Chapter 1:Singly Reinforced Sections

Ex1.1:pg-27

In [2]:
 #let the depth of neutral axis be x
import math
b=200 #width, in mm
d=350 #effective depth, in mm
m=18.66  #modular ratio
sigma_cbc=5 #in MPa
sigma_st=140 #in MPa
x=d/(1+sigma_st/(m*sigma_cbc)) #in mm
print "The depth of neutral axis = ",(x)," mm\n",
 #to find area of steel
Ast=b*x*sigma_cbc/(2*sigma_st) #in sq mm
print "Area of steel = ",(Ast)," mm**2\n"
 #to find percentage steel
pst=Ast*100/(b*d) #in %
print "Percentage of steel = ",(pst)," percent\n", 

The depth of neutral axis =  139.969995714  mm
Area of steel =  499.892841835  mm**2

Percentage of steel =  0.714132631192  percent

Ex1.2:pg-27

In [3]:
 #let the depth of neutral axis be x
import math
b=150.0 #width, in mm
d=400 #effective depth, in mm
Ast=804 #area of steel, in sq mm
m=18.66 #modular ratio
 #b(x**2)/2=mAst(d-x)-->this becomes a quadratic equation of form px**2+qx+r=0
p=b/2.0
q=m*Ast
r=-m*Ast*d
 #solving the quadratic equation
x=(-q+math.sqrt(q**2-4*p*r))/(2*p) #in mm
print "The depth of neutral axis = ",(x)," mm"

The depth of neutral axis =  200.011732416  mm

Ex1.3:pg-28

In [5]:
import math
#assume d = 400 mm and b = 200 mm
b=200 #in mm
d=400 #in mm
sigma_cbc=5 #in MPa
sigma_st=140 #in MPa
m=18.66 #modular ratio
Xc=d/(1+sigma_st/m/sigma_cbc) #in mm
z=d-Xc/3 #in mm
Mr=b*Xc*sigma_cbc/2*z #in N-mm
Ast=b*Xc*sigma_cbc/2/sigma_st #in sq mm
pt=Ast*100/b/d #in %
print "When d is assumed as 400 mm and b as 200 mm\n(a) Position of neutral axis=",(Xc)," mm\n(b) Lever arm=",(z)," mm\n(c) Moment of resistance=",(Mr/10**6)," kN-m\n(d) Percentage of steel=",(pt)," percent"

When d is assumed as 400 mm and b as 200 mm
(a) Position of neutral axis= 159.965709387  mm
(b) Lever arm= 346.678096871  mm
(c) Moment of resistance= 27.7283038475  kN-m
(d) Percentage of steel= 0.714132631192  percent

Ex1.4:pg-28

In [7]:
import math
b=250.0 #width, in mm
d=500.0 #effective depth, in mm
sigma_cbc=5 #in MPa
sigma_st=140 #in MPa
m=18.66 #modular ratio
#to find critical depth of neutral axis
Xc=d/(1+sigma_st/(m*sigma_cbc)) #in mm
z=d-Xc/3 #lever arm, in mm
Mr=b*Xc*sigma_cbc*z/2 #in N-mm
print "Moment of resistance of the beam = ",(Mr/10**6)," kN-m"

Moment of resistance of the beam =  54.1568434521  kN-m

Ex1.5:pg-29

In [10]:
import math
b=250 #width, in mm
D=550.0 #overall depth, in mm
Ast=1521 #area of steel, in sq mm
cover=25 #in mm
d=D-cover #effective depth, in mm
sigma_cbc=7 #in MPa
sigma_st=140 #in MPa
m=13.33 #modular ratio
 #to find critical depth of neutral axis
Xc=d/(1+sigma_st/(m*sigma_cbc)) #in mm
 #to find actual depth of neutral axis using b(x**2)/2=mAst(d-x)--> this will become of the form px**2+qx+r=0
p=b/2
q=m*Ast
r=-m*Ast*d
x=(-q+math.sqrt(q**2-4*p*r))/(2*p) #in mm
 #x>Xc; hence beam is over-reinforced
Mr=b*x*sigma_cbc/2*(d-x/3) #in N-mm
print "Moment of resistance of the beam=",(Mr/10**6)," kN-m"

Moment of resistance of the beam= 87.5318457534  kN-m

Ex1.6:pg-30

In [11]:
import math
b=200 #width, in mm
d=450 #effective depth, in mm
Ast=3*.785*16**2 #three 16 dia bars, in sq mm
sigma_cbc=5 #in MPa
sigma_st=140 #in MPa
m=18.66 #modular ratio
 #to find critical depth of neutral axis
Xc=d/(1+sigma_st/(m*sigma_cbc)) #in mm
 #to find actual depth of neutral axis using b(x**2)/2=mAst(d-x), which becomes of form px**2+qx+r=0
p=b/2
q=m*Ast
r=-m*Ast*d
x=(-q+math.sqrt(q**2-4*p*r))/(2*p) #in mm
 #as x<Xc, beam is under-reinforced
Mr=Ast*sigma_st*(d-x/3) #in N-mm
print "Moment of resistance of the beam = ",(Mr/10**6)," kN-m"

Moment of resistance of the beam =  33.0389810897  kN-m

Ex1.7:pg-31

In [12]:
import math
b=300 #width, in mm
D=700.0 #overall depth, in mm
Ast=3*.785*20**2 #3-20mm dia bars, in sq mm
cover=50 #in mm
d=D-cover #effective depth, in mm
sigma_cbc=7 #in MPa
sigma_st=190 #in MPa
m=13.33 #modular ratio
l=6 #span, in m
w=25 #unit weight of concrete, in kN/m**3
 #to find critical depth of neutral axis
Xc=d/(1+sigma_st/(m*sigma_cbc)) #in mm
 #to find actual depth of neutral axis using b(x**2)/2=mAst(d-x), which becomes of the form px**2+qx+r=0
p=b/2
q=m*Ast
r=-m*Ast*d
 #solving quadratic equation
x=(-q+math.sqrt(q**2-4*p*r))/(2*p) #in mm
 #x<Xc, hence beam is under-reinforced
Mr=sigma_st*Ast*(d-x/3) #in N-mm
UDL=(Mr/10.0**6)*8/l**2 #in kN/m
self_weight=w*b*D/10**6 #in kN/m
net_weight=UDL-self_weight #in kN/m
print "Moment of resistance=",(Mr/10**6)," kN-m\nSafe uniformly distributed load that the beam can carry=",(net_weight)," kN/m"

Moment of resistance= 104.695224175  kN-m
Safe uniformly distributed load that the beam can carry= 18.0156053722  kN/m

Ex1.8:pg-32

In [1]:
import math

b=250.0 #width, in mm
D=500 #overall depth, in mm
Ast=4*.785*22**2 #four 22 mm dia bars, in sq mm
cover=25 #in mm
d=D-cover #effective depth, in mm
l=5 #effective span, in m
sigma_cbc=5 #in MPa
sigma_st=190 #in MPa
m=18.66 #modular ratio
 #to find critical depth of neutral axis
Xc=d/(1+sigma_st/(m*sigma_cbc)) #in mm
 #to find actual depth of neutral axis using b(x**2)/2=mAst(d-x), which becomes of the form px**2+qx+r=0
p=b/2
q=m*Ast
r=-m*Ast*d
x=(-q+math.sqrt(q**2-4*p*r))/(2*p) #in mm
 #as x>Xc, beam is over-reinforced
Mr=b*sigma_cbc*x/2*(d-x/3) #in N-mm
self_weight=25*(b/10**3)*(D/10**3) #in kN/m
M=Mr/10**6-self_weight*l**2/8 #moment of resistance available for external load, in kN-m
W=4*M/l #in kN
print "The concentrated load the beam can support at centre=",(W)," kN"

The concentrated load the beam can support at centre= 46.4305432264  kN

Ex1.9:pg-33

In [14]:
import math
d=120.0 #effective depth of slab, in mm
 #consider 1 m strip of slab
b=1000.0 #in mm
s=80.0 #spacing of 12mm dia bars centre-to-centre, in mm
Ast=1000*.785*12**2/s #in sq mm
l=3.2 #span, in m
sigma_cbc=7 #in MPa
sigma_st=140 #in MPa
m=13.33 #modular ratio
 #to find critical depth of neutral axis
Xc=d/(1+sigma_st/(m*sigma_cbc)) #in mm
 #to find actual depth of neutral axis using b(x**2)/2=mAst(d-x), which becomes of the form px**2+qx+r=0
p=b/2
q=m*Ast
r=-m*Ast*d
x=(-q+math.sqrt(q**2-4*p*r))/(2*p) #in mm
 #as x>Xc, the beam is over-reinforced
Mr=b*sigma_cbc*x/2*(d-x/3)/10**6 #in kN-m
UDL=Mr*8/l**2 #in kN/m
self_weight=25*(d/10.0**3)*(b/10**3) #in kN/m
W=UDL-self_weight #in kN/m
print "The safe load for slab=",(W)," kN/m"

