Introduction to Simulation - Lecture 2. Equation Formulation Methods. Jacob White. Thanks to Deepak Ramaswamy, Michal Rewienski, and Karen Veroy

Transcription

1 Introduction to Simulation - Lecture Equation Formulation Methods Jacob White Thanks to Deepak Ramaswamy, Michal Rewienski, and Karen Veroy

2 Outline Formulating Equations rom Schematics Struts and Joints Example Matrix Construction From Schematics Stamping Procedure Two Formulation Approaches Node-Branch More general but less eicient Nodal Derivable rom Node-Branch

3 Formulating Equations rom Schematics Y x, y Struts Example Identiying Unknowns x, y X,, hinged Assign each joint an X,Y position, with one joint as zero. SMA-HPC 3 MIT Given a schematic or the struts, the problem is to determine the joint positions and the strut orces. Recall the joints in the struts problem correspond physically to the location where steel beams are bolted together. The joints are also analogous to the nodes in the circuit, but there is an important dierence. The joint position is a vector because one needs two (X,Y) (three (X,Y,Z)) coordinates to speciy a joint position in two (three) dimensions. The joint positions are labeled x,y,x,y,..xj,yj where j is the number o joints whose positions are unknown. Like in circuits, in struts and joints there is also an issue about position reerence. The position o a joint is usually speciied with respect to a reerence joint. Note also the symbol This symbol is used to denote a ixed structure ( like a concrete wall, or example). Joints on such a wall have their positions ixed and usually one such joint is selected as the reerence joint. The reerence joint has the position, (,, in three dimensions). 3

4 Formulating Equations rom Schematics x, y x, y 3 3 x, y Struts Example Identiying Unknowns 4 4 x, y load Assign each strut an X and Y orce component. SMA-HPC 3 MIT The second set o unknowns are the strut orces. Like the currents in the circuit examples, these orces can be considered branch quantities. There is again a complication due to the two dimensional nature o the problem, there is an x and a y s component to the orce. The strut orces are labeled,,...,, where s is the number o struts. s x y x y 4

5 Formulating Equations rom Schematics Struts Example Aside on Strut Forces Y (,) L x y x, y X ( ) = EAc = ε L L L x y L x = L y = L L L = x + y SMA-HPC 3 MIT The orce,, in a stretched strut always acts along the direction o the strut, as shown in the igure. However, it will be necessary to sum the orces at a joint, individual struts connected to a joint will not all be in the same direction. So, to sum such orces, it is necessary to compute the components o the orces in the X and Y direction. Since one must have selected the directions or the X and Y axis once or a given problem, such axes are reerred to as the global coordinate system. Then, one can think o the process o computing x, y shown in the igure as mapping rom a local to a global coordinate system. The ormulas or determining x and y rom ollow easily rom the geometry depicted in the igure, one is imply projecting the vector orce onto coordinate axes. 5

6 Formulating Equations rom Schematics 3 x x x 3 y y y + + = + + = 3 x, y x, y Struts Example Conservation Law 4 + = 4 3 x x load + = 4 3 y y load load x y,, Force Equilibrium Sum o X-directed orces at a joint = Sum o Y-directed orces at a joint = SMA-HPC 3 MIT The conservation law or struts is usually reerred to as requiring orce equilibrium. There are some subtleties about signs, however. To begin, consider that the sum o X-diirected orces at a joint must sum to zero otherwise the joint will accelerate in the X-direction. The Y-directed orces must also sum to zero to avoid joint acceleration in the Y direction. To see the subtlety about signs, consider a single strut aligned with the X axis as shown below x, x,, then the strut will exert orce in attempt to contract, as I the strut is stretched by shown below x, x +, a b 6

7 The orces a and b, are equal in magnitude but opposite in sign. This is because a points in the positive X direction and b in the negative X direction. I one examines the orce equilibrium equation or the let-hand joint in the igure, then that equation will be o the orm Other orces + a = whereas the equilibrium equation or the right-hand joint will be Other orces + b = Other orces - a = In setting up a system o equations or the strut, one need not include both a and b as separate variables in the system o equations. Instead, one can select either orce and implicitly exploit the relationship between the orces on opposite sides o the strut. As an example, consider that or strut 3 between joint and joint on the slide, we have selected to represent the orce on the joint side o the strut and labeled that orce 3. Thereore, or the conservation law associated with joint, orce 3 appears with a positive sign, but or the conservation law associated with joint, we need the opposite side orce, - 3. Although the physical mechanism seems quite dierent, this trick o representing the equations using only the orce on one side o the strut as a variable makes an algebraic analogy with the circuit sum o currents law. That is, it appears as i a strut s orce leaves one joint and enters another. 7

