Practical question for mechanical/Civil engineers

[cyrille de brébisson]

Hello,

I find myself faced with a mechanical problem at home that I am unsure how to resolve and since I am pretty sure that some of you can solve stuff like that in your sleep, decided to post it here. Thanks for your help in advance.

Basically, I am trying to calculate the point of maximum deflection on a board set vertically and used to hold concrete in place while it is being poured in a vertical shaft.

Imagine a 'U' channel, but set vertically, 20cm*20cm by 3meters high. 3 sides are already enclosed in concrete, but the last one is 'open'. My plan is to place a board to 'close' the open face, fix it on top, bottom and brace it, if possible at the point of maximum pressure, which I imagine is NOT going to be the middle!

I would also like to get an idea as to how much pressure will be there.

Thanks,
Cyrille

ps: My thinking was that the point of maximum deflection would be at the point where 1/2 of the maximum pressure is exerted.
Assuming that the pressure at any 'height', h is proportional to the sum of the weight of the concrete above it, it means that the pressure at h i sum(0 to h, of p*h), ie p/2*h² where p is a constant based on concrete weight (2.4t/m^3) and how much of that weight actually does act to press.
so, pressure at the bottom is p/2*9=4.5p.
1/2 of this is 2.25p.
h where 1/2 the pressure is exerted would then be sqrt(2.25*2/p) around 2.1meter down from the top...

Am I any close to the solution of just blowing bubbles?

[CR Haeger]

Ill briefly take a shot at this.

I think you are correct in that the max poured concrete pressure will be at the bottom of the 3m column of concrete (~0.71 bar).

I think you are also correct in assuming that if you fix the top and bottom of the board, then the location of max (unfixed) board deflection will be lower than the 1.5m point, due to the linearly increasing pressure being placed on the board per meter of depth. At 2m, the estimated pressure will be about 0.47 bar.

As to the total force from the wet concrete pressing against the board, it seems you may generate about 21kN total, with the majority being borne by the middle and lower board "fixtures".

I am sure there are experts out there who can check my math and probably give you more detailed advice.

   

PS - Here is a fun link that may help directly Beam Calculator

[Alex Mark]

Hi cyrille,

Assuming that one can model the vertical board as a beam with both ends fixed (meaning that both ends have to be firmly connected to the structure and/or the soil), then the board will be loaded with a triangular load, ranging from 0 at the top, to γ*h at the bottom (where γ is the specific weight of concrete (~24 KN/m^3) and h is the beam's height (h = 3m). So, we have 0 at top, 72 KN/m^2 at bottom. This load is statically equivalent to (can be replaced by) a horizontal force loading the beam at a height of h/3 from bottom (the height of the triangle's centre of gravity). This force F has a measure equal to the area of the loading triangle, i.e. F = 0.5*γ*h^2.

But the loading point of the statically equivalent force F does not coincide with the point of maximum deflection, nor the point of maximum bending moment. Using Reinforced Concrete Designer's Handbook, Eleventh Edition, By Charles E. Reynolds, James C. Steedman, Anthony J. Threlfall, one can find the necessary equations to solve the problem. So, the maximum deflection is expected at 0.4753*h = 1.41 m from bottom, while the maximum bending moment at (1-sqrt(0.3))*h = 1.35 m from bottom.

For practical purposes, this means a bit lower the the beam's mid point. But you must also take into consideration that all these calculations are for static liquid pressure, i.e. after the concrete has been cast. While concrete casting, the beam load will be larger due to the concrete "falling down" into the channel with some velocity. The height of 3 m is not negligible, so the board has to be firmly fixed at both ends, and/or the concrete has to be cast carefully and not at once. I am afraid this calculation is beyond my mechanics knowledge, but as a typical engineer, I would just multiply the static figures by 2 or even 3, to be on the safe side.

[Edit]: In the link above, E is the elastic modulus (Young's modulus) of the board's material (8-9 GPa for pine wood), and I is the moment of inertia of the board's cross section, which for a rectangular cross section is I = b*d^3/12, where b is the cross section's width and d is the cross section's height. In your case, b >= 20 cm and d is the board's thickness (I guess a few cm).

Finally, do not forget to add some reinforcement to that. It will minimise the concrete flaking/cracking.

