san suon toan khoix (1)

Chng 6Sn sn ton khiSn sn ton khi

(Tnh theo cng )

6.1 Khi nim chung

6.2 Sn sn ton khi c bn loi dm

6.3 Sn sn ton khi c bn k 4 cnh

6.4 Cc loi sn khc

6.1.1 nh ngha v phn loi

6.1.2 Phn bit bn loi dm v bn k 4 cnh

6.1. Khi nim chung6.1.1. nh ngha v phn loi

6.1.3 Ti trng tc dng ln cc phng ca bn

Phn loi.- Theo phng php thi cng:+ Sn bn lp ghp

Khi nim chungSn l mt kt cu chu lc trc tip ca ti trng s dng tc dng ln cng trnh, ti trng nyc truyn ln dm ri t dm truyn ln ct ri xung mng.

2

+ Sn bn lp ghp+ Sn lp ghp+ Sn ton khi.- Theo s kt cu+ Bn c dm - Bn mt phng

- Bn hai phng- Sn c

+ Bn khng c dm (sn nm).

6.1 Khi nim chung



6.4 Cc loi sn khc



6.1.2. Phn bit bn chu un mt phng (bn loi dm) vi bn chu un hai phng (bnk 4 cnh)Tnh cht lm vic ca bn ph thuc vo kch thc bn v kiu lin kt.


Bn chu un mt phng (bn loi dm)Khi bn ch c lin kt 1 cnh hoc 2 cnh i din, ti trng tc dng ln bn ch c truyntheo phng c lin kt hay l bn ch lm vic theo mt phng ta gi l bn loi dm.

3Hnh 6.1 S tnh ca bn loi dm

6.1 Khi nim chung



6.4 Cc loi sn khc



Bn chu un hai phng (bn k 4 cnh)Khi bn c lin kt c 4 cnh (hoc 2 hoc 3 cnh khng i din) ti trng c truyn theoc hai phng. Ta gi loi bn ny l bn k 4 cnh (lm vic theo 2 phng).


4

Hnh 6.2 S tnh ca bn k 4 cnh

6.1 Khi nim chung



6.4 Cc loi sn khc




6.1.3 Xc nh ti trng truyn theo mi phng- Vi bn lm vic theo 1 phng, vic xc nh ti trng, ni lc c tin hnh nh trong dm(bng cch ct mt di bn rng 1m ri tin hnh dn ti v tnh ton).- Bn k 4 cnh: xt mt bn k 4 cnh chu ti trng phn b u q,gi ti trng truyn theo phng cnh b l1 l q1gi ti trng truyn theo phng cnh ln l2 l q2Ta c q1 + q2= q (1)

Trng hp cc cnh k

5

=

41 1

15

384q lfEJ

=

42 2

25

384q lfEJ

Trng hp cc cnh kCt mt di bn c b rng bngn v ti chnh gia bn theo haiphng. vng ti chnh giami di

+ theo phng l2

+ theo phng l1

6.1 Khi nim chung



6.4 Cc loi sn khc




4 42 1 1 1 2 2 (2)f f q l q l= =

qll

lq 42

41

42

1+

= qll

lq 42

41

41

2+

=

42

1 241

lq ql

=

= =

21max 2

2max 1

9M lM l

im gia hai bn c cng vng

T (1), (2)

Khi l2 > l1 th q1 > q2. Nu l2/l1 > 3 q1 > 81q2 hay

Chng t ti trng ch yu truyn theo phng cnh ngn l1 (gy un theo phng cnh ngn l1),M kh b so vi M , c th b qua s lm vic theo cnh di v tnh ton nh bn mt phng.

6

M2 kh b so vi M1, c th b qua s lm vic theo cnh di v tnh ton nh bn mt phng.Trong tnh ton thc hnh c th tnh ton theo bn mt phng khi l2/l1 2. Khi l2/l1 < 2 cntnh bn lm vic theo hai phng (bn k bn cnh).

Trng hp cc cnh c lin kt bt kTng tng ly hai di bn vung gc vi nhau v tnh vng ca hai di bn im giaonhau. Dng iu kin vng ti im giao nhau tnh theo hai di bn l bng nhau tm titrng truyn theo mi phng.

6.1 Khi nim chung



6.4 Cc loi sn khc

6.2.1 S kt cu

6.2.2 Tnh ton cc b phn ca bn sn

6.2. Sn sn ton khi c bn loi dm6.2.1. S kt cuSn c th c dm chnh t theo phng dc hocphng ngang (tu theo s b tr chung ca cngtrnh v yu cu thng gi v chiu sng).

1 2

34

5

Hnh 6.3 Mt s s sn1. Bn; 2. Dm ph; 3. Dmchnh; 4. Ct; 5. Tng.

7

14

2 3

chnh; 4. Ct; 5. Tng.

1 2

6.1 Khi nim chung



6.4 Cc loi sn khc

6.2.1 S kt cu


=

11 1

25 35bh l

Cc b phn chnh ca sn (1) Bn, (2) Dm ph, (3) Dm chnh, (4) Ct, (5) Tng.Sn gm bn sn v h dm (sn) c lin khi, bn k ln dm ph, dm ph k ln dm chnh,dm chnh k ln ct v tng.

Yu cu cu to: h 6 cm i vi sn nh dn dng, h 7 cm i vi sn nh cng nghip.

Khong cch dm ph l1 = (1 4) m thng thng l1 = (1,7 2,8) m.Khong cch dm chnh l2 = (4 10) m thng thng l1 = (5 8) m.Chiu dy bn

8

=

1 112 20dp nhip

h LChiu cao dm ph

Yu cu cu to: hb 6 cm i vi sn nh dn dng, hb 7 cm i vi sn nh cng nghip.

Chiu cao dm chnh =

1 18 12dc nhip

h L

B rng dm bd = (0,2 0,5) hdNu chu vi ca sn c k ln tng gch th chiu di on k S (12 cm, hb) i vi bn; 22cm i vi dm ph; 34 cm i vi dm chnh.

