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Chng 3

TRNG THI NG SUT

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Chng 3. Trng thi ng sut

3.1. Khi nim v trng thi ng sut ti mt im3.2. Trng thi ng sut phng

3.3. Vng trn Mohrng sut

3.4. Trng thi ng sut phng c bit3.5. Trng thi ng sut khi

3.6. Quan h ng sut bin dng.

nh lut Hooke3.6.iu kin bn cho phn t TTS phc tp

Cc thuyt bn

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3.1. Khi nim v trng thi ngsut ti mt im (1)

a. Khi nim ng sut

im K(x,y,z) mt ct (php tuyn n)

Mt ct bt k i qua K ng sut php s ng sut tip t

Qua K: v s mt ctTrng thi ng sut ti mt im l tp hp

tt c nhng thnh phn ng sut trn ttc cc mt i qua im

K

x

y

z

n


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nghin cu TTSti mt im => tch raphn t lp phng vcng b cha im => gn h trc xyz =>trn mi mt vunggc vi trc c 3 thnh

phn ng sut: 1 tpng sut php v 2thnh phn ng suttip


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Chn thnh phn ng sut tc dngtrn 3 cp mt vung gc vi ba trcto thnh ten-x ng sut

s

s t t

t s t

t t s

x xy xz

yx y yz

zx zy z

T


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b. Mt chnh ng sut chnhphng chnh Mt chnh: L mt khng c

tc dng ca ng sut tip.

Phng chnh: l phngphp tuyn ca mt chnh.

ng sut chnh: l ng sutphp tc dng trn mtchnh.

Phn t chnh: ng sut tiptrn cc mt bng 0


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d) Qui c gi tn cc ng sut chnh:Ti 1 im lun tn ti ba mt chnh

vung gc vi nhau vi ba ng sut

chnh tng ng k hiu lTheo qui c:e) Phn loi TTS

- TTS n- TTS phng- TTS khi

1 2 3s s s 1 2 3, ,s s s

Nghin cu trng thi ng sut phng

3

1

2


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- Trng thi ng sut n: Haitrong 3 ng sut chnh bng 0

- Trng thi ng sut phng: Mttrong 3 ng sut chnh bng 0

- Trng thi ng sut khi: C 3ng sut chnh khc 0


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3.2. TTS phng (1)

Mt vung gc vi trc z l mt chnh c ngsut chnh = 0 => Ch tn ti cc thnh phnng sut trong xOy

sxtxy

sy

x

y

z

tyx

x

y

sxtxy

sy

O

tyx


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3.2. TTS phng (2)

Qui c du ng sut php dng khi c chiu i ra khi phn t

ng sut tip c chiu dng khi i vng quanhphn t theo chiu kim ng h

a) nh lut i ng ca ng sut tip

ng sut tip trn hai mt bt k vung gc vi nhauc tr s bng nhau, c chiu cng i vo cnh chung

hoc cng i ra khi cnh chung.

TTS phng xc nh bi: sx ,sy, txy

zM 0 |txy| = |tyx|C t xyt yx


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3.2. TTS phng (3)

b) ng sut trn mt nghing (//z)Mt nghing c php tuyn u hpvi phng ngang x gc a (a > 0:t x quay n u theo chiu ngc

chiu kim ng h)

dy ds a xy u

sy

sxtxy tyx

tuvsu

yz

y

xsx

sy

sz

txytxz

tyxtyztzytzx

vz

u

sx

sy

txy

tyx

sutuv


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Qui c du a>0 - chiu ngc kim ng h

0uF 2

2

cos cos sin

sin sin cos 0

u x xy

y yx

A A A

A A

s s a t a a

s a t a a

su >0 - hng ra t uv - thun chiu kim ng h

2

uv xy x

2

yx y

A - Acos - Acossin

+ Asin + Asincos = 0

vF 0

3.2. TTS phng (*)

sy

x

y

sxtxy

sy

O

u

sytyxsx

v

u

a

AAcosa

Asina

su

tu va

txy


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3.2. TTS phng (4)

TTS phngx y x y

u xycos sins s s s

s a t a

2 22 2

x y

uv xysin 2 cos 22

s st a t a

tyxsyx

su

tu v

a

sy

sxtxy


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3.2. TTS phng (5)

c) ng sut php cc tr l cc ng sut chnh ng sut php cc tr khi:

Cc ng sut chnh (phng chnh) xc nh t k:

a) T (1) v (2):

uv 0 =>t

(1)

0 (d.p.c.m)a a

xyu

x y

2d0 => tg2 =-

d

tsa

a s s

xy

0

x y

2

tg2 =-

t

a s s(2)


