Civil Estimate

    

ESTIMATE OF A RCC BOX, SPAN 6 M, CLEAR HEIGHT 6.0M, FORMATION HEIGHT ABOVE BED 8 M

                               

RCC box is a highly populated small span bridge being constructed in Indian Railway and in many road bridges also. Most of the RCC box in Indian Railway is provided with single opening. There are few bridges with 02 openings may be 10% of the minor Boxes. Multi span boxes having 03 or more openings are banned now.

RESTRICTION IN THE USE OF RCC BOX ON INDIAN RAILWAY PERSPECTIVE.

1. RCC Box type minor bridges are opted using as balancing culvert in Indian Railway.

2. It is a minor bridge with Shallow foundation. Depth of M15 leveling course is only 1.35m from bed level.

3. In flowing Stream/ Nalliah,  bed is subjected to scour, hence RCC box bridge can not be opted, even in occasional flow where  scour is apprehended.

4. RCC box is placed on top soil generally with a poor bearing capacity. Required bearing capacity as suggested by the designer has to be ensured by excavating the top soil, 500mm to 1000mm even more,  below the levelling course and filled with  sand in layers and consolidated. 

5. Being Top soil, a poor bearing capacity, subjected to settlement/unequal  settlement, chances of failure is more than usual in multiple boxes.

 As per Railway standard and RDSO DRAWING an estimated cost of a RCC box bridge is tabulated below.

DETAILS OF GIVEN DATA/ DESIGNED DATA

1. Opening 6m x 6.15m (Inner height)

2. Top Slab, Bottom Slab and Vertical wall thickness: 700mm

3. Haunch: 0.6m x 0.6m

4. Height of Bank above Bed= 8m

5. Formation width of single track=7.85m

6. Boulder Backing thickness = 0.6m

7. Curtain Wall/Drop Wall thickness = 0.3m 

8. Wing wall thickness at top= 0.30m

9. Wing wall thickness at bottom= 0.80m

10. Base footing of Wing wall,  thickness: 0.8m

11. Base footing of Wing wall, wide: 6.9m.

12. Barrel Length of the Box: 13m

13. Sand filling below Box/Wing wall base 0.5m

14. Schedule of Rates followed DSR'21, IRSSOR'2019

Based on the data mentioned above  Estimated quantity and costing of a typical RCC box is tabulated below.

END OF RCC BOX

 Steel required for 1000 sqft Building

Considering a RCC column/beam structure, steel is provided fo
1. Footing, 2. Plinth Beam, 3. Lintel (continuous ), 4. Beam, 5. Slab, 6. Column. 7. Stair case. and other features like septic tank etc not considered.

 Considering height of the building= 3.34m

ntity of   Steel required for

1. Slab= 650 kg

2. Footing=340 Kg

3. Column=650 kg

4. Plinth Beam=520 kg

5. Continuous Lintel= 400 kg

6. Beam= 760 kg

7. Stair case=65 kg

8. Stair case Roof=60Kg

Total=3445 kg i/c wastage 5%.

A. M20 CONCRETE- MATERIAL REQUIREMENT 10' X 10' X 5'' THK.=41.5 CFT=1.18CUM.

1. Given Data:

2. Wet Volume of concrete =10' x 10' x 0.4166'(5'') =41.7 cft.=1.18cum.

3. Increasing 54%, dry volume of the concrete=1.18 x 1.54=1.82cum.

4. Quantity Calculation:

a. Quantity of Cement = Dry volume x (a/{a+b+c})

=1.82 x (1/{1+1.5+3}.

= 1.82 x (1/5.5}) = 0.331cum= 0.331 x 1440= 476.5 kg =9.53 bags approx.

b. Quantity of sand = Dry volume x (b/{a+b+c}

∴ Quantity of sand = Dry volume x (1.5/5.5)

= 1.82 x (1.5/5.5}) = 0.496 cu.m=0.496 x 35.31= 17.5 cft approx.

c. Quantity of Stone Aggregate:

quantity of aggregates = Dry volume x (c/{a+b+c} =1.82x (3/5.5) .

= 0.9927 cum= 0.9927 x 35.31= 35 cft approx.

