Civil Engineering Reference
In-Depth Information
(
w/
4) tan
α
r
=
(5
/
12)
w
=
(1
/
4)
F
s
. Then the top stringer force at point c is (7
/
8)
F
s
-
(1
/
4)
F
s
=
(5
/
8)
F
s
. Similar logic would give the top stringer force at point d to be (5
/
8)
F
s
-
(3
4)
F
s
, and at point e to be 0. These forces in top stringer at points a, b, c, d and e
are connected by straight lines in Figure 8.2(f). In actuality, of course, the top stringer forces
should change at the discrete points of the stirrups. In the region c to d, for example, each
stirrup force increment is (3
/
8)
F
s
=
(1
/
40)
F
s
.
In the local 'fanning' region between d and e, however, each discrete vertical force at the
stirrup locations is (
w/
8)(1
/
5) tan
α
r
=
(1
/
8)
w
=
(3
/
w/
2)(1
/
5)
=
w/
10 and the increments of the top stringer force are
(
w/
10) tan
α
r
, depending on the variable angle
α
r
. The values of tan
α
r
changes from d to e in
/
/
,
/
/
,
/
/
,
/
/
/
/
the following sequence: (9
10)(5
3)
(7
10)(5
3)
(5
10)(5
3)
(3
10)(5
3) and (1
10)(5
3).
/
/
Therefore, the stirrup force increments change from d to e as follows: (9
100)
F
s
,(7
100)
F
s
,
(5
100)
F
s
.
The force diagram for the top stringer derived from the plasticity truss model is compared
with the force diagram calculated from the conventional moment diagram in Figure 8.2(f). It
can be seen that the former is smaller than the latter due to the effect of shear. This shear effect
is frequently ignored in design practice.
Now, let us find the forces in the bottom stringers according to the plasticity truss model.
Looking at the bottom stringer in Figure 8.2(a), the vertical force from a
to b
is zero.
Therefore, no increment of force will occur in the bottom stringer between these two points
and the stringer force remains a constant of
F
s
from a
/
100)
F
s
,(3
/
100)
F
s
and (1
/
to b
(Figure 8.2e). In the next region
from b
to c
, however, the vertical force is
w/
8. The force increment in the bottom stringer is
8)
F
s
. Therefore, the bottom stringer force at c
is
F
s
-(1
(
w/
8) tan
α
r
=
(5
/
24)
w
=
(1
/
/
8)
F
s
8)
F
s
. Similar reasoning gives the bottom stringer forces at points d
and e
to be (5
=
(7
/
/
8)
F
s
and (1
4)
F
s
, respectively.
Figure 8.2(e) compares the forces in the bottom stringer obtained from plasticity truss model
with those calculated from the conventional moment diagram. It can be seen that the shear
force increases the bottom stringer force. Therefore, the 'shift rule' given in Section 8.1.1.3
for detailing of bottom reinforcement remains valid.
/
8.1.2.3 Comments on Plasticity Truss Model
The beam example in Sections 8.1.1 and 8.1.2, analyzed by the plasticity truss model, provides
a very clear concept of stress flow in the beam and the resulting forces in the stirrups,
stringers and concrete struts. This model is quite valid for the case of a beam subjected to
midspan concentrated load as shown in Section 8.1.1. However, the solution may not be
conservative in the case of a beam subjected to uniformly distributed load as shown in Sec-
tion 8.1.2.
The plasticity truss model satisfies only the force equilibrium condition, not the strain
compatibility condition. Because the second assumption of infinite plastic deformation can
not be guaranteed for concrete, a design based on the staggered shear diagram (Section 8.1.2.1)
may not be conservative. In Section 8.2, a compatibility truss model will be presented, based
on the elastic behavior of concrete and steel. As far as the material property is concerned, the
plasticity truss model provides an upper bound solution, while the compatibility truss model
yields the lower bound solution. For a nonlinear material of limited plasticity, such as concrete,
the solution should lie in between these two limits.
