lemn vedrasco

7252019 Lemn Vedrasco

httpslidepdfcomreaderfulllemn-vedrasco 117

Ministerul Educaţiei Ştiinţei şi Sport

Facultatea Cadastru Geodezie şi Construcţie

Catedra Constructii si Mecanica Structurilor

Memoriu explicativ La disciplina Constructii din lemn si materiale compozite

Tema Calculul si proiectarea constructiilor din lemn

eeictuat st r CC$ampamp Musurivsc(i )

ictor

veriicat conuniv +urculet M

C(işinu amp-

0



Date initiale

- Localitatea constructiei or Chisinau- Deschiderea constructiei l=10 m- Traveia constructiei B=51 m

- Inaltimea stilpului H=58 m- Lungimea cladirii L=306 m- Tipul acoperisului rece(Astereala din 2 straturi de

scandura)

1 Calculul asterelii in doua straturi

Pentru calculu asterelii simple s-a utilizaIndicati metodica nr210(p20) Astereala poate fi confec983395ionată sub formă de plăci dimensionate 3x1m

utilizicircnd scicircnduri cu grosimea de 16cm şi lă983395imea de 25 cm Distan983395adintre pane este de 15 m

Fig1 Astereala in doua straturi

5

5

1



Pana independenta

140x200x5100

Strat de protectie16x100

Strat util30x100x5000

2 5 0 0

2 5 0 0

Sarcina pe 1 m2

de acoperi Pa

Calculăm valorile sarcinilor provenite din greutatea proprie a plăcii deicircnchidere şi a icircnvelitorii de la 1 m2 de acoperiş

Valoarea greută983395ii normate a stratului de zăpadă pe 1 m2 de protec983395ieorizontală a suprafe983395ei terestre pentru or Chisinau S0=500 Pa=05KNmPanta acoperisului o adopt 120 Pentru α=6ordmlt25ordm coeficientul μ = 1 Valoarea sarcinii normate totale provenite din zăpadă pe 1 m2 deprotec983395ie orizontală a acoperişului

0 500 1 500nS S )a micro = times times = times times =

Valorile sarcinii normate si respectiv de calcul ce revine pe 1m2 deproiectie orizontala a acoperisuluisi cele perpendiculare axei de calcul se

considera egale deoarece cos6ordmasymp1

Sarcina Valoareanormata

Coeficientde

siguranta f

Valoareadecalcul

1Greutatea invelitorii din3 straturi de ruberoid

90 13 117

2Strat de nivelare dinasfalt δ=3 cm

360 13 468

3Material termoizolant(stabilit δ=15 cm)

650 13 849

4Bariera de vapori 30 13 39

5Placa asterialei cugrosimea medie δ=46 cm

150 11 165

Total 1199 1634

2



11991199 112361cos 097

16341634 116592cos 097

n

)a

)a

α

α

= = =

= = =

0

121752435 08500

n S = = gt

Coeficientul de siguranţă a sarcinii din zăpadă γf = 14 pentru raportul Valoarea sarcinii de calcul din zăpadă

500 14 700n

S S )aγ = times = times times

Valoarea sarcinii normate totale12361 500 17361n n n S )a= + = + =

Valoarea sarcinii de calcul totale16592 700 23592 S )a= + = + =

11Calculul pentru combinarea intiala a sarcinilor

1 cos 17361 1 097 11684017n n m

α = times times = times times =

1 cos 23592 1 097 2288424 m


Momentul sarcinilor de calcul2 2

32288424 2513845 1385 10

8 8 x

l M 0 m M mminustimes

= = = = times

Modulul de rezistenta necesar al sectiunii scindurei stratului util 3

3 31385 100106 10

13nec

Mx1 m

2i

minusminustimes

= = = times

3

2 2

6 6 0106 101022

0025

necnec

1 3 m

(

minustimes timesΣ = = =

Nu poate i 1 m lungime sa incapa 11 m de scindura Adoptam bxh=01x003m pentru stratul util

3

2 2

6 6 0106 10071

003

necnec

1 3 m

(


Utilizind opt scinduri cu latimea sumara8 01 08 071nec3 m 3 mΣ = times = gt Σ =

gasim marimea interspatiului dintre ele

0

1 08003

7a m

minus= =

Verificam sageata relativa a asterelei6 3

4 8

213 1684017 10 25 1

146384 10 225 10

l

minus

minus

times times times= =

times times times unde

3 38 408 003

225 1012 12 x

3(

mminusΣ times

= = = times

3



12Calculul pentru combinarea a doua a sarcinilor

Componentele normale ale sarcinilor1 cos 16592 097 1609424

mα = times times = times =

12 1000 12 097 1164 ) ) = times = times times = Momentul de incovoere

2 2 6007 021 2 007 1609424 25 021 2 1164 25 1315 10 Mx l )l M mminus= times + times = times times + times times times = times Modulul de rezistenta al sectiuni utile

2

6 308 003120 10

61e mminustimes

= = times

Tensiunile normale la icircncovoiere6

6

1315 101096 13 12 156

120 10 i s

Mx 2 m M)a

1e σ

minus

minus

times= = = lt = times =

times

2 Calculul panelor continuie compuse din doua scinduri

Pentru calculul penei continuie s-a utilizatIndicati metodica nr210

Panele sunt amplasate pe construc983395iile portante ale acoperişului cupanta de 115 Pasul construc983395iilor portante este de 51m pasul panelor ndashde 25 m Valorile sarcinii permanente respectiv normate şi de calcul pe 1 m2 deprotec983395ie orizontală a acoperişului sunt

(1 08 003 1 1 0016) 500 10 1 1 200

200 11 220

n

n

G )a

G G )aγ

= times times + times times times times times =

= times = times =1249 1249

12876cos 097

n )aα

= = =

1685 16851737

cos 097n )a

α = = =

Sarcinile respectiv normate şi de calcul din zăpadă pe 1m2 de protec983395ieorizontală

0 500 1 500nS S )a micro = times = times times = 500 14 700

n S S )aγ = times = times =

Sarcinile totale respectiv normate şi de calcul pe unitate de lungime apanelor

( ) cos (12876 500) 25 097 4335n n n

S a mα = + times times = + times times =

4



( ) cos (1737 700) 25 097 5425 S am

α = + times times = + times times =

Momentul icircncovoietor2 2cos 5425 097 51

11406

12 12 x

l M m

α times times times times= = = times

Momentul de rezistenta necesar6

6 311406 10 8774 10

13nec x i1x M 2 m

minusminustimes

= = = times

In urma consultarii sortimentului pentru scindurile dimensionate 006 x02m gasim

2

6 32 007 02933 10

6 x nec1 m 1xminustimes times

= = times asymp

Panele de capat se intaresc cu o a 3 scindura

Sageata relativa in deschiderile de capat

( )6 3

3

4 4

25 4335 10 51 1 1 25 384 0005

374 200384 10 14 10

n x l l E

minus

minustimes times times

= = = lt =times times times

Unde

34 4007 3 02

14 1012

x mminustimes times= = times - momentul de inertie al sectiunii panei de capat

compuse din 3 scinduri cu sectiunea totala 021 x 0200mSageata relativa in deschiderile intermediare

( )6 3

3

4 6

1 4335 10 51 1 1 1 384 0005

623 200384 10 933 10

n x l l E

minus

minus

times times times= = = lt =

times times times

Unde3

6 4007 2 02933 10

12 x mminustimes times= = times - momentul de inertie al sectiunii panei

compuse din 2 scinduri cu sectiunea totala 014 x 02 m

Imbinarea scindurilor

Distanta de la axa rostului pina la axa reazemului 021 021 51 1071a l mprime = = times =Distanta de la centrul frontului de cuie pina la axa reazemului

225 1071 225 05 106 x a d cm= minus = minus times = unde d = 05 cm ndash diametru cuielor

Forta taietoare in rost 6611406 10

2 53802 102 106

x4 M x M minus

minustimes= = = times

timesCapacitatea portanta Tmin a unui cui in imbinarea asimetrica cu un

singur plan de alunecare o vom determina din relatiile2 2 2 21) 25 001 25 05 001 7 1115

2) 035 035 7 05 1225

i

s

+ d a 0

+ cd 0

= + = times + times =

= = times times =

unde a = c = 7 cm ndash grosimea scindurilorCalculam numarul necesar de cuie

5



min

538 6

1115n 4 + piese= = = deci utilizam 6 cuie

3Calculul grinzii din scindurii si placaj cu sectiunea dublu T

cheson

Se cere de proiectat grinda cu sectiunea cheson in doua pante incleiate dinlemn si placaj cu deschiderea de calcul de 158 m si panta talpii superioare 115 Lungimea totala a grinzii este de 130m Valorile sarcinilor sint Sarcina normata qn= 945 KNm

Sarcina de calcul q = 1269KNm Acceptam inaltimea sectiunii grinzii la mijlocul deschiderii dupa raportulrecomandat hl=18-112

hm=l12= 09mInaltimea sectiunii de reazem pentru i=110 =00667

09 05 10 00667 06r ( m= minus times times =

Pentru construirea utilizam la talpi scanduri de 33x24 si pentru inimaplacaj cu grosimea de 24grinzii

Latimea sectiuni initiala cheson va fii1 4 33 2 12 1563 cmδ δ = Σ + Σ = times + times =

Inimile se vor realiza din 10 foi de placaj Lungimea taeturii oblice( )152 13 1020 12 7967 10 12 12 z cm= times minus = gt times =

Distanta intre axele rosturilor intermediare ale inimii

1 152 7967 7233a lp z cm= minus = minus =

In cazul nervurilor cu latimea de 100 mm lungimea unui panouintermediar al inimii

1 10 7233 10 6233a a cm= minus = minus =

Lungimea panoului marginal al inimii152 3 10 122ma cm= minus times =

Distanta dintre centrele de greutate al talpilor in sectiunea de reazem 06 0144 0456r r t ( ( ( m= minus = minus =

Sectiunea de calcul se afla la distanta x de la reazem[ (1 ) ] 10 ( 068 (1 068) 068) 389 x l mγ γ γ = times times + minus = times times + minus =

Unde ( ) 0456 (10 00667) 068r ( l iγ = times = times = Inaltimea sectiuni de calcul

06 00667 389 086 x r ( ( i x m= + times = + times =

6



Distanta dintre centrele de giratie ale talpilor 086 02 066 x( m= minus =

Inaltimea inimii in lumina066 02 046ix( m= minus =

Momentul incovoitor in sectiunea de calcul( ) 2 1269 46 (13 389) 2 1783 M x l x 5n m= times times minus = times times minus =

Momentul de inertie al grinzii pentru raportul90100009000 == E Ep

3 2

3 3 2

3 4 4

2 ( 2 ) 12 ( 2 ) ( 2)

2 (12 ) 2 (156 2 12) 20 12 (156 2 12) 20 (66 2)

2 12 76 09 12 12 73 10

e t t x

x

p Ep 5p E 3 ( 3 ( (

( Ep 5p E

cm

δ δ

δ

= + times times = minus times times + minus times times times + + times times times times times = minus times times + minus times times times

+ times times times times = times

Modulul de rezistenta4 3 3 3 32 73 10 2 86 172 10 172 10e e x1 ( cm mminus= times = times times = times = times

Rezistenta talpii inferioare la intindere3 3 1783 10 172 10 ) 2073 9 095 947 x M 1e 2t d M)aσ γ minus minus= = times times = lt = =

Unde Rt=9Mpa-rezistenta de calcul a lemnului incleiat de calitatea a 2-ala intindere γ d = 095 ndashcoificient de siguranta in destinatie Marim sectiunile si folosim la constructia fermei 6x33 cu 24 240(δ = =

3 2

3 3 2

3 4 4

2 ( 2 ) 12 ( 2 ) ( 2)

2 (12 ) 2 (213 2 24) 24 12 (213 2 24) 24 (62 2)

2 24 86 09 12 12 1074 10

e t t x

x

p Ep 5p E 3 ( 3 ( (

( Ep 5p E

cm

δ δ

δ

= + times times = minus times times + minus times times times +

+ times times times times times = minus times times + minus times times times


Modulul de rezistenta4 3 3 3 32 2 1074 10 2 86 25 10 25 10e e x1 ( cm mminus= times = times times times = times = times


Tensiunile de intindere in inima3 3) ( ) 1783 10 09 12 (25 10 ) 77

14 08 095 118

x p e pt p d M E 5p 1 E 2 m

M)a

δ γ minus minus= times times times = times times times times = lt times

= times =Rpt=14MPa-rezistenta de calcul a placajului la intindere Mp= 08-coificient care tine cont de micsorarea rezistentei placajului in

rostul incleiat prin taetura oblica Verificam stabilitatea talpii superioare In cazul fixarilor peste fiecare15m flexibilitatea talpii din planul grinzii

(0289 ) 150 0289 213 244 70 6 pl 3λ = times = times = lt

Coificientul de flambaj2 2

1 08( 100) 1 08(244 100) 08 6ϕ λ = minus = minus =

7



Tensiunile de compresiune in talpa superioara3 3

178 10 (25 10 ) 7132 094 12 095 1187 x e 6 pc d M 1 2 M)aδ ϕ γ minus minus= = times times = lt times = times =

Rpc=12 MPa-rezistenta de calcul a placajului la compresiune Verificam stabilitatea inimii in sectiunea periculoasa X Calculam

valorilor momentelor de inertie si static ale talpii reduse fata de placaj

3 1 2 3

3 2 3

4 4

2 ( 2 ) 12 ( 2 ) ( 2) ( ) 2 12

2 (213 2 24) 24 12 (213 2 24) 24 (62 2) 1 (09 12) 2 24 86 12

10535 10

xp t t x p x 3 ( 3 ( ( E E 0p (

cm

δ δ δ = times minus times times + minus times times times times times + times times =

= times minus times times + minus times times times times times + times times

= times1 1

3 3

( 2 ) ( 2) ( ) 2 ( 2) (213 2 24) 24 (62 2) 1 (09 12)

2 16 24 (62 2) 149 10

tip x p t xS 3 ( E E 5p ( (

cm

δ δ = minus times times times + times times times = minus times times times times

+ times times times = times

Pentru sectienea periculoasa cu parametrii 38 24 158ix( t = =

129 38 34ix

a (γ = = = din fig89 gasim

12 5i

= 20 5

τ

=

Eforturile taietoare in sectiunea periculoasa( 2 ) 1269 (10 2 389) 14 x4 l x 5 = times minus = times minus =

Tensiunile normale si tangetiale in inima la nivelul marginilor interioareale talpilor

3 405 1783 10 05 038 10535 10 ) 321i x ix xp M ( M)aσ minus minus= times times = times times times times =

3 3 4 ( ) 14 10 149 10 (10535 10 2 0024) 041i x tip xp4 S M)aτ δ minus minus minus= times Σ = times times times times times times =

Conditia de stabilitate conform formulei (816)

2 2

1

(100 ) (100 )

i i

i

i e

0 0 ( (

τ

σ τ

+ le

part part

2 2

321 041017 1

100 10024( ) 20( )

60 60

+ = lt

Verificam rezistenta inimii la actiunea tensiunilor principale de intinderein zona rostului intii al inimii deci in sectiunea primei nervurii de lareazemul grinzii la distanta 1 10 5 122 10 5 137m x a cm= + + = + + =

Momentul incovoitor si efortul taietor in aceasta sectiune

1 1 1( ) 2 1269 137 (10 137) 2 75 x M x l x 5 m= times times minus = times times minus =

1 1( 2 ) 1269 (10 2 137) 464x l x 5 = times minus = times minus =

Inaltimea sectiunii in acest loc al deschideri

1 1 06 137 00667 069 x r ( ( x i m= + times = + times =Inaltimea inimii in lumina in sectiunea 1

x

1 1 2 069 2 024 021ix x t ( ( ( m= minus times = minus times =

8



Distanta dintre centrele de giratie ale talpilor

1 1 069 024 045 x x t ( ( ( m= minus = minus =

Momentul de inertie redus fata de placaj al sectiunii3 1 2 3

1 1 1

3 2 3

4 4

2 ( 2 ) 12 ( 2 ) ( 2) ( ) 2 12

2 (213 2 24) 24 12 (213 2 24) 24 (45 2) (09 12) 2 24 69 12

503 10

x p t t x x 3 ( 3 ( ( E Ep 5p (

cm

δ δ δ = times minus times times + minus times times times times times + times times

= times minus times times + minus times times times times + times times = timesMomentul static redus fata de placaj al talpii

1 11 1 1

3 3

( 2 ) ( 2) ( ) 2 ( 2) (213 2 24) 24 (45 2) (09 12)

2 24 24 (45 2) 1084 10

tx p t x t xS 3 ( ( E Ep 5p ( (

m

δ δ

minus

= minus times times times times times + times times = minus times times times


Tensiunile normale si tangentiale in inima la nivelul marginii interioare altalpii intinse

3 4

1 1 1

(2 ) 75 10 021 (2 503 10 ) 156i x ix x p

M ( M)aσ minus minus= times times = times times times times =3 3 4

1 1 1 ( ) 46 10 1084 10 (503 10 0024) 206i x tx p x p4 S M)aτ δ minus minus minus= times times Σ = times times times times times =

Tensiunile principale de intindere in inima2 2 2 2

05 (05 ) 05 156 (05 156) 206 369

45 08 095 379

it i i i p d 2pta m

M)a

σ δ δ τ γ = times + times + = times + times + = lt times

= times =

unde Rpta=45Mpa ndashrezistenta de calcul a placajului la intindere subungiul a fata de fibre

05 (2 ) 05 (2 206 156) 3463i iarct arct α τ δ = times times = times times =

Verificarea stabilitati inimii la mijlocul primului panou de la capatulgrinziiSectiunea de verificare se afla la distanta de la reazem

2 10 122 2 71 x cm= + =

Inaltimea sectiunii

2 2 06 071 00667 0647 x r ( ( x i m= + times = + times =

Inaltimea inimii in lumina in acest loc al deschiderii

2 0647 2 024 0167ix( m= minus times =

Parametri primului panou

2 0167 0024 696 50ix( δ = = gtStabilitatea locala a inimii este asigurata (valoarea la mijlocul primului

panou)

Verificarea rezistentei inimii la taiere in sectiunea de reazem conformformulei (814)Momentul de inertie al sectiunii de reazem redus fata de placaj

9



3 2 3

3 2

3 4 4

2 ( 2 ) 12( 2 ) ( 2) ( ) 12

2 (213 2 24) 24 12 (213 2 24) 24 (36 2) 1 (09 12)

2 24 60 12 3592 10

ep t t r r 3 ( 3 ( ( E Ep 5p (

cm

δ δ δ = times minus times times minus times times times times times + times

= times minus times times + minus times times times times times

+ times times = times

Momentul static1 2

2 3 3

( 2 ) ( 2) ( ) 2 8 (213 2 24) 24 (36 2) 1 (09 12)

2 24 60 8 876 10

ep t r r S 3 ( ( E Ep 5p (

cm

δ δ = minus times times times times times + times times = minus times times times times


Eforturile taietoare in sectiunea de reazemQ = ql 2 =126910 2 = 6345KN

Rezistenta inimi la taiere3 3 4

90

( ) 6345 10 876 10 (3592 10 2 0024) 322

6 095 63

ep ep

p

4 S M)a

2 M)a

τ δ minus minus minus= times times Σ = times times times times times times =

lt = =

Verificam rezistenta rosturilor dintre talpa si inima la forfecare aplicindformulaMomentul static redus al talpii in sectiunea de reazem

1 3 3( 2 ) ( 2) ( ) (213 2 24) 24 (36 2) 1 (09 12) 66 10tep t r S 3 ( ( E Ep 5p cmδ = minus times times times times times = minus times times times times times =

Rezistenta rosturilor3 3 4

( ) 6345 10 66 10 3592 10 2 024)

024 08 095 084

tep ep t 4 S (

2op n M)a

τ

γ

minus minus minus= times times Σ = times times times times times times

= lt = = Verificam regiditatea grinzii conform formulii 420Momentul de inertie al

sectiunii de la mijlocul deschiderii3 2 3

3 2 3 4 4

2 ( 2 ) 12 ( 2 ) ( 2) 2 (12 )

2 (213 2 24) 24 12 (213 2 24) 24 (833 2) 2 24 90 09 12 (12 1) 17268 10

e t t m m p Ep 5p E 3 ( 3 ( ( ( Ep 5p E

cm

δ δ δ = times times times = times minus times times + minus times times times + times times times times

= times minus times times + minus times times times + times times times times times =

Valorile coeficientelor K si C (vanexa4tab3)04 06 04 06 06 09 08

(453 69 ) (453 69 06 09) 099 49

(213 2 24) 24 (833 2 24) 091

5

C

t i

β

β γ

γ

= + times = + times =

= minus times times = minus times times =

= = minus times times times times =

Calculam4 3 4 4 4

0

2 20

5 (384 ) 5 945 10 10 (384 10 17268 10 ) 0007

1 ( ) 0007 1 49 (09 10) 08 0012

0012 10 1 883 [ ]lim1 300 00033

n

m

l E e m

C ( l 5 m

l l

minus minus= times times times times = times times times times times times = = times + times = times + times =

= = lt =

10



4 Calculul sticirclpilor din lemn icircncleiat

H

f i

x

q1 q2

1

2

c

x

f i

Sticirclpul este expus ac983395iunii următoarelor sarciniSarcina din greutatea proprie de icircnchidere a elementelor a peretelui P=

2 828 r s inc(

) l G G 5 = times + + =

Sarcina temporară provenită din zăpadă

0

1005 16 51 204

2 2

l S S 7 5 γ = times times times = times times times =

Sarcinile orizontale concentrate de calcul provenite din vicircnt aplicate lacapetele superioare ale sticirclpilor

17136 09 154224 p p r

1 8 ( = times = times = 8568 09 77112

cp cp1 8 (r = times = times =

Sarcinile de calcul provenite din vicircnt distribuite uniform pe lungimea

sticirclpilor0

03 14 08 51 17136 p

0 c t m

ω ω γ = times times times = times times times =

0 3 0225 14 04 51 8568

cp c t 0 mω ω γ = times times times = times times times =

Reac983395iunea la nivelul grinzii cadrului3 ( ) 3 58(17136 8568) 154224 77112

13173316 2 16 2

p cp p cp 9 1 1

x 0 ω ω times minus minus times minus minus

= + = + =

Eforturile maximale icircn baza sticirclpuluin din sticircngaMomentul de icircncovoiere

2 2

1

17136 58( ) (154224 131733) 58 301272 2

p

p

9

M 1 x 9 0 m

ω times times

= minus times + = minus times + = times

11



efortul de tăiere

1 152424 131733 17136 58 10146

p p4 1 x 9 0 ω = minus + times = minus + times =

for983395a de comprimare axială

1 2 828 ) 0 = = =

efortul maximal icircn sticirclpul din dreapta2 2

2

8568 58( ) (71112 131733) 58 265

2 2

cp

cp

9 M 1 x 9 0 m

ω times times= + times + = + times + = times

2 77112 131733 8568 58 706

cp cp4 1 x 9 0 ω = + + times = + + times =

Aplicicircnd rela983395ia hs= (18 115)H determinăm icircn prealabil icircnăl983395imeasec983395iunii sticirclpului

n

s(

= (18 115)H = (18 115)middot4 = 0725 0387m Adoptăm icircnăl983395imea medie

0725 0387

5562

n

s( cm

+

= =

Pachetul poate fi icircncleiat din scicircndură cu grosimea de 345cm şi decinumărul lor necesar pentru formaea acestui pachet

556161 16

345n 3ucati= = rArr

Utilizicircnd 14 scicircnduri aflăm icircnăl983395imea reală a sec983395iunii16 345 552 0552

s 9 cm m= = =

Avicircnd icircn vedere respectarea condi983395ilor de reazem ale panourilor de peretelă983395imea sec983395iunii sticirclpului poate fi adoptată 243 cm= Deci caracteristicilegeometrice ale sec983395iunii 0552x024m dimensionate

4 224 552 13248 10 s

3 ( mminus= times = times = times 3 3

8 424 5523363932210

12 12

s x

3 ( mminustimes times

= = =

2 26 324 552

1218816 106 6

s x

3 (1 mminustimes times

= = = times

Flexibilitatea sticirclpului22 22 58

80 1200289 0289 0552

s

l 9

i (λ

times times= = = = lt

times times

1) Verificarea rezisten983395ei sticirclpului din sticircnga

c 2

1

M

e

d

e

le+=σ

2 2 3

4

1 1 80 828 100887

3000 3000 13248 10 15e c

2

λ ξ

minus

minus

minus times minus times times= = =

times times times times times

1 30127

0340887

d

M M M m

ξ = = = times

3

4 6

828 10 03434 15

13248 10 1218816 10 c M) 2 M)aσ

minus

minus minus

times= + = lt =

times times

12



2) Verificarea stabilită983395ii formei plane a icircncovoierii

1le

sdotsdot+

sdotsdot

n

ce m

d

ce 21

M

2

ϕ ϕ

Stabilitatea in forma plana este asigurataFlexibilitatea sticirclpului icircn afara cadrului

58836 120

0289 0289 024

l l

i 3λ = = = = lt

times times

Utilizicircnd la mijlocul icircnăl983395imii sticirclpului o riglă vom aveaCalculăm φ 2 21 08( 100) 1 08(836 100) 044ϕ λ = minus = minus =

2 2140 ( ) 140 024 232 (58 0552) 584

m e 3 0 l (ϕ = times times times = times times times =

Icircn jumătatea superioară a sticirclpului epurei e similar celei triunghiulare inaşa caz 322= 0

22 2

2

06 58 11 075 006 ( ) 1 1 075 006( ) 1 248

1 0552 2

p p

)

s s

l l m0

( ( m

α times times = + + times + minus times = + + minus times = +

2

2

0142 176 58 0142 142 0552 11 14 1 1 1 083

1 0552 4 2

p s )M

s p

l ( m0

( l mα

times times times times = + + + minus times = + + minus times = +

unde 0=α Aşadar vom avea

044 248 109 1 )

0 φ times = times = gt584 083 485 1

M )M 0 φ times = times = gt

Şi deci calculul stabilităţii formei plane de icircncovoiere este inutil

3) Verificăm rezisten983395a sec983395iunii icircn tăiere (la forfecare) aplicicircnd formula

e

23

4S lesdot

=τ 12

8

101458 914112 10022 15

33639322 10 0144

M)a 2 M)aτ minus

minus

times times= = lt =

times times

1 101458

11440887

44 0

ξ = = =

2 2

324 552 9141128 8

s3 (S cmtimes times= = = 3 3

424 55233639322

12 12

s3 (

cmtimes times

= = =

06 06 024 0144e

3 3 m= times = times =

Rezistenţa secţiunii icircn tăiere (la forfecare) este asigurata

4) Calculul prinderii sticirclpului de funda983395ii 1

828 204 624 ) ) S 0 = minus = minus =

109 09 624

5111 11

)

0

times= = =

13



2 2 3

4

80 51 101 1 099

3000 3000 13248 10 15c

2

λ ξ

minus

minus

times times times= minus = minus =


Momentul icircncovoietor de calcul icircn sec983395iunea inferioară a sticirclpului1 30431 00304

d

M M M m

ξ = = = times

3

4 6

51 10 0030421 15

13248 10 1218816 10t c

M)a 2 M)aσ minus

minus minus

times= minus + = lt =

times times

3

4 6

51 10 00304288 15

13248 10 1218816 10c c

M)a 2 M)aσ minus

minus minus

times= minus minus = minus lt =

times times

Ordinea epiurilor de icircntindere şi comprimare288 0552

03221 288

c sc

t c

(( m

σ

σ σ

times times= = =

+ +

21 0552023

288 21

t st

t c

(( m

σ

σ σ

times times= = =

+ +

Utilizicircnd pentru fixarea pe funda983395ie eclise cu grosimeaa = 12cm şialcătuind ecua983395ia de echilibru a for983395elor fa983395ă de centrul de greutate al epureitensiunilor de comprimare determinăm efortul de icircntindere icircn buloanele deancoraj

300304 51 0064 10008

04 04

d an t

M t M

e e

minustimes times times= = minus = minus =

2 012 2 032023 04

2 3 2 3

ct

(ae ( m

times times= + + = + + =

012 0552040 00642 2

sa (t e m+ += minus = minus =

Aria necesară totală a sec983395iunii buloanelor de ancoraj2008

123120 08 085

annec

3a a

cm

2 m m= = =

times times times times unde

80=am şi 850=m R ba=120Mpa Adoptăm doi tiran983395i de ancoraj cu diametrul d = 36mm fiecare şi ariatotală netă

2 22 759 1518 123n nec

cm cm= times = gt =

5) Calculăm icircmbinarea ecliselor cu corpul sticirclpuluiReeşind din faptul că lă983395imea eclisei trebuie să fie egală cu lă983395imea

sec983395iunii sticirclpului şi 983395inăicircnd cont de prescrip983395iile de amplasarea buloanelordeterminăm numărul lor necesar

3 35 3 24d d d cm+ + = 24

252695

d cm= =

Utilizăm deci buloane cu d=26mmCalculăm capacitatea portantă Tmina bulonului pentru un plan de

alunecare

14



08 08 12 26 2496a

+ ad 0 = = times times = 035 035 552 26 502

c s+ ( d 0 = = times times =

2 2

mx 25 25 26 169

i+ d 0 = = times = 2 2 2 218 002 18 26 002 12 1505

i+ d a 0 = + = times + times =

Prin urmare Tmin = Ti = 1505kNNumărul necesar de buloane

3

min

008531 6

1505 10

an

n piese+ minus= = = =

times

Adoptăm 6 buloane

Lungimea necesară a eclisei avicircnd icircn vedere interspa983395iul minimal deamplasare a buloanelor icircn lungimea fibrelor egală cu 7middotd

7 4 3 0026 4 0312e

l d m= times times = times times =

Adoptăm lungimea ecliseile = 32cm Icircntre eclisă şi funda983395ii este necesarde prevăzut un rost de 2 cm pentru a asigura lucrul bazei sticirclpului icircnconformitate cu schema de calcul primită

B I B L I O G R A F I E

1 II ndash 25 ndash 80 Construc983395ii din lemn Norme de proiectare

2 20107-85

3Indica983395ie metodică nr 210 Proiectarea elementelor de icircnchidere şi a

sticirclpilor clădirilor cu schelet

15



4Indica983395ie metodică nr 396 Construc983395ii din lemn şi mase plastice

(Arce)


(Construc983395ii de icircnchidere partea a III-a)

16



Date initiale

- Localitatea constructiei or Chisinau- Deschiderea constructiei l=10 m- Traveia constructiei B=51 m

- Inaltimea stilpului H=58 m- Lungimea cladirii L=306 m- Tipul acoperisului rece(Astereala din 2 straturi de

scandura)

1 Calculul asterelii in doua straturi

Pentru calculu asterelii simple s-a utilizaIndicati metodica nr210(p20) Astereala poate fi confec983395ionată sub formă de plăci dimensionate 3x1m

utilizicircnd scicircnduri cu grosimea de 16cm şi lă983395imea de 25 cm Distan983395adintre pane este de 15 m

Fig1 Astereala in doua straturi

5

5

1



Pana independenta

140x200x5100



2 5 0 0

2 5 0 0

Sarcina pe 1 m2

de acoperi Pa







Coeficientde

siguranta f

Valoareadecalcul


90 13 117


360 13 468


650 13 849



150 11 165

Total 1199 1634

2



11991199 112361cos 097

16341634 116592cos 097

n

)a

)a

α

α

= = =

= = =

0

121752435 08500

n S = = gt


500 14 700n





1 cos 17361 1 097 11684017n n m


1 cos 23592 1 097 2288424 m



32288424 2513845 1385 10

8 8 x


= = = = times


3 31385 100106 10

13nec

Mx1 m

2i

minusminustimes

= = = times

3

2 2

6 6 0106 101022

0025

necnec

1 3 m

(



3

2 2

6 6 0106 10071

003

necnec

1 3 m

(




0

1 08003

7a m

minus= =


4 8

213 1684017 10 25 1

146384 10 225 10

l

minus

minus



3 38 408 003

225 1012 12 x

3(

mminusΣ times

= = = times

3








2

6 308 003120 10

61e mminustimes

= = times


6

1315 101096 13 12 156

120 10 i s

Mx 2 m M)a

1e σ

minus

minus


times




(1 08 003 1 1 0016) 500 10 1 1 200

200 11 220

n

n

G )a

G G )aγ


= times = times =1249 1249

12876cos 097

n )aα

= = =

1685 16851737

cos 097n )a

α = = =





( ) cos (12876 500) 25 097 4335n n n


4



( ) cos (1737 700) 25 097 5425 S am



11406

12 12 x

l M m



6 311406 10 8774 10

13nec x i1x M 2 m

minusminustimes

= = = times


2

6 32 007 02933 10


= = times asymp



( )6 3

3

4 4

25 4335 10 51 1 1 25 384 0005

374 200384 10 14 10

n x l l E

minus



Unde

34 4007 3 02

14 1012



( )6 3

3

4 6

1 4335 10 51 1 1 1 384 0005

623 200384 10 933 10

n x l l E

minus

minus


times times times

Unde3

6 4007 2 02933 10







2 53802 102 106

x4 M x M minus




2) 035 035 7 05 1225

i

s

+ d a 0

+ cd 0


= = times times =


5



min

538 6



cheson











1 10 7233 10 6233a a cm= minus = minus =






6







3 2

3 3 2

3 4 4

2 ( 2 ) 12 ( 2 ) ( 2)

2 (12 ) 2 (156 2 12) 20 12 (156 2 12) 20 (66 2)

2 12 76 09 12 12 73 10

e t t x

x

p Ep 5p E 3 ( 3 ( (

( Ep 5p E

cm

δ δ

δ






3 2

3 3 2

3 4 4

2 ( 2 ) 12 ( 2 ) ( 2)

2 (12 ) 2 (213 2 24) 24 12 (213 2 24) 24 (62 2)

2 24 86 09 12 12 1074 10

e t t x

x

p Ep 5p E 3 ( 3 ( (

( Ep 5p E

cm

δ δ

δ







14 08 095 118


M)a






1 08( 100) 1 08(244 100) 08 6ϕ λ = minus = minus =

7







3 1 2 3

3 2 3

4 4

2 ( 2 ) 12 ( 2 ) ( 2) ( ) 2 12

2 (213 2 24) 24 12 (213 2 24) 24 (62 2) 1 (09 12) 2 24 86 12

10535 10

xp t t x p x 3 ( 3 ( ( E E 0p (

cm



= times1 1

3 3

( 2 ) ( 2) ( ) 2 ( 2) (213 2 24) 24 (62 2) 1 (09 12)

2 16 24 (62 2) 149 10


cm




129 38 34ix


12 5i

= 20 5

τ

=






2 2

1

(100 ) (100 )

i i

i

i e

0 0 ( (

τ

σ τ

+ le

part part

2 2

321 041017 1

100 10024( ) 20( )

60 60

+ = lt







x


8




1 1 069 024 045 x x t ( ( ( m= minus = minus =


1 1 1

3 2 3

4 4

2 ( 2 ) 12 ( 2 ) ( 2) ( ) 2 12

2 (213 2 24) 24 12 (213 2 24) 24 (45 2) (09 12) 2 24 69 12

503 10

x p t t x x 3 ( 3 ( ( E Ep 5p (

cm



1 11 1 1

3 3

( 2 ) ( 2) ( ) 2 ( 2) (213 2 24) 24 (45 2) (09 12)

2 24 24 (45 2) 1084 10

tx p t x t xS 3 ( ( E Ep 5p ( (

m

δ δ

minus




3 4

1 1 1

(2 ) 75 10 021 (2 503 10 ) 156i x ix x p




05 (05 ) 05 156 (05 156) 206 369

45 08 095 379

it i i i p d 2pta m

M)a


= times =




2 10 122 2 71 x cm= + =

Inaltimea sectiunii



2 0647 2 024 0167ix( m= minus times =



panou)


9



3 2 3

3 2

3 4 4

2 ( 2 ) 12( 2 ) ( 2) ( ) 12

2 (213 2 24) 24 12 (213 2 24) 24 (36 2) 1 (09 12)

2 24 60 12 3592 10

ep t t r r 3 ( 3 ( ( E Ep 5p (

cm




Momentul static1 2

2 3 3

( 2 ) ( 2) ( ) 2 8 (213 2 24) 24 (36 2) 1 (09 12)

2 24 60 8 876 10

ep t r r S 3 ( ( E Ep 5p (

cm





90

( ) 6345 10 876 10 (3592 10 2 0024) 322

6 095 63

ep ep

p

4 S M)a

2 M)a


lt = =




( ) 6345 10 66 10 3592 10 2 024)

024 08 095 084

tep ep t 4 S (

2op n M)a

τ

γ




3 2 3 4 4

2 ( 2 ) 12 ( 2 ) ( 2) 2 (12 )

2 (213 2 24) 24 12 (213 2 24) 24 (833 2) 2 24 90 09 12 (12 1) 17268 10

e t t m m p Ep 5p E 3 ( 3 ( ( ( Ep 5p E

cm




(453 69 ) (453 69 06 09) 099 49

(213 2 24) 24 (833 2 24) 091

5

C

t i

β

β γ

γ




Calculam4 3 4 4 4

0

2 20

5 (384 ) 5 945 10 10 (384 10 17268 10 ) 0007

1 ( ) 0007 1 49 (09 10) 08 0012

0012 10 1 883 [ ]lim1 300 00033

n

m

l E e m

C ( l 5 m

l l


= = lt =

10




H

f i

x

q1 q2

1

2

c

x

f i


2 828 r s inc(

) l G G 5 = times + + =


0

1005 16 51 204

2 2



17136 09 154224 p p r

1 8 ( = times = times = 8568 09 77112



sticirclpilor0

03 14 08 51 17136 p

0 c t m


0 3 0225 14 04 51 8568



13173316 2 16 2

p cp p cp 9 1 1


= + = + =


2 2

1

17136 58( ) (154224 131733) 58 301272 2

p

p

9

M 1 x 9 0 m

ω times times


11



efortul de tăiere

1 152424 131733 17136 58 10146



1 2 828 ) 0 = = =


2

8568 58( ) (71112 131733) 58 265

2 2

cp

cp

9 M 1 x 9 0 m


2 77112 131733 8568 58 706



n

s(


0725 0387

5562

n

s( cm

+

= =


556161 16

345n 3ucati= = rArr


s 9 cm m= = =


4 224 552 13248 10 s


8 424 5523363932210

12 12

s x


= = =

2 26 324 552

1218816 106 6

s x


= = = times


80 1200289 0289 0552

s

l 9

i (λ


times times


c 2

1

M

e

d

e

le+=σ

2 2 3

4

1 1 80 828 100887

3000 3000 13248 10 15e c

2

λ ξ

minus

minus



1 30127

0340887

d

M M M m

ξ = = = times

3

4 6

828 10 03434 15

13248 10 1218816 10 c M) 2 M)aσ

minus

minus minus

times= + = lt =

times times

12




1le

sdotsdot+

sdotsdot

n

ce m

d

ce 21

M

2

ϕ ϕ


58836 120

0289 0289 024

l l

i 3λ = = = = lt

times times


2 2140 ( ) 140 024 232 (58 0552) 584



22 2

2

06 58 11 075 006 ( ) 1 1 075 006( ) 1 248

1 0552 2

p p

)

s s

l l m0

( ( m


2

2

0142 176 58 0142 142 0552 11 14 1 1 1 083

1 0552 4 2

p s )M

s p

l ( m0

( l mα



044 248 109 1 )

0 φ times = times = gt584 083 485 1




e

23

4S lesdot

=τ 12

8

101458 914112 10022 15

33639322 10 0144

M)a 2 M)aτ minus

minus

times times= = lt =

times times

1 101458

11440887

44 0

ξ = = =

2 2

324 552 9141128 8


424 55233639322

12 12

s3 (

cmtimes times

= = =

06 06 024 0144e




828 204 624 ) ) S 0 = minus = minus =

109 09 624

5111 11

)

0

times= = =

13



2 2 3

4

80 51 101 1 099

3000 3000 13248 10 15c

2

λ ξ

minus

minus




d

M M M m

ξ = = = times

3

4 6

51 10 0030421 15

13248 10 1218816 10t c

M)a 2 M)aσ minus

minus minus


times times

3

4 6

51 10 00304288 15

13248 10 1218816 10c c

M)a 2 M)aσ minus

minus minus


times times


03221 288

c sc

t c

(( m

σ

σ σ

times times= = =

+ +

21 0552023

288 21

t st

t c

(( m

σ

σ σ

times times= = =

+ +


300304 51 0064 10008

04 04

d an t

M t M

e e


2 012 2 032023 04

2 3 2 3

ct

(ae ( m


012 0552040 00642 2



123120 08 085

annec

3a a

cm

2 m m= = =



2 22 759 1518 123n nec

cm cm= times = gt =



3 35 3 24d d d cm+ + = 24

252695

d cm= =


alunecare

14



08 08 12 26 2496a

+ ad 0 = = times times = 035 035 552 26 502


2 2

mx 25 25 26 169

i+ d 0 = = times = 2 2 2 218 002 18 26 002 12 1505



3

min

008531 6

1505 10

an


times

Adoptăm 6 buloane


7 4 3 0026 4 0312e





2 20107-85



15




(Arce)



16



Pana independenta

140x200x5100



2 5 0 0

2 5 0 0

Sarcina pe 1 m2

de acoperi Pa







Coeficientde

siguranta f

Valoareadecalcul


90 13 117


360 13 468


650 13 849



150 11 165

Total 1199 1634

2



11991199 112361cos 097

16341634 116592cos 097

n

)a

)a

α

α

= = =

= = =

0

121752435 08500

n S = = gt


500 14 700n





1 cos 17361 1 097 11684017n n m


1 cos 23592 1 097 2288424 m



32288424 2513845 1385 10

8 8 x


= = = = times


3 31385 100106 10

13nec

Mx1 m

2i

minusminustimes

= = = times

3

2 2

6 6 0106 101022

0025

necnec

1 3 m

(



3

2 2

6 6 0106 10071

003

necnec

1 3 m

(




0

1 08003

7a m

minus= =


4 8

213 1684017 10 25 1

146384 10 225 10

l

minus

minus



3 38 408 003

225 1012 12 x

3(

mminusΣ times

= = = times

3








2

6 308 003120 10

61e mminustimes

= = times


6

1315 101096 13 12 156

120 10 i s

Mx 2 m M)a

1e σ

minus

minus


times




(1 08 003 1 1 0016) 500 10 1 1 200

200 11 220

n

n

G )a

G G )aγ


= times = times =1249 1249

12876cos 097

n )aα

= = =

1685 16851737

cos 097n )a

α = = =





( ) cos (12876 500) 25 097 4335n n n


4



( ) cos (1737 700) 25 097 5425 S am



11406

12 12 x

l M m



6 311406 10 8774 10

13nec x i1x M 2 m

minusminustimes

= = = times


2

6 32 007 02933 10


= = times asymp



( )6 3

3

4 4

25 4335 10 51 1 1 25 384 0005

374 200384 10 14 10

n x l l E

minus



Unde

34 4007 3 02

14 1012



( )6 3

3

4 6

1 4335 10 51 1 1 1 384 0005

623 200384 10 933 10

n x l l E

minus

minus


times times times

Unde3

6 4007 2 02933 10







2 53802 102 106

x4 M x M minus




2) 035 035 7 05 1225

i

s

+ d a 0

+ cd 0


= = times times =


5



min

538 6



cheson











1 10 7233 10 6233a a cm= minus = minus =






6







3 2

3 3 2

3 4 4

2 ( 2 ) 12 ( 2 ) ( 2)

2 (12 ) 2 (156 2 12) 20 12 (156 2 12) 20 (66 2)

2 12 76 09 12 12 73 10

e t t x

x

p Ep 5p E 3 ( 3 ( (

( Ep 5p E

cm

δ δ

δ






3 2

3 3 2

3 4 4

2 ( 2 ) 12 ( 2 ) ( 2)

2 (12 ) 2 (213 2 24) 24 12 (213 2 24) 24 (62 2)

2 24 86 09 12 12 1074 10

e t t x

x

p Ep 5p E 3 ( 3 ( (

( Ep 5p E

cm

δ δ

δ







14 08 095 118


M)a






1 08( 100) 1 08(244 100) 08 6ϕ λ = minus = minus =

7







3 1 2 3

3 2 3

4 4

2 ( 2 ) 12 ( 2 ) ( 2) ( ) 2 12

2 (213 2 24) 24 12 (213 2 24) 24 (62 2) 1 (09 12) 2 24 86 12

10535 10

xp t t x p x 3 ( 3 ( ( E E 0p (

cm



= times1 1

3 3

( 2 ) ( 2) ( ) 2 ( 2) (213 2 24) 24 (62 2) 1 (09 12)

2 16 24 (62 2) 149 10


cm




129 38 34ix


12 5i

= 20 5

τ

=






2 2

1

(100 ) (100 )

i i

i

i e

0 0 ( (

τ

σ τ

+ le

part part

2 2

321 041017 1

100 10024( ) 20( )

60 60

+ = lt







x


8




1 1 069 024 045 x x t ( ( ( m= minus = minus =


1 1 1

3 2 3

4 4

2 ( 2 ) 12 ( 2 ) ( 2) ( ) 2 12

2 (213 2 24) 24 12 (213 2 24) 24 (45 2) (09 12) 2 24 69 12

503 10

x p t t x x 3 ( 3 ( ( E Ep 5p (

cm



1 11 1 1

3 3

( 2 ) ( 2) ( ) 2 ( 2) (213 2 24) 24 (45 2) (09 12)

2 24 24 (45 2) 1084 10

tx p t x t xS 3 ( ( E Ep 5p ( (

m

δ δ

minus




3 4

1 1 1

(2 ) 75 10 021 (2 503 10 ) 156i x ix x p




05 (05 ) 05 156 (05 156) 206 369

45 08 095 379

it i i i p d 2pta m

M)a


= times =




2 10 122 2 71 x cm= + =

Inaltimea sectiunii



2 0647 2 024 0167ix( m= minus times =



panou)


9



3 2 3

3 2

3 4 4

2 ( 2 ) 12( 2 ) ( 2) ( ) 12

2 (213 2 24) 24 12 (213 2 24) 24 (36 2) 1 (09 12)

2 24 60 12 3592 10

ep t t r r 3 ( 3 ( ( E Ep 5p (

cm




Momentul static1 2

2 3 3

( 2 ) ( 2) ( ) 2 8 (213 2 24) 24 (36 2) 1 (09 12)

2 24 60 8 876 10

ep t r r S 3 ( ( E Ep 5p (

cm





90

( ) 6345 10 876 10 (3592 10 2 0024) 322

6 095 63

ep ep

p

4 S M)a

2 M)a


lt = =




( ) 6345 10 66 10 3592 10 2 024)

024 08 095 084

tep ep t 4 S (

2op n M)a

τ

γ




3 2 3 4 4

2 ( 2 ) 12 ( 2 ) ( 2) 2 (12 )

2 (213 2 24) 24 12 (213 2 24) 24 (833 2) 2 24 90 09 12 (12 1) 17268 10

e t t m m p Ep 5p E 3 ( 3 ( ( ( Ep 5p E

cm




(453 69 ) (453 69 06 09) 099 49

(213 2 24) 24 (833 2 24) 091

5

C

t i

β

β γ

γ




Calculam4 3 4 4 4

0

2 20

5 (384 ) 5 945 10 10 (384 10 17268 10 ) 0007

1 ( ) 0007 1 49 (09 10) 08 0012

0012 10 1 883 [ ]lim1 300 00033

n

m

l E e m

C ( l 5 m

l l


= = lt =

10




H

f i

x

q1 q2

1

2

c

x

f i


2 828 r s inc(

) l G G 5 = times + + =


0

1005 16 51 204

2 2



17136 09 154224 p p r

1 8 ( = times = times = 8568 09 77112



sticirclpilor0

03 14 08 51 17136 p

0 c t m


0 3 0225 14 04 51 8568



13173316 2 16 2

p cp p cp 9 1 1


= + = + =


2 2

1

17136 58( ) (154224 131733) 58 301272 2

p

p

9

M 1 x 9 0 m

ω times times


11



efortul de tăiere

1 152424 131733 17136 58 10146



1 2 828 ) 0 = = =


2

8568 58( ) (71112 131733) 58 265

2 2

cp

cp

9 M 1 x 9 0 m


2 77112 131733 8568 58 706



n

s(


0725 0387

5562

n

s( cm

+

= =


556161 16

345n 3ucati= = rArr


s 9 cm m= = =


4 224 552 13248 10 s


8 424 5523363932210

12 12

s x


= = =

2 26 324 552

1218816 106 6

s x


= = = times


80 1200289 0289 0552

s

l 9

i (λ


times times


c 2

1

M

e

d

e

le+=σ

2 2 3

4

1 1 80 828 100887

3000 3000 13248 10 15e c

2

λ ξ

minus

minus



1 30127

0340887

d

M M M m

ξ = = = times

3

4 6

828 10 03434 15

13248 10 1218816 10 c M) 2 M)aσ

minus

minus minus

times= + = lt =

times times

12




1le

sdotsdot+

sdotsdot

n

ce m

d

ce 21

M

2

ϕ ϕ


58836 120

0289 0289 024

l l

i 3λ = = = = lt

times times


2 2140 ( ) 140 024 232 (58 0552) 584



22 2

2

06 58 11 075 006 ( ) 1 1 075 006( ) 1 248

1 0552 2

p p

)

s s

l l m0

( ( m


2

2

0142 176 58 0142 142 0552 11 14 1 1 1 083

1 0552 4 2

p s )M

s p

l ( m0

( l mα



044 248 109 1 )

0 φ times = times = gt584 083 485 1




e

23

4S lesdot

=τ 12

8

101458 914112 10022 15

33639322 10 0144

M)a 2 M)aτ minus

minus

times times= = lt =

times times

1 101458

11440887

44 0

ξ = = =

2 2

324 552 9141128 8


424 55233639322

12 12

s3 (

cmtimes times

= = =

06 06 024 0144e




828 204 624 ) ) S 0 = minus = minus =

109 09 624

5111 11

)

0

times= = =

13



2 2 3

4

80 51 101 1 099

3000 3000 13248 10 15c

2

λ ξ

minus

minus




d

M M M m

ξ = = = times

3

4 6

51 10 0030421 15

13248 10 1218816 10t c

M)a 2 M)aσ minus

minus minus


times times

3

4 6

51 10 00304288 15

13248 10 1218816 10c c

M)a 2 M)aσ minus

minus minus


times times


03221 288

c sc

t c

(( m

σ

σ σ

times times= = =

+ +

21 0552023

288 21

t st

t c

(( m

σ

σ σ

times times= = =

+ +


300304 51 0064 10008

04 04

d an t

M t M

e e


2 012 2 032023 04

2 3 2 3

ct

(ae ( m


012 0552040 00642 2



123120 08 085

annec

3a a

cm

2 m m= = =



2 22 759 1518 123n nec

cm cm= times = gt =



3 35 3 24d d d cm+ + = 24

252695

d cm= =


alunecare

14



08 08 12 26 2496a

+ ad 0 = = times times = 035 035 552 26 502


2 2

mx 25 25 26 169

i+ d 0 = = times = 2 2 2 218 002 18 26 002 12 1505



3

min

008531 6

1505 10

an


times

Adoptăm 6 buloane


7 4 3 0026 4 0312e





2 20107-85



15




(Arce)



16



11991199 112361cos 097

16341634 116592cos 097

n

)a

)a

α

α

= = =

= = =

0

121752435 08500

n S = = gt


500 14 700n





1 cos 17361 1 097 11684017n n m


1 cos 23592 1 097 2288424 m



32288424 2513845 1385 10

8 8 x


= = = = times


3 31385 100106 10

13nec

Mx1 m

2i

minusminustimes

= = = times

3

2 2

6 6 0106 101022

0025

necnec

1 3 m

(



3

2 2

6 6 0106 10071

003

necnec

1 3 m

(




0

1 08003

7a m

minus= =


4 8

213 1684017 10 25 1

146384 10 225 10

l

minus

minus



3 38 408 003

225 1012 12 x

3(

mminusΣ times

= = = times

3








2

6 308 003120 10

61e mminustimes

= = times


6

1315 101096 13 12 156

120 10 i s

Mx 2 m M)a

1e σ

minus

minus


times




(1 08 003 1 1 0016) 500 10 1 1 200

200 11 220

n

n

G )a

G G )aγ


= times = times =1249 1249

12876cos 097

n )aα

= = =

1685 16851737

cos 097n )a

α = = =





( ) cos (12876 500) 25 097 4335n n n


4



( ) cos (1737 700) 25 097 5425 S am



11406

12 12 x

l M m



6 311406 10 8774 10

13nec x i1x M 2 m

minusminustimes

= = = times


2

6 32 007 02933 10


= = times asymp



( )6 3

3

4 4

25 4335 10 51 1 1 25 384 0005

374 200384 10 14 10

n x l l E

minus



Unde

34 4007 3 02

14 1012



( )6 3

3

4 6

1 4335 10 51 1 1 1 384 0005

623 200384 10 933 10

n x l l E

minus

minus


times times times

Unde3

6 4007 2 02933 10







2 53802 102 106

x4 M x M minus




2) 035 035 7 05 1225

i

s

+ d a 0

+ cd 0


= = times times =


5



min

538 6



cheson











1 10 7233 10 6233a a cm= minus = minus =






6







3 2

3 3 2

3 4 4

2 ( 2 ) 12 ( 2 ) ( 2)

2 (12 ) 2 (156 2 12) 20 12 (156 2 12) 20 (66 2)

2 12 76 09 12 12 73 10

e t t x

x

p Ep 5p E 3 ( 3 ( (

( Ep 5p E

cm

δ δ

δ






3 2

3 3 2

3 4 4

2 ( 2 ) 12 ( 2 ) ( 2)

2 (12 ) 2 (213 2 24) 24 12 (213 2 24) 24 (62 2)

2 24 86 09 12 12 1074 10

e t t x

x

p Ep 5p E 3 ( 3 ( (

( Ep 5p E

cm

δ δ

δ







14 08 095 118


M)a






1 08( 100) 1 08(244 100) 08 6ϕ λ = minus = minus =

7







3 1 2 3

3 2 3

4 4

2 ( 2 ) 12 ( 2 ) ( 2) ( ) 2 12

2 (213 2 24) 24 12 (213 2 24) 24 (62 2) 1 (09 12) 2 24 86 12

10535 10

xp t t x p x 3 ( 3 ( ( E E 0p (

cm



= times1 1

3 3

( 2 ) ( 2) ( ) 2 ( 2) (213 2 24) 24 (62 2) 1 (09 12)

2 16 24 (62 2) 149 10


cm




129 38 34ix


12 5i

= 20 5

τ

=






2 2

1

(100 ) (100 )

i i

i

i e

0 0 ( (

τ

σ τ

+ le

part part

2 2

321 041017 1

100 10024( ) 20( )

60 60

+ = lt







x


8




1 1 069 024 045 x x t ( ( ( m= minus = minus =


1 1 1

3 2 3

4 4

2 ( 2 ) 12 ( 2 ) ( 2) ( ) 2 12

2 (213 2 24) 24 12 (213 2 24) 24 (45 2) (09 12) 2 24 69 12

503 10

x p t t x x 3 ( 3 ( ( E Ep 5p (

cm



1 11 1 1

3 3

( 2 ) ( 2) ( ) 2 ( 2) (213 2 24) 24 (45 2) (09 12)

2 24 24 (45 2) 1084 10

tx p t x t xS 3 ( ( E Ep 5p ( (

m

δ δ

minus




3 4

1 1 1

(2 ) 75 10 021 (2 503 10 ) 156i x ix x p




05 (05 ) 05 156 (05 156) 206 369

45 08 095 379

it i i i p d 2pta m

M)a


= times =




2 10 122 2 71 x cm= + =

Inaltimea sectiunii



2 0647 2 024 0167ix( m= minus times =



panou)


9



3 2 3

3 2

3 4 4

2 ( 2 ) 12( 2 ) ( 2) ( ) 12

2 (213 2 24) 24 12 (213 2 24) 24 (36 2) 1 (09 12)

2 24 60 12 3592 10

ep t t r r 3 ( 3 ( ( E Ep 5p (

cm




Momentul static1 2

2 3 3

( 2 ) ( 2) ( ) 2 8 (213 2 24) 24 (36 2) 1 (09 12)

2 24 60 8 876 10

ep t r r S 3 ( ( E Ep 5p (

cm





90

( ) 6345 10 876 10 (3592 10 2 0024) 322

6 095 63

ep ep

p

4 S M)a

2 M)a


lt = =




( ) 6345 10 66 10 3592 10 2 024)

024 08 095 084

tep ep t 4 S (

2op n M)a

τ

γ




3 2 3 4 4

2 ( 2 ) 12 ( 2 ) ( 2) 2 (12 )

2 (213 2 24) 24 12 (213 2 24) 24 (833 2) 2 24 90 09 12 (12 1) 17268 10

e t t m m p Ep 5p E 3 ( 3 ( ( ( Ep 5p E

cm




(453 69 ) (453 69 06 09) 099 49

(213 2 24) 24 (833 2 24) 091

5

C

t i

β

β γ

γ




Calculam4 3 4 4 4

0

2 20

5 (384 ) 5 945 10 10 (384 10 17268 10 ) 0007

1 ( ) 0007 1 49 (09 10) 08 0012

0012 10 1 883 [ ]lim1 300 00033

n

m

l E e m

C ( l 5 m

l l


= = lt =

10




H

f i

x

q1 q2

1

2

c

x

f i


2 828 r s inc(

) l G G 5 = times + + =


0

1005 16 51 204

2 2



17136 09 154224 p p r

1 8 ( = times = times = 8568 09 77112



sticirclpilor0

03 14 08 51 17136 p

0 c t m


0 3 0225 14 04 51 8568



13173316 2 16 2

p cp p cp 9 1 1


= + = + =


2 2

1

17136 58( ) (154224 131733) 58 301272 2

p

p

9

M 1 x 9 0 m

ω times times


11



efortul de tăiere

1 152424 131733 17136 58 10146



1 2 828 ) 0 = = =


2

8568 58( ) (71112 131733) 58 265

2 2

cp

cp

9 M 1 x 9 0 m


2 77112 131733 8568 58 706



n

s(


0725 0387

5562

n

s( cm

+

= =


556161 16

345n 3ucati= = rArr


s 9 cm m= = =


4 224 552 13248 10 s


8 424 5523363932210

12 12

s x


= = =

2 26 324 552

1218816 106 6

s x


= = = times


80 1200289 0289 0552

s

l 9

i (λ


times times


c 2

1

M

e

d

e

le+=σ

2 2 3

4

1 1 80 828 100887

3000 3000 13248 10 15e c

2

λ ξ

minus

minus



1 30127

0340887

d

M M M m

ξ = = = times

3

4 6

828 10 03434 15

13248 10 1218816 10 c M) 2 M)aσ

minus

minus minus

times= + = lt =

times times

12




1le

sdotsdot+

sdotsdot

n

ce m

d

ce 21

M

2

ϕ ϕ


58836 120

0289 0289 024

l l

i 3λ = = = = lt

times times


2 2140 ( ) 140 024 232 (58 0552) 584



22 2

2

06 58 11 075 006 ( ) 1 1 075 006( ) 1 248

1 0552 2

p p

)

s s

l l m0

( ( m


2

2

0142 176 58 0142 142 0552 11 14 1 1 1 083

1 0552 4 2

p s )M

s p

l ( m0

( l mα



044 248 109 1 )

0 φ times = times = gt584 083 485 1




e

23

4S lesdot

=τ 12

8

101458 914112 10022 15

33639322 10 0144

M)a 2 M)aτ minus

minus

times times= = lt =

times times

1 101458

11440887

44 0

ξ = = =

2 2

324 552 9141128 8


424 55233639322

12 12

s3 (

cmtimes times

= = =

06 06 024 0144e




828 204 624 ) ) S 0 = minus = minus =

109 09 624

5111 11

)

0

times= = =

13



2 2 3

4

80 51 101 1 099

3000 3000 13248 10 15c

2

λ ξ

minus

minus




d

M M M m

ξ = = = times

3

4 6

51 10 0030421 15

13248 10 1218816 10t c

M)a 2 M)aσ minus

minus minus


times times

3

4 6

51 10 00304288 15

13248 10 1218816 10c c

M)a 2 M)aσ minus

minus minus


times times


03221 288

c sc

t c

(( m

σ

σ σ

times times= = =

+ +

21 0552023

288 21

t st

t c

(( m

σ

σ σ

times times= = =

+ +


300304 51 0064 10008

04 04

d an t

M t M

e e


2 012 2 032023 04

2 3 2 3

ct

(ae ( m


012 0552040 00642 2



123120 08 085

annec

3a a

cm

2 m m= = =



2 22 759 1518 123n nec

cm cm= times = gt =



3 35 3 24d d d cm+ + = 24

252695

d cm= =


alunecare

14



08 08 12 26 2496a

+ ad 0 = = times times = 035 035 552 26 502


2 2

mx 25 25 26 169

i+ d 0 = = times = 2 2 2 218 002 18 26 002 12 1505



3

min

008531 6

1505 10

an


times

Adoptăm 6 buloane


7 4 3 0026 4 0312e





2 20107-85



15




(Arce)



16








2

6 308 003120 10

61e mminustimes

= = times


6

1315 101096 13 12 156

120 10 i s

Mx 2 m M)a

1e σ

minus

minus


times




(1 08 003 1 1 0016) 500 10 1 1 200

200 11 220

n

n

G )a

G G )aγ


= times = times =1249 1249

12876cos 097

n )aα

= = =

1685 16851737

cos 097n )a

α = = =





( ) cos (12876 500) 25 097 4335n n n


4



( ) cos (1737 700) 25 097 5425 S am



11406

12 12 x

l M m



6 311406 10 8774 10

13nec x i1x M 2 m

minusminustimes

= = = times


2

6 32 007 02933 10


= = times asymp



( )6 3

3

4 4

25 4335 10 51 1 1 25 384 0005

374 200384 10 14 10

n x l l E

minus



Unde

34 4007 3 02

14 1012



( )6 3

3

4 6

1 4335 10 51 1 1 1 384 0005

623 200384 10 933 10

n x l l E

minus

minus


times times times

Unde3

6 4007 2 02933 10







2 53802 102 106

x4 M x M minus




2) 035 035 7 05 1225

i

s

+ d a 0

+ cd 0


= = times times =


5



min

538 6



cheson











1 10 7233 10 6233a a cm= minus = minus =






6







3 2

3 3 2

3 4 4

2 ( 2 ) 12 ( 2 ) ( 2)

2 (12 ) 2 (156 2 12) 20 12 (156 2 12) 20 (66 2)

2 12 76 09 12 12 73 10

e t t x

x

p Ep 5p E 3 ( 3 ( (

( Ep 5p E

cm

δ δ

δ






3 2

3 3 2

3 4 4

2 ( 2 ) 12 ( 2 ) ( 2)

2 (12 ) 2 (213 2 24) 24 12 (213 2 24) 24 (62 2)

2 24 86 09 12 12 1074 10

e t t x

x

p Ep 5p E 3 ( 3 ( (

( Ep 5p E

cm

δ δ

δ







14 08 095 118


M)a






1 08( 100) 1 08(244 100) 08 6ϕ λ = minus = minus =

7







3 1 2 3

3 2 3

4 4

2 ( 2 ) 12 ( 2 ) ( 2) ( ) 2 12

2 (213 2 24) 24 12 (213 2 24) 24 (62 2) 1 (09 12) 2 24 86 12

10535 10

xp t t x p x 3 ( 3 ( ( E E 0p (

cm



= times1 1

3 3

( 2 ) ( 2) ( ) 2 ( 2) (213 2 24) 24 (62 2) 1 (09 12)

2 16 24 (62 2) 149 10


cm




129 38 34ix


12 5i

= 20 5

τ

=






2 2

1

(100 ) (100 )

i i

i

i e

0 0 ( (

τ

σ τ

+ le

part part

2 2

321 041017 1

100 10024( ) 20( )

60 60

+ = lt







x


8




1 1 069 024 045 x x t ( ( ( m= minus = minus =


1 1 1

3 2 3

4 4

2 ( 2 ) 12 ( 2 ) ( 2) ( ) 2 12

2 (213 2 24) 24 12 (213 2 24) 24 (45 2) (09 12) 2 24 69 12

503 10

x p t t x x 3 ( 3 ( ( E Ep 5p (

cm



1 11 1 1

3 3

( 2 ) ( 2) ( ) 2 ( 2) (213 2 24) 24 (45 2) (09 12)

2 24 24 (45 2) 1084 10

tx p t x t xS 3 ( ( E Ep 5p ( (

m

δ δ

minus




3 4

1 1 1

(2 ) 75 10 021 (2 503 10 ) 156i x ix x p




05 (05 ) 05 156 (05 156) 206 369

45 08 095 379

it i i i p d 2pta m

M)a


= times =




2 10 122 2 71 x cm= + =

Inaltimea sectiunii



2 0647 2 024 0167ix( m= minus times =



panou)


9



3 2 3

3 2

3 4 4

2 ( 2 ) 12( 2 ) ( 2) ( ) 12

2 (213 2 24) 24 12 (213 2 24) 24 (36 2) 1 (09 12)

2 24 60 12 3592 10

ep t t r r 3 ( 3 ( ( E Ep 5p (

cm




Momentul static1 2

2 3 3

( 2 ) ( 2) ( ) 2 8 (213 2 24) 24 (36 2) 1 (09 12)

2 24 60 8 876 10

ep t r r S 3 ( ( E Ep 5p (

cm





90

( ) 6345 10 876 10 (3592 10 2 0024) 322

6 095 63

ep ep

p

4 S M)a

2 M)a


lt = =




( ) 6345 10 66 10 3592 10 2 024)

024 08 095 084

tep ep t 4 S (

2op n M)a

τ

γ




3 2 3 4 4

2 ( 2 ) 12 ( 2 ) ( 2) 2 (12 )

2 (213 2 24) 24 12 (213 2 24) 24 (833 2) 2 24 90 09 12 (12 1) 17268 10

e t t m m p Ep 5p E 3 ( 3 ( ( ( Ep 5p E

cm




(453 69 ) (453 69 06 09) 099 49

(213 2 24) 24 (833 2 24) 091

5

C

t i

β

β γ

γ




Calculam4 3 4 4 4

0

2 20

5 (384 ) 5 945 10 10 (384 10 17268 10 ) 0007

1 ( ) 0007 1 49 (09 10) 08 0012

0012 10 1 883 [ ]lim1 300 00033

n

m

l E e m

C ( l 5 m

l l


= = lt =

10




H

f i

x

q1 q2

1

2

c

x

f i


2 828 r s inc(

) l G G 5 = times + + =


0

1005 16 51 204

2 2



17136 09 154224 p p r

1 8 ( = times = times = 8568 09 77112



sticirclpilor0

03 14 08 51 17136 p

0 c t m


0 3 0225 14 04 51 8568



13173316 2 16 2

p cp p cp 9 1 1


= + = + =


2 2

1

17136 58( ) (154224 131733) 58 301272 2

p

p

9

M 1 x 9 0 m

ω times times


11



efortul de tăiere

1 152424 131733 17136 58 10146



1 2 828 ) 0 = = =


2

8568 58( ) (71112 131733) 58 265

2 2

cp

cp

9 M 1 x 9 0 m


2 77112 131733 8568 58 706



n

s(


0725 0387

5562

n

s( cm

+

= =


556161 16

345n 3ucati= = rArr


s 9 cm m= = =


4 224 552 13248 10 s


8 424 5523363932210

12 12

s x


= = =

2 26 324 552

1218816 106 6

s x


= = = times


80 1200289 0289 0552

s

l 9

i (λ


times times


c 2

1

M

e

d

e

le+=σ

2 2 3

4

1 1 80 828 100887

3000 3000 13248 10 15e c

2

λ ξ

minus

minus



1 30127

0340887

d

M M M m

ξ = = = times

3

4 6

828 10 03434 15

13248 10 1218816 10 c M) 2 M)aσ

minus

minus minus

times= + = lt =

times times

12




1le

sdotsdot+

sdotsdot

n

ce m

d

ce 21

M

2

ϕ ϕ


58836 120

0289 0289 024

l l

i 3λ = = = = lt

times times


2 2140 ( ) 140 024 232 (58 0552) 584



22 2

2

06 58 11 075 006 ( ) 1 1 075 006( ) 1 248

1 0552 2

p p

)

s s

l l m0

( ( m


2

2

0142 176 58 0142 142 0552 11 14 1 1 1 083

1 0552 4 2

p s )M

s p

l ( m0

( l mα



044 248 109 1 )

0 φ times = times = gt584 083 485 1




e

23

4S lesdot

=τ 12

8

101458 914112 10022 15

33639322 10 0144

M)a 2 M)aτ minus

minus

times times= = lt =

times times

1 101458

11440887

44 0

ξ = = =

2 2

324 552 9141128 8


424 55233639322

12 12

s3 (

cmtimes times

= = =

06 06 024 0144e




828 204 624 ) ) S 0 = minus = minus =

109 09 624

5111 11

)

0

times= = =

13



2 2 3

4

80 51 101 1 099

3000 3000 13248 10 15c

2

λ ξ

minus

minus




d

M M M m

ξ = = = times

3

4 6

51 10 0030421 15

13248 10 1218816 10t c

M)a 2 M)aσ minus

minus minus


times times

3

4 6

51 10 00304288 15

13248 10 1218816 10c c

M)a 2 M)aσ minus

minus minus


times times


03221 288

c sc

t c

(( m

σ

σ σ

times times= = =

+ +

21 0552023

288 21

t st

t c

(( m

σ

σ σ

times times= = =

+ +


300304 51 0064 10008

04 04

d an t

M t M

e e


2 012 2 032023 04

2 3 2 3

ct

(ae ( m


012 0552040 00642 2



123120 08 085

annec

3a a

cm

2 m m= = =



2 22 759 1518 123n nec

cm cm= times = gt =



3 35 3 24d d d cm+ + = 24

252695

d cm= =


alunecare

14



08 08 12 26 2496a

+ ad 0 = = times times = 035 035 552 26 502


2 2

mx 25 25 26 169

i+ d 0 = = times = 2 2 2 218 002 18 26 002 12 1505



3

min

008531 6

1505 10

an


times

Adoptăm 6 buloane


7 4 3 0026 4 0312e





2 20107-85



15




(Arce)



16



( ) cos (1737 700) 25 097 5425 S am



11406

12 12 x

l M m



6 311406 10 8774 10

13nec x i1x M 2 m

minusminustimes

= = = times


2

6 32 007 02933 10


= = times asymp



( )6 3

3

4 4

25 4335 10 51 1 1 25 384 0005

374 200384 10 14 10

n x l l E

minus



Unde

34 4007 3 02

14 1012



( )6 3

3

4 6

1 4335 10 51 1 1 1 384 0005

623 200384 10 933 10

n x l l E

minus

minus


times times times

Unde3

6 4007 2 02933 10







2 53802 102 106

x4 M x M minus




2) 035 035 7 05 1225

i

s

+ d a 0

+ cd 0


= = times times =


5



min

538 6



cheson











1 10 7233 10 6233a a cm= minus = minus =






6







3 2

3 3 2

3 4 4

2 ( 2 ) 12 ( 2 ) ( 2)

2 (12 ) 2 (156 2 12) 20 12 (156 2 12) 20 (66 2)

2 12 76 09 12 12 73 10

e t t x

x

p Ep 5p E 3 ( 3 ( (

( Ep 5p E

cm

δ δ

δ






3 2

3 3 2

3 4 4

2 ( 2 ) 12 ( 2 ) ( 2)

2 (12 ) 2 (213 2 24) 24 12 (213 2 24) 24 (62 2)

2 24 86 09 12 12 1074 10

e t t x

x

p Ep 5p E 3 ( 3 ( (

( Ep 5p E

cm

δ δ

δ







14 08 095 118


M)a






1 08( 100) 1 08(244 100) 08 6ϕ λ = minus = minus =

7







3 1 2 3

3 2 3

4 4

2 ( 2 ) 12 ( 2 ) ( 2) ( ) 2 12

2 (213 2 24) 24 12 (213 2 24) 24 (62 2) 1 (09 12) 2 24 86 12

10535 10

xp t t x p x 3 ( 3 ( ( E E 0p (

cm



= times1 1

3 3

( 2 ) ( 2) ( ) 2 ( 2) (213 2 24) 24 (62 2) 1 (09 12)

2 16 24 (62 2) 149 10


cm




129 38 34ix


12 5i

= 20 5

τ

=






2 2

1

(100 ) (100 )

i i

i

i e

0 0 ( (

τ

σ τ

+ le

part part

2 2

321 041017 1

100 10024( ) 20( )

60 60

+ = lt







x


8




1 1 069 024 045 x x t ( ( ( m= minus = minus =


1 1 1

3 2 3

4 4

2 ( 2 ) 12 ( 2 ) ( 2) ( ) 2 12

2 (213 2 24) 24 12 (213 2 24) 24 (45 2) (09 12) 2 24 69 12

503 10

x p t t x x 3 ( 3 ( ( E Ep 5p (

cm



1 11 1 1

3 3

( 2 ) ( 2) ( ) 2 ( 2) (213 2 24) 24 (45 2) (09 12)

2 24 24 (45 2) 1084 10

tx p t x t xS 3 ( ( E Ep 5p ( (

m

δ δ

minus




3 4

1 1 1

(2 ) 75 10 021 (2 503 10 ) 156i x ix x p




05 (05 ) 05 156 (05 156) 206 369

45 08 095 379

it i i i p d 2pta m

M)a


= times =




2 10 122 2 71 x cm= + =

Inaltimea sectiunii



2 0647 2 024 0167ix( m= minus times =



panou)


9



3 2 3

3 2

3 4 4

2 ( 2 ) 12( 2 ) ( 2) ( ) 12

2 (213 2 24) 24 12 (213 2 24) 24 (36 2) 1 (09 12)

2 24 60 12 3592 10

ep t t r r 3 ( 3 ( ( E Ep 5p (

cm




Momentul static1 2

2 3 3

( 2 ) ( 2) ( ) 2 8 (213 2 24) 24 (36 2) 1 (09 12)

2 24 60 8 876 10

ep t r r S 3 ( ( E Ep 5p (

cm





90

( ) 6345 10 876 10 (3592 10 2 0024) 322

6 095 63

ep ep

p

4 S M)a

2 M)a


lt = =




( ) 6345 10 66 10 3592 10 2 024)

024 08 095 084

tep ep t 4 S (

2op n M)a

τ

γ




3 2 3 4 4

2 ( 2 ) 12 ( 2 ) ( 2) 2 (12 )

2 (213 2 24) 24 12 (213 2 24) 24 (833 2) 2 24 90 09 12 (12 1) 17268 10

e t t m m p Ep 5p E 3 ( 3 ( ( ( Ep 5p E

cm




(453 69 ) (453 69 06 09) 099 49

(213 2 24) 24 (833 2 24) 091

5

C

t i

β

β γ

γ




Calculam4 3 4 4 4

0

2 20

5 (384 ) 5 945 10 10 (384 10 17268 10 ) 0007

1 ( ) 0007 1 49 (09 10) 08 0012

0012 10 1 883 [ ]lim1 300 00033

n

m

l E e m

C ( l 5 m

l l


= = lt =

10




H

f i

x

q1 q2

1

2

c

x

f i


2 828 r s inc(

) l G G 5 = times + + =


0

1005 16 51 204

2 2



17136 09 154224 p p r

1 8 ( = times = times = 8568 09 77112



sticirclpilor0

03 14 08 51 17136 p

0 c t m


0 3 0225 14 04 51 8568



13173316 2 16 2

p cp p cp 9 1 1


= + = + =


2 2

1

17136 58( ) (154224 131733) 58 301272 2

p

p

9

M 1 x 9 0 m

ω times times


11



efortul de tăiere

1 152424 131733 17136 58 10146



1 2 828 ) 0 = = =


2

8568 58( ) (71112 131733) 58 265

2 2

cp

cp

9 M 1 x 9 0 m


2 77112 131733 8568 58 706



n

s(


0725 0387

5562

n

s( cm

+

= =


556161 16

345n 3ucati= = rArr


s 9 cm m= = =


4 224 552 13248 10 s


8 424 5523363932210

12 12

s x


= = =

2 26 324 552

1218816 106 6

s x


= = = times


80 1200289 0289 0552

s

l 9

i (λ


times times


c 2

1

M

e

d

e

le+=σ

2 2 3

4

1 1 80 828 100887

3000 3000 13248 10 15e c

2

λ ξ

minus

minus



1 30127

0340887

d

M M M m

ξ = = = times

3

4 6

828 10 03434 15

13248 10 1218816 10 c M) 2 M)aσ

minus

minus minus

times= + = lt =

times times

12




1le

sdotsdot+

sdotsdot

n

ce m

d

ce 21

M

2

ϕ ϕ


58836 120

0289 0289 024

l l

i 3λ = = = = lt

times times


2 2140 ( ) 140 024 232 (58 0552) 584



22 2

2

06 58 11 075 006 ( ) 1 1 075 006( ) 1 248

1 0552 2

p p

)

s s

l l m0

( ( m


2

2

0142 176 58 0142 142 0552 11 14 1 1 1 083

1 0552 4 2

p s )M

s p

l ( m0

( l mα



044 248 109 1 )

0 φ times = times = gt584 083 485 1




e

23

4S lesdot

=τ 12

8

101458 914112 10022 15

33639322 10 0144

M)a 2 M)aτ minus

minus

times times= = lt =

times times

1 101458

11440887

44 0

ξ = = =

2 2

324 552 9141128 8


424 55233639322

12 12

s3 (

cmtimes times

= = =

06 06 024 0144e




828 204 624 ) ) S 0 = minus = minus =

109 09 624

5111 11

)

0

times= = =

13



2 2 3

4

80 51 101 1 099

3000 3000 13248 10 15c

2

λ ξ

minus

minus




d

M M M m

ξ = = = times

3

4 6

51 10 0030421 15

13248 10 1218816 10t c

M)a 2 M)aσ minus

minus minus


times times

3

4 6

51 10 00304288 15

13248 10 1218816 10c c

M)a 2 M)aσ minus

minus minus


times times


03221 288

c sc

t c

(( m

σ

σ σ

times times= = =

+ +

21 0552023

288 21

t st

t c

(( m

σ

σ σ

times times= = =

+ +


300304 51 0064 10008

04 04

d an t

M t M

e e


2 012 2 032023 04

2 3 2 3

ct

(ae ( m


012 0552040 00642 2



123120 08 085

annec

3a a

cm

2 m m= = =



2 22 759 1518 123n nec

cm cm= times = gt =



3 35 3 24d d d cm+ + = 24

252695

d cm= =


alunecare

14



08 08 12 26 2496a

+ ad 0 = = times times = 035 035 552 26 502


2 2

mx 25 25 26 169

i+ d 0 = = times = 2 2 2 218 002 18 26 002 12 1505



3

min

008531 6

1505 10

an


times

Adoptăm 6 buloane


7 4 3 0026 4 0312e





2 20107-85



15




(Arce)



16



min

538 6



cheson











1 10 7233 10 6233a a cm= minus = minus =






6







3 2

3 3 2

3 4 4

2 ( 2 ) 12 ( 2 ) ( 2)

2 (12 ) 2 (156 2 12) 20 12 (156 2 12) 20 (66 2)

2 12 76 09 12 12 73 10

e t t x

x

p Ep 5p E 3 ( 3 ( (

( Ep 5p E

cm

δ δ

δ






3 2

3 3 2

3 4 4

2 ( 2 ) 12 ( 2 ) ( 2)

2 (12 ) 2 (213 2 24) 24 12 (213 2 24) 24 (62 2)

2 24 86 09 12 12 1074 10

e t t x

x

p Ep 5p E 3 ( 3 ( (

( Ep 5p E

cm

δ δ

δ







14 08 095 118


M)a






1 08( 100) 1 08(244 100) 08 6ϕ λ = minus = minus =

7







3 1 2 3

3 2 3

4 4

2 ( 2 ) 12 ( 2 ) ( 2) ( ) 2 12

2 (213 2 24) 24 12 (213 2 24) 24 (62 2) 1 (09 12) 2 24 86 12

10535 10

xp t t x p x 3 ( 3 ( ( E E 0p (

cm



= times1 1

3 3

( 2 ) ( 2) ( ) 2 ( 2) (213 2 24) 24 (62 2) 1 (09 12)

2 16 24 (62 2) 149 10


cm




129 38 34ix


12 5i

= 20 5

τ

=






2 2

1

(100 ) (100 )

i i

i

i e

0 0 ( (

τ

σ τ

+ le

part part

2 2

321 041017 1

100 10024( ) 20( )

60 60

+ = lt







x


8




1 1 069 024 045 x x t ( ( ( m= minus = minus =


1 1 1

3 2 3

4 4

2 ( 2 ) 12 ( 2 ) ( 2) ( ) 2 12

2 (213 2 24) 24 12 (213 2 24) 24 (45 2) (09 12) 2 24 69 12

503 10

x p t t x x 3 ( 3 ( ( E Ep 5p (

cm



1 11 1 1

3 3

( 2 ) ( 2) ( ) 2 ( 2) (213 2 24) 24 (45 2) (09 12)

