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The Finite Element Method:Its Basis and Fundamentals
Sixth Edition
Problem Solutions
O.C. Zienkiewicz, CBE, FRSUnesco Professor of Numerical Methods in Engineering
International Centre for Numerical Methods in Engineering, Barcelona
Previously Director of the Institute of Numerical Methods in Engineering
University of Wales, Swansea
R.L. Taylor ∗
Professor in the Graduate School
Department of Civil and Environmental Engineering
University of California at Berkeley
Berkeley, California
J.Z. Zhu †
Senior Scientist
ESI US R & D Inc.
5850 Waterloo Road, Suite 140
Columbia, Maryland
March 29, 2005
∗e-mail: [email protected]†e-mail: [email protected]
ii
1
Introduction
This document presents solutions to many of the problems contained in The Finite Ele-ment Method: Its Basis and Fundamentals. The problems included in the text range fromthose whose solution are quite simple to some whose solutions are quite involved. The orderin each chapter is random to force the reader to distinguish between the possible levels ofcomplexity. The problems presented are not an exhaustive list of those which may be givento readers but should provide sufficient scope that an instructor can devise many additionalexamples on their own.
Many of the problems included in the text are posed such that the reader needs to supplytheir own interpretation as well as a choice for particular data to be used. This is intentionalas learning a new subject from a text should require the reader to inject some effort. Thus,some of the problem solutions included in this manual provide a choice for data that is notunique. This is especially true for problems involving construction of shape functions andtheir derivatives.
At this time there are no solutions to problems requiring use or writing of a computer pro-gram to complete. While suggestions to use MATLAB©R are included in the text, an instructormay wish to use another programming language to complete the assigned programming as-signments. It is the belief of the authors, however, that mastery of the finite element methodcan only be achieved by programming the methods presented in the text. The FEM is asubject that requires a computer to solve realistic problems!
It is expected that some interpretations provided in this manual can be improved upon.For such situations the authors solicit comments and suggestions. Please send such commentsby e-mail to: [email protected].
OCZ, RLT, JZZ
2
Errata:
Problem Description of change1.1 Change solution at node 4 from 30.81 to 39.01. Alternatively,
change the head at node 6 to 20.0.3.1 Part (b) add subscripts to g. Let u(0) = g0 and
(a du/dx+ ku)|x=1 = g1.3.10 Instructor to assign formulation used for the solution. Can be any
of those developed in Problems 3.2 to 3.9.4.12 The edge node is located at (0.5, 1, 1) not (0.75, 1, 1).5.3 The interior node for transition element B is not needed.
5.15 Strength of singularity shoulde be 1/r1/2 not 1/r2.6.10 Figure 6.24(a) should not have a node at the center.
3
Chapter 1: Solutions
1.1 For the network shown in the figure one must choose an order for the nodes on elements1 and 4. For the solution given here, the node orders are (1, 2, 3) for element 1 and(6, 5, 4) for element 4. The problem is given as a standard discrete form by
K V = J .
(a) Using the Using the two types of element arrays for the network shown, theassembled matrix is given by
K =
3 −2 −1 0 0 0−2 5 −2 0 −1 0−1 −2 4 −1 0 0
0 0 −1 7 −4 −20 −1 0 −4 9 −40 0 0 −2 −4 6
There are no element loads, but removal at nodes 2 and 4 gives
JT =[0 −30 0 −10 0 0
].
(b) Modifying rows 1 and 6 of K and J for the known boundary heads gives
K =
1 0 0 0 0 0−2 5 −2 0 −1 0−1 −2 4 −1 0 0
0 0 −1 7 −4 −20 −1 0 −4 9 −40 0 0 0 0 1
and JT =
[100 −30 0 −10 0 30
].
(c) Solution gives VT =[100 69.47 69.49 39.01 38.89 30.00
]. If head at node 6
is lowered to 20, solution is given by VT =[100 66.34 65.87 30.81 29.95 20.00
](corresponding to answer in book).
1
2
3 4
5 6
(1)
(2)
(3)
(4)
Fluid network for Problem 1.1.
4
1.2 The solution of a plane truss as a standard discrete problem uses arrays from theproblem statement together with Eqs (1.23) and (1.25) to compute the individualdiscrete elements. Accordingly,
(a) Arrays are given by
q =
cos θ 0sin θ 0
0 cos θ0 sin θ
u′1u′2
and
K =EA
L
cos θ 0sin θ 0
0 cos θ0 sin θ
[1 −1−1 1
] [cos θ sin θ 0 0
0 0 cos θ sin θ
]
=EA
L
cos2 θ sin θ cos θ − cos2 θ − sin θ cos θ
sin θ cos θ sin2 θ − sin θ cos θ − sin2 θ− cos2 θ − sin θ cos θ cos2 θ sinθ cos θ− sin θ cos θ − sin2 θ sin θ cos θ sin2 θ
(b) If nodes are reversed then first two rows and columns interchange with last two
rows and columns. Due to double symmetry there is no change in the finalstiffness. When θ = 30o, the final stiffness is given by:
K =EA
4L
3
√3 −3 −
√3√
3 1 −√
3 −1
−3 −√
3 3√
3
−√
3 −1√
3 1
.
x (u,U)
y (v,V) x′
y′
u2′
u1′
θ
1
2
u
vu′
θ
(a) Truss member description (b) DisplacementsTruss member for Problem 1.2.
5
1.3 The fill pattern for the global problem is to be determined for a plane truss with nodesnumbered as shown in parts (a) and (b) of the figure.
(a) Let • denote terms in K which are non-zero. After inserting each of the elementsfor Fig. 1.6(a), we obtain:
K =
• • 0 0 0 0 • 0 0 0• • • 0 0 0 • 0 0 00 • • • 0 0 • • • 00 0 • • • 0 0 • • •0 0 0 • • • 0 0 0 •0 0 0 0 • • 0 0 0 •• • 0 • 0 0 • • 0 00 0 • • 0 0 • • • 00 0 • • 0 0 0 • • •0 0 0 • • • 0 0 • •
Maximum bandwidth is 7 (including diagonal).
1 2 3 4 5 6
7 8 9 10
(a)
1 2 4 6 8 10
3 5 7 9
(b)Truss for Problems 1.3 and 1.4.
6
(b) After inserting each of the elements for Fig. 1.6(b), we obtain:
K =
• • • 0 0 0 0 0 0 0• • • • 0 0 0 0 0 0• • • • • 0 0 0 0 00 • • • • • • 0 0 00 0 • • • • • 0 0 00 0 0 • • • • • • 00 0 0 • • • • 0 • 00 0 0 0 0 • 0 • • •0 0 0 0 0 • • • • •0 0 0 0 0 0 0 • • •
Maximum bandwidth is 4 (including diagonal). Thus the ordering in figure (b)produces the smallest bandwidth.
1.4 A sample MATLAB©R program for truss analysis is given by a main program:
% Matlab program for truss.m analysis
% This program requires following data arrays% coord - x,y coordinate for each node% element - node connections and material data for each element% forces - nodal forces% displ - nodal displacement
clear all % Clear workspace.format compact
% Input coordinate array
% Number nodes per element
nel = 2;
% Input data
[x,y,ix,matl,force,displ] = input2d(nel);
% Global solution
u = gstiff(x,y,ix,matl,force,displ,nel);
% Output nodal displacements
u_node = sprintf(’ %0.5e %0.5e\n’,u)
% Plot undeformed and deformed mesh
plot_mesh(x,y,u,ix,1e5)
7
An input routine (input2d.m)
function[x,y,ix,matl,force,displ] = input2d(nel)
% Input of data for 2-d problems
% Input coordinate array
load coordx = coord(:,1); % Store x-coordsy = coord(:,2); % Store y-coords
% Specify number of nodes on each element: 3 = triangle
load elementnum_dat = length(element(1,:));ix = element(:,1:nel); % Connection listmatl = element(:,nel+1:num_dat); % Material data
% Input force and displacement arrays
load force % Columns: 1= node, 2=dof, 3=value
load displ % Columns: 1= node, 2=dof, 3=value
A global stiffness routine (gstiff.m):
function[u] = gstiff(x,y,ix,matl,force,displ,nel)
% Set problem size
num_np = length(x)num_el = length(ix(:,1))num_fc = length(force(:,1))num_di = length(displ(:,1))
% Allocate sparse matrix to hold all stiffness coefficients
nzmax = num_el*(nel*2)^2;neq = num_np*2;K = spalloc(neq,neq,nzmax);b = zeros(neq,1);
% Loop through elements, compute stiffness and assemble
for n = 1:num_el[k_el,ld] = el_stif(ix(n,:),x,y,matl(n,:));K(ld,ld) = K(ld,ld) + k_el;
end
% Modify for boundary conditions
for n = 1:num_di
8
ne = 2*(displ(n,1)-1) + displ(n,2);b = b - K(1:neq,ne)*displ(n,3);b(ne) = displ(n,3);K(1:neq,ne) = 0;K(ne,1:neq) = 0;K(ne,ne) = 1;
end
% Add nodal forces
for n = 1:num_fcne = 2*(force(n,1)-1) + force(n,2);b(ne) = b(ne) + force(n,3);
end
% Solve equations
u = K\b;
An element stiffness routine (k el.m):
function[k_el,ld] = el_stif(ixl,x,y,matl)
% Function to compute element stiffness and assembly vector for triangle
for i = 1:2ld(2*i-1) = 2*ixl(i) - 1; % Assembly vectorld(2*i ) = 2*ixl(i);
end
% Direction cosine matrix of truss
cn = x(ixl(2)) - x(ixl(1));sn = y(ixl(2)) - y(ixl(1));L = sqrt(cn*cn + sn*sn); % Truss length
bmat(1,1) = cn/L;bmat(1,2) = sn/L;bmat(2,3) = cn/L;bmat(2,4) = sn/L;
% Material moduli times thickness
dmat(1,1) = matl(1)*matl(2)/L; % matl(1) = e; matl(2) = admat(1,2) = -dmat(1,1);dmat(2,1) = dmat(1,2);dmat(2,2) = dmat(1,1);
% Stiffness
k_el = bmat’*dmat*bmat;
A plot routine (plot mesh.m)
9
function void = plot_mesh(x,y,u,ix,f)
% Set number of elements to plot
num_el = length(ix(:,1));neq = length(u);
% Get deformed coordinates
xd = x + f*u(1:2:neq-1);yd = y + f*u(2:2:neq );
% Plot undeformed (dashed) and deformed (solid) shapes
figure(1)clfhold onaxis offaxis equalfor n = 1:num_elplot([ x(ix(n,:))’ x(ix(n,1))],[ y(ix(n,:))’ y(ix(n,1))], ...
’k--’,’linewidth’,2)plot([xd(ix(n,:))’ xd(ix(n,1))],[yd(ix(n,:))’ yd(ix(n,1))], ...
’k-’ ,’linewidth’,2)end
The input data for the truss problem shown in part (b) of the figure for the specifiedvalues is given by
coord file element file0.0 0.0 1 2 200e9 0.0010.5 0.0 1 3 200e9 0.0010.5 0.8 2 3 200e9 0.0011.0 0.0 2 4 200e9 0.0011.0 0.8 3 4 200e9 0.0011.5 0.0 3 5 200e9 0.0011.5 0.8 4 5 200e9 0.0012.0 0.0 4 6 200e9 0.0012.0 0.8 4 7 200e9 0.0012.5 0.0 5 6 200e9 0.001
5 7 200e9 0.001force file 6 7 200e9 0.0016 2 -100.0 6 8 200e9 0.001
6 9 200e9 0.001displ file 7 9 200e9 0.0011 1 0.0 8 9 200e9 0.0011 2 0.0 8 10 200e9 0.001
10 2 0.0 9 10 200e9 0.001
The solution from FEAPpv for the same data is given by
Node 1 Coord 2 Coord 1 Displ 2 Displ1 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00
10
2 5.0000E-01 0.0000E+00 6.2500E-08 -5.4996E-073 5.0000E-01 8.0000E-01 4.6013E-07 -5.4996E-074 1.0000E+00 0.0000E+00 1.2500E-07 -1.0218E-065 1.0000E+00 8.0000E-01 3.3513E-07 -1.1018E-066 1.5000E+00 0.0000E+00 2.8125E-07 -1.2667E-067 1.5000E+00 8.0000E-01 1.7888E-07 -1.1867E-068 2.0000E+00 0.0000E+00 3.7500E-07 -6.9193E-079 2.0000E+00 8.0000E-01 -8.6188E-09 -6.9193E-07
10 2.5000E+00 0.0000E+00 4.6875E-07 0.0000E+00
Executing the MATLAB©R program produces the same answers as well as the plot of thetruss and deformed shape.
1.5 A standard discrete problem for a variable cross-section axially loaded elastic bar hasindividual elements given by:
Ke =EA
h
[1 −1−1 1
].
Thus, the individual element matrices are given by:
K1 =200 · 25
37.5
[1 −1−1 1
]=
[133.33 −133.33−133.33 133.33
]K2 =
[160 −160−160 160
]and K3 =
[192 −192−192 192
].
Assembling, inserting the loads and boundary restraint gives the matrix problem.1.00 0.00 0.00 0.00
−133.33 293.33 −160.00 0.000.00 −160.00 352.00 −192.000.00 0.00 192.00 192.00
u1
u2
u3
u4
=
0.0
10.0−3.5
6.0
u1
u2
u3
u4e=1 e=2 e=3
L1
L2
L3
E1,A
1 E2,A
2 E3,A
3
(a) Bar geometry
u1=0 P
2P
3P
4
(b) Problem 1.5
u1=0 P
2P
3u
4=0
(c) Problem 1.6Elastic bar descriptions for Problems 1.5 and 1.6.
11
which has the solution
uT =[0.000 0.0938 0.1094 0.1406
]cm.
1.6 The individual element matrices for loading shown in part (c) of the figure are givenby:
K1 =200 · 30
37.5
[1 −1−1 1
]=
[160 −160−160 160
]K2 =
[133.33 −133.33−133.33 133.33
]and K3 =
[80 −80−80 80
].
Assembling, inserting the loads and boundary restraint gives the matrix problem.1.00 0.00 0.00 0.00
−160.00 293.33 −133.33 0.000.00 −133.33 213.33 −80.000.00 0.00 0.00 1.00
u1
u2
u3
u4
=
0.0
−10.03.50.0
which has the solution
uT =[0.000 0.0372 0.0068 0.0000
]cm.
1.7 A tapered bar is loaded by an end load P and a uniform loading b as shown in figurepart (a). The area varies as A(x) = Ax/L when the origin of coordinates is located asshown in the figure.
The problem is converted into a standard discrete system by dividing it into equallength segments of constant area as shown in part (b) of the figure. The array for eachsegment is determined from
qe = Keue + f e
where Ke and ue are defined in Problem 1.5 and
f e = 12b h
11
.
x
y
L L
A 2AP b
u1
u2
u3
u4
u5=0
h=L/4
e=1 e=2 e=3 e=4
(a) Tapered bar geometry (b) Approximation by 4 segmentsTapered bar description for Problem 1.7.
12
For the properties L = 100 cm, A = 2 cm2, E = 104 kN/cm2 P = 2 kN, b =−0.25 kN/cm and u(2L) = 0, the displacement from the solution of the differentialequation is u(L) = −0.03142513 cm.
A MATLAB©R program to solve the problem is given by
% Solution of tapered bar. Problem 1.7
clear allformat short e
% Data
A = 2; L = 100; E = 10e3; P = 2; b = -0.25;
% Exact solution
c1 = b*L*L/(E*A); c2 = P*L/(E*A);x = linspace(L,2*L,101);u = c1*(2- x/L + log(x/2/L)) - c2*log(x/2/L);
% Approximate solution (jj = number of refinements)
jj = 7; ee = zeros(3,jj);
% Start with 1 element (n)
n = 1;for j = 1:jj
kg = zeros(n+1);bg = zeros(n+1,1);h = L/n;for i = 1:n
% Element stiffness and load
c3 = E*A*(1+i*h/L - h/2/L)/h;ke = [ c3 -c3
-c3 c3];be = b*h/2*[ 1 1]’;
% Assemble to global arrays
kg(i:i+1,i:i+1) = kg(i:i+1,i:i+1) + ke;bg(i:i+1,1) = bg(i:i+1,1) + be;
end
% Modify for fixed end condition (u = 0)
kg(n+1 ,1:n+1) = 0; kg(1:n+1,n+1 ) = 0;kg(n+1 ,n+1 ) = 1; bg(n+1 ,1) = 0;
13
% Add applied end force and solve equations
bg(1) = bg(1) + P;ug = kg\bg;
% Save solution and error (displayed later)
ee(1,j) = n;ee(2,j) = ug(1);ee(3,j) = abs(ug(1)-u(1));
% Double number of elements, repeat solution jj times
n = n*2;endee’ % Display solutionu_exact = u(1)
figure(1)clfplot(x,u,’k-’,’linewidth’,2)hold onhh = gca;set(hh,’fontsize’,14,’fontweight’,’bold’,’linewidth’,2)axis([L 2*L -0.035 0])xlabel(’x/L’)ylabel(’u - displacement’)grid onbox onxg = linspace(L,2*L,n/2+1); ! Plot of last solutionplot(xg,ug,’ko’) ! for jj refinement
The results from running the program are given in the table.
Elements u1 |u(L)− u1
1 -0.035000 3.5749e-032 -0.032529 1.0034e-034 -0.031685 2.6018e-048 -0.031491 6.5695e-05
16 -0.031442 1.6465e-0532 -0.031429 4.1190e-0664 -0.031426 1.0299e-06
From the table the error is less than 10−5 cm when 32 elements are used.
14
Chapter 2: Solutions
1 2
3
a
b
1 2
34
a
b
(a) Triangle (b) RectangleElement description for Problems 2.1 to 2.8.
2.1 For the triangle shown in part (a) of the figure with the coordinate origin placed atnode 1 using Eq. 2.4 we can write
u1
u2
u3
=
1 0 01 3 01 0 4
α1
α2
α3
.
Inverting the matrix givesα1
α2
α3
=1
12
12 0 0−4 4 0−3 3 0
u1
u2
u3
.
This gives the interpolation
u = 112
[(12− 4x− 3y)u1 + 4x u2 + 3y u3]
and thus the shape functions are given by
N1 = 112
(12− 4x− 3y) ; N2 = 112
4x and N3 = 112
3y .
2.2 For the rectangular element shown in part (b) of the figure with the origin at node 1,using Eq. 2.9 we can write (letting x′ = x and y′ = y)
u1
u2
u3
u3
=
1 0 0 01 6 0 01 6 4 241 0 4 0
α1
α2
α3
α4
.
15
Inverting the matrix givesα1
α2
α3
α4
=1
24
24 0 0 0−4 4 0 0−6 0 0 6
1 −1 1 −1
u1
u2
u3
u4
.
This gives the interpolation
u = 124
[(24− 4x− 6y + xy) u1 + (4x− xy) u2 + xy u3 + (6y − xy) u4]
and thus the shape functions are given by
N1 = 124
(24− 4x− 6y + xy) = 124
(6− x)(4− y)N2 = 1
24(4x− xy) = 1
24x(4− y)
N3 = 124xy
N4 = 124
(6y − xy) = 124
(6− x)y .
2.3 The strain-displacement matrix is given by Eq. 2.15 and after inserting the derivativesof the interpolations computed in Problem 2.1 for the triangular element shown in part(a) of the figure we obtain
B =1
12
−4 0 4 0 0 00 −3 0 0 0 3−3 4 0 4 3 0
.
2.4 The strain-displacement matrix is given by Eq. 2.15 and after inserting the derivativesof the interpolations computed in Problem 2.2 for the rectangular element shown inpart (b) of the figure we obtain
B =1
24
−(4− y) 0 (4− y) 0 y 0 −y 00 −(6− x) 0 −x 0 x 0 (6− x)
−(6− x) −(4− y) −x (4− y) x y (6− x) −y
.
2.5 The body force vector for an element is computed from Eq. 2.28(b) as
f e = −∫
Ωe
Na(x, y)b dΩ .
For the triangle shown in part (a) of the figure, it is necessary to evaluate the integralsfor the polynomials [1, x, y]. Accordingly, for the dimensions specified we can computethe integrals as∫ 4
0
(∫ 3−0.75y
0
[1, x, y] dx
)dy =
∫ 4
0
[(3−0.75y), 12(3−0.75y)2, (3−0.75y)y] dy = [6, 6, 8] .
16
Using this result for each of the shape functions defined in Problem 2.1 gives∫Ωe
Na dx dy = 2 for a = 1, 2, 3 .
For the body force bT = [5, 0] we obtain
f e =[−10 0 −10 0 −10 0
]T
and for the body force bT = [0, −30]
f e =[0 30 0 30 0 30
]T.
2.6 The body force vector is computed from Eq. 2.28(b) as
f e = −∫
Ωe
Na(x, y)b dΩ .
For the rectanglular element shown in part (b) of the figure, it is necessary to evaluatethe integrals for the polynomials [1, x, y, xy]. Accordingly, for the dimensions specifiedwe compute the integrals as∫ 4
0
(∫ 6
0
[1, x, y, xy] dx
)dy =
∫ 4
0
[6, 18, 6y, 18y] dy = [24, 72, 48, 144] .
Using this result for each of the shape functions defined in Problem 2.1 gives∫Ωe
Na dx dy = 6 for a = 1, 2, 3, 4 .
For the body force bT = [5, 0] we obtain
f e =[−30 0 −30 0 −30 0 −30 0
]T.
2.7 Using integrals from Problem 2.6 for the body force vector bx = 0 and by = −30 gives
f e =[0 120 0 120 0 120 0 120 0 120
]T.
2.8 Let un be the normal displacement and us the tangential displacement to the edge. Ifθ is the angle the normal makes with the x-axis, the transformation of displacementsis given by
uv
=
[cos θ − sin θsin θ cos θ
] un
us
= LT u′ .
Thus for the triangle shown we have cos θ = 0.8 and sin θ = 0.6 which gives
LT =
[0.8 −0.60.6 0.8
].
17
1 2
a
F
h
1 2
q
a
h
(a) Point loading (b) Hydrostatic loadingTraction loading on boundary for Problems 2.9 and 2.10.
2.9 The solution for a concentrated load, F , applied to the edge of a two-dimensional planestrain problem is constructed as follows:
(a) Let F1 and F2 be the forces at nodes 1 and 2, respectively. We may write momentequilibrium equivalence about node 1 as
F · a ≡ F2 · h giving F2 =a
hF .
Similarly by equivalence about node 2 we obtain
F · (h− a) ≡ F1 · h giving F1 =h− ah
F .
Note that F1 + F2 = F gives final equivalence of vertical forces.
