Solución de Resis - Wil

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  • 7/24/2019 Solucin de Resis - Wil

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    SOLUCIN:

    PROPERTIES OF THE STEEL SHAFT (s)

    Gs=80GPad 1=70mm

    Allowable shear stress:s=110MPa .

    IPS=

    32d

    1

    4=2.3572106 m4 .

    GsIPS=188.574103N .m

    2

    PROPERTIES OF THE BRASS SLEEVE (b)

    Gb=40GPad2=90mm d1=70mm

    Allowable shear stress:b=70MPa .

    IPB=

    32( d2

    4d1

    4 )=4.0841106m4

    GsIPB=163.363103N . m

    2

    TORQUES IN THE COPOSITE BAR AB

    TS= Tor!"e #$ the Steel sha%t AB

    Tb= Tor!"e #$ the brass slee&e AB

    Fro' E! (**a): TS=T( G SIps

    GSIPS+GbIpb)

    TS=T(0.53582) (E! +)

    Tb=TTS=T(0.46418) (E! ,)

    AN-LE OF T.IST OF THE COPOSITE BAR AB

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    AB=TS(

    L

    2)

    GSIPS=

    TS(L

    2)

    G SIPS=

    Tb (L

    2)

    GbIPb=(5.6828106 ) T (E! )

    U$#/a/es:T=N .m=rad

    AN-LE OF T.IST OF PART BC OF THE STEEL SHAFT

    BC=T(

    L

    2)

    GSIPS=(10.6059106 ) T (E! *)

    AN-LE OF T.IST OF THE ENTIRE SHAFT ABC

    =AB+BC (E!s 0 *)

    =(16.2887106 ) T

    U$#/a/es: 1 2 ra/ T 2 N'

    a) Allowable tor!"eT

    1 base/ "3o$ a$4le o% tw#st

    allow=8.0=0.13963rad

    =(16.2887106 ) T=0.13963 rad

    T1=8.57 kN .m

    b) Allowable tor!"eT

    2 base/ "3o$ shear stress #$ the brass

    slee&e

    b=Tb(d2/2)

    Ipb=70MPa

    Tb=0.46418T(deeq.2)

    70MPa=(0.46418T)(0.045m)4.0841106 m4

    Sol"5#6$ 3ara T (E!"al toT

    2 ):T

    2=13.69 k N . m

    5) Allowable tor!"eT

    3 base/ "3o$ shear stress #$ the steel sha%t

    BC

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    S=T(d2/2)IPS

    =110 MPa

    110MPa= T(0.035m)

    2.3572106m4

    Sol"5#6$ 3ara T (E!"al toT

    3 ):

    T3=7.41 kN .m

    /) a7#'"$ allowable tor!"eShear stress #$ steel 4o&er$s

    Tmx=7.41kN .m