giao trinh co ket cau - cĐgtvt ii

7/28/2019 Giao trinh Co Ket Cau - CGTVT II

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Trng Cao ng Giao thng Vn Ti II Bi ging : Chc kt cu

Bin son: Th.S Nguyn Ph Th Trang 1

CHNG 1 : MU

1.1Gii thiu, nhim v, i tng mn hc.

Chc kt cu l mn khoa hc thc nghim, nghin cu cch tnh bn, cng v n nh ca cng trnh.

bn l m bo cho cng trnh c kh nng chu ng c tc dng cacc ti trng v cc nguyn nhn khc m khng b ph hoi.

cng nhm m bo cho cng trnh khng c chuyn v v rung ng lnngn cn s lm vic bnh thng ca cng trnh.

n nh nhm tm hiu kh nng bo ton v tr v hnh dng ban u trong ccdng cn bng ca trng thi bin dng.

Vy nhim v ch yu ca Chc kt cu l xc nh ni lc trong cng trnh.T s tnh ton xc nh kch thc, cu to ca cng trnh.

i tng mn hc : Bao gm nhng h thanh lin kt vi nhau.

1.2 S cng trnh , s tnh ton, cc gi thuyt tnh ton.Trong s cng trnh cc yu t ch yu c gi li, cn cc yu t ph th

c th b qua. Vy s cng trnh l hnh nh n gin ha ca cng trnh c th,trong ch k n cc s liu cbn xc nh phm cht ca cng trnh.

tnh ton cng trnh ta cn phi a cng trnh thc t v s tnh

Trong s cng trnh cc thanh c thay bng ng trc, mt ct ngangc thay bng cc i lng c trng nh din tch F, mmen qun tnh J,...Vys tnh l s cng trnh c n gin ha.

Th d hnh v 1.1 v 1.2

P1

P2

1P P

2

1 2PP

Cc gi thuyt tnh ton :

Gi thuyt 1 : Vt liu lm vic n hi v tun theo nh lut Huck.(Giabin dng v ni lc c quan h bc nht)


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Gi thuyt 2 : Bin dng v chuyn v trong cng trnh rt b (Di tc dngca ti trng hnh dng ca cng trnh thay i rt t)

Vi nhng cng trnh cho php s dng c c hai gi thuyt trn th cngtrnh cng cho php p dng nguyn l cng tc dng.

1.3 Cc nguyn nhngy ra ni lc.

C nhiu nguyn nhn gy ra ni lc trong cc cng trnh, cc nguyn nhnthng gp l ti trng, s thay i ca nhit v chuyn v ca cc lin kt.Titrng gy ra ni lc, bin dng v chuyn v trong tt c cc cng trnh, cn sthay i nhit v chuyn v ca cc lin kt ch gy ra ni lc trong h siu tnh,gy ra chuyn v trong c h tnh nh v h siu tnh. mn hc ny ta ch xt tcdng ca ti trng.

Phn loi ti trng theo v tr tc dng :

- Ti trng c nh l nhng ti trng c v tr khng thay i, nh trng lngbn thn, trng lng ca cc thit b c nh t trn cng trng.v.v..

- Ti trng di ng l nhng ti trng c v tr lun lun thay i, nh ontu, t v.v...

Phn loi ti trng theo tnh cht tc dng :

- Ti trng tnh l ti trng c tr s tng t t n gi tr cui cng ca n.

- Ti trng ng l nhng ti trng khi tc dng ln cng trnh c gy ra lcqun tnh, v d ti trng ca ba trn cc.v.v...


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CHNG 2 : CU TO H PHNG

2.1 Khi nim cbn.

1. Kt cu bt bin hnh, bin hnh v bin hnh tc thi

Kt cu bt bin hnh l kt cu khi chu tc dng ca mi loi ti trng, dng

hnh hc ban u ca n khng b thay i (b qua thay i do bin dng n hi)

P

Th d hnh 2.1 : Kt cu bin hnh l kt cu khi chu tc dng ca ti trng,dng hnh hc ban u ca n b thay i hu hn mc d ta vn xem tng cu kincu n l cng tuyt i.

Th d hnh 2.2 : Ngoi hai loi kt cu bt bin hnh v bin hnh trn cnmt loi kt cu khi chu ti trng dng hnh hc ban u ca n b thay i mtlng v cng nh mc d ta xem tng cu kin cu n l tuyt i cng. Loi nyc gi l kt cu bin hnh tc thi.

Th d hnh 2.3 : Kt cu bin hnh tc thi.

Tm li khi xy dng cng trnh nht thit phi cu to n thnh h bt binhnh, khng c cu to lm cho cng trnh b bin hnh hoc bin hnh tc thi.

2.Ming cng(tm cng) : Tm cng c th l mt thanh thng, thanh cong,thanh gp khc hoc mt h thanh bt bin hnh nh hnh v 2.4a,b

a) b) c)

Ta qui c biu din cc tm cng nh hnh 2.4 c

3. Bc tdo ca im, ca tm cng

Bc t do ca h l s thng s hnh hc c lp, xc nh v tr ca h ivi mt h khc c xem l c nh.

Trong mt phng : mt im c hai bc t do cn mt tm cng c ba bc tdo. T khi nim trn ta nhn thy mun im hay tm cng c nh (khng chuynng) ta phi tm cch kh bc t do ca n v ch khi no kh ht c tt c cc

bc t do ca im hay tm cng th im hay tm cng y mi c nh c.

kh ht cc bc t do ca im hay tm cng trong k thut ngi ta dng


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lin kt.

2.2 Cc loi lin kt trong h phng

1. Lin kt thanh, lin kt khp v lin kt hn.

a. Lin kt thanh( lin kt loi 1)

Cu to lin kt ny l mt thanh thng c khp l tng hai u lin ktvi tm cng. (hnh 2.5a)

a) b) c)

Lin kt thanh ch kh c 1 bc t do v trong thanh ch pht sinh mt phn

lc lin kt c phng dc theo trc thanh (hnh 2.5 b)Ch : V bn cht c th xem thanh cong hay tm cng c hai u khp nh

mt lin kt thanh c phng qua hai khp y (hnh 2.5c)

b. Lin kt khp

Khp ni 2 tm cng c gi l khp n (hnh 2.6 a), cn khp ni t 3 tmcng trln c gi l khp bi hay khp phc tp (hnh 2.6 b,c)

d) e)

a) b) c)

Khp n kh c 2 bc t do nn n tng ng 2 lin kt thanh.Ngc li 2 lin kt thanh khng song song cng tng ng vi 1 khp n t giao im ca 2 lin kt ( khp ny l khp gi) xem hnh 2.6 d,e.

Nu khp bi ni D tm cng th tr 1 tm cng xem l c nh, i vi mitm cn li khp bi kh c 2 bc t do nn i vi D-1 tm cng cn li khp skh c 2(D-1) bc t do. Ngha l nu khp bi ni D tm cng th n s tngng vi K khp n, trong K = D-1

c. Lin kt hn


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Lin kt hn dng ni 2 tm cng A v B nh hnh v 2.7a c gi l linkt hn n

Cn lin kt ni trn hai tm cng nh hnh 2.7c,d th gi l lin kt hn bihay lin kt hn phc tp.

d)

a)

b)

c)

Nh vy lin kt hn n kh c 3 bc t do. Trong lin kt hn c 3 thnh

phn : 2 phn lc c phng cho trc v 1 phn lc mmen.V kh c 3 bc t do nn 1 lin kt hn n tng ng vi 3 lin kt

thanh hoc 1 lin kt thanh v mt lin kt khp.

Tng t lin kt bi, lin kt hn bi c s bc t do kh c tng ngvi K =D-1 ln s bc t do ca lin kt hn n.

2. Cc lin kt gi ni t.

ni cc cu kin vi t ngi ta dng cc loi lin kt gi l gi ding, gi c nh, ngm trt, ngm cng...S lin kt thanh tng ng vi miloi gi theo bng 2.1

1 2 3

2.3 Cch sdng cc loi lin kt ni cc tm cng.

Khi ni cc tm cng thnh h bt bin hnh ta cn phi xem xt 2 iu kinsau :

- iu kin cn : s lng lin kt phi s dng.

- iu kin : Cch sp xp b tr v s dng cc lin kt .


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1. Ni 2 tm cng: Xem 1 ming cng l c nh, ta cn phi lin kt mingcng cn li. M 1 ming cng tn ti 3 bc t do.

Vy iu kin cn ni 2 tm cng thnh h khng bin hnh l phi dng slin kt kh c 3 bc t do. Ngha l ta phi dng 3 lin kt thanh hoc 1 linkt thanh v 1 lin kt khp hoc 1 lin kt hn nh hnh 2.8 a,b,c

a) c)

iu kin : Lin kt hn th lun lun c 1 h bt bin hnh

Dng 3 lin kt thanh hoc 1 lin kt thanh v 1 lin kt khp th khng bao gicng c 1 h bt bin hnh. V

Khi dng 3 lin kt thanh ni 2 tm cng ta s gp cc trng hp sau :- 3 thanh c phng bt k hnh 2.8a

- 3 thanh c phng ng qui hnh 2.9a

- 3 thanh c phng song song hnh 2.9b,c

a) b) c)

Nu 3 thanh ng qui th tm cng B quay quanh I, lc ny h bin hnh tcthi. Nu 3 lin kt song song th khng c thanh no ngn cn B chuyn ng tnhtin theo phng vung gc vi cc thanh. Khi cc thanh di bng nhau hnh 2.9bth h bin hnh v tm cng B b dch chuyn ln hu hn. khi 3 thanh song songkhng di bng nhau hnh 2.9c th h bin hnh tc thi v B ch b chuyn dch vcng b.

Nu 3 thanh c phng bt k hnh 2.8a h s bt bin hnh.

Khi dng 1 lin kt thanh v 1 lin kt khp ta ch gp: trong 2 trng hp sau.

- Thanh c phng khng qua khp hnh 2.8b : h s khng bin hnh

- Thanh c phng qua khp hnh 2.10:lc ny h s bin hnh tc thi

Tm li: iu kin cn v ni 2 tm cng thnh hbt bin hnh l: phi dng 1 lin kt hn, hoc

- Dng 3 lin kt thanh khng c song song hocng qui

- Dng 1 lin kt thanh v 1 lin kt khp th thanh


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phi c phng khng c i qua khp

2. Ni 3 tm cng:

Nu ta xem 1 tm cng c nh cn li 2 tm t do chng ta cn phi lin kt 2tm cng cn li do ta phi kh c 6 bc t do. kh c 6 bc t do ta cnhiu cch.

a) b) c) d)

e) f) g)

1

2 3

h)

1

23

i) k)

- Dng 6 lin kt thanh hnh 2.11 a,b,c

- Dng 3 lin kt khp hnh 2.11 d

- Dng 2 lin kt hn hnh 2.11 e

- Dng 4 lin kt thanh, 1 lin kt khp hnh 2.11 f,g

- Dng 2 lin kt thanh, 2 lin kt khp hnh 2.11 h,i

- Dng 1 lin kt thanh, 1 lin kt khp v 1 lin kt hn hnh 2.11 k,...

Khi s dng cc lin kt k trn, nu s sp xp b tr cc lin kt khng hp lth h vn c th b bin hnh hoc bin hnh tc thi do ta cn xt iu kin .

- Trng hp 2 trong 3 tm cng ni vi nhau thnh 1 tm cng nh hnh2.11 b,c,e,f,i,k lc ny ni 3 tm cng trv bi ton ni 2 tm cng nh ni trn.


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I

a)

I

b) c)

- Trng hp tng cp 2 tmcng ni vi nhau bng 1 khp ththoc gi th ta ch gp 1 trong 2 trnghp hoc khp khng thng hngnh cc hnh 2.11 a,d,g,h hoc 3 khp thng hng nh cc hnh 2.12 a,b.

Nu 3 khp tht hoc gi thng hngth h b bin hnh tc thi.

