FACTORS EFFECTING STRENGTH OF CORE OF CONCRETE...

FACTORS   WHICH   EFFECT   STRENGTH   OF   CORE   OF   CONCRETE   OR PRECAUTION   TAKEN   DURING   CORE   OF   CONCRETE   IS   GIVEN   BELOW :


1) Concrete which core is needed must be 14 days old in the case of M25 grade and if the grade of concrete more than M25 than it can be extracted earlier, but concrete must be set and hard before taking core.

2) core of concrete must be taken from the middle of wall and core cutting machine must be make right angle triangle with wall i.e machine must be kept in horizontal direction.

3) If the core of concrete taken vertically i.e for slab than in this case 15 to 20% of concrete core length from the top must be not calculated during the measuring of length i.e it must be trimmed off.( Due to uneven distribution of aggregate.)

4) If the core of concrete is not taken fully length of element of concrete than in this case 10 to 15% length of core of concrete from the bottom  is trimmed off due to cracks develop at the end of concrete.

5) Ratio between diameter of core and maximum aggregate size used in concrete is not less than 3. Generally diameter of core must be 100 mm. If we not take diameter of core 100 mm than in this case following factor is apply which is given below:
         
Diameter of core 70 mm to 80 mm = 1.03 * compressive strength of core

Diameter of core less than 70 mm  =  1.06 * compressive strength of core

6) Length or Diameter ratio of core must be 2 , following factor apply in the case when ratio is less than  2 .

 Factor of safety equation = 0.11 N + 0.78
     
 where  N is ratio of length and diameter of core.

7) core of concrete must not contain any traverse reinforcement it may cause reduction of strength of core.

8) If the capping of core is done by calcium aluminate  cement and sand than in this case ratio is 3:1 where 3 is cement and 1 is part of sand most of sand paased from 300 micron seive, water cement ratio must be not more than 0.35  .

9) Core of concrete must test in saturated condition for this kept core in water not less 40 hours and not more than 48 hours. If it is testes in dry condition than core must be kept in oven for minimum 40 hours and not more than 48 hr. Strength of core is more in the case of dry condition.

10) During the grinding of core, core must be not remove from water more than 1 hr , if grinding take  more time than in this case submerged core of concrete in water again and left it for 1 hr and after 1 hr you can do further grinding for only 1 hr.

11) During compressive strength pace rate of CTM must be 140 kg/cm2/minute.



                 

  

ESTIMATE QUANTITY OF BRICK WORK......

ESTIMATION  OF  QUANTITY  OF  INGREDIENTS  DURING  BRICKWORK  FOR  A  WALL OF  L*B*T  IS   GIVEN BELOW:

LENGTH       =       300 CM
BREATH       =        300 CM
THICKNESS =       10 CM

STANDARD SIZE OF BRICKS IS USED FOR CONSTRUCTION WORK.

LET START WITH SIZE OF STANDARD BRICK =19*9*9 CM

VOLUME OF WALL =3*3*0.1 = 0.9 M3

WE CALCULATE FIRST FOR  1 m3 BRICKWORK

Volume of one brick without mortar = 19 *9*9 = 0.001539 m3

volume of one brick with mortar   = 20*10*10 = 0.002 m3

Number of bricks for 1 m3 =  1/0.002 = 500 bricks

so volume of bricks without mortar = 500* 0.001539 = 0.769 m3

then volume of mortar = 1-0.769 = 0.2305m3

volume of dry mortar = 1.28* 0.2305 = 0.295 m3

ratio of mortar  = 1:4

number of ingredients in above ratio = 1+4 =5

Quantity of cement  = 1/5*0.295 = 0.059
                                =  0.059/ 0.0347 = 1.7 bags
                                = 1.7*50 = 85 kg

Quantity of sand    =  4/5 *0.295 = 0.236 m3 = 401.2 kg ( Density of sand 1700 kg/m3)

Number of bricks = 500 NOS  ( ABOVE CALCULATION FOR 1 M3 BRICKWORK)

Quantity of water = 85*0.45 = 38.25 kg
( 0.45 is water cement ratio)

IF WE WANT TO CALCULATE FOR 0.9 M3 BRICKWORK THAN IT IS :

CEMENT = 85*0.9 = 76.5 KG
SAND      = 401.2*0.9 = 361.08 KG
BRICKS   = 500*0.9 = 450 NOS .
WATER   =  38.25 *0.9 = 34.42 KG

HOW TO ESTIMATE QUANTITY OF INGREDIENT IN MORTAR.

