EXAMPLE 3.2.2. ESTIMATING MORTAR MATERIALS IN A CONCRETE STRUCTURE

From the previous example 3.1.2, estimate the materials for the mortar using class B mixture.

Volume method:

Block Laying: The small structure's height is 3m exposed and 0.90m embedded, making a total of 3.90m. This is to be divided by the height of each CHB unit which is 0.20m.

                          The total perimeter of the wall is also necessary in order to compute the volume properly. From example 3.1.2., the total perimeter of the wall is 27.60m.

                      
                 
                      

CLASS	PROPORTION	CEMENT	CEMENT	SAND
		40kg	50kg
A	1:2	18.0	14.5	1.0
B	1:3	12.0	9.5	1.0
C	1:4	9.0	7.0	1.0
D	1:5	7.5	6.0	1.0

                      

                      

Cell Filler. From the previous example, it was computed that the structure would require 1216pcs of 6" CHB.

                     

                     

                     


Take the summation of materials' quantities:







Area method: The table for Area method accounts for both the block laying and cell filler materials. This method would only require the area of the wall.

CHB SIZE	CHB NUMBER	A	B	C	D	SAND
		(40kg)	(40kg)	(40kg)	(40kg)
cm	per sqm	bags	bags	bags	bags	cu.m.
10 x 20 x 40	12.5	0.792	0.522	0.394	0.328	0.0435
15 x 20 x 40	12.5	1.526	1.018	0.763	0.633	0.0844
20 x 20 x 40	12.5	2.260	1.500	1.125	0.938	0.1250

             






Hundred Block Method:

CHB SIZE	A	B	C	D	SAND
	(40kg)	(40kg)	(40kg)	(40kg)
cm	bags	bags	bags	bags	cu.m.
10 x 20 x 40	6.336	4.176	3.152	2.624	0.3480
15 x 20 x 40	12.150	8.104	6.072	5.064	0.6750
20 x 20 x 40	18.072	12.000	9.000	7.504	1.0000

                

Back to 3.0. Estimating Masonry