SGChapt02 - II. Fluid Statics From a force analysis on a...

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Unformatted text preview: II. Fluid Statics From a force analysis on a triangular fluid element at rest, the following three concepts are easily developed: For a continuous, hydrostatic, shear free fluid: 1. Pressure is constant along a horizontal plane, 2. Pressure at a point is independent of orientation, 3. Pressure change in any direction is proportional to the fluid density, local g, and vertical change in depth. These concepts are key to the solution of problems in fluid statics, e.g. 1. Two points at the same depth in a static fluid have the same pressure. 2. The orientation of a surface has no bearing on the pressure at a point in a static fluid. 3. Vertical depth is a key dimension in determining pressure change in a static fluid. If we were to conduct a more general force analysis on a fluid in motion, we would then obtain the following: ∇ P = ρ{ g − a} + µ ∇ V 2 Thus the pressure change in fluid in general depends on: effects of fluid statics (ρ g), Ch. II inertial effects (ρ a), Ch. III viscous effects ( µ ∇ 2 V ) Chs VI & VII Note: For problems involving the effects of both (1) fluid statics and vv (2) inertial effects, it is the net g − a acceleration vector that controls both the magnitude and direction of the pressure gradient. | v v II-1 | e-Text Main Menu | Textbook Table of Contents | Study Guide This equation can be simplified for a fluid at rest (ie., no inertial or viscous effects) to yield ∇p = ρ g ∂p ∂x = 0; ∂p ∂y =0; ∂p ∂z = dp dz = −ρ g 2 P2 − P1 = − ∫ ρ g d Z 1 Free surface Pressure = Pa For liquids and incompressible fluids, this integrates to z P2 P1 – P2 = -ρg (Z2 – Z1) h =Z 2 Z2 Note: P Z 1 Z 1 1 y Z2 – Z1 is positive for Z2 above Z1. but x P2 – P1 is negative for Z2 above Z1. We can now define a new fluid parameter useful in static fluid analysis: γ = ρg ≡ specific weight of the fluid With this, the previous equation becomes (for an incompressible, static fluid) P2 – P1 = - γ (Z2 – Z1) The most common application of this result is that of manometry. | v v II-2 | e-Text Main Menu | Textbook Table of Contents | Study Guide Consider the U-tube, multifluid manometer shown on the right. If we first label all intermediate points between A & a, we can write for the overall pressure change PA - Pa = (PA- P1) + (P1 - P2) + (P2 - Pa ) This equation was obtained by adding and subtracting each intermediate pressure. The total pressure difference now is expressed in terms of a series of intermediate pressure differences. Substituting the previous result for static pressure difference, we obtain PA - PB = - ρ g(ZA- Z1) – ρ g (Z1 – Z2) – ρ g (Z2 - ZB ) Again note: Z positive up and ZA > Z1 , Z1 < Z2 , Z2 < Za . In general, follow the following steps when analyzing manometry problems: 1. On manometer schematic, label points on each end of manometer and each intermediate point where there is a fluid-fluid interface: e.g., A – 1 – 2 - B 2. Express overall manometer pressure difference in terms of appropriate intermediate pressure differences. PA - PB = (PA- P1) + (P1 – P2) + (P2 - PB ) 3. Express each intermediate pressure difference in terms of appropriate product of specific weight * elevation change (watch signs) PA - PB = - ρ g(zA- z1) – ρ g (z1 – z2) – ρ g (z2 - zB ) 4. Substitute for known values and solve for remaining unknowns. | v v II-3 | e-Text Main Menu | Textbook Table of Contents | Study Guide When developing a solution for manometer problems, take care to: 1. Include all pressure changes 2. Use correct ∆Z and γ with each fluid 3. Use correct signs with ∆ Z. If pressure difference is expressed as PA – P1, the elevation change should be written as ZA – Z1 4. Watch units. Manometer Example: a Given the indicated manometer, determine the gage pressure at A. Pa = 101.3 kPa. The fluid at A is Meriam red oil no. 3. S.G. = .83 H0 A 18 cm ρgw = 9790 N/m 3 10 cm 1 ρg A = S.G.