8 Formulating Equations rom Schematics Struts Example Conservation Law x = Fx ( x, y ) y = Fy( x, y ) 3x = Fx ( x x, y y) 3y = Fy( x x, y y) 3 4 load x = Fx ( x, y ) y = Fy( x, y ), = F ( x, y ) 4 x x = F ( x, y ) 4 y y Use Constitutive Equations to relate strut orces to joint positions. It is worth examining how the signs o the orce are determined. Again consider a single strut aligned with the X axis. x, x, The X- axis alignment can be used to simpliy the relation between the orce on the x side and x and x to x x L x x x x L x = Note that there are two ways to make x negative and point in the negative x direction. Either x- x>, which corresponds to lipping the strut, or x- x < L which corresponds to compressing the strut. 8

9 Formulating Equations rom Schematics Struts Example Summary Unknowns or the Strut Example Joint positions (except or a reerence or ixed joints) Strut orces Equations or the Strut Example One set o conservation equations or each joint. One set o constitutive equations or each strut. Note that the # equations = # unknowns SMA-HPC 3 MIT 9

10 SMA-HPC 3 MIT Strut Example To Demonstrate Sign convention Two Struts Aligned with the X axis x, y = x, y = Conservation Law At node : + = x x At node : - + = x L L

11 Strut Example To Demonstrate Sign convention Two Struts Aligned with the X axis L x, y = x, y = Constitutive Equations x x = ε ( L x ) x SMA-HPC 3 MIT x x = L x x x ε ( ) x x

12 Strut Example To Demonstrate Sign convention Two Struts Aligned with the X axis Reduced (Nodal) Equations x x x ε( L x ) + ε( L x x ) = x x x x x x ε ( L x x ) + L = x x x SMA-HPC 3 MIT

13 Strut Example To Demonstrate Sign convention Two Struts Aligned with the X axis Solution o Nodal Equations = (orce in positive x direction) L x = L + x = x+ L + ε ε x, y = x, y = L SMA-HPC 3 MIT 3

14 Strut Example To Demonstrate Sign convention Two Struts Aligned with the X axis L x, y = x, y = Notice the signs o the orces x = (orce in positive x direction) x = (orce in negative x direction) SMA-HPC 3 MIT 4

15 Formulating Equations rom Schematics Examples rom last time x, y x, y 4 Circuit Modeling VLSI Power Distribution 3, Struts and Joints Modeling a Space Frame SMA-HPC 3 MIT 5

16 Formulating Equations rom Schematics Two Types o Unknowns Circuit - Node voltages, element currents Struts - Joint positions, strut orces Two Types o Equations Conservation Law Circuit - Sum o Currents at each node = Struts - Sum o Forces at each joint = Constitutive Relations Circuit branch (element) current proportional to branch (element) voltage Struts - branch (strut) orce proportional to branch (strut) displacement 6

17 Generating Matrices rom Schematics Assume Linear Constitutive Equations... Circuit Example One Matrix column or each unknown N columns or the Node voltage B columns or the Branch currents One Matrix row or each equation N rows or KCL B rows or element constitutive equations (linear!) SMA-HPC 3 MIT 7

18 Generating Matrices rom Schematics Assume Linear Constitutive Equations Struts Example in -D One pair o Matrix columns or each unknown J pairs o columns or the Joint positions S pairs o columns or the Strut orces One pair o Matrix rows or each equation J pairs o rows or the Force Equilibrium equations S pairs o rows or element constitutive equations (linear!) SMA-HPC 3 MIT 8

19 Generating Matrices rom Schematics Circuit Example Conservation Equation i5 R5 V R V i i is is is3 R R4 R3 i4 V 4 i3 V 3 SMA-HPC 3 MIT To generate a matrix equation or the circuit, we begin by writing the KCL equation at each node in terms o the branch currents and the source currents. In particular, we write signed branch currents = signed source currents where the sign o a branch current in the equation is positive i the current is leaving the node and negative otherwise. The sign o the source current in the equation is positive i the current is entering the node and negative otherwise. 9

20 Generating Matrices rom Schematics Circuit Example Conservation Equation i5 R5 R V R i i is is is3 V R4 R3 SMA-HPC 3 MIT i4 V 4 i3 i + i = is i i5 = i 3 = is3 i + i = i + i is is 3 4 s s 3 V 3

21 Generating Matrices rom Schematics Circuit Example Conservation Equation One Row or each KCL Equation N SMA-HPC 3 MIT 3 4 Matrix Form or the Equations A i is i is i s3 i3 = i s3 i4 is + i 5 s i B One column or each branch current The matrix A is usually not square Right Hand Side or Source Currents