6.1 Khi nim chung



6.4 Cc loi sn khc

6.2.1 S kt cu


6.2.2. Tnh ton cc b phn ca snTnh bn sn theo s khp dol2/l1 2, bn thuc loi bn dm. Ct dibn rng 1m theo phng song song vil1 (Hnh 3), ta c s di bn v s tnh ca di bn nh trn hnh 6.4

Di bn tnhdm ph

Di b

n

tn

hdm

ch

nh

Hnh 6.4 S mt bng kt cu ca sn

9

Di b

n

tn

hdm

ch

nhD

i b

n

tn

h b

n

Hnh 6.4 S mt bng kt cu ca sn sn ton khi c bn loi dm

a. Dm chnh t theo phng ngang nh

6.1 Khi nim chung



6.4 Cc loi sn khc

6.2.1 S kt cu


Bn Dm chnh Dm ph

Hnh 6.4 S mt bng kt cu ca sn sn ton khi c bn loi

10

ca sn sn ton khi c bn loi dm

b. Dm chnh t theo phng dc nh

6.1 Khi nim chung



6.4 Cc loi sn khc

6.2.1 S kt cu


11

pb

gb

qb=pb+gb

Hnh 6.5 S tnh ton ca di bn

6.1 Khi nim chung



6.4 Cc loi sn khc

6.2.1 S kt cu


Nhp tnh ton:Vi cc gi gia, cng ca di bn trong phm vi gi ta ln nn khp do hnh thnh cc mp gi ta. Vi gi bin, v tr gi ta trng vi im t phn lc ca tng, v tr ny c quy c cch mp trong ca tng mt on bng 0,5 hb. T :

2221btdp

bhbbll += lg = l1 bdp

Ti trng:Ti trng tc dng trn bn sn gm tnh ti (k hiu g) v hot ti (k hiu p).Tnh ti: Tnh ti trn bn c xc nh theo cu to thc t ca bn sn. Gi g c l tnh ti tiu

*

12

Tnh ti: Tnh ti trn bn c xc nh theo cu to thc t ca bn sn. Gi gic l tnh ti tiuchuNn trn 1m2 sn ca lp th i cu to nn bn v i l h s tin cy tng ng ca lp , tac tnh ti tnh ton trn 1m2 bn l: g = igic.

11

2bb lqM =

16

2bb lqM =

Hot ti: Theo TCVN 2737-1995, hot ti tiu chuNn phn b u trn 1m2 sn (k hiu pc) cxc nh theo loi phng v loi cng trnh.

+ i vi nhp bin v gi th hai:

+ i vi cc nhp gia v cc gi gia:

Ni lc: Xc nh ni lc trong di bn theo s khp do. Khi s chnh lch gia cc nhpbng (lg-lb)/lg100% = ... < 10%, bng phng php phn phi m men trn nguyn tc m boiu kin cn bng tnh hc, xc nh c gi tr m men cc nhp v cc gi nh sau:

6.1 Khi nim chung



6.4 Cc loi sn khc

6.2.1 S kt cu

6.2.2 Tnh ton cc b phn ca bn sn*

qb.lb2

11qb.lg16

2qb.lg16

2qb.lg16

2

qb.lb2

11

qb.lg2

16

qb.lg2

16

qb.lg2

16

Biu m men c dng nh trn hnh 6.6:

13

11 16 16 16

Tnh ton ct thp:Tnh ton ct thp trong di bn nh i vi cu kin chu un tit din ch nht t ct n b h = 100 hb. Ch vi tiet din tnh theo s khp do th iu kin m pl phi c thamn, nu khng th hoc phi tng chiu dy bn, hoc phi tng cp bn ca b tng.Sau khi tnh ton ct thp, tin hnh tnh hm lng ct thp.i vi bn % nm trong khong 0,3 0.9 l hp l. Nu % nm ngoi khong trn, nn thayi hb v tnh ton li. Trng hp % < min m khng th gim chiu dy bn th chn AS min 100ho.i vi cc bn c c bn cnh lin kt vi dm, do nh hng ca hiu ng vm cho phpgim khng qu 20% lng ct thp so vi kt qu tnh ton. Nh vy trn Hnh 6.7 cc bntrong khu vc gch cho c gim ct thp.

6.1 Khi nim chung



6.4 Cc loi sn khc

6.2.1 S kt cu


Hnh 6.6 Ch dn vngc gim thp

(Vng gim thp cgch cho trn cc hnh

a,b)

14

B tr ct thp:Ct thp chu lc:Cn c kt qu tnh ton c trn, chn ng knh ct thp, sau xc nh khong cchgia cc thanh ct thp.

6.1 Khi nim chung



6.4 Cc loi sn khc

6.2.1 S kt cu


ng knh ct thp trong bn thng c chn trong khong 610mm (c th ln hn), nnchn d hb/10 v mi vng chu lc:+ Hoc chn cng mt loi ng knh (ph bin).+ Nu dng hai loi ng knh th trnh nhm ln v m bo cho cc thanh ct thp lmvic tng i ng u trong di bn, chn d = 2mm.Khong cch gia cc thanh ct thp cnh nhau a tnh theo yu cu chu lc ng thi:

+ m bo thi cng d dng, nhanh chng, yu cu a 7cm

15

a 20cm khi hb 15cm1,5 hb khi hb > 15cm

+ m bo thi cng d dng, nhanh chng, yu cu a 7cm+ m bo cho b tng v ct thp kt hp lm vic tt vi nhau, yu cu:

Ct thp c th c b tr mt cch n gin thun li cho thi cng, th d nh Hnh 6.8

6.1 Khi nim chung



6.4 Cc loi sn khc

6.2.1 S kt cu


B

16

B

AA

6.1 Khi nim chung



6.4 Cc loi sn khc

6.2.1 S kt cu


17

A-a

Hnh 6.8 B tr thp trong bna) Mt bng; b,c : Cc mt ct; + =0.25 khi p/g3

6.1 Khi nim chung



6.4 Cc loi sn khc

6.2.1 S kt cu


18

Hnh 6.8 B tr thp trong bna) Mt bng; b,c : Cc mt ct; + =0.25 khi p/g3

b - b

6.1 Khi nim chung



6.4 Cc loi sn khc

6.2.1 S kt cu


Khi khong cch gia cc ct thp b, a < 15cm, tit kim, c th gim bt ct thp bng mttrong cc cch: - t cc thanh di ngn xen k nhau (Hnh 6.9a); - Dng cc thanh ngn hnbnh thng t so le nhau ((Hnh 6.9b).

