pressure for ces
* Fluid statics -
> fluids at rest Pressure Forces on any Plane Surface
no shear forces
-
-
only stress is normal Stress (pressure) is
variation of pressure (p ggh) =
& irregular
Introduction Shape
↳ tilted plane
↳ pressure forces normal to surfaces
act
they act on
- Sphere
to a
of fluid
larger body of
is stationary relative
same fluid
Consider
water surfaceis at
a plane surface
an
situated
angle of O to
some distance
that surface
below the
-
>
-
all action is reaction pressure forces ·
snow that the magnitude of the resultant force is :
are in equilibrium (no motion)
F =
09 (sin O) Al
even as
d - > 0 ·
distance from origin to line of action is :
↳ The pressure at any point acts
equally in all directions l = i +
Factors that influence the distribution depthSha
:
,
in ↳ for a vertical surface : Y' =
j +
BL3
Pressure Forces on Submerged Plane Surfaces ·
Where Io =
12
for rectangle of breadth i
EXAMPLE 1 What We know :
3 height L
R
surface atmospheric P ·
Only normal forces to surface are
↳ Io = for circle of radius R
considered * DERIVATION
·
Pressure will vary linearly with depth
(incompressible fluids -
density won't vary)
EXAMPLE :
Hydrostatic force on a dam
5m
& >
Q Consider horizontal plate of width b Im
: 43 m
:
a =
.
9
,
000
length a =
2 5m .
situated at a depth of
y
=
2m , com" 8
000 880
what is the pressure force (F PA) ?= V 00 do 000
A: Fr =
(Po +
ggh) ab =
(101 .
4 FN/m2 + 1000 x 9 81 +
.
2)((x2 . 5) A rockfill dam is to have the cross-section illustrated .
=
49 . 3 kN The resevoir design depth is to be 18m .
Estimate :
·
resultant force with a line of action through the centre 1. Force on dam per unit width
of pressure i. e
. Centroid (can be proved by taking moments) .
2 locationis direction of pressure force
=
.
cos
EXAMPLE 2 :
What we know : A : 1 . F =
pgsin OAE nw Wetted height :
10
1000 (9 81) (sin 45 ) (14 14) (7 07)
%
magnitude of resultant force COS 45
°
is the =
average
-
=
height
.
wetted
.
.
of the prism pressure force distribution
=
693 5 . kN/m nw = 14 14 .
m
-
line of action is
through centroid of
.
2 l = +
e Area into' page
prism pressure force distribution
= ot
↑
nu
↓
-
Tilted 3 irregular shape (14 14) (7 07)
. .
= /m ->
↳ =
9 43m
generic equation that can be
.
i = y =
7 07m
applied to plane submerged
.
any
surface
A =
(14 14) (1)
.
=
14 14 ma.
* Fluid statics -
> fluids at rest Pressure Forces on any Plane Surface
no shear forces
-
-
only stress is normal Stress (pressure) is
variation of pressure (p ggh) =
& irregular
Introduction Shape
↳ tilted plane
↳ pressure forces normal to surfaces
act
they act on
- Sphere
to a
of fluid
larger body of
is stationary relative
same fluid
Consider
water surfaceis at
a plane surface
an
situated
angle of O to
some distance
that surface
below the
-
>
-
all action is reaction pressure forces ·
snow that the magnitude of the resultant force is :
are in equilibrium (no motion)
F =
09 (sin O) Al
even as
d - > 0 ·
distance from origin to line of action is :
↳ The pressure at any point acts
equally in all directions l = i +
Factors that influence the distribution depthSha
:
,
in ↳ for a vertical surface : Y' =
j +
BL3
Pressure Forces on Submerged Plane Surfaces ·
Where Io =
12
for rectangle of breadth i
EXAMPLE 1 What We know :
3 height L
R
surface atmospheric P ·
Only normal forces to surface are
↳ Io = for circle of radius R
considered * DERIVATION
·
Pressure will vary linearly with depth
(incompressible fluids -
density won't vary)
EXAMPLE :
Hydrostatic force on a dam
5m
& >
Q Consider horizontal plate of width b Im
: 43 m
:
a =
.
9
,
000
length a =
2 5m .
situated at a depth of
y
=
2m , com" 8
000 880
what is the pressure force (F PA) ?= V 00 do 000
A: Fr =
(Po +
ggh) ab =
(101 .
4 FN/m2 + 1000 x 9 81 +
.
2)((x2 . 5) A rockfill dam is to have the cross-section illustrated .
=
49 . 3 kN The resevoir design depth is to be 18m .
Estimate :
·
resultant force with a line of action through the centre 1. Force on dam per unit width
of pressure i. e
. Centroid (can be proved by taking moments) .
2 locationis direction of pressure force
=
.
cos
EXAMPLE 2 :
What we know : A : 1 . F =
pgsin OAE nw Wetted height :
10
1000 (9 81) (sin 45 ) (14 14) (7 07)
%
magnitude of resultant force COS 45
°
is the =
average
-
=
height
.
wetted
.
.
of the prism pressure force distribution
=
693 5 . kN/m nw = 14 14 .
m
-
line of action is
through centroid of
.
2 l = +
e Area into' page
prism pressure force distribution
= ot
↑
nu
↓
-
Tilted 3 irregular shape (14 14) (7 07)
. .
= /m ->
↳ =
9 43m
generic equation that can be
.
i = y =
7 07m
applied to plane submerged
.
any
surface
A =
(14 14) (1)
.
=
14 14 ma.
