Repercussions on the Shear Force of an Internal Beam–Column Connection from Two Symmetrical Uniformly Distributed Loads at Different Positions on the Beam ^†

Abstract

The beam–column connection is an important element in frame construction. Despite numerous studies, there is still no uniform procedure for shear force design across countries. We continue to witness serious problems and even collapse of buildings under seismic activity caused by failures in the beam–column connection of the frame. During the last 60 decades, a large number of experimental studies have been carried out on frame assemblies, where various parameters and their compatibility under cyclic activities have been investigated. What remains misunderstood is the magnitude and distribution of the forces passing through the joint and their involvement in the magnitude of the shear force. Here, the creation of a new mathematical model for the beam and column contributes significantly to our understanding of the flow of forces in the frame connection. For this purpose, the full dimensions of the beam and its material properties are taken into account. All investigations were carried out before crack initiation and after crack propagation along the face of the column, where it separates from the beam. In the present work, the beam is subjected to two symmetrical, transverse, uniformly distributed loads. Expressions are derived to determine the magnitudes of the support reactions from the beam, as a function of the height of its lateral edge. The load positions corresponding to the extreme values of the support reactions are determined. Numerical results are presented for the effect over the magnitudes of the support reactions from different strengths of concrete and steel on the beam. The results are compared with those given in the Eurocode for shear force calculation. It is found that the shear force determined by the proposed new model exceeds the force calculated by Eurocode by 4–62.5%, depending on the crack development stage and the beam materials.

1. Introduction

The transfer of forces in framed structures from beams to columns under dynamic loads can threaten the integrity of the connection. Determining the shear force in the connection continues to preoccupy researchers.

The first definition of shear force in a joint was given in [1] by Hanson and Connor (1967) as the horizontal force acting at the mid-height of the beam–column connection. Experimental and analytical studies conducted since the introduction of this definition have observed the influence of various variables on the response of the beam–column joint [2,3,4,5,6,7,8,9,10,11].

In Eurocode 8 [12], the beam forces contributing to the shear force are determined based on capacity design. These are the forces absorbed by the longitudinal reinforcing bars when the steel yields. The same assumption is also made in [13]. Nowadays, capacity design methods give additional consideration to the involvement of the concrete section and stirrups in the joint [14,15,16]. In [17], a graphical method called monograms is proposed to determine the geometric dimensions of the beam–column connection and the number of stirrups. In [18], the experimental results are compared with a 3D model, allowing the researchers to vary the characteristics of the construction materials used.

However, all these approaches do not answer the question of how large the shear force actually is. In [19,20,21,22,23], a beam and cantilever model is proposed to determine the magnitude of the shear force by considering the contribution of the concrete section, the beam dimensions, and the material characteristics of the constituent members. In this study, the forces transferred from the beam are determined as a result of two symmetrical, uniformly distributed loads. Determining the exact expressions of the beam’s forces before the beam–column connection allows the exact magnitude of the shear force to be calculated. The appearance of a crack between the beam and the column and its growth allows us to track the change in shear force in a limit stage. The results obtained for the magnitude of the shear force are compared with those given in the literature and prescribed in Eurocode 8 (2004) [12].

2. Materials

In their study [1], Hanson and Connor defined the shear force in an internal beam–column connection from Figure 1, by Equation (1).

$$V_{jh}=T+{C^{\prime }}_S+{C^{\prime }}_C-V_C=T+T^{\prime }-V_C,$$

(1)

where $C_C$ and $C_S$ are the compressive forces in concrete and in the longitudinal reinforcing bars in the beam passing through the connection; $T$ and $T^{\prime }$ are the tensile forces in the longitudinal reinforcing bars in the beam; and $V_C$ is the shear force of the column.

The difficulty encountered in determining the forces $T$ and $T^{\prime }$ from Equation (1) leads to the adoption in the literature of:

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

(2)

where $M_b$ and ${M^{\prime }}_b$ are the moments at the column face and $j_b$ and ${j^{\prime }}_b$ are the lever arms corresponding to these moments. The lever arms are assumed to remain constant during the deformation process.

