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7/30/2019 SBVL1 - Ch3 - TTUS (2012)
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Chng 3
TRNG THI NG SUT
7/30/2019 SBVL1 - Ch3 - TTUS (2012)
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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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3.1. Khi nim v trng thi ngsut ti mt im (2)
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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3.1. Khi nim v trng thi ngsut ti mt im (3)
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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3.1. Khi nim v trng thi ngsut ti mt im (4)
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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3.1. Khi nim v trng thi ngsut ti mt im (5)
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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3.1. Khi nim v trng thi ngsut ti mt im (6)
- 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)
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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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3.6. Quan h ng sut bin dng (2)
( 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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3.6. Quan h ng sut bin dng (3)
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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3.6. Quan h ng sut bin dng (3)
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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3.6. Quan h ng sut bin dng (4)
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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3.6. Quan h ng sut bin dng (5)
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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3.6. Quan h ng sut bin dng (6)
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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3.6. Quan h ng sut bin dng (7)
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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3.7. Cc thuyt bn (2)
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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3.7. Cc thuyt bn (3)
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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3.7. Cc thuyt bn (4)
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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3.7. Cc thuyt bn (5)
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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3.7. Cc thuyt bn (6)
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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3.7. Cc thuyt bn (7)
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
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