Shear Force of Interior Beam–Column Joints under Symmetrical Loading with Two Transverse Forces on the Beam ^†

Abstract

The beam-to-column connection is a particularly vulnerable element in frame structures under seismic action and is often responsible for building damages. Experimental investigations carried out over the past six decades on shear strength in frame joints have not led to the establishment of a uniform procedure in the design codes of different countries. The reason lies probably in the varied nature of the investigated parameters and in the varied configurations of beam–column connections. A good knowledge of the forces passing through the frame joints in the beam–beam and column–column direction would allow both their adequate computation in new buildings and the verification of existing ones without requiring experimental studies. In the design codes of the leading countries in seismic engineering, the shear force is determined by the capacitive method, considering only the area of the longitudinal reinforcement of the beam passing through the column. This method shows us how much shear force the beam reinforcement can take, but not what the magnitude of the resulting forces actually is as a result of the acting loads. In addition, the method of the codes does not indicate the contribution of the concrete to the total magnitude of the shear force in the beam–column connection. In the proposed mathematical model for calculating the forces that leave the beam, the full dimensions of the cross-section of the beam were taken into account. The material properties and cross-sectional shape were also taken into account. A determining factor for the magnitude of forces entering the beam–column joint is the acting load on the beam. In this paper, the load of two transverse forces was considered. The forces are applied in different possible positions, while remaining symmetrically located on the beam. The calculations are based on Menabrea’s theorem to determine the hyperstatic unknowns. The results of the proposed method for the considered beam show that the magnitude of the shear force differs from that accepted in the literature and the norms by 2% to 27%, depending on the stage of development of the crack. In comparison, the Eurocode-recommended method shows differences in the order of 27% to 40% for the adopted beam under static loads.

1. Introduction

The beam–column connection is a fundamental element in frame structures. Damage in the beam–column joints leads to failure of part of buildings or even to their destruction. Over the past few decades, many frame structures have experienced sudden failure due to joint shear during cyclic loading, such as in earthquakes.

The first quantitative definition of shear strength was given by Hanson and Connor in [1]. In their report on the test results of RC interior beam–column connections, the researchers defined joint shear as the horizontal force transferred to the mid-horizontal plane in a beam–column connection. They suggest that joint shear failure can be prevented by limiting the joint shear stress to the level at which joint shear failure occurs.

Design codes are created to provide a limiting value of the joint shear stress. A number of parameters have been introduced that affect joint shear strength. Researchers rely on different combinations of parameters [2,3,4,5,6,7,8,9,10,11,12,13]. This does not make it possible to create a uniform procedure for designing shear force in the beam–column joint.

An experimental and analytical investigation of RC beam–column joints of existing substandard RC structures subjected to seismic loading was conducted in [14]. “Four sub-assemblages were designed with poor details to simulate the equivalent of structural members found in existing substandard RC structures”. The study examined anchorage length and lap splicing and their effect on the hysteresis curve of the joint.

Parametric studies on the influence of different parameters affecting joint shear strength were carried out in [15]. The results of numerous studies on five parameters with respect to different countries were summarized. One of the main conclusions that emerged was that “concrete cylinder strength increases the joint shear strength”.

“A number of RC/ECC composite beam–column joints have been tested under reversed cyclic loading to study the effect of substitution of concrete with ECC in the joint zone on the seismic behaviors of composite members”. This research was conducted in [16]. “The substitution of concrete with ECC in the joint zone was experimentally proved to be an effective method to increase the seismic resistance of beam–column joint specimens”.

In [17], Shiohara proposed a new model for calculating and detailing the beam–column connection, based on the capacitive design. The study shows an irrationality in the joint shear model adopted in the most current design codes of reinforced concrete beam–column joints. The conclusions are based on the data of tests from twenty reinforced concrete interior beam-to-column-joint that failed due to joint shear. The analysis indicated that joint shear stress increased in most specimens, even after apparent joint shear failure began. The proposed new capacitive model takes into account the participation of both the concrete and the column stirrups in the joint area.

In [18], the authors noted that various structural design codes and researchers have focused on predicting the nominal shear capacity of external reinforced concrete assemblies. These approaches do not capture the essence of the complex shear response of reinforced concrete assemblies. This is because most of the available models tend to ignore key influencing parameters such as beam and column geometry, concrete and steel material properties, longitudinal and transverse reinforcement and column axial loads.

In [19], seismic tests on reinforced concrete beam–column joint sub-assemblages subjected to lateral and long-term vertical loads were reported. The influence of an additional transverse force on the beam applied near the support of the cantilever beam was observed. It was reported that such a force did not significantly affect to the frame joint. Observations showed that such a force did not significantly affect the beam–column connection.

A procedure is needed to determine the magnitudes of the forces along the height of the lateral edge of the beam. These forces will correspond to the reinforcing bars and the concrete cross-section, accounting for the entire geometry of the beam, as well as the location of the steel bars relative to the axis of the beam. The material characteristics of the construction materials composing it, steel and concrete, must also be taken into account. In this way, it will be possible to track how the variation in a selected parameter affects the magnitudes of the forces leaving the beam.

A mathematical model of the beam is created to correspond to a real beam in a frame structure. The introduction of the new support reactions makes the beam statically indeterminate. Menabrea’s theorem is used to determine the hyperstatic unknowns.

In the prescribed procedure in the norms of different countries, the forces are determined capacitively. They are the product of the area of the longitudinal reinforcement of the beam passing through the beam–column connection and the yield strength of the reinforcement. These are the forces that can be absorbed by the available reinforcement. However, this does not address the question of how big the forces actually are, which depends on a given load, as well as the geometric and material characteristics of the beam. All these parameters can vary. This will allow for the contribution of each variable to the total shear force of the joint to be accounted for.

