# Ductility Calculation of Prefabricated Shear Wall with Rabbet-Unbond Horizontal Connection

Authors: Chongfang Sun; Shuting Liang; Xiaojun Zhu; Yamin Song; Dongyue Wu
Journal: Advances in Civil Engineering
 (2018)

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## Abstract

Considering the effect of confined end-zone and unbonded reinforcement, the yield curvature and ultimate curvature of rabbet-unbond horizontal connection (RHC) shear wall were calculated. Based on the curvature calculation result, the yield displacement and ultimate displacement were calculated using displacement superposition, which were compared with test values. The result showed that theoretical values were slightly smaller than test values; however, on the whole, both were in good agreement. The author studied the effect of unbonded reinforcement on yield displacement and ultimate displacement, and the result showed that the effect on ultimate displacement is more pronounced than yield displacement. The present work could be useful for the design of new prefabricated shear wall.


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## Body

## 1. Introduction

In recent years, due to high industrialized level, good quality, convenient installation, energy conservation, and environment protection, the precast concrete structure is widely used. In particular, the prefabricated shear wall structure with large rigidity and great bearing capacity is the first choice for high-rise housings. In this regard, China has made policies to promote the development of the prefabricated shear wall structure [1], and many institutes and universities have carried out the relevant research [2, 3]; meanwhile, they cooperate with the national housing industrialization bases [4, 5] and applies the structure in some estate development projects as demonstration [6, 7]. What is more, the implementing subjects continue rise, for example, the prefabricated concrete shear wall structure system of Zhongnan Group [8]. For the prefabricated shear wall structure, the horizontal connection in the structure is the key to ensure structural integrity and seismic performance [9], and for this purpose, the authors invent a new horizontal connection [10] whose bearing capacity has been studied; however, the ductility which unbonded reinforcement can improve has not been studied.To date, addressing the ductility calculation of different shear wall structures, many engineers or scholars have made contributions. For the conventional cast-in-situ shear wall, Si et al. [11] proposed displacement ductility calculation method considering the effect of axial compression ratio, shear span ratio, confined end-zone, and stirrup eigenvalue. Park et al. [12] deduced the crack curvature, yield curvature, peak curvature, and ultimate curvature of flexure-dominated shear wall with concentrated end-zone reinforcements and uniformly web reinforcements along the cross section. For high-performance concrete shear wall, Liang et al. [13] deduced the crack-bending moment and curvature, yield-bending moment and curvature, peak-bending moment and curvature, and ultimate-bending moment and curvature. For prefabricated shear wall, Chen [14] deduced the formula of crack-bending moment and curvature, yield-bending moment and curvature, peak-bending moment and curvature, and ultimate-bending moment and curvature.Therefore, here, firstly, for the newly horizontal connection, considering the effect of confined end-zone and unbonded reinforcement, yield curvature and ultimate curvature are calculated, which prepares for displacement calculation. And then, yield displacement and ultimate displacement are calculated using displacement superposition, which is verified by the test values. Finally, the effects of unbonded reinforcement on yield displacement and ultimate displacement are studied.


## 2. Rabbet-Unbond Horizontal Connection

The arrangement of the rabbet-unbond horizontal connection (RHC) is shown in Figures1(a) and 1(b). The reinforcements reserved in the lower wall are connected with those in the upper wall by a well-developed grouting-connection technology. The two walls are integrated by filling high-performance mortar into the space between a pair of trapezoidal rabbets. Compared with the conventional connection (Figure 1(c)), a segment reinforcement of RHC in the lower wall is left unbonded to increase the ductility and energy dissipation of the assembled structure, and the trapezoidal rabbets are used to enhance the connection’s shear capacity.Figure 1
(a) Prefabricated shear wall, (b) the RHC details, and (c) conventional connection details. Note: For the sake of clarity, the lateral web reinforcements in the upper and lower walls are not shown in (b) and (c).
(a)
(b)
(c)


## 3. Curvature Calculation of RHC Shear Wall

This section calculates yield curvature and ultimate curvature considering effect of confined end-zone and unbonded reinforcement, meanwhile, which prepares for calculation of yield displacement and ultimate displacement in Section4.


### 3.1. Deformation Coordination Equation

#### 3.1.1. Strain Lag of Unbonded Reinforcement

According to test phenomenon of literature [15], compared with the bonded specimen, unbonded specimen’s crack width and length are greater under the same loading, which states that the unbonded cross section’s neutral axis is closer to compressed zone, as shown in Figure 2. Assuming that bending moments M of the two sections in Figure 2 are equal and the tensile reinforcements do not yield, we can get AsEsεs1Z1=AsEsεs2Z2=M, where As and Es are the cross section’s area and Young’s modulus of the tensile reinforcement, respectively, εs1,εs2 are the tensile reinforcement strains in the bonded cross section and the unbonded cross section, respectively, Z1andZ2 are the lever of the bonded cross section and the unbonded cross section, respectively, the subscripts 1 and 2 denote the bonded and unbonded. Apparently, under a constant moment M,Z2>Z1 leads to εs2<εs1, and this is called “strain lag.”Figure 2
Force in (a) bonded cross section and (b) unbonded cross section. Note:C1 and C2 represent the resultant force of compressive concrete, respectively.Regarding the “strain lag,” the unbonded cross section’s average strain distribution is no longer linear but bilinear (Figure3).Figure 3
Average strain distributions in (a) bonded cross section and (b) unbonded cross section.


#### 3.1.2. Deformation Coordination Region

For the bonded shear wall, the ductility’ calculation is based on the plane section assumption and the deformation coordination assumption between reinforcement and concrete. Because the maximum moment locates at the bottom of the shear wall, the plane section assumption regards a rectangular region at the bottom as the basic deformation coordination region (Figure4(a)), and the deformation coordination equation is(1)xc1h0−xc1=εc1lc1εs1ls1,where xc1 is the height of the compressed zone in the cross section of the bonded shear wall (Figure 3(a)). h0 is the effective height of the cross section. εc1 and εs1 are the edge strain of the compressed concrete and the strain of the tensile reinforcement in the cross section of the bonded shear wall, respectively (Figure 3(a)). lc1 and ls1 are the heights of the compressed zone and the tensile zone of the deformation coordination region of the bonded shear wall, respectively (see Figure 4(a)). Because lc1=ls1, the deformation coordination (1) is reexpressed as(2)xc1h0−xc1=εc1εs1.Figure 4
Deformation coordination region of (a) bonded shear wall and (b) unbonded shear wall.
(a)
(b)For the unbonded shear wall, we assume that in the compressed zone of the wall, the deformation of the unbonded reinforcement is consistent with the concrete deformation, but in the tensile zone of the wall, the deformation of the unbonded reinforcement is not consistent with the concrete deformation. In this regard, the traditional plane section assumption is no longer applicable, and the shape of the deformation coordination region is changed from the rectangular to the trapezoidal (Figure4(b)); however, the strain of the tensile reinforcement in the trapezoidal is uniformly distributed.


