17–19
Solutions Solutions to Exercises Exercises
Homework Exercises for Chapter 17: Isoparametric Quadrilaterals Solutions This exercise can be done using the Mathematica script listed in Figure E17.5. Running that code produces the results shown in Figure E17.6.
EXERCISE EXERCISE 17.1
ClearAll[Em, ν ,a,b,h]; Em=48; h=1; a=4; b=2; ν =0; ncoor={{0,0},{a,0},{a,b},{0,b}}; Emat=Em/(1- ν ^2)*{{1, ν ,0},{ ν ,1,0},{0,0,(1- ν )/2}}; For [p=1, p<=4, p++, Ke= Quad4IsoPMembraneStiffness[ncoor,Ema t,h,{True,p}]; Print["Gauss integration rule: ",p," x ",p]; Print["Ke=",Chop[Ke]//MatrixForm]; Print["Eigenvalues of Ke=",Chop[Eigenvalues[N[Ke]]]] ]; Figure E17.5.
Gauss integration rule: 1 x 1 18. 6. 6.
Ke =
Script to do Exercise 17.1.
− 6.
− 18.
− 6.
− 6.
6.
6. 6. − 6. 18.
27. 6. 21. − 6.
6. 1 8. − 6. − 6.
2 1. − 6. 2 7. 6.
− 6. − 6. 6. 18.
− 27. − 6. − 21. 6.
− 6. − 18. 6. 6.
− 21. 6. − 27. − 6.
− 6. − 6. 6.
− 27. − 6. − 21.
− 6. − 18. 6.
− 21. 6. − 27.
6. 6. − 6.
27. 6. 21.
6. 18. − 6.
21. − 6. 27.
− 6. − 18. − 6.
− 12. − 6. − 12.
6. − 30. 6.
6. 0 6.
− 18. − 6. 12.
24. − 6.
− 6. 36.
− 12. − 6. − 12. 6.
6. − 30. 6. − 18.
0 6.
− 6. 12.
24. − 6.
− 6. 36.
Eigenvalues of Ke= {896., 60., 24., 0, 0, 0, 0, 0} Gauss integration rule: 2 x 2 24. 6. 0 − 6. − 12. 6. 36. 6. 1 2. − 6. 0 6. 24. − 6. − 12.
− 6. 12. − 6. 3 6. 6. − 30. − 6. − 12. 6. 24. 6. − 12. − 6. − 18. − 6. − 30. 6. 36. − 6. − 12. 6. 0 6. − 12. 6. − 30. 6. − 18. − 6. 12. Eigenvalues of Ke= {896., 60., 36., 24., 24., 0, 0, 0} Gauss integration rule: 3 x 3 24. 6. 0 − 6. − 12. − 6. 6. 36. 6. 1 2. − 6. − 18. 0 6. 24. − 6. − 12. − 6. − 6. 12. − 6. 3 6. 6. − 30. Ke = − 12. − 6. − 12. 6. 24. 6. − 6. − 18. − 6. − 30. 6. 36. − 12. − 6. − 12. 6. 0 6. 6. − 30. 6. − 18. − 6. 12. Eigenvalues of Ke= {896., 60., 36., 24., 24., 0, 0, 0} Gauss integration rule: 4 x 4 24. 6. 0 − 6. − 12. − 6. 6. 36. 6. 12. − 6. − 18. 0 6. 24. − 6. − 12. − 6. − 6. 12. − 6. 36. 6. − 30. Ke = − 12. − 6. − 12. 6. 24. 6. − 6. − 1 8. − 6. − 30. 6. 36. − 12. − 6. − 12. 6. 0 6. 6. − 3 0. 6. − 18. − 6. 1 2. Eigenvalues of Ke= {896., 60., 36., 24., 24., 0, 0, 0} Figure E17.6.
− 12.
6.
− 6. − 12. 6.
− 30. 6. − 18.
0 6. 24.
− 6. 12. − 6.
− 6.
36.
Results from running the script of Figure E17.5.
