cơ kết cấu 1- bài giảng

8/6/2019 c kt cu 1- bi ging

1/83

PGS. TS. KIN QUC

KHOA K THUT XY DNG

BI GING CHC KT CU

nh nghi:

C hc kt cu (CHKC) l mn khoa hc L thuyt Thc nghim trnh by cc phng php tnhton kt cu v bn, cng v n nh do

cc nguyn nhn khc nhau: ti trng, nhit ,ln, ch to khng chnh xc.

1.KHI NIM MN HC

MU 2


2/83

Phng php nghin cu:

L thuyt T hc nghim:

L thuyt (LT): d bo khnng lm vic ca kt cu.

Thc nghim (TN): phthin tnh cht vt liuvk imtra l thuyt.

1.KHI NIM MN HC (TT)

MU 3

TN

LT

LT

LT

C s xy dngl thuyt

Kim tra lthuyt

Nhim v ch yu:

Xy dng cc phng php tnh ton ni lc,lm c s kim tra cc iu kin bn, cngv n nh (hin i: tui th, tin cy).


MU 4


3/83

V tr mn hc:

Qu trnh thit k cng trnh bao gm:


MU 5

Tnh ni lc Tnh tit din Kim tra bn, cng, n nh

CHKC & chuyn mn CHKC Chuyn mn CHKC & chuyn mn

Khu kh khn v quan trng nht

S kt cu

S tnh = S cng trnh + cc gi thit ngin ho.

2. S TNH CA CNG TRNH

MU 6

E, A, I


4/83

Cc gi thit gm:- Thay thanh bng trc thanh; bn & v bng mttrung gian.- Tit din E, A, I- Lin kt L tng (khng ma st, cng, nhi).- Ti trng a v trc thanh.- Thm gi thit ph nu cn (nt khp, tnggch, sn btng).

2. S TNH CA CNG TRNH

MU 7

E, A, I

Lu : Lc chn s tnh cn phn nh t t slm vic ca cng trnh tht v ph hp vi khnng tnh ton.

2. S TNH CA CNG TRNH (TT)

MU 8

E, A, I

Hnh 1


5/83

3. PHN LOI CNG TRNH

MU 9

c) Khungd) Vm

a) Dm b) Dn

Theo s tnh:

3. PHN LOI CNG TRNH (TT)

MU 10

Theo s tnh (tt):

H phng: cu kin v lc u nm trong mtphng.

H khng gian: Khng phng

Trong thc t ch yu l h khng gian: dmtrc giao, dn khng gian, kt cu tm v thd: nh cao tng, cu, dn khoangNhiu biton khng gian khi tnh ton c a v s h phng.


6/83


MU 11

Theo phng php tnh ni lc

Phng php lc:

H tnh nh: ch dng phng trnh cn bngl tm ni lc.

H siu tnh: phi b sung iu kin hnh hc(chuyn v, bin dng)


MU 12

Theo phng php tnh ni lc (tt)

Phng php chuyn v:Hxcnh ng: xc nh c bin dng ca cc

phn t thuc h ch t iu kin ng hckh ih bchuyn v cng bc.

H siu ng: khi h chu chuyn v cng bc, nu ch

dng iu kin ng hc (hnh hc) th khng xcnh bin dng ca cc phn t.

a) Hxcnh ng

b) Hsiu ng


7/83

Ti trng:Gy ra ni lc, chuyn v cho mi h. Mt s cch phnloi:

Theo v tr : bt ng

di ng

Theo tnh cht t c dng: tnh: gia tc nh, bqua lc qun tnh khixt cn bng.

ng: phi xt n lc

qun tnh trongphng trnh cn bng.

Theo kh nng nhn bit: tin nh: P = P(t)

ngu nhin: ch bit theoqui lut xc sut

4. CC NGUYN NHN GY NI LC V

CHUYN V

MU 13

Nhit

Ln

Hai nguyn nhn ny gy ni lc, chuyn v trong hsiu tnh, nhng ch gy chuyn v trong h tnh nh.

4. CC NGUYN NHN GY NI LC VCHUYN V (TT)

MU 14


8/83

Cc gi thit nhm n gin ho tnh ton:

1- Vt liu n hi tun theo nh lut Hooke.