The formulas are derived under the limit state conditions. This makes it possible to determine the forces from the beam both before the opening of a crack between the beam and the column and in the process of its growth. These formulas can also be applied to steel structures to determine the forces in bolted connections.

This paper is part of a comprehensive study to determine the forces acting on the beam that contribute to shear force [20,21]. Here, we consider a beam of a frame construction loaded with two transverse forces, applied at different possible positions on the beam and remaining symmetrical with respect to the supports. Their position can be set near the supports, in accordance with [19].

2. Materials

In [1], Hanson and Connor defined the joint shear $V_j$ in an interior beam–column connection, shown in Figure 1, as the force that is determined by the given Equation (1). The joint shear $V_j$ from Equation (1) is an internal force acting on the free body along the horizontal plane at the mid-height of the beam–column connection.

$$V_j=T+{C^{\prime }}_S+{C^{\prime }}_C-V_C=T+T^{\prime }-V_C$$

(1)

where $C_S$ and ${C^{\prime }}_S$ are the compressive forces in the bottom and top longitudinal reinforcing bars in the beam passing through the connection;

$C_C$ and ${C^{\prime }}_C$ are the compressive forces in concrete on the bottom and top edge of the beam;

$T$ and $T^{\prime }$ are the tensile forces in the top and bottom reinforcing bars in the beam passing through the connection;

$V_C$ is the column shear force.

The difficulty encountered in determining the forces from Equation (1) leads to the adoption of another way of writing the expression for shear force in the literature. Usually, $T$ and $T^{\prime }$ are defined by Equation (2).

$$T={\displaystyle \frac{M_b}{j_b}} \mathrm{and} T^{\prime }={\displaystyle \frac{{M^{\prime }}_b}{j_b}}$$

(2)

where $M_b$ and ${M^{\prime }}_b$ represent the moments at the column face;

$j_b$ and ${j^{\prime }}_b$ represent the lengths of the bending moment arms at the column face. They are assumed to be constant and unchanging in the process of deformation.

Then, Equation (1) is rewritten to represent the shear force with the moment in the beam section at the column faces as Equation (3).

$$V_j={\displaystyle \frac{M_b}{j_b}}+{\displaystyle \frac{{M^{\prime }}_b}{{j^{\prime }}_b}}-V_C$$

(3)

The assumption in Equation (2) obliges us to assume equal forces in the bottom and top reinforcement of the beam at the face of the column. In the author’s previous publications, these values were shown to differ substantially [22,23].

In this article, the following tasks are set: 1. to determine the expressions for the forces from Figure 1, at the column face, when the beam is loaded with two symmetrical transverse forces; 2. to compare the obtained results with the results from Equations (2) and (3).

3. Methods

3.1. Mathematical Model of Beam

A beam from a frame structure is considered, as depicted in Figure 2. Linear spring supports act along the vertical edge of the beam, taking into account the connection between the concrete of the beam and that of the column. The forces in all springs are reduced to one force, $H_1 \left[\mathrm{kN}\right]$. In the case of large deformations, part of the vertical edge is destroyed. The unbroken edge has length of $2b \left[\mathrm{cm}\right]$. The reaction, $H_1\hspace{0.17em}\left[\mathrm{kN}\right]$, which is symmetrically located with respect to the intact lateral edge, moves along the height of the beam as the crack length increases. For convenience, it was transferred $H_1\hspace{0.17em}\left[\mathrm{kN}\right]$ to the support along the bottom edge (support one), after applying Poinsot’s theorem concerning the transfer of forces in parallel to their directrix. This necessitated the introduction of compensating moments $H_{1}b\hspace{0.17em}\left[\mathrm{kN}.\mathrm{cm}\right]$. The coefficient of the linear spring is $k_1$. It is set as the reduced tensile/compressive stiffness of the concrete section by the multiplier ${\zeta }_1$.

$$k_1={\zeta }_1{\displaystyle \frac{E_1{A}_1}{L}}.$$

(4)

where $L \left[\mathrm{cm}\right]$ is the length of the beam;

$A_1 \left[{\mathrm{cm}}^2\right]$ is the area of the cross-section of the concrete;

$E_1 \left[{\mathrm{kN}/\mathrm{cm}}^2\right]$ is the modulus of elasticity of the concrete.

At the level of the reinforcing bars, the elastic supports 2 and 3, with linear spring coefficients $k_2$ and $k_3$, are introduced. They are set as the reduced tension/compression stiffness of the reinforcing bar by the multipliers ${\zeta }_2$ and ${\zeta }_3$, respectively.

$$k_2={\zeta }_2{\displaystyle \frac{E_2{A}_2}{L}} \mathrm{and} k_3={\zeta }_3{\displaystyle \frac{E_3{A}_3}{L}}$$

(5)

where $L \left[\mathrm{cm}\right]$ represents the length of the beam;

$A_2 \left[\mathrm{c}{\mathrm{m}}^2\right]$ and $A_3 \left[\mathrm{c}{\mathrm{m}}^2\right]$ represent the cross-sectional areas of the bottom and top longitudinal reinforcing bars in the beam passing through the connection;

$E_2 \left[\mathrm{kN}/\mathrm{c}{\mathrm{m}}^2\right]$ and $E_3 \left[\mathrm{kN}/\mathrm{c}{\mathrm{m}}^2\right]$ represent the moduli of elasticity of the bottom and top longitudinal reinforcing bars in the beam passing through the connection.

The supporting reactions that occur here are $H_2 \left[\mathrm{kN}\right]$ and $H_3 \left[\mathrm{kN}\right]$.

$EA=E_1{A}_1+E_2{A}_2+E_3{A}_3$ represents the tensile (compressive) stiffness of the composite section;

$EI=E_1{I}_1+E_2{I}_2+E_3{I}_3$ represents the bending stiffness of the composite section;

$I_1 \left(I_{y1}\right)$, $I_2 \left(I_{y2}\right)$ and $I_3 \left(I_{y3}\right)$ are the moment of inertia of the concrete cross-section and of the top and bottom reinforcing bars relative to the principal axis of inertia, у.

