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UNIT 6 EXAMPLES WITH DIFFERENT TYPES OF STRUCTURE
Structure 6.1 Introduction
Objectives
6.2 Quantification Principles of Various Items of Work 6.2.1 Earthwork for Foundation 6.2.2 Earthwork for Canals and Roads
6.2.3 Masonry 6.2.4 Plain and Reinforced Cement Concrete
6.2.5 Structural Steel 6.2.6 Woodwork 6.2.7 Painting 6.2.8 Plumbing and other Lump-sum Items
6.3 Rate Analysis
6.4 Examples 6.4.1 A Complete Example - Compound Wall 6.4.2 Single-bed Roomed Residential Building 6.4.3 Steel Shnchion with Grillage Foundation 6.4.4 Steel Roof Truss 6.4.5 Arched Culvert 6.4.6 Roads
6.5 Computer-aided Estimating
6.6 Summary
6.7 Answers to SAQs
6.1 INTRODUCTION
In Unit 5, estimating was defined as "careful and correct measurement of various quantities involved in a construction project as reflected in contract documents and drawings and the subsequent oalculation and application of accurate unit prices to these quantities". Principles involved in such estimating were also described in detail therein.
In this unit, typical examples of structures will be taken up for quantification. One exampk of a structure is given a complete treatment including costing whereas in other examples, only quantification of items of work (without costing) is given. This curtailing is made to primarily make the size of the unit concise; the secondary reason, of course, is that while quantities of items of work remain unaltered as long as the drawings1 specifications are not altered, cost varies from site to site and time to time.
It is emphasised that the availability of detailed, precise and well-dimensioned drawiqgs form an essential prerequisite for accurate estimating on the one hand. On the other, the estimator or the quantity surveyor is expected to possess adequate knowledge to correctly read the drawings, interpret abbreviations and specifications, infer any missing data and correctly apply appropriate formulae using principles of mensuration. At times, the estimator may have to make independent decisions without consulting the architect or the engineer in order to complete the job on hand in time. There is a vast array of codes, standards, regulations and legislation which apply to construction processes and cause either opportunities or restraints in the use of specific materials, methods or systems. A general knowledge of these is invaluable to the estimator in making independent decisions. More importantly, the estimator should have intimate knowledge of the working procedufes, strengths and limitations of hisher company or organisation in order to have access to reliable data on labour productivity, equipment utilisation and material costs including transportation so that the cost estimates prepared by himher are accurate.
Objectives After studying this unit, you should be able to
apply the principles learnt in previous unit to estimate actual structures,
Estimating describe precisely the procedure of costing structures as a "three-component" exercise,
determine the cost of any structurelproject,
conceptualise that the degree of accuracy of the estimate critically depends upon the degree of detail with which drawings, specifications and cost data are furnished,
gain confidence to take up any structure for estimating,
seek any missing relevant drawingsldata, if you are given an exercise of estimating,
, appreciate the commonly stated principle that "practice makes things perfect" and try out more examples on your own, and
have a feel for the applications of programs designed for construction estimating using personal computers (PCs).
6.2 QUANTIFICATION PRINCIPLES OF VARIOUS ITEMS OF WORK
6.2.1 Earthwork for Foundation (a) Any earthwork item involves one or more of the following activities :
(i) Cutting
(ii) Filling
(iii) Conveying
(b) Foundation trenches are quantified to the exact dimensions as given in the drawings. Actual width of excavation may be more for construction convenience but not provided for in the estimate except in cases such as in deeper trenches where benching is required.
(c) Excavations of areas wider than 1.5 m and 10 sq m in area and depth less than 30 cm are quantified in sq m and termed as surface excavation and different from excavation over area or excavation of trenches. The latter two are quantified in cu m.
(d) Excavation quantities shall be separately done foi different types of soils, viz.
(i) Loose/soji Soil : Any soil that yields to the ordinary application of pick & shovel or ordinary digging implements, e.g. organic soil (loam), sand, silt etc.
(ii) Harddense Soil : Soil which requires close application of pick or jumpers or scarifiers to loosen, e.g. stiff clay, gravel etc.
(iii) Mud : A mixture of soil and water in fluid or weak solid state.
(iv) So8 Disintegrated Rock (not requiring blasting) : Split with crow bars, e.g. laterite, hard conglomerate etc.
(v) Hard Rock (requiring blasting).
(vi) Hard Rock (blasting prohibited) : Prohibited for any reason, e.g. safety etc. and excavati6n has to be carried out by chiselling, wedging etc.
To identify the nature of soil, details are taken from trial pits or bore holes, which are normally made to assess safe bearing capacity of soil.
(e) Lead and Lift
Lead: Horizontal distance through which the excavated soil is to be conveyed in order to be dumped or filled.
Gff - Vertical distance through which the excavated soil is to be conveyed in order to be dumped oa filled.
The quantities of earth that require additiaul lead and lift should be separately prepxed. Noqnally, initial lead and liftace taken as 10 m and 2 m. respectively Every 10 m or part thereof over 30 m will constitute an additional lead.
Similarly, every 1 m or part thereof over the initial lift of 2 m will constitute an additional lift.
(f) Filling in the sides of trenches with excavated earth is not separately quantified unless done with soil other than excavated earth.
(g) Earthwork in plinth filling is calculated as a separate item (even if done with the excavated soil).
I
(h) Extra earth required or such filling is brought from outside and surplus earth, if any is carted away. (After using for filling low lying areas in order to raise the general level of ground etc.)
Example 6.1
The trench to be excavated is shown in Figure 6.1. Length of the trench is 50 m. Earth is to be deposited 25 m away from the site of excavation.
Figure 6.1 : Excavated Trench
t Solution Quantification is as follows :
1 I
6.2.2 Earthwork for Canals and Roads Depending upon the terrain, alignment of canal or road and the proposed formationlbed level, the nature of earthwork will be either in cutting or in filling or partially in cutting and filling. Filling is called banking or embankment. Principles of conveying earth (lead and lift) and earthwork in different types of soils are the same as for earthwork excavation in foundation as described in Section 6.2.1.
50 cu m
125 cu m
175 cu m
50 cu m 1
25 cu m
In order that the excavated or the banked sections remain without collapsing, side slopes depending upon the nature of the soil (angle of repose) are provided. Therefore, most cross-sections (except in rocks) are trapezoidal in geometry.
The quantity of excavation or banking can be computed by using various mensuration techniques out of which three are given below :
(a) Mid-section Area Method,
50 x 1.0 x 1.0
50 x 1.0 x 2.5
125 + 50
50 x 1.0 x 1.0
50 x 1.0 x 0.5
a.
b.
c.
d.
e.
(b) Mean Sectional Area Method or Trapezoidal Method, and
(c) Prismoidal Method.
Earthwork in Sandy Soil
Earthwork in Hard Stiff Clay
Extra for 2 leads
Extra for 1 lift
Extra for 2 lifts
Note : Use of tabular statement is advocated for calculation of earthwork using any of the three methods.
Examples with Diierent Types of Stmcture
Mid-section Area Method In this method, the mean depth is to be calculated first by averaging the depth of two consecutive sections, From the mean depth, the area of the mid-section is to be worked out and the volume of earthwork to be computed by multiplying the area of mid-section by the distance between the two original sections.
Volume of earthwork = A, x L
where. A, = area of mid-section, and
L = length of distance between two consecutive sections.
Referring to Figure 6.2,
(dl + d2) we get, dm = 2
2 A, = Area of mid-section + area of two sides ,= Bd, + Sd,
Therefore, V - (Bd, + sdm2) x L
Mean Sectional Area Method or Trapezoidal Method
This method is based on the assumption that the mid area of a pyramid is half the average area of the ends and the end sections are in parallel planes. If A1 andA2 are the areas of the ends and L is the length between two sections, the volume of the prismoid is given by
( A r + A 2 ) or V = L x A ,
where, A, = mid-sectional area.
Considering a number of consecutive sections, having areas A l , A2, . . . , A,-1, A,. L
Therefor'e, the total volume of earthwork = 2 (Al + Z42 + U3 + . . . + + A 3
Prismoidal Method ,
From mensuration, volume of a prism having end faces that are'in parallel planes
where A1 and 42 are the cross-sectional areas at the ends of a portion of embankment or cutting of length L and A , is the mid-sectional area.
Example 6.2
Calculate the quantity of earthwork in theireach of a road 300 m long over an uniform ground. The heights of formation at two sections at the beginning and end of the reach respectively are 1.5 m and 2.2 m. The formation width is 12 m and side slopes are 2H : 1V. Assume that there is no transverse slope.
Solution
Here, we have B = 12 m, S = 2, L = 300 m, d l = 1.5 m, d2= 2.2 m
Mid-section Area Method
(d l + d2) - (1.5 + 2.2) Mean depth, d = 2 - 2 = 1.85 m
Quantity, V = (Bd, + sdm2) x L = ( 1 2 x 1.85) + ( 2 x 1 .852) x 300
= 29.045 x 300 = 8713.5 cum. c
Mean Sectional Area Method
Al = Sectional area at one end = (Bd, + sd12) = (12 x 1.5) + (2 x 1.5~) = 22.5 sq m.
A2 = Sectional area at other end = (Bd2 + sd22) = (12 x 2.2) + (2 x 2.22) = 36.08 sq m.
(22.5 + 36.08) Mean sectional area =
2 = 29.29 sq m.
. Quantity = Mean sectional area x Length = 29.29 x 300 = 8787 cu m.
Prismoidal Method d
~uantity. V = ( ) x (Al + A2 + 41,)
Here, A, = 22.5 sq m, A2 = 36.08 sq m.
