able of Contents CHPER INRODUCION O HE FINIE EEMEN MEHOD ND BR EEMENS.... Simple springs.... Bar elements....3 System analysis... 4.3. Eample ially loae bars with varying cross-sections... 5.4 Properties of the stiffness matri... 8.4. Sparsity... 8.4. Symmetry... 8.4.3 Only positive iagonal elements... 9.4.4 Positive efinite... 9.4.5 Singularity... 9.5 rbitrary orientation of the stiffness matri... 9.5. pplication to bars....6 Etene eample....6. Step Rotate the element stiffness matrices... 3.6.. Element... 3.6.. Element... 3.6..3 Element 3... 4.6. Step - ugment the iniviual element stiffness matrices... 4.6.. ugmentation of element stiffness matri for element... 5.6.. ugmentation of element stiffness matri for element... 5.6..3 ugmentation of element stiffness matri for element 3... 5.6.3 Step 3 Implement bounary conitions... 6.6.4 Step 4 Establish the loa vector... 7.6.5 Solve for isplacements... 7.7 Eercises... 8.7. Matlab script... 8.7. NSYS script... CHPER N INRODUCION O ENERGY MEHODS.... n introuction to energy methos..... Potential energy in a bar..... he principle of minimum potential energy... 3..3 Equilibrium of a bar revisite... 4..4 Etene eample Bar Frame II... 7..4. Rotation of element stiffness matrices... 8..4. ugmentation of the element stiffness matri to a global system... 9..4.3 Summation to global stiffness matri... 3 Page iii of vi
..4.4 Implementation of bounary conitions... 33..4.5 Establish a loa vector... 35..4.6 Solution... 35. Eercises... 4.. Continuation of etene eample..4... 4.. 4..3 4 3 BSIC ESICIY HEORY... 43 3. Equilibrium... 43 3. Kinematics... 47 3.3 Material law... 48 3.3. Eample - bar... 48 3.3. Eample Membrane... 48 3.4 Graphic representation... 5 3.5 Weak form... 5 3.5. Choice of primary variable... 5 3.5. Weak an strong formulations... 5 3.5.3 Variational principle... 5 3.6 he principle of minimum potential energy... 5 3.7 Eamples... 57 3.7. he ifferential equation for a beam uner istribute loaing... 57 3.7. Displacement of a simply supporte bar... 58 3.7.3 ransverse eflection of a cantilever beam... 59 3.7.4 Displacement of a simply supporte beam... 6 3.8 Eercises... 6 4 HE RYEIGH-RIZ MEHOD ND BEM EEMENS... 64 4. Generalise methoology to establish the stiffness matri... 64 4. Eample Deuction of the stiffness matri for a bar element... 67 4.3 imitations to the Rayleigh-Ritz metho... 68 4.4 he stiffness matri for a beam element... 68 4.5 Etene eample Beam frame... 7 4.5. he problem... 7 4.5. Meshing... 7 4.5.3 ocalisation... 7 4.5.4 Element formulation... 7 4.5.5 Formulation of element loa vectors... 73 4.5.6 Globalisation... 75 4.5.7 ssembly... 77 Page iv of vi
4.5.8 Bounary conitions... 79 4.5.9 Solution... 8 4.6 Eercises... 83 4.6. Manatory assignment in MEK455... 83 4.6.. Eercise... 83 4.6.. Eercise... 84 5 INER RINGE ND BIINER RECNGE... 85 5. Displacement assumption... 85 5. Coorinate systems an coorinate transformations... 86 5.3 Constant strain triangle... 87 5.3. Eample Stiffness matri for a constant strain triangle element... 89 5.3. ifferent approach to etermine the stiffness matri of a constant strain triangle element... 9 5.3.3 Eample Simplifie analysis of a triangular stiffener... 9 5.4 he bi-linear rectangle... 97 5.4. Natural coorinates... 98 5.4. Choice of isplacement functions... 99 5.4.3 Development of the stiffness matri... 5.5 Eercises... 3 5.5. Eercise... 3 5.5. Eercise... 3 5.5.3 Eercise 3... 4 5.5.4 Eercise 4... 4 5.5.5 Eercise 5... 5 6 SSEMBY... 6 6. Backgroun... 6 6. opology matrices... 7 6.. Eample of a topology matri... 7 6.3 ugmentation of the stiffness matri... 8 6.3. Etene eample Revisit of the triangular stiffener... 8 6.3.. Element stiffness matrices... 9 6.3.. Bounary conitions... 6.3..3 ssembly... 6.3..4 Solution... 6.3..5 Matlab coe... 6.3..6 NSYS coe... 5 6.4 Eercises... 7 6.4. Eercise... 7 6.4. Eercise... 7 Page v of vi
7 INERPOION ND CHOICE OF DISPCEMEN FUNCIONS... 8 BOUNDRY CONDIIONS ND KINEMIC COMPIBIIY... 9 ISO-PRMERIC EEMENS ND GEOMERIC MPPING... 3 SIC CONDENSION... 4 REFERENCES... 5 Page vi of vi
CHPER INRODUCION O HE FINIE EEMEN MEHOD ND BR EEMENS Cook :.,.,.4 (for bars),.5 an.6. Simple springs o establish the systematic elastic behaviour of bar elements we shall etermine equilibrium equations for a bar element by two separate methoologies. First we assume the bar is uniform an behaves like a spring, which is equivalent to the most funamentally simple epression of a uniform an uniformly loae bar. F F Figure.- - Forces an egrees of freeom in a spring Displacement is enote, forces are enote F, an the spring stiffness is enote k. From Figure.- we fin that force equilibrium may be escribe accoring to equation (). F F () Furthermore we fin that the isplacements an may be epresse accoring to the forces F an F as escribe in equations (); k k F F () Page of 5
If we organise the system of equations () in a matri equation, we fin the following relation (3); k F F (3) On element level it is common to generalise equation (3) with the following notation (4); k r, where (4) k k, F r F, In equation (4), k is known as the element stiffness matri, is the element isplacement vector an r is the element loa vector. he system of equations (3) makes little sense on its own, since the equilibrium equations () show that F an F are oppositely equal forces. Consequently an are also oppositely equal isplacements if we only consier the case shown in Figure.-. herefore it is initially no real reason to introuce the matri form of the equilibrium equations since we coul more easily calculate the elongation of the spring by the following epression (5); N / k (5) Furthermore, if we trie to solve equation (3) we woul fin that the matri equation i not have a unique solution in its current state. he eplanation is that equation (3) only has a solution if consistent bounary conitions are applie, which is an issue we will come back to later in this chapter. he interesting question is thus; why are we organising our equations in matri form, as state in equation (3)? For single springs, there is no real reason to use a matri equation in orer to calculate isplacements. However, for systems of springs we may utilise the general matri equilibrium formulation systematically to achieve a single matri equation for the entire system of springs. he reason we are looking at the matri formulations is at the centre of the finite element metho. his entire course is eicate to eplaining why an how we shall utilise such matri formulations in orer to establish simple linear systems of equations for larger structures compose of bars, beams an membranes. Now we shall interchange the spring with a bar an inclue more etails in section... Bar elements Page of 5
In Figure.-, a bar element is shown. t each en of the bar the position is enote as noes. When elements are subject to bounary conitions or elements are connecte to one another, the applications of bounary conitions or connections are always in the noes of the element. Interactions between elements may in specialise cases be applie between noes, but for the purposes of this course, bounary conitions an connections between elements will always be place in noes. For bar elements the noes are always at either en of the bar. Figure.- bar element Displacements an forces are enote an F respectively. It is a convention in structural mechanics that; noal forces an noal isplacements within an element are always efine as positive in the same irection. Figure.- Sectional equilibrium for the bar element With reference to Figure.- an Figure.- we will establish equilibrium much in the same manner as for the spring in section.. Equilibrium is euce in equations (6) an (7). F F E (6) If we substitute strain for stress into equations (6) we fin two equilibrium equations for the bar (7); E F E F E F E F (7) Page 3 of 5
s for the spring element, we may organise the equilibrium equations in matri form E F F (8) E If we substitute the tensile stiffness of the bar with the spring stiffness k as given in section., we fin that equations (8) an (3) are in fact the same equation. s for the spring element, it is customary to generalise equation (8) with the following terminology k r, where (9) E k, F r F, k is the element stiffness matri, is the element isplacement vector an r is the element loa vector. From equation (9) we shoul also make the following important observation; column of k is the vector of loas that must be applie to an element at its noes to maintain a eformation state in which the corresponing noal egree of freeom has unit value while all other noal egrees of freeom are zero.(given that the stiffness matri is subject to consistent bounary conitions) For bar elements, each noe has only one egree of freeom, which makes the above observation normally fairly obvious. However, the result is in fact general for all element formulations, an plays an important part in how kinematic compatibility is achieve when using the finite element metho. We shall return to this observation at a later stage when we iscuss beam elements..3 System analysis In sections. an. we looke at the elastic properties of springs an bars, an we foun matri equations for equilibrium of single elements, assuming linear elastic material properties. he system analysis is concerne with connecting elements to one another, an to apply bounary conitions an loaing. For any structural mechanics problem the following conitions must apply; Kinematic compatibility Equilibrium material law Page 4 of 5
Page 5 of 5 In sections. an. the two latter bullets were consiere. Bounary conitions an continuity in the structure is covere uner kinematic compatibility, an furthermore loaing as further consieration on equilibrium. o illustrate how systems are combine in the finite element metho, we will look at a simple system of two connecte bars..3. Eample ially loae bars with varying cross-sections Figure.3- ially loae, simply supporte bar with varying cross-section In Figure.3-, there are three noes,, an 3. Note that noe number is share by both bar elements. From section. we know that the element equilibrium equation for a single bar element is given by equation (9). he two iniviual element stiffness relations may therefore be escribe by the following equations; F F l E F F l E However, in the global system we choose suffies for the global noe numbers, an not the iniviual element. In the global system we have three noes, an we may reformulate the equilibrium equations for each iniviual element; Element 3 D D D l E
Element E l D D D 3 D i enotes isplacements in the global system, an the suffi i marks the relevant noe. From the above equations it shoul be note that the reunant isplacements at noe 3 for element an noe for element 3 have been inclue in the stiffness relations. he noes are not connecte to the relevant elements so there is no relation between force an isplacement. his has been inicate by rows an columns of zeroes. Since we have inclue all the reunant equilibrium relations in the iniviual element stiffness matrices, they may now easily be combine into a global stiffness matri since all the element stiffness matrices have the same imension; k k E l K Note that the element stiffness matrices are enote by small letter bolface k, with suffi equal to the element number. he global stiffness matri is enote by a capital bolface K. Now we have establishe the stiffness matri for the two connecte bars. he global isplacement vector is trivially given as; D D D D 3 Note that the global isplacement vector is enote by a capital bolface D. Since we have establishe the stiffness an the isplacement, we are left with applying the forces to the system. In this case, we are face with a istribute loa. his is a problem we shall return to at a later stage in the course. For the purposes of this lecture we shall only assume that there are two concentrate loas, R an R 3 at noes an 3 both irecte aially an to the right relative to Figure.3-. In general, the forces must be balance at each iniviual noe. Page 6 of 5
Page 7 of 5 Figure.3- Noal force equilibrium he noal force R i for a noe i must balance all the forces s i j entering the noe from neighbouring elements. Formally the relation may be epresse as follows (); m e e R i s i () R enotes the noal force an m is the number of elements with a bounary to noe i. With the loaing conitions in place, we are able to complete the epression for the global equilibrium equation; 3 3 R R R D D D l E R KD In the process of establishing the equilibrium equations for the global system of two aially loae bars, we have complete the following main steps of the algorithm we shall know as the finite element metho; Meshing (we create two elements from a system) Establish local stiffness matrices ssemble local elements for a global stiffness matri pply loaing he net step in the algorithm is to inclue bounary conitions. In our case, there is a bounary conition at noe no. which requires that the isplacement at noe no. is zero. his can easily be inclue in our equilibrium equation by simply emaning that D is zero. We perform this by zeroing out the rows an columns in the global stiffness matri which are governe by the isplacement D. 3 3 R R D D l E KD
Now we are reay to solve our system. We invert our global stiffness matri an multiply with the loa vector. his process yiels the global isplacements; D K R For larger systems it may be more or less impossible to compute K by han, so for the remainer of this course we shall use computational ais in Matlab an NSYS in orer to compute the actual solutions. Only for very simplifie systems will there be a nee for manual calculation of inverses. ().4 Properties of the stiffness matri he global stiffness matri of a structure has several useful properties which are relevant both for computational efficiency in solving large systems as well as properties which are important in orer to prove eistence an uniqueness of solutions. he global stiffness matri K is/has; Sparse Symmetric Only positive iagonal elements Positive efinite Singular.4. Sparsity he global stiffness matri is almost always sparse, since only contributions from neighbouring elements are inclue for iniviual element entries in the matri. his has little theoretical value, but a huge importance for effective solution methoologies..4. Symmetry Symmetry of the global stiffness matri follows from Betti-Mawells theorem; If two sets of loas act on a linearly elastic structure then work one by the first set of loas in acting through isplacements prouce by the secon set of loas is equal to the work one by the secon set in acting through isplacements prouce by the first set. More precisely, if loas R an R are prouce isplacements D an D, then; R D RD If we substitute for the loa vectors R an R we fin; KD D KD D D K D DK D K is an n n matri, an the isplacement vectors D an D are n vectors. hus both proucts on either sie of the equation are scalar quantities. Since they are scalars they may be transpose without isturbing the equality; D K K KD D DK D DKD DK D D Since neither D nor D are zero vectors (as that woul mean zero loaing) the epression insie the parentheses must vanish. his conclues the proof that K K Page 8 of 5
.4.3 Only positive iagonal elements It is physically obvious that iagonal elements must be positive. If all egrees of freeom ecept one (arbitrarily chosen one) is constraine, a negative iagonal element woul imply a negative isplacement for a positive force. his is of course impossible..4.4 Positive efinite he global stiffness matri is positive efinite, which by efinition means that; K, R / n he consequence of a positive efinite stiffness matri is that it is possible to use various types of matri factorisations on K, which is highly useful when etracting Eigen-values an for optimise algorithms. thorough eplanation of why the stiffness matri is positive efinite will be given chapter 4..4.5 Singularity he stiffness matri is singular before bounary conitions are applie. When no bounary conitions are applie, the system is a mechanism rather than a static system in equilibrium. herefore the eformations are unetermine, an the consequence is that the stiffness matri is singular..5 rbitrary orientation of the stiffness matri he stiffness matri is normally establishe in the local coorinate system of an element, but when applie to a structure, the stiffness matri must often be rotate in orer to match the global coorinate system, see for instance Figure.5-. Page 9 of 5
Figure.5- Bars rotate relative to the global coorinate system If we assume a set of forces r an a corresponing set of isplacements, we may epress the forces an isplacements in an arbitrary coorinate system. If we assume two consistent coorinate systems (which means they must be complete an have a basis) a an b, we assemble the loa an isplacement vectors in the two coorinate systems respectively; r a, a an r b, b. he work one by the loaing r is not epenent on the coorinate system in which it is epresse. herefore; r r a a b b If we assume that there eists a linear transformation which transforms a vector in one coorinate system to a vector in the other; b ab a If we apply the transformation on the isplacement vector a we fin; b a If this relation is inserte into equation () we fin; r r a a b a Since the isplacement an loaing is arbitrarily chosen, this equation must apply for any a, which leas to; r a r b If we insert our new foun relations into the equilibrium equation for an element we fin the following; r k r r a r b e e e e k r k r k r b b b b a b a a a () (3) Page of 5
Since the isplacements an loas were chosen arbitrarily the relation must be vali for any choice of ; r a, a an r b, b, we fin that; k a k, r b a r b (4).5. pplication to bars he stiffness matri foun in sections.an. is euce base on the notion that the isplacement is only aial, an we have place the coorinate system with the -ais in the aial irection. hus the eformation has only one component at each en. In two imensional space the isplacement component in aial irection for the bar oes not necessarily align itself with one of the aes of the coorinate system. In that case, the isplacement is aial along an arbitrary line in two imensional space. Displacement along such a line has components in both aes of the coorinate system, an therefore isplacement components for a bar in two imensional space will generally increase to four rather than two compare to the one-imensional case. In Figure.5- a bar is epresse in two separate coorinate systems. he coorinate system marke by an asterisk ( ) is the oriente along the ais of the bar. he other coorinate system is a simple Cartesian coorinate system with an origin at the first noe. Figure.5- Bar element in local an rotate coorinate systems We want to epress the equilibrium equation for the bar in the Cartesian coorinate system, such that we may combine it with other bars in the same coorinate system. In orer to achieve this, we must fin the isplacement components of u in an y. his is one by simple trigonometry; u is the -component of u v is the y-component of u Page of 5
u is the -component of u v is the y-component of u he transformation matri is thus; cos sin cos sin From equation (4) we fin that the stiffness matri an the loa vector may be escribe as follows (5); k k' r r' (5).6 Etene eample frame of bars consisting of three members is shown in Figure.6-. In this eercise we shall solve for isplacements in the three bars base on analytical calculations. he material an structural parameters are foun in the list below; E = 7 GPa (ypical for harene steel) =.5 m (5cm 5cm rectangular cross-section) =.5 m (3cm 5cm rectangular cross-section) We alreay know the element stiffness relations from equation (9). What we nee to o may be summarize by the following list;. Rotate the element stiffness matrices such that they are all represente in the same coorinate system. ugment the iniviual element stiffness matrices such that they may be summe to a global stiffness matri 3. Implement bounary conitions an eliminate all rows an columns in the global stiffness matri relate to constraine egrees of freeom 4. Establish a loa vector 5. Invert the global stiffness matri an solve for isplacements Page of 5
Page 3 of 5 Figure.6- Frame of three bars.6. Step Rotate the element stiffness matrices.6.. Element Element is efine between noes an, an is rotate 9 egrees relative to the -ais. hus the cosine is zero an the sine is. From section.5., we know the form of the transformation matri, an thus we fin; he rotate element stiffness matri is foun accoring to equation (5); E E e k k.6.. Element Element is efine between noes an 3, an is rotate relative to the -ais. We coul establish the angle of rotation before we transform the stiffness matri, but it is easier to compute sines an cosines irectly from the triangle. he sine of the angle is the height of the element, which is m, ivie by the length which is m 5. Similarly we fin the cosine;
Page 4 of 5 5 5 cos 5 5 sin his leaves us with a transformation matri; 5 5 gain we employ equation (5); 4 4 4 4 5 5 5 E E e k k.6..3 Element 3 Element 3 is rotate relative to the -ais an we perform practically the same calculation as for element. Note that the cosine is now negative, since the angle to the -ais is greater than 9 egrees. 5 5 cos 5 5 sin he trigonometric calculations give us the transformation matri; 5 5 he transformation matri allows us to use equation (5); 4 4 4 4 5 5 5 3 E E e k k.6. Step - ugment the iniviual element stiffness matrices We have two egrees of freeom in each noe (Ref. Figure.5-), isplacement in -irection an y-irection respectively. When we rotate the stiffness matri, using the transformation matri, we committe to using the same isplacements as we use when we evelope the transformation matri. hus the new element isplacement an element loa vectors shoul rea;
Page 5 of 5,,,,, y y F F F F v u v u r We observe that we have two egrees of freeom for each noe, an we have four noes in our system. hus we nee to have 8 egrees of freeom in our global system (before bounary conitions, where some of these will be eliminate). When we have 8 egrees of freeom, we nee an 88 global stiffness matri, an 8 isplacement vector an an 8 loa vector..6.. ugmentation of element stiffness matri for element he first element is connecte to the first 4 egrees of freeom, lateral an vertical isplacements in noes an respectively. he remaining 4 egrees of freeom are however not inclue for the first element, so for these egrees of freeom the element stiffness matri shoul have entries of zero; E k.6.. ugmentation of element stiffness matri for element he secon element is connecte to noes an 3, which results in a relation to egrees of freeom 3 through 6 (as an are relate to noe an 7 an 8 are relate to noe 4). he augmente stiffness matri becomes the following; 4 4 4 4 5 E k.6..3 ugmentation of element stiffness matri for element 3 Element 3 is connecte to noes an 4, which relates element 3 to egrees of freeom 3, 4, 7 an 8. he augmente stiffness matri becomes the following;
Page 6 of 5 4 4 4 4 5 3 E k.6.3 Step 3 Implement bounary conitions Noes 3 an 4 are constraine, which means that the 4 egrees of freeom 5 through 8 are constraine. his implies that the rows 5 through 8 an columns 5 through 8 in the element stiffness matrices may be eliminate. he resulting global stiffness matri may be foun by the following epression; 5 5 4 5 5 5 5 5 5 4 5 4 5 * 3 * * E E E E K k k k K Note that the global element stiffness matri has zero entries in its first row an its first column. his happens since the bar has no stiffness in -irection (we o not consier bening). Since there are only zero entries in the relevant row an column, the stiffness matri is not invertible, an the solution for this egree of freeom is irrelevant. herefore we must eliminate the first row an first column in the global stiffness matri. he isplacement vector is inclue in the emonstration in orer to avoi confusion on which egrees of freeom are solve for an which are not. 5 5 4 5 5 5 5 5 5 5 5 4 5 5 5 5 5 5 v u v E v u v u E KD
Page 7 of 5.6.4 Step 4 Establish the loa vector We have vertical loaing equal to 3 MN in noes an. he irection is opposite to the y-ais, so the loa vector must invert the values for the loas; F F R.6.5 Solve for isplacements If we insert the values for areas, Young moulus an forces we achieve the following system of equations; 5.5498.373.5875.373.746.5875.5875 6 8 v u v If we invert the stiffness matri an solve for the isplacements we fin the following solution;.6.8.33 v u v
.7 Eercises In this eercise we shall reo the work one in section.6. he following tasks shall be performe;. With the below foun matlab script an the ansys coe, confirm that the solution in section.6 is correct. Note that NSYS gives a vertical isplacement in noe, eplain why the solutions are still consistent.. Move noe from position (,) to position (, ), pply a horizontal loa in positive -irection, F, in noe equal to MN, an reo the eample in section.6. When you are finishe, confirm your results using NSYS an Matlab.7. Matlab script % Declaration of stiffness, forces an geometric parameters E=7; =.5; =.5; =; =sqrt(5); F=3; F=3; % Calculation of element stiffness matrices %------------------------------------------- % ocal stiffness matri kloc=[ -; - ] % --- Element --- =[ ; ]; k_e_='*kloc* % --- Element --- Page 8 of 5
=(sqrt(5)/5)*[ ; ]; k_e_='*kloc* % --- Element 3 --- =(sqrt(5)/5)*[- ; - ]; k_e_3='*kloc* % ugmentation of element stiffness matrices to prepare for assembly % --- Element --- k=zeros(8,8); % Create 88 matri of only zero entries k(:4,:4)=k_e_; % he first 44 matri in the upper quarant of k is substitute for the local rotate stiffness matri % --- Element --- k=zeros(8,8); % Create another 88 matri of only zero entries k(3:6,3:6)=k_e_; % row 3 to row 6 an column 3 to column 6 is a 44 matri which relates to egrees of freeom 3 to 6, which in turn relate to noes an 3 % --- Element 3 --- k3=zeros(8,8); k3(3:4,3:4)=k_e_3(:,:); k3(3:4,7:8)=k_e_3(:,3:4); k3(7:8,3:4)=k_e_3(3:4,:); k3(7:8,7:8)=k_e_3(3:4,3:4) % ssembly of stiffness matrices K_tot=E*/*k+E*/*k+E*/*k3; % Bounary conitions - We know that the egrees of freeom in noes 3 an Page 9 of 5
% 4 are constraine. his means that egrees of freeom 5 through 8 are % zero. K=zeros(3,3); K(:3,:3)=K_tot(:4,:4); % Forces are acting oppositely to the y-ais an therefore they must be % inverte in the global loa vector. R=[-F; ; -F]; D=inv(K)*R.7. NSYS script /BCH,IS /FINM,e4 /IE, ineã r statisk analyse rett stav /PREP7 E,,! INK elementer R,,.5! verrsnitts areal til staven R,,.5 MP,EX,,7e9! E-moulen!Geometri ("soli moelling") K,,,! Punkt er origo K,,,! Punkt B er i =, y= K,3,,4! Punkt C i =, y=4 K,4,-,4! Punkt D i =-, y=4,,! inje B,,3! inje BC,,4! inje BD!Inneling i elementer ESIZE,,,,! Deklarer at linje B skal inneles i ett element ESIZE,,,,! Deklarerer at linje BC skal inneles i ett element ESIZE,3,,,! Deklarerer at linje BD skal inneles i ett element RE,! Bruk tverrsnittsareal nr. for innelingen (neste to linjer) Page of 5
MESH,! Inneling av linje MESH,! Inneling av linje RE,! Bruk tverrsnittsareal nr. for innelingen (neste linje) MESH,3!Inneling av linje 3 FINISH! Ut av Preprossessoren /SOU! øsningsprossessoren NYPE, SIC! Statisk analyse (efault) DK,3,all! Ingen forskyvninger i punkt C DK,4,all! Ingen forskyvninger i punkt D FK,,fy,-3e6! Belastning i punkt FK,,fy,-3e6! Belastning i punkt B DRN! Overfører grensebetetingelser til elementmoell SBCRN! og belastningen SOVE! øsningsproseyren FINISH! Ut av øsningsprossessoren /POS! Postprossessoren SE! ast inn analyseresultatene PDISP,! Deformert konstruksjon PRNSO,U,COMP! Utskrift av forskyvningene (global akse) OC,,,,,,53.3! okalt aksesystem RSYS,! aktiveres og brukes til å lese PRNSO,U,COMP! forskyvningene og PRESO, F! kreftene PRESO,EEM! a ut tilgjengelige elementresultater (aksialkrefter) Page of 5
CHPER N INRODUCION O ENERGY MEHODS Cook : 4., 4., 4.3, 4.4 (parts of section 4.4, will be revisite in chapters 4 an 5 of this compenium). n introuction to energy methos In classic soli mechanics it is common to euce equations of motion by equilibrium calculations, particularly for bars, beams an plates. In a finite element contet it is however not common to use equilibrium to euce equations of motion. he finite element metho is base on assuming isplacement functions between noes. he nature of the finite element metho will therefore rener energy methos much more applicable, since energy methos are also base on assuming a set of eformations in whichever irections are relevant (i.e. aial irection for bars, vertical an aial irection for beams, all irections for solis etc.). When we assume a set of isplacement functions in the element metho, we may use energy methos to use these assume isplacements in orer to etermine element equilibrium equations on the form k = r. Specifically, energy methos are the most efficient manner of etermining both the stiffness matri k, an the only general manner of etermining a consistent loa vector r... Potential energy in a bar If we eamine the aially loae, simply supporte bar in Figure.-, we fin that the applie aial force is F, the Young moulus is E, the cross-sectional area is, an the length of the bar is. Figure.- ially loae, simply supporte bar We know from Section. that the isplacement of the bar is linear in F, an we may epress the force isplacement in terms of a function (Ref.Figure.-) Page of 5
Force (F) oa isplacement curve for a bar Displacement (u) Figure.- oa isplacement curve for an aially loae bar If we want to etermine the potential energy in the bar, we can integrate the loa isplacement curve (6); E F u u( F) U F uu E E u E F F F F F E E u In the above equation, U has been efine as the total potential energy. he work one by the aial force is simply the force times the isplacement (7). F In the book by Cook et. al., the work one by eternal forces has been given the symbol. It is also common to use the symbol H, which may be foun in several other references, however since Cook et. al. is our main reference, we shall use. he ifference between the internal energy, which in this case is store elastic energy, an the eternal work one is calle the total potential energy functional; U he total potential energy is given the symbol... he principle of minimum potential energy he principle of minimum potential energy may be state as the following; From all amissible eformations, the system which fulfils the equilibrium equations of a conservative system is the system which has the least potential energy By a conservative system it is meant that the potential energy of an arbitrary eformation configuration is path inepenent, i.e. that the potential energy oes not epen on the loa eformation history. linearly elastic eformation of a bar is an eample of a conservative system. plastically eforme bar is an eample of a non-conservative system, since a (6) (7) (8) Page 3 of 5