Hnh 6.9 Mt s cch t ct thp trong bn

19

Hnh 6.9a

Hnh 6.9b

6.1 Khi nim chung



6.4 Cc loi sn khc

6.2.1 S kt cu


- Khi chiu dy bn hb 8cm c th un bt mt s thanh chu m men dng nhp ln chum men m gi ((Hnh 6.9c). Thng thng cch mt thanh un mt thanh. Sau khi un thp tnhp ln, nu thp trn gi cn thiu th t thm cc ct m. Gc un ct thp thng l 300, khichiu dy bn ln c th un gc 450.Sau khi ct hoc un ct thp, s ct thp mt di i vo gi ta c din tch khng b hnmt phn ba so vi tit din gia nhp v khong cch gia cc thanh khng qu 330 mm. Ccthanh ny phi c neo chc vo gi ta mt on khng nh hn 15 ln ng knh thanh.

20

Hnh 6.9c Mt s cch t ct thp trong bn

6.1 Khi nim chung



6.4 Cc loi sn khc

6.2.1 S kt cu


Ct thp chu m men m theo cu to: n gin tnh ton, khi xc nh ni lc trong bn b qua s ngn cn chuyn v xoay khibn b chn cng vo tng v b qua s lm vic ca bn theo phng cnh di. Thc t dctheo chu vi bn khi bn b chn cng vo tng v ti khu vc ln cn dm chnh m men mxut hin vi tr s ng k. trnh cho bn khng b nt do cc m men v lm tng cng tng th cho bn, ngi ta t ct thp m theo cu to dc theo lin kt gia bn vi tngv dc theo cc dm chnh vi mt lng khng t hn 56 trong mi mt v cng khng t hn50% ct thp chu lc tnh ton cc gi gia (Hnh 6.10),

21

50% ct thp chu lc tnh ton cc gi gia (Hnh 6.10),

Thp cu to

v khng t hn 3 thanh m bocho li thp khng b xc xch

Thp m cu to

Tng bin

Thp m cu to

Dm chnh

Hnh 6.10 Ct thp chu m men m theo cu to

6.1 Khi nim chung



6.4 Cc loi sn khc

6.2.1 S kt cu


Thp phn b - cu to: gi ng v tr cho cc ct chu lc cn phi t ct phn b vung gc vi ct chu lc vlin kt chng vi cc ct chu lc bng dy thp buc 0,8 1mm (hoc hn). Ct phn bthng c s dng nhm CI (CII t dng) v t gn trc trung ha hn so vi ct chu lc,ng knh b hn hoc bng ct chu lc (thng dng 6) khong cch 25 30cm.Ch rng vi nhng ct phn b t mt di bn song song vi phng l2 (th d thp s 5trn Hnh 6.8), ngoi chc nng nh v ct dc n cn chu m men dng theo phng l m

22

trn Hnh 6.8), ngoi chc nng nh v ct dc n cn chu m men dng theo phng l2 mkhi tnh ton b qua. Din tch tit din ngang cc ct ny tnh cho mi mt b rng bn khngt hn 20% AS khi 2l1 < l2 3l1 v khng t hn 15% AS khi l2 > 3l1, trong AS - din tch ctthp chu lc theo tnh ton.

6.1 Khi nim chung



6.4 Cc loi sn khc

6.2.1 S kt cu


Tnh dm ph theo s khp do: S tnhHnh 6.11. Cc kch thc hnh hc ca dm ph

23

Tng ng ct di bn c b rng l1 theo phng song song vi l2 sao cho trc di bn trngvi trc dm ph (Hnh 6.4). (Trng hp bn c nhp khng u nhau th b rng di bn cxc nh t trc dm ph sang tri v sang phi l l1T/2 v l1p/2). T c s tnh dm cth hin trn Hnh 6.12 Pd

gd

qd=pd+gd

Hnh 6.12 S tnh dm ph

6.1 Khi nim chung



6.4 Cc loi sn khc

6.2.1 S kt cu


Nhp tnh ton:2222abbll dctb += lg = l2 bdc

Ti trng:* Hot ti (pd) pd = pb l1* Tnh ti (gd) gd = gb l1 + g0Trong : g0 - Trng lng bn thn phn sn ca dm ph; g0 = bdp (hdp hb) 1 1,1 bt y: 1 - 1m dm; 1,1 - H s tin cy i vi trng lng bn thn dm; bt Trng lng ring ca b tng * Tng ti trng trn dm: qd = pd + gdCh : Trng hp nhp bn khng bng nhau th trong cc cng thc trn l c thay bng

24

211 phailtrail +

Ni lc:Xc nh ni lc trong dm theo s khp do. Khi cc nhp cnh nhau khng chnh lch qu10%, ngi ta xc nh c: - Tung ca biu bao nhnh max: Mmax = 1 qd l2;

- Tung ca biu bao nhnh min: Mmin = 2 qd l2Trong :i vi nhp bin dng lb. i vi nhp gia dng lg. i vi cc gi, trng hp nhpbn tri v nhp bn phi khng bng nhau th m men ti gi c xc nh theo nhp c trs ln. Chia mi nhp thnh 5 phn bng nhau. ng vi mi im chia tr s 1 c cho trnHnh 6.13, tr s 2 ph thuc t s pd/gd v c cho trong Bng tra trong gio trnh. M menm trit tiu ti nhp bin cch gi ta th 2 mt on klb, trong k cng c cho trong Bngtra trong gio trnh.

Ch : Trng hp nhp bn khng bng nhau th trong cc cng thc trn l1 c thay bng

6.1 Khi nim chung



6.4 Cc loi sn khc

6.2.1 S kt cu


0

0,2l

1 2 3 4 5 5 6 7 8 9 10 10 11 12

0,2l 0,2l 0,2l 0,2l 0,2l 0,2l 0,2l 0,2l 0,2l 0,2l 0,2l 0,1l

=0,

065

=0 ,

090

=0,

091

=0,

075

=0,

020

=0,

075

=0,

018

=0,

058

=0,

0625

=0,

058

=0,

018

=0,

018

=0 ,

058

=0,

0625

25

Hinh 6.13; 6.14 Hnh dng biu bao m men v lc ct

Q =0,4.q .la d b Q =0,5.q .lb d gp

Q =0,5.q .lb d gt

Q =0,5.q .lc d gt

Q =0,5.q .lc d gp

6.1 Khi nim chung



6.4 Cc loi sn khc

6.2.1 S kt cu


Tnh ton ct thp:* Tnh ct dc:- Dng m men c tr s tuyt i ln nht mi nhp v trn tng gi tnh ton cho tng nhpv gi tng ng.- Vi tit din chu m men m, cnh tit din nm trong vng ko, tnh nh tit din ch nht bd hd- Tnh ton theo s khp do. Nn phi kim tra theo iu kin m pl. Trong :pl = 0,3 tng ng pl = 0,37 vi b tng c cp bn t B 22,5 tr xung. = 0,25 tng ng = 0,3 vi b tng c cp bn t B25 tr ln.

