TRNG I HC BCH KHOA H NI

VIN C KH
B mn C hc vt liu v kt cu
a ch: C3-201. in thoi: (04) 3.868.0103; PTN: C3-101

SC BN VT LIU

TS. HONG S TUN

tuan.hoangsi@hust.vn

Edited by Hoang Sy Tuan

Ni dung

Chng 1. Cc khi nim c bn

Chng 2. L thuyt v ni lc
Chng 3. Ko, nn ng tm
Chng 4. Trng thi ng sut
Chng 5. Cc thuyt bn
Chng 6. c trng hnh hc ca mt ct ngang
Ni dung

Chng 7. Xon thun ty thanh thng MCN trn

Chng 8. Un ngang phng nhng thanh thng
Chng 9. Sc chu phc tp

Chng 10. n nh
Chng 11. Tnh chuyn v ca h thanh
Chng 12. Gii h siu tnh
Ti liu tham kho
1. Sc bn vt liu, tp 1&2.
L Quang Minh, Nguyn Vn Vng.
2. Sc bn vt liu, tp 1&2.
ng Vit Cng, Nguyn Nht Thng, Nh Phng Mai.
3. L thuyt v bi tp sc bn vt liu.
Nh Phng Mai.
4. Bi tp Sc bn vt liu.
Bi Trng Lu, Nguyn Vn Vng.
5. Bi tp Sc bn vt liu.
Thi Th Hng, ng Vit Cng, Nguyn Nht Thng,
Nh Phng Mai, Hong Th Bch Thy, Trn nh Long
 Chng I
CC KHI NIM C BN

Edited by Hoang Sy Tuan

1.1. i tng, mc ch nghin cu ca mn hc

i tng nghin cu: vt rn bin dng (thanh).

Mc ch: Tnh ton bn, cng v n nh ca thanh di

tc dng ca ngoi lc.

iu kin bn: C th chu c lc ln nht m khng b ph

hy.

iu kin cng: Bin dng ln nht khng vt qu gii hn

cho php m bo kt cu hot ng tt.

iu kin n nh: C th khi phc li trng thi cn bng

ban u, sau khi b tc ng ca lc bn ngoi.
1.2. M hnh ha kt cu v gi thuyt v vt liu
3 m hnh hnh hc c bn: Thanh, Tm-V, Khi.

Cu treo, M Nh ht ln, c p chn nc, M

Gi thuyt v vt liu:
Lin tc: Ti mi im u c vt liu.
ng nht: Tnh cht c hc ti mi im u nh nhau.
ng hng: Tnh cht c hc ti mi hng u nh nhau.
n hi: Bin dng l v cng b so vi kch thc ca vt th.
1.3. Quan h gia SBVL v cc mn hc khc
L thuyt n hi.
C hc kt cu.
L thuyt do.

L thuyt t bin.
 Chng II
L THUYT V NI LC

Edited by Hoang Sy Tuan

2.1. nh ngha v thanh v cc lin kt
Mt ct ngang Trc

Thanh thng tit din khng i

Khung
Trc bc

Y Y Y
X X
M
Gi ta c nh Gi ta di ng Ngm c nh
Y Y
M X
X
X=X
M Y=Y
Gi ta chng xoay Ngm trt Khp ni
2.2. Ni lc v phng php mt ct
Mt ct
P1 P2

Pn P3
(A) (B)

Pn-1 P4

P1

Pn H ni
(A) lc

Pn-1
2.3. Cc thnh phn ni lc

Nz - Lc dc
Mz Qy y
Qx, Qy - Lc ct
Mx
0
Nz z Mx, My - M men un
Qx My
x Mz - M men xon

My Qx y
Nz
Qy 0 z
Mx
Mz
x
2.4. Biu ni lc
L ng biu din gi tr ni lc ca tt c cc mt ct trn
ton b thanh.

