Civil Engineering Reference
In-Depth Information
locally available aggregates. Recommendations related to exposure condi-
tions discussed earlier should be followed.
Tables 7.11 and 7.12 are used for proportioning concrete mixes by
weight and volume, respectively. The tables provide ratios of components,
with a sum of one unit. Therefore, the required total weight or volume of the
concrete mix can be multiplied by the given ratios to obtain the weight or
volume of each component. Note that for proportioning by volume, the com-
bined volume is approximately two-thirds of the sum of the original bulk
volumes of the components, since water and fine materials fill the voids be-
tween coarse materials.
Sample Problem 7.5
Determine the required weights of ingredients to make a 3500-lb batch of non-air-en-
trained concrete mix with a maximum gravel size of 1/2 in.
Solution
From Table 7.11:
Weight of
Weight of wet fine
Weight of wet coarse
Weight of
cement
=
3500
*
0.185
=
647.5 lb
aggregate
=
3500
*
0.363
=
1270.5 lb
aggregate
=
3500
*
0.377
=
1319.5 lb
water
=
3500
*
0.075
=
262.5 lb
Sample Problem 7.6
0.5
-
m
3
Determine the required volumes of ingredients to make a
batch of air-entrained
concrete mix with a maximum gravel size of 19 mm.
Solution
Sum of the original bulk volumes of the
0.75 m
3
.
components
=
0.5
*
1.5
=
From
Table 7.12:
0.12 m
3
Volume of
Volume of wet fine
Volume of wet coarse
Volume of water
cement
=
0.75
*
0.160
=
0.27 m
3
aggregate
=
0.75
*
0.360
=
0.3 m
3
aggregate
=
0.75
*
0.400
=
0.06 m
3
=
0.75
*
0.080
=
Search WWH ::
Custom Search