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## CE 3130 Week 5 Notes from Lecture with In-class problem solutions

by: Aaron Bowshier

143

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# CE 3130 Week 5 Notes from Lecture with In-class problem solutions CIVILEN 3130

Marketplace > Ohio State University > CIVILEN 3130 > CE 3130 Week 5 Notes from Lecture with In class problem solutions
Aaron Bowshier
OSU
GPA 3.52
FLUID MECHANICS
Colton Conroy

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The fifth week of notes for Fluid Mechanics (lectures 11b, 12, and 14). This includes every single fill-in for the blanks in the notes as well as ALL the homework problem solutions done in class.
COURSE
FLUID MECHANICS
PROF.
Colton Conroy
TYPE
Class Notes
PAGES
13
WORDS
KARMA
25 ?

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This 13 page Class Notes was uploaded by Aaron Bowshier on Friday February 13, 2015. The Class Notes belongs to CIVILEN 3130 at Ohio State University taught by Colton Conroy in Spring2015. Since its upload, it has received 143 views.

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Date Created: 02/13/15
CIVIL EN 3130 LECTURE 11 14 The pressure prism 8 Lecture 11 Continued 2915 14 The pressure prism 2 f 2 l T M l p h E F 7b a b For submerged surfaces of relatively simple shape such as a rectangle a useful graphical approach can be used in lieu of the equations derived in the previous sec tions to determine the magnitude and location center of pressure of the resultant force acting on the submerged surface To illustrate this approach consider the situation depicted in the gure above where we wish to determine the resultant force acting on the side of the tank Using Equation 3 of section 12 the resultant force can be computed as 152 YEA where we have used the fact that ft y sin 90 g h 2 for this case The pressure distribution shown in gure a runs across t e entire width w of the tank as shown in gure b and the magnitude of the resultant force is equal to the volume of this shape the so called pressure prism that is l l Volume of pressure prism c9 W 4 A where wh is the area A of the side of the tank CIVIL EN 3130 LECTURE 11 14 The pressure prism 9 l p T a YR The resultant force acts through the centroid of the pressure prism not the cen troid of area A For the volume of gure b of the previous page this occurs at a distance h 3 from the bottom of the tank since the centroid of a triangle is located a distance h 3 above its base It can easily be shown that this result is consistent with that obtained from Equations 4 derived in the previous section The pressure prism approach can also be used for cases where the submerged surface does not extend all the way to the free surface and or the submerged surface is inclined at some arbitrary angle 6 as illustrated in gures a and b above In both cases shown above the pressure prisms are trapezoidal instead of triangular but the magnitude of the resultant force is still equal to the volume of the pressure prism and the pressure center can be computed as the centroid of the pressure prism Resultant Force Vol of pressure prism Pressure Center centroid of the pressure prism CIVIL EN 3130 SPRING 2015 LECTURE 12 Contents Reading Section 26 of the text book Homework Problems 284 289 and 299 from the text book CIVIL EN 3130 LECTURE 12 FORCE COMPONENTS ON CURVED SURFACES 1 Force components on curved surfaces 0 When the submerged surface is curved rather than planar as shown in the gure above the differential pressure forces dF on the surface vary in Direction 0 At any particular point on the curved surface the associated differ ential force dF may have both Horizonal de and Vertical dFy force components 0 Thus to nd the Magnitude of the resultant force we must rst sum integrate the horizontal and vertical differential force components separately 0 We denote the total sum of these horizontal and vertical component forces on the curved surface by FX and FX respectively 0 These component forces can then be used to nd the Magnitude of the total resultant force via the stan dard equation F 1 W W 0 Let s look at how we nd these component forces beginning with the horizontal force Fm CIVIL EN 3130 LECTURE 12 11 Horizontal force component 11 Horizontal force component 3 L i dAx dA cos6 V 9 yravrzc Hgy Referring to the gure above the component of the differential force in the c direction is given by ampFx9 JFLCMB 3 use Summing integrating these components over the entire surface gives Fx A remade The product dA cos6 E dAm is the horizontal projection of the the dif ferential area dA and the pressure p equation above as l so we can write the Fx 4 4 73le I qx A The integral expression in the above equation is equal to the