B.MATERIAL REQUIREMENTS OF  CEMENT MORTAR IN BRICK WORK (1:6)

1. Given Data:

a. Quantity of Brick Work= 5 cum

b. Ratio of Cement Sand Mortar=1:6

c. Mortar thickness generally =10mm.

d. Size of Bricks(modular)= 190 x 90 x 90 mm.

e. Bulk density of Cement= 1440 Kg/cum.

2. Size of Brick with Mortar= 200 x 100x 100mm.

3. Volume of one brick including mortar=0.2 x 0.1 x 0.1=0.002cum.

4. No. of bricks with mortar alround=Volume of Brick work(5 cum)/Volume of one brick with mortar

    =5 / 0.002=2500nos.

5. Total volume of bricks only=2500 x 0.19 x 0.09 x 0.09=3.8475cum.

6. Wet Volume of Mortar= Volume of Brick work - Volume of 2500nos. of bricks.=5 - 3.8475=1.1525      cum.

7. Increasing 15% for frog filling, wastage etc, wet volume comes as=1.15 x 1.1525 = 1.3254cum.

8. Increasing 33%,  Dry Volume of Mortar will be= 1.33 x 1.3254 = 1.7627 cum.

9. Quantity Calculation-cement

a. Cement

b. Ratio of Cement Sand Mortar-1:6

c. Sum of the Ratio of Cement and Sand=1+6=7

d. Quantity of Cement=(Vol. of Dry Mortar x Ratio of cement)/Sum of the Ratio=(1.7627 x 1)/7=0.252cum.

= 1440 x 0.252=363kg.=7.26bags

10. Quantity Calculation-Sand

a. Quantity of Sand= (Volume of Dry Mortar x Ratio of sand) / Sum of the Ratio=(1.7627 x 6)/7=1.511cum.

=1.511 x 35.31=53.35cft.

B.MATERIAL REQUIREMENTS OF  CEMENT MORTAR IN BRICK WORK (1:4)

1. Given Data:

a. Quantity of Brick Work= 5 cum

b. Ratio of Cement Sand Mortar=1:4

c. Mortar thickness generally =10mm.

d. Size of Bricks(modular)= 190 x 90 x 90 mm.

e. Bulk density of Cement= 1440 Kg/cum.

2. Size of Brick with Mortar= 200 x 100x 100mm.

3. Volume of one brick including mortar=0.2 x 0.1 x 0.1=0.002cum.

4. No. of bricks with mortar alround=Volume of Brick work(5 cum)/Volume of one brick with mortar

    =5 / 0.002=2500nos.

5. Total volume of bricks only=2500 x 0.19 x 0.09 x 0.09=3.8475cum.

6. Wet Volume of Mortar= Volume of Brick work - Volume of 2500nos. of bricks.=5 - 3.8475=1.1525      cum.

7. Increasing 15% for frog filling, w

astage etc, wet volume comes as=1.15 x 1.1525 = 1.3254cum.

8. Increasing 33%,  Dry Volume of Mortar will be= 1.33 x 1.3254 = 1.7627 cum.

9. Quantity Calculation-cement

a. Cement

b. Ratio of Cement Sand Mortar-1:4

c. Sum of the Ratio of Cement and Sand=1+4=5

d. Quantity of Cement=(Vol. of Dry Mortar x Ratio of cement)/Sum of the Ratio=(1.7627 x 1)/5=0.3525cum.

= 1440 x 0.3525=508kg.=10 bags nearly.

10. Quantity Calculation-Sand

a. Quantity of Sand=

 (Volume of Dry Mortar x Ratio of sand) / Sum of the Ratio=

(1.7627 x 4)/5=1.411cum.

=1.511 x 35.31=49.8 cft.

C. MATERIAL REQUIREMENTS OF  CEMENT-SAND PLASTER IN BRICK WALL.

1. Given Data:

a. Quantity of plaster-100m2

b. Ratio of Cement Sand-1:6

c. Plaster thickness=12mm.

d. Bulk density of Cement- 1440 kg/cum.

e. Weight of cement bags- 50 kg

2. Volume of plaster= 100m2 x 0.012=1.2cum.

3. Adding 20% for filling unevenness, joints etc wet volume of mortar comes to-1.2 x 1.2=1.44cum.

Dry volume of mortar(increasing 33%)=1.44 X 1.33=1.92 cum.