Quantity = (?Ix (22.5 + 36.08 + 4 x 29.045) = - x 174.76= 8738 cu m. (3:)
Example 6.3
The profile of a road alignment in reach ABC is shown in Figure 6.3. The formation (top) level of the road at A is 30.0. In order to save on earthwork quantity, following two gradients are proposed :
(a) 1 in 250 rising between A and B
(b) 1 in 400 falling between B and C
Top width of formation is 10 m, with side slopes of 2H : 1 V. Estimate the quantities of earthwork in cutting and filling.
Solution
Here, s = 2.0 and B = 10 m. Referring to Figure 6.3, from chainage 1600 to 1800, there is a changeover from filling to cutting.
Fsgure 6.3 : Earthwork for Road Fomtion
Changeover point : 140 x 200 = 175 m. (1.60)
Similarly, there is a change over from chainage 1800 to 2000.
Examples with Dierent Types of St~c(ure
Changeover point : @ x 200 = 47 m. [ 0.85)
Estimating Ouantification is as follows
Ear
Thus, we get,
quantity of cutting = 75.44 cu m, and
quantity of filling = 22547.07 cu m.
.thwork for Canals 1 The bed level of the canal is generally decided by the command level required and the bed slope. Depending upon the terrain, alignment of canal and the proposed bed level, the nature of earthwork will be either fully in cutting or fully in filling or partially in cutting and filling.
Fully in Cutting
-*7 Figure 6.4 : Fully in Cutting
Referring to Figure 6.4,
Quantity of earthwork,
Q = ( ~ d + sd2) x Length
Note : In this case, the depth d may be equal to or more than the required designed depth.
Partly in Excavation and Partly in Embankment
Figure 6 5 : Partly in Cutting and Partly in Filling
Referring to Figure 6.5,
Quantity of earthwork in cutting,
Ql = ( ~ d + sd2) x Length
Quantity of earthwork in filling,
Q2 = (blh + slh2) x Length + (b2h + s2h2) x Length
As far as possible, the excavated earth if found suitable is used for fillling to achieve economy. In some cases, this may involve conveyance of excavated earth over long distances, making it uneconomical. Under such circumstances, earth for filling is taken from borrow pits.
Fully in Banking
Figure 6.6 : Fully in Banking
Referring to Figure 6.6, quantity of earthwork may be calculated considering the whole area as a solid block and deducting the channel portion.
Quantity = (Whole sectional area - Sectional area of channel) x Length
= [(BD + s ~ D ~ ) - (bH + SIH~) ] x Length
where B = b+bl+b2+ZIi lH and D = d + H
Here, it may be seen that the bed of the canal is above G.L. Such a case in practice is always avoided as any fresh earth is likely to be eroded away. But wherever it is necessary, this is provided and the bed may be lined.
Example 6.3
Calculate the quantity of earthwork required for an irrigation channel whose longitudinal section (LS) and typical cross-section are given in Figure 6.7. The bet width of the channel is 5 m and top width of the banks are 3 m for the left bank and 1.5 m for the right hank. Side slopes of excavation is 1 : 1 and of bank 1; : 1.
Examples with Dierent Types of Structure
! Height of banks from bed is 2.55 m throughout. The longitudinal slope of channel 1 is 1 in 5000. t
figure 6 7 (a) : Cross-section of Channel
Solution
Quantity of earthwork in chailnel
Quantity of excess earthwork required for = 1651.30m3 banks to be taken from borrow pits - Grand total = 6559.25 m3
Table 6.1 : Earthwork Calculation Using Mid-Section Area Method for the Irrigation Channel in Example 6.3
1 I EXCAVATION I BANIUNG . 1 I
I Total 1 4375.80 1 1 4375.80 1 1651.30 1
Figom. 6.7 (b) : Loagitudinal Section Showing Bed Level, G w n d Level &Top of Bund Level
Abstract
(Refer Table 6.1) Quantity Rate Per Amount
Earthwork in channels and banks 6559.25 m3 Rs.200 100 m3 Rs. 13118.50
Permanent Land
As soon as the alignment for the proposed road or canal is finalised, area required to be acquired for the project is also to be worked out along with the quantity of earthwork.
For a typical section shown in Figure 6.5,
Area of Permanent Land = [b + bl + b2 + 2S1H + 2S2D + extra width] x Length
The extent of extra width required is dictated by the
(a) Nature of the soil - easily erodable or hard
(b) Height of banking - higher the height of banking more the area
(c) Value of land on either side
(dl Whether the bank slopes a e protected or not.
If very steep banks are provided as for approaches to over-bridge and high embankments, or if the soil used for banking is oor against erosion, the banks are S to be protected. As the depth of banking varies rom section to section, the width of permanent land also will vary at different sections. In practice, uniform land width, equal to the maximum width in a particular reach is proposed for acquisition.
For the above problem,
Area of permanent land
= Width of permanent land x Length
= [b + b1 + b2 + 2S1 H + 2S2h + extra width] x Length
Width of permanent land, W = W1 + 2S2h
We know that W1 = b + bl + b2 + 2S1K + extra width
Therefore, W = 19.15 + 2S2h
From earlier tabular statement given in Table 6.1, Col.10 (h = H - 6) is multiplied b,y 2S2 as shown in Table 6.2. I
- Examples with Different
Types of Structure
Estimating Table 6.2
Maximum width in the reach between chainage 10 and 20 = 24.31 m
Area of permanent land = 24.31 x 500 = 12155 sq m or 1.2155 Hectares.
6.2.3 Masonry Masonry is a composite material used in construction of walls, both load bearing and non-load bearing. The composite consists of masonry unit, which can be brick, block (hollow or solid), rubble, dressed stone etc. laid in mortar. The mortar can be mud mortar, lime mortar, cement-sand mortar or any other combination mortar. Masonry is quantified by volume (in cu m). Sometimes, partition walls constructed of half brick thick masonry or brick laid on edge are quantified on area basis. Masonry in foundation and basement is distinguished as a separate item as compared to masonry in superstructure. The main reason for this is that construction of masonry in superstructure requires scaffolding. In addition, openings for doors, windows, ventilators etc. have to be left.
Deduction for Openings, Bearing etc. in Masonry
The quantity of masonry 1s first calculated considering it as a solid mass. Openings are subsequently deducted. There are certain conventions for such deductions. The principle is that very small openings and openings with unconventional shape are not deducted in order to compensate the additional involved in leaving such openings during construction.
No deduction is made for the following :
Opening each up to 1000 sq cm or 0.1 sq m (1 sq ft).
End of beams, purlins, posts, rafters etc. up to 500 sq cm or 0.05 sq m (72 sq inches) in section.
Bed plate, waIl plate, bearing of chajjas and the like up to 10 cm (4") depth.
Bearings of floor and roof slabs are de 'ucted from masonry. For other openings deductions are made in the following manner:
Rectangular Openings
Full deduction is made as under :
Deduct (b x h x thickness of wall)
where, b is the width of opening and h is the height of opening.
Segmental Arch Openings
For small segmental arches, deduction is made for the rectangular portion only up to the springing point. The segmental portion is considered as solid.
For large segmental arch openings, deductions is made for the whole openings - the rectangular portion, as well as the segmental portion.
The area of segmental portion is obtained approximately by taking 2/3 of the span x rise x thickness of wall.
Examples with Diirent Types of Straeture
Thus, the total deduction = (1 x h) + - x 1 x r x thickness of wall [ i: I1 where, I is the span of the arch/width of opening.'
Semi-circular Arch Openings
The area of semicircular portion = $2
The area of the semi-circular portion may also be obtained approximately by 314 x span x rise.
Thus, the total deduction = x thickness of wall
Elliptical arches may be considered as semi-circular and may be dealt in same manner. For large arches, the actual area of opening shall be calculated correctly by mensuration formulae, and deducted.
Masonry Arches
Openings in masonry are spanned by flat lintels or arches. Arches can be constructed of any suitable material. Masonry can also be used to construct arches. There are different types of masonry arches. Masonry arches are quantified as separate items distinct from normal masonry walls in view of the fact that arch masonry requires special masonry units and skills for construction.
For any arch,
Quantity = mean length x thickness of masonry wall x ,depth of arch
For example, in the case of a culvert with masonry arch,
Quantity = the width of the culvert from face to face x mean length x depth of arch
Case I : Segmental Arch with Span and Angle Given
Figure 6.8 : Arch Masonry
The arch is shown in Figure 6.8.
Here, s = span of arch
0 = angle subtended at the centre
r = radius
rm = meanradius
lm = mean length of arch
t = thickness of arch
h = breadth of wall
Therefore, the quantity of arch masonry = 1, x b x t
Estimating Normally, arches over doors and windows are segmental arches with 60° at the centre. Thus, in this case,
r = s (in case of equilateral triangle) and
Therefore, quantity, q = 3
Case 2 : Segmental Arch with span and Rise Given
%ore 6.9 : Arch Mmary - SepmenfBI Arch
Referring to Figure 6.9,
Here, r = radius of intrados
1 = length of intrados
To find radius r,
To find length 1,
From menauration, 1 = and 3
Therefore, quantity q = x b x t
For a semi-circular arch,
7t Im = Wm = - X ( S + t )
2
Case 3: Flat Arches
Flat arches may be used over openings for doors and windows having small spans. Flat arch csually subtends an angle of 60' at the centre and forms an equilateral triangle.
Figure 6.10 : Arch Masonry -Flat Arch
Referring to Figure 6.10, inclined thickness of arch at the s ~ r i n ~ i n ~ mint,
t l=~-i- - - 1.15 t sin 60' 0.866
1 Mean length of arch, 1, = s + - 2
Quantity, q = 1, x b x t
I Serni7elliptical arches may be considered as segmental arches
2 h3 Area of the segmental portion = -sh + -
3 2s
2 h3 Quantity, q = -sh 4 - x thickness of wall
3 2s
But in practice,
2 Ouantitv = rsh x thickness of wall
I h3 . (the term - is small and may be omitted)
2s I Example 6.4
The arch of a culvert, subtends an angle of 135' at the centre. The span of the arch I is 6.00 m and the thickness of the arch is 60 cm. The length of arch from face to
face is 9 m. Calculate the quantity of arch masonry work.
