# A Conceptual Design Approach of Coupled Shear Walls

Authors: Dipendu Bhunia; Vipul Prakash; Ashok D. Pandey
Journal: ISRN Civil Engineering
 (2013)

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## Abstract

Earthquake causes considerable damage to a large number of RCC high-rise buildings and tremendous loss of life. Therefore, designers and structural engineers should ensure to offer adequate earthquake resistant provisions with regard to planning, design, and detailing in high-rise buildings to withstand the effect of an earthquake and minimize disaster. As an earthquake resistant system, the use of coupled shear walls is one of the potential options in comparison with moment resistant frame (MRF) and shear wall frame combination systems in RCC high-rise buildings. Furthermore, it is reasonably well established that it is uneconomical to design a structure considering its linear behavior during earthquake. Hence, an alternative design philosophy needs to be evolved in the Indian context to consider the postyield behavior wherein the damage state is evaluated through deformation considerations. In the present context, therefore, performance-based seismic design (PBSD) has been considered to offer significantly improved solutions as compared to the conventional design based on linear response spectrum analysis.


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## Body

## 1. Introduction

The growth of population density and shortage of land in urban areas are two major problems for all developing countries including India. In order to mitigate these two problems, the designers resort to high-rise buildings, which are rapidly increasing in number, with various architectural configurations and ingenious use of structural materials. However, earthquakes are the most critical loading condition for all land based structures located in the seismically active regions. The Indian subcontinent is divided into different seismic zones as indicated by IS 1893 (Part 1) [1], facilitating the designer to provide adequate protection against earthquake. A recent earthquake in India on January 26th, 2001 caused considerable damage to a large number of RCC high-rise buildings (number of storey varies from 4 to 15) and tremendous loss of life. The reasons were (a) most of the buildings had soft and weak ground storey that provided open space for parking, (b) poor quality of concrete in columns, and (c) poor detailing of the structural design (http://www.nicee.org/eqe-iitk/uploads/EQR_Bhuj.pdf). Therefore, this particular incident has shown that designers and structural engineers should ensure to offer adequate earthquake resistant provisions with regard to planning, design, and detailing in high-rise buildings to withstand the effect of an earthquake to minimize disaster.As an earthquake resistant system, the use of coupled shear walls is one of the potential options in comparison with moment resistant frame (MRF) and shear wall frame combination systems in RCC high-rise buildings. MRF system and shear wall frame combination system are controlled by both shear behavior and flexural behavior; whereas, the behavior of coupled shear walls system is governed by flexural behavior. However, the behavior of the conventional beam both in MRF and shear wall frame combination systems is governed by flexural capacity, and the behavior of the coupling beam in coupled shear walls is governed by shear capacity. During earthquake, infilled brick masonry cracks in a brittle manner although earthquake energy dissipates through both inelastic yielding in beams and columns for MRF and shear wall frame combination systems; whereas, in coupled shear walls, earthquake energy dissipates through inelastic yielding in the coupling beams and at the base of the shear walls. Hence, amount of dissipation of earthquake energy and ductility obtained from both MRF and shear wall frame combination systems are less than those of coupled shear walls system in the high-rise buildings [2–11]. However, the Indian codes of practice governing the earthquake resistant design, such as IS 1893 (Part 1) [1] and IS 4326 [12], do not provide specific guidelines with regard to earthquake resistant design of coupled shear walls. On the other hand, IS 13920 [13] gives credence to the coupled shear walls as an earthquake resistant option but it has incorporated very limited design guidelines of coupling beams that are inadequate for practical applications. It requires further investigations and elaborations before practical use.Further, it is reasonably well established that it is uneconomical to design a structure considering its linear behavior during earthquake as is recognized by the Bureau of Indian Standards [1, 12, 13] currently in use. Hence, an alternative design philosophy needs to be evolved in the Indian context to consider the postyield behavior, wherein the damage state is evaluated through deformation considerations.In the present context, therefore, performance-based seismic design (PBSD) can be considered to offer significantly improved solutions as compared to the conventional design based on linear response spectrum analysis. Performance-based seismic design (PBSD) implies design, evaluation, and construction of engineered facilities whose performance under common and extreme loads responds to the diverse needs and objectives of owners, tenants, and societies at large. The objective of PBSD is to produce structures with predictable seismic performance. In PBSD, multiple levels of earthquake and corresponding expected performance criteria are specified [16]. This aspect emphasizes nonlinear analyses for seismic design verification of any structure. This procedure gives some guidelines for estimating the possible local and global damages of structures. A retrofitted structure can be evaluated with the help of PBSD. Similarly, economics in the form of life-cycle cost along with construction cost of the structure is inherently included in PBSD [21].On the basis of the aforesaid discussion, an effort has been made in this paper to develop a comprehensive procedure for the design of coupled shear walls.


## 2. Investigation of Coupling Beam

Coupled shear walls consist of two shear walls connected intermittently by beams along the height. The behavior of coupled shear walls is mainly governed by the coupling beams. The coupling beams are designed for ductile inelastic behavior in order to dissipate energy. The base of the shear walls may be designed for elastic or ductile inelastic behaviors. The amount of energy dissipation depends on the yield moment capacity and plastic rotation capacity of the coupling beams. If the yield moment capacity is too high, then the coupling beams will undergo only limited rotations and dissipate little energy. On the other hand, if the yield moment capacity is too low, then the coupling beams may undergo rotations much larger than their plastic rotation capacities. Therefore, the coupling beams should be provided with an optimum level of yield moment capacities. These moment capacities depend on the plastic rotation capacity available in beams. The geometry, rotations, and moment capacities of coupling beams have been reviewed based on previous experimental and analytical studies in this paper. An analytical model of coupling beam has also been developed to calculate the rotations of coupling beam with diagonal reinforcement and truss reinforcement.


### 2.1. Geometry of Coupling Beam

The behavior of the reinforced concrete coupling beam may be dominated by (1) shear or by (2) flexure as per ATC 40 [16], FEMA 273 [14], and FEMA 356 [15]. Shear is dominant in coupling beams when ϕ≤2 [5] or Lb/db≤4.There are various types of reinforcements in RCC coupling beams described as follows.


#### 2.1.1. Conventional Reinforcement

Conventional reinforcement consists of longitudinal flexural reinforcement and transverse reinforcement for shear. Longitudinal reinforcement consists of top and bottom steel parallel to the longitudinal axis of the beam. Transverse reinforcement consists of closed stirrups or ties. If the strength of these tiesVs1≥3/4 required shear strength of the beam and the spacing these ties ≤d/3 over the entire length of the beam, then stable hysteresis occurs, and such transverse reinforcement is said to be conforming to the type now mandatory for new construction. If the transverse reinforcement is deficient either in strength or spacing then it leads to pinched hysteresis, and such reinforcement is said to be nonconforming [14–16].According to IS 13920 [13], the spacing of transverse reinforcement over a length of 2d at either end of a beam shall not exceed (a) d/4 and (b) 8 times the diameter of the smallest longitudinal bar; however, it need not be less than 100 mm. Elsewhere, the beam shall have transverse reinforcement at a spacing not exceeding d/2. Whereas, the shear force to be resisted by the transverse reinforcement shall be the maximum of (1) calculated factored shear force as per analysis and (2) shear force due to formation of plastic hinges at both ends of the beam plus the factored gravity load on the span.


#### 2.1.2. Diagonal Reinforcement

Diagonal reinforcement consists of minimum four bars per diagonal. It gives a better plastic rotation capacity compared to conventional coupling beam during an earthquake whenLb/db<1.5 as per Penelis and Kappos [5]. The provisions for diagonal reinforcement according to different codes have already been shown in Table 1. If the diagonal reinforcement is subjected to compressive loading it may buckle, in which case it cannot be relied on to continue resisting compressive loading. Under the action of reversing loads, reinforcement that buckles in compression with loading in one direction may be stressed in tension with loading in the opposite direction. This action may lead to low-cycle fatigue failure, so that the reinforcement cannot continue to resist tensile forces. For this reason, it is necessary to ensure that this reinforcement does not buckle. To prevent buckling due to compressive loading, the spacing between two adjacent diagonal bars should be greater than (Ld/12) [based on buckling condition of a column]. It has also been noticed from Englekirk [3] that diagonal reinforcement should not be attempted in walls that are less than 16 in. (406.4 mm) thick. Unless strength is an overriding consideration, diagonally reinforced coupling beams should not be used.Table 1
Rotation capacities for coupling beams controlled by flexure as per FEMA 273 [14] and FEMA 356 [15].
Type of coupling beam
Conditions
Plastic Rotation Capacity (Radians)
Shear
/
t
w
L
w
f
c
′
IO
LS
CP
Conventional longitudinal reinforcement with conforming transverse reinforcement
≤3
0.006
0.015
0.025
≥6
0.005
0.010
0.015
Conventional longitudinal reinforcement with non-conforming transverse reinforcement
≤3
0.006
0.012
0.020
≥6
0.005
0.008
0.010
Diagonal Reinforcement
NA
0.006
0.018
0.030
Flexure dominant steel coupling beam
b
f
/
2
t
f
≤
52
/
F
y and h/tw′≤418/Fy
1
θ
y
6
θ
y
8
θ
y
b
f
/
2
t
f
≥
65
/
F
y and h/tw′≥640/Fy
0.25
θ
y
2
θ
y
3
θ
y


#### 2.1.3. Truss Reinforcement

Truss reinforcement represents a significant and promising departure from traditional coupling beam reinforcements. The primary load transfer mechanism of the system is represented by the truss taken to its yield capacity. A secondary load path is created by the global strut and tie. The load transfer limit state will coincide with the yielding of all of the tension diagonals, provided the so-produced compression loads do not exceed the capacity of the concrete compression strut. The yield strength of the primary truss is governed by the tensile strength of its diagonal; whereas, the primary truss transfer mechanism must include the shear travelling along the compression diagonal. According to Penelis and Kappos [5] and Galano and Vignoli [17], when 1.5≤Lb/db≤4, truss reinforcement offers best seismic performance in comparison with conventional and diagonal reinforcements. This type of reinforcement has not been used till now. Detailing and placement problems must be carefully studied if their use is contemplated. Clearly, additional experimentation is required because the system appears to have merit, especially in thin walls [3].When the postyield rotational level is much higher compared to rotational level for truss reinforcement, then steel beam can be provided as a coupling beam. There are two types of steel beams which are provided as coupling beams based on the following factors as per Englekirk [3], AISC [26], AISC [27], and AISC [28].(a) Shear Dominant. In this beam, the shear capacity Vsp is attained, and the corresponding bending moment is equal to Vsp×e/2 which must be less than 0.8Mp. The postyield deformation is accommodated by shear and it is presumed e≤1.6Mp/Vsp; where e= clear span of the coupling beam + 2 × concrete cover of shear wall, Mp= moment capacity of coupling beam, and Vsp= shear capacity of coupling beam.(b) Flexure Dominant. In this beam, the bending moment capacity Mp is attained, and the corresponding shear force is equal to 2Mp/e which must be less than 0.8Vsp. The postyield deformation is accommodated by flexure and it is presumed e≥2.6Mp/Vsp.


### 2.2. Moment Capacity of Coupling Beam

The bending moment capacity of coupling beam depends on the geometry and material property of coupling beam. Bending moment capacity and shear force capacity of the coupling beam are related with each other. Englekirk [3], Park, and Paulay [4], Paulay [29], Harries et al. [30], AISC [26], AISC [27] and AISC [28] describe these capacities as follows.


#### 2.2.1. Reinforced Concrete Coupling Beam

Shear capacity of coupling beam with conventional reinforcement can be calculated as(1)Vsp=2MpLb=2Asfy(d-d′)Lb.
Whereas, shear capacity of coupling beam with diagonal reinforcement can be calculated as
(2)Vsp=Vf+Vd1=2MpLb+2T1sinθ=2Asfy(d-d′)Lb+4As′′fy((db/2)-d′)Ld,
and shear capacity of coupling beam with truss reinforcement is as
(3)Vsp=2V1+V2=2Td(d-d′Ld1)+2T′(d-d′Ld2)=2As′′fy(d-d′)Ld1+2(As-As′′)fy(d-d′)Ld2,
where, Ld1=(Lb/2)2+(d-d′)2 and Ld2=(Lb)2+(d-d′)2. All three shear capacities must be less than equal to [(0.35fc′bbw(d-d′)/Ldλ0] or [(0.35fc′bb(0.25Ld)(d-d′)/Ldλ0] or [(0.08fc′bb(d-d′)/λ0] which is based on the statement, that is, capacity of a concrete strut in cylindrical elements will diminish to a level of 30 to 35% of fc′ as cracking increases, where, λ0 is member over strength factor of 1.25.


#### 2.2.2. Shear Dominant Steel Coupling Beam

For I-section type of steel coupling beam, shear capacity (permissible shear resisted by web only) for shear dominant steel coupling beam is denoted asVsp=0.6Fytw′(D-2tf) and moment capacity is Mp=ZpFy; where, Fy is yield stress of structural steel, tw′ is web thickness, D is the overall depth of the section, tf is flange thickness, and Zp is plastic section modulus.


#### 2.2.3. Flexure Dominant Steel Coupling Beam

The transferable shear force (Vnf) for flexure dominant steel coupling beam is the lesser of 2Mp/e and Vsp; where, Mp is the moment capacity which is ZpFy.


### 2.3. Rotational Capacity of Coupling Beam

The rotation capacity in coupling beams depends upon the type of coupling beam. When the rotational demand is greater than rotational capacity of RCC coupling beam with conventional flexural and shear reinforcement then diagonal or truss reinforcement type of coupling beam could be provided depending on theLb/db ratio. The steel coupling beam could be used when the rotational limit due to lateral loading exceeds the rotation capacity of RCC coupling beam with truss reinforcement. Various research works conducted by Paulay [31, 32], Hindi, and Sexsmith [33], FEMA356 [15], Xuan et al. [34] describe these capacities. ATC 40 [16], FEMA 273 [14], FEMA 356 [15], Galano and Vignoli [17], Chao et al. [35] and Englekirk [3] describe the following rotational capacities for various types of coupling beams considering the behavior controlled by flexure and shear during earthquake. Tables 1, 2, 3, 4, 5, and 6 show these different rotational capacities for various coupling beams.Table 2
Rotation capacities for coupling beams controlled by shear as per FEMA 273 [14] and FEMA 356 [15].
Type of coupling beam
Conditions
Plastic Rotation Capacity (Radians)
Shear
/
t
w
L
w
f
c
′
IO
LS
CP
Conventional longitudinal reinforcement with conforming transverse reinforcement
≤3
0.006
0.012
0.015
≥6
0.004
0.008
0.010
Conventional longitudinal reinforcement with non-conforming transverse reinforcement
≤3
0.006
0.008
0.010
≥6
0.004
0.006
0.007
Shear dominant steel coupling beam
0.005
0.11
0.14Table 3
Rotation capacities for coupling beams controlled by flexure as per ATC 40 [16].
Type of coupling beam
Conditions
Plastic Rotation Capacity (Radians)
Shear
/
b
w
d
f
c
′
IO
LS
CP
Conventional longitudinal reinforcement with conforming transverse reinforcement
≤3
0.006
0.015
0.025
≥6
0.005
0.010
0.015
Conventional longitudinal reinforcement with non-conforming transverse reinforcement
≤3
0.006
0.012
0.020
≥6
0.005
0.008
0.010
Diagonal reinforcement
NA
0.006
0.018
0.030Table 4
Rotation capacities for coupling beams controlled by shear as per ATC 40 [16].
Type of coupling beam
Conditions
Plastic Rotation Capacity (Radians)
Shear
/
b
w
d
f
c
′
IO
LS
CP
Conventional longitudinal reinforcement with conforming transverse reinforcement
≤3
0.006
0.012
0.015
≥6
0.004
0.008
0.010
Conventional longitudinal reinforcement with non-conforming transverse reinforcement
≤3
0.006
0.008
0.010
≥6
0.004
0.006
0.007Table 5
Rotation capacities for coupling beams as per Galano and Vignoli [17].
Type of coupling beam
Aspect ratio
Rotation Capacity (Radians)
L
b
/
d
b
θ
Lu
Conventional reinforcement
1.5
0.051
Diagonal reinforcement
1.5
0.062
Truss reinforcement
1.5
0.084Table 6
Rotation capacities for coupling beams as per Englekirk [3].
Type of coupling beam
Aspect ratio
Rotation Capacity (Radians)
L
b
/
d
b
θ
u
,
max
Conventional reinforcement
1.5
0.02
Diagonal reinforcement
1.5
0.04
Truss reinforcement
1.5
0.06Shear
/
t
w
L
w
f
c
′ or Shear/bwdfc′≤3 or ≥6 is based on the aspect ratio (Lb/db) of coupling beam and bf/2tf≤52/Fy and h/tw′≤418/Fy or bf/2tf≥65/Fy and h/tw′≥640/Fy are the conditions of the flexure dominant steel coupling beam to prevent local buckling.Specifications in Tables1, 2, 3, and 4 can be questioned on the basis of the following observations:(1)
As per Tables1 and 2, the rotational capacities of beam depends on size of wall (tw,Lw) which is illogical.(2)
When shear span to depth ratioϕ≤2 or aspect ratio Lb/db≤4, the behavior of RCC coupling beams is controlled by shear. For this reason, as aspect ratio (Lb/db) of diagonally reinforced beam is less than 1.5, it means that the behavior of diagonally reinforced beam is controlled by shear. Whereas, Tables 1 and 3 show that diagonally reinforced coupling beam behavior is controlled by flexure which is not acceptable.(3)
Conventional longitudinal reinforcement with nonconforming transverse reinforcement is not accepted for new construction.(4)
If the behavior of coupling beam is controlled by flexure [aspect ratio(Lb/db) is greater than 4], the length of the coupling beam is quite larger. According to Munshi and Ghosh [36], weakly coupled shear walls can be obtained for larger span of the coupling beam and the design results of this type of coupled shear walls are inconsistent with regard to the ductility and energy dissipation during earthquake motion. Hence, it can be said that rotational capacity of coupling beams controlled by flexure as per ATC 40 [16], FEMA 273 [14], and FEMA 356 [15] cannot be accepted.Similarly, specifications in Tables5 and 6 can also be questioned on the basis of following observation.For aspect ratioLb/db=1.5, Galano and Vignoli [17] show different results regarding the ultimate rotation of various RCC coupling beams in comparison with the results made by Englekirk [3].