3.2. Support Reactions

The beams in Figure 3 and Figure 4 are considered. Due to the symmetry of the beam with respect to the vertical midsection, the horizontal forces on the left side are equal to those on the right side. The beam is three times statically indeterminate.

The solution is based on Menabria’s theorem about statically indeterminate systems in first-order theory.

The potential energy of deformation in special bending, combined with tension (compression) and with the effects of linear springs, taken into account, will be as follows:

$$\Pi ={\displaystyle \frac{1}{2}}{\displaystyle \underset{0}{\overset{\mathrm{L}}{\int }}\frac{M^2\left(x\right)}{EI}}dx+{\displaystyle \frac{1}{2}}{\displaystyle \underset{0}{\overset{L}{\int }}\frac{N^2\left(x\right)}{EA}dx+\frac{H_1^2}{k_1}}+{\displaystyle \frac{H_2^2}{k_2}}+{\displaystyle \frac{H_3^2}{k_3}}$$

(6)

It is a well-known fact that, according to Menabria’s theorem, the desired hyperstatic unknown is determined by the minimum potential energy condition with respect to it, or

$${\displaystyle \frac{\partial \Pi }{\partial H_1}}=0;{\displaystyle \frac{\partial \Pi }{\partial H_2}}=0;{\displaystyle \frac{\partial \Pi }{\partial H_3}}=0.$$

(7)

3.2.1. Mathematical Model of Beam with Symmetrical Cross-Section

The beam with symmetrical cross-section from Figure 3 is considered.

The following notations are also introduced:

$h \left[\mathrm{cm}\right]$—the height of the beam;

$e\left[\mathrm{cm}\right]$ and $a\left[\mathrm{cm}\right]$—offset of the reinforcing bars from the top and bottom edges of the beam and from the axis of the beam, respectively.

The vertical support reactions are

$$A={\displaystyle \frac{P}{2}} \mathrm{and} B={\displaystyle \frac{P }{2}}.$$

(8)

The bending moments for the three parts of the beam will be as follows:

$$M_1={\displaystyle \frac{P}{2}}x-H_{3}a-H_{2}a-H_1\left({\displaystyle \frac{h}{2}}-b\right);$$

(9)

$$M_2={\displaystyle \frac{P }{2}}\left(g+x\right)-{\displaystyle \frac{P}{2}}x-H_{3}a-H_{2}a-H_1\left({\displaystyle \frac{h}{2}}-b\right);$$

(10)

$$M_3={\displaystyle \frac{P }{2}}\left(g-x\right)-H_{3}a-H_{2}a-H_1\left({\displaystyle \frac{h}{2}}-b\right);$$

(11)

and the normal force, respectively,

$$N=-H_1-H_2+H_3.$$

(12)

Substitute Equations (9)–(12) in Equation (6). We apply Equation (7). A system of three linear equations with respect to the three unknowns is obtained. The solutions give the formulas of the horizontal support reactions, provided below:

$$H_1={\displaystyle \frac{-Pg{k}_1\left(L-g\right)\left\{2EA{h}_1+L{k}_2{n}_1-L{k}_3{n}_2\right\}}{2EA\left[8EI+L{D}_1\right]+8EILK+L{D}_2}};$$

(13)

$$H_2={\displaystyle \frac{Pg{k}_2\left(L-g\right)\left\{4EAa+L{k}_1{n}_1+L{k}_{3}4a\right\}}{2EA\left[8EI+L{D}_1\right]+8EILK+L{D}_2}};$$

(14)

$$H_3={\displaystyle \frac{Pg{k}_3\left(L-g\right)\left\{4EAa+L{k}_1{n}_2+L{k}_{2}4a\right\}}{2EA\left[8EI+L{D}_1\right]+8EILK+L{D}_2}};$$

(15)

where $h_1=2b-h$; $n_1=2a+h_1$; $n_2=2a-h_1$;

$K=k_1+k_2+k_3$;

$D_1=\left(k_2+k_3\right)4a^2+k_1{h}_1^2$;

$D_2=L\left[k_1{k}_2{\left(2a+h_1\right)}^2+k_1{k}_3{\left(2a-h_1\right)}^2+k_2{k}_{3}16a^2\right]$.

Neglecting the normal force in the strain potential energy expression, the support reactions become

$$H_1={\displaystyle \frac{-Pg{k}_1\left(L-g\right)\left(2b-h\right)}{8EI+L{D}_1}};$$

(16)

$$H_2={\displaystyle \frac{2Pag{k}_2\left(L-g\right)}{8EI+L{D}_1}};$$

(17)

$$H_3={\displaystyle \frac{2Pag{k}_3\left(L-g\right)}{8EI+L{D}_1}}.$$

(18)

The substitution of the expression $g=L/2$ in Equations (13)–(15) and Equations (16)–(18) leads to midsection force loading formulas in Equations (А1)–(А3) and Equations (А4)–(А6), respectively.

3.2.2. Mathematical Model of Beam with Asymmetrical Cross-Section

The beam with asymmetrical cross-section from Figure 4 is considered.