#### 3.1.3. Deformation Coordination Equation

For the sake of convenience, regard the concrete edge compressive strain of the maximum bending moment cross section as the calculated value, correspondingly, the equivalent compressive height oflc2 [16] is denoted as le. It is worth mentioned that the traditional plane section assumption is still applicable beyond trapezoidal deformation region, and the boundary condition of the region is linear. Thus, assuming that the deformation of the trapezoidal region also obeys the linear distribution along the cross section height, we can get the deformation coordination equation:(3)xc2h0−xc2=εc2leεs2ls2,where xc2 is the height of the compressed zone in the cross section of the unbonded shear wall (Figure 3(b)), εc2 and εs2 are the edge strain of the compressed concrete and the strain of the tensile reinforcement in the cross section of the unbonded shear wall, respectively (Figure 3(b)), ls2 is the height of the tensile zone of the deformation coordination region of the unbonded shear wall, and it equals the unbonded length, and le is the equivalent compressive height of the unbonded shear wall, derived from the loading tests of partial unbonded reinforced concrete beams with pure bending regions [16], as shown in Figure 5.Figure 5
Partial unbonded reinforced concrete beam.The formula ofle is obtained in the following method: In tests of partial unbonded reinforced concrete beam, strain values of the top concrete under each level load are recorded, and the total deformation Δlc of top concrete corresponding to unbonded length can be calculated. The average strain value of the top concrete under the same level load within pure bending region is expressed as εc. Divide Δlc by εc, and we can get the equivalent compressive height. Take the average value of equivalent compressive heights under all level loads as the final equivalent compressive height. By fitting, the equivalent compressive height formula is expressed as(4)le=0.74ls2,90mm<ls2<900mm32.81+0.09ls2,900mm<ls2<1575mm.Since the shear wall can be regarded as the cantilever beam, the author employsle to calculate the ductility of RHC shear wall. Here, ignore the effect of axial force acting on the shear wall.


### 3.2. Assumption

The assumptions to calculate the ductility of RHC shear wall are stated as follows:(1)
The concrete is based on the plane section assumption and its tensile strength is not considered.(2)
The compressed stress-strain curve of concrete and stress calculation of the vertical reinforcement refer to China code “Code for the design of concrete structures (GB 50010-2010, 2010).” The ultimate tensile strain of the vertical reinforcement is 0.01.(3)
In compressed zone, the unbonded reinforcement deformation is coordinated with the concrete deformation.(4)
The deformation coordination region in the unbonded segment is from rectangular to trapezoidal. The tensile reinforcement strain in the trapezoidal region is uniformly distributed. Deformation in the trapezoidal region obeys linear distribution.


### 3.3. Yield Curvature

In the yield state, the lateral deformation of compressive concrete is small, so stirrup stress is small, and stirrup restraint in the confined end-zone is not considered. The stress-strain relationship of concrete adopts unconfined stress-strain relationship. Assume that when bonded reinforcement segments in confined end-zone yield, the shear wall yields. Stress of the compressive concrete is assumed to be the linear triangle distribution. Deformation of reinforcement and concrete in compressive zone is assumed to be consistent. Stress and strain distribution in the yield state is shown in Figure6.Figure 6
Stress and strain distribution profile in the yield state.According to the force equilibrium, the following equation can be obtained as follows:(5)N1+Ty+Tswy=Cc+T′sw+T′sy,where N1 is the axial force acting on the shear wall’s cross section in the yield state. Ty is the yield force of vertical reinforcement in the tensile zone which can be defined as Ts=fyAs, where fy is the yield stress of tensile reinforcement. Tswy is the tensile force of web reinforcements in the tensile zone which can be defined as Tswy=ρwbwhw0−xyfyw/2, where ρw is the reinforcement ratio of web reinforcement; bw is the cross section width; hw0 is the effective height of the cross section; xy is the depth of compressive zone; and fyw is the yield stress of web reinforcement. Cc is the compressive force of concrete in the compressive zone which can be defined as Cc=bwEcxy2φy/2, where Ec is the elastic modulus of concrete. T′sw is the compressive force of web reinforcements in the compressive zone which can be defined as Tsw′=ρwbwxy−as′2Esφy/2, where as′ is the distance of compressed reinforcement resultant force to compressed edge. Tsy′ is the compressive force of vertical reinforcement in compressive constraint zone which here can be defined as Tsy′=Esxy−as′φyAs′, where As′ is the compressed reinforcement cross section area.According to assumption (1), the deformation coordination equation is(6)xyhw0−xy=εcmleεyls2,where εcm is the concrete compressive strain and εy is the yield strain of reinforcement.Solve above equations andφy=εcm/xy can be calculated according to the concrete strain profile in Figure 6.


### 3.4. Ultimate Curvature

In the ultimate state, the lateral deformation of compressive concrete is great, so stirrup restraint in confined end-zone should be considered. The stress-strain relationship of concrete adopts the confined stress-strain relationship. Assume that when the compressive concrete strain in confined, end-zone reaches ultimate compressive strain and shear wall reaches the ultimate state. Deformation of reinforcement and concrete in compressive zone is assumed to be consistent. What is more, compressive reinforcements are assumed to yield. The compression effect of unconfined concrete is not considered. Stress and strain distribution in ultimate state is shown in Figure7.Figure 7
Stress and strain distribution profile in the ultimate state.According to the force equilibrium, the following equation can be obtained:(7)N2+Ts+Tswu=Ccc+Cuc+T′su,where N2 is the axial force acting on the shear wall’s cross section in the ultimate state. Ts is the tensile force of vertical reinforcement in the tensile constraint zone which can be defined as Ts=σsAs, where σs is the tensile stress of vertical reinforcements in the constraint zone which can be defined as σ=sEsεccule/ls2hw0β/x−1 from (6), where εccu is the ultimate compressive strain of confined concrete; β is the coefficient corresponding to the rectangular stress block, refer to China code “Code for design of concrete structures (GB 50010-2010, 2010)”; x is the depth of the compressive zone in rectangular stress block which can be defined as x=xuβ1. Tswu is the tensile force of web reinforcements in the tensile zone which can be defined as Tswu=fywbwρwhw0−1.5x. Ccc is the compressive force of confined concrete. Cuc is the compressive force of unconfined concrete. Ccc and Cuc have different formulas for different eccentric compression states. Tsu′ is the compressive force of vertical reinforcement in the compressive constraint zone which here can be defined as Tsu′=σs′As′, where σs′ is the compressive stress of vertical reinforcements.The constitutive law of confined concrete referring to literature [17] is as follows:(8)εcc=1+3.5λvεc0,εccu=2.34+2.49λv0.73εcc,fcc=1+1.79λvfc,where εcc is the peak compressive strain of confined concrete, λv is the stirrup eigenvalue, εc0 is the peak compressive strain of unconfined concrete and it is set as 0.0018, fcc is the axial compressive strength of confined concrete, and fc is the axial compressive strength of unconfined concrete.According to assumption (1), the deformation coordination equation is(9)xuhw0−xu=εcculeεsls2.From (7) and (9), x can be calculated. Different x leads to different eccentric compression states and different ultimate curvatures; then, the explanations are as follows:(1)
Whenx≤xb, the wall is in large eccentric compression, Ts=Ts′=fyAs, where xb is the boundary depth of the compressive zone. This situation consists of two states: xb>x>lc and x≤lc.(a)
Whenxb>x>lc,N2=Ccc+Cuc−Tsw, where Ccc=αfccbwlc, α is the coefficient corresponding to constraint zone length and it is set as 0.8 and lc is the confinement depth. Cuc=fcbwx−lc.(b)
Whenx≤lc,N2=Ccc−Tsw, where Cuc=0,Ccc=αfccbwx.(2)
Whenx>xb, the wall is in small eccentric compression, N2=Ccc+Cuc+Ts′−Ts, where here Ts′=fy′As′. This situation consists of two states: x>lc and xb<x≤lc.(a)
Whenx>lc,Ccc=αfccbwlc,Cuc=fcbwx−lc, the equation about x can be obtained as:(10)fcbwx2+αfccbwlc−fcbwlc+f′yA′s+AsEsεcculels2−N2⋅x−AsEsεccuhw0β1lels2=0.Whenxb<x≤lc,N2=Ccc+Ts′−Ts, where Cuc=0,Ccc=αfccbwx, the equation about x can be obtained as(11)αfccbwx2+fy′As′+AsEsεcculels2−N2⋅x−AsEsεccuhw0β1lels2=0.Solve the above equations, and finally we can obtain the ultimate curvatureφu=εccu/xu.


