Chapter 3:T and L Beams

Ex3.1:pg-72

In [3]:
import math
Bf=1300 #width of flange, in mm
Df=80.0 #thickness of flange, in mm
d=600 #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
Xc=240 #round-off, in mm
 #to find Ast
Ast=Bf*Df*(Xc-Df/2)/(m*(d-Xc)) #in sq mm
print "Neutral axis depth=",(Xc)," mm\nArea of steel=",Ast," mm**2"

Neutral axis depth= 240  mm
Area of steel= 4334.41693757  mm**2

Ex3.2:pg-72

In [5]:
import math
Bf=1500 #width of flange, in mm
Bw=300 #breadth of web, in mm
Df=100 #thickness of flange, in mm
d=700 #effective depth, in mm
m=18.66 #modular ratio
Ast=8*0.785*25**2 #eight 25 mm dia bars, in sq mm
 #assume depth of neutral axis is less than or equal to thickness of flange; find x using Bf(x**2)/2=mAst(d-x), which becomes of the form px**2+qx+r=0
p=Bf/2.0
q=m*Ast
r=-m*Ast*d
 #solving quadratic equation
x=(-q+math.sqrt(q**2-4*p*r))/(2*p) #in mm
 #x>Df; hence our assumption is incorrect; equating moments of area on compression and tension sides about N.A.
x=(m*Ast*d+Bf*Df**2/2)/(m*Ast+Bf*Df) #in mm
x=263 #round-off, in mm
print "Neutral axis depth=",(x)," mm"

Neutral axis depth= 263  mm

Ex3.3:pg-73

In [8]:
import math
Bf=1200 #width of flange, in mm
Bw=200 #breadth of web, in mm
Df=100 #thickness of flange, in mm
d=400 #effective depth, in mm
m=13.33 #modular ratio
Ast=4*0.785*18**2 #four 18mm dia bars, in sq mm
 #assume x > Df; ; equating moments of area on compression and tension sides about N.A.
x=(m*Ast*d+Bf*Df**2/2.0)/(m*Ast+Bf*Df) #in mm
 #as x < Df; our assumption was incorrect
#x < Df; find x using Bf(x**2)/2=mAst(d-x), which becomes of the form px**2+qx+r=0
p=Bf/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<Df; hence our assumption is correct
print "Neutral axis depth=",round(x,2)," mm"

Neutral axis depth= 84.45  mm

Ex3.4:pg-74

In [9]:
import math
Bf=1500 #width of flange, in mm
Bw=300 #breadth of web, in mm
Df=100 #thickness of flange, in mm
d=700 #effective depth, in mm
sigma_cbc=5 #in MPa
sigma_st=140 #in MPa
m=18.66 #modular ratio
Ast=8*0.785*25**2 #eight 25 mm dia bars, in sq mm
 #assume x < Df; find x using Bf(x**2)/2=mAst(d-x), which becomes of the form px**2+qx+r=0
p=Bf/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 > Df; hence our assumption is incorrect; equating moments of area on compression and tension sides about N.A.
x=(m*Ast*d+Bf*Df**2/2)/(m*Ast+Bf*Df) #in mm
 #to find critical depth of neutral axis
Xc=d/(1+sigma_st/(m*sigma_cbc)) #in mm
 #as x < Xc, beam is under-reinforced
sigma_cbc=sigma_st/m*x/(d-x) #in MPa
sigma_cbc_dash=sigma_cbc*(x-Df)/x #in MPa
 #to find lever arm
z=d-(sigma_cbc+2*sigma_cbc_dash)/(sigma_cbc+sigma_cbc_dash)*Df/3 #in mm
Mr=Bf*Df*(sigma_cbc+sigma_cbc_dash)*z/2 #in N-mm
print "Moment of resistance of the beam=",Mr/10**6,"kN-m"

Moment of resistance of the beam= 359.322310166 kN-m

Ex3.5:pg-76

In [10]:
import math
Bf=1200 #width of flange, in mm
Bw=200 #breadth of web, in mm
Df=100 #thickness of flange, in mm
d=400 #effective depth, in mm
sigma_cbc=7 #in MPa
sigma_st=190 #in MPa
m=13.33 #modular ratio
Ast=4*0.785*18**2 #four 18 mm dia bars, in sq mm
#assume x < Df; find x using Bf(x**2)/2=mAst(d-x), which becomes of the form px**2+qx+r=0
p=Bf/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 < Df; hence our assumption is correct
#to find critical depth of neutral axis
Xc=d/(1+sigma_st/(m*sigma_cbc)) #in mm
#as x < Xc, beam is under-reinforced
sigma_cbc=sigma_st/m*x/(d-x) #in MPa
#taking moments about tensile steel
Mr=Bf*x*sigma_cbc*(d-x/3)/2 #in N-mm
print "Moment of resistance of the beam=",Mr/10**6,"kN-m"

