Want Know how to calculate those triangless in red.
J124 was included in this example，but didn't explain about J524 clearly.the last one is about J542I would appreciate it if you could answerReally want to know how to solve it!!!
J124 was included in this example，but didn't explain about J524 clearly.the last one is about J542
I would appreciate it if you could answer
Really want to know how to solve it!!!
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2.3 Two-dimensional finite element approximation Strain potential energy and kinetic energy of a two-dimensional soft robot are formulated by integrals over two-dimensional region S, which is often described by an irregular shape, making analytical calculation of integrals difficult or impossible. Let us approximate two- dimensional region S (Fig. 2.1(a)) by a set of small triangles (Fig. 2.1(b)). Then, integral over two-dimensional region 5 can be approximated by the sum of integrals over small triangles: AP,P,PL Here we apply piecewise linear approximation so that we can analytically or numerically calculate individual integrals over small triangles. Finite element approximation of kinetic energy Let us calculate kinetic energy of a two-dimensional soft robot given by eq. (1.22), that is: T= [/pu¹unds 19 (2.30)
which directly yields P4 P1 1 First, we calculate integral over triangle region AP,P,P₂: Tijk = p.pp, püühas AP,P, (2.31) Piecewise linear approximation of function u over triangle region AP,P,P, is described as follows: Mijk = P5 P6 Figure 2.2: Example of rectangle region u=u; Nijk+uj Nj,ki + uz Nej Noting that u,. u. u depend on time while N N N do not, we have = Nijk+uj Nika + N pha 12 P2 P3 καθε Δ 12 NjNl2x2 [[:] where I2x2 represents 2 x 2 identical matrix. For sake of simplicity, assume that density p and thickness h are constants. Then, (A/6)12x2 (A/12) 12x2 (A/12) 12x2 Tij.x = []ph (A/12)/2x2 (A/6)/2x2 (4/12)/2x2 (A/12)2x2 (A/12) 12x2 (A/6)12x2 ][E] (NkI2x2 NjNjk2x2 NjNajl2x2 (N₁)²12x2 NjN₁12x2 NijNijl2x2 NjNjI22 (N)²12x2 where A = AP,P;P& (see Problem 6). Matrix 212x2 12×2 12x2 12x2 212x2 12x2 12x2 12x2 212x2 2/2x2 12x2 12x2 12x2 212x2 12x2 12x2 12x2 212x2 M₁2.4 M2,35 M5.4.2 = M6,5,3 = (2.32) 2/2x2 12x2 12x2 12x2 212x2 12x2 12x2 12x2 2/2x2 (2.33) 20 14, is referred to as partial inertia matriz. Note that the sum of all blocks of matrix Mis equal to ph 12x2, which denotes the mass of triangle region. Let us calculate the total kinetic energy over rectangle region P,PPP, shown in Fig. 2.2. This region consists of four triangle regions: AP,P₂P, AP₂PP. APP₂P₂, and APP&P3. For sake of simplicity, assume that phA/12 is constantly equal to 1. Then, partial inertia matrices are given as: (2.34) (2.35)
