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Shear walls are specially designed structural walls incorporated in buildings to resist lateral forces that are produced in the plane of the wall due to wind, earthquake and other forces. Shear walls are usually provided in tall buildings and have been found of immense use to avoid total collapse of buildings under seismic forces. It is always advisable to incorporate them in buildings built in regions likely to experience earthquake of large intensity or high winds. Shear walls for wind are designed as simple concrete walls. The design of these walls for seismic forces required special considerations, as they should be saf3e under repeated loads. Shear walls for wind or earthquakes are generally made of concrete or masonry. They are usually provided between columns, in stairwells,

Toilets, utility shafts, etc. Initially shear walls were used in RCC buildings to resist wind forces. These came into general practice only as slate as 1940. With the introduction of shear walls, concrete construction can be used for tall buildings. Earlier tall buildings were made only of steel, as bracings to take lateral load could be easily provided in steel constructions. However, since recent observations as shown consistently the excellent performance of buildings with shear walls even under seismic forces, such walls is now extensively used for all earthquake resistant design.

1.CLASSIFICATION OF SHEAR WALLS:

1. Simple rectangular types and the flanged walls (called the bar bell type walls with boundary elements)

2. Coupled shear walls 3. Rigid frame shear walls

4. Framed walls with in filled frames 5. Column supported shear walls 6. Core type shear walls

2.DESIGN OF RECTANGULAR AND FLANGED SHEAR WALLS:

IS 13920-1993 clause 9(7) deals with requirements and design of simple free standings shear walls.

2.1General Dimensions:

The following factors determine the general dimensions of the wall. 1. The thickness of the wall (t) should not be less than 150mm

2. If it is flanged wall, the effective extension of the flange width beyond the face of web to be considered in design, is to be lee of the following

(a) ½ distance to an adjacent shear wall web (b) 1/10th _{of the total wall height}

(c) Actual width

3. Where the extreme fiber compressive stresses in the wall due to all loads (the gravity loads and the lateral forces) exceed 0.2 fck boundary elements are to be

provided along the vertical boundaries of the walls. Boundary elements are portions along the wall edges specially enlarged and strengthened by longitudinal and transverse steel as in columns. These elements can be discontinued when the compressive stresses are less than 0.15 fck. Boundary elements are also not

required if the entire wall section is provided with special confirming steel reinforcements.

2.2 Reinforcements

The following rules are to be observed for detailing of steel:

1. Walls are to be provided with reinforcement in two orthogonal directions in the plane of the wall. The minimum steel ratios for each of the vertical and horizontal directions should be 0.0025.

[As/Ac (gross)]=ρ ≥ 0.0025

This reinforcement is distributed uniformly in the wall.

2. If the factored shear stress (v) exceeds 0.25√ fck or if the thickness of the wall

exceeds 200mm the bars should be provided as two mats in the plane of the wall one on each face. (This adds to ductility of wall by reducing the fragmentation and premature deterioration on reversal of loading).

3. The diameter of the bars should not exceed 1/10th _{of the thickness of the part of}

the wall.

4. The maximum spacing should not exceed, L/5, 3t or 450mm, where L is the length of the wall.

2.3 Reinforcements for shear

The nominal shear is calculated by the formula V=Vu/td

Where

D = effective width (= 0.8 for rectangular sections) Vu = the factored shear

The nominal shear should not exceed the maximum allowable beam shear given in IS 456-2000 Table 20.

Table 62 of SP 16 can be used for determining the diameter of shear steel and its spacing. The vertical steel provided in the wall for shear should not be less than the horizontal shear.

PROBLEM 1.

(Design of simple shear wall with enlarged ends)

Design a shear wall of length 4.16m and thickness 250mm is subjected to the following forces. Assume fck = 25 and fy =415N/mm2 and the wall is a high wall with the following

loadings:

Loading Axial force (KN) Moment (KN-m) Shear (KN) 1. DL+LL 2. Seismic load 1950 250 600 4800 20 700 Step 1. Determine design loads

P1(Case1)=(0.8*1950)+(1.2*250)=1860KN

P2(Case2)=1.2(1950+250)=2640KN

Moment =1.2(4800+600)=6480KN Shear =1.2(700+20)=864KN

Step 2. Check whether boundary elements are required

(Extreme stresses are more than 4N/mm2 _{boundary elements are to be provided)}

Assuming uniform thickness L=4160mm; t=250mm I=1.5x1012 _mm4

A= 4160x250=1.04x106_mm2

fc=[P/A]±[My/I]

=[(2.64)106_/(1.04)106_]±[(6.48)109_{*4160/2(1.5)10}12_]

0.2fck=0.2*25=5N/mm2

As extreme stresses are high, boundary elements are needed. Also there is tension in one end due to bending moment.

