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Chng 11. Tnh h thanh siu tnh bng phng php lc
11.1
Chng 11. TNH H THANH SIU TNH BNG PHNG PHP LC
I. KHI NIM C BN H tnh nh (HT): s lin kt = s phng trnh cn bng tnh hc. H siu tnh (HST) l h c s lin kt nhiu hn s phng trnh cn
bng tnh hc. H siu tnh l h bt bin hnh v c cc lin kt tha. Bc siu tnh ca h c tnh bng s lin kt tha. S lin kt tha ca mt h c th l lin kt ngoi (lin kt cn thit gi cho h c c nh) hay lin kt ni (lin kt gia cc phn i vi nhau trong cng mt h)
So vi h tnh nh, HST c nhng c im sau: Ni lc trong HST phn b u hn, ng sut v bin dng nh hn
so vi HT c cng kch thc v ti trng. HST c nhc im l d pht sinh cc ng sut khi nhit thay i, khi c ln cc gi ta, gia cng lp ghp khng chnh xc.
Khi nhng lin kt tha b h hng th h vn khng b ph loi, v khi h vn bt bin hnh hc.
V d: Hnh 11.1a,e: h tha 2 lin kt ngoi: bc siu tnh ca h l 2. Hnh 11.1b: h tha 1 lin kt ngoi: bc siu tnh ca h l 1. Hnh 11.1c:
h tha 3 lin kt ngoi v 3 lin kt ni: bc siu tnh l 6. Hnh 11.1d: h tha 3 lin kt ni, bc siu tnh ca h l 3.
Khung khp kn (hnh 1.1f) siu tnh bc ba. V mun ni phn (A) v (B), cn 3 lin kt n hoc 1 khp v 1 lin kt n hay thay ba lin kt n bng mi hn cng (hnh 11.1g,h).
Khi nim lin kt tha ch c tnh qui c. Bi v m bo cho h bt bin hnh th chng l tha, nhng s c mt ca chng s to cho kt cu c cng cao hn v do , lm vic tt hn so vi h tnh nh. Sau y ta gii HST bng phng php lc.
a)
e)
b) c) d)
f) g) h) (A) (B) (A) (B)
Hnh 11.1
Chng 11. Tnh h thanh siu tnh bng phng php lc
11.2
II. GII H SIU TNH BNG PHNG PHP LC 1. H c bn ca HST l mt HT c c t HST cho bng cch b bt cc lin kt tha.
HST c th c nhiu h c bn (hnh 11.2).
Cn ch rng: Sau khi b cc lin
kt tha, h phi m bo tnh bt bin hnh ca n. Ch c php gim
bt cc lin kt n ch khng c php thm lin kt n vo mt mt ct bt k.
V d: h trn hnh 11.3b, c khng phi l h c bn ca h trn hnh 11.3a, v n s bin hnh. 2. HT tng ng HT tng ng vi HST cho khi bin dng v chuyn v ca
chng hon ton ging nhau. HT tng ng l h c bn chn ca HST: cc lin kt tha biu
din phn lc lin kt (hnh 11.4). Phn lc lin kt c xc nh vi iu kin bin dng v chuyn v ca HT hon ton ging nh HST cho.
Hnh 11.4
3. Thit lp h phng trnh chnh tc xc nh cc phn lc lin kt Vi mi phn lc lin kt Xi ta c mt iu kin chuyn v:
l
q
l
Hnh 11.2(a) (b) (c)
a) b) c)
l
l
Hnh 11.3
(a) (b) (c)
Chng 11. Tnh h thanh siu tnh bng phng php lc
11.3
Gi i l chuyn v ca im t ca Xi theo phng ca Xi , gy ra do ti trng Pi v tt c cc Xj (j = 1, 2, , n), vi n l bc siu tnh ta c:
i = i (i = 1, 2, , n) (11.1) y i l chuyn v ti im t ca Xi v theo phng Xi do ti trng
cho gy ra trong HST, du (+) ly khi chiu chuyn v ca i cng chiu vi chiu ca lc Xi v ly du (-)khi chiu chuyn v ca i ngc chiu vi chiu ca lc Xi. Trong cc trng hp thng gp nh gi c nh, di ng, ngm th ta c i = 0. Tuy nhin c nhng trng hp i 0, chng hn gi ta n hi. Nu HST c n bc siu tnh n phng trnh (11.1) h phng
trnh chnh tc xc nh cc phn lc lin kt Xi (i = 1, 2, .., n):
1 11 1 12 2 1n n 1p
2 21 1 22 2 2n n 2p
n n1 1 n2 2 nn n np
X X ... X 0
X X ... X 0
.............................................................
X X ... X 0
= + + + + = = + + + + = = + + + + = (11.2)
trong : ip l chuyn v theo phng i ca h c bn do ti trng gy nn. ik l chuyn v n v theo phng i ca h c bn do lc n v t theo phng k gy nn. Ta c th tnh c ip v ik theo cng thc Mo sau:
= = = = + + + i i i
l l ln n nzi zk xi xk zi zk
iki 1 i 1 i 1x p0 0 0
N N M M M Mdz dz ... dz
EF EJ GJ
= = =
= + + + i i il l ln n n
zi zp xi xp zi zpip
i 1 i 1 i 1x p0 0 0
N N M M M Mdz dz ... dz
EF EJ GJ
Nu b qua nh hng ca ko-nn v xon so vi un, th ip v ik tnh theo cng thc Mo sau (b qua ch s x, y trong cng thc):
iln
i kik
i 1 0
M Mdz
EJ= = ; i
lni p
ipi 1 0
M Mdz
EJ= = (11.3)
Sau khi xc nh c cc phn lc lin kt Xi, t cc phn lc lin kt Xi cng vi ti trng ln h c bn mt HT tng ng. Gii HST bng phng php lc ta c cc bc sau: Bc 1. Xc nh bc siu tnh v chn h c bn Bc 2. Xc nh HT tng ng bng cch t vo h c bn cc phn
lc lin kt tng ng vi cc lin kt tha b i. Bc 3. Thit lp h phng trnh chnh tc
Chng 11. Tnh h thanh siu tnh bng phng php lc
11.4
V d 11.1: V biu ni lc ca khung nh hnh v 11.5a
Gii: Khung c hai bc siu tnh, h c bn c chn nh hnh 11.5b.
