steel = 10 mm dia bars @ ",s1," mm c/c\nDistribution steel = 8 mm dia @ ",s2," mm c/c"
 #answer in textbook for spacing of 10 mm dia bars is incorrect

Summary of design
Slab thickness= 155  mm
Cover = 25 mm
Main steel = 10 mm dia bars @  110  mm c/c
Distribution steel = 8 mm dia @  270  mm c/c

Ex19.9:pg-1022

In [23]:
import math
b=0.2 #column width, in m
D=0.3 #column depth, in m
fck=15 #in MPa
fy=415 #in MPa
P1=600 #load on column, in kN
P2=0.05*P1 #weight of footing, in kN
P=P1+P2 #in kN
Pu=1.5*P #in kN
q=150 #bearing capacity of soil, in kN/sq m
qu=2*q #ultimate bearing capacity of soil, in kN/sq m
A=Pu/qu #in sq m
L=math.sqrt(A) #assuming footing to be square, in m
L=1.8 #round-off, in m
p=P1*1.5/L**2 #soil pressure, in kN/sq m
p=277.8 #round-off, in kN/sq m
bc=b/D
ks=0.5+bc #>1
ks=1
Tc=0.25*math.sqrt(fck)*10**3 #in kN/sq m
Tv=Tc
#let d be the depth of footing in metres
#case I: consider greater width of shaded portion in Fig. 19.6 of textbook
d1=L*(L-b)/2*p/(Tc*L+L*p) #in m
#case II: refer Fig. 19.7 of textbook; we get a quadratic equation of the form e d**2 + f d + g = 0
e=p+4*Tc
f=b*p+D*p+2*(b+D)*Tc
g=-(L**2-b*D)*p
d2=(-f+math.sqrt(f**2-4*e*g))/2/e #in m
d2=0.35 #round-off, in m
 #bending moment consideration, refer Fig. 19.8 of textbook
Mx=1*((L-b)/2)**2/2*p #in kN-m
My=1*((L-D)/2)**2/2*p #in kN-m
d3=math.sqrt(Mx*10**6/0.138/fck/10**3) #<350 mm, hence OK
#steel
#Xu=0.87*fy*Ast/0.36/fck/b = a*Ast
a=0.87*fy/0.36/fck/10**3
#using Mu=0.87 fy Ast (d-0.416 Xu), we get a quadratic equation
p=0.87*fy*0.416*a
q=-0.87*fy*d2*10**3
r=Mx*10**6
Ast=(-q-math.sqrt(q**2-4*p*r))/2/p #in sq mm
Ast=L*Ast #steel required for full width of 1.8 m
 #provide 12 mm dia bars
dia=12 #in mm
n=Ast/0.785/dia**2 #no. of 12 mm dia bars
n=12 #round-off
Tbd=1.6 #in MPa
Ld=dia*0.87*fy/4/Tbd #in mm
Ld=677 #assume, in mm
 #this length is available from the face of the column in both directions
D=d2*10**3+dia/2+100 #in mm
print "Summary of design:\nOverall depth of footing=",D," mm\nCover=100 mm\nSteel-",n," bars of 12 mm dia both ways"

Summary of design:
Overall depth of footing= 456.0  mm
Cover=100 mm
Steel- 12  bars of 12 mm dia both ways