The bending moments for the three parts of the beam will be as follows:

$$M_1={\displaystyle \frac{P}{2}}x-H_{3}d-H_{2}a-H_1\left(z-b\right);$$

(19)

$$M_2={\displaystyle \frac{P}{2}}\left(g+x\right)-{\displaystyle \frac{P}{2}}x-H_{3}d-H_{2}a-H_1\left(z-b\right);$$

(20)

$$M_3={\displaystyle \frac{P }{2}}\left(g-x\right)-H_{3}d-H_{2}a-H_1\left(z-b\right);$$

(21)

and the normal force, respectively,

$$N=-H_1-H_2+H_3.$$

(22)

Substitute Equations (19)–(22) in Equation (6). We apply Equation (7). A system of three linear equations with respect to the three unknowns is obtained. The formulas of the horizontal support reactions are given below:

$$H_1={\displaystyle \frac{Pg{k}_1(L-g)\left[2AE\left(z-b\right)+L{k}_2\left(c-b\right)+L{k}_3\left(z-b+d\right)\right]}{2\left\{2EA\left[2EI+L{k}_1{\left(z-b\right)}^2+L{k}_2{a}^2+L{k}_3{d}^2\right]+2EILK+L{D}_3\right\}}};$$

(23)

$$H_2={\displaystyle \frac{Pg{k}_2(L-g)\left[2AEa+L{k}_1\left(a+b-z\right)+L{k}_3\left(a+d\right)\right]}{2\left\{2EA\left[2EI+L{k}_1{\left(z-b\right)}^2+L{k}_2{a}^2+L{k}_3{d}^2\right]+2EILK+L{D}_3\right\}}};$$

(24)

$$H_3={\displaystyle \frac{Pg{k}_3(L-g)\left[2AEd+L{k}_1\left(d+z-b\right)+L{k}_2\left(a+d\right)\right]}{2\left\{2EA\left[2EI+L{k}_1{\left(z-b\right)}^2+L{k}_2{a}^2+L{k}_3{d}^2\right]+2EILK+L{D}_3\right\}}};$$

(25)

where $h_1=2b-h$; $n_1=2a+h_1$; $n_2=2a-h_1$;

$K=k_1+k_2+k_3$; $z=a+c$;

$D_3=L\left[k_1{k}_2{\left(c+b\right)}^2+k_1{k}_3{\left(z-b+d\right)}^2+k_2{k}_3{\left(a+d\right)}^2\right]$.

Neglecting the normal force in the strain potential energy expression, the support reactions become

$$H_1={\displaystyle \frac{Pg{k}_1\left(L-g\right)\left(z-b\right)}{2\left[2EI+L{k}_1{\left(z-b\right)}^2+L\left(k_2{a}^2+k_3{d}^2\right)\right]}};$$

(26)

$$H_2={\displaystyle \frac{Pag{k}_2\left(L-g\right)}{2\left[2EI+L{k}_1{\left(z-b\right)}^2+L\left(k_2{a}^2+k_3{d}^2\right)\right]}};$$

(27)

$$H_3={\displaystyle \frac{Pdg{k}_3\left(L-g\right)}{2\left[2EI+L{k}_1{\left(z-b\right)}^2+L\left(k_2{a}^2+k_3{d}^2\right)\right]}}.$$

(28)

By substituting $d=a$, $c=e$ and $z=a+c=h/2$ in Equations (23)–(25) and Equations (26)–(28), the formulas for the symmetric section Equations (13)–(15) and Equations (16)–(18) are obtained, respectively.

The substitution of $g=L/2$ in Equations (23)–(25) and Equations (26)–(28) leads to midsection force loading formulas in Equations (А7)–(А9) and Equations (А10)–(А12).

The study of the extremum of the support reactions from Equations (13)–(15), Equations (16)–(18), Equations (23)–(25) and Equations (26)–(28) provides the value for the distance, $g=L/2$.

The solution was performed in the symbolic environment of the MATLAB R2017b program [24].

4. Results and Discussion

The numerical results shown in Section 4 and Section 5 are for two cross-sections of the beam, both symmetrical and asymmetrical indicated in Figure 5.

For both cross-sections, the following is accepted:

$A_2 =A_3=25 \mathrm{c}{\mathrm{m}}^2$ and $E_2 =E_3=\mathrm{39,000} \mathrm{kN}/\mathrm{c}{\mathrm{m}}^2$ represent the areas of the bottom and top reinforcement and the modulus of elasticity, respectively;

$e=3 \mathrm{cm}$ and $c=3 \mathrm{cm}$—the cover of the reinforcement;

$L=700 \mathrm{cm}$—the length of the beam;

$E_1=1700 \mathrm{kN}/\mathrm{c}{\mathrm{m}}^2$—the modulus of elasticity for normal concrete and $E_1=3700 \mathrm{kN}/\mathrm{c}{\mathrm{m}}^2$ for high-strength concrete;

${\zeta }_1=1$—multiplier for reduced tensile/compressive stiffness of the concrete section;

${\zeta }_2=1$ and ${\zeta }_3=1$—multiplier for reduced tensile/compressive stiffness of the reinforcing bars;

$P=50 \mathrm{kN}$—for all numerical results.

4.1. Symmetrical Cross-Section

А beam with a cross-section of $25/25 \mathrm{cm}$, as shown in Figure 5a, is introduced.

The distance $b \left[\mathrm{cm}\right]$ varies in the interval $\left[12.5, 0\right)$ and is monitored by the ratio $h/b$.

Figure 6 shows the variation in the parameters of the three support reactions, ${\displaystyle \frac{H_1}{P}}, {\displaystyle \frac{H_2}{P}} \mathrm{and} {\displaystyle \frac{H_3}{P}}$, calculated by Equations (13)–(15), while the crack between the beam and the column grows, and the two loading forces are in the middle vertical section of the beam, $g=L/2=350 \mathrm{cm}$. Before the appearance of a crack, $H_2=H_3=2.12P$ and $H_1=0$. After the appearance of a crack, $H_2$ decreases, while $H_3$ slightly increases to $b/h=2.9$ and then decreases. The support reaction parameter ${\displaystyle \frac{H_1}{P}}$ increases to a value of 1.04P, which corresponds to a support reaction $H_1$ equal in magnitude to the loading force. Using high-strength concrete (Figure 6b) leads to smaller values of $H_2$ and $H_3$, along with a significant increase in $H_1$, compared to Figure 6a, which depicts normal concrete.

