Calcul Buloane de Ancoraj (4 Ipoteze) Rev.1 [PDF]

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CALCUL BULOANE DE ANCORAJ kN  1000 newton

combin  3

N  1 newton

i  1  combin

REACTIUNI PE PLACA DE BAZÃ 1.Combinatia 35 (nod 40)

3.Combinatia 40 (nod 23)

2.Combinatia 33 (nod 6) Ny  i

M x 

M z 

i

430kN 252kN 1kN

i

11  kN m 390kN m 1kN m

0 kN m 1kN m 1kN m

DIMENSIUNI PLACA DE BAZÃ

Ny

T

T f

Mx,z

l(b)

l  850 mm

b  520 mm

l0  750 mm

b 0  420 mm

N

Rb  9

2

mm Efortul în bulon dupã directia x M ax  M x  Ny   l0  i i i

  2 l

 150.75  M ax   308.1   kN m    0.675 

2

r0x  i

l 0  b  Rb 2 M ax

i

r0x  i

r0x i

 2.955  r0x   2.067     44.159 

αx  1.5  i

2.25  3    2   r0xi     

Nboltx  i

0.119     0.256 αx      4  5.129  10 

αx  l0  b  Rb i 2

 Ny i

 639  Nboltx   197   kN    0 

Efortul în bulon dupã directia z

0 M az   1   kN m   1

M az  M z i i

αz  i

0 if M az = 0 i

 1.5  2.25     

Nboltz  i

 0  αz   0.001     0.001 

 otherwise  b 0 2 l  R   b    2 Maz   i   3

 0  Nboltz   2.382   kN    2.382 

αz  b 0  l Rb i 2

Numar buloane - nb n bx  3

Nboltx  i

n bz  2 Nboltx i n bx

 213.107  Nboltx   65.703   kN    0.033 

daN  10N

Nboltz  i

A'b  i

σa

Nb  Nboltx  Nboltz i i i

n bz

 0  Nboltz   1.191   kN    1.191  σa  210

Material bulon grupa 6.6

Nb i

Nboltz i

 21310.736  Nb   6689.422   daN    115.778  N 2

mm

Ab  i

A'b if A'b  0 i i 0 otherwise

 10.148  2 Ab   3.185   cm    0.055 

d  i

4 Ab π

i

 35.946  d   20.139   mm    2.649 

=> 6xM42