<?xml version="1.0" encoding="UTF-8"?><!DOCTYPE article  PUBLIC "-//NLM//DTD Journal Publishing DTD v3.0 20080202//EN" "http://dtd.nlm.nih.gov/publishing/3.0/journalpublishing3.dtd"><article xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink" dtd-version="3.0" xml:lang="en" article-type="research article"><front><journal-meta><journal-id journal-id-type="publisher-id">MSCE</journal-id><journal-title-group><journal-title>Journal of Materials Science and Chemical Engineering</journal-title></journal-title-group><issn pub-type="epub">2327-6045</issn><publisher><publisher-name>Scientific Research Publishing</publisher-name></publisher></journal-meta><article-meta><article-id pub-id-type="doi">10.4236/msce.2021.96002</article-id><article-id pub-id-type="publisher-id">MSCE-110218</article-id><article-categories><subj-group subj-group-type="heading"><subject>Articles</subject></subj-group><subj-group subj-group-type="Discipline-v2"><subject>Chemistry&amp;Materials Science</subject></subj-group></article-categories><title-group><article-title>
 
 
  A Simple Elasto-Plastic Iterative Method by Fem for the Analysis of Plane Articulated Truss: Case of a 3-Bar Truss
 
</article-title></title-group><contrib-group><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Djomi</surname><given-names>Rolland</given-names></name><xref ref-type="aff" rid="aff1"><sup>1</sup></xref><xref ref-type="corresp" rid="cor1"><sup>*</sup></xref></contrib><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Fokam</surname><given-names>Bopda Christian</given-names></name><xref ref-type="aff" rid="aff1"><sup>1</sup></xref></contrib><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Kenmeugne</surname><given-names>Bienvenu</given-names></name><xref ref-type="aff" rid="aff1"><sup>1</sup></xref></contrib><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Kanmogne</surname><given-names>Abraham</given-names></name><xref ref-type="aff" rid="aff1"><sup>1</sup></xref></contrib><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Mbobda</surname><given-names>Gerard</given-names></name><xref ref-type="aff" rid="aff1"><sup>1</sup></xref></contrib></contrib-group><aff id="aff1"><addr-line>Laboratoire Engineering Civil et Mécanique, National Advanced School of Engineering (ENSPY/UY1), University of Yaounde 1, Yaoundé, Cameroon</addr-line></aff><pub-date pub-type="epub"><day>29</day><month>06</month><year>2021</year></pub-date><volume>09</volume><issue>06</issue><fpage>12</fpage><lpage>23</lpage><history><date date-type="received"><day>14,</day>	<month>April</month>	<year>2021</year></date><date date-type="rev-recd"><day>26,</day>	<month>June</month>	<year>2021</year>	</date><date date-type="accepted"><day>29,</day>	<month>June</month>	<year>2021</year></date></history><permissions><copyright-statement>&#169; Copyright  2014 by authors and Scientific Research Publishing Inc. </copyright-statement><copyright-year>2014</copyright-year><license><license-p>This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/</license-p></license></permissions><abstract><p>
 
 
  This paper presents a simplified numerical simulation tool for the elasto-plastic calculation of plane articulated truss by the finite element method (FEM) in MATLAB. The simplified approach consists of linearizing isotropic strain-hardening (to obtain a bilinear material law). The numerical implementation is built on the basis of the incremental and iterative FEM algorithms. The numerical resolution technique used is based on the projection methods of the modified Newton-Raphson solution. The MATLAB program is developed for the application of a 3-bar truss under monotonous quasi-static loading. Different values of the approximation error of the convergence criterion are used to study its impact on the quality of the algorithm. Numerical simulations have shown the reliability and quality of our simplified approach regardless of the approximation error.
