- Introduction to “
**Analysis of Rates**” **Analysis of Materials**- Analysis of
**Cement Concrete** - Analysis of
**Brick Work** - Analysis of
**Mortar**

- Analysis of

- Introduction to “

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Download Quantity Survey & Estimation (Lecture # 7) PDF

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Quantity Survey & Estimation

Course Instructor

**Engr. Shad Muhammad**

**Lecturer, Department of Civil Engineering**

**Qurtuba University of Science & IT, D. I. Khan.**

**Review of Lecture # 6**

Example #1: Estimate of **R.C.C Column**

Example #2: Estimate of **R.C.C Retaining Wall**

**Content of Lecture # 7**

Introduction to “**Analysis of Rates**”

**Analysis of Materials**

vAnalysis of **Cement Concrete**

vAnalysis of **Brick Work**

vAnalysis of **Mortar**

**Analysis of Rates**

ØThe **determination of rate per unit** of a particular item of work, from the cost of quantities of materials, the cost of laborers and other miscellaneous petty expenses required for its completion is known as “Analysis of Rates”.

ØA reasonable profit, usually **10% for the contractor **is also wp-wp-includessd in the analysis of rates.

ØRates of materials are usually taken as the **rates delivered at site **of work and if the materials are to be carried from a distant place, **more than 8 km (5 miles), then cost of transport is also added. **

**Analysis of Rates**

The rates of a particular item of work depend on the following:

1.**Specification of Works and Materials**, quality of materials, the proportion of mortar, method of construction operations, etc.

2.**Quantities of materials **and their rates, number of different types of laborers and their rates.

3.**Location of the site **of work and its distances from the source of materials and the rate of transport, availability of water.

4.**Profits** and overhead expenses of contractor.

**Analysis of Rates**

__ Overhead Costs:__ Overhead costs wp-wp-includess general office expenses, rents, taxes, supervision and other costs which are

**Analysis of Rates**

**Contractor’s Profits:**

- The contractor may be allowed a net profit of 6 to 8%, and the miscellaneous overhead expenses may come to about 5 to 10%.
- For overhead expenses and contractor’s profit 15% of the actual cost may be reasonable amount, but it is usual practice to add 10% for all these under the head profit.
- For small works, overhead cost may be very little.

**Analysis of Rates**

The analysis of rates is usually worked out for the unit of payment of the particular item of work under two heads:

1.**Materials**

2.**Labors**

- And their costs added together give the cost of items of work.
- The
**costs of materials**are taken as delivered at site, inclusive of the transport, local taxes, and other charges. - For
**tools & plants (T. & P.)**and**miscellaneous petty items (sundries)**which cannot be accounted in details, lump-sum provision is made. - A provision for
**water charges**@ 1.5% of the total cost is made in the rate. - Adding 10%
**Contractor’s Profit**, the rate per unit of the item of work is obtained.

**Analysis of Rates ****(Conditions & Situations)**

- If transport of materials is to be done from a distant place more than 8 km (5 miles)
**analysis of transport work**may be done separately. - If cement and steel are supplied by the department and the contractor is not to invest any money on these, 10% profit is not allowed on cement and steel.
- The
**cost of carriage**of cement and steel**from the go down to the site of work**should be allowed to the contractor. - If cement and steel are to be arranged by the contractor, then 10% should be added as profit on these materials too.
- 10% profit may be added over the whole cost of laborers and materials including cement and steel, if it is not specified that these will be supplied by the department.

**Analysis of Rates ****(Labor Performance)**

**Task or Out-Turn Work:**

- The out-turn of work per artisan varies to some extent, according to the nature, size, height, situation, location, etc.
- In bigger cities, where specialized and experienced labor is available, the out-turn is greater than the small towns and country sides.
- In well-organized work less labor is required.

