Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations SDETERS on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

Net average moment of inertia in CMU 2

Status
Not open for further replies.

Buleeek

Structural
Sep 5, 2017
98
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.
 
Replies continue below

Recommended for you

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.
 
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.)
 
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.

Structural Engineering Software: Structural Engineering Videos:
 
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,
 
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.
 
I'll work through your example.

Masonry_Wall_-_Grouted_24in_x6gyg1.png


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.)
 
Thank you for your help, it makes sense now.
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor