1. Reinforced Concrete Design-II
Lecture 11: Load transfer to the foundation in Columns
Semester – Fall 2020
Dr.Tahir Mehmood
Reference
Design of Concrete Structures 14th Ed. by Nilson, Darwin and Dolan.
2. Bearing-Load transfer for column to footing
At the base of the column,
the permitted design
bearing strength is
2
A
A
A
A
A
)
f
85
.
0
(
P
1
2
1
2
1
c
n
A1 is column cross-sectional
area
A2 is the area of the portion
of the supporting footing
At least four dowels having
area not less that 0.005
times the x-sectional area
of the column and the
diameter of dowel should
not exceed the diameter of
the column reinfrocement
by more than 0.15”.
3. Development Length
Dowel should have sufficient development length
For compression bars, that is, b
y
c
b
y
d d
f
0003
.
0
f
d
f
02
.
0
l
The lapped length must be at least that required for a lap
splice in compression. i.e. the length of lap must not be less
than the usual development length in compression and must
not be less than 0.0005 fydb
Where bars of different sizes are la spliced, the splice length
should be the larger of the development length of larger bar or
the splice length of the smaller bar.
Splice length for compression
For bar with 0.0005 fydb
psi
000
,
60
fy
For bar with fy > 60,000 psi (0.0009fy-24)db
Not less than 12”
For fc’ <3000 psi lap is increased by one third.
4. Problem
Design a square column footing to carry a column dead load of
197 k and a live load of 160 k from a 16” squired tied column
containing # 11 bars as the principal column steal. The
allowable soil pressure is 4.5 k/ft2. Use fc’=3000 psi fy=40000
psi , 4’ below final grade.
Solution
Assume 2’ thick column footing
qe = 4500 – (2)(150) - (2)(100) = 4000 lb/ft2
Required area 2
ft
25
.
89
4
160
197
Let 9’ – 6” = 90.5 ft2
Pu = (1.2) (197) + (1.6) (160) = 492.4 k
2
u ft
k/
07
.
6
25
.
90
8
.
547
q
Let depth required for punching shear d=20”
bo=16+20=36”=3’
492.4 5.45
5. Vu = (qu) (Area)
=(5.45){(9.5)2-32} = 442.81 k
βc=1
k
3
.
536
98
.
630
85
.
0
V
k
98
.
630
1000
1
20
36
4
3000
4
d
b
f
4
V
c
o
c
c
Solution
The other equations are checked and give larger value so
this value is valid and for this øVc
øVc > Vu
Therefore no shear reinforcement is required.
9
3
.
18
)
36
(
)(4)
3000
)(
4
)(
85
.
0
(
)
1000
)(
19
.
493
(
b
f
4
V
d
o
c
u
So d=20” is O.K. w.r.t two-way shear.
0.75 473.23
442.81
0.75
18.71’’
6. For one-way shear
Solution
k
55
.
139
5
.
9
42
.
2
07
.
6
V
2
4
.
2
12
20
2
167
.
8
12
16
5
.
9
u
According to ACI code
k
3
.
212
76
.
249
85
.
0
V
k
76
.
249
1000
1
20
12
5
.
9
3000
2
d
b
f
2
V
c
w
c
c
Since øVc>Vu therefore no shear reinforcement is required
5
1
.
13
)
12
5
.
9
)(
3000
)(
2
)(
85
.
0
(
1000
55
.
139
b
f
2
V
d
w
c
u
= 8.167
8.167
2.42
0.75 187.32
5.45 125.3 k
125.3
0.75
13.34’’
7. Therefore depth d=20” is O.K. w.r.t. one-way shear.
Bending strength
083
.
4
ft
.
k
66
.
480
2
167
.
8
2
16
5
.
9
)
5
.
9
(
2
083
.
4
)
083
.
4
)(
07
.
6
(
Mu
Solution
2
s
c
y
y
n
2
2
u
n
in
76
.
8
20
114
00384
.
0
bd
A
quired
Re
00384
.
0
000
,
40
81
.
148
69
.
15
2
1
1
69
.
15
1
69
.
15
3000
85
.
0
000
,
40
f
85
.
0
f
m
f
mR
2
1
1
m
1
psi
81
.
148
)
20
)(
114
)(
85
.
0
(
)
12000
)(
66
.
480
(
bd
M
R
quired
Re
5.45
432.56
432.56
0.75
151.78
151.78
8. Use As = 0.0041 ( 114)x( 20) =9.34 in2
Try 16 # 7 having As = 9.6 in2
Solution
C = 0.85fc’ba = (0.85)(3)(9.5)(12)(a) = 290.7 a
T = Asfy = (9.6)(40) = 384 k
"
2384
.
1
7
.
290
360
a
ft
.
k
66
.
480
M
M
ft
.
k
29
.
523
12
1
2
2384
.
1
20
40
9
9
.
0
2
a
d
f
A
9
.
0
M
u
n
y
s
n
So use 16 # 7 bars each way
384 1.32
1.32
557
Ok
9.6
9. Bearing strength
øPn = ø(0.85fc’) Ag
= (0.7)(0.85 x 3000)(16)2
= 424320 lb = 424.32 k
Pu = 492.4 k
Solution
øPn < Pu thus column load can not be transferred by
concrete bearing alone and dowels bars are required.
Excess Pu = 492.4 – 424.32 = 68.08 k
2
y
u
s in
25
.
3
)
40
)(
7
.
0
(
84
.
90
f
P
Excess
A
Required
2
c
y
u
s
in
47
.
3
)
3
85
.
0
40
)(
7
.
0
(
84
.
90
)
f
85
.
0
f
(
P
Excess
A
More correctly Required
0.65
68.08
0.65
2.62
0.65
68.08
2.80
10. Use 4 # 8 bars As = 3.16 in2 Min As=(0.005)(16)2=1.28 in2
The # 8 dowels must be developed below and above the
junction of column and footing.
Development length in compression
7
4
.
16
3000
128
.
1
000
,
40
02
.
0
f
d
f
02
.
0
c
b
y
db
Bearing strength
Solution
If should not be less than ldb = 0.0003 fydb
=(0.0003)(40000)(1)=12”
24” thick footing is adequate for straight dowels.
1
14.60
11. 24-3.0(cover)-2x0.875(footing bars)-1(dowels)= 18.25” >
14.60”
alternate solutions (i) thicker footing(ii) Larger number of
smaller bars(iii) pedestal
Development length of main bars
for # 7 bars from table ld = 18”
Available length )
er
(cov
3
2
16
)
12
)(
5
.
9
( "
If bends are provided in dowels then the available length is
Solution
=46” > ld=18”
