# Calculation Model of Shear Capacity of Multiple Composite Core Column Joints Based on Softened Tension-Compression Bar Model

Authors: Xu Wentao; Yang Chengyu
Journal: Advances in Civil Engineering
 (2021)

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## Abstract

This paper aims to study the seismic performance of multiple composite core column joints. The influence of the stress mechanism, axial compression ratio, and shear span ratio on the failure mode, hysteretic performance, and shear capacity of the multiple composite core column joints was studied through the low-reversed cyclic loading tests of three specially designed and manufactured multiple composite core column joints. The angle ratio method is used to calculate the effective area of the vertical tie bar, and based on the mechanism of the softening tension-compression bar, the formula for calculating the shear capacity of the joint with multiple composite core column is established. In addition, it is also verified by the test data in this paper. The experimental results show that when the axial compression ratio increases from 0.26 to 0.45, the number and width of cracks at the beam end decrease. When the shear span ratio increases from 1.67 to 2.22, the number and width of cracks at the joint beam end increase. The average value and standard deviation of the ratio between the measured value and the calculated value of the shear capacity are 0.97 and 0.16, indicating that the proposed calculation method has a high agreement with the actual value and strong engineering application.


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## Body

## 1. Introduction

Beam-column joints play an important role in the transmission and distribution of internal forces and are also the key to the seismic performance of the structure in the frame structure. Most of the ordinary reinforced concrete beam-column joints show brittle failure, which is not conducive to the seismic resistance of the structure [1–3]. To explore a new type of reinforcement and improve the transmission and distribution of internal forces in the frame, much attention has been paid by researchers. In recent years, multiple composite reinforced core columns can effectively guarantee the realization of “strong column and weak beam” and have good shear bearing capacity and seismic performance under earthquake action, which has attracted increasing attention of researchers. As the core index of seismic shear capacity is particularly important, many researchers have done a series of studies on it [4–12].You et al. [13] comprehensively analyzed the seismic tests of numerous international frame joints, found the important correlation between the stress state of longitudinal bars after yield and the number of longitudinal bars that affect the shear capacity of joints, and proposed the shear capacity formula of such failure-type joints. Xing et al. [14–16] proposed a modified calculation model for reinforced concrete joints with orthogonal beams on the basis of the model of softened tension-compression bars [17, 18]. Yang [19] studied the influence of axial compression ratio, stirrup spacing, stirrup ratio, and stirrup form on the seismic performance of multiple composite core columns and proposed the corresponding calculation formula of shear capacity and design suggestions. Wakabayshi et al. [20] and Omine et al. [21] found that the seismic performance of cored columns was significantly better than that of ordinary reinforced concrete columns in their studies under low-cyclic reciprocating loads. Kadoya et al. [22] found that the initial stiffness and strength of columns with core columns are about 1.2 times that of ordinary reinforced concrete columns. At present, most of the joint researches at home and abroad are made of reinforced concrete beam-column joints in the form of ordinary reinforcement. Although the research on the seismic performance of composite core columns has made some progress, there are few reports on the seismic performance of composite core columns, especially the calculation of its shear capacity needs to be further studied.Based on the above analysis, this paper carried out pseudostatic tests on the beam-column joints composed of multiple composite core columns, analyzed the effects of axial compression ratio and shear span ratio on the seismic performance of the joints, and proposed a formula for calculating the shear capacity of the multiple composite core column joints based on the mechanism of softening the tension and compression rods. It provides reference for the popularization and use of such nodes in engineering construction.


