Civil Engineering Reference
In-Depth Information
Example
Consider a beam with span
l
=3
h
/2, reinforcement strength
y
o
=
h
/
v
=
h
/3 and
level of reinforcement
c
=
h
/12 corresponding to
d
=0.92/
h
. From Eqn (8.16)
we find the solution:
V
/
bvf
c
=
x
=
h
/6=0.167h and Eqn (8.19) gives
s
l
=
h
/2. The
geometry of the considered beam is shown in
Figure 8.7
, and we note that
attainment of the flexural capacity requires a substantial length of the
support platen.
If more realistically we assume
s
=
t
=
h
/6, the solution is found from Eqn
(8.20):
V
/
bvf
c
=
x
=0.108
h
corresponding to a reduction by 35%. If, on the
other hand, the level of the reinforcement is increased to
c
=
h
/6,
corresponding to
d
=0.83
h
, we find from Eqn (8.16):
V
/
bvf
c
=
x
=0.148
h
. This
is a reduction of 11% only, and the attainment of this flexural capacity
requires no oversize support platen, as we now have
s
l
=
x
by Eqn (8.13).
F
8.4.2
Upper bound analysis
The kinematically admissible failure mechanism shown in Figure 8.10
consists of a straight yield line running at the inclination ¦ from the edge of
the load platen to the edge of the support platen. The relative displacement
rate is v, inclined at the angle
a
to the yield line.
Figure 8.10
Failure mechanism for beam with point loading.
We assume that the reinforcement is not compressed, i.e.
a‡p
/2-¦ or
a
+¦
‡p
/2. The rate of external work done by the load is
W
E
=
V
sin (
a
+¦). The
rate of internal work dissipated in the mechanism is:
