Truss by Direct Stiffness

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    DIRECT STIFFNESS METHOD FOR TRUSSES:

    3.1 INTRODUCTION

    In the previous Lectures the procedure for obtaining thestructure stiffness matrix was discussed. The structure stiffness matrixwas established by the following equation:

    [K] = [T]T [kc] [T] -----------(2.21)

    However if a large and complicated structure is to be analyzedand if more force components are to be included for an element thenthe size of composite stiffness matrix [kc] and deformationtransformation matrix [T] will be increased. Therefore the procedureoutlined in the preceding chapter for formation of structure stiffnessmatrix appears to be inefficient, furthermore this procedure is notsuitable to automatic computation.

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    In this chapter an alternative procedure called the direct stiffnessmethod is introduced. This procedure provides the basis for mostcomputer programs to analyze structures. In this method eachindividual element is treated as a structure and structure stiffnessmatrix is obtained for this element using the relationship:

    [K]m = [T]T

    m[k]m [T]m -----------(3.1)

    Where

    [K]m= Structure stiffness matrix of an individual element.

    [T]m = Deformation Transformation matrix of an individualelement.

    [k]m = Member Stiffness matrix of an individual element.

    Total structure stiffness matrix can be obtained by superimposing thestructure stiffness matrices of the individual elements..

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    As all members of a truss are not in the same direction i.e.

    inclination of the longitudinal axes of the elements varies, thereforestiffness matrices are to be transformed from element coordinatesystem to structure or global coordinate system. When the matrices forall the truss elements have been formed then adding or combiningtogether the stiffness matrices of the individual elements can generatethe structure stiffness matrix [K] for the entire structure, because ofthese considerations two systems of coordinates are required.

    i ) Local or member or element coordinate system:

    In this coordinate system x-axis is collinear with the longitudinalaxis of the element or member. Element stiffness is calculated with

    respect to this axis. This system is illustrated in figure 3.1

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    Element 'B'Element 'A'

    A B

    Structure or

    globel axes

    Local or member

    or element axes for

    element B

    Local or member

    or element axes

    for element 'A'

    Figure 3.1

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    ii) Structure or global coordinate system:

    A single coordinate system for the entire structure is chosen, with

    respect to which stiffness of all elements must be written.

    3.2 PROCEDURE FOR THE FORMATION OF TOTAL STRUCTURESTIFFNESS MATRIX FOR AN ELEMENT USING DIRECT STIFFNESS

    METHOD:

    Following is the procedure for the formation of structure

    stiffness matrix:

    i) Formation of the element stiffness matrix using equation 2.16.

    11

    11

    L

    AEk m ----------- (2.16)

    ii) Formation of the deformation transformation

    matrix [T] for a single element:

    []m = [T]m []m -----------(3.2)

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    where

    []m= Element or member deformation matrix.[]m= Structure deformation matrix of an element or member[T]m= Element or member deformation transformation matrix.

    (iii)Formation of a structure stiffness matrix [K] or an element using the

    relation

    [K]m = [T]T

    m[k]m [T]m -----------(3.1)

    The above mentioned procedure is discussed in detail in the subsequent

    discussion.

    3.2.1 The formation of element sti ffness matr ix in local co-ordinates:

    It has already been discussed in the previous chapter. However it

    is to be noted that for horizontal members the structure stiffness matrix

    and element stiffness matrix are identical because both membercoordinate systems and structure coordinate systems are identical. But

    for inclined members deformation transformation matrices are to be

    used because member coordinate system and structure coordinate

    system are different; therefore their structure stiffness matrix and

    element stiffness matrix will also be different.

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    3.2.2 The formation of deformation transformation matrix:

    As the main difference between the previously discussed method and directstiffness method is the formation of the deformation transformation matrix.In this article deformation transformation matrix for a single element will be

    derived.Before the formation of deformation transformation matrix followingconventions are to be established in order to identify joints, members,element and structure deformations.

    1) The member is assigned a direction. An arrow is written along the member,with its head directed to the far end of the member.

    2) i, j, k, and l, are the x and y structure deformations at near and far (tail& head) ends of the member as shown in figure 3.2. These are positive in theright and upward direction.

