Giaotrinh co hoc ket cau

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C HOC KT CU I Page 4

Goi kS la ai lng S do ring Pk = 1 gy ra. Tc la S(Pk) = kS .PkVy S(P1, P2,..... Pn) = nn PSPSPS ........ 2211 ++

Chu y:Nguyn ly cng tac dung ch ap dung cho h tuyn tnh vt ly cung nhtuyn tnh hnh hoc.

4. PHN LOAI CNG TRNHI. Phn loai theo s tnh:1. H phng: khi tt ca cac

cu kin cung thuc mt mt phngva tai trong tac dung cung nmtrong mt phng o.

Cac loai h phng:

- Dm (H.6)- Dan (H.7)- Vom (H.8)

- Khung (H.9)

- H lin hp (H.10)

2. H khng gian: khi cac cu kin khng cung nm trong mt mt phng, hoccung nm trong mt mt phng nhng tai trong tac dung ra ngoai mt phng o.

Cac loai h khng gian:

- H dm trc giao (H.11)- Khung khng gian (H.12)- Dan khng gian (H.13)- Ban (H.14)- Vo (H.15)

H.6bH.6a

H.7a H.7b

H.10bH.10a

H.11

H.12 H.13 H.15H.14

H.9aH.9b

H.8a H.8b


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II. Phn loai theo phng phap tnh:1. Da vao s cn thit hay khng phai s dung iu kin ng hoc khi xac

nh toan b cac phan lc va ni lc trong h, ngi ta chia ra hai loai h:a. H tnh nh: la loai h ma ch bng cac iu kin tnh hoc co th xac nh

c toan b ni lc va phan lc trong h. V du cac h trn hnh a t (H.6) n (H.10).b. H siu tnh: la loai h ma ch bng cac iu kin tnh hoc th cha u xacnh toan b cac ni lc va phan lc ma con phai s dung thm iu kin ng hoc vaiu kin vt ly. V du cac h trn hnh b t (H.6) n (H.10).

2. Da vao s cn thit hay khng phai s dung iu kin cn tnh hoc khixac nh bin dang trong h khi h chu chuyn v cng bc, ngi ta chia ra hailoai h:

a. H xac nh ng: la loai hkhi chu chuyn v cng bc, co thxac nh bin dang cua h ch bng caciu kin ng hoc (hnh hoc). V duhcho trn hnh (H.16).

b. H siu ng: la loai h khi chu chuyn v cng bc, nu ch bng cac iukin ng hoc th cha th xac nh c bin dang cua h ma cn phai s dung thmiu kin tnh hoc.V duh cho trn hnh (H.17).

III. Phn loai theo kch thc tng i cua cac cu kin:- Thanh: nu kch thc mt phng kha ln hn hai phng con lai (H 18a).

- Ban: nu kch thc cua hai phng kha ln hn phng con lai (H.18b).- Khi: nu kch thc cua ba phng gn bng nhau (H.18c)

IV. Phn loai theo kha nng thay i hnh dang hnh hoc:- H bin hnh.

- H bin hnh tc thi.- H bt bin hnh.

H.18aH.18b H.18c

D

B C

A'A

H.16

D D

A B

C

DC' D'

H.17


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5. CAC NGUYN NHN GY RA NI LC, BIN DANG VACHUYN V

I. Tai trong: gy ra ni lc, bin dang va chuyn v trong tt ca cac loai h.Phn loai tai trong:- Theo thi gian tac dung: tai trong lu dai (nh trong lng ban thn cng

trnh...) con c goi la tnh tai va tai trong tam thi (nh tai trong do gio, do con ngii lai khi s dung..) con c goi la hoat tai.

- Theo s thay i v tr tac dung: tai trong bt ng va tai trong di ng.- Theo tnh cht tac dung co gy ra lc quan tnh hay khng: tai trong tac dung

tnh va tai trong tac dung ng.Ngoai ra, con phn loai tai trong theo hnh thc tac dung cua tai trong: tai trong

tp trung, tai trong phn b...

II. S thay i nhit : chnh la s thay i nhit tac dung ln cng trnh khilam vic so vi luc ch tao ra no.

i vi h tnh nh, tac nhn nay ch gy ra bin dang va chuyn v, khng gyra ni lc, con i vi h siu tnh th gy ra ng thi ca ba yu t trn.

III. Chuyn v cng bc cua cac gi ta (lun) va do ch tao lp rap khngchnh xac.

i vi h tnh nh, tac nhn nay ch gy ra chuyn v, khng gy ra bin dangva ni lc; con i vi h siu tnh th gy ra ng thi ca ba yu t trn.


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CHNG 1PHN TCH CU TAO HNH HOC CUA H PHNG

1. CAC KHAI NIM

I. H bt bin hnh (BBH): la h khng co s thay i hnh dang hnh hoc ditac dung cua tai trong nu xem cac cu kin cua h la tuyt icng.

V du: Phn tch h hnh ve (H.1.1a)Nu quan nim AB, BC, trai t la tuyt i cng, tc la

lAB, lBC, lCA = const th tam giac ABC la duy nht, nn h a chola h BBH.

- Mt h BBH mt cach ro rt goi chung la ming cng (tm cng)- Cac loai ming cng: (H.1.1b)

- Ky hiu ming cng: (H.1.1c)

* Chu y: Do h BBH co kha nng chu lc tac dung nn no c s dung lam cackt cu xy dng va thc t la chu yu s dung loai h nay.

II. H khng bt bin hnh:

1. H bin hnh (BH): la h co s thay ihnh dang hnh hoc mt lng hu han di tac dungcua tai trong mc du xem cac cu kin cua h latuyt i cng.

V du: H ABCD cho trn hnh (H.1.2a) coth thanh h AB'CD, nn h a cho la h BH.

* Chu y: Do h BH khng co kha nng chu tai trong tac dung nn cackt cu xy dng khng s dung loai h nay.

H BH trn hnh (H.1.2b) cho phep s dung v theo phng ng, taitrong tac dung ln h trang thai cn bng.2. H bin hnh tc thi (BHTT): la h co s thay i hnh dang hnh

hoc mt lng v cung be di tac dung cua tai trong mc du xem cac cukin cua h la tuyt i cng.

V du: H ABC co cu tao nh trn hnh (H.1.3a), khp A co th i xung mtoan v cung bed, nn h a cho la h BHTT.

*Chu y: Cac kt cu xy dng khng s dungh BHTT hay h gn BHTT (la h ma ch cn thay imt lng v cung be hnh dang hnh hoc se tr thanhh BHTT, v du h BA'C trn hnh (H.1.3a) v ni lc

B C

A

H.1.1a

H.1.1bH.1.1c

A D

B

B'

C H.1.2a

H.1.2b

H.1.3aB

C

A

A'd


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trong h gn BHTT rt ln.Tht vy, xet h trn hnh (H.1.3b).

Lc doc trong hai thanh AB va AC la N.

N =asin.2

P- .

Khi a 0, h BAC tin n h gnBHTT.

N = -

)sin.2

(lim0 aa

P .

III. Bc t do: la s cac thng s c lp u xac nh v tr cua mt h so vimt h c nh khac.

Trong h phng, mt chtim co bc t do bng 2 (

H.1.4a); mt ming cng co bct do bng 3 (H.1.4b).

2. CAC LOAI LIN KT VA TNH CHT CUA LIN KTI. Lin kt n gian: la lin kt ni hai ming cng vi nhau.Cac loai lin kt n gian

1. Lin kt thanh: (lin kt loai mt)a. Cu tao: Gm mt thanh thng khng chu tai trong cohai khp ly tng hai u. (H.1.5a)

b. Tnh cht cua lin kt:+ V mt ng hoc: lin kt thanh khng cho ming

cng di chuyn theo phng doc truc thanh, tc la kh cmt bc t do

+ V mt tnh hoc: tai lin kt ch co th phat sinh mtthanh phn phan lc theo phng doc truc thanh (H.1.5b).

* Kt lun: lin kt thanh kh c mt bc t do va lam phat sinh mt thanhphn phan lc theo phng lin kt.

* Trng hp c bit: mt mingcng co hai u khp va khng chu taitrong th co th nh mt lin kt thanh,co truc thanh la ng ni hai khp(H.1.5c).

* Chu y: lin kt thanh la m rng cua khai nim gi di ng ni t (H.1.5d).

2. Lin kt khp: (lin kt loai 2)a. Cu tao: Gm hai ming cng ni vi nhau bng mt khp ly tng (H.1.6a).

O x

yM (xo,yo)

xo

yo

H.1.4a H.1.4b

yo

O

y

MC(xo,yo,a

)xo x

a

H.1.5a(A) (B)

H.1.5b

N

N(A) (B)

H.1.5cTruc thanh(A) (B)

(A)

RH.1.5d

a a

a

H.1.3b

P

A

B CAN N

P


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V mt cu tao, ch co lin kt khp phc tap (H.1.8a) va han phc tap (H.1.8b).* phc tap cua lin kt: la s lin kt n gian cung loai, tng ng vi

lin kt a cho. Ky hiu p.* Cng thc xac nh phc tap: (1 - 1)

D: s ming cng quy tu vao lin kt.* V du: Xac nh phc tap cua lin kt han trn hnh(H.1.8c)

p = D - 1 = 4 - 1 = 3.Co ngha la lin kt han phc tap a cho tng ng vi

ba lin kt han n gian.

.3 CACH NI CAC MING CNG THANH H BT BINHNHI. Ni mt im (mt) vao mt ming cng:a. iu kin cn: ni mt im vao ming cng cn phai kh hai bc t do

cua no. Ngha la cn dung hai lin ktthanh (H.1.9a).

b. iu kin u: hai lin ktthanh khng c thng hang.

Hai lin kt thanh khng thnghang ni mt im vao ming cng goila b i (H.1.9a).

* Tnh cht cua b i: khi thm hay bt ln lt cac b i th tnh cht nghoc cua h khng thay i. Tnh cht nay c s dung phn tch cu tao hnh hoccua h, va phn tch theo hai hng sau:

+ Phng phap thu hep ming cng: t h ban u, ln lt loai bo dn cac bi a v h n gian cui cung. Nu h thu c la BBH hay BH th h ban ucung BBH hay BH. V du h trn hnh (H.1.9c)

+ Phng phap phat trin ming cng: t ming cng ban u, thm ln lt cacb i th cui cung thu c ming cng. V du h trn hnh (H.1.9d)

p = D - 1

(A)(B)

(D)

Mi han

(C)

H.1.8c

H.1.9a

(A)

Mt

B i

H.1.9bH BHTT

Loai bo cac b i1

12

23

34

4

5

5

H BBHH BBH

H.1.9c


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Vy iu kin cn h BBH la(1 - 3)

* Cac trng hp cua n: tng t nh trn* Cac loai lin kt ni t (H.1.12a):

2. iu kin u:Thng s dung tnh cht cua b i, cach ni hai hoc ba ming cng nhm thu

hep hoc phat trin h n mc ti a cho phep. Nu kt qua thu c:

+ Mt ming cng: h a cho la BBH.+ Hai hoc ba ming cng: s dung iu kin u cua bai toan ni hai, ba ming

cng a bit phn tch tip.V du:

* Ngoai ra con s dung phng phap tai trong bng khng hoc phng phapng hoc khao sat.Xem giao trnh mn C hoc kt cu - Lu Tho Trnh.

V. Trng hp c bit: H dan.H dan la h gm nhng thanh thng lin kt vi nhau ch bng cac khp hai

u mi thanh.

* i vi h dan cung cho phep ap dung cng thc (1 - 2) hoc (1 - 3) khao satiu kin cn. Tuy nhin, trong h dan, cac lin kt khp thng la khp phc tap cnquy i v khp n gian. Cach lam nh vy thng d nhm ln. Di y se trnh bay

mt cach khac thun li hn ma khng phai quan tm n phc tap cua cac lin ktkhp.

n = T + 2.K + 3.H + C - 3.D 0

Gi di ngC =1

Gi c nhC = 2

Ngam trtC = 2

NgamC = 3

H.1.12a

H.1.12d

Khng phai h dan

Mt

H dan

H.1.12b

(A)

(C)

(B)

(D

(Phat trin ming cng)H.1.12c

Loai bo cac b i

5 2 1

H BBH

4

53

43 12

H BBH(Thu hep ming cng)


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1. Trng hp h dan khng ni t:Xet h dan khng ni t gm D thanh dan va M mt.Xem mt thanh dan la ming cng c nh, con lai (D - 1) thanh. Va i ni (M -

2) mt con lai vao ming cng c nh, ngha la cn phai kh 2.(M - 2) bc t do.

Xem cac thanh dan la cac lin kt thanh. Nh vy, (D -1) thanh con lai co khanng kh c (D -1) bc t do.Vy iu kin cn h BBH la:

(1 - 4)

2. Trng hp h dan ni t:Xet h dan gm D thanh dan va M mt. Ngoai ra h dan con ni t bng s lin

kt tng ng C lin kt loai mt. Ni M mt vao ming cng c nh. Ngha la cnkh 2.M bc t do.

Xem cac thanh dan la cac lin kt thanh. Nh vy, D thanh dan co kha nng khc D bc t do. Ngoai ra cac lin kt ni t kh c C bc t do.

Vy iu kin cn h BBH la:(1 - 5)

* Chu y: - Cac trng hp cua n va iu kin u vn nh trng hp tng quat.

