DamsDams (cont.)Fluid StaticsDam DesignSlide 5Example (cont.)Slide 7DamsArch DamGravity DamArch & Gravity DamDams (cont.)Hydrostatic forcesEnergy conversionBernoulli equationTurbineHydrostatic upliftFluid StaticsWhen a surface is submerged in a fluid at rest, hydrostatic forces develop on the surface due to the fluid pressure. These forces must be perpendicular to the surface since there is no shear action present. These forces can be determined by integrating the static pressure distribution over the area it is acting on.Example: What is the force acting on the bottom of the tank shown?hTank area AFluid with density Dam DesignDesign concern: (Hydrostatic Uplift) Hydrostatic pressure above the heel (upstream edge) of the dam may cause seepage with resultant uplift beneath the dam base (depends largely on the supporting material of the dam). This reduces the dams stability to sliding and overturning by effectively reducing the weight of the dam structure. (Question: What prevents the dam from sliding?)Determine the minimum compressive stresses in the base of a concrete gravity dam as given below. It is important that this value should be greater than zero because (1) concrete has poor tensile strength. Damage might occur near the heel of the dam. (2) The lifting of the dam structure will accelerate the seeping rate of the water underneath the dam and further increase hydrostatic uplift and generate more instability.Catastrophic breakdown can occur if this factor is not considered: for example, it is partially responsible for the total collapse of the St. Francis Dam in California, 1928.Dam Design30 m40 m20 mconcrete=2.5 waterFirst, calculate the weight of the dam (per unit width): W=Vg=(2.5)(1000)(20)(40)(1)(9.8)=19.6106 (N)The static pressure at a depth of y: P(y)=wgyyFree surface3022 6w w0The total resultant force acting on the dam by the water pressure is:R= P(y)dy= (1000)(9.8)(1/ 2)(30) 4.4 10 ( )2hhgydy g N        Example (cont.)3303w2w w w0The resultant force, R, is acting at a depth h below the free surface so that23Rh= P(y)ydy= ( ) , 20( )3 3Assume the load distribution under the dam is linear (it mighhgh hgy ydy g y dy g h mR        max minminht not be linear if the soildistribution is not uniform)Therefore, the stress distribution can be written as(x)=20x  RW20 mx dam,x20y , max min0In order to reach equilibrium, both the sum of forces and the sum of moments have to balance to zeroF 0, R=F (frictional force and the air drag force)F 0, ( ) 10( )1.96 1dam yW F x dx      6max min0 ( )N  minmaxFree surfaceExample (cont.)20O020 206 6 2max minmax0 06max minThe sum of moments has to be zero also: Taking moment w.r.t. the heel of the damM 0, (10) (10) ( ) 0(10)(4.4 10 19.6 10 )20240 10 133.3 66.7Solve: R W x xdxxdx x dx             6 6max min1.64 10 ( ), 0.32 10 ( )The minimum compressive stress is significantly lower than the maximum stressThe hydrostatic lift under the dam (as a result of the buoyancy induced by water seeping unN N   liftder the dam structure) can induce as high as one half of the maximumhydrostatic head at the heel of the dam and gradually decrease to zero at the other end. 1That is ( ) (0.5)(1000)(9.8)(302wgh  66) 0.147 10 ( )Therefore, the effective compressive stress will only be 0.173(=0.32-0.147) 10 ( ).NN