 
</p></abstract><kwd-group><kwd>Articulated Truss</kwd><kwd> Elasto-Plastic</kwd><kwd> Newton-Raphson</kwd><kwd> Finite Element Method</kwd></kwd-group></article-meta></front><body><sec id="s1"><title>1. Introduction</title><p>Structures consisting of truss elements are widely used in many engineering fields [<xref ref-type="bibr" rid="scirp.110218-ref1">1</xref>]. These trusses structures are fast to build, lightweight and allow large spans. The design of trusses structures has been the subject of numerous studies [<xref ref-type="bibr" rid="scirp.110218-ref2">2</xref>] [<xref ref-type="bibr" rid="scirp.110218-ref3">3</xref>] [<xref ref-type="bibr" rid="scirp.110218-ref4">4</xref>]. Many methods have been developed for the dimensioning of trusses. For small structures, classical analytical methods such as the node method (Culmann method) and the Ritter method are often used [<xref ref-type="bibr" rid="scirp.110218-ref5">5</xref>] [<xref ref-type="bibr" rid="scirp.110218-ref6">6</xref>]. For large structures, the graphical method was one of the first adopted techniques.</p><p>As a result of the increasing performance of computers, finite element software is now widely used for the calculation of trusses. Using numerical simulation tools, some authors have worked on methods for structural optimization of trusses [<xref ref-type="bibr" rid="scirp.110218-ref7">7</xref>] [<xref ref-type="bibr" rid="scirp.110218-ref8">8</xref>] [<xref ref-type="bibr" rid="scirp.110218-ref9">9</xref>]. McKeown [<xref ref-type="bibr" rid="scirp.110218-ref7">7</xref>] presented a displacement optimization approach for the design of linear elastic trusses. The mechanical modeling of truss structures is simple under ordinary conditions to which the classical assumption of linear elasticity applies. However, modern applications of truss structures often require plasticity material analysis, which can accurately predict the structural responses. This observation is justified by structural failures, such as the collapse of the I-35W Bridge over the Mississippi River in Minneapolis [<xref ref-type="bibr" rid="scirp.110218-ref10">10</xref>] [<xref ref-type="bibr" rid="scirp.110218-ref11">11</xref>]. Fan and et al. [<xref ref-type="bibr" rid="scirp.110218-ref12">12</xref>] studied the static and dynamic elasto-plastic stability of cross-linked hulls, while Ferrari et al. [<xref ref-type="bibr" rid="scirp.110218-ref13">13</xref>] numerically studied the non-linear elasto-plastic response of the historical Pardermo truss bridge. All these approaches are strongly non-linear.</p><p>In the present work, we propose a numerical simulation approach by FEM for the plane articulated truss. Our modeling will be based on a linearized material law of isotropic plastic strain-hardening (linearization of the elasto-plastic behavioral law: bilinear law) and on a projection method using the modified Newton-Raphson approach.</p></sec><sec id="s2"><title>2. Simplified Behaviour Model</title>Selecting a Template<p>In this study, we are interested in the elasto-plastic behaviour of articulated truss structures. A simplified linear isotropic strain-hardening model of the Johson Cook type [<xref ref-type="bibr" rid="scirp.110218-ref14">14</xref>] is implemented. Thus, the simplified model is bilinear and characterized by a Young’s modulus E and a tangent modulus E<sub>T</sub>, the yield strength σ 0 , yield stress projection σ ′ 0 (see <xref ref-type="fig" rid="fig1">Figure 1</xref>). The material constituting our truss is made of a metal alloy, therefore the actual behaviour law is shown in <xref ref-type="fig" rid="fig1">Figure 1</xref>(a). The corresponding simplified model is shown in <xref ref-type="fig" rid="fig1">Figure 1</xref>(b). In this study, we do not consider the phenomena of damage and failure [<xref ref-type="bibr" rid="scirp.110218-ref15">15</xref>].</p><p>This approximation of the real material behaviour law (<xref ref-type="fig" rid="fig1">Figure 1</xref>(a)) by a bilinear behaviour law of elasto-plastic type with linear strain hardening (<xref ref-type="fig" rid="fig1">Figure 1</xref>(b)) permits to write the following incremental plastic flow law (Equation (1)):</p><p>d σ = E T d ε (1)</p><p>With d ε = d ε e + d ε p ,</p><p>d ε e : elastic deformation increment;</p><p>d ε p : plastic deformation increment.</p><p>We deduce the evolution law of linear strain-hardening (Equation (2)):</p><p>σ = σ s ( σ 0 , E ) + H d ε p (2)</p><p>With H = E T 1 − E T E , H is called the strain-hardening modulus, ( σ 0 , E ) : Current</p><p>yield strength.