a) Ko nn ng tm

4P P
2P

a a a a

d Nz
Biu lc dc Nz: Nz =n z
dz
- Trn on c n(z) = 0 Nz l hng s.
- Trn on c n(z) = n0 = const Nz l bc nht.
- Ti mt ct c lc tp trung Nz c bc nhy, ln bc
nhy bng ln lc tp trung (lc gy BD ko th nhy v pha
+, cn gy BD nn th nhy v pha ca biu ).
b) Xon thun ty
M2=3m.a
M1=m.a
m
B
A C

2a 2a

d Mz
Biu m men xon Mz: Mz =m z
dz
- Trn on c m(z) = 0 Mz l hng s.
- Trn on c m(z) = m0 = const Mz l bc nht.
- Ti mt ct c m men xon tp trung Mz c bc nhy,
ln bc nhy bng ln m men tp trung (m men cng chiu
KH th nhy v pha + ca biu ).
 c) Un ngang phng
M=q2
q Qy Qy
B 0 x
A C
2
Mx Mx
y
Biu lc ct Qy v m men un Mx:
Lc ct Qy (quy c + trn, - di) Qy q z
- Trn on c q = 0 Qy l hng s.
- Trn on c q = q0 = const Qy l bc nht.
- Ti mt ct c lc tp trung Qy c bc nhy, ln bc nhy bng
ln lc (ngoi lc lm thanh quay theo chiu KH th bc nhy v pha +).
M men un Mx (quy c + di, - trn) Mx =q z =Qy
- Trn on c q = 0 Mx l bc nht.
- Trn on c q = q0 = const Mx l bc hai, cong hng ly chiu ca lc
phn b v t cc tr ti mt ct c Qy=0.
- Ti mt ct c m men tp trung Mx c bc nhy, ln bc nhy
bng ln m men (m men lm cng th no th bc nhy v pha ).
 Chng III
KO, NN NG TM

Edited by Hoang Sy Tuan

3.1. Khi nim
Mt thanh gi l ko (nn) ng tm khi trn mt ct ngang ca
thanh ch c 1 thnh phn ni lc l lc dc (Nz).

3P B D P
A 2d C d

P
Nz
2P

3.2. ng sut trn mt ct ngang

Nz
z
F
3.3. Bin dng nh lut Hc
nh lut Hc: z E. z
N
z dz z
dz z dz
0 0 E 0 EF

Nz n
N zi i n
SiN z
const : EF i const :
EF i i 1 Ei Fi i 1 Ei Fi

3.4. c trng c hc ca vt liu

Ptl Pch P P
tl ch PB
F0 F0 PB
PB Pch
B Ptl
F0
1 0
100%
0
F0 F1 0 l 0 l
100% Vt liu do Vt liu gin
F0
3.5. ng sut cho php, h s an ton -
Ba bi ton c bn.
ng sut gii hn nguy him: Do: 0 ch Gin: 0 B
0
ng sut cho php: n: h s an ton (n>1).
n
Ba bi ton c bn: - Kim tra bn: max
Nz max
- Xc nh kch thc mt ct ngang: F

- Xc nh ti trng cho php: N z .F

V d: Cho mt thanh chu lc v lin kt nh hnh v. V biu

lc dc Nz, ng sut z v chuyn v u. Cho P, a, d, E.

5P 2d P
d P

a a a a
3.6. Bi ton siu tnh
Nu s n s (ni lc/phn lc lin kt) > s phng trnh cn
bng tnh hc, ta phi b sung thm cc phng trnh tng thch
bin dng. K

P 2d EF l P
d
2a C
B
a a a a
A
EF l
A B C 450
H
EF
EF
EF 450 h
600

D
P
 Chng IV
TRNG THI NG SUT

Edited by Hoang Sy Tuan

4.1. Khi nim ng sut
x
dR
p lim dR
dF 0 dF
dF
p z
n v: N/m2 (Pa), kN/cm2,
M
p zx .i zy . j z .k
y

ng sut ng sut
tip php
zx
zy y y
n n 0
0
z z z zy z
zx x x

TTS l tp hp tt c cc vc t ng sut p ca tt c cc mt ct
qua im M.
4.2. ng sut trn mt ct nghing bt k
cos , x ; m cos , y ; n cos , z z
A
(OAC): px x , xy , xz

