sbvl1 - ch1 - noi luc thanh (2012)

7/30/2019 SBVL1 - Ch1 - Noi Luc Thanh (2012)

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Chng 1

NI LC TRONG BI TON THANH


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Chng 1. Ni lc trong bi ton thanh

NI DUNG1.1. Cc thnh phnnoi lc trn mt ct ngang

1.2. Biu ni lc PP mt ct bin thin

1.3. Lin h vi phn gia m men un, lc ct v ti trng phn b

1.4. Phng php v biu ni lc theo im

c bit1.5. Biu ni lc ca dm tnh nh nhiu nhp

1.6. Biu ni lc ca khung phng

University of Architechture


3/41

1.1. Cc thnh phn noi lc trn mtct ngang (1)

Trong trng hp tng qut trn mtct ngang ca thanh chu tc dng cangoi lc c 6 thnh phnnoi lc:

y

z

xMx

My

Mz Qx

NZ

Qy



4/41


Bi ton phng: Ngoi lc nm trong mtphng i qua trc z (yOz) => Ch tn ticc thnh phn ng lc trong mt phng

ny: Nz, Mx, Qy

Nz - lc dc; Qy - lc ct; Mx m men un

y

z

xMx

NZ

Qy



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Qui c du cc thnh phn ng lc Lc dc: N>0 khi c chiu i ra khi mt ct

Lc ct: Q>0 khi c chiu i vng quanh phnthanh ang xt theo chiu kim ng h

M men un: M>0 khi lm cng cc th di

N

N

xc nh cc thnh phn ni lc: PHNG PHP MT CT



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N

Q Q

M M

1

1



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Cch xc nh cc thnh phn noi lc Gi thit chiu cc thnh phn M, N, Q theo chiu

dng qui c

Thit lp phng trnh hnh chiu ln cc trc z, yv phng trnh cn bng m men vi trng tmO ca mt ct ngang

0 => N= ...Z 0 => Q= ...Y 0 => M= ...OM



8/41

1.2. Biu ni lc (1)

Khi tnh ton => cn tm v tr mt ctngang c tr snoi lc ln nht =>

biu Biu ni lc - l th biu din sbin thin ca cc thnh phnnoi lc

theo to mt ct ngangCc bc v biu ni lc Phngphp mt ct bin thin



9/41

1.2. Biu ni lc (2)

a. Xc nh phn lc ti cc lin ktb. Phn on sao cho tren moi oan khong

colc tap trung & momen tap trungc. Vit biu thc xc nh cc thnh phn nglc N, Q, M theo to mt ct ngang

bng phng php mt ctd. V biu cho tng on cn c vo

phng trnh nhn c t bc (c)e. Kim tra biu nh vo cc nhn xt

mang tnh trc quan, tnh kinh nghim.



10/41

1.2. Biu ni lc (3)

Biu lc dc, lc ct v theo qui cv mang du

Biu m men lun v v pha thcng

N, Q

z

M

z



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V d 1.1 (1)

V biu noi lc trn cc mtctngang ca thanh chu titrng nh hnh v

GII:

1. Xc nh phn lc VAVB

F

a b

C

0A BM V a b Fa

0B AM V a b Fb

B

FaV

a b

A

FbV

a b

Th li: 0YUniversity of Architechture


12/41

V d 1.1 (2)

F

a bVA VB

C

1

1

Mt ct 1 1:

VAz1

Q

M

N

0N

10 z a

0A A

FbY Q V Q V

a b

Mt ct 2 2:

1

0 1 10

A A

FbzM M V z M V z

a b

0N

20 z b

0B B

FaY Q V Q V

a b

2

0 2 20

B B

FazM M V z M V z

a b

2

2

VB

z2Q

M

N

on AC

on BC

A B



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V d 1.1 (3)

Nhn xt 1 Ti mt ct c lc tptrung => biu lcct c bc nhy, ln bc nhy bnggi tr lc tp trung, xtt tri qua phi, chiubc nhy cng chiulc tp trung

F

a bVA VB

Fb

a+b

a+b

Fa

+

N

M

Q

M

Fab

a+b

F

C :

FbAC Q a b

:

FaBC Q

a b

1:Fbz

AC M

a b

2:

FazBC M

a b



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V d 1.2 (1)

L

q

VA VB

V biu cc thnh phnnoi lctrn cc mt ct ngang ca thanhchu ti trng nh hnh v

GII1. Xc nh cc phn lc lin kt

2

. 02

A B

qlM V l

.2

Aq lV

2. 0

2B A

qlM V l

.

2B

q lV

Bi ton i xng:.

2A B

q lV V

Hoc:

2. Biu thc ni lc

Xt mt ct 1-1(0 zL)

.

2

qlQ q z

2. .2 2

ql qM z z

1

1

Qz

VA

M

Nq

0AY Q qz V

2

00

2A

qzM M V z



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V d 1.2 (2)

Nhn xt 2

Ti mt ct c lc ctbng 0, biu mmen t cc tr

L

q

VA VBqL/2

qL/2

+

Q

L/2

qL2/8

M

.2qlQ q z

2. .2 2

ql qM z z

0 2AqL

z Q

2B

qLz L Q

0 0Az M

0Bz L M

2

qLM ' qz 0

2

LM ' z

0M '' q

2

2

8

max z L /

qLM M



16/41

V d 1.3 (1)

1. Xc nh phn lc:

.( ) 0A BM V a b M

.( ) 0B AM V a b M

B

MV

a b

A

MV

a b 2. Lp cc biu thc noi lc: AC: Xt mt ct 1-1 ( 0 z1 a)

y A

MQ V

a b

VA VBa b

C

M

.x AM V z

QVA

M

z1 VB

M

Q z2

1

1

2

2

BC: Xt mt ct 2-2 ( 0 z2 b)y A

MQ V

a b

2.x BM V z



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V d 1.3 (2)

a bVA VB

M(a+b) M

(a+b)Ma

(a+b)

Mb(a+b)

M

Q

M

C

M

y A

MQ V

a b

1.x AM V z

AC: ( 0 z1 a)

y A

MQ V

a b

BC: ( 0 z2 b)

2.x BM V z

Nhn xt 3

Ti mt ct c m men tptrung, biu m men c bcnhy, ln bc nhy bnggi tr m men tp trung, xt ttri qua phi, mmen tp trungquay thun chiu kim ng hth bc nhy i xung


1 3 Li h i h i


18/41

1.3. Lin h vi phn gia m men un,lc ct v ti ngang phn b (1)

Xt dm chu ti phn bq(z)>0: hng ln

Tch on thanh c chiudi dz gii hn bi 2 mtct ngang 1-1 v 2-2

q(z)1

1

2

2dz

Q Q+dQ

M M+dM

dz

o hm bc hai ca m men un bng o hm bc nht ca

lc ct v bng cng ti trng ngang phn b

( ) 0Y Q dQ Q q z dz ( )

dQq z

dz ( ) 0

2 2

dz dz M M dM M Q dQ Q

dMQ

dz 2

2( )

d M dQq z

dz dz



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ng dng Nhn dng cc biu Q, M khi bit qui lut

phn b ca ti trng q(z). Nu trn mt onthanh biu thc ca q(z) bc n th biu thc

lc ct Q bc (n+1), biu thc m men M bc(n+2)

Ti mt ct c Q=0 => M cc tr

Tnh cc thnh phn Q, M ti mt ct bt kkhi bit gi tr ca chng ti mt ct xc nh

Qphi = Qtri + Sq ( Sq Dtch biu q)

Mphi = Mtri + SQ ( SQ Dtch biu Q)



20/41

q

z

q(z)

A B

( )

B B

A A

dQ q z dz B A qQ Q SSq

Q

z

Q(z)

A B

SQ

( )

B B

A A

dM Q z dz B A QM M S




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1.4. V biu ni lc theo im c bit (1)

C s: Da vo mi lin h vi phngia Q, M v q(z)

Bit ti trng phn b =>nhn xt dng

biu Q, M => xc nh s im cnthit v c biu

q=0 => Q=const => QA=? (hoc QB)

M bc 1 => MA

=? v MB

=?

q=const => Q bc 1 => QA=? QB=?

M bc 2 => MA=?; MB=?; cc tr?

tnh li, lm,..?



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1.4. V biu ni lc theo im c bit (2)

Cc gi tr QA, QB, MA, MB, cc tr - l gi trcc im c bit. c xc nh bi:

Quan h bc nhy ca biu

Phng php mt ct Qphi = Qtri + Sq (Sq - Dtch biu q)

Mphi = Mtri + SQ (SQ - Dtch biu Q)

Vd



24/41

V d 1.4 (1)

Xc nh phn lc qF=qa

VA VB

.3 2 .2 . 0B AM V a qa a F a 5

3AV qa

.3 2 . .2 0A BM V a qa a F a 4

3BV qa

Xt on AC:

2a a

C

q=const Q bc 1

QA=VA

QC=VA+Sq=5qa/3-2qa=-qa/3

M bc 2: MA=0

MC=MA+SQ=4qa2/3; Mmax=25qa2/18

5

3 qa

1

3qa

+

5a/3

Mmax=25qa2/18

4qa2/3



25/41

V d 1.4 (2)

2a aVA VB

C

5

3

qa

4

3qa

1

3qa

+

5a/3

Mmax=25qa2/18

4qa2/3

Xt on BC: q

F=qa

q= 0Q = const

QB= - VB

M bc 1:

MB=0

MC=MB-SQ=4qa2/3

Q

M



26/41

4.5. Biu ni lc dm tnh nh nhiu nhp

nh ngha: L h tnh nh gm tp hp cc dm, ni vi nhaubng cc lin kt khp

Cch v biu :

- Phn bit dm chnh v dm ph

- Dm chnh l dm khi ng c lp vn chu c ti trng- Dm ph l dm khi ng c lp khng chu c titrng, phi ta ln dm chnh mi chu c ti trng-Ti trng t ln dm chnh khng nh hng ti dm ph,

ti trng t trn dm ph s truyn ti dm chnh thngqua phn lc lin kt

- V biu cho dm ph trc ri n dm chnh, sau ghp li vi nhau



27/41

V d 1.5 (1)

V d 1.5: V biu ni lc cho dm

ghp tnh nh sau:

Bi gii:

H dm ABCD l h dm ghp gm:

+ Dm ph BCD

+ Dm chnh AB

1) Dm ph BCD:

- Xc nh phn lc:

F

a aa

A B

CD

2D

FV R

a aa

A R

B

BC

D

VD

F

R



28/41

0 ( )

2 2C B Q

F FaM M S a

(Q)2

F

2

F

(M)

2

Fa

0 ( )2 2

C D Q

Fa FaM M S

F

a aa

A B

CD

BC

D

VD

F

R

a. on BC: q(z)=0

=> Q=const => QB= R = F/2

=> M bc nht 0BM

b. on CD: q(z)=0

=> Q=const => QD= -VD = - F/2

=> M bc nht 0D

M

V d 1.5 (2)



29/41

2.) Dm chnh AB:

2

F

(Q)

2

Fa

(M)

F

a aa

A B

CD

A R

B

V d 1.5 (3)



30/41

a aa

A B

CD

2

Fa

2

Fa

2

F

2

F

2

F

(Q)

(M)

3.) Biu noi lc ton h dm ghp

V d 1.5 (4)



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4.6. Biu ni lc khung phng

Khung phng l h phng gm nhng thanh ninhau bng cc lin kt cng (l lin kt m gcgia cc thanh ti im lin kt khng thay ikhi khung chu lc)

i vi cc on khung nm ngang, biu ccthnh phn ng lc v nh qui c vi thanhthng

i vi cc on khung thng ng, biu N,Q v v pha ty v mang du. Biu m men

v v pha th cng kim tra biu ta cn kim tra iu kin cn

bng cc mt khung: Ti mt khung, ni lc vngoi lc tho mn iu kin cn bng tnh hc.