Ex19.10:pg-1023

In [24]:
import math
fck=15 #in MPa
fy=415 #in MPa
phi=30 #angle of repose, in degrees
H=5 #height of wall, in m
B=0.6*H #assume, in m
T=B/4 #assume toe to base ratio as 1:4, in m
W=16 #density of retained earth, in kN/cu m
Wu=1.5*W #factored load, in kN/cu m
P=Wu*H**2/2*(1-math.sin(phi))/(1+math.sin(phi)) #in kN
M1=P*H/3 #in kN-m
M1=167 #round-off, in kN-m
 #bending moment at 2.5 m below the top
h=2.5 #in m
M2=Wu*h**2/2*(1-math.sin(phi))/(1+math.sin(phi))*h/3 #in kN-m
M2=21 #round-off, in kN-m
 #thickness of stem (at the base)
d=math.sqrt(M1*10**6/0.138/fck/1000) #in mm
d=285 #round-off, in mm
dia=20 #assume 20 mm dia bars
D1=d+dia/2+25 #in mm
D2=200 #thickness at top, in mm
D3=D2+(D1-D2)*h/H #thickness at 2.5 m below top, in mm 
d3=math.sqrt(M2*10**6/0.138/fck/1000) #in mm
D3=d3+dia/2+25 #< 260 mm (provided), hence OK
D3=260 #in mm
d3=D3-dia/2-25 #in mm
#main steel
#(a) 5 m below the top
#Xu=0.87*fy*Ast/0.36/fck/b = a*Ast
a=0.87*fy/0.36/fck/10**3
#using Mu=0.87 fy Ast (d-0.416 Xu), we get a quadratic equation
p=0.87*fy*0.416*a
q=-0.87*fy*d
r=M1*10**6
Ast=(-q-math.sqrt(q**2-4*p*r))/2/p #in sq mm
pt=Ast/1000/d*100 #in %
 #provide 20 mm dia bars
s1=1000*0.785*20**2/Ast #in mm
s1=155 #round-off, in mm
#(b) 2.5 m below the top
#Xu=0.87*fy*Ast/0.36/fck/b = a*Ast
a=0.87*fy/0.36/fck/10**3
#using Mu=0.87 fy Ast (d-0.416 Xu), we get a quadratic equation
p=0.87*fy*0.416*a
q=-0.87*fy*d3
r=M2*10**6
Ast=(-q-math.sqrt(q**2-4*p*r))/2/p #in sq mm
Astmin=0.12/100*10**3*D3 #in sq mm
Ast=max(Ast,Astmin) #in sq mm
#provide 12 mm dia bars
s2=1000*0.785*12**2/Ast #in mm
s2=360 #round-off, in mm
#distribution steel
Ads=0.12/100*10**3*D3 #in sq mm
#provide 8 mm dia bars
s3=1000*0.785*8**2/Ads #in mm
s3=160 #round-off, in mm
#check for shear
Vu=P #in kN
Tv=Vu*10**3.0/10**3/d #in MPa
#for M15 grade concrete and pt=0.71
Tc=0.54 #in MPa
#as Tc > Tv, no shear reinforcement required
#development length
 #(a) At the base of stem
dia=20 #in mm
Tbd=1.6 #in MPa
Ld=dia*0.87*fy/4/Tbd #in mm
Ld=1130 #round-off, in mm
 #(b) At 2.5 m below the top
dia=12 #in mm
Ld=dia*0.87*fy/4/Tbd #in mm
Ld=680 #round-off, in mm
print "Summary of design:\nThickness of stem (at base) = ",D1," mm\nThickness of stem at top = ",D2," mm\nRefer Fig. 19.10 of textbook for reinforcement details"

Summary of design:
Thickness of stem (at base) =  320  mm
Thickness of stem at top =  200  mm
Refer Fig. 19.10 of textbook for reinforcement details

Ex19.11:pg-1024

In [25]:
import math
P=1000 #in kN
Pu=1.5*P #in kN
fck=15 #in MPa
fy=415 #in MPa
l=3.5 #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=420 #in mm
emin=l*10**3/500+L/30 #in mm
ep=0.05*L #=emin, hence OK
Asc=0.01*L**2 #in sq mm
 #provide 6-20 mm dia bars
Asc=6*0.785*20**2 #in sq mm
print "Summary of design:\nColumn size - ",L," x ",L," mm\nSteel-main = 6-20 mm dia bars"

Summary of design:
Column size -  420  x  420  mm
Steel-main = 6-20 mm dia bars

Ex19.12:pg-1025

In [26]:
import math
P=500 #in kN
Pu=1.5*P #in kN
fck=15 #in MPa
fy=250 #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=315 #in mm
emin=l*10**3.0/500+L/30 #<20
emin=20 #in mm
ep=0.05*L #<emin, hence the column is to be checked for bending
Mu=Pu*10**3*emin #in N-mm
a=Pu*10**3/fck/L/L
b=Mu/fck/L/L**2 #b=0.032
d1=40 #cover(assume), in mm
c=d1/L #c=d'/D
 #for d'/D = 0.15
p=0.07*fck #in %
Asc=p/100*L**2 #in sq mm
 #provide 4-20 mm dia bars
Asc=4*0.785*20**2 #in sq mm
print "Summary of design:\nColumn size - ",L," x ",L," mm\nSteel-main = 4-20 mm dia bars"

Summary of design:
Column size -  315  x  315  mm
Steel-main = 4-20 mm dia bars

Ex19.13:pg-1026

In [27]:
import math
P=500 #in kN
Pu=1.5*P #in kN
fck=15 #in MPa
fy=250 #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
b=250 #in mm
D=Ag/b #in mm
D=400 #round-off, in mm
emin1=l*10**3/500+D/30 #in direction of Y axis, in mm, < 20 mm
emin1=20 #in mm
ep1=0.05*D #=emin, hence no moment is required to be considered in this direction
emin2=l*10**3/500+b/30 #in direction of X axis, in mm, < 20 mm
emin2=20 #in mm
ep2=0.05*b #<emin, hence moment in this direction needs to be considered
 #interaction diagram
b=400 #in mm
D=250 #in mm
Mu=Pu*10**3*emin2 #in N-mm
m=Pu*10**3/fck/b/D
n=Mu/fck/b/D**2 #b=0.032
d1=40 #cover(assume), in mm
c=d1/D #c=d'/D
 #referring to Fig. 19.12
p=0.08*fck #in %
Asc=p/100*b*D #in sq mm
 #provide 6-16 dia bars
Asc=6*0.785*16**2 #in sq mm
print "Summary of design:\nColumn size - ",D," x ",b," mm\nSteel-main = 6-16 mm dia bars"

