Hydrostatic Force on Surfaces
3. Hydrostatic Force on Surfaces
3.1. INTRODUCTION
Engineers must calculate forces exerted by fluids in order to design constraining structures satisfactorily. In this chapter all t hree characteristics of hydrostatic forces will be evaluated: magnitude, direction, and sense. In addition, locations of forces will be found. 3.2. FORCE EXERTED BY A LIQUID ON A PLANE AREA
The force F exerted exerted by a liquid on a plane area A is equal to the product of the specific weight γ of the liquid, depth of the center of gravity of the area hcg, and the area.
(1) Note that the product of specific weight and depth of the center of gravity of t he area yield s the intensity of pressure at the area's center of gravity. The line of action of the force passes through the center of pressure, which can be located by applying the formula
(2) where I cg is the moment of inertia of the area about its center of gravity gravity axis (see Fig. 3-1). Distances γ are measured along along the plane from an axis located at the intersection of the plane and the liquid surface, both extended if necessary.
Figure 3-1.
3.3. FORCE EXERTED BY A LIQUID ON A CURVED SURFACE
The horizontal component of the hydrostatic force on a curved surface is equal to the normal force on the vertical projection of the surface. The component acts through the center of pressure for the vertical projection. The vertical component of of the hydrostatic force on a curved surface is equal to the weight of the volume of liquid above the area, real or imaginary. The force passes through the center of gravity of the volume. 3.4. HOOP OR CIRCUMFERENTIAL TENSION
Hoop tension or circumferential tension is created in the walls of a cylinder subjected to internal pressure. For thin-walled cylinders ( t < 0.1d ), ),
(3) typical units being lb/in2 (psi) or Pa. 3.5. LONGITUDINAL STRESS IN THIN-WALLED CYLINDERS
Longitudinal stress in thin-walled cylinders closed at the ends is equal to half the hoop tension. 3.6. HYDROSTATIC FORCES ON DAMS
Large hydrostatic forces to which dams are subjected tend to cause a dam to (1) slide horizontally along its base and (2) overturn about its downstream edge (which is known as the toe of the dam). Another factor that may affect dam stability is hydrostatic uplift along the bottom of the dam, caused by water seeping under the dam. Checks for dam stability are made by finding (1) the factor of safety against sliding, (2) the factor of safety against overturning, and (3) the pressure intensity on the base of the dam. The factor of safety against sliding is determined by dividing sliding resistance by sliding force. The factor of safety against overturning is computed by dividing the total righting (resisting) moment by the total overturning moment, all moments being taken about the toe of the dam. Pressure intensity on the base of the dam can be calculated using the flexural formula
(4) where
Equation ( 4) gives the pressure distribution along the dam's base if the resultant reaction on the base acts within its middle third. 3.6.1. Solved Problems
3.1. (a) Develop the equation for the hydrostatic force acting on a plane area, and ( b) locate the force. Solution:
a. Let trace AB represent any plane area acted upon by a fluid and making an angle θ with the horizontal, as shown in Fig. 3-2. Consider an element of area such that every particle is the same distance h below the surface of the liquid. The strip shown crosshatched is such an area (dA), and the pressure is uniform over this area. Then the force acting on the area dA is equal to the uniform intensity of pressure p times the area dA, or below the surface of the liquid. The strip shown crosshatched is such an area ( dA), and the pressure is uniform over this area. Then the force acting on the area dA is equal to the uniform intensity of pressure p times the area dA, or
Figure 3-2.
