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BE BA1 VA HUGNG DAN GIAI

BA1 TAP L(')’N

SUC BEN VAT LIEU — CO’ HQC KET CAU
LEU MQC LAN — NGUYEN VU VIET NGA

DE BAI VA HU(’)’NG DAN GIAI
BAI TAP LON

sU’C BEN VAT LIEU — CO H(_)C KET CAU

NXB-. ..
LOI G161 THIEU

Ta‘i li_éu tham khdo “Dd bdi vi:  hmfng ddn gidi bdi top [671 Stir bén vfit li_éu - C0‘ hoc
két cdu“ dujorc...
CAC YEU CAU CHUNG

I -YIAEU CAU vi‘:  TRlNH BAY
S Trang bia trlnh bay theo mau qui djnh (xem phan Phu luc cfia tai Iiéu na...
PHANI
DE VA HU’O’NG DAN GIAI
BAI TAP LON SL'. l’C BEN VAT LIEU
BAI TAP LON so 1
TlNH DAC TRU’NG HlNH HOC CUA HlNH PHANG

BANG so LIEU BAI TAP LON so 1

D(cm) Bxbxd (mm)
180x110x10
250x1...
2. Tinh ca'c me man quan tinh chinh trung tam: 

-" Chon he true trung tam XCY (di qua trong tam C Va song song Vévi he tr...
HlNH DANG MAT cAT NGANG

     
     

      
 
       

/ .4

X0

     

8

b 2D

    
       

N

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I IN”
.  
. ‘

 
...
C

 

C

  
 
  

 
   

   

     
   

   

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B 3 a D B X0



   

X0

 
    

    

V-, : m: s

:  
" 1
‘s
...
Vl Dl_J THAM KHAO
De bai: 

Xac djnh cac me men quan tinh chinh trung t j tri he truc quan tinh
chinh trung tam cua hinh p...
Bai lam: 
1. Xac djnh trong tam: 
Chon he truc ban dau xoyo nhu hinh Ve:  xem hinh 1.2.

Chia hinh phang da cho thanh 3 hi...
Bang ket qua tinh toan

5;,  (m3) 8;,  (cm

12 000,000 36 000,000
13 651,162 6 243,877
2 222,335 3 756,225

Teng 1730,66 2...
a.  T061 <76 trong tém C08 mung hinh thanh phén 076/ vL’7i hé truc trung tém XCY

/ E31.’
Hinh a,  (cm) bi (cm)
1 3,420 - ...
4 2
Ta 06: Jf) =  i R—— 
4“ 8 3.71 3.7: 4

Jfgyz =  :(0,125R4—0,14154R4) =  :0,01654R4
Trwérng hcyp nay 1g am < 0 nén Jgzg...
c.  Tinh mé men quén tinh trung tém c1Ja toén hinh: 
Jx =   =  84 739,883 + 107 854,23 + 16 678,602

4 Jx =  209 272,715 c...
Jmx =  600580,67 cm‘

   

 

 
   

JXY=82164,210 , 

_]Y =  583 328,384 1‘

Hinh 1.4

V1 tri hé truc qua'n tinh chinh tr...
BAI TAP LON so 2
 

BANG so LIEU BAI TAP LON so 2

PKN

l
2
3
4
5
6
7
8
9
1

3

 

6: Sinh vién chon nhfyng s6 Iiéu trong ...
2. Kiém tra / ai d/ éu k/ _én bén khi co’ ké dén trong / wqng bén thén: 
-' Vé biéu d6 ncfni luc trong trunfmg hop cc’) ké...
3. Xéc djnh dung suét chinh: 

—' Tinh LJ’ng suét chinh va phwcmg chinh tai 5 diém dzfzc biét trén mejt cét cc’)
MX V2‘;  ...
S0’ D0 TiNH
M q q M
12:3»;  P1” % P 2;. ”
71a}:  b 1 “At 71/“),  b 4;“ ,1:

22
V1 DU THAM KHAO

Iaé bai: 
Chon sé hiéu thép chip I (N°l) caa mat cét ngang dam dwdi day, 
Biét:  [0] =  210 MN/ m2, (xem ...
*TaiZ1=0(taiC):  Qy= -10KN;  Mx=0

* TaiZ1= 1 m (tai gifra doan):  Qy =  — 30 KN;  Mx =  — 20 KNm

*Te_Ii 21 =2m(tai A):  ...
Vé biéu dc“) ncf)i Iuxc

 

     

10 | IIIII| ||| r'e

. .‘
.4

25.71 :25 71 5

1.4. So’ bé chon mat cat theo diéu kién b...
1 31:). 
q=20‘315 KN/  m M=40KNm q =  20,315KN/  m
0,315 KN/ m

  
  

  
  

18,6
(KNm)

58,68
Hinh 2.3 Mmax =  75.25 KNm...
- Ccjng biéu d6 vtra vé vévi biéu d6 trén hinh 2.2 sé dwqc biéu d6 nhu trén
hinh 2.3.

2. 2. Viét phwong trinh néi / l_. P...
2.3. Xa’c djnh vj tri co’ Mmax: 

- Cho phucyng trinh Qy =  0 (6 doan DB),  ta tim dwqc toa d6 mét cét cc’) Mmax: 
(mat cé...
cs, ,a, = 202,9 MN/ m2 < [0] =  210 MN/ m2
Thoé mén diéu kien bén tai bién trén va bién dwcyi caa mét cét. 
6 Kiém tra cho...
CE =  151 000 KN/ m25 CE =  151 MN/ m2
.  . .  C vé C-
Taldlém Eco.  S = SX—d7 véb —d

(Xem hinh 2.5)

 

Do do: 
2 2
Q0 E...
A Diém E:  6 trén (16 cc’)
6E =  - 6E =151 MN/ m2
T5 =  T}:  = — 27,5 MN/ m2

2 2
151 151
6,, ,,, ,,, ,,, = E:   +122 =  T...
2
Tai dufyng trung hoé 06: cmamin =   +12 =  : 35,4 MN/ m2

3 : 
2
Tai diém 0 cc’) céc [mg suét chinh la: 
61 =  35,4 MN/ ...
4. Viét phuwng trinh dwbvng dén h5i cfla truc dém: 

p :10 KN q=20,315KN/ m
Q=0,315 KN/  m

q=20,315KN/ m

  

3 m
vA=77,54...
1023 + 20,3152‘ _ 77,54(2—2)3 _ 2o. (2—2)4

%Y2(Z)= Vo+ (Po-2+

3!EJ 4!EJ 3!EJ 4!EJ
_ 77,54(2—2)2 20.(2—2)3
2 — 2 - T — T
...
10.93 + 20,315.94 _ 77,54(9—2)3 _ 20.(9—2)4 _40(9—5)3

” V“ + "’°' 9 + 3!EJ 4!EJ 3!EJ 4!EJ 2!EJ
+ 20.(9—5)4 =  0
4!EJ
1215...
36

28,719 + 10.53 + 20,315.53 _ 77,54(5-2)3 _ 20.(5-2)3
EJ 2!EJ 3!EJ 2!EJ 3!EJ

1 80 58
(p2(Z) =  EJ(— 28,719 +125 + 423,...
BAI TAP LON so 3
TTNH COT CHIU LUC PHUC TAP

BANG SO LIEU BAI TAP LON SO 3

b(m) A(m) 1(m) P(KN) KN/ m3 q(KN/ m)

ON

>-‘>...
" Xéc djnh céc mé men quan tinh chI'nhtrungté1m: Jx, Jy

" Xéc djnh céc bén kI'nh quén tinh chinh trung tém:  ix,  iv. 

3...
S0 D0 MAT CAT COT

 

@ ® a 221 4
3b
3b
3b
1.5b
E F E F
in 7a T1‘
3 3a a
@1,5a a 3a a 1,53
3b
3b
2],  2b
E F E F

39
2b

321

_#b

3b

2b

_
7

, »

221

 

2a

2a

2a

40
V1 DU THAM KHAO

Dé bai: 
Cho mat cét cét chiu | L_. vc nhu hinh vé,  biét céc 1L_J’C Iéch tém P1=P2=P3=500KN, 
a’p | L_J’...
CfXa'1c djnh céc mé men qua’n tinh chinh trung tém: 

Hé truc XCY trén hinh 3.3 Ia hé true quén tinh chinh trung tam.  Chi...
2 48.203 8.183

Jx + (2895 —10)2.48.2O + + (28,195 — 29)2.18.8 +

+(2o+18+12—28,95). 

+[48.363 2 48.36}

—> Jx =  32 000 ...
F J, 
Béng két qué t1’nh toén: 

X,  y,  11515,741 + 143175 1 |  i 1120001 )4 GZ
(cm) ‘T1968 - 3256753 V 268032 (KN/ cmz)
...
TL‘r hinh vé vé béng két qué tinh toén,  ta théy ngay 2 diém nguy hiém Ia E & I

= ?Xa'1c d_inh biéu J6 Ling sua"t phép ta...
a——i=7—= —17,2021cm a= —17,2cm

XK 7,9
-12 -20 52 _)

b= —X= —'= —14,75cm = -14,7cm
YK 28,5

Toe 66 DTH 15: (—17,2; -14,7)...
+ Dwdng A1:
Cét trL_1c Y tai diém I(0; 45,05) nén cc’):  b1 =  45,05 cm
Cét trL_1c X tai diém A(xA;  0) nén sé cc’):  a1 =...
2%
'4 
/ 

 

Hinh 3.9
Tu» hinh 3.6 ta théy:  diém dét | L_rc doc Iéch tam K ném ngoai |6i nén biéu (16

(mg suét phép tré...
BAI TAP LON so 4
TiNH DAM TREN NEN DAN HOI

BANG so LIEU BAI TAP LO'N so 3

0,03516
0,01800
0,04267
0,02289
0,01800
0,0285...
- Lap diéu kién bién va giéi hé phwong trinh dé tim ra cac én sé yo Va 00.

-' Lap béng két qué tinh toén n©i | |,J’C (mé ...
SO 90 TINH DAM TREN NEN DAN HOI

6) PM 2: (2). , M 
/ # / #
4134 b 4 C J’ #34“ b 4' C 2}’
@P q M 2', ‘ ® P q M 2P
A!  At
+...
52

q M P M
2P
%f<aj%b%+£c / #
4/ %kaj~€bfi%C%/ 
P
+a ”“ M
fi%b0Pc%1/ fiéa / #
‘4§bfif%c§y
P / /V’
q 2P .1_6./  p 2P
M
Ta W‘. ...
Vi DU THAM KHAO

A

De bz‘-Ji: 

TI'nh gia' tri néi luc trén céc mef1tcét(ca'ch nhau 1 m) vé vé biéu d6 néi | L_rc cho
dém...
Dean 1 Dean 2 Doan 3
Tai A (Z =  0) T61‘ B (Z =3m) T61‘ C (Z =7 m)

 

2. Viét phwomg trinh n<f>i | l_PC cho t| ‘.Png doan...
c.  Thay céc gié tri tinh s§n vac phworng trinh n1_3i / LPG cda dém: 
M1: -1500,5313.. . + 31,9753.10“. yo.  cm.  + 73,815...
Thay diéu kién bién tai D véo phworng trinh ncfai Igyc (a) 61 doan 3, ta cc’): 
M3 =  -1500,531.B, ,,_11+ 31 ,9753.104.yo....
_ 932973485
_ 936,5597.104

Thay gia' tri yo V3 goo vao,  ta cc’) phwang trinh néi luc trong 3 clean cfla dém dé
cho nhw sa...
BANG KET QUA TiNH Ll_J’C CAT TAI cAc MAT cAT YEU cAu

90311 2 mZ -650.A, ,,Z 1379,804.13mz -970,278.C, ,,Z m(Z-3) —650A, ,...
BANG KET QUA T1NH MO MEN TREN CAC MAT CRT YEU CAU

Dean Z m Z -1500.531B, ,,z 3|85.315C, ,,z - 2239.893D, ,,z m. (Z-3) - |...
60

Biéu d6 néi | l.1’CZ

650 KN 650 KN

3m

14m1

3 486.735

 
 

Biéu 86 n61 Iuc (dé tham khéo): 

KET QUA TINH TOAN BAN...
10.0 0.0026692 0.1309270 -6948760000 -1755120164
11.0 0.0061471 0.3602250 —8000000000 00000000

550 KN 650 KN

 

   

0

...
62
PHAN II

are VA HU’C)’NG DAN GI/3.1
BAI TAP LON CO’ HOC KET CAU
BAI TAP LON so 1

T1NH HE THANH PHANG TTNH DINH

BANG so LIEU BAI TAP LON so 1

 Sinh vién chc_>n nhtrng 36 liéu trong bén...
1.5. Tim vj tri bét | c_ri nhét cL’1a dean téi trong gém 4 | I,J’C tap trung di déng trén
hé khi co mét truyén 10:6 86 m6 ...
S0 D0 TINH HE TTNH D! NH
P 15,,  2F,  a= L1/4;b= L2/4;c= L3/4.

  
 

      

-b--j-'b--

’q

F1 1 1 W31 1 4: 1 {”"1'°'1~°...
vi DL_J THAM KHAO

x

De bai: 
S6 aé:  4. 5. 3

4 P7 86 thL'r tL_}’ ctla so 66 két céu

5 w 36 liéu V6 kich 111066 hinh ho...
2MB= YA,8-P2-YN.11=YA.8'120.2-150.11=09YA=236,25KN

Kiém tra Iai két qué tinh YA va YB béng phuong trinh ZY =  0 9 Cho ta ...
1.2.1. Vé biéu 66 m6 men M (Hinh 1.3). 

4400

  
 
 

  
     
  
    

9'»
vgwo,  2175
. -.‘. ‘§‘: ’:‘: ’:‘: ’o"/  --
L ...
475,625

  
         
  
  

325,625

“ 104,375

309,375

362,5
I

   

609,375

 

(9

(KN)

4. Kiém tra cén béng céc nut...
ZY =  465. 0,8 - 30. 0,8 -104,375.0,6 — 475,625. 0,6 =  0
o Kiém tra nut L (Hinh 1.66): 

EX =  362,5 - 104,375. 0,8 - 465...
2. Dung dah dé kiém tra lai céc tri 56 RA_ MB,  QB Va Q,  dé tinh bang giai ti’ch: 
1,375 - 6

RA:  50. +120-0,25 :  236,2...
4. Tim Vi trl’ bét Iqi nhét 003 he 4 | L_PC tap trung P1; P2; P3; P4 di dcfmg trén he
khi cc’) mét truyén Iuc aé MK cc’) g...
Lén Iuvcjt cho doan tai trong di déng 102 trai qua phai sao cho cac | L_. vc tap
trung Ién lwqt dejt V30 céc dinh I,  II, ...
F
HJAKJ =  F'3.tga1 + P4. tgag =  —180.0,25 + 240. 0,5 =  75 > 0
Z

Ta nhén théy ciao ham 551 déu nén P4 dat o d1nh I I5 P...
+ Khi P3 dét c’v bén tréi dinh II ta cé: 

dM T
[ d K} =  P1- t90t1+ (P2 + P3)-1902+ P4- T9063

Z

=  -1200,25 + (120 +180...
max | MK|  = 450 KNm. 
2. Tinh chuyén vi trong hé tinh dinh
Theo yéu céu cL’ia dé bai ta phéi xéc dinh chuyén V1 géc xoay ...
— 1 4350~12 2
‘PR(P)= (MP)-(Mk):  {T A — -2]

3EJ 2 3

+ T 4050.5<2*T'5+35°'51,5+T.0,5 +2125-5435
2EJ 2 2 3 3 2

+ 1 2075....
BAI TAP LON co HOC KET CAU so 2
TTNH KHUNG SIEU TTNH THEO PHUTOTNG PHAP LUC

BANG so LIEU CHUNG VE KicH THUO'C VA TAI TRQN...
6. Vé biéu G6 luc cat Qp Va | u’c doc Np trén hé siéu tTnh da cho. 
1.2. Xéc dinh chuyén v_i ngang cda diam I hoac géc xoa...
S0 D0 T1NH KHUNG SIEU TTNH

@

®
Vi Dl_. l THAM KHAo
De bai: 
s6 Be:  10.5.8
10 w=  S6 thu» tu cfia so 06 két céu
5 rr se lieu ve kich thuoc hinh hoc (hang...
3. Lap he phwong trinh chinh tac dang chfrz
511X1"’ 512 X2 T’ 513 X3 ‘T’ A1p =  0
521X1+ 522 X2 T’ 523 X3 T A2):  = 0

531...
-1- Dung céng thL’: =c Mécxoen- Mo vé phép nhén biéu <36 dé tinh céc hé s6 vé
ca’c s6 hang tu do cfla hé phwong trinh chinh...
:1_10O-1O.2.12_2_200_10.6+i960-12‘ 6+2” :2088O
2EJ 3 2 3 EJ

2 3 3EJ

1 960-12 19200
A = . @= ,_. _.10=, _
2" X 3EJ 2 EJ

...
1 .12+12_8+ 1 ‘2+12‘10.12+ 1112-12‘2.12=11sa
3EJ 2 2EJ 2 EJ 2 3 EJ

r Kiém tra tat ca cac he s6 cfla he phwomg trinh chinh ...
Mex 13,35 0 0 0 13,35

MCH 13,35 419,14 124,8 ’ 0 307,69
MHC 40,05 419,14 0 960 ‘ —5o0,31
Ma;  0 419,14 124,8 ' 0 294,34
M...
+ 1 124,310 2+2“) _294,34.10 2+1_10 = 
2EJ 2 3 2 3

i(2705,04 — 3922,03) :  — 121794 =  — 
EJ EJ 2.10 .10" .10

(m)

=  — ...
OCH =  0 (500,81 + 67,
QHC 2 307,69)/12 375
QCB =  0 - (294,34 + —
QBC 0 124,8)/10 41,914
QBA =  10.
0 124 8/12
QM,  2 ’ 4...
XX =  NKE.  0,8 - 92,67. 0,6 - 2,25 =  0
4 NKE =  72,315 KN
EU =  NKC.  0,8 + 2,25. 0,6 + 92,67 =  0 —> NKC =  -117,525 KN...
..  III
| IIlllII1.a1lIll| |I| ||II. . 1*

       

( KN)
Hinh 2.6

 

1.2. Tinh chuyén v, i géc xoay tai K: 
,_ 8 2 -5 4 ...
2. TI'nh hé siéu tinh chiu téc dung déng thévi cua ba nguyén nhén (téi
trgng,  Sl_. |' thay d6i nhiét 61} vs‘:  g6i twa db...
cé:  E: -1,35KN;  E= N__-, =0

10’5~8

AB =10-5-32(—1,35-10)+ (%=  4368-10‘5 =0,044

A21:A31=0

Phén | l._J’C dulng tai li...
2. Céch tinh chuyén vi géc xoay tai K: 
9 Lap trang théi phL_1 "k" nhu trén (Hinh 2.3

9 TI'nh hé finh dinh dé chc_>n 6 tra...
BAI TAP LCVN CO’ HOC KET CAU SO 3

T1NH KHUNG SIEU TTNH THEO PHUONG PHAP
CHUYEN VI VA PHU’O’NG PHAP PHAN PHOI MO MEN. 