Let us be a collective vector consisting of all displacement vectors at nodal points: UN = 141 U₂ us which is referred to as nodal displacement vector. The total kinetic energy is then described by a quadratic form with respect to us: T=; where M is referred to as inertia matriz. Noting that (1,2,3) x (1.2.3) blocks of M₁2,4 contribute to (1.2.4) x (1.2.4) blocks of M. 12x2 namely, (1, 1), (1,2), (1,3) blocks of M₁,2,4 corresponds to (1, 1), (1, 2), (1,4) blocks of M, (2.1), (2, 2), (2.3) blocks of M₁,2,4 corresponds to (2,1). (2.2), (2, 4) blocks of M. (3.1), (3,2), (3,3) blocks of M₁2, corresponds to (4,1), (4,2), (4,4) blocks of M. we find contribution of M₁,2,4 to M as follows: 212x2 12x2 12x22/2x2 12x2 12x2 12x2 2/2x2 Similarly. (1,2,3) x (1,2,3) blocks of Ms.4.2 contribute to (5,4.2) x (5,4, 2) blocks of M. (2.36) 2/2x2 12x2 12x2 namely, (1, 1), (1,2). (1.3) blocks of M5,42 corresponds to (5,5). (5. 4), (5,2) blocks of M. (2, 1), (2, 2), (2,3) blocks of M₁,4,2 corresponds to (4,5), (4,4), (4,2) blocks of M. (3, 1), (3,2), (3, 3) blocks of M₁,42 corresponds to (2,5), (2, 4). (2.2) blocks of M. we find contribution of M5,4,2 to M as follows: 12x2 12x2 2/2x2 12x2 12x2 2/2x2
Summing up all contributions, we finally have 212x2 12x2 12x2 12x2 612x2 12x2 212x2 212x2 12x2 412x2 12x2 212x2 212x2 M = This inertia matrix M consists of 62 2 x 2 blocks and is a sparse matriz. We simply describe the above calculation as: M = M1,2,4 M2,3,5 ⒸM5.4.2 ⒸM6.5.3- Operator works block-wise. where Finite element approximation of strain potential energy We apply the above cal- culation to strain potential energy. First, let us calculate strain potential energy stored in small triangle region AP,P,P: du ər Uisk = r.PP, (Mx + ₂)ch ds. (2.38) Piecewise linear approximation of function u over triangle region AP,P,P is described as u=₁N₁.jk+uj Nj.k+u N... Introducing collective vectors = [u, uy, and 4, 5, ¹, we find = a Yu a= 1 24 412x2 12x2 212x2 12x2 612x2 12x2 12x2 12x2 212x2 du Əy Yjyk yky {]- Ij - Ik Ik - Ii 2 IiIj (see Problem 2). Then, strain vector is given as -=[1²7² Substituting the above equation into eq. (2.38), we have aa ab ba bbT 2aa +bb aft yi-yj au + 2/2x2 12x2 =b²% dv dr (see Problem 7). Then, we have U₁=¹ (AHx+pH₂) Y a Ye U₁=[²] ][*] ba 2bb + aa de Əy (2.37) =b²% T] ^ [*]. (2.39) (2.40)
where 7= - [ J hΔ. Ημπ - [ T aa ab babb H₂ = x=jk The above equation is a quadratic form with respect to 7 = [u. . . . . . Let us permutate rows and columns of H₂ and H, so that U is described by a quadratic form with respect to ujk = [₁,₁,₁, , . That is H₂ = 2aa +bb ab Permutating rows and columns so that 1.4.2.5.3.6 rows and columns of H₂ be 1.2.3.4.5.6 rows and columns of J, we obtain matrix J. Similarly, permutating rows and columns so that 1, 4, 2,5, 3,6 rows and columns of H, be 1,2,3,4,5,6 rows and columns of Jijk, we obtain matrix J. Finally, we have Uis.x = −µ², (Aſk + µ£²5k) Uz zik where J and J are referred to as partial connection matrices. Once coordinates of P₁, P₁, P₁ are given, we can calculate partial connection matrices and J 1-10 1 0-1 -1 1 0-1 0 1 0 0 1 0 0 0 0 <-1 0 1 0 0 0 0 -1 1 0 -1 Summing up all strain potential energies over small triangle regions, we obtain the total strain potential energy described as U=(Jx+J) un where matrices J, and J, are referred to as connection matrices Matrix K = XJ₁ + HJ₂ is referred to as stiffness matriz. 