eformation may be achieve by either elastic or inelastic loa isplacement history. permanent plastic eformation has ifferent energy than a linear elastic eformation, an these two eformations may be equal. he work one in these two cases is obviously ifferent, an thus loa eformation is path epenent. If we insert the two epressions we have euce, for internal potential energy (6) an work one by eternal forces (7) respectively into the equation for the total potential energy (8), we get an algebraic epression for the total potential energy; E u Fu We observe that the total potential energy (functional) is a secon orer equation in u, which means it has a global minimum as the secon orer term is positive. ccoring to the principle of minimum potential energy, we must fin the configuration with minimum potential energy in orer to fin the system which fulfils the equilibrium equations. Obviously, the function has a global minimum, an we may generally fin that global minimum by fining the stationary value of the erivative (Ref. ()); E E uu Fu u F u u In equation (9) we have the benefit that the function is a secon egree polynomial in u, which means we know that there is only one stationary value, an since the secon orer term is positive, that stationary value is a global minimum. For a more general system, it is not obvious however that the stationary value is unique, nor that it is a global minimum. For the purposes of this course however, all systems we shall investigate will only have one unique stationary value for the potential energy functional, an that stationary value shall be the a global minimum, which means the principle of minimum potential energy applies, an we may fin the stationary value by a ifferentiation/variation operation. Note that this oes not apply for nonlinear systems, an in those cases we nee to sort through ifferent possible solutions in orer to fin the physically relevant one. Nonlinear analyses will be investigate in moule of this course. F E (9) ()..3 Equilibrium of a bar revisite he equilibrium for a simply supporte bar was establishe in equation (). If we o not have any bounary conitions, an we wish to establish the potential energy in a generally supporte an generally loae bar, we may simply wait to impose bounary conitions until after we have establishe the equilibrium equations. Consier the bar from Figure.-, given again here as Figure.-3 for ease of reference. Page 4 of 5
Figure.-3 bar element We choose two isplacement functions (); u u () he functions have the properties that u is unity at noe an zero at noe. u is unity at noe an zero at noe. he functions are plotte in Figure.-4. Figure.-4 Shape functions u an u he importance of choosing the given isplacement functions cannot be overstate. he benefits of choosing shape functions in this manner are the following; If u or u are set to zero, either a bounary conition at noe or noe of zero isplacement will be automatically fulfille for any en isplacement on the other noe. his is a formal requirement an a necessary conition for a shape function if it is implemente using the principle of minimum potential energy. inear functions are chosen for a bar, since the isplacement of a uniform bar is always linear (as long as the material is linearly elastic, an large eformations are not Page 5 of 5
consiere), as such they are capable of returning eact eformations relative to the theory applie Continuity at each noe is assure Rotational continuity is not assure, but bar theory oes not require rotational continuity between iniviual members. (For beams for instance, we cannot use linear functions since rotational continuity is require). Before we start to introuce the assume isplacement functions, we shall revisit the euction for equilibrium of a bar element using the principle of minimum potential energy. he work one by eternal forces an the store potential energy in the resulting isplacements accoring to the assume isplacement functions may be calculate base on equation (8); E U u u Fu F u E u u Fu Fu s we have foun the total potential energy function, we may invoke the principle of minimum potential energy an ifferentiate to fin the stationary value; E E u u u u u u F u u u F u F u E F u Since isplacements at either en of the bar are amissible, the sum of the two ifferentials u an u can only be zero for arbitrarily chosen u an u if u an u are zero iniviually (ater we shall iscover that this statement is calle the funamental theorem of variational calculus). his in turn means we can euce two equations from the ifferential of the total potential energy functional; E E u u u u F E F u F u F We conclue that the equilibrium equation for a bar element may be euce using the principle of minimum potential energy. When we introuce shape functions, we can no longer view the isplacement relate to each noe. We must instea integrate the store elastic energy along the length of the element, since any combination of isplacement functions coul theoretically be applie. he epression for the potential energy store as elastic energy in a bar may be epresse on the following form; U u E If we introuce our assume isplacement functions we may integrate in, assuming that E an are constant along the length of the bar. Page 6 of 5
Page 7 of 5 E E U E E U Now we have our total potential energy epresse in terms of two unetermine coefficients which give the amplitues of our assume isplacement polynomials. he remaining issue is to etermine the work one by eternal forces; F F In orer to etermine these coefficients we invoke the principle of minimum potential energy; F F E Since the above equation is vali for amissible an, we may iniviually set an to zero; F F E F E F E o repeat the process of assembly an a practical approach to applying bounary conitions we shall have another etene eample of a bar frame...4 Etene eample Bar Frame II We consier a vertically loae bar frame, shown in Figure.-5 Figure.-5 vertically loae frame of bars
he noal coorinates are given in the following list, along with the material properties an cross-sectional area of the bars; E= = F=. Noe : (,) Noe : (,) Noe 3: (,4) Noe 4: (,) Noe 5: (3,) We follow the same proceure as we chose for the bar frame in the etene eample of section.6;. Rotate the element stiffness matrices such that they are all represente in the same coorinate system. ugment the iniviual element stiffness matrices such that they may be summe to a global stiffness matri 3. Implement bounary conitions an eliminate all rows an columns in the global stiffness matri relate to constraine egrees of freeom 4. Establish a loa vector 5. Invert the global stiffness matri an solve for isplacements..4. Rotation of element stiffness matrices he basic stiffness matri for a bar element is; E k Elements, an 7 have the same orientation, length an aial stiffness (E). hus they have ientical element stiffness matrices. he irection of each element is parallel to the -ais, which means the sine is zero an the irection cosine is. hus the transformation matri is simply; We fin the element stiffness matri epresse in the global coorinate system; e k k k e k e 7 k e Page 8 of 5
Page 9 of 5 Elements 6 an 4 have the same angle to the global -ais an the same length, an therefore they as well have the same element stiffness matri (an thus the same transformation matri). e e e 6 4 6 6 6 6 k k k k Elements 3 an 5 have the same angle to the global -ais an the same length. his means the thir an final element stiffness matri configuration may be calculate for element 3 an 5 both; e e e 5 3 5 5 5 5 k k k k..4. ugmentation of the element stiffness matri to a global system he system in Figure.-5 is a two-imensional frame of bars, which means each noe has two egrees of freeom. he total number of egrees of freeom in the system (incluing those constraine by bounary conitions) is. he global isplacement vector may therefore be written as follows (); 5 5 4 4 3 3 v u v u v u v u v u a D () he small suffi a on D has been inclue to show that this is the augmente global isplacement vector, to istinguish it from the full global isplacement vector in which we have inclue bounary conitions. he sub inices on u an v inicate noe number an u inicates
Page 3 of 5 isplacement in -irection as v inicates isplacement in y-irection. Since a D is a vector, a i k (i.e. the augmente element stiffness matrices) have imension. ugmentation of element stiffness matri for element Element has two egrees of freeom in noes an. this means Element relates to egrees of freeom through 4. he augmente element stiffness matri becomes; E a k ugmentation of element stiffness matri for element Element relates to egrees of freeom 3 through 6 via noes an 3; E a k ugmentation of element stiffness matri for element 3 Element 3 has egrees of freeom in noes 3 an 5, which means element 3 relates to egrees of freeom 5, 6, 9 an ;
Page 3 of 5 3 E a k ugmentation of element stiffness matri for element 4 Element 4 has egrees of freeom in noes an 5, which means egrees of freeom 3, 4, 9 an are relate to the augmente stiffness matri; 4 E a k ugmentation of element stiffness matri for element 5 Element 5 is relate to noes an 4, which means element 5 is relate to egrees of freeom 3, 4, 7 an 8; 5 E a k ugmentation of element stiffness matri for element 6
Page 3 of 5 Element 6 is relate to noes an 4, which means element 6 is relate to egrees of freeom,, 7 an 8; 6 E a k ugmentation of element stiffness matri for element 7 Element 7 has egrees of freeom in noes 4 an 5 which relates to egrees of freeom 7 through. 7 E a k..4.3 Summation to global stiffness matri Note that when the matrices have been summe, it has been inclue that the element lengths for the lateral elements are whereas the element lengths for the angle elements is.
Page 33 of 5 K a..4.4 Implementation of bounary conitions From the constraint in noe we must fi egrees of freeom an. From the constraint in noe 3 we must fi egrees of freeom 5 an 6. We achieve this by eliminating rows an columns,, 5 an 6 from the global stiffness matri, an entries,, 5 an 6 from the global loa vector;
Page 34 of 5 he resulting global stiffness matri becomes; K he relevant egrees of freeom are the initial augmente global isplacement vector, where egrees of freeom,, 5 an 6 have been eclue;
Page 35 of 5 5 5 4 4 v u v u v u D..4.5 Establish a loa vector he global loa vector consists of a single point loa in noe, with irection along the y-ais. his is relate to isplacement v an must therefore have the same place in the loaing vector as the isplacement v. here are no other loas, an therefore the global loa vector may be written as follows;. R..4.6 Solution he isplacement vector may be foun as by the following equation;.44..44..488 R K D Matlab script for the solution of this eercise is given below; E=; =; _straight=; _angle=sqrt(); 6=[sqrt()/ sqrt()/ ; sqrt()/ sqrt()/]; 5=[-sqrt()/ sqrt()/ ; -sqrt()/ sqrt()/];
ke=(e*/_straight)*[ - ; ; - ; ]; ke=ke; k7e=ke; k=[ -; - ]; k6e=(e*/_angle)*6'*k*6; k5e=(e*/_angle)*5'*k*5; k4e=k6e; k3e=k5e; K=zeros(,); % EEMEN K(:4,:4)=ke; % EEMEN K(3:6,3:6)=K(3:6,3:6)+ke; % EEMEN 3 K(5:6,5:6)=K(5:6,5:6)+k3e(:,:); K(5:6,9:)=K(5:6,9:)+k3e(:,3:4); K(9:,5:6)=K(9:,5:6)+k3e(3:4,:); K(9:,9:)=K(9:,9:)+k3e(3:4,3:4); % EEMEN 4 K(3:4,3:4)=K(3:4,3:4)+k4e(:,:); K(3:4,9:)=K(3:4,9:)+k4e(:,3:4); K(9:,3:4)=K(9:,3:4)+k4e(3:4,:); K(9:,9:)=K(9:,9:)+k4e(3:4,3:4); % EEMEN 5 K(3:4,3:4)=K(3:4,3:4)+k5e(:,:); K(3:4,7:8)=K(3:4,7:8)+k5e(:,3:4); Page 36 of 5
K(7:8,3:4)=K(7:8,3:4)+k5e(3:4,:); K(7:8,7:8)=K(7:8,7:8)+k5e(3:4,3:4); % EEMEN 6 K(:,:)=K(:,:)+k6e(:,:); K(:,7:8)=K(:,7:8)+k6e(:,3:4); K(7:8,:)=K(7:8,:)+k6e(3:4,:); K(7:8,7:8)=K(7:8,7:8)+k6e(3:4,3:4); % EEMEN 7 K(7:8,7:8)=K(7:8,7:8)+k7e(:,:); K(7:8,9:)=K(7:8,9:)+k7e(:,3:4); K(9:,7:8)=K(9:,7:8)+k7e(3:4,:); K(9:,9:)=K(9:,9:)+k7e(3:4,3:4); K_glob=zeros(6,6); K_glob(:,:)=K(3:4,3:4); K_glob(:,3:6)=K(3:4,7:); K_glob(3:6,:)=K(7:,3:4); K_glob(3:6,3:6)=K(7:,7:); R=zeros(6,); R()=-.; D=inv(K_glob)*R Page 37 of 5
n nsys script is given below for the solution of this eercise; /BCH,IS /FINM,e4 /IE, ineã r statisk analyse rett stav /PREP7 E,,! INK elementer R,,! verrsnitts areal til staven MP,EX,,! E-moulen!Geometri ("soli moelling") K,,,! Punkt er i origo K,,,! Punkt B er i =, y= K,3,4,! Punkt C i =4, y= K,4,,! Punkt D i =, y= K,5,3,! Punkt E i =3, y=,,! inje,,3! inje 3,3,5! inje 35,,5! inje 5,,4! inje 4,,4! inje 4,4,5! inje 45!Inneling i elementer ESIZE,,,,! Deklarer at alle linjer skal inneles i ett element RE,! Bruk tverrsnittsareal nr. for innelingen (neste to linjer) MESH,! Inneling av alle linjer FINISH! Ut av Preprossessoren /SOU! à sningsprossessoren NYPE, SIC! Statisk analyse (efault) DK,,all! Ingen forskyvninger i punkt DK,3,all! Ingen forskyvninger i punkt C FK,,fy,-.! Belastning i punkt B DRN! Overfà rer grensebetetingelser til elementmoell SBCRN! og belastningen SOVE! à sningsproseyren Page 38 of 5
FINISH! Ut av à sningsprossessoren /POS! Postprossessoren SE! ast inn analyseresultatene PDISP,! Deformert konstruksjon PRNSO,U,COMP! Utskrift av forskyvningene (global akse) OC,,,,,,53.3! okalt aksesystem PRESO,EEM! a ut tilgjengelige elementresultater (aksialkrefter) Page 39 of 5
. Eercises.. Continuation of etene eample..4.. a) Before bounary conitions are applie to the global stiffness matri, constraine egrees of freeom are still present in the stiffness matri. If we multiply the global stiffness matri with the noal isplacements we get the forces in each noe, which allows us to fin the reaction forces in noes an 3. Fin the reaction forces. b) pply cross-sectional area =.5 an Young moulus E=7 GPa. Substitute the force F with ---. Recalculate the noal isplacements an reaction forces. c) he steel is of type X65, which has a specifie minimum yiel stress of 45 MPa. However, bars are inaccurate (no bening for instance), an therefore the allowe aial stress is well below the yiel limit. he allowable aial stress is 5 MPa. Fin the maimum loa F which keeps the aial stress below 5 MPa anywhere in the structure ) Use NSYS to eperiment on the consequences of allowing lateral isplacement in noe no. 3 (i.e. change the bounary conition to only constrain the vertical egree of freeom in noe 3). In the figure below a system of springs is shown. Fin the global stiffness matri for the system of springs; Page 4 of 5
..3 Page 4 of 5
