A. Hydrostatic Force on Plane Surface 1. Formula π = πΈππ¨ π = π·ππ π¨ π°π π= π¨π Where: β = depth of liquid above the centroid of the submerged area πππ = pressure at the center of gravity cg = center of gravity cp = the center of pressure πΌπ = Inertia with respect to an axis passing through its centroid and parallel to the x axis e = eccentricity, distance between the center of pressure and center of gravity 2.Pressure Diagram F = Volume of Pressure Diagram Z = h/3 (rectangular section) Z = h/4 (triangular section) 3. Integration Force: π π = π·ππ β π π¨ ∫ π π = ∫ πΈπ β ππ π Location: Using Varignon’s Theorem πΉ β π¦ = ∫ ππΉ β π₯ π= ∫ π π β π π B. Hydrostatic Force on Curved Surface ππ― = πΈππ¨ = π·ππ π¨ ππ½ = πΈ β π½ πΉ = √ππ― π + ππ½ π π π Location: πππ§ π½ = πΊ = π π½ π― ADS Problems: A. Hydrostatic Force on Plane Surface 1. A vertical rectangular plane of height 3 m and base 1.5 m is submerged in water with its top edge at the surface. Determine the total force F acting on one side and its location from the liquid surface using a) Formula b) Pressure Diagram and c) Integration 2. A vertical triangular plate whose height is 3.6 m has its base horizontal and vertex upper most in the water surface. a) Find the depth to which it must be lowered from its vertex so that the difference in level between the center of gravity and the center of pressure shall be 20 cm. b) What is the base width of the plate of the total hydrostatic force acting on the plate is 76.28 kN. 3. Find the location of the resultant force F of the water on the triangular gate and the minimum force P necessary to hold the gate of it is to be applied at top of the gate. Neglect the weight of the gate. 4. The semi-circular gate shown below is hinged at B. Determine the force P required to hold the gate in position. 5. Determine the magnitude and location of the total hydrostatic force acting on the 2m x 4 m gate shown in the figure. 6. Water in a tank is pressurized to 85 cm Hg. Determine the total force per meter width on panel AB. 7. In the figure shown, the gate is 1.2 m square and weighs 6690 N. Neglecting the thickness of the gate and the weight of the chain. Compute a) the normal force acting on the gate due to oil above the gate b) normal force acting on the gate due to water below the gate and c) minimum force required to open the gate. Problems: 1. Problem in Gillesania 2. The crest gate shown consists of a cylindrical surface of which AB is the base supported by a structural frame hinged at O. The length of the gate is 10 m. Compute the magnitude and location of the horizontal and vertical components of the total pressure on AB. 3. In the figure shown, determine the horizontal and vertical components of the total force acting on the cylinder per m of its length. Dams 1. Definition - Structures that provide safe retention and storage of water - Block the flow of a river, stream, or other waterway 2. Purpose of Dams 1) Water Supply and Irrigation - Stores rainfall with an average of 2.5m/yr from catchment area 2) Flood Control 3) River Regulation - Controls the flow from upstream to downstream side a) When water flow to ocean, it becomes saline which is generally not good for drinking b) To have water supply during summer 4) Hydroelectric Power Generation - Dams harness the energy of flowing water to generate electric power - Angat Dam can produce 270 MW if dam is at full capacity 5) Navigation and Roadways - For highway inspection and recreation 3. Location of Dams - Lowest point of river to store maximum rain - Good catchment area - Low slope to minimize runoff since sediments (solids due to runoff) can be stored in dam occupying space for storage which may require desilting (removal of sediments) 4. Design Team Professionals involved in Dams 1) Hydraulic Engineer - Analyze flow of water - Design of open channels, pipelines, etc. 2) Hydrologist - Determines the amount of rainfall 3) Structural Engineer - Design of dam itself 4) Geotechnical Engineer - Determines the property of soil where dam will be constructed - Determines the soil condition suitable for which type of dam should be constructed 5. Issues 1) Environmental - No aquatic life Solution: placing of ladder to preserve salmon reproduction