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  • Dept. of Civil and Environmental Eng., SNU

    Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr

    Structural Analysis I

    Spring Semester, 2018

    Hae Sung Lee

    Dept. of Civil and Environmental Engineering Seoul National University

    yf

    zδ zf

    xf

    yM yθ

    zM zθ xM xθ

  • Dept. of Civil and Environmental Eng., SNU

    Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr

    This page is intentionally left blank.

  • Dept. of Civil and Environmental Eng., SNU

    Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr

    Contents

    1. Introduction

    2. Reactions & Internal Forces by Free Body Diagra

    ms

    3. Principle of Virtual Work

    4. Analysis of Statically Indeterminate Beams

    5. Analysis of Statically Indeterminate Trusses

    6. Analysis of Statically Indeterminate Frames

    7. Influence Lines for Determinate Structures

    8. Influence Lines for Indeterminate Structures

  • Dept. of Civil and Environmental Eng., SNU

    Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr

    This page is intentionally left blank.

  • Dept. of Civil and Environmental Eng., SNU

    Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr

    1

    Chapter 1

    Introduction

  • Dept. of Civil and Environmental Eng., SNU

    Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr

    2

    1.1 Mechanics of Material - Structural Mechanics

    Problem

    Calculate the reaction force at each support and draw the moment and shear force dia

    gram for the two-span beam shown in the figure.

    Solution

    Equilibrium Equation

    qLRRRF cbay 20 =++→=

    qLRRLRLRLqLM cbcba 220220 =+→=×−×−×→=

    0022

    0 =−→=×−×+×+×−→= cacab RRLRLRL

    qLL

    qLM

    qLRRLRLRLqLM babac 220220 =+→=×+×+×−→=

    Since there are three unknowns in two independent equations, we cannot determine a unique

    solution for the given structure, and thus we need one more equation to solve this problem.

    The main issue of this class is how to build additional equations to analyze statically inde-

    terminate structures.

    EI EI

    q

    Ra Rb Rc

    L L

    q

  • Dept. of Civil and Environmental Eng., SNU

    Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr

    3

    Fundamentals of Differential Equations

    Governing Equations

    The governing equations of a (engineering) system are usually defined by a system of diffe-

    rential equations, which governs behaviors of the given system within a domain. Notice that

    the domain does not include boundaries.

    11 qwEI =′′′′ for lx

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    Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr

    4

    1.2 Mechanics of Material

    Governing Equation

    – Left span

    112

    13

    1

    4

    1''''

    1 24dxcxbxa

    EI

    qxwqEIw ++++=→=

    – Right Span

    222

    23

    2

    4

    2''''

    2 24dxcxbxa

    EI

    qxwqEIw ++++=→=

    Boundary Conditions

    – Left support

    0)0()0( , 0)0( 111 =′′−== wEIMw

    – Center support

    )()( , )()( , 0)()( 212121 LwLwLwLwLwLw ′′=′′′−=′==

    – Right support

    0)0()0( , 0)0( 222 =′′−== wEIMw

    Since there are eight unknowns with eight conditions, we can solve this problem.

    Determination of Integration Constant – Left Support

    xcxaEI

    qxwbwdw 1

    31

    4

    11111 2402)0( , 0)0( ++=→==′′==

    – Right Support

    xcxaEI

    qxwbwdw 2

    32

    4

    22222 2402)0( , 0)0( ++=→==′′==

    x x

    y

    z

    z

    y

    q

    w1 w2

  • Dept. of Civil and Environmental Eng., SNU

    Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr

    5

    – Center Support

    ==

    −==→

    +=+

    −−−=++

    =++=

    =++=

    EI

    qLcc

    EI

    qLaa

    LaEI

    qLLa

    EI

    qL

    cLaEI

    qLcLa

    EI

    qL

    LcLaEI

    qLLw

    LcLaEI

    qLLw

    48

    48

    3

    62

    62

    36

    36

    024

    )(

    024

    )(

    3

    21

    21

    2

    2

    1

    2

    22

    2

    3

    12

    1

    3

    23

    2

    4

    2

    13

    1

    4

    1

    )32(48

    33421 xLLxxEI

    qww +−=≡

    83

    , 8

    32 11

    211

    qLqxwEIVx

    qLx

    qwEIM +−=′′′−=+−=′′−=

    Moment Diagram

    Shear Diagram

    Reactions

    0.375qL

    L83

    +

    -

    +

    0.625qL

    -

    0.375qL 0.375qL 1.25qL

    0.125qL2

    0.070qL2

    L8

    3

    +

    -

    +

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    Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr

    6

    1.3 Mechanics of Material + α

    1.3.1 Main idea

    Original Problem

    Case I (Removal of the center support)

    Case II (Application of the reaction force)

    Original Problem = Case I + Case II (compatibility condition)

    δ0+ δR=0

    1.3.2 Calculation of δ0

    Bending Moment

    qLxqx

    M +−=2

    2

    q

    q

    δ0

    δR

    Rb

    q

    qL2/2

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    Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr

    7

    Governing Equation

    baxqLxqx

    EIwMwEI ++−=→−=′′624

    34

    00

    Boundary (support) Conditions

    – Left Support : 0)0(0 =w

    – Right Support : 0)2(0 =Lw

    Determination of Integration Constant

    – Left Support

    0 0)0(0 =→= bEIw

    – Right Support

    EI

    LqaLa

    LqLqLEIw

    24

    )2(0)2(

    6

    )2(

    24

    )2()2(

    334

    0 =→=+−=

    Deflection

    ))2()2(2(24

    3340 LxLxxEI

    qw +−=

    EI

    LqLLLLL

    EI

    qLw

    384)2(5

    ))2()2(2(24

    )(4

    33400 =+−==δ

    1.3.3 Calculation of δR

    Bending Moment

    22,

    22

    21

    1

    LRxRM

    xRM bbb −=−=

    Governing Equation

    +++−=

    ++=→

    −=′′−=′′

    222

    22

    32

    2

    111

    31

    1

    22

    11

    412

    12

    bxaLxRxR

    EIw

    bxaxR

    EIw

    MwEI

    MwEI

    bbR

    bR

    R

    R

    1x 2x

    RbL/2

    Rb

  • Dept. of Civil and Environmental Eng., SNU

    Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr

    8

    Boundary (support and mid-span) Conditions

    – Left Support & Right Support

    0)( , 0)0( 21 == Lww RR

    – Mid-span

    )( )0()()( , )0()( 221121 LwLwLwLw RRRRRR θ=′=′=θ=

    Determination of Integration Constant

    – Left Support

    0 0)0( 11 =→= bwR

    – Right Support & Mid-span

    −=

    =

    −=

    ′==+=′

    ==+=

    =+++−=

    6

    04

    (0)4

    )(

    (0)12

    )(

    0412

    )(

    3

    2

    2

    2

    1

    221

    2

    1

    221

    3

    1

    22

    33

    2

    LRb

    a

    LRa

    wEIaaLR

    LwEI

    EIwbLaLR

    LEIw

    bLaLRLR

    LEIw

    b

    b

    Rb

    R

    Rb

    R

    bbR

    Deflection

    )23(12

    )3(12

    322

    322

    123

    11

    LLxxEI

    Rw

    xLxEI

    Rw

    bR

    bR

    +−−=

    −=

    EI

    RLwLw bRRR 48

    )2()0()(

    3

    21 −===δ

    Compatibility Condition

    δ0+ δR=0 → 048

    )2(

    384

    )2(5 34 =−EI

    RL

    EI

    Lq b → qLRb 810=

  • Dept. of Civil and Environmental Eng., SNU

    Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr

    9

    1.4 Structural Mechanics

    Original Problem

    Case I (Removal of the center support)

    Case II (Application of the reaction force)

    Original Problem = Case I + Case II

    δ0+ δR=0

    Principle of Virtual Work

    EI

    Lqdx

    EI

    MML R384

    )2(5 42

    0

    00 −== δ , EI

    LRdx

    EI

    MM bL

    RRR 48

    )2( 32

    0

    == δ

    Solution

    δ0+ δR=0 → 048

    )2(384

    )2(5 34 =+−EI

    RL

    EI

    Lq b → qLRb 810=

    RbL/2

    Rb

    q

    q

    qL2/2

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    Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr

    10

    – Moment

    – Shear

    +

    -

    +

    -

    +

    + -

    +

    -

    =

    0.070qL2

    5qL2/8

    Rb

    +

    =

    0.125qL2 L

    8

    3

    +

    -

    +

    qL2/2

  • Dept. of Civil and Environmental Eng., SNU

    Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr

    11

    1.5 Supports

    – Supports are used to maintain the position of the structure by constraining free mo-

    tions at some discrete points.

    – If you want to constrain a free motion of a structure at any part of the structure, some

    kind of forces should be applied.

    – Since a support prevent free motion of a structure, a force called as a reaction force is

    developed in the direction of a constraints.

    – The support conditions provide boundary conditions for the governing equation(s).

    – The reaction forces are unknown, and should be calculated through the structural

    analysis.

    – The most popular support types are a fixed support, a hinge support and a roller sup-

    port.

    δ0δ

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    12

    Fixed Support(고정단)

    – Both the horizontal and vertical motion and rotational motion are prevented.

    – Reaction forces are developed in three directions.

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    13

    Hinged Support(회전단)

    – The horizontal and vertical motion are prevented, but the rotational motion is allowed.

    – The horizontal and vertical reaction forces are developed in all three directions

    Roller Support(이동단)

    – Only the vertical motion is prevented, and the horizontal and the rotational motion is

    allowed.

