Truss Analysis Using The Stiffness Chapter 14 Truss Analysis Using The Stiffness Method. Stiffness Method

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  • 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 N AE

    q d L

     'F N AE

    q d L

     

    ''N F AE

    q d L

      ''F F AE

    q d L

    N N F

    AE AE q d d

    L L  

    F N F

    AE AE q 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

    AE k

    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 x x x

    L  

      

    cos F Ny y y y

    L  

      

        2 2

    cos F Nx x F N F N

    x x

    x x y y  

      

      

        2 2

    cos F Ny y F 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 TD or

    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 q or 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 TDand

    'q k T D 

    TQ T q 'TQ T k TD  Q kD

    'Tk T k T 0.

    0. 0. 0.1 1

    0. 0. 0.1 1

    0.

    x

    y x y

    x x y

    y

    AE k

    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

    NAE k

    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 0 1

    3 x

      

    0 0 0

    3 y

      

    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

    NAE k

    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

    0.333 3

    x

    L

        

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

    3 0 0.6

    5 x

      

    4 0 0.8

    5 y

      

    2 2

    2 2

    2 2

    2 2

    xx x y x x y

    yy