HT tng ng nh trn hnh 11.5c. Phng trnh chnh tc c dng:
11 1 12 2 1p
21 1 22 2 2p
X X 0
X X 0
+ + = + + = Biu mmen un do ti trng (Mp) nh hnh 11.5d. p dng phng php nhn biu Versaghin ta c:
3n
1 111
x x xi 1 0
M M 1 1 2 4adz .a.a. a a.a.aEJ EJ 2 3 3EJ=
= = + = il
3n
1 212
x x xi 1 0
M M 1 1 adz .a.a.aEJ EJ 2 2EJ=
= = = il
3n
2 222
x x xi 1 0
M M 1 1 2 adz .a.a. aEJ EJ 2 3 3EJ=
= = = il
2 2 4n
p 11p
x x xi 1 0
M M 1 1 a 3 a 5qadz .aq. . a q. .a.aEJ EJ 3 2 4 2 8EJ=
= = + = il
2 4n
p 22p
x x xi 1 0
M M 1 1 a qadz .aq. .aEJ EJ 2 2 4EJ=
= = = il
Hnh 11.5
X2
X1
(c)
X2=1
2M
(f)
a
C B
A
a
a
(a)
q
(e)X1=1
1M
a
C B
A
MP
qa2/2
(d)
(b)
H c bn
Chng 11. Tnh h thanh siu tnh bng phng php lc
11.5
Thay vo phng trnh chnh tc, ta c:
+ = = = + =
3 3 4
1 2 1x x x
3 3 4
21 2x x x
3 a 1 a 5 qa 3X X 0 X qa4 EJ 2 EJ 8 EJ 731 a 1 a 1 qa X qaX X 0
282 EJ 3 EJ 4 EJ
v biu M, N, Q ta t cc lc X1, X2 vo h c bn vi lc X1 c chiu ngc li v kt qu mang du m. Biu M, N, Q nh hnh 11.6.
Hnh 11.6
III. TNH H SIU TNH I XNG 1. nh ngha : H i xng l h khi c t nht mt trc i xng. H i xng chu ti trng i xng khi ti trng t ln phn ny l
nh ca ti trng t ln phn kia qua gng phng t ti trc i xng v vung gc vi mt phng ca h. Nu ti trng ca phn ny l nh ca phn kia nhng c chiu ngc
li th ta gi l h i xng chu ti trng phn i xng. Hnh (11.7a,b,c) - HST
i xng, h chu ti trng i xng, h chu ti trng phn i xng. 2. Tnh cht (mnh ) Tng t, ni lc
cng c tnh cht i xng hoc phn i xng. Trong mt phng: Nz , Mx c tnh
i xng, Qy c tnh phn i xng Trong khng gian: Nz, Mx, My l i
xng, Qx, Qy v Mz phn i xng.
a) b) c)Hnh 11.7
Mx
Mx
Qy
Qy
Hnh 11.8
Chng 11. Tnh h thanh siu tnh bng phng php lc
11.6
Tnh cht ca HST i xng: Nu mt h i xng chu ti trng i xng th ni lc phn i xng trn mt ct trong mt phng i xng ca h l bng khng. Ngc li nu ti trng l phn i xng th ni lc i xng phi bng khng. Ch cc nhn xt sau: Khi h
l i xng chu ti trng i xng th biu mmen l i xng, ngc li nu ti trng phn i xng th biu mmen l phn i xng. Php nhn Vrsaghin gia biu i xng v phn i xng l bng khng.
Chng minh. Gi s c HST i xng chu ti phn i xng (hnh 11.10b). Chn h c bn bng cch ct i khung. Phi chng minh cc thnh phn ni lc i xng X1 v X2 trn mt ct l bng khng.
X1 , X2 , X3 l nghim ca phng trnh chnh tc:
11 1 12 2 13 3 1p
21 1 22 2 23 3 2p
31 1 32 2 33 3 3p
X X X 0
X X X 0
X X X 0
+ + + = + + + = + + + = (11.4)
Biu 1M , 2M l i xng cn biu 3M l phn i xng nn: 13 = 31 = 23 = 32 =1p = 2p = 0 Do h phng trnh chnh tc trn thu gn li nh sau:
11 1 12 2
21 1 22 2
33 3 3p
X X 0
X X 0
X 0
+ = + = + = (11.5)
Hai phng trnh u l mt h thng phng trnh thun nht 2 n s nh thc khc khng X1 = X2 = 0.
Mx
Qx
Nz
xy
z
yx
zMyQy
Mz
Hnh 11.9
X3 X3
X2 X2
X1 Pk=1
Hnh 11.10
(a)
Pk=1 P
(b)
P
l l Pl
kM Mm l l l
Pl
l
Chng 11. Tnh h thanh siu tnh bng phng php lc
11.7
Tng t, khung chu lc i xng nh hnh v 11.10a. Lc biu ti trng l i xng nn: 13 = 31 = 23 = 32 = 3P = 0
H phng trnh chnh tc: 11 1 12 2 1p
21 1 22 2 2p
33 3
X X 0
X X 0
X 0
+ + = + + = = (11.6)
T phng trnh th 3 ta c X3 = 0 pcm. Trng hp h i xng nhng ti trng bt k tng tc dng ca h
c ti trng i xng v h chu ti trng phn i xng (hnh 11.11).
IV. HST C CC LIN KT CHU CHUYN V CNG BC tnh ton nhng HST c cc gi ta chu chuyn v cng bc ta
cng s s dng nhng l lun va m t trn. Ni lc trong h c cc lin kt chu chuyn v cng bc l do cc gi ta chu cc chuyn v cng bc. p dng h phng trnh chnh tc (11.2) vo trng hp ny ta phi
ch khi chn h c bn, khng nn loi b cc lin kt c chuyn v cng bc m phi ct cc lin kt y. Ngoi ra c th la chn h c bn bng cch loi b cc lin kt tha khng c chuyn v cng bc.
