CALCULATION OF MATERIALS

A Vision To Clear Solutions.

a)For 1 m3 f concrete Mix 1:2:4 (M15)
Add 50% for wet concrete=1.50m3
=1.50/(1+2+4)=0.214 m3
For 1m3    =30 bags of cement required (1440/50)Say 30 bags
Cement     =0.214X30=6.42 bags
Sand         =0.214X2=0.428 m3
Aggregate =0.214X4=0.856 m3

b)Wall Plastering in CM 1:4 of 12mm thick for 100m2
Volume =100 X (12/1000)=1.20 m3
Add 30 to 35% as bulking sand=0.36 m3
Add 20 as wastage of sand =0.312 m3
        Total=1.872 m3
        =(1.872/1+4)=0.374 m3
Cement = 0.374X30 =10.77 bags
Sand = 0.374X4 = 1.496 m3

c)For 100 m3 of solid Block masonry in CM 1:6 mix 8" thick

Volume =100 X 0.2 (Thickness of wall) = 20 m3
No.of blocks required =20/(0.4 X 0.2 X 0.2)=12502 No's
Volume of mortar = 20-(0.39 X 0.19 X 0.19 X1250)
                             =20-17.598
                             =2.40 m3
Note:200mm - 10mm for mortar thickness = 190mm

Blocks = 17.598/(0.4 X 0.2 X 0.2)
            =1100 No's
Add 2 % Wastage =22
Total=1122 No's

Increase by 25 % for dry mortar = 3 m3
     =3/(1+6)
     =0.429 m3
Cement =0.429 X 30 =12.50 bags
Sand     =0.429 X 6 =2.57 m3
Blocks =1122 No's

STEEL:
Weight of rod per meter length = d2/162 where d in mm
"L" for column main rod in footing is minimum of 300mm
Hook for stirrups is 9D for one side
For Cantilever anchorage length for main steel is 69D
No.of Stirrups=(Clear span/Spanning)+1
Design strength is M40,target strength is 1.25 X 40 = 50 Mpa

TRAPEZOIDAL FOOTING FORMULA:

Volume V -(L X B X D) =H/3 (A1 + A2 + sq.rt (A1+A2)
where A1= L X B
           A2=l X b B
           H = D-d (Overall depth of footing - depth of rectangular                          footing)

THEORETICAL WEIGHT:

Cement	1440 kg/m3
Steel	7850 Kg/m3
Bricks	1600-1920 Kg/m3
1 HP	745.7 watts

WEIGHT OF ROD PER METER LENGTH:

6mm	0.222
8mm	0.395
10mm	0.616
12mm	0.888
16mm	1.578
20mm	2.466
25mm	3.853
32mm	6.313
40mm	9.865

CEMENT REQUIREMENTS:

M10	210 Kg
M20	320 Kg
M25	340 Kg
M30	380 Kg
M35	410 Kg
M40	430 Kg
M45	450 Kg
M50	450 +M.S 7.5%