QUANTITY   OF   INGREDIENT   IN   MORTAR.....

Let start with ratio of mortar 1:6 and area for plaster is 200 m2  and thickness of plaster is 12 mm.

Area = 200 m2
Thickness =12 mm

we convert 12 mm in meter it is 12/1000 = 0.012 m

So wet volume of mortar = 200*0.012 = 2.4 m3
Total ingredient in mortar = 1+6  =7

Dry volume of mortar  = 2.4* 1.28 =3.072 m 3 ( we increase 28% volume in wet mortar because voids are present in cement and sand which may be reach 25 % to 28% and causing reduction in specified quantity Here we assume it 28%. )

Cement quantity = 1/7 *3.072 = 0.438 m3  
                        
                           = 0.438/0.0347 = 12.64 Bags (volume of 1 bag of cement = 0.0347)
                           = 12.64*50   =  632  kg          (weight of 1 cement bag =50 KG )


sand Quantity     =  6/7 *3.072 = 2.63 m3 

                           = 4471 kg  ( Assume density of sand 1700 kg/ m3 . Density = mass / volume)
 ( Than 1700= mass/ 2.63 = 1700*2.63 = 4471 kg)


water quantity let we assume water cement ratio is 0.45 

weight of water / weight of cement = w/C

0.45*632 = 285 kg


Cement = 632 kg
Sand     =  4471 kg
Water    =  285 kg


  

REACTION OF TRICALCIUM ALUMINATE DURING HYDRATION.

Tricalcium aluminate during hydration...

Tricalcium aluminate reacts with Gypsum in presence of water any form calcium aluminate trisulphate hydrate which is known as entringite,  which is unstable product which reacts with C3A in the absence of gypsum and create Calcium aluminate monosulphate hydrate crystal which is stable only in absence of sulphate  but in the presence of sulphate mono sulphate crystal again convert in trisulphate hydrate crystal which size is more than two or 2.5  times of monosulphate crystal due to these large size crystal crack are developed in concrete and cement known as sulphate attack. 

HOW TO CALCULATE EFFICIENCY OF ADMIXTURE.

EFFICIENCY  OF ADMIXTURE

EFFICIENCY OF ADMIXTURE DETERMINE BY MARSH CONE TEST.

Marsh cone : Marsh cone is a brass cone with a stand which is used to measure efficiency of admixture as well as to determine optimum dose of an admixture. 

PROCEDURE IS GIVEN BELOW:
1) Take 2000gm of cement and 0.60 water cement ratio it will come 1200gm, mix both with a suitable mixture for uniform paste without lumps, seive it through 1.18mm seive to remove any lumps present in paste after mixing, pour the paste in the brass cone keep the aperture of marsh cone closed with one finger, after pouring 1 liter of paste in the cone remove finger and make the paste free from cone to jar which is placed under the Marsh cone,  simultaneously start the stop watch to determine the time taken by the paste come from cone to jar.

2) Repeat the procedure again with 1% of admixture dose and same quantity of water and determine the time taken by paste to come from cone to jar.

3) Repeat the procedure again with 1% admixture dose and with 1100 gm water and determine the time taken by paste for complete the test.

4) Repeat the test with 1000gm and 900 gm of water with 1% of admixture dose for both water content and determine the time.