since salmons tend to go to higher places when laying eggs 2) Economic - Displaces communities - Solution: building of mini dams (3 mini dams = 1 big dam) 6. Types of Dams 1) Gravity Dam Stability depends on weight. Hence, force of gravity is used to resist water pressure from pushing the dam downstream or tipping it over - Key – for additional shear resistance other than friction - Critical part is base since that is where hydrostatic pressure is maximum 2) Arch Dam - Transmits most horizontal thrust of water behind to the abutments by arch action - Thinner cross-section - Used in narrow canyons (sides should have high geologic foundations) - May rest on mountains, rocks, canyons - Critical part are the connections on the sides 3) Buttress Dam Each buttress has its own footing to decrease the bearing pressure exerted by the structure to the soil - Spread Footing: used when bearing capacity of soil is uniform Matt Footing: used when bearing capacity of soil is not uniform so that structure will settle as a whole - Sloping membrane which is the slab transmits the water load to a series of buttresses - Buttresses keep the slab from collapsing - Sometimes called hollow dams 4) Embankment - Gravity dam formed out of loose rock, earth, or combination - Downstream and upstream are flatter (21.5O maximum slope) to prevent runoff Disadvantage: seepage may take place since embankment is porous because it is made of soil and rocks - Solution: placing of impermeable core made of concrete to prevent seepage - Rock-fill-toe – placing of big rock so that soil will not flow with water 7.Parts of Dam Failure Modes 1) Rotation about the toe - Causes overturning of the dam - Dangerous due to possible spillage of water - Compressive stress occurs at toe – critical since dam might sink if too compressed 2) Sliding - Less critical than overturning - Occurs at horizontal plane - Placement of key to add shear resistance Forces in Dams Step 1 Consider 1 unit length (1 m length) of dam perpendicular to the cross section. Step 2 Determine all the forces acting: 1. Vertical forces W = Weight of dam, γW_c=γγ_c V Fv = Weight of water in the upstream side (if any), γF_v=W_w=γγ_w V U = Hydrostatic uplift, U= γV Weight of permanent structures on the dam 2. Horizontal forces FH = Horizontal component of total hydrostatic force, F_H=γ¯h A Wind pressure, wave action, floating bodies, earthquake load, etc. Step 3 Solve for the reaction Horizontal component of the reaction R_x=∑βγF_H=γ F_H Vertical component of the reaction R_y=∑βγF_v=γ W+F_v-U Step 4 Moment about the toe • Righting moment, RM RM = Sum of all rotation towards the upstream side RM=Wγβxγ_1+F_vβx_2 Overturning moment, OM OM = Sum of all rotation towards the downstream side OM=F_H γβyγ_1+Uβy_2 Factor of Safety Factor of safety against sliding, FSs γFSγ_s=(μR_y)/R_x =(Sliding Resistance (Friction))/(Sliding Force) γFSγ_s>1 Where: μ = coefficient of friction between the base of the dam and the foundation. Factor of safety against overturning, FSo γFSγ_O=RM/OM=Capacity/Actual>1.5 Foundation Pressure/ Pressure Intensities @ the Base Location of Resultant, x Μ R_yβ x Μ =RM-OM Eccentricity, e e=|B/2-x Μ | 1. for e≤ B⁄6 Ry is within the middle third and the foundation pressure is trapezoidal acting from heel to toe. If e is exactly B/6, the shape of foundation pressure is triangular also acting from heel to toe. q=-R_y/B (1±6e/B) Note: +q = Tensile stress - q = Compressive stress 2. for e> B⁄6 In foundation design, soil is not allowed to carry tensile stress, thus, any +q will be neglected in the analysis. If e>B/6, Ry is outside the middle ππ = ππΉπ Μ ππ Problems: 1. A concrete dam retaining water is shown in the figure. If the specific weight of concrete is 23.5 kN/cu.m., find the factor of safety against sliding and overturning, and the pressure intensity of the base. Assume hydrostatic uplift varies uniformly from full hydrostatic pressure at the heel of the dam to zero at the toe and that the coefficient of friction between dam and foundation soil is 0.45. 