Download 2-Fluid Statics-2022

Topics
Fluid statics
• Fluid statics: forces applied by fluids at rest, there are no shear stress and
only normal pressure present
Hydrostatics-> Fluid is liquid
Hydrostatics
•Pressure (absolute, gauge pressure)
CE2201 Fundamentals of Fluid Mechanics
•Pressure at a point, pressure variation with
depth
Dr. B. M. L. A. Basnayake
•Pressure measuring devices
Department of Civil and Environmental Engineering
•Hydrostatic forces on submerged bodies
University of Ruhuna
•Buoyant force on submerged and floating
bodies
E2020Batch-2022
•Stability of submerged and floating bodies
Pressure
Pressure
• Pressure: Normal force exerted by a fluid per unit area
Principles described by Blaise Pascal
𝑃=
𝐹
𝐴
Unit: N/m2, Pa (Pascal)
1. Pressure acts uniformly in all directions
on a small fluid volume
2. Pressure acts perpendicular to the boundary
1 𝑏𝑎𝑟 = 105 𝑃𝑎
1 𝑎𝑡𝑚 = 101325 𝑃𝑎
• Absolute pressure : Actual pressure at a point
• Gauge pressure: Pressure relative to the atmospheric pressure
Gauge pressure = Absolute pressure- Atmospheric pressure
Direction of fluid pressure on boundaries
1
Pressure variation with depth
Consider a cylindrical fluid element dV having cross sectional area dA and length ds in any s
direction inclined at an angle 𝜽 to horizontal
𝐒𝐮𝐦 𝐨𝐟 𝐭𝐡𝐞 𝐟𝐨𝐫𝐜𝐞𝐬 𝐢𝐧 𝐬 𝐝𝐢𝐫𝐞𝐜𝐭𝐢𝐨𝐧
𝜕𝑝
(𝑝 + (𝜕𝑝/𝜕𝑠). 𝑑𝑠 )𝑑𝐴
𝑝 𝑑𝐴 − 𝑝 +
. 𝑑𝑠 𝑑𝐴 = ρg dV Sin𝜃
𝜕𝑠
−
θ
𝑝. 𝑑𝐴
ρg dV
𝜕𝑝
. 𝑑𝑠 𝑑𝐴 = ρg dV Sin𝜃
𝜕𝑠
𝜕𝑝
= −ρgSin𝜃
𝜕𝑠
𝑝 = 𝑝(𝑥, 𝑦, 𝑧)
𝜕𝑝 𝜕𝑝
Pressure in a static fluid
increases vertically
downwards and remains
𝝏𝒑
When 𝜃 = 90°;
= −𝝆𝒈 constant horizontally.
𝝏𝒛
When 𝜃 = 0;
,
𝜕𝑥 𝜕𝑦
=0
• Pressure increases with depth due to ‘extra weight’ on deeper layers
Pressure variation
𝜕𝑝
= −𝜌𝑔; 𝑝 = 𝑝 𝑧
𝜕𝑧
In a static fluid constant pressure
surfaces and constant density
surfaces are horizontal.
In an incompressible fluid (𝜌𝑔 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡)
𝜕𝑝
= −𝜌𝑔
𝜕𝑧
𝑝 + 𝜌𝑔𝑧 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
Pressure measurement
Free surface open to
atmospheric pressure
Interface between
liquids
In gauge pressure 𝑝 = 𝜌𝑔𝐻
As a pressure head 𝐻 =
𝑃
𝜌𝑔
• Relation between the height of a column of liquid and the pressure at its
base
𝑃𝑒𝑖𝑧𝑜𝑚𝑒𝑡𝑟𝑖𝑐 ℎ𝑒𝑎𝑑 ℎ =
𝑃
𝜌𝑔
*For measuring small gauge pressures of static liquids simple piezometer
tubes may be adequate
𝑝1 = 𝑝𝑐 + 𝜌𝑔 𝑧𝑐 − 𝑧1 = 𝑝𝑐 + 𝜌𝑔ℎ1
𝑝0 = 𝑝𝑐 + 𝜌𝑔 𝑧𝑐 − 𝑧𝑜 = 𝑝𝑜 + 𝜌𝑔ℎ0
2
Manometers
Micromanometer
• Equating the pressure in the same liquid horizontally
U tube manometer
Differential manometer
𝑦
2
𝑦
= 𝑝2 + 𝜌𝐴 𝑔 ℎ − ∆𝑧 + 𝜌𝐵 𝑔 𝑧 + ∆𝑧 −
+ 𝜌𝑐 𝑔𝑦
2
𝑝1 + 𝜌𝐴 𝑔 ℎ + ∆𝑧 + 𝜌𝐵 𝑔 𝑧 − ∆𝑧 +
Inclined manometer
Amount of liquid B on each side remains constant
𝑎1 ∆𝑧 = 𝑎2
𝑝 = 𝜌𝑚 𝑔𝑆 𝑆𝑖𝑛𝜃
𝑦
2
𝑝1 − 𝑝2 = 𝑔𝑦 𝜌𝐶 − 𝜌𝐵 1 −
𝑝1 = 𝑝2 = 𝑝𝑎𝑡𝑚 + 𝜌𝑔ℎ
𝑎2
𝑎2
− 𝜌𝐴
𝑎1
𝑎1
𝑎2 ≪ 𝑎1
𝑝1 − 𝑝2 = 𝜌𝐶 − 𝜌𝐵 𝑔𝑦
𝑝1 + 𝜌1 𝑔 ℎ + 𝑎 = 𝑝2 + 𝜌1 𝑔𝑎 + 𝜌2 𝑔ℎ
When 𝜌𝐶 , 𝜌𝐵 are closely similar a reasonable value of y
can be achieved for a small pressure difference
Example
Pressure variation in a variable density fluid
The water in a tank is pressurized by air, and the pressure is measured by a multi-fluid
manometer. The tank is located on a mountain at an altitude of 1400 m where the atmospheric
pressure is 85.6 kPa. Determine the air pressure in the tank if h1= 0.1 m, h2=0.2 m, and h3=0.35
m. Take the densities of water, oil, and mercury to be 1000 kg/m3, 850 kg/m3, and 13,600
kg/m3, respectively.
