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Eng. 6002 Ship Structures 1

Matrix Analysis UsingMATLAB

Example

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Introduction Last class we divided the FEM into five

steps:1. construction of the element stiffness matrix in

local coordinates,

2. transformation of the element stiffness matrixinto global coorindinates,

. assembl! to the global stiffness matrix usingtransformed element stiffness matrices,

". application of the constraints to reduce theglobal stiffness matrix,

#. determination of un$nown nodaldisplacements.

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The Method %he stiffness matrix for each element is

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The Method this matrix describes the stiffness of an

elastic element that relates the load vectorto the displacement vector

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The Method the element stiffness matrix transformed to

global coordinates is defined as:

{ } [ ]{ }uf k= [ ] [ ] [ ][ ] kk T=

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The Method %he transformation matrix is given b!:

{ } [ ]{ } { } [ ]{ }uuff == and

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Example cont. &n order to easil! combine the element stiffness

matrices, each element stiffness matrix is storedin a matrix the si'e of the global stiffness matrix,with the extra spaces filled with 'eros.

&n this example, the element stiffness matrix forelement 1 is stored in the portion of the globalstiffness matrix that involves nodes 1 and 2, i.e., the upper ( x ( portion of the matrix. %hus, the

expanded stiffness matrix that describes element 1 isgiven b!:

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Example cont.

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Example cont. %he element stiffness matrix for element 2 is

stored in the portion of the global stiffness matrixthat involves nodes 2 and , i.e., the lower ( x (portion of the matrix. %he expanded stiffness

matrix that describes element 2 is given b!:

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Example cont. %he expanded individual stiffness matrices can

now be added together so that the globalstiffness matrix for the two)element structure isgiven as:

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Next ClassExample 1: two beam elements * simple structure made of two beams has the following

parameters L1 + 12 in, L2 + 1 in, and E + - x 1-(psi. %he cross)

section of the beams are -.# in x -.# in and + "#/. &f

a downward vertical force of 1--- lb is applied at node2, find the displacement at node 2.

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Step 1 %he first step in the solution procedure is to

discreti'e the domain, i.e. select the number ofelements. *s a first approximation, the problem ismodeled with two beam elements. %he elementstiffness matrix for each of the beam elements isof the form:

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Step 1 (code m + 20 number of elements b + -.# -.#30 width 4in5 h + -.# -.#30 height 4in5 a + b.6h0 cross)sectional area & + b.6h.7.8120 moment of inertia l + 12 130 length of each element 4in5 e + -61-7( -61-7(30 modulus of elasticit! 4psi5 theta + - )"#6pi81-30 orientation angle f + )1---0 force 4lbs5 9tep 1: onstruct element stiffness matrices $ + 'eros4(,(,m50 for n + 1:m

$11 + a4n56e4n58l4n50 $22 + 126e4n56&4n58l4n570 $2 + (6e4n56&4n58l4n5720 $+ "6e4n56&4n58l4n50 $( + 26e4n56&4n58l4n50 $4:,:,n5 + $11 - - )$11 - -0- $22 $2 - )$22 $20- $2 $ - )$2 $(0)$11 - -

$11 - -0- )$22 )$2 - $22 )$20 - $2 $( - )$2 $30 end

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Step 1 ;ote: we are building < matrices 4i.e. a

series of 2< matrices = with si'e+m5

9ee $ in code

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Step 2 %he second step is to transform each element

stiffness matrix in local coordinates to the globalcoordinate s!stem.

%his transformation is accomplished b! .

%he transformation angle for element 1 is + -,

and, thus, the transformation matrix is simpl! theidentit! matrix. >ecause the rotation angle ismeasured countercloc$wise, the transformationangle for element 2 is + )"#/.

{ } [ ]{ }uf k= [ ] [ ] [ ][ ] kk T=

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Step 2 for n + 1:m

c + cos4theta4n550 s + sin4theta4n550

lamda + c s - - - -0 )s c - - - -0 - - 1 - - -0

- - - c s -0 - - - )s c -0 - - - - - 130 $bar 4:,:,n5 + lamda?6$4:,:,n56lamda0

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Step ! %he third step is to assemble the global

stiffness matrix that describes the entirestructure b! properl! combining the

individual element stiffness matrices. Elements stiffness matrices are stored in

the proper portion of the matrix, i.e. eachmatrix is shifted b! three rows and three

columns.

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Step ! 9tep : ombine element stiffness

matrices to form global stiffness matrix

for i + 1:( for @ + 1:( Ae4iBshift,@Bshift,n5 + $bar4i,@,n50 end end shift + shift B 0

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Step ! Ce then add the matrices together to form

the global stiffness matrix

A + sum4Ae,5

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Step " %he fourth step is to appl! the constraints and reduce the

global stiffness matrix so that the specific problem ofinterest can be solved. For this problem, the displacementsat node 1 and node are $nown, i.e. U1 + U2 + U + UD+ U + U + -, and the loads are applied to node 2 are

specified, i.e. F" + -, F# + )1--- lb, and F( + -. %hus, thes!stem of euations can be written as:

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Step " %his problem has three un$nowns, U", U#,

and U(, and, thus, reuires threeindependent euations.

Matrix algebra allows three independenteuations to be constructed b! theremoval of the rows and columns thatcorrespond to the $nown displacements,

i.e. the global stiffness matrix reduces to:

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Step "

the stiffness matrix reduces to:

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Step " 9tep ": Geduce global stiffness matrix

with constraints

Ar + A0 Ar4:,5 + 30 Ar4,:5 + 30 Ar4:,5 + 30

Ar4,:5 + 30 Ar4:,D5 + 30 Ar4D,:5 + 30 Ar4:,5 + 30 Ar4,:5 + 30 Ar4:,25 + 30

Ar42,:5 + 30 Ar4:,15 + 30 Ar41,:5 + 3

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Step " %he numerical values from the code

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Step # %he final step is simpl! to solve the reduced

s!stem of euations for the un$nowndisplacements. Fr + -0 f0 -30

Hr + ArIFr

%he numerical solution from the M*%L*> code is U" + )-.--1( in,

U# + )-.--(" in, and

U( + )-.--- radians.

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$iscussion %his problem is staticall! indeterminate,

and the solution is not trivial. *dditional information can be found with

the addition of nodes. For example, thedeflection at the midpoint of each elementcan be found if two additional nodes areadded.

Jf course, then a x s!stem ofeuations must be solved. ;ote that theaddition of nodes at the midpoints doesnot change the solution at node 2.

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%ssignment Hsing the approach outlined in this lecture,

develop a matlab routine to find the deflection atthe midpoint of each element shown.

L1 + 12 in, L2 + 1 in, and E + - x 1-(psi. %hecross)section of the beams are -.# in x -.# in and + "#/. &f a downward vertical force of 1--- lb isapplied at node 2