26

pl plpl = 0,25 tng ng pl = 0,3 vi b tng c cp bn t B25 tr ln.+ m pl : t ct n* Tnh ct ai: - Dng tr s tuyt i ln nht ca lc ct trn mi tnh ct ai cho dm. Bqua phn cnh, tnh nh tit din ch nht.

6.1 Khi nim chung



6.4 Cc loi sn khc

6.2.1 S kt cu


Tnh dm chnh theo s n hiS tnhDm chnh cng vi ct to thnh khung. Thng thng ni lc trong dm chnh c xc nht vic tnh ton khung vi t hp cc ti trng ng v ngang tc dng vo khung.Trng hp dm chnh k ln ct, hoc khi cng n v ca dm ln hn bn ln cng nv ca ct, m men phn phi vo ct nh c th b qua v tnh dm chnh nh mt dm lin tck ln cc ct v tng.Tng tng ct di bn c b rng l2 theo phng song song vi l1 sao cho trc dm trng vi

27

Tng tng ct di bn c b rng l2 theo phng song song vi l1 sao cho trc dm trng vitrc di bn (Hnh 6.4).T c s tnh dm chnh nh trn Hnh 6.15.Nhp tnh ton:+ Nhp gia ly bng khong cch gia cc trc ct.+ Nhp bin ly bng khong cch t trc ct n trc tng.Khi kch thc cc nhp chnh lch nhau di 10% th c th coi cc nhp u nhau bng l v lytr s ln hn trong cc nhp.Xc nh ti trngTi trng trn sn truyn v dm chnh c qui v thnh lc tp trung t ti v tr dm ph gcln dm chnh. Vi pd v gd l hot ti v tnh ti phn b u trn dm ph xc nh c phn trn. Ta c:

6.1 Khi nim chung



6.4 Cc loi sn khc

6.2.1 S kt cu


28

P

G

P

G

P

G

P

G

* Hot ti tp trung P: P = 0,5pdl2T + 0,5pdl2p. Khi l2T = l2p th P = pdl2.

* Tnh ti tp trung G: G = G1 + GO (kN)

6.1 Khi nim chung



6.4 Cc loi sn khc

6.2.1 S kt cu


Trong :+ G1 - lc tp trung do dm ph truyn v G1 = 0,5 gd (l2T + l2P)+ G0 - lc tp trung do trng lng bn thn phn sn dm chnh G0 = b(h hb) l1 bt 1,1

+

y b, h - l b rng v b cao tit din chnh

Xc nh ni lc theo s n hi:Ni lc trong dm chnh c xc nh bng cc phng php ca c hc kt cu. Khi dm cnhp u nhau hoc c nhp chnh lch nhau khng qu 10% c th dng cc bng vi cc cngthc lp sn. C hai cch: cch trc tip v cch t hp. Cch trc tip n gin nn dng khi

29

thc lp sn. C hai cch: cch trc tip v cch t hp. Cch trc tip n gin nn dng khitnh ton thit k. Cch t hp cho ta thy r hn bn cht ca biu bao ni lc v rn luyncho ngi s dng k nng t hp ni lc.* Cch trc tip:Tung nhnh dng v nhnh m ca biu bao m men:Mmax = 0 Gl + 1 PlMmin- = 0 Gl 2 PlTung nhnh dng v nhnh m ca biu bao lc ct:Qmax = 0G + 1PQmin = 0G 2PTrong : i; i cho trong cc bng lp sn ph thuc vo s nhp dm v s t ti trn minhp.

6.1 Khi nim chung



6.4 Cc loi sn khc

6.2.1 S kt cu

6.2.2 Tnh ton cc b phn ca bn sn+

* Cch t hp: xy dng biu bao m men ta tin hnh theo hai bcBc 1:+ V ring biu ni lc do tnh ti MG. Tnh ti G c cht trn ton b dm (Hnh 6.17a)+ V ring tng biu ni lc do cc trng hp bt li ca hot ti Mpi i = 1,2,3... (Hnh6.17b,c,d....)Hot ti c th thay i v v tr. c c cc trng hp bt li, cn ch :+ Hot ti xp cch hai nhp s cho m men dng bt li gi khng cht ti.

30

+ Hot ti xp cch hai nhp s cho m men dng bt li gi khng cht ti.+ Hot ti xp cch nhp s cho m men dng bt li nhp xp ti.+ Hot ti xp hai nhp k gi s cho m men m v lc ct bt li ti gi .ng thi cn ch ti tnh cht i xng ca h v cn c nhng nhn xt loi bt cctrng hp khng bt li. V d vi dm bn nhp v li dng tnh cht i xng ta cn xt sutrng hp ca P nh trn Hnh 6.17a,b,c,... h. c MG v Mpi dng cng thc:MG = Gl ; Mpi = Pl c QG v Qpi dng cng thc:QG = G ; Qpi = P v cho trong bng IV ca Ph lc (hoc trong cc bng lp sn ca gio trnh v cc cNmnang kt cu BTCT).

6.1 Khi nim chung



6.4 Cc loi sn khc

6.2.1 S kt cu


G G G G G G G G

P P P P

P P P P

P P P P P P

MG

MP1

MP2

31

P P P P P P

P P P P

P P P P

P P

MP3

MP3

MP4

MP5

6.1 Khi nim chung



6.4 Cc loi sn khc

6.2.1 S kt cu


Trong bng ch cho gi tr v ti mt s tit din quan trng. Mun c c biu ni lctrong tng nhp ca dm cn ch :+ i vi m men cn thc hin cc php tnh b tr theo phng php ca c hc kt cu. emct nhp dm ra nh mt dm tnh nh k trn hai gi t do, t thm m men gi tnh cri tnh dm v v biu m men. Hoc bng cch treo biu (Hnh 6.17)i vi lc ct on gia nhp, xc nh c bng phng php mt ct vi ch rng ti titdin c lc tp trung, biu lc ct c bc nhy bng ng tr s ca lc tp trung .Ngoi ra vn cn phi c bit ch ti tnh cht i xng ca h.