    3) r and s are the element deformations at near and far (tail & head) ends ofthe member as shown in figure 3.2.

    4)The member axis (x-axis of member coordinate system) makes an angle x,with the x-axis of the structure coordinate system as shown in figure 3.2.

    5)The member axis (x-axis of member coordinate system) makes an angle y,with the y-axis of the structure co-ordinate system as can be seen from figure3.2.

    6) The cosines of these angles are used in subsequent discussion Letters l andm

    represent these respectively.

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    8Figure 3.2

    Near

    end

    Far

    end

    Memberdirection

    x

    y

    (b)

    r

    s

    i

    j

    k

    Nearend

    Farend

    Memberdirection

    (a)

    x

    y

    l

    i

    j

    k

    r

    s

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    2122

    12

    1212

    YYXX

    XX

    L

    XX

    = Cosxand m = Cos

    y

    = Cos

    x=

    m = Cosy=

    2

    12

    2

    12

    1212

    YYXX

    YY

    L

    YY

    The algebraic signs ofs will be automatically accounted for the members

    which are oriented in other quadrants of X-Y plan.

    In order to form deformation transformation matrix, once again considerthe member of a truss shown in figure 3.1.

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    s=0

    r

    j

    i

    s=0

    j

    r=1

    (b)(a)

    =1

    j

    rjc

    osy

    =mr

    cos

    xl

    x

    i

    i

    i

    y

    x

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    r=0

    r=0

    y

    x

    Figure 3.3

    (d)(c)

    =1k

    =1

    s1cos

    y=m

    1

    l

    kk

    s

    kco

    sx=

    1

    s

    s

    x

    y

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    Following four cases are considered to form deformation transformation matrix.

    Case-1: Introduction of horizontal deformation to the structure i = 1, while farend of the member is hinged (restrained against movement). From the geometry

    of figure 3.3(a)

    r = i Cosx = 1. Cosx = Cosx = lr =l ---------- (3.3)s = 0 ---------- (3.4)

    Case-2: I ntroduction of vertical deformation to the structurej = 1, while nearend of the member is hinged (restrained against movement). F rom the geometry of

    f igure 3.3 (b) r =j Sinx =j Cosy = 1. Cosy = mr = m ---------- (3.5)s = 0 ---------- (3.6)

    Case-3:Int roduc t ion of hor izontal deformat ion to the structurek = 1, whi lenear end o f the member is hinged (restrained against m ovement).From th e

    geometry of f igure 3.3(c)

    s =k Cosx = 1. Cosx = Cosx = lr = 0 ---------- (3.7)s = l ---------- (3.8)

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    Case-4: Introduction of vertical deformation to the structure L = 1, whilenear end is hinged

    From the geometry of figure 3.3(d)

    L = 1r = 0 ---------- (3.9)s = LCosy = m ---------- (3.10)

    Effect of four structure deformations and two member deformations can bewritten as

    r =i Cosx +j Cosy +k .0 + L .0s =i .0 +j .0 +k Cosx + L Cosy

    r= l.

    i+ m.

    j+ 0.

    k+ 0.

    L ----------- (3.11)

    s= 0.

    i+ 0.

    j+ l.

    k+ m.

    l----------- (3.12)

    Arranging equations 3.11 and 3.12 in matrix from

    L

    k

    j

    i

    s

    r

    mm

    0000

    ----------- (3.13)

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    Comparing this equation with the following equation

    []m = [T]m []m ---------------(3.2)After comparing equation (3.14) and (3.2) following deformation

    transformation matrix is obtained.

    mm

    T m

    0000 - ---------- (3.14)

    This matrix [T]m transforms four structure deformation (i, j, k, L )into two element deformation (r and s).3.2.3 Formation of structure stiffness matrix:

    Structure Stiffness matrix of an individual member is first to be

    transformed from member to structure coordinates. This can be done by

    using equation 3.1.

