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Chapter I PROPERTIES OF FLUIDS

1. If certain gasoline weighs 7,000 N/m, what are the values of its density, specific volume, and specific gravity relative to water at 15°C? Given: Specific weight (w) =7kN/m³ Required: density ( ), volume of gasoline ( ) and specific gravity (s.g.) of gasoline. Solution: ;

`

2. A certain gas weighs 16 N / at a certain temperature and pressure. What are the values of its density, specific volume, and specific gravity relative to air weighing 12 N / ? Given: Weigh of gas ( ) Weigh of air ( ) Required: density of gas ( ), Volume of gas ( ) and specific gravity (s.g.) of gas. Solution: ; 1.6310 kg/m3

0.6131m3/kg

3. If 5.30 of a certain oil weighs 43,860 N. Calculate the specific weight,density and specific gravity of this oil. Given: Required: specific weight (w) =? Density ( =? Specific gravity (Sg) =? Solution:

)

4. The density of alcohol is 790 and specific volume.

. Calculate its specific weight, specific gravity

Required: specific weight (w) =? Specific gravity (Sg) =? specific volume (v) =? Given: Solution: For specific weight(w)

For specific gravity (sg)

For specific volume (v)

5. A cubic meter of air at 101.3 KPa and 150C weighs 12 N/m3. What is its specific volume? Given: Required: Solution: ; =

6. At a depth of 8 km in the ocean the pressure is 82.26 MPa. Assume the specific weight at the surface to be 10.10kN/m3 and that the average bulk modulus is 2344 MPa for that pressure range. (a) What will be the change in specific volume between that at the surface and at the depth? (b) What will be the specific volume at that depth? (c) What will be the specific weight at that depth?

Solution: (a).

(

)

7. To two significant figures what is the bulk modulus of water in kN/m2 at 50°C under a pressure of 30 MN/m2? Given: pressure of water ( ) = 30 MN/m2 Required: bulk modulus=? Solution: By Interpolation 2

Pressure, P (MN/m ) 10.340 31.030 30 Bulk modulus= 2.49 kN/m2

Bulk Modulus, Ev (MN/m2) 2560 2500 2493.03

8. If the dynamic viscosity of water at 20˚ C is 1x10-3 N-s/m2, what is the kinematic viscosity in the English Units? Required: Solution:

9. The kinetic viscosity of 1ft2/sec is equivalent to how many strokes? (1 strokes = 1cm2/sec) Solution: Convert ft2/sec to cm2/sec 1 ft2/sec x ((1 cm)2/(0.0328ft)2) = 929.0304 cm2/sec or 929 cm2/sec 1ft2/sec = 929 strokes

10. A volume of 450 litres of a certain fluid weighs 3.50 kN. Compute the mass density. (1m3 = 1000 litres). Given:

volume of fluid ( = 450 L Weight = 3.5 kN

Required: mass density of fluid Solution: 450 L x (1m3/1000L) = 0.45m3 ⁄ =

⁄

11. Compute the number of watts which are equivalent to one horsepower. (1 HP = 550 ft-lb/s; 1 W = 107 dyne-cm/s; 1 lb = 444,800 dynes. Solution:

=

Watts

12. A city of 6000 population has an average total consumption per person per day of 100 gallons. Compute the daily total consumption of the city in cubic meter per second. (1 ft3 = 7.48 gallons) Solution: = =

13. Compute the conversion factor for reducing pounds to Newtons. Solution: =

Chapter 2 Principles of Hydrostatics

1. If the pressure 3 m below the free surface of a liquid is 140 kPa, calculate its specific weight and specific gravity. Solution:

2. If the pressure at a point in the ocean is 1400 kPa, what is the pressure 30 m below this point? The specific gravity of salt water is 1.03 Solution:

3. An open vessel contains carbon tetrachloride (s = 1.50) to a depth of 2 m and water above this liquid to a depth of 1.30 m. What is the pressure at the bottom?

Solution:

4. How many meters of water are equivalent to a pressure of 100 kPa? How many cm of mercury? Solution:

5. What is the equivalent pressure in kPa corresponding to one meter of air at 15° C under standard atmospheric conditions? Solution:

6. At sea level a mercury barometer reads 750 mm and at the same time on the top of a mountain another mercury barometer reads 745 mm. The temperature of the air is assumed constant at 15° C and its specific weight assumed uniform at 12 N/m3. Determine the height of the mountain. Solution:

7. At ground level the atmospheric pressure is 101.3 kPa at 15° C. Calculate the pressure at a point 6500 m above the ground, assuming (a) no density variation, (b) an isothermal variation of density with pressure.

Solution: a)

b)

8. If the barometer reads 755 cm of mercury, what absolute pressure corresponds to a gage pressure of 130 kPa?

Solution: +

9. Determine the absolute pressure corresponding to a vacuum of 30 cm of mercury when the barometer reads 750 mm of mercury.

Solution:

10. Figure A shows two closed compartments filled with air. Gage (1) reads 210 kPa, gage (2) reads 25 cm of mercury. What is the reading at gage (3)? Barometric pressure is 100 kPa. Solution: P3 = P1 + P2 + Pb P1 = 210 kPa P2 = 25cmHg *10mmHg/1cmHg*101.325 kPa/760mmHg P2 =33.25 kPa 11. If the pressure in a gas tank is 2.50 atm, find the pressure in kPa and the pressure head in meter of water.

Solution:

12. The gage at the suction side of a pump shows a vacuum of 25 cm of mercury. Compute (a) pressure head in meter of water, (b) pressure in kPa, (c) absolute pressure in kPa if the barometer reads 755 cm of mercury.

Solution: a) b)

(

)

13. Oil of sp. gravity 0.80 is being pumped. A pressure gage located downstream of the pump reads 280 kPa. What is the pressure head in meter of oil? Solution:

14. The pressure of the air inside a tank containing air and water is 20 kPa absolute.

Determine the gage pressure at a point 1.5 m below the

water surface. Assume standard atmospheric pressure. Solution:

15. A piece of timber 3 m long and having a 30 cm by 30 cm section is placed in a body of water in a vertical position. If the timber weighs 6.5 kN/m3, what vertical force is required to hold it with its upper end flush with the water surface?

Solution:

[

]

16. A glass tube 1.60 m long and having a diameter of 2.5 cm is inserted vertically into a tank of oil (sp. gravity 0.80) with the open end down and the closed end uppermost. If the open end is submerged 1.30 m from the oil surface, determine the height to which the oil will rise in the tube. Assume barometric pressure is 100 kPa and neglect vapour pressure. Solution:

Before the glass tube was inserted:

After the glass was inserted:

(

)

*

+

17. A gas holder at sea level contains illuminating gas under a pressure equivalent to 5 cm of water. What pressure in cm of water is expected in a distributing pipe at a point 160 m above sea level? Consider standard atmospheric pressure at sea level and assume the unit weights of air and gas to be constant at all elevations with values of 12 N/m3 and 6 N/m3, respectively. Solution:

18. If the barometric pressure is 758 mm of mercury, calculate the value of h of Figure B. Solution:

19. The manometer of Figure C is tapped to pipeline carrying oil (sp. gravity 0.85). Determine the pressure at the center of the pipe.

Solution:

20. Determine the gage reading of the manometer system of Figure D. Solution:

21. In Figure E, calculate the pressure at point m. Solution:

22. In Figure F, find the pressure and pressure head at point m: fluid A is oil (s = 0.90), fluid B is carbon tetrachloride (s = 1.50) and fluid C is air.

Solution:

23. Compute the gage and absolute pressures at point m of Figure G: fluids A and C are air, fluid B is mercury.

Solution:

24. The pressure at point m is increased from 70 kPa to 105 kPa. This causes the top level of mercury to move 20 cm in the sloping tube. What is the inclination Ɵ? Figure H. Solution:

[

]

25. In Figure I, determine the elevation of the liquid surface in each piezometer Solution: .

26. In Figure J, fluid A is water, fluid B is oil (s = 0.85). Determine the pressure difference between points m and n.

Solution:

27. In Figure K determine ⍴m-⍴n. Solution:

28. In Figure L, fluid A has a specific gravity of 0.90 and fluid B has a specific gravity of 3.00. Determine the pressure at point m.

Solution:

29. If the pressure at m obtained from Problem 28 is increased by 7 kPa, how many cm will fluid B in the 12 mm tube? Solution: 7kPA = 3(9.81)h 7kPa = 29.4h h = 24cm

30. In Figure M, fluid A is a gasoline (s = 0.70), fluid B is mercury. Find the pressure head difference between points m and n.

31. In Figure N, x = 25 cm initially. If the pressure at m is increased by 35 kPa while maintaining the pressure at n constant, calculate the new value of x. Solution:

32. The diameters of the cylinders of the hydraulic jack of Figure O are 7.5 cm and 60 cm, respectively. What force F is required to maintain equilibrium if the load weighs 35 kN? Solution:

33. A 5 cm pipe is connected with the end of a cylinder having a diameter of 50 cm. There is a piston in the pipe and a piston in the cylinder, the space between being filled with oil (s = 0.80). The larger piston is connected by a rod with a 5 cm piston in a third pipe, the two pipes and cylinder having their axes horizontal and collinear. A force of 100 N is applied to the piston in the first pipe. What pressure is affected in the third pipe?

34.Assuming normal barometric pressure, how deep is the ocean at appoint where an air bubble, upon reaching the surface, has six times the volume that it had at the bottom? Specific gravity of sea water is 1.03. Solution: l

35. A bottle consisting of a cylinder 30 cm in diameter and 30 cm high has a neck 5 cm in diameter and 30 cm long. If the bottle, filled with air under normal atmospheric condition, is inverted and submerged in water until the neck is just filled with water, find the depth to which the open end is submerged. Neglect vapour pressure. Solution: Before the bottle was inserted:

When the bottle is inserted:

(

)

depth to which the open end is submerged =

CHAPTER IV ACCELERATED LIQUIDS IN RELATIVE EQUILIBRIUM

1. A car travelling on a horizontal road has a rectangular crosssection, 6m long by 2.40m wide by 1.50m high. If the car is half full of water, what is the maximum acceleration it can undergo without spilling any water? Neglecting the weight of the car, what force is required to produce maximum acceleration? Solution: F= Ma F= ρVa F= (1)(0.75)(6)(2.4)(2.45) = 14°

F= 26.5 KN

a= 2.45 m/s²

2.

A cylindrical bucket is accelerated upward with an acceleration of gravity. If the bucket is 0.60m in diameter and 1.20 m deep, what is the force on the bottom of the bucket if it contains 0.90 m depth of wet concrete whose specific weight is 22,000 N/m3? What is force on the bottom idf the bucket is accelerated downward at 9.81 m/s2? Solution: ∑

3. A rectangular car is 3 m long by 1.5 m wide and 1.5 m deep. If friction is neglected and the car rolls down a plane with an inclination of 20°, what is the inclination of the water surface if the car contained 0.60 m depth of water when the car was horizontal?

Note:

4.

An open tank, 9.15 m long is supported on a car moving on a level track and uniformly accelerated from rest to 48 km/hr. When at rest the tank was filled with water to within 15 cm of its top. Find the shortest time in which the acceleration may be accomplished without spilling over the edge. Solution:

= 1.87788°

(

)(

)(

)

5. A rectangular tank, 60 cm long and containing 20 cm of water is given an acceleration of a quarter of the acceleration of gravity along the length. How deep will the water be at the rear end? At the front end? What is the pressure force at the rear end if it is 45 cm wide? Solution:

= 14.0362° a.) =27.5 cm b.) 12.5 cm

6. Fig. GG shows a container having a width of 1.50 m. Calculate the total forces on the ends and bottom of the container when at rest and when being accelerated vertically upward at 3 m/s². Solution: Sin 60°

z= 1.5011 m

x = 0.7506 When at rest,

Force at the bottom,

Vertically upward, (

)

(

)

7. A closed rectangular tank 1.20 m high by 2.40 m long by 1.50 m wide is filled with water and the pressure at the top is raised to 140 KPa. Calculate the pressures in the corners of this tank when it is accelerated horizontally along its length at 4.60 m/s². Solution :

= 25.1223°

At

At

+ 140

8. A pipe 2.50 cm in diameter is 1.0 m long and filled with 0.60 m of water. If the pipe is capped at both ends and rotated in a horizontal plane about a vertical axis through one end of the pipe, what is the pressure at the other end of the pipe when it is rotating at 200 RPM?

9. An open vertical cylindrical tank 0.60 m in diameter and 1.20 m high is half full of water. If it is rotated about its vertical axis so that water just reach the top, find the speed of rotation. What will then be the maximum pressure in the tank? If the water were 1.0 m deep, what speed will cause the water to just reach the top? What is the depth of water at the center? Required: Depth of water=?

Solution:

h=

w = 16.17 kN (

)

h = 0.4

w =9.34 rad/s Depth= 1.2m – 0.4 m Depth D epth Depth=0.8m Depth 10. If the tank of Prob. 9 is half full of oil (sp.gr. 0.75) what speed of rotation is necessary to expose one- half of the bottom diameter? How much oil is lost in attaining this speed? Solution:

Y= 0.4 m

w = 18. 68 rad/s = 0.169646

V= 41.41 L

11. The U-tube of figure HH is given a uniform acceleration of 1.22 m/s² to the right. What is the depth in AB and the pressures at B, G and D? Solution: (

)

(

(

)

)

12. The U-tube of fig. HH is rotated about an axis through HG so that the velocity at B is 3 m/s. What are the pressures at B and G?

13. The U-tube of fig. HH is rotated about HG. At what angular velocity does the pressure at G become zero gage? What angular velocity is required to produce a cavity at G? Solution: h= 0.3 =

14. The tank of Prob. 9 is covered with a lid having a small hole at the center and filled with water. If the tank is then rotated about its vertical axis at 8 rad/s, what is the pressure at any point on the circumference of the upper cover? Of the lower cover?

15. The tank of Prob. 9 contains 0.60 m of water covered by 0.30 m of oil (sp. Gr. 0.75) What speed of rotation will cause the oil to reach the top? What is then the pressure at any point on the circumference of the bottom? Solution: h= 0.6 = w = 11.44 rad/s P= w P= (0.75)(9.81)(0.3)+(9.81)(0.9) P= 11.04 kPa 16.The tube of figure II is rotated about axis AB. What angular velocity is required to make the pressures at B and C equal? At that speed where is the location of the minimum pressure along BC? Solution : h= 0.3 =

17. A vessel 30 cm in diameter and filled with water is rotated about its vertical with such a speed that the water surface at a distance of 7.50 cm from its axis makes an angle of 45° with the horizontal?