32/41

V d 1.6 (1)

K

a

a a a

M F

q

A

D

B

C

Vd 5: V biu khung phng sau:

Bit M=qa2, F=2qa

Bi gii:

1. Xc nh cc phn lc:

T iu kin cn bng ca khung ta c

0AM

2

0

2 2 2

1

.2 2

1.2 2 0

2

K

K

V a Fa M qa

V a qa qa qa

VK

VA

HA

7

4KV qa

0 AX H qa



33/41

V d 1.6 (2)

0

2 2 2 2

3.2 .2 .

2

3.2 2 2 0

2

A A

A

aV a H a qa M Fa

V a qa qa qa qa

0KM

A1

4V qa

2. Nhn xt dng biu cc thnhphn ng lc trn tng on:

+ Biu lc dc:Bng phng php mt ct d dng xc nh:

4

0

AB BC A

DK CD

qaN N V

N N

a

a a a

M F

q

A

KD

B

C

VK

VA

HA

1

4qa

N



34/41

V d 1.6 (3)

on AB: q=const

Biu Q bc nht

Cn xc nh: QA = HA = qa

QB = QA+Sq = qa+(-q).a = 0

Biu M bc haiCn xc nh: MA = 0MB = MA+SQ = 0 + qa.a/2 = qa

2/2;

ti B c Q = 0 => Mmax=qa2/2

a

a a a

M F

q

A

KD

B

C

VK

VA

HA

qa

Q(kN)M(kNm)

2

2

qa



35/41

V d 1.6 (4)

on BC: q=0

Biu Q=const

Cn xc nh QB=0

2 / 2;AB

B BM M qa 2 2/ 2 0 / 2C B QM M S qa qa

Biu M bc nht

Cn xc nh

a

a a a

M F

q

A

KD

B

C

VK

VA

HA

qa

Q(kN)

M(kNm)

21

2qa

2

2

qa



36/41

V d 1.6 (5)

Trn on CD: q=0

Biu Q=const => Cn xc nh7 1

24 4

D KQ F V qa qa qa

27

4D KM V a qa 2 27 1 3

4 4 2C D QM M S qa qa a qa

Biu M bc nht => Cn xc nh

a

a a a

M F

q

A

KD

B

C

VK

VA

HA

qa

Q(kN)

27

4qa

232

qa

2

2

qa

M(kNm)

2

2

qa

1

4qa



37/41

V d 1.6 (6)

Trn on DK: q=0

Biu Q=const => Cn xc nh7

4K KQ V qa

0KM ( ) 27

4

CD

D DM M qa a

a a a

M F

q

A

KD

B

C

VK

VA

HA

Biu M bc nht => Cn xc nh

Q(kN)

27

4qa

M(kNm)

232

qa

2

2

qa

2

2

qa

1

4qa

qa

7

4qa



38/41

V d 1.6 (7)

4. Xt cn bng cc mt khungTi mt C, biu din cc ngoi lc, cc thnh phn ng lc trn hai mt

ct ngay st C thuc on BC v CD theo chiu thc (cn c vo cc biu

). Kim tra iu kin cn bng: Ti mt khung tng ni lc v ngoi lc

bng khng.

0; 0; 0CX Y M

1

4

qa

23

2

qa

1

4qa

2

2

qa

2qa



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1

4

qa

23

2

qa

1

4qa

2

2

qa

2qa

Biu ni lc ca khung

1

4qa

N

kN

1

4qa

7

4qa

qa

Q

kN

27

4qa

23

2 qa

2

2

qa

2

2

qa

M

kNm

V d 1.6 (8)



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CC VN SINH VIN CN NM

Qui c du cc gi tr ni lc khi xt

Thnh tho cc phng php v biu

Hc thuc lng v vn dng tt cc ntrong v biu

Hc thuc lng biu ni lc ca c

n gin sau y


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sbvl1 - ch1 - noi luc thanh (2012)

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