240𝑔
𝑑𝑝
= 𝜌𝑔
𝑑𝑧
𝜌 = 800 + 30ℎ1/2
𝑃1 + 𝜌𝑤 𝑔ℎ1 + 𝜌𝑜𝑖𝑙 𝑔ℎ2 = 𝑃2 + 𝜌𝑚 𝑔ℎ3
𝑊𝑎𝑡𝑒𝑟
𝑃1 = −𝜌𝑤 𝑔ℎ1 − 𝜌𝑜𝑖𝑙 𝑔ℎ2 + 𝑃2 + 𝜌𝑚 𝑔ℎ3
𝑃1
= −1000 × 9.81 × 0.1 − 850 × 9.81 × 0.2 + 85.6 × 1000
+ 13600 × 9.81 × 0.35 = 130 𝑘𝑃𝑎
𝜌 = 𝜌(𝑧)
3600𝑔
7200𝑔
A
B
C
4
𝑃𝑜 + න 800 + 30ℎ1/2 𝑔 𝑑ℎ = 3.6𝜌𝑔
0
D
E
2
× 30 × 43/2 𝑔 = 3.6𝜌𝑔
3
𝑃𝑜 + 3200𝑔 + 160𝑔 = 3.6𝜌𝑔
𝑃𝑜 = 240g
𝑃𝑜 + 800𝑔 × 4 +
3
Pressure and temperature variation in the atmosphere
Standard Atmosphere
Real atmosphere
Pressure variation in gasses
At an altitude of 11000 m, the atmospheric temperature T is -56.6 °C and the pressure is 22.4 kNm-2.
Assuming that the temperature remains the same at higher altitudes, calculate the density of the air
Isothermal
-56 °C
at an altitude of 15000 m. Assume R= 287 J kg-1 K-1.
Lapse rate 0.0065 K/m
𝑃
𝜌𝑛
= Constant;
𝑝2 𝜌2
𝑔
=
= 𝑒𝑥𝑝 −
𝑧 − 𝑧1
𝑝1 𝜌1
𝑅𝑇 2
n=1.238
Pressure variation in gasses
With altitude at constant temperature (isothermal)
• For gases 𝜌 = 𝜌 𝑝, 𝑇
𝑝 = 𝜌𝑅𝑇 Perfect gas equation
𝑑𝑝
= −𝜌𝑔 = −𝑝𝑔/𝑅𝑇
𝑑𝑧
Temperature of the gas is constant
𝑑𝑝
𝒈
=−
𝑑𝑧
𝑝
𝑹𝑻
Integrate from (z1,p1) to (z2,p2)
𝑝2
𝑔
= 𝑒𝑥𝑝 −
𝑧 − 𝑧1
𝑝1
𝑅𝑇 2
𝑅𝑇 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡; 𝑃 = 𝜌𝑅𝑇
𝑝2 𝜌2
𝑔
=
= 𝑒𝑥𝑝 −
𝑧 − 𝑧1
𝑝1 𝜌1
𝑅𝑇 2
Pressure variation in gasses
With altitude for linear decrease of temperature
𝑇 = 𝑇𝑜 − 𝑎𝑧; 𝑇 = 𝑇0 , 𝑧 = 0
𝑑𝑝
𝒈
=−
𝑑𝑧
𝑝
𝑹(𝑻 − 𝒂𝒛)
ln
𝑝2
𝑔 𝑇0 − 𝑎𝑧2
=
ln
𝑝1 𝑎𝑅 𝑇0 − 𝑎𝑧1
As 𝑃 = 𝜌𝑅𝑇
𝑝2
𝑇2
=
𝑝1
𝑇1
𝑔/𝑎𝑅
𝑝2 𝜌2 𝑇2
=
𝑝1 𝜌1 𝑇1
𝜌2
𝑇2
=
𝜌1
𝑇1
𝑔
−1
𝑎𝑅
4
Barometer
Hydrostatic force on a submerged surface
• Atmospheric pressure is measured
ℎ = 760 𝑚𝑚
• Atmospheric pressure is referred to as the
barometric pressure
𝑝=𝑝 𝑧
𝑑𝐹 = 𝑝 𝑑𝐴
• Atmospheric pressure can be measured by
inverting a mercury-filled tube into a mercury
container that is open to the atmosphere
(Evangelista Torricelli )
Varies in magnitude and direction over
the surface
Resultant force
𝒅𝑭
𝒑
𝐹 = ෍ 𝑑𝐹 = න 𝑝 𝑑𝐴
𝒅𝑨
𝐴
Centre of pressure: the point on the surface through which the
resultant force acts 𝑃 ≡ 𝑋𝑝 , 𝑌𝑝
Line of action of F
𝐹𝑋𝑝 = න 𝑥 𝑝 𝑑𝐴
𝐴
𝐹𝑌𝑝 = න 𝑦 𝑝 𝑑𝐴
𝐴
The centroid and the centroidal moments of inertia for some common geometries.