5. CC GI THIT V NGUYN L CNG

TC DNG

MU 15

Cc gi thit nhm n gin ho tnh ton (tt):

5. CC GI THIT V NGUYN L CNGTC DNG

MU 16

2- Bin dng v chuyn v b (c dng nh khinim v cng b trong ton hc). Cho php dngs khng bin dng. Dng c cc xp x:

sin tan , cos = 1 T dn ti nguyn l cng tc dng:

P1 P2

P1

1

P2

2= +

Hnh 5

(P1, P2) = (P1) + (P2)


9/83

3. PHN LOI CNG TRNH

MU 17

3. PHN LOI CNG TRNH

MU 18


10/83

3. PHN LOI CNG TRNH

MU 19

3. PHN LOI CNG TRNH

MU 20


11/83

3. PHN LOI CNG TRNH

MU 21

3. PHN LOI CNG TRNH

MU 22


12/83

3. PHN LOI CNG TRNH

MU 23

3. PHN LOI CNG TRNH

MU 24


13/83

3. PHN LOI CNG TRNH

MU 25

3. PHN LOI CNG TRNH

MU 26


14/83

3. PHN LOI CNG TRNH

MU 27

3. PHN LOI CNG TRNH

MU 28


15/83

3. PHN LOI CNG TRNH

MU 29

3. PHN LOI CNG TRNH

MU 30


16/83

3. PHN LOI CNG TRNH

MU 31

3. PHN LOI CNG TRNH

MU 32


17/83

3. PHN LOI CNG TRNH

MU 33

3. PHN LOI CNG TRNH

MU 34


18/83

3. PHN LOI CNG TRNH

MU 35

3. PHN LOI CNG TRNH

MU 36


19/83

3. PHN LOI CNG TRNH

MU 37

3. PHN LOI CNG TRNH

MU 38


20/83

3. PHN LOI CNG TRNH

MU 39

3. PHN LOI CNG TRNH

MU 40


21/83

PGS. TS. KIN QUC

KHOA K THUT XY DNG

BI GING CHC KT CU

CHNG 1

1. H bt bin hnh (BBH)

nh ngha: H BBH l h khi chu ti trngbt kvn gic hnh dng ban u nu bqua bin dng n hi.

Tnh cht: c kh nng chu lc trn hnh

dng ban u p ng c yu cu sdng.

1.1 CC KHI NIM

Chng 1: Cu to hnh hc ca hphng 2


22/83

2. H bin hnh (BH)nh ngha: l h khi chu ti trng bt k s

thay i hnh dng hu hn nu coi cc phnt cng tuyt i.

Tnh cht: Khng c kh nng chu lc bt ktrn hnh dng ban u khng dng cnh l 1 kt cu.

1.1 CC KHI NIM (TT)


3. H bin hnh tc thi (BHTT)nh ngha: l h thay i hnh dng hnh hc

v cng b nu coi cc phn t cng tuyti (chnh xc hn: b qua lng thay i vcng b bc cao).

Th d: vi hnh bn ta c dn di L == VCB bc cao 0.. Tnh cht: kt cu mm, ni lc rt ln, nn

khng dng trong thc t.

1.1 CC KHI NIM (TT)


P

L L

2

2L


23/83

4. Ming cng (MC)nh ngha: MC l h phng BBH.

Th d:

1.1 CC KHI NIM (TT)


H BBH

Ming cng

ngha: gip kho st tnh cht hnh hc ca1 h phng d dng hn (ch quan tm tnhcht cng, khng quan tm cu to chi tit).

5. Bc t do (BTD)- Bc t do ca 1 h l s thng sc lp xcnh v tr 1 h so vi mc cnh.

- Bc t do cu 1 h l s chuyn v kh dc lpso vi mc cnh.

Trong mt phng, 1 im c 2 BTD (2 chuyn vthng), 1 m/c c 3 BTD (2 chuyn v thng, 1 gcxoay).

H BBH l h c BTD bng 0, h BH c BTD khc0. V vy, khi nim BTD c th dng k/s cuto hnh hc.

1.1 CC KHI NIM (TT)



24/83

1. Lin kt n gin Lin kt thanh: l thanh c khp 2 u.

1.2 CC LOI LIN KT (TT)


Tng nglin kt thanh

Tnh cht: kh 1 bc t do, pht sinh 1 phn lc(ni 2 khp).1 m/c c 2 khp th tng ng 1 lin ktthanh

1. Lin kt n gin (tt) Lin kt khp:

Tnh cht: kh 2 BTD, phtsinh 2 thnh phn phn lctheo 2 phng xc nh.

V mt ng hc, 1 khptng ng vi 2 lin ktthanh.

Giao ca 2 thanh tngng vi khp gi to. Vtr ca khp gi to K thayi khi B dch chuyn sovi A khp tc thi.

1.2 CC LOI LIN KT



25/83

1. Lin kt n gin (tt)

Lin kt hn:

Ni cng 2 ming cng vi nhau thanh 1ming cng ln. n gin vic kho stcu to hnh hc, nn gom li t s mingcng nht v ch nn quan nim lin kt chgm thanh v khp. V vy phn sau skhng bn n lin kt hn na v ch lmphc tp.

