Quantity Survey & Estimation – Lecture # 8

Table of Content

- Analysis of Materials
  - Analysis of Plastering
  - Analysis of Pointing
  - Analysis of Masonry Mortar
- Introduction to CSR (Composite Schedule of Rates) & MRS (Market Rate System)
- Unit of Analysis (MRS-2016)
  - Carriage
  - Mortar
  - Concrete
- Software Applications to Rate Analysis

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Quantity Survey & Estimation – Lecture # 8 – PDF

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Quantity Survey & Estimation – Lecture # 8 – Excerpt

Quantity Survey & Estimation

Course Instructor

Engr. Shad Muhammad

Lecturer, Department of Civil Engineering

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

Review of Lecture # 7

Introduction to “Analysis of Rates”

Analysis of Materials

vAnalysis of Cement Concrete

vAnalysis of Brick Work Masonry

vAnalysis of Mortar

Content of Lecture # 8

Analysis of Materials

vAnalysis of Plastering

vAnalysis of Pointing

vAnalysis of Masonry Mortar

Introduction to CSR (Composite Schedule of Rates) & MRS (Market Rate System)

Unit of Analysis (MRS-2016)

Carriage

Mortar

Concrete

Software Applications to Rate Analysis

Analysis of Plastering

Problem: Determine the quantities of various material used in 12 mm thick plastering for wall of 100 m2 with cement mortar (1:6).

Solution:

Quantity of mortar for uniform thickness = Area*Thickness = 100 m2 * 0.012 m = 1.2 m3

Adding 30% to fill up joints, uneven surfaces, etc,

The quantity of mortar comes = 1.2 + 0.30 * 1.2 = 1.56 m3

Increasing 25%, the dry volume = 1.56 + 0.25 * 1.56 = 1.95 m3

For 1:6 Mortar,

Cement = 1.95/ (1+6) = 0.28 m3

Sand = 6 * 0.28 = 1.68 m3

No. of bags of cement for 1 m3 = 30 bags

No. of bags of cement for 0.28 m3 = 30 * 0.28 = 8 ½ bags

Analysis of Plastering

üFor ceiling plastering, unevenness of surfaces will be less; So, 20% of extra mortar may be taken to get even surface.

üFor plastering of floors over concrete, same quantity of mortar may be taken as walls for sufficient unevenness in surface.

Analysis of Plastering

Ceiling Plastering Problem: Determine the quantities of various materials used in 12 mm thick plastering for ceiling of 100 m2 with cement mortar (1:6).

Solution:

Quantity of wet mortar = Area * Thickness = 100 * 0.012 = 1.2 m3

Add 20% for uneven surface = 20% *1.2 = 0.24 m3

Volume of wet mortar = 1.2 + 0.24 = 1.44 m3

Increase 25% the dry volume = 1.44 + 25% * 1.44 = 1.80 m3

For different mortar ratios

vFor mortar (1:2), quantity of cement = 1.80/(1+2) = ___________ m3

vFor mortar (1:1.5), quantity of cement = 1.80/(1+1.5) = _________ m3

Analysis of Pointing

For pointing in brickwork, the total dry volume of material is taken as 0.60 m3 for 100 m2.

Analysis of Masonry Mortar

Problem: Determine the quantities of various materials used in 100 ft3, brick work in cement mortar (1:5).

Solution:

Quantity of brick work = 100 ft3

Number of Brick work = 1350

Sum of Ratio = 1+5 =6

Note : For 100 ft3 of masonry 30 ft3 of wet mortar is used.

Quantity of dry Cement = [1.25 x Ratio of Cement/Sum of Ratio] × Quantity of Mortar

=1.25 x 1/6×30 = 6.25 ft3

NOTE: Volume of Cement bag = 1.25 ft3

=6.25/1.25 = 5 bags

Quantity of Sand = [1.25*Ratio of Sand/Sum of Ratio] × Quantity of Mortar

=1.25* 5/6×30 = 31.25 ft3

Bricks = 1350, Cement = 5 bags, Sand = 31.25 ft3 Answer

Analysis of Masonry Mortar

Problem: Determine the quantities of various materials used in 1 m3 brick work in cement mortar (1:6)

Solution:

Quantity of Brick work = 1 m3

Number of Bricks = 500

Note: For 1 cum of work , 500 bricks are required of standard size.

Ratio of Cement Mortar = 1:6

Sum of Ratio = 1+6 =7

Note: Quantity of Wet Mortar in one cubic meter of brickwork is 30%.