Moment of resistance of the beam= 71.8778860857 kN-m

Ex3.7:pg-77

In [12]:
import math
Bf=1500 #width of flange, in mm
Bw=200 #breadth of web, in mm
Df=100 #thickness of flange, in mm
d=400 #effective depth, in mm
sigma_cbc=5 #in MPa
sigma_st=140 #in MPa
m=18.66 #modular ratio
Ast=2190 #in sq mm
 #assume x>Df
x=(m*Ast*d+Bf*Df**2/2)/(m*Ast+Bf*Df) #in mm
 #to find critical depth of neutral axis
Xc=d/(1+sigma_st/(m*sigma_cbc)) #in mm
 #as x<Xc, beam is under-reinforced
sigma_cbc=sigma_st/m*x/(d-x) #in MPa
sigma_cbc_dash=sigma_cbc*(x-Df)/x #in MPa
 #to find lever arm
z=d-(sigma_cbc+2*sigma_cbc_dash)/(sigma_cbc+sigma_cbc_dash)*Df/3 #in mm
 #taking moments about tensile steel
Mr=Bf*Df*(sigma_cbc+sigma_cbc_dash)*z/2 #in N-mm
print "Moment of resistance of the beam=",Mr/10**6," kN-m"

Moment of resistance of the beam= 110.719528403  kN-m

Ex3.8:pg-78

In [13]:
import math
Bf=1200 #width of flange, in mm
Bw=300 #breadth of web, in mm
Df=120 #thickness of flange, in mm
d=500 #effective depth, in mm
sigma_cbc=7 #in MPa
sigma_st=190 #in MPa
m=13.33 #modular ratio
Ast=5*0.785*20**2 #five 20 mm dia bars, in sq mm
l=6 #span, in m
#assume depth of neutral axis is less than or equal to thickness of flange; find x using Bf(x**2)/2=mAst(d-x), which becomes of the form px**2+qx+r=0
p=Bf/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 < Df; hence our assumption is correct
#to find critical depth of neutral axis
Xc=d/(1+sigma_st/(m*sigma_cbc)) #in mm
#as x<Xc, beam is under-reinforced
sigma_cbc=sigma_st/m*x/(d-x) #in MPa
#taking moments about tensile steel
Mr=Bf*x*sigma_cbc*(d-x/3)/2 #in N-mm
W=(Mr/10**6)*8.0/l**2 #in kN/m
print "Moment of resistance of the beam=", Mr/10**6," kN-m\nCapacity to take uniformly distributed load(including self-weight)=",W," kN/m"

Moment of resistance of the beam= 137.638859204  kN-m
Capacity to take uniformly distributed load(including self-weight)= 30.5864131564  kN/m

Ex3.9:pg-79

In [15]:
import math
Bf=1400 #width of flange, in mm
Df=120 #thickness of flange, in mm
d=600.0 #effective depth, in mm
m=18.66 #modular ratio
Ast=4000 #in sq mm
M=160*10**6 #in N-mm
 #Assume x>Df; equating moments of area on compression and tension sides about N.A.
x=(m*Ast*d+Bf*Df**2/2)/(m*Ast+Bf*Df) #in mm
 #let sigma_cbc_dash=a*sigma_cbc
a=(x-Df)/x
 #to find lever arm
z=d-(1+2*a)/(1+a)*Df/3 #in mm
sigma_cbc=2*M/(Bf*Df*(1+a)*z) #in MPa
sigma_st=m*sigma_cbc*(d-x)/x #in MPa
print "Stress in concrete=",sigma_cbc," N/mm**2\nStress in tension steel=",sigma_st," N/mm**2"
 #answer given in textbook is incorrect

Stress in concrete= 2.36901404247  N/mm**2
Stress in tension steel= 73.0962080852  N/mm**2

Ex3.10:pg-80

In [16]:
import math
Bf=1250 #width of flange, in mm
Df=120 #thickness of flange, in mm
d=700 #effective depth, in mm
m=13.33 #modular ratio
Ast=5500 #in sq mm
W=60 #UDL including self-weight, in kN/m
l=8 #span, in m
M=W*l**2/8.0*10**6 #in N-mm
 #Assume x>Df. Equating moments of area on compressiona and tension sides about N.A.
x=(m*Ast*d+Bf*Df**2.0/2)/(m*Ast+Bf*Df) #in mm
 #let sigma_cbc_dash=a*sigma_cbc
a=(x-Df)/x
 #to find lever arm
z=d-(1+2*a)/(1+a)*Df/3 #in mm
sigma_cbc=2*M/(Bf*Df*(1+a)*z) #in MPa
sigma_st=m*sigma_cbc*(d-x)/x #in MPa
print "Stress in concrete=",sigma_cbc," N/mm**2\nStress in tension steel=",sigma_st," N/mm**2"