2. A masonry dam (s=2.4) of a trapezoidal cross section with one face vertical and horizontal base is 24 m high. Its crest width is 2 m and the width of the base is 12 m. The uplift pressure will vary from 60% at the heel to 75% at the toe. a.) what should be the depth of water on the vertical side if the resultant pressure intersects the base at the outer edge of the middle third? Other keywords: resultant force falls within the middle thirds; maximum intensity of pressure at the toe is twice the average pressure at the base; resultant force will cut the extreme edge of the middle third near the toe; resulting soil pressure diagram is triangular (qmin=0) b.) If the angle of friction between the dam’s base and the soil is 21.8 degrees, determine the factor of safety against sliding in this condition. c.) what is the minimum coefficient of friction to prevent sliding? d.) Find the unit horizontal shearing stress at the base Buoyancy Archimedes' principle of buoyancy or Law of hydrostatics states that the upward buoyant force that is exerted on a body immersed in a fluid (gas or liquid), whether fully or partially submerged, is equal to the weight of the fluid that the body displaces and acts in the upward direction at the centroid of the displaced fluid or the center of buoyancy. Buoyancy = weight of displaced fluid Illustration of Archimedes’ principle of buoyancy. Here a 5-kg object immersed in water is shown being acted upon by a buoyant (upward) force of 2 kg, which is equal to the weight of the water displaced by the immersed object. The buoyant force reduces the object’s apparent weight by 2 kg—that is, from 5 kg to 3 kg Body is partially or fully submerged in fluid: • Loss in weight of the body = Weight of water displaced by the body = Buoyant Force exerted by water on the body. • Volume of the water displaced = Volume of the body immersed in water. A net upward vertical force results because pressure increases with depth and the pressure forces acting from below are larger than the pressure forces acting from above. BF= γ_fβV_D For floating body in a homogeneous liquid at rest: BF=W W=γ_OβV_O For completely submerged: BF= W_(O in air)- W_(O in fluid) Where: BF = buoyant force γ_f = unit weight of the fluid V_D = volume displaced by the body W = weight of the object γ_O = unit weight of the object V_O = volume of the object W_(O in air) = actual weight of object measured in air W_(O in fluid) = apparent weight of object measured when submerged in fluid Problems: 1. A stone weighs 468 N in air. When submerged in water it weighs 298 N. Compute the volume and the unit weight of the stone? 2. A tank with vertical sides is 1.5 m square, 3.5 m depth is filled to a depth of 2.8 m of a liquid having a sg = 0.80. A cube of wood having a sg = 0.60 measuring 1 m on an edge is placed on the liquid. a) Find the weight of the volume of liquid displaced. b) By what amount will the liquid rise on the tank? c) What will be the forced increased on the side of the tank? 3. A block of wood 0.60 m x 0.60 m x h meters in dimension was thrown into the water and floats with 0.18 m projecting above the water surface. The same block was thrown into a container of a liquid having a specific gravity of 0.90 and it floats with 0.14 m projecting above the surface. Determine the following: a) value of h and b) weight of the block 4. A cube 2.2 feet on an edge has its lower half of sg = 1.6 and upper half of sg = 0.7. It rests in a two layer fluid, with lower sg = 1.4 and upper sg = 0.8. Determine the height h of the top of the cube above the interface. 5. A hollow cylinder 1.0 m in diameter and 2 m long weighs 3825 N. a) How many kN of lead weighing 110 kN/m^3 must be fastened to the outside bottom to make the cylinder float vertically with 1.50 m submerged in fresh water. b) How many kN of lead weighing 110 kN/m^3 must be placed inside the cylinder to make the cylinder float vertically with 1.50 m submerged in fresh water c) What additional load must be placed inside the cylinder to make the top of the cylinder flush with the water surface? 6. A ship having a displacement of 24, 000 tons and a draft of 34 feet in ocean enters a harbor of fresh water. If the horizontal section of the ship at the waterline is 32,000 sq. ft. what depth of fresh water is required to float the ship? Assume the marine ton is 2240 lb and the sea water and fresh water weight 64 pcf and 62.2 pcf, respectively.
