Civil Engineering Reference
In-Depth Information
= 0.57 + 67
t
0.90
Spiral
0.70
0.65
= 0.48 + 83
t
Other
Compression
controlled
Transition
Tension
controlled
Figure 10.25
Variation of
with
t
and
c
/
d
t
for Grade 60
reinforcement and for prestressing
steel. (Figure R9.3.2 in ACI Code.)
(Permission of American Concrete
Institute.)
net tensile
t
= 0.002
t
= 0.005
c
c
= 0.600
= 0.375
d
t
d
t
Interpolation on
c/d
t
: Spiral
= 0.37 + 0.20/(
c/d
t
)
Other
= 0.23 + 0.25/(
c/d
t
)
t
.
The procedure described here can be used to make a long-hand determination of
As a beginning we assume
c
/
d
t
t
0.60 where
0.002, as shown in Figure 10.25. With
c
t
,
f
s
, and for our column. Then with reference to
Figure 10.26, moments can be taken about the centerline of the column and the result
solved for
M
n
and
e
determined.
c
,
this value we can calculate
c
,
a
,
f
d
t
d
d
t
d
h
2
a
2
M
n
T
s
C
s
C
c
2
2
t
As the next step,
c
/
d
t
can be assumed equal to 0.375 (where
0.005 as shown in
t
determined. If the
e
or our column falls between the
two
e
values we have just calculated, the column falls in the transition zone for
Figure 10.25) and another value of
. To de-
termine its value we can try different
c
/
d
t
values between 0.600 and 0.375 until the calcu-
lated
e
equals the actual
e
of the column.
b
d
t
- d'
d'
d
t
h
P
n
M
n
T
s
C
c
C
s
Figure 10.26