    – A reaction force is developed only in the vertical direction.

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    14

    1.6 Two-Dimensional Idealization of a Structure

    Real 3-D Structures

    – 3 force components and 3 moment components

    – 3 displacement components and 3 rotational components

    Main Structures : A part of a structure designed mainly to resist external loads.

    – Beams, Truss, Frame, Arch…

    Secondary Structure : A part of a structure designed to transfer external loads to

    main structures or increase stability and/or stiffness of a structure

    – Stringer : Small beam that runs parallel to a main structure to support the floor

    system.

    – Cross beam : Small beam that runs perpendicular to the main structure to support

    stringers.

    – Bracing (tie) : member that connects two or more main structures to increase mainly

    lateral stability of the main structures.

    xδ xf

    ym yθ zδ

    x

    y

    z

    yδ yf

    zδ zf

    xδ xf ym yθ

    zm zθ xm xθ

  • Dept. of Civil and Environmental Eng., SNU

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    15

    Load transfer path

    External load → Slab → Stringer → Cross Beam → Main Structure → Support →

    Foundation

    – External loads applied on a slab are supported by stringers, which transfer the carried

    loads to cross beams. The cross beams, in turn, transfer the carried-over loads from

    the stringers to main structures.

    주 구조물 (Main Structure)

    가로 보 (Cross Beam)

    세로 보 (Stringer )

    Cross Bracing (Wind Bracing)

    지 점 (Support)

  • Dept. of Civil and Environmental Eng., SNU

    Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr

    16

    Two-dimensional Idealization of a Structures

    All members are idealized as one-dimensional members, which are represented by

    lines passing the centroids of the cross sections, because dimensions in cross-

    sectional directions are very small compared to their lengths (usually less than 1/20).

    Suppose we somehow know portions of the external loads transferred to each main

    structure.

    Suppose all members in each main structure lie in one plane, and external loads car-

    ried-over from substructures are applied on the same planes.

    Then, all structural responses are developed in the same plane, and main load resist-

    ing actions occur in the same plane.

    If this the case, each main structure in the same plane can be separated from the

    whole structure by neglecting three-dimensional effects, and analyzed in the plane

    where each main structure is defined for the simplicity of a structural analysis.

    Two-dimensional structures are often referred to as plane structures.

    However, fundamentals employed in the 2-D analysis of structures are exactly the

    same as those in the 3-D analysis.

  • Dept. of Civil and Environmental Eng., SNU

    Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr

    17

    Plane truss idealization

    – Structure composed of a number of bars carrying axial forces that lie in one plane.

    – Vertical and horizontal force

    – Vertical and horizontal displacement

    절점

    xf xδ

    yδ yf

  • Dept. of Civil and Environmental Eng., SNU

    Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr

    18

    Plane beam idealization

    – Structure composed of a number of beams carrying bending moments and shear

    forces that lies in one plane.

    – Shear force and Moment on z-axis

    – Vertical displacement and rotational angle with respect to z-axis

    11 , wV 22 , wV

    33 , wV

    11 , θM 22 , θM 33 , θM

  • Dept. of Civil and Environmental Eng., SNU

    Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr

    19

    Plane Frame idealization

    – Structure composed of a number of beams carrying bending moments, shear forces

    and axial forces that lie in one plane.

    – Vertical, horizontal force and moment w.r.t. z-axis

    – Vertical, horizontal displacement rotational angle w.r.t. z-axis

    xδ xf

    yf yδ zM zθ

  • Dept. of Civil and Environmental Eng., SNU

    Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr

    20

    1.7 Stability of Structures

    Internal Stability (내적 안정)

    A structure maintains its shape under the actions of all possible loads within a certain limit.

    External Stability (외적 안정)

    A structure maintains its position under the actions of all possible loads within a certain limit.

  • Dept. of Civil and Environmental Eng., SNU

    Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr

    21

    This page is intentionally left blank.

  • Dept. of Civil and Environmental Eng., SNU

    Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr

    22

    Chapter 2

    Reactions & Internal Forces

    by Free Body Diagrams

  • Dept. of Civil and Environmental Eng., SNU

    Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr

    23

    2.1 Free Body Diagram

    Any portion of a structure can be isolated by passing any desired section through structure.

    A free body sketch is a diagram drawn showing this portion acted upon by the external loads

    and reactions, together with any forces that may be act on the faces of the members cut by the

    isolating section. Or, a diagram drawn by removing some part(s) of a structure and replacing

    the removed part(s) with unknown forces.

  • Dept. of Civil and Environmental Eng., SNU

    Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr

    24

    It is impossible to draw too many free-body diagrams.

    Time spent in doing so is never wasted

    - C. H. Norris & J. B. Wilbur & S. Utku -

  • Dept. of Civil and Environmental Eng., SNU

    Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr

    25

    2.2 Reactions

    Beams

    +=−→=

    =−+→=

    ) (clockwise 00

    00

    LRPaM

    PRRF

    BA

    BAVP

    L

    aRB = , PL

    bRA =

    +=−+→=

    =+→=

    ) (clockwise 0)(0

    0

    LRaLPM

    PRRF

    BA

    BAVP

    L

    aRB )1( += , PL

    aRA −=

    Truss

    PHPHF AAH −=→=+→= 00 ,

    00 =−+→= PRRF BAV

    0)3(0 =−+→= aRPaPaM BA (Clockwise +)

    PRA 31= , PRB 3

    2=

    P a b

    L

    P

    RAL

    RBL

    b a

    P a

    L

    P a

    L RAL

    RBL

    RAL

    RB P

    P

    HAL

    P

    P

    [email protected]

    a

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    26

    Frame

    +=

    −=→

    =−−=

    =−−=

    =+=

    =−→=−+=−+=

    42

    2

    0422

    02

    082

    24

    2 02

    qLHR

    HR

    LqLLH

    LRM

    LHL

    RM

    HHF

    qLHH

    qLH

    qLH

    qLRRF

    AA

    BB

    AALh

    BBRh

    BAH

    BABABAV

    What if there is no internal hinge ?

    042

    )(2

    )(

    0422

    02

    0

    02

    =−+−−→

    =−−+−=

    =−−=

    =+=

    =−+=

    LqLLHH

    LRR

    LqLLH

    LRMM

    LHL

    RMM

    HHF

    qLRRF

    ABBA

    AALh

    BBRh

    BAH

    BAV

    HB

    RA RB

    HA

    L

    L

    h

    q

    →−=

    =

    16

    16qL

    H

    qLH

    B

    A

    8

    8

    3

    qLR

    qLR

    B

    A

    =

    =

    HB

    RA RB

    HA

    q

    M M

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    27

    2.3 Internal Forces in Framed Structures

    Axial Force

    Shear Force

    Bending Moment

    Torsion

    +

    +

    +

    +

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    28

    2.4 Internal Forces in Simple Beams

    Reactions

    q

    RA=qL/2 Rb= qL/

    q

    RA=qL/2 RB= qL/2

    RA RB

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    29

    Free Body Diagram for Shear and Moment

    qxqL

    qxRVVqxRF AxxAV −=−=→=−−= 20

    22

    02

    2qxx

    qLMM

    xqxxRM xxAx −=→=−−=

    Differential Equation

    baxxq

    Mqdx

    Md ++−=→−= 22

    2

    2

    Boundary Conditions: 2222

    ,00)()0( xq

    xqL

    MqL

    abLMM −=→==→==

    Shear Force and Moment Diagrams

    Deflected Shape

    Remember 2

    2

    dx

    wdEIM −= !!!

    RA

    x

    RB

    Mx

    Vx

    +

    qL/2

    qL2/8

  • Dept. of Civil and Environmental Eng., SNU

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    30

    2.5 Gerber Systems

    A structural system that an unstable structural component is supported by a stable structure.

    Sometimes there may be several unstable structures are supported by a stable structure, and

    vice versa. An external load applied to an unstable structure is transferred to a stable struc-

    ture, but an external load applied to a stable structure is not transferred to an unstable structure.

    In the later case, a unstable structure is just deflected as a rigid body without deformation.

  • Dept. of Civil and Environmental Eng., SNU

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    31

  • Dept. of Civil and Environmental Eng., SNU

    Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr

    32

    2.5.1 Internal Forces in a Gerber Beam - I

    Reactions

    PRL

    PL

    RM HHC 3

    20

    24

    30 =→=×−×→=

    PRPRRF CCHv 3

    100 =→=−+→=

    PRL

    RLRM BHBA 6

    50

    4

    50 =→=×+×−→=

    PRPRRF ABAv 6

    10

    3

    20 −=→=−+→=

    L/4

    P

    RH

    RA RB

    RC

    P

    P/6 5P/6 P/3

    P

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    33

    Shear Force

    i) Lx ≤≤0

    ii) LxL2

    3≤≤

    iii) LxL 22

    3 ≤≤

    Bending Moment

    i) Lx ≤≤0

    P/6

    V= -P/6

    P/6 5P/6

    V= 2P/3

    P/6 5P/6

    P

    V= -P/3

    +

    - -

    2P/3

    P/3 P/6

    P/6 P/6

    Mx= Px6

    1−

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    34

    ii) LxL2

    3≤≤ , Lxx −=′

    iii) LxL 22

    3 ≤≤ xLx −=′ 2

    Deflected Shape

    qdx

    wdEI =

    4

    4

    or 2

    2

    dx

    wdEIM −=

    P/6 5P/6

    2P/3

    Mx=

    xPPL ′+− 2

    P/6 5P/6

    P

    P/3 P/3

    Mx= xP ′3

    1

    +

    -

    PL/6

    PL/6

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    35

    2.5.2 Internal Forces in a Gerber Beam - II

    Free Body diagram

    Shear

    q

    2

    ql

    2

    ql

    2

    ql

    2

    ql

    2

    2ql

    2

    2ql

    2

    ql

    2

    ql

    +

    L

    q

    L L

    2

    ql

    2

    ql

    x

    qxql

    Vx −= 2 2

    ql

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    Moment

    Deflected Shape

    8

    2ql

    2

    2ql 2

    2ql

    +

    xqlql

    Mxqlql

    M xx 220

    22

    22

    +−=→=−+

    2

    ql

    2

    2ql

    2

    ql

    x

    2

    ql

    220

    2

    2qxx

    qLMM

    xqxxR xxA −=→=++−

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    2.6 Truss (트러스)

    Structures composed of a number of straight bars carrying axial forces.