17–19
17–20
Chapter 17: ISOPARAMETRIC QUADRILATERALS
The K e given by the 1 × 1 Gauss rule has a rank deficiency of two because it has five zero eigenvalues instead of three. This behavior is explained in Chapter 19. The 2 × 2, 3 × 3 and 4 × 4 rules produce the same stiffness matrix. This matrix has three zero eigenvalues, which correspond to the three independent rigid body modes in two dimensions. The reason for the repeating stiffness matrices is that the integrand h BT EB J is at most quadratic in ξ and η because h and E are constant, B is linear in ξ and η , and for a rectangle J is constant over the element. A 2-point product Gauss rule is exact for up to cubic polynomials in the ξ and η directions, so it does quadratics exactly. The verification canbe done in severalways: (1) doing it allby hand, (2) using the suggested Mathematica script of Figure E17.3 and visually comparing to (E17.5) entry by entry, or (3) doing a fully automatic verification. The script shown in the top cell of Figure E17.7 takes the latter approach. (Any of the 3 ways gets credit, but the last one is quicker.) EXERCISE 17.2
ClearAll[Em, ν,a,b,γ ,h,ψ 1,ψ 2,ψ 3,ψ 4,ψ 5,ψ 6,ψ 7,ψ 8]; ψ 1=(1+ ν)*γ ; ψ 2=(1-3* ν)*γ ; ψ 3=2+(1- ν)*γ ^2; ψ 4=2*γ ^2+(1- ν); ψ 5=(1- ν)*γ ^2-4; ψ 6=(1- ν)*γ ^2-1; ψ 7=4*γ ^2-(1- ν); ψ 8=γ ^2-(1- ν); b=a/γ ; kfac=Em*h/(24*γ *(1- ν^2)); Kr=kfac*{{4ψ 3, 3ψ 1, 2ψ 5,-3ψ 2,-2ψ 3,-3ψ 1,-4ψ 6, 3ψ 2}, {0, 4ψ 4, 3ψ 2, 4ψ 8,-3ψ 1,-2ψ 4,-3ψ 2,-2ψ 7}, {0,0, 4ψ 3,-3ψ 1,-4ψ 6,-3ψ 2,-2ψ 3, 3ψ 1}, {0,0,0, 4ψ 4, 3ψ 2,-2ψ 7, 3ψ 1,-2ψ 4}, {0,0,0,0, 4ψ 3, 3ψ 1, 2ψ 5,-3ψ 2}, {0,0,0,0,0, 4ψ 4, 3ψ 2, 4ψ 8}, {0,0,0,0,0,0, 4ψ 3,-3ψ 1}, {0,0,0,0,0,0,0, 4ψ 4}}; For[i=2,i<=8,i++,For[j=1,j<=i-1,j++,Kr[[i,j]]=Kr[[j,i]]]]; Emat=Simplify[Em/(1- ν^2)*{{1, ν,0},{ ν,1,0},{0,0,(1- ν)/2}}]; ncoor={{0,0},{a,0},{a,b},{0,b}}; Ke=Quad4IsoPMembraneStiffness[ncoor,Emat,h,{False,2}]; dK=Simplify[Kr-Ke]; Print["This should be the null matrix:", dK//MatrixForm];
This should be the null matrix:
Figure E17.7.
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
Script to do Exercise 17.2 and results.
The code of Figure E17.7 builds the alleged exact stiffness (E17.5) in Kr (the 2 For loops symmetrize that matrix), then invokes Quad4IsoPMembraneStiffness with symbolic inputs placing the re turned stiffness in Ke. The difference Ke-Kr is printed upon simplification. If the two matrices are identical for any symbolic input the difference should be the null matrix of order 8, which can be easily inspected. As shown in the bottom cell of Figure E17.7, the matrices match. EXERCISE 17.3
The verification (a) is immediate. The script shown in Figure E17.8 does (b), computes the energy ratio r for (c), and plots it for Poisson’s ratios ν = 0, 14 , 12 . The plots show that r never exceeds 1. 17–20
17–21
Solutions to Exercises
ClearAll[Em, ν,a,b,h,Iz,M]; b=γ *a; Iz=h*b^3/12; kappa=M/(Em*Iz); ncoor={{-a/2,-b/2},{a/2,-b/2},{a/2,b/2},{-a/2,b/2}}; Emat=Em/(1- ν^2)*{{1, ν,0},{ ν,1,0},{0,0,(1- ν)/2}}; Ke= Quad4IsoPMembraneStiffness[ncoor,{Emat,0,0},{h},{False,2}]; Ke=Simplify[Ke]; u=kappa*a*b*{-1,0,1,0,-1,0,1,0}/4; Uquad=Simplify[u.Ke.u/2]; Ubeam=Simplify[M*kappa*a/2]; Print["Uquad=",Uquad, ", Ubeam=",Ubeam]; r=Simplify[Ubeam/Uquad]; Print["r=Ubeam/Uquad=",r]; Plot[{r/. ν->0,r/. ν->1/4,r/. ν->1/2},{γ ,0.1,10}]; 3 M 2 1 2 Γ 2 Ν Uquad , a2 Em h Γ 5 1 Ν2
6 M2 Ubeam a2 Em h Γ 3
2 Γ 2 1 Ν2 rUbeamUquad 1 2 Γ 2 Ν
1 0.8 0.6 0.4 0.2
2
4
6
Figure E17.8.
EXERCISE 17.4
Not assigned.
EXERCISE 17.5
Not assigned.
EXERCISE 17.6
Not assigned.
EXERCISE 17.7
Not assigned.
8
10
Script to do Exercise 17.3 and results.
17–21
