Fluid Mechanics: Calculating Force Required

1. Sep 19, 2015

Sheezey

1. The problem statement, all variables and given/known data

Gate AB in Fig. P2.62 is 15 ft long and 8 ft wide into the paper and is hinged at B with a stop at A. The water is at 20°C. The gate is 1-in-thick steel, SG =7.85. Compute the water level h for which the gate will start to fail.

2. Relevant equations
F = δhcgA

Ycp = Ixxsinθ/hcgA

Ixx = bd3/12

Moment = F*D

3. The attempt at a solution
F = (62.3lb/ft3)(h/2)(120ft2)

I am getting stuck as far as how to put together the equation to find the hydrostatic force. Mainly, what height value to use for hcg and what value to use for area.

According to a similar example in the book, hcg represents the center of gravity for h so we would have h/2 but I'm getting confused as to what the area is for this equation. Is it the area of the submerged portion of the plate? If so, I am not sure how to compute that. Any help is greatly appreciated.

2. Sep 19, 2015

billy_joule

Yes it's the submerged area, that is what you'll need to find. hwater depth is related to A, can you show this relationship?

Start with the easy part, what is the torque about B due to the self weight of the gate and the hanging mass? This is the torque the water must apply to open the gate.

3. Sep 19, 2015

Sheezey

Hi billy_joule,
Thank you for responding. The torque about B due to the self weight would be

10,000lb* 15ft = 150,000lbft

4. Sep 19, 2015

billy_joule

That's the clockwise torque due to the hanging mass, you need to find the anticlockwise torque from the self weight of the gate itself too.
That's why you're given the density of the steel.

5. Sep 19, 2015

Sheezey

Right. You must mean the weight force of the steel plate which would act along the center of the plate.

So we would have...
(7.85)(62.3lb/ft3)(15ft)(8ft)(1/12ft)(7.5cos(60)) = 18339.6lb

Does that look right?

6. Sep 19, 2015

Sheezey

Correction
(7.85)(62.3lb/ft3)(15ft)(8ft)(1/12ft)(7.5ftcos(60)) = 18339.6lbft

7. Sep 19, 2015

Sheezey

Thinking more about it there would be a force due to the water acting on to the steel plate. I believe this force would also act in the center of the plate. I think this is what I am looking for to insert for the area in my original equation.

8. Sep 19, 2015

SteamKing

Staff Emeritus
Does the hydrostatic force act at the center of the plate? Remember, water pressure varies with depth.

9. Sep 19, 2015

Sheezey

No it wouldn't. It does vary and will be become greater the deeper you go.

10. Sep 20, 2015

billy_joule

You'll also need to express other quantities in terms of hwater depth too. e.g. the length of the lever arm that the hydrostatic force acts at.

You won't actually find the submerged area directly, though you could find it after you've found hwater depth, at which point there's no need.

Ultimately you want an expression:
M=Fd
where the LHS is the moment about B due to the gate weight and hanging weight (which you've almost found) and the only unknown on the RHS is hwater depth.
(The knowns on the RHS will be 60o, water density and 8ft..)

11. Sep 20, 2015

Staff: Mentor

If z is the vertical elevation above point B and h is the total depth of water, what is the water depth at elevation z? What is the pressure at elevation z? What is the direction of the resultant force per unit area (pressure force) acting on the gate at elevations between elevation z and elevation z + dz? What is the differential area of the gate between elevations z and z + dz? What is the moment arm of this force about line B?

Chet

12. Sep 20, 2015

Sheezey

Ok so the submerged area of the plate is going to be (8ft)*(the submerged length of A to B)........

If I create a triangle on the drawing and call the unknown submerged length 'x' I get x=h/sin(60).

Now I can plug that into the original equation I was asking about, hydrostatic force. So I now have F = (62.3lb/ft3)(h/2)(8*h*csc(60)) = 287.8h2

So looking at my moments I have my weight force, my 10,000 force and my hydrostatic force F acting on the plate. I should be able to sum up the moments to find h.

I have

M = -(150,000lbft) + (18339.6lbft) + 287.8h2(some distance from b) = 0

Does that look like the right direction so far?

13. Sep 20, 2015

Sheezey

Ok got it! The "some distance from b" in my moment equation is = to Ycp. Which requires me to find the moment of inertia.

Ixx = (1/12)(8)(hcsc(60))3

Ycp = (Ixx(8)(hcsc(60))3sin(60))/((h/2)(8hcsc(60))

M = -150000+18339.6+287.8h2(Ycp) = 0 solve for h

∴ h = 10.59 ft

14. Sep 20, 2015

Staff: Mentor

I really hate the idea of using memorized equations, like those for Ycp and Ixx. Using the simple method I outlined in post #11, you get a moment about axis B of $$M=\frac{ρgwh^3}{6}\csc^2(60)$$
where w = 8 ft. This leads to the same value for h that you obtained.

Chet

15. Sep 20, 2015

Sheezey

Interesting. I understand what you mean. That is a great way to look at it. Good advice. Thank you for your help.