Gi s cho mt dm nh hnh 11.12a, nu chn h c bn bng cch loi b lin kt gi ta B c chuyn v cng bc th iu kin bin dng theo phng ca n s X1 do cc n s Xk nu c (trn hnh 11.12a khng ch ra nhng n s ny) v chuyn v cng bc gy ra s khng bng khng. C th l:
1X 1 2 n(X ,X ,...,X ) 0 =
By gi nu ta chn h c bn bng cch ct lin kt c chuyn v cng bc B th iu kin chuyn v theo phng lin kt y vn bng khng. V lc ny iu kin va ni l iu kin m ta chuyn v tng i ca hai mt ct ca lin kt va b ct:
P/2P/2P/2P/2P
Hnh 11.11
= +
Chng 11. Tnh h thanh siu tnh bng phng php lc
11.8
1X 1 2 n(X ,X ,...,X ) 0 =
Nu chn h c bn bng cch ct cc lin kt tha c chuyn v cng bc th phng trnh th k c dng:
+ + + + + + =k1 1 k2 2 kk k kn n kX X ... X ... X 0 (k=1,n) (11.7) Cc h s kj tnh nh i vi trng hp h chu ti trng.
k l chuyn v theo phng ca lc Xk do chuyn v cng bc gy ra trong h c bn. N c xc nh theo cng thc sau:
= =
= n nk i i i ii 1 i 1
R M (11.8)
Trong i iR ,M l phn lc theo phng lin kt th i do lc kX = 1 gy ra trong h c bn. i l chuyn v thng theo phng lin kt th i v
i l gc xoay ti lin kt th i trong h siu tnh cho.
V d 11.2: Tnh mmen un ln nht trong trc c cho trn hnh
11.12a, nu khi ch to tm ca lch i mt on . Gii H c bn chn nh hnh 11.12b. Phng trnh chnh tc c dng: + =11 1 1X 0 Trong = = = 1 1R . 1. Biu 1M cho trn hnh 11.12c, nhn biu ny vi chnh n ta c:
3
111 1 1 2 12 .
EJ 2 2 3 2 6EJ = =
ll. l. l
Thay 1 v 11 va tm c vo phng trnh chnh tc ta c:
12
l
A B
g
l l
C
X1
a)
b)
c)1X 1=
2
3EJ l
d)
1M
M
Hnh 11.12
Chng 11. Tnh h thanh siu tnh bng phng php lc
11.9
1 36EJX = l
Biu mmen un v gi tr mmen un ln nht trn hnh 11.12d.
IV. TNH HST CHU NHIT THAY I
Vic tnh HST chu nhit thay i cng tng t nh tnh h chu tc dng ca ti trng, ch khc y l s bin thin ca nhit l nguyn nhn gy ra ni lc trong h. V th s hng kp thay bng kt l chuyn v theo phng Xk do s bin thin nhit gy ra trong h c bn. C th l
= + + + + = = + + + + = = + + + + =
1 11 1 12 2 1n n 1t
2 21 1 22 2 2n n 2t
n n1 1 n2 2 nn n nt
X X ... X 0
X X ... X 0
.............................................................
X X ... X 0
(11.9)
Trong cc h s kt c xc nh nh sau: 2 1kt c k k
0 0
t - tt N dZ M dZh
= + l l Hay ( ) ( ) 2 1kt k c k t - tN t M h = + Trong ( )kN v ( )kM l din tch ca biu lc dc v mmen
un do lc kX 1= gy ra trong h c bn; 1 2c t + tt 2= ; l h s dn n nhit ca vt liu ca h; h l chiu cao MCN; t1 v t2 l bin thin ca nhit hai pha ca MCN.
Cc h s kj c xc nh nh trng hp h chu ti trng. Sau khi thit lp v gii h phng trnh chnh tc ta s tm c cc n
s X1, X2, X3, Vic v cc biu ni lc c tin hnh theo cc phng php bit.
Chng 11. Tnh h thanh siu tnh bng phng php lc
11.10
IV. TNH DM LIN TC 1. nh ngha Dm lin tc l dm
siu tnh t trn nhiu gi ta n, trong c mt gi ta c nh (hnh 11.13a). Khong cch gia hai gi ta gi l nhp. Bc siu tnh ca dm bng s nhp tr mt. 2. Phng trnh ba mmen Chn h c bn ca
dm bng cch t ln mi gi ta mt khp chia dm thnh nhiu dm n (hnh 11.13b). Nhng lc tc dng ln
mt nhp no ch nh hng n chuyn v ca nhp bn cnh khi xt chuyn v mt gi ta bt k, ch cn xt hai nhp lin tip nhau v cc n s ch l cc mmen un ni lc Mi (hnh 11.13c) (Mi>0 lm cng th di). Phng trnh chnh
tc (phng trnh ba mmen) vit theo iu kin gc xoay tng i gia hai mt ct ti gi ta phi bng khng. V d ti gi ta th i: 11M1 + 12M2 ++ i,i-1Mi-1 + i,iMi + i,i+1Mi+1 ++ 1nMn + ip = 0 Cc h s i1 = i2 = = i(i-2) = = 0, do lc tc dng trn hai nhp
trn hai gi ta th i ch nh hng n gc xoay ca gi ta trn hai nhp . Phng trnh chnh tc ca h c dng sau:
i,i-1Mi-1 + i,iMi + i,i+1Mi+1 + ip = 0 (11.10) Cc h s v s hng t do trong (11.10) tnh theo phng php nhn
biu Vrsaghin, ta c:
1
ibi+1 ai+1 bi
P
li+1 li
M1 M2 M3 M3
P
M2 M1
ll ll
q
M4 M0q
0 1 2 3 4
Mi-1 Mi Mi+1 Mi+1Mi Mi-1q
i-1 i i+1
aii-1 i+1
C C
i i+1
Mi-1=1
Mi=1
1
1
Mi+1=1
i 1M
iM
i 1M +
Mp
a)
b)
c)
d)
e)
f)
g)
Hnh 11.13
Chng 11. Tnh h thanh siu tnh bng phng php lc
11.11
i 1
i 1 i ii,i 1 i
i i i0
M M 1 1 1dz . .1. .EJ EJ 2 3 6EJ
= = =l ll
i
i i i i 1i,i i i 1
i i 1 i i 10
M M 1 1 2 1 1 2dz . .1. . . .1. .EJ EJ 2 3 EJ 2 3 3EJ 3EJ
+++ +
= = + = +l l ll l
i 1
i i 1 i 1i,i 1 i 1
i 1 i 1 i 10
M M 1 1 1dz . .1. .EJ EJ 2 3 6EJ
+ + ++ ++ + +
= = =l ll
i
p i i i 1ip i i 1
i i i 1 i 10
M M a b1 1dz . . . .EJ EJ EJ
+++ +
= = + l
l l
trong : li, li+1 : di ca nhp th i v th (i+1). i, i+1: Din tch ca biu mmen do ti trng gy nn trn hai nhp th i v th (i+1). ai, bi+1: Khong cch t trng tm ca cc din tch n gi ta th (i-1) v (i+1).
Thay cc tr s vo phng trnh chnh tc, ta c: i i i 1 i 1 i i i 1 i 1
i 1 i i 1i i i 1 i 1 i i i 1 i 1
.a .bM M M 06EJ 3EJ 3EJ 6EJ EJ EJ
+ + + + ++ + + +
+ + + + + = l l l l
l l (11.11)
Nu cng EJ khng i trn sut chiu di ca dm, ta c :
( ) i i i 1 i 1i i 1 i i 1 i i 1 i 1i i 1
.a .bM 2 M M 6 0+ + + + ++
+ + + + + = l l l l
l l (11.12)
V d :V biu lc ct, mmen un ca dm lin tc nh hnh v (11.14a)
l
M1 M2 M3
P=ql
M2 M1
l l l
q
M0
0 1 2 3
a)
b)
l l
ql2/8 ql2/4
Mpc)
Hnh 11.14 5ql/8
M2 M2
1M
2M
M1M1
d)
ql2/40
e)
3ql2/20
ql/40ql2/20
3ql2/20 7ql2/20
3ql/813ql/20
7ql/20
Mx
Qy
f)
g)
Chng 11. Tnh h thanh siu tnh bng phng php lc
11.12
Gii: y l HST bc 2. H c bn nh hnh 11.14b. Biu mmen un Mp nh hnh 11.14c.
Phng trnh ba mmen i vi gi ta th 1, 2 v 3 l:
+ + + + = + + + + =
2
0 1 2
2 2
1 2 3
2 q 1M 2( + )M M 6 0 . . . 0
3 8 2
2 q 1 1 q 1M 2( + )M M 6 . . . . . . 0
3 8 2 2 4 2
ll l l l l
l ll l l l l l
Trong M0 = M3 = 0 (do cc khp khng c mmen ngoi lc tp trung). Gii h phng trnh trn ta c:
M1 = 2ql
40 ; M2 =
23ql20
; Du (-) ch cc mmen c chiu ngc vi chiu chn. Cng cc biu Mp, M1, M2 ta c biu Mx (hnh 11.14f). Sau khi
tnh phn lc cc gi ta ca biu Mp, M1, M2 v cng cc vect phn lc, ta thu c biu Qy nh trn hnh 11.14g. 3. Trng hp c bit Trng hp dm lin tc c u tha v u ngm th cch gii ca
chng ta nh sau: Tng tng b u tha v thu gn tt c ngoi lc t trn on
v gi ta cui cng. Mmen un thu gn c th xem l mmen lin kt ti mt ct ca gi ta cui cng (mmen c tr s dng khi n lm cng th di v c tr s m khi n lm cng th trn) hoc c xem l mmen un ngoi lc tc ng ln dm. Cn lin kt ngm th c thay bng mt nhp t trn mt gi ta c nh v mt lin kt n. cng EJx ca on nhp ny c xem l ln v cng v chiu di ca nhp c xem l bng khng (hnh 11.15).
Hnh 11.15
Phng trnh ba mmen c p dng i vi tng nhp cnh nh phn trn.
l0=0
P
P
l
M=Pl
Chng 11. Tnh h thanh siu tnh bng phng php lc
11.13
V d: V biu ni lc ca dm chu lc nh hnh v (11.15a).
Gii: H c bn v th t cc nhp, cc gi ta c nh s nh hnh
11.15b. Biu Mp do ti trng gy nn trn h c bn nh hnh v (11.15c). Mmen thu gn gi ta cui cng c xem l mmen lin kt trn mt
ct ca gi ta . V vy trn biu mmen Mp khng c mmen . Vi cc gi ta (1), (2), ta thit lp c cc phng trnh ba mmen nh sau:
+ + + = + + + =
ll l l l l
ll l l l l
2
0 1 2
2
1 2 3
2 q 1M 2( + )M M 6 . . . 0
3 8 2
2 q 1M 2( + )M M 6 . . . 0
3 8 2
Gii h trn vi M0 = 0 v M3 = 0.5Pl = 0.5ql2 ta c:
M1 = 25ql
28 ; M2 = 23ql
28 ; Mmen M10 c
ngha l M2 lm cng th di. Biu mmen un v lc ct cho trn hnh 11.15f,g.
