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

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written 7.8 years ago by

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

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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.

$\Box$ = 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 $C_L-No. \ of \ T {\times}\frac{B}{2} \\ =5.2-10{\times}\frac{0.9}{2} \\ =48.3 \ m$	1	(calc),48.3	0.9,(P.C.C width)	1.4,[H from G.L]	$L{\times} B {\times} H,60.86$	$m^3$
b)	P.C.C (1:2:4) for foundation bed Net length = Total $C_L-No. \ of \ T \ junction {\times} \frac{B}{2} \\ =52.8-10{\times}\frac{0.9}{2} \\ =48.3 \ m$	1	48.3	0.9	HH of Pcconly,0.2	$L {\times} B {\times} H,8.69$	$m^3$
c)	D.P.C (2.5 cm thick) in 1:3:6 (At P.L),Note:- No. height in DPC,Net length =$T_{c1}- N_T{\times}\frac{B}{2} \\ =52.8-10{\times} \frac{0.3}{2} \\ =51.3$	1	51.3	0.3	-	$L{\times}B \\ 15.39$	$m^3$
d)	$1^{st}$ class brickwork in superstructure,Net length = $T_{c1}-N_T{\times}\frac{B}{2} \\ =52.8-10{\times}\frac{0.3}{2} \\ 51.3$	1	51.3	0.3	Floor to floor ht.,3	$L{\times} B {\times}H,46.17$	$m^3$
-	Total,quantity	-	-	-	-	46.17	$m^3$
-	Deduction:-,(-),	-	-	-	-	-	-
-	1) Door (Analysis fromdiagram & given data)	-	-	-	-	$L{\times}B{\times}H$	-
-	D	-2	1.2	0.3	2.1	-1.512	$m^3$
-	$D_1$	-2	1	0.3	2.1	-1.26	$m^3$
-	2) Windows	-	-	-	-	$L{\times}B{\times}H$	-
-	$W_1$	-1	2	0.3	1.3	-0.7	$m^3$
-	$W_2$	-4	1.5	0.3	1.3	-2.34	$m^3$
-	V	-2	0.6	0.3	0.9	-0.32	$m^3$
-	O	-2	1.2	0.3	2.1	-1.5	$m^3$
-	-	-	-	-	-	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.

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