Hydrostatic force
Consider an inclined plane surface completely submerged in a liquid
𝐹𝑅 = න 𝑝 𝑑𝐴 = න
𝐴
𝑝0 + 𝜌𝑔𝑦𝑆𝑖𝑛𝜃 𝑑𝐴
𝐴
= 𝑝0 𝐴 + 𝜌𝑔𝑆𝑖𝑛𝜃 න 𝑦 𝑑𝐴
𝐴
First moment of
area
Y- coordinate of the centroid of the area 𝑦𝑐
1
𝑦c = න 𝑦 𝑑𝐴
𝐴 𝐴
𝐹𝑅 = 𝑝0 𝐴 + 𝜌𝑔𝐴𝑦𝑐 𝑆𝑖𝑛𝜃
= Pressure at the centroid of the surface x Area
𝐼𝑓 𝑝0 = 0; 𝐹𝑅 = 𝜌𝑔𝐴𝑦𝑐 𝑆𝑖𝑛𝜃
5
Pressure prism method
Center of pressure
𝐹𝑅 𝑦𝑝 = න 𝑦 𝑝 𝑑𝐴
𝐴
Volume of the pressure prism =‫𝐴𝑑 𝑃 𝐴׬‬
= න 𝑦 𝑝𝑜 + 𝜌𝑔𝑦𝑆𝑖𝑛𝜃 𝑑𝐴
𝐴
𝐹𝑅 = Volume of the pressure prism
= 𝑝𝑜 න 𝑦 𝑑𝐴 + 𝜌𝑔𝑆𝑖𝑛𝜃 න 𝑦 2 𝑑𝐴
𝐴
𝐴
𝐹𝑅 𝑦𝑝 = 𝑝𝑜 𝐴𝑦𝑐 + 𝜌𝑔𝑆𝑖𝑛𝜃𝐼𝑥𝑥,𝑜
Parallel axis theorem to find the 𝐼𝑥𝑥,𝑜
𝐼𝑥𝑥,𝑜 = 𝐼𝑥𝑥,𝑐 + 𝐴𝑦𝑐 2
𝐹𝑅 = 𝑝0𝐴 + 𝜌𝑔𝐴𝑦𝑐 𝑆𝑖𝑛𝜃
𝑦𝑝 = 𝑦𝑐 +
𝐼𝑥𝑥,𝑐 𝜌𝑔𝑆𝑖𝑛𝜃
𝐹𝑅
𝐼𝑓𝑝0 = 0; 𝐹𝑅 = 𝜌𝑔𝐴𝑦𝑐 𝑆𝑖𝑛𝜃
2𝑛𝑑 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑎𝑟𝑒𝑎
𝑦𝑝 =
1𝑠𝑡 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑎𝑟𝑒𝑎
Pressure prism method
Special case: A constant pressure 𝑃𝑐 acting on a submerged surface
1st moment of volume of pressure prism
න 𝑦 𝑑𝑉 = න 𝑦 𝑝𝑑𝐴 = න 𝑦 𝑑𝐹 = 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑓𝑜𝑟𝑐𝑒 𝐹 = 𝑦𝑝 𝐹
Resultant force
𝐹 = න 𝑝 𝑑𝐴 = න 𝑃𝑐 𝑑𝐴
𝐴
𝐴
= 𝑷𝒄 𝑨
Centre of pressure
𝑦𝑝 =
‫𝑉𝑑 𝑦 ׬‬
= 𝑐𝑒𝑛𝑡𝑟𝑜𝑖𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑝𝑟𝑖𝑠𝑚
𝑉
𝐹𝑦𝑝 = න 𝑦𝑑𝐹 = න 𝑦𝑃𝑐 𝑑𝐴
𝐴
𝐴
𝑃𝑐 𝐴𝑦𝑝 = 𝑃𝑐 න 𝑦 𝑑𝐴
𝐴
𝑦𝑝 =
‫𝐴𝑑 𝑦 𝐴׬‬
= 𝑦𝑐
𝐴
𝑪𝒆𝒏𝒕𝒓𝒆 𝒐𝒇 𝒑𝒓𝒆𝒔𝒔𝒖𝒓𝒆 𝑷 ≡ 𝑪𝒆𝒏𝒕𝒓𝒐𝒊𝒅 𝒐𝒇 𝒂𝒓𝒆𝒂 (𝑪)
6
Moments of a triangular area
Example 1
The moment of area of a triangular area with mid points of sides at depth ℎ1, ℎ2,
and ℎ3 below 𝑂𝑥
Resultant force
Top edge at 𝒑 = 𝟎
𝐹 = න 𝑝 𝑑𝐴
𝑏
𝑝=0
𝑎
𝐴
= න 𝜌𝑔𝑦𝑏𝑑𝑦
𝑥
0
𝑂
𝒉𝟏
𝑑𝐹
𝒉𝟐
𝒉𝟑
1st
moment about Ox=
2nd moment about Ox=
1
𝐴
3
1
𝐴
3
𝑎