Vy trng hp tng cp 2 tm cng ni vi nhau bng 1 khp tht hoc gi th

3 khp tht hoc gi khng c thng hng.3. Ni im vi tm cng- B i.

Mt im tn ti 2 bc t do, chng ta cn phi kh 2 bc t do ny. vy iukin cn ti thiu phi c 2 lin kt thanh.

iu kin 2 lin kt thanh ny khng c thng hng hnh 2.13a, hai thanhlin kt khng c song song nh hnh 2.13 b,c.

Hai lin kt thanh khng thng hng ni im vi tm cng l mt b i.

Khi chng ta thm hoc bt 1 b i vo 1 h th tnh cht ng hc ca hkhng thay i. nu h bt bin hnh th sau khi thm (hay bt) b i h vn bt

bin hnh. Ngc li nu h bin hnh th sau khi thm (hay bt) b i h vn binhnh.

Da vo tnh cht ny ta c th phn tch cu to hnh hc ca h.

- T mt tm cng ban u tathm ln lt cc b i c 1 tmcng hnh 2.14

- T h ban u ta ln lt loi bcc b i cho n cui cng l h bt

bin hnh nu ban u h l bt binhnh v l h bin hnh nu ban u l h

bin hnh2.4 Cch phn tch ng hc ca kt cu.

1. Bc tdo ca kt cu - iu kin cn:

a. H kt cu khng ni vi t

Kt cu c D tm cng th 1 tm cng xem nh c nh, (D-1) tm cng cn lic 3(D-1) bc t do.

Kt cu c T lin kt thanh, K khp n (k c s khp bi i ra khp

12

3

a)

32

a)

1


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n), H lin kt hn n (k c s lin kt hn bi i ra lin kt hn n) thtng s bc t do kh c l T+2K+3H

Do iu kin cn: n = T+2K+3H 3(D-1) 0 (2.1)

b. H ni vi t.

Xem tri t l 1 tm cng c nh th ta cn phi lin kt D tm cng cn li.Do ta cn 1 t hp lin kt kh c 3D bc t do v nu cng dng T lin ktthanh, k lin kt khp n, H lin kt hn n ni D cu kin vi nhau v dngC0 lin kt tng ng lin kt thanh ni cc cu kin vi t th iu kin cn kt cu c D cu kin ni t thnh mt kt cu bt bin hnh l :

T + 2K + 3H + C0 3D 0 (2.2)

S lin kt thanh tng ng vi mi loi gi C0 xem bng 2.1

1. Cch phn tchng hc kt cu - iu kin .

Nu kt cu ch tha mn iu kin cn nh h thc 2.1 v 2.2 trn m cchsp xp cc lin kt khng hp l th nhng lin kt dng trong kt cu vn khngc kh nng kh ht c bc t do ca kt cu, do kt cu vn c th bin hnh.Vy iu kin kt cu bt bin hnh l cc lin kt phi c sp xp hp l.

phn tch ng hc kt cu xem cc lin kt b tr c hp l hay khng tadng khi nim b i hoc cch ni 2,3 tm cng ri p dng iu kin cn v xt.

Th d 2.1: Phn tch ng hc kt cu chonh hnh 2.15

Gii:

Xt iu kin cn: Xem mi thanh l mt cu

kin th kt cu cho trn hnh 2.15 c D cu kinni vi nhau bng cc khp n v phc tp (khp 1,4 v 7 lin kt n, khp 2,6 l khp phctp c phc tp l 2, khp 3,5 l khp phc tpc phc tp l 3).

Theo hnh v ta c D =10, K = 13 , T = 0, H =0. Theo 2.1 ta c

T+2K+3H 3(D-1) 0 => 0 + 2.13 + 0 3(10-1) = -1< 0 Khng tha mn iu kin cn.Khng cn xt iu kin

Kt lun : H bin hnh v thiu lin kt.

Th d 2.2 : Phn tch ng hc kt cu chohnh 2.16

Gii :

Xt iu kin cn : H cho l kt cu ni t ta dng h thc 2.2 xt iukin cn.

C nhiu cch quan nim khc nhau p dng h thc 2.2


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- Nu cho rng cc thanh gy khc l tm cng th h thc 2.2 ta c :D = 4 , K =3, T= 0, H= 0 v C0 = 6 (theo bng 2.1)

Ta c : T + 2K + 3H + C0 3D 0

0 + 2.3 + 3.0 + 6 3.4 = 0. Vy tha mn iu kin cn

- Nu cho rng mi thanh thng l 1 tm cng th : D = 8, T = 0, K =3,H = 4, v C0 = 6 nh vy ta c kt qu : 0 + 2.3 + 3.4 + 6 - 3.8 = 0 Vy cng thamn iu kin cn.

Cng c th dng h thc 2.1 gii nu ta xem tri t cng l 1 tm cng thlc ny D = 5, T = 0, K = 6, H = 0 vy :T+2K+3H 3(D-1) 0

0 + 2.6 + 3.0 3(5-1) = 0 Tha mn iu kin cn.

iu kin : Cng c nhiu cch xt.

- Nu xem tri t l 1 tm cng, thanh gy AD v DB ni D vi t l mtb i th t v b i ADB l mt tm cng, tng t DE v EC l 3 tm cng nivi nhau bng 3 khp D,E,C khng thng hng nn h bt bin hnh.

- Nu xem tng cp 2 thanh DE, EC v AD,DB l nhng b i. p dngtch cht loi b b i DEC ri ADB, cui cng cn li tri t l 1 tm cng. Ktlun h bt bin hnh.

Th d 2.3 Phn tch ng hc kt cu cho hnh 2.1........

Xt iu kin cn : H cho l kt cu ni t ta dng h thc 2.2 xt iukin cn.

- Nu cho rng cc thanh thng l tm cng th h thc 2.2 ta c : D = 6, K=3, T= 0, H= 2 v C0 = 6 (theo bng 2.1)

Ta c : T + 2K + 3H + C0 3D 00 + 2.3 + 3.2 + 6 3.6 = 0. Vy tha mn iu kin cn

iu kin :

- Nu xem tri t l 1 tm cng , thanh EFG l ming cng 2, xem thanhADE v GHC l lin kt thanh. Vy h l 2 tm cng lin kt vi nhau bng 3 linkt thanh ng qui ti i. vy h bin hnh tc thi. Nu 3 lin kt thanh khng ngqui th h bt bin hnh.


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CHNG 3 : NG NH HNG

3.1 Khi nim v ti trng di ng, ti trng tiu chun.

Ti trng ng khi tc dng gy ra lc qun tnh v do thay i v tr nngi tr ca yu t xt cng thay i theo. Phng php gii quyt nhsau:

-Xem ti trng ng nhti trng tnh di dng bng cnh nhn gi tr cati trng vi h s xung kch (1 + ). (cha nghin cu)

-Sdi ch ca ti trng cho thy gi tr ca yu t xt s thay i theo vtr ca n. Qu trnh di chuyn ti trng s c ch cc i, gi l v tr bt linht, ta dng li tnh. tm v tr bt li ca ti trng t c cgi tr cci ca yu tang xt ta dng phng php ng nh hng.

Ti trng di ng trn cng trng l on tu, t,... nhng ti trngny rt phc tp, do khi thit k ly on ti trng tiu chun no tnhton l do nhim v thit k quy nh.

Hnh 3.1 gii thiu mt son ti trng tiu chun trong quy trnh thit k cu

hin ang dng nc ta.

60kN

120kN

120kN

6m 1,610m

60kN

120kN

120kN

1,66m

60kN

120kN

120kN

6m 1,610m 10m 10m

4m8m

30kN

70kN

4m8m

35kN

95kN

4m 4m

30kN

70kN

8m

30kN

70kN

4m

1,2m

200kN

1,2m1,2m

200kN

200kN

200kN

0,8m

2,7m

0,8m

5m2,6m

0,7m 0,7m

q=60

l1,5mx5

Z Z Z Z Z 0,36Z

a, on t tiu chun H.30;


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z

l

z=l/2

z=l

Az

10,5

b, on t tiu chun H.10. Vi on t tiu chun H.8 v H.13 th khongcch gia cc trc xe cng nh khong cch gia cc xe hon ton nh trn, cn trs ti trng bng ti trng ca on t H.10 nhn vi 0,8 cho on H.8, nhn vi1,3 cho on H.13;

c, Xe bnh tiu chun XB 80 (ch c mt xe);

d, Xe xch tiu chun X 60 (ch c mt xe);e, Tu ha tiu chun T Z, trong Z l ti trng trc ca u my. Th d T

10 th mi trc ca u my ;l 100 kN (10T), toa xe l 0,36.100 = 36 kN/m

3.2 ng nh hng

1. Khi nim Cch v:

a.Khi nim:

Theo nguyn l cng tc dng gi tr ca mt yu t no do nhiu ti trngng thi tc dng ln kt cu sinh ra bng tng cc gi tr ca yu t do tng titrng tc dng ring r sinh ra. Mt khc gi tr ca ti trng tng ln hoc gim i

bao nhiu ln th gi tr cu yu t do ti trng sinh ra cng ng ln hoc gim iby nhiu ln. Do tnh tc dng ca ti trng di dng u tin ta ch xt vi 1ti trng tp trung c tr s bng n v, c phng chiu thng ng hng xungdi dng trn kt cu.

V tr ca ti trng P = 1 c xc nh bngmt ta z, nh hnh 3.2 nu ly A l gc to th

Khi P =1 ti A (z = 0) c VA = 1

Khi P =1 gia dm (z = l/2) c VA = 0,5

Khi P =1 gia ti gi B (z = l) c VA = 0Lc ny ta vc th biu din mi quan

h gia gi tr ca yu t xt l VA vi v tr titrng (ta z) th ny c c gi lng nh hng ca yu t xt.

Vy ng nh hng ca yu t no l th biu din quy lut bin i gitr ca yu t xut hin ti mt v tr xc nh trn cng trnh (Phn lc gi,mmen un, lc ct... mt ct no ) theo v tr ca ti trng n v di ng trncng trnh.

Tung ng nh hng ca yu t no (phn lc, ni lc...) l gi tr cayu t khi ti trng P= 1 v tr tng ng vi tung sinh ra.

Th nguyn ca tung ng nh hng l t s gia th nguyn ca yutang xt vi th nguyn ca lc P, nh vy tung ng nh hng phn lc,lc ct l i lng khng th nguyn, cn tung ng nh hng mmen c thnguyn chiu di, thng l mt.

b.Cch vngnh hng.

C nhiu cch vng nh hng, y ta ch xt cch vng nh


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ng theo phng php tnh.

-Chn h trc c trc z song song vi trc dm biu th v tr ca ti trng nv, trc vung gc vi trc z biu th gi tr ca yu t xt. Gc ta thng lyng ng vi gi tri ca mi dm.

-Vit biu thc quan h gia gi tr ca yu t xt vi v tr ca ti trng n v

z ta c S = f(z) gi l phng trnh ng nh hng.-n c vo biu thc S = f(z) vng nh hng ca yu t S, sau khi v

cn nh du (+), (-), ghi tung nhng v tr c bit v ghi tn ng nhng.

2.ngnh hng dmn gin.

a.ngnh hng phn lc (VA, VB, HA =0).

Chn gc ta O tng ng vi gi tri ca dm, trc z hng sang phi,trc V biu th gi tr ca phn lc hng ln trn.

- ng nh hng VA :

mB = - l.VA + P(l-z) = 0, m P = 1 =>l

zlVA

=

M VA = f(z) l bc nht ca z nn ng nh hng VA l ng thng nnch cn xc nh hai tung .

Khi z = 0 c VA = 1

Khi z = l c VA = 0

Ni hai m (0,1) v (l,0) ta cng nh hng VA nh hnh 3.3b

- ng nh hng VB :mA = l.VB P.z = 0, m P =

1 =>l

zVB =

Khi z = 0 c VB = 0Khi z = l c VB = 1

Ni hai m (0,0) v (1,1) ta cng nh hng VB nh hnh 3.3c.

