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The precast concrete stairs at the home I recently purchased do not have a footing. Lack of a footing coupled with drainage issues has caused them to sink over time. I'm going to raise them and get them properly supported underneath.

To do that I want to clamp them near their center of gravity (along the length), so that I can jack them up back to where they belong and minimize the risk of breaking the concrete (I'll continuously add temporary support underneath them as I slowly jack them up).

I have dug out the bottom of the steps and determined that they are monolithically poured and solid.

What is the measurement along the length of the steps where the center of gravity will be?

enter image description here

Edit: This is not homework. Here are the stairs in question:

enter image description here

You can see how they are pulling away from the foundation and have rotated forward and dropped. The sidewalk slab in front of it has also sunk and needs to be raised, so it is worse than it seems in this photo.

I attempted to solve this by hand because it seems trivial. I'm glad I asked, because the solution I came up with was wrong. I assumed I could reduce it to working with the ratio of cross sections which are simply three rectangles (since they all have the same width).

If I take the area of each step individually I get:

$$\begin{align} 17.5 \times 18 &= 315 \\ 14 \times 11 &= 154 \\ 47.5 \times 3 &= 142.5 \end{align}$$

I noticed that if I add the bottom two steps together, they are a smaller combined area (and thus volume) than the large step [pretty sure this is where I went wrong ... when you multiply by width this is no longer the case]. So it appeared to me at the time that the COG would be somewhere in the large step. I'll denote the COG as $c$.

So I then used this to give me $c$:

$$18 \times c = 18 \times (17.5 - c) + 154 + 142.5$$

That worked out to 16.98 (17 inches for all practical purposes).

However, after seeing @Chris Johns solution, I realize that what I had in my brain is not correct, and anybody with sinking stairs should ignore everything above.

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  • $\begingroup$ This looks like a "homework question". In order for such questions to be answered in this site, we need you to add details describing the precise problem you're having. What have you tried to solve this yourself? Please edit your question to include this information. $\endgroup$ – Wasabi Apr 24 '16 at 19:53
  • $\begingroup$ It's not homework. I haven't sat in a college course in over 20 years. $\endgroup$ – alfreema Apr 24 '16 at 20:03
  • $\begingroup$ Please note the quotation marks. "Homework questions" don't have to be actual homework. They are questions wherein the asker simply throws a numerical exercise (that's all this really is) and asks for an answer. The equation to calculate the centroid of a shape which is just a composition of rectangles is quite simple and can be found on Wikipedia. So I ask you: what are you having trouble with in using these methods? That's what you need to tell us. $\endgroup$ – Wasabi Apr 24 '16 at 20:16
  • $\begingroup$ Well for starters, I don't even know what a "centroid" is, so I wouldn't have known to Google that. I'll add what I did to the question, but it's just gibberish at this point and won't help a home owner fix this very common problem. $\endgroup$ – alfreema Apr 24 '16 at 20:18
  • $\begingroup$ For an object made of a single material, centroid is equal to center of gravity. You could also look at the Wikipedia page on centers of gravity to get the same equations. $\endgroup$ – Wasabi Apr 24 '16 at 20:21
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The equation to find the center of gravity of an object which can be subdivided into smaller parts is the following:

$$\text{CG} = \dfrac{\sum \overline{y}_iA_i\rho_i}{\sum A_i\rho_i}$$

where $\overline{y}_i$, $A_i$ and $\rho_i$ are the centroid, area and specific weight of each part, respectively.

In your case, the specific weight of all parts is equal, and so the equation can be simplified to

$$\text{CG} = \overline{y} = \dfrac{\sum \overline{y}_iA_i}{A}$$

Therefore, in such a case, the center of gravity of the object is equal to its centroid ($\overline{y}$). Also, to be clear, the $A$ in the denominator is the object's total area.