= 𝜌𝑔𝑏𝑎2 /2
𝑭 = 𝑷𝒓𝒆𝒔𝒔𝒖𝒓𝒆 𝒂𝒕 𝒕𝒉𝒆 𝒄𝒆𝒏𝒕𝒓𝒐𝒊𝒅 × 𝑨𝒓𝒆𝒂
Centre of pressure
ℎ1 + ℎ2 + ℎ3
𝒂
𝐹𝑦𝑝 = න 𝑦𝑑𝐹 = න 𝜌𝑔𝑦 2 𝑏𝑑𝑦
2
2
ℎ1 + ℎ2 + ℎ3
𝐴
2
𝟎
𝜌𝑔𝑏𝑎2 /2 𝑦𝑝 = 𝜌𝑔𝑏𝑎3 /3
2
𝑦𝑝 = 𝑎
3
𝑰𝑶𝑿
𝒚𝒑 =
=
Example 2: Method 1
Example 2: Method 2
Top edge is at 𝒂 𝒅𝒆𝒑𝒕𝒉 𝒆 𝒃𝒆𝒍𝒐𝒘
3
𝑒
𝑏
𝑝𝑐 = 𝜌𝑔𝑒
𝑒
𝑝=0
𝑝𝑐 = 𝜌𝑔𝑒
𝑏
𝑎/2
𝑑𝐹
𝑎
≡
𝑑𝐹
𝑎
𝑦𝑝 = 𝑒 +
𝑎
𝑦 = 𝜌𝑔𝑏𝑎2 /2
2 𝑝
𝑎 3𝑒 + 2𝑎
3 2𝑒 + 𝑎
1st moment o𝐟 𝐀𝟏 =
𝟏𝟏
𝒃𝒂
𝟑𝟐
=
𝑨𝟐
𝑨𝟏
𝑎
1st moment o𝐟 𝐀𝟐 =
+
2
𝑎
𝑎 + 𝜌𝑔𝑒 𝑎𝑏
3
2
𝑏𝑎
3𝑒 2 + 3𝑒𝑎 + 𝑎2
𝑦𝑝 = 3
𝑏𝑎
2𝑒 + 𝑎
2
𝑦𝑝 = 𝑒 +
𝑎 3𝑒 + 2𝑎
3 2𝑒 + 𝑎
𝒆 + 𝒂/𝟐 + 𝒆 + 𝒂 + 𝒆 + 𝒂/𝟐
𝒃𝒂
𝟑𝒆 + 𝟐𝒂
𝟔
𝟏𝟏
𝒃𝒂
𝟑𝟐
=
𝑭 = 𝑷𝒄 𝑨
𝜌𝑔𝑏𝑎 𝑒 +
1
1
2
2
2
1st moment about Ox= 𝐴 ℎ1 + ℎ2 + ℎ3 2nd moment about Ox= 3 𝐴 ℎ1 + ℎ2 + ℎ3
𝑝=0
𝑝=0
𝑏
𝑨𝒀𝑮
𝒃𝒂𝟑 /𝟏𝟐 + 𝒃𝒂𝒂𝟐 /𝟒 𝟐
𝑰𝑶𝒄 + 𝑨𝒚𝟐
=
= 𝒂
𝑨𝒚𝑮
𝒃𝒂𝒂/𝟐
𝟑
𝑏𝑎
2𝑒 + 𝑎
2
𝒃𝒂
𝟑𝒆 + 𝒂
𝟔
𝟏𝟏
2nd moment o𝐟 𝐀𝟏 = 𝟑 𝟐 𝒃𝒂 𝒆 + 𝒂/𝟐 𝟐 + 𝒆 + 𝒂
𝒃𝒂
=
𝟑𝒆𝟐 + 𝟒𝒆𝒂 + 𝟑𝒂𝟐 /𝟐
𝟔
2nd
𝑇𝑜𝑡𝑎𝑙 =
𝒆 + 𝒂/𝟐 + 𝒆 + 𝒆 + 𝒂/𝟐
𝟐
+ 𝒆 + 𝒂/𝟐
𝟐
𝑏𝑎
𝑇𝑜𝑡𝑎𝑙 =
3𝑒 2 + 3𝑒𝑎 + 𝑎2
3
𝟏𝟏
moment o𝐟 𝐀𝟐 = 𝒃𝒂 𝒆 + 𝒂/𝟐 𝟐 + 𝒆 𝟐 + 𝒆 + 𝒂/𝟐 𝟐
𝟑𝟐
𝒃𝒂
=
𝟑𝒆𝟐 + 𝟐𝒆𝒂 + 𝒂𝟐 /𝟐
𝟔
7
Example 2: Pressure prism method
Example 2: Method of superposition
𝑝=0
𝑏
𝑏
𝑒
𝑎+𝑒
=
𝑒
𝑎
2
𝜌𝑔𝑒𝑎 + .5𝜌𝑔𝑎𝑎 𝑦𝐺 = 0.5𝜌𝑔𝑎𝑎 × 𝑎 + 𝜌𝑔𝑒𝑎 × 0.5𝑎
3
𝑑𝐹
𝑏
𝑝=0
−
𝑒
Resultant force= Volume of the pressure prism
𝐹 = 𝜌𝑔𝑒𝑎 + .5𝜌𝑔𝑎𝑎 𝑏 = 𝜌𝑔𝑏𝑎 𝑒 + 0.5𝑎
𝑝=0
𝑏
𝑝=0
𝑎
𝑦𝐺 =
𝑭𝟏 = 𝝆𝒈
𝑎 3𝑒 + 2𝑎
3 2𝑒 + 𝑎
𝒚𝒑𝟏 =
𝒂+𝒆
𝒃(𝒂 + 𝒆)
𝟐
𝟐
𝒂+𝒆
𝟑
𝑭𝟐 = 𝝆𝒈
𝒚𝒑𝟐 =
𝐹𝑦𝑝 = 𝐹1 𝑦𝑝1 − 𝐹2 𝑦𝑝2
𝜌𝑔𝑏𝑎 𝑒 +
Example 3
𝒆
𝒃𝒆
𝟐
𝟐
𝒆
𝟑
𝑎
𝑎+𝑒
2
𝑒
2
𝑦 = 𝜌𝑔
𝑏 𝑎+𝑒
𝑎 + 𝑒 − 𝜌𝑔
𝑏𝑒 𝑒
2 𝑝
2
3
2
3
𝑎 3𝑒 + 2𝑎
𝑦𝑝 = 𝑒 +
3 2𝑒 + 𝑎
Example 3
𝑝=0
Vertical gate HJ hinged at H to water tank is just kept closed by a
horizontal force P at the lower edge J. Find P when water surface is (a)
at H (b) 0.4 m above H
𝑯