### 2.4. Analytical Program

The above study shows the inconsistent modeling parameters and inconsistent evaluative parameters. However, the behavior of coupled shear walls is controlled by the characteristics of various coupling beams. These characteristics depend on the following parameters:(1)
Beam span to depth ratio.(2)
Reinforcement details.For this reason, more study is required to investigate into the limitations on behavior of coupling beams. Since computer programme ATENA2D [18] has some advantages in comparison with other software packages like SAP V 10.0.5 [23], ATENA2D [18] was considered to carry out this study. The advantages as well as disadvantage of ATENA2D [18] are as follows.


### 2.5. Advantages of ATENA2D Are

(i)
Material, element, and reinforcement can be modeled individually, and
(ii)
Geometric and material nonlinearity can be provided.


### 2.6. Disadvantage of ATENA2D Is

(i)
Only static loading in one direction can be applied.


### 2.7. Reinforcement Layouts

There were eighteen RCC coupling beams and three different reinforcement layouts considered in the analytical program using ATENA2D [18]: (a) longitudinal with conforming transverse ties, (b) diagonal with conforming transverse ties around the main bars, and (c) truss with conforming transverse ties around the main bars. For each layout, the cross section of the coupling beam was considered as 600 mm (depth,db)×300 mm (width,bb) and the beam span-depth ratio (Lb/db) was considered as 1, 1.5, and 2.


### 2.8. Materials

The concrete (M20 grade) and steel (Fe 415 grade) were considered as two materials to model the coupled shear walls. The Poisson’s ratio was considered as 0.2. The unit-weight of concrete was considered as 23 kN/m3 and the unit-weight of steel was considered as 78.5 kN/m3. Both coupling beam and shear wall elements were assigned as 4-noded quadrilateral elements; material in coupling beam was assigned as SBeta (inelastic), whereas, material in shear wall was assigned as plane stress elastic isotropic.


### 2.9. Investigative Model

Figure1 and Table 7 describe the investigative models considered for ATENA2D [18] analysis. The behaviors of all eighteen coupling beams were governed by shear. The load (F) was calculated based on the shear force in beam and other parameters according to the provisions of FEMA 273 [14] and FEMA 356 [15].Table 7
(a) investigative model of coupling beam in ATENA2D [18] as per IS 13920 [13], IS 456 [19], SP-16 [20], FEMA 273 [14], and FEMA 356 [15]. (b) investigative model of coupling beam in ATENA2D  [18] as per IS 13920 [13], IS 456 [19], SP-16 [20], FEMA 273 [14], and FEMA 356 [15]. (c) investigative model of coupling beam in ATENA2D [18] as per IS 13920 [13], IS 456 [19], SP-16 [20], FEMA 273 [14], and FEMA 356 [15].
(a)
Coupling beam
Type
L
b (m)
Shear
/
t
w
L
w
f
c
′
F (kN)
Reinforced steel
Longitudinal
Transverse
Conventional beam with longitudinal and transverse conforming reinforcement
0.6
≤3
585.4
8–10ϕ
2-legged 16ϕ@200 c/c
≥6
1171
8–20ϕ
2-legged 25ϕ@200 c/c
0.9
≤3
623.5
8–10ϕ
2-legged 16ϕ@200 c/c
≥6
1247
8–20ϕ
2-legged 25ϕ@200 c/c
1.2
≤3
661.7
8–10ϕ
2-legged 16ϕ@200 c/c
≥6
1323
8–20ϕ
2-legged 25ϕ@200 c/c
(b)
Coupling beam
Type
L
b (m)
Shear
/
t
w
L
w
f
c
′
F (kN)
Reinforced steel
Longitudinal
Transverse
Beam with diagonal reinforcement
0.6
≤3
585.4
8–10ϕ+4–20ϕ as  one diagonal
2-legged 16ϕ@200 c/c
≥6
1171
8–20ϕ+4–30ϕ as one diagonal
2-legged 25ϕ@200 c/c
0.9
≤3
623.5
8–10ϕ+4–25ϕ as one diagonal
2-legged 16ϕ@200 c/c
≥6
1247
8–20ϕ+4–30ϕ as one diagonal
2-legged 25ϕ@200 c/c
1.2
≤3
661.7
8–10ϕ+4–25ϕ as one diagonal
2-legged 16ϕ@200 c/c
≥6
1323
8–20ϕ+4–35ϕ as one diagonal
2-legged 25ϕ@200 c/c
(c)
Beam
Type
L
b (m)
Shear
/
t
w
L
w
f
c
′
F (kN)
Reinforced steel
Longitudinal
Transverse
Beam with truss reinforcement
0.6
≤3
585.4
8–10ϕ+4–30ϕ as one truss
2-legged 16ϕ@200 c/c
≥6
1171
8–20ϕ+4–45ϕ as one truss
2-legged 25ϕ@200 c/c
0.9
≤3
623.5
8–10ϕ+4–30ϕ as one truss
2-legged 16ϕ@200 c/c
≥6
1247
8–20ϕ+4–40ϕ as one truss
2-legged 25ϕ@200 c/c
1.2
≤3
661.7
8–10ϕ+4–30ϕ as one truss
2-legged 16ϕ@200 c/c
≥6
1323
8–20ϕ+4–40ϕ as one truss
2-legged 25ϕ@200 c/cFigure 1
Initial sketch of the analytical model.The depth of the wall is considered asLw=4 m, thickness of the wall is considered as tw=300 mm, and minimum reinforcement in the shear wall is taken as 0.25% of its gross area @450 c/c.Here, Young’s modulus for concrete in beam=Ecb=2.24×104 MPa, Young’s modulus for steel in beam =Esb=2.1×105 MPa, Young’s modulus for concrete in wall =Ecw=2.24×104 MPa, and Young’s modulus for steel in wall =Esw=2.1×105 MPa.


### 2.10. Results and Discussions

The results using ATENA2D [18] have been tabulated in Table 8. It shows the comparison of rotational limit at CP level among FEMA 273 [14], FEMA 356 [15], and ATENA2D [18]. There are a lot of differences among the results of FEMA 273 [14], FEMA 356 [15], and ATENA2D [18]. The comparison has also been extended for considering ATC 40’s [16] provisions. There are also big differences between the results of ATC 40 [16] and ATENA2D [18] shown in Table 9. It may be because of the limitations of ATENA2D [18] software. However, it is unexpected in FEMA 273 [14], FEMA 356 [15], and ATC 40 [16] that the rotational limit is more or less same, whereas, the parameters considered for calculation of shear strength are different. Therefore, it can be said that the parameters given in FEMA 273 [14], FEMA 356 [15], and ATC 40 [16] are questionable which have already been discussed in this paper. It has also been observed from Tables 8 and 9 that crack width in beam is quite significant although the rotational values in ATENA2D [18] are unexpectedly varying with FEMA 273 [14], FEMA 356 [15], and ATC 40 [16].Table 8
Compare the Modeling Parameters and Numerical Acceptance Criteria with FEMA 273 [14] and FEMA 356 [15].
Longitudinal reinforcement and transverse reinforcement
Shear
/
t
w
L
w
f
c
′
Rotational limit at collapse prevention level (CP) in radians
Crack width in coupling beam  at CP level in meters by    ATENA2D [18]
Member controlled   by flexure
Member controlled   by shear
ATENA2D [18]
FEMA 273 [14]
FEMA356 [15]
FEMA 273 [14]
FEMA 356 [15]
L
b
=
0.6 m
L
b
=
0.9 m
L
b
=
1.2 m
L
b
=
0.6 m
L
b
=
0.9 m
L
b
=
1.2 m
Conventional longitudinal reinforcement with conforming transverse reinforcement
≤3
0.025
0.025
0.015
0.020
0.000881
0.00104
0.002325
0.000263
0.000306
0.000559
≥6
0.015
0.02
0.010
0.016
0.00348
0.00528
0.00886
0.0007125
0.001726
0.003124
Diagonal
≤3
0.03
0.03
—
—
0.00235
0.011
0.0111
0.000494
0.004315
0.00372
≥6
0.03
0.03
—
—
0.00292
0.00833
0.00978
0.0005724
0.002961
0.003228
Truss
≤3
NA
NA
NA
NA
0.001176
0.000422
0.00093
0.0003144
0.0001066
0.000204
≥6
NA
NA
NA
NA
0.001413
0.00297
0.0029
0.000344
0.0007514
0.00066Table 9
Compare the Modeling Parameters and Numerical Acceptance Criteria with ATC 40 [16].
Longitudinal reinforcement and transverse reinforcement
Shear
/
b
w
d
f
c
′
Rotational limit at collapse prevention level (CP) in radians
Crack width in coupling beam  at CP level in meters by    ATENA2D [18]
Member controlled by flexure
Member controlled by shear
ATENA2D [18]
ATC 40 [16]
ATC 40 [16]
L
b
=
0.6 m
L
b
=
0.9 m
L
b
=
1.2 m
L
b
=
0.6 m
L
b
=
0.9 m
L
b
=
1.2 m
Conventional longitudinal reinforcement with conforming transverse reinforcement
≤3
0.025
0.018
0.0001023
0.000784
0.00198
0.0000001308
0.0005
0.001613
≥6
0.015
0.012
0.0002423
0.001944
0.00344
0.00163
0.00136
0.00297
Diagonal
≤3
0.03
—
0.00012
0.000416
0.00055
0.0000194
0.0002184
0.00021
≥6
0.03
—
0.000415
0.000422
0.001533
0.0001795
0.0001483
0.00093Hence, the results obtained from the above study using ATENA2D [18] were found unsatisfactory. Therefore, a new model has been created with some assumptions in the manner shown in Figure 2 to carryout further study.Figure 2
Schematic diagram of coupling beam.


### 2.11. Assumptions

(i)
The effect of gravity loads on the coupling beams has been neglected.
(ii)
Deflection of the coupling beam occurs due to lateral loading.
(iii)
Contra flexure occurs at the mid-span of the coupling beam.
(iv)
The confined concrete, due to the confining action is provided by closely spaced transverse reinforcement in concrete, is assumed to govern the strength.Total elongation in the horizontal direction (Figure2) due to lateral loading can be written as
(4)ΔLb=db×θb,
and strain in the concrete
(5)εc=ΔLbLb.
Hence, considering (4) and (5) the following equation can be written as
(6)coupling  beam  rotation,θb=εc×Lbdb.
The results, considering (6) with maximum strain in confined concrete (εcu) of 0.02 (Confining action is provided by closely spaced transverse reinforcement in concrete as per ATC 40 [16]), have been tabulated in Table 10.Table 10
Maximum rotations in radians.
Type of eeinforcement
L
b
/
d
b
Value as per (6)
Galano and Vignoli [17]
Englekirk [3]
ATC40 [16], FEMA 273 [14] and FEMA 356  [15]
Diagonal
<1.5
<0.03
0.062
0.04
0.03
Truss
1.5 to 4.0
0.03 to 0.08
0.084
0.06
—It can be observed from Table10 that the values obtained as per (6) have similar trend with the values specified by ATC 40 [16], FEMA 273 [14], FEMA 356 [15], Galano and Vignoli [17], and Englekirk [3].Based on the above study, Table11 has been prepared containing modified parameters governing the coupling beam characteristics, which are also considered for the developments of the design technique discussed below. As design technique is based on collapse prevention (CP) level of structure, plastic rotation capacity given in Table 11 is for CP level only.Table 11
Modified Parameters governing the coupling beam characteristics controlled by shear.
Type of coupling beam
Shear Span to Depth Ratio
L
b
/
d
b
Type of detailing
Plastic Rotation Capacity (Radians)
Shear
/
b
w
d
f
c
′
CP
Reinforced concrete coupling beam
ϕ
≤
2
No limit
Conventional longitudinal reinforcement with conforming transverse reinforcement
≤3
0.015
≥6
0.010
<1.5
Diagonal Reinforcement (strength is an overriding consideration and thickness of wall should be greater than 406.4 mm)
—
<0.03
1.5 to 4.0
Truss Reinforcement (additional experimentation is required)
—
0.03–0.08
Steel coupling beam
e
≤
1.6
M
p
/
V
s
p
Shear dominant
—
0.15
/
L
b


## 2.1. Geometry of Coupling Beam

The behavior of the reinforced concrete coupling beam may be dominated by (1) shear or by (2) flexure as per ATC 40 [16], FEMA 273 [14], and FEMA 356 [15]. Shear is dominant in coupling beams when ϕ≤2 [5] or Lb/db≤4.There are various types of reinforcements in RCC coupling beams described as follows.


### 2.1.1. Conventional Reinforcement

Conventional reinforcement consists of longitudinal flexural reinforcement and transverse reinforcement for shear. Longitudinal reinforcement consists of top and bottom steel parallel to the longitudinal axis of the beam. Transverse reinforcement consists of closed stirrups or ties. If the strength of these tiesVs1≥3/4 required shear strength of the beam and the spacing these ties ≤d/3 over the entire length of the beam, then stable hysteresis occurs, and such transverse reinforcement is said to be conforming to the type now mandatory for new construction. If the transverse reinforcement is deficient either in strength or spacing then it leads to pinched hysteresis, and such reinforcement is said to be nonconforming [14–16].According to IS 13920 [13], the spacing of transverse reinforcement over a length of 2d at either end of a beam shall not exceed (a) d/4 and (b) 8 times the diameter of the smallest longitudinal bar; however, it need not be less than 100 mm. Elsewhere, the beam shall have transverse reinforcement at a spacing not exceeding d/2. Whereas, the shear force to be resisted by the transverse reinforcement shall be the maximum of (1) calculated factored shear force as per analysis and (2) shear force due to formation of plastic hinges at both ends of the beam plus the factored gravity load on the span.