(OAB): p y yx , y , yz
0 y
(OBC): pz zx , zy , z C
B
(ABC): p X ,Y , Z x
X x yx m zx n x yx zx

Y xy y m zy n T
xy y zy - Ten-x ng sut
xz yz z
Z xz yz m z n
T
, m, n p T .

xy yx ; yz zy ; zx xz
4.3. ng sut chnh, phng chnh
T
p . , m, n Phng
chnh
T .E 0 v 0
det T .E 0 Mt
chnh
3 I1. 2 I 2 . I 3 0
I1, I2, I3 l cc bt bin ca Ten-x ng sut T.
I1 x y z x xy y yz z xz
I2
I 3 det T xy y yz z xz x
2 2

1 2=0 1 1 1 1 1
3=0 3=0
3

2 2
TTS n TTS phng TTS khi
 4.4. Vng trn Mo ng sut y
v u
Bi ton S phng: x , y , xy
u
x y x y uv
u cos 2 xy sin 2
2 2 x
x y xy
uv sin 2 xy cos 2 x
2 0 xy
2 2
x y 2

x y 2
y
uv uv xy
2 2
uv
M u , uv C,R - vng trn Mo
2 uv M
x y x y 2
C ,0; R xy xy P M0
2 2
min max u
x y y u C x
max/ min R 0
1
2
2 xy
tan 2 max/ min 2
x y
 Chng V
CC THUYT BN

Edited by Hoang Sy Tuan

5.1. Khi nim v thuyt bn
2

1 1 t t

3

2
iu kin bn: td

5.2. Cc thuyt bn thng dng

a) Thuyt bn ng sut tip ln nht (Thuyt bn th 3)
Hai phn t A v B c coi l c bn tng ng khi m
STLN ca 2 phn t bng nhau.
td 1 3
Ph hp vi vt liu do.
b) Thuyt bn th nng bin i hnh dng (Thuyt bn th 4)
Nguyn nhn gy nn s ph hng ca vt liu l do th nng
bin i hnh dng.
td 12 22 32 1 2 2 3 3 1
Ph hp vi vt liu do.
 c) Thuyt bn Mo (Thuyt bn th 5)

B
A
K C
L
ch ch ch
3 ch 1
0 02 0 01 03

Gii hn bi ng bao tip xc vi cc vng trn ng sut.

td 1 3
K
vi
N
Ph hp vi c 2 loi vt liu do v gin.
 Chng VI
C TRNG HNH HC
CA MT CT NGANG

Edited by Hoang Sy Tuan

6.1. M men tnh (m men din tch cp 1)
y Y
i vi 0: S 0 rdF
F
dF M
i vi cc trc ta : y X
G

S x ydF S y xdF r F
F F
x
G gi l trng tm ca mt ct ngang:
0 x
V d: Xc nh trng
S G GM dF 0
F tm ca hnh ch L?
Cng thc xc nh ta trng tm mcn: y
Sy Sx a
xG yG F1
F F F2
Trc i qua trng tm ca mt ct ngang 2a G1
a
gi l trc trung tm. G2 x
0 4a
6.2. M men qun tnh (m men din tch cp 2)
M men qun tnh c cc i vi 0: y

I p r 2 dF M
F y
i vi cc trc ta :
r F
I x y 2 dF I y x 2 dF
F F
x
I p Ix I y 0 x
y
M men qun tnh li tm:
I xy xydF
F dF dF
Mt h trc ta c Ixy = 0 th c gi l
h trc qun tnh chnh. x
Mt h trc ta qun tnh chnh i qua 0
trng tm G ca mt ct ngang th c
gi l h trc qun tnh chnh trung tm.
6.3. M men qun tnh ca mt s mcn n gin
y y
bh3 d4
Ix Ip
h
0 x 12 0 x 32
hb3 d4
Iy Ix I y
12 64
b d

