import math
b=0.2 #column width in m
D=0.3 #column depth in m
fck=15 #in MPa
sigma_cbc=5 #in MPa
sigma_st=230 #in MPa
P1=600 #load on column in kN
P2=0.05*P1 #weight of footing, in kN
P=P1+P2 #in kN
q=150 #bearing capacity of soil in kN/sq m
A=P/q #in sq m
L=math.sqrt(A) #assuming footing to be square
L=2.1 #assume, in m
p=P1/L**2 #soil pressure, in kN/sq m
p=136 #assume, in sq m
bc=b/D
ks=0.5+bc #>1
ks=1
Tc=0.16*math.sqrt(fck)*10**3 #in kN/sq m
Tv=Tc
#let d be the depth of footing in metres
#case I: consider greater width of shaded portion in Fig. 11.3 of textbook
d1=L*(L-b)/2*p/(Tc*L+L*p) #in m
#case II: refer Fig. 11.4 of textbook; we get a quadratic equation of the form e d**2 + f d + g = 0
e=p+4*Tc
f=b*p+D*p+2*(b+D)*Tc
g=-(L**2-b*D)*p
d2=(-f+math.sqrt(f**2-4*e*g))/2/e #in m
d2=0.362 #assume, in m
#bending moment consideration, refer Fig. 11.5 of textbook
Mx=1*((L-b)/2)**2/2*p #in kN-m
My=1*((L-D)/2)**2/2*p #in kN-m
d3=math.sqrt(Mx*10**6/0.65/10**3) #<362 mm, hence OK
z=0.9*d2*10**3 #lever arm, in mm
Ast1=(Mx*10**6/sigma_st/z) #in sq mm
Ast=L*Ast1 #steel required for full width of 2.1 m, in sq mm
#provide 12 mm dia bars
dia=12 #in mm
n=Ast/0.785/dia**2 #no. of 12 mm dia bars
n=16 #assume
Tbd=0.84 #in MPa
Ld=dia*sigma_st/4/Tbd #in mm
Ld=825 #assume, in mm
c=50 #side cover, in mm
La=(L-D)/2*10**3-c #>Ld, hence OK
D=d2*10**3+dia/2+100 #in mm
print "Summary of design:\nOverall depth of footing=",(D)," mm\nCover=100 mm bottom; 50 mm side\nSteel-",(n)," bars of 12 mm dia both ways"
import math
b=0.4 #column width, in m
D=0.4 #column depth, in m
fck=15 #in MPa
sigma_cbc=5 #in MPa
sigma_st=140 #in MPa
P1=1000 #load on column, in kN
P2=0.05*P1 #weight of footing, in kN
P=P1+P2 #in kN
q=200 #bearing capacity of soil, in kN/sq m
A=P/q #in sq m
L=math.sqrt(A) #assuming footing to be square
L=2.3 #assume, in m
p=P1/L**2 #soil pressure, in kN/sq m
p=189 #assume, in kN/sq m
bc=b/D
ks=0.5+bc #>1
ks=1
Tc=0.16*math.sqrt(fck)*10**3 #in kN/sq m
Tv=Tc
#let d be the depth of footing in metres
#case I: consider greater width of shaded portion in Fig. 11.7 of textbook
d1=L*(L-b)/2*p/(Tc*L+L*p) #in m
#case II: refer Fig. 11.8 of textbook; we get a quadratic equation of the form e d**2 + f d + g = 0
e=p+4*Tc
f=b*p+D*p+2*(b+D)*Tc
g=-(L**2-b*D)*p
d2=(-f+math.sqrt(f**2-4*e*g))/2/e #in m
d2=0.425 #assume, in m
d=max(d1,d2) #in m
#bending moment consideration, refer Fig. 11.9 of textbook
Mx=1*((L-b)/2)**2/2*p #in kN-m
d3=math.sqrt(Mx*10**6/0.87/10**3) #<425 mm, hence OK
z=0.87*d*10**3 #lever arm, in mm
Ast1=(Mx*10**6/sigma_st/z) #in sq mm
Ast=L*Ast1 #steel required for full width of 2.3 m, in sq mm
#provide 18 mm dia bars
dia=18 #in mm
n=Ast/0.785/dia**2 #no. of 18 mm dia bars
n=15 #assume
Tbd=0.6 #in MPa
Ld=dia*sigma_st/4/Tbd #in mm
c=50 #side cover, in mm
La=(L-D)/2*10**3-c #in mm
#providing hook at ends
La=La+16*dia #>Ld, hence OK
D=d2*10**3+dia/2+100 #in mm
print "Summary of design:\nOverall depth of footing=",(D)," mm\nCover=100 mm bottom; 50 mm side\nSteel-",(n)," bars of 18 mm dia both ways"
import math
B=0.5 #column diameter, in m
fck=20 #in MPa
sigma_cbc=7 #in MPa
sigma_st=230 #in MPa
P1=1600 #load on column, in kN
P2=0.05*P1 #weight of footing, in kN
P=P1+P2 #in kN
q=300 #bearing capacity of soil, in kN/sq m
A=P/q #in sq m
L=math.sqrt(A) #assuming footing to be square
L=2.4 #assume, in m
p=P1/L**2 #soil pressure, in kN/sq m
p=278 #assume, in kN/sq m
bc=1
ks=0.5+bc #>1