CAC V DU*V du 1:Phn tch cu tao hnh hoc cua h cho trn hnh H.1.13a1. iu kin cn: H a cho thuc h bt ky ni t nn iu kin cn s dung

biu thc (1 - 3). Co th giai bai toan theo nhiu quannim khac nhau:

a. Quan nim mi oan thanh thng la mtming cng:

Luc nay D = 5, T = 0, K = 1, H = 3, C = 4. Thayvao (1 - 3)

n = T + 2.K + 3.H + C - 3.D = 0 + 2.1 + 3.3 + 4 - 3.5 = 0 H a cho co kha nng BBH.b. Quan nim mi thanh gay khuc la mt ming cng (quan nim s ming

cng ti thiu):Luc nay D = 2 (ab, bce), T = 0, K = 1, H = 0, C = 4. Thay vao (1 - 3)

n = 0 + 2.1 + 3.0 + 4 - 3.2 = 0 H a cho co kha nng BBH.c. Quan nim trai t la mt ming cng:

Luc nay xem h la khng ni t nn iu kin cn s dung biu thc (1 - 2).Luc nay D = 3 (ab, bce va trai t), T = 2, K = 2, H = 0. Thay vao (1 - 2)

n = (D - 1) - 2.(M - 2) = D - 2.M + 3 0

H.1.13a

a

b

cd

e

n = D - 2M + C


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e

a

H.1.13d

b

cd (I)

H.1.13e

a

b

c

(II)

d

e

O

N = 2 + 2.2 + 3.0 - 3.(3 - 1) = 0 H a cho co kha nng BBH.* Nhn xet: - Co nhiu cach quan nim ming cng khac nhau, va co anh hng

n s lng ming cng va cac lin kt.

- Nn quan nim s ming cng ti thiu v s lng D, T, K, H se tnht.2. iu kin u: Co nhiu cach quan nim.a. a h v thanh bai toan ni hai ming cng: trai t (II) va bce (I). Hai

ming cng nay ni vi nhau bng ba thanh ab, cd, ef (H.1.13b). Ba thanh nay khngng quy hay song song nn h a cho la h BBH (h tnh nh).

b. a h v thanh bai toan ni ba ming cng:Trai t (II), bce (I) va ab (III). Ba ming cng nay ni nhau bng ba khp (1,2

xa v cung), (2,3), (3,1). Ba khp nay khng thng hang nn h a cho la h BBH(H.1.13c).

* Lu y: Khi khao sat iu cn va u cho mt h, ch cn s dung mt quan nimla u.

* V du 2:Ni dung ging v du 1 nhng thanh e-f nghing i 45o (hnh H.1.13d).

1. iu kin cn: khng thay i so vi v du 1.2. iu kin u: a h v thanh bai toan ni hai ming cng:o la trai t (II) va bce (I). Hai ming cng nay ni vi nhau bng ba thanh ab,

cd, ef (H.1.13e). Ba thanh nay ng quy tai O nn h a cho la h BHTT.

*V du 3:Phn tch cu tao hnh hoc cua h cho trn hnh H.1.13f.

(I)

H.1.13c

a

b

c

(II)

d

e(III) (3,1)

(2,3

(1,2)

e

a

H.1.13b

b

c d

(II)

(I)


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1. iu kin cn: H a cho thuc h bt ky ni t. S dung biu thc (1 - 3) khao sat iu kin cn.

Quan nim h gm cac ming cng: (A), (B), (C), (D), (E), (F).Vy D = 6, T = 4, K = 3, C = 8, H = 0. Thay vao (1 - 3)

n = T + 2.K + 3.H + C - 3.D = 4 + 2.3 + 3.0 + 8 - 3.6 = 0.H a cho co kha nng BBH.2. iu kin u: a h v thanh bai toan ni ba ming cng (I), (II) & (III) nh

trn hnh (H.1.13g). Ba ming cng nay ni vi nhau bng ba khp (1,2), (2,3) & (3,1)khng thng hang nn h a cho la BBH (tnh nh).

*V du 4:Phn tch cu tao hnh hoc cua h cho trn hnh (H.1.13h).1. iu kin cn: H a cho thuc h

bt ky ni t. S dung biu thc (1-3) khao sat iu kin cn.

Quan nim h gm cac ming cng(A), (B), (C).

Vy D = 3, T = 2, K = 1, H = 0, C = 5.Thay vao (1 - 3)

n = T + 2.K + 3.H + C - 3.D = 2 + 2.1 + 3.0 + 5 - 3.3 = 0.H a cho co kha nng BBH.2. iu kin u:Dung phng phap phat trin ming cng:

Vy h a cho la h BBH (h tnh nh).* V du 5:Phn tch cu tao hnh hoc cua h cho trn hnh (H.1.13i).

1. iu kin cn: H a cho thuc h bt ky ni t. S dung biu thc (1-3) khao sat iu kin cn.

(A) + Trai tngam (1)

MC + (B)khp 2

thanh 3

MC + (C)

3 thanhMC duy nht.

(4) va (5)

H.1.13f

(E) (F)

(D)

(C)(B)

(A)

H.1.13g

(II) (III)

(I)

(1,2) (3,1)(2,3)

H.1.13h

12 4

53

A B

C


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Quan nim h gm cac ming cng (af), (eb), (bg), (fh), (hc).Vy D = 5, T = 0, K = 5, H = 0,

C = 5. Thay vao (1 - 3)n = T + 2.K + 3.H + C - 3.D

= 0 + 2.5 + 3.0 + 5 - 3.5 = 0.H a cho co kha nng BBH.2. iu kin u:Dung phng phap phat trin ming cng:

Vy h a cho la h BBH (h tnh nh).* V du 6:Phn tch cu tao hnh hoc cua h cho trn hnh (H.1.13j).

1. iu kin cn: H a cho thuc h dan ni t. S dung biu thc (1 - 5) khao sat iu kin cn.

Vy D = 11, M = 7, C = 3. Thay vao (1 - 5)n = D - 2.M + C = 11 - 2.7 + 3 = 0

H a cho co kha nng BBH.

2. iu kin u:Dung phng phap phat trin ming cng (H.1.13k).

Tng t, (2-3-7-5) la ming cng (II)

H

H.1.13i

e

dc

ga

bk

h

(af) + Trai tngam (a)

MC + (eb)khp e

MC + (fh)

khp fMC + (hc)

(thanh bg)

MC duy nht.khp h

thanh cd

thanh k

(1-4-6)b i (1-2) & (2-4)

MC (I)

(I) + (II)khp 2

MC + Trai tkhp A

MC duy nht

thanh Bthanh 4-5

1

6

4

2 3

7

5H.1.13j

1

6

4

2 3

7

5(I) (II)H.1.13kA

BA

B


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a cho la h BHTT (h tnh nh).* V du 7:Phn tch cu tao hnh hoc cua h cho trn hnh (H.1.13l)1. iu kin cn: H a cho thuc h dan ni t. S dung biu thc (1 - 5)

khao sat iu kin cn.

Vy T = 18, M = 10, C = 4. Thay vao (1-5)n = D -2.M + C = 18 - 2.10 + 4 = 2 > 0.H a cho co kha nng BBH va tha lin kt.

2. iu kin u:a h v thanh bai toan ni ba ming cng.+ Trai t: (I).

+ (1, 2, 5, 6, 9): (II). D thy (II) tha mt lin kt thanh.+ Tng t (3, 4, 7, 8, 10) la ming cng (III) cung tha mt lin kt thanh.Ba ming cng nay ni vi nhau bng ba khp (1,2 xa v cung), (2,3), (3,1). Ba

khp nay thng hang (H.1.13m).Vy h a cho la BHTT.

1 2 3 4

5 6 7 8

109A B

H.1.13l

(1,2)

(I) (II)

(III)

(3,1) (2,3)H.1.13m


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CHNG M U 1. I TNG NGHIN CU VA NHIM VU CUA MN HOC

I. i tng nghin cu va nhim vu cua mn hoc:

1. i tng nghin cu: la vt rn bin dang an hi, tc la co th thay i hnhdang di tac dung cua cac nguyn nhn bn ngoai.2. Pham vi nghin cu:Pham vi nghin cu cua mn C hoc kt cu la ging mn Sc bn vt liu

nhng gm nhiu cu kin lin kt lai vi nhau. Do vy, trong kt cu hay dung tn goila h kt cu.

II. Nhim vu cua mn hoc:Nhim vu chu yu cua mn C hoc kt cu la i xac nh ni lc, bin dang va

chuyn v trong cng trnh nhm xy dng cng trnh thoa man cac yu cu:- iu kin v bn: am bao cho cng trnh khng b pha hoai di tac dung

cua cac nguyn nhn bn ngoai- iu kin v cng: am bao cho cng trnh khng co chuyn v va bin dang

vt qua gii han cho phep nhm am bao s lam vic bnh thng cua cng trnh.- iu kin v n nh: am bao cho cng trnh co kha nng bao toan v tr va

hnh dang ban u cua no di dang cn bng trong trang thai bin dang.Vi yu cu v bn, cn i xac nh ni lc; vi yu cu v cng, cn i

xac nh chuyn v; vi yu cu v n nh, cn i xac nh lc ti han ma kt cu co th

chu c.III. Cac bai toan mn hoc giai quyt:1. Bai toan kim tra: bai toan nay, ta a bit trc hnh dang, kch thc cu

th cua cac cu kin trong cng trnh va cac nguyn nhn tac ng.Yu cu: kim tra cng trnh theo ba iu kin trn ( bn, cng & n nh)

co am bao hay khng? Va ngoai ra con kim tra cng trnh thit k co tit kim nguynvt liu hay khng?

2. Bai toan thit k: bai toan nay, ta mi ch bit nguyn nhn tac ng bn

ngoai. Yu cu: Xac nh hnh dang, kch thc cua cac cu kin trong cng trnh mtcach hp ly ma vn am bao ba iu kin trn. giai quyt bai toan nay, thng thng, da vao kinh nghim hoc dung

phng phap thit k s b gia thit trc hnh dang, kch thc cua cac cu kin. Sauo tin hanh giai bai toan kim tra nh a noi trn. Va trn c s o ngui thit k iuchnh lai gia thit ban u cua mnh, tc la i giai bai toan lp.

IV. V tr cua mn hoc:La mn hoc ky thut c s lam nn tang cho cac mn hoc chuyn nganh nh: kt

cu b tng, kt cu thep & g, ky thut thi cng...Trang b cho ngi lam cng tac xy dng nhng kin thc hu ch.


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2. PHNG PHAP NGHIN CUI. S cng trnh:1. Khai nim: S cng trnh la hnh anh n gian hoa ma vn am bao phan

anh c chnh xac s lam vic thc t cua cng trnh va phai dung tnh toan c.

2. Cac yu t anh hng n vic chon s tnh:- Hnh dang, kch thc cua cng trnh.- Ty l cng cua cac cu kin.- Tm quan trong cua cng trnh.- Kha nng tnh toan cua ngi thit k.- Tai trong va tnh cht tac dung cua no.- v.v.v3. Cac bc la chon s tnh:

a. Bc 1: a cng trnh thc v s cng trnh:- Thay cac thanh bng ng truc thanh.- Thay cac ban va vo bng cac mt trung gian.- Thay tit din, vt liu bng cac ai lng c trng: din tch (F), mmen quan

tnh (J), mun an hi (E), h s dan n v nhit (a) ...- Thay thit b ta bng cac lin kt ly tng.- a tai trong tac dung ln mt cu kin v truc cu kin.V du:

b. Bc 2: a s cng trnh v s tnh:Trong mt s trng hp, s cng trnh a v cha phu hp vi kha nng tnh

toan, ta loai bo nhng yu t th yu n gian bai toan va a v s tnh, tnhc.

V du:

H.1 E, J, F, h, a...

(Bc 1)

(Bc 2)

H.2


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3. CAC GIA THIT TNH TOAN VA NGUYN LY CNGTAC DUNG

I. Cac gia thit tnh toan:1. iu kin vt ly cua bai toan:Gia thit rng vt liu la an hi tuyt i va tun theo

nh lut Hook, ngha la quan h gia ni lc va bin dang la

quan h tuyn tnh (E

s

e = ).

Chu y: Nu chp nhn gia thit nay th bai toan goi laan hi tuyn tnh (tuyn tnh vt ly). Nu khng chp nhn gia thit nay th bai toan goila an hi phi tuyn (phi tuyn vt ly).

2. iu kin hnh hoc cua bai toan:

Chuyn v va bin dang c xem nh la nhng ailng v cung be. Do vy khi tnh toan, xem cng trnh lakhng co bin dang.

Chu y: Nu chp nhn gia thit nay th bai toan goi latuyn tnh hnh hoc. Nu khng chp nhn gia thit nay th baitoan goi la phi tuyn hnh hoc.

II. Nguyn ly cng tac dung:1. Phat biu: Mt ai lng nghin cu S (ni lc, phan lc, chuyn v...) do mt

s cac nguyn nhn ng thi tac dung gy ra se bng tng ai s hay hay tng hnh hoccua ai lng S do tng nguyn nhn tac dung ring re gy ra.

Ly tng ai s khi ai lng S la ai lngv hng, ly tng hnh hoc khi ai lng S la ailng vec t.

V du: Xet dm chu tac dung cua 2 lc P1 &P2 va ai lng nghin cu S la phan lc VA trnhnh (H.5a)

Xet chnh dm o nhng chu tac dung ring

re cua 2 lc P1, P2 trn hnh (H.5b) & (H.5c).Theo nguyn ly cng tac dung:

21 AAA VVV += .Va nu xet toan din, th h (H.5a) bng

tng cua hai h (H.5b) & (H.5c).2. Biu thc giai tch cua nguyn ly cng tac dung:

S(P1, P2,..... Pn) = S(P1) + S(P2) +.....+ S(Pn)- S(P1, P2,..... Pn): la ai lng S do cac nguyn nhn P1, P2,..... Pn ng thi tac

dung ln h gy ra.- S(Pk): la ai lng S do ring Pk tac dung ln h gy ra.

H.3O e

s

H.4

D 0D 0

VA

VA1

P1

VA2

A P2 B

BA

A BP1 P2

H.5a

H.5b

H.5c


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Goi kS la ai lng S do ring Pk = 1 gy ra. Tc la S(Pk) = kS .PkVy S(P1, P2,..... Pn) = nn PSPSPS ........ 2211 ++

Chu y:Nguyn ly cng tac dung ch ap dung cho h tuyn tnh vt ly cung nhtuyn tnh hnh hoc.

4. PHN LOAI CNG TRNHI. Phn loai theo s tnh:1. H phng: khi tt ca cac

cu kin cung thuc mt mt phngva tai trong tac dung cung nmtrong mt phng o.

