Calculation of Trusses System in MATLAB - Multibody

1. Introduction

We fully agree with the statement in the editorial [1,2] that Computer-Aided technologies (CAx) encompass the use of computer technology to assist in designing, analyzing, and manufacturing products. Considerable progress has been made in this field, thanks to rapid advancements in science and technology. Advanced CAx tools combine various aspects of Product Lifecycle Management (PLM), including design, Finite Element Method (FEM), production, production planning, and the product itself. In terms of transitioning to the Industry 4.0 concept, the idea of a digital twin is coming to the forefront, and existing CAx systems must adapt to this trend.

In the past period, our team focused on research and development of the Universal Rotary Module (URM). Specifically, we explored the control method of the URM [3], which fully aligns with the concept of Industry 4.0. This module can serve as a fundamental building block for modular robot structures. In addressing various tasks to improve stiffness optimization of the URM construction, we arrived at the need for calculating rod systems. Therefore, we are publishing our findings in this article.

The present paper examines a statically indeterminate trusses system developed through analytical computation and subsequent use of software tools MATLAB and SOLIDWORKS. MATLAB – Simscape Multibody [4,5,6,7] has in its library – Flexible Bodies Library elements named Flexible Beam [6,7] with various cross – sections. These are flexible, slender beams capable of elastic deformation. The deformation involves additions due to extension, bending and torsion. The stiffness and inertia of the body (Flexible Beam) are calculated based on the geometry and material properties. The calculation principle employed complies with the Euler – Bernoulli beam theory [8] (pp. 174-220), [9] (pp. 8 – 66), [10]; hence it is suitable for the trusses system. Numerous programs for calculating such systems exist, many in different programming languages and differing in the purpose for which they were designed. The market offers a diverse range of options. References [11,12] present an example of a statically indeterminate planar trusses system calculated using the finite element method (FEM). This is a standalone program, with the calculation verified in ANSYS [13,14,15]. In the present paper, a model developed in the SOLIDWORKS environment using the Simulation tool is intended to verify the results. The whole design was created using the Weldments tool. The calculation also includes the visualization of the results using the specified input parameters. The analytical calculations are performed in a linear domain subject to the well – known Hooke's law. The term bar refers to a body of the same cross – section fixed at both ends using a rotary link, i.e. joints. An axis of the bar passes through the center of the joints. The point of connection of at least two bars in the center of the joints (forming a rotational link) is called a gusset or node. The gussets are the nodes where the transfer of forces to the rest of the structure occurs. The bar transmits the force in its axis through tension and compression, subject to the aforementioned Bernoulli – Navier hypothesis. In the case of compression, however, this is true only to a certain extent. This hypothesis assumes the preservation of the flatness of the cross – sections before and after deformation. In other words, after deformation, the cross – sections remain perpendicular to the deformed axis of symmetry of the cross – section. The same properties apply to the material of the beam in tension and compression. The normal stress is distributed uniformly; thus, there is no shear inside the bar. Certainly, there are several methods for the identification of yield strength in materials. More information on this can be found, for example, in this article [16]. For a higher level of material utilization, it is advisable to consider composite materials, which can also be beneficial. The production technology of such materials is described [17].

It should also be mentioned that the Bernoulli – Navier hypothesis is only accurate for rather slender bars. Under compressive loading, the so – called buckling [18] (pp.483 – 507) occurs when the bar may lose its stability and buckle out of its axis. The buckling stability of trusses systems is difficult to handle [19,20,21], especially if several parameters are monitored and optimized simultaneously [17].

2. Example of Analytical Design Based on in – Plane Deformation Conditions

The system in Figure 1 is composed of five bars. It is externally and internally statically 1x indeterminate. When one degree of freedom looseness is suitably removed at node A or D, which binds the system to the world frame, the system does not turn into a mechanism. Similarly, after removing any bars 1 to 5, the structure remains unchanged. This means that the given system is externally and internally 1x statically indeterminate. Thus, we partially relax the system and express the unknown reactions in terms of the deformation conditions.

The example mainly involves comparing theoretical results with simulations in MATLAB – Multibody and then conducting verification in SOLIDWORKS. The structure of the trusses system loaded by the force at node B after release from the bonds at node(s) A, D and release of the bar between nodes AB is displayed in Figure 2.

For the energy change of the whole system under the N_AB force [18] p. 204, [23] p. 160 in the bar, the following applies:

$${\displaystyle \frac{\partial U}{\partial N_{AB}}}=0$$

(1)

To determine the forces in individual nodes or gussets we use the Method of joint [24] p.70, [25,26]. By successively releasing the nodes, we denote the forces acting in the bars by the index of the nodes where the bar is attached.

Node – A:

Figure 3. Reactions after release from the bonds at node A and release of bar AB, AC.

Figure 3. Reactions after release from the bonds at node A and release of bar AB, AC.

In general, the following applies to the equilibrium of forces at a node:

$$\sum \mathbf{F}=0$$

(2)

By decomposing into the individual axes of the coordinate system, the following applies:

$$\sum F_x=0;F_{Ax}+{\displaystyle \frac{F_{AC}}{\sqrt{2}}}-N_{AB}=0$$

(3)

$$\sum F_y=0;F_{Ay}+{\displaystyle \frac{F_{AC}}{\sqrt{2}}}=0$$

(4)

Node – B:

Figure 4. Reactions after release from the bonds at node B and release of bar AB, BD, BC.

Figure 4. Reactions after release from the bonds at node B and release of bar AB, BD, BC.

$$\sum F_x=0;N_{AB}-F_{Bx}-{\displaystyle \frac{F_{BD}}{\sqrt{2}}}=0$$

(5)

$$\sum F_y=0;F_{BC}-F_{By}+{\displaystyle \frac{F_{BD}}{\sqrt{2}}}=0$$

(6)

Node – C:

Figure 5. Reactions after release from the bonds at node C and release of bar AC, BC, CD.

Figure 5. Reactions after release from the bonds at node C and release of bar AC, BC, CD.

$$\sum F_x=0;-F_{CD}-{\displaystyle \frac{F_{AC}}{\sqrt{2}}}=0$$

(7)

$$\sum F_y=0;-F_{BC}-{\displaystyle \frac{F_{AC}}{\sqrt{2}}}=0$$

(8)

Node – D:

Figure 6. Reactions after release from the bonds at node D and release of bar BD, CD.

Figure 6. Reactions after release from the bonds at node D and release of bar BD, CD.

$$\sum F_x=0;F_{Dx}+F_{CD}+{\displaystyle \frac{F_{BD}}{\sqrt{2}}}=0$$

(9)

$$\sum F_y=0;F_{Dy}-{\displaystyle \frac{F_{BD}}{\sqrt{2}}}=0$$

(10)

Consider the system of given equations (5), (6), (7), (8), where the variables without reactions at nodes A and D are searched for:

$$N_{\mathrm{A}\mathrm{B}}-{\displaystyle \frac{\sqrt{2}\hspace{0.17em}{F}_{\mathrm{B}\mathrm{D}}}{2}}=F_{Bx}$$

(11)

$$F_{\mathrm{B}\mathrm{C}}+{\displaystyle \frac{\sqrt{2}\hspace{0.17em}{F}_{\mathrm{B}\mathrm{D}}}{2}}=F_{By}$$

(12)

$$-F_{\mathrm{C}\mathrm{D}}-{\displaystyle \frac{\sqrt{2}\hspace{0.17em}{F}_{\mathrm{A}\mathrm{C}}}{2}}=0$$

(13)

$$-F_{\mathrm{B}\mathrm{C}}-{\displaystyle \frac{\sqrt{2}\hspace{0.17em}{F}_{\mathrm{A}\mathrm{C}}}{2}}=0$$

(14)

And its transcription into matrix entry for automated computer calculation:

$$\left[\begin{array}{ccccc}1& 0& 0& -{\displaystyle \frac{\sqrt{2}}{2}}& 0\\ 0& 0& 1& {\displaystyle \frac{\sqrt{2}}{2}}& 0\\ 0& -{\displaystyle \frac{\sqrt{2}}{2}}& 0& 0& -1\\ 0& -{\displaystyle \frac{\sqrt{2}}{2}}& -1& 0& 0\end{array}\right]·\left[\begin{array}{c}\begin{array}{c}{N}_{\mathrm{A}\mathrm{B}}\\ F_{\mathrm{A}\mathrm{C}}\end{array}\\ \begin{array}{c}{F}_{\mathrm{B}\mathrm{C}}\\ F_{\mathrm{B}\mathrm{D}}\\ F_{\mathrm{C}\mathrm{D}}\end{array}\end{array}\right]=\left[\begin{array}{c}\begin{array}{c}{F}_{\mathrm{B}\mathrm{x}}\\ F_{\mathrm{B}\mathrm{y}}\end{array}\\ \begin{array}{c}0\\ 0\end{array}\end{array}\right]$$

(15)

From system (15). The unknown forces in the bars expressed in terms of the known variables F_Bx, F_By, N_AB, are as follows:

$$F_{\mathrm{A}\mathrm{C}}=-\sqrt{2}·\left(F_{\mathrm{B}\mathrm{x}}-N_{\mathrm{A}\mathrm{B}}+F_{\mathrm{B}\mathrm{y}}\right)$$

(16)

$$F_{\mathrm{B}\mathrm{C}}=F_{\mathrm{B}\mathrm{x}}-N_{\mathrm{A}\mathrm{B}}+F_{\mathrm{B}\mathrm{y}}$$

(17)

$$F_{\mathrm{B}\mathrm{D}}=\sqrt{2}·\left(N_{\mathrm{A}\mathrm{B}}-F_{\mathrm{B}\mathrm{x}}\right)$$

(18)

$$F_{\mathrm{C}\mathrm{D}}=F_{\mathrm{B}\mathrm{x}}-N_{\mathrm{A}\mathrm{B}}+F_{\mathrm{B}\mathrm{y}}$$

(19)

Based on the deformation condition (1) regarding the energy change in the whole system under the N_AB force in the bar, the following applies:

$$\sum {\displaystyle \frac{F_i·l_i}{E_i·S_i}}·{\displaystyle \frac{\partial F_i}{\partial N_{AB}}}=0$$

(20)

The cross – sections S_i and the Young's modulus E_i are the same in the present case. The lengths l_i can be expressed as multiples of the uniform length l. Subsequently, based on the deformation condition (1) regarding the energy change in the whole system under the force N_AB in the bar AB, the following applies:

$${\displaystyle \frac{l}{E·S}}\sum F_i·{\displaystyle \frac{\partial F_i}{\partial N_{AB}}}=0$$

(21)

$$N_{\mathrm{A}\mathrm{B}}-\sqrt{2}·2·\left(F_{\mathrm{B}\mathrm{x}}-N_{\mathrm{A}\mathrm{B}}+F_{\mathrm{B}\mathrm{y}}\right)-\left(F_{\mathrm{B}\mathrm{x}}-N_{\mathrm{A}\mathrm{B}}+F_{\mathrm{B}\mathrm{y}}\right)+\sqrt{2}·2·\left(N_{\mathrm{A}\mathrm{B}}-F_{\mathrm{B}\mathrm{x}}\right)-\left(F_{\mathrm{B}\mathrm{x}}-N_{\mathrm{A}\mathrm{B}}+F_{\mathrm{B}\mathrm{y}}\right)=0$$

(22)

$$N_{\mathrm{A}\mathrm{B}}=2·{\displaystyle \frac{\left(2·\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{x}}+\left(\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{y}}}{\left(4·\sqrt{2}+3\right)}}$$

(23)

In general, the equations for external equilibrium are as follows:

$$\sum \mathbf{F}=0$$

(24)

$$\sum \mathbf{M}=0$$

(25)

After decomposing:

$$\sum F_x=0;F_{Ax}+F_{Dx}=F_{Bx}$$

(26)

$$\sum F_y=0;F_{Ay}+F_{Dy}=F_{By}$$

(27)

$$\sum M_A=0;{-F}_{Dx}·\mathcal{l}=F_{By}·\mathcal{l}$$

(28)

Alternatively, its transcription into matrix entry for automated computer calculation is as follows:

$$\left[\begin{array}{cccc}1& 0& 1& 0\\ 0& 1& 0& 1\\ 0& 0& -1& 0\end{array}\right]·\left[\begin{array}{c}\begin{array}{c}{F}_{\mathrm{A}\mathrm{x}}\\ F_{\mathrm{A}\mathrm{y}}\end{array}\\ \begin{array}{c}{F}_{\mathrm{D}\mathrm{x}}\\ F_{\mathrm{D}\mathrm{y}}\end{array}\end{array}\right]=\left[\begin{array}{c}{F}_{\mathrm{B}\mathrm{x}}\\ F_{\mathrm{B}\mathrm{y}}\\ F_{\mathrm{B}\mathrm{y}}\end{array}\right]$$

(29)

From system (29). The unknown forces in the bars expressed in terms of the known variables F_Bx, F_By, F_Ay are as follows:

$$F_{\mathrm{A}\mathrm{x}}=F_{\mathrm{B}\mathrm{x}}+F_{\mathrm{B}\mathrm{y}}$$

(30)

$$F_{\mathrm{D}x}=-F_{\mathrm{B}\mathrm{y}}$$

(31)

$$F_{\mathrm{D}\mathrm{y}}=F_{\mathrm{B}\mathrm{y}}-F_{\mathrm{A}\mathrm{y}}$$

(32)

The force F_Ay can be calculated from the equilibrium of forces of node A using equation (4) as follows:

$$F_{Ay}=-{\displaystyle \frac{F_{AC}}{\sqrt{2}}}$$

(33)

After substituting equation (16) into (33):

$$F_{Ay}=-{\displaystyle \frac{-\sqrt{2}·\left(F_{\mathrm{B}\mathrm{x}}-N_{\mathrm{A}\mathrm{B}}+F_{\mathrm{B}\mathrm{y}}\right)}{\sqrt{2}}},$$

(34)

and substituting (23) into (34), equation yields:

$$F_{Ay}=F_{\mathrm{B}\mathrm{x}}-2·{\displaystyle \frac{\left(2·\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{x}}+\left(\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{y}}}{\left(4·\sqrt{2}+3\right)}}+F_{\mathrm{B}\mathrm{y}}$$

(35)

The deflections at nodes B and C of the trusses system are determined by Castigliano's first theorem [18] (pp. 202 – 204), [23] (36) for node B and (53) for node C. This will be determined by the partial derivative of the external component of the force vector F_e acting in the direction of the deflection sought. It's important to note that the situation is simplified assuming the same cross – sections S, Young's modulus E and multiples of uniform length l.

Node – B:

Figure 7. Structure of the trusses system loaded by the force vector F_B at node B and the additional external force vector F_eB.

Figure 7. Structure of the trusses system loaded by the force vector F_B at node B and the additional external force vector F_eB.

The deflection vector δ_B of node B derived from the first Castigliano theorem can be calculated using the following equation:

$${\mathit{\delta }}_B={\displaystyle \frac{l}{E·S}}\left[\begin{array}{c}\sum F_i·{\displaystyle \frac{\partial F_i}{\partial F_{eBx}}}\\ \sum F_i·{\displaystyle \frac{\partial F_i}{\partial F_{eBy}}}\end{array}\right]$$

(36)

Each component of the force F_i is expressed as a function of the component of the external force vector F_eB as follows:

$${\mathit{\delta }}_B={\displaystyle \frac{l}{E·S}}\left[\begin{array}{c}{F}_{eBx}+N_{\mathrm{A}\mathrm{B}}·{\displaystyle \frac{\partial N_{\mathrm{A}\mathrm{B}}}{\partial F_{eBx}}}+{\sqrt{2}·F}_{\mathrm{A}\mathrm{C}}·{\displaystyle \frac{\partial F_{\mathrm{A}\mathrm{C}}}{\partial F_{eBx}}}+F_{\mathrm{B}\mathrm{C}}·{\displaystyle \frac{\partial F_{\mathrm{B}\mathrm{C}}}{\partial F_{eBx}}}+{\sqrt{2}·F}_{\mathrm{B}\mathrm{D}}·{\displaystyle \frac{\partial F_{\mathrm{B}\mathrm{D}}}{\partial F_{eBx}}}+F_{\mathrm{C}\mathrm{D}}·{\displaystyle \frac{\partial F_{\mathrm{C}\mathrm{D}}}{\partial F_{eBx}}}\\ F_{eBy}+N_{\mathrm{A}\mathrm{B}}·{\displaystyle \frac{\partial N_{\mathrm{A}\mathrm{B}}}{\partial F_{eBy}}}+{\sqrt{2}·F}_{\mathrm{A}\mathrm{C}}·{\displaystyle \frac{\partial F_{\mathrm{A}\mathrm{C}}}{\partial F_{eBy}}}+F_{\mathrm{B}\mathrm{C}}·{\displaystyle \frac{\partial F_{\mathrm{B}\mathrm{C}}}{\partial F_{eBy}}}+\sqrt{2}·F_{\mathrm{B}\mathrm{D}}·{\displaystyle \frac{\partial F_{\mathrm{B}\mathrm{D}}}{\partial F_{eBy}}}+F_{\mathrm{C}\mathrm{D}}·{\displaystyle \frac{\partial F_{\mathrm{C}\mathrm{D}}}{\partial F_{eBy}}}\end{array}\right]$$

(37)

By substituting the corresponding component of the external force vector F_eB, into equation (23) derived from the deformation condition, the equation reads:

$$N_{\mathrm{A}\mathrm{B}}=2·{\displaystyle \frac{\left(2·\sqrt{2}+1\right)·\left(F_{\mathrm{B}\mathrm{x}}+F_{\mathrm{e}\mathrm{B}\mathrm{x}}\right)+\left(\sqrt{2}+1\right)·\left(F_{\mathrm{B}\mathrm{y}}+F_{\mathrm{e}\mathrm{B}\mathrm{y}}\right)}{\left(4·\sqrt{2}+3\right)}},$$

(38)

Substituting N_AB into equations (16) to (19) yields the equation as follows:

$$F_{\mathrm{A}\mathrm{C}}=-\sqrt{2}·\left(F_{\mathrm{B}\mathrm{x}}+F_{\mathrm{e}\mathrm{B}\mathrm{x}}-2·{\displaystyle \frac{\left(2·\sqrt{2}+1\right)·\left(F_{\mathrm{B}\mathrm{x}}+F_{eBx}\right)+\left(\sqrt{2}+1\right)·\left(F_{\mathrm{B}\mathrm{y}}+F_{\mathrm{e}\mathrm{B}\mathrm{y}}\right)}{\left(4·\sqrt{2}+3\right)}}+F_{\mathrm{B}\mathrm{y}}+F_{\mathrm{e}\mathrm{B}\mathrm{y}}\right)$$

(39)

$$F_{\mathrm{B}\mathrm{C}}=F_{\mathrm{B}\mathrm{x}}+F_{\mathrm{e}\mathrm{B}\mathrm{x}}-2·{\displaystyle \frac{\left(2·\sqrt{2}+1\right)·\left(F_{\mathrm{B}\mathrm{x}}+F_{\mathrm{e}\mathrm{B}\mathrm{x}}\right)+\left(\sqrt{2}+1\right)·\left(F_{\mathrm{B}\mathrm{y}}+F_{\mathrm{e}\mathrm{B}\mathrm{y}}\right)}{\left(4·\sqrt{2}+3\right)}}+F_{\mathrm{B}\mathrm{y}}+F_{\mathrm{e}\mathrm{B}\mathrm{y}}$$

(40)

$$F_{\mathrm{B}\mathrm{D}}=\sqrt{2}·\left(2·{\displaystyle \frac{\left(2·\sqrt{2}+1\right)·\left(F_{\mathrm{B}\mathrm{x}}+F_{\mathrm{e}\mathrm{B}\mathrm{x}}\right)+\left(\sqrt{2}+1\right)·\left(F_{\mathrm{B}\mathrm{y}}+F_{\mathrm{e}\mathrm{B}\mathrm{y}}\right)}{\left(4·\sqrt{2}+3\right)}}-F_{\mathrm{B}\mathrm{x}}-F_{\mathrm{e}\mathrm{B}\mathrm{x}}\right)$$

(41)

$$F_{\mathrm{C}\mathrm{D}}=F_{\mathrm{B}\mathrm{x}}+F_{\mathrm{e}\mathrm{B}\mathrm{x}}-2·{\displaystyle \frac{\left(2·\sqrt{2}+1\right)·\left(F_{\mathrm{B}\mathrm{x}}+F_{\mathrm{e}\mathrm{B}\mathrm{x}}\right)+\left(\sqrt{2}+1\right)·\left(F_{\mathrm{B}\mathrm{y}}+F_{\mathrm{e}\mathrm{B}\mathrm{y}}\right)}{\left(4·\sqrt{2}+3\right)}}+F_{\mathrm{B}\mathrm{y}}+F_{\mathrm{e}\mathrm{B}\mathrm{y}}$$

(42)

The individual elements of the matrix in equation (37) after deriving by F_eBx are as follows:

$$N_{\mathrm{A}\mathrm{B}}·{\displaystyle \frac{\partial N_{\mathrm{A}\mathrm{B}}}{\partial F_{eBx}}}=4·{\displaystyle \frac{\left(2·\sqrt{2}+1\right)·\left(F_{\mathrm{B}\mathrm{x}}+F_{\mathrm{e}\mathrm{B}\mathrm{x}}\right)+\left(\sqrt{2}+1\right)·\left(F_{\mathrm{B}\mathrm{y}}+F_{\mathrm{e}\mathrm{B}\mathrm{y}}\right)}{\left(4·\sqrt{2}+3\right)}}·{\displaystyle \frac{\left(2·\sqrt{2}+1\right)}{\left(4·\sqrt{2}+3\right)}}$$

(43)

$$F_{\mathrm{A}\mathrm{C}}·{\displaystyle \frac{\partial F_{\mathrm{A}\mathrm{C}}}{\partial F_{eBx}}}=2·\left(F_{\mathrm{B}\mathrm{x}}+F_{\mathrm{e}\mathrm{B}\mathrm{x}}-2·{\displaystyle \frac{\left(2·\sqrt{2}+1\right)·\left(F_{\mathrm{B}\mathrm{x}}+F_{\mathrm{e}\mathrm{B}\mathrm{x}}\right)+\left(\sqrt{2}+1\right)·\left(F_{\mathrm{B}\mathrm{y}}+F_{\mathrm{e}\mathrm{B}\mathrm{y}}\right)}{\left(4·\sqrt{2}+3\right)}}+F_{\mathrm{B}\mathrm{y}}+F_{\mathrm{e}\mathrm{B}\mathrm{y}}\right)·\left(1-2·{\displaystyle \frac{\left(2·\sqrt{2}+1\right)}{\left(4·\sqrt{2}+3\right)}}\right)$$