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    Assumption

    1. All joints are hinges.

    2. All members are straight.

    3. Small deformatiom

    4. The external loads are applied only at joints.

    The 1st assumption seems to be unreasonable, but it is very reasonable assumption !!!

    Because the length of each member is short compared to the total length of a truss, the mo-

    ment induced by rigid joints is negligibly small in an engineering sense.

    Characteristics of truss

    – By the 2nd , 3rd and 4th assumptions

    02

    2

    2

    2

    =→−=dx

    Mdq

    dx

    Md→ baxM +=

    – By the 1st assumption

    0 , 000)()0( ≡≡→==→== VMbaLMM

    – No bending moment and shear force are induced in all members in a truss structure.

    – Only axial forces are the internal forces in a truss.

    – The axial force is constant along a member by assumption 1.

    – There is only one unknown per member.

    – As the equilibirium of each member is always satisfied, the equilibrium conditions at

    joints are to be considered.

    – Determinancy

    – The number of unknowns : RM NN +

    – The number of Equilibrium Equations : JN2

    – 02 =−+ JRM NNN : Determinant, 02 >−+ JRM NNN : Indeterminant

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    39

    2.6.1 Internal Forces in Howe Truss

    At U1 and U3

    At L1

    2 ,

    22

    022

    2

    02

    2

    23

    3

    23 PFPF

    PF

    FF=−=→

    =+

    =+

    At L2

    2 , 0- , 0 265625

    PFFPFFFPF ===→=+=−

    F1 F3

    F2

    P/2

    F5

    F2 F6

    P

    F4=0

    F1=0

    F8 =0

    F9=0

    9+3-2×6=0: determinant truss

    L2

    7

    9 1

    4

    2

    3

    U1 U2 U3

    6

    5

    L1

    P/2 P/2

    L3

    8

    P

    [email protected]

    a

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    At U2

    =+−

    −==→

    =−−−

    =++−−

    02

    1

    2

    12

    2

    02

    2

    2

    2

    02

    2

    2

    237

    753

    8734

    PPP

    PFF

    FFF

    FFFF

    At L3

    Because we already have 3 equilibrium equations to calculate the reactions

    Axial Force Diagram

    2

    P

    2

    P

    F5

    F4 F8

    F3 F7

    2

    P 0

    P/2

    P/2

    P/2 P/2

    P 0

    0 0

    0

    P P/2 P/2

    Tension Compression

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    Equilvalent Beam Action

    Axial forces in members generate bending moments and shear forces just like in a beam !!!

    Deflected Shape

    2.6.2 Internal Forces in Warren Truss

    V=P/2

    P/2

    1 P

    3P

    32P

    2 3

    4

    5

    6

    7

    8

    9

    10

    11

    L1 L2 L3

    U1 U2 U3

    L4

    x

    PL/4 M=Px/2

    [email protected]

    l

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    At L1

    PFPFP

    F

    FF

    32

    , 3

    22

    03

    222

    022

    12

    2

    12

    =−=→

    =+

    =+

    At U1

    PFPFF

    FF

    FFF

    34

    ,232

    022

    22

    02

    2

    2

    2

    423

    32

    432

    −==−=→

    =+

    =++−

    At L2

    PFPF

    FFFF

    PFF==→

    =+−+

    =−+65

    5361

    35

    ,32

    022

    22

    -

    022

    22

    F2

    F1

    2P/3

    F4

    F2 F3

    F3

    F1 F6

    P

    F5

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    At U2

    PFPF

    FF

    FFFF

    32

    ,32

    022

    22

    02

    2

    2

    2

    87

    75

    7584

    −=−=→

    =−−

    =+−+−

    At L3

    3,

    32

    022

    22

    022

    22

    109

    97610

    97 PFPF

    FFFF

    FF==→

    =+−−

    =+

    At U3

    PF

    FF

    FFF

    32

    022

    22

    022

    22

    11

    119

    8119

    −=→

    =−−

    =−+−

    F8

    F5 F7

    F4

    F7

    F6 F10

    F9

    F11

    F8

    F9

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    At L1

    OK

    Deflected Shape

    Equivalent Beam Action

    1 2 3 4 5 6

    1. PxM32=

    2. PxPlM32

    32 +=

    3. PxPlxlPPlM31

    34

    )(31 −=−+=

    4. PxPlxlPPlM31

    )(31

    32 −=−+=

    5. PxPlxlPPlM31

    32

    )(31

    31 −=−+=

    6. PxPlxlPM31

    31

    )(31 −=−=

    P/3

    P/3

    P32

    P

    Pl3

    4

    PxPlM3

    2

    3

    2 += 2P/3

    2P/3

    4P/3=2P/3+2P/3

    2P/3 x

    x

    2P/3

    PxM3

    2=

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    2.6.3 Method of Sections

    PFL

    FLPM L 34

    023

    2442 −=→=×+×=

    PFL

    FL

    PLPMU =→=×−×−×= 662 02223

    32

    PFFPPFV 32

    022

    32

    55 =→=×+−=

    P

    3P

    P32

    1

    2 3

    4

    5

    6

    7

    8

    9

    10

    11

    L1 L2 L3

    U1

    U

    U2 U3

    L4

    Cut out

    P

    P32

    F5

    F4

    F6

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    2.7 프레임 (Frame)

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    2.7.1 Internal Forces in a Frame

    Reactions

    Freebody Diagram

    H

    qH

    L

    L

    qH

    2

    2

    q

    L

    qH

    2

    2

    2

    2qH

    qH

    L

    qH

    2

    2

    L

    qH

    2

    2

    2

    2qH

    L

    qH

    2

    2

    LqH

    2

    2

    L

    qH

    2

    2

    L

    qH

    2

    2

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    Axial, Shear and Moment diagram

    Deflected Shape

    L

    qH

    2

    2

    + - Axial

    L

    qH

    2

    2

    qH

    -

    + Shear

    L

    qH

    2

    2

    2

    2qH

    +

    Moment

    +

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    2.7.2 Internal Forces in a 3-hinged Frame

    Reactions (+:Clockwise for mement)

    02

    =+

    =+

    BA

    BA

    HH

    qLRR

    = 0RhM : BBBB HRLHLR 202 −=→=−−

    = 0LhM : 420422qL

    HRLqL

    LHL

    R AAAA +=→=−−

    082

    24

    2

    =+

    =−→=−+

    BA

    BABA

    HH

    qLHH

    qLH

    qLH

    16

    16qL

    H

    qLH

    B

    A

    −=

    = →

    8

    83

    qLR

    qLR

    B

    A

    =

    =

    L

    L

    HA HB

    RA RB

    h

    q

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    Freebody Diagram

    Axial, Shear and Moment diagram

    16qL

    16qL

    8qL

    16qL

    83qL

    16qL

    8qL

    83qL

    8qL

    16qL

    16qL

    16

    2qL 16

    2qL

    16

    2qL 16

    2qL

    16qL

    16qL

    - +

    - 8

    qL

    +

    83qL

    Shear

    83qL

    8qL

    16qL

    - -

    -

    Axial

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    283

    16

    028

    316

    22

    2

    xqqLx

    qLM

    xqxqLx

    qLM

    −+−=

    =−+−−

    Deflected Shape

    - -

    -

    Moment

    - -

    2

    16qL 2

    16qL

    83qL

    16

    2qL

    M

    V

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    2.8 Arches

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    2.8.1 Three Hinged Arch

    Arch Curve : )( 22

    2xl

    l

    hy −=

    Reactions

    2

    2P

    R

    PR

    B

    A

    =

    = ,

    h

    PlH

    h

    PlH

    B

    A

    2

    2

    =

    =

    Freebody Diagram

    h

    PlH

    PV

    2

    2

    =

    =

    0)( =++−− xLRyHM AA

    )(2

    )(2

    )(2

    )(22

    2

    222

    xlxl

    P

    xlP

    xll

    h

    h

    Pl

    xlP

    yh

    PlM

    +=

    ++−−=

    ++−=

    8maxPl

    M −=

    HA

    RA

    HB

    RB

    P

    h

    2l

    x

    y

    HA

    RA

    V

    H

    M

    y

    l+x

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    Axial force and Shear Force

