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Unformatted text preview: ASE321K Final Exam
ME. Mear Fall 2006 .5/ Problem 1: The truss shown in the ﬁgure is comprised of bars with axial rigidity
AE. Let the global degrees of freedom (and associated nodal forces) be numbered as indicated. 3L 3 :l r
l ( l T Z :'>
+
:___’ #55
7;, .1, / (a) Consider the three ‘typical’ elements shown below, and give the stiffness matrix
L, for each of these elements. 4 4‘.
z ‘t 3
4—3 5
I B93 5’
l "" l
2. ‘7’ (b) With reference to these three elements, state the connectivity vector for each of
the ﬁve bars. (0) Recall that the global system of equations can be partitioned such that K f f U f =
P; — Kchc and Pc = chUf + KCCUC. Here P; (is a vector which) con
tains the applied forces and U , contains corresponding displacements (viz. the
“free degrees of freedom”), while UC contains prescribed displacements and PC
contains associated forces (viz. the reactions). For the problem under consid~ eration: H (1) State explicitly the vectors U f, UC, P f and PC. 13 (ii) Follow an ‘elementbyelement’ assembly process to construct K f f. "’7 Problem 2: Consider a solid circular bar subjected to a uniformly distributed torque to as indicated in the ﬁgure. The bar is linearly elastic with shear modulus G and to polar moment of the area J. / —:v ~—:9 «9 —»
/ Nﬁ: Earth/wok ingm Equilibrium of the bar dictates that x T: 63" 3% dﬁb
d—i(GJd—m)+t =0 , O<x<L in which ¢(:I:) is the angle of twist. The boundary conditions for the problem under
consideration are that 95(0) = 0 and d¢/dx(L) = 0. Establish a symmetric weakform statement of the boundary value problem. Make sure that you explicitly state the essential and natural boundary conditions. Problem 3: An axial bar is connected to rigid supports by means of Springs at its
ends as shown. The bar has length L and axial rigidity AE, and the springs have
stiffness P6 = AE/(4L). A concentrated force Po acts at the center of the bar as indicated. WSW
hams—Vs! 4 j A symmetric weakform statement of the problem is as follows. Let dudu“
W*E—nu(0)u*(O)—nu(L )u *(L)+Pu*)(L/2 / 14de d3: A’x The actual displacement ﬁeld (Le. the one that satisﬁes equilibrium) is the one for
which W* = 0 for every test function u*(:r) Use Galerkin’s method to obtain an approximate solution to the boundary value
problem. To do so, take the trial ﬁeld as uN(:t) = 051 + a2(E)2 in which {1511,le2} are constants to be determined. L
._—
0—» a %_ a
J." ;/'
11. Problem 4: Consider again the boundary value problem introduced in Problem 3.
We wish to obtain an approximate solution by means of the Finite Element Method.
Consider two (linear) elements of equal length, and construct the system of equations
K U = P. You need not solve for the free degrees of freedom. 2 L.
éifiﬂax +J’rodpﬁcbx =0
Ax Ax o L
“‘20) 4m) — a3$(o)¢*(o) —;6I L
A M u ,0
ﬁfganx “Rap x E: Wiajc 1’0”“
L A A PK L ’6 r ﬁst
—CTIii° (04970 afar}; 4x 441,4: Ax :0 1} my
k“, 4):...4/ 9 I a
“Liﬂﬁﬁ “E. Emmh«J 13c: Mo): 0 Naﬁxrd M : TR. = o mm 3 = L QC ’ X
a“ “1+ “L 5;) J “N“: Z“??—
' L... f x
L0; = ﬁn 33105:) ) uﬁ = 27”? .9, w"  J<o<1rsl — «(99+XL)(FI+FL) + Pﬁsl L) JAE(2¢<2:_‘ )(2‘2.LT>:;)AX
[Max— «(x1+o<L)+P]F,+[1c(a<1+o«)+P
W‘*=o jaw ﬁrﬁrg MN: §“j<’<"' K(o<,+°<L)+P =0
3C(’<++°<L)+ .3 _4RE°‘2_ ._ 4A5 “LJFL .2
4 ..__.... 4 31. R} aue 3C: ite— P1 4L
uié _ CE = i “ 41. L H)” O —’5§(a<4+°g +fﬁ_4ﬁ.ﬁx
““— ') 4 3L
2&1"qu jg?!"— "'—I—«..__. . 5‘5 ﬁE QEFDUJM‘}
Arf’rbeMtLtk u = :W‘q‘ro‘)
N'
2W?”
[Lb43 M. WNW: jrorw‘:
M: _ ocZu c? (o) 2 Edna) —0cZLL.JML)ZH¢ (9+ Findja')
L
jéfz ”4de 2:?3'43 3/)A7‘
N 0 L
: Zﬁ‘{ 3<Zu1¢ mam.) tiwamgwg + Yw‘b (.2) [ZWSAEMAJ
f4 11 1 0 :5
= 2:? HZ”? mm» (o) mmudvay @543:de “3+ wig/J} u
2‘95“"? m“ ”N
L''. L
WM Kit): [mﬂﬁcb (o) + d>k(L)a>(1_)‘5 ENE 4chb Ax
P3“: ?Dd:i 1;)
TL krm 5/0435); $1! A»: f; a 6er dimiard L'M'ar XJQRW;
. 1 —1
Kg: 6;; A ‘] Am} HM ﬁrm «3(d?(°)bj°'( J+M¢((L)4’J(L) it. ’rEJ (.mh’ikuh‘m [A
W 5V ”‘3” +3 ﬁdeR, “MW; MMMV Cpr by EM k 'Hgk’r'wwzi. W ' E? 1 + «L: :1 ...
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 Spring '06
 Dr.MarkMear

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