EAS207-2009-lec33

EAS207-2009-lec33 - L€Cx A > . 3 Q) FLUID...

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Unformatted text preview: L€Cx A > . 3 Q) FLUID STR‘HCS (ox Haamwgcs) Px éflm‘ci Gk Test in WAch “to suppox‘t SMY . ’§O‘fC€-‘~W\us)q OJC “fest con/\EXQYJU 0M? WOXMOX «fomen cm G‘VbOmxcfiiwg Swofiqg. Pascw {CAN > ‘ . ‘ . m pesswe 0* 0W8 'CXUQW 530%“. {W 0 m $0M '\Y\ QM‘ngEC/Q‘fim . ‘5nonwwa‘ob cfivfexk ’jj” m “g F...— ‘UAK p‘x'esswce AU) O\ fiumckm ’- 47 PM L I tm ‘(sii C&\‘ P (‘5’ <i-—- 0/3 \92 \ WM¥M $9 P v "PZRWW *9 *7 ==h+fz I ‘ K WW; E: Py. on m guféokce O} JIM wkgm I ' - 23220 4 5} R (I) dug 4:0 akymsflxm‘c PmeATRJ’Mm F2 €3‘Z2TZ := 030.31% meWrQ) LP:£E:WW ’— Emmg‘ln PYOHW 2 i What is the simplest resultant force from the water and where does it act on the 60-m-high 800-m-long straight earthfill dam? (Water weighs 9,806 N/m3.) ‘ 9 3" Water level 5% l \ (60%) L71“) U306) .5 l - fl FR’TKP’ILQC'YZ’ng‘f/l fiyéarywolo Zi-é‘WXlOm N "Zl'é‘flfly kN. N cm! A At what height I: will the water cause the door to rotate clockwise? The door is 3 m wide. Neglect friction and the weight M of the door. PM Jew MW 4; mm cw emf MM (xxme GROOY JED Yéjtoie dOckw‘EQ‘ Sokmfirim : ' I (603% ‘03 N W Q” k «#21359. 53 wésrufis‘wxca _ C0530“ dockwigz- E:‘><Q\ ma 3.14 M V/ Determine the magnitude and location of the resultant hydrostatic force ' acting on the submerged rectangular plate {113 shown in Fig. 9~28a. The plate has a width of1.5 m; pW = 1000 kg/mo. SOLUTlON l The water pressures at depths A and B are 19A 2 pwng '= (1000 kg/m3)(9.81 m/52)(2 m) 2 19.62 kPa p3 —-= png3 = (1000 kg/m3)(9.81 m/82)(5 m) = 49.05 kPa ‘ Since the plate has a constant width, the pressure loading can be Viewed in two dimensions as shown in Fig. 9—281). The intensities of the load at A and B are W = pr = (1.5 m)(19.62 kPa) = 29.43 kN/m 14/8 2 pr = (1.5 m)(49.05 kPa) = 73.58 kN/rn From the table on the inside back cover, the magnitude of the resultant force FR created by this distributed load is FR = area ofa trapezoid = %(3)(29.4 + 73.6) = 154.5 kN Ans. This force acts through the centroid of this area, I i 1 2(29.43) + 73.58 ‘ = §<m><3> = 1-29 m Am- measured upward from B, Fig. 9—3112. SOLUTlQN it The same results can be obtained by considering two componentsi'ofji FR, defined by the triangle and rectangle shown in Fig. 9~28c. Each-7' force acts through its associated centroid and has a magnitudeof 1 * FRE = (29.43 kN/m)(3 m) = 88.3 kN Pr = 544.15 kN/m)(3 m) 2 66.2 kN Hence, FR = FRe + E = 88.3 + 66.2 = 154.5 kN ; Ans; The location of FR is determined by summing moments abbut Fig. 9—28!) and c,i.e., ‘ ' C+(MR)B = EMB;(154.5)}1 = 88.3(15) + 66.2(1) h = 1.29 m NQTE: Using Eq. 9&4, the resultant force can be calculated. as} FD = m4 = (9810 N/m3w35 mi(3 mms ml 2 154.5 kN. ' * 442;? Determine the magnitude of the resultant hydrostatic force acting on the surface of a seawall shaped in the form of a parabola as shown in Fig. 9—290.The wall is 5 m long; pw = 1020 kg/m3. wB : 150.1kN/rn <b> Fig. ma smiuraom The horizontal and vertical components of the resultant force will be calculated, Fig. 9—2919. Since 193 = pwng = (1020 kg/m3)(9.81 m/s2)(3 m) = 30.02 kPa then W}; = pr = 5 m(30.02 kPa) = 150.1 kN/m Thus, F11 H %<3 m)(150.1kN/m) = 225.1 kN The area of the parabolic sector ABC can be determined using the table on the inside back cover. Hence, the weight of water Within this 5 m long region is F2) 2 (pwgb>(areaABC) (1020 kg/m3)(9.8l yin/52x5 m)[§(1m)(3 rm] 2 50.0 kN The resultant force is therefore FR = \/F% + F?) = \/(225.1kN)2 + (50.0 kN)2 = 231 kN ' ' Ans. ...
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EAS207-2009-lec33 - L€Cx A &amp;gt; . 3 Q) FLUID...

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