(b) The vertical loading is given by q(x) = F δ(x− a) where δ(y) is a delta functionacting at point y. Using virtual work we have∫ x2
x1
δv q(x) dx =[δv1 δv2
] ∫ h
0
[N1(x)N2(x)
]q(x) dx =
[δv1 δv2
] [N1(a)N2(a)
]F
where N1(a) and N2(a) are the values of the shape functions at x = a. It is easyto show that the shape functions are given by
N1(x) =h− xh
and N2(x) =x
h
and thus the equivalent forces by
F1 =h− ah
F and F2 =a
hF .
2.10 The solution for a triangular traction load applied to the edge of a two-dimensionalplane strain problem (shown in part (b) of the figure) is computed as follows:
18
(a) Let F1 and F2 be the forces at nodes 1 and 2, respectively. The resultant of thetriangular loading is F = q · a/2 and acts at x = a/3. We may write momentequilibrium equivalence about node 1 as
F · a3≡ F2 · h giving F2 =
a
3hF =
1
6hq · a2 .
Similarly by equivalence about node 2 we obtain
F · (h− a/3) ≡ F1 · h giving F1 =3h− a
3hF =
3h− a6h
q · a .
Note that F1 + F2 = q · a/2 gives final equivalence of vertical forces.
(b) The vertical loading is given by q(x) = q · (1 − x/a) for 0 ≤ x ≤ a. Thus, thevirtual work expression for loads may be written as∫ x2
x1
δv q(x) dx =[δv1 δv2
] ∫ 1−x/a
0
[N1(x)N2(x)
]q (1− x/a) dx .
The expressions for shape functions are given in Problem 2.9 and when used inthe above integral gives the result∫ x2
x1
δv q(x) dx =1
6h
[δv1 δv2
] q (3h− a) a
q a2
which is identical to the equilibrium equivalence given above.
1 2 3
54
6 cm. 3 cm.
4 cm.
Element assembly for Problem 2.11.
2.11 The stiffness coefficients for nodes a and b are computed from
Kab =
∫Ωe
BTa DBb dΩ
and thus we need to compute B2 and B5 for the two elements
19
The strain displacement matrix for node a is given by
Ba =
∂Na
∂x0
0∂Na
∂y∂Na
∂y
∂Na
∂x
Let the origin of local coordinates for the rectangle be at node 1 and for the triangleat node 2, then, using the shape function expressions from Problem 2.1 we have (withsuitable permutation of the node numbering)
N2(x, y) = 124x (4− y)
N5(x, y) = 124x y
, for the rectangle
N2(x, y) = 112
(12− 4x− 3y)
N5(x, y) = 112
3y, for the triangle .
Thus, the strain-displacement arrays are
B2 =1
24
(4− y) 00 xx (4− y)
and B5 =1
24
y 00 xx y
for the rectangle, and
B2 =1
12
−4 00 −3−3 −4
and B5 =1
12
0 00 33 0
for the triangle. The elastic modulus matrix is given by
D =E
(1− ν2)
1 ν 0ν 1 00 0 (1− ν)/2
.
Thus for the rectangular element with specified properties we have
D =1
15
16000 4000 04000 16000 0
0 0 6000
and for the triangular element
D =
1200 0 00 1200 00 0 600
.
20
Factoring the common term from each Ba and D arrays and noting the thickness is0.2, the stiffness coefficients for the rectangle are:
K22 =10
432
∫ 4
0
∫ 6
0
[(4− y) 0 −x
0 −x (4− y)
]16 4 04 16 00 0 6
(4− y) 00 −x−x (4− y)
dx
dy
=
[187.41 −33.33−33.33 124.44
]
K55 =10
432
∫ 4
0
∫ 6
0
[y 0 x0 x y
]16 4 04 16 00 0 6
y 00 xx y
dx
dy
=
[187.41 33.3333.33 124.44
]
K25 =10
432
∫ 4
0
∫ 6
0
[(4− y) 0 −x
0 −x (4− y)
]16 4 04 16 00 0 6
y 00 xx y
dx
dy
=
[−16.30 −6.67
6.67 −97.78
]Similarly, factoring common terms from the Ba and D arrays and noting the area andthickness are 6 and 0.2, the stiffness coefficients for the triangle are:
K22 =60
144
[4 0 30 3 4
]12 0 00 12 00 0 6
4 00 33 4
=
[205 6060 170
]
K55 =60
144
[0 0 30 3 0
]12 0 00 12 00 0 6
0 00 33 0
=
[45 00 90
]
K25 = − 60
144
[4 0 30 3 4
]12 0 00 12 00 0 6
0 00 33 0
= −[45 060 90
].
The final assembly adds the two contributions to K22, K55 and K25 together.
2.12 A one dimensional elasticity problem formulation may be developed by simplifying theequations to be expressed in terms of u(x), ε(x) and σ(x).
(a) For a two-node finite element with the origin of the x-axis placed at node 1, wehave
ue1
ue2
=
[1 01 h
] α1
α2
where h = xe
2 − xe1. Solving for the αa we obtain
α1
α2
=
1
h
[h 0−1 1
] ue
1
ue2
21
t1 2 3
1 2
a a
One dimensional elasticity description for Problem 2.12.
which gives the interpolation
ue =(1− x
h
)ue
1 +x
hue
2 = N1(x) ue1 +N2(x) u
e2 .
(b) The strain matrix is determined by differentiating ue in an element and gives
εe =due
dx=
1
h
[−1 1
]ue
1
ue2
.
(c) Using (2.25) we obtain (2.28a) for the stiffness matrix as
Ke =
∫ h
0
BTDB dx =1
h2
∫ h
0
[−1
1
]E
[−1 1
]dx =
E
h
[1 −1−1 1
].
From (2.28b) the force vector is given by (noting ε0 and σ0 are zero)
f e = −∫ h
0
1− x
hx
h
b dx = − 12
bhbh
.
(d) For h = a = 5, b = 2 and E = 200 the stiffness and load vector for each elementare given by
Ke =
[40 −40−40 40
]and f e = −
55
.
Assembling for two elements and imposing the boundary conditions gives thematrix problem 1 0 0
−40 80 −400 −40 40
ue
1
ue2
ue3
=
010
5 + 4
Inverting the matrix gives
ue1
ue2
ue3
=1
1600
1600 0 01600 40 401600 40 80
010
5 + 4
=
0.0000.4750.700
.
22
(e) Substituting the stress-strain relation into the equilibrium equation gives the or-dinary differential equation
Ed2u
dx2+ b = 0
which has the general solution
u = β1 + β2 x−b x2
2E.
Differentiating gives the expression for stress
σ = E β2 − b x .
Satisfying the boundary conditions gives β1 = 0 and β2 = 0.12, thus, the solutionsare
u = 0.12x− 0.005x2 and σ = 24− 2x .
(f) The value of the exact solution at the two nodes gives u(5) = 0.475 and u(10) =0.700 – thus nodal solutions are exact! However, as seen in the plot there is anerror in displacement for each element that is maximum at the element center.For the first element the error is
Eu = |uexact − ufe| = |0.26875− 0.2375| = 0.03125
Similarly for the second element the error is
Eu = |0.62875− 0.5 ∗ (0.7 + 0.475)| = 0.03125 .
The stresses from the finite element solution are given by
σ1fe = 200 ∗ (0.475)/5 = 19.0 and σ2
fe = 200 ∗ (0.7− 0.475)/5 = 9.0 .
The stress error is maximum at the ends of elements and for each element is thesame and given by (see figure)
Eσ = |σexact − σfe| = 5.0 .
(g) For four elements, h = a/2 = 2.5 with remaining data the same. Thus the elementarrays become
Ke =
[80 −80−80 80
]and f e = −
2.52.5
.
Assembling and imposing boundary conditions gives the matrix problem1 0 0 0 0
−80 160 −80 0 00 −80 160 −80 00 0 −80 160 −800 0 0 −80 80
ue
1
ue2
ue3
ue4
ue5
=
0555
2.5 + 4
23
0 2 4 6 8 100
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
x − coordinate
u −
disp
lace
men
t
0 2 4 6 8 104
6
8
10
12
14
16
18
20
22
24
x − coordinate
σ −
stre
ss
Solution for displacement and stress for Problem 2.12(e).
Inverting and evaluating the solution givesue
1
ue2
ue3
ue4
ue5
=
0.000000.268750.475000.618750.70000
.
Again, all nodal values are exact. The error in each element will be the same,thus, for the first element we have
Eu = |0.14218875− 0.5 ∗ (0.26875)| = 0.0078125 .
Similarly, for the stress error we obtain Eσ = 2.5, and thus observe the error isdecreasing linearly, whereas the displacement error is decreasing quadratically.
2.13 Download the program FEAPpv and user manual from a web site given in Chapter 19.Note that both source code for the program and an executable version for Windowsbased systems are available at the site.
If source code is used it is necessary to compile the program to obtain an executableversion.
2.14 Use FEAPpv (or any available program) to solve the rectangular beam problem givenin Example 2.3 – verify results shown in Table 2.1. A simple mesh of triangles forFEAPpv is given by
feap * * Example 2.3 - Rectangular beam0 0 0 2 2 3
material 1solid
elastic isotropic 1000 0.25
24
plane stress
parametersn1 = 4 ! Number elements in heightn2 = 10 ! Number elements in lengthc = 10 ! Half heightl = 100 ! Lengthp = 80 ! Loadt = 3*p/(4*c) ! Shear traction at centerty = 6 ! Crossed triangles
blockcart n1 n2 0 0 0 0 ty
1 0 c2 0 -c3 l -c4 l c
ebou1 l 1 0
cbounode l 0 1 1
csurfacetangentialquadratic
1 0 c 02 0 -c 03 0 0 -t
endinterstop
Consult user manual for command description.
2.15 Use FEAPpv (or any available program) to solve the curved beam problem given inExample 2.4 – verify results shown in Table 2.2. A simple mesh of triangles for FEAPpvis given by
feap * * Example 2.4 - Curved beam0 0 0 2 2 3
material 1solid
elastic isotropic 10000 0.25plane stress ! Default thickness is 1.0
parametersn1 = 4n2 = 6
25
a = 5b = 10u0 = -0.01ty = 1
blockpolar n1 n2 0 0 0 0 ty
1 a 02 b 03 b 904 a 90
ebou1 0 1 02 0 1 0
cbounode 0 a 1 1
edis2 0 2 u0 0
endinterstop
Consult user manual for command description.
x
y
q0
h
L
Uniformly loaded cantilever beam description for Problem 2.16.
2.16 The uniformly loaded cantilever beam shown in the figure has properties
L = 2 m ; h = 0.4 m ; t = 0.05 m and q0 = 100 N/m.
Use FEAPpv (or any available program) to perform a plane stress analysis of theproblem assuming linear isotropic elastic behavior with E = 200 GPa and ν = 0.3.
In your analysis:
26
(a) Use 3-node triangular elements with an initial mesh of 2-elements in the depthand 10-elements in the length directions.
(b) Compute consistent nodal forces for the uniform loading.
(c) Compute nodal forces for a parabolically distributed shear traction at the re-strained end which balances the uniform loading q0.
(d) Report results for the centerline displacement in the vertical direction and thestored energy in the beam.
(e) Repeat the analysis 3 additional times using meshes of 4 × 20, 8 × 40 and 16 ×80 elements. Tabulate the tip vertical displacement and stored energy for eachsolution.
(f) If the energy error is given by
∆E = En − En−1 = C hp
estimate C and p for your solution.
(g) Repeat the above analysis using rectangular 4-node elements.
No solution provided in this version of the solution manual.
1 2
34
σx = 2
x
y
6
5
Figure 1: Problem 2.17: Patch test for triangles.
2.17 The program to solve a plane stress problem using 3-node triangular elements canuse the same routines developed in Problem 1.4 for the truss to perform many of thesolution steps. In particular the same ’input2d’ and ’plot mesh’ modules are useddirectly. Thus, the only additions needed are a main module, the module to computethe triangular element array, and the stress output module. It is now, however, crucialto check that the sequence of the nodes on each element produces positive area. Thisis accomplished below by computing the area and if negative swapping the order onthe connection array (ix).
% Matlab program for plane stress analysis
% This program requires following data arrays% coord - x,y coordinate for each node
27
% element - node connections and material number for each element% forces - nodal forces% displ - nodal displacement
clear all % Clear workspace.format compact
% Number nodes per element
nel = 3;
% Input data
[x,y,ix,matl,force,displ] = input2d(nel);
% Check triangle area is positive
n_el = length(ix(:,1));for n = 1:n_el
area = x(ix(n,1))*(y(ix(n,2)) - y(ix(n,3))) ...+ x(ix(n,2))*(y(ix(n,3)) - y(ix(n,1))) ...+ x(ix(n,3))*(y(ix(n,1)) - y(ix(n,2)));
if area < 0 % If negative swap 2 and 3 ixii = ix(n,2);ix(n,2) = ix(n,3);ix(n,3) = ii;
endend
% Global solution
u = gstiff(x,y,ix,matl,force,displ,nel);
% Output nodal displacements
u_node = sprintf(’ %0.5e %0.5e\n’,u)
% Plot undeformed and deformed mesh
plot_mesh(x,y,u,ix,10000)
The element module is given by:
function[k_el,ld] = el_stif(ixl,x,y,matl)
% Element stiffness vector for 3-node triangle
for i = 1:3ld(2*i-1) = 2*ixl(i) - 1; % Assembly vectorld(2*i ) = 2*ixl(i);
endxl = x(ixl(1:3)); % Element x coordinateyl = y(ixl(1:3)); % Element y coordinate
28
% Material moduli times thickness
e = matl(1);nu = matl(2);t = matl(3);dmat(1,1) = e*t/(1-nu*nu);dmat(1,2) = nu*dmat(1,1);dmat(2,1) = dmat(1,2);dmat(2,2) = dmat(1,1);dmat(3,3) = e*t/2/(1+nu);
% Shape function derivatives for triangle
jj = [2 3 1]; % Permutation arrayskk = [3 1 2]; % Permutation arrays
ai = xl(jj).*yl(kk) - xl(kk).*yl(jj);area = 0.5*sum(ai); % Triangle areabi = (yl(jj) - yl(kk))/(area*2);ci = (xl(kk) - xl(jj))/(area*2);
% Strain-displacement matrix
bmat(1,1) = bi(1);bmat(2,2) = ci(1);bmat(3,1) = ci(1);bmat(3,2) = bi(1);
bmat(1,3) = bi(2);bmat(2,4) = ci(2);bmat(3,3) = ci(2);bmat(3,4) = bi(2);
bmat(1,5) = bi(3);bmat(2,6) = ci(3);bmat(3,5) = ci(3);bmat(3,6) = bi(3);
% Stiffness
k_el = bmat’*dmat*bmat*area;
2.18 To add a graphics capability to the program developed in Problem 2.17 to plot contoursof the computed finite element displacements it is only necessary to add a contourmodule and to insert the following into the main module
% Plot contours of u
len_u = length(u);tricont(ix,x,y,u(1:2:len_u-1),2) % Names figure 2tricont(ix,x,y,u(2:2:len_u ),3) % Names figure 3
A simple contour plot module is given by
29
function [] = tricont(ix,x,y,v,n_plot)
% Plot triangular and rectangular element results (v)
figure(n_plot)clfhold onaxis offaxis equal
e = length(ix(1,:)); % Number nodes on element (3 or 4)for n = 1:length(ix(:,1))
xl(1,1) = x(ix(n,1)); % Following remaps to rectanglexl(1,2) = x(ix(n,2)); % for plot surfacexl(2,1) = x(ix(n,3));xl(2,2) = x(ix(n,e));
yl(1,1) = y(ix(n,1));yl(1,2) = y(ix(n,2));yl(2,1) = y(ix(n,3));yl(2,2) = y(ix(n,e));
vl(1,1) = v(ix(n,1));vl(1,2) = v(ix(n,2));vl(2,1) = v(ix(n,3));vl(2,2) = v(ix(n,e));
surf(xl,yl,vl)endshading interp % Convert to shaded interpolation
This module will also work for rectangular elements.
Executing the matlab program with the input data produces the contours for the uand v displacements. Results for one analysis are shown in the figure for the mesh andcontour of u.
(a) Deformed mesh (b) u-contoursResults for Problems 2.18.
30
Chapter 3: Solutions
3.1 A Weak form for the problems is constructed as follows:
1. For the first order equation
adu
dx+ c u+ q = 0 ; u(0) = g
multiply by the arbitrary function v and integrate over domain∫ L
0
v[adu
dx+ c u+ q] dx = 0 .
In this case it is not possible to reduce the order of derivatives so no integrations byparts are perfomed. It is thus necessary to impose the boundary condition u(0) = gexplicitly.
2. For the second order equation
d
dx
(a
du
dx
)+ q = 0 ; u(0) = g & a
du
dx+ k u = g .
multiply by v and integrate over domain∫ 1
0
v
[d
dx
(a
du
dx
)+ q
]dx = 0 .
Integrate by parts to obtain∫ 1
0
v
[− dv
dxa
du
dx+ v q
]dx+ v
(a
du
dx
)∣∣∣∣10
= 0 .
We can impose the boundary condition u(0) = g0 and set v(0) = 0. Similarly we canimpose the other boundary condition and write the weak form as∫ 1
0
v
[− dv
dxa
du
dx+ v q
]dx+ v(1)
(g1 − k u(1)
)= 0 .
Note that it is necessary to include the term k u(1) since it does not appear otherwise.
3. For the second order equation
− d
dx
(a
du
dx
)+ b
du
dx+ q = 0 ; u(0) = g0 ; u(1) = g2 .
multiply by v, integrate over the domain 0 < x < 1 and integrate first term by partsgives ∫ 1
0
v
[− dv
dxa
du
dx+ v
(b
du
dx+ q
)]dx+ v
(a
du
dx
)∣∣∣∣10
= 0 .
The boundary term may be omitted by imposing the exact boundary conditions andsetting v(0) = v(1) = 0.
31
4. For the third order equation
d
dx
(a
d2u
dx2
)+ f = 0 ; u(0) = g0 ;
du
dx
∣∣∣∣x=0
= h0 & u(1) = g1 .
multiplying by v and integrating once by parts gives∫ 1
0
v
[− dv
dx
(a
d2u
dx2
)+ v f
]dx+ v
(a
d2u
dx2
)∣∣∣∣10
= 0
Last term is dropped by imposing the boundary conditions for u(0) and u(1) andsetting v(0) = v(1) = 0. Integrating once again by parts gives finally∫ 1
0
v
[(a
d2v
dx2
)du
dx+ v f
]dx−
(adv
dx
)du
dx
∣∣∣∣10
= 0
which upon using the boundary condition for du/dx may be rewritten as∫ 1
0
v
[(a
d2v
dx2
)du
dx+ v f
]dx− adv
dx
∣∣∣∣x=1
)du
dx
∣∣∣∣x=1
+
(adv
dx
∣∣∣∣x=0
)h0 = 0
5. For the second order equation
− ∇T (k∇u) + c bT (∇u) + q = 0 in Ω ; u = g on Γ .
multiply by the arbitrary function v and using Green’s theorem gives∫Ω
[(∇v)T k∇u+ v
(c bT (∇u) + q
) ]dΩ−
∫Γ
v nT (k∇u) dΓ = 0 .
Enforcing the boundary condition on Γ and setting v = 0 on Γ permits the last termto be omitted.
3.2 Multiplying the differential equations for bending of a beam by the arbitrary functionsspecified gives
(1) ±∫
Ωδw
[dV
dx+ q
]dx = 0 (2) ±
∫Ωδθ
[dM
dx+ V
]dx = 0
(3)∫
ΩδM
[dθ
dx− M
EI
]dx = 0 (4)
∫ΩδV
[dw
dx− θ − V
GA
]dx = 0
In order to observe symmetry we integrate the first term of the first two equations byparts to obtain
(1) ±∫
Ω
[− dδw
dxV + δw q
]dx± δw V
∣∣∣∣ΓV
= 0
(2) ±∫
Ω
[− dδθ
dxM + δθ V
]dx± δθ M
∣∣∣∣ΓM
= 0 .
32
Comparing these with the third and fourth equations shows symmetry if we select aminus sign for the weighting functions. This gives the weak forms
(1)∫
Ω
[dδw
dxV − δw q
]dx− δw V
∣∣∣∣ΓV
= 0 (2)∫
Ω
[dδθ
dxM − δθ V
]dx− δθ M
∣∣∣∣ΓM
= 0
(3)∫
ΩδM
[dθ
dx− M
EI
]dx = 0 (4)
∫ΩδV
[dw
dx− θ − V
GA
]dx = 0 .
3.3 To obtain all boundary conditions in the problem we integrate the first term of thethird and fourth equation by parts to obtain
(3)∫
Ω
[− dδM
dxθ − δM M
EI
]dx+ δM θ
∣∣∣∣Γθ
= 0
(4)∫
Ω
[− dδV
dxw − δV
(θ +
V
GA
)]dx+ δV w
∣∣∣∣Γw
= 0 .
To restore symmetry with the first two equations in Problem 3.2 we can integrate theequations by parts again to obtain
(3)∫
Ω
[δM
dθ
dx− δM M
EI
]dx + δM (θ − θ)
∣∣∣∣Γθ
= 0
(4)∫
Ω
[δV
dw
dx− δV
(θ +
V
GA
)]dx+ δV (w − w)
∣∣∣∣Γw
= 0 .
Thus, using (1) and (2) from Problem 3.2 with these gives a weak form with all bound-ary conditions.
3.4 To construct a variational theorem which gives the weak form we write the four equa-tions in the operator form Lu + b = 0 we obtain
u =
wθMV
; L =
0 0 0 − d
dx
0 0 − d
dx−1
0d
dx− 1
EI0
d
dx−1 0 − 1
GA
; b =
−q
000
To obtain the variational theorem, use (3.84) to obtain (we omit all boundary termsfor simplicity)
Π =
∫Ω
[1
2uT (Lu)uTb
]dΩ .
Inserting the above arrays and expanding gives
Π =
∫Ω
[− 1
2w
dV
dx− 1
2θ
dM
dx+ 1
2V
dw
dx+ 1
2M
dθ
dx− V θ − M2
2EI− V 2
2GA
]dx .
33
Integrating the first two terms by parts gives (again omitting boundary terms)
Π =
∫Ω
[V
dw
dx+M
dθ
dx− V θ − M2
2EI− V 2
2GA
]dx .
The quivalence with the weak form may be verified by constructing δΠ.
3.5 From (4) with GA =∞ we obtain
θ =dw
dx
which when inserted into (3) gives
M = EId2w
dx2.