Hnh 11.15
0
M1
l0=0
M0=0
l/2
M2
P=ql
M3=-ql2/2 M2
l l
q
1 2 3
a)
b)
l l
ql2/8
Mpc)
17ql/28
1M
2M
M1d)
5ql2/28
e)
3ql2/28
11ql/14
5ql2/28ql2/2
3ql/14
ql
Mx
Qy
f)
g)
3ql2/28
Chng 12. Ti trng ng
12-1
Chng 12. ti trng ng
I. Khi nim 1. Ti trng tnh, ti trng ng Ti trng tnh tc l nhng lc hoc ngu lc c t ln
m hnh kho st mt cch t t, lin tc t khng n tr s cui cng v t tr i khng i, hoc bin i khng ng k theo thi gian. Ti trng tnh khng lm xut hin lc qun tnh. Ti trng tc dng mt cch t ngt hoc bin i theo thi
gian, v d nhng ti trng xut hin do va chm, rung ng, v.v... nhng ti trng ny c gi l ti trng ng. Mt cch tng qut, ta gi nhng ti trng gy ra gia tc c
tr s ng k trn vt th c xt, l nhng ti trng ng. 2. Phn loi ti trng ng Bi ton chuyn ng c gia tc khng i w=const, v d,
chuyn ng ca cc thang my, vn thang trong xy dng, nng hoc h cc vt nng, trng hp chuyn ng trn vi vn tc gc quay hng s ca cc v lng hoc cc trc truyn ng. Bi ton c gia tc thay i v l hm xc nh theo thi
gian w = w(t). Trng hp gia tc thay i tun hon theo thi gian, gi l dao ng. V d bn rung, m di, m bn lm cht cc vt liu, bi ton dao ng ca cc my cng c, ... Bi ton trong chuyn ng xy ra rt nhanh trong mt
thi gian ngn, c gi l bi ton va chm. V d phanh mt cch t ngt, ng cc bng ba, sng p vo p chn, 3. Cc gi thit khi tnh ton. Ta chp nhn nhng gi thit sau:
a) Tnh cht vt liu khi chu ti trng tnh v ti trng ng l nh nhau.
b) Chp nhn cc gi thit v tnh cht bin dng ca thanh nh khi chu ti trng tnh, chng hn cc gi thit v tit din phng, gi thit v th dc khng tc dng tng h.
S dng cc kt qu, cc nguyn l v ng lc hc, chng hn: - Nguyn l DAlembert: qtF mw=
r r (12.1)
- Nguyn l bo ton nng lng: T + U = A (12.2) - Nguyn l bo ton xung lng: ng lng ca h trc v
sau khi va chm l mt tr s khng i.
Chng 12. Ti trng ng
12-2
II. Chuyn ng vi gia tc khng i 1. Bi ton ko mt vt nng ln cao Xt mt vt nng P
c ko ln theo phng thng ng vi gia tc khng i bi mt dy cp c mt ct F. Trng lng bn thn ca dy khng ng k so vi trng lng P (hnh 8.1). p dng nguyn l
almbe (dAlembert) v phng php mt ct, chng ta d dng suy ra ni lc trn mt ct ca dy cp:
N = P + Pqt
N = P + P wg = w
1g
+ P = KP (12.3)
Vi K = 1 + wg
Khi gia tc w = 0, th K = 1 v N = Nt = P. Ti trng Nt (khi khng c gia tc) l ti trng tnh, ti trng
N (khi c gia tc) l ti trng ng: N = KNt. ng sut mt ct ca dy khi khng c gia tc t, khi c gia
tc l ng sut ng . V dy chu ko ng tm, nn: t
t
N NK K
F F = = = (12.4)
Cc cng thc (12.3) v (12.4) cho thy: bi ton vi ti trng ng tng ng nh bi ton vi ti trng tnh ln hn K ln. H s K c gi l h s ng hay h s ti trng ng. Kt lun: Nh vy, ni chung, nhng yu t khc nhau gia
ti trng ng v ti trng tnh c xt n bng h s ng v vic gii cc bi ton vi ti trng ng quy v vic xc nh cc h s ng .
P
1 1
z
l
Hnh 8.1
Chng 12. Ti trng ng
12-3
2. Chuyn ng quay vi vn tc khng i Xt v lng c b dy t rt b so vi ng knh trung bnh D = 2R
quay vi vn tc gc khng i (hnh 12-2a). V lng c din tch mt ct ngang F, trng lng ring ca vt liu l . Tnh ng sut ng ca v lng. n gin, ta b qua nh hng ca
cc nan hoa v trng lng bn thn v lng. Nh vy, trn v lng ch c lc ly tm tc dng phn b u q V v lng quay vi vn tc gc =
const, nn gia tc gc & = 0. Vy gia tc tip tuyn wt = & R = 0 v gia tc php tuyn wn = 2R Trn mt n v chiu di c khi
lng F, cng ca lc ly tm l: q =
2 2n
F F FRW Rg g g = =
Ni lc trn mt ct ngang: tng tng ct v lng bi mt ct xuyn tm. Do tnh cht i xng, trn mi mt ct ngang ch c thnh phn ni lc l lc dc N, ng sut php c coi l phn b u (v b dy t b so vi ng knh). (hnh 12-2b) Lp tng hnh chiu cc lc theo phng y, ta c:
= = = x x2 22 2 0 0
FR FR2.N q .ds.sin d . sin d 2 .