32

Ngoi ra vn cn phi c bit ch ti tnh cht i xng ca h.

M (

bi

t)B

M (

bi

t)C

P P

M =Pl/3o

M2.1=Mo-X1; M2.2=Mo-X2;

Hnh 6.17 Treo biu tm m men ti tit din cha bit

6.1 Khi nim chung



6.4 Cc loi sn khc

6.2.1 S kt cu


MG + Mp1 = M1MG + Mp2 = M2. . . . . . . . . . . .

V chung M1, M2, ... Mi trn cng mttrc theo cng mt t l. ng baongoi cng chnh l biu bao m

Bc 2:T hp v v biu bao ni lc.Cch 1:V chung cc biu ni lc thnh phn ln cng mt trc. ng bao ngoi cng chnh l biu bao ni lc. V d vi m men:

33

. . . . . . . . . . . .

MG + Mpi = Mi

ngoi cng chnh l biu bao mmen.

Cch 2:T hp v v biu bao ni lc theo tng tit din, th d ti tit din K, tung biu baonhnh max v nhnh min c xc nh nh sau:+ i vi m men:M-max(K) = MG(K) + max Mpi(K)M-min(K) = MG(K) + min Mpi(K)Ni cc tung max vi nhau ta c nhnh max ca biu bao m men. Ni cc tung minvi nhau ta c nhnh min ca biu bao m men. (Hnh 6.18).+ i vi lc ct cng lm tng t nh m men ta c nhnh max v nhnh min ca biu bao lc ct (Hnh 6.19).

6.1 Khi nim chung



6.4 Cc loi sn khc

6.2.1 S kt cu


Nhnh Min

Nhnh Max

Hnh 6.18 Hnh bao m men dm chnh 4 nhp

34

Hnh 6.18 Hnh bao m men dm chnh 4 nhpNhnh Max (ng nt lin)

Nhnh Min (ng nt t)

Q 2Q3

T

Q 2P

Q 1

Q g

Q g

Hnh 6.19 Hnh bao lc ct ca dm chnh 4 nhp

6.1 Khi nim chung



6.4 Cc loi sn khc

6.2.1 S kt cu


Tnh ton ct thp:* Tnh ton ct dc:Vi cc tit din chu m men dng tnh ton theo tit din ch T do cnh (bn sn) thuc vngnnVi cc tit din chu m men m tnh ton theo tit din ch nht bh vi ch rng dm chnhc xc nh ni lc theo s n hi. Tr s m men m ti cc gi ta chnh l tr s mmen tng ng ti cc tit din qua trc ca gi. D dng nhn thy rng trong phm vi gi ta(t mp ct tri n mp ct phi), tit din tham gia chu m men gi bao gm c phn chiu

35

(t mp ct tri n mp ct phi), tit din tham gia chu m men gi bao gm c phn chiucao ct, nn chiu cao lm vic ca cc tit din trong phm vi gi ta l rt ln. Do vy vic tnhton ct thp dc chu m men m ti gi ch cn thc hin vi cc tit din i qua mp gi vitr s m men tng ng (k hiu Mmg). Trng hp m men hai bn mp gi no khngbng nhau th ly m men c tr s tuyt i ln tnh ton v b tr ct dc cho gi .* Tnh ct ai:- Dng tr s tuyt i ln nht ca lc ct trn mi tnh ct ai cho dm. B qua phn cnh,

tnh nh tit din ch nht.- Ch tnh ct treo chu ti tp trung ca dm ph truyn ln dm chnh. m bo iu kinkhng b git t th ni dm ph gc ln dm chnh cn phi gia cng cho dm chnh bng ctai hoc ct xin, gi l ct treo. Nu dng ct ai th tng din tch tit din ct ai cn thitc xc nh theo cng thc: ( )1 w w , , sins s s inc s incP G R A R A + +

6.1 Khi nim chung



6.4 Cc loi sn khc

6.2.1 S kt cu


(P+G1) l tng ti trng tp trung ti im tnh ct treoRsw l cng ca ct ai khi tnh chu lc ct

As,inc l tng din tch tit din ct ai m thp chc thng ct quaRs,inc l cng ca ct vai b khi tnh chu lc ctAsw l tng din tch tit din ct vai b m thp chc thng ct qua

l gc nghing ca ct vai bG1+P

36

Lng ct treo tnh ton phi c b trhai bn mp dm ph trong khong hs theos Hnh 6.20. Thng thng ch dngct treo di dng ct ai. Khi khng kch thc th c th tng ng knh ctai hoc dng ct xin dng vai b ltngc.

Hnh 6.20 S b tr ct treo (trng hpdng ct ai)

6.1 Khi nim chung



6.4 Cc loi sn khc

6.3.1 S kt cu v s lm vic ca bn

6.3.2 Cu to ct thp

6.3.3 Tnh ton bn sn

6.3.4 Tnh dm ca bn k 4 cnh

6.3. Sn sn ton khi c bn bn k bn cnh (chu un 2 phng)6.3.1.S kt cu v s lm vic ca bn.Sn gm bn sn v h sn c lin khi; T l cc cnh ca bn l2/l1 2 (thng ly 1-1.5);Kch thc cc cnh l1, l2 = 4 - 6m.; Chiu dy bn hb ly khong l1/40 n l1/50

37

Hnh 6.21 S kt cu ca bn k 4 cnh

6.1 Khi nim chung



6.4 Cc loi sn khc


6.3.2 Cu to ct thp

6.3.3 Tnh ton bn sn


Xt mt bn k 4 cnh chu ti trng phn b u tng dn, quan st thy bin dng ca bnnh sau (Hnh 6.22):+ Mt di ca bn: xut hin cc vt nt theo phng ng phn gic cc gc, cn gia bnc cc vt nt theo phng cnh di.+ Mt trn: nu cc cnh l ngm cng th c cc vt nt chy vng theo chu vi, nu k t do thcc gc bn s b vnh ln.