    [K] = [T]Tm [k]m [T]m --------(3.1)

    ml

    mlmT

    00

    00 ----- (3.15)

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    1111

    L

    AEk m ------(2.16)

    Lkji

    m

    m

    m

    m

    L

    k

    j

    i

    Km

    00

    00

    11

    11

    0

    0

    0

    0

    Lkji

    mm

    mm

    L

    AE

    m

    m

    L

    k

    ji

    Km

    0

    0

    00

    L

    k

    j

    iLkji

    mmmm

    mm

    mmmm

    mm

    L

    AEKm

    22

    22

    22

    22

    --------------------------------(3.16)

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    3.2.4 ASSEMBLING OF THE INDI VIDUAL STRUCTURE ELEMENTSTI FFNESS MATRICES TO FORM TOTAL STRUCTURESTIFFNESS MATRIX

    Combining the stiffness matrix of the individual members can generatethe stiffness matrix of a structure However the combining processshould be carried out by identifying the truss joint so that matrixelements associated at particular member stiffness matrices arecombined. The procedure of formation of structure stiffness matrix isas follows:

    Step 1: Form the individual element stiffness matrices for each member.

    Step 2: Form a square matrix, whose order should be equal to that ofstructure deformations.

    Step 3: Place the elements of each individual element stiffness matrix

    framed into structure in the corresponding rows and columns ofstructure stiffness matrix of step-2.

    Step 4: If more than one element are to be placed in the same location ofthe structure stiffness matrix then those elements will be added.

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    1

    2 3

    2

    3

    2

    1

    Structurestiffness

    matrix formember-2

    Structurestiffness

    matrix formember-1

    Structure

    stiffness matrix

    Foll owing f igure shows the above mentioned process.

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    18

    2

    121

    kk

    kkK

    11

    11

    1

    3

    2

    32

    kk

    kkK

    22

    22

    2

    21 KKK

    3

    2

    1

    kk

    kkkk

    kk

    K

    321

    22

    2211

    11

    0

    0

    In the above matrix the element in the second row and

    second column is k1+k2 where k1 is for member 1 and k2

    is for member 2.This is because both k1 and k2 have

    structure coordinates 2-2.

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    Illustrative Examples Regarding the Formation of [K] Matrix:

    L/2

    (x1,y1) = (0,0)

    (x1,y1) = (L/2,2L/3)(x2,y2) = (L/2,2L/3)

    (x2,y2) = (L,0)

    (x1,y1) = (L,0)(x2,y2) = (0,0)

    (a)

    (b)

    (c)

    (Structure forces and deformations)

    (Free body diagram indicating structure forces

    and coordinates)

    2L/31 1

    2 2

    6 4

    5 3

    6 4

    5 3

    1

    5

    6

    3

    4

    2

    12

    11.5

    1

    1

    2

    23

    3

    L/2

    Example 3.1: Form the structure stiffness matrix for the following

    truss by direct stiffness method.

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    Member Length Near

    End

    X1 Y1

    Far End

    X2, Y2

    i j k L l=Cosx =X2-X1/

    Length

    m=CosyY2-Y1/

    Length

    1 5L/6 0 0 L/2 2L/3 5 6 1 2 0.6 0.8

    2 5L/6 L/

    2

    2L/3 L 0 1 2 3 4 0.6 -0.8

    3 L L 0 0 0 3 4 5 6 -1 0

    l

    k

    ji

    mlmmlm

    lmllmlmlmmlmlmllml

    lkji

    L

    AEK m

    22

    22

    22

    22

    According to eq. (3.16) we get

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    2165

    768.0576.0768.0576.0576.0432.0576.0432.0768.0576.0768.0576.0576.0432.0576.0432.0

    2165

    1

    L

    AE

    K

    432

    1

    768.0576.0768.0576.0576.0432.0576.0432.0 768.0576.0768.0576.0

    576.0432.0576.0432.0

    4321

    2

    LAEK

    6

    5

    43

    0000

    000.1000.1

    0000000.1000.1

    6543

    L

    AE3K

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    22

    6

    5

    4

    3

    2

    1

    768.0576.0000.0000.0768.0576.0

    576.0432.1000.0000.1576.0432.0

    000.0000.0768.0576.0768.0576.0

    000.0000.1576.0432.1576.0432.0

    768.0576.0768.0576.0536.1000.0

    576.0432.0576.0432.0000.0864.0

    654321

    L

    AEK

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    Example 3.2: Form the structure stif fness matrix of the following truss.