</p><p>Equation (2) can also be expressed as a σ ′ 0 function as follows (Equation (3)):</p><p>σ = σ ′ 0 ( σ 0 , E ) + E T ε (3)</p><p>The linearized elasto-plastic evolution allows a simplified writing of the discretized elementary equilibrium, obtained analytically by the FEM an articulated truss (Equation (4)):</p><p>( ∫ V ( [ B ] T E T [ B ] d V ) ︸ [ K e ] ) { q e } = [ F ] e − ∫ V [ B ] T σ ′ 0 d V (4)</p><p>[ K e ] : local stiffness matrix, [ B ] : strain-displacement relationship matrix, { q e } : Nodal displacement field.</p></sec><sec id="s3"><title>3. Elasto-Plastic Numerical Resolution</title><p>The numerical solving technique used in our elasto-plastic problem is based on the classical incremental, iterative approach and on the minimization of a convergence criterion. The modified Newton Raphson-type projection method, which uses the elastic stiffness matrix at each iteration. <xref ref-type="fig" rid="fig2">Figure 2</xref> presents the algorithm followed in MATLAB program [<xref ref-type="bibr" rid="scirp.110218-ref2">2</xref>].</p></sec><sec id="s4"><title>4. Projection Method</title><p>The projection method on the solution sought is the modified Newton-Raphson method. The linear approximation of the plasticity law allows an easy calculation of the hardening, thus saving the CPU processing time. <xref ref-type="fig" rid="fig3">Figure 3</xref> shows the principle of projection on the boundary of the current elasticity domain for an increase in dε deformation. We have represented the most general case of an element being plasticized. We consider a history of initial stress equal to σ and the state of current strain hardening is denoted σ<sub>h</sub>.</p><p>The linear approximation of the strain hardening allows to obtain by a simple calculation, the plastic evolution law. From <xref ref-type="fig" rid="fig3">Figure 3</xref>(b), are geometrically deduced the relations presented in equations 5 to 8. First, we define a quantity R, as follows (Equation (5)):</p><p>R = A B A C = σ + d σ e − σ h d σ e = 1 + σ − σ h d σ e (5)</p><p>We deduce from the <xref ref-type="fig" rid="fig3">Figure 3</xref>(b) (Equation (6)):</p><p>X = R d ε ⇒</p><p>| Plasticslope   E T = B σ ′ X → B σ ′ = R E T d ε B C = A C − A B → B C = ( 1 − R ) d σ e (6)</p><p>The law of evolution (Equation (7)):</p><p>σ ′ = σ + ( 1 − R ) E d ε + R E T d ε ︸ d σ (7)</p><p>The increments of elastic and plastic deformations are deduced from the stress increment as follows (Equation (8)):</p><p>‖ ε e = 1 E d σ → d ε e = ( 1 − R + R E T E ) d ε d ε p = d ε − d ε e → d ε p = R ( 1 − E T E ) d ε (8)</p><p>The numerical implementation is done through the projection algorithm of <xref ref-type="fig" rid="fig4">Figure 4</xref>.</p></sec><sec id="s5"><title>5. Case of Application</title><p>The application case structure in this article is a truss consisting of three (03) bars of identical cross-section S (see <xref ref-type="fig" rid="fig5">Figure 5</xref>). The inclined bar is of length 2 L and the other of length L. The material behaviour law described is the one presented in Section 2. The numerical values considered are: L = 500 mm, Young’s modulus E= 70.000 MPa, tangent modulus E<sub>T</sub> = 510.8 MPa, yield strength σ<sub>0</sub> = 281.559 MPa, bar section S = 50 mm<sup>2</sup>. <xref ref-type="table" rid="table1"><xref ref-type="table" rid="table">Table </xref>1</xref> shows the table of connectivities of the 3-bar truss structure.</p><p>The resolution by FEM involves either a matrix or a discrete formulation of the problem. In the displacement approach, the main unknowns of the problem are the nodal displacements of the structure. In this article, the stiffness matrix is used to construct the FEM model. The elementary stiffness matrix (for each bar element) is written. In the case of the articulated trusses we consider 1 nodal degree of freedom in the local reference and 2 nodal degrees of freedom in the global reference.</p><p>The global stiffness matrix is obtained after assembling the elementary stiffness matrices. Thus, the main unknown, which is the nodal displacement vector is calculated by solving the global matrix equilibrium equation [K]{Q} = {F}. In this equation, [K] is the global stiffness matrix, {F} the nodal force vector and {Q} the nodal displacement vector. When the bar element is in the plastic behaviour phase, the local matrix equilibrium equation is modified as shown in Equation (4). Once the nodal displacement vector has been determined, the internal forces in the bars can be deduced. In the application case of 3-bar truss in this article, the global stiffness matrix is presented in Equation (9). A computational code is developed in MATLAB for the application case.