38

Hnh 6.22 Vt nt trong bn k 4 cnh chu ti trng phn b u

6.1 Khi nim chung



6.4 Cc loi sn khc


6.3.2 Cu to ct thp

6.3.3 Tnh ton bn sn


6.3.2.Cu to ct thp.Ct thp chu lc c t theo c hai phng. Thp gia nhp t theo tnh ton, vo gn gi(dy bin lk) c th gim. Thp trn gi xc nh theo m men un. C th un 1/2 -> 2/3 lngthp nhp ln v t thm ct m xen k yu cu.

39

6.1 Khi nim chung



6.4 Cc loi sn khc


6.3.2 Cu to ct thp

6.3.3 Tnh ton bn sn


6.3.3.Tnh bn sn:Theo s khp do: Quan st s lm vic ca k 4 cnh, khi trng thi cn bng gii hn,theo cc khe nt s hnh thnh khp do, chia bn thnh cc ming cng. Nh vy c th xembn nh gm cc ming cng ni vi nhau bi cc khp do (Hnh 6.23).M men khp do: Mkd = Ra.Fa.Z; Mkd l m menkhp do trn 1 n v di; Fa din tch ct thp trn1 n v di; Z l cnh tay n ni lc (Z 0.9h0).Nu cnh k t do th m men trn cnh =0.Tnh m men trong bn da trn nguyn l cn bng

40

Hnh 6.23 S khp do trn bn k 4 cnh chu ti trng

phn b u

Tnh m men trong bn da trn nguyn l cn bngcng kh d ca ni v ngoi lc, ta c cc cngthc sau:- Trng hp ct thp lp di c b tr u:

2' '1 2 1

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

12 I I II IIl l lq M M M l M M M l = + + + + +

- Trng hp ct thp lp di c b tr khngu ct thp chu m men dng gia nhp gp idi bin nn:

2' '1 2 1 1

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

12 2I I II IIl l l lq M M M l M M M l M M = + + + + + +

6.1 Khi nim chung



6.4 Cc loi sn khc


6.3.2 Cu to ct thp

6.3.3 Tnh ton bn sn


Ta ch c 1 phng trnh m cha 6 m men cn tm. V vy c th ly M1 lm Nn s, cn cc mmen cn li c biu din qua M1 vi cc h s c chn da trn cc hng dn thc hnh vd nh Bng 6.1

1,0 - 1,5 1,0 - 0,3 2,5 - 1,5 2,5 - 0,8

2

1

ll

= 221

Ma

M=

'

'

1 11 1

;I IM M

a aM M

= =

'

'

2 21 1

;II IIM M

a aM M

= =

41

1,0 - 1,5 1,5 - 2,0

1,0 - 0,3 0,5 - 0,15

2,5 - 1,5 2,0 - 1,0

2,5 - 0,8 1,3 - 0,3

Bng 6.1 Cc t s khi tnh bn k bn cnh theo s khp do

6.1 Khi nim chung



6.4 Cc loi sn khc


6.3.2 Cu to ct thp

6.3.3 Tnh ton bn sn


Theo s n hiTnh bn n:Dng l thuyt n hi xc nh ni lc nh trong hng dn cc ti liu chuyn v tm mngn hi.Bn c xem l ta khp khi n gi vo tng hoc k t do trn dm.Mp bin ca bn c xem l ngm khi mp bin nm trn gi ca mt bn lin tc, cchuyn v xoay kh nh.Khi bn c lin khi vi dm bin, ln ca m men m ca bn st mp dm ph thuc

42

Khi bn c lin khi vi dm bin, ln ca m men m ca bn st mp dm ph thuc cng chng xon ca dm bin. Dm c cng chng xon ln s cn tr bin dng xoayca bn lm m men m trong bn ln (gn vi lin kt ngm). Trong tnh ton thc hnh, c thcoi mp bn lin kt vi dm bin l lin kt khp nhng vn t thp chu m men m.Khi bn k ln dm, cho vng ca bn gi ta bng khng ch l gn ng. Thc ra iukin bin phi l vng bin ca bn bng vng ca dm.C th dng cc phn mm tnh ton kt cu (nh SAP, SAFE...) xc nh ni lc trong bnvi cu kin tm (plate) hoc tm v (shell) (phn t tm v khc phn t tm l c k thm nilc ko hoc nn trong mt phng v). Hoc dng cc bng tra sn gi tr m men cho 9 loi bn n vi cc lin kt l tng khp hoc ngm.

6.1 Khi nim chung



6.4 Cc loi sn khc


6.3.2 Cu to ct thp

6.3.3 Tnh ton bn sn


1 1 1 11 2 1 2( . . ). .kM q q l l = +( . . ). .M q q l l = +

Tnh bn lin tcC th dng cc phn mm tnh ton kt cu xc nh ni lc trong bn lin tc vi ccphng n ti trng ri t hp ni lc.Trong tnh ton thc hnh, c th ch k n 2 phng n cht hot ti cch nhp v s dng ccbng tra ca bn n tnh ton nh sau:t q1 = g + 0.5p; q2 = 0.5p (g v p l tnh ti v hot ti phn b u trn bn sn).

;

M men dng gia nhp tnh theo cng thc sau:

,

,

43

2 2 1 12 2 1 2( . . ). .kM q q l l = +

1 1 2.( ). .I kM g p l l= +'

2 1 2.( ). .I kM g p l l= +

k1, k2 l h s 1, 2 trong bng tra ng vi bn c lin kt dng k

M men m trn gi :

k1, k2 l h s 1 v 2 trong bng tra ng vi bn c lin kt dng k. Trng hp bin k tdo th m men m bng 0.

,

11, 12 l h s 1, 2 trong bng tra ng vi bn c lin kt dng 1 (4 bin k t do)

6.1 Khi nim chung



6.4 Cc loi sn khc


6.3.2 Cu to ct thp

6.3.3 Tnh ton bn sn


6.3.4.Tnh dm ca bn k bn cnh: Cc bc v phng php tnh dm ny cng ging nhcch tnh dm ca sn sn ton khi c bn loi dm ch khc nhau phn phn ti t bn vodm. Hnh 6.24 th hin s phn ti t bn vo dm (cha k trng lng bn thn dm).Nn ti trng dng hnh thang hochnh tam gic tnh ni lc cho dm.Nu cn tnh n gin ha th c thquy v ti trng phn b u theocng thc sau:

44

Hnh 6.24 S phn ti cho dm bn k bn cnh

cng thc sau:Vi dng tam gic: qt = 5/6.qd;Vi dng hnh thang: qt = (1 - 2. 2 + 3)qd

Trong qd l ti trng nh tam gic hayhnh thang. Lu : s quy i ti trng nych l s tng ng ca m men ln nhttrong dm. Cc gi tr khc ca ni lc vchuyn v khng tng ng do khngnn dng ti trng phn b u tngng tnh ton cc bc tip theo.