BA...
13181: E =  2.108 (KN/ m2);  J =  106. 1., “ (m4). 

Chtfl )2.‘
1. Vé xong biéu 115 m6 men uén Mp cén Kiém tra can béng cé...
vi Dl_. l THAM KHAo
4
s6 08: 10.7.5

10 <9: S6 thu=  tu» cua so 06 két céu

7 =8 $61180 vé kich thuoc hinh hc_>c(hé1ng 111...
1.1. Xéc djnh s6 éin s6:
n= ng+n, =1+1=2.
1.2. Lép he‘ 00 ba’n (HCB): 

Thém V60 nut B m6t lién két m6 men Va m<f)t Iién k...
Hinh 3.5

      

Hinh 3.4 01°47“

0,6EJ I'11=1,1EJ
0,0945%

r11

0,5EJ
R1 p

r12 =  r21 =  - 0,094EJ
160
00 V22

R1,, = —...
1.6. Giéi h_é phuvng trinh chinh téc: 
1,1EJ Z1- 0,094EJ Z2 - 240 =  O {Z1= 200,712/EJ
| :C>
- 0,094EJ Z1 + 0,03EJ Z2 + 25...
2. Dung phwomg phép phan phéi m6 men (PPMM) vé biéu d5 m6 men
u6n MP; 
Hé siéu tTnh dé Cho (:6 mcf>t nut cL'rng B cc’) chu...
:-1- Lap so d6 PPMM dé vé (Hinh 3.8::  (7 day chung téi sir dung két qué biéu

tra bén do Z2 =1 (7 trén,  dc’) la (Hinh 3....
2.5. Xa’c djnh ca’c hé s6 cda phwang trinh chinh téc: 

B 711

       
   
   

P R
B 11)
T11: 0,021EJ R19 =  4525

0,015E...
3. Vé biéu (16 | |_. Pc cét Qp vé biéu 66 | L_rc doc Np: 

3.1 Biéu d5I1rc cét Qp (Hinh 3.11) dwqc suy ra 10’ biéu (16 Mp....
Q98 10 (40.10.0,8)/2 279,573/10 187,957

QC“ 10 (40.10.0,8)/2 270573/10 132,043

QCE
_ 4 0 95,196/4 23,799
‘ QEC

QED 4 O ...
2 x =  NOB.  0,8 + 132,043. 0,6 + 23,62 =  0
NOB=  -123,557 KN

2 u =  NOB.  0,8 + 23,62. 0,6 + 132,043 =  0
NOB =  — 182,...
_ 1 16032 169,395-8 19,5736
0K‘”’_x@=2EJ1 2 '1j+EJ1 2 0‘ 2 '1]

(pm) =  - 0,00015 rad

Vey tiét dien K se 1:1 xoay met géc...
PhL_1 | L_1c:  Méu Trang bia

TRUONG DAI HQC THUY LOI HA NOI
BO MON sfrc BEN - CO KET CAU

BAI TAP LO’N . ... . . .
SO 

T...
ML_JC LL_JC

L6-i giévi thieu
Céc yeu céu chung
Phén I: 
sue BEN VAT LIEU
Béi tap | c')’n S6 1:
Dec trwng hinh hgc cua hin...
co HQC KET CAU
Bel tep lén s61:
Tinh he thanh phéng tinh dinh
Béng $6 lieu 65
Vi du tham khéo 68
Bai tep ldn $6 2:
Tinh kh...
Đề Bài và Hướng dẫn giải bài tập lớn sức bền vật liệu
Đề Bài và Hướng dẫn giải bài tập lớn sức bền vật liệu
Đề Bài và Hướng dẫn giải bài tập lớn sức bền vật liệu
Đề Bài và Hướng dẫn giải bài tập lớn sức bền vật liệu
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Đề Bài và Hướng dẫn giải bài tập lớn sức bền vật liệu

