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This 10 page Class Notes was uploaded by Jackeline Wuckert on Friday October 23, 2015. The Class Notes belongs to ME512 at University of Louisville taught by RogerBradshaw in Fall. Since its upload, it has received 61 views. For similar materials see /class/228364/me512-university-of-louisville in Mechanical Engineering at University of Louisville.
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Date Created: 10/23/15
Displacement Field 2 Node 1D Bar Element gt Recast ux in terms of nodal displacements u1 and u2 Express ux at nodal locations u1 at x1 u2 at x2 Solve simultaneous equations to find B1 and B2 in terms of ul u2 L 7 M 499 l quot 3 1329 X 0 am A 5 51YZ e w p a w 3961 r latTa a Substitute to solve for B1 3 F 2 I I I 5 quoti in 1 a z l 5 1quot XML X2 a2gtltf L L 29 Displacement Field 2 Node 1D Bar Element gt Substitute solutions for 31 and 32 into ux wefv 2133 OW5139 iil w meg4quotqu 562L 1 Lrwmmw 4 NJ MW wh r3 37x l 4 X X a ggo w U 1 39 1 Ll xiii gt Typical FEA form eld expressed in terms of nodal values Can repeat procedure for many cases parabolic 2D 3D etc gt Check that we have amt K A e u the expected behavior Note that these ux forms are identical L 3 quot quot fV l z W2 10 01 a L41 Shape Functions 2 Node 1D Linear Bar gt Consider the uX form expressed in terms of ul u2 x2 x L gt We can rewrite this as x xl L ux ju1 142 14xNlxulN2xM2 Nlxx2L xNZltxgt x1xlj gt N1 and N2 are called the element shape functions Consider their behavior over the element domain X 6 X1 X2 Each equals 1 at its associated node 0 at other node i I t x Ah I 4 91quot n a z o p gt were K 5 vb 76 lav e qv damnk 1me Shape Functions General gt Shape functions are used for all elements in FEA gt General expression of field quantities using shape functions Field quantities uX vXy temperature magnetic eld etc Summation over the number of nodes in the element i l 7 N no MOO N 20 0 29 ub i N 2amp7 ULL ED M Vt 739 1 7amp3 i gt Key feature of shape functions Equal l at their associated node ie NiXi l Equals 0 at all other nodes ie NiXJ 0 when i at j Equals function values at locations away from nodes gt This ensures that the field quantity in question always equals the value prescribed at a given node We will consider 2D and 3D shape functions later Formal Method To Find Stiffness Matrix K gt Formal method to determine K always applicable k fg aaw Jva 5v may Vdew Ma aval Ph ml WWW Lawzquot Shani dvspl uanu Maui6 E 7 constitutive matrix stress strain relationship B 7 element strain expressed in terms of nodal displacement Integrate quantity over domain of the elment Velem gt This approach is based on energy methods Principle of minimum potential energy PMPE Application of variational calculus 7 find minimum PE Stiffness matrix K results as above We will derive this equation later gt Use this to find K for 1D bar element Formal Method To Find K 1D Bar Element gt Express displacement eld in terms of nodal DOFs Use shape functions obtained earlier 39 ra U I Ma Q X F gt 39 39 gtF u x f 2 z 1 g 2 N 3 t Rewrite this in matrix form as x f 10 5344 were gabE 2 g y slam9amp7 Kindaer Inaw UL cl r qu DGF VCL VV Differentiate to find the strain at any point in the element cLt d e 77 5mm 40 ex 39 D 7 kl 4 JIMj Lm E i r i I 397 f Uquot r lt 39quot I i v c39 if LA 1 I r 91L I u 14 Formal Method To Find K 1D Bar Element gt Check strain using strain