## 3.1. Deformation Coordination Equation

### 3.1.1. Strain Lag of Unbonded Reinforcement

According to test phenomenon of literature [15], compared with the bonded specimen, unbonded specimen’s crack width and length are greater under the same loading, which states that the unbonded cross section’s neutral axis is closer to compressed zone, as shown in Figure 2. Assuming that bending moments M of the two sections in Figure 2 are equal and the tensile reinforcements do not yield, we can get AsEsεs1Z1=AsEsεs2Z2=M, where As and Es are the cross section’s area and Young’s modulus of the tensile reinforcement, respectively, εs1,εs2 are the tensile reinforcement strains in the bonded cross section and the unbonded cross section, respectively, Z1andZ2 are the lever of the bonded cross section and the unbonded cross section, respectively, the subscripts 1 and 2 denote the bonded and unbonded. Apparently, under a constant moment M,Z2>Z1 leads to εs2<εs1, and this is called “strain lag.”Figure 2
Force in (a) bonded cross section and (b) unbonded cross section. Note:C1 and C2 represent the resultant force of compressive concrete, respectively.Regarding the “strain lag,” the unbonded cross section’s average strain distribution is no longer linear but bilinear (Figure3).Figure 3
Average strain distributions in (a) bonded cross section and (b) unbonded cross section.


### 3.1.2. Deformation Coordination Region

For the bonded shear wall, the ductility’ calculation is based on the plane section assumption and the deformation coordination assumption between reinforcement and concrete. Because the maximum moment locates at the bottom of the shear wall, the plane section assumption regards a rectangular region at the bottom as the basic deformation coordination region (Figure4(a)), and the deformation coordination equation is(1)xc1h0−xc1=εc1lc1εs1ls1,where xc1 is the height of the compressed zone in the cross section of the bonded shear wall (Figure 3(a)). h0 is the effective height of the cross section. εc1 and εs1 are the edge strain of the compressed concrete and the strain of the tensile reinforcement in the cross section of the bonded shear wall, respectively (Figure 3(a)). lc1 and ls1 are the heights of the compressed zone and the tensile zone of the deformation coordination region of the bonded shear wall, respectively (see Figure 4(a)). Because lc1=ls1, the deformation coordination (1) is reexpressed as(2)xc1h0−xc1=εc1εs1.Figure 4
Deformation coordination region of (a) bonded shear wall and (b) unbonded shear wall.
(a)
(b)For the unbonded shear wall, we assume that in the compressed zone of the wall, the deformation of the unbonded reinforcement is consistent with the concrete deformation, but in the tensile zone of the wall, the deformation of the unbonded reinforcement is not consistent with the concrete deformation. In this regard, the traditional plane section assumption is no longer applicable, and the shape of the deformation coordination region is changed from the rectangular to the trapezoidal (Figure4(b)); however, the strain of the tensile reinforcement in the trapezoidal is uniformly distributed.


### 3.1.3. Deformation Coordination Equation

For the sake of convenience, regard the concrete edge compressive strain of the maximum bending moment cross section as the calculated value, correspondingly, the equivalent compressive height oflc2 [16] is denoted as le. It is worth mentioned that the traditional plane section assumption is still applicable beyond trapezoidal deformation region, and the boundary condition of the region is linear. Thus, assuming that the deformation of the trapezoidal region also obeys the linear distribution along the cross section height, we can get the deformation coordination equation:(3)xc2h0−xc2=εc2leεs2ls2,where xc2 is the height of the compressed zone in the cross section of the unbonded shear wall (Figure 3(b)), εc2 and εs2 are the edge strain of the compressed concrete and the strain of the tensile reinforcement in the cross section of the unbonded shear wall, respectively (Figure 3(b)), ls2 is the height of the tensile zone of the deformation coordination region of the unbonded shear wall, and it equals the unbonded length, and le is the equivalent compressive height of the unbonded shear wall, derived from the loading tests of partial unbonded reinforced concrete beams with pure bending regions [16], as shown in Figure 5.Figure 5
Partial unbonded reinforced concrete beam.The formula ofle is obtained in the following method: In tests of partial unbonded reinforced concrete beam, strain values of the top concrete under each level load are recorded, and the total deformation Δlc of top concrete corresponding to unbonded length can be calculated. The average strain value of the top concrete under the same level load within pure bending region is expressed as εc. Divide Δlc by εc, and we can get the equivalent compressive height. Take the average value of equivalent compressive heights under all level loads as the final equivalent compressive height. By fitting, the equivalent compressive height formula is expressed as(4)le=0.74ls2,90mm<ls2<900mm32.81+0.09ls2,900mm<ls2<1575mm.Since the shear wall can be regarded as the cantilever beam, the author employsle to calculate the ductility of RHC shear wall. Here, ignore the effect of axial force acting on the shear wall.


## 3.1.1. Strain Lag of Unbonded Reinforcement

According to test phenomenon of literature [15], compared with the bonded specimen, unbonded specimen’s crack width and length are greater under the same loading, which states that the unbonded cross section’s neutral axis is closer to compressed zone, as shown in Figure 2. Assuming that bending moments M of the two sections in Figure 2 are equal and the tensile reinforcements do not yield, we can get AsEsεs1Z1=AsEsεs2Z2=M, where As and Es are the cross section’s area and Young’s modulus of the tensile reinforcement, respectively, εs1,εs2 are the tensile reinforcement strains in the bonded cross section and the unbonded cross section, respectively, Z1andZ2 are the lever of the bonded cross section and the unbonded cross section, respectively, the subscripts 1 and 2 denote the bonded and unbonded. Apparently, under a constant moment M,Z2>Z1 leads to εs2<εs1, and this is called “strain lag.”Figure 2
Force in (a) bonded cross section and (b) unbonded cross section. Note:C1 and C2 represent the resultant force of compressive concrete, respectively.Regarding the “strain lag,” the unbonded cross section’s average strain distribution is no longer linear but bilinear (Figure3).Figure 3
Average strain distributions in (a) bonded cross section and (b) unbonded cross section.


## 3.1.2. Deformation Coordination Region

For the bonded shear wall, the ductility’ calculation is based on the plane section assumption and the deformation coordination assumption between reinforcement and concrete. Because the maximum moment locates at the bottom of the shear wall, the plane section assumption regards a rectangular region at the bottom as the basic deformation coordination region (Figure4(a)), and the deformation coordination equation is(1)xc1h0−xc1=εc1lc1εs1ls1,where xc1 is the height of the compressed zone in the cross section of the bonded shear wall (Figure 3(a)). h0 is the effective height of the cross section. εc1 and εs1 are the edge strain of the compressed concrete and the strain of the tensile reinforcement in the cross section of the bonded shear wall, respectively (Figure 3(a)). lc1 and ls1 are the heights of the compressed zone and the tensile zone of the deformation coordination region of the bonded shear wall, respectively (see Figure 4(a)). Because lc1=ls1, the deformation coordination (1) is reexpressed as(2)xc1h0−xc1=εc1εs1.Figure 4
Deformation coordination region of (a) bonded shear wall and (b) unbonded shear wall.
(a)
(b)For the unbonded shear wall, we assume that in the compressed zone of the wall, the deformation of the unbonded reinforcement is consistent with the concrete deformation, but in the tensile zone of the wall, the deformation of the unbonded reinforcement is not consistent with the concrete deformation. In this regard, the traditional plane section assumption is no longer applicable, and the shape of the deformation coordination region is changed from the rectangular to the trapezoidal (Figure4(b)); however, the strain of the tensile reinforcement in the trapezoidal is uniformly distributed.