18. A cylindrical vessel, 0.30 m deep, is half filled with water. When it is rotated about its vertical axis with a speed of 150 RPM, the water just rises to the rim of the vessel. Find the diameter of the vessel. Required:

Solution: w = 150 RPM( ) w = 15.70796 rad/sec 0.3 = r = 0.15445 D= 2r D=0.3089 m D= 30.89 cm 19. A conical vessel with vertical axis has an altitude of 1 m and is filled with water. Its base, 0.60 m in diameter, is horizontal and uppermost. If the vessel is rotated about its axis with a speed of 60 RPM, how much water will remain in it? Solution: 60 x h= h= h = 0.18 m V= =

= = = 0.094 m³

20. A cylindrical bucket, 35 cm deep and 30 cm in diameter, contains water to a depth of 30 cm. A man swings this bucket thru a vertical plane, the bottom of the bucket describing a circle having a diameter of 2.15 m. What is the minimum speed of rotation that the bucket can have without permitting water to spill? Solution: CF= W CF= ma CF= W= r= r = 0.925 W= g= 9.81 =

21. If the water which just fills a hemispherical bowl of 1.0 m radius be made to rotate uniformly about the vertical axis of the bowl at the rate of 30 RPM, determine the amount of water that will spill out? Solution:

30 RPM x h= h= h =0.503 m V= =

=

22. The open cylindrical tank of fig. JJ is rotated about its vertical axis at the rate of 60 RPM. If initially filled with water, how high above the top of the tank will water rise in the attached piezometer? Solution:

w =60 RPM x

2.012 m

y = 2.012- 0.85 y = 1.16 m 23. A closed cylindrical tank with axis vertical, 2 m high and 0.60 m in diameter is filled with water, the intensity of pressure at the top being 140 KPa. The metal making up the side is 0.25 cm thick. If the vessel is rotated about its vertical axis at 240 RPM, compute a) total pressure on the side wall, b) total pressure against the top, c) maximum intensity of hoop tension in pascals.

24. A small pipe, 0.60 m long, is filled with water and capped at both ends. If placed in a horizontal position, how fast must it be rotated about a vertical axis, 0.30 m from one end, to produce maximum pressure of 6,900 KPa? Solution:

3450= 9.81h h = 351.68 m

w = 276.89 rad/s

25. A vertical cylindrical tank 2 m high and 1.30 m in diameter, two – thirds full of water, is rotated uniformly about its axis until it is on the point of overflowing. Compute the linear velocity at the circumference. How fast will it have to rotate in order that 0.170 m³ of water will spill out? Solution:

h= h=

h= 1.3333 =

26. A steel cylinder, closed at the top, is 3 m high and 2 m in diameter. It is filled with water and rotated about its vertical axis until the water pressure is about to burst the sides of the cylinder by hoop tension. The metal is 0.625 cm thick and its ultimate strength is 345 MPa. How fast must the vessel be rotated? Solution:

345x10³ =

(

)

2156.25 = 9.81(1)

√

27. A conical vessel with axis vertical and sides sloping at 30° with the same is rotated about another axis 0.60 m from it. What must be the speed of rotation so that water poured into it will be entirely discharged by the rotative effect? Given: x= 0.6 m g= 9.81 m/s² =90°-30°

Tan = Tan60°= w= √ w = 5.32 rad/s

CHAPTER III HYDROSTATIC FORCE ON SURFACES

1. A rectangular plate 4m by 3m is immersed vertically with one of the longer sides along the water surface. How must a dividing line be drawn parallel to the surface so as to divide the plate into two areas, the total forces upon which shall be equal. Given: ω=1 ħ = = 1.5 Req’d: h=?

A = 4 x 3 = 12

Solution: FT = ω ħ A FT = (9.81) ( 1.5) (12) FT = 176.58 KN FT = F1 + F2 FT = 2F1 176.58 = 2(9.81)(h/2)(4)(h) h = 2.12m

2. A triangle of height h and base b is vertically submerged in a liquid. The base b coincides with the liquid surface. Derive the relation that will give the location of the center of pressure. Req’d: =? Solution:

Ig= bh3/36 e = (bh3/36) / (1/2bh* 1/3h) e = h/6 Yp = h + (h/6) Yp = h/2 from the Lower Surface

3. The composite area shown in Figure A is submerged vertically in a liquid with specific gravity 0.85. Determine the magnitude and location of the total hydrostatic force on one face of the area.

Given: Sg = 0.85 Req’d:

=? F=? Solution: F1 = whA F1 = (9.81)(0.85)(3.25)(1.5)(3.5) F1 = 142.28kN F2 = whA F2 = (9.81)(0.85)(4.25)(1.5)(1.5) F2 = 79.74kN F = F1 + F2 F = 142.28kN + 79.74kN F = 222.02kN yp = ̅ yp= (1.75 +

)+ 1.5

yp= 2.33 + 1.5 yp = 3.83m 4. The gate in Figure B is subjected to water pressure on one side and to air pressure on the other side. Determine the value of x for which the gate will rotate counter clockwise if the gate is (a) rectangular, 1.5 m by 1.0 m base and 1.0 m high. Given: V = 1.5 x 1.0 x 1.0 ħ = 0.5 Req’d: x=? Solution: Pair – PH2O = (1.5) (1.0) (9.81) (x+0.5+e) 1.5(1) – 30 kN/m2 = (1.5)(1)(9.81)(x+.5+.17) 30=9.81(x+0.67) 30=9.81x = 6.54 9.81x = 23.46 X= 2.39 ≈2.40

b. Pair = PH2O = ½ (1.5) (1.0) (9.81) (x+.05+e) ½ (1.5) (1) – 30 = 9.81 (x+²/ (0.5) + 0.17) 30= 9.81 (x+.5) 9.81x = 30 – 4.905 9.81x = 25.095 X = 2.558 ≈ 2.60

5. A vertical circular gate 1 m in diameter is subjected to pressure of liquid of specific gravity 1.40 on one side. The free surface of the liquid is 2.60 m above the uppermost part of the gate. Calculate the total force on the gate and the location of the center of pressure.

Given: D=1m sg = 1.4 y = 2.6 m Req’d: F=? e=? Solution:

(

𝐞

)

( )

𝟎 𝟎𝟐 𝐛𝐞𝐥𝐨𝐰 𝐭𝐡𝐞 𝐜𝐞𝐧𝐭𝐫𝐨𝐢𝐝

6. A horizontal tunnel having a diameter of 3m is closed by a vertical gate. When the tunnel is (a) ½ full (b) ¾ full, of water, determine the magnitude and location of the total force. Given: D = 3m Req’d: F=? Solution: a. h1/2 = h1/2 = h1/2 = 0.6366 A=

=

F =9.81(0.6366)( ) F = 22.10 kN

b. h3/4=( + )/2 = + = 0.954929 A= =

)/2

F = 9.81(0.95429) ( F = 33.10 kN

)

7. In Figure C is a parabolic segment submerged vertically in water. Determine the magnitude and location of the total force on one face of the area. Req’d: F=? yp = ? Solution: F = ЋA F = 9.81(

(

F = 9.81( )( )(3)(3) F = 106 kN e= e= e=

= 0.34

Ῡ+e= + .34 yp = 2.14 below water surface 8. A sliding gate 3 m wide by 1.60 m high is in a vertical position. The coefficient of friction between the gate and guides is 0.20. If the gate weighs 18 kN and its upper edge is 10 m below the water surface, what vertical force is required to lift it? Neglect the thickness of the gate. Given: A = (3m)(1.6) = 4.8 μ = 0.20 Req’d: =? Solution: F = AωЋ Ћ = 0.8 + 0.8 + 10 Ћ = 10.8 F = 4.8 (9.81) (10.8) F= 508.55

W = 18 kN Ћ = 10 m

Ff = μ N Ff = (0.20) (508.55) Ff = 101.71 ∑ Fy = 0 = W + Ff = 18 + 107. 71 = 119.71 ≈ 120

9. The upper edge of a vertical trapezoidal gate is 1.60 m long and flush with the water surface. The two edges are vertical and measure 2 m and 3 m, respectively. Calculate the force and location of the center of pressure on one side of the gate. Given: b = 1.60m = 2m = 3m Req’d: F=? Solution:

( )

(

)(

)

10. How far below the water surface is it necessary to immerse a vertical plane surface, 1 m square, two edges of which are horizontal, so that the center of pressure will be located 2.50 cm below the center of gravity. Given: e = 0.025 m A = 1m square Req’d: Ῡ=? Solution: e = Ig / AῩ ; Ig=bh3/ 12 = 0.08 0.025 Ῡ = 0.08 / 1Ῡ 0.025 Ῡ = 0.08 Ῡ = 3.33 - .5 Ῡ = 2.83 m

11. The gate shown in Figure D is hinged at B and rests on a smooth surface at A. If the gate is 1.60 m wide perpendicular to the paper, find BH and BV. Given: w = 1.60m Req’d: BH = ? BV = ? Solution: Pω = 9.81 (2.8) Pω = 27.468 P = 27.468 (1.6) (3.6) P = 158.22 kN P = ωЋA 158.22 = 9.81h (1.6) (3.6) 158.22 = 56.5056h h = 2.8 Ῡ = Ћ / sin 56.31 Ῡ = 3.37 e = Ig/ AῩ =

⁄

e = 0.32 x = 1.8 + e x = 2.12 ∑ MB = 0 158.22 (2.12) – F (2) 2F = 335.4264 F = 167. 71 kN ∑Fy = 0 = 167.71 – 158.22 cos 56.31 – Bv Bv = 79.95 ≈ 80 kN ∑Fx = 0 BH = 158.22 sin 56.31 BH = 131.65 ≈132 kN 12. In Figure E gate AB is 2m wide perpendicular to the paper. Determine FH to hold the gate in equilibrium. Req’d: FH = ? Solution: P = (9.81) (0.8) (0.6) P = 4.7 P2= P1 + wh = 4.7 + 9.81 (2) P2= 24.4 h = 2.44 / 9.81 h = 2. 49 3.2 sin60 = 2.77 h of gate = 2.49 / sin 60 h of gate = 2. 88 Ћ= 2.49 / 2 Ћ = 1. 245 P = AωЋ = 5.75 (9.81) (1.245) P = 70. 23 e= e=

⁄

e = 0.48 ∑M = 0 2.77 Fh = P (1.44 – 0.48)

Fh = 24.34 kN

13. Calculate the resultant of the hydrostatic forces on the gate of Figure F. What vertical force P is necessary to lift the gate at point A? Width of gate is 1.60 m. Given: w = 1.60 Req’d: P=? Solution: Ћ1 = 3/2 + 1/3 Ћ = 2.8 A1 = 1.6 (3.6) A1 = 5. 76 m2 F1 = ωhA F1 = (9.81) (2.8) (5.76) F1 = 158.22 Ћ =1.5 A2 =5.76 F2 =Aωh F2 =84.76 KN a= a= a =2.1 = (3.437) = 0.314 b= b = 2.4 Ῡ =Ћ Ῡ =2.36/sin 56.31 Ῡ =3.437 ∑ Mb =0 2P = F1(a)- F2(b) 2P =158.22(2.12)-84.76(2.4)

2P = 132KN P = 66KN P=66 kN to open the gate

14. The width of the gate of Figure G normal to the paper is 3m. What vertical force must be applied at “a” to prevent collapse when h = 6 m? Neglect weight of the gate. What is the stress in strut BC? Given: w = 3m h = 6m Req’d: =? ς=? Solution:

15. The gate of Figure H is hinged at A and rests on a smooth surface at B. The gate is circular having a diameter of 3 m, determine the value of the vertical having a diameter of 3 m. Determine the value of the vertical force P that will open the gate at B. Given: D = 3m Req’d: P=? Solution: Y = 3sin 45 Y =2.12 h = 2.12/2+1.16 h =2.66m x =3cos 45 x =2.12 A =π =π =7.06m2 F = ЋA

=9.81(.8)(2.66)(7.06) =147.382 m e= = =1.5 a= 3/2 + e = 1.5 +.15 = 1.65 Ῡ =h/sinθ Ῡ = 2.66/sin 45 Ῡ =3.76 ∑Ma=0 2.12 P = F(a) 2.12 P =147.382(1,67) P = 114.707 P-7 =107.707 KN P=107.707 KN to open the gate

16. A triangular gate having a horizontal base of 1.30 m and an altitude of 2 m is inclined 45° from the vertical with the vertex pointing upward. The base of the gate is 2.60 m below the surface of oil (s = 0.80). What normal force must be applied at the vertex of the gate to open it? Given: b = 1.30m h = 2m

ϴ = 45° s = 0.80

Req’d: N=? Solution: Sin 45= c = 2.8284 F = gЋ F = 9.81(0.8)(1/3(2))(1/2(1.3)(2.8284)) F = 9.6188 N = 9.6188KN sin 45 N = 6.80 kN

17. What depth of water will cause the rectangular gate of Figure I to fall? Neglect weight of the gate. Req’d: d=? Solution:

( )

(

( (

)

)

)(

)

∑ (

) (

)

18. Determine the horizontal and vertical components of the total force on the gate of Figure J. The width of the gate normal to the paper is 2 m. Given: w = 2m Req’d: FH = ? FV = ? Solution: h=6 FH= whA FH= (9.81) (3) (6) (2)

FH = 353.16 kN FV= wV FV= (9.81){[(60/360)(6)2(π)+-[(1/2)(6)(6sin60°)]}(2) FV= (9.81) (6π – 15.6) (2) FV = 63.76 kN

19. The corner of a floating body has a quarter cylinder AB having a length normal to the paper of 3 m. Calculate the magnitude and location of each of the components of force on AB. Figure K. Given: L = 3m Req’d: FT = ? FV = ? Solution: FT = whA FT = (9.81) (3.25) (1.5) (3) FT = 143.47 kN FV= wV FV= (9.81) [(4)(1.5)-((π) (1.52))/4](3) FV = 124.57 kN

20. The cylindrical gate of Figure L is 3m long. Find the total force on the gate. What is the minimum weight of the gate to maintain equilibrium of the system? Given: L = 3m Req’d: F=? =? Solution: x = 1.5sin30° = 0.75m y = 1.5cos30° = 1.3m

FH = FH1 + FH2 FH = (1.5)(3)(9.81)(1.5/2) – (0.2)(3)(2/2)(9.81) FH = 33.11 – 0.5886 FH = 32.52 KN A = (1/4) πr2 + (1/2)bh + (θ/360)πr2 A = 1.77 + 04875 + 0.6 A = 2.86 m2 V = 2.86(3) = 8.58m3 Fv = wV Fv = (9.81)(8.58) Fv = 84.17 √ F= = 90.23 KN 21. Determine the magnitude and direction of the force on the radial gate of Figure M. AB has a length of 2 m normal to paper. Given: l = 2m Req’d: F=? ϴ=? Solution: FH= whA FH= (9.81) (1.35) (2.7) (2) FH = 71.51 KN FV= wV FV= (9.81){[(1/2)(1.53)(2.70)] + *(π)(30)(6)2/360]- [(1/2)(3.10)(5.8)]}(2) FV = 48.30 KN √ √

ϴ = tan-1(48.30/71.51) ϴ= 34° from the horizontal

22. End AB of Figure N has a section in the shape of a quadrant. If the tank has a length of 3 m, determine the total force acting on the end. Given: l = 3m Req’d: FH = ? F=? Solution: Fv = wV Fv = (9.81)(3.1416)(3) Fv= 92.46 KN A = (1/4) πr2 A = 3.1416 m2 √ √

e = IG/Ay = (2/6)(3) Ig= bh3/36 = 2 e = 0.11m FH = whA FH = (9.81)(3)(2)(3) FH = 176.58 KN

23. The gate of Figure O is 3 m long. Find the magnitude and location of the horizontal and vertical of the forces components on the gate AB.

Req’d: FH = ?

FHy = ?

FV = ?

FVx = ?

