[94 Fluid Mechanics ] where [= MOL of gate = 24 = 50X12" 2 2 72m AS 4 4.576 = 013 + 4.576 = 4.589 m Depth of centre of pressure h* = 6x4576 a But from Fig. 3.23 (a), * asin ase ig. (a), OH sin Distance, 589 x JF = 6489 m Distance, Bo=—>_ = 5x J2=70T1m ana Distance, BH = BO ~ OH = 7.071 ~ 6489 = 0.582 m Distance 2 - 0.582 = 0.618 m Taking the moments aout the hinge A P x AB = F x (AH) where Ps the fore nomal tthe gate applied at B Px 12 209383 x 0518 2683430618 ~ 12 Problem 3.19 A gate supporting water is shown n Fig. 3.24 Find th eight ofthe water 50 that the pte tps about ihe hinge, Tak he wlth o the gate ani Solution. Given: 0 =60° hh * sin 60° V3. P = 138708 N. Ans. Distance, where h = Depth of water. The gate will stat tipping about hinge B if the resultant pressure force acts at B. If the resultant pressure force passes through a point which i lying from B to C anywhere on the gate, the gate will ip ‘over the hinge. Hence limiting ease is when the resultant force passes through B. But the resultant force passes through the centre of pressure. Hence for the given position, point B becomes the centre of pressure. Hence depth of centre of pressure, hv = (h—3)m FREE WATER SURFACE = losin? @ 5 Ah HINGE But h* is also given by ‘Taking width of gate unity. Then Area, AzACxI Fig. 3.24 bd? _1xAC? 82h? 2 12 12 12x33 9x3[ Hydrostatic Forces on Surfaces 95] 3 2 2h x in? 60° he 4 Equating the two values of h*, 7 he ++ Height of water for tipping the gate = 9m. Ans. Problem 3.20 A rectangular sluice gate AB, 2 m wide and 3 m long is hinged at A as shown in Fig. 3.25. Itis kept closed by a weight fixed to the gate. The total weight of the gate and weight fixed to the gate is 343350 N. Find the height of the water ‘h’ which will just cause the gate to open. The centre of gravity of the weight and gate is at G. Solution. Given : Width of gate, b= 2m; Length of gate L= 3m Area, x3=6m? x3=9m Weight of gate and Let / is the required height of water. Depth of C.G. of the gate and weight From Fig. 3.25 (a), ~ ED = h~(AD~ AE) ~ (AB sin @~ EG tan 8) = h~G sin 45° - 0.6 tan 45°) ~ 2.121 ~ 0.6) = (h~ 1.521) m ‘The total pressure force, F is given by F = pgAh = 1000 x9.81 x6 x (h 1.521) = 58860 (lr ~ 1.521) N. ‘The total force F is acting at the centre of pressure as shown in Fig. 3.25 (b) at H. The depth of H from free surface is given by h* which is equal to 45 xsin? 45° 0375 “ —— ——— + (h- 1.521 6x(h— 1521) * isa *4 X[96 Fluid Mechanics ] HINGE ons" @ Fig. 3.25 Now taking moments about hinge A, we get 343350 x EG = F x AH AK or 343350 x 0. sin 45° [Fm AAKH, Fig.325 (b) AK = AH sin @= AH sin 45°. A ae 58860 (ht ~ 1.521) x AK ‘sin 45° 343350 x0.6 xsin 45°_ 03535 x7 AK fi 38860 (1521) (1521) 2 37s But Ak =h -ac= — 375 _ 4 (41.521) -AC wolf 1320 * 7 oy But AC = CD ~ AD = h~AB sin 45° = h-3 xin 45° = h-2.121 Substituting this value in (i), we get 315 $= 1521) h- 2.121 nisi * ~« , 315 315 +2421 - 1.521 +06 Ait 521 uu) Equating the two values of AK from (i) and (ii) L521[ Hydrostatic Forces on Surfaces 97| 0.3535x7 _ _ 0.375 h=152 = 1521 or 375 + 0.6 (h~ 1.521) = 0.375 + 0.6 h~ 0.6 x 1.521 or . ATAS — 375 + 0.6 x 1.521 = 2.0995 + 0.9126 = 3.0121 +06 m. Ans. Problem 3.21 Find the total pressure and position of centre of pressure on a triangular plate of base 2 mand height 3 m which is immersed in water in such a way that the plane of the plate makes an angle of 60° with the free surface of the water. The base