9. EXERCISES ON THE FINITE-ELEMENT METHOD

Transcription

1 9. EXERCISES O THE FIITE-ELEMET METHOD

2 Exercise Thickness: t=; Pane strain proem (ν 0): Surface oad Voume oad; 4 p f ( x, ) ( x ) 0 E D ( ) 4 p F( xy, ) Interna constrain: rigid rod etween D and B This exercise can e sove in two different ways: i. expressing the deformed configuration of the eement as a function of the two independent aowed dispacements; the eight noda dispacements are reduced to two, ecause there are 5 externa constrains and one interna constrain. ii. sseming the stiffness matrix without considering the interna constrain; we consider it ony in the end. In this way, the terms of the stiffness matrix, which we need to cacuate, are the usua ones of the four nodes pane eements. Remark: the numers in rackets (),(),... refer to a noda quantities, whie the simpe numer,,... to the degrees of freedom (DOF).

3 Procedure I: We consider the stiffness interna constrain from the eginning and we express the dispacement fied s as a function of the dispacement U 6 and U 8. x( y) xy y( x) ( x, y) ; ( x, y) ; 4( x, y) ; u(x,y)=0 s = U4U8 v(x,y)= U U U U U U + U () 4 () 6 (4) 8 () (4) 8 () 6 (8) 8 (6) 6 ( ) ( ) (8)(, ) x y y x y x ; (6)( x, y) xy ; The matrix B is: x u x 0 0 U6 U6 x xu6 ε 0 0 6, y 8, y y v y 6 8 U 8 U 8 U 8 6, x 8, x y x y x y y B We cacuate the stiffness matrix K, taking into account that, in this case, B i =0 and D =D =0 k ij tb ri D rs B sj d t B i D B j D B j D B j Bi D B j DB j D Bi D B D j B DB d t B DB B DB d j j i j i j B B j

4 88 sx/ t y/ E x y k B D B D d dxdy k 00 E s t dsdt 00 E 6 sx/ E 66 BD BD d s t dsdt t y/ 00 E 6 sx/ ty/ E 68 BD B BD B d s s t t dsdt 00 E k ow we cacuate the equivaent noda oads. The surface oads are: t y/ 6 6 x x dx p dx p t t dt ()(4) 0 0 x x x p p (, )f (, ) 4 4 ty/ x x p 8 8 x x dx p dx p t t dt ()(4) 0 0 p (, )f (, ) 4 4 The voume oads are: P 0 0 F d t d t d 0 F F 6 6 T t F 8 8 sx/ ty/ xy 4p 4 p P dxdy p stdsdt sx/ ty/ x( y) y( x) y 4 p 4 P dxdy p t( s) s( t) dsdt p Then, the governing system of the proem is: K66 K68 U6 p6 P6 K K U p P Repacing the found vaues, we can cacuate the soution of the proem: E (- ) U6 p ( ) U 8 4

5 8 p ( ) p ( ) 6 8 U ; U. E ( ) E ( ) Procedure II: We sove the exercise, cacuating ony the term of the stiffness matrix which are associated to the free DOFs 4,6,8. The presence of the interna constrain reduces to two the independent inear dispacement ut it introduces an unknown interna reaction given y the tension in the rigid rod. If we want to cacuate this rod reaction, we have to sove the system of three equations in three unknowns; whie we sove a system of two equations in two unknown dispacements. x( y) xy y( x) ( x, y) ; ( x, y) ; 4( x, y) ; u u u y 0 y 0 y 0 u4 0 x 0 x 0 x ε u5 x y x y x y u 6 B u 7 u 8 5

6 We cacuate the stiffness matrix coefficients, taking into account that D =D =0 and for even index i B i =0. kij tbri Drs Bsjd t BiD B j B idb j d (i even) sx/ / t y E x y 44 B4D B4D d dxdy 00 E k E In the same way, we otain k k k k k E 6 s 00 t dsdt 6 sx/ E 46 t B4D B6 B4D B6 d s t t dsdt t y/ 00 E E 6 sx/ ty/ E E 68 t B6DB 8 B6DB 8 d s s t dsdt sx/ t y/ E 48 t B4D B8 B4D B8 d s s t t dsdt The surface oads are: 6 () 6 ()(4) 0 E x x x p p p ( x, )f ( x, ) dx 4 p dx and p, whie the voume oads are: xy 4 p p P6 ()( x, y)f( x, y) d dxdy ; 00 p p P8 (4)( x, y)f( x, y) d ; P4 ()( x, y)f( x, y) d. nd so, the governing system is: 6