Solution: FH= whA FH= (9.81) (6.36) (2.12/2)

FH = 66.14 KN FHy = 2.12 ( y + e) ; e = IG/Ay FHy = 2.12 [1.06 + (3)(2)(12)3/(3)(2.12)(1.06)] FHy = 2.12 – 1.41 FHy = 0.71 above O FV= wV FV= (9.81) (0.58) (3) FV = 17.15 KN ∑ (66.14)(0.71) = (17.07)(FVx) FVx = 2.75m left of O

24. A pyramidical object having a square base (2 m on a side) and 1.50 m high weighs 18 kN. The base covers a square hole (2 m on a side) at the bottom of a tank. If water stands 1.50 m in the tank, what force is necessary to lift the object off the bottom? Assume that atmospheric pressure acts on the water surface and underneath the bottom of the tank. Given: A=4 h = 1.50 m Wobj= 18kN Req’d: BF = ?

Solution: V = 1/3Abaseh V = 1/3 (1.5) V=2 Fv = wV Fv= (9.81)(2) Fv = 19.62kN BF = Wobj + Fv BF = 18kN + 19.62kN BF = 37.62kN

25. The hemispherical dome of Figure P surmounts a closed tank containing a liquid of specific gravity 0.75. The gage indicates 60 kPa. Determine the tension holding the bolts in place. Given: s g = 0.75 = 60 kPa Req’d: F=? Solution: F =9.81(0.75)((π(1.5)2(2)-(2/3)π(1.5)3 F = 52.0071 KN (2) F = 104.0142 Heq= P/Y = 60 KPA/9.81(0.75) = 8.15499-1.6 H = 6.5549 F = y Vw F = π(1.5)2(6.5549)+ 1/2*(4/3)π(1.5)3 V = 53.402 m3 F = 9.81(0.75)(53.402) F = 392.91 KN F = 392.91 KN – 104.0142KN F =288.89 KN 26. Figure Q shows a semi-conical buttress. Calculate the components of the total force acting on the surface of this semi-conical buttress. Req’d: F=? Solution FH = whA FH = (9.81)(1.3+1.5)(1.5)(3) FH = 123.606kN Fv = wV Fv = (9.81)(1/2)(1.5)(3)(1.5) Fv = 33.103kN F= F=√ F = 127.963kN

27. In Figure R a circular opening is closed by a sphere. If the pressure at B is 350 kPa absolute, what horizontal force is exerted by the sphere on the opening? Given: P = 350kPa Req’d: F=? Solution: F = PA F = 350(0.2)(0.25) F = 17.5 KN 28. Calculate the force required to hold the cone of Figure S in position. Req’d: F=? Solution:

PA = 3.5kPa – (9.81)(0.8)(1.5) PA = -8.272kPa F1 = PAA F1 = 8.272(π)( F1 = 3.65kN F2 = (9.81)(0.8) F2 = 1.156kN w=π w = 8.67kN

(2.5)(9.81)(0.8)

∑ F + F1 + F2 = w F + 3.65 + 1.156 =8.67 F = 3.864kN

29. The section of a gate at the gate at the end of a tank is shown in Figure T. It has a length of 3 m normal to the paper and hinged at O. If the depth of the water is 2.50 m, calculate P to maintain equilibrium.

Given: l = 3m h = 2.50m Req’d: P=? Solution: FH = (2.5)(3)(1.25)(9.81) FH = 91.969 kN Fv = 9.81 [(1.45)(2.5)-⅔(1.45)(2.5)](3) Fv= 35.56 kN 30. A pipe having a diameter of 30 cm is exposed to an atmospheric pressure of 70 cm of mercury, while the pressure inside is 33 cm of mercury. What is the maximum allowable internal pressure in the pipe? Given: D = 30cm = 70 kPa = 33 kPa Req’d: P =? Solution: Patm = 70(101.3/76) Patm = 93.303 kPa Pg = 33(101.3/76) Pg = 43.986 kPa FB = (93.303-43.986)(0.3) FB = 14.795 N/m FB = PiD 14.795 = Pi (0.3) Pi = 49.317 kPa

31. A steel pipe having a diameter of 15 cm and a wall thickness of 9.50 mm has an allowable stress of 140,000 kPa. What is the maximum allowable internal pressure in the pipe? Given: d = 15cm t = 9.50mm ς = 140,000kPa Req’d: P=? Solution:

Pi = wh S = (PD/ 2t) 140,000 KPa = P(0.15)/2(9.5E-3) P = 17,733.33 KPa or 17.73 MPa

32. A pipe carrying steam at a pressure of 7,000 kPa has an inside diameter of 20 cm. If the pipe is made of steel with an allowable stress of 400,000 kPa, what is the factor of safety if the wall thickness is 6.25 mm? Given: l = 3m h = 2.50m Req’d: P=? Solution:

FB = (Pi – Pe)D FB = (7000-0)(0.2) FB = 1400kN T = FB/2 T = 1400kN/2 T = 700kN S = T/t S=

S = 112000kPa FS = SA/S FS = FS = 3.57 33. A 60 cm cast-iron main leads from a reservoir whose water surface is at El. 1590 m. In the heart of the city the main is at EL 1415m. What is the stress in the pipe wall if the thickness of the wall is 12.5 mm and the external soil pressure is 520 kPa? Assume static conditions. Solution: S = [ (1716.75-520) (0.6) / 2 ] / 0.0125 S = 28.7 MPa

34. Compute the stress in a 90 cm pipe with wall thickness of 9.50 mm if the water fills it under a head of 70 m. Solution: S = (9.81)(70)(900) / (2)(9.5) S = 32527.89474 KPa or 32,530 KPa S = 32.53 MPa 35. A wood stave pipe, 120 cm in inside diameter, is to resist a maximum water pressure of 1,200 kPa. If the staves are bound by steel flat bands (10 cm by 2.50 cm), find the spacing of the bands if its allowable stress is 105 MPa. Solution: S = [(2)(10.5E-3)(10)(2.5)] / (1200)(120) S = 36.46cm

36. A continuous wood stave pipe is 3 m in diameter and is in service under a pressure head of 30 m of water. The staves are secured by metal hoops 2.50 cm in diameter. How far apart should the hoops be spaced in order that the allowable stress in the metal hoop of 105 MPa be not exceeded? Assume that there is an initial tension in the hoops of 4.50 kN due to cinching. Given: D = 3m S = 2.5 cm Allowable = 105 MPa Solution:

T = FB / 2 T = 98.1 / 2 T = 49.05 S = [(105) (2.5)] / (49.05) S = 10.6 cm 37. A vertical cylindrical container, 1.60 m in diameter and 4 m high, is held together by means of hoops, one at the top and the other at the bottom. A liquid of specific gravity 1.40 stands 3 m in the container. Calculate the tension in each hoop. Solution: ∑ 2T2 (4) = F (3) F = whA F = (9.81)(1.04)(1.5)(3)(1.6) F = 98.88 KN 2T2(4) = (98.88)(3) T2= 37.08 KN at the bottom of the hoop) 38. An open cylindrical wood stave tank contains three liquids with specific gravities 2, 3, and 4, respectively. The depth of the bottom liquid is 2 m, while the other two has a depth of 1 m each. If the diameter of the container is 2 m, determine the tension in the top and bottom hoops which are holding the container in place. Solution: p1=0 F1= (1/2)(p2)(2)(1) p2= p1+ωAЋA = 2ω = 0 + (2ω)(1) F2= p2(2)(1) = 2ω = 4ω p3= p2+ωBЋB F3= (1/2)(p3- p2)(1)(2) = 2ω + (3ω)(1) = 3ω = 5ω F4= p3(2)(2) P4= p3+ωCЋC = 20ω = 5ω + (4ω)(2) F5=(1/2)(p4- p3)(2)(2) = 13ω = 16ω ∑ Top=0 2T2(4) = F1(2/3) + F2(1.5) + F3(5/3) + F4(3) + F5(10/3) 8T2= T2=

=

T2= 154.10kN

39. An open tank in the shape of a frustum of a cone is bound by two hoops, one at the top and the other at the bottom. The diameters at the top and bottom are 2 m and 3 m, respectively. If the total bursting force is to be taken care of by the hoops, find the tension in each hoop if water stands 2 m in the tank. Height of tank is 3 m. Solution:

40. A masonry dam has a trapezoidal section: one face is vertical, width at the top is 60 cm and at the bottom is 3 m. The dam is 7 m high with the vertical face subjected to water pressure. If the depth of water is 5m, where will the resultant force intersect the base? Determine the distribution of pressure along the base, (a) assuming there is no uplift pressure; (b) assuming that the uplift pressure varies uniformly from the full hydrostatic at the heel to zero at the toe. Specific weight of masonry is 23.54 kN/m3. Required:x, x, &

& if there’s no uplift pressure if there’s an uplift pressure

Solution: (a) Vertical forces

Horizontal forces

(from toe)

(b).with uplift pressure

Vertical forces

Horizontal forces

(from toe)

41. The masonry dam of Problem 40 has its inclined face subjected to pressure due to a depth of 5 m of water. If there is no uplift pressure, where will the resultant intersect the base? Specific weight of concrete is 23.54 kN/m3. SOLUTION:

42. A masonry dam of trapezoidal cross section, with one face vertical has a thickness of 60 cm at the top, 3.70 m at the base, and has height of 7.40 m. What is the depth of water on the vertical face? What is the minimum length of the base of the dam such that the resultant will intersect the base at the downstream edge of the middle third? Assume that the uplift pressure varies uniformly from full hydrostatic at the heel to zero at the toe. SOLUTION:

43. A concrete dam is triangular in cross section and 30 m high from the horizontal base. If water reaches a depth of 27 m on the vertical face, what is the minimum length of the base within the middle third? What minimum coefficient of friction is required to prevent sliding? Determine the pressure distribution along the base. SOLUTION: P= ΥhA P=(9.81)(13.5)(27)(1) P=3575.75 kN W_1= 30B/2 (23.5) W_1=352.5 B R= W= 352.5 B RM= W_1 ( 2/3 B) RM= 352.5B( 2/3 B)

RM= 235B^2 OM= 3575.75( 27/3) OM=32181.75 R_x=RM-OM 352.5B( B/3)= 235B^2- 32181.75 117.5B^2= 32181.75 B= 16.55 m Min. coeff. of friction FS= μR/P 1= (μ(352.5)(16.55))/3575.75 μ=0.61 qT = (-R_y)/B [1+ 6e/B] = (-15(23.43)(16.55))/((16.55)) [1+ (6(2.755))/16.55] qT = 706.2kPa at the toe

44. The section of a masonry dam is shown in Figure U. If the uplift pressure varies uniformly from full hydrostatic at the heel to full hydrostatic at the toe, but acts over 2/3 of the area of the base, find: (a) The location of the resultant, (b) factor of safety against overturning, (c) factor of safety against sliding if the coefficient of friction between base and foundation is 0.60.

SOLUTION: Vertical Forces: W1 = ωCV1 = (23.5)(1/2)(21)(7)(1) = 1727.25kN

W4 = ωH20V4 = (9.81)(5)(8)(1) = 392.40kN W5 = ωH20V5

W2 = ωCV2 = (9.81)(1/2)(5)(10)(1) = (23.5)(21)(5)(1) = 245.25kN = 2467.50kN W6 = ωH20V6 W3 = ωCV3 = (9.81)(1/2)(5/3)(5)(1) = (23.5)(1/2)(5)(10)(1) = 40.875kN = 587.50kN U1 = ωH20VUplift U2 = ωH20VUplift = (9.81)(2/3)(5)(1)(16) = (9.81)(2/3)(1/2)(13)(10.6667) = 523.20kN = 453.44kN Horizontal Forces: FH1=ωЋA FH2=ωЋA = (9.81)(9)(1)(18) = (9.81)(2.5)(5)(1) = 1589.22kN = 122.625kN Rx=∑▒Fx Rx = FH1 – FH2 = 1589.22kN – 122.625kN = 1466.595kN Ry =∑▒Fy Ry = W1 +W2 +W3 +W4 +W5 -U1 -U2 = 1727.25 +2467.5 + 587.5 + 392.4 + 245.25 + 40.875 – 784.8 – 680.16 = 3995.815kN RM = W1x1 + W2x2 + W3x3 + W4x4 + W5x5 + W6x6 + FH2y2 = (1727.25)(4) + (2467.5)(8.5) + (587.5)(13.5) + (245.25)(14.33) + (40.875)(0.56) + (122.625)(1.67) = 44365.88kN-m OM = U1z1 + U2z2 + FH1y1 = (523.2)(8) +(453.44)(7.11) + (1589.22)(6) = 16944.88kN-m X =( RM-OM)/Ry = ( 44365.88 -16944.88)/3995.815 = 6.86m FSO = RM/OM =(44365.88kN-m)/(16944.88kN-m) = 2.62 > 1, safe from overturning FSS = μRy/Rx = ((0.6)(3995.815))/1466.595 = 1.63 > 1, safe from sliding

45. Shown in Figure V is an overflow dam. If there is no uplift pressure, determine the location of the resultant. SOLUTION:

46. The base of a solid metal cone (sp. gr. 6.95) is 25 cm in diameter. The altitude of the cone is 30 cm. If placed in a basin containing mercury (sp. gr. 13.60) with the apex of the cone down, how deep will the cone float? SOLUTION:

(

(

)

)

47. If the metal sphere 60 cm in diameter weighs 11,120 N in the air, what would be its weight when submerged in (a) water? (b) Mercury? SOLUTION: Vsphere= (4/3)πr3 (4/3)π(0.30)

3

0.1130973355 m3

Vsphere= Vsphere=

Wair =

Wwater= ? BF = Wair– Wwater 11,120 – Wwater = 9810 (0.1130973355) Wwater = 10,010.51514 N

11,120 N There will be no answer since sphere will float in Hg. Sgog Hg is greater than metal. 48. A rectangular solid piece of wood 30 cm square and 5 cm thick floats in water to a depth of 3.25 cm. How heavy an object must be placed on the wood (sp. gr. 0.50) in such a way that it will just be submerged? Solution: ΣF_V=0 BF-W_obj-W_W=0 W_obj=BF-W_W ΣF_V=0 W_W=BF W_W=wV_D =9.81(1000)(0.3×0.3×0.0325) W_W=28.69 N BF=wV_O

=9.81(1000)(0.3×0.3×0.05) BF=44.145 N W_obj=BF-W_W =44.145-28.69 W_obj=15.455 N

49. A hollow vessel in the shape of a paraboloid of revolution floats in fresh water with its axis vertical and the vertex down. Find the depth to which it must be filled with a liquid (sp. gr. 1.20) so that its vertex will be submerged at 45 cm from the water surface. SOLUTION: r^2/0.45= x^2/h x^2= (r^2 h)/0.45 BF = W ω VD = ωV (9.81)(1)(1/2)(πr^2)(0.45) = (9.81)(1.20)(1/2)πx^2h ½(0.45) r^2 = ½ (1.20)((r^2 h)/0.45) h h = 0.41m = 41cm 50. A barge is 16 m long by 7 m wide by 120 cm deep, outside dimensions. The sides and bottom of the barge are made of timber having thickness of 30 cm. The timber weighs 7860 N/m3. If there is to be a freeboard of 20 cm in fresh water how many cubic meters of sand weighing 15700 N/m3 may be loaded uniformly into the barge? SOLUTION: W = BF Volume of Barge (VB) VB = (7x 16 x 1.2) – (6.40 x 15.4 x0.9) VB = 134.4 – 88.703 VB = 45.696 m3 Volume of Submerged Barge ( VSB) VSB = 1 x 7 x 16

VSB = 112 m3 BF = Ws + WB (9.81)(112) = (7.860)(45.696) + (15.7)(Vs) 1098.72 = 359.17 + 15.7Vs Vs= 47.10 m3

51. A brass sphere (sp. gr. 8.60) is placed in a body of mercury. If the diameter of the sphere is 30 cm (a) what minimum force would be required to hold it submerged in mercury? (b) What is the depth of floatation of the sphere when it is floating freely?