ofthe plate is parallel to water surface and at 4@ depth of 2.5 m from water surface. SURFACE, Solution. Given : Base of plate, m Height of plate, h=3m Area, Ae PRL DS 3 yt 2 2 Inclination, 6 = 60° Depth of centre of gravity from free surface of water, hi =2.5 + AG sin 60° Fig. 3.26 1 BB 1 sebeaxtl —— {sacLortiatuia} = 2.5 + 866 m= 3.366 m (0 Total pressure force (F) F = peAl = 1000 x 9.81 x 3 x3.366 = 99061.38 N. Ans. (Go Centre of pressure (h). Depth of centre of pressure from free surface of water i given by 29 where I,= ME = 2%3 3 3% 6D 1S xsin? 60° 3X 2366 + 3.366 = O.111 + 3.366 = 3.477 m. Ans. > 3.6 CURVED SURFACE SUB-MERGED IN LIQUID Consider a curved surface AB, sub-merged in a static fluid as shown in Fig. 3.27. Let dA is the arca of a stall strip at a depth of fr from water surface. ‘Then pressure intensity on the area dA is = pgh and pressure force, dF = p x Arca = pghx dA GAD This force dF acts normal to the surface. Hence total pressure force on the curved surface should be F: J pghaa (3.12)[98 Fluid Mechanics WATER n SURFACE _C |5 7 3 iar io ancoso E ——— ‘AREA dA 8 0) (a) a Fig. 327 But here as the direction of the forces on the small areas are not in the same direction, but varies from point to point. Hence integration of equation (3.11) for curved surface is impossible. The problem can, however, be solved by resolving the force dF in two components dF, and dF, in the x and y directions respectively. The total force in the x and y directions, i.e., F, and F, are obtained by integrating dF, and dF, Then total force on the curved surface is +R B.13) FE and inclination of resultant with horizontal is tan @ = — B14) Resolving the force dF given by equation (3.11) in x and y directions : dF, = dF sin = pghdA sin 8 : dF = pghdA} and DahdA cos 8 “Total orcs in th x andy dition ae Fy J dP. J pati sin 0 pe f ht sin 0 19 and Fy= | aFy= J penta cos 0= pe tas oor 1) Fig. 3.27 (b) shows the enlarged area dA. From this figure, ie., AEFG, ‘Ths inequation (3.15). sin 9 PE Verte pojectinf the rca dt nd hene the exprsin oe J isn ® repens th press fore on te pote asa ofthe cued suc on he vere plane, Ths F, = Total pressure force on the projected area of the curved surface on vertical plane. (3.17) Also dA cos 6 horizontal projection of dA and hence MA cos @ isthe volume of the liquid contin nthe elena ae dA upto fee suace ofthe Lig Ths [A cos is the ol volume conte betwen the cured surface extend ut five src, Hence pf td cos is th ttl wight uppity the carved surface. Thus Fy=pe [nas cos 0 = weight of lid supported bythe ved surface upto fie surface of ii ous)[ Hydrostatic Forces on Surfaces 99 In Fig. 3.28, the curved surface AB is not supporting any fluid. In such cases, F, is equal to the weight of the imaginary liquid supported by AB upto free surface of liquid. The direction of F, will be taken in upward direction, Problem 3.22 Compute the horizontal and vertical components of the total force acting on a curved surface AB, which is in the form of a quadrant of a circle of radius 2 m as shown in Fig. 3.29. Take the width of the gate as unity. Solution. Given : Width of gate Radius of the gate Distance ‘equation (3.17) as, F, = Total pressure force on the projected area of curved surface AB on vertical plane rotal pressure force on OB {projected area of curved surface on vertical plane = OB x 1) = ped wo 981 x21 (+2) WATER SURFACE Fig. 3.28 {Area of OB = A= BOx1=2x1=2, Ti = Depth of C.G. of