</p><table-wrap id="table1" ><label><xref ref-type="table" rid="table1"><xref ref-type="table" rid="table">Table </xref>1</xref></label><caption><title> <xref ref-type="table" rid="table">Table </xref>of connectivity</title></caption><table><tbody><thead><tr><th align="center" valign="middle" >Element</th><th align="center" valign="middle" >Node 1</th><th align="center" valign="middle" >Node 2</th><th align="center" valign="middle" >Length</th><th align="center" valign="middle" >Section</th><th align="center" valign="middle" >Inclination</th></tr></thead><tr><td align="center" valign="middle" >Bar 1</td><td align="center" valign="middle" >1</td><td align="center" valign="middle" >2</td><td align="center" valign="middle" >2 L</td><td align="center" valign="middle" >S</td><td align="center" valign="middle" >135˚</td></tr><tr><td align="center" valign="middle" >Bar 2</td><td align="center" valign="middle" >1</td><td align="center" valign="middle" >3</td><td align="center" valign="middle" >L</td><td align="center" valign="middle" >S</td><td align="center" valign="middle" >90˚</td></tr><tr><td align="center" valign="middle" >Bar 3</td><td align="center" valign="middle" >1</td><td align="center" valign="middle" >4</td><td align="center" valign="middle" >2 L</td><td align="center" valign="middle" >S</td><td align="center" valign="middle" >45˚</td></tr></tbody></table></table-wrap><p>[ K ] = [ E S 2 L 2 + E S 2 L 2 0 − E S 2 L 2 E S 2 L 2 0 0 − E S 2 L 2 − E S 2 L 2 0 E S 2 L 2 + E S 2 L 2 + E S L E S 2 L 2 − E S 2 L 2 0 − E S L − E S 2 L 2 − E S 2 L 2 − E S 2 L 2 E S 2 L 2 E S 2 L 2 − E S 2 L 2 0 0 0 0 E S 2 L 2 − E S 2 L 2 − E S 2 L 2 E S 2 L 2 0 0 0 0 0 0 0 0 0 0 0 0 0 − E S L 0 0 0 E S L 0 0 − E S 2 L 2 − E S 2 L 2 0 0 0 0 E S 2 L 2 E S 2 L 2 − E S 2 L 2 − E S 2 L 2 0 0 0 0 E S 2 L 2 E S 2 L 2 ] (9)</p></sec><sec id="s6"><title>6. Results and Validation</title><p>The results presented in this section are obtained from the finite element computation code developed under MATLAB and, having the architecture of the numerical resolution algorithm presented in <xref ref-type="fig" rid="fig2">Figure 2</xref> and <xref ref-type="fig" rid="fig4">Figure 4</xref>. The solving method uses an incremental and iterative method based on minimizing a convergence criterion. Different values of the approximation error of the convergence criterion are used to study its impact on the algorithm used. The MATLAB program is illustrated in the case of the articulated truss shown in <xref ref-type="fig" rid="fig5">Figure 5</xref>. A load F = 35 kN is applied to the structure, the load increment dF is chosen to be equal to 100N.</p><sec id="s6_1"><title>6.1. Simulation and Validation</title><p>In this paragraph, examples of numerical simulation are presented in order to show the reliability of our algorithm to correctly simulate an elasto-plastic behavior of an articulated truss. Then, the validation is made by comparing with the results obtained from the ABAQUS/CAE computation code.</p><p>Tables 2-4 show the main results (N<sub>1</sub>, N<sub>2</sub>, N<sub>3</sub>: internal forces in each bar element and V<sub>1</sub>: vertical displacement of node 1, see <xref ref-type="fig" rid="fig5">Figure 5</xref>) for some load increments obtained from the calculation code developed in MATLAB. The results are presented in each table according to the selected value of the approximation error ξ of the convergence test. The ξ values shown in the tables are: 1, 0.01, and 0.00001.</p><p>The loads chosen as examples in these different tables correspond to the 3 phases of behaviour of the structure: elastic structure (elastic bars), partial plastic structure (bar 2 in the plastic domain), plastic structure (all bars in the plastic domain).</p><table-wrap id="table2" ><label><xref ref-type="table" rid="table">Table </xref>2</label><caption><title> Internal forces (N<sub>1</sub>, N<sub>2</sub> and N<sub>3</sub>) in the bars and vertical displacement V<sub>1</sub> (of node 1) for different loads (approximation error ξ = 1)</title></caption><table><tbody><thead><tr><th align="center" valign="middle"  colspan="5"  >dF = 100 N and convergence test ξ = 1</th></tr></thead><tr><td align="center" valign="middle" >F (N)</td><td align="center" valign="middle" >N<sub>1</sub> = N<sub>3</sub> (N)</td><td align="center" valign="middle" >N<sub>2</sub> (N)</td><td align="center" valign="middle" >Displacement V<sub>1</sub> (mm)</td><td align="center" valign="middle" >N˚ of bar-element in the plastic zone</td></tr><tr><td align="center" valign="middle" >100</td><td align="center" valign="middle" >29.28</td><td