trus Equations

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Transcript of trus Equations

  • 1MS5019 FEM 1

    MS5019 FEM 2

    3.1. Definition of the Stiffness Matrixz We will consider now the derivation of the stiffness matrix for the

    linear-elastic, constant-cross-sectional area (prismatic) bar element shown in Figure 3-1.

    1

    2T

    y

    ux ,

    L

    Txx fd 11 ,

    xx fd 22 ,

    y

    xFigure 3-1 Bar subjected to tensile forces T; positive nodal displacements and forces

  • 2MS5019 FEM 3

    z The bar element is assumed to have constant cross-sectional area A, modulus elasticity E, and initial length L. The nodal d.o.f are local axial displacements (longitudinal displacements directed along the length of the bar).

    )(0

    obtain we, respect to with atingdifferenti and (c)in (a) thenand (a)in (b) Eq. Usingbar. on the acting load ddistribute nofor

    )(constanthave wem,equilibriu force From

    )(

    )(have weip,relationshplacement strain/dis theand law sHooke' From

    dxdudAE

    xdd

    x

    cTA

    bdxud

    aE

    x

    =

    ==

    ==

    MS5019 FEM 4

    z The following assumptions are used in deriving the bar elements stiffness matrix:

    .by strain axial torelated is stress axial is, that applies; law sHooke' 3.

    ignored. isnt displaceme e transversofeffect Any 2..0 and 0 is that force;shear sustain cannot bar The 1. 21

    xxx

    x

    yy

    E

    ff

    =

    ==

    The steps previously outlined in Chapter 1 are now used to derive the stiffness matrix for bar element.

  • 3MS5019 FEM 5

    z Step 1 Select Element TypeRepresent the bar by labeling nodes at each end and in general by labeling the element number (see Figure 3-1).

    z Step 2 Select a Displacement Functions

    Assume a linear displacement variation along the local axis of the bar because a linear function with specified endpointshas a unique path.

    )1.1.3( 21 xaau +=with the total number of coefficients ai always equal to the total number of d.o.f associated with the element. Using the same procedure as in Section 2.2 for the spring element, we express Eq. (3.1.1) as

    )2.1.3(

    112 xxx dxLddu +

    =

    MS5019 FEM 6

    [ ]

    )4.1.3(

    and

    1

    bygiven functions shapewith

    )3.1.3(

    becomes (3.1.2) Eq. form,matrix in or

    21

    2

    121

    LxN

    LxN

    ddNNu

    x

    x

    ==

    =

    The linear displacement function plotted over the length of the bar element os shown in Figure 3-2. The bar is shown with the same orientation as in Figure 3-1.

  • 4MS5019 FEM 7

    Figure 3-2 Linear displacement plotted over the length of the element

    1

    2u x

    L

    T

    xd1

    xd2

    y

    x

    MS5019 FEM 8

    z Step 3 Define the Strain-displacement and Stress-strain Relationships

    The strain-displacement relationship is.

    )6.1.3(

    is iprelationshain stress/str theand(3.1.5), Eq.obtain tousedbeen have (3.1.4) and (3.1.3) Eqs. where,

    )5.1.3(

    12

    xx

    xxx

    E

    Ldd

    xdud

    =

    ==

  • 5MS5019 FEM 9

    z Step 4 Derive the Element Matrix and Equations

    ( ) )10.1.3(becomes (3.1.9) Eq. (3.1.8), Eq. usingby or,

    )9.1.3(1,-3 Figure of conventionsign force nodal by the Also,

    )8.1.3(

    obtain we(3.1.7), Eq.in (3.1.6) and (3.1.5) Eqs. Using)7.1.3(

    havewemechanics, elementary From

    211

    1

    12

    xxx

    x

    xx

    x

    ddL

    AEf

    Tf

    LddAET

    AT

    =

    =

    =

    =

    MS5019 FEM 10

    ( )

    element. spring afor constant spring the toanalogous iselement bar afor (3.1.14), Eq.In

    element.bar or trussafor matrix stiffness therepresents (3.1.14) Eq.

    )14.1.3(1111

    (3.1.3) Eq. from have, we, because Now,

    )13.1.3(

    1111

    have weform,matrix in together (3.1.12) and (3.1.10) Eqs. Expressing

    )12.1.3(

    becomes (3.1.11) Eq. (3.1.8), Eq. usingby or,)11.1.3(Similarly,

    2

    1

    2

    1

    122

    2

    kLAE

    LAE

    dd

    LAE

    ff

    ddL

    AEf

    Tf

    x

    x

    x

    x

    xxx

    x

    ==

    =

    =

    =

    k

    dkf

  • 6MS5019 FEM 11

    z Step 5 Assemble the Element Equations to Obtain the Total/Global Equations

    Assemble the global stiffness and forces vectors and global equations using the direct stiffness method described in Chapter 2 can still be adopted in this case.The method applies for structures composedd of more than one element such that