OB from free surface = 1.5+ 3) Fy 81 x 2000 x 2.5 = 49050 N. Ans. +2 2x25 75 = 0.1333 + 2.5 = 2.633 m from free surface. Vertical force, F,, exerted by water is given by equation (3.18) F, = Weight of water supported by AB upto free surface Weight of portion DABOC Weight of DAOC + Weight of water AOB 1g [Volume of DAOC + Volume of AOB] = 1000 x 9.81 [apxaoxi+ = (aoy x1][100 Fluid Mechanics ] “| = 1000 x 9.81 [3.0 + RIN = 60249.1 N. Ans. Problem 3.23 Fig. 3.30 shows a gate having a quadrant shape of radius 2 m. Find the resultant force due to water per metre length of the gate. Find also the angle at which the total force will act. Solution. Given : a Radius of gate Width of gate Horizontal Force = 1000 x 9.81 [1sx20x1+ 2x Force on the projected area of the curved surface on vertical plane = Force on BO = peal 1 where A= Areaof BO=2x1=2m’,h astm F, = 1000 x 9.81 x2. 1 = 19620 N Thi wile at dn of 22 = m fom ssi i Vertical Foree, F, F, = Weight of water (imagined) supported by AB 1g x Area of AOB x 1.0 = 1000 9.81 x E (2)4x 1.0 = 30819 N ‘This will act ata distance of 28 = 4%20 anon Resultant force, Fis given by Fs (haP = 196205 30BT9 = [5859445004 DABBIOTOT = 368344 N. Ans. ‘The angle made by the resultant with horizontal is given by F, _ 30819 F, 19620 . @ = tan"! 1.5708 = $7° 31’. Ans. Problem 3.24 Find the magnitude and direction of the resultant force due to water acting on a roller gate of cylindrical form of 4.0 m diameter, when the gate is placed on the dam in such a way that water is just going t0 spill. Take the length of the gate as 8 m. Solution. Given : Dia. of gate Radius, Length of gate, = 0.848 m from OB. tan 5708[ Hydrostatic Forces on Surfaces 101) Horizontal force, F, acting on the gate is WATER, SURFACE F, = peAli = Force on projected area of curved surface ‘ACB on vertical plane = Force on vertical area AOB where A = Area of AOB = 4.0 x 8.0 = 32.0 m? I; = Depth of CG. of AOB = 4/2 = 2.0m F, = 1000 9.81% 32.0 2.0 627840 N aa Vertical force, F, is given by F, = Weight of water enclosed or supported (actually or imaginary) by the curved surface ACB = pg x Volume of portion ACB 1g Area of ACB XI = 1000 x 9.81 x F(R) x8.0 = 9810 x 52)" x 81 It will be acting in the upward direction. Resultant force, = «[F2 + Fe = 6270+ AVSION = 798328 N. Ans. F, _ 493104 Direction of resultant force is given by tan = 0.785 ia wlan L a F, 627840 ee 0=31°8' Ans. Problem 3.25 Find the horizontal and vertical component of water pressure acting on the face of a tainter gate of 90° sector of radius 4 m as shown in Fig. 3.32. Take width of gate unity. Solution. Given : Radius of gate, R=4m Horizontal component of force acting on the gate is F, = Force on area of gate projected on vertical plane = Force on area ADB mesure = pea where A= ABx Width of gate =2KADx1 (cs AB = 2AD) Fig. 3.32 x 4x sin 45° = 8 x .707 = 5.656 m? (1 AD = 4 sin 45°} AB 25.856 <9 998m 2. 2 F, = 1000 x 9.81 x 5.656 x 2.828 N = 186911 N. Ans. Vertical component F, = Weight of water supported or enclosed by the curved surface Weight of water in portion ACBDA = pg x Area of ACBDA x Width of gate 1000 x 9.81 x [Area of sector ACBOA ~ Area of ABO] x 1[102 Fluid Mechanics ] AOx BO 2 : AAOB is a right angled] Kp? = 9810 x| ER? s10%|% = 9810 [4° 224) «44796 N. Ans ao Problem 3.26 Calculate the horizontal and vertical components of the water pressure exerted on a tainter gate of radius 8 m as shown in Fig. 3.33. Take width of gate