Prof. Dr. Hafzullah Aksoy
Fluid Mechanics Week 04
Hydrostatics
Numerical Example
Draw diagram of pressure action on the spherical
solid body within water.
Water
Determine the weight of the body in water.
Solid
Body
Solution
𝑭𝒘𝒂𝒕𝒆𝒓
water
Solid
Body
Solid
Body
𝑭𝒔𝒐𝒍𝒊𝒅
𝐅𝐧𝐞𝐭 = 𝐅𝐬𝐨𝐥𝐢𝐝 − 𝐅𝐰𝐚𝐭𝐞𝐫
𝐕𝐨𝐥𝐮𝐦𝐞
= (𝜸𝑠𝑜𝑙𝑖𝑑 ∀𝑠𝑜𝑙𝑖𝑑 ) − (𝜸𝑤𝑎𝑡𝑒𝑟 ∀𝑤𝑎𝑡𝑒𝑟 ) {
∀𝐬𝐨𝐥𝐢𝐝 =
Fwater : Uplift force (Archimedes buoyancy law)
Fsolid : Weight of solid body
For a spherical solid body
𝑭𝒏𝒆𝒕 = 𝒈
𝟒 𝟑
𝝅𝒓 (𝝆𝑠𝑜𝑙𝑖𝑑 − 𝝆𝑤𝑎𝑡𝑒𝑟 )
𝟑
𝟒
𝝅. 𝒓𝟑
𝟑
Prof. Dr. Hafzullah Aksoy
Numerical Example
Gate AB has a rectangular geometry. It is
inclined with horizontal axis at angle α. Its
width (third dimension) is b = constant.
Calculate pressure force by
𝒉𝑩
B
Liquid
a) Resulting pressure diagram
b) Components’ pressure diagrams
c) Centroid method
𝒉𝑨
𝑳𝑨𝑩
𝜶
A
Solution
a)
Resulting Pressure Diagram
Liquid
𝒑𝑩
F
𝒉𝑩
B
𝒉𝑨
𝒑𝑨
𝑳𝑨𝑩
𝜶
A
þ𝐀 = (𝛄 𝐡𝑨 ) = (𝝆𝒈 𝐡𝑨 )
þ𝐁 = (𝛄 𝐡𝑩 ) = (𝝆 𝒈 𝐡𝑩 )
Resulting Pressure Force
𝟏
𝑭 = þ𝐁 (𝐋𝑨𝑩 𝐛) + (þ𝑨 − þ𝐁 )(𝐋𝑨𝑩 𝐛)
𝟐
Prof. Dr. Hafzullah Aksoy
b)
Components’ Pressure Diagrams
𝑭
Liquid
𝒑𝑩
𝒑𝑨
𝑭𝒉
𝑭𝒙
𝒑𝑩
𝒉𝑩
B
𝒉𝑨
𝑳𝑨𝑩
𝜶
𝒑𝑨
A
𝒃 : Width of the figure
𝟏
(
)
𝑭𝒙 = þ𝐁 𝐡𝑨 − 𝐡𝐁 𝒃 + (þ𝑨 − þ𝐁 )(𝐡𝑨 − 𝐡𝐁 ) 𝒃
𝟐
Horizontal Pressure Force
Vrtical Pressutre Force = Weight of liquid contained in the volume above the surface AB
𝑭𝒉 = (𝒈𝝆)∀𝑨𝑩 = 𝜸(𝑨𝐴𝐵 . 𝒃)
Resulting Pressure Force
𝑭 = √𝑭𝟐𝒙 + 𝑭𝟐𝒉
c)
Centroid Method
𝑭 = þ𝐆 𝑨𝒔𝒖𝒓𝒇𝒂𝒄𝒆 = (𝛄𝐡G ) 𝑨𝒔𝒖𝒓𝒇𝒂𝒄𝒆
𝑨𝑘𝑎𝑝𝑎𝑘 = 𝐋AB . 𝒃
𝟏
𝑭 = 𝜸 (𝐡𝐁 + (𝐡𝑨 − 𝐡𝐁 )) 𝑨𝑠𝑢𝑟𝑓𝑎𝑐𝑒
𝟐
Prof. Dr. Hafzullah Aksoy
Numerical Example
Calculate pressure acting on the surface
AB of the dam (width b = constant)
y
B
water
Dam
𝒉𝑨
𝒚 = 𝒇(𝒙)
A
x
Solution: We can determine pressure force on a surface with a form of a curve by using each component
y
separately.