1.2 CC LOI LIN KT


2. Khp phc tp L khp ni nhiu ming cng vi nhau.

phc tp ca khp phc tp l s khp n gintng ng v mt lin kt.

p = D - 1

p phc tp ca khp tng ng s khpn gin

D s ming cng ni vo khp K. Mcch: qui i tt c lin kt dng trong h thanh

thnh s lin kt thanh tng ng.

1.2 CC LOI LIN KT


A

BC

B

A

C =K K1K2


26/83

1.2 CC LOI LIN KT


1.2 CC LOI LIN KT



27/83

1.2 CC LOI LIN KT


1.2 CC LOI LIN KT



28/83

1.2 CC LOI LIN KT


1.2 CC LOI LIN KT



29/83

1.2 CC LOI LIN KT


1. iu kin cn:L in kin v s lng lin kt ni cc ming cngthnh 1 h BBH.

a)H bt kH gm D ming cng, ni vi nhau bng T thanh v Kkhp n gin.

S bc t do: Coi 1 ming cng l cnh th cn khi 3(D-1) = BTD bc t do.

S lin kt thanh qui i: T + 2K = LKLp hiu s:

n = LK BTD = T + 2K 3(D-1)

n < 0 : khng lin kt BHn = 0 : lin kt

n > 0 : d lin kt

1.3 NI CC MING CNG THNH H BBH


Phi xt thm iukin kt lun.


30/83

1. iu kin cn (tt):b) H ni t

H c D ming cng ni vi t bng C thanh (qui i).

S BTD = 3DS lin kt qui i: LK = T + 2K + C

Hiu s:

n = T + 2K + C 3Dn < 0 : khng lin kt BHn = 0 : lin kt

n > 0 : d lin ktQuii lin kt thanh :

1.3 NI CC MING CNG THNH H BBH (TT)


Phi xt thm iukin kt lun.

1 32

1. iu kin cn (tt):c) H dn Gm cc thanh thng, ni khp 2 u. Gi s dn c D thanh v M mt. Coi 1 thanh l mingcng cnh th ch cn li D 1 lin kt thanh, khc2(M 2) bc t do. Nh vy:

n = D -1 - 2(M - 2) = D + 3 - 2M Nu h nit th:n = D + C - 2M



< 0 : BH

0 : Xt iu kin

< 0 : BH 0 : Xt iu kin

D thanhM mt


31/83

2. iu kin :

a) H gm 2 ming cng

Cn : dng s lin kt qui i ti thiu tngng 3 thanh.

: + 3 thanh khng ng qui hoc songsong.

+ 1 thanh khng i qua khp.



2. iu kin (tt):

b) H gm 3 ming cng

Cn : dng s lin kt qui i ti thiu tngng 6 thanh

: 3 khp thc hoc gi to khng thng

hng.




32/83

2. iu kin (tt):

c) Bi

nh ngha : bi l 2 lin kt thanh khngthng hng, ni 1 im vo 1 h cho.

Tnh cht : thm hoc bt bi khng lmthay i tnh cht hnh hc ca h. Do , kho st tnh cht hnh hc c th dngphng phppht trin bi hoc loi tr bi..



2. iu kin (tt):

d) Cch kho st tnh cht hnh hc ca 1 h

C gng gom v t ming cng nht (2 hoc3) v dng iu kin cn v kt lun. Vihn gin, c th dng ngay iu kin , cgng li dng tnh cht ca bi.

Nu s ming cng nhiu hn 3 th phidng phng php tng qut (v cng phctp hn) nh ti trng bng 0, ng hc, thayth lin kt.




33/83

3. Mt s th d



K

a) BHTT

I II

III

(1,2)

c) BHTT (gn BHTT: khng tt)

(1,3)

f) BHTT

I II

III

B i

b) BBH

(2,3)

e) BHTT


34/83

PGS. TS. KIN QUC

KHOA K THUT XY DNG

BI GING CHC KT CU

CHNG 2

1. H n gin

H dm: thanh thng, chu un l ch yu(thng N = 0).

2.1 PHN LOI V C IM CHU LC CAH KT CU

Chng 2: Xcnh ni lc do ti trng btng 2


35/83

1. H n gin

H dm:

2.1 PHN LOI V C IM CHU LC CA

H KT CU


1. H n gin

H dm:




36/83

1. H n gin (tt)


H KT CU


H khung: thanh gy khc, ni lc gm M, Q, N.