g g
ng sut ko trong v lng l: 2 2
R
g
= (12.5) Nhn xt: ng sut trong v lng tng rt nhanh nu tng hay R. iu kin bn khi tnh v lng l: [ ] =
2 2
k
R
g
trong []k: ng sut cho php khi ko ca vt liu Ghi ch :Chu k T l khong thi gian thc hin mt dao ng (s). Tn
s f l s dao ng trong 1 giy (hertz). Tn s vng (tn s ring): s dao
ng trong 2 giy: 2 2 fT = =
y
x
t
Hnh 12-2
R
q (N/cm)
a)
d
ds dP=q.ds
N=.F N=.F
b)
Chng 12. Ti trng ng
12-4
III. DAO NG CA H N HI 1. Khi nim chung v dao ng Khi nghin cu v dao ng ca h n hi, trc tin ta cn c khi
nim v bc t do: bc t do ca mt h n hi khi dao ng l s thng s c lp xc nh v tr ca h. V d: hnh 12-3a, nu b qua
trng lng ca dm th h c 1 bc t do (ch cn bit tung y ca khi lng m xc nh v tr ca vt m). Nu k n trng lng ca dm h c v s bc t do v cn bit v s tung y xc nh mi im trn dm. Trc truyn mang hai puli (hnh
12-3b). Nu b qua trng lng ca trc 2 bc t do (ch cn bit hai gc xon ca hai puli ta s xc nh v tr ca h). Khi tnh phi chn s tnh,
da vo mc gn ng cho php gia s tnh v h thc ang xt. V d: nu khi lng m >> so vi khi lng ca dm lp s
tnh l khi lng m t trn dm n hi khng c khi lng h mt bc t do. Nu trng lng ca khi lng m khng ln so vi trng lng dm, ta phi ly s tnh l mt h c v s bc t do bc t do ca mt h xc nh theo s tnh chn, ngha l ph thuc vo s gn ng m ta chn khi lp s tnh. Dao ng ca h n hi c chia ra: Dao ng cng bc: dao ng ca h n hi di tc dng ca ngoi
lc bin i theo thi gian (lc kch thch). P(t) 0 Dao ng t do: dao ng khng c lc kch thch P(t)=0: Dao ng t do khng c lc cn: h s cn = 0; P(t) = 0 Dao ng t do c n lc cn ca mi trng: 0 ; P(t) = 0
Trng lng ca khi lng m c cn bng vi lc n hi ca dm tc ng ln khi lng.
m
y
Hnh 12.3
a)
2
1
b)
Chng 12. Ti trng ng
12-5
2. Dao ng ca h n hi mt bc t do a) Phng trnh vi phn biu din dao ng Dm mang khi lng m
(b qua trng lng dm). Lc kch thch P(t) bin i theo thi gian tc dng ti mt ct ngang c honh z. Tm chuyn v y(t) ca khi lng m theo thi gian t. Vn tc v gia tc ca khi
lng ny l:
2
2
dy d yv y(t) ; a y(t)dt dt
= = = =& && Chuyn v ca m do nhng lc sau y gy ra: Lc kch thch P(t), lc
cn ngc chiu chuyn ng v t l vi vn tc: Fc = -y& ; ( - h s cn), lc qun tnh: Fqt = - m y&& Gi l chuyn v gy ra do lc bng mt n v ti v tr m chuyn
v do lc P(t) gy ra l .P(t), chuyn v do lc cn gy ra l .Fc = - . y(t)& , chuyn v do lc qun tnh gy ra l -.m y(t)&& Chuyn v do cc lc tc dng vo h gy ra l [ ]y(t) P(t) y(t) my(t)= & && (12.6) Chia (12.6) cho m. v t: 2 m
= ; 2 1m. =
Do ta c: 2 P(t)y(t) 2 y(t) y(t) m+ + =&& & (12.7) y l phng trnh vi phn ca dao ng. H s biu din nh
hng ca lc cn ca mi trng n dao ng v < . b) Dao ng t do khng c lc cn Dao ng t do khng c lc cn: P(t) = 0, = 0. Phng trnh vi phn ca dao ng c dng: + =&& 2y(t) y(t) 0 (12.8) Nghim ca phng trnh ny c dng: y(t) = C1cost + C2sint Biu din C1 v C2 qua hai hng s tch phn mi l A v bng cch t: C1 = A sin ; C2 = A cos Ta c phng trnh dao ng t do: y(t) = A sin(t + ) (12.9) iu kin ban u t = 0 => y(0) = y0; 0y(0) y=& & xc nh C1 v C2
z
a
m
y(t)
z
Hnh 12.4
P(t)
Chng 12. Ti trng ng
12-6
Phng trnh (12-9) cho thy: Chuyn ng t do khng lc cn l mt dao ng iu ho c bin A
v chu k T = 2 . th dao ng hnh
sin nh trn hnh 12-5.
Tn s dao ng f = 1T 2
= . Tn s gc hay tn s dao ng
ring: = 2f ;
0
1 g g
m mg y = = = (Hert = 1/s) c) Dao ng t do c k n lc cn V P(t) = 0, 0, khi phng trnh vi phn ca dao ng l: + + =&& & 2y(t) 2 y(t) y(t) 0 (12.10) Vi iu kin hn ch < (lc cn khng qu ln), nghim c dng:
t1y(t) Ae sin( t )
= + (12.11) Dao ng l hm tt dn theo thi gian vi tn s gc:
2 21 = <
Chu k dao ng: = =
1 21
2
2 2 1T
1
Dng dao ng c biu din trn hnh 12.6, bin dao ng gim dn theo thi gian, bi vy ta gi l dao ng t do tt dn. Khi lc cn cng ln, tc l h s cng ln th s tt dn cng nhanh.
Sau mi chu k T1, bin dao ng gim vi t s:
1
1
tT
(t T )
e e conste
+ = = tc l gim theo cp s nhn Hnh 12.6
Chng 12. Ti trng ng
12-7
3. Dao ng cng bc - hin tng cng hung Dao ng cng bc: xt lc P(t) bin thin tun hon theo thi gian:
P(t) = Posint Lc cng bc bt k c th khai trin theo chui Fourier trng
hp ring m ta nghin cu khng lm gim tnh tng qut ca kt qu. Phng trnh vi phn dao ng c dng khng thun nht:
2 0Py(t) 2 y(t) y(t) sin tm
+ + = && & (12.12) Nghim tng qut ca phng trnh ny c dng: y(t) = y1(t) + y2(t) Nghim tng qut ca phng trnh vi phn thun nht l biu thc: y1 = e-t C sin(1t + 1) (12.13) Cn nghim ring y2(t) c dng: y2(t) = C1sint + C2cost Thay y2 vo (12.12), sau mt s bin i ta tm c: y2 = A1sin(t + ) (12.14)
vi k hiu 01 22 2 22 4
PA
41
= +
; ( )2 2
22 2 2 2
arcos4
= +
Nghim tng qut ca dao ng cng bc: y(t) = e-t C sin(1t + 1) + A1sin(t + ) (12.15) S hng th nht tt dn theo thi gian, sau mt thi gian ln h ch
cn li s hng th hai vi tn s ca lc cng bc , bin A1: y(t) = A1sin(t + ) = 022 2 2
2 4
sin( t )P
41
+ +
(12.16)
Lng P0 tng ng vi gi tr chuyn v gy ra bi mt lc tnh yt, c tr s bng bin lc cng bc v c phng theo phng dao ng:
y(t) = t t22 2 22 4
sin( t )y k (t)y
41
+ = +
(12.17)
trong k(t) l h s ng, hm ny t cc tr K khi sin(t + ) = 1. Chuyn v cc tr tng ng, k hiu bng y: y(t) = K. yt (12.18) K = +
22 2 2
2 4
1
41
(12.19)
Chng 12. Ti trng ng
12-8
C th gii bi ton ng bng cch gii bi ton tnh ri nhn vi h s ng k . ng sut c dng: = = t tk . ; k . (12.20) H s ng cc tr K cng ln th hiu ng ng cng ln. H s ny
ph thuc vo t s /. th quan h gia K v / ng vi cc gi tr khc nhau ca h s cn nht c trnh by trn hnh 12.7. tnh bn khi
ng sut thay i c th dng v theo (12.20). Nu trn h cn c ti trng tnh tc dng th tp l tng ng sut do ti trng tnh v ng sut ng , . + Hin tng cng hng: th K - (/) cho
thy: khi / 1, ngha l khi tn s lc cng bc trng vi tn s dao ng ring ca h y rt ln, c th bng v cng nu khng c lc cn. l hin tng cng hng.
