Analysis & Design of RCC shear walls in the model using ETABS

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Analysis & Design of RCC shear walls in the model using ETABS

The etabs file for a G+6 building is provided. Run the analysis and design the RCC shear walls (show in the etabs model). The following challenge deals specifically with the shear wall of size 300x1500 at grid A-11 in the structural model.

• Extract values of longitudinal steel ratio, boundary element length and shear reinforcing ratio from the design results.
• Provide cross sectional steel arrangement of the shear wall from its base up to terrace, in compliance with provisions of IS 13920 – 2016. AutoCAD application can be used for this exercise. Alternatively, it can be drawn by hand as well.
• Provide elevation drawings of the reinforcement details of the shear wall from base up to terrace.

Ans.

Aim:-

• To run th analysis and sedign the RCC shear wall for given ETABS model.
• To extract the values of longitudnal steel ratio, boundary elemets length and shear reinforing ration from the design results.
• To provide the cross-sectional arrangement of shear wall from its case up to terrace, in compliance with provision of IS 3920-2016.
• To provide elevation drawing of reinforcement details of shear wall from base to terrace.

Introduction:-

A shear wall consists of web and boundary elements at either side. The boundary elements in shear wall are subjected to alternate compression and tension when subjected to alteral forces (EQ), hece it is necessary to provide special reinfrocement deatilas of shear walla from base to terrace.

IS 3920-2016 provides guidelines on several aspect of shear wall deatiling.

• Thickness of wall (Clause 10.1.2)
• Minimum ratio of wall length to thickness (Clause 10.1.3)
• Largest dia of longitudnal steel bar (Clause 10.1.8)
• Maximum spacing of horizontal and vertical reinforcement (Clause 10.1.9)
• Condition for base of shear wall (Clause 10.1.10)
• Minimum reinforcement in RC shear wall (Clause 10.1.6)

From clause 10.1.4, walls can be cateriosed as squat walls, intermediate and slender wall as given below :

a. Squat walls = h_w / Lw <1,

b. Intermediate walls = 1≤ h_w / Lw ≤ 2 and

c. Slender walls = h_w / Lw > 2

Now,

Min. Shear reinforcement in boundary element of RC shear wall can be determined from formula

Procedure:-

• Open given model in ETABS
• Now, analysis and check the model

• Run analysis

• Once analysis is completed we need to carry out the design of shear wall, from this go to shear wall design.
• Shear wall ----> View/ revise prefernces

• Now we need to start designing
• Once the design is completed go to display design option
• Then we have checked for any failures in the design
• Now we need to extract certain values from the design
• Go to display design info ---> pier longitudnal reinfrocing.

• Go to display design info ---> pier reinfrocing ration

• Go to display design info ---> pier edge/ boundary zone width

• Go to display design info ---> pier shear reinforcing

Since, we gathered all the information, now we need to draft the deatils of RC wall in AutoCAD

A. Shear wall detailing calculations :

Here, the total height of wall is 21000mm and width of wall is 1500mm.

Hence, ratio is obtained as 21000/1500 = 14.........which is greater than 2, hence it can be considered as slender wall

Max. dia of bar that can be used from IS code = 1/10 thickness of that part = 30mm

Hence, we consider dia of bar =25mm

Max. spacing between bars shold not exceed the minimum of the following

a. 1/5 = 300mm

b. 3X300 = 900mm

c. 450mm

Hence min. spacing is limited to 300mm.

B. Longitudnal reinforcement calculations :

• From ETABS results the maximum longitudnal reinforcement obtained is 13128 sq.mm

1. Since we are using 25mm dia, the no. of bars required will be 13128/490 = 28 bars. But we are providing two additional bars in the boundary elements for proper confinement i.e. total 30 longitudinal bars.
2. Hence the total area of reinforcement provided is 13720 sq.mm.........Hence, it is safe.

• Now from code the minimum vertical reinforcement required is 0.0025+0.01375 (tw/ Lw) = 0.00525

i.e. 0.00525 X 300 X 1500 = 2362.5 sq.mm, which is less than actually provided........Hence, safe.

• Now minimum vertical reinforcement to be provided in boundary region = 0.008X 1500 X300 = 3600 sq.mm

1. The reinforcement provided for boundary element in right side = 4410 sq.mm and
2. At left side = 5390 sq.mm.............Both values are greater than required.......... Hence, safe.

• In the web area the min. rrinforcement provided is 0.0025 X 1500 X300 = 1125sq.mm and provided amount of reinforcement in web area is 3290 sq.mm .......... Hence, safe.

C. Horizontal reinforcement calculations:

• From ETABS the shear reinforcement amount is obtained as 750 sq.mm
• So let us provide 8mm dia bars @ 100mm c/c, then provided area of reinforcement will be 1000 sq.mm............Hence, safe.
• This shear reinforcement can be provided for entire shear wall
• Now, we need to check for boundary element
• From, clause 10.4.4 the shear reinforcement required in boundary element can be obtained from the equation

1. Ash is the area of shear reinforcement required as per the link
2. Sv = spacing between shear reinforcement = 100mm
3. h = spacing between exterme longitudnal reinforcement in the boundary element i.e.600-80 =520mm
4. fck = 25 N/sq.mm and fy =500N/sq.mm
5. Hence, its is obtained as Ash =0.05 X100X 520 X (25/415) = 156 sq.mm...........this value is for link
6. One link has 2 leg........hence, area of each leg = 156/2 = 78sq.mm
7. Hence, we can choose 10mm dia bars for shear reinforcement in the boundary elelment width whose area is around 78.5sq.mm..........Provide on left side of wall
8. Now, on right side of boundary element width is 450 sq.mm......So, h=450-80 =370mm.......hence, Ash is required
9. Ash = 0.05X 100X370 X (25/4115) = 111sq.mm.........for one link..........hence for one leg =55 sq.mm
10. Hence, we can provide 8mm dia bars for shear reinforcement in the boundary elelment width whose area is around 50 sq.mm................Provide ths on right side of wall.
11. i.e. for boundary element width special confining reinforcement to be provided is 10mm dia bars@ 100 mm c/c at left side BE and 8mm dia bars @100mm c/c BE.

Result:-

• The value of longitudnal steel ratio, boundary element length and shear reinforing from the design results have been extracted successfully.
• The c/s steel arrangement of shear wall from ts base up to terrace, in compliance with previous as per IS-13920-2016 has been drafted appropriately by calculations.
• The elevation drawings of reinforcement details of shear walls from base to terrace has been drafted.