### 2.1.2. Diagonal Reinforcement

Diagonal reinforcement consists of minimum four bars per diagonal. It gives a better plastic rotation capacity compared to conventional coupling beam during an earthquake whenLb/db<1.5 as per Penelis and Kappos [5]. The provisions for diagonal reinforcement according to different codes have already been shown in Table 1. If the diagonal reinforcement is subjected to compressive loading it may buckle, in which case it cannot be relied on to continue resisting compressive loading. Under the action of reversing loads, reinforcement that buckles in compression with loading in one direction may be stressed in tension with loading in the opposite direction. This action may lead to low-cycle fatigue failure, so that the reinforcement cannot continue to resist tensile forces. For this reason, it is necessary to ensure that this reinforcement does not buckle. To prevent buckling due to compressive loading, the spacing between two adjacent diagonal bars should be greater than (Ld/12) [based on buckling condition of a column]. It has also been noticed from Englekirk [3] that diagonal reinforcement should not be attempted in walls that are less than 16 in. (406.4 mm) thick. Unless strength is an overriding consideration, diagonally reinforced coupling beams should not be used.Table 1
Rotation capacities for coupling beams controlled by flexure as per FEMA 273 [14] and FEMA 356 [15].
Type of coupling beam
Conditions
Plastic Rotation Capacity (Radians)
Shear
/
t
w
L
w
f
c
′
IO
LS
CP
Conventional longitudinal reinforcement with conforming transverse reinforcement
≤3
0.006
0.015
0.025
≥6
0.005
0.010
0.015
Conventional longitudinal reinforcement with non-conforming transverse reinforcement
≤3
0.006
0.012
0.020
≥6
0.005
0.008
0.010
Diagonal Reinforcement
NA
0.006
0.018
0.030
Flexure dominant steel coupling beam
b
f
/
2
t
f
≤
52
/
F
y and h/tw′≤418/Fy
1
θ
y
6
θ
y
8
θ
y
b
f
/
2
t
f
≥
65
/
F
y and h/tw′≥640/Fy
0.25
θ
y
2
θ
y
3
θ
y


### 2.1.3. Truss Reinforcement

Truss reinforcement represents a significant and promising departure from traditional coupling beam reinforcements. The primary load transfer mechanism of the system is represented by the truss taken to its yield capacity. A secondary load path is created by the global strut and tie. The load transfer limit state will coincide with the yielding of all of the tension diagonals, provided the so-produced compression loads do not exceed the capacity of the concrete compression strut. The yield strength of the primary truss is governed by the tensile strength of its diagonal; whereas, the primary truss transfer mechanism must include the shear travelling along the compression diagonal. According to Penelis and Kappos [5] and Galano and Vignoli [17], when 1.5≤Lb/db≤4, truss reinforcement offers best seismic performance in comparison with conventional and diagonal reinforcements. This type of reinforcement has not been used till now. Detailing and placement problems must be carefully studied if their use is contemplated. Clearly, additional experimentation is required because the system appears to have merit, especially in thin walls [3].When the postyield rotational level is much higher compared to rotational level for truss reinforcement, then steel beam can be provided as a coupling beam. There are two types of steel beams which are provided as coupling beams based on the following factors as per Englekirk [3], AISC [26], AISC [27], and AISC [28].(a) Shear Dominant. In this beam, the shear capacity Vsp is attained, and the corresponding bending moment is equal to Vsp×e/2 which must be less than 0.8Mp. The postyield deformation is accommodated by shear and it is presumed e≤1.6Mp/Vsp; where e= clear span of the coupling beam + 2 × concrete cover of shear wall, Mp= moment capacity of coupling beam, and Vsp= shear capacity of coupling beam.(b) Flexure Dominant. In this beam, the bending moment capacity Mp is attained, and the corresponding shear force is equal to 2Mp/e which must be less than 0.8Vsp. The postyield deformation is accommodated by flexure and it is presumed e≥2.6Mp/Vsp.


## 2.1.1. Conventional Reinforcement

Conventional reinforcement consists of longitudinal flexural reinforcement and transverse reinforcement for shear. Longitudinal reinforcement consists of top and bottom steel parallel to the longitudinal axis of the beam. Transverse reinforcement consists of closed stirrups or ties. If the strength of these tiesVs1≥3/4 required shear strength of the beam and the spacing these ties ≤d/3 over the entire length of the beam, then stable hysteresis occurs, and such transverse reinforcement is said to be conforming to the type now mandatory for new construction. If the transverse reinforcement is deficient either in strength or spacing then it leads to pinched hysteresis, and such reinforcement is said to be nonconforming [14–16].According to IS 13920 [13], the spacing of transverse reinforcement over a length of 2d at either end of a beam shall not exceed (a) d/4 and (b) 8 times the diameter of the smallest longitudinal bar; however, it need not be less than 100 mm. Elsewhere, the beam shall have transverse reinforcement at a spacing not exceeding d/2. Whereas, the shear force to be resisted by the transverse reinforcement shall be the maximum of (1) calculated factored shear force as per analysis and (2) shear force due to formation of plastic hinges at both ends of the beam plus the factored gravity load on the span.


## 2.1.2. Diagonal Reinforcement

Diagonal reinforcement consists of minimum four bars per diagonal. It gives a better plastic rotation capacity compared to conventional coupling beam during an earthquake whenLb/db<1.5 as per Penelis and Kappos [5]. The provisions for diagonal reinforcement according to different codes have already been shown in Table 1. If the diagonal reinforcement is subjected to compressive loading it may buckle, in which case it cannot be relied on to continue resisting compressive loading. Under the action of reversing loads, reinforcement that buckles in compression with loading in one direction may be stressed in tension with loading in the opposite direction. This action may lead to low-cycle fatigue failure, so that the reinforcement cannot continue to resist tensile forces. For this reason, it is necessary to ensure that this reinforcement does not buckle. To prevent buckling due to compressive loading, the spacing between two adjacent diagonal bars should be greater than (Ld/12) [based on buckling condition of a column]. It has also been noticed from Englekirk [3] that diagonal reinforcement should not be attempted in walls that are less than 16 in. (406.4 mm) thick. Unless strength is an overriding consideration, diagonally reinforced coupling beams should not be used.Table 1
Rotation capacities for coupling beams controlled by flexure as per FEMA 273 [14] and FEMA 356 [15].
Type of coupling beam
Conditions
Plastic Rotation Capacity (Radians)
Shear
/
t
w
L
w
f
c
′
IO
LS
CP
Conventional longitudinal reinforcement with conforming transverse reinforcement
≤3
0.006
0.015
0.025
≥6
0.005
0.010
0.015
Conventional longitudinal reinforcement with non-conforming transverse reinforcement
≤3
0.006
0.012
0.020
≥6
0.005
0.008
0.010
Diagonal Reinforcement
NA
0.006
0.018
0.030
Flexure dominant steel coupling beam
b
f
/
2
t
f
≤
52
/
F
y and h/tw′≤418/Fy
1
θ
y
6
θ
y
8
θ
y
b
f
/
2
t
f
≥
65
/
F
y and h/tw′≥640/Fy
0.25
θ
y
2
θ
y
3
θ
y


## 2.1.3. Truss Reinforcement

Truss reinforcement represents a significant and promising departure from traditional coupling beam reinforcements. The primary load transfer mechanism of the system is represented by the truss taken to its yield capacity. A secondary load path is created by the global strut and tie. The load transfer limit state will coincide with the yielding of all of the tension diagonals, provided the so-produced compression loads do not exceed the capacity of the concrete compression strut. The yield strength of the primary truss is governed by the tensile strength of its diagonal; whereas, the primary truss transfer mechanism must include the shear travelling along the compression diagonal. According to Penelis and Kappos [5] and Galano and Vignoli [17], when 1.5≤Lb/db≤4, truss reinforcement offers best seismic performance in comparison with conventional and diagonal reinforcements. This type of reinforcement has not been used till now. Detailing and placement problems must be carefully studied if their use is contemplated. Clearly, additional experimentation is required because the system appears to have merit, especially in thin walls [3].When the postyield rotational level is much higher compared to rotational level for truss reinforcement, then steel beam can be provided as a coupling beam. There are two types of steel beams which are provided as coupling beams based on the following factors as per Englekirk [3], AISC [26], AISC [27], and AISC [28].(a) Shear Dominant. In this beam, the shear capacity Vsp is attained, and the corresponding bending moment is equal to Vsp×e/2 which must be less than 0.8Mp. The postyield deformation is accommodated by shear and it is presumed e≤1.6Mp/Vsp; where e= clear span of the coupling beam + 2 × concrete cover of shear wall, Mp= moment capacity of coupling beam, and Vsp= shear capacity of coupling beam.(b) Flexure Dominant. In this beam, the bending moment capacity Mp is attained, and the corresponding shear force is equal to 2Mp/e which must be less than 0.8Vsp. The postyield deformation is accommodated by flexure and it is presumed e≥2.6Mp/Vsp.


## 2.2. Moment Capacity of Coupling Beam

The bending moment capacity of coupling beam depends on the geometry and material property of coupling beam. Bending moment capacity and shear force capacity of the coupling beam are related with each other. Englekirk [3], Park, and Paulay [4], Paulay [29], Harries et al. [30], AISC [26], AISC [27] and AISC [28] describe these capacities as follows.


### 2.2.1. Reinforced Concrete Coupling Beam

Shear capacity of coupling beam with conventional reinforcement can be calculated as(1)Vsp=2MpLb=2Asfy(d-d′)Lb.
Whereas, shear capacity of coupling beam with diagonal reinforcement can be calculated as
(2)Vsp=Vf+Vd1=2MpLb+2T1sinθ=2Asfy(d-d′)Lb+4As′′fy((db/2)-d′)Ld,
and shear capacity of coupling beam with truss reinforcement is as
(3)Vsp=2V1+V2=2Td(d-d′Ld1)+2T′(d-d′Ld2)=2As′′fy(d-d′)Ld1+2(As-As′′)fy(d-d′)Ld2,
where, Ld1=(Lb/2)2+(d-d′)2 and Ld2=(Lb)2+(d-d′)2. All three shear capacities must be less than equal to [(0.35fc′bbw(d-d′)/Ldλ0] or [(0.35fc′bb(0.25Ld)(d-d′)/Ldλ0] or [(0.08fc′bb(d-d′)/λ0] which is based on the statement, that is, capacity of a concrete strut in cylindrical elements will diminish to a level of 30 to 35% of fc′ as cracking increases, where, λ0 is member over strength factor of 1.25.


### 2.2.2. Shear Dominant Steel Coupling Beam

For I-section type of steel coupling beam, shear capacity (permissible shear resisted by web only) for shear dominant steel coupling beam is denoted asVsp=0.6Fytw′(D-2tf) and moment capacity is Mp=ZpFy; where, Fy is yield stress of structural steel, tw′ is web thickness, D is the overall depth of the section, tf is flange thickness, and Zp is plastic section modulus.


### 2.2.3. Flexure Dominant Steel Coupling Beam

The transferable shear force (Vnf) for flexure dominant steel coupling beam is the lesser of 2Mp/e and Vsp; where, Mp is the moment capacity which is ZpFy.


## 2.2.1. Reinforced Concrete Coupling Beam

Shear capacity of coupling beam with conventional reinforcement can be calculated as(1)Vsp=2MpLb=2Asfy(d-d′)Lb.
Whereas, shear capacity of coupling beam with diagonal reinforcement can be calculated as
(2)Vsp=Vf+Vd1=2MpLb+2T1sinθ=2Asfy(d-d′)Lb+4As′′fy((db/2)-d′)Ld,
and shear capacity of coupling beam with truss reinforcement is as
(3)Vsp=2V1+V2=2Td(d-d′Ld1)+2T′(d-d′Ld2)=2As′′fy(d-d′)Ld1+2(As-As′′)fy(d-d′)Ld2,
where, Ld1=(Lb/2)2+(d-d′)2 and Ld2=(Lb)2+(d-d′)2. All three shear capacities must be less than equal to [(0.35fc′bbw(d-d′)/Ldλ0] or [(0.35fc′bb(0.25Ld)(d-d′)/Ldλ0] or [(0.08fc′bb(d-d′)/λ0] which is based on the statement, that is, capacity of a concrete strut in cylindrical elements will diminish to a level of 30 to 35% of fc′ as cracking increases, where, λ0 is member over strength factor of 1.25.


## 2.2.2. Shear Dominant Steel Coupling Beam

For I-section type of steel coupling beam, shear capacity (permissible shear resisted by web only) for shear dominant steel coupling beam is denoted asVsp=0.6Fytw′(D-2tf) and moment capacity is Mp=ZpFy; where, Fy is yield stress of structural steel, tw′ is web thickness, D is the overall depth of the section, tf is flange thickness, and Zp is plastic section modulus.


## 2.2.3. Flexure Dominant Steel Coupling Beam

The transferable shear force (Vnf) for flexure dominant steel coupling beam is the lesser of 2Mp/e and Vsp; where, Mp is the moment capacity which is ZpFy.


## 2.3. Rotational Capacity of Coupling Beam

The rotation capacity in coupling beams depends upon the type of coupling beam. When the rotational demand is greater than rotational capacity of RCC coupling beam with conventional flexural and shear reinforcement then diagonal or truss reinforcement type of coupling beam could be provided depending on theLb/db ratio. The steel coupling beam could be used when the rotational limit due to lateral loading exceeds the rotation capacity of RCC coupling beam with truss reinforcement. Various research works conducted by Paulay [31, 32], Hindi, and Sexsmith [33], FEMA356 [15], Xuan et al. [34] describe these capacities. ATC 40 [16], FEMA 273 [14], FEMA 356 [15], Galano and Vignoli [17], Chao et al. [35] and Englekirk [3] describe the following rotational capacities for various types of coupling beams considering the behavior controlled by flexure and shear during earthquake. Tables 1, 2, 3, 4, 5, and 6 show these different rotational capacities for various coupling beams.Table 2
Rotation capacities for coupling beams controlled by shear as per FEMA 273 [14] and FEMA 356 [15].
Type of coupling beam
Conditions
Plastic Rotation Capacity (Radians)
Shear
/
t
w
L
w
f
c
′
IO
LS
CP
Conventional longitudinal reinforcement with conforming transverse reinforcement
≤3
0.006
0.012
0.015
≥6
0.004
0.008
0.010
Conventional longitudinal reinforcement with non-conforming transverse reinforcement
≤3
0.006
0.008
0.010
≥6
0.004
0.006
0.007
Shear dominant steel coupling beam
0.005
0.11
0.14Table 3
Rotation capacities for coupling beams controlled by flexure as per ATC 40 [16].
Type of coupling beam
Conditions
Plastic Rotation Capacity (Radians)
Shear
/
b
w
d
f
c
′
IO
LS
CP
Conventional longitudinal reinforcement with conforming transverse reinforcement
≤3
0.006
0.015
0.025
≥6
0.005
0.010
0.015
Conventional longitudinal reinforcement with non-conforming transverse reinforcement
≤3
0.006
0.012
0.020
≥6
0.005
0.008
0.010
Diagonal reinforcement
NA
0.006
0.018
0.030Table 4
Rotation capacities for coupling beams controlled by shear as per ATC 40 [16].
Type of coupling beam
Conditions
Plastic Rotation Capacity (Radians)
Shear
/
b
w
d
f
c
′
IO
LS
CP
Conventional longitudinal reinforcement with conforming transverse reinforcement
≤3
0.006
0.012
0.015
≥6
0.004
0.008
0.010
Conventional longitudinal reinforcement with non-conforming transverse reinforcement
≤3
0.006
0.008
0.010
≥6
0.004
0.006
0.007Table 5
Rotation capacities for coupling beams as per Galano and Vignoli [17].
Type of coupling beam
Aspect ratio
Rotation Capacity (Radians)
L
b
/
d
b
θ
Lu
Conventional reinforcement
1.5
0.051
Diagonal reinforcement
1.5
0.062
Truss reinforcement
1.5
0.084Table 6
Rotation capacities for coupling beams as per Englekirk [3].
Type of coupling beam
Aspect ratio
Rotation Capacity (Radians)
L
b
/
d
b
θ
u
,
max
Conventional reinforcement
1.5
0.02
Diagonal reinforcement
1.5
0.04
Truss reinforcement
1.5
0.06Shear
/
t
w
L
w
f
c
′ or Shear/bwdfc′≤3 or ≥6 is based on the aspect ratio (Lb/db) of coupling beam and bf/2tf≤52/Fy and h/tw′≤418/Fy or bf/2tf≥65/Fy and h/tw′≥640/Fy are the conditions of the flexure dominant steel coupling beam to prevent local buckling.Specifications in Tables1, 2, 3, and 4 can be questioned on the basis of the following observations:(1)
As per Tables1 and 2, the rotational capacities of beam depends on size of wall (tw,Lw) which is illogical.(2)
When shear span to depth ratioϕ≤2 or aspect ratio Lb/db≤4, the behavior of RCC coupling beams is controlled by shear. For this reason, as aspect ratio (Lb/db) of diagonally reinforced beam is less than 1.5, it means that the behavior of diagonally reinforced beam is controlled by shear. Whereas, Tables 1 and 3 show that diagonally reinforced coupling beam behavior is controlled by flexure which is not acceptable.(3)
Conventional longitudinal reinforcement with nonconforming transverse reinforcement is not accepted for new construction.(4)
If the behavior of coupling beam is controlled by flexure [aspect ratio(Lb/db) is greater than 4], the length of the coupling beam is quite larger. According to Munshi and Ghosh [36], weakly coupled shear walls can be obtained for larger span of the coupling beam and the design results of this type of coupled shear walls are inconsistent with regard to the ductility and energy dissipation during earthquake motion. Hence, it can be said that rotational capacity of coupling beams controlled by flexure as per ATC 40 [16], FEMA 273 [14], and FEMA 356 [15] cannot be accepted.Similarly, specifications in Tables5 and 6 can also be questioned on the basis of following observation.For aspect ratioLb/db=1.5, Galano and Vignoli [17] show different results regarding the ultimate rotation of various RCC coupling beams in comparison with the results made by Englekirk [3].