    θθθ−θ

    =

    θ+θ=θ−θ=

    S

    A

    V

    H

    SAV

    SAH

    cossin

    sincos

    cossin

    sincos

    θθ−θθ

    =

    V

    H

    S

    A

    cossin

    sincos

    224224

    2

    2

    4

    2sin ,

    4cos

    2tan

    xhl

    hx

    xhl

    l

    l

    hxy

    +

    −=θ+

    −=′=θ

    )2

    (4

    cos2

    sin2

    )2

    (4

    sin2

    cos2

    2

    224

    23

    224

    llx

    xhl

    PP

    h

    PlS

    h

    xhl

    xhl

    PP

    h

    PlA

    −−+

    =θ−θ=

    +−=θ−θ−=

    Deflected Shape

    V

    H

    A S θ

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    2.8.2 Zero Moment Arch I

    Reactions

    2

    2P

    R

    PR

    B

    A

    =

    = ,

    h

    PlH

    h

    PlH

    B

    A

    2

    2

    =

    =

    Freebody Diagram

    h

    PlH

    PV

    2

    2

    =

    =

    0)( =++−− xLRyHM AA

    )(

    0)(22

    xll

    hy

    xlP

    yh

    PlM

    +=

    =++−=

    HA

    RA

    HB

    RB

    P

    h

    2l

    x

    y

    H

    HA

    RA

    V

    M

    y

    l+x

    x

    y

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    2.8.3 Zero Moment Arch II

    Reactions

    qlR

    qlR

    B

    A

    == ,

    h

    qlH

    h

    qlH

    B

    A

    2

    22

    2

    =

    =

    Freebody Diagram

    02

    )()()( =++−++−− xlxlqxlRyHM AA

    )(

    02

    )()()(

    2

    222

    2

    xll

    hy

    xlxlqxlqly

    h

    qlM

    −=

    =++−++−=

    HA

    RA

    HB

    RB

    q

    h

    2l

    x

    y

    HA

    RA

    V

    H

    M

    y

    l+x

    q

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    Chapter 3

    Principle of Virtual Work

    The principle of virtual work is the most important subject in the area of the structural analysis !!!!

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    3.1 Beam Problems

    3.1.1 Governing Equations

    Equilibrium for vertical force

    qdx

    dVqdxVdVV −=→=+−+ 0)(

    Equilibrium for moment

    Vdx

    dMdxqdxVdxMdMM =→=+−−+ 0

    2)(

    Elimination of shear force

    qdx

    Md −=2

    2

    Strain-displacement relation

    A plane section before deformation remains a plane after deformation.

    A section perpendicular to the neutral axis before deformation maintain the right

    angle to the neutral axis after deformation

    ydx

    wd2

    2

    −=ε

    Stress-strain relation (Hooke’s law)

    ydx

    wdEE

    2

    2

    −=ε=σ

    Definition of Moment

    2

    22

    2

    2

    dx

    wdEIdAy

    dx

    wdEydAEydAM

    AAA

    −=−=ε=σ=

    Beam Equation

    qdx

    wdEI =

    4

    4

    M M+dM

    V V+dV

    q

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    3.1.2 Limit Process

    Equilibrium for vertical force

    qdx

    dVq

    x

    Vq

    x

    VxqVVV

    x−=→=+

    ∆∆→=+

    ∆∆→=∆+−∆+

    →∆0)(lim00)(

    0

    Equilibrium for moment

    0))()(1

    (lim0)()()(0

    00

    =ξξ−∆ξ+∆

    +−∆

    ∆→=ξξ−∆ξ++∆−−∆+ ∆

    →∆

    dxxqx

    Vx

    MdxxqxVMMM

    x

    x

    x

    00)(),()0,(),(

    lim)(1

    lim)(lim

    )(),(

    00

    00

    00

    =×−=ξ

    ξ−=∆

    −∆−=ξξξ+∆

    −ξξ+

    ξξξ+=ξ

    =ξ→∆

    →∆

    →∆

    xqd

    xdQ

    x

    xQxxQdxq

    xdxq

    dxqxQ

    x

    x

    x

    x

    x

    3.1.3 Modelling of Concentrate loads - Dirac delta functions

    0lim→ε

    = = )( ξ−δ x

    122

    1lim)0

    2

    10(lim)(

    00

    00

    =εε

    =+ε

    +=ξ−δ→ε

    ε+ξ

    ε+ξ

    ε−ξ

    ε−ξ

    →ε ll

    dxdxdxdxx

    )()(2

    )()(lim)(

    2

    1lim

    )0)(2

    1)(0)((lim)()(

    00

    00

    0

    ξ=ξ′=ε

    ε−ξ−ε+ξ=ε

    =

    +=ξ−δ

    →ε

    ε+ξ

    ε−ξ→ε

    ε+ξ

    ε+ξ

    ε−ξ

    ε−ξ

    →ε

    fFFF

    dxxf

    dxxfdxxfdxxfdxxxfll

    ξ

    ξ

    ε21

    ξ

    M M+∆M

    V V+∆V

    q

    ∆x

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    3.1.4 Principle of Virtual Work (Beam)

    The real(actual) load case A virtual(imaginary) load case

    For the real load case

    02

    2

    =+ qdx

    Md,

    2

    2

    dx

    wdEIM −= for lx

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    Principle of virtual work

    dxEI

    MMdxM

    dx

    wd ll

    −=00

    2

    2

    dxqwdxEI

    MM ll =00

    → extWW δ=δ int

    The internal virtual work is equal to the external virtual work if a beam satisfy the

    equilibrium for the real and virtual load.

    Equilibrium equation for load case q

    0)(0

    2

    2

    =+ dxqdxMd

    wl

    Virtual work expression

    dxqwdxEI

    MM ll

    =00

    Betti-Maxwell’s Reciprocal Theorem

    dxEI

    MMdx

    EI

    MM ll

    =00

    → dxqwdxqwll

    =00

    Calculation of displacement for the load case q

    dxqwdxEI

    MM ll

    =00

    In case q system represents a single unit concentrated load applied at the position where

    you want to calculate the displacement for q system.

    )()( 00

    0

    0

    xwdxxxwdxEI

    MM ll =−δ= → dxEIMM

    xwl

    =0

    0 )(

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    3.1.5 Example

    A simple beam subject to an uniform load

    – Moment of load case q

    – Moment of load case q

    – Deflection at the center of the span

    EI

    ql

    EI

    qllqlql

    EI

    dxxq

    xql

    EIdxx

    qx

    qlx

    EI

    dxxq

    xqlxl

    EIdxx

    qx

    qlx

    EIdx

    EI

    MMlw

    ll

    l

    l

    ll

    4443

    2/

    0

    322/

    0

    2

    2/

    22/

    0

    2

    0

    384

    5)

    256

    1

    96

    1(

    2))

    2(

    4

    1

    4)

    2(

    3

    1

    4(

    2

    )44

    (2

    )22

    (2

    2

    )22

    )(22

    (1

    )22

    (2

    1)

    2(

    =−=−=

    −=−=

    −−+−==

    or from the integration table,

    EI

    qlqll

    EI

    lMM

    l

    ab

    EI

    ldx

    EI

    MMlw

    l

    3845

    84)

    41

    1(3

    )1(3

    )2

    (42

    3120

    =+=+==

    l/4

    1

    q

    ql2/8

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    Values of Product Integrals L

    LU dxMM0

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    3.1.6 Conservation of Energy

    Equilibrium and Conservation of Energy

    – Equilibrium Equation

    qdx

    wdEI =

    4

    4

    – External work

    int

    ll

    lll

    lll

    ll

    ext

    WdxEI

    Mdx

    dx

    wdEI

    dx

    wd

    MwVdxdx

    wdEI

    dx

    wd

    dx

    wdEI

    dx

    dw

    dx

    wdwEIdx

    dx

    wdEI

    dx

    wd

    dxdx

    wdwEIwqdxW

    ===

    θ+−=

    −+=

    ==

    0

    2

    02

    2

    2

    2

    000

    2

    2

    2

    2

    02

    2

    03

    3

    02

    2

    2

    2

    04

    4

    0

    2

    1

    2

    1

    ][2

    1

    ][2

    1

    2

    1

    2

    1

    Conservation of Energy in each load case

    dxwqdxEI

    M ll

    =00

    2

    21

    21

    , dxqwdxEI

    M ll

    =00

    2

    21

    21

    Two load cases are applied simultaneously.

    dxqwqwdxEI

    MM

    dxwqqwqwqwdxEI

    Mdx

    EI

    MMdx

    EI

    M

    dxqqwwdxEI

    MM

    ll

    llll

    ll

    +=

    +++=++

    ++=+

    00

    00

    2

    00

    2

    00

    2

    )(2

    1

    )((2

    1

    2

    1

    2

    1

    ))((2

    1)(

    2

    1

    w

    q

    w

    q

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    65

    External work for sequential loading (q first)

    =

    dxwqdxwqdxqwqwdxwq

    dxwqqwqwqwdxqwqdxwwqdx

    dxqqwwdxqwdxwqwqdx

    llll

    lll l

    lll l

    =→+=

    +++=++

    ++=++

    0000

    000 0

    000 0

    )(21

    )((21

    21

    21

    ))((21

    21

    21

    Principle of Virtual work

    dxqwdxqwdxqwqwdxEI

    MM llll

    ==+=0000

    )(2

    1

    3.1.7 General Conservation and Equilibrium

    Conservation in General

    =+⋅−S V

    fdVdS 0nv

    v n

    dS

    w

    q

    w

    q

    ww +

    w

    q

    w

    q

    ww +qq + qq +

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    – By divergence theorem,

    ⋅∇−=⋅−VS

    dVdS vnv where ),,(),,(321 xxxzyx ∂

    ∂∂∂

    ∂∂=

    ∂∂

    ∂∂

    ∂∂=∇

    =+⋅−∇=+⋅∇−=+⋅−VVS VV

    dVffdVdVfdVdS 0)( vvnv

    – Since the integral equation should hold for all systems,

    0=+⋅∇− fv

    Potential Problems

    – The vector field of a system is defined by a gradient of a scalar function referred to as

    a potential function

    Φ∇⋅−= kv , ),,(zyx ∂Φ∂

    ∂Φ∂

    ∂Φ∂=Φ∇

    – The famous Laplace equation for a conservative system.