Figure 7 shows the variation in the parameters of the three support reactions, ${\displaystyle \frac{H_1}{P}}, {\displaystyle \frac{H_2}{P}} \mathrm{and} {\displaystyle \frac{H_3}{P}}$, calculated by Equations (13)–(15), while the crack between the beam and the column grows, and the two loading forces are in the vertical section offset from the ends of the beam, at a distance $g=200 \mathrm{cm}$. Before the appearance of a crack, $H_2=H_3=1.73P$, and $H_1=0$. After the appearance of a crack $H_2$ decreases, while $H_3$ slightly increases to $b/h=3.3$ and then decreases. The support reaction parameter ${\displaystyle \frac{H_1}{P}}$ increases to a value of 0.85P. The use of high-strength concrete (Figure 7b) leads to smaller values of $H_2$ and $H_3$ and a significant increase in $H_1$, compared to normal concrete, as depicted in Figure 7a. The magnitudes of all force parameters are smaller than those shown in Figure 6a,b, when the two loading forces are in the middle vertical section of the beam, $g=L/2=350 \mathrm{cm}$.

Figure 8 shows the variation in the parameter of the support reaction $H_1$, calculated by Equations (13)–(15), while the crack between the beam and the column grows, and the two loading forces are applied in the vertical section, offset from the ends of the beam at different distances, with values of $g=100 \mathrm{cm},\text{} g=200 \mathrm{cm}, g=300 \mathrm{cm}$ and $g=L/2=350 \mathrm{cm}$. As was noted in the study of the extremum of the expressions for the support reactions, the largest values of $H_1, H_2 \mathrm{and} H_3$ occur at $g=L/2=3.5 m$. A rapid increase in $H_1$ is observed for smaller values of g, and a slower increase in $H_1$ for large values of g, i.e., at loading forces near the mean vertical section of the beam. The use of high-strength concrete (Figure 8b) leads to a significant increase in $H_1$ as compared to normal concrete, as depicted in Figure 8a.

Figure 9 shows the variation in the parameters of the three support reactions, ${\displaystyle \frac{H_1}{P}}, {\displaystyle \frac{H_2}{P}} \mathrm{and} {\displaystyle \frac{H_3}{P}}$, calculated by Equations (16)–(18), while the crack between the beam and the column grows, and the two loading forces are in the middle vertical section of the beam, $g=L/2=350 \mathrm{cm}$. Figure 9 shows the complete matching of $H_2 \mathrm{and} H_3$. After the appearance of a crack $H_2 \mathrm{and} H_3$ decrease, while a serious increase in $H_1$ is observed. The use of high-strength concrete (Figure 9b) leads to smaller values of $H_2 \mathrm{and} H_3$ and a significant increase in $H_1$ compared to normal concrete (Figure 9a).

4.2. Asymmetrical Cross-Section

The numerical results are calculated for beams with a trapezoidal cross-section, as shown in Figure 5b. The shape of the cross-section is tentatively adopted, to meet the asymmetric cross-section condition. The short base has a size equal to the height $h\hspace{0.17em}=25\text{} \mathrm{cm}$ of the trapezoid, and the long base has a size of $\left(2\times {\displaystyle \frac{1}{5}}+1\right)h$. The position of the support, in which the support reaction $H_1$ occurs, varies in the interval $\left[b=z_c={\displaystyle \frac{\left(38h-60\right)h}{72h-120}}; 0\right)$ measured from the bottom edge of the beam.

Figure 10 shows the variation in the parameters of the three support reactions ${\displaystyle \frac{H_1}{P}}, {\displaystyle \frac{H_2}{P}} \mathrm{and} {\displaystyle \frac{H_3}{P}}$ for asymmetric cross-sections, calculated by Equations (23)–(25), while the crack between the beam and the column grows, and the two loading forces are in the middle vertical section of the beam, $g=L/2=350 \mathrm{cm}$. Different values of $H_2=2.16P$ and $H_3=1.95P$ before the appearance of the crack, along with a negative value of $H_1=-0.03P$ are observed. Therefore, in an asymmetric section, $H_1$ is on a tensile state before the appearance of a crack. After the appearance of a crack, $H_2=1.53P$, while $H_3$ slightly increases to $b/h=3.0$, $H_3=2.00P$ and then decreases to $H_3=1.84P$. A significant increase in $H_1=1.23P$ is observed. The use of high-strength concrete (Figure 10b) leads to smaller values of $H_2 \mathrm{and} H_3$ and a significant increase in $H_1=1.89P$ compared to normal concrete (Figure 10a).

5. Shear Force

The magnitudes of the forces $H_3, H_2^{\prime } \mathrm{and} H_1^{\prime }$ are already known (Figure 11). Then, the determination of the shear force in RC internal beam–column connections will be determined by Equation (29) instead of Equation (1).

$$V_j=H_3+H_2^{\prime }+H_1^{\prime }-V_c.$$

(29)

If the frame is symmetric and other conditions are equal, we will achieve the equality of $H_1=H_1^{\prime }$, $H_2=H_2^{\prime }$ and $H_3=H_3^{\prime }$. Then, Equation (29) becomes Equation (30)

$$V_j=H_3+H_2+H_1-V_c.$$

(30)

The advantage of the proposed solution is that the exact magnitudes of shear forces are considered, taking into account a significant number of parameters affecting the magnitudes of the support reactions entering the beam–column joint.

To verify the proposed solution, a comparison with the approximate solution recommended in the literature and codes is made.

The comparison of Equations (30) and (3) is expressed in the comparison shown in Equation (31).

$$H_3+H_2+H_1 \stackrel{?}{=} {\displaystyle \frac{M_b}{j_b}}+{\displaystyle \frac{{M^{\prime }}_b}{{j^{\prime }}_b}}.$$

(31)

where $M_b=Pg\left(L-g\right)/2L$ is the moment of the beam on the face of the column. The maximum value is at $g=L/2$, and then $M_b=PL/8$.