## 2.4. Analytical Program

The above study shows the inconsistent modeling parameters and inconsistent evaluative parameters. However, the behavior of coupled shear walls is controlled by the characteristics of various coupling beams. These characteristics depend on the following parameters:(1)
Beam span to depth ratio.(2)
Reinforcement details.For this reason, more study is required to investigate into the limitations on behavior of coupling beams. Since computer programme ATENA2D [18] has some advantages in comparison with other software packages like SAP V 10.0.5 [23], ATENA2D [18] was considered to carry out this study. The advantages as well as disadvantage of ATENA2D [18] are as follows.


## 2.5. Advantages of ATENA2D Are

(i)
Material, element, and reinforcement can be modeled individually, and
(ii)
Geometric and material nonlinearity can be provided.


## 2.6. Disadvantage of ATENA2D Is

(i)
Only static loading in one direction can be applied.


## 2.7. Reinforcement Layouts

There were eighteen RCC coupling beams and three different reinforcement layouts considered in the analytical program using ATENA2D [18]: (a) longitudinal with conforming transverse ties, (b) diagonal with conforming transverse ties around the main bars, and (c) truss with conforming transverse ties around the main bars. For each layout, the cross section of the coupling beam was considered as 600 mm (depth,db)×300 mm (width,bb) and the beam span-depth ratio (Lb/db) was considered as 1, 1.5, and 2.


## 2.8. Materials

The concrete (M20 grade) and steel (Fe 415 grade) were considered as two materials to model the coupled shear walls. The Poisson’s ratio was considered as 0.2. The unit-weight of concrete was considered as 23 kN/m3 and the unit-weight of steel was considered as 78.5 kN/m3. Both coupling beam and shear wall elements were assigned as 4-noded quadrilateral elements; material in coupling beam was assigned as SBeta (inelastic), whereas, material in shear wall was assigned as plane stress elastic isotropic.


## 2.9. Investigative Model

Figure1 and Table 7 describe the investigative models considered for ATENA2D [18] analysis. The behaviors of all eighteen coupling beams were governed by shear. The load (F) was calculated based on the shear force in beam and other parameters according to the provisions of FEMA 273 [14] and FEMA 356 [15].Table 7
(a) investigative model of coupling beam in ATENA2D [18] as per IS 13920 [13], IS 456 [19], SP-16 [20], FEMA 273 [14], and FEMA 356 [15]. (b) investigative model of coupling beam in ATENA2D  [18] as per IS 13920 [13], IS 456 [19], SP-16 [20], FEMA 273 [14], and FEMA 356 [15]. (c) investigative model of coupling beam in ATENA2D [18] as per IS 13920 [13], IS 456 [19], SP-16 [20], FEMA 273 [14], and FEMA 356 [15].
(a)
Coupling beam
Type
L
b (m)
Shear
/
t
w
L
w
f
c
′
F (kN)
Reinforced steel
Longitudinal
Transverse
Conventional beam with longitudinal and transverse conforming reinforcement
0.6
≤3
585.4
8–10ϕ
2-legged 16ϕ@200 c/c
≥6
1171
8–20ϕ
2-legged 25ϕ@200 c/c
0.9
≤3
623.5
8–10ϕ
2-legged 16ϕ@200 c/c
≥6
1247
8–20ϕ
2-legged 25ϕ@200 c/c
1.2
≤3
661.7
8–10ϕ
2-legged 16ϕ@200 c/c
≥6
1323
8–20ϕ
2-legged 25ϕ@200 c/c
(b)
Coupling beam
Type
L
b (m)
Shear
/
t
w
L
w
f
c
′
F (kN)
Reinforced steel
Longitudinal
Transverse
Beam with diagonal reinforcement
0.6
≤3
585.4
8–10ϕ+4–20ϕ as  one diagonal
2-legged 16ϕ@200 c/c
≥6
1171
8–20ϕ+4–30ϕ as one diagonal
2-legged 25ϕ@200 c/c
0.9
≤3
623.5
8–10ϕ+4–25ϕ as one diagonal
2-legged 16ϕ@200 c/c
≥6
1247
8–20ϕ+4–30ϕ as one diagonal
2-legged 25ϕ@200 c/c
1.2
≤3
661.7
8–10ϕ+4–25ϕ as one diagonal
2-legged 16ϕ@200 c/c
≥6
1323
8–20ϕ+4–35ϕ as one diagonal
2-legged 25ϕ@200 c/c
(c)
Beam
Type
L
b (m)
Shear
/
t
w
L
w
f
c
′
F (kN)
Reinforced steel
Longitudinal
Transverse
Beam with truss reinforcement
0.6
≤3
585.4
8–10ϕ+4–30ϕ as one truss
2-legged 16ϕ@200 c/c
≥6
1171
8–20ϕ+4–45ϕ as one truss
2-legged 25ϕ@200 c/c
0.9
≤3
623.5
8–10ϕ+4–30ϕ as one truss
2-legged 16ϕ@200 c/c
≥6
1247
8–20ϕ+4–40ϕ as one truss
2-legged 25ϕ@200 c/c
1.2
≤3
661.7
8–10ϕ+4–30ϕ as one truss
2-legged 16ϕ@200 c/c
≥6
1323
8–20ϕ+4–40ϕ as one truss
2-legged 25ϕ@200 c/cFigure 1
Initial sketch of the analytical model.The depth of the wall is considered asLw=4 m, thickness of the wall is considered as tw=300 mm, and minimum reinforcement in the shear wall is taken as 0.25% of its gross area @450 c/c.Here, Young’s modulus for concrete in beam=Ecb=2.24×104 MPa, Young’s modulus for steel in beam =Esb=2.1×105 MPa, Young’s modulus for concrete in wall =Ecw=2.24×104 MPa, and Young’s modulus for steel in wall =Esw=2.1×105 MPa.


## 2.10. Results and Discussions

The results using ATENA2D [18] have been tabulated in Table 8. It shows the comparison of rotational limit at CP level among FEMA 273 [14], FEMA 356 [15], and ATENA2D [18]. There are a lot of differences among the results of FEMA 273 [14], FEMA 356 [15], and ATENA2D [18]. The comparison has also been extended for considering ATC 40’s [16] provisions. There are also big differences between the results of ATC 40 [16] and ATENA2D [18] shown in Table 9. It may be because of the limitations of ATENA2D [18] software. However, it is unexpected in FEMA 273 [14], FEMA 356 [15], and ATC 40 [16] that the rotational limit is more or less same, whereas, the parameters considered for calculation of shear strength are different. Therefore, it can be said that the parameters given in FEMA 273 [14], FEMA 356 [15], and ATC 40 [16] are questionable which have already been discussed in this paper. It has also been observed from Tables 8 and 9 that crack width in beam is quite significant although the rotational values in ATENA2D [18] are unexpectedly varying with FEMA 273 [14], FEMA 356 [15], and ATC 40 [16].Table 8
Compare the Modeling Parameters and Numerical Acceptance Criteria with FEMA 273 [14] and FEMA 356 [15].
Longitudinal reinforcement and transverse reinforcement
Shear
/
t
w
L
w
f
c
′
Rotational limit at collapse prevention level (CP) in radians
Crack width in coupling beam  at CP level in meters by    ATENA2D [18]
Member controlled   by flexure
Member controlled   by shear
ATENA2D [18]
FEMA 273 [14]
FEMA356 [15]
FEMA 273 [14]
FEMA 356 [15]
L
b
=
0.6 m
L
b
=
0.9 m
L
b
=
1.2 m
L
b
=
0.6 m
L
b
=
0.9 m
L
b
=
1.2 m
Conventional longitudinal reinforcement with conforming transverse reinforcement
≤3
0.025
0.025
0.015
0.020
0.000881
0.00104
0.002325
0.000263
0.000306
0.000559
≥6
0.015
0.02
0.010
0.016
0.00348
0.00528
0.00886
0.0007125
0.001726
0.003124
Diagonal
≤3
0.03
0.03
—
—
0.00235
0.011
0.0111
0.000494
0.004315
0.00372
≥6
0.03
0.03
—
—
0.00292
0.00833
0.00978
0.0005724
0.002961
0.003228
Truss
≤3
NA
NA
NA
NA
0.001176
0.000422
0.00093
0.0003144
0.0001066
0.000204
≥6
NA
NA
NA
NA
0.001413
0.00297
0.0029
0.000344
0.0007514
0.00066Table 9
Compare the Modeling Parameters and Numerical Acceptance Criteria with ATC 40 [16].
Longitudinal reinforcement and transverse reinforcement
Shear
/
b
w
d
f
c
′
Rotational limit at collapse prevention level (CP) in radians
Crack width in coupling beam  at CP level in meters by    ATENA2D [18]
Member controlled by flexure
Member controlled by shear
ATENA2D [18]
ATC 40 [16]
ATC 40 [16]
L
b
=
0.6 m
L
b
=
0.9 m
L
b
=
1.2 m
L
b
=
0.6 m
L
b
=
0.9 m
L
b
=
1.2 m
Conventional longitudinal reinforcement with conforming transverse reinforcement
≤3
0.025
0.018
0.0001023
0.000784
0.00198
0.0000001308
0.0005
0.001613
≥6
0.015
0.012
0.0002423
0.001944
0.00344
0.00163
0.00136
0.00297
Diagonal
≤3
0.03
—
0.00012
0.000416
0.00055
0.0000194
0.0002184
0.00021
≥6
0.03
—
0.000415
0.000422
0.001533
0.0001795
0.0001483
0.00093Hence, the results obtained from the above study using ATENA2D [18] were found unsatisfactory. Therefore, a new model has been created with some assumptions in the manner shown in Figure 2 to carryout further study.Figure 2
Schematic diagram of coupling beam.


## 2.11. Assumptions

(i)
The effect of gravity loads on the coupling beams has been neglected.
(ii)
Deflection of the coupling beam occurs due to lateral loading.
(iii)
Contra flexure occurs at the mid-span of the coupling beam.
(iv)
The confined concrete, due to the confining action is provided by closely spaced transverse reinforcement in concrete, is assumed to govern the strength.Total elongation in the horizontal direction (Figure2) due to lateral loading can be written as
(4)ΔLb=db×θb,
and strain in the concrete
(5)εc=ΔLbLb.
Hence, considering (4) and (5) the following equation can be written as
(6)coupling  beam  rotation,θb=εc×Lbdb.
The results, considering (6) with maximum strain in confined concrete (εcu) of 0.02 (Confining action is provided by closely spaced transverse reinforcement in concrete as per ATC 40 [16]), have been tabulated in Table 10.Table 10
Maximum rotations in radians.
Type of eeinforcement
L
b
/
d
b
Value as per (6)
Galano and Vignoli [17]
Englekirk [3]
ATC40 [16], FEMA 273 [14] and FEMA 356  [15]
Diagonal
<1.5
<0.03
0.062
0.04
0.03
Truss
1.5 to 4.0
0.03 to 0.08
0.084
0.06
—It can be observed from Table10 that the values obtained as per (6) have similar trend with the values specified by ATC 40 [16], FEMA 273 [14], FEMA 356 [15], Galano and Vignoli [17], and Englekirk [3].Based on the above study, Table11 has been prepared containing modified parameters governing the coupling beam characteristics, which are also considered for the developments of the design technique discussed below. As design technique is based on collapse prevention (CP) level of structure, plastic rotation capacity given in Table 11 is for CP level only.Table 11
Modified Parameters governing the coupling beam characteristics controlled by shear.
Type of coupling beam
Shear Span to Depth Ratio
L
b
/
d
b
Type of detailing
Plastic Rotation Capacity (Radians)
Shear
/
b
w
d
f
c
′
CP
Reinforced concrete coupling beam
ϕ
≤
2
No limit
Conventional longitudinal reinforcement with conforming transverse reinforcement
≤3
0.015
≥6
0.010
<1.5
Diagonal Reinforcement (strength is an overriding consideration and thickness of wall should be greater than 406.4 mm)
—
<0.03
1.5 to 4.0
Truss Reinforcement (additional experimentation is required)
—
0.03–0.08
Steel coupling beam
e
≤
1.6
M
p
/
V
s
p
Shear dominant
—
0.15
/
L
b


## 3. Proposed Design Technique

In this paper an attempt has been made to develop a technique to design coupled shear walls considering its ideal seismic behavior (stable hysteresis with high earthquake energy dissipation). For preparing this design technique, symmetrical coupled shear walls have been considered. Design/capacity curve of coupled shear walls is obtained at the collapse mechanism of the structure based on this technique. This technique is applied to both fixed base and pinned base coupled shear walls. To start with, this technique is useful in selecting the preliminary dimensions of symmetrical coupled shear walls and subsequently arrives at a final design stage. Further, this technique is particularly useful for designer, consultant and practicing engineer who have no access to sophisticated software packages. A case study has been done implementing the technique with the help of Microsoft Excel Spreadsheet and the results have also been validated.


### 3.1. Proposed Formulation

In Figure3, the coupled shear walls are subjected to a triangular variation of loading with amplitude F1 at the roof level. The value of F1 is obtained corresponding to the CP level of structure. Subsequently, the base shear and roof displacement can be determined. The procedure involving Figure 3, the assumptions, steps, and mathematical calculation with initial value of F1 as unity have been illustrated as in Figure 3.(a) Coupled shear walls. (b) Free body diagram of coupled shear walls.
(a)
(b)


### 3.2. Assumptions

The following assumptions are adopted for the design technique to obtain the ideal seismic behavior of coupled shear walls.(1)
The analytical model of coupled shear walls is taken as two-dimensional entity.(2)
Coupled shear walls exhibit flexural behavior.(3)
Coupling beams carry axial forces, shear forces, and moments.(4)
The axial deformation of the coupling beam is neglected.(5)
The effect of gravity loads on the coupling beams is neglected.(6)
The horizontal displacement at each point of wall 1 is equal to the horizontal displacement at each corresponding point of wall 2 due to the presence of coupling beam.(7)
The curvatures of the two walls are same at any level.(8)
The point of contra flexure occurs at mid-point of clear span of the beam.(9)
The seismic design philosophy requires formation of plastic hinges at the ends of the coupling beams. All coupling beams are typically designed identically with identical plastic moment capacities. Being lightly loaded under gravity loads they will carry equal shear forces before a collapse mechanism is formed. All coupling beams are, therefore, assumed to carry equal shear forces.(10)
In the collapse mechanism for coupled shear walls, plastic hinges are assumed to form at the base of the wall and at the two ends of each coupling beam. In the wall the elastic displacements shall be small in comparison to the displacements due to rotation at the base of the wall. If the elastic displacements in the wall are considered negligible then a triangular displaced shape occurs. This is assumed to be the distribution displacement/velocity/acceleration along the height. The acceleration times the mass/weight at any floor level gives the lateral load. Hence, the distribution of the lateral loading is assumed as a triangular variation, which conforms to the first mode shape pattern.


### 3.3. Steps

The following iterative steps are developed in this thesis for the design of coupled shear walls.(1)
Selection of a particular type of coupling beam and determining its shear capacity.(2)
Determining the fractions of total lateral loading subjected on wall 1 and wall 2.(3)
Determining shear forces developed in coupling beams for different base conditions.(4)
Determining wall rotations in each storey.(5)
Checking for occurrence of plastic hinges at the base of the walls when base is fixed. For walls pinned at the base this check is not required.(6)
Calculating coupling beam rotation in each storey.(7)
Checking whether coupling beam rotation lies at collapse prevention level.(8)
Calculating base shear and roof displacement.(9)
Modifying the value ofF1 for next iteration starting from Step (2) if Step (7) is not satisfied.


