PLAIN AND CIVIL: EXAMPLE 2.7.4. ESTIMATING AN IRREGULAR PAVEMENT

Saturday 20 July 2019

Using class B, estimate the concrete components for the 4" thick park pavement as shown.

Solve for the total area:

   $Area_{rectangle}=8.0m\left ( 20.0m \right )=160m^{2}$

   $Area_{circle}=\frac{\pi \left ( 15.0m \right )^{2}}{4}=176.7146m^{2}$

$Area_{trapezoid}=\frac{1}{2}\left ( 4.0m \right )\left ( 5.00m+15.00m \right )=40m^{2}$

$Area_{rectangle}=9.00m\left ( 15.00m \right )=135m^{2}$

   $Area_{total}=511.7146m^{2}$

From the table:

SLAB THK	A	B	C	A	B	C	SAND	GRAVEL
	(40KG)	(40KG)	(40KG)	(50KG)	(50KG)	(50KG)
cm	bags	bags	bags	bags	bags	bags	cu.m.	cu.m.

5.0	0.450	0.375	0.300	0.350	0.300	0.350	0.0250	0.050
7.5	0.675	0.563	0.450	0.525	0.450	0.375	0.0375	0.075
10.0	0.900	0.750	0.600	0.700	0.600	0.500	0.0500	0.100
12.5	1.125	0.938	0.750	0.875	0.750	0.625	0.0630	0.125
15.0	1.350	1.125	0.900	1.050	0.900	0.750	0.0750	0.150
17.5	1.575	1.313	1.050	1.225	1.050	0.875	0.0880	0.175
20.0	1.800	1.500	1.200	1.400	1.200	1.000	0.1000	0.200
22.5	2.030	1.688	1.350	1.575	1.350	1.125	0.1125	0.225
25.0	2.250	1.875	1.500	1.750	1.500	1.250	0.1250	0.250
27.5	2.475	2.063	1.650	1.925	1.650	1.375	0.1380	0.275
30.0	2.700	2.250	1.800	2.100	1.800	1.500	0.1500	0.300

Coefficients are as follows:

Cement (40kg) = 0.750
Sand = 0.050
Gravel = 0.100

$Cement=0.750\left (511.7146m^{2} \right )=383.7860\simeq {\color{Red} \mathbf{384bags}}$

$Sand=0.050\left (511.7146m^{2} \right )=25.5857\simeq {\color{Red} \mathbf{26m^{3}}}$

$Gravel=0.100\left (511.7146m^{2} \right )=51.1715\simeq {\color{Red} \mathbf{51.5m^{3}}}$

Back to 2.0 Estimating Concrete
2.7. Estimating Concrete Slabs by the Area Method

PLAIN AND CIVIL

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Saturday 20 July 2019

EXAMPLE 2.7.4. ESTIMATING AN IRREGULAR PAVEMENT

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