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Lecture 13: Trusses & Grids Stiffness Method Washkewicz College of Engineering Plane Truss Stiffness Matrix The distinguishing feature of a plane truss is that loads are applied in the plane of the structure whereas in a space truss they are not. We now wish to outline the procedure of formulating the joint stiffness matrix [S J ] for a plane truss structure. The extension to a three dimensional space truss will be intuitively obvious. 1

Plane Truss Stiffness Matrix · Lecture 13: Trusses & Grids –Stiffness Method Washkewicz College of Engineering 2 Consider an arbitrary member, i. isolated from a generalized plane

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  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

    Plane Truss – Stiffness Matrix

    The distinguishing feature of a plane truss is that loads are applied in the plane of the

    structure whereas in a space truss they are not.

    We now wish to outline the procedure of formulating the joint stiffness matrix [SJ] for a

    plane truss structure. The extension to a three dimensional space truss will be intuitively

    obvious. 1

  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

    2

    Consider an arbitrary member, i. isolated from a

    generalized plane truss depicted below:

    The joints at the end of truss member i are denoted j and k. The plane truss lies in the x-y

    plane. The joint translations are the unknown displacements and these displacements are

    expressed in terms of their x and y components.

    We will relax the requirement that truss members are two force members. This allows for

    loads that are applied between joints to a truss member, and it allows consideration of the

    weight of the truss member.

    i

    k

    j

    x

    y

    Primary load carrying capabilities

    are in the plane of the truss –

    hence the need for shear walls.

  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

    The positive directions of the four displacement components (two translations at either

    end) of member i are depicted in the figure below

    It will be convenient to utilize the direction cosines associated with this arbitrary

    member. In terms of the joint coordinates the direction cosines are

    L

    xx

    C

    jk

    X

    1cos

    L

    yy

    C

    jk

    Y

    2cos

    22 jkjk yyxxL with

    3

  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

    The beam member stiffness matrix developed in the previous section of notes can be

    easily adapted for use in the case of a plane truss. The member stiffness matrix [SM] for

    an arbitrary truss member with member axes Xm and Ym oriented along the member and

    perpendicular to the member can be obtained by considering Case #1 and Case #7 from

    the previous section of notes.

    Using the numbering joint numbering system and the member axes depicted in the

    following figure

    then the member stiffness matrix for

    a truss member is as follows

    4

  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

    Note that [SJ] is based on axes oriented to the

    structure. Truss member stiffnesses may be

    obtained in one of two ways. Either the

    stiffnesses are directly computed using the figure

    to the left, or the second method consists of first

    obtaining the stiffness matrix relative to the

    member oriented axes and then imposing a

    suitable matrix transformation that transforms

    these elements to axes relative to the structure.

    We will focus on the direct method first to help

    develop an intuition of how the structure

    behaves. Unit displacement in both the x and y

    directions are applied at each end of the member.

    If a unit displacement in the x direction is applied

    to the j end of the member, the member shortens

    and an axial compression force is induced. The

    magnitude of the force is

    xx C

    L

    EA

    5

  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

    Restraint actions at the ends of the truss member in the x and y directions are required.

    They are equal to the components of the axial force induced in the member, and are

    identified here as elements of the [SMD] matrix in order to distinguish them from elements

    of the [SM] matrix. The numbering of these elements are shown in the previous figure.

    Thus

    yxxMDMD

    xx

    MDMD

    yxx

    MD

    xx

    MD

    CCL

    EASS

    CL

    EASS

    CCL

    EAS

    CL

    EAS

    2141

    2

    1131

    21

    2

    11

    6

  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

    In a similar fashion, a unit displacement in the y direction at the j end of the member yields

    22242

    1232

    2

    22

    12

    yx

    MDMD

    yxx

    MDMD

    yx

    MD

    yxx

    MD

    CL

    EASS

    CCL

    EASS

    CL

    EAS

    CCL

    EAS

    7

  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

    In a similar fashion, a unit displacement in the x direction at the k end of the member yields

    yxxMDMD

    xx

    MDMD

    yxx

    MD

    xx

    MD

    CCL

    EASS

    CL

    EASS

    CCL

    EAS

    CL

    EAS

    2343

    2

    1333

    23

    2

    13

    8

  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

    In a similar fashion, a unit displacement in the y direction at the k end of the member yields

    22444

    1434

    2

    24

    14

    yx

    MDMD

    yxx

    MDMD

    yx

    MD

    yxx

    MD

    CL

    EASS

    CCL

    EASS

    CL

    EAS

    CCL

    EAS

    9

  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

    We have just developed the four rows of the [SMD] matrix, i.e.,

    10

  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

    Example 13.1

    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 L1, L2, L3 and

    L4 and axial rigidities EA1, EA2, EA3 and EA4

    The loads consist

    of two

    concentrated

    forces P1 and P2action at joint E.

    We will consider

    the bar weights

    identified 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. 11

  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

    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

    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, thus

    We next consider the restrained

    structure shown at the right. Here joint

    E is fixed with a pin support that

    produce loads ADL1 and ADL2 associated

    with D1 and D2.

    2

    1

    P

    PAD

    12

  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

    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

    correspond 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.

    13

  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

    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.e.,

    2

    0

    2222

    0

    44332211 WLwLwLwLwADL

    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

    MLi

    LwA sin

    2

    444

    333

    222

    111

    sin

    sin

    sin

    sin

    2

    1

    Lw

    Lw

    Lw

    Lw

    AMLor

    14

  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

    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.15

  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

    When the upper joint of the element

    moves to the right, the lower joint stays

    fixed.

