Civil Engineering Reference
In-Depth Information
Equation (2.87) shows that the interaction of
V
and
T
is also circular for a constant
M
in the second mode of failure, Figure 2.10. This equation represents an interaction surface
which intersects the vertical
V
-
M
plane to form an interaction curve expressed by Equation
(2.62) and intersects the horizontal
T
-
M
plane to form an interaction curve expressed by
Equation (2.73).
The intersection of the two failure surfaces for the first and second modes of failure will
form a peak interaction curve between
V
and
T
. The expression of this peak curve can be
obtained by solving Equations (2.87) and (2.84) to eliminate the
M
term:
V
V
o
2
T
T
o
2
1
+
R
+
=
(2.88)
2
R
The location of this curve can be obtained by solving Equations (2.87) and (2.84) to eliminate
the
V
and
T
terms:
M
M
o
=
1
−
R
(2.89)
2
Equation (2.89) is, of course, the same as Equation (2.64). The plane formed by the peak
curve at
M
/
M
o
=
(1
−
R
)
/
2 will be designated as the peak plane. Equation (2.88) for
R
=
0.25, 0.5 and 1 on the peak planes is plotted as a series of dotted curves in Figure 2.13.
2.2.3.3 Third Failure Mode
In the third failure mode, failure is caused by yielding in the
top bar
(not the top stringer), in
the
bottom bar
and in the
transverse reinforcement
, all on the side where shear flows due to
shear and torsion are additive (i.e. Left-hand side).
Taking moments about the right side wall, Figure 2.12, where shear flows due to shear and
torsion are subtractive, will furnish the following equilibrium equation:
1
2
(
N
b
+
b
2
−
b
2
0
=
N
t
)
b
v
−
q
w
d
v
tan
α
w
b
v
−
q
t
w
b
v
tan
α
t
w
q
b
w
b
v
tan
α
b
w
(2.90)
Substituting
q
w
,
q
t
w
,
q
b
w
from Equations (2.74)-(2.76) and tan
α
w
,tan
α
t
w
,tan
α
b
w
from
Equations (2.78)-(2.80) into Equation (2.90) and simplifying gives
V
2
d
v
2
T
2
A
o
2
N
b
+
N
t
d
n
t
T
2
A
o
b
n
t
=
+
+
(2.91)
2
When the square term is multiplied in Equation (2.91) we notice the appearance of a mixed
term for
VT
. Grouping the two
T
2
terms and dividing the whole equation by
N
t
results in:
V
2
d
v
2
T
2
A
o
2
2
V
2
d
v
T
2
A
o
d
N
t
N
b
+
N
t
2
N
t
d
N
t
1
n
t
+
(
b
v
+
d
v
)
1
n
t
+
1
n
t
=
(2.92)
N
t
Assume the yielding of the top bar, the bottom bar and the transverse steel in the left wall,
then
N
t
=
N
t
y
,
N
b
=
N
b
y
, and
n
t
=
n
ty
. Substituting the definitions of
V
o
and
T
o
into