unity. Solution. The horizontal component of water pressure is given by Ah “orce on the vertical area of BD where A= BD x Width of gate = 4.0 1 = 4.0 m ‘orce on the area projected on vertical plane WATER SURFACE ¢ = 1000 x 9.81 x 4.0 x 2.0 = 78480 N. Ans. Fig. 333 Vertical component of the water pressure is given by F, = Weight of water supported or enclosed (imaginary) by curved surface CB = Weight of water in the portion CBDC 1g x [Area of portion CBDC] x Width of gate = pg x [Area of sector CBO ~ Area of the triangle BOD] x 1 30. BDx be = 010981 xan = 360 2 2 (+ DO = BO cos 30° = 8 x cos 30°) 9810 x [16.755 — 13.856] = 28439 N. Ans. Problem 3.27 A cylindrical gate of 4 m diameter 2 m long has water on its both sides as shown in Fig. 3.34. Determine the magnitude, location and direction of the resultant force exerted by the water ‘on the gate. Find also the least weight of the cylinder so that it may not be lifted away from the floor. Solution. Given : WATER SURFACE A Dia. of gate Radius m (@ The forces acting on the left side of the cylinder are = ‘The horizontal component, F,, where F,, = Force of water on area projected on vertical plane = Force on area AOCHydrostatic Forces on Surfaces 103) F,, = weight of water enclose by ABCOA = 1000 9.81 x[ER| 200 = 9810 x © x 2? x20 = 123276 N. Right Side of the Cylinder = ptA;fz = Force on verical area CO = 1000 9.81 «2%23{ Ay=COx1=2«1=2 mh: 9240 N Weight of water enclosed by DOCD = pax [Ee] x Width of gate = 1000 x 9.81 x © x 2?x2 = 61638 N Resultant force in the direction of x, F,= F,,~ Fe, = 156960 ~ 39240 = 117720 N Resultant force in the direction of y, F, =F, + Fy, = 123276 + 61638 = 184914 N (H Resultant force, F is given as, F= JF +F = ¥(117720)' +(184914)° = 219206 N. Ans. (id Direction of resultant force is given by F, _tsdo1a an 5707 an O= 117720 0 =57° 31" Ans. (iii) Location of the resultant force 24 — 2.67 m from the top surface of water on left side, while F,, 33 m from free surface on the right side of the cylinder. The resultant Force, F,, acts ata distance of 2 scsaradisaceot2 2 force F, in the direction of x will act at a distance of y from the bottom as F,xy= F, [4-267] ~ Fy, [2-133] or 117720 x y = 156960 x 1.33 ~ 39240 x .67 = 208756.8 - 26290.8 = 182466 12466 _ 155 m from the bottom. 11720 ae) 8488 m from AOC towards Force Facts ata distance 28 from AOC ora distance " an leat 0c Aso a8 ata distance #2 088 m om AOC toast ip of AOC. Te eat force F, will act ata distance x from AOC which is given by[104 Fluid Mechanics ] Fy xx= Fy, X 8488 — Fx 8488 or 184914 x x= 123276 x 8488 - 61638 x .8488 = 8488 [123276 - 61638] = 52318.4 318.4 Teaoqa 702829 m from AOC. (iv) Least weight of cylinder. The resultant force in the upward direction is F, = 184914 N ‘Thus the weight of cylinder should not be less than the upward force F,. Hence least weight of cylinder should be at least. 184914 N. Ans. Problem 3.28 Fig. 3.35 shows the cross-section of a tank full of water under pressure. The length of the tank is 2 m. An empty cylinder lies along the length of the tank on one of its corner as shown. Find the horizontal and vertical components of the force acting on the curved surface ABC of the oylinder. Solution. Radius, R= 1m o2 katen? Length of tank, 1=2m Pressure, 2 keflem’ {962 Nlem? = 1.962 x 10* Nh _p 196210" pg 1000x981 Free surface of water will be at a height of 2 m from the top of the tank. Fig. 3.36 shows the equivalent free