Civil Engineering Reference
In-Depth Information
the shears and moments are calculated. For strength design, the footing must be propor-
tioned for the effects of these loads as required in ACI Section 9.2.
EXAMPLE 12.8
Determine the width needed for a wall footing to support loads:
D
18 k/ft and
L
12 k/ft. In ad-
dition, a lateral load of 6 kt/ft is assumed to be applied 5 ft above the top of the footing. Assume the
footing is 18 in. thick, its base is 4 ft below the final grade, and
q
a
4 ksf.
SOLUTION
First Trial
18
12
30
12
Neglecting moment
q
e
4000
(150)
(100)
3525 psf
Width required
18
12
3.525
8.51
Try 9
0
M
6
distance to footing base
6
6.5
39 ft- k
A
(9)(1)
9 ft
2
1
12
I
(1)(9)
3
60.75 ft
4
9
(39)(4.5)
P
A
Mc
30
q
max
I
60.75
6.22 ksf
3.525 ksf
No good
9
(39)(4.5)
P
A
Mc
30
q
min
0.44 ksf
I
60.75
Second trial
Assume 14-ft-wide footing (after a check of a 13-ft-wide footing proves it to be insufficient)
A
(14)(1)
14 ft
2
1
12
(1)(14)
3
228.7 ft
4
I
(39)(7)
228.7
30
14
q
max
3.34 ksf
3.525 ksf
14
(39)(7)
30
q
min
0.95 ksf
Use 14
0
footing
228.7
12.13
TRANSFER OF HORIZONTAL FORCES
When it is necessary to transfer horizontal forces from walls or columns to footings, the
shear friction method previously discussed in Section 8.12 of this text should be used.
Sometimes shear keys (see Figure 13.1) are used between walls or columns and footings.
This practice is of rather questionable value, however, because appreciable slipping has to
occur to develop a shear key. A shear key may be thought of as providing an additional
mechanical safety factor, but none of the lateral design force should be assigned to it.
The following example illustrates the consideration of lateral force transfer by the
shear friction concept.