Cac loai h phng:

- Dm (H.6)- Dan (H.7)- Vom (H.8)

- Khung (H.9)

- H lin hp (H.10)

2. H khng gian: khi cac cu kin khng cung nm trong mt mt phng, hoccung nm trong mt mt phng nhng tai trong tac dung ra ngoai mt phng o.

Cac loai h khng gian:

- H dm trc giao (H.11)- Khung khng gian (H.12)- Dan khng gian (H.13)- Ban (H.14)- Vo (H.15)

H.6bH.6a

H.7a H.7b

H.10bH.10a

H.11

H.12 H.13 H.15H.14

H.9aH.9b

H.8a H.8b


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II. Phn loai theo phng phap tnh:1. Da vao s cn thit hay khng phai s dung iu kin ng hoc khi xac

nh toan b cac phan lc va ni lc trong h, ngi ta chia ra hai loai h:a. H tnh nh: la loai h ma ch bng cac iu kin tnh hoc co th xac nh

c toan b ni lc va phan lc trong h. V du cac h trn hnh a t (H.6) n (H.10).b. H siu tnh: la loai h ma ch bng cac iu kin tnh hoc th cha u xacnh toan b cac ni lc va phan lc ma con phai s dung thm iu kin ng hoc vaiu kin vt ly. V du cac h trn hnh b t (H.6) n (H.10).

2. Da vao s cn thit hay khng phai s dung iu kin cn tnh hoc khixac nh bin dang trong h khi h chu chuyn v cng bc, ngi ta chia ra hailoai h:

a. H xac nh ng: la loai hkhi chu chuyn v cng bc, co thxac nh bin dang cua h ch bng caciu kin ng hoc (hnh hoc). V duhcho trn hnh (H.16).

b. H siu ng: la loai h khi chu chuyn v cng bc, nu ch bng cac iukin ng hoc th cha th xac nh c bin dang cua h ma cn phai s dung thmiu kin tnh hoc.V duh cho trn hnh (H.17).

III. Phn loai theo kch thc tng i cua cac cu kin:- Thanh: nu kch thc mt phng kha ln hn hai phng con lai (H 18a).

- Ban: nu kch thc cua hai phng kha ln hn phng con lai (H.18b).- Khi: nu kch thc cua ba phng gn bng nhau (H.18c)

IV. Phn loai theo kha nng thay i hnh dang hnh hoc:- H bin hnh.

- H bin hnh tc thi.- H bt bin hnh.

H.18aH.18b H.18c

D

B C

A'A

H.16

D D

A B

C

DC' D'

H.17


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CHNG 1PHN TCH CU TAO HNH HOC CUA H PHNG

1. CAC KHAI NIM

I. H bt bin hnh (BBH): la h khng co s thay i hnh dang hnh hoc ditac dung cua tai trong nu xem cac cu kin cua h la tuyt icng.

V du: Phn tch h hnh ve (H.1.1a)Nu quan nim AB, BC, trai t la tuyt i cng, tc la

lAB, lBC, lCA = const th tam giac ABC la duy nht, nn h a chola h BBH.

- Mt h BBH mt cach ro rt goi chung la ming cng (tm cng)- Cac loai ming cng: (H.1.1b)

- Ky hiu ming cng: (H.1.1c)

* Chu y: Do h BBH co kha nng chu lc tac dung nn no c s dung lam cackt cu xy dng va thc t la chu yu s dung loai h nay.

II. H khng bt bin hnh:

1. H bin hnh (BH): la h co s thay ihnh dang hnh hoc mt lng hu han di tac dungcua tai trong mc du xem cac cu kin cua h latuyt i cng.

V du: H ABCD cho trn hnh (H.1.2a) coth thanh h AB'CD, nn h a cho la h BH.

* Chu y: Do h BH khng co kha nng chu tai trong tac dung nn cackt cu xy dng khng s dung loai h nay.

H BH trn hnh (H.1.2b) cho phep s dung v theo phng ng, taitrong tac dung ln h trang thai cn bng.2. H bin hnh tc thi (BHTT): la h co s thay i hnh dang hnh

hoc mt lng v cung be di tac dung cua tai trong mc du xem cac cukin cua h la tuyt i cng.

V du: H ABC co cu tao nh trn hnh (H.1.3a), khp A co th i xung mtoan v cung bed, nn h a cho la h BHTT.

*Chu y: Cac kt cu xy dng khng s dungh BHTT hay h gn BHTT (la h ma ch cn thay imt lng v cung be hnh dang hnh hoc se tr thanhh BHTT, v du h BA'C trn hnh (H.1.3a) v ni lc

B C

A

H.1.1a

H.1.1bH.1.1c

A D

B

B'

C H.1.2a

H.1.2b

H.1.3aB

C

A

A'd


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trong h gn BHTT rt ln.Tht vy, xet h trn hnh (H.1.3b).

Lc doc trong hai thanh AB va AC la N.

N =asin.2

P- .

Khi a 0, h BAC tin n h gnBHTT.

N = -

)sin.2

(lim0 aa

P .

III. Bc t do: la s cac thng s c lp u xac nh v tr cua mt h so vimt h c nh khac.

Trong h phng, mt chtim co bc t do bng 2 (

H.1.4a); mt ming cng co bct do bng 3 (H.1.4b).

2. CAC LOAI LIN KT VA TNH CHT CUA LIN KTI. Lin kt n gian: la lin kt ni hai ming cng vi nhau.Cac loai lin kt n gian

1. Lin kt thanh: (lin kt loai mt)a. Cu tao: Gm mt thanh thng khng chu tai trong cohai khp ly tng hai u. (H.1.5a)

b. Tnh cht cua lin kt:+ V mt ng hoc: lin kt thanh khng cho ming

cng di chuyn theo phng doc truc thanh, tc la kh cmt bc t do

+ V mt tnh hoc: tai lin kt ch co th phat sinh mtthanh phn phan lc theo phng doc truc thanh (H.1.5b).

* Kt lun: lin kt thanh kh c mt bc t do va lam phat sinh mt thanhphn phan lc theo phng lin kt.

* Trng hp c bit: mt mingcng co hai u khp va khng chu taitrong th co th nh mt lin kt thanh,co truc thanh la ng ni hai khp(H.1.5c).

* Chu y: lin kt thanh la m rng cua khai nim gi di ng ni t (H.1.5d).

2. Lin kt khp: (lin kt loai 2)a. Cu tao: Gm hai ming cng ni vi nhau bng mt khp ly tng (H.1.6a).

O x

yM (xo,yo)

xo

yo

H.1.4a H.1.4b

yo

O

y

MC(xo,yo,a

)xo x

a

H.1.5a(A) (B)

H.1.5b

N

N(A) (B)

H.1.5cTruc thanh(A) (B)

(A)

RH.1.5d

a a

a

H.1.3b

P

A

B CAN N

P


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b. Tnh cht:+ V mt ng hoc: lin kt khp khng cho ming cng

chuyn v thng (nhng co th xoay), tc la kh c hai bc tdo.

+ V mt tnh hoc: tai lin kt co th phat sinh mt thanhphn phan lc co phng cha bit. Phan lc nay thng cphn tch thanh hai thanh phn theo hai phng xac nh(H.1.6b).

* Kt lun: lin kt khp kh c hai bc t do va lamphat sinh hai thanh phn phan lc.

* Trng h p c bit: hai lin kt thanh co th xem lamt lin kt khp (khp gia tao), co v trtai giao im ng ni hai truc thanh(H.1.6c).

* Chu y: lin kt khp la m rngcua khai nim gi c nh ni t(H.1.6d)

3. Lin kt han: (lin kt loai 3)a. Cu tao: Gm hai ming cng ni vi nhau bng

mt mi han (H.1.7a).b. Tnh cht:

+ V mt ng hoc: lin kt hankhng cho ming cng co chuyn v, tc lakh c 3 bc t do.

+ V mt tnh hoc: lin kt co th lamphat sinh mt thanh phn phan lc co phngva v tr cha bit. Thng a phan lc nayv tai v tr lin kt va phn tch thanh bathanh phn ( yx RRM ,, )(H.1.7b)

* Kt lun: lin kt han kh c ba bc t do va lam phatsinh ba thanh phn phan lc.

* Chu y:- Lin kt han tng ng vi ba lin kt thanh hoc mt

lin kt thanh va mt lin kt khp c sp xp mt cach hp ly.- Lin kt han la m rng cua khai

nim lin kt ngam ni t (H.1.7c)II. Lin kt phc tap: la lin kt ni

nhiu ming cng vi nhau, s ming cngln hn hai.

(B)(A)

Mi han

H.1.7a

Khp

(A) (B)

H.1.6a

(A) Rx

RyR

H.1.6byx RRR +=

(A)

(B

(C)

H.1.8a

(A)(B)

(D)

Mi han

(C)

H.1.8b

H.1.6c (B)(A)

Khp gia (A)

RyRx

H.1.6d

Ry(A)

H.1.7b R

Rx

R

M = R.d

dRM

RRR

MRRR

yx

yx

.

,,

=

+=

d

Rx

Ry


29/139


V mt cu tao, ch co lin kt khp phc tap (H.1.8a) va han phc tap (H.1.8b).* phc tap cua lin kt: la s lin kt n gian cung loai, tng ng vi

lin kt a cho. Ky hiu p.* Cng thc xac nh phc tap: (1 - 1)

D: s ming cng quy tu vao lin kt.* V du: Xac nh phc ta p cua lin kt han trn hnh(H.1.8c)

p = D - 1 = 4 - 1 = 3.Co ngha la lin kt han phc tap a cho tng ng vi

ba lin kt han n gian.

.3 CACH NI CAC MING CNG THANH H BT BINHNHI. Ni mt im (mt) vao mt ming cng:a. iu kin cn: ni mt im vao ming cng cn phai kh hai bc t do

cua no. Ngha la cn dung hai lin ktthanh (H.1.9a).

b. iu kin u: hai lin ktthanh khng c thng hang.

Hai lin kt thanh khng thnghang ni mt im vao ming cng goila b i (H.1.9a).

* Tnh cht cua b i: khi thm hay bt ln lt cac b i th tnh cht nghoc cua h khng thay i. Tnh cht nay c s dung phn tch cu tao hnh hoccua h, va phn tch theo hai hng sau:

+ Phng phap thu hep ming cng: t h ban u, ln lt loai bo dn cac bi a v h n gian cui cung. Nu h thu c la BBH hay BH th h ban ucung BBH hay BH. V du h trn hnh (H.1.9c)

+ Phng phap phat trin ming cng: t ming cng ban u, thm ln lt cacb i th cui cung thu c ming cng. V du h trn hnh (H.1.9d)

p = D - 1

(A)(B)

(D)

Mi han

(C)

H.1.8c

H.1.9a

(A)

Mt

B i

H.1.9bH BHTT

Loai bo cac b i1

12

23

34

4

5

5

H BBHH BBH

H.1.9c


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II. Cach ni hai ming cng:1. iu kin cn: Xem mt ming cng la c nh. ni ming cng con lai

vao ming cng c nh cn kh ba bc t do cua no, ngha la cn s dung t hp caclin kt:

+ Ba lin kt thanh (H.1.10a).+ Mt lin kt thanh cng mt lin kt khp (H.1.10b).+ Mt lin kt han (H.1.10c).

2. iu kin u:a. Nu s dung ba lin kt thanh: yu cu ba thanh khng c ng quy hoc

song song (H.1.10d, H.1.10e & H.1.10f).

b. Nu s dung mt lin kt thanh cng mt lin kt khp: yu cu khpkhng c nm trn ng truc thanh (H.1.10g).

c. Nu s dung lin kt han: th o cung la iu kinu.

III. Cach ni ba ming cng:1. iu kin cn: xem mt ming cng la c nh.

ni hai ming cng con lai vao ming cng c nh cn phaikh sau bc t do, ngha la cn phai s dung t hp cac linkt:

+ Ba lin kt khp (H.1.11a).+ Sau lin kt thanh (H.1.11b).+ Hai lin kt han (H.1.11c).

+ Mt lin kt thanh cng mt lin kt khp cng mt lin kt han (H.1.11d) .+ v.v.v.

H.1.9d Thm cac b i

1

1

22

33

4

4

5

5

H BBH H BBH

H.1.10a(A) (B) H.1.10b(A) (B) H.1.10c

Mi han(A) (B)

H.1.10d

(A) (B)

H.1.10e

(A) (B) (B)(A)

H.1.10f

(BHTT) (BHTT) (BH)

H BHTTH.1.10g

(A)

(B)


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2. iu kin u:+ Nu cac ming cng ni ln lt vi nhau: tr v

lai bai toan ni hai ming cng. V du (H.1.11e).+ Nu cac ming cng ni ng thi vi nhau (nu

loai bo mt ming cng bt ky, hcon lai b bin hnh): luc nay h cn

s dung ba lin kt khp (thc hocgia tao) tng h (H.1.11f). Va yucu cac lin kt khp khng cungnm trn mt ng thng (H.1.11g).

IV. Cach ni nhiu mingcng:

1. iu kin cn:a. Trng hp h bt ky khng ni t:

Xet mt h khng ni t gm D ming cng. Cac lin kt gia cac ming cngla: T lin kt thanh, K lin kt khp a quy v khp n gian va H lin kt han a quyv han n gian.

Xem mt ming cng la c nh. Ni (D - 1) ming cng con lai vao ming cngc nh, ngha la cn phai kh 3.(D-1) bc t do. o la yu cu.

V kha nng: T , K, H kh c T + 2.K + 3.H bc t do.Nh vy, iu kin cn h BBH la

(1 - 2)

* Cac trng hp cua n:+ n = 0 va h a cho la h BBH th h la h tnh.+ n > 0 va h a cho la h BBH th h la h siu tnh.+ n < 0 th h la h BH.b. Trng hp h bt ky co ni t:Xet mt h ni t gm D ming cng. Cac lin kt gia cac ming cng la: T

lin kt thanh, K lin kt khp a quy v khp n gian va H lin kt han a quy v hann gian. Lin kt gia h va trai t gm C lin kt a quy v lin kt loai mt.