D4
d x Ix
64
1 4
d

D4 D
Ip
32
1 4
D
6.4. Cng thc chuyn trc song song
Oxy l h trc trung tm (OG) v y

Iu I x a2 F
x
2
Iv I y b F OG
a F
I uv I xy abF u
01 b
V d: Ix ? y y
y
t
01 x1 t
0 x 0 x 3t
4t 0 x
02 x2

t
d
2t
3t
 Chng VII
XON THUN TY THANH
THNG MT CT NGANG TRN

Edited by Hoang Sy Tuan

7.1. Khi nim
Ni lc: m men xon Mz.

M Mz M

Qui c du: Biu ni lc:

Mz
y
4kNm
0 2kNm
z 5kNm/m
B C D
2d A 3d
x
y Mz 1m 0,5m 0,5m
z
0

x
7.2. Quan h gia m men xon vi cng sut v
s vng quay ca trc truyn

N: cng sut [W] n

N M z .
Mz: m men xon [Nm] 30
n: tc quay [vng/pht] 30 N N
: vn tc gc [rad/s] Mz 9, 55
n n
7.3. ng sut trn mt ct ngang
max ng sut tip r ti v tr im M c:
Mz
r - Phng vi OM.
M - Chiu xc nh theo chiu Mz.
r 0 Mz
- ln: r r
max Jp
ng sut tip ln nht:
max Mz Mz Jp
max R vi Wp
Jp Wp R
max
Mz
Hnh trn c:
d4 d3
0
Jp 0,1d 4 Wp 0, 2d 3
32 16
max Hnh trn rng (vnh khn): d D
D4
max
Jp
32
1 4
0,1 D 4
1 4

D3
Wp
16
1 4
0, 2 D 3
1 4

7.4. Bin dng ca thanh trn chu xon
Gc xon t i (gc xon trn mt n v chiu di):
d M z
[rad/c.di]
dz GJ p
G: m un trt; GJp: cng chng xon
l Gc xon tng i gia 2 mt ct:
Mz
dz

GJ p
M
Mz M z
Thanh c 1 on v const

GJ p GJ p
Mz n
M
Thanh c n on v const zi i
GJ p i 1 GJ pi
i 1, n
n
SiM z
Thanh c n on v GJ
p i 1, n const
i 1 GJ pi
7.5. iu kin bn, iu kin cng
iu kin bn:
- ng sut tip ln nht (STLN)
2

- Th nng bin i hnh dng (TNBHD)
max

3
iu kin cng:

max
- Gc xon t i cho php.

V d: Cho mt trc bc mt ct 7M M

ngang trn c kch thc v lin

kt nh hnh bn. A B
2d
C
D d
H

1) V biu ni lc.
3M
2) Xc nh gi tr m men xon
cho php [M] trc tha mn iu a a a a

kin bn theo TB STLN. Cho [].

3) V biu gc xon ca trc.
Cho G.
7.6. Bi ton siu tnh
PP kh: B sung thm cc phng trnh tng thch bin dng
hoc chuyn v.

V d 1: Cho mt trc bc mt ct V d 2: Tnh max trong trc AC

ngang trn c kch thc v lin kt v z trong cc thanh 1 v 2.
nh hnh bn.
Cho bit M=6000Nm;
V biu m men xon, ng sut l=1,5m; D=8cm; d=1,6cm;
tip ln nht v gc xon ca cc mt E=2.107N/cm2; G=8.106N/cm2.
ct trong ton trc. Cho G.
m M=ma

A B D
2d C d

2a a a
 Chng VIII
UN NGANG PHNG
NHNG THANH THNG

Edited by Hoang Sy Tuan

 Mt phng
8.1. Khi nim Mt phngti trng
ti trng

Mt phng
qun tnh chnh
trung tm

0
x
z
y
Ni lc: lc ct Qy v m men un Mx.
Un thun ty Qui c du cc ni lc:
(Mx) Mx Mx
Phn loi
z
Un ngang phng 0