Superposition
1.2 𝑚
𝑱
0.2 𝑚
0.2 𝑚
𝑷
0.6 𝑚
≡
0.288𝜌𝑔𝑦𝑐 = 0.216𝜌𝑔 × 0.8 + 0.072𝜌𝑔 × 0.4
𝑦𝑝 = 0.7 𝑚
+
+0.6 𝑚
+
0.3 𝑚
𝐹𝑦𝑝 = 𝐹1 𝑦𝑝1 + 𝐹2 𝑦𝑝2
0.3 𝑚
≡
𝑭𝟏 = 𝟎. 𝟔𝝆𝒈 × 𝟎. 𝟑 × 𝟏. 𝟐
= 𝟎. 𝟐𝟏𝟔𝝆𝒈
𝟐
𝒚𝒑𝟏 = × 𝟏. 𝟐 = 𝟎. 𝟖
𝟑
𝐹𝑦𝑝 = 𝑃 𝐻𝐽
0.288 𝜌𝑔 × 0.7 = 𝑃 1.2
𝑃 = 0.168𝜌𝑔
𝑭𝟐 = 𝟎. 𝟑𝝆𝒈 × 𝟎. 𝟒 × 𝟎. 𝟔
= 𝟎. 𝟎𝟕𝟐𝝆𝒈
𝒚𝒑𝟐 =
𝟐
× 𝟎. 𝟔 = 𝟎. 𝟒
𝟑
𝑭𝟏 = 𝟎. 𝟔𝝆𝒈 × 𝟎. 𝟑 × 𝟏. 𝟐
= 𝟎. 𝟐𝟏𝟔𝝆𝒈
𝟐
𝒚𝒑𝟏 = × 𝟏. 𝟐 = 𝟎. 𝟖
𝟑
𝑭𝟐 = 𝟎. 𝟑𝝆𝒈 × 𝟎. 𝟒 × 𝟎. 𝟔
= 𝟎. 𝟎𝟕𝟐𝝆𝒈
𝑭𝟑 = 𝑷𝒄 𝑨
= 𝟎. 𝟒𝝆𝒈𝑨
𝟐
𝒚𝒑𝟐 = × 𝟎. 𝟔 = 𝟎. 𝟒
𝟑
𝐹 × 𝑦𝑝 = 0.216𝜌𝑔 × 0.8 + 0.072𝜌𝑔 × 0.4 + 0.4𝜌𝑔 × 0.6 × 0.36 + 0.4𝜌𝑔 × 0.3 × 0.24
= 𝑃 𝐻𝐽
𝑃 = 0.264 𝜌𝑔
8
Example 3: Pressure prism method
Example 4
𝑝=0
Vertical gate HJ hinged at H to water tank is just kept closed by a
horizontal force P at the lower edge J. Find P when water surface is
(a) at H
𝑯
1𝑚
1𝑚
2𝑚
(b) 0.5 m above H
1𝑚
1𝑚
𝑱
𝑷
𝑨𝟏
𝑨𝟐
1
1st moment about Ox= 3 𝐴 ℎ1 + ℎ2 + ℎ3
11
=2×
2 0.5 + 1.5 + 1 = 2
32
1
2nd moment about Ox= 𝐴 ℎ1 2 + ℎ2 2 + ℎ3 2
3
11
=2×
2 0.52 + 1.52 + 12 = 7/3
32
7/3
𝑦𝑐 =
= 7/6
2
Example 4: Pressure prism method
𝐹 = 𝜌𝑔 × 1 𝑠𝑡 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑎𝑟𝑒𝑎 = 2𝜌𝑔
𝐹 × 𝑦𝑐 = 𝑃 𝐻𝐽
𝑃 = 0.5 𝜌𝑔
𝑃2 =
5
𝜌𝑔
3
Example 5
Rectangular gate AB (0.8 m wide, negligible weight)
smoothly hinged at A. It is kept closed by a vertical
force P applied at B.
a) Draw pressure diagram on AB
b)Find force P and horizontal and vertical reactions
at the hinge A.
9
Example 5
Example 6
A trapezoidal opening in the vertical wall of a tank
is closed by a flat plate hinged at upper edge.
Calculate the moment about the hinge line required
to keep the plate closed.
𝑃𝐴 = 750 × 𝑔 × 1.2
= 900𝑔
𝑃𝐵 = 𝑃𝐴 + 750 × 𝑔 × 1.2