SOLUTION: BF = WBody = F F = (9810)(13.6)(4/3)( π)(0.15)3 – (9810)(8.60)(4/3)(π)(0.15)3 F = 693.43N BF = W (9810)(13.6) VS = (9810)(8.60)(4/3)(π)(0.15)3 VS = 8939.679cm3 VS = π/3 D2 (3r-D) 8939.679cm3 = π/3 D2 (3(15cm)-D) D = 17.75cm 52. A spherical balloon weighs 3115 N. How many Newtons of helium have to be put in the balloon to cause it to rise, (a) at sea level? (b) At an elevation of 4570 m ? SOLUTION:

53. The specific gravity of rock used as concrete aggregate is often desirable to know. If a rock weighed 6.15 N in the air and 3.80 N when submerged in water, what would be the specific gravity of the rock? Given:

Required: Specific gravity=? Solution:

BF

54. A piece of wood weighs 17.80 N in air and a piece of metal weighs 17.80 N in water. Together the two weigh 13.35 N in water. What is the specific gravity of the wood? SOLUTION: BF = WWood and Metal in H20 BF = 13.35N BF = WWood + WMetal 13.35N = WWood + 17.80N WWood = -4.45N s.g = s.g = s.g = 0.8

55. In Figure W is shown a thin-walled inverted box 1.60 m long and 30 cm square, which was full of air before being submerged. In position (a), it is being held by 0.0280 m3 of concrete anchor weighing 23540 N/m3. Determine the depth D1. At what other depth D2 (position B) is the system in equilibrium? SOLUTION: Wair + Wrock = BF air + BFrock (0.3)(0.3)(1.6)(0.012) + (23.54)(0.028) = (0.3)(0.3) d (9.81) + (9.81)(0.028) 0.6608 – 0.2747 = 0.8829 d d = 0.44 m

56. In Figure X a gate 1.30 m square and 30 cm thick is hinged at A. It is subjected to water pressures on both sides. What force T is required to open the gate if it weighs 1780 N?

SOLUTION: F = whA F = (9.81)(1.8 - 0.3)(1.3)(0.3) F = 5.73885 KN F2 = (9.81)(1.2 – 0.3)(1.3)(0.3) F2 = 3.44331 KN

BF = 5.73885 + 1.78 BF = 7.51385 KN 1.3 T = 5.74885(1.3) + 3.44331(1.3) + 1.78(1.3) – 7.51385(1.3) T = 3.45 KN

W = 1.78 KN 57. The timber AB of Figure Y is 7.20 m long by 15 cm square. Find the specific weight of the timber and the total weight of the anchor (sp. gr. 2.40).

SOLUTION:

a.) ∑MA = 0 3.6 W – 2.6 BF = 0 (3.12)(ω)(0.15)(0.15)(7.2) – (2.6)(9.81)(6)(0.15)(0.15) = 0 ω = 6.44

sg = 6.44 / 9.81 sg = 0.7 b.) Wt + Wa = BFt + BFa (0.7)(0.15)(0.15)(7.2) + (2.4)(9.81)(VD) = (9.81)(6)(0.15)(0.15) + 9.81 (VD) VD = 0.15429 W = (2.4)(9.81)(0.15429) W = 363.25 N

58. A sphere 1 m in diameter floats half submerged in a tank of liquid (sp. gr. =0.80) (a) what is the weight of the sphere? (b) What is the minimum weight of the anchor (sp. gr. 2.40) that will require to submerge the sphere completely? Solution:

(

)

In half submerged:

(

)

59. Figure Z shows a hemispherical shell covering a circular hole 1.30 m in diameter at the vertical side of a tank. If the shell weighs 12,450 N, what vertical force is necessary to lift the shell considering a friction factor of 0.30 between the wall and the shell?

60. An iceberg has a specific gravity of 0.92 and floats in salt water (sp. gr. 1.03). If the volume of ice above the water surface is 700 m3, what is the total volume of the iceberg? SOLUTION: s.g. (iceberg) = 0.92 s.g. (saltwater) = 1.03 VD = (s.g.body/s.g. fluid)(V) VD = VT -700 m3 ∑ BF = W WliquidVD = wbodyVT (1.03)(9.81)(VT – 700) = 9.81(0.92) VT VT = 6554.545455 m3 61. A concrete cube 60 cm on each edge (sp. gr. 2.40) rests on the bottom of a tank in which sea water stands to a depth of 5 m. The bottom edges of the block are sealed off so that no water is admitted under the block. Find the vertical pull required to lift the block. SOLUTION: P = ,ρghseawater –ρghconcrete} Area = ,*(1000kg⁄m3 )(1.03)(9.81m/s2)(5m)] – *(1000kg⁄m3)(2.4)(9.81m/s2)(0.6m)]}(0.6m)2 P =13,102.236N ≈13.1022kN

62. A 15 cm by 15 cm by 7 m long timber weighing 6280 N/m3 is hinged at one end as shown in Figure AA. If the anchor weighs 23540 N/m3, determine the minimum total weight it must have. SOLUTION: [W1 + W2]V = Wtotal [6280N/m3 + 23540 N/m3](0.15m × 0.15m × 7m) = Wtotal [29820N/m3](0.1575m3) = Wtotal 4696.65N = Wtotal Wtotal = 4696.65N ≈4700N 63. A cylinder weighing 445 N and having a diameter of 1 m floats in salt water (sp. gr. 1.03) with its axis vertical as in Figure BB. The anchor consists of 0.280 m3 of concrete weighing 23540N/m3. What rise in the tide r will be required to lift the anchor off the bottom?

Solution: ∑ BF = W 445 N + 23540(0.280) = 0 BF = 7036.2 N W = (9.81)(1.03)(Vc + 0.280m2) W = 10.1043 (0.3156 + 0.785r)

W = 5.21 + 7.93r BF = W 7.0362 KN = 5.21 + 7.93r r = 0.23 m or 23cm

64. A timber 15 cm square and 5 m long has a specific gravity of 0.50. One end is hinged to the wall and the other is left to float in water (Figure CC). For a = 60 cm, what is the length of the timber submerged in water? Solution: P = wh P = (9.81)(1.30) P = 13.2435 KN Wwood = wV Wwood = (9.81)(0.5)(0.270) Wwood = 1.05948 P = F/A P = 1.05948/(o.6)2 P = 2.94 KN

65. A metal block 30 cm square and 25 cm deep is allowed to float on a body of liquid which consists of 20 cm layer of water above a layer of mercury. The block weighs 18,850 N/m3. What is the position of the upper level of the block? If a downward vertical force of 1110 N is applied to the centroid of the block, what is the new position of the upper level of the block? Solution:

66. Two spheres, each 1.2 m in diameter, weigh 4 and 12 kN, respectively. They are connected with a short rope and placed in water. What is the tension in the rope and what portion of the lighter sphere protrudes from the water? What should be the weight of the heavier sphere so that the lighter sphere will float halfway out of the water? Solution: Tension in the hoops T+BF_2=12 kN T+(π(1.2)^3)/6(9.79)=12 kN T=3.14 kN Portion of the lighter sphere protrudes from the water BF1=4kN+3.14 kN BF1=7.14 kN BF2=(π(1.2)^3)/6 (9.79)=8.56 kN By proportion: (8.56-7.14)/8.56= V_1/V V1= (8.56-7.14)/8.56 V V1=0.1659 V1=16.59 % of V above the water surface

67. A flat-bottomed scow is built with vertical sides and sloping ends. Its length on deck is 24.40 m, on the bottom 19.80 m, its width 6.10 m and its vertical depth 3.70 m. What is its depth of flotation in sea water (sp. gr. 1.03) if the scow weighs 2,220 kN? Solution: BF = W ωVD= 2220kN (9.81)(1.03) [((24.40+19.80))/2] (6.10)(x)=2220kN X = 1.63m 68. A ship with cargo weighs 44,480 kN and draws 7.60 m of salt water. On crossing a bar at the entrance to a river her draft is decreased by 30 cm by the discharge of 2670 kN of water ballast. In going up the river to fresh water, 445 kN of coal burned. What will her draft be then and how much ballast must be required to increase it by 30 cm? Assume that the sides of the vessel near the water surface are vertical. Solution:

69. A tank with vertical sides is 1.30 m square, 3 m deep and is filled to a depth of 2.70 m with water. By how much, if at all, will the pressure on one side of the tank be changed if a cube of wood (sp. gr. 0.50), measuring 60 cm on an edge, be placed in the water so as to float with one ace horizontal? Solution: P = 9.81(1.35) P = 13.2435 Wwood= ωV = 9.81(. = 2.12 kN

)

70. A cask which weighed 270 N was placed on platform scales and then nearly filled with water. A total load on the scales of 1425 N was read. Should the net weight of water as computed from these data be corrected by reason of the fact that a 7.5 cm diameter vertical shaft suspended from the ceiling above has its lower end submerged in the water to a depth of 30 cm? If so, by what amount? Solution:

71. . If the specific gravity of a body is 0.80, what proportional part of its total volume will be submerged below the surface of a liquid (sp. gr. 1.20) upon which it floats? Solution: BF = W ωVD = ωV (9.81)(1.20)(V-x) = (9.81)(0.8)(V) 11.772V – 11.772x = 7.848V 3.924V/11.772 = 11.772x/11.772 X = 1/3 V VSubmerged = V – x = V - 1/3 V VSubmerged= V

72. .A vertical cylindrical tank, open at the top, contains 45.50 m3 of water. It has a horizontal sectional area of 7.40 m2 and its sides are 12.20 m high. Into it is lowered another similar tank, having a sectional are of 5.60 m2 and a height of 12.20 m. The second tank is inverted so that its open end is down and it is allowed to rest on the bottom of the first. Find the maximum hoop tension in the outer tank. Neglect the thickness of the inner tank. Solution: P=ωh P = (9.81) (6.15) P = 60.32 kPa FB = 60.32 (5.60) FB = 337.79 T = FB / 2 T = 337.79 / 2

T = 168. 896 kN/m 73. . A small metal pan of length 1 m, width 20 cm, and depth 4 cm floats in water. When a uniform load of 15 N/m is applied as shown in Figure DD, the pan assumes the figure shown. Find the weight of the pan and the magnitude of the righting moment developed. Solution:

74. .A ship of 39,140 kN displacement floats in sea water with the axis of symmetry vertical when a weight of 490 kN is midship. Moving the weight 3 m toward one side of the deck causes a plumb bob, suspended at the end of a string 4 m long, to move 24 cm. Find the metacentric height. Solution: Tanθ= Θ= 3.43 RM= w(MGsinθ) MG= MG= MG= 0.627m 75. .A rectangular scow 9.15 m wide by 15.25 m long and 3.65 m high has a draft of 2.44 m in sea water. Its center of gravity is 2.75 m above the bottom of the scow. (a) Determine the initial metacentric height. (b) If the scow tilts until one of the

longitudinal side is just at the point of submergence, determine the righting couple or the overturning couple. Solution: A) MG= MGo + GBo MBo=

(1+

MBo=

) ; θ=0 (1+

)

GBo= 2.75- (2.44-2) GBo= 1.53m

MBo= 2.82m MG= 2.84 – 1.53 MG= 1.33m

B) OM= BF. X

GBo= 2.75-1.22

X= MGsinθ

= 1.53

MG= MBo-GBo MBo=

(1+

Θ= tan (

)

MG=2.96-1.53

)

= 1.43m

Θ= 14.81°

x= MGsinθ x= 1.43 sin (14.81)

MBo=

(1+

)

x= 0.37m

MBo= 2.96 OM= BF.x = (9.81) (1.03)(9.15)(2.44)(15.25)(0.37) OM= 1256.9KN.m 76. A cylindrical caisson has an outside diameter of 6 m and floats in fresh water with its axis vertical. Its lower end is submerged to a depth of 6 m below the water surface. Find: (a) the initial metacentric height; (b) the righting couple when the caisson is tipped through an angle of 10°. Solution:

77. A rectangular scow 9.15 m wide by 15.25 m long has a draft of 2.44 m in fresh water. Its center of gravity is 4.60 m above the bottom. Determine the height of the scow if, with on side just at the point of submergence, the scow is in unstable position. Solution: GBo = 4.60 -1.22 GBo = 3.38 MBo = [(9.15^2) / 12(2.44)] [1 + (tan0 ^2) / 2] MBo = 2.86 MG = 3.38 – 2.86 MG = 0.52 m y = 4.6 – 1.22 y = 3.38 h = 3.38 + 2.44 – 0.52 h = 5.3 m 78. A rectangular scow 9.15 m wide by 15. 25 m long and 4.60 m high has a draft of 2.75 m. Its center of gravity, transversely and longitudinally, is at the center of the scow. If the sow is tipped transversely until one end is at the point of submergence, find the righting couple. Solution:

(

)

79. .A rectangular raft 3 m wide and 6 m long has a thickness of 60 cm and is made of solid timbers (sp. gr. 0.60). If a man weighing 890 N steps on the edge of the raft at the middle of one side, how much will the original water line on that side be depressed below the water surface?

Solution:

80. .The timber shown in Figure EE is 30 cm square and 6 m long, having a specific gravity of 0.50. A man standing at a point 60 cm from one end causes that end to be just submerged. Find the weight of the man. Solution:

81. .A submarine of 10, 700 kN displacement has its center of gravity 30 cm from its center of volume. What is the righting couple when it is submerged in sea water and the angle of heel is 5˚. Solution:

82. .A log 30 cm in diameter weighs 4900 N/m3. What is its shortest length so that it may float in water with the axis horizontal? Solution: Wlog = 4.9 (0.30)3 = 0.1323 kN BF = Wwood 9.81 (1) *π (0.15)2 h] = 4.9 h = 0.1908 m h = 19.08 cm

83. .A block of wood is 15 cm square and 30 cm long. What is the specific gravity of the wood if the metacentre is at the same point as the center of gravity of the wood when the block is floating in water on its side? Would it be stable floating on its end? Explain. Solution: W = BF 9.81(0.15)(0.15)(.3)(S) = 9.81(1)(0.15)(0.3)(.10) S = 0.667

84. .Figure FF shows a scow equipped with a derrick with a boom 5.50 m long. What maximum weight could be picked by the boom at its end along the longitudinal side in order that water will not enter into the scow? Assume the weight of the boom to be negligible and consider its position to be always horizontal.

