Modulus of Elasticity for CMU

[Mike1144]

How do you find the modulus of elasticity of CMU?

I have a couple possibilities i'd like some input on.

1) UBC 2106.2.12 says its E = 750 * f`m

2) NCEES sample exam #539 says its E = 2.2 - (100/500)*0.6 for f`m = 1500psi. I have no idea what those values are, but if you recognize the format of the formula that could be a clue.

 

[scottiesei]

UBC is not a reference on the exam list. You should be in ACI530.

Section 1.8.2.2.1

Em(clay)=700*f'm

EM(concrete)=900*f'm

The NCEES sample exam that I have uses 900*f'm? Maybe you have an old version?

 

[Mike1144]

UBC is not a reference on the exam list. You should be in ACI530.Section 1.8.2.2.1

Em(clay)=700*f'm

EM(concrete)=900*f'm

The NCEES sample exam that I have uses 900*f'm? Maybe you have an old version?

Thanks for the info.

The exact text of the solution says :

"f`m = 1,500psi Net area compressive strength = 1,900 (ACI Table 2)

Em = 2.2 - (100/500)*(0.6) = 2.08 * 10 ^6 "

You can see how that seems a little cryptic. Is the "ACI Table 2" part of ACI530?

 

[scottiesei]

Yeah, you must have an old version. Mine is 04', those equations that you are writing do not match what I have in my sample exam nor what the code has. With that being said you may want to pick up ACI530, it is small and will take about a week to get familar with.

 

[Mike1144]

OK, for what its worth I now know what they were doing.

I have ACI530-95, they went to Table 2 of the "Specification For Masonry Structures", and got the 1900psi based on f`m = 1500 and Type M or S mortar. Then went to Table 5.5.2.3 and interpolated to find E. Of course it still makes little sense why they simplified the interpolation equation, instead of just showing the values they used.