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Estimate For Grouted Riprap





Estimate :Stone Masonry I.EARTHWORKS 1.)Excavation a.)@ section “A” Volume of excavation=1.5(12)(3.5)+4. excavation=1.5(12)(3.5)+4.85(12)(2.9)+3.9(12)(2.3)+3 85(12)(2.9)+3.9(12)(2.3)+3(1.8)(12) (1.8)(12) =404.22cum b.)@ section “B” Volume of excavation=6.2(3.5)(24)+12( excavation=6.2(3.5)(24)+12(5.7)(3.4)+12(5.2)(3.10) 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(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(3.2)(1.9)+12(2.7)(1.6)+12(2.2)(1.3)+12(1.7)(1)+ 12(2.2)(1.3)+12(1.7)(1)+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)+ Excavation=27(6.2)(3.5)+12(5.7)(3.4)+12(4.7)(2.8)+ 12(5.7)(3.4)+12(4.7)(2.8)+12(3.7)(2.2)+ 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)+ Excavation=6(6.2)(3.5)+6(4.8)(2.9)+6(3.4)( 6(4.8)(2.9)+6(3.4)(1.3) 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)( 12(.5)(.25+3.5)(6.2)+12(0.2 5+2.9)(.5)(4.85)+12(3.9)(0.25+2.30)(0.5)+12(0.25+1. 0.25+2.30)(0.5)+12(0.25+1.8)(0.5)(3) 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” Volume of Backfill=20%(240.24cum) =48.04 cum Total Volume Backfill=76.49cum+341.28cum+244.96cum+48.04 cum =710.77 cum II.MASONRY WORKS 1.)@ section “A” Total Volume=12(.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) =327.73cum 2. @Section “B” Total Volume={24(6.2)(0.25+3.5)+12(5.7)(3.4+0.25)+24(5.2)(3.10+.25)+12(4.7)(2.8+0.25)+12(4.2)(2.5+0.25)+1 2(3.7)(2.2+0.25)+12(3.2)(1.9+0.25)+12(2.7)(1.6+0.25)+12(2.2)(1.3+0.25)+12(1.7)(1+0.25)+12(1.2)(0.7+0. 25)+12(0.70)(0.40+0.25)}{0.50} =936.6cum 3. @Section “C” Total Volume={27(6.5)(3.5+0.25)+12(5.7)(3.4+0.25)+12(4.7)(2.8+0.25)+12(3.7)(2.2+0.25)+12(3.7)(2.2+0.25)+1 2(2.7)(1.6+0.25)+12(1.7)(1+0.25)}{0.50} =691.40 cum 4. @Section “D” Total Volume=0.5(6){6.2(3.5+0.25)+4.8(0.25+2.9)+3.4(1.3+0.25)} =130.92cum Total Volume of Stone Masonry=327.73cum+936.6cum+691.40 cum+130.92cum =2086.65 cum Using Class C Rock with Class B Grout Mixture: No. of bags cement=2086.65x1.5 =3129.98 say 3130 bags cement No. of cum sand=2086.65x 0.124 =258.75 say 258 cum sand For Base Coarse: 1.@ Section “A” Volume={12}{0.15}(3.5+2.9+2.3+1.8) =18.9 cum 2.@ section “B”   Volume={12}{0.15}(3.4+2.8+2.5+2.2+1.9+1.6+1.3+1+0.70+0.30)+24(0.15)(3.5+3.10) =55.62 cum 3. .@ section “C” Volume=12(0.15)(3.4+2.8+2.2+2.2+1.6+1)+27(0.15)(3.5) =37.94 cum 4.)@ section “D” Volume=6(0.15)(3.5+2.9+1.3) =6.93 cum Total Volume of Base Coarse=18.9 cum+55.62 cum+37.94 cum+6.93 cum =119.39 cum For Gravel Bedding 1.@ Section “A” Volume={12}{0.10}(3.5+2.9+2.3+1.8) =12.6cum 2.@ section “B” Volume={12}{0.10}(3.4+2.8+2.5+2.2+1.9+1.6+1.3+1+0.70+0.30)+24(0.10)(3.5+3.10) =37.08 cum 3. .@ section “C” Volume=12(0.10)(3.4+2.8+2.2+2.2+1.6+1)+27(0.10)(3.5) =25.29 cum 4.)@ section “D” Volume=6(0.10)(3.5+2.9+1.3) =4.62cum Total Volume of gravel bedding=12.6cum+37.08 cum+25.29 cum+4.62 cum =79.59 cum For Beam Total Length=345 m Using 12mmx 6 mm, 345/5.5 =63 pcsx4 =252 pcs 12 mm x 6 m bars For stirrups, Spacing=0.20m Total no of Stirrups=345/0.20 =1725 stirrups Length=4(.30)-4(.05)+.2 =1.2m Total no of pcs per commercial le ngth=6/1.2 =5 pcs Total no of bars=1725/5 =use 350 pcs 10 mm x 6 m bars Volume of beam=.30(.30)(345) =31.05cum No of bags=31.05x 9=288 bags cement No of cum sand=16 cum No of cum gravel=32 cum gravel FOR FENCING Total area=1.3(345) =448.5 sqm No of CHB=448.5*12.5 =5606.25 say 5620 pcs for breakage Total Volume of Mortar=3(5620)(0.003) =50.58 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=448.5(0.11) =50 bags P.Cx 2 ==100 bags NO of cum sand=0.006 x 448.5 =2.69 say 3x2==6 cum For reinforcements; Total length=345 Spacing=0.6 m Total no of Bars=345/0.6 =575 pcs No of pcs per commercial length=6/1.3 =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=448.5 sqm Total Length=448.5x2.7(Spaced every 2 layers) =1210.95 m Total no of pcs 10 mm x 6 m bars =1210.95/6 =201.83 say 202 pcs bars For Weep holes: Average length of weep hole=1.5 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(.20)(.20) =14 cum gravel filter For Formworks Total area of beam=.30(4)(345) =414sqm Total no of plywood=414/2.88 =144 pcs plywood NO of studs and wales=345/4.85 =70 pcsx 4 280 pcs 2x2x16 Computation for no of days in concreting and masonry works No of hours=5620/100 =56.2 days/4 =14 days for laying chb Finishing of chb joints=5620/420 =13.38 days/4 =3.35 days say 5 days Concreting of beam No of days=31.05/0.50 =62.1days/4 =15.53 say 16 days Reinforcement Total no of kgs=396(5.33)+552(3.7) =4154 kgs Total no of hours=9 man hours/kl(4154)/100 =373.86 hours/8 =47/8 =6 days For plastering Total area=448.5x2 =897sqm No of days=897/8 =113/4 =29 days