# Chapter 17:Shear and Development Length¶

## Ex17.1:pg-905¶

In [1]:
import math
b=230 #width, in mm
d=500 #effective depth, in mm
l=4.5 #span, in m
Ast=4*0.785*20**2 #four 20 mm dia bars, in sq mm
fck=20 #in MPa
W=24 #in kN/m
Vu=Wu*l/2 #in kN
Tv=Vu*10**3/b/d #in MPa
Tcmax=2.8 #for M20, in MPa
#Tv<Tcmax, hence OK
p=Ast/b/d*100 #p=1.1, approximately
#for p=1.1 and M20 grade concrete
Tc=0.64 #in MPa
#Tv>Tc, hence shear reinforcement required
print "Nominal shear stress=",Tv," MPa\nShear strength of concrete=",Tc," MPa"

Nominal shear stress= 0.704347826087  MPa
Shear strength of concrete= 0.64  MPa


## Ex17.2:pg-906¶

In [2]:
import math
b=300 #width, in mm
d=1010 #effective depth, in mm
l=7 #span, in m
Ast=round(6*0.785*22**2) #six 22 mm dia bars, in sq mm
fck=15 #in MPa
fy=250 #in MPa
W=45 #in kN/m
Vu=Wu*l/2 #in kN
Tv=Vu*10**3/b/d #in MPa
#Tv<Tcmax, hence OK
p=Ast/b/d*100 #p=0.75, approximately
#for p=0.75 and M15 grade concrete
Tc=0.54 #in MPa
#Tv>Tc, hence shear reinforcement required
Vus=Vu-Tc*b*d/10**3 #in kN
#provide 6 mm dia stirrups
Sv=0.87*fy*2*0.785*6**2*d/Vus/10**3 #in mm
Sv=171 #approximately, in mm
Svmin=2*0.785*6**2*fy/b/0.4 #in mm
Svmin=118 #approximately, in mm
Sv=min(Sv,Svmin) #in mm
print "Provide 6 mm dia stirrups at ",(Sv)," mm c/c as shear reinforcement"

Provide 6 mm dia stirrups at  118  mm c/c as shear reinforcement