Solution: MG = Mbo-Gbo Mbo = I/V =0.1188 in. I = bh3/12 =(10)3(16)/12 = 16000 m4 V = bhw = 10(16)(2.6) = 46 m3

Gbo = 3= 1.7 m MG = Mbo + Gbo MG = .12 + 1.7 MG = 1.82

W = ѡV W = 9.81(1)(416) W = 4080.86 KN

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1. If certain gasoline weighs 7,000 N/m, what are the values of its density, specific volume, and specific gravity relative to water at 15°C? Given: Specific weight (w) =7kN/m³ Required: density ( ), volume of gasoline ( ) and specific gravity (s.g.) of gasoline. Solution: ;

`

2. A certain gas weighs 16 N / at a certain temperature and pressure. What are the values of its density, specific volume, and specific gravity relative to air weighing 12 N / ? Given: Weigh of gas ( ) Weigh of air ( ) Required: density of gas ( ), Volume of gas ( ) and specific gravity (s.g.) of gas. Solution: ; 1.6310 kg/m3

0.6131m3/kg

3. If 5.30 of a certain oil weighs 43,860 N. Calculate the specific weight,density and specific gravity of this oil. Given: Required: specific weight (w) =? Density ( =? Specific gravity (Sg) =? Solution:

)

4. The density of alcohol is 790 and specific volume.

. Calculate its specific weight, specific gravity

Required: specific weight (w) =? Specific gravity (Sg) =? specific volume (v) =? Given: Solution: For specific weight(w)

For specific gravity (sg)

For specific volume (v)

5. A cubic meter of air at 101.3 KPa and 150C weighs 12 N/m3. What is its specific volume? Given: Required: Solution: ; =

6. At a depth of 8 km in the ocean the pressure is 82.26 MPa. Assume the specific weight at the surface to be 10.10kN/m3 and that the average bulk modulus is 2344 MPa for that pressure range. (a) What will be the change in specific volume between that at the surface and at the depth? (b) What will be the specific volume at that depth? (c) What will be the specific weight at that depth?

Solution: (a).

(

)

7. To two significant figures what is the bulk modulus of water in kN/m2 at 50°C under a pressure of 30 MN/m2? Given: pressure of water ( ) = 30 MN/m2 Required: bulk modulus=? Solution: By Interpolation 2

Pressure, P (MN/m ) 10.340 31.030 30 Bulk modulus= 2.49 kN/m2

Bulk Modulus, Ev (MN/m2) 2560 2500 2493.03

8. If the dynamic viscosity of water at 20˚ C is 1x10-3 N-s/m2, what is the kinematic viscosity in the English Units? Required: Solution:

9. The kinetic viscosity of 1ft2/sec is equivalent to how many strokes? (1 strokes = 1cm2/sec) Solution: Convert ft2/sec to cm2/sec 1 ft2/sec x ((1 cm)2/(0.0328ft)2) = 929.0304 cm2/sec or 929 cm2/sec 1ft2/sec = 929 strokes

10. A volume of 450 litres of a certain fluid weighs 3.50 kN. Compute the mass density. (1m3 = 1000 litres). Given:

volume of fluid ( = 450 L Weight = 3.5 kN

Required: mass density of fluid Solution: 450 L x (1m3/1000L) = 0.45m3 ⁄ =

⁄

11. Compute the number of watts which are equivalent to one horsepower. (1 HP = 550 ft-lb/s; 1 W = 107 dyne-cm/s; 1 lb = 444,800 dynes. Solution:

=

Watts

12. A city of 6000 population has an average total consumption per person per day of 100 gallons. Compute the daily total consumption of the city in cubic meter per second. (1 ft3 = 7.48 gallons) Solution: = =

13. Compute the conversion factor for reducing pounds to Newtons. Solution: =

Chapter 2 Principles of Hydrostatics

1. If the pressure 3 m below the free surface of a liquid is 140 kPa, calculate its specific weight and specific gravity. Solution:

2. If the pressure at a point in the ocean is 1400 kPa, what is the pressure 30 m below this point? The specific gravity of salt water is 1.03 Solution:

3. An open vessel contains carbon tetrachloride (s = 1.50) to a depth of 2 m and water above this liquid to a depth of 1.30 m. What is the pressure at the bottom?

Solution:

4. How many meters of water are equivalent to a pressure of 100 kPa? How many cm of mercury? Solution:

5. What is the equivalent pressure in kPa corresponding to one meter of air at 15° C under standard atmospheric conditions? Solution:

6. At sea level a mercury barometer reads 750 mm and at the same time on the top of a mountain another mercury barometer reads 745 mm. The temperature of the air is assumed constant at 15° C and its specific weight assumed uniform at 12 N/m3. Determine the height of the mountain. Solution:

7. At ground level the atmospheric pressure is 101.3 kPa at 15° C. Calculate the pressure at a point 6500 m above the ground, assuming (a) no density variation, (b) an isothermal variation of density with pressure.

Solution: a)

b)

8. If the barometer reads 755 cm of mercury, what absolute pressure corresponds to a gage pressure of 130 kPa?

Solution: +

9. Determine the absolute pressure corresponding to a vacuum of 30 cm of mercury when the barometer reads 750 mm of mercury.

Solution:

10. Figure A shows two closed compartments filled with air. Gage (1) reads 210 kPa, gage (2) reads 25 cm of mercury. What is the reading at gage (3)? Barometric pressure is 100 kPa. Solution: P3 = P1 + P2 + Pb P1 = 210 kPa P2 = 25cmHg *10mmHg/1cmHg*101.325 kPa/760mmHg P2 =33.25 kPa 11. If the pressure in a gas tank is 2.50 atm, find the pressure in kPa and the pressure head in meter of water.

Solution:

12. The gage at the suction side of a pump shows a vacuum of 25 cm of mercury. Compute (a) pressure head in meter of water, (b) pressure in kPa, (c) absolute pressure in kPa if the barometer reads 755 cm of mercury.

Solution: a) b)

(

)

13. Oil of sp. gravity 0.80 is being pumped. A pressure gage located downstream of the pump reads 280 kPa. What is the pressure head in meter of oil? Solution:

14. The pressure of the air inside a tank containing air and water is 20 kPa absolute.

Determine the gage pressure at a point 1.5 m below the

water surface. Assume standard atmospheric pressure. Solution:

15. A piece of timber 3 m long and having a 30 cm by 30 cm section is placed in a body of water in a vertical position. If the timber weighs 6.5 kN/m3, what vertical force is required to hold it with its upper end flush with the water surface?

Solution:

[

]

16. A glass tube 1.60 m long and having a diameter of 2.5 cm is inserted vertically into a tank of oil (sp. gravity 0.80) with the open end down and the closed end uppermost. If the open end is submerged 1.30 m from the oil surface, determine the height to which the oil will rise in the tube. Assume barometric pressure is 100 kPa and neglect vapour pressure. Solution:

Before the glass tube was inserted:

After the glass was inserted:

(

)

*

+

17. A gas holder at sea level contains illuminating gas under a pressure equivalent to 5 cm of water. What pressure in cm of water is expected in a distributing pipe at a point 160 m above sea level? Consider standard atmospheric pressure at sea level and assume the unit weights of air and gas to be constant at all elevations with values of 12 N/m3 and 6 N/m3, respectively. Solution:

18. If the barometric pressure is 758 mm of mercury, calculate the value of h of Figure B. Solution:

19. The manometer of Figure C is tapped to pipeline carrying oil (sp. gravity 0.85). Determine the pressure at the center of the pipe.

Solution:

20. Determine the gage reading of the manometer system of Figure D. Solution:

21. In Figure E, calculate the pressure at point m. Solution:

22. In Figure F, find the pressure and pressure head at point m: fluid A is oil (s = 0.90), fluid B is carbon tetrachloride (s = 1.50) and fluid C is air.

Solution:

23. Compute the gage and absolute pressures at point m of Figure G: fluids A and C are air, fluid B is mercury.

Solution:

24. The pressure at point m is increased from 70 kPa to 105 kPa. This causes the top level of mercury to move 20 cm in the sloping tube. What is the inclination Ɵ? Figure H. Solution:

[

]

25. In Figure I, determine the elevation of the liquid surface in each piezometer Solution: .

26. In Figure J, fluid A is water, fluid B is oil (s = 0.85). Determine the pressure difference between points m and n.

Solution:

27. In Figure K determine ⍴m-⍴n. Solution:

28. In Figure L, fluid A has a specific gravity of 0.90 and fluid B has a specific gravity of 3.00. Determine the pressure at point m.

Solution:

29. If the pressure at m obtained from Problem 28 is increased by 7 kPa, how many cm will fluid B in the 12 mm tube? Solution: 7kPA = 3(9.81)h 7kPa = 29.4h h = 24cm

30. In Figure M, fluid A is a gasoline (s = 0.70), fluid B is mercury. Find the pressure head difference between points m and n.

31. In Figure N, x = 25 cm initially. If the pressure at m is increased by 35 kPa while maintaining the pressure at n constant, calculate the new value of x. Solution:

32. The diameters of the cylinders of the hydraulic jack of Figure O are 7.5 cm and 60 cm, respectively. What force F is required to maintain equilibrium if the load weighs 35 kN? Solution:

33. A 5 cm pipe is connected with the end of a cylinder having a diameter of 50 cm. There is a piston in the pipe and a piston in the cylinder, the space between being filled with oil (s = 0.80). The larger piston is connected by a rod with a 5 cm piston in a third pipe, the two pipes and cylinder having their axes horizontal and collinear. A force of 100 N is applied to the piston in the first pipe. What pressure is affected in the third pipe?

34.Assuming normal barometric pressure, how deep is the ocean at appoint where an air bubble, upon reaching the surface, has six times the volume that it had at the bottom? Specific gravity of sea water is 1.03. Solution: l

35. A bottle consisting of a cylinder 30 cm in diameter and 30 cm high has a neck 5 cm in diameter and 30 cm long. If the bottle, filled with air under normal atmospheric condition, is inverted and submerged in water until the neck is just filled with water, find the depth to which the open end is submerged. Neglect vapour pressure. Solution: Before the bottle was inserted:

When the bottle is inserted:

(

)

depth to which the open end is submerged =

CHAPTER IV ACCELERATED LIQUIDS IN RELATIVE EQUILIBRIUM

1. A car travelling on a horizontal road has a rectangular crosssection, 6m long by 2.40m wide by 1.50m high. If the car is half full of water, what is the maximum acceleration it can undergo without spilling any water? Neglecting the weight of the car, what force is required to produce maximum acceleration? Solution: F= Ma F= ρVa F= (1)(0.75)(6)(2.4)(2.45) = 14°

F= 26.5 KN

a= 2.45 m/s²

2.

A cylindrical bucket is accelerated upward with an acceleration of gravity. If the bucket is 0.60m in diameter and 1.20 m deep, what is the force on the bottom of the bucket if it contains 0.90 m depth of wet concrete whose specific weight is 22,000 N/m3? What is force on the bottom idf the bucket is accelerated downward at 9.81 m/s2? Solution: ∑

3. A rectangular car is 3 m long by 1.5 m wide and 1.5 m deep. If friction is neglected and the car rolls down a plane with an inclination of 20°, what is the inclination of the water surface if the car contained 0.60 m depth of water when the car was horizontal?

Note:

4.

An open tank, 9.15 m long is supported on a car moving on a level track and uniformly accelerated from rest to 48 km/hr. When at rest the tank was filled with water to within 15 cm of its top. Find the shortest time in which the acceleration may be accomplished without spilling over the edge. Solution:

= 1.87788°

(

)(

)(

)

5. A rectangular tank, 60 cm long and containing 20 cm of water is given an acceleration of a quarter of the acceleration of gravity along the length. How deep will the water be at the rear end? At the front end? What is the pressure force at the rear end if it is 45 cm wide? Solution:

= 14.0362° a.) =27.5 cm b.) 12.5 cm

6. Fig. GG shows a container having a width of 1.50 m. Calculate the total forces on the ends and bottom of the container when at rest and when being accelerated vertically upward at 3 m/s². Solution: Sin 60°

z= 1.5011 m

x = 0.7506 When at rest,

Force at the bottom,

Vertically upward, (

)

(

)

7. A closed rectangular tank 1.20 m high by 2.40 m long by 1.50 m wide is filled with water and the pressure at the top is raised to 140 KPa. Calculate the pressures in the corners of this tank when it is accelerated horizontally along its length at 4.60 m/s². Solution :

= 25.1223°

At

At

+ 140

8. A pipe 2.50 cm in diameter is 1.0 m long and filled with 0.60 m of water. If the pipe is capped at both ends and rotated in a horizontal plane about a vertical axis through one end of the pipe, what is the pressure at the other end of the pipe when it is rotating at 200 RPM?

9. An open vertical cylindrical tank 0.60 m in diameter and 1.20 m high is half full of water. If it is rotated about its vertical axis so that water just reach the top, find the speed of rotation. What will then be the maximum pressure in the tank? If the water were 1.0 m deep, what speed will cause the water to just reach the top? What is the depth of water at the center? Required: Depth of water=?

Solution:

h=

w = 16.17 kN (

)

h = 0.4

w =9.34 rad/s Depth= 1.2m – 0.4 m Depth D epth Depth=0.8m Depth 10. If the tank of Prob. 9 is half full of oil (sp.gr. 0.75) what speed of rotation is necessary to expose one- half of the bottom diameter? How much oil is lost in attaining this speed? Solution:

Y= 0.4 m

w = 18. 68 rad/s = 0.169646

V= 41.41 L

11. The U-tube of figure HH is given a uniform acceleration of 1.22 m/s² to the right. What is the depth in AB and the pressures at B, G and D? Solution: (

)

(

(

)

)

12. The U-tube of fig. HH is rotated about an axis through HG so that the velocity at B is 3 m/s. What are the pressures at B and G?

13. The U-tube of fig. HH is rotated about HG. At what angular velocity does the pressure at G become zero gage? What angular velocity is required to produce a cavity at G? Solution: h= 0.3 =

14. The tank of Prob. 9 is covered with a lid having a small hole at the center and filled with water. If the tank is then rotated about its vertical axis at 8 rad/s, what is the pressure at any point on the circumference of the upper cover? Of the lower cover?

15. The tank of Prob. 9 contains 0.60 m of water covered by 0.30 m of oil (sp. Gr. 0.75) What speed of rotation will cause the oil to reach the top? What is then the pressure at any point on the circumference of the bottom? Solution: h= 0.6 = w = 11.44 rad/s P= w P= (0.75)(9.81)(0.3)+(9.81)(0.9) P= 11.04 kPa 16.The tube of figure II is rotated about axis AB. What angular velocity is required to make the pressures at B and C equal? At that speed where is the location of the minimum pressure along BC? Solution : h= 0.3 =

17. A vessel 30 cm in diameter and filled with water is rotated about its vertical with such a speed that the water surface at a distance of 7.50 cm from its axis makes an angle of 45° with the horizontal?