Solution S 1
Radius, r = - + - 2 sm (e!)
where,
s = span = 6.00 m
0 = angle subtended at the centre = 135'
r = =+ 2 sin (13.h)
= 4.08 m
0 6 Mean radius, r, = 4.08 + - = 4.38 m.
2
0 135 Mean length, 1, = 2nr, x- = 2n x4.38 x- = 10.32 m. 3 60' 360
I Length of arch from face to face, L = 9 m.
Quantity of arch masonry, q = L x 1, x t = 9 x 10.32 x 0.6 = 55.73 cu m.
6.2.4 Plain and Reinforced Cement Concrete While cast-in-place plain cwcrete is laid either with or without formwork, most reinforced cement concrete construction involve some form of centering and shuttering, which in short is called the formwork. Moulds are used largely in prefabricated construction.
Quantification of any reinforced concrete work can be split in to three components. They are as follows :
(a) Formwork,
(b) Reinforcing Grill Work, and
(c) Concrete.
Formwork
The purpose of formwork is to mould the plastic concrete into specific shapes in specific positions as required by the structural design. Centering or falsework supports the formwork that provides a level area or an area of the required shape to receive concrete. Centering consists of a framework of verticals, horizontals and inclined supports, either of timber or mild steel sections or pipes. Concrete receiving formwork is made of wooden planks, mild steel plates or plywood sheets. The surface is oiled and the gaps are properly filled up. Reinforcement grills are placed on the surface over suitably placed cover blocks and concrete is poured and compacted. After the concrete has gained sufficient strength, the formwork is struck off. Falsework and formwork are then inspected for damage, if any, repaired, if necessary and stored carefully for future reuse.
Formwork is quantified in area basis. It is the inner surface area of the box (formwork:) that receives the concrete.
Slab
Formwork area = Bottom area + Area of sides all around
Bottom area is generally the clear size of rooms or between beams and area retaining the sides is arrived at as Perimeter x Depth of slab.
Beam
Formwork area = Clear span of the beam x [width of beam + (2 x rib depth of beam)] + (area of end faces given by 2 x width x rib depth).
Lintel with Sunshude
Lintels over openings in the external walls are usually cast integrally (monolithically) with sunshades, as shown below
Formwork area = Bottom surface (It should be seen whether the sunshade has bearing over the entire length of lintel or out of opening only) + sides and ends of lintel + 2 x sides of sunshade.
(The front surface of sunshade compensates for the side of lintel at the junction).
Column
Formwork area = Perimeter x height of column
Normally, moulds of the standard size of columns of a particular height will be used.
Reinforcing Grill Work
Reinforcing grills are fabricated out of mostly mild steel rods of suitable (design) grade and estimated as quintals (100 kg) or tomes. There are following two methods to estimate the quantity of reinforcing grill :
Approximate Method
ExactMethod
Approximate Method
In the approximate method, the quantity of concrete is first arrived at. Based on tables aviulable in handbooks, giving the reinfoicement quantity per cubic metre of concrete, for different structural members, the quantity of reinforcing grills is arrived at. In the absence of any handbook, the estimator can use hislher own previous experience.
The estimator is expected to use his judgement to select and use a particular value nr;ih;n rho r-n'no n ; r r o n ;n TohIo A 1
Table 6.3 : Quantity of Reinforcing Grill Examples with DitPemt Types of Structure
Exact Method
The exact method involves actual quantification of the reinforcement grills. This presupposes the availability of detailed design of the structural concrete members and detailed drawings.
The designer provides the structural drawings along with detailing of reinforcement as per design requirements and according to the provisions of the codes oi-' practice. Normally, this is made available in the form of a Bar Bending Schedule (BBS) using which the estimxtor prepares a Bar Requirement Schedule (BRS). The total quantity of reinforcer~~ent is arrived at by multiplying the length of bars with its corresponding linear weight. Table 6.4 gives the linear weight of different diameters of commonly available reinforcing steel.
Table 6.4 : Linear Weight of Different Diameters
Note : The linear weight can also be calculated using unit weight of steel as 7850 kg/m3. To calculate the linear weight per metre length (kglm) of any diameter rbd, multiply the cross-sectional area expressed in sq m by 7850.
Example 6.5 : Slab
A RCC two way slab 150 mm thick covers a hall of clear size 3.6 m x 7.0 m. The supporting walls all around are 230 mm thick.
Reinforcement along short direction - 10 mm diameter HYSD bars @ 180 mm clc
Reinforcement along long direction - 8 mm diameter HYSD bars @ 150 mm clc.
Altemate bars are c r e e d up at 0.2L (115 span) from face of support. Clear cover to reinforcement and end cover are 20 mm and 25 mm, respectively. Quantify the reinforcing steel for the slab. The structural details are given in Figure 6.1 1.
The bar bending schedule is as follows :
Figure 6.11 : Reinforcement Details (Two-way Slab)
Solution
BAR REQUIREMENT SCHEDULE
Location Mark Type/ Total Length Total Linear Total I 1 m / No. ( (m) I Length I Weight ( Weight ( SLAB - I ' I lOHYSD ( 40 1 4.24 / 169.6m 1 0.620 1 105.2kp I Short Span
SLAB - 'b' 8 HYSI) 2.S 7.63 190.75 m 0.395 75.3 kg Long Span
At Cranked 'c' 8 HYSD 6 4.16 24.96 m 0.395 9.9 kg up portion --- At Cranked 'd' 8 HYSL) 4 7.56 30.24 m 0.395 11.9 kg up portioi~
Total 202 kg
1 Add 5% (approximate) for Wastage and Laps (rounded up) 1 9.70 1 1 TOTAL 1 2 12.0 kg 1
Example 6.6 : Beam
The main reinforcement of a RCC beam of size 30 cm x 60 cm consists of 3 nos. of 20 mm dia HYSD bars at bottom out of which one middle bar is cranked up over support. The clear span of beam is 5.0 m with 25 cm bearing on each end. The hanger (top) bars are 2 nos. of 16 nun dia. HYSD. The crank point is 1/5 span (clear). There are 5 such beams.
Clear coier to main reinforcement - 25 mm
Clear end cover to reinforcement - 2d of bar (d - diameter of bar)
Shear reinforcement - 8 rnm dia. HYSD - 2 legged stirrups @ 15 cm c/c at end quarter spans and @ 25 cm c/c in the middle half of the span. Estimate the quantity of steel per cu m of concrete. Figure 6.12 shows the structural details.
Figure 6.12 : Reinforcement Details (Beam)
The bar bending schedule is as tollows : Examples with Different Typs of Stmcture
Location & Mark me/ Number Length 1 sire I orsets I (em) 1 2~E:;:t (
Solution
BEAM - Bottom - Straight (Mark p)
BEAM -Bottom - Cranked (Mark q)
BEAM - Top - (Mark r)
BEAM - Stirrups (Mark s)
/ BEAM-cranked / 20 HYSD I 5 I 6.24 I 31.20 m ( 2.46 ) 76.8 kg ) marked 'q'
20 HYSD
20 HYSD
16 HYSD
8 HYSD
BAR REQUIREMENT SCHEDULE
1 Total / 411.12kg 1
5
5
5
5
Total Weight
142.7 kg
At top portion marked 'r '
Stimps nmked 's'
Linear Weight
2.46
Concrete
In quantifying concrete, different types of concrete, viz. lime concrete, cement concrete, concrete using stone jelly, concrete using brick jelly etc. shall be quantified as separate items. Similarly, concrete to be made with different proportions of materials, methods of mixing, curing etc, shall be quantified as separate items. Further, concrete used in different component, viz. foundation, roof, wall, columns may have to be classified and measured under separate heads.
580.00
623.90
572.40
Location
BEAM -bottom straight marked 'p'
16 HYSD
8 HYSD
Add 5% (approximate) for Wastage and Laps (rounded up)
TOTAL
Concrete shall be quantified in cu m and measurements of length, breadth and height or thickness shall be takeII' to the nearest lcm except that the thickness of slabs, partitions, posts, beams and the like shall be measured to the nearest 0.5 cm. No deduction shall be made for openings up to 0.1 sq m.
2
1
2
Length (m)
5.80
20.80
432.0 kg
Fair finish to the exposed surface of ccncrete or hacking or roughening surfaces of concrete shall be included in the description and the thickness of finishing shall not be measured with the concrete. Special finishes except in precast concrete shall be measured separately in sq m. In RCC work, the volume occupied by reinforcement shall not be deducted from the measured concrete volume.
176.80 29
Total Length
58.00 m
Type/ Size
20HYSD
10
145
6.2.5 Structural Steel
Total No.
10
There are a large number of major buildings where steel is extensively used. They are
5.72
1.768
Industrial and very tall residential buildings,
Large span structures, e.g. suspension bridges, railway bridges etc.,
57.20m
256.40 m
Towers and water tanks, and
- Large span roofs for auditoria and warehouses.
1.58
0.395
It is fairly easy to estimate the quantity of structural steel due to the following reasons : (a) Only standard sections of limited sizes are manufactured and their linear weiiht
is known.
90.4 kg
101.3 kg
(b) InvaAably, the structural drawings are made to a high degree of accuracy and detail, enabling the estimator to directly work out the quantity of structural steel used.
Standard strucb~ral steel sectlons consist of angles (equal and unequal), channels, beams, plates, flats and ropes. Steel structures are fabricated by cutting and shaping these sections and joining them by welding or bolting or riveting. This process is known as "wrought and put up" in estimating parlance. The estimator arrives at the total weight (tonnage) of steel used. This is done by just multiplying the length of the structural steel sections by their respective linear weights as done in the case of reinforcing steel. In case of plates - (Area of plates used x weight/unit area) will give the quantity.