align="center" valign="middle" >58.57</td><td align="center" valign="middle" >−0.0084</td><td align="center" valign="middle" >-</td></tr><tr><td align="center" valign="middle" >500</td><td align="center" valign="middle" >146.44</td><td align="center" valign="middle" >292.89</td><td align="center" valign="middle" >−0.0418</td><td align="center" valign="middle" >-</td></tr><tr><td align="center" valign="middle" >5000</td><td align="center" valign="middle" >1464.46</td><td align="center" valign="middle" >2928.93</td><td align="center" valign="middle" >0.4184</td><td align="center" valign="middle" >-</td></tr><tr><td align="center" valign="middle" >20000</td><td align="center" valign="middle" >5857.86</td><td align="center" valign="middle" >11,715.72</td><td align="center" valign="middle" >1.6736</td><td align="center" valign="middle" >-</td></tr><tr><td align="center" valign="middle" >23600</td><td align="center" valign="middle" >6912.28</td><td align="center" valign="middle" >13,824.55</td><td align="center" valign="middle" >−1.9749</td><td align="center" valign="middle" >-</td></tr><tr><td align="center" valign="middle" >24000</td><td align="center" valign="middle" >7029.43</td><td align="center" valign="middle" >14,058.87</td><td align="center" valign="middle" >−2.0084</td><td align="center" valign="middle" >-</td></tr><tr><td align="center" valign="middle" >24100</td><td align="center" valign="middle" >7085.55</td><td align="center" valign="middle" >14,078.62</td><td align="center" valign="middle" >−2.0244</td><td align="center" valign="middle" >2</td></tr><tr><td align="center" valign="middle" >24400</td><td align="center" valign="middle" >7293.92</td><td align="center" valign="middle" >14,081.67</td><td align="center" valign="middle" >−2.0840</td><td align="center" valign="middle" >2</td></tr><tr><td align="center" valign="middle" >30000</td><td align="center" valign="middle" >11,183.47</td><td align="center" valign="middle" >14,138.43</td><td align="center" valign="middle" >−3.1952</td><td align="center" valign="middle" >2</td></tr><tr><td align="center" valign="middle" >34200</td><td align="center" valign="middle" >14,087.37</td><td align="center" valign="middle" >14,199.52</td><td align="center" valign="middle" >−4.3913</td><td align="center" valign="middle" >1, 2 and 3</td></tr><tr><td align="center" valign="middle" >34600</td><td align="center" valign="middle" >14,203.36</td><td align="center" valign="middle" >14,431.50</td><td align="center" valign="middle" >−8.9327</td><td align="center" valign="middle" >1, 2, and 3</td></tr></tbody></table></table-wrap><table-wrap id="table3" ><label><xref ref-type="table" rid="table">Table </xref>3</label><caption><title> Internal forces (N<sub>1</sub>, N<sub>2</sub> and N<sub>3</sub>) in the bars and vertical displacement V<sub>1</sub> (of node 1) for different loads (approximation error ξ = 0.01)</title></caption><table><tbody><thead><tr><th align="center" valign="middle"  colspan="5"  >dF = 100 N and convergence test ξ = 0.01</th></tr></thead><tr><td align="center" valign="middle" >F (N)</td><td align="center" valign="middle" >N<sub>1</sub> = N<sub>3</sub> (N)</td><td align="center" valign="middle" >N<sub>2</sub> (N)</td><td align="center" valign="middle" >Displacement V<sub>1</sub> (mm)</td><td align="center" valign="middle" >N˚ of bar-element in the plastic zone</td></tr><tr><td align="center" valign="middle" >100</td><td align="center" valign="middle" >29.28</td><td align="center" valign="middle" >58.57</td><td align="center" valign="middle" >−0.0084</td><td align="center" valign="middle" >-</td></tr><tr><td align="center" valign="middle" >500</td><td align="center" valign="middle" >146.44</td><td align="center" valign="middle" >292.89</td><td align="center" valign="middle" >−0.0418</td><td align="center" valign="middle" >-</td></tr><tr><td align="center" valign="middle" >5000</td><td align="center" valign="middle" >1464.46</td><td align="center" valign="middle" >2928.93</td><td align="center" valign="middle" >−0.4184</td><td align="center" valign="middle" >-</td></tr><tr><td align="center" valign="middle" >20,000</td><td align="center" valign="middle" >5857.86</td><td align="center" valign="middle" >11,715.72</td><td align="center" valign="middle" >−1.6736</td><td align="center" valign="middle" >-</td></tr><tr><td align="center" valign="middle" >23,600</td><td align="center" valign="middle" >6912.28</td><td align="center" valign="middle" >13,824.55</td><td align="center" valign="middle" >−1.9749</td><td align="center" valign="middle" >-</td></tr><tr><td align="center" valign="middle" >24,000</td><td