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3 BSIC ESICIY HEORY Curriculum hese notes, in aition to Cook et. al. chapters 4., 4., 4.3, 4.4, 4.7 an 4.9 his chapter concerns itself with the basic conitions of elasticity theory. he equations apply for linear static loa an response. Dynamic effects of inertia are thus not consiere. he chapter covers the following topics; Equilibrium Kinematics isplacements an strains Material law Bounary conitions he combine consierations of the four above mentione topics allows for consistent static analyses. We shall also use to types of notation; Vector notation, an ensor notation Vector notation is known from linear algebra. ensor notation is typical for continuum mechanics theory, but not all stuents know this notation. he course will therefore only inclue this notation when absolutely necessary. 3. Equilibrium he starting point for our analyses is the principle of conservation of momentum, which in the static case reuces to a consieration of a balance of forces an moments. If we ignore inertia forces, this is simply Newton s secon law. We consier a linearly elastic boy B, which covers part of the plane V R. he boy is circumscribe by a smooth bounary S, with unit normal vector n, which points out from the boy. he boy is loae by a set of static loas:, which is a set of tractions which affect the bounary of the boy, an F, which constitute volume forces which affect the interior of the boy. ypical volume forces are gravitational forces Page 43 of 5
Figure 3.- he boy B with a volume V, bounary S epose to tractions an forces ypically we know tractions as istribute loas. more rigorous efinition may be foun in equation (3); lim f Continuum mechanics hols three main principles;. Conservation of mass. Conservation of force equilibrium an moment equilibrium 3. Conservation of energy he sum of forces is epresse accoring to the following integral; FV S F (4) V S (3) ractions an boy forces may be ecompose along unit vectors. In a Cartesian coorinate system for instance, the tractions an boy forces may be epresse by their iniviual components (5). F,F y Fy (5) In the efinition of traction, a surface is involve. raction is force per unit of surface, (which incientally is the same efinition as stress). he surface is efine by the surface normal n. he same type of efinition is mae for stress, an the term stress tensor is introuce such that we may link the stress to an arbitrary surface, just like the surface tractions. Page 44 of 5
tensor is a generalisation of the vector concept. While a vector has two components in imensions, a tensor may have many. he stress tensor, for instance, has 4 components in imensions. he imensional stress tensor may be written on the following form; y y yy he stress tensor consists of four stresses, each of which relate to a surface in the plane, an a irection on that surface. he physical interpretation is liste as follows; he stress is classe by surface an irection. he secon ine gives the unit normal of the surface. he first ine gives the irection of the stress on that surface o establish the equilibrium equations on local form, we must fin a relation between tractions which epen on the surfaces, an the stresses. In orer to establish this relation, we will perform equilibrium calculations on a small ifferential element as shown in Figure 3.-. (6) Figure 3.- Stresses an tractions on a small ifferential element Cauchy s law may be formulate base on equilibrium calculations Page 45 of 5
y s y s y y yy y o establish the relation in terms of Cartesian coorinates we use the following relations; y n s n s y If we combine equations (7) an (8) we fin a more convenient epression for equilibrium of the small element; y y n yy ny y his epression may be written on total form or by ine notation an Einstein s summation rule; n or n ij j i By applying equations (4) an (9) we fin the ifferential equation for equilibrium of an elastic boy through use of the ivergence theorem; V V V F V S (3) FV S S F V ns We have evelope the ifferential equation for an arbitrary volume of soli, which in turn means the equation must be vali for any volume. his implies (3) F On component form, for the two-imensional problem, equation (3) may be rewritten (3);, y, y, y yy, y F F y Moment equilibrium may be epresse similarly an leas to the useful conclusion that the stress tensor is symmetric, i.e.; Since the stress tensor is symmetric, we o not nee to know all the elements in orer to uniquely escribe its contents. In two imensions we only nee three components, since the fourth component follows from symmetry. Generally the stress tensor is written as a vector in finite element contets, where only the necessary components to completely escribe the stress tensor are inclue. In two imensions the following stresses are relevant; yy y (7) (8) (9) (3) (3) Page 46 of 5
he equilibrium equation in two imensions may thereby be rewritten as well; 3. Kinematics When a system is epose to loaing it will eform. he relation between isplacement an strains is essential to allow us to use equilibrium to balance forces an kinematics (isplacements) through a material law. We ivie strains into two categories, illustrate in Figure 3.-; Volumetric strains, an Shape changing strains (i.e. shear strains) Figure 3.- Volumetric strain to the left an shear strain to the right Change in length yiels the aial strain; u u u u Change in angle yiels the shear strain. Note that we assume small isplacements. From vector calculus we can calculate the angle between two vectors; a b cos a b v u y y y v u y y his yiels the epression for the shear strains if the following assumptions are mae; Page 47 of 5
y v u sin cos y he general epression for the strain tensor is; ui ij j u j i Note that the strain tensor is symmetric. ike the stresses, strains may also be epresse on vector form, an the symmetry of the strain tensor is also taken avantage of; yy y y u y y u y v Note that. his is the ifference between tensorial an engineering strains, an ha no physical significance. It is only a matter of convenience in terms of interpretation an epression of strains. 3.3 Material law In this course we shall only consier linear elastic materials. If we use the vectorise forms of stress an strain we thus have a linear relation calle the generalise Hooke s law; E We may also efine an epane Hooke s law, where initial stresses an strains are consiere; E In both cases, E is a matri/tensor. 3.3. Eample - bar In one-imensional analyses, say for bars, there is only one isplacement component, which is the aial one. hus there is only coorinate in the epressions for stresses an strains; u u o fin the strains, we insert into the efinition; ui j 3.3. Eample Membrane u j u u u i membrane has two isplacement components; Page 48 of 5
Page 49 of 5 v u u he strains are therefore escribe by two irections, an the stress tensor has 4 components. Since the stress tensor is symmetric, only three of them are inepenent; u v u y y y yy For plane stress we get the generalise Young moulus; v v v v E E For plane strain we get a ifferent Young moulus; v v v v v v v E E
3.4 Graphic representation In Figure 3.4-, the relation between isplacements, forces, stresses, tractions an strains are shown. Figure 3.4- Relation between known an unknown entities in a finite element solution Page 5 of 5
3.5 Weak form Previously, in this chapter, we investigate the basic equations for a linearly elastic material uner static loaing an response. ll equations were on strong form (eplicit ifferential equations). In their strong form the equations apply for all interior points in the omain of the unknown functions. he equations we have summarize are;. Strain-isplacement relations. Stress-strain relations 3. Equilibrium on the interior of an elastic boy 4. Equilibrium on the bounary of an elastic boy 5. Continuity on the bounary In the element metho we shall not apply all these equations irectly, since this is inconvenient computationally. We transfer these equations into a weak form. We shall erive the governing equations from a ifferent stanpoint, which we introuce in Chapter, using the principle of minimum potential energy. hen the equilibrium equations are a consequence of a stationary value of a functional which epresses the energy in an elastic system. his is a variational formulation. he variational formulation is also an integral formulation, where, if we choose, we can require that the governing equations hol true over an integral of a selection of the omain of the equations. In that case, the solution is accurate as an average value over a finite element, rather than over the entire omain as they woul be on a strong form. Further etails on variational principles may be foun in Cook et. al, Chapters 4., 4., 4.3, 4.4, 4.7 an 4.9. When structuring a variational principle, we must follow certain rules;. Choice of primary variable. Choice of weak an strong formulations 3. Choice of variational principle/rule 3.5. Choice of primary variable In the finite element, the isplacements are the primary variable, an stresses an strains are immeiately following from the isplacements. hus all solutions we shall be solving for are isplacements, from which stresses, strains, reaction forces etc. will be euce afterwars. 3.5. Weak an strong formulations We want to establish the equilibrium equations on weak form. he weak equations are thus alternative formulations for;. Equilibrium insie the boy. Equilibrium on the bounary he strong formulations are all the equations that remain; Strain- isplacement relations Page 5 of 5
Stress-strain relations Continuity on the bounary he main ifference between the weak an the strong formulations are; Strong formulations are satisfie pointwise on the omain of the functions Weak formulations are satisfie as integrals, or mean values, piece-wise over the omain 3.5.3 Variational principle Basically we have two stanar choices (among others, but these are ominating); he principle of virtual isplacements he principle of minimum potential energy Virtual isplacements are often intuitive, an stuents often have a basic unerstaning of the principle. However, the principle of minimum potential energy is governing in the fiel of the finite element metho, as the most literature, be it instruction manuals or tetbooks, are formulate base on the PMPE. hus this course shall apply the principle of minimum potential energy even though the principle of virtual isplacements is more intuitive for some, an equally applicable. 3.6 he principle of minimum potential energy he principle of minimum potential energy may be iscusse in further etail than we i in Chapter. Now we introuce the strain energy ensity (33); u he strain energy ensity is the prouct of stress an strain for any material point in the continuum. he concept of strain energy ensity will be illustrate with the eamples of volumetric strain, an shape changing strain, i.e. shear strain. Strain energy ensity is the strain energy store in an elastic boy per unit of volume. hus when we integrate the strain energy ensity over the volume, as will be one later in this Section, the total strain energy in the boy is the result. hereby: u = energy/volume We start by showing that equation (33) applies for volumetric strains. In Figure 3.6-, a cubic volume of an elastic material epose to an aial force F is shown, where the sies of the volume have length a, an the resulting isplacement is δ. (33) Page 5 of 5
Figure 3.6- Volume of elastic material subject to a force F he stress on the surface where F is applie is trivially: F a F a he strain in the volume is foun from the isplacement δ: a a he elastic strain energy U store in the volume follows from follows from Figure.-, an is simply the area uner the loa isplacement curve. his is epresse as follows: U F If we insert for the stress an strain relations, we fin the store elastic strain energy as a function of stresses an strains: U 3 a he strain energy ensity is efine as the energy per unit of volume, so the strain energy ensity may therefore be foun: u U V 3 a 3 a he strain energy ensity thus conforms to equation (33) for volumetric strains. he same argument also applies to strain energy ensity for shear strains. In Figure 3.6- a cubic volume of elastic material is subject to shear stresses τ y. Page 53 of 5
Figure 3.6- Cubic volume of elastic material subject to a shear stress τ y in un-eforme configuration he eforme configuration for the volume in Figure 3.6- is shown in Figure 3.6-3: Figure 3.6-3 Cubic volume of elastic material subject to shear stress in eforme configuration Base on Figure 3.6-3, an the assumption of small isplacements, the isplacement may be calculate as follows: a y he force acting on the surface of the shear stress is simply the stress multiplie with the area on which it acts: F y a y he elastic energy may be calculate: U 3 F y a y y Page 54 of 5
Fining the strain energy ensity only leaves iviing the total elastic energy with the volume of the cube: u U V y y he strain energy ensity thus conforms to equation (33) for shear strains also. If we integrate the strain energy ensity over the volume of the boy, we necessarily achieve the strain energy elastically store in the boy. he total strain energy is efine in Chapter an terme U; U V V V EV If we also inclue initial stresses an strains to the equation we may now reformulate the total potential energy functional; V E E V u FV V S u S Equilibrium equations may now be foun by etermining the stationary value of this functional (35). he process of etermining the stationary value of the energy functional may be unerstoo in the following manner; If we have an elastic system we may se what happens to the potential energy close to an equilibrium position. his is illustrate in Figure 3.6-4. (34) (35) Figure 3.6-4 Variation about an equilibrium position hus, if you wish to fin how heavy something is, try to lift it. We se what happens to the potential energy when we change the isplacement fiel in a small omain close to the equilibrium position. he nee to solve ifferential equations may be avoie by applying the Rayleigh Ritz metho on a functional such as. With this metho, we establish an approimation to the unknown functions which in the potential energy functional are isplacements, u. he result is a problem with a finite number of unknowns, an is escribe by algebraic equations rather than ifferential equations. he element metho is a special form of the Rayleigh-Ritz metho, which we shall iscuss in etail in the net chapter. Much time in MEK455 shall be evote to etermine epressions for U, since the strain energy functional is the basis for evelopment of stiffness matrices. he strong an weak formulations are illustrate in Figure 3.6-5 Page 55 of 5
Figure 3.6-5 - Weak an strong equations Page 56 of 5
3.7 Eamples In this section we shall employ the principle of minimum potential energy accoring to the Rayleigh metho. he Rayleigh metho is base on assuming that the isplacements of a system may be escribe by a single function. We shall make three eamples of this. But first we shall use the principle of minimum potential energy to euce the ifferential equation for a beam subject to istribute loaing. 3.7. he ifferential equation for a beam uner istribute loaing Figure 3.7- finite length beam with uniformly istribute loa an uniform crosssection, mae from an isotropic material When we have a beam of uniform cross-section with a uniformly istribute loa, if the isplacements are small an shear is negligible we may assume Euler-Bernoulli beam theory. With linear theory an small isplacements, the isplacement fiel across a beam may be escribe accoring to the following equations; w u z w w() For beams we only consier aial strains, thus; u w z We insert the strains into the total potential energy functional; U V V EV EV V V u FV EV V S u S w Ez V When we integrate over the volume, we assume that the beam ais goes is positione in the centre of area for the beam cross-section, thus the function w is not epenent on the y an z coorinates. herefore we may isolate the integration; z I his allows us to simplify the epression for the internal strain energy; Page 57 of 5
