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CHAPTER 3 Moment Distribution Objectives 1. เพื่อให้เข้าใจวิเคราะห์ของโครงสร้าง Statically Indeterminate ด้วยวิธี Moment Distribution (MOM-DIS) 2. เพื่อให้ทราบและเข้าใจสมการ joint equilibrium 3. เข้าใจพฤติกรรมของคานและเสา ที่มี sidesway, no sidesway Method of Analyzing Indeterminate Structure Moment Distribution Developed by Prof Hardy Cross 1930 ใช้การหาค่า joint moment ในคานต่อเนื่องและโครงสร้างแข็ง Principles and Definition เป็นวิธีการประมาณอย่างต่อเนื่อง โดยสมมุติฐานว่า joint เป็ น fixed ด้วย unlocking และ locking แต่ละ joint แล้ว moment ภายในมีการกระจายจนกว่า joint จะหมุนตัวไปในตําแหน่ง สุดท้าย

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  • CHAPTER 3Moment Distribution

    Objectives1. Statically Indeterminate Moment Distribution (MOM-DIS)2. joint equilibrium 3. sidesway, no sidesway

    Method of Analyzing Indeterminate Structure

    Moment Distribution Developed by Prof Hardy Cross 1930 joint moment

    Principles and Definition joint

    fixed unlocking locking joint moment joint

  • General Principles

    Sign Convention slope-deflection

    Fixed-End Moments (FEMS)

    Member Stiffness Factor

    Joint stiffness factorLEIK 4

    Derivation of Slope-Deflection Eqs

    Moment due to angular displacement @A, A

    MAB disp conjugate bm method

    4

    2

    AB A

    BA A

    EIMLEIML

  • General Principles Distribution Factor (DF)

    moment

    KKDF

    KK

    MMDF

    i

    iii

    General Principles Member relative stiffness factor

    modulus

    Carry-over factor

    LIKR

    AA 2 ; 4

    LEIM

    LEIM BAAB

    ABBA MM 5.0

  • Example 1

    DF

    kNmwLFEM

    kNmwLFEM

    CB

    BC

    800012

    )(

    800012

    )(

    2

    2

    Example 1 contd

    assume Joint B fixed locked [email protected] span BC apply 8000 N.m joint joint

  • Example 2

    internal moment @ support

  • Example 2 contd

    Stiffness-Factor Modification

    MOM DIS fixed support carry over moment

    factor MOM DIS

    pin support

    3AB A

    EIML

  • Stiffness-Factor Modification

    (contd) member pin support Stiffness factor

    Carry over factor = 0 pin support moment

    fixed pin support stiffness factor 3/4

    3EIKL

    Stiffness-Factor Modification

    Symmetric BM & Loading

    BMD internal MOM @ B, C M span BC

  • Stiffness-Factor Modification

    Symmetric BM & Loading (contd)

    MOM stiffness factor

    LEIK

    LEIM

    EIMLV

    LLEIMLVM

    B

    BC

    2

    22

    '

    02

    )(' - 0'

    Stiffness-Factor Modification

    Symmetric BM & Asymmetric Loading

    BM internal MOM @ B MOM @C

  • Stiffness-Factor Modification

    Symmetric BM & Asymmetric Loading (contd)

    MOM = MLEIK

    LEIM

    EIMLV

    LLEIMLL

    EIMLV

    M

    B

    B

    C

    6

    66

    '

    0622

    16

    522

    1)(' -

    0'

    Moment Distribution for Frames:No Sidesway

    BM

    Example 3 internal moment @joint, @E pinned, @D fixed

  • Example 3 contd

    Moment Distribution for Frames: Sidesway internal moment @joint moment

    distribution principle of superposition frame joint support

    R sidesway

  • Moment Distribution for Frames: SideswayMulti-story Frames Multistory frameworks may have several independent joints disp Consequently, the moment distribution analysis using the above

    techniques will involve more computation The structure shown can have 2 independent joint disp since the

    sidesway of the first story is independent of any disp of the second story

    Moment Distribution for Frames: SideswayMulti-story Frames These disp are not known initially The analysis must proceed on the basis of superposition 2 restraining forces R1 and R2 are applied The fixed end moments are determined & distributed Using the eqn of eqm, the numerical values of R1 and R2 are then determined The restraint at the floor of the first story is removed & the floor is given a disp This disp causes fixed end moment (FEMs) in the frame which can be assigned

    specific numerical values By distributing these moments & using the eqn of eqm, the associated numerical

    values of R1 and R2 can be determined In a similar manner, the floor of the second story is then given a disp With reference to the restraining forces we require equal but opposite application of

    R1 and R2 to the frame such that:

    111

    222

    "'''

    "'''

    RCRCR

    RCRCR

  • Moment Distribution for Frames: SideswayMulti-story Frames

    Simultaneous solution of these eqn yields the values of C and C These correction factors are then multiplied by the internal joint

    moments found from moment distribution The resultant moments are found by adding these corrected

    moments to those obtained for the frame

    Example 4

    moment @joint, EI

  • Example 4 contd

    moment @joint, EI

    Example 4 contd

    moment @joint, EI

  • Example 4 contd

    moment @joint, EI

    Example 4 contd