  1. 1. BE BA1 VA HUGNG DAN GIAI BA1 TAP L(')’N SUC BEN VAT LIEU — CO’ HQC KET CAU
  2. 2. LEU MQC LAN — NGUYEN VU VIET NGA DE BAI VA HU(’)’NG DAN GIAI BAI TAP LON sU’C BEN VAT LIEU — CO H(_)C KET CAU NXB-. ..
  3. 3. LOI G161 THIEU Ta‘i li_éu tham khdo “Dd bdi vi: hmfng ddn gidi bdi top [671 Stir bén vfit li_éu - C0‘ hoc két cdu“ dujorc bién soon theo du'ng dd cimrng "Chmrng trinh gidng doy mzin SBVL vd CHKC“ do tiéu ban mdn hoc Clio bé gido doc vd dda too soon thdo . SB VL vd CHKC cung cdp mot phdn kién thu'rc co sot cho czic k_}7 su theo hoc trong cdc trudng doi hoc k_}7 thudt nhu : thuy l_o'i , xdy dung , gioo thdng Hai mzin hoc ndy trong b_i cho cdc sinh vién vd cdc 197 so nhfmg kién thiic cdn thiét dd gidi quyét cdc bdi todn thuc té tit cling viéc thiét ké , thdm d_inh . dén thi cling vd Id co so cha viéc nghién cifm cdc min /97 thudt thuéc cdc chuyén ngdnh khdc. Trong chmmg trinh ddo too hai mén h_0L’ ndy , ngodi cdc bdi tdp nho' bd tri sou mdi chmmg ciia gido trinh , cdc sinh vién Lon buds phdi hodn thdnh mét s6 bdi tip [611 , Cd tinh chdt tong hop cdc kién thiic co bdn nhdt , vo duos b6 tn’ theo tftng hoc phdn czia mén hoc. Dd gizip cdc sinh vién cling cd cdc kién thllt‘ ciia mén hoc vzi ndm wing timg bzm"c gidi quyét cdc yéu cdu cda cdc bdi tdp ldn trong chuorng trinh ddo too Clio hai mén hoc, chdng téi bién soon 2121' liéu [ham khdo ndy vdi ddy dd cdc bdi tdp Ion cu'o hai mén SBVL vd CHKC . T dz’ liéu ndy bao gdm hai phdn , tuong u"ng vdi hat" mén hoc . Phzin céng bién soon nhu sou . ' 0Phdn I do C6 gido Nguyén V1? Viét Ngo bién soon , boo gdm 4 btii tgip lo"n SB VL. 9Phdn 11 do ca gido Léu Méc Lon bién soon , boo gdm 3 bdi tfip lo"n CHKC‘. Cdc bdi trip lo"n ndy yéu cdu cdc sinh vién phdi hodn thdnh theo dung yéu cdu czia gido vién phu trdch rmin hoc , phi) hop véi tirng giai doon . Trong mdi phdn ctia an liéu ndy, déu boo gdm : pin? » are‘ bdi vd phdn bdi gicii mdu. Trongphdn bdi gidi mdu , tdi liéu ndy se" gio"i thiéu cho cdc bon doc cdc budc gidi cling nhu‘ cdch trinh bdy mot bdi trip [on , nhdm cling Cd cdc kién thu"c co bdn tru'o’c khi [hi hét mon hoc . Tuy dd Cd nhiéu Cd gdng trong qud trinh bién soon , nhzrng do trinh do vd thdi gion cé hon nén khéng trdnh khdi nhfing soi sot . Chdng téi mong nhdn duoc nhiéu kién ddng gop Clio cdc bon ddng nghiép , co’: bon sinh vién vd cdc bon doc , dd tdi liéu ndy ngdy Cong dwoc hodn thién hon . Xin chdn thdnh com on so’ quart tdm vd nh1Z‘ngy' kién dong gép quy' bdu czia tdt Cd cdc ddng nghiép dd gulp dd chzing téi rd! nhiéu trong qud trinh bién soon tdi liéu ndy . CAC TAC GIA
  4. 4. CAC YEU CAU CHUNG I -YIAEU CAU vi‘: TRlNH BAY S Trang bia trlnh bay theo mau qui djnh (xem phan Phu luc cfia tai Iiéu nay); LT Bai lam trinh bay trén kh6 giay A4; ff Cac hinh vé lrong bai lam phai r6 rang, phai ghi day dfi cac kich thLrc’vc va tai trong as cho bang s6 lén so (16 tinh; 3 Cac bwoc tinh toan, cac két qua tinh toan, Ezac biéu dE‘)‘n<f)i lL_l’C v. .v. .. can phai duoc trinh bay r6 rang, sach sé va theo bai mau (xem phan vi du tham khao cua tai liéu nay). II —YEU CAU VE NOI DUNG 3 Mon Strc bén vat lieu co 4 bai tap Ion sau: 1. Tinh dac trung hinh hoc cfia hlnh phang 2. Tinh dam thép 3. Tinh cot chju luc pl‘| U’C tap 4. Tinh dam trén nan dan hoi. as Mon Co hoc két céu co 3 ball tap | c'rn sau : 1. Tinh he tinh djnh 2. Tinh khung siéu tinh theo phuong phap Iuzc 3. Tinh khung siéu tinh theo phuong phap chuyén vi Va phwoing phap phan phéi momen
  5. 5. PHANI DE VA HU’O’NG DAN GIAI BAI TAP LON SL'. l’C BEN VAT LIEU
  6. 6. BAI TAP LON so 1 TlNH DAC TRU’NG HlNH HOC CUA HlNH PHANG BANG so LIEU BAI TAP LON so 1 D(cm) Bxbxd (mm) 180x110x10 250x160x20 125x80x7 125x80x10 140x90x8 140x90x10 160x100x9 l60xl00xl2 180x110x12 200x125X16 250xl60x18 250x160x20 1 2 3 4 5 6 7 8 9 L: Sinh Vién chon nhfrng 36 liéu trong bang s6 liéu phu hop Vdi hinh Vé cua mlnh. YEU cAu VA THU’ Tl_l’ THl_. l’C HIEN Yéu cau: Xac dinh cac mo men quan tinh chinh trung tam Va phuong cua cac truc quan tinh chinh trung tam cfia hinh phang da cho. Giai bang hai phuong phap: gia’i tich Va dd giai. Céc bu'c'>'c giai: 1. Xac djnh toa do trong tam cda hinh phéng: -' Chon he truc ban dau xoyotuy y —' Xac dinh toa do trong tam Va tinh cac dién tich, cac mé men tinh cfia timg hinh thanh phan val hé truc ban dau as chon, -' Dung cong thtrc xac dinh trong tam C(x¢, yC): ZSYO . yc: ZSXO ZF ’ ZF X9:
  7. 7. 2. Tinh ca'c me man quan tinh chinh trung tam: -" Chon he true trung tam XCY (di qua trong tam C Va song song Vévi he truc ban dau). Xac dinh toe do trong tam cua ttrng hinh thanh phan déi V0l he truc trung tam XCY. ' Tinh céc mo men quan tinh trung tam cL‘Ia tL‘I'ng hinh thanh phan (Jl , J‘Y Va J‘XY) lay Véil he truc XCY bang cach dung cong thuc chuyen truc song song. To do tinh ca'c m6 men quan tinh trung tam cfia toan hinh (Jx, Jy, Jxv). -' Tinh me men quan tinh chinh trung tam Jmax_ min bang hai phwong pha'p: a) Phuovng phap giai tich: Dung cong thuic xoay trL_Ic de xéc djnh mo men quan tinh chinh trung tam Va VI tri cua he truc quan tinh chinh trung tam (Jmax, Jmin Va O. ”-Iax) _JX+JY+ JX—JY2 2 ‘ 2 Jmaxmin JXY Jxv t mx= ’ g a a Jmax _JY ‘JX _‘Jmin b) Phuong phap de giai: DL_ra Vao cac gia trj Jx, Jy, Jxy as tinh ducyc (3 tren, Vé Va sir dung Vong tron Mo quan tinh de xac djnh me men quan tinh chinh trung tam Va VI tri cfla he truc quan tinh chi nh trung tam (Jmax, Jmin Va (Xmax).
  8. 8. HlNH DANG MAT cAT NGANG / .4 X0 8 b 2D N C I IN” . . ‘ R D D C
  9. 9. C C “ h °0 X0 co B 3 a D B X0 X0 V-, : m: s : " 1 ‘s 10
  10. 10. Vl Dl_J THAM KHAO De bai: Xac djnh cac me men quan tinh chinh trung t j tri he truc quan tinh chinh trung tam cua hinh phang cho tren hinh 1.1, b" : Thép géc Bxbxd: 250x160x20(mm); D = 20 cm; = m; R = 24cm. R=24 cm $§V l Tra ban the dc co: B = 250 mm F = 78,5 cmz X0 = 3,85 cm = 49 b=1 m Jx=498 4 yo=8,31cm =0, 5 d = 20 Jy = 161 3 85 yo 10.191 1T’fit
  11. 11. Bai lam: 1. Xac djnh trong tam: Chon he truc ban dau xoyo nhu hinh Ve: xem hinh 1.2. Chia hinh phang da cho thanh 3 hinh (xem hinh 1.2), kich thwoc Va loa do trong tam CUB tiling hinh thanh phan lay Vol he truc ban dau la: - Hinh 1 (chCv nhet): b1=b+D+R=16+20+24=60cm; h1 = 20 cm; X1 = 30 cm; y1 = 10 cm; 0, ( 30,10 ); F1: b1.h1=1200 cmz; s; “ = F1.y1 = 1200.10 = 12000 cm3 sy = F1. x1 = 1200. 30 = 36 000 cm3 - Hinh 2 (1/4 tron): R = 24 cm; Tea do trong tem cL’1a ‘/4 tron Vol he truc di qua trong tam hinh tron la: 4 * 4R 4.24 =10,191cm —> x2 = R—x"2 = 24 — 10,191 = 13,809 cm —> Y2 = c + y*2 = 20 + 10,191 = 30,191 cm 02 ( 13,809; 30,191); F2 = r1:. R2/ 4 = 452,16cm2; sf’ = F2. y2 = 452,16. 30,191 =13 651,162 cm3 3.1,” = F2. x2 = 452,16.13,809 = 5 243,877 cm3 Hinhl.2o — Hinh 3 (thép goc): s1’II dung cac gia tri tra bang thép o tren, ta co x‘3 = 3,85 cm y*3 = 8,31 cm X3: R+D+ X; =24+20+3,85cm=47,85cm ya = c + y; = 20 + 8,31= 28,31 cm 03 ( 47,85; 28,31); F3 = 78,5 cmz. sf’ = F3. y3 = 78,5. 28,31 = 2 222,335 cm3 sly“ = F3. x3 = 78,5. 47,85 = 3 756,225 cm3 12
  12. 12. Bang ket qua tinh toan 5;, (m3) 8;, (cm 12 000,000 36 000,000 13 651,162 6 243,877 2 222,335 3 756,225 Teng 1730,66 27 873,497 46 000,102 Toa de trong tam: zsiyo : 36000 + 6243,877 + 3756,225 : 46000,102 > X9 = + 26,58cm ): F, 1200 + 452.16 + 78,5 1730,66 X13: 28:0 =12000 +13651,162 + 2222,3315 = 27873,497 ZF1 1200 + 452,16 + 78,5 1730,66 Toa do trong tam trong he truc ban dau xoyo la: C(+26,58; +16,106) Y. ;= >Yc= +16,106cm 2. Tinh cac mo men quan tinh trung tém: Chon he trI. _Ic trung tam XCY nhw hinh Vé: Xem hinh 1.3. . Y 5 3 52 g *3 § ’ 111111. 2’ / /// ,,/2 12.771 3.42 26,5 8 cm A 21.27 1 H1nh1.3 13
  13. 13. a. T061 <76 trong tém C08 mung hinh thanh phén 076/ vL’7i hé truc trung tém XCY / E31.’ Hinh a, (cm) bi (cm) 1 3,420 - 6,106 2 —12,771 14,085 3 21,270 12,204 b. Tinh mé men quén tinh caa tfrng hinh thanh phén 1761' vc'7i hé true trung tém XCY: Dung céng thU’C chuyén true song song. — Hinh 1: chip nhét JQ’ = J9) + b1ZF = 50203 + (—6,106)2.120O = 40 000 + 44 738,883 4 J)! ’ = 84 739,883 cm“ JQ’ = J, “ + a,2F, = 20503 +(3,42)21200 = 360 000 + 14 035,68 4 J9’ = 374 035,68 cm“ J9; = a1b1F1 = (3,42)(- 6,106) 1200 = — 25 059,024 cm“ —» J9; = - 25 059,024 cm“ - Hinh 2: 1/4 tr<‘)n Tinh mé men qua'n tinh J9] vé Jfl léy vdi hé trL_1c trung tém cfla hinh 1.2 J12) : J92) = [fi_[4Rj2 7:52] 16 E 4 —» J? ‘ = J? ) = 0,19625R“— 0,14154R“ = 0,05471R“ Véy: J‘? = J? ) + 52215: o,05471R4 + b22F2 J? = o,05471. 244 + (14,085)2. 452,16 =18151,464 + 89 702,765 —» J5) = 107 854,23 cm“ . Y Twang tu: J? ’ = J(y2) + a22F2 max J? ) = 005471. 244 + (—12,771)2. 452,16 ~§‘ = .x 18151,464 + 73 746,59 4 JG’ = 91 898,054 cm‘ 10.191 Ap dung céng thL'rc: J‘X2Y> = J3“ + a2b2F2 HirLh1.3u 14
  14. 14. 4 2 Ta 06: Jf) = i R—— 4“ 8 3.71 3.7: 4 Jfgyz = :(0,125R4—0,14154R4) = :0,01654R4 Trwérng hcyp nay 1g am < 0 nén Jgzgyz = 0,01654R4, Iéy déu > 0: J<X2Y>= 0,01654R4 + a2b2F2 = 0,014654.244 + (14,085). (—12,771).452,16 Jgfg = 5 487,575 — 81 334,328 = — 75 846,753 cm‘ — Hinh 3: thép géc J? = 4987 + b§. F3 = 4987 + (12,204)2.78,5 = 4987 + 11 691,606 4 J33’ = 16 678,602 m4 J? ’ = 1613 + a§. F3 =1613 + (21,27)2.78,5 =1613 + 35 514,412 3,85 4 JG’ = 37127,412 cm4 J(x3v’ = J§3s’ys ’' asbs F3 Ap dung céng thL'rc: tg Uvmax = # ‘* Jxy = (Jmin ‘ JX) tg amax min x v1 tg am > 0 nén J1?“ < 0, do do J43“ cue thép géc la: J2)“ = (949 — 4987). 0,405 = - 1 635,39 cm4 J9; = J1?“ + a3b3F3 = — 1 635,39 + (21,27). (12,204).78,5 4 J9; = - 1 635,39 + 20 376,957 = 18 741,567 cm Béng két qué tinh toén Hinh 11» <cm‘> 1 40 000 360 000 2 18151,464 18151,464 54117575 —12,771 3 4 987 1613 1635,39 21,27 84 739 883 374 035,68 — 25 059 024 107 854 23 91 898,054 -75 846 753 16 678,602 37 127,412 18 741,567 15
  15. 15. c. Tinh mé men quén tinh trung tém c1Ja toén hinh: Jx = = 84 739,883 + 107 854,23 + 16 678,602 4 Jx = 209 272,715 cm4 Jy = ZJ; = 374 035,68 + 91 898,054 + 37 127,412 4 Jy = 583 328,384 cm4 Jxy = XJXY = -25 059,204 4 75 846,753 + 18 741,567 4 Jxy = 4 82 164,210 cm4 3. Tinh céc m6 men quén tinh chinh trung tém: J J J J 2 Jmax, min= X; Y i [XE YJ +J§(Y _ 209272.715 + 583328.384 Jmax, min ‘ 2 J[209272,715 4 583328,384 2 + 2 J +(—82164,21O)2 Jma><. min ‘ 792601,099 + [—374055,669 2 _ 2 Jmax = 39630055 + 204280,12 = 600580.67 Jmin = 39630055 - 204280,12 =192020,43 JXY = _ 482164,210 = _ 482164,210 JW 4 JY 600580,67 4 583328,384 17252,29 2 J +(—82164,210)2 tgumax= ‘ = 4,7625 4. Két qué tinh toén: Jmax = 600580,67 cm4 Jmm = 192020,43 cm4 am 2 780085” Vbng Mo trén hinh 1.4 duwyc vé vé1i: _ Tam; C ( .3,“ k, ,,, , R = ; 0 ) 4 C (396300,55; 0) 2 2 J +(—82164,210)2 4 R = 204280,12 - Cuc: P (Jy, Jxv) —> P ( 583 328,384; - 82164,210) 16
  16. 16. Jmx = 600580,67 cm‘ JXY=82164,210 , _]Y = 583 328,384 1‘ Hinh 1.4 V1 tri hé truc qua'n tinh chinh trung tém duqc biéu dién trén hinh 1.5 Y Max Min / // // % 9 ¢ 2 16,106 cm 17
  17. 17. BAI TAP LON so 2 BANG so LIEU BAI TAP LON so 2 PKN l 2 3 4 5 6 7 8 9 1 3 6: Sinh vién chon nhfyng s6 Iiéu trong béng s6 lieu phi. ) hC_)’p vc’yi hinh vé cfla minh. YEU CAU VA THU’ TU’ THL_rc HIEN Yéu céu: Héy chon s6 hiéu mat cét cho dém lam béng thép chfr I (INO) dé thoé man diéu kién bén cfla dém, biét [6] = 210 MN/ m2. Tinh chuyén vj tai mzfat cét D. Céc bwévc giéi: 1. Chan so bcf) m_ét cét: -' Vé biéu di‘) ncf>i luc cfla so (16 tinh vdi téi trong dé cho (Mx, Qy) -' TU» biéu d6 Mx vé dwcyc, chcpn mat cét nguy hiém cé MX max -' Chc_>n kich thuévc mat cét theo diéu kién bén cfla (mg suét phép: WX Z lMXlmax [6] T0 dc’: tra béng thép dé dwqc s6 hiéu thép (N° I) cén fim. 18
  18. 18. 2. Kiém tra / ai d/ éu k/ _én bén khi co’ ké dén trong / wqng bén thén: -' Vé biéu d6 ncfni luc trong trunfmg hop cc’) ké dén trong lwcjng bén than dém. - Chon céc mét cét nguy hiém: tn: biéu d6 Mx Va Qv chon ra 3 loai mat cét sau: * Mat cét cé | Mxmax * Mat cét co’ | QY| max * Métcét cé Mx Va Qy cung | c’vn (déi khi 3 Ioai miat cét néy trung nhau). -' Kiém tra bén cho dém tai céc diém sau: * Diém cc’) (mg suét pha'p ldn nhét (tai céc diém trén bién cfla mat cét cc’) Wlxlmax) * Diém cc’) (mg suét tiép | c’rn nhét (tai céc diém trén dwdng trung hoé cfla mat cét cé Qymax ): _ °v Tmax ‘ SP] max X Jxbc s [r] Theo thuyét bén (mg suét tiép cue dai thi: [1] = % Theo thuyét bén thé néng bién déi hinh dang thi: [T] = % * Diém co [mg suét phép va L'rng suét tiép déu khé Idn (diém tiép giép gifra than va ca'nh trén mat cét cc’) Mx va Qy cung lérn): Theo thuyét bén (mg suét tiép CL_J’C dai thi: on = 1 (522 +41% < [0] Theo thuyét bén thé néng bién d6i hinh da'ng rm: 6“ = 3 522 +31% S [0] - Néu mot trong ca'c diéu kién bén trén khéng thoé man thi phéi chon Iai sé hiéu thép, Va kiém tra bén Iai cho dém. 19