Xem trai t la c nh. Ni D ming cng con lai vao tr t, ngha la phai kh3.D bc t do. o la yu cu.

V kha nng: T, K, H, C kh c T + 2.K + 3.H + C bc t do.

n = T + 2.K + 3.H - 3.(D -1) 0

(C)(A) (B)

H.1.11e

(A)

(B)

(C)

H.1.11fH BHTT

(A)

(B)

H.1.11g

(C)

(B)(A)

(C)

H.1.11a H.1.11b(A)

(B)

(C)

H.1.11c

(A) (B)

(C)(C)

(A) (B)

H.1.11d


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Vy iu kin cn h BBH la(1 - 3)

* Cac trng hp cua n: tng t nh trn* Cac loai lin kt ni t (H.1.12a):

2. iu kin u:Thng s dung tnh cht cua b i, cach ni hai hoc ba ming cng nhm thu

hep hoc phat trin h n mc ti a cho phep. Nu kt qua thu c:

+ Mt ming cng: h a cho la BBH.+ Hai hoc ba ming cng: s dung iu kin u cua bai toan ni hai, ba ming

cng a bit phn tch tip.V du:

* Ngoai ra con s dung phng phap tai trong bng khng hoc phng phapng hoc khao sat.Xem giao trnh mn C hoc kt cu - Lu Tho Trnh.

V. Trng hp c bit: H dan.H dan la h gm nhng thanh thng lin kt vi nhau ch bng cac khp hai

u mi thanh.

* i vi h dan cung cho phep ap dung cng thc (1 - 2) hoc (1 - 3) khao satiu kin cn. Tuy nhin, trong h dan, cac lin kt khp thng la khp phc tap cnquy i v khp n gian. Cach lam nh vy thng d nhm ln. Di y se trnh bay

mt cach khac thun li hn ma khng phai quan tm n phc tap cua cac lin ktkhp.

n = T + 2.K + 3.H + C - 3.D 0

Gi di ngC =1

Gi cnh

Ngam trtC = 2

NgamC = 3

H.1.12a

H.1.12d

Khng phai h dan

Mt

H dan

H.1.12b

(A)

(C)

(B)

(D

(Phat trin ming cng)H.1.12c

Loai bo cac b i

5 2 1

H BBH

4

53

43 12

H BBH(Thu hep ming cng)


33/139


1. Trng hp h dan khng ni t:Xet h dan khng ni t gm D thanh dan va M mt.Xem mt thanh dan la ming cng c nh, con lai (D - 1) thanh. Va i ni (M -

2) mt con lai vao ming cng c nh, ngha la cn phai kh 2.(M - 2) bc t do.

Xem cac thanh dan la cac lin kt thanh. Nh vy, (D -1) thanh con lai co khanng kh c (D -1) bc t do.Vy iu kin cn h BBH la:

(1 - 4)

2. Trng hp h dan ni t:Xet h dan gm D thanh dan va M mt. Ngoai ra h dan con ni t bng s lin

kt tng ng C lin kt loai mt. Ni M mt vao ming cng c nh. Ngha la cnkh 2.M bc t do.

Xem cac thanh dan la cac lin kt thanh. Nh vy, D thanh dan co kha nng khc D bc t do. Ngoai ra cac lin kt ni t kh c C bc t do.

Vy iu kin cn h BBH la:(1 - 5)

* Chu y: - Cac trng hp cua n va iu kin u vn nh trng hp tng quat.

CAC V DU*V du 1:Phn tch cu tao hnh hoc cua h cho trn hnh H.1.13a1. iu kin cn: H a cho thuc h bt ky ni t nn iu kin cn s dung

biu thc (1 - 3). Co th giai bai toan theo nhiu quannim khac nhau:

a. Quan nim mi oan thanh thng la mtming cng:

Luc nay D = 5, T = 0, K = 1, H = 3, C = 4. Thayvao (1 - 3)

n = T + 2.K + 3.H + C - 3.D = 0 + 2.1 + 3.3 + 4 - 3.5 = 0 H a cho co kha nng BBH.b. Quan nim mi thanh gay khuc la mt ming cng (quan nim s ming

cng ti thiu):Luc nay D = 2 (ab, bce), T = 0, K = 1, H = 0, C = 4. Thay vao (1 - 3)

n = 0 + 2.1 + 3.0 + 4 - 3.2 = 0 H a cho co kha nng BBH.c. Quan nim trai t la mt ming cng:

Luc nay xem h la khng ni t nn iu kin cn s dung biu thc (1 - 2).Luc nay D = 3 (ab, bce va trai t), T = 2, K = 2, H = 0. Thay vao (1 - 2)

n = (D - 1) - 2.(M - 2) = D - 2.M + 3 0

H.1.13a

a

b

cd

e


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e

a

H.1.13d

b

cd (I)

H.1.13e

a

b

c

(II)

d

e

O

N = 2 + 2.2 + 3.0 - 3.(3 - 1) = 0 H a cho co kha nng BBH.* Nhn xet: - Co nhiu cach quan nim ming cng khac nhau, va co anh hng

n s lng ming cng va cac lin kt.

- Nn quan nim s ming cng ti thiu v s lng D, T, K, H se tnht.2. iu kin u: Co nhiu cach quan nim.a. a h v thanh bai toan ni hai ming cng: trai t (II) va bce (I). Hai

ming cng nay ni vi nhau bng ba thanh ab, cd, ef (H.1.13b). Ba thanh nay khngng quy hay song song nn h a cho la h BBH (h tnh nh).

b. a h v thanh bai toan ni ba ming cng:Trai t (II), bce (I) va ab (III). Ba ming cng nay ni nhau bng ba khp (1,2

xa v cung), (2,3), (3,1). Ba khp nay khng thng hang nn h a cho la h BBH(H.1.13c).

* Lu y: Khi khao sat iu cn va u cho mt h, ch cn s dung mt quan nimla u.

* V du 2:Ni dung ging v du 1 nhng thanh e-f nghing i 45o (hnh H.1.13d).

1. iu kin cn: khng thay i so vi v du 1.2. iu kin u: a h v thanh bai toan ni hai ming cng:o la trai t (II) va bce (I). Hai ming cng nay ni vi nhau bng ba thanh ab,

cd, ef (H.1.13e). Ba thanh nay ng quy tai O nn h a cho la h BHTT.

*V du 3:Phn tch cu tao hnh hoc cua h cho trn hnh H.1.13f.

(I)

H.1.13c

a

b

c

(II)

d

e(III) (3,1)

(2,3

(1,2)

e

a

H.1.13b

b

c d

(II)

(I)


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1. iu kin cn: H a cho thuc h bt ky ni t. S dung biu thc (1 - 3) khao sat iu kin cn.

Quan nim h gm cac ming cng: (A), (B), (C), (D), (E), (F).Vy D = 6, T = 4, K = 3, C = 8, H = 0. Thay vao (1 - 3)

n = T + 2.K + 3.H + C - 3.D = 4 + 2.3 + 3.0 + 8 - 3.6 = 0.H a cho co kha nng BBH.2. iu kin u: a h v thanh bai toan ni ba ming cng (I), (II) & (III) nh

trn hnh (H.1.13g). Ba ming cng nay ni vi nhau bng ba khp (1,2), (2,3) & (3,1)khng thng hang nn h a cho la BBH (tnh nh).

*V du 4:Phn tch cu tao hnh hoc cua h cho trn hnh (H.1.13h).1. iu kin cn: H a cho thuc h

bt ky ni t. S dung biu thc (1-3) khao sat iu kin cn.

Quan nim h gm cac ming cng(A), (B), (C).

Vy D = 3, T = 2, K = 1, H = 0, C = 5.Thay vao (1 - 3)

n = T + 2.K + 3.H + C - 3.D = 2 + 2.1 + 3.0 + 5 - 3.3 = 0.H a cho co kha nng BBH.2. iu kin u:Dung phng phap phat trin ming cng:

Vy h a cho la h BBH (h tnh nh).* V du 5:Phn tch cu tao hnh hoc cua h cho trn hnh (H.1.13i).

1. iu kin cn: H a cho thuc h bt ky ni t. S dung biu thc (1-3) khao sat iu kin cn.

(A) + Trai tngam (1)

MC + (B)khp 2

thanh 3

MC + (C)

3 thanhMC duy nht.

(4) va (5)

H.1.13f

(E) (F)

(D)

(C)(B)

(A)

H.1.13g

(II) (III)

(I)

(1,2) (3,1)(2,3)

H.1.13h

12 4

53

A B

C


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Quan nim h gm cac ming cng (af), (eb), (bg), (fh), (hc).Vy D = 5, T = 0, K = 5, H = 0,

C = 5. Thay vao (1 - 3)n = T + 2.K + 3.H + C - 3.D

= 0 + 2.5 + 3.0 + 5 - 3.5 = 0.H a cho co kha nng BBH.2. iu kin u:Dung phng phap phat trin ming cng:

Vy h a cho la h BBH (h tnh nh).* V du 6:Phn tch cu tao hnh hoc cua h cho trn hnh (H.1.13j).

1. iu kin cn: H a cho thuc h dan ni t. S dung biu thc (1 - 5) khao sat iu kin cn.

Vy D = 11, M = 7, C = 3. Thay vao (1 - 5)n = D - 2.M + C = 11 - 2.7 + 3 = 0

H a cho co kha nng BBH.

2. iu kin u:Dung phng phap phat trin ming cng (H.1.13k).

Tng t, (2-3-7-5) la ming cng (II)

H.1.13i

e

dc

ga

bk

h

(af) + Trai tngam (a)

MC + (eb)khp e

MC + (fh)

khp fMC + (hc)

(thanh bg)

MC duy nht.khp h

thanh cd

thanh k

(1-4-6)b i (1-2) & (2-4)

MC (I)

(I) + (II)khp 2

MC + Trai tkhp A

MC duy nht

thanh Bthanh 4-5

1

6

4

2 3

7

5H.1.13j

1

6

4

2 3

7

5(I) (II)H.1.13kA

BA

B


37/139


H a cho la h BHTT (h tnh nh).* V du 7:Phn tch cu tao hnh hoc cua h cho trn hnh (H.1.13l)1. iu kin cn: H a cho thuc h dan ni t. S dung biu thc (1 - 5)

khao sat iu kin cn.

Vy T = 18, M = 10, C = 4. Thay vao (1-5)n = D -2.M + C = 18 - 2.10 + 4 = 2 > 0.H a cho co kha nng BBH va tha lin kt.

2. iu kin u:a h v thanh bai toan ni ba ming cng.+ Trai t: (I).

+ (1, 2, 5, 6, 9): (II). D thy (II) tha mt lin kt thanh.+ Tng t (3, 4, 7, 8, 10) la ming cng (III) cung tha mt lin kt thanh.Ba ming cng nay ni vi nhau bng ba khp (1,2 xa v cung), (2,3), (3,1). Ba

khp nay thng hang (H.1.13m).Vy h a cho la BHTT.

1 2 3 4

5 6 7 8

109A B

H.1.13l

(1,2)

(I) (II)

(III)

(3,1) (2,3)H.1.13m


38/139

C HOC KT CU 1 Page 19

CHNG 2XAC NH NI LC TRONG H PHNG TNH NH CHU

TAI TRONG BT NG

1. CAC KHAI NIM.I. Ni lc:1. Khai nim: Ni lc la bin thin lc lin kt cua cac phn t bn trong

cu kin khi cu kin chu tac dung cua ngoai lc va cac nguyn nhn khac.* Chu y: Khai nim v ni lc va phan lc la co th ng nht vi nhau nu

quan nim tit din la mt lin kt han hoc lin kt tng ng ni hai ming cng hai bn tit din. V vy, sau nay ta co th ng nht vic xac nh ni lc vi vic xacnh phan lc trong cac lin kt.

2. Cac thanh phn ni lc: Mn C hoc kt cu chu yu xac nh 3 thanh phnni lc:

- Mmen un. Ky hiu M.- Lc ct. Ky hiu Q.- Lc doc. Ky hiu N.3. Quy c du cac thanh phn ni lc:- Mmen un quy c xem la dng khi no lam cng th di va ngc lai

(H.1a).

- Lc ct quy c xem la dng khi no lam cho phn h xoay thun chiu kimng h va ngc lai (H.1b).

- Lc doc quy c xem la dng khi no gy keo va ngc lai (H.1c).

*Chu y:

- Cach quy c du ni lc la ging mn Sc bn vt liu.- Quy c chon v tr ngi ng quan sat co hng nhn t di ln i vi

thanh ngang; t phai sang trai i vi thanh ng va thanh xin khi xet du ni lc(H.1d).

4. Cach xac nh ni lc (phan lc):

M > 0M > 0 M < 0 M < 0

H.1aQ > 0Q > 0 Q < 0 Q < 0

H.1b

N > 0 N > 0 N < 0N < 0

H.1c H.1d


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Ni lc (phan lc) c xac nh bng phng phap mt ct. Cac bc tinhanh nh sau:

* Bc 1: Thc hin mt mt ct qua tit din cn xac nh ni lc (qua lin ktcn xac nh phan lc). Mt ct phai chia h thanh hai phn c lp. Gi lai mt phn

bt ky.* Bc 2: Thay th tac dung cua phn h b loai bo bng cac thanh phn ni lc

(phan lc) tng ng. Cac thanh phn nay co chiu cha bit, co th gia thit co chiudng, va chung cung la cac ai lng cn tm.

* Bc 3: Thit lp cac iu kin cn bng di dang cac biu thc giai tch.Xem bang cac iu kin cn bng.

Dang h lcDang iukin cn bng H lc ng quy tai O H lc song song H lc bt ky

Dang I

X = 0; Y = 0.Yu cu: Truc Xkhng c songsong vi truc Y

X = 0; Y = 0; MA= 0.Yu cu: Truc Xkhng c song songvi truc Y

Dang II

X = 0; MA = 0.Yu cu: Truc Xkhng c vung

goc vi OA.