## 2. Materials and Methods

To avoid the contingency caused by single data, three reinforced concrete beam joints with multiple composite core columns were prepared for parallel tests. The beam section size of each node is 150 mm × 380 mm, the column section size is 400 mm × 400 mm, and the concrete protective layer thickness of the beam and column is 20 mm. The specimens BCJ-1, BCJ-2, and BCJ-3 were designed and manufactured with axial compression ratio and shear span ratio as variables. Specific joint dimensions and reinforcement can be found in Figures1 and 2 and Table 1.Figure 1
Detailed drawing of reinforcement for BCJ-1 (BCJ-2) joints.Figure 2
Detailed drawing of reinforcement for BCJ-3 joints.Table 1
The detailed parameters of the frame node in test.
ProjectBCJ-1BCJ-2BCJ-3MandrelRectangular frame columnSection (mm × mm)400 × 400400 × 400400 × 400Angle steelUnilateral longitudinal barsStirrupOuter mandrel XZ-1Section (mm × mm)Longitudinal barsStirrupInner mandrel XZ-2Section (mm × mm)140 × 140 (circle)140 × 140 (circle)140 × 140 (circle)Longitudinal barsStirrupStrength of concrete (MPa)C45C45C45Axial pressure (kN)156027501560Lower column height2.42.43.2Height of reverse bending point of lower column1.21.21.6Height of upper column1.91.91.9Height of upper column reverse bending point1.01.01.0Experimental axial compression ratio0.260.450.26Shear span ratio1.671.672.22BeamSection (mm × mm)150 × 380150 × 380150 × 380Concrete strength (MPa)C45C45C45StirrupBeam length (m)3.753.753.75Length of reverse bending point (m)1.871.871.87Midspan compression barsNegative bending regionsLower longitudinal barsLongitudinal barsThe measured compressive strength of concrete cube used in the test is 58.6 MPa, and the measured material properties of steel bars with different diameters are shown in Table2.Table 2
Property of steel bars.
Diameter (mm)Yield strengthfy (N/mm2)Ultimate strengthfu (N/mm2)Elastic modulusEs (N/mm2)6434.60605.202.0 × 1058539.70706.702.0 × 10510490.24659.452.0 × 10512484.20624.902.0 × 10514455.10639.452.0 × 105The static loading method is used as shown in Figure3 in the experiment of this paper. The loading process is mainly divided into the following two steps:(1)
Vertical load was applied to the top of the node column by a hydraulic Jack.The first application of axial pressure is 0.4 times to ensure that the device is not loosened, and then two further cycles of 0.4 times. Finally, the axial pressure applied at the top of the column is 1 time and remains constant.(2)
Horizontal displacement load is applied to the joint column end through the actuator.Before the formal loading, 5 kN force was used to push and pull the joint specimen three times for trial loading to ensure the normal working of each device, acquisition system, and fixed device. After the complete system is tested to work normally, the horizontal load is returned to 0. The step size of initial loading of horizontal displacement is 2 mm, 4 mm, 6 mm, 8 mm, and 10 mm in turn, and then the step size remains 10 mm. Each step size is repeated three times until the specimen is damaged or the horizontal load is less than 0.85 times the peak load, and the loading is finished. The vibration frequency of the load is the simulated seismic load, as shown in Figure4.Figure 3
Schematic diagram of loading device. Note: ① reaction wall, ② gantry mounting, ③ MTS actuator, ④ horizontal rolling system, and ⑤ test specimen.Figure 4
The vibration frequency of the load.


