import math
sigma_cbc=5 #in MPa
sigma_st=230 #in MPa
phi=30 #angle of repose, in degrees
H=5 #height of wall, in m
B=0.6*H #assume, in m
T=B/4 #assume toe to base ratio as 1:4
W=16 #density of retained earth, in kN/cu m
P=W*H**2/2*(1-math.sin(phi))/(1+math.sin(phi)) #in kN
P=67 #assume, in kN
M1=P*H/3 #in kN-m
M1=112 #assume, in kN-m
#bending moment at 2.5 m below the top
h=2.5 #in m
M2=W*h**2/2*(1-math.sin(phi))/(1+math.sin(phi))*h/3 #in kN-m
M2=14 #in kN-m
#thickness of stem (at the base)
d=math.sqrt(M1*10**6/0.65/1000) #in mm
d=415 #in mm
dia=20 #assume 20 mm dia bars
D1=d+dia/2+25 #in mm
D2=200 #thickness at top, in mm
D3=D2+(D1-D2)*h/H #in mm
d3=math.sqrt(M2*10**6/0.65/1000) #in mm
D3=d3+dia/2+25 #< 325 mm (provided), hence OK
D3=325 #in mm
d3=D3-dia/2-25 #in mm
#main steel
#(a) 5 m below the top
Ast=M1*10**6/sigma_st/0.9/d #in sq mm
#provide 20 mm dia bars
s1=1000*0.785*20**2/Ast #in mm
s1=240 #assume, in mm
#(b) 2.5 m below the top
Ast=M2*10**6/sigma_st/0.9/d3 #in sq mm
Astmin=0.12/100*10**3*D3 #in sq mm
Ast=max(Ast,Astmin) #in sq mm
#provide 12 mm dia bars
s2=1000*0.785*12**2/Ast #in mm
s2=290 #assume, in mm
#distribution steel
Ads=0.12/100*10**3*D3 #in sq mm
#provide 8 mm dia bars
s3=1000*0.785*8**2/Ads #in mm
s3=125 #assume, in mm
#check for shear
V=P #in kN
Tv=V*10**3.0/10**3/d #in MPa
#for M15 grade concrete and pt=0.31
Tc=0.22 #in MPa
#as Tc > Tv, no shear reinforcement required
#development length
#(a) At the base of stem
dia=20 #in mm
Tbd=0.84 #in MPa
Ld=dia*sigma_st/4/Tbd #in mm
Ld=1370 #assume, in mm
#(b) At 2.5 m below the top
dia=12 #in mm
Ld=dia*sigma_st/4/Tbd #in mm
Ld=825 #assume, in mm
#check for stability
D4=500 #thickness of base, in mm (assume)
V1=1/2.0*(D1-D2)/10**3*H*25 #in kN
V2=(D2/10**3)*H*25 #in kN
V3=(D4/10**3)*B*25 #weight of base, in kN
V4=(B-T-D1/10**3)*H*W #weight of soil, in kN
V=V1+V2+V3+V4 #in kN
M=V1*(T+2/3*(D1-D2)/10**3)+V2*(T+(D1-D2)/10**3+D2/10**3.0/2)+V3*B/2+V4*(B-(B-T-D1/10**3)/2) #in kN-m
x=M/V #in m
x=1.8 #assume, in m
#factor of safety
#for overturning
F1=V*x/P/(H/3) #> 1.5, hence OK
mu=0.5
#for sliding
F2=mu*V/P #> 1.5, hence OK
print "Summary of design:\nThickness of stem (at base) = ",(D1)," mm\nThickness of stem at top = ",(D2)," mm\nRefer Fig. 12.4 of textbook for reinforcement details"
import math
sigma_cbc=5 #in MPa
sigma_st=140 #in MPa
phi=35 #angle of repose, in degrees
H=6 #height of wall, in m
B=0.4*H #assume, in m
T=B/4 #assume toe to base ratio as 1:4
W=18 #density of retained earth, in kN/cu m
P=W*H**2/2*(1-math.sin(phi))/(1+math.sin(phi)) #in kN
P=88 #assume, in kN
M1=P*H/3 #in kN-m
#bending moment at 3 m below the top
h=3 #in m
M2=W*h**2/2*(1-math.sin(phi))/(1+math.sin(phi))*h/3 #in kN-m
M2=22 #in kN-m
#thickness of stem (at the base)
d=math.sqrt(M1*10**6/0.87/1000) #in mm
d=450 #in mm
dia=20 #assume 20 mm dia bars
D1=d+dia/2+25 #in mm
D2=200 #thickness at top, in mm
D3=D2+(D1-D2)*h/H #in mm
d3=math.sqrt(M2*10**6/0.87/1000) #in mm
D3=d3+dia/2+25 #< 342.5 mm (provided), hence OK
D3=342.5 #in mm
d3=D3-dia/2-25 #in mm
#main steel
#(a) 6 m below the top
Ast=M1*10**6/sigma_st/0.87/d #in sq mm
#provide 20 mm dia bars
s1=1000*0.785*20**2/Ast #in mm
s1=95 #assume, in mm
#(b) 3 m below the top
Ast=M2*10**6/sigma_st/0.87/d3 #in sq mm
#provide 10 mm dia bars
s2=1000*0.785*10**2/Ast #in mm
s2=130 #assume, in mm
#distribution steel
Ads=0.15/100*10**3*D3 #in sq mm
#provide 10 mm dia bars
s3=1000*0.785*10**2/Ads #in mm
