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Aim:- Analysis and design of RCC shear wall in the model by using ETABS Introduction:- The wall which can carry havy lateral loads are called shear walls Shear walls are mandatory in the zone 4 and zone 5 regions because it resists the earthquake loads It will helps to avoid the sudden failure of the structure…
Rajendra Prasad
updated on 12 Sep 2022
Aim:- Analysis and design of RCC shear wall in the model by using ETABS
Introduction:-
Procedure:-
Step1:-
Step2:-
Step3;-
Step4:-
Step5:-
Step6:-
Step7:-
Provide the 600mm boundary zone width at the left and provide 450mm boundary zone width at the right uniform throughout the height
Step8:-
Step8:-
Detailing guidelines for the shear wall
Which is greater we should draw in the detailing
hW / LW = 21000/1500 = 14>2
1.4>2, so its a slender wall
(ρv,net)min = 0.0025+0.01375*(300/1500) = 2363mm^2 (ρv,be)min (ρv,web)min (ρv,net)min
600mm width at the left
450mm width at the right
(ρh) = 750/(300*1000) = 0.0025
hW = 21,000mm
LW = 1500mm
(ρh)min = 0.0025+0.5(21/15 -2)*(0)
(ρh)min = 0.0025
1/10th of wall thickness = 1/10 * 300 = 30mm
30mm dia rebar is the maximum rebar but we will use 25mm dia bar which is a commercial rebar
From the ETABS we get the longitudinal reinforcement value as Ast = 13128mm^2
No of bars = 13128/490 = 26.79 = 28bars
(490mm^2 is the area of 25mm dia bar)
Confining reinforcement
Confining reinforcement spacing for the boundary element as per the IS 13920-2016 clause 10.4.4
maximum spacing of confining reinforcement in boundary element is 150mm
We use a 8mm dia bar for the confining reinforcement
No of bars = Ast / Area of 8mm bar 4 Legs =
= 750/4*50
= 3.75= 4 No of bars (per meter )
Spacing = 1000/No of bars
= 1000/4 = 250mm c/c (per meter)
Use 4 Nos of bar at 250mm c/c
Ast = 750mm^2
No of bars = Ast / Area of 8mm bar 2 Legs =
= 750/2*50
= 7.5 = 8 No of bars (per meter)
Spacing = 1000/No of bars
= 1000/8 = 125mm c/c (per meter)
Use 8 Nos of bars at 125mm c/c
Confinig reinforcement within boundary element as per IS 13920 - 2016 clause 10.4.4
sv = 150mm , h = 1000mm (because of the horizontal reinforcement ), fck = 25, fy = 415
Ash = 0.05*150*1000*(25/415) = 451.80 = 452mm^2 (minimum)
For the deatiling we used
AutoCAD deatiling ;-
This is the cross section of the shear wall
This is the longitudinal section of the shear wall
This is the cross section and longitudinal section of the shear wall
Result:-
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