## 3. Results and Discussion

### 3.1. Phenomena and Failure Modes

For multiple composite core column joints, the frame columns are always in the elastic stage, and the core areas of the columns and joints have no obvious damage under the reciprocating action of horizontal displacement loads. With the loading process, penetration cracks appear at the beam-column junction, and cross oblique cracks appear at the beam end away from the edge of the column. Concrete in this area gradually spall off with the subsequent loading, and obvious plastic failure occurs at the edge of the beam away from the column. The failure modes of each specimen are shown in Figures5 to 7.Figure 5
BCJ-1 node failure pattern diagram.Figure 6
BCJ-2 node failure pattern diagram.Figure 7
BCJ-3 node failure pattern diagram.It can be found that the number and width of cracks at the beam end decrease as the axial compression ratio increases from 0.26 (node BCJ-1) to 0.45 (node BCJ-2) when the specimen is damaged. However, the horizontal displacement of failure is just the opposite. The horizontal displacement of BCJ-1 node is 70 mm, while that of BCJ-2 node is 50 mm. This is because the increase of axial compression ratio will cause the concrete at the core of the joint to be subjected to greater compression, so as to delay the generation of cracks at the beam and column ends at the core of the joint. It is shown that increasing the axial compression ratio can delay the development of the beam end cracks but reduce the ductility of the joint under seismic load. Similarly, when the shear-span ratio increases from 1.67 (node BCJ-1) to 2.22 (node BCJ-3), the number and width of fractures at the joints increase. This is because when the shear span ratio increases, the second-order effect of joints becomes more obvious, which intensifies the generation and development of beam-end cracks. For details, see the red mark in the revised manuscript. Increasing the shear-span ratio will accelerate the generation of fractures at the joints.


### 3.2. Hysteretic Behavior

The hysteretic curves of each specimen are shown in Figures8 to 10. In the early stage of loading, the hysteretic loop area and residual deformation of each joint are small, and the stiffness degradation of the specimen is not obvious. The area and residual deformation of the hysteresis loop gradually increase with the increase of the load. When the horizontal displacement load reaches its peak, the stiffness of the joint degrades obviously. With the continuous loading, the hysteresis loop area of the curve increases, the maximum load at all levels begins to decrease, and the rising part of the curve tends to be gentle. The hysteresis curves of the three specimen nodes all show a spindle shape without obvious pinching, and the load strength and specimen stiffness do not significantly degrade when the load of the same level is loaded each time. The cumulative damage of the frame joints under cyclic loads at all levels is small, and the residual deformation of the joints under cyclic loads at all levels is uniform after the peak load. However, the load-displacement hysteresis curves of the beam-column joints with common reinforcement form mostly present bow and inverse “S” shape [23–25], which proves that the multicomposite core column joints have better seismic performance than the beam-column joints with common reinforcement form.Figure 8
BCJ-1 load-displacement hysteresis curve.Figure 9
BCJ-2 load-displacement hysteresis curve.Figure 10
BCJ-3 load-displacement hysteresis curve.Skeleton curve of the node refers to the line between the first loading curve of the node and the peak points of hysteretic loops in the hysteretic loop curve under the action of low-cycle reciprocating load. The skeleton curve drawn by the line of the peak points of each hysteretic loop under low-cycle reciprocating load is shown in Figure11. The initial stiffness of the frame joints increases by about 21.6 kN/mm, and the peak load increases by about 103.5 kN, when the axial compression ratio increases from 0.26 to 0.45. As for the effect of the shear to span ratio, the initial stiffness of the node decreases by about 1.3 kN/mm and the peak value by about 22.9 kN, when the shear to span ratio increases from 1.67 to 2.22. The results show that the increase of axial compression ratio can significantly increase the initial stiffness and peak load of the composite core column joints, while the shear span ratio is just the opposite.Figure 11
Comparison diagram of skeleton curve.