1. H n gin (tt)



H khung:


37/83

1. H n gin (tt)


H KT CU


H khung:

1. H n gin (tt)

2.1 PHN LOI V C IM CHU LC CAH TNH NH


H khung:


38/83

1. H n gin (tt)

H dn:


H KT CU(TT)


t Mt Bin trn

Bin di

Thanh xin

Thanh ng

Nhp

Hnh 2.3

Trong thc t, mt dn l nt cng h siu tnh phc tp. n gin ho, dng cc gi thit sau: Mt dn l khp l tng.

Ti trng ch tc dng mt dn.

Trng lng khng ng k ( b qua un thanh).

u im: tit kim vt liu kt cu nh, vt nhp ln.

Ni lc ch clc dc N 0

1. H n gin (tt)

H dn:

2.1 PHN LOI V C IM CHU LC CAH TNH NH (TT)



39/83

1. H n gin (tt)

H dn:


H KT CU(TT)


1. H n gin (tt)

H dn:

2.1 PHN LOI V C IM CHU LC CAH KT CU(TT)



40/83

1. H n gin (tt)H 3 khp


H KT CU(TT)


Ni lc: M, Q, N; Lc dc nn: dng vt liu dn.

Phn lc: c lc x nn kt cu mng bt li hn.

2. H ghp

c ni bi cc h n gin. Thng c 2 loitrong thc t:



Dm tnh nh nhiu nhp

Khung tnh nh nhiu nhp


41/83

2. H ghp (tt)


H KT CU(TT)


Dm tnh nh nhiu nhp


V cu to: gm h chnh v ph.

Chnh : BBH hoc c kh nng chu lc khi bkt cu bn cnh.

Ph : BH hoc khng c kh nng chu lc khib qua kt cu bn cnh.

2. H ghp (tt)



Dm tnh nh nhiu nhp


Cch tnh: t ph chnh; truyn lc t ph sang chnh.


42/83

3. H lin hp (Xem sch)

Lin hp cc dng kt cu khc nhau nh dm vm, dm dy xch, dn vm


H KT CU(TT)


3. H lin hp




43/83

3. H lin hp


H KT CU(TT)


3. H lin hp




44/83

4. H c mt truyn lc

Mt truyn lcctcdng c nh v tr ti trngtc dng vo kt cu chnh.


H KT CU(TT)


H thng dm truyn lc

Nhp

Mt truyn lc

4. H c mt truyn lc




45/83

4. H c mt truyn lc


H KT CU(TT)


4. H c mt truyn lc




46/83

1. Ni lc:M, Q, N

M : v theo th cng.

Q & N : ghi du ( qui c nh SBVL).

2.2 NI LC TRONG H DM & KHUNG N GIN.


Q

N

M

Hnh 2.7

2. Phng php v:

Phng php mt ct :

Tnh phn lc.

Chia on (ph thuc q, P, trc thanh).

Lp biu thc tng on. V


2.2 NI LC TRONG H DM & KHUNG N GIN (TT)


47/83

2. Phng php v (tt):

Phng php c bit :

Tnh phn lc.

Chia on.

Nhn xt dng biu & im c bit.

Tnh im c bit v v biu .



3. Th d:


Cho h c lin kt v chu lc nh hnh v. Hyv biu M, Q, N.

q

P= qa

a

a

2qa

2



48/83

3. Th d (tt):


q

P= qa

a

aVD = qaVA = 0

HA = qa

Phn lc:HA = P = qa

2qa

2

Ni lc:

qa2

M Q N

qa

qaqa

qa

2qa

2 2qa

8

Ch : ntcn bng

P = qaqa

qaqa

Hnh 2.10

qa2

2qa

2

2qa

2


1. Phng php tch mt: Ni dung:

Ln lt tch mt v vit phng trnh cnbng lc thu c cc phng trnh tm ni lc.


d d dd

P

h

A B

12

3

N1

N2

2.3 TNH TON H DN (TT)


49/83

1. Phng php tch mt (tt): Trnh t & th thut:

Trnh t: tch mt sao cho mt mt ch c2 lc dc cha bit.

Th thut: lp 1 phng trnh cha 1 n:loi b lck iabng cch chiu ln phngtrnh vung gc vi n.



d d dd

P

h

A B

12

3

N1

N2

A

1

N2

N1

y

x

1. Phng php tch mt (tt):

Th d:

Cho h dn c lin kt v chu ti trng nhhnh v. Hy xc nh ni lc thanh N1, N2



d d dd

P

h

12

3

N1

N2


50/83

1. Phng php tch mt (tt): Th d (tt): Gii



d d dd

P

h

A B

12

3

N1

N2

A PY 0: N sin A 0 N - -

2 2 sin 2sin+ = = =

1 2 1 2

PX 0: N N cos 0 N -N cos - cotg

2= + = = =

A =

1

N2

N1

y

x

P

2

1. Phng php tch mt (tt):

Nhn xt:

Mt c 2 thanh, khng c ti trng: N1=N2=0.