7 k44 k46 k48 u4 p4 X / k64 k66 k68 u6 p6 P 6 k k k u p P X / where the to the axes. is the vertica component of the traction X in the rigid rod incined at 45 reative ow, we impose the interna constrain: in order to carify the procedure, we rewrite the system in the form of principe of virtua work, from which it derives: T T L L u u Ku u P u i e k44 k46 k48 u4 P4 X / u4 u6 u8k64 k66 k68 u6 u4 u6 u8 p6 P 6 u k k k u p P X / Considering that δu 4 =δu 8, we otain k44 k46 k48 u4 P4 X / u8 u6 u8 k64 k66 k68 u6 u8 u6 u8 p6 P 6 u k k k u p P X / u4 k64 k66 k68 p6 P6 u u u u u u k44 k84 k46 k86 k48 k88 P4 p8 P8 ow, we do the same passages for u 4 =u 8. u 8 7

8 u8 k64 k66 k68 p6 P6 u u u u u u k44 k84 k46 k86 k48 k88 P4 p8 P8 u 8 k66 k64 k68 u6 p6 P6 u u u u u k46 k86 k44 k84 k48 k88 u8 P4 p8 P8 k66 k64 k68 u6 p6 P6 u k k k k k k u P p P E u6 p u 8 This governing system is exacty identica to the system otained y the first procedure. The effect of the constrain (u 4 =u 8 ) consists of adding together the respective rows and coumns of the stiffness matrix. Known u 6 and u 8 =u 4, the unknown X can e cacuated with first and third equation of the system. 8

9 Exercise Finite eement (FE) of Euer-Bernoui; Inextensie rods E. Stiffness matrix of the generic Euer Bernoui FE We consider the Euer Bernoui FE, shown in the foowing figure: The dispacement fied of the FE can e express as: u(x)= u u 4 4 v(x)= u u u + u expressing its in vectoria form: u U v T where U u u u u u u The shape functions have the foowing expressions: ( x) x, 9

10 x 4( x) x x ( x) x x 5( x) ( x) x x x x 6( x) From the dispacement fied, through the compatiiity operator, we can write the expression of the deformations: ( x) u( x) U. ( x) v( x) The Stiffness matrix of the eement is: E 0 T k B DBdx; D 0 0 EI E E EI EI EI EI EI EI EI EI k E E EI EI EI EI EI EI EI EI B 0

11 Resoution of the exercise : To sove the proem, we numerate, as shown in foowing figure, the goa degrees of freedom (DOFs). These are the free DOFs of the structure, which they aow to determine the deformed configuration, and the DOFs associated to constrain reactions that we want to know. The soution of the exercise is dividend in five parts: i. Cacuation of the stiffness matrix of the first FE; ii. Cacuations of the stiffness matrix of the second FE; iii. sseming of the goa stiffness matrix; iv. Cacuation of the equivaent noda oads; v. Soution of the inear system. The stiffness matrix of the first FE Having in mind the stiffness matrix of the generic Euer Bernoui FE, we can simpy cacuate the stiffness matrix of the first FE. We consider ony the oca DOFs associated to the goa DOFs, in this case,5,6; we note that the ending stiffness of the first FE is EI. 6 4 EI The stiffness matrix of the second FE In this case, we consider the oca DOFs, and 5.

12 6 EI sseming of the goa stiffness matrix In order to asseme the goa stiffness matrix, we compose the connection matrix of the proem: DOF FE / / / / / 4 / In the first row there are the oca DOFs of the eements, whie in the first coumn there are the FEs of the structure. Each numer in the tae indicates the goa DOF that correspond to the oca one, j (coumn) of the FE i (row). In this way, we find the connection etween goa and oca DOFs. For exampe the eement (5,6) of the stiffness matrix of the first FE is a contriution of the term (,) of the goa stiffness matrix. Then, we otain the foowing matrix: V 4 6 V EI V 6 0 V 4 In the previous expression we normaize the unit dimension of the unknowns and, to simpify the understanding, we write separatey the contriutions of the eements. The equivaent noda oads In order to determinate the equivaent noda oads, we can proceed in two ways: i. with the cassica method, i.e. ppying the formua which defined the equivaent noda oad associated to the component that we want to know.

