of 20/20
Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD Example Consider the plane truss with four bars meeting at a common joint E. This truss only has two degrees of freedom from a kinematic standpoint. It is a convenience to identify the bars of the truss numerically. The bars have lengths L 1 , L 2 , L 3 and L 4 and axial rigidities EA 1 , EA 2 , EA 3 and EA 4 The loads consist of two concentrated f P d P forces P 1 and P 2 action at joint E. We will consider the bar weights identified here as w 1 , w 2 , w 3 and w 4 (force/length). The unknown displacements at joint E are identified as D 1 and D 2 . We seek to calculate member end actions A M1 , A M2 , A M3 and A M4 .

Lecture 13 Trusses and Grids.pptLecture 13: TRUSSES & GRIDS – STIFFNESS METHOD Example Consider the plane truss with four bars meeting at a common joint E. This truss only has two

  • View
    4

  • Download
    0

Embed Size (px)

Text of Lecture 13 Trusses and Grids.pptLecture 13: TRUSSES & GRIDS – STIFFNESS METHOD Example Consider...

  • Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD

    ExampleConsider the plane truss with four bars meeting at a common joint E. This truss only has two degrees of freedom from a kinematic standpoint. It is a convenience to identify the bars of the truss numerically. The bars have lengths L1, L2, L3 and L4 and axial rigidities EA1, EA2, EA3 and EA4

    The loads consist of two concentrated f P d Pforces P1 and P2action at joint E. We will consider the bar weights gidentified here as w1, w2, w3 and w4(force/length).

    The unknown displacements at joint E are identified as D1 and D2. We seek to calculate member end actions AM1, AM2, AM3 and AM4.

  • Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD

    Because the weight of each truss member is included, the axial forces at either end of a truss member will be different at joints A, B, C and D then the axial force at joint E. The axial forces at joint E could be computed as well as the shear stresses joint E. The axial forces at joint E could be computed as well as the shear stresses at the end of each truss member, however they are omitted in this example for simplicity.

    The loads P1 and P2 correspond to unknown displacements D1 and D2, thus1 2 p p 1 2,

    { }

    =2

    1

    PP

    AD

    We next consider the restrained structure shown at the right. Here joint E i fi d ith i t th tE is fixed with a pin support that produce loads ADL1 and ADL2 associated with D1 and D2.

  • Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD

    Each truss member can be considered loaded as shown below. The points of support are indicated as A and E for the purpose of discussion and do not

    d l j i i l b l d i h i i l O ld h G kcorrespond to actual joints in labeled in the original truss. One could use the Greek alphabet, but the nomenclature should be transparent given the context where it used.

  • Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD

    Since the weights of the truss members produce no horizontal reactions, the actions ADL1 must be zero and ADL2 must be equal to half the weight of all the truss elements, i ei.e.,

    { }

    =

    =

    00

    A{ }

    =

    +++

    =

    2222244332211 WLwLwLwLw

    ADL

    The quantity W is the total weight of the truss. For the purpose of calculating end actions for the vector AML, consider that from the previous figure

    iii

    MLiLwA γsin

    2−

    = { }

    −=333

    222

    111

    sinsinsin

    21

    γγγ

    LwLwLw

    AMLor2

    444

    333

    sin γγ

    Lw

  • Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD

    The next step is formulating the stiffness matrix by imposing unit displacement associated with D1 and D2 on the restrained structure as indicated below

    To obtain the stiffness values it is necessary to compute the forces in the truss elements when the unit displacements are applied to joint E.

  • Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD

    When the upper joint of the element moves to the right, the lower joint stays fixedfixed.

    When the upper joint of the element moves up, again the lower joint stays fixed. Both actions elongate the truss gelements. The geometry of the elongation is determined by the translation of joint E.

  • Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD

    EA

    When joint E is subjected to a unit translation to the right the truss element elongates an amount

    γcosL

    EA

    When joint E is subjected to a unit translation vertically the truss element elongates

    γsinL

    EAan amount

    The formulas given above are suitable for use in analyzing this plane truss. In a later lecture a more systematic approach to the development of member stiffnesses is developed that works for trusses and all types of structures.

    The stiffness S11 is composed of contributions from various elements of the truss. Consider the contribution to S11 from member 3, i.e.,

    A3

    2

    3

    311

    3 cos γL

    EAS =

  • Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD

    Thus

    42

    4

    43

    2

    3

    32

    2

    2

    21

    2

    1

    1

    114

    113

    112

    111

    11

    coscoscoscos γγγγL

    EAL

    EAL

    EAL

    EASSSSS

    +++=

    +++=

    ( ) ( )4

    43

    2

    3

    32

    2

    2

    2

    1

    1

    4321

    0coscos1 γγ

    EAEAEAL

    EAL

    EAL

    EAL

    EA+++=

    32

    3

    32

    2

    2

    2

    1

    1 coscos γγL

    EAL

    EAL

    EA++=

    The final expression results from the fact that truss element 1 is horizontal and truss element 4 is vertical.

  • Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD

    Similarly the stiffness S21 is composed of contributions from various elements of the truss. Consider the contribution to S21 from member 3, i.e.,

    333

    321

    3 sincos γγL

    EAS =

    Thus

    214

    213

    212

    211

    21 SSSSS +++=

    Thus

    EAEAEAEA

    ( )( ) ( )( )43332221

    444

    433

    3

    322

    2

    211

    1

    1

    10sincossincos01

    sincossincossincossincos

    γγγγ

    γγγγγγγγ

    EAEAEAEAL

    EAL

    EAL

    EAL

    EA

    +++=

    +++=

    ( )( ) ( )( )

    333

    322

    2

    2

    433

    322

    21

    sincossincos γγγγ

    γγγγ

    LEA

    LEA

    LLLL

    +=

  • Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD

    By an analogous procedure S12 and S22 are

    333

    322

    2

    212 sincossincos γγγγ L

    EAL

    EAS +=

    EAEAEA

    4

    43

    2

    3

    32

    2

    2

    222 sinsin L

    EAL

    EAL

    EAS ++= γγ

    The two expressions on this page as well as the two from the previous page constitute the stiffness matrix [S]. The next step would be inverting this matrix and performing the following matrix computation to find the displacement D1 and D2.

    { } [ ] { } { }{ }DLD AASD −= −1

    Th {A } d h i {A } bli h d liThe vector {AD} and the matrix {ADL} were established earlier.

  • Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD

    Since the vector {AML} was determined earlier as well, we need only identify the elements of the matrix {AMD}. This matrix contains the member end-actions due to unit displacements associated with the displacements D1 and D2, but the end actions are computed using the restrained structure. Thus for ith member using a previous figure

    ii

    iMDi L

    EAA γcos1 = ii

    iMDi L

    EAA γsin2 =i i

    11

    11

    1

    1 sincos γγL

    EAL

    EAthus

    { }

    =3

    33

    3

    22

    22

    2

    2

    11

    sincos

    sincos

    γγ

    γγ

    EAEAL

    EAL

    EA

    AMD

    4

    4

    44

    4

    4

    33

    33

    sincos

    sincos

    γγ

    γγ

    LEA

    LEA

    LL

    And we can now solve

    { } { } { }{ }DAAA MDMLM +=

  • Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD

    ExampleThe grid shown below consists of two members (AB and BC) that are rigidly joined at B. Each member is assumed to have flexural rigidity EI and torsional rigidity GJ. Kinematically, the only unknowns are the displacements at B. Since axial rigidities of the members is assumed to be quite large relative to EI and GJ, the displacements at B consist of one translation (D1) and two rotations (D2 and D3) Determine theseconsist of one translation (D1) and two rotations (D2 and D3). Determine these unknown displacements.

  • Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD

    When analyzing a grid by the stiffness method, an artificial restraint is provided at joint B, i.e.,

    It is easier to see what the reactions are if we break the structure above into two substructures such that

  • Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD

    From the last figure it is easy to see that

    PLP8

    02 321

    PLAAPA DLDLDL −=′=′=′

    or

    000 321 =′′=′′=′′ DLDLDL AAA

    80

    2 321PLAAPA DLDLDL −===

    d i t i f t

    { }

    =

    PADL 04

    8

    and in a matrix format

    { }

    − L

    DL 8

  • Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD

    The vector {AD} represents actions in the unrestrained structure associated with the unknown displacement D1, D2 and D3. Since there are no loads

    0

    p 1, 2 3associated with these displacements {AD} is a null vector and in a matrix format

    { }

    =

    00DA

    { } [ ] { } { }{ }1

    We have {ADL} and {AD} the next step is the solution of the superposition expression

    { } [ ] { } { }{ }DLD AASD −= −1

    for the unknown displacements. To do that we need to formulate the stiffness matrix and find its inversematrix and find its inverse.

  • Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD

    The stiffness matrix is found by analyzing the restrained structure for the effects of unit translations and rotations associated with the unknown displacements. In the following figure the grid structure is once again split into two substructuresfigure the grid structure is once again split into two substructures.

    From the figures above

    231213116012LEISS

    LEIS −=′=′=′ 23 LL

    0612 31221311 =′′=′′=′′ SLEIS

    LEIS

    2312213116624LEIS

    LEIS

    LEIS −===

  • Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD

    To obtain the second column of the stiffness matrix utilize the following figure

    From the figures above

    00 322212 =′=′=′ SLGJSS

    From the figures above

    046 3222212 =′′=′′=′′ SLEIS

    LEIS

    046 3222212 =+== SLGJ

    LEIS

    LEIS

  • Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD

    To obtain the third column of the stiffness matrix utilize the following figure

    From the figures aboveFrom the figures above

    LEISS

    LEIS 406 3323213 =′=′−=′

    JGJSSS =′′=′′=′′ 332313 00

    LGJ

    LEISS

    LEIS +==−= 406 3323213

  • Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD

    Define

    EIGJ

    [ ] ( )

    +−

    = 2 0466624

    LLLL

    EIS η

    then

    [ ] ( )( )

    +−

    +=2

    3

    406046

    LLLL

    LS

    ηη

    and inverting this stiffness matrix leads to

    [ ]( )

    ( )

    +−

    =− 2

    21

    2

    1 04666

    241 LL

    LLCL

    CEICS η

    ( ) +−221 406

    24LL

    CEICη

    η4 +=Cwhere

    ηηη

    2514

    3

    2

    1

    +=+=+=

    CCC

  • Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD

    { } [ ] { } { }{ }DLD AASD −= −1Solving

    { } [ ] { } { }{ }DLD

    ( ) ( )

    ++− 2542 ηηL

    leads to

    { } ( ) ( )

    ( ) ( )( )

    −+

    ++=

    18256

    4196

    2

    ηηη

    ηηEIPLD