Stress in concrete= 6.37094400445  N/mm**2
Stress in tension steel= 135.157527269  N/mm**2

Ex3.11:pg-81

In [17]:
import math
Bf=1300 #width of flange, in mm
Df=100 #thickness of flange, in mm
d=500 #effective depth, in mm
sigma_cbc=5 #in MPa
sigma_st=275 #in MPa
m=18.66 #modular ratio
Ast=1570 #in sq mm
Asc=1256 #in sq mm
top_cover=30 #in mm
 #to find critical depth of neutral axis
Xc=d/(1+sigma_st/(m*sigma_cbc)) #in mm
 #assume x>Df; equating moments of area on compression and tension sides about N.A.
x=(m*Ast*d+Bf*Df**2/2+(1.5*m-1)*Asc*top_cover)/(m*Ast+Bf*Df+(1.5*m-1)*Asc) #in mm
 #as x<Xc, beam is under-reinforced
sigma_cbc=sigma_st/m*x/(d-x) #in MPa
sigma_cbc_dash=sigma_cbc*(x-top_cover)/x #stress in concrete at level of compression steel, in MPa
sigma_cbc_double_dash=sigma_cbc*(x-Df)/x #stress in concrete at the underside of the slab, in MPa
 #to find lever arm
z=round(d-(sigma_cbc+2*sigma_cbc_double_dash)/(sigma_cbc+sigma_cbc_double_dash)*Df/3) #in mm
 #taking moments about tensile steel
Mr=Bf*Df*(sigma_cbc+sigma_cbc_double_dash)*z/2+(1.5*m-1)*Asc*sigma_cbc_dash*(d-top_cover) #in N-mm
print "Moment of resistance of the beam=",Mr/10**6," kN-m"

Moment of resistance of the beam= 200.669336976  kN-m

Ex3.12:pg-82

In [19]:
import math
Bf=1500 #width of flange, in mm
Df=150 #thickness of flange, in mm
d=600 #effective depth, in mm
sigma_cbc=7 #in MPa
sigma_st=230 #in MPa
m=13.33 #modular ratio
Ast=1964 #in sq mm
Asc=1140 #in sq mm
top_cover=50 #in mm
#to find critical depth of neutral axis
Xc=d/(1+sigma_st/(m*sigma_cbc)) #in mm
#assume x>Df; equating moments of area on compression and tension sides about N.A.
x=(m*Ast*d+Bf*Df**2/2+(1.5*m-1)*Asc*top_cover)/(m*Ast+Bf*Df+(1.5*m-1)*Asc) # in mm
#we find that x<Df, hence our assumption that x>Df is wrong
#to find x using Bf(x**2)/2 + (1.5m-1)Asc(x-d')=mAst(d-x), which becomes of the form px**2+qx+r=0
p=Bf/2
q=m*Ast+(1.5*m-1)*Asc
r=-(m*Ast*d+(1.5*m-1)*Asc*top_cover)
#solving quadratic equation
x=(-q+math.sqrt(q**2-4*p*r))/(2*p) #in mm
#as x<Xc, beam is under-reinforced
sigma_cbc=sigma_st/m*x/(d-x) #in MPa
sigma_cbc_dash=sigma_cbc*(x-top_cover)/x #stress in concrete at level of compression steel, in MPa
#taking moments about tensile steel
Mr=Bf*x*sigma_cbc*(d-x/3)/2+(1.5*m-1)*Asc*sigma_cbc_dash*(d-top_cover) #in N-mm
print "Moment of resistance of the beam=",Mr/10**6," kN-m"
 #answer given in textbook is incorrect