Apologies for not going through all the calculations in full detail, but my work load does not allow me to do so. I sincerely hope this has helped a bit.

Kind regards,

Alexandros Markatis
Greece

PS: I hate disclaimers, but I have to point out that all the above are just a part of a friendly conversation, and not a thorough study of the specific problem. If necessary, kindly consult an engineer.

[cyrille de brébisson]

Hello,

Thanks a lot guys, I will keep you posted on the 'pour' which, if all goes well should happen this WE.
I have decided to hold the board in 3 points using 3 threaded rods and do the pour in 2 passes (bottom and then top)... to reduce pressure on my board.

Cyrille

[Paul Dale]

That amount of cement will take a while to cure properly.

- Pauli

[Claudio L.]

(05-18-2015 05:37 AM)cyrille de brébisson Wrote:  Hello,
I find myself faced with a mechanical problem at home that I am unsure how to resolve and since I am pretty sure that some of you can solve stuff like that in your sleep, decided to post it here. Thanks for your help in advance.

Basically, I am trying to calculate the point of maximum deflection on a board set vertically and used to hold concrete in place while it is being poured in a vertical shaft.

Imagine a 'U' channel, but set vertically, 20cm*20cm by 3meters high. 3 sides are already enclosed in concrete, but the last one is 'open'. My plan is to place a board to 'close' the open face, fix it on top, bottom and brace it, if possible at the point of maximum pressure, which I imagine is NOT going to be the middle!

I would also like to get an idea as to how much pressure will be there.

The lateral pressure of concrete depends on how fluid they make the mix. The worst case is a very fluid mix, in which case it asymptotically approaches the same pressure diagram as water.
So it's safe to say the pressure at any height is no more than 2.4t/m3*h, with h=height of pour above.

Your worst-case pressure for 3 m high pour is then 7200 kgf/m3 or 70.6 kPa, so you got that right.

(05-18-2015 05:37 AM)cyrille de brébisson Wrote:  ps: My thinking was that the point of maximum deflection would be at the point where 1/2 of the maximum pressure is exerted.

...

Am I any close to the solution of just blowing bubbles?

Regarding anchorage: Do NOT anchor it only at 3 points. Use many small concrete screws (like Tapcons, for example) closely spaced along the vertical sides.
If you anchor at 3 points only, your board will also deflect between anchors, and deflection = leaks!. I don't know what kind of board you mean (I think plywood perhaps?), but you need lots of small anchors to make sure the board is tight to the other concrete at all points. With your proposal You'll end up with a mess of concrete on the floor, and your board surface will be "bulged" (despite the fact your anchorage will be sufficient to hold it), and so will the concrete when it hardens. Even small bulging makes it look bad. From an engineering point of view, you could calculate the deflection, of course, but being the owner you want NO deflection.

What I'd recommend you do, is attach to the sides, not top and bottom. That gives you a small span = very small deflections.
Since you are in the US, use 1/2 plywood or OSB with 1/4" Tapcons spaced 6 in. on center, 1 1/2" minimum embedment into the concrete (so you need 2 1/4" long screws). At least in the lower third, then you can increase the spacing to 8 in above, but keep in mind too much spacing means more chance for possible leaks.
If you think Tapcons is too much work, you can use what concrete workers use: an air gun with nails for concrete (I'd go 4 in spacing if using nails) or if you like firearms, use powder actuated fasteners (more fun).

Remember you only have one shot at this: once the concrete is there, if it bulges it will be a mess that's really hard to cleanup, so it pays to put some extra effort in preparations.
If you want a smooth finish (will it be exposed to the view?), use plywood with phenolic finish on one side, and very fluid concrete.

The pour height is important, but not to reduce the pressure (unless you wait several hours between pours). Do it in 2 passes, one right after the other, it won't reduce the pressure, but will help prevent segregation of the concrete and give you an overall better finish. Use a hammer to hit the board (ideally a concrete vibrator) to help the finish.

Good luck with your pour.
Claudio

[Claudio L.]

(05-18-2015 05:37 AM)cyrille de brébisson Wrote:  Basically, I am trying to calculate the point of maximum deflection on a board set vertically and used to hold concrete in place while it is being poured in a vertical shaft.