### 3.4. Mathematical Calculation

The steps which are described above have been illustrated in this section as follows.Step 1.
The type of coupling beam can be determined as per Table11, and shear capacity can be calculated as per Section 2.2.Step 2.
In Figure3(b), free body diagram of coupled shear walls has been shown; α and β are fractions of total lateral loading incident on wall 1 and wall 2, respectively, such that,
(7)α+β=1.0.
For symmetrical coupled shear walls, moments of inertias of two walls are equal for equal depths and thicknesses at any level. Further, curvatures of two walls are equal at any level. Hence based on the Assumption (7), equation (7) can be written as
(8)α=β=0.5.Step 3.
In this step, it is explained how to calculate the shear force developed in the coupling beams for different types of boundary conditions. CSA [25] and Chaallal et al. [37] defined the degree of coupling which is written as
(9)DC=T×lMot,
where, l=Lw+Lb; T is the axial force due to lateral loading and Mot is total overturning moment at the base of the wall produced due to lateral loading. For fixed base condition DC varies from 0 to 1 and (9) can also be written as
(10)DC=k′(db)a(Lw)b×(Lb)c.
The above equation (10) is proposed by Chaallal et al. [37]; N is the total number of storeys, k′ is constant, and a, b, and c are exponents which are given in Table 12.
So based upon the above criteria and considering (9) and (10), shear force developed in the coupling beam could be determined as follows.
For fixed base condition following equation can be written as(11)C=T=∑i=1NVi=Motl×k′(db)a(Lw)b×(Lb)c,
where, Mot is total overturning moment at the base due to the lateral loading.
Therefore, based on the Assumption (9) shear force in coupling beam at each storey is
(12)V=∑i=1NViN.Table 12
Values of constantk′ and exponents a, b and c.
N
k
′
a
b
c
6
2.976
0.706
0.615
0.698
10
2.342
0.512
0.462
0.509
15
1.697
0.352
0.345
0.279
20
1.463
0.265
0.281
0.190
30
1.293
0.193
0.223
0.106
40
1.190
0.145
0.155
0.059Pinned Base Condition. In this study, pinned base condition has been introduced as one of the possible boundary conditions for coupled shear walls. It can be constructed by designing the foundation for axial load and shear force without considering bending moment. It is expected that stable hysteresis with high earthquake energy dissipation can be obtained for considering this kind of base condition.DC is 1 for pinned base condition from (9). Hence, the equation can be written as
(13)C=T=∑i=1NVi=Motl.
Therefore, based on the Assumption (9) shear force in coupling beam at each storey is
(14)V=∑i=1NViN.Step 4.
After obtainingα, β, and V at each storey for the particular value of F1, bending moment values in each storey could be determined for each wall. Subsequently, curvature diagram for each wall is generated by using moment area method as adopted in the Microsoft excel spreadsheet; which is required to determine the wall rotation in each storey. The following equations are considered to calculate the wall rotation.
Overturning moment at a distance “x” from base with respect to each wall can be written as
(15)Mot(x)=∑j=0N-i{0.5×F1H(H-jhs)(H-x-jhs)},
where, i is storey number and it is considered from the base as 0,1,2,3,…,N.
Resisting moment in wall due to shear force in the coupling beam at a distance “x” from base can be written as
(16)Mwr(x)=(Lw2+Lb2)∑j=iNVj,
where, net moment in the wall at a distance “x” from base, generated due to overturning moment and moment due to shear force in the coupling beam, can be written as
(17)Mnet(x)=Mot(x)-Mwr(x).
Wall rotation at ith storey for fixed base can be written as
(18)θwi=∫0ihsMnet(x)dxEcI,
where,
(19)I=tw×Lw312.
For plastic hinge rotation at the fixed base of wall or rotation at the pinned base of wall, (18) could be written as
(20)θwi=∫0ihsMnet(x)dxEcI+θw0,
where, θw0 is the plastic hinge rotation at the fixed base of wall or rotation at the pinned base of wall.Step 5.
Consider (i) Tensile forces at the base of wall 1 (T) as well as compressive forces at the base of wall 2 (C) are calculated due to lateral loading.
(ii) Compressive loads at the bases of wall 1 and wall 2 are calculated due to gravity loading.
(iii) Net axial forces at the bases of wall 1 and wall 2 are calculated, that is, Net axial force= Tensile or Compressive force due to lateral loading (T or C) ± Compressive load due to gravity loading.
(iv) Then, according to these net axial forces for the particular values offck, bb, d, and p, the yield moment values at the bases of wall 1 and wall 2 can be determined from P-M interaction curve [2, 19]. Where fck, bb, d, and p are yield strength of concrete, breadth of a section, depth of that section and percentage of minimum reinforcement in that particular section, respectively; and P is the axial force and M is the moment; here, net axial force is considered as P in the P-M interaction curve.
(v) Therefore, if calculated bending moment value at any base of the two walls is greater than yield moment value, plastic hinge at that base would be formed, otherwise no plastic hinge would be formed.Step 6.
The rotation of coupling beam in each storey is determined in Figure4.
Rotation of coupling beam atith storey for symmetrical walls [3] as per Figure 4 is given by
(21)θbi=θwi(1+LwLb),
where, θwi is rotation of wall at ith storey and can be calculated as per (18), Lw= depth of wall, Lb= length of coupling beam.
For plastic hinge rotation at the fixed base of wall or real hinge rotation at the pinned base of wall, (21) could be written as:
(22)θbi=Lwb{θwi},
where, θwi can be calculated as per (20) for fixed base of wall or for pinned base of wall and
(23)Lwb=(1+LwLb).Figure 4
Deformed shape of aith storey symmetrical coupled shear walls.Step 7.
The rotational limit for collapse prevention level of different types of RCC coupling beams and steel beams are given in Table11. The task was to check whether the rotations of beams attained their rotational limit of CP level at the collapse mechanism of the structure simultaneously.Step 8.
The roof displacement can be calculated as per the following equations(24)Δroof=hs×(∑i=0Nθwi),
where, displacement at ith storey can be calculated as
(25)Δi=hs×(∑j=0iθwj).
The base shear can be calculated as follows:
(26)VB=F1×(N+1)2.Step 9.
TheF1 is modified as follows when the condition of Step 7 is not satisfied.
To obtain the collapse mechanism of the structure, it is required to increaseF1 with equal increment until all coupling beams attain their rotation limit of CP level simultaneously.


### 3.5. Validation of the Proposed Design Technique

The following numerical example has been considered to validate the propose design technique. In this study, plan and elevation with dimensions and material properties of the coupled shear walls have been adopted as given in Chaallal et al. [37].


### 3.6. Numerical Example

The coupled shear walls considered here are part of a 20-storey office building (Figure5). It is subjected to triangular variation of lateral loading. The dimension and material properties are tabulated in Table 13. Dead loads and live loads are discussed in the following section. A comparison of the results regarding design/capacity curve (Figure 7) and ductility (27) obtained from the proposed design technique with the results obtained in SAP V 10.0.5 [23] and DRAIN-3DX [22] software packages may, thus, be required. For obtaining more perfection about the results, these two softwares have been considered in Table 13 simultaneously.Table 13
Dimensions and material properties of coupled shear walls for validation of proposed design technique.
Depth of the wall (Lw)
4 m
Length of coupling beam (Lb)
1.8 m
Depth of coupling beam (db)
600 mm
Number of storeys (N)
20
Wall thickness (tw)
300 mm
Width of coupling beam (bb)
300 mm
Storey height (hs)
3.0 m
Modulus of concrete (Ec)
27.0 GPa
Modulus of steel (Es)
200.0 GPa
Steel yield strength (fy)
415 Mpa(a) Plan view of building. (b) Coupled shear walls at section “a-a”.
(a)
(b)Figures5(a) and 5(b) show the plan and sectional elevation of the coupled shear wall building, respectively.


#### 3.6.1. Loading Consideration

Dead loads (DL) of 6.7 kN/m2 and live loads (LL) of 2.4 kN/m2 have been considered as suggested in Chaallal et al. [37]. Total gravity loading on coupled shear walls at section “a-a” has been calculated as the sum of dead load plus 25% LL as per IS 1893 (part 1) [1] for floor; however, in case of roof only dead load is considered.


#### 3.6.2. Modeling of Coupled Shear Walls in Proposed Design Technique

The modeling of coupled shear walls involving Figure3, assumptions, and steps with mathematical calculation are already described in Section 3.1.


#### 3.6.3. Modeling of Coupled Shear Walls in SAP and DRAIN-3DX

Wide column frame analogy [38] has been considered for modeling of coupled shear walls in SAP V 10.0.5 [23] and DRAIN-3DX [22] as given in Figure 6. In this analogy, shear walls are represented as two line elements (centre line of shear wall), and beams are represented as line elements (centre line of beam) by joining with each other with rigid link. Beam column elastic element (Type-17) and inelastic element (Type-15) are considered for modeling.Figure 6
Modeling in SAP V 10.0.5 [23] and DRAIN-3DX [22].Figure 7
Bilinear representation for Capacity Curve.


#### 3.6.4. Calculation of Ductility

The obtained design/capacity curve from the proposed design technique, SAP V 10.0.5 [23], and DRAIN-3DX [22] is bilinearized. The bilinear representation is prepared in the manner shown in Figure 7 based on the concepts given in ATC 40 [16].It can be seen from Figure7 that bilinear representation can be due to the basis of initial tangent stiffness and equal energies (Area a1 = Area a2). Subsequently, ductility of the coupled shear walls has been calculated as
(27)μΔ=Δroof,CPΔroof,yield,
where, Δroof,CP and Δroof,yield can be calculated from (24); μΔ is the ductility which represents how much earthquake energy dissipates during an earthquake.


### 3.7. Results and Discussions

Coupled shear walls at section “a-a” as shown in Figure5 are considered for conducting the study.


### 3.8. RCC Coupling Beam with Conventional Longitudinal Reinforcement and Conforming Transverse Reinforcement

RCC coupling beam with Conventional longitudinal reinforcement and conforming transverse reinforcement in each storey has been selected as per Step1 for the study. The results of this study for fixed base as well as pinned base conditions have been shown in Figure 8 and Table 14.Table 14
Ductility of coupled shear walls considering different approaches.
Method
Ductility
Fixed base
Pinned base
Proposed Design Technique
7
7.5
DRAIN-3DX [22]
6.75
7.45
SAP V 10.0.5 [23]
6.92
7.47(a) Capacity curve for fixed base condition. (b) Capacity curve for pinned base condition.
(a)
(b)


#### 3.8.1. Discussions of Numerical Results

Figure8(b) shows that the results obtained from proposed design technique for pinned base conditions are almost similar with the results obtained from DRAIN-3DX [22] and SAP V 10.0.5 [23]. Whereas, Figure 8(a) is showing a bit differences about the results obtained from proposed design technique, DRAIN-3DX [22], and SAP V 10.0.5 [23] although same dimensions, same material properties, and same loading were considered in all the three techniques. However, the differences were not very high (5–10%). Table 14 is showing the results about ductility obtained for fixed and pinned base conditions with the help of the Figures 8(a) and 8(b) and Section 3.6.4. It is noticed that ductility for pinned base condition is greater than fixed base conditions. It means that stable hysteresis with high earthquake energy dissipation can be obtained for coupled shear walls with pinned base.The results obtained from the proposed design technique are satisfactory. However, it is necessary to find the limitations of the proposed design technique. Therefore, in the following section, parametric study is elaborately discussed to detect the limitations of the proposed design technique.


## 3.1. Proposed Formulation

In Figure3, the coupled shear walls are subjected to a triangular variation of loading with amplitude F1 at the roof level. The value of F1 is obtained corresponding to the CP level of structure. Subsequently, the base shear and roof displacement can be determined. The procedure involving Figure 3, the assumptions, steps, and mathematical calculation with initial value of F1 as unity have been illustrated as in Figure 3.(a) Coupled shear walls. (b) Free body diagram of coupled shear walls.
(a)
(b)


## 3.2. Assumptions

The following assumptions are adopted for the design technique to obtain the ideal seismic behavior of coupled shear walls.(1)
The analytical model of coupled shear walls is taken as two-dimensional entity.(2)
Coupled shear walls exhibit flexural behavior.(3)
Coupling beams carry axial forces, shear forces, and moments.(4)
The axial deformation of the coupling beam is neglected.(5)
The effect of gravity loads on the coupling beams is neglected.(6)
The horizontal displacement at each point of wall 1 is equal to the horizontal displacement at each corresponding point of wall 2 due to the presence of coupling beam.(7)
The curvatures of the two walls are same at any level.(8)
The point of contra flexure occurs at mid-point of clear span of the beam.(9)
The seismic design philosophy requires formation of plastic hinges at the ends of the coupling beams. All coupling beams are typically designed identically with identical plastic moment capacities. Being lightly loaded under gravity loads they will carry equal shear forces before a collapse mechanism is formed. All coupling beams are, therefore, assumed to carry equal shear forces.(10)
In the collapse mechanism for coupled shear walls, plastic hinges are assumed to form at the base of the wall and at the two ends of each coupling beam. In the wall the elastic displacements shall be small in comparison to the displacements due to rotation at the base of the wall. If the elastic displacements in the wall are considered negligible then a triangular displaced shape occurs. This is assumed to be the distribution displacement/velocity/acceleration along the height. The acceleration times the mass/weight at any floor level gives the lateral load. Hence, the distribution of the lateral loading is assumed as a triangular variation, which conforms to the first mode shape pattern.


## 3.3. Steps

The following iterative steps are developed in this thesis for the design of coupled shear walls.(1)
Selection of a particular type of coupling beam and determining its shear capacity.(2)
Determining the fractions of total lateral loading subjected on wall 1 and wall 2.(3)
Determining shear forces developed in coupling beams for different base conditions.(4)
Determining wall rotations in each storey.(5)
Checking for occurrence of plastic hinges at the base of the walls when base is fixed. For walls pinned at the base this check is not required.(6)
Calculating coupling beam rotation in each storey.(7)
Checking whether coupling beam rotation lies at collapse prevention level.(8)
Calculating base shear and roof displacement.(9)
Modifying the value ofF1 for next iteration starting from Step (2) if Step (7) is not satisfied.