X = 0; MA = 0.Yu cu: Truc Xkhng c vung

goc vi phng caclc

X = 0; MA = 0;MB = 0;Yu cu: Truc X

khng c vung gocvi AB.

Dang III

MA = 0; MB = 0.Yu cu: A, B, Okhng c thnghang

MA = 0; MB =0.Yu cu: A, Bkhng c songsong vi phngcac lc

MA = 0; MB = 0;MC = 0.Yu cu: A, B, Ckhng c thnghang

Bang 1. Bang cac iu kin cn bng.* Bc 4: Giai h phng trnh cac iu kin cn bng se xac nh c cacthanh phn ni lc (phan lc). Nu kt qua mang du dng th chiu cua ni lc (phanlc) ung chiu a gia nh va ngc lai.

* V du: Xac nh cac thanh phn phan lc va ni lc tai tit din k (H.2a).1. Xac nh cac thanh phn phan lc:{ }CAA VHV ,,

X = 0 HA + P = 0 HA = -P = -2(T) < 0.MI = 0 4.VA + 4.P - 4.q.2 = 0.

4.VA + 4.2 - 4.1,2.2 = 0 VA = 0,4(T) > 0.MA = 0 -4.VC + 4.P + 4.q.2 = 0


40/139


-4.VC + 4.2 + 4.1,2.2 = 0 VC = 4,4(T) > 0* Kim tra: Y = 0 VA + VC - 4.q = 0 0,4 + 4,4 - 4.1,2 = 0 (ung)

2. Xac nh ni lc tai tit din k:{ }kkk NQM ,, Thc hin mt ct (1-1), gi lai va xet cn bng phn bn phai (H.2b).

Mk = 0 Mk + 2.q.1 - 2.VC = 0 Mk = 2.VC - 2.q.1 = 2.4,4 - 2.1,2.1 = 6,4(T.m) > 0.

Y = 0 Qk -2.q + VC = 0 Qk = 2.q - VC = 2.1,2 - 4,4 = -2(T) < 0.

X = 0 Nk = 0(T)II. Biu ni lc:1. Khai nim: Biu ni lc la th biu din quy lut bin thin cua ni lc

doc theo chiu dai cu kin.2. Cac thanh phn cua biu ni lc:- ng chun: la h truc dung dung cac tung .- Tung : tung cua biu ni lc tai mt v tr nao o la biu th cho ni

lc tai tit din tng ng.- ng biu : la ng ni cac tung .3. Cac quy c khi ve biu ni lc:

- ng chun: thng chon la ng truc thanh.- Tung phai dng vung goc vi ng chun.- Biu mmen: tung dng dng v pha di, tung m dng ln trn

ng chun. iu nay co ngha la tung dng v pha th cng.- Biu lc ct: tung dng dng ln trn ng chun va ngc lai.- Biu lc doc: tung dng thng dng ln trn dng chun va ngc

lai.- Ghi ky hiu , (Q ) vao mim dng (m) cua biu lc ct va lc doc.

- Ghi tn va n v trn cac biu a ve c.4. Cach ve biu ni lc:

A

HA

P = 2T

kB

4m

x

VAI

q = 1,2T/m

O

y VC

C

2m 2m

1

1

H.2a

k

VC = 4,4

C

Mk

QkNk

q = 1,2T/m

H.2b


41/139


Theo mn C hoc kt cu, ve biu ni lc tin hanh theo cac bc sau:* Bc 1: Xac nh cac thanh phn phan lc (nu cn).* Bc 2: Xac nh ni lc tai cac tit din c trng.- Tit din c trng: la nhng tit din chia h thanh nhng oan thanh thng

sao cho trn oan thanh o hoc la khng chu tai trong hoc la ch chu tai trong phnb lin tuc.

Nh vy, v tr cac tit din c trng thng la: nut (ni giao nhau cac thanh), v tr lc tp trung, hai u tai trong phn b, tai v tr cac gi ta....V du, vi hcho trn hnh (H.3a & H.3b), v tr cac tit din c trng la ni ghi ky hiu bng cacch hoa A, B, C, E, F.

- Xac nh ni lc: tin hanh theo nguyn tc a trnh bay. Tuy nhin, sau khiphn tch cac iu kin cn bng, ta thy co th xac nh nh sau:

+ Mmen un tai tit din k (Mk): co gia tr c xac nh bng tng mmencua tai trong tac dung ln phn h gi lai ly i vi trong tm tit din k.

+ Lc ct tai tit din k (Qk): co gia tr c xac nh bng tng hnh chiu cuacac tai trong tac dung ln phn h c gia lai ln phng vung goc vi tip tuyntruc thanh tai tit din k (phng cua Qk).

+ Lc doc tai tit din k (Nk): co gia tr c xac nh bng tng hnh chiu cuacac tai trong tac dung ln phn h c gi lai ln phng tip tuyn vi truc thanh taitit din k (phng cua Nk).

- Du cua cac ai lng trong biu thc xac nh ni lc:+ Tai trong gy cng th di tai tit din k se cho Mk mang du dng va

ngc lai.+ Tai trong tac dung ln phn bn trai co chiu hng ln hay phn bn phai co

chiu hng xung se cho Qk mang du dng va ngc lai.+ Tai trong gy keo tai tit din k se cho Nk mang du dng va ngc lai.* Bc 3: Ve biu ni lc.S dung cac lin h vi phn ve. Chi tit se c trnh bay sau bc 4.* Bc 4: Kim tra lai kt qua.Ging mn hoc Sc bn vt liu.* V du: Xac nh ni lc tai cac tit din k, m, n cua h cho trn hnh (H.4a).

D

B C

qP

A

H.3a H.3b

qP

A B

C

D E

F


42/139


43/139


Mi quan h ni lc 2 u oan thanh: trphtrph

phtrNN

s

MMQQ =

-== ;

3. Trng hp trn oan thanh chu tai phn b u: (H.5c)Tc la q = const. Nh vy, (Q) & (N) trn oan nay se la oan ng thng

c ve qua hai im; (M) se la ng parabol c ve qua ba im.

8

. 2lqf= (goi la tung treo); f treo vung goc vi ng chun va theo chiu q.

Mi quan h gia mmen va lc ct tai hai u thanh:

aa cos...2

1;cos...

2

1lq

s

MMQlq

s

MMQ

trphph

trphtr -

-=+

-=

4. Trng hp trn oan thanh chu tai trong phn b hnh tam giac: (H.5c& H.5d)

Tc la q co dang bc nht. Nh vy, (Q) & (N) trn oan nay se la oan ng

parabol c ve qua ba im; (M) se la ng bc ba, cho phep ve qua ba im.

-16

. 2lqfM = ; fM treo vung goc vi ng chun va treo theo chiu q.

- acos.8

.lqfQ = , fQ treo vung goc vi ng chun va co chiu sao cho tai v tr

q = 0, tip tuyn vi ng biu song song vi ng chun.

- asin.8

.lqfN = , fN treo vung goc vi ng chun va co chiu sao cho tai v tr

q = 0, tip tuyn vi ng biu song song vi ng chun.* Mi quan h gia mmen va lc ct tai hai u thanh:

Qtr

/2

Qph

l

Mtr

a

fMMph

/2

fQ

/2

/2

q

Q

M

Ntr

Nph

/2

/2fN

N

H.5c

Qph

Mph

/2

/2

/2Qtr

/2

fQ

l

q

Mtr

fM

a

Q

M

Nph

/2Ntr

fN/2

N

H.5d


44/139


- Khi tai phn b tam giac co ay bn phai (H.5c):

;cos..3

1;cos..

6

1aa lq

s

MMQlq

s

MMQ

trphph

trphtr -

-=+

-=

- Khi tai phn b tam giac co ay bn trai (H.5d):

;cos..61;cos..

31 aa lq

sMMQlq

sMMQ

trph

ph

trph

tr --=+-= 5. Trng hp trn oan thanh chu tai trong phn b hnh thang:Dang ng cua cac biu khng thay i so vi trng hp tai phn b hnh

tam giac. Co th a v thanh tng cua hai bai toan a bit (H.5e).

6. Trng hp trn oan thanh chu tai trong phn b quy lut bt ky:Dung cach treo biu (H.5f).- i vi (Q), (N), cach thc hin tng t.

* Cac chu y:- Trng hp tai trong

phn b theo chiu dai xincua truc thanh, co th a vtheo phng ngang bngcach chia tai trong o chocosa (H.6a).

- Tai v tr chu taitrong tp trung, ni lc co s thay i:

+ Mmen tp trung (H.6b & H.6c)

q2

a

l

q1

l

a

q1

l

a

(q2 - q1)

= +

H.5e

M M1

q

MtrNtr

Qtr

MphNph

Qph

Ntr Mtr

Qtr

Qph

NphMph q

= +

MtrMph

MtrMph

=+

H.5M2

q

b

a a

acos'

qq =

b

H.6a

a


45/139


+ Lc tp trung co phng vung goc vi truc thanh (H.6d).

+ Lc tp trung co phng trung truc thanh (H.6e).+ Lc tp trung co phng bt ky: co th a v tng cua hai bai toan (H.6f).

2. DM, KHUNG N GIAN.I. Dm n gian:1. Phn tch cu tao h:a. nh Ngha: Dm n gian la h gm mt thanh thng ni vi trai t bng

s lin kt tng ng vi ba lin kt loai mt tao thanh h BBH.b. Phn loai:- Dm n gian hai u khp. (H.7a)- Dm n gian co u tha. (H.7b)- Dm cng xn. (H.7c)

2. Xac nh cac thanh phn phan lc:Trong h dm n gian, tn lai ba thanh phn phan lc. Cach xac nh a c

trnh bay trong phn xac nh phan lc. Tuy nhin, tranh vic giai h phng trnhtoan hoc, nn thit lp sao cho trong mi phng trnh ch co mt n s. Cach thchin nh sau:

- Nu hai n con lai ng quy tai mt im I, phng trnh cn thit lp la tng

mmen toan h i vi im I bng khng. (SMI = 0)

M

a

a

Qph

Nph

Qtr

Ntr

M

Nph

Qph

Ntr

Qtr

a

Ma

M

H.6b H.6c

Qtr = Qph = tga; Ntr = Nph

H.6d H.6e

Ntr

Qtr

Qph

Nph

b

P

aP

Ntr

NphP

P

Qtr

Qph

Mph

Mtr

H.7a H.7b H.7c

H.6f

P

P2

P1 P1 P2= +

P1^ truc thanhP2 truc thanh


46/139


- Nu hai n con lai song song nhau, phng trnh cn thit lp la tng hnhchiu toan h ln phng vung goc phng hai n song song bng khng. (SZ = 0, Zco phng vung goc vi phng hai n song song)

- Nu hai n con lai la mt lc va mt mmen, phng trnh cn thit lp la

tng hnh chiu ln phng vung goc cua n lc bng khng. (SZ = 0, Z co phngvung goc vi phng n lc)

* Minh hoa:1. Xac nh phan lc cua h cho trn

hnh (H.7d):Cac thanh phn phan lc gm{ }BAA VHV ,,

- HA: X = 0 f1(HA) = 0 HA.- VA :MI = 0 f2(VA) = 0 VA.

- VB :MA = 0 f3(VB) = 0 VB.2. Xac nh phan lc cua h cho trn hnh (H.7e):Cac thanh phn phan lc gm{ }AAA MHV ,,

- HA: X = 0 f4(HA) = 0 HA.- MA :MA = 0 f5(MA) = 0 MA.- VA :Y = 0 f6(VA) = 0 VA.3. Xac nh va ve cac biu ni lc:- Xac nh ni lc tai cac tit din c trng: a trnh bay- Dng tung biu tai cac tit din c trng.- Ve biu ni lc trn tng oan thanh theo cac lin h vi phn gia ni lc

va ngoai lc.4. Kim tra lai biu ni lc: a trnh bay.

CAC V DU V DM N GIAN* V du 1:Ve cac biu ni

lc cua dm cho trn hnh (H.8a)

1. Xac nh cac thanhphn phan lc:{ }BAA VHV ,,

- SX = 0 HA = 0.- SMA = 0 8.VB - 6.30 - 10.4.2 = 0 VB = 32,5 (> 0)- SMB = 0 8.VA - 2.30 - 10.4.6 = 0 VA = 37,5 (> 0)

* Kim tra: SY = 0 - q.4 - P + VA + VB = 0

y

xH.7d

P

A

B

HAVA

VB

I

x

y

P

VA

A

BHA

MA

H.7e

q = 10kN/mP = 30kN

4m 2m 2m

A

DC B

VA

HA

VB

Q

N

M

37,52,5

32,5

657020

(kN)

(kN)

(kN.m)

y

O x

H.8a


47/139


-4.10 - 30 + 37,5 + 32,5 = 0 (ung)2. Xac nh ni lc tai cac tit din c trng:Tai A: MA = 0; QA = +VA = 37,5; NA = HA = 0.Tai B: MB = 0; QB = - VB = -32,5; NB = 0.

Tai C: MC = VA.4 - q.4.2 = 37,5.4 - 10.4.2 = 150 - 80 = 70.QC = VA - q.4 = 37,5 - 10.4 = 2,5; NC = 0.

Tai D: MD = +VB.2 = 32,5.2 = 65.Tai D co lc tp trung nn biu (Q) co bc nhay.QDC = +VA - q.4 = - 2,5; QDB = -VB = 32,5.

3. Ve cac biu ni lc cui cung:a. Biu mmen (M):- Trn oan AC co q phn b u nn co tung treo:

208

4.10

8

. 22

===lq

f - Trn cac oan con lai la nhng oan thng.b. Biu lc ct (Q):- La nhng oan ng thng.c. Biu lc doc (N):- La nhng oan ng thng.4. Kim tra lai cac biu a ve: T kim tra.* V du 2:Ve cac biu ni lc cua dm cho trn hnh (H.9a)

Quy tai trong phn b u v tacdung trn ng nm ngang:

qt = 309,230cos

2

cos==

o

q

a.

