Application of singular functions in bending deformation of material mechanics

Abstract

In the mechanics of materials, the integral method is often used to calculate the bending deformation of the beam. It is relatively simple to use this method under a single load. However, in real life, the load conditions of most beams are complex and diverse, resulting in more segmentation of the bending equation, too many integral constants, complicated equations involved, cumbersome calculation process, and heavy calculation. In order to solve this problem, the concept of singular function is introduced in this paper. By using a bending equation to express the bending moment internal force of the whole beam, and there are only two integral constants after integration, which greatly simplifies the calculation process and reduces the calculation amount. This paper mainly introduces the properties of the singular function and how to use the singular function to express the bending equation of the beam, and solves the bending deformation of the beam with the boundary conditions, and discusses the application of the singular function in the calculation of the statically indeterminate beam.

Keywords

Singular function material mechanical bending deformation statically determinate beam statically indeterminate beam

1. Introduction

The fundamental technique for resolving the bending deformation of beams in material mechanics is the integral method. Although more straightforward, the complexity of the beam’s load situation, the segmentation of the bending moment equation, and the differential equation of the deflection curve increase with complexity. Consequently, the number of integral constants increases. If the beam has n bending moment equations, there are 2n integral constants, and 2n simultaneous equations must be resolved. The calculation process is time-consuming, the amount of calculation is significant, and practical application is considerably inconvenient [1].

To address this issue, this paper introduces the singular function as a solution to the bending deformation problem in material mechanics. Through a single functional equation, the internal force of the entire beam can be expressed, allowing for integral calculation utilizing the unique properties of the singular function. This method remains effective regardless of the complexity of the beam’s load situation. There are typically only two integral constants, which can be solved efficiently by using the boundary conditions to derive the rotation angle and deflection equations for the beam. This simplifies the calculation process significantly and makes it much easier to understand [2].

2. The nature of singular functions

Singular functions, also known as impulse functions, were developed by the German mathematician A. Clebsch, who first suggested that they had a wide range of applications in solving discontinuous problems. Setting up singular functions

$\displaystyle{F}_{n}(x)=\langle x-a\rangle^{n}$

where $n$ is an arbitrary integer and a is a constant [3].

When $n$ is equal to 0, there is the following:

$\displaystyle\left\langle{x-a}\right\rangle^{0}=\left\{{\begin{array}[]{l}1,x% \geqslant a\\ 0,x<a\\ \end{array}}\right.$

Singular functions have the following properties:

(1)
For $n\geqslant 0$

rule:

$\displaystyle\langle x-a\rangle^{n}=\left\{{\begin{array}[]{c}\left({x-a}% \right)^{n}\\ 0\\ \end{array}}\right.\begin{array}[]{l}\left({x\geqslant a}\right)\\ \left({x<a}\right)\\ \end{array}$
(2)
For $n\geqslant 0$

rule:

$\displaystyle\int{\left\langle{x-a}\right\rangle^{n}\text{d}x=}\frac{1}{n+1}% \langle x-a\rangle^{n+1}+C$
(3)
For $n\geqslant 1$

rule:

$\displaystyle\frac{\text{d}}{\text{dx}}\langle x-a\rangle^{n}=n\langle x-a% \rangle^{n-1}$

The concept of singular function results from the efforts and contributions of numerous mathematicians in the field of mathematics. It holds significant application and research value within analytic number theory, integral transformation, physics application, and function approximation. It plays a crucial role in solving complex problems and profoundly comprehending the essence of mathematical and mechanical phenomena. The application and research value of singular function in mathematics and physics cannot be ignored. This paper introduces the use of singular function in solving the bending deformation problem in material mechanics. The unique properties of the singular function provide an advantage in the solution of complex discontinuous problems.
3. Use a singular function to express the bending moment equation of the beam

In the mechanics of materials, the integral method –sometimes referred to as the section method– is commonly applied to derive the bending moment equation for a beam. This method is advantageous when dealing with a single load’s smooth variation on a beam. However, in practical scenarios, the beam typically experiences various concentrated forces, concentrated couples, and distributed loads concurrently, necessitating the writing of multiple bending moment equations for different sections and obtaining several integral constants after integration. This calculation can be arduous [4].

The singular function exhibits a natural advantage in addressing the issue of discontinuity, by introducing a singular function based on the load of the beam, the internal force of the entire beam can be expressed directly through a function equation. After integration, at most two integral constants remain, resulting in a simplified calculation process and faster acquisition of the internal force of the beam.

This paper takes the cantilever beam as an example, and gives the following bending moment equation of the beam under common loads, which can be solved in the same way if it is an extended beam or a simply supported beam [5].