Page 58 of 5 w EI V w Ez U V We fin the variation of the energy functional, but we shall vary the internal strain energy separately for simplicity. w w EI V w Ez U V, When the variation inclues erivatives of the isplacements, we use integration by parts to transform the varie entities to variations in the functions themselves an not their erivatives. We perform integration by parts twice on the internal strain energy; w w EI w w w w w w EI w w w w EI 4 4 3 3,, 3 3,, Variation of a functional at a fie point is equivalent to a constant in terms of ifferentiation, i.e. there is no potential for change in a function that has a fie evaluation point. w w EI U 4 4 We complete the epression for the total potential energy functional, an use the principle of minimum potential energy; EI q w w q w EI w q w w EI qw w EI 4 4 4 4 4 4 We recognize the ifferential equation for a transversely loae beam. 3.7. Displacement of a simply supporte bar
We assume that the bar starts at the origin an that the -coorinate is parallel to the bar ais. We further assume that the lateral isplacement may be epresse accoring to the following function; u( ) he aial strain is foun by ifferentiating the isplacement with regar to. observe that the isplacement is zero at the origin, hence the bounary conitions for the system are fulfille by the assume isplacement function. We insert the assume isplacement function into the epression for the total potential energy; F E EV u S V S V E F E V F E E F F We observe that the assume isplacement function yiels the eact result. his is simply because the chosen isplacement function is the eact solution. 3.7.3 ransverse eflection of a cantilever beam u We Figure 3.7- Cantilever beam with en loa he bounary conitions for a cantilever beam are that both the isplacement an rotation at the origin is zero. simple function which fulfils both of the bounary conitions is. If we assume that the isplacement function be epresse as follows; w( ) c, we fin that the total potential energy may Page 59 of 5
w M cf EI cf 4EIc 3 4EIc F c F c 3 4EI cf 4 EIc 3 cf 3 F We fin that the en eflection is fairly well approimate since the eact solution is, thus 3EI the isplacement is unerestimate by 5%. We shall unerstan in following chapters that the Rayleigh metho (an the Rayleigh-Ritz metho) unerestimates isplacements in loa controlle conitions in general. One important observation is that the isplacements are estimate with fairly goo accuracy. he moments on the other han are not. When the isplacements are secon orer functions, the moments are constant functions accoring to the ifferential equation for a beam. hus the moments are; F M he eact solution for the moment istribution is F(-), which is only a match to our solution in the mile of the beam. he important lesson is that the isplacements are more accurately preicte than the moments. Page 6 of 5
3.7.4 Displacement of a simply supporte beam Figure 3.7-3 Simply supporte beam with istribute loa he bounary conitions for the simply supporte beam is zero isplacements at the origin an at position =. function which fulfils the bounary conitions is w( ) csin. If we insert this isplacement assumption into the total potential energy functional, we achieve the following result; M 4 c EI sin 4 4 c EI qw 3 w EI 4 q 4q c 5 EI qw qcsin c 4 EI 4 3 q c Compare to the eact solution, the mi isplacement eviates by less than %. he eact 5q 4 solution for the mi isplacement is. 384EI Page 6 of 5
3.8 Eercises Page 6 of 5
Page 63 of 5
4 HE RYEIGH-RIZ MEHOD ND BEM EEMENS From the previous three chapters in this compenium, we have become acquainte with bars, the stiffness matri for a bar an we have stuie variational principles, specifically the principle of minimum potential energy. In this chapter we shall aopt the principle of minimum potential energy on bars an beams, an we shall introuce new notation an new means of formulating isplacement functions in orer to fin stiffness matrices efficiently. 4. Generalise methoology to establish the stiffness matri In section..3 we foun the stiffness matri for a bar element using two assume isplacement functions. his le to fairly lengthy calculations involving the noal isplacements an, as well as their respective ifferentials. For more comple elements, the calculations are even more lengthy, an in the en so lengthy that they are inconvenient. o establish a methoology for generation of stiffness matrices for general elements, a stanar notation which generalises the variational operation is necessary. his methoology an the associate notation will be introuce in this section. We start with the equation for the total potential energy of a system without initial stresses or strains; U EV u FV u V V S We know from section 3. that the relation between isplacements an strains coul be epresse accoring to the following formula; u he finite element metho is base on assuming a isplacement fiel u, after which unetermine coefficients in the isplacement fiel are establishe using the principle of minimum potential energy. In section 3.7 we introuce the Rayleigh metho, where we assume a single isplacement function as a solution to various systems. In this chapter we generalise this methoology to inclue more than one isplacement function an the new/augmente metho is calle the Rayleigh-Ritz metho. For the purposes of the finite element metho in structural mechanics, the isplacement fiel is always base on polynomials. Generally these are polynomials in three imensions, i.e. epenent on, y an z, however we shall consier only polynomials in one imension in this chapter, i.e. polynomials in. In later chapters, we will introuce aitional egrees of freeom to our polynomials. When the isplacement fiel is a polynomial, we may epress it generally; u c c c... c n n In FE problems, we reorganise the general isplacement fiel to associate iniviual polynomials with noal egrees of freeom; u S,,.. P,,.... P,,.. P n n If we epress u as a vector equation, we may place the iniviual polynomials in one vector an the noal egrees of freeom in another; Page 64 of 5
u P P... P n. n We name the vector of polynomials N, an the polynomials epress shape functions. he noal isplacement vector is name. hus; u N, N P. n P... P n, It is not reaily eplaine how a general assumption that the isplacements may be epresse in terms of polynomials leas to the conclusion that the iniviual elements of the unetermine coefficients epress noal egrees of freeom. For now, we shall simply state that this is the case. he eplanation lies with completeness an continuity, but also convenience. he etails of how an why the polynomials are aapte to represent iniviual noal egrees of freeom shall be the subject of much attention in this course. he shape functions N is a row vector (when we introuce more variables in the polynomial assumptions it is convenient to turn N into a matri) while the noal isplacement vector is a column vector. he prouct N is therefore a scalar quantity. If we use the relation between strain an isplacements we fin the following relation; u N he noal isplacement vector oes not epen on, an therefore we only nee to ifferentiate the shape functions in orer to fin the strains. he unit isplacement strain matri B, may thus be epresse as; B N B If we substitute the strain epressions in the total potential energy functional, we fin; B EB V N FV N V V S Now we shall attempt to evelop a general epression for the stiffness matri k an the loa vector r. In orer to o so, we have to use one general result on the symmetry of the strain energy ensity as well as a result from calculus in linear algebra. he strain energy ensity is only a number for each volumetric point; B EB f (, y, z) his entails that the strain energy ensity is a scalar function (as oppose to a vector function). his is physically obvious, since the strain energy in a point cannot have a irection. Since the strain energy ensity is a scalar function, it is equal to its own inverse; S (36) (37) Page 65 of 5
B EB B EB We also nee the following result from linear algebra; Now we invoke the principle of minimum potential energy; B EB N FV N S V V V S he variation operation shoul relate to the unetermine coefficients, since B, N an E are matrices of known functions. he variation operation may therefore be epresse as follows; B EB V N FV N V V S B EB B EB V N FV V V S B EB B EB ; We use the relation S N S B EB B EBV N FV B EBV N FV V V S V V S B EBV N FV N V V S N S S N S By the funamental theorem of variational calculus, we conclue that the variation is amissible, an therefore we may set the content of the parenthesis equal to zero; B EBV N FV N V V S S We may group the terms accoring to physical interpretation; V V B EB V k N FV S N S r hus the integral over the strain energy ensity is interprete as the element stiffness matri an the accumulate work one by volume forces an tractions give rise to the element loa vector. he variational operation may thus be conclue by the following equation; k r, which is the familiar equilibrium equation. We have establishe element equilibrium equations by means of the Rayleigh-Ritz metho. We have assume a set of isplacement polynomials an applie the principle of minimum potential energy to establish equilibrium. here are ifferent means of establishing element equilibrium, one eample is weighte resiuals methos Page 66 of 5
(introuce in Cook et. al. in chapter 5), others are irect physical erivation an the metho of virtual isplacements. For structural mechanics problems, the Rayleigh-Ritz metho covers all necessary applications since the principle of minimum potential energy is sufficient to escribe static problems. hus we shall evote ourselves to this metho in the following parts of this course. 4. Eample Deuction of the stiffness matri for a bar element We have euce the stiffness matri for a bar element several times uring this course. We shall perform this operation one last time, in orer to show that the new notation an the new result for the generalisation of the variational operation results in a smooth an simple general metho to establish stiffness matrices. We assume the familiar isplacement functions; u he isplacement functions may be epresse in our new notation; N, he only strains we are concerne with are aial strains, an they are assume to be constant over any perpenicular cross-section. he strains are thus simply u. he ifferential operator is therefore. We calculate the B-matri by multiplying the ifferential operator with the shape functions; B N When we know the B-matri we may euce the stiffness matri; k V B EB V E k E E We fin that the new notation an technique to establish the stiffness matri is convenient an much simpler than previous efforts. Page 67 of 5
4.3 imitations to the Rayleigh-Ritz metho We have seen how to use the Rayleigh metho an the Rayleigh-Ritz metho to approimate isplacement behaviour by assuming isplacement functions an calculating their amplitue(s) by means of the principle of minimum potential energy. However, this metho introuces some limitations an some important problems which shoul be unerstoo when using the metho. When we introuce shape functions to ictate the isplacement behaviour of a system we constrain its natural freeom of isplacement which generally is total. We force the solution to eist in a solution space which we have confine by the etent/applicability of our assume isplacement functions. Whenever we constrain our system, we must re-evaluate the principle of minimum potential energy. he isplacements in a natural unconstraine system will fin a state of equilibrium for which the internal strain energy in the system is minimize. If we constrain the solution space, we will always, at least to some etent, reuce the systems inherent ability to fin the least complicate state of equilibrium, where as little strain energy as possible is store. hus the consequence of restraining the solution space is to heighten the stiffness of our systems. When we heighten the stiffness of our systems this has two essential effects; In loa controlle conitions we generally unerestimate the isplacements In isplacement conitions we generally overestimate the inuce stresses nother important aspect is the requirement that the isplacement functions must fulfil the bounary conitions of our physical problems. Otherwise, strain energy will be store where it is not suppose to. 4.4 he stiffness matri for a beam element In orer to establish the stiffness matri for a beam, we nee isplacement functions relate to vertical isplacements an rotations at both noes of an element. he isplacement functions for beams are calle the beam equations, an they are liste as follows; N 3 graphic representation is given in Figure 4.4-. 3 3 3 3 3 Page 68 of 5
Page 69 of 5 Figure 4.4- Shape functions for beam elements (the beam equations) We note that by constraining either function we impose a bounary conition of fie isplacement an/or rotation at the relevant noe in a beam element. his is how bounary conitions are introuce practically we impose them by eliminating the assume isplacement functions which relate to the relevant egree of freeom. From section 3.7. we know that the strain energy in a beam may be epresse as the prouct of moment an curvature integrate over the length of the beam; w EI M U If we assume that the secon moment of inertia an the Young moulus is constant over the length of the beam, we may fin a ifferential operator for a beam element; Using the relation N B, we fin; 6 6 4 6 6 3 3 3 3 3 3 N B Now we may insert into the epression for the general calculation of the stiffness matri; EI EI 3 4 6 6 6 6 6 4 6 6 6 B B k (38)
Noal loas are couple to the relevant noal isplacements. Noal moments are couple to the relevant noal rotations. Distribute loas are transforme to noal loas an noal moments by integration of the isplacement functions. Eample by istribute loa for beam element; r qn q q q q (39) 4.5 Etene eample Beam frame In this Section we shall thoroughly investigate a beam frame, inspire by an eample given by Bell, (Bell, K. Matrisestatikk, apir Forlag, 994). his eample is also intene to emonstrate the process of a finite element analysis by iviing the analysis into a local element analysis an a system analysis. he local element analysis is characterize by the following steps: Uncoupling (in noes)/ Meshing ocalisation (apply local coorinate systems) Element formulation (Stiffness relation / Material law) Element loa vector formulations he system analysis applies all the local element analyses to combine into a global stiffness relation by the following steps: Globalisation (Global coorinate system) ssembly (Combine the iniviual local element stiffness matrices to a global stiffness matri) Bounary conitions Solution (Determine noal isplacements) Derive solutions (Stresses, strains, reaction forces an/or force resultants an moment resultants etc.) o make the process of transforming local element analyses to the system analysis smooth, we set two requirements: he system egrees of freeom an the element egrees of freeom are epresse in the same coorinate system he system egrees of freeom have a one-to-one relation with the element egrees of freeom 4.5. he problem In Figure 4.5-, the beam frame which we shall solve for is shown. Page 7 of 5