13 ii. using the anaogy etween Euer Bernoui equivaent noda oads and the constrain reactions of the doue camped eam. Method I The vector of the equivaent noda oads for the FE is: Q T fdx 0 In our case f T =[0 p(x)], where p(x) is the distriuted oad in y-direction. Repacing the definition of the matrix, we otain: T Q 0 ( x) p( x) dx ( x) p( x) dx 0 5( x) p( x) dx 6( x) p( x) dx In this way, we can cacuate the equivaent noda oad components of the first FE (p(x)=- (-x/)p). s done for the stiffness matrix, we focus the attention ony on the free DOFs (,5,6), which aow to determinate the deformed configuration. Q Q Q x x ( x) p( x) dx p x dx p x x x 6 6( x) p( x) dx p dx p x x x ( x) p( x) dx p dx p In the previous expressions the suscript represent the oca DOF, whie the apex is the numer of the eement. The same operation can e repeated for the second FE, with p(x)=p and considering ony the DOFs,,5. Method II If we know the constrain reactions of the doue camped eam under the distriuted oad of the proem, we can simpify the cacuation, ecause these reactions are the same, uness the sign, to the Euer Bernoui equivaent noda oads. The reactions for p(x) inear and constant, are:

14 sseming fter determining the equivaent noda oads, we do an asseming operation, as made for the stiffness matrix, in which the oca contriutions are summed to asseme the goa vector of known terms Q. The contriutions of the moment p is added to the third component of the vector. p Q 0 p Q 5+Q p 0 Q Q 6+Q p p p p 0 Q p 5 Resoving system ow, we can write the inear governing systems in the adimensiona form: EI V 6 6 V p V V 4 0 Using the Gauss eimination method, the soution is: 4

15 65 V 4 69 V p. 0 EI 0 V 9 V 4 dimensionayzing of the governing system Generay, the terms of a stiffness matrix are not in the same dimension. In this section, we riefy review how otain an adimensiona system. We have the system: F F V F L, F M M where V are the unknown dispacement, θ=v/l the unknown noda rotations, F are the forces and M are the moments. First we homogenize the dimension of the unknowns F F V F L L. F FL M In this way the coefficient of the same row have the same dimension; secondy, dividing to L the rows of equations in rotation, we arrive at a homogeneous matrix. F F V F L L M. F F L L L L 5

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17 Exercise Determinate the dispacement of the point, using two constant strain triange (CST) FEs. ν=0 First, we numerate the FEs and the goa DOFs, as shown in foowing figure. We introduce the numeration of the oca DOFs, and the connection tae. ote that the sign - indicates that the goa DOF has opposite verse respect to oca one. DOF FE / / / / / / / / - / 7

18 We can note that the FE coincide with FE, at ess than a rigid rotation cockwise y 80. In this way, we can use the stiffness matrix of the first FE aso for the contriution of the second one to the goa stiffness matrix. ow, we consider the first FE; we introduce the reference system. In the foowing figure, there are the equations of the trianguar sides, ecause we use them for the determination of the shape functions. The shape functions of the first FE are: x y ( x, y) ( x y); ( x, y) ; ( x, y) ; y Then, we can derive the first, third and fifth coumns of the matrix B: 8

19 0 0 0 B ; DE We cacuate the Stiffness matrix terms (the suscripts indicate row and coumn of the matrix, whie the superscripts the FE). k B D B d r rr r 9 E E; 8 k B D B d r rr r k B D B d r rr r E E; EE; k k B D B d r5 rr r5 E E ; 8 The vector of the equivaent noda oads is simpy to cacuate. I fact the concentrated force is directy appied to the DOF ; whie the first component of the vector is cacuated expoiting the anaogy with a doue camped eam. Then the resuting of the oad PL is distriuted eveny on oth ends of the oaded side n aternative method to determinate the first component of equivaent noda oad vector is the cacuation of the foowing integra: P P ()( o, y) p( y) dy y dy P. 0 0 sseming the quantities, we can write the resoving system: 9-8 u u P. E 8 9 E 8 P 8 - u u 9

20 Exercise ppying the Ritz method, determine the diagrams of ending moment and shear. We adopt the poynomia of minima degree, separatey in two parts 0,. k=ei/ The reference system of the proem is shown in the foowing figure. The five oundary conditions are: v (0) 0, v(0) 0, v ( ) 0, v ( ) v (0), v( ) v (0) We use two poynomia of second degree, six parameters in tota. Five of these are determinate y compatiiity conditions and one y imposing the stationary of the tota potentia energy. x x x x v( x) a c ; v( x) d e f x e x v ( x) c ; v ( x) f ; We impose the oundary conditions: v (0) 0 a 0; v(0) 0 0; 0

