Concrete
There are different grades of design mix ratios are used in concrete
Nominal grade
- M5 = (1:5:10)
- M10 = (1:3:6)
- M15 = (1:2:4)
- M20 = (1:1.5:3)
Standard grade
- M25 = (1:1:2)
- M30 = design mix
- M35 = design mix
- M40 = design mix
- M45 = design mix
High strength grade
- M50 = design mix
- M55 = design mix
- M60 = design mix
- M65 = design mix
- Above M70 = design mix
For example; M25 means
M – mix and compressive strength of concrete cube after 28 days of curing
25 – compressive value (25 N/mm3)
Cube dimension = 150mm x 150mm x 150mm (or) 100mm x 100mm x 100mm
(1:1:2) – 1 part of cement, 1 part of fine aggregate, 2 parts of coarse aggregate
How to calculate 1m3 material quantity for M25 concrete mix
M25 = 1:1:2 (cement : fine aggregate : coarse aggregate )= ((part of material)/(sum of ratio)) x total volume required
Cement
= ((part of cement) / (sum of ratio)) x total volume required
Part of cement = 1
Sum of ratio = (1+1+2) = 4
The total volume required = Wet volume x dry volume
Wet volume = Actual required volume, Dry volume = 1.54 (for concrete)
What is dry volume? why 1.54 in concrete? CLICK HERE
Total volume = ( 1 x 1.54 )
= 1.54 m3so, = ( 1/4 ) x 1.54= 0.385 m3
Unit weight of cement = 1440 kg/m3
= 0.385 x 1440
= 554.4 kg
What is the unit weight? 1440? CLICK HERE
1 bag cement = 50kg = 554.4 / 50
= 11.088 ~ 11 bag
Fine aggregate
= ((part of fine aggregate) / (sum of ratio)) x total volume required
= ( 1/4 ) x 1.54= 0.385 m3 (or) 13.6 cubic feet
Unit conversions? CLICK HERE
Unit weight of F.A = 1600 to 1840 kg3
take 1600 kg/m3= 0.385 x 1600=616 kg of fine aggregate required
Coarse aggregate
= ((part of coarse aggregate) / (sum of ratio)) x total volume required
= ( 2/4 ) x 1.54 = 0.77 m3
Unit weight of C. A = 1200 to 1750kg/m3 ( density of coarse aggregate may go higher if void spaces decrease. 5mm coarse aggregate has more density when compared to 20mm size coarse aggregate )
= 1450 to 1750kg/m3 ( ASTM ) American society for testing and materials
so, assume = 1560 kg/m3
= 0.77 x 1560
= 1201.2 kg of coarse aggregate required
Water
Assume water-cement ratio = 0.50 ( 50% of cement )
= 554.4 x 0.50 = 277.2 kg
= 277.2 liters ~ 277 (or) 278 litre of water required
Summary
Materials required for 1m3 of M25 concrete
- Cement = 11 bags (1 bag = 50kg)
- Fine aggregate = 616 kg
- Coarse aggregate = 1201.2 kg
- water = 278 litre
For more details about in concrete design mix ratio and materials required for all mixes, READ MORE
Cement Mortar Materials Calculator ( Cement, Fine aggregate, and Water. ) – ” Click Here “
Mortar
Cement mortar is also used in different grades
- CM 1:1
- CM 1:2
- CM 1:3
- CM 1:4
- CM 1:5
- CM 1:6
- CM 1:8
Example; CM 1:1 means
CM – cement mortar (cement + fine aggregate + water)
1:1 – 1 part of cement and 1 part of fine aggregate
Materals required for 1m3 mortar in CM 1:6
CM 1:6 = Cement mortar 1:6 (cement: fine aggregate)
Cement
= ((part of cement) / (sum of ratio)) x total volume required
Part of cement = 1
Sum of ratio = (1+6) = 7
The total volume of required = Wet volume x dry volumeWet volume
= Actual required Dry volume = 1.33 (for mortar)
What is dry volume?1.33? CLICK HERE
Total volume required = 1×1.33 = 1.33
= (1/7) x 1.33= 0.19 m3
Unit weight of cement = 1440 kg/m3
= 0.19 x 1440= 273.6 kg
= 273.6 / 50 (1 bag cement = 50kg)
= 5.472 bags of cement required
Fine aggregate
= ((part of fine aggregate) / (sum of ratio)) x total volume required
Part of Fine aggregate = 6
Sum of ratio = (1+6) = 7
Total volume required = 1×1.33 = 1.33
= (6/7) x 1.33= 1.14 m3
Unit weight of F.A = 1600 to 1840 kg/m3
take 1600= 1.14 x 1600 = 1824 kg
Water
Assume water-cement ratio = 0.50 (50% of cement)
= 273.6 x 0.50= 136.8 kg
= 136.8 liters ~ 137 liters of water required
Summary
Material required for 1m3 mortar CM 1:1
- Cement = 5.472 bags (1 bag is 50kg)
- Fine aggregate = 1824 kg
- Water = 137 litre
For more details about in mortar design mix ratio and materials required for all mixes, READ MORE
Cement Mortar Materials Calculator ( Cement, Fine aggregate, and Water. ) – ” Click Here “
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