Figure 4.5- Beam frame with a point loa an a istribute loa he first task in the local analysis is to ecouple the elements in noes, or to mesh the system. 4.5. Meshing he meshing may be one in an arbitrary number of ways, as we coul ivie each iniviual member of the frame into any number of elements. For the present analyses we shall choose three elements, an we shall introuce noes at positions, B, C, an D. his leas to the following elements: Figure 4.5- Meshing of the system into 3 elements he following global noe numbering is chosen: Noe in point Noe in point B Noe 3 in point C Noe 4 in point D Page 7 of 5
4.5.3 ocalisation he element noal enumeration an corresponing irectionalities of egrees of freeom are given as follows: Figure 4.5-3 ocal egrees of freeom an orientation of beam aes for the three ecouple elements In Figure 4.5-3, the isplacements u wi are transverse noal isplacements perpenicular to the beam ais, u ui are aial isplacements parallel to the beam ais an u θi are noal rotations. Each beam element has its own beam ais, an orients its local coorinate system using the beam ais as the -ais. 4.5.4 Element formulation he stiffness matri for a beam element where aial isplacements are isregare is given in equation (4). In the present case, however, we shall inclue aial isplacements also. ial isplacements of beams are governe by the same behaviour as aial isplacements of bars. his will be shown in Section 7, when a more rigorous euction of the element stiffness matri for beams shall be performe. Combining the element stiffness relations of bars an beams results as follows: E k r 3 I 6I I 6I 6I 4I 6I I I 6I I 6I u 6I u I u u 6I u 4I u u w u w r r r r r r he element stiffness relation shown above is simply the linear combination of the stiffness relations for beams an bars, assuming that bar an beam behaviours are linearly inepenent. s the stiffness relation is etermine, the remaining part of the localisation process is to etermine the local noal loa vectors for the iniviual elements. u w u w Page 7 of 5
4.5.5 Formulation of element loa vectors oa vector for element : Element is epose only to a noal loa at noe, in transverse irection relative to the beam ais. he loa vector for the element is thus: e r oa vector for element : Element is epose to a noal loa at global noe, which is the local noe number for the element. his loa also applies to element, which leaves us with the choice of whether we shoul apply the loa in the element loa vector for element or. We have inclue it for element, an therefore we cannot apply it for element as well. he other loa acting on element is the uniformly istribute loa q. his loa acts at an angle to the beam ais, ifferent from the perpenicular angle, an may therefore be ecompose into two separate irections. In the figure above, the angle of attack of q is perpenicular to the otte line, an q w is perpenicular to element. hus the angle between these lines must be equal. Base on the imensions of element, shown in Figure 4.5-, we fin the following: 6 sin 8 cos he perpenicular an parallel components of the loaing may therefore be epresse as functions of q: q q w u qcos qsin he perpenicular component q w is a istribute loa corresponing to the beam behaviour of element. he element loas for this loa is escribe in equation (4). he result is therefore: Page 73 of 5
6 r qw qcos 6 Note that the transverse noal loas are negative, since the positive y-ais is irecte in the opposite irection of the loaing. For the horizontal component q u, the loa must be calculate accoring to the below epression: N FV N S V S r he assume isplacement functions for bars are given in Section 4.: N Combining the two equations allows a calculation of the corresponing loa vector: r qu 9 qcos qcos 9 Note that the aial loaing vector is set negative, since the irection of q u is the opposite of the positive beam ais. Epaning the r qu vector to the right imension, an combining with r qw yiels the element loa vector: e e rqu rqw r 9 9 oa vector for element 3: Element 3 is only subject to an uniformly istribute loa, perpenicular to the beam ais, which means that equation (4) yiels a consistent loa vector: Page 74 of 5
Page 75 of 5 5 5 5 5 6 6 3 q e r 4.5.6 Globalisation t the present stage in the analyses, all elements an loa vectors are represente in their own local coorinate system. he net step in the analysis is to epress all the local element stiffness matrices an the element loa vectors in a common global coorinate system. For this process we nee first to choose a global coorinate system, an seconly to fin a transformation matri for each of the elements. For the present analysis, the global coorinate system is chosen to have its origin at position, an the y-ais is parallel to element while the -ais is parallel to element 3. he net step is to fin transformation matrices for each iniviual element. he transformation of aial an transverse isplacement components was illustrate in Section.5. for bars. he same approach applies for beams in terms of rotation of translational egrees of freeom. Rotations, however, are rotationally invariant an therefore the rotations are not subject to transformation when the coorinate system changes. he resulting transformation matri is foun below: cos sin sin cos cos sin sin cos Now it only remains to transform the element stiffness matrices an the loa vectors. ransformation of element Element has its beam ais 9 egrees rotate from the -ais. he resulting transformation matri is given:
k k. 6 3. 6 5 5 6 4 6. 6. 6 5 5 6 6 4 ransformation of element Element has its beam ais at an angle θ relative to the -ais. he resulting transformation matri is given:.8.6 k k.6.8.8.6 644 474 36 3 644 474 36.6.8 474 368 48 474 368 48 36 48 4 36 48 644 474 36 644 474 36 474 368 48 474 368 48 36 48 36 48 4 he loa vector is also rotate at an angle θ relative to the -ais, so the loaing vector must also be rotate: r r e 6 6 ransformation of element 3 Element 3 is parallel to the -ais, an therefore there is no nee to rotate neither the stiffness matri nor the loa vector: Page 76 of 5
Page 77 of 5 4 6 6 6. 6. 5 5 6 4 6 6. 6. 5 5 3 k 3 k e 3 r 3 r 4.5.7 ssembly Prior to any application of bounary conitions, the number of egrees of freeom is a simple function of the number of noes. In each noe there are three egrees of freeom, consisting of two translations an a rotation. In the entire system we have chosen 4 noes, which result in a total of egrees of freeom. he global isplacement vector may therefore be formulate as follows: 4 4 4 3 3 3 v u v u v u v u g D Since the system has egrees of freeom, the global stiffness matri must be a matri. his means we must augment all the local element stiffness matrices to matrices, an all local element loa vectors to vectors. his process is best performe for each iniviual element, an will be iscusse in this fashion. Element Element has global noes an, an since there are three egrees of freeom for each noe, the first 6 egrees of freeom are associate with element. hus the contribution from element to the global stiffness matri is given as follows: 4 6 6 5 5 6. 6. 6 4 6 5 5 6. 6. 3 g k he loa vector is associate with the same egrees of freeom an becomes as follows: e r Element
Page 78 of 5 Element has global noes an 3, an since there are three egrees of freeom for each noe, egrees of freeom 4 through 9 are associate with element. hus the contribution from element to the global stiffness matri is given as follows: 4 48 36 48 36 48 368 474 48 368 474 36 474 644 36 474 644 48 36 4 48 36 48 368 474 48 368 474 36 474 644 36 474 644 3 g k he loa vector is associate with the same egrees of freeom an becomes as follows: 5 5 e r Element 3 Element 3 has global noes 3 an 4, an since there are three egrees of freeom for each noe, the last 6 egrees of freeom are associate with element 3. hus the contribution from element 3 to the global stiffness matri is given as follows: 4 6 6 6. 6. 5 5 6 4 6 6. 6. 5 5 3 3 g k he loa vector is associate with the same egrees of freeom an becomes as follows: 5 5 5 5 3 e r
Having etermine the element stiffness matrices an loa vectors in the global coorinate system, an epane their imensions, the net step is to assemble the global stiffness matri an the global loa vector: K K g g k g k g. 6. 6 3 k g 3 5 5 6 4 6. 6 646 474 3 644 474 36 5 474 868 48 474 368 48 6 3 48 44 36 48 644 474 36 644 474 36 5 474 368 48 474 38. 6 36 48 36 8 6 5 5. 6. 6 6 6 4 We observe that the global stiffness matri is symmetric, an that all the iagonal elements are positive. he global loa vector is calculate net: R g g g g r r r 5 3 5 5 5 3 4.5.8 Bounary conitions We have etermine all the components of the global stiffness relation; K D R g g. 6. 3 6 g 5 5 6 4 6. 6 646 474 3 644 474 36 5 474 868 48 474 368 48 6 3 48 44 36 48 644 474 36 644 474 36 5 474 368 48 474 38. 6 36 48 36 8 6 5 5. 6. 6 u v u v 5 u 3 6v3 3 3 5 u 4 6 v4 5 4 4 5 he beam fram is fie in noes an 4 which means that egrees of freeom,, 3,, an must be fie. hus we eliminate the relevant rows an columns: Page 79 of 5
he remaining system is therefore: KD R 3 645 474 3 644 474 36 474 868 48 474 368 48 3 48 44 36 48 644 474 36 644 474 36 474 368 48 474 368 36 u 48 v 5 36u3 v 3 3 8 3 5 4.5.9 Solution Inverting the above relation, an solving for the isplacements yiels the following global isplacement vector: 9.5.68.69 D 3.7 3.46.48 he eact same solution may be etermine using the following Matlab script: E=5*^6; E=^7; EI=^5; EI=^6; =; a=e/; b=ei/^3; Page 8 of 5
k=[a -a ; *b -6*b* -*b -6*b*; -6*b* 4*b*^ 6*b* *b*^; -a a ; -*b 6*b* *b 6*b*; -6*b* *b*^ 6*b* 4*b*^]; a=e/; b=ei/^3; k=[a -a ; *b -6*b* -*b -6*b*; -6*b* 4*b*^ 6*b* *b*^; -a a ; -*b 6*b* *b 6*b*; -6*b* *b*^ 6*b* 4*b*^]; =[ ; - ; ; ; - ; ]; kr='*k*; =[.8.6 ; -.6.8 ; ;.8.6 ; -.6.8 ; ]; kr='*k*; K=kr; K(:3,:3)=K(:3,:3)+kr(4:6,4:6); K(4:6,4:6)=K(4:6,4:6)+k(:3,:3); R=[ -5-3 5]'; inv(k)*r* he solution has also been implemente in NSYS, resulting also in the eact same isplacements: /BCH,IS /FINM,Eample4_5 /IE, inear static beam frame analysis /PREP7! Preprossessoren E,,3! BEM3 elementer R,,.5e-5,5e-7,.! Height, I-value an E-value for element R,,5e-5,5e-6,.!Height, I-value an E-value for elements an 3 MP,EX,,e! E-moulen! Define all 4 noes N,,,, N,,,, N,3,8,6, N,4,8,6,! Set the real constant set for element RE,! Define Element from noe to noe E,,! Set the real constant set for element an 3 RE,!Define elements an 3 from noe to 3, an 3 to 4 respectively E,,3 E,3,4! Bounary conitions in noe D,,u D,,uy D,,rotz! Bounary conitions in noe 4 Page 8 of 5
D,4,u D,4,uy D,4,rotz! pplie loaing in noe! Note that the efault irection of the z-ais is out of the plane of the screen. hus the moments! must be negative rather than positive. F,,f, F,,fy,-5 f,,mz,-! pplie loaing in noe 3, comment on moment irection from noe applies here also f,3,fy,-3 f,3,mz,-5! Meshing, element efinitions, bounary conitions an loaing has been complete. FINISH /SOU SOVE Page 8 of 5
4.6 Eercises 4.6. Manatory assignment in MEK455 4.6.. Eercise In this eercise we shall look at an en loae cantilever beam. We will use the Rayleigh-Ritz metho to evelop a solution to the isplacements. Generally we shall assume; w m n cn n a) Show that the isplacement assumption fulfils the bounary conitions for the beam b) Vectorise the isplacement assumption, an give it on the form u=n. c) Fin the loa vector of the system ) Fin the stiffness matri of the system e) Implement the stiffness matri an the loa vector for general m in Matlab f) ssume EI=, =, F=. Use the analytical solution for comparison; How many terms must we inclue to achieve less than % error on the isplacements? How accurately may we escribe the moments? g) We have create an element with particular bounary conitions an a set of isplacement polynomials to achieve % accuracy on isplacements. May this element be combine with other elements, if yes, how may continuity an kinematic compatability be ensure? h) Woul you characterise the propose solution as a goo solution? i) Recalculate the loa vector instea with a constant istribute loa q over the length of the cantilever j) How many terms are necessary now, in orer to achieve a goo solution (% accuracy) on the isplacements? k) How i the change of loa conition influence the accuracy of the moment calculations? Page 83 of 5
4.6.. Eercise EI=, =, = We assume that bening stiffness, length an area is equal for all three elements. a) he maimum loa from the istribute loa is q. Determine the loa vector for the beam using the beam equations. b) Establish the global isplacement vector c) Introuce the bounary conitions an re-establish the isplacement vector ) Establish the stiffness matri for the system with bounary conitions reaily implemente. e) ssume q= an F= an solve for isplacemenets f) Use NSYS to verify your results with BEM3 elements g) Use matlab to calculate the reaction forces in the two constraints an use NSYS to confirm the results. Page 84 of 5
5 INER RINGE ND BIINER RECNGE Cook Sections 3., 3.4, 3.8 an 3.9 he elements we shall work with in this chapter are calle membrane elements. Membrane behaviour is classifie by pure in-plane behaviour. his means a membrane only eforms within its plane. n analogy to beams an bars woul be that bars are to beams what membranes are to plates. bar consiers only aial eformation while the beam also inclues rotation about the ais. membrane only consiers aial an in-plane shear eformation within its plane, but no rotations about any line in the membrane s plane. We shall start by investigating the two most funamental membrane elements. he linear triangle element, in the literature often referre as the constant strain triangle, an the bilinear rectangle. First we shall look at a mathematical escription of the membrane behaviour. 5. Displacement assumption Membrane elements only consier in-plane eformations, an all other eformations are consiere negligible. hus we have a two-imensional problem. he isplacement assumption is given in equation (4); u u u v v, y, y he strain fiel is base on plane stress or plane strain. hus we only inclue in-plane strains in our stress-strain relation. he strain fiel is given in equation (4); yy y he relation between the isplacement an the strain fiels is given by the general ifferential operator ; y y Before we make assumptions on the isplacement fiel, i.e. insert polynomial assumptions for the isplacements, we nee to unerstan which requirements we place on the isplacement polynomials; Convergence requirement: he isplacements an their erivatives are continuous within the element borers. On the bounary the isplacements are continuous but the strains are m not. he general level of continuity is therefore C where m is the orer of the polynomials in the polynomial assumptions. (4) (4) (4) Page 85 of 5