### 3.3. Calculation of Horizontal Shear Stress on Nodes

The force analysis of nodes is shown in Figure12.Figure 12
Schematic diagram of force of nodes: (a) diagram of external forces on node and (b) analysis of the internal forces on node.
(a)(b)The upper half of the node is taken as the isolation body, and the horizontal shear force of the node has the following equilibrium relationship:(1)Vjh=Tb1+Cb2−Vc1.According to the geometric relationship, the vertical shear force of the node is(2)Vjv≈hb′hc′×Vjh,(3)Tb1=Mb2hb′,(4)Cb2=Mb1hb′,where Vjh is the horizontal shear force of the node, Tb1 is the tensile force of the reinforcement of the beam on the right side of the node, and Cb2 is the pressure of the beam on the left side.The orientation ofTb1 and Cb2 is assumed to be consistent although not necessarily in the actual force, for simplicity of calculation. Vc1 is the horizontal shear force at the column end at the node, which is equal to the load value of horizontal loading. Mb1 and Mb2 are the bending moments at the end of the beam; hb′ and hc′ are the internal moment arms of the beam and column, respectively, the approximate distance between the center lines of upper and lower longitudinal bars of the beam is taken to calculation the hb′.According to equations (1), (3), and (4), we can get(5)Vjh=Mb1+Mb2hb′−Vc1.Balanced by the bending moment of the node, the following equation can be obtained:(6)Mc1+Mc2=Mb1+Mb2.Among them,(7)Mc1=Vc1h1,Mc2=Vc2h2,(8)Vc1=Vc2.Equations (6) to (8) can be deduced as follows:(9)Vc1=Mb1+Mb2h1+h2=Mb1+Mb2h−hb,where Mc1 and Mc2 are bending moments at the upper and lower column ends; h is the distance between the reverse bending points of the upper and lower columns; h1 and h2 are the distance between the reverse bending points of the upper and lower columns and the core area of the joint, respectively; and hb is the height of the cross section of the beam.By combining equations (9) and (5), we can get(10)Vjh=Mb1+Mb2hb′1−hb′h−h0.The shear forces on the three nodes in the horizontal direction are finally calculated, as shown in Table3.Table 3
The shear on the three nodes in the horizontal direction.
SpecimenBCJ-1BCJ-2BCJ-3VjhKN668.711161.32708.58


## 3.1. Phenomena and Failure Modes

For multiple composite core column joints, the frame columns are always in the elastic stage, and the core areas of the columns and joints have no obvious damage under the reciprocating action of horizontal displacement loads. With the loading process, penetration cracks appear at the beam-column junction, and cross oblique cracks appear at the beam end away from the edge of the column. Concrete in this area gradually spall off with the subsequent loading, and obvious plastic failure occurs at the edge of the beam away from the column. The failure modes of each specimen are shown in Figures5 to 7.Figure 5
BCJ-1 node failure pattern diagram.Figure 6
BCJ-2 node failure pattern diagram.Figure 7
BCJ-3 node failure pattern diagram.It can be found that the number and width of cracks at the beam end decrease as the axial compression ratio increases from 0.26 (node BCJ-1) to 0.45 (node BCJ-2) when the specimen is damaged. However, the horizontal displacement of failure is just the opposite. The horizontal displacement of BCJ-1 node is 70 mm, while that of BCJ-2 node is 50 mm. This is because the increase of axial compression ratio will cause the concrete at the core of the joint to be subjected to greater compression, so as to delay the generation of cracks at the beam and column ends at the core of the joint. It is shown that increasing the axial compression ratio can delay the development of the beam end cracks but reduce the ductility of the joint under seismic load. Similarly, when the shear-span ratio increases from 1.67 (node BCJ-1) to 2.22 (node BCJ-3), the number and width of fractures at the joints increase. This is because when the shear span ratio increases, the second-order effect of joints becomes more obvious, which intensifies the generation and development of beam-end cracks. For details, see the red mark in the revised manuscript. Increasing the shear-span ratio will accelerate the generation of fractures at the joints.