(44)

$$F_{\mathrm{B}\mathrm{C}}·{\displaystyle \frac{\partial F_{\mathrm{B}\mathrm{C}}}{\partial F_{eBx}}}=\left(F_{Bx}+F_{\mathrm{e}\mathrm{B}\mathrm{x}}-2·{\displaystyle \frac{\left(2·\sqrt{2}+1\right)·\left(F_{\mathrm{B}\mathrm{x}}+F_{\mathrm{e}\mathrm{B}\mathrm{x}}\right)+\left(\sqrt{2}+1\right)·\left(F_{\mathrm{B}\mathrm{y}}+F_{\mathrm{e}\mathrm{B}\mathrm{y}}\right)}{\left(4·\sqrt{2}+3\right)}}+F_{\mathrm{B}\mathrm{y}}+F_{\mathrm{e}\mathrm{B}\mathrm{y}}\right)·\left(1-2·{\displaystyle \frac{\left(2·\sqrt{2}+1\right)}{\left(4·\sqrt{2}+3\right)}}\right)$$

(45)

$$F_{\mathrm{B}\mathrm{D}}·{\displaystyle \frac{\partial F_{\mathrm{B}\mathrm{D}}}{\partial F_{eBx}}}=2·\left(2·{\displaystyle \frac{\left(2·\sqrt{2}+1\right)·\left(F_{\mathrm{B}\mathrm{x}}+F_{\mathrm{e}\mathrm{B}\mathrm{x}}\right)+\left(\sqrt{2}+1\right)·\left(F_{\mathrm{B}\mathrm{y}}+F_{\mathrm{e}\mathrm{B}\mathrm{y}}\right)}{\left(4·\sqrt{2}+3\right)}}-F_{Bx}-F_{\mathrm{e}\mathrm{B}\mathrm{x}}\right)·\left(2·{\displaystyle \frac{\left(2·\sqrt{2}+1\right)}{\left(4·\sqrt{2}+3\right)}}-1\right)$$

(46)

$$F_{\mathrm{C}\mathrm{D}}·{\displaystyle \frac{\partial F_{\mathrm{C}\mathrm{D}}}{\partial F_{eBx}}}=\left(F_{\mathrm{B}\mathrm{x}}+F_{\mathrm{e}\mathrm{B}\mathrm{x}}-2·{\displaystyle \frac{\left(2·\sqrt{2}+1\right)·\left(F_{\mathrm{B}\mathrm{x}}+F_{\mathrm{e}\mathrm{B}\mathrm{x}}\right)+\left(\sqrt{2}+1\right)·\left(F_{\mathrm{B}\mathrm{y}}+F_{\mathrm{e}\mathrm{B}\mathrm{y}}\right)}{\left(4·\sqrt{2}+3\right)}}+F_{\mathrm{B}\mathrm{y}}+F_{\mathrm{e}\mathrm{B}\mathrm{y}}\right)·\left(1-2·{\displaystyle \frac{\left(2·\sqrt{2}+1\right)}{\left(4·\sqrt{2}+3\right)}}\right)$$

(47)

The individual elements of the matrix in equation (37) after deriving by F_eBy are as follows:

$$N_{\mathrm{A}\mathrm{B}}·{\displaystyle \frac{\partial N_{\mathrm{A}\mathrm{B}}}{\partial F_{eBy}}}=4·{\displaystyle \frac{\left(2·\sqrt{2}+1\right)·\left(F_{\mathrm{B}\mathrm{x}}+F_{\mathrm{e}\mathrm{B}\mathrm{x}}\right)+\left(\sqrt{2}+1\right)·\left(F_{By}+F_{\mathrm{e}\mathrm{B}\mathrm{y}}\right)}{\left(4·\sqrt{2}+3\right)}}·{\displaystyle \frac{\left(\sqrt{2}+1\right)}{\left(4·\sqrt{2}+3\right)}}$$

(48)

$$F_{\mathrm{A}\mathrm{C}}·{\displaystyle \frac{\partial F_{\mathrm{A}\mathrm{C}}}{\partial F_{eBy}}}=2·\left(F_{\mathrm{B}\mathrm{x}}+F_{\mathrm{e}\mathrm{B}\mathrm{x}}-2·{\displaystyle \frac{\left(2·\sqrt{2}+1\right)·\left(F_{\mathrm{B}\mathrm{x}}+F_{\mathrm{e}\mathrm{B}\mathrm{x}}\right)+\left(\sqrt{2}+1\right)·\left(F_{\mathrm{B}\mathrm{y}}+F_{\mathrm{e}\mathrm{B}\mathrm{y}}\right)}{\left(4·\sqrt{2}+3\right)}}+F_{\mathrm{B}\mathrm{y}}+F_{\mathrm{e}\mathrm{B}\mathrm{y}}\right)·\left(1-2·{\displaystyle \frac{\left(\sqrt{2}+1\right)}{\left(4·\sqrt{2}+3\right)}}\right)$$

(49)

$$F_{\mathrm{B}\mathrm{C}}·{\displaystyle \frac{\partial F_{\mathrm{B}\mathrm{C}}}{\partial F_{eBy}}}=\left(F_{Bx}+F_{\mathrm{e}\mathrm{B}\mathrm{x}}-2·{\displaystyle \frac{\left(2·\sqrt{2}+1\right)·\left(F_{\mathrm{B}\mathrm{x}}+F_{\mathrm{e}\mathrm{B}\mathrm{x}}\right)+\left(\sqrt{2}+1\right)·\left(F_{\mathrm{B}\mathrm{y}}+F_{\mathrm{e}\mathrm{B}\mathrm{y}}\right)}{\left(4·\sqrt{2}+3\right)}}+F_{\mathrm{B}\mathrm{y}}+F_{\mathrm{e}\mathrm{B}\mathrm{y}}\right)·\left(1-2·{\displaystyle \frac{\left(\sqrt{2}+1\right)}{\left(4·\sqrt{2}+3\right)}}\right)$$

(50)

$$F_{\mathrm{B}\mathrm{D}}·{\displaystyle \frac{\partial F_{\mathrm{B}\mathrm{D}}}{\partial F_{eBy}}}=2·\left(2·{\displaystyle \frac{\left(2·\sqrt{2}+1\right)·\left(F_{\mathrm{B}\mathrm{x}}+F_{\mathrm{e}\mathrm{B}\mathrm{x}}\right)+\left(\sqrt{2}+1\right)·\left(F_{\mathrm{B}\mathrm{y}}+F_{\mathrm{e}\mathrm{B}\mathrm{y}}\right)}{\left(4·\sqrt{2}+3\right)}}-F_{Bx}-F_{\mathrm{e}\mathrm{B}\mathrm{x}}\right)·\left(2·{\displaystyle \frac{\left(\sqrt{2}+1\right)}{\left(4·\sqrt{2}+3\right)}}\right)$$

(51)

$$F_{\mathrm{C}\mathrm{D}}·{\displaystyle \frac{\partial F_{\mathrm{C}\mathrm{D}}}{\partial F_{eBy}}}=\left(F_{\mathrm{B}\mathrm{x}}+F_{\mathrm{e}\mathrm{B}\mathrm{x}}-2·{\displaystyle \frac{\left(2·\sqrt{2}+1\right)·\left(F_{\mathrm{B}\mathrm{x}}+F_{\mathrm{e}\mathrm{B}\mathrm{x}}\right)+\left(\sqrt{2}+1\right)·\left(F_{\mathrm{B}\mathrm{y}}+F_{\mathrm{e}\mathrm{B}\mathrm{y}}\right)}{\left(4·\sqrt{2}+3\right)}}+F_{\mathrm{B}\mathrm{y}}+F_{\mathrm{e}\mathrm{B}\mathrm{y}}\right)·\left(1-2·{\displaystyle \frac{\left(\sqrt{2}+1\right)}{\left(4·\sqrt{2}+3\right)}}\right)$$

(52)

In equations (37), (43) to (52), under the theorem, zero value is substituted for the vector F_eB (F_eB = 0) to obtain the resulting deflection δ_B of node B.