In case you still want to know the point of maximum deflection, here's a formula:

\[ x=h * \left( 1 - \sqrt{1-\sqrt{ 8 \over 15 } } \right) \approx 0.48 * h\]

This is the height measured from the bottom at which maximum deflection would happen if you only had supports top and bottom.

[Claudio L.]

(05-19-2015 01:45 PM)Claudio L. Wrote:  

(05-18-2015 05:37 AM)cyrille de brébisson Wrote:  Basically, I am trying to calculate the point of maximum deflection on a board set vertically and used to hold concrete in place while it is being poured in a vertical shaft.

In case you still want to know the point of maximum deflection, here's a formula:

\[ x=h * \left( 1 - \sqrt{1-\sqrt{ 8 \over 15 } } \right) \approx 0.48 * h\]

This is the height measured from the bottom at which maximum deflection would happen if you only had supports top and bottom.

And full formulae derivation, for those who care about the math:

\[ q=qmax-k*x \] (with x=height from the floor)

\[ qmax = \gamma * h * b \] ( \( \gamma \) = density of concrete, h = total height, b = width of the new pour)
and of course \( k= { qmax \over h } \) to make the pressure go to zero for x=h.

Now to get deflections we need to integrate a number of times:

\[ v = { \int_{}^{} p \cdot dx } = qmax \cdot x - k \cdot { x^2 \over 2 } + C1\]
\[ m = { \int_{}^{} v \cdot dx } = qmax \cdot { x^2 \over 2 } - k \cdot { x^3 \over 6 } + C1 \cdot x + C2 \]
\[ \phi = - { \int_{}^{} { m \over EI } \cdot dx } = - { 1 \over EI } \cdot \left[ qmax \cdot { x^3 \over 6 } - k \cdot { x^4 \over 24 } + C1 \cdot { x^2 \over 2 } + C2 \cdot x + C3 \right] \]
and finally:
\[ y = { \int_{}^{} \phi \cdot dx } = - {1 \over EI } \cdot \left[ qmax \cdot { x^4 \over 24 } - k \cdot { x^5 \over 120 } + C1 \cdot { x^3 \over 6 } + C2 \cdot { x^2 \over 2 } + C3 \cdot x + C4 \right] \]

The only thing left to do is to apply boundary conditions:

\[ \left. m=0 \right|_{x=0} \]
\[ \left. m=0 \right|_{x=h} \]

which allows to solve for C1 and C2 (C2 is trivially zero).

\[ 0 = qmax \cdot { h^2 \over 2 } - k \cdot { h^3 \over 6 } + C1 \cdot h \]
\[ C1 = k \cdot { h^2 \over 6 } - qmax \cdot { h \over 2 } \]

and at the supports:
\[ \left. y=0 \right|_{x=0} \]
\[ \left. y=0 \right|_{x=h} \]

to solve for C3 and C4 (C4=0 as well).

\[ 0 = - {1 \over EI } \cdot \left[ qmax \cdot { h^4 \over 24 } - k \cdot { h^5 \over 120 } + \left( k \cdot { h^2 \over 6 } - qmax \cdot { h \over 2 } \right) \cdot { h^3 \over 6 } + C3 \cdot h \right] \]

\[ C3 = k \cdot { h^4 \over 120 } - qmax \cdot { h^3 \over 24 } - \left( k \cdot { h \over 6 } - { qmax \over 2 } \right) \cdot { h^3 \over 6 } \]

and finally, the equation for deflection:

\[ y = - {1 \over EI } \cdot \left[ qmax \cdot { x^4 \over 24 } - k \cdot { x^5 \over 120 } + \left( k \cdot { h^2 \over 6 } - qmax \cdot { h \over 2 } \right) \cdot { x^3 \over 6 } + \left( k \cdot { h^4 \over 120 } - qmax \cdot { h^3 \over 24 } - \left( k \cdot { h \over 6 } - { qmax \over 2 } \right) \cdot { h^3 \over 6 } \right) \cdot x \right] \]

EI is the product of the elastic modulus of the material and the moment of inertia of the cross section of the board. I'll leave it for the reader to verify that \( y'=0 \) will result in a value of X per the formula given above (I didn't check).