Mt c 3 thanh: N1 = N2 = 0; N3 = 0



N1

N2

N3

N1 N2

Nhc im:

D b sai s truyn


51/83

2. Phng php mt ct n gin Ni dung:

Ct dn ( khng nhiu hn 3 thanh). Lp 3phng trnh cn bng gii 3 n.



d d dd

P

h

A= B

N1

N3

N2

J

P

IP

2

2. Phng php mt ct n gin (tt)

Th thut:

Lp phng trnh cha 1 n, bng cch loi i2 lc cha cn tm.

Nu 2 thanh song song: chiu ln phngvung gc.

Nu 2 thanh ct nhau: ly mmen vi imct.



d d dd

P

h

A= B

N1

N3

N2

J

P

IP

2


52/83

2. Phng php mt ct n gin (tt) Th d:

Cho h dn c lin kt v chu ti trng nhhnh v. Hy xc nh ni lc trong thanh N1,N2, N3.



d d dd

P

h

N1

N3

N2

J

P

I

2. Phng php mt ct n gin (tt)

Th d: (Gii)



d d dd

P

h

A= B

N1

N3

N2

J

P

IP

2

1

2

03

.20

0 sin sin

d

I

I

d

J

J

d

MAdM N

h h

MA dM N

h h

QA

Y N

= = = = = =

= = =

Nhn xt:- Thanh bin : du v tr s - Thanh xin : duvtr s Qd

dM

h


53/83

3. Phng php mt ct phi hp Ni dung:

Khi s n ln hn 3 dng 1 s mt ct phi hp to s phng trnh. Trong thc tthng dng nhiu lm l 2 mt ct.



A= B

N2

N1

P

1

1

2-2

P

2

3. Phng php mt ct phi hp (tt)



N2

N1

P

1

1

2-2

Th d:

Cho h dn c lin kt v chu ti trng nh hnhv. Hy xc nh ni lc trong thanh N1, N2, N3


54/83

3. Phng php mt ct phi hp (tt)



A= B

N2

N1

P

1

1

2-2

P

2

1 2 2 1

A PY 0 N cos N cos A 0 N N

cos 2cos

+ = = =1 2 1 2

X 0 N sin N sin 0 N N + = = M/c 1-1:M/c 2-2 (tch mt):

1

PN

4cos

2

PN

4cos=

Th d (tt): Gii

1. Tnh phn lc

Phn tch phn lc nh hnh v. Mi phng trnhcn bng ch cha 1 n:


2.4 TNH TON H 3 KHP

d

B A

d

A B

Trai

C A

Phai

C A

M 0 V

M 0 V

M 0 Z

M 0 Z

= = = =

B

A

VdAVA

HA

ZA

ZB

VBVdB

HB

P3

P1

P2C

Sau , c th phn tch phn lc theo phngng v ngang. Nu ti trng thng ng th:HA = HB = H Lc x ca h 3 khp


55/83

2. Tnh ni lc

-Vm 3 khp: thit lp biu thc ni lc theo ta z. Biu M,Q, N v theo trc chun nmngang. Ring vm th qui c N>0 l nn.

-Khung 3 khp: v biu ni lc theo im cbit.


2.4 TNH TON H 3 KHP (TT)

B

A

VdAVA

HA

ZA

ZB

VBVdB

HB

P3

P1

P2C

3. Th d:

Cho h c lin kt v chu ti trng nh hnh v.Hy v biu M, Q, N



q

a

a a

A B

C


56/83

3. Th d (tt): Gii



q

a

a

qa

H= qa/2H

qaa

A B

C

A B

C

M

A B

C

Q

A B

CN

qa

qa

qa/2 qa/2

qa/2

qa qa

2qa

2

2qa

2

Trnh t tnh

Tch h ghp ra cc h n gin.

Tnh h ph.

Truyn lc t h ph sang chnh v tnh hchnh.

Ghp cc biu li.


2.5 TNH TON H GHP


57/83

Th d:

Cho h ghp c lin kt v chu ti trng nh hnhv. Hy v biu M, Q


2.5 TNH TON H GHP (TT)

3 3 2 8 m

P = 40 kN q = 10 kN/m

Th d:



3 3 2 8 m

P = 40 kN q = 10 kN/m

P = 40 kN

q = 10 kN/m20 kN

20 kN20 kN

Q20

20

45

35

6040

80

60

M

(kN/m)

(kN)


58/83



3 3 2 8 m

P = 40 kN q = 10 kN/m

6040

80

60

M

(kNm)

40 kN q = 10 kN/m

75 80

Th d (tt) So snh vi dm n gin:

Trnh t tnh

Truyn lc t dm ph xung dm chnh.