In this article, the following tasks are set: 1. To derive expressions for the forces from Figure 1 and from Equation (1); 2. To use the derived expressions to calculate the forces for selected sections and determine the magnitude of the shear force; 3. To compare the resulting shear force values with those prescribed in Equation (2) and in Eurocode 8 [12]; 4. To investigate how much the magnitude of shear force changes when the material characteristics of some sections are changed.

3. Method

3.1. Support Reactions with Axial Force in the Strain Energy Expression

A frame structure beam is considered below. The beam is statically indeterminate, prismatic, and symmetrical. The beam is subjected to special bending and tension/compression, and the Bernoulli–Euler hypothesis is applied (Figure 2).

The solution is based on Menabria’s theorem for 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, is as follows:

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

(3)

where $M_i$ is the bending moment in particular sections;

$N$ is the axial force of the beam;

$H_1$ is the support reaction at support 1, corresponding to a spring with a spring constant of $k_1={\zeta }_1{\displaystyle \frac{E_1{A}_1}{L}}$. It is set as the reduced tensile/compressive stiffness of the concrete section by the multiplier ${\zeta }_1$;

In supports 2 and 3, the support reactions $H_2$ and $H_3$ occur. The corresponding springs have spring constants $k_2={\zeta }_2{\displaystyle \frac{E_2{A}_2}{L}}$ and $k_3={\zeta }_3{\displaystyle \frac{E_3{A}_3}{L}}$. They are set as the reduced tensile/compressive stiffness of the steel section by the multiplier ${\zeta }_2$ and ${\zeta }_3$;

$L\hspace{0.33em}$ is the length of the beam;

$E_1{A}_1$, $E_2{A}_2$, and $E_3{A}_3$ are the tensile (compressive) stiffness of the concrete cross-section and the reinforcing bars, respectively;

$E_1{I}_1$, $E_2{I}_2$, and $E_3{I}_3$ are the bending stiffness of the concrete cross-section and the reinforcing bars, respectively;

$EA=E_1{A}_1+E_2{A}_2+E_3{A}_3\hspace{0.33em}$ are the tensile (compressive) stiffness of the composite section;

$EI=E_1{I}_1+E_2{I}_2+E_3{I}_3\hspace{0.33em}$ are the bending stiffness of the composite section.

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

$${\displaystyle \frac{\mathit{\partial }\Pi }{\mathit{\partial }{H}_1}}=0;\hspace{1em}{\displaystyle \frac{\mathit{\partial }\Pi }{\mathit{\partial }{H}_2}}=0;\hspace{1em}{\displaystyle \frac{\mathit{\partial }\Pi }{\mathit{\partial }{H}_3}}=0$$

(4)

A mathematical model of a beam with a symmetrical cross-section is given in Figure 2a. The cross-section of the beam is shown in Figure 2b.

The equilibrium conditions of statics give, respectively:

$$A=ql; B=ql$$

(5)

The axial force and bending moment for the individual sections of the beam are substituted into Equation (3). We apply the solution according to Equation (4). The resulting horizontal support reactions are, respectively:

$$\begin{gathered} H_1={\displaystyle \frac{\left(-q{k}_1\right)\left[Z\right]\left[2EA{h}_1+L{k}_2{n}_1-L{k}_3{n}_2\right]}{3\left\{2EA\left[8EI+L{D}_1\right]+8EILK+L{D}_2\right\}}}; \\ {H}_2={\displaystyle \frac{\left(q{k}_2\right)\left[Z\right]\left[4EAa+L{k}_1\left(2a+2b-h\right)+4L{k}_{3}a\right]}{3\left\{2EA\left[8EI+L{D}_1\right]+8EILK+L{D}_2\right\}}}; \\ {H}_3={\displaystyle \frac{\left(q{k}_3\right)\left[Z\right]\left[4EAa+L{k}_1\left(2a-2b+h\right)+4L{k}_{2}a\right]}{3\left\{2EA\left[8EI+L{D}_1\right]+8EILK+L{D}_2\right\}}}. \end{gathered}$$