## 3.1.3. Deformation Coordination Equation

For the sake of convenience, regard the concrete edge compressive strain of the maximum bending moment cross section as the calculated value, correspondingly, the equivalent compressive height oflc2 [16] is denoted as le. It is worth mentioned that the traditional plane section assumption is still applicable beyond trapezoidal deformation region, and the boundary condition of the region is linear. Thus, assuming that the deformation of the trapezoidal region also obeys the linear distribution along the cross section height, we can get the deformation coordination equation:(3)xc2h0−xc2=εc2leεs2ls2,where xc2 is the height of the compressed zone in the cross section of the unbonded shear wall (Figure 3(b)), εc2 and εs2 are the edge strain of the compressed concrete and the strain of the tensile reinforcement in the cross section of the unbonded shear wall, respectively (Figure 3(b)), ls2 is the height of the tensile zone of the deformation coordination region of the unbonded shear wall, and it equals the unbonded length, and le is the equivalent compressive height of the unbonded shear wall, derived from the loading tests of partial unbonded reinforced concrete beams with pure bending regions [16], as shown in Figure 5.Figure 5
Partial unbonded reinforced concrete beam.The formula ofle is obtained in the following method: In tests of partial unbonded reinforced concrete beam, strain values of the top concrete under each level load are recorded, and the total deformation Δlc of top concrete corresponding to unbonded length can be calculated. The average strain value of the top concrete under the same level load within pure bending region is expressed as εc. Divide Δlc by εc, and we can get the equivalent compressive height. Take the average value of equivalent compressive heights under all level loads as the final equivalent compressive height. By fitting, the equivalent compressive height formula is expressed as(4)le=0.74ls2,90mm<ls2<900mm32.81+0.09ls2,900mm<ls2<1575mm.Since the shear wall can be regarded as the cantilever beam, the author employsle to calculate the ductility of RHC shear wall. Here, ignore the effect of axial force acting on the shear wall.


## 3.2. Assumption

The assumptions to calculate the ductility of RHC shear wall are stated as follows:(1)
The concrete is based on the plane section assumption and its tensile strength is not considered.(2)
The compressed stress-strain curve of concrete and stress calculation of the vertical reinforcement refer to China code “Code for the design of concrete structures (GB 50010-2010, 2010).” The ultimate tensile strain of the vertical reinforcement is 0.01.(3)
In compressed zone, the unbonded reinforcement deformation is coordinated with the concrete deformation.(4)
The deformation coordination region in the unbonded segment is from rectangular to trapezoidal. The tensile reinforcement strain in the trapezoidal region is uniformly distributed. Deformation in the trapezoidal region obeys linear distribution.


## 3.3. Yield Curvature

In the yield state, the lateral deformation of compressive concrete is small, so stirrup stress is small, and stirrup restraint in the confined end-zone is not considered. The stress-strain relationship of concrete adopts unconfined stress-strain relationship. Assume that when bonded reinforcement segments in confined end-zone yield, the shear wall yields. Stress of the compressive concrete is assumed to be the linear triangle distribution. Deformation of reinforcement and concrete in compressive zone is assumed to be consistent. Stress and strain distribution in the yield state is shown in Figure6.Figure 6
Stress and strain distribution profile in the yield state.According to the force equilibrium, the following equation can be obtained as follows:(5)N1+Ty+Tswy=Cc+T′sw+T′sy,where N1 is the axial force acting on the shear wall’s cross section in the yield state. Ty is the yield force of vertical reinforcement in the tensile zone which can be defined as Ts=fyAs, where fy is the yield stress of tensile reinforcement. Tswy is the tensile force of web reinforcements in the tensile zone which can be defined as Tswy=ρwbwhw0−xyfyw/2, where ρw is the reinforcement ratio of web reinforcement; bw is the cross section width; hw0 is the effective height of the cross section; xy is the depth of compressive zone; and fyw is the yield stress of web reinforcement. Cc is the compressive force of concrete in the compressive zone which can be defined as Cc=bwEcxy2φy/2, where Ec is the elastic modulus of concrete. T′sw is the compressive force of web reinforcements in the compressive zone which can be defined as Tsw′=ρwbwxy−as′2Esφy/2, where as′ is the distance of compressed reinforcement resultant force to compressed edge. Tsy′ is the compressive force of vertical reinforcement in compressive constraint zone which here can be defined as Tsy′=Esxy−as′φyAs′, where As′ is the compressed reinforcement cross section area.According to assumption (1), the deformation coordination equation is(6)xyhw0−xy=εcmleεyls2,where εcm is the concrete compressive strain and εy is the yield strain of reinforcement.Solve above equations andφy=εcm/xy can be calculated according to the concrete strain profile in Figure 6.


## 3.4. Ultimate Curvature

In the ultimate state, the lateral deformation of compressive concrete is great, so stirrup restraint in confined end-zone should be considered. The stress-strain relationship of concrete adopts the confined stress-strain relationship. Assume that when the compressive concrete strain in confined, end-zone reaches ultimate compressive strain and shear wall reaches the ultimate state. Deformation of reinforcement and concrete in compressive zone is assumed to be consistent. What is more, compressive reinforcements are assumed to yield. The compression effect of unconfined concrete is not considered. Stress and strain distribution in ultimate state is shown in Figure7.Figure 7
Stress and strain distribution profile in the ultimate state.According to the force equilibrium, the following equation can be obtained:(7)N2+Ts+Tswu=Ccc+Cuc+T′su,where N2 is the axial force acting on the shear wall’s cross section in the ultimate state. Ts is the tensile force of vertical reinforcement in the tensile constraint zone which can be defined as Ts=σsAs, where σs is the tensile stress of vertical reinforcements in the constraint zone which can be defined as σ=sEsεccule/ls2hw0β/x−1 from (6), where εccu is the ultimate compressive strain of confined concrete; β is the coefficient corresponding to the rectangular stress block, refer to China code “Code for design of concrete structures (GB 50010-2010, 2010)”; x is the depth of the compressive zone in rectangular stress block which can be defined as x=xuβ1. Tswu is the tensile force of web reinforcements in the tensile zone which can be defined as Tswu=fywbwρwhw0−1.5x. Ccc is the compressive force of confined concrete. Cuc is the compressive force of unconfined concrete. Ccc and Cuc have different formulas for different eccentric compression states. Tsu′ is the compressive force of vertical reinforcement in the compressive constraint zone which here can be defined as Tsu′=σs′As′, where σs′ is the compressive stress of vertical reinforcements.The constitutive law of confined concrete referring to literature [17] is as follows:(8)εcc=1+3.5λvεc0,εccu=2.34+2.49λv0.73εcc,fcc=1+1.79λvfc,where εcc is the peak compressive strain of confined concrete, λv is the stirrup eigenvalue, εc0 is the peak compressive strain of unconfined concrete and it is set as 0.0018, fcc is the axial compressive strength of confined concrete, and fc is the axial compressive strength of unconfined concrete.According to assumption (1), the deformation coordination equation is(9)xuhw0−xu=εcculeεsls2.From (7) and (9), x can be calculated. Different x leads to different eccentric compression states and different ultimate curvatures; then, the explanations are as follows:(1)
Whenx≤xb, the wall is in large eccentric compression, Ts=Ts′=fyAs, where xb is the boundary depth of the compressive zone. This situation consists of two states: xb>x>lc and x≤lc.(a)
Whenxb>x>lc,N2=Ccc+Cuc−Tsw, where Ccc=αfccbwlc, α is the coefficient corresponding to constraint zone length and it is set as 0.8 and lc is the confinement depth. Cuc=fcbwx−lc.(b)
Whenx≤lc,N2=Ccc−Tsw, where Cuc=0,Ccc=αfccbwx.(2)
Whenx>xb, the wall is in small eccentric compression, N2=Ccc+Cuc+Ts′−Ts, where here Ts′=fy′As′. This situation consists of two states: x>lc and xb<x≤lc.(a)
Whenx>lc,Ccc=αfccbwlc,Cuc=fcbwx−lc, the equation about x can be obtained as:(10)fcbwx2+αfccbwlc−fcbwlc+f′yA′s+AsEsεcculels2−N2⋅x−AsEsεccuhw0β1lels2=0.Whenxb<x≤lc,N2=Ccc+Ts′−Ts, where Cuc=0,Ccc=αfccbwx, the equation about x can be obtained as(11)αfccbwx2+fy′As′+AsEsεcculels2−N2⋅x−AsEsεccuhw0β1lels2=0.Solve the above equations, and finally we can obtain the ultimate curvatureφu=εccu/xu.


