Monday 17th January 2011
Lecture 3:
Review of basic hydrostatics equation
Applications:- a) incompressible fluids (- review again ..... )
barometers
manometers
other pressure gauges
b) compressible fluids (atmospheric behaviour, gases)
isothermal behaviour
pressure variation in an isothermal gas
temperature lapse rates
pressure variation in an atmosphere with a constant lapse rate
approximations for small z
Principles of assessing forces, lines of action, and moments on immersed objects
coffer dam in a river
rectangular gate in a tank
other cases ....
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CIVIL, ENVIRONMENTAL and GEOMATIC ENGINEERING DEPARTMENT
1st YEAR FLUIDS
2.5
CEGE1009 MECHANISMS
Forces and moments on plane immersed surfaces
The aim of this section is to outline a general method for determining the total force on a plane
immersed surface, the position of its line of action, and the moment exerted by that force about
some appropriate axis (or hinge) for any specified depth of immersion in a liquid. The procedure
can be summarised as follows:
i) draw a sketch - to large scale,
ii) adopt appropriate axes (usually selecting any hinge that might exist as the origin),
iii) define a typical "element" in this coordinate system,
iv) evaluate an expression for the force on that element,
v) evaluate an equivalent expression for the moment about the chosen axis,
vi) integrate both expressions between limits to determine the total force and moment,
vii) determine where the resultant force acts.
2.5.1 Coffer dam:
Considering first the case of a coffer dam
designed to allow the construction of a bridge
pier in a river. Such structures are normally
constructed of interlinking sheet piles driven
into the river bed.
Patm
y
Taking the y origin at the water surface and y
positive downwards, the absolute pressure
acting from the water on a small element dy is
{patm +  g y } and the force acting on horizontal
width b is:
y
air
Patm
water
{patm +  g y} b. dy
River bed
Hence, the hydrostatic force acting on the full
wetted height of the dam is:
h
F =
( p
atm
+  g y) b dy = b h patm + 21  gbh 2
0
However, the air on the dry side of the coffer dam also exerts a force on the small element:
{ patm } b. dy
and hence the total force on the dry face is: b h patm
This leaves a net force on the wall of: ½ g b h2
Similarly, the moment exerted about the water surface by the wetted side of the small horizontal
element is:
{ patm +  g y } b y dy
Hence the moment generated by the full wetted face is:
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h

M =
( patm +  g y ) b y dy =
1
2
patm bh 2 + 13  g bh 3
0
and that by the air on the other side:
h

2
1
M a = patm b y . dy  patm bh
2
0
giving a resultant net moment M = 13  g bh 3
If the resultant force acts at a depth y' below the water surface, then - as it must also generate the
same net moment, M, - we have:
1
2
 g bh 2 y  = 13  g bh 3 and y  =
2
3
h
that is, the centre of pressure is at a depth 2h/3 below the water surface for a rectangular coffer
dam wall of length b.
In this example, the effect of atmospheric pressure on the two sides of the coffer dam wall cancels
out - and illustrates why gauge pressure is often used by engineers to assess hydrostatic forces on
submerged structures. However, if the problem had involved a pressurised container, then it
would have been necessary stay with absolute pressures.
Incidentally, if you were required to calculate the overturning moment of the coffer dam wall
about level of the river bed (where it might be likely to fail in bending), it would be wise to adopt
axes with an origin at the river bed.
2.5.2 Rectangular gate in the side of a tank:
Now consider the forces and moments acting on a rectangular gate, height a and width b, hinged
s
s
liquid
hinge
Air at Patm
y
F1
F2
a
b
about its upper edge and located in the side of a tank with water to a depth s above the top of the
gate. Taking y positive downwards from hinge level (at the top of the gate), the net force dF
acting on the element d y is:
dF = dF1 - dF2 = {patm + g(s + y)} b dy - patm b dy = g (s + y) b dy
and the net moment:
dM = dM1 - dM2 = {patm + g(s + y)} b y dy - patm b y dy =  g (s + y) b y dy
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The total force on the gate can then be found by integration (as for the coffer dam):
a
2
y a
a
F =   g (s + y ) b dy =  g b [s y + ] =  g b a [ s + ]
2 0
2
0
The total moment about the hinge is:
a
3
y a
1
s a
M =   g (s + y ) b y dy =  g b [ s y 2 + ] =  g b a 2 [ + ]
2
3 0
2 3
0
Let the line of action of F be at y' below the hinge O:
 y =
F y' = M
s
2
a
3 = a (3s  2a)
a
3 (2s + a)
 g ba [ s  ]
2
 g b a2 [ + ]
For very deep immersion with s >>a , y' = a / 2
In other words, the line of action tends to the centroid of the gate for large depths. On the other
hand, if the gate is just submerged and s = 0, then y' = 2 a / 3
The same procedure is followed for gates of other shapes (circular, triangular, etc.), inclined gates,
and problems with multiple liquids - as will become apparent when you tackle the tutorial sheets
and exam questions on this topic. [One of the commonest errors is to forget that b is also a
function of y if the width of the gate is not constant.]
2.6
Forces on submerged curved surfaces:
y
y

l
The horizontal component of force acting on any curved surface immersed in a fluid is
independent of the shape of that surface. Consider the forces on the two tanks illustrated here:
The horizontal force H acting on the small element dl (of width b) is:
{ g b y dl} cos 
But dl cos is simply the vertical projection of the element dy. Therefore, H =  g b y dy which
is the same as the force on a vertical strip of height dy experienced by the plane vertical wall at the
other end of the tank.
Of course, if the horizontal force component was not independent of shape, then the ends of the
tanks illustrated could experience different forces and (friction allowing) accelerate sideways.
The vertical component of force is dependent on shape of the surface on which it acts. It can be
calculated by integrating the vertical components of force on each element down the face.
Alternatively (and more conveniently), it is equal to the weight of the column of liquid directly
above the surface.
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CIVL1009 – Mechanisms (Fluids) - Problems – week 3
1. Two alternative dike designs are schematically shown in figures a) and b). Compare loads on the base
(force and torque) caused by water pressure in each case. Which of the designs is more efficient and why?
2.
Two identical buckets are placed on the scales as shown in the figure, and filled with the same amount of
water. The force due to water pressure acting on each scale plate is pressure times the bottom area. The
bottom area of the left bucket is 4 times larger then that of the right one. Therefore, the force acting on the
left plate should be 4 times larger then the force acting on the right one. However, the weight of both
buckets is the same, and the scales are in the equilibrium. What is missing in the consideration? Resolve
the contradiction.
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3.
Two buckets on the scales (see the figure) are filled with water up to the edge. A solid object is placed in
the left bucket. How will the equilibrium of the scales be change if the density of the object is
(i) larger then the water density,
(ii) smaller the water density?
Explain your reasoning.
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