Masonry Wall - Question on radius of gyration value / section properties

[Target Builder]

I'm working through the NCEES SE sample questions book and one of the questions (Q133 in the AM section) is on masonry walls, and I can't work out how they get their values in the solutions.

Wall = 8" thick, reinforced at 48" O.C.

The NCEES solutions use the following: Area = 161 in^2 (manually calculated), r = 2.66" (no calculation).

How is their r value calculated and is their method for calculating the area correct?

Going by MDG (4) Appendix A Table 8 the properties should be: Area = 57 in^2/ft, Ixx = 377 in^4/ft so a 48" wide section has the following: Area = 228 in^2, r = 2.57"

The NCEES uses a manually calculated area, using the same section to manually calculate the section properties gives Ixx = 1324 in^4, suggesting a radius of gyration of 2.86"

The errata on their website doesn't show any issues for this question.

Which is the correct way to calculate the required wall properties?

Any help on this would be greatly appreciated - I rarely work in masonry.

 

[ADB]

ncees uses r = sqrt(I/a)

 

[Target Builder]

Where do they get their I value from?

In their answer they use:

r = 2.66 in

A = 166 in^2

Which implies I = 1139 in^4

But the section dimensions they use give I = 1324 in^4 (48*7.625^3 - 40*5.125^ 3)/12

And neither value matches up with the properties in the MDG (4) - did the MDG values change?

 

[ItzmeJ0e]

The NCMA publishes a document titled "Section Properties of Concrete Masonry Walls" TEK 14-1B. You can find it online for free with a quick google search. The document provides net and average cross-sectional properties for grouted and partially grouted CMU walls. I believe that this is where NCEES is getting the value of 2.66" for r.

I would recommend printing out this document are bringing it with you to the exam in a binder (unless you're taking the exam in Illinois in which case you're out of luck). That way, you won't need to calculate any masonry section properties by hand during the exam. Make sure you understand how to read the tables--use the average cross-section properties for deflection and stability calculations and use the net cross-section properties for determining stress and strain. Note that the NCMA table gives a net area for the wall in this question of 40.7 in^2/ft (very close to what the NCEES solution calculates by hand).

On a related note, I think NCEES may be incorrect to include the contribution from the steel reinforcing when calculating the axial capacity of the wall. Per ACI 530 Sec. 2.3.2.2.1, "The compressive resistance of steel reinforcement shall be neglected unless lateral reinforcement is provided in compliance with the requirements of Section 1.14.1.3." The question statement does not mention lateral ties, and typically they are not provided in a masonry wall, so I would have ignored the contribution from the steel.

Hope this helps!

 

[McEngr]

ItzmeJ0e is right. It is impossible to get two curtains of reinforcement in an 8" wall. I think it would require at least a 10" wall to have a second curtain to actually tie to. I think the compressive equation becomes more simple for 8" walls without ties as it is.

 

[steelhead]

I agree, NCEES is incorrect on the compression steel in the wall. I solved the problem incorrectly and got it right (just like NCEES did it) but after reviewing my Sue Frey class notes I see it's supposed to be neglected.

 

[carls88]

Hey ItzmeJ0e - what did you mean when you wrote "unless you're taking the exam in Illinois in which case you're out of luck"?

I'm sitting for the S.E. in October and based on the NCEES candidate agreement it seems fine as long as it’s in a binder .....

 

[Vinh Hien Nguyen]

I do the NCEES problem in 2017, which is 4 year later since this topic had started in 2013, I struggle which An and r of masonry wall, the value is same , A=161in2 and r=2.66, nothing has been change since 2013. i try to use TEK 14-B, table 3a and 3b, but not get the correct number. They use the masonry column concept to apply for masonry wall. If anyone can find the correct solution, please post

 

[StandardPractice]

NCEES is using r(avg) which is based on I(avg) and A(avg). They don't do a very good job explaining this at all. A(avg) is the average between A and Anet and I is the same. I plan to use average values on the exam or hopefully they just provide the r, I, A they want used in masonry problems. You can find a table of the average values online pretty easily and they are also located in the "Reinforced Masonry Engineering Handbook) 7th edition table GN-8

 

[StructuralVFL]

Using the average values provided in TEK 14-1B, Table 3a for 8" walls grouted at 48" o.c., I = 351.7 in^3 and A = 49.6 in^2 which gives an r = 2.66 in.