22 Generating Matrices rom Schematics Circuit Example Conservation Equation n n k A SMA-HPC 3 MIT How each resistor contributes to the matrix KCL at n KCL at n ik n n i i other other Rk = i A has no more than two nonzeros per column + i i k k = i s s What happens to the matrix when one end o a resistor is connected to the reerence ( or the zero node). ik n In that case, there is only one contribution to the kth column o the matrix, as shown below k n

23 Generating Matrices rom Schematics Circuit Example Conservation Equation How each current source contributes to the Right Hand Side n n isother + isb isother isb SMA-HPC 3 MIT RHS KCL at n KCL at n isb n n i = i + i 's other b b s s i = i i 's other b b s s 3

24 Generating Matrices rom Schematics SMA-HPC 3 MIT Circuit Example Conservation Matrix Equation Generation Algorithm For each resistor i ( n i ( n Set Is = zero vector For each current source i ( n i ( n Conservation Equation ik n n > ) A( n, b) = > ) A( n, b) = > ) > ) Is( n isb n n Is( n) = Is( n) is ) = Is( n) + is b b 4

25 Generating Matrices rom Schematics i 5 R 5 Circuit Example Conservation Equation R i R i is is is 3 R 4 SMA-HPC 3 MIT i i 3 R 3 A i i i3 i 4 i5 i i i i i = is is is i i 3 s + s3 i s 5

26 Generating Matrices rom Schematics Circuit Example Constitutive Equation i = Vb = ( V ) R R 4 i4 = Vb 4 = ( V 4 ) i3 = Vb3 = ( V 3 V 4) R4 R4 R3 R3 First determine Voltages across resistors (Branch Voltages) Second relate Branch currents to Branch Voltages SMA-HPC 3 MIT i i 5 i i 4 i 3 i5 = Vb5 = ( V R5 R5 i = Vb = ( V3 R R V 4 3 ) ) The current through a resistor is related to the voltage across the resistor, which in turn is related to the node voltages. Consider the resistor below. i V V R The voltage across the resistor is V-V and the current through the resistor is i = ( V V ) R Notice the sign, i, is positive i V > V. In order to construct a matrix representation o the constitutive equations, the irst step is to relate the node voltages to the voltages across resistors, the branch voltages. 6

27 Generating Matrices rom Schematics Circuit Example Constitutive Equation SMA-HPC 3 MIT i i = V b = ( R R V Examine Matrix Construction i 5 ) i i 4 i 3 i 5 = V b 5 = ( V R 5 R 5 i = Vb = ( V R R V 3 4 i 4 = V b 4 = ( V 4 ) i 3 = V b 3 = ( V 3 V 4) R 4 R 4 R 3 R 3 V V V V V b b b 3 b 4 b 5 = V V V V 3 4 ) ) To generate a matrix equation that relates the node voltages to the branch voltages, one notes that the voltage across a branch is just the dierence between the node voltages at the ends o the branch. The sign is determined by the direction o the current, which points rom the positive node to the negative node. Since there are B branch voltages and N node voltages, the matrix relating the two has B rows and N columns. 7

28 8 SMA-HPC 3 MIT Generating Matrices rom Schematics Constitutive Equation Circuit Example = b b b b b T V V V V V V V V V A = Is i i i i i A KCL Equations Node to Branch Relation The node-to-branch matrix is the transpose o the KCL Matrix A relation exists between the matrix associated with the conservation law (KCL) and the matrix associated with the node to branch relation. To see this, examine a single resistor. For the conservation law, branch k contributes two non- zeros to the k th column o A as in Note that the voltage across branch k is Vl -Vm, so the k th branch contributes two non-zeros to the k th row o the node- branch relation as in l V m V ik Rk : : : s B I I I = l A m k

29 k l m V V b : : = : V N V bb It is easy to see that each branch element will contribute a column to the incidence matrix A, and will contribute the transpose o that column, a row, to the node-tobranch relation. 9

30 Generating Matrices rom Schematics R i R4 is i4 i5 4 α SMA-HPC 3 MIT i R5 i3 R is R3 The k th resistor contributes Circuit Example The matrix relates branch voltages to branch currents. - One row or each unknown current. - One column or each associated branch voltage. The matrix α is square and diagonal. is3 Rk 3 to i Constitutive Equation b i R V b R i3 = Vb 3 R3 i4 Vb4 R 4 i5 Vb 5 5 α( kk, ) α R V 3