= 1800𝑔
𝐹1 = 900𝑔 × 1.5 × 0.8
= 1080 𝑔
1
𝐹2 = × 900𝑔 × 1.5 × 0.8
2
= 540 𝑔
𝐴𝐶1 = 0.75 𝑚; 𝐴𝐶2 = 1 𝑚
Taking moments around A
𝐴𝐶1 𝐹1 + 𝐴𝐶2 𝐹2 = 𝑃 𝐴𝐵
0.75 × 1080𝑔 + 1 × 540𝑔 = 0.9 × 𝑃
𝑃 = 1500𝑔
𝑅𝑥 + (𝐹1 +𝐹2 )𝐶𝑜𝑠𝜃 = 0
𝑅𝑥 = − 1080𝑔 + 540𝑔
4
= 1296 𝑔
5
𝑅𝑦 + (𝐹1 +𝐹2 )𝑆𝑖𝑛𝜃 − 𝑃 = 0
𝑅𝑦 = 1500𝑔 − (1080𝑔 + 540𝑔)
Hydrostatic forces on curved surfaces
Consider the equilibrium of the element ACB
→ 𝑃 = 𝑅ℎ
And should act on the same line, center of
pressure of CA
3
= 528 𝑔
5
Example 1
𝑝=0
𝑶
1.5 𝑚
5𝜌𝑔
AC is the vertical projection of AB
𝑱
𝑦=
𝑭𝑽
Rv = Weight of the fluid above AB act vertically
through the center of gravity G of ABDE
𝑭𝑯
4.5𝜌𝑔
Find force per meter width on HJO
3
𝐻𝐽 = 𝑦 = 𝑥 2
8
JO vertical
3 2
𝑥 ; 𝑥 = 2𝑚, 𝑦 = 1.5 𝑚
8
?𝑚
𝛼
𝑹
𝐹𝐻 =
𝑯
1
3𝜌𝑔 × 3 = 4.5𝜌𝑔
2
2
3𝑚
2𝑚
𝐹𝑉 = 𝜌𝑔𝑉 = 𝜌𝑔 න
0
3
3 − 𝑥 2 𝑑𝑥 = 5𝜌𝑔
8
𝑅 = 6.73 𝜌𝑔 𝑡𝑎𝑛𝛼 = 4.5/5
𝐹
4.5𝜌𝑔
3𝜌𝑔
1𝑚
10
Example 2
Example 3
A sluice gate is in the form of circular arc of radius 6
m. Calculate the magnitude and direction of the
resultant force on the gate, and the location with
respect to O
PQ= depth of water= 2 × 6𝑆𝑖𝑛30°
𝑅ℎ = 𝐹𝑜𝑟𝑐𝑒 𝑜𝑛 𝑃𝑄 = 𝜌𝑔ℎ 2/2= 176.58 kN/m
𝑅𝑣 = 𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑤𝑎𝑡𝑒𝑟 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑 𝑏𝑦 𝑡ℎ𝑒 𝑠𝑒𝑔𝑚𝑒𝑛𝑡 𝑃𝑄
60
= 𝜋𝑅2 360 −
= 32 𝑘𝑁/𝑚
1
𝑃𝑄
2
𝑝=0
𝑵
𝑴
Cylindrical drum gate of width b is hinged at H
a) Find the force on the drum when water surface
is at level M
b) Find the force on the drum when water surface
is at N
𝑯
× 𝑃𝑂 𝑆𝑖𝑛30°
𝑅=
ℎ
+
Buoyancy
Equilibrium of floating bodies
Forces
Consider immersed an object
Horizontal forces are equal and opposite=>