Summing all the forces acting on the area and considering that h = y sinθ,
where γ and θ are constants and, from statics, ∫ y d A = ycgA. Since hcg sin θ,
(1)
b. To locate this force F , proceed as in static mechanics by taking moments. Axis O is chosen as the intersection of the plane area and the water surface, both extended if necessary. All distances y are measured from this axis, and the distance to the resultant force is called ycp, which is the distance to the center of pressure. Since the sum of the moments of all the forces about axis O = the moment of the resultant force, we obtain
In more convenient form, from the parallel axis theorem,
(2) Note that the position of the center of pressure is always below the center of gravity of the area, or ycp – ycg is always positive because I cg is always positive. 3.2. Locate the lateral position of the center of pressure. Refer to Fig. 3-2. Solution:
While in general the lateral position of the center of pressure is not required to solve most engineering problems concerning hydrostatic forces, occasionally this information may be needed. Using the sketch in the preceding problem, area dA is chosen as ( dx dy) so that the moment arm x is properly used. Taking moments about any nonintersecting axis Y 1 Y ,1,
Using values derived in Problem 3.1,
or
(3) since h = y sin θ. The integral represents the product of inertia of the plane area about the X and Y axes chosen, designated by I xy . Then
(4) Should either of the centroidal axes be an axis of symmetry of the plane area, I xy becomes zero, and the lateral position of the center of pressure lies on the Y axis, which passes through the center of gravity (not shown in the figure). Note that the product of inertia about the center of gravity axes, (l Xy)c, may be positive or negative, so that the lateral position of the center of pressure may lie on either side of the centroidal y axis.
3.3. Determine the resultant force F due to water acting on the 3 m by 6 m rectangular area A B shown in Fig. 3-3.
Figure 3-3. Solution:
This resultant force acts at the center of pressure which is at a distance ycp from axis O1 and
3.4. Determine the resultant force due to water acting on the 4 m by 6 m triangular area CD shown in Fig. 3-3. The apex of the triangle is at C. Solution:
This force acts at a distance ycp from axis O2 and is measured along the plane of the area CD.
3.5.
Water rises to level E in the pipe attached to tank ABCD in Fig. 3-4. Neglecting the weight of the tank and riser pipe, ( a) determine
and locate the resultant force acting on area AB, which is 8 ft wide; ( b) compute the total force on the bottom of the tank; and ( c) compare the total weight of the water with the result in ( b) and explain the difference.
Figure 3-4. Solution:
A free body of the lower part of the tank (cut by a horizontal plane just above level BC ) will indicate a downward force on area BC of 179,700 lb, vertical tension in the walls of the tank, and the reaction of the supporting plane. The reaction must equal the total weight of water or 60,700 lb. The tension in the walls of the tank is caused by the upward force on the top AD of the tank, which is
An apparent paradox is thus clarified since, for the free body considered, the sum of the vertical forces is zero, i.e.,
and hence the condition for equilibrium is satisfied.
3.6. Gate AB in Fig. 3-5(a) is 4 ft wide and is hinged at A. Gage G reads —2.17 psi, and oil of specific gravity 0.750 is in the right-hand tank.
What horizontal force must be applied at B for equilibrium of gate AB?
Figure 3-5. Solution:
The forces acting on the gate due to the liquids must be evaluated and located. For the right-hand side,
acting
It should be noted that the pressure intensity acting on the right-hand side of rectangle AB varies linearly from zero gage to a value due to 6 ft of oil ( p = γh is a linear equation). Loading diagram ABC indicates this fact. For a rectangular area only, the center of gravity of this loading diagram coincides with the center of pressure. The center of gravity is located ( )(6) = 4 ft from A, as above. For the left-hand side, it is necessary to convert the negative pressure due to the air to its equivalent in feet of the liquid, water.
This negative pressure head is equivalent to having 5.01 ft less of water above level A. It is convenient and useful to employ an imaginary water surface (IWS) 5.01 ft below the real surface and solve the problem by direct use of basic equations. Thus,
For the submerged rectangular area,
from O, or the center of pressure is (10.29 – 6.99) – 3.30 ft
from A. In Fig. 3-5( b), the free-body diagram of gate AB shows the forces acting. The sum of the moments about A must equal zero. Taking clockwise as plus,
3.7. The tank in Fig. 3-6 contains oil and water. Find the resultant force on side ABC , which is 4 ft wide.
Figure 3-6. Solution:
The total force on ABC is equal to ( F AB + F BC). Find each force, locate it, and, using the principle of moments, determine the position of the total force on side ABC . a. F AB = (0.800 × 62.4)(5)(10 × 4) = 9980 lb acting at a point ( ) (10) ft from A or 6.67 ft down. The same distance can be obtained by formula as follows:
b. Water is acting on area BC , and any superimposed liquid can be converted into an equivalent depth of water. Employ an imaginary water surface (IWS) for this second calculation, locating the IWS by changing 10 ft of oil to 0.800 × 10 = 8 ft of water. Then
The total resultant force = 9980 + 16,470 = 26,450 lb acting at the center of pressure for the entire area. The moment of this total force = the sum of the moments of its two parts. Using A as a convenient axis,
Other methods of attack may be employed, but it is believed that the method illustrated will greatly reduce mistakes in judgment and calculation.