*ρgw = 0.83*9790 N/m 3 1 ρg A = 8126 N/m3 ρgair = 11.8 N/m3 With the indicated points labeled on the manometer, we can write PA - Pa = (PA- P1) + (P1 – P2) + (P2 - Pa ) Substituting the manometer expression for a static fluid, we obtain PA - Pa = - ρgA(zA- z1) – ρgw(z1 – z2) – ρga(z2 - za ) Neglect the contribution due to the air column. Substituting values, we obtain PA - Pa = - 8126 N/m3 * 0.10 m – 9790 N/m3 * -0.18 = 949.6 N/m2 Note why: (zA- z1) = 0.10 m and (z1 – z2) = -0.18 m, & did not use Pa Review the text examples for manometry. v v II-4 | | 2 2 e-Text Main Menu | Textbook Table of Contents | Study Guide Hydrostatic Forces on Plane Surfaces Consider a plane surface of arbitrary shape and orientation, submerged in a static fluid as shown: If P represents the local pressure at any point on the surface and h the depth of fluid above any point on the surface, from basic physics we can easily show that the net hydrostatic force on a plane surface is given by (see text for development): F = ∫ PdA = Pcg A A The basic physics says that the hydrostatic force is a distributed load equal to the integral of the local pressure force over the area. This is equivalent to the following: The hydrostatic force on one side of a plane surface submerged in a static fluid equals the product of the fluid pressure at the centroid of the surface times the surface area in contact with the fluid. Also: Since pressure acts normal to a surface, the direction of the resultant force will always be normal to the surface. Note: In most cases since it is the net hydrostatic force that is desired and the contribution of atmospheric pressure Pa will act on both sides of a surface, the result of atmospheric pressure Pa will cancel and the net force is obtained by F = ρ gh cg A F = Pcg A | v v II-5 | e-Text Main Menu | Textbook Table of Contents | Study Guide Pcg is now the gage pressure at the centroid of the area in contact with the fluid. Therefore, to obtain the net hydrostatic force F on a plane surface: 1. 2. 3. Determine depth of centroid hcg for the area in contact with the fluid Determine the (gage) pressure at the centroid Pcg Calculate F = PcgA. The following page shows the centroid, and other geometric properties of several common areas. It is noted that care must be taken when dealing with layered fluids. The required procedure is essentially that the force on the plane area in each layer of fluid must be determined individually for each layer using the steps listed above. We must now determine the effective point of application of F. This is commonly called the “center of pressure - cp” of the hydrostatic force. Define an x – y coordinate system whose origin is at the centroid, c.g, of the area. The location of the resultant force is determined by integrating the moment of the distributed fluid load on the surface about each axis and equating this to the moment of the resultant force. Therefore, for the moment about the x axis: F y cp = ∫ y P dA A Applying a procedure similar to that used previously to determine the resultant force, and using the definition (see text for detailed development), for Ixx defined as the ≡ moment of inertia, or 2nd moment of area we obtain Ycp = − ρ gsin θ I xx Pcg A ≤ 0 Therefore, the resultant force will always act at a distance ycp below the centroid of the surface ( except for the special case of sin θ = 0 ). | v v II-6 | e-Text Main Menu | Textbook Table of Contents | Study Guide PROPERTIES OF PLANE SECTIONS Geometry Moment of Inertia Ix x Centroid y x , 2 Area bL3 L b L Product of Inertia Ix y 2 0 b ⋅L 0 12 πR b πR y x 4 2 4 b y x 3, L bL3 3 2 − 36 bL b⋅L 2 2 72 b y 0 ,a = x a R 4R R 3π 4 π 8 − 9 π 8 y a= L x π R2 0 2 s L bL b ( b − 2s) L 1 3 36 72 2 2 3 b⋅ L b R y a= x 4R π 3π 16 4 − R 9π 4 4 R − 8 9π 4 1 π R2 4 b 1 y h x a= h( b + 2b 1 ) h 3 (b 2 + 4bb 1 + b 1 2 ) 36 ( b + b1 ) 3( b + b 1 ) ( b + b 1) 0 b Fluid Specific Weight 3 1bf /ft 3 N /m Seawater 64.0 10,050 8,996 Glycerin 78.7 12,360 62.4 9,790 Mercury 846. 