  19. 19. 3. Xéc djnh dung suét chinh: —' Tinh LJ’ng suét chinh va phwcmg chinh tai 5 diém dzfzc biét trén mejt cét cc’) MX V2‘; Qy cung | c'm (diém trén 2 bién, dié-m trén dwdng trung hoé, diém tiép giép gifya thén Va ca’nh) béng phuvng phép giéi tI'ch -' Xéc djnh (mg suét chinh Vé phuong chinh tai 5 diém dc’) béng phwdng phép vé vc‘)ng Mo. 4. Tinh chuyén v_i: -' Viét phwcmg trinh dc) Véng Va gc’)c xoay cho toén dém béng phwcrng phép théng s6 ban déu. ' TI’nh chuyén Vj dang Va géc xoay tai mat cét D. 20
  20. 20. S0’ D0 TiNH
  21. 21. M q q M 12:3»; P1” % P 2;. ” 71a}: b 1 “At 71/“), b 4;“ ,1: 22
  22. 22. V1 DU THAM KHAO Iaé bai: Chon sé hiéu thép chip I (N°l) caa mat cét ngang dam dwdi day, Biét: [0] = 210 MN/ m2, (xem hinh 2.1). TI’nh dcf) Vc‘)ng Va géc xoay tai mat cat D. P=10KN q=2OKN/ m M=4OKNmm q=2OKN/ m 75,714 KN 54.286 KN Hinh 2.1 Bai lam: 1. Chc_)n so bf) mat cat theo diéu kién bén cfia (mg suat phép: 1.1. Xéc djnh phan Iuvc géi tu= a.- —2o—4o+4o+4o0 2MA= V13.7+P.2+q.2.1—M—q.4.5=0—>VB= 7 —> VB = 54,286 KN 90+320—40+160 2MB= -VA.7+P.9+q.2.8-M+q.4.2=0—>VA= 7 -> VA = 75,714 KN Kiém tra Iai phan | l_. PCI 2Y= VA+VB—P—q.2—q.4=75,714+54,286—10—20.2—20.4=0 » VA Va VB da tinh dung. 1.2. Viétphuvng trinh néi / l_J’C. ' Chia dam lam 3 doan — Doan CA: Chc_)n géc toa dé tai C Va true 2 hucyng sang phai (O 5 21 S 2 m ) Qy = - P — q. Z1—> Qy = - 10 — 20.21 —> Phwong trinh bac nhét Z12 MX = - P. Z1— q.7 —> MX = -10.21 — 10.212 —> Phucmg trinh bac hai 23
  23. 23. *TaiZ1=0(taiC): Qy= -10KN; Mx=0 * TaiZ1= 1 m (tai gifra doan): Qy = — 30 KN; Mx = — 20 KNm *Te_Ii 21 =2m(tai A): Qy= -50 KN; Mx= —60 KNm — Doan AD: Chc_)n géc toa dtf) tai Ava true 2 hLrc’)ng sang phai (O S 22 S 3 m) Qy= - P—q.2+VA= -10—20.2+75,714 —> Q1 = 25,714 KN —> Phwcyng trinh hang s6 Mx = - P. (2+Z2) — q.2.(1+Z2) + VA. Z2 = - 10(2+Z2) — 40.(1+Z2) +75,714.Z2 —> Mx = 25,714.22 - 60 —> Phwdng trinh bac nhét * Tai Z2 = 0 (tai A): Qy = 25,714 KN; Mx = - 60 KNm * Tai Z2 = 3 m (tai D): Qy = 25,714 KN; Mx = 17,142 KNm - Dean DB: Chc_)n géc toa dtf) tai B Va true 2 hwcvng sang trai (0 S 23 S 4 m) Qy = — VB + q. 23 —> Qy = - 54,286 + 20.23 —> Phwcrng trinh bac nhat 2 Mx = V3.23 — q. % a MX = 54,286.23 -10 232 » Phwong trinh bac hai * Tai Z3 = 0 (tai B): Qy = - 54,286 KN; Mx = 0 * Tai 23 = 2 m (tai gicva doan): Qy = —14,286 KN; Mx = 14,286 KNm * Tai Z3 = 4 m (tai D): Qy = 25,714 KN; Mx = 57,14 KNm 1.3. Xa'c djnh I/ _i tri cé Mmax: Cho phwcvng trinh Qy = 0 (E) doan DB), ta tim ducyc toa dcf) mat cat cc’) Mmax: (mat cat E) 54286 — 54,286 + 20.23 = 0 —> 23 = 20 —> 23 = 2,714 m Tinh gia tri Mmax: Mmax = 54,286. 2,714 — 10. 2,7142 —> Mmax = 73,67 KNm Bang két qua tinh toan ncf)i | L_.1’CZ CA 0 - 10 0 2 -60 -60 24
  24. 24. Vé biéu dc“) ncf)i Iuxc 10 | IIIII| ||| r'e . .‘ .4 25.71 :25 71 5 1.4. So’ bé chon mat cat theo diéu kién bén cda I}’ng suat phap: Tai mat cat E trén hinh Vé cc’) mc‘) men ldn nhat: MX| = + 73,67 KNm (xem hinh 2.2) nén: max max _ 73,67KNm [6] _ 210.103KN/ m2 Tra bang thép chfy I, chon | No:27 cc’) Wx = 371 cm3 thoa man diéu kién trén Va cc’) cac dac trung sau: qm = 31,5 Kg/ m = 315 N/ m = 0,315 KN/ m; h = 27 cm; b = 12,5 cm; d = 0,6 cm; t= 0,98 cm; F = 40,2 cm2;Wx = 371 cm3; sx = 210 cm3; Jx = 5010 cm4; =3,5.1o“ m3 4 w 2 350 cm3 2. Kiém tra lai diéu kién bén: (khi ké dén trong lwqng ban than dam) Sc) 66 tinh cfla dam khi cc’) ké dén trc_)ng lwcmg ban than nhw sau: 25
  25. 25. 1 31:). q=20‘315 KN/ m M=40KNm q = 20,315KN/ m 0,315 KN/ m 18,6 (KNm) 58,68 Hinh 2.3 Mmax = 75.25 KNm 2.1. Xac djnh phan / l_l’C géi tu’a. ' (xem hinh2.3) EMA = V5.7 + P 2 + q.2.1+ qb1.2.1— M — q. 4. 5 — q1,1.7.3,5 = 0 ~> V3 = 55,3 KN 2MB= -VA.7+ P. 9+q.2.8-M +q4.2+qb1.9.4,5=0 a VA = 77,54 KN Kiém tra lai phan | L_PCI EY = VA + VB — P—q.2 — q.4 — q1,1.9 = 77,54 + 55,3 -10 — 20.2 — 20.4 — 0,315.9 = 0 a VA Va VB da tinh dung. : - TI'nh trLrc‘)ng hcyp do riéng trc_)ng lwcyng ban than gay ra (xac dinh phan | L_l’C, vé biéu dc? ) ncf)i | L_. I’C) 26
  26. 26. - Ccjng biéu d6 vtra vé vévi biéu d6 trén hinh 2.2 sé dwqc biéu d6 nhu trén hinh 2.3. 2. 2. Viét phwong trinh néi / l_. PC. ' Chia dém lam 3 (Joan, chon true 2 Va géc toa d<’_5 cho m5i doan twang tu nhu trén: - Doan CA: 0 S Z1 3 2 m (g6c toa dc) tai C) Qy= -P—(q+qm). Z1 —>Qy= -10—20,315.Z1 Mx = — P. Z1 — (q + q. ,.). Z7‘2 4) Mx = -10.2, — 2052515 2,? *Ta_1iZ1=0(te_1i C): Qy= —10KN; Mx=0 * TaiZ1 = m (tai gifra doan): Qy = - 30,315 KN; Mx = -20.15? KNm * Tai Z1 = 2m (tai A): Qy = - 50,63 KN; Mx = - 60,63 KNm - Dean AD: 0 S Z2 5 3 m (géc toa dé tai A) QY = - P — q.2 — qbt. (2+ Z2) + VA = - 10 — 20.2 — 0,315.(2 + Z2) + 77,54 a Qy = 26,91 — 0,315.22 (2+z2)2 Mx = - F’-(2+Z2) - q-2-(1+Z2) + VA-Z2 - qm- 2 2 —> Mx = - 0,315. Z2—2+ 26,91 . Z2 — 60,63 * Tai Z2 = 0 (tai A): Qy = 26,91 KN; Mx = - 60,63 KNm * Tai Z2 = 2m: Qy = 26,28 KN; Mx = - 7,43 KNm * Tai Z2 = 3m (tai D): QY = 25,965 KN; Mx = 18,68 KNm -Doan DB:0 : Z3 5 4m(g6ctoad6taiB) Qy = — VB + (q + qbg). Z3 ~> Qy = - 55,3 + 20,315.23 Z2 20,315 Mx = VB-Z3 - (Q +qbt)-—3 —> MX = 55,323 - . z 2 2 2 3 * Tai Z3 = O (tai B): Qy = — 55,3 KN; Mx = 0 * Tai Z3 = 2 m (tai gifra doan): Qy = -14,67 KN: MX = 69,93 KNm * Tai Z3 = 4 m (tai D): Qy = 25,96 KN; Mx = 58,68 KNm 27
  27. 27. 2.3. Xa’c djnh vj tri co’ Mmax: - Cho phucyng trinh Qy = 0 (6 doan DB), ta tim dwqc toa d6 mét cét cc’) Mmax: (mat cét E) _ 55,3 ' 20,315 — 55,3 + 20,315.Z = O a —) Z3 = 2,72 m — TI'nh gié tri Mmax 2 Mmax = 55,3. 2,72 — 20,315. 2'72 a Mmax = 75,25 KNm Béng két qué tinh toén: OWN) 0 ~10 0 .505 50.63 AD 0 26,91 — 60,63 3 25,96 18,68 4 25,96 58,68 2,72 0 75,25 0 55,3 0 Vé biéu <16 ncfai Iuc: Céc biéu <16 noi luc Mx va Qy biéu dién trén hinh 2.3 2.4. Chon mat cét nguy hiém vé kiém tra bén: — Chen ba mét cét nguy hiém sau: 6 Mint cét H co M, = + 75,25 KNm 5 kiém tra diéu kién bén theo tvng suét pha’p om, cho ca’c diém trén bién. 5 Mat cét B co QY = +55,3 KN —» kiém tra diéu kién bén theo Llvng suét tiép rmax cho ca’c diém trén dwdng trung hoé. 6 Mét cét A ( tréi ) cé MX = - 60,63 KNm Va‘! Qy = - 50,63 KN 4) kiém tra theo thuyét bén thé néng hoéc thuyét bén (mg suét tiép cho céc diém tiép giép gifra lbng vé dé. - Kiém tra bén: 6 Kiém tra cho céc diém trén bién (diém I hoac K)te_1i mét cét H: _ Mx 5max = ' Gmin ‘ max = 202 900 KN/ m2 g[, ,] _, ,, W=7i5 Wx 371.1055 28
  28. 28. cs, ,a, = 202,9 MN/ m2 < [0] = 210 MN/ m2 Thoé mén diéu kien bén tai bién trén va bién dwcyi caa mét cét. 6 Kiém tra cho céc diém trén dwérng trung hoé (diém O - cc’) (mg suét tiép) tai mat cét B theo thuyét bén thé néng: _ IQY Tmax ‘ max Jxbc 3;] s [1] = M / § trong céng thtrc trén, ta léy be = d trong béng, they 36 ta (mac: -5 5 cm, - 555'3'21°‘1° = 38 600 KN/ m2 _ 5010.10'8.O,6.10'2 «max: 38,6 MN/ m2< M = 121 MN/ m2 J5 Thoé mén diéu kién bén tai céc diém trén truc trung hoa cfla mat cét. Biéu d6 L'rng suét ctla mat cét A (tréi) 6 Kiém tra cho céc diém tiép giép gifra than vé ca'nh (diém E hoéc F) tai mat cét A tréi theo thuyét bén TNBDHD: Gm = / G2 +312 S [6] Ta 06: M —e h — 2 GE: X. yE = °'63_8.—{ 4) = 6°'63_8.—[ 7—O,98j .10’? JX 5010.10 2 5010.10 2 29
  29. 29. CE = 151 000 KN/ m25 CE = 151 MN/ m2 . . . C vé C- Taldlém Eco. S = SX—d7 véb —d (Xem hinh 2.5) Do do: 2 2 Q0 Es, —dV? EJ —50,63.[210—0,6.12’52 J10-5 TE = — = = — 27500 KN/ m2 Jx. d 501O.10’8.0,6.1O'2 5 «. -E = — 27,5 MN/ m2, 5 (Sm = (1s1)° +3(— 27,5)° = 156,33 MN/ m2 6., = 156,33 MN/ m2 < [a] = 210 MN/ m2 Thoé mén theo diéu kién bén cI. ’Ia thuyét bén TNBDHD. Két luénz Chon mat cét | N° 27 dém béo diéu kién bén cho toén dém. 3. Xéc dinh (mg suét chinh: DI, ra véo biéu G6 L’rng suét trén hinh 2.4, tinh céc Iivng suét chinh vé phwong chinh cho céc diém déc biét trén mét cét A (tréi) 30 3.1. Being gia’i tich: - Diém trén bén (I vé K) 5max, min= i = + -5 = i 163 500 KN/ m2 = 11635 MN/ m2 Tai I c6: 61 = cm, = 163,5 MN/ m2 (phén t6 kéo dun) (xlmax = 0° Tai K 06: G3 = 6,, " = -163,5 MN/ m2 (phén t6 nén dorn) 6;, = 0° 2 - Diém tiép giép gifya long va dé (E Va F): cr, ,,, ,,, ,,, ,I, , = g: E—J +12
  30. 30. A Diém E: 6 trén (16 cc’) 6E = - 6E =151 MN/ m2 T5 = T}: = — 27,5 MN/ m2 2 2 151 151 6,, ,,, ,,, ,,, = E: +122 = T: +(—27,5)2 Do dc’): 61E: 75,5+ 60,352 MN/ m2 = 155,852 MN/ m2 63E: 75,5 - 60,352 MN/ m2 = - 4,852 MN/ m2 —27,5 - tgumax = ‘GE §Ecmin = '151_(_4,852) = 0,177 ) 0LrEnax =10o6 I Diém F2 2 5max, min = ‘ 1:1 i + (‘ 27v5)2 61F: -75,5+ 60,352 MN/ m2 = 4,852 MN/ m2 63F: -75,5 - 80,352 MN/ m2 = — 155,852 MN/ m2 tga; aX= _ 6, = _ .275 6,—6m, , —151~(-155,852) Véy: Tai diém E cé céc Iivng suét chinh la: 61 = 155,35 MN/ m2 C73 = - 4,852 MN/ m2 65,, = 10%’ Tai diém F cc’) ca’c Ifrng suét chinh la: 61 = 4,852 MN/ m2 G3 = - 155,852 MN/ m2 6;, = 60° 5 Diém trén duévng trung hofa (O): on = 0 _ Q$~IS§ -50,63.210.10‘6 = 5,67 > 65,3, = 80° 6,, — — = 5 = — 35 400 KN/ m2 = —35,4 MN/ m2 JX. b° 5010.10-“.0610-2 31
  31. 31. 2 Tai dufyng trung hoé 06: cmamin = +12 = : 35,4 MN/ m2 3 : 2 Tai diém 0 cc’) céc [mg suét chinh la: 61 = 35,4 MN/ m2 G3 = - 35,4 MN/ m2 “flax = 450 Vi phén t6 tai DTH Ié phén té truort thuén tuy. 3.2. Biéu dié'n phén to‘ t_ai 5 diém déc biét trén m_a‘t cét ngang va‘ vé vo‘ng Mo ting suét cho 5 diém db: = = ‘ (1. =00 %| :'+¢T1_€5.. ..x C; |:| _‘>71 Um max (7 01 1 er I 1 1 6 2 “max = 1005 0' Z ‘CE E N53 0' G) G) QY= ‘-50.63 KN @ __ Ext x<>7r"‘ 3a, .u; =4s° 6) ® Mx=75,25 KNm Hinh 2.6 32
  32. 32. 4. Viét phuwng trinh dwbvng dén h5i cfla truc dém: p :10 KN q=20,315KN/ m Q=0,315 KN/ m q=20,315KN/ m 3 m vA=77,54 KN ‘ v5=55,3 K Hinh 2.7 Béng théng 56 ban déu Dean CA (a=0) Doan AD a=2 Dean DB (a=5 A Y 4.1. Phwang trinh dé véng cfia tfrng doan dém: ( EJ = const ) AMa_. (Z — a)2 _ AQa_. (Z —a)3 y"” = y” + Aye + M" (Z_ 3) ' 2!EJ 3!EJ — Thay céc gié tri véo phwcrng trinh trén: *Doan CA: (0 : Z 3 2) 3 4 + 10Z + 20,3152 N2) = Yo + <Po- Z 3!EJ 4!EJ _ 1022 2031523 M2)" ‘°° + 2!EJ + E *E)oan AD: (2 s Z 3 5) _ 77,54(z — 2)3 _ 2o. (z — 2)“ z = Z M ) y‘( ) 3!EJ 4!EJ 33
  33. 33. 1023 + 20,3152‘ _ 77,54(2—2)3 _ 2o. (2—2)4 %Y2(Z)= Vo+ (Po-2+ 3!EJ 4!EJ 3!EJ 4!EJ _ 77,54(2—2)2 20.(2—2)3 2 — 2 - T — T <P2() (P1() ZEJ 3!EJ _NP2(Z) = (PM 1023 + 20.31523 _ 77,54(2—2)3 _ 20.(2—2)3 2!EJ 3!EJ 2!EJ 3!EJ *Doan DB: (5 5 2 s 9) 40(z—5)3 + 20.(2—5)“ 2 = 2 — Y3() Y2() ZEJ 4!EJ _ 1023 2031524 77,54(2—2)3 2o. (z—2)4 _’y3(Z)_y°+ ‘P°'Z+ 3!EJ + 4!EJ ' 3!EJ ' 4!EJ ' _ 40(2—5)2 + 20.(2—5)4 2!EJ 4!EJ _ 40(2—5) 2o. (2—5)3 <P3(Z) - <I>2(Z) ' T "' T3!EJ 2 3 2 3 awz) = ¢0+ 102 + 20,3152 _ 77,54(2—2) _ 20.(2—2) 2!EJ 3!EJ 2!EJ 3!EJ _ 40(2—5) + 20.(2—5)“ EJ 4!EJ 4.2. Xéc djnh yo va‘ (p 0 tL‘P ca’c diéu kién bién sau: -TaiA(Z=2)céy1=y2=0,thays6: 1023 + 20,3152“ _ 77,54(2—2)3 _ 2o. (2—2)4 = 0 312(2) = Y0 ’' <Po~ 2* 3!EJ 4!EJ 3!EJ 4!EJ 10.23 20315.24 2 = _2 *2 :0 M ) y°+ W + 3151 + 4!EJ 2687 .2 ' -0 —> Yo+ (P0 + E] -TaiD(Z=9)céy3=0 1023 + 20,3152‘ _ 77,54(2—2)3 _ 20.(z—2)4 _ 40(2—5)3 3!EJ 4!EJ 3!EJ 4!EJ 2!EJ Yo+<I>0-Z+ 4 + 20.(2—5) =0 4! EJ 34
  34. 34. 10.93 + 20,315.94 _ 77,54(9—2)3 _ 20.(9—2)4 _40(9—5)3 ” V“ + "’°' 9 + 3!EJ 4!EJ 3!EJ 4!EJ 2!EJ + 20.(9—5)4 = 0 4!EJ 1215 5553,613 4432,703 2000,833 320 213,333 _ yo + (pg. 9 + + — — — + — 0 EJ EJ EJ EJ EJ EJ —> Yo"' (P0-9+ 22341 =0 Ta co’ hé hai phuong trinh sau: 26,87 28,791 Y0"'(P0-2"’ EJ :0 #00:‘? 228,41 30,713 Yo+<Po-9+ EJ :0 Yo: EJ 4.3. Phwong trinh 60 véng vé géc xoay toan dém: 3 4 NZ): 30,713 _ 28.791 2+ 102 + 20,3152 EJ EJ 3154 4!EJ 2 7 1 1 23 2 1 23 M2); 8,9 + 0 + 0,3 5 EJ 2!EJ 3!EJ 30,713 _ 28,719Z+ 1023 + 2031524 _ 77,54(2—2)3 _20.(2—2)4 Z = M ) EJ EJ 3!EJ 4!EJ 3!EJ 4!EJ 2 71 3 2 1 23 — 3 . — 3 (p2(Z)= _ 8, 9+102 + 0,3 5 _77,54(2 2) _20(2 2) EJ 2!EJ 3!EJ 2!EJ 3!EJ 3 4 3 M2): 30,713 _ 28,719z+ 102 + 20,3152 _ 77,54(2—2) _ EJ EEJ 3!EJ 4!EJ 3!EJ _ 20.(z—2)4 _ 40(2—5)3 + 20.(2—5)4 4!EJ 2!EJ 4!EJ 271 123 2 123 7742-23 (P3(z)= _ 8, 9+ 0 + 0,35 _ ,5( )_ EJ 2!EJ 3!EJ 2!EJ _ 20.(2—2)3 _ 40(2—5) + 20.(2—5)4 3!EJ EJ 4!EJ 4.4. Tinh dz} v6ng va‘ go’c xoay tai m_a't cét D: Tai mat cét D cc’) Z = 5 m (thuéc doan 2), do dc’) thay véo phuong trinh <p2(Z) va y2(Z) ta C6 35
