Chapter 4:Shear and Development Length¶

Ex4.1:pg-134¶

In :
import math
b=250 #width, in mm
d=500.0 #effective depth, in mm
W=20 #UDL including self-weight, in kN/m
Pt=1 #percentage tensile steel
l=6 #span, in m
V=W*l/2.0 #in kN
Tv=(V*10**3)/(b*d) #in MPa
#for Pt=1% and for M15 grade concrete
Tc=0.37 #in MPa
#as Tv>Tc, shear reinforcement is required
print "Nominal shear stress in beam=",Tv," MPa\nShear strength of concrete=",Tc," MPa"
Nominal shear stress in beam= 0.48  MPa
Shear strength of concrete= 0.37  MPa

Ex4.2:pg-135¶

In :
import math
b=230 #width, in mm
d=500 #effective depth, in mm
W=24 #UDL including self-weight, in kN/m
Ast=4*0.785*20**2 #four 20 mm dia bars, in sq mm
Pt=Ast/(b*d)*100 #percentage tensile steel
l=4.5 #span, in m
V=W*l/2.0 #in kN
Tv=(V*10**3)/(b*d) #in MPa
#for Pt=1.1% and for M20 grade concrete
Tc=0.40 #in MPa
#as Tv>Tc, shear reinforcement is required
print "Nominal shear stress in beam=",Tv," MPa\nShear strength of concrete=",Tc," MPa"
Nominal shear stress in beam= 0.469565217391  MPa
Shear strength of concrete= 0.4  MPa

Ex4.3:pg-135¶

In :
import math
b=300 #width, in mm
d=600 #effective depth, in mm
W=100 #UDL including self-weight, in kN/m
Pt=2 #percentage tensile steel
l=7.2 #span, in m
sigma_cbc=7 #in MPa
sigma_st=190 #in MPa
m=13.33 #modular ratio
V=W*l/2.0 #in kN
Tv=(V*10**3)/(b*d) #in MPa
Tcmax=1.8 #in MPa
#as Tv>Tcmax, section is to be redesigned so that Tv becomes less than Tcmax
print "Nominal shear stress in beam=",Tv," MPa\nFor given grade of concrete, Tcmax=1.8 MPa and as Tv > Tcmax, section is to be redesigned so that Tv becomes less than Tcmax"
Nominal shear stress in beam= 2.0  MPa
For given grade of concrete, Tcmax=1.8 MPa and as Tv > Tcmax, section is to be redesigned so that Tv becomes less than Tcmax

Ex4.4:pg-136¶

In :
import math
b=1000 #consider 1 m width of slab
D=100 #depth of slab, in mm
cover=20 #in mm
d=D-cover #effective depth, in mm
W=7 #uniformly distributed load, in kN/m**2
dia=10 #in mm
s=100 #spacing of 10 mm dia bars, in mm
l=4 #span, in m
V=W*l/2.0 #in kN
Pt=1000*.785*dia**2/(s*b*d)*100 #in %
Tv=(V*10**3)/(b*d) #in MPa
#for given Pt and M15 grade concrete
Tc=0.37 #in MPa
#and for solid slabs
k=1.3
Tc=k*Tc #in MPa
print "Nominal shear stress in slab, Tv=",Tv," MPa\nShear strength of slab, Tc=",Tc," MPa. As Tc > Tv, no shear reinforcement is required"
Nominal shear stress in slab, Tv= 0.175  MPa
Shear strength of slab, Tc= 0.481  MPa. As Tc > Tv, no shear reinforcement is required

Ex4.5:pg-136¶

In :
import math
b=300 #width, in mm
d=1010 #effective depth, in mm
W=45 #UDL including self-weight, in kN/m
Ast=6*0.785*22**2 #six 22 mm dia bars, in sq mm
l=7 #span, in m
sigma_cbc=5 #in MPa
sigma_sv=140 #in MPa
Fy=250 #in MPa
V=W*l/2.0 #in kN
Tv=(V*10**3)/(b*d) #in MPa
Tcmax=1.6 #in MPa
#Tv<Tcmax; OK
Pt=Ast/(b*d)*100 #percentage tensile steel
#for given Pt and for M15 grade concrete
Tc=0.34 #in MPa
Vs=V-Tc*b*d/10**3 #in kN
#providing 6 mm dia stirrups
dia=6 #in mm
Asv=2*0.785*dia**2 #in sq mm
Sv1=Asv*sigma_sv*d/(Vs*10**3) #in mm
Sv1=145 #round-off, in mm
#Sv<0.75d or 450 mm, whichever is less; hence OK
#calculating minimum spacing of shear reinforcement
Sv2=Asv*Fy/(b*0.4) #in mm
Sv2=118 #round-off, in mm
Sv=min(Sv1,Sv2)
print "Provide 6 mm dia bars at ",Sv," mm c/c throughout the length of the beam, as shear reinforcement"
Provide 6 mm dia bars at  118  mm c/c throughout the length of the beam, as shear reinforcement