Tng nh hng VA v VB suy ra

cch v nhanh dng nh hng phn lcca dm n gin nh sau: khi vngnh hng phn lc ca gi no trnng chun v tr tngng vi gi dng tung bng +1, ni nh tung ny vi m c tung bng khng tngng vi gi cn li bngng thng.

b. ng nh hng ni lc.(M,Q,N=0)

la b

dah Vab)

dah Vb

1

1

dah Mc

a.bl

a

b

lb

al

1

1

dah Qc

c)

d)

e)

z


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vng nh hng ni lc (mmen un, lc ct) ca mt mt ct ta tngng ct dm ti mt ct . Mi phn dm u cn bng di tc dng ca ngoilc v ni lc tc dng ln phn .

- Cho P = 1 di chuyn phn dm bn tri (0 z a), xt s cn bng caphn dm bn phi (phn c t lc hn), t vit c phng trnh cho nhnh tri.

MCph =0 -MC + b.VB = 0 =>MC = b.l

z

Y=0 QC = -VB = -l

z

Khi z = 0 c MC= 0; QC = 0;

Khi z = a c MC=l

ba.; QC =

l

a

Vy nhnh tri ca ng nh hng mmen un ca mt ct C l on thng

i qua hai im (0, 0) v (a, lba.

). Trn ng chun v tr di mt ct C (z = a)

dng tung bngl

ba., ni nh tung ny vi m c tung 0 di gi A (z

= 0) c nhnh tri ca ng nh hng (hnh 3.3d), nu ko di nhnh ny nv tr tng ng vi gi B ta c tung ng bng b (khong cch t gi B nmt ct C)

Ti v tr tng ng vi mt ct C trn dng chun ta dng tung bng

l

a , ni nh tung ny vi n c tung bng 0 ng vi gi tri ta c nhnh

ca ng nh hng QC, nu ko di nhnh ny n v tr tng ng vi gi phi ta

ng tung bng -1. (hnh 3.3 e)- Khi P = 1 di chuyn phn dm bn phi (a z l), xt s cn bng ca

phn dm bn tri.

MCtr=0 MC - a.VA = 0 => MC = a.VA = a.l

zl

Y=0 QC - VA =0 => QC = VA =l

zl

Khi z = a c MC =l

ba.; QC =

l

b;

Khi z = l c MC = 0; QC = 0

Vy nhnh phi ca ng nh hng mmen un ca mt ct C l on

thng i qua hai m (a,l

ba.) v (l, 0); Cn c vo cc gi tr tnh ta vc

nhnh phi ca ng nh hng QC (hnh 3.3e), nu ko di nhnh ny n v trng ng vi gi tri ta c tung bng +1.

Tng nh hng ca ni lc ta suy ra cch v nhanh ng nh hngca ni lc nh sau:


15/69



- ngnh hng ca mmen un: Trn dng chun ti nhng v tr tngng vi cc gi ta dng cc tung bng khong cch tgi n mt ct, ni nhcc tung ny vi cc m c tung bng 0 trn ng chun gi bn kiabngng thng, ngnh hng mmen un l tam gic c nh v tr dimt ct v tung ti nh bng tch cc khong cch tmt ctn hai gi chiacho chiu di dm.

- ngnh hng ca lc ct: Trn dng chun ti nhng v tr tngngvi gi tri v phi ta ln lt dng cc tung bng 1 v -1, ni cc nh tung ny vi cc m c tung bng 0 tngng vi cc gi phi v tri bngng thng, t mt ct hng dng, ngnh hng lc ct l hai tam gicvung c nh nm tngng vi cc gi.

3.ngnh hng dm mt tha

Gi chiu di mt tha bn tri l l1, mt tha bn phi l l2, im gc l gitri, vy z bin i (- l1 z l + l2)

a. ngnh hng phn lc.

- Phn lc gi A: MB = VA.l + p(l - z) = 0 VA =l

zl

- Phn lc gi B: MA = 0 VB =l

z

Thay cc gi tr z

Khi z = - l1 c VA =l

zl=

l

l

l

ll 11 1 +=+

; VB =l

l1

Khi z = 0 c VA = 1; VB = 0

Khi z = l c VA = 0; VB = 1

Khi z = l + l2 c VA =l

zl=

l

l2 ; VB =l

l21+

Ta c ng nh hng ca VA, VB hnh 3.5

* Cch v nhanh : ngnh hng phn lc dm mt tha ta ch cn vngnh hng ca dm n gin c chiu di bng khong cch 2 gi ( l ), sau ko di on mt tha.

b. ngnh hng ni lc (mmen un v lc ct) : Chia mt ct lm 2loi: mt ct trong nhp (gia 2 gi) v mt ct ngoi nhp (ngoi 2 gi)

- Mt ct trong nhp ti C

+ Khi P = 1 di ng pha tri mt ct C (-l1[ z [ a). Xt cn bng phn phi ta

c MC = b.VB = b.l

zv QC = - VB =

l

z

Khi z = -l1 MC = - b.l

l1 ; QC =

l

l1

Khi z = 0 MC = 0 ; QC = 0


16/69



Khi z = a MC =l

ab; QC =

l

a

Vng nh hng nhnh tri.

la.b

blal

1

1

l1h1

l2h2

EBD A

l

C

a b

dah Vab)

a)

c)

d)

e)

g)

h)

i)

k)

m)

n)

p)

q)

r)

l)

dah Vb

dah Mc

dah Qc

dah Md

dah Qd

dah Me

dah Qe

dah Ma

dah Mb

dah Qph

B

trdah QA

trdah QB

Adah Qph

1+l1l

l1l

l1b.l

l1l

l1l

l1l

1h

1

1l

1

l2l

l2l

1+ h2

1

l2

1

l2l

l

l2

l

l2

a.l

l2

+ Khi P = 1 di ng pha phi MC (a [ z [ l + l2). Xt cn bng pha tri: MC

= a.VA = a.l

zl v QC = VA =l

zl

Khi z = a th MC =l

ab; QC =

l

b

Khi z = l th MC = 0 ; QC = 0

Khi z = l + l2 th MC = -a.l

l2 ; QC =

l

l2

Vng nh hng nhnh phi

Tcch vMC, QC suy ra cch vnhanh ngnh hng mmen, lc ctnhsau:

+ u tin vngnh hng ca ni lc ca mt ct tngng trn dmn gin c chiu di bng khong cch 2 gi, sau ko di ng nh hngcho phn mt tha.

+ p dng v nhanh ng nhng QA

tr, QAph, QB

tr, QBph

- ng nh hng mt ct

l

zz

l1

h1z z

h2

l2


17/69



ngoi nhp (ngoi 2 gi -on mt tha). Chn gc ta trng vi mt ct D vchiu dng trc z hng ra u mt tha bn tri (hnh 3.4).

+ Khi P = 1 di ng trn phn tri mt ct D (0 [ z [ h1), xt cn bng phnphi (t lc hn): MD = 0 MD = -z.P = -z v QD = -1 =Const

Khi z = 0 th MD = 0

Khi z = h1 th MD = -h1

+ Khi P = 1 di ng trn phn phi mt ct D (h1 [ z [ l + l2), xt cn bngphn tri (t lc hn): MD = 0 MD = -z.P = -z v QC = 0

Khi z = 0 th MD = 0

Khi z = h2 th MD = -h2

T ta c cch v nhanh ng nh hng mmen v lc ct ngoi nhp:

-ngnh hng mmen un: Trn ng chun tng ng vi u mttha (n mt tha c cha mt ct) dng tung -h (vi h l khong cch tmt

tha ti mt ctang xt). Sau , ni nh tung ny vi m c tung bng 0ngng vi mt ct.

-ngnh hng lc ct: Trn ng chun v tr tngng vi mt ctdng tung -1 nu mt ctn mt tha bn tri, +1 nu mt ctn mttha bn phi. Tnh tung ny kng song songng chun cho n mttha n cha mt ct. Nhnh cn li ca ng nh hng trng vi ngchun.

4.ngnh hng ca dm tnh nh nhiu nhp

a. Khi nim vdm tnh nh nhiu nhp: L h gm nhiu dm ghp li vinhau bng cc khp v t trn nhiu gi ta sao cho h l bt bin hnh v khng

c lin kt tha. Th d cc dm hnh 3.6

a)

b)

c)

m bo cho h bt bin hnh th trong mt nhp khng c c qu haikhp nm trn mt ng thng.

Trong dm nhiu nhp u mt ca dm ny l gi ta ca dm khc, dm c gi l dm chnh, cn cc dm gi ln n l dm phi vi n.

Trn hnh 3.6a c dm D1, D3, D5 l dm chnh, D2 l dm phi vi dm


18/69



D1 v D3, D4 l dm phi vi dm D3 v D5.

Hnh 3.6b c dm D1 l dm chnh, D2 l dm phi vi dm D1 nhng ldm chnh i vi dm D3, D3 l dm phi vi D2 v D1 nhng li l dmchnh i vi D4, cn D4 l dm phi vi c D3,D2 v D1.

Hnh 3.6c c dm D1 v D4 l dm chnh, D2 l dm phi vi dm D1

nhng l dm chnh i vi dm D3, D3 l dm phi vi c D2,D1 v D4.Nh vy c nhng dm lun l dm chnh hoc ph, nhng cng c

nhng dm i vi dm ny n l chnh nhng i vi dm kia n l ph. Tanhn bit theo nhn xt sau:

-Dm c s lin kt vi t tng ng vi thai lin kt thanh tr lnth n lun lun l dm chnh.

-Dm khng c lin kt vi t th n lun l dm ph.

-Dm c 1 lin kt ni vi t th khi ngoi cng n s lun l dm ph,khi trong n s l va chnh, va ph.

u : Qu trnh tnh ton c tin hnh tdm ph, ri n dm chnh. Titrng tc dng ln dm chnh skhng gy ra phn lc trong dm phi vi n,ngc li khi c ti trng tc dng ln dm ph n sgy ra phn lc v ni lctrong tt c cc dm chnh i vi n.

b.ngnh hng ca dm tnh nh nhiu nhp.

Khi vng nh hng c th xy ra 2 trng hp: Yu txt thuc dm ph,dm chnh hay dm va chnh va ph.

a)

b)

K

V I

dah Vb

dah Mk

dah Vh

dah Qk

dah MI

dah QI

dah Mv

dah Qv

c)

d)

e)

g)

h)

i)

k)

l)

l1 l2 l3 l4 l5 l6

h a bc d

1

a.bl3

3la

3lb

1

5ld.l4

c.dl5

c.ll5

6

l4l5

5ld

l5

c

l

5l6

h

1

- Yu t vng nh hng thuc dm ph: Vng nh hng VB v MK: Khi P =1 di ng trn dm BC, ta xem nh dm BC l mt dm n gin v ddng vc ng nh hng ca dm ny. Nh phn tch khi P=1 di chuyntrn cc dm khc khng nh hng g n VB nn ng nh hng trng ving chun. (hnh 3.7c,d)


19/69



- Yu t vng nh hng thuc dm chnh hoc dm va chnh va ph:Vng nh hng VH ca dm CD.

+ Khi P = 1 di chuyn trn CD lc ny ch c Dm CD chu lc, cc dm khckhng lm vic, Xem nh dm CD l dm c lp, d dng vc ng nhng ca dm CD.

+ Khi P = 1 di chuyn trn BC l dm ph ca CD, lc ny ti trng truynqua dm CD qua khp C. Ta chng minh n ng nh hng tng ng vi dmBC l mt n thng i qua nh hai tung :tung tng ng di khp C ( ni dm phv chnh) ng bng tung khi ti trngP=1 di chuyn trn dm chnh v tung khng tng ng di gi cn li ( gi B) cadm ph.

