# Concrete Masonry Units Problems



## passpepasspe (Oct 5, 2010)

I was wondering if someone can help me with where I can find problems asking for the number of concrete masonry units (CMU) needed given dimensions, etc?


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## pmblair (Oct 5, 2010)

Typical CMU's are 8" high and that is standard for 8" and 12" CMU.

Number of courses is equal to H(ft)x12/8". Thats it


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## passpepasspe (Oct 5, 2010)

pmblair said:


> Typical CMU's are 8" high and that is standard for 8" and 12" CMU.
> Number of courses is equal to H(ft)x12/8". Thats it



http://www.tpub.com/content/construction/1...s/14045_145.htm

where can i find more examples like the one here, i googled it and found this.


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## maximus808 (Oct 5, 2010)

Great example! Does the CERM have anything about this topic?


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## passpepasspe (Oct 6, 2010)

maximus808 said:


> Great example! Does the CERM have anything about this topic?


That's what I am trying to find out, whether the CERM or the practice problem book have these type of problems. I am not sure in a typical problem what would be provided, what assumptions would need to be made and how to arrive at an answer.


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## pmblair (Oct 6, 2010)

In there example its ..............Given: A building 20-feet long by 8-feet wide by8-feet high

8' high x12"/8" per block = 12 blocks per foot x the perimeter =20+20+8+8=56'x12"/16" per block= 42

So.... 12(42)=504 blocks of 8"x16" cmu.

I think you guys are over thinking this. There is no need for equations, this is a standard geometry question. Typical block are 8"x8"x16".


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## dastuff (Oct 6, 2010)

And no, i never was able to find this type of problem in the CERM.

Another common one is:

A gallon of paint covers 400 sq ft. For a 20'x40'x12' room how many gallons are needed for 2 coats.

They're just geometry/unit problems.


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## passpepasspe (Oct 7, 2010)

dastuff said:


> And no, i never was able to find this type of problem in the CERM.
> Another common one is:
> 
> A gallon of paint covers 400 sq ft. For a 20'x40'x12' room how many gallons are needed for 2 coats.
> ...


Thanks for the help guys.. I thought about this a lot..

It seems like CERM chapter 67 had some key things about masonry. I am an environmental engineering with 0 background in most of the topics for Civil PE. so i apologize if this is too easy for u guys. lol. but its totally not for me.

ok, so for the building of 20 ft long by 8 ft wide and 8 ft high, how many CMU?

So if you do the permiter for the building, you get , 20+20+8+8 = 56 ft, multiply by the height to get 56X 8 ft =448 sf.

now, the CMUs are given in W X LX H, so 8X 16X 8 for one CMU

taking the face dimension to be 8 X 16 = 128/144 = .88 sf per CMU. So dividing 448/.88 = 504 CMU

OK. As easy as this problem might be to you guys, have i assumed correctly, and solved correctly?


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## pmblair (Oct 8, 2010)

Yah you got it. If it were 12" CMU you would use 12x16/144.

I didnt mean to insult as to the easiness of the problem.

just meant to infer that is was a geometry problem, and no real equations are needed.

I feel you on the test. I have a lot of experience in structures, but the rest is Greek to me.


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## humner (Oct 8, 2010)

am I missing something here? I figure that you would need the same amount of blocks regardless if they were 4", 8" or 12", that number would be the thickness of the block and would not be considered for the count.


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## maximus808 (Oct 8, 2010)

When dimensions of a cmu is given as 8,8,16 this means height, width, and length? Or width, height, length?


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## pmblair (Oct 9, 2010)

Hummer your absolutly right. Sorry about the confusion.

Maximums808 its Width Height length. Typical you only call out the first dimension as most blocks are 8" high by 16" long.

So someone says 8" or 12" block they are talking about the width.


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## JKwon (Oct 11, 2010)

Should we not consider thickness of Mortar joints?

What about breakage and waste factor?


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## maximus808 (Oct 11, 2010)

I believe we can ignore it, can someone confirm?


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## dastuff (Oct 11, 2010)

maximus808 said:


> I believe we can ignore it, can someone confirm?


I'll confirm.. Mainly because I think that would make the problem too difficult for a morning type question.

Also the class I took had one of these "red brick" type questions as an example and he didn't include mortar in his calcs.


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## pmblair (Oct 11, 2010)

I have designed over 100 masonry buildings, and only a handful of times has mortar joints come into the equation.... and it wasn't to calculate heights or count of block.


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## maximus808 (Oct 11, 2010)

Thanks! I knew it wasn't too complicated  But when given the dimensions of a block X, X, X, which represents the length, width, and height. I know the longest is typically the length, but depth is usually not a factor when counting the amount of blocks to build a wall correct? It is based on the surface area divided by the area of one block (length x height) not depth or thickness of the block? I just don't know what is typical when they describe the 3 dimensions of the block, thanks.


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## pmblair (Oct 11, 2010)

Typical CMU will be 8" wide 8" tall 16" long.

The 16" is that you can have a running bond instead of a stack bond.----- So the seams dont line up, when i say seams i mean the end of one block isnt directly above the end of the ones above and below, thats a stacked bond, and it is weaker for obvious reasons such as only the mortar resist the force at that point instead of a grout filled cell.

So the 16" is you length 99.99% of the time, its 2 cells.

The height is standard as well its 8" about 99% of the time, this has to do with a couple different reasons, weight of the cmu/ leveling/ and others.

The depth or width is usually 8" most of the time like 75%. Even here in Florida with hurricanes dominating design of the components and cladding system, which masonry usually is, 8" is typical. For shelters and such some times you use 12" block so you have a greater effective depth. Similar to concrete beam design your moment arm is larger to your steel.

The question really isnt about masonry so much as a geometry question. As passpepasspe put it in his solution, which is really good way to look at it.

"ok, so for the building of 20 ft long by 8 ft wide and 8 ft high, how many CMU?

So if you do the permiter for the building, you get , 20+20+8+8 = 56 ft, multiply by the height to get 56X 8 ft =448 sf.

now, the CMUs are given in W X LX H, so 8X 16X 8 for one CMU

taking the face dimension to be 8 X 16 = 128/144 = .88 sf per CMU. So dividing 448/.88 = 504 CMU"


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