Q.NO. DESIGN A CONCRETE FOR STRENGTH M30 (i.e.30 �

���)

SPECIFICATION OF MATERIALS:-

a) GRADE DESIGNATION : M30

b) TYPE OF CEMENT : OPC 53 GRADE

c) MAX NOMINAL SIZE OF : 20mm DOWN

AGGREGATE

d) Min. CEMENT CONTNT : 320 ��

��

e) WATER CEMENT RATIO : 0.45

f) WORKABILITY : 75mm SLUMP

g) EXPOSER CONDITION : SEVERE (R.C.C)

h) METHOD OF CONCRETE : MANUAL

PLACING

i) Max. CEMENT CONTENT : 480 ��

��

j) CHEMICAL ADMIXTURE : Nil

k) FINE AGGREGATE ZONE : ZONE 1

SPECIFICATION OF INGRADIENTS:

A. CEMENT:

a) TYPE OF CEMENT : OPC 53 GRADE

b) SPECIFIC GRAVITY : 2.857

B. COURSE AGGREGATE :

a) SPECIFIC GRAVITY : 2.75

b) WATER ABSORPTION : 1.123%

c) FREE SURFACE MOISTURE : Nil.

C. FINE AGGREGATE :

a) SPECIFIC GRAVITY : 2.486~2.5

b) WATER ABSORPTION : 5.48%

c) FREE SURFACE MOISTURE : Nil.

D. CHEMICAL ADMIXTURES : Nil.

STEP 1:TARGETED STRENGTH:

F’ck=Fck+1.65xS

=30+1.65x5

=38.25 �

��� (Where, Fck -is mean strength,

S-standard deviation,)

STEP 2: SELECTION OF WATER CEMENT RATIO:

Maximum Water CEMENT ratio is=0.45

Based on experience adopting W/C ratio to 0.42 .Here, 0.42<0.45 hence, O.K.

STEP 3:SELECTION OF WATER CONTENT:

From table no.2 Max. water content is=186 lit.(20mm aggregate) for 50mm slump.

To achieve slump value 75mm slump 191.5 litre of water is necessary.

(NOTE:if slump value is more than 50mm then every 25mm slump water is increased

by 3% i.e.=�

�� + � =191.5 litre).

There is no chem. Admixture hence no change in water content.so 191.5 lit.is O.K

STEP 4:CALCULATION OF CEMENT CONTENT:

Water cement ratio(W/C) = 0.42

Therefore,

Cement content = �.�

.��

= 456 ��

��

STEP 5: PROPRTION OF VOLUME OF COURSE & FINE AGGREGATES:

From table no.3 Volume of C.A corresponding to 20mm size aggregate & Fine aggregate

(Zone1)

For, W/C ratio of 0.42=0.6

In present case W/C ratio is 0.42 therefore vol.of course aggregate is required to be

increased, to decrese fine aggregate content

As, W/C ratio lowered by 0.01 the proportion of vol.of course aggregate is increased

by 0.006 therefore corrected proportion of vol.of course aggregate for W/C ratio of 0.42 is

0.606

Therefore volume of Course aggregate is=0.606X1=0.606

Volume of Fine aggregate is=1-0.606=0.394

STEP 6: CAlCULATION OF MIX PROPORTINS.

A)VOLUME OF CONCRETE:= 1 ��

B) VOLUME OF CEMENT = ��������������

� .!"�#$�%�����������

=��

�.��&�

=0.160 ��

C) VOLUME OF WATER = ��������'���"�

� .!"�#$�%����'���"�

=�.�

�

=0.192 ��

D)VOLUME OF CHEMICAL ADMIXTURES= Nil.

E) VOLUME OF ALL IN AGGREGATE=[A-(B+C+D)] =[1-(0.160+0.192+0.000)]

=0.648 �� (say value as ‘e’)

F) MASS OF COURSE AGGREGATE= e�X�vol. of�C. A. X�Sp. gravity�of�C. A. X�1000

=0.648 X 0.606 X 2.75 X 1000

=1079.892 Kg

~=1080 ��

��

G) MASS OF FINE AGGREGATE= e X vol.of F.A. X Sp.gravity of F.A X 1000

=0.648 X 0.394 X 2.5 X 1000

=638.28 Kg ~638 ��

��

STEP 7: CHANGES IN CONTENT DUE TO CHANGES IN CONDITION OF INGRADIENTS:

1) WATER ABSORPTION:

QUANTITY OF WATER ABSORBED BY

COURSE AGGREGATE =%����'���"��;��"<�$��

X�MASS�OF�C. A

=.��

���1080

=12.1 ��

�� ~12

��

��

MASS OF C.A IN FIELD CONDITION =1080-12

=1068 ��

��

QUANTITY OF WATER ABSORBED BY

FINE AGGREGATE= %����'���"��;��"<�$���

X�MASS�OF�F. A

= �.��

X�638

=35 ��

�� ~ 35

��

��

MASS OF F.A IN FIELD CONDITION = 638-35

= 603 ��

��

WITH REGARDS TO WATER = 12+35= 47 ��

�� IS ABSORBED BY F.A & C.A SO THIS

MUCH OF WATER IS ADDED TO TOTAL WATER =191.5+47

=238.5 ��

��

STEP 8:QUANTITIES OF MATERIALS:

1) CEMENT = 456 ��

��

2) FINE AGGREGATE =603 ��

��

3) COURSE AGGREGATE =1068 ��

��

4) WATER =238.5 ��

��

5) WATER CEMENT RATIO =0.42

TO EXPRESS PROPORTION IN USUAL WAY CEMENT : FA : CA : Water

456 : 603 : 1068 : 238.5

1 : 1.322 : 2.342 : 0.523

REFERANCE TABLES AS PER IS 10262-2009

Selection of Water Cement Ratio