LET

In first case when we not use admixture time is = 32 seconds

In second case when we use both adx. and water (1200gm) and 1% admixture time is =26 seconds

In third case time is = 28 seconds ( adx dose 1% and water 1100gm)

In fourth case time is = 30 seconds ( adx dose 1% and water 1000gm)

In fifth case time is = 32 seconds ( adx dose 1% and water is 900 gm)

THAN CALCULATION IS

       (1200 - 900)/1200 = 25%

SO EFFICIENCY OF ADMIXTURE IS   25% I.E
IT WILL REDUCE WATER UP TO 25% IN CONCRETE.

Aperture size not less than 6mm to avoid any blockage .

MIX DESIGN OF CONCRETE AS PER IS 10262:2019

MAKING A MIX DESIGN ACCORDING TO INDIAN STANDARD 10262:2019

For grade M 80 for severe condition as per IS 10262:2019

1) Let start with target mean strength :  fck+1.65*standard deviation  or fck +8 (whichever greater)

  80 +1.65*6 =89.9n/mm2         or                          80  +8 =88n/ mm2    

                                                   

  we take target mean strength : 89.9 n/mm2

2)water cement ratio from table no. 8 in is 10262 :2019 = 0.28 ( for max. aggregate size 10mm)

0.28 is less than 0.45 

Hence ok

3) water content for aggregate MAS 10mm is : 200 kg/m3 (from table number 7) for 50mm slump.

    ( required slump is =200mm)

    for every 25mm slump we increse 3% of        water.

    so for 200mm  slump we require to               increase 18% of water

   so total water =200+ 200*(18/100)

                         =200 +36 = 236 kg/m3

we use high water reducing admixture who reduce water up to 27% as per test perform from marsh cone.

 so final total water = 236 - 236*(27/100)

                                =236 - 63.72 =172.28 kg/m3

4) cementitious content  = 172/0.28

                                       =   614.28 = 614 kg/m3

614 is more than 320 kg/m3 

Hence ok.

But according to experience and trial we increase 20% of cementitious material.

than cementitious material     614+614*(20/100)= 736.8 = 737 kg/m3 

and water/cement ratio :  172/737= 0.233

5) From table no 10 of IS 10216:2019 

    volume of coarse aggregate for second zone of c/sand is :  0.54 ( for w/c = 0.30)

    

   (for every decreament of 0.05 of water cement ratio volume of coarse aggregate  increased by 0.01

   and for every increment of 0.05 of water cement ratio volume of coarse aggregate decresed by      0.01  here water cement ratio from we make comparison is 0.30 come from table no 10 of 10262)

 But in previous our actual water cement ration is  : 0.233 ( 4 th step)

here water cement ratio decreased by 0.067 so the coarse aggregate volume increase.

for 0.05 water cement ratio voulme of coarse aggregate increase = 0.01

for   1      water cement ratio volume of coarse aggregate increase  = 0.01/0.05= 0.2

for  0.067 water cement ratio volume of coarse aggregate increase by = 0.2*0.067 = 0.0134m3

so the volume of coarse aggregate is:  0.54+0.0134 =  0.5534m3

and fine aggregate voulme = 1-0.5534=  0.4466m3

6) Let cementitious content is = 737 kg

we use 8% as microsilica , 10 % alcofine and 30 percent GGBFS from this calculation is below:

for microsilica  = (737*8/100)

                         =  58.96 kg/m3

 for alcofine     =  (737*10/100)

                         =  73.7 kg/m3

for GGBFS     =    ( 737*30/100)

                       =      221.10 kg/m3

so cement content = 737-(58.96+73.7+221.10)

                              =737 - 353.76

                              =383.24 kg/m3

7) Entrapped  air  % = 1%        or               admixture  is 1% of cementitious =7.37kg/m3

 volume of concrete  =1 m3

than   volume of cement = 383.24  /   3.15*1000 = 0.121m3        ( 3.15=specific gravityof cement

        volume of microsilica = 58.96  /  2.25*1000 =0.026 m3         (2.25 = specific gravity of microsilica)

        volume of GGBFS   =  221.10  /  2.9*1000                                             =  0.0762 m3

        volume of alcofine   =  73.7  /  2.9*1000                                                 =  0.0254 m3

