0
1.5kviews
Figure shows plan and sectional details of a load learning structure work out the quantities of the following items of work from figure:

(a) UCR (1:5) masonry in substructure - (b) DPC(2.5cm thk) in 1:2:4 - (c) 1st class brickwork in super structure in CM (1:4) - (d) Internal plaster in CM (1:4) -

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

Marks: 6 M

Year: Dec 2012

1 Answer
0
4views

Step 1:- Calc $C_L$ & no. T junction Draw centre line diagram & analyse further.

enter image description here

T =2 x 4.8 + 1 + 2 x 5.3 + 2 x 5.8 + 3.8 + 0.4 +2 x 3.8 + 2 x 4.1 + 2 x 1.9 + 2 x 3.1 + 3.2 + 2.3 + 3.5 + 2.8 = 75.6 m.

Note: Cross the no. with the diagram and complete all the sides.

$\Box$ = No. of T junctions = 12

Step 2:- calc what is the question given in the paper.

Sr No. Description of work Nos L B H Quantity Unit
a) UCRmasonry in sub str = plinth + 3rd footing,+ 2nd footing + 1stfooting [No. P.C. C] - - - - - -
- $Plinth :-T_{C_1}-N_T{\times}\frac{B}{2} \\Total \ length-75.6-12{\times}{0.3}{2} \\ =53.8$ 1 73.8 0.3 0.7 $L{\times} B {\times} H,\\ 15.49$ $m^3$
- $3_{rd}$ footing (below G.L),Total length =$T_{C_1}-N_T{\times}{B}{2}\\ =75.6-12{\times}\frac{0.5}{2} \\ =72.6$ 1 72.6 0.5 0.3 $L{\times} B {\times} H,\\ 10.89$ $m^3$
- $2_{nd}$ footing :-Total length =$T_{C_1}-N_T{\times}{B}{2}\\=75.6-12{\times}\frac{0.7}{2} \\ =71.4$ 1 714 0.7 0.3 $L{\times} B {\times} H,\\ 14.99$ $m^3$
- 1st,footing:-$T_{C_1}-N_T{\times}{B}{2}$ Total Lenght=$75.6-12{\times}{1}{2} \\ 69.9$ 1 69.6 1 0.3 $L{\times} B {\times} H,\\ 20.88$ $m^3$
b) DPC (2.5cm thk) (At P.L) in1:2:4,Total length$=T_{C_1}-N_T{\times}{B}{2} \\ 75.6-12{\times}\frac{0.3}{2} \\=73.8$ 1 73.8 0.3 - $L {\times} B {\times} H,\\ 22.14$ $m^3$
c) $1^{st}$ class brickwork in superstructure,Net length = $T_{c1}-N_T{\times}\frac{B}{2} \\ =75.6{\times}\frac{0.3}{2} \\ =73.8$ 1 73.8 0.3 3.1 $L{\times} B {\times}H,\\ 68.63$ $m^3$
- Deduction:-,(-), - - - - - -
- 1) Door (Analysis fromdiagram & given data) - - - - $L{\times}B{\times}H$ -
- D -1 1.4 0.3 2.1 -0.88 $m^3$
- $D_1$ -3 1.2 0.3 2.1 -2.27 $m^3$
- $D_2$ -2 1 0.3 2.1 -1.26 $m^3$
- O -1 1.3 0.3 2.1 -0.82 $m^3$
- 2) Windows - - - - $L{\times}B{\times}H$ -
- W -3 1.2 0.3 1.5 -1.62 $m^3$
- $W_1$ -4 1.5 0.3 1.5 -2.7 $m^3$
- $W_2$ -2 1 0.3 0.8 -0.48 $m^3$
- - - - - - 58.6 $m^3$

d) Internal Plaster:

Note:-

1) No width

2)Internal dimension to be taken

3) In deductions, count both side of doors, if it is present inside the plan

4) count only internal side for windows as it is an internal plaster

Deduction:-

Opening area i.e L x B < $0.5m^2$ no deduction . 0.5m < opening < $3m^2$ -50% deduction

No separate measurement for jamb lining.

Opening > $3m^2$ = 100% deduction

Seperate measurement for jamb lining.

enter image description here

Please log in to add an answer.