    0)( =+Φ∇⋅⋅∇=+⋅∇− ff kv

    – If the system properties are homogeneous and isotropic, Φ∇−= kv

    0)( 2 =+Φ∇=+Φ∇⋅∇=+⋅∇− fkfkfv or 02

    2

    2

    2

    2

    2

    =+∂

    Φ∂+∂

    Φ∂+∂

    Φ∂f

    zyx

    3.1.8 Equilibrium in General

    Force Equilibrium : === 0zyx FFF or 0=F

    =+S V

    dVdS 0bT or =+S V

    ii dVbdST 0 for i = 1,2,3

    – Suppose nT ⋅= σ or =

    ⋅=σ=3

    1jijiji nT nσ ,

    =

    σσσσσσσσσ

    =

    3

    2

    1

    333231

    232221

    131211

    σσσ

    σ ,

    =

    3

    2

    1

    n

    n

    n

    n

    TT=

    E, ν

    uu = Su

    St

    V

    x

    y

    z

    bx

    by

    bz

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    67

    – Divergence Theorem

    0)( =+⋅∇=

    +⋅∇=+⋅=+

    V

    ii

    V V

    ii

    S V

    ii

    S V

    ii

    dVb

    dVbdVdVbdSdVbdST

    σ

    σσ n for i = 1,2,3

    Since the integral equation should hold for all systems in equilibrium,

    03

    1

    321 =+∂σ∂

    =+∂σ∂

    +∂σ∂

    +∂σ∂

    =+⋅∇ =

    ij j

    iji

    iiiii bx

    bzyx

    bσ for i = 1,2,3 or

    0

    0

    0

    3333231

    2232221

    1131211

    =+∂σ∂

    +∂σ∂

    +∂σ∂

    =+∂σ∂

    +∂σ∂+

    ∂σ∂

    =+∂σ∂

    +∂σ∂+

    ∂σ∂

    bzyx

    bzyx

    bzyx

    Moment Equilibrium : 0= iM for i=1, 2, 3 or 0=M

    0=+×+× VVS

    dVdVdS mfxvx

    211231133223 , , σ=σσ=σσ=σ

    – In case elasticity problems are potential problems

    ii

    i uC ∇⋅=σ

    – However, to maintain symmetry condition of stress,

    uC ∇= :σ or = = ∂

    ∂=σ3

    1

    3

    1k l l

    kijklij u

    uC

    – Equilibrium equation in terms of the potential functions

    0):( =+∇⋅∇ buC

    – What is σ, and why is σ related to a potential function in such a way?

    out of scope of this class !

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    3.1.9 Displacement on boundaries

    Virtual Work Expression

    llll

    MVwdxqwdxEI

    MM00

    00

    θ−+=

    llll

    MVwdxqwdxEI

    MM00

    00

    θ−+=

    )0()0()()()0()0()()(0

    0000

    MlMlVwlVlwdxqwMVwdxqwdxEI

    MM lllll θ+θ−−+=θ−+=

    By coinciding the positive direction of forces and displacement

    )0()0()()()0()0()()(00

    MlMlVwlVlwdxqwdxEI

    MM ll θ+θ+++=

    Deflection of a cantilever beam subject to an end load

    0)0( , 0)0( , 0)( , 1)( =θ=== wlMlV

    q (real) system q (virtual) system

    )0()0()()()0()0()()(00

    MlMlVwlVlwdxqwdxEI

    MM ll θ+θ+++=

    EI

    PlPll

    EI

    lMM

    EI

    ldx

    EI

    MMlw

    l

    3))((

    33)(

    3

    31

    0

    =−−===

    Or, you can obtain the same answer by assuming the unit concentrate load is applied at

    just left side of the boundary.

    Rotation of a cantilever beam subject to an end load

    0)0( , 0)0( , 1)( , 0)( =θ=== wlMlV

    q (real) system q (virtual) system

    )0()0()()()0()0()()(00

    MlMlVwlVlwdxqwdxEI

    MM ll θ+θ+++=

    EI

    PlPl

    EI

    lMM

    EI

    ldx

    EI

    MMl

    l

    2)(1

    22)(

    2

    31

    0

    −=−××===θ

    P

    -Pl

    1

    -l

    P

    -Pl

    1

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    Rotation in the a body (or a structure)

    – Modeling of a unit moment applied at x0

    )0()0()()()0()0()()(00

    MlMlVwlVlwdxqwdxEI

    MM ll θ+θ+++=

    )()]2

    ()2

    ([1

    lim

    ))]2

    ((1

    ))2

    ((1

    [lim

    0000

    0

    000

    0

    0

    xdx

    dwxwxw

    dxxxxxwdxEI

    MM

    xx

    ll

    θ−=−=ε+−ε−ε

    =

    ε+−δε

    −ε−−δε

    =

    =→ε

    →ε

    – by coinciding the positive direction of the rotational angle with that of the applied

    moment.

    dxEI

    MMx

    l

    =θ0

    0 )(

    ε 1/ε 1/ε

    x0

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    3.2 Principle of Virtual Work in General

    3.2.1 3-Dimensional Elastic Body

    Rigid Body

    Rigid body: A body that does not deform at all. Displacement is constant over the body.

    Deformation of a body is caused by the variation of displacements in a body.

    If a real q-force system is acting on a rigid body is in equilibrium and remains in

    equilibrium as the body is given any small displacement, the virtual work done by the q-

    force system is equal to zero.

    0)( =+⋅=⋅+⋅=δS VS V

    ext dVdSdVdSW bTwbwTw

    Deformable Body

    If a deformable body is in equilibrium under a real q force system while it is subjected to

    small and compatible displacement caused by a virtual q force system, the external

    virtual work done by the real q force system is equal to internal virtual work done by

    the internal q stress !!!

    q-Force System

    q -Force System

    w

    T

    T

    dVdSij V

    ijij

    S

    σε=⋅ Tw

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    71

    ⋅+⋅=δS V

    ext dVdSW bwTw

    =δV

    dVW σε :int

    = == =

    = == ==

    ∂σ∂

    +σ∂∂=σ

    ∂∂=

    σ=σ==⋅

    3

    1

    3

    1

    3

    1

    3

    1

    3

    1

    3

    1

    3

    1

    3

    1

    3

    1

    )()(i j V j

    ijiij

    j

    i

    i j V

    ijij

    i j S

    jijii S j

    jijii S

    ii

    S

    dVx

    wx

    wdVw

    x

    dSnwdSnwdSTwdSTw

    int

    3

    1

    3

    1

    3

    1

    3

    1

    3

    1

    3

    1

    3

    1

    3

    1

    3

    1

    3

    1

    3

    1

    3

    1

    3

    1

    3

    1

    3

    1

    3

    1

    3

    1

    3

    1

    3

    1

    3

    1

    3

    1

    )(2

    1

    )(2

    1

    )(2

    1

    )(

    )(

    WdVdVx

    w

    x

    w

    dVx

    wdV

    x

    w

    dVx

    wdV

    x

    wdV

    x

    w

    dVbx

    wdVx

    w

    dVbwdVx

    wx

    wW

    i j V

    ijiji j V

    iji

    j

    j

    i

    i j V

    iji

    j

    i j V

    ijj

    i

    j i V

    jii

    j

    i j V

    ijj

    i

    i j V

    ijj

    i

    i V ji

    j

    iji

    i j V

    ijj

    i

    i V

    iii j V j

    ijiij

    j

    iext

    δ=σε=σ∂∂

    +∂∂=

    σ∂∂

    +σ∂∂=

    σ∂∂

    +σ∂∂=σ

    ∂∂=

    +∂σ∂

    +σ∂∂=

    +∂σ∂

    +σ∂∂=δ

    = == =

    = == =

    = == == =

    = == =

    == =

    =⋅+⋅S VV

    dVdVdS σε :bwTw

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    3.2.2 General Framed Structures

    )( τγ+σε=τγ+σε=σ⋅εe VVVVij V

    ijij dVdVdVdVdVee

    Internal virtual work by normal stress – bending moment

    =−−=

    ==

    ==−−=σε

    ee

    ee

    e

    e

    eeee

    ll

    ll

    A

    l

    AVVV

    dxEI

    MMdx

    EI

    MEI

    EI

    M

    dxdx

    wdEI

    dx

    wddx

    dx

    wddAEy

    dx

    wd

    dAdxydx

    wd

    dx

    wdEdVy

    dx

    wd

    dx

    wdEdVy

    dx

    wdEy

    dx

    wddV

    00

    02

    2

    2

    2

    02

    22

    2

    2

    0

    22

    2

    2

    22

    2

    2

    2

    2

    2

    2

    2

    2

    )()(

    )(

    )()(

    Internal virtual work by normal stress – Axial Force

    ===σεee

    eee

    ll

    AVV

    dxEA

    FFdx

    A

    FdA

    EA

    FdV

    A

    F

    EA

    FdV

    00

    )(

    Internal virtual work by shear stress

    QyIb

    V

    )(=τ and Q

    yGIb

    V

    )(=γ where =

    a

    y

    ydAQ

    ===τγee

    eee

    ls

    l

    AVV

    VdxVGA

    fdAdx

    ybI

    Q

    G

    VVQdV

    yIb

    VQ

    yGIb

    VdV

    0022

    2

    )()()(

    Total displacement

    ++=e

    l ll

    s

    e ee

    dxEA

    FFdx

    GA

    VVfdx

    EI

    MMxw

    0 00

    0 )()(

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    73

    3.2.3 Effect of Shear Deformation

    For simple beam with a uniform load case

    EI

    qllwM 384

    5)