  35. 35. 36 28,719 + 10.53 + 20,315.53 _ 77,54(5-2)3 _ 20.(5-2)3 EJ 2!EJ 3!EJ 2!EJ 3!EJ 1 80 58 (p2(Z) = EJ(— 28,719 +125 + 423,23 — 348,93 — 90) = (Rad) <I>2(Z) = - 30,713 28,719 10.53 20,315.54 77,545-23 20.5-24 y2<z)= - + + - 4 ’- ‘ 3 EJ EJ 3!EJ 4!EJ 3!EJ 4!EJ 1 y2(Z) = E—J(30,713) (m) Két gué: yo = EiJ(30,713) (m) 1 8.3 = E(s0,5s) (Rad)
  36. 36. BAI TAP LON so 3 TTNH COT CHIU LUC PHUC TAP BANG SO LIEU BAI TAP LON SO 3 b(m) A(m) 1(m) P(KN) KN/ m3 q(KN/ m) ON >-‘>-‘D0OlO<. h-J3UJIJ>—- Ho OOU‘IOl<7Vl'J10l0O Ghi chL'1: Sinh vién chon nhfyng S6 Iiéu trong béng 56 Iiéu phfl hqp vdi hinh vé cfla minh. YEU cAu VA THU’ TU’ THl_J’C HIEN Yéu céu: - Xéc djnh néi Iuc tai mét cét déy cét. - Vé biéu d6 Evng suét phép tai mat cét déy cét - Vé |6i cfla mat cét déy cét. Biét réng m6i so 06 ccf)t 06 3 Iuc doc Iéch tém (P, trén hinh vé ky hiéu diém eat I3 (8) ), I I3 chiéu cao cét, y la trong Iuqng riéng cfia cét, q (KN/ m2) I3 Iuc phén b6 déu vuéng géc vc’7i mat phéng chflla canh EF. Céc bwévc giéi: 1. Vé hinh chiéu truc do caa cr_3t: ' TU’ so d6 hinh chiéu béng dé cho, vé hinh chiéu trL_1c do cfla c(_‘)t trén hé truc toa dc} Dé Céc. ' ChL'1 y ghi déy GL1 kich thwérc vé téi trong dé cho. 2. Xa’c dinh 0:520 1750 trwng hinh hoc cda m_a‘t cét ngang: -' Xéc djnh toa G6 trong tém cL’1a mét cét ngang C(Xc, YC) 37
  37. 37. " Xéc djnh céc mé men quan tinh chI'nhtrungté1m: Jx, Jy " Xéc djnh céc bén kI'nh quén tinh chinh trung tém: ix, iv. 3. X530 djnh m_5i lgrc va‘ L'mg suét t_ai mat cét déy cét: -' Xéc djnh tea dc) céc diém dét lgrc doc léch tém F’; (X‘ , Y‘K ). - Tinh céc gia' tri ncfai Iuc tai mét cét déy ctfat, Ién luzcyt do céc téi trong dé cho géy ra. -' Xa’c dinh dLr<‘7ng trung hoé vé (mg suét pha’p | c’yn nhét vé nhé nhét tai mat cét day ccfnt, tu’ do vé biéu dc‘) (mg suét phép phéng tai mét cét déy c©t. -' Xéc djnh vi tn’ diém dét ILPC doc léch tém K ( XK, yK ) twang duong tai mat cét déy ccfnt —'‘’'v- = & XK NZ, YK NZ 4. Xéc djnh /6i cda mét cét day c_6t an | <‘g. ~< II . 5' | ><_r§> 5. Biéu dién: v Biéu dién nci Iuc tai mét cét da’y cét béng hinh chiéu trL_1c do ' Biéu dién vj trl’ hé truc quén tinh chinh trung tém - Biéu dién diém dat | L_rc doc léch tém tai mat cét déy cot -' Biéu di§n dudng trung hoé tai mét cét déy ccf>t -' Vé biéu 65 (mg suét phép phéng tai mét cét déy cét -' Vé | c'>i cfla mat cét. 38
  38. 38. S0 D0 MAT CAT COT @ ® a 221 4 3b 3b 3b 1.5b E F E F in 7a T1‘ 3 3a a @1,5a a 3a a 1,53 3b 3b 2], 2b E F E F 39
  39. 39. 2b 321 _#b 3b 2b _ 7 , » 221 2a 2a 2a 40
  40. 40. V1 DU THAM KHAO Dé bai: Cho mat cét cét chiu | L_. vc nhu hinh vé, biét céc 1L_J’C Iéch tém P1=P2=P3=500KN, a’p | L_J’C phén b6 déu trén mét EF Ié q = 15 KNI m2‘ trgng Iucyng riéng cfla ccfwt la y = 20 KN/ m3 vé cét cao L = 4m (xem hinh 3.1). Yéu céu: — Vé biéu d6 L'mg suét phép tai mét cét déy ccfnt — Vé |6i cfla mét cét da’y cL’1a ccfnt 20cm 8cm 20cm q Hinh 3.1 Béi Iém 1. Vé hinh chiéu truc do cfla c{>t: TU’ mét cét c4_5t dé cho trén hinh 3.1, ta vé dwoc hinh chiéu truc do cfla cét trong hé truc toa G6 Dé Céc nhu trén hinh 3.2. 2. Xéc dinh céc déc trwng hinh hgc cfla mat cét ngang cét: ~'= =Xéc djnh toa dé trong tém cL’Ia mat cét déy cét: Chon hé truc toe d0 ban déu Ia xoyo nhw hI‘nh3.2. ZS 43.2o.1o+s.13.(2o+9)+36'43.(2o+1s+12) YC= X0 = 2F 48.20 + 18.8 + 3648 _, yC= =@ _, YC=2395cm 960 +144 + 864 1968 41
  41. 41. CfXa'1c djnh céc mé men qua’n tinh chinh trung tém: Hé truc XCY trén hinh 3.3 Ia hé true quén tinh chinh trung tam. Chia mefat cét ngang cét thanh 3 hinh thanh phén: 2 hinh chfy nhét Va 1 hinh tam giéc. 45,05 42
  42. 42. 2 48.203 8.183 Jx + (2895 —10)2.48.2O + + (28,195 — 29)2.18.8 + +(2o+18+12—28,95). +[48.363 2 48.36} —> Jx = 32 000 + 344 774,785 + 3 888 + 0,346 + (62 208 + 382 804,186) 8 Jx = 825 675,3 cm4 _ 20.483 18.83 [36.243 — +2. Jy + 12 12 8 JY = 268 032 cm4 F = 1968 cm2 J =184 320 + 768 + 2.41472 6=Xa'c djnh ca’c ba’n kinh qua'n tinh chinh trung tém: 1% = / 8 ix = 20,5 cm; 1% = 2:332? 8 iv = 11,7 cm 3. Xéc dinh n¢i luc va (mg suét tai mét cét déy c¢t: ix iv 6’Xa’c djnh toa d_6 ca’c tfiém d_a‘t I1,. rc doc I_éch tém P1: TU hinh 3 ta cc’): P1(24; 9,05); P2 (24; -895); P3 (-24; -28,95) 3 Xéc d_inh m_3i / l_. FC. ' Mx = 2Pi. yiK + qFL§ = (—500).9,05 + (—500). (—8,95) + + (-500). (-28,95) -15. 48.10'2.4.200 Mx = - 4 525 + 4 475 +14 475 — 288.200 —> Mx = - 43175 KN. cm My = 2P; .xiK = (-500).24 + (-500).24 + (-500). (-24) —> My = -12 000 KN. cm N; = —(z P, + y. L.2F) = 3.(—500) — 20.4.1968.10'4 8 N; = -1 515,74 KN 6’Xa'c djnh (mg suér tai mét cét déy 1:61: TI'nh (mg suét tai céc diém géc trén mat cét ngang theo céng thL'rc: 43
  43. 43. F J, Béng két qué t1’nh toén: X, y, 11515,741 + 143175 1 | i 1120001 )4 GZ (cm) ‘T1968 - 3256753 V 268032 (KN/ cmz) — 8,951 — 0 7702 1.51385 107450 1.81815 ,9510 — 0,7702 0.46810 1,07450 0.77240 — —8,9510 — 0,7702 0.46810 0,17910 —0.1230 — 0,1791 -0.4812 W J, ‘ M i IX1 D16 5 - 8 _’>IJI. > 1,07450 -0.1689 1 [Q OJ; -B 0 —3.1258 —1,07450 —2.3179 -0,17910 -1.4225 0,1791O -1.0643 —1,07450 —1.3766 3 A D 1 H N P G F E2882 —1,07450 —0.33085 Tai E: em, = 1,81815 KN / cmz; Tai 1: am = — 3,1258 KN / cm? Phén vung (mg suét tai mat cét déy cét do My vé My gay ra: 44
  44. 44. TL‘r hinh vé vé béng két qué tinh toén, ta théy ngay 2 diém nguy hiém Ia E & I = ?Xa'1c d_inh biéu J6 Ling sua"t phép tai mat cét déy cét: a) Biéu dién néi Iuc tai mét cét déy cot: Hinh 3.5 b) Xéc dinh diém dét | L_PC doc Inf. -ch tém K(xK, yK) X= &_ 812000 _ 9 K NZ 81515,74 ’ yK= &=i75=285cm NZ —1515,74 ‘ Diém dét | L_rc doc Inf. -ch tém biéu dién trén hinh 3.6: K(7,9; 28,5) / » 7,9 cm Z - P= l515,74 KN ‘ . YoEY c) Xéc dinh dLrc‘mg trung hoa: Phwong trinh DTH: 5 + % =1 5 45
  45. 45. a——i=7—= —17,2021cm a= —17,2cm XK 7,9 -12 -20 52 _) b= —X= —'= —14,75cm = -14,7cm YK 28,5 Toe 66 DTH 15: (—17,2; -14,7)cm Biéu dién 6160 06 uvng suét pha’p phéng trén hinh 3.7 2 4 -Xéc dinh |6i cfla m§'t cét L818" KN/ °'" Chqn 3 du’<‘7ng trung hoé gié thiét: A1, A2, A3 nhu hinh 3.8 duéai déy, I Y0 EY D H A B X A1 A3 E Hinh 3.8 46
  46. 46. + Dwdng A1: Cét trL_1c Y tai diém I(0; 45,05) nén cc’): b1 = 45,05 cm Cét trL_1c X tai diém A(xA; 0) nén sé cc’): a1 = XA. TI'nh XA theo tinh chét déng dang cfla 2 tam gia’c Ia: ACI vé HJI JH IJ 24 36 _ 24.45,049 T = — 4) T = 9 XA - 5 CA IC XA 45.049 Vef1y cc’): a1 = 30,03 cm; b1 = 45,05 cm = 30,033 cm + Dwang A2: song song véri truc Y nén cé: a2 = 24 cm; [32 = 00 + Dwdng A3: Song song vdi truc X nén c6: 33: co; b3 = - 28,95 cm -2 -2 Dung céng thwc: XK = - ll, YK = - ifi; V61: ix = 20,483 cm, iy =11,67 cm a dé xéc djnh céc diém trén |6i la 3 diém dét Iwc doc Iech tém twang (mg: K1, K2, K3. Tw hinh vé trén, ta lap dwac béng sau: DTHgie'1thiét a; (cm) b1(cm) Diém XK; (cm) YK, (cm) A 30,03 45,05 K1 — 4,54 — 9,31 24 K2 - 5,67 — 28,95 K3 14,49 Vi mat cét déi xL'mg qua trL_1c Y, nén ta léy thém 2 diém déi xwng vai 2 diém dét lwc doc léch tém do dwang A1 V8 dwang A2 5 trén, dwac them 2 diém K4 vé K5. N615 diém (36 I831, ta dwac |6i cfla mét cét cc’: dang nhw hinh vé 9 dwdi déy. . 47
  47. 47. 2% '4 / Hinh 3.9 Tu» hinh 3.6 ta théy: diém dét | L_rc doc Iéch tam K ném ngoai |6i nén biéu (16 (mg suét phép trén hinh 3.7 cé hai déu. Muén biéu <36 (mg suét phép chi co déu (-) thi | L_l’C doc Iéch tém tai mét cét déy ccjt phéi dét véo |6i. 48
  48. 48. BAI TAP LON so 4 TiNH DAM TREN NEN DAN HOI BANG so LIEU BAI TAP LO'N so 3 0,03516 0,01800 0,04267 0,02289 0,01800 0,02858 0,04267 0,051 18 0,03652 0,02843 0,040 I 6 1 2 3 4 5 6 7 8 9 10 ll ouJl. »Jl. oJJ><. »JI)U1I>-J><. »3 '-‘O>-‘U1-J31»)-J>IJU1IJ-J> U-)'—OI)I)J>J>bJbJJ>UJ 0: Sinh vién chgn nhfrng s6 Iiéu trong béng s6 Iiéu phi. ) hcyp vc’yi hinh vé cfla minh. Yéu cAu VA THU’ TU’ THl_J’C HIEN Yéu céu: Vé biéu di‘) néi | L_J’C (biéu d6 M vé biéu G6 Q) cfla dém dét trén nén Winkler. SCV as tinh cho trén hinh vé, médun dén h6i cfla dém Ia E = 107 KN/ m2. Yéu céu lap béng két qué tinh toén n(f)i Iuc cho céc mist cét Iién tiép céch nhau 1 m. Ca’c bmfvc giéi: 1. Tinh s5n céc tr_i s6 cén S0’ dung: Dc} cL'mg: EJ, hé sé cfla dém trén nén dén h6i: m, m2, m3, m4 2. L_a? p béng théng s6 ban déu: Lap béng vdi 6 théng $6 cho céc dean dém. 3. Viét phwang trinh mé men uén va Igrc cét cL’Ia dém: -' Viét phu’c7ng trinh néi luc cfla dém. -' Lép béng céc hé sé Crulép tai ca’c mét cét cén tinh toa'n. -' Lap phuang trinh n<f)i | L_J’C cfla toén dém dufyi dang s6. 49
  49. 49. - Lap diéu kién bién va giéi hé phwong trinh dé tim ra cac én sé yo Va 00. -' Lap béng két qué tinh toén n©i | |,J’C (mé men va Iuc cét) tai céc mét cét cén tinh toa'n. 4. Vé b/ éu d6 néi / _u: c: DL_ra vao béng két qué tinh toén ér trén, vé céc biéu d6 ncfni Iuc. 50
  50. 50. SO 90 TINH DAM TREN NEN DAN HOI 6) PM 2: (2). , M / # / # 4134 b 4 C J’ #34“ b 4' C 2}’ @P q M 2', ‘ ® P q M 2P A! At +d4~w+~e~+ 4a~T~w%c~+ @" 2P®P q 2P M /1; / # fl> P 2P P M 2P q PT-»/ ”e+ ra+bT”c+ +a4~b4:eav %H%b—icH
  51. 51. 52 q M P M 2P %f<aj%b%+£c / # 4/ %kaj~€bfi%C%/ P +a ”“ M fi%b0Pc%1/ fiéa / # ‘4§bfif%c§y P / /V’ q 2P .1_6./ p 2P M Ta W‘. /0 b ,1 / ‘L / {(3% b7|/
  52. 52. Vi DU THAM KHAO A De bz‘-Ji: TI'nh gia' tri néi luc trén céc mef1tcét(ca'ch nhau 1 m) vé vé biéu d6 néi | L_rc cho dém dét trén nén Winkler nhw so d6 cho trén hinh 4.1, cho biét: q=8OKN/ mb=1m M = 800 KNm J = 0,0425 m4 P = 650 KN E =10’ KN/ m2 Hé s6 nén K0 = 6.104 KN/ m3 650 KN 650 KN 100KNm 80 KNlm 0 1. Tinh sin céc tri 56 cén sir dung: EJ = O,0426.107 = 426.103 KNm2 TI'nh céc hé s6 cfla dém trén nén dan héi 4: Kob : 8104.1 4.EJ 4.426.103 m = 4./0,035 = O,433182 m2 = 0,‘l8764 m3 = 0,081285 m = 0,0352 Béng théng 56 ban déu: 53
  53. 53. Dean 1 Dean 2 Doan 3 Tai A (Z = 0) T61‘ B (Z =3m) T61‘ C (Z =7 m) 2. Viét phwomg trinh n<f>i | l_PC cho t| ‘.Png doan: a. Viét phwang trinh m0 men u6n V51 /1140 cét cda dém déi havu han dudéi dang chL'P. ' P Ky K0) M1 = i mZ +i0'Cn1Z+f~,0'])mZ In IT] III P <1 M2 = M1 +E‘Bm(Z—3) 4'? -Cm(z—3) M3 = M2 +MAm(z—7) K. K. Q1: P. Amz +fl. Bmz +i’2°. cmZ m m E m. Q3 = Q3 —4m. M.Dm(Z_7) b. Tinh sa'n céc hé s6 cda ca’c s6 hang trong céc phwang trinh trén: Q2 = Q1 +P. A,, ,(Z_3) + Bm(Z_3) 4 _ E = _ 550 = _1500,531K—° = 610 =13,851.104 m 0.43318 m 0.43318 4 _ E = _ 8° = _ 184.6807 K43 = 1 = 31.975310“ m 0.43318 m2 0.433182 4 — 12 = i2 = -426.3372 K—‘; = ix = 73.815210“ m 0,4331B m 0.43318 4.m. M = 4. O,43318.(— 100) = —173,272 54
  54. 54. c. Thay céc gié tri tinh s§n vac phworng trinh n1_3i / LPG cda dém: M1: -1500,5313.. . + 31,9753.10“. yo. cm. + 73,8152.10“. mom. M2 = - 1500531. 3.. .. + 31,9753.10“. yo. cm. + 73,8152.10“. mom. — (8) — 1500.531.Bm, (z.3; — 426,3372.c. ..z.3. M3 = -1500,5313.. . + 31.9753.10“. yo. Cmz + 73,8152.10“. ¢70.DmZ — — 1500531 . B.. _.z_3. — 426.3372.Cm_(z_3) — 100.Am_(z.7) Q1 = — 650.Amz + 13.851 .10“. y0.B. .z +31 .9753.10“. mom. Q2 = - 650A. .. + 13.851 .10“. y0.B. .. +31.9753.10“. ¢po. cm. — 650.Am(z.3, (a) — 184.6807. B. .(z_3. Q3 = - 650A. .. + 13,851 .10“. yo. BmZ +31,9753.104.(po. Cmz - 650.Am(z.3) — 184.6807. B. ,_(Z.3, — (—173,272). Dm_. Z.7, 3. viét aiéu kién bién: Tai D (Z =11m)ta cé: M3=—800KNm va Q3=0 4. TI'nh sin céc ham CrLP|6p: (Lap béng tinh sén céc ham Crwlép cho céc mét cét cén tinh toén, céch nhau 1 m, vc’7i hé s6 m = 043318) 3 N H 3 N H 0 0 0.43318 0.43267 0.86636 0, 8501 1.29954 1.1764 1.73272 — 0.4702 2.16590 - 2.4770 2.59909 — 5.7919 3.03227 - 10.3342 3.46545 -15.1787 -10.1222 - 2,5426 2.5134 3.89863 — 17.9374 — 17.4354 — 8.4670 0.2436 4.33182 -14.1328 - 24.7268 -17.6567 - 5.2994 4.76501 3.0853 — 27.7519 — 29.2901 — 15.4184 55