Using (2) we can also express V in terms of displacements as
V = − d
dx
(EI
d2w
dx2
)which when substitued into (1) and (2) gives the irreducible equation
− d2
dx2
(EI
d2w
dx2
)+ q = 0 .
The boundary conditions are obtained immediately from the above as
(1) − d
dx
(EI
d2w
dx2
)= V or w = w
(2) EId2w
dx2= M or
dw
dx= θ .
3.6 For the irreducible differential equation given in Problem 3.5 it is only necessary toconsider one arbitrary function, namely δw. Accordingly, we may write∫
Ω
δw
[− d2
dx2
(EI
d2w
dx2
)+ q
]dx = 0 .
This form may be integrated by parts to give∫Ω
dδw
dx
[d
dx
(EI
d2w
dx2
)]dx+
∫Ω
δw q dx+ δw V
∣∣∣∣ΓV
= 0
in which we have inserted the shear boundary condition on V and also assumed δw = 0on the displacement boundary where w = w. Integating a second time reduces allderivatives to a minimum and gives
−∫
Ω
d2δw
dx2
(EI
d2w
dx2
)dx+
∫Ω
δw q dx+ δw V
∣∣∣∣ΓV
+dδw
dxM
∣∣∣∣ΓM
= 0
34
where we now use the moment boundary condition on M and also assume dδw/dx = 0on the rotation boundary where dw/dx = θ.
Often the sign on the whole form given above is reversed.
Since the highest derivative in the weak form is now two (2), it is necessary to useapproximating functions that are C1. that is, the function and its first derivative mustbe continuous in Ω.
3.7 Using Eq. (3.84) the variational theorem for the irreducible form is given immediatelyby
Π = −∫
Ω
12
d2w
dx2EI
d2w
dx2dx+
∫Ω
w q dx+ w V
∣∣∣∣ΓV
+dw
dxM
∣∣∣∣ΓM
= 0 .
Again, changing the sign gives a variational principle whose solution gives a minimum(known as the Minimum Potential Energy Principe) value of Π (whereas the formabove would give a maximum).
3.8 When GA =∞ a pair of equations in terms of w and M can be obtained by combiningthe two equilibrium equations (1) and (2) to obtain
d2M
dx2− q = 0
and the two constitutive equations (3) and (4) to obtain
d2w
dx2+M
EI= 0 .
The boundary conditions may also be given as
(1) − dM
dx= V or w = w
(2) EI
dodif 2wx2 = M ordw
dx= θ .
3.9 A weak form for Problem 3.8 is given by∫Ω
δw
[d2M
dx2− q
]dx = 0 and
∫Ω
δM
[d2w
dx2+M
EI
]dx = 0
and after integrating the first term in both by parts we obtain a form which containsonly first derivatives as
−∫
Ω
dδw
dx
dM
dxdx−
∫Ω
δw q
]dx− δw V
∣∣∣∣ΓV
= 0
−∫
Ω
dδM
dx
dw
dxdx+
∫Ω
δMM
EIdx+ δM θ
∣∣∣∣Γθ
= 0
35
In the above form we have used the boundary conditions for V and θ and also assumedthat δw = 0 where w = w and δM = 0 where M = M .
If we approximate
w =2∑
a=1
Nawa and M =2∑
a=1
NaMa
we obtain the matrix form
−2∑
a=1
2∑b=1
[δwa δMa
] ∫Ω
0dNa
dx
dNb
dx
dNa
dx
dNb
dx− 1
EINaNb
dx
wb
Mb
+2∑
a=1
∫Ω
Na q dx− δwcVc + δMdθd = 0 .
It is obvious that the stiffness matrix is symmetric in the above form.
For an element of length h with EI and q constant in the element the element stiffnessis given by
Ke = − 1
h
0 1 0 −11 0 −1 00 −1 0 1−1 0 1 0
− h
6EI
0 0 0 00 2 0 10 0 0 00 1 0 2
in which the unknowns are given in the order
ue =[w1 M1 w2 M2
]T.
Similarly the force vector is given by
f e = 12q h
[1 0 1 0
]T.
3.10 A simply supported beam of length 10, constant cross section EI = 3, uniform loadof q = 1, and boundary conditions w = M = 0 at each end is solved using the arraysdeveloped in Problem 9. The exact solution for moment M and displacement w isgiven by
M(x) = 12q (L x− x2)
w(x) = 124q (2 L x3 − x3 − L3x) .
The maximum occurs at the center of M(L/2) = qL2/8 and w(L/2) = −qL4/(384 EI).The numerical solution is carried out using a MATLAB©R program for 2, 4, 8, 16, 32 and64 elements. Results for the moment and displacement are shown in parts (a) and (b)of the figure. The convergence of the error in energy is shown in part (c) of the figure.
36
0 2 4 6 8 100
2
4
6
8
10
12
14
x
M −
Mom
ent
Exact2 elements4 elements8 elements
0 2 4 6 8 10−45
−40
−35
−30
−25
−20
−15
−10
−5
0
x
w −
Dis
plac
emen
t
Exact2 elements4 elements8 elements
(a) Moment (b) Displacement
10−2 10−1 10010−4
10−3
10−2
10−1
100
log10 h/h1
log 10
(Err
or =
| E h −
Eex
|/Eex
)
(c) Convergence of EnergySolution of beam moment, displacement and energy error for Problem 3.10.
3.11 For the four element problem considered in Example 3.5 with boundary conditionsimposed by a penalty method the matrix form is given by (after dividing the force bythe coefficent of K)
(1 + k) −1 0 0 0−1 2 −1 0 0
0 −1 2 −1 00 0 −1 2 −10 0 0 −1 (1 + k)
φ1
φ2
φ3
φ4
φ5
=Q0L
2
192
00165
.
Using MATLAB©R to solve for k = 2 × 10p, the boundary error is less than 10−6 φ4 forp = 7.
3.12 The variation of the specified functional is given by
δΠ(u) =
∫ b
a
2
[dδu
dx
(EI
du
dx
)− Pδu u
]dx− δu g
∣∣∣∣x=b
= 0 .
To obtain the Euler equation it is necessary to integrate the first term by parts. Ac-
37
cordingly, we obtain
δΠ(u) = −∫ b
a
2 δu
[d
dx
(EI
du
dx
)+ P u
]dx+ δu
(2 EI
du
dx− g
)∣∣∣∣x=b
= 0 .
We set δu(a) = 0 where the boundary condition u(a) = 0 must be enforced. Thus, theEuler equation is
− d
dx
(EI
du
dx
)− P u = 0
and boundary conditons are
u = 0 for x = a
2 EIdu
dx= g for x = b .
Taking the second variation gives
δ2Π = 2
∫ b
a
[EI
(dδu
dx
)2
− P (δu)2
]dx
hence if EI > 0 then Π is a saddle point unless P < 0 when it is a minimum principle.
3.13 The variation of the specified functional gives
δΠ(u) = 2
∫ b
a
[dδu
dx
(EA
du
dx
)+ δu k u− δu q
]dx+2 α [δu(a) u(a) + δu(b) u(b)] = 0
Integrating by parts gives
δΠ(u) = 2
∫ b
a
δu
[− d
dx
(EA
du
dx
)+ k u− q
]dx+ 2 δu
(EA
du
dx
)∣∣∣∣ba
+ 2 α [δu(a) u(a) + δu(b) u(b)] = 0
thus the Euler equation is
− d
dx
(EA
du
dx
)+ k u− q = 0
and the boundary conditions as α→∞ are u(a) = u(b) = 0.
3.14 The first variation of the specified functional is given by
δΠ(u) = 2
∫ b
a
[dδu
dx
(EA
du
dx
)+ δu k u− δu q
]dx
+ δλa u(a) + δu(a) λa + δλb u(b) + δu(b) λb = 0
38
Integrating by parts gives
δΠ(u) = 2
∫ b
a
δu
[− d
dx
(EA
du
dx
)+ k u− q
]dx+ 2 δu
(EA
du
dx
)∣∣∣∣ba
+ δλa u(a) + δu(a)
(λa + 2 EA
du
dx
)+ δλb u(b) + δu(b) λb
(+ 2 EA
du
dx
)= 0
thus the Euler equation is
− d
dx
(EA
du
dx
)+ k u− q = 0
and the boundary conditions associated with δλa and δλb are u(a) = u(b) = 0. Theremaining terms define the Lagrange multipliers λa and λb.
3.15 For the transient heat equation in one-dimension given by
− ∂
∂x
(k∂φ
∂x
)+Q+ c
∂φ
∂t= 0
with boundary conditions
φ = φ on Γ1 or q = − k∂φ
∂x= q on Γ2
the probleml solution is given by:
(a) The weak form ∫Ω
δφ
[− ∂
∂x
(k∂φ
∂x
)+Q+ c
∂φ
∂t
]dx = 0
which upon integration by parts of the first term gives∫Ω
δφ
[∂δφ
∂xk∂φ
∂x+ δφ
(Q+ c
∂φ
∂t
)]dx+ δφ q
∣∣∣∣Γ2
= 0
where we have assumed δφ = 0 on Γ1 such that we must enforce φ = φ there.
(b) Using the specified shape functions the semi discrete form for a typicall elementis given by
[δφ1 δφ2
]∫ xe2
xe1
dN1
dxdN2
dx
k
[dN1
dx
dN2
dx
]dx
φ1
φ2
+
∫ xe2
xe1
[N1
N2
]c
[N1 N2
]dx
˙φ1
˙φ2
+
∫ xe2
xe1
[N1
N2
]Q dx
= 0 .
39
(c) For a region of length 10, with properties k = 5, c = 1, Q = 0 the elementmatrices for the semi-discrete form are
Ke =k
h
[1 −1−1 1
]and Ce = 1
6c h
[2 11 2
].
For the properties given above the assembled equations for four elements is givenby
2 −2 0 0 0−2 4 −2 0 0
0 −2 4 −2 00 0 −2 4 −20 0 0 −2 2
φ1
φ2
φ3
φ4
φ5
+
5
12
2 1 0 0 01 4 1 0 00 1 4 1 00 0 1 4 10 0 0 1 2
˙φ1
˙φ2
˙φ3
˙φ4
˙φ5
=
00000
Note that no boundary conditions have been imposed.
(d) For a finite difference time approximation of the equations we obtain2 −2 0 0 0−2 4 −2 0 0
0 −2 4 −2 00 0 −2 4 −20 0 0 −2 2
φn1
φn2
φn3
φn4
φn5
+
5
12 ∆t
2 1 0 0 01 4 1 0 00 1 4 1 00 0 1 4 10 0 0 1 2
φn1 − φn−1
1
φn2 − φn−1
2
φn3 − φn−1
3
φn4 − φn−1
4
φn5 − φn−1
5
=
00000
The equations may be regrouped for solution as
[K +5
12 ∆tC]φ
n= − 5
12 ∆tC φ
n−1.
The solution is displayed in the figure at the requested times. Part (a) uses the(consistent) C array while part (b) uses a diagonal C array. The results are quiteclose, however, we note that the use of a diagonal form for C results in answerswhich are always greater than zero at every nodal location. Indeed, this is also aproperty of an exact solution and, thus, it is often advocated to use the diagonalform whenever possible. One should note that it is not possible to find a diagonalform for many element types, especially those of higher order.
40
0 2 4 6 8 10−2
0
2
4
6
8
10
x
φ(t)
t = 0.01t = 0.10t = 1.00t = 10.0
0 2 4 6 8 100
1
2
3
4
5
6
7
8
9
10
x
φ(t)
t = 0.01t = 0.10t = 1.00t = 10.0
(a) Consistent C (b) Diagonal CSolution of transient heat conduction for Problem 3.15.
41
Chapter 4: Solutions
1 2
34
5
689
1 2
34
5
89
1 2
34
5
7
8
(a) Problem 4.1 (b) Problem 4.2 (c) Problem 4.3Quadrilateral element for Problems 4.1 to 4.3.
The shape functions for the 9 nodes appearing in the above elements may be written inhierarchical form as:
N1 = 14(1− ξ)(1− η) N2 = 1
4(1 + ξ)(1− η)
N3 = 14(1 + ξ)(1 + η) N4 = 1
4(1− ξ)(1 + η)
N5 = 12(1− ξ2)(1− η) N6 = 1
2(1− ξ)(1− η2)
N7 = 12(1 + ξ2)(1 + η) N8 = 1
2(1− ξ)(1− η2)
N9 = (1− ξ2)(1− η2)
The coordinates of the nodes for the coordinate range −1 ≤ ξ, η ≤ 1 are given by:
Node numbersf 1 2 3 4 5 6 7 8 9
1 1 1 1 1 1 1 1 1 1ξ -1 1 1 -1 0 1 0 -1 0η -1 -1 1 1 -1 0 1 0 0ξ2 1 1 1 1 0 1 0 1 0ξη 1 -1 1 -1 0 0 0 0 0η2 1 1 1 1 1 0 1 0 0ξ2η -1 -1 1 1 0 0 0 0 0ξη2 -1 1 1 -1 0 0 0 0 0ξ2η2 1 1 1 1 0 0 0 0 0
1− ξ2 0 0 0 0 1 0 1 0 11− η2 0 0 0 0 0 1 0 1 1
N9 0 0 0 0 0 0 0 0 1
42
4.1. The standard shape functions at nodes 1, 3 and 6 for the element shown are given by
N1 = N1 − 12(N5 + N8) + 1
4N9
N3 = N3 − 12N6
N6 = N6 − 12N9
The shape functions may be expanded and written as
N1 = 14(1− ξ)(1− η) ξη
N3 = 14(1 + ξ)(1 + η) η
N6 = 12(1 + ξ)(1− η2) ξ
To show the polynomials in the Pascal triangle in ξ and η we need all the shapefunctions. Accordingly,
N2 = N2 − 12(N5 + N6) + 1
4N9
N4 = N4 − 12N9
N5 = N5 − 12N9
N8 = N8 − 12N9
N9 = N9
with the remainder given above. Writing these in a Matrix form as:
N1
N2
N3
N4
N5
N6
N8
N9
=
1
4
4 0 0 0 −2 0 −2 10 4 0 0 −2 −2 0 10 0 4 0 0 −2 0 10 0 0 4 0 0 −2 10 0 0 0 4 0 0 −20 0 0 0 0 4 0 −20 0 0 0 0 0 4 −20 0 0 0 0 0 0 4
N1
N2
N3
N4
N5
N6
N8
N9
To show which polynomials are present we use
f(ξ, η) =∑
a
faNa(ξ, η)
in which we use specified fa from the list shown above. We start by considering the
43
individual polynomials up to quadratic order
1 = N1 + N3 + N3 + N4 ≡ 1
ξ = −N1 + N3 + N3 − N4 ≡ ξ
η = −N1 − N3 + N3 + N4 ≡ η
ξη = N1 − N3 + N3 − N4 ≡ ξη
1− ξ2 = N5 + 12N9 6= 1− ξ2
1− η2 = N6 + N8 ≡ 1− η2
ξ2η = −N1 − N2 + N3 + N4 + N5 − 12N9
= η + N5 − 12N9 6= ξ2η
ξη2 = −N1 + N2 + N3 − N4 − N6 + N8
= ξ − ξ (1− η2) ≡ ξη2
Since 1 and 1− η2 are included the polynomial η2 is obtained. While we do always getN9 correctly, since the polynomial ξ2 is not present the polynomial ξ2η2 is also missing.Thus the Pascal triangle of complete polynomials is given by:
1ξ η
ξη η2
ξη2
4.2. The standard shape functions at nodes 2, 3 and 9 for the element shown are given by
N2 = N2 − 12N5
N3 = N3 − 14N9
The shape functions may be expanded and written as
N2 = 14(1 + ξ)(1− η) ξ
N3 = 14(1 + ξ)(1 + η) (ξ + η − ξη)
N9 = (1− ξ2)(1− η2)
To show which polynomials are present we use
f(ξ, η) =∑
a
faNa(ξ, η)
in which we use specified fa from the list shown above. We start by considering the
44
individual polynomials up to quadratic order
1 = N1 + N3 + N3 + N4 ≡ 1
ξ = −N1 + N3 + N3 − N4 ≡ ξ
η = −N1 − N3 + N3 + N4 ≡ η
ξη = N1 − N3 + N3 − N4 ≡ ξη
1− ξ2 = N5 + 12N9 6= 1− ξ2
1− η2 = N8 + 12N9 6= 1− η2
ξ2η = −N1 − N2 + N3 + N4 + N5 − 12N9
= η + N5 − 12N9 6= ξ2η
ξη2 = −N1 + N2 + N3 − N4 − N8 − 12N9
= ξ + N8 − 12N9 6= ξη2
Thus the Pascal triangle of complete polynomials is given by:
1ξ η
ξη
4.3. The standard shape functions at nodes 1, 2 and 5 for the element shown are given by
N1 = N1 − 12(N5 + N8)
N2 = N2 − 12N5
The shape functions may be expanded and written as
N1 = 14(1− ξ)(1− η) (−1− ξ − η)
N2 = 14(1 + ξ)(1− η) ξ
N5 = 12(1− ξ2)(1− η)
To show which polynomials are present we use
f(ξ, η) =∑
a
faNa(ξ, η)
in which we use specified fa from the list shown above. We start by considering the
45
individual polynomials up to quadratic order
1 = N1 + N3 + N3 + N4 ≡ 1
ξ = −N1 + N3 + N3 − N4 ≡ ξ
η = −N1 − N3 + N3 + N4 ≡ η
ξη = N1 − N3 + N3 − N4 ≡ ξη
1− ξ2 = N5 + N7 ≡ 1− ξ2
1− η2 = N8 6= 1− η2
ξ2η = −N1 − N2 + N3 + N4 + N5 − N7
= η − (1− ξ2)η ≡ ξ2η
ξη2 = −N1 + N2 + N3 − N4 − N8
= ξ + N8 6= ξη2
Thus the Pascal triangle of complete polynomials is given by:
1ξ η
ξ2 ξηξ2η
1 2
34
5 6
7
8
1 2
34
5 6
78
(a) Problem 4.5 (b) Problem 4.6Quadrilateral elements for Problems 4.5 and 4.6.
4.5. To develop an explicit form of the standard shape functions at nodes 1, 2 and 5 forthe element shown in part (a) of the figure the solution uses the procedure described
46
in Sec. 4.6. We begin by defining the basic shape functions Na for all the nodes as:
Na = 14
(1 + ξaξ) (1 + ηaη) for a = 1, 2, 3, 4
Na = 932
(1− ξ2) (1− 3ξ) (1− η) for a = 5
Na = 932
(1− ξ2) (1 + 3ξ) (1− η) for a = 6
Na = 12
(1− ξ2) (1 + η) for a = 7
Na = 12
(1− ξ) (1− η2) for a = 8 .
Now constructing the shape function from the steps shown in Figs 4.10 and 4.11 weobtain:
N1 = N1 − 13
(2 N5 + N6)− 12N8
N2 = N2 − 13
(2 N6 + N5)
N5 = N5
Using the usual procedure in which we introduce in turn polynomials in ξ and η wecan show that the polynomials present include only:
1ξ η
ξ2 ξηξ2η
4.6. Develop an explicit form of the standard shape functions at nodes 1, 5 and 7 for theelement shown in part (b) of the figure.
The solution uses the procedure described in Sec. 4.6. We begin by defining the basicshape functions Na for all the nodes as:
Na = 14
(1 + ξaξ) (1 + ηaη) for a = 1, 2, 3, 4
Na = 932
(1− ξ2) (1− 3ξ) (1− η) for a = 5
Na = 932
(1− ξ2) (1 + 3ξ) (1− η) for a = 6
Na = 12
(1− ξ) (1− η2) for a = 7
Na = (1− ξ2) (1− η2) for a = 8 .
Now constructing the shape function from the steps shown in Figs 4.10 and 4.11 weobtain:
N1 = N1 − 13
(2 N5 + N6)− 12N7 + 9
32N9
N5 = N5 − 932N9
N7 = N7 − 12N9
The factor 9/32 results from N1(0, 0) = 1/4, N5(0, 0) = N6(0, 0) = 9/32 and N7(0, 0) =1/2 which when combined in the proportions shown for N1 gives −9/32.
47
Using the usual procedure in which we introduce in turn polynomials in ξ and η wecan show that the polynomials present include only:
1ξ η
ξη
The highest polynomial is only that present in the 4-node element.
1 2 3 4 5
8
11109
6 7
6 cm. 3 cm.
4 cm.
Quadratic rectangle and triangle for Problem 4.7.
4.7 The mesh for a problem contains an 8-node quadratic serendipity rectangle adjacent toa 6-node quadratic triangle as shown in the figure. Show that the coordinates computedfrom each element satisfy C0 continuity along the edge 3− 7− 11.
Using the shape functions given in Eq. (4.22)(a) and (b) we obtain the following shapefunctions for nodes 3, 7, and 11 in the rectangular element:
N3(ξ, η) = 14
(1 + ξ) (1− η) (ξ − η − 1)
N11(ξ, η) = 14
(1 + ξ) (1 + η) (ξ + η − 1)
N7(ξ, η) = 12
(1 + ξ) (1− η2) .
Evaluation for ξ = 1 gives the form:
N3(1, η) = 12
(η2 − η)N11(1, η) = 1
2(η2 + η)
N7(1, η) = (1− η2) .
Repeating for the triangular element with node 3 at (L1, L2, L3) = (1, 0, 0) node 11 at(0, 0, 1) and node 7 at (1/2, 0, 1/2) and using quadratic functions from Sec. 4.7.2 wehave
N3(L1, L2, L3) = (2 L1 − 1) L1
N11(L1, L2, L3) = (2 L3 − 1) L3
N7(L1, L2, L3) = 4 L1 L3 .
48
For the edge 3-7-11, L2 = 0 and thus L1 +L3 = 1 which permits the shape function tobe written in terms of L1 as
N3(L1, 0, L3) = (2 L1 − 1) L1
N11(L1, 0, L3) = (1− 2 L1) (1− L1)
N7(L1, 0, L3) = 4 L1 (1− L1) .
Along the edge we have −1 ≤ η ≤ 1 for the rectangle and 0 ≤ L1 ≤ 1 for the triangle.We can express one in terms of the other using linear interpolation in which
L1 = 12(1− η)
which obviously satisfies the range for L1 using the range for η. Substituting this weobtain
N3(L1, 0, L3) ≡ (−η) 12(1− η) = 1
2(η2 − η)
N11(L1, 0, L3) ≡ (+η) 12(1 + η) = 1
2(η2 + η)
N7(L1, 0, L3) ≡ 2 (1− η) 12(1 + η) = (1− η2) .
The two interpolations agree and thus the two elements produce C0 compatibilitydespite the fact that two different interpolation schemes are used within each element.