Node – C:

Figure 8. Structure of the trusses system loaded by the external force vector F_eC at node C.

Figure 8. Structure of the trusses system loaded by the external force vector F_eC at node C.

The deflection vector δ_C of node C derived from the first Castigliano theorem can be calculated using the following equation:

$${\mathit{\delta }}_C={\displaystyle \frac{l}{E·S}}\left[\begin{array}{c}\sum F_i·{\displaystyle \frac{\partial F_i}{\partial F_{eCx}}}\\ \sum F_i·{\displaystyle \frac{\partial F_i}{\partial F_{eCy}}}\end{array}\right]$$

(53)

Each component of the force F_i is expressed as a function of the component of the external force vector F_eC as follows:

$${\mathit{\delta }}_C={\displaystyle \frac{l}{E·S}}\left[\begin{array}{c}{F}_{eCx}+N_{\mathrm{A}\mathrm{B}}·{\displaystyle \frac{\partial N_{\mathrm{A}\mathrm{B}}}{\partial F_{eCx}}}+\sqrt{2}·F_{AC}·{\displaystyle \frac{\partial F_{AC}}{\partial F_{eCx}}}+F_{BC}·{\displaystyle \frac{\partial F_{BC}}{\partial F_{eCx}}}+{\sqrt{2}·F}_{BD}·{\displaystyle \frac{\partial F_{BD}}{\partial F_{eCx}}}+F_{CD}·{\displaystyle \frac{\partial F_{CD}}{\partial F_{eCx}}}\\ F_{eCy}+N_{\mathrm{A}\mathrm{B}}·{\displaystyle \frac{\partial N_{\mathrm{A}\mathrm{B}}}{\partial F_{eCy}}}+{\sqrt{2}·F}_{AC}·{\displaystyle \frac{\partial F_{AC}}{\partial F_{eCy}}}+F_{BC}·{\displaystyle \frac{\partial F_{BC}}{\partial F_{eCy}}}+\sqrt{2}·F_{BD}·{\displaystyle \frac{\partial F_{BD}}{\partial F_{eCy}}}+F_{CD}·{\displaystyle \frac{\partial F_{CD}}{\partial F_{eCy}}}\end{array}\right]$$

(54)

Consider again the system of the given equations (5), (6), (7), (8) with the additional vector of the external force F_eC:

$$N_{\mathrm{A}\mathrm{B}}-{\displaystyle \frac{\sqrt{2}\hspace{0.17em}{F}_{\mathrm{B}\mathrm{D}}}{2}}=F_{Bx}$$

(55)

$$F_{\mathrm{B}\mathrm{C}}+{\displaystyle \frac{\sqrt{2}\hspace{0.17em}{F}_{\mathrm{B}\mathrm{D}}}{2}}=F_{By}$$

(56)

$$-F_{\mathrm{C}\mathrm{D}}-F_{\mathrm{e}\mathrm{C}\mathrm{x}}-{\displaystyle \frac{\sqrt{2}\hspace{0.17em}{F}_{\mathrm{A}\mathrm{C}}}{2}}=0$$

(57)

$$-F_{\mathrm{B}\mathrm{C}}-F_{\mathrm{e}\mathrm{C}\mathrm{y}}-{\displaystyle \frac{\sqrt{2}\hspace{0.17em}{F}_{\mathrm{A}\mathrm{C}}}{2}}=0$$

(58)

And its transcription into matrix entry for automated computer calculation as follows:

$$\left[\begin{array}{ccccccc}1& 0& 0& -{\displaystyle \frac{\sqrt{2}}{2}}& 0& 0& 0\\ 0& 0& 1& {\displaystyle \frac{\sqrt{2}}{2}}& 0& 0& 0\\ 0& -{\displaystyle \frac{\sqrt{2}}{2}}& 0& 0& -1& -1& 0\\ 0& -{\displaystyle \frac{\sqrt{2}}{2}}& -1& 0& 0& 0& -1\end{array}\right]·\left[\begin{array}{c}{N}_{\mathrm{A}\mathrm{B}}\\ F_{\mathrm{A}\mathrm{C}}\\ F_{\mathrm{B}\mathrm{C}}\\ F_{\mathrm{B}\mathrm{D}}\\ F_{\mathrm{C}\mathrm{D}}\\ F_{\mathrm{e}\mathrm{C}\mathrm{x}}\\ F_{\mathrm{e}\mathrm{C}\mathrm{y}}\end{array}\right]=\left[\begin{array}{c}\begin{array}{c}{F}_{\mathrm{B}\mathrm{x}}\\ F_{\mathrm{B}\mathrm{y}}\end{array}\\ \begin{array}{c}0\\ 0\end{array}\end{array}\right]$$

(59)

From system (59). The unknown forces in the bars expressed in terms of the known variables F_Bx, F_By, N_AB, F_eCx , F_eCy are as follows:

$$F_{\mathrm{A}\mathrm{C}}=-\sqrt{2}·\left(F_{\mathrm{B}\mathrm{x}}-N_{\mathrm{A}\mathrm{B}}+F_{\mathrm{B}\mathrm{y}}+F_{\mathrm{e}\mathrm{C}\mathrm{y}}\right)$$

(60)

$$F_{\mathrm{B}\mathrm{C}}=F_{\mathrm{B}\mathrm{x}}-N_{\mathrm{A}\mathrm{B}}+F_{\mathrm{B}\mathrm{y}}$$

(61)

$$F_{\mathrm{B}\mathrm{D}}=\sqrt{2}·\left(N_{\mathrm{A}\mathrm{B}}-F_{\mathrm{B}\mathrm{x}}\right)$$

(62)

$$F_{\mathrm{C}\mathrm{D}}=F_{\mathrm{B}\mathrm{x}}-N_{\mathrm{A}\mathrm{B}}+F_{\mathrm{B}\mathrm{y}}-F_{\mathrm{e}\mathrm{C}\mathrm{x}}+F_{\mathrm{e}\mathrm{C}\mathrm{y}}$$

(63)

Based on the deformation condition (1), where the cross – sections S_i = S, Young's modules E_i = E, and the lengths l_i are expressed as multiples of a uniform length l, the following is true for the energy change in the whole system under the N_AB force:

$${\displaystyle \frac{l}{E·S}}\sum F_i·{\displaystyle \frac{\partial F_i}{\partial N_{\mathrm{A}\mathrm{B}}}}=0$$

(64)

$$N_{\mathrm{A}\mathrm{B}}-\sqrt{2}·2·\left(F_{\mathrm{B}\mathrm{x}}-N_{\mathrm{A}\mathrm{B}}+F_{\mathrm{B}\mathrm{y}}+F_{\mathrm{e}\mathrm{C}\mathrm{y}}\right)-\left(F_{\mathrm{B}\mathrm{x}}-N_{\mathrm{A}\mathrm{B}}+F_{\mathrm{B}\mathrm{y}}\right)+\sqrt{2}·2·\left(N_{\mathrm{A}\mathrm{B}}-F_{\mathrm{B}\mathrm{x}}\right)-\left(F_{\mathrm{B}\mathrm{x}}-N_{\mathrm{A}\mathrm{B}}+F_{\mathrm{B}\mathrm{y}}-F_{\mathrm{e}\mathrm{C}\mathrm{x}}+F_{\mathrm{e}\mathrm{C}\mathrm{y}}\right)=0$$

