Question

The proportioning of a concrete mixture has resulted in the following SSD weights.

Material WSSD (Ib/yd^3) Other Info
Cement 739 Gs=3.15
Rock (coarse) 1550 Gs(SSD)= 2.7, % tot= 6%, % abs= 0.8%
Sand (fine) ? SSD =2.6, % tot=1.4%, % abs= 1.4%
Water 325
Air ___ % Air =7%

1. What is the volume of sand required?

a. 6.45 ft^3
b. 6.94 ft^3
c. 8.83 ft^3
d. 9.66 ft^3

2. What is the batch water weight?

a. 230 lbs
b. 245 lbs
c. 325 lbs
d. 405 lbs

Answer 1

1) 8.84 ft ^3 ( c )

2) 325 Ibs

Step-by-step explanation:

1) Determine the volume of sand required

Volume of sand = [ 1 - volume of cement - volume rock - volume air - volume water ] * Gs of sand -------- ( 1 )

Note : volume = [ mass / (density of water * Gs) ]

Volume of cement = [ 739 / ( 1685.55 * 3.15 ) ] = 0.139 yd^3

Volume of rock = [ 1550 / ( 1685.55 * 2.7 ) ] = 0.340 yd^3

volume of air = 0.07 yd^3

volume of water = 0.325

back to equation ( 1 )

[ 1 - 0.139 - 0.340 - 0.07 - 0.325 ] * 2.6 * 27 = 8.84 ft ^3

2) Determine Batch water weight

= 325 Ib/yd^3 * 1 yd^3

= 325 Ibs