Total Quantity of Wet Mortar = (30/100) *1 = 0.30 m3

Analysis of Masonry Mortar

Quantity of Dry Cement = 1.25*Ratio of Cement/Sum of Ratio × Quantity of Mortar

= 1.25 * 1/7 * 0.30 = 0.0514 m3

= 0.0514/0.035 = 1.47 bags

Note: Volume of Cement bag in m3 is 0.035.

Quantity of Sand = 1.25 * (Ratio of Sand/Sum of Ratio) × Quantity of Mortar

= 1.25 * 6/7 ×0.30 =0.308 m3

Bricks = 500, Cement =1.47 bags, Sand = 0.308 m3 Answer

Analysis of Rates

üThe 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 and plants (T. and 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% to this cost as Contractor’s Profit, the rate per unit of the item of work is obtained.

Introduction to CSR & MRS

v‘CSR’ stands for Composite Schedule of Rates, while ‘MRS’ stands for Market Rate System.

üExisting system of Composite Schedule of Rates (CSR) is discarded.

üA committee was formed under the Chairmanship of Secretary C&W Department.

ü1600 construction items were collected from the market and verified.

üThe complete MRS has been prepared, analyzed, compared with Punjab rates and was submitted to Finance Department on 15/08/2013.

üFinance Department issued approval vide SO (Dev-IV)/ FD/CSR/4-16/2012-13 dated 10/09/2013.

üMRS document will be updated by Bi-Annually in a year.

Website Address:

1.https://www.cwd.gkp.pk/downloads.php

2.https://communication_works.kp.gov.pk/page/market_rate_system_mrs/page_type/citizen

CSR & MRS Links

MRS 2016_Jan-Mar_(Qtr-1)

CSR-2014-Khyber-Pakhtunkhwa (NHA)

CSR 2016 Pakistan Railways

Unit of Analysis (MRS-2016)-Carriage

Unit of Analysis (MRS-2016)-Dismantling (Demolition)

Unit of Analysis (MRS-2016)-Mortar

Unit of Analysis (MRS-2016)-Concrete

Analysis of Masonry

Problem: Prepare analysis of rates for first class bricks work in cement mortar (1:4). Cement will be brought from a distance of 2 miles, bricks from 6 miles and sand from 3.5 miles.

Solution:

Unit of Rate = 100 ft3

Number of Bricks = 1350

Ratio of cement mortar = 1:4

Sum of Ratio = 1+4 =5

Total quantity of Wet mortar =30 ft3

Quantity of cement = 1.2*Ratio of cement/sum of ratio × Quantity of Mortar

= 1.2* 1/5×30 = 7.2 ft3

No. of bags of cement = 7.2/1.25 =5.76 bags

Analysis of Masonry

Quantity of Sand =1.2*Ratio of sand/Sum of ratio × Quantity of Mortar

=1.2*4/5×30 = 28.8cft

Materials Required; Bricks =1350, Cement = 5.76 bags, Sand = 28.8 ft3

Cost of Materials

5.76 bags @Rs.300.00/bag =5.76 × 300 =Rs. 1728/-

28.8 ft3 sand @Rs.300/100cft =28.8×300/100 =Rs.86.4/-

1350 bricks@Rs.1200/1000 =1350×1200/1000 =Rs.1620/-

Total =Rs.3434/-

Analysis of Masonry (1 Slide is Missing)*******

4th mile = 1350×19.70/1000 =Rs. 26.60/-

5th mile = 1350×17.70/1000 =Rs. 23.90/-

6th mile = 1350×15.75/1000 =Rs. 21.26/-

Total =Rs.311.65/-

Sand 28.8 ft3

1st mile =28.8×129.55/100 =Rs.37.31/-

2nd mile =28.8×25.75/100 =Rs.7.344 /-

3rd mile =28.8×22.40/100 =Rs.6.45 /-

4th mile =28.8×19.70/100 =Rs.5.67 /-

Total =Rs.56.774/-

Note: Half mile or more will be considered full mile while less than half will be ignored.

Analysis of Masonry

Total Cost of carriage =Rs.386.66/-

Cost of Labor

2.5 mason @ Rs.200/no/day = 2.5×200 =Rs.500/-

4 ⅚ coolies@ Rs.100/no/day = 4.83×100 =Rs.483/-

¾ waterman@ Rs. 120/no/day = 0.75×120 =Rs.90/-

Total = Rs.1073/-

Total Cost = Cost of materials + Cost of carriage + Cost of labor

= 2652.00 + 386.66.83 + 1073.00 = Rs.4111.66/-

Add 10% contractor’s profit = 4100.83 = Rs.411.16/-

Total =Rs.4522.82/- Answer

Rate Analysis Software Application

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