wood pillar

[odisseo 18]

hi, an information if I can: how much can a wood pillar 100x100 , x 3 meters high in berries loaded with tip? I'm supposed to do a provisional droop.

 

[Betoniera]

hi, an information if I can: how much can a wood pillar 100x100 , x 3 meters high in berries loaded with tip? I'm supposed to do a provisional droop.

the maximum load of a wooden dot cm 10x10 high 3 m is about 1800 kg
hypothesis:
- constraints must be fixed
- the load must be centered
- wood must not have imperfections
if there are geometric imperfections these must be put into account (acting on mdx and mdy moments different from 0)
I'll give you the verification according to dm 2008.
Hi.

slu design load = 1800*1.5 = 2700 kg

verification of wood rod to itxion slu according to ntc 2008 #4.4.8.2.2
wood beam verification
solid wood c16 (uni en 338-2004)
bending fm,k = kg/cm2 160
parallel traction fiber ft,0,k = kg/cm2 100
perpendicular traction fiber ft,90,k = kg/cm2 5
parallel compression fiber fc,0,k = kg/cm2 170
perpendicular compression fiber fc,90,k = kg/cm2 22
cut fv,k = kg/cm2 18
medium fiber parallel elastic module e0,mean = kg/cm2 80000
parallel elastic module fiber e0,05 = kg/cm2 54000
elastic module perpend. medium fiber e90,mean = kg/cm2 2700
cutting module (medium) gmean = kg/cm2 5000
volume mass rk = kg/m3 310
medium volume mass rmean = kg/m3 370
Material safety coefficient Gammam = 1.5
service class service class 1
short load duration
axial action slu nd = cm 2700
moment y slu mdy = kgm 0
moment z slu mdz = kgm 0
length rod lx = cm 300
width section b = cm 10
Height section h = cm 10
bond coefficient y betay = 1
constraint coefficient z betaz = 1
area section area = cm2 100
resistance module wy=b*h^2/6 wly = cm3 166,66
resistance module wz=h*b^2/6 wlz = cm3 166,66
radius inertia iy=h/(12^0,5)
iy = cm3 2,88
radius inertia iz=b/(12^0,5)
iz = cm3 2,88
Snellezza y lamday=betay*l/iy lamday = 103,92
lamdaz=betaz*l/iz lamdaz = 103,92
lamdarel=lamda/3,14*(fc0k/e0)^0,5 [4.4.14]lamdarely = 1.8
lamdarelz = 1.8
reductive coefficients of critical tension kcy and kcz [4.4.15]imperfection coefficient bc = 0.2
== sync, corrected by elderman ==
== sync, corrected by elderman =
kcy=1/ (ky^2-lamdarely^2)^0,5 kcy = 0.27
(kz^2-lamdarelz^2)^0,5 kcz = 0.27
model coefficient kmod = 0.9
resistances of bending [4.4.1]fm,d=kmpd*fm,k/gammam fm,d = kg/cm2 96
compression resistance [4.4.1]
fc,0,d=kmod*fc,0,k/gammam fc,0,d = kg/cm2 102
tensione compressione sc0d=nd/area sc,0,d = kg/cm2 27
tensione flessioney smyd=myd/wy sm,y,d = kg/cm2 0
tensione flessionez smzd=mzd/wy sm,z,d = kg/cm2 0

verifica instabilità [4.4.13] sc0d/(kcz * fc0d) + 0,7*smyd/fmd + smzd/fmd < 1
0.95 + 0 + 0 = 0.95 < 1 ok

 

[odisseo 18]

thanks you were very precious I don't know how to thank you, the next good to make