  55. 55. Thay diéu kién bién tai D véo phworng trinh ncfai Igyc (a) 61 doan 3, ta cc’): M3 = -1500,531.B, ,,_11+ 31 ,9753.104.yo. c,. ,_1, + 73,8152.104. ¢2o. D,, ._11 - 1500,531.Bm_(11.3) — 426,3372.Cm(11.3) — 1O0.Amm(11_7) = - 800 (b) Q3 = - 65o. Am_, , + 13,851.104.yg. Bm,11 + 31,9753.1o4. ¢o. c,, ,_, , — 650.A, ,,_(11.3) — 184,6807.Bm, (11.3) + 173,272.D, ,,, ,,.7, = 0 Thay gia' tri cfla céc ham s6 Crwlép léy to bang trén vao phwang trinh (b), ta C02 M3: — 1500,531.(— 27,7519) + 31 ,9753.104.y. ,.(— 292901) + + 73,8152.104.(po. (- 15,4184) — 1500,531.(— 10,1222) + (C) — 426,3372.(- 2,5426) — 10o. (—o,4702) = — 800 Q3 = - 650.(3,0853) + 13,851 . yo.104.(-27,7519) + + 31 ,9753.104.(po. (-29,2901) — 650.(-15,1787) — 184,68o7.(-10,1222) + 173,272.o,8229 = o d){ 936,5597.1o4.y. , + 1138,1123.104.q; ,, = 58762.286 384,392.104.yg + 936,5597.104. (p 0 = 98726705 5. Giéi hé phworng trinh: Ttv phuorng trinh thL’r nhét cfla hé phuong trinh (d) 6 trén, ta co: = 58762,286—1138,1123.104 y° 936,5597.104 Thay véo phu1oIngtr‘InhthL'r hai cfla hé (d), ta C6: 04 58762,286—113s,1123.1o4 384,392.1 4 936,5597.10 + 936,5597.1O4.q>o = 9872,6705 241177926 — 467,1152.104 + 9365597104. 60 = 98726705 5 469,4445. 104.80 = —14245.1221 60 = - 30.3446.104Rad = 58762,286 —1138,1123.104-(—3o,3446.1o*4) = 58762,286 + 345355625 9365597104 936,5597.104 V0 56
  56. 56. _ 932973485 _ 936,5597.104 Thay gia' tri yo V3 goo vao, ta cc’) phwang trinh néi luc trong 3 clean cfla dém dé cho nhw sau: M1 = —1500,531.B, ,z + 31,9753.104.99,618.10'4 Cmz + + 73,8152.104.(—30,3446.10‘4). Dmz Q1 = - 650.Amz + 13,851.104.99,618.10'4.Bmz + + 31 ,9753.104.(-30,3446.10‘4). Cmz M2 = — 1500,531. Bmz + 31,9753.104.99,618.104 cm; + + 73,8152.104.(—30,3446.10'4). Dmz— 1500,531.B, ,._(z.3) — 426,3372.C, ,._(z.3) Q2 = - 650.A, ,.z + 13,851.104.99,618.10‘4.B, ,.z + 31 ,9753.1o4.(—30,3446.104). cmz — 650.A. .._(z.3) — 184,6807. B, ,_z.3) M3 = —1500,531.B, ,z + 31,9753.104.99,618.10‘4 Cmz + + 73,8152.1o4.(—30,3446.104). Dmz + 1500,531.B, ,1,(z.3) — 426,3372.C, ,.,1z.3) — 100.A, ,._(z.7) Q; = - 650.Amz + 13,851.104.99,618.10'4.Bmz + + 31 ,9753.1o4(— 30,3446.1o“)c. ..z — - 65O. A., ._(z.3) — 184,6807. Bm_z.3) + 173,272.Dm_1z.7) Ta 06 phwong trinh ncji | L_rc cL’1a toén dém nhw sau: M1 = — 1500,5151. Bmz + 3185.315 Cmz — 2239.893.D, ,.z Q1 = - 650.Amz + 1379,804.Bmz - 970.278.0111; —» yo= 99.61810“‘m M2 = — 1500,5151. Bmz + 3185.315 Cmz — 2239.893.Dmz — 150O,531.Bm(z.3) — 426,3372.Cm(z.3) Q2 = - 650.Amz + 1379,804.Bmz - 970.273.Cmz - 650.Am(z.3) — 184.6807 Bm(z.3) M3 = -1500,531.Bmz + 3185.315 Cmz — 2239.893.Dmz — 1500,531.Bm1z3, - 426,3372.Cm(z.3) — 100.A, ,,1z.1, Q3 = - 650.Amz + 1379,804.Bmz — 970.278.Cmz — 65o. A., ,1z.3, — 184,6807 8,112.3, + 173,272 . Dm(z.7) 57
  57. 57. BANG KET QUA TiNH Ll_J’C CAT TAI cAc MAT cAT YEU cAu 90311 2 mZ -650.A, ,,Z 1379,804.13mz -970,278.C, ,,Z m(Z-3) —650A, ,,Z__1, ~184,6807B, ,,(Z__1, m. (Z-7) 173,272D, ,,Z_1, Q(KN) 0 0 -650 0 0 - - — - - -650 1 0.43318 — 646.165 597,000 — 90,915 — — — — — -140.08 1 2 0,86636 - 589,030 1172971 - 361,817 - - — . - 222.124 3 1.29954 — 343,070 1623,201 — 793,396 — — — — — 486.735 3 129954 — 343,070 1623,201 — 793,396 0 — 650 0 — - -163.265 4 1,73272 305,630 l680,877 — 1311.622 0,43318 — 646,165 — 79,906 — — -51.186 H 5 21659 1610,05 859,618 —1729,035 0.86436 —589,03 —156,997 — — -5.394 6 259909 3764735 — 1543.035 — 1675,185 1.29954 — 343,07 — 217,258 — — -13.1313 7 303227 6717230 — 6313,983 — 547,625 1.73272 305,63 — 224,978 — — -63.726 7 303227 6717230 — 6313,983 — 547,625 1,73272 305,63 — 224,978 0 0 —63.726 8 3,46545 9866,155 . 13966,652 2467029 2,1659 1610,05 — 115,056 04331:; 2,339 436.135 111 9 3.89863 11659.31 —24057.435 8215.344 2.59909 3764,735 206,528 0.86636 18,713 -192.805 10 433182 9186320 — 34118,138 17131908 3.03227 6717,23 845,099 1,29954 62.517 —175.064 11 476501 — 2005,444 — 38292,l83 28419540 3.46545 9866,155 1869,375 173272 142.586 0.029 58
  58. 58. BANG KET QUA T1NH MO MEN TREN CAC MAT CRT YEU CAU Dean Z m Z -1500.531B, ,,z 3|85.315C, ,,z - 2239.893D, ,,z m. (Z-3) - |500,531B, ,,(z.3) - 426,3372C. ,,(z.3) m. (Z-7) - l00A, ,,(z.7, M(KNm) 0 0 0 0 0 - - - - - 0 I 1 0.43318 - 649,235 298.464 - 30.239 - - - ~ - - 381.01 2 0.86636 - 1275.601 1187.804 - 241.908 ~ ~ - ~ - - 329.705 3 1,29954 - 1765.225 2604.632 - 808.153 - - - - - 31.254 3 1.29954 - 1765,225 2604.632 - 808.153 0 0 0 - - 31,254 II 4 1.73272 - 1827.947 4305.909 - 1858.950 0.43318 - 649,235 - 39,948 - — - 70.171 5 2.16659 - 934,831 5676.231 - 3399.934 0.86636 - 1275.601 - 158.981 - - - 93.116 6 2.59909 1678.044 5499.446 - 5162.729 1,29954 - 1765.225 - 348,616 - - - 99.08 7 3,03227 6866.430 1797.792 - 6395.118 1.73272 - 1827.947 - 576.323 - - - 135.166 7 3.03227 6866.430 1797.792 - 6395.118 1.73272 - 1827.947 - 576.323 0 -100 - 235.166 In 8 3.46545 15188,675 -8098.982 - 5629.747 2.16659 - 934,831 - 759.733 0.43218 - 99,41 - 334.028 9 3.89863 26162.358 - 26970.062 - 545.638 2.59909 1678.044 - 736.071 0.86636 - 90.62 - 501.989 10 4.33182 37103.33 - 56242.151 11870.089 3.03227 6866.430 - 240.625 1.29954 - 52,78 - 695.707 11 4,7650] 41642,586 - 93298195 34535566 3.46545 15188,675 1084.005 1,73272 47,02 - 800.343 59
  59. 59. 60 Biéu d6 néi | l.1’CZ 650 KN 650 KN 3m 14m1 3 486.735 Biéu 86 n61 Iuc (dé tham khéo): KET QUA TINH TOAN BANG MAY TINH: M NMC Y0 Tetao 0.43318 23 0.00996 -0.00303 Z Y Tata M O 0.0 0.0099615 -0.0030343 0.0000000 -650.0000000 1.0 0.0071266 -0.0027936 -3808450000 -140.1870775 2.0 0.0051018 -0.0040241 -3297900000 2220708631 3.0 0.0037909 -0.0093300 31.1228000 4867868019 3.0 0.0037909 -0.0093300 31 .1228000 —163.2131981 4.0 0.0026245 —0.0198178 -70.1688000 —51.1123330 5.0 0.0016077 -0.0348689 -932355000 -5.1574182 6.0 0.0008065 -0.0516563 -987306000 -138467519 7.0 00002431 -0.0624000 -1346650000 -636694719 7.0 0.0002431 -0.0624000 -2346650000 -636694719 8.0 0.0001258 -0.0519507 -3336200000 -1356515721 9.0 0.0008060 0.0026625 -5014760000 -1927289675 4m 800 KNm (KN) 800,343 6) (KNm)
  60. 60. 10.0 0.0026692 0.1309270 -6948760000 -1755120164 11.0 0.0061471 0.3602250 —8000000000 00000000 550 KN 650 KN 0 31.12 Biéu 86 Q: Hinh 4.4 61
  61. 61. 62
  62. 62. PHAN II are VA HU’C)’NG DAN GI/3.1 BAI TAP LON CO’ HOC KET CAU
  63. 63. BAI TAP LON so 1 T1NH HE THANH PHANG TTNH DINH BANG so LIEU BAI TAP LON so 1 Sinh vién chc_>n nhtrng 36 liéu trong béng 56 liéu phL‘1 hqp vdi hinh vé ctla minh. I. Xéc dinh néi lL_1’c trong hé ghép tinh dinh: 1.1. Xéc djnh phén ILPC tai céc géi tua. YEU cAu VA THU’ TL_l’ THL_l’C HIEN Kich thudc hinh hoe (m) T81 tr(_)ng hu m L1 L2 L3 N/ m§l(K (KN) P (KNm])/ I 8 12 10 30 80 50 10 8 12 40 100 20 12 10 8 50 120 00 4 8 10 12 20 100 50 ‘ 10 12 8 40 80 50 12 8 10 30 120 20 ' 8 8 10 50 I00 50 10 10 8 20 80 00 12 12 10 40 I20 50 10 12 12 30 100 20 0 1.2. V6 céc biéu d6 néi I010: mé men uén M. |l_. PC cét Q va Iuc doc N. 1.3. Vé céc duzdng énh huxérng: dahRA, dahMB. dahQB vé dahQ, khi | L_rc théng dtrng P = 1 di dcfmg trén hé khi chua cé hé théng mét truyén | L_vc. Dung dah dé kiém tra lai céc trj s6 RA, MB. QB, Q, d511nh duqc béng giéi tich. 1.4. Vé Iai céc duzdng énh huzévngz dahRA. dahMB. dahQB vé dahQ, khi IL_rc théng dtmg P = 1 di dcfang trén he khi cc’) he théng mét truyén |1,vc.
  64. 64. 1.5. Tim vj tri bét | c_ri nhét cL’1a dean téi trong gém 4 | I,J’C tap trung di déng trén hé khi co mét truyén 10:6 86 m6 men uén tai tiét dién K 66 gié trj tuyét 801 ldn nhat ll. Xéc dinh mét trong céc chuyén vi sau cfla hé tinh dinh: Chuyén vj dL'mgte_1i F, Chuyén vi ngang tai H, Chuyén vj géc xoay tai tiét dién R do téc dung déng théyi cfla hai nguyén nhén téi trong vé chuyén vi cL. vc"7ng bL’rc cL’1a gc">i tua (xem hinh vé). Biét: .11 = 23; J, = 31; E = 2.108(KN/ m2); .1 =10'6.L.4(m4); A = 001. L1 (m); 6 = A/ L2_
  65. 65. S0 D0 TINH HE TTNH D! NH P 15,, 2F, a= L1/4;b= L2/4;c= L3/4. -b--j-'b-- ’q F1 1 1 W31 1 4: 1 {”"1'°'1~°%'°. d*)1= B
  66. 66. vi DL_J THAM KHAO x De bai: S6 aé: 4. 5. 3 4 P7 86 thL'r tL_}’ ctla so 66 két céu 5 w 36 liéu V6 kich 111066 hinh hoc (héng 1110 5): L. = 10m; L2 = 12m; L3 = 8m 3 G’ S6 liéu vé téi trong (hang thi: ’ 3): q = 50KN/ m; P =120 KN; M =100 KNm. V1"yi céc s6 liéu dé cho, so d6 tinh cfla két céu dmjyc vé Iai nhw sau (Hinh 1): q=50KN/ m _ P=120KN P=120KN | (2,5)| ( 4m )| e4m ‘ 5m _1(3m_)| 6m +3m+ 4m+2+2+2% Hinh 1.1 Trinh tw tinh toén: 1. Xéc dinh néi | l_. |’C trong hé tinh dinh 1.1 Xa’c djnh céc pha’n Iwc g6i twa: 1? Bait tén céc géi tua va céc nut cfla khung (Hinh 1.1). 1'47‘ Phén tI'ch hé chinh phu: Lép so (10 téng (Hinh 1.2) *3 Lén Iwqt tinh toén tL‘P hé phL_J dén hé Ch1nh1he01hU’1L_P sau: 1 YM = 150 KN 9 Truyén phén Iuc xuéng khung GEM YN = 150 KN 9 Truyén phén | L_rc xuéng dém AB 2.T1’nh dém AB: 2 MA= - YB_8 + P.6 - YN_3 = - YB_8 + 120.6 - 150.3 = 09 Y3 = 33,75 KN
  67. 67. 2MB= YA,8-P2-YN.11=YA.8'120.2-150.11=09YA=236,25KN Kiém tra Iai két qué tinh YA va YB béng phuong trinh ZY = 0 9 Cho ta két qué dL'1ng. 3. TI'nh khung GEM: q=50KN/ m P=120KN M N M=100KN L m YM = 150 YN = 150 P=120KN Hinh 1.2 2 M5 = - XE.6 + q.3. 6,5 + YM_8 = - X5. 8 + 50.3.6,5 + 150.8 = 0 9 XE = 362,5 KN EX = 0 9 X5 = 362,5 KN 2Y=09YG=300 KN Truyén phén | l._J’C XG va YG sang khung chinh CD (Iwu )7 déi chiéu cfla phén | l_PC) 4. TI'nh khung CD: 2Mc= -YD.8-P.2,5+q.5.2+M+YG_8+XG_6=0 9 YD = 609,375 KN EX = 0 9 Xe = 362,5 KN ZY = O 9 Yc= 60,625 KN 1.2. Dung phwang phép mat cét xéc d_inh néi lwc trong hé:
  68. 68. 1.2.1. Vé biéu 66 m6 men M (Hinh 1.3). 4400 9'» vgwo, 2175 . -.‘. ‘§‘: ’:‘: ’:‘: ’o"/ -- L . .0.0IIA.4§: "'1 -. ‘aw Hinh 1.3 1.2.2. Vé biéu (I6 lwc 0511 Q: DL_ra véo céc Iién hé vi phan gifra mt‘) men M Va AM luc cét Q, dung céng thl'J= c: QAB : Q23 1 | L 1 biéu G6 | l,l’C cét Q (Hinh 1.4) dwqc suy tL‘r biéu d6 mé men M. 30 1fl/ "WVE h 300 :0: | Il| |EI>ll| | 1 3. Vé biéu db Iwc doc N: Biéu d6 luc doc N (Hinh 1.5) dwqc suy tL‘r biéu d6 II_. vc 051’: Q béng céch téch céc not Va xét can béng vé | l,J’C.
  69. 69. 475,625 325,625 “ 104,375 309,375 362,5 I 609,375 (9 (KN) 4. Kiém tra cén béng céc nut: S; T; L; G; R cfla khung CD khung GEM. 6 V5 mt‘) men: Nut G khéng cén kiém tra vi cc’) ca’c m6 men ncf)i | L_PC, ngoai | L_PC D509 0- 4050 T 4400 2 M3 = 4350 — 300 - 4050 = 0 5 440° mfgx (D 2 MT = 4400 - 4400 = 0 20,5 565 1500 R L 2M| _=2075+100—2175=O £22100 1 6:75 21 zMR=1500+575—2175=0 2175 75 ové | L_rc: T02 kich thuéyc hinh hc_>c cfla khung ta cé: Sina = 0, 6; Cosa = 0, 8 Y1 c) 104,375 ‘V IL 465 ~, m—--»x -6 X , 362,5 30 104,375 309,375 Hinh 1.6 - Kiém tra nut s: (Hinh 1.6a) 2x = 325,625. 0,8 — 362,5 + 170. 0,6 = 0 zv = 50,525 -120 -170.03 + 325,625. 0,6 = 0 o Kiém tra nut T: (Hinh 1.6b) 2x = 30. 0,6 - 475,625. 0,8 + 104,375. 0,8 + 465. 0,6 = 0
  70. 70. ZY = 465. 0,8 - 30. 0,8 -104,375.0,6 — 475,625. 0,6 = 0 o Kiém tra nut L (Hinh 1.66): EX = 362,5 - 104,375. 0,8 - 465. 0,6 = 0 ZY = 309,375 +104,375.U,6 - 465. 0,8 = 0 - Kiém tra nut G (Hinh 1.71»): EX = 362,5 - 362,5 = O ZY = — 309,375 - 300 + 609,375 = 0 9 Kiém tra téng hqp mét phén cfla khung (Hinh 1.7a): V1 (Z) P=120KN b) 1309 375 1 362,5 4-1’- M=100KNm G 3523 X 4350KNm 2175KNm T 300 60,625KN 309,375KN 609-375 Hinh 1.7 EX = 362,5 - 362,5= 0 ZY = 60,625 + 309,375 -120 — 50. 5 = 0 2MS= 4350 - 120. 2,5 + 50.5.2 +100 - 2175 - 309,375. 8 = 0 1.3. Vé céc du= c‘7ng énh hufmg (dah) RA, MB, QB, Q, _- Khi | l_J’C théng dang P =1 di déng trén hé khi chwa cc’) mét truyén | L_. vc (Hinh 1.8) ta nhan théy céc tiét dién cén vé dah déu thutfnc hé phu cfla CD nén khi P = 1 di déng trén khung chinh CD thi dah sé trung vdi dudng chuén do dc’) ta chi quan tém va vé dah thuéc hé MN V3 AB. 1. Vé céc dahRfl dahMg, dahQBT, dahQBF V3 dahQ. khi | L_1’C théng dtvng P= 1 di d6ng trén he khi chwa cé mét truyén Iuc (Hinh 1.8b, c,d, e,f): P=120KN
  71. 71. 2. Dung dah dé kiém tra lai céc tri 56 RA_ MB, QB Va Q, dé tinh bang giai ti’ch: 1,375 - 6 RA: 50. +120-0,25 : 236,25KN; ME = 0 Q, ,T= 50‘ °'375'5 —1200,75 = — 33,75KN; QBF = 0 0,7 = 50»°‘375'6 - 120 . 0,25 = 86,25KN; of = 50.5 — 120 0,75 : — 33,75KN; So sanh vc'ri két qua tinh theo giai ti’ch cho ta théy két qua tinh theo hai each 1.5 béng nhau. 3. Vé lai céc dahRA, dahMB dahQBT dahQBF, dahQ, Va dahM; _( khi Iuc théng dtvng P = 1 di 50ng trén he khi co mét truyén 10¢ (Hinh 1.9): 5,5m 5.5m 12
  72. 72. 4. Tim Vi trl’ bét Iqi nhét 003 he 4 | L_PC tap trung P1; P2; P3; P4 di dcfmg trén he khi cc’) mét truyén Iuc aé MK cc’) gié trj tuyét 061 Ian nhét. Ta nhan théy dahMK (Hinh 1.10a) gm 4 doan théng —> tinh céc tri s5 tga, ‘ (mg Vdi céc doan théng lén luqt tu= tréi qua phéi: tga1 = - 0,25,‘ tgaz = 0,5,‘ tgas = 0; tga:4 = - 0, 5.
