3 spread footing
TRANSCRIPT
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Dr.: Youssef Gomaa Youssef
CE 402: Part AShallow Foundation Design
Lecture No. (3): Spread Footing
Fayoum UniversityFaculty of Engineering
Department of Civil Engineering
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Design of Spread Footing
• Plain concrete footing (P.C.)
.* F La
P Area A B
q Eq.(1)
Assume thickness of P.C.:
t = (0.25 to 0.50)
Dim. of P.C. = A * B * tA
A1
t 1
t
P F.L
P G.S
q a
G.S
( ) ( ) A a B b Eq.(2) optional
Solve Eqs (1)&(2) to get A and B
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Design of Spread Footing• Reinforced concrete footing (R.C.)
1 2 A A X
21( ) / 22 I n
A a M p
.
1 1*
G s
n
P p
A B
* cu
M d C
b F
1 covt d er
Steel cover=5.0 to 7.0cm
Dim. of R.C. = A1 * B 1 * t 1
A1
t 1
P G.S
p n
G.S
a
+
+ +
+
B.M.D
1 2 B B X
= 0.80 → 1.00 ∗
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Design of Spread Footing
• Shear Stress:
1*2 s n
A aQ p d
0.75 cu suc
f q
* s
s su
Qq q
b d
Qs: shear force at critical sec. (II).
qs: shear stress.
qsu: ultimate shear strength.
If q s > q su , Increase d
Notes:
• No shear RFT in Footing .
t 1
A1
P G.S
(II)
G.S
a
(II)
(II) p n
d
+
-
S.F.D
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Design of Spread Footing
• Punching Stress:
. * ( )*( ) p G s nQ P p a d b d
*2* ( ) ( ) p A d a d b d
p p
p
Qq
A
0.5 ( / ) / /cup cu c cu cq a b f f
If q p > q cup , Increase d
t 1
A1
P G.S
G.S
a
p n
d/2 a d/2
d/2
d/2
b
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Design of Spread Footing
• Footing Reinforcement:
Which is required?
Top or bottom RFT
why?
* * I
s y
M A
f d j
Notes:
• Minimum number of bars per meter is five .
• Minimum diameter for main RFT is 12mm.
• Number of bars may be taken 5 to 8.
• Diameter of bars may be selected from 12 to 18mm.
t 1
G.S
a
+
+ +
+
B.M.D
G.S
A1
Main RFT5∅12/ ′
Main RFT5∅12/ ′
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Design of Spread Footing
• Example(1):
Make a complete design for a footing supporting a 30cm X 60cm
column load of 230t at ground surface (G.S.). The foundation
level is 1.5 m below G.S. and the net allowable bearing capacity is1.75kg/cm2. Make the design considering the following two cases:
1- with plain concrete base
2- without Plain concrete base
a = 0.60m. B= 0.30m
p G.S = 230 t/m’
q a = 1.75kg/cm 2 = 17.5t/m 2 .
f cu = 250kg/cm 2 .
f y =3600kg/cm 2
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Design of Spread Footing
• Plain concrete footing (P.C.)2. 230*1.15* 15.11
17.5 F L
a
P Area A B m
q
Assume thickness of P.C.: t = 0.30m
Dim. of P.C. = 4.05 * 3.75 * 0.30
A
A1
t 1
t
P F.L
P G.S
q a
G.SSolve Eqs (1)&(2) to get A and B
( ) ( ) A a B b
2
( 0.30) * 15.11 B B m
2 0.30* 15.11 0.0 B B
20.30 0.30 4*1*15.113.75
2*1 B m
3.75 0.30 4.05 A m
Eq.(1)2* 15.11 A B m
( ) 0.30 A B Eq.(2)0.30 A B
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Design of Spread Footing• Reinforced concrete footing (R.C.)
1 4.05 2*0.30 3.45 A m
0.30 X m
21( ) / 22 I n
A a M p
2.
1 1
230*1.5031.75 /* 3.45*3.15
G sn
P p t m A B
532.23*105 56.77 60
* 100*250cu
M d C cm
b F
1 60 5 65t cm
Dim. of R.C. = 3.45 * 3.15* 65
A1
t 1
P G.S
p n
G.S
a
+
+ +
+
B.M.D
1 3.75 0.60 3.15 B m
2(3.45 0.60)/ 231.75 32.23 . / '
2 I M m t m
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Design of Spread Footing
• Shear Stress:1*2 s n
A aQ p d
2500.75 0.75 9.68 /
1.50cu
suc
f q t m
3226.19*10 4.37 /
* 100*60 s
sQ
q t mb d
Qs: shear force at critical sec. (II).
qs: shear stress.
qsu: ultimate shear strength.
Notes:
• No shear RFT in Footing .
t 1
A1
P G.S
(II)
G.S
a
(II)
(II) p n
d
+
-
S.F.D
3.45 0.6031.75( 0.60) 26.19 / '
2 sQ t m
s suq q safe shear
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Design of Spread Footing
• Punching Stress:
. * ( )*( ) p G s nQ P p a d b d
*2* ( ) ( ) p A d a d b d
3
24317.10*10 12.58 /2.52*10
p p p
Qq kg cm A
0.5 ( / ) / 250 /1.50 12.91cup cu cq b a f
t 1
A1
P G.S
G.S
a
p n
d/2 a d/2
d/2
d/2
b
230*1.50 31.75* (0.60 0.50)* (0.30 0.50) 317.1 pQ t
20.60* 2* (0.60 0.60) (0.30 0.60) 2.52 p A m
p cupq q safe punching
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Design of Spread Footing
• Footing Reinforcement:
5232.23*10 21.50
* * 3600*60*0.695 I
s y
M A cm
f d j
As (short dir.)= 9 φ 18 mm/m’
As (long dir.)= 9 φ 18 mm/m’
0.65
G.S
0.60
+
+ +
+
B.M.D
G.S
3.45
(9 18/ ')m
(9 18/ ')m
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Design of Spread Footing
• Problem(1):
Make a complete design for a footing supporting a 25cm X 50cm
column load of 150 t at ground surface (G.S.). The foundation
level is 1.5 m below G.S. and the net allowable bearing capacity is
1.50kg/cm2
. The column location with respect to neighbors isshown in Figure.
1.50m 0.50
0.25
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Design of Spread Footing
• Problem(2):
Make a complete design for a footing supporting a 25cm X 50cm
column load of 120 t at ground surface (G.S.). The foundation
level is 1.5 m below G.S. Soil Profile under footing is shown in
the Figure. Allowable settlement is 2.0cm.
Cu=1.00kg/cm 2
mv= 0.10cm 2/kg
Cu=0.50kg/cm 2
mv =0.06cm 2/kg
3.00
5.00
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Design of Spread Footing• Problem(3):
Prove that the unconfined compressive strength of saturated clay
can be considered as the allowable net bearing capacity of this
clay.
1 2 (ECP:202/3) Eq. (3-8)ult c c c f q q qq CN i D N i BN i
*5.14*1.30*1.0 6.68ult u c c c u uq C N i C C
F.S. from Table (3-3) ECP:202/3
6.682.60 2
. . 2.50ult u
a u u uq C
q C C q F S
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Design of Spread Footing• Problem(4):
A large footing settles more than a small one resting on the same
ground profile and subjected to the same stresses. Give reasons
using neat sketches.