In measuring jointed frame, full length of structural sections out of which site lengths are fabricated shall be considered. Sirmlarly, in case of plates, e.g. gusset plates, full size of plates (rectangular or square), out of which odd shapes are cut, shall be measured.
Normally, all structural steel is painted with a coat of red oxide primer - number of coats depending upon the degree of protection required. if needed one or two coats of enamel/ aluminium paint is given as further protection from corrosion. Painting is measured on area (m2) bass.
6.2.6 Woodwork Woodwork is required for doors, windows, ventilators, cupboards, partition etc. While estimating woodwork items, the type is specified as to whether it is panelled, glazed, partly panelled and partly glazed, louvered, venetian, etc. and the area is measured in (m2). It may not be required to take out the quantities of the woodwork components on volume basis - if standardised doors, windows are adopted for which standard data is available. In major buildings, large number of doors and windows of standard size and design are used. In such cases, doors 'uld windows can be quantified simply as "so many" numbers. If standardised doors/windows are not adopted, it is necessary to arrive at the quantities of woodwork components individually -as shown in the example below for a door with panelled shutter.
In measuring framed timber, length of tenons, scrafs shall be added to site length of framed timber.
Referring to Figure 6.13, Table 6.5 gives quantification for the woodwork of a door with panelled shutter.
Table 6.5
Item Particulars No. L B D Quantity No.
1. Door Frame Verticals out of 2m-3m long 2 2.10 0.10 0.075 0.3 15 scantlings
Horizontals out of 2m-3m long I 1.20 0.10 0.075 0.0090
c. Lock Rail & Bottom Rail
(*) Height = 2.10 - 0.075 + 0.02 - 0.025 (floor covering)
[1.20- (0 .075~ 2)] (+) Width =
2 + 0.01 5 + 0.02
2.02 - (0.075 + 0.1 + 0. I) (#) Height =
2 + (2 x 0.012) = 0.897 (Shutter inset)
($) Width = 0.56 - (2 x 0.075) + (2 x 0.012) = 0.434
Items to be Priced
(a) Teak wood or country wood scantlings 2 m to 3 m long
Door Frame 0.0315 mi Shutters 0.0230 m
Total 0.0545 mr
(b) Teak wood or country wood scantlings below 2 m long
Door Frame 0.0090 m3 Shutters 0.0032 m" Shutters 0.0085 m"
Total 0.0207 m"
(c) Teak woodcountry woodplywood planks more than 30 cm wide for panels - 0.0389 m3.
(d) Teak wood or country wood reepers for fixing panels - 21.30 m. \
blmens~ons in m n .
Figure 6.13 : Panelled Door
Fittings
Fittings made of anodised aluminium or steel
100 mm - Butt hinges - 6 nos.
250 rnm - Aldrop bolt - 1 no.
100 mm - Tower bolt - 1 no.
150 mm - Tower bolt - 1 no.
Handle - 1 no. Labour
Making frames, fixing, holdfasts complete
= 0.031 5 + 0.0090 = 0.0405 m3
Making Shutters with moulded panels and fixing fittings complete
Quantities are similarly taken for Glazed Panels or Louvers.
The above quantities are multiplied by corresponding costs and added up to get at the total cost of one door with panelled shutters.
6.2.7 Painting 'aintingtwhite washing/colour washing of walls/ceiling including sunshades etc. shall be leasured in sq m. Area can be suitably taken out from the corresponding plastering
Ekamples with Different Types of Structure
Estimating quantity that has been already calculated. Deduction shall be dealt in the same way as for plastering. Preparation of surface as cleaning, brooming, scraping, etc. shall be included in the item.
Paintings for doors/windows or any wood work follow certain guidelines as it is rather time consuming to ascertain areas of small cuts, grooves etc. Coefficients are used to multiply the over all area of doors/windows/ventilators in order to arrive at the actual area of painting both faces. Table 6.6 gives the painting coefficients that are normally adopted.
Table 6.6 : Painting Coefficients
I I
1. I Batten doors and windows 1 - 2.25
1 2. 1 Panel doors and windows 1 2.25 1 I 1 3. 1 Panel and.venetian doors 1 3.25 1 ) 4. / Panel and venetian doors with glazed top . 1 3.00' I 1 5, ] Iron barred doors 1 1.50 1 1 6 . 1 Iron barred doors with batten and sheet 1 3.75 1
1 10. 1 Venetian windows with glazed top and iron b k s 1 4.50 1 I 11. I Venetian windows with iron bars and glass shutter& / 5.00 /
1
1 12, 1 Glazed windows with iron bars ,I 2.00 1
- -- - -
Batten windows with iron bars . Venetian windows
Venetian windows with iron bars
-
7.
8.
/ 13. ' Glazed windows without iron bars I ' 1.50 1
2.75
3.50
4.00 I 9.
Coal tarring and varnishing etc. shall be measured similar to painting.
14.
15. -
Painting Corrugation
Corrugated surface shall be measured for painting as flat and not girthed in sq m and the quantity so measured shall be increased by the following percentages and the resultant shall be included in general areas.
(a) Corrugated steel sheets - 14%
(b) Corrugated asbestos cement sheets - 20%
- - -
Glazed shutters (measured over shutters only)
Welded mesh
(c) Semi-corrugated asbestos cement sheets - 10%
1.00
0.75
Painting Structural Sections
Painting area for any structural sections is taken as follows :
Painting area = 'iength of the section x perimeter.
The perimeter is calculated as given below :
Angle = 4 times the size of the leg. (In case of unequal angles, sum of 2 times of each leg)
Channel = 2 times the size of the web + 4 times the size of the flange.
Beam = 2 times the size of the web + 4 times the size of the flange.
6.2.8 Plumbi~ig and other Lump-sum Items Plumbing an4 other Services
While preparing estimates of civil works, estimates relating to structural and finishing items are prepared first with suitable lump-sum provisions towards services like plumbing, electrification and other non-structural items. Detailed estimates are subsequently prepared after consulting the concerned engineers, who
specialise in services. This is due to the fact that services in buildings especially large ones, require elaborate planning and detailing and is inter-disciplinary in nature, involving a fair amount of knowledge of electrical and mechanical engineering aspects.
Other Lump-sum Items
Suitable percentages of the cost of civil works are marked out as lump-sum for items for many other items. These percentages are approximate and similar detailed estimates are prepared if needed.
Discretion should be used in arriving at the percentages depending upon the individual cases.
Usually adopted percentages are given below :
Internal & external electrification = 10%
Internal & external water supply = 7.5% and sanitary arrangements
Architectural features = 1%
Lawns and roads = 2%
Fluctuation in marked rates of labour = 10% and material
Unforeseen items = 2%
Petty supervision = 3%
In case of hospitals, where provision for electrification and water supply & sanitary arrangements will be on large scale, the L.S amounts should be higher by 12.5% and 10% respectively.
In large projects, adequate provision towards purchase of special equipment such as concrete mixers, vibrators, mortar mills, lorries etc. has to be made. Also provision for construction of temporary quarters for the construction of staffs, godowns should be made.
6.3 RATE ANALYSIS
Rate analysis or pricing refers to the process of arriving at the price per unit item of work. An item of work is a part of the total construction project which can be distinguished from all other parts, and for which costs can be determined separately, e.g. plastering as distinct from reinforcing grill work in RCC.
The principles involved in pricing any item of work was dealt with in detail in Unit 5. The four elements of pricing any item of work are as follows :
(a) Materials (c) Equipment
(b) Labour (d) Overheads An item of work may involve some or all these four elements. A contractor while quoting his bid to any project adds up the profit margin to the rate arrived at considering the above four elements.
Consider the following two examples with the pricing data :
(A) Concrete mix 1 : 2 : 4 (by volume) using 20 mm size broken stone in cement mortar for reinforced concrete works - lcu m.
Materials
Labour
Equipment
Broken stone 20 mm size Sand Cement
Mason - Semi-skilled Mazdoor I Mazdoor II
Mixer Machine
0.90 cum. 0.45 cum. 324 kg
0.35 number 2.12 number 3.53 number
30 minutes
Estimating (B) Random rubble masonry in cement-sand mortar 1 : 6 - 10 cu m
Bond stone
Cement
Labour Mason - Skilled Mason - Semi-skilled Mazdoor I Mazdoor II
10 c u m 1 cu m 3.4 cu m 816 kg
7.10 number 10.6 number 14.1 number 14.6 number
On a cursory observation of the pricing data relating to the above two items, the following points can be noted :
(a) The items to be priced are described with reference to their salient specifications.
(b) The unit of measurement, i.e. 1 cu m or 10 cu m as the case may be is mentioned.
(c) The quantities of materialAabourIequipment time are given.
(d) The quantities of material/labour/equipment time are not necessarily in full numbers.
(e) Overheads are not mentioned.
(f) Profit margin is not specified.
(g) While in Example (A), three elements (materials, labour and equipment) are involved, in Example (B), only two elements (materials and labour) are involved.
It is to be further noted that the tabulated quantity of materials required for the unit of work will remain unchanged whoever carries out the work. But the labour/equipment requirements can change either way (increase or decrease) depending upon the productivity levels. Similarly, overheads and profit margins also can be significantly different with different contractors.
The data books available with the Central Public Works Department and Public Works Departments of different States give average labour/equipment requirements, which can be used for preparation of estimates. The equipment usage can be arrived at based on the principles given in Unit 5 . Arriving at labour requirements for a given unit of work is a difficult exercise. A large number of factors are involved in deciding on the crew size and its productivity for each and every item of work. Observations taken from a large number of sites are averaged out and used as labour data for rate analysis. Major contractors have their own data based on the productivity of their own labour force or on their subc'ontractors' rates. These are made available in a "Standard Data Book" which includes a large number of items of work covering most Civil Engineering works.