21 e f v ( ) 0 f ; c e v ( ) v ( ) e c; v ( ) v (0) c d. But v and v ony depend from parameter c: x x x v ( x) c ; v ( x) c c c ; x c x c c v ( x) c ; v ( x) c ; v ( x) ; v ( x). We write the tota potentia energy as a function of parameter c: EPT( v, v ) EIv dx+ EIv dx+ kv ( ) p( x) v ( x) dx EPT( v, v ) EIv dx+ EIv dx+ kv ( ) p( x) v ( x) dx c c x EPT( c) EI dx+ EI dx+ k( c) p( x) c dx 4c 4c x EPT( c) EI dx+ EI dx+ kc p( x) c dx We impose the stationary of the tota potentia energy: EPT EI EI x 0 4 c dx c dx 4 + kc p ( x ) dx 0 c EI EI p x 8EI p 4c x 4 c x + kc 0 k c c p 8 / / EI k Repacing the vaue k=ei/, we otain: c 4 p / p EI EI EI 8 / / 7

22 4 4 p x p x x v( x) ; v ( x) ; 7EI 7EI p x p x p p v ( x) ; v ( x) ; v ( x) ; v ( x). 7EI 7EI 7EI 7EI The foowing figure shows the deformed configuration. In ascissa there is x/ and in the ordinate v/ c. ow, we cacuate the ending moment and the shear: M EIv ; T EIv ; p M( x) EIv ( x) ; T ( x) EIv ( x) 0 7 p M ( x) EIv ( x) ; T ( x) EIv ( x) 0 7 The shear is nu, ecause we chose the second degree poynomia. The moment is piecewise constant and discontinuous in x=. In the foowing figure, the diagram of moment is represented in the part of the fiers stretched. It is different from the exact soution, which is paraoic in the first part and inear in the second one.

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24 Exercise 4 Determine the interna actions of the rods of the frame, as shown in the figure, using the Euer Bernoui FEs, with the hypothesis of axia inextensiiity. The Young moduus E, and the geometrica characteristics and I are the same in a rods. The structure is symmetric for geometric and oads respect to the vertica axis; then, the centra node doesn t rotate and move horizontay. However, the vertica rod is inextensie and consequenty it can t move verticay aso. We can study the reduced structure, composed y one rod with two unknown dispacements. 4

25 The governing system wi e in the foowing form: K K u P KU P, K K u P where the goa DOFs are connected to the oca DOFs (this numeration is shown in foowing figure) from the connection tae: DOF FE / / / / We can cacuate the stiffness matrix coefficients of a FE from the reaction of a doue camped eam with a unitary noda dispacement (yieding or rotation) imposed. nd so, we otain: EI EI EI K k55= ; K k56= 6 ; K k66=4 ; where k ij indicates the stiffness matrix coefficients of a FE and K ij indicates the goa stiffness matrix coefficients. We cacuate the shape functions that we need for the determination of the equivaent noda oad. x x x x x i( x) a c d ; i( x) c d. 5

26 Cacuation of 5 : (0) 0; (0) 0; ( ) ; ( ) 0; c d d 5(0) 0 a 0; 5 (0) 0 0; 5 ( ) 0 0 c= ; 5( ) c d d d d c. x x 5( x). Cacuation of 6 : (0) 0; (0) 0; ( ) 0; ( ) ; (0) 0 a 0; (0) 0 0; ( ) 0 c d 0 c d; ( ) c d d d d c. x x 6( x). The distriuted oad is p(+x/); then, we otain: tx x x x x p5 5( x) p dx p dx p t t tdt t t. p5 p t t t dt p t p tx x x x x p6 6( x) p dx p dx p t t tdt t t p6 p t t dt p p ternative method to cacuate the equivaent noda oads is to use the constrain reactions of a doue camped eam under constant distriuted oad and with trianguar one.. 6

27 p 7 p 7 p p p 5 ; 6. p p p The vector of the known terms is: 7 p p P 5 0, P p 6 p 5 where P i indicates the equivaent noda oad of the goa DOFs, whie p i the equivaent noda oad of the oca DOFs. ow, repacing the terms found in the resoving system, we can find the dispacement. EI EI u p 6 u 4 u 4 0 p 0 p 60. EI EI EI EI u p 6 4 u u We cacuate the deformation and their derivatives: v( x) ( x) u ( x) u ( x) u ( x) u ; ( ) p x x p x v x x v( x) p x x ; 60 EI 4 EI EI 0 0 p 4 x 7 x p 4 7 x 7 p v ( x) ; v ( x) ; v ( x). EI EI EI We cacuate shear and ending moment: p x 7 M ( x) EI ( x) EIv ( x) 5 4 ; T(x)=M (x) EIv ( x) p We have found constant shear unike the exact soution of the eastic ine EIv which is paraoic shear. p 7