## 3.2. Hysteretic Behavior

The hysteretic curves of each specimen are shown in Figures8 to 10. In the early stage of loading, the hysteretic loop area and residual deformation of each joint are small, and the stiffness degradation of the specimen is not obvious. The area and residual deformation of the hysteresis loop gradually increase with the increase of the load. When the horizontal displacement load reaches its peak, the stiffness of the joint degrades obviously. With the continuous loading, the hysteresis loop area of the curve increases, the maximum load at all levels begins to decrease, and the rising part of the curve tends to be gentle. The hysteresis curves of the three specimen nodes all show a spindle shape without obvious pinching, and the load strength and specimen stiffness do not significantly degrade when the load of the same level is loaded each time. The cumulative damage of the frame joints under cyclic loads at all levels is small, and the residual deformation of the joints under cyclic loads at all levels is uniform after the peak load. However, the load-displacement hysteresis curves of the beam-column joints with common reinforcement form mostly present bow and inverse “S” shape [23–25], which proves that the multicomposite core column joints have better seismic performance than the beam-column joints with common reinforcement form.Figure 8
BCJ-1 load-displacement hysteresis curve.Figure 9
BCJ-2 load-displacement hysteresis curve.Figure 10
BCJ-3 load-displacement hysteresis curve.Skeleton curve of the node refers to the line between the first loading curve of the node and the peak points of hysteretic loops in the hysteretic loop curve under the action of low-cycle reciprocating load. The skeleton curve drawn by the line of the peak points of each hysteretic loop under low-cycle reciprocating load is shown in Figure11. The initial stiffness of the frame joints increases by about 21.6 kN/mm, and the peak load increases by about 103.5 kN, when the axial compression ratio increases from 0.26 to 0.45. As for the effect of the shear to span ratio, the initial stiffness of the node decreases by about 1.3 kN/mm and the peak value by about 22.9 kN, when the shear to span ratio increases from 1.67 to 2.22. The results show that the increase of axial compression ratio can significantly increase the initial stiffness and peak load of the composite core column joints, while the shear span ratio is just the opposite.Figure 11
Comparison diagram of skeleton curve.


## 3.3. Calculation of Horizontal Shear Stress on Nodes

The force analysis of nodes is shown in Figure12.Figure 12
Schematic diagram of force of nodes: (a) diagram of external forces on node and (b) analysis of the internal forces on node.
(a)(b)The upper half of the node is taken as the isolation body, and the horizontal shear force of the node has the following equilibrium relationship:(1)Vjh=Tb1+Cb2−Vc1.According to the geometric relationship, the vertical shear force of the node is(2)Vjv≈hb′hc′×Vjh,(3)Tb1=Mb2hb′,(4)Cb2=Mb1hb′,where Vjh is the horizontal shear force of the node, Tb1 is the tensile force of the reinforcement of the beam on the right side of the node, and Cb2 is the pressure of the beam on the left side.The orientation ofTb1 and Cb2 is assumed to be consistent although not necessarily in the actual force, for simplicity of calculation. Vc1 is the horizontal shear force at the column end at the node, which is equal to the load value of horizontal loading. Mb1 and Mb2 are the bending moments at the end of the beam; hb′ and hc′ are the internal moment arms of the beam and column, respectively, the approximate distance between the center lines of upper and lower longitudinal bars of the beam is taken to calculation the hb′.According to equations (1), (3), and (4), we can get(5)Vjh=Mb1+Mb2hb′−Vc1.Balanced by the bending moment of the node, the following equation can be obtained:(6)Mc1+Mc2=Mb1+Mb2.Among them,(7)Mc1=Vc1h1,Mc2=Vc2h2,(8)Vc1=Vc2.Equations (6) to (8) can be deduced as follows:(9)Vc1=Mb1+Mb2h1+h2=Mb1+Mb2h−hb,where Mc1 and Mc2 are bending moments at the upper and lower column ends; h is the distance between the reverse bending points of the upper and lower columns; h1 and h2 are the distance between the reverse bending points of the upper and lower columns and the core area of the joint, respectively; and hb is the height of the cross section of the beam.By combining equations (9) and (5), we can get(10)Vjh=Mb1+Mb2hb′1−hb′h−h0.The shear forces on the three nodes in the horizontal direction are finally calculated, as shown in Table3.Table 3
The shear on the three nodes in the horizontal direction.
SpecimenBCJ-1BCJ-2BCJ-3VjhKN668.711161.32708.58