So, in this example, the solution is:

$$\begin{align} \overline{x}_1A_1 &= \left(\dfrac{17.5}{2}\right) \times \left(18 \times 17.5\right) \\ \overline{x}_2A_2 &= \left(17.5 + \dfrac{14}{2}\right) \times \left(10.5 \times 14\right) \\ \overline{x}_3A_3 &= \left(17.5 + 14 + \dfrac{47.5}{2}\right) \times \left(3.0 \times 47.5\right) \\ A &= \left(18 \times 17.5\right) + \left(10.5 \times 14\right) + \left(3.0 \times 47.5\right) \\ \color{red}{\overline{x}} &= \color{red}{\dfrac{\sum \overline{x}_iA_i}{A} = 23.5} \\ \overline{y}_1A_1 &= \left(\dfrac{18}{2}\right) \times \left(18 \times 17.5\right) \\ \overline{y}_2A_2 &= \left(\dfrac{10.5}{2}\right) \times \left(10.5 \times 14\right) \\ \overline{y}_3A_3 &= \left(\dfrac{3.0}{2}\right) \times \left(3.0 \times 47.5\right) \\ \color{red}{\overline{y}} &= \color{red}{\dfrac{\sum \overline{y}_iA_i}{A} = 6.3} \end{align}$$

Your center of gravity is 23.5" from the left-hand side of your image (from the exterior face of the upper step) and 6.3" from the bottom of your image.

That being said, don't think that just because you raise the steps from their center of gravity that they won't get damaged. If you just do that, then the steps may become like a dual-cantilever, in which case your thin "step" (if you can call it that) will risk breaking apart due to its low height and therefore low moment of inertia. Now, if what you actually want is help on raising your steps without damaging them, I suggest you ask another question referencing this one.

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  • $\begingroup$ Ahh, this is awesome. I have to switch the answer over to this one. So $\overline{x_i}$ is just the offset from the left side to the midpoint of the "next section". Now if I chop the large thin slab off leaving just a couple of inches due to the width of the circular saw, I'll know how to recalculate! (Thanks for cleaning my question up too!) $\endgroup$ – alfreema Apr 24 '16 at 21:50
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From the left hand side in your drawing I make it x = 23.5" y = 6.32" ie more or less in the centre of the second step.

Also note that as you have done a CAD drawing of it already there is probably a tool in the software to calculate the CoG for you (that's how I did it).

If it's going to break anywhere the inside corner between the first and second steps is probably the most likely place. So try to avoid putting to much bending load on the bottom step.

For academic interest the way to do this by hand is to break the areas down into convenient blocks as you have done and calculate moments of each individual centre of mass about the CoG. By definition the sum of the moments about the CoG will be zero. Here even though the bottom step is thin, it has significant effect because it extends a long way from the CoG.

EDIT

You may also be interested in foam injection leveling this is a method of leveling, jacking and stabilising concrete by injecting expanding PU foam underneath it. Here the gas pressure produced by the foaming reaction lifts the slab and the foam itself creates a stable base. I can't vouch for the usefulness of this method as I've only come across it in passing but it seems like it is worth considering in this situation.

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  • $\begingroup$ Ugh, I was off on the weeds then. I threw this together in Sketchup and I doubt the free version will calculate COG. I doubt even the Pro version will. $\endgroup$ – alfreema Apr 24 '16 at 19:58
  • $\begingroup$ Thanks for the answer, I was way off in the weeds. I thought I could take the cross section area of the three steps and go from there. But ignoring the width (and hence volume) screwed me totally up. I'm actually considering using a masonry blade and sawing off the thin first step near the second step, just so I minimize the risk of cracking this monster. Then I can level those two piece independently and the cut will just look like a typical concrete relief cut. $\endgroup$ – alfreema Apr 24 '16 at 20:01
  • $\begingroup$ Well apparently I can ignore the width after all, but somehow still managed to screw this up by hand. $\endgroup$ – alfreema Apr 24 '16 at 20:57
  • $\begingroup$ In a nutshell you need to consider each block as a point mass and calculate the point where they are connected such that the net moment (ie weight x distance from point) is zero. $\endgroup$ – Chris Johns Apr 24 '16 at 21:03

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