Cracked Moment of Inertia When Top and Bot Steel in Tension (kd > c): 2

[Struct123ure]

Hi, I have a 200mm slab and based on the covers the d=167.5mm and d’=54.5mm (distance to centre of bottom steel and top steel respectively). Based on the loading and strain diagram I can calculate a c (distance to N.A.) of 43.3mm.
When calculating deflection though following the CSA A23.3-14 on pg 6-5 (Table 6.2a) there are formulas to calculate the “Moment of Inertia of Cracked Transformed Section - with compression steel” which gives you a kd distance to natural axis.
This kd=52.1mm which is larger than the c=43.3mm calculated previously?

1. I am trying to figure out what I should use for Icr, because it doesn’t make sense to use a larger cracked section then what I know will crack under loading?
The kd equations are also used in the ACI 435R except there they call it “a”.
I suspect that this is because my calculated c is for slab failure and the reinforcement has yielded. While the deflection kd is for serviceability so the slab is not expected to crack as much under service load?
2. Does calculating Icr based on bh3/12+Ad2 (as shown in attached) make any sense?
3. If kd=52.1mm < d’=54.5mm am I justified in using the “no compression steel” formula since this would mean the top steel is actually in tension?

To get the full picture here are some other things:
I can’t increase the depth of the slab.
The spans are pretty short, but the load kind of high that’s why I think I will be able to get the deflection to work.

 

[IDS]

Note that I don't use the Canadian concrete codes.

Any single formula for a cracked moment of inertia will be very much an approximation, so it makes sense to be conservative if deflections are critical to the design.

Yes, the depth to the NA will be greater under service loads than under ultimate loads, but if the top reinforcement is close to the NA it will have very little effect on deflections, and it is conservative to ignore the "compression" steel, so it makes sense to ignore it for deflection calculations.


Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[bugbus]

Don't calculate the neutral axis depth at SLS based on a rectangular stress block. The stress block should be triangular and should extend all the way down to the neutral axis. So, the 52.1 mm appears to be correct. Also, the maximum compressive strain isn't necessarily 0.0035 at SLS, this is only at ULS.

You don't appear to have accounted for tension stiffening anywhere, which might be OK if your slab is heavily loaded, but I suspect the slab will be much stiffer than the fully cracked Icr. Use the formula for effective second moment of area in Eurocode / Australian Standard (by Bischoff) (and not the one in ACI by Branson).

 

[Struct123ure]

Thank you bugbus, I started by creating a moment curvature plot and found the SLS moment to be half of the ULS moment. At SLS moment the N.A. is approx 53mm which is very close to the kd=52.1mm.

Thanks IDS, I will ignore the "compression" steel since even under SLS load it isn't in compression 53mm < d’=54.5mm.