28 To cacuate the constant axia action in the vertica ar, we superimpose the equivaent noda oad of oca DOF due to constant and trianguar distriuted oad. ( x) p p p

29 Exercise 5 Determine the dispacement of the node, using a FE with four nodes. The oad f(x) is paraoic [F/L], whie the F f / is constant [F/L ]. 0. We can note that the proem is symmetric for oads and geometry. Then, we anaysis ony haf the proem, as shown in figure. But in this case the haved proem has more DOFs than the compete proem; instead of. For this reason, we consider a square FE of the side, with the numeration of DOFs iustrated in foowing figure. However, we expoit the symmetry of the proem, noting that the diagona terms of the stiffness matrix, the equivaent noda oads due to surface and voume oads, are equa for the unknowns DOFs 6 and 8. We actuay trace ack to a proem in one unknown. 9

30 We cacuate the shape functions at node and 4: ( ) ()(, ) xy ; (4)(, ) y x y x y x. Using these shape functions, we cacuate the sixth and eighth coumns of matrix B: x x B ; D E 0 0. y y We can cacuate the eements k 66 and k 88 of stiffness matrix, using the expression of matrix D, with ν=0. k sx/ ty/ x E y E E dxdy sx/ ty/ x x E y E k E dxdy s t 0 0 We cacuate the voume equivaent noda oads (g 8 =g 6 )... E E dsdt ; s s t 6 () xy f f x y f g ( x, y) F( x, y) dxdy dxdy ; 4 E dsdt 0. 6 and the surface equivaent noda oads (p 8 =p 6 )... 0

31 x x x f 6 ()(, ) ( ) 4 4 4, p x p x dx f dx f s s ds f where we cacuate the expression p(x), knowing that it has a paraoic shape and imposing the condition of transition for three known points: p(0) 0 xx p( ) 0 p( x) 4 f. p( / ) f We note that, expoiting the symmetry of the proem, we coud cacuate the equivaent noda oad p 6 ike as haf of the resutant of the distriuted oad on the side (i.e. as reaction of douy supported eam under the same oad): x x f ( ) 4 4 4, p x dx f dx f s s ds f Being k 68 =0, we have two identica equations uncouped. Then we find the soution of the proem: E f f 7 k66u6 p6 g6 u f 6 u8 u6. 4 6E

32 Exercise 6 Determine the vertica dispacement of the node C, using two Euer Bernoui FE. The Young moduus E, and the Geometrica characteristics of the section, and I, are the same for a the ars. The area of the section is I/. The figure shows the DOfs of Euer Bernoui FE of side. Its stiffness matrix is: E E EI EI EI EI EI EI EI EI k E E EI EI EI EI EI EI EI EI

33 We can cacuate the stiffness matrix coefficients of a FE from the reactions of a doue camped eam under a unitary noda dispacement (yieding or rotation) imposed. Otherwise we cacuate the foowing integra: k E 0 T B DB dx; i.e. kij Bsi Dst Btjdx; with D, EI where the matrix B is the connection etween the noda parameters and the generaized strains. ( x) u( x) U. ( x) v( x) For exampe, to cacuate k, we use and ; note that B =B =0 x x (0) ; ( ) ( ) (0) 0. x x (0) ; ( ) ( ) (0) 0. B 6 x x 6 x x x x ( x) ; (x) ; ( x) 4 ; ( x) ; 6 x x k B D B dx B D B dx ( x) D ( x) dx EI dx s st t t 7 6 k EI t t dt EI t t t EI. 0 0 We indicate the goa DOFs with V B, V C, and θ B, which are connected to the oca DOFs from the connection tae. The numeration and the verse are shown in the foowing figure. From the verse of each FE, we can know the numeration and the verse of the oca DOFs.