  73. 73. Lén Iuvcjt cho doan tai trong di déng 102 trai qua phai sao cho cac | L_. vc tap trung Ién lwqt dejt V30 céc dinh I, II, III cfla dahMK (theo 5 so (16 trong hinh 1.10b. c.d. e.f). Tim Vi tn’ co dao ham FMK1 061 déu dé xéc djnh 106 P, ,,_ 122'“ . _ 4m _1% 4m%| P, = P2 =120KN; P3 =180KN; P4 = 240KN IIIIIIIKOJIII I 4 /1 __: >—" 1“ Hinh1.10 .1. Tha» Ian 1: Cho P4 dét vao dinh I caa dahMK (so 06 1) + Khi P4 5510 bén tréi dinh I ta co: T FMK1 : (P3 + P4). tga1 = — (130 + 240). 0,25 = -105 < 0 + Khi P465161 bén phéi dinh I ta ct’):
  74. 74. F HJAKJ = F'3.tga1 + P4. tgag = —180.0,25 + 240. 0,5 = 75 > 0 Z Ta nhén théy ciao ham 551 déu nén P4 dat o d1nh I I5 P1.. . TI'nh MK 0»ng voi so (36 1: MK = -180. 0,5 — 240.15 = — 450 KNm .1. T110 lén 2: Cho P3 551 Vac dinh I cua crahMK (so 56 2) + Khi P3 (1511 6 bén tréi dinh I ta 00: T Hm] : (P; + P3). tga1+ P4. {gag = (120 +180).0,25 + 240. 0,5 = 45 > Z + Khi P365116 bén phai dinh I ta (:0: dM F = (P1 + P2). tga1+ (P3 '1' P4). tgag z = — (120 +120). 0,25 + (180 +240). 0,5 =15O > 0 Ta nhan théy dao ham khéng 061 déu nén khéng cho gié 111 MK cue tri. Tiép tuc dich chuyén dean téi trong sang bén phai. -1- Thfy Ian 3: Cho P4 dét vao dinh II cfIa dahMK (30 C16 3) + Khi P465116 bén tréi dinh II ta cc’): dM T [ d K1 = 031* P2)- T9111 1' (P3 ‘I’ P4)- 79052 Z = - (120 +120). 0,25 + (180 + 240). 0,5 =150 > 0 + Khi P4 6231 (3 bén phéi dinh II ta cé: dM F [ d K: | 2 031*‘ P2). tga1+ P3. tga2+ P4. tgag Z = - (120 +120).0,25 +130.0,5 + 240.0 = 30 > 0 Ta nhan théy dao ham khéng d6i déu nén khéng cho gié trj MK cuvc tri. Tiép tuc dich chuyén dean tai trong sang bén phai. .1. T110 lén 4: Cho P3551 vao dinh ll caa dahMK (so 56 4)
  75. 75. + Khi P3 dét c’v bén tréi dinh II ta cé: dM T [ d K} = P1- t90t1+ (P2 + P3)-1902+ P4- T9063 Z = -1200,25 + (120 +180). 0,5 + 240.0 =120 > 0 + Khi P3 dét 6 bén phéi (Sinh II ta ct’): dMK dz F : | = P1. tgt/ Z1+ P2. tg(l2+ P3. {gag + P4. ITQDK4 = —120.025 +120.0,5 + 180.0 — 240. 0,5 = — 90 < 0 Ta nhan théy dao ham d6i déu nén P3 dét :3 dinh II I3 Pm. TI'nh Mk (mg vc'vi so d6 4 ta cc’): Mk = -120. 1,25 -120. 1 +180. 1 + 24o.1=15o KNm 4. Tha» Ian 5: Cho P3 dét véo dinh III caa dahMK (sci d6 5) + Khi P3 dét G bén tréi dinh III ta cé: dM T [ d K] = (P1 + P2). tgag+ P3.tga3+ P4.tga4 z = (120 +120).0,5 + 180.0 — 240. 0.5 = 0 + Khi P3 dét 6 bén phéi dinh | || ta C6: dM T [ d K: | = P1. tga2+ P2. tga3+ P3. IQOI4 Z =120.0,5 + 120.0 +180.0,5 = — 30 < 0 Ta nhafm théy dao ham déi tu 0 sang dwcrng nén P3 dét 6 dinh III Ia P1.. . TI'nh Mk L'1=ng vc'yi so (36 5 ta cé: Mk=0+120.1+180.1—240.1=60KNm Néu djch chuyén tiép, dean téi trong sé ra ngoai dahMk, qué trinh thlfr cc’) thé dlimg Iai ducyc. So sénh hai tri s6: MKTTTTT = - 450 KNm MKTTTTX = 150 KNm Ta cé thé két Iuén: Vi tn’ bét Icyi nhét cfla hé | L_rc tzfap trung di dcfmg trén hef: khi cc’) mét truyén Iuc dé mé men uén tai tiét dién K cé gié tri tuyét déi Ion nhét Ié vi tri dét téi theo so <16 1. U’ng viii so 66 nay ta cc’):
  76. 76. max | MK| = 450 KNm. 2. Tinh chuyén vi trong hé tinh dinh Theo yéu céu cL’ia dé bai ta phéi xéc dinh chuyén V1 géc xoay tai tiét dién R do hai nguyén nhén Ia téi trgng véi géi tua C dich chuyén sang phéi mcf>t doan Ia A. Vdi: J1= 24; J2 = 3.1; E = 2. 108 (KN/ m2); J = 10'“. L, ‘ (m4) = 10"? 10‘ = 1o‘2(m“) A = 0,01. L1 (m) = 0,01. 10 = 0,1 (m). 2.1. Lép trang théi phu “k”: 1. Bat mcf>t mc‘) nen tap trung Mk = 1 véo tiét dién R cén xéc dinh chuyén vi géc xoay. 2. Tinh hé 61 trang théi "k": Ta cé nhén xét Mk = 1 duorc flat véo hé khung GEM nén né chi énh huéing dén néi Iuc cfia khung GEM vé khung chinh CD cfia né ChLTJ’ khéng énh huémg dén néi | L_J’C trong céc hé phu MNAB cfla né, vi véy khi tinh he 61 trang théi “k” ta chi cén quan tam dén néi | L_rc E7 phén khung CDGEM. + Xéc dinh céc phén | L_PCI XE = X0 = Y0 = YD = (chiéu cfia phén | u'c 1. 1 6 ’ 8 xem hinh 10). + Vé biéu d6 (Mk ): (Hinh 1.11). 2.2. Tinh hé 5» trang ma; "p ". - Dung két qué as tinh (7 phén trén, dé dé theo dc'>i trong qué trinh nhén biéu d6 ta vé Iai phén biéu dc‘) (MP) trong khung CDGEM (Hinh 1.11). 2.3. Dang céng thwc Mécxoen-Mo tinh chuyén v_i cén tim: 1. TI'nh chuyén vj géc xoay tai R do téi trong géy ra: (Mp): Van dung céng thtrc nhén biéu G6 tinh chuyén vi géc xoay tai nut R do téi trong géy ra vc'7i luu )7 trong hé dém khung cc’) thé b6 qua énh hu’c'1ng cfia | l,l’C cét vé | L_PC doc.
  77. 77. — 1 4350~12 2 ‘PR(P)= (MP)-(Mk): {T A — -2] 3EJ 2 3 + T 4050.5<2*T'5+35°'51,5+T.0,5 +2125-5435 2EJ 2 2 3 3 2 + 1 2075.5-2,5+2325.51+2l0!5 + 1 2175.6‘2‘1+ 1 12175.6-2-1 2EJ 2 2 3 3EJ 2 3 2EJ 2 3 : EiJ(1240O +17718,75 -1- 729,167 + 364,583 + ($484,375 -1- 3875 +1450 + 2175) — %45T95'875 = 0,023 rad. T 2.10“.10“‘.104 2. Tinh chuyén vj géc xoay tai R do géi tuna C djch chuyén sang phéi: (0,313) N 7. . 1 (pR(A) = — ZRk. A'm = — (—EAj = 0,017 (rad) 1 3. Tinh chuyén vi géc xoay tai R do cé hai nguyén nhén déng thc‘7i téc dung: (PR = (| )R(p) + (pR(A) = 0,023 + 0,017 = 0,04 (rad). Két qué mang déu duzcrng cho ta két luéin tiét dién R dL. vc’yi ta’c dung cfia hai nguyén nhén trén sé bi xoay di 1 géc 0,04 (rad) thuzfan chiéu kim déng h6 (cung chiéu vai Mk = 1 as gié thiét).
  78. 78. BAI TAP LON co HOC KET CAU so 2 TTNH KHUNG SIEU TTNH THEO PHUTOTNG PHAP LUC BANG so LIEU CHUNG VE KicH THUO'C VA TAI TRQNG Kich thuéc hinh ‘ Th1'r hoe km) Téu trong S°°ES°°58°° 2 3 4 5 6 7 8 9 »_4 O »_«»_. Oh. ) YEU cAu VA THU’ TU’ THU’C HIEN 1. TI'nh hé siéu tinh do tai trc_>ng tac dung: 1.1. Vé ca’c biéu d6 ncji Iirc: M6 men uén Mp, |L_rc cat Qp, |L_rc doc Np trén he siéu tinh da cho. Biét F = 10J/ L12 (m2). 1. Xac dinh bac siéu finh Va chon hé co ban (HCB). 2. Thanh lap hé phtmng trinh chinh téc dang Chfl’. 3. Xac dinh cac hé s6 Va s6 hang tL_P do cfla hé phuctng trinh chinh téc, kiém tra cac két qua da tinh ducjc. 4. Giai hé phucyng trinh chinh tac. 5. Vé biéu d6 m6 men trén hé siéu tinh da cho do iai trong tac dung Mp_ Kiém tra can bang cac nut Va kiém tra theo diéu kién chuyén Vi.
  79. 79. 6. Vé biéu G6 luc cat Qp Va | u’c doc Np trén hé siéu tTnh da cho. 1.2. Xéc dinh chuyén v_i ngang cda diam I hoac géc xoay cifia tiét dién K. Biét: E = 2.108 (KN/ m2); J = 105. L14 (m4). 2. Tinh hé siéu tinh chiu tac dung déng thévi cfla ba nguyén nhan (tai trong, nhiét dé thay déi Va g6i tL_ra dari ch6). 2.1. Viét hé phworng trinh chinh téc dang sé. 2.2. Trinh bay: 1. Cach Vé biéu (16 m6 men uén Mm do 3 nguyén nhan déng thbri tac dung trén hé siéu tTnh 05 cho Va céch kiém tra. 2. Cach tinh cac chuyén Vida néu E1 muc trén. Biét: $7 Nhiét dé thay déi trong thanh xién: + 0 thév trén I5 7,, = +36” + Cr tho dwéri is Td = +2a°. 3 Thanh xién cc’) chiéu cao tiét dién h = 0,1 m. E Hé sé dan n0 dai Vi nhiét do a = 105. @ Chuyén Vi géi tua: + Géi D dich chuyén sang phai mét doan A1 = 0,001. L, (m). + Géi H bi Iun xuc”3ng mcf)t doan A2 = 0,001. L2 (m).
  80. 80. S0 D0 T1NH KHUNG SIEU TTNH @ ®
  81. 81. Vi Dl_. l THAM KHAo De bai: s6 Be: 10.5.8 10 w= S6 thu» tu cfia so 06 két céu 5 rr se lieu ve kich thuoc hinh hoc (hang mu 5): L1= 10 m; L2: 12 m. 8 P“ S6 lieu ve tai trong (hang thu 8): q = 20 KN/ m; P = 80 KN; M =100 KNm. Voi cac $6 lieu Ga cho, so dc‘) tinh ctia két cau duoc Vé Iai nhu sau: (Hinh 2.1). T em M=100 KNm q=2OKN/ m i P=80KN .1. Trinh tw tinh toan: gm 10m Hinh 2.1 1. Tinh he siéu tinh chiu tac dung cfla tai trqng 1.1. Ve cac bieu d6 net‘ Iirc: me men uén Mp, luc cat Qp Va luc doc Np_ 1. Xéc dinh bacsiéu tinh: n = 3V- K= 3.2 -3 = 3. 2. Chen he co ban (HCB): La he tinh dinh (Hinh 2.2a) duoc suy 10 he siéu tinh 55 cho bang cach loai b6 bot 3 lien két thua (2 lien két tai A; 1 lien két ngan can chuyen Vi ngang tai D), sau dé them Vao D Va A ba an | L_. vc X1; X2; X3.
  82. 82. 3. Lap he phwong trinh chinh tac dang chfrz 511X1"’ 512 X2 T’ 513 X3 ‘T’ A1p = 0 521X1+ 522 X2 T’ 523 X3 T A2): = 0 531X1+ 532 X2 T’ 533 X3 + A3p : 0 4. Xac dinh cac he sf’) 8km Va cac sf’) hang tu’ do Akp cga 11¢ phuJQ'ng mm; -1- Vé cac bieu dc’) mc“> men don Vi: @, @va @3013» luotcac an luc X1 = 1 (Hinh 2.2b)_ X2 = 1 (Hinh 2.2c) va X3 = 1 (Hinh 2.2d) tae dung tren HCB. .1. Vé 1,1151) 55 mo men do tai trcpng tac dung tren HCB (Hinh 2.2e).
  83. 83. -1- Dung céng thL’: =c Mécxoen- Mo vé phép nhén biéu <36 dé tinh céc hé s6 vé ca’c s6 hang tu do cfla hé phwong trinh chinh téc: _ _ 1 1212 2 1 12-10 2 811—@><@—E[T~§+12]+E[T>§~12]+ 1 18-18 2 1464 +— j~—~18 = T 3EJ 2 3 EJ 1 18 6 430 W521‘@*@"@1%““°1“E _ _; _ V 110-102 _17o0_56e,67 522'@X@_ 3EJ(1O1210)+2EJ[ 2 31O]_3EJ_ EJ 1 12.12 1 10.10 540 523=532=@X@= _3EJ[ 2 '101’2EJ[ 2 ‘121=’EJ 5ss= @*@= 1 _12-122‘ 112.122 1488 12+ 1 (12.1o‘12)= — 2EJ EJ 12+ 3 EJ 2 3 3EJ 2 813=631=@x@= %[1%(e+%-12H=2ELJ° A, p=*@=
  84. 84. :1_10O-1O.2.12_2_200_10.6+i960-12‘ 6+2” :2088O 2EJ 3 2 3 EJ 2 3 3EJ 1 960-12 19200 A = . @= ,_. _.10=, _ 2" X 3EJ 2 EJ _ _ 1 96042 1 7680 A3v'*@'fiT‘5"2=fi -1- Kiém tra két qua tinh céc he 56 5km cfla hé phuong trinh chinh téc: > Vé biéu 616 m6 men don vi téng céng (Hinh 2.3): @= @@@ Hinh 2.3 > Kiém tra cac he 36 thucf>c hang mu» nhét cfla hé phucmg trinh: . . 2 . _112.122 2a+m=311+a12+o13=@x@—E[T~3.12j+ 1 [1210212] 1 [66 2 6+(6+218_12_8H:1224 + + . . 2E. J 2 3 3EJ 2 3 EJ > Kiém tra cac hé $6 thucf>c hang thfy hai cfla hé phwcrng trinh: 282m: 521+ 522 "' 523 = @X@= 1 ——.1o.12.s—i T 2+1.10 : —453'33 3EJ 2EJ 2 3 EJ > Kiém tra cac he s6 thucfac hang mu» ba cfla hé phucrng trinh: 263m = 531''' 532 "' 533 = @X@=
  85. 85. 1 .12+12_8+ 1 ‘2+12‘10.12+ 1112-12‘2.12=11sa 3EJ 2 2EJ 2 EJ 2 3 EJ r Kiém tra tat ca cac he s6 cfla he phwomg trinh chinh tac: 1.12.12_2_ + 1 ‘12.1o_2.12+ 1 3-s_2‘6+8_12_8 EJ232EJ233EJ23 + 1 2.10.2+12+10-10 2+2“) +1.12«12‘2‘12=1958,75 2EJ 2 2 3 EJ 2 3 EJ > Kiém tra cac s6 hang tu do cfia he phwong trinh chinh tac: 2A«»= Aw+A2»+As»= x@= 1 [ 100-10 2 2 J 1 96012 9360 +_. 3:: — —: ._.12——-200.10-6 2EJ 2 3 3 2 EJ 5.Giai hé phucrng trinh chinh tac: 1464X1 - 480X2 + 240X3 + 20880 = 0 X1 = — 2,225 KN —430x1 + 566,67X2 — 54ox3 — 19200 = 0 : > X2 = 41,914 KN 240X1 — 540X2 +1488X3 + 7680 = 0 X3 = 10,4 KN 6. Vé biéu dc‘) m6 men trén hé siéu tTnh da cho do tai trong téc dung: @= @X1+@X2+@Xs+ Khi céng céc biéu d6 ta can phai cc’) sL_P théng nhét chung vé déu cL'1a cac ncf>i Iuc gifra cac biéu <10. Ba (16 nham Ian ta cc’) thé tL_r qui um M > 0 khi cang duoi véri cac thanh ngang; cang phai véri cac thanh dang va ngwqc lai. C)’ day chung t6i lap bang t1’nh céc mt‘) men tai céc dau thanh vdi qui u’é1c: ngu’<‘; vi quan sat dlfrng 6 trong khung HCBA; M > 0 cang vé phI’a ngu’c‘Jri quan sat; M < 0 cang vé phia ngworc Iai. Déiu M1.X1 M . X M3.X3 o thanh 2 2 Mp MP(KNm) MED 26,7 0 0 0 26,7 MEK 26,7 0 0 100 126,7
  86. 86. Mex 13,35 0 0 0 13,35 MCH 13,35 419,14 124,8 ’ 0 307,69 MHC 40,05 419,14 0 960 ‘ —5o0,31 Ma; 0 419,14 124,8 ' 0 294,34 Mac 0 0 124,8 ‘ 0 . 124,3 MBA 0 0 124,8 ‘ 0 —124,s * Kiém tra can bang not C vé m6 men: ZMC = 307,69 - 13,35 - 294,34 = O =1 Kiém tra theo diéu kién chuyén vi: 1_20,7.12212 EME =1OO + 26,7 - 126,7 = 0 2MB=124,8-124,8= 0 2+ EJ 2 3 721:1 2 1 [126,7+10‘2_1 2-200-10-6:‘ 3 + 3EJ 2 3 2 1 [113,35-62.6130169-12_8+500,:1~12‘8] 1[124,s.12v2.12 + EJ 2 3 >1 Kiém tra can bang nut E vé m6 men: * Kiém tra can bang n1'1t B V6 mc‘) men: 1
  87. 87. + 1 124,310 2+2“) _294,34.10 2+1_10 = 2EJ 2 3 2 3 i(2705,04 — 3922,03) : — 121794 = — EJ EJ 2.10 .10" .10 (m) = — 0,61 (mm). |A+B1 TI'nh sai s6 theo biéu thL'1’c: 5 2 (°o) 1 2705,04 — 3922,08 1 2705,04 ta cc’) thé coi chuyén vj da 11’nh la bang 0, diéu dc’) chfyng té biéu d6 Mp da vé dung. ta co 5 = = 0,45 < 5% la sai 56 trong gic'yi han cho phép 7. Vé biéu d6 Iuic cat Qp Va biéu (16 |1_. vc dc_>c Np: -1- Biéu C16 |1,yc 6310,, (Hinh 2.5) (11100 suy ra 10» biéu d6 Mp dL, ya vao méi Iién A 2 A . . hé vi phan gifra M va Q: Dung céng thL'1’c QAB = Q33 : 1 LM‘ de lap bang t1nh uc cat tai cac 3311 thanh véri sina = 0,6; cosa = 0,8. Dau Q0 i AM Qp thanh (m) AB - (KN) QDE = 2,2 0 26 7 - 0 /12 QED 2 ( ’ ) 25 QCK : - 0 - 13 35 - 0 /6 QKC ‘ * ) 2,225 20.10.0,8 / 67, QEK 0 (2 ) 126,7/10 33 Q ’ 126 /10 ’ “E 0 (20.10.0,8)/2 ’7 92,67 QC” : 0 (500,81+ 67, QHC 2 307,69)/12 375