1. Xac nh cac thanh phn phanlc: { }BAA VHM ,,

- SX = 0 HA = 0.- S 0=

I

M

MA + M - 2.P - qt.2.1 = 0. MA + 3,5 - 2.3 - 2,309.2.1 = 0. MA = +7,118 (> 0)- SY = 0 VB - P - q

t.2 = 0 VB - 3 - 2,309.2 = 0 VB = +7,618 (> 0)

2. Xac nh ni lc tai cac titdin c trng:- Tai A: MA = 7,118; Q A = 0;

NA = 0.

a = 30o

HA

MA A VBI

2m 2m 2m

q = 2T/m

M = 3,5T.m

q' = 2,309T/m

P = 3T

C Dx

y

O

H.9a

(T)N

(T)

Q

(T.m)M

1,154

10,6187,118

6,5972,598

3,809

1,5

10,618


48/139


- Tai B: MB = 0; QB = -VB.cosa= -7,618.cos30o = -6,597.

NB = VB.sina = 7,618.sin30o = 3,809.- Tai C: MCA = 7,118; MCD = MA + M = 7,118 + 3,5 = 10,618.

QCA = QCD = 0, NCA = NCD = 0.- Tai D: MD = MA + M = 10,618; QDC = 0; QDB = -P.cosa = -3.cos30o = -2,598.

NCD = 0; NDB = P.sina = 3.sin30 = 1,53. Ve cac biu ni lc cui cung:a. Biu mmen (M):- Trn oan DB co qt phn b u nn co tung treo:

154,18

2.309,2

8

. 22===

lqf

td

.

- Trn cac oan con lai la nhng oan thng.

b. Biu lc ct (Q):- La nhng oan ng thng.c. Biu lc doc (N):- La nhng oan ng thng.4. Kim tra lai cac biu a ve: T kim tra.* V du 3:Ve cac biu ni lc cua dm cho trn hnh H.10* Nhn xet rng nu ta gi lai phn bn

phai khi xet cn bng mt phn h th khng

cn quan tm n phan lc.1. Xac nh ni lc tai cac tit din

c trng:Tai C: MC = 0; QC = P1 + P2.sin45

oQC = 2.(1 + sin45

o) = 3,414;NC = - P2.cos45

o = - 2.cos45o = -1,414.Tai B: MBC = -(P1 + P2.sin45

o).2 == -2.(1 + sin45o).2 = -6,828;

MBA = MBC + M = -3,828;QB = P1 + P2.sin45o = 3,414;

NB = - P2.cos45o = -1,414.

Tai A: MA = -(P1 + P2.sin45o).4 + M +

+ 2.q.1 = -(2 + 2.sin45o).4 + 3 + 2.2.1 == -6,656;

QA = P1 + P2.sin45o -2.q = 2 + 2.sin45o - 2.2 = -0,586; NA = -P2.cos45

o = -1,414.2. Ve biu ni lc cui cung:a. Biu mmen:- Trn oan AB co tai trong q phn b u, co tung treo:

N

(T.m)

Q(T)

M

B

2m 2m

P1 = 2T

P2

= 2T

45o

M = 3T.m

q = 2T/m

A C

3,414

1,4141,4141,414

0,586 3,414

16,656 6,828

3,828

(T)

H.10


49/139


50/139


- q.4 - P + VA + VD = 0 -2.4 - 3 + 8,5 + 2,5 = 0 (ung)

2. Xac nh ni lc tai cac tit din c trng:Tai A: MA = 0; QA = -HA = 0; NA = -VA = -8,5.

Tai B: MB = 0;QB = -P = -3; NB = 0.Tai C: MCA = -3.HA = 0; MCB = -2.P = -6; MCD = -2.P -3.HA = -6;

QCA = -HA = 0; QCB = -P = -3; QCD = -P + VA = -3 + 8,5 = 5,5.NCA = -VA = -8,5; NCB = 0; NCD = -HA = 0.

Kim tra s cn bng mmen nut C (H.13b).Tai D: MD = 0, QD = -VD = -2,5; ND = 0.3. Ve cac biu ni lc cui cung:

a. Biu mmen (M):

- Trn oan CD co q phn b u nn co tung treo:

48

4.2

8

. 22===

lqf

- Trn cac oan con lai la nhng oan thng.b. Biu lc ct (Q):- La nhng oan ng thng.c. Biu lc doc (N):

- La nhng oan ng thng.4. Kim tra lai cac biu a ve: T kim tra.

* V du 2:Ve cac biu ni lc cua dm cho trn hnh H.14a.

1. Xac nh cac thanh phn phan lc:{ }BCA HVV ,, - SX = 0 HB = -P

HB = -2 (< 0).- SMI = 0 -8.VC + 4.P +

+ 4.q.2 +

+ 434.

2.4 q = 0

(T.m)M

H.13c

6

4

(T)

Q

H.13d

2,5

5,5

3

(T)

N

H.13e 8,5

0

66C

H.13b


51/139


-8.VC + 4.2 + 4.1,2.2 +

+

+ 43

4.

2

2,1.4= 0

VC = 3,8 (> 0)

- SMC = 0 8.VA + 4.P -4.q.6 -

4.

3

2.

2

.4 q= 0

8.VA + 4.2 -

4.1,2.6 -

4.

3

2.

2

2,1.4= 0.

VA = 3,4 (> 0)

* Kim tra: SY = 0 VA + VC - 4.q -2

.4 q= 0

3,4 + 3,8 - 4.1,2 -22,1.4 = 0 (ung)

2. Xac nh ni lc tai cac tit din c trng:Tai A: MA = 0; QA = VA = 3,4; NA = -P = -2.Tai B: MB = 0; QB = -HB = -(-2) = 2; NB = 0Tai D: MDA = 4.VA - 4.q.2 = 4.3,4 - 4.1,2.2 = 4.

MDB = -4.HB = -4.(-2) = 8.MDE = 4.VA - 4.q.2 - 4.HB = 4.3,4 - 4.1,2.2 - 4.(-2) = 12

* Kim tra cn bng mmen nut D: (t kim tra)QDA = VA - 4.q = 3,4 - 4.1,2 = -1,4QDB = -HB = -(-2) = 2; QDE = VA - 4.q = -1,4.

NDA = -P= -2; NDB = 0; NDE = -P - HB = 0.Tai E: MED = 0; MEC = 0; QED = -VC = -3,8; QEC = 0.

NED = 0; NEC = -VC = -3,8.3. Ve cac biu ni lc cui cung:a. Biu mmen (M):- Trn oan AD co q phn b u nn co

tung treo:

4,28

4.2,1

8

. 22===

lqf

- Trn oan DE co q phn b tam giacnn co tung treo:

2,116

4.2,1

16

. 22===

lqf

- Trn cac oan con lai la nhng oanthng.

b. Biu lc ct (Q):

H.14a

4m

O

HB

B C

VC

x

q = 1,2T/m

D

y

E

4m 4m

VA

A

P = 2T

I

(T)

Q

H.14c

3,81,4

2

3,4

0,6

(T.m)M

H.14b1,2

2,4

8

4

12


52/139


- Trn oan DE co q phn b tam giac nn co tung treo:

6,01.8

4.2,1cos.

8

.=== a

lqf

- Trn cac oan con lai la nhng oan thng.

c. Biu lc doc (N):- Trn oan DE co q phn b tamgiac nn co tung treo:

00.8

4.2,1sin.

8

.=== a

lqf

- Trn cac oan con lai la nhngoan thng.

4. Kim tra lai cac biu a ve:T kim tra.

* V du 3:Ve cac biu ni lc cua dm cho trn hnh H.15a

Quy tai trong phn b u vtac dung trn ng nm ngang:

qt = 346,130cos

2,1

cos==

o

q

a.

1. Xac nh cac thanh phnphan lc:{ }CAA VHV ,,

- SX = 0 HA

= 0.- SMI = 0 6.VA - 4.qt.4 + 2.P = 0 6.VA - 4.1,346.4 + 2.2 = 0. VA = 2,922 (> 0)- SMA = 0 -6.VC + 4.q

t.2 + 8.P = 0 -6.VC + 4.1,346.2 + 8.2 = 0. VC = 4,461 (> 0)

* Kim tra: SY = 0 - qt.4 - P + VA + VC = 0 -1,346.4 - 2 + 2,922 + 4,461 = 0 (ung)

2. Xac nh ni lc tai cac tit din c trng:Tai A: MA = 0; QA =VA.cosa = 2,922.cos30o = 2,530; NA = -VA.sina = -1,461.Tai B: MB = 2.VC - 4.P = 2.4,461 - 4.2 = 0,922.

QBA = VA.cosa - 4.qtq.cosa = -2,132; QBC = P - VC = -2,461.NBA = -VA.sina +

4.qt.sina = 1,231; NBC = 0.Tai C: MC = -2.P = -2.2 = -4;

QCB = - VC + P = -

2,461;QCD = P = 2; NC = 2.

N(T) H.14d

2

3,8

H.15a

4m

A

HA

VA

I

qt = 1,346T/m

O

y

P = 2T

D

x

VC

C

2m 2m

B

q = 1, 2T/m

30

o

M(T.m)

H.15b2,692

0,922

4


53/139


Tai D: MD = 0; QD = P = 2; ND = 0.Kim tra s cn bng mmen nut B: T kim tra3. Ve cac biu ni lc cui cung:a. Biu mmen (M):

- Trn oan AB co qt

phn b u nn co tung treo:692,2

8

4.346,1

8

. 22===

lqf

td

.

- Trn cac oan con lai la nhng oan thng.b. Biu lc ct (Q):- La nhng oan ng thng.c. Biu lc doc (N):- La nhng oan ng thng.

4. Kim tra lai cac biu a ve: T kim tra.

3. H BA KHP.I. Phn tch cu tao h:1. nh ngha: H ba khp la h gm hai ming cng

ni vi nhau bng mt khp va lin kt vi trai t bng haikhp (gi c nh) tao thanh h BBH.

2. Tnh cht cua h ba khp:- Trong h lun tn tai thanh phn phan lc nm ngang ngay ca khi tai trong ch

tac dung theo phng thng ng.- Ni lc trong h ba khp (mmen un va lc ct) noi chng la nho hn trong

h n gian cung nhp, cung chu tai trong.3. Phn loai h ba khp:a. Vom ba khp: Khi cac ming cng cua h la nhng thanh cong (H.17a).

Trong vom ba khp, noi chung phat sinh y u ba thanh phn ni lc.b. Khung ba khp: Khi cac ming cng cua h la cac thanh gay khuc (H.17b).

Trong khung ba khp, noi chung phat sinh y u ba thanh phn ni lc.

c. Dan ba khp: Khi cac ming cng cua h la nhng dan phng tnh nh(H.17c). Trong dan ba khp, cac thanh ch tn tai lc doc.

Q(T)

H.15c

2,530

2

2,4612,132

(T)N H.15d

1,461

1,231

(B)(C)

H.16


54/139


55/139


Trong trng hp tai trong ch tac dung theo phng thng ng, d thy rngcac thanh phn dB

d

A VV , ging nh cac phan lc trong dm n gian tng ng. V vy

cac phan lc nay goi la phan lc dm va c ky hiu nh trn.b. Xac nh BA ZZ , :

Ct qua C, gi lai phn bn traiva vit phng trnh cn bng mmen i vi C.

SMC = 0 -ZA.h +tr

CM = 0 ZA =h

Mtr

C .

Trong o, trCM la tng mmen cua cac lc tac dung l phn h bn trai C, khng

k ZA. Trong biu thc xac nhtr

CM , cac ngoai lc lam cho phn h xoay thun chiu

kim ng h quanh C ly du dng.

Tng t, xet cn bng mmen cho phn h bn phai C ZA =h

Mph

C . phCM la

tng mmen cua cac lc tac dung ln phn h bn phai C, khng k ZB. Trong biuthc xac nh phCM , cac ngoai lc lam cho phn h xoay ngc chiu kim ng h

quanh C, ly du dng.- Cac thanh phn BA ZZ , goi la phan lc vom.

2. Phn tch theo phng thng ng va phng ngang:AAA HVR & ; BBB HVR &

a. Xac nh BA HH , :T quan h hnh hoc trn hnh ve. HA = ZA.cosb; HB = ZB.cosb.

- HA, HB goi la cac lc x.- Trong trng hp tai trong tac dung theo phng ng, d thy H A = HB = H,

nn ZA = Z B = Z.b. Xac nh BA VV , :

T quan h hnh hoc trn hnh ve.VA =

d

AV + ZA.sinb; VB =d

BV - ZB.sinb.

Hay VA =d

AV + HA.tgb; VB =d

BV - HB.tgb.

3. Xac nh cac phan lc toan phn BA RR , :

AAAdAA HVZVR +=+=

BBB

d

BB HVZVR +=+=

V gia tr ( ln): 22 AAA HVR += ;22

BBB HVR +=

* Chu y:- b > 0 nu gi B cao hn gi A va ngc lai.- Co th xac nh c BBAA HVHV &,, thng qua giai h phng trnh:

+ Vit phng trnh cn bng mmen toan h i vi gi B:SMB = 0 f1(VA, HA) = 0 (a)

+ Tach qua C, vit phng trnh cn bng mmen cua na h bn trai i vi C:


56/139


S trCM = 0 f2(VA, HA) = 0 (b)

Giai h phng trnh (a), (b) se c { }AA HV ,

+ Tng tSMA = 0 f3(VB, HB) = 0 (c)

S phCM = 0 f4(VB, HB) = 0 (d)Giai h phng trnh (c), (d) se c { }BB HV ,

- Nu h ba khp s dung cac khp gia tao, phn tch cac phan lc xut hin taicac lin kt vit phng trnh cn bng hp ly. Xem v du h trn hnh (H.19).

- SU = 0 f1(VA, HA) = 0 (1)- S trY = 0 f2(VA, HA) = 0 (2)

Giai h (1), (2) xacnh c (VA, HA). Va xac nh (RB, MB), phn tch phanlc thit lp cp phng trnh tng t.

III. Xac nh ni lc tai cac tit din c trng va ve biu ni lc:

y i trnh bay cho h vom va khung ba khp.

