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1 ! Fundamentals of the Stiffness Method ! Member Local Stiffness Matrix ! Displacement and Force Transformation Matrices ! Member Global Stiffness Matrix ! Application of the Stiffness Method for Truss Analysis ! Trusses Having Inclined Supports, Thermal Changes and Fabrication Errors ! Space-Truss Analysis TRUSSES ANALYSIS

TRUSSES ANALYSIS · Trusses Having Inclined Supports, Thermal Changes and Fabrication Errors! Space-Truss Analysis TRUSSES ANALYSIS. 2 2-Dimension Trusses. 3 Fundamentals of the Stiffness

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Text of TRUSSES ANALYSIS · Trusses Having Inclined Supports, Thermal Changes and Fabrication Errors!...

  • 1

    ! Fundamentals of the Stiffness Method! Member Local Stiffness Matrix! Displacement and Force Transformation Matrices! Member Global Stiffness Matrix! Application of the Stiffness Method for Truss

    Analysis! Trusses Having Inclined Supports, Thermal Changes

    and Fabrication Errors! Space-Truss Analysis

    TRUSSES ANALYSIS

  • 2

    2-Dimension Trusses

  • 3

    Fundamentals of the Stiffness Method

    Node and Member Identification

    Global and Member Coordinates

    Degrees of Freedom

    12

    3 4

    2

    1

    3

    (x1, y1)

    (x3, y3)

    (x2, y2)

    (x4, y4)

    1

    23

    4

    56

    78

    34

    56

    78

    x

    y

    Known degrees of freedom D3, D4, D5, D6, D7 and D8 Unknown degrees of freedom D1 and D2

  • 4

    AE/LAE/L

    x djAE/L x d´j

    AE/L

    x diAE/LAE/L x d´i

    Member Local Stiffness Matrix

    x´y´

    i

    j

    q´i

    q´j

    AE/L

    AE/L

    jii dLAEd

    LAEq ''' −=

    AE/LAE/Ld´ i =

    1

    d´ j = 1

    x di

    x dj

    jij dLAEd

    LAEq ''' +−=

    −=

    j

    i

    j

    i

    dd

    LAE

    qq

    ''

    1111

    ''

    −=

    1111

    ]'[L

    AEk

    [q´] = [k´][d´] ----------(1)

    q´j

    q´ix´

    x´y´

  • 5

    x´y´

    m

    i

    j

    (xi,yi)

    (xj,yj)x

    y

    Displacement and Force Transformation Matrices

    θyθx

    22 )()(cos

    ijij

    ijijxx

    yyxx

    xxL

    xx

    −+−

    −=

    −== θλ

    22 )()(cos

    ijij

    ijijyy

    yyxx

    yyL

    yy

    −+−

    −=

    −== θλ

  • 6

    x

    y

    Global

    m

    i

    jdjx

    djy

    dix

    diy

    djx

    djyx´y´

    m

    i

    j

    Local

    d´i

    d´j

    dix

    diy

    d´j

    d´i

    Displacement Transformation Matrices

    yiyxixi ddd θθ coscos' +=

    =

    jy

    jx

    iy

    ix

    yx

    yx

    j

    i

    dddd

    dd

    λλλλ00

    00''

    θyθx

    yjyxjxj ddd θθ coscos' +=

    =

    yx

    yxTλλ

    λλ00

    00][

    λx λy

    [d´] = [T][d] ----------(2)

  • 7

    x

    y

    θyθx

    m

    i

    j

    x

    y

    Global

    x´y´

    m

    i

    j

    Local

    q´i

    q´j

    Force Transformation Matrices

    θyθx

    xiix qq θcos'=

    yiiy qq θcos'=

    xjjx qq θcos'=

    yjjy qq θcos'=

    =

    j

    i

    y

    x

    y

    x

    jy

    jx

    iy

    ix

    qq

    qqqq

    ''

    00

    00

    λλ

    λλ

    =

    y

    x

    y

    x

    TT

    where

    λλ

    λλ

    00

    00

    ][

    λx

    λy

    [q] = [T]T[q´] ----------(3)

    qjx

    qjy

    qix

    qiy

  • 8

    Member Global Stiffness Matrix

    [ q ] = [ T ]T ([k´][d´] + [q´F] ) = [ T ]T [ k´ ][T][d] + [ T ]T [q´F] = [k][d] + [qF]

    [ k ] [ k ] = [ T ]T[ k´ ][T]

    [qF] = [ T ]T [q´F]

    [q] = [T]T[q´] ----------(3)

    Substitute ( [q´] = [k´][d´] + [q´F] ) into Eq. 3, yields the result,

    λyλx

    λxλx

    −λyλx

    −λxλx

    λyλy

    λxλy

    −λyλy

    −λxλyλyλx

    λxλx

    −λyλx

    −λxλx

    λyλy

    λxλy

    −λyλy

    −λxλy

    V U VU

    [ k ] = AELVUV

    U

    00λyλxλyλx00

    -11

    1-1

    AEL[ k

    ] =

    00

    λy

    λx

    λy

    λx

    00

  • 9

    [Qa] = [K][D] + [QF]

    [Qk] = [K11][Du] + [K12][Dk] + [QF]

    Reaction Boundary Condition

    Unknown DisplacementJoint Load

    Equilibrium Equation:

    Partitioned Form:

    Application of the Stiffness Method for Truss Analysis

    [Du] = (([Qk] - [QF]) - [K12][Dk])[Ku]-1

    Qk

    Qu

    Du

    Dk=

    K12

    K22

    K11

    K21

    +QFk

    QFu

  • 10

    +

    −=

    jF

    iF

    j

    i

    j

    i

    qq

    dd

    LAE

    qq

    ''

    ''

    1111

    ''

    jy

    jx

    iy

    ix

    yx

    yx

    dddd

    λλλλ00

    00

    +

    −=

    jF

    iF

    jy

    jx

    iy

    ix

    yx

    yx

    j

    i

    qq

    DDDD

    LAE

    qq

    ''

    0000

    1111

    ''

    λλλλ

    Member Forces

    x

    y

    θyθx

    x´y´

    m

    i

    j

    q´i

    q´j

  • 11

    +

    −−

    −−=

    jF

    iF

    jy

    jx

    iy

    ix

    yxyx

    yxyx

    j

    i

    qq

    DDDD

    LAE

    qq

    ''

    ''

    λλλλλλλλ

    Dyi

    Dxi

    Dyj

    Dxjq´j =

    AEL −λx −λy λx λy qj´

    F+

    x

    y

    θyθx

    x´y´

    m

    i

    j

    Member Forces

    q´i

    q´j

  • 12

    Member Forces

    Dyi

    Dxi

    Dyj

    Dxjqm =

    AEL −λx −λy λx λy qj´

    F+

    x

    y

    θyθx

    x´y´

    m

    i

    j

    Member Forces

    qm

  • 13

    3 m

    3 m

    4 m 4 m

    80 kN

    50 kN

    5 m

    5 m5 m

    Example 1

    For the truss shown, use the stiffness method to:(a) Determine the deflections of the loaded joint.(b) Determine the end forces of each member and reactions at supports.Assume EA to be the same for each member.

