Hoja1
Verificacin rpida de una seccin para viga de polipasto
(Secciones W, HEA, HEB, I y similares), y polipasto de 4
ruedas.Entrar datos en color
azul.http://sites.google.com/site/antolintinez
PROYECTO: La Esmeralda.ELEMENTO: Viga carrilera polipasto en
estacin de bombeo.INGENIERO: AMAFECHA: May-12OBSERVACIONES:Usar
perfil W 12x22 Acero A-36.
Verificacin de la seccin
PerfilW12x14Acero A36
Datos
rea
Antolin: Antolin:rea de la seccin transversal del
perfil.26.84cm2d
Antolin: Antolin:d, bf, tf, tw de acuerdo al dibujo de la
derecha.30.25cmbf10.08cm (dato no
utilizado)tf0.57cmtw0.51cmIxx3,687.81cm4Iyy98.23cm4 (dato no
utilizado)fy2,533kgf/cm2Sxx243.81cm3PP viga21.07kgf/mCarga
izable1,000kgfPP polipasto40kgfImpacto25%FS1.15Factor de seguridad
adicional.Carga diseo1,495kgfCarga/canal (P)747.5kgf
Verificacin de la viga, caso de voladizoRecomendacin de Covenin
1618-82 sobre las flechas (condicin de servicio).L voladizo2.00mM
mx.3,032.14kgfxmTensin flexin1,243.64kgf/cm2 = 0.49fyTensin
admisible0.60fyOkTensin corte100.02kgf/cm2 = 0.04fyTensin
admisible0.40fyOkDeformacin voladizoDeformacin mx.0.52cm = L /
384Flecha alta
Verificacin de la viga, caso de viga de 1 tramo simplemente
apoyada
L tramo3.50mM mx.1,340.39kgfxmTensin flexin549.76kgf/cm2 =
0.22fyTensin admisible0.60fyOkTensin corte51.04kgf/cm2 =
0.02fyTensin admisible0.40fyOkDeformacin tramo isostticoDeformacin
mx.0.18cm = L / 1,969Ok
Verificacin de la tensin en el alma
P/2373.75kgfk15.13cmTensin27.79kgf/cm2 = 0.01fyft = P/2A =
P/(2tw)(3.5k) = P/(7k)tw
Antolin: Antolin:Vase fundamento terico ms abajo.Ok
Verificacin de la tensin en el ala
Tensin3,432.97kgf/cm2 = 1.36fyfb = M/S = 3eP/(4e)(tf)2 =
0.75P/(tf)2No verifica
Ref.:http://www.modernsteel.com/steelinterchange_details.php?id=499
Esta hoja de clculo est intencionada para verificaciones rpidas,
tiles para el predimensionamiento de una viga carrilera que
soportar un polipasto tpico, como el mostrado en la figura de
arriba, y no constituye un anlisis sofisticado que, segn el caso,
se podra requerir. Queda a juicio del ingeniero el utilizar mtodos
ms precisos, o no. El autor no se responsabiliza por daos derivados
del empleo de esta hoja de clculo. Puerto Ordaz, mayo 12.
Steel Interchange
Bottom Flange Bending Capacity
How do you calculate the lower flange loading capacity of a
steel beam to be used to support an underhung crane? Are there any
published ASD or LRFD design procedures?James F. Jendusa, P.E.MSI
General CorporationOconomowoc, WI
Answer
The bottom part of the crane beam must be checked for:1. Tension
in the web.2. Bending of the bottom flange.Most underslung cranes
will have each end supported by 2 pairs of wheels. Each individual
wheel load will include a portion of the lifted load (in its most
critical position), the dead loads, and impact. Impact is usually
about 25% of the lifted load but will depend on the speed and
braking ability of the hoist. Allowable stresses must be reduced
due to the cyclical nature of the applied load.The wheels must be
purchased to suit the profile of the supporting crane beam, either
an S-shape or a W-shape. The web tension at each pair of wheels is
checked at the intersection of the web and fillet (at the "k"
distance).Referring to Figure 1 below, the length of resistance is
seen to be3.5k. The 30 years. Assuming 4 wheels (2 pair) at each
end of the crane, each wheel will supportP/4delivered to the
supporting crane beam. In Figure 1, two wheels cause the web
tension, so the load isP/2.The tensile stress in the web
becomes:ft= P/2A = P/(2tw)(3.5k) = P/(7k)tw
Flange bending depends on the location of the wheels with
respect to the beam web. Referring to Figure 2, this is dimensione.
As stated previously, each wheel load isP/4.
The longitudinal length of flange participating in the bending
resistance can be taken as 2e per yield-line analysis. See Figure
3.
The section modulus at the plane of bending is(bd2)/6which
translates toe(tf)2/3. From Figure 2 the bending moment iseP/4. The
bending stress is:fb= M/S = 3eP/(4e)(tf)2= 0.75P/(tf)2Local
loadings such as this often result in biaxial and triaxial
stresses. These stress combinations are quite common, and designers
must design accordingly. For more information on crane loading,
refer to my paper in theEngineering Journal, 4th quarter 1982,
called "Tips for Avoiding Crane Runway Problems."David T. Ricker,
P.E.Javelina ExplorationsPayson, AZ
Posted on December 1, 1999
Flange bending depends on the location of the wheels with
respect to the beam web. Referring to Figure 2, this is dimensione.
As stated previously, each wheel load isP/4.
The longitudinal length of flange participating in the bending
resistance can be taken as 2e per yield-line analysis. See Figure
3.
The section modulus at the plane of bending is(bd2)/6which
translates toe(tf)2/3. From Figure 2 the bending moment iseP/4. The
bending stress is:fb= M/S = 3eP/(4e)(tf)2= 0.75P/(tf)2Local
loadings such as this often result in biaxial and triaxial
stresses. These stress combinations are quite common, and designers
must design accordingly. For more information on crane loading,
refer to my paper in theEngineering Journal, 4th quarter 1982,
called "Tips for Avoiding Crane Runway Problems."David T. Ricker,
P.E.Javelina ExplorationsPayson, AZ
Posted on December 1, 1999