(65)

$$N_{\mathrm{A}\mathrm{B}}={\displaystyle \frac{2·\left(\left(2·\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{x}}+\left(\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{y}}\right)-F_{\mathrm{e}\mathrm{C}\mathrm{x}}+\left(2·\sqrt{2}+1\right)·F_{\mathrm{e}\mathrm{C}\mathrm{y}}}{\left(4·\sqrt{2}+3\right)}}$$

(66)

By substituting N_AB from equation (66) into the system of equations (60) to (63), the following is derived:

$$F_{\mathrm{A}\mathrm{C}}=-\sqrt{2}·\left(F_{\mathrm{B}\mathrm{x}}-{\displaystyle \frac{2·\left(\left(2·\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{x}}+\left(\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{y}}\right)-F_{\mathrm{e}\mathrm{C}\mathrm{x}}-\left(2·\sqrt{2}+2\right)·F_{\mathrm{e}\mathrm{C}\mathrm{y}}}{\left(4·\sqrt{2}+3\right)}}+F_{\mathrm{B}\mathrm{y}}\right)$$

(67)

$$F_{\mathrm{B}\mathrm{C}}=F_{\mathrm{B}\mathrm{x}}-{\displaystyle \frac{2·\left(\left(2·\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{x}}+\left(\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{y}}\right)-F_{\mathrm{e}\mathrm{C}\mathrm{x}}+\left(2·\sqrt{2}+1\right)·F_{\mathrm{e}\mathrm{C}\mathrm{y}}}{\left(4·\sqrt{2}+3\right)}}+F_{\mathrm{B}\mathrm{y}}$$

(68)

$$F_{\mathrm{B}\mathrm{D}}=\sqrt{2}·\left({\displaystyle \frac{2·\left(\left(2·\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{x}}+\left(\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{y}}\right)-F_{\mathrm{e}\mathrm{C}\mathrm{x}}+\left(2·\sqrt{2}+1\right)·F_{\mathrm{e}\mathrm{C}\mathrm{y}}}{\left(4·\sqrt{2}+3\right)}}-F_{\mathrm{B}\mathrm{x}}\right)$$

(69)

$$F_{\mathrm{C}\mathrm{D}}=F_{\mathrm{B}\mathrm{x}}-{\displaystyle \frac{2·\left(\left(2·\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{x}}+\left(\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{y}}\right)+\left(4·\sqrt{2}+2\right)·F_{\mathrm{e}\mathrm{C}\mathrm{x}}-2·\left(\sqrt{2}+1\right)·F_{\mathrm{e}\mathrm{C}\mathrm{y}}}{\left(4·\sqrt{2}+3\right)}}+F_{\mathrm{B}\mathrm{y}}$$

(70)

The individual elements of the matrix in equation (54) after deriving by F_eCx are as follows:

$$N_{\mathrm{A}\mathrm{B}}·{\displaystyle \frac{\partial N_{\mathrm{A}\mathrm{B}}}{\partial F_{eCx}}}=-{\displaystyle \frac{2·\left(\left(2·\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{x}}+\left(\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{y}}\right)-F_{\mathrm{e}\mathrm{C}\mathrm{x}}+\left(2·\sqrt{2}+1\right)·F_{\mathrm{e}\mathrm{C}\mathrm{y}}}{{\left(4·\sqrt{2}+3\right)}^2}}$$

(71)

$$F_{AC}·{\displaystyle \frac{\partial F_{AC}}{\partial F_{eCx}}}=\left(F_{\mathrm{B}\mathrm{x}}-{\displaystyle \frac{2·\left(\left(2·\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{x}}+\left(\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{y}}\right)-F_{\mathrm{e}\mathrm{C}\mathrm{x}}+\left(2·\sqrt{2}+1\right)·F_{\mathrm{e}\mathrm{C}\mathrm{y}}}{\left(4·\sqrt{2}+3\right)}}+F_{\mathrm{B}\mathrm{y}}+F_{\mathrm{e}\mathrm{C}\mathrm{y}}\right)·{\displaystyle \frac{2}{\left(4·\sqrt{2}+3\right)}}$$

(72)

$$F_{BC}·{\displaystyle \frac{\partial F_{BC}}{\partial F_{eCx}}}=\left(F_{\mathrm{B}\mathrm{x}}-{\displaystyle \frac{2·\left(\left(2·\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{x}}+\left(\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{y}}\right)-F_{\mathrm{e}\mathrm{C}\mathrm{x}}+\left(2·\sqrt{2}+1\right)·F_{\mathrm{e}\mathrm{C}\mathrm{y}}}{\left(4·\sqrt{2}+3\right)}}+F_{\mathrm{B}\mathrm{y}}\right)·{\displaystyle \frac{1}{\left(4·\sqrt{2}+3\right)}}$$

(73)

$$F_{\mathrm{B}\mathrm{D}}·{\displaystyle \frac{\partial F_{\mathrm{B}\mathrm{D}}}{\partial F_{eCx}}}=-\left({\displaystyle \frac{2·\left(\left(2·\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{x}}+\left(\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{y}}\right)-F_{\mathrm{e}\mathrm{C}\mathrm{x}}+\left(2·\sqrt{2}+1\right)·F_{\mathrm{e}\mathrm{C}\mathrm{y}}}{\left(4·\sqrt{2}+3\right)}}-F_{\mathrm{B}\mathrm{x}}\right)·{\displaystyle \frac{2}{\left(4·\sqrt{2}+3\right)}}$$

(74)

$$F_{\mathrm{C}\mathrm{D}}·{\displaystyle \frac{\partial F_{\mathrm{C}\mathrm{D}}}{\partial F_{eCx}}}=-\left(F_{\mathrm{B}\mathrm{x}}-{\displaystyle \frac{2·\left(\left(2·\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{x}}+\left(\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{y}}\right)+\left(4·\sqrt{2}+2\right)·F_{\mathrm{e}\mathrm{C}\mathrm{x}}-2·\left(\sqrt{2}+1\right)·F_{\mathrm{e}\mathrm{C}\mathrm{y}}}{\left(4·\sqrt{2}+3\right)}}+F_{\mathrm{B}\mathrm{y}}\right)·{\displaystyle \frac{\left(4·\sqrt{2}+2\right)}{\left(4·\sqrt{2}+3\right)}}$$

(75)

The individual elements of the matrix in equation (54) after deriving by F_eCy read as follows:

$$N_{\mathrm{A}\mathrm{B}}·{\displaystyle \frac{\partial N_{\mathrm{A}\mathrm{B}}}{\partial F_{eCy}}}=\left({\displaystyle \frac{2·\left(\left(2·\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{x}}+\left(\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{y}}\right)-F_{\mathrm{e}\mathrm{C}\mathrm{x}}+\left(2·\sqrt{2}+1\right)·F_{\mathrm{e}\mathrm{C}\mathrm{y}}}{\left(4·\sqrt{2}+3\right)}}\right)·{\displaystyle \frac{\left(2·\sqrt{2}+1\right)}{\left(4·\sqrt{2}+3\right)}}$$