Table 1
Several common loads and bending moment equations

Serial number	Load type	Bending moment equation
1	Concentrated couple	${M(x)}=-{M}\langle x-a\rangle^{0}$
2	Concentrated force	${M(x)}=-{F}\langle x-a\rangle^{1}$
3	Uniform loading	${M(x)}=-\frac{q}{2}\langle x-a\rangle^{2}+\frac{q}{2}\langle x-{b}\rangle^{2}$
4	Linearly distributed load	${M}(x)=-\frac{k^{\prime}}{3!}\langle x-a\rangle^{3}$
5	Variable load	${M(x)}=-\frac{k^{n}}{(n+2)!}\langle x-a\rangle^{n+2}$

It is worth noting that in this paper, the equation assumes the origin of coordinates to be at the left end of the beam, and the equation follows the load sequence from left to right. The positive and negative signs in the formula only indicate the direction; if the external forces F and the distributed load q point upwards, they are positive signs and vice versa; if the external force couple M rotates clockwise, it is a positive sign and vice versa [6].

Note the third case in the table, for the uniform load $q$ , should be continuous from the left to the right end of the beam, if there is a gap in the middle, or only act on a certain interval length in the beam, the uniform load should be extended to the far right end of the beam, and the reverse uniform load q should be added at the same time in the extension part, and then the superposition principle is applied to write the bending moment equation, see Fig. 1 [7].

Figure 1.

Diagram of the calculation method of the singular function of the uniform load.

4. The singular function calculates the deformation of the beam

In mechanics of materials, the calculation of bending of beams is very important. On any section of any beam, the differential equation of bending moment $M$ , the shear force $F_{Q}$ , and load concentration $q\left(x\right)$ is:

$\displaystyle{q(x)}=\frac{{dF}_{Q}}{{dx}}=\frac{{d}^{2}{M(x)}}{{dx}^{2}}$ (1)

The differential equation of the deflection curve during its deformation is [8]:

$\displaystyle\text{EIy}{{{}^{\prime\prime}}}{=M(x)}$ (2)

In the formula, EI represents the bending stiffness of the beam while $y$ represents the deflection. By integrating the differential equation of the deflection curve, it is easy to obtain the rotation angle equation and the deflection equation. When using the singular function to express the bending moment equation of the beam, only one equation is necessary to express the bending equation of the entire beam, resulting in two integral constants obtained through integration. These constants can be calculated by using boundary conditions, which greatly simplifies the calculation process.

Example 1: Find the angle and deflection of the C section of the simple support beam in Fig. 2.

Figure 2.

Simply supported beam.

Solution: Find the support reaction force of the beam as:

$\displaystyle{F}_{\textit{RA}}=\frac{5qa}{6},{F}_{\textit{RD}}=\frac{7qa}{6}$

Then the equation for the bending moment of the beam expressed by the singular function is:

$\displaystyle{M(x)}=\frac{5qa}{6}{x}-qa^{2}\langle x-a\rangle^{0}-qa\langle x-% 2a\rangle^{1}-\frac{q}{2}\langle x-2a\rangle^{2}$ (3)

Substituting Eq. (3) into Eq. (2), integrate once to get the angle equation:

$\displaystyle{\textit{EI}\theta}=\frac{5qa}{12}x^{2}-qa^{2}\langle x-a\rangle^% {1}-\frac{qa}{2}\langle x-2a\rangle^{2}-\frac{q}{6}\langle x-2a\rangle^{3}+C$ (4)

Integrating Eq. (4) again gives the deflection equation :

$\displaystyle\textit{EIy}=\frac{5qa}{36}x^{3}-\frac{qa^{2}}{2}\langle x-a% \rangle^{2}-\frac{qa}{6}\langle x-2a\rangle^{3}-\frac{q}{24}\langle x-2a% \rangle^{4}+\textit{Cx}+D$ (5)

where $C$ and $D$ are the integration constant to be determined, which can be determined by the boundary conditions.

The boundary conditions are:

$\displaystyle x=0,{y}=0$ (6) $\displaystyle x=3a,y=0$

Substituting Eq. (6) into Eq. (5), the properties of the singular function can be calculated as follows:

$\displaystyle\textit{EI}\cdot 0=\frac{5qa}{36}\cdot 0-\frac{qa^{2}}{2}\cdot 0-% \frac{qa}{6}\cdot 0-\frac{q}{24}\cdot 0+C\cdot 0+D$ (7) $\displaystyle\textit{EI}\cdot 0=\frac{5qa}{36}\cdot(3a)^{3}-\frac{qa^{2}}{2}% \cdot(2a)^{2}-\frac{qa}{6}\cdot(a)^{3}-\frac{q}{24}\cdot a^{4}+C\cdot 3a+D$ (8)