displacement matrix B S c p A k llquot 2 4L 5131 gt 6 s V J a Cx N l B7 7 I 3 L L gt Substitute into integral and solve E is simply modulus E L riffssz H bigH Vi X V D 1 AE l quotl E rlt ll gt Same as direct method K Will always turn out this way gt Form BT E B With symmetric E guarantees K is symmetric 2715 Model of Several Elements Global Stiffness gt Actual structure typically consist of many elements Above 7 K for a single element often called Kelem or K81 Assemble all element stiffness matrices to describe the entire structure Label this collection as the global stiffness matrix K gt One approach augment element stiffness matrices Include all model DOFs place 0 values in slots not in each element Simply add all element matrices together to find global stiffness matrix Relates all DOFs in problem to all nodal loads in problem quot1 quot2 quot3 i 2 quot3 Examgle om textbook 11 41 0 0 0 0 7k k1 0 and 0 k2 42 l5 3 0 0 o 0 4 k2 Element 1 Elemcm 2 Figure 223 3 Structure formed by two bar elements k1 4 1 0 i F f These arethe kl kllk2 42 H2 F or KllDl R external nodal loads 0 42 12 143 F3 appliedto the model Assembly of Elements Nodal Loads gt Recall example from previous page One item that may occur to you 7 what are F1 F2 F3 Element stiffness matrices gt forces are elernent nodal loads What about assembly Speci cally what about node 2 F2 below k1 4 l 0 1 F1 k1 klk2 42 42 F2 1 2 3 1quot ka quot3 F 3 0 k2 k2 gt DL Logan A First Course in the Finite Element Method supplemental reading handout Consider the sum of all elernent nodal loads at a given node By equilibrium this is equal to the externally applied nodal load Thus only need to consider external nodal load for K d r I D 3 2 FL In 13 lei fizz 29 FH Figure 2 7 Nodal forces consistent with element force sign convention Spring Models Examples gt Consider two spring model examples in detail Additional examples in reading handout F 5 9565 25 Iii gt1 Zaao I75 MW wt 00 55 4 2 395 t Lug K Xvijocks y glaml 3917 am quot 4 I nd E gxi W W gquot 51 jODrS Myc 2 5 w I 200 Ix M grab Ir Jimhawquot em f ch Fawn a gt c 5 I Lu 0 A ML 2 4 ND 14 2 Q I E I A W waft 402 1 7 i LL54 30 pa Zr aw PJ MAX General FEA Formal Solution Method gt Preceding spring FEA examples Solved by breaking equations into parts Found unknown DOFs a led to unknown nodal forces gt Formal method same approach but formally stated Assume there are m DOFs A g c c C 1 M Rearrange K d r into two sets of equations mpx m W76 r k K kl W1 D017 m 4 N Maw 137 lt IIZLL quot P Umkwa DoFJ M AJ where 11 px r9 Iquot Px p 52 F quotl SQ le v 2 AX 51 quot 390 may DoFx Kmun Grabf lt11 N x A an Cam an DoFr quotLN Wquot 0 J 219 gt Last slide reorganize equations it c r Ff 52quot f N N n l I K2 in I39 2 x kl A gt Rewrite this as two matrixvector equations c 67 ff 91X EIL dc 1 WA 12 313 139 Ell Elf 7 VX gt If a sufficient number of boundary conditions DOFs are provided to prevent rigid body motion gt K11 not singular Can solve equation VV for unknown vector of DOFs dx v 9 quot c fn 2 gt 1 f39f 57 25 KWWH gt Substitute dx into VVV to find unknown nodal forces rx Solution is complete 2720 Spring Models Example Repeat gt Apply formal solution method to earlier example Key step is rearranging K d r 7 not just swapping rows 7on9 Lana n 72ch FER Examefe 2T i mL H377 x0e Is Fm tarLEV we x a z I o 07 F a2 Drag 2900 2 1127 0 gt7 MaoO o 7 I Ulgj my goo we can wow 62 111552 EguzLwtj 45 