18. A cylindrical vessel, 0.30 m deep, is half filled with water. When it is rotated about its vertical axis with a speed of 150 RPM, the water just rises to the rim of the vessel. Find the diameter of the vessel. Required:

Solution: w = 150 RPM( ) w = 15.70796 rad/sec 0.3 = r = 0.15445 D= 2r D=0.3089 m D= 30.89 cm 19. A conical vessel with vertical axis has an altitude of 1 m and is filled with water. Its base, 0.60 m in diameter, is horizontal and uppermost. If the vessel is rotated about its axis with a speed of 60 RPM, how much water will remain in it? Solution: 60 x h= h= h = 0.18 m V= =

= = = 0.094 m³

20. A cylindrical bucket, 35 cm deep and 30 cm in diameter, contains water to a depth of 30 cm. A man swings this bucket thru a vertical plane, the bottom of the bucket describing a circle having a diameter of 2.15 m. What is the minimum speed of rotation that the bucket can have without permitting water to spill? Solution: CF= W CF= ma CF= W= r= r = 0.925 W= g= 9.81 =

21. If the water which just fills a hemispherical bowl of 1.0 m radius be made to rotate uniformly about the vertical axis of the bowl at the rate of 30 RPM, determine the amount of water that will spill out? Solution:

30 RPM x h= h= h =0.503 m V= =

=

22. The open cylindrical tank of fig. JJ is rotated about its vertical axis at the rate of 60 RPM. If initially filled with water, how high above the top of the tank will water rise in the attached piezometer? Solution:

w =60 RPM x

2.012 m

y = 2.012- 0.85 y = 1.16 m 23. A closed cylindrical tank with axis vertical, 2 m high and 0.60 m in diameter is filled with water, the intensity of pressure at the top being 140 KPa. The metal making up the side is 0.25 cm thick. If the vessel is rotated about its vertical axis at 240 RPM, compute a) total pressure on the side wall, b) total pressure against the top, c) maximum intensity of hoop tension in pascals.

24. A small pipe, 0.60 m long, is filled with water and capped at both ends. If placed in a horizontal position, how fast must it be rotated about a vertical axis, 0.30 m from one end, to produce maximum pressure of 6,900 KPa? Solution:

3450= 9.81h h = 351.68 m

w = 276.89 rad/s

25. A vertical cylindrical tank 2 m high and 1.30 m in diameter, two – thirds full of water, is rotated uniformly about its axis until it is on the point of overflowing. Compute the linear velocity at the circumference. How fast will it have to rotate in order that 0.170 m³ of water will spill out? Solution:

h= h=

h= 1.3333 =

26. A steel cylinder, closed at the top, is 3 m high and 2 m in diameter. It is filled with water and rotated about its vertical axis until the water pressure is about to burst the sides of the cylinder by hoop tension. The metal is 0.625 cm thick and its ultimate strength is 345 MPa. How fast must the vessel be rotated? Solution:

345x10³ =

(

)

2156.25 = 9.81(1)

√

27. A conical vessel with axis vertical and sides sloping at 30° with the same is rotated about another axis 0.60 m from it. What must be the speed of rotation so that water poured into it will be entirely discharged by the rotative effect? Given: x= 0.6 m g= 9.81 m/s² =90°-30°

Tan = Tan60°= w= √ w = 5.32 rad/s

CHAPTER III HYDROSTATIC FORCE ON SURFACES

1. A rectangular plate 4m by 3m is immersed vertically with one of the longer sides along the water surface. How must a dividing line be drawn parallel to the surface so as to divide the plate into two areas, the total forces upon which shall be equal. Given: ω=1 ħ = = 1.5 Req’d: h=?

A = 4 x 3 = 12

Solution: FT = ω ħ A FT = (9.81) ( 1.5) (12) FT = 176.58 KN FT = F1 + F2 FT = 2F1 176.58 = 2(9.81)(h/2)(4)(h) h = 2.12m

2. A triangle of height h and base b is vertically submerged in a liquid. The base b coincides with the liquid surface. Derive the relation that will give the location of the center of pressure. Req’d: =? Solution:

Ig= bh3/36 e = (bh3/36) / (1/2bh* 1/3h) e = h/6 Yp = h + (h/6) Yp = h/2 from the Lower Surface

3. The composite area shown in Figure A is submerged vertically in a liquid with specific gravity 0.85. Determine the magnitude and location of the total hydrostatic force on one face of the area.

Given: Sg = 0.85 Req’d:

=? F=? Solution: F1 = whA F1 = (9.81)(0.85)(3.25)(1.5)(3.5) F1 = 142.28kN F2 = whA F2 = (9.81)(0.85)(4.25)(1.5)(1.5) F2 = 79.74kN F = F1 + F2 F = 142.28kN + 79.74kN F = 222.02kN yp = ̅ yp= (1.75 +

)+ 1.5

yp= 2.33 + 1.5 yp = 3.83m 4. The gate in Figure B is subjected to water pressure on one side and to air pressure on the other side. Determine the value of x for which the gate will rotate counter clockwise if the gate is (a) rectangular, 1.5 m by 1.0 m base and 1.0 m high. Given: V = 1.5 x 1.0 x 1.0 ħ = 0.5 Req’d: x=? Solution: Pair – PH2O = (1.5) (1.0) (9.81) (x+0.5+e) 1.5(1) – 30 kN/m2 = (1.5)(1)(9.81)(x+.5+.17) 30=9.81(x+0.67) 30=9.81x = 6.54 9.81x = 23.46 X= 2.39 ≈2.40

b. Pair = PH2O = ½ (1.5) (1.0) (9.81) (x+.05+e) ½ (1.5) (1) – 30 = 9.81 (x+²/ (0.5) + 0.17) 30= 9.81 (x+.5) 9.81x = 30 – 4.905 9.81x = 25.095 X = 2.558 ≈ 2.60

5. A vertical circular gate 1 m in diameter is subjected to pressure of liquid of specific gravity 1.40 on one side. The free surface of the liquid is 2.60 m above the uppermost part of the gate. Calculate the total force on the gate and the location of the center of pressure.

Given: D=1m sg = 1.4 y = 2.6 m Req’d: F=? e=? Solution:

(

𝐞

)

( )

𝟎 𝟎𝟐 𝐛𝐞𝐥𝐨𝐰 𝐭𝐡𝐞 𝐜𝐞𝐧𝐭𝐫𝐨𝐢𝐝

6. A horizontal tunnel having a diameter of 3m is closed by a vertical gate. When the tunnel is (a) ½ full (b) ¾ full, of water, determine the magnitude and location of the total force. Given: D = 3m Req’d: F=? Solution: a. h1/2 = h1/2 = h1/2 = 0.6366 A=

=

F =9.81(0.6366)( ) F = 22.10 kN

b. h3/4=( + )/2 = + = 0.954929 A= =

)/2

F = 9.81(0.95429) ( F = 33.10 kN

)

7. In Figure C is a parabolic segment submerged vertically in water. Determine the magnitude and location of the total force on one face of the area. Req’d: F=? yp = ? Solution: F = ЋA F = 9.81(

(

F = 9.81( )( )(3)(3) F = 106 kN e= e= e=

= 0.34

Ῡ+e= + .34 yp = 2.14 below water surface 8. A sliding gate 3 m wide by 1.60 m high is in a vertical position. The coefficient of friction between the gate and guides is 0.20. If the gate weighs 18 kN and its upper edge is 10 m below the water surface, what vertical force is required to lift it? Neglect the thickness of the gate. Given: A = (3m)(1.6) = 4.8 μ = 0.20 Req’d: =? Solution: F = AωЋ Ћ = 0.8 + 0.8 + 10 Ћ = 10.8 F = 4.8 (9.81) (10.8) F= 508.55

W = 18 kN Ћ = 10 m

Ff = μ N Ff = (0.20) (508.55) Ff = 101.71 ∑ Fy = 0 = W + Ff = 18 + 107. 71 = 119.71 ≈ 120

9. The upper edge of a vertical trapezoidal gate is 1.60 m long and flush with the water surface. The two edges are vertical and measure 2 m and 3 m, respectively. Calculate the force and location of the center of pressure on one side of the gate. Given: b = 1.60m = 2m = 3m Req’d: F=? Solution:

( )

(

)(

)

10. How far below the water surface is it necessary to immerse a vertical plane surface, 1 m square, two edges of which are horizontal, so that the center of pressure will be located 2.50 cm below the center of gravity. Given: e = 0.025 m A = 1m square Req’d: Ῡ=? Solution: e = Ig / AῩ ; Ig=bh3/ 12 = 0.08 0.025 Ῡ = 0.08 / 1Ῡ 0.025 Ῡ = 0.08 Ῡ = 3.33 - .5 Ῡ = 2.83 m

11. The gate shown in Figure D is hinged at B and rests on a smooth surface at A. If the gate is 1.60 m wide perpendicular to the paper, find BH and BV. Given: w = 1.60m Req’d: BH = ? BV = ? Solution: Pω = 9.81 (2.8) Pω = 27.468 P = 27.468 (1.6) (3.6) P = 158.22 kN P = ωЋA 158.22 = 9.81h (1.6) (3.6) 158.22 = 56.5056h h = 2.8 Ῡ = Ћ / sin 56.31 Ῡ = 3.37 e = Ig/ AῩ =

⁄

e = 0.32 x = 1.8 + e x = 2.12 ∑ MB = 0 158.22 (2.12) – F (2) 2F = 335.4264 F = 167. 71 kN ∑Fy = 0 = 167.71 – 158.22 cos 56.31 – Bv Bv = 79.95 ≈ 80 kN ∑Fx = 0 BH = 158.22 sin 56.31 BH = 131.65 ≈132 kN 12. In Figure E gate AB is 2m wide perpendicular to the paper. Determine FH to hold the gate in equilibrium. Req’d: FH = ? Solution: P = (9.81) (0.8) (0.6) P = 4.7 P2= P1 + wh = 4.7 + 9.81 (2) P2= 24.4 h = 2.44 / 9.81 h = 2. 49 3.2 sin60 = 2.77 h of gate = 2.49 / sin 60 h of gate = 2. 88 Ћ= 2.49 / 2 Ћ = 1. 245 P = AωЋ = 5.75 (9.81) (1.245) P = 70. 23 e= e=

⁄

e = 0.48 ∑M = 0 2.77 Fh = P (1.44 – 0.48)

Fh = 24.34 kN

13. Calculate the resultant of the hydrostatic forces on the gate of Figure F. What vertical force P is necessary to lift the gate at point A? Width of gate is 1.60 m. Given: w = 1.60 Req’d: P=? Solution: Ћ1 = 3/2 + 1/3 Ћ = 2.8 A1 = 1.6 (3.6) A1 = 5. 76 m2 F1 = ωhA F1 = (9.81) (2.8) (5.76) F1 = 158.22 Ћ =1.5 A2 =5.76 F2 =Aωh F2 =84.76 KN a= a= a =2.1 = (3.437) = 0.314 b= b = 2.4 Ῡ =Ћ Ῡ =2.36/sin 56.31 Ῡ =3.437 ∑ Mb =0 2P = F1(a)- F2(b) 2P =158.22(2.12)-84.76(2.4)

2P = 132KN P = 66KN P=66 kN to open the gate

14. The width of the gate of Figure G normal to the paper is 3m. What vertical force must be applied at “a” to prevent collapse when h = 6 m? Neglect weight of the gate. What is the stress in strut BC? Given: w = 3m h = 6m Req’d: =? ς=? Solution:

15. The gate of Figure H is hinged at A and rests on a smooth surface at B. The gate is circular having a diameter of 3 m, determine the value of the vertical having a diameter of 3 m. Determine the value of the vertical force P that will open the gate at B. Given: D = 3m Req’d: P=? Solution: Y = 3sin 45 Y =2.12 h = 2.12/2+1.16 h =2.66m x =3cos 45 x =2.12 A =π =π =7.06m2 F = ЋA

=9.81(.8)(2.66)(7.06) =147.382 m e= = =1.5 a= 3/2 + e = 1.5 +.15 = 1.65 Ῡ =h/sinθ Ῡ = 2.66/sin 45 Ῡ =3.76 ∑Ma=0 2.12 P = F(a) 2.12 P =147.382(1,67) P = 114.707 P-7 =107.707 KN P=107.707 KN to open the gate

16. A triangular gate having a horizontal base of 1.30 m and an altitude of 2 m is inclined 45° from the vertical with the vertex pointing upward. The base of the gate is 2.60 m below the surface of oil (s = 0.80). What normal force must be applied at the vertex of the gate to open it? Given: b = 1.30m h = 2m

ϴ = 45° s = 0.80

Req’d: N=? Solution: Sin 45= c = 2.8284 F = gЋ F = 9.81(0.8)(1/3(2))(1/2(1.3)(2.8284)) F = 9.6188 N = 9.6188KN sin 45 N = 6.80 kN

17. What depth of water will cause the rectangular gate of Figure I to fall? Neglect weight of the gate. Req’d: d=? Solution:

( )

(

( (

)

)

)(

)

∑ (

) (

)

18. Determine the horizontal and vertical components of the total force on the gate of Figure J. The width of the gate normal to the paper is 2 m. Given: w = 2m Req’d: FH = ? FV = ? Solution: h=6 FH= whA FH= (9.81) (3) (6) (2)

FH = 353.16 kN FV= wV FV= (9.81){[(60/360)(6)2(π)+-[(1/2)(6)(6sin60°)]}(2) FV= (9.81) (6π – 15.6) (2) FV = 63.76 kN

19. The corner of a floating body has a quarter cylinder AB having a length normal to the paper of 3 m. Calculate the magnitude and location of each of the components of force on AB. Figure K. Given: L = 3m Req’d: FT = ? FV = ? Solution: FT = whA FT = (9.81) (3.25) (1.5) (3) FT = 143.47 kN FV= wV FV= (9.81) [(4)(1.5)-((π) (1.52))/4](3) FV = 124.57 kN

20. The cylindrical gate of Figure L is 3m long. Find the total force on the gate. What is the minimum weight of the gate to maintain equilibrium of the system? Given: L = 3m Req’d: F=? =? Solution: x = 1.5sin30° = 0.75m y = 1.5cos30° = 1.3m

FH = FH1 + FH2 FH = (1.5)(3)(9.81)(1.5/2) – (0.2)(3)(2/2)(9.81) FH = 33.11 – 0.5886 FH = 32.52 KN A = (1/4) πr2 + (1/2)bh + (θ/360)πr2 A = 1.77 + 04875 + 0.6 A = 2.86 m2 V = 2.86(3) = 8.58m3 Fv = wV Fv = (9.81)(8.58) Fv = 84.17 √ F= = 90.23 KN 21. Determine the magnitude and direction of the force on the radial gate of Figure M. AB has a length of 2 m normal to paper. Given: l = 2m Req’d: F=? ϴ=? Solution: FH= whA FH= (9.81) (1.35) (2.7) (2) FH = 71.51 KN FV= wV FV= (9.81){[(1/2)(1.53)(2.70)] + *(π)(30)(6)2/360]- [(1/2)(3.10)(5.8)]}(2) FV = 48.30 KN √ √

ϴ = tan-1(48.30/71.51) ϴ= 34° from the horizontal

22. End AB of Figure N has a section in the shape of a quadrant. If the tank has a length of 3 m, determine the total force acting on the end. Given: l = 3m Req’d: FH = ? F=? Solution: Fv = wV Fv = (9.81)(3.1416)(3) Fv= 92.46 KN A = (1/4) πr2 A = 3.1416 m2 √ √

e = IG/Ay = (2/6)(3) Ig= bh3/36 = 2 e = 0.11m FH = whA FH = (9.81)(3)(2)(3) FH = 176.58 KN

23. The gate of Figure O is 3 m long. Find the magnitude and location of the horizontal and vertical of the forces components on the gate AB.

Req’d: FH = ?

FHy = ?

FV = ?

FVx = ?