align="center" valign="middle" >7029.43</td><td align="center" valign="middle" >14,058.87</td><td align="center" valign="middle" >−2.0084</td><td align="center" valign="middle" >-</td></tr><tr><td align="center" valign="middle" >24,100</td><td align="center" valign="middle" >7086.16</td><td align="center" valign="middle" >14,078.63</td><td align="center" valign="middle" >−2.0246</td><td align="center" valign="middle" >2</td></tr><tr><td align="center" valign="middle" >24,400</td><td align="center" valign="middle" >7296.11</td><td align="center" valign="middle" >14,081.70</td><td align="center" valign="middle" >−2.0846</td><td align="center" valign="middle" >2</td></tr><tr><td align="center" valign="middle" >30,000</td><td align="center" valign="middle" >11,215.07</td><td align="center" valign="middle" >14,138.89</td><td align="center" valign="middle" >−3.2043</td><td align="center" valign="middle" >2</td></tr><tr><td align="center" valign="middle" >34,200</td><td align="center" valign="middle" >14,109.89</td><td align="center" valign="middle" >14,244.57</td><td align="center" valign="middle" >−5.2732</td><td align="center" valign="middle" >1, 2 et 3</td></tr><tr><td align="center" valign="middle" >34,600</td><td align="center" valign="middle" >14,227.04</td><td align="center" valign="middle" >14,478.86</td><td align="center" valign="middle" >−9.8600</td><td align="center" valign="middle" >1, 2 et 3</td></tr></tbody></table></table-wrap><table-wrap id="table4" ><label><xref ref-type="table" rid="table">Table </xref>4</label><caption><title> Internal forces (N<sub>1</sub>, N<sub>2</sub> and N<sub>3</sub>) in the bars and vertical displacement V<sub>1</sub> (of node 1) for different loads (approximation error ξ = 0.00001)</title></caption><table><tbody><thead><tr><th align="center" valign="middle"  colspan="5"  >dF = 100N and convergence test ξ = 0.00001</th></tr></thead><tr><td align="center" valign="middle" >F (N)</td><td align="center" valign="middle" >N<sub>1</sub> = N<sub>3</sub> (N)</td><td align="center" valign="middle" >N<sub>2</sub> (N)</td><td align="center" valign="middle" >Displacement V<sub>1</sub> (mm)</td><td align="center" valign="middle" >N˚ of bar-element in the plastic zone ˚</td></tr><tr><td align="center" valign="middle" >100</td><td align="center" valign="middle" >29.28</td><td align="center" valign="middle" >58.5786</td><td align="center" valign="middle" >−0.0084</td><td align="center" valign="middle" >-</td></tr><tr><td align="center" valign="middle" >500</td><td align="center" valign="middle" >146.44</td><td align="center" valign="middle" >292.8932</td><td align="center" valign="middle" >−0.0418</td><td align="center" valign="middle" >-</td></tr><tr><td align="center" valign="middle" >5000</td><td align="center" valign="middle" >1464.46</td><td align="center" valign="middle" >2928.93</td><td align="center" valign="middle" >−0.4184</td><td align="center" valign="middle" >-</td></tr><tr><td align="center" valign="middle" >20,000</td><td align="center" valign="middle" >5857.86</td><td align="center" valign="middle" >11,715.72</td><td align="center" valign="middle" >−1.6736</td><td align="center" valign="middle" >-</td></tr><tr><td align="center" valign="middle" >5200</td><td align="center" valign="middle" >1523.04</td><td align="center" valign="middle" >3046.08</td><td align="center" valign="middle" >−0.4352</td><td align="center" valign="middle" >-</td></tr><tr><td align="center" valign="middle" >23,600</td><td align="center" valign="middle" >6912.28</td><td align="center" valign="middle" >13,824.55</td><td align="center" valign="middle" >−1.9749</td><td align="center" valign="middle" >-</td></tr><tr><td align="center" valign="middle" >24,000</td><td align="center" valign="middle" >7029.43</td><td align="center" valign="middle" >14,058.87</td><td align="center" valign="middle" >−2.0084</td><td align="center" valign="middle" >-</td></tr><tr><td align="center" valign="middle" >24,100</td><td align="center" valign="middle" >7086.17</td><td align="center" valign="middle" >14,078.63</td><td align="center" valign="middle" >−2.0246</td><td align="center" valign="middle" >2</td></tr><tr><td align="center" valign="middle" >24,400</td><td align="center" valign="middle" >7296.13</td><td align="center" valign="middle" >14,081.70</td><td align="center" valign="middle" >−2.0846</td><td align="center" valign="middle" >2</td></tr><tr><td align="center" valign="middle" >30,000</td><td align="center" valign="middle" >11,215.48</td><td align="center" valign="middle" >14,138.90</td><td align="center" valign="middle" >−3.2044</td><td align="center" valign="middle" >2</td></tr><tr><td