Modulus of Rupture 1

[HardyParty]

For the slender wall method (Chapter 14.8 in ACI 318-08) the modulus of rupture is listed as 7.5*lambda*sqrt(f'c).

For structurally plain concrete (Chapter 22 of the same code), it is given as 5 * lambda * sqrt(f'c).

Why are these values different? Why is the modulus stronger for the slender wall method?

 

[dik]

I would think that for plain concrete you might want to be a little more conservative, where with slender walls, you have reinforcing that can mitigate an overload and also maybe some compressive load... also with pavements, depending on the application, you could be looking at 8 or 9 as the factor.

Dik

 

[Ron]

Agree with dik. Axial compression of the slender wall likely plays into that as well.

 

[FlashSet]

I have also had this question in the past.

I had thought about conservatism, but there are phi factors present for this. The code also requires plain concrete to be in overall compression.

To add more to this, if I pour my little beam on the job site and test it in three or four point bending I will be testing a plain concrete beam, regardless if my pour is actually reinforced. Is the resulting modulus of rupture from the lab analogous to ACI 14.8 or ACI chapter 22?

 

[Lion06]

I don't think that Chapter 22 states that the modulus of rupture is 5*sq rt f'c. What it says is that the nominal moment capacity is S*(5*sq rt f'c). You can make the case that the implication is that the modulus of rupture used is 5*sq rt f'c, but I'm not sure I see it that way. It's likely just additional conservatism built into the equations.

 

[Gumpmaster]

The 7.5*sq rt f'c is for deflections only. It isn't a strength calculation.

MacGregor & Wight note that the mean modulus of rupture is 8.3*sq rt f'c, so the 7.5 makes sense for deflections. 5*sq rt f'c is closer to the lower bound value, which would make sense for strength calculations. MacGregor & Wight have a 3 page explanation of it if you're interested.