Hand Calculation for 6" CMU Block With 1" Face Shell 1

[TylerM94]

I am trying to prepare for the PE exam and wanted to do a hand calc of a typical 6" hollow CMU block with 1" face shell.I am assuming that the CMU block is hollow and does not have any grout. NCMA TEK 14-1B provides Inet of 130.3 in^4/ft for a hollow 6 inch CMU block wall with 1 inch face shell. I calculated the net moment of inertia as 514.87 in^4 so clearly I am way off. Could someone help me figure out where I am going wrong? My approach was to subtract the moment of inertia of the two hollow cores from the gross section and I tried to include a PDF of my work below.

 

[dold]

Set the x-axis at the centerline of the shape. Then just subtract: Ix1 - Ix2 - Ix3 = 231.74 - 25.06 - 25.06 = 181.62 in^4 (per block).

Then consider that NCMA gives the value in in^4/ft. So: 181.62 in^4 * (12" / 15.625") = 139.3 in^4/ft (not sure if you had a typo on the 130.3in^4/ft? or if they used some other trickery... or if I did this all wrong) Still pretty close.

 

[hokie66]

dold has it right. Nothing complicated about it, as the OP seemed to think.

 

[TylerM94]

Perfect, thank you so much! Selecting the neutral axis to be the center of the shape makes everything so much easier.

not sure if you had a typo

The reference lists 139.3 for I[sub]average[/sub] and then they have a separate column for what they call I[sub]net[/sub] listed as 130.3 in[sup]4[/sup] per foot... I am not sure what the trickery is to get the I[sub]net[/sub] but I am super happy I was able to get the I [sub]average[/sub] with your help. Thank you!

 

[Celt83]

I[sub]net[/sub] comes from consideration of face shell only bedding and is computed neglecting the webs. You’ll notice in the NCMA table the line with full bed I[sub]net[/sub] = I[sub]avg[/sub]

 

[TylerM94]

Thanks Celt83, I will re-run the calc to exclude the web area and see if I get the Inet to match the table value.