(Qy, Mx) y Qy
Qy
 8.2. Un thun ty
M
P P
B C
A B A D
a a
M
Mx Mx
Pa Pa
Mx Mx M
z y z max y max x W x M men chng un.
Ix Ix Wx

Trc trung ha l trc trung tm.

b d
min min
3
bh d4
Ix Ix
0 x 12 0 x 64
h z z
bh 2 d3
Wx Wx
6 32
max=|min| max=|min|
y y
8.3. Un ngang phng
Mx
M men un Mx ng sut php z: z y
Ix
y
Lc ct Qy ng sut tip zy (Jurapxki):
Fc
Qy .S xc M x
zy 0
I x .bc
b d

0 x max 0 x max
h
y y

y
y
3Qy 4 y 2 3Qy 4Qy y2 4Qy
1 2 max 1 max
2bh h 2F 3 R 2 R 2 3F
8.4. iu kin bn
(Thng thng b qua nh hng ca lc ct)
- Xc nh mt ct nguy him c |Mx|max.
- Xc nh ng sut ln nht: Mx
max max

Wx
k
k
iu kin bn: max
max (Do)
n (Gin)
max n
V d: Cho mt dm lin kt v chu lc nh hnh v.
1. V biu cc ni lc.
2. Xc nh ng knh cho php [d] dm tha mn iu kin
bn. Cho [] v khi tnh b qua nh hng ca lc ct.
q P=ql d

A C D

B
2l l l
8.5. Chuyn v ca dm chu un
P
y(z) vng. z
0
P
(z) = y(z) gc xoay. ng
y
n hi
Ptvp ng n hi: y
Mx
y EJ cng chng un.
EJ
Cc pp gii:

- Tch phn khng nh hn.

- Thng s ban u (t c).

- Dm gi to.

- Hm gin on.
a) Phng php tch phn khng nh hn
Mx Mx
y dz C y dz dz Cz D
EJ EJ
Cc hng s tch phn c xc nh t cc iu kin bin v cc
iu kin lin tc. P
EJ
V d 1: Xc nh vng ti A B
u t do ca dm cng-xn.
l

V d 2: Xc nh vng P
ti B v gc xoay ti mt
ct A ca dm chu lc A C
nh hnh v. Bit cng B
ca dm l EJ. l/2 l/2

Nhc im: Nu dm c n on th s tch phn n ptvp v c 2n

hng s TP cn c xc nh.
b) Phng php dm gi to
Mx
qgt y Qgt v y M gt
EJ
Dm thc Dm gi to
y 0 y 0 Qgt 0 Qgt 0
y0 y0 M gt 0 M gt 0

y 0 y 0 Qgt 0 Qgt 0
y0 y0 M gt 0 M gt 0

y 0 y 0 y 0 Qgt 0 Qgt 0 Qgt 0

y0 y0 y0 M gt 0 M gt 0 M gt 0

y 0 y 0 y 0 y 0 Qgt 0 Qgt 0 Qgt 0 Qgt 0

y0 y0 y0 y0 M gt 0 M gt 0 M gt 0 M gt 0

Ch : t qgt ln dm gi to tng ng, chiu lun hng ra xa dm.

 1 1 2
S h S h S h
2 3 3
Bc 2 Bc 2
h C h h C
C

l/3 l/4 3l/8

l l l

P
V d 1: Xc nh vng v EJ
gc xoay ti u t do ca A B
dm cng-xn chu lc nh
l
hnh bn.
V d 2: Cho mt dm chu lc nh hnh v. Bit cng chng
un ca dm l EJ.
Xc nh vng ti mt ct C v gc xoay ti mt ct A.
q P=ql

A C
B
2l l
c) Phng php hm gin on
nh ngha: n 0 nu z< a. Ch :
za n
z a nu z a. <0>0 = 1
Tnh cht: n 1
n n 1 n za
1) z a n z a 2) z a dz C
n 1
Biu din m men un theo hm gin on:
M
- M men tp trung: z
0
0
Mx M z a a