## 4. Calculation of Displacement Ductility

The displacement ductilityμΔ is defined as μΔ=Δu/Δy, where Δu is the ultimate displacement and Δy is the yield displacement.Assume that wall’s displacement consists of two parts approximately: one part is the displacement that the wall generates resisting external force when all reinforcements are bonded, and assume the wall is not a rigid body and another part is the displacement that elongation of unbonded reinforcements leads the wall to generate when the wall is RHC wall, and assume the wall is a rigid body.(12)Δy=Δy1+Δy2,Δu=Δu1+Δu2,where Δy1andΔu1 are yield displacement and ultimate displacement that the wall generates resisting external force when all reinforcements are bonded, and assume the wall is not a rigid body, respectively. Δy2andΔu2 are yield displacement and ultimate displacement that elongation of unbonded reinforcements leads the wall to generate when the wall is RHC wall, and assume the wall is a rigid body, respectively.


### 4.1. Calculation ofΔy1,Δu1

The shear wall can be regarded as a cantilever member. In a cantilever member, the curvature distribution of plastic hinge region is very irregular (Figure8(b)). The height of plastic hinge region can be determined by the curvature distribution along the wall height.Figure 8
(a) Bending moment, (b) curvature, and (c) lateral displacement of cantilever shear wall.
(a)
(b)
(c)Based on the displacement equivalent principle in maximum bending moment region, the curvature distribution of plastic hinge region can be simplified, which consists of two parts: the plastic deformation region and the elastic deformation region.Curvature is defined as the corner angle unit height, so the corner angle relative to the base section at any heighth can be expressed as θh=∫0hφdx.Therefore, the lateral displacement relative to the base section at any heighth can be expressed as Δh=∫0hxφdx.Define that when the base section of cantilever shear wall yields, shear wall yields. The yield curvature is linear distribution along the wall heightH (Figure 8(b)). The curvature in the top section is zero, and the curvature in the base section is the yield curvature φy (Figure 9), and the resulting equation can be expressed as x/H=φ/φy.Figure 9
Curvature of cantilever shear wall.Applying above equations, the yield displacement can be derived as follows:(13)Δy1=∫0Hφyx2Hdx=φyH23.Assume that when shear wall reaches the ultimate state, there is a equivalent plastic hinge region heightlp at the wall base, in which plastic curvature is equal to the largest plastic curvature φp, and the resulting equation can be expressed as φp=φu−φy.Applying above equations, the ultimate displacement can be derived as follows:(14)Δu1=∫0Hφyx2Hdx+∫H−lpHφu−φyxdx=φyH23+φu−φy2Hlp−lp22,where lp is the equivalent plastic hinge region height which can be defined as lp=0.20+0.044H/hwhw, where hw is the cross section length. The formula of lp is generally a good estimate for RC shear walls [9].


### 4.2. Calculation ofΔy2,Δu2

#### 4.2.1. Calculation ofΔy2

When calculatingΔy2, as shown in Figure 10(a), assume that the shear wall is a rigid body and the deformation of bonded reinforcement is ignored, the yield displacement can be expressed approximately as Δy2=θy×H, where θy is the wall’s corner angle, as shown in Figure 10(b), which can be expressed as θy=δy/hw/2−xy±e, where xy is the compressive zone depth in the yield state obtained by Section 3.3 approximately, e is the distance of unbonded reinforcement to the cross section’s geometric center, and δy is the elongation of unbonded reinforcement.Figure 10
(a) Rigid body rotation and (b) unbonded reinforcement deformation of RHC shear wall.
(a)
(b)When unbonded reinforcements in tensile confined end-zone yield, elongation of unbonded reinforcements can be expressed asδy=εy×ls.Applying above equations, the yield displacement can be derived as follows:(15)Δy2=εylsHhw/2−xy±e.


#### 4.2.2. Calculation ofΔu2

When calculatingΔu2, as shown in Figure 10(a), assume that the shear wall is a rigid body and the elongation of bonded reinforcement is ignored, the ultimate displacement can be expressed approximately as Δu2=θu×H, where θu is the corner angle at wall base, which can be expressed as θu=δu/hw/2−xu±e, where xu is the compressive zone depth in the ultimate state obtained by Section 3.4 approximately and δu is the elongation of unbonded reinforcement.Forδu, according to the provision 4.2.4 in China code “Code for design of concrete structures (GB 50010-2010, 2010),” the gross elongation of HRB400 ordinary reinforcement under the largest force should not be less than 7.5%. When calculating Δu2, we assume that the deformation of unbonded reinforcements in tensile confined end-zone reaches the minimum elongation percentage δgt=7.5%; the elongation can be expressed as δu=δgt×ls.Applying above equations, the ultimate displacement can be expressed as(16)Δu2=δgtlsHhw/2−xu±e.


## 4.1. Calculation ofΔy1,Δu1

The shear wall can be regarded as a cantilever member. In a cantilever member, the curvature distribution of plastic hinge region is very irregular (Figure8(b)). The height of plastic hinge region can be determined by the curvature distribution along the wall height.Figure 8
(a) Bending moment, (b) curvature, and (c) lateral displacement of cantilever shear wall.
(a)
(b)
(c)Based on the displacement equivalent principle in maximum bending moment region, the curvature distribution of plastic hinge region can be simplified, which consists of two parts: the plastic deformation region and the elastic deformation region.Curvature is defined as the corner angle unit height, so the corner angle relative to the base section at any heighth can be expressed as θh=∫0hφdx.Therefore, the lateral displacement relative to the base section at any heighth can be expressed as Δh=∫0hxφdx.Define that when the base section of cantilever shear wall yields, shear wall yields. The yield curvature is linear distribution along the wall heightH (Figure 8(b)). The curvature in the top section is zero, and the curvature in the base section is the yield curvature φy (Figure 9), and the resulting equation can be expressed as x/H=φ/φy.Figure 9
Curvature of cantilever shear wall.Applying above equations, the yield displacement can be derived as follows:(13)Δy1=∫0Hφyx2Hdx=φyH23.Assume that when shear wall reaches the ultimate state, there is a equivalent plastic hinge region heightlp at the wall base, in which plastic curvature is equal to the largest plastic curvature φp, and the resulting equation can be expressed as φp=φu−φy.Applying above equations, the ultimate displacement can be derived as follows:(14)Δu1=∫0Hφyx2Hdx+∫H−lpHφu−φyxdx=φyH23+φu−φy2Hlp−lp22,where lp is the equivalent plastic hinge region height which can be defined as lp=0.20+0.044H/hwhw, where hw is the cross section length. The formula of lp is generally a good estimate for RC shear walls [9].


