Hydraulics Solution Sheet 3 Forces on submerged …

1 HydraulicsSolution Sheet 3 Forces on submerged concrete dam with a vertical upstream face impounds water of density to a total depth ofd,with gravitational a graph showing the variation of pressure with the force on the dam per unit have the formula for the force on any planeFx= gAx this case, the area of a rectangle,Ax=d 1, as we are calculating per unit width, and h=d/2, as the centroid is at the centre of the rectangle, henceFx=12 is a well-known and commonly-used the depth of the centre of pressure, and add it to your sketch, showing the resultantof the can use the formula that we obtained in the additional sheetMx=hCPFx= gI,whereIis the second moment of area about an axis in the surface. As the axis is about one side ofa rectangle of width1and depthd, we use the formula from Table 1 on the Revision Sheet ofMomentshCP=IAx hx=2nd Moment of area about axis through surface1st Moment of area about axis at the surface=13 1 d312 1 d3= the centre of pressure lies2d/3below the surface, which is also a well-known and the resultant force and the centre of pressure vertical square plate , the centre of the the surface.

2 (Ans.:) , gA h= 1000 h=IGA h= m,hCP= + vertical circular plate , the centre of the the (Ans. , m.)F= gA h= 1000 /4 h=IGA h= m,hCP= + cylindrical boom shown is required to retain a spill of oil of depthdoand relative density obyfloating in seawater of relative density that the seawater on the other side of the boom rises to a depth ofdw=do o/ wabovethe oil-water pressure on the underside of the oil (pressure equals density offluid g depth) is the hydrostatic pressure equation between a point in the seawater just under the oiland a point on the other side of the boom at the same elevation, ogdo=pressure due to seawater= wgdw, dw=do o/ that the horizontal force on the boom per unit length is12 ogd2o(1 o/ w),where is the density of fresh calculate the force per unit length we exploit our knowledge (actually about to be obtainedin the next question) that the force per unit length on a depthvertical rectangular shape, here theprojection of the boom, is1/2 density g depth2.

3 force to right due to oil above interface=12 ogd2oForce to left due to seawater above level of interface=12 the level of the interface the problem is symmetric, hence, the net force isF=12 g od2o wd2w ,however using the result fordwfrom part (a),F=12 g od2o wd2o 2o/ 2w =12 g od2o(1 o/ w). which direction is the force ? What implication does this have for the stability of suchflexiblebooms in retaining oil spills? What plan shape do you think aflexible boom would take up?The force is to the right, as o/ wis less than 1. This means that the boom will will be pushedoutwards by the oil, so that the configuration is generally stable, and we expect it to tend to acircle in vertical bulkhead in a ship has a door which has to be designed such that it will not be forcedopen if part of the shipfills with seawater of density1025 kg m 3. The door is2mhigh and1mwide, and the sea surface is assumed to be1mabove the top of the the force on the door and its total force on the door isF= gA h= 1025 2 1 (1 + 2/2) = position of the centre of pressure ishCP h=IGA h=1 23/122 1 (1 + 2/2)= m,hence it mbelow the top of the door.

4 By symmetry, it is in the left-right centre of door is fastened by two hinges A and B on one vertical edge,15 cmfrom top and bottom,and by a latch C in the centre of the other vertical edge. Calculate the Forces on each hinge andon the latch when one face of the bulkhead is subject to water pressure (Ans. , , ).Now take moments about the axis AB - the moment arm of the force at C is twice that of thewater force , henceFC= take moments about a horizontal axis through B. The moment arm ofFAis2 2 = The moment arm ofFCis1 = m. The water force mbelow thecentroid, hence it has a moment arm = +FC = 0FA + evaluate the remaining force we just consider horizontal equilibrium and obtainFB= mixing tank with trapezoidal ends , at the top, at the base. If the tank is completelyfilled with water, total weight force of water in the ( + ) = m2 Hence the volume m3, and the total weight is gV= 1000 total force exerted by water on the of water which would occupy the volume between base and surface= gV= 1000 total force exerted on one end and where it acts.

5 To do this it will be necessary to calculatethefirst and second moments of area of the trapezium which forms the face about an axis alongthe top of the trapezium, the water hcan be calculated by using known properties of rectan-gles and triangles. It is suggested that you calculate it that way (MO= m3). , to check and to prepare for the next part set up and evaluate an integral for thatfirst moment of area in terms of the widthbof the trapezium as a function of , set up and evaluate a similar integral for the second moment of areaIO,sothatyoucan use the expressionhCP= :i. To do this wefirst have to calculateA h, which is thefirst moment of area of the trapeziumabout the surface. We can use the fact that the centroid of a triangle is1/3of its height fromits h=A h square+2 A h triangle= 12 +2 12 13 m3ii. The width of the trapezium athbelow the surface isb= 2 dh= ( 2h/3)hdh= m3 HenceF= gA h= 1000 = Now to calculate its position we usesCP= gsin IO gsin A s=IOA s,10so we must also calculateIOof the ( 2h/3)h2dh= m4,and sohCP= mfrom the sluice gate consists of a radial gate of radius2mpivoted at its centre O, as shown in the magnitude and direction of the resultant force on the gate due to the water, and thenet moment required to open the = 2mD = 3mGate B = 2m wideFThe horizontal component on the gate is equal to the horizontal force on the projection onto a ver-tical plane of heightRand widthBwith its top edgeDbelow the surface.