    L L

    P

    1

    2

    3

    4

    5

    6

    7

    8

    (L,L)

    (0,0)

    (L,0)

    (2L,0)

    (L,L)

    (L,L)

    y

    x

    1

    2

    32 2

    1

    1

    3

    4

    65

    7

    8

    1

    2

    Free body diagram showing structure forces andcoordinates

    Structure forces and deformations

    Structure to be analysed

    L

    1

    2

    3

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    Member Near End Far End Length =Cosx m=Cosy i j k L

    X1 Y1 X2 Y2

    1 0 0 L L 2L 0.707 0.707 7 8 1 2

    2 L L L 0 L 0 -1 1 2 5 6

    3 L L 2L 0 2L 0.707 -0.707 1 2 3 4

    22

    22

    22

    22

    mlmmlm

    lmllml

    mlmmlm

    lmllml

    L

    AEmK

    So

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    25

    2

    1

    8

    7

    0.50.5.5.5

    0.50.5.5.5

    .5.50.50.5

    .5.50.50.5

    2187

    1

    2L

    AEK

    6

    5

    2

    1

    1010

    0000

    1010

    0000

    6521

    2

    L

    AE

    K 4

    3

    2

    1

    .5.5.5.5

    .5.5.5.5

    .5.5.5.5

    .5.5.5.5

    3

    4321

    2L

    AE

    K

    Now the matrix can be written as [k]2:

    6

    5

    2

    1

    414.10414.10

    0000

    414.10414.10

    0000

    6521

    22

    L

    AEK

    2

    1

    8

    7

    5.5.5.5.

    5.5.5.5.

    5.5.5.5.

    5.5.5.5.2187

    21

    L

    AEK

    So [k] = [k]1+ [k]

    2+ [k]

    3

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    8

    7

    6

    54

    3

    2

    1

    5.5.00005.5.

    5.5.00005.5.

    00414.1000414.10

    0000000000005.5.5.5.

    00005.5.5.5.

    5.5.414.105.5.414.20

    5.5.005.05.01

    87654321

    2

    L

    AEK

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    3.4 Force transformation matrix:

    The axial force (wm) in the members of a truss can befound by using the relationship between member forces

    and member deformations [equation (2.15)] and betweenelement and structure deformations [equation (3.1)]

    [w]m= [k]m []m ------------ (2.15)[]m = [T]m []m ------------ (3.3)

    Substituting value of []m from equation 3.1 into equation2.15

    [w]m = [k]m [T]m []m ------------ (3.18)Substituting value of [k]m and [T]m from equation 2.16 and

    3.15 respectively

    L

    k

    j

    i

    mmmm

    L

    AEmw 11

    11

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    Lk

    j

    i

    mmmm

    L

    AE

    wsrw

    In above Equations, wr= Force at near end

    See figure below:

    ws

    = Force at far end

    As wr = -ws so:

    l

    k

    j

    i

    smlml

    L

    AEw -------------- (3.19)

    Sign conventions

    wr

    ws

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    If

    wris positive then member is in compression.

    wsis negative then member is in compression.

    wris negative then member is in tension.

    wsis positive then member is in tension.

    3.5 ANALYSIS OF TRUSSES USING DIRECT STI FFNESS METHOD:

    Basing on the derivations in the preceding sections of this chaptera truss can be completely analyzed. The analysis comprises ofdetermining.

    i) Joint deformations.

    ii) Support reactions.

    iii) Internal member forces.

    As the first step in the analysis is the determination of unknown jointdeformation. Using the equations can do this.

    [W] = [K] []The matrices [W], [K] and [] can be divided in submatrices in the

    following form

    k

    u

    u

    k

    KK

    KK

    W

    W

    2221

    1211 ----------- (3.20)

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    Where

    Wk = Known values of loads at joints.

    Wu = Unknown support reaction.

    u = Unknown joint deformation.k = Known deformations, generally zero due to supportconditions.

