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Chapter 14 Truss Analysis Using The Stiffness Method

Truss Analysis Using The Stiffness Methodsite.iugaza.edu.ps/marafa/files/Chapter-14-2019.pdf · Chapter 14 Truss Analysis Using The Stiffness Method. Stiffness Method • Fundamentals

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Text of Truss Analysis Using The Stiffness Methodsite.iugaza.edu.ps/marafa/files/Chapter-14-2019.pdf ·...

  • Chapter 14

    Truss Analysis

    Using The Stiffness Method

  • Stiffness Method

    • Fundamentals of the stiffness method

    – There are essentially two ways in which structures can be analyzed using

    matrix methods.

    • Flexibility method

    • Stiffness method

    – The stiffness method can be used to analyzed both statically determinate

    & indeterminate structures, whereas the flexibility method required a

    different procedure for each of these cases.

    – The stiffness method yields the displacements & forces directly, whereas

    with the flexibility method the displacements are not obtained directly.

  • – Application of this method requires subdividing the structure into a series

    of discrete finite elements & identifying their end points as nodes.

    – The force-displacement properties of each element are determined & then

    related to one another using the force equilibrium equations written at the

    nodes

    – These relationships, for the entire structure, are then grouped together into

    what is called the structure stiffness matrix K.

    – Once the K is established, the unknown displacements of the nodes can

    then be determined for any given loading on the structure.

    – When the displacements are known, the external & internal forces in the

    structure can be calculated using the force-displacement relations for each

    member.

    Stiffness Method

  • Preliminary Definitions & Concepts

    • Member & node identifications– We will specify each member by a number enclosed within a square,

    & use a number enclosed within a circle to identify the nodes.

    – Also the ‘near’ & ‘far’ ends of the member must be identified, this will

    be done using an arrow written along the member with the head of the

    arrow directed toward the far end.

  • • Global & member coordinates. – We will use two different type of coordinate systems, global or

    structure coordinate system and local or member coordinate system.

    – Global system x, y, used to specify the sense of each of the external

    force & displacement components at the nodes.

    – Local system x’,y’ used to specify the sense of direction of members

    displacements & internal loadings.

    Preliminary Definitions & Concepts

  • • Degrees of freedom – The unconstrained truss has two degree of freedom or two possible

    displacements for each joint (node).

    – Each degree of freedom will be specified on the truss using a code

    number, shown at the joint or node, & referenced to its positive global

    coordinate direction using an associated arrow.

    • For example– The truss has eight degree of freedom or eight possible displacements.

    – 1 through 5 represent unknown or

    unconstrained degree of freedom.

    – 6 through 8 represent constrained

    degree of freedom

    Lowest code numbers will always be used

    to identify the unknown displacements.

    Highest code numbers will be used to

    identify the known displacements

    Preliminary Definitions & Concepts

  • Member Stiffness Matrix

    • Case I

    – Positive displacement dN on the near end

    • Case II

    – Positive displacement dF on the far end

    • Case I + Case II

    – Resultant forces caused by both displacements are

    'N NAE

    q dL

    'F NAE

    q dL

    ''N FAE

    q dL

    ''F FAE

    q dL

    N N F

    AE AEq d d

    L L

    F N F

    AE AEq d d

    L L

  • – These load-displacement equations written in matrix form

    1 1

    1 1

    N N

    F F

    q dAE

    q dL

    'q k d

    1 1'

    1 1

    AEk

    L

    or

    where

    The matrix, k` is called the member stiffness matrix.

    It is of the same form for each member of the truss.

    Member Stiffness Matrix

  • Transformation Matrices• Displacement & force transformation

    – We will now develop a method for

    transforming the member forces q and

    displacements d defined in local coordinate

    to global coordinates

    cos F Nx xx x

    L

    cos F Ny yy y

    L

    2 2

    cos F Nx xF N F N

    x x

    x x y y

    2 2

    cos F Ny yF N F N

    y y

    x x y y

  • Transformation Matrices• Displacement transformation matrix

    – In global coordinate each end of the member can have two independent

    displacements.