    2(

    4

    =

    Shear Effect

    ▬ Shear force of load case q

    ▬ Shear force of load case q

    ▬ Deflection by shear force

    GA

    qlfVV

    l

    GA

    fVdxV

    GA

    fVdxV

    GA

    fw ss

    ls

    ls

    S

    e

    822

    122 22/

    00

    ====

    GA

    EI

    l

    f

    EIql

    GAqlf

    w

    w ss

    M

    s

    40384

    384/58/

    24

    2

    ==

    for a rectangle section of bh× with steel

    2

    2

    2

    3

    5.21240

    6.238456

    l

    h

    bhl

    bh

    w

    w

    M

    s =×

    ××=

    For small h/l, the effect of shear deformation can be neglected.

    1/2

    ql/2

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    3.3 Truss Problems

    3.3.1 Principle of Virtual Work

    From General principle

    =++=

    α+=α=⋅

    ii

    ii

    e

    l ll

    s

    iii

    S

    lEA

    FFdx

    EA

    FFdx

    GA

    VVfdx

    EI

    MM

    vuwdS

    e ee

    0 00

    22

    )(

    coscoswq

    From equilibrium equation

    0 , 0)(

    1

    )(

    1

    =+−=+− ==

    iim

    j

    ij

    iim

    j

    ij YVXH for njni ,,1L= (

    ij

    ij

    ij

    ij

    ij

    ij FVFH θ=θ= sin , cos )

    where m(i), ijH and ijV are the number of member connected to joint i, the horizontal

    component and the vertical component of the bar force of j-th member connected to joint

    i, respectively.

    0])( )[(1

    )(

    1

    )(

    1

    =+−++− = ==

    njn

    i

    iiim

    j

    ij

    iiim

    j

    ij vYVuXH

    == ==

    = ==

    +=θ+θ

    =+θ−++θ−

    njn

    i

    iiiinjn

    i

    im

    j

    ij

    ij

    iim

    j

    ij

    ij

    i

    njn

    i

    iiim

    j

    ij

    ij

    iiim

    j

    ij

    ij

    vYuXFvFu

    vYFuXF

    11

    )(

    1

    )(

    1

    1

    )(

    1

    )(

    1

    ) ()sin cos(

    0))sin( )cos((

    Each member has two ends, and each end should be connected to two different joints.

    Therefore adding all member end forces in a joint-wise fashion is equivalent to adding all

    member end forces in a member-wise fashion because the member end forces of a

    member

    iY

    iX

    iF1−

    ijF−

    i imF )(−

    iY

    iX

    pqF−

    ijF ijF−

    pqF

    i-th joint

    p-th joint

    k-th member

    1

    2

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    75

    ==

    == ==

    +=−θ+−θ

    +=θ+θ

    njn

    i

    iiiinmb

    ikkkkkkkk

    njn

    i

    iiiinjn

    i

    im

    j

    ij

    ij

    iim

    j

    ij

    ij

    i

    vYuXvvFuuF

    vYuXFvFu

    11

    1212

    11

    )(

    1

    )(

    1

    ) ())(sin )(cos(

    ) ()sin cos(

    ==

    ====

    =+

    ==∆=−θ+−θ

    nmb

    k k

    kkknjn

    i

    iiii

    nmb

    k k

    kkk

    k

    kknmb

    kk

    nmb

    kkk

    nmb

    kkkkkkkk

    EA

    lFFvYuX

    EA

    lFF

    EA

    lFFlFvvuuF

    11

    1111

    1212

    ) (

    ))(sin )((cos

    Since α represnts the angle between the applied unit load and the displacement vector, αcosiu are the displacement of the joint i in the direction of the applied unit load.

    For vertical displacement For Horizontal displacement

    iu αcosiu

    kkk vv θ− sin)(12

    kk uu − cos)(12

    )( 12 kk uu −

    )( 12 kk vv −By applying Betti-Maxwell reciprocal

    theorem,

    ==

    =+nmb

    i k

    kkknjn

    i

    iiii

    EA

    lFFvYuX

    11

    ) (

    The displacement of a joint i in a truss is

    obtained by applying a unit load at a joint i

    in an arbitrary direction.

    =

    =α=

    α=+nmb

    k k

    kkki

    iiiii

    EA

    lFF

    vYuX

    1

    cos

    cos

    u

    uX

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    3.3.2 Example

    Real System Virtual System

    Table for calculation of the deflection of a truss

    Member EA

    l F F EA

    lFF

    1 1 30 0.75 22.5

    2 2 -30 2 -0.75 2 45 2

    3 1 30 0.75 22.5

    4 1 40 0.50 20

    5 2 -10 2 0.25 2 -5 2

    6 1 -30 -0.75 22.5

    7 1 20 0 0

    8 1 40 0.5 20

    9 2 -10 2 -0.25 2 5 2

    10 1 -30 -0.25 7.5

    11 1 30 0.25 7.5

    12 1 30 0.25 7.5

    13 2 -30 2 -0.25 2 15 2

    130+60 2

    EA

    l

    EA

    L215)260130( =+=δ

    1

    2

    3

    4

    5

    6

    7

    8

    9

    10

    11

    12

    13

    20 20 20

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    3.3.3 Conservation of Energy (Truss)

    Equilibrium and Conservation of Energy

    ▬ Equilibrium Equation

    0 , 0)(

    1

    )(

    1

    =+−=+− ==

    iim

    j

    ij

    iim

    j

    ij YVXH for njni ,,1L=

    ▬ External work

    = ====

    +=⋅∆=+=njn

    i

    iim

    j

    ij

    iim

    j

    ij

    njn

    iii

    njn

    i

    iiiiext vVuHvYuXW

    1

    )(

    1

    )(

    111

    ) (21

    21

    )(21

    P

    int

    nmb

    k k

    kkknmb

    kkk

    nmb

    kkkkkkkkk

    njn

    i

    im

    j

    ij

    ij

    iim

    j

    ij

    ij

    i

    njn

    i

    iim

    j

    ij

    ij

    iim

    j

    ij

    ij

    njn

    i

    iim

    j

    ij

    iim

    j

    ijext

    WEA

    lFFlF

    vvFuuF

    FvFu

    vFuFvVuHW

    ==∆=

    −θ+−θ=

    θ+θ=

    θ+θ=+=

    ==

    =

    = ==

    = === ==

    11

    1

    1212

    1

    )(

    1

    )(

    1

    1

    )(

    1

    )(

    11

    )(

    1

    )(

    1

    2

    1

    2

    1

    ))(sin )(cos(2

    1

    )sin cos(2

    1

    )sin cos(2

    1) (

    2

    1

    intext WW =

    Conservation of Energy in each load case

    ==

    ⋅∆=njn

    iii

    nmb

    k k

    kk

    EA

    lF

    11

    2

    21

    21

    P , ==

    ⋅∆=njn

    iii

    nmb

    k k

    kk

    EA

    lF

    11

    2

    2

    1

    2

    1P

    Two load cases are applied simultaneously

    ====

    ⋅∆+⋅∆=→+⋅∆+∆=+njn

    iiiii

    nmb

    k k

    kkknjn

    iiiii

    nmb

    k k

    kkk

    EA

    lFF

    EA

    lFF

    1111

    2

    )(2

    1)()(

    2

    1)(

    2

    1PPPP

    External work for sequential loading (P first)

    ====

    ====

    ⋅∆=⋅∆→⋅∆=⋅∆+⋅∆

    ∆+⋅∆+⋅∆=+⋅∆+∆

    njn

    iii

    njn

    iii

    njn

    iii

    njn

    iiiii

    njn

    iii

    njn

    iii

    njn

    iii

    njn

    iiiii

    1111

    1111

    )(21

    21

    21

    )()(21

    PPPPP

    PPPPP

    ===

    ⋅∆=⋅∆=njn

    iii

    njn

    iii

    nmb

    k k

    kkk

    EA

    lFF

    111

    PP

    ===

    δ=δ=njn

    kkk

    njn

    kkk

    nmb

    k k

    kkk lFlFEA

    lFF

    111

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    78

    3.4 Frame Problems

    ++=∆e

    l ll

    s

    e ee

    dxEA

    FFdx

    GA

    VVfdx

    EI

    MM

    0 00

    )(

    where ∆ is the displacement in the direction of applied unit concentrate load in the virtual system.

