1. Northern Technical University
Engineering Technical College of Mosul
Building & Construction
Technology Engineering Dept.
Estimation, Specifications and
Contracts
SECOND CLASS
Lecturer:
Dr. Muthanna Adil Najm ABBU
2017-2018
5. Estimation, Specifications
& Contracts General rules, units Lecture .4
4
Dr. Muthanna Adil Najm Northern Technical University
س o.36
0.9
4.2
ص
ع
3.6
Methods of working quantities for various items
excavation and fill works for wall footings
Example (1):- calculate the amount of work to excavate the wall footing for a
room with internal clear dimensions of 4.2x3.60 m. note that the wall
thickness is 36 cm and the width and depth of the footing 90 cm and 1 m
respectively.
Item
No.
description
unit
number
dimensions
quantityNotes and calculations
Length
width
height
1
2
األسس ياترحف
الشكل (أنظر
)أعاله
م3117.040.901.0015.33الجدار عن األساس بروز مسافة
=س(90,0-36,0÷)2=0.27م
=ص92,4+27,0×2=46,5م
=ع60,3-27,0-27,0=06,3م
طول=األساس2)(ص+ع
=2(46,5+06,3)=04,17م
15.33
6. Estimation, Specifications
& Contracts General rules, units Lecture .4
5
Dr. Muthanna Adil Najm Northern Technical University
In the above calculations we used method of stripping, although the method of
centerlines can be used too.
Example (2):- calculate the quantities of the following sub-items to perform the
wall footing shown in Figure:
1. excavation works.
2. placing a layer of broken bricks beneath the base.
3. placing concrete (1: 2: 4).
4. construction of walls
using bricks and (1: 3)
cement-sand mortar up
to D.P.C. level.
0.90
0.48
0.08
0.25
0.24
0.24
0.29
0.19
0.48
0.36
1.10م
أ مقط-أ
0.24
0.240.24 3.000.24 3.00
0.24
0.24
3.00 أأ
plan
8. Estimation, Specifications
& Contracts General rules, units Lecture .4
7
Dr. Muthanna Adil Najm Northern Technical University
Example (3): Compute the quantity of fill for wall footing in previous example
Solution: here we shall calculate the quantity of earth filling up to original or
natural ground level (G.L.)
- Compute V0.24 up to G. L.
V0.24 = 22.44*0.290.24
= 1.56 m3
V const = 1.56+1.93+2.56
= 6.05 m3
V total = (0.24+0.24+0.29)0.921.78
= 15.09 m3
V fill = V total - V const
= 15.09-6.05= 9.04 m3