Completeness he element must be capable of escribing rigi boy isplacements an rigi boy rotations as well as a constant strain fiel eactly. In orer to achieve complete polynomials we apply Pascal s triangle: Pascal s triangle for polynomials: 5. Coorinate systems an coorinate transformations 3 y y Our stuy on the funamental linear membrane elements changes the perspective of coorinate systems relative to bars an beams. When we stuy uni-aial elements, we only nee to unerstan the orientation of the element relative to the -ais, an to unerstan the length of each element. hereby we may assemble global stiffness matrices fairly simply. When we take the step into a two-imensional reality, we may no longer apply this simpler approach. twoimensional element may take any size an within the confines of its efinition also shape. hus the stiffness matri for an iniviual element may vary greatly from other elements. here is a vast array of methos to hanle the issues of coorinate systems an inee how to etermine stiffness matrices an the relation between their local an global coorinates. In this chapter we shall investigate two methos, which are among the simpler approaches. s an introuction to coorinate systems an coorinate transformations in two imensions we will stuy: Generalise coorinate systems, an Natural coorinates he systems we introuce are general an applicable to most element configurations. hey are not confine to two imensions either. For the constant strain triangle we shall apply generalise coorinates, an for the bi-linear rectangle we shall apply natural coorinates. Natural coorinates are not reaily applicable to triangular elements, but generalise coorinates apply to any kin of element geometry. pplying generalise coorinates to the bi-linear rectangle will be the subject of eercises. y y y y 3 Page 86 of 5
Page 87 of 5 5.3 Constant strain triangle Figure 5.3- hree noes with respective - an y coorinates he number of egrees of freeom for the constant shear triangle is si. We have two isplacements in each noe, which means si linearly inepenent isplacements. In each irection however, we only have three egrees of freeom. With three egrees of freeom we nee three isplacement polynomials. Base on Pascal s triangle, we fin the three first polynomials to use as a isplacement assumption for the constant strain triangle. y P P,, For the constant strain triangle we shall apply generalise coorinates. he principle of generalise coorinates is illustrate by eample applie to the constant strain triangle. First we write our isplacement assumption on vector matri form: v u v u q q N N u q q (43) he quantities q N an q are escribe by the following equations: y q q q q q q v v v v u u u u, 3 3 N q q q In equation (43) we have thus rewritten the isplacement assumption on a vector matri form. Generally, we want the unetermine coefficients of each of the polynomials to represent the magnitue of a specific noal isplacement rather than being general coefficients of a
Page 88 of 5 polynomial, in orer to conveniently apply bounary conitions an kinematic compatibility. hus we want the shape of our equation on the form: N u (44) Currently we have an equation on the form; N q u q (45) In orer to transfer the equation over to a form we want, we have to perform some transformation of the prouct q N q. he transformation is achieve by simply evaluating the isplacement polynomials at the noes of the element: 3 3 y y y v u v u q q q By inverting the above relation, we fin an epression for q; q By combining equations (44) an (45) with the epression for q, we fin; N N q N u q q his leas to our conclusion for N: N N q (46) Since q u an q v are orere in a specific fashion, we have introuce a restriction on, which efines the orering of the noal egrees of freeom. he vector is thus: 3 3 v v v u u u (47) ssuming that the inverse of eists, which is not always the case, it is straightforwar to etermine the epression for the stiffness matri. he B-matri may be foun by applying the ifferential operator efine in equation (4) to the epression for N as shown in equation (46). N N B q On block-matri form, we may clarify further how the B-matri is caculate:,,,, N N N N N N N B q q y q y q q q q y y (48)
he erivatives of N q are easily etermine: N N q, q, y s iscusse in Section 3.3., the material law for a membrane is either plane stress or plane strain. he generalise Young s moulus E, foun in the epression for the stiffness matri will in this case thus be either that of plane stress or plane strain. pplying the epression for the B- matri with a suitable material law yiels the final epression for the element stiffness matri; B EBV k V Since the ifferentiation operation on the assume polynomial isplacements result in constant values for the B-matri, the integration over the volume of the triangle simply results in a factor on the prouct B EB which is the volume, assuming that the thickness of the membrane element is constant over the area. hus: k hb EB In equation (49), is the area of the triangle an h is the assume constant thickness of the triangle. From linear algebra, we may apply a general result to etermine the area of the triangle: et hus we arrive at the final epression for k: h et k B EB (49) (5) 5.3. Eample Stiffness matri for a constant strain triangle element We shall fin the stiffness matri for a triangle with the following properties: E = 7 GPa v =.3 Uniform thickness of membrane mm Plane strain shall be assume he same element will be reinvestigate in section 5.3.3, an in that eample subject to static analysis. he triangle is shown in Figure 5.3-. Page 89 of 5
Figure 5.3- three noe triangular element, units are in millimetres First we compute :..4.3..5.5 Note that millimetres have been transforme to metres. his has been one since the Young s moulus is efine in terms of force per square metre. From equation (48) we recognize that the inverse of is necessary to compute: he proucts N N q, q, y.5749 8.574 8.574 q, an N 8.574 8.574.4857 4.8574 57.49.749 4.857 85.749 q, are compute: N y 4.8574 57.49 4.857 85.749 Base on the above calculations, the B-matri is foun: 8.574 B 8.574 4.8674 57.49 4.857 85.749 8.574 8.574 57.49 4.8674 85.749 4.857 he remaining part is to compute the prouct of B-matrices an integrate over the volume to etermine the stiffness matri k. Page 9 of 5
he prouct h et k B EB B EB is compute: 8.57 8.57 4.87 57.4.3 8.57 E 4.9 85.7 B EB.3.9 8.57 8.57.35 8.57 57.4 4.87 85.74 4.9 78 439 583 67 65 78 374 557 4 749 55 439 557 35 87 68 875 B EB E 583 4 87 33 534 67 79 68 33 485 568 65 55 875 534 568 85 4.87 57.4 4.9 85.7 Finally we may calculate k by multiplying the remainer of equation (5): h et k B 78 439 EB 365 583 67 65 78 374 557 4 79 55 439 557 35 87 68 875 583 4 87 33 534 8.57 8.57 67 749 68 33 485 568 57.4 4.87 65 55 875 534 568 85 85.7 4.9 Since lengths were transforme to metres, the unit of the stiffness matri is N/m. Question: If rotational egrees of freeom are involve, will all the entries of the stiffness matri still have the same unit? 5.3. ifferent approach to etermine the stiffness matri of a constant strain triangle element s a note, an alternative calculation proceure for the stiffness matri will be shown. he two methoologies, i.e. the methoology escribe previously in this chapter an the metho which will be shown in the following, o not result in any particular ifferences in terms of computational efficiency or computational stability, so the choice of methoology is generally a choice of convenience. Instea of transforming the N q -matri to to an N-matri via the -matri, the B q -matri may be calculate instea: Page 9 of 5
B q N q y Nq y N q Nq Nq,, y N N q, y q, Base on the B q -matri, the stiffness matri is calculate in two steps: k q h et BqEBqV BqEB V q From equation (5) the k q matri is transforme to the correct global coorinate system via the -matri: k k q o prove the relation between the k an the k q matri is left to the eercises. 5.3.3 Eample Simplifie analysis of a triangular stiffener (5) (5) Figure 5.3-3 Part of a larger structure, containing a stiffener he triangle shown in Figure 5.3-3 is a steel stiffener in a larger structure. he structure is fie in noes an 3, while the arm along the y-ais is fleible. he steel arms are conservatively consiere to be so fleible that the stiffener absorbs all the loaing. he loaing along the y-ais shall be simplifie to inclue only point loas in noe. F = kn, F y = 8 kn. he specifications for the stiffener are liste as follows: E = 7 GPa, v =.3 h = 6 mm,.4,, y.,.4,.4. i Plane stress shall be assume i Maimum allowable local principal stress is 4 MPa he eample shall etermine if the stress in the triangular stiffener is within the allowable limit. Page 9 of 5
Page 93 of 5 First the -matri is etermine:.4.4.4.4.4.4 he inverse of is foun by simple calculation:.5.5.5.5.5.5.5.5 N q, an N q,y are foun by ifferentiation of N q :,, y q q N N he B q -matri is simply a function of N q, an N q,y :,,,, q y q y q q q N N N N B he stiffness matri in generalise coorinates, k q may now be calculate base on equation (5):.99.376.38.38.38.38.376.99 et 8 q q q h EB B k From equation (5), the element stiffness matri in global coorinates may be foun: 9.6.3885 6.84 4.4357.437.3885.3885.3885.3885.3885 6.84 6.84.473.473 4.4357.3885.473 9.6 6.84.3885.473.473 6.84 6.84.3885.3885.3885.3885 8 k k q
We now know the element stiffness matri for the stiffener. he net step is naturally to implement the bounary conitions. he element is fie in noes an 3, which leaves only the egrees of freeom u an v. hese are entries an 4 in the noal isplacement vector, an therefore the global stiffness matri, an global isplacement vector may be foun as follows: k K k D 4 u v k k 4 44 8.3385 6.84 he loas are point loas, an therefore we may apply them irectly in the global loa vector: F R F y 8 o etermine the noal isplacements in noe is now only a matter of algebra: D K R.54 3.7 In orer to etermine the element stress vector, it is easiest to etermine the strains first. From equation (37), we know that ε = B. Using the same methoology as in Eample 5.3., the B matri is foun: B.5.5.5.5.5.5.5.5 he noal isplacement vector is the result of the global isplacement vector inserte into the element isplacement vector: 3.54.7 he prouct of these quantities yiels the strain: B yy y 3.93.56 Note that the compressive strain in -irection is zero. Since the strains are known, it is a trivial matter to insert the strains into the material law to etermine the stresses: E yy 67 MPa y Question: How can the compressive stress in -irection be non-zero when the strain is zero? he acceptance criterion was base on the principal stresses. he two principal stresses are the Eigen-values of the stress tensor. he stress tensor is state for ease of reference: Page 94 of 5
ij y y yy he Eigen-values of the stress tensor are in this case -46 MPa an 59 MPa respectively. he st principal stress is therefore much lower than the acceptable 4 MPa reference, an thus the stiffener is not epose to ecessive loaing base on an analysis with only element. In the eercises, the consequences of increasing the number of elements will be investigate. he valiity of the analyses is checke using NSYS, with a linear triangular element calle PNE4. he isplacement fiel is shown in the figure below: he isplacement fiel in the figure above is the vector sum of the lateral an transverse isplacements. With the following NSYS coe, an eact match with the analytical results is foun: /BCH, IS /FIENME, ES /IE, nalyse av triangulær stiver /PREP7! Element type PNE4! Keyopts = (global coorinate system), = (no etra isplacement shapes), 3=3 (Plane stress! with thickness input), 5=, 6= (No refinement on stress solution). E,,4,,,3,,,! Real constant set, is the thickness of the plate Page 95 of 5
R,,.6! ssigning the real constants to element type RE,! Material properties MP,EX,,7E9 MP,EY,,7E9 MP,NUXY,,.3! Definition of noes N,,,, N,,.4,.4, N,3,,.4,! Defining an element with noes, an 3 E,,,3! Bounary conitions D,,UX, D,,UY, D,3,UX, D,3,UY,! pplication of loas F,,FX, F,,FY,8! Finish preprocessor FINISH! Solution processor /SOU SOVE he Matlab coe for the present eample is given below: % Establish constants Emo=7*^9; v=.3; h=.6; F=; Fy=8; % Plane stress material law E=Emo/(-v^)*[ v ; v ; (-v)/]; %------------------------------------------- % Calculate stiffness matri %------------------------------------------- % -matri =[ ;.4.4;.4]; (:3,:3)=; (4:6,4:6)=; % Determine inverse of an i=inv(); i=inv(); % Differentiate Nq-vectors Nq=[ ]; Page 96 of 5
Nqy=[ ]; % Determine Bq-matri Bq=[Nq ; Nqy; Nqy Nq]; % kq-matri kq=.5*et()*h*bq'*e*bq % ransform kq to k k=i'*kq*i; % Calculate B-matri for later use with % Stress calculations Nqi=Nq*i; Nqyi=Nqy*i; B=[Nqi ; Nqyi; Nqyi Nqi]; %------------------------------------------ % Determine global stiffness matri % an global loa vector %------------------------------------------ K=zeros(,); K(,)=k(,); K(,)=k(,4); K(,)=k(4,); K(,)=k(4,4); R=[F; Fy]; %------------------------------------------ % Solve for the isplacements %------------------------------------------ D=inv(K)*R %------------------------------------------ % Determine stresses %------------------------------------------ =zeros(6,); (,)=D(,); (4,)=D(,); Epsilon=B*; Sigma=E*Epsilon Sigma_ensor=[Sigma(,) Sigma(3,); Sigma(3,) Sigma(,)]; % Principal stresses eigs(sigma_ensor) 5.4 he bi-linear rectangle In Figure 5.4-, a rectangular membrane element with 8 egrees of freeom is shown. he egrees of freeom are (u i, v i ) an they are efine in each of the 4 corners of the rectangle. Page 97 of 5
Figure 5.4- rectangular membrane with sies a an b If we wish to efine the stiffness matri for general, rectangular membrane elements, there are an infinite number of possibilities to o so, since the imensions of the rectangles have infinite possibilities for variation. We shall introuce natural coorinates to epress all possible rectangles in the same unit type formulation, such that the stiffness matri, an loa vector epressions, may be epresse generally. he other alternative is to etermine the stiffness matri iniviually for any rectangular membrane element as part of the numerical process in the finite element solution methoology, but this is costly in terms of computational efficiency an therefore it is esirable to etermine general epressions. he natural coorinates allow for a general methoology for rectangular elements. 5.4. Natural coorinates We introuce two new coorinates: a y b if efine on the element an on its bounaries. It is also convenient to efine the relation between the stanar Cartesian coorinate representation an the natural coorinates on vector-matri form: he two new coorinates will take the intervals,,, a y a, b b he coorinate transformation matri is most often calle the Jacobi-matri, an generally enote J in stanar mathematical tets. he inverse relation is trivially foun by inverting the coorinate transformation matri: a y b Page 98 of 5