4.8 Determine an explicit expression for the shape function of node 1 of the linear triangularprism shown in Fig. 4.20(a).
This is a 6-node prism with triangular and rectangular faces. If we let the trianglularfaces have interpolation in terms of (L1, L2, L3) area coordinates with node 1 locatedat (1, 0, 0) and use linear Lagrange interpolation in ζ in the direction of rectangularfaces with ζ = 1 nn the 1− 2− 3 face and ζ = −1 nn the 4− 5− 6 face we may writethe interpolation directly for node 1 as
N1(L1, L2, L3, ζ) = 12L1 (1 + ζ) .
Setting the area coordinate to (1, 0, 0) and ζ = 1 gives N1 = 0. Setting L1 = 0 givesN2 = N3 = 0. Setting ζ = −1 gives N4 = N5 = N6 = 0. Thus, the interpolation issatisfied at nodes. Similarly, evaluated for ζ = ± 1 gives a linear interpolation in L1
on each face. Finally, on the rectangular faces we obtain a bilinear interpolation forfaces adjacent to node 1 and zero on the other face. Thus the interpolation satisfiesall the conditions for C0 requirements.
4.9 Determine an explicit expression for the hierarchical shape function of nodes 1, 7 and10 of the quadratic triangular prism shown in Fig. 4.20(b).
Here standard lagrangian and serendipity interpolation is used for the hierarchic func-tions. We let the triangular faces with vertex nodes 1−2−3 and 4−5−6 be expressedin terms of quadratic order area coordinates (L1, L2, L3) with nodes 1 and 4 locatedat (1, 0, 0). Similarly the trianglular plane at nodes 7 − 8 − 9 is defined by a linearform in the area coordinates with node 7 located at (1, 0, 0). Let the other direction
49
be expressed in a ζ coordinate between −1 and +1. In hierarchical form we start withlinear interpolation for the vertex nodes and thus node 1 has the solution given forProblem 4.8 above as
N1 = 12L1 (1 + ζ) .
Similarly, node 7 is defined by a linear form in area coordinates and a quadratic termin ζ as
N7 = 12L1 (1− ζ2) .
Finally, the node 10 is generated for the serendipity face and a quadratic triangularface as
N10 = 4 L1 L212
(1 + ζ) = 2 L1 L2 (1 + ζ) .
4.10 Determine an explicit expression for the shape function of nodes 1, 7, 13 and 25 of thecubic triangular prism shown in Fig. 4.20(c).
Here the top and bottom faces are given in (L1, L2, L3) area coordinates and faces byserendipity functions for the ζ (vertical) direction. We can construct the interpolationby taking first the shape functions for the cubic triangular faces with linear interpola-tion in the ζ direction. This is then corrected using the functions for nodes which havenon-zero values. Thus, for node 1 we have
N1 = 12
(3L1 − 1) (3L1 − 2) L112
(1 + ζ)
= 14
(3L1 − 1) (3L1 − 2) L1 (1 + ζ)
N7 = L198
(1− ζ2) (3ζ + 1)
= 98L1 (1− ζ2) (3ζ + 1)
N13 = 92L1 L2 (3L1 − 1) 1
2(1 + ζ)
= 94L1 L2 (3L1 − 1) (1 + ζ)
Examining the values at other nodes we find at node 7 N1(1, 0, 0,+1/3) = 2/3 and atnode 10 N1(1, 0, 0,−1/3) = 1/3, thus,
N1 = N1 − 13
(2N7 +N8) .
Similarly evaluating the shape functions N7 and N13 at all other nodes shows that
N7 = N7 and N13 = N13 .
4.11 On the sketch below the location of the 35 nodes of a quartic order tetrahedron areshown.
An explicit expression for the shape function of the vertex node located at (L1, L2, L3, L4) =(1, 0, 0, 0) may be constructed following the procedure given in Eq. 4.30 generalized as
Na = lII (L1) lJJ (L2) l
KK(L3) l
LL(L4)
50
(a) Nodes at L1 = 1 (b) Nodes at L1 = 0.75
(c) Nodes at L1 = 0.5 (d) Nodes at L1 = 0.25 (e) Nodes at L1 = 0.0Location of nodes for quartic tetrahedron of Problem 4.11.
where I + J + K + L = M . For the vertex node defined above for a quartic orderinterpolation this simiplifies to
Na = l44(L1) =(4L1 − 3) (4L1 − 2) (4L1 − 1) L1
(4− 3) (4− 2) (4− 1) 1= 1
6(4L1−3) (4L1−2) (4L1−1) L1 .
For a mid-edge node located at (0.25, 0.75, 0, 0) the formula specializes to
Na = l11(L1) l33(L2) =
4L1
1
(4L2 − 2) (4L2 − 1) 4L2
(3− 2) (3− 1) 3=
16 L1 L2 (4L2 − 1) (4L2 − 2)
(1) (2) (3)
=8
3L1 L2 (4L2 − 1) (4L2 − 2) .
4.12 The location of the nodes for the quartic member of the serendipity family is shown inthe figure. In part (a) of the figure an isometric view of the element is shown togetherwith the location of the visible nodes. In parts (b) to (f) of the figure the position ofthe nodes at different elevations of the ζ coordinate is shown. The ξ, η coordinates liein the plane of each level. There are a total of 50 nodes on the element (compared to
51
(a) Isometric view (b) Nodes at ζ = −1 (c) Nodes at ζ = −0.5
(d) Nodes at ζ = 0.0 (e) Nodes at ζ = 0.5 (f) Nodes at ζ = 1.0Location of nodes for quartic serendipity brick of Problem 4.12.
125 if a lagrangian interpolation of quartic order were assumed). Note that no interiornodes are needed to obtain the quartic order interpolation. The nodes where shapefunctions are to be determined are shown as solid bullets, other nodes as open ones.
We assign the nodes as follows:
N1(ξ, η, ζ) ; located at (1, 1, 1). N2(ξ, η, ζ) ; located at (1, 1, 0.5).N3(ξ, η, ζ) ; located at (1, 1, 0). N4(ξ, η, ζ) ; located at (1, 1,−0.5).N5(ξ, η, ζ) ; located at (1, 1,−1). N6(ξ, η, ζ) ; located at (1, 0, 0).N7(ξ, η, ζ) ; located at (0, 1, 0). N8(ξ, η, ζ) ; located at (0, 0, 1).
There is a typographical error for the location of the edge node. The coordinatesshould read (0.5, 1, 1). For this location we can construct a shape function by firstconstructing the quartic lagrange function in ζ and multiplying by the bi-linear functionin the plane. Accordingly, we have
N2(ξ, η ζ) =(2ζ + 2)(2ζ + 1)ζ(2ζ − 2)
(3)(2)(0.5)(−1)(1 + ξ) (1 + η)
= −1
3(2ζ + 2)(2ζ + 1)ζ(2ζ − 2) (1 + ξ) (1 + η) .
52
This function is zero at all the other nodes and, hence N2 = N2.
For the vertex node located at (1, 1, 1) we start with the tri-linear function
N1(ξ, η, ζ) = 18(1 + ξ) (1 + η) (1 + ζ)
which is zero for all nodes located at ξ = −1, η = −1 and ζ = −1. However it isnon-zero at all the mid-side nodes where ξ = η = 1 and the mid-face node locatedat (1, 0, 0), the mid-face node located at (0, 1, 0) and the mid-face node located at(0, 0, 1) For the mid-face node the function Na has the value
N6(1, 0, 0) = N7(0, 1, 0) = N8(0, 0, 1) =1
4.
For the edge nodes we have
N2(1, 1, 0.5) =3
4; N3(1, 1, 0) =
1
2; N4(1, 1,−0.5) =
1
4.
We can note that, except for N3, all the correction shape functions are similar inbehavior to N2 such that Na ≡ Na at these nodes. At node-2 we need to apply thecorrections as
N3 = N3 − 12
(N6 + N7)
= N3 − 12
(N6 +N7) .
Thus, the final interpolation function is given by
N1 = N1 − 14
(3N2 + 2N3 +N4 +N6 +N7 +N8) .
Alternatively, it may be expressed in the Na as
N1 = N1 − 14
(3N2 + 2N3 + N4 + N8) .
The shape functions for the mid-faces have the form
N6 = N6 = 12
(1− ξ2) (1− η2) (1 + ζ)
with similar forms for N7 and N8 by permutation of the coordinates. The shapefunction for N4 is similar in form to that for N2 with only the lagrange polynomial inζ adjusted for the different locatin of the node.
4.13 The location of the nodes for the quartic member of the serendipity wedge familyis shown in the figure. In part (a) of the figure an isometric view of the elementis shown together with the location of the visible nodes. In parts (b) to (f) of thefigure the position of the nodes at different elevations of the ζ coordinate is shown.The area coordinates La lie in the plane of each level. There are a total of 42 nodeson the element. Note that no interior nodes are needed to obtain the quartic order
53
(a) Isometric view (b) Nodes at ζ = −1 (c) Nodes at ζ = −0.5
(c) Nodes at ζ = 0.0 (d) Nodes at ζ = 0.5 (e) Nodes at ζ = 1.0Location of nodes for quartic serendipity wedge of Problem 4.13.
interpolation. The nodes where shape functions are to be determined are shown assolid bullets, other nodes as open ones.
We define the location of the nodes as follows:
N1(L1, L2, L3, ζ) ; located at (1, 0, 0,−1).N2(L1, L2, L3, ζ) ; located at (1, 0, 0,−0.5).N3(L1, L2, L3, ζ) ; located at (0.75, 0.25, 0,−1).
We construct the hierarchical vertex shape function as the product of linear area co-ordinates times a linear lagrange polynomial in ζ. Thus, for ζ = −1 we use the areacoordinates for the linear triangle. From Sec. 4.7.2 we obtain
N1 = 12L1 (1− ζ) .
For the edge node on the triangle we use a cubic area coordinate function from Sec.4.7.2 and the linear ζ function as
N3 = 94L1 L2 (3L1 − 1) (1− ζ) .
54
(a) Isometric view (b) Nodes at ζ = −1
(c) Nodes at ζ = 0.0 (d) Nodes at ζ = 1.0Location of nodes for quadratic lagrangian wedge of Problem 4.14.
For the edge of the rectangular element we use the cubic ζ function with the lineararea coordinate to
N2(L1, L2, L3, ζ) = L1(2 + 2ζ) (1 + 2ζ) (2− 2ζ)
(3)(2)(1)
= 23L1 (1− ζ2) (1 + 2ζ) .
4.14 The location of the nodes for the quadratic member of the lagrangian wedge familyis shown in the figure. In part (a) of the figure an isometric view of the element isshown together with the location of the visible nodes. In parts (b) to (d) of the figurethe position of the nodes at different elevations of the ζ coordinate is shown. Thearea coordinates La lie in the plane of each level. There are a total of 18 nodes on theelement. Note that no interior node is needed to give full quadratic order interpolation.The nodes where shape functions are to be determined are shown as solid bullets, othernodes as open ones.
55
We define the location of the nodes as follows:
N1(L1, L2, L3, ζ) ; located at (1, 0, 0,−1).N2(L1, L2, L3, ζ) ; located at (1, 0, 0, 0).N3(L1, L2, L3, ζ) ; located at (0.5, 0.5, 0,−1).
The interpolation for all shape functions may be constructed by taking the (tensor)product of the quadratic area coordinates times the quadratic Lagrange interpolationin ζ. Thus for the vertex node 1 the interpolation is given as
N1(L1, L2, L3, ζ) = 12L1 (2L1 − 1) (1− ζ) .
Similarly for the mid-edge of the rectangular faces for node-2 the interpolation is
N2(L1, L2, L3, ζ) = L1 (2L1 − 1) (1− ζ2)
and for the mid-edge of the triangular face for node-3 the interpolation is
N3(L1, L2, L3, ζ) = 2 L1 L2 (1− ζ) .
4.15 The location of the nodes for the cubic member of the lagrangian wedge family isshown in the figure. In part (a) of the figure an isometric view of the element is showntogether with the location of the visible nodes. In parts (b) to (e) of the figure theposition of the nodes at different elevations of the ζ coordinate is shown. The areacoordinates La lie in the plane of each level. There are a total of 40 nodes on theelement. Note that two interior nodes are needed in the lagrangian form. The nodeswhere shape functions are to be determined are shown as solid bullets, other nodes asopen ones.
We define the location of the nodes as follows:
N1(L1, L2, L3, ζ) ; located at (1, 0, 0,−1).N2(L1, L2, L3, ζ) ; located at (1, 0, 0,−1/3).N3(L1, L2, L3, ζ) ; located at (2/3, 1/3, 0,−1).N4(L1, L2, L3, ζ) ; located at (1/3, 1/3, 1/3,−1).N5(L1, L2, L3, ζ) ; located at (2/3, 1/3, 0,−1/3).N6(L1, L2, L3, ζ) ; located at (1/3, 1/3, 1/3,−1/3).N7(L1, L2, L3, ζ) ; located at (1/3, 1/3, 1/3, 1/3).
The interpolation for all shape functions may be constructed by taking the (tensor)product of the quadratic area coordinates times the quadratic Lagrange interpolationin ζ. Thus for the vertex node 1 the interpolation is given as
N1(L1, L2, L3, ζ) = 14L1 (3L1 − 1)(3L1 − 2) (ζ2 − ζ) .
Similarly for the edge of the rectangular face for node-2 the interpolation is
N2(L1, L2, L3, ζ) = 12L1 (3L1 − 1)(3L1 − 2) −9
16(1− ζ2) (3ζ − 1)
= − 932L1 (3L1 − 1)(3L1 − 2) (1− ζ2) (3ζ − 1)
56
(a) Isometric view (b) Nodes at ζ = −1
(c) Nodes at ζ = −1/3 (d) Nodes at ζ = 1/3 (e) Nodes at ζ = 1.0Location of nodes for cubic lagrangian wedge of Problem 4.15.
and for the edge of the triangular face for node-3 the interpolation is
N3(L1, L2, L3, ζ) = 94L1 L2 (3L1 − 1) (ζ2 − ζ) .
For the mid-face node on the triangular facet we have the interpolation
N4 = 54 L1 L2 L3 (ζ2 − ζ)
and for the mid-face node on the rectangular face
N5(L1, L2, L3, ζ) = 92L1 L2 (3L1 − 1) −9
16(1− ζ2) (3ζ − 1)
= − 8132L1 L2 (3L1 − 1) (1− ζ2) (3ζ − 1) .
For the two internal nodes we have
N6(L1, L2, L3, ζ = 27 L1 L2 L3−916
(1− ζ2) (3ζ − 1)
= − 24316
L1 L2 L3 (1− ζ2) (3ζ − 1)
similarly for the other internal node we have
N7(L1, L2, L3, ζ = 27 L1 L2 L3916
(1− ζ2) (3ζ + 1)
= 24316
L1 L2 L3 (1− ζ2) (3ζ + 1) .
57
Chapter 5: Solutions
ξ
N
1 1
1 3 2x
f
a b
1 3 2
(a) Parent element (b) Mapped elementQuadratic element for Problem 5.1.
5.1 A quadratic one dimensional element is shown in the figure in parent form and in themapped configuration. Let a+ b = h the total length of the mapped element.
(a) The shape functions Na(ξ) for the three nodes may be determined using anisoparametric mapping in which
x =3∑
a=1
Na(ξ) xa .
Using the lagrange interpolation given by Eq. (4.18) we may write the Na as
N1(ξ) = 12(ξ2 − ξ) ; N2(ξ) = 1
2(ξ2 + ξ) and N3(ξ) = (1− ξ2) .
−1 −0.5 0 0.5 1−0.2
0
0.2
0.4
0.6
0.8
1
1.2
ξ
x/h
0.20.30.40.50.60.70.8
Problem 5.1. x vs. ξ.
58
(b) Using the interpolation for x above with the origin at node-1, the nodal coordi-nates xa may be taken as
x1 = 0 ; x2 = h and x3 = a = α h .
Thus,
x = 12h (ξ2 + ξ) + α h (1− ξ2) =
[(1
2− α) ξ2 + 1
2ξ + α
]h .
For a ranging from 0.2h to 0.8h in increments of 0.1h (i.e., 0.2 ≤ α ≤ 0.8) thebehavior of x vs ξ is given by:
0 0.2 0.4 0.6 0.8 1−0.2
0
0.2
0.4
0.6
0.8
x/h
N 1
0.20.30.40.50.60.70.8
0 0.2 0.4 0.6 0.8 1−0.2
0
0.2
0.4
0.6
0.8
x/h
N 2
0.20.30.40.50.60.70.8
N1 vs. x. N2 vs. x.
0 0.2 0.4 0.6 0.8 1−0.2
0
0.2
0.4
0.6
0.8
x/h
N 3
0.20.30.40.50.60.70.8
0 0.2 0.4 0.6 0.8 1−4
−3
−2
−1
0
1
2
x/h
dN1/d
x
0.20.30.40.50.60.70.8
N3 vs. x. dN1/dx vs. x.
0 0.2 0.4 0.6 0.8 1−2
−1
0
1
2
3
4
x/h
dN2/d
x
0.20.30.40.50.60.70.8
0 0.2 0.4 0.6 0.8 1−6
−4
−2
0
2
4
6
x/h
dN3/d
x
0.20.30.40.50.60.70.8
dN2/dx vs. x. dN3/dx vs. x.Problem 5.1. Na and dNa/dx vs. x.
59
(c) The plot of Na vs. x is shown in the figures.
(d) The plot of dNa/dx vs. x is shown in the figures
5.2 For the one dimensional problem in the domain 0 ≤ x ≤ 1 the weak form is given by:∫ 1
0
[dδu
dx
du
dx− δu q
]dx− δu σ
∣∣∣∣x=1
= 0 with u(0) = 0
with q = σ = 1.
(a) Integrating by parts the weak form becomes
−∫ 1
0
δu
[d2u
dx2+ q
]dx+ δu
(du
dx− σ
) ∣∣∣∣x=1
= 0
with δu(0) = u(0) = 0. Thus, the Euler differential equation and boundaryconditions for the problem are
d2u
dx2+ q = 0 ; for 0 < x < 1
du
dx− σ = 0 ; for x = 1
u = 0 ; for x = 0 .
(b) An exact solution to the differential equation may be constructed by integrationand after applying the boundary conditions and specified values gives:
u(x) = σ x+ q (x− 12x2) = 2 x− 1
2x2
σ(x) =du
dx= σ + q (1− x) = 2− x .
(c) For the single quadratic order element shown the require shape functions are:
i. Lagrange interpolation in x
N1 = 15
(5− 16x)(1− x) ;dN1
dx= 1
5(32x− 1)
N2 = − 111x (5− 16x) ;
dN2
dx= 1
11(32x− 5)
N3 = 25655
x (1− x) ;dN3
dx= 256
55(1− 2x)
ii. Isoparametric interpolation for Na(ξ) with x = Na(ξ)xa.
N1(ξ) = 12
(ξ2 − ξ) ;dN1
dξ= ξ − 1
2
N2(ξ) = 12
(ξ2 + ξ) ;dN2
dξ= ξ + 1
2
N3(ξ) = (1− ξ2) ;dN3
dξ= −2 ξ
60
with the interpolation for x given by
x(ξ) = 516
(1− ξ2) + 12
(ξ2 + ξ)
= 116
(5 + 8ξ + 3ξ2) .
The interpolation for x gives
dx
dξ= 1
8(4 + 3ξ)
thus, the derivatives of the shape functions are given by
dNa
dx= 8
dNa
dξ/(4 + 3ξ) .
Evaluate all integrals using two-point gaussian quadrature.
0 0.2 0.4 0.6 0.8 1−0.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
x
u
LagrangeIsoparametric
0 0.2 0.4 0.6 0.8 1−1
−0.5
0
0.5
1
1.5
2
x
u
LagrangeIsoparametric
(a) Displacement (b) StressProblem 5.2. Plot of displacements and stresses
(d) A plot for the u and σ = du/dx for the two solution methods is shown in thefigure. It is evident that when the fully quadratic displacements using Lagrangeinterpolation in x are used the answers are exact (since the exact solution foru involves no higher than quadratic terms). However, when the isoparametricsolution is used with severe distortion for the location of node-3 the solution haslarge errors. Use of additional elements will, however, lead to convergent solutions.
5.3 It is proposed to create transition elements for use with 4-node quadrilateral elementmeshes as shown in the figure.
(a) The layout for a 5-node element is shown in part (a) of the figure for transitionelement A. The element is divided into two parts so that the piecewise linearbehavior along the right side of the composite element may be obtained. The
61
A
B
Problem 5.3. Transition elements for use with 4-node quadrilaterals.
1 2
34
5
a
b
1
1 2
34
5
(a) Element A layout (b) N5 shape function.
Problem 5.3. Transition element A.
shape function for N5 is constructed from two piecewise polynomials as shown inpart (b) of the figure. This function is defined by
N5(ξ, η) =
12
(1 + ξ) (1 + η) ; for − 1 ≤ η ≤ 012
(1 + ξ) (1− η) ; for 0 ≤ η ≤ 1
; for − 1 ≤ ξ ≤ 1 .
Using hierarchic form for the shape functions at nodes 1 to 4 we have
Na = 14
(1 + ξa ξ) (1 + ηa η) ; for a = 1, 2, 3, 4 .
The final shape functions are then given by N5 and
N1 = N1
N2 = N2 − 12N5
N3 = N3 − 12N5
N4 = N4 .
62
1 2
34
5
6
7
a
b
c
1 2
34
5
6
a
b c
1 2
34
5
6
a b
cd
(a) Layout 1 (b) Layout 2 (c) Layout 3
Problem 5.3. Transition element B.
The functions for N1, N2 and N5 are all discontinuous in the element, thus, it isnecessary to perform any integrals by computation on the two sub-elements.
(b) In the figure we show 3 possible configurations for a composite transition elementof the type B. In the first layout we show 3-quadrilaterals with a center node(labeled 7). For this layout we have severely distorted elements and the needto eliminate the interior node, however, the formulation may be achieved usingstandard 4-node isoparametric elements for each sub-element. In the second lay-out we elminate the interior node by using two triangles. The two triangles havelower accuracy than quadrilaterals and in addition the remaining quadrilateralis highly distorted. We could divide the quadrilateral into two triangles and,thus, use standard 3-node triangular elements for each sub-element, however, thiswould give even poorer results. The best choice is to use the layout shown inpart (c) of the figure. This uses two pairs of rectangular sub-elements of the typeintroduced above for transition type A. Moreover, if necessary, we could continuethis construction to form a transition element with three sides subdivided.