chuk and 2 71 M o I i o M Pill o 1 o Fzooo TAU I N 7 4mm heach Ar 7 g mk IKEhaul 2721 1D Beam Elements gt 2 node lD beam elements quotIquot Uniform area crosssection 1 9d 5 1 2922 Constant modulus E Constant moment of inertial I 39 2 gt 2 DOFs for each node 4 total m a Lateral displacement V1 V2 I M at 2 M2 Rotation 61 92 1 x gt 2 loads for each node 4 total 5 L F Lateral force F1 F2 I I Moment M1 M2 quotgm 139 Ell i if ii39 i 5122352355 quot quot39 gt No axial DOFs loads later 7 simply add lD bar to element gt Theoretical aspects limitations EulerBernoulli beam theory 7 ME 323 no transverse shear effects Beam must be symmetric about xy plane Centroid of beam must pass through line y z 0 7 Can use offset in some beam elements to avoid this not discussed 2 22 ME 323 Symmetric Member In Pure Bending gt Straight beam with single moment 0 Beam symmetric about Xy plane Moment M acts in xy plane MZ gt Consider section in beam Incremental area dA 0 Normal stress 6X 0 Shear stresses CXZ cyZ O 0 M only causes 6X gt Neutral axis 0 Passes through centroid of section Integral is the moment of inertia relative to centroidal axis in y direction M c lesz a amz A I i 0 Can also de ne usmg sectlon modulus S c C M l aquot S 7 a 0m 2 Neutral surface A A 0 Written as I strictly speaking should be IZ 1 M3 Stiffness Matrix 1D Beam Element v M V 213139 gt Beam stiffness matrix M g g g H i 39 e 4 V1 61 V2 J D 239 239 4 loads F1 M1 F2 M2 Wmmmfm wwai 0 K is 4gtlt4 as at ri ht g K11 K12 K13 K14 V1 F1 K K K K 9 M gt Direct method 21 22 23 24 1 1 Cantilever beam equations K31 K32 K33 K34 V2 F2 0 Combine de ectionsslopes K41 K 42 K43 K44 92 M2 I Appmdlx D Beam De ections and Slopes Burn and Loading Elam eum atop in End Equa m at Ehl c cum F 3 Pl P 39 quot5 quot23 FEW 3L 7W 39 Mir2 ML M E quota 3 L M Mechanics of Materials Fourth Edition FP Beer EFt Johnston Jr JT DeWolt 23924 Beam Stiffness Matrix 3 Column F1 K11 K12 K13 K14 V1 VI 3 MI V1 3122 M1 2 K21 K22 K23 K24 91 M 9 92 3H2 F2 K31 K32 K33 K34 V2 aw M2 K41 K42 K43 K44 92 L gt Example 3rd column of K gt Evaluate nodal loads Use the direct method Calculate loads necessary 3 d DOF v2 7 set equal to 1 i39h nodal load is equal to slot KB Other DOFs v1 61 62 7 set to 0 Values shown below gt Combine two cases Cantilever with a load b moment Find values to satisfy DOFs use equilibrium F Fl K11 K12 K13 K14 0 Fl K13 1K1 M1 2 K21 K22 K23 K24 0 M1 2 K23 w i F2 K31 K32 K33 K34 1 F2 K33 7 F33 M2 K41 K42 K43 Km 0 M2 K43 V11 2725 Beam Stiffness Matrix 3 Column gt Cantilever with end load P I W 7 P L3 4 l V V 13916 2 4 Je I P Ptquot L Q i y L 2E gt Cantilever with end moment M M v J i 39 t M L1 44 LT 9 FL ZE I M L r T 39 elm a gt Combine cantilever with end load P and end moment M m M v Vi H5119 I a 2 i P ML par av Jr 2 71 Beam Stiffness Matrix 3rd Column gt Substitute P 2 K33 and M 2 K43 gt Enforce DOF conditions VxL 2 V2 2 1 and 9XL 92 0 z 393 2 v1 Lu i F3 le 33 2 21 3839 25 9 r M 92 o 143 53L 95 23 8 gt Solve for K33 and K43 r z Kn L3 La 0 L ALE Koal 361 2 zen 24 12531 lZEE A i LE 4kg K53 5 33 2 A Lil L1 a 72375 Beam Stiffness Matrix 3rd Column gt Recall original diagram 1445 K33 K43 now known he 5 gt Two terms remains Use static equilibrium f 39 Sum of forces gt K13 Sum of moments gt KM V13 3M0 0 9 lt23 K3gtQ K45 0 ET 1221 i A i 25 L L7 L3 3 L K23 gt Entire 3r01 column of K is now known K13 K23 K33 K43 12E 6E1 12E 6E1 T L2 7 L2 K13 K23 K33 K13 as