Question

How much water, graval, sand and concrete do I need for 90.32 Yards (complete yards of cement house interior/exterior) for slab.

any grade or concrete, or the concrete normally used, its for an estimate of a house.

Answer 1

Considering 4 ksi concrete and usual mix proportioning

Amount of material roughly required for 1 yd3 would be (may change based on material properties and concrte specifications),

Cement = 550 lb

sand = 1300 lb

graval = 1800 lb

water = 290 lb

Now,

for 90.32 yards slab of 6 inch thickness,

Volume of concrete = 90.32 * 0.167 = 15.1 yd3

Adding 10% extra for wastage,

Volume of concrete required = 16.6 yd3

Therefore, amount of materials required,

Cement = 9135 lb

sand = 21600 lb

graval = 29900 lb

water = 4820 lb