(76)

$$F_{\mathrm{A}\mathrm{C}}·{\displaystyle \frac{\partial F_{\mathrm{A}\mathrm{C}}}{\partial F_{eCy}}}=\left(F_{\mathrm{B}\mathrm{x}}-{\displaystyle \frac{2·\left(\left(2·\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{x}}+\left(\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{y}}\right)-F_{\mathrm{e}\mathrm{C}\mathrm{x}}-\left(2·\sqrt{2}+2\right)·F_{\mathrm{e}\mathrm{C}\mathrm{y}}}{\left(4·\sqrt{2}+3\right)}}+F_{\mathrm{B}\mathrm{y}}\right)·{\displaystyle \frac{2·\left(2·\sqrt{2}+2\right)}{\left(4·\sqrt{2}+3\right)}}$$

(77)

$$F_{\mathrm{B}\mathrm{C}}·{\displaystyle \frac{\partial F_{\mathrm{B}\mathrm{C}}}{\partial F_{eCy}}}=-\left(F_{\mathrm{B}\mathrm{x}}-{\displaystyle \frac{2·\left(\left(2·\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{x}}+\left(\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{y}}\right)-F_{\mathrm{e}\mathrm{C}\mathrm{x}}+\left(2·\sqrt{2}+1\right)·F_{\mathrm{e}\mathrm{C}\mathrm{y}}}{\left(4·\sqrt{2}+3\right)}}+F_{\mathrm{B}\mathrm{y}}\right)·{\displaystyle \frac{\left(2·\sqrt{2}+1\right)}{\left(4·\sqrt{2}+3\right)}}$$

(78)

$$F_{\mathrm{B}\mathrm{D}}·{\displaystyle \frac{\partial F_{\mathrm{B}\mathrm{D}}}{\partial F_{eCy}}}=\left({\displaystyle \frac{2·\left(\left(2·\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{x}}+\left(\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{y}}\right)-F_{\mathrm{e}\mathrm{C}\mathrm{x}}+\left(2·\sqrt{2}+1\right)·F_{\mathrm{e}\mathrm{C}\mathrm{y}}}{\left(4·\sqrt{2}+3\right)}}-F_{\mathrm{B}\mathrm{x}}\right)·{\displaystyle \frac{2·\left(2·\sqrt{2}+1\right)}{\left(4·\sqrt{2}+3\right)}}$$

(79)

$$F_{\mathrm{C}\mathrm{D}}·{\displaystyle \frac{\partial F_{\mathrm{C}\mathrm{D}}}{\partial F_{eCy}}}=\left(F_{\mathrm{B}\mathrm{x}}-{\displaystyle \frac{2·\left(\left(2·\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{x}}+\left(\sqrt{2}+1\right)·F_{\mathrm{B}\mathrm{y}}\right)+\left(4·\sqrt{2}+2\right)·F_{\mathrm{e}\mathrm{C}\mathrm{x}}-2·\left(\sqrt{2}+1\right)·F_{\mathrm{e}\mathrm{C}\mathrm{y}}}{\left(4·\sqrt{2}+3\right)}}+F_{\mathrm{B}\mathrm{y}}\right)·{\displaystyle \frac{2·\left(\sqrt{2}+1\right)}{\left(4·\sqrt{2}+3\right)}}$$

(80)

In equations (54), (71) to (80), following the theorem, a zero value is substituted into for the vector F_eC (F_eC = 0) to obtain the resulting deflection δ_C of the node C.

3. Building a Trusses Structure in MATLAB – Simscape Multibody

The Simscape Multibody environment [4,5,6,7] enables simulations of mechanical systems in space. It formulates and solves the equations of motion for the entire mechanical system and also provides visualization of the system dynamics. This allows for the examination of various dynamic and static processes on the model. Examining kinematic and dynamic quantities on advanced models can be quite challenging. Dealing with spatial assignments can be tough without an understanding of vector calculus and rotation matrix representations. The probability of error in derivation and computation is considerably high. Utilizing software tools with predefined blocks representing physical solids can assist in resolving such assignments. There are different programs available for calculating trusses systems and various tools for simulating the behavior of structures made from various materials. In such simulation program environments, there are different blocks such as "Flexible Beams". These blocks simulate a slender bar with constant cross – section with small and linear deformations, including extension, bending and torsion. Thus, a flexible body can be represented as a system composed of multiple bodies (multibody) [27]. In the following Figure 9, the model of the trusses system from Figure 1 is loaded with the external force vector F_B at node B with the calculation results. The following values have been defined:

F_B = [-2, -10]^T kN, bar lenghts l = 1 m, l_AB = l, l_BC = l, l_CD = l, l_AC = √2∙l m, l_BD = √2∙l m, bar diameters are the same with d = 0.01 m. Young’s modulus E is 210 GPa, shear modulus G is 79 GPa and density is 7800 kg/m³. Poisson’s ratio is 0.28. The values of proportional or Rayleigh damping are left "default" with respect to computational practicality b_m = 0.01 s^-1, b_k = 0.01 s. The Rayleigh model [28,29,30] defines the damping matrix C as a linear combination of the mass matrix M and the stiffness matrix K as follows:

$$\mathbf{C}=b_m·\mathbf{M}+b_k·\mathbf{K},$$

(81)

with b_m being a mass coefficient and b_k being a stiffness coefficient. These are the scalar coefficients. The simulation time is also "default" in the model settings and equals t = 10 s.

As already mentioned, for the sake of simplicity, so that we do not have to adjust and adapt the values of t, b_m, b_k to the level of external load in the form of force F_B and cross – section S_i, the so – called multiplication constant – con, is introduced. Its role is to proportionally shift the simulation to the order of units depending on the value of the external force vector F_B and also the uniform cross – section of the bar S_i. Practically, we can also start from the normal stress in the most stressed bar if the cross – sections of the S_i bars of a given structure would be different. This avoids unnecessary changes in the defined coefficients b_m for the mass coefficient M, b_k for the stiffness coefficient K in equation (81) in the Rayleigh damping model. For the multiplication constant, the following applies:

$$con={\displaystyle \frac{‖{\mathbf{F}}_{\mathrm{B}}‖}{1200{·S}_i}}$$

(82)

This simple modification ensured an increase of two to four orders of magnitude in the accuracy of the calculation in accordance with theory, within an unlimited range of external F_B force loading. This multiplication constant can be optimized for best accuracy based on equations (5) and (6), provided that the following holds with the smallest possible deviation:

$$F_{Bx}=N_{\mathrm{A}\mathrm{B}}-{\displaystyle \frac{F_{BD}}{\sqrt{2}}}$$

(83)

$$F_{By}=F_{BC}+{\displaystyle \frac{F_{BD}}{\sqrt{2}}},$$

(84)

The magnitudes of the components in the force vector F_B are known to us, and the other forces are computational outputs.

Figure 9. Structure of the trusses system loaded by the external force vector F_B at node B in MATLAB – Simscape Multibody, based on the Figure 1.

Figure 9. Structure of the trusses system loaded by the external force vector F_B at node B in MATLAB – Simscape Multibody, based on the Figure 1.

Figure 10 shows a visualization of the trusses system model in the Mechanics Explorers window of the MATLAB – Simscape Multibody environment. It is a running visualization in the case of a properly constructed model.

Figure 10. Visualization of the MATLAB – Simscape Multibody model of the trusses system, in the Mechanics Explorers window: (a) Unloaded; (b) Loaded with visible deformation.

Figure 10. Visualization of the MATLAB – Simscape Multibody model of the trusses system, in the Mechanics Explorers window: (a) Unloaded; (b) Loaded with visible deformation.

The individual systems forming a particular trusses system should be explained. The World Frame subsystem is formed by combining the World Frame and Solver Configuration library blocks, these blocks being in the "default" configuration.

Figure 11. The World Frame subsystem is formed by combining the World Frame and Solver Configuration library blocks.

Figure 11. The World Frame subsystem is formed by combining the World Frame and Solver Configuration library blocks.