4. Quantity Calculation:

a. Sum of the Ratio of Cement and Sand= 1 +6 =7

b. Quantity of Cement=(Volume of Dry Mortar x Ratio of cement)/ Sum of the Ratio =(1.92 x 1)/7=0.2736 cum

c. Quantity of Cement in Kg.=0.2736 x 1440=394kg.

d. Quantity of Cement in bags=394/50=7.88 bags

e. Quantity of Sand= (Volume of Dry Mortar x Ratio of sand) / Sum of the Ratio =(1.92 x 6)/7=1.6416 cum.= 1.6416 x 35.31=58 cft.

D. REQUIREMENTS OF BRICKS IN 100 SFT WALL, 9''THK.

1. Given Data:

a. Size of wall 10' x 10' x 9'' thk.

b. Size of one brick 9''x 4.5'' x 3''

c. Thickness of mortar= 10mm=10/25.4=0.393''

2. Area of wall face= 

10' x10'=100sft.=100x12x12=14400 sq inch.

3. Due to mortar 02 sides i.e 4.5'' and 3'' each will be increased  by 10mm or 0.3937inch.

4. Size of exposed face with mortar of a brick will be=4.8937'' x 3.3937''

5. face area representing one brick with mortar=4.8937'' x 3.3937'' =16.61sq.inch.

6. No. of bricks= Total face area of wall/exposed face area of one brick with mortar.

=Sl no.2 / Sl no.5=14400/16.61=867nos.

E. REQUIREMENTS OF BRICKS IN 100 SFT WALL, 4''THK.

1. Given Data:

a. Size of wall 10' x 10' x 4'' thk.

b. Size of one brick 9''x 4''x 3''

c. Thickness of mortar= 10mm=10/25.4=0.393''

2. Area of wall face= 

10' x10'=100sft.=100x12x12=14400 sq inch.

3. Due to mortar 02 sides i.e 9'' and 3'' each side will be increased  by 10mm or by 0.3937inch.

4. Size of exposed face with mortar of a brick will be=9.3937'' x 3.3937''

5. face area representing one brick with mortar=9.3937'' x 3.3937'' =31.88 sq.inch.

6. No. of bricks= Total face area of wall/exposed face area of one brick with mortar. =Sl no.2 / Sl no.5=14400/31.88=452nos.

F-Estimating and costing of Rubble Stone Masonry with cement mortar 1:6: 

1. Let quantity of Stone masonry=10 cum.

2. Quantity to be brought at site taking into account wastage, voids etc, extra 30%.

3. Quantity of stone boulder=1.3 x 10=13cum.

4. Volume of wet Cement mortar required=30% of 10 cum, quantity of Masonry=3.0 cum.(In brick work this is nearly 22 % of brick work)

5. Volume of Dry Cement mortar, increasing 33 %= 3 x 1.33 =4cum.

6. Quantity of cement=Dry volume x ratio of cement/sum of the ratios=4 x1/(1+6)=4/7=0.57cum.=0.57 x 1440=820kg=17bags

7. Quantity of sand=Dry volume x ratio of cement/sum of the ratios=4 x6/(1+6)=24/7=3.429 cum.=3.429 x 35.31=121 cft.

Costing: materials cost shall be taken from local market. However effort is given to have a rough costing in kolkata area.

1. Rubble stone 13Cum @ 

2. Cement 17 bags @ Rs375/- per bag=6460/-

3. Sand 121 cft @ 47/- =5687/-

4. Add L/S cost 2 % for water, electricity

5. Add labour Cost @ 15%, considering material available within a lead of 10m and lift up to 3.5m

6. Add contractors profit @ 12.5%.

G. How much does 1,000 sq ft slab work cost?

Assuming residential house up to 02 story:

1. Area of the slab in m2=93 sqm.

2. Thickness of Slab assumed=125mm

3. Volume of weight concrete for slab= 93 x 0.125=11.625 cum.

4. Volume of concrete for beam projection 32% of Slab quantity= 

    0.32 x  11.625=3.6cum.

