4 – Hydrostatic Forces on Curved Surfaces, Slides of Fluid Mechanics

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ENGR. BON RYAN ANIBAN

Case I: FLUID IS ABOVE THE CURVED SURFACE

𝑣

𝑒

𝜃

𝐹 𝐻

𝑐𝑔

𝐴𝑟𝑒𝑎 = 𝐴

Vertical projection of the

curved surface

ത ℎ

𝑣

𝑐𝑔 𝑜𝑓 𝑣𝑜𝑙𝑢𝑚𝑒

Curved surface

𝑉

𝐻

; Volume of water above

; A = Area of vertical projection

𝑉

2

• 𝐹 𝐻

2

Case III: FLUID ARE BELOW AND ABOVE THE CURVED

SURFACE

𝑣

𝐹 𝐻

Curved surface

𝑣

𝐹 𝐻

𝐹 𝐻 𝑒

𝐴

𝑣

c𝑔 𝑜𝑓 𝑣𝑜𝑙𝑢𝑚𝑒

Net vertical force 𝐹

𝑐𝑔

Net vertical projection

of area

Net horizontal force

𝑉

𝑉 2

𝑉 1

; Volume of water above

; A = Area of vertical projection

𝑉

2

• 𝐹 𝐻

2

𝐻

𝐻 2 −

𝐻

2

1

2

2

1

1

The submerged curve AB is one quarter of a circle or radius 2m. and is located on the lower corner of the water tank as

shown. The length of the tank perpendicular to the sketch is 4m. Find the magnitude and location of the horizontal and

vertical component of the total force acting on AB.

B

A

4m

2m

2m

2m

O

The submerged curve AB is one quarter of a circle or radius 2m. And is located on the lower corner of the water tank as

shown. The length of the tank perpendicular to the sketch is 4m. Find the magnitude and location of the horizontal and

vertical component of the horizontal and vertical component of the total force acting on AB.

B

A

4m

2m

2m

F

2m

F

H

F

V

g

5 m

y

D C

𝑉

𝐴𝐵𝐶𝐷

Solution

𝑉

𝐴𝐵𝐶𝐷

𝐴𝐵𝐶𝐷

A

1

A

2

1

+ A

2

2

2 𝜋

= 3. 142 m

2

1

X

= 8 m

2

𝐴 = 11. 142 m

2

𝐴𝐵𝐶𝐷

= 44. 566 m

3

𝑉

𝑉

= 437. 196 kN

Location of Fv

𝐴തx = 𝐴 1

തx 1

2

xത 2

X 2

X 1

11. 142 xത = 8

തx = 0. 957 m

O

The submerged curve AB is one quarter of a circle or radius 2m. And is located on the lower corner of the water tank as

shown. The length of the tank perpendicular to the sketch is 4m. Find the magnitude and location of the horizontal and

vertical component of the horizontal and vertical component of the total force acting on AB.

B

A

4m

2m

2m

F

2m

F

H

F

V

y

D C

Solution

A

1

A

2

X

Location of Fv

X 2

X 1

1. 067 m

Since Total pressure is always normal to the surface

and normal to the circle passes is center,

then the total force F shall also pass through the center of circle O,

hence the momnet about O due to F or due to Fh and Fv is zero.

O

0

𝑉

xത − 𝐹 𝐻

തy = 0

437. (^196) (തx) − 392. (^4) ( 1. 067 )

392. 4 kN

തx = 0. 957 m

For the curved surface AB: (a) Determine the magnitude, direction and line of action of vertical component of hydrostatic force

actin on the surface. (b) Determine the magnitude, direction and line of action of vertical component of hydrostatic force actin

on the surface. Use l =1m.

F F v

F H

1 m

1 m

1 m

𝐻

𝐻

𝐻

= 14. 715 kN

𝑔

3

𝑒 = 0. 056 m

𝑦 = 0. 5 + 0. 056 m

∴ F

H

is acting 0. 556 m below B

Solution

For the curved surface AB: (a) Determine the magnitude, direction and line of action of vertical component of hydrostatic force

actin on the surface. (b) Determine the magnitude, direction and line of action of vertical component of hydrostatic force actin

on the surface. Use l =1m.

F F v

F H

1 m

𝐻

= 14. 715 kN ∴ F H

is acting 0. 556 m below B

Solution

𝑉

1

+ A

2

2

2 𝜋

= 0. 785 m

2

1

= 1 ( 1 ) = 1 m

2

𝐴 = 1. 785 m

2

𝐴𝐵𝐶𝐷

= 1. 785 m

3

𝑉

𝑉

= 17. 515 kN

0

𝑉

xത − 𝐹 𝐻

yത = 0

17.51 (^5) (xത) − 14. (^715) ( 0. 556 )

x ത = 0. 467 m