# Chapter 7:Slab Design One Way¶

## Ex7.1:pg-298¶

In [3]:
import math
sigma_cbc=5 #in MPa
sigma_st=140 #in MPa
MF=1.6 #modification factor
#let a be span to depth ratio
l=4 #span, in m
a=MF*20
D=l*1000/a #in mm
self_weight=25*(D/10**3) #in kN/m
finish=1 #in kN/m
lef=l+D/1000 #in m
M=W*lef**2/8.0 #in kN-m
#check for depth
d=round((M*10**6/(0.87*1000))**0.5) #in mm
#assume 12 mm dia bars
D=d+12/2+15 #in mm
#the calculated value of D is more than its assumed value
D=150 #revised value of depth, in mm
self_weight=25*(D/10**3) #in kN/m
finish=1 #in kN/m
lef=l+D/1000 #in m
M=W*lef**2/8.0 #in kN-m
#check for depth
d=round((M*10**6/(0.87*1000))**0.5) #in mm
D=d+12/2+15 #in mm
Ast=round(M*10**6/(sigma_st*0.87*d)) #in sq mm
s1=1000*0.785*12**2/Ast #which is less than 3d= 387 mm
s1=120 #approximately, in mm
Ads=0.15/100*1000*D #distribution steel, in sq mm
#assume 8 mm dia bars
s2=1000*0.785*8**2/Ads #which is less than 5d= 645 mm
s2=220 #approximately, in mm
#to calculate development length
w=0.345 #support width, in m
lef=l+w #in m
R=W*lef/2 #reaction at support, in kN
M1=R*w/2-W*w**2/2 #bending moment at the face of wall, in kN-m
sigma_st=M1*10**6/(Ast/2*0.87*d) #in MPa
Tbd=0.6 #in MPa
Ld=12*sigma_st/(4*Tbd) #in mm
La=w*1000-25 #available length for bar over wall, which is greater than development length
#check for shear
V=W*4.15/2 #in kN
Tv=V*10**3/(1000*d) #in MPa
Tc=0.33 #permissible shear in concrete for p=0.71 and M15, in MPa
Tc=1.3*Tc #permissible shear for slabs, in MPa
#Tc>Tv; hence no shear reinforcement is required
print "Summary of design\nSlab thickness=",(D)," mm\nCover=15 mm\nMain steel = 12 mm dia @ ",(s1)," mm c/c\nAlternate bars are bent up @ 45-degree at support at a distance l/7 from support face\nDistribution steel=8 mm dia @ ",(s2)," mm c/c"

Summary of design
Slab thickness= 104.0  mm
Cover=15 mm
Main steel = 12 mm dia @  120  mm c/c
Alternate bars are bent up @ 45-degree at support at a distance l/7 from support face
Distribution steel=8 mm dia @  220  mm c/c


## Ex7.2:pg-299¶

In [4]:
import math
sigma_cbc=5 #in MPa
sigma_st=230 #in MPa
MF=1.4 #modification factor
#let a be span to depth ratio
l=4.5 #span, in m
a=MF*20
D=l*1000.0/a #in mm
D=160 #approximately, in mm
self_weight=25*(D/10**3) #in kN/m
finish=1 #in kN/m
partitions=1 #in kN/m
lef=l+D/1000 #in m
M=W*lef**2.0/8 #in kN-m
#check for depth
d=(M*10**6/(0.9*sigma_cbc/2*0.29*1000))**0.5 #in mm
#assume 12 mm dia bars
D=d+12/2+15 #in mm
#the calculated value of D is more than its assumed value
D=1.1*D #revised value of depth, in mm
D=250 #assume, in mm
self_weight=25*(D/10**3) #in kN/m
finish=1 #in kN/m
partitions=1 #in kN/m
lef=l+D/1000 #in m
M=W*lef**2/8 #in kN-m
#check for depth
d=round((M*10**6/(0.9*sigma_cbc/2*0.29*1000))**0.5) #in mm
D=d+12/2+15 #in mm
D=250 #approximately, in mm
Ast=round(M*10**6/(sigma_st*0.9*d)) #in sq mm
s1=1000*0.785*12**2/Ast #which is less than 3d= 690 mm
s1=155 #approximately, in mm
pt=Ast/1000/d*100 #in %
Ads=0.12/100*1000*D #distribution steel, in sq mm
#assume 8 mm dia bars
s2=1000*0.785*8**2/Ads #which is less than 5d= 1150 mm
s2=165 #approximately, in mm
#to calculate development length
w=0.23 #support width, in m
l=l+w #in m
R=W*l/2.0 #reaction at support, in kN
M1=R*w/2-W*w**2/2 #bending moment at the face of wall, in kN-m
sigma_st=M1*10**6/(Ast/2*0.9*d) #in MPa
Tbd=0.6 #in MPa
Ld=12*sigma_st/(4*Tbd) #in mm
La=w*1000-25 #available length for bar over wall, which is greater than development length
#check for shear
V=W*lef/2 #in kN
Tv=V*10**3/(1000*d) #in MPa
Tc=0.2212 #permissible shear in concrete for p=0.315 and M15, in MPa
Tc=1.15*Tc #permissible shear for slabs, in MPa
#Tc>Tv; hence no shear reinforcement is required
print "Summary of design\nSlab thickness=",D," mm\nCover=15 mm\nMain steel = 12 dia @ ",s1," mm c/c\nAlternate bars are bent up at 45-degree at support at a distance of l/7 from support face\nDistribution steel=8 dia @ ",s2," mm c/c"