  • 14

    Ljyy

    Lixx ijij

    ij

    )()( −+−

    cosθx = λx cosθy = λy

    1

    2

    3

    434

    56

    78

    1

    2

    (0,0)(-4,-3)

    (-4,3)

    (4,-3)

    31

    23 m

    3 m

    4 m 4 m

    80 kN

    50 kN

    5 m

    5 m5 m

    Member

    #1

    #2

    λx λy

    #3

    λyλx

    λxλxλyλy

    λxλy

    λyλx

    λxλxλyλy

    λxλy

    −λyλx

    −λxλx−λyλy

    −λxλy

    −λyλx

    −λxλx−λyλy

    −λxλy

    Vi Uj VjUi

    [ k ]m = AEL

    ViUjVj

    Ui

    31

    2

    -4/5 = -0.8

    -4/5 = -0.8

    4/5 = 0.8

    -3/5 = -0.6

    3/5 = 0.6

    -3/5 = -0.6

  • 15

    -0.48

    1.08-0.48

    1.92-0.48

    0.64

    0.36

    -0.481

    2

    1 2

    [K] = AE5

    1

    2

    3

    434

    56

    78

    1

    2

    31

    2

    0.48

    0.64

    0.36

    0.48-0.48

    -0.64

    -0.36

    -0.48

    -0.48

    -0.64

    -0.36

    -0.48[ k ]1 =

    2 3 41

    AE5

    23

    4

    1

    [ k ]2 =

    2 5 61

    AE5

    25

    6

    1

    -0.48

    0.64

    0.36

    -0.480.48

    -0.64

    -0.36

    0.48

    0.48

    -0.64

    -0.36

    0.48

    [ k ]3 =

    2 7 81

    AE5

    27

    8

    1

    -0.48

    0.64

    0.36

    -0.480.48

    -0.64

    -0.36

    0.48

    0.48

    -0.64

    -0.36

    0.48

    31

    2

    34

    56

    78

    Member

    #1

    #2

    #3

    λx

    -0.8

    - 0.8

    0.8

    λy

    -0.6

    0.6

    -0.6

    λx2 λx λy λy2

    0.64 0.48 0.36

    0.64 -0.48 0.36

    0.64 -0.48 0.36

    0.48

    0.64

    0.36

    0.48

    -0.48

    0.64

    0.36

    -0.48

  • 16

    34

    56

    78

    1

    2

    31

    2

    Global

    Q1 = -50

    Q2 = -80

    D1

    D2

    D1

    D2 =-250.65/AE

    -481.77/AE

    3 m

    3 m

    4 m 4 m

    80 kN

    50 kN

    5 m

    5 m5 m

    1

    2

    1

    -0.48

    2

    -0.48

    1.08

    1.92

    = AE5

    80 kN

    50 kN

    0

    0+1

    2

  • 17

    1

    2

    3

    434

    56

    78

    1

    2

    31

    2

    Local

    = -97.9 kN (C)

    [q´F]1 = AE5 0.8 0.6 -0.8 -0.6

    D2=D1=

    D4=D3=

    -481.77/AE-250.65/AE

    0.00.0

    = +17.7 kN (T)

    D2=D1=

    D6=D5=

    -481.77/AE-250.65/AE

    0.00.0

    = -17.7 kN (C)

    80 kN

    50 kN 36.87o

    97.9 kN

    17.7 kN

    17.7 kN

    #1 -0.8 -0.6#2 -0.8 0.6#3 0.8 -0.6

    [q´F]2 = AE5 0.8 -0.6 -0.8 0.6

    [q´F]3 = AE5 -0.8 +0.6 +0.8 -0.6

    D2=D1=

    D8=D7=

    -481.77/AE-250.65/AE

    0.00.0

    Member

    #1#2#3

    λx λy

    [ ] [ ]F

    yj

    xj

    yi

    xi

    xxyxmF q

    DDDD

    LAEq ']'[ +

    −−= λλλλ

  • 18

    80 kN

    50 kN 36.89o

    97.9 kN

    17.7 kN

    17.7 kN

    80 kN

    50 kN

    Member

    #1#2#3

    λx-0.8-0.80.8

    λy-0.60.6-0.6

    1

    2

    3

    434

    56

    78

    1

    2

    31

    2

    97.9(0.8)=78.32 kN

    97.9(0.6)=58.74 kN

    17.7(0.8)=14.16 kN

    17.7(0.6)=10.62 kN

    17.7(0.8)=14.16 kN

    17.7(0.6)=10.62 kN

    ΣFx ´ = 0:+ 17.7 + 17.7 +50cos 36.89 - 97.9cos73.78 - 80cos53.11 = 0, O.K

    Check :

  • 19

    Example 2

    For the truss shown, use the stiffness method to:(b) Determine the end forces of each member and reactions at supports.(a) Determine the deflections of the loaded joint.The support B settles downward 2.5 mm. Temperature in member BDincrease 20 oC. Take α = 12x10-6 /oC, AE = 8(103) kN.

    A

    BC

    D

    8 kN

    4 kN

    4 m

    3 m+20

    o C

    ∆B = 2.5 mm

  • 20

    λyλx

    λxλxλyλy

    λxλy

    λyλx

    λxλxλyλy

    λxλy

    −λyλx

    −λxλx−λyλy

    −λxλy

    −λyλx

    −λxλx−λyλy

    −λxλy

    ∆B = 2.5 mm

    A

    BC

    D

    8 kN

    4 kN

    4 m

    3 m+20

    o C

    Vi Uj VjUi

    [ k ]m = AEL

    ViUjVj

    Ui

    12

    3 4

    (0,0)

    (-4,-3)

    (-4,0)

    (0,-3)

    2

    1

    3

    1

    23

    4

    56

    78

    Member

    #1

    #2

    λx λy

    #3

    -4/4 = -1

    -4/5 = -0.8

    0

    0

    -3/5 = -0.6

    -3/3 = -1

    Ljyy

    Lixx ijij

    ij

    )()( −+−

    cosθx = λx cosθy = λy

  • 21

    0.096

    0.128

    0.072

    0.096

    0

    0

    0.333

    0

    0

    0.25

    0

    0

    0

    0.25

    0

    00

    -0.25

    0

    0

    0

    -0.25

    0

    0[k]1 = 8x103

    2 3 41

    234

    1

    [k]3 = 8x103

    2 7 81

    278

    1

    0

    0

    0.333

    00

    0

    -0.333

    0

    0

    0

    -0.333

    0

    Member

    #1

    #2

    #3

    λx

    -1

    - 0.8

    0

    λy

    0

    -0.6

    -1

    λx2/L λx λy/L λy2/L

    0.25 0 0

    0.128 0.096 0.072

    0 0 0.333

    0.096

    0.128

    0.072

    0.096

    -0.096

    -0.128

    -0.072

    -0.096-0.096

    -0.128

    -0.072

    -0.096

    [k]2 = 8x103

    2 5 61

    256

    1

    12

    3 4

    (0,0)

    (-4,-3)

    (-4,0)

    (0,-3)

    2

    1

    3

    1

    23

    4

    56

    78

    1

    23

    4

    56

    78

    [K] = 8x103

    21

    21

    0.096

    0.378

    0.405

    0.096

  • 22

    ∆B = 2.5 mm

    1.536 kN

    1.152 kN

    2+20oC1.536 kN

    1.152 kN

    2+20oC

    1.92 kΝ =α(∆T1)AE = (12x10-6)(20)(8x103)

    1.92 kN

    1.536 kN

    1.152 kN

    1.536 kN

    1.152 kN

    12

    34

    (0,0)

    (-4,-3)

    (-4,0)

    (0,-3)