34 DOF FE / / / V B / -θ B / V B - θ B / V C / The sign - in the connection tae indicates that the goa DOF has contrary verse respect to the corresponding oca one. ow we cacuate the contriution of each FE to the goa stiffness matrix. They are sumatrices of the stiffness matrix of the generic Euer-Bernoui FE, where we change the sign to the rows and coumns associated to the oca DOFs that has contrary verse respect to the correspondent goa one. k EI EI EI 6 V B E 0 VB EI EI EI ; B ; EI k 0 4 B EI EI EI 6 V C We can cacuate the goa stiffness matrix of the system, repacing with I/ : EI E EI EI EI EI EI 6 V 4 6 B EI EI EI EI EI EI K ; B EI EI EI EI EI EI 6 V 6 C whie the equivaent noda oad vector is: 4

35 F T p p 0; The first component of the vector is simpy to cacuate, ecause the oad p is directy appied to the DOF V B, whie the vaue and the verse of the second component can e cacuated expoiting the anaogy etween the equivaent oads of the Euer Bernoui FE and the reactions of the doue camped eam under the same oads. Otherwise we have to resove the integra to find the equivaent noda oads: P 0 T n dx, p where n is the axia oad vector, p the transversa one and the matrix connects the dispacements to the noda parameters. s the exampe, we cacuate the second component of the vector F: from the connection tae, we have (the superscript indicated the eement and the suscript the component of the vector): B 6 P6 6 0 F P P ( x) p( x) dx where p(x)=-p in the chosen reference and 6 is: 6( ) x x x 6 ( ) ; 6( ) 6 (0) 6 (0) 0. Then 4 x x t t p F pdx t t pdt p. B ow, we resove the system and we cacuate the dispacement V C : 5

36 EI 4 6 VB VB p p p B B VC. 0 EI 0 EI 6 V C 0 V C 6

37 Exercise 7 Determine the governing system using two CST FEs 0; g ; E. 4 In figure there are the oca and goa DOF numerations of the system. In the second FE, the numeration is chosen in order to make it coincide with the first FE ess than a rigid rotation of 90 countercockwise. DOF FE / 5-5 / / / 7

38 The sign - in the connection tae indicates that the goa DOF has opposite verse respect to the corresponding oca one. We cacuate the shape functions and the stiffness matrix of the first FE. From the previous figure, the shape functions are: () ( x, y) x ; ()( x, y) y x; ()( x, y) y x, and so the matrix B is: B ; D E ow, we can cacuate the coefficient of the stiffness matrix; for exampe, we report the cacuation of k 5 : 5 r rr r5 0 k B D B d E E with the same procedure, we can otain the stiffness matrix of the first FE: 8

39 k E ote that the second FE stiffness matrix is equa to first one, at ess than a rigid rotation; the geometry and the numeration of the DOFs are the same. Expoiting the connection tae (changing the sign to the stiffness matrix coefficients, where in the connection tae there is - ), we report each FE contriution to the goa stiffness matrix. Outside the matrix coumns, we indicate the correspondent goa DOFs. k E 4 ; E k ; k E spring E In the previous expressions, there is aso the contriution of the spring. ow we cacuate the equivaent noda oads: first, we consider the voume oad of the first FE (the superscript indicated the eement, whie the suscript the oca DOF): 9

40 We cacuate the voume equivaent noda oad associated to the sixth DOF of the first FE. In the chosen reference g( x, y). 0 x 0 x 0 s 6 () x x s g ( x, y) g( x, y) dxdy ( x y)( ) dxdy ( t s) dtds 0 s 0 0 t s 6 s g st ds s ds. Remark: the voume oad is constant, and so the voume equivaent noda oads 6 4 g, g, g are equa to the resutant of the voume force, divided y three. The other components are cacuated in the same way. Instead, to cacuate the voume equivaent noda oads of the second FE, we note that they are the same of one due to the horizonta voume oad of -γ in the first FE. In this way we don t cacuate the shape functions of the second FE. Expoiting the connection tae, we can find the goa vector of voume equivaent noda oads: g T go To write the system, we have to cacuate the equivaent noda oads associated to the trianguar distriuted oad on the side D. The ony contriution to the governing system is due to equivaent noda oad of DOF of the first FE (p ). t ()(, ) ( ) ( ) ( ) ( )( ) p y p y dy x y dy t t dt t p ow, we can write the governing system of the proem: u u 6 E 4 u 8 0 u4 6 u5 6 40

41 Then, the soution of the previous system is: u u 5 u 4. 6 E u4 0 u5 40 Exercise 8 Determine the defection of the ends,using the Ritz method with the poynomia of minimum degree. Euer Bernoui eam mode; width of the section is constant; Specific weight γ [F][L] - ; α. The inear variation of the section height aong the eam axis invoves a different weight per unit ength and different moment of inertia to vary ascissa. First, we cacuate the aw of eam height variation with position x on the axis, knowing that it has inear aw. h( x) Bx 4

