AME-309 Exam1-soln

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Unformatted text preview: ................................... .............................................. .... ... .... .......... .................................... .......................................... ............................. ......... ......................................... w (a) Clearly the center fluid exerts the same hydrostatic force Fc on both gates There are two additional hydrostatic forces acting viz the force on the left gate from the left fluid F and the force on the right gate from the right fluid Fr Since the gate is rectangular with height 2H its centroid is at a distance H below the top of the gate So for the fluids to the left center and right respectively the centroid locations relative to the free surface are z = 2H + H = 3H, ¯ zc = H + h, ¯ zr = H + H = 2H ¯ Also the area and moment of inertia for the gate are A = 2Hw and I= 1 2 w(2H )3 = wH 3 12 3 Therefore the hydrostatic forces on the gate from left center and right fluids are F = ρg z A = ρg (3H )(2Hw) = 6ρgH 2 w ¯ ¯ Fc = 2ρg zc A = 2ρg (h + H )(2Hw) = 4ρgH (H + h)w 3 3 ρg zr A = ρg (2H )(2Hw) = 6ρgH 2 w ¯ Fr = 2 2 (b) The centers of pressure are 2 3 28 I 3 wH = 3H + =H zA ¯ (3H )(2Hw) 9 2 3 1 H2 I 3 wH =H +h+ =H +h+ = zc + ¯ zc A ¯ (H + h)(2Hw) 3H +h 2 3 13 I 3 wH = 2H + =H = zr + ¯ zr A ¯ (2H )(2Hw) 6 ( zcp) = z + ¯ (c zcp) (r zcp) (c) There are four forces acting on each gate In addition to the hydrostatic forces from each side there are reaction forces at the top of the gate Ft and at the stop Fs We seek the depth of the center fluid h at which each gate is on the brink of opening i e the depth that yields Fs = 0 To find the depth we must balance moments It is simplest to take moments about the top of the gate (which obviates the need to determine Ft ) Left Gate: F ( (c zcp) − 2H + Fs (2H ) = Ft (0) + Fc zcp ) − h 3 Now, the centers of pressure are ( zcp) = 28 H 9 and (c zcp) = H + h + 1 H2 3H +h Thus, substituting into the moment-balance equation with Fs = 0 yields 6ρgH 2 w 10 H 9 = 4ρgH (H + h)w H + 1 H2 3H +h or, 3H + 3h + H 20 ρgH 3 w = 4ρgH 2 w 3 3 Solving for h, there follows 1 h= H 3 =⇒ 5H = 4H + 3h Right Gate: (c (r Fc zcp) − h + Fs (2H ) = Ft (0) + Fr zcp ) − H Now, the centers of pressure are (c zcp) = H + h + 1 H2 3H +h and (r zcp) = 13 H 6 Thus,...
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This note was uploaded on 02/01/2013 for the course AME 309 taught by Professor Phares during the Fall '06 term at USC.

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