The exercise of rate analysis is to associate costs to the listed quantum of materials, labour, equipment for different items of work. While associating costs to materials, the estimator has to work out the cost of various materials to be used at site of work.
Cost of materials at site = Cost of materials at source + Transportation cost
In a particular locality, though the cost of a material may be same at the source of availability, the distance of different work sites in the locality from that source may be different and hence, the landed cost per unit of material may be different. General guidelines are given below for working out transportation costs of materials. In working out transportation costs, the principle of "Erne is money" is used.
The following data are needed :
(a) Mode of conveyance - lorry, cart etc.
(Most of the urban work sites are connected by good, black topped roads and conveyance of building materials is generally by lorries. In rural areas, where road conditions are poor, bullock carts are used.)
(b) Capacity of conveying vehicle for different materials.
(c) . Distance of site from source of availability of material.
I (e) Speed linlits, if any. Examples with Dierent Types d S t ~ c t ~ u - e
(If different speed limits are specified as in cities, then split up distances outside and inside city limits are to be noted.)
(f) Hire charges per day of the conveying vehicle.
(g) Loading and unloading time of material.
(h) Loading and unloading charges of material per trip. The following example will illustrate the procedure nf working out "cost of material at site".
Example 6.7 The site is located in an urban area. Material to be transported is river sand. Cost at source is Rs. 15 per cu m.
Data given is as follows :
Hire charges for lorry is Rs. 1440 per day of 8 hours.
The capacity of lorry is 4 cu m for sand.
Permissible speed limits : 45 kmlh outside the town and 40 kmlh inside the town.
Loading and unloading time - 20 minutes each.
Loading and unloading charges - Rs. 50 each operation.
Lead - 40 km out of which 30 km is outside town and 10 km is inside town
Solution It is assumed that the lorry makes 2 trips every time to bring 1 lorry load of material to the site (1 empty trip to the source and 1 loaded trip to the site).
1440 Cost per minute use of lorry = - = Rs. 3.00
(8 x 60)
Time spent for bringing 1 lorry load of sand to site
= loading time at source + travel time to site with load + unloading time at site + travel time to source without load
= 150 min.
Cost of conveyance of 1 lorry load of sand to site
= Cost of lorry use + Loading and unloading charges
= (150 x 3.00) + 50 + 50 = Rs. 550
Cost of conveyance of 1 cu m of sand
- Cost of 1 lorry load - .- - Capacity of lorry
550 = Rs. 137.50 - 4
Cost of 1 cu m of sand at site = Cost at source + Cost of conveyance
= 15 + 137.50 = 152.50
i . .: $, \%%at is i?c palntir:g urea fore srrrck of ~t~uct:.irr,l s t ~ ~ l CC,IIS~S!~ :~K 0f,1::'~('9. YSh.Is 49(f (I:mgc width 23'3 mm) and 3 nos. of !SA 150 'F, 1 1.5 x i 2 '!
EPtimating 6.4 EXAMPLES
In the earlier sections, you were given an insight into the nuances of quantification involved in individual items of work. In the following sections, you will be exposed to quantification of actual structures. Different types of structures are chosen so that, by themselves, these together cover a fairly wide range of civil engineering projects. Nevertheless, these examples alone cannot be considered exhaustive. This is in view of the limitations of this unit and you are advised to refer to good textbooks suggested in the section "Further Reading" given at the end of this block.
The first example given below is complete by itself in the sense that it includes all the three components, viz. quantity estimate, rate analysis and abstract estimate. In other examples, only quantity estimates are furnished. The reason is that while quantities of items of work remain unaltered as long as the drawings/specifications are not altered, cost changes from site to site and time to time.
6.4.1 A Complete Example - A Compound Wall The example details below will give a complete picture as to how you should proceed in costing any work. The example chosen is simple with a small number of items of work. The type and nurnber of items of work may vary with other projects for which costing needs to be done, but essentially, the procedure will be identical to the one described below.
Example 6.8
Prepare a detailed estimate for the consm' tion of a compound wall. F
Fig- 6.14 : Compound (Boundary) Wall
Specifications
. The site is located in a town.
The foundation is with PCC 1 : 4 : 8 using 40 mm size broken stone jelly and 230 mm thick wall built with stock bricks - I sort of size 23 x 11.5 x 7.5 cm in cement-sand mortar 1 : 5.
The superstructure consists of 115 mrn thick wall built with stock bricks - I sort of size 23 x 11.5 x 7.5 cm size in cement-sand mortar 1 : 4, plastered on both faces using cement-sand mortar 1 : 5 - 12 mm thick, extending 20 crn below ground level; white wash with 3 coats up to G.L.
The super~tructure is strengthened with 230 x 230 rnm size pilasters at 2.4 m cic, built with stock bricks - I sort of size 23 x 11.5 x 7.5 cm in cement-sand mortar 1 : 4; plastered and whitewashed.
Total number of bays - 80 (2.4 m length each).
Solution QUANTITY ESTIMATE -per bay of 2.4 m length
RATE ANALYSIS
Item No.
1.
2.
3.
The following data are available from the records of the contractor's office :
Examples with Different T y p d S t m e t a n
Particulars
Earthwork excavation for foundation in sandynoose soil (including refilling the sides)
Concrete in foundation
Brickwork in foundation and pilaster
Foundation Pilaster
The contractor has his ownlorry for transportation. According to the costing data, the running cost per km is Rs. 5.00. Loading/unloading charges are Rs. 30 per each operation.
Rates of labow Earthwork ex'cavation :
. Open excavation - Rs. 70 per m3
Excavation for foundation in loose soil - Rs. 120 per m3 Refilling the sides - Rs. 10 per m3
0.421 m3
3.69 m2
0.05 1.17
0.25 7.38
8.850 m2
0.05 1.04
0.25 6.51
7.850 m2
Mazdoor I - Rs. 80 per day and Mazdoor I1 - Rs. 75 per day z 57,.
N
1
1
1 1
4.
5.
L
2.40
2.40
2.4 0.23
B
0.65
0.65
0.23 0.23
Brick wall - 115 mm thick
Plastering 12 mm thick
Pilaster - top - sides all round
115 mm thick wall -top - sides
1
1 1
1 1 x 2
6.
D
1.00
0.20
0.60 1.70
2.17
0.23 0.69
2.17 2.17
Quantity
1.560 m3
0.312 m3
0.331 m3 0.090 m3
White washing 3 coats
Pilaster -top - sides all round
115 mmthick wall - top - sides
-
0.23 -
0.115 -
1.70
- 1.70
- 1.70
1 1
1 1 x 2
0.23 0.69
2.17 2.17
0.23 -
0.115 -
- 1.50
- 1 .SO
Mason (skilled) - Rs. 130 per day Mason (semi-skilled) - Rs. 110 per day
Mortar mixing per m3 - 1/3 Mazdoor I1 = 7513 = Rs. 25 . t
Cost of Materials at Site
Cement
Conveyance cost per lorry load = 2 x 5 x 5 + 60 = Rs. 110.
[ l io) = Rs. 22 Conveyance cost per tonne = -
Cost at site = Rs. 3400 + Rs. 22 = Rs. 3422
Bricks
Conveyance cost per lorry load = 2 x 30 x 5 + 60 = Rs. 360.
Conveyance cost per 1000 nos. = (E] = Rs. 240
Cost at site = Rs. 1750 + Rs. 240 = Rs. 1990
Stone jelly
Cor.veyance cost per lorry load = 2 x 10 x 5 + 60 = Rs. 160.
[ = Rs. 40 Conveyance cost per m = -
Cost at site = Rs. 150 + Rs. 40 = Rs. 190
Sand
Conveyance cost per lorry load = 2 x 30 x 5 + 60 = Rs. 360.
(3:0]= Rs.90 Conveyance cost per m = -
Cost at site = Rs. 15 + Rs. 90 = Rs. 105
Shell lime
Conveyance cost per lorry load = 2 x 15 x 5 + 60 = Rs. 210.
Conveyance cost per m = (:O) - = R s 52.50
Cost at site = Rs. 900 + Rs. 52.50 = Rs. 952.50
Rate Analysis
Cement-sand mortar 1 : 4 - 1 m3
PCC 1 : 4 : 8 using 40 mm size broken stone - 1 m3
.
Quan'tity
l m3
360 kg
1 m3
Per
1 m3
1 tonrle
1 m3
Detaii
Sand for mortar
Cement
Mixing mortar
Amount (Rs.)
105.00
1231.92
25.00
Total
Rate (Rs.)
105
3422
25
Rs. 136 1.92
Cement-sand mortar 1 : 5 - 1 m' Examples with Different Types of Stmetare
Brickwork in cement-sand mortar 1 : 5 using I sort-stock bricks -' 100 cft
Quantity
1 m3
288 kg
1 m3
Brickwork in cement-sand mortar 1 : 4,
= 4436.80+ 0.708 x (1361.92- 1115.54)
= Rs. 461 1.24 per 100 cft or Rs. 1629.41 per m3
Brickwork in wall (1 15 mm) using stock bricks in cement-sand mortar 1 : 4 - 10 m2
Details
Sand for mortar
Cement
Mixing mortar
Plastering with cement-sand mortar 1 : 5 - 12 mm thick - 10 m2
Total
Quantity
1.15 m3
0.5 nos.
Rate (Rs)
105
3422
25
Rs. 1115.54
White washing - 3 coats - 100 m2
Details
Brickwork in cement-sand mortar1 : 4
Mason semi-skilled - extra
Per
1 m3
1 tonne
1 n13
Total for 10 m2
Quantity
0.14 m3
1.1 nos.
0.5 nos.
1.1 nos.
Amount (Rs.)