Tnh dm chnh.

Th d:


2.5 TNH TON H C MT TRUYN LC

q


59/83

PGS. TS. KIN QUC

KHOA K THUT XY DNG

BI GING CHC KT CU

CHNG 3

1. Ti trng di ng v phng php tnh

Ti trng ding: c v tr thay i gy rani lc thay i.

Th d: Xe la, t, ngi, dm cu chy

Vn cn gii quyt: Cnt mSmax (ni lc, phnlc )

3.1 PHNG PHP NG NH HNG

Chng 3: Xcnh ni lc do ti trng ding 2

z

K

Hnh 3.1


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1. Ti trng di ng v phng php tnh (tt) Cc phng php gii quyt:

Gii tch: lp biu thc gii tch S(z) v khost cc tr: phc tp khng dng.Th d:

ng vi 5 v tr ca ti trng

ng nh hng: dng nguyn l cng tcdng. c dng trong thc t.

3.1 PHNG PHP NG NH HNG (TT)


1

2

k

5

S ( )...

S

Sz

S

=

2. Phng php ng nh hng

nh ngha:

th ca i lng S theo v tr mt lc tptrung P=1 (khng th nguyn) c phngchiu khng i, di ng trn cng trnh.

K hiu: ah S hoc S




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2. Phng php ng nh hng (tt)

Trnh tv S:

t P=1 ti v tr Z; coi nh lc bt ng.

Lp biu thc S=S(z), thng gm nhiubiu thc khc nhau cho nhiu onkhc nhau.

Cho z bin thin v v th S=S(z).




Quic:

ng chun vung gc P=1 (hoc //trc thanh)

Trung vung gc ng chun.

Trung (+) dng theo chiu ca P.




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2. Phng php ng nh hng (tt) Ch

Phn bit s khc nhau gia ah S vbiu S.

Th nguyn tung ah =

Th d :



[S]

[P][M] F-L

["M"]= = =L[P] F


Th d:

V ng nh hng A, B, Mk, Qk



a b

P = 1

K

A B

L

z


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2. Phng php ng nh hng (tt) Th d (tt):

Phn lc:



L-zA=

L

zB=

L

A1

B1

a b

P = 1

K

A B

L

z

z


Ni lc:ah gm 2 on: ng tri v ng phi.Xt cn bng phn t lc n gin hn(phn khng c lc P=1).



tk

tk

z= -B = -Q

L

b= B.b = zM

L

0 z a

b

Qk

t

K B

Mk

t

ng tri

a b

P = 1

KA B

L

z

z


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Ni lc:



p

k

pk

L-z=A=Q

L

a=A.a= (L-z)M

L

a z L ng phi

a

QkpK

A

Mk

p

a b

P = 1

KA B

L

z

z


Ni lc (tt):



a b

P = 1

K

A B

L

z

Mk

b. tri . phia


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Ni lc (tt) :



1

Qk1. tri

. phi

a b

P = 1

K

A B

L

z

Xt dm n gin c u tha v l tr ng hptng qut ca dm n gin v dm cng xn.

3.2 NG NH HNG TRONG DM TNH NHN GIN


P = 1

L


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1. ng nh hng phn lc

3.2 NG NH HNG TRONG DM TNH NH

N GIN


B

A

L-zM 0 : A=

L

zM 0 : B=

L

=

V ah vi 2 tung ti Av B, tc l z= 0 v z= L

bc 1

P = 1

A B

L

1

1

A

B

2. ng nh hng ni lc (tt)

Tit din trong nhp:

Mk1: tri giao phi di k1 cch v nhanh.

3.2 NG NH HNG TRONG DM TNH NHN GIN (TT)


Mk1

a . tri . phi

ab

K1A B

L

K3K2

c


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2. ng nh hng ni lc (tt)

Tit din trong nhp (tt):

Qk1: tri song song phi v nhanh.


N GIN (TT)


Qk1

1

1. tri

. phi

ab

K1A B

L

K3K2

c

2. ng nh hng ni lc (tt):

Tit din trong nhp (tt):

Ch : v

3.2 NG NH HNG TRONG DM TNH NHN GIN (TT)


p

AQt

BQ

QAP

1

1. tri

. phi

ab

K1A B

L

K3K2

c


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2. ng nh hng ni lc (tt):

Tit din u tha:

Ch : ging dm cngxn.


N GIN (TT)


ab

K1A B

L

K3K2

c

Mk2b

Qk31

Mk3c

Qk2 1

v ah thuc h chnh, thc hinccbcsau:

1) V ah, coi P=1 di ng trc tip trn h chnh.

2) Gi li tung di mt truyn lc.

