0
3.1kviews
Figure shows plan and sectional details of a load bearing structure work out

the quantities of the following items of work from figure:- - (a) Earthwork in excavation for foundation - (b) P.C.C (1:2:4) for foundation bd. - (c) 1st class brick work in superstructure in cement mortor (1:4) - (d) D.P.C (2.5 cm thick) in 1:3:6 -

Mumbai University > CIVIL > Sem 7 > Quantity survey, Estimation and valuation

Marks: 12 M

Year: May 2014

1 Answer
0
33views

Calc CL & no. T junction first of all calculate the total centre line of the plan given in the question bt drawing the centre line diagram

enter image description here

Total C = 3 x 5.6 x + 6 x 3.8 + 2 x 4.3 + 2.8 + 1.8 =52.8 m

Note:- see the shape in the figure and count it as 1 segment.

= No. of T junctions marked

No. of T junctions = 10

Step 2:- To calculate what is asked.

Sr No. Description of work Nos L B H Quantity Unit
a) Earthquake in excavation for foundation:-,Net length=total CLNo. of T×B2=5.210×0.92=48.3 m 1 (calc),48.3 0.9,(P.C.C width) 1.4,[H from G.L] L×B×H,60.86 m3
b) P.C.C (1:2:4) for foundation bed Net length = Total CLNo. of T junction×B2=52.810×0.92=48.3 m 1 48.3 0.9 HH of Pcconly,0.2 L×B×H,8.69 m3
c) D.P.C (2.5 cm thick) in 1:3:6 (At P.L),Note:- No. height in DPC,Net length =Tc1NT×B2=52.810×0.32=51.3 1 51.3 0.3 - L×B15.39 m3
d) 1st class brickwork in superstructure,Net length = Tc1NT×B2=52.810×0.3251.3 1 51.3 0.3 Floor to floor ht.,3 L×B×H,46.17 m3
- Total,quantity - - - - 46.17 m3
- Deduction:-,(-), - - - - - -
- 1) Door (Analysis fromdiagram & given data) - - - - L×B×H -
- D -2 1.2 0.3 2.1 -1.512 m3
- D1 -2 1 0.3 2.1 -1.26 m3
- 2) Windows - - - - L×B×H -
- W1 -1 2 0.3 1.3 -0.7 m3
- W2 -4 1.5 0.3 1.3 -2.34 m3
- V -2 0.6 0.3 0.9 -0.32 m3
- O -2 1.2 0.3 2.1 -1.5 m3
- - - - - - 38.53 -

Note:- Brickwork in super structure possess deductions. Deductions are done because of doors & windows present in the wall Information given in the diagram helps us for the deduction.

Please log in to add an answer.