no horizontal force on the body
Upthrust (buoyancy force)= Weight of fluid
over ACB- Weight of fluid over ADB
• Upthrust R acting on the centre of the buoyancy
• Weight W on the centre of gravity
For equilibrium;
R and W should be equal and act in the same line
= Weight of fluid displaced by the body
= 𝜌𝑔𝑉
V- volume of the object
11
Example 1
Example 2
K
𝑝=0
N
G
H
M
B
𝜃
J
Solid triangular prism HJK of uniform thickness b and density
𝜎floats in a liquid of density 𝜌 as shown unsupported with edge
H at free surface p=0.
JH=JK=a, HJK=𝜃 Prove that JK is vertical and 𝜎 = 𝜌𝐶𝑜𝑠𝜃
W and F should be equal and in same line
𝑀𝐺: 𝐺𝐾 = 1: 2
𝑀𝐵: 𝐵𝑁 = 1: 2
BG //𝑁𝐾
BG is vertical ∴ 𝐽𝐾 𝑖𝑠 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙
𝑊=𝐹
𝜎𝑔0.5 × 𝐽𝐾 × 𝐻𝑁 × 𝑏 = 𝜌𝑔 × 0.5 × 𝐽𝑁 × 𝐻𝑁
𝜎=
𝜎=
𝐽𝑁
𝜌
𝐽𝐾
M
A rectangular prism of thickness 0.2 m and density
𝜎 has cross section shown and floats supported by a
vertical force P as shown. HM= 0.3 m, HN= 0.4 m. Show
34
that 𝜎 = 𝜌
P
0.3 𝑚
𝑝=0
G
H
75
L K
𝜃 J
B
𝑊 𝐺𝐽 = 𝐹 𝐿𝐽
0.4 𝑚
𝜎𝑔 0.4 × 0.3 × 0.2 × 0.25 = 𝜌𝑔 0.5 × 0.4 × 0.3 × 0.2 × 0.68/3
N
𝐺𝐵: 𝐵𝑁 = 1: 2; 𝐺𝐿: 𝐿𝐾 = 1: 2
𝐽𝐾 = 0.3 𝐶𝑜𝑠𝜃 = 0.18
𝐺𝐾 = 𝐺𝐽 − 𝐽𝐾 = 0.25 − 0.18 = 0.07
2
𝐿𝐽 = 𝐽𝐾 + 𝐿𝐾 = 0.18 + 𝐺𝐾 = 0.68/3
3
𝐽𝐻 𝑐𝑜𝑠𝜃
𝜌
𝐽𝐾
𝜎=
1 1
𝜌 × 0.68/3
2 0.25
𝜎=
0.34
𝜌
0.75
𝜎=
34
𝜌
75
𝜎 = 𝜌𝐶𝑜𝑠𝜃
Example 3
Stability of floating bodies
A rectangular pontoon has a width B of 6 m, a length 𝑙 of 12 m, and a draught D of 1.5 m in fresh water.
Calculate
a) Weight of the pontoon
b) Its draught in sea water
c) The load that can be supported by the pontoon in freshwater if the maximum draught permissible is 2
m.