Subsystems of nodes labelled Node_A and Node_D are formed as follows:

Figure 12. Detailed view of nodes A, D: (a) Node_A; (b) Node_D.

Figure 12. Detailed view of nodes A, D: (a) Node_A; (b) Node_D.

The blocks "Point" – Graphic sphere and "Point Cloud" – Frame visualisation are used only for visualisations. Physically they are not involved in the given model. The F_A force vector value is measured directly from the "Revolute Joint" – Total force sensing block. The bars are connected using "Revolute Joint" joints in the "default" setting. The block "Interpreted MATLAB Function" – u(1:2) defines a two – dimensional force vector in the F_A, F_B plane, whose z – component is of zero value. The block "Gain" – con provides the multiplication by the multiplication constant – con. An explanation of constant – con has been provided in equations (82) to (84).

Subsystems of nodes labelled Node_B and Node_C are formed as follows:

Figure 13. Detailed view of nodes B, C: (a) Node_B; (b) Node_C.

Figure 13. Detailed view of nodes B, C: (a) Node_B; (b) Node_C.

Similarly, the bars are connected using "Revolute Joint" joints in the "default" setting. The member subsystems labelled Beam_AB, Beam_AC, Beam_BC, Beam_BD, Beam_CD are identical in structure and are formed as follows:

Figure 14. An example of the bar subsystem labelled Beam_AB.

Figure 14. An example of the bar subsystem labelled Beam_AB.

The "Rigid Transform" blocks ensure that the bar is rotated into the desired plane around the y – axis of the world frame. The N_AB force vector value is measured directly from the "Weld Joint" block – Total force sensing. The block "Interpreted MATLAB Function" – u(3) selects the z – component of the force vector in the N_AB bar. The block physically representing the bar is a "Flexible Cylindrical Beam" block type. It defines the individual parameters such as bar length l_AB, bar diameter d, Young's modulus E, shear modulus G, density and proportional Rayleigh damping values b_m, b_k.

Subsystems labelled Node_Deflection_B, Node_Deflection_C are designed to measure the deflection of nodes B, C and are formed as follows:

Figure 15. Detailed view of the node deflection measurement subsystems labelled Node_deflection _B and Node_deflection_C: (a) Node_B node deflection; (b) Node_C node deflection.

Figure 15. Detailed view of the node deflection measurement subsystems labelled Node_deflection _B and Node_deflection_C: (a) Node_B node deflection; (b) Node_C node deflection.

The "Transform Sensor" – Translation [x, y, z] blocks provide deflection measurement relative to the world frame. The remaining blocks select only those components of the vector that define the deflection in the x, y plane relative to the original position.

Figure 16 illustrates the control measurement of the reaction forces to optimize the multiplication constant – con based on equations (83), (84).

Figure 16. Subsystem of control measurement of reaction forces for optimizing the multiplication constant – con based on equations (83), (84).

Figure 16. Subsystem of control measurement of reaction forces for optimizing the multiplication constant – con based on equations (83), (84).

4. Building a Trusses Structure in the SOLIDWORKS – Weldments Tool

To verify the calculation of the given trusses system, the SOLIDWORKS environment was used and a similar model was developed using the "Weldments" tool, which allows the welded structure to be designed as a single multibody part. This tool yielded the most accurate results when simulated on bars under bending, torsion, tension – compression and continuous loads against the theory.

Load F_B = [-2, -10]^T kN, bar lengths l = 1 m, l_AB = l, l_BC = l, l_CD = l, l_AC = √2∙l m, l_BD = √2∙l m, bar diameters are the same with d = 0.01m full profile. The material used is "Plain Carbon Steel" with Young's modulus E is 210 GPa, shear modulus G of 79 GPa and density of 7800 kg/m3. Poisson's ratio is 0.28.

A detailed procedure for developing such a model is not provided in this section as it is available in [31,32,33,34]. It should be noted that the node where the diagonal bars intersect is not a fixed joint. This is also evident when loaded with the visible deformation in Figure 17b. A comparison of the individual results from MATLAB – Multibody, SOLIDWORKS – Weldments tool and the theoretical calculations are summarized in Table 1, Chapter 5. Results.

Figure 17. The figure displays a similar trusses structure developed with the Weldments tool in SOLIDWORKS: (a) Unloaded with indicated reactions at nodes A, D; (b) Loaded with visible deformation of the bars.

Figure 17. The figure displays a similar trusses structure developed with the Weldments tool in SOLIDWORKS: (a) Unloaded with indicated reactions at nodes A, D; (b) Loaded with visible deformation of the bars.

5. Results

A comparison of the individual results from MATLAB – Multibody, SOLIDWORKS – Weldments tool and the theoretical calculation is presented in Table 1 overview.

Firstly, the respective values are substituted into the individual equations and the results are quantified:

5.1. Deriving from the Theory

By substituting the respective values into the individual equations for N_AB, F_AC, F_BC, F_BD, F_CD, the following results can be derived:

$$N_{\mathrm{A}\mathrm{B}}=2·{\displaystyle \frac{\left(2·\sqrt{2}+1\right)·\left(-2000\right)+\left(\sqrt{2}+1\right)·\left(-10000\right)}{\left(4·\sqrt{2}+3\right)}}=-7.346546206455528{·10}^{+03}$$

(85)

From the equilibrium in the nodes, the reaction forces F_i in the bars are derived:

$$F_{\mathrm{A}\mathrm{C}}=-\sqrt{2}·\left(-2000-F_{\mathrm{A}\mathrm{B}}-10000\right)=6.580977466707121{·10}^{+03}$$

(86)

$$F_{\mathrm{B}\mathrm{C}}=-2000-F_{\mathrm{A}\mathrm{B}}-10000=-4.653453793544472{·10}^{+03}$$

(87)

$$F_{\mathrm{B}\mathrm{D}}=\sqrt{2}·\left(F_{\mathrm{A}\mathrm{B}}+2000\right)=-7.561158157023830{·10}^{+03}$$

(88)

$$F_{\mathrm{C}\mathrm{D}}=-2000-F_{\mathrm{A}\mathrm{B}}+10000=-4.653453793544472{·10}^{+03}$$

(89)

Consequently, the N_i forces in the bars are as follows:

$$N_{\mathrm{A}\mathrm{C}}={-F}_{\mathrm{A}\mathrm{C}}=-6.580977466707121{·10}^{+03}$$

(90)

$$N_{\mathrm{B}\mathrm{C}}={-F}_{\mathrm{B}\mathrm{C}}=4.653453793544472{·10}^{+03}$$

(91)

$$N_{\mathrm{B}\mathrm{D}}={-F}_{\mathrm{B}\mathrm{D}}=7.561158157023830{·10}^{+03}$$

(92)

$$N_{\mathrm{C}\mathrm{D}}={-F}_{\mathrm{C}\mathrm{D}}=4.653453793544472{·10}^{+03}$$

(93)

5.2. The MATAB and SOLIDWORKS Results

For ease of overview, the individual results are arranged in a table.

It can be noticed that the accuracy in MATLAB – Multibody is guaranteed to six decimal places, while in SOLIDWORKS – Weldments tool, it is guaranteed to two decimal places. It is important to note that SOLIDWORKS also offers many other deliverables and extensions that would have to be completed in MATLAB. Additionally, the calculation in MATLAB has been optimized using – con to ensure the calculation is not tedious and delivers interesting results. If the simulation had not been proportionally shifted to lower values using the – con constant, the computation would have been tedious, and the accuracy would have been approximately four decimal places. Alternatively, it would be necessary to increase the values of damping b_k and adjust the simulation time accordingly.

In MATLAB, for a more complex structure, individual blocks representing a repeating part can be created, and the entire structure can be further developed for static or dynamic assessment. See Figure 18.