42 h h(0) h h x h( x) h h () h h h B h B The inertia moment as function of the position x, is: x I( x) h ( x) h, and in the same way the distriuted oad, considering the weight of the eam infinitesima part dx: dv h( x) dx p( x) dx p( x) h( x) x p( x) h. Our proem is to cacuate the defection of the Euer- Bernoui eam under a distriuted oad and variae inertia moment with known aw. First, we write the tota potentia energy for this case: EPT ( v) EI ( x) v( x) dx p( x) v( x) dx. 0 0 We have to choose the approximant function ( ) the compatiiity, i.e. v( x) C 0, and v(0) v( x) 0. vx for the deformation, which it respect We take the second degree poynomia, which has ony one free parameter in addition to the two one that are necessary to satisfy the oundary conditions. x x B x C v( x) B C ; v( x) C ; v ( x) ; v(0) 0 0 x. c. B v( x) C. v(0) 0 B 0 We write the tota potentia energy in function of the free parameter C: 4

43 C x EPT ( C) EI( x) dx p( x) C dx 0 0 / x C x x tx E h 4 dx h C dx C Eh t dt C h 4 tt dt 0 0 t dt C h t 6 t dt 0 0 C Eh We impose the stationary of the tota potentia energy: EPT ( C) Eh C t dt h t t dt 0. C 0 0 We cacuate the integras in the previous equation: 4 4 t ; 4( ) 4( ) 4 t dt t t dt 0 0 t t 4, 4 4 and repacing them in the equation of stationary of EPT: EPT ( C) Eh C h C 4 0; from which we can find the expression of C: dept ( C) dc C E h C 4 he 4. h. The deformation of the eam and the dispacement in x= are: 4 v x C x x h E ( ) ; 4 v( ) C C. h E 4

44 We can do a dimensiona verification in order to confirm that the parameter C has the dimension of a distance (α is adimensiona): F L E FL L h L ; ; ; ; 4 4 F L L C he F L L L. 44

45 Exercise 9 nayze the foowing structure using FEs for pane proems ssume 0 and pane stress proem. We use two CST FEs and one FE with four nodes; the proem has four unknowns, i.e. the vertica dispacements v B, v C, v E, v F of nodes B, C, E, F. But for the symmetry of the proem (oth geometry and oads), v B =v C, v E =v F, the proem has ony two independent unknowns. If we expoit the symmetry to reduce the computation, we can sove haved structure, as shown in the foowing figure. This figure iustrate aso the numeration of the eements and the goa DOfs: 45

46 Therefore, in this case, the numer of unknown are four; this ecause we impicity adopt for the compete structure, a discretization richer that the previous one ( FE of four nodes + CST in comparison to FE of four nodes + CST). priori for the pane proem v H v B =v C and v G v E =v F, whie discretizing a the structure with ony FE of four nodes and CST, eing the dispacement of the sides EF and BC inear, we have: v v v /v v and / G E F E F v v v v v. H B C B C Then, we can use different approaches for the soution of this proem: i. Sove the compete structure with FE of four nodes and CST FEs; ii. Sove the proem with ony two independent unknowns, considering ony haved structure; we otain the same soution of method (i). iii. Consider haf of the structure with four independent unknowns, otaining a soution more accuratey that one of point (i) and (ii). We choose to sove the proem using the approach (ii), ecause the more pecuiar than the others and ecause we have ony two goa DOFs simutaneousy, and the advantage of anayzing ony haf structure. Of course aso other approaches are vaid. In the end, there is the soution otained with the approach (iii), which is different from the soution (ii) ony for assemy. In the appendix, there is an aternativey approach to sove the proem with method (ii). Then, we consider the haved proem, and we reduce the proem to one with two DOFs. To do this, we impose v B =v H and v G =v E. The foowing figure iustrates the two FEs and their oca numeration. 46

47 We impose aso in the connection tae the equaities v B =v H and v G =v E. This condiction in term of DOFs ecames u =u 4 and u =u. DOF FE / / / / 4 / / / / / 4 / 4 We cacuate the stiffness matrix of the first FE. In foowing figure there is the reference system, the node coordinates and the equations of the trianguar sides to cacuate the shape functions. From the previous figure, we cacuate the shape functions: () ( x, y) x ; ()( x, y) x y ; ()( x, y) y, ow, we get the second, fourth and sixth coumn of the matrix B B,, 0 ; 0. y y, y E D,, 0 x x, x We cacuate the coefficients of the stiffness matrix; this goes ack to the integra of the sum of two terms, expoiting that D =D =0 and that the first eement of the even coumns of the matrix B is nu (B ii =0, i even). 47

