Civil Engineering Reference
In-Depth Information
0.25, 0.5 and 1. A series of complete
interaction curves for shear and bending is now clearly illustrated in Figure 2.8.
Take, for example, the case of
R
Equation (2.62) is also plotted in Figure 2.8 for
R
=
0.5 where the top stringer has one-half the yield capacity
of the bottom stringer. The moment will then vary from
M
=
/
M
o
=−
0.5 to 1 depending on the
magnitude of
V
0.5, this negative moment introduces a tensile force in
the top stringer equal to the yield force. Consequently, the shear strength
V
which is based on
the top stringer strength becomes zero. When the moment is increased (in an algebraic sense
toward the right), the tensile force in the top stringer decreases. The remaining tensile capacity
in the top stringer is now available to resist shear, resulting in an increase of the shear strength.
When the moment is increased to zero (i.e.
M
/
V
o
. When
M
/
M
o
=−
/
M
o
=
0), the full capacity of the top stringer
is available to resist shear, and the shear strength
V
becomes the pure shear strength
V
o
(i.e.
V
1).
When the moment becomes positive, it induces a compressive stress in the top stringer,
which can be used to reduce the tensile stress due to shear. As a result, the shear strength
continues to increase until the second mode of failure (yielding of top stringer) is changed into
the first mode of failure (yielding of bottom stringer). When
M
/
V
o
=
0.25, the condition is
reached where the bottom stringer and the top stringer yield simultaneously. This peak point
provides the highest possible shear strength. Beyond this point, the shear strength decreases
with increasing moment, because failure is now caused by the tensile yielding of the bottom
stringer. In the bottom stringer the tensile stresses due to shear and bending are additive. When
the moment reaches the pure bending strength (
M
/
M
o
=
1), the bottom stringer will yield
under the moment itself, leaving no capacity for shear. The shear strength then becomes zero
(
V
/
M
o
=
0).
The point of maximum shear, which corresponds to the simultaneous yielding of top and
bottom stringers, is the intersection point of the two curves for the first and second modes
of failure. The locations of these peak points can be obtained by solving the two interaction
equations, Equations (2.58) and (2.62). Multiplying Equation (2.62) by
R
and adding it to
Equation (2.58), we derive
V
/
V
o
=
/
V
o
by eliminating
M
/
M
o
:
1
V
V
o
=
+
R
(2.63)
2
R
/
Multiplying Equation (2.62) by
R
and subtracting it from Equation (2.58), we derive
M
M
o
by eliminating
V
/
V
o
:
M
M
o
=
1
−
R
(2.64)
2
For the case of
R
=
0.5, Equat
ion
s (2.64) and (2.63) illustrate that the peak point is located
V
o
=
√
1
at
M
/
M
o
=
0.25 and
V
/
.
5. These values are indicated in Figure 2.8.
2.2.2 Torsion-Bending Interaction
A model of beam subjected to torsion and bending is shown in Figure 2.9. The moment
M
creates a tensile force
M
d
v
in the bottom stringer and an equal compressive force in the
top stringer. The torsional moment, however, induces a total tensile force of (
Tp
o
/
/
α
r
in the longitudinal steel. Due to symmetry, the top and bottom stringers should each resist
2
A
o
)tan