    2

    1

    3

    1

    23

    4

    56

    78

    ∆B = 2.5 mm

    +20o C

    -1.536

    1.536-1.152

    1.152

    +

    1

    2

    6

    5

    q1q2

    q1

    q5

    q2

    q6

    -1.536

    -1.152+1

    2

    0

    -2.5x10-35

    6-0.096-0.128

    -0.072

    -0.096+ 8x1035 6

    = 8x103

    21

    21

    0.096

    0.128

    0.072

    0.096 d1d2

    d1

    d5 = 0d2

    d6 = -2.5x10-3

    0.096

    0.128

    0.072

    0.096

    = 8x103

    2 5 61

    256

    1

    0.096

    0.128

    0.072

    0.096

    -0.096

    -0.128

    -0.072

    -0.096-0.096

    -0.128

    -0.072

    -0.096

    [q] = [k]m[d] + [qF]Member 2:

    q1q2

    -1.536

    -1.152+1

    2= 8x103

    21

    0.096

    0.128

    0.072

    0.096 d1d2

    1.92

    1.44+

  • 23

    [Q] = [K][D] + [QF]

    Global:

    A

    BC

    D

    8 kN

    4 kN

    4 m

    3 m+20

    o C

    ∆B = 2.5 mm

    12

    3 4

    (0,0)

    (-4,-3)

    (-4,0)

    (0,-3)

    2

    1

    3

    1

    23

    4

    56

    78

    Q1 = -4Q2 = -8

    = 8x103

    21

    21

    0.096

    0.378

    0.405

    0.096 D1D2

    -1.536

    -1.152+1.92

    1.44+

    D1D2

    -0.8514x10-3 m

    -2.356x10-3 m=

  • 24

    Member

    #1

    #2

    #3

    λx λy

    Local12

    3 4

    2

    1

    3

    1

    23

    4

    56

    78

    [q´F]1 = 8x103 1.0 0.0 -1.0 0.04

    = -1.70 kN (C)

    D2=D1=

    D4=D3=

    0.00.0

    -0.8514x10-3

    -2.356x10-3

    -1.92+[q´F]2 = 8x103 0.8 0.6 -0.8 -0.65

    = -2.87 kN (C)

    D2=D1=

    D6=D5=

    -0.00250.0

    -0.8514x10-3

    -2.356x10-3

    = -6.28 kN (C)

    D2=D1=

    D8=D7=

    0.00.0

    -0.8514x10-3

    -2.356x10-3[q´F]3 = 8x103 0.0 1.0 0.0 -1.0

    3

    2+20oC

    1.92 kN

    1.92 kN

    -1 0

    - 0.8 -0.6

    0 -1

    [ ] [ ]F

    yj

    xj

    yi

    xi

    xxyxmF q

    DDDD

    LAEq ']'[ +

    −−= λλλλ

  • 25

    4 kN

    8 kN

    6.28 kN

    1.70 kN

    2.87 kN

    12

    3 4

    2

    1

    3

    1

    23

    4

    56

    78

    Member

    #1

    #2

    #3

    cosθx

    -1

    - 0.8

    0

    cosθy

    0

    -0.6

    -1

    [q´]m

    -1.70

    -2.87

    -6.28

    4 kN

    8 kN

    2

    1

    3

    1.70 kN

    6.28 kN

    2.87(0.8) = 2.30 kN

    2.87(0.6) = 1.72 kN

  • 26

    ∆ AD = +

    3 mm

    AB

    C

    3 m

    D

    8 kN

    4 kN

    4 m 4 m∆ = - 4 m

    m

    Example 3

    For the truss shown, use the stiffness method to:(a) Determine the end forces of each member and reactions at supports.(b) Determine the displacement of the loaded joint.Take AE = 8(103) kN.

  • 27

    ∆ AD = +

    3 mm

    AB

    C

    D

    8 kN

    4 kN

    3 m

    4 m 4 m

    ∆ = - 4 mm

    21

    4

    3

    5

    1

    2

    34

    56

    78

    1

    2 3 4

    (0,0)

    (-4,-3) (4,-3)(0,-3)

    λyλx

    λxλxλyλy

    λxλy

    λyλx

    λxλxλyλy

    λxλy

    −λyλx

    −λxλx−λyλy

    −λxλy

    −λyλx

    −λxλx−λyλy

    −λxλy

    Vi Uj VjUi

    [ k ]m = AEL

    ViUjVj

    Ui

    Member

    #1

    #2

    λx λy

    #3

    #4

    #5

    -4/5 =-0.8

    0

    4/5 = 0.8

    4/4 = 1

    4/4 = 1

    -3/5 = -0.6

    -3/3 = -1

    -3/5 = -0.6

    0

    0

    Ljyy

    Lixx ijij

    ij

    )()( −+−

    cosθx = λx cosθy = λy

  • 28

    [k]1 = 8x103

    2 3 41

    234

    1

    0.096

    0.128

    0.072

    0.096

    -0.096

    -0.128

    -0.072

    -0.096-0.096

    -0.128

    -0.072

    -0.096

    0.096

    0.128

    0.072

    0.096

    [k]2 = 8x103

    2 5 61

    256

    1

    0

    0

    0.333

    00

    0

    -0.333

    0

    0

    0

    0.333

    0

    0

    0

    -0.333

    0

    [k]3 = 8x103

    2 7 81

    278

    1

    -0.096

    0.128

    0.072

    -0.0960.096

    -0.128

    -0.072

    0.096

    0.096

    -0.128

    -0.072

    0.096-0.096

    0.128

    0.072

    -0.096

    Member

    #1

    #2

    λy

    #3

    λx-0.8

    0

    0.8

    -0.6

    -1

    -0.6

    λx2/L λy2/Lλxλy/L

    0.128 0.0720.096

    0 0.3330

    0.128 0.072-0.096

    21

    4

    3

    5

    1

    2

    34

    56

    78

    1

    2 3 4

    (0,0)

    (-4,-3) (4,-3)(0,-3)

    5 m

    3 m 5 m

    4 m 4 m

  • 29

    21

    4

    3

    5

    1

    2

    34

    56

    78

    1

    2 3 4

    (0,0)

    (-4,-3) (4,-3)(0,-3)

    5 m

    3 m 5 m

    4 m 4 m

    Member

    #4

    #5

    λyλx1

    1

    0

    0

    λx2/L λy2/Lλxλy/L

    0

    0.25

    0

    00

    -0.25

    0

    0

    0

    0.25

    0

    0

    0

    -0.25

    0

    0

    0

    0.25

    0

    00

    -0.25

    0

    0

    0

    0.25

    0

    0

    0

    -0.25

    0

    0[k]5= 8x103

    6 7 85

    678

    5

    0.25 00

    0.25 00

    Global Stiffness Matrix

    [K] = 8x103

    2 5 71

    257

    1

    34 6

    8

    [k]4= 8x103

    4 5 63

    456

    3

  • 30

    Global Stiffness Matrix

    0.2560.0 0.477

    0.0 0.00.0

    0.0 0.0-0.128 0.096

    -0.1280.096

    0.50-0.25

    -0.250.378

    [K] = 8x103

    2 5 71

    257

    1

    [k]1 = 8x103

    2 3 41

    234

    1

    0.096

    0.128

    0.072

    0.096

    -0.096

    -0.128

    -0.072

    -0.096-0.096

    -0.128

    -0.072

    -0.096

    0.096

    0.128

    0.072

    0.096

    [k]2 = 8x103

    2 5 61

    256

    1

    0

    0

    0.333

    00

    0

    -0.333

    0

    0

    0

    0.333

    0

    0

    0

    -0.333

    0

    [k]3 = 8x103

    2 7 81

    278

    1

    -0.096

    0.128

    0.072

    -0.0960.096

    -0.128

    -0.072

    0.096

    0.096

    -0.128

    -0.072

    0.096-0.096

    0.128

    0.072

    -0.096

    0

    0.250

    -0.25

    0

    00

    0

    0

    0.25

    0

    -0.250

    0

    0

    0[k]4= 8x103

    4 5 63

    456

    3

    0

    0.250

    -0.25

    0

    00

    0

    0

    0.25

    0

    -0.250

    0

    0

    0[k]5= 8x103

    6 7 85

    678

    5

  • 31

    ∆AE/L = 10.67 kN

    ∆ AD = +

    3 mm1

    3.84 kN

    2.88 kN

    3.84 kN

    2.88 kN

    ∆ AD = +

    3 mm

    AB

    C

    D

    8 kN

    4 kN

    3 m

    4 m 4 m

    ∆ = - 4 mm

    1

    2

    21

    4

    3

    534

    56

    78

    Global Fixed end forces

    0.00.0

    257

    1∆AE/L = 4.8 kN

    4.8 kN

    ∆AE/L = 10.67 kN

    10.67 kN∆ = -4 m

    m

    2 Fixed End3.84 kN

    2.88 kN-3.84-2.88 + 10.67 = 7.79

  • 32

    ∆ AD = +

    3 mm

    AB

    C

    D

    8 kN

    4 kN

    3 m

    4 m 4 m

    ∆ = -4 mm

    Global:

    1

    2

    21

    4

    3

    534

    56

    78

    [Q] = [K][D] + [QF]

    Q1 = 4

    Q5 = 0Q2 = -8

    Q7 = 0

    = 8x103

    2 5 71

    257

    1

    -0.250.500.00.0

    0.378-0.250.096

    -0.128

    -0.1280.00.0

    0.256

    0.0960.0

    0.4770.0 D1

    D5

    D2

    D7

    -3.84

    0.07.79

    0.0

    +

    D1

    D5D2

    D7

    6.4426x10-3 m

    2.6144x10-3 m-5.1902x10-3 m

    5.2288x10-3 m

    =

    8 kN

    4 kN

  • 33

    D1

    D5D2

    D7

    6.4426x10-3 m

    2.6144x10-3 m-5.1902x10-3 m

    5.2288x10-3 m

    =

    -4.8+D2

    D1

    00

    = -1.54 kN (C)

    1

    2

    21

    4

    3

    534

    56

    78

    Member forces

    = -3.17 kN (C)

    10.67+

    [q´F]1 = 8x103 0.8 0.6 -0.8 -0.65

    [q´F]2 = 8x103 0.0 1.0 0.0 -1.03

    D2

    D1

    0D5

    Member

    #1

    #2

    λiyλix

    -0.8 -0.6

    0 -1

    1

    4.8 kN

    4.8 kN∆ AD

    = + 3 m

    m

    10.67 kN

    10.67 kN

    2

    [ ] [ ]F

    yj

    xj

    yi

    xi

    xxyxmF q

    DDDD

    LAEq ']'[ +

    −−= λλλλ

  • 34

    D1

    D5D2

    D7

    6.4426x10-3 m

    2.6144x10-3 m-5.1902x10-3 m

    5.2288x10-3 m

    =

    1

    2

    21

    4

    3

    534

    56

    78

    00

    0D5

    [q´F]4 = 8x103 -1.0 0.0 1.0 0.04

    = 5.23 kN (T)

    0D5

    0D7

    [q´F]5 = 8x103 -1.0 0.0 1.0 0.04

    = 5.23 kN (T)

    D2

    D1

    0D7

    [q´F]3 = 8x103 -0.8 0.6 0.8 -0.65

    = -6.54 kN (C)

    Member λx λy#3#4

    #5

    0.8 -0.61 0

    1 0

    [ ] [ ]F

    yj

    xj

    yi

    xi

    xxyxmF q

    DDDD

    LAEq ']'[ +

    −−= λλλλ

  • 35

    1

    2

    21

    4

    3

    534

    56

    78

    -0.8

    0

    0.81

    1

    -0.6

    -1

    -0.60

    0

    Member

    #1

    #2

    λx

    #3#4

    #5

    λy [q´]

    4 kN8 kN

    3.17 kN

    3.17 kN

    6.54 kN

    6.54 kN

    1.54 kN

    1.54 kN

    5.23 kN5.23 kN

    4 kN8 kN

    0.92 kN

    4 kN

    3.17 kN 3.92 kN

    -1.54

    -3.17

    -6.545.23

    5.23

  • 36

    Special Trusses (Inclined roller supports)

  • 37

    λix = cos θiλiy = sin θi

    λjx = cos θjλjy = sin θj

    [ q* ] = [ T ]T[ q´ ] 1

    2

    3*

    4*5

    6

    78

    3

    2

    4

    5

    1

    x *

    y *

    θi

    x

    y

    θj

    1

    i

    j

    q´i

    q´j

    q*3

    q1

    q*4

    q2

    q´iq´j

    [T]T

    00λiyλixλjyλjx00

    [ T ] = [[ T ]T]T =

    =00

    λiy

    λix

    λjy

    λjx

    00

    1

    i

    j1

    2

    3*

    4*

    Transformation Matrices

  • 38

    [ k ] = [ T ]T[ k´ ][T]

    λjyλjx

    λjxλjx

    −λiyλjx

    −λixλjx

    λjyλjy

    λjxλjy

    −λiyλjy

    −λixλjyλiyλix

    λixλix

    −λjyλix

    −λjxλix

    λiyλiy

    λixλiy

    −λjyλiy

    −λjxλiy

    Vi Uj VjUi

    [ k ]m = AEL

    ViUjVj

    Ui

    00λiyλixλjyλjx00

    -11

    1-1

    AEL[ k

    ]m =

    00

    λiy

    λix

    λjy

    λjx

    00

  • 39

    Example 5

    For the truss shown, use the stiffness method to:(b) Determine the end forces of each member and reactions at supports.(a) Determine the displacement of the loaded joint. AE is constant.

    3 m

    4 m

    45o

    30 kN

  • 40

    3 m

    4 m45o

    2

    1

    3

    1

    2

    5

    63*

    4*

    Member 1:

    15

    63*

    4*

    [q*]θi = 0,λix = cos 0 = 1,λiy = sin 0 = 0

    θij = -45o = 135o,λix = cos (-45o) = 0.707,λiy = sin(- 45o) = -0.707

    45o

    [q*] = [T*]T[q´] + [T*]T[q´F]

    q5

    q3*

    q6

    q4*

    q´iq´j

    [T*]T

    =0001

    -0.7070.707

    00

    q´i q´ji j

    1

  • 41

    15

    63*

    4*

    [q*]θi = 0 ;λix = cos 0 = 1,λiy = sin 0 = 0

    θij = -45o = 135o,λix = cos (-45o) = 0.707,λiy = sin(- 45o) = -0.707

    45o

    0-0.707

    00.707

    00

    10

    AE4

    1-1

    -11[k*]1 =

    0001

    -0.7070.707

    00

    Member 1:

    [k*]1 = AE63*4*

    53*

    -0.1250.125

    0-0.1768

    4*

    0.125-0.125

    00.1768

    5

    0.1768-0.1768

    00.25

    6

    0000

    q´i q´ji j

    1

    3 m

    4 m45o

    2

    1

    3

    1

    2

    5

    63*

    4*

    [ ]TL

    AETk T

    −=

    1111

    ][*][

  • 42

    Member 2:q´i

    q´j

    i

    j

    22

    1

    2

    3*4*

    θi = -90o = 270o,λix = cos(-90o) = 0,λiy = sin(-90o) = -1

    90o+45o =135o

    90o

    θj = -135o = 215o ,λix = cos (-135o) = -0.707,λiy = sin(- 135o) = -0.707

    0-0.707

    0-0.707

    -10

    00

    AE3

    1-1

    -11[k*]2 =

    00-10

    -0.707-0.707

    00

    [k*]2 = AE

    2 4*1

    23*4*

    1

    0.16670.1667-0.2357

    0

    0.16670.1667-0.2357

    0

    0000

    -0.2357-0.23570.3333

    03*

    3 m

    4 m45o

    2

    1

    3

    1

    2

    5

    63*

    4*

    [ ]TL

    AETk T

    −=

    1111

    ][*][

  • 43

    [ ]TL

    AETk T

    −=

    1111

    ][][

    36.87o

    Member 3:

    θi = θj = 36.87o ;λix = λjx = cos (36.87o) = 0.8,λiy = λjy = sin(36.87o) = 0.6

    00.6

    00.8

    0.60

    0.80

    AE5

    1-1

    -11[k]3 =

    00

    0.60.8

    0.60.800

    3

    1

    2

    5

    63

    q´i

    q´j

    i

    j36.87o

    [k]3 = AE

    6 25

    612

    5

    0.0960.128-0.096-0.128

    0.0720.096-0.072-0.096

    -0.096-0.1280.0960.128

    -0.072-0.0960.0720.096

    1

    3 m

    4 m45o

    2

    1

    3

    1

    2

    5

    63*

    4*

  • 44

    3 m

    4 m45o

    2

    1

    3

    1

    2

    5

    63*

    4*

    [k*]1 = AE63*4*

    53*

    -0.1250.125

    0-0.1768

    4*

    0.125-0.125

    00.1768

    5

    0.1768-0.1768

    00.25

    6

    0000

    5

    6 4*

    Global Stiffness:

    [k*]2 = AE

    2 4*1

    23*4*

    1

    0.16670.1667-0.2357

    0

    0.16670.1667-0.2357

    0

    0000

    -0.2357-0.23570.3333

    03*

    [k]3 = AE

    6 25

    612

    5

    0.0960.128-0.096-0.128

    0.0720.096-0.072-0.096

    -0.096-0.1280.0960.128

    -0.072-0.0960.0720.096

    1[K] = AE 2

    3*

    11

    00.0960.128

    2

    -0.23570.40530.096

    0.2917-0.2357

    03*

  • 45

    Global :

    [Q] = [K][D] + [QF]

    Q1 = 30

    Q3*= 0

    Q2 = 0D1

    D3*D2 = AE 2

    3*

    11

    00.0960.128

    2

    -0.23570.40530.096

    0.2917-0.2357

    03*

    D1

    D3*D2 =

    352.5

    -127.3-157.5AE

    1

    3 m

    4 m45o

    30 kN

    3 m

    4 m45o

    2

    1

    3

    1

    2

    5

    63*

    4*

    5

    6 4*

  • 46

    [q´F]1 = AE00

    0D3*

    -1 0 0.707 -0.7074

    = -22.50 kN, (C)

    D2

    D1

    0D3*

    [q´F]2 = AE 0 1 -0.707 -0.7073

    = -22.50 kN, (C)

    00

    D2

    D1[q´F]3 = AE -0.8 -0.6 0.8 0.6

    5 = 37.50 kN, (T)

    D1

    D3*D2 =

    352.5

    -127.3-157.5AE

    1

    Member Forces :

    Member

    #1

    λiyλix λjx λjy

    #2

    #3

    1 0 0.707 -0.707

    0 -1 -0.707 -0.707

    0.8 0.6 0.8 0.6

    3 m

    4 m45o

    2

    1

    3

    1

    2

    5

    63*

    4*

    5

    6 4*

    [ ] [ ]F

    yj

    xj

    yi

    xi

    xxyxmF q

    DDDD

    LAEq ']'[ +

    −−= λλλλ

  • 47

    36.87o 45o45o

    3 m

    4 m45o

    2

    1

    3

    1

    2

    5

    63*

    4*

    Member

    Member Force (kN)

    [q´]2[q´]1 [q´]3

    -22.50 -22.50 37.50

    22.50 kN

    22.50 kN

    37.50 kN

    22.50 kN

    7.50 kN31.82 kN

    Reactions :

    3 m

    4 m45o

    30 kN

  • 48

    Example 6

    For the truss shown, use the stiffness method to:(b) Determine the end forces of each member and reactions at supports.(a) Determine the displacement of the loaded joint. AE is constant.

    3 m

    4 m

    45o

    30 kN

    4 m

  • 49

    15

    63*

    4*

    [q*]θi = 0o,λix = cos 0o = 1,λiy = sin 0o = 0

    θij = -45oλix = cos (-45o) = 0.707,λiy = sin(- 45o) = -0.707

    45o

    [q*] = [T*]T[q´] + [T*]T[q´F]

    q5

    q3*

    q6

    q4*

    q´iq´j

    [T*]T

    =0001

    -0.7070.707

    00

    q´i q´ji j

    1

    3 m

    4 m 4 m

    2

    1

    3

    4

    5

    12

    5

    6 3*4*

    78Member 1:

  • 50

    3 m

    4 m 4 m

    15

    63*

    4*

    [q*]θi = 0o,λix = cos 0o = 1,λiy = sin 0o = 0

    θij = -45o = 315o,λix = cos (-45o) = 0.707,λiy = sin(- 45o) = -0.707

    45o

    0-0.707

    00.707

    00

    10

    AE4

    1-1

    -11[k*]1 =

    0001

    -0.7070.707

    00

    [k*]1 = AE63*4*

    53*

    -0.1250.125

    0-0.1768

    4*

    0.125-0.125

    00.1768

    5

    0.1768-0.1768

    00.25

    6

    0000

    q´i q´ji j

    1

    2

    1

    3

    4

    5

    12

    5

    6 3*4*

    78Member 1:

    [ ]TL

    AETk T

    −=

    1111

    ][*][

  • 51

    q´i

    q´j

    i

    j

    22

    1

    2

    3*4*

    θi = -90o,λix = cos(-90o) = 0,λiy = sin(-90o) = -1

    90o

    θj = -135o = 215o,λix = cos (-135o) = -0.707,λiy = sin(- 135o) = -0.707

    0-0.707

    0-0.707

    -10

    00

    AE3

    1-1

    -11[k*]2 =

    00-10

    -0.707-0.707

    00

    [k*]2 = AE

    2 4*1

    23*4*

    1

    0.16670.1667-0.2357

    0

    0.16670.1667-0.2357

    0

    0000

    -0.2357-0.23570.3333

    03*

    3 m

    4 m 4 m

    2

    1

    3

    4

    5

    12

    5

    6 3*4*

    78Member 2:

    [ ]TL

    AETk T

    −=

    1111

    ][*][

    90o+45o =135o

  • 52

    36.87o

    Member 3:

    θi = θj = 36.87o ;λix = λjx = cos (36.87o) = 0.8,λiy = λjy = sin(36.87o) = 0.6

    00.6

    00.8

    0.60

    0.80

    AE5

    1-1

    -11[k]3 =

    00

    0.60.8

    0.60.800

    3

    1

    2

    5

    63

    q´i

    q´j

    i

    j36.87o

    [k]3 = AE

    6 25

    612

    5

    0.0960.128-0.096-0.128

    0.0720.096-0.072-0.096

    -0.096-0.1280.0960.128

    -0.072-0.0960.0720.096

    1

    3 m

    4 m 4 m

    2

    1

    3

    4

    5

    12

    5

    6 3*4*

    78

    [ ]TL

    AETk T

    −=

    1111

    ][*][

  • 53

    3 m

    4 m 4 m

    θi = θij = 0o; λix = λjx = cos 0o = 1, λiy = λjy = sin 0o = 0

    00

    01

    00

    10

    AE4

    1-1

    -11[k]1 =

    0001

    0100

    q´i q´ji j

    1

    2

    1

    3

    4

    5

    12

    5

    6 3*4*

    78Member 4:

    47

    8

    [q]1

    2

    [k]4 = AE278

    17

    00.25

    0-0.25

    8

    0000

    1

    0-0.25

    00.25

    2

    0000

    [ ]TL

    AETk T

    −=

    1111

    ][*][

  • 54

    Member 5:

    θi = - 8.13o;λix = cos (- 8.13o) = 0.9899,λiy = sin(- 8.13o) = -0.1414

    00.6

    00.8

    -0.14140

    0.98990

    AE5

    1-1

    -11

    5

    q´i

    q´j

    i

    j36.87o

    3 m

    4 m 4 m

    2

    1

    3

    4

    5

    12

    5

    6 3*4*

    78

    53*4*

    7

    8

    [k*]5 =00

    -0.14140.9899

    0.60.800

    [k*]5 = AE

    4* 83*

    4*78

    3*

    0.0960.128

    0.02263-0.1584

    0.0720.096

    0.01697-0.1188

    -0.1188-0.1584-0.0280.196

    0.016970.022630.004-0.028

    7

    θ j = 36.