    – Joint N has displacements DNx & DNy in global coordinate

    cos cosN Nx x Ny yd D D

    N Nx x Ny yd D D

    or

  • – Also joint F has displacements DFx & DFy

    – Displacements at N & F

    cos cosF Fx x Fy yd D D

    F Fx x Fy yd D D

    or

    N Nx x Ny yd D D

    F Fx x Fy yd D D

    0. 0.

    0. 0.

    Nx

    x y NyN

    x y FxF

    Fy

    D

    Dd

    Dd

    D

    d TDor

    0. 0.

    0. 0.

    x y

    x y

    T

    where

    Transformation Matrices

  • • Force transformation matrix

    – Force qN applied to the near end of the member

    – Force qF applied to the far end of the member

    – Rewrite in a matrix form:

    cosNx N xQ q or

    cosNy N yQ q

    Nx N xQ q Ny N yQ q

    cosFx F xQ q or

    cosFy F yQ q

    Fx F xQ q Fy F yQ q

    0.

    0.

    0.

    0.

    Nx x

    Ny y N

    Fx x F

    Fy y

    Q

    Q q

    Q q

    Q

    TQ T qor where

    0.

    0.

    0.

    0.

    x

    yT

    x

    y

    T

    Transformation Matrices

  • Member Global Stiffness Matrix

    • Stiffness matrix

    – We will determine the stiffness matrix for a member which relates

    the member’s global force components Q to its global

    displacements D. 'q k d d TDand

    'q k T D

    TQ T q 'TQ T k TD Q kD

    'Tk T k T0.

    0. 0. 0.1 1

    0. 0. 0.1 1

    0.

    x

    y x y

    x x y

    y

    AEk

    L

  • Member Global Stiffness Matrix

    2 2

    2 2

    2 2

    2 2

    xx x y x x y

    yy x y y x y

    xx x y x x y

    yy x y y x y

    N

    NAEk

    FL

    F

    x y x yN N F F

    'Tk T k T

  • Example 1 Determine the structure stiffness matrix for the two-member truss

    shown. AE is constant

    Solution:

    Establish the x, y global system

    Identify each joint & member numerically.

  • Member 1:

    Determine x & y, where L = 3ft

    3 01

    3x

    0 00

    3y

    2 2

    2 2

    2 2

    2 2

    xx x y x x y

    yy x y y x y

    xx x y x x y

    yy x y y x y

    N

    NAEk

    FL

    F

    x y x yN N F F

    1

    0.333 0. 0.333 0. 1

    0. 0. 0. 0. 2

    0.333 0. 0.333 0. 3

    0. 0. 0. 0. 4

    k AE

    1 2 3 4

    Dividing each element by L = 3ft

    22 1

    0.3333

    x

    L

  • Member 2:Determine x & y, where L = 5ft

    3 00.6

    5x

    4 00.8

    5y

    2 2

    2 2

    2 2

    2 2

    xx x y x x y

    yy x y y x y

    xx x y x x y

    yy x y y x y

    N

    NAEk

    FL

    F

    x y x yN N F F

    2

    0.072 0.096 0.072 0.096 1

    0.096 0.128 0.096 0.128 2

    0.072 0.096 0.072 0.096 5

    0.096 0.128 0.096 0.128 6

    k AE

    1 2 5 6

    Dividing each element by L = 5ft 0.8 0.6

    0.0965

  • Structure stiffness matrix

    K = k1 + k2

    0.333 0 0.333 0 0 0 1 0.072 0.096 0 0 0.072 0.096

    0. 0 0. 0 0 0 2 0.096 0.128 0 0 0.096 0.128

    0.333 0 0.333 0 0 0 3 0. 0. 0 0 0. 0.

    0. 0 0. 0 0 0 4 0. 0. 0 0 0. 0.