Step3. Adopt the dimensions of boundary elements.

Adopt a bar bell type wall with a central 3400mm portion and two ends 380X760mm giving a total length of (3400+380)=4160mm.

Step 4. Check whether two layers of steel are required. Two layers are required if

(a) Shear stress is more than 0.25√ fck.

(b) The thickness of section is more than 200mm.

a) Depth of section= center to center boundary elements=3400+380=3780mm V=V/bd=864x103_{/(250x3780)=0.92N/mm}2

0.25√ fck =0.25√20=1.11N/mm2

b) Thickness of 250 is more than 200mm Use two layers of steel with suitable cover. Step 5.Determine steel required

Let us put min-required steel and check the safety of wall (p=0.0025) As (min)=0.0025*250*1000 per meter length=625mm2 is two layers

Provide 10φ @250mm,provide the same vertical and horizontal steel. Step 6. Calculate Vs to be taken by steel

V=0.92N/mm2

τ c (for 0.25% steel and fck=20)=0.36 N/mm2

Maximum shear allowed=3.1 N/mm2

Designed steel is necessary for Vs

Vs=(0.92-0.36) bd=0.56*250*3780=529.2KN

Step7. Calculate steel necessary to take Vs

As the wall is high, horizontal steel is more effective. Therefore Vs/d =0.87fy(Asv/Sv), d=3780mm

Required Asv/Sv = [(529.2*103)/(0.87*415*3780)] = 0.3888

Consider 1m height = Sv

Horizontal steel area = 628mm2 _{= A} sv

Available Asv/Sv = 628/1000=0.628

The nominal steel provided will satisfy shear requirements. Step 7(a). Find flexural strength of web part of wall

Assuming it as a UDL over the area the axial load for the central part beams =Pw

Pw= 1860{3400*250/[(3400*250)+2(380*760)]}=1860*0.595=1107KN ---(1)

Step8. Calculate the parameters λ ,φ ,β , x/L (As per code IS 13920)

λ = Pw/( fck*t*L)= 1107*103/(20*250*3400)=0.065 φ = 0.87 fy p/ fck = 0.87*415*0.0025/20=0.045, β =0.516 x/L=(λ +φ )/(0.36+2φ )=0.24< 0.5 Mu/( fck*t*L2)= φ (1+λ /φ )(0.5-0.42 x/L)= 0.041 Mu= 2370kN-m< 6480(required)

Step 9. Calculate compression and tension in the boundary elements due to M1=6480-

M1=6480-2370=4110KN

Step 10. Calculate compression and tension in the boundary elements due to M1

Distance between boundary elements=3480+380=3.86m=c Axial load=M1/c=4110/3.86=1065KN

This load acts as tension at one end and compression at the other end. Step 11. Calculate compression due to the load at these ends

Fraction of area at each end= (1-0.595)/2 = 0.2025 (from step 7a point 1) P2 = 0.2025*2640=535KN

Factored compression at tension end (taking P1)= 0.2025*1860=377KN

Compression at compression end= 1065+535=1600KN Tension at the tension end=-1065+377=-688KN

Step 12. Design the boundary elements compression (a) Design one end as column with details (b) Check laterals for confinement

(c) Check for anchorage and splice length

Step 13. Design the reinforcement around openings, if any, of the wall

Openings are provided in the main body of the wall. Assume opening size as 1200x1200 Area of reinforcement cut off by the opening

=1200 (thickness)% of steel/100=1200*250*0.0025=750mm2

4no’s of 16mm bar area=804mm2

Provide 2no’s of 16mm dia one each face of the wall, on all the sides of the hole to compensate for the steel cut off by the hole.