Xt dm BC v CD nh hnh 3.8 ta c

Dm BC c mB = l1.VC z.P = 0 VC = 1lz

(1)

Dm CD c mF = l3.VH + VC.l2 = 0 VH =3

2

l

l VC (2)

Thay (1) vo (2) c VH =3

2

l

l .

1l

z(3)

Vy VH l hm bc 1 ca z nn tng ng vi an BC ng nh hng VHl ng thng.

Khi z = 0 ( P = 1 trn gi B), VH = 0

Khi z = l1 ( P = 1 trn khp C), VH =3

2

l

l

ng bng tung khp C lc P=1 di chuyn trn dm chnh CD.

+ Khi P = 1 di chuyn trn dm chnh AB, r rng dm CD khng chulc nn an ng nh hng ca VH ng ng vi dm CD trng vi ngchun ( hnh 3.7g)

ng t ta vc ng nh hng MI, MV QV

Tm li vng nh hng ca dm tnh nh nhiu nhp ta thc hin.

- Khi P = 1 di ng trn dm cha yu t cn vng nh hng, xem ynh 1 dm c lp, ngha l ch c n lm vic, cn li cc dm khc khng lmvic, ng nh hng ca cc dm ny trng ng chun ( thng y l dmmt tha).

- Khi P = 1 di ng trn dm chnh i vi dm cha yu t cn vng nhng th ng nh hng trng vi ng chun ( v khng gy ra phn lc vni lc).

- Khi P = 1 di ng trn dm phi vi dm cha cc yu t cn vngnh hng th ng nh hng tng ng l on thng i qua nh tung ti

l1

z

P=1B

VC

l4l3l2

VC

F H D


20/69



ch ni dm chnh vi dm ph v im 0 tng ng vi gi cn li ca dm ph.

* Cch v nhanh ng nh hng dm tnh nh nhiu nhp.

+ Khi P = 1 di ng trn dm cha yu tcn vngnh hng th xem nl c lp, vngnh hng dm ny.

+ Ln lt vngnh hng qua cc dm k tip khi cho P = 1 di ng.Nu l dm chnh so vi dmang xt th ngnh hng trngng chun.Nu l dm ph so vi dmang xt th vngnh hng l on thng ktipthuc dm ph ca dm ang xt qua cc tung bng 0 tngng vi cc gi cadm ph hoc ng vi khp gp u tin ca dm chnh khc.

5.ngnh hng ca dm chu ti gin tip

Khi tnh ton cng trnh ta thng gp ti trng khng tc dng trc tip lndm chnh m tc dng gin tip ln dm thng qua cc mt truyn lc t nhngim nht nh trn dm. Hnh 3.9.

Trc khi vng nh

ng ta nghin cu 2 cim ca dm chu ti trngtruyn qua mt truyn lc.

- Khi P = 1 t ngvo mt truyn lc ca dmth thc t l ti trng ttrc tip ln dm chnh. Vytung ng nh hng cc v tr tng ng vi mttruyn lc ng bng tung khi ti trng di ng trc tipln dm chnh. c bit khi P= 1 t mt truyn lc trnt hay ngoi dm th titrng khng nh hngn dm nn tng ng vimt ny th ng nh hnglun bng 0.

- Khi P = 1 di ngtrong khong gia 2 mttruyn lc lin tip (trong

phm vi 1 dm) th ngnh hng l 1 on thngni nh 2 tung tng ngvi 2 mt truyn lc 2 u.

* Cch vngnh hng cho dm chu ti trng gin tip:

- Vng nh hng cho ti trng di ng tc dng trc tip ln dm chnh.

- Ti v tr tng ng vi cc mt truyn lc ly tung bng tung ngnh hng v

d)

(P=1 truyn qua mt)

(P=1 t trc tip trn dm chnh)

(P=1 truyn qua mt)


h)

Cdah Q

Cdah Q

g)

dah MC

e)

dah M

C

(P=1 truyn qua mt)


c)Adah V

b)

Adah V

a)


21/69



dah S

PnPP1 2

y

a) C

1 2y

ny

P3

y3

- Ni nh tung cc ng nh hng bng cc n thng c ngnh hng ca dm chu ti trng gin tip.

p dng phng php nu ta vc cc ng nh hng ca dm chuti trng truyn qua mt nh trn hnh 3.9 c,e,h.

3.3 S dng ng nh hng tnh yu t xt di tc dng ca cc

dng ti trng khc nhau.1. Ti trng cnh

a. Ti trng cnh tp trung.

Gi s trn cng trnh c ti trng cnh tp trung P1, P2, ..., Pn tc dng. Cnphi tnh gi tr cc yu t xt S do h ti trng sinh ra.

Trn ng nh hng ca yu t S, tung tng ng vi cc ti trng P1,P2, ..., Pn l y1, y2, ..., yn (hnh 3.10)

Gi tr ca yu t S do h ti trng P1, P2, ..., Pn gy ra l:

S1 = P1.y1S2 = P2.y2

..............

Sn = Pn.yn

Gi tr S do h ti trng ng thigy ra l:

S = S1 + S2+....+ Sn = P1.y1 + P2.y2 + ...+Pn.yn = in

iyP.

1 (3.3)

Vy: Gi tr ca yu tS do ti trng cnh tp trung gy ra bng tngisca tch cc lc vi tungngnh hng tngng

* Ch :

- Lc Pi ly du (+) khi cng chiu P = 1

- Lc Pi ly du (-) khi ngc chiu P = 1

- Ly du tung yi theo du ng nhng .

- Nu Pi t ti v tr tung ng nhng c bc nhy th gi tr ca yu t S c 2gi tr tng ng vi Pi l : hnh 3.11

Siphi = Pi.yi

tri

Sitri = Pi.yi

phi

V d 3.1: S dng ng nh hng, tnh MC v QC cho dm chu ti trng cnh nh hnh 3.12a. Kim tra li kt qu bng cch v biu lc ct v mmenun. Cho bit P1 = 0,5T, P2 = 1T, P3 = 0,8T.

yph

i

tr

iy

Pi


22/69



C

Biu M

e)3034

26

Biu Q

8m

1m2m

2

13

d)

c) 8

8

b)10

85

2

8

10

dah Q83

8

C

8

Cdah M3

815

3m

9

2m

a)

PPP1 2 3Gii:

Vng nh hng MC v QC hnh3.12b,c. Theo cng thc (3.3) ta c

MC = P1.y1 + P2.y2 + P3.y3

= Tmxxx 4,3898,081518105,0 =++

Vi QC ta phi xt 2 trng hp.

- Khi P2 bn tri mt ct C :

QC = Tmxxx 2,08

38,0)

8

3(1)

8

2(5,0 =++

- Khi P2 bn phi mt ct C :

QC = Tmxxx 8,08

38,0)

8

5(1)

8

2(5,0 =++

Bi tp V d 3.1: S dng ngnh hng, tnh MC v QC cho dm chuti trng cnh nh hnh 3.12a. Kim trali kt qu bng cch v biu lc ct vmmen un. Cho bit P1 = 1,5T, P2 =2,5T, P3 = 0,6T.

Ti trng c nh phn bu:

Gi s ti trng phn bu c cng tc dng l q tc dng ln cng trnh.Cn tnh gi tr yu t S do ti trng ny gy ra

Gi honh m u v im cui ca ti trngphn bu l a v b. Xt 1 on dz thay n bng lctp trung l q.dz (hnh 3.13)

Do : dS = q.dz.y

S = q. Trong :

q: cng ti trng phn b

q > 0 khi cng chiu P = 1

q < 0 khi ngc chiu P = 1

: din tch ng nh hng tng ng. Du ca ly theo du ca ng nh hng. Nu ng nh hng gm nhiu b phn cdu khc nhau th l tng i s din tch ca cc phn . = (+) + (-)

* Vy: gi tr ca yu t xt do ti trng phn bu gy ra bng tch cang ti trng phn bu vi din tch ngnh hng tngng

V d 3.2: Dm mt tha chu tc dng ca ti trng phn bu q = 10 kN/mnh hnh 3.14a. Tnh mmen v lc ct mt ct C gia dm.

Gii:

a

b

q

dz

qdz

y ya yb


23/69



Tnh MC v QC theo ng nh hng hoc Sc bn vt liu.

Theo ng nh hng :

0,5

8m

26m

0,3125

c)dah QC

2,5C

0,5

dah M

b) 4

5m

a) C

8m

0,3125

2,5

5m

MC = qM = 10

+

2

5.5,2

2

16.4= 10(32-12,5) = 10.19,5 = 195 kNm

QC = 0 v phn dng v m bng nhau.

Theo Sc bn vt liu:

VA = VB = kN1302

26.10=

MC = 130.8 - kNm1302

13.13.10=

QC = 130 - 10.13 = 0

b. Ti trng cnh l mmen tp trung:

Gi s trn cng trnh chu tc dng ti trng l mmen tp trung M

Gi tr ca yu t xt tnh theo cng thc:

S = =

n

i

iitgm

1

.

Trong :

mi: mmen tp trung

mi (+): khi quay thun chiu kim ng h.

mi (-): khi quay ngc chiu kim ng h.

i : gc nghing ca ng nh hng v tr tng ng mi

tgi(+) :khi ng nh hng ng bin


24/69



tgi (-) :khi ng nh hng nghch bin

Nu ti v tr tng ng vi mmen tp trung ng nh hng c gy khchoc c bc nhy th khi M bn tri ly tgitr,M bn phi ly tgiph. (hnh 3.15)

Stri = =n

i

ii tgm1 . tr

Sphi = =

n

i

iitgm

1

. ph

Vy: Gi tr yu t S do h ti ti trng cnh l cc mmen tp trung gy ra bng tngcc tch s gia mmen vi tg gc nghing cangnh hng v tr tngng.

V d 3.3: S dng ng nh hng tnh MA, QA, MC, QC ca dm trn hnh 3.16a.

Bit P1 = 8kN; P2 = 12kN; q = 4kN/m;M = 10kNm.

Gii:

Vng nh hng MA, QA,MC, QC nh hnh v 3.16b,c,d,e

Theo cng thc (3.3), (3.4), (3.5)ta c

S = P1.y1 + P2.y2 + q. + M.tg

MA = 8.(-1) +12(-3) + 4 2)4.(4

-10.(-1) = -66kNm

QA = 8.1 +12.1 +4.4.1 -10.0 =36kN

Tnh MC cn phn bit 2trng hp

- Khi M t bn tri mt ct C:tgtr= 0

MC = 12.(-1) + 4 2

)2.(2

=-20kN.m

- Khi M t bn phi mt ct C: tgph = 12

2=

MC = 12.(-1) + 42

)2.(2 - 10.(-1) =-10kN.m

QC = 12.(1) +4.1.2 = 20kN.

2. Ti trng di ng

dah S

ph

i

dah S

tr

i

Pi

tri

iph

mi

a)

e)

dah Q

d)

dah M

dah Q

2

c)

dah M

1

b) 1

C

1

1 2

4

1

4

q1P M 2P


25/69



tnh tc dng ca ti trng di ng cn xc nh v tr bt li nht sau dng ti trng v tr ny v xem ti trng l cnh tnh gi tr yu t xt.

Gi Smax l gi tr dng ln nht

Smin l gi tr m ln nht

a. Ti trng di ng l lc phn bu q:- Trng hp chiu di ti trng phn bu ln hn hoc bng chiu di t

ti ca ng nh hng (d l),(hnh 3.17). Lc ny ta st ti trng ph knng nh hng, chnh l v tr bt li nht. Ta C Smax = q. nu ng nhng (+) hoc Smin = q. nu ng nh hng (-). Gi Stnh l ni lc chn tnh th: Stnh = q. Trong :

q: cng ti trng phn b ( ly du theo P =1).

: din tch ng nh hng

l

d

dah S

b)

dzdzl

2y1y t p

a)

d

q

- Trng hp chiu di ti trng phn bu nhn hn chiu di t ti ca

ng nh hng (d < l),(hnh 3.18). V tr bt li l v tr c tung ng nhng tng ng vi u tri (ytr) v u phi (yph) ca ti trng phn b bng nhau.(ytr= yph)

Ch : Nu ng nh hng c hai du th phi t ring cho tng phn cdu dng v du m tnh Smax v Smin

Th d: 3.4 Cho dm n gin c chiu di l =8m.