105.00
985.54
25.00
Rate
1629.41
110
Rs. 1928.82
Details
Cement-sand mortar 1 : 5
Mason skilled
Mazdoor I
Mazdoor II
Rate (Rs)
1115.54
130.00
80.00
75.00
Quantity
0.09 m3
2.2 nos.
0.5 nos.
3.8 nos.
L. S.
Total for 10 m2
Per
1 m3
Each
Rs. 418.68
Per
1 rn3
Each
Each
Each
Details
Fine screened shell lime
Mason semi-skilled
Mazdoor I
Mazdoor I1
Brushes, Gum etc.
Amount (Rs.)
1873.82
55.00
Amount (Rs.)
156.18
143.00
40.00
82.50
Total for 100 m2 Rs. 702.73
Amount (Rs)
85.73
242.00
40.00
285.00
50.00
Rate (Its.)
952.50
110.00
80.00
75.00
-
Per
1 m3
Each
Each
Each
-
Estimating ABSTRACT ESTIMATE
LEGEND< 0 - Opening - 100 x 200 J-RCJali -180x120
6. 1 7.85 m2 1 White washing 1 702.73 1 100m2
Total for 1 bay
Total for 80 bays
TOILET
rJ 260 X 160 C
D
DRAlJlNt K I T ~ M c W W( 0
480 x 360 , - 360x360 ..W
b- V E R A N DAH
8 E D 4- r 200
% 3 6 0 ~ 8 9 0 - D J
55.16
Rs. 2228.14
Rs. 1,78,251
LIME PLAM-
( ~ h o d n j clear m t e r d dimenstons of rooms @ f 10or Iebel)
All dimensions in centimetres
Figure 6.15 : One Bed Roomed House
6.4.2 Single Bed-roomed Residential Building A Note on Centre Line Method
In case of most residential buildings, the planner initially visualises the requirements as a line plan. The line plan includes the arrangement of rooms with theit internal clear dimensions at floor level. Once the line plan is frnalised it should be possible for the estimator to proceed with quantification of most of the items even without a detailed plan. Kmay be seen that the load bearing walls in
,. -
super structure, basement and foundatian are, in most cases concentric; the centre line length remaining the same.
The centre line length of walls of a plan form can be visualised as the perimeter of the rectangle enclosing the same. The length of all cross walls will then be measured as their individual centre line lengths minus the widthtbreadth under quantification. Following example illustrates the usage of centre line method.
Example 6.9
Prepare a quantity estimate for thk construction of a one bed-roomed residential building. Refer to drawings given in Figure 6.15. Use Centre Line Method for quantification.
Specifications * Foundation concrete PCC - 1 : 4 : 8. * RR Masonry in CM - 1 : 5 in foundation and basement.
Brick Masonry in CM - 1 : 6 in superstructure.
All RCC items (lintels, sunshades, roof slab etc.) in cement concrete 1 : 2:4.
Plastering of walls in CM - 1 : 4.
Weathering course with brick jelly rime concrete with one course of pressed clay tiles.
Flooring with heavy tiles 2 cm thick over PCC - 1 : 4 : 8 - 10 cm thick.
All woodwork in the best country wood.
Painting woodwork with one coat of primer and 2 coats of ready mixed enamel paint.
All walls whitelcolour washed.
Additional Specifications
The sidewalls for the cupboard in the bed room, for the hearth slab in kitchen and in front of the toilet area are assumed to be built with'half brick thick walls and constructed over the floor.
All lintels over openings are 120 rnrn thick with bearing of 200 mrn on either side.
Solution F
Details of Measurement and Calculation of Quantities
ExanQles with Dittermt Types of Strudlllp
Quantity
20.72
4.48
0.78
1.68
1.12
0.22
29.0 A3
D
0.7
0.7
0.7
0.7
0.7
Item No.
1.
Add for steps etc. - ,Lump sum
Particulars .
Earthwork excavation for foundation Walls all round : 2 x (L + B)
L=(0 .1+3 .6+0 .2+4 .8+0 .1 ) = 8.8m B=(0 .1+3 .9+0 .2+3 .6+0 .2+ 1.6+0.1)=9.7rn
Cross wall between bed and hawing (0.1 +3.6+0.2+4.8+0.1)-0.8=8.00m
Cross wall between bed and verandah (2.2-0.8)= 1.4m .
Cross wall between kitchen and drawing 3.8 - 0.8 = 3.00 m
Cross wall between kitchen and toilet 2.8 - 0.8 = 2.00 m
L
37.0
8.0
1.4
3.0
2.0
Total
N
1
1
1
1
1
B
0.8
0.8
0.8
0.8
0.8
I)
0.1
Quantity
4.11
-
L
51.4
N
1
Item No.
2.
0.14
4.25 m3
11.10
4.68
B
0.8
Add for steps etc.
-
Particulars
Concrete in foundation
Total length as in item 1
Lump sum
* The sum of centre line lengths of all 4 cross walls is 17.6 m. As the width uhder quantification is 0.5 m, 4 times 0.5 m should be deducted from this length to get the net length. Alternately, four times the difference between the width under quantification in item 1 or 2 can be added to the net length of cross walls shown in item 1.
3.
2nd footing :
Walls all round
All cross walls (15.60 + 4 x 0.1)
Total
37.0
16
1
1
RR masonry in foundation and basement
1st footing :
Walls all round
All cross walls (14.40 + 4 x 0.30)'
Total
0.4
0.4
25.32 m3
1
1
0.45
0.45
, , I I ,
Walls all round
All cross walls (16.0 + 4 x 0.2)
6.66
2.88
RC Jali
0.6
0.6
37 0
15.6
0.5
0.5
1
Net brickwork (32.52 m3 - 6.14 m3)
2.20
6 14 m3
26.38 m3
0.23 0.12 0.06
Item No. Particulars N L B D Quantity
5. Concrete for RCC works
Lintels as above 0.6 1
Lintel over toilet door 1 1.20 0.12 0.10 0.01
Roof slab 2 8.00 4.00 0.12 7.68
1 5.60 1.00 0.12 0.67
1 41.6 0.75 0.12 3.74
Hearth slab 1 2.60 0.75 0.07 0.14
Cup board slabs 3 2.60 0.50 0.03 0.04
Total 12.89 m3 - 6. Reinforcement grill work (following approximate method)
Lintels 0.62 m3 @ 80 kg/m3 49.6 kg
Roof slab 12 09 m3 @ 50 kg/m3 604.5
Hearth slab and Cup board slabs 0.18 m3 @ 30 kg/m3 5.4
Total 659.5 kg
7. Fonnwork for concrete
Roof slab - Bed room 1 3.6 3.9 - 14.04 rn2
* Drawing 1 4.8 3.6 - 17.28
* Verandah 1 4.8 2.0 - 9.6
* Kitchen 1 3.6 3.6 - 12.96
* Toilet 1 3.6 1.6 - 5.76
* Sides all around 1 44.6 - 0.12 5.35
1 41.6 0.75 - 31.2
Lintels - Door bottom 1 3 9 0.23 ' - 0.90
- Door sides 2 5.5 - 0.12 1.32
- Ends 8 - 0.23 0.12 0.22
Lintels -Window bottom 1 7.5 0.23 - 1.73
-Window sides 2 8.5 - 0.12 2.04
- Ends 10 - 0.23 0.12 0.28
Lintels - Openings bottom 1 3.0 0.23 - 0.69
- Openings sides 2 4.2 - 0.12 1.01
- Ends 6 - 0.23 0.12 0.17
Lintels - Jali bottom 1 1.8 0.23 - 0.41
- Jali sides 2 2.2 - 0.12 0.53
- Ends 2 - 0.23 0.12 0.06
- Total 105.55 m2
Note : Precast lintels are used over toilet door and ventilator openings.
Examples with Dierent Typos of Structure
Item No. Particulars N L B D Quantity
8. Filling in basement with sand
Baj room 1 3.4 3 7 0.35 4.40 m3
Drawing 1 4.6 3.4 0.35 5.47 m3
Verandah 1 4.6 1.8 0.35 2.90m3
Kitchen 1 3.4 3.4 0.35 4.05 m3
Toilet 1 3.4 1.4 0.35 1.67m3
Bed room 2 7.5 3 45.00 m2 -
Drawing 2 8.4 - 3 50.4
Verandah 2 6.8 - 3 40.8
Kitchen 2 7.2 - 3 43.2
Toilet 2 5.2 - 3 31.2
External plastering 1 37.8 - 3 113.4
1 37.2 0.05 - 1 9
1 37.4 - 0.45 16.8
Total 342.7 m2
Deduct for all openings once I
Doors 3 1.00 - 2.00 6.0
1 090 - 2.00 1.8 - Windows 3 1.50 - 1.20 5.4
2 1.00 - 1.20 2.4
Ventilators 1 0.60 - 0.50 0.3
Openings 3 1.00 - 2.00 6.0
RC Jali 1 1.80 - 1.20 2.2
Total deductions 24.1 m2
Net plastering area 318.6 m2 (342.7 m2 - 24.1 m2) \ I
14. Woodwork
Par~elled doors - 100 x 200 cm 3 3 nos.
Panelled doors - 90 x 200 cm 1 1 no.
Panelled doors - 80 X 200 cm 1 1 no.
Panelled doors - 90 x 200 cm
Glazed windows - 150 x 120 cm
Slab projection (41.6 x 0.75)
Sides of slab etc. 6.7 mZ
Total 420.0 m2
6.4.3 Steel Stanchion with Grillage Foundation Example 6.10
Prepare the quantity estimate for a steel stanchion with grillage foundation shown in Figure 6.16.
Specifications
Foundation
The square grillage foundation is of size 3 m x 3 m (size of concrete encasement). The area of grillage is 2.75 m x 2.75 m. It consists of 3 beams of ISMB 350 in top tier tied with 2 nos. of 10 mm dia spacer bolts and 8 beams of ISLB 250 in bottom tier tied with 2 nos. of ISA ,
50 x 50 x 6 on either side.
