Showing posts with label "How to calculate circular slab reinforcement". Show all posts
Showing posts with label "How to calculate circular slab reinforcement". Show all posts

Wednesday, August 18, 2021

How to calculate circular slab reinforcement

 

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

 



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