    Moment Shear Axial

    +=δ

    2/

    00

    2 ll

    M dxdxEI

    EI

    PllPlllPll

    EIM 16

    3)

    446443(

    2 =××+××=δ

    l

    l

    HA HB=P/4

    RA=P/2 RB=P/2

    P

    -

    Pl/4

    +

    P/2

    -

    - +

    P/4

    - -

    -

    P/2 P/4

    ×

    Pl/4 l/4

    ×

    Pl/4 l/4

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    79

    +=δ

    2/

    00

    2 lls

    S dxdxGA

    f

    GA

    Plf)

    PlPl(

    GA

    fdx

    GA

    VVf ss

    V

    sS 8

    3

    2

    1

    224

    1

    4

    2 =××+××== δ

    +=

    2/

    00

    2 lls

    A dxdxEAδ

    EA

    PlPlPl

    EAdx

    EA

    AA

    V

    A 16

    9)

    4

    1

    422

    1

    2(

    2 =××+××==δ

    ))(75.0)(56.11(16

    )961(16

    223

    22

    3

    l

    h

    l

    h

    EI

    Pl

    EAl

    EI

    GAl

    EIf

    EI

    Pl sASM

    ++=

    ++=δ+δ+δ=δ

    In most cases, the deformation caused by the shear force and the axial force negligibly

    small compared to that caused by the bending moment. If this is the case, the

    displacement of a frame can be approximated by considering only the bending moment.

    =∆e

    le

    dxEI

    MM

    0

    ×

    P/4 1/4

    ×

    P/2 1/2

    ×

    P/4 1/4

    ×

    P/2 1/2

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    This page is intentionally left blank.

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    81

    Chapter 4

    Analysis of Statically Indeterminate Beams

    Rxθ L

    1

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    82

    4.1 Propped Cantilever Beam

    Equilibrium equation

    02

    0

    =−+

    =+−−

    lRl

    qlM

    qlRR

    BA

    BA

    4.1.1. The first idea

    =

    +

    Compatibility condition

    00 =δ+δ xBR

    – The end displacements of the cantilever beam for two loads cases are calculated by

    the principle of virtual work.

    EI

    qll

    qll

    EIdx

    EIdx

    EI

    MM ll

    8))(

    2(

    4

    1

    1 42

    00

    0 =−−===δ

    EI

    lll

    l

    EIdx

    EIdx

    EI

    MM ll

    x 3))((

    3

    1

    1 3

    00

    =−−===δ

    q

    RA RB

    MA

    δ0

    R

    xBR δ

    -ql2/2

    1

    -l

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    Compatibility condition and the final solution

    qlREI

    lR

    EI

    qlBB 8

    30

    38

    34

    −=→=+ (up)

    281

    ,85

    qlMqlR AA −==

    Moment Diagram

    Deflected shape

    4.1.2. The second idea

    =

    +

    Compatibility condition

    00 =θ+θ xAM

    -ql2/2

    3ql2/8

    -ql2/8 -

    2

    1289

    ql

    3l/8

    + =

    θ0

    MA

    xAM θ

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    – Rotional Angle at the fixed end

    EI

    qlqll

    EIdx

    EIdx

    EI

    MM ll

    241

    83

    1

    1 32

    00

    0 =×===θ

    EI

    ll

    EIdx

    EIdx

    EI

    MM ll

    x 311

    3

    1

    1

    00

    =××===θ

    Compatibility condition and the final solution

    23

    81

    0324

    qlMEI

    lM

    EI

    qlAA −=→=+

    Other reactions by a free body diagram

    =

    ql2/8 1

    1

    ql/2 ql/2 ql/8

    ql2/8

    ql/8

    +

    3ql/8 5ql/8

    ql2/8

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    4.2 Cantilever Beam with Spring Support

    Robin BC (The third type BC)

    )()( lkwlwEIV −=′′′−=

    Primary structure

    Compatibility Condition

    wbeam(l)=δspring → spring0 δ−=δ+δ xBR

    EIkl

    klql

    kEI

    lEI

    ql

    Rk

    R

    EI

    lR

    EI

    qlB

    BB 38

    31

    3

    838 3

    3

    3

    4

    34

    +−=

    +−=→−=+

    As qlRk B 8

    3, →∞→ , and As 0,0 →→ BRk

    Deflected Shape for 3

    100l

    EIk =

    δ0

    BR

    xBR δ

    )(lkw

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    86

    4.3 Support Settlement

    Primary structure

    Compatibility condition

    ∆=→∆=δ+δ30

    3

    l

    EIRR BxB

    Deflected Shape

    RB

    xBR δ

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    87

    4.4 Temperature Change

    Primary structure

    Curvature due to temperature change

    dxTTdxTThd )()( 0102 −α−−α=θ

    2

    212 )(

    dx

    wd

    h

    TT

    dx

    d −=−α

    baxxh

    TTw ++

    −α−= 212

    2

    )(

    For simple beam, 0)()0( == lww → lh

    TTa

    2

    )( 12 −α=

    Comaptibility condition

    00 =θ+θ xAM → 032)( 12 =+

    −αEI

    lMl

    h

    TTA → EIh

    TTM A

    )(

    2

    3 12 −α−=

    T1

    T2

    θ0

    MA

    xAM θ

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    88

    4.5 Shear Effect

    Primary Structures – Shear force diagram

    Compatibility condition

    0)()( 000 =δ+δ+δ+δ=δ+δSx

    MxB

    SMxB RR

    GA

    fqlql

    l

    GA

    fdx

    GA

    fdx

    GA

    VVf

    llS

    2)1)((

    2

    2

    00

    0 ====δ

    GA

    fll

    GA

    fdx

    GA

    fdx

    GA

    VVf

    llSx ====δ )1)(1(

    00

    2

    2

    2

    2

    3

    24

    7801

    0411

    8

    3

    31

    41

    8

    3

    3

    28

    )l/h(.

    )l/h(.ql

    GAl/fEI

    GAl/fEIql

    GA/flEI/ql

    GA/fqlEI/qlRB +

    +−=++−=

    ++−=

    – For retangular section

    ( ) 222

    3

    2260

    125

    30126

    12

    12

    5

    6)

    l

    h(.)

    l

    h(

    .

    )v(/Ehbl

    /Ebh

    GAl

    fEI =×

    +×=+

    =

    For 201=

    l

    h

    8

    300071

    002001

    002601

    8

    3

    7801

    0411

    8

    32

    2 ql.

    .

    .ql

    )l/h(.

    )l/h(.qlRB −=+

    +−=++−= (0.07 % error)

    1

    ql 1

    + +

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    89

    For 101=

    l

    h

    8

    300261

    007801

    010401

    8

    3

    7801

    0411

    8

    32

    2 ql.

    .

    .ql

    )l/h(.

    )l/h(.qlRB −=+

    +−=++−= (0.26 % error)

    For 51=

    l

    h

    8

    301011

    031201

    041601

    8

    3

    7801

    0411

    8

    32

    2 ql.

    .

    .ql

    )l/h(.

    )l/h(.qlRB −=+

    +−=++−= (1.0 % error)

    You may neglect the effect of the shear deformation in most cases !!

    4.6 2-Span Continuous Beam

    Primary structure

    Compatibility

    )( 00RxB

    RLxB

    L MM θ+θ−=θ+θ → 0)(00 =θ+θ+θ+θ RxLxBRL M

    EI EI q

    ql

    Rxθ

    Lxθ

    1

    4

    2ql 82ql

    q

    ql

    R0θ L0θ

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    90

    EI

    qlqlql

    EI

    qllqll

    EI

    dxEI

    dxEI

    dxEI

    MM

    ll

    lRL

    33322

    00

    2

    0

    00

    48

    5)

    2416(

    1)1

    831

    4)

    2

    11(

    6(

    1

    1

    1

    =+=××+×+=

    +=

    =θ+θ

    EI

    ldx

    EIdx

    EI

    dxEI

    MM

    ll

    lRx

    Lx

    3

    2

    1

    1

    00

    2

    0

    =+=

    =θ+θ

    200

    325

    qlMRx

    Lx

    RL

    B −=++−=

    θθθθ

    Deflected shape

    1

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    4.7 Fixed-Fixed End Beam

    4.7.1. Primary Structure type I

    0-th load case (X0)

    1st load case (X1)

    2nd load case (X2)

    ijδ : the deflection in the direction of the unit concentrated load in load case i caused by

    load case j. In most of engineering problem, the first subscript denotes results while the

    second subscript indicates causes.

