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# Introduction to Fluid Mechanic CE 321

MSU

GPA 3.85

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This 18 page Class Notes was uploaded by Jolie Shields on Saturday September 19, 2015. The Class Notes belongs to CE 321 at Michigan State University taught by Roger Wallace in Fall. Since its upload, it has received 23 views. For similar materials see /class/207369/ce-321-michigan-state-university in Civil Engineering at Michigan State University.

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Date Created: 09/19/15

Lecture 4 Forces on Submerged Bodies Forces on Plane Surfaces Important in the design of storage tanks ships dams etc For a fluid at rest force is perpendicular to the surface no shear inside P varies linearly with depth hydrostatic What if the surface is incHned He I l f CC Some Concepts from Mechanics The moment of a quotge force is a measure of its turning effect A1 F 7 sin H A1 F 7 for a 15pm lt4 354410quot What can this person do differently to open the door easily The First Moment of Areal Qw39Jc a 1 UAW fydA yOA A yC is the y coordinate of the centroid measured from the X axis Second Moment 0 The second moment of area also known as 2 the second moment of IQ fy dA inertia is a property of a shape that is used to predict its resistance to bending and deflection The Parallel Axis Theorem The parallel axis rule can be used to A determine the moment x of inertia ofa rigid object 39 about any axis given the moment of inertia of the object about the arallel axis throuV h the object39s center of mass 2 and the perpendicular Z X distance between the axes Fvee swim 0 Force on an Inclined Surface FR fwhdA fwysin 0dA h y Sin 9 A F 39 9 11 I R 78111 1 7 11 70A 482W L M cens ii aiifi sm NOTE The point of application ofthis force is NOT the centroid Force on an Inclined Surface We now have an equation for the resultant force We are now ready to ask the next question Where does this force act In other words what are the coordinates of the point of application of the resultant force Point of Application of FR The moment of the resultant force must be equal to the moment of the distributed force FRyR fde fypdA fy7ydA vsin 0f ygdA 7A sin 0 f yfdA f filA A A A R 7 Sin 9f ydA 90A 9014 A can calculate XR in a similar way Using the parallel axis theorem we can express lX as IX IXC Ayc2 20 yR yCA yo m can calculate XR in a similar way Geometric Properties of Common Shapes Ai 70 PATH Pgaggq lbl chle my Rectangle The point through which the resultant force acts is called the Center of pressure m1 In 7 Wu 7 2m 3 quot T3 m 5mm CentrOIdal coordinates Inertia TOI39 some common shapes lcl Quanev wcle Jr Vial A7 I 5 Pressure Prism Average P at depth h2 a b Force FR EMA VL JA Volume of prism length x width x height 2 7 A Magnitude of Resultant Force NOTE The centroid that we are referring to here is that of the triangular distribution of pressure Do not confuse this with the centroid of the object we showed earlier Pressure Prism The resultant force passes through the centroid of the pressure prism This result can be verified using the formulas for the coordinates of FR For a triangle the centroid is located at a distance h3 above the base Problem 229 2 24 A large open tank contains water and is connected to a 6ft diameter conduit as shown in Fig P211 A circular plug is used to seal the conduit Determine the magnitude direction and location of the force of the water on the plug Problem 229 Solution FR m A MIAWm w a 1 IX 77 J M W when In T 3 5 5 IZH zH R umWm The 14kg of all200M 404 Z 7 75 emd 7 wafer sumac and 439s Pero ndfcuqr 7 e pu Sur ace 45 Show Problem 230 p 60 230 A homogeneous 4ftwido 8tftlong rectangular gate weighing 300mll3 is held in place by a horizontal exible cable as shown in Fig P230 Water acts against the gate which is hinged at point A Friction in the hinge is negligi ble Determine the tension in the cable i 6 sin60 Cable 360 230 Solution Pg XAcA UAW Ac f39 fn a I f quot552 2 4 3 Qf sMbQMtx 4145 33701 230 Solution 7 Made FE 3 71 ffc 1015 7a5 50 7711119 74 4mm 3 6 4mm 37 644 IE

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