Project on Concrete Mix Design for various grades of Concrete

1)Step1 :

• AIM :- To Calculate the Mix Design for M35 grade of Concrete with fly ash.
• Procedure :-This section contains specifications for the materials and requirements for the concrete to be manufactured. The specific gravity of Fly ash is 2.1 to 3, while the specific gravity of GGBS is 2.85.

• A	Grade of Designation	M35
  B	Type of Cement	OPC 53
  C	Type of admixture	Fly Ash
  D	Maximum Nominal Size of Aggregate	20mm
  E	Minimum Cement Content	340 kg	Table 5-IS 456
  F	Maximum Cement Content	450 kg	As per code
  G	Partial replacement of cement by Fly Ash	30%	As per requirement
  H	Maximum W/C ratio	0.45	Table 5-IS 456
  I	Workability	100mm
  J	Chemical Admixture	PCS type

Step 2:-

Materials used :-

A	Cement Used	OPC 53	Confirming to IS 8112
B	Specific Gravity of cement	3.15	Test - IS:2720
C	Mineral admixture	Fly Ash	Confirming to IS 12089
D	Specific gravity of fly ash	2.9
E	Chemical admixture	0.6%	IS 9103
F	Specific Gravity of
	Coarse Aggregate	2.74
	Fine Aggregate	2.74
	Admixture PCE based	1.12
G	Water Absorption		Test - IS 2386
	Coarse Aggregate	0.5%
	Fine Aggregate	1%
H	Sieve Analysis		Test - IS 2386
	Coarse aggregate		Confirming to Table 2- IS383
	Fine aggregate	Zone 2	Confirming to Table 4- IS383

Step3:-

Target mean strength

Fck	=Fck+1.65s	"S" from Table 1 -IS10262
Fck	=35+1.65x5
Fck	=43.25N/mm^2

Step 4:-

Slection of Water Cement Ratio :_

The maximum water-cement ratios for different grades of concrete are given in Table 5-IS456.

Maximum W/C ratio	=0.45
Adopted W/C ratio	=0.45

Step 5:-

Selection of Water Content:-

Maximum Water COntent

=186 L

Table 2-IS1026

The water content above is for a slump of 25 to 50 mm, but according to the specification our concrete requires a slump of 100 mm. Therefore, to get the desired slump, the water content must be increased. Clause 4.2 of IS1026 recommends increasing the water content by 3% to obtain an additional 25 mm slump. Therefore, to obtain a slump of 50 mm in addition to our concrete, the water content must be increased by 6%.

Water-Content for 100mm Slump	=186+(0.06x186)
	=197.16 L

As we use super plasticizers we can reduce the water content to gain strength in the concrete. The amount of water-content to be reduced depends on the type of super plasticizer.

Percentage of Reduction in water content	=20%
Water content requirement after Super plasticizer	=198-(0.20x198)
	=157.73 L
Adopted Water Content	=160 L

Step 6:-

Calculation of cement and fly ash content

Adopted water cement ratio	0.45
Total cement+fly ash req.	=Water content/water cement ratio
	=160/0.45
	=355.55 kg

Since flying ash is used, the amount of cementitious content (cement + fly ash) should be increased to get good performance and strength. The percentage of increasing cementitious content depends on experience and research. Here we have increased the cementitious content by 10%.

Final content (Cement + Fly ash)	=356x1.1
	=391.6 kg
Minimum Cement content required for severe exposure	=340 kg	Table 5-IS456
Adopted (Cement + Fly ash) content	391.6 kg	>340 kg
Fly ash content requirement	=391.6x0.3	30% replacement
	=117.48=118 kg
Cement content requirement	=(Cement+fly ash)- Fly ash
	=391.6 - 117.48
	=274.12 kg

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

Step 7:-

Calculation of coarse and fine aggregate:-

Coarse aggregate ratio	=0.62	Table 3- IS10262 (for W/C =0.5)
Coarse Aggregate artio for 0.45 w/c ratio	=0.63	(adding 0.01 for every decrease 0.05 and vice versa)
Final fine aggregate artio	=0.37	(1-0.63)=0.37

Step 8:-

Mix Calculations of M35 Grade of concrete

A	Volume of Concrete	= 1 cum
B	Total Volume of cement	=Cement/(S.G X1000)
		=275/(3.15 x 1000)
		=0.087 cum
C	Total Volume of Fly Ash	=Fly ash/(S.G x 1000)
		=118/(2.5 x 1000)
		=0.041 cum
D	Volume of Water	=Water/(S.G x 1000)
		=160/(1 x 1000)
		=0.16 cum
E	Volume of Super Plasticizers	=Superplasticizers/(S.G x 1000)
		=2.35/(1.12x1000)
		=0.002 cum
F	Total Aggregates requirement	= A – (B+C+D+E)
		=1 -(0.087 +0.041+0.160+0.002)
		=0.71 cum.
G	Coarse Aggregate	= F * C.A ratio * S.G * 1000
		= 0.71*0.63 *2.74 *1000
		= 1225.6 kg
I	Fine Aggregate	= F * F.A ratio * S.G * 1000
		= 0.71*0.37*2.74 *1000
		=719.8 kg

Step 9:-

Sr. 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 Super Plasticizer	2 L

RESULT :-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

2)M50 grade without flyash :-

The mix design M-50 grade (Using Admixture –Sikament) provided here is for reference purpose only. Actual site conditions vary and thus this should be adjusted as per the location and other factors.

Parameters for mix design M50

Grade Designation = M-50
Type of cement = O.P.C-43 grade
Brand of cement = Vikram ( Grasim )
Admixture = Sika [Sikament 170 ( H ) ]
Fine Aggregate = Zone-II

Sp. Gravity
Cement = 3.15
Fine Aggregate = 2.61
Coarse Aggregate (20mm) = 2.65
Coarse Aggregate (10mm) = 2.66

Minimum Cement (As per contract) =400 kg / m³
Maximum water cement ratio (As per contract) = 0.45

Mix Calculation: –

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 /m^3

^{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