5. Total volume of concrete= 11.625 + 3.6=15.225 Cum.

6. Dry volume of concrete= 1.53 x 15.225=23.3 cum

7. Assuming 1: 1.5: 3 concrete, 

a. Volume of cement=4.24 cum=122 bags.

b. Volume of sand= 6.36 cum.=225 cft

c. Volume of Stone chips=12.72cum=450 cft.

8. Reinforcement

a. Slab Reinforcement 62 Kg per 100 sqft= (62/100) x 1000=620 Kg.

b. Beam reinforcement: 76 Kg per 100 sft=(76/100) x 1000=760 Kg.

c. Total Steel = 620 + 720= 1380 Kg.

9. Costing:

1. Cement: 122 bags x 375/-=45750/-

2. Sand: @47/- per cft for 225 cft x 47=10,575/-

3. Stone chips: @ 58/- per_ cft., for 450 cft cost comes to:26100/-

4. Reinforcement 1380 kg x 68/- = 93840/- 

5. Labour/mason cost for casting i/c ply wood shuttering, bamboo props, excluding vibrator etc: @ 70/- per sq ft=1000/-x 70/-=70000/- 

6. Hire charges of vibrator @ 1200/- per day

Total cost for 1000 sqft slab including Beam

=45750/- + 10,575/- + 26100/- +  93840/- + 70000/- + 1200/- = Rs.2,47,465/-

Hence Cost per Cum of Concrete= 2,47,465/- / 15.25= Rs16227/-

Add 2% for wastage and miscellaneous cost.

H. REQUIREMENTS OF BRICKS IN 100 SFT WALL, 10"THK.

1. Given Data:

a. Size of wall 10' x 10' x 10'' thk.

b. Size of one brick 9.75''x 4.75'' x 2.75'' (In west Bengal, this size is adopted in general)

c. Thickness of mortar= 10mm=10/25.4=0.393''

2. Area of wall face= 

10' x10'=100sft.=100x12x12=14400 sq inch.

3. Due to mortar 02 sides i.e 4.75'' and 2.75'' each will be increased  by 10mm or 0.3937inch.

4. Size of exposed face with mortar of a brick will be=5.1437'' x 3.1437''

5. face area representing one brick with mortar=5.1437''  x 3.1437'' =16.17 sq.inch.

6. No. of bricks= Total face area of wall/exposed face area of one brick with mortar.

=Sl no.2 / Sl no.5=14400/16.17=890nos.

7. No. of Bricks per cum of above brick wall

Total Volume=100sft. x 0.833=83.3 cft.=2.36 cum.

Therefore 2.36 cum brick work contains- 891 bricks

Bricks required per cum= 891/2.36=377 nos.

J. BAR BENDING FOR STIRRUPS FOR BEAM AND COLUMNS

   Cutting Length of Rectangular Stirrups:

1. Assumed Data:

Diameter of stirrups: 8mm

Size of the beam= B x D= 250 x 400 mm

Clear cover= 40mm.

2. Effective cover= clear cover + 0.5 dia of stirrups= 40 + 4=44mm.

3. Hook Length=10d or 75mm whichever is maximum, d=dia of stirrups.= 10 x 8=80mm.

4. Deduction of each 45 deg bend=1d, d=dia of stirrups.= 8mm.

5. Deduction of each 90 deg bend=2d= 2 x 8= 16mm.

6. Deduction for each 135 deg bend=3d= 3 x 8= 24mm.

7. Width of stirrups, b=B - 2x clear cover - 2*half dia of stirrup, b=250-2x40-2x4=162 mm.

8. Depth of stirrups, d=D- 2x clear cover- 2*half dia of bar, d=400-2x40-2x4=312 mm.

9. Cutting length of stirrups =2*width + 2*depth +2*hook lenth - 3nos of 90 deg bend -2 

nos of 135 deg. bend= 2 x 162 + 2x 312 + 2 x 80 - 3 x 16 - 2 x 24= 1012 mm.