6 The force isFx= gAx hx= gBR D+R2 .The position of the centre of pressure is that of the centre of pressure of the projection onto a verticalplane, which is a rectangle of dimensionsB R, however the top edge is stillDbelow the surface,so wecannotjust use the formula13bd3. We could use the parallel axes theorem, however it is moreintuitive here to calculate the net second moment of a large rectangle of total depthD+Rwithedge at the surface minus that of a rectangle of depthDbut also with edge at the surface:I=13B(D+R)3 13BD3=BR 13R2+RD+D2 .In case anybody is disappointed that we didn t use the Parallel Axes Theorem, here it is, using thesecond moment of area of a rectangle about an axis through the centroid112BR3plus the area timesthe square of the distance between surface axis and centroidBR (D+R/2)2:I=112BR3+BR D+R2 2=BR 13R2+RD+D2 showing that the intuitive method ,hCP=IAx hx=13R2+RD+D2D+ the vertical force is equal to the weight force on thefluid which would occupy the regionbetween the level of the free surface and the surface on which we are to calculate the force .

7 In thiscase this is the volume occupied by the quadrant of the gateplus the rectangular region above gV= gB 4R2+DR = gBR 4R+D 11 This force acts through the centre of gravity of that volume and is this we takemoments of volume about an axis through O (to do this we need to know that the distance of thecentroid of a semi-circle or quadrant is4/(3 )times the radius from the centre of the circle):Total moment of vertical force about O= gB 14 R2 4R3 +DR 12R = gBR2 R3+D2 Thus, the vertical force acts a horizontal distance of xfrom O: x=Moment of vertical forceFz= gBR2 R3+D2 gBR 4R+D =R R3+D2 4R+DNow we calculate the net moment of the horizontal and vertical Forces about O:Net moment= gBR2 R3+D2 Fx(hCP D)= gBR2 R3+D2 gBR D+R2 13R2+RD+D2D+R2 D!=0!As it should be in this radial gate we put numbers in:B=2m,D=3m,R=2m,and = 1000 kg m 3andg= 2:Fx= gBR D+R2 = 1000 2 2 (3 + 2/2) = 157 kNhCP=13R2+RD+D2D+R2= mFz= gBR 4R+D =179kN x=R R3+D2 4R+D= mThus, a horizontal force of157 kNacts to the right, a distance 3= mbelowthe pivot, and the vertical force of179 kNacts vertically upwards (the gate is above the water) ahorizontal distance mfrom the resultant force is 1572+ 1792= 238 kN, at an angle to the horizontal ofarctan (179/157) = 49.

8 Net moment of pressure Forces about pivotNet moment of pressure Forces about pivot= 157 179 0to our order of accuracy. This result is expected, as every element of the pressure force is perpendic-ular to the arc, so that the line of action of the resultant is through the pivot, and there is no moment,the reason for using radial spherical container is made up of two hemispheres, the joint between the two halves being hor-izontal. The sphere is completelyfilled with water through a small hole in the top. It is found that50kg of water are required for this purpose. If the two halves of the container are not secured to-gether, what must be the mass of the upper hemisphere if it just fails to lift off the lower hemisphere?(Ans. ).12 The force in the direction of gravity is equal to the weight force on thefluid which would occupy avolume between the surface on which the force is to be calculated and the plane of the free is, it is the region between the upper hemisphere, and the level of the water surface, whichforms a cylinder of radiusrand heightr.

9 The direction of the force is volume of this region= r2 r|{z}Vol. of cylinder 12 43 r3|{z}Vol. of hemisphere=13 r3,hence the mass of water in that region ism= 13 r3. However the mass of water whichfills thesphere is50 kg, hence 43 r3=50,andm=50/4= corner of a tank is bevelled by equal dimensionsdas shown in thefigure. It isfilled tohabovethe top of the bevel. What is the force on the triangular corner?xzydddOYXZd+hForce inxdirection= force on projection OYZ= gAOYZ hOYZ= g 12d2 h+23d = gd26(3h+2d)= force inydirection by symmetryForce inzdirection=Weight force on prism above XYZ= g Vo l u m e= g d22 13(h+d+h+d+h)= gd26(3h+2d).Thus, each of the force components is the same. This is what we would expect for such a surfacewhose direction cosines are the same for all 3