## 4. Calculation Formula of Shear Capacity Based on Softening Tension-Compression Bar Model

The complex mechanical behavior of the whole structure is abstracted into a relatively simple strut-and-tie model by taking the compression concrete as the compression struts; and the steel bar in the tension zone as the tie bar. The simplified softened strut-and-tie model is a model proposed by Professor Huang Shijian to calculate the shear capacity of beam-and-column joints on the basis of the softened strut-and-tie model [26, 27]. The schematic diagram of the softened strut-and-tie model is shown in Figure 13. The model is composed of oblique, horizontal, and vertical parts. The oblique part is composed of a baroclinic bar, and the included angle between the baroclinic bar and the horizontal axis is θ; the horizontal part is composed of a horizontal tie bar and two gentle compression struts, which are composed of stirrups at the nodes; the vertical part consists of a vertical tie bar and two steep compression struts, which are composed of the longitudinal bars of the columns [15]. The specific calculation method and calculation process are as follows:Figure 13
The strut-and-tie model of the node: (a) oblique part, (b) horizontal part, and (c) vertical part.
(a)(b)(c)The horizontal shear bearing capacity of this model,Vj is calculated as follows:(11)Vj=Kζfc′Astrcosθ,where θ is the included angle between the baroclinic bar and the horizontal direction, which can be determined by the following equation:(12)θ=tan−1hb″hc″.For the convenience of calculation,b and c in the equation are assumed to be the distances between the center lines of the outermost reinforcement of the beam and column, respectively, and the inclination direction and magnitude of the baroclinic bar in the joint area and the main compressive stress of the concrete in the core area are considered to be the same.The areaAstr of the baroclinic bar can be calculated as follows:(13)Astr=as×bs,where as and bs are the effective height and width of the baroclinic bar, respectively, determined by the following equation:(14)as=ab2+ac2,(15)bs=minbb+2×ac6,bc,where ab is the height of the compression zone of the beam cross section and ac is the height of the compression zone of the column cross section, which can be determined by the following equation:(16)ab=hb5,(17)ac=0.25+0.85NAgfc′hc,where bb is the beam cross-section width, bc is the column cross-section width, hb is the beam cross-section height, hc is the column cross-section height, N is the axial pressure, determined according to the test, Ag is the column cross-section area, and fc′ is the compressive strength of the concrete cylinder. According to the conversion formula of “Standard for test methods of concrete physical and mechanical properties” (GB/T50081-2019) [28], the compressive strength of the concrete cube is converted into the compressive strength of the cylinder. K is the tension/compression bar coefficient. Professor Huang Shijian [26] suggested the following equation for calculation:(18)K=Kh+Kv−1,where Kh is the horizontal tie bar coefficient and Kv is the vertical tie bar coefficient. The values of the two are shown in the following equation:(19)Kh=1+Kh¯−1AthfyhFh¯≤Kh¯,Kv=1+Kv¯−1AtvfyvFv¯≤Kv¯,where Athfyh is the product of the area of the horizontal tie bar and the yield strength of the corresponding steel bar and Atvfyv is the product of the area of the vertical tie bar and the yield strength of the corresponding reinforcing bar. If the tie bar consists of different types of reinforcement, then Athfyh and Atvfyv shall be the sum of the products of different types.Kh¯ is the balance coefficient of the horizontal tie bar and Kv¯ is the balance coefficient of the vertical tie bar, which can be calculated as follows:(20)Kh¯≈11−0.2γh+γh2,Kv¯≈11−0.2γv+γv2,where γh is the ratio of the horizontal tie bar to the horizontal shear stress of the node and γv is the ratio of the vertical tie bar to the horizontal shear stress of the node, which can be calculated as follows:(21)γh=2tanθ−13,γv=2cotθ−13,Fh¯ is the balance tension of the horizontal tie bar, and Fv¯ is the balance tension of the vertical tie bar. They can be calculated by the following equation:(22)Fh¯=γh×Kh¯ζfc′Astr×cosθ,Fv¯=γv×Kv¯ζfc′Astr×sinθ,ζ is the softening coefficient of concrete under compression and can be calculated as follows:(23)ζ≈3.35f′.The main calculation steps of the modified strut-and-tie model are as follows :(1)