ks=1
Tc=0.16*math.sqrt(fck)*10**3 #in kN/sq m
Tv=Tc
#let d be the depth of footing in metres
#case I: refer Fig. 11.11 of textbook
d1=L*(L-B)/2*p/(Tc*L+L*p) #in m
#case II: refer Fig. 11.12 of textbook; we get a quadratic equation of the form e d**2 + f d + g = 0
e=math.pi/4*p+math.pi*Tc
f=2*math.pi/4*B*p+math.pi*B*Tc
g=-(L**2-math.pi/4*B**2)*p
d2=(-f+math.sqrt(f**2-4*e*g))/2/e #in m
d2=0.57 #assume, in m
d=max(d1,d2) #in m
#bending moment consideration, refer Fig. 11.13 of textbook
M=1*((L-B)/2)**2/2*p #in kN-m
d3=math.sqrt(M*10**6/0.88/10**3) #<570 mm, hence OK
z=0.9*d*10**3 #lever arm, in mm
Ast1=(M*10**6/sigma_st/z) #in sq mm
Ast=L*Ast1 #steel required for full width of 2.4 m
#provide 20 mm dia bars
dia=20 #in mm
n=Ast/0.785/dia**2 #no. of 20 mm dia bars
n=9 #assume
Tbd=1.12 #in MPa
Ld=dia*sigma_st/4/Tbd #in mm
Ld=1030 #assume, in mm
c=50 #side cover, in mm
La=(L-B)/2*10**3-c #in mm
#bend bar at right angle and provide length, l
l=Ld-La #in mm
D=d*10**3+dia/2+100 #in mm
print "Summary of design:\nOverall depth of footing=",(D)," mm\nCover:100 mm bottom; 50 mm side\nSteel:",(n),"-20 mm dia bars both ways"
#answer in textbook is incorrect
import math
b=0.3 #column width in m
c1=0.4 #column depth in m
fck=20 #in MPa
sigma_cbc=7 #in MPa
sigma_st=275 #in MPa
P1=1200 #load on column, in kN
P2=0.05*P1 #weight of footing, in kN
P=P1+P2 #in kN
q=200 #bearing capacity of soil, in kN/sq m
A=P/q #in sq m
L1=2 #in m
L2=A/L1 #assuming footing to be square
L2=3.2 #assume, in m
p=P1/L1/L2 #soil pressure, in kN/sq m
bc=b/c1
ks=0.5+bc #>1
ks=1
Tc=0.16*math.sqrt(fck)*10**3 #in kN/sq m
Tv=Tc
#let d be the depth of footing in metres
#case I, refer Fig. 11.15 of textbook
#short direction
d1=L1*(L2-c1)/2*p/(Tc*L1+L1*p) #in m
#long direction
d2=L2*(L1-b)/2*p/(Tc*L2+L2*p) #in m
#case II: refer Fig. 11.16 of textbook; we get a quadratic equation of the form e d**2 + f d + g = 0
e=p+4*Tc
f=b*p+c1*p+2*(b+c1)*Tc
g=-(L1*L2-b*c1)*p
d3=(-f+math.sqrt(f**2-4*e*g))/2/e #in m
d3=0.47 #assume, in m
d=max(d1,d2,d3) #in m
#bending moment consideration, refer Fig. 11.17 of textbook
Mx=1*((L1-b)/2)**2/2*p #in kN-m
My=1*((L2-c1)/2)**2/2*p #in kN-m
d4=math.sqrt(My*10**6/0.8/10**3) #in mm
d4=480 #>470 mm (provided for shear)
d=d4 #in mm
z=0.92*d #lever arm, in mm
#short direction
Ast1=(Mx*10**6/sigma_st/z) #in sq mm
Ast=L2*Ast1 #steel required for full width of 3.2 m, in sq mm
b1=L1 #central band width, in m
beta=L2/L1
Astc=L1/(beta+1)*Ast #in sq mm
#provide 12 mm dia bars
dia=12 #in mm
n1=Astc/0.785/dia**2 #no. of 12 mm dia bars
n1=13 #assume
Astr=Ast-Astc #steel in remaining width, in sq mm
n2=Astr/0.785/dia**2
n2=4 #assume
n2=n2/2 #on each side
Tbd=1.12 #in MPa
Ld=dia*sigma_st/4/Tbd #in mm
c=50 #side cover, in mm
La=(L1-b)/2*10**3-c #>Ld, hence OK
#long direction
Ast1=(My*10**6/sigma_st/z) #in sq mm
Ast=L1*Ast1 #steel required for full width of 2 m, in sq mm
#provide 18 mm dia bars
dia=18 #in mm
n=Ast/0.785/dia**2 #no. of 18 mm dia bars
n=12 #assume
Ld=dia*sigma_st/4/Tbd #in mm
c=50 #side cover, in mm
La=(L2-c1)/2*10**3-c #>Ld, hence OK
D=d+dia/2+100 #in mm
D=590 #assume, in mm
print "Summary of design:\nOverall depth of footing=",(D)," mm\nCover=100 mm bottom; 50 mm side\nSteel-long direction\n",(n)," bars of 18 mm dia in ",(L1)," m width equally spaced\nShort direction\nCentral band ",(L1)," m:",(n1),"-12 mm dia bars equally spaced\nRemaining sides:",(n2),"-12 mm dia bars on each side"