The safe load for slab= 11.3607587975  kN/m

Ex1.10:pg-34

In [16]:
import math
b=300.0 #width, in mm
D=700 #overall depth, in mm
Ast=4*.785*25**2 #four 25mm dia bars, in sq mm
cover=30 #in mm
d=D-cover #effective depth, in mm
M=130*10**6 #bending moment, in N-mm
m=18.66 #modular ratio
 #to find actual depth of neutral axis using b(x**2)/2=mAst(d-x), which becomes of the form px**2+qx+r=0
p=b/2.0
q=m*Ast
r=-m*Ast*d
x=(-q+math.sqrt(q**2-4*p*r))/(2*p) #in mm
z=d-x/3 #lever arm, in mm
 #assuming under-reinforced section, Mr=Ast*sigma_st(d-x/3) and equating Mr to M
sigma_st=M/(Ast*z) #in MPa
sigma_st=116 #round-off, in MPa
sigma_cbc=(sigma_st/m)*x/(d-x) #in MPa
sigma_cbc=5 #round-off, in MPa
print "Stress in steel=",(sigma_st)," N/mm**2\nStress in concrete=",(sigma_cbc)," N/mm**2"

Stress in steel= 116  N/mm**2
Stress in concrete= 5  N/mm**2

Ex1.11:pg-35

In [2]:
import math

b=350.0 #width, in mm
D=650 #overall depth, in mm
Ast=4*.785*22**2 #four 22mm dia bars, in sq mm
cover=25 #in mm
d=D-cover #effective depth, in mm
W=20 #UDL, in kN/m
l=7 #span, in m
M=W*l**2/8.0*10**6 #bending moment, in N-mm 
m=13.33 #modular ratio
 #to find actual depth of neutral axis using b(x**2)/2=mAst(d-x), which becomes of the form px**2+qx+r=0
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 #lever arm, in mm
 #assuming under-reinforced section, Mr=Ast*sigma_st(d-x/3) and equating Mr to M
sigma_st=M/(Ast*z) #in MPa
sigma_cbc=(sigma_st/m)*x/(d-x) #in MPa
print "Stress in steel=",(sigma_st)," N/mm**2\nStress in concrete=",(sigma_cbc)," N/mm**2"

Stress in steel= 145.869819292  N/mm**2
Stress in concrete= 5.83077431505  N/mm**2

Ex1.12:pg-35

In [3]:
import math

b=250.0 #width, in mm
sigma_cbc=5.0 #in MPa
sigma_st=190.0 #in MPa
m=280.0/(3*sigma_cbc) #modular ratio
M=75*10.0**6 #bending moment, in N-mm
 #critical depth of neutral axis, Xc=d/(1+sigma_st/(m*sigma_cbc))=a*d
a=1/(1+sigma_st/(m*sigma_cbc))
d=(M/(b*sigma_cbc*a*(1-a/3.0)/2.0))**0.5 #in mm
d=640 #round-off, in mm
Xc=a*d #in mm
Ast=b*Xc*sigma_cbc/(2*sigma_st) #in sq mm
print "Effective depth=",round(d)," mm\nArea of steel=",(Ast)," mm**2",

Effective depth= 640.0  mm
Area of steel= 693.498452012  mm**2

Ex1.13:pg-36

In [4]:
import math

#b=d/2 (given)
sigma_cbc=5 #in MPa
sigma_st=140.0 #in MPa
m=18.66 #modular ratio
M=65*10.0**6 #bending moment, in N-mm
 #critical depth of neutral axis, Xc=d/(1+sigma_st/(m*sigma_cbc))=a*d
a=1/(1+sigma_st/(m*sigma_cbc))
d=(M/(sigma_cbc*a*(1-a/3.0)/4))**(1/3.0) #in mm
d=530.0 #round-off, in mm
Xc=a*d #in mm
b=d/2.0 #in mm
Ast=M/sigma_st/0.87/d #in sq mm
Ast=1007.0 #round-off, in sq mm
print "Dimensions of section=",(b)," x ",(d)," mm\nArea of steel=",(Ast)," mm**2"

Dimensions of section= 265.0  x  530.0  mm
Area of steel= 1007.0  mm**2