    0. 0 0. 0 0 0 5 0.072 0.096 0 0 0.072 0.096

    0. 0 0. 0 0 0 6 0.096 0.128 0 0 0

    K AE AE

    1

    2

    3

    4

    5

    .096 0.128 6

    1 2 3 4 5 6 1 2 3 4 5 6

    0.405 0.096 0.333 0. 0.072 0.096

    0.096 0.128 0. 0. 0.096 0.128

    0.333 0. 0.333 0. 0. 0.

    0. 0. 0. 0. 0. 0.

    0.072 0.096 0. 0. 0.072 0.096

    0.096 0.128 0. 0. 0.096 0.128

    K AE

  • Example 2

    Determine the structure stiffness matrix for the truss shown. AE is constant

  • Member 1:

    Determine x & y, where L = 10ft

    10 01

    10x

    0 00

    10y

    2 2

    2 2

    2 2

    2 2

    xx x y x x y

    yy x y y x y

    xx x y x x y

    yy x y y x y

    N

    NAEk

    FL

    F

    x y x yN N F F

    1

    0.1 0 0.1 0 1

    0 0 0 0 2

    0.1 0 0.1 0 6

    0 0 0 0 5

    k AE

    1 2 6 5

    Dividing each element by L = 10ft

    2

    10.1

    10

  • Member 2:

    Determine x & y, where L = 14.14ft10 0

    0.70714.14

    x

    10 0

    0.70714.14

    y

    2 2

    2 2

    2 2

    2 2

    x x y x x y

    y x y y x y

    x x y x x y

    y x y y x y

    AEk

    L

    2

    0.035 0.035 0.035 0.035 1

    0.035 0.035 0.035 0.035 2

    0.035 0.035 0.035 0.035 7

    0.035 0.035 0.035 0.035 8

    k AE

    1 2 7 8

    Member 3:

    Determine x & y, where L = 10ft0 0

    0.10

    x

    10 0

    110

    y

    3

    0 0 0 0 1

    0 0.1 0 0.1 2

    0 0 0 0 3

    0 0.1 0 0.1 4

    k AE

    1 2 3 4

  • Member 4:Determine x & y, where L = 10ft

    10 01

    10x

    10 100.

    10y

    2 2

    2 2

    2 2

    2 2

    x x y x x y

    y x y y x y

    x x y x x y

    y x y y x y

    AEk

    L

    4

    0.1 0 0.1 0 3

    0 0 0 0 4

    0.1 0 0.1 0 7

    0 0 0 0 8

    k AE

    3 4 7 8

    Member 5:

    Determine x & y, where L = 14.14ft10 0

    0.70714.14

    x

    0 10

    0.70714.14

    y

    5

    0.035 0.035 0.035 0.035 3

    0.035 0.035 0.035 0.035 4

    0.035 0.035 0.035 0.035 6

    0.035 0.035 0.035 0.035 5

    k AE

    3 4 6 5

  • Member 6:Determine x & y, where L = 10ft

    10 100.