(6)

$$\begin{gathered} where h_1=2b-h; n_1=2a+h_1; n_2=2a-h_1; K=k_1+k_2+k_3 \\ D_1=\left(k_2+k_3\right)4a^2+k_1{h}_1^2; D_2=L\left[k_1{k}_2{n}_1^2+k_1{k}_3{n}_2^2+16k_2{k}_3{a}^2\right]; \\ Z=L^3-4l^3+3Lg\left(4l-L\right)+2g^2\left(2g+3l\right)+6Ll\left(2l-L\right) \end{gathered}$$

(7)

3.2. Support Reactions Without Axial Force in Strain Energy Expression

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

$$H_1={\displaystyle \frac{\left(-q{k}_1\right)\left[Z\right]h_1}{3\left[8EI+L{D}_1\right]}}; H_2={\displaystyle \frac{\left(q{k}_2\right)\left[Z\right]2a}{3\left[8EI+L{D}_1\right]}}; H_3={\displaystyle \frac{\left(q{k}_3\right)\left[Z\right]2a}{3\left[8EI+L{D}_1\right]}};$$

(8)

Introducing $g=0$ and $l=L/2$ into Equations (6) and (8) leads to the equations for a beam loaded uniformly along its entire length, based on [24,25].

The solution was performed in the symbolic environment of MATLAB R2017b [26].

4. Results and Discussion

For the numerical results, a beam with a cross-section of $25/25\hspace{0.33em}\mathrm{cm}$ was introduced. The transverse load is $q=0.5\hspace{0.33em}\mathrm{kN}/\mathrm{cm}\prime$ with $l=200\hspace{0.33em}\mathrm{cm}$; $e=3\hspace{0.33em}\mathrm{cm}$, $a=9.5\hspace{0.33em}\mathrm{cm}$, and $A_2=A_3=6.75\hspace{0.33em}{\mathrm{cm}}^2$. The distance $b\hspace{0.33em}\left[\mathrm{cm}\right]$ varies in the interval $\left[12.5\hspace{0.33em};\hspace{0.33em}0\right)$ and is monitored by the ratio $h/b$. The length of the beam is $L=700\hspace{0.33em}\mathrm{cm}$.

When we consider rigid support between structural elements through the static schemes, we assume that the connections between them do not allow the sections to move or rotate. The rigid support has a larger $k_i$, and from there, a larger ${\zeta }_i$. It was shown in [19] that it can be assumed with sufficient accuracy that ${\zeta }_1=10$ and ${\zeta }_2={\zeta }_3=20$. Three examples are considered, each with a different moduli of elasticity for concrete and steel. The first example, Case I, is calculated with moduli of elasticity $E_1=1700\hspace{0.33em}{\mathrm{kN}/\mathrm{cm}}^2$ and $E_2=E_3=18500\hspace{0.33em}{\mathrm{kN}/\mathrm{cm}}^2$. The second example, Case II, is for moduli of elasticity $E_1=3310\hspace{0.33em}{\mathrm{kN}/\mathrm{cm}}^2$ and $E_2=E_3=18500\hspace{0.33em}{\mathrm{kN}/\mathrm{cm}}^2$ [27]. The third example, Case III, is for $E_1=1700\hspace{0.33em}{\mathrm{kN}/\mathrm{cm}}^2$ and $E_2=E_3=39000\hspace{0.33em}{\mathrm{kN}/\mathrm{cm}}^2$.

4.1. Comparison of Support Reaction Magnitude Results for Case I, Case II, and Case III

It can be seen from Figure 3 that the movement of the uniformly distributed loads towards the mid-section of the beam, with an increase of $g$, leads to an increase in the magnitudes of the three support reactions. The growth of the crack between the beam and the column also leads to an increase in the support reaction $H_1$ and a decrease in the support reaction $H_2$, while the support reaction $H_3$ increases to $h/b=2.9$, and then decreases. This is observed in all three examples considered. The percentage differences in the magnitudes of the three support reactions for some characteristic values of $h/b$ for the three examples are shown in

Table 1. Percentage differences in the magnitudes of the three support reactions for the three examples.