- Lc tp trung: P
z
1 0
Mx P z a
a
q
- Lc phn b:
2 z
0
q za
Mx a
2
V d: Cho mt dm chu lc nh hnh v. Bit cng chng
un ca dm l EJ.
Xc nh vng ti mt ct B v gc xoay ti mt ct C.
M=Pl 3P P
EJ D
A C
B
2l l l

q P=ql

A C D

B
2l l l
 Chng IX
SC CHU PHC TP

Edited by Hoang Sy Tuan

 9.1. Un xin
Ni lc: M men un Mx (Qy) v My (Qx).

M u M x M y khng nm trn mt phng qun tnh chnh trung tm.
z
ng sut php ti M(x,y):
My Mx
Mx My
My z y x


+
y Ix Iy
+

ng
Mu Mx Phng trnh ng trung ha:
trung ha x My
Mx
y x0
z Ix Iy

Mx Gc nghing gia ng trung ha v trc x:

My
y Ix M y Ix M y
y x tg .x vi tg
x Iy Mx Iy Mx
 z
im nguy him ti u? A2 My Mx

Mx My
A1: max yk xk +
z max max y
Ix Iy +
A1
Mx My ng
A2: min yn xn trung ha x
z max max
Ix Iy
Mt ct ngang i xng:
max min Mx My
z z
Wx Wy
iu kin bn:
- Gin: zmax k zmin n
- Do: z max

Chuyn v:
f f x2 f y2 x2 y2
V d: Dm cng-xn chu tc dng ca lc theo phng thng ng
v ngang nh hnh v.
Xc nh v tr ng trung ha ti mt ct nguy him, max v vng
ton phn ti u t do ca dm. Cho E = 2.104 kN/cm2.

P=2,4kN

20 cm
P

2 cm
1
2a a=0,5m
9.2. Ko v un ng thi (ko-nn lch tm)

Ni lc: Lc dc Nz, m men un Mx v My.

z P
z
b
Nz
y

Mx x z
My
P
y Pa/2
a
x
Pb/2 y

x
9.2. Ko v un ng thi (ko-nn lch tm)

z ng sut php ti M(x,y):

Nz Nz M x My
z y x
F Ix Iy
Mx
My Phng trnh ng trung ha:
y
Nz M x My
x y x0
F Ix Iy
Mt ct ngang hnh trn
z z z

Mu My Nz v Nz v
A B
Mu Mu
y u u
Mx Mu B A
x
u

Nz Mu Nz Mu
2
Mu M M
x
2
y
zmax zmin
F Wx F Wx
im A l im nguy him c
Nz Mu
z max

F Wx
Mt ct ngang hnh ch nht
z z
Nz
Nz
A Mx B Mx
+ +
+ +
My
y My y
B A

x x

max Nz M x My min Nz M x My
z z
F Wx Wy F Wx Wy

im A l im nguy him c
Nz Mx My
z max

F Wx Wy
 P1=16kN
V d 1: Ct c mt ct ngang hnh y
ch nht b=12 cm, h=16cm, chiu z
cao l=2 m, chu tc dng ca lc x
P1=16 kN, P2=4 kN v ti trng
phn b u q=2kN/m.
Tnh max, min v xc nh v tr l/2
ng trung ha chn ct.

q P2=4kN

l/2

b
V d 2: Trc mt ct ngang hnh trn ng knh d=10 cm, chu lc
nh hnh v.
Tnh max, min?