 

[TehMightyEngineer]

Agreed with the above. They use r_avg = 2.66 in to calculate h/r = 54.1. Then An = 40.7 in^2/ft x 4 ft = 162.8 in^2. The solution has An = 161 in^2 so slightly different but close enough. Then they calculate Pa per the MSJC.

Everyone definitely needs a copy of TEK 14-01B in their exam notes.

 

[ZEZO4]

I found errors not only in NCEES Practice, I found even an errors (most of them typo errors) in the AISC Seismic Manual and ASCE7, so Imagine!

 

[TehMightyEngineer]

I found errors not only in NCEES Practice, I found even an errors (most of them typo errors) in the AISC Seismic Manual and ASCE7, so Imagine!

Always check your errata, many problems have (hopefully) already been found and reported.

 

[ZEZO4]

Always check your errata, many problems have (hopefully) already been found and reported.

That is correct, Thanks.

 

[polarnaut]

I think that the answers here still aren't very clear.  Per ACI 530, "radius of gyration shall be calculated using the average net cross-sectional area of the member considered."  Think of your assemblage as an I-beam where the face shells are your flanges.  The web of the I-beam would have an equivalent width equal to the total of your CMU block web thickness divided by width of the CMU block plus the width of one grouted cell divided by spacing of the grouted cells.  So symbolically, it would look something like this (I use a 12" standard block with 24" grout spacing in the horizontal plane as an example):

bavg = 3*bw/bcmu,nom +bcell/s = 5.59 in/ft = net average thickness of equivalent web

Aavg = 2*tfs + bavg*tcell = 81.0 in^2/ft = net average cross-sectional area

Iavg = 2*((tcmu^3-tcell^3)/12) + bavg*tcell^3/12 = 1165.4 in^4/ft = net average moment of inertia

r = sqrt(Iavg/Aavg) = 3.79 in = net average radius of gyration

Where:

tfs = 1.25" = thickness of CMU face shells

bw = 1.125" = thickness of the CMU webs

bcmu,nom = 16" = nominal width of one block

tcmu = 11.625" = actual thickness of block

tcell = 9.125" = actual thickness of one cell (this would be the depth of the equivalent web)

bcell = 6.125" = actual width of one cell

s = 24" = grout spacing

 

[Phil Clark]

Some folks keep talking about ACI, but the original post was about masonry.  TMS should be used not ACI.  Having studied for these exams recently, I would agree that some of the solutions in the NCEES practice exam are not very clear.  The PPI practice exam is much better in terms of showing where there solutions came from for all steps of the problems.

 

[StructEngBrah]

Some folks keep talking about ACI, but the original post was about masonry.  TMS should be used not ACI.  Having studied for these exams recently, I would agree that some of the solutions in the NCEES practice exam are not very clear.  The PPI practice exam is much better in terms of showing where there solutions came from for all steps of the problems.

 I was under the impression that ACI 530 is the same thing TMS 402.

 

[EBAT75]

Old post. New member. So let me see.

I don't see any issue with the NCEES solution in this case. Areas - Net is used for strength calcs, Average is used for stiffness, deflection.

Net area is used in the first portion of the allowable load equation - axial compressive strength only.  Average area is the basis for the r value because stiffness, deflection are based on it, not on net values. The second component is stiffness related, so r from average area is the one to use.

Yes, the code says contribution of steel can be neglected. This is based on larger column and pilaster loads. Walls are treated as a series of columns stacked side by side. The NCEES solution treats it like a reinforced column. Only thing is columns must have lateral ties. The code says if the R/F does not have lateral ties as in columns, the contribution of steel can be neglected. The difference in this case is only about 10%.

There is another aspect that may control in some cases. That is the 1/4 of Euler buckling load. That seems to be not checked for. In general terms, masonry columns and walls have factor of safety of 4 built in. So in real terms omitting the steel would not make a big dent.