## 4.2. Calculation ofΔy2,Δu2

### 4.2.1. Calculation ofΔy2

When calculatingΔy2, as shown in Figure 10(a), assume that the shear wall is a rigid body and the deformation of bonded reinforcement is ignored, the yield displacement can be expressed approximately as Δy2=θy×H, where θy is the wall’s corner angle, as shown in Figure 10(b), which can be expressed as θy=δy/hw/2−xy±e, where xy is the compressive zone depth in the yield state obtained by Section 3.3 approximately, e is the distance of unbonded reinforcement to the cross section’s geometric center, and δy is the elongation of unbonded reinforcement.Figure 10
(a) Rigid body rotation and (b) unbonded reinforcement deformation of RHC shear wall.
(a)
(b)When unbonded reinforcements in tensile confined end-zone yield, elongation of unbonded reinforcements can be expressed asδy=εy×ls.Applying above equations, the yield displacement can be derived as follows:(15)Δy2=εylsHhw/2−xy±e.


### 4.2.2. Calculation ofΔu2

When calculatingΔu2, as shown in Figure 10(a), assume that the shear wall is a rigid body and the elongation of bonded reinforcement is ignored, the ultimate displacement can be expressed approximately as Δu2=θu×H, where θu is the corner angle at wall base, which can be expressed as θu=δu/hw/2−xu±e, where xu is the compressive zone depth in the ultimate state obtained by Section 3.4 approximately and δu is the elongation of unbonded reinforcement.Forδu, according to the provision 4.2.4 in China code “Code for design of concrete structures (GB 50010-2010, 2010),” the gross elongation of HRB400 ordinary reinforcement under the largest force should not be less than 7.5%. When calculating Δu2, we assume that the deformation of unbonded reinforcements in tensile confined end-zone reaches the minimum elongation percentage δgt=7.5%; the elongation can be expressed as δu=δgt×ls.Applying above equations, the ultimate displacement can be expressed as(16)Δu2=δgtlsHhw/2−xu±e.


## 4.2.1. Calculation ofΔy2

When calculatingΔy2, as shown in Figure 10(a), assume that the shear wall is a rigid body and the deformation of bonded reinforcement is ignored, the yield displacement can be expressed approximately as Δy2=θy×H, where θy is the wall’s corner angle, as shown in Figure 10(b), which can be expressed as θy=δy/hw/2−xy±e, where xy is the compressive zone depth in the yield state obtained by Section 3.3 approximately, e is the distance of unbonded reinforcement to the cross section’s geometric center, and δy is the elongation of unbonded reinforcement.Figure 10
(a) Rigid body rotation and (b) unbonded reinforcement deformation of RHC shear wall.
(a)
(b)When unbonded reinforcements in tensile confined end-zone yield, elongation of unbonded reinforcements can be expressed asδy=εy×ls.Applying above equations, the yield displacement can be derived as follows:(15)Δy2=εylsHhw/2−xy±e.


## 4.2.2. Calculation ofΔu2

When calculatingΔu2, as shown in Figure 10(a), assume that the shear wall is a rigid body and the elongation of bonded reinforcement is ignored, the ultimate displacement can be expressed approximately as Δu2=θu×H, where θu is the corner angle at wall base, which can be expressed as θu=δu/hw/2−xu±e, where xu is the compressive zone depth in the ultimate state obtained by Section 3.4 approximately and δu is the elongation of unbonded reinforcement.Forδu, according to the provision 4.2.4 in China code “Code for design of concrete structures (GB 50010-2010, 2010),” the gross elongation of HRB400 ordinary reinforcement under the largest force should not be less than 7.5%. When calculating Δu2, we assume that the deformation of unbonded reinforcements in tensile confined end-zone reaches the minimum elongation percentage δgt=7.5%; the elongation can be expressed as δu=δgt×ls.Applying above equations, the ultimate displacement can be expressed as(16)Δu2=δgtlsHhw/2−xu±e.


## 5. Verification of Ductility

Two full-scale specimens (RHC-1, RHC-2) are designed and tested. The RHC specimen is composed of a base and a wall. The wall has a height, length, and thickness of 3.4 m, 1.7 m, and 0.2 m, as shown in Figure11. The base has a length, width, and height of 2.3 m, 0.69 m, and 0.82 m. The grouting region is located in the wall. The unbonded segment at the vertical reinforcements is located in the base. Based on the finite element simulation results of specimens before test, select the unbonded length at the vertical reinforcement as 300 mm. The horizontal connection is located between the wall and the base. Considering the vertical reinforcement intensity of confined end-zones and construction of metal bellows, seven rabbets are set up symmetrically at connection. The rabbet’s size is shown in Figure 11. High-performance mortar is grouted in metal bellows and connection. Two specimens have the same details of reinforcement and size, and the difference is the axial compression ratio with 0.1 of RHC-1 and 0.2 of RHC-2.Figure 11
Dimension and reinforcement details of the RHC shear wall specimen (unit: mm).


### 5.1. Material Properties

All of the materials (concrete, steel, and mortar) employed in the tests are selected based on China code “Code for design of concrete structures (GB 50010-2010, 2010).” The grade of concrete is C35, which denotes that the ultimate compressive strength of the cubic concrete specimens (15 cm × 15 cm × 15 cm) cured in standard conditions is 35 MPa, and its Poisson’s ratio is 0.2. Here, six cubic concrete specimens are tested.The grade of steel in the tests is HRB400, where HRB denotes hot-rolled ribbed-steel bar and the number 400 denotes the yield strength. Its elastic modulus is 200 GPa and Poisson’s ratio is 0.3, and three sets of steels with different diameters (d = 10 mm, 12 mm, and 16 mm) are tested.The high-performance mortar is tested (160 mm × 40 mm × 40 mm) with a type of H-80, which is a characteristic of early strength, high strength, no shrinkage, and high fluidity.The mechanical parameters of all materials are presented in Tables1–3.Table 1
Mechanical property of concrete.
Specimen number
Failure load (kN)
Compressive strength (MPa)
Average compressive strength (MPa)
1
880
39.11
39.86
2
910
40.44
3
822
36.53
4
960
42.67
5
950
42.22
6
860
38.22Table 2
Mechanical property of steel.
Specimen number
Steel diameter
Yield strength (MPa)
Ultimate strength (MPa)
Ratio of tensile strength to yield strength
Elongation (%)
1-1
10
445
625
1.4
24
1-2
455
615
1.35
24
Average value
450
620
1.375
24
2-1
12
445
575
1.29
25
2-2
460
585
1.27
25
Average value
452.5
580
1.28
25
3–1
16
465
600
1.29
25
3-2
460
600
1.30
25
Average value
462.5
600
1.295
25Table 3
Mechanical property of high-performance mortar.
Test item
Time
Measured value
Compressive strength (MPa)
1 d
24.96
28 d
93.52
Rupture strength (MPa)
1 d
5.29
28 d
10.44