P=3,14kN
4P

z
d
P
x
d 3d d

y
9.3. Un v xon ng thi thanh MCN trn

Ni lc: M men un Mx, My v m men xon Mz.

z
M u M x2 M y2
v
im nguy him ti 2 im A1 v A2:
A2
Mu Mz
max max
Mz Wx Wp
Mu
u Theo thuyt bn STLN:
A1 1
tdmax max
2 2
4 max M x2 M y2 M z2
Wx
Theo thuyt bn TNBHD:
max 2 2 1
td max 3 max M x2 M y2 0.75M z2
Wx
V d 1: V biu m men P
xon, m men un Mx, My. M=30 kNm d=10cm
0,2 m TC
Tnh ng sut tng
ng ln nht theo thuyt
2P
bn STLN.
0,5 m 0,5 m 0,5 m

0,125 m

V d 2: Trc AB mt ct ngang 3P
hnh trn ng knh d. Thanh
D M=Pa
CD cng tuyt i hn vung a
gc vi AB. d
A C K B
Xc nh ng knh cho
php ca trc theo thuyt bn P
a a a
TNBHD. Bit []. Tnh gc
xoay ti K. Cho bit E v G.
9.4. Ko, un v xon ng thi

Ni lc: Lc dc Nz, m men un Mx, My v m men xon Mz.

z v
Nz M u M x2 M y2
Mz

im A l im nguy him c:
Mu
u Nz Mu Mz
max
A max
F Wx Wp

Theo thuyt bn STLN:

z v
Nz 2 2
A Mz td max 4 max
Mu Theo thuyt bn TNBHD:
u
2 2
td max 3 max
Chng X
N NH

Edited by Hoang Sy Tuan

10.1. Khi nim
P
R Pth

10.2. Bi ton le
Mx Pth
y M x Pth y z h y
EJ x
2 Pth
2
y z y z 0 b
x
EJ x
l
y z C1 sin z C2 cos z y(z)

z
Cc iu kin bin:
z 0, y 0 C2 0 Pth
y
z l , y 0 sin l 0 x
k
l k hay , k 1, 2, 3,...
2 2
l
k EJ x
Pth
l2
2 EJ x
Thanh b mt n nh khi k=1: Pth
l2 Pth
 10.3. Cc trng hp lin kt khc
Lc ti hn theo le:
2 EJ x
Pth 2
l h s ph thuc lin kt hai u thanh.

l
Pth
Pth Pth
Pth

/4
/2
0,7

/2

/2
=1 =2 /4 =0,5
=0,7
10.4. ng sut ti hn, gii hn p dng
cng thc le
Pth 2 EJ min 2 E l J min
th 2
2 imin
F l F imin F
th
iu kin n nh: od
kod
iu kin p dng cng thc le: th tl
2E th
0
tl ng Iaxinxki
ch
0 p dng cng thc Iaxinxki:
Hypecbn le
tl
th a b
1 1 a ch b
th ch (do) th B (gin) 0 1 0
 Chng XI
TNH CHUYN V CA
H THANH

Edited by Hoang Sy Tuan

Gii thiu

Tnh chuyn v ca cc thanh c dng bt k, nh khung, h

thanh, chu lc bt k.

Cc gi thit:
Ti trng tc dng tnh.
Chuyn v tun theo nguyn l cng tc dng.

Cc phng php:
Nguyn l bo ton nng lng.
Nguyn l cng kh d.
11.1. Xc nh chuyn v theo cng kh d

a) Cng kh d ca ngoi lc Pk
k dz
Xt 2 trng thi:
+ Trng thi th nht gi l
trng thi k chu lc Pk.
Pm
+ Trng thi th hai gi l dz
trng thi m chu lc Pm. m
Angkm Pk . km
km
km
hoc Ang Pik .ikm
i

b) Nguyn l cng kh d
km km
Angkm Ankm 0 ik i n 0
P A
i
c) Cng kh d ca ni lc tb
d
Qk
Mk Nm Qm
Nk Nm Qm
Mm Mm
Nk
Mk
Qk

dz dz ds
dz+dz dz

N m dz dz M m dz Qm dz
dz dz d ds
EF EJ GF
dAngkm N k dz M k d Qk ds
km km N k N m dz M k M m dz Qk Qm dz
dA dA
GF
n ng
EF EJ
N N dz M M dz Q Q dz
Ankm k m k m k m
EF EJ GF
N k N m dz M k M m dz Qk Qm dz
i Pik km EF EJ GF
d) Cc nh l tng h
- nh l tng h v cng kh d ca ngoi lc:
P
i
ik km Pjm mk
j
- nh l tng h v cc chuyn v n v:
km mk