31 Generating Matrices rom Schematics Circuit Example Constitutive Equation R i is R4 i4 i5 4 i R5 i3 R is R3 is3 3 i V i T V i3 α A V 3 i4 V 4 5 i = VS The node voltages can be related to branch currents. - A T relates node voltages to branch voltages. α - relates branch voltages to branch currents. SMA-HPC 3 MIT 3

32 Generating Matrices rom Schematics B N T I α A Ib = A VN Is B T Ib α A VN A I = b = N N = Number o Nodes with unknown voltages B = Number o Branches with unknown currents I s Circuit Example Node-Branch Form Constitutive Relation Conservation Law 3

33 Generating Matrices rom Schematics Struts Example In -D One pair o columns or each unknown - J pairs o columns or the Joint positions - S pairs o columns or the Strut positions One pair o Matrix Rows or each Equation - J pairs o rows or the orce equilibrium equations - S pairs o rows or the Linearized constitutive relations. 33

34 Generating Matrices rom Schematics Struts Example Follow Approach Parallel to Circuits ) Form an Incidence Matrix, A, rom Conservation Law. ) Determine strut deormation using A T. 3) Use linearized constitutive equations to relate strut deormation 4) Combine (),(), and (3) to generate a node-branch orm. 34

35 Generating Matrices rom Schematics, 3 4 x y x, y Struts Example l Conservation Equation, SMA-HPC 3 MIT, x+ x+ 3x = y+ y+ 3y = = x x lx + = 3y 4y l y As a reminder, the conservation equation or struts is naturally divided in pairs. At each joint the sum o X-directed orces = and the sum o Y-directed orces =. Note that the load orce is known, so it appears on the right hand side o the equation. 35

36 Generating Matrices rom Schematics x y x y, x SMA-HPC 3 MIT 3 x, y x, y 4 l y x, y Struts Example A 3x 3 y 4 x Conservation Equation Stamping Approach Load pair o columns per strut Load right side or load 4 y x y x y = 3x lx 3y l y 4x FL 4y Note that the incidence matrix, A, or the strut problem is very similar to the incidence matrix or the circuit problem, except the two dimensional orces and positions generate x blocks in the incidence matrix. Consider a single strut xj, yj xj, yj s The orce equilibrium equations or the two joints at the ends o the strut are At joint j At joint j j j + = x s l other x x + = y s l other y y j j j j = x s l other x x = y s l other y y j j 36

37 Examining what goes in the matrix leads to a picture xj yj xj yj sx sy Note that the matrix entries are x blocks. Thereore, the individual entries in the matrix block or strut S s contribution to j s conservation equation need speciic indices and we use jx, jy to indicate the two rows and Sx, Sy to indicate the two columns. 37

38 Generating Matrices rom Schematics For each strut I ( j is not ixed) I ( j is not ixed) For each load I ( j is not ixed) A has at most non-zeros / column Struts Example Conservation Matrix Generation Algorithm Conservation Equation A( jx, bx) = A( jy, by) = A( jx, bx) = A( jy, by) = xj, yj load F j F j L ( ) = ( ) x L x loadx L ( ) = L( ) load F j F j y x y 38

39 Generating Matrices rom Schematics Y (,) L x, y X Struts Example Constitutive Equation First linearize the constitutive relation I x, y are close to some x, y, x + y = L SMA-HPC 3 MIT F F x x ( x, y) ( x, y) x x y ux = y Fy F y u y ( x, y) ( x, y) x y ux = x x uy = y y As shown beore, the orce through a strut is where L= x + y x x = Fx( x, y) = ε L L L y y = Fy( x, y) = ε L L L and x, y are as in ( ) ( ) y L X I x and y are perturbed a small amount rom some x, y such that x + y = L, then since F x (x,y ) = Fx Fx x ( x, y) ( x x) + ( x, y)( y y) x y and a similar expression holds or y. One should note that rotating the strut, even without stretching it, will violate the small perturbation conditions. The Taylor series expression will not give good approximate orces, because they will point in an incorrect direction. 39

40 Generating Matrices rom Schematics Struts Example Constitutive Equation, u, u x y 3 x 4, u, u y l x y x α ux y α y T u A = 3x α33 ux 3y α44 uy 4 α x 4 y SMA-HPC 3 MIT 4

41 Generating Matrices rom Schematics Struts Example Constitutive Equation α (,) ss SMA-HPC 3 MIT The α(,) s s block ux s, uy ux, uy Initial position Initial position, y x, y Fx Fx ( x x, y y ) ( x x, y y ) x y = Fy F y ( x x, y y) ( x x, y y) x y x 4

42 Generating Matrices rom Schematics S J T I α A s = A u L S J S =Number o Struts s J = Number o unixed Joints T = α A u = A s = Struts Example Node-Branch From Constitutive Equation Conservation Law 4