### 5.2. Test Setup

The quasistatic cyclic tests of the RHC shear wall are performed. Axial load is applied by tensioning prestressed reinforcement strands, and the lateral load is applied by the 150 t hydraulic actuator. The loading protocol employed consists of a load control procedure first and a displacement control procedure then. Moreover, four steel supports are employed on both sides of the wall symmetrically to keep the lateral stability of the specimen during the testing process. The lateral displacements of the wall are monitored by seven displacement transducers. The testing system is shown in Figure12.Figure 12
Test setup: (a) schematic and (b) photograph.
(a)
(b)


### 5.3. Crack Pattern and Failure Mode

For the RHC-1 specimen, the first crack observed at 210 kN was horizontal flexural cracks when the specimen was pulled, which occurred near the bottom of the wall. The crack on the top interface of rabbet was generated at 240 kN when the specimen was pushed. However, for the RHC-2 specimen, the first crack was observed at 270 kN when the specimen was pushed, which occurred near the bottom of the wall. The crack on the top interface of rabbet was generated at 300 kN when the specimen was pushed.The yield state was defined as when bonded reinforcement segments in the confined end-zone yielded, the specimen yielded. The increase of the applied load resulted in the propagation of cracks and initiation of new flexural cracks along the specimen. Further load increase extended the existing flexural cracks into flexure shear cracks and caused initiation of vertical cracks. In addition, the cracks in the center of the test specimen propagated to the compressed zone, and the crack width and the displacement increased. Finally, when the lateral resistance degenerated to 85% of the peak load, defined as the failure of the specimen, the failure mode was a typical flexural failure for each specimen, which was controlled by crushing concrete at the plastic region near the bottom of the wall. The final crack pattern and failure mode are shown in Figure13.Figure 13
Final crack pattern and failure mode of (a) RHC-1 specimen and (b) RHC-2 specimen.
(a)
(b)From Figure13(b), it was noted that the failure mode was not symmetrical. The reason was that the push force and pull force of hydraulic actuator during test were not symmetrical, but had little effect.


### 5.4. Load-Displacement Relationship

The lateral load versus top displacement curves of the two specimens is shown in Figure14.Figure 14
Hysteresis curves and skeleton curves of RHC-1 and RHC-2. (a) Hysteresis curve of RHC-1. (b) Skeleton curve of RHC-1. (c) Hysteresis curve of RHC-2. (d) Skeleton curve of RHC-2.
(a)
(b)
(c)
(d)It is observed in Figure14 that the number of hysteresis loop is greater and the curve is plumper in the RHC-1 specimen than in the RHC-2 specimen. The bearing capacity of the specimen is decreased by reducing the axial compression ratio, whereas the yield displacement and ultimate displacement are increased at the same time.


### 5.5. Bearing Capacity and Deformation

From the tests, we obtain the crack loadFcr, crack displacement Δcr, yield load Fy, yield displacement Δy, peak load Fm, peak displacement Δm, ultimate load Fu, and ultimate displacement Δu of two specimens, as shown in Table 4.Table 4
Test results of two specimens.
Specimen
F
cr (kN)
Δcr (mm)
F
y (kN)
Δy (mm)
F
m (kN)
Δm (mm)
F
u (kN)
Δu (mm)
RHC-1
210
12.28
330
18.30
610
85.44
518.50
137.70
RHC-2
270
11.74
420
13.60
718
46.49
610.30
75.50Table4 shows that Fcr,Fy,Fm,andFu increase, whereas Δcr,Δy,Δm,andΔu decrease with increase of the axial compression ratio.


### 5.6. Verification

Applying (13) and (15), the yield displacement can be expressed as(17)Δy=φyH23+εylsHhw/2−xy±e.Applying (14) and (16), the ultimate displacement can be expressed as(18)Δu=φyH23+φu−φy2Hlp−lp22+δgtlsHhw/2−xu±e.From (17) and (18), the yield displacement and ultimate displacement can be calculated, which are compared with test values (Table 5).Table 5
Comparison of theoretical and test values of displacement ductility.
Specimens
Yield displacement (mm)
Ultimate displacement (mm)
Displacement ductility
Theoretical
Test
Theoretical
Test
Theoretical
Test
Theoretical/test
RHC-1
15.23
18.30
133.57
137.70
8.77
7.52
1.16
RHC-2
11.74
13.60
68.52
75.50
5.83
5.60
1.04Table5 shows that theoretical values are a little smaller than test values. There are two reasons. One is that the elongation of bonded segment is ignored when calculating Δy2 and Δu2, which leads to the theoretical values smaller. Another reason is that there exist some errors and virtual displacements in test inevitably, which leads to the test values larger. However, on the whole, both are in good agreement.


## 5.1. Material Properties

All of the materials (concrete, steel, and mortar) employed in the tests are selected based on China code “Code for design of concrete structures (GB 50010-2010, 2010).” The grade of concrete is C35, which denotes that the ultimate compressive strength of the cubic concrete specimens (15 cm × 15 cm × 15 cm) cured in standard conditions is 35 MPa, and its Poisson’s ratio is 0.2. Here, six cubic concrete specimens are tested.The grade of steel in the tests is HRB400, where HRB denotes hot-rolled ribbed-steel bar and the number 400 denotes the yield strength. Its elastic modulus is 200 GPa and Poisson’s ratio is 0.3, and three sets of steels with different diameters (d = 10 mm, 12 mm, and 16 mm) are tested.The high-performance mortar is tested (160 mm × 40 mm × 40 mm) with a type of H-80, which is a characteristic of early strength, high strength, no shrinkage, and high fluidity.The mechanical parameters of all materials are presented in Tables1–3.Table 1
Mechanical property of concrete.
Specimen number
Failure load (kN)
Compressive strength (MPa)
Average compressive strength (MPa)
1
880
39.11
39.86
2
910
40.44
3
822
36.53
4
960
42.67
5
950
42.22
6
860
38.22Table 2
Mechanical property of steel.
Specimen number
Steel diameter
Yield strength (MPa)
Ultimate strength (MPa)
Ratio of tensile strength to yield strength
Elongation (%)
1-1
10
445
625
1.4
24
1-2
455
615
1.35
24
Average value
450
620
1.375
24
2-1
12
445
575
1.29
25
2-2
460
585
1.27
25
Average value
452.5
580
1.28
25
3–1
16
465
600
1.29
25
3-2
460
600
1.30
25
Average value
462.5
600
1.295
25Table 3
Mechanical property of high-performance mortar.
Test item
Time
Measured value
Compressive strength (MPa)
1 d
24.96
28 d
93.52
Rupture strength (MPa)
1 d
5.29
28 d
10.44


## 5.2. Test Setup

The quasistatic cyclic tests of the RHC shear wall are performed. Axial load is applied by tensioning prestressed reinforcement strands, and the lateral load is applied by the 150 t hydraulic actuator. The loading protocol employed consists of a load control procedure first and a displacement control procedure then. Moreover, four steel supports are employed on both sides of the wall symmetrically to keep the lateral stability of the specimen during the testing process. The lateral displacements of the wall are monitored by seven displacement transducers. The testing system is shown in Figure12.Figure 12
Test setup: (a) schematic and (b) photograph.
(a)
(b)


