# Chapter 14:Limit State Met¶

## Ex14.1:pg-761¶

In [8]:
import math
b=250 #width, in mm
d=500 #effective depth, in mm
Ast=4*0.785*20**2 #four 20 mm dia bars, in sq mm
fck=15 #in MPa
fy=250 #in MPa
Xu=round(0.87*fy*Ast/0.36/fck/b) #in mm
Xc=0.531*d #in mm
#as Xu<Xc, it is under-reinforced section, hence OK
Mu=0.87*fy*Ast*(d-0.416*Xu)/10**6 #in kN-m
print "Moment of resistance of the beam=",Mu," kN-m"

Moment of resistance of the beam= 113.63413824  kN-m


## Ex14.2:pg-761¶

In [9]:
import math
b=300 #width, in mm
d=600 #effective depth, in mm
fck=15 #in MPa
fy=500 #in MPa
Xc=0.456*d #in mm
Mu=0.36*fck*b*Xc*(d-0.416*Xc)/10**6 #in kN-m
Ast=round(0.36*fck*b*Xc/0.87/fy) #in sq mm
print "Moment of resistance of the beam=",Mu," kN-m\nSteel required=",Ast," sq mm"

Moment of resistance of the beam= 215.491597517  kN-m
Steel required= 1019.0  sq mm


## Ex14.3:pg-762¶

In [10]:
import math
b=300 #width, in mm
d=600 #effective depth, in mm
fck=20 #in MPa
fy=415 #in MPa
Xc=0.479*d #in mm
Mu=0.36*fck*b*Xc*(d-0.416*Xc)/10**6 #in kN-m
Ast=round(0.36*fck*b*Xc/0.87/fy) #in sq mm
print "Moment of resistance of the beam=",Mu," kN-m\nSteel required=",(Ast)," sq mm"
#answer does not match with textbook because of round-off error

Moment of resistance of the beam= 298.250458214  kN-m
Steel required= 1719.0  sq mm


## Ex14.3:pg-763¶

In [11]:
import math
b=300 #width, in mm
d=650 #effective depth, in mm
Ast=942 #in sq mm
lef=6 #in m
fck=20 #in MPa
fy=340 #in MPa
Xu=round(0.87*fy*Ast/0.36/fck/b) #in mm
Xc=0.5*d #in mm
#as Xu<Xc, it is under-reinforced beam, hence OK
Mu=0.87*fy*Ast*(d-0.416*Xu)/10**6 #in kN-m
Wu=Mu*8/lef**2 #in kN/m
self_weight=25*(b/1000)*(d/1000) #in kN/m
W=Wu/1.5-self_weight #in kN/m
print "Safe load on the beam=",W," kN/m"

Safe load on the beam= 24.6170681259  kN/m


## Ex14.4:pg-764¶

In [12]:
import math
b=1000 #width, in mm
d=120 #effective depth, in mm
Ast=1412 #in sq mm
lef=3.2 #in m
fck=20 #in MPa
fy=250 #in MPa
Xu=0.87*fy*Ast/0.36/fck/b #in mm
Xc=0.531*d #in mm
#as Xu<Xc, it is under-reinforced section, hence OK
Mu=0.87*fy*Ast*(d-0.416*Xu)/10**6 #in kN-m
Wu=Mu*8/lef**2 #in kN/m
self_weight=25*(b/1000)*(d/1000) #in kN/m
W=Wu/1.5-self_weight #in kN/m
print "Safe load on the slab=",W," kN/m"

Safe load on the slab= 16.3561454229  kN/m


## Ex14.5:pg-765¶

In [13]:
import math
fck=15 #in MPa
fy=250 #in MPa
#b=d/2
M=65 #in kN-m
Mu=1.5*M #factored moment, in kN-m
d=(Mu*10**6/(0.149*fck*0.5))**(1/3) #in mm
d=445 #approximately, in mm
b=d/2 #in mm
Xc=0.531*d #in mm
Ast=round(0.36*fck*b*Xc/0.87/fy) #in sq mm
print "b=",b," mm\nd=",d," mm\nAst=",Ast," sq mm"
#answer does not match with textbook because of round-off error

b= 222  mm
d= 445  mm
Ast= 1302.0  sq mm


## Ex14.6:pg-766¶

In [14]:
import math
b=300 #width, in mm
d=500 #effective depth, in mm
fck=20 #in MPa
fy=500 #in MPa
Mu=175 #in kN-m
Mulim=0.133*fck*b*d**2/10**6 #in kN-m
#as Mu<Mulim, beam is under-reinforced
#using Cu=Tu, Xu=0.87 fy Ast/(0.36 fck b); let Xu= a Ast
a=0.87*fy/(0.36*fck*b)
#Mu=0.87 fy Ast (d-0.416 Xu), putting Xu = a Ast, we get p Ast**2 + q Ast + r =0
p=0.87*0.416*fy*a
q=-0.87*fy*d
r=Mu*10**6

Area of steel required= 959.0  sq mm