PLAIN AND CIVIL: EXAMPLE 3.2.1. ESTIMATING MORTAR MATERIALS IN EXAMPLE 3.1.1.

Determine the materials for mortar to construct a wall of height=3m by length= 4.8m. CHB thickness used is 4". Use class A mortar.

Block Laying. There is only one procedure in estimating the mortar for block laying.

   $Layers=\frac{3.0m}{0.20m}=15layers$

$Volume=15layers\left [ 0.012m\left ( 0.10m \right ) \left ( 4.80m \right )\right ]=0.0864m^{3}$

$Cement=0.0864m^{3}\left ( 18.0 \right )={\color{Red} \mathbf{1.5552bags\left ( 40kg \right )}}$

$Sand=0.0864m^{3}\left ( 1.0 \right )={\color{Red} \mathbf{0.0864m^{3}}}$

Cell Filler.

Volume method. Based on the previous example, the wall requires 180pcs 4" CHB.

   $Vol_{chb}=4 cells\left ( length_{cell} \right )\left ( width_{cell} \right )\left ( height_{cell} \right )$

   $Vol_{chb}=4 cells\left ( 0.075m \right )\left ( 0.05m \right )\left ( 0.20m \right )=0.003m^{3}$

   $Vol_{total}=180\left (0.003m^{3}\right )=0.54m^{3}$

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

$Cement=0.54m^{3}\left ( 18.0 \right )={\color{Red} \mathbf{9.72bags\left ( 40kg \right )}}$

$Sand=0.54m^{3}\left ( 1.0 \right )={\color{Red} \mathbf{0.54m^{3}}}$

$Cement_{total}=1.5552+9.72=11.2752\simeq {\color{Red} \mathbf{12 bags\left ( 40k \right )}}$

$Sand_{total}=0.0864+0.54=0.6264\simeq {\color{Red} \mathbf{1m^{3}}}$

Area Method.

$Area_{wall}=3.0m\left ( 4.80m \right )=14.40m^{2}$

CHB NUMBER	A	B	C	D	SAND
	(40kg)	(40kg)	(40kg)	(40kg)
per sqm	bags	bags	bags	bags	cu.m.

12.5	0.792	0.522	0.394	0.328	0.0435
12.5	1.526	1.018	0.763	0.633	0.0844
12.5	2.260	1.500	1.125	0.938	0.1250

$Cement=14.40m^{2}\left ( 0.792 \right )=11.4048\simeq {\color{Red} \mathbf{12 bags\left ( 40k \right )}}$

$Sand=14.40m^{2}\left ( 0.0435 \right )=0.6264\simeq {\color{Red} \mathbf{1m^{3}}}$

Hundred Block Method.

$Factor=\frac{180blocks}{100}=1.80$

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

$Cement=1.80\left ( 6.336 \right )=11.4048\simeq {\color{Red} \mathbf{12bags\left ( 40kg \right )}}$

$Sand=1.80\left ( 0.348 \right )=0.6264\simeq {\color{Red} \mathbf{1m^{3}}}$

Back to 3.0. Estimating Masonry

PLAIN AND CIVIL

Pages

Monday 22 July 2019

EXAMPLE 3.2.1. ESTIMATING MORTAR MATERIALS IN EXAMPLE 3.1.1.

No comments:

Post a Comment