Ex4.6:pg-137¶

In :
import math
Bf=1600 #width, in mm
Df=100 #thickness of slab, in mm
d=400 #effective depth, in mm
Bw=225 #breadth of web, in mm
b=Bw
W=30 #UDL including self-weight, in kN/m
Ast=5*0.785*22**2 #five 22 mm dia bars, in sq mm
l=9.2 #span, in m
sigma_cbc=5 #in MPa
sigma_sv=230 #in MPa
Fy=415 #in MPa
V=W*l/2.0 #in kN
Tv=(V*10**3)/(b*d) #in MPa
Tcmax=1.6 #in MPa
#Tv<Tcmax; OK
Pt=Ast/(b*d)*100 #percentage tensile steel
#for given Pt and for M15 grade concrete
Tc=0.44 #in MPa
Vs=V-Tc*b*d/10**3 #in kN
#providing bent-up bars
Asv=0.785*22**2 #in sq mm
Vs1=Asv*sigma_sv*math.sin(45.0)/10**3 #in kN
#but shear taken up by bent-up bar is limited to Vs/2
Vs1=Vs/2 #in kN
#providing 6 mm dia stirrups, which will take up remaining shear force
Vs2=Vs-Vs1 #in kN
dia=6 #in mm
Asv=2*0.785*dia**2 #in sq mm
Sv=Asv*sigma_sv*d/(Vs2*10**3) #in mm
Sv1=105 #round-off, in mm
#Sv<0.75d or 450 mm, whichever is less; hence OK
#calculating minimum spacing of shear reinforcement
Sv2=Asv*Fy/(b*0.4) #in mm
Sv2=260 #round-off, in mm
#to calculate distance 'x' from support where shear stress in concrete is equal to Tc
x=Tc/Tv*l/2.0 #in m
print "Provide 6 mm dia stirrups at ",(Sv1)," mm c/c upto ",(l/2.0-x)," m from both ends\nFor the remaining portion, provide 6 mm dia stirrups at ",(Sv2)," mm"
Provide 6 mm dia stirrups at  105  mm c/c upto  3.28  m from both ends
For the remaining portion, provide 6 mm dia stirrups at  260  mm

Ex4.7:pg-138¶

In :
import math
D=100 #thickness of slab, in mm
l=3 #span of slab, in m
s=0.23 #thickness of support, in m
Lef=l+s #effective span, in m
W=5 #UDL, in kN/m
cover=15 #in mm
R=W*Lef/2 #in kN
M=(R*s/2-W*s**2/2)*10**6 #bending moment at face of wall, in N-mm
#10 mm dia bars at 145 mm c/c as main steel
dia=10 #in mm
c=145 #spacing of reinforcement, in mm
Ast=1000*0.785*dia**2/c #in sq mm
#as alternate bars are bent up
Ast=Ast/2 #available steel reinforcement at face of wall, in sq mm
d=D-10/2.0-cover #in mm
#assuming balanced section
z=0.87*d #in mm
sigma_st=M/(Ast*z) #in MPa
Tbd=0.6 #bond stress, in MPa
Ld=dia*sigma_st/(4*Tbd) #in mm
Ld=177 #round-off, in mm
print "Development length required from the face of the support = ",(Ld)," mm"
#answer given in textbook is incorrect
Development length required from the face of the support =  177  mm

Ex4.8:pg-139¶

In :
import math
b=230 #width, in mm
d=500 #effective depth, in mm
l=6 #span, in m
s=0.3 #thickness of support, in m
Lef=l+s #effective span, in m
W=60 #UDL, in kN/m
Ast=6*0.785*20**2 #six 20 mm dia bars at bottom, in sq mm
Asc=2*0.785*20**2 #two 20 mm dia bars at top, in sq mm
dia=20 #in mm
sigma_cbc=5 #in MPa
sigma_st=230 #in MPa
m=18.66 #modular ratio
R=W*l/2.0 #in kN
M=(R*s/2-W*s**2/2.0)*10**6 #bending moment at face of wall, in N-mm
#assuming balanced section
z=0.87*d #in mm
sigma_st1=M/(Ast*z) #in MPa
Tbd=0.6*1.4 #bond stress in MPa for deformed steel and M15
Ld1=dia*sigma_st1/(4*Tbd) #in mm
#to find critical depth of neutral axis
Xc=d/(1+sigma_st/(m*sigma_cbc)) #in mm
Xc=144 #round-off, in mm
#at face of support
sigma_cbc=sigma_st1/m*Xc/(d-Xc) #in MPa
sigma_sc=1.5*m*sigma_cbc #in MPa
Tbd=1.68 #bond stress in MPa for M15 and deformed steel in compression
Ld2=dia*sigma_sc/(4*Tbd) #in mm
print "Development length required from the face of the support for tension steel = ",(Ld1)," mm\nDevelopment length required from the face of the support for compression steel = ",(Ld2)," mm"
Development length required from the face of the support for tension steel =  176.492736343  mm
Development length required from the face of the support for compression steel =  53.542740239  mm

Ex4.9:pg-139¶

In :
import math
D=120 #thickness of slab, in mm
l=1.5 #span of slab, in m
s=0.23 #thickness of support, in m
Lef=l+s #effective span, in m
W1=3 #UDL, in kN/m**2
cover=15 #in mm
sigma_cbc=5 #in MPa
sigma_st=140 #in MPa
m=18.66 #modular ratio
W=W1+W2 #in kN/m
M=W*l**2/2*10**6 #bending moment at face of wall, in N-mm
#10 mm dia bars at 145 mm c/c as main steel
dia=10 #in mm
d=D-dia/2-cover
c=100 #spacing of reinforcement, in mm
Ast=1000*0.785*dia**2/c #in sq mm
#assuming balanced section
z=0.87*d #in mm
sigma_st=M/(Ast*z) #in MPa
Tbd=0.6 #bond stress in MPa
Ld=dia*sigma_st/(4*Tbd) #in mm
Ld=412 #round-off,in mm
print "Development length required from the face of the support = ",(Ld)," mm"
Development length required from the face of the support =  412  mm