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Chapter 10

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SC BN VT LIU 2SC BN VT LIU 2Trn Minh Ti hc Xy dng H ni

B mn Sc bn Vt liuKhoa Xy dng Dn dng & Cng nghip

zx

zy

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Tran Minh Tu University of Civil Engineering

E-mail: [email protected](20)Chapter 10

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Chng 10

Tnh bn kt cu theo ti trng gii hn

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10.1. Cc khi nim chung

10.2. Phng php tnh bn theo ti trnggii hn

10.3. Tnh h thanh chu ko (nn) ng tm

10.4. Tnh thanh chu un thun ty phng10.5. Tnh thanh chu un ngang phng

Tnh bn kt cu theo ti trng gii hn

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1. Cc quan im tnh ton kt cu Mi cc nh gi bn u km theo cc quan nim,

cc tiu chun. C hai quan im chnh tnh tonkt cu: quan im tnh theo ng sut cho php vquan im tnh theo ti trng gii hn.

a. Tnh bn theo ng sut cho php

Ch cho php vt liu thanh lm vic trong min nhi, khi mt im bt k hay mt mt ct no thucvt th xutt hin bin dng do (ng sut t ti sch)

=> h b ph hoi iu kin bn:

[ ] 0maxn

= [ ] 0maxn

=

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ch u im: n gin, ch cho phpbin dng b ( 0,2%) Nhc im: qu thin v an tonnn lng ph vt liu, cha xem xtn s lm vic ca ton b kt cu

b. Tnh bn theo ti trng gii hn

=> cn c mt phng php khc

khc phc nhc im

- i vi vt liu do, khi xut hin bin dng do mt vi

im (un, xon), mt vi mt ct ngang (h siu tnh) h vncha b ph hoi (vn cn kh nng chu lc)

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F

ch

ch

P

1 2 3

=>nh gi bn ca kt cu cn phi xt n kh nng chu lc ca c h.

=> Cn xt xem hp ng hay khng p ng c cc yu cu t ra vmt chu lc

Trng thi trung gian gia hai trng thi: p ng v khng p ng c

cc yu cu chu lc gi l trng thi gii hn, ti trng tng ng gi l titrng gii hn = > K hiu Fgh Tnh bn theo ti trng gii hn cho php pht sinh bin dng do, h trng thi gii hnkhi bin dng do pht trin ti mc ton kt cu mt khnng chu lc.

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10.2. Phng php tnh bn theo ti trng gii hn

ch

T th ko vt liu do: bindng do >> bin dng n hi

C th quan nim th ch

gm 2 giai on: n hi v do=> th Prandtl

ch

Biu qui c (Prandtl)

iu kin bn :

ghgh

PP Pn

=

u im: tit kim vt liu

Nhc im: cho php bindng ln => khng ph hp chongnh c kh chnh xc

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Thanh chu ko (nn) ng tm: z=const Khi bt kim no c ng sut php t tich => c tit din u t ti ch.=> Lc dc trn mt ct ngang gi l Nd

d chN A=

1. Thanh n hoc h thanh tnh nh: tnh theo

SCP v TTGH l nh nhau

Khi 1 im no trn mt ct ngang c ng sut t ti ch th theoSCP y l trng thi nguy him. ng thi do =const nn ton bmt ct ngang u t ti ch => bin dng ca thanh l ty => thanhmt kh nng chu lc: TTGH

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2. H siu tnh: s lng lin kt nhiu hn s lng cn thit khi1 thanh xut hin bin dng do th h vn cn kh nng chu lc,cho n khi h c (n+1) thanh b chy do h mi hon ton mtkh nng chu lc

Phng php gii theo ti trng gii hn

Phng php n hi:

- Xc nh ni lc trong tt c cc thanh => xc nh ng sut

- Ln lt cho (n+1) thanh c tr s ng sut ln nht xuthin chy do (lc dc Nd=ch.A)- Khi thanh th (n+1) thanh b chy do: h TTGH => xcnh ti trng gii hn tng ng

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H gm dm tuyt i cng BCD,chu ti trng phn bu q. Dm clin kt khp ti A v treo bi 2 thanh1 v 2 c cng chiu di v cngEA. Xc nh ti trng cho php theophng php USCP v TTGH, bitch ca vt liu thanh treo

BC D

1 2

a a

q

B C D

N1 N2q

Bi gii

- Gii theo SCP2

2 12 2 0

B M N a N a qa= + =

l1 l2

2 1 2 12 2l l N N = => =

1 2

2 4;

5 5 N qa N qa = =

iu kin bn : [ ]224

5

N qa

A A

= = [ ][ ]5 5

4 4

chdh

Aq

A nA

= =


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Khi thanh 1 b chy do: N1=Nd=chA=> h TTGH