o etermine the stiffness matri we must apply the general ifferential operator, which means we must ifferentiate isplacement assumptions. If we epress our isplacement assumptions in terms of natural coorinates it is therefore necessary to efine ifferentiation of functions epresse in natural coorinates. his is, however, a simple matter of application of the chain rule: a y y y b his formulation also allows for a vector-matri representation of ifferentiation: y In aition to ifferentiation, we must also integrate functions epening on the natural coorinates, which means we must epress the relation between integration in Cartesian coorinates an natural coorinates: b a ba f, yy g, ab For stanar coorinate transformations, the factor separating y from ξη is the eterminant of the coorinate transformation matri, i.e. the eterminant of the Jacobi matri. his factor represents the ifference in the area for an incremental part of the integration omains for the two coorinate representations. (53) 5.4. Choice of isplacement functions For the triangular element we ha three noes, an three polynomial functions to choose from when etermining the assume isplacement fiel. For the rectangle, we have 4 noes, an subsequently we must apply 4 polynomial functions. We choose the basic form of the functions from Pascal s triangle: Figure 5.4- Four isplacement functions chosen from Pascal s triangle Note that either or y coul have been chosen instea of y as the fourth term, as both of these functions are represente in the same row of Pascal s triangle. Either of the functions woul generally work, but empirically it is foun that the best general results are achieve if the Page 99 of 5
functions are chosen such that the balance between - an y-terms is as even as possible. We woul for instance favour either - or y-terms if one of the quaratic or y terms were chosen instea. Physically this woul be interprete as having a ominance of secon orer isplacement behaviour in only one of the coorinate irections. In orer to fulfil kinematic compatibility an to ease the implementation of bounary conitions, our isplacement fiel must consist of functions which have unit value at one noe an zero at the remaining noes. s for the triangular elements, we shall use the same isplacement polynomials in orer to approimate isplacements in u- an v-irections both. hus we wish to etermine four polynomials which have the properties shown in Figure 5.4-3. Figure 5.4-3 Displacement polynomials for a bi-linear rectangle In Section 7, interpolation techniques an methos to erive such polynomials will be iscusse in etail. For now, the polynomial epressions will simply be presente: Ni i 4 i In the above equation, ξ i is the coorinate value of ξ in the i-th noe, an corresponingly η i is the coorinate value of η in the i-th noe. If we enumerate the noes accoring to Figure 5.4-4, we may fin epressions for the isplacement polynomials. (54) Figure 5.4-4 Enumeration of noes for the bi-linear rectangle For noe, for instance, the ξ-coorinate is an the η-coorinate is -. hus the isplacement polynomial becomes: Page of 5
Page of 5 4 N We also observe that the polynomial has the correct form accoring to the isplacement polynomials shown in Figure 5.4-. 5.4.3 Development of the stiffness matri For the evelopment of the stiffness matri it is necessary to formalise the isplacement assumption. he isplacement polynomials erive in the previous sub-section are applie for isplacements in both u- an v-irections. hus the isplacement assumption may be written as follows: v u N N N u (55) In equation (55), the variables may be escribe as follows:,,3,4,, i N i i v v i u u N N i are taken from equation (54), an i represents the relevant noe number. he full epression for N is given below: 4 N Base on the epression for N, we may erive the unit isplacement strain matri B: B N However, the isplacement polynomials are epresse in natural coorinates, so the generalise ifferential operator must be re-assesse: y y (56) Combining equations (53) an (56) yiels the following epression for the ifferential operator:
Page of 5 a b b a pplying the ifferential operator to the isplacement polynomials yiels the B-matri:,,,, N N N N N N N B a b b a a b b a he partial erivatives of the N-matri are calculate as follows: 4 4,, N N he remaining part is compute the element stiffness matri, base on the epression for the B- matri, assuming that the thickness of the element is uniform: V ab h V EB B EB B k he result is an 88 stiffness matri, an each term is fairly lengthy. For instance, assuming plane stress: 6 v ab b v a Eh k
5.5 Eercises 5.5. Eercise Use the relation between N an N q in equation (46) to prove that the transformation in equation (5) yiels the correct relation between k an k q. 5.5. Eercise three noe triangular element is shown in the figure below; ssume that.,.,.5, y.,.5,.5. i a) Determine N i b) Evaluate the isplacement polynomials in all three noes, i.e. etermine that N has unit value in noe, an zero in noes an 3 etc. for all three isplacement polynomials. c) Determine k ssume u = u = v =, an point loes in noe 3: F y = F =. Convert the Matlab eample in Section 5.3.3 to: ) Determine the noal isplacements e) Determine the stress an the strain tensors for the element f) Convert the NSYS file in Section 5.3.3 to confirm the Matlab results g) Using nsys, what is the relative ifference in noal isplacements if you apply 4, 6 or 64 elements instea of just one, compare to the results from )? Page 3 of 5
5.5.3 Eercise 3 h) Using nsys, what is the relative ifference in maimum stress in the triangle if you apply 4, 6 or 64 elements instea of just one, compare to the results from e)? i) Shoul inaccuracies in isplacements theoretically be smaller or larger than inaccuracies in stresses an strains? Does this fit with the finings from g) an h)? What o these finings tell you about the results in Eample 5.3.3? he 4-noe bi-linear rectangle shown in Figure 5.4-4 will be stuie in this eercise. Use Matlab to etermine an analytical epression for the stiffness matri. Confirm that k becomes the same epression as shown in Section 5.4.3. 5.5.4 Eercise 4 (Note Eercise 3 must be solve first). Base on the stiffness matri evelope in Eercise 3, the problem shown in the Figure below shall be investigate: he rectangle is fie in noe, an in noe v is fie. he material properties are given as follows: 9 E 7 v =.3 a) Solve for noal isplacements using Matlab, an confirm using NSYS with a 4-noe PNE4 element. b) Fin the strain vector using Matlab, an etermine the maimum principal strain on the omain of the element an confirm against the NSYS solution. Page 4 of 5
5.5.5 Eercise 5 For a 4-noe bi-linear rectangle, place noe at the origin of a Cartesian coorinate system. et the length be.4, an the with be.4 for the element. a) pply generalise coorinates to etermine the stiffness matri for the element (using Matlab for the ifferentiations an integrations) b) Compare the stiffness matri to the element stiffness matri etermine in Eercise 4, to show that the two ifferent methoologies prouce the same element stiffness matri Page 5 of 5
6 SSEMBY When implementing solutions for a single element, or systems compose of only a few elements, implementation of bounary conitions an subsequently assembling the global stiffness matri an global loa vector is fairly simple. he systems we have analyse so far in this course inee are only compose of one or just a few elements. In this Section we shall euce a formal methoology for the implementation of bounary conitions an the subsequent assembly process. he formal methoology allows us to epan our scope such that we may analyse larger systems of elements. It shoul be note that the methoologies euce in this Section are not the same as the methos applie in professional finite element software, since the methos euce here are unfavourable in terms of computational efficiency. From a mathematical perspective, however, the methos are instructive an transparent. 6. Backgroun In Section 3, we foun the epression for the total potential energy functional: u U EV u FV u V V S he total potential energy functional is efine over the entire volume V of the elastic boy, an the tractions act on the close surface S which circumscribes the volume. For the purpose of the finite element metho, we choose to partition the volume V into a set of n elements such that: n V V e e In other wors, the volume V may be partitione into a set of element volumes V e such that the union of all the element volumes becomes the total volume. n important feature of the partition is that it must satisfy the conition that all the iniviual element volumes are isjoint, i.e. no part of the volume is inclue in more than one element of the partition. More precisely: V V, i j, when i, j,..., n i j Each iniviual part of the partition is calle an element. subset of the partition inclues elements which have a surface bounary inclue in S. his implies that: n S S e e In the above equation, S e is zero when the element oes not have a surface contact with the outer surface bounary, an when the element has a surface bounary inclue in S, S e is the area of that surface bounary. Since integration is aitive we may reformulate the epression for the total potential energy functional: n u EVe u FVe e V V S e e e u S e S Page 6 of 5
Page 7 of 5 In other wors, the integration over the total potential energy functional may be performe iniviually for each element, an the sum must be equal to the integral over the entire volume. In Section 4., the concept of element isplacement vectors an element isplacement polynomials was introuce, an introucing these concepts to the present iscussion yiels: n e V V S e e e e e e e e e e e e e e S V V N F N EB B (57) In equation (57), e is the element isplacement vector, N e is the matri of element isplacement polynomials an B e is the element unit isplacement strain matri. 6. opology matrices opology matrices are generally introuce in orer to augment stiffness matrices an to sort local from global egrees of freeom. Simply put, there eists a matri a e such that: D a e e (58) In equation (58), D is the global isplacement vector an R is the global loa vector. Generally speaking, if an element has k egrees of freeom, an the total number of global egrees of freeom is m, then a e is a k m matri. If the element egrees of freeom correspon to global egrees of freeom (GDOF) accoring to the following sequence: i a k a a GDOF,...,, then a e is a matri of zeros with the eception of {k,a,k,a, k k,ak } which have the value. he generation of a e is illustrate by an eample below. 6.. Eample of a topology matri Say we have a 8 DOF system, an an element with 4 DOF. If the element egrees of freeom correspon to the global egrees of freeom, 5, 8 an 7, then a e is a 4 8 matri an it becomes: a e By irect calculation this can easily be verifie: 7 8 5 8 7 6 5 4 3
6.3 ugmentation of the stiffness matri he concept of the topology matri may be introuce to the general euction of the stiffness matri an loa vector, by simple alteration of the euction in Section 4.. he euction will be given in part, an the remaining arithmetic is left to the eercises. Inserting for equation (58) into (57) leaves: n D ae Be EBeaeDVe D a e e N e FV e V V S e e e D a e N e S Performing the variation operation an applying the principle of minimum potential energy results in the following epression: D n a e B e e e e e e e e e EB V a D a N FV a V e n e V S From Section 4. we know the following ientities: V V e e B N e e EB V e FV e e k S e N e e S e r e e e N e S Combining equations (6) an (6) yiels the epression for augmentation of local stiffness matrices an summation into a global stiffness matri as well as an epression for augmentation of element loa vectors an summation into a global loa vector: K R n e n e a a e e k r e e a e k R k In equation (6), the global force vector has been ae to k 6.3. Etene eample Revisit of the triangular stiffener R k e e (59) (6) (6) (6). R k are noal loas/moments. In eample 5.3.3, a triangular stiffener epose to a point loa was stuie. In this eample, the same stiffener will be analyse, only this time with 4 membrane elements rather than. he triangle is shown again, in a meshe state: Page 8 of 5
Figure 6.3- Part of a larger structure, containing a stiffener In Figure 6.3-, the noes are given outsie of the structure an the element enumeration is shown in the interior of each element. hus, the meshe structure has 6 noes an 4 elements. he structure is fie in noes 4 an 5 an 6, while the arm along the y-ais is fleible. he steel arms are conservatively consiere to be so fleible that the stiffener absorbs all the loaing. he loaing along the y-ais shall be simplifie to inclue only point loas in noe. F = kn, F y = 8 kn. he specifications for the stiffener are liste as follows: E = 7 GPa, v =.3 h = 6 mm,,.,.,.,.4, y.,.,.,.4,.4,.4. i Plane stress shall be assume i Maimum allowable local principal stress is 4 MPa he eample shall etermine if the stress in the triangular stiffener is within the allowable limit. 6.3.. Element stiffness matrices Generalise coorinates have been applie. Element he global an local noal enumerations have been chosen as equal.... N B N q, q, y N N q, y q, Page 9 of 5
k h et( ) B EB Element Element noes, an 3 have been chosen as global noes, 3 an 4 respectively. k Element 3....4 h et( ) B EB Element noes, an 3 have been chosen as global noes 3, 4 an 5 respectively k Element 4 3....4 h et( ) B EB Element noes, an 3 have been chosen as global noes 3, 5 an 6 respectively...4..4.4 k 4 h et( ) B EB 6.3.. Bounary conitions he global isplacement vector is chosen as: g D u v u v u3 v3 u4 v4 u5 v5 u6 v6 pplying a fie conition in global noes 4, 5 an 6 reuces the global isplacement vector to: D u v u v u3 v3 6.3..3 ssembly he topology matrices are etermine for each iniviual element: Element Page of 5
6 element egrees of freeom shall match with 6 global egrees of freeom. he topology matri becomes: a Since the a-matrices are large, an full of zeroes, they will not be shown in full for the remainer of the eample. Instea, the entries of the relevant a-matri that are non-zero, i.e., will be given on ine form. So for a, the following ine pairs yiels entries of : Element,,3 3,5 4, 5,4 6,6 he following ine pairs yiels entries of : Element 3,3,5 3,7 4,4 5,6 6,8 he following ine pairs yiels entries of : Element 4,5,7 3,9 4,6 5,8 6, he following ine pairs yiels entries of : Summing up,5,9 3, 4,6 5, 6, s all the element stiffness matrices are etermine an augmente, the global stiffness matri may be assemble: K a ka aka a3k3a3 a4k4a4 pplying the bounary conition simply consists of eliminating rows an columns 4, 5 an 6. he final result is given below:.388.388 K 9.388.388.684.47.684.47.388.47.845.3648.388.684.845.4777.47.3648.845.388.4777.845 here is little work in establishing the loa vector, since we only have point loas: Page of 5
R 8 6.3..4 Solution Solving for the isplacements yiels the following isplacement vector: D K R 3.5.46.886.89..6 With the below NSYS coe (in Section 6.3..6), this isplacement vector is replicate eactly. he isplacement vector yiels the following isplace structure (epicte in NSYS): From the figure it is observe that the linear isplacement along the y-ais in Eample 5.3.3 is replace with a non-linear one, an that the larger part of the isplacements are foun in noe. Compare to Eample 5.3.3, u in noe is also ouble, i.e. the system of four elements is significantly less stiff than the system of element. Calculation of stresses 6.3..5 Matlab coe % Establish constants Emo=7*^9; v=.3; h=.6; F=; Fy=8; % Plane stress material law E=Emo/(-v^)*[ v ; v ; (-v)/]; %------------------------------------------- Page of 5