surface of water. Pressure head, (® Horizontal Component of Force F,= peak where A= Area projected on vertical plane 1.5 2.0 = 3.0 m? Ls 5 1000 x 9.81 x 3.0 x 2.75 = 80932.5 N. Ans. (ii) Vertical Component of Force ‘eight of water enclosed or supported actually or imaginary by curved surface ABC ‘eight of water in the portion CODE ABC ‘Weight of water in CODFBC ~ Weight of water in AEFB But weight of water in CODFBC ‘= Weight of water in [COB + ODFBO] = os 7 + nox 00] 2 = 1000%9.81 Fat st0x25] x2 4458.5 N 9g [Area of AEFB] x 2.0 Weight of water in AEFBHydrostatic Forces on Surfaces 105 = 1000 x 9.81 [Area of (AEFG + AGBH ~ AHB)] x 2.0 In AAHO, sing = AH _ 05 10 ‘AO BH @= 30° (0 ~ HO = 1.0 - AO cos @ = 1.0 ~ 1x 00s 30° = 0. ABH = Area ABO ~ Area AHO = ni 3 AHXHO _ AR? 05% 866 _ 5455 30020 Weight of water in AEF 810 x [AE x AG + AG x AH — 0.0453] x 2.0 = 9810 x [2.0 x .134 + .134 x .5 ~ 0453] x 2.0 18.10 x [.268 + .067 — .0453] x 2.0 = 5684 N : F, = 6458.5 ~ 5684 = 587745 N, Ans. Problem 3.29 Find the magnitude and direction of the resltant water pressure acting on a curved Area, face of adam which is shaped according to the elation y = as shown in Fig. 3.37. The height ofthe water retained by the dam is 10 m. Consider the width of the dam as unity. Solution, Equation of curve AB is Height of water, Width, ‘The horizontal component, F, is given by Fig. 3.37 ressure due to water on the curved area projected on vertical plane ressure on area BC seh where A=BCx1=10x1m,h=4+x10=5m F, = 1000 «9.81 x 10 x5 = 490500 N Vertical component, F, is given by F, = Weight of water supported by the curve AB ‘eight of water in the portion ABC = palArea of ABC] x Width of dam [[xe]1.0 {arora ay asaase [0] ar) 000 x 9.81 x [3yF ay = 2043022)" 2pep -m00 5] 20402 fy"? = 19620 110% 19620 x 31.622 = 620439 N[106 Fluid Mechanics ] Resultant water pressure on dam [F? + F} = y(490500)° +(620439)" = 790907 N = 790.907 KN. Ans. tiven by F, _ 620439 F, 490500 = 51° 40", Ans. F Direction of the resultant tan @ = 1.265 Problem 3.30 A dam has a parabolic shape y (z) as shown in Fig, 3.38 blow having x =6m and yo = 9m. The fluid is water with density = 1000 kg/m’. Compute the horizontal, vertical and the resultant thrust exerted by water per metre length of the dam. Solution. Given : Equation of the curve OA is yi Width of dam, _ ( Horizontal thrust exerted by water F, = Force exerted by water on vertical surface OB, ie., the surface obtained by projecting ont [53h 6 m the curved surface on vertical plane = pet Fig. 3.38 = 1000981 (9 x 1) x5 (ii) Vertical thrust exerted by water F, = Weight of water supported by curved surface OA upto free surface of water ‘eight of water in the portion ABO 9g x Area of OAB x Width of dam = 10009381 x( fx dy] x10 1 ) = 1909.81 [faa] 10 vey ap A = 19620 x|2—_| = x= 9] sox] 9620 x 5 I 19620227 = 8160, As[ Hydrostatic Forces on Surfaces 107) (iii) Resultant thrust exerted by water F= JF) +82 = (3973054 353160 = 531574 N. Ans. Direction of resultant is given by 353160 _ 5 agg F, 397305 O= tan”! 