Project on Concrete Mix Design for various grades of Concrete

1.Calculate the Concrete Mix Design for M35 grade concrete with fly ash

Aim:

To determine the concrete mix design for M35 grade concrete with fly ash

Procedure:

TARGET STRENGTH FOR MIX PROPORTIONS OF M35 GRADE CONCRETE

The strength of the concrete may deviate based on the quality of the material, method of mixing and construction practices. Hence target strength (Fck) should be higher than the required strength of the concrete (fck). The value of standard deviation(s) for different grades of concrete is given in Table 1 of IS456.

Fck	= fck+1.65s	“S” from Table1-IS10262
Fck	= 35+1.65×5
Fck	= 43.25 N/mm2

1. SELECTION OF WATER-CEMENT RATIO

The maximum water-cement ratio for different grades of concrete is given in Table 5-IS456. You can reduce the water-cement ratio based on your requirements. The Water-cement ratio directly affects the strength of the concrete.

Maximum W/C ratio	=0.45	Table 5-IS456
Adopted W/C ratio	=0.45

1. SELECTION OF WATER CONTENT

Maximum Water Content

=186 L

Table 2-IS1026

Above water-content is for 25 to 50mm slump, but we need 100mm slump for our concert as per specification. So we need to increase the water-content to get the required slump. Clause 4.2 of IS1026 suggests to increases 3% of water-content to get an extra 25mm slump. Hence we to increase water content by 6% to get an additional 50mm slump for our concrete.

Water-content for 100mm Slump	=186+(0.06*186)
	=197.16 L

Since we are using superplasticizers water-content can be reduced inorder to get strength in concrete. The amount of water-content to be reduced depends on the type of superplasticizer.

Percentage of Reduction in water content	=20%
Water content requirement after SP	=198 – (0.20*198)
	=157.73
Adopted Water Content	=160 L

1. CALCULATION OF CEMENT AND FLY ASH CONTENT

Adopted Water Cement Ratio	=0.45
Total Cementitious (Cement +Fly ash) requirement	= Water Content / WaterCement ratio
	=160/0.45
	=355.55 kg

Since Fly ash is used the amount of Cementitious content (Cement+Fly ash) shall be increased to get good workability and strength. The percentage of cementitious content to be increased depends on the experience and research. Here we have increased the cementitious content by 10%.

Final Cementitious Content (Cement+Fly ash)	= 356*1.10
	=391.6 kg
Minimum Cement Content required for ‘severe’ exposure	=340kg	Table 5-IS456
Adopted Cementious content	=391.6 kg	> 340kg

Fly ash content requirement	= 391.6 * 0.30	30% replacement
	= 117.48 ~ 118 kg
Cement Content requirement	=Cementious content – Fly ash
	=391.6-117.48
	= 274.12

Calculation of Super Plasticiers @ 0.6 percent to the mass of cementitious content = = 391.6*(0.6/100) = 2.35 kg

VII COARSE AND FINE AGGREGATE

Coarse Aggregate ratio	= 0.62	Table3-IS10262 (For W/C =0.5)
Coarse Aggregate ratio for 0.45 w/c ratio	= 0.63	(adding 0.01 for every decrease .05 and vice versa)
Final Fine Aggregate ratio	= 0.37	(1-0.63) = 0.37

VIII MIX CALCULATIONS OF M35 GRADE CONCRETE

A	Volume of Concrete	= 1 cu.m
B	Total Volume of Cement	= Cement/(S.G*1000)
		= 275/(3.15*1000)
		= 0.087 cu.m
C	Total Volume of Fly ash	= Flyash/(S.G*1000)
		= 118/(2.5*1000)
		= 0.041 cu.m
D	Volume of Water	=Water /(S.G*1000)
		= 160/(1*1000)
		= 0.160 cu.m
E	Volume of Superplasticizers	= Superplasticizers /(S.G*1000)
		= 2.35/(1.12*1000)
		= 0.002 cu.m
F	Total Aggregates requirement	= A – (B+C+D+E)
		=1 -(0.087 +0.041+0.160+0.002)
		=0.71 cu.m
G	Coarse Aggregate (C.A)	= F * C.A ratio * S.G * 1000
		= 0.71*0.63 *2.74 *1000
		= 1225.6 kg
I	Fine Aggregate (F.A)	= F * F.A ratio * S.G * 1000
		= 0.71*0.37*2.74 *1000
		=719.8 kg

Cement Flyash Sand Aggregate Water and Superplasticizer needed for M35 grade concrete

S.No.	MATERIALS	QUANTITY
1	Cement	275 kg
2	Flyash	118 kg
3	Fine Aggregate	720 kg
4	Coarse Aggregate (20mm)	1226 kg
5	WATER	160 L
6	PCE based SuperPlasticizer	2 L

To find the design mix ratio divide the calculated value of all materials by the weight of cement. Therefore Mix Ratio of M35 Grade concrete by weight is Cement : Flyash: F.A : C.A : Water = 1:0.4:2.6:4.46:0.58

1.Calculate the Concrete Mix Design for M50 grade concrete without fly ash

Aim:

To determine the concrete mix design for M50 grade concrete without fly ash

Procedure:

1. Target Mean Strength = 50 + ( 5 X 1.65 ) = 58.25 Mpa
2. Selection of water cement ratio:-
   Assume water cement ratio = 0.35
3. Calculation of water: –
   Approximate water content for 20mm max. Size of aggregate = 180 kg /m³(As per Table No. 5 , IS : 10262 ). As plasticizer is proposed we can reduce water content by 20%.

Now water content = 180 X 0.8 = 144 kg /m³

4. Calculation of cement content:-
   Water cement ratio = 0.35
   Water content per cum of concrete = 144 kg
   Cement content = 144/0.35 = 411.4 kg / m³
   Say cement content = 412 kg / m³(As per contract Minimum cement content 400 kg / m³ )
   Hence O.K.
5. Calculation for C.A. & F.A.: [ Formula’s can be seen in earlier posts]-

Volume of concrete = 1 m³
Volume of cement = 412 / ( 3.15 X 1000 ) = 0.1308 m³
Volume of water = 144 / ( 1 X 1000 ) = 0.1440 m³
Volume of Admixture = 4.994 / (1.145 X 1000 ) = 0.0043 m³
Total weight of other materials except coarse aggregate = 0.1308 + 0.1440 +0.0043 = 0.2791 m³

Volume of coarse and fine aggregate = 1 – 0.2791 = 0.7209 m³
Volume of F.A. = 0.7209 X 0.33 = 0.2379 m³ (Assuming 33% by volume of total aggregate )

Volume of C.A. = 0.7209 – 0.2379 = 0.4830 m³

Therefore weight of F.A. = 0.2379 X 2.61 X 1000 = 620.919 kg/ m³

Say weight of F.A. = 621 kg/ m³

Therefore weight of C.A. = 0.4830 X 2.655 X 1000 = 1282.365 kg/ m³

Say weight of C.A. = 1284 kg/ m³

Considering 20 mm: 10mm = 0.55: 0.45
20mm = 706 kg .
10mm = 578 kg .
Hence Mix details per m³
Increasing cement, water, admixture by 2.5% for this trial

Cement = 412 X 1.025 = 422 kg
Water = 144 X 1.025 = 147.6 kg
Fine aggregate = 621 kg
Coarse aggregate 20 mm = 706 kg
Coarse aggregate 10 mm = 578 kg
Admixture = 1.2 % by weight of cement = 5.064 kg.

Water: cement: F.A.: C.A. = 0.35: 1: 1.472: 3.043