dw 2 3(w) x 4 x 4 =L x 3 x 1 =0 x 2 El #1 El #2 El #3 Potential energy of element 3: Total potential energy Potential energy of element 1:

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1 MAE 44 & CIV 44 Introduction to Finite Elements Reading assignment: ecture notes, ogan. Summary: Pro. Suvranu De Finite element ormulation or D elasticity using the Rayleigh-Ritz Principle Stiness matri and nodal load vectors or D elasticity prolem Aially loaded elastic ar y F A() cross section at () ody orce distriution (orce per unit length) E() Young s modulus Potential energy o the aially loaded ar corresponding to the eact solution u() du Π(u) EA d u d Fu( ) d Potential energy o the ar corresponding to an admissile displacement w() Π(w) EA d d w d Fw( ) Finite element idea: Step : Divide the truss into inite elements connected to each other through special points ( nodes ) Π(w) 4 El # El # El # otal potential energysum o potential energies o the elements EA d d w d Fw( )

2 otal potential energy Π(w) EA d d 4 El # El # El # Potential energy o element : Π(w) EA d w d d Potential energy o element : Π (w) EA d w d d w d Fw( ) 4 El # El # El # Potential energy o element : 4 4 Π (w) EA d w d Fw( ) d otal potential energysum o potential energies o the elements Π(w) Π(w) + Π (w) + Π (w) Step : Descrie the ehavior o each element In the direct stiness approach, we derived the stiness matri o each element directly (See lecture on Springs/russes). ow, we will irst approimate the displacement inside each element and then show you a systematic way o deriving the stiness matri (sections. and. o ogan). ASK : APPROXIMAE HE DISPACEME WIHI EACH EEME ASK : APPROXIMAE HE SRAI and SRESS WIHI EACH EEME ASK : DERIVE HE SIFFESS MARIX OF EACH EEME (this class) USIG HE RAYEIGH-RIZ PRICIPE Summary Inside an element, the three most important approimations in terms o the nodal displacements (d) are: Displacement approimation in terms o shape unctions w() d Strain approimation in terms o strain-displacement matri ε() B d () Stress approimation in terms o strain-displacement matri and Young s modulus σ EB d () ()

3 he shape unctions or a D linear element For a linear element - () El # Within the element, the displacement approimation is - - w() d + d - () Displacement approimation in terms o shape unctions Strain approimation Stress approimation - w() - dw d ε [ ] d d E d σ Eε [ ] d d d Why is the approimation admissile? 4 El # El # El # For the entire ar, the displacement approimation is w() w () () + w () () + w Where w (i) () is the displacement approimation within element (i). et use set d. hen, can you seen that the aove approimation does satisy the two conditions o eing an admissile unction on the entire ar, i.e., () w( ) dw () d () () eists ASK : DERIVE HE SIFFESS MARIX OF EACH EEME USIG HE RAYEIGH-RIZ PRICIPE Potential energy o element : Π (w) σ ε Ad w d ets plug in the approimation w() d ε() B d σ EB d ( ) ( ) A Π (d) d B EB d d d d

4 ets see what the matri BEB Ad is or a D linear element Recall that Hence B B E B [ ] E E [ ] ( ) [ ] E ( ) B EB Ad AEd ( AEd ) ( ) ( ) ow, i we assume E and A are constant ( ) ( ) ( ) AE( ) BEB Ad AEd AE ( ) Rememering that ( - ) is the length o the element, this is the stiness matri we had derived directly eore using the direct stiness approach!! hen why is it necessary to go through this complicated procedure??. Easy to handle nonuniorm E and A. Easy to handle distriuted loads For nonuniorm E and A, i.e. E() and A(), the stiness matri o the linear element will O e EA ( ) But it will AWAYS e k BEB Ad ow lets go ack to Π (d) d B EB d d d d A k d k d d Element stiness matri k BEB Ad Element nodal load vector due to distriuted ody orce d 4

5 Apply Rayleigh-Ritz principle or the D linear element Π(d) Π(d) Π (d) Recall rom linear algera (ecture notes on inear Algera) Π(d) d kd d Π(d) kd Hence Π (d) k d Eactly the same equation that we had eore, ecept that the stiness matri and nodal orce vectors are more general Recap o the properties o the element stiness matri k BEB Ad. he stiness matri is singular and is thereore non-invertile. he stiness matri is symmetric. Sum o any row (or column) o the stiness matri is zero! Why? k Sum o any row (or column) o the stiness matri is zero Consider a rigid ody motion o the element d d k k d k k + k d k k and k + k Element strain ε B d ( A ) k d B EB d d ( ) BEBd Ad 5