133,100 49.2 7,733 Carbon 99.1 15,570 1bf/ft 3 N /m 3 Air .0752 11.8 Oil 57.3 Water Ethyl II-7 | v v L 0, 0 2R | e-Text Main Menu | Textbook Table of Contents | Study Guide h 2 Proceeding in a similar manner for the x location, and defining Ixy = product of inertia, we obtain X cp = − ρ g sinθ I xy Pcg A where Xcp can be either positive or negative since Ixy can be either positive or negative. Note: For areas with a vertical plane of symmetry (e.g., squares, circles, isosceles triangles, etc.) through the centroid, i.e. the ( y - axis), the center of pressure is located directly below the centroid along the plane of symmetry, i.e., Xcp = 0. Key Points: The values Xcp and Ycp are both measured with respect to the centroid of the area in contact with the fluid. Xcp and Ycp are both measured in the plane of the area; i.e., o Ycp is not necessarily a vertical dimension, unless θ = 90 . Special Case: For most problems where (1) we have a single, homogeneous fluid ( i.e., not applicable to layers of multiple fluids) and (2) the surface pressure is atmospheric, the fluid specific weight γ cancels in the equation for Ycp and Xcp and we have the following simplified expressions: F = ρ g h cg A Ycp = − I xx sinθ X cp = − h cg A I xy sin θ h cg A However, for problems where we have either (1) multiple fluid layers, or (2) a container with surface pressurization > Patm , these simplifications do not occur and the original, basic expressions for F , Ycp , and Xcp must be used; i.e., take care to use the approximate expressions only for cases where they apply. The basic equations always work. | v v II-8 | e-Text Main Menu | Textbook Table of Contents | Study Guide Summary: 1. The resultant force is determined from the product of the pressure at the centroid of the surface times the area in contact with the fluid 2. The centroid is used to determine the magnitude of the force. This is not the location of the resultant force 3. The location of the resultant force will be at the center of pressure which will be at a location Ycp below the centroid and Xcp as specified previously 4. Xcp = 0 for areas with a vertical plane of symmetry through the c.g. Example 2.5 Seawater 64 lbf/ft3 Given: Gate, 5 ft wide Hinged at B Holds seawater as shown Find: a. Net hydrostatic force on gate b. Horizontal force at wall - A c. Hinge reactions - B 9’ h c.g. 15’ A • B c. g. θ 8’ a. By geometry: θ = tan-1 (6/8) = 36.87 o Since plate is rectangular, hcg = 9 ft + 3ft = 12 ft Neglect Patm A = 10 x 5 = 50 ft2 Pcg = γ hcg = 64 lbf/ft3 * 12 ft = 768 lbf/ft2 Fp = Pcg A = 768 lbf/ft2 * 50 ft2 = 38,400 lbf II-9 | v v ∴ | e-Text Main Menu | Textbook Table of Contents | Study Guide 6’ b. Horizontal Reaction at A Must first find the location, c.p., for Fp ycp = − ρ g sinθ Ixx I sin θ = − xx Pcg A hcg A P θ Fw For a rectangular wall: Bz 3 Ixx = bh /12 θ y c.p. • c.g. • c.p. Bx Ixx = 5 * 103/12 = 417 ft4 6 ft 8 ft Note: The relevant area is a rectangle, not a triangle. Note: Do not overlook the hinged reactions at B. y cp = − 417 ft 4 0.6 417 ft 4 0.6 = − 0.417 ft y cp = − = − 0.417 ft 12 ft 50 ft 2 12 ft 50 ft 2 below c.g. xcp = 0 due to symmetry ∑ MB = 0 P (5 − 0.417) ⋅38,400 − 6 P = 0 Bz ← P = 29,330 lbf θ Fw Bx θ y c.p. • c.g. • c.p. 6 ft 8 ft | v v II-10 | e-Text Main Menu | Textbook Table of Contents | Study Guide c. ∑ Fx = 0, Bx + F sinθ − P = 0 Bx + 38,400*0.6 - 29,330 = 0 → Bx = 6290 lbf ∑ Fz = 0, Bz − F cosθ = 0 Bz = 38,400 * 0.8 = 30,720 lbf ↑ Note: Show the direction of all forces in final answers. Summary: To find net hydrostatic force on a plane surface: 1. 2. 3. 4. 5. Find area in contact with fluid. Locate centroid of that area. Find hydrostatic pressure Pcg at centroid, typically = γ hcg ( generally neglect Patm ). Find force F = Pcg A. Location will not be at c.g., but at a distance ycp below centroid. ycp is in the plane of the area. Review all text examples for forces on plane surfaces. | v v II-11 | e-Text Main Menu | Textbook Table of Contents | Study Guide Forces on Curved Surfaces Since this class of surface is curved, the direction of the force is different at each location on the surface. Therefore, we will evaluate separate x and y components of net hydrostatic