calculul la forta taiatoare pereti mulati + dimensionare spraituri

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  • Verificarea la forta taiatoareAsl As 34.56 cm

    2==

    Ved 419.60kN=

    VRd.c CRd.c k 100l fck( )13

    bw d=

    c 1.5=

    CRd.c0.18c

    0.12==

    k 1 200mmd

    + 1.626== < 2

    => k 1.626=

    lAsl

    bw d0.007== < 0.02

    => l 0.0068=

    VRd.c CRd.c k 100 l fck( )13

    bw d 245.42 kN== > 0.035 k

    32

    fck

    12

    bw d 185.05 kN=

    => VRd.c 245.42 kN=

    VRd.c 245.42 kN= < Ved 419.6 kN=

    => se impune armatur la = 45 i = 90

    z 0.9 d 459 mm==

    AswS

    =

    Vedfywd z

    3.047 mm2

    mm=

    nec

    n 6= 14mm= swpi

    2

    41.539 cm2==

    Asw n sw 9.24 cm2

    == => SmaxAswVed

    fywd z

    303.1 mm==

    S 300mm=

  • Verificare armturiilor supuse la for tietoare:

    AswS

    3.079 mm2

    mm= >

    AswS

    2.878 mm2

    mm=

    eff nec

    VRd.sAsw

    S

    3.079 mm2

    mm==

    eff

    VRd.max 0.5 bw 1fcd

    fywd=

    1 0.6 1fck

    250 N

    mm2

    0.54==

    VRd.max 0.5 bw 1fcd

    fywd 15 mm

    2

    mm== > VRd.s 3.079

    mm2

    mm=

    => etrieri 14/30

  • Dimensionarea praiului:NEd 1600kN=

    L 7.10m=M0 1.0=M1 1.0=

    - Oel folosit S275 => fy 275N

    mm2

    =

    Predimensionare:

    NEd Nc.Rd=

    => AnecNEd M0

    fy58.18 cm2==

    Nc.RdAnec fy

    M0=

    => CHS 324x8: d 324mm=t 8mm=

    h 324 mm=b 324 mm=iy 103.5mm=

    Aextpi d2

    4824.48 cm2==

    Aintpi d 2 t( )2

    4745.06 cm2==

    A Aext Aint 79.419 cm2

    ==

    Clasa seciunei:

    235 N

    mm2

    fy0.924==

    dt

    40.5= < 50 2 42.727= => seciune de clasa 1

  • Verificarea de rezisten:

    NEdNc.Rd

    1.0

    Nc.RdA fy

    M02184 kN==

    NEd 1600 kN=

    NEdNc.Rd

    0.733= Seciunea este aleas corect din punct de vedere al rezistenei

    Verificarea la flambaj:NEd

    Nb.Rd1

    Lcr.yy 1 L cm==

    E 210000 N

    mm2

    =

    1 piEfy

    86.815==

    baratLcr.yy

    iy

    11

    0.79==

    seciune finisat la cald

    Din tabelul 6.1 i 6.2, SR EN 1993-1-1:2006 => 0.21=

    0.5 1 barat 0.2( )+ barat2+ 0.874==

    1

    2

    barat2

    +

    0.801==

    Nb.Rd A fy

    M11750 kN== NEd 1600 kN=

    NEdNb.Rd

    0.91= Seciunea este aleas corect