s3=150 #assume, in mm
#check for shear
V=P #in kN
Tv=V*10**3.0/10**3/d #in MPa
#for M15 grade concrete and pt=0.71
Tc=0.34 #in MPa
#as Tc > Tv, no shear reinforcement required
#development length
#(a) At the base of stem
dia=20 #in mm
Tbd=0.6 #in MPa
Ld=dia*sigma_st/4/Tbd #in mm
Ld=1170 #assume, in mm
#(b) At 3 m below the top
dia=10 #in mm
Ld=dia*sigma_st/4/Tbd #in mm
Ld=590 #assume, in mm
#check for stability
D4=500 #thickness of base, in mm (assume)
V1=1.0/2*(D1-D2)/10**3*H*25 #in kN
V2=(D2/10**3)*H*25 #in kN
V3=(D4/10**3)*B*25 #weight of base, in kN
V4=(B-T-D1/10**3)*H*W #in kN
V=V1+V2+V3+V4 #in kN
M=V1*(T+2/3*(D1-D2)/10**3)+V2*(T+(D1-D2)/10**3+D2/10**3/2)+V3*B/2+V4*(B-(B-T-D1/10**3)/2) #in kN-m
x=M/V #in m
#factor of safety
#for overturning
F1=V*x/P/(H/3) #> 1.5, hence OK
mu=0.5
#for sliding
F2=mu*V/P #< 1.5, hence it is not safe against sliding
print "Summary of design:\nThickness of stem (at base) = ",(D1)," mm\nThickness of stem at top = ",(D2)," mm\nRefer Fig. 12.7 of textbook for reinforcement details"
#answers in textbook for factor of safety against overturning and sliding are incorrect
import math
sigma_cbc=5 #in MPa
sigma_st=230 #in MPa
phi=30 #angle of repose, in degrees
H=5 #height of wall, in m
B=0.6*H #assume, in m
T=B/4 #assume toe to base ratio as 1:4
t=450 #thickness of wall, in mm
W=16 #density of retained earth, in kN/cu m
P=W*H**2/2*(1-math.sin(phi))/(1+math.sin(phi)) #in kN
P=67 #assume, in kN
y=1.8 #in m
P=67 #in kN
Wt=223 #in kN
D=0.5 #thickness of base, in m
x=1.8-P*(H/3+D/10**3)/Wt #in m
x=1.15 #in m
e=B/2-x #in m
q1=Wt/B+Wt*e/(1*B**2.0/6) #maximum pressure, in kN/sq m
q2=Wt/B-Wt*e/(1*B**2.0/6) #minimum pressure, in kN/sq m
Pa=q1-(q1-q2)/B*T #pressure at A, in kN/sq m
Pa=100 #assume, in kN/sq m
Pb=q1-(q1-q2)/B*(T+t/10**3) #pressure at B, in kN/sq m
Pb=85 #assume, in kN/sq m
Ma=Pa*T**2/2+1/2*(q1-Pa)*T*2/3*T-T*D*25*T/2 #bending moment at A, in kN-m
Ma=30 #round-off, in kN-m
Mb=(B-T-t/10**3)**2*H*W/2+(B-T-t/10**3)**2*D*25/2-q2*(B-T-t/10**3)**2/2-(Pb-q2)*1/3*(B-T-t/10**3)**2/2 #bending moment at B, in kN-m
Mb=80 #in kN-m
#design of toe
d=math.sqrt(Ma*10**6/0.65/10**3) #in mm
D=d+10/2+70 #<500 mm (provided), hence OK
D=500 #in mm
d=D-70 #in mm
Ast=Ma*10**6/sigma_st/0.9/d #in sq mm
Astmin=0.12/100*10**3*D #in sq mm
Ast=max(Ast,Astmin) #in sq mm
s1=1000*0.785*10**2/Ast #in mm
s1=130 #assume, in mm
#distribution steel is same as above
#check for shear
V=(q1+Pa)/2*T #in kN
Tv=V*10**3/10**3/d #in MPa
#for M15 grade concrete and pt=0.32
Tc=0.2368 #in MPa
#as Tc > Tv, no shear reinforcement required
#development length
dia=10 #in mm
Tbd=0.84 #in MPa
Ld=dia*sigma_st/4/Tbd #in mm
Ld=685 #assume, in mm
#design of heel
d=math.sqrt(Mb*10**6/0.65/10**3) #< 430 mm (provided), hence OK
d=430 #in mm
Ast=Mb*10**6/sigma_st/0.9/d #in sq mm
s2=1000*0.785*10**2/Ast #in mm
s2=85 #assume, in mm
#distribution steel: 0.12% of Ag, hence provide 10 mm dia bars @ 130 mm c/c
V=(B-T-t/10**3)*H*W-(Pb+q2)/2*(B-T-t/10**3) #in kN
Tv=V*10**3.0/10**3/d #in MPa
#for M15 grade concrete and pt=0.32
Tc=0.2368 #in MPa
#as Tc > Tv, no shear reinforcement required
#development length
dia=10 #in mm
Tbd=0.84 #in MPa
Ld=dia*sigma_st/4/Tbd #in mm
Ld=685 #assume, in mm
print "Summary of design:\nThickness of base slab=",(D)," mm. Refer to Fig. 12.11 of textbook for reinforcement details."
#answer in textbook for spacing of 10 mm dia bars for main steel in toe and distribution steel is incorrect