$\mathit{g}=150\hspace{0.33em}\mathbf{cm}$	$\mathit{h}/\mathit{b}$	Case I	Case II		Case III
		Value [kN]	Value [kN]	% Relative to Case I	Value [kN]	% Relative to Case I
$H_1$	0	0	0	-	0	-
	2.9	311.31	410.50	31.86	205.30	−34.05
	50	569.90	668.46	17.29	433.35	−23.96
$H_2$	0	705.43	643.31	−8.81	745.37	5.66
	2.9	525,38	434.53	−17.29	616.77	17.39
	50	169.60	82.75	−51.21	308.21	81.73
$H_3$	0	705.43	643.31	−8.81	745.37	5.66
	2.9	772.82	716.62	−7.27	795.49	2.93
	50	622.57	542.10	−12.93	685.45	10.10

.

The results in Table 1 are for $g=150\hspace{0.33em}\mathrm{cm}$ and show that as the concrete modulus $E_1$ increases, the support reaction $H_1$ increases, and the support reactions $H_2$ and $H_3$ decrease. Larger steel moduli $E_2$ and $E_3$ lead to a decrease in $H_1$ and an increase in $H_2$ and $H_3$.

A model to determine the shear force in an RC interior beam–column connection is proposed on Figure 4.

$$V_{jh}=H_3+H_2^{\prime }+H_1^{\prime }-V_c,$$

(9)

If the frame is symmetrical and all other conditions being equal, we have equality for $H_1=H_1^{\prime }$, $H_2=H_2^{\prime }$, and $H_3=H_3^{\prime }$. Then (9) becomes:

$$V_{jh}=H_3+H_2+H_1-V_c.$$

(10)

4.2. Comparison of the Results from Equation (2), Equation (10), and Eurocode

The values of $M_b$ were determined using the RuckZuck 4.0 software [28]. The beam has two rigid supports.

All results are converted into parametric form by dividing the load by 2ql.

Figure 5 and Figure 6 provide information about the increase in shear force with the change in distance $g$. For the specific material values, the magnitude of $H_1+H_2+H_3$ significantly exceeds the Eurocode limit. The percentage differences for the three considered examples at $g=150\hspace{0.33em}\mathrm{cm}$ are shown in

Table 2. Comparison of the parameters of the three support reactions for distance g = 150 cm with Equation (2) and with Eurocode.

$\mathit{g}=150\hspace{0.33em}\mathit{c}\mathit{m}$	$\mathit{h}/\mathit{b}$	$\frac{\left(\frac{{\mathit{M}}_{\mathit{b}}}{{\mathit{j}}_{\mathit{b}}}+\frac{{{\mathit{M}}^{\prime }}_{\mathit{b}}}{{{\mathit{j}}^{\prime }}_{\mathit{b}}}\right)-\left({\mathit{H}}_3+{\mathit{H}}_2+{\mathit{H}}_1\right)\hspace{0.33em}\hspace{0.33em}}{\left({\mathit{H}}_3+{\mathit{H}}_2+{\mathit{H}}_1\right)}100\mathit{\%}$	$\frac{1.2\times {\mathit{f}}_{\mathit{y}\mathit{d}}\left({\mathit{A}}_{\mathit{s}1}+{\mathit{A}}_{\mathit{s}2}\right)-\left({\mathit{H}}_3+{\mathit{H}}_2+{\mathit{H}}_1\right)\hspace{0.33em}\hspace{0.33em}}{\left({\mathit{H}}_3+{\mathit{H}}_2+{\mathit{H}}_1\right)}100\mathit{\%}$
Case I	0	16.35	−56.94
	3.3	1.66	−62.38
	50	20.52	−55.40
Case II	0	27.59	−52.78
	3.3	4.48	−61.34
	50	26.93	−53.03
Case III	0	10.12	−59.25
	3.3	1.28	−62.52
	50	15.03	−57.43

$f_{yd}=37.5\hspace{0.33em}{\mathrm{kN}/\mathrm{cm}}^2$ is the design value of the yield strength of steel [27].

.

5. Conclusions

The expressions for the horizontal support reactions for the beam of a frame structure with two uniformly distributed loads are derived. The obtained forces are used to calculate the shear force in the beam–column connection. The method allows us to take into account the participation of the concrete cross-section in the total magnitude of the shear force. The dimensions of the beam and the material characteristics of the composite elements are also taken into account and have an impact on the shear force. The expressions allow us to select different combinations of data and to calculate the shear force as it is at the specific load. Joint designs should be carried out based on these determined forces.