* Column and base The column consists of 2 nos. of ISWB 400 with a battened plate of 10 mm thick at the top end and 7 nos. of intermediate plates.
The column is supported by cleat angles 2 nos. of ISA 75 x 75 x 8 at the web side. Ths size of the base plate is 750 x 750 x 22 mm and that of gusset plates - 2 nos. of 750 x 350 x 12 mm, Packing plates - 2 nos. of 750 x 350 x 10 mm, Gusset angle - 2 nos. of ISA 150 x 115 x 12.
* Minimum cover of concrete is 125 mm.
S e t i o n A A . 7. .-
7 5 ~ ~ . brick FB S O / id3
beams ISLB 259 .
figurn 6.16 (a) : Grillage Foundation for Steel Staochi~n
Examplea with Different Types of Structure
Figore 6.16 (b) : Steel Staaehion figure 6.16 (c) : Details of Gusseted Base
Solution
Details of Measurement and Calculation of Quantities.
3.
Note : No deduction is to be made for the volume occupied by the grillage beams. This practice is similar to that adopted for reinforced concrete.
Cement concrete in foundation ( 1 2 4 ) Lower portion Upper portion 1
1
Total
3.00 3.00
7.00 m3
3.00 1.05
0.55 0.65
4.95 m3 2.05 m3
Item I No. / ~arription of item I *a ( L ( B ( H / Quanti@ / pr weisht udt 1 itl~upntity I
(a) / Bottom tier ISLB ( 8 1 2.75 / 1 1 22 1 27.9 ( 613.8kg / 250
4. STEELWORK I I I
(b)
(c)
(e) 1 B W ~ plate 33 mm I I 1 0.75 10.75 1 1 0.5625 / 172.7 1 97.14kg ( thick
I
Bottom tier cleat I S A 5 0 ~ 5 0 ~ 6
(d) I I I
Top tier ISMB 350
(g) Packing plate 10 I ! 1 2 / 0.75 1 0.35 / I 0.525 1 78.5 1 41.21 kg I mm thick
2
Top tier 10 rnm spacer bolts
( f )
. 3
6.4.4 Steel Roof Truss Example 6.11
Prepare a quantity estimate of the roof of an industrjal shed.
2.75
I
2
Gusset plate 12 mm thick
(h)
. (i)
(j)
(h)
Specifications
2.75
The span is 12.0 m and the spacing of trusses is 4.0 m.
The inside length of the shed% 24 m.
5.5
0.75
2
Gusset angle ISA 15ox 115 x 12
Column section ISWB 400
Cleat angle (web) lSA75 ~ 7 5 x 8 -- Batten plates 10 mm thick End plate Int~:rmediatc:plate
At the ends of the shed, there are gable walls to support the purlis.
8.25 52.4
Rivets 22 mm diameter - 4% of the total weight of all members (kg)
Total quantity of steel required
The AC sheet roofig is supported on steel trusses and purlins.
4.5
432.3 kg
1.5
0.75
2
2
2
1 7
All connections are riveted and cleats for purlins and fixing plates are of ISA 60 x 60 x 6 rnm. All steel works shall be painted with one coat of red oxide primer and two coats of aluminium paint.
i
24.75 kg
2077.97 82.03
2160 kg
Principal rafter and ties : each 2 nos. ISA 50 x 50 x 6 @ 4.5 kglm
3.0
0.35
0.75
5.0
0.15
0.525 0.525
')
Purlins : ISA 125 x 75 x 8 @J 12.1 kglm.
4.50 kg
Verticals marked 3 and 4 : ISA 40 x 40 x 6 @ 3.5 kglm.
0.525
0.4 0.32
All others : ISA 60 x 60 x 6 @J 5.4 kglm.
uss set plate sizes - all 8 mrn thick.
94.2
- - - -
49.46 kg .
1.5
10.0
0.3
-
0.2 1 1.176
23.8
66.7
8.9
78.5 78.5
35.70kg
667.00 kg
2.67 kg
L
16.49 kg 92.32 kg
1
\
At A-400x250 D-300x300 E-200x100
F 2207 180 C -180 x 280 G, H, K, L - 200 x 200
Fixing plate is 400 x 400 x 12 mm thick.
Cleats for purlins - 10 nos. ISA 60 x 60 x 6 @5.4 kglm
Cleats for fixing plates - 4 nos. ISA 60 x 60 x 6 85.4 kglm
The actual length of members are given in Figure. 6.17.
Note : Ty P ~ C A L sc>lt4T
The members at a joint do not join at a point The Gussct Plate joins tbm and its size is cut out from (L x B).
Figure 6.1 7 : Steel Roof TNSS
Solutlon '
Details of Measurement and Calculation of Quantities
I t~ I Description of item 1 No' 1 I / QusntiQ 1 Weight
1 (a) Principal rafter ties 1 1 : I 2:: 1 1 z:: 1 2: 1
Examples wiUm Different Types of Strocture
Total 235.8 kg
Details of Measurement and Calculation of Area of Painting '1 i
QuantiQ
0.7 1.4
B
Item No.
2.
2 (a)
2 (b)
I
Total we.$t j
2.45 4.90
Item NO.
1 (b)
7.35 kg
11.34 7.56 14.58 16.20 17.82
67.5 kg
No.
2 2
Oescription of item
Verticles marked 3 Verticles marked 4
1 (c)
1 (d) Gussets 8 mm thick - at A 2 0.40 0.25 0.20 m2 E 2 0.20 0.10 0.04 F 2 0.22 0.18 0.08, (3, H, K, L 8 0.20 0.20 0.32 D 1 0.30 0.30 0.09 C 1 0.18 0.28 0.05
L
0.35 0.7
Quantity Description of item
Total
Painting the structural steel with 1 coat of red oxide primer and 2 coats of enamel paint
2.1 1.4 2 7 3.0 3.3
B No.
Total
1.05 1.40 1.35 1.50 1.65
All other members marked 5 6 7 8 9
L
Principal rafter Ties
2 1 2 2 2
4 4
6.9 6.2
10.48 m2
0.112 0.224
0.20 0.20
Total
5.52 4.96
0.16 0.16
Total
0.35 0.7
Verticles marked 3 verticles marked 4
0.336 m2
2 2
-
Examples with Different T~l~esofstmctwe
Figwe 6.18 : Arched Masonry Culvert
Item No.
2 (c)
.L P L A N
No.
1 2 2 2 2
Description of item
All other members marked 5 marked 6 marked 7 marked 8 marked 9
L
1.40 1.05 1.35 1.50 1.65
2.976 m2
1.56 m2
0.64
1.259 0.384
0.643 m2
16.635 83.175
116.64 m2
3.69 m2
203.51 m2
335.35 m2
24.30 m
49.6 m
18 m
2 (d)
2 (e)
2(f)
B
0.24 0.24 0.24 0.24 0.24
Total c
Quantity J
0.336 0.48 0.648 0.72 0.792
Painting 6 r 1 truss = (10.48 + 0.336 + 2.976 + 1.56 + 0.64 + 0.643) Painting for 5 truss
0.40
0.24 0.24
Gussets - vide item 1 Area = (2 x 0.78)
Fixing plate 12 mm thick
Cleats for purlins Cleats for fixing plates
Total
2 (g)
2 (h)
4
10 4
24.3
24.3
0.4
0.108 0.40
Purlins
Wind ties
0.4
0.076
Total
12
2
3.
4.
5.
6.
AC Sheet roof including all fittings
AC Sheet ridging
AC Gutters
A@ Raiowater pipes 150 dia.
2
1
2
6
24.30
24.30
24.30
3 .O
0.4
I Estimating
Examples with D i i e m t Types d Structure
4.
5.
6.
Description of Item
Deduct
Arch optking segmental portion
Arch masonry
Triangular portion above abutment
Triangular portion above parapet
No.
1
2
2
6.8
6.8
21.18 m3
38.76 m3
4.478 m3
Total Deductions
Net masonry (54.94 m3 - 21.18 m3)
1st class brick work in arch
Breadth Height or (m) I D ~ ~ &
(% x 3 x 0.45)
(U x 4.2 x 0.35)
( M X 4.2 X 0.35 x 0.40)
Arch calculations :
r = (';)+(s) = (y ) + ("1 8 x 0.45 = 2.725 m
, = r + (t 1 = 2.725 + (y ) = 2.825 m
b = @Z = d(l.s2 + 0.45~ = 1.566 m
I = ((8b ; 2a) ) = ((8 x 1.566) - (2 x 1.5) ) = 3.176 m
3 . = lx (+ )= 3 . 1 7 6 X ( s ) = 3.293-
Q = L x In x t = 6.8 X 3.293 x 0.2 = 4.478 m3 *
1
at stone work laid in coping
Cement Pointing
Facewall upto top of parapet
Inner face of parapets above road level
Ends of parapets
Ends of parapets
Inner face of abu trnents
Soffits of arch
Quantity
6.120
4.478
9.996
0.588
Remark
2/3 span x rise
As in Item 4.
Area of triangle X
breadth of wall
6.8
2
2
2
4
4
2
1
Total
3.293
10.6
10.5
10.5
- -
6.8
6.8
118.89 mZ
0.2
1= 3.176
0.35
-
-
0.40
0.40
-
3.176
0.075
3.00
0.75
0.20
0.45
1.3
0.5565
63.00
15.75
0.32
0.54
17.68
- 21.60
Estimating
6.4.6 Road Example 6.13
Remark
Rectan- gular and segmental opening are to be deducted
A
Prepare a q-ty estimate for the construction of a new road with the , cross section as given in Figure 6.19.
Qua.6~
7.80
1.80
13.40
23 .OO m2
95.89 m2
Figure 6.19 : Cnms-section &Road
Specifications
The length of road is 1 km.
The formation width of the road is 10 metre.