48 k ij B ri D rs B sj d B i Ds Bsj Bi D B j DB j D k ij B i DB j B i DB j d k 44 B 4 D B 4 D d k 66 B 4 D d k 46 B 4 D B 6 d k 4 B D B 4 d B B D j i E E ; 4 E E ; E E ; E E. 4 B j D B j DB j d The even rows of the fourth coumn of the stiffness matrix correspond to the noda vertica reactions due to the unitary dispacement imposed to the DOF four; Their sum, for equiirium in y-direction must e nu. This cacuation provides a usefu verification of the correctness of the cacuations. E k4 k44 k We consider the second FE: in foowing figure there are the noda numeration and coordinates, the oca DOFs, the oca reference system and the equations of the sides. We cacuate the shape functions: () ( x, y) x y ; ()( x, y) x y ; 48

49 ()(, ) x y ; (4)(, ) x x y x y y ; and the even coumns of the matrix B are: x x x x B,,, 4 y y y, y ; y y y y,,, 4 x x x, x so in this case, eing invoved ony the vertica DOFs, aso for ν 0, the cacuation of the stiffness matrix terms is simpified: Bi 0 i even kij BiD B j Bi DB j d D D 0 However, expoiting the symmetry and the meaning of the stiffness matrix terms such as reactions to an unitary dispacements (see foowing figure), we can further reduce cacuations y oserving that: k k k k ; k k ; k k ; k k ;

50 In the foowing integras, we use the transformation of coordinates: x s dx ds y t dy dt k B D B D d E s 4t dsdt 00 E E k k44 k66 k88 ; E k B D B B D B d s s 4 t dsdt E 4 E k4 k68 4 ; E k B D B B D B d s s 4 t t dsdt E E k6 k48 4 ; E k B D B B D B d s 4t t dsdt 00 E E E k8 k46 4 ;

51 s verification of the previous cacuations, we contro that the sum of the vertica reactions, for equiirium in y-direction, due to the unitary dispacement of the DOF is nu: E 4 k k k k Then, we assemy the stiffness matrix, using the connection tae: k k k k k k k k k k K k k k k k k k k k k In the stiffness matrix, the apex indicates the numer of eement and the suscript the row and the coumn of its stiffness matrix. Being the oad vector otainae without any cacuation, we can write the governing system and its soution: E 5 4 u 0 u P 4 P. 4 u u E This soution is equa to one otain with the approach (i), i.e. discretizing the entire structure with CST FEs and FE of four nodes. If we do not impose any equaity of DOFs in the connection tae, we can sove the proem using a four DOFs. It 's the case of the approach (iii) descried at the eginning: the soution is more accurate ecause we are impicity using 4 eements instead of one to discretize the entire structure. In this case the connection tae ecomes: DOF FE / / / / 4 / / / / / / 4 and the stiffness matrix is: 5

52 k44 k k4 k6 k46 k k k44 k46 k 48 E K. k k k k k46 k k48 k68 k66 k 88 nd so the resoving system is: E u u 0 P ; u u 0 u 8 8 u P. u E u 6 4 If we are interested in cacuating staticay indeterminates associated to the constraint of equaity of the dispacements, we can use this system (the stiffness matrix is unchanged): u X E u X ; u Y u 4 P Y X P Y 4 and u P 4. u E 4 5 ppendix In the appendix, we riefy introduce an aternative way to sove the proem with the approach (ii). 5

53 Rather than take into account the constraint of equaity of the dispacements in the connection tae, we can do it from the description of the dispacement fied of the FE with four nodes. s ( x, y) u ( x, y) u ( x, y) u ( x, y) u y () () () (4) 4 where the u i are the goa DOFs. Since u = u and u = u 4 we can introduce the matrix : sx 0 0 u ; s u y () () () (4) 4 and the matrix B is: 0 0 x s x x 0 0 u ε 0 0 y s y y () () () (4) u 4 y x y x 0 0 () () () (4) 0 0 u u ε., y, yu 4 u4 () () () (4) 0 0, x, x B The stiffness matrix is cacuated very easiy using the usua formua k ij ri rs sj B D B d E u k ; u 4 To otain the resoving matrix, we assemy to k the contriution of the trianguar FE. E E / E 5 4 k So we have otained the same matrix with a different procedure. B 5

54 54