CALCULATING
CONCRETE MATERIALS
Concrete materials are calculated based on the
working plans and specifications. By determining
the volume of the concrete needed in cubic
meters and the class of concrete, you can easily
calculate the number of cement bags, and the
amounts of sand and gravel required.
Example:
Supposing a secondary road is to be paved with 8” thick,
6 meter wide and 800 meters in length concrete.
Determine the number of cement, sand and gravel in
cubic meter using class A concrete.
Solution:
1. Determine the given figures: Thickness, Width and
Length
Thickness = 8” = 0.2 meters
Width = 6.0 meters
Length = 800 meters
(Note: if the given is not in meters, you must convert it first.)
Eg. 1000mm = 1m
100cm = 1m
0.001km = 1m
Solution:
2. Compute the volume: V=LxWxT
V = 0.2 m x 6.0 m x 800 m
V = 960 m³
3. Determine the Class of Concrete to be used.
(for this figure Class A is the example)
Solution:
4. Compute Cement, Sand and Gravel Using Class A
Cement = V x 9cu.m
= 960cu.m x 9bags
= 8640 bags of cement
Sand = V x 0.5 cu.m
= 960cu.m x 0.5 cu.m
= 480 cu.m of sand
Gravel = V x 1 cu.m
= 960cu.m x 1 cu.m
= 960 cu.m of Gravel
Therefore:
Cement = 8640 bags of cement
Sand = 480 cu.m of sand
Gravel = 960 cu.m of Gravel
Example No. 2
A concrete path walk from the exit door to the gate is to be constructed or paved. The
path walk should be 6” thick, 48” wide and 3 meters long. Solve for the quantity of
cement in bags, sand and gravel in cubic meters using class B concrete.
Solution :
Given: T = 6” = 0.15 m
W = 48” = 1.20 m
L = 3 meters
Volume = 0.15 m x 1.20 m x 3 m
= 0.54 m³
Using Class B cement has 7.5 bags, sand 0.5, gravel of 1
C = 0.54 m³ X 7.5 bags / m³
= 4.05 bags; say 4 bags or 5 bags
S = 0.54 m³ x 0.5
= 0.27 m³ or 0.5 m³
G = 0.54 x 1
= 0.54 m³ or 1 m³
Since bags of sand and gravel are already available at the hardware or
construction supplies, 0.27 m³ of sand can be converted into bags ;
No. of bag of sand = 0.27 m³
0.027 m³ / bags – volume of 1 bag of cement
= 10 bags
Just make the quantity or number of sand double to get the quantity of gravel.
No. of gravels on bags = 0.54
0.027
= 20 bags
Example No. 3 :
A two story residential building has 6 concrete columns having 250 mm x 250 mm cross
– section and height of 6.8 meters with corresponding concrete footings of 8” thick x 1.0
meter wider x 1.0 m on length. Find / solve for the quantity of cement in bags, sand and
gravel in cubic meter using class A – A concrete for the concrete footing and column.
Solution :
Given : 6 columns and 6 footings
Footings x 6
Given T = 8” = 0.2m
W = 1.0m
L = 1.0m
V = 0.2m x 1m x 1m
= 0.2 m³ x 6 footings
= 1.2 m³
Using class AA
C = 12, Sand = 0.5, Gravel = 1
Cement = 1.2 m³ x 12 = 14.4 bags or 15 bags
Sand = 1.2 x 0.5 = 0.6m³ or 1m³
Gravel = 1.2 x 1 = 1.2 or 2m³
Or to be exact – Sand bags = 0.6
= 23 bags
0.027
Gravel bags = 1.2
= 45 bags or 46 bags
0.027
Column x 6
Given :
TxW = 250mm x 250m
= 0.25m x 0.25m
Height = 6.80m
Volume = 0.25m x 0.25m x 6.80
= 0.425 m³ x 6 columns
= 2.55 m³
Using class AA C has 12 bags, S = 0.5, G = 1
Cement = V x 12 bags / m³
= 2.55 m³ x 12 bags / m³
= 31 bags
Sand = 2.55 m³ x 0.5
= 1.3 m³ say 2m³
Gravel = 2.55 m³ x 1
= 2.55 or 2.6 say 3m³ or 4³
Example No.4
A lot area of 6 meters x 12 meters needs to be secured by a 6” thick 60” high concrete perimeter
fence from footings. The concrete fence has a continuous footing of 6” thick and 18” wide. Compute
the required quantity of cement in bags, sand and gravel in cubic meters using class A concrete.
Solution No. 4
Concrete Perimeter Fence
Given :
Area = 6m x 12m
Height = 60” = 1.5m
Thickness = 6” = 0.15m
Perimeter = (2 x 6) + (2 x 12)
= 12m + 24m
= 36m
Volume = 0.15m x 1.5m x 36m
= 8.1 m³
Using class A = C = 9 ; S = 0.5, G = 1
Cement = 8.1m³ x 9
= 72.9 bags or 73 bags
Sand = 8.1m³ x 0.5
= 4.05m³ or 5m³
Gravel = 8.1m³ x 1
= 8.1m³ or 9m³ or 10m³
Continuous Concrete Wall Footing
Given :
T x W = 6” x 18”
= 0.15m x 0.45m
Area = (6m + 0.3m) + (12m + .3m)
= 6.3m x 12.3m
Perimeter = 2 (6.3m) + 2 (12.3m)
= 12.6m x 24.6m
= 37.2m
Volume = 0.15 x 0.45m x 37.2m
= 2511m³
Using class A = C = 9 ; S = 0.5 ; G = 1
Cement = 2.511m³ x 9
= 22.6 bags or say 23 bags
Sand = 2.511m³ x 0.5
= 1.5m³ or say 2m³
Gravel = 2.511m³ x 1
= 2.511 or say 3m³ or 4m³
ACTIVITY :
Problem 1:
A rectangular parking lot is to be paved with concrete that is 0.15 meters thick, 10
meters wide, and 50 meters long. Using Class A concrete (requiring 9 bags of
cement, 0.5 m³ of sand, and 1 m³ of gravel per cubic meter), calculate:
1.The volume of the parking lot in cubic meters.
