**Introduction of Rate Analysis for Concrete**

**1**– estimation of labour, materials, equipments & miscellaneous items^{st}Step**2**– check the component of structure to which the RCC rate analysis is needed,^{nd}Step

**Data required for RCC Rate Analysis**

## Step 1: Estimation of Materials

Principal Components are: cement, sand, coarse aggregate & steel. Let mix design is M20 (1:1.5:3). Total volume of substances required is 1.54 m^{3} for 1 m^{3 }of wet concrete.

**For 1 m**^{3}of m15 (1:2:4) grade- Sub total cost = cement + sand + aggregate = Rs. (2520 +775 + 1860) = Rs. 5155.

**For 1 m**^{3}of m7.5 (1:4:8) grade- Sub total cost = cement + sand + aggregate = Rs. (1360 +837 + 2008 )= Rs. 4205.

**For 1 m**^{3}of m10 (1:3:6) grade- Sub total cost = cement + sand + aggregate = Rs. (1776 +815 + 1956) = Rs. 4547.

**1.1 Bags of cement required:**

- Cement density = 1440 kg/m
^{3} - 1 bag cement weight = 50 kg
- For 1 m
^{3}of Concrete, Cement needed =0.28 m^{3} - Therefore, number of bags of cement required (volume of 1 bag of cement = 0.0347 m
^{3}= 8.07 bags of cement. **For 1m**^{3 }of M15 (1:2:4) grade- Amount of cement = 1/7 × 1.54 m3 × 1440 kg/m
^{3}= 316.8 kg, - For 1m
^{3}m15 concrete, number of cement bags needed = 316.8/50 = 6.3. - If rate of cement = Rs. 430/bag,
- Hence, cost of 5 bags cement = 5 ×430 = Rs. 2150.

- Amount of cement = 1/7 × 1.54 m3 × 1440 kg/m

**For 1 m**^{3}of M7.5 (1:4:8) grade- Amount of cement = 1/13 × 1.54 m
^{3}× 1440 kg/m^{3}= 170 kg, - For 1m
^{3}5 concrete, number of cement bags needed = 170/50 = 3.4. - If rate of cement = Rs. 440/bag,
- Hence, cost of 4 bags cement = 4 ×440 = Rs. 1760.

- Amount of cement = 1/13 × 1.54 m
**For 1 m**^{3 }of M10 (1:3:6) grade- Amount of cement = 1/10 × 1.54 m3 × 1440 kg/m3 = 222 kg,
- For 1m
^{3}m10 concrete, number of cement bags needed = 222/50 = 4.44. - If rate of cement = Rs. 450 per bag,
- Hence, cost of 5 bags cement = 5 ×450 = Rs. 2250.

**1.2 Volume of Sand required:**

- Volume of sand required for 1m
^{3}of concrete = 0.42 m^{3}of sand. - Conventionally sand is measured in CFT & Sand Density = 35.3147
**For 1 m**^{3 }of M15 (1:2:4) grade- Amount of sand = 2/7 ×1.54 × 35.3147 = 15.5 cft,
- If rate of sand is Rs. 50/cft,
- Hence, cost of 15 cft sand = 15 × 50 = Rs. 750,

**For 1 m**^{3 }of M7.5 (1:4:8) grade- Amount of sand = 4/13 ×1.54 × 35.3147 = 16.73 cft,
- If rate of sand is Rs. 55/cft,
- Hence, cost of 17 cft sand = 17 × 55 = Rs. 935,

**For 1 m**^{3}of M10 (1:3:6) grade- Amount of sand = 3/10 ×1.54 × 35.3147 = 16.3 cft,
- If rate of sand is Rs. 60/cft,
- Hence, cost of 19 cft sand = 19 × 60 = Rs. 1140,

**1.3 Volume of Coarse Aggregate Required**

For 1 m^{3 }of concrete = 0.84 m^{3} of coarse aggregates.

