Civil Engineering Reference
In-Depth Information
After time
t
m
, the two velocities are equal and
the block comes to rest with respect to the base,
that is, the relative velocity,
u
(a)
W
tan
0. The value of
t
m
is calculated by equating the ground velocity
to the velocity of the block to give the following
expression for the time
t
m
:
=
y
(
t
)
W
=
Mg
v
b
g
tan
φ
Acceleration
(b)
t
m
=
(6.36)
ag
The displacement
δ
m
of the block relative to the
ground at time
t
m
is obtained by computing the
area of the shaded region on Figure. 6.12(c) as
follows:
a
y
=
g
tan
t
0
Time
1
1
Velocity
(c)
δ
m
=
2
vt
m
−
2
vt
0
V
g
=
agt
v
2
2
g
tan
φ
−
v
2
2
ag
=
V
=
agt
0
V
b
=
gt
tan
1
v
2
2
g
tan
φ
tan
φ
a
=
−
(6.37)
t
0
t
m
Time
Equation (6.37) gives the displacement of the
block in response to a single acceleration pulse
(duration
t
0
, magnitude
ag
) that exceeds the yield
acceleration
(g
tan
φ)
, assuming infinite ground
displacement. The equation shows that the dis-
placement is proportional to the square of the
ground velocity.
While equation (6.37) applies to a block on
a horizontal plane, a block on a sloping plane
will slip at a lower yield acceleration and show
greater displacement, depending on the direction
of the acceleration pulse. For a cohesionless sur-
face where the factor of safety of the block FS is
equal to
(
tan
φ/
tan
ψ
p
)
and the applied accelera-
tion is horizontal, Newmark shows that the yield
acceleration
a
y
, is given by
Figure 6.12
Displacement of rigid block on rigid
base (Newmark, 1965): (a) block on moving base;
(b) acceleration plot; (c) velocity plot (Newmark,
1965).
and is of magnitude
(W
tan
φ)
, corresponding
to a yield acceleration
a
y
of
(g
tan
φ)
, as shown
by the dashed line on the acceleration plot (Fig-
ure 6.12(b)). The shaded area shows that the
ground acceleration pulse exceeds the accelera-
tion of the block, resulting in slippage.
Figure 6.12(c) shows the velocities as a func-
tion of time for both the ground and the block
accelerating forces. The maximum velocity for
the ground accelerating force has a magnitude
v
which remains constant after an elapsed time of
t
0
. The magnitude of the ground velocity
v
g
is
given by
a
y
=
(
FS
−
1
)g
sin
ψ
p
(6.38)
v
g
=
agt
0
(6.34)
where
φ
is the friction angle of sliding surface,
and
ψ
p
is the dip angle of this surface. Note that
for
ψ
p
while the velocity of the block
v
b
is
g
tan
φ
. Also equation (6.38)
shows that for a block on a sloping surface, the
yield acceleration is higher when the acceleration
=
0,
a
y
=
v
b
=
gt
tan
φ
(6.35)
