Numerical problem on concrete mix design by is 10262(2009) method
4,447 views
18 slides
Apr 12, 2020
Slide 1 of 18
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
About This Presentation
Design of M25 grade concrete mix as per IS:10262-2009, Concrete mix proportioning guidelines with given design stipulations
Size: 801.9 KB
Language: en
Added: Apr 12, 2020
Slides: 18 pages
Slide Content
Subject: Concrete Technology
Numerical Problem on Concrete Mix
Design by IS:10262(2009) method
by
Dnyaneshwar. D. MoreDnyaneshwar. D. More
(Assistant Professor)DEPARTMENT OF CIVIL ENGINEERING
SanjivaniRural Education Society’s
SANJIVANI COLLEGE OF ENGINEERING, KOPARGAON –423 603
SAVITRIBAI PHULE PUNE UNIVERSITY, PUNE
vii. Exposure condition : Mild
viii. Degree of supervision : Good
ix. Type of aggregate : Crushed angular aggregate
x. Maximum cement content : 450 kg/m
3
xi. Chemical admixture : Not recommended
2. Test Data For Materials2. Test Data For Materials
p) Cement used : OPC 43 grade confirming to IS 8112
q) Specific gravity of cement : 3.15
r) Specific gravity of (SSD Condition):
i.Coarse aggregate : 2.68
ii.Fine aggregate : 2.65
s) Water absorption:
i.Coarse aggregate : 0.6 percent
ii.Fine aggregate : 1.0 percent
t) Free (surface) moisture:
i.Coarse aggregate : Nili.Coarse aggregate : Nil
ii.Fine aggregate : Nil
u) Sieve analysis result (Conforming IS: 383)
i.Coarse aggregate : Table 2 of IS: 383
ii.Fine aggregate : Zone I of IS: 383
4.SelectionofWaterCementRatio
FromTable5ofIS:456-2000,maximumwatercement
ratio=0.55(Mildexposure),
adoptwatercementratioas0.50.
So,0.5<0.55,henceok.
5.SelectionofWaterContent5.SelectionofWaterContent
From Table-2, IS:10262, maximum water content =
186 liters (for 25mm to 50mm slump range and for 20
mm aggregates)
Estimated water content for 75 mm slump = 186
+3/100x186=191.6liters
6. Calculation of Cement Content
Water cement ratio = 0.50
Cement content = W.C./w-c ratio=191.6/0.5=383 kg/m3 i.e.
> 320 kg/m3
also, mini. cement content as per Table 5 of
IS: 456 for mild exposure condition =300 kg/m
3
and
also < 450 kg/m
3
Hence, Ok.
7.ProportionofVolumeofCoarseAggregateandFine7.ProportionofVolumeofCoarseAggregateandFine
AggregateContent:
From Table 3, of IS:10262 corresponding to 20 mm size
aggregate and fine aggregate (Zone I) for water-cement
ratio of 0.50
volume of coarse aggregate =0.60
Correctionforfreesurfacemoistureandwaterabsorption
(AggregatesotherthanSSD):
(In our case the coarse and fine aggregates are in dry condition as
water absorption is given)
a)Fine Aggregate (Dry) = (Mass of SSD F.A.) / {1+((water absorp. by F.A.)/100)}
=
727 /{1+((1)/100)}
= 719.802 kg/m
3
b)Coarse Aggregate (Dry) = (Mass of SSD C.A.) /{1+((water absorp. C.A.)/100)} b)Coarse Aggregate (Dry) = (Mass of SSD C.A.) /{1+((water absorp. C.A.)/100)}
= 1103 /{1+((0.6)/100)}
=1096.42 kg/m3
c)The extra water to be added for absorption:
Fine Aggregate = (Mass of FA in SSD condition –Mass of FA in dry condition)
=727–719.802
=7.2kg
coarse Aggregate = (Mass of CA in SSD condition –Mass of CA in dry condition)
=1103–1096.42
=6.58kg
Trial mix No.1234
Final mix
proportions
Qty. per bag
of cement (kg)IngredientQuantity per cu. meter
Water-cement ratio
Cement .
Water
C.A.-I ( %)
C.A.-II ( %)
F.A.
W
Mix Proportions Summary
W
O
R
K
ability
Slump(mm)
C.F.
Flow Table
Visual Observations
Comp.
Strength
(N/mm
2
)
7
Days
28
Days
Remark
17