5.1. Determination of the Reduction Multipliers for the Stiffnesses of the Linear Springs Relative to the Tension/Compression Stiffness of the Beam for the Three Supports

The spring coefficients $k_1$, $k_2$ and $k_3$ of the three supports are reduced to the tension/compression stiffness of the beam by the ${\zeta }_1$, ${\zeta }_2$ and ${\zeta }_3$ multipliers. When we consider the rigid support between structural elements using static schemes, we assume that the connections between them do not allow the sections to move and to rotate. The rigid support will have a larger $k$ and therefore a larger $\zeta$. We will now consider how large values to choose for ${\zeta }_1$, ${\zeta }_2$ and ${\zeta }_3$.

The following results are for a beam with a symmetrical cross-section (Figure 5а).

$A_2 =A_3=25 \mathrm{c}{\mathrm{m}}^2$ and $E_2 =E_3=\mathrm{18,500} \mathrm{kN}/\mathrm{c}{\mathrm{m}}^2$—the areas of the bottom and top reinforcement and the moduli of elasticity, respectively;

$e=3 \mathrm{cm}$ and $c=3 \mathrm{cm}$—the cover of reinforcement;

$L=700 \mathrm{cm}$—the length of the beam;

$E_1=3310 \mathrm{kN}/\mathrm{c}{\mathrm{m}}^2$—modulus of elasticity for normal concrete;

${\zeta }_1$, ${\zeta }_2$ and ${\zeta }_3$—multiplier for reduced tensile/compressive stiffness of the concrete section and of the reinforcing bars, respectively, and they have different values.

The data on the material properties are reported from [19].

Figure 12 shows the parameters of the three horizontal support reactions, $H_1/P$; $H_2/P$ and $H_3/P$ are calculated by Equations (13)–(15), while the crack between the beam and the column grows, and the two loading forces are in the middle of the vertical section of the beam, $g=L/2=350 \mathrm{cm}$. Multipliers for reduced tensile/compressive stiffness values—${\zeta }_1$, ${\zeta }_2$ and ${\zeta }_3$—are set. They vary from 0 to infinity (10,000). In Figure 12a, the plots for ${\zeta }_1=5, 10 \mathrm{and} \mathrm{10,000}$ almost coincide. It is observed that the interval of variation of $H_2$ and $H_3$, following ${\zeta }_2=20$ and ${\zeta }_3=20$, narrows significantly (Figure 12b,c). The curves $\zeta =0 \mathrm{to} 5$ for the three parameters in Figure 12 diverge. Then, for the rigid support modelling, we will assume ${\zeta }_1=10$, ${\zeta }_2=20$ and ${\zeta }_3=20$.

5.2. Comparison of the Results of the Derived Formulas $H_3+H_2+H_1$ and the Prescribed Simplification ${\displaystyle \frac{M_b}{j_b}}+{\displaystyle \frac{{M^{\prime }}_b}{{j^{\prime }}_b}}$—Equation (31)

The numerical results are shown for two cross-sections of the beam—symmetrical and asymmetrical (Figure 5).

A size of 25/25 cm is accepted for both sections:

$A_2 ={А}_3=12.5 \mathrm{c}{\mathrm{m}}^2$ and $E_2 ={Е}_3=\mathrm{18,500} \mathrm{kN}/\mathrm{c}{\mathrm{m}}^2$—the areas of the bottom and top reinforcement and moduli of elasticity, respectively; the bending design for the adopted beam gives the values for $A_2 \mathrm{and} {А}_3$;

$e=3 \mathrm{cm}$ and $c=3 \mathrm{cm}$—the cover of reinforcement;

$L=700 \mathrm{cm}$—the length of the beam;

$E_1=3310 \mathrm{kN}/\mathrm{c}{\mathrm{m}}^2$; $E_1=4000 \mathrm{kN}/\mathrm{c}{\mathrm{m}}^2$—moduli of elasticity for concrete;

${\zeta }_1=10$—multiplier for reduced tensile/compressive stiffness of the concrete section;

${\zeta }_2=20$ and ${\zeta }_3=20$—multipliers for reduced tensile/compressive stiffness of the reinforcing bars.

Figure 13 shows the variation in the sum with respect to the parameters of the three support reactions, ${\displaystyle \frac{H_1}{P}}+ {\displaystyle \frac{H_2}{P}} + {\displaystyle \frac{H_3}{P}}$, calculated by Equations (13)–(15), while the crack between the beam and the column grows, and the two loading forces are in the middle vertical section of the beam, $g=L/2=350 \mathrm{cm}$. The comparison is made with $\left(M_b/j_b+{M^{\prime }}_b/{j^{\prime }}_b\right)/P$. The two graphs have the closest values at h/b = 3.3.

The graphs show that the proposed new model for calculating shear force gives us not only its most unfavorable value but also makes the shear force traceable throughout the crack development process.

The results of

Table 1. Comparison of the results of Equations (3) and (30) for the symmetrical cross-section.

