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Lecture 2: One‐dimensional
Problems
APL705 Finite Element Method
Steps in FEM Problems
1. System idealization –the given actual problem is broken
down into idealized elements
2. Element equilibrium – the equilibrium requirements are
established in terms of displacements or the state
variables (or primary unknowns)
3. Assembly of elements –the element interconnections
are used to develop a set of simultaneous equations in
the unknown state variables
44. CalculaCalculattionion ofof rreesponsesponse –– byby solvingsolving thethe sysysstteemm ofof
equations, the response of each element and thereby
that of the entire system under consideration is
determined
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Direct Stiffness Method
The direct aprroach
ThThe process off obtbtaiiniing ththe tttotall or systtem stiftifffness matitrix bby
summation of individual element stiffness matrices is called
direct stiffness method.
n
K=∑k(j)
i=1
(j)
Here k are element stiffness matrices. Through the examples
discussed later, we will see that this approach is general and can
be applied to other non‐structural problems also. This method is
simple and gives a basic idea about obtaining the behaviour of a
finite element of a continuum.
Some Examples
A single element model
HerHeree wewe cacann coconnssiiddeerr aa rorodd, springspring, trusstruss membermember, beambeam, pipepipe oror
any such simple structural element for analysis. Consider the
following single linear spring model as an example.
i k j
U U
i f j
fi j
HerHeree ii andand jj araree thethe twtwoo nodesnodes wherwheree thethe
displacements and forces are present and k is
the stiffness (spring constant) of the linear
spring element under consideration
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Simple Examples
Force – displacement relation
Let us consider the equilibrium of forces at nodes iand j:
f =k(u −u )=ku −ku
i i j i j
f =k(u −u)=ku −ku
j j i j i
Expressing these in a matrix form:
⎡ ⎤⎧ u ⎫ ⎧ f ⎫
k −k ⎪ i ⎪=⎪ i ⎪⇒ku= ff
⎢⎢ −kk kk ⎥⎥⎨ u ⎬ ⎨ ff ⎬
⎣⎣ ⎦⎦⎪⎪ j ⎪⎪ ⎪⎪ j ⎪⎪
⎩ ⎭ ⎩ ⎭
Here [k] –element stiffness matrix, {u}‐ nodal
displacement vector and {f}‐ element force vector
Direct Approach
Two‐element model
NowNow lelett usus coconnssiiddeerr thethe ffoollowingllowing linearlinear elaselastictic springssprings ccoonnectnnecteded endend
to end as follows:
1 x K
1 K 3
U 2 U
f 1 (1) 2 (2) 3
1 U f3
f 2
2
HerHeree agagainain thethe nomenclanomenclaturturee isis similarsimilar atat nodesnodes
and elements. Now we have 3 nodes and two
elements with their stiffness as k1 and k2
respectively.
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Direct Approach
Simple Two‐element Example
Let us extend the idea of to a two element system now:
For Element 1 ⎡ ⎤⎧ ⎫ ⎧ (1) ⎫
k −k ⎪ u ⎪ ⎪ f ⎪
⎢ 1 1 ⎥⎨ 1 ⎬=⎨ 1 ⎬
⎢ −k k ⎥⎪ u ⎪ ⎪ f(1) ⎪
For Element 2 ⎣ 1 1 ⎦⎩ 2 ⎭ ⎩ 2 ⎭
⎡ k −k ⎤⎧ u ⎫ ⎧ f(2) ⎫
⎢ 2 2 ⎥⎪ 2 ⎪=⎪ 1 ⎪
⎢⎢ −kk kk ⎥⎥⎨ u ⎬ ⎨ ((22)) ⎬
⎣ 2 2 ⎦⎪⎪ 3 ⎪⎪ ⎪⎪ f2 ⎪⎪
⎩ ⎭ ⎩ ⎭
For assembling the system or total stiffness matrix,
there are two approaches: (i) Equilibrium of forces
(ii) Superposition of element matrices
Assembling Stiffness Matrix
(i) Considering Equilibrium of forces
NNodde11 ff ((11)) = RR ⇒kk u −kk u
1 1 1 1 1 2
Node2 f(1)+ f(2) = R ⇒ku +(k +k )u −k u
2 1 2 1 1 1 2 2 2 3
Node3 f(2) = R ⇒k u +k u
2 3 2 2 2 3
⎡ k −k 0 ⎤⎧ u ⎫ ⎧ R ⎫
⎢ 1 1 ⎥⎪ 1 ⎪ ⎪ 1 ⎪
⎢⎢ −k k +k −k ⎥⎥⎨⎨ u ⎬⎬ = ⎨⎨ R ⎬⎬
⎢ 1 1 2 2 ⎥⎪ 2 ⎪ ⎪ 2 ⎪
0 −k k u R
⎢ 2 2 ⎥⎪ 3 ⎪ ⎪ 3 ⎪
⎣ ⎦⎩ ⎭ ⎩ ⎭
(ii) Superposition of element matrices
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