    10x

    10 01

    10y

    2 2

    2 2

    2 2

    2 2

    x x y x x y

    y x y y x y

    x x y x x y

    y x y y x y

    AEk

    L

    6

    0 0 0 0 6

    0 0.1 0 0.1 5

    0 0 0 0 7

    0 0.1 0 0.1 8

    k AE

    6 5 7 8

    Structure stiffness matrix

    K = k1 + k2 + k3 + k4 + k5 + k6

  • 0.1 0 0 0 0 0.1 0 0 1

    0 0 0 0 0 0 0 0 2

    0 0 0 0 0 0 0 0 3

    0 0 0 0 0 0 0 0 4

    0 0 0 0 0 0 0 0 5

    0.1 0 0 0 0 0.1 0 0 6

    0 0 0 0 0 0 0 0 7

    0 0 0 0 0 0 0 0 8

    K AE

    1 2 3 4 5 6 7 8

    0.035 0.035 0 0 0 0 0.035 0.035 1

    0.035 0.035 0 0 0 0 0.035 0.035 2

    0 0 0 0 0 0 0 0 3

    0 0 0 0 0 0 0 0 4

    0 0 0 0 0 0 0 0 5

    0 0 0 0 0 0 0 0 6

    0.035 0.035 0 0 0 0 0.035 0.035 7

    0.035 0.035 0 0 0 0 0.035 0.035 8

    AE

    1 2 3 4 5 6 7 8

  • 0 0 0 0 0 0 0 0 1

    0 0.1 0 0.1 0 0 0 0 2

    0 0 0 0 0 0 0 0 3

    0 0.1 0 0.1 0 0 0 0 4

    0 0 0 0 0 0 0 0 5

    0 0 0 0 0 0 0 0 6

    0 0 0 0 0 0 0 0 7

    0 0 0 0 0 0 0 0 8

    AE

    0 0 0 0 0 0 0 0 1

    0 0 0 0 0 0 0 0 2

    0 0 0.1 0 0 0 0.1 0 3

    0 0 0 0 0 0 0 0 4

    0 0 0 0 0 0 0 0 5

    0 0 0 0 0 0 0 0 6

    0 0 0.1 0 0 0 0.1 0 7

    0 0 0 0 0 0 0 0 8

    AE

    0 0 0 0 0 0 0 0 1

    0 0 0 0 0 0 0 0 2

    0 0 0.035 0.035 0.035 0.035 0 0 3

    0 0 0.035 0.035 0.035 0.035 0 0 4

    0 0 0.035 0.035 0.035 0.035 0 0 5

    0 0 0.035 0.035 0.035 0.035 0 0 6

    0 0 0 0 0 0 0 0 7

    0 0 0 0 0 0 0 0 8

    AE

    0 0 0 0 0 0 0 0 1

    0 0 0 0 0 0 0 0 2

    0 0 0 0 0 0 0 0 3

    0 0 0 0 0 0 0 0 4

    0 0 0 0 0.1 0 0 0.1 5

    0 0 0 0 0 0 0 0 6

    0 0 0 0 0 0 0 0 7

    0 0 0 0 0.1 0 0 0.1 8

    AE

    1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8

    1 2 3 4 5 6 7 81 2 3 4 5 6 7 8

  • 0.135 0.035 0 0 0 0.1 0.035 0.035

    0.035 0.135 0 0.1 0 0 0.035 0.035

    0 0 0.135 0.035 0.035 0.035 0.1 0

    0 0.1 0.035 0.135 0.035 0.035 0 0

    0 0 0.035 0.035 0.135 0.035 0 0.1

    0.1 0 0.035 0.035 0.035 0.135 0 0

    0.035 0.035 0.1 0 0 0 0.1

    K AE

    1

    2

    3

    4

    5

    6

    35 0.035 7

    0.035 0.035 0 0 0.1 0 0.035 0.135 8

    1 2 3 4 5 6 7 8

    0. 0. 0.1 0. 0.035 0. 0.135

    0. 0. 0. 0. 0.035 0. 0.035

  • Application of Stiffness Method• Truss analysis

    – The global force components Q acting on the truss can be

    related to its global displacements D using

    Qk, Dk = know external loads & displacements. The

    loads here exist on the truss as part of the

    problem

    Qu, Du = unknown loads & displacements. The loads

    here represent the unknown support reactions

    K = structure stiffness matrix

    11 12

    21 22

    k u k

    u u k

    Q K D K D

    Q K D K D

    k 11 12 u

    u 21 22 k

    =

    Q KD

    Q K K D

    Q K K D

  • Application of Stiffness Method

    – The member forces can be determined using

    0. 0.1 1

    0. 0.1 1

    Nx

    x y NyN

    x y FxF

    Fy

    D

    Dq AE

    Dq L

    D

    'q k TD

    Nx

    Ny

    F x y x y

    Fx

    Fy

    D

    DAEq

    DL

    D

  • Example 3

    • Determine the force in each member of the two member truss

    shown. AE is constant.

  • Structure stiffness matrix: from previous example

    0.405 0.096 0.333 0. 0.072 0.096

    0.096 0.128 0. 0. 0.096 0.128

    0.333 0. 0.333 0. 0. 0.

    0. 0. 0. 0. 0. 0.

