How to calculate cement sand and aggregate quantity in concrete

Quantities of materials for concrete such as cement, sand, and aggregates for production of required quantity of concrete of given mix proportions such as 1:2:4 (M15), 1:1.5: 3 (M20), 1:1:2 (M25).

We assume that we have 30 cft volume of concrete and the mixing ratio is 1:2:4 (M15).
So we know that the unit weight of concrete is 150 lbs per cubic foot.  so, first of all, we need to find the weight of 30 cft concrete.

Weight of 30 cft concrete 

30 x 150 = 4500 lbs

We know that this is the weight of wet concrete so now we will find the dry weight of concrete.

for converting wet concrete weight to dry concrete weight the weight of wet concrete multiply with 1.54.

(If you want to know why we multiplied it with 1.54 then Read this : 

      Quantity Of Cement And Sand Calculation In Mortar CFT )

Dry weight of concrete 

4500 x 1.54 = 6930 lbs

Now we will calculate the quantity of cement, sand, coarse aggregate and water.

Calculation

First of all, we will find the sum of ratio

Sum of ratio = 1+2+4 = 7

Sum of ratio = 7

Now

Quantity of cement = 

( 1 ÷ 7 ) x 6930 = 990 lbs

We know that 1 Kg = 2.204 lb so

990 ÷ 2.204 = 449.18 kg  say 150 kg

We also know that  the weight of1 bag of cement is 50 Kg so

150 ÷ 50 = 3 bags

Quantity of cement = 3 Bags

Quantity of sand = 

( 2 ÷ 7 ) x 6930 = 1980 lbs

Quantity of sand in kg = 

1980 ÷ 2.204 = 898.36 kg say 900 kg

Quantity of sand = 900kg

Quantity of coarse aggregate = 

( 3 ÷ 7 ) x 6930 = 2970 lbs

Quantity of coarse aggregate in kg = 

2970 ÷ 2.204 = 1347.54 say 1348 kg

Quantity of coarse aggregate= 1348 kg

(Note = 0.4 to 0.6 water-cement ratio used in M15 concrete. we will take o.5 w/c ratio)

Quantity of water = 

0.5 x 150 = 75 liter 

(150 is the weight of cement)

Quantity of water = 75 liters 

Quantity of material in 30 cft:

Cement = 3 bags

Sand = 900 kg

Coarse aggregate = 1348 kg

Water = 75 liters

What is Soil? Formation and types of Soil?

What is soil?

The term "Soil" has different meanings depending on the field of general professional consideration.
To an agriculturist, Soil is the material on the surface of the earth which grows and produces plant life.
To the Geologist, Soil is the substance in the relatively thin surface area within which there are roots.
To an engineer, Soil is an unconsolidated agglomerate of minerals found on or near the earth's surface with or without organic matter through which and on which engineers build structures.
It covers the entire thickness of the earth's crust that is accessible and feasible for practical use as foundation support or building material.
The behavior of soil as a foundation or building material is heavily influenced by the following:

• Moisture content present in soil pores
• The fluctuation of the groundwater table
• Freezing and thawing phenomena
• Presence of organic matter
• History of the formation of soil
• Seismicity of area

Mechanical or attractive forces bind the particles of the soil together.
The binding strength of the soil is very low compared to the rocks.
The type of soil will differ between clay and gravel and even between cobble and boulders.
The topsoil, normally two feet deep, contains organic matter and is generally considered unacceptable for use in civil engineering.

Formation of soil:

Soil is generally formed by rock decomposition or disintegration (rock weathering) on or near the earth's surface through the action of many natural, physical, and chemical agents that often break them into smaller and smaller particles.

The weathering of rocks may be the following: 

1. Physical/mechanical weathering
2. Chemical weathering 

1.Physical/mechanical weathering:

It is the disintegration of rocks caused by changes in temperature, freezing and thawing, swelling, flowing water erosion, natural disasters (earthquake, land sliding, etc.) and animal activities including people. 

Soils formed by physical weathering retain parent rock minerals.

Coarse-grained soils (gravels, sands and their mixtures) are the physical weathering products

2. Chemical weathering:

Weathering caused by oxidation, hydration, carbonation, desilication, and leaching of rock minerals is known as chemical weathering.

The common soils formed by chemical weathering are different types of clay and organic soils (peat, muck, hummus, etc.).

Types of Soil Based on Particle Size:

Types of soil based on engineering considerations depend on particle size. As the engineering properties of soil change with particle size change, different names are given for different particle size ranges. The range of particle size defined for each soil type varies among agencies.

clay:

Composed of fine particles with a size of less than 0.002 mm. Flaky in the form with a large area of the surface. Have a strong attraction of inter-particles and therefore have sufficient cohesion. Poor permeability is vulnerable to swelling and shrinking. Typically the color is brown.

silt:

Composed of particles between 0.002 and 0.06 mm in size. Have high capillarity and dry strength very low. The intermediate particle size between clay and sand, therefore, has sand and clay properties, i.e. it shows slight friction and friction as well. The silty soil color is mostly brown.

