# Chapter 10:Staircase¶

## Ex10.1:pg-517¶

In [1]:
import math
l=1 #span, in m
t=0.27 #tread in m
sigma_cbc=5 #in MPa
sigma_st=140 #in MPa
MF=1.6
a=MF*7
D=l*10**3/a #in mm
D=100 #assume, in mm
W1=D/10**3*t*25 #in kN/m
M1=W1*l/2 #in kN-m
M2=t*3*l/2 #in kN-m
M3=1.3*l #in kN-m
M=M1+max(M2,M3) #in kN-m
d=math.sqrt(M*10**6/0.87/t/10**3) #in mm
d=83 #in mm
#assume 8 mm dia bars
dia=8 #in mm
D=d+dia/2+15 #this is slightly more than assumed value, hence OK
D=100 #in mm
z=0.87*d #in mm
Ast=M*10**6/sigma_st/z #in sq mm
n=Ast/0.785/8**2
n=4 #assume
Ads=0.15/100*D*t*10**3 #distribution steel, in sq mm
#provide 6 mm dia bars
s=415 #in mm
Tbd=0.6 #in MPa
Ld=dia*sigma_st/4/Tbd #in mm
Ld=470 #assume, in mm
print "Summary of design\nThickness of steps=",(n)," mm\nCover from top=15 mm\nMain steel = 8 mm dia, ",(n)," in each step with development length of ",(Ld)," mm\nDistribution steel = 6 mm dia @ ",(s)," mm c/c"

Summary of design
Thickness of steps= 4  mm
Cover from top=15 mm
Main steel = 8 mm dia,  4  in each step with development length of  470  mm
Distribution steel = 6 mm dia @  415  mm c/c


## Ex10.2:pg-518¶

In [2]:
import math
l=2.7+1 #span, in m
R=0.15 #rise, in m
t=0.27 #tread, in m
sigma_cbc=5 #in MPa
sigma_st=230 #in MPa
#assuming 50 mm per 1 m of span
D=50*l #in mm
D=200 #assume, in mm
W1=D/10**3*25*math.sqrt(R**2+t**2)/t #slab load on plan, in kN/m
W2=1/2.0*R*t*25/t #load of step per metre, in kN/m
W3=3 #live load, in kN/m
W=W1+W2+W3 #in kN/m
M=W*l**2/8 #in kN-m
d=math.sqrt(M*10**6/0.65/10**3) #in mm
d=170 #in mm
#assume 10 mm dia bars
dia=10 #in mm
D=d+dia/2+25 #which is equal to assumed value, hence OK
z=0.9*d #in mm
Ast=M*10**6/sigma_st/z #in mm
s1=1000*0.785*dia**2/Ast #spacing of 10 mm dia bars
s1=150 #assume, in mm
Ads=0.12/100*D*10**3 #distribution steel, in sq mm
#provide 8 mm dia bars
s2=210 #in mm
#let span-to-depth ratio be 'a'
a=l*10**3/D
#for Fe415 grade steel and pt=.32
MF=1.2
b=20*MF #permissible span-to-depth ratio
#as a<b, hence OK
print "Summary of design\nSlab thickness=",(D)," mm\nCover = 25 mm\nMain steel = 10 mm dia bars @ ",(s1)," mm c/c\nDistribution steel = 8 mm dia @ ",(s2)," mm c/c"

Summary of design
Slab thickness= 200  mm
Cover = 25 mm
Main steel = 10 mm dia bars @  150  mm c/c
Distribution steel = 8 mm dia @  210  mm c/c


## Ex10.3:pg-519¶

In [3]:
import math
l=2.5+1.5 #span, in m
R=0.15 #rise, in m
t=0.25 #tread in m
sigma_cbc=7 #in MPa
sigma_st=275 #in MPa
#assuming 50 mm per 1 m of span
D=50*l #in mm
W1=D/10**3*25*1.5*math.sqrt(R**2+t**2)/t #slab load on plan, in kN/m
W2=1/2*R*t*1.5*25/t #load of step per metre, in kN/m
W3=1.5*5 #live load, in kN/m
W=W1+W2+W3 #in kN/m
M=W*l**2/8 #in kN-m
d=math.sqrt(M*10**6/0.81/1.5/10**3) #in mm
d=177 #in mm
#assume 10 mm dia bars
dia=10 #in mm
D=d+dia/2+25 #which is slightly more than assumed value, hence OK
D=200 #in mm
d=D-dia/2-25 #in mm
z=0.92*d #in mm
Ast=M*10**6/sigma_st/z #in sq mm
s1=1500*0.785*dia**2/Ast #spacing of 10 mm dia bars, in mm
s1=130 #assume, in mm
Ads=0.12/100*D*1.5*10**3 #distribution steel, in sq mm
#provide 8 mm dia bars
s2=140 #in mm
#let span-to-depth ratio be 'a'
a=l*10**3/D
pt=Ast/1500/D*100 #pt=0.3
#for Fe500 grade steel and pt=.3
MF=1.2
b=20*MF #permissible span-to-depth ratio
#as a<b, hence OK
print "Summary of design\nSlab thickness=",D," mm\nCover = 25 mm\nMain steel = 10 mm dia bars @ ",s1," mm c/c\nDistribution steel = 8 mm dia @ ",s2," mm c/c"