Summary of design:
Column size -  250  x  400  mm
Steel-main = 6-16 mm dia bars

Ex19.14:pg-1027

In [28]:
import math
P=500.0 #in kN
Pu=1.5*P #in kN
fck=15.0 #in MPa
fy=250.0 #in MPa
l=5.0 #effective length, in m
lex=5.0 #in m
ley=5.0 #in m
L=315.0 #column dimension in mm (square column)
Asc=1256.0 #in sq mm
m=lex*10**3/L #>12
n=ley*10**3/L #>12
 #hence the column is slender on both the axes
Max=Pu*10**3*L/2000*(lex/(L/10**3))**2.0/10**6 #in kN-m
May=Max
Puz=(0.45*fck*(L**2-Asc)+0.75*fy*Asc)/10**3 #in kN
c=40 #cover, in mm
 #to find Pb
xu=(L-c)/(1+0.002/0.0035) #in mm
Pb=0.36*fck*L*xu/10**3 #in kN
k=(Puz-Pu)/(Puz-Pb) #>1
Max=k*Max #in kN-m
Mu=15 #in kN-m
Mu=Mu+Max #in kN-m
a=Pu*10**3/fck/L/L
b=Mu*10**6/fck/L/L**2 #b=0.047
d1=c/L #d1=d'/D
 #for d'/D = 0.1
p=0.095*fck #in %
Asc=p/100*L**2 #in sq mm
 #provide 4-18 mm + 4-12 mm dia bars
Asc=4*0.785*18**2+4*0.785*12**2 #in sq mm
print "Summary of design:\nColumn size - ",L," x ",L," mm\nSteel-main = 4-18 mm + 4-12 mm dia bars"

Summary of design:
Column size -  315.0  x  315.0  mm
Steel-main = 4-18 mm + 4-12 mm dia bars

Ex19.15:pg-1028

In [29]:
import math
Pu=2000.0 #in kN
Mux=50.0 #in kN-m
Muy=Mux
fck=20.0 #in MPa
fy=415.0 #in MPa
 #assume 2% steel
p=2 #in %
Ag=Pu*10**3/(0.4*fck*(1-p/100)+0.67*fy*p/100) #in sq mm
L=math.sqrt(Ag) #assuming a square column
L=400 #in mm
m=Pu*10**3/fck/L/L
n=p/fck
c=50 #cover (assume), in mm
d1=c/L #d1=d'/D
 #from Fig. 19.21, for d'/D = 0.15 and Pu / fck b D = 0.625
f=0.046
Mux1=f*fck*L*L**2.0/10**6 #in kN-m
Muy1=Mux1
Puz=(0.45*fck*(1-p/100)*L**2+0.75*fy*p/100*L**2)/10**3 #in kN
a=Pu/Puz #>0.8
an=2
b=(Mux/Mux1)**an+(Muy/Muy1)**an #>1
 #assume 2.5% steel
p=2.5 #in %
n=p/fck
 #from Fig. 19.21, for d'/D = 0.15 and Pu / fck b D = 0.625
f=0.08
Mux1=f*fck*L*L**2/10**6 #in kN-m
Muy1=Mux1
Puz=(0.45*fck*(1-p/100)*L**2+0.75*fy*p/100*L**2)/10**3 #in kN
a=Pu/Puz #<0.8
an=1+1/0.6*(a-0.2)
b=(Mux/Mux1)**an+(Muy/Muy1)**an #<1, hence OK
Asc=p/100*L**2 #in sq mm
 #provide 12-22 mm dia bars
Asc=12*0.785*22**2 #in sq mm
print "Summary of design:\nColumn size - ",L," x ",L," mm\nSteel-main = 12-22 mm dia bars placed equally on four faces of the column"

Summary of design:
Column size -  400  x  400  mm
Steel-main = 12-22 mm dia bars placed equally on four faces of the column