    K11, K12, K21, K22are the sub-matrices of [K]

    Expanding equation 3.20

    Wk = K11u + K12

    k ----------- (3.21)

    Wu = K21u + K22k ----------- (3.22)If the supports do not move, then k = 0 therefore equation 3.21

    & 3.22 can be written as

    Wk = K11u ----------- (3.23)Wu = K21u ----------- (3.24)

    By pre-multiplying equation 3.23 by [K11]-1 following equation isobtained

    [K11]-1 [WK]=[K11]-1 [K11][U][u] = [K11]-1 [Wk] ----------- (3.25)

    Substituting value u from equation 3.25 into equation 3.24[Wu] = [K21] [K11]-1 [Wk] ----------- (3.26)

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    Using equation 3.25, 3.26 and 3.19, joint deformations, support reactionsand internal member forces can be determined respectively.

    As this method does not depend upon degree of indeterminacy so it canbe used for determinate as well as indeterminate structures.

    Using the basic concepts as discussed in the previous pages, followingare the necessary steps for the analysis of the truss using stiffnessmethod.

    1-Identify the separate elements of the structure numerically and specifynear end and far end of the member by directing an arrow along thelength of the member with head directed to the far end as shown in thefig. 3.2

    2-Establish the x,y structure co-ordinate system. Origin be located at oneof the joints. Identify all nodal co-ordinates by numbers and specifytwo different numbers for each joint (one for x and one for y). Firstnumber the joints with unknown displacements.

    3-Form structure stiffness matrix for each element using equation 3.16.

    4-Form the total structure stiffness matrix by superposition of the element

    stiffness matrices.5-Get values of unknown displacements using equation 3.25.

    6-Determine support reaction using equation 3.26.

    7-Compute element or member forces using equation 3.19.

    3 6 Ill t ti E l R di C l t A l i f T

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    3.6- Illustrative Examples Regarding Complete Analysis of Trusses:

    Example 3.3: Solve truss in example 3.1 to find member forces.

    L/2

    (x1,y1) = (0,0)

    (x1,y1) = (L/2,2L/3)(x2,y2) = (L/2,2L/3)

    (x2,y2) = (L,0)

    (x1,y1) = (L,0)(x2,y2) = (0,0)

    (a)

    (b)

    (c)

    (Structure forces and deformations)

    (Free body diagram indicating structure forces and

    coordinates)

    2L/31 1

    2 2

    6 4

    5 3

    6 4

    5 3

    1

    5

    6

    3

    4

    2

    12

    11.5

    1

    1

    2

    23

    3

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    The [K] matrix for the truss as formed in example 3.1 is:

    6

    5

    4

    3

    2

    1

    768.0576.0000.0000.0768.0576.0

    576.0432.1000.0000.1576.0432.0

    000.0000.0768.0576.0768.0576.0

    000.0000.1576.0432.1576.0432.0

    768.0576.0768.0576.0536.1000.0

    576.0432.0576.0432.0000.0864.0

    654321

    L

    AEK

    Using the relation [W] = [K] [] we get

    k

    u

    u

    k

    KK

    KK

    W

    W

    2221

    1211

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    6

    5

    4

    3

    2

    1

    654321

    6

    5

    4

    3

    2

    1

    768.0576.0000.0000.0768.0576.0

    576.0432.1000.0000.1576.0432.0

    000.0000.0768.0576.0768.0576.0

    000.0000.1576.0432.1576.0432.0

    768.0576.0768.0576.0536.1000.0

    576.0432.0576.0432.0000.0864.0

    L

    AE

    W

    W

    W

    W

    W

    W

    From equation (3.25) [u] = [K11]-1 [Wk]

    2

    1

    536.1000.0

    000.0864.0

    2

    11

    W

    W

    AE

    L

    125.11

    651.0000.0000.0157.1

    2

    1AEL

    812.7

    31.13

    2

    1

    AE

    L

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    Using the relation:

    [Wu] = [K21] [u]

    812.7

    310.13

    768.0576.0

    576.0432.0

    768.0576.0

    576.0432.0

    6

    5

    4

    3

    W

    W

    W

    W

    1.66

    1.25

    13.66

    10.25

    6W5W4W3W

    Now the member forces can be found using the relation:

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    36

    L

    k

    j

    i

    mmL

    AEwr

    09.2

    812.7

    31.13

    00.0

    00.0

    8.06.08.06.05

    61

    AE

    L

    L

    AEw kips

    09.17

    00.0

    00.0

    812.7

    31.13

    8.06.08.06.05

    62

    AE

    L

    L

    AEw kips

    00.0

    00.0

    00.0

    00.0

    00.0

    0.00.10.00.13

    AE

    L

    L

    AEw

    kips

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    11.5k

    12k

    1.25k 10.25k

    13.66k

    17.09kip

    2.09

    kip

    s

    1.66k

    0 kips

    Final Resul ts & sketch

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    Example 3.4:

    L L

    P

    1

    2

    3

    4

    5

    6

    7

    8

    (L,L)

    (0,0)

    (L,0)

    (2L,0)

    (L,L)

    (L,L)

    y

    x

    1

    2

    32 2

    1

    1

    3

    4

    65

    7

    8

    1

    2

    (Structure forces and deformations)

    (Structure to be analysed)

    L

    1

    2

    3

    (Free body diagram showing structure forces and

    coordinates)

    Solve the truss in example 3.2 to find member forces.

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    The [K]matrix for the truss as formed in example 3.2 is as follows:

    8

    7

    65

    4

    3

    2

    1

    5.5.00005.5.

    5.5.00005.5.

    00414.1000414.1000000000

    00005.5.5.5.

    00005.5.5.5.

    5.5.414.105.5.414.20

    5.5.005.05.01

    87654321

    2

    L

    AEK

    Now as W = k

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    8

    7

    6

    5

    4

    3

    2

    1

    8

    7

    6

    5

    4

    3

    2

    1

    5.5.00005.5.

    5.5.00005.5.

    00414.1000414.10

    0000000000005.5.5.5.

    00005.5.5.5.

    5.5.414.105.5.414.20

    5.5.005.05.01

    87654321

    2LAE

    W

    W

    W

    WW

    W

    W

    W

    k

    u

    u

    k

    kk

    kk

    W

    W

    2221

    1211

    As k= 0 or 3, 4, 5, 6, 7,8, all are zero due to supports and using[W

    k] = [k

    11] [

    u] we get

    W

    W

    AE

    L

    1

    2

    1

    22

    1 0

    0 2 414

    .

    Which gives

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    W1= 0

    W2= P

    Also using [Wu] = [k21] [u] we have

    0

    2

    1 0

    0 2 414

    1

    2P

    AE

    L

    .

    Or

    PAE

    L 0

    414.20

    0121

    2

    1

    2 1

    2 4142 414 0

    0 10L

    AE P.

    ..

    Or

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    42

    1

    2

    2

    2 414

    0

    L

    AE P.

    Or

    22

    2 414 LP

    AE.Or

    2 05858 .PL

    AE

    Now Wu= [K

    21] [

    u]

    W

    W

    WW

    W

    W

    AE

    L

    3

    4

    5

    6

    7

    8

    1

    22

    5 5

    5 5

    0 00 1414

    5 5

    5 5

    . .

    . .

    .

    . .

    . .

    W

    W

    W

    W

    W

    W

    P

    3

    4

    5

    6

    7

    8

    1

    2 414

    5 5

    5 5

    0 0

    0 1414

    5 5

    5 5

    0

    .

    . .

    . .

    .

    . .

    . .

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    43

    W

    W

    W

    W

    W

    W

    P

    P

    P

    P

    P

    3

    4

    5

    6

    7

    8

    1

    2 414

    0 5

    0 5

    0

    1414

    0 5

    0 5

    .

    .

    .

    .

    .

    .

    W3= 0.207 P

    W4= -0.207 P

    W5= 0

    W6= -0.5858 PW

    7= -0.207 P

    W8= -0.207 P

    Now as

    l

    k

    j

    i

    s

    r

    mlml

    mlml

    L

    AE

    w

    w

    So

    2

    1

    8

    7

    l

    k

    j

    s

    i

    mlmlL

    AEw

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    So, for member # 1:

    AE

    LP

    L

    AE

    wsw5858.0

    0

    0

    0

    707.0707.0707.0707.01

    Or

    ws= 0.707 x 0.5858 P

    w1= 0.414 P

    Similarl y, for member # 2:

    6

    5

    2

    1

    0

    0

    5858.

