Challenge

Project on Concrete Mix Design for various grades of Concrete

VVenkatesan Shanmughanathan5 Jun 2022

Design a concrete mix for grade M35 using fly ash as substitute material in cement with following data

STEP 1: Find target mean strength (f’_ck) @ 26days compressive strength in Mpa …clause 4.2 IS 10262:2019

S=5.0 N/m³, X=6.5

f’_ck= f_ck+1.65×S

f’_ck= 35+1.65×5

f’_ck= 43.27 Mpa

f’_ck= f_ck+X

f’_ck= 35+6.5

f’_ck= 41.5 Mpa

therefore, designing for a target mean strength of 43.27 Mpa

STEP 2: Approximate air content…clause 5.2 SI 10262: 2019

Entrapped air for nominal maximum size of 20mm is 1% of volume of concrete.

STEP 3: selection of water- cementitious ratio

The water / cementitious ratio for target mean strength and OPC 43= 0.45

According to table 5 IS 456: 2000; Maximum water cement ratio for severe exposure=0.50

STEP 4: selection of water content…. Clause 5.3 SI 10262: 2019

This table gives water for angular coarse aggregates to give a slump of 50mm

Wc= 186, x=120

wc_x=wc₅₀+× ×wc₅₀

wc₁₂₀=186 +× ×186

wc₁₂₀=201.626 kg

for water reducing admixtures reduce water content by 5-10%

for superplasticisers reduce water content by 20-30%

Reducing water content by 23% for 1% superplasticizer used.

Wc₁₂₀=201. 624×0.77

Wc₁₂₀=155.2504 kg 156kg

STEP 5: calculation of cementations content

cementations content = ≥minimum cementitious content table 5 IS 456:2000

cementations content =

cementations content = 346.66 kg/m³ 346.66 kg/m³ < 340 kg/m³ (minimum cementations content)

increased cementations content =346.66×1.1

increased cementations content = 381.326 kg/m³ 382 kg/m³

Revised w/c ratio =

Revised w/c ratio = 0. 406

Fly ash content = 30% × increased cementations content

Fly ash content = 30% × 382

Fly ash content = 114.6 kg/m³

Cement content = increased cementations content - Fly ash content

Cement content = 382 -114

Cement content = 268 kg/m³ <450 kg/m³ (maximum cemetent)

STEP 6: proportion of volume of coarse aggregate & fine aggregate… clause 5.5 SI 10262: 2019

V_ca@w = V_{ca@ 0.5} + _×_0.01

V_ca@w = 0.62 + _×_0.01

V_ca@w= 0.6546m³

Revised volume of coarse aggregate per unit volume

V_{[email protected]}= 0.6546×0.9

V_{[email protected]}= 0.5891 58.91%

Revised volume of fine aggregate per unit volume

V_{[email protected]}=1- 0.5891

V_{[email protected]}= .04109 41.09%

STEP 7: mxi calculations

Total Volume	1m³
Quantities	Mass	Sq. Gravity			Volume
Volume of entrapped air (a)	-	-			1%×1 = 0.01m³
Volume of water (b)	156kg	1.00			156/(1×1000) = 0.156m³
Volume of cement (c)	268 kg	3.15			268/(3.15×1000)=0.085m³
Volume of fly ash (d)	114.6 kg	2.20			114.6/(2.20×1000)=0.0521m³
Volume of chem.adm. (e)	1%(268+114.6)=3.82kg		1.145		3.82/(1.145×1000)=0.0033m³
Total volume of aggregates (f)	1-(a+b+c+d+e) = 1- (0.01+0.085+0.0521+0.0033) = 0.8496m³
	Total Volume (f)	% of Aggregate		Sq. Gravity		Mass of Aggregate
Mass of coarse aggregate	0.6546m³	0.5891		2.74		0.6546×0.5891×2.74×1000=1056.6 kg
Mass of fine aggregate	0.6546m³	0.4109		2.65		0.6546×0.4109×2.65×1000=712.78 kg

STEP 8:Mix proportion for trial 1

_{Required Quantities}
_Cement	268 kg/m³
_Water	156 kg/m³
_{Water/cementitious ratio}	0. 406
_{Fly Ash}	114.6
_{Superplasticizer}	3.82kg/m³
_{coarse Aggregates (SSD)}	1056.6 kg/m³
_{fine Aggregates (SSD)}	712.78 kg/m³

STEP 9: Adjustment in mix design

fine aggregates (dry) =

fine aggregates (dry) = =741.666kg/m³

Extra water required for fine aggregate simulate a SSD condition

A = (³

coarse aggregates (dry) = = 1072.69 kg/m³
Extra water required for coarse aggregate simulate a SSD condition

B = ( 1072.69) = -16.09 kg/m³

Total water required =water requirement +A+B

Total water required = 156-16.09 = 111.02 kg/m³

Design a concrete mix for grade M50 using in cement with following data