2.The number of bags of cement required.
3.The quantities of sand and gravel needed in cubic meters.
Problem 2:
A concrete driveway is to be constructed with a thickness of 0.25 meters, a width of 4
meters, and a length of 100 meters. Using Class B concrete (requiring 7.5 bags of
cement, 0.50 m³ of sand, and 1 m³ of gravel per cubic meter), determine:
1.The total volume of the driveway in cubic meters.
2.The number of bags of cement needed.
3.The quantities of sand and gravel required in cubic meters.
ACTIVITY :
Problem 3:
A concrete slab for a building foundation is 0.3 meters thick, 12 meters wide, and 30
meters long. Using Class C concrete (requiring 6 bags of cement, 0.5 m³ of sand,
and 1 m³ of gravel per cubic meter), compute:
1.The volume of the concrete slab in cubic meters.
2.The number of cement bags required.
3.The quantities of sand and gravel needed in cubic meters.
Calculating Masonry Materials
(Concrete Hollow Block)
Laying and Plastering
Masonry materials for CHB laying including plastering are
the concrete hollow blocks, CHB mortar joints, CHB cell
mortar or filler, mortar plaster and the concrete materials.
For class A mortar, it requires 1 part of cement plus 2 parts of sand
and for every cubic meter quantity of mortar, it requires 18 bags of
cement and 1 cubic meter of sand.
Example 1 :
Supposing a CHB perimeter fence is needed to secure a lot area of 8 meter x 10 meters. The height of
the CHB fence is 1.5 meters from footing with the use of 6” thick CHB. If the CHB continuous footing is
150mm thick and 450mm wide solve for the quantity of masonry materials using class A.
CHB Footing requires concrete materials, so used the concrete proportion.
Given : T x W = 150mm x 450mm
= 0.15m x 0.45m
Perimeter = (8m + 0.30m) 2 + (10m + 0.30m) 2
= 16.6m + 20.6m
= 37.2 meters
Volume = 0.15m x 0.45m x 37.2m
= 2.511m³
Using Class A concrete cement in 9 bags ; sand = 0.5 ; gravel = 1
Cement = 2.511m³ x 9
= 23 bags
Sand = 2.511 x 0.5
= 1.3m³ say 1.5m³ or 2m³
Gravel = 2.511 x 1
= 3m³ or 4m³
Quantity of CHB 6”
1. CHB 6 / m² = Quantity of CHB per square meter
= 12.5 pcs say 13pcs/m²
Base on the standard size of the CHB
= Height x Length
= 8” x 16”
= 0.2m x 0.4m
2. Perimeter
P = (8m)² + (10m)²
= 16m + 20m
= 36m
Q = 2.5 pcs x 5pcs
= 12.5 or 13 pcs
4. Height = 1.5m
5. Area of wall / fence
= 1.5m x 36m
= 54m²
6. CHBQ = Area x CHB /m²
= 54m² x 13 pcs / m² = 702 pcs
1. T x W = 0.02m x 0.15m
2. Perimeter = 36m
3. No. of Layers = Height
CHB height
H ( h of CHB ) = 1.5m
0.2m
= 7.5 or 8 layers
4. Volume of Mortar
V = T x W x P x No. of Layers
= 0.02m x 0.15m x 36m x 8
= 0.864m³
5. Using class A mortar: C is 18 bags / m²
S is 1 m³ / m³
6. Cement = 0.864m³ x 18
= 16 bags
7. Sand = 0.864m³ x 1
= 1m³
1. Volume of 1 cell = 0.09m x .075m x 0.2m
= 0.00135m
CHB No. of Cells = 4
2. volume for 1 CHB = 0.00135m x 4 cells
= 0.0054 or 0.006m³
3. Quantity of CHB = 702 pcs
4. Volume = 702 pcs x 0.006m³
= 4.212 m³
5. Using class A mortar C = 18
S = 1
6. Cement = 4.212m³ x 18
= 76 bags
7. Sand = 4.212m³ x 1
= 5 m³
E. Plaster
1. Assume thickness of plaster say 15mm = 0.015m
2. Area of wall x 2 (for interior and exterior plaster)
Pxhx2
= 36m x 1.5m x 2
= 108m
3. Volume of Mortar = Area x Thickness of Plaster
= 108m³ x 0.015m
= 1.62m³
4. Using class A mortar
5. Cement = 18 x 1.62m³
= 30 bags
6. Sand = 1 x 1.62m³
= 2m³
Note:
• Quantity of cement in bags has no decimal point. It’s always
in whole number.
• Quantity of aggregates with decimal should be as 0.5
intervals only.
• Conversions should have rough equivalents.