K. BAR BENDING SCHEDULE FOR COLUMN FOOTING: 

1. Assumed Data:

a. Dia of reinforcement= 12mm

b. Size of the footing= L x B=2000 x 1000mm.

c. Depth of the footing=D=350mm.

d. Clear cover=50mm.

2. Spacing of bars along X- direction=150 mm c/c.

3. Spacing of bars along y- direction=200 mm c/c.

4. Horizontal part of each bar along X-direction= L - 2*clear cover =2000 - 2 x 50= 1900mm

5. Horizontal part of each bar along y-direction= B - 2*clear cover.

= 1000 - 2 X 50= 900 mm.

6. Vertical part of each bar in fig. above =D-2*clear cover = 350- 2 x 50 = 250mm

7. Deduction for 90 deg bend=2d, d= dia of footing bar = 2 x 12= 24mm

8. No. of bars along X- direction=((B-2*clear cover)/150)+1=(900/150) +1=7 nos

9. No. of bars along Y- direction=((L-2*clear cover)/200)+1=(1900/200) +1=11 nos

10. Cutting length of each bar along X-direction= L- 2*clear cover + 2*vertical part - 2*90 deg bend

= 2000 - 2x 50 + 2 x 250 - 2 x 24 = 2352mm.

11. Total length of all bars along X-direction= No. of Bars x each cutting length = 7 x 2352=16464mm.=16.464m.

12. Cutting length of each bar along Y-direction= B - 2*clear cover + 2*vertical part - 2*90 deg bend

= 1000 - 2x 50 + 2 x 250 - 2 x 24 = 1352mm.

13. Total length of all bars along Y-direction= No. of Bars x each cutting length = 11 x 1352=14872mm.=14.872m

14. Requirement of 12mm dia Steel for one footing (SL no11 + SL no.13)=16.464m + 14.872m=31.336m @0.89 Kg/m=  27.85kg.

L. REQUIREMENT OF CEMENT IN TILE FLOORING:

Tile flooring is constructed with 22mm thick tiles laid on a bed of 20mm thick 1:4 cement mortar. Cement slurry prepared @ 44 kg/10sqm is sprayed on the mortar before laying the tiles. After laying tiles tapped lightly to remove the air void. Then joints are filled with white cement with colour pigment matching with colour of tiles. 

Cement required for 10 sqm area of floor

1. 20mm thk cement mortar

a. Volume of wet mortar=10 x 0.02=0.2       cum.

b. Volume of dry mortar=1.33 x 0.2=0.266cum.

c. Sum of the ratio=4+1=5

c. Volume of cement=0.266/5=0.0532cum=77kg.

2. Grey cement slurry @ 44 kg/10sqm=44 kg

3.White cement for filling joint@ 44kg per 10 Sqm=44kg.

Total Cement=77 + 44 + 44=165 kg for 10 sqm.

COST OF TRACK MATERIAL

Date: 11.09.21

A: COST OF 1 KM TRACK MATERIAL STRAIGHT TRACK
Description of Item	Unit	qty	Basic cost of Material	Transp, inspecn and overhead 26% on C-4	GST-18%  on C-5 +C-4	Unit rate	Amount
1	2	3	4	5	6	7	8
Supply of 52 kg rail 90 uts	MT	105	62500	16250	14175	92925	9757125
Supply of PSC sleeper BG Mono Block	No.	1540	1920	499.2	89.9	2830	4357553
Supply of elastic Rail Clips	No.	5544	60	15.6	2.8	87	484087
Supply of grooved rubber sole plate	No.	3080	30	7.8	1.4	44	134378
Supply of Metal liner	No.	5544	50	13	2.3	68	376257
Supply of fish plate	Set	154	2460	639.6	115	3636	559907
Supply of fish bolts and nuts + washer=55.51	No.	616	50	13	2.3	75	45985
Supply of end cut liner	No.	616	20	5.2	0.9	32	19574
Supply of J type clips	No.	616	55	14.3	2.6	78	48161
Total Cost of 1 KM Track materials=						Rs	15783026
Cost per M track=						Rs	15783