Civil Engineering Reference
In-Depth Information
ties in the web. The concrete compression struts are inclined at an angle of
, because the
diagonal cracks due to shear is assumed to develop at this angle with respect to the longitudinal
axis. Each cell of the truss, therefore, has a longitudinal length of
d
v
cot
α
, except at the local
regions near the concentrated loads where the longitudinal length is (
d
v
cot
α
2.
The forces in the struts and ties of this idealized truss can be calculated from the equilibrium
conditions by various procedures. According to the sectional method, a cut along the section
A-A on the right-hand side of the truss will produce a free body, as shown in Figure 1.2(e).
Equilibrium assessment of this free body shows that the top and bottom stringers are each sub-
jected to a force of (1/2)
V
cot
α
)
/
α
and (3/2)
V
cot
α
, respectively. The force in the compression
/
α
strut is
V
. From the vertical equilibrium of the node point
a
, the force in the vertical tie
is
V
. The results from similar calculations are recorded on the left-hand side of the truss for
all the struts and ties.
Figure 1.3 gives a much simplified struts-and-ties model for a beam to resist torsion. The
longitudinal and hoop bars are assumed to have the same cross-sectional area and are both
spaced at a constant spacing of s. The concrete compression struts are inclined at an angle of
45
◦
. Since each hoop bar is treated as a series of straight ties of length
s
, a long plane truss is
formed in the longitudinal direction between two adjacent longitudinal bars. A series of this
kind of identical plane trusses is folded into a space truss with an arbitrary cross-section, as
shown in Figure 1.3(b). Because each plane truss is capable of resisting a force
F
, a series of
F
thus form a circulatory shear flow, resulting in the torsional resistance
T
. It has been proven by
Rausch (1929), using the equilibrium conditions at the node points, that the force
F
is related
to the torsional moment
T
by the formula,
F
sin
2
A
o
, where
A
o
is the cross-sectional area
within the truss or the circulatory shear flow (see derivation of Equation (2.46) in Section
2.1.4, and notice the shear flow
q
=
Ts
/
=
F
/
s
).
1.4.3 Struts-and-ties Model for Knee Joints
The knee joint, which connects a beam and a column at the top left-hand corner of the frame,
as shown in Figure 1.1, will be used to illustrate how the struts-and-ties model are utilized to
help design the reinforcing bars (rebars). Under gravity loads the knee joint is subjected to a
closing moment, as shown in Figure 1.4(a). The top rebars in the beam and the outer rebars in
the column are stretched by tension, while the bottom portion of the beam and the inner portion
of the column are under compression. If the frame is loaded laterally, say by earthquake forces,
the knee joint may be subjected to an opening moment, as shown in Figure 1.4(b). In this case
the bottom rebars of the beam and the inner rebars of the column are stretched by tension,
while the top portion of the beam and the outer portion of the column are under compression.
For simplicity, the small shear stresses on the boundaries of the knee joint are neglected.
Three rebar arrangements are shown in Figure 1.4 (c-e). Figure 1.4(c) gives a type of rebar
arrangement frequently utilized to resist a closing moment. In this type of arrangement the top
and bottom rebars of the beam are connected to form a loop in the joint region. A similar loop is
formed by the outer and inner rebars of the column. This type of arrangement is very attractive,
because the separation of the beam steel from the column steel makes the construction easy.
Unfortunately, tests (Swann, 1969) have shown that the strength of such a joint can be as low as
34% of the strength of the governing member (i. e. the beam or the column, whichever is less).
Two examples of incorrect arrangements of rebars to resist an opening moment are shown
in Figure 1.4(d) and (e). In Figure 1.4(d) the bottom tension rebars of the beam are connected
