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Net average moment of inertia in CMU
2

Net average moment of inertia in CMU

Net average moment of inertia in CMU

(OP)
Hello,

Can someone write a formula for the net average moment of inertia (Iavg) of the CMU wall ? I have been trying to come up with a formula for Iavg but it is always way off the NCMA charts.

For example:
12" wall (11.625")
face shell = 1.25"
webs = 1.125"
spacing = 24"

Iavg = 1165.4 in2/ft

I appreciate your help. Thanks.

RE: Net average moment of inertia in CMU

Use Steiner´s rule. If the section is unsymmetric (centroid not obvious), use balance of first moment of area (centroid = sum(A*d) / sum(A) for all discrete parts at distance "d" from a chosen reference line) to calculate the centroid first and then apply Steiner´s rule.

RE: Net average moment of inertia in CMU

For hollow walls I use the minimum face shell values and ignore the web contribution.

Check the spec, but I think the minimum face shell thickness for 12” CMU is 1 3/8”. That’s the value I’d use. (1 1/4 for 8”CMU.)

RE: Net average moment of inertia in CMU

I agree with JLNJ to ignore the web contribution.

BA

RE: Net average moment of inertia in CMU

Do you know how to determine the net moment of inertia (not the average moment of inertia)? The average moment of inertia is done the exact same way, but you get to also include the contribution from every web even if they do not have mortar at the joints. This is because the average properties consider the horizontal plane at every height and not just at the height of the mortar joint where the wall is weakest.

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RE: Net average moment of inertia in CMU

(OP)
ProgrammingPE,

Yes, I got Inet formula and it works just fine. How do I include the contribution of other horizontal planes ?

centondollar,

I see the section as symmetric - like an I-beam with two equal size flanges and a web (filled core and two cmu webs). That way the centroid is obvious and is located at center (1/2 thicknes). Please correct me if I'm wrong.

JLNJ,

The specs I am looking at (NCMA) face shell thickness is 1.25". What formula do you use if you ignore the web contribution?

Thanks,

RE: Net average moment of inertia in CMU

I haven't looked at ASTM C90 in a while, sad to say. As of 2011, minimum face shells for CMU greater than 8" are 1 1/4". That's the number I'd use.

As far as the "formula" goes, simply compute the I per foot using two 1 1/4" face shells 12" long, 11 5/8" out-to-out. As I said before, I'd ignore any web contribution. The webs don't contribute too much and I'm not sure you get full bedding on the webs anyway.

RE: Net average moment of inertia in CMU

2
I'll work through your example.



For a 12" masonry wall grouted 24" o.c. the grouting pattern repeats every 3 blocks. Over that width you get to consider 2 grouted cells and 9 webs for the average properties (net properties only consider the 4 webs that are adjacent to the grout).

Cell Depth = Block Depth - 2*Face Shell Thickness = 11.625in - 2*1.25in = 9.125in
Cell Width = (Block Width - 3*Web Thickness) / 2 cells = (15.625in - 3*1.125in) / 2 = 6.125in

Iavg, face shell = 1/12 * 48in * ((11.625in)³ - (9.125in)³) = 3244.8in^4
Iavg, webs = 1/12 * (9*1.125in) * (9.125in)³ = 641.1in^4
Iavg, grout = 1/12 * (2*6.125in) * (9.125in)³ = 775.6^4

Iavg = (3244.8in^4 + 641.1in^4 + 775.6in^4) / (48in / 12in/ft) = 1165.4 in^4/ft

(Also, the current ASTM C90 uses a minimum web thickness of 3/4". It changed at some point, so if you are using a larger value then you need to confirm the actual block dimensions being used.)

RE: Net average moment of inertia in CMU

(OP)
Thank you for your help, it makes sense now.

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