align="center" valign="middle" >34,200</td><td align="center" valign="middle" >14,110.1953</td><td align="center" valign="middle" >14,245.1695</td><td align="center" valign="middle" >−5.2848</td><td align="center" valign="middle" >1, 2 et 3</td></tr><tr><td align="center" valign="middle" >34,600</td><td align="center" valign="middle" >14,227.3526</td><td align="center" valign="middle" >14,479.4840</td><td align="center" valign="middle" >−9.8720</td><td align="center" valign="middle" >1, 2 et 3</td></tr></tbody></table></table-wrap><p>The results presented in Tables 2-4 are compared with the results obtained using the ABAQUS/CAE calculation code. For example, we choose two load cases of 20 kN and 30 kN corresponding to partial and total plasticization phases of the structure. <xref ref-type="fig" rid="fig6">Figure 6</xref> shows the stress states in the truss for the two load cases obtained by ABAQUS. The internal forces in the bars are respectively the couple (N<sub>1</sub> = N<sub>3</sub> = 5858 N, N<sub>2</sub> = 11715.5 N) and (N<sub>1</sub> = N<sub>3</sub> = 11216 N, N<sub>2</sub> = 14138.5 N) for loads F of 20 kN and 30 kN respectively. Whatever the load considered, the results are quite close to ours (see the corresponding row in Tables 2-4).</p><p><xref ref-type="fig" rid="fig7">Figure 7</xref> shows the evolution of the internal forces as a function of the load. It can be seen that each curve obtained has 3 areas of linearity corresponding to the 3 phases of behaviour of the following truss structure: elastic phase (bars in the elastic), partial plastic phase (bar 2 in the plastic) and complete plasticization phase (bars 1, 2 and 3 in the plastic).</p><p><xref ref-type="fig" rid="fig8">Figure 8</xref> shows the evolution of the vertical displacement V<sub>1</sub> of node 1 as a function of the load F. As in <xref ref-type="fig" rid="fig7">Figure 7</xref>, a tri-linear curve of the displacement V<sub>1</sub> is observed. An almost exponential increase of the displacement can be seen when all the bars are plasticized. This is explained by the low value of the tangent modulus (E<sub>T</sub> ≈ 510 MPa) of the bilinear material behaviour law used (see <xref ref-type="fig" rid="fig1">Figure 1</xref>).</p></sec><sec id="s6_2"><title>6.2. Influence of the Value of Approximation Error x</title><p>In this section, we are interested in the influence of the ξ value on the final result obtained. <xref ref-type="fig" rid="fig9">Figure 9</xref> and <xref ref-type="fig" rid="fig1">Figure 1</xref>0 illustrate respectively the influence of the ξ value on the solution of the internal forces in the bars and the vertical displacement V<sub>1</sub> of node 1. The illustrative load cases are: 5000, 24,400 and 34,600 N corresponding to the 3 phases of behaviour of the structure. Whatever the load level, it can be seen that the value of ξ has little influence on the final result. Nevertheless, for high loads (F = 34,600 N), a non-negligible influence of the ξ value on the displacement V<sub>1</sub> is observed (<xref ref-type="fig" rid="fig1">Figure 1</xref>0). This weak influence is explained by the bilinear nature of the implemented behaviour law in our study.</p></sec></sec><sec id="s7"><title>7. Conclusion</title><p>This article presented the development of a FEM calculation code for the simple calculation of an articulated truss structure. A finite element program was developed in the MATLAB environment for a three-bar truss structure. The special feature of the numerical resolution of our computational code is the use of a modified Newton-Raphson projection method, by using a linearized elastoplastic material law. Simulations on a 3-bar articulated truss structure and a comparison with a similar model under ABAQUS have proven the reliability of our approach.</p></sec><sec id="s8"><title>Conflicts of Interest</title><p>The authors declare no conflicts of interest regarding the publication of this paper.