Civil engineer

 Depending use of grade & Flooring thickness Assume General M20 grade use & 150 mm thickness M 20 per M3 =8 bag 1 sqm area 1 sqm X 0.150 m thickness =0.150 M3 0.150 X 8 = 1.2 bag = 1.2X 50 =60 kg required  Any your calculation required pls comment 

  Cement 1 part  Sand 1.5 part  Aggregate 3 part Total dry volume of Material Required = 1.57 cu.m 1+1.5+3 =5.5 Volume of cement needed = Ratio of cement x 1.57/(1+1.5+3) = 1 x 1.57/5 .5 = 0.285 cu.m X 1440 kg/m3 =410 kg =410/50 =8.2 bags Cement density =1440 kg/m3 Volume of sand needed = Ratio of Sand x 1.57/(1+1.5+3) = 1.5 x 1.57/5 .5 = 0.427 cu.m X 1680 kg/m3 =717.36 kg/m3 Average density of sand =1680 kg/m3 Volume of aggregate needed = Ratio of agg x 1.57/( 1+1.5+3) = 3 X 1.57/5.5 = 0.854 cu.m X 1800 kg/M3 =1537.27 kg/M3 Aggregate density valu 1600–1800 kg/M3 Density value as site metarial available Cement required of bag =8.2 bag Sand required =717.36 kg/M3 Aggregate required =1537.27 kg/m3  Below blog question answer for civil engineer https://civilbilling.blogspot.com/2023/03/civil-engineering-gpsc-questions-paper.html

Quantity of steel in concrete depending on type of structure and load of structure. For Foundation :- 0.5 to 0.8 % steel are required Let we take minimum 0.5 % of steel is required on 1 m3 concrete Then quantity of steel. = (0.5/100)×1× 7850 = 39.25 kg 7850 = density of steel If we take maximum 0.8% steel is required For maximum 6% = (6/100)× 7850 = 471 kg 3. For Beam :- 1- 2 % For minimum 1% = (1/100) ×7850 = 78.5 kg For maximum 2 % = (2/100) × 7850 = 157 kg 4. Slab and lintel :- 0.7 – 1.0 % For minimum :- (0.7/100)× 7850 = 54.95 kg For maximum 1% :- (1/100)× 7850 = 78.5 kg Overall we assume Average 2.5 % of steel required in 1 m3 concrete work Then = (2.5/100) × 1 ×7850 = 196.25 kg Approx 200 kg steel 1m3 = 200 kg steel 1m3 = 2 Quintal steel required Note :- As per site Analysis not according to Is code

 Density of aggregate 1600- 1800 kg/m3 1 cuft =3.28 X 3.28 X 3.28 =35.28 cuft 1 cum required aggregate =1800 kg/m3 1 cum =1 cuft 1800 = 35.28 cuft Cuft =1800 /35.28 1 cuft =51 kg Around 51 kg required But required proportional multiple cross and find quantity Density test at laboratory and find of your coarse aggregate density

 TMT bar weight calculation formula  =D2 /162 D = dia of bar =16 mm (for example 16 mm dia bar)  16 X 16 = 256 (D x D ) Steel per meter kg=256/162 =1.58 per kg