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2.31 Project 6 Getting ready for nonlinearity Team 3: De Marco, Abao, Jeunnette Due November 21 We are going to start by modeling the same channel considered for Assignment 10, but this week we will use beam elements. The geometry, boundary conditions, and loading are the same (σσσσtop= 100 MPa, R=2m, H=4 m, W=0.5 m, t=0.5 m). This week, lets make the material stiffer, and change the Young’s modulus to 200 GPa. Will this change the resulting stress? Why? (if you are not sure, rerun your continuum model). Model the channel using quadratic beam elements (B32: Timoshenko shear flexible beams ) with a seed of 0.5. Think carefully about the appropriate boundary conditions on the supported end of the beam. Make sure you click on the ÊDuring analysis option in the section integration, so as to have stress contours available in the visualization module. As a default, the stress will be available at the four outer corners of the cross section (points 1,3,21,25 on Figure 6-2 of Handout 12). You can require additional points for the stress output: looking at Figure 6-2, figure out what is the number of the section point in the middle of the section, then in the in the Step module, use the Field Output Request Manager to edit the requests: right underneath the variable list, change the Output at shell, beam, and layered section points to ÊSpecify, and in the field enter the numbers for the four corners and of the middle point. This saves stress values at all 5 section points to the .odb file. Also, in the same panel, add Section Forces and Moments to the required forces/reactions output quantities. Now run the job and compare the results of the beam model to the results for the continuum model and of the shell model that you obtained in Assignment 10. I would like to see comparisons for the axial stress, and for section forces and moments. For the beam elements, the axial section force is SF1 and it represents the total axial force on the beam, SF2 is the total transverse shear force in the local 2-direction, and SF3 is the total transverse shear force in the local 1-direction (SF2 and SF3 are not available for Euler-Bernoulli beams). The Moments are SM1 : bending moment about the local 1-axis, SM2: bending moment about the local 2-axis, and SM3: the twisting moment. The stress components are S11, the axial stress, and S12, the shear stress. The S12 component only accounts for the shear stress due to torsion, not for the shear stress from the transverse shear force, so it will not be of much use to us here. Get ready to give a presentation of this work to the class to help everybody understand better the output from beam elements. 2RHσσσσtop tWNow to get ready for nonlinearity, I would like you to review what you learned in 2.002 on elastic-plastic beam bending. To help refresh your memory and provide fresh material for your nightmares, I am attaching a 2.002 handout on elastic-plastic beam bending. I will give a brief review of this stuff on Wed before your presentation. In the following calculations you should focus on the flat left part of the beam, which, in the fully loaded condition, is subjected to a constant bending moment, Mloaded=112.5MNm, and neglect the effects of the superposed axial force. Assume that the material is elastic-perfectly plastic with a yield stress σy=4GPa. What is the elastic limit moment for this beam? What will be the extension of the elastic core in the fully loaded condition? What will be the curvature of the beam in the fully loaded configuration? What permanent curvature will the beam spring back to when we unload it? How many integration points through the thickness do you think we will need in a beam model or a shell model to capture this behavior properly? How many quadratic continuum elements through the thickness would you use for a continuum model of the structure? Get ready to discuss these issues in your presentation.ρMMBeam Bending -Review--Plane sections remain plane.xyρ : radius of curvature; k = 1/ρ :curvature; M: momentCompatibility : εxx= - y/ρ Independent of material behaviorEquilibrium: M = -σxx y dA Independent of material behavioryneutralaxisεxxtensioncompressionMyσxx(1)AM = -σxx y dA = E /ρ y dAyneutralaxistensioncompressionElastic beam bendingelastic material behavior : σxx =E εxxE1εσAσxx =E εxx = -E (y/ρ)A2I : second moment of areaM=EI / ρσxx=-M y/Ilinear stress distributionEI11/ρΜlinear moment-curvature relationshipσxxI = y dAA2for rectangular beamsfor I-beamsBHBhHb/2 b/2I = BH /12I = (BH -bh )/12333(2)Elastic -plastic beam bendingelastic-perfectly plastic material behaviorE1εσEI11/ρMσyMpeM<Me : elastic regime: |σ| <σy at all pointsyM>Me : elastic-plastic regime: |σ|=σy at outer fibers of beamσxxyσxxσyM ->Mp :fully plastic regime:|σ|=σy at all fibers of beamyσxxσynonlinear Moment-curvature relationshipIn the elastic regime σxx=-M y /I -> | σmax | =M ymax /I Yielding initiates at the outer fibers of the beam when | σmax |= σy. This corresponds to a bending moment Me :Me = (σy I )/ymaxfor rectangular beams : Me = (σy BH )/6(3)2Elastic -plastic beam bendingyFor M>Me the beam is in the elastic-plastic regime: the core of the beam (between y=- c and y=c)is in the elastic regime, while the outer fibers are in the plastic regime (σ=σy).εxx = -y/ρyσxxσyfor rectangular beams : Mp = (σy BH ) / 4 = 1.5 Me2εσσyεy = σy/Eelastic-perfectly plasticmaterial behaviorεy−εyc-cyσxxσy−σyc-c for -c<y<c ε=σ/EFor |y| = c , ε=-σy/E --> -y/ρ = −c/ρ=-σy/E--> the extension of the elastic region is given by:The moment-curvature relationship is then given by:For a rectangular beam :c= ρ (σy/E)M = σy (3H -4c ) B/1222For very large curvatures c-->0 and the moment approaches thelimit moment Mp, where the entire section is in the plastic regimeA2-cc-ymax-ccymax M = -σxx y dA = E /ρ y dA + -σy y dA + -σy y dAM p = -σy y dAA(4)Elastic -plastic beam bending . Unloading /springbackmaterial behavior: elastic unloadingE1εσyσxxyσxxσyE1∆σ∆ε∆σ = E ∆εThe beam unloads elastically∆(1/ρ)=∆M / EI =-M loaded /EI(1/ρ)unloaded =(1/ρ)loaded-M loaded /EIEI11/ρEI1M(1/ρ)(1/ρ)loadedunloadedMloaded∆M∆(1/ρ)Stress distribution upon unloadingyσxxσystress distribution in theloaded configuration σloadedChange in stressupon unloading ∆σ=(Mloaded


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MIT 2 31 - Nonlinearity

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