product gmAm Am Where gm is the y coordinate of the centroid of the horizontally projected area CIVIL EN 3130 LECTURE 12 11 Horizontal force component Thus we can calculate the horizontal component of force on a curved area by calculating the force on the horizontal PPOJeCtlon of the area that is Fquot t KX Ax 1 Where 561 2 321 is the depth of the uid at the centroid of the horizontal pro m jection Am 39 In a similar way it can be shown that the line of action of the horizontal force component passes through the Pressure center of the projected area Ax CIVIL EN 3130 LECTURE 12 12 Vertical force component 12 Vertical force component i dAy dA cos L Referring to the gure above the component of the differential force in the y direction is given by JFcos L05 Again summing integrating these components over the entire surface gives h P as The product dA cos E dA is the vertical projection of the the differential area dA and the pressur so the equation above can be written as I F r IdAV JV l 39 In 39 39 I v Where we have de ned 395 l cl A of a prism of height y With a base39 of area dAy Which is equal to the volume CIVIL EN 3130 LECTURE 12 12 Vertical force component Z T S 8 Liquid B 73 hi 2 12 O I Vl mm 7A Liquid A 7A Thus the vertical force component is given by W Fy YV39 2 where V is the T0131 VOIDme between the curved sur face and the free surface Note that this expression is equivalent to the Weight of the liquid in the Volume V39 A special case to consider is when the liquid is BGIOW the curved surface and the pressure is known at some point say 0 see gure above In that case we use the following procedure 1 An imaginary or equivalent free surface s s is constructed some height h 1 Fp Y above 0 where p0 is the pressure at point 0 and y is the speci c weight of the liquid in contact with the curved surface 2 The vertical component of the pressure force on the curved surface is then the Weight of the imaginary volume vertically above the curved surface see page 68 of your text book Finally it can be easily shown see page 69 of your text book that the line of action of the vertical force component passes through the CentrOid of the volume V real or imaginary CIVIL EN 3130 SPRING 2015 LECTURE 14 Contents Reading Sections 27 and 28 of the text book Homework Problems 2104 and 2107 from your text book CIVIL EN 3130 LECTURE 14 BUOYANT FORCE 2 1 Buoyant force 0 The resultant force exerted on a oating or submerged body by a static uid is called the Buoyant Force which we denote by F B o The buoyant force Always acts vertically Upward o It is equal to the Specific Weight of the uid times the Volume of the uid displaced by the body or equivalently the Weight of the uid displaced by the body O In equation form we have Fbth 1 0 To understand how we arrive at this equation let s consider the following NET HORIZONTAL FORCE IS ZERO CIVIL EN 3130 LECTURE 14 BUOYANT FORCE 3 Elt AK Cv Tn V I I H YVU FL I YV39L NET WEI LT t LL JfL Fd FluioL 7 YWL Via W Submerged D Buoyant quot Force FB 7 CIVIL EN 3130 LECTURE 14 STABILITY OF FLOATING AND SUBMERGED BODIES 4 o It can easily be shown that the Buoyant Force acts through the CentPOid of the ispla edmelmeof uid see pages 73 and 74 of your text book o This holds for both Smeng ed and Floating obj ects o The centroid of the displaced volume of uid is called the center of Buoyancy 2 Stability of oating and submerged bodies 0 Another important thing to consider with submerged or oating bodies is their Stability o A submerged or oating body is said to be in Stable equilibrium if when displaced it Returns to its equilibrium position 0 Conversely a body is in UnStable equilibrium if when displaced even slightly it Moves to a new equilibrium position o The stability of a body is determined by the relative positions of its centers of Buoyancy and Gravity 0 Consider the example below of a completely submerged body Center of Buoyancy Center of Gravity Restoring Couple Stable CIVIL EN 3130 LECTURE 14 STABILITY OF FLOATING AND SUBMERCED BODIES 5 V co 0 q CB Overturning Couple Unstable 0 Thus a completely submerged body is in Stable equilibrium if its center of gravity falls BGIOW the center of buoyancy 0 Conversely a completely submerged body is in Unstable equi librium if its center of gravity is Above the center of buoyancy o For Floating bodies the stability problem is more complicated since as the body rotates the location of the center of buoyancy Which is the centroid of the displaced volume May change o For example consider a oating body such as a barge that rides low in the wa ter as shown below Which can be Stable even though the center of gravity lies Above the center of buoyancy V l A c El 5 I 1 I It I If I

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