-Xc nh v tr bt li v tnh mmen ln nht mt ct K

-Tm Qmax v Qmin mt ct K.

Cho bit ti trng di ng phn bu c cng q = 120 kN/m v chiu di

d = 5m; a = 3m; b = 5m.Gii:

Chiu di t ti ca ng nh hng MK l 8m di hn chiu di ti trng d= 5m nn c v tr bt li cn t ti trng sao cho ytr= yph . hnh 3.19.


26/69



Tnh z ta c :a

z

y

y

m

t = (1)

b

zdl

y

y

m

ph = v ytr= yph nn

a

z

b

zdl

=

l

adlz

)( = (2)

Thay (2) vo (1) cl

dlyt

= my.

l

baym

.=

8

15=

Din tch ng nh hng trong phmvi ytrv yph

= )(2

)(2

adzyy

zayy mpmt +

++

+

Thay yt , yp, v z vo ta c

=l

dl

.2

.2 dym .. = 8.2

58.2 .

8

15.5 =

128

825

v MKmax = q. = 120. 128825

773,4kNm.

- Tnh QKmax : chiu di t ti b =5mng bng chiu di t ti phn bu nnv tr bt li l v tr ti trng ph kn ng nh hng (hnh 3.19c)

QKmax = q.+ = =2

85.5

.120 187,5 kN

- Tnh QKmin : Phn ng nh hng QK c du m c chiu di t ti a =3m, v tr bt li c biu th nh (hnh 3.19c)

QKmin = q.- = =2

83.5.120 67,5 kN

b. Ti trng di ng l lc tp trung:

- Trng hp ch c 1 lc tp trung P: Ta nhn thy: Smax ch cn t ti trngtp trung ti v tr c tung ymax v Smax = P.ymax . Khi tnh Smin ch cn t ti trngtp trung ti v tr c tung ymin v Smin = P.ymin (hnh 3.20)

- Trng hp c nhiu ti trng tp trung: Khi xc nh v tr bt li nht ivi ng nh hng a gic c th da vo 2 tnh cht sau:

a bKdah M

8

5

Kdah M38

c)

a) K

my

yy

z

t p

db)

l

q

ba

a)

max

d.a.h S miny

yc)

b)

P

P


27/69



+ Tnh cht 1: V tr bt li nht ch c th xy ra khi c 1 ti trng tp trungtrng vi 1 nh no ca ng nh hng.

+ Tnh cht 2: Khi c t nht 1 ti trng tp trung 1 nh no ca ngnh hng S. Nu v tr S c cc tr th khi dch ti trng sang tri hay phi 1on dz cn phi tha mn iu kin

> 0 qua phi = 0 (1)

= 0

i

i

i

i

V d 3.5: Nh hnh 3.22 gi s P4t ti C ca ng nh hng . Khi dchchuyn sang tri 1 n dz ta c:

R1 = P1

R2 = P2 + P3 + P4

R3 = P5

R4 = P6 + P7

Khi dch chuyn sang phi 1on dz ta c:

R1 = P1

R2 = P2 + P3

R3 = P4 + P5

R4 = P6 + P7

Ttnh cht ta suy ra:

1

EA

C

2

B

4

D

3

321 PPP 7654 PPP P


28/69



+ Di chuyn ti trng 1 ti trng ri vo 1 nh no ca ng nhng

+ Dch ti trng sang tri 1 n dz v tnh =

n

i

iitgR1

+ Dch ti trng sang phi 1 n dz v tnh =n

i

iitgR1

+ So snh 2 ln dch chuyn trn xem k (1) c tho mn hay khng. Nukhng tho mn th chc chn n khng phi l v tr bt li. Nu tho mn th tnh

yu t xt S theo cng thc: S = in

i

iyP.

1

=

- Ch : C nhiu v tr tho mn k (1) nn cn phi lm nhiu ln sau sosnh cc kt qu ca S v chn gi tr c tr tuyt i ln nht l Stnh

* Khi xv tr bt li nhti vi ngnh hng l tam gic: (hinh 3.23)

Bin i u kin (1) thnh iu kin n gin hn:+ Khi dch chuyn sang tri 1 n:

>

a

PRtht

+ >

b

Rph (2)

=

+ Khi dch chuyn sang phi 1 n:

N9-8 = VA = 30kN

Bi tp Th d 4.2 : Bng phng php tch mt. Tnh ni lc cc thanh cadn trn hnh 4.5 a Bit P = 10kN. gc = 300

4.3 ng nh hng ni lc cc thanh trong dn.


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Dn n gin l dn ch c gi ta hai u. Trong dn ti trng ch truynqua mt nn xem nh ti trng truyn gin tip, n ng nh hng trong phmvi hai mt lin tip trn ng xe chy l on thng.

ng nh hng phn lc ca dn n gin ging ng nh hng phnlc ca dm n gin cng nhp.

i tc dng ca ti trng cnh ni lc ca cc thanh c xc nh theophng php no th khi vng nh hng ta cng dng phng php .

1. Vng nh hngni lc thanh 2-3: Dng phng php tm mmen vng nh hng hnh 4.6

Tng tng mt ct I-I ct dn lm 2 phn, tr thanh 2-3 cc thanh cn ling qui ti 11.

- Khi P = 1 di ng trn phn tri dn (0 [ z [ d), xt cn bng phn phihnh 4.6b ta c : m11 = N2-3.r1 + VB.4d = 0

N2-3 = 1

.4

r

dVB

= lr

dz

1

4

+ Khi z = 0 (P=1 gi A) th N2-3 = 0

a)1

2

3 4 5

6

7

89101112

k

1

I

I I II

I II II

IIP=1

6d

b)

11 10 9

3 4 5

7

8 B

6

V

2-3N

N2-11

K

N

K

c)

2-3N

11

2-11

A

V

12

2

1r

1r

ZK

4.d3.r1

2.d13.r

2.dr1 1r

1zd)

. a .h 2-3N


41/69



zrk

k6r

z +6.d

3r2.z

k

k

k

ke)

g)

1sin 1

sin1

1

3 sin11

2 sin11

N . a .h2-11

N . a .h 4-9

h)

sink1

sink13 sin

1k

k2 sin1

N . a .h 4-11

i) 1

N . a .h 4-10N

Hnh 4.6

3-43

N3-11

N

2-3N

k)

i)

3-11N . a .h3r

1

4d.cos

+ Khi z = d (P=1 mt 12) th N2-3 =dr

dd

.6

.4

1

=13

2

r

d

T s liu c vc nhnh tri ng nh hng N2-3

- Khi P = 1 di ng pha phi tm mmen (2d[ z [ l), xt cn bng phntri.( hnh 4.6c ta c : m11 = - N2-3.r1 - VA.2d = 0

N2-3 =1

.2r

dVA =dr

zld

6.)(2

1

+ Khi z = 2d (P=1 gi A) th N2-3 =dr

ddd

6.

)26(2

1

=

dr

d

3.

4

1

+ Khi z = 6d (P=1 gi B) th N2-3 = 0

T s liu c vc nhnh phi ng nh hng N2-3

Vng ni cc nh tugn tng ng vi cc mt ct hai bn khoang b


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ct ( mt 12 v 11) c ng nh hng N2-3 (hnh 4.6d)

2. Vng nh hng N2-11. Vn dng mt ct I-I, tr thanh 2-11 cn vng nh hng, cc thanh cn li ng quy ti K nn vng nh hng N2-11 ta dng phng php tm mmen tng t nh N2-3 ta vc ng nh hng

N2-11 nh hnh 4.6e . T suy ra cch v nhanh ng nh hng ni lc bng

phng php tm mmen nh sau :+ Dngng chun.

+ Ti v tr tngng vi gi tri trn ng chun dng tung k

k

r

z.

Trong:

Zk: l khong cch ttm mmen n gi tri, rk: l khong cch tni lccn tm ti tm mmen

k

k

r

z: ly du (+) khi xt phn dn bn tri, phn lc gi tri v ni lc cn v

ngnh hng quay ngc chiu nhau quanh tm mmen.

k

k

r

z: ly du (-) khi ngc li.

+ Ni nh tung vi tung bng 0 tngng vi gi phi trn ngchun, ging t tm mme xung gp ng va vti 1 im. T im n gi

phi l ng phi, n gi tri l ng tri.

+ Tmt bn phi ca khoang b ct nm trn ng xe chy ging xungng tri ctng tri 1 im, t im n gi tri gi l nhnh tri. Ninhnh tri v nhnh phi bngng thng.

3. Vngnh hng N4-9. Dng mt ct II-II ct dn thnh hai phn hnh4.6a, tr thanh N4-9 cn v ng nh hng, hai thanh b ct cn li song song vinhau nn ta v ng nh hng N4-9 bng phng php chiu ln trc vung gcvi hai thanh 4-5 v 9-10 (trc y)

- Khi P = 1 di ng trn phn tri dn (0 [ z [ 3d), xt cn bng phn phi tac : Y = N4-9.sin1 + VB = 0

N4-9 =1sin

BV =1sin. l

z

+ Vi z = 0 (P=1 gi tri ) th N4-9 = 0

+ Vi z = 3d (P=1 mt 10 ) th N4-9 =1sin.2

1

T s liu c vc nhnh tring nh hng N4-9

- Khi P = 1 di ng trn phn phi dn (4d [ z [ l), xt cn bng phn tri tac : Y = -N4-9.sin1 + VA = 0


43/69



N4-9 =1sin

AV =1sin. l

zl

+ Vi z = 4d (P=1 mt 9 ) th N4-9 =1sin.6

46

ddd =

1sin.3

1

+ Vi z = l (P=1 gi B ) th N4-9 = 0T s liu c vc nhnh phing nh hng N4-9

V ng ni nh cc tung tng ng vi cc mt hai bn khoang bct (mt 9 v 10) ta c ng nh hng N4-9 (hnh 4.6 g)


44/69



CHNG V : KHUNG TNH NH VM 3 KHP

5.1 Khung tnh nh.

1. Tnh ni lc bng phng php gii tch.

a. Khi nim: Khung l 1 hbt bin hnh gm cc thanh thng hoc cong ni

vi nhau bng cc nt cng hoc bng khp. Nu khung bt bin hnh v c linkt (bc t do bng 0) th khung l khung tnh nh. (hnh 5.1)

Hnh 5.1

Khung tnh nh c th tm ni lc, phn lc tt c cc b phn ca khungbng cc phng trnh cn bng tnh hc.

b. Tnh ni lc trn mt ct bt k ca khung:

Trn mt ct bt k ca khung c 3 thnh phn ni lc: M, Q, N

- Mmen un M: Mment un ca mt ct bt k bng tng i s mmen cacc ngoi lc 1 bn ca mt ct ly i vi trng tm mt ct . Biu M cakhung lun lun v v pha chu ko. Nn khi tnh M mt ct bt k khng cn kn du m ch xt xem mmen un lm thno chu ko.

- Lc ct Q: Lc ct Q mt ct bt k ca khung bng tng i s hnh chiuca cc lc 1 bn mt ct ln trc vung gc vi trc thanh v tr mt ct. Duca Q c qui c ging nh trong Sc bn Vt liu

- Lc dc N: Lc dc N mt ct bt k ca khung bng tng i s hnhchiu ca cc lc 1 bn mt ct ln trc tip tuyn vi trc thanh v tr mt ct.Du ca lc dc qui c nh trong Sc bn Vt liu.

Th d: 5.1 Xc nh mmen un, lc ct v lc dc ca cc mt ct C, D, Eca khung tnh nh hnh 5.2. Cho bit q = 15kN/mt ct.