       

        volume of admixture = 7.37 /  1.1*1000                                                  = 0 .0067m3

       volume of water         =  172/1*1000                                                        =0.172m3

volume of all in aggregate  = 1-(0.01+0.121+0.026+0.0762+0.0254+0.0067+0.172)

         

                                            =   1- 0.4273

                                            = 0.5727m3

8) weight of coarse aggregate =  0.5727 *2.9*0.5534*1000 = 919.10 kg /m3 (2.9 s.gravity)

     weight of fine aggregate    =  0.5727*2.67*0.4466*1000= 683 kg/m3 (2.67 s.gravity)

So final mix design  for M80 grade of concrete as per IS 10262  :  2019

cement =383 kg /m3

microsilica = 58.96 kg/m3

GGBFS    =  221.10kg/m3

Alcofine  =   73.7 kg/m3

10mm  =   919.10 kg/m3

c/ sand   = 683 kg/m3

water =   172 kg/m3

admixture = 7.37 kg/m3

YIELD = 1.00003   Hence ok 

    

    

THUMB RULE OF MAKING A MIX DESIGN.

THUMB  RULE   OF MAKING  A   MIX   DESIGN

It is very simple to understand:

for example we need to making of mix design of grade M40 for severe conditions. 

So in this case first we start from cement content 

AS PER INDIAN STANDARD 456:2000 MINIMUM CEMENT CONTENT FOR M40 GRADE FOR SEVERE CONDITION IS  : ( 320 KG/M3)

AND MAXIMUM WATER CEMENT RATIO IS : ( 0.45)

So here we got cement content and water cement ratio. 

w/c  = weight of water/weight of cementitious content

HERE WE DESIGN MIX FOR PURE OPC

Than we arrive weight of water just put value of cement and water in formula above.

we got weight of water  :  144 kg

If we convert weight of cement and water into volume than we get:

volume of cement  =  320/3.15  = 101.58 

volume of  water    =  144/ 1      =  144 

Than volume of Aggregate:  1000 - (101.58+144)

                                                :   754.42

Let specific gravity of 20mm: 2.90

       specific gravity of 10mm: 2.86

       specific gravity of c/sand: 2.70

THAN weight of aggregate :

for 20 mm= 754.42/ (% of 20mm/ specific gravity )

for 10 mm= 754.42/ (% of 10mm/ specific gravity )

for c / sand = 754.42/ (% of sand/ specific gravity ) 

We derive % of aggregate by combined grading of aggregate which is explain in my previous blog check this if you miss this 

Let % of 20mm  :   40%

       % of 10mm  :   25%

       % of c/ sand :   35%

IF WE PUT THIS IN ABOVE FORMULA WE GET :

754.42÷(0.40/2.90)+ 754.42÷(0.25/2.86)+754.42÷(0.35/2.7) = 2180 KG 

Here 2180kg is the weight of All aggregate. 

so weight of 20mm :  0.40×2180 = 872 kg

      weight of 10mm : 0.25× 2180= 545 kg

      weight of c / sand : 0.35 × 2180= 763 kg

THAN THE MIX DESIGN IS  :

CEMENT :  320 KG ( MINIMUM)

WATER    :  144 KG ( ACCORDING TO CEMENTITIOUS     CONTENT)

20 MM     : 872 KG

10MM      :  545 KG

C / SAND :   763 KG

FOR CUBIC METER CONCRETE.

This mix design is only for knowledge purpose not applicable for high grade concrete.

THIS IS ONLY THUMB RULE TO MAKE A MIX DESIGN FASTLY OTHER WISE WE MAKE MIX DESIGN ACCORDING TO INDIAN STANDARD 10262:2019

REACTION BETWEEN CEMENT AND WATER AND INTERFACIAL TRANSITION ZONE...

Reaction between cement and water and interfacial transition zone.

Reaction between cement and water is known as heat of hydration it is exothermic reaction .

cement contains four components known as Bouge components.