The average height of embankment is 1 m with the side slope of 2 : 1.
Height or Deptl, etb(m)
1.3
The metalled width is 3.70 m and three coats of metalling are to be provided.
The surface shall be finished with two coats of black topping. Solution
Details of Measurement and Calculation of Quantities
Breadth D-rip.onofItem
Deduct -
Rectangular opening
Arch opening segmental portion
Triangular '
portion below earth slope in face walls
No.
2
Total deduction
Net Pointing (118.89 &- 23.Q0m2)
L;y
3.0
2
4
Item No.
1.
2.
3.
?,5 x 3 x0.45
lh x 2.35 x 2.85
Quantity
l k m
12000m2
40000 m2
Remarks
B = lOm d = l m
Depthof borrow pit = 0.3 m
Description of Item
Surveying, degbelling
Earthwork (EW) in
Land acquisition temporary
No.
1
Q = ( ~ d + sd2) x L Q = (10 x 1 + 2 x 12) x 1000
- Quantity of EW in embankment -
Depth of borrow pit
12000 --- - - 40000 0.3
1 km /
,,'
6.
7.
8.
9.
10.
BLACK
1
1
1
1
1
METALLING
Preparation of subgrade
Soling coat
Stone boulders 15 cm size
Laying and consolidation of boulders including binding with sandy soil
Inter coat
Stone ballast 50mmsize
Laying and consolidation of ballast including binding with sandy soil
Top coat
Stone ballast 40 mm gauge
Laying and consolidation of ballast including binding with sandy soil
Beam or patri dressing
TOP SURFACE
lo00
1000
(SAME
1000
(SAME
1 0 0
(SAME
1 km
50 m3
8.14 tonne
3700 m2
3.70m
3.70m
3.70 m
11.
30 cm wider
30 cm wider
12cmto loose layer
compacted to 8 cm
12cmto loose layer
compacted to 8 cm
4.00
4.00
AS
3.70
AS
3.70
AS
' 1.35 100
220 100
1
1
1
1st coat with roat tar No. - 3
Stone grit 20 mm gauge @ 1.35 m3 per 100 m2
Painting or binding Road tar No.4 @ 220 kg per 100 m2
Laying
1000
1000
1000
- -
0.15
ABOVE)
0.12
ABOVE)
0.12
ABOVE)
4000 m2
600 m3
600m3
444m3
444 m3
444m3
444 m3
l k m
Estimating
-
6.5.1 Introduction We have seen that estimating involves handling of an enormous amount of compilation of measurements, calculations of quantities using mostly simple arithmetic and the use of a fairly large volume of cost data, which are mostly standard. Witb the advent of microcomputers and their ability to rapidly accept, file, store, sort and retrieve or delete large amount of complex data with minimal error and in desired and advantageous patterns, it is realised that an estimator needs to educate himself on the use of personal computers just in the same way as electronic calculators were substituted for manual . calculations a few decades back. Computers significantly reduce the time required to prepare an estimate, improve the accuracy of computations and enhance the quality of the various analyses required. I
Applications \ In the particular fields of constrbction estimating and cost accounting, there are already hundreds of specialist programs with more beiqg developed every year. All of these exhibit varying degrees of quality and utility, and each has been to serve some-real or imaginary specific need of the construction estimator or cost accountant, either on an individual basis or on an industry-wide basis. Programs of most interest to construction estimators broadly fall into the categories of spreadsheets, databases, CADD and speci-d estimating packages.
I
I MISCELLANEOUS LUMP-SUM lTEMS (for which estimates are to be subsequently prepared)
Hei$t Depa (m)
0.75/100
120/100
4 .
~ u m t i $
27.75 m3
4.44 tonne
3700 m2
1 km
Itern No.
12.
13.
, <
R . ~ l u
14.
15.
16.
17.
'(m)
1000
1000
1000
l k m
l k m
1 km
1 km
l k m
Breadth (m)
3.70 m
3.70 m
3.70m
Description of Itgm
2nd coat with Asphalt
Stone grit 12 mm gauge @ 0.75 m3 per 100 m2
Painting or binding Asphalt @ 120 kg per 100 mZ
Laying
Brick edging on both sides including bricks and labour
-
Bridges (minor) and culverts
b h a l f k m boundary stones
Formation level pillers
Road direction posts, caution
No.
1
1
1
1
--- l k m
l k m
I 18.
1 19.
1
1
1
1
signs etc ---- Traffic diversion, service roat, etc
Arbonculture
l k m
l k m
1 km
1 kin
1
1
1 km
-p-p---
1 km
A spreadsheet is nothing more than a two-dimensional matrix which consists of khmplos with Differed
vertical columns connecting with horizontal rows, to produce small boxes or Types dStmdure
"fields" into which data can be inserted. The ordinary quantity take-off paper used by estimators is a good example of a spreadsheet. Most of the examples illustrated in this unit can be easily implemented on a spreadsheet. The columns are usually identified by letters and the rows by numbers. As all columns and lines chat appear on the computer console screen at once, the program must be able to permit "scrolling" from side to side and forward and backward to permit all parts of the spreadsheet to be viewed and modified. The main advantage of the computer spreadsheet is that the program can command the CPU to manipulate data inserted into the various field cells; for example, lengths can be multiplied by widths to produce areas, and numbers of areas can be summed instantly and correctly. - Reflected dimensions can be passed automatically to one or more cells, as
I required. If a change is made in any one cell, the results can also be automatically amended to reflect the effects of the change. Some of the popular spreadsheet packages available in the market include Microsoft Excel, Lotus 1-2-3, etc.
Apart from measurement and summarising, there are many programs available to suit the needs of estimators involved with cost control, scheduling, purchasing payroll and other features involved in the proper management of any construction company. Filling and mailing programs are also very popular and useful for companies with large numbers of customers or clients, or any other extensive data base requiring storage, manipulation, and retrieval, such a labour, material, and equipment output and cost data. In the design field, there are computer programs intended to assist with the graphical aspects of design, in which the data are pictorial or graphical, as distinct from vertical or numerical. Such programs are generally known by the acronym CADD, which stands for "computer aided design and drafting". CADD prograrns can also be used to generate data of use to the design estimator, m the form of the perimeter lengths of walls, floor areas, building volumes, quant~ties of specified elements, heating and cooling roads, and the like, which when used with other programs can generate probable costs of proposed solutions to design problems with in seconds.
%,4$12 :, 5,
r? % r ii. Irni: pl:lt.: -,I a building wl fh clear intertlal dirnlcnsinns (in cmj at fli~ur level is ~tt!'..;ii in !71yur.r 6 . X (a). Ahr-; ih!wn i s !he cross-section of the founclati!)n for the
. *a
u:a . . in F;gi:r.i: b.20 (I-.). Nnte that the supersti-ccwe wall is placed eccentric i:~~:. ii the ft,oriug:; zxr: crjncenric) with rcfereilcc rc; foundation concrete. Estimate ;i!.: ii:!:!!i:!iry :.'f ~ ~ . ~ ~ ~ c : I ~ ! c in l'c:u~ldatir!~l and masonry ill the t%o for:lings:
- . SAQ3 m e line plan of a proposed residential building, showing clear intkmal dimensions (in cm) of rooms, is given m Figure 6.21 (a). The eastem outer wall is to be built to coincide with the property limit of the plot and hence, is provided with eccentric footing [Figure 6.21 (b)l. All other remaining walls have concentric footings
\ [Figure 6.21 (c)]. Make a quantity estimate of concrete in foundation and , brickwork in foundation arid basement. I
LET
FIgPrr 6.21 : Line Am
SAQ 4 Work out lhe area of formwork needed for concreting of the beams and slab in the room shown in Figure 6.22. Also calculate the quantity of reinforcing grill work needed for both the slab and the beams assuming approximate quantities of 60 kg/cu m of concrete for slab and 110 kglcu m of concrete for beams.
Figure 622 : Plan of the Room and Section of Beam
6.6 SUMMARY Examples with D i i m t Types of SLmCan
In this unit, an overview of procedures involved in quantification of various items of civil construction works are given. To the extent possible, all essential items of work normally encountered by the estimator are covered. The principles given in Unit 5 and the procedures described in this unit are simple to comprehend but only by practise (by workmg out a number of examples), a student can imbibe well all these principles and procedures. It is emphasised that the three components of estimating, viz. quantification, rate analysis and abstract estimating sh$l always be carried out as three separate and independentunits as illustrated in the "compound wall" example. Only quantity estimates are given for other examples. The student is advised to try out rate analysis using standard data given in text books or collected from Public Works Departments.
With the advent of computers, the job of estimating has been made a lot more simpler and less time consuming. The student is also advised to farniliarise himself with the use of computers for esttmating.
6.7 ANSWERS TO SAQs
SAQ 1
(a) 26.6 sq m.
(b) 49.1 sq m. SAQ 2
Concrete in foundation = 6.07 cu m.
M~SONY in two footings = 15.11 cu m.
SAQ 3
Concrete in foundation = 8.18 cu m.
Brick work in foundation and basement = 19.93 cu m.
SAQ 4
Area of formwork = 92.9 sq m.
Reinforcing grill work = 801 kg.
FURTHER READING
Glenn M. Hardie, Construction Estimagng Techniques, Prentice-Hall Inc., Englewood Cliffs, New Jersey, 1986.
B. N. Dutta, Estimating and Costing in Civil Engineering : Theory and Practice, UBS Publishers' Distributors Ltd., New Deihi, 1996.
M. Chakraborti, Estimating, Costing, Specification and Valuation in Civil Engineering, Calcutta, 1997.
IS : 1200-1974, Parts 1 to 25, Methods of Measurement'of Building and Civil Engineering Works, Bureau of Indian Standards, New Delhi.
"Standard Data Books" of Central Public Works Department (CPWD) and Public Works Departments of States.
