Two Slab Supported by 4 walls 2

[Colts56]

What method of analysis would be used when a two way slab is being supported continuously on 4 sides? An example would be a 20'x20' masonry structure with a concrete roof. Due to building limitations, the slab has to be one thickness. I'd assume a two way slab design would be best, but when looking at ACI 318, there is no analysis information when a slab is supported by continuous walls. It seems to only discusses two way slabs for column grids.

 

[JedClampett]

There's tools (Roark; Bureau of Reclamation Tables, finite element) to analyze these cases. A two way slab design is likely the best, but you're going to have to do your own analysis. ACI two way slab design is oriented to slabs on beams with thickened sections at the columns.
For a case you're talking about, model the edges as pinned, do your analysis, and add reinforcing as necessary. For a lot of these, the slabs are designed in one direction and the controlling reinforcing is provided in both directions.

 

[DETstru]

Look into yield line analysis for a simply supported slab on all sides. The yield lines will form an "X" in the middle of your slab (since it's a square).

I would just toss it in RAM Concept though...
If you really want to shortcut it... you could design it as one way and put that reinforcing in both directions...

 

[BAretired]

If the slab is square in plan, symmetry helps to simplify the design. If the slab is rectangular (but not square), you could use the Hillerborg Strip Method or Yield Line Analysis. If the slab is tied down to the wall at the corners, don't forget to design for "corner levers" because the slab will have a tendency to lift off the supports near the corners. Holding it down will produce negative moments.

BA

 

[Colts56]

I was wanting to use RAM Concept, but didn't know how to do the design strips since there are no columns to help know what the design strip width should be. I thought about just making a bunch of 1' wide strips each direction, but didn't know if that would be correct.

 

[DETstru]

Just run one strip in each direction. Centerline of wall to the opposite side.
If you want, run another model with three in each direction and take the worst case.

 

[dik]

and use the same moment resistance in both directions... slab behaves isotropically for yield line... also make sure you have top reinforcing in the corners... you can develop 'corner levers and have a negative moment at 45degrees.

Good Luck

Dik

 

[ShawThing]

If the slab is a square and the slab is discontinuous at the walls, in the Australian Concrete Code the bending moment would be:
M* = 0.056FdL² where Fd is your uniformly distributed design load and L is the slab span. You would need to check deflection also.

 

[P1ENG]

See the attached for Roark's equations. This is from a project with steel plate and it calculates the stresses and deflection for multiple (4) edge support conditions. I don't know off-hand if the stresses use the elastic or plastic section modulus. I could figure that out though if you want to use this approach, or maybe someone else can chime in before I get back to you.

Juston Fluckey, SE, PE, AWS CWI
Engineering Consultant

 

[BAretired]

If the slab is square and simply supported at all four walls, it seems appropriate to consider half the load carried in each direction. A good approximation for positive moment is M = wL²/16 in each direction, slightly conservative because of the corner moments. If corner reinforcement is omitted, cracks may be expected in the corners. It is customary to provide top corner bars at the same spacing as the positive moment, placed parallel to the diagonal line between corners. Positive moment at the corners is satisfied if the bottom bars are continued to the support in each direction.

BA

 

[dik]

If simply supported on 'knife edges', with no attachment, you can still develop corner levers due to the corners wanting to lift. You need to provide top steel in the corners...

Dik

 

[hokie66]

Agree with BA, except the top bars are usually placed orthogonnally.

 

[dik]

PIENG:
Can you post a complete pdf copy of your mathcad program; I'd like to try to enter it on my pocket calculator. You may not want to, and, I fully understand.

Thanks Dik

 

[P1ENG]

dik,
I don't understand your request. The rest of the calculation checks beams, columns, and connections for a stair assembly. I don't have any other options in the program other than the (4) I included:
1.) (4) edges simply supported
2.) (2) short edges fixed, (2) long edges simply supported
3.) (2) long edges fixed, (2) short edges simply supported
4.) (4) edges fixed

I have uploaded the actual Mathcad (v15) portion of the Roark's formulas. Nothing was hidden, so all it does is save anyone some typing if they want to use it. Table 11.4 of Roark's 7th Ed. has several other conditions for rectangular plates and loading conditions.

Also, to answer my question earlier about elastic/plastic section modulus; it is the elastic section modulus. Proof is found by taking a/b = infinity for the (4) edges simply supported. We know stress would be equal to M/S or M/Z. With beta=0.75, we find it is equal to M/S. Now you can calculate the maximum moment in the plate instead of stress.

Juston Fluckey, SE, PE, AWS CWI
Engineering Consultant

 

[dik]

Thanks... I often use bd^2/4 for plates... almost never bd^2/6...

Dik

 

[P1ENG]

dik,
I would too for steel if I needed it. In my applications, the deflection controls so it doesn't matter if it is elastic or plastic. That is just the way the stress equations are given by Roark's. Knowing now that it is the elastic, I will probably adjust my beta values so that I calculate a moment instead of stress to match the Mn/Ω vs M comparisons of the last (2) releases of the AISC SCM.

Juston Fluckey, SE, PE, AWS CWI
Engineering Consultant