examp partii 13
TRANSCRIPT
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Forces on submerged surfaces
2.1
Obtain an expression for the depth of the centre of pressureof a plane surface wholly submerged in a fluid and inclined
at an angle to the free surface of the liquid.
A horizontal circular pipe, 1.25 m diameter, is closed by a butterfly disk which rotates about a horizontal axis through
its centre. Determine the torque which would have to be
applied to the disk spindle to keep the disk closed in a
vertical position when there is a 3m head of fresh water
above the axis.
[1176 Nm]
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The question asks what is the moment you have to apply to the
spindle to keep the disc vertical i.e. to keep the valve shut?
So you need to know the resultant force exerted on the disc by thewater and the distance x of this force from the spindle.
We know that the water in the pipe is under a pressure of 3m head of
water (to the spindle)
Diagram of the forces on the disc valve, based on an imaginary water surface
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, the depth to the centroid of the disc
h’ = depth to the centre of pressure (or line of action of the force)
Calculate the force:
= 36.116 kN
Calculate the line of action of the force, h’
m3h =
AhgF ρ=
2)225.1(381.91000 π×××=
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Ah
I00=
By the parallel axis theorem 2nd
moment of area about O (in thesurface) where IGG is the 2nd moment of area about a
line through the centroid of the disc and IGG = πr 4/4.
So the distance from the spindle to the line of action of the force is
2
GG00 hAII +=
hhA
I
h
GG'+=
33)r (4
r 2
4
+π
π
=
m0326.3312
r 2=+=
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m0326.03-0326.3h-hx '===
And the moment required to keep the gate shut is
moment = Fx = 36.116 x 0.0326 = 1.176 kNm
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2.2
A dock gate is to be reinforced with three horizontal
beams. If the water acts on one side only, to a depth of 6m,find the positions of the beams measured from the water
surface so that each will carry an equal load. Give the load
per meter.
[58860 N/m, 2.31m, 4.22m, 5.47m]
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First of all draw the pressure diagram, as below:
The resultant force per unit length of gate is the area of the pressure
diagram. So the total resultant force is
(per m length) N176580681.910005.0gh
2
1R 22
=×××=ρ=
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Alternatively the resultant force is, R = Pressure at centroid x Area,
(take width of gate as 1 m to give force per m)
(per m length)
This is the resultant force exerted by the gate on the water.
The three beam should carry an equal load, so each beam carries the
load f, where
If we take moments from the surface,
DR = fd1 + fd2 + fd3
N176580)1h(2
h
gR =××ρ=
N588603
R f ==
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D(3f) = f (d1 + d2 + d3)
12 = d1 + d2 + d3
Taking the first beam, we can draw a pressure diagram for this,
(ignoring what is below),
We know that the resultant force, so,gH2
1F 2ρ= g
F2H ρ=
m46.381.91000
588602
g
F2H =
×
×=
ρ=
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And the force acts at 2H/3, so this is the position of the 1st beam
position of 1st beam
Taking the second beam into consideration, we can draw the
following pressure diagram,
The reaction force is equal to the sum of the forces on each beam, so
as before
m31.2H3
2==
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m9.481.91000
)588602(2
g
F2H =
×
××=
ρ=
The reaction force acts at 2H/3, so H = 3.27m. Taking moments fromthe surface,
(2 x 58860) x 3.27 = 58860 x 2.31 + 58860 x d2
depth to second beam d2 = 4.22m
For the third beam, from before we have,
12 = d1 + d2 + d3
depth to third beam d3 = 12 – 2.31 – 4.22 = 5.47m
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2.3
The profile of a masonry dam is an arc of a circle, the arc having a
radius of 30m and subtending an angle of 60 at the centre ofcurvature which lies in the water surface. Determine (a) the load on
the dam in N/m length, (b) the position of the line of action to this
pressure.
[4.28 106 N/m length at depth 19.0m]
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Draw the dam to help picture the geometry,
h = 30 sin 60 = 25.98m
a = 30 cos 60 = 15.0m
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Calculate Fv = total weight of fluid above the curved surface (per m
length)
Fv = ρg (area of sector – area of triangle)
= 2711.375 kN/m
Calculate Fh = force on projection of curved surface onto a vertical
plane
])2
1598.25(-)
360
6030([81.91000 2
××π××=
2
h gh21F ρ=
m/kN681.331098.2581.910005.0 2=×××=
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The resultant,
= 4279.27 kN/m
acting at angle
As this force act normal to the surface, it must act through the
centre of radius of the dam wall. So the depth to the point where the
force acts is,
y = 30sin39.31 = 19m
222
h
2
vR 375.2711681.3310FFF +=+=
819.0F
Ftan
h
v==θ
032.39=θ
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2.4
The arch of a bridge over a stream is in the form of a semi-circle of
radius 2m the bridge width is 4m. Due to a flood the water level isnow 1.25m above the crest of the arch. Calculate (a) the upward
force on the underside of the arch, (b) the horizontal thrust on one
half of the arch.
[263.6 kN, 176.6 kN]
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The bridge and water level can be drawn as:
The upward force on the arch = weight of (imaginary) water above the
arch.
ter volumeofwagR v ×ρ=
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3
2
m867.264]2
24)225.1[( =×
π×+=volume
(b)
The horizontal force on half of the arch, is equal to the force on
the projection of the curved surface onto a vertical plane.
Fh = pressure at centroid x area
= ρg (1.25 + 1) x (2 x 4)= 176.58 kN
kN568.263861.2681.91000R v
=××=
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2.5
The face of a dam is vertical to a depth of 7.5m below the water
surface then slope at 30° to the vertical. If the depth of water is 17mwhat is the resultant force per meter acting on the whole face?
[1563.29 kN]
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h2 = 17.0 m, so h1 = 17.0 – 7.5 = 9.5
x = 9.5/tan 60 = 5.485 m.
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vertical force = weight of water above the surface,
= 659.123 kN/m
The horizontal force = force on the projection of the surface on to a
vertical plane.
The resultant force is
xh5.0xhgF 12v ×+×ρ=
( )485.55.95.0485.55.79810 ××+××=
2
h gh2
1F ρ=
m/kN545.1417
1781.910005.0 2
=
×××=
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222
h
2
vR 545.1417123.659FFF +=+=
m/kN29.1563=
And acts at the angle
465.0
F
Ftan
h
v==θ
094.24=θ
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2.6
A tank with vertical sides is square in plan with 3m long
sides. The tank contains oil of relative density 0.9 to a depth of 2.0mwhich is floating on water a depth of 1.5m. Calculate the force on
the walls and the height of the centre of pressure from the bottom of
the tank.
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Consider one wall of the tank. Draw the pressure diagram:
density of oil ρoil = 0.9 ρwater = 900 kg/m3
Force per unit length, F = area under the graph = sum of the three
areas = f 1 + f 2 + f 3
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N5297432
2)281.9900(f 1 =×
×××=
( ) N7946135.1281.9900f 2 =
××××=
N3310932
5.1)5.181.91000(f 3 =×
×××=
To find the position of the resultant force F, we take moments fromany point. We will take moments about the surface.
DF = f 1 d1 + f 2 d2 + f 3 d3
D = 2.347 m (from surface)
= 1.153 m (from base of wall)
)5.13
22(333109)
2
5.12(794612
3
252974D165544 +×++×+×=