force. Consider curved surface, a-b. Force balances in x & y directions yields Fh = FH Fv = Wair + W1 + W2 From this force balance, the basic rules for determining the horizontal and vertical component of forces on a curved surface in a static fluid can be summarized as follows: Horizontal Component, Fh The horizontal component of force on a curved surface equals the force on the plane area formed by the projection of the curved surface onto a vertical plane normal to the component. hcg a’ The horizontal force will act through the c.p. (not the centroid) of the projected area. ycp Fh b’ a Projected vertical plane Curved surface cp b | v v II-12 | e-Text Main Menu | Textbook Table of Contents | Study Guide Therefore, to determine the horizontal component of force on a curved surface in a hydrostatic fluid: 1. Project the curved surface into the appropriate vertical plane. 2. Perform all further calculations on the vertical plane. 3. Determine the location of the centroid - c.g. of the vertical plane. 4. Determine the depth of the centroid - hcg of the vertical plane. 5. Determine the pressure - Pcg = g hcg at the centroid of the vertical plane. 6. Calculate Fh = Pcg A, where A is the area of the projection of the curved surface into the vertical plane, ie., the area of the vertical plane. 7. The location of Fh is through the center of pressure of the vertical plane , not the centroid. Get the picture? All elements of the analysis are performed with the vertical plane. The original curved surface is important only as it is used to define the projected vertical plane. Vertical Component - Fv The vertical component of force on a curved surface equals the weight of the effective column of fluid necessary to cause the pressure on the surface. The use of the words effective column of fluid is important in that there may not always actually be fluid directly above the surface. ( See graphic that follows.) This effective column of fluid is specified by identifying the column of fluid that would be required to cause the pressure at each location on the surface. | v v II-13 | e-Text Main Menu | Textbook Table of Contents | Study Guide Thus to identify the effective volume - Veff: 1. 2. 3. 4. Identify the curved surface in contact with the fluid. Identify the pressure at each point on the curved surface. Identify the height of fluid required to develop the pressure. These collective heights combine to form Veff. Veff Veff a P b P P P a fluid P P b No fluid actually above surface Fluid above the surface These two examples show two typical cases where this concept is used to determine Veff. The vertical force acts vertically through the centroid (center of mass) of the effective column of fluid. The vertical direction will be the direction of the vertical components of the pressure forces. Therefore, to determine the vertical component of force on a curved surface in a hydrostatic fluid: 1. Identify the effective column of fluid necessary to cause the fluid pressure on the surface. 2. Determine the volume of the effective column of fluid. 3. Calculate the weight of the effective column of fluid - Fv = ρgVeff. 4. The location of Fv is through the centroid of Veff. | v v II-14 | e-Text Main Menu | Textbook Table of Contents | Study Guide Finding the Location of the Centroid A second problem associated with the topic of curved surfaces is that of finding the location of the centroid of Veff. Recall: Centroid = the location where the first moment of a point area, volume, or mass equals the first moment of the distributed area, volume, or mass, e.g. xcgV1 = ∫ x dV V1 This principle can also be used to determine the location of the centroid of complex geometries. For example: If Veff = V1 + V2 V1 a then V 2 xcgVeff = x1V1 + x2V2 b or VT = V1 + Veff Veff b V1 xTVT = x1V1 + xcgVeff fluid a Note: In the figures shown above, each of the x values would be specified relative to a vertical axis through b since the cg of the quarter circle is most easily specified relative to this axis. | v v II-15 | e-Text Main Menu | Textbook