0.888 = 41.63". Ans. Problem 3.31 A cylinder 3 m in diameter and 4 m long retains water on one side. The cylinder is supported as shown in Fig. 3.39. Determine the horizontal reaction at A and the vertical reaction at B The cylinder weighs 196.2 kN. Ignore friction. WATER SURFACE Solution. Given : Dia. of cylinder Length of cylinder Weight of cylinder, W= 196.2 KN = 196200 N Horizontal force exerted by water F,,= Force on vertical area BOC = pedi eal tx 2 F, = 1000 x 9.81 x 12 x 1.5 = 176580 N ‘The vertical force exerted by water F, = Weight of water enclosed in BDCOB where A=BOCxI=3x4= 12m’, = pe (Ea?) x1 10003 9.81«% x 4= 138684 N Force Fis acting in the upward direction, For the equilibrium of eylinder Horizontal reaction at A "76580 N = Weight of cylinder ~ F, 196200 ~ 138684 = S7516 N. Ans. Vertical reaction at B > 3.7. TOTAL PRESSURE AND CENTRE OF PRESSURE ON LOCK GATES Lock gates are the devices used for changing the water level in a canal or a river for navigation. Fig. 3.40 shows plan and elevation of a pair of lock gates. Let AB and BC be the two lock gates. Each gate is supported on two hinges fixed on their top and bottom at the ends A and C. In the closed position, the gates meet at B. Let F = Resultant force due to water on the gate AB or BC acting are right angles to the gate R = Reaction at the lower and upper hinge P = Reaction at the common contact surface of the two gates and acting perpendicular to the contact surface. Let the force P and F meet at O. Then the reaction R must pass through O as the gate AB is in the ‘equilibrium under the action of three forces. Let @ is the inclination of the lock gate with the normal to the side of the lock.[108 Fluid Mechanics ] In ZABO, ZOAB = ZABO = 8. Resolving all forces along the gate AB and putting equal to zero, we get R cos 0 P cos = OorR=P (3.19) WATER SURFACE UPSTREAM SIDE DOWNSTREAM SIDE Fig. 3.40 Resolving forces normal to the gate AB Rsin 0+ P sin @- F=0 o F=Rsin 8+ P sin 9=2P sin 0 (eRe?) F 6.20 Zaind ae To calculate P and R In equation (3.20), P can be calculated if F and @ are known. The value of 8 is calculated from the angle between the lock gates. The angle between the two lock gate is equal to 180° — 28. Hence @ can be calculated. The value of F is calculated as : me hot water onthe apt sie igh wa om dw ie aren onthe pon ota ie rps nh on Sora ie of ep i wisn oe Now 8A i 1 x Hy x 1x a 2 i ea oat : simu, Fess = pact i al 1H? _pgltt? | — Fy Pal pale Substituting the value of @ and F in equation (3.20, the value of P and R can be calculated. Reactions at the top and bottom hinges Let R, = Reaction of the top hinge Resultant force F=[ Hydrostatic Forces on Surfaces 109) R, = Reaction of the bottom hinge Then R=R, +R, ‘The resultant water pressure F acts normal to the gate. Half of the value of F is resisted by the hinges of one lock gates and other half will be resisted by the hinges of other lock gate. Also Fy acts at A Hy a distance of 5 from bottom while F, ats at a distance of “> from botiom. ‘Taking moments about the lower hinge R,xsinOxH 0 where H/= Distance between two hinges Resolving forees horizontally R,sin 0 +R, sin = LB alii) 2 From equations (i) and (i), we can find R, and Ry. Problem 3.32 Each gate of a lock is 6 m high and is supported by two hinges placed on the top and bottom of the gate. When the gates are closed, they make an angle of 120°. The width of lock is 'm. Ifthe water levels are 4 m and 2 mon the upstream and downstream sides respectively, determine the magnitude of the forces on the hinges due to water pressure Solution. Given : HINGE Ry, Height of lock Width of lock Width of each lock gate Angle between gates “ 6 2 = 30° Height of water on upstream side Ay and " Fig. 3.41 Total water pressure on upstream side F, = pgAyh where Ay = Hy x1= 40 «2.887 m? = 1000 x 9.81 x 4 x 2.887 x 2.0 = 