## 3.4. Mathematical Calculation

The steps which are described above have been illustrated in this section as follows.Step 1.
The type of coupling beam can be determined as per Table11, and shear capacity can be calculated as per Section 2.2.Step 2.
In Figure3(b), free body diagram of coupled shear walls has been shown; α and β are fractions of total lateral loading incident on wall 1 and wall 2, respectively, such that,
(7)α+β=1.0.
For symmetrical coupled shear walls, moments of inertias of two walls are equal for equal depths and thicknesses at any level. Further, curvatures of two walls are equal at any level. Hence based on the Assumption (7), equation (7) can be written as
(8)α=β=0.5.Step 3.
In this step, it is explained how to calculate the shear force developed in the coupling beams for different types of boundary conditions. CSA [25] and Chaallal et al. [37] defined the degree of coupling which is written as
(9)DC=T×lMot,
where, l=Lw+Lb; T is the axial force due to lateral loading and Mot is total overturning moment at the base of the wall produced due to lateral loading. For fixed base condition DC varies from 0 to 1 and (9) can also be written as
(10)DC=k′(db)a(Lw)b×(Lb)c.
The above equation (10) is proposed by Chaallal et al. [37]; N is the total number of storeys, k′ is constant, and a, b, and c are exponents which are given in Table 12.
So based upon the above criteria and considering (9) and (10), shear force developed in the coupling beam could be determined as follows.
For fixed base condition following equation can be written as(11)C=T=∑i=1NVi=Motl×k′(db)a(Lw)b×(Lb)c,
where, Mot is total overturning moment at the base due to the lateral loading.
Therefore, based on the Assumption (9) shear force in coupling beam at each storey is
(12)V=∑i=1NViN.Table 12
Values of constantk′ and exponents a, b and c.
N
k
′
a
b
c
6
2.976
0.706
0.615
0.698
10
2.342
0.512
0.462
0.509
15
1.697
0.352
0.345
0.279
20
1.463
0.265
0.281
0.190
30
1.293
0.193
0.223
0.106
40
1.190
0.145
0.155
0.059Pinned Base Condition. In this study, pinned base condition has been introduced as one of the possible boundary conditions for coupled shear walls. It can be constructed by designing the foundation for axial load and shear force without considering bending moment. It is expected that stable hysteresis with high earthquake energy dissipation can be obtained for considering this kind of base condition.DC is 1 for pinned base condition from (9). Hence, the equation can be written as
(13)C=T=∑i=1NVi=Motl.
Therefore, based on the Assumption (9) shear force in coupling beam at each storey is
(14)V=∑i=1NViN.Step 4.
After obtainingα, β, and V at each storey for the particular value of F1, bending moment values in each storey could be determined for each wall. Subsequently, curvature diagram for each wall is generated by using moment area method as adopted in the Microsoft excel spreadsheet; which is required to determine the wall rotation in each storey. The following equations are considered to calculate the wall rotation.
Overturning moment at a distance “x” from base with respect to each wall can be written as
(15)Mot(x)=∑j=0N-i{0.5×F1H(H-jhs)(H-x-jhs)},
where, i is storey number and it is considered from the base as 0,1,2,3,…,N.
Resisting moment in wall due to shear force in the coupling beam at a distance “x” from base can be written as
(16)Mwr(x)=(Lw2+Lb2)∑j=iNVj,
where, net moment in the wall at a distance “x” from base, generated due to overturning moment and moment due to shear force in the coupling beam, can be written as
(17)Mnet(x)=Mot(x)-Mwr(x).
Wall rotation at ith storey for fixed base can be written as
(18)θwi=∫0ihsMnet(x)dxEcI,
where,
(19)I=tw×Lw312.
For plastic hinge rotation at the fixed base of wall or rotation at the pinned base of wall, (18) could be written as
(20)θwi=∫0ihsMnet(x)dxEcI+θw0,
where, θw0 is the plastic hinge rotation at the fixed base of wall or rotation at the pinned base of wall.Step 5.
Consider (i) Tensile forces at the base of wall 1 (T) as well as compressive forces at the base of wall 2 (C) are calculated due to lateral loading.
(ii) Compressive loads at the bases of wall 1 and wall 2 are calculated due to gravity loading.
(iii) Net axial forces at the bases of wall 1 and wall 2 are calculated, that is, Net axial force= Tensile or Compressive force due to lateral loading (T or C) ± Compressive load due to gravity loading.
(iv) Then, according to these net axial forces for the particular values offck, bb, d, and p, the yield moment values at the bases of wall 1 and wall 2 can be determined from P-M interaction curve [2, 19]. Where fck, bb, d, and p are yield strength of concrete, breadth of a section, depth of that section and percentage of minimum reinforcement in that particular section, respectively; and P is the axial force and M is the moment; here, net axial force is considered as P in the P-M interaction curve.
(v) Therefore, if calculated bending moment value at any base of the two walls is greater than yield moment value, plastic hinge at that base would be formed, otherwise no plastic hinge would be formed.Step 6.
The rotation of coupling beam in each storey is determined in Figure4.
Rotation of coupling beam atith storey for symmetrical walls [3] as per Figure 4 is given by
(21)θbi=θwi(1+LwLb),
where, θwi is rotation of wall at ith storey and can be calculated as per (18), Lw= depth of wall, Lb= length of coupling beam.
For plastic hinge rotation at the fixed base of wall or real hinge rotation at the pinned base of wall, (21) could be written as:
(22)θbi=Lwb{θwi},
where, θwi can be calculated as per (20) for fixed base of wall or for pinned base of wall and
(23)Lwb=(1+LwLb).Figure 4
Deformed shape of aith storey symmetrical coupled shear walls.Step 7.
The rotational limit for collapse prevention level of different types of RCC coupling beams and steel beams are given in Table11. The task was to check whether the rotations of beams attained their rotational limit of CP level at the collapse mechanism of the structure simultaneously.Step 8.
The roof displacement can be calculated as per the following equations(24)Δroof=hs×(∑i=0Nθwi),
where, displacement at ith storey can be calculated as
(25)Δi=hs×(∑j=0iθwj).
The base shear can be calculated as follows:
(26)VB=F1×(N+1)2.Step 9.
TheF1 is modified as follows when the condition of Step 7 is not satisfied.
To obtain the collapse mechanism of the structure, it is required to increaseF1 with equal increment until all coupling beams attain their rotation limit of CP level simultaneously.


## 3.5. Validation of the Proposed Design Technique

The following numerical example has been considered to validate the propose design technique. In this study, plan and elevation with dimensions and material properties of the coupled shear walls have been adopted as given in Chaallal et al. [37].


## 3.6. Numerical Example

The coupled shear walls considered here are part of a 20-storey office building (Figure5). It is subjected to triangular variation of lateral loading. The dimension and material properties are tabulated in Table 13. Dead loads and live loads are discussed in the following section. A comparison of the results regarding design/capacity curve (Figure 7) and ductility (27) obtained from the proposed design technique with the results obtained in SAP V 10.0.5 [23] and DRAIN-3DX [22] software packages may, thus, be required. For obtaining more perfection about the results, these two softwares have been considered in Table 13 simultaneously.Table 13
Dimensions and material properties of coupled shear walls for validation of proposed design technique.
Depth of the wall (Lw)
4 m
Length of coupling beam (Lb)
1.8 m
Depth of coupling beam (db)
600 mm
Number of storeys (N)
20
Wall thickness (tw)
300 mm
Width of coupling beam (bb)
300 mm
Storey height (hs)
3.0 m
Modulus of concrete (Ec)
27.0 GPa
Modulus of steel (Es)
200.0 GPa
Steel yield strength (fy)
415 Mpa(a) Plan view of building. (b) Coupled shear walls at section “a-a”.
(a)
(b)Figures5(a) and 5(b) show the plan and sectional elevation of the coupled shear wall building, respectively.


### 3.6.1. Loading Consideration

Dead loads (DL) of 6.7 kN/m2 and live loads (LL) of 2.4 kN/m2 have been considered as suggested in Chaallal et al. [37]. Total gravity loading on coupled shear walls at section “a-a” has been calculated as the sum of dead load plus 25% LL as per IS 1893 (part 1) [1] for floor; however, in case of roof only dead load is considered.


### 3.6.2. Modeling of Coupled Shear Walls in Proposed Design Technique

The modeling of coupled shear walls involving Figure3, assumptions, and steps with mathematical calculation are already described in Section 3.1.


### 3.6.3. Modeling of Coupled Shear Walls in SAP and DRAIN-3DX

Wide column frame analogy [38] has been considered for modeling of coupled shear walls in SAP V 10.0.5 [23] and DRAIN-3DX [22] as given in Figure 6. In this analogy, shear walls are represented as two line elements (centre line of shear wall), and beams are represented as line elements (centre line of beam) by joining with each other with rigid link. Beam column elastic element (Type-17) and inelastic element (Type-15) are considered for modeling.Figure 6
Modeling in SAP V 10.0.5 [23] and DRAIN-3DX [22].Figure 7
Bilinear representation for Capacity Curve.


### 3.6.4. Calculation of Ductility

The obtained design/capacity curve from the proposed design technique, SAP V 10.0.5 [23], and DRAIN-3DX [22] is bilinearized. The bilinear representation is prepared in the manner shown in Figure 7 based on the concepts given in ATC 40 [16].It can be seen from Figure7 that bilinear representation can be due to the basis of initial tangent stiffness and equal energies (Area a1 = Area a2). Subsequently, ductility of the coupled shear walls has been calculated as
(27)μΔ=Δroof,CPΔroof,yield,
where, Δroof,CP and Δroof,yield can be calculated from (24); μΔ is the ductility which represents how much earthquake energy dissipates during an earthquake.


## 3.6.1. Loading Consideration

Dead loads (DL) of 6.7 kN/m2 and live loads (LL) of 2.4 kN/m2 have been considered as suggested in Chaallal et al. [37]. Total gravity loading on coupled shear walls at section “a-a” has been calculated as the sum of dead load plus 25% LL as per IS 1893 (part 1) [1] for floor; however, in case of roof only dead load is considered.


## 3.6.2. Modeling of Coupled Shear Walls in Proposed Design Technique

The modeling of coupled shear walls involving Figure3, assumptions, and steps with mathematical calculation are already described in Section 3.1.


## 3.6.3. Modeling of Coupled Shear Walls in SAP and DRAIN-3DX

Wide column frame analogy [38] has been considered for modeling of coupled shear walls in SAP V 10.0.5 [23] and DRAIN-3DX [22] as given in Figure 6. In this analogy, shear walls are represented as two line elements (centre line of shear wall), and beams are represented as line elements (centre line of beam) by joining with each other with rigid link. Beam column elastic element (Type-17) and inelastic element (Type-15) are considered for modeling.Figure 6
Modeling in SAP V 10.0.5 [23] and DRAIN-3DX [22].Figure 7
Bilinear representation for Capacity Curve.


## 3.6.4. Calculation of Ductility

The obtained design/capacity curve from the proposed design technique, SAP V 10.0.5 [23], and DRAIN-3DX [22] is bilinearized. The bilinear representation is prepared in the manner shown in Figure 7 based on the concepts given in ATC 40 [16].It can be seen from Figure7 that bilinear representation can be due to the basis of initial tangent stiffness and equal energies (Area a1 = Area a2). Subsequently, ductility of the coupled shear walls has been calculated as
(27)μΔ=Δroof,CPΔroof,yield,
where, Δroof,CP and Δroof,yield can be calculated from (24); μΔ is the ductility which represents how much earthquake energy dissipates during an earthquake.


## 3.7. Results and Discussions

Coupled shear walls at section “a-a” as shown in Figure5 are considered for conducting the study.


## 3.8. RCC Coupling Beam with Conventional Longitudinal Reinforcement and Conforming Transverse Reinforcement

RCC coupling beam with Conventional longitudinal reinforcement and conforming transverse reinforcement in each storey has been selected as per Step1 for the study. The results of this study for fixed base as well as pinned base conditions have been shown in Figure 8 and Table 14.Table 14
Ductility of coupled shear walls considering different approaches.
Method
Ductility
Fixed base
Pinned base
Proposed Design Technique
7
7.5
DRAIN-3DX [22]
6.75
7.45
SAP V 10.0.5 [23]
6.92
7.47(a) Capacity curve for fixed base condition. (b) Capacity curve for pinned base condition.
(a)
(b)


### 3.8.1. Discussions of Numerical Results

Figure8(b) shows that the results obtained from proposed design technique for pinned base conditions are almost similar with the results obtained from DRAIN-3DX [22] and SAP V 10.0.5 [23]. Whereas, Figure 8(a) is showing a bit differences about the results obtained from proposed design technique, DRAIN-3DX [22], and SAP V 10.0.5 [23] although same dimensions, same material properties, and same loading were considered in all the three techniques. However, the differences were not very high (5–10%). Table 14 is showing the results about ductility obtained for fixed and pinned base conditions with the help of the Figures 8(a) and 8(b) and Section 3.6.4. It is noticed that ductility for pinned base condition is greater than fixed base conditions. It means that stable hysteresis with high earthquake energy dissipation can be obtained for coupled shear walls with pinned base.The results obtained from the proposed design technique are satisfactory. However, it is necessary to find the limitations of the proposed design technique. Therefore, in the following section, parametric study is elaborately discussed to detect the limitations of the proposed design technique.


## 3.8.1. Discussions of Numerical Results

Figure8(b) shows that the results obtained from proposed design technique for pinned base conditions are almost similar with the results obtained from DRAIN-3DX [22] and SAP V 10.0.5 [23]. Whereas, Figure 8(a) is showing a bit differences about the results obtained from proposed design technique, DRAIN-3DX [22], and SAP V 10.0.5 [23] although same dimensions, same material properties, and same loading were considered in all the three techniques. However, the differences were not very high (5–10%). Table 14 is showing the results about ductility obtained for fixed and pinned base conditions with the help of the Figures 8(a) and 8(b) and Section 3.6.4. It is noticed that ductility for pinned base condition is greater than fixed base conditions. It means that stable hysteresis with high earthquake energy dissipation can be obtained for coupled shear walls with pinned base.The results obtained from the proposed design technique are satisfactory. However, it is necessary to find the limitations of the proposed design technique. Therefore, in the following section, parametric study is elaborately discussed to detect the limitations of the proposed design technique.


## 4. Parametric Study

It has been observed from the CSA [25] and Chaallal et al. [37] that the behavior of the ductile coupled shear walls depend on degree of coupling, where degree of coupling depends upon depth and length of the coupling beam as well as depth and height of the coupled shear walls [4, 10].Therefore, this study has been restricted on length of the coupling beam and number of stories as basic variables and other parameters are considered as constant. These parameters have been considered in proposed method to make out effect on the behavior of coupled shear walls. Further, modifications to achieve ideal seismic behavior according to the proposed method have been included in this study.


### 4.1. Model for Parametric Study

A typical building with symmetrical coupled shear walls is shown in Figures9(a) and 9(b). Coupled shear walls at section “a-a” have been considered to carry out the parametric study.(a) Plan view of building with symmetrical coupled shear walls. (b) Coupled shear walls at section “a-a”.
(a)
(b)


### 4.2. Loading Consideration

Dead loads (DL) of 6.7 kN/m2 and live loads (LL) of 2.4 kN/m2 have been considered as per the suggestions made by in Chaallal et al. [37]. Total gravity loading on coupled shear walls at section “a-a” has been calculated as the sum of dead load plus 25% LL as per IS 1893 (part 1) [1] for floor; however, in case of roof only dead load is considered.


### 4.3. Parameters

Table15 mentions the different parameters with dimensions and material properties which have been considered to carry out the parametric study.Table 15
Dimensions and material properties of coupled shear walls for parametric study.
Depth of the wall (Lw)
4 m
Length of beam (Lb)
1 m, 1.5 m and 2 m
Depth of beam (db)
800 mm
Number of stories (N)
10, 15 and 20
Wall thickness (tw)
300 mm
Width of coupling beam (bb)
300 mm
Storey height (hs)
3.6 m
Modulus of concrete (Ec)
22.4 GPa
Yield strength of steel (fy)
415 MPa


### 4.4. Analysis Using Proposed Design Technique

The above mentioned building has been studied by the design technique. The results for different parameters have been described in this section.


### 4.5. Observed Behavior

To study the influence of length of the coupling beam (Lb) on the behavior of coupled shear walls, length of the coupling beam is considered as 1 m, 1.5 m and 2 m for both fixed and pinned base conditions. RCC coupling beam with conventional longitudinal reinforcement with conforming transverse reinforcement has been selected. Shear capacity in the coupling beam is calculated as per Step 1. The rotational limit of coupling beam has been selected as per Step 7. The study has been performed for coupled shear walls with number of stories 20, 15, and 10 for both fixed and pinned base conditions.


#### 4.5.1. For Number of StoriesN=20

For more details, see Figures10, 11, 12, and 13.(a) Storey displacement for fixed base condition at CP level. (b) Storey displacement for pinned base condition at CP level.
(a)
(b)(a) Wall rotation for fixed base condition at CP level. (b) Wall rotation for pinned base condition at CP level.
(a)
(b)(a) Beam rotation for fixed base condition at CP level. (b) Beam rotation for pinned base condition at CP level.
(a)
(b)(a) Capacity curve for fixed base condition. (b) Capacity curve for pinned base condition.
(a)
(b)


#### 4.5.2. Discussion of Results forN=20

The deflection for the case of pinned base condition is much higher than the case of fixed base (Figure10); however, the base shear for the case of pinned base condition is lower than the case of fixed base (Figure 13). It shows satisfactory results based on the behavior of coupled shear walls. Because, coupled shear walls with pinned base deflected more subjected to lesser lateral loading in comparison with the coupled shear walls with fixed base and base shear is directly varying with the lateral loading (26). Since wall rotation is directly varying with the length of the beam (Figure 11) and deflection is the summation of the wall rotation (25), deflection is directly varying with the length of the beam (Figure 10). It has been also observed that all beams reach to their rotational limit of CP level for pinned base condition; however, very few beams reach to their rotational limit of CP level for fixed base condition (Figure 12). Hence, it can be said that coupled shear walls are behaving as a rigid body motion for pinned base condition; which is expected. The explanations for fixed base condition (Figure 12) are given in the following manner:(i)
The rotation of the cantilever wall is maximum at the free end of the wall. This rotation decreases towards the base of the wall and is zero at the base for fixity.(ii)
Fixed base coupled shear walls with short span coupling beam is behaving as a cantilever wall (Lb=1 m of Figure 11). It is also one of the behaviors of a coupled shear walls. However, fixed base coupled shear walls with long span coupling beam does not show cantilever wall (Lb=1.5 m and Lb=2 m of Figure 11) behavior.(iii)
Beam rotation is proportional to the wall rotation.Therefore, it can be said from the above observations that coupled shear walls with short span coupling beam (Lb=1 m) can be acceptable in comparison with the long span coupling beam (Lb=1.5 m and Lb=2 m) although the behavior of all three coupling beams is governed by shear according to Table 11.With the help of Section3.6.4 and Figure 13, ductility for pinned base condition and fixed base condition has been calculated in Table 16.Table 16
Ductility of coupled shear walls forN=20.
Base condition
Length of the coupling beam (Lb)
Values
Fixed
1 m
3.33
1.5 m
4.8
2 m
6.3
Pinned
1 m
5.11
1.5 m
6.35
2 m
7.1It has been observed from Table16 that ductility is more for pinned base condition in comparison with the fixed base condition and ductility increases with increase in length of the coupling beam ((24) and (27), Figures 10, 11, and 13).