Using the third option we can specify the shape function for node 6 as
N6(ξ, η) =
12
(1 + ξ) (1 + η) ; for − 1 ≤ ξ ≤ 012
(1− ξ) (1 + η) ; for 0 ≤ ξ ≤ 1
; for − 1 ≤ η ≤ 1 .
Using the interpolations for N5 and N6 with the standard bilinear interpolationsfor Na, a = 1, 2, 3, 4 we may write the corrected vertex interpolations as:
N1 = N1
N2 = N2 − 12N5
N3 = N3 − 12
(N5 + N6)
N4 = N4 − 12N5
63
1 2
34
5 6
7
8
910
11
1213 14
1516
ξ
η
1 2
3=4
5 6
7
811
1213
(a) 16-node quadrilateral (b) Degenerate 10-node triangleCubic elements for Problems 5.4, 5.5 and 5.6.
(c) The location of the quadrature points is shown by plus (+) signs on the sketchof each composite element. Note that it is necessary to split the integral overcontinuous parts of each interpolation.
(d) As an alternative to transition elements, 4-node elements may be used for allelements and constraints imposed to maintain compatibility. For the mesh shownin the figure, number all nodes and write the constraint equations necessary tomaintain compatibility. The interior node of element B is not needed and can beignored.
5.4 Hierarchical interpolation functions in ξ, η coordinates are to be developed for the16-node cubic order quadrilateral shown in part (a) of the figure in the form
f(ξ, η) =4∑
a=1
Na(ξ, η)fa +12∑
a=5
Na(ξ, η)∆fa +16∑
a=13
Na(ξ, η)∆∆fa .
For the first four functions we merely use the linear functions
Na = 14
(1 + ξa xi) (1 + ηa η) ; a = 1, 2, 3, 4 .
For the edge nodes we use the serendipity functions
Na = 916
(1 + 9ξa ξ) (1− ξ2) (1 + ηa η) ; a = 5, 6, 9, 10Na = 9
16(1 + 9ηa η) (1− η2) (1 + ξa ξ) ; a = 7, 8, 11, 12 .
Finally for the interior nodes we use the bubble functions given by
Na = (1 + 9ξa ξ) (1 + 9ηa η) (1− ξ2) (1− η2) ; a = 13, 14, 15, 16 .
The values for the ξa and ηa are given by values in the table.
64
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16ξ -1 1 1 -1 −1
919
1 1 19−1
9-1 -1 −1
919
19−1
9
η -1 -1 1 1 -1 -1 −19
19
1 1 19−1
9−1
9−1
919
19
5.5 The hierarchical interpolation functions in L1, L2, L3 area coordinates for the 10-nodecubic order triangle shown in part (b) of the figure are given in the form
f(L1, L2, L3) =3∑
a=1
Na(L1, L2, L3)fa +8∑
a=5
Na(L1, L2, L3)∆fa +12∑
a=11
Na(L1, L2, L3)∆fa
+N13(L1, L2, L3)∆∆f13 .
The vertex node shape functions are given by the linear functions
Na = La ; a = 1, 2, 3 .
The edge functions are given by
N5 = L1 L2 (2L2 − L1) N6 = L1 L2 (2L1 − L2)N7 = L2 L3 (2L3 − L2) N8 = L2 L3 (2L2 − L3)N11 = L3 L1 (2L1 − L3) N12 = L3 L1 (2L3 − L1) .
Finally, the internal function is given by the bubble function
N13 = 27 L1 L2 L3 .
5.6 Using the shape functions developed in Problem 5.4, determine the modified shapefunctions to degenerate the cubic 16-node quadrilateral into the cubic 10-node trian-gular element using numbering as shown in part (b) of the figure. The final elementmust be completely consistent with the shape functions developed in Problem 5.5.
5.7 The shape functions for an 8-node hexahedral element may be degenerated to forma 5-node pyramid with a rectangular base by merging 4-nodes with any ξ, η or ζ =±1. We choose to merge nodes 5 to 8 to form the 5-node pyramid. Starting from thefunctions
Na = 18
(1 + ξa ξ) (1 + ηa η) (1 + ζa ζ) ; a = 1, 2, . . . , 8
we add the last 4 functions as
Np5 =
8∑a=5
Na
where the superscript p denotes the function for the pyramid and can note that
8∑a=5
14
(1 + ξa ξ) (1 + ηa ηa) ≡ 1 .
Thus, the final 5 functions are
Npa = Na ; a = 1, 2, 3, 4
Np5 = 1
2(1 + ζa ζ) .
65
5.8 The shape functions for the triangular element shown in the figure are given in areacoordinates by
N1 = L1 (2 L1 − 1) ; N2 = L2 (2 L2 − 1) ; N3 = L3 (2 L3 − 1)N4 = 4 L1 L2 ; N5 = 4 L2 L3 ; N6 = 4 L3 L1
For the origin of the coordinates placed as shown we have
x = 0 ·N1 + 12 ·N2 + 0 ·N3 + 6 ·N4 + 6 ·N5 + 0 ·N6
= 6 · (2 N2 +N4 +N5) .
Substituting the shape function definitions we obtain
2 N2 +N4 +N5) = 4 L22 − 2 L2 + 4 L1 L2 + 4 L2 L3
= 4 L22 − 2 L2 + 4 L2 (L1 + L3)
= 4 L22 − 2 L2 + 4 L2 (1− L2)
= 2 L2
Thus x = 12 L2. Similarly, we have
y = 0 ·N1 + 0 ·N2 + 18 ·N3 + 9 ·N4 + 9 ·N5 + 9 ·N6
= 9 · (2 N3 +N5 +N6) .
Using the same construction as above for the x coordinate we obtain
2 N3 +N5 +N6 = 2 L3
and thus y = 18 L3. The final form for the coordinates is given by
x =6∑
a=1
Na(Lb) xa =
12L2
18L3
.
x
y
1 4 2
5
3
6
6 6
9
9
Quadratic triangle. Problems 5.8 to 5.10
66
5.9 For the triangular element shown in the figure the integrals∫
∆N2N3 d∆ and
∫∆N2N4 d∆
are given by: ∫∆
N2N3 d∆ =
∫∆
L2 (2L2 − 1) L3 (2L3 − 1) d∆∫∆
N2N4 d∆ =
∫∆
L2 (2L2 − 1) 4 L2 L3 d∆ .
From the figure ∆ = 108.
(a) Expanding the polynomials and using Eq. (4.31) we obtain∫∆
N2N3 d∆ =
∫∆
(2L22 − L2) (2L2
3 − L3) d∆
=
∫∆
[4L22L
23 − 2(L32L2 + L2
2L3) + L2L3] d∆
=
[4
2! 2!
6!− 2
2! + 2!
5!+
1
4!
]2 180 = − 0.6∫
∆
N2N4 d∆ =
∫∆
[8 L32L3 − 4 L2
2L3] d∆
=
[8
3!
6!− 4
2!
5!
]2 180 = 0 .
(b) The polynomials to be integrated involve quartic terms. Consequently, from Table5.3 we need to use the 7-point formula to obtain correct answers. Using the valuesfrom the table the integrals are given by∫
∆
N2N3 d∆ = 1807∑
l=1
L2(l) (2L2(l)− 1) L3(l) (2L3(l)− 1)w(l) = −0.60000075
∫∆
N2N4 d∆ = 1807∑
l=1
L2(l) (2L2(l)− 1) 4 L2(l) L3(l) w(l) = 0 .
The first integral gives a small error due to the finite precision of the numbers givenin the table. The second integral will give a very small number – say O(10−8).(To avoid tedius sums the solution should be carried out by performing the sumwith a small computer program.)
5.10 For the triangular element shown in the figure the integrals required are given by∫∆Na d∆ =
∫∆
(2L2a − La) d∆ ; for a = 1, 2, 3∫
∆Na d∆ =
∫∆
4La−3Lb−3 d∆ ; for a = 4, 5, 6
In the above b is the cyclic permutation 5, 6, 4 for a = 4, 5, 6. We only need to computetwo of the integrals.
67
(a) Using Eq (4.31) the integral for a vertex node becomes∫∆
Na d∆ =
∫∆
(2L21 − L1) d∆
= 22!
4!− 1
3!2 180 = 0 .
The second integral is given by∫∆
Na d∆ =
∫∆
4L1L2 d∆ = 41
4!2 180 = 60 .
Thus the vertex nodes have no values while the mid-edge nodes receive one-thirdof the area.
(b) For this case the integrals involve quadratic terms only, hence the 3-point formulain Table 3 is adequate to compute the integral. Accordingly,∫
∆
(2L21 − L1) d∆ = ∆
[2
(12
)2 − 12
+ 2(
12
)2 − 12
]13
= 0 .
The second integral is given by∫∆
4L1L2 d∆ = ∆∑
l
[4 L1(l)L2(l) w(l)] = ∆ 4 12
12
13
= 13
∆ = 60.
In the present case we are able to perform the sums very easily due to the locationof the points and the values for the weight as a fraction.
x
y
6
3
3
1 2
34
Quadrilateral for Problem 5.11
5.11 A 4-node quadrilateral element is used in solve a problem in which the dependent vari-able is a scalar, u. For the geometry shown in the figure we can write the isoparametricmap using the bilinear shape functions
Na(ξ, η) = 14
(1 + ξaξ) (1 + ηaη) .
68
(a) Inserting the shape functions and noting the location of the nodal points theexpression for the coordinates is given by
x = 14
(1 + ξ) (9− 3 η)
y = 32
(1 + η) .
(b) The area of the element is A = 13.5. The location of the centroid is given by
A x = 9 · 1.5 + 4.5 · 4 = 31.5A y = 9 · 1.5 + 4.5 · 1 = 18.0
which gives
x = 31.513.5
= 73
and y = 1813.5
= 43.
Using the expressions for the coordinates we have
y = 43
= 32
(1 + η)
x = 73
= 14
(1 + ξ) (9− 3 η) .
Solving gives
η = − 19
and ξ = 0 .
(c) The expression for the jacobian transformation J of the element is given by
J =
∂x
∂ξ
∂y
∂ξ∂x
∂η
∂y
∂η
=
14
(9− 3 η) 0
−34
(1 + ξ) 32
Thus, at the centroid the value is given by
J = 14
[9 0
−3 6
].
(d) The derivatives of the shape function N3 with respect to the parent coordinatesξ and η are given by
∂N3
∂ξ= 1
4(1 + η) and
∂N3
∂η= 1
4(1 + ξ) .
The derivatives with respect to x and y are given by
J
∂N3
∂x∂N3
∂y
=
∂N3
∂ξ∂N3
∂η
.
69
Substituting the values for the jacobian and computing the inverse times the parentderivatives gives
∂N3
∂ξ∂N3
∂η
=1
6(9− 3η)
[6 0
3(1 + ξ) (9− 3η)
] (1 + η)
(1 + ξ)
=
1
(9− 3η)
(1 + η)
2(1 + ξ)
.
Evaluating the derivative at the centroid gives∂N3
∂ξ∂N3
∂η
=1
42
4
9
.
x
yη
ξ
1 2
3 = 4
a
b
Degenerate triangle for Problem 5.12
5.12 The triangular element formed by degenerating a 4-node quadrilateral element as shownin the figure has nodal coordinates
xa =[10, 30, 10
]ya =
[8, 8, 38
].
The three degenerate shape functions are given by
N1 = 14
(1− ξ) (1− η) ; N2 = 14
(1 + ξ) (1− η) and N3 = 12
(1 + η) .
(a) The expressions for x and y in terms of ξ and η are given by
x = 14
(1− η) (10− 10ξ + 30 + 30ξ) + 12
(1 + η) 10
= 15− 5 η + 5 ξ (1− η)y = 1
4(1− η) (1− ξ + 1 + ξ) 8 + 1
2(1 + η) 38
= 23 + 15 η .
70
(b) Using the above coordinates the jacobian J(ξ, η) for the element is given by
J =
∂x
∂ξ
∂y
∂ξ∂x
∂η
∂y
∂η
=
5 (1− η) 0
−5 (1− η)
.
(c) The jacobian determinant is given by
J(ξ, η) = 75 (1− η) .
(d) If we define a one-point quadrature formula by
I =
∫ 1
−1
∫ 1
−1
f(ξ, η) dξ dη = f(ξi, ηi)Wi
then to exactly integrate the jacobian J and also any integral of a constant wehave ∫ 1
−1
∫ 1
−1
1 dξ dη = 4 and
∫ 1
−1
∫ 1
−1
η dξ dη = 0 .
Thus using quadrature we have∫ 1
−1
∫ 1
−1
1 dξ dη = 1 W1 = 4∫ 1
−1
∫ 1
−1
η dξ dη = η1 W1 = 0
which has the solution η1 = 0 and W1 = 4 (which is identical to the single pointquadrature for any quadrilateral element also). While there are no integrals in ξwe also would use ξ1 = 0 for any case where linear terms existed, such as in bodyforces.
(e) The physical point in the element is obtained by substituting the quadrature pointlocation into the expressions for x and y. This gives
x = 15 and y = 23 .
If we used area coordinates instead we would have
x =3∑
a=1
La xa and y =3∑
a=1
La ya
with the one point integral location at (L1, L2, L3) = (1/3, 1/3, 1/3) to give thephysical location at
x = 13
3∑a=1
xa = 50/3 6= 15 and y = 13
3∑a=1
ya = 64/3 6= 23
and thus the point is different. The reason for the difference is the variablejacobian arising from the ξ, η form as opposed to the constant one in the areacoordinate form.
71
5.13 Gauss–Lobatto quadrature is defined by∫ 1
−1
f(ξ) dξ = [f(−1) + f(1)]W0 +n∑1
f(ξn)Wn
and may be used to integrate exactly polynomials of the form
f(ξ) =1, ξ, ξ2, . . . , ξn
∫ 1
−1
f(ξ) dξ =
2, 0,
2
3, . . . ,
1
n+ 1[(1)n+1 − (−1)n+1]
If we substitute the function into the quadrature formula and compare to exact answerswe can find the points. Consider:
(a) Starting with the three point formula we have∫ 1
−1
f(ξ) dξ = [f(−1) + f(1)]W0 + f(ξ1)W1 .
Inserting our functions up to the order where only 3 unique equations exist weobtain
∫ 1
−1
1ξξ2
ξ3
dξ =
20
2/30
=
(1 + 1) W0 + 1 W1
(−1 + 1) W0 + ξ1 W1 = ξ1 W1
(1 + 1) W0 + ξ1 (ξ1 W1) = 2 W0
(−1 + 1) W0 + (ξ1)2 (ξ1 W1) = 2 W0
From the integral of ξ2 we obtain W0 = 1/3 then from the integral or 1we findW0 = 4/3. Finally, both of the other equations are satisfied for ξ1 = 0. Werecognize the result as the classical Simpson’s rule. If we try any higher orderpolynomial the quadrature formula produces an error.
(b) For the four point formula we have∫ 1
−1
f(ξ) dξ = [f(−1) + f(1)]W0 + f(ξ1)W1 + f(ξ2)W2 .
Repeating the process for the polynomials up to ξ5 we have the equations
2 = 2 W0 +W1 +W2
0 = ξ1 W1 + ξ2 W2
2/3 = 2 W0 + ξ1(ξ1W1) + ξ2(ξ2W2)
0 = ξ21(ξ1W1) + ξ2
2(ξ2W2)
2/5 = 2 W0 + ξ31(ξ1W1) + ξ3
2(ξ2W2) .
It is evident that these equations are non-linear, however we can solve them exactly.Inserting the second equation into the fourth gives
(ξ21 − ξ2
2) (ξ1W1) = 0
72
and thus the quadrature point is symmetrically located around ξ = 0 and we mayuse now ξ2 = −ξ1 in the remaining equations. From the second equation we find thatW1 = W2. Now the remaining three equations may be rewritten in the remainingunknowns as (after deleting the common factor 2)
1 = W0 +W1
1/3 = W0 + ξ21W1
1/5 = W0 + ξ41W1 .
These are now easily solved by substituting W0 = 1−W1 in the last two equations andtaking their ratio to obtain
4
5
3
2=
6
5=
(1− ξ41) W1
(1− ξ21) W1
= 1 + ξ2 .
The solution gives ξ1 = −ξ2 = 1/√
5. Substituting this into the equations givesW0 = 1/6 and W1 = 5/6. Thus the final points and weights for the quadrature are
ξl =[−1, − 1/
√5, 1/
√5, 1
]Wl = 1
6
[1, 5, 5, 1
].
4
1
2 3
π/4
Problem 5.14: Blending function 3× 3 mesh.
5.14 The region to be defined by a blending function mapping is shown in the figure as solidlines. The procedure to define a blending function mapping is defined in Sec. 5.13and shown schematically in Fig. 5.20. It is assumed that the curved edge is located atξ = 1. Thus, starting from Eq. (5.57) we write the blending function mapping as
x = Pη x+ Pξ x− PξPηx
y = Pη y + Pξ y − PξPηy .
73
We need to define each of the interpolations. Starting from the last term which ismerely the 4-node bilinear function we have
PξPηx = a 14
[(1− ξ)(1− η) 1√
2+ (1 + ξ)(1− η) + (1 + ξ)(1 + η) 2 + (1− ξ)(1 + η)
√2
]PξPηy = a 1
4
[(1− ξ)(1− η) 1√
2+ (1− ξ)(1 + η)
√2
]=
a
4√
2(1− ξ)(3 + η) .
Next we consider Pξx and Pξy which are defined by
Pξx = 12
[(1− ξ) x(−1, η) + (1 + ξ) x(1, η)]
Pξy = 12
[(1− ξ) y(−1, η) + (1 + ξ) y(1, η)] .
However on the edges ξ = ±1 the boundaries are straight lines and defined by thelinear interpolations
x(−1, η) = 12
[(1− η) a+ (1 + η) 2a = 32a
y(−1, η) = 12
[(1− η) · 0 + 2 (1 + η) · 0 = 0
and
x(1, η) = 12
[(1− η) a√2
+ (1 + η)
√2a
2=
a
2√
2(3 + η)
y(1, η) = x(1, η)
Combining all the parts to define Pξ we find that
Pξx ≡ PξPηx and Pξy ≡ PξPηy .
Thus, the final interpolation is given by Pηx and Pηy defined by
Pηx = x(ξ,−1) 12
(1− η) + x(ξ, 1) 12
(1 + η)
Pηy = y(ξ,−1) 12
(1− η) + y(ξ, 1) 12
(1 + η) .
At η = −1 we use the linear interpolations
x(ξ,−1) = 12
[(1− ξ) a+ (1− ξ) a√2]
y(ξ,−1) = 12
[(1− ξ) · 0 + (1− ξ) a√2]
= 12
(1− ξ) a√2.
At ξ = 1 we have a circular edge for which
θ(η) = 12
(1 + η)π
2
74
which, for −1 ≤ η ≤ 1, ranges from 0 ≤ θ ≤ π/2. Thus, an interpolation along theedge is given by
x(1, η) = 2a cos θ(η)
y(1, η) = 2a sin θ(η) .
Thus, the final interpolation is given by
x(ξ, η) = 12
[(1− ξ) x(−1, η) + (1 + ξ) x(1, η)]
y(ξ, η) = 12
[(1− ξ) y(−1, η) + (1 + ξ) y(1, η)]
with the functions defined above. The region may be divided into a 3 × 3 mesh byusing equal increments of the ξ and η defined at (1, −1/3, 1/3, 1). The final mesh isshown as dashed lines in the figure.
a/4 3a/4
b/4
3b/4
1
2
3
4
5
6
Problem 5.15: Quarter point singularity
5.15 The shape functions for a 6-node triangular element are given by
Na = La (2La − 1) ; for a = 1, 2, 3Na = 4 La−3 Lb−3 ; for a = 4, 5, 6
Consider an element with straight edges and nodes 4 and 6 placed at the quarter pointsas shown in the figure. The x and y coordinates are given by
x = a [L2(2L2 − 1) + L1L2 + L2L3]
= a[2L2
2 − L2 + L2(1− L2)]
= a L22
y = b [L3(2L3 − 1) + L2L3 + L3L1]
= b[2L3
2 − L3 + L3(1− L3)]
= b L23 .
We can solve these expressions for L2 and L3 to obtain
L2 =
√x
aand L3 =
√y
b
75
thus from L1 + L1 + L1 = 1 we have
L1 = 1−√x
a−
√y
a
The interpolation for a dependent variable u is given by
u =6∑
a=1
Na(Lc) ua
and we want to show that ∂u/∂x and ∂u/∂x at node 1 are of order 1/r1/2. (N.B. prob-lem statement gives 1/r2 which is not correct) From the interpolation for displacementswe may consturct the two dimensional form of equation (5.18) as
∂rx
∂x
∂rx
∂y∂ry
∂x
∂ry
∂y∂r1∂x
∂r1∂y
=
1 0
0 1
0 0
=
0 2aL2 0
0 0 2bL3
1 1 1
∂L1
∂x
∂L1
∂y∂L2
∂x
∂L2
∂y∂L3
∂x
∂L3
∂y
which solves to give
∂L1
∂x
∂L1
∂y∂L2
∂x
∂L2
∂y∂L3
∂x
∂L3
∂y
=1
4abL2L3
−2bL3 −2aL2
2bL3 00 2aL2
=
− 1
2√ax
− 1
2√by
1
2√ax
0
01
2√by
.
Substituting into the expression for ∂u/∂x we obtain
∂u
∂x= (4L1 − 1)
∂L1
∂xu1 + (4L2 − 1)
∂L2
∂xu2 + (4L3 − 1)
∂L3
∂xu3
+ 4 (L2∂L1
∂x+ L1
∂L2
∂x) u4 + 4 (L3
∂L2
∂x+ L2
∂L3
∂x) u5
+ 4 (L1∂L3
∂x+ L3
∂L1
∂x) u6
which when evaluated as L1 → 1 and L2, L3 → 0 gives
∂u
∂x→ 3
∂L1
∂xu1 + 4
∂L2
∂xu4 + 4
∂L3
∂xu6 = 3 u1 +
1
2√ax
(u4 − u1) .
Noting that along the x-axis we have x ≡ r yields the specified reciprocal square rootsingularity strength.
76
x
y
z1
7
413
5
14
6
9
3
11
2
8
10
12
15
a
b
c
Problem 5.16: Quadratic triangular prism geometry.
5.16 The quadratic triangular prism to be considered is shown in the figure. The particularwedge to be considered has straight edges in which the nodes are uniformly placed.Thus, the geometry can be defined by the subparametric interpolation
x =6∑
a=1
Na(L1, L2, L3, ζ) xa
y =6∑
a=1
Na(L1, L2, L3, ζ) ya .
The shape functions are defined by
Na = 12La (1 + ζaζ)
where parent and cartesian coordinates are defined in the table.