Gii:- Tnh phn lc: mA = 6.VB - 4. 8q = 0

VB =6

.32 q= 80kN

mB = 6.VA - 4. 8q = 0


45/69



A =6

.32 q= 80kN

X = 8q - HA= 0 HA= 120 kN

- Tnh ni lc:

+ Ti mt ct C: Xt cn bngphn AC:

MC = HA.4 - 4q. 2

MC = 120.4 4.15.2 = 360 kNm

QC = HA 4.q = 60kN

NC = VA = 80 kN

+ Ti mt ct D: Xt cn bngphn DHEB:

MD = 3.VB = 240 kNmQD = VB = -80kN

ND = 0

+ Ti mt ct E: Xt cn bng phn EB:

ME = 0

QE = 0

NE = - VB = -80 kN

2. Vbiu ni lc khung.

a. Biu mmen un M:

- Khi khung c nhiu on thanh ta v tng on, vi mi on cch v honton ging nh trong Sc bn Vt liu.

- Biu mmen lun v v thchu ko, khng cn du.

- cc nt cng lun lun c s cn bng mmen (ngha l tng mmen ngoilc v ni lc bng 0) (hnh 5.3)

Hnh 5.3

a) b)

M1

2M

M1

2M

- cc nt khp nu khng c mmen ngoi lc th mmen ti khp bng 0,

6m

Hnh 5.2

q=15kNm

4m

3m

4m

3m

4m


46/69



ti biu ct ng chun.

b. Biu lc ct Q:

C 2 cch v:

- Theo Sc bn Vt liu.

- V theo biu mmen un:

Qph =2

ql

l

MM trph

(5.1)

Qtr=2

ql

l

MM trph+

(5.2)

Trong :

Qph, Qtr : lc ct u tri v u phi on thanh.

L : chiu di on thanh.

M

ph

, M

tr

: ln lt l mmen u tri v u phi ca on cn v biu ivi ngi quan st ng sao cho lc phn b hng xung di.

Trng hp q =0

Qtr= Qph =l

MMtrph

c. Biu lc dc N:

C 2 cch v:

- Theo Sc bn Vt liu.

- Tch nt cng, kho st s cn bng ca nt

* Ch :

Nu lc ct (+) th chiu ca lc ct c khuynh hng lm nt quay thunchiu kim ng h v ngc li

Th d 5.2: V M, Q, N ca khung cho nh hnh v: 5.4a

Gii:

- V biu mmen un

o Thanh AK.

MA = 0; MC = 360 kNm (thphi chu ko);

MK= 8.HA - 8.q = 8.120 -32.15 = 480 kNm (thphi chu ko)

o Thanh ngang KH.

MK= 480 (thdi chu ko);

MH = 0

o Thanh ng HB. Ta c MB = MH = 0


47/69



Hnh 5.4

kNm480

4m

4m

6m

3m

q=15mkN

3m

4m

a)

b) d)

kNm80 kNm80

kNm

c)

120

80kNm

T cc kt qu c ta v biu M (hnh 5.4b)

- V biu lc cto Thanh AK. Theo cng thc 5.1 v 5.2

QK = Qph = )(

1AK

MMl

-2

.lq=(480 - 0)/8 - 15.8/2 = 60 - 60 = 0

Mt = MA = 0 ; Mph = MK = 480 kNm (hnh 5.5a)

QA = Qt = )(

1AK MM

l +

2

.lq=(480 - 0)/8+15.8/2 = 60 + 60 = 120 kN

8m

6m

a)

b)

Hnh 5.5


48/69



o Thanh ngang AH. (hnh 5.5b)

QK = QH =l

1(Mph - Mt)=

6

1(0 - 480) = -80 kN

o Thanh ng HB.

QB = QH = 0T cc kt qu c ta v biu Q (hnh 5.4c)

- V biu lc dc. Tch nt K v H (hnh 5.6)

Hnh 5.6

NKA

NKH

KQ =80kN

KHN

Q =80kNK

NHB

nt K : X = NKH = 0 ; Y = 80 - NKA = 0 suy ra NKA = 80 kN

nt H: Y = -NHB - 80 = 0 do NHB = -80 kN

T cc kt qu c ta v biu N (hnh 5.4d)

5.2 Vm 3 khp

1 . Khi nim: Vm 3 khp l h gm 2 thanh cong ni vi nhau bng 1khp C v t trn 2 gi ta cnh khc l A v B. (hnh 5.6a)

- C 2 dng vm 3 khp:

+ Vm ngang mc (hnh 5.6a)+ Vm khc mc (hnh 5.6b)

Hnh 5.6

a) b)

* Ch : trong vm 3 khp c pht sinh phn lc ta nm ngang ngay c khi

c ti trng thng ng.( chng li lc x ngang) (hnh 5.7)2. Tnh ton vm 3 khp chu ti trng cnh bng phng php gii

tch:

a. Xc nh cc phn lc: (hnh 5.7)

Phn lc ta vm 3 khp c 4 thnh phn VA, VB, HA, HB. Xc nh nh sau:

+ Vit phng trnh MA = 0 tm c VB+ Vit phng trnh MB = 0 tm c VA


49/69



Hnh 5.7

H

V R

K

R

P

R

RV

H

+ Vit phng trnh MCtr = 0 tm c HA

+ Vit phng trnh MCph = 0 tm c HBMC

tr, MCph, ln t l tng i s mmen pha tri (k c VA) i vi m C

v mmen ca cc lc na vm bn phi (k c VB )i vi m C, f: ng tn

ca vmTa c:

HA = f

Mtr

C

HB = f

MphC

- Trng hp vm ch c ti trng thng ng tc dng, ta c:

HA = HB = H = f

Mtr

C = f

MphC

Th d 5.3 Tnh cc phn lc trn hnh 5.8

Gii : mA = -12.q - 40.16 - 40.20 + 24.VB = 0

Suy ra : VB = 120 kN

mB = 0 Suy ra : VA = 200 kN

Trn vm ch c cclc thng ng tc dng nncc phn lc ngang c tr s

bng nhau.

HA = HB =

f

Mtrc =f

Vq A126..12 + =

6

200.126.20.12 +=200 kN

b. Tnh ton ni lc trn mt ct ngang ca vm

- Biu thc mmen un: Xt cn bng 1 phn vm tri mt ct K

RHnh 5.8

V

H

V

H

6m

kNq=20m

4m 4m 4m12m

P=40 kNkNP=40


50/69



MK = VA.zK- P1(zK- a1) - HA.yK

So snh khu vm vi khu dm c cng kch thc, c ti trng tngng vi nhau, ta nhn thy: (hnh 5.9a,b)

H

P K

H

V

PK

zKyKK

P

P

V

V

V

l

a1

Hnh 5.9

b)

a)

VAd = VA

; VB = VBd

MKd = VA.zk - P1(zK- a1)

Vy: MK= MKd - HA.yK (5.3)

Trng hp ch c lc thng ng tc dng.

HA = HB = H nn cng thc (5.3) vit li l:

MK

= MK

d - H.yK

(5.4)

* Nhn xt: Mmen un trong vm nh hn mmen un trong dm n ginng ng. Nu kho chn hnh dng vm sao cho tch s ( H.yK) = MK

d th mmenun ca mi mt ct ca vm bng 0. Lc ny vm khng chu un m ch chu nn.Vy s tit kim c vt liu v tn dng vt liua phng tt lm vm nhgch, , ...

- Biu thc lc ct: Xt cn bng phn vm bn tri mt ct K

QK= VA.cosK- P1.cosK- H.sinKQK= (VA - P1)cosK- H.sinK

QK= QKd

.cosK- H.sinK (5.5)Vi (VA - P1) = QK

d

- Biu thc lc dc: Xt cn bng phn vm bn tri mt ct K

NK= VA.sinK- P1.sinK+ HcosK

NK= (VA - P1).sinK+ HcosK

NK= QKd.sinK+ HcosK (5.6)

Ch : vm chu nn nn qui c lc ko l lc (+)


51/69



Th d : 5.4 : Vm c ng trc bin thin theo phng trnh y =

9

)12( zzy

= . Xc nh ni lc ti mt ct K c honh zk = 3m. Bit P= 40kN, q

=20m

kN(hnh 5.10a)

V

H z

a)

b)

VV

K

K

P P P

Hnh 5.10

q=20m P

1,53m

Hnh 5.10

kNP

P

1,5 1,5 1,53m

4m

H

V

Gii :

- Tnh cc s liu ng ng zk= 3m ta c myk 393)312(

=

=

z = 6 m, f =9

6)612( = 4m

tg = y=9

).212( z

tgk =9

).212( kz =9

)3.212( =

3

2

Tra lng gic : k = 3304130 ; sink = 0,5549 ; cosk = 0,8321

- Tnh phn lc : mA = 12VB - 6.q.3 -7.5.P - 9.P -10,5.P = 0

Suy ra VA = dAV = 120kN

mB = -12VA + 6.q.9 +4.5.P + 3.P + 1,5.P = 0Suy ra VB = dBV = 120kN

Trn vm ch c ti trng thng ng nn

HA = HB = H =f

MtrC =4

3.20.66.120 =90kN

- Tnh MK: dkM =3.d

AV -3.q.1,5 = 3.120-3.20.1,5 = 270kNm


52/69



Theo cng thc (5.4) ta c MK= MKd - H.yK= 270- 9.30 = 0

- Tnh QK:d

kQ =d

AV -3.q = 120-3.20 = 60kN

Theo cng thc (5.5) ta c QK= QKd.cosK- H.sinK

= 60.0,8321 -90.0,5549 = 0

- Tnh NK: Theo cng thc (5.6) ta c NK= QKd.sinK+ HcosK= 60.0,5549 + 90.0,8321 = 108,1kN

Ti mt ct K mmen un v lc ct u bng khng, lc dc l lc nn vitr s l 108,1kN

3. Khi nim v trc hp l ca vm.

a. nh ngha: Trc hp l ca vm l trc c chn sao cho mmen untrn tt c cc mt ct ca vm bng 0 (do Q = 0), lc ny trong vm ch c lcdc. (trc hp l ca vm phi trng vi ng p lc), nh vy trc hp l cavm ph thuc vo ti trng trn vm.

b. Xc nh trc hp l ca vm khi chu ti trng thngng v ti trngny khng ph thuc vo dng vm.

Theo cng thc 5.4 ta c: MK= MKd - H.yK= 0 suy ra yK=

H

MdK

V chu ti trng thng ng H = const nn dng trc hp l ca vm chnh ldng ca biu mmen un trong dm n gin tng ng, ngha l c cng titrng v khu vi vm, trong cc tung ca biu mmen gim i H ln.


53/69



CHNG 6 :

TNH KT CU SIU TNH BNG PHNG PHP LC

6.1 Khi nim v kt cu siu tnh

1.nh ngha:

Hc gi l h siu tnh nu trong ton h hoc mt vi b phn ca h takhng th dng cc phng trnh cn bng tnh hc xc nh tt c cc phn lcv ni lc.

H siu tnh l hbt bin hnh v c lin kt tha.