Table of Contents | Study Guide Example: A • Gate AB holds back 15 ft of water. Neglecting the weight of the gate, determine the magnitude (per unit width) and location of the hydrostatic forces on the gate and the resisting moment about B. Water Width - W F H 15 ft F V B • γ = ρg = 62.4 lbf/ft3 a. Horizontal component Rule: Project the curved surface into the vertical plane. Locate the centroid of the projected area. Find the pressure at the centroid of the vertical projection. F = Pcg Ap a h A • cg Pcg Note: All calculations are done with the projected area. The curved surface is not used at all in the analysis. B b • The curved surface projects onto plane a - b and results in a rectangle, (not a quarter circle) 15 ft x W. For this rectangle: Pcg = γhcg = 62.4 lbf/ft3 * 7.5 ft = 468 lbf/ft2 hcg = 7.5, Fh = Pcg A = 468 lbf/ft2 * 15 ft*W= 7020 W lbf Location: Ixx = bh3/12 = W * 153 /12 = 281.25 W ft4 Ixx sin θ 281.25W ft 4 sin 90o =− = − 2.5 ft hcg A 7.5 ft 15W ft2 The location is 2.5 ft below the c.g. or 10 ft below the surface, 5 ft above the bottom. II-16 | v v ycp = − | e-Text Main Menu | Textbook Table of Contents | Study Guide b. Vertical force: A Rule: Fv equals the weight of the effective column of fluid above the curved surface. b C • • c.g. F v B • Q: What is the effective volume of fluid above the surface? What volume of fluid would result in the actual pressure distribution on the curved surface? Vol = A - B - C Vrec = Vqc + VABC, VABC = Vrec - Vqc VABC = Veff = 152 W - π 152/4*W = 48.29 W ft3 Fv = ρg Veff = 62.4 lbf/ft3 * 48.29 ft3 = 3013 lbf Note: Fv is directed upward even though the effective volume is above the surface. c. What is the location? A C • • Rule: Fv will act through the centroid of the “effective volume causing the force. b We need the centroid of volume A-B-C. c.g. F v B • How do we obtain this centroid? Use the concept which is the basis of the centroid, the “first moment of an area.” | v v II-17 | e-Text Main Menu | Textbook Table of Contents | Study Guide Since: Arec = Aqc + AABC Mrec = Mqc + MABC MABC = Mrec - Mqc Note: We are taking moments about the left side of the figure, ie., point b. WHY? (The c.g. of the quarter circle is known to be 4 R/ 3 π w.r.t. b.) xcg A = xrec Arec - xqc Aqc xcg {152 - π*152/4} = 7.5*152 - {4*15/3/π}* π*152/4 xcg = 11.65 ft { distance to rt. of b to centroid } Q: Do we need a y location? Why? d. Calculate the moment about B needed for equilibrium. ∑M B =0 clockwise positive. MB + 5 Fh + ( − xv )Fv = 0 15 MB + 5 × 7020 W + (15 − 11.65 ) 3013 W = 0 MB + 5 × 7020 W + (15 − 11.65 ) 3013 W = 0 Pa ≠ ρ g y G≠g MB + 35,100 W + 10,093.6 W = 0 MB = − 45,194 W ft − lbf Why negative? The hydrostatic forces will tend to roll the surface clockwise relative to B, thus a resisting moment that is counterclockwise is needed for static equilibrium. Always review your answer (all aspects: magnitude, direction, units, etc.) to determine if it makes sense relative to physically what you understand about the problem. Begin to think like an engineer. | v v II-18 | e-Text Main Menu | Textbook Table of Contents | Study Guide Buoyancy An important extension of the procedure for vertical forces on curved surfaces is that of the concept of buoyancy. The basic principle was discovered by Archimedes. P atm It can be easily shown that (see text for detailed development) the buoyant force Fb is given by: Vb Fb = ρ g Vb Fb where Vb is the volume of the fluid displaced by the submerged body and ρ g is the specific weight of the fluid displaced. Thus, the buoyant force equals the weight of the fluid displaced, which is equal to the product of the specific weight times the volume of fluid displaced. The location of the buoyant force is: Through a vertical line of action, directed upward, which acts through the centroid of the volume of fluid displaced. Review all text examples and material on buoyancy. | v v II-19 | e-Text Main Menu | Textbook Table of Contents | Study Guide Pressure distribution in rigid body motion All of the problems considered to this point were for