1. Biu thc mmen un (Mk):Gia s cn xac nh mmen un tai tit din k cua vom ba khp chu tai trong

tac dung thng ng nh trn hnh ve. (H.20a)

Dung mt ct qua k, gi lai va xet cn bng phn bn trai.Mk =d

AV .zk - P1.a - ZA.hk (a)

B

A

VAHA MB

RB

P1

u

H.19aVA

HA A

P1

H.19b

C

HC

MC

y

a

A

AZA

P1

P1

Bh

P2

H.20a

b

P3

B

P2C

P3

dAV

d

BV

ZB

y

x

f

k

k

a

yk hkb

ak

d

AV

d

BV

k

A

A

ZA

P1

d

AV

dAV

bhkyk

P1 k

b x

ak

y

d

kM

d

kQ

Mk

QkNk

H.20b

zk zk

d

kN


57/139


Qk =d

kQ .cosak - H.(sinak - tgb.cosak)

Mk =d

kM - H.yk.

Nk = -d

kQ .sinak - H.(cosak + tgb.sinak)

Mk =Mk =d

kM - H.yk.

Qk =d

kQ .cosak - H.sinak.

Nk = -d

kQ .sinak - H.cosak.

Goi dkM la mmen un tai tit din k trn dm n gian cung nhp, cung chu tai

trong.D thy

d

kM =d

AV .zk - P1.a (b)

Ta bit: ZA =bbb coscoscos

HHH BA == ; ma yk =bcos

kh

Suy ra ZA.hk = H.yk (c)T (a), (b), (c) suy ra:Biu thc chng to rng mmen un trong vom ba khp nho hn mmen un

trong dm n gian cung nhp, cung chu tai trong mt lng H.yk. Va nu kheo chonhnh dang cua vom (yk) sao cho

d

kM = H.yk th mmen un tai moi tit din u bng

khng. Luc nay trong vom ch tn tai lc doc nn tit kim vt liu.

2. Biu thc lc ct (Qk):Tng t nh trn nhng i thit lp phng trnh hnh chiu ln phng Qk

(phng vung goc vi tip tuyn truc vom tai tit din k).

Trong o- dkQ : lc ct tai tit din k trong dm n gian tng ng cung nhp, cung chu

tai trong.

- Qk: lc ct trong vom tai tit din k.- a: goc hp bi tip tuyn vi truc vom tai tit din k vi phng ngang.

3. Biu thc lc doc (Nk):Tng t nh xac nh lc ct nhng i thit lp phng trnh hnh chiu ln

phng Nk (phng cua tip tuyn truc vom tai tit din k).

* Chu y:- b > 0 khi gi B cao hn gi A va ngc lai.- ak > 0 khi y'(zk) > 0 va ngc lai.- yk = y(zk) - zk.tgb.- Khi b = 0 (gi A & B cung cao )

- Cac biu thc trn c thit lp cho tai trong tac dung theo phng thngng.

3. Ve cac biu ni lc cui cung:- i vi khung ba khp: Tin hanh ging h dm, khung n gian.


58/139


- i vi vom ba khp: Sau khi chon va xac nh ni lc tai cac tit din trnkt cu. Cac tit din thng chon la cac tit din c trng va mt s tit din trunggian tng tnh chnh xac. Biu ni lc c ve gn ung bng cach ni cac tung lin tip bng cac oan thng.

Qua trnh tnh toan co th lp thanh bang sau: (Bang tham khao)Titdin

zk yk tgak = y'(zk) sinak cosakd

kM d

kQ Mk Qk Nk

- - - - - - - - - - -

Bang 2. Bang phn tch ni lc trong vom ba khp.* Chu y: Co th chon ng chun la ng nm ngang khi ve bii ni lc.* Chu thch: i vi h dan vom ba khp, cach tnh c thc hin nh sau:+ Xac nh phan lc tai cac gi ta theo cach a trnh bay trn.

+ Ni lc trong cac thanh dan ch la lc doc. Xem cach xac nh trong bai hdan.

CAC V DU V H BA KHP.* V du 1: Xac nh ni lc tai tit din k cua vom ba khp cho trn hnh ve (H21)

Cho bit phng trnh truc vom

la parabol y(z) = mlzzl 10;).(5

1=-

1. Cac s liu suy ra t bai:

-z

k =3m; yk = y(3) = 4,2m; b = 0.- tgak = y'(3) = 3).2(

5

1=- Zzl = 0,8.

sinak =

= 624,08,01

8,0

1 22=

+=

+ k

k

tg

tg

a

a

cosak =

= 780,08,01

1

1

1

22=

+=

+ktga

2. Xac nh cac thanh phn phan lc:{ }BABABAdBdA HHVVZZVV ,,,,,,, Do b = 0 va tai trong ch tac dung theo phng thng ng nn

BABAB

d

BA

d

A HHZZVVVV ===== ,, .

a. Xac nh VA, HA:SMB = 0 10.VA - 5.8 - 5.2.2,5 = 0 VA = 6,5 (> 0).S trCM = 0 5.VA - 5.HA - 5.3 = 0 5.HA = 5.6,5 - 5.3 HA = 3,5 (> 0).

b. Xac nh VB, HB:SMA = 0 -10.VB + 5.2 + 5.2.7,5 = 0 VB = 8,5 (> 0).S phCM = 0 -5.VB + 5.HB + 5.2.2,5 = 0 5.HB = 5.8,5 - 5.2.2,5.

H.21

2m 5m2m1m

10m

y

zVA

HA

A

VB

HBB

P = 5T

akk

q = 2T/m

C

5m


59/139


HB = 3,5 (> 0).* Kim tra: SX = 0 HA - HB = 0 3,5 - 3,5 = 0 (ung)

SY = 0 VA + VB - P - 5.q = 0 6,5 - 8,5 - 5 - 5.2 = 0 (ung)* Kt lun: 5,6== dAA VV ; 5,8==

d

BB VV ; ZA = ZB = HA = HB = H = 3,5.

3. Xac nh ni lc tai cac tit k:Mk =Mk =

d

kM - H.yk = 6,5.3 - 5.1 - 3,5.4,2 = - 0,2 (d

kM = 6,5.3 - 5.1).

Qk =d

kQ .cosak - H.sinak = (6,5 - 5).0,780 - 3,5.0,624 = -1,014.

Nk = -d

kQ .sinak - H.cosak = -(6,5 - 5).0,624 - 3,5.0,780 = -3,666.

* V du 2: Ni dung va hnh ve ging v du 1 nhng P nghing 1 goc 450(H.22a)

Cac s liu suy ra t bai toanging v du trn.

1.Xac nh cac thanh phnphan lc:

{ }BABABAdBdA HHVVZZVV ,,,,,,, Do b = 0 nn

BBAAB

d

BA

d

A HZHZVVVV ==== ;,,

. SMB = 0 10.VA - P.sin45o.8 ++ P.cos45o.3,2 - 5.q.2,5 = 0.

10.VA - 5.sin45o.8 +

+ 5.cos45o

.3,2 - 5.2.2,5 = 0 VA = 4,197(>0)

S trCM = 0 5.VA - 5.HA - P.sin45o.3 -P.cos45o.(5 - 3,2) = 0.

5.4,197 - 5.HA - 5.sin45o.3 - 5.cos45o(5 - 3,2) = 0 HA = 0,802.

SMA = 0 -10.VB + P.sin45o.2 + P.cos45o.3,2 + 5.q.7,5 = 0. 10.VB + 5.sin45o.2 + 5.cos45o.3,2 + 5.2.7,5 = 0. VB = 9,338 (> 0).

Sph

CM = 0 -5.VB + 5.HB + 5.q.2,5 = 0.

-5.9,338 + 5.HB + 5.2.2,5 = 0 HB = 4,338.

* Kim tra: SX = 0 HA - HB + P.cos45o = 0 0,802 - 4,338 - 5.cos45o = 0

(ung)SY = 0 VA + VB - P.sin45

o - 5.q = 0.

5m

2m

10m

ak

HAVA

A HB

2m1m 5m

H.22a

VB

Bz

P = 5Ty

q = 2T/m

Ck45o

VB.cosak

B

HB.sinak

kak

MkNk

Qk

qtqt.cosak

qt.sinak

HB

HB.cosak

VB.sinakVB

H.22b


60/139


61/139


Titdin

zk(m)

yk(m)

tgak =y'(zk)

sinak cosakd

kM

(T.m)

d

kQ

(T)

Mk(T.m)

Qk(T)

Nk(T)

A 0 0 1,60 0,848 0,529 0 3,75 0 0,66 -4,01

1 1,0 1,44 1,28 0,788 0,615 3,75 3,75 1,50 1,08 -3,92

2 2,0 2,56 0,96 0,721 0,692 7,50 3,75 3,50 1,47 -3,783tr 2,5 3,00 0,80 0,624 0,781 9,37 3,75 4,69 1,95 -3,56

3ph 2,5 3,00 0,80 0,624 0,781 9,37 -1,25 4,69 -1,95 -0,44

4 3,0 3,36 0,64 0,539 0,642 8,75 -1,25 3,50 -1,88 -0,65

5 4,0 3,84 0,32 0,305 0,952 7,50 -1,25 1,50 -1,68 -1,10

C 5,0 4,00 0 0 1,000 6,25 -1,25 0 -1,25 -1,56

6 6,0 3,84 -0,32 -0,305 0,952 5,00 -1,25 -1,00 -0,71 -1,87

7 7,0 3,36 -0,64 -0,539 0,842 3,75 -1,25 -1,50 -0,81 -2,00

8 8,0 2,56 -0,96 -0,721 0,692 2,50 -1,25 -1,50 -0,25 -2,00

9 9,0 1,44 -1,28 -0,788 0,615 1,25 -1,25 -1,00 0,46 -1,95

B 10 0 -1,60 -0,848 0,529 0 -1,25 0 0,67 -1,89

B3. Bang minh hoa tnh toan* V du 4: Ve cac biu ni lc cua h trn hnh ve (H.24a).1. Xac nh cac thanh phn

phan lc:{ }BBAA HVHV ,,, SMB = 0 -1.HA + 5.VA -

- 4.q.5 - 3.P + M = 0 -1.HA + 5.VA -

- 4.2.5 - 3.4 + 3,2 = 0 -HA +5.VA -48,8 = 0 (a)S trCM = 0 -4.HA + 2.VA -

- 4.q.2 = 0 -4.HA + 2.VA - 4.2.2 = 0.

-4.HA + 2.VA -16 = 0 (b)T (a), (b) giai ra VA = 9,955; HA = 0,977.SMA = 0 -1.HB - 5.VB + 2.P + M = 0

-1.HB - 5.VB + 2.4 + 3,2 = 0 -HB - 5.VB + 11,2 = 0 (c)

S phCM = 0 3.HB - 3.VB + M = 0

3.HB - 3.VB + 3,2 = 0.(d)T (c), (d) giai ra VB = 2,044; HB = 0,977.

* Kim tra:SX = 0 HA - HB = 0 (ung)

2m

A

HAVA

ED

O

y

xVB

F

q = 2T/m P = 4T

4m

2m 3m

3mC

B

M = 3,2T.m

HBH.24a


62/139


SY = 0 VA + VB - P - 4.q = 0 (ung)2. Xac nh ni lc tai cac tit din c trng:Tai A: MA = 0; QA = -HA = - 0,977; NA = -VA = -9,955.Tai D: MD = 0; QD = 0; ND =0.

Tai E: MED = -2.q.1 = -2.2.1 = -4; MEA = -4.HA = -4.0,977 = -3,911.MEC = -2.q.1 - 4.HA = -2.2.1 - 4.0.977 = -7,908.QED = -2.q = -2.2 = -4; QEA = -HA = -0,977;QEC = -2.q + VA = -2.2 + 9,955 = 5,955.

NED = 0; NEA = -VA = -9,955; NEC = -HA = -0,977.Tai C: MC = 0; QCE = VA - 4.q = 9,955 - 2.4 = 1,955; QCF = -VB = -2,044.

NC = -HA = -0,977.Tai B: MB = 0; QB = HB = 0,977; NB = -VB = -2,044.Tai F: M

FB= 3.H

B= 3.0,977 = 2,91; M

FC= - 3.H

B- M = -3.0,977 - 3,2 = -6,11.

QFB = HB = 0,977; QFC = -VB = -2,044.NFB = -VB = -2,044; NFC = -HB = -0,977.

3. Ve cac biu ni lc cuicung:

a. Biu mmen (M):- Trn oan DE & EC co q phn b

u nn co tung treo:

18

2.2

8

. 22

===

lq

f - Trn cac oan con lai la nhng

oan thng.b. Biu lc ct (Q):- La nhng oan ng thng.

c. Biu lc doc (N):- La nhng oan ng thng.4. Kim tra lai cac biu a ve: T kim tra.

&. Phn tch trng hp h ba khp co thanh cng:1. Trng hp tai trong khng tac dung ln thanh cng: Trong thanh cngch xut hin lc doc.

6,11

2,91

7,9084

3,911

1 1

(T.m)

M

H.24b

0,977 0,977

1,9555,955

2,0444

(T)Q

H.24c 9,955 2,044

0,977

(T)N

H.24d


63/139


a. Xac nh phan lc:- Cac thanh phn VA, VB HA: Xac nh nh trong trng hp dm, khung n

gian.- Thanh phn lc doc NDE: Dung mt ct qua khp C va thanh cng. Xet cn

bng mmen cua mt phn bt ky i vi khp C.S trCM = 0 hay S

ph

CM = 0

*Nhn xet:- VA, VB ong vai tro nh

d

B

d

A VV , .

- NED ong vai tro nh ZA, ZB.- H ging h ba khp A*CB*.b. Xac nh ni lc:Theo nguyn tc chung xac nh ni lc

a trnh bay.* Chu y: Nu thanh cng la ming cng

gay khuc: Lc xut hin trong thanh cng lalc i qua hai khp hai u, ngc chiunhau, bng nhau v gia tr.