3.8. In Fig. 3-7, gate ABC is hinged at B and is 4 m long. Neglecting the weight of the gate, determine the unbalanced moment due to the water
acting on the gate.
Figure 3-7. Solution:
F AB =
(9.79)(4)(9.24 × 4) = 1447 kN, acting ( ) (9.24) = 6.16 m from A. F BC = (9.79)(8)(3 × 4) = 940 kN, acting at the center of gravity of BC since the pressure on BC is uniform. Taking moments about B (clockwise plus),
3.9. Determine the resultant force due to the water acting on the vertical area shown in Fig. 3-8(a), and locate the center of pressure in the x and
y directions.
Figure 3-8. Solution:
Divide the area into a rectangle and a triangle. The total force F 1 acting is equal to force the rectangle plus force F 2 acting on the triangle. a.
The resultant force F = 7990 + 7490 = 15,480 lb. Taking moments about axis XX ,
b. To locate the center of pressure in the X direction (seldom required), use the principle of moments after having located x1 and x2 for the rectangle and triangle, respectively. For the rectangle, the center of pressure for each horizontal strip of area dA is 2 ft from the YY axis; therefore its center of pressure is 2 ft from that axis. For the triangle, each area dA has its center of pressure at its own center; therefore the median line contains all these centers of pressure, and the center of pressure for the entire triangle can now be calculated. Referring to Fig. 38(b) and using similar triangles, x2 /2 = 3.80/6, from which x2 = 1.27 ft from YY . Taking moments,
An alternative method can be used to locate the center of pressure. Instead of dividing the area into two parts, calculate the center of gravity position for the entire area. Using the parallel-axis theorem, determine the moment of inertia and the product of inertia of the entire area about these center of gravity axes. The values of ycp and xcp are then calculated by formulas (2) and (4), Problems 3.1 and 3.2. Generally this alternative method has no particular advantage and may involve more arithmetic.
3.10. The 2-m-diameter gate AB in Fig. 3-9 swings about a horizontal pivot C located 40 mm
below the center of gravity. To what depth h can
the water rise without causing an unbalanced clockwise moment about pivot C?
Figure 3-9. Solution:
If the center of pressure and axis C should coincide, there would be no unbalanced moment acting on the gate. Evaluating the center of pressure distance,
Then
from which h = 5.25 m above A. 3.11. Determine and locate the components of the force due to the water acting on curved area AB in Fig. 3-10, per meter of its length.
Figure 3-10. Solution:
acting through the center of gravity of the volume of liquid. The center of gravity of a quadrant of a circle is located at a distance (4/3) × ( r/π ) from either mutually perpendicular radius. Thus
Note: Each force dP acts normal to curve AB and would therefore pass
through hinge C upon being extended. The total force should also pass through C. To confirm this statement, take moments of the components about C, as follows.
3.12.
The 6-ft-diameter cylinder in Fig. 3-11 weighs 5000 lb and is 5 ft long. Determine the reactions at A and B, neglecting friction.
Figure 3-11. Solution:
a. The reaction at A is due to the horizontal component of the liquid force acting on the cylinder or
to the right. Hence the reaction at A must be 4490 lb to the left. b. The reaction at B is the algebraic sum of the weight of the cylinder and the net vertical component of the force due to the liquid. The curved surface CDB acted upon by the liquid consists of a concave-downward part CD and a concave-upward part DB. The net vertical component is the algebraic sum of the downward force and the upward force.
