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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
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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
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1.1. Cc thnh phn noi lc trn mtct ngang (2)
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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1.1. Cc thnh phn noi lc trn mtct ngang (3)
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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1.1. Cc thnh phn noi lc trn mtct ngang (4)
N
Q Q
M M
1
1
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1.1. Cc thnh phn noi lc trn mtct ngang (5)
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
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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
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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.
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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
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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
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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
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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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1.3. Lin h vi phn gia m men un,lc ct v ti ngang phn b (2)
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)
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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
1.3. Lin h vi phn gia m men un,lc ct v ti ngang phn b (3)
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1.3. Lin h vi phn gia m men un,lc ct v ti ngang phn b (4)
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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
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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
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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
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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
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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
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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)
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2.) Dm chnh AB:
2
F
(Q)
2
Fa
(M)
F
a aa
A B
CD
A R
B
V d 1.5 (3)
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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.
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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
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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
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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
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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
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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
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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
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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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E- mail:[email protected]