Suppose
we have a circular slab having internal diameter of 11'-0" and external
diameter as 12'-6" (including both walls) and RCC slab having 1/2"
dia bars @ 6" c/c both ways.
So
solving this example we will use following formula
Spacing
of bars will be started from point
"A" and will be gradually increased adding each bar from point
"A" towards Points "C" .This process will provide us half
reinforcement of single layer and 1/4th of two layers of reinforcement.
L = 2 (( 2R-S)S))^1/2
L = Length of
each bar with respect of positioning of bar from points ''A" and
"B"
S =
Spacing of each bar from point "A" towards point "C" OR Spacing of each bar from
point "B" towards point "C".
R =Effective
radius slab (Clear span of slab + two bearings for MS bars.
L = 2 (( 2R-S)S))^1/2
Before
putting values in above listed formula let me check the correctness of formula by checking the length of central bar
adding 2 bearings of 4.5"as it will be 11'-0" + 2(4.5") =
11'-9"
Trial
for central bar will be done as follows
L = 2 (( 2R-S)S))^1/2
L= 2x
((2x5.875-5.875)5.875))^1/2
=2x
((34.5156))^1/2
=2x
(5.875) = 11.75'
So
the formula is correct to calculate the length of all bars
Bar-1
at 6" = L= 2x ((2x5.875-0.5) 0.5))^1/2
=4.743'
Bar-2
at 1'= L= 2x ((2x5.875-1) 1))^1/2 =6.557'
Bar-3
at 1.5'= L= 2x ((2x5.875-1.5) 1.5))^1/2 =7.842'
Bar-4
at 2'= L= 2x ((2x5.875-2.0) 2.0))^1/2 =8.831'
Bar-5
at 2.5'= L= 2x ((2x5.875-02.5) 02.5))^1/2 =9.617'
Bar-6
at 3'= L= 2x ((2x5.875-3.0) 3.0))^1/2 =10.25'
Bar-7
at 3.5'= L= 2x ((2x5.875-3.5) 3.5))^1/2 =10.75'
Bar-8
at 4'= L= 2x ((2x5.875-4.0) 4.0))^1/2 =11.135'
Bar-9
at 4.5'= L= 2x ((2x5.875-4.5) 4.5))^1/2 =11.423'
Bar-10
at 5'= L= 2x ((2x5.875-5.0) 5.0))^1/2 =11.618'
Bar-11
at 5.5'= L= 2x ((2x5.875-5.5) 5.5))^1/2 =11.726'
Common
Bar-12 at 5.875'= L= 2x ((2x5.875-5.875) 5.875))^1/2 =11.75
Length of one side bars from Point
"C" (Excluding central bar) = 104.50 Ft
Length of one mesh (Including central bar) = 2x104.50 Ft+11.75 =220.75
Ft
Length of 2- Meshes (Main and Distribution
bars) = 2x220.75=441.50 Ft
Weight of bars = Total Length of Bars x Unit Weight
in Unit length
Weight of 1/2" dia bar in One Ft length =
0.302 Kg/Ft
Weight of bars = Total Length of Bars x Unit Weight
in Unit length
= 441.50 x 0.302 = 133.33
Kgs
Add 3% for wastage margin =133.33x1.03 = 137.33
say 138 KGS
