Estimate :Stone MasonryI.EARTHWORKS 1.)Excavation a.)@ section “A” Volume of excavation=1.5(12)(3.5)+4.85(12)(2.9)+3.9(12)(2.3)+3(1.8)(12) =404.22cum b.)@ section “B” Volume of excavation=6.2(3.5)(24)+12(5.7)(3.4)+12(5.2)(3.10) 12(5.2)(3.10)+12(4.7)(2.8)+12(4.2)(2.5)+12(3.7)(2.2)+12(3.2)(1.9)+12(2.7)(1.6)+12(2.2)(1.3)+12(1.7)(1)+1 2(1.2)(0.70)+12(0.7)(0.40) =1706.4cum c.)@ section “C” Volume of Excavation=27(6.2)(3.5)+12(5.7)(3.4)+12(4.7)(2.8)+12(3.7)(2.2)+ 12(3.7)(2.2)+12(1.7)(1.6)+12(1.7)(1) =1224.78 cum d.)@ section “D” Volume of Excavation=6(6.2)(3.5)+6(4.8)(2.9)+6(3.4)(1.3) =240.24 Total Volume of Excavation=404.22cum+1706.4cum+1224.78 cum+240.24 =3575.64cum B.Backfill a.) @ section “A” Volume of backfill=404.2212(.5)(.25+3.5)(6.2)+12(0.25+2.9)(.5)(4.85)+12(3.9)(0.25+2.30)(0.5)+12(0.25+1.8)(0.5)(3) =76.49cum %Volume of Backfill=76.49/404.22 =19 % say 20% b.)@ section “B” Volume of Backfill=20%(1706.4cum) =341.28cum c.)@ section “C” Volume of Backfill=20%(1224.78 cum) =244.96cum d.)@ section “D” 7)(2.4(1.3+0.40+0.25)+12(5.7)(3.3+0.25)+1 2(3.25)+12(3.2)+12(0.40 cum+130.6+0.9)(0.9 cum 2.MASONRY WORKS 1. @Section “D” Total Volume=0.25)+12(2.25)+12(3.40 cum 4.98 say 3130 bags cement No.65x 0.7)(2.4+0.25)+12(1.2)(1.6+0.2+0.25)+12(4. @Section “B” Total Volume={24(6.5)+12(0.2)(0.85)+12(3.77 cum II.25+1.7)(2.25)+12(4.7)(2.25)+4.70)(0. of bags cement=2086.5)(3) =327.92cum Total Volume of Stone Masonry=327.25+3. @Section “C” Total Volume={27(6.5+0.50} =936.8(0. of cum sand=2086.8+0.5)(.25)+1 2(2.5+0.7)(1+0.5)+12(5.4+0.5)(6.25+2.6cum 3.24cum) =48.5+0.2(3.25+3.25)}{0.25)}{0.8) =18.25+2.7)(3.04 cum =710.9)(.25)+12(4.25)+12(1.7)(2.2)(2.124 =258.10+.)@ section “A” Total Volume=12(.8)(0.25)+12(1.92cum =2086.5+2.5)(4.5(6){6.6cum+691.2+0.25)+12(2.73cum 2.25)+12(3.04 cum Total Volume Backfill=76.96cum+48.9+0.25)+24(5.25+2.2)(1. 25)+12(0.7+0.5 =3129.5)(3.2+0.7)(1+0.2)(0.8+0.28cum+244.75 say 258 cum sand For Base Coarse: 1.65x1.Volume of Backfill=20%(240.@ section “B” .@ Section “A” Volume={12}{0.2)(3.25)} =130.3+1.49cum+341.50} =691.9)+3.30)(0.7)(1.15}(3.7)(1.65 cum Using Class C Rock with Class B Grout Mixture: No.73cum+936.9+2. 5) =37.30)+24(0. 345/5.Volume={12}{0.6cum 2.6+1.08 cum+25.62 cum =79.8+2.10)(3.4+2.10)(3. .8+2.5+2.2+1.70+0.8+2.3) =6.2+2.3) =4.2+1. Spacing=0.10) =37.3+1+0.9+2.5+3.8) =12.5+2.@ Section “A” Volume={12}{0.10)(3.15)(3.70+0.5 =63 pcsx4 =252 pcs 12 mm x 6 m bars For stirrups.2+2.6cum+37.15}(3.3+1.6+1)+27(0.62cum Total Volume of gravel bedding=12.62 cum 3.2+1.9 cum+55.20m .5+3.8+2.)@ section “D” Volume=6(0.6+1)+27(0.2+1.39 cum For Gravel Bedding 1.10)(3.94 cum 4.5) =25.@ section “C” Volume=12(0. .5+2.@ section “C” Volume=12(0.)@ section “D” Volume=6(0.94 cum+6.4+2.15)(3.@ section “B” Volume={12}{0.4+2.62 cum+37.6+1.5+2.5+2.15)(3.4+2.15)(3.10) =55.9+1.59 cum For Beam Total Length=345 m Using 12mmx 6 mm.29 cum+4.9+1.08 cum 3.93 cum =119.93 cum Total Volume of Base Coarse=18.9+1.9+1.3+1+0.30)+24(0.29 cum 4.10}(3.10}(3. Cx 2 ==100 bags NO of cum sand=0.30)(345) =31.11) =50 bags P.05)+.5 =5606.58 say 51 cum No of bags=51x9=459 bags PC cement No of cum sand=26 cum sand No of cum Gravel=51 cum gravel For plaster No of cement=448.30)-4(.05cum No of bags=31.006 x 448.25 say 5620 pcs for breakage Total Volume of Mortar=3(5620)(0.Total no of Stirrups=345/0.5 =2.3(345) =448.003) =50.30(.5 sqm No of CHB=448.5(0.2m Total no of pcs per commercial length=6/1.69 say 3x2==6 cum For reinforcements.5*12.2 =1.05x 9=288 bags cement No of cum sand=16 cum No of cum gravel=32 cum gravel FOR FENCING Total area=1. .2 =5 pcs Total no of bars=1725/5 =use 350 pcs 10 mm x 6 m bars Volume of beam=.20 =1725 stirrups Length=4(. 6 =575 pcs No of pcs per commercial length=6/1.95/6 =201.5x2.83 say 202 pcs bars For Weep holes: Average length of weep hole=1.3 =4 pcs No of 12 mm x 6 m bars=575/4 =144 pcs 12 mm x 6 bars(Vertical Bars) For Horizontal Bars Total area=448.20)(.6 m Total no of Bars=345/0.7(Spaced every 2 layers) =1210.5 m No of pcs per length=2 pcs Total no of pc =345/2 =173 pcs 3” diam pipe PVc Volume of Gravel Filter=345(.Total length=345 Spacing=0.20) =14 cum gravel filter For Formworks Total area of beam=.85 =70 pcsx 4 280 pcs 2x2x16 Computation for no of days in concreting and masonry works .88 =144 pcs plywood NO of studs and wales=345/4.95 m Total no of pcs 10 mm x 6 m bars =1210.5 sqm Total Length=448.30(4)(345) =414sqm Total no of plywood=414/2. 05/0.33)+552(3.2 days/4 =14 days for laying chb Finishing of chb joints=5620/420 =13.53 say 16 days Reinforcement Total no of kgs=396(5.7) =4154 kgs Total no of hours=9 man hours/kl(4154)/100 =373.No of hours=5620/100 =56.1days/4 =15.35 days say 5 days Concreting of beam No of days=31.5x2 =897sqm No of days=897/8 =113/4 =29 days .86 hours/8 =47/8 =6 days For plastering Total area=448.38 days/4 =3.50 =62.