Conventionally aggregate measured in CFT & 1m3 = 35.3147

**For 1 m**^{3}of M15 (1:2:4) grade- Amount of aggregate = 4/7 ×1.54 × 35.3147 = 31 cft,
- If rate of aggregate is Rs. 60/cft,
- Hence, cost of 30 cft aggregate = 30 × 60 = Rs. 1800,

**For 1 m**^{3 }of M7.5 (1:4:8) grade- Amount of aggregate = 8/13 ×1.54 × 35.3147 = 33.47 cft,
- If rate of aggregate is Rs. 65/cft,
- Hence cost of 32 cft aggregate = 32 × 65 = Rs. 2080,

**For 1 m**^{3}of M10 (1:3:6) grade- Amount of aggregate = 6/10 ×1.54 × 35.3147 = 32.6 cft,
- If rate of aggregate is Rs. 70/cft,
- Hence, cost of 34 cft aggregate = 34 × 70 = Rs. 2380 ,

**1.4 Estimation of Reinforced Steel:**

- Entire weight of steel required / total volume of concrete for various components.
- Assuming the proportion of reinforcement for different components.
- For slabs = 1 % of concrete
- For Beam = 2 % concrete
- For column = 2.5 % of concrete
- For RCC Roads, 0.6% concrete

## Step2: Labor Requirement for 1m^{3} of RCC

- Mason: 1 mason required for 0.37 days.
- Labors: 1 Unskilled labor required for 3.5 days.
- Water carrier: 1 water-carrier required for 1.39 days.
- Bar Bender: 1 bar-bender required for 100 kg steel for 1 day.
- Mixer Operator: 1 mixer-operator required for 0.0714 days.
- Vibrator Operator: 1 vibrator-operator required for 0.0714 days.
- Other charges 7% (approx)

Let 1 mason & a pair of helpers completed 1 m^{3} of work in 8 hour,

Mason rate Rs. 550

Helper Rate 450,

Hence, cost of labour in concreting work = Rs. 550 +(2×450) = Rs. 1450.

**For 1m**^{3 }of M15 (1:2:4) grade- Sub total cost = Rs. (5155 + 1450) = Rs. 6605

**For 1m**^{3}of M7.5 (1:4:8) grade- Sub total cost = Rs. (4205 + 1450) = Rs. 5655

**For 1m**^{3 }of M10 (1:3:6) grade- Sub total cost = Rs. (4547 + 1450) = Rs. 5997

**Step 3: Equipments and sundries**

Let it be 7.5%.

**For 1m**^{3}of M15 (1:2:4) grade- Let cost of apparatus is 1.5 %,then 1.5% of 6605 = Rs. 99.075
- Sub total cost = Rs. (6605 + 99.075) = Rs. 6704.07

**For 1m**^{3}of M7.5 (1:4:8) grade- Let cost of apparatus is 1.5 %,then 1.5% of 5655 = Rs. 84.825
- Sub total cost = Rs. (5655 + 84.825) = Rs. 5739.82

**For 1m**^{3 }of M10 (1:3:6) grade- Let cost of apparatus is 1.5 %,then 1.5% of 5997 = Rs. 89.955
- Sub total cost = Rs. (5997 + 89.955) = Rs. 6086.95

**Step 4: Contractor’s Profit**

It ranges from 10 – 20%. Let it be 15% of total cost of materials, labours & equipments.

**For 1m**^{3 }of M15 (1:2:4) grade- Let contractor’s profit is 10%, then 10% of 6605 = Rs. 660.5
- Total cost = Rs. (6605 + 660.5) = Rs. 7265.50.

**For 1m**^{3}of M7.5 (1:4:8) grade- Let contractor’s profit is 10%, then 10% of 5655 = Rs. 565.5
- Total cost = Rs. (5655 + 565.5) = Rs. 6220.50.

**For 1m**^{3}of M10 (1:3:6) grade- Let contractor’s profit is 10%, then 10% of 5997 = Rs. 599.7
- Total cost = Rs. (5997 + 599.7) = Rs. 6596.70.

## Step 5: Rate Analysis for Floors

A floor has the subsequent modules.

- Sub Base
- Base.
- Topping

**Sub Base**

**Sub Base**is the lower part of the floor.- Sub Base is developed by Lime Concrete or Lean Cement Concrete.
- Concrete (1:4:8) or (1:3:6) is employed.
- Sub Base isn’t required if the bottom is solid & strong.

**Base**

**The base**is developed directly over the sub-grade on the ground.- This part is formed with concrete (1:2:4).

**Topping**

**The topping**is the topmost part of the ground.- The topping is sometimes developed with Terrazzo, Mosaic, Cement Concrete, and Tiles.

Also Read: Building Layout | How to Building Layout | Construction Layout Techniques

## Step6: Rate Analysis for White Washing and Painting

- Unit Area is chosen.
- 10 sq m & 100 sq ft are considered as Unit Area in meters & feet respectively.
- 2 kgs of white fat lime are going to be required.
- 06 kgs of adhesive are going to be required into the blend.

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