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3.2. TTS phng (6)

ng sut php cc tr l cc ng sut chnh

Hai phng chnh vung gc vi nhau

0

01,02 0

090

aa

a

0

21

2

xy

x y

arctgt

as s

2

2

1,2(3)2 2 xy

x y x y

max, min

s s s s s s t

22

xy

x y

tgt

a

s s

Hoc:1

max

xy

y

tgt

as s

2min

xy

ytg

t

a s s


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3.2. TTS phng (7)

d) ng sut tip cc tr: mt c ng sut tipcc tr hp vi mt chnh gc 450

e) Bt bin ca TTS phng: tng cc ng sutphp trn hai mt bt k vung gc vi nhau timt im c gi tr khng i

2

2

2

s st t

xy

x y

max,min

s s s s x y u v const

x y 0

0

xy

d0 => tg2 = => = 45

d 2

s st a

a t


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3.3. Vng trn Mohr ng sut (1)

Bit TTS ti mt im => cc thnh phn ng suttrn mt nghing, ng sut chnh, phng chnhtheo cng thc : PHNG PHP GII TCH

Bng th => vng trn Mohrng sut

Pt ng trn

Tm bn knh

- cos2 sin 22 2

sin 2 cos22

x y x yu xy

x y

uv xy

s s s ss a t as st a t a

( )2 ( )2

( )2 ( )2

2 2

2 2-2 2

x y x y

u uv xys s s s s t t

,02

x yI

s s 2

2

2

x y

xyRs s t


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3.3. Vng trn Mohr ng sut (2) Cch dng vng trn Mohr

OB AI

s1sx

sy

s2

Mtxysu

tuv

au

sx

sy

sytxy

tyx

tyxtxy

sx au

su

tuv

tmin

tmax

u1

u2

R

K

` a01a02

im cc

TTS phng

( ,y xyM s t

,02

x yI

s s 2

2

2

x y

xyRs s t


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3.4. TTS phng c bit (1)

F

z

y

x

C

c

y xxz zy

x x c

Q SM y,I I b

s t

Thanh chu un ngang phng

t zyt yz

s zz

TTS phng m 1 trong 2 thnh

phn ng sut php sx, sy bng 0 =>k hiu cc thnh phn ng sut: sv t

st

tt

s

ts

t

I

smaxmin

max,min 3

2

2

1,

2 2

s s t

s s 2 21ax

3m

2 2

s s s t t


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3.4. TTS phng c bit (2)

M

C t xyCt yx

txytyx

Thanh chu xon

x ys s 0z

xy

P

M

Wt t

ts

s minmax

t

smin smax

TTS trt thunty: trn cc mt caphn t ch c ngsut tip


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3.5. TTS khi (1)

TTS khi c c 3 thnh phn ng sutchnh s1, s2, s3 0ng vi mi cp ng sut (s1, s2), (s1, s3),

(s2, s3) ta v c 3 vng trn c tm C1, C2,C3.

LTH chng minh: ng sut trong mt ct nghing bt k (khng //

vi mt chnh no) tng ng vi 1 im nm

trong vng gch cho Cc im nm trn chu vi ng trn C1(s1, s2),

tng ng vi cc thnh phn ng sut trn trnmt // vi phng chnh cn li s3


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3.5. TTS khi (2)

s1x

y

z

s2

s3b

a s

t

123

C1C2 C3


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Cho phn t TTS phng c ccthnh phn ng sut trn cc mtnh hnh v. Tm phng chnh,ng sut chnh ca TTS ti im. Bit =60o

GiI

210 / ;u kN cms

u

6KN/cm2

4KN/cm2

10KN/cm2

6KN/cm2

4KN/cm

2

x

y

Gn h trc xy cho phn t nh hnh v

Php tuyn u ca mt nghing to viphng ngang gc a

aTa c: 24 / ;y

kN cms 26 / ;xy kN cmt

150oa

V d 3.1 (1)


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V d 3.1 (1)


atassss

s 2sin2cos22

xyyxyx

u

218,928 /x kN cms

Phng chnh:

22

xy

x y

tgt

a

s s

Li c:u

6KN/cm2

4KN/cm2

x

y

a

1 2 119,4 ; 90 109,4o o oa a a

2

2

min

max22

xyyxyx

t

ssss

s

ng sut chnh: 2max /041,21 cmKN s

2min /887,1 cmKN s

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3.6. Quan h ng sut bin dng (1)

1. Trng thi ng sut n

E

x

x

s xy

Es

xzEs

2. Trng thi ng sut trt thun ty

Gxy

xy

t 0zxyz

x

y

z

sx

x

y

z

t x yUniversity of Architechture

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( zyx

zyx

x

E

EEE

s

ss

1

sy

x

y

z

sz txysx

3. Trng thi ng sut tng qut

- Theo nguyn l cng tc dng

- Gt: bin dng di ch sinh ra ngsut php, bin dng gc lm phtsinh ng sut tip


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a. Quan h ng sut php bin dng di

b. Quan h ng sut tip bin dng gc

vi E, , G l m un n hi ko (nn), h sPoisson, m un n hi trt, lin h vi nhaubi cng thc:

( 1

x x y zE

s s s

( 1

y y x z

E

s s s

( 1

z z x yE

s s s

xy

xyG

t xzxzG

t yz

yzG

t

( 2 1E

G

( 1322

1 sE (

1233

1 sE

( 3211

1 sE


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Trng thi ng sut phng:1

x x yE

s s

1y y xE

s s

1 1 21

E s s

2 2 11E s s

xy

xy

G

t


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c. Quan h ng sut php bin dng th tch

321aaaV

)(a)(a)(aV3322111

111 321

1 V

VV

x y z( ) ( )E E

s s s s s s

1 2 3

1 2 1 2


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3. Th nng bin dng n hi

s2s3

s 1u u us t

s t 1 1

2 2

Xt phn t chnh: t=0 =>

3322112

1

2

1

2

1u

( 312321

2

3

2

2

2

12

2

1 sE


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s2s3

s 1

ac

s3-s tb

s 1-s tb

s2-s tb

s tb

b

s tb

s tb

=

Bin dng => thay i Hnh dngTh tch

tb( )s s s s

1 2 3

1

3


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hd tt u u u=

ttuE

s s s

21 2 31 2

( )6

( ( ( hduE

s s s s s s

2 2 2

1 2 2 3 3 1

1

6

TNBDH ring u TNB hnh dng uh d

TNB th tch ut t


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3.7. Cc thuyt bn (1)

A. Khi nim v cc thuyt bn- TTS n (ko nn ng tm):iu kin bn:

P

M

- TTS trt thun ty:

0

max

k

k n

ss s

0

min

n

n n

ss s

0maxn

tt t

- Gi tr cc ng sut cho php xcnh theo ng sut nguy him => tthc nghim

TTS phc tp: cn phi thc nghim xc nh nhng ngsut nguy him cho TTS tng ng => khng thc hin c

???


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L do: S lng th nghim ln ( p ng c cc t l gia cc

ng sut chnh c th xy ra trong thc t)

K thut th nghim cha thc hin c

Khng tin hnh thc nghim c =>Khng bit nguyn nhn gy ra s ph hoivt liu => Gi thit

Thuyt bn: Cc gi thit v nguyn nhn

gy ra s ph hoi vt liu Cc nguyn nhn c th: ng sut, bin

dng, th nng bin dng n hi,


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B. Cc thuyt bna. Thuyt bn 1 - Thuyt bn ng sut

php ln nht

Nguyn nhn vt liu bph hoi l do ng sutphp ln nhtca phn t TTS phc tp tti ng sut php nguy him ca phn t TTS n

iu kin bn

Hn ch: Chph hp vi vt liu gin, v TTS n

1 1t ks s 1 3t ns s


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b. Thuyt bn 2 - Thuyt bn bin dng ditng i ln nht (M a r i o t t e )

Nguyn nhn vt liu b ph hoi l do bindngdi tng i ln nhtca phn t TTS

phc tp t ti bin dngdi tng i trngthi nguyhim ca phn t TTS n

iu kin bn:

Chph hp vi vt liu gin

( 2 1 2 3t ks s s s ( 2 3 1 2t ns s s s


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c. Thuyt bn 3 - Thuyt bn ng suttip ln nht (T r es c a -S ai n t V en a n t )

Nguyn nhn vt liu bph hoi l do ng sut

tip ln nht ca phn t TTS phc tp tti ng sut tip nguy him ca phn t TTSn

iu kin bn:

Ph hp vi vt liu do. Thng s dng trongngnh c kh

3 1 3t s s s s


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d. Thuyt bn 4 - Thuyt bn th nng bin ihnh dng cc i (H u b e r V o n M i s e s )

Nguyn nhn vt liu bph hoi l do th nngbin i hnh dng ca phn t TTS phctp t ti th nng bin i hnh dng trngthi nguy him nguy him ca phn t TTSn

iu kin bn: Ph hp vi vt liu do. S dng trong ngnh

k thut xy dng v c kh ch to.

2 2 2

4 1 2 3 1 2 1 3 2 3t ks s s s s s s s s s s


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e. Thuyt bn 5 - Thuyt bn Mohr Da vo kt qu th nghim => V vng trn

ng sut gii hn => V ngbao => Xcnhmin an ton ca vt liu

iu kin bn:

Chph hp vt liu gin

5 1 3

kt k

n

s

s s s s

s

[s] ns u

u v

[s ]kO2

O3 O1


(O.Mohr),1835 1918

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