[TASP]

Hope you get a good seal, too. Even if form doesn't deflect (much) seeing the mix oozing around it is annoying.

[BruceH]

(05-19-2015 02:58 PM)Claudio L. Wrote:  

(05-19-2015 01:45 PM)Claudio L. Wrote:  In case you still want to know the point of maximum deflection, here's a formula:

\[ x=h * \left( 1 - \sqrt{1-\sqrt{ 8 \over 15 } } \right) \approx 0.48 * h\]

This is the height measured from the bottom at which maximum deflection would happen if you only had supports top and bottom.

And full formulae derivation, for those who care about the math:

\[ q=qmax-k*x \] (with x=height from the floor)

\[ qmax = \gamma * h * b \] ( \( \gamma \) = density of concrete, h = total height, b = width of the new pour)
and of course \( k= { qmax \over h } \) to make the pressure go to zero for x=h.

Now to get deflections we need to integrate a number of times:

\[ v = { \int_{}^{} p \cdot dx } = qmax \cdot x - k \cdot { x^2 \over 2 } + C1\]
\[ m = { \int_{}^{} v \cdot dx } = qmax \cdot { x^2 \over 2 } - k \cdot { x^3 \over 6 } + C1 \cdot x + C2 \]
\[ \phi = - { \int_{}^{} { m \over EI } \cdot dx } = - { 1 \over EI } \cdot \left[ qmax \cdot { x^3 \over 6 } - k \cdot { x^4 \over 24 } + C1 \cdot { x^2 \over 2 } + C2 \cdot x + C3 \right] \]
and finally:
\[ y = { \int_{}^{} \phi \cdot dx } = - {1 \over EI } \cdot \left[ qmax \cdot { x^4 \over 24 } - k \cdot { x^5 \over 120 } + C1 \cdot { x^3 \over 6 } + C2 \cdot { x^2 \over 2 } + C3 \cdot x + C4 \right] \]

The only thing left to do is to apply boundary conditions:

\[ \left. m=0 \right|_{x=0} \]
\[ \left. m=0 \right|_{x=h} \]

which allows to solve for C1 and C2 (C2 is trivially zero).

\[ 0 = qmax \cdot { h^2 \over 2 } - k \cdot { h^3 \over 6 } + C1 \cdot h \]
\[ C1 = k \cdot { h^2 \over 6 } - qmax \cdot { h \over 2 } \]

and at the supports:
\[ \left. y=0 \right|_{x=0} \]
\[ \left. y=0 \right|_{x=h} \]

to solve for C3 and C4 (C4=0 as well).

\[ 0 = - {1 \over EI } \cdot \left[ qmax \cdot { h^4 \over 24 } - k \cdot { h^5 \over 120 } + \left( k \cdot { h^2 \over 6 } - qmax \cdot { h \over 2 } \right) \cdot { h^3 \over 6 } + C3 \cdot h \right] \]

\[ C3 = k \cdot { h^4 \over 120 } - qmax \cdot { h^3 \over 24 } - \left( k \cdot { h \over 6 } - { qmax \over 2 } \right) \cdot { h^3 \over 6 } \]

and finally, the equation for deflection:

\[ y = - {1 \over EI } \cdot \left[ qmax \cdot { x^4 \over 24 } - k \cdot { x^5 \over 120 } + \left( k \cdot { h^2 \over 6 } - qmax \cdot { h \over 2 } \right) \cdot { x^3 \over 6 } + \left( k \cdot { h^4 \over 120 } - qmax \cdot { h^3 \over 24 } - \left( k \cdot { h \over 6 } - { qmax \over 2 } \right) \cdot { h^3 \over 6 } \right) \cdot x \right] \]

EI is the product of the elastic modulus of the material and the moment of inertia of the cross section of the board. I'll leave it for the reader to verify that \( y'=0 \) will result in a value of X per the formula given above (I didn't check).

I entered all of that into my Prime and the answer came out as... 3 rolls of Duck tape. ;-)

[cyrille de brébisson]

Hello,

Quote:I entered all of that into my Prime and the answer came out as... 3 rolls of Duck tape. ;-)

You must have set it up using US units. What is that in SI?

Cyrille