1 2 3 4 5 6L1 1 0 0 1 0 0L2 0 1 0 0 1 0L3 0 0 1 0 0 1ζ 1 1 1 -1 -1 -1xa 0 0 a 0 0 aya 0 b 0 0 b 0za c c c 0 0 0
With these definitions we have
x = a L3
y = b L1
z = 12c (1 + ζ)
77
We can define the 15 shape functions for interpolation of the variable u in an hierar-chical manner. Accordingly, the mid-edge functions are given by
Na = La−6 (1− ζ2) ; for a = 7, 8, 9Na = 2 La−9Lb−9 (1 + ζ) ; for a = 10, 11, 12Na = 2 La−12Lb−12 (1− ζ) ; for a = 13, 14, 15
in which the cyclic permutation for a = 1, 2, 3 gives b = 2, 3, 1. These edge functionsalso define the serendipity interpolations for the wedge. The linear interpolations atthe vertex nodes need to be corrected to define the final shape functions. For example
N1 = N1 − 12
(N7 +N10 +N12) .
To compute the x and y derivatives for the shape function of nodes 1, 7 and 10 we usethe chain rule
∂Na
∂x=∂Na
∂L1
∂L1
∂x+∂Na
∂L2
∂L2
∂x+∂Na
∂L3
∂L3
∂x+∂Na
∂ζ
∂ζ
∂x.
Following the procedure given in Eq. (5.17) we can write
rx = x−∑6
a=1Na xa = 0
ry = y −∑6
a=1Na ya = 0
rz = z −∑6
a=1Na za = 0
r1 = 1− L1 − L2 − L3 = 0
which differentiates to give
1 0 00 1 00 0 10 0 0
=
6∑a=1
∂Na
∂L1
xa
6∑a=1
∂Na
∂L2
xa
6∑a=1
∂Na
∂L3
xa
6∑a=1
∂Na
∂ζxa
6∑a=1
∂Na
∂L1
ya
6∑a=1
∂Na
∂L2
ya
6∑a=1
∂Na
∂L3
ya
6∑a=1
∂Na
∂ζya
6∑a=1
∂Na
∂L1
za
6∑a=1
∂Na
∂L2
za
6∑a=1
∂Na
∂L3
za
6∑a=1
∂Na
∂ζza
1 1 1 0
∂L1
∂x
∂L1
∂y
∂L1
∂z∂L2
∂x
∂L2
∂y
∂L2
∂z∂L3
∂x
∂L3
∂y
∂L3
∂z∂ζ
∂x
∂ζ
∂y
∂ζ
∂z
.
Using the definition for the coordinates assumed the 4 × 4 coefficient matrix givenabove simplifies to
A =
0 0 a 0b 0 0 00 0 0 c/21 1 1 0
78
which has the inverse
A−1 =
0 1/b 0 0−1/a −1/b 0 1
1/a 0 0 00 0 2/c 0
and thus
∂L1
∂x
∂L1
∂y
∂L1
∂z∂L2
∂x
∂L2
∂y
∂L2
∂z∂L3
∂x
∂L3
∂y
∂L3
∂z∂ζ
∂x
∂ζ
∂y
∂ζ
∂z
=
0 1/b 0−1/a −1/b 0
1/a 0 00 0 2/c
.
The derivatives are now easily computed. For N1 = L3 (1 + ζ)/2 we have
∂N1
∂x=∂N1
∂L3
∂L3
∂x+∂N1
∂ζ
∂ζ
∂x= 1
2(1 + ζ)
1
a+ 1
2L3 · 0 =
1
2a(1 + ζ)
∂N1
∂y=∂N1
∂L3
∂L3
∂y+∂N1
∂ζ
∂ζ
∂y= 1
2(1 + ζ) · 0 + 1
2L3 · 0 = 0 .
For N7 = L1 (1− ζ2) we have
∂N7
∂x=∂N7
∂L1
∂L1
∂x+∂N7
∂ζ
∂ζ
∂x= (1− ζ2)
1
a− 2 L1 ζ · 0 =
1
a(1− ζ2)
∂N7
∂y=∂N7
∂L1
∂L1
∂y+∂N7
∂ζ
∂ζ
∂y= (1− ζ2) · 0− 2 L1 ζ · 0 = 0 .
For N10 = 2 L1 L2 (1 + ζ) we have
∂N10
∂x=∂N10
∂L1
∂L1
∂x+∂N10
∂L2
∂L2
∂x+∂N10
∂ζ
∂ζ
∂x
= 2 L2 (1 + ζ) · 0 + 2 L1 (1 + ζ)−1
a+ 2 L1 L2 ζ · 0 = − 2
aL1 (1 + ζ)
∂N10
∂y=∂N10
∂L1
∂L1
∂y+∂N10
∂L2
∂L2
∂y+∂N10
∂ζ
∂ζ
∂y
= 2 L2 (1 + ζ) · 0 + 2 L1 (1 + ζ)−1
b+ 2 L1 L2 ζ · 0 = −2
bL1 (1 + ζ) .
79
For N12 = 2 L3 L1 (1 + ζ) we have
∂N12
∂x=∂N12
∂L1
∂L1
∂x+∂N12
∂L3
∂L3
∂x+∂N12
∂ζ
∂ζ
∂x
= 2 L3 (1 + ζ) · 0 + 2 L1 (1 + ζ)1
a+ 2 L1 L2 ζ · 0 =
2
aL1 (1 + ζ)
∂N12
∂y=∂N12
∂L1
∂L1
∂y+∂N12
∂L3
∂L3
∂y+∂N12
∂ζ
∂ζ
∂y
= 2 L3 (1 + ζ)1
b+ 2 L1 (1 + ζ) · 0 + 2 L1 L2 ζ · 0 ==
2
bL3 (1 + ζ) .
The final form for the derivatives of N1 are computed from
∂N1
∂x=∂N1
∂x− 1
2
(∂N7
∂x+∂N10
∂x+∂N12
∂x
)∂N1
∂y=∂N1
∂y− 1
2
(∂N7
∂y+∂N10
∂y+∂N12
∂y
).
5.17 Program development project: Extend the program system started in Problem 2.17 topermit mesh generation using as input a 4-node isoparametric block and mapping asdescribed in Sec. 5.13. The input data should be the coordinates of the block verticesand the number of subdivisions in each direction.
Include as an option generation of coordinates in r, θ coordinates that are then trans-formed to x, y cartesian form.
Hint: Once coordinates for all node points are specified, MATLAB©R can generate anode connection list for 3-node triangles using DELAUNAY.† A plot of the mesh may beproduced using TRIMESH.
Use your program to generate a mesh for the rectangular beam described in Example2.3 and the curved beam described in Example 2.4. Note the random orientation ofdiagonals which is associated with degeneracy in the Delaunay algorithm (viz. Chap.8).
5.18 Program development project: Extend the mesh generation scheme developed in Prob-lem 5.17 to permit specification of the block as a blending function. Only allow twocases: (i) Lagrange interpolation which is linear or quadratic; (ii) Circular arcs withspecified radius and end points.
Test your program for the beam problems described in Examples 2.3 and 2.4.
†In Chapter 8 we discuss mesh generation and some of the difficulties encountered with the Delaunaymethod.
80
Chapter 6: Solutions
6.1 The stress-strain relation for shear may be written as
τx′y′ = Gγx′y′ .
The stress and strain in the rotated x′− y′ coordinates are given by (6.23) and (6.24).Using these transformations for θ = 45o where the transformation array T is given by
T =1√2
1 1 0− 1 1 0
0 0√
2
.
we obtain
τx′y′ = 12(σy − σx) and γx′y′ = εy − εx .
Using the stress-strain relation (6.28) we obtain
σy − σx =E
d(1− 2ν) (εy − εx) =
E
1 + ν(εy − εx)
and thus we may write
τx′y′ =E
2 (1 + ν)γx′y′
which yields the desired result G = E/[2(1 + ν)].
6.2 Using (6.28) we obtain
p = (σx + σy + σz)/3 =E
3d(1 + ν) (εx + εy + εz) =
E
3(1− 2 ν)ε
thus from p = K εv we obtain the desired result K = E/[3(1− 2ν)].
6.3 For the strain displacement equations given by
ε =
εrr
εθθ
γrθ
=
∂ur
∂r1
r
∂uθ
∂θ+ur
r1
r(∂ur
∂θ− uθ) +
∂uθ
∂r
.
with displacements expressed as
ur =∑
n
un(r) cosnθ and uθ =∑
n
vn(r) sinnθ .
we can introduce a standard finite element representation and let:
81
(a) The displacement expansion for each harmonic be given by
un(r) =∑
a
Na(r) una and vn(r) =
∑a
Na(r) vna
thus, the strain-displacement matrix for a harmonic n is given by
Bna =
cosnθ 0 00 cosnθ 00 0 sinnθ
dNa
dr0
Na
r
nNa
r
− nNa
r
(dNa
dr− Na
r
)
= Tn(θ) Bn
a
(b) The stiffness matrix for a linear elastic material is deduced from
Kmnab =
∫r
(∫ 2π
0
BmTa DBn
b dθ
)r dr
=
∫r
BmTa
(∫ 2π
0
TTmDTn dθ
)Bn
b r dr .
The integrals are∫ 2 π
0dθ = 2π ; for m = n = 0∫ 2 π
0cosnθdθ = 2π ; for m = 0, n > 0∫ 2 π
0sinnθdθ = 2π ; for m = 0, n > 0∫ 2 π
0cosmθ cosnθ dθ =
π for m = n > 00 for m 6= n∫ 2 π
0sinmθ sinnθ dθ =
π for m = n > 00 for m 6= n∫ 2 π
0sinmθ cosnθ dθ = 0 ; for all m, n
Thus, the double sum reduces to a single sum with∑n
(Knnab un
b + fna ) = 0
with
Knnab =
∫r
BnTa DnBn
b dr
in which
Dn = πD for n > 0
= 2 πT0DT0 for n = 0 T0 =
11
0
.
82
6.4 Solution of spherically symmetric problems in one dimension use the strain-displacementrelations
εrr =∂ur
∂r; εθθ = εφφ =
ur
r
γrφ =∂uφ
∂r− uφ
r; γrθ =
∂uθ
∂r− uθ
r; γθφ = 0 .
(a) A rigid body mode can exist if: (a) a constant displacement causes no strain or(b) if a rotation based on a linear displacement causes no strain. From the abovestrain displacement relations we observe that
uθ = α r and uφ = β r
both yield zero strains in all components. Hence there are two rigid body modes.
(b) The expansion for displacements may be given by If we letur(r)uθ(r)uφ(r)
=∑
a
Na(r)
ua
r
uaθ
uaφ
where Na(r) are one-dimensional C0 shape functions and ua
r , uaθ , u
aφ are nodal
parameters.
(c) Inserting the above expansion of displacements into the strain-displacement rela-tions gives
ε =
εrr
εθθ
εφφ
γrθ
γrφ
=
dNa
dr0 0
Na
r0 0
Na
r0 0
0
(dNa
dr− Na
r
)0
0 0
(dNa
dr− Na
r
)
ua
r
uaθ
uaφ
The 5× 3 coefficient matrix defines the strain displacement matrix Ba.
(d) For a linear elastic material the stiffness matrix is given by the standard form
Kab =
∫Ω
BTa DBb dΩ
where, for isotropy the elasticity matrix is given by the 5× 5 form
D =E
d
(1− ν) ν ν 0 0ν (1− ν) ν 0 0ν ν (1− ν) 0 00 0 0 (1− 2ν)/2 00 0 0 0 (1− 2ν)/2
=
D11
GG
83
where d = (1 + ν)(1− 2ν), dΩ = 4π r2 dr and D11 is the upper 3× 3 block of D.Expanding the expression for the stiffness gives
Kab =
Krab
Kθab
Kφab
where
Krab =
∫Ω
[dNa
dr
Na
r
Na
r
]D11
dNb
drNb
rNb
r
dΩ
(e) For the ur displacement alone, for coordinates r1 ≤ r ≤ r2, the linear shapefunctions are given by
N1 =r2 − rr2 − r1
and N2 =r − r1r2 − r1
.
A one-point quadrature at r = (r1 + r2)/2 gives
N1 = N2 =1
2
dN2
dr= − dN1
dr=
1
h
where h = r2 − r1. The stiffness is given by
Ke = 4 π
−1
h
1
2
1
21
h
1
2
1
2
D11
− 1
h
1
h1
h
1
21
h
1
2
r2 h .
A correct rank of Ke is two. Numerical verification of the one-point form doesgive rank two.
6.5 For a linear elastic isotropic material the stiffness matrix may be computed by numer-ical integration using Eq. (6.68). Alternatively, the stiffness matrix may be computedin indicial form as indicated in Appendix B.
For plane strain the matrices involved are
Ba =
Na,x 00 Na,y
Na,y Na,x
and D =
D11 D12 0D12 D11 00 0 D33
where the third row of each array has been removed since all the operations wouldinvolve zeros. Assuming a 2 × 2 quadrature we will have the following operationsrepeated 4 times.
84
(a) Using Equation (6.68) we can compute first for each quadrature point l
D(ξl)← D ∗ J(ξl) ∗Wl
which requires one multiply to get J ∗Wl and 3 multiplies to obtain the remainingunique entries. Next we compute the product of DBb for each node, expandingthis becomes
Qb = D Bb =
D11B11 D12B22
D12B11 D22B22
D33B31 D33B32
which involves 6 multiplies, or for the 4-nodes a total of 24 multiplies. Finally,for each pair of a and b we need to compute BT
a Qb, expanding this becomes
Kab ← Kab + BTa + Qb
=
[Kab
11 +B11 ∗Q11 +B31 ∗Q31 Kab12 +B11 ∗Q12 +B31 ∗Q32
Kab21 +B22 ∗Q21 +B32 ∗Q31 Kab
22 +B22 ∗Q22 +B32 ∗Q32
]The inclusion of the Kab is necessary to account for the sum appearing in Eq.(6.68). For each nodal pair the above requires 8 multiplies and 8 additions. Wecan now summarize all the operations involved for each quadrature point.
Step Times Multiples AdditionsJ ∗W 1 1 0D ∗ J ∗W 1 3 0D Bb 4 6 0BT
a Qb 10 8 8Total 108 80
In the above we have noted symmetry of K and only computed the terms above orbelow the diagonal. The others may be formed using symmetry, thus Kba = KabT .Accounting for the 4 quadrature points gives a total of 432 multiplies and 320additions.
(b) We now repeat the above calculation by first computing integrals of the derivativesas
Wab =4∑
l=1
[Na,x
Na,y
] [Nb,x Nb,y
]Jl Wl .
This is followed by multiplying the result by the elasticity matrix D (which isconstant for the whole element) in terms of its two unique Lame parameters λand µ as given in Eq. (B.54) as
Kabik = λ W ab
ik + µ[W ab
ki + δik Wabjj
].
The operations are as follows: Compute first the product of c = J ∗ Wl foreach quadrature point. Then for each node compute the product with the shapefunctions as
Qb =[Nb,x ∗ c Nb,y ∗ c
]
85
which involves only 2 multiplies for each node. Next we compute the Wab as
Wab =
[W ab
11 +Na,x ∗Q1 W ab12 +Na,x ∗Q2
W ab21 +Na,y ∗Q1 W ab
22 +Na,y ∗Q2
]which involves 4 multiplies and 4 additions for each of the nodal pairs a and b(again, with symmetry there are 10 pairs). The assembly step for each nodal pairis given by, computing W ab = W ab
11 +W ab22 ,[
Kab11 Kab
12
Kab21 Kab
22
]=
[λ ∗W ab
11 + µ ∗ (W ab11 +W ab) λ ∗W ab
12 + µ ∗W ab21
λ ∗W ab21 + µ ∗W ab
12 λ ∗W ab22 + µ ∗ (W ab
22 +W ab)
]This involves 1 addition to get W ab and 8 multiplies plus 6 additions to do theassembly, thus for the 10 pairs there are 90 multiplies and 60 additions. The totaloperations involved are:
Step Times Multiples AdditionsJ ∗W 1 1 0NT
b ∗ c 4 2 0Na Qb 10 4 4Total 49 40
For the 4 quadrature points this gives a total of 196 multiplies and 160 additions.To this we must add the assembly operations giving a total of 286 multipliesand 220 additions. Thus, there is considerable savings using this approach.Note, however, that it only can be used when the D is constant in the element.
6.6 In the classical plane strain problem the strain normal to the plane of deformation(i.e., εz was assumed to be zero. The problem may be ‘generalized’ by assuming εz isconstant over the entire analysis domain. The constant strain may then be related toa resultant force Fz applied normal to the deformation plane.
(a) Following the steps given in Sec. 6.3, we can write the strain-displacement rela-tions as
ε =
εx
εy
εz
γxy
= Ba ua + εz
where all the arrays are defined in Eq. (6.57), however, now εz is a parametersuch that
δε = Ba δua + δεz .
Thus, the virtual work expression is given by∫Ω
(δuT
a BTa + δεT
z
)D (Bbub + εz) dΩ−
∫Ω
δuTaNa b dΩ−
∫Γt
δuTaNat dΓ−
∫Ω
δεz tz dΩ = 0 .
86
It is assumed the expressions are written for a unit thick slab (in the z direction)so that the relative displacement in the z direction is equal to εz. Since εz isconstant over the entire domain Ω (as is its variation). We have∫
Ω
tz dΩ = Fz
the force resultant on the z-face.
(b) To obtain a finite element form the evaluation of the remaining integrals gives thematrix problem [
Kab Ca
CTb Kzz
]ub
εz
=
faFz
where
Kab =∫
ΩBT
a D Bb dΩ Ca =∫
ΩBT
a D3 dΩKzz =
∫ΩD33 dΩ fa =
∫ΩNa b dΩ +
∫ΓtNat dΓ
(c) From the above form we find immediately that the force Fz is related to theconstant strain εz by
Fz = Kzz εz + CTb ub .
For a well posed problem we must specify either the resultant force or the amountof strain εz. We also see that if εz = 0 the above form reduces to the usualproblem for plane strain.
1 3 2
a
F
h/2 h/2
q
a
1 3 2
h/2 h/2
(a) Point loading (b) Hydrostatic loadingTraction loading on boundary for Problems 6.7 and 6.8.
6.7 Placing the origin of the x-axis at node 1, the shape functions for a quadratic edge asshown in the figure are given by Lagrange interpolation which may be written as
N1 =(2x− h)(x− h)
h2; N2 =
x(2x− h)h2
and N3 =4x(h− x)
h2.
The equivalent nodal forces are computed from
Fa =
∫ h
0
Na(x) t(x) d x
87
which for a single concentrated load placed at x = a gives
Fa = Na(a)F .
We can evaluate the shape functions to be
N1 =(2a− h)(a− h)
h2; N2 =
a(2a− h)h2
and N3 =4a(h− a)
h2
which after multiplying by F give the nodal forces
6.8 Equivalent nodal forces for triangular traction load.
(a) For the hydrostatic loading shown in the figure the traction is given by
t(x) =q(a− x)
afor 0 ≤ x ≤ a .
Evaluating the force relation
Fa =
∫ a
0
Na(x) t(x) d x
gives the nodal forces. Evaluating the integrals gives
F1 =qa (3h2 − 3 , ah+ a2)
6h2; F2 =
qa2 (a− h)6h2
and F3 =qa2 (2h− a)
3h2.
Summing forces gives∑
a Fa = q a/2 which is the correct resultant force. Settinga = h (triangular load over entire element) gives
F1 = 16qh ; F2 = 0 and F3 = 1
3qh
(b) To perform numerical integration we must have continuous functions over therange of ξ. Accordingly, we let −1 ≤ ξ ≤ 1 correspond to 0 ≤ x ≤ a so that
x = 12(1 + ξ) a .
Now, the integrals become∫ a
0
f(x) dx =
∫ 1
−1
f(ξ) 12a dξ .
Since the integrands for the nodal forces give f(x) which have polynomial terms upto cubic order, we need 2-point Gaussian quadrature and, thus, exact evaluationsoccur for ∫ 1
−1
f(ξ) 12a d = 1
2a
2∑l=1
f(ξl)wl .
88
The two points are obtained as ξl = ±1/√
3 with Wl = 1. Thus, the polynomialpoints xl are
xl = 12(1± 1√
3) .
Integrals are: ∫ 1
−11 dξ = 2
∫ 1
−1x2 dξ = 2
3a2∫ 1
−1x dξ = a
∫ 1
−1x3 dξ = 1
2a3 .
The force expressions are
F1 =q
2h2
2∑l=1
[−2x3
l + (3h+ 2a)x2l − (h2 + 3ah)xl + ah2
]=
q
2h2
[−a3 + (3h+ 2a) 2
3a2 − (h2 + 3ah)a+ 2ah2
]=
qa
6h2
[3h2 − 3ha+ a2
]F2 =
q
2h2
2∑l=1
[−2x3
l + (h+ 2a)x2l − ahxl
]=
q
2h2
[−a3 + (h+ 2a) 2
3a2 − aha
]=
q
6h2
[a3 + ha2
]F3 =
q
2h2
2∑l=1
[4x3
l − 4(h+ a)x2l + 4haxl
]=
q
2h2
[2a3 − 4(h+ a)a2 2
3+ 4haa
]=qa2
3h2[2h− a]
which are exact. If one-point integration (i.e., ξ1 = 0 and w1 = 2) is used theintegrals are, upon noting that x1 = a/2∫ 1
−11 dξ = 2
∫ 1
−1x2 dξ = 1
2a2∫ 1
−1x dξ = a
∫ 1
−1x3 dξ = 1
4a3 .
Thus the force expressions are
F1 =q
2h2
1∑l=1
[−2x3
l + (3h+ 2a)x2l − (h2 + 3ah)xl + ah2
]2
=q
2h2
[−1
2a3 + (3h+ 2a) 1
2a2 − (h2 + 3ah)a+ 2ah2
]=
qa
4h2
[2h2 − 3ha+ a2
]F2 =
q
2h2
1∑l=1
[−2x3
l + (h+ 2a)x2l − ahxl
]=
q
2h2
[−a3 + (h+ 2a) 1
2a2 − aha
]= − qa2
4h2[a− h]
F3 =q
2h2
1∑l=1
[4x3
l − 4(h+ a)x2l + 4haxl
]=
q
2h2
[a3 − 4(h+ a)a2 1
2+ 4haa
]=qa2
4h2[4h− 2a] .
89
The sum is correct, however, the distribution is in error.
x
y
13
2
c
F
a(ξ)
h h
Concentrated normal load on a curved boundary for Problem 6.9.