3) Ni cc tung bng cc on thng.

3.3 NG NH HNG CA H C MTTRUYN LC


P = 1


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Chng minh:

3.3 NG NH HNG CA H C MT

TRUYN LC (TT)


d

i i+1

d-z zR = ,R =

d d

Mk =Riyi + Ri+1yi+1= bc 1 ng thng.Khi z=0 Mk = yi

z=d Mk = yi+1

a

yi yi+1Mk

1

1

Qk

RiaK Ri+1

i i+1

P = 1z

1. ng nh hng thuc h ph

Khi P=1 di ng trn h ph: v ah nh i vih n gin.

Khi P=1 trn h chnh: ah = 0.

3.4 NG NH HNG CA H GHP


K2 K1K3

Mk1


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2. ng nh hng thuc h chnh Khi P=1 trn h chnh: h ph khng lm vic

xt ring h chnh.

Khi P=1 trn h ph: ah l ng thng i quatung ng di khp ni h chnh vi ph, vtung =0 ng di gi ta t ca dm ph(lin kt thng ng).

3.4 NG NH HNG CA H GHP (TT)


K2 K1K3

Qk2

Mk3

K

O

(0,1)

(1,2)

(0,2) (2,3)

L

I II III(0,3) M

k

Ch :

Nu h ghp phc tp, c th dng phng phpng v dng ah, sau tnh 1 tung cbit v suy ra cc tung khc.




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K

O

(0,1)

(1,2)

(0,2) (2,3)

L

I IIIII

(0,3) Mk

Ch : Th d:



Phng php ng v ah:

3 khp tng h ca 3 ming cng ca 1 h BHthng hng: (1,2) + (2,3) = (1,3).

Tung ng vi khp ni vi t th bng 0(khng c chuyn v ng)

1. ng nh hng phn lcPhn lc c tnh tng t nh trong dn dm.

3.5 NG NH HNG TRONG DN DM


h

N1

N3 N4N2

L = 4d 2d2dA B

C DE

1

1

A B

A B

A

B

B A

L-z zM 0 : A= M 0 : B=

L L= =


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2. ng nh hng ni lc bng phng phpmt ct n gin

M/c trong nhp: N1 v N2Ct t cha N1 v N2.

1/ P=1 bn tri t b ct: xt cn bng phnphi ( t lc)

2/ P=1 bn phi t b ct: xt phn tri.

3/ P=1 trong t ct: ng ni.

3.5 NG NH HNG TRONG DN DM (TT)


h N1

N3 N4N2

L = 4d 2d2d A BC DE

2. ng nh hng ni lc bng phng phpmt ct n gin (tt)

M/c trong nhp: N1



h

N1

N3 N4N2

L = 4d 2d2d A BC DE

. tri . phi

. ni

A BC DN1

d

h3d

h


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2. ng nh hng ni lc bng phng phpmt ct n gin (tt) M/c trong nhp: N2



h N1

N3 N4N2

L = 4d 2d2dA B

C DE

. tri. phi

. ni

ABC

DN2

1

cos

1

cos

2. ng nh hng ni lc bng phng phpmt ct n gin (tt)

M/c u tha: N31/ P=1 bn tri t b ct

2/ P=1 bn phi t b ct




h N1

N3 N4N2

L = 4d 2d2d A BC DE

. ni. tri

. phi = 0AEN3

1

cos


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3. ah ni lc bng phng php tch mtLp biu thc ni lc khi:

1/ P=1 t ti mt

2/ P=1 ngoi t ct


Minh ha N4



h N1

N3 N4N2

L = 4d 2d2d A BC DE

1

P=1 ngoi t ct. ni

P=1 ti mtAE C B

N4

1cos

P = 1N4 = 0

A = 1

P=1 ti mt

N4 = -A

A

P = 1

P=1 ngoi t ct

1. Ti trng tp trung

Dng nguyn l cng tc dng

3.6 XC NH I LNG S BNG AH


n

i

i i

1

S P y=

= Pi PnP1

y1 yi yn

S


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1. Ti trng tp trung (tt)

3.6 XC NH I LNG S BNG AH


K

Pyt

yp

Qk

Ch : Nu S c bc nhy:

St = P.ypSp = P.yt

2. Ti trng phn b

Trng hp thng gp: q = const

3.6 XC NH I LNG S BNG AH (TT)


b b

a aS yqdz q ydz q= =

S q

S

q

ba

dz


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3. Momen tp trung

Th M bng ngu lc



M

Pdz

=S ( ) - .

.

