Files-2-Lectures LEC 06 Shear Stresses for Beams in Bending

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    Chapter 4

    Load and Stress AnalysisLoad and Stress Analysis

    A. Bazoune

    Shear Stresses for BeamsShear Stresses for Beams in Bendingin Bendingecec..

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    4.84.8Shear Stresses forShear Stresses forBeams in BendingBeams in Bending

    Most beams have both shear forces and bending momentpresent.

    It is only occasionally that we encounter beams subjected, .

    The flexural formula was developed on the assumption ofpure bending in order to eliminate complicating effects ofshear force.

    For engineering purposes, the flexure formula is valid nomatter whether the shear force is present or not.

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    Fig. 4.20 showsa beam ofconstant crosssection

    subjected to ashear force Vand a bendingmoment M

    Figure 4-20 Beam section isolation.Note: Only forces in x-direction are

    shown on the dx element.

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    Associate the hollowAssociate the hollowvector with yourvector with yourright hand.right hand.

    If you place theIf you place thethumb of our ri htthumb of our ri ht

    Figure 4-20 Beam section isolation.Note: Only forces in x-direction are

    shown on the dx element.

    hand in thehand in the negativenegativezz-- directiondirection, then, thenyour fingers, whenyour fingers, whenbent, will indicatebent, will indicatethethe direction ofdirection of

    moment Mmoment M..

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    Shear force andShear force andbending moment arebending moment arerelated by:related by:

    dMV =

    Figure 4-20 Beam section isolation.Note: Only forces in x-direction are

    shown on the dx element.

    dx

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    At some pointalong the beam,we cut transverse

    section dx at a

    distancey1 abovethe neutral axis.

    We remove thissection to studythe forces that acton it.

    Figure 4-20 Beam section isolation.Note: Only forces in x-direction are

    shown on the dx element.

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    Because shear

    force is present,the B.M. ischanging as wemove along the x-

    .

    We can designatethe B.M. as M onthe Near Side of

    the section and as(M + dM) on theFar Side.

    Figure 4-20 Beam section isolation.Note: Only forces in x-direction are

    shown on the dx element.

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    The moment M

    produces a normalstress , and themoment (M + dM),a normal stress

    + d).

    The compressive force on the far side being

    greater than on the near side.

    The Resultant of these two would cause thesection to tend to slide in the negative x-direction.

    These normalstresses producenormal forces onthe vertical faces of

    the element.

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    This resultant mustbe balanced by a

    shear force actingin the positive x-direction on thebottom of the

    For the near face

    (b)

    where the limits indicate that we integrate from y = y1 to y = c.

    sec on.

    This shear forceresults in a shearstress.

    1

    c

    Ny

    F dA=

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    For ,

    Eq. (b) becomes

    (b) (c)

    y I=

    c

    N

    MF y dA=

    For the Far side

    1y

    ( )1 1

    c c

    Fy y

    M dMF d dA y dA

    I

    += + = (d)

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    The force on thebottom face is theshear stress timesthe area of thebottom face

    (e)BF bdx=

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    In this equation, the integralis the first moment of thearea of the isolated verticalface about the neutral axis.

    1

    c

    yy dA Q=

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    The moment is usually designated Q

    ==c

    yAydAyQ

    1

    ''(4-30)

    V Q

    I b = (4-31)

    where, for the isolated area fromy1 to c, is the distance from they-axis to the area centroid andA is the area.

    Finally, we write

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    Shear Stresses in StandardShear Stresses in Standard--SectionSectionBeamsBeams

    The shear stress distribution in a beam depends on how Q/b

    varies as a function of y1.

    g. - s ows a por on o a eam w a rec angu ar cross-

    section.

    It is subjected to shear force V and a bending Moment M.

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    Figure 4-22Shear Stress in a rectangular beam

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    A normal stress is developed on a cross-section such as A-A due toM.

    Section A-A is in compression above the N.A. and in tension below.

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    Consider an element of area dA at a distancey above the N.A. such that

    dA bdy=

    Eq. (4-30) becomes

    ( )1 1

    2

    2 2

    1

    12 2

    cc c

    y yy

    by bQ y dA b y dy c y

    = = = =

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    Substitute this value into Eq.(4-31)

    From Table A-18 and for a rectangular cross-section, we haveSubstitute and gives

    ( )2 212

    Vc y

    I = (4-32)

    2

    3

    AcI=

    3

    12

    bh

    I=A bh= 2h c=

    (b)

    s ng s va ue or n q. - g ves

    The maximum shear stress exists when y=0, which is at the bendingneutral axis. Thus

    2

    1

    2

    31

    2

    V y

    c

    =

    (4-33)

    (4-34)max

    3

    2

    V =

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    As we move away from the neutral axis, the shear stress

    decreases parabolically until it is zero at the outer surfacewhere y = c .

    Notice that the shears ress s max mum a

    the bending neutralaxis, where thenormal stress due tobending is zero andthat the shear stress

    is zero at the outersurfaces, where thebending stress is amaximum.

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    Particular Case:Particular Case:Shear Stresses in Standard Section BeamsShear Stresses in Standard Section Beams

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    Example 4Example 4--7 Textbook7 Textbook

    AA beambeam 1212 inin longlong isis toto supportsupport aa loadload ofof 488488 lblb..ff actingacting 33 inin fromfrom thethe leftleftsupportsupport asas shownshown inin FigFig-- 44--2121 aa.. BasingBasing thethe designdesign onlyonly onon bendingbending stress,stress, aa

    designerdesigner hashas selectedselected aa 33--inin columncolumn channelchannel withwith thethe crosscross--sectionalsectionaldimensionsdimensions asas shownshown.. IfIf thethe directdirect shearshear isis neglected,neglected, thethe stressstress beambeam maymay bebeactuallyactually higherhigher thanthan thethe designerdesigner thinksthinks.. DETERMINEDETERMINE thethe principalprincipal stressesstressesconsideringconsidering bendingbending andand directdirect shearshear andand comparecompare thenthen withwith thatthat consideringconsideringbendingbending onlyonly..

    Figure 4Figure 4--2121..

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    Example 4Example 4--7 Contd7 Contd

    TheThe loading,loading, shearshear forceforce andandbendingbending momentmoment diagramdiagram areare

    shownshown inin FigureFigure44--2121(b)(b)..

    IfIf thethe directdirect shearshear forceforce isisincludedincluded inin thethe analysis,analysis, thethe

    andand bottombottom ofof thethe beambeam willwillbebe thethe samesame asas if if onlyonlybendingbending werewere consideredconsidered..TheThe maximummaximum stressesstresses areare::

    Figure 4Figure 4--21 (b)21 (b)..

    ( )1098 1.5992

    1.66psi

    Mc

    I

    =

    = =

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    Example 4Example 4--7 Contd7 Contd

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    Example 4Example 4--7 Contd7 Contd

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    yy QQ ss tt ss11,s,s

    22

    inin inin33 psipsi psipsi psipsi

    00 0.6530.653 00 847847 847,847, --847847

    .. .. ,, --

    0.50.5 0.6310.631 331331 818818 670,670, --10001000

    0.750.75 0.6050.605 496496 785785 575,575, --10711071

    11 0.5680.568 661661 737737 477,477, --11381138

    1.2271.227 0.5250.525 812812 681681 387,387, --12001200

    1.51.5 00 992992 00 0,0, --992992