I need you to do the specific calculations for the best two admixtures which are
ID: 1713040 • Letter: I
Question
I need you to do the specific calculations for the best two admixtures which are silica fumes and superplasticizer from the table, i think that the calculations goes like this ( its an example ):If this is not all the required calculations please provide us with more and thanks in advance
Materials Properties Type of Materials 2.68 0.5 2.68 0.5 0.9 Sieve 25 100 100 100 100 14 10 100 89 100 100 97 100 100 100 100 100 100 36 1.18 0.6 0.3 25 46 0.075 FM or fine Aggregates 2.8 Rodded Unit Weight for Aggregates-1.57 Mg/m Refer to BS 882 or ASTM C33 for Combined Aggregates Specifications Additives/Admixtures Properties Other Materials S.G % of cementitious Materials t(OPC) 3.15 2.9 36-80% 3-8% 0.5-1.5Kg/m3 0.5-1.5Kg/m3 0.8-2.5 lit per 100kg of cementitious micro 0.9 macro0.9 1.05 r Entraining Agent 1.01 0 06-04 % by weight of cement
Explanation / Answer
STEP 1: GIVEN DATA :
step 2 : Targeted mean strength (or) Required Strength :
fm = fck + 1.65 x Standard deviation
let us take 5% of std.deviation
fm = 30+1.65X5 = 38.25 Mpa , (1 pa = 1 N/m2)
STEP3 : water cement ratio:
Referring to Table 7.8 for OPC, uncrushed aggregate, we get the W.C. ratio = 0.5, but the maximum W.C.ratio is =0.45 ,so take W.C. as 0.45
STEP 4 : determination of water content :
now assume the slump as 75 mm for (20 mm C.A.) As per B.S.1988
let us asssume Water demand for F.A = 200 Ltrs
water demand for C.A. = 225 ltrs
as per Formula Water content = (2/3)X WFA + (1/3) X WCA = 2/3 X200 + 1/3 X 225 = 206.25 kg/m3
STEP 4 : Determination of Cement Content :
Cement content = ((Water content) / (W.c. ratio)) = 206.25/0.45 =458.33 kg
Which is more than 350 kg (As per Table No. 9.2 of BS 8110 : Part I : 1985 ) Hence safe..
STEP 5 : Determination of admixture content :
assume air content as 6-8 %
Admixture content = unit weight x aircontent(cement/100)= 0.157x6x(458.33/100)
=4.31 kg/m3
STEP 6 : TOTAL WEIGHT OF THE AGGREGATES:
density of fresh concrete from from bs code for water content of 206.25 kg/m3 and 20mm C.A of sp.gravity =2.6
weight of wet aggregates = 2375 kg/m3
Weight of Total Aggregate =Wet density- ( weight of cement + weight of water ) =2375-( 458+200 )
= 1717 kg/m3.
step 8 : Weight of F.A :
For 20 mm aggregate size, W/C ratio of 0.45,Slump of 75 mm, for 50% fines passing through 1.18mm sievepercentage of % of F.A. passing =40%refer the 1st table of the given question for this.
Weight of F.A. = 1717 X (40/100) =686.8 kg/m3
Weight of CA =1717 X(59/100) = 1013.03 kg/m3
STEP 9 : Adjustment for Field Condition
The proportions are required to be adjusted for the field conditions. Fine Aggregate has surface moisture of 2 %
weight of F.A. = 686.6 +(2% x686.6) = 700.332 kg/m3
Course Aggregate absorbs 1% water
weight of C.A = 1013.03 + ( 1% of 1013.03) = 1023.16 kg/m3
STEP10 : Final design properties :
Water 200 ltrs cement 458.33kg Fine aggregates 700.332 kg/m3 Coarse aggregates 1023.16 kg/m3 Admixture 4.31 kg/m3