anonymous
  • anonymous
A gate in a salt water canal is hinged at A and is held closed by the pin-connected assembly CDF. (SEE FIGURE)
Engineering
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
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mathmate
  • mathmate
|dw:1449354075120:dw|
mathmate
  • mathmate
First, the hydrostatic force can be reduced to a single equivalent force by integrating hydrostatic pressure over the depth of the salt water. The magnitude is Fh and should be applied at a point between A and B such that it gives the same moment as the hydrostatic pressure, kind of finding the centroid of the hydrostatic pressure. Call this depth d. |dw:1449354258204:dw|

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mathmate
  • mathmate
For the two arms CD and DE, they are two force members, therefore forces are axial. By symmetry, each will have an axial force of Fs (s for strut, since they are both in compression). Consider member AB, hinged at A. Fs and Fh create opposing moments about hinge A, such that the sum of moments =0, or Fs*((1.1+0.4)cos(20)=Fh*d |dw:1449354906308:dw|
mathmate
  • mathmate
Now having found Fs, draw the FBD for strut DE, recall that DE is a two force member. |dw:1449355102781:dw| We can solve for T/2 using Lami's theorem: (T/2)/sin(160)=Fs/sin(90). and the problem is solved. If you would like to check your answer, please post your result.
mathmate
  • mathmate
Correction to previous diagram for calculation of moment on member AB |dw:1449355525267:dw| The lever arm for hydrostatic force Fh is actually d+0.4, since d was measured from the liquid surface.
anonymous
  • anonymous
The tension that I calculated is 114 KN.
anonymous
  • anonymous
@mathmate
mathmate
  • mathmate
I double-checked my work, 114.244 is what I got also. Congrats!
rvc
  • rvc
YAY! thanks @mathmate
anonymous
  • anonymous
is it ok to ask question on a old topic? I was reviewing hydrostaticforce. I just want to know how I can find d?
mathmate
  • mathmate
@acuben |dw:1454553100269:dw| As we know, the hydrostatic pressure at depth y equals rho*y. d is basically the centre of gravity of the triangular hydrostatic pressure, or two-thirds from the fluid surface. If you'd like a more rigorous expression, we take moments of the forces about the water surface: \(F_h = \int_0^h \rho y dy = (1/2)\rho h^2\) \(F_h*d = \int_0^h y*\rho ydy=(1/3)\rho h^3 \) Substitute \(F_h\) and solve for d: d=(2/3)h

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