    87o ;

    λ jx = cos

    (36.87

    o ) = 0.8,

    λ jy = sin

    (36.87

    o ) = 0.6

    [ ]TL

    AETk T

    −=

    1111

    ][*][

    8.13o

  • 550 -0.2357-0.2357

    0

    [k*]1 = AE63*4*

    53*

    -0.1250.125

    0-0.1768

    4*

    0.125-0.125

    00.1768

    5

    0.1768-0.1768

    00.25

    6

    0000

    [k*]3 = AE

    6 25

    612

    5

    0.0960.128-0.096-0.128

    0.0720.096-0.072-0.096

    -0.096-0.1280.0960.128

    -0.072-0.0960.0720.096

    1

    3 m

    4 m 4 m

    2

    1

    3

    4

    5

    Global Stiffness:1

    2

    5

    6 3*4*

    78

    5

    6 4*

    78

    [k*]2 = AE

    2 4*1

    23*4*

    1

    0.16670.1667-0.2357

    0

    0.16670.1667-0.2357

    0

    0000

    -0.2357-0.23570.3333

    03*

    [k]4 = AE278

    17

    00.25

    0-0.25

    8

    0000

    1

    0-0.25

    00.25

    2

    0000

    [k*]5 = AE

    4* 83*

    4*78

    3*

    0.0960.128

    0.02263-0.1584

    0.0720.096

    0.01697-0.1188

    -0.1188-0.1584-0.0280.196

    0.016970.022630.004-0.028

    7

    [K] = AE 23*

    11 2

    0.0960.378

    0.40530.096

    3*

    0.4877

  • 56

    Global :30 kN

    4 m 4 m

    3 m2

    1

    3

    4

    5

    12

    5

    6 3*4*

    78

    5

    6 4*

    78

    [Q] = [K][D] + [QF]

    Q1 = 30

    Q3*= 0

    Q2 = 0D1

    D3*D2

    D1

    D3*D2 =

    86.612

    -13.791-28.535AE

    1

    = AE 23*

    11

    00.0960.378

    2

    -0.23570.40530.096

    0.4877-0.2357

    03*

  • 57

    3 m

    4 m 4 m

    2

    1

    3

    4

    5

    12

    5

    6 3*4*

    78

    5

    6 4*

    78

    D1

    D3*D2 =

    86.612

    -13.791-28.535AE

    1

    [q´F]1 = AE00

    0D3*

    -1 0 0.707 -0.7074

    = -2.44 kN, (C)

    D2

    D1

    0D3*

    [q´F]2 = AE 0 1 -0.707 -0.7073

    = -6.26 kN, (C)

    00

    D2

    D1[q´F]3 = AE -0.8 -0.6 0.8 0.6

    5 = 10.43 kN, (T)

    Member Forces :

    Member

    #1

    λiyλix λjx λjy

    #2

    #3

    1 0 0.707 -0.707

    0 -1 -0.707 -0.707

    0.8 0.6 0.8 0.6

    [ ] [ ]F

    yj

    xj

    yi

    xi

    xxyxmF q

    DDDD

    LAEq ']'[ +

    −−= λλλλ

  • 58

    Member

    #4

    λiyλix λjx λjy

    #5

    3 m

    4 m 4 m

    2

    1

    3

    4

    5

    12

    5

    6 3*4*

    78

    5

    6 4*

    78

    D1

    D3*D2 =

    86.612

    -13.791-28.535AE

    1

    D2

    D1

    00

    [q´F]4 = AE -1 0 1 04

    = -21.65 kN, (C)

    0D3*

    00

    [q´F]5 = AE -0.9899 0.141 0.8 0.65

    = 2.73 kN, (T)1 0 1 0

    0.9899 -0.141 0.8 0.6

    Member Forces :

    [ ] [ ]F

    yj

    xj

    yi

    xi

    xxyxmF q

    DDDD

    LAEq ']'[ +

    −−= λλλλ

  • 59

    36.87o 45o45o

    81.87o

    36.87o

    Member

    Member Force (kN)

    [q´]1 [q´]2 [q´]3 [q´]4 [q´]5

    -2.44 -6.26 10.43 -21.65 2.73

    Reactions :30 kN

    4 m 4 m

    3 m2

    1

    3

    4

    5

    12

    5

    6 3*4*

    78

    5

    6 4*

    78

    6.26 kN10.43 kN

    21.65 kN

    2.73 kN

    2.44 kN 2.44 kN5.90 kN

    6.26 kN

    19.47 kN

    1.64 kN

    6.54 kN

  • 60

    AB

    C

    3 m

    D

    8 kN

    4 kN

    4 m36.87o

    4 m

    λjyλjx

    λjxλjx

    −λiyλjx

    −λixλjx

    λjyλjy

    λjxλjy

    −λiyλjy

    −λixλjyλiyλix

    λixλix

    −λjyλix

    −λjxλix

    λiyλiy

    λixλiy

    −λjyλiy

    −λjxλiy

    Vi Uj VjUi

    [ k *]m = AEL

    ViUjVj

    Ui

    Example 7

    For the truss shown, use the stiffness method to:(b) Determine the end forces of each member and reactions at supports.(a) Determine the displacement of the loaded joint.Take AE = 8(103) kN.

  • 61

    Member 1:

    λix = cos 73.74o = 0.28,λiy = sin 73.74o = 0.96

    [q*] = [T*]T[q´] + [T*]T[q´F]

    λjx = cos 36.87o = 0.8,λjy = sin 36.87o = 0.6

    AB

    C

    3 m

    D

    8 kN

    4 kN

    4 m

    36.87o

    4 m

    x *

    y *

    73.74o

    x

    y

    36.87o

    1

    i

    j1

    2

    3*

    4*1

    i

    j

    q´i

    q´jq3*

    q1

    q4*

    q2

    q´iq´j=

    00

    0.960.28

    0.60.8

    00

    [T*]T

    3

    2

    4

    5

    1

    1

    2

    56

    78

    3*

    4*

  • 62

    [k]1 = 8x103

    4* 1 23*

    4*12

    3*

    0.0960.128

    -0.1536-0.0448

    0.0720.096

    -0.1152-0.0336

    -0.0336-0.04480.053760.01568

    -0.1152-0.15360.184320.05376

    000.960.280.60.800

    [k]1 =00

    0.960.28

    0.60.8

    00

    8x1035

    1-1

    -11

    AB

    C

    3 m

    D

    8 kN

    4 kN

    4 m

    36.87o

    4 m

    3

    2

    4

    5

    1

    1

    2

    56

    78

    3*

    4*

    [ ]TL

    AETk T

    −=

    1111

    ][][

  • 63

    x *

    y*

    36.87x

    yMember 2:

    λix = cos 36.87o = 0.8,λiy = sin 36.87o = 0.6

    [q*] = [T*]T[q´] + [T*]T[q´F]

    AB

    C

    3 m

    D

    8 kN

    4 kN

    4 m

    36.87o

    4 m

    λjx = cos 0o = 1,λjy = sin 0o = 0

    q3*

    q5q4*

    q6

    q´iq´j

    [T*]T

    [k] = [TT] AEL

    [T]1-1

    -11

    q´i q´ji j

    2

    =00

    0.60.8

    0100

    i j2 5

    6

    3*

    4*

    [k]2 = 8x103

    4* 5 63*

    4*56

    3*

    00.25

    -0.15-0.2

    0000

    0-0.20.120.16

    0-0.150.090.12

    3

    2

    4

    5

    1

    1

    2

    56

    78

    3*

    4*

  • 64

    x

    y

    270o 1

    2

    784

    Member 3:

    AB

    C

    3 m

    D

    8 kN

    4 kN

    4 m

    36.87o

    4 m

    λx = cos 323.13o = 0.8,λy = sin 323.13o = -0.6

    Member 4:

    [k]4 = 8x103

    2 7 81

    278

    1

    -0.0960.1280.096

    -0.128

    0.072-0.096-0.0720.096

    0.096-0.128-0.0960.128

    -0.0720.0960.072

    -0.096

    λx = cos 270o = 0, λy = sin 270o = -1

    [k]3 = 8x103

    0000

    0.3330

    -0.3330

    00

    00

    2 5 61

    256

    10.333

    0

    -0.3330

    x

    y

    323.13o3

    1

    2

    56

    3

    2

    4

    5

    1

    1

    2

    56

    78

    3*

    4*

  • 65

    x

    yMember 5:

    AB

    C

    3 m

    D

    8 kN

    4 kN

    4 m

    36.87o

    4 m

    λx = cos 0o = 1,λy = sin 0o = 0

    00.25

    0-0.25

    0000

    00.25

    0-0.25

    00

    00

    [k]5 = 8x103

    6 7 85

    678

    5

    5 78

    56

    3

    2

    4

    5

    1

    1

    2

    56

    78

    3*

    4*

  • 660.17568

    -0.2-0.20.5

    00

    0 0

    -0.0336-0.0448

    -0.0448 -0.03360.0

    0.2560.4740.0

    [k]1 = 8x103

    4* 1 23*

    4*12

    3*

    0.0960.128

    -0.1536-0.0448

    0.0720.096

    -0.1152-0.0336

    -0.0336-0.04480.053760.01568

    -0.1152-0.15360.184320.05376

    [k]4 = 8x103

    2 7 81

    278

    1

    -0.0960.1280.096

    -0.128

    0.072-0.096-0.0720.096

    0.096-0.128-0.0960.128

    -0.0720.0960.072

    -0.096

    00.25

    0-0.25

    0000

    00.25

    0-0.25

    00

    00

    [k]5 = 8x103

    6 7 85

    678

    5

    [k]2 = 8x103

    4* 5 63*

    4*56

    3*

    00.25

    -0.15-0.2

    0000

    0-0.20.120.16

    0-0.150.090.12

    [k]3 = 8x103

    0000

    0.3330

    -0.3330

    00

    00

    2 5 61

    256

    10.333

    0

    -0.3330

    [K] = 8x103

    2 3* 51

    23*5

    1

    3

    2

    4

    5

    1

    12

    56

    78

    3*

    4* 6

    784

    *

  • 67

    3

    2

    4

    5

    1

    2

    3*

    4*5

    6

    78A

    BC

    3 m

    D

    8 kN

    4 kN

    4 m

    36.87o

    4 m

    1

    Global: [Q] = [K][D] + [QF]

    D1

    D3*D2

    D5

    1.988x10-3 m

    1.996x10-4 m-2.0824x10-3 m

    7.984x10-5 m

    =

    Q1 = 4

    Q3*= 0

    Q2 = -8

    Q5 = 0

    D1

    D3*

    D2

    D50.17568

    -0.2-0.20.5

    00

    0 0

    -0.0336-0.0448

    -0.0448 -0.03360.0

    0.2560.4740.0

    = 8x103

    2 3* 51

    23*

    5

    1

    8 kN

    4 kN

  • 68

    AB

    C

    3 m

    D

    8 kN

    4 kN

    4 m

    36.87o

    4 m

    3

    2

    4

    5

    1

    2

    3*

    4*5

    6

    78

    1

    Member forces

    D1

    D3*D2

    D5

    1.988x10-3 m

    1.996x10-4 m-2.0824x10-3 m

    7.984x10-5 m

    =

    0D3*

    D2

    D1[q´F]1 = 8x103 -0.28 -0.96 0.8 0.6

    5 = 0.46 kN, (T)

    0D3*

    0D5

    [q´F]2 = 8x103 -0.8 -0.6 1 04

    = -0.16 kN, (C)

    Member

    #1

    #2

    λiyλix λjx λjy

    0.28 0.96 0.8 0.6

    0.8 0.6 1 0

    [ ] [ ]F

    yj

    xj

    yi

    xi

    xxyxmF q

    DDDD

    LAEq ']'[ +

    −−= λλλλ

  • 69

    D1

    D*3D2

    D5

    1.988x10-3 m

    1.996x10-4 m-2.0824x10-3 m

    7.984x10-5 m

    =

    3

    2

    4

    5

    1

    2

    3*

    4*5

    6

    78

    1

    [q´F]3 = 8x103 D2

    D1

    0D5

    0 1 0 -13

    = -5.55 kN

    D2

    D1

    00

    [q´F]4 = 8x103 -0.8 0.6 0.8 - 0.65

    = -4.54 kN

    0D5

    00

    [q´F]5 = 8x103 -1 0 1 04

    = -0.16 kN

    Member

    #3

    λiyλix λjx λjy

    #4

    #5

    0 -1 0 -1

    0.8 -0.6 0.8 -0.6

    1 0 1 0

    [ ] [ ]F

    yj

    xj

    yi

    xi

    xxyxmF q

    DDDD

    LAEq ']'[ +

    −−= λλλλ

  • 70

    y*

    x *

    36.87o

    36.87o 36.87o

    AB

    C

    3 m

    D

    8 kN

    4 kN

    4 m

    36.87o

    4 m

    3

    2

    4

    5

    1

    2

    3*

    4*5

    6

    78

    1

    0.46 -0.16 -5.55 -4.54 -0.16

    0.46 kN

    5.55 kN0.36 kN

    4.54 kN

    3.79 kN

    2.72 kN

    Member

    Member Force (kN)

    [q]2[q]1 [q]3 [q]4 [q]5

    0.16 kN 0.16 kN

    5.55 kN

  • 71

    Space-Truss Analysis

  • 72

    +

    −=

    Fj

    Fi

    j

    i

    j

    i

    qq

    dd

    LEA

    qq

    ''

    ''

    1111

    ''

    [ ] [ ]F

    jz

    jy

    jx

    iz

    iy

    ix

    q

    dddddd

    TL

    EA '1111

    +

    −=

    [ ]

    =

    zyx

    zyxT

    where

    λλλλλλ000

    000,

    [q´] = [k´][d´] + [q´F]

    = [k´][T][d] + [q´F]

    Member Local Stiffness [k´]:

  • 73

    Member Global Stiffness [km]:

    [km]= [T]T[k´] [T]

    [ ]

    =zyx

    zyx

    z

    y

    x

    z

    y

    x

    m LEAk

    λλλλλλ

    λλλ

    λλλ

    000000

    1111

    000

    000

  • 74

    Global equilibrium matrix:

    [Q] = [K][D] + [QF]