#### 4.5.3. For Number of StoriesN=15

For more details, see Figures14, 15, 16, and 17.(a) Storey displacement for fixed base condition at CP level. (b) Storey displacement for pinned base condition at CP level.
(a)
(b)(a) Wall rotation for fixed base condition at CP level. (b) Wall rotation for pinned base condition at CP level.
(a)
(b)(a) Beam rotation for fixed base condition at CP level. (b) Beam rotation for pinned base condition at CP level.
(a)
(b)(a) Capacity curve for fixed base condition. (b) Capacity curve for pinned base condition.
(a)
(b)


#### 4.5.4. Discussion of Results forN=15

With the help of Section3.6.4 and Figure 17, ductility for pinned base condition and fixed base condition has been calculated in Table 17.Table 17
Ductility of coupled shear walls forN=15.
Base condition
Length of the coupling beam (Lb)
Values
Fixed
1 m
2.93
1.5 m
4.0
2 m
5.9
Pinned
1 m
4.5
1.5 m
5.85
2 m
6.87It has been observed from Figures14 to 17 and Table 17 that the results obtained for N=15 are similar with the results of N=20 for fixed base condition and pinned base condition.


#### 4.5.5. For Number of StoriesN=10

For more details, see Figures18, 19, 20, and 21.(a) Storey displacement for fixed base condition at CP level. (b) Storey displacement for pinned base condition at CP level.
(a)
(b)(a) Wall rotation for fixed base condition at CP level. (b) Wall rotation for pinned base condition at CP level.
(a)
(b)(a) Beam rotation for fixed base condition at CP level. (b) Beam rotation for pinned base condition at CP level.
(a)
(b)(a) Capacity curve for fixed base condition. (b) Capacity curve for pinned base condition.
(a)
(b)


#### 4.5.6. Discussion of Results forN=10

Figures20 and 21 show that beam rotation and capacity curve reach CP level for the case of Lb=1 m with pinned base condition only. However, beam rotation and capacity curve do not reach the CP level for the other cases while shear capacities in all coupling beams have been achieved. It means that ideal seismic behavior (stable hysteresis with high earthquake energy dissipation) of coupled shear walls has only been achieved for Lb=1 m with pinned base condition. Proposed design technique does not show ideal seismic behavior of coupled shear walls for Lb=1 m, 1.5 m and 2 m with fixed base condition and Lb=1.5 m, and 2 m with pinned base condition. Now, remedial action has been considered in the following manner to obtain the ideal seismic behavior.


#### 4.5.7. Remedial Action forN=10

The remedy for the cases ofLb=1 m, 1.5 m, and 2 m with fixed base condition and Lb=1.5 m and 2 m with pinned base condition to achieve CP level is mentioned in (Figures 22, 23, 24 and 25). To obtain the CP level, it is required to increase the wall rotation. Since wall rotation ((18) and (19)) is inversely varying to the Lw3, it is required to decrease the Lw. It has been observed from Figure 25 that the ideal seismic behavior of coupled shear walls has been achieved.(a) Storey displacement for fixed base condition at CP level. (b) Storey displacement for pinned base condition at CP level.
(a)
(b)(a) Wall rotation for fixed base condition at CP level. (b) Wall rotation for pinned base condition at CP level.
(a)
(b)(a) Beam rotation for fixed base condition at CP level. (b) Beam rotation for pinned base condition at CP level.
(a)
(b)(a) Capacity curve for fixed base condition. (b) Capacity curve for pinned base condition.
(a)
(b)


#### 4.5.8. Discussion of the Above Results

Figures24 and 25 show that beam rotation and capacity curve reach CP level for all cases although the results are not satisfactory for fixed base condition according to the explanations given in Section 4.5.1.Hence, it can be said from the above results that proposed design technique is useful to design the coupled shear walls during earthquake motion. To confirm it more, nonlinear static analysis is considered in the following manner to assess the proposed design technique.


## 4.1. Model for Parametric Study

A typical building with symmetrical coupled shear walls is shown in Figures9(a) and 9(b). Coupled shear walls at section “a-a” have been considered to carry out the parametric study.(a) Plan view of building with symmetrical coupled shear walls. (b) Coupled shear walls at section “a-a”.
(a)
(b)


## 4.2. Loading Consideration

Dead loads (DL) of 6.7 kN/m2 and live loads (LL) of 2.4 kN/m2 have been considered as per the suggestions made by in Chaallal et al. [37]. Total gravity loading on coupled shear walls at section “a-a” has been calculated as the sum of dead load plus 25% LL as per IS 1893 (part 1) [1] for floor; however, in case of roof only dead load is considered.


## 4.3. Parameters

Table15 mentions the different parameters with dimensions and material properties which have been considered to carry out the parametric study.Table 15
Dimensions and material properties of coupled shear walls for parametric study.
Depth of the wall (Lw)
4 m
Length of beam (Lb)
1 m, 1.5 m and 2 m
Depth of beam (db)
800 mm
Number of stories (N)
10, 15 and 20
Wall thickness (tw)
300 mm
Width of coupling beam (bb)
300 mm
Storey height (hs)
3.6 m
Modulus of concrete (Ec)
22.4 GPa
Yield strength of steel (fy)
415 MPa


## 4.4. Analysis Using Proposed Design Technique

The above mentioned building has been studied by the design technique. The results for different parameters have been described in this section.


## 4.5. Observed Behavior

To study the influence of length of the coupling beam (Lb) on the behavior of coupled shear walls, length of the coupling beam is considered as 1 m, 1.5 m and 2 m for both fixed and pinned base conditions. RCC coupling beam with conventional longitudinal reinforcement with conforming transverse reinforcement has been selected. Shear capacity in the coupling beam is calculated as per Step 1. The rotational limit of coupling beam has been selected as per Step 7. The study has been performed for coupled shear walls with number of stories 20, 15, and 10 for both fixed and pinned base conditions.


### 4.5.1. For Number of StoriesN=20

For more details, see Figures10, 11, 12, and 13.(a) Storey displacement for fixed base condition at CP level. (b) Storey displacement for pinned base condition at CP level.
(a)
(b)(a) Wall rotation for fixed base condition at CP level. (b) Wall rotation for pinned base condition at CP level.
(a)
(b)(a) Beam rotation for fixed base condition at CP level. (b) Beam rotation for pinned base condition at CP level.
(a)
(b)(a) Capacity curve for fixed base condition. (b) Capacity curve for pinned base condition.
(a)
(b)


### 4.5.2. Discussion of Results forN=20

The deflection for the case of pinned base condition is much higher than the case of fixed base (Figure10); however, the base shear for the case of pinned base condition is lower than the case of fixed base (Figure 13). It shows satisfactory results based on the behavior of coupled shear walls. Because, coupled shear walls with pinned base deflected more subjected to lesser lateral loading in comparison with the coupled shear walls with fixed base and base shear is directly varying with the lateral loading (26). Since wall rotation is directly varying with the length of the beam (Figure 11) and deflection is the summation of the wall rotation (25), deflection is directly varying with the length of the beam (Figure 10). It has been also observed that all beams reach to their rotational limit of CP level for pinned base condition; however, very few beams reach to their rotational limit of CP level for fixed base condition (Figure 12). Hence, it can be said that coupled shear walls are behaving as a rigid body motion for pinned base condition; which is expected. The explanations for fixed base condition (Figure 12) are given in the following manner:(i)
The rotation of the cantilever wall is maximum at the free end of the wall. This rotation decreases towards the base of the wall and is zero at the base for fixity.(ii)
Fixed base coupled shear walls with short span coupling beam is behaving as a cantilever wall (Lb=1 m of Figure 11). It is also one of the behaviors of a coupled shear walls. However, fixed base coupled shear walls with long span coupling beam does not show cantilever wall (Lb=1.5 m and Lb=2 m of Figure 11) behavior.(iii)
Beam rotation is proportional to the wall rotation.Therefore, it can be said from the above observations that coupled shear walls with short span coupling beam (Lb=1 m) can be acceptable in comparison with the long span coupling beam (Lb=1.5 m and Lb=2 m) although the behavior of all three coupling beams is governed by shear according to Table 11.With the help of Section3.6.4 and Figure 13, ductility for pinned base condition and fixed base condition has been calculated in Table 16.Table 16
Ductility of coupled shear walls forN=20.
Base condition
Length of the coupling beam (Lb)
Values
Fixed
1 m
3.33
1.5 m
4.8
2 m
6.3
Pinned
1 m
5.11
1.5 m
6.35
2 m
7.1It has been observed from Table16 that ductility is more for pinned base condition in comparison with the fixed base condition and ductility increases with increase in length of the coupling beam ((24) and (27), Figures 10, 11, and 13).


### 4.5.3. For Number of StoriesN=15

For more details, see Figures14, 15, 16, and 17.(a) Storey displacement for fixed base condition at CP level. (b) Storey displacement for pinned base condition at CP level.
(a)
(b)(a) Wall rotation for fixed base condition at CP level. (b) Wall rotation for pinned base condition at CP level.
(a)
(b)(a) Beam rotation for fixed base condition at CP level. (b) Beam rotation for pinned base condition at CP level.
(a)
(b)(a) Capacity curve for fixed base condition. (b) Capacity curve for pinned base condition.
(a)
(b)


### 4.5.4. Discussion of Results forN=15

With the help of Section3.6.4 and Figure 17, ductility for pinned base condition and fixed base condition has been calculated in Table 17.Table 17
Ductility of coupled shear walls forN=15.
Base condition
Length of the coupling beam (Lb)
Values
Fixed
1 m
2.93
1.5 m
4.0
2 m
5.9
Pinned
1 m
4.5
1.5 m
5.85
2 m
6.87It has been observed from Figures14 to 17 and Table 17 that the results obtained for N=15 are similar with the results of N=20 for fixed base condition and pinned base condition.


### 4.5.5. For Number of StoriesN=10

For more details, see Figures18, 19, 20, and 21.(a) Storey displacement for fixed base condition at CP level. (b) Storey displacement for pinned base condition at CP level.
(a)
(b)(a) Wall rotation for fixed base condition at CP level. (b) Wall rotation for pinned base condition at CP level.
(a)
(b)(a) Beam rotation for fixed base condition at CP level. (b) Beam rotation for pinned base condition at CP level.
(a)
(b)(a) Capacity curve for fixed base condition. (b) Capacity curve for pinned base condition.
(a)
(b)


### 4.5.6. Discussion of Results forN=10

Figures20 and 21 show that beam rotation and capacity curve reach CP level for the case of Lb=1 m with pinned base condition only. However, beam rotation and capacity curve do not reach the CP level for the other cases while shear capacities in all coupling beams have been achieved. It means that ideal seismic behavior (stable hysteresis with high earthquake energy dissipation) of coupled shear walls has only been achieved for Lb=1 m with pinned base condition. Proposed design technique does not show ideal seismic behavior of coupled shear walls for Lb=1 m, 1.5 m and 2 m with fixed base condition and Lb=1.5 m, and 2 m with pinned base condition. Now, remedial action has been considered in the following manner to obtain the ideal seismic behavior.


### 4.5.7. Remedial Action forN=10

The remedy for the cases ofLb=1 m, 1.5 m, and 2 m with fixed base condition and Lb=1.5 m and 2 m with pinned base condition to achieve CP level is mentioned in (Figures 22, 23, 24 and 25). To obtain the CP level, it is required to increase the wall rotation. Since wall rotation ((18) and (19)) is inversely varying to the Lw3, it is required to decrease the Lw. It has been observed from Figure 25 that the ideal seismic behavior of coupled shear walls has been achieved.(a) Storey displacement for fixed base condition at CP level. (b) Storey displacement for pinned base condition at CP level.
(a)
(b)(a) Wall rotation for fixed base condition at CP level. (b) Wall rotation for pinned base condition at CP level.
(a)
(b)(a) Beam rotation for fixed base condition at CP level. (b) Beam rotation for pinned base condition at CP level.
(a)
(b)(a) Capacity curve for fixed base condition. (b) Capacity curve for pinned base condition.
(a)
(b)


### 4.5.8. Discussion of the Above Results

Figures24 and 25 show that beam rotation and capacity curve reach CP level for all cases although the results are not satisfactory for fixed base condition according to the explanations given in Section 4.5.1.Hence, it can be said from the above results that proposed design technique is useful to design the coupled shear walls during earthquake motion. To confirm it more, nonlinear static analysis is considered in the following manner to assess the proposed design technique.


## 4.5.1. For Number of StoriesN=20

For more details, see Figures10, 11, 12, and 13.(a) Storey displacement for fixed base condition at CP level. (b) Storey displacement for pinned base condition at CP level.
(a)
(b)(a) Wall rotation for fixed base condition at CP level. (b) Wall rotation for pinned base condition at CP level.
(a)
(b)(a) Beam rotation for fixed base condition at CP level. (b) Beam rotation for pinned base condition at CP level.
(a)
(b)(a) Capacity curve for fixed base condition. (b) Capacity curve for pinned base condition.
(a)
(b)


## 4.5.2. Discussion of Results forN=20

The deflection for the case of pinned base condition is much higher than the case of fixed base (Figure10); however, the base shear for the case of pinned base condition is lower than the case of fixed base (Figure 13). It shows satisfactory results based on the behavior of coupled shear walls. Because, coupled shear walls with pinned base deflected more subjected to lesser lateral loading in comparison with the coupled shear walls with fixed base and base shear is directly varying with the lateral loading (26). Since wall rotation is directly varying with the length of the beam (Figure 11) and deflection is the summation of the wall rotation (25), deflection is directly varying with the length of the beam (Figure 10). It has been also observed that all beams reach to their rotational limit of CP level for pinned base condition; however, very few beams reach to their rotational limit of CP level for fixed base condition (Figure 12). Hence, it can be said that coupled shear walls are behaving as a rigid body motion for pinned base condition; which is expected. The explanations for fixed base condition (Figure 12) are given in the following manner:(i)
The rotation of the cantilever wall is maximum at the free end of the wall. This rotation decreases towards the base of the wall and is zero at the base for fixity.(ii)
Fixed base coupled shear walls with short span coupling beam is behaving as a cantilever wall (Lb=1 m of Figure 11). It is also one of the behaviors of a coupled shear walls. However, fixed base coupled shear walls with long span coupling beam does not show cantilever wall (Lb=1.5 m and Lb=2 m of Figure 11) behavior.(iii)
Beam rotation is proportional to the wall rotation.Therefore, it can be said from the above observations that coupled shear walls with short span coupling beam (Lb=1 m) can be acceptable in comparison with the long span coupling beam (Lb=1.5 m and Lb=2 m) although the behavior of all three coupling beams is governed by shear according to Table 11.With the help of Section3.6.4 and Figure 13, ductility for pinned base condition and fixed base condition has been calculated in Table 16.Table 16
Ductility of coupled shear walls forN=20.
Base condition
Length of the coupling beam (Lb)
Values
Fixed
1 m
3.33
1.5 m
4.8
2 m
6.3
Pinned
1 m
5.11
1.5 m
6.35
2 m
7.1It has been observed from Table16 that ductility is more for pinned base condition in comparison with the fixed base condition and ductility increases with increase in length of the coupling beam ((24) and (27), Figures 10, 11, and 13).