Noting that square DOCE upward less area DEC downward equals quadrant of circle DOC , the net vertical component is
Finally,
In this particular problem, the upward component (buoyant force) equals the weight of the displaced liquid to the left of the vertical plane COB.
3.13. Referring to Fig. 3-12, determine the horizontal and vertical forces due to the water acting on the 6-ft-diameter cylinder per foot of its
length.
Figure 3-12. Solution:
The hatched area (volume) is contained in each of the above volumes, one force being upward and the other downward. Thus they cancel, and
Dividing this volume into convenient geometric shapes,
If it is desired to locate this resultant vertical component, the principle of moments is employed. Each part of the 2500-lb resultant acts through the center of gravity of the volume it represents. By static mechanics, the centers of gravity are found and the moment equation is written (see Problems 3.7 and 3.9).
3.14. In Fig. 3-13, an 8-m-diameter cylinder plugs a rectangular hole in a tank that is 3 m long. With what force is the cylinder pressed against
the bottom of the tank due to the 9 m of water?
Figure 3-13. Solution:
3.15. In Fig. 3-14, the 8-ft-diameter cylinder weighs 500 lb and rests on the bottom of a tank that is 3 ft long. Water and oil are poured into the
left- and right-hand portions of the tank to depths of 2 and 4 ft, respectively. Find the magnitudes of the horizontal and vertical components of the force that will keep the cylinder touching the tank at B.
Figure 3-14. Solution:
3.16. The half-conical buttress ABE shown in Fig. 3-15 is used to support a half-cylindrical tower ABCD. Calculate the horizontal and vertical
components of the force due to water acting on buttress ABE .
Figure 3-15. Solution:
3.17.
A 1.2-m-diameter steel pipe, 6 mm thick, carries oil of sp gr 0.822 under a head of 120 m of oil. Compute ( a) the stress in the steel
and (b) the thickness of steel required to carry a pressure of 1.72 MPa with an allowable stress of 124 MPa. Solution:
3.18.
A wooden storage vat, 20 ft in outside diameter, is filled with 24 ft of brine, sp gr 1.06. The wood staves are bound by flat steel
bands, 2 in wide by
in thick, whose allowable stress is 16,000 psi. What is the spacing of the bands near the bottom of the vat, neglecting any
initial stress? Refer to Fig. 3-16.
Figure 3-16. Solution:
Force P represents the sum of all horizontal components of small forces dP acting on length y of the vat, and forces T represent the total tension carried in a band loaded by the same length y. Since the sum of the forces in the X direction must be zero, 2T (lb) – P (lb) = 0, or
Then
and
3.19. Refer to Fig. 3-17. What is the minimum width b for the base of a dam 100 ft high if upward pressure beneath the dam is assumed to vary
uniformly from full hydrostatic head at the heel to zero at the toe, and also assuming an ice thrust F 1 of 12,480 lb per linear foot of dam at the top? For this study make the resultant of the reacting forces cut the base at the downstream edge of the middle third of the base (at O) and take the weight of the masonry as 2.50 γ.
Figure 3-17. Solution:
Shown in the diagram are the H and V components of the reaction of the foundation, acting through O. Consider a length of 1 ft of dam, and evaluate all the forces in terms of γ and b, as follows:
To find the value of b for equilibrium, take moments of these forces about axis O. Considering clockwise moments positive,
3.20.
A concrete dam retaining 6 m of water is shown in Fig. 3-18(a). The unit weight of the concrete is 23.5 kN/m .3 The foundation soil
is impermeable. Determine ( a) the factor of safety against SS sliding, ( b) the factor of safety against overturning, and ( c) the pressure intensity on the base of the dam. The coefficient of friction between the base of the dam and the foundation soil is 0.48.