Summary of design
Slab thickness= 250  mm
Cover=15 mm
Main steel = 12 dia @  155  mm c/c
Alternate bars are bent up at 45-degree at support at a distance of l/7 from support face
Distribution steel=8 dia @  165  mm c/c


## Ex7.3:pg-300¶

In [6]:
import math
sigma_cbc=7 #in MPa
sigma_st=230 #in MPa
MF=1.22 #modification factor
#let a be span to depth ratio
l=5 #span, in m
a=MF*26
D=l*1000/a #in mm
D=160 #assume, in mm
self_weight=25*(D/10**3) #in kN/m
finish=0.75 #in kN/m
partitions=1 #in kN/m
lef=5.15 #effective span, in m
M1=Wd*lef**2/12+Wl*lef**2/10 #bending moment at mid-span, in kN-m
M2=Wd*lef**2/10+Wl*lef**2/9 #bending moment at support next to end support, in kN-m
#check for depth
d=(M2*10**6/(0.89*1000))**0.5 #in mm
dia=12 #assume 12 mm dia bars
D=d+12/2+15 #>160, hence depth not suitable
D=1.1*D #in mm
D=210 #assume, in mm
self_weight=25*(D/10**3) #in kN/m
Wd=self_weight #in kN/m
M1=Wd*lef**2/12+Wl*lef**2/10 #bending moment at mid-span, in kN-m
M2=Wd*lef**2/10+Wl*lef**2/9 #bending moment at support next to end support, in kN-m
#check for depth
d=round((M2*10**6/(0.9*sigma_cbc/2*0.29*1000))**0.5) #in mm
D=d+12/2+15 #<210, hence OK
D=200 #assume, in mm
d=D-dia/2-15 #in mm
#main steel at mid-span
Ast1=round(M1*10**6/(sigma_st*0.91*d)) #in sq mm
s1=1000*0.785*12**2/Ast1 #in mm
s1=175 #approximately, in mm
#main steel at support
Ast2=round(M2*10**6/(sigma_st*0.91*d)) #in sq mm
#alternate bars from mid-span are available at the central support as bent up bars; assuming same amount of steel is available from another adjoining mid-span steel
Ast2=Ast2-Ast1 #which is nominal, hence no separate steel is required
Ads=0.12/100*1000*D #distribution steel, in sq mm
#assume 8 mm dia bars
s2=200 #approximately, in mm
print "Summary of design\nSlab thickness=",D," mm\nMain steel = 12 mm dia @ ",s1," mm c/c\nAlternate bars are bent up at support\nDistribution steel=8 mm dia @ ",s2," mm c/c"
#answer given in textbook is incorrect

Summary of design
Slab thickness= 200  mm
Main steel = 12 mm dia @  175  mm c/c
Alternate bars are bent up at support
Distribution steel=8 mm dia @  200  mm c/c


## Ex7.4:pg-301¶

In [7]:
import math
sigma_cbc=5 #in MPa
sigma_st=230.0 #in MPa
MF=1.4 #modification factor
#let a be span to depth ratio
l=1 #span, in m
a=MF*7
D=l*1000/a #in mm
D=105 #assume, in mm
self_weight=25*(D/10**3)*1.5 #in kN/m
finish=0.5*1.5 #in kN/m
lef=l+0.23/2 #effective span, in m
M=W*lef/2 #in kN-m
#check for depth
d=(M*10**6/(0.65*1500))**0.5 #in mm
dia=12 #assume 12 mm dia bars
D=d+12/2+15 #<105, hence OK
D=100 #assume, in mm
d=D-dia/2-15 #in mm
#main steel at mid-span
Ast=M*10**6/(sigma_st*0.9*d) #in sq mm
s1=1500*0.785*12**2/Ast #>3d = 237 mm
s1=235 #assume, in mm
Ads=0.12/100*1000*D #distribution steel, in sq mm
#assume 6 mm dia bars
s2=235 #assume, in mm
Tbd=0.84 #in MPa
Ld=dia*sigma_st/4/Tbd # in mm
Ld=821 #round-off, in mm
Tv=W*10**3/1500/d #in MPa
As=1500*0.785*12**2/235 #in sq mm
pt=As/1500/d*100 #in %
Tc=0.316 #in MPa
#as Tc>Tv, no shear reinforcement required
print "Summary of design\nThickness of slab = ",D," mm\nCover = 15mm\nMain steel = 12 mm dia @ ",s1," mm c/c\nProvide development length of ",Ld," mm in the beam from face of beam\nDistribution steel = 6 mm dia @ ",s2," mm c/c"

Summary of design
Thickness of slab =  100  mm
Cover = 15mm
Main steel = 12 mm dia @  235  mm c/c
Provide development length of  821  mm in the beam from face of beam
Distribution steel = 6 mm dia @  235  mm c/c