    EI

    qlqll

    l

    EIdx

    EI

    MMl

    8)

    2()(

    4

    1 42

    0

    0110 =−×−×==δ

    EI

    qlqll

    EIdx

    EI

    MMl

    6)

    2(1

    3

    1 32

    0

    0220 −=−××==δ

    EI

    lll

    l

    EIdx

    EI

    MMl

    3)()(

    3

    1 3

    0

    1111 =−×−×==δ

    EI

    ll

    l

    EIdx

    EI

    MMl

    21)(

    2

    1 2

    0

    212112 −=×−==δ=δ

    EI

    ll

    EIdx

    EI

    MMl=××==δ 11

    1

    0

    2222

    q

    RA RB

    MA MB

    -ql2/2

    M0

    1

    -l

    M1

    1 1

    M2

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    Compatibility condition (Flexibility equation)

    =+−−

    =−+→

    =δ+δ+δ=δ+δ+δ

    026

    0238

    0

    0

    21

    23

    2

    2

    1

    34

    22212120

    21211110

    XEI

    lX

    EI

    l

    EI

    ql

    XEI

    lX

    EI

    l

    EI

    ql

    XX

    XX

    21ql

    X −= , 12

    2

    2

    qlX −=

    4.7.2. Primary Structure type II

    EI

    qlqll

    EIdx

    EI

    MMl

    2481

    3

    1 42

    0

    0110 =××==δ

    EI

    qlqll

    EIdx

    EI

    MMl

    2481

    3

    1 32

    0

    0220 =××==δ

    EI

    ll

    EIdx

    EI

    MMl

    3)1()1(

    3

    1

    0

    1111 =−×−×==δ

    EI

    ll

    EIdx

    EI

    MMl

    611

    61

    δδ

    0

    212112 =××===

    EI

    ll

    EIdx

    EI

    MMl

    311

    3

    1

    0

    2222 =××==δ

    M0

    M2 1

    1

    M1

    1

    1 8

    2ql

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    Compatibility condition (Flexibility equation)

    =++

    =++→

    =δ+δ+δ=δ+δ+δ

    03624

    06324

    0

    0

    21

    3

    21

    3

    22212120

    21211110

    XEI

    lX

    EI

    l

    EI

    ql

    XEI

    lX

    EI

    l

    EI

    ql

    XX

    XX

    12

    2

    21

    qlXX −==

    Reactions and Moment Diagrams

    Deflected Shape

    24

    2ql 12

    2ql−

    2

    ql 2

    ql

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    94

    4.8 3-Span Continuous Beam

    4.8.1. Uniform load case

    Primary structure

    )1(248

    13

    1

    81

    3

    1

    2

    1

    1

    32

    2

    2

    1

    2

    0

    0110 I

    I

    EI

    qlqll

    EI

    qll

    EIdx

    EI

    MMl+=××+××==δ

    )1(248

    13

    1

    81

    3

    1

    2

    1

    1

    32

    1

    2

    2

    2

    0

    0220 I

    I

    EI

    qlqll

    EI

    qll

    EIdx

    EI

    MMl+=××+××==δ

    )1(3

    113

    111

    3

    1

    2

    1

    121

    2

    0

    1111 I

    I

    EI

    ll

    EI

    l

    EIdx

    EI

    MMl+=××+××==δ

    220

    212112 6

    116

    1δδ

    EI

    ll

    EIdx

    EI

    MMl =××===

    )1(3

    113

    111

    3

    1

    2

    1

    112

    2

    0

    2222 I

    I

    EI

    ll

    EI

    l

    EIdx

    EI

    MMl+=××+××==δ

    EI1 EI2

    q

    EI1

    M0

    1

    M1

    1

    M2

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    Compatibility condition (Flexibility equation)

    =++++

    =++++→

    =δ+δ+δ=δ+δ+δ

    0)1(36

    )1(24

    06

    )1(3

    )1(24

    0

    0

    22

    1

    11

    22

    1

    1

    3

    22

    12

    1

    12

    1

    1

    3

    22212120

    21211110

    XI

    I

    EI

    lX

    EI

    l

    I

    I

    EI

    ql

    XEI

    lX

    I

    I

    EI

    l

    I

    I

    EI

    ql

    XX

    XX

    2

    1

    2

    1

    221

    5.11

    1

    8

    1

    I

    II

    I

    qlXX+

    +−==

    In case 21 II = , 2

    21 101

    qlXX −==

    4.8.2. Complicated Load Case

    Primary structure

    M0

    EI EI

    q

    EI

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    96

    EI

    qlqll

    EIdx

    EI

    MMl

    2481

    3

    1 32

    0

    0110 =××==δ

    EI

    ql

    EI

    ql

    EI

    qlqll

    EI

    qll

    EIdx

    EI

    MMl

    48

    5

    162441)

    2

    11(

    6

    1

    81

    3

    1 333222

    0

    0220 =+=××++××==δ

    Compatibility condition (Flexibility equation)

    =++

    =++→

    =δ+δ+δ=δ+δ+δ

    03

    2

    648

    5

    063

    2

    240

    0

    21

    3

    21

    3

    22212120

    21211110

    XEI

    lX

    EI

    l

    EI

    ql

    XEI

    lX

    EI

    l

    EI

    ql

    XX

    XX

    2

    1 401

    qlX −= , 22 406

    qlX −=

    Compatibility Condition (Flexibility Equation) in General

    =

    ∆=δ+δn

    jijiji X

    10

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    97

    Chapter 5

    Analysis of Statically Indeterminate Trusses

    1 1

    1

    21− 2

    1−

    21−

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    5.1 Various Types of Trusses

    Determinate Truss

    Externally Indeterminate Truss

    Internally Indeterminate Truss

    Mixed Indeterminate Truss

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    5.2 A Simple Truss

    5.2.1 Method - I

    +

    P

    X

    P

    P

    =

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    Primary structure

    At L1

    0

    75.06.0

    25.1

    18.0

    1

    31

    3

    ==+

    ==

    F

    FF

    F

    F

    P -1.0

    1.25

    0

    -0.75 0

    1

    0.75 0.75

    F1 F3

    0.75

    1

    ①,②,④:0.5A

    ③,⑤:A

    P ②

    L

    0.75L

    L1 L2

    U1 U2

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    At L2

    75.0 , 25.1

    18.0

    06.0

    45

    5

    54

    −===

    =+

    FF

    F

    FF

    At U1

    1 , 75.0

    08.0

    06.0

    21

    52

    51

    −=−==+=+

    FF

    FF

    FF

    1

    -1.0

    1.25

    1.25

    -0.75

    1

    -0.75

    F4 F5

    1

    F1

    F2

    F5

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    102

    Axial force table for primary structure

    Mem 0F xF A L LEA

    FF x00 =δ

    LEA

    Fxx

    2

    1 0 -0.75 0.5 0.75L 0 LEA5.0

    75.0 3

    2 -P -1.0 0.5 L EA

    PL

    5.0 L

    EA5.01

    3 1.25 P 1.25 1.0 1.25L EA

    PL325.1

    EA

    L325.1

    4 -0.75 P -0.75 0.5 0.75L EA

    PL

    5.075.0 3

    LEA5.0

    75.0 3

    5 0 1.25 1.0 1.25L 0 EA

    L325.1

    EA

    PL79.4

    EA

    L59.7

    Compatibility Condition

    00 =δ+δ xX PX 63.0−=→

    Final Solution

    Mem 0F xF xXF xXFF +0

    1 0 -0.75 0.47P 0.47P

    2 -P -1.0 0.63P -0.37P

    3 1.25 P 1.25 -0.79P 0.46P

    4 -0.75 P -0.75 0.47P -0.28P

    5 0 1.25 -0.79P -0.79P

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    103

    5.2.2 Method - II

    P

    +

    P

    X

    X

    =

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    104

    Compatibility condition

    EA

    XLX x −=δ+δ0

    Primary structure

    -1.0P P

    1.25P -0.75P

    P

    0

    0.75P 0.75P

    -0.8

    1.0 -0.6

    0.8

    -0.6

    0.8

    1

    1

    X

    = +

    F0 Fx

    EA

    XL

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    105

    Axial force table for primary structure

    Mem 0F xF A L LEA

    FF x00 =δ LEA

    FF xxx =δ

    1 0 -0.6 0.5 0.75L 0 EA

    L54.0

    2 -P -0.8 0.5 L EA

    PL6.1 EA

    L28.1

    3 1.25 P 1.0 1.0 1.25 L EA

    PL56.1 EA

    PL25.1

    `4 -0.75 P -0.6 0.5 0.75 L EA

    PL68.0

    EA

    PL54.0

    5 - - 1.0 - - -

    EA

    PL84.3 EA

    L61.3

    EA

    XLX x −=δ+δ0 → XEA

    LX

    EA

    L

    EA

    PL 25.161.384.3 −=+

    PPX 79.086.4

    84.3 −==

    PXH 63.08.02 −==

  • Dept. of Civil and Environmental Eng., SNU

    Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr

    106

    5.3 A Truss with 1 Roller Support

    Primary Structures

    1

    2

    3

    4

    5

    6

    7

    8

    9

    10

    3P

    P 2P 3P

    -P

    2P P 2P

    P 3P - 2P - 2P - 22 P

    1

    1

    1

    2

    1− 2

    1−

    2

    1−

    2

    1−

  • Dept. of Civil and Environmental Eng., SNU

    Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr

    107

    Axial force table for primary structure

    Mem F 0 xF EAL

    LEA

    FF x00 =δ LEA

    FF xxx =δ

    1 P 0 1 0 0

    2 P2− 0 2 0 0

    3 P 2

    1− 1 2P−

    2

    1

    4 P2 2

    1− 1 22P−

    2

    1

    5 P2− 1 2 P2− 2

    6 - - 2 - -

    7 P− 21− 1

    2

    P

    2

    1

    8 P3 2

    1− 1 23P−

    2

    1

    9 P2 0 1 0 0

    10 P22− 0 2 0 0

    -5.54P 3.41

    Compatibility condition

    LEA

    XXx 20 −=δ+δ

    PXXXP 15.141.141.354.5 =→−=+−

    X

  • Dept. of Civil and Environmental Eng., SNU

    Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr

    108

    Temperature change and fabrication error

    )22(0 fx LTLEA

    XX ∆+∆α+−=δ+δ

    In case of no external loads

    )2(21.0)2(82.4 ff LTL

    EAXLTX

    EA

    L ∆+∆−=→∆+∆−= αα

    5.4 Truss with Two Hinge Supports

    Primary structure and compatibility condition

    0

    2

    22212120

    121211110

    =δ+δ+δ

    −=δ+δ+δ

    XX

    LEA

    XXX

    3P 2P P

    X1

    1

    2

    3

    4

    5

    6

    7

    8

    9