43 Generating Matrices rom Schematics Struts Example Comparison S J B N T I α A s = A u L S J T I α A Ib Vs = A VN Is B N 43

44 Nodal Formulation Generating Matrices Circuit Example i + V + ( V V ) = s R R V V R is is R R5 i + i + ( V V) + V = s s3 R R5 is 3 R4 V 4 R3 V 3 i i + V + ( V V ) = s s R4 R3 ) Number the nodes with one node as. ) Write a conservation law at each node. except () in terms o the node voltages! SMA-HPC 3 MIT 44

45 Nodal Formulation R V R V i 4 i i is 3 is is R 4 i 5 R 5 i 3 R 3 Generating Matrices Circuit Example One row per node, one column per node. For each resistor n R n V 3 + R R R v + i = s i R R R s3 5 v i R 3 R3 3 s3 + v R 4 3 R3 R4 i s +i s G SMA-HPC 3 MIT V 4 v i s Is Examining the nodal equations one sees that a resistor contributes a current to two equations, and its current is dependent on two voltages. i k V n V n R k KCL at node n i + ( V V ) = i Rk KCL at node n others ( n n i V V ) = is Rk So, the matrix entries associated with R k are others n n s n n n Rk Rk n Rk Rk 45

46 Nodal Formulation n i ( ) & ( ) else i ( ) else > n > SMA-HPC 3 MIT Nodal Matrix Generation Algorithm n > Gn (, n) = Gn (, n), Gn (, n) = Gn (, n) R R Gn (, n) = Gn (, n) +, Gn (, n) = Gn (, n) + R R Gn n = Gn n + R (, ) (, ) Gn n Gn n (, ) (, ) = + R Generating Matrices Circuit Example 46

47 Nodal Formulation Generating Matrices N G Vn = I s (Resistor Networks) N J G uj = F J L (Struts and Joints) 47

48 Nodal Formulation Comparing to Node-Branch orm Constitutive Conservation Law Node-Branch Matrix T I α A Ib = A VN Is Nodal Matrix [ G ][ VN] = [ Is] 48

49 Nodal Formulation G matrix properties Diagonally Dominant. Gii Gij G = j i G Symmetric.. ij ji Smaller.. N N J J << << ( N + B) ( N + B) (J + S) (J + S) 49

50 Nodal Formulation Node-Branch orm Node-Branch ormulation T I α A I b = A V n I s M x b Not Symmetric or Diagonally Dominant Matrix is (n+b) x (n+b) SMA-HPC 3 MIT 5

51 Nodal Formulation Deriving Formulation From Node- Branch I T b α A VN = T A ( Ib α A VN) = A AI b = I s α = G T A A VN Is 5

52 Nodal Formulation Problem element Voltage Source Vs Voltage source is + Vn Vn Constitutive Equation is + Vn Vn = Vs SMA-HPC 3 MIT 5

53 53 SMA-HPC 3 MIT Nodal Formulation Voltage Source Can orm Node-Branch Constitutive Equation with Voltage Sources R i R i R3 i3 R4 i4 3 4 R5 i5 + s s T V V V V V V A i i i i i i V = α i6 Vs 5 α = R R 3 R 4 R 5 R Problem Element

54 Nodal Formulation Problem Element Voltage Source Resistor currents Voltage source currents missing SMA-HPC 3 MIT Cannot Derive Nodal Formulation IbR T α AV N = (Constitutive Equation) IbR T A Aα A V N = (Multiply by A) A Ib= Is (Conservation Law) Ai IbR Cannot Eliminate! Nodal Formulation requires Constitutive relations in the orm Conserved Quantity = F ( Node Voltages)! 54

55 Nodal Formulation Problem Element Rigid Rod Rigid rod x, y L x, y ( ) ( ) L x x x + y y ( y y) ( x x) + = y SMA-HPC 3 MIT 55

56 Nodal Formulation Resistor Grid Comparing Matrix Sparsity Example Problem V V V 3 V 4 V 99 V V V V 3 V V 9 V 9 V 93 V SMA-HPC 3 MIT 56

57 Nodal Formulation Comparing Matrix Sparsity Example Problem Matrix non-zero locations or x Resistor Grid Node-Branch Nodal 57

58 Summary o key points... Developed algorithms or automatically constructing matrix equations rom schematics using conservation law + constitutive equations. Looked at two node-branch and nodal orms. 58

59 Summary o key points Node-branch General constitutive equations Large sparser system No diagonal dominance Nodal Conserved quantity must be a unction o node variables Smaller denser system. Diagonally dominant & symmetric. 59