Sand:

Particle size from 0.06 to 2 mm, in shape, can be rounded to angular in color brown. No plasticity, high resistance in a confined environment and significant resistance to friction. angular particles have a high resistance to friction compared to rounded particles. It is extremely permeable and has low capillarity.

gravel:

The particle size ranges between 2 and 60 mm. Form good material for the foundation. Show high resistance to friction. Angular particles have a high resistance to friction compared to rounded particles. The gravel produced by rock crushing is angular in shape, while those taken from the riverbeds are sub-rounded to rounded.

Cobbles and Boulders:

Particles larger than gravel are commonly referred to as cobbles or boulders. Cobbles vary between 60 and 200 mm in length. Boulders are the material greater than 200 mm.

Soil types According to ASTM and AASHTO:

Sieve Analysis of Coarse Aggregate ASTM : Explanation and Procedure of TEST Step by Step

TO FIND THE GRADATION OF COARSE AGGREGATE BY SIEVE ANALYSIS (ASTM C136-05).

THEORY & IMPORTANCE:

This experiment is conducted to find and check coarse aggregate gradation i.e. crush.  The main bulky component of the concrete is crush; it is used together with its strength-giving properties to increase the volume of the concrete. it calculates its fineness module to find the water ratio to be used, along with the value of the specific crush gravity and its maximum grain size.

The maximum size of the coarse aggregate will be calculated from the modulus fineness table in this experiment. For example, if its maximum size is 3⁄4 inches, it means that all grain size is less than 3⁄4 inches. This can also be expressed by writing it as a crush: 3⁄4 inch down.

APPARATUS DETAILS:

For this gradation, the sieve used is 1 1⁄2 in, 1 in, 3⁄4 in, 1⁄2 in, 3/8 in, and #4 sieve. The distinction is that #4 sieve means that in one linear inch of the sieve there are four holes, while 1 1⁄2 inch sieve means the one side dimension of the sieve hole is 1 1⁄2 inches (not that in one linear inch of the sieve there are one and a half inches). The sieves are stacked with their lengths in ascending order. The crushed sample here is 3000 g.

APPARATUS:

• Sieve Apparatus or sieve set.
• Electronic Balance.
• Brittle brush.
• Empty plate.

Electronic Scale

Sieve Set & shaker

MATERIAL:

Sample of Coarse aggregate

PROCEDURE: 

Ø  Took the digital balance and set the scale reading to zero

Ø  Took and measured the weight of pan.

Ø  Put some crush in the pan and measure 3000 grams of crush with the help of electronic balance.

Ø  Put the crush in a sieve of 1. 5 inches and began shaking until no more seeds could escape it readily. Using the electronic balance, measured the weight retained on the sieve and noted it in the table.

Ø  Some stone had fallen out of the reduced sieve placed it in it and begins to shake the sieve for sufficient time. Noted the sieves read of the retained weight.

Ø  Similarly, placed the entire remaining crush in the below sieve and began shaking for enough moment and noticed the weight retained in all the sieves.

Ø  Calculated the percentage of each sieve's weight retained.

Ø  The percentage of the weight that had passed through each sieve was found in the next phase.  The complete quantity entered will be 3000 grams for 1.5 inch sieve.

Ø  But it won't be the same for the 1 inch sieve because the upper sieve had retained some weight, so the proportion of the passing was calculated relative to the quantity that enters that sieve.

Ø  For a sieve of 3⁄4 inches, the total quantity entering the sieve will be less the value for the upper sieve, similarly, for other sieves, the same method should be used to calculate the percentage of the passed weight

Ø  The cumulative proportion was calculated in the next column.  This was the weight proportion that would be retained if the crush were placed directly on the sieve.  It will be the same as it was for 1. 5 inch sieve, but for 1-inch sieve, it would be the sum of the proportion retained by 1. 5 inch sieve plus the one retained.  Similarly, the cumulative proportion was calculated for other sieves, this was denoted by a1, a2, a3, etc

Ø Then at the end to find the crush's fineness module add each sieve's cumulative proportion and divide it by 100.

Observations and Calculations:

Total weight of natural coarse aggregate = 3kg.

Sieve No	Weight Retained on Sieve (kg)	Percentage of weight Retained (%)	Percentage Of weight Passed (%)	Cumulative Percentage of Retained (%)
1 in	0	0	100	0
3/4 in	0.3594	11.98	88.02	11.98
1/2 in	1.1526	38.42	49.6	50.40
3/8 in	0.9732	32.44	12.16	82.84
#4	0.489	16.30	0.86	99.14

Finess modulus formula= a1+a2+a3+a4+a5 / 100

(Note:- a is the Cumulative Percentage of Retained (%) )

Finess modulus= 244.36/100=2.44