226571 N Foxe Fw beating at adsanee ot 133m om oom Similarly, total water pressure on the downstream side F = pgAzhia, where A= Hz x= 2% 2.887 m? = 1000 x 9.81 x 2x 2.887 x 1.0[110 Fluid Mechanics ] = 56643 N Hy 2 FF, will act ata distance of S*= > = 0.67 m from bottom, Resultant water pressure on each gate F = Fy ~ F, = 26571 ~ 56643 = 169928 N. Let xis height of F from the bottom, then taking moments of F,, F and F about the bottom, we have Fxx= F,x133- Fx 0.67 or 169928 x x = 226571 x 1.33 ~ 56643 x 0.67 226571 x 133 — 56643 x 0.67 _ 30139-37950 _ | 551, 169928 169928 F__ 169928 From equation (3.20), P 169928 N 2sind 2sin30 From equation (3.19), R= P = 169928 N, If Ry and Rp are the reactions atthe top and bottom hinges, then Ry + Ry= R= 169928 N. ‘Taking movements of hinge reactions R, Ry and R about the bottom hinges, we have Ry X60 + RyX0= RX 155 ~ Ry = 169928 ~ 43898 = 126030 N. Ans. Problem 3.88 The end gates ABC of a lock are 9 m high and when closed include an angle of 120°. The width of the lock is 10 m. Each gate is supported by two hinges located at I m and 6 m above the bottom of the lock. The depths of water on the two sides are 8 m and 4 m respectively. Find: (i) Resultant water force on each gate, (ii) Reaction between the gates AB and BC, and (iii) Force on each hinge, considering the reaction of the gate acting in the same horizontal plane as resultant water pressure. Solution. Given : Height of gate Inclination of gate PUN ) ELEVATION Fig. 3.42[ ‘Hydrostatic Forces on Surfaces 111] Width of lock =10m 5 Width of each lock = S30 8 I= 577m Depth of water on upstream side, Hy Depth of water on downstream side, Hy (0 Water pressure on upstream side F,= pea where Ay = DH, = 5.773 x 8 = 46.184 m, hy = BE $=40m F, = 1000 x 9.81 x 46.184 x 4,0 = 1812260 N = 1812.26 kN Water pressure on downstream side, Fre peAshs where Ay = 1% Hy 773 x4 = 23.092 m, Tn = 4 = 20 1000 x 9.81 x 23.092 x 2.0 = 453065 N = 453.065 kN Resultant water pressure 1 ~ Fy = 1812.26 ~ 453,065 = 1359.195 kN (i Reaction between the gates AB and BC. The reaction (P) between the gates AB and BC is given by equation (3.20) as pe P= 1389195 = sin 2xsin 30° = 1359.195 kN. Ans. (ii) Force on each hinge. If Ry and Ry are the reactions at the top and bottom hinges then Ry + Ry But from equation (3.19), R = . Ry + Ry 359.195 359.195 epee 3 ‘The resultant force F will act at a distance x from bottom is given by Fx 2.67 - F, x 133 or 1812.26%2.67-453.065%133 “7359195 Hence R is also acting at a distance 3.11 m from bottom, ‘Taking moments of Ry and R about the bottom hinge Ry X [6.0 ~ 1.0] = Rx (x ~ 1.0) Rx(e=L0)_ 1359195211 r 50 50 Ry= R~ Ry = 1359.195 ~ 573.58 85.615 kN. Ans. 67m om toma Ft Hn 1.3m fom tom.[112 Fluid Mechanics ] > 3.8 PRESSURE DISTRIBUTION IN A LIQUID SUBJECTED TO CONSTANT HORIZONTAL/VERTICAL ACCELERATION In chapters 2 and 3, the containers which contains liquids, are assumed to be at rest. Hence the liquids are also at rest. They are in static equilibrium with respect to containers. But if the container ‘containing a liquid is made to move with a constant acceleration, the liquid particles initially will move relative to each other and after some time, there will not be any relative motion between the liquid particles and boundaries of the container. The liquid will take up a new position under the effect of acceleration imparted to its container. The liquid will come to rest in this new position