PLAIN AND CIVIL: EXAMPLE 3.1.3 ESTIMATING CHB OF AN IRREGULAR WALL

Determine the number of 6" thk. CHB required to construct the building firewall.

Fundamental method:

Area A:

$Vert_{layers}=\frac{4.00m}{0.40m}=10layers$

$Horiz_{layers}=\frac{3.40m}{0.20m}=17layers$

   $CHB_{area}=10\left ( 17 \right )=170pcs$

Area B: For irregular shaped wall such as Area B, estimating area will be by taking the average of the sloping height. For this case 1.00 is divided into two and then added to the height of 5.00.

$Vert_{layers}=\frac{7.00m}{0.40m}=17.50layers$

$Horiz_{layers}=\frac{5.00m+\frac{1.00m}{2}+0.40m}{0.20}=29.50layers$

$CHB=17.50\left ( 29.50 \right )=516.25pcs$

Area C:

   $Vert_{layers}=\frac{5.00m}{0.40m}=12.50layers$

$Horiz_{layers}=\frac{1.50m+0.40m}{0.20}=9.50layers$

$CHB=12.50\left ( 9.50 \right )=118.75pcs$

$CHB^{total}=170+516.25+118.75pcs={\color{Red} \mathbf{805pcs}}$

Area method:

$Area_{A}=4.00m\left ( 3.40m \right )=13.60m^{2}$

$Area_{B}=7.00m\left ( 5.90m \right )=41.30m^{2}$

   $Area_{C}=5.00m\left ( 1.90m \right )=9.50m^{2}$

$Area_{total}=13.60m^{2}+41.30m^{2}+9.50m^{2}=64.40m^{2}$

$CHB_{total}=64.40m^{2}\left ( 12.50\frac{pcs}{m^{2}} \right )={\color{Red} \mathbf{805pcs}}$

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Monday, 22 July 2019

EXAMPLE 3.1.3 ESTIMATING CHB OF AN IRREGULAR WALL

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