    QI

    QII

    Du

    Dk

    KI,II

    KII,II

    KI,I

    KII,I

    Reaction Support Boundary Condition

    Unknown DisplacementJoint Load

    QFI

    QFII= +

    Fixed End Forces

  • 75

    q´j

    q´ii

    j

    m

    i

    j

    m

    =EAL

    λxλx

    λzλx

    λyλx

    λxλyλyλyλzλy

    λxλzλyλzλzλz

    λxλx

    λzλx

    λyλx

    λxλyλyλyλzλy

    λxλzλyλzλzλz

    −λxλx

    −λzλx

    −λyλx

    −λxλy−λyλy−λzλy

    −λxλz−λyλz−λzλz

    −λxλx

    −λzλx

    −λyλx

    −λxλy−λyλy−λzλy

    −λxλz−λyλz−λzλz

    diydix

    djxdiz

    djydjz

    qiyqix

    qjx

    qiz

    qjyqjz

    qjy

    qjx

    qjzqiy

    qix

    qiz

    +

    −=

    Fj

    Fi

    j

    i

    j

    i

    qq

    dd

    LEA

    qq

    ''

    ''

    1111

    ''

    [ ] [ ] [ ]F

    jz

    jy

    jx

    iz

    iy

    ix

    zyxzyxmjq

    dddddd

    LEAq '' +

    −−−= λλλλλλ

  • 76

    x

    z

    y

    4 m4 m

    3 m3 m

    10 m

    80 kN60 kN

    O

    Example 6

    For the truss shown, use the stiffness method to:(b) Determine the end forces of each member.(a) Determine the deflections of the loaded joint.Take E = 200 GPa, A = 1000 mm2.

  • 77

    λ1 = (-4/11.18)i + (3/11.18)j + (-10/11.18)k

    = -0.3578 i + 0.2683 j - 0.8944 k

    λ2 = (+4/11.18)i + (3/11.18)j + (-10/11.18)k= +0.3578 i + 0.2683 j - 0.8944 k

    λ3 = (+4/11.18)i + (-3/11.18)j + (-10/11.18)k

    = +0.3578 i - 0.2683 j - 0.8944 k

    = -0.3578 i - 0.2683 j - 0.8944 k

    λ4 = (-4/11.18)i + (-3/11.18)j + (-10/11.18)k

    Member

    #1

    #2

    #3

    λx λy λz

    #4

    -0.3578 +0.2683 -0.8944

    +0.3578 +0.2683 -0.8944

    +0.3578 -0.2683 -0.8944

    -0.3578 -0.2683 -0.8944

    λm = λxi + λyj + λzk

    12

    34

    12

    3

    (0, 0, 10)

    (-4, 3, 0)

    (4, 3, 0)

    (4, -3, 0)

    (-4, -3, 0)4

    1

    2

    53

    x

    z

    y

    4 m4 m

    3 m3 m

    10 m

    80 kN60 kN

    O

  • 78

    Member

    #1#2

    #3

    λx λy λz

    #4

    -0.3578 +0.2683 -0.8944+0.3578 +0.2683 -0.8944

    +0.3578 -0.2683 -0.8944-0.3578 -0.2683 -0.8944

    Member Stiffness Matrix [k]6x6

    [k]m =[k12]3x3

    [k22]3x3

    [k11]3x3

    [k21]3x3

    2 31

    23

    1

    +0.80-0.240

    +0.320

    +0.320-0.096

    +0.128

    -0.240+0.072

    -0.096

    [k11]1 =AEL

    2 31

    23

    1

    +0.80-0.240

    -0.320

    -0.320+0.096

    +0.128

    -0.240+0.072

    +0.096

    [k11]2 =AEL

    2 31

    23

    1

    +0.80+0.240

    -0.320

    -0.320-0.096

    +0.128

    +0.240+0.072

    -0.096

    [k11]3 =AEL

    2 31

    23

    1

    +0.80+0.240

    +0.320

    +0.320+0.096

    +0.128

    +0.240+0.072

    +0.096

    [k11]4 =AEL

    2 31

    23

    1

    3.20.0

    0.0

    0.00.0

    0.512

    0.00.288

    0.0

    [KI,I] =AEL

  • 79

    12

    34

    12

    3

    (0, 0, 10)

    (-4, 3, 0)

    (4, 3, 0)

    (4, -3, 0)

    (-4, -3, 0)4

    1

    2

    53

    x

    z

    y

    4 m4 m

    3 m3 m

    10 m

    80 kN60 kN

    O

    0.0-80

    60

    D3

    D2

    D1

    0.00.0

    0.0+

    2 31

    23

    1

    3.20.0

    0.0

    0.00.0

    0.512

    0.00.288

    0.0

    =AEL

    [Q] = [K][D] + [QF]

  • 80

    Global equilibrium matrix:

    [Q] = [K][D] + [QF]

    QI

    QII

    Du

    Dk

    KI,II

    KII,II

    KI,I

    KII,I

    Reaction Support Boundary Condition

    Unknown DisplacementJoint Load

    QFI

    QFII= +

    Fixed End Forces

    (AE/L) = (1x10-3)(200x106)/(11.18) = 17.89x103 kN

    0.0-80

    60

    D3

    D2

    D1

    0.00.0

    0.0+

    2 31

    23

    1

    3.20.0

    0.0

    0.00.0

    0.512

    0.00.288

    0.0

    =AEL

    0.0 mm-15.53 mm

    6.551 mm=

    D3

    D2

    D1= LAE

    0.0-277.8

    +117.2

  • 81

    Member forces:

    q´F+

    Member

    #1#2

    #3

    λx λy λz

    #4

    -0.3578 +0.2683 -0.8944+0.3578 +0.2683 -0.8944

    +0.3578 -0.2683 -0.8944-0.3578 -0.2683 -0.8944

    dyi

    dxi

    dxj

    dzi

    dyjdzj

    [q´j]m = AEL −λx −λy −λz λx λy λz

    D3D2D1

    [ 0 ]

    = +116.5 kN (T)

    +0.3578 -0.2683 +0.8944[q´j]1 =AEL

    0.0

    117.2

    -277.8L

    AE

    x

    z

    y

    4 m4 m

    3 m3 m

    10 m

    80 kN60 kN

    O

  • 82

    Member

    #1#2

    #3

    λx λy λz

    #4

    -0.3578 +0.2683 -0.8944+0.3578 +0.2683 -0.8944

    +0.3578 -0.2683 -0.8944-0.3578 -0.2683 -0.8944

    = +32.61 kN (T)-0.3578 -0.2683 +0.8944[q´j]2 =AEL

    0.0

    117.2

    -277.8L

    AE

    = -116.5 kN (T)-0.3578 +0.2683 +0.8944[q´j]3 =AEL

    0.0

    117.2

    -277.8L

    AE

    = -32.61 kN (T)+0.3578 +0.2683 +0.8944[q´j]4 =AEL

    0.0

    117.2

    -277.8L

    AE

    x

    z

    y

    4 m4 m

    3 m3 m

    10 m

    80 kN60 kN

    O

  • 83

    5

    32.6 kN116.5 kN

    116.5 kN

    32.6 kN

    Member

    #1#2

    #3

    λx λy λz

    #4

    -0.3578 +0.2683 -0.8944+0.3578 +0.2683 -0.8944

    +0.3578 -0.2683 -0.8944-0.3578 -0.2683 -0.8944

    [q´j]m116.5 32.6

    -32.6-116.5

    12

    34

    80 kN60 kN

    R5x = (-32.6)(-0.3578) = 11.66 kN

    R5y = (-32.6)(-0.2683) = 8.75 kN

    R5z = (-32.6)(-0.8944) = 29.16 kN