It is easy to calculate the constant of integration as follows,

$\displaystyle\left\{{\begin{array}[]{l}C=-\frac{37}{72}qa^{3}\\ D=0\\ \end{array}}\right.$ (9)

Substituting the results of Eq. (9) into Eq. (4) and Eq. (5), we can obtain any section of the angle equation and deflection equation of the beam as follows,

$\displaystyle{\textit{EI}\theta}=\frac{5qa}{12}x^{2}-qa^{2}\langle x-a\rangle^% {1}-\frac{qa}{2}\langle x-2a\rangle^{2}-\frac{q}{6}\langle x-2a\rangle^{3}-% \frac{37}{72}qa^{3}$ (10) $\displaystyle\textit{EIy}=\frac{5qa}{36}x^{3}-\frac{qa^{2}}{2}\langle x-a% \rangle^{2}-\frac{qa}{6}\langle x-2a\rangle^{3}-\frac{q}{24}\langle x-2a% \rangle^{4}-\frac{37}{72}qa^{3}x$ (11)

From the above Eq. (3), Eq. (10), and Eq. (11), the specific numerical magnitude of the internal force and deformation of any section of the beam can be obtained, which is very simple.

Then to find the angle and deflection of the $C$ section, simply substitute $x=$ 2a into Eqs (10) and (11) to obtain,

$\displaystyle\theta|_{x=2a}=\frac{11qa^{3}}{72\textit{EI}}$ (12) $\displaystyle{y|}_{x=2a}=-\frac{5qa^{4}}{9\textit{EI}}(\downarrow)$ (13)

It can be inferred from the example that deploying the singular function to address the bending deformation of static beams necessitates only one bending moment equation, irrespective of the complexity of the applied load. This equation fully encompasses the entire beam’s internal force. After integration, only two integral constants remain, which can be swiftly acquired by amalgamating the boundary conditions. This simplifies the calculation process significantly and reduces the amount of computation necessary compared to the integral method and the section method.

5. Singular function calculates statically indeterminated beam

The statically indeterminate beam also referred to as a geometrically invariant system with redundant constraints, presents a challenge for problem-solving. The solution process cannot solely rely on the static equilibrium equation, but must also consider the structure’s deformation. To solve this problem, the deformation coordination equation and geometric relationship must be combined. This entails using several equations and can be a challenging task to resolve.

In material mechanics, the support reaction can be solved by the static equilibrium equation for statically determinated beams, and the main difficulty lies in calculating their rotation angle and deflection. In real life, most beams are statically indeterminated beams and their support reaction forces often cannot be solved by static equilibrium equations alone, which is very complicated [9].

In this paper, we introduce the singular function to simplify the calculation process of statically indeterminate beams and provide instructions on how to use it for calculation. It primarily addresses the issue of determining the support for statically indeterminate beams and subsequently resolves their internal forces and bending deformations. This highlights the efficacy of singular functions in tackling statically indeterminate beam problems [10].

Example 2: Solve the support reaction force of the super statically indeterminated beam shown in Fig. 3.

Figure 3.

Free-body diagram of statically indeterminated beam.

Solution: Take beam AB as the basic research system, and replace the reaction force with $M_{A}$ , $H_{\textit{RA}}$ , $F_{\textit{RA}}$ , and $F_{\textit{RB}}$ , as shown in Fig. 3(b) shown.

Then the equation for the bending moment of the beam expressed by the singular function is:

$\displaystyle{M}(x)=M_{A}\langle x\rangle^{0}-F_{\textit{RA}}\langle x\rangle^% {1}+F\langle x-a\rangle^{1}$ (14)

Integrate Eq. (14) once to get the angle equation,

$\displaystyle\textit{EI}\theta=M_{A}\langle x\rangle^{1}-\frac{F_{\textit{RA}}% }{2}\langle x\rangle^{2}+\frac{F}{2}\langle x-a\rangle^{2}+C$ (15)

Integrate Eq. (15) again to get the deflection equation ,

$\displaystyle\textit{EIy}=\frac{M_{A}}{2}\langle x\rangle^{2}-\frac{F_{\textit% {RA}}}{6}\langle x\rangle^{3}+\frac{F}{6}\langle x-a\rangle^{3}+\textit{Cx}+D$ (16)

where $C$ and $D$ are the integration constants to be determined, which can be determined by the boundary conditions.