BC D

Nd=chAqgh Nd=chA

1 2 2gh ch ch N q a A A = =

3

2

chgh

qa

=

BC D

N1Nd=chA

q1

1 2

2 4;

5 5 N qa N qa = =

=> thanh 2 chy do trc => s:

2

1 12 2 0

B chM a N a q a= + =

1 12 2 ch N q a A =

3

2

gh chgh

qq

n na

= =

[ ][ ]5 5

4 4

chdh

Aq

A nA

= =

6

5

gh

dh

q

q

=

- Gii theo TTGH


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Phng php ng:- Gi thit (n+1) thanh bt k b chy do => H TTGH gi thit =>Xc nh ti trng gii hn gi thit tng ng Fgh

i

- Gi tr nh nht trong cc ti trng gii hn gi thit l ti trng giihn ca kt cu

Fgh=min{Fghi}

V d: Cho h thanh chu ti trng nh

hnh v. Tm [Fgh] bit A1=A2=A3=A, giihn chy ca vt liu ch, h s an ton n

F

1

2

3300 450

Nhn xt: - Cc thanh 1, 2, 3 u chu ko- H siu tnh bc 1

=> h TTGH khi 2 trong 3 thanh b chy do- Thanh 1 v 2 b chy do => Fgh1

Cc trng thi gii hn gi thit - Thanh 1 v 3 b chy do => Fgh2

- Thanh 2 v 3 b chy do => Fgh3

=> Loi

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TH1: Thanh 1 v 2 b chy do

Fgh1

3300450

N2=chA

N1=chA

u

0 1 0os30 os30 0

ch ch ghu A Ac F c = + =1 2 3 3

2,153gh ch chF A A

+

=

TH3: Thanh 2 v 3 b chy do

1 300450

Fgh3

N2=chA

N3=chA

v

0 3 0os45 sin 30 0

ch ch gh

v A Ac F = + =

( )3 1 2 2, 41gh ch chF A A = + Fgh=min{Fgh

i}

12,15

gh gh chF F A= =

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10. 4. Tnh dm chu un thun ty

Xt dm chu un thun ty, vt liudm l n hi tuyn tnh

max=ch

min=ch

- Biu ng sut l ng bcnht, iu kin bn:

maxW

xch

x

M =

Khi ti trng tng n gi tr :

max=lminl=ch => dm trng thinguy him

, ,W

x dh ch x dhM =

max=ch

min=ch

Ti trng tip tc tng, min dolan rng dn v min n hi thuhp li

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Ti trng tng n lc ton b mtct ngang b chy do hon ton, dmmt kh nng chu lc => TRNGTHI GII HN

max=ch

min=ch

trng thi gii hn, ng phn cchgia 2 min ko v nn gi l ng trungho chy do.ng trung ho chy do chia mt ct

ngang lm 2 phn c din tch bng nhau x

x1

min=ch

max=ch

Ak

An

Ak = An

trng thi gii hn, m men un nilc trn mt ct ngang gi l m menun do Mx,d

, ,d ch x d W=

Wx,d m men chng un do

1 1

( ) ( )

,k nA A

x d x xW S S= +

1

( )kA

xS- m men tnh caA

ki vi x

1

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Khi mt ct ngang c 2 trc i xng th x x1

h

b

2

,6

x dh

bhW =

2

,4

x d

bhW =

D

- Mt ct ngang ch nht

- Mt ct ngang trn

3

,32

x dh

DW

=

3

,6

x d

DW =

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10. 5. Tnh dm chu un ngang phng

Xt dm chu un ngang phng F- Do Mx const => cc mt ctngang c mc chy do khcnhau, khng ging nhau nh unthun ty phng

Mmax

ch

ch

- Khi Mmax=Mx,h: xut hin bindng do u tin ti mp trn vdi ca mt ct ngang im tlc F

- Khi Mmax>Mx,h: bin dng do landn vo trong v ra hai bn dc

theo chiu di dm- Khi Mmax = Mx,d: tit din im t lc b chy do hon ton, trong lccc tit din ln cn cha b chy do hon ton. Min chy do c hnhdng nh hnh v

Do

n hin hi

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10. 5. Tnh dm chu un ngang phng

F

Mmax

ch

ch

Do

n hin hi

Hai phn thanh n hi haipha tri, phi lin kt vi nhau ch mt im. im ni ny ngvai tr nh l khp v gi l

khp do. Khp tht c th xoay t do vc hai pha, v c Mx=0

Khp do ch c th xoaychuyn ng v pha th cng, vc Mx=Mx,d

Mx,d Mx,d- Vi dm tnh nh, khi xut hinkhp do, dm tr thnh c cu=> mt kh nng chu lc =>TTGH => Fgh

- Vi dm siu tnh bc n, h TTGH khi hnh thnh (n+1) khp do.

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???

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