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ME471 Assignment 6 Solutions
1.) Textbook p.195, problem 6.1. In one dimension, the master
element (M = [−1, 1]) quadraticisoparametric shape
functions are H (s) = [s(−1 + s)/2, 1 − s2, s(1
+ s)/2]T , and the mapping fromM to the
element [x1, x3] = [2, 6] with interior node x2 = 4 is
given by
x = x̂(s) = ( 2 4 6 ) H
with Jacobian (or derivative in 1D)
J = dx̂
ds = ( 2 4 6 )
(−1 + 2s)/2−2s
(1 + 2s)/2
= 2.
The derivative of the shape functions are found from the
definition Ĥ (x̂) = H (s), i.e.,
d Ĥ/dx =(dH/ds)J −1 so that d
Ĥ 1/dx = (−1 + 2s)/4. Utilizing this
information we have
K 11 = 1
−1 J −2
dH 1
ds 2
+ H 21 Jds = 1
1 1
4
(−1 + 2s)2
4 +
s2(−1 + s)2
4 2ds.Integrating we find K 11 = 67/60
2.) Textbook p.195, problem 6.3. The master element for bilinear
shape functions on rectanglesis the square M =
{(s, t) : −1 ≤ s, t ≤ 1}.
The bilinear master element shape functions areH (s, t) = [(1−
s)(1− t)/4, (1 + s)(1− t)/4, (1 + s)(1+ t)/4, (1− s)(1+
t)/4]T and the isoparametrictransformation from the master
element to the rectangle R (physical element) shown in
Fig.P6.3may be written as
xy
=
x̂(s, t)ŷ(s, t)
=
−1 3 5 −2−1 −2 4 5
H ≡ XH
The Jacobian matrix of this transformation is
J =
∂ ̂x/∂ s ∂ ̂y/∂t∂ ̂y/∂ s ∂ ̂y/∂t
= X ( ∂H/∂s ∂H/∂t ) = 1
4X −(1 − t) −(1 − s)
(1 − t) −(1 + s)(1 + t) (1 + s)−(1
+ t) (1 − s)
or
J = 1
4
11 + 3t 1 + 3s−2 12
.
The inverse of this matrix is
J −1 = 2
67 + 3s + 18t
12 −1 − 3s2 11 + 3t
.
The matrix of partial derivatives of the shape functions for the
physical element with respect to
x, y is given by (Recall that the physical element shape
functions ˆH j(x, y) are related to the masterelement
shape functions by Ĥ j(x̂(s, t), ŷ(s, t)) =
H j(s, t).)
[∂ Ĥ
∂x ,
∂ Ĥ
∂y ] = [
∂H
∂s ,
∂H
∂t ]J −1.
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Since all the requisite matrices have been computed, the
derivatives may be found by matrix algebra.The problem requires
∂ Ĥ i/∂x and
∂ Ĥ i/∂y for i = 1, 2.
These are computed as follows:
∂ Ĥ 1/∂x = 2 [−3(1 − t) − (1 − s)/2]
/(67 + 3s + 18t),∂ Ĥ 1/∂y = 2 [(1 −
t)(1 + 3s)/4 − (1 − s)(11 + 3t)/4] /(67 + 3s +
18t),∂ Ĥ 2/∂x = 2 [3(1− t) − (1
+ s)/2] /(67 + 3s + 18t),∂ Ĥ 2/∂y
= 2 [−(1 − t)(1 + 3s)/4 − (1 + s)(11 + 3t)/4] /(67 + 3s +
18t).
It is clear that numerical integration of the
element K 12 is the best approach to the
computation. Onepossibility is to compute the required integral
manually, i.e.,
M
F (s, t) dsdt =n
i,j=1 F (si, tj)wiwjwhere sk, tk, wk k
= 1, . . . , n are the Gauss-Legendre points and weights
listed in the text, and fora hand computation n = 2
would be a reasonable choice. An efficient approach, and certainly
thepreferred one if more than one stiffness element were to be
computed, is to modify the file ex661.mprovided by with the
text. An appropriate modification is shown below. (The initial
commentssupplied with the file have been deleted to save
space.)
function k=prob6_3(xpts, ypts, ngx, ngy)
nnel=4; % number of nodes per element
ndof=1; % degrees of freedom per node
edof=nnel*ndof; % degrees of freedom per element
nglx=ngx; ngly=ngy; % use ngx x ngy integration rule
xcoord=xpts; % x coordinate values
ycoord=ypts; % y coordinate values
[point2,weight2]=feglqd2(nglx,ngly); % sampling points &
weights
%--------------------------------
% numerical integration
%--------------------------------
k=zeros(edof,edof); % initialization to zero
for intx=1:nglx
x=point2(intx,1); % sampling point in x-axis
wtx=weight2(intx,1); % weight in x-axis
for inty=1:ngly
y=point2(inty,2); % sampling point in y-axis
wty=weight2(inty,2) ; % weight in y-axis
[shape,dhdr,dhds]=feisoq4(x,y); % compute shape functions
and
% derivatives at sampling point
jacob2=fejacob2(nnel,dhdr,dhds,xcoord,ycoord); % compute
Jacobian
detjacob=det(jacob2); % determinant of Jacobian
invjacob=inv(jacob2); % inverse of Jacobian matrix
[dhdx,dhdy]=federiv2(nnel,dhdr,dhds,invjacob); % derivatives
w.r.t.
% physical coordinate
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%------------------------------
% element matrix loop
%------------------------------
for i=1:edof
for
j=1:edofk(i,j)=k(i,j)+(dhdx(i)*dhdx(j)+dhdy(i)*dhdy(j))*wtx*wty*detjacob;
end
end
end
end
%-----------------------------------------------------------------
This file produces the entire stiffness matrix as follows:
EDU>> prob6_3([-1,3,5,-2],[-1,-2,4,5],4,4)
ans =
0.8529 -0.3175 -0.3951 -0.1404
-0.3175 0.7163 -0.1092 -0.2896
-0.3951 -0.1092 0.6344 -0.1302
-0.1404 -0.2896 -0.1302 0.5602
From which we read that K 12 = −0.3175
3.) Textbook p.195, problem 6.5. The bilinear shape functions,
H i(s, t), i = 1, . . . , 4, on the mastersquare
are given on p.163 of the textbook. The isoparametric map between
the master square andthe element shown in Fig. P6.5 is given by
x̂ŷ
=
1 2 2.5 1.51 1 2 2
H = X H
with H = (H 1, . . . , H 4)T .
The Jacobian of the mapping is
J = X ( ∂H/∂s ∂H/∂t ) = 1
4X
−(1 − t) −(1 − s)(1 − t) −(1 + s)(1 + t)
(1 + s)−(1 + t) (1 − s)
= 0.5 0.25
0 0.5
3
