Method of findingthe dry ingredient quantity of 1 m3
concrete.
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Answer Posted / oretan omowunmi olatunde
volume of concrete considering 1m3
ration 1:2:4
1+2+4= 7
cement
1/7+40%(compaction)+5%(waste)
0.142+0.0572+0.00715=0.20735m3
density of cement = 1440kg/cm3
1m3 of cement = 1440kg/cm3
xm3 contains 50kg = 50/1440 = 0.035m3
no of bags of cement =0.20735/0.035= 5.92bags
Sand
2/7 + 40% + 5%
0286 + 0.1144 + 0.143=0.4147m3
3.33m3 of lorry = 5cubic yard
no of load = 0.414/3.33 =0.125load/m3
Granite
4/7 + 40 + 5
0.572 + 0.229 + 0.0286 = 0.8296m3
3.33m3 of lorry = 5cubic yard
no of load = 0.8296/3.33 =0.25load/m3
cements = 6bags
Sand = 0.125load/m3
Granite = 0.250load/m3
| Is This Answer Correct ? | 52 Yes | 25 No |
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