    [ ] { }

    3.4). and3.3 Sectionsin described is nsnsformatiomatrix tra stiffness and

    coordinate ofconcept (This (3.1.15). Eq.by indicated as appliedis methodditrect thebefore matrices stiffnesselement global to

    ed transformbemust matrices stiffnesselement local all now where

    )15.1.3( and 1

    )(

    1

    )(

    kk

    fFkK ==

    ====N

    e

    eN

    e

    e FK

    MS5019 FEM 12

    z Step 6 Solve for the Nodal Displacements

    Determine the displacement by imposing boundary conditions and simultaneously solving a system of equations, F = K d.

    z Step 7 Solve for the Element Forces

    Finaly, determine the strains and stress in each element by back-substitution of the displacement into equations similar to Eqs. (3.1.5) and (3.1.6).

    Example 3.1

  • 7MS5019 FEM 13

    3.2. Selecting Approximation Functions for Displacement

    Consider the following guidelines, as they relate to the one-dimensional bar element, when selecting a displacement function. Further discussion will be provided in Chapter 4 (for the beam element).

    1. Common approximation functions (AF) are usually polynomialssuch as that given by Eq. (3.1.1) or equivalently by Eq. (3.1.3), where the functionis expressed in terms of the shape functions.

    2. The AF should be continuous within the bar element. The simple linear function of Eq. (3.1.1) certainly is continuous within the element.

    3. The AF should provide interelement continuity for all d.o.f at each node for discrete line elements, and along common boundary lines and surfaces for two- and three-dimensional elements.

    2k

    MS5019 FEM 14

    For the bar element, we must ensure that nodes common to two or more elements remain common to the these elements upon deformation and thus prevent overlaps or voids between elements.For the two-bar structure (Figure 3-3), the linear function for displacement within each element will ensure that elements 1 and 2 remain connected; that is, the displacement at node 2 for element 1 will equal the displacement at the same node 2 for element 2. The linear function is then called a conforming (or compatible) functionfor the bar element because it ensure both the satisfaction of continuity between adjacent elements and of continuity within the element.

    Figure 3-3 Interelement continuity of a two-bar structure

    L1

    L

    2 31 2

  • 8MS5019 FEM 15

    4. The approximation function should allow for rigid-body motiondisplacement and for a state of constant strain within the element. The 1D displacement function, Eq. (3.1.1), satisfies these criteria because a1 term allows for rigid-body motion and the term allows for constant strain since is a constant. (This state of constant strain in the element can, if fact, occur if elements are chosen small enough).The simple polynomial Eq. (3.1.1) satisfying this fourth guidelines is said to be complete for the bar element. The completeness of a function is a necessary condition for convergence to the exact answer, for instant, for displacement and stresses.

    The idea that the interpolation (approximation) function must allow for a rigid-body displacement means that the function must be capable of yielding a constant value (say, a1), because such a value can, in fact, occur.

    xa22 axdudx ==

    MS5019 FEM 16

    occurs.nt displacemebody -rigid awhen alueconstant v a yield will that soelement in thepoint withevery at unity to

    addmust functionsion interpolatnt displaceme that theshows (3.2.5) Eq. Thus,

    )5.2.3(1

    obtain we(3.2.4), Eq.by Therefore,

    )4.2.3()(

    have then we(3.2.3), and (3.2.1) Eqs. From

    )3.2.3()(

    have we(3.1.3), Eq.in (3.2.2) Eq. Using)2.2.3(or

    )1.2.3(case heconsider tmust weTherefore,

    21

    1211

    1212211

    211

    1

    u

    NN

    aNNau

    aNNdNdNu

    ddaau

    xx

    xx

    =+

    +==

    +=+=

    ===

  • 9MS5019 FEM 17

    3.3 Transformation of Vectors in Two Dimensionz In many problem it is convenient to introduce both local and global

    coordinates. Local coordinates are always chosen to convenientlyrepresent the individual element. Global coordinates are chosen to be convenient for the whole structures.

    z Given the nodal displacement of an element, represented by the vector d in Figure 3-4, we want to relate the components of this vector in one coordinate system to components in another.

    Figure 3-4 General displacement vector di

    d

    x

    y

    j

    ij

    xy

    MS5019 FEM 18

    z For general purposes, we will assume that d is not coincident with neither the local nor the global axes. In this case, we want to relate global displacement components to local ones.

    z In doing that, we will develop a transformation matrix that willsubsequently be used to develop the global stiffness matrix for a bar element.

    5.-3 Figure of use through and to and relate now will We.directions and in vectorsunit are and ;directions and in the rsunit vecto are and where

    )1.3.3( systems coordinateboth in