    0.072 0.096 0. 0. 0.072 0.096

    0.096 0.128 0. 0. 0.096 0.128

    K AE

    Displacements and loads

    Q KD

    1 1

    2 2

    3 3

    4 4

    5 5

    6

    0.405 0.096 0.333 0. 0.072 0.096

    0.096 0.128 0. 0. 0.096 0.128

    0.333 0. 0.333 0. 0. 0.

    0. 0. 0. 0. 0. 0.

    0.072 0.096 0. 0. 0.072 0.096

    0.096 0.128 0. 0. 0.096 0.128

    Q D

    Q D

    Q DAE

    Q D

    Q D

    Q

    6D

  • The known external displacements are D3 = D4 = D5 = D6 = 0.

    Determine the unknown displacements by

    The known external loads are Q1 = 0., Q2 = -2k

    Rewrite the matrix

    1

    2

    3

    4

    5

    6

    0 0.405 0.096 0.333 0. 0.072 0.096

    2 0.096 0.128 0. 0. 0.096 0.128

    0.333 0. 0.333 0. 0. 0. 0

    0. 0. 0. 0. 0. 0. 0

    0.072 0.096 0. 0. 0.072 0.096 0

    0.096 0.128 0. 0. 0.096 0.128 0

    D

    D

    QAE

    Q

    Q

    Q

    1

    2

    0 0.405 0.096 0

    2 0.096 0.128 0

    DAE

    D

    1 20 0.405 0.096AE D D

    1 22 0.096 0.128AE D D 1

    4.505D

    AE

    2

    19.003D

    AE

  • The support reactions are now obtained by

    The force in each member obtained by

    3

    4

    5

    6

    0.333 0 04.05

    0 0 0

    0.072 0.096 19.003 0

    0.096 0.128 0

    Q

    Q AEAE

    Q

    AEQ

    3 0.333(4.505) 1.5Q K

    4 0Q

    5 0.072(4.505) 0.096( 19.003) 1.5Q K

    6 0.096(4.505) 0.128( 19.003) 2.0Q K

    Nx

    Ny

    F x y x y

    Fx

    Fy

    D

    DAEq

    DL

    D

  • Member 1:

    1

    4.505

    19.0031 0 1 0 1.5

    3

    0

    0

    AE

    AEq k

    AE

    1x 0y 3L ft

    Member 2:

    2

    4.505

    19.0030.6 0.8 0.6 0.8 2.5

    5

    0

    0

    AE

    AEq k

    AE

    0.6x 0.8y 5L ft

  • Example 4

    • Determine the support reactions and the force in member 2 of

    the truss shown in figure. AE is constant

  • Example 4

    • The Stiffness matrix has been determine in Example 2 using the

    same notation as shown.

  • Example 5

    Determine the force in member 2 of the assembly if the support at joint 1

    settles downward 25mm. AE = 8103 kN

  • Member 1:

    1

    0, 1, 3

    0

    0.002580000 1 0 1 8.333

    0.005563

    0.021875

    x y L m

    q kN

    Member 2:

    2

    0.8, 0.6, 5

    0.00556

    0.02187580000.8 0.6 0.8 0.6 13.9

    05

    0

    x y L m

    q kN

    Member 3:

    3

    1, 0, 4

    0

    080001 0 1 0 11.11

    0.005564

    0.021875

    x y L m

    q kN

    Members Forces

  • Nodal Coordinate

    To solve this problem

    A set of nodal coordinate system x’’, y’’ located at the inclined

    support will be used

    If the support is an inclined roller

    The zero deflection cannot be defined using single horizontal and

    vertical global coordinate system

  • Nodal Coordinate

    '' ''

    '' ''

    0.

    0.

    0.

    0.

    Nx x

    Ny y N

    Fx x F

    Fy y

    Q

    Q q

    Q q

    Q

    The Nodal Forces

    '' '' ''

    ''

    0. 0.

    0. 0.