C2S- DI CALCIUM SILICATE

C3S- TRI CALCIUM SILICATE

C3A - TRI CALCIUM ALUMINATE

C4AF- TETRA CALCIUM ALUMINO FERRITE.

when water reacts with cement they formed calcium silicate hydrate,calcium hydroxide and calcium aluminte trisulphate hydrate( known as ettringite)or calcium aluminate monosulphate hydrate which is final product of cement and water reaction.

C- S-H is cementitious product useful for concrete but Calcium hydroxide and ettringite is harmful for concrete.

when C3S reacts with water is generate less C-S-H  and more calcium hydroxide or more quantity of heat in comparison of C2S.

C3S causing initial strength for concrete where as C2S is causing gaining strength gradually for long time.

INTERFACIAL TRANSITION ZONE:

CONCRETE CONSIST MAINLY TWO ZONES

: PASTE PHASE  ( LIQUID PHASE)

: SOLID PHASE

Third phase is known as Transition zone which is lies between above these phase.

when aggregate of concrete settle down water is trapped underneath of aggregate due to which water cement ratio of that perticular area increase this zone is known as transition zone which is weaker part of concrete  causing cracks when load is applied and crackson is propagate from this zone to another zone and causing break of cube.

To minimise this zone we use POZZALANIC materials in concrete they are:

FLYASH

MICROSILICA

ALCOFINE

GGBFS

COMPRESSIVE STRENGTH OF WHITE CEMENT...

               

   COMPRESSIVE   STRENGTH   OF   WHITE   CEMENT  

As per IS 8042 compressive strength of white cement not less than

90% of specified strength of OPC 33.

Specified strength of OPC 33 after 28 days: 33N/mm2

If we calculate it is come 29.7 N/ mm2 (90%  of 33N/mm2.)


so the strength of white cement not less than 29.7 N/mm2....

EFFECT OF SPECIFIC GRAVITY ON CONCRETE...

Effect of specific gravity on concrete is given below:

• If the specific gravity of aggregate which we use to produce concrete is less than aggregate which we use in trial before to produce concrete in this case water requirement for concrete is more than trial to meet slump criteria due to which water cement ratio high and result less strength.

• In the case of low specific gravity we require more quantity of cement in mix to meetc strength criteria explaination : Let we have two types of aggregate 1) Low specific gravity 2) High specific gravity.

    fill 1m3 box with these both aggregate,

    we see that for same mass number of 

   aggregate in the case of low specific          

   gravity    is more than high specific gravity aggregate .

Hence number of aggregate more surface area of aggregate more due to this cement paste require to cover the surface of aggregate is more and water requirement more.

• If specific gravity of aggregate less than aggregate use in trial than in this case volume of concrete more I. e yield of concrete more because in this case volume of water maximum.
• If specific gravity of aggregate less it require more space in stock yard in comparison of high specific gravity of aggregate. 
• If specific gravity less density of concrete also less and vise versa.  

    

     

COMBINED GRADING OF AGGREGATE...

GENERAL QUESTIONS RELATED FOR CONCRETE

General Question:

Pace rate of C TM  for 150 mm cube size      -                 140 kg/ cm 2 /minute  -  5.25 kn/s

Pace rate of CTM for  100 mm cube size       -                 140 kg/cm 2/ minute  -   2.33 kn/s

Pace of     CTM   for cement mortar cube       -                350 kg/cm 2/ minute  -    2.9 kn/s

Pace rate of CTM for crushing value of aggregate   -      40 ton /10 minute  -        0.67 kn/s


DIFFERENT INDIAN STANDARDS .

O P C 33  -    IS 269
O P C 43  -     IS 8112
O P C 53  -     IS 12269
P S C        -     IS 455
P P C        -     IS 1489 PART 1ST ( FLY ASH BASED)
               -     IS 1489 PART 2ND( CALCINED CLAY BASED)

MASS CONCRETE WORK - IS 457

FLY ASH - 3812 PART 1ST USED AS          POZZALANIC MATERIAL.

                 -3812 PART 2ND  USED AS MINERAL ADMIXTURE.
         
                 -3812 PART 3RD  USED AS FINER MATERIAL IN CONCRETE.