𝑭𝒉
B
𝑭
liquid
Dam
𝑭𝒙
𝒑𝑨
𝒚 = 𝒇(𝒙)
x
𝒑𝑨
A
Horizontal Pressure Force
𝟏
𝟏
𝑭𝒙 = þ𝑨 (𝐡𝑨 𝐛) = (𝜸𝐡𝐴 )(𝐡𝑨 𝐛)
𝟐
𝟐
Vertical Pressure Force : Weight of liquid contained in the volume above the surface AB
𝑨
𝑭𝒉 = 𝜸∀𝑨𝑩 = 𝜸 (∫ 𝒇(𝒙)𝒅𝒚) 𝒃
𝑩
Resulting Pressure Force :
𝑭 = √𝑭𝟐𝒙 + 𝑭𝟐𝒉
Prof. Dr. Hafzullah Aksoy
Numerical Example
Determine pressure force on the
quarter-circle surface AB with contant
width b. Use each component
separately.
Liquid
hB
B
hA
A
Solution
Liquid
𝑭𝒉
hB
𝑭
𝒑𝑩
𝑭𝒙
B
hA
𝒑𝑨
Hoizontal Pressure Force
𝑭𝒙 =
𝟏
𝟏
𝟏
þ𝑨 (𝐡𝑨 𝐛) − þ𝑩 (𝐡𝑩 𝐛) = (𝝆𝒍𝒊𝒒𝒖𝒊𝒅 𝒈)(𝒉𝟐𝑨 − 𝒉𝟐𝑩 )𝒃
𝟐
𝟐
𝟐
OR
𝟏
𝑭𝒙 = þ𝑩 𝑨𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 + (þ𝑨 − þ𝑩 )𝑨𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙
𝟐
𝟏
= þ𝑩 {(𝐡𝑨 − 𝐡𝑩 )(𝐛)} + (þ𝑨 − þ𝑩 ){(𝐡𝑨 − 𝐡𝑩 )(𝐛)}
𝟐
Vertical Pressure Force : Weight of liquid contained in the volume above the surface AB
𝑭𝒉 = (𝒈𝝆𝒍𝒊𝒒𝒖𝒊𝒅 )∀𝑨𝑩 = 𝜸𝒍𝒊𝒒𝒖𝒊𝒅 {𝐡𝑨 (𝐡𝑨 − 𝐡𝑩 ) − 𝛑(𝐡𝑨 − 𝐡𝑩 )𝟐 }𝐛
Resulting Force :
𝑭 = √𝑭𝟐𝒙 + 𝑭𝟐𝒉
Prof. Dr. Hafzullah Aksoy
Numerical Example
Determine pressure force on the
quarter-circle surface AB with contant
width b. Use each component
separately.
atmosphere
𝒉𝑩
Liquid
B
𝟏
(𝒉 − 𝒉𝑨 )
𝟐 𝑩
A
Solution
B’
A’
𝑭𝒉
Liquid
𝒑𝑩
atmosphere
𝒑𝑩
B
𝒑𝑨
A
𝑭𝒙
𝑭
𝒑𝑨
Horizontal Pressure Force
𝟏
𝑭𝒙 = þ𝑩 𝑨𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 + (þ𝑨 − þ𝑩 )𝑨𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙
𝟐
𝟏
= þ𝑩 {(𝐡𝑨 − 𝐡𝑩 )(𝐛)} + (þ𝑨 − þ𝑩 ){(𝐡𝑨 − 𝐡𝑩 )(𝐛)}
𝟐
Vertical Pressure Force : Weight of liquid contained in the volume above the surface AB
𝑭𝒉 = (𝒈𝝆𝒍𝒊𝒒𝒖𝒊𝒅 )∀𝐴𝐴′𝐵𝐵′ = 𝜸𝒍𝒊𝒒𝒖𝒊𝒅 {(𝑨𝒍𝒂𝒏𝐴𝐴′𝐵𝐵′ )𝐛}
Resulting Pressure Force
𝑭 = √𝑭𝟐𝒙 + 𝑭𝟐𝒉
𝒉𝑨
Prof. Dr. Hafzullah Aksoy
Numerical Example
Calculate pressure force acting on surface AB (constant width b ≠ f(h)).
Determine the acting point of the resulting force
Draw presure diagram and use
a)
Resulting force diagram
b)
Centroid method
(ρwater=1000 kg/m3 ; 𝑰𝑮 =
𝒃.𝑳𝟑
𝟏𝟐
𝑓𝑜𝑟 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒) .
Sonuç : Bileşke Basınç Kuvveti , Fsu = 90644 N ; x M = 4,716 m ; hM = 4,084 m
60o
B =1 m
hA=3 m
water
A
HB=5 m
B
Solution
45o
water
B =2 m
hA=2 m
þ𝐀 = 𝟐 𝜸𝒘𝒂𝒕𝒆𝒓
þ𝐁 = 𝟒 𝜸𝒘𝒂𝒕𝒆𝒓
A
𝑳𝑨𝑩 = 𝟐√𝟐 =
𝒑𝑨
hB=4 m
𝟐
𝒔𝒊𝒏𝟒𝟓
G
M
𝑭
B
𝑨 = 𝐋𝑨𝑩 . 𝒃 = (𝟐√𝟐) 𝟐 = 𝟒√𝟐 𝒎𝟐
𝒑𝑩
𝟏
𝐡 𝐆 = 𝐡 𝑨 + (𝐡 𝑩 − 𝐡 𝐀 ) = 𝟑 𝒎
𝟐
Resulting Pressure Force
𝟏
𝑭 = þ𝐀 (𝐋𝐴𝐵 𝐛) + (þ𝑩 − þ𝐀 )(𝐋𝐴𝐵 𝐛) = (𝟏𝟐 √𝟐 𝜸𝒘𝒂𝒕𝒆𝒓 ) N (Newton)
𝟐
Centroid Method
𝑭 = þ𝐆 𝑨 = 𝜸𝒘𝒂𝒕𝒆𝒓 (𝐡𝑮)𝑨 = 𝜸𝒘𝒂𝒕𝒆𝒓 (𝟑)(𝟒√𝟐) = (𝟏𝟐√𝟐𝜸𝒘𝒂𝒕𝒆𝒓 ) = 𝟏𝟔, 𝟗𝟕 𝜸𝒘𝒂𝒕𝒆𝒓 = 𝟏𝟔𝟔𝟒𝟖𝟏 N
Prof. Dr. Hafzullah Aksoy
Numerical Example
Surface AB with a variable width (b = f(h)) is consdiered.
Calculate pressure force anad find its acting point M.
ρwater
𝑏.L3
=1000 kg/m3 , 𝐼𝐺 =
36
𝑓𝑜𝑟 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 .
60o
B =1
m
hA=3 m
water
A
G
hB=5 m
B
𝑭
Solution
M
Centroid method
𝟐
𝟒√𝟑
𝐋𝑨𝑩 = 𝒔𝒊𝒏𝟔𝟎 = (
𝟑
𝟏
)𝑚
𝟐√𝟑
𝑨 = 𝟐 𝐋𝑨𝑩 . 𝒃 = (
𝟑
) 𝑚2
𝑭 = þ𝐆 𝑨
𝑭 = 𝜸𝒔𝒖 (𝐡𝑮 )𝑨 = (
𝐡
𝑮
xG = 𝒔𝒊𝒏𝟔𝟎
=
𝟐𝟐√𝟑
𝟐𝟐√𝟑
𝟗
𝟗
) 𝜸𝒔𝒖
𝟏
𝐡𝐆 = 𝐡𝑨 + (𝐡𝑩 − 𝐡𝐀 ) = 𝟑. 𝟔𝟕 𝒎
𝟑
= 𝟒. 𝟐𝟑𝟒 𝜸𝒔𝒖 N (Newton)
= 𝟒. 𝟐𝟑𝟒 m
𝑏. L3𝐴𝐵
𝐼𝐺 =
=
36
𝐱𝐌 =
4√3
1 .( 3 )
36
𝟏𝟔√𝟑
( 𝟖𝟏 )
3
𝟏𝟔√𝟑
=(
)
𝟖𝟏
𝐈𝐆
𝟐𝟐√𝟑
+ 𝐱𝐆 =
+(
) = 𝟒. 𝟑𝟎𝟒 𝒎
𝐱 𝐆. 𝐀
𝟗
𝟐𝟐√𝟑 𝟐√𝟑
(
)(
)
𝟗
𝟑
Depth of acting point of the resulting force
𝐡𝑴 = xM 𝒔𝒊𝒏𝟔𝟎 = 3.73 m
Prof. Dr. Hafzullah Aksoy
Unsolved Example
Quarter-circle surface AB has a fixed-width b=2 m.
a)
Show horizontal and vertical pressure diagrams.
b)
Calculate resulting force.
water
2m
A
1m
B
Example Pressure Diagrams