    Nx

    x y NyN

    x y FxF

    Fy

    D

    Dd

    Dd

    D

    The Nodal Displacements

  • Nodal Coordinate

    ''

    ''

    2

    '' ''

    2

    '' ''

    2

    '' '' '' '' ''

    2

    '' '' '' '' ''

    xx x y x x x y

    yy x y y x y y

    x x y x x x y x

    x y y y x y y y

    N

    NAEk

    FL

    F

    '' ''x y x yN N F F

    'Tk T k T

    The Stiffness Matrix

    '' '' ''

    ''

    0.

    0. 0. 0.1 1

    0. 0. 0.1 1

    0.

    x

    y x y

    x x y

    y

    AEk

    L

  • – The member forces can be determined using

    '' '' ''

    ''

    0. 0.1 1

    0. 0.1 1

    Nx

    x y NyN

    F x y Fx

    Fy

    D

    Dq AE

    Dq L

    D

    'q k TD

    ''

    ''

    '' ''

    Nx

    Ny

    F x y x y

    Fx

    Fy

    D

    DAEq

    DL

    D

    Nodal Coordinate

    The Member Forces

  • Example 6

    • Determine the support reaction for the truss shown

  • Member 1:

    '' ''

    1

    1, 0, 0.707, 0.707, 4

    0

    011 0 0.707 0.707 22.5

    127.34

    0

    x y x yL m

    EAq kN

    EA

    Member 2:

    '' ''

    2

    0, 1, 0.707, 0.707, 3

    352.5

    157.510 1 0.707 0.707 22.5

    127.33

    0

    x y x yL m

    EAq kN

    EA

    Member 3:

    3

    0.8, 0.6, 5

    0

    010.8 0.6 0.8 0.6 37.5

    352.55

    157.5

    x y L m

    EAq kN

    EA

    Members Forces

  • Thermal Changes and Fabrication Errors

    Thermal Effects

    If a truss member of length L is subjected to a temperature

    increase T, The member undergo an increase in length of

    L TL

    0

    0

    N

    F

    q AE T

    q AE T

    Then

    Transforming into global coordinate

    0

    0

    0

    0

    0.

    0. 1

    0. 1

    0.

    Nxx x

    Ny y y

    x xFx

    y yFy

    Q

    QAE T AE T

    Q

    Q

  • Thermal Changes and Fabrication Errors

    Fabrication Errors

    If a truss member is made too long by an amount L, then the force

    q0 needed to keep the member at its design length L

    0

    0

    N

    F

    AE Lq

    L

    AE Lq

    L

    In global coordinates

    0

    0

    0

    0

    Nxx

    Ny y

    xFx

    yFy

    Q

    Q AE L

    LQ

    Q

  • Thermal Changes and Fabrication Errors

    0Q KD Q

    Q0 : is the column matrix for the entire truss of theinitial fixed-end force caused by the temperaturechanges and fabrication errors of the members

    defined in previous equations

    kk 11 12 u 0

    u 21 22 k u 0

    =QQ K K D

    Q K K D Q

    Matrix Analysis

  • Thermal Changes and Fabrication Errors

    The Member forces

    '

    0q k TD q

    0

    Nx

    Ny

    F x y x y F

    Fx

    Fy

    DAE T

    DAEq q

    DLLAED L

  • Example 7

    • Determine the force in member 1 and 2 of the pin-connected

    assembly if the member 2 was made 0.01m too short before it

    was fitted into place. Take AE=8(103)

  • Example 8

    Member 2 of the truss shown is subjected to an increase in temperature

    of 150 F. Determine the force developed in member 2. E=29(106)lb/in2.

    Each member has across sectional area of A=0.75 in2

  • Springs Structures

    1 1'

    1 1

    Where:

    :is the spring stiffness

    s

    s

    k k

    k

    'q k dks

    ks q2q1

    d2d1

  • Example 9

    Given: For the spring system shown,

    K1 = 100 N/mm, K2 = 200 N/mm,

    K3 = 300 N/mm

    P = 500 N.