static fluids. We will now consider an extension of our static fluid analysis to the case of rigid body motion, where the entire fluid mass moves and accelerates uniformly (as a rigid body). The container of fluid shown below is accelerated uniformly up and to the right as shown. From a previous analysis, the general equation governing fluid motion is ∇ P = ρ( g − a ) + µ ∇ V 2 For rigid body motion, there is no velocity gradient in the fluid, therefore µ∇2 V = 0 The simplified equation can now be written as ∇ P = ρ( g − a ) = ρ G G = g − a ≡ the net acceleration vector acting on the fluid. where | v v II-20 | e-Text Main Menu | Textbook Table of Contents | Study Guide This result is similar to the equation for the variation of pressure in a hydrostatic fluid. However, in the case of rigid body motion: * ∇ P = f {fluid density & the net acceleration vector- G = g − a } * ∇ P acts in the vector direction of G = g − a * Lines of constant pressure are perpendicular to G . The new orientation of the free surface will also be perpendicular to G . The equations governing the analysis for this class of problems are most easily developed from an acceleration diagram. Acceleration diagram: Free surface For the indicated geometry: θ -a ax ax θ = tan−1 θ = tan−1 g + az g + az dP ds and 2 } 2 1 θ P2 2 P 1 g G P2 − P1 = ρ G(s 2 − s 1 ) Note: P2 − P1 ≠ ρ g(z 2 − z 1 ) Note: s is the depth to a given point perpendicular to the free surface or its extension. s is aligned with G . and s2 – s1 is not a vertical dimension v v II-21 | | e-Text Main Menu az ax s where G = {a x + (g + a z ) = ρG a | Textbook Table of Contents | Study Guide In analyzing typical problems with rigid body motion: 1. Draw the acceleration diagram taking care to correctly indicate –a, g, and θ, the inclination angle of the free surface. 2. Using the previously developed equations, solve for G and θ. 3. If required, use geometry to determine s2 – s1 (the perpendicular distance from the free surface to a given point) and then the pressure at that point relative to the surface using P2 – P1 = ρ G (s2 – s1) . Key Point: Do not use ρg to calculate P2 – P1, use ρ G. ax Example 2.12 Given: A coffee mug, 6 cm x 6 cm square, 10 cm deep, contains 7 cm of coffee. Mug is accelerated to the right with ax = 7 m/s2 . Assuming rigid body motion. ρc = 1010 kg/m3, 6 cm ∆z θ 10 cm Determine: a. Will the coffee spill? 7 cm b. Pg at “a & b”. c. Fnet on left wall. a. First draw schematic showing original orientation and final orientation of the free surface. ρc = 1010 kg/m3 a ax = 7m/s2 b g = 9.8907 m/s2 az = 0 Have a new free surface angle θ where -a θ = tan−1 ax a g + az θ 7 = 35.5° θ = tan 9.807 −1 ∆z = 3 tan 35.5 = 2.14 cm G g | v v II-22 | e-Text Main Menu | Textbook Table of Contents | Study Guide hmax = 7 + 2.14 = 9.14 cm < 10 cm ∴ Will not spill. ax b. Pressure at “ a & b.” 6 cm Pa = ρ G ∆ sa 2 ∆z 2 .5 2 θ 2 .5 G = {a x + g } = { 7 + 9.807 } G = 12.05 m/s2 10 cm ∆ sa = {7 + z} cos θ ∆ sa = 9.14 cm cos 35.5 = 7.44 cm 3 7 cm ∆s a θ 2 Pa = 1010 kg/m *12.05m/s *0.0744 m Pa = 906 (kg m/s2)/m2 = 906 Pa a b Note: Pa ≠ ρ g y G ≠ g Q: How would you find the pressure at b, Pb? c. What is the force on the left wall? ax We have a plane surface, what is the rule? 6 cm ∆z Find cg, Pcg, F = Pcg. A Vertical depth to cg is: θ θ ∆ s cg zcg = 9.14/2 = 4.57 cm 10 cm cg • ∆scg = 4.57 cos 35.5 = 3.72 cm 7 cm Pcg = ρ G ∆scg Pcg = 1010 kg/m3*12.05 m/s2* 0.0372 m Pcg = 452.7 N/m2 a b F = Pcg A = 452.7 N/m2*0.0914*0.06m2 F = 2.48 N ← | v v II-23 | e-Text Main Menu | Textbook Table of Contents | Study Guide What is the direction? Horizontal, perpendicular to the wall; i.e., Pressure always acts normal to a surface. Q: How would you find the force on the right wall? | v v II-24 | e-Text Main Menu | Textbook Table of Contents | Study Guide ...
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This note was uploaded on 04/21/2009 for the course MECH 314 taught by Professor White during the Spring '09 term at Clarkson University .

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