Figure 3-18. Solution:
3.6.2. Supplementary Problems 3.21. For an 8-ft length of gate AB in Fig. 3-19, find the compression in strut CD due to water pressure {B, C , and D are pins). Ans. 15.850 lb
Figure 3-19. 3.22. A 3.7-m high by 1.5-m wide rectangular gate AB is vertical and is hinged at a point 150 mm below its center of gravity. The total depth of
water is 6.1 m. What horizontal force F must be applied at the bottom of the gate for equilibrium? Ans. 15 kN
3.23. Find dimension z so that the total stress in rod BD in Fig. 3-20 will be not more than 18,000 lb, using a 4-ft length perpendicular to the
paper and considering BD pinned at each end. Ans. 5.87 ft
Figure 3-20. 3.24. A dam 20 m long retains 7 m of water, as shown in Fig. 3-21. Find the total resultant force acting on the dam and the location of the center
of pressure. Ans. 5541 kN, 4.667 m below water surface.
Figure 3-21.
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Schaum's Fluid Mechanics and Hydraulics Problem 3-24: Resultant Force on a Dam
This video illustrates how to calculate the resultant force acting on a solid surface due to the presence of a static fluid, as well as the determination of its location. Thom Adams, Ph.D., Professor, Mechanical Engineering, Rose-Hulman Institute of Technology 2013 Copy Link
3.25. Oil of specific gravity 0.800 acts on a vertical triangular area whose apex is in the oil surface. The triangle is 9 ft high and 12 ft wide. A
vertical rectangular area 8 ft high is attached to the 12-ft base of the triangle and is acted upon by water. Find the magnitude and position of the resultant force on the entire area. Ans. 83,300 lb; 12,18 ft down 3.26. In Fig. 3-22, gate AB is hinged at B and is 1.2 m wide. What vertical force, applied at the center of gravity of the 20-kN gate, will keep it in
equilibrium? Ans. 54 kN
Figure 3-22. 3.27. A tank is 20 ft long and of cross section shown in Fig. 3-23. Water is at level AE . Find ( a) the total force acting on side BC and (b) the total
force acting on end ABCDE in magnitude and position. Ans. 200.000 lb; at 11.17 ft
Figure 3-23. 3.28. Given the vertical rectangular gate with water on one side, shown in Fig. 3-24, find the total resultant force acting on the gate and the
location of the center of pressure. Ans. 84.59 kN, 3.633 m below water surface
Figure 3-24.
3.29. In Fig. 3-25, the 4-ft-diameter semicylindrical gate is 3 ft long. If the coefficient of friction between the gate and its guides is 0.100, find
the force F required to raise the 1000-lb gate. Ans. 347 lb
Figure 3-25. 3.30. A tank with vertical sides contains 0.914 m of mercury and 5.029 m of water. Find the total force on a square portion of one side 0.61 m by
0.61 m in area, half of this area being below the surface of the mercury. The sides of the square are horizontal and vertical. Ans. 21.8 kN, 5.069 m down 3.31. An isosceles triangle, base 18 ft and altitude 24 ft, is immersed vertically in oil of specific gravity 0.800 with its axis of symmetry
horizontal. If the head on the horizontal axis is 13 ft, determine the total force on one face of the triangle and locate the center of pressure vertically. Ans. 140,400 lb; 14.04 ft 3.32. How far below the water surface should a vertical square, 1.22 m on a side with two sides horizontal, be immersed so that the center of
pressure will be 76 mm below the center of gravity? What will be the total force on the square? Ans. 1.01 m, 23.7 kN 3.33. In Fig. 3-26, the 4-ft-diameter cylinder, 4 ft long, is acted upon by water on the left and oil of sp gr 0.800 on the right. Determine ( a) the
normal force at B if the cylinder weighs 4000 lb and ( b) the horizontal force due to oil and water if the oil level drops 1 ft. Ans. 1180 lb, 3100 lb to right
Figure 3-26. 3.34. For the inclined circular gate 1.0 m in diameter with water on one side, shown in Fig. 3-27, find the total resultant force acting on the gate
and the location of the center of pressure. Ans. 14.86 kN, 2.260 m below water surface measured along inclination of gate
Figure 3-27.
