Influence of Axial & Shear

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    INFLUENCE OF AXIAL FORCE ON THE VALUE

    OF THE PLASTIC MOMENT

    If an axial load is applied to the CrossSection of a short

    column, then the load will give rise to a uniform compressive

    stress over the section. Addition of small bending moment will

    then produce a linear variation of elastic stress across the

    section. Further increase of B.M. with axial load remainingconstant will cause yield on one face followed by yield on the

    other face and eventually yield on the whole Cross-Section.

    During this process, the zero-stress axis shifts progressively

    towards the final position in the fully plastic state.

    To explain this, take a rectangular section on which an

    axial load P is acting at the centre line of cross section, in

    addition to a bending moment Mp sufficient to cause the whole

    section to be fully plastic. Due to force P across the Cross-

    Section, the zero stress axes must shift from the centre line by

    plastic

    Fully

    tension

    Yield in

    in CompYield

    B.M.Medium

    B.M.smallNo B.M.

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    an amount d. It will be seen that the fully plastic stress

    distribution may be regarded as being composed of two parts,

    i.e. on the original fully plastic distribution in the absence of

    axial load may be superimposed a fictitious distribution

    involving stress of magnitude 2 fL. This distribution must be

    equivalent to a total force P and it will reduce the value of the

    full plastic moment by an amount P

    d2

    1

    P = 2bdfL = o

    Mp = Mpo P

    d2

    1

    = Mpo 2

    12d. Po

    Mpofbdbdfd2

    1dPo2

    1L

    2

    L

    2 ===

    Mp = Mpo (1 2)

    2

    C

    2 Bddtc

    Bd

    2t1

    tt

    +=

    T

    CC

    T

    Bd

    b

    2d

    c c

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    o is the squash load of the Cross-Section in the absence

    of bending moment and Mpo is the full plastic moment (bd2fL)

    in the absence of axial load.

    So2

    2

    Po

    P11

    Mpo

    M

    ==

    => 1Po

    P

    Mpo

    M2

    =

    +

    This gives the doubly symmetric curve. This is true for

    tensile load also. In a system of Cartesian coordinates with

    Po

    Pas abscissa and (M/Mpo) as ordinate the complete plot is

    as shown.

    For the

    general case of a cross section having at least one axis of

    symmetry, it may be supposed that the axial load P causes the

    zero-stress axis to shift so that an area a is transferred from

    tension to compression (or vice-versa).

    If the centre of the transferred area a is at distance

    y

    from the original zero-stress axis.

    P = 2fLa

    3

    oP/P

    po/ MpM

    +1

    +1

    -1

    -1

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    Mp = Mpo - P

    y

    i.e. Zp

    = Zpo

    - 2a

    y

    I Section subjected to bending with respect to its centroidal

    axis of maximum moment of inertia (strong axis)

    A

    da2

    Po

    P=

    If the neutral layer is in the web i.e.,

    N/Np < AW/A, AW = area of web

    4

    =+

    t ctc

    2tc

    y

    ha

    h

    y

    hX X

    b

    t

    M

    N

    d

    a

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    A = total area

    Zpx = Zp along X-X axis

    So that Zd = ad2

    pxZ

    Zd1

    Mp

    M= 1

    2

    Po

    P

    = 1

    2

    L

    L

    fA

    adf2

    = 1 -2

    22

    A

    da4

    = 1 -Zp

    Zd1

    ah

    ad42

    2

    =

    aZp4

    A

    yA

    yda41

    Zp

    ad1

    Mp

    M 2

    22

    2222

    ==

    = 1 -aZp4

    A

    Po

    P 22

    forA

    Av

    Po

    P 0.15 Po, the linear interaction as given

    may be used.

    15.0Po

    Pfor)

    Po

    P1(18.1

    Mp

    M=

    For bending about the weak axis, M approximately equals

    Mp as long as P < 0.4 Po and

    40.0Po

    P

    for)Po

    P

    1(19.1Mp

    M2

    =

    Influence of the shear force on the value of the plastic

    moment

    The effect of shear force on a beam of general cross-

    section is much more complex than that of the axial load. With

    the latter, the resulting stresses could be superimposed

    directly on the bending stresses, since they were all

    6

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    longitudinal. Shear combined with bending gives rise to a two-

    dimensional stress system. Here, the special case of I

    section with approximate method will be discussed. It is

    commonly assumed in elastic design that the shear stresses

    are uniformly distributed over the web of an I-section and the

    flanges not contributing at all to the carrying of the shear

    force.

    If this same assumptions are made in the plastic analysis

    of the problem then an empirical solution may be obtained

    which gives good agreement with experimental results.

    Full plastic moment of an I section is

    Mp = BT (D T) fL + L2

    ftT2

    D

    = Mf (flange) = BT (D T) fL

    WW (Web) = L2

    tT2

    Df

    If the shearing force F acts on the web causing an

    uniform shear stress in the web, so

    F = (D 2T) t.