Th d hnh 6.1a l dm siu tnh. Dm ny c mt lin kt tha, nu b 1 linkt tha i th ta c cc dm nh hnh 6.1 b,c,d l cc dm tnh nh v bt binhnh.

a)

b)

c)

P

A B C D

A B DC

P

A B DC

P

d)

A B

P

C D

2. Bc siu tnh

Bc siu tnh ca h siu tnh l s lin kt tha tng ng loi 1 ngoi slin kt cn thit cho h bt bin hnh.

a. H ni vi t: n = T + 2K + 3H + C0 3D

S lin kt thanh tng ng vi mi loi gi C0 xem bng 2.1 ( nghin

cu kchng 2)Th d 6.1: Xc nh bc siu tnh ca dm hnh 6.2

A B C

Gii: H c : T=0; K=1;H=0; C0=6 ; D=2, vy bc siu tnh ca h


54/69



n = 0 + 2.1 + 3.0 + 6 3.2 = 2

b. H kt cu khng ni vi t: n = T+2K+3H 3(D-1)

Th d 6.2: Xc nh bc siu tnh ca dm hnh 6.3a,b

A B

CD

a) b)

D C

A BE

Gii: Hnh 6.3a h c : T=0; K=0; H=4; D=4, vy bc siu tnh ca h

n = 0 + 2.0 + 3.4 3(4 - 1) = 3

Hnh 6.3b h c 1 lin kt khp ti E do : K=1; T=0; H=4; D=5, vy bcsiu tnh ca h n = 0 + 2.1 + 3.4 3(5 - 1) = 2

Nhn xt:Mt chu vi kn c 3 bc siu tnh, nu thm vo chu vi kn 1 khpn th bc siu tnh gim i mt

c.i vi h khung : n = 3C- K (C: chu vi kn, K : S khp n)

Th d 6.3: Xc nh bc siu tnh ca khung trnhnh 6.4

Gii: Khung c 3 chu vi kn,C =3, K= 6 vy bcsiu tnh ca khung. n = 3C- K = 3.3 - 6 = 3 .

3. Cc c m ca h siu tnh:So vi h tnh nh h siu tnh c cc c m

sau:

- Chuyn v v bin dng ni chung trong h siutnh ni chung nh hn trong h tnh nh c cng kchthc v ti trng. V vy, h siu tnh tit kim vt liun so vi h tnh nh.

Th d 6.4. Dm tnh nh trn 6.5a c: Mmax =8

2ql

v vng

ymax=EJ

ql

384

5 4

Dm siu tnh trn 6.5b c: Mmax ti mt ct ngm =12

2ql

v vng ln

nht ti mt ct gia nhp ymax=EJ

ql

384

4

A B

C


55/69



q

a)

l

q

l

b)

- Ni lc trong h siu tnh ph thuc vo kch thc tit din cc cu kinca h EJ, EF,... ( trong h tnh nh khng ph thuc kch thc ny).

- Trong h siu tnh ni lc c th xut hin do s thay i nhit , chuynv ca cc gi ta hay do ch to v lp ghp khng chnh xc.

* Nhn xt: Qua c m trn ta thy ni lc trong h siu tnh c th sinhra do: ti trng, nhit , chuyn v gi hay lp ghp khng chnh xc, trong phm vicng trnh ta ch ch yu tnh ni lc do ti trng sinh ra.

6.2. Ni dung ca phng php lc tnh kt cu siu tnh.1.Trnh ttnh ton vbiu ni lc ca kt cu siu tnh bng phng

php lc:

- Xc nh bc siu tnh (s lin kt tha)

- Chn h cbn: bng cch loi b lin kt tha.

- Vit phng trnh chnh tc theo phng php lc.

- Gii phng trnh chnh tc, tm gi tr cc lin kt tha.

- Tm gi tr ni lc cui cng.

2.Thc hin:a. Bc siu tnh:

- Bc siu tnh ca h siu tnh l s lin kt tha tng ng loi 1 ngoi slin kt cn thit cho hbt bin hnh

- Gi n l s bc siu tnh (s lin kt tha ngoi s lin kt hbt binhnh)

+ Kt cu siu tnh ni t:n = T + 2K + 3H + C0 - 3D

+ Kt cu siu tnh khng ni t: n = T + 2K + 3H - 3(D - 1)

+i vi dn phng nu trong dn c T thanh, C0 lin kt ni t, M mt dn

th: n = T + C0 - 2M+ Kt cu khng ni t: n = 3C - K

Vi C: chu vi kn K: s khp n.

* Ch :

+ Gi di ng c xem l 1 khp n.

+ Gi cnh c xem l 2 khp n.


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+ Tri t l 1 ming cng h( chu vi h)

b. Chn h cbn cho phng php lc:

- H cbn: L hbt bin hnh suy ra t h siu tnh bng cch loi bimt s hoc tt c cc lin kt tha. Mt kt cu siu tnh c th c nhiu h cbn.

y ta ch xt h cbn c suy ra bng cch loi bi tt c cc lin kttha nn h cbn l kt cu tnh nh.

Th d h siu tnh hnh 6.6a

a) b) c)

1X

2X

d) b) c)

X1

2X

* Skhc nhau gia h siu tnh v h cbn:

Trn h siu tnh v tr loi b lin kt s c phn lc theo phng ca linkt cn trn h cbn khng c lin kt nn khng c phn lc. V vy, cho h

bn lm vic ging h siu tnh cho theo cc phng ca lin kt loi bcn t cc phn lc tng ng l X1, X2, ...., Xn

c. Phng trnh chnh tc ca phng php lc: h siu tnh lm vicging h cbn cn xt cc phng trnh m bo cho chuyn v tng ng vi cclin kt bi bng 0, phng trnh ny gi l phng trnh chnh tc.

- Phng trnh chnh tc ca kt cu siu tnh c 1 bc siu tnh:

11.X1 + 1p = 0 (1)

- Phng trnh chnh tc ca kt cu siu tnh c 2 bc siu tnh:

11.X1 + 12.X2 + 1p = 0

21.X1 + 22.X2 + 2p = 0 (2)

- Phng trnh chnh tc ca kt cu siu tnh c n bc siu tnh:

11.X1 + 12.X2 + ....+ 1p.Xp + 1n.Xn + 1p = 0

21.X1 + 22.X2 + ....+ 2p.Xp + 2n.Xn + 2p = 0

........................ (n)

n1.X1 + n2.X2 + ....+ np.Xp + nn.Xn + np = 0


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Trong :

km: l chuyn v theo phng ca lc XKdo Xm= 1 gy ra h cbn.

+ Nu k m (vd: 12, 21, 34., ....): gi l h s ph.

+ Nu k = m (vd: 11, 21, 33., ....): gi l h s chnh

kp: l chuyn v theo phng ca lc Xk do cc ti trng trn h cbn gyra. S hng ny l s hng t do.

* Tnh cc h s chnh kk, h s ph km, s hng t do kp:

+ t ln h cbn lc Xk = 1, v biu kM

+ t ln h cbn lc Xm = 1, v biu mM

d.Nhn biu Vrsaghin xc nh km theo cng thc:

km = mkMEJ

M. (3)

Tnh kk= mk MEJ

M. (4)

* Nhn xt:

T (3) v (4) ta thy:kM mM = .y

kk > 0, km c th > 0, < 0, = 0

: din tch biu mmen c hnh dng bt k

y: tung ca biu mmen c dng ng thng ly tng ng vi

trng tm Mk, Mm: biu mmenn v do Xk= 1 gy ra h cbn.

- p dng cng thc: kp =EJ

MM pk.

Gii phng trnh chnh tc cho phng php lc: xc nh c X1, X2,.., Xn

e. Vbiu ni lc cui cng:

-V biu mmen un

M = X1. 1M + X2. 2M + ......+ Xn. nM + Mp

* Ch :

+ Nu biu c dng ng thng, tnh y gy khc chia ra nhiu n tnh.

+ Nu biu phc tp chia ra nhiu hnh n gin tnh ton.

- V biu Q

+ V theo biu mmen un nghin cu chng 5)


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Qph =2

ql

l

MM trph

(5)

Qtr =2

ql

l

MMtrph

+

(6)

Trong :

Qph, Qtr : lc ct u tri v u phi on thanh.

L : chiu di on thanh.

Mph, Mtr: ln lt l mmen u tri v u phi ca on cn v biu i

vi ngi quan st ng sao cho lc phn b hng xung di. (hnh 6.7, 6.8)

Trng hp q =0

Qtr= Qph =l

MM trph (7)

tr

tr

tr

ph

ph

ql

tr

trtr

Hnh 6.8

q

ph

ph

ph

ph

tr

l

tr

q

tr

ph

ph

Hnh 6.7

l

ph

Th d 6.5: Dm siu tnh hnh 6.9a c biu mmen un hnh 6.9b. Vbiu lc ct ca khung

Gii : Xt hai on AB v BC

- on AB : QA, MA tng ng Qtr, Mtrv QB, MB tng ng Q

ph, Mph

Theo cng thc (6.6) ta c : QA =2

.)( AB

AB

AB lq

l

MM+

=2

2.142

)04( + = 12 kN

Theo cng thc (6.5) ta c : QB =2

.)( AB

AB

AB lq

l

MM

=2

2.14

2

)04(

= -16 kN

Trn on BC : B l u tri, C l u phi, theo cng thc 6.7 ta c :


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Qtr = Qph =l

MMtrph =

BC

BC

l

MM =

2

)4(2 =3 kN

T v biu hnh 6.9c

b) c)

Hnh 6.9

q=14

2m

kNm

2m

a)

- Biu lc dc N: Tch nt cng, kho st s cn bng ca nt* Ch :

Nu lc ct (+) th chiu ca lc ct c khuynh hng lm nt quay thunchiu kim ng h v ngc li

Th d 6.6: V biu lc dc theo biu lc ct hnh 6.9c

Gii : Tch nt B xt cn bng ta c

Hnh 6.10

NBKA

Q

BCN

BC

a) QBA b)

QBC = 3 kN,

QBA = 16 kN,

QBC vung gc vi BC, QBA vung gc vi BA cn chiu nh hnh 6.10b

X = NBC + QBA = 0, suy ra NBC = - QBA = -16 kN

Y = - NBA - QBC = 0, suy ra NBA = - QBC = -3 kN

V trn thanh BA v BC ch c lc vung gc vi trc thanh nn lc dc trnton thanh khng i v biu lc dc vc nh hnh 6.10b (du tr chng tthanh b nn)

Th d 6.7 V biu mmen un v lc ct ca dm siu tnh trn hnh6.11a. Cho bit P = 20 kN, m = 5 kNm.


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Gii: Chn h cbn bng cch loi b lin kt tha gi B

Phng trnh chnh tc:

11.X1 + 1p = 0

V 1M v MP

t X1 = 1 nh hnh 6.11b t vc 1M (hnh 6.11c). Trn h c bn cht cc ti trng vc MP (hnh 6.11e)

Tnh 11 =EJ

MM 11. =

=EJ

1. 4.

3

2.

2

4.4=

EJ3

64

Tnh 1p cn nhn biu 1M vi

biu MP,

do ta chia biu 1M thnh 3din tch.

1 = 22

2.2= ; 2 =2.2. = 4;

3 = 22

)24(2=

c trng tm l C1,

C2 , C3 ly trn biu Mp ln lt l

y1 = 5 ; y2 = 252

455=

+ ;

y3 = 395)545(325 =+ (hnh 6.11e)

1p = EJ

MM p.1 = )...(1

332211 yyyEJ

++

= )3

95.225.45.2(

1++

EJ= -

EJ3

520

Thay 11 v 1p vo phng trnh chnh tc ta c

EJ3

64X1 -

EJ3

520= 0 suy ra X1 = kN125,8

64

520=

X1 c du dng chng t chiu ginh hnh 6.11b l ng

v biu mmen un ta p dng cng thc

M = X1. 1M + Mp

Ta ch cn xc nh tung tng ng vi m A,C cn BD biu gingMp v n ny biu 1M trng ng chun.

