co hoc ket cau

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


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)


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


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


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


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


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

= +


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


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.


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


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)


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


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

co hoc ket cau

Documents