3.35. In Fig. 3-28, for a length of 8 ft determine the unbalanced moment about the hinge O due to water at level A. Ans. 18,000 ft-lb clockwise
Figure 3-28. 3.36. The tank whose cross section is shown in Fig. 3-29 is 1.2 m long and full of water under pressure. Find the components of the force
required to keep the cylinder in position, neglecting the weight of the cylinder. Ans. 14 kN down, 20 kN to left
Figure 3-29. 3.37. Determine, per foot of
length, the horizontal and vertical components of water pressure acting on the Tainter-type gate shown in Fig. 3-30.
Ans. 3120 and 1130 lb
Figure 3-30. 3.38. Find the vertical force acting on the semicylindrical dome shown in Fig. 3-31 when gage A reads 58.3 kPa. The dome is 1.83 m long. Ans.
113 kPa
Figure 3-31. 3.39. If the dome in Problem 3.38 is changed to a hemispherical dome of the same diameter, what is the vertical force acting? Ans. 60 kPa
3.40. Referring to Fig. 3-32, determine ( a) the force exerted by water on bottom plate AB of the 1-m-diameterriser pipe and ( b) the total force on
plane C Ans. 38.45 kN, 269 kN
Figure 3-32.
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Schaum's Fluid Mechanics and Hydraulics Problem 3-40: Resultant Force Versus Weight
This video demonstrates how to calculate the total force due to the contact of water with a plane surface. This force is compared to the total weight of the the water. Thom Adams, Ph.D., Professor, Mechanical Engineering, Rose-Hulman Institute of Technology 2013 Copy Link
3.41. The cylinder shown in Fig. 3-33 is 10 ft long. Assuming a watertight condition at A and no rotation of the cylinder, what weight of cylinder
is required to impede motion upward? Ans. 12,700 lb
Figure 3-33. 3.42. A wood stave pipe, 48 in inside diameter, is bound by flat steel bands 4 in wide and
in thick. For an allowable stress of 16,000 psi in the
steel and an internal pressure of 160 psi, determine the spacing of the bands. Ans. 12.5 in 3.43. For the parabolic seawall shown in Fig. 3-34, what moment about A per ft of wall is created by the 10-ft depth of water? ( γ = 64.0 lb/ft3)
Ans. 25,200 ft-lb counterclockwise
Figure 3-34. 3.44. The tank shown in Fig. 3-35 is 10 ft long, and sloping bottom BC is 8′ wide. What depth of mercury will cause the resultant moment about
C due to the liquids to be 101,300 ft-lb clockwise? Ans. 2 ft
Figure 3-35. 3.45. The gate shown in Fig. 3-36 is 6.10 m long. What are the reactions at hinge O due to the water? Check to see that the torque about O is
zero. Ans. 136 kN, 272 kN
Figure 3-36.
3.46. Refer to Fig. 3-37. A flat plate hinged at C has a configuration satisfying
the equation x2 + 1.5y = 9. What is the force of the oil on the plate, and what is the torque about hinge C due to the oil? Ans. 6240 lb; 16,400 ft-lb
Figure 3-37. 3.47. In Fig. 3-38, parabolic gate ABC is hinged at A and is acted upon by oil weighing 50 lb/ft 3. If the gate's center of gravity is at B, what must
the gate weigh per ft of length (perpendicular to the paper) in order for equilibrium to exist? Vertex of parabola is at A. Ans. 408 lb/ft
Figure 3-38. 3.48. In Fig. 3-39, automatic gate ABC weighs 1.50 tons/ft of length and its center of gravity is 6 ft to the right of hinge A. Will the gate turn
open due to the depth of water shown? Ans. Yes
Figure 3-39. 3.49. Referring to Fig. 3-40, calculate the width of concrete wall that is necessary to prevent the wall from sliding. The unit weight of the concrete is 23.6 kN/m 3, and the coefficient of friction between the base of the wall and the foundation soil is 0.42. Use 1.5 as the factor of safety
against sliding. Will it also be safe against overturning? Ans. 3.09 m, yes
Figure 3-40. 3.50. Solve Problem 3.20 assuming there is hydrostatic uplift that varies uniformly from full hydrostatic head at the heel of the dam to zero at
the toe. Ans. (a) 1.02; ( b) 1.81; ( c) p A = 173.5 kPa, pB = 14.5 kPa