    7

    D

    TB

    t

    fc

    fL

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    If the web remains fully plastic then the longitudinal

    stress in the web available to resist the B.M. will be reduced

    below the value fL. There are several yield criterion for

    combined stress system, but the Von-Mises yield condition fits

    best in this case.

    L

    22 fZ3f =+

    The stress distribution over the cross section under

    combined shear and bending sufficient to cause full plasticityis as shown.

    The contribution from the flanges remains unaltered, but

    the contribution from the web is reduced.

    Experiments indicate that the influence of the shear force

    is negligible for a beam with rectangular cross section and a

    ratio of span l to depth h of at least 10. The influence of the

    shear force V is also negligible for an I beam with a similar

    ratio as long as V does not exceed fL Aw/ 3 which

    corresponds to full plastification of the web.

    Example

    8

    Z

    shear stressBending stress

    f

    Lf

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    Suppose that the ISHB 300 is subjected to a shearing

    force of 300 kN. fL = 250 N/mm2. Find the moment carrying

    capacity.

    =

    2mm/N5.114

    4.9)26.10300(

    000,300=

    2

    L

    L

    )f/(31f

    f =

    =2

    250

    5.11431

    = 0.608

    Zpo = Zf + Zw Zf + ZW

    Zp = Zf + Zw.

    Lf

    fZf + Zw

    Lf

    f

    = Zpo Zw

    Lf

    f1

    So the original plastic modules Zpo reduces by

    Zw (1 f/fL) = Zw (0.392)

    Zw = 4.9x)4.139(tT2

    D 2w

    2

    =

    Effect of

    axial load on

    I section.

    9

    9.4

    250

    10.6

    300

    D

    TB

    t

    fL

    BD

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    Mp = Mpo 2D2 t.fL

    P = 2tDfL

    If the mean axial stress is denoted by p so that P = p.A

    where A is the total cross-sectional area and if n = p/fL then

    equation

    Zp = Zpo 2D2t

    2tD2 = 2L

    2

    L

    222

    f

    f.

    t4

    Dt4

    =t4

    1.

    f

    Ap

    tf4

    P2

    L

    22

    2

    L

    2

    =

    = 22

    n.t4

    A

    So Zp = Zpo 22

    n.t4

    A

    Combined effects of shear and axial load. This can be

    found by considering the figure as given.

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    Zw = Zwo (1 2) .L

    f

    f

    P = Po

    Lf

    f

    Where Po is the squash load of the web alone. From yield

    equation

    Zw = Zwo)f/(31

    )Po/P()f/(31

    2

    L

    22

    L

    In this equation, the mean web shearing stress is

    computed, as before, assuming that none of the shear force is

    carried by the flanges of the I-section.

    Biaxial Bending

    11

    Z

    shear stressBending stress

    f

    Lf

    DBD

    consined with

    axial stress

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    When a section is subjected to bending about both axes

    such that full plasticity is produced, the neutral axis changes

    its inclination. However, if no overall axial thrust or tension is

    present, the neutral axis must continue to divide the cross-

    section into two equal are as

    A section of dimension b x d is subjected simultaneously

    to bending moments Mx about XX and My about yy, causing

    tensile and compressive plastic zones as shown on wither side

    of neutral axis NN. For purposes of calculation, this stress

    distribution may be considered as the sum of the distribution

    plastic bending about XX only and zones OAC with stress + 2 y

    and OBD with -2 y. If Y is used to define the inclination of the

    neutral axis.

    Mx = 3/Y2y2

    b

    2

    12.

    4

    bdyy

    2

    My = 22

    b

    3

    22y

    2

    b

    2

    1y

    12

    stress

    Zero

    A

    C

    y

    d/2

    d/2

    b/2b/2

    +=

    N

    N

    XX

    Y

    Y

    y-2- y

    y+2y+

    B

    D

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    These expressions are valid provided Y< d/2. The

    dimension Y may be eliminated by using expressions for the

    plastic module of the section about XX and YY

    ==

    4

    dbZpy&

    4

    bdZpx

    22

    1Zpy

    My

    4

    3

    Zpx

    Mx2

    yy

    =

    +

    0 < 1Zpx

    Mx

    3

    2

    Zpy

    My

    yy

    If the N.A. is assumed to intersect the edges of dimension

    b, it is found that

    1Zpy

    My

    Zpx

    Mx

    4

    3

    y

    2

    y

    =

    +

    1Zpy

    My

    3

    2

    Zpx

    Mx0

    yy

    Empirically, it is found that the relationship.

    1Zpy

    My

    Zpx

    Mx71.1

    y

    71.1

    y

    =

    +

    and gives a very close approximation to the accurate

    result.

    Hollow, square and rectangular sections gives the

    identical with those for all solid rectangular sections.

    Circular sections (both solid and hollow) give the

    relationship.

    1Zpy

    My

    Zpx

    Mx2

    y

    2

    y

    =

    +

    *****

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