## 5.3. Crack Pattern and Failure Mode

For the RHC-1 specimen, the first crack observed at 210 kN was horizontal flexural cracks when the specimen was pulled, which occurred near the bottom of the wall. The crack on the top interface of rabbet was generated at 240 kN when the specimen was pushed. However, for the RHC-2 specimen, the first crack was observed at 270 kN when the specimen was pushed, which occurred near the bottom of the wall. The crack on the top interface of rabbet was generated at 300 kN when the specimen was pushed.The yield state was defined as when bonded reinforcement segments in the confined end-zone yielded, the specimen yielded. The increase of the applied load resulted in the propagation of cracks and initiation of new flexural cracks along the specimen. Further load increase extended the existing flexural cracks into flexure shear cracks and caused initiation of vertical cracks. In addition, the cracks in the center of the test specimen propagated to the compressed zone, and the crack width and the displacement increased. Finally, when the lateral resistance degenerated to 85% of the peak load, defined as the failure of the specimen, the failure mode was a typical flexural failure for each specimen, which was controlled by crushing concrete at the plastic region near the bottom of the wall. The final crack pattern and failure mode are shown in Figure13.Figure 13
Final crack pattern and failure mode of (a) RHC-1 specimen and (b) RHC-2 specimen.
(a)
(b)From Figure13(b), it was noted that the failure mode was not symmetrical. The reason was that the push force and pull force of hydraulic actuator during test were not symmetrical, but had little effect.


## 5.4. Load-Displacement Relationship

The lateral load versus top displacement curves of the two specimens is shown in Figure14.Figure 14
Hysteresis curves and skeleton curves of RHC-1 and RHC-2. (a) Hysteresis curve of RHC-1. (b) Skeleton curve of RHC-1. (c) Hysteresis curve of RHC-2. (d) Skeleton curve of RHC-2.
(a)
(b)
(c)
(d)It is observed in Figure14 that the number of hysteresis loop is greater and the curve is plumper in the RHC-1 specimen than in the RHC-2 specimen. The bearing capacity of the specimen is decreased by reducing the axial compression ratio, whereas the yield displacement and ultimate displacement are increased at the same time.


## 5.5. Bearing Capacity and Deformation

From the tests, we obtain the crack loadFcr, crack displacement Δcr, yield load Fy, yield displacement Δy, peak load Fm, peak displacement Δm, ultimate load Fu, and ultimate displacement Δu of two specimens, as shown in Table 4.Table 4
Test results of two specimens.
Specimen
F
cr (kN)
Δcr (mm)
F
y (kN)
Δy (mm)
F
m (kN)
Δm (mm)
F
u (kN)
Δu (mm)
RHC-1
210
12.28
330
18.30
610
85.44
518.50
137.70
RHC-2
270
11.74
420
13.60
718
46.49
610.30
75.50Table4 shows that Fcr,Fy,Fm,andFu increase, whereas Δcr,Δy,Δm,andΔu decrease with increase of the axial compression ratio.


## 5.6. Verification

Applying (13) and (15), the yield displacement can be expressed as(17)Δy=φyH23+εylsHhw/2−xy±e.Applying (14) and (16), the ultimate displacement can be expressed as(18)Δu=φyH23+φu−φy2Hlp−lp22+δgtlsHhw/2−xu±e.From (17) and (18), the yield displacement and ultimate displacement can be calculated, which are compared with test values (Table 5).Table 5
Comparison of theoretical and test values of displacement ductility.
Specimens
Yield displacement (mm)
Ultimate displacement (mm)
Displacement ductility
Theoretical
Test
Theoretical
Test
Theoretical
Test
Theoretical/test
RHC-1
15.23
18.30
133.57
137.70
8.77
7.52
1.16
RHC-2
11.74
13.60
68.52
75.50
5.83
5.60
1.04Table5 shows that theoretical values are a little smaller than test values. There are two reasons. One is that the elongation of bonded segment is ignored when calculating Δy2 and Δu2, which leads to the theoretical values smaller. Another reason is that there exist some errors and virtual displacements in test inevitably, which leads to the test values larger. However, on the whole, both are in good agreement.


## 6. Effect of Unbonded Reinforcement on Yield Displacement and Ultimate Displacement

In order to study the effect of unbonded reinforcement on yield displacement and ultimate displacement, set RHC-2 as example, Table6 lists the effect of unbonded reinforcement on yield displacement and ultimate displacement.Table 6
Effect of unbonded reinforcement on yield displacement and ultimate displacement.
Yield displacement (mm)
Ultimate displacement (mm)
Δy1
Δy1/Δy
Δy2
Δy2/Δy
Δu1
Δu1/Δu
Δu2
Δu2/Δu
9.22
78.53%
2.52
21.47%
21.06
30.74%
47.46
69.26%Table6 shows that the effect of unbonded reinforcement on ultimate displacement is more pronounced than that of yield displacement. The reason is that in the yield state, deformation of unbonded reinforcement is small which leads to small effect; however, in the ultimate state, unbonded reinforcement yields and its deformation are large which leads to large effect.


## 7. Conclusions and Suggestions

### 7.1. Conclusions

In this paper, the curvature and displacement of RHC shear wall are analyzed, and the following conclusions can be drawn:(1)
Considering the effect of confined end-zone and unbonded reinforcement on curvature, yield curvature and ultimate curvature are calculated.(2)
In calculating displacement, assume that the displacement consists of two parts approximately: one part is that the wall generates resisting external force when all reinforcements are bonded, and assume the wall is not a rigid body and another part is that elongation of unbonded reinforcements leads the wall to generate when the wall is RHC shear wall, and assume the wall is a rigid body. Overlay the two parts linearly, and obtain the calculation equation of yield displacement and ultimate displacement. The theoretical and test values are compared. Also, the result shows that theoretical values are slightly smaller than test values, but both are in good agreement on the whole.(3)
According to the theoretical equation, effects of unbonded reinforcement on yield displacement and ultimate displacement are studied. The result shows that the effect of unbonded reinforcement on ultimate displacement is more pronounced than that of yield displacement.


### 7.2. Suggestions

Only two specimens were applied to verify the displacement ductility in the paper, which was inadequate to convince. Numerous tests considering different parameters should be performed to verify the displacement ductility.


## 7.1. Conclusions

In this paper, the curvature and displacement of RHC shear wall are analyzed, and the following conclusions can be drawn:(1)
Considering the effect of confined end-zone and unbonded reinforcement on curvature, yield curvature and ultimate curvature are calculated.(2)
In calculating displacement, assume that the displacement consists of two parts approximately: one part is that the wall generates resisting external force when all reinforcements are bonded, and assume the wall is not a rigid body and another part is that elongation of unbonded reinforcements leads the wall to generate when the wall is RHC shear wall, and assume the wall is a rigid body. Overlay the two parts linearly, and obtain the calculation equation of yield displacement and ultimate displacement. The theoretical and test values are compared. Also, the result shows that theoretical values are slightly smaller than test values, but both are in good agreement on the whole.(3)
According to the theoretical equation, effects of unbonded reinforcement on yield displacement and ultimate displacement are studied. The result shows that the effect of unbonded reinforcement on ultimate displacement is more pronounced than that of yield displacement.


## 7.2. Suggestions

Only two specimens were applied to verify the displacement ductility in the paper, which was inadequate to convince. Numerous tests considering different parameters should be performed to verify the displacement ductility.



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*Source: 5197125-2018-04-03.xml*