6.1 Khi nim chung



6.4 Cc loi sn khc

6.4.1 Sn sn lp ghp

6.4.2 Sn c

6.4.3 Sn nm

6.4. Cc loi sn khc6.4.1.Sn sun lp ghpS kt cu: + Panen k ln dm hoc tng.

+ Khong cch gia cc nhp ca panen l lp = (2,8 6,8) m.+ Nhp dm ld = (4 7,2) m.

Phn loi panen.Chia lm 3 loi: Panen c, panen c l v panen sna. Panen c+ C th cu to mt lp hoc nhiu lp (mt lp btng ct thp chu lc v cc lp cch m,

45

+ C th cu to mt lp hoc nhiu lp (mt lp btng ct thp chu lc v cc lp cch m,cch nhit).+ Chiu dy h = 8 15 cm.u im: D sn xut, nhanh, lin kt n gin, chiu dy sn b.Nhc im: Tn vt liu, cch m km.

6.1 Khi nim chung



6.4 Cc loi sn khc

6.4.1 Sn sn lp ghp

6.4.2 Sn c

6.4.3 Sn nm

b. Panen c l:+ Cu to c th c mt l hoc nhiu l, mt ct ngang cc l c th hnh thang, hnh ch nht,hinh bu dc...+ Chiu cao ph thuc vo chiu di (nhp).+ Chiu di (nhp) t (2,5 4,5) m.+ B rng (45 60) cm loi mt l; (90 120) cm loi nhiu l.+ B dy cnh (2 3) cm tu thuc vo vng nn hay vng ko.

+ B dy sn (2,5 5) cm.

46

+ B dy sn (2,5 5) cm.u im: To c trn v sn phng, cch nhit v cch m tt, tn t vt liu.

Nhc im: Kh ch toc. Panen c sn+ Cu to bao gm bn v sn: Thng c hai loi sn dc v sn ngang cch nhau khongt (1,5 2,0) m.+ Sn ngang c kch thc b hn sn dc, sn c th nm pha trn hoc pha di (nmpha di s hp l v mt chu lc, sn nm pha trn s c sn phng).+ Chiu dy cnh (5 6) cm khi sn pha di.+ Chiu dy cnh (3 3,5) cm khi sn pha trn.+ B rng panen (40 80) cm.

6.1 Khi nim chung



6.4 Cc loi sn khc

6.4.1 Sn sn lp ghp

6.4.2 Sn c

6.4.3 Sn nm

Tnh ton panen.+ Tnh un tng th+ Tnh un cc b+ Kim tra vnga. Tnh un tng th+ S tnh: Coi panen nh mt dm n gin k t do ln hai u+ Nhp tnh ton: Ly bng khong cch trng tm cc gi+ Ti trng: Gm tnh ti v hot ti phn b trn sn ca din tch b mt panen ang xt (av ti trng phn b).

47

v ti trng phn b).+ Tit din tnh ton: c quy i v cc dng tit din n gin nh tit din ch I, ch T...+ Tnh ton ct thp: Bao gm vic tnh ton ct dc chu mmen b tr trong min chu ko vct ai b tr trong sn ca panen (tnh ton theo kh nng chu ct ca btng).b. Tnh un cc b (vi panen sn hoc panen l)+ Tnh bn chu un: Xem bn lin kt n hi vi dm (tnh nh bn k 4 cnh hoc bn loidm). + Tnh sn ngang: Nh dm n gin k t do ln cc sn dc.c. Kim tra vng+ Tnh ton nh mt cu kin chu un+ Tnh vi tit din quy i thnh dng ch I, ch T tng ng theo quy tc sau: Cc l trnc quy i thnh l vung, l bu lc thnh l hnh ch nht. Gi nguyn v tr trng tm titdin, din tch v mmen qun tnh ca tit din.

6.1 Khi nim chung



6.4 Cc loi sn khc

6.4.1 Sn sn lp ghp

6.4.2 Sn c

6.4.3 Sn nm

Cu to v tnh dm.+ Tu theo yu cu chu lc, cch gc panen m chn tit din ch nht hay ch T cnh dihay cnh trn.+ Ti trng gm ti t panen truyn xung (vi panen c v panen hp l ti phn b, panensn l ti tp trung ti cc v tr t sn dc), trng lng bn thn dm.+ Cu to v tnh ton ct thp nh dm ca sn ton khi+ Vi cu kin lp ghp cn kim tra kh nng chu lc khi vn chuyn, cNu lp.

48

6.1 Khi nim chung



6.4 Cc loi sn khc

6.4.1 Sn sn lp ghp

6.4.2 Sn c

6.4.3 Sn nm

49

6.1 Khi nim chung



6.4 Cc loi sn khc

6.4.1 Sn sn lp ghp

6.4.2 Sn c

6.4.3 Sn nm

Khi nim chungSn nm l sn khng c dm, bn sn da trc tip trn ct(hnh 1). Dng sn nm s gim cchiu cao kt cu, vic lm vn khun n gin v d dng b tr ct thp. Sn nm c c mtdi phng nn vic chiu sng v thng gi tt hn sn c dm. Ngoi ra vic ngn chia ccphng trn mt sn cng s linh hot v rt thch hp vi cc bc tng ngn di ng. v.v...Khi chu ti trng thng ng, bn sn c th b ph lm v ct theo kiu b ct m thng. tng cng kh nng chu ct, c th to ra m ct (Hnh 6.25a) hoc to bn ng ct c chiudy ln hn (Hnh 6.25b).

