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  • 7/25/2019 CE302 Beams Ch6

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    MECHANICS OF

    MATERIALS

    CHAPTER

    6 Shearing Stresses in Beams and Thin-Walled MembersPROF. AHMED B. SHURAIM

    STRUCTURAL ENGINEERING

    CIVIL EN GINEERING DEPT.

    KING SAUD UNIVERSITY

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    6 - 2

    Introduction

    Distribution of normal and shearing

    stresses satisfies

    Transverse loading applied to a beam

    results in normal and shearing stresses intransverse sections.

    When shearing stresses are exerted on the

    vertical faces of an element, equal stressesmust be exerted on the horizontal faces

    Longitudinal shearing stresses must exist

    in any member subected to transverse

    loading.

    ( )

    ( ) !!

    !

    !!

    ====

    ===== ===

    xzxzz

    xyxyy

    xyxzxxx

    yMdAF

    dAzMVdAF

    dAzyMdAF

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    shear formula

    6 - 3

    olving for

    oting that

    hear formula

    Area#

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    &etermination of the Shearing Stre'' in a(eam

    The averageshearing stress on the horizontal

    face of the element is obtained by dividing theshearing force on the element by the area of

    the face.

    "n the upper and lo#er surfaces of the beam,

    yx$ !. %t follo#s that xy$ ! on the upper and

    lo#er edges of the transverse sections.

    %f the #idth of the beam is comparable or large

    relative to its depth, the shearing stresses at

    edges &D'andD() are significantly higher than

    at the middle &D.

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    the di'tri)ution of 'hearing 'tre''e' in a tran'ver'e 'ection of arectangular )eam i'parabolic. *a+ingy # 0 in E,. 6./ e o)tain thevalue of the ma1imum 'hearing 'tre'' in a given 'ection of a narrowrectangular beam:

    6 - 6

    Shearing Stre''e' xyin *ommon Types of +eams

    or -merican tandard &/beam)

    and #ide/flange &W/beam) beams

    web

    ave

    A

    V

    It

    VQ

    =

    =

    max

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    6 -

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    6 -

    C 302

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    6 -

    E1am4le 6.05

    A )eam i' made of three4lan+' nailed together.noing that the '4acing

    )eteen nail' i' 2% mm andthat the vertical 'hear in the)eam i'V# %00 ! determine the'hear force in each nail.

    "L0T%"12

    Determine the horizontal force per

    unit length or shear flo# qon the

    lo#er surface of the upper plan3.

    *alculate the corresponding shear

    force in each nail.

    CE 302

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    6 - 50

    E1am4le 6.05

    "L0T%"12

    Determine the horizontal force per

    unit length or shear flo# qon the

    lo#er surface of the upper plan3.

    *alculate the corresponding shear

    force in each nail for a nail spacingof (4 mm.

    ( ) ( )

    ( ) ( )

    ( ) ( )

    ( ) ( )

    56

    (

    '('

    '(

    '

    6

    m'!(!.'6

    7m!6!.!m'!!.!m!(!.!

    m!(!.!m'!!.!8(

    m'!!.!m!(!.!

    m'!'(!

    m!6!.!m'!!.!m!(!.!

    =

    +

    +

    =

    =

    ==

    I

    yAQ

    m19!5

    m'!'6.(!

    )m'!'(!)&14!!&

    56/

    6

    =

    ==

    I

    VQq

    mNqF 9!5)&m!(4.!&)m!(4.!& ==

    16.:(=F

    CE 302

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    6 - 55

    Sam4le Pro)lem 6.2

    A tim)er )eam i' to 'u44ort thethree concentrated load''hon. noing that for thegrade of tim)er u'ed

    determine the minimumre,uired de4th dof the )eam.

    "L0T%"12

    Develop shear and bending moment

    diagrams. %dentify the maximums.

    Determine the beam depth based on

    allo#able normal stress.

    Determine the beam depth based on

    allo#able shear stress.

    ;equired beam depth is equal to the

    larger of the t#o depths found.MPaMPa

    allall

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    Sam4le Pro)lem 6.2

    "L0T%"12

    Develop shear and bending momentdiagrams. %dentify the maximums.

    mkNM

    kNV

    .:4.'!

    4.'5

    max

    max

    ==

    CE 302

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    6 - 53

    Sam4le Pro)lem 6.2

    Determine the beam depth based on allo#able

    normal stress.

    Determine the beam depth based on allo#ableshear stress.

    ;equired beam depth is equal to the larger of the t#o.

    ;

    .6((mmd =

    CE 302

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    6 - 5$

    7urther &i'cu''ion of the &i'tri)ution ofStre''e' in a !arro 8ectangular (eam

    *onsider a narro# rectangular cantilever beam

    subected to loadPat its free end2

    hearing stresses are independent of the distance

    from the point of application of the load.

    1ormal strains and normal stresses are unaffected

    by the shearing stresses.

    rom aint/=enant>s principle, effects of the load

    application mode are negligible except in immediate

    vicinity of load application points.

    tress?strain deviations for distributed loads are

    negligible for typical beam sections of interest.