Solution: FH= whA FH= (9.81) (6.36) (2.12/2)

FH = 66.14 KN FHy = 2.12 ( y + e) ; e = IG/Ay FHy = 2.12 [1.06 + (3)(2)(12)3/(3)(2.12)(1.06)] FHy = 2.12 – 1.41 FHy = 0.71 above O FV= wV FV= (9.81) (0.58) (3) FV = 17.15 KN ∑ (66.14)(0.71) = (17.07)(FVx) FVx = 2.75m left of O

24. A pyramidical object having a square base (2 m on a side) and 1.50 m high weighs 18 kN. The base covers a square hole (2 m on a side) at the bottom of a tank. If water stands 1.50 m in the tank, what force is necessary to lift the object off the bottom? Assume that atmospheric pressure acts on the water surface and underneath the bottom of the tank. Given: A=4 h = 1.50 m Wobj= 18kN Req’d: BF = ?

Solution: V = 1/3Abaseh V = 1/3 (1.5) V=2 Fv = wV Fv= (9.81)(2) Fv = 19.62kN BF = Wobj + Fv BF = 18kN + 19.62kN BF = 37.62kN

25. The hemispherical dome of Figure P surmounts a closed tank containing a liquid of specific gravity 0.75. The gage indicates 60 kPa. Determine the tension holding the bolts in place. Given: s g = 0.75 = 60 kPa Req’d: F=? Solution: F =9.81(0.75)((π(1.5)2(2)-(2/3)π(1.5)3 F = 52.0071 KN (2) F = 104.0142 Heq= P/Y = 60 KPA/9.81(0.75) = 8.15499-1.6 H = 6.5549 F = y Vw F = π(1.5)2(6.5549)+ 1/2*(4/3)π(1.5)3 V = 53.402 m3 F = 9.81(0.75)(53.402) F = 392.91 KN F = 392.91 KN – 104.0142KN F =288.89 KN 26. Figure Q shows a semi-conical buttress. Calculate the components of the total force acting on the surface of this semi-conical buttress. Req’d: F=? Solution FH = whA FH = (9.81)(1.3+1.5)(1.5)(3) FH = 123.606kN Fv = wV Fv = (9.81)(1/2)(1.5)(3)(1.5) Fv = 33.103kN F= F=√ F = 127.963kN

27. In Figure R a circular opening is closed by a sphere. If the pressure at B is 350 kPa absolute, what horizontal force is exerted by the sphere on the opening? Given: P = 350kPa Req’d: F=? Solution: F = PA F = 350(0.2)(0.25) F = 17.5 KN 28. Calculate the force required to hold the cone of Figure S in position. Req’d: F=? Solution:

PA = 3.5kPa – (9.81)(0.8)(1.5) PA = -8.272kPa F1 = PAA F1 = 8.272(π)( F1 = 3.65kN F2 = (9.81)(0.8) F2 = 1.156kN w=π w = 8.67kN

(2.5)(9.81)(0.8)

∑ F + F1 + F2 = w F + 3.65 + 1.156 =8.67 F = 3.864kN

29. The section of a gate at the gate at the end of a tank is shown in Figure T. It has a length of 3 m normal to the paper and hinged at O. If the depth of the water is 2.50 m, calculate P to maintain equilibrium.

Given: l = 3m h = 2.50m Req’d: P=? Solution: FH = (2.5)(3)(1.25)(9.81) FH = 91.969 kN Fv = 9.81 [(1.45)(2.5)-⅔(1.45)(2.5)](3) Fv= 35.56 kN 30. A pipe having a diameter of 30 cm is exposed to an atmospheric pressure of 70 cm of mercury, while the pressure inside is 33 cm of mercury. What is the maximum allowable internal pressure in the pipe? Given: D = 30cm = 70 kPa = 33 kPa Req’d: P =? Solution: Patm = 70(101.3/76) Patm = 93.303 kPa Pg = 33(101.3/76) Pg = 43.986 kPa FB = (93.303-43.986)(0.3) FB = 14.795 N/m FB = PiD 14.795 = Pi (0.3) Pi = 49.317 kPa

31. A steel pipe having a diameter of 15 cm and a wall thickness of 9.50 mm has an allowable stress of 140,000 kPa. What is the maximum allowable internal pressure in the pipe? Given: d = 15cm t = 9.50mm ς = 140,000kPa Req’d: P=? Solution:

Pi = wh S = (PD/ 2t) 140,000 KPa = P(0.15)/2(9.5E-3) P = 17,733.33 KPa or 17.73 MPa

32. A pipe carrying steam at a pressure of 7,000 kPa has an inside diameter of 20 cm. If the pipe is made of steel with an allowable stress of 400,000 kPa, what is the factor of safety if the wall thickness is 6.25 mm? Given: l = 3m h = 2.50m Req’d: P=? Solution:

FB = (Pi – Pe)D FB = (7000-0)(0.2) FB = 1400kN T = FB/2 T = 1400kN/2 T = 700kN S = T/t S=

S = 112000kPa FS = SA/S FS = FS = 3.57 33. A 60 cm cast-iron main leads from a reservoir whose water surface is at El. 1590 m. In the heart of the city the main is at EL 1415m. What is the stress in the pipe wall if the thickness of the wall is 12.5 mm and the external soil pressure is 520 kPa? Assume static conditions. Solution: S = [ (1716.75-520) (0.6) / 2 ] / 0.0125 S = 28.7 MPa

34. Compute the stress in a 90 cm pipe with wall thickness of 9.50 mm if the water fills it under a head of 70 m. Solution: S = (9.81)(70)(900) / (2)(9.5) S = 32527.89474 KPa or 32,530 KPa S = 32.53 MPa 35. A wood stave pipe, 120 cm in inside diameter, is to resist a maximum water pressure of 1,200 kPa. If the staves are bound by steel flat bands (10 cm by 2.50 cm), find the spacing of the bands if its allowable stress is 105 MPa. Solution: S = [(2)(10.5E-3)(10)(2.5)] / (1200)(120) S = 36.46cm

36. A continuous wood stave pipe is 3 m in diameter and is in service under a pressure head of 30 m of water. The staves are secured by metal hoops 2.50 cm in diameter. How far apart should the hoops be spaced in order that the allowable stress in the metal hoop of 105 MPa be not exceeded? Assume that there is an initial tension in the hoops of 4.50 kN due to cinching. Given: D = 3m S = 2.5 cm Allowable = 105 MPa Solution:

T = FB / 2 T = 98.1 / 2 T = 49.05 S = [(105) (2.5)] / (49.05) S = 10.6 cm 37. A vertical cylindrical container, 1.60 m in diameter and 4 m high, is held together by means of hoops, one at the top and the other at the bottom. A liquid of specific gravity 1.40 stands 3 m in the container. Calculate the tension in each hoop. Solution: ∑ 2T2 (4) = F (3) F = whA F = (9.81)(1.04)(1.5)(3)(1.6) F = 98.88 KN 2T2(4) = (98.88)(3) T2= 37.08 KN at the bottom of the hoop) 38. An open cylindrical wood stave tank contains three liquids with specific gravities 2, 3, and 4, respectively. The depth of the bottom liquid is 2 m, while the other two has a depth of 1 m each. If the diameter of the container is 2 m, determine the tension in the top and bottom hoops which are holding the container in place. Solution: p1=0 F1= (1/2)(p2)(2)(1) p2= p1+ωAЋA = 2ω = 0 + (2ω)(1) F2= p2(2)(1) = 2ω = 4ω p3= p2+ωBЋB F3= (1/2)(p3- p2)(1)(2) = 2ω + (3ω)(1) = 3ω = 5ω F4= p3(2)(2) P4= p3+ωCЋC = 20ω = 5ω + (4ω)(2) F5=(1/2)(p4- p3)(2)(2) = 13ω = 16ω ∑ Top=0 2T2(4) = F1(2/3) + F2(1.5) + F3(5/3) + F4(3) + F5(10/3) 8T2= T2=

=

T2= 154.10kN

39. An open tank in the shape of a frustum of a cone is bound by two hoops, one at the top and the other at the bottom. The diameters at the top and bottom are 2 m and 3 m, respectively. If the total bursting force is to be taken care of by the hoops, find the tension in each hoop if water stands 2 m in the tank. Height of tank is 3 m. Solution:

40. A masonry dam has a trapezoidal section: one face is vertical, width at the top is 60 cm and at the bottom is 3 m. The dam is 7 m high with the vertical face subjected to water pressure. If the depth of water is 5m, where will the resultant force intersect the base? Determine the distribution of pressure along the base, (a) assuming there is no uplift pressure; (b) assuming that the uplift pressure varies uniformly from the full hydrostatic at the heel to zero at the toe. Specific weight of masonry is 23.54 kN/m3. Required:x, x, &

& if there’s no uplift pressure if there’s an uplift pressure

Solution: (a) Vertical forces

Horizontal forces

(from toe)

(b).with uplift pressure

Vertical forces

Horizontal forces

(from toe)

41. The masonry dam of Problem 40 has its inclined face subjected to pressure due to a depth of 5 m of water. If there is no uplift pressure, where will the resultant intersect the base? Specific weight of concrete is 23.54 kN/m3. SOLUTION:

42. A masonry dam of trapezoidal cross section, with one face vertical has a thickness of 60 cm at the top, 3.70 m at the base, and has height of 7.40 m. What is the depth of water on the vertical face? What is the minimum length of the base of the dam such that the resultant will intersect the base at the downstream edge of the middle third? Assume that the uplift pressure varies uniformly from full hydrostatic at the heel to zero at the toe. SOLUTION:

43. A concrete dam is triangular in cross section and 30 m high from the horizontal base. If water reaches a depth of 27 m on the vertical face, what is the minimum length of the base within the middle third? What minimum coefficient of friction is required to prevent sliding? Determine the pressure distribution along the base. SOLUTION: P= ΥhA P=(9.81)(13.5)(27)(1) P=3575.75 kN W_1= 30B/2 (23.5) W_1=352.5 B R= W= 352.5 B RM= W_1 ( 2/3 B) RM= 352.5B( 2/3 B)

RM= 235B^2 OM= 3575.75( 27/3) OM=32181.75 R_x=RM-OM 352.5B( B/3)= 235B^2- 32181.75 117.5B^2= 32181.75 B= 16.55 m Min. coeff. of friction FS= μR/P 1= (μ(352.5)(16.55))/3575.75 μ=0.61 qT = (-R_y)/B [1+ 6e/B] = (-15(23.43)(16.55))/((16.55)) [1+ (6(2.755))/16.55] qT = 706.2kPa at the toe

44. The section of a masonry dam is shown in Figure U. If the uplift pressure varies uniformly from full hydrostatic at the heel to full hydrostatic at the toe, but acts over 2/3 of the area of the base, find: (a) The location of the resultant, (b) factor of safety against overturning, (c) factor of safety against sliding if the coefficient of friction between base and foundation is 0.60.

SOLUTION: Vertical Forces: W1 = ωCV1 = (23.5)(1/2)(21)(7)(1) = 1727.25kN

W4 = ωH20V4 = (9.81)(5)(8)(1) = 392.40kN W5 = ωH20V5

W2 = ωCV2 = (9.81)(1/2)(5)(10)(1) = (23.5)(21)(5)(1) = 245.25kN = 2467.50kN W6 = ωH20V6 W3 = ωCV3 = (9.81)(1/2)(5/3)(5)(1) = (23.5)(1/2)(5)(10)(1) = 40.875kN = 587.50kN U1 = ωH20VUplift U2 = ωH20VUplift = (9.81)(2/3)(5)(1)(16) = (9.81)(2/3)(1/2)(13)(10.6667) = 523.20kN = 453.44kN Horizontal Forces: FH1=ωЋA FH2=ωЋA = (9.81)(9)(1)(18) = (9.81)(2.5)(5)(1) = 1589.22kN = 122.625kN Rx=∑▒Fx Rx = FH1 – FH2 = 1589.22kN – 122.625kN = 1466.595kN Ry =∑▒Fy Ry = W1 +W2 +W3 +W4 +W5 -U1 -U2 = 1727.25 +2467.5 + 587.5 + 392.4 + 245.25 + 40.875 – 784.8 – 680.16 = 3995.815kN RM = W1x1 + W2x2 + W3x3 + W4x4 + W5x5 + W6x6 + FH2y2 = (1727.25)(4) + (2467.5)(8.5) + (587.5)(13.5) + (245.25)(14.33) + (40.875)(0.56) + (122.625)(1.67) = 44365.88kN-m OM = U1z1 + U2z2 + FH1y1 = (523.2)(8) +(453.44)(7.11) + (1589.22)(6) = 16944.88kN-m X =( RM-OM)/Ry = ( 44365.88 -16944.88)/3995.815 = 6.86m FSO = RM/OM =(44365.88kN-m)/(16944.88kN-m) = 2.62 > 1, safe from overturning FSS = μRy/Rx = ((0.6)(3995.815))/1466.595 = 1.63 > 1, safe from sliding

45. Shown in Figure V is an overflow dam. If there is no uplift pressure, determine the location of the resultant. SOLUTION:

46. The base of a solid metal cone (sp. gr. 6.95) is 25 cm in diameter. The altitude of the cone is 30 cm. If placed in a basin containing mercury (sp. gr. 13.60) with the apex of the cone down, how deep will the cone float? SOLUTION:

(

(

)

)

47. If the metal sphere 60 cm in diameter weighs 11,120 N in the air, what would be its weight when submerged in (a) water? (b) Mercury? SOLUTION: Vsphere= (4/3)πr3 (4/3)π(0.30)

3

0.1130973355 m3

Vsphere= Vsphere=

Wair =

Wwater= ? BF = Wair– Wwater 11,120 – Wwater = 9810 (0.1130973355) Wwater = 10,010.51514 N

11,120 N There will be no answer since sphere will float in Hg. Sgog Hg is greater than metal. 48. A rectangular solid piece of wood 30 cm square and 5 cm thick floats in water to a depth of 3.25 cm. How heavy an object must be placed on the wood (sp. gr. 0.50) in such a way that it will just be submerged? Solution: ΣF_V=0 BF-W_obj-W_W=0 W_obj=BF-W_W ΣF_V=0 W_W=BF W_W=wV_D =9.81(1000)(0.3×0.3×0.0325) W_W=28.69 N BF=wV_O

=9.81(1000)(0.3×0.3×0.05) BF=44.145 N W_obj=BF-W_W =44.145-28.69 W_obj=15.455 N

49. A hollow vessel in the shape of a paraboloid of revolution floats in fresh water with its axis vertical and the vertex down. Find the depth to which it must be filled with a liquid (sp. gr. 1.20) so that its vertex will be submerged at 45 cm from the water surface. SOLUTION: r^2/0.45= x^2/h x^2= (r^2 h)/0.45 BF = W ω VD = ωV (9.81)(1)(1/2)(πr^2)(0.45) = (9.81)(1.20)(1/2)πx^2h ½(0.45) r^2 = ½ (1.20)((r^2 h)/0.45) h h = 0.41m = 41cm 50. A barge is 16 m long by 7 m wide by 120 cm deep, outside dimensions. The sides and bottom of the barge are made of timber having thickness of 30 cm. The timber weighs 7860 N/m3. If there is to be a freeboard of 20 cm in fresh water how many cubic meters of sand weighing 15700 N/m3 may be loaded uniformly into the barge? SOLUTION: W = BF Volume of Barge (VB) VB = (7x 16 x 1.2) – (6.40 x 15.4 x0.9) VB = 134.4 – 88.703 VB = 45.696 m3 Volume of Submerged Barge ( VSB) VSB = 1 x 7 x 16

VSB = 112 m3 BF = Ws + WB (9.81)(112) = (7.860)(45.696) + (15.7)(Vs) 1098.72 = 359.17 + 15.7Vs Vs= 47.10 m3

51. A brass sphere (sp. gr. 8.60) is placed in a body of mercury. If the diameter of the sphere is 30 cm (a) what minimum force would be required to hold it submerged in mercury? (b) What is the depth of floatation of the sphere when it is floating freely?

