Civil Engineering Reference
In-Depth Information
experimental maximum shear forces equal to or greater than those of the
corresponding beams with higher
r
t
(0.0245).
In view of the theory and the above tests, it seems reasonable to state that
the effectiveness of transverse reinforcement decreases when
a/h
ratio
decreases from 2 to 0.5. When
a/h
0.5 an increase of transverse
reinforcement beyond the ACI Code minimum requirement,
£
r
t
=0.25%, is not
effective in increasing the shear strength of deep beams.
Table 7.3
Effect of tensverse reinforcement at low
a/h
ratios
7.5
Explicit shear strength equation
7.5.1
Derivation of equation
The accuracy of the theoretical results confirms the usefulness of the
theoretical model. However, the solution procedure is too complicated to be
used in design. An explicit formula suitable for practical design is presented
in this section. The formula is derived from the three equilibrium Eqns
(7.6a-c) alone. Recognising that the three quantities
r
t
f
t
may be
estimated, the shear capacity v may be expressible in terms of these three
quantities by eliminating the other unknowns
s
t
,
r
l
f
t
, and
s
d
and
a
from Eqns (7.6a-c).
This is achieved by the following manipulation.
Utilizing the identity sin
2
a
+cos
2
a
=1, one may rewrite Eqns (7.6a and
b) as
(
s
d
-
s
r
)cos
2
a
=-
r
l
f
l
-
s
r
(7.15a)
(7.15b)
(
s
d
-
s
r
)sin
2
a
=-
K
v
-
r
t
f
t
-
s
r
Eqn (7.6c) may be squared to become
(
=
v
2
Multiplying Eqn (7.15a) by Eqn (7.15b) and subtracting the result from Eqn
(7.15c) gives:
s
d
-
s
r
)
2
sin
2
a
cos
2
a
(7.15c)
v
2
-(
r
l
f
l
+
s
r
)(
K
v
+
r
t
f
t
+
s
r
)=0
(7.16)