0 -1 0 0 0 1 H₁ = Example Let us calculate partial connection matrices of triangle P P₂P, shown in Fig. 2.2. Vectors a, b are given by a = [-1, 1, 01 and b=[-1.0, 1]. Assuming h = 2, we have 3 -2-1 -2 ba 2bb + aa -1 1 -1 0 220000 101 ]h.A. (2.41) 10 1 -1 0 0 0 0 0 LO -1 1 <-1 3 -1 -1 1 -2 0 (2.42) ON (2.43) 2 (2.44) -2 0
Permulating rows and columns of the above matrices, we find 0 0 1 1 -1 namely, -1 0 0 1 1 -1 3 1 0 1 1 -1 0 -1 0 -1 0 1-2-1-1 0 3 0 -1 -1 -2 -2 0 -1 -1 0 -1 -1 0 -2 -1 -1 -1 1 -1 0 -1 1 0 0 1 Let us calculate partial connection matrices of triangle P.PP, shown in Fig. 2.2. Vectors a, b are given by a = [-1, 1, 0] and b= -1, 0, 1]. Thus, assuming h= 2. we find J42 J124 and J5.4.2 J124, Partial connection matrices are invariant with respect to translation displacement. As a result, unde the same assumption, we have = J124 12.35 J54.276.5,3 0 Let us calculate connection matrices J, and J, of rectangle region IP, PPP, shown in Fig. 2.2. Noting that (1,2,3) x (1,2,3) blocks of J24 contribute to (1.2.4) x (1,2,4) blocks of Ja. 11-1 0 0 121 olo ole0 namely, (1, 1), (1,2), (1,3) blocks of J2 corresponds to (1, 1), (1,2), (1,4) blocks of Jx, (2,1), (2,2), (2,3) blocks of J24 corresponds to (2.1), (2,2), (2,4) blocks of J (3,1), (3,2), (3, 3) blocks of J24 corresponds to (4.1). (4,2), (4,4) blocks of J. we obtain contribution of J124 to J₁. Noting that (1.2.3) x (1.2.3) blocks of J2 contribute to (5.4.2) x (5.4.2) blocks of J₁, 0 2 0 0 1 0 1 0 0 = (1,1), (1,2), (1,3) blocks of J2 corresponds to (5,5), (5, 4), (5,2) blocks of Ja. (2,1), (2.2), (2,3) blocks of J42 corresponds to (4,5), (4,4), (4.2) blocks of JA (3.1). (3.2), (3, 3) blocks of 4.2 corresponds to (2.5). (2.4). (2.2) blocks of J₁. we obtain contribution of J42 to J. Summing up all contributions, we finally have 1000- 000000 1-1 2-1 -1 -1 1 0 0 1 0 1 1 -1 -1 0-1 0 -1-2 1 0 0 0 0 2 1 0 0 1 000- 1 0-1 0 0-1 0-1 olo- 0 0-1 1 0 -1 -2 --- 10 1 -1 0 0 1 0 -1 0 1 0 1 0 0 1 0-1 -1 1021 -1 10 1-1 -1 NDO 1-0 2 0 -1 0 1 -1 0 1 1 1
We simply describe the above calculation as: Jx=J¹24 J23,5J5.4.26.5.3 71,2,4 Operator works block-wise. Similarly, we have which yields Jμ = 3 1 -2 -1 Ⓡ 1 -2 -1 3 0 -1 -1 -1 0 -1 0 0-2 1 LL 6 1 -2 -1 -1 -N 16T 0 HT 1 0 -2 -1 177 -1 -4 -2 2030 el 0 1 -1 -1 A -1 Matrices Jx and J, are sparse matrices. 103 -1 -1 -2 0 1 3 MONO 0 1 -2 0 0 -2 0210 0 1-2 -1 om 21012- 0 -2 -1 -1 -1 3 -1 6 1 IN O 177 0 01 0 -1 -1 -1 0-2 1 -2 0 6 -1 -1 1 1 3 -2 -1 3 -1 (2.45) (2.46)
contribution of J54,2 to Jx 0 0 0 0 0 1 10 0-1 0-1 0 0 OH 1 10 0 OO 0 OO 0 0 -1 0 OO -1 0 0 0 -1 -1 -1 0 1 TL 1 -1 - 0 1 1