IS      -   456      ( FOR P C C AND R C C)
IS      -  4925     ( SPECIFICATION FOR BATCHING CONCRETE AND MIXING PLANT)
IS      -  4926     ( SPECIFICATION FOR READY MIX CONCRETE)
IS      -   10262  ( CONCRETE MIX DESIGN)
IS      -   9103    ( SPECIFICATION FOR ADMIXTURE)
IS      -    9013    ( FOR ACCELERATED CUBE/CONCRETE STRENGTH TEST )
IS      -    650     ( STANDARD SAND)
IS      -   1343    (SPECIFICATION FOR PRE STRESSED CONCRETE)
IS      -   1344    ( SPECIFICATION FOR CALCINED CLAY FOR MAKING PPC)
IS      -   4031    (PART IST ( FINESS OF CEMENT BY SEIVE)
IS      -   4031    (PART 2 ND (FINESS BY BLAIN AIR PERMEABILITY METHOD)
IS      -   4031     (PART 3RD    ( SOUNDNESS OF CEMENT)
IS      -   4031     (PART  4TH ( FOR CONSISTENCY OF CEMENT)
IS      -   4031     ( PART  5 TH ( FOR INITIAL AND FINAL TEST OF CEMENT)3
IS      -   4031    ( PART  6TH ( FOR CEMENT MOTAR CUBE TEST)


SURFACE AREA

O PC  - NOT LESS THAN 225 M 2/KG
P SC  - NOT LESS THAN   225 M 2/KG
P PC  - NOT LESS THAN   300 M 2/ KG

MICRO SILICA  -15000 -20000 M2/KG
AL CO FINE       -1200 M 2 /KG
G G B F S          -  NOT LESS THAN 275 M 2/KG

SPECIFIC GRAVITY

O PC                               :          3.10 - 3.15
P PC                                :          2.9  -3.0
G G B F S                       :          2.9
MICRO SILICA             :          2.2
WATER                           :         1.0
AGGREGATE  COARSE :       2.7 - 2.9
FINE AGGREGATE         :       2.6 -  2.7
AL CO FINE                   :         2.9

P S C                                :        2.9-3.0

ACCEPTANCE CRITERIA FOR STRENGTH OF CONCRETE

Acceptance criteria for strength of concrete.

1st) variation in percentage of stregth of single cube not more than and not less than  15% of average strength percentage of sample.

Let start  with one example:
For M 35 grade of concrete the following results are arrives of a sample which is given below:

weight           load           strength         
kg                  kn               N/mm 2                       

8.720  -     1020.1   -        45.33
8.700  -     980.4     -        43.57         
8.724  -     965.1     -        42.89

Average strength  of sample: 43.93 N/mm2


If we convert individual strength in to percentage than we
N/mm 2             PERCENTAGE
45.33     -             129.51%                                                       
43.57     -              124.48%                                   calculation for % (45.33/35)=129.51%                                     
42.89     -             122.54%

And for average strength it will be come - 43.93- 125.51%

If we calculate variation of single strength percentage of cube to average strength percentage of sample than it will be following:

129.51     -      125.51    =     4%
124.48     -      125.51    =   -1.03%
122.54     -      125.51    =   -2.97%

So it is satisfy first criteria.



2nd) Strength of four non overlapping consecutive test result not less than

fck+3 or f ck +0.825* Standard deviation
( which ever greater)
and for individual it is f ck -3.


where f ck - characteristic of compressive strength of concrete.

if the sampling of concrete is not done by IS 456

Than it is

                         f ck+4  or   f ck+ 0.825*standard deviation (which is more)
                   
                         and for individual it is   f ck-2


WHAT IS THE MEANING OF FOUR CONSECUTIVE NON OVERLAPPING TEST RESULT IS GIVEN BELOW:


LET START WITH  EXAMPLE .