Thc t tn ti min cng hng, nm trong khong 0,75 1,25 ; h s ng trong min ny t tr s kh ln. Trnh hin tng cng hng, cn cu to h sao cho tn s dao ng
ring ca h khng gn vi tn s ca lc cng bc, chng hn thay i khi lng ca h hoc thay i kt cu bng cch thm cc thit b gim chn nh l xo, cc tm m n hi. + Kt lun chung v tnh ton kt cu chu dao ng cng bc i vi h n hi, vt liu tun theo nh lut Hc, ta c th vit biu
thc (12.18) cho i lng nghin cu bt k: S = K.St (12.21) v S = S0 + S = S0 + K.St (12.22)
trong S - i lng nghin cu c th l chuyn v, ng sut, bin dng ca h, S0 - i lng tng ng trong bi ton tnh do tc ng ca trng lng m t sn trn h, St - i lng tng ng trong bi ton tnh do tc ng ca mt lc tnh, tr s bng bin ca lc cng bc v c phng theo phng dao ng, K - h s ng cc tr, tnh theo biu thc (12.19).
Hnh 12.7
Chng 12. Ti trng ng
12-9
V d 12.1: Mt mt trng lng 6kN t ti chnh gia dm n gin (hnh 12.8) c chiu di nhp 4,5m lm t thp I s 30, c tc quay ca trc n = 600 vng/ph. Trc c trng lng 50 N, c lch tm e = 0,5 cm. B qua lc cn, tnh ng sut php ln nht pht sinh trn tit din ca dm.
Bi gii
Tc gc ca trc quay: 2 n 2 .600
62,85rad / s60 60
= = = . Lc ly tm pht sinh khi trc quay lch tm:
2 2
0
1 1 50P me . 0,5.62,85 5038N
2 2 9,80= = =
Lc cng bc c dng: P(t) = P0 sint = 5,038 sin62,85 kN. Theo bng thp nh hnh Jx=7080 cm4; Wx=472 cm3; E=2,1.104 kN/cm2. vng ban u, do trng lng mt P t sn gy ra:
= = =3 3
0 4
P 6.(450)y 0,0766 cm
48EJ 48.2,1.10 .7080
l
Tn s dao ng ring ca dm: = = =0
g 980113 (1/s)
y 0,0766
H s ng, khi b qua lc cn:
K =
2 2
2 2 2 2 222
22
1 1 1131,448
113 62,8511
= = = =
Mmen un ln nht ti tit din chnh gia nhp bng:
M =M0+M=M0+K Mt= + = + =0 PP 6.4,5 5,038.4,5K 1,448 14,957 kNm4 4 4 4ll
ng sut php ln nht trn tit din:
2
max
M 1495,73,17kN / cm
W 472 = = =
A B
Hnh 12.8
l/2 l/2
P0 50N e
N0 30
Chng 12. Ti trng ng
12-10
IV. BI TON TI TRNG VA CHM 1. Va chm ng ca h mt bc t do Va chm: hin tng hai vt tc
dng vo nhau trong thi gian rt ngn. Cc gi thuyt sau: a) Khi chu va chm vt liu vn tun
theo nh lut Hc b- Mun n hi E ca vt liu khi
chu ti trng tnh v khi chu va chm l nh nhau.