A small displacement
is given
a) Weight of the pontoon = Upthrust = Weight of the fluid displaced
𝑊 = 𝜌𝑔𝑉 = 1000 × 9.81 × 6 × 12 × 1.5 = 1059.5 𝑘𝑁
b) Weight of the pontoon = Upthrust = Weight of the fluid displaced
𝑊 = 𝜌𝑔𝑉 = 1025 × 9.81 × 6 × 12 × 𝐷 = 1059.5 𝑘𝑁
𝐷 = 1.46 𝑚
c) Weight of the pontoon + load = Upthrust = Weight of the fluid
displaced
𝑊 + 𝑤 = 𝜌𝑔𝑉 = 1000 × 9.81 × 6 × 12 × 2 = 𝑤 + 1059.5 × 103
𝑤 = 353.1 𝑘𝑁
G remain same
Floating body in
equilibrium
Immersed volume changed, B moved to B’
R and W are not at the same line.
Turning moment 𝑊 𝐺𝑀 𝜃
M- Metacentre
GM- Metacentric height
12
Stability of floating bodies
Stability of floating bodies
M lies above G, a righting moment is
produced
Come back to initial position
M lies below G, an overturning
moment is produced
Move further
Equilibrium is unstable
M coincides G
Equilibrium is neutral
Equilibrium is stable
Determination of the metacentric height: Experimental
Measuring angle of tilt 𝜃 caused by moving a load 𝑤 a known distance 𝑥 across the deck
Determination of the metacentric height: Theoretical
For small angle of tilt
𝐵𝑀 =
Overturning moment= 𝑤𝑥
Righting moment = 𝑊 𝐺𝑀 𝜃
For equilibrium;
𝑤𝑥 = 𝑊 𝐺𝑀 𝜃
𝑥=
𝑊
𝑤
𝐺𝑀 𝜃
𝐵𝐵′
𝜃
Change of centre of buoyancy from B to B’
is due to the change in immersed area AOA’
and COC’
Taking moment around waterline OO
‫𝐵𝐵 𝑅 = 𝑥 𝜃𝑥 𝐴𝑑𝑔𝜌 ׬‬′
𝜌𝑔𝜃 න 𝑥 2 𝑑𝐴 = 𝜌𝑔𝑉 𝐵𝐵′
𝜃𝐼 = 𝑉 𝐵𝑀 𝜃
𝑰
𝑩𝑴 =
𝑽
V - immersed volume
I - Second moment of area of
waterline plane about OO
Cross-section
Plan
𝑩𝑴 = 𝑩𝑮 + 𝑮𝑴
𝑮𝑴 = 𝑩𝑴 − 𝑩𝑮
13
Example 1
Example 2
A uniform rectangular block of specific gravity 0.6
floats in water. Block has width b height a and length l.
Show that block is in stable equilibrium if b> 1.2 a.
𝑙
𝑝=0
𝑎
𝑂𝐵 =
G
B
𝐼
=
𝑉
𝑙𝑏3 /12
𝑙𝑏𝑑
=
𝑏2
12 × 0.6𝑎
𝐺𝑀 = 𝐵𝑀 − 𝐵𝐺
𝑏2
− 0.2𝑎
7.2𝑎
For stable eqm 𝐺𝑀 > 0; 𝑏 > 1.2𝑎
𝐺𝑀 =
River barge 2.4 m wide 4m long with rectangular plan has
uniform vertical section as shown. It is loaded with sand.
Loaded barge weights 56.5 kN and CG is at center of water
0.6 𝑚 line. Find GM and righting moment for 1°.
0.6 𝑚
𝑑
𝑎
= 0.3𝑎; 𝑂𝐺 = ; 𝐵𝐺 = 0.2𝑎
2
2
𝐵𝑀 =
G
B
O
𝑑
𝑊 = 0.6𝜌𝑔𝑏𝑎𝑙
𝑅 = 𝜌𝑔𝑏𝑑𝑙
For equilibrium, 𝐹 = 𝑊; 𝑑 = 0.6𝑎
𝑏
𝑑
2.4 𝑚
𝑝=0
𝐹=𝑤
𝜌𝑔𝑉 = 56500; 𝑉 = 5.76 𝑚 3
Volume of triangular area= 0.5 × 2.4 × 0.6 × 4 = 2.88𝑚 3
1.2 𝑚
Immersed rectangular area= 2.88 m3
If the immersed depth is d, volume=2.4 × 𝑑 × 4 = 2.880
𝑑 = 0.3 𝑚
0.3 × 2.4 × 0.15 + 0.5 × 2.4 × 0.6 × 0.5 = 0.3 × 2.4 + 0.5 × 2.4 × 0.6 𝐺𝐵
𝐵𝐺 = 0.325 𝑚
𝐵𝑀 =
𝐼 4 × 2.43 /12
=
= 0.8 𝑚
𝑉
5.76
𝐺𝑀 = 𝐵𝑀 − 𝐵𝐺 = 0.475 𝑚
Righting moment= 𝑊 𝐺𝑀 𝑆𝑖𝑛𝜃 = 468.4 Nm
14