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Diag de Fza Cortante y Mto Flexor

Apr 06, 2018

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    Diagramas de FuerzaDiagramas de Fuerzacortante y Momento Flexorcortante y Momento Flexor

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    BeamsBeams areare usuallyusually long,long, straightstraight prismaticprismatic membersmembersdesigneddesigned toto supportsupport loadsloads appliedapplied atat variousvarious pointspoints alongalong

    thethe membermember.. InIn general,general, thethe loadsloads areare perpendicularperpendicular toto

    thethe axisaxis of of thethe beambeam andand produceproduce onlyonly shearshear andand

    bendingbending inin thethe beambeam.. TheThe loadsloads maymay bebe either either

    concentratedconcentrated atat specificspecific points,points, oror distributeddistributed alongalong

    thethe entireentire lengthlength oror aa portionportion ofof thethe beambeam.. TheThe beambeam

    itselfitself maymay bebe supportedsupported inin variousvarious waysways..

    SinceSince onlyonly staticallystatically determinatedeterminate beamsbeams areare consideredconsideredinin thisthis text,text, wewe limitlimit ourour analysisanalysis toto thatthat of of simplysimply

    supportedsupported beamsbeams,, overhangingoverhanging beamsbeams,, andand

    cantileveredcantilevered beamsbeams..

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    ToTo obtainobtain thethe shearshearVV andand bendingbending momentmomentMMatat aa givengiven pointpoint CC ofof aa beam,beam, wewe firstfirst determinedetermine

    thethe reactionsreactions atat thethe supportssupports byby consideringconsidering thethe

    entireentire beambeam asas aa freefree bodybody.. WeWe thenthen cutcut thethe

    beambeam atat CCandand useuse thethe freefree--bodybody diagramdiagram ofof oneoneofof thethe twotwo portionsportions obtainedobtained inin thisthis fashionfashion toto

    determinedetermine VVandand MM..

    C

    C

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    TheThe signsign conventionconvention forfor positivepositive shearshear forceforce andand

    bendingbending momentmoment isis asas shownshown.. OnceOnce thethe valuesvalues

    ofof shearshear andand bendingbending momentmoment areare establishedestablished atat

    severalseveral selectselect pointspoints alongalong thethe beam,beam, itit isisusuallyusually possiblepossible toto drawdraw thethe shearshear diagramdiagram andand

    bendingbending--momentmoment diagramdiagram forfor thethe entireentire beambeam..

    V

    M

    V

    M

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    Dibuje los diagramas de fuerza cortante yDibuje los diagramas de fuerza cortante y

    momento flexor para la viga mostrada.momento flexor para la viga mostrada.

    L

    w0

    3

    2

    0Lw

    2

    0Lw

    L

    w0

    2/3L

    2

    0Lw

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    x

    3

    3

    0Lw

    2

    0Lw

    2/3x

    x

    L

    xwA

    ! 0

    2

    1

    L

    xww

    0!

    M

    V

    220

    00

    2

    02

    1

    2

    0

    xL

    L

    wV

    VxL

    xwLw

    Fy

    !

    !

    !7

    3230

    00

    2

    0

    326

    03

    1)(

    2

    1)(

    23

    0

    xxLLL

    wM

    MxxL

    xwx

    LwLw

    M

    !

    !

    !7

    Proponemos un nuevo

    corte a una distancia x

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    2202

    xLL

    wV !

    3230 326

    xxLLL

    wM !

    V

    x

    M

    x

    3

    3

    0Lw

    2

    0Lw0w

    M

    V

    2

    0Lw

    3

    3

    0Lw

    L

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    EjemploEjemplo

    Trace los diagramas de fuerzas cortantesTrace los diagramas de fuerzas cortantes

    y de momentosy de momentos flexionantesflexionantes de las vigas yde las vigas y

    la carga que se muestra.la carga que se muestra.

    60 kN

    2m1m2m

    A B

    C D

    25 kN/m60 kN

    2m1m2m

    A B

    C D

    25 kN/m

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    60 kN

    2m1m2m

    A B

    C D

    60 kN

    2m1m2m

    A B

    C D

    50 kN

    Primero determinar las fuerzas exteriores don el DCL

    MA=0 B(5m)-(60kN)(2m)-(50kN)(4m)=0

    B=64kN

    Fy=0 Ay+64kN-60kN-50kN=0

    Ay=46kN

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    Proponemos un corte:

    x M

    V46kN

    J

    De A a C

    Fy=0 46kN -V=0V= 46kN

    MJ=0 M-(46kN)(x)=0

    M=46 x (kN m)

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    M

    De C a D

    Fy=0 46kN 60kN -V=0V= -14kN

    MJ=0 M-(46kN)(x)+60kN (x-2)=0

    M=(120-14x) [kN m]

    xV46kN

    J

    60kN

    2m

    Cuando x=2m MC = 92kN m

    Cuando x=3m MD = 78 kN m

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    u

    V

    64kN

    (25)(u)u/2

    De D a B

    M

    De D a B

    Fy=0 V+64-25u=0V= (25u-64)kN

    MJ=0 64u-(25u)(u/2)-M=0

    M=(64u-12.5u2) [kN m]

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    Para graficar:

    AB

    C D

    -14

    -64

    46

    92

    V

    M 78.0

    x

    x