6.9 A concentrated force of F = 100N is applied normal to a circular arc of 2 θ = 30o withradius R = 10.
Let F = 100 N , R = 10 cm and θ = 15o. For ξ = 0, 0.25, 0.50, 0.75, 1.0 determine:
(a) If the force is approximated by a normal to quadratic interpolations given by:
x = (1 + ξ)h and y = (1− ξ2) c
then for each ξ the angle of the normal is given by αh = tan−1(−dy/dx). Fromthe interpolation we obtain dx = h dξ and dy = −2 ξ c. Using the values forξ =
[0.0, 0.25, 0.50, 0.75, 1.0
]we obtain the forces[
F hx
F hy
]=
[0.0, 6.5684, 13.0526, 19.3737, 25.4626
100.0, 99.7840, 99.1445, 98.1053, 96.7040
]The equivalent forces acting on nodes 1, 2 and 3 for this case will be:
ξ 0.0 0.25 0.50 0.75 1.00
F 1x 0.0 -0.6158 -1.6316 -1.8163 0.0000F 1
y 0.0 -9.3548 -12.3931 -9.1974 0.0000F 2
x 0.0 1.0263 4.8947 12.7140 25.4626F 2
y 0.0 15.5913 37.1792 64.3816 96.7040
F 3x 0.0 6.1579 9.7895 8.4760 0.0000F 3
y 100.0 93.5475 74.3584 42.9211 100.0
x-Tot 0.0 6.5684 13.0526 19.3737 25.4626y-Tot 100.0 99.7840 99.1445 98.1053 96.7040
The totals are also indicated in the table and give the correct applied load as acheck.
(b) The normal to the boundary (which is still approximate by the quadratic segment)is given by computing the angle each point makes as αc = tan−1(∆x/∆y) where
90
∆x is the horizontal distance from node 2 and ∆y is the height from the centerof the circular arc. These are given by ∆x = ξ h and ∆y = R− ξ2 c. Accordingly,at the same locations as in part (a) we obtain the force components[
F cx
F cy
]=
[0.0 6.4707, 12.9424, 19.4142, 25.8819
100.0, 99.7904, 99.1589, 98.0973, 96.5926
]The equivalent nodal forces using the normal to the circular boundary are thengiven by:
ξ 0.0 0.25 0.50 0.75 1.00
F 1x 0.0 -0.6066 -1.6178 -1.8201 0.0000F 1
y 0.0 -9.3554 -12.3949 -9.1966 0.0000F 2
x 0.0 1.0110 4.8534 12.7406 25.8819F 2
y 0.0 15.5923 37.1846 64.3764 96.5926F 3
x 0.0 6.0663 9.7068 8.4937 0.0000F 3
y 100.0 93.5535 74.3692 42.9176 0.0000
x-Tot 0.0 6.4707 12.9424 19.4142 25.8819y-Tot 100.0 99.7904 99.1589 98.0973 96.5926
(c) The error between the two forms is obtained by computing absolute value of thedifference between parts (a) and (b). A plot of the error in the force is shown inthe figure.
0 0.5 1 1.5 2 2.5 3−0.5
−0.4
−0.3
−0.2
−0.1
0
0.1
0.2
x−Coordinate
E =
Ffe
− F
ex −
Err
or
Fx
Fy
Error in forces for Problem 6.9
6.10 A mesh for a plane strain problem contains the quadratic order rectangular elementsshown in the figures. The elements are subjected to a constant body force b =(0,−ρ g)T where ρ is mass density and g is acceleration of gravity.
91
The nodal values for the body force are computed from
fa =
∫ 1
−1
∫ 1
−1
Na(ξ, η)b ab dξ dη
where ab is the jacobian of the 2a× 2b rectangle.
(a) The standard shape functions for the 8-node serendipity element are given by
Na(ξ, η) = 14(1 + ξaξ)(1 + ηaη)(ξaξ + ηaη − 1) for a = 1, 2, 3, 4
Na(ξ, η)12(1− ξ2)(1 + ηaη) for a = 5, 7
Na(ξ, η)12(1 + ξaξ)(1− η2) for a = 6, 8
The nodal forces require the evaluation of the integrals
Ia =
∫−1
1
∫−1
1Na(ξ, η) dξ dη .
and these become
Ia = 13
−1 ; for a = 1, 2, 3, 4
4 ; for a = 5, 6, 7, 8.
Thus, the body force vectors are:
Fa = −13ρ g ab
0−1
for a = 1, 2, 3, 4
Fa = −13ρ g ab
04
for a = 5, 6, 7, 8
x
y
a a
b
b
1 2
34
5
7
68x
y
a a
b
b
1 2
34
5
7
689
Quadrilateral 8 and 9 node elements for Problems 6.10 and 6.11.
92
For the 9-node lagrangian element the shape functions are products of 1-d La-grange interpolations and thes may be written as
Na(ξ, η) = 14(ξ2 + ξaξ)(η
2 + ηaη) for a = 1, 2, 3, 4
Na(ξ, η) = 12(1− ξ2)(η2 + ηaη) for a = 5, 7
Na(ξ, η) = 12(ξ2 + ξaξ)(1− η2) for a = 6, 8
N9(ξ, η) = (1− ξ2)(1− η2)
These have the integrals
Ia = 19
1 ; for a = 1, 2, 3, 44 ; for a = 5, 6, 7, 816 ; for a = 9
.
Thus, the nodal forces are given by
Fa = −19ρ g ab
01
for a = 1, 2, 3, 4
Fa = −19ρ g ab
04
for a = 5, 6, 7, 8
Fa = −19ρ g ab
016
for a = 9 .
(b) Use hierarchical shape functions for Na and develop a closed form expression forthe nodal forces in terms of a, b and ρ g.
6.11 The thermal behavior for a linear elastic material is given by
σ = D (ε−α ∆T )
thus, ε0 = α ∆T and we may use the relations given in Eq. (6.62) where
f ea = −
∫Ωe
BTa D α ∆T dΩ .
For an elastic material we can define a thermal stress array as
β = D α
and for an isotropic material obtain
β = 3 K α m
where m =[1, 1, 1, 0
]T. Thus, the thermal force for a constant ∆T is given by
f ea = −3 K α ∆T
Na,x
Na,y
dΩ .
93
For the elements shown the Cartesian coordinates may be expressed in terms of parentcoordinates as
x = a ξ and y = b η .
Thus derivatives are trivial given by
∂
∂x=
1
a
∂
∂ξand
∂
∂y=
1
b
∂
∂η.
Also, the jacobian is given simply as dx dy = ab dξ dη so that the integrals neededbecome
f ea = −3 K α ∆T
b Na,ξ
a Na,η
dξ dη .
It only remains to compute the derivatives for the shape functions and evaluate theresulting integrals.
(a) The standard shape functions for the 8-node element shown in part (a) of thefigure are given by
Na(ξ, η) = 14(1 + ξaξ)(1 + ηaη)(ξaξ + ηaη − 1) for a = 1, 2, 3, 4
Na(ξ, η)12(1− ξ2)(1 + ηaη) for a = 5, 7
Na(ξ, η)12(1 + ξaξ)(1− η2) for a = 6, 8
Derivatives are given by
Na,ξ = 14ξa (1 + ηaη) [2 ξaξ + ηaη]
Na,η = 14ηa (1 + ξaξ) [(ξaξ + 2 ηaη]
for nodes a = 1, 2, 3, 4. For nodes 5 and 7 we obtain
Na,ξ = −ξ (1 + ηaη)
Na,η = 12ηa (1− ξ2)
and for nodes 6 and 8
Na,ξ = 12ξa (1− η2)
Na,η = −η (1 + ξaξ) .
Performing the integrals gives the results
f ea = K α ∆T
ξa bηa a
; for a = 1, 2, 3, 4
f ea = 4K α ∆T
0ηa a
; for a = 5, 7
f ea = 4K α ∆T
ξa b0
; for a = 6, 8
.
Use standard shape functions for Na and develop a closed form expression for thenodal forces in terms of a, b and the elastic properties.
94
(b) Use hierarchical shape functions for Na and develop a closed form expression forthe nodal forces in terms of a, b and the elastic properties.
6.12 Use the program FEAPpv (or any other available program) to solve the rectangularbeam problem given in Example 6.4 and verify the results shown in Table 6.2.
6.13 Use the program FEAPpv (or any other available program) to solve the curved beamproblem given in Example 6.5 and verify the results shown in Table 6.3.
x
y
q0
h
L
Uniformly loaded cantilever beam for Problem 6.14.
6.14 The uniformly loaded cantilever beam shown in the figure has properties
L = 2 m ; h = 0.4 m ; t = 0.05 m and q0 = 100 N/m.
Use FEAPpv or any other available program to perform a plane stress analysis of theproblem assuming linear isotropic elastic behavior with E = 200 GPa and ν = 0.3.
In your analysis:
(a) The mesh for quadratic lagrangian elements is given by
feap * * Problem 6.14 - Uniform load on a beam0 0 0 2 2 9
parametersf = 1 ! Multiplier on mesh sizen1 = 2*fn2 = 10*fh = 0.4l = 2.0t = 0.05q0 = 100s = 1.5*q0*L/hc = h/2e = 200e9nu = 0.3ty = 9
material
95
solidelastic isotropic e nuthickness slice tplane stress
blockcart n1 n2 0 0 0 0 ty
1 0 c2 0 -c3 l -c4 l c
eboun1 0 1 0
cbounnode 0 0 1 1
csurflinear ! Compute loads for q0
1 l c -q02 0 c -q0
tangential ! Compute loads for shearquadratic
1 0 c 02 0 -c 03 0 0 -s
endinterstop
(b) Using the CSURFace command, consistent nodal forces for the uniform loadingare computed by the program, however, students should compute at least one byhand. The shape functions for a quadratic edge are given by
N1 = 12ξ (ξ − 1)
N2 = 12ξ (ξ + 1)
N3 = (1− ξ2) .
For the top we let a local origin for each element be located at the mid-side edge(node 3), thus, the nodal forces are vertical (in the y-direction) and given by
f ea = − 1
2q0 h
∫ 1
−1
Na dξ
= −12
100 0.4
∫ 1
−1
12ξ(ξ − 1)
12ξ(ξ + 1)(1− ξ2)
dξ = −dfrac203
114
.
(c) Similarly to the previous part, nodal forces for a parabolically distributed sheartraction at the restrained end which balance the uniform loading are computed
96
by the program using the parameters specified. However, here we show the nodalforces for the 1×5 element mesh. The total shear force at the left end is 100N/m×2m = 200N . The edge average traction times thickness is 200/0.4 = 500N/m.Assuming the end shear is parabolic and given by s (1 − ξ2) we obtain s =3(500)/2 = 750N/m. Thus, the nodal foces are obtained from
f ea = 1
2s h
∫−1
1Na (1− ξ2) dξ
=300
15
118
=
2020160
.
(d) The centerline displacement in the vertical direction at each node is given by
x-Coord v-displ.0.0 0.0000e+000.2 -9.6962e-080.4 -3.1261e-070.6 -6.2208e-070.8 -1.0031e-061.0 -1.4357e-061.2 -1.9040e-061.4 -2.3929e-061.6 -2.8954e-061.8 -3.4006e-062.0 -3.9054e-06
The stored energy in the beam is 1.6017288e-04 (= uKu/2).
(e) Repeat the analysis 3 additional times using meshes of 2× 10, 4× 20 and 8× 40elements. The tip vertical displacement and stored energy for each solution isgiven in the table (FEAPpv prints the work which is twice the stored energy).
Mesh v-displ. 2× E1× 5 -3.9054e-06 3.203457590e-04
2× 10 -3.9267e-06 3.231039817e-044× 20 -3.9285e-06 3.233425011e-048× 40 -3.9286e-06 3.233649763e-04
16× 80 -3.9286e-06 3.233671141e-0432× 160 -3.9286e-06 3.233673062e-04
(f) If the energy error is given by
∆E = En − En−1 = C hp
a solution may be constructed by taking the (natural) logarithm of the relationto write
log ∆E = logC + p log h
97
and writing the set of linear equations in terms of logC and p as
log ∆E = b = [1g] logCp = A x
where g = log h. Since there are more equations than unknowns, using a leastsquares relation gives the solution from
ATA x = ATb .
Using the solutions above to estimate C and p gives logC = −5.57 or C = 0.0038and p = 3.44. Expected p was 4 for an element with full quadratic terms.
(a) 2× 10 Mesh
10−1 10010−10
10−9
10−8
10−7
10−6
10−5
h/h1
Ener
gy In
crem
ent −
| E n −
En−
1|
(b) Incremental energyMesh and energy behavior for Problem 6.14.
6.15 A circular composite disk is restrained at its inner radius and free at the outer radius.The disk is spinning at a constant angular velocity ω as shown in part (a) of the figure.The disk is manufactured by bonding a steel layer on top of an aluminum layer asshown in part (b) of the figure.
When spinning at an angular velocity of 50 RPM it is desired that the top surface beflat. This will be accomplished by milling the initial shape of the top to a specifiedlevel. Your task is to determine the profile for milling. To accomplish this
98
A A
15 85
x
y
10
10
(a) Plan of disk (b) Cross-section A–ASpinning composite disk for Problem 6.15.
(a) An analysis for an initially flat top surface is performed using the dimensionsgiven in the figure (lengths given in mm.) The elastic properties for steel areE = 200 GPa, ν = 0.3 and ρ = 7.8 µg/mm3; those for aluminum are E = 70 GPa,ν = 0.35 and ρ = 2.6 µg/mm3 (where µ = 10−6). Consistent units in mm, sec.,and µg are used for the properties. With these units the elastic properties havethe same numerical values as in pascals. Other units are obtained by droppingthe µ. Finally, ω is converted to radians per second as 50× 2 π/60.
The inner radius of the disk is be restrained in the radial direction (i.e., u(15, z) =0) and xxial restraint by v(15, 0) = 0.
(b) Using the results for the vertical displacements computed in (a), the locationof the nodal coordinates at the top of the disk are repositioned using the userfunction:
subroutine umacr1(lct,ctl,prt)
c * * F E A P * * A Finite Element Analysis Program
c.... Copyright (c) 1984-2005: Robert L. Taylorc All rights reservedc-----[--.----+----.----+----.-----------------------------------------]c Purpose: User function to reposition nodal coordinates and toc output current values.c Inputs:c lct - Command character parametersc ctl(3) - Command numerical parametersc prt - Flag, output if truec Outputs:c - new position for coordinatesc-----[--.----+----.----+----.-----------------------------------------]
99
implicit none
include ’iofile.h’include ’umac1.h’include ’pointer.h’include ’comblk.h’logical pcomp,prtcharacter lct*15integer ndir,ndofreal*8 ctl(3), xval
save
c Set command wordif(pcomp(uct,’mac1’,4)) then ! Usual form
uct = ’xadj’ ! Coordinate adjustelseif(urest.eq.1) then ! Read restart dataelseif(urest.eq.2) then ! Write restart dataelse ! Perform user operation
ndof = nint(ctl(1))ndir = nint(ctl(2))xval = ctl(3)if(pcomp(lct,’out’,3)) then
call xoutput(hr(np(43)),hr(np(40)), ndof,ndir,xval)else
call xadjust(hr(np(43)),hr(np(40)), ndof,ndir,xval)endif
endif
end
subroutine xadjust(x,u, ndof,ndir,xval)
implicit none
include ’cdata.h’include ’sdata.h’integer ndof,ndir, nreal*8 x(ndm,*),u(ndf,*), xval,sign, tol
data tol /1.d-08/
if(ndof.gt.0) thensign = 1.0d0
elsesign = -1.0d0
endifndof = abs(ndof)do n = 1,numnp
if(abs(x(ndir,n) - xval).lt.tol) thenx(ndir,n) = x(ndir,n) - sign*u(ndof,n)
endifend do ! n
100
end
(c) Performing a new analysis with the coorinates repositioned and outputing thefinal positions using the user function with subprogram:
subroutine xoutput(x,u, ndof,ndir,xval)
implicit none
include ’cdata.h’include ’sdata.h’include ’iofile.h’integer ndof,ndir, nreal*8 x(ndm,*),u(ndf,*), xval,sign, tol
data tol /1.d-02/
if(ndof.gt.0) thensign = 1.0d0
elsesign = -1.0d0
endifndof = abs(ndof)do n = 1,numnp
if(abs(x(ndir,n) - xval).lt.tol) thenwrite(iow,2000) n, x(ndir,n) + sign*u(ndof,n)
endifend do ! n
2000 format(’node = ’,i6,’ Level = ’,1p,1e19.10)end
gives deformed positions to about 9 digits of accuracy. The solution commandsto achieve the solution are given by:
batchtang,,1xadj coords 2 2 10tang,,1xadj output 2 2 10
Note that a tolerance of 1.d-02 is used to find the location of coordinates on thenew top surface.
Any additional improvement should involve an analysis that includes finite de-formation effects (i.e., change in geometry and stress effects). Then an iterativemethod involving the successive correction of initial shape can be performed.
6.16 A rectangular region with a circular hole is shown in the figure. The traction on thecircular hole is zero. The region is used for the solution of an infinitely extending planestress problem in which the stress at infinity is given by a uniformly distributed normalstress σ0 acting in the x-direction. To match the infinite plate using a finite region,
101
x
y
RA B
CD
E
θ
r
Rectangular region with circular hole for Problem 6.16.
the traction along the boundary BCD is subjected to traction loads from the exactsolution given by
σr = 12σ0
[1− (
a
r)2
]+
[1 + 3(
a
r)4 − 4 (
a
r)2
]cos 2θ
σθ = 1
2σ0
[1 + (
a
r)2
]−
[1 + 3(
a
r)4
]cos 2θ
τrθ = −1
2σ0
1− 3(
a
r)4 + 2(
a
r)2
sin 2θ
In addition, to obtain the energy stored in the region the displacements given by
ur =σ0r
2E
[1 + (
a
r)2
]+
[1− (
a
r)4 + 4(
a
r)2
]cos 2θ + ν
[1− (
a
4)2
]− ν
[1− (
a
r)4
]cos 2θ
uθ =
σ0r
2E
[1 + (
a
r)4 + 2(
a
r)2
]+ ν
[1 + (
a
4)4 − 2(
a
r)2
]sin 2θ
are used to compute
E =
∫BCD
(u tx + v ty) t dΓ .
which is twice the stored energy in a slice of thickness t.
A subprogram for FEAPpv is written to compute the consistent nodal forces and Eon the boundary BCD and is given by
subroutine elmt04(d,ul,xl,ix,tl,s,r,ndf,ndm,nst,isw)
c * * F E A P * * A Finite Element Analysis Programc - - - -c.... Copyright (c) 1984-2005: Robert L. Taylorc All rights reservedc-----[--.----+----.----+----.-----------------------------------------]c Purpose: Plane boundary stress loading for circular hole problem
c Control data:
102
c nel = 2: 1 o - - - - - - - o 2
c nel = 3 1 o - - - o - - - o 2c 3c nel = 4 1 o - - o - - o - - o 2c 3 4c-----[--.----+----.----+----.-----------------------------------------]
implicit none
include ’eldata.h’include ’iofile.h’
character text*15logical errck, tinput, pcompinteger ndf,ndm,nst,isw, i, ix(*)real*8 d(*),xl(ndm,*),tl(*),ul(ndf,*),s(nst,*),r(ndf,*),td(2)
c Output descriptorif(isw.eq.0 .and. ior.lt.0) then
write(*,*) ’ Elmt04: Boundary loading’
c Input/Output material set dataelseif(isw.eq.1) then
text = ’start’do while(.not.pcomp(text,’ ’,4))
errck = tinput(text,1,td,2)if (pcomp(text,’stre’,4)) then
d(1) = td(1)elseif(pcomp(text,’radi’,4)) then
d(2) = td(1)elseif(pcomp(text,’quad’,4)) then
d(3) = td(1)elseif(pcomp(text,’elas’,4)) then
d(4) = td(1)d(5) = td(2)
endifend do ! while
c Output problem parameterswrite(iow,2000) d(1),d(2),nint(d(3)), d(4),d(5)
c Compute residual and tangentelseif(isw.eq.3 .or. isw.eq.6) then
call pboun2h(d, xl, r, ndf,ndm, isw)
c Compute output and energyelseif(isw.eq.2) then
call pboun2h(d, xl, r, ndf,ndm, isw)endif
c Format statement2000 format(5x,’2-D B o u n d a r y L o a d i n g’//
& 10x,’Axial stress :’,1p,1e12.4/
103
& 10x,’Radius of hole :’,1p,1e12.4/& 10x,’Quadrature :’,i8/& 10x,’Elastic Modulus :’,1p,1e12.4/& 10x,’Poisson Ratio :’,1p,1e12.4/)end
subroutine pboun2h(d, xl,p, ndf,ndm, isw)
implicit none
include ’eldata.h’include ’elflag.h’include ’iofile.h’
integer ndf,ndm, isw, i, l,l1real*8 d(*),xl(ndm,*),p(ndf,*)real*8 sg(2,10),shp(2,5),sig(4), t(2), u(2)real*8 nx,ny, xx,yy, rr,th, engy
c Set start value for energyif(n.eq.elstart(ma)) then
engy = 0.0d0endif
c Set quadrature orderl1 = nint(d(3))call int1dg(l1,sg)
c Do quadrature of sidedo l = 1,l1
c Form shape functions and natural derivativescall shp1dn(sg(1,l),shp,nel)
c Compute direction cosines times ds and coordinatesnx = 0.0d0ny = 0.0d0xx = 0.0d0yy = 0.0d0do i = 1,nel
nx = nx + shp(1,i)*xl(2,i)ny = ny - shp(1,i)*xl(1,i)xx = xx + shp(2,i)*xl(1,i)yy = yy + shp(2,i)*xl(2,i)
end do ! i
c Scale by quadrature weightnx = nx*sg(2,l)ny = ny*sg(2,l)
c Compute stresses: sig_x, sig_y, tau_xyrr = sqrt(xx*xx + yy*yy)th = atan2(yy,xx)
104
call shole2d(d, rr,th, sig)call dhole2d(d, rr,th, u)
c Compute traction and energyt(1) = nx*sig(1) + ny*sig(4)t(2) = ny*sig(2) + nx*sig(4)if(isw.eq.2) then
engy = engy + t(1)*u(1) + t(2)*u(2)write(iow,*) ’ X,U,T =’,xx,yy,u,t
c Compute nodal forceselse
do i = 1,nelp(1,i) = p(1,i) + shp(2,i)*t(1)p(2,i) = p(2,i) + shp(2,i)*t(2)
end do ! iendif
end do ! l
c Output energy for last elementif(isw.eq.2. and. n.eq.ellast(ma)) then
write( *,2000) engywrite(iow,2000) engy
endif
c Format2000 format(5x,’Energy for region =’,1p,1e25.15/)
end
subroutine shole2d(d, rr,th, sig)
implicit nonereal*8 d(*), rr,th, sig(4)real*8 ar,ar2,ar4, sig0,sigr,sigt, taur, cn2,sn2
c Form geometric factors (a/r) and stress coefficientar = d(2)/rrar2 = ar*arar4 = ar2*ar2sig0 = d(1)*0.5d0
c Trigonometric functions for double anglecn2 = cos(2.d0*th)sn2 = sin(2.d0*th)
c Polar coordinate stressessigr = sig0*(1.d0 - ar2 + (1.d0 + 3.d0*ar4 - 4.d0*ar2)*cn2)sigt = sig0*(1.d0 + ar2 - (1.d0 + 3.d0*ar4)*cn2)taur = -sig0*(1.d0 - 3.d0*ar4 + 2.d0*ar2)*sn2
c Cartesian coordinate stressessig(1) = 0.5d0*((sigr+sigt) + (sigr - sigt)*cn2) - taur*sn2
105
sig(2) = 0.5d0*((sigr+sigt) - (sigr - sigt)*cn2) + taur*sn2sig(3) = 0.0d0sig(4) = 0.5d0*(sigr - sigt)*sn2 + taur*cn2
end
subroutine dhole2d(d, rr,th, u)
implicit nonereal*8 d(*), rr,th, u(2)real*8 ar,ar2,ar4, c1,c2, sig0,ur,ut, cn2,sn2
c Form geometric factors (a/r) and stress coefficientar = d(2)/rrar2 = ar*arar4 = ar2*ar2sig0 = d(1)*0.5d0
c Form displacement coefficientsc1 = sig0*rr/d(4)c2 = d(5)*c1
c Polar coordinate displacementsur = c1*(1.d0 + ar2 + (1.d0 - ar4 + 4.d0*ar2)*cos(2.d0*th))
& - c2*(1.d0 - ar2 - (1.d0 - ar4)*cos(2.d0*th))ut = -c1*(1.d0 + ar4 + 2.d0*ar2)*sin(2.d0*th)
& - c2*(1.d0 + ar4 - 2.d0*ar2)*sin(2.d0*th)
c Cartesian coordinate displacementsu(1) = ur*cos(th) - ut*sin(th)u(2) = ur*sin(th) + ut*cos(th)
end
A plane stress solution to the problem is computed using FEAPpv for the input data:hole radius R = 10 cm.; slice thickness t = 0.1R; elastic modulus E = 200 GPa.;Poisson ratio ν = 0.3 and σx(∞) = σ0 = 1. The boundary BC is placed at x = 30 cm.and CD at y = 30 cm. Assume a unit value for the stress σ0.