The boundary conditions are:

$\displaystyle x=0,\theta=0$ $\displaystyle x=0,y=0$ (17) $\displaystyle x=l,y=0$

Similarly, according to the solution method of the statically determinate beam, Eq. (5) is substituted into Eqs (15) and (16), and the following calculation is carried out according to the properties of the singular function,

$\displaystyle\textit{EI}\cdot 0=M_{A}\cdot 0-\frac{F_{\textit{RA}}}{2}\cdot 0+% \frac{F}{2}\cdot 0+C$ (18) $\displaystyle\textit{EI}\cdot 0=\frac{M_{A}}{2}\cdot 0-\frac{F_{\textit{RA}}}{% 6}\cdot 0+\frac{F}{6}\cdot 0+C\cdot 0+D$ (19) $\displaystyle\textit{EI}\cdot 0=\frac{M_{A}}{2}\cdot l^{2}-\frac{F_{\textit{RA% }}}{6}\cdot l^{3}+\frac{F}{6}\cdot(l-a)^{3}+C\cdot l+D$ (20)

It is easy to calculate the constant of integration as follows,

$\displaystyle\left\{{\begin{array}[]{l}D=0\\ C=0\\ M_{A}l^{2}=\frac{F_{RA}}{3}l^{3}-\frac{F}{3}(l-a)^{3}\\ \end{array}}\right.$ (21)

The static equilibrium conditions are as follows,

$\displaystyle\sum F_{x}=0,H_{\textit{RA}}=0$ (22) $\displaystyle\sum F_{y}=0,F-F_{\textit{RA}}-F_{\textit{RB}}=0$ (23) $\displaystyle\sum M_{A}=0,\textit{Fa}-{F}_{\textit{RB}}l-M_{A}=0$ (24)

According to Eqs (21), (22), (23), and (24), the supporing reaction are obtained as follows,

$\displaystyle\left\{{\begin{array}[]{l}F_{\textit{RB}}=\frac{F}{2}(3\frac{a^{2% }}{l^{2}}-\frac{a^{3}}{l^{3}})\\ F_{\textit{RA}}=F(1-\frac{3a^{2}}{2l^{2}}-\frac{a^{3}}{2l^{3}})\\ M_{A}=\frac{\textit{Fl}}{2}(2\frac{a}{l}-3\frac{a^{2}}{l^{2}}+\frac{a^{3}}{l^{% 3}})\\ \end{array}}\right.$ (25)

The support reaction force of the superstatic beam can be determined [11].

By replacing Eq. (25) with Eqs (14), (15), and (16), we can derive the bending, rotation angle, and deflection equations of a statically indeterminate beam. This enables us to determine the internal force and bending deformation at any section of the beam. Explanation of technical abbreviations will be provided upon their first use.

In material mechanics, the loads of statically indeterminate beams are often complex and diverse, which makes it very difficult to solve the stress and bending deformation of statically indeterminate beams, and the calculation process is long and the calculation amount is large. In this paper, the singular function is used to solve the statically indeterminate beam, which can avoid solving the equations and reduce the difficulty of calculation, to quickly solve the bearing reaction force, and then get the internal force and bending deformation of the beam, which greatly simplifies the calculation process and is very simple [12].

6. Conclusion

In summary, by introducing the singular function into the calculation of bending deformation in material mechanics, the difficulty of calculation can be greatly reduced and the results can be calculated quickly. For statically determinate beams, the coordinate origin must be selected first, and then the bending moment equation is listed from left to right, i.e. the force of the whole beam can be expressed by a bending moment equation. The integral calculation is performed according to the properties of the singular function. Combined with the boundary conditions of the beam, the angle of rotation and deflection of the beam can be quickly obtained, which greatly simplifies the calculation process. Especially when the load on the beam is complex, only one bending moment equation is needed to represent the internal force of the whole beam. At the same time, the bending moment equation can be integrated at most twice to quickly determine the bending deformation of any section of the beam. In solving statically indeterminate beam, according to the characteristics of singular function, this paper lists the bending moment equation, rotation angle equation and deflection equation with unknown parameters, and then combines the static equilibrium equation to quickly solve the support reaction force of the beam, and then determine the internal force and bending deformation of the beam, which is very simple.

The cross-application of singular functions in mechanics is of great significance, as their use allows for more precise analysis and calculation of the mechanical properties of systems. Through the unique characteristics of singular functions, one can gain a greater comprehension and description of unusual behaviour in physical systems. This information is highly applicable in the describing of specific phenomena, such as mechanical discontinuity and mutation, which can aid a deeper understanding and analysis of the underlying mechanics.

Footnotes

Acknowledgments

This work was jointly supported by the Mechanics Innovation Fund at Hubei University of Technology and the Innovation Demonstration Base of Ecological Environment Geotechnical and Ecological Restoration of Rivers and Lakes (Grant No. 2020EJB004). The author expresses sincere gratitude to them.

Conflict of interest

The authors have no relevant financial or non-financial interests to disclose. The authors have no competing interests to declare that are relevant to the content of this article. All authors certify that they have no affiliations with or involvement in any organization or entity with any financial interest or non-financial interest in the subject matter or materials discussed in this manuscript. The authors have no financial or proprietary interests in any material discussed in this article.

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