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Table 4-3.Average Concrete Masonry Units and Mortar Per 100 Sq Ft of wall

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Table 4-3.—Average Concrete Masonry Units and Mortar Per 100 Sq Ft of wall required   for   every   100   square   feet   of   a   concrete length by the number of block in height, which, in this masonry wall. NOTE:  Actual wall length is measured from the outside  edge  to  outside  edge  of  units  and  equals  the nominal length minus 3/8” (one mortar joint). NOTE: For concrete masonry units 7 5/8” and 3 5/8” in height laid with 3/8” mortar joints. Height is measured from center to center of mortar joints. NOTE:  Mortar  is  based  on  3/8”  joint  with  a face-shell mortar bed and 10% allowance for waste. As a Builder, you might find yourself in the field without  the  tables  handy.  To  solve  that  problem,  we will   cover   two   methods   for   estimating   concrete masonry  units  (CMUs)  without  the  tables.  The  first method is CHASING THE BOND, which is using the 3/4 rule and the 3/2 rule. Remember when estimating, example, is 75 CMUs in length times 15 courses high, which equals 1,125 (FB). Let’s take another example: Given: A building 20-feet long by 8-feet wide by 8-feet high .75 x 20 x 2 (sides)  = 30 (8" x 8“ x 16" block) .75 x 8 x 2 (sides)  = 12 (8" x 8“ x 16" block) Or  you  can  find  the  total  linear  feet  (LF)  of  the building and multiply by .75. 20 x 2 (sides) + 8 x 2 (sides) = 56 LF 56 x .75 =42  FB 1.5 x 8 = 12 courses  high 42 FB x 12 courses = 504 total FB always  use  OUTSIDE  measurements  to  calculate  the The  second  method  of  estimating  CMU  is  the number of blocks required per course. In most Seabee SQUARE FOOT METHOD. It is usually the quickest construction,  8-inch  by  8-inch  by  16-inch  block  is used. Using the 3/4 rule (three full block per 4 feet in length) or .75, multiply the length of the wall by .75. For  example,  a  retaining  wall  that  is  10  feet  high  by 100 feet in length (1,000 sf) will require 75 block for the first course. Length of course in feet  x rule 3/4 = number  of  CMU per course Using the 3/2 rule (three full block per 2 feet in  height), multiply the height of the wall by 1.5. For example, the height of the retaining wall is 10 feet. Multiply (10) by the  rule  3/2  (1.5)  which  will  equal  15  block  high (courses  high).  See  the  following  formula: Height of wall in feet x  rule 3/2 = courses high Then, to find the total number of full block (FB) in the retaining wall, multiply the number of block in and   simplest   method   but   NOT   the   most   accurate. However,  you,  the  estimator,  will  use  this  method quite  frequently.  Remember  in  the  first  example,  the retaining wall was 10 feet high and 100 feet in length. All you do is multiply L x H = SF in this example; the answer is 1,000 square feet (SF). To find the number of 8" x 8" x 16" block required, you must determine the  square  footage  of  one  CMU  which  is  .89  SF  per block.  Next,  you  divide  1,000  SF  by  .89  SF/CMU which equals 1,124 FB. You calculated the block for 1,000 SF and the difference was (1) less block figuring by the SF method. See the following formula: Total  SF  divided  by  SF/CMU  =  total  number  of CMU Now calculate the 20 ft x 20 ft x 8 ft building: 20 x 8 = 160 SF x 2 (sides) = 320 SF 4-3



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