MA = - 45 + 4.8,8125 = -12,5 kNm

a)

2m 2m 1m

P m

A BC DEJ =Const

X =1b)

1

c)

1

42 12

3

d)

P m

e)12

3

g)5

11,25

12,5

h)

8,125

11,875

p

Hnh 6.11


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MC = - 5 + 2.8,8125 = 11,25 kNm

n c vo cc tung ta vc biu M nh hnh 6.11g

Trn dm khng c ti trng phn b nn v Q theo biu M ta p dng

cng thc (7) Qtr= Qph =l

MMtrph cho c 3 on AC,CB v BD

QAC =AC

AC

l

MM )( =

[ ]2

)5,12(25,11 = 11,875kN

QCB =CB

CB

l

MM )( =

2

)25,11(5 = -8,125kN

QBD =BD

BD

l

MM )( = [ ]

2

)5(5 = 0KN (lu :hnh v sai)

T cc gi tr v biu lc ct Q nh hnh 6.11h

(Gii li thbng phn mm FBTW, SAP 2000

Kt qu ni lc ca thanh do ti trng tnh gy ra

Tn PT n t i n t j N Mi Mj Qi Qj

(Tn) (T.m) (T.m) (Tn) (Tn)

1 1 2 0.000 -12.500 11.250 11.875 11.875

2 2 3 0.000 11.250 -5.000 -8.125 -8.125

3 3 4 0.000 -5.000 -5.000 0.000 0.000

4 4 5 0.000 -5.000 -5.000 -0.000 -0.000


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CHNG 7 : DM LIN TC

7.1 Khi nim chung

1.nh ngha: Dm lin tc l dm ch c 1 thanh thng t trn nhiu gi ta( s gi ta > 2)

- C 3 loi dm lin tc:a)

b)

c)

+ Dm lin tc n gin (hnh 7.1 a)

+ Dm lin tc c u tha (hnh 7.1 b)

+ Dm lin tc c u ngm (hnh 7.1 c)

2. Bc siu tnh ca dm lin tc.

Cng thc tnh: n = G0 - 3 (7-1)Trong G0: s lin kt tng ng lin kt loi 1

n: bc siu tnh ca dm lin tc

Th d 7.1: Dm trn hnh 7.1 a c G0 = 5 (gi cnh tng ng 2 lin ktloi 1) nn n = 5-3 = 2 ; hnh 7.1 b c G0 = 4; hnh 7.1c c G0 = 6 (lin kt ngmng ng 3 lin kt loi 1)

3. H cbn: Dm lin tc l h siu tnh nn ta dng phng php lc tnh. hnh 7.2 c nhiu cch la chn h cbn, tuy nhin ta phi c cch chnh cbn sao cho hp l gii bi ton chn h cbn theo hnh 7.2d l hp lnht.

7.2 Phng trnh ba mmen ca dm lin tc.

C th chn h c bn bng cch t khp vo cc mt ct trn gi trunggian, gi ny cho php 2 mt ct st 2 bn khp chuyn v gc tng i. V vy, nlc tha trong trng hp ny l cc mmen X1, X2. Chn nh th sn gin hnkhi tnh h s chnh, h s ph v s hng t do. Png trnh chnh tc ca phng

php lc cho h siu tnh bc 3 c dng:


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c)

b)

a)

X1

X2

X1 X2

X1 2X

11.X1 + 12.X2 + 13.X3 + 1p = 0 (1)

21.X1 + 22.X2 + 23.X3 + 2p = 0 (2)

31.X1 + 32.X2 + 33.X3 + 3p = 0 (3)

Tnhc bng cch nhn biu n v hnh v 7.3:

11 = =

3

1

11.

i EJ

MM=

1

1

EJ.

2

1.

1

1.1

l.

3

2.1 +

2

1

EJ.

2

1.

1

1.1

l.

3

2.1 =

1

1

3EJ

l +2

2

3EJ

l

12 = 21 =2

2

6EJ

l

22 =3

3

2

2

33 EJ

l

EJ

l

+

23 = 32 =3

3

6EJ

l

2p =2

1

EJ.2.y2 +

3

1

EJ.3.y3

Trong :

y2 =2

2

l

a; y3 =

3

3

l

bdo

2p =22

22.

EJl

a+

33

33

.

.

EJl

b

Tng qut:

ip =ii

ii

EJl

a.+

11

11

.

.

++

++

ii

ii

EJl

b


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a)

P1 1P' 2P 3P q

10 2 3 4EJ1 2EJ EJ 3 EJ4

l1 l2 l3 l4

b)

X1 X2 X3

c)X =11

d)1

M1

e)1

2M

g)1

3M

h)

2PP'11P 3P q

i)

m

m

C1C2 C3

C4

a1 b1 a2 b2 a3 b3 a4 b4

Thay tt c vo phng trnh 3 mmen cho gi th i trong trng hp chu titrng tc dng:

i

i

EJ

l

6.Mi-1 + (

1

1

33 +

++i

i

i

i

EJ

l

EJ

l ).Mi +1

1

6 +

+

i

i

EJ

l .Mi+1 +ii

ii

EJl

a. +11

11

.

.

++

++

ii

ii

EJl

b = 0 (7-2)

Trong :

li, li+1: chiu di ca nhp th i v i + 1

EJi, EJi + 1: cng chng un nhp i v i + 1

i, i + 1: din tch biu Mp nhp i v i + 1ai: khong cch t trng tm din tch in gi tri nhp i

bi+1: khong cch t trng tm din tch i+1n gi phi nhp i + 1

Trong dm lin tc n gin c n nhp ta s vit (n -1) phng trnh 3 mmencho gi ta trung gian. i vi dm lin tc u tha hay u ngm a v dm lintc bng cch sau:

1. Trng hp dm lin tc cu tha:


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a dm lin tc c u tha v dm lin tc n gin bng cch bu thai v thay tc dng ca ca ti trng phn u tha bng nhng ngoi lc t gi2 u ca dm lin tc n gin theo nguyn l di lc nh hc cl thuyt.Th d nh cc dm lin tc c u tha trn hnh 7.4 a v 7.4c c tha v dmlin tc n gin tng ng nh 7.4 b v 7.4 d

3

C1

0a)

M =P C

P21P1 2

q

C2

b)

P1 P2

o 1 1 M =P C3 22

c) 01 2 3

a

m

M =d)

o

2

P =q.a1

M =m3

P2

q.a2

2. Trng hp dm lin tc cu ngm:

a dm lin tc c ngm cng v dm lin tc n gin bng cch thayngm bng 1 nhp c 1 u l gi cnh, 1 u l gi di ng. Nhp ny c di

bng 0, cng l v cng. Th d nh cc dm lin tc c u ngm trn hnh 7.5 a

v 7.4c c tha v dm lin tc n gin tng ng nh 7.5 b v 7.5 d

a)10 2

a)210-1

Hoc EJ =

l = 00

c)210 3

0d) 1 2 3 4

0

4Hoc EJ =

l = 04

7.3 Cch v biu M, Q cho dm lin tc.


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1. Trnh ttnh ton:

h cbn chn bng cch t khp vo cc mt ct trn gi ta trunggian cht cc ti trng cho. Xem mi nhp nh dm n gin v Mp ri xcnh i, ai, bi cho cc nhp ca dm. Nu gp biu phc tp th chia thnh nhiu

phn n gin tnh ton

i.ai = =

n

j

ijija

1

.

i.bi = =

n

j

ijijb

1

.

ij: din tch th j trong nhp i

aij, b j: khong cch t trng tm din tch th j ca nhp i n gi triv gi phi.

- Vit phng trnh 3 mmen cho cc gi trung gian

- Gii cc phng trnh 3 mmen xc nh mmen gi.- V M, Q

2. Cch vM, Q.

v biu M cui cng v v biu Q, trn h cbn t cc ti trng cho v cc mmen gi tnh c khi gii phng trnh 3 mmen.

Khi v biu mmen un cui cng theo nguyn l cng tc dng: mmenun mt ct no (M) bng tng mmen do cc ti trng cho sinh ra vmmen do cc gi sinh ra ti chnh mt ct. Mmen un do ti trng cho sinhra ly trn biu Mp v k hiu l Mp, cn mmen un do cc mmen gi sinh raly trn biu mmen gi k hiu l Mg. Biu mmen gi vc bng cch cc gi 2 u mi nhp dng tung bng mmen gi tnh khi gii phng trnh3 mmen, sau ni nh cc tung ny bng ng thng. Nh vy cng thctnh mmen v biu cui cng l:

M = Mp + Mg (7.3)

Biu Q v theo biu M nh nghin cu chng 6

- Vi dm lin tc c u tha th v thm phn biu ca n u tha.

Th d: 7.2 V biu M v Q cho dm lin tc n gin trn hnh 7.6 a

Cho bit P = 10kN, q = 4 kN/m, l1 = l3 = 8m, l2 = 6m, EJ = Const.

Gii:Trn h cbn cht cc ti trng cho (hnh 7.6 b), coi mi nhp nh mt

dm n gin ta vc biu Mp (hnh 7.6 c). biu Mp c.

1 = 2802

8.20kNm= , a1 = b1= 4m, 2 = 2726.18

3

2kNm= , a2 = b2= 3m,3 = 0


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E

E K

q

a) 31 2

l1 l2 l3

l1/2 l1/2

P

2b)

P

1

q

3

a1 b1 b2a2

0

0

c)

20kNm18kNm

pM

15,34kNm

Mg4,43kNm

d)

M

e)

13,33kNm

15,34kNm

8,12kNm

4,43kNm

g)

3,08kNm

6,92kNm

13,82kNm

10,18kNm

0,55kNm

Q

Phng trnh 3 mmen ca gi 1 c dng:

01

1

6M

EJ

l+ )

33(

2

2

1

1

EJ

l

EJ

l+ M1+ 2

2

2

6M

EJ

l+

11

11

EJl

a+

22

22

EJl

b=0 (1)

Thay M0 = 0 v cc gi tr bit vo phng trnh (1) ta c :

)3

6

3

8(

EJEJ+ M1+ 26

6M

EJ+

EJ8

4.80+

EJ6

3.72=0

EJ314 M1+ 21 M

EJ+

EJ76 =0

14M1 +3M2 +228 = 0 (2)

Phng trnh 3 mmen ca gi 2 c dng:

12

2

6M

EJ

l+ )

33(

3

3

2

2

EJ

l

EJ

l+ M2+ 3

3

3

6M

EJ

l+

22

22

EJl

a+

33

33

EJl

b= 0 (3)

Thay M3 = 0 v cc gi tr bit vo phng trnh (3) ta c :


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EJ6

6M1+ )

3

8

3

6(

EJEJ+ M2 +

EJ6

3.72= 0

3M1 +14M2 +108 =0 (4)

Nhn phng trnh (2) vi 3 v phng trnh (4) vi -14 ta c :

42M1 + 9M2 + 684 =0 (5)

-42M1 -196M2 - 1512 =0 (6)

- 187M2 - 828 =0

M2=-828/187 - 4,3kNm

Thay vo phng trnh (4) ta c

3M1 +14(- 4,3) +108 =0

M1 =-2868/187 -15,3kNm

Trn ng chun v tr tng ng vi gi 1 dng v pha trn tung 15,3kNm, v tr tng ng gi 2 cng dng v pha trn tung 4,3kNm, ni nhny vi nhau v vi m khng cc gi u cnh n ta c biu Mg (hnh7.6d)

v biu M cui cng ta s dng cng thc (7.3)

Trong nhp 1:

M0 = Mp0 + Mg0 = 0 + 0 = 0 ;

ME = MpE + MgE = 20 - 7,67 = 12,33kNm; (E l mt ct gia nhp)

M1 = Mp1 + Mg1 = 0 - 15,3 = - 15,3kNm ;

Trong nhp 2: 3,151 =phM kNm

MK= MpK+ MgK = 18 - 9,883 = 8,12 kNm (K l mt ct gia nhp 2)

=trM2 Mp2 + Mg2 = 0 - 4,43 = - 4,43kNm ;

Nhp 3 : Biu Mp trng ng chun nn biu M han ton ging biu Mg

Da vo cc gi tr c ta v M cui cng hnh 7.6e

on OE : Q =l

MM trph )( =

4

)033,12( = 3kN

on E1 : Q =4

)33,1234,15( = - 6,92kN

on 12 : Qtr =l

MM trph )( +

2

.lq=

[ ]6

)34,15(43,4 +

2

6.4= 13,82kN

Qph =l

MM trph )( -

2

.lq=

[ ]6

)34,15(43,4 -

2

6.4= -10,18kN


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on 23 : Q =8

))43,4(0( = - 0,55kN

Da vo cc gi tr c ta v Q cui cng hnh 7.6g

giao trinh co ket cau - cĐgtvt ii

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