6 he nodal load vector () d d ( ) ( ) d d d d ( ) ( ) ( ) d ( ) d d Consistent nodal loads () /unit length Replaced y d d A distriuted load is represented y two nodal loads in a consistent manner e.g., i ( ) d ( ) d ( ) d ( ) d Divide the total orce into two equal halves and lump them at the nodes What happens i ()? d d Summary: For each element Displacement approimation in terms o shape unctions w() d Strain approimation in terms o strain-displacement matri ε() B d Stress approimation σ EB d Element stiness matri k BEB Ad Element nodal load vector d What happens or element #? 4 4 Π (w) EA d w d Fw( ) d For element d w() 4 4 d 4 w( ) d 4 he discretized orm o the potential energy ( ) ( ) 4 A 4 Π (d) d B EB d d d d Fd4 6

7 What happens or element #? ow apply Rayleigh-Ritz principle Π (d) Step:Assemly eactly as you had done eore, assemle the gloal stiness matri and gloal load vector and solve the resulting set o equations y properly taking into account the displacement oundary conditions kd + F Hence there is an etra load term on the right hand side due to the concentrated orce F applied to the right end o the ar. OE that whenever you have a concentrated load at AY node, that load should e applied as an etra right hand side term. Prolem: 6 Pl 4 E 6 psi ρ.86 l/in hickness o plate, t Model the plate as inite elements and ()Write the epression or element stiness matri and ody orce vectors ()Assemle the gloal stiness matri and load vector ()Solve or the unknown displacements (4)Evaluate the stress in each element (5)Evaluate the reaction in each support Solution () ode-element connectivity chart Finite element model Element # ode ode El # El # Pl Stiness matri o El # () E k BEB d A ( A( ) d () ) A( ) d t(6.5 ) d t (6.5 ) d 6 in () E 6 k ( 6).5 () 7

8 Stiness matri o El # ( ) () 4 E 4 k BEB Ad A( ) d () A( ) d t(6.5 ) d t (6.5 ) d 45 in () E 6 k ( 45) 9.75 () ow compute the element load vector due to distriuted ody orce (weight) d For element # () ρ A d () ( ) t () ( ) A( ) ( ρ ) d A d ρ (6.5 ) d.86 l l 8.58 () () () ( ) () ( ) ( ) ( ) Superscript in parenthesis indicates element numer El # For element # () ρ A d () 4 ( ) t () ( ) A ( ) ( ρ ) d A d ρ (6.5 ) d 4.86 l l () ( ) () ( ) () () El # El # 4 ( ) ( ) Solution () Assemle the system equations K concentrated load l l concentrated load l

9 Solution () Hence we need to solve.5.5 d R d d R is the reaction at node. otice that since the oundary condition at (d ) has not een taken into account, the system matri is not invertile. Incorporating the oundary condition d we need to solve the ollowing set o equations d d Solve to otain 5 d.996 in 5 d Solution (4) Stress in elements otice that since we are using linear elements, the stress within each element is constant. In element # () σ () () EB d E d [ ] d 6 d d.99 psi In element # () σ EB () () d E d [ ] d.588 psi 6 d -d ( ) Solution (5) Reaction at support Go ack to the irst line o the gloal equilirium equations.5.5 d R d d R.688 l (he ve sign indicates that the orce is in the ve -direction) R Check 6 he reaction at the wall rom orce equilirium in the -direction 4 R P+ 4 ρa() d Pl 4 ρt + (6.5 ) d.688 l 9

10 Prolem: Can you solve or the displacement and stresses analytically? Check out Comparison o displacement solutions. -4 Analytical solution u anal + < + + Stress or or 4 du du σ ( ) anal 6 anal anal E d d Displacement (in) Finite element solution 5 5 (in) otice:. Slope discontinuity at (why?). he inite element solution does not produce the eact solution even at the nodes. We may improve the solution y () Increasing the numer o elements () Using higher order elements (e.g., quadratic instead o linear) Comparison o stress solutions Stress (psi) 5 5 Finite element solution Analytical solutions (in) he analytical as well as the inite element stresses are discontinuous across the elements

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