Moment of resistance of the beam= 252.263730672  kN-m

Ex3.13:pg-83

In [20]:
import math
Bf=1450 #width of flange, in mm
Df=120 #thickness of flange, in mm
d=400 #effective depth, in mm
m=13.33 #modular ratio
Ast=1800 #in sq mm
Asc=450 #in sq mm
top_cover=30 #in mm
M=200*10**6 #in N-mm
#assume x>Df; equating moments of area on compression and tension sides about N.A.
x=(m*Ast*d+Bf*Df**2/2+(1.5*m-1)*Asc*top_cover)/(m*Ast+Bf*Df+(1.5*m-1)*Asc) #in mm
#we find that x<Df, hence our assumption that x>Df is wrong
#to find x using Bf(x**2)/2 + (1.5m-1)Asc(x-d')=mAst(d-x), which becomes of the form px**2+qx+r=0
p=Bf/2
q=m*Ast+(1.5*m-1)*Asc
r=-(m*Ast*d+(1.5*m-1)*Asc*top_cover)
#solving quadratic equation
x=(-q+math.sqrt(q**2-4*p*r))/(2*p) #in mm
#as x<Xc, beam is under-reinforced; let stress in concrete at level of steel be equal to 'a' times the stress in concrete at top
a=(x-top_cover)/x
#taking moments about tensile steel
sigma_cbc=M/(Bf*x*(d-x/3)/2+(1.5*m-1)*Asc*a*(d-top_cover)) #in MPa
sigma_st=m*sigma_cbc*(d-x)/x #in MPa
sigma_sc=1.5*m*a*sigma_cbc #in MPa
print "Stress in concrete=",sigma_cbc," N/mm**2\nStress in tension steel=",sigma_st," N/mm**2\nStress in compression steel=",sigma_sc," N/mm**2"
 #answer in textbook is incorrect

Stress in concrete= 7.17910912778  N/mm**2
Stress in tension steel= 301.875837251  N/mm**2
Stress in compression steel= 98.8192837935  N/mm**2

Ex3.14:pg-84

In [22]:
import math
Bf=500 #width of flange, in mm
Bw=250 #breadth of web, in mm
Df=100 #thickness of flange, in mm
d=500 #effective depth, in mm
sigma_cbc=5 #in MPa
sigma_st=140 #in MPa
m=18.66 #modular ratio
Ast=2000.0 #in sq mm
 #to find critical depth of neutral axis
Xc=d/(1+sigma_st/(m*sigma_cbc)) #in mm
 #assume x>Df
x=(m*Ast*d+Bf*Df**2/2.0)/(m*Ast+Bf*Df) #in mm
 #as x>Xc, beam is over-reinforced
sigma_cbc_dash=sigma_cbc*(x-Df)/x #in MPa
 #to find lever arm
z=d-(sigma_cbc+2*sigma_cbc_dash)/(sigma_cbc+sigma_cbc_dash)*Df/3 #in mm
 #taking moments about tensile steel
Mr=Bf*Df*(sigma_cbc+sigma_cbc_dash)*z/2 #in N-mm
print "Moment of resistance of the beam=",round(Mr/10.0**6,2)," kN-m"

Moment of resistance of the beam= 90.15  kN-m

Ex3.15:pg-85

In [24]:
import math
Bf=750 #width of flange, in mm
Bw=250 #breadth of web, in mm
Df=100 #thickness of flange, in mm
d=700 #effective depth, in mm
sigma_cbc=7 #in MPa
sigma_st=190 #in MPa
m=13.33 #modular ratio
M=460*10**6 #in N-mm
 #to find critical depth of neutral axis
Xc=d/(1+sigma_st/(m*sigma_cbc)) #in mm
sigma_cbc_dash=sigma_cbc*(Xc-Df)/Xc #in MPa
 #to find lever arm
z=d-(sigma_cbc+2*sigma_cbc_dash)/(sigma_cbc+sigma_cbc_dash)*Df/3 #in mm
 #taking moments about tensile steel
Ast=M/(sigma_st*z) #in sq mm
Ast=3699 #round-off, in sq mm
print "Area of steel required=",(Ast),"mm**2"

Area of steel required= 3699 mm**2

Ex3.16:pg-86

In [26]:
import math
Df=120 #thickness of flange, in mm
Bw=200 #breadth of web, in mm
d=550 #effective depth, in mm
l=6 #span, in m
Bf=l*1000.0/12+Bw+3*Df #in mm
m=13.33 #modular ratio
Ast=3200 #in sq mm
M=190*10**6 #in N-mm
#assume x>Df; equating moments of area on compression and tension sides about N.A.
x=(m*Ast*d+Bf*Df**2/2)/(m*Ast+Bf*Df) #in mm
#we find that x>Df, hence our assumption that x>Df is correct
#as x<Xc, beam is under-reinforced; let stress in concrete at underside of slab be equal to 'a' times the stress in concrete at top
a=(x-Df)/x
#to find lever arm
z=d-(1+2*a)/(1+a)*Df/3 #in mm
z=500 #round-off, in mm
 #taking moments about tensile steel
sigma_cbc=M/(Bf*Df*(1+a)*z/2) #in MPa
sigma_st=m*sigma_cbc*(d-x)/x #in MPa
print "Stress in concrete=",sigma_cbc," N/mm**2\nStress in tension steel=",sigma_st," N/mm**2"

Stress in concrete= 4.44406201666  N/mm**2
Stress in tension steel= 118.75  N/mm**2