SOLUTION: BF = WBody = F F = (9810)(13.6)(4/3)( π)(0.15)3 – (9810)(8.60)(4/3)(π)(0.15)3 F = 693.43N BF = W (9810)(13.6) VS = (9810)(8.60)(4/3)(π)(0.15)3 VS = 8939.679cm3 VS = π/3 D2 (3r-D) 8939.679cm3 = π/3 D2 (3(15cm)-D) D = 17.75cm 52. A spherical balloon weighs 3115 N. How many Newtons of helium have to be put in the balloon to cause it to rise, (a) at sea level? (b) At an elevation of 4570 m ? SOLUTION:

53. The specific gravity of rock used as concrete aggregate is often desirable to know. If a rock weighed 6.15 N in the air and 3.80 N when submerged in water, what would be the specific gravity of the rock? Given:

Required: Specific gravity=? Solution:

BF

54. A piece of wood weighs 17.80 N in air and a piece of metal weighs 17.80 N in water. Together the two weigh 13.35 N in water. What is the specific gravity of the wood? SOLUTION: BF = WWood and Metal in H20 BF = 13.35N BF = WWood + WMetal 13.35N = WWood + 17.80N WWood = -4.45N s.g = s.g = s.g = 0.8

55. In Figure W is shown a thin-walled inverted box 1.60 m long and 30 cm square, which was full of air before being submerged. In position (a), it is being held by 0.0280 m3 of concrete anchor weighing 23540 N/m3. Determine the depth D1. At what other depth D2 (position B) is the system in equilibrium? SOLUTION: Wair + Wrock = BF air + BFrock (0.3)(0.3)(1.6)(0.012) + (23.54)(0.028) = (0.3)(0.3) d (9.81) + (9.81)(0.028) 0.6608 – 0.2747 = 0.8829 d d = 0.44 m

56. In Figure X a gate 1.30 m square and 30 cm thick is hinged at A. It is subjected to water pressures on both sides. What force T is required to open the gate if it weighs 1780 N?

SOLUTION: F = whA F = (9.81)(1.8 - 0.3)(1.3)(0.3) F = 5.73885 KN F2 = (9.81)(1.2 – 0.3)(1.3)(0.3) F2 = 3.44331 KN

BF = 5.73885 + 1.78 BF = 7.51385 KN 1.3 T = 5.74885(1.3) + 3.44331(1.3) + 1.78(1.3) – 7.51385(1.3) T = 3.45 KN

W = 1.78 KN 57. The timber AB of Figure Y is 7.20 m long by 15 cm square. Find the specific weight of the timber and the total weight of the anchor (sp. gr. 2.40).

SOLUTION:

a.) ∑MA = 0 3.6 W – 2.6 BF = 0 (3.12)(ω)(0.15)(0.15)(7.2) – (2.6)(9.81)(6)(0.15)(0.15) = 0 ω = 6.44

sg = 6.44 / 9.81 sg = 0.7 b.) Wt + Wa = BFt + BFa (0.7)(0.15)(0.15)(7.2) + (2.4)(9.81)(VD) = (9.81)(6)(0.15)(0.15) + 9.81 (VD) VD = 0.15429 W = (2.4)(9.81)(0.15429) W = 363.25 N

58. A sphere 1 m in diameter floats half submerged in a tank of liquid (sp. gr. =0.80) (a) what is the weight of the sphere? (b) What is the minimum weight of the anchor (sp. gr. 2.40) that will require to submerge the sphere completely? Solution:

(

)

In half submerged:

(

)

59. Figure Z shows a hemispherical shell covering a circular hole 1.30 m in diameter at the vertical side of a tank. If the shell weighs 12,450 N, what vertical force is necessary to lift the shell considering a friction factor of 0.30 between the wall and the shell?

60. An iceberg has a specific gravity of 0.92 and floats in salt water (sp. gr. 1.03). If the volume of ice above the water surface is 700 m3, what is the total volume of the iceberg? SOLUTION: s.g. (iceberg) = 0.92 s.g. (saltwater) = 1.03 VD = (s.g.body/s.g. fluid)(V) VD = VT -700 m3 ∑ BF = W WliquidVD = wbodyVT (1.03)(9.81)(VT – 700) = 9.81(0.92) VT VT = 6554.545455 m3 61. A concrete cube 60 cm on each edge (sp. gr. 2.40) rests on the bottom of a tank in which sea water stands to a depth of 5 m. The bottom edges of the block are sealed off so that no water is admitted under the block. Find the vertical pull required to lift the block. SOLUTION: P = ,ρghseawater –ρghconcrete} Area = ,*(1000kg⁄m3 )(1.03)(9.81m/s2)(5m)] – *(1000kg⁄m3)(2.4)(9.81m/s2)(0.6m)]}(0.6m)2 P =13,102.236N ≈13.1022kN

62. A 15 cm by 15 cm by 7 m long timber weighing 6280 N/m3 is hinged at one end as shown in Figure AA. If the anchor weighs 23540 N/m3, determine the minimum total weight it must have. SOLUTION: [W1 + W2]V = Wtotal [6280N/m3 + 23540 N/m3](0.15m × 0.15m × 7m) = Wtotal [29820N/m3](0.1575m3) = Wtotal 4696.65N = Wtotal Wtotal = 4696.65N ≈4700N 63. A cylinder weighing 445 N and having a diameter of 1 m floats in salt water (sp. gr. 1.03) with its axis vertical as in Figure BB. The anchor consists of 0.280 m3 of concrete weighing 23540N/m3. What rise in the tide r will be required to lift the anchor off the bottom?

Solution: ∑ BF = W 445 N + 23540(0.280) = 0 BF = 7036.2 N W = (9.81)(1.03)(Vc + 0.280m2) W = 10.1043 (0.3156 + 0.785r)

W = 5.21 + 7.93r BF = W 7.0362 KN = 5.21 + 7.93r r = 0.23 m or 23cm

64. A timber 15 cm square and 5 m long has a specific gravity of 0.50. One end is hinged to the wall and the other is left to float in water (Figure CC). For a = 60 cm, what is the length of the timber submerged in water? Solution: P = wh P = (9.81)(1.30) P = 13.2435 KN Wwood = wV Wwood = (9.81)(0.5)(0.270) Wwood = 1.05948 P = F/A P = 1.05948/(o.6)2 P = 2.94 KN

65. A metal block 30 cm square and 25 cm deep is allowed to float on a body of liquid which consists of 20 cm layer of water above a layer of mercury. The block weighs 18,850 N/m3. What is the position of the upper level of the block? If a downward vertical force of 1110 N is applied to the centroid of the block, what is the new position of the upper level of the block? Solution:

66. Two spheres, each 1.2 m in diameter, weigh 4 and 12 kN, respectively. They are connected with a short rope and placed in water. What is the tension in the rope and what portion of the lighter sphere protrudes from the water? What should be the weight of the heavier sphere so that the lighter sphere will float halfway out of the water? Solution: Tension in the hoops T+BF_2=12 kN T+(π(1.2)^3)/6(9.79)=12 kN T=3.14 kN Portion of the lighter sphere protrudes from the water BF1=4kN+3.14 kN BF1=7.14 kN BF2=(π(1.2)^3)/6 (9.79)=8.56 kN By proportion: (8.56-7.14)/8.56= V_1/V V1= (8.56-7.14)/8.56 V V1=0.1659 V1=16.59 % of V above the water surface

67. A flat-bottomed scow is built with vertical sides and sloping ends. Its length on deck is 24.40 m, on the bottom 19.80 m, its width 6.10 m and its vertical depth 3.70 m. What is its depth of flotation in sea water (sp. gr. 1.03) if the scow weighs 2,220 kN? Solution: BF = W ωVD= 2220kN (9.81)(1.03) [((24.40+19.80))/2] (6.10)(x)=2220kN X = 1.63m 68. A ship with cargo weighs 44,480 kN and draws 7.60 m of salt water. On crossing a bar at the entrance to a river her draft is decreased by 30 cm by the discharge of 2670 kN of water ballast. In going up the river to fresh water, 445 kN of coal burned. What will her draft be then and how much ballast must be required to increase it by 30 cm? Assume that the sides of the vessel near the water surface are vertical. Solution:

69. A tank with vertical sides is 1.30 m square, 3 m deep and is filled to a depth of 2.70 m with water. By how much, if at all, will the pressure on one side of the tank be changed if a cube of wood (sp. gr. 0.50), measuring 60 cm on an edge, be placed in the water so as to float with one ace horizontal? Solution: P = 9.81(1.35) P = 13.2435 Wwood= ωV = 9.81(. = 2.12 kN

)

70. A cask which weighed 270 N was placed on platform scales and then nearly filled with water. A total load on the scales of 1425 N was read. Should the net weight of water as computed from these data be corrected by reason of the fact that a 7.5 cm diameter vertical shaft suspended from the ceiling above has its lower end submerged in the water to a depth of 30 cm? If so, by what amount? Solution:

71. . If the specific gravity of a body is 0.80, what proportional part of its total volume will be submerged below the surface of a liquid (sp. gr. 1.20) upon which it floats? Solution: BF = W ωVD = ωV (9.81)(1.20)(V-x) = (9.81)(0.8)(V) 11.772V – 11.772x = 7.848V 3.924V/11.772 = 11.772x/11.772 X = 1/3 V VSubmerged = V – x = V - 1/3 V VSubmerged= V

72. .A vertical cylindrical tank, open at the top, contains 45.50 m3 of water. It has a horizontal sectional area of 7.40 m2 and its sides are 12.20 m high. Into it is lowered another similar tank, having a sectional are of 5.60 m2 and a height of 12.20 m. The second tank is inverted so that its open end is down and it is allowed to rest on the bottom of the first. Find the maximum hoop tension in the outer tank. Neglect the thickness of the inner tank. Solution: P=ωh P = (9.81) (6.15) P = 60.32 kPa FB = 60.32 (5.60) FB = 337.79 T = FB / 2 T = 337.79 / 2

T = 168. 896 kN/m 73. . A small metal pan of length 1 m, width 20 cm, and depth 4 cm floats in water. When a uniform load of 15 N/m is applied as shown in Figure DD, the pan assumes the figure shown. Find the weight of the pan and the magnitude of the righting moment developed. Solution:

74. .A ship of 39,140 kN displacement floats in sea water with the axis of symmetry vertical when a weight of 490 kN is midship. Moving the weight 3 m toward one side of the deck causes a plumb bob, suspended at the end of a string 4 m long, to move 24 cm. Find the metacentric height. Solution: Tanθ= Θ= 3.43 RM= w(MGsinθ) MG= MG= MG= 0.627m 75. .A rectangular scow 9.15 m wide by 15.25 m long and 3.65 m high has a draft of 2.44 m in sea water. Its center of gravity is 2.75 m above the bottom of the scow. (a) Determine the initial metacentric height. (b) If the scow tilts until one of the

longitudinal side is just at the point of submergence, determine the righting couple or the overturning couple. Solution: A) MG= MGo + GBo MBo=

(1+

MBo=

) ; θ=0 (1+

)

GBo= 2.75- (2.44-2) GBo= 1.53m

MBo= 2.82m MG= 2.84 – 1.53 MG= 1.33m

B) OM= BF. X

GBo= 2.75-1.22

X= MGsinθ

= 1.53

MG= MBo-GBo MBo=

(1+

Θ= tan (

)

MG=2.96-1.53

)

= 1.43m

Θ= 14.81°

x= MGsinθ x= 1.43 sin (14.81)

MBo=

(1+

)

x= 0.37m

MBo= 2.96 OM= BF.x = (9.81) (1.03)(9.15)(2.44)(15.25)(0.37) OM= 1256.9KN.m 76. A cylindrical caisson has an outside diameter of 6 m and floats in fresh water with its axis vertical. Its lower end is submerged to a depth of 6 m below the water surface. Find: (a) the initial metacentric height; (b) the righting couple when the caisson is tipped through an angle of 10°. Solution:

77. A rectangular scow 9.15 m wide by 15.25 m long has a draft of 2.44 m in fresh water. Its center of gravity is 4.60 m above the bottom. Determine the height of the scow if, with on side just at the point of submergence, the scow is in unstable position. Solution: GBo = 4.60 -1.22 GBo = 3.38 MBo = [(9.15^2) / 12(2.44)] [1 + (tan0 ^2) / 2] MBo = 2.86 MG = 3.38 – 2.86 MG = 0.52 m y = 4.6 – 1.22 y = 3.38 h = 3.38 + 2.44 – 0.52 h = 5.3 m 78. A rectangular scow 9.15 m wide by 15. 25 m long and 4.60 m high has a draft of 2.75 m. Its center of gravity, transversely and longitudinally, is at the center of the scow. If the sow is tipped transversely until one end is at the point of submergence, find the righting couple. Solution:

(

)

79. .A rectangular raft 3 m wide and 6 m long has a thickness of 60 cm and is made of solid timbers (sp. gr. 0.60). If a man weighing 890 N steps on the edge of the raft at the middle of one side, how much will the original water line on that side be depressed below the water surface?

Solution:

80. .The timber shown in Figure EE is 30 cm square and 6 m long, having a specific gravity of 0.50. A man standing at a point 60 cm from one end causes that end to be just submerged. Find the weight of the man. Solution:

81. .A submarine of 10, 700 kN displacement has its center of gravity 30 cm from its center of volume. What is the righting couple when it is submerged in sea water and the angle of heel is 5˚. Solution:

82. .A log 30 cm in diameter weighs 4900 N/m3. What is its shortest length so that it may float in water with the axis horizontal? Solution: Wlog = 4.9 (0.30)3 = 0.1323 kN BF = Wwood 9.81 (1) *π (0.15)2 h] = 4.9 h = 0.1908 m h = 19.08 cm

83. .A block of wood is 15 cm square and 30 cm long. What is the specific gravity of the wood if the metacentre is at the same point as the center of gravity of the wood when the block is floating in water on its side? Would it be stable floating on its end? Explain. Solution: W = BF 9.81(0.15)(0.15)(.3)(S) = 9.81(1)(0.15)(0.3)(.10) S = 0.667

84. .Figure FF shows a scow equipped with a derrick with a boom 5.50 m long. What maximum weight could be picked by the boom at its end along the longitudinal side in order that water will not enter into the scow? Assume the weight of the boom to be negligible and consider its position to be always horizontal.

Solution: MG = Mbo-Gbo Mbo = I/V =0.1188 in. I = bh3/12 =(10)3(16)/12 = 16000 m4 V = bhw = 10(16)(2.6) = 46 m3

Gbo = 3= 1.7 m MG = Mbo + Gbo MG = .12 + 1.7 MG = 1.82

W = ѡV W = 9.81(1)(416) W = 4080.86 KN

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