**Analysis of Rates**

**Task or Out-Turn Work:**

** **

**Preparation of Analysis of Rates**

- The number of Mazdors may be adopted from the general ideas and different operations of construction of the particular item of work.
- For mortar and concrete, the ingredient materials-
**lime, cement and sand have voids varying from 40% to 50%**and the finer ingredient to fill up the voids in the coarse ones. - For rich mortar or rich concrete the
**finer ingredients are in excess**of the volume of voids in the coarser ones, hence**the volume of the finished mortar or concrete will increase.** **Dry volume of materials**of mortar concrete, as taken in the calculation of analysis of rates,**means total volume of each ingredient added together.**

**Preparation of Analysis of Rates**

- In working out Analysis of Rates, the laborer has been on
**daily wages basis for 8 hours**working a day. - When full day for a particular laborer is not required one laborer has to work part of a day, in such cases part labor of laborer has been taken into account.
**For example,***1 laborer for half day is equivalent to half laborer per day*.**In working out the analysis of rates, bigger units have been taken**and then rates have been deducted for unit of payment.

**Analysis of Cement Concrete**

**Problem # 1.****Find the material for 10 m****3**** of cement concrete of 1:4:8 proportion**.

vSum total quantity of determining the quantity of materials for 10 m3 concrete is to divide 15.2 by the sum of the numerals of the proportion of the materials which gives the quantity of cement in m3.

**Illustration:**

For 10 m3, volume increase =( )*10,Suppose 52

Cement = m3

Fine Aggregate = 4 *1.17 = 4.68 m3

Coarse Aggregate = 8 * 1.17 = 9.36 m3

**Now, we calculate No. of Bags of Cement**

1 m3 of Portland cement = 30 bags

1.17 m3 of Portland cement = 1.17 * 30 = 35 ½ bags

**Analysis of Cement Concrete**

**Analysis of Cement Concrete**

**Analysis of Cement Concrete**

**Problem # 2:****Determine the quantities of various materials to prepare 100 ft****3**** concrete (1:4:8)**

**Solution:**

Quantity of Wet material = 100 ft3

Increase in volume for dry material = (1+ 54%)*100 = 154 ft3

Sum of ratio of concrete = 1+4+8 =13

Quantity of Cement = 154/13 =11.78 ft3

No. of bags of cement = 11.78/1.25 = 9.42 bags

Quantity of sand = (Ratio of sand / Sum of ratio) × Dry Material =(4/13) × 154 = 47.12 ft3

Quantity of coarse aggregate = (Ratio of Course aggregate / Sum of ratio) × Dry Material

=8/13 × 154 =94.24 ft3

**Analysis of Brickwork ****(Standard Modular Brick Wall)**

**Problem # 1:**** A wall of 1½ brick thick, 30 cm nominal thickness of 20 m length and 5 m height. Determine the quantities of no. of bricks and the quantity of mortar (1:6).**

**Solution:**

Nominal volume of wall = L*B*H = 20*0.30*5 = 30 m3

For 1 cm mortar joint, then actual thickness of wall will be 29 cm.

Actual Volume of Wall = L*B*H = 20*0.29*5 = 29 m3

No. of Standard Modular Bricks = = = 14500 nos.

No. of bricks per m3 (nominal) = =

**Considering 5% breakages/wastages **= 484*5%+484 = 500 nos. per m3

For 1 m3 of brickwork, nos. of bricks required = 500

For 10 m3 of brickwork, nos. of bricks required = 500*10 = 5000 nos. per m3

**Analysis of Brickwork ****(Traditional Brick Wall)**

**Analysis of Mortar ****(Previous Example Continued)**

**Mortar Requirement **= Actual Vol. of wall – Net Vol. of Bricks

= 29 – (0.19*0.09*0.09*14500) = 6.685 m3

**For frog filling, use of cut bricks, for bonding, for uniform joints, wastages, etc. ****15% extra mortar ****is required. **

Vol. of mortar = 6.685 + 6.685 * 0.15 = 7.688 m3

For dry volume, increase by **25% **dry volume of mortar =7.688 + 0.25*7.688 = 9.61 m3

For 30 m3 of brickwork, dry vol. of mortar = 9.61 m3

For 1 m3 of brickwork, dry vol. of mortar = 9.61/30 m3

For 10 m3 of brickwork, dry vol. of mortar = (9.61/30)*10 = 3.2 m3

**Calculation of Cement & Sand for 3.2 m****3**** mortar having ingredients ratio 1:6.**

Cement = 3.2 / (1+6) = 0.46 m3

Sand = 6*0.46 = 2.76 m3

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