    Find:

    (a) The global stiffness matrix

    (b) Displacements of nodes 1 and 2

    (c) The reaction forces at nodes 3 and 4

    (d) The force in the spring 2

    1 23 41 2 3

  • Solution

    (a) The global stiffness matrix

    1

    100 0 100 0 1

    100 100 3 0 0 0 0 2

    100 100 1 100 0 100 0 3

    0 0 0 0 4

    k

    2

    200 200 0 0 1

    200 200 1 200 200 0 0 2

    200 200 2 0 0 0 0 3

    0 0 0 0 4

    k

    3

    0 0 0 0 1

    300 300 2 0 300 0 300 2

    300 300 4 0 0 0 0 3

    0 300 0 300 4

    k

  • Solution

    The global stiffness matrix

    100 0 100 0 1 200 200 0 0 1 0 0 0 0 1

    0 0 0 0 2 200 200 0 0 2 0 300 0 300 2

    100 0 100 0 3 0 0 0 0 3 0 0 0 0 3

    0 0 0 0 4 0 0 0 0 4 0 300 0 300 4

    K

    K = k1 + k2 + k3

    300 200 100 0 1

    200 500 0 300 2

    100 0 100 0 3

    0 300 0 300 4

    K

    1 23 41 2 3

  • Application of Stiffness Method– The global force components Q acting on the Structure can

    be related to its global displacements D using

    k 11 12 u

    u 21 22 k

    =

    Q KD

    Q K K D

    Q K K D

    Qk, Dk = know external loads & displacements. The loads

    here exist on the truss as part of the problem

    Qu, Du = unknown loads & displacements. The loads here

    represent the unknown support reactions

    K = structure stiffness matrix

    11 12

    21 22

    k u k

    u u k

    Q K D K D

    Q K D K D

  • (b) Displacements of nodes 1 and 2

    D3 = D4 = 0., Q1 = 0 and Q2 = P = 500

    1 1

    2 2

    10

    300 200 0. 11

    200 500 500 15

    11

    D D

    D D

    1 1

    2 2

    3 3

    4 4

    300 200 100 0

    200 500 0 300

    100 0 100 0

    0 300 0 300

    Q KD

    Q D

    Q D

    Q D

    Q D

    1

    2

    3

    4

    0 300 200 100 0

    500 200 500 0 300

    100 0 100 0 0

    0 300 0 300 0

    D

    D

    Q

    Q

  • 3

    3 4

    4

    10

    0300 200 100 0 11 1000

    500200 500 0 300 15 11

    11100 0 100 0 4500

    0 110 300 0 300

    0

    Q

    Q Q

    Q

    (c) The reaction forces at nodes 3 and 4

    (d) The force in the spring 2

    22 2 1 1

    22 2 2 2

    2 2

    1 1

    2 2

    2 2

    10 1000

    200 200 11 11

    200 200 15 1000

    11 11

    k k d q

    k k d q

    q q

    q q

  • Space Truss Analysis

    2 2 2

    2 2 2

    2 2 2

    cos

    cos

    cos

    F N F Nx x

    F N F N F N

    F N F Ny y

    F N F N F N

    F N F Nz z

    F N F N F N

    x x x x

    L x x y y z z

    y y y y

    L x x y y z z

    z z z z

    L x x y y z z

    0 0 0

    0 0 0

    x y z

    x y z

    T

  • Stiffness Matrix

    2 2

    2 2

    2 2

    2 2

    2 2

    2 2

    x y z x y z

    x x y x z x x y x z x

    y x y y z y x y y z y

    x z y z z x z y z z z

    x y x x z x x y x z x

    x y y y z x y y y z y

    x z y z z x z y z z z

    N N N F F F

    N

    N

    NAEk

    FL

    F

    F

    0

    0

    0 0 00 1 1

    0 0 00 1 1

    0

    0

    x

    y

    x y zz

    x y zx

    y

    z

    AEk

    L

  • Stiffness Matrix

    2

    2

    2

    xyz xyz

    xyz xyz

    x x y x z

    xyz x y y y z

    x z y z z

    C Ck

    C C

    AEC

    L