Summary of design
Slab thickness= 200  mm
Cover = 25 mm
Main steel = 10 mm dia bars @  130  mm c/c
Distribution steel = 8 mm dia @  140  mm c/c


## Ex10.4:pg-520¶

In [4]:
import math
R=0.15 #rise, in m
t=0.3 #tread, in m
sigma_cbc=5 #in MPa
sigma_st=230 #in MPa
l1=1.8+1.5 #span for flight AB, in m
l2=1.2+1.5+1.5 #span for flight BC, in m
l3=1.8+1.5 #span for flight CD, in m
#assuming 50 mm slab thickness per 1 m of span
D=50*l2 #slab thickness, in mm
W1=D/10**3*25*1.5*math.sqrt(R**2+t**2)/t #slab load on plan, in kN/m
W2=1/2*R*t*1.5*25/t #load of step per metre, in kN/m
W3=1.5*5 #live load, in kN/m
W=W1+W2+W3 #in kN/m
#bending moment
#(a) flight AB and CD, refer Fig. 10.9
Rb=(W/2*1.5*(1.8+1.5/2)+W*1.8**2/2)/(1.5+1.8) #in kN
Ra=W/2*1.5+W*1.8-Rb #in kN
x=Ra/Rb #point of zero shear force from Ra, in m
M1=Ra*x-W*x**2/2 #maximum bending moment, in kN-m
#(b) flight BC, refer Fig. 10.10
Rb=(W/2*1.5**2/2+W*1.2*(1.2/2+1.5)+W/2*1.5*(1.5+1.2+1.5/2))/(1.5+1.2+1.5) #in kN
Rc=Rb #in kN
#maximum bending moment will be at centre
M2=Rb*(1.5+1.2/2)-W/2*1.5*(1.5/2+1.2/2)-W*(1.2/2)**2/2 #maximum bending moment, in kN-m
M=max(M1,M2) #in kN/m
d=math.sqrt(M*10**6/0.65/1.5/10**3) #in mm
#assume 10 mm dia bars
dia=10 #in mm
D=d+dia/2+25 #< 210 mm (assumed value)
D=210 #in mm
d=D-dia/2-25 #in mm
#steel
#flight AB and CD
z=0.9*d #in mm
Ast=M1*10**6/sigma_st/z #in sq mm
s1=1500*0.785*dia**2/Ast #spacing of 10 mm dia bars, in mm
s1=210 #round-off, in mm
Ads=0.12/100*D*1.5*10**3 #distribution steel, in sq mm
#provide 6 mm dia bars
s2=70 #round-off, in mm
#flight BC
Ast=M2*10**6/sigma_st/z #in sq mm
s3=1500*0.785*dia**2/Ast #spacing of 10 mm dia bars, in mm
s3=130 #round-off, in mm
#distribution steel is same as flights AB and CD
#let span-to-depth ratio be 'a'
a=l2*10**3/D
#for Fe415 grade steel and pt=.32
MF=1.2 #modification factor
b=20*MF #permissible span-to-depth ratio
#as a < b, hence OK
print "Summary of design\nSlab thickness=",D," mm\nCover = 25 mm\n(a)Flight AB and CD\nMain steel = 10 mm dia bars @ ",s1," mm c/c\nDistribution steel = 6 mm dia @ ",s2," mm c/c\n(b)Flight BC\nMain steel = 10 mm dia bars @ ",s3," mm c/c\nDistribution steel = 6 mm dia @ ",s2," mm c/c"
#answer in textbook is incorrect

Summary of design
Slab thickness= 210  mm
Cover = 25 mm
(a)Flight AB and CD
Main steel = 10 mm dia bars @  210  mm c/c
Distribution steel = 6 mm dia @  70  mm c/c
(b)Flight BC
Main steel = 10 mm dia bars @  130  mm c/c
Distribution steel = 6 mm dia @  70  mm c/c