Fig. shows a plan and section of a small house. Workout the
quantities of the following items and enter them on standard measurement form
with full description of the items.
a)
Excavation in murum for foundation trenches.
b)
U.C.R. masonry in CM (1:6) in foundation.
Solution.
Figure represents the foundation trench plan.
By the Centre line method –
The length of wall A & B = 4.3 m each
The length of wall C
= 8.1 m
The length of wall D & E = 7.1 m each
The length of wall F
= 3.8 m
The length of wall G
= 2.65 m
Therefore,
The total length of wall
= 2 ( 4.3 ) + 8.1 + 2 ( 7.1 ) + 3.8 + 2.65 m
= 37.35 m
In the problem there are three junction walls, hence
deduction is to be made for them.
If the total centre length of wall is multiplied by the
breadth and depth, the junctions shown in hatched lines come twice, and we get
the quantity in excess, by these portions and therefore these excesses shall
have to be deducted. The deduction may be effected reducing the centre length by
half width for each junction.
Thus, the quantity of earth work in excavation centre,
= ( Total length – 3 x ½ x breadth ) x ( breadth ) x ( depth
)
= ( 37.35 – 3/2 x 1.0 ) ( 1.0 ) ( 1.15 )
= 43.53 cu.m.
Details of Measurement and Calculation of Quantities.
Item
No.
|
Particulars of Item
|
No.
|
Length
In m.
|
Breadth in m.
|
Height or Depth in m.
|
Qty in
Cu.m.
|
Explanatory
Notes.
|
1.
|
Excavation in murum for foundation trenches
|
1
|
35.85
|
1.0
|
1.15
|
43.53 cu.m.
|
Total length = 37.35 m
L = 37.35 – 3 x (1.0/2)
= 35.85 m
|
2.
|
IInd class brick masonry in CM (1:6)
In plinth
|
1
|
36.60
|
0.5
|
0.9
|
16.47 cu.m.
|
L = 37.35 – 3 x (0.5/2)
= 37.35 – 0.75
= 36.60
|