e) Cng thc Maxwell-Morh

Cho Pk = 1: N k N m dz M k M m dz Qk Qm dz
km
EF EJ GF
- xc nh chuyn v thng (hoc gc xoay) ti mt v tr no
ta t lc (m men) tp trung n v ti .
- xc nh chuyn v (hoc gc xoay) tng i gia 2 mt
ct th ta t 2 lc (hoc m men) tp trung n v ngc chiu
nhau ti 2 mt ct .
11.2. Phng php nhn biu Vrsaghin

I F z G z dz G,F
G(z)
l

az b G z dz d

C
l

az b d z
O
z dz
zC
a zd b d F(z)=az+b

azC b F(zC)

azC b O
z

L
I F zC
 Bc 2
Bc 2
h h

z1 z2 z1 z2

l l

1 1 3 2 3 5
hl , z1 l , z2 l hl , z1 l , z2 l
3 4 4 3 8 8

Bc 3 Bc n
h h

z1 z2 z1 z2

l l

1 1 4 1 1 n 1
hl , z1 l , z2 l hl , z1 l , z2 l
4 5 5 n 1 n2 n2
 Chng XII
GII H SIU TNH

Edited by Hoang Sy Tuan

12.1. H siu tnh
Nu s lin kt nhiu hn s phng trnh cn bng tnh hc th h
gi l h siu tnh.
Bc siu tnh = S n S phng trnh cn bng
P P

P P
Kh h siu tnh:
- Bc 1: Xc nh bc siu tnh v chn h c bn tng ng ca
h siu tnh cho (sao cho phi m bo tnh bt bin hnh ca h).

q
- Bc 2: Xc nh h tnh nh tng ng (bng cch a vo
h c bn cc phn lc lin kt tng ng vi cc lin kt tha
b i).
X2
- Bc 3: Thit lp h phng trnh chnh tc
X1

11 X1 12 X 2 ... 1n X n 1 p 0

21 X1 22 X 2 ... 2 n X n 2 p 0

........................................................
X X ... X 0
n1 1 n 2 2 nn n np

trong
1 1
ij Mi M j ip Mi Mp
EJ EJ
Biu ni lc: N z X 1 N 1 X 2 N 2 ... X n N n N p

Qy X 1 Q1 X 2 Q 2 ... X n Q n Q p

M x X 1 M 1 X 2 M 2 ... X n M n M p
12.2. H siu tnh i xng
a) nh ngha M M

EJ EJ

P P

2EJ 2EJ
EJ = const EJ = const

b) Tnh cht X2 X2 X2=0 X2=0

X1 X1 X1=0
X3=0 X3=0 X3 X3

P P P P
12.3. Dm lin tc
a) nh ngha P q
Mi-1 Mi Mi Mi+1
L dm t trn nhiu gi ta n,
trong c mt gi ta c nh. i i+1

P q Mi=1
Mi

1
Mi-1=1
Mi-1
Bc siu tnh = S nhp - 1
1
Mi+1=1
b) Phng trnh 3 m men Mi+1

P q 1
i i+1
M0 M1 M1 M2 M2 M3
MP
ai bi ai+1 bi+1
0 1 2 3
 li li li 1 li 1
M i 1 Mi M i 1
6 Ei J i 3Ei J i 3Ei 1 J i 1 6 Ei 1 J i 1
i ai i 1bi 1
0
li Ei J i li 1 Ei 1 J i 1
Nu cng EJ khng i
i ai i 1bi 1
li M i 1 2 li li 1 M i li 1 M i 1 6 0
li li 1
q P=q
V d: V biu ni lc.
Cho EJ = const.
/2 /2

Ch :
q P q
M=P/2
0=0

/2
EJ0=