## 4.5.3. For Number of StoriesN=15

For more details, see Figures14, 15, 16, and 17.(a) Storey displacement for fixed base condition at CP level. (b) Storey displacement for pinned base condition at CP level.
(a)
(b)(a) Wall rotation for fixed base condition at CP level. (b) Wall rotation for pinned base condition at CP level.
(a)
(b)(a) Beam rotation for fixed base condition at CP level. (b) Beam rotation for pinned base condition at CP level.
(a)
(b)(a) Capacity curve for fixed base condition. (b) Capacity curve for pinned base condition.
(a)
(b)


## 4.5.4. Discussion of Results forN=15

With the help of Section3.6.4 and Figure 17, ductility for pinned base condition and fixed base condition has been calculated in Table 17.Table 17
Ductility of coupled shear walls forN=15.
Base condition
Length of the coupling beam (Lb)
Values
Fixed
1 m
2.93
1.5 m
4.0
2 m
5.9
Pinned
1 m
4.5
1.5 m
5.85
2 m
6.87It has been observed from Figures14 to 17 and Table 17 that the results obtained for N=15 are similar with the results of N=20 for fixed base condition and pinned base condition.


## 4.5.5. For Number of StoriesN=10

For more details, see Figures18, 19, 20, and 21.(a) Storey displacement for fixed base condition at CP level. (b) Storey displacement for pinned base condition at CP level.
(a)
(b)(a) Wall rotation for fixed base condition at CP level. (b) Wall rotation for pinned base condition at CP level.
(a)
(b)(a) Beam rotation for fixed base condition at CP level. (b) Beam rotation for pinned base condition at CP level.
(a)
(b)(a) Capacity curve for fixed base condition. (b) Capacity curve for pinned base condition.
(a)
(b)


## 4.5.6. Discussion of Results forN=10

Figures20 and 21 show that beam rotation and capacity curve reach CP level for the case of Lb=1 m with pinned base condition only. However, beam rotation and capacity curve do not reach the CP level for the other cases while shear capacities in all coupling beams have been achieved. It means that ideal seismic behavior (stable hysteresis with high earthquake energy dissipation) of coupled shear walls has only been achieved for Lb=1 m with pinned base condition. Proposed design technique does not show ideal seismic behavior of coupled shear walls for Lb=1 m, 1.5 m and 2 m with fixed base condition and Lb=1.5 m, and 2 m with pinned base condition. Now, remedial action has been considered in the following manner to obtain the ideal seismic behavior.


## 4.5.7. Remedial Action forN=10

The remedy for the cases ofLb=1 m, 1.5 m, and 2 m with fixed base condition and Lb=1.5 m and 2 m with pinned base condition to achieve CP level is mentioned in (Figures 22, 23, 24 and 25). To obtain the CP level, it is required to increase the wall rotation. Since wall rotation ((18) and (19)) is inversely varying to the Lw3, it is required to decrease the Lw. It has been observed from Figure 25 that the ideal seismic behavior of coupled shear walls has been achieved.(a) Storey displacement for fixed base condition at CP level. (b) Storey displacement for pinned base condition at CP level.
(a)
(b)(a) Wall rotation for fixed base condition at CP level. (b) Wall rotation for pinned base condition at CP level.
(a)
(b)(a) Beam rotation for fixed base condition at CP level. (b) Beam rotation for pinned base condition at CP level.
(a)
(b)(a) Capacity curve for fixed base condition. (b) Capacity curve for pinned base condition.
(a)
(b)


## 4.5.8. Discussion of the Above Results

Figures24 and 25 show that beam rotation and capacity curve reach CP level for all cases although the results are not satisfactory for fixed base condition according to the explanations given in Section 4.5.1.Hence, it can be said from the above results that proposed design technique is useful to design the coupled shear walls during earthquake motion. To confirm it more, nonlinear static analysis is considered in the following manner to assess the proposed design technique.


## 5. Assessment of Proposed Design Technique Using Nonlinear Static Analysis

In this paper, nonlinear static analysis is carried out to determine the response reduction factors of coupled shear walls at different earthquake levels.


### 5.1. Design Example

The following design example is presented for carrying out the nonlinear static analysis of coupled shear walls. These walls are subjected to triangular variation of lateral loading. The base of the walls is assumed as fixed. Table18 mentions the different parameters with dimensions and material properties which have been considered to carry out the study. Figures 26(a) and 26(b) show the plan and sectional elevation of the coupled shear wall building, respectively. The place considered for this study is Roorkee and soil type for this place is medium (Type II); maximum considered earthquake (MCE) level and design basis earthquake level (DBE) are considered for the study.Table 18
Dimensions and material properties of coupled shear walls for nonlinear static analysis.
Depth of the wall (Lw)
4 m
Length of beam (Lb)
1 m
Depth of beam (db)
800 mm
Number of stories (N)
20 and 15
Wall thickness (tw)
300 mm
Width of coupling beam (bb)
300 mm
Storey height (hs)
3.6 m
Modulus of concrete (Ec)
22.4 GPa
Modulus of steel (Es)
200.0 GPa
Steel yield strength (fy)
415 MPa(a) Plan view of building with coupled shear walls. (b) Coupled shear walls at section “a-a”.
(a)
(b)


### 5.2. Loading Consideration

Dead loads (DL) of 6.7 kN/m2 and live loads (LL) of 2.4 kN/m2 have been considered as given in Chaallal et al. [37]. Total gravity loading on coupled shear walls at section “a-a” has been calculated as the sum of dead load plus 25% LL as per IS 1893 (part 1) [1] for floor; however, in case of roof only dead load is considered.


### 5.3. Results and Discussions

The results and discussions are described in Figure27.(a) Capacity curve forN=20. (b) Capacity curve for N=15.
(a)
(b)


#### 5.3.1. Calculation of Performance Point

Place considered here is Roorkee which belongs to the seismic zone IV and Z is 0.24 as per IS 1893 (part 1) [1]. 5% damped elastic response spectra as per IS 1893 (part 1) [1] are considered here as demand curve. DBE and MCE levels are considered for calculation of performance point (pp). Capacity curves are already obtained in Figure 27. The performance point has been calculated with the help of capacity spectrum method of ATC 40 [16] which is shown in  Figure 28.Figure 28
Performance point at the MCE level forN=20.In this case, modal mass co-efficientα1=0.616 and Mode participation factor PF1=1.51 derived with the help of modal analysis in SAP V 10.0.5 [23]. Figure 28 shows that pp is the performance point. The base shear at the performance point (pp), Vb,pp=1173.1 kN and roof displacement at the performance point (pp), Δroof,pp=0.31 m.In this case, modal mass co-efficientα1=0.616 and Mode participation factor PF1=1.51. Figure 29 shows that pp is the performance point. The base shear at the performance point (pp), Vb,pp=957.6 kN and roof displacement at the performance point (pp), Δroof,pp=0.097 m.Figure 29
Performance point at the DBE level forN=20.In this case, modal mass co-efficientα1=0.644 and Mode participation factor PF1=1.485. Figure 30 shows that pp is the performance point. The base shear at the performance point (pp), Vb,pp=1455.3 kN and roof displacement at the performance point (pp), Δroof,pp=0.259 m.Figure 30
Performance point at the MCE level forN=15.In this case, modal mass co-efficientα1=0.644 and Mode participation factor PF1=1.485. Figure 31 shows that pp is the performance point. The base shear at the performance point (pp), Vb,pp=1251.5 kN and roof displacement at the performance point (pp), Δroof,pp=0.101 m.Figure 31
Performance point at the DBE level forN=15.


#### 5.3.2. Calculation of Response Reduction Factor at the Performance Point

Table19 shows different response reduction factors for MCE and DBE levels. These are calculated at different performance points (Figures 28–31).Table 19
Response Reduction Factors for DBE and MCE levels.
Parameters
μ
Δ
e
1[24]
μ
Δ
e
2 [24]
R
μ
ξ [24]
R
μ
IDRS
[First method of Energy-Ductility Based Response Reduction] [24]
R
μ
IDRS
[Second method of Energy-Ductility Based Response Reduction] [24]
R
d as per CSA [25]
N
=
20
DBE
1.04
1.004
1.02
1.04
1.004
1.5 or 2 for coupled shear walls with conventional reinforced coupling beam
MCE
2.05
1.2
1.58
2.05
1.34
N
=
15
DBE
1.01
1.00
1.002
1.01
1.00
MCE
1.87
1.13
1.39
1.87
1.22From Table19, response reduction factor of coupled shear walls is varying between 1.22 to 2.05 for maximum considered earthquake (MCE) level; which is almost same as the provision of CSA [25] for coupling beam with conventional reinforcement.


## 5.1. Design Example

The following design example is presented for carrying out the nonlinear static analysis of coupled shear walls. These walls are subjected to triangular variation of lateral loading. The base of the walls is assumed as fixed. Table18 mentions the different parameters with dimensions and material properties which have been considered to carry out the study. Figures 26(a) and 26(b) show the plan and sectional elevation of the coupled shear wall building, respectively. The place considered for this study is Roorkee and soil type for this place is medium (Type II); maximum considered earthquake (MCE) level and design basis earthquake level (DBE) are considered for the study.Table 18
Dimensions and material properties of coupled shear walls for nonlinear static analysis.
Depth of the wall (Lw)
4 m
Length of beam (Lb)
1 m
Depth of beam (db)
800 mm
Number of stories (N)
20 and 15
Wall thickness (tw)
300 mm
Width of coupling beam (bb)
300 mm
Storey height (hs)
3.6 m
Modulus of concrete (Ec)
22.4 GPa
Modulus of steel (Es)
200.0 GPa
Steel yield strength (fy)
415 MPa(a) Plan view of building with coupled shear walls. (b) Coupled shear walls at section “a-a”.
(a)
(b)


## 5.2. Loading Consideration

Dead loads (DL) of 6.7 kN/m2 and live loads (LL) of 2.4 kN/m2 have been considered as given in Chaallal et al. [37]. Total gravity loading on coupled shear walls at section “a-a” has been calculated as the sum of dead load plus 25% LL as per IS 1893 (part 1) [1] for floor; however, in case of roof only dead load is considered.


## 5.3. Results and Discussions

The results and discussions are described in Figure27.(a) Capacity curve forN=20. (b) Capacity curve for N=15.
(a)
(b)


### 5.3.1. Calculation of Performance Point

Place considered here is Roorkee which belongs to the seismic zone IV and Z is 0.24 as per IS 1893 (part 1) [1]. 5% damped elastic response spectra as per IS 1893 (part 1) [1] are considered here as demand curve. DBE and MCE levels are considered for calculation of performance point (pp). Capacity curves are already obtained in Figure 27. The performance point has been calculated with the help of capacity spectrum method of ATC 40 [16] which is shown in  Figure 28.Figure 28
Performance point at the MCE level forN=20.In this case, modal mass co-efficientα1=0.616 and Mode participation factor PF1=1.51 derived with the help of modal analysis in SAP V 10.0.5 [23]. Figure 28 shows that pp is the performance point. The base shear at the performance point (pp), Vb,pp=1173.1 kN and roof displacement at the performance point (pp), Δroof,pp=0.31 m.In this case, modal mass co-efficientα1=0.616 and Mode participation factor PF1=1.51. Figure 29 shows that pp is the performance point. The base shear at the performance point (pp), Vb,pp=957.6 kN and roof displacement at the performance point (pp), Δroof,pp=0.097 m.Figure 29
Performance point at the DBE level forN=20.In this case, modal mass co-efficientα1=0.644 and Mode participation factor PF1=1.485. Figure 30 shows that pp is the performance point. The base shear at the performance point (pp), Vb,pp=1455.3 kN and roof displacement at the performance point (pp), Δroof,pp=0.259 m.Figure 30
Performance point at the MCE level forN=15.In this case, modal mass co-efficientα1=0.644 and Mode participation factor PF1=1.485. Figure 31 shows that pp is the performance point. The base shear at the performance point (pp), Vb,pp=1251.5 kN and roof displacement at the performance point (pp), Δroof,pp=0.101 m.Figure 31
Performance point at the DBE level forN=15.


### 5.3.2. Calculation of Response Reduction Factor at the Performance Point

Table19 shows different response reduction factors for MCE and DBE levels. These are calculated at different performance points (Figures 28–31).Table 19
Response Reduction Factors for DBE and MCE levels.
Parameters
μ
Δ
e
1[24]
μ
Δ
e
2 [24]
R
μ
ξ [24]
R
μ
IDRS
[First method of Energy-Ductility Based Response Reduction] [24]
R
μ
IDRS
[Second method of Energy-Ductility Based Response Reduction] [24]
R
d as per CSA [25]
N
=
20
DBE
1.04
1.004
1.02
1.04
1.004
1.5 or 2 for coupled shear walls with conventional reinforced coupling beam
MCE
2.05
1.2
1.58
2.05
1.34
N
=
15
DBE
1.01
1.00
1.002
1.01
1.00
MCE
1.87
1.13
1.39
1.87
1.22From Table19, response reduction factor of coupled shear walls is varying between 1.22 to 2.05 for maximum considered earthquake (MCE) level; which is almost same as the provision of CSA [25] for coupling beam with conventional reinforcement.


## 5.3.1. Calculation of Performance Point

Place considered here is Roorkee which belongs to the seismic zone IV and Z is 0.24 as per IS 1893 (part 1) [1]. 5% damped elastic response spectra as per IS 1893 (part 1) [1] are considered here as demand curve. DBE and MCE levels are considered for calculation of performance point (pp). Capacity curves are already obtained in Figure 27. The performance point has been calculated with the help of capacity spectrum method of ATC 40 [16] which is shown in  Figure 28.Figure 28
Performance point at the MCE level forN=20.In this case, modal mass co-efficientα1=0.616 and Mode participation factor PF1=1.51 derived with the help of modal analysis in SAP V 10.0.5 [23]. Figure 28 shows that pp is the performance point. The base shear at the performance point (pp), Vb,pp=1173.1 kN and roof displacement at the performance point (pp), Δroof,pp=0.31 m.In this case, modal mass co-efficientα1=0.616 and Mode participation factor PF1=1.51. Figure 29 shows that pp is the performance point. The base shear at the performance point (pp), Vb,pp=957.6 kN and roof displacement at the performance point (pp), Δroof,pp=0.097 m.Figure 29
Performance point at the DBE level forN=20.In this case, modal mass co-efficientα1=0.644 and Mode participation factor PF1=1.485. Figure 30 shows that pp is the performance point. The base shear at the performance point (pp), Vb,pp=1455.3 kN and roof displacement at the performance point (pp), Δroof,pp=0.259 m.Figure 30
Performance point at the MCE level forN=15.In this case, modal mass co-efficientα1=0.644 and Mode participation factor PF1=1.485. Figure 31 shows that pp is the performance point. The base shear at the performance point (pp), Vb,pp=1251.5 kN and roof displacement at the performance point (pp), Δroof,pp=0.101 m.Figure 31
Performance point at the DBE level forN=15.


## 5.3.2. Calculation of Response Reduction Factor at the Performance Point

Table19 shows different response reduction factors for MCE and DBE levels. These are calculated at different performance points (Figures 28–31).Table 19
Response Reduction Factors for DBE and MCE levels.
Parameters
μ
Δ
e
1[24]
μ
Δ
e
2 [24]
R
μ
ξ [24]
R
μ
IDRS
[First method of Energy-Ductility Based Response Reduction] [24]
R
μ
IDRS
[Second method of Energy-Ductility Based Response Reduction] [24]
R
d as per CSA [25]
N
=
20
DBE
1.04
1.004
1.02
1.04
1.004
1.5 or 2 for coupled shear walls with conventional reinforced coupling beam
MCE
2.05
1.2
1.58
2.05
1.34
N
=
15
DBE
1.01
1.00
1.002
1.01
1.00
MCE
1.87
1.13
1.39
1.87
1.22From Table19, response reduction factor of coupled shear walls is varying between 1.22 to 2.05 for maximum considered earthquake (MCE) level; which is almost same as the provision of CSA [25] for coupling beam with conventional reinforcement.


## 6. Conclusions

From the above studies, the following recommendations have been made for the design of coupled shear walls under earthquake motion.(i)
Design technique should be adopted for fixing the dimensions of coupled shear walls.(ii)
Coupled shear walls withN≥15 with equal storey height hs=3.6 m can be designed with an optimum ratio of Lb/Lw=0.25 for Lb/db=1.25 and Ib/I=8×10-03 to obtain consistency between the behavior with respect to the wall rotation and earthquake energy dissipations.(iii)
Pinned base condition can be provided at the base of the shear wall as this type of base condition offers better nonlinear behavior in compare to the fixed base condition.(iv)
The behavior of coupling beam should be governed by shear.



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*Source: 161502-2013-10-09.xml*