Project on Concrete Mix Design for various grades of Concrete

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

Step 6:- Calculation of cement content

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

Step 7:- calculation of course and fine aggregate

Coarse Aggregate ratio	= 0.62	From 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

Step 8:- material calculation for M35 grade o

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

Result:- Materil needed for M35 grade of 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

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

Q B) :-

Answer:-

Aim:-The Concrete Mix Design for M50 grade concrete with fly ash. And 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. Use the appropriate specific gravity value based on the type of admixture used.

Procedure:-

Step 1:- following data required for concrete mix design calculation

A	Grade Designation	M35
B	Type of Cement	OPC 53
C	Type of Admixture	No
D	Maximum Nominal Size of Aggregate	20mm
E	Minimum Cement Content	340kg	Table 5-IS 456
F	Maximum Cement Content	450kg	As per code IS: 456
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:- Test data for naterial

A	Cement Used	OPC 53	Confirming to IS8112
B	Specific Gravity of Cement	3.15	Test – IS: 2720
C	Mineral Admixture	Fly ash	Confirming to IS12089
D	Specific Gravity of Fly ash	2.90
E	Chemical Admixture	@0.6%	IS9103
F	Specific Gravity of		Test – IS 2386
	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 II	Confirming to Table 4-IS383

step 3:- The target mean strength fck is calculated as shown below.

Fck	= fck+1.65s	Form IS-10262
Fck	= 50+1.65×5
Fck	=58.25 N/mm2

Step 4:-Selection of w/c ratio.

Selection of w/c ratio should be done in proper manner. if it is not. It affecr the various property of concrete and it may lead to develop porous and weak concrete. and it directly affect inversly to the strength property of concrete.

Maximum W/C ratio	=0.45	Table 5-Is 456
Adopted W/C ratio	=0.45

Step 5:- Slection of water content:-

the selection of maximum water content based on the nominal size of aggregate. as shown below but the 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 IS10262 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+(0.06*186)
	=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 SP	=198 – (0.20*198)
	=157.73
Adopted Water Content	=160 L

Step 6:- Calculation of cement content

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

Calculation of Super Plasticiers @ 0.6 percent to the mass of cementitious content = = 355.55*(0.6/100) = 2.1333 kg

Step 7:- calculation of course and fine aggregate

Coarse Aggregate ratio	= 0.62	From 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

Step 8:- material calculation for M35 grade o

A	Volume of Concrete	= 1 cu.m
B	Total Volume of Cement	= Cement/(S.G*1000)
		= 355.55/(3.15*1000)
		= 0.113 cu.m
C	Volume of Water	=Water /(S.G*1000)
		= 160/(1*1000)
		= 0.160 cu.m
D	Volume of Superplasticizers	= Superplasticizers /(S.G*1000)
		= 2.133/(1.12*1000)
		= 0.002 cu.m
E	Total Aggregates requirement	= A – (B+C+D)
		=1 -(0.113+0.160+0.002)
		=0.725cu.m
F	Coarse Aggregate (C.A)	= E * C.A ratio * S.G * 1000
		= 0.725*0.63 *2.74 *1000
		= 1251.495 kg
G	Fine Aggregate (F.A)	= E * F.A ratio * S.G * 1000
		= 0.725*0.37*2.74 *1000
		=735.005 kg

Result:- Material needed for M50 grade of concrete.

S.No.	MATERIALS	QUANTITY
1	Cement	355.55 kg
2	Flyash	no
3	Fine Aggregate	735.005 kg
4	Coarse Aggregate (20mm)	1251.495 kg
5	WATER	160 L
6	PCE based SuperPlasticizer	2 L

Therefore Mix Ratio of M35 Grade concrete by weight is Cement : F.A : C.A : Water = 1:2.067:3.195:0.45