thiet lap cong thuc tinh br tru va br thang theo do ben tiep xuc va do ben uon

* - tinh banh rang thang va tru theo do ben tiep xuc

us tiep xuc lon nhat @H duoc tinh theo cong thuc hec'

@H=Zm.can(qn/2p)

( p la ro)

qn: cuong do tai trong phap tuyen

p ban kinh duong cong tuong duong cua be mat

Zm he so xet den co tinh cua vat lieu

Zm= 2.E1.E2/[pi.[(1-luy2^2).E1+(1-luy1^2).E2]]

E1,E2la mondun dan hoi

luy1 , luy2 he so poat xong ( luy la gan giong chu m)

voi banh rang thep la 275.(NPa)^1/2   

"A la anpha"

luy = luy1.luy2/(luy1+luy2)

luy1= N1.w=sin (aw.dw1)/2

luy2=N2.w=sin(aw.dw2)/2=u.dw1.sin(aw)/2

suy ra luy=u.dw1.sinaw/[(u+1).2]

co' qn= Fn/LH

" H viet nho"

LH= 3.bw/(4-epa)

" w viet nho, ep la epsilon, a la anpha"

epa la he so trung khop

Fn= Ft/cosaw = 2T1/(d1.cosaw)

Fn: luc phap tuyen

LH tong chieu dai tiep xuc

ke den tai trong Fn= 2T.KHpeta.KHV/(d1.cosaw)

" Hpeta, HV viet nho"

suy ra cong thuc kiem nghiem

@H=(Zm.Zep.ZH/dw1).can[ 2T1.(u+1).KHpeta.KHV/(u.bw)] <= [@H]

cong thuc thiet ke

dw1 >= cambac3( Zm^2.zep^2.Zh^2). canbac3(2T1.(u+1).KHbeta.KHV/(u.bw))

suy ra dw1 >= 77.canbac3(2T1.(u+1).KHpeta.KHV/(#bd.u.[@H]))

" #bd la xi , 1 duong cong va 1 duong thang"

aw>50.(u+1).canbac3(2T1.KHbeta.KHV/(#bd.u.[@H]^2))

- theo Us uon

@F<=[@F]

@u= Fn.cosa'.h/wu

"wu , u viet nho"

cac he so tai trong

KF=KFa.Kpbeta.Kfv=Kpbeta.Kpv

suy ra 

@u= Fn.cosa'.h.KF/wu

co' wu= bw.S^2/6

" bw, w viet nho "

@n=Fn.sina'.Kf/A

A=bw.S

tai vi tri nguy hiem

@=@u-@n

@= Fn.cosa'.h.Kp.6/(bw.s^2) -(Fn.sina'.kp/(bw.s))

thay s= gm, h=lm, Fn= Ft/cosaw=2T/(d.cosaw)

suy ra @=(2T.Kf/d.cosaw.bw).[6h.cosa'/(s^2) -sina'/s]

suy ra

@= (2T.Kf.K@/d.bw.m).(6.l.cosa'/(g^2.cosaw) - sina'/(g.cosaw))

dat ( 6.l.cosa'.................cosaw)) =YF

suy ra cong thuc kiem nghiem

@F= 2T.KF.YF/(m.bw.dw)<=[ @F]

suy ra @F1=2T1.Kp.YF1.Yep/(m.dw1.bw) <=[@F1]

@F2= YF2.@F1/YF1<=[@F2]

cong thuc thiet ke

m >= 1,4 canbac3( T1.Kp.Yp1/(#bd.Z1^2.[@F1])

"#bd la xi , 1 duong cong o tren va 1 dung thang dung"