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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 indirect expenses and non-productive expenses on the job.
Analysis of Rates
Contractor’s Profits:
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
Analysis of Rates (Conditions & Situations)
Analysis of Rates (Labor Performance)
Task or Out-Turn Work:
Analysis of Rates
Task or Out-Turn Work:
Preparation of Analysis of Rates
Preparation of Analysis of Rates
Analysis of Cement Concrete
Problem # 1. Find the material for 10 m3 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 ft3 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 m3 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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