Cc giai on va chm: a) Giai on th nht: trng lng Q ri va chm trng lng P: vn tc
v0 ca trng lng Q trc lc va chm b gim t ngt cho n lc c hai trng lng P v Q cng chuyn ng vi vn tc v. Theo nh lut bo ton
ng lng: 0 0Q Q P Q
v v v vg g Q P
+= = + b) Giai on th hai: c hai trng lng Q v P gn vo nhau v cng
chuyn ng vi vn tc v n lc c hai dng li do sc cn ca h n hi. on ng m Q v P va thc hin chnh l chuyn v y ln nht ti mt ct va chm. Trong giai on ny ng nng ca h l:
( )+ + = = = + +
2
2 20 0
1 Q P 1 Q P Q 1 QT . v T . v v
2 g 2 g Q P 2 g 1 P / Q
Khi P v Q cng di chuyn mt on y, th nng ca h: = (Q +P)y Nu gi U l th nng bin dng n hi ca h nhn c do va chm
th theo nh lut bo ton nng lng ta c: U = T + Th nng bin dng n hi c tnh nh sau: lc u trn dm c t
sn trng lng P, th nng bin dng n hi lc : 1 t1U P.y2
= trong : yt l chuyn v tnh ti mt ct va chm do P gy ra, yt = P. ( chuyn v tnh do lc bng mt n v gy ra)
2t
1y1U
2=
Khi va chm, chuyn v ton phn mt ct va chm l (yt + y). Theo cc gi thuyt trn, th nng bin dng n hi lc :
+= 2
t 2
(y y )1U
2
Nh vy th nng bin dng n hi do va chm l:
+= = = + = + 2 2 2 2
t t t 2 1
(y y ) y y y y y1 1U U U P.y
2 2 2 2
P
Q
P
Q
y
yt
H
Hnh 12.9
Chng 12. Ti trng ng
12-11
Do U = T + ( )+ = + + +2
2 0
y 1 QP.y v (Q P)y
2 2 g 1 P / Q
hay ( ) =+
22 0
Qvy 2 Qy 0
g 1 P / Q (12.23)
Gi t l chuyn v tnh ca h n hi ti mt ct va chm do trng lng Q c t mt cch tnh ln h gy ra th tng t nh trn ta c:
t = Q. tQ =
Th vo (12.23) ta c: = +
22 t 0 t
vy 2 y 0
Pg 1
Q
Ch ly nghim dng ca phng trnh: 2
2 t 0 t t
vy
Pg 1
Q
= + + + >0
Thay 20v 2gH= , ta c: tt
2Hy (1 1 )
P1
Q
= + + + (12.24)
H s ng k, tc l s ln ln hn ca chuyn v ng (do va chm) i vi chuyn v tnh do trng lng Q t mt cch tnh ln h:
= = tt
yk y k .y
y t
2Hk 1 1
P1
Q
= + + + (12.25)
Cc trng hp t bit: 1. Nu trn dm khng c khi lng P t sn th h s ng:
t
2Hk 1 1= + + (12.26)
2. Nu trng lng Q tc dng t ngt vo h, tc l: H = 0, th k = 2, tc l chuyn v ng, ng sut ng ln gp hai ln so vi bi ton tnh. ng sut php v tip do ti trng va chm: = k.t ; = k.t Nu trn h cn c ti trng tnh th ng sut ng v chuyn v ng: = (Q) + t(P); y = y(Q) + yt(P); Nhn xt: trong cng thc ca h s ng, ta thy nu chuyn v tnh yt
ln, tc l h c cng nh th h s ng k nh. Vy mun gim h s
Chng 12. Ti trng ng
12-12
ng ta phi gim cng ca h hay t ti mt ct va chm nhng b phn c cng nh nh l xo, ... tng yt. Khi xc nh h s ng k ta b qua trng lng bn thn ca h
n hi. Ngi ta chng minh c rng nu k n trng lng bn thn ca h th h s ng cng khng thay i nhiu. Do trong khi tnh vi ti trng va chm, ta khng xt n trng lng bn thn ca h. 2. Va chm ngang ca h mt bc t do Va chm ngang nh hnh 12.10. Qu trnh va chm
vn thc hin qua hai giai on nh trong va chm ng. V cc khi lng u di chuyn theo phng ngang nn th nng = 0. vy theo nh lut bo ton nng lng:
T = U
ng nng T: 20
1 QT v
2 Pg 1
Q
= +
Th nng bin dng n hi m h nhn c sau va chm c tnh nh sau: tuy c trng lng P t trc trn dm, nhng P khng lm dm bin dng ngang nn: U1 = 0. Khi va chm, chuyn v ca mt ct va chm l y nn lc th nng bin dng n hi:
= 2
2
y1U
2 2
2 2 20 0
y1 Q 1 Qv y v
2 2P Pg 1 g 1
Q Q
= = + + (12.27)
Nu gi yt l chuyn v tnh theo phng ngang mt ct va chm do lc c gi tr bng trng lng va chm Q tc dng tnh ln phng ngang:
t = Q. =
tQ
Do ta c th vit biu thc (12.27) li nh sau: = +
2 2t 0y vP
g 1Q
Gi tr y ch ly du dng, do y = k.t
Vi = +
20
t
vk
Pg 1
Q
(12.28)
PQ
Hnh 12.10
Chng 12. Ti trng ng
12-13
V d 12.2: Xc nh ng sut php ln nht trn tit din mt ct chu va chm theo phng thng ng cho trn hnh 12.11. B qua trng lng ca ct. Cho bit Q = 600 N; H = 6cm; E = 103 kN/cm2.
Gii Chuyn v tnh bng bin dng di ca ct do
trng lng Q t tnh trn ct l:
yt = t = l = + = 31 21 2
Q. Q.3,4.10 cm
EF EF
l l
H s ng: t
2Hk 1 1
P1
Q
= + + + =
= t
2H1 1+ + = 3
2.61 1 60,41
3,4.10+ + =
ng sut pht ln nht trn tit din:
= = = = 2 t 2
Q 0,6k . k . 60,41. 1,82kN / cm
F 20
V d 12.3: Xc nh h s ng ca dm thp ch I s 14 (hnh 12.12) chu va chm bi vt c trng lng 100 N chuyn ng theo phng ngang vi vn tc v0 = 20km/h khi khng k v khi c k n trng lng ca dm.
Gii Thp ch I s 14 ta c cc c trng:
trng lng trn 1m di l 137N, Jx = 572 cm4, E = 2,1.104 kN/cm2. Chuyn v tnh:
3 32
t 4x
Q 0,1.400y 1,1.10 cm
48EJ 48.2,1.10 .572= = =l
- Khi khng k n trng lng bn thn
2 20
2t
v 555,5k 169
gy 980.1,1.10= = =
Khi k n trng lng bn thn, ta thu gn trng lng v tit din va chm chnh gia dm vi h s thu gn l 17/35 v c trng lng thu gn l P = (17/35).137.4 = 266 N
2 20
2
t
v 555,5k 88
266P980. 1 .1,1.10g 1 y
100Q
= = = ++
Nh th trng lng bn thn lm gim nh hng ca va chm. Vic khng k n trng lng bn thn khin php tnh thin v an ton.
6 cm
80 cm
60 cmF2=20cm2
F1=30cm2
Q
Hnh 12.11
Q
Hnh 12.12
N0 14
v0
Qyt
l=4m