2. Trng hp tai trong tac dung lnthanh cng: Trong thanh cng xut hin ca bathanh phn ni lc M, Q, N.

a. Xac nh phan lc:- Cac thanh phn VA, VB, HA: ging nh

trong dm, khung n gian.- Cac thanh phn tai khp D, E: c

xac nh qua hai bc.+ Bc 1: Tach ring ED ra khoi h va

xet cn bng no.Phn tch phan lc tai D va E theo hai

phng: phng DE (ZD, ZE) va phng ng

(VD, VE).Vit cac phng trnh cn bng mmen

i vi D va E se giai ra c VD, VE.+ Bc 2: Xet cn bng cua phn h

ABC (sau khi a tach bo thanh DE).Truyn phan lc tai D va E tai hai u

thanh cng vao (co chiu ngc lai).Thc hin mt ct qua khp C. Gi lai va xet cn bng tng phn cua h.

Str

CM = 0 ZD vaSph

CM = 0 ZE.b. Xac nh & ve biu ni lc: Theo nguyn ly chung.

A D

H.25b

EB

P1C P2

VAHA VB

NDENDE

1

1

A*B*

A B

C

DE

P1 P2

VBVAHA

H.25a

A DE B

P1 CP2

P3

VBVA

HA

EP3D

VD

VEZE

ZD

H.26a

H.26b

P1

DA

HAVA

B1

H.26c VB

1 P2C

E

ZDZE

VEVD


64/139


* V du 5: Ve cac biu ni lc cua h trn hnh ve (h.27a).

1. Xac nh cac thanh phn phan lc:{ }BAA VHV ,, - SX = 0 HA = 0.- SMB = 0 8.VA - 3.4 -2.4.6 = 0 VA = 7,5.(> 0)

- SMA = 0 -8.VB + 3.4 + 2.4.2 = 0 VB = 3,5 (> 0).* Kim tra: SY = 0 - q.4 - P + VA + VB = 0 -2.4 - 3 + 7,5 + 3,5 = 0

(ung)* Xac nh lc doc trong

thanh DE: NDES phCM = 0

2.NDE - 4.VB = 0 NDE = 1,75

(> 0, gy keo)2. Xac nh ni lc tai cac

tit din c trng:Trong thanh DE ch tn tai

lc doc NDE = 1,75.Tai A: MA = 0; QA = VA = 7,5; NA = 0.Tai F: MFA = 2.VA - 2.q.1 = 11; MFD = -2.NDE = -3,5.

MFC = 2.VA - 2.q.1 - 2.NED = 7,5.QFA = VA -2.q = 7,5 - 2.2 = 3,5; NFA = 0.QFD = - NED = -1,75; NFD = 0.QFC = VA - 2.q = 3,5; NFC = -NDE = -1,75.

Tai C: MC = 0; QCF = -VB + P = -3,5 + 3 = -0,5; QCG = -VB = -3,5.NC = - NED = -1,75.

Tai G: MGC = 2.VB - 2.NDE = 3,5.2 - 1,75.2 = 3,5; MGE = 2.NED = 3,5.MGB = 2.VB = 7.QGC = - VB = -3,5; NGC = -NED = -1,75; QGE = NED = 1,75; NGE = 0.QGB -VB =3,5; NGB = 0.

Tai D: MD = 0; QDF = -NED = -1,75; NDF = 0.Tai E: ME = 0; QEG = NED = 1,75; NEG = 0;Kim tra s cn bng mmen nut F & G: d thy thoa man.3. Ve cac biu ni lc cui cung:

a. Biu mmen (M):- Trn oan AF & FC co q phn b u nn co tung treo:

B

2m2m

VA

2m2m

VB

P = 3T

O

q = 2T/m

HA

A

y

F

x

C G2m

EDH.27a

q = 2T/m

VAHA

A

D E

C

P = 3T

F G

VB

B

1

1

NDE NDEH.27b


65/139


18

2.2

8

. 22===

lqf

- Trn cac oan con lai la nhngoan thng.

b. Biu lc ct (Q):- La nhng oan ng thng.

c. Biu lc doc (N):

- La nhng oan ng thng.4. Kim tra lai cac biu a ve: T kim tra.V du 5: Ve cac biu ni lc cua h trn hnh ve (H.28a)1. Xac nh cac thanh

phn phan lc:- D thy VA = VB = 6; HA

= 0.- Xac nh cac phan lc tai

v tr ming cng ni thanh cng.Tach thanh cng GH. Cac thanhphn phan lc gm: (VG, HG, VH,HH).

+ Xet cn bng thanh cngGH:

D thy VG = VH = 6.+ Xet cn bng phn h bn

di:Dung mt ct 1 - 1, xet cn

bng phn bn trai:S trCM = 0 4.VA -

2.VG - 2.HG = 0 HG = 6.Tng t, xet cn bng

phn bn phai:S phCM = 0 HH =

6.2. Xac nh ni lc tai cac tit din c trng:

3,5 3,5

7,5113,5 71 1

H.27c

(T.m)

M

VA

A

O

y

x

C

P = 2T

D E

2m

VB

B

2m2m

HA

2m 2m

P = 2T q = 2T/m

KHGF

H.28a

G

HAVA

AD

P = 2T

F

C

VB

E B

q = 2T/m

H

P = 2T

KG HHG VG VHHH

VG

HG

VH

HH

1

1

H.28b

1,75

1,75H.27e N

(T)

3,53,5

7,5

3,50,5

1,75H.27d

(T)

Q


66/139


thun li cho vic xac nh ni lc, ta tach ring h ra lam hai phn c lpnh luc xac nh phan lc lin kt trong cac thanh cng.

Tai A: MA = 0; QA = VA = 6; NA = 0.Tai F: MF = 0; QF - P = -2; NF = 0.

Tai G: MGF = -2.P= -2.2 = -3; MGH = -2.P= -4; MGD = 0.QGF = -P = -2; QGH = VG - P = 6 - 2 = 4; Q GD = -HD = -6.

NGF = 0; NGH = -HG -6; NGD = -VG = -6.Tai D: MDA = 2.VA = 2.6 = 12; MDH = 2.HG = 2.6 =12; MDC = 2.VA - 2.HG =0.

QDA = VA =6; QDG = -HG = -6; QDC = VA - VG =0.NDA = 0; NDG = -VG = -6; NDC = HG = 6.

Tai C: MC = 0; QC = VA - VG = 0; NC = HG = 6.Tng t cho cac tit din con lai.

3. Ve cac biu ni lc cuicung:a. Biu mmen (M):- Trn oan GH co q phn b u

nn co tung treo:

48

4.2

8

. 22===

lqf

- Trn cac oan con lai la nhng oan thng.b. Biu lc ct (Q):

- La nhng oan ng thng.

c. Biu lc doc (N):- La nhng oan ng thng.4. Kim tra lai cac biu a ve: T kim tra.IV. Truc hp ly cua vom ba khp:1. t vn : Mmen un tai tit din k trong vom ba khp c xac nh

bng biu thc:Mk(z) =

d

kM - H.yk.

Nu ta kheo chon hnh dang cua vom (yk) sao cho H.yk =d

kM th Mk(z) = 0. Va

Qk(z) = 0 v Qk(z) = 0)(

=dz

zdMk.

6

4

6

6

2

4 2

(T)

Q

H.28d

6

6

6 6

(T)

N

H.28e

44 4

12 12

(T.m)M

H.28c


67/139


Luc nay trong vom ch tn tai lc doc Nk nn tit kim vt liu ch tao vom.Tht vy, bng cach so sanh s phn b ng sut phap trn tit din vom trong haitrng hp.

+ Trng hp M = 0, ng sut phn b u. Vt liu tai moi tit din c s

dung nh nhau nn phat huy ht kha nng lam vic cua vt liu. Ngha la tit kim vtliu.

Ngoai ra, lc doc trong vom thng gy nen nn nu ch co N 0, ta co th sdung nhng loai vt liu chu nen tt nhng re tin nh b tng, gach, a...*Kt lun:Truc hp ly cua vom la truc c chon sao cho mmen un tai tt ca

cac tit din cua vom u bng khng.2. Cac dang truc vom hp ly:a. Trng hp tai trong thng ng khng phu thuc vao hnh dang cua

vom:Khi ma s thay i cua trong lng ban

thn va cac tac dung bn ngoai khi truc vom

thay i la khng ang k . V du h trn hnhH.30, cac lc P1, P2, q truyn qua h thng cacthanh ng ln vom la khng thay i khi trucvom thay i.

Nh vy, dkM se khng thay i. Suy ra:

yk =H

Md

k (a)

b. Trng hp tai trong thng ng phu thuc dang truc vom:

Luc nay khng th s dung biu thc (a) v dkM phu thuc vao yk. Do yk thayi th tai trong thay i nn dkM se thay i.

Ly vi phn hai ln biu thc (a)

H

q

dz

yd=

2

2

(b)

Giai phng trnh (b) se xac nh c yk nhng noi chung la rt phc tap.c. Trng hp tai trong tac dung vung goc vi truc vom: Nh cac cng

trnh lam vic trong mi trng cht long, cht kh.

i khao sat cn bng mt oan v cung be cua truc vom hp ly.

M

N1

smax

s

N2

smax

smaxN1< N2 H.29

s < smax

H.30B

C

A

P2q

P1


68/139


Ly tng mmen i vi tm OMo = 0 N.r - (N + dN).r = 0 dN = 0

N = const.Ly tng hnh chiu ln phng U:

U = 0 0.2sin.

2sin. =-+ dsqdNdN aa

Do da la VCB nn22

sinaa dd

=

N.da - q.ds = 0.

Mc khac: ds = r.da. Nnq

N=r .

Trng hp q = const (phn b u)r = const (cung tron)

ds

Oda

r

u

N

N+dN

H.31


69/139


4. H DANI. Phn tch h:1. nh ngha: la h gm cac thanth thng lin kt vi nhau ch bng cac khp

ly tng hai u mi thanh tao thanh h BBH.2. Cu tao cua dan:- Khoang cach gia hai gi ta goi la nhp dan.- Cac khp cua dan goi la cac mt dan.- Cac thanh dan nm trn ng bin dan goi la cac thanh bin (gm bin trn

va bin di).- Cac thanh dan nm bn trong bin goi la cac thanh bung (gm thanh ng va

thanh xin).- Khoang cach gia hai mt dan thuc cung mt ng bin goi la t.

2. Cac gia thit tnh dan:- Mt dan phai nm tai giao im cua truc cac thanh dan va la khp ly tng (coth xoay t do, khng ma sat).

- Bo qua trong lng ban thn cua cac thanh dan.- Tai trong ch tac dung ln mt dan.Vy cac thanh dan lam vic nh cac lin kt thanh, ngha la ch tn tai lc doc.4. c im cua h dan:- Tit kim vt liu.

- Trong lng ban thn be.- Co th vt qua c nhng nhp ln.- Kho thi cng, lp dng.II. Xac nh ni lc trong cac thanh dan:Co nhiu phng phap khac nhau. y ch trnh bay phng phap giai tch va

phng phap giai.1. Phng phap giai tch:a. Phng phap tach mt: Ni dung cua phng phap la i khao sat s cn

bng cua tng mt c tach ra khoi dan. Thc ra, y la trng hp c bit cua

phng phap mt ct vi h lc khao sat la h lc ng quy.

H.32

nhp dan

t

Mt Bin trn

Bin di

Thanh ng

Thanh xin


70/139


Cac bc tin hanh nh sau:- Xac nh cac thanh phn phan lc (nu cn).- Ln lt tach cac mt ra khoi dan bng cac mt ct quanh mt.- Thay th tac dung cua thanh dan b ct bng lc doc trong thanh o. Luc u,

cac lc doc cha bit, gia thit co chiu dng (ve hng ra ngoai mt).- Khao sat s cn bng cua tng mt: Lc tac dung ln mt gm ngoai lc tp

trung (nu co) va lc doc trong cac thanh dan. y la h lc ng quy nn thng sdung hai phng trnh hnh chiu theo hai phng khng song song.

=S

=S

0

0

Y

X

- Khao sat cn bng cho tt ca cac mt, se c h thng cac phng trnh. Giaih phng trnh se xac nh c cac lc doc cn tm. Nu kt qua mang du dng th

lc doc gy keo (ung chiu a gia nh) va ngc lai.* Minh hoa: Tach va xet cn bng mt s 7 cua h dan trn hnh (H.33a).

Hai phng trnh cn bng hnh chiu theo hai phng co th thit lp:X = 0 N7-8 + N7-4.cosa - N7-6 - N7-2.cosa = 0.X = 0 N7-3 + N7-2.sina + N7-4.sina - P = 0.

* tranh giai h phng trnh toan hoc, ta i thit lp iu kin cn bng saocho trong mi phng trnh ch cha mt n s. Cach tin hanh nh sau:

- Tach mt theo th t sao cho tai mi mt ch co ti a hai n s cha bit.- tm lc doc trong thanh cha bit th nht, ta thit lp phng trnh cn

bng hnh chiu ln phng vung goc vi thanh cha lc doc cha bit th hai.

* Minh hoa: Tr lai v du cho trn hnh(H.33a) ta co th tach mt theo th t: 1 6 2 3...Chn han, tach mt 1:

X = 0 N1-2.sina + 1,5.P = 0

N1-2 =asin

5,1- (< 0, gy nen).

Z = 0 N1-6.sina - 1,5.P.cosa = 0

N1-6 =a

a

sin

cos..5,1 P(> 0, gy keo).

* Cac h qua rut ra t phng phap tach mt:

3

2

1

6 78

4

5B

A PPP 1 1

VA =1,5P

HA =0

VB =1,5P

P7N7-8

N7-4N7-3

N7-2

N7-6

a a

y

O xH.33aH.33b

VA =1,5P

HA = 0

1N1-6

N1-2a

O

z

y

H.33c

A

(Z ^ thanh 1-2)


71/139


+ H qua 1: Nu mt mt ch co hai thanh khng thng hang (b i) va khngchu tai trong tac dung th lc doc trong hai thanh o bng khng (Cn bng mt trnhnh H.34a, N1 = N2 = 0).

+ H qua 2: Nu

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