(a) Using a 4 × 8 mesh of 4-node quadrilateral elements as shown in part(a) of thefigure, the problem is solved. Subsequently, the mesh is subdivided in half andfive additional solutions are obtained.
(b) The displacements around the hole boundary are plotted in part (b) of the figurefor the 4× 8 mesh and in part(d) of the figure for the 8× 16 mesh (shown in part(c) of the figure). Results for the exact solution are also shown.
(c) The work performed by several meshes is given in the table and plotted in part(c) of the figure.
106
0 10 20 30 40 50 60 70 80 90−1
0
1
2x 10
−10
θ − angle
u −
dis
pla
cem
ents
Tension Strip − ν = 0.3
ur − radial
uθ − tangential
(a) 4× 8 Mesh (b) Displacements at hole
0 10 20 30 40 50 60 70 80 90−1
0
1
2x 10
−10
θ − angle
u −
dis
pla
cem
ents
Tension Strip − ν = 0.3
ur − radial
uθ − tangential
(c) 8× 16 Mesh (d) Displacements at holeMesh and displacements for 4-node elements for Problem 6.16.
Mesh Work (E)4× 8 4.8057892014824E-098× 16 4.8579696908747E-0916× 32 4.8756782891852E-0932× 64 4.8806241461361E-0964× 128 4.8819016628505E-09128× 256 4.8822238361990E-09
Exact 4.8823314976138E-09
(d) Results for energy behavior using 8-node serendipity elements are given in thetable and plotted in part(e) of the figure.
107
10−1
100
10−4
10−3
10−2
10−1
h
E
Tension Strip − ν = 0.3
10−1
100
10−8
10−7
10−6
10−5
10−4
10−3
10−2
h
E
Tension Strip − ν = 0.3
Q8 ElementQ9 Element
(e) 4-node elements (f) 8- and 9-node elementsEnergy convergence for 4-, 8- and 9-node elements for Problem 6.16.
Mesh Work (E)8-Node Quads. 9-Node Quads.
2× 4 4.8532102331947E-09 4.8541058087301E-094× 8 4.8770470777973E-09 4.8771300769207E-098× 16 4.8817537308566E-09 4.8817566246168E-0916× 32 4.8822864204817E-09 4.8822864808224E-0932× 64 4.8823284663016E-09 4.8823284674541E-0964× 128 4.8823312907864E-09 4.8823312908135E-09Exact 4.8823314976138E-09 4.8823314976138E-09
(e) Results for energy behavior using 9-node lagrangian elements are given in thetable and plotted in part(f) of the figure. The 2 × 4 mesh of 9-node elements(same number of nodes as the 4×8 mesh of 4-node elements) is shown in part (g)of the figure and in part (h) the displacement around the hole is given. Similarlyfor a 4× 8 mesh [part (i) of figure] the displacements are shown in part (j) of thefigure. It is easy to see that convergence using the higher order elements is muchmore rapid.
6.17 Program development project: Extend the program developed in Problem 2.17 toconsider plane strain and axisymmetric geometry.
No solution available at this time.
6.18 Program development project: Extend the program developed in Problem 2.17 tocompute nodal forces for specified boundary tractions which are normal or tangentialto the element edge. Assume tractions can vary up to quadratic order (i.e., constant,linear and parabolic distributions) and use numerical integration to compute values.
Test your program for an edge with constant normal stress. Then test for linearnormal and finally quadratic tangential values. Compare results with those computed
108
0 10 20 30 40 50 60 70 80 90−1
0
1
2x 10
−10
θ − angle
u −
dis
pla
cem
ents
Tension Strip − ν = 0.3
ur − radial
uθ − tangential
(g) 2× 4 Mesh (h) Displacements at hole
0 10 20 30 40 50 60 70 80 90−1
0
1
2x 10
−10
θ − angle
u −
dis
pla
cem
ents
Tension Strip − ν = 0.3
ur − radial
uθ − tangential
(i) 4× 8 Mesh (j) Displacements at holeMesh and displacements for 9-node elements for Problem 6.16.
by FEAPpv (or any available program).
No solution available at this time.
6.19 Program development project: Extend the program developed in Problem 2.17 tocompute nodal values of stress and strain. Follow the procedure given in Sec. 6.4 toproject element values to nodes.
Test your program using (a) the patch test of Problem 2.17 and (b) the curved beamproblem shown in Fig. 2.11.
No solution available at this time.
6.20 Program development project: Add a module to the program developed in Problem2.17 to plot contours of stress and strain components for plane stress, plane strain and
109
axisymmetric solids. Use the capability developed in Problem 6.18 to obtain nodalvalues and the contour routine developed in Problem 2.18.
Test your program system by plotting contours of stress components for the curvedbeam meshes described in Problem 2.18.
No solution available at this time.
6.21 Program development project: Add a 4-node quadrilateral element to the the programsystem developed in Problem 2.17. Use shape functions and numerical integration tocompute the element stiffness matrix. Also include the force vector from a constantelement body force (you may need to add b to your input module).
Test your program on the curved beam problems described in Problem 2.18. Comparethe accuracy to that obtained using triangular elements.
No solution available at this time.
110
Chapter 7: Solutions
x
y
x′
y′
z = z′
θ
Orientation of axes for Problem 7.1.
7.1 For the anisotropic properties kx′ = 0.4, ky′ = 2.1 and kz′ = 1.0 and axes as shown inthe figure with an angle θ = 30 the transformation array is given by
T =
cos θ sin θ 0− sin θ cos θ 0
0 0 1
=1
2
√3 1 0
−1√
3 00 0 2
Applying Eq. (7.10b) we have
k = TTk′T =1
4
√3 −1 0
1√
(3) 00 0 2
0.4 0 00 2.1 00 0 1
√3 1 0
−1√
(3) 00 0 2
=
0.825 −0.425√
3 0
−0.425√
3 1.675 00 0 1
7.2 A two-dimensional heat equation has a surface heat convection from the surrounding
region given byQ(x, y) = −β[φ(x, y)− φ0] .
If we assume steady heat flow (transient terms may be added based on results in Sec.7.4) the weak form given by (7.13) may be used in a two dimensional form. Substitutingthe heat source above, with sign reversed to reflect the behavior for thermal body, givesthe relation∫
Ω
[(∇δφ)T k∇φ+ δφ 2 β (φ− φ0)
]dΩ +
∫Γq
δφ [q +H (φ− φ0)] dΓ = 0
where the factor 2 results from radiation from both surface and we assume that theboundary condition for φ = φ will be imposed on Γφ.
Using the finite element approximations
φ =∑
a
Na(x, y) φa
δφ =∑
a
Na(x, y) δφa
111
the tangent matrix is given by
Kab =
∫Ω
[(∇Na)
T k∇Nb +Na 2 β Nb
]dΩ +
∫Γq
NaH Nb dΓ
and the force vector by
fa =
∫Γq
Na [q −H φ0] dΓ−∫
Ω
Na 2 β φ0 dΓ
In the above dΩ = h dx dy where h is the slab thickness. The problem solution is givenby ∑
b
Kab φb + fa = 0 ; for a = 1, 2, · · · , n
7.3 For an 8-node serendipity element with unit side lengths and k = 1, the number ofzero eigenvalues for the stiffness matrix computed using quadrature are shown in thetable below
Order Eigenvalues1× 1 2 2 0 0 0 0 0 02× 2 5.333 2.667 1.854 1.854 0.667 0.479 0.479 03× 3 5.333 2.667 2.090 2.090 0.667 0.510 0.510 04× 4 5.333 2.667 2.090 2.090 0.667 0.510 0.510 0
(a) The lowest order quadrature that gives correct rank is 2× 2 Gauss quadrature.
(b) The lowest order quadrature that exactly evaluates H is 3× 3 Gauss quadrature.This can be predicted from the fact that the derivatives of the shape functionsalways leave one direction with a polynomial of order 2. Thus, products of shapefunctions in volve polynomials up to order 4 and the first quadrature order toreach an order greater than this is n = 3 (since we know the polynomials of order2n− 1 are then exactly integrated).
7.4 For a 9-node lagrangian element with unit side lengths and k = 1, the number of zeroeigenvalues for the stiffness matrix computed using quadrature are shown in the tablebelow
Order Eigenvalues1× 1 2 2 0 0 0 0 0 0 02× 2 5.333 2.667 1.854 1.854 0.667 0.479 0.479 0 03× 3 6.605 2.667 2.090 2.090 0.861 0.667 0.510 0.510 04× 4 6.605 2.667 2.090 2.090 0.861 0.667 0.510 0.510 0
(a) The lowest order quadrature that gives correct rank is 3× 3 Gauss quadrature.
112
(b) The lowest order quadrature that exactly evaluates H is also 3× 3 Gauss quadra-ture. This can be predicted from the fact that the derivatives of the shape func-tions always leave one direction with a polynomial of order 2. Thus, products ofshape functions in volve polynomials up to order 4 and the first quadrature orderto reach an order greater than this is n = 3 (since we know the polynomials oforder 2n− 1 are then exactly integrated).
Using the eigenvector for the zero eigenvalue of the fully integrated element array Hdetermine and sketch the shape of eigenvectors from any additional (spurious) zeroeigenvalues. (Note: The fully integrated element has one zero eigenvalue, v0).
Hint: For the case where two zero eigen-vectors v1 and v2 exist they may be expressedin terms of v0 and another orthogonal unit vector w0 as:
v0 = α1v1 + α2v2 where αi = vT0 vi
w0 = α2v1 − α1v2 .
number of non-zero eigenvalues λi (a zero being values below the computer round-off).The vectors v0, w0 and the vectors v1, w2 are both eigen-vectors of the same subspace.
+ -
+
-
+-
+
-
+
-
-
+
-
+
+
-
+
-+
-
(a) a/b = 1 (b) a/b = 1.25 (c) a/b = 2Warping function for torsion of rectangular bar. Problem 7.5.
7.5 A mesh of 40×40 9-node lagrangian elements which gives a mesh of 6561 nodes and 1600elements is used to analyze one quadrant of the rectangular bar. The mesh occupiesthe region 0 ≤≤ 0.5 and 0 ≤ 0.5 ∗ r with the nodes at x = 0 and y = 0 restrainedto give a warping function of zero along the coordinate axes. The value of r is variedbetween 1.2 and 1.5 in increments of 0.1 to bracket the ratio of the transition. It isdisovered that the solution along x = a = 0.5 has both positive and negative values upto 1.4 and all the same sign at 1.5 (with the sign agreeing with that at y = 0.5r).
The problem is then repeated using increments of r of 0.01 for 1.4 ≤ r ≤ 1.5. Theprocess is repeated using smaller increments until 6 significant figures are obtained,giving r = 1.45126. This problem is discussed in A.E.H. Love, A Treatise on theMathematical Theory of Elasticity, 4th edition, Cambridge University Press, 1927. Lovegives an answer of r = 1.4513 that corresponds well to the computed value using afinite element solution.
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0.05
0.05
0.025 0.075 0.05
A
B
C
D
Thermal analysis of composite section for Problem 7.6.
7.6 A cross-section of a long prismatic section is shown in the problem figure and subjectedto a constant uniform temperatures 370 Co on the left boundary and 66 Co on the rightboundary. The top and bottom edges are assumed to be insulated so that qn = 0.
The cross-section is a composite of fir (A), concrete (B), glass wool (C) and yellow pine(D). The thermal conductivity for each of the parts is: kA = 0.11, kB = 0.78 kC = 0.04and kD = 0.147 in consistent units for the geometry of the section shown.
(a) Estimate the heat flow through the cross section assuming qy = 0 and qx constantin each part. Let the temperatures at each junction be T (0) = 370, T (0.025) = T1,T (0.10) = T2 and T (0.15) = 66. Hint: Assume T is a function of x only.
(b) Use FEAPpv (or any other available program) to compute a finite element solutionusing 4-node and 9-node quadrilateral elements. First perform a solution on acoarse mesh and use this to design a mesh using a finer discretization. Let yourfinal 4-node element mesh have nodal locations which coincide with those usedfor the corresponding 9-node element mesh.
Plot a distribution of heat flow qn across each of the internal boundaries.
7.7 The cross section of two tubular sections is shown in the problem figure. The partsare to be assembled by heating the outer part until it just passes over the inner partas shown in the figure. Let ri = 10 cm., t = 5cm. and h = 10 cm. and take elasticproperties as E = 200 Gpa, ν = 0.3 and α = 12 × 10−6 per Co. The parts are stressfree at the room temperature 20 Co. The parts just fit when the outer bar is heatedto 220 Co (while the inner part is maintained at room temperature).
(a) What is the correct inner radius of the outer part at room temperature?
(b) Solve the problem using FEAPpv (or any other available program). Use a meshof 4-node quadrilateral elements to compute the final solution for the assembled
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r
z
ri t t
h
h
Thermal assembly of tubular sections for Problem 7.7.
part at room temperature assuming complete contact at the mating surface andno slip during cooling. Plot the radial displacement at the interaction surface.
(c) Compute an estimate of the traction components at the interaction surface. Doyou think there will be slip? Why?
x
y
z
a
a
b
x
y
qn = H (T− T
∞)
qn = H (T− T
∞)
qn = 0
qn = 0
(a) 3-D block region (b) 2-D analysis regionThermal quench in 2 and 3 dimensions for Problem 7.8.
7.8 Company X&Y plans to produce a rectangular block which needs to be processed bya thermal quench in a medium which is 100C above room temperature. The blockshown in part (a) of the figure a = 10 and b = 20 (i.e., the block is 10 × 10 × 20). Ithas been determined that the thermal properties of the block may be specified by anisotropic Fourier model in which k = 1 and c = 1. The surface convection constant His 0.05.
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The quench must be maintained until the minimum temperature in the block reaches99C above room temperature. Use FEAPpv (or any other available program) toperform a transient analysis to estimate the required quench time.
(a) First perform a 2-d plane analysis on a 10×10 cross section using a uniform meshof 4-node quadrilateral elements. Use symmetry to reduce the size of the domainanalyzed. The surface convection will be modeled by 2-node line elements alongthe outer perimeter. The analysis region is shown in part (b) of the figure withthe boundary conditions to be imposed. Locate the node where the minimumtemperature occurs and plot the behavior vs. time (a good option is to useMATLAB©R to perform plots).
Estimate the duration of time needed for the minimum temperature to reach thedesired value. (Hint: One approach to selecting time increments is to select avery small value, e.g. ∆t = 10−8 and perform 10 steps of the solution. Multiplythe time increment by 10 and perform 9 more steps. Repeat the multiplicationuntil the desired time is reached.
(b) Using the time duration estimated in (a) perform a 3-d analysis using a uniformmesh of 8-node hexahedral elements. Use symmetry to reduce the size of theregion analyzed. Note: The convection condition applies to all outer surfaces.
Estimate the duration of quench time needed for the minimum temperature toreach the desired value.
(c) What analyses would you perform if the block was 10× 10× 5?
(d) Comment on use of a 2-d solution to estimate the required quench times for othershaped parts.
z
y
PA
A
x
y
w w
h
h
(a) Cantilever beam (b) Section A–AEnd loaded cantilever beam for Problem 7.9.
7.9 The distribution of shear stresses on the cross section of a cantilever beam shown inpart (a) of the figure may be determined by solving the quasi-harmonic equation
∂2φ
∂x2+∂2φ
∂y2= 0
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with boundary condition
φ =P
2I
[ ∫y2 dx− ν
3(1 + ν)y3
]where P is the end load, I is the moment of inertia of the cross section, ν is Poissonratio of an isotropic elastic material and φ is a stress function. The shear stresses aredetermined from
τxz = −∂φ∂y
and τyz =∂φ
∂x+P
2I
[ν
1 + νx2 − y2
].
(a) Show that the stress function satisfies the equilibrium equation when the bendingstress is computed from
σz = − P (L− z) yI
.
and σx = σy = τxy = 0. L is the length of the beam.
(b) Develop a weak form for the problem in terms of the stress function φ.
(c) For a finite element formulation develop the relation to compute the boundarycondition for the case when either 3-node triangular or 4-node quadrilateral ele-ments are used.
(d) Write a program to determine the boundary values for the cross-section shown inpart (b) of the figure. Let w = 2 and h = 3. Use the quasi-harmonic thermal ele-ment in FEAPpv (or any other available program) to solve for the stress functionφ. Plot the distribution for φ on the cross section.
(e) Modify the expressions in FEAPpv (or any other program for which source codeis available) to compute the stress distribution on the cross-section. Solve andplot their distribution. Compare your results to those computed from the classicalstrength of materials approach.
Hint: Normalize your solution by the factor P/2I to simpify expressions.
7.10 A long sheet pile is placed in soil as shown in the problem figure. The anisotropicproperties of the soil are oriented so that x = x′, and y = y′. The governing differentialequation is given in Sec. 7.5.3. The soil has the properties kx = 2 and ky = 3. UseFEAPpv (or any other available program) to determine the distribution of head andthe flow in the region shown. Solve the problem using a mesh of 4-node, 8-node, and9-node quadrilateral elements. Model the problem so that there are about 4 times asmany 4-node elements as used for the 8- and 9-node models (and thus approximatelyan equal number of nodes for each model). Compare total flow obtained from eachanalysis.
7.11 An axisymmetric sheet pile is placed in soil as shown in the problem figure. Theanisotropic properties of the soil are oriented so that r = r′, and z = z′. The governingequation for plane flow is given in Sec. 7.5.3. Deduce the Euler differential equation for
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Pile
30
50
50
90 110
H=20
H=0
qn=0
qn=0
qn=0
x
y
Data for seepage under a sheet pile for Problem 7.10.
Pile
30
50
50
90 110
H=0
H=40
qn=0
qn=0
r
z
Data for seepage under an axisymmetric sheet pile for Problem 7.11.
the axisymmetric problem from the weak form given in Secs 7.2.3 and 7.3.2 suitablymodified for the seepage problem.
Assuming isotropic properties with k = 3, use FEAPpv (or any other available pro-gram) to determine the distribution of head and the flow in the region shown. Solvethe problem using a mesh of 4-node, 8-node, and 9-node quadrilateral elements. Modelthe problem so that there are about 4 times as many 4-node elements as used for the 8-and 9-node models (and thus approximately an equal number of nodes for each model).Compare total flow obtained from each analysis.
7.12 A membrane occupies a region in the x−y plane and is stretched by a uniform tensionT . When subjected to a transient load q(x, y, t) acting normal to the surface the
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governing differential equation is given by
−T[∂2u
∂x2+∂2u
∂y2
]+m
∂2u
∂t2= q(x, y, t) .
(a) Construct a weak form for the differential equation for the case when boundaryconditions are given by u(s, t) = 0 for s on Γ.
(b) Show that the solution by a finite element method may be constructed using C0
functions.
(c) Approximate the u and δu by C0 shape functions Na(x, y) and determine thesemi-discrete form of the equations.
(d) For the case of steady harmonic motion, u may be replaced by
u(x, y, t) = w(x, y) exp iωt
where i =√−1 and ω is the frequency of excitation.
Using this approximation, deduce the governing equation for w. Construct a weakform for this equation. Using C0 approximations for w determine the form of thediscretized problem.
7.13 Program development project: Modify the program developed for solution of linearelasticity problems to solve problems described by the quasi-harmonic equation forheat conduction. Include capability to solve both plane and axisymmetric geometry.
Specify the material properties by anisotropic values k′x, k′y and ψ (where ψ is the angle
x′ makes with the x-axis.
Use your program to solve the problem described in Problem 7.6. Plot contours fortemperature and heat flows qx and qy.
7.14 Program development project: Extend the program developed in Problem 7.13 to solvetransient problems.
Include an input module to specify the initial temperatures.
Also add a capability to consider time dependent source terms for Q.
Test your program by solving the problem described in Example 7.6 of Sec. sec7.5.2.
7.15 Program development project: Extend the program developed in Problem 2.17 tocompute nodal values of for fluxes from the quasi-harmonic equation. Follow theprocedure given in Sec. 6.4 to project element values to nodes.
Test your program using (a) a patch test of your design and (b) the problem describedin Example 6.6.