    When the upper joint of the element

    moves up, again the lower joint stays

    fixed. Both actions elongate the truss

    elements. The geometry of the elongation

    is determined by the translation of joint E.

    cos

    AE

    PL

    L

    EAP cos

    16

  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

    cosL

    EA

    sinL

    EA

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

    an amount

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

    an 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.,

    3

    2

    3

    311

    3 cos L

    EAS

    17

  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

    Thus

    3

    2

    3

    32

    2

    2

    2

    1

    1

    4

    43

    2

    3

    32

    2

    2

    2

    1

    1

    4

    2

    4

    43

    2

    3

    32

    2

    2

    21

    2

    1

    1

    11

    4

    11

    3

    11

    2

    11

    1

    11

    coscos

    0coscos1

    coscoscoscos

    L

    EA

    L

    EA

    L

    EA

    L

    EA

    L

    EA

    L

    EA

    L

    EA

    L

    EA

    L

    EA

    L

    EA

    L

    EA

    SSSSS

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

    element 4 is vertical.

    18

  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

    21

    4

    21

    3

    21

    2

    21

    1

    21 SSSSS

    Similarly the stiffness S21 is composed of contributions from various elements of the

    truss. Consider the contribution to S21 from member 3, i.e.,

    33

    3

    321

    3 sincos L

    EAS

    Thus

    33

    3

    322

    2

    2

    4

    433

    3

    322

    2

    2

    1

    1

    44

    4

    433

    3

    322

    2

    211

    1

    1

    sincossincos

    10sincossincos01

    sincossincossincossincos

    L

    EA

    L

    EA

    L

    EA

    L

    EA

    L

    EA

    L

    EA

    L

    EA

    L

    EA

    L

    EA

    L

    EA

    19

  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

    By an analogous procedure S12 and S22 are

    33

    3

    322

    2

    212 sincossincos

    L

    EA

    L

    EAS

    4

    43

    2

    3

    32

    2

    2

    222 sinsin

    L

    EA

    L

    EA

    L

    EAS

    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

    The vector {AD} and the matrix {ADL} were established earlier.

    20

  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

    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

    i

    i

    iMDi

    L

    EAA cos1 i

    i

    iMDi

    L

    EAA sin2

    4

    4

    44

    4

    4

    3

    3

    33

    3

    3

    2

    2

    22

    2

    2

    1

    1

    11

    1

    1

    sincos

    sincos

    sincos

    sincos

    L

    EA

    L

    EA

    L

    EA

    L

    EA

    L

    EA

    L

    EA

    L

    EA

    L

    EA

    AMD

    thus

    And we can now solve

    DAAA MDMLM 21

  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

    22

    Example 13.2

  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

    23

    Structural Grids (Need a theory section on - Stiffness Method as it applies to grids

    Examples of two dimensional grids are depected below:

    Grids behave more like frames. However, loads are typically applied perpendicular

    to the plane of the structure. Moments are allowed in the plane of the structure and

    perpendicular to the structure.

  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

    24

    A space grid – the structure is not in a plane, but the loads are in essence perpendicular to

    the structure.

  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

    Example 13.3

    The 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 these

    unknown displacements.

    25

  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

    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

    26

  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

    From the last figure it is easy to see that

    or

    L

    PADL 0

    4

    8

    80

    2321

    PLAA

    PA DLDLDL

    and in a matrix format

    80

    2321

    PLAA

    PA DLDLDL

    000 321 DLDLDL AAA

    27

  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

    0

    0

    0

    DA

    The vector {AD} represents actions in the unrestrained structure associated

    with the unknown displacement D1, D2 and D3. Since there are no loads

    associated with these displacements {AD} is a null vector and in a matrix

    format

    DLD AASD 1

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

    expression

    for the unknown displacements. To do that we need to formulate the stiffness

    matrix and find its inverse.

    28

  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

    From the figures above

    23121311

    60

    12

    L

    EISS

    L

    EIS

    0612

    31221311 S

    L

    EIS

    L

    EIS

    231221311

    6624

    L

    EIS

    L

    EIS

    L

    EIS

    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 substructures.

    29

  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

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

    00 322212 SL

    GJSS

    046

    3222212 S

    L

    EIS

    L

    EIS

    046

    3222212 S

    L

    GJ

    L

    EIS

    L

    EIS

    From the figures above

    30

  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

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

    From the figures above

    L

    EISS

    L

    EIS

    40

    63323213

    J

    GJSSS 332313 00

    L

    GJ

    L

    EISS

    L

    EIS

    40

    63323213

    31

  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

    Define

    EI

    GJ

    2

    2

    3

    406

    046

    6624

    LL

    LL

    LL

    L

    EIS

    2

    2

    2

    1

    2

    21

    1

    406

    046

    66

    24

    1

    LL

    LL

    LLCL

    CEICS

    then

    and inverting this stiffness matrix leads to

    25

    1

    4

    3

    2

    1

    C

    C

    C

    where

    32

  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

    DLD AASD 1

    18

    256

    254

    4196

    2

    L

    EI

    PLD

    Solving

    leads to

    33

  • Lecture 13: Trusses & Grids – Stiffness Method

    Washkewicz College of Engineering

    34

    Example 13.4