Rate Analysis for Concrete

Introduction of Rate Analysis for Concrete

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

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 m³ for 1 m³of wet concrete.

For 1 m³ of m15 (1:2:4) grade
- Sub total cost = cement + sand + aggregate = Rs. (2520 +775 + 1860) = Rs. 5155.
For 1 m³ of m7.5 (1:4:8) grade
- Sub total cost = cement + sand + aggregate = Rs. (1360 +837 + 2008 )= Rs. 4205.
For 1 m³ 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/m³
1 bag cement weight = 50 kg
For 1 m³of Concrete, Cement needed =0.28 m³
Therefore, number of bags of cement required (volume of 1 bag of cement = 0.0347 m³= 8.07 bags of cement.
For 1m³of M15 (1:2:4) grade
- Amount of cement = 1/7 × 1.54 m3 × 1440 kg/m³ = 316.8 kg,
- For 1m³ 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.

For 1 m³ of M7.5 (1:4:8) grade
- Amount of cement = 1/13 × 1.54 m³ × 1440 kg/m³ = 170 kg,
- For 1m³5 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.
For 1 m³of M10 (1:3:6) grade
- Amount of cement = 1/10 × 1.54 m3 × 1440 kg/m3 = 222 kg,
- For 1m³ 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 1m³ of concrete = 0.42 m³ of sand.
Conventionally sand is measured in CFT & Sand Density = 35.3147
For 1 m³of 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 m³of 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 m³ 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 m³of concrete = 0.84 m³ of coarse aggregates.

Conventionally aggregate measured in CFT & 1m3 = 35.3147

For 1 m³ 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 m³of 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 m³ 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

Also Read: What Is Lap Length? | Lap Length in Beam | Why Lapping Is Provided? | How to Calculate Lap Length? | Lap Length as Per Is Code 456 | What Are the General Rules for Lap Length? | Lapping Zone

Step2: Labor Requirement for 1m³ 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 m³ 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 1m³of M15 (1:2:4) grade
- Sub total cost = Rs. (5155 + 1450) = Rs. 6605
For 1m³ of M7.5 (1:4:8) grade
- Sub total cost = Rs. (4205 + 1450) = Rs. 5655
For 1m³of M10 (1:3:6) grade
- Sub total cost = Rs. (4547 + 1450) = Rs. 5997

Step 3: Equipments and sundries

Let it be 7.5%.

For 1m³ 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 1m³ 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 1m³of 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 1m³of 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 1m³ 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 1m³ 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 toppingis 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.

FAQ

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Petrographic testing (astm c856) is a method that uses microscopes to examine the mineralogical and chemical characteristics of concrete and minerals. Each of these characteristics gives the engineer a sense of what is going on with the concrete and its physical properties.

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