    =(

    (

    '(

    c

    y

    A

    Pxy

    I

    Pxyx +=

    CE 302:

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    6 - 5%

    9ongitudinal Shear on a (eam Elementof Ar)itrar Sha4e

    We have examined the distribution of

    the vertical components xyon a

    transverse section of a beam. We

    no# #ish to consider the horizontal

    components xzof the stresses.

    *onsider prismatic beam #ith anelement defined by the curved surface

    *DD>*>.

    @xcept for the differences inintegration areas, this is the same

    result obtained before #hich led to

    ( ) +==a

    dAHF CDx !

    I

    VQ

    x

    Hqx

    I

    VQH =

    ==

    CE 302:

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    6 - 56

    E1am4le 6.0$

    A ',uare )o1 )eam i'con'tructed from four 4lan+' a''hon. noing that the'4acing )eteen nail' i' 5.% in.and the )eam i' 'u)ected to avertical 'hear of magnitude V#600 l) determine the 'hearing

    force in each nail.

    "L0T%"12

    Determine the shear force per unit

    length along each edge of the upper

    plan3.

    +ased on the spacing bet#een nails,determine the shear force in each

    nail.

    CE 302:

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    6 - 5

    E1am4le 6.0$

    7or the u44er4lan+

    7or the overall )eam cro''-'ection

    "L0T%"12

    Determine the shear force per unitlength along each edge of the upper

    plan3.

    +ased on the spacing bet#een nails,

    determine the shear force in eachnail.

    ( ) ( ) ( )

    in((.5

    .in

    CE 302:

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    6 - 5

    Shearing Stre''e' in

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    6 - 5

    Shearing Stre''e' in

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    6 - 20

    Shearing Stre''e' in

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    6 - 25

    The section becomes fully plastic &y!$ !) at

    the #all #hen

    orP"BM! , yield is initiated at$and$.

    or an elastoplastic material, the half/thic3ness

    of the elastic core is found from

    Pla'tic &eformation'

    ;ecall2

    or M % P" & M! , the normal stress does

    not exceed the yield stress any#here along

    the beam.

    Caximum load #hich the beam can support is

    '! MMP" ==(

    =

    (

    (

    ''

    (

    c

    yMPx !!

    momentelasticmaximum== !!c

    IM

    "

    MP

    '=max

    CE 302:

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    Pla'tic &eformation'

    Areceding discussion #as based on

    normal stresses only

    *onsider horizontal shear force on an

    element #ithin the plastic zone,

    Therefore, the shear stress is zero in theplastic zone.

    hear load is carried by the elastic

    core,

    -sAdecreases, maxincreases and

    may exceed

    !

    ( ) ( ) !=== dAdAH !!DC

    A

    P

    byA

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    y

    A

    P!

    !

    xy

    =

    =

    =

    (

    (#here'

    (

    max

    (

    (

    CE 302:

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    6 - 23

    Sam4le Pro)lem 6.3

    noing that the vertical'hear i' %0 +i4' in a =5016rolled-'teel )eam determinethe hori?ontal 'hearing 'tre''in the to4 >ange at the 4ointa.

    "L0T%"12

    or the shaded area,

    The shear stress at a,

    ( ) ( ) ( )

    in:

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    6 - 2$

    @n'mmetric 9oading of

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    6 - 2%

    When the force A is applied at a distance e to theleft of the #eb centerline, the member bends in

    a vertical plane #ithout t#isting.

    @n'mmetric 9oading of

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    6 - 26

    E1am4le 6.0%

    Determine the location for the shear center of the

    channel section #ith b$ 5 in., ($ 6 in., and t$ !.'4 in.

    #here

    *ombining,

    I

    (Fe=

    I

    Vt(b

    d)(

    )t

    I

    Vd)

    I

    VQd)qF

    b bb

    5

    ((

    ! !!

    =

    ===

    ( )(bt(

    (btbtt(III fla*geweb

    +

    ++=+=

    6

    ('(

    '(

    '(

    '(

    ('('

    (

    ( ).in5

    .in6(

    in.5

    ( +

    =+

    =

    b

    (

    be .in6.'=e

    CE 302:

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    Prof.Ahmed

    Shuraim

    6 - 2

    E1am4le 6.06

    Determine the shear stress distribution for

    V$ (.4 3ips.

    hearing stresses in the flanges,

    hearing stress in the #eb,

    ItVQ

    tq ==

    ( )

    ( )( ) ( )( ) ( )

    ( ) ( ) ( )3si((.(

    in6in56in6in'4.!

    in53ips4.(6

    66

    6(

    ((

    ('('

    =+

    =

    +=

    +=

    ===

    (bt(Vb

    (bt(

    V(b

    )I

    V(()t

    It

    V

    It

    VQ

    $

    ( )( )( )

    ( )

    ( )

    ( ) ( )

    ( ) ( ) ( )3si!6.

    in6in66in6in'4.!(

    in6in553ips4.(

    6(

    5

    6

    5

    ('('