relative to the container. The entire fluid mass moves as a single unit, Since the liquid after attaining a new position is in static condition relative to the container, the laws of hydrostatic can be applied to determine the liquid pressure. As there is no relative motion between the liquid particles, hence the shear stresses and shear forces between liquid particles will be zero. The pressure will be normal (0 the surface in contact with the liquid. ‘The following are the important cases under consideration : (Liquid containers subject to constant horizontal acceleration, (i) Liquid containers subject to constant vertical acceleration, 3.8.1 Liquid Containers Subject to Constant Horizontal Acceleration. Fig. 3.43 (a) shows a tank containing a liquid upto a certain depth. The tank is stationary and free surface of liquid ishorizontal, Let this tank is moving with a constant acceleration ‘a’ in the horizontal direction towards right as shown in Fig. 3.43 (b). The initial free surface of liquid which was horizontal, now takes the shape as shown in Fig. 3.43 (b). Now AB represents the new free surface of the liquid. Thus the free surface of liquid due to horizontal acceleration will become a downward sloping inclined plane, with the liquid rising at the back end, the liquid falling at the front end. The equation for the free liquid surface can be derived by considering the equilibrium of a fluid element C lying on the free surface. ‘The forces acting on the element C are : Free surface of y que Origa igi surface Free surface of fue |e . — Tank (staionay) @ ) ‘Tank moving Fig. 3.43 (D the pressure force P exerted by the surrounding fluid on the element C. This force is normal to the free surface. (id, the weight of the fluid element i.e., mx g acting vertically downward. (iii) accelerating force ie., m x a acting in horizontal direction.[ Hydrostatic Forces on Surfaces 113) Resolving the forces horizontally, we get Psin@+mxa=0 or P sin @=~ ma oli) Resolving the forces vertically, we get P cos @~ mg =0 or P cos O=mxg ii) Dividing (0) by (i), we get tn o=- 4 (onan) 11.204) ee ‘The above equation, gives the slope ofthe free surface ofthe liquid which i contained in a tank which is subjected to horizontal constant acceleration. The term (a/g) isa constant and henee tan 8 will be constant The -ve sign shows thatthe fee surface of ligud is sloping downwards. Hence the tree surface isa stright plane inlined down at an angle @ along the direction of acceleration. Now let us find the expression forthe pressure at any point D in the liquid mass subjected to horizontal acceleration. Let the point Dis ata depth of “Wr fiom the fee surface. Consider an elementary pris DE of height "F"and eros-secional area dA as shown in ig. 3-44 [Unes of constant pressure Consider the equilibrium of the elementary prism DE. The forces acting on this prism DE in the vertical direction are ()_ the atmospheric pressure force (pp x dA) at the top end of the prism acting downwards, (ii) the weight of the element (p x g x hx dA) at the C.G. of the element acting in the downward direction, and (ii) the pressure force (p x dA) at the bottom end of the prism acting upwards. Since there is no vertical acceleration given to the tank, hence net force acting vertically should be ze10. px dA ~ px dA~pgh dA =0 or P~Po~peh=0 or p=po+psh or P~Po=pah fof gauge pressure at point D is given by P=pgh or pressure head at point D, 2 =h. Pa