Ex19.16:pg-1029

In [30]:
import math
b=400 #in mm
D=500 #in mm
Pu=1600 #in kN
Mux=90 #in kN-m
Muy=50 #in kN-m
fck=15 #in MPa
fy=415 #in MPa
p=1.5 #assume 1.5% steel, placed on four sides
m=p/fck
c=50 #cover (assume), in mm
 #to find Mux1
n=c/D #n=d'/D
l=Pu*10**3/fck/b/D
 #referring to Fig.19.20, for Pu/ fck/ b/ D = 0.53 and p/ fck = 0.1
f=0.09
Mux1=f*fck*b*D**2/10**6 #in kN-m
 #to find Muy1
b=500 #in mm
D=400 #in mm
n=c/D #n=d'/D
l=Pu*10**3/fck/b/D
 #referring to Fig.19.21, for Pu/ fck/ b/ D = 0.53 and p/ fck = 0.1
f=0.08
Muy1=f*fck*b*D**2/10**6 #in kN-m
Puz=(0.45*fck*(1-p/100)*b*D+0.75*fy*p/100*b*D)/10**3 #in kN
a=Pu/Puz #<0.8
an=1+1/0.6*(a-0.2)
r=(Mux/Mux1)**an+(Muy/Muy1)**an #<1
Asc=p/100*b*D #in sq mm
 #provide 6-16 mm + 6-20 mm dia bars
Asc=6*0.785*16**2+6*0.785*20**2 #in sq mm
print "Summary of design:\nColumn size - ",D," x ",b," mm\nSteel-main = 6-16 mm + 6-20 mm dia bars"
 #answer in textbook is incorrect

Summary of design:
Column size -  400  x  500  mm
Steel-main = 6-16 mm + 6-20 mm dia bars

Ex19.17:pg-1030

In [31]:
import math
b=300 #in mm
Pu=1500 #in kN
Mux=100 #in kN-m
Muy=70 #in kN-m
fck=15 #in MPa
fy=250 #in MPa
p=1.5 #assume 1.5% steel, placed on four sides
Ag=Pu*10**3/(0.4*fck*(1-p/100)+0.67*fy*p/100) #in sq mm
D=Ag/b #in mm
D=600 #assume, in mm
m=p/fck
c=60 #cover (assume), in mm
 #to find Mux1
n=c/D #n=d'/D
l=Pu*10**3/fck/b/D
 #referring to Fig.19.17, for Pu/ fck/ b/ D = 0.56 and p/ fck = 0.1
f=0.038
Mux1=f*fck*b*D**2/10**6 #in kN-m
 #to find Muy1
b=600 #in mm
D=300 #in mm
n=c/D #n=d'/D
l=Pu*10**3/fck/b/D
 #referring to Fig.19.19, for Pu/ fck/ b/ D = 0.56 and p/ fck = 0.1
f=0.038
Muy1=f*fck*b*D**2.0/10**6 #in kN-m
Puz=(0.45*fck*(1-p/100)*b*D+0.75*fy*p/100*b*D)/10**3 #in kN
a=Pu/Puz #>0.8
an=2
r=(Mux/Mux1)**an+(Muy/Muy1)**an #>1
p=4 #assume 4% steel, second trial
m=p/fck
 #to find Mux1
b=300 #in mm
D=600 #in mm
 #referring to Fig.19.17, for Pu/ fck/ b/ D = 0.56 and p/ fck = 0.26
f=0.15
Mux1=f*fck*b*D**2/10**6 #in kN-m
 #to find Muy1
b=600 #in mm
D=300 #in mm
n=c/D #n=d'/D
 #referring to Fig.19.19, for Pu/ fck/ b/ D = 0.56 and p/ fck = 0.26
f=0.15
Muy1=f*fck*b*D**2.0/10**6 #in kN-m
Puz=(0.45*fck*(1-p/100)*b*D+0.75*fy*p/100*b*D)/10**3 #in kN
a=Pu/Puz #<0.8
an=1+1/0.6*(a-0.2)
r=(Mux/Mux1)**an+(Muy/Muy1)**an #<1, hence OK
 #but steel can be reduced
p=3 #assume 3% steel, second trial
m=p/fck
 #to find Mux1
b=300 #in mm
D=600 #in mm
 #referring to Fig.19.17, for Pu/ fck/ b/ D = 0.56 and p/ fck = 0.2
f=0.12
Mux1=f*fck*b*D**2.0/10**6 #in kN-m
 #to find Muy1
b=600 #in mm
D=300 #in mm
n=c/D #n=d'/D
 #referring to Fig.19.19, for Pu/ fck/ b/ D = 0.56 and p/ fck = 0.2
f=0.12
Muy1=f*fck*b*D**2.0/10**6 #in kN-m
Puz=(0.45*fck*(1-p/100)*b*D+0.75*fy*p/100*b*D)/10**3 #in kN
a=Pu/Puz #<0.8
an=1+1/0.6*(a-0.2)
r=(Mux/Mux1)**an+(Muy/Muy1)**an #<1, hence OK
Asc=p/100*b*D #in sq mm
 #provide 12-25 dia bars
Asc=12*0.785*25**2 #in sq mm
print "Summary of design:\nColumn size - ",B," x ",b," mm\nSteel-main = 12-25 mm dia bars"

Summary of design:
Column size -  3.0  x  600  mm
Steel-main = 12-25 mm dia bars