  88. 88. OCH = 0 (500,81 + 67, QHC 2 307,69)/12 375 QCB = 0 - (294,34 + — QBC 0 124,8)/10 41,914 QBA = 10. 0 124 8/12 QM, 2 ’ 4 K 2,25 NEK E 3 ------------ -->X I I ' X K ‘o‘ 0 ‘”"9‘§'6'7""‘ 1 2,25 % ' 0! x 2,25 1' X 1 -. _- NKE % “Kc _. ( KN ) Hinh 2.5 -1- Biéu <36 luc doc Np (Hinh 2.6) ducyc suy tu’ biéu db | L_rc cét Qp béng céch xét cén béng hinh chiéu céc ngoai Iuc vé néi Iwc tai céc nut E, K, C vé B véi sina = 0,6; 00501 = 0,8. 41,914 -1- Xet can bang nut E: (_"NE2c_ / B 2x = NEK. 0,8 + 87,33. 0,8 — 2,25 = o X 4 | E0 N = — 47,885 KN ‘"525 fir EK NBAl1o'4 zu = NED. 0,8 + 87,33 — 2,25. 0,6 = o 2125 ‘L 41,914 ; y NED = — 82,475 KN 67 3: | —4/T9-yé-—-J‘ -1- Xét cén béng nut K: ' <1- NCH
  89. 89. XX = NKE. 0,8 - 92,67. 0,6 - 2,25 = 0 4 NKE = 72,315 KN EU = NKC. 0,8 + 2,25. 0,6 + 92,67 = 0 —> NKC = -117,525 KN -1. Xét can béng nut B: :x = NBB— 80 + 10,4 = 0 4 NBC = 69,6 KN EY = NBA + 41,914 = 0 —> NBA = — 41,914 KN -1- Xét can béng nut C: EX = 69,6 - 67,375 — 2,25 = 0 EY = NBH + 117,525 -41,914 = 0 —» NBB = - 75,811 KN
  90. 90. .. III | IIlllII1.a1lIll| |I| ||II. . 1* ( KN) Hinh 2.6 1.2. Tinh chuyén v, i géc xoay tai K: ,_ 8 2 -5 4 4 -6 4 -2 4 Vc11E=2.10 KN/ m ; J=10 . L, (m )=10 .10 =10 (m) 1. Laptr3_1ngthéiphL_1“k”trén hé tTnh dinh duqc suy ra ttr hé siéu tTnh dé cho béng céch loai b6 3 Iién két thfra. 0’ day chung téi chon giéng HCB (Hinh 2.2). 2. Vé biéu (16 m6 men 67 trang théi phu “k” (Hinh 2.7). 3. Dung c8ng thL'rc nhén 8180 88 tinh «PB: 1 13,35 »6 ¢PK(P): ‘ 307,69»12 500,81»12 1 + — ~1 3EJ J 2 2 2 1 = E4J(13,35 + 615,38 —10o1,82)= - 37289 EJ : $57289 : — 0,00019 rad -2.103 .1016 .104 Vay tiét dién K b1 xoay mét géc 0,00019 rad thuén chiéu kim déng hi‘).
  91. 91. 2. TI'nh hé siéu tinh chiu téc dung déng thévi cua ba nguyén nhén (téi trgng, Sl_. |' thay d6i nhiét 61} vs‘: g6i twa dbai ch5): 2.1. Viét 11,5 phuvang trinh chinh téc dang s6: 1. Chan hé co bén giéng nhu’ trén (Hinh 2.8). 2. Lap hé phwong trinh chinh téc dang chfr: 511 X1 + 512 X2 + 513 X3 + A1p + A11 + Am = A1 521 X1 + 522 X2 + 523 X3 + A2p+ A21 + A2A = 0 531 X1 "' 532 X2 + 533 X3 + Asp "' A31 +A3A = 0 3. Xa’c djnh céc s6 hang my do AA, Va‘ AAA; AKA = ' ' Aim | A h Vdi A1= 0,001.L1= 0,001. 10 = 0,01 (m) Ak, =20-1, ~Qfi+Z : 0t~ ‘A-10W] A2 = 0,O01.L2 = 0,001. 12 = 0,012 (m) tc = (36 + 28)/2 = 320; 1A, | = 3°; 01 =1o'5; h = 0,1 (m) s0» dung céc két qué tinh nofai Iuc don vi cua thanh xién EK dé tinh E1 trén ta
  92. 92. cé: E: -1,35KN; E= N__-, =0 10’5~8 AB =10-5-32(—1,35-10)+ (%= 4368-10‘5 =0,044 A21:A31=0 Phén | l._J’C dulng tai lién két H duqc ghi trong céc (Hinh 2.3), (Hinh 2.4) vé (Hinh 2.5). A, A = — 2,25. AB = — 2,25. 0,012 = - 0,027 A2A= — 1. AB: —0,012 A3A = 0 4. Lap hé phuong trinh chinh téc dang sf}: 1464 X, - 480 X2 + 240 X3 + 20880 + (0,044 - 0,027)EJ = 0,01EJ -480 X, + 566,67 X2 — 540 X3 — 19200 + 0 - 0,012EJ = 0 240X1- 540X2+1488X3+7680+ 0+0 =0 1464 x, — 480 X2 + 240 X3 + 34880 = 0 :9 480 x, + 566,67 X2 — 540 X3 — 43200 = 0 (**) 240 X, — 540 X2 + 1488 XB + 7680 = 0 2. 2. Trinh bay céch tinh: 1. M6 men uén M55 trén hé siéu tTnh dé cho do téc dung déng thfri cua 3 nguyén nhén: téi trgng, su’ thay d6i nhiét dtf) trong thanh xién EK via su’ dc‘7i ch5 cua géi 11,131 D va H. 9 Giéi hé phuong trinh (**) ta dwoc céc nghiém X1, X2, X3 4 @= @x, +@x2+@x3+ o Kiém tra theo diéu kién chuyén v1 ta dung biéu thfrc: 3 3 @@ = — 2A“ — ZAKA +vé phéi cua hé phwong trinh k:1 k:1 = — A, ,— A, A — A2, + A, = 0,005 4 Néu két qué nhén biéu 06 théa man biéu thL'rc trén 1111 0180 08 Mcc ducyc xem I3 dung.
  93. 93. 2. Céch tinh chuyén vi géc xoay tai K: 9 Lap trang théi phL_1 "k" nhu trén (Hinh 2.3 9 TI'nh hé finh dinh dé chc_>n 6 trang théi "k": Xéc djnh phén |1_1Ic tai D, H; vé biéu 131‘) m6 men vé xéc djnh luc doc trong thanh xién EK. 0 0 ‘ ‘PK(P:10;A) = X + An +AkA 0 day véri trang tha'i “k” (35 chon dé tinh géc xoay tai K ta ct’) phén Iuc tai géi tua D; phén |1_yc dung tai H; V3 néi |1_yc m6 men, |1_yc doc trong thanh xién EK béng 0 nen AND = ABAO = O.
  94. 94. BAI TAP LCVN CO’ HOC KET CAU SO 3 T1NH KHUNG SIEU TTNH THEO PHUONG PHAP CHUYEN VI VA PHU’O’NG PHAP PHAN PHOI MO MEN. BANG so LIEU CHUNG VE KicH THUOC VA TAI TRQNG Kich 111m’-8 hinh hoe (m) T31 trqng YEU CAU VA THU’ TU’ THL_PC HIEN: 1. Vé biéu G6 mt“) men u6n Mp cua khung siéu tTnh dé cho theo phwong pha’p chuyén V1. 2. Vé biéu (16 mo men uén Mp cua khung siéu tTnh dé cho theo phwong pha’p phén phéi mt“) men. 3. Vé 0180 88 10c cét QB, 10c doc Np trén he siéu tTnh 05 cho. 4. Xéc dinh chuyén vj ngang cua diém I hoéc géc xoay cua tiét dién K.
  95. 95. 13181: E = 2.108 (KN/ m2); J = 106. 1., “ (m4). Chtfl )2.‘ 1. Vé xong biéu 115 m6 men uén Mp cén Kiém tra can béng céc nut va cén béng hinh chiéu cho céo biéu df“) | L_.1’C cét Qp, |l_. PC doc Np. 2. Cén so sénh két qué tinh ncf>i | L_J’C gifra hai phwcvng phép. 3. can hiéu r6 y nghTa cua céng 1h0»c11nh chuyén v1 va céch lap trang 11151 phu "k“ dé t1'nh chuyén vi. so no T1NH KHUNG s1Eu TTNH (B31 tap Iévn s8 3)
  96. 96. vi Dl_. l THAM KHAo 4 s6 08: 10.7.5 10 <9: S6 thu= tu» cua so 06 két céu 7 =8 $61180 vé kich thuoc hinh hc_>c(hé1ng 1110 7): L1: 8 m; L2: 8 m. 5 =3’ S6 Iiéu V6 téi trong (hang 1110» 5): q = 4OKN/ m; P = 80 KN. Vc'>’i 0310 $6 lieu dé cho, so d6 tinh cua két céu dwcyc vé lai nhw sau (Hinh 3.1): P — 80 KN Hinh 3.1 }‘_2n1a| (B Sm B)‘ Trinh tw tinh toén: 1. Dung phwong phép chuyén vi vé biéu (16 m6 men u6n MB do téi trgng téc dung trén hé siéu tinh dé cho:
  97. 97. 1.1. Xéc djnh s6 éin s6: n= ng+n, =1+1=2. 1.2. Lép he‘ 00 ba’n (HCB): Thém V60 nut B m6t lién két m6 men Va m<f)t Iién két | l,J’C, twang umg V61 chung I3 0130 6n chuyén vi Z, V3 Z2 (Hinh 3.2). 1.3. Lép he‘ phwang trinh chinh téc: [r11Z1+ r12Z2 + Rm: 0 r21Z1+ '22 22 + R2p = 0 1.4. Dung béng tra Vé ca’c biéu d6 don 1/,1’: @ @3 0 lén luzqt céc én Z, = 1 (Hinh 3.3), Z2 = 1 (Hinh 3.4) V31 téi trong (Hinh 3.5) géy ra trén HCB. 1.5. Tinh ca’c hé 36: r11; r12; r22 V31 céc 56 hang tu do R, p; RZB; + Téch nut B 6 céc biéu d6,@ @/3 @ét can béng V6 m6 men d6 xéc dinh céc phén Iuc m6 men F11. F12 Vél R1,, trong lien két m6 men dunyc them V310 B trén HCB. + Xét cén béng vé 10c cfla thanh BC 6 biéu (36 @a xéc dinh céc phén luc théng F22 V3 R2,, trong Iién két Iuc ducyc thém V30 B trén HCB.
  98. 98. Hinh 3.5 Hinh 3.4 01°47“ 0,6EJ I'11=1,1EJ 0,0945% r11 0,5EJ R1 p r12 = r21 = - 0,094EJ 160 00 V22 R1,, = — 240 1: R2,) 0,024EJ I'22=0,03EJ 0,006EJ R2,: 25 c 25
  99. 99. 1.6. Giéi h_é phuvng trinh chinh téc: 1,1EJ Z1- 0,094EJ Z2 - 240 = O {Z1= 200,712/EJ | :C> - 0,094EJ Z1 + 0,03EJ Z2 + 25 = 0 Z2 = - 204,436/EJ 1.7. Vé biéu 66 m6 men trén hé siéu tinh dé cho (Hinh 3. 6): = @Z«+@z2+ Khi céng céc biéu d6 ta cén phéi cc’) Sl_. P théng nhét chung vé déu cfla céc ncf>i luc gifra céc biéu d6. Dé dc")! nhém | §n ta cé thé tL_. I’ qui woo M > 0 khi céng dwfri vc'yi céc thanh ngang; céng phéi v<’7i céc thanh dL’rng vé ngwoc lai. C)’ day chL’mg téi lap béng tinh m6 men tai céc déu thanh vc'yi qui Lrdc: ngLrc‘7i quan sét dL'rng fr trong khung khi dc’) M > 0 céng vé phia ngudi quan sét Va M < 0 céng vé phI’a ngwoc lai. 951] M1. Z1 M2- 22 Mp Mp thanh MBK 0 0 - 160 - 160 M _ _ _ BA 100,356 19,217 0 119,573 M _ _ BC 120,427 0 400 279573 MAB 50,178 19,217 0 69,395 M _ _ _ DC 0 9,608 120 129,608 119 573 Kiém tra can béng nL’1t B vé mc‘) men: EMS = 279,573 -119,573 -160 = O
  100. 100. 2. Dung phwomg phép phan phéi m6 men (PPMM) vé biéu d5 m6 men u6n MP; Hé siéu tTnh dé Cho (:6 mcf>t nut cL'rng B cc’) chuyén vi théng, trinh tL_r tinh nhu’ sau: 2.1. Xéc djnh s6 an S6: n = nt= 1. 2. 2. Lép h_é ca bén (HCB) (Hinh 3. 7). 2.3. hé phunng trinh chinh téc: @ r11 Z1 + R”) = 0 2.4. biéu 06 mm v_i @ do 2, = 1 géy ra trén HCB (Hinh 3.8). Khéc vc’ri phwong phép chuyén vi 61 day mé men tai not B (151 cén béng sau khi thL, rc hién so as PPMM (Hinh 3.8). -1-Xéc dinh dé cL'yng don vi qui um: pm: pBA = iBA = O,125EJ; 3- 3 2EJ pm = Z|BC = Z»W= O,15EJ -1- Xéc dinh céc hé sé phén phéi m6 men pkji 0125EJ 015EJ : Z : 0,455; 2 ‘Z 2 “BA 0,125EJ + 0,15EJ 1'1 3° 0,125EJ + 0,15EJ 0,545 4- Kiém tra céc hé s6 PPMM: 2 113,- = MBA *' NBC = 0,455 + 0,545 =1
  101. 101. :-1- Lap so d6 PPMM dé vé (Hinh 3.8:: (7 day chung téi sir dung két qué biéu tra bén do Z2 =1 (7 trén, dc’) la (Hinh 3.4) trco 051 J tI'Z1=1 0 phuong +0.05lEJ F4 phép c -0,051EJ fF0,094EJ} 054 + ‘ ------- -‘ -n ’ —0,043EJ Q : «= ' I : 1:1 I 1 ; C I , -_€’—_°_2_%_h_J A + 40,0940 +0,072EJ .1. Lap so dé PPMM dé vé (Hinh 3.9) Str dung két qué biéu df“) tra béng do téi trong téc dung E7 trén, do la Mpo (Hinh 3. 5) trong phén tinh theo phwong phép chuyén vi. >269,2 .1303 r----: + ;400;
  102. 102. 2.5. Xa’c djnh ca’c hé s6 cda phwang trinh chinh téc: B 711 P R B 11) T11: 0,021EJ R19 = 4525 0,015EJ _, 20,475 C C 0,O06EJ __ 25 2.6. Gia’i phuong trinh chinh téc: 0,02‘1EJ. Z1 + 4,525 = 0 =5 Z1 = - 215,476/EJ 2. 7. Vé biéu d6 mé men trén hé siéu tinh: (Hinh 3.10). —@3z«+ Lap béng tinh céc m6 men tai céc déu thanh vdi qui u»t’yc: ngwdi quan sét dtrng (3 trong khung; M > 0 céng vé phia ngwfyi quan sét; M < 0 céng vé phfa ngucyc Iai. Sau do so sénh két qué tinh Mp gifya phtycvng phép chuyén vj va phuong phép phén phéi mé men (sai $6 két qué tinh gifra 2 phwang phép dwcyc ghi trong béng). Diiu 1v| '1_ Z1 0 M Sai s5 gifia hai thanh Mp p PP MM 0 — 160 — 150 0% MB‘ 10 989- 109 2 _ 120 189 05% M” 10 989- 269 2 _ 280 189 032% MAB 15,514 54,6 70,114 0,1% MD“ 10,127‘ "120 130,127 0‘4%
  103. 103. 3. Vé biéu (16 | |_. Pc cét Qp vé biéu 66 | L_rc doc Np: 3.1 Biéu d5I1rc cét Qp (Hinh 3.11) dwqc suy ra 10’ biéu (16 Mp. 0’ day chL’1ng téi dung két qué tI’nh MP theo phwomg phép chuyén vi (Hinh 3.6) dé tinh | L_PC cét tai céc déu thanh dua véo méi Iién he vi phén gifra M vé Q: 3. 2. Dung céng thLi’c: 1 4M1 L _ 0 Q1121‘ QAB 0-“ Két qué tinh luc cét tai céc déu thanh dwqc ghi trong béng sau: Déu 0 1 fll thanh L(m) QAB - L QP : Q QBK 2 0 - (160 - 0)/2 - 80 KB _ Q” 8 0 ' ' — Q31 (69,395+119,573)/8 23,621
  104. 104. Q98 10 (40.10.0,8)/2 279,573/10 187,957 QC“ 10 (40.10.0,8)/2 270573/10 132,043 QCE _ 4 0 95,196/4 23,799 ‘ QEC QED 4 O - 029,608 1* -56 201 = QDE 95,196)/4 ’ 3.3. Biéu (76 I111: doc Np (Hinh 3.12) duqc suy ttv biéu 66 Iuc cét Q1, béng céch xét cén béng hinh chiéu céc ncf>i |1_. vc vé ngoai | L_PC tai céc n1JtB vé C vévi sina = 0,6; cosa = 0,8. Hinh 3.11 -1- Xét cén béng not B: NBK = 0 0 13 187957 2 x = NBC. 0,8 — 187,957. 0,6 + 23,62 = 0 “ a ,1 " "" 3? NBC = 111,443 KN 23.62 , NBC 2 U = NBA. 0,8 + 187,957 + 80. 0,8 - 23,62. 0,6 = 0 U , ‘ NBA —> NBA = - 297,183 KN -1- Xét can béng nL’1t C: N03
  105. 105. 2 x = NOB. 0,8 + 132,043. 0,6 + 23,62 = 0 NOB= -123,557 KN 2 u = NOB. 0,8 + 23,62. 0,6 + 132,043 = 0 NOB = — 182,769 KN 4. TI'nh chuyen vi géc xoay tai K: V61 E = 2. 10” KN/ m2; MK=1 J =10'0. L14(m4) = 4096. 100 (m4) 4.1. Lep tr-fang tha'i phu “k”tren he finh dinh duqc suy ra 10 he siéu finh dé cho béng céch loai b6 lien két khc'rp tai C (Hinh 3.13). 4.2. Vé biéu L16 m6 men 6* trang théi phu "k” (Hinh 3.14}. 279,573 Hinh 3.13 Hinh 3.14 Hinh 3.6 4.3. Dung céng thuvc nhén bieu (76 tinh 112,1:
  106. 106. _ 1 16032 169,395-8 19,5736 0K‘”’_x@=2EJ1 2 '1j+EJ1 2 0‘ 2 '1] (pm) = - 0,00015 rad Vey tiét dien K se 1:1 xoay met géc 0,00015 rad thuen chieu kim deng 1.6.
  107. 107. PhL_1 | L_1c: Méu Trang bia TRUONG DAI HQC THUY LOI HA NOI BO MON sfrc BEN - CO KET CAU BAI TAP LO’N . ... . . . SO TiNH HE TTNH D1NH s6aé:4.5.3 H0 vé tén sinh vien : L<'7p : _ Ngwfii hwfyng dén : Hé Nei -2006
  108. 108. ML_JC LL_JC L6-i giévi thieu Céc yeu céu chung Phén I: sue BEN VAT LIEU Béi tap | c')’n S6 1: Dec trwng hinh hgc cua hinh phéng Béng s6 lieu Vi du tham khéo Béi tep lc’7n $6 2: Tinh dém thép Béng $6 lieu Vi du tham khéo Béi tap léin s6 3: Tinh cet chiu Iu’c phfrc tap Béng s6 lieu Vi du tham Khéo Bel tep | <')’n s6 4: Tinh dém trén nen den hei Béng s6 lieu Vi du tham khéo Phén ll: Trang 11 18 23 37 41 49 53
  109. 109. co HQC KET CAU Bel tep lén s61: Tinh he thanh phéng tinh dinh Béng $6 lieu 65 Vi du tham khéo 68 Bai tep ldn $6 2: Tinh khung siéu tinh theo phuuyng phép Iuic Béng s6 lieu 81 Vi di_i tham khéo 84 Bei rep i6n s6 3: Tinh khung siéu finh theo phwong phép chuyen V1 V6 phwomg phép phén ph6i m6 men Béng $6 lieu 96 Vi du tham Khéo 98 Phl, I | l_. lc 108 Muc | L_Ic 109
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Đề bài và Hướng dẫn giải BTL sức bền vật liệu

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