Section		$\mathit{g}$	$\mathit{h}/\mathit{b}$	$\frac{{\mathit{H}}_3+{\mathit{H}}_2+{\mathit{H}}_1}{\mathit{P}}$	$\frac{\frac{{\mathit{M}}_{\mathit{b}}}{{\mathit{j}}_{\mathit{b}}}+\frac{{{\mathit{M}}^{\prime }}_{\mathit{b}}}{{{\mathit{j}}^{\prime }}_{\mathit{b}}}}{\mathit{P}}$	$\frac{\left({\mathit{H}}_3+{\mathit{H}}_2+{\mathit{H}}_1\right) - \left(\frac{{\mathit{M}}_{\mathit{b}}}{{\mathit{j}}_{\mathit{b}}}+\frac{{{\mathit{M}}^{\prime }}_{\mathit{b}}}{{{\mathit{j}}^{\prime }}_{\mathit{b}}}\right)}{\left({\mathit{H}}_3+{\mathit{H}}_2+{\mathit{H}}_1\right)}100\mathit{\%}$
25/25	E1 = 3310 kN/cm2	L/2 = 350	2.0	7.90	9.21	16.60
			3.3	9.06		1.62
			50	7.63		20.72
	E1 = 4000 kN/cm2	L/2 = 350	2.0	7.73	9.21	19.12
			3.3	9.02		2.14
			50	7.53		22.37

show the differences between the exact method ($H_3+H_2+H_1$) and the approximate method $\left({\displaystyle \frac{M_b}{j_b}}+{\displaystyle \frac{{M^{\prime }}_b}{{j^{\prime }}_b}}\right)$ used in Equation (3) for the symmetrical section. For sections with E₁ = 3310 kN/cm² for concrete, the differences between the two methods exceed 10% before the opening of a crack and at its final size. The smallest difference is at h/b = 3.3–1.62%. For concrete with E₁ = 4000 kN/cm², the smallest difference is at h/b = 3.3–2.14%.

Figure 14 shows the variation in the sum with respect to the parameters of the three support reactions, ${\displaystyle \frac{H_1}{P}}+ {\displaystyle \frac{H_2}{P}} + {\displaystyle \frac{H_3}{P}}$, calculated by Equations (16)–(18), neglecting the axial force in the expression of the potential energy of the deformation, while the crack between the beam and the column grows, and the two loading forces are in the middle vertical section of the beam, $g=L/2=350 \mathrm{cm}$. The comparison is made with $\left({\displaystyle \frac{M_b}{j_b}}+{\displaystyle \frac{{M^{\prime }}_b}{{j^{\prime }}_b}}\right)/P$. The two graphs have the closest values at h/b = 2.9. The results of Figure 14 are shown in

Table 2. Comparison of the results of Equations (3) and (30) for symmetrical cross-section, neglecting the axial force in the expression of the potential energy of the deformation.

Section		$\mathit{g}$	$\mathit{h}/\mathit{b}$	$\frac{{\mathit{H}}_3+{\mathit{H}}_2+{\mathit{H}}_1}{\mathit{P}}$	$\frac{\frac{{\mathit{M}}_{\mathit{b}}}{{\mathit{j}}_{\mathit{b}}}+\frac{{{\mathit{M}}^{\prime }}_{\mathit{b}}}{{{\mathit{j}}^{\prime }}_{\mathit{b}}}}{\mathit{P}}$	$\frac{\left({\mathit{H}}_3+{\mathit{H}}_2+{\mathit{H}}_1\right) - \left(\frac{{\mathit{M}}_{\mathit{b}}}{{\mathit{j}}_{\mathit{b}}}+\frac{{{\mathit{M}}^{\prime }}_{\mathit{b}}}{{{\mathit{j}}^{\prime }}_{\mathit{b}}}\right)}{\left({\mathit{H}}_3+{\mathit{H}}_2+{\mathit{H}}_1\right)}100\mathit{\%}$
25/25	E1 = 3310 kN/cm2	L/2 = 350	2.0	7.90	9.21	16.60
			2.9	9.86		−6.60
			50	7.54		22.19
	E1 = 4000 kN/cm2	L/2 = 350	2.0	7.73	9.21	19.12
			2.9	9.90		−6.96
			50	7.58		21.55

.

The results of Table 2 show the differences between the exact method ($H_3+H_2+H_1$) and the approximate method $\left({\displaystyle \frac{M_b}{j_b}}+{\displaystyle \frac{{M^{\prime }}_b}{{j^{\prime }}_b}}\right)$ for symmetrical section, calculated by Equations (16)–(18), neglecting the axial force in the expression of the potential energy of the deformation. For sections with E₁ = 3310 kN/cm² for concrete, the differences between the two methods exceed 16% before the opening of a crack and at its final size. The smallest difference is at h/b = 2.9–6.6%. For concrete with E₁ = 4000 kN/cm², the smallest difference is at h/b = 2.9–6.96%.

Figure 15 shows the variation in the sum of the parameters of the three support reactions, ${\displaystyle \frac{H_1}{P}}+ {\displaystyle \frac{H_2}{P}} + {\displaystyle \frac{H_3}{P}}$, calculated by Equations (23)–(25) for an asymmetric cross-section, while the crack between the beam and the column grows, and the two loading forces are in the middle vertical section of the beam, $g=L/2=350 \mathrm{cm}$. The comparison is made with $\left({\displaystyle \frac{M_b}{j_b}}+{\displaystyle \frac{{M^{\prime }}_b}{{j^{\prime }}_b}}\right)/P$. The two graphs have the closest values at h/b = 3.0.

The results of

Table 3. Comparison of the results of Equations (3) and (30) for asymmetrical cross-section.