Determine relevant parameters of nodes:bb, bc, hb, hc, hb″, hc″, Ath, Atv, and fc′.Considering that all the stirrups at the joints do not yield when the joints fail, the joint area is simply divided by force. All the area of the stirrups at the middle half participates in the force, while only half of the area of the reinforcing bars at both sides participates in the force [14]. See Figure 14 for specific zones.The vertical tie bar is composed of the longitudinal bar of the column, and the reinforcement partition of the longitudinal bar participating in the force is shown in Figure15. It is composed of longitudinal bars of core column and longitudinal bars of rectangular column. According to the stress zone, the calculation method of Atv is as follows:(24)Atv=Atvj+Atvx,cosα=h″/4R,Atvx=m×π−2απ×Atvx1+n×Atvx2,whereAtv is the effective area of the vertical tie bar, Atvj is the effective area of the longitudinal bars of the rectangular column, Atvx is the effective area of the longitudinal reinforcement of the core column, Atvx1 is the effective surface of the longitudinal reinforcement of the outer core column, and Atvx2 is the effective area of the longitudinal reinforcement of the inner core column.(2)
Calculate the heightab of the compression zone of the joint beam cross section (equation (16)).(3)
Calculate the height of the compression zone of the joint column cross sectionac (equation (17)).(4)
Calculate the effective height of the baroclinic bar in the joint area (equation (14)).(5)
Calculate the effective width of the baroclinic bar in the joint area is calculated,bs (equation (15)).(6)
Calculate the softening coefficientζ of concrete (equation (23)).(7)
Calculateγh and γv (equation (21)).(8)
CalculateKh¯ and Kv¯ (equation (20)).(9)
CalculateFh¯ and Fv¯ (equation (22)).(10)
CalculateKh and Kv (equation (19)).(11)
CalculateK (equation (18)).(12)
CalculateVj (equation (11)).Figure 14
The area partition diagram of the node.Figure 15
Diagram of columnAtv calculation method.The comparison between the calculation results and the test results of the shear capacity of the three multiple composite core column joints (calculated according to the above steps) is shown in Table4. The average value and standard deviation of the ratio between the calculated value and the measured value are 0.965 and 0.16 respectively. It can be seen that the calculated value of the shear capacity based on the softened tension-compression bar model is in good agreement with the experimental value. The results indicate that it is feasible to use the angle ratio to calculate the effective area of the longitudinal reinforcement of the core column.Table 4
The shear force calculation results of each joint are compared with the test results.
Node numberVjVjhVj/VjhAverageStandard deviationBCJ-1707.25668.711.0580.9650.16BCJ-2975.611161.320.840BCJ-3707.25708.580.998


## 5. Conclusion

(1)
According to the experimental study, the three specimens of multiple composite core column joints were all damaged by plastic hinge at the beam end, conforming to the seismic design of “strong column and weak beam” and had better seismic performance compared with ordinary beam-column joints.(2)
The hysteresis curves of the joint specimens all show full spindle shape, and the hysteresis loops are relatively full. Increasing the axial compression ratio can significantly improve the initial stiffness and peak load of the joint, while increasing the shear span ratio will lead to a slight decrease in the initial stiffness and peak load.(3)
When the modified softened tension-compression bar model is used to calculate the shear at the joints, a calculation method for the effective area of the vertical tension-compression bar suitable for multiple composite core columns is proposed. Compared with the test results, the model value is slightly lower than the test value, which indicates that the calculation results of this method are safer when designing the shear capacity of frame joints and can provide a theoretical basis for the structural design of practical engineering.



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