Civil Engineering Reference
In-Depth Information
1778
2 33 0 9 60
A
s
,
14 13
.
in
2
(
9116 11
.
mm
2
)
min
=
=
.
×
.
×
With three parallel ties, the area of reinforcement required to ensure
the factored flexural resistance is at least 1.2
M
cr
, which
is equal to
14.13/3 = 4.71 in
2
(3038.7 mm
2
).
The amount of reinforcement required to resist 1.33 times the factored
loads is
2
2
A
st
=
1 33 3 52 4 68
.
×
.
=
.
in
(
3019 35
.
mm
)
The amount of reinforcement required to resist 1.33 times the factored loads
is less than the amount required to resist 1.2
M
cr
; therefore, this smaller
amount will be checked against the amount provided by the original design
of the footing. There are presently 18 no. 8 bars provided in the lower mat
in each direction. This equals 18 × 0.79 = 14.22 in
2
(9174.18 mm
2
). This
results in 14.22/3 = 4.74 in
2
(3058.06 mm
2
) per tie zone. This reinforce-
ment is distributed across the full width of the footing and not the limits
of the nodes. Even though the total amount of reinforcement is greater
than the 4.68 in
2
(3019.35 mm
2
) required, it is not placed within the
region defined by the nodes and therefore does not meet the requirements
of STM.
13.2.2.2
Check the capacity of struts
Take the representative strut AJ (Figure 13.10c).
Using node A as representative of all the corner nodes, the area of the
vertical projection of the strut may be calculated as
25 45 6 88
2
.
.
25 45 19 8
2
.
.
×
+
×
A
srut
=
+ ×
2 25 45
.
+
2
87 55 50 90
.
.
45 25 1 185 10
.
=
.
×
5
mm
2
(
183 7
.
in
2
)
=
+
+
Because this value is a vertical section of the strut, the cross-sectional area
perpendicular to the axis of the strut can be calculated by cos(28.4°) ×
183.70 = 1.0426 × 10
5
mm
2
(161.6 in
2
).
The limiting compressive stress
f
cu
in the strut depends on the principal
strain, ε
1
, in the concrete surrounding the tension ties.
The tensile strain in tie AB is
P
A E
190
4 74 29 000
u
1 382 10
3
−
ε
s
=
=
=
.
×
.
×
,
st
s
In
accordance with AASHTO LRFD C5.6.3.3.3, the strain will be approxi-
mately equal to
1 382 10 2 0 691 10
.
×
−
3
/
=
.
×
−
3
at the midpoint of the strut. Using
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