50

Hnh 6.25 M ct v bn u ct

6.1 Khi nim chung



6.4 Cc loi sn khc

6.4.1 Sn sn lp ghp

6.4.2 Sn c

6.4.3 Sn nm

Bn c chiu dy ln hn trn u ct cn c tc dng tng cng kh nng chu momen, v tit din st n ct, momen un trong bn t gi tr ln nht.Chiu rng nhp thch hp vi sn nm, thng l 4 n 8 mt i vi b tng ct thp thng,khi nhp ca bn t 7m tr ln nn c ct thp ng li trc c th gim chiu dy bn vgim vng.Chiu dy cc bn sn nm khng c ng lc trc, c th ly khong 1/30 nhp;Bn u ct phi c b dy c tng thm t nht bng 1/4 chiu dy ca bn gia v brng ca di nn phi khng nh hn 1/3 khong cch gia hai trc ct (hai trc ca bn u ct

51

rng ca di nn phi khng nh hn 1/3 khong cch gia hai trc ct (hai trc ca bn u cttrng vi trc ca ct).i vi bn sn nm c ct thp ng lc trc, chiu dy ca bn c th s b gi thit khngnh hn 1/42 cnh ln ca bc ct i vi bn sn c khng di hai nhp.Chiu dy ca bn hoc chiu dy ca bn u ct phi c tnh ton kim tra loi tr khnng bn b m thng.Trong tnh ton v cu to bn sn nm, ngi ta thng chia bn ra thnh di bn trn u ct vgii gia nhp, hai gii ny u c chiu rng bng 1/2 bc ct nh Hnh 6.26

6.1 Khi nim chung



6.4 Cc loi sn khc

6.4.1 Sn sn lp ghp

6.4.2 Sn c

6.4.3 Sn nm

52Hnh 6.26 Hnh nh bin dng v m men trong cc di bn

6.1 Khi nim chung



6.4 Cc loi sn khc

6.4.1 Sn sn lp ghp

6.4.2 Sn c

6.4.3 Sn nm

Gi s ti trng trn bn l phn b u, xem xt bin dng ca di trn u ct A1B ta thy ti vtr u ct ( A, B ) vng ca bn bng khng, ti v tr gi nhp (1) vng l ln nht. Tng n hi ( vng) ta suy ra dng ca biu momen un di trn u ct nh Hnh 6.26trong MA v MB l momen m, M1 l momen dng. i vi di gia nhp 324 vng ti vtr 3 l f3 s nh hn vng ti v tr 2 l f2. C th tng tng rng di gia nhp 324 gingnh mt dm lin tc k ln cc gi ta l cc di trn u ct A3D, B4C, v.v..T suy ra dngca biu momen un nh trn hnh 4c, trong M2 l momen dng M3, M4 l momen m. Hon ton tng t, c th suy ra hnh nh bin dng v momen un ca di trn u ct v digia nhp ca phng vung gc.

53

gia nhp ca phng vung gc.

Tnh ton n lc. tnh c cc gi tr ni lc mt tit din no ca bn c th dng nhiu cch khc nhauda trn l thuyt n hi hoc cn bng gii hn, c th dng phng php gii tch hocphng php s.Vn t ra l cn phi tnh c cc gi tr momen un trong cc di bn trn u ct v dibn gia nhp theo c hai phng ca h li ct. Trong thit k thc hnh, ngi ta thng sdng phng php phn phi trc tip v phng php khung thay th.

6.1 Khi nim chung



6.4 Cc loi sn khc

6.4.1 Sn sn lp ghp

6.4.2 Sn c

6.4.3 Sn nm

Phng php phn phi trc tip l xc nh trc tip cc gi tr ni lc ca cc di gia nhp vgii trn u ct. Cc nc khc nhau cho cc h s phn phi khc nhau tu theo quan nim vs phn phi li ni lc trong kt cu, tnh cht lm vic n hi do ca vt liu.Phng php khung thay th c dng xc nh ni lc (momen un v lc ct) cho bn snv ct khi chu ti trng thng ng v ti trng ngang, nhp ca bn c th u hoc khng u.Ngi ta coi sn nh ghp t hai h khung phng vung gc vi nhau tnh ton ni lc mtcch ring bit, ct khung l ct nh cn x ngang khung l bn sn vi chiu rng bng khongcch gia hai trc ca hai bn ln cn vi ct. Hnh 6 cho mt s v d v vic xc nh b rngca bn tham gia vo x ngang ca khung thay th theo hai phng x v y. C th dng cc

54

ca bn tham gia vo x ngang ca khung thay th theo hai phng x v y. C th dng ccphng php c hc kt cu khc nhau xc nh momen un trong bn v ct. Ti trng trnmi khung thay th l ton b ti tc dng ln sn. Vic phn chia cc gi tr momen tnh ccho cc di trn u ct v cc di gia nhp c th theo cc bng hng dn.

Tnh ton ct thp dc trong bn snT cc gi tr momen trong cc di bn trn u ct v di bn gia nhp c th xc nh cdin tch ct thp dc trong bn sn theo cc cng thc chung phn cu kin c bn. xt nnhng sai lch thin v an ton trong tnh ton ni lc v tnh ton tit din, c th gim bt ctthp dc trong bn theo cng thc :

6.1 Khi nim chung



6.4 Cc loi sn khc

6.4.1 Sn sn lp ghp

6.4.2 Sn c

6.4.3 Sn nm

Cn phn bit chiu cao h0 ca bn i vi phng c ct thp t di v phng c ct thpt trn khi c bn mi ct, chiu cao h0 ly theo chiu dy ca bn v bn mi ct. Ct thp chumomen m ca di trn u ct s c t hai phn ba (2/3) trn bng chy qua nh ct cchiu rng bng 1/2 chiu rng ca di trn u ct, 1/3 cn li t sang hai bn.B tr ct thp trong bn sn nmVic b tr ct thp v ct ct thp i vi bn chu ti trng phn b u c th theo quy tc ngin v an ton th hin trn Hnh 6.27

55Hnh 6.27 B tr ct thp trn bn sn nm

6.1 Khi nim chung



6.4 Cc loi sn khc

6.4.1 Sn sn lp ghp

6.4.2 Sn c

6.4.3 Sn nm

B tr ct thp trong bn m ct v bn u ctB tr ct thp trong m ct v bn u ct c th hin trn Hnh 6.28

56

Hnh 6.28 B tr ct thp trn m ct v bn u ct

san suon toan khoix (1)

Documents