</p></sec><sec id="s9"><title>Cite this paper</title><p>Rolland, D., Christian, F.B., Bienvenu, K., Abraham, K. and Gerard, M. (2021) A Simple Elasto-Plastic Iterative Method by Fem for the Analysis of Plane Articulated Truss: Case of a 3-Bar Truss. Journal of Materials Science and Chemical Engineering, 9, 12-23. https://doi.org/10.4236/msce.2021.96002</p></sec></body><back><ref-list><title>References</title><ref id="scirp.110218-ref1"><label>1</label><mixed-citation publication-type="other" xlink:type="simple">Miyachi, K., Nakamura, S. and Manda, A. (2012) Progressive Collapse Analysis of Steel Truss Bridges and Evaluation of Ductility. Journal of Constructional Steel Research, 78, 192-200. https://doi.org/10.1016/j.jcsr.2012.06.015</mixed-citation></ref><ref id="scirp.110218-ref2"><label>2</label><mixed-citation publication-type="other" xlink:type="simple">Oudin, H. (2009) Introduction à la plasticié. Ecole d’ingénieur. Ecole Centrale de Nantes, 49. https://cel.archives-ouvertes.fr/cel-00483603</mixed-citation></ref><ref id="scirp.110218-ref3"><label>3</label><mixed-citation publication-type="other" xlink:type="simple">Dos Reis, F. and Ganghoffer, J.F. (2014) Homogenized Elastoplastic Response of Repetitive 2D Lattice Truss Materials. Computational Materials Science, 84, 145-155. https://doi.org/10.1016/j.commatsci.2013.11.066</mixed-citation></ref><ref id="scirp.110218-ref4"><label>4</label><mixed-citation publication-type="other" xlink:type="simple">Felipe, T.R.C., Leonel, E.D. and Haach, V.G., (2019) A Comprehensive Ductile Damage Model for 3D Truss Structures. International Journal of Non-Linear Mechanics, 112, 13-24. https://doi.org/10.1016/j.ijnonlinmec.2019.02.010</mixed-citation></ref><ref id="scirp.110218-ref5"><label>5</label><mixed-citation publication-type="other" xlink:type="simple">Kurrer, K.-E. (2018) The History of the Theory of Structures: Searching for Equilibrium. 2nd Edition, John Wiley &amp; Sons, Berlin, 1212 p. https://doi.org/10.1002/9783433609163</mixed-citation></ref><ref id="scirp.110218-ref6"><label>6</label><mixed-citation publication-type="other" xlink:type="simple">Rebielak, J. (2019) Simple Method of Approximate Calculation of Statically Indeterminate Trusses. International Journal of Computational Methods, 16, Article No. 1840026. https://doi.org/10.1142/S0219876218400261</mixed-citation></ref><ref id="scirp.110218-ref7"><label>7</label><mixed-citation publication-type="other" xlink:type="simple">McKeown, J.J. (1989) The Design of Optimal Trusses via Sequences of Optimal Fixed Displacement Structures. Engineering Optimization, 14, 159–178.https://doi.org/10.1080/03052158908941209</mixed-citation></ref><ref id="scirp.110218-ref8"><label>8</label><mixed-citation publication-type="other" xlink:type="simple">Missoum, S., Gurdal, Z., Hernandez, P. and Guillot, J. (2002) A Displacement Based Optimization for Truss Structures Subjected to Static and Dynamic Constraints. AIAA Journal, 40.</mixed-citation></ref><ref id="scirp.110218-ref9"><label>9</label><mixed-citation publication-type="other" xlink:type="simple">Madah, H. and Amir, O. (2017) Truss Optimization with Buckling Considerations Using Geometrically Nonlinear Beam Modeling. Computers &amp; Structures, 192, 233-247.https://doi.org/10.1016/j.compstruc.2017.07.023</mixed-citation></ref><ref id="scirp.110218-ref10"><label>10</label><mixed-citation publication-type="other" xlink:type="simple">Subramanian, N. (2008) I-35W Mississippi River Bridge Failure—Is It a Wakeup Call? Indian Concrete Journal, 82, 29-38.</mixed-citation></ref><ref id="scirp.110218-ref11"><label>11</label><mixed-citation publication-type="other" xlink:type="simple">Salem, H.M. and Helmy, H.M. (2014) Numerical Investigation of Collapse of the Minnesota I-35W Bridge. Engineering Structures, 59, 635-645. https://doi.org/10.1016/j.engstruct.2013.11.022</mixed-citation></ref><ref id="scirp.110218-ref12"><label>12</label><mixed-citation publication-type="other" xlink:type="simple">Fan, F., Cao, Z. and Shen, S. (2010) Elasto-Plastic Stability of Single-Layer Reticulated Shells. Thin-Walled Structures, 48, 827-836.https://doi.org/10.1016/j.tws.2010.04.004</mixed-citation></ref><ref id="scirp.110218-ref13"><label>13</label><mixed-citation publication-type="other" xlink:type="simple">Ferrari, R., Cocchetti, G. and Rizzi, E. (2018) Computational Elastoplastic Limit Analysis of the Paderno d’Adda Bridge (Italy, 1889). Archives of Civil and Mechanical Engineering, 18, 291-310. https://doi.org/10.1016/j.acme.2017.05.002</mixed-citation></ref><ref id="scirp.110218-ref14"><label>14</label><mixed-citation publication-type="other" xlink:type="simple">Remache, D., Semaan, M., Rossi, J.M., Pithioux, M. and Milan, J.L. (2020) Application of the Johnson-Cook Plasticity Model in the Finite Element Simulations of the Nanoindentation of the Cortical Bone. Journal of the Mechanical Behavior of Biomedical Materials, 101, Article No. 103426. https://doi.org/10.1016/j.jmbbm.2019.103426</mixed-citation></ref><ref id="scirp.110218-ref15"><label>15</label><mixed-citation publication-type="other" xlink:type="simple">Lemaitre, J., Chaboche, J.L., Benallal, A. and Desmorat, R. (2008) Mécanique des matériaux solides. 3ième edn, Dunod, Paris, 580.</mixed-citation></ref></ref-list></back></article>