Here we describe it with M 35 grade of concrete strength, result of samples are given  below:

wt.           Load        strength        Average
Kg              KN          N/ mm2        N/mm2
8.510       980            43.55
8.512       921            40.93             41.81
8.628       922            40.97                         


8.610        898             39.11             
8.598        878             39.02           39.22                     
8.602        890             39.55

8.600       1010            44.88     
8.624         990            44.00           43.55           
8.630         940            41.77             


8.600       900.1           40.00             
8.498       878.4           39.04           39.38 
8.598       880.1           39.11


8.604       940.1            41.78         
8.502       840.4            37.35          39.34 
8.558       875.4            38.90


8.648      899.4             39.97
8.548     1000.1            44.44            42.21 
8.595     950.4              42.24

For calculaing non overlapping consecutive test result: we must calculate results of strength of  four consecutive samples which is non overlap i.e in above six samples results we add average strength  of 1 2 3 4 sample for non overlapping consecutive result but we can't add 1 2 3 5 because in this case overlapping occure and non consecutive result.

If we need four consecutive test result for non overlapping cube from the above given example it must be

41.81+39.22+43.55+39.38 (average result of samples) = 40.99 N/mm 2

40.99 N/mm 2 is greater than f ck +3 i.e (35+3= 38 N/mm 2 ) than it is satisfy the 2nd criteria.

And for individual result it is come :35 N/mm 2 - 3 N/ mm 2 =  32 N/ mm 2  which is more than above any single result of sample.


HERE GRADE OF CONCRETE M 35...

FLAKINESS AND ELONGATION INDEX

FLAKINESS AND ELONGATION  TEST OF AGGREGATE

FLAKINESS INDEX: Percentage of aggregate by weight whose least dimension is less than 3/5 times of mean seive size is known as flakiness index

ELONGATION INDEX:Percentage of aggregate by weight whose greater dimension is more than 9/5 times of mean seive size is known as elongation index.

THIS TESTS IS NOT APPLICABLE FOR AGGREGATE SIZE LESS THAN 6.3 mm.

According To IS 383 :2016 COMBINED FLAKINESS AND ELONGATION VALUE IS NOT MORE THAN 40%.

PROCEDURE OF TEST IS GIVEN IN IS 2386 PART 1ST.

Meaning of 3/5 of mean seive size is:
mean seive size = aggregate passing from seive + aggregate retained on sieve ÷2

Then: 3/5× (20+16/2)= 10.8 mm

Here 20 mm is seive size where aggregate pass and 16 mm is seive where aggregate retained.


So here, when you see Flakiness gauge the gap between these particular seive 20 mm and 16 mm is 10.8 mm.
And in Elongation it is : 9/5× (20+16/2)= 32.4 mm.

Example for flakiness and elongation is given below:

Let we take 200 pieces of 20 mm after sampling and weight it
let it come : 2000 gm

seive size              Retained wt.          Passing
                                                             on gauge
                                     gm                        gm
20-16                          220                         21
16-12.5                       822                         220
12.5-10                       792                         190
10-6.3                         120                         16
                          Total  1954 gm               447 gm

% of Flakiness = 447/1954= 22.87%

Here 46 gm passing from 6.3 mm seive and retained on pan so it is not counted.


sieve size           Retained wt              Retained
                                                             on gauge

20-16                      199                         08
16-12.5                   602                         120
12.5-10                   602                         104
10-6.3                     104                         07
             Total        1507                         239

% Of Elongation = 15.85%


so combined Flakiness and Elongation value  : 22.87+15.85 = 38.72%

PACE RATE OF COMPRESSIVE TESTING MACHINE

As per Indian standard 516 pace rate of C. T. M must be 140 kg/cm 2/minute.  so what is meaning of this given below:

surface area of cube 15 cm dimension= 225 cm 2

so in one minute load applied on cube :140*225=31500 kg/minute

Than in one second it is : 31500/60=525 kg/second

one kilogram = 9.81 N (10 Newton)

so                  = 525*10= 5250 N/second

one newton= 0.001 kn

                  =5250*0.001=5.25 kn/second

If we use 10 cm cube than in this case surface area change and we calculate it will be come 2.33 kn/second.