P y dy Py P dy

M dydy M M tg

dz dz

= + == = =

dz

M > 0

yi y + dy

Nu c nhiu momen

1

n

i

i

S M tg =

= 0 : ham tangtg >NuSb gy:

St= MtgpSp = Mtgt pt

M

Th d: Tnh Mk, v bng phng php ah



t

kQ kQ

q P = qL

L L

K


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Th d (tt):Tnh Mk



q P = qL

L L

K

P = 1

L/2

Mk

230.5 0.5 0.5

4k k M Py q qL L q L L qL+ = + =

Th d (tt):

Tnh



q P = qL

L L

K

Pyt

yp

Qk

Hnh 3.11t

k R

p

k L

Q Py q qL q L qL

Q Py q qL q L qL

10.5 0.5 ( 0.5)

4

3( 0.5) 0.5 ( 0.5)

4

= + = + == + = + =

t

kQ kQ


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Th d (tt):

Kim tra li



t

k

p

k

Q qL qL qL

Q qL qL qL

5 1

4 4

1 3

4 4

= == =

5

4V qLA

=

q P = qL

L L3

4V qLB

=

5 3 2

4 2 4

L M qL L qL qL

k= =


3.7 AH GM CC ON THNG


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Tnh cht:C th thay tc dng ca cc ti trng trn tngphn thng ca ah bng hp lc ca chng.

Chng minh:

3.7 AH GM CC ON THNG


n n

i i i i i i

i=1 i=1

S = P y = P tg.z = tg P z Theo nh l Varinhng

v zotg = yo S = RyoCh : vi ti trng phnb cng chng minhtng t.

i i oP z = Rz Pi PnP1

y1 yi yn

S

yo

Ozo

ziR

1. on ti trng tiu chun v v tr bt li

L on ti trng dng thit k kt cu, tuntheo qui phm v ti trng, khong cch

V tr bt li l v tr ca on ti trng gy racc tr Smax(min)

3.8 DNG AH XC NH V TR BT LI



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2. Biu hin gii tch ca v tr bt li

Vi ah S v on ti trng tiu chun c th lpc biu thc gii tch ca S(z). V tr cho cc trca S nh sau:

Nu S(z) l hm trn:

iu kin: dSdz

= 0

3.8 DNG AH XC NH V TR BT LI (TT)


2. Biu hin gii tch ca v tr bt li (tt) Nu S(z) l hm khng trn v cc tr ti im gy th

biu hin cc tr nh hnh v di y:

iu kin cn Nu c cc i ti im ang xt th:

Tng t, nu cc tiu th:

Cc tr:



' ' '

t max

S =S -S 0 Sp < 0 St > 0

Sp = 0 St = 0

Sp < 0

Cc i


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3. ng nh hng a gic1- Cc tr ca S ch c th xy ra khi c t nht mt ti trngtp trung t ti nh ca ng nh hng.

S = Riyi(z)S = Riyi(z)S = Ritg i , tg i = const

cho cc tr th cn thit phi c St Sp, do Ri phi cthay i, tc l c t nht 1 lc tp trung t ti 1 nh cang nh hng. Lc gi l lc ti hn Pth.



12 3

S

R1

Ri

Rn

y1 yiyn

3. ng nh hng a gic (tt)

2- Nu Pth t ti nh li th c th cho Smax; ngc li, tti nh lm th c th cho Smin.

St = Ritg i + Pthtg tSp = Ritg i + PthtgpS= Pth(tgp - tg t)S= Pthtg 0, nu nh lm Smin



lilm t p

R1 RnRpRt Pth

S


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4. Cch tm Smax hoc Smin trong thc t

Nu on ti trng ngt c th ch t ln ngnh hng 1 du (du (+) tm Smax, du (-) tm Smin).

t ti trng ln ln cc tung ln, thng tPmax ln tung ymax (v S =Piyi).

Nu cn c th th 1 s phng n t ti.



5. Khi nim biu bao

nh ngha: l biu th hin ni lc ln nht v nh nhtti mi tit din, do ng thi tnh ti vhot ti gy ra.

Th d: Xc nh cc tit din cn tnh ni lc: 0, 1, , 6. V biu do tnh ti.

V ng nh hng cc tit din.Tnh ni lc do hot ti.

= P.y2max

= P.y2min

Xc nh cc gi tr bao

= Mtnh +

= Mtnh +



hoat tai

2maxM

hoat tai

2minM

bao

maxM

bao

minM

hoat tai

maxMhoat tai

minM

0 1 2 3 4 5 6

P (di ng)q


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5. Khi nim biu bao (tt)

Th d (tt):



0 1 2 3 4 5 6

P (di ng)q

M2t

Mtnh

Mbao

bao

maxM

bao

minM

M1y1

max

y1minPP

y2max

P y2min

P

M2

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cơ kết cấu 1- bài giảng