Section		$\mathit{g}$	$\mathit{h}/\mathit{b}$	$\frac{{\mathit{H}}_3+{\mathit{H}}_2+{\mathit{H}}_1}{\mathit{P}}$	$\frac{\frac{{\mathit{M}}_{\mathit{b}}}{{\mathit{j}}_{\mathit{b}}}+\frac{{{\mathit{M}}^{\prime }}_{\mathit{b}}}{{{\mathit{j}}^{\prime }}_{\mathit{b}}}}{\mathit{P}}$	$\frac{\left({\mathit{H}}_3+{\mathit{H}}_2+{\mathit{H}}_1\right) - \left(\frac{{\mathit{M}}_{\mathit{b}}}{{\mathit{j}}_{\mathit{b}}}+\frac{{{\mathit{M}}^{\prime }}_{\mathit{b}}}{{{\mathit{j}}^{\prime }}_{\mathit{b}}}\right)}{\left({\mathit{H}}_3+{\mathit{H}}_2+{\mathit{H}}_1\right)}100\mathit{\%}$
25/25	E1 = 3310 kN/cm2	L/2 = 350	2.0	7.41	9.21	24.28
			3.0	9.00		2.29
			50	7.39		24.57
	E1 = 4000 kN/cm2	L/2 = 350	2.0	7.20	9.21	27.85
			3.0	8.94		3.05
			50	7.28		26.58

show the differences between the exact method ($H_3+H_2+H_1$) and the approximate method $\left({\displaystyle \frac{M_b}{j_b}}+{\displaystyle \frac{{M^{\prime }}_b}{{j^{\prime }}_b}}\right)$ used in Equation (3) for the asymmetrical section. For sections with E₁ = 3310 kN/cm² for concrete, the differences between the two methods exceed 10% before the crack opening and at its final size. The smallest difference is at h/b = 3.0–2.29%. For concrete with E₁ = 4000 kN/cm², the smallest difference is at h/b = 3.3–3.05%.

5.3. Comparison of Eurocode Results and Prescribed Simplification $\left({\displaystyle \frac{M_b}{j_b}}+{\displaystyle \frac{{M^{\prime }}_b}{{j^{\prime }}_b}}\right)$

Equation (5.22) of [25] gives us the magnitude of the shear force:

$$V_{jhd}={\gamma }_{Rd}\left(A_{s1}+A_{s2}\right)f_{yd}-V_c.$$

(32)

where $f_{yd}=21 \mathrm{kN}/\mathrm{c}{\mathrm{m}}^2$—design value of the yield strength of steel;

${\gamma }_{Rd}$ should not be taken less than 1.2.

For the accepted data in our example, we obtain ${\gamma }_{Rd}\left(A_{s1}+A_{s2}\right)f_{yd}/P=1.2\times 25\times 21/50=12.6$. Compared to $\left({\displaystyle \frac{M_b}{j_b}}+{\displaystyle \frac{{M^{\prime }}_b}{{j^{\prime }}_b}}\right)/P=9.21$, we obtain a difference of 36.80%. Comparing the Eurocode results with those of Table 1, Table 2 and Table 3 gives us the differences, respectively,

Table 1: for ${\displaystyle \frac{\left(H_3+H_2+H_1\right)}{P}}=9.06$—39.07%

- for ${\displaystyle \frac{\left(H_3+H_2+H_1\right)}{P}}=9.02$—39.69%

Table 2: for ${\displaystyle \frac{\left(H_3+H_2+H_1\right)}{P}}=9.86$—27.79%

- for ${\displaystyle \frac{\left(H_3+H_2+H_1\right)}{P}}=9.90$—27.27%

Table 3: for ${\displaystyle \frac{\left(H_3+H_2+H_1\right)}{P}}=9.00$—40.00%

- for ${\displaystyle \frac{\left(H_3+H_2+H_1\right)}{P}}=8.94$—40.94%

The comparison shows that the calculated exact magnitudes of the forces for the considered static loading forces are smaller than those predicted by Eurocode. Designing according to Eurocode includes several calculated safety factors.

6. Conclusions

A solution for a “simple” beam with a special arrangement of the supports was developed. The actual dimensions of the beam were taken into account. The beam was loaded with two transverse forces located symmetrically to the mid-section of the beam.

The derived expressions for the reactions of the horizontal supports yielded results that clearly show the distribution of the forces along the height of the beam. The formulas were derived for the limit stage and enabled us to determine the distribution of forces before and after the appearance of a crack between the beam and the column.

The resulting expressions for the support reactions take into account the influence of both the geometry of the beam and the material properties of all its components. This makes it possible to trace the variation in the forces acting on the beam and entering the beam–column connection with different data combinations of the included quantities.

The selection of the reduction coefficients for the stiffness of the springs in the supports relative to the stiffness of the tension/compression beam was shown. These coefficients must be chosen carefully, due to their significant sensitivity to the final results. These $\zeta$ coefficients can help us to model the bond of reinforcement or different grades of concrete in the joint.

A comparison was made between the contribution of beam forces to the shear force value in RC internal beam–column connections and those known from the literature and from Eurocode.

The results showed that the proposed exact method gives results that differ from the adopted one by 2% to 27%, depending on the stage of crack development. The difference between the new exact method and that of the Eurocode ranges from 27% to 40%, based only on the largest shear force value determined by the exact method. The Eurocode result is loaded with multiple safety factors, and this is probably what sets the results so far apart. Eurocode specifies the force that the intended reinforcement could withstand, while the new formulas show how big the forces actually are. The new formulas can protect us from choosing an inappropriate combination of data, as well as oversizing the reinforcing bars, as this will overestimate the horizontal forces at the joint. The new approach will prevent us from displacing the reinforcing bars towards the axis of the beam, as we will obtain smaller forces in the reinforcement, whereas the forces in the concrete will increase [22,23]. The new formulas allow us to calculate the contribution of concrete to the shear force and the capacity of the concrete section.

The static solution for the shear strength confirms the observations in [15] “Concrete cylinder strength increases the joint shear strength”. After the appearance of a crack on the face of the column, the support reactions in the beam increase significantly, and the shear force increases accordingly. This corresponds with the observations reported in [17] about “joint shear stress had increased in the most specimens, even after apparent joint shear failure starts”.

The results show that the values of all parameters included in the formulas have a serious impact on the magnitudes of the support reactions. The model allows to track how and by how much the change in just one parameter will affect the final results.

The derived expressions for the support reactions can also be used for steel structures in the design of bolted connections.

The obtained results can be useful for both researchers and practicing engineers. The research from this paper can help to interpret the results obtained from construct analyses and experimental studies.