    0

    10102

    AE

    PL

    L

    AEwws

    Or w2= 1 x 0.5858 P

    Or w2= 0.5858 P

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    For member # 3:

    4

    3

    2

    1

    l

    k

    js

    i

    mlmlL

    AEW

    0

    0

    5858.0

    707.707.707.707.3 AE

    PL

    LAEw

    Orw3= 0.707 x 0.5858 P

    w3= 0.414 P

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    46

    P

    0.207P 0.59P 0.207P

    0.207P

    0.207P

    0.41

    4P0.414P

    0.5

    86P

    w1= 0.414 P

    w2= 0.586 P

    w3= 0.414 P

    Example 3 5 Analyze the truss shown in the f igure

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    Example 3.5 Analyze the truss shown in the f igure.

    5k2kW1,1

    W7,7

    W8,8

    W5,5W6,

    6

    W3,3

    W4,4

    W2,2

    12

    3

    4

    5

    8' 6'

    8'

    Truss to be analysedStructure forces and

    displacements

    A and E are constant for each member.

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    48

    Member Length l m i j k l

    1 10 -0.6 -0.8 7 8 1 2

    2 11.34 -0.7071 0.7071 1 2 3 4

    3 8 1 0 3 4 5 6

    4 6 1 0 5 6 7 8

    5 8 0 1 1 2 5 6

    Using the properties given in the above table we can find the structures

    stiffness matrices for each element as follows.

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    Element forces and displacements

    w , 5

    w , 4

    5

    4

    w , w ,

    2

    3 3

    w , 9 9

    6w ,

    3

    6w ,

    5

    w , 10

    7

    10

    7

    1

    2 2

    w ,

    w ,

    4

    8

    1

    8

    1

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    50

    6

    5

    4

    3

    0000

    0125.00125.0

    0000

    0125.00125.0

    k

    6543

    4

    3

    2

    1

    044.0044.0044.0044.0

    044.0044.0044.0044.0

    044.0044.0044.0044.0

    044.0044.0044.0044.0

    k

    4321

    21

    8

    7

    064.0048.0064.0048.0048.0036.0048.0036.0

    064.0048.0064.0048.0

    048.0036.0048.0036.0

    k

    2187

    3

    2

    1

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    51

    8

    7

    4

    3

    125.00125.00

    0000

    125.00125.00

    0000

    k

    8743

    8

    7

    6

    5

    0000

    0167.00167.0

    0000

    0167.00167.0

    k

    8765

    5

    4

    Using equation [K] = [K1]+[K2]+[K3]+[K4]+[K5] we get the structure

    stiffness matrix of (8x8) dimensions. Partitioning this matrix with respectto known and unknown deformations we get [K11] and [K12] portions as

    follows.

    21

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    52

    665.9

    983.62

    2

    5

    233.0004.0

    004.008.01

    2

    1

    2

    1

    233.0004.0

    004.008.0

    21

    11

    K

    8

    7

    6

    5

    4

    3

    064.0048.0

    048.0036.0

    125.00

    00

    044.0044.0

    044.0004.0

    21

    12

    K

    Using equation [Du]=[K11]1[Wk] we get the

    Putting the above value in equation [Wu]=[K21][Du]

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    53

    405.2803.1

    208.1

    0197.3

    197.3

    665.9

    983.62

    064.0048.0048.0036.0

    125.00

    00044.0044.0

    044.0004.0

    6

    5

    4

    3

    2

    1

    WW

    W

    WW

    W

    Element forces can be calculated using equation as follows.

    0

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    54

    4.54kips

    3.197 K

    3.197 K

    1.2

    08

    kips

    -3.006

    kip

    s

    2 K

    5 K

    1.803 K

    2.405 K

    0 K

    1.208 K

    kipw

    kipw

    kipw

    kipw

    kipw

    208.1

    0

    0

    665.9

    983.62

    10108

    1

    0

    0

    0

    0

    0

    01016

    1

    0

    0

    0

    0

    0

    01018

    1

    540.4

    0

    0

    665.9

    983.62

    707.0707.0707.0707.0314.11

    1

    006.3

    665.9

    983.62

    08.06.08.06.0

    10

    1

    5

    4

    3

    2

    1