2 24 24 (45 2) 1084 10

tx p t x t xS 3 ( ( E Ep 5p ( (

m

δ δ

minus




3 4

1 1 1

(2 ) 75 10 021 (2 503 10 ) 156i x ix x p




05 (05 ) 05 156 (05 156) 206 369

45 08 095 379

it i i i p d 2pta m

M)a


= times =




2 10 122 2 71 x cm= + =

Inaltimea sectiunii



2 0647 2 024 0167ix( m= minus times =



panou)


9



3 2 3

3 2

3 4 4

2 ( 2 ) 12( 2 ) ( 2) ( ) 12

2 (213 2 24) 24 12 (213 2 24) 24 (36 2) 1 (09 12)

2 24 60 12 3592 10

ep t t r r 3 ( 3 ( ( E Ep 5p (

cm




Momentul static1 2

2 3 3

( 2 ) ( 2) ( ) 2 8 (213 2 24) 24 (36 2) 1 (09 12)

2 24 60 8 876 10

ep t r r S 3 ( ( E Ep 5p (

cm





90

( ) 6345 10 876 10 (3592 10 2 0024) 322

6 095 63

ep ep

p

4 S M)a

2 M)a


lt = =




( ) 6345 10 66 10 3592 10 2 024)

024 08 095 084

tep ep t 4 S (

2op n M)a

τ

γ




3 2 3 4 4

2 ( 2 ) 12 ( 2 ) ( 2) 2 (12 )

2 (213 2 24) 24 12 (213 2 24) 24 (833 2) 2 24 90 09 12 (12 1) 17268 10

e t t m m p Ep 5p E 3 ( 3 ( ( ( Ep 5p E

cm




(453 69 ) (453 69 06 09) 099 49

(213 2 24) 24 (833 2 24) 091

5

C

t i

β

β γ

γ




Calculam4 3 4 4 4

0

2 20

5 (384 ) 5 945 10 10 (384 10 17268 10 ) 0007

1 ( ) 0007 1 49 (09 10) 08 0012

0012 10 1 883 [ ]lim1 300 00033

n

m

l E e m

C ( l 5 m

l l


= = lt =

10




H

f i

x

q1 q2

1

2

c

x

f i


2 828 r s inc(

) l G G 5 = times + + =


0

1005 16 51 204

2 2



17136 09 154224 p p r

1 8 ( = times = times = 8568 09 77112



sticirclpilor0

03 14 08 51 17136 p

0 c t m


0 3 0225 14 04 51 8568



13173316 2 16 2

p cp p cp 9 1 1


= + = + =


2 2

1

17136 58( ) (154224 131733) 58 301272 2

p

p

9

M 1 x 9 0 m

ω times times


11



efortul de tăiere

1 152424 131733 17136 58 10146



1 2 828 ) 0 = = =


2

8568 58( ) (71112 131733) 58 265

2 2

cp

cp

9 M 1 x 9 0 m


2 77112 131733 8568 58 706



n

s(


0725 0387

5562

n

s( cm

+

= =


556161 16

345n 3ucati= = rArr


s 9 cm m= = =


4 224 552 13248 10 s


8 424 5523363932210

12 12

s x


= = =

2 26 324 552

1218816 106 6

s x


= = = times


80 1200289 0289 0552

s

l 9

i (λ


times times


c 2

1

M

e

d

e

le+=σ

2 2 3

4

1 1 80 828 100887

3000 3000 13248 10 15e c

2

λ ξ

minus

minus



1 30127

0340887

d

M M M m

ξ = = = times

3

4 6

828 10 03434 15

13248 10 1218816 10 c M) 2 M)aσ

minus

minus minus

times= + = lt =

times times

12




1le

sdotsdot+

sdotsdot

n

ce m

d

ce 21

M

2

ϕ ϕ


58836 120

0289 0289 024

l l

i 3λ = = = = lt

times times


2 2140 ( ) 140 024 232 (58 0552) 584



22 2

2

06 58 11 075 006 ( ) 1 1 075 006( ) 1 248

1 0552 2

p p

)

s s

l l m0

( ( m


2

2

0142 176 58 0142 142 0552 11 14 1 1 1 083

1 0552 4 2

p s )M

s p

l ( m0

( l mα



044 248 109 1 )

0 φ times = times = gt584 083 485 1




e

23

4S lesdot

=τ 12

8

101458 914112 10022 15

33639322 10 0144

M)a 2 M)aτ minus

minus

times times= = lt =

times times

1 101458

11440887

44 0

ξ = = =

2 2

324 552 9141128 8


424 55233639322

12 12

s3 (

cmtimes times

= = =

06 06 024 0144e




828 204 624 ) ) S 0 = minus = minus =

109 09 624

5111 11

)

0

times= = =

13



2 2 3

4

80 51 101 1 099

3000 3000 13248 10 15c

2

λ ξ

minus

minus




d

M M M m

ξ = = = times

3

4 6

51 10 0030421 15

13248 10 1218816 10t c

M)a 2 M)aσ minus

minus minus


times times

3

4 6

51 10 00304288 15

13248 10 1218816 10c c

M)a 2 M)aσ minus

minus minus


times times


03221 288

c sc

t c

(( m

σ

σ σ

times times= = =

+ +

21 0552023

288 21

t st

t c

(( m

σ

σ σ

times times= = =

+ +


300304 51 0064 10008

04 04

d an t

M t M

e e


2 012 2 032023 04

2 3 2 3

ct

(ae ( m


012 0552040 00642 2



123120 08 085

annec

3a a

cm

2 m m= = =



2 22 759 1518 123n nec

cm cm= times = gt =



3 35 3 24d d d cm+ + = 24

252695

d cm= =


alunecare

14



08 08 12 26 2496a

+ ad 0 = = times times = 035 035 552 26 502


2 2

mx 25 25 26 169

i+ d 0 = = times = 2 2 2 218 002 18 26 002 12 1505



3

min

008531 6

1505 10

an


times

Adoptăm 6 buloane


7 4 3 0026 4 0312e





2 20107-85



15




(Arce)



16







3 2

3 3 2

3 4 4

2 ( 2 ) 12 ( 2 ) ( 2)

2 (12 ) 2 (156 2 12) 20 12 (156 2 12) 20 (66 2)

2 12 76 09 12 12 73 10

e t t x

x

p Ep 5p E 3 ( 3 ( (

( Ep 5p E

cm

δ δ

δ






3 2

3 3 2

3 4 4

2 ( 2 ) 12 ( 2 ) ( 2)

2 (12 ) 2 (213 2 24) 24 12 (213 2 24) 24 (62 2)

2 24 86 09 12 12 1074 10

e t t x

x

p Ep 5p E 3 ( 3 ( (

( Ep 5p E

cm

δ δ

δ







14 08 095 118


M)a






1 08( 100) 1 08(244 100) 08 6ϕ λ = minus = minus =

7







3 1 2 3

3 2 3

4 4

2 ( 2 ) 12 ( 2 ) ( 2) ( ) 2 12

2 (213 2 24) 24 12 (213 2 24) 24 (62 2) 1 (09 12) 2 24 86 12

10535 10

xp t t x p x 3 ( 3 ( ( E E 0p (

cm



= times1 1

3 3

( 2 ) ( 2) ( ) 2 ( 2) (213 2 24) 24 (62 2) 1 (09 12)

2 16 24 (62 2) 149 10


cm




129 38 34ix


12 5i

= 20 5

τ

=






2 2

1

(100 ) (100 )

i i

i

i e

0 0 ( (

τ

σ τ

+ le

part part

2 2

321 041017 1

100 10024( ) 20( )

60 60

+ = lt







x


8




1 1 069 024 045 x x t ( ( ( m= minus = minus =


1 1 1

3 2 3

4 4

2 ( 2 ) 12 ( 2 ) ( 2) ( ) 2 12

2 (213 2 24) 24 12 (213 2 24) 24 (45 2) (09 12) 2 24 69 12

503 10

x p t t x x 3 ( 3 ( ( E Ep 5p (

cm



1 11 1 1

3 3

( 2 ) ( 2) ( ) 2 ( 2) (213 2 24) 24 (45 2) (09 12)

2 24 24 (45 2) 1084 10

tx p t x t xS 3 ( ( E Ep 5p ( (

m

δ δ

minus




3 4

1 1 1

(2 ) 75 10 021 (2 503 10 ) 156i x ix x p




05 (05 ) 05 156 (05 156) 206 369

45 08 095 379

it i i i p d 2pta m

M)a


= times =




2 10 122 2 71 x cm= + =

Inaltimea sectiunii



2 0647 2 024 0167ix( m= minus times =



panou)


9



3 2 3

3 2

3 4 4

2 ( 2 ) 12( 2 ) ( 2) ( ) 12

2 (213 2 24) 24 12 (213 2 24) 24 (36 2) 1 (09 12)

2 24 60 12 3592 10

ep t t r r 3 ( 3 ( ( E Ep 5p (

cm




Momentul static1 2

2 3 3

( 2 ) ( 2) ( ) 2 8 (213 2 24) 24 (36 2) 1 (09 12)

2 24 60 8 876 10

ep t r r S 3 ( ( E Ep 5p (

cm





90

( ) 6345 10 876 10 (3592 10 2 0024) 322

6 095 63

ep ep

p

4 S M)a

2 M)a


lt = =




( ) 6345 10 66 10 3592 10 2 024)

024 08 095 084

tep ep t 4 S (

2op n M)a

τ

γ




3 2 3 4 4

2 ( 2 ) 12 ( 2 ) ( 2) 2 (12 )

2 (213 2 24) 24 12 (213 2 24) 24 (833 2) 2 24 90 09 12 (12 1) 17268 10

e t t m m p Ep 5p E 3 ( 3 ( ( ( Ep 5p E

cm




(453 69 ) (453 69 06 09) 099 49

(213 2 24) 24 (833 2 24) 091

5

C

t i

β

β γ

γ




Calculam4 3 4 4 4

0

2 20

5 (384 ) 5 945 10 10 (384 10 17268 10 ) 0007

1 ( ) 0007 1 49 (09 10) 08 0012

0012 10 1 883 [ ]lim1 300 00033

n

m

l E e m

C ( l 5 m

l l


= = lt =

10




H

f i

x

q1 q2

1

2

c

x

f i


2 828 r s inc(

) l G G 5 = times + + =


0

1005 16 51 204

2 2



17136 09 154224 p p r

1 8 ( = times = times = 8568 09 77112



sticirclpilor0

03 14 08 51 17136 p

0 c t m


0 3 0225 14 04 51 8568



13173316 2 16 2

p cp p cp 9 1 1


= + = + =


2 2

1

17136 58( ) (154224 131733) 58 301272 2

p

p

9

M 1 x 9 0 m

ω times times


11



efortul de tăiere

1 152424 131733 17136 58 10146



1 2 828 ) 0 = = =


2

8568 58( ) (71112 131733) 58 265

2 2

cp

cp

9 M 1 x 9 0 m


2 77112 131733 8568 58 706



n

s(


0725 0387

5562

n

s( cm

+

= =


556161 16

345n 3ucati= = rArr


s 9 cm m= = =


4 224 552 13248 10 s


8 424 5523363932210

12 12

s x


= = =

2 26 324 552

1218816 106 6

s x


= = = times


80 1200289 0289 0552

s

l 9

i (λ


times times


c 2

1

M

e

d

e

le+=σ

2 2 3

4

1 1 80 828 100887

3000 3000 13248 10 15e c

2

λ ξ

minus

minus



1 30127

0340887

d

M M M m

ξ = = = times

3

4 6

828 10 03434 15

13248 10 1218816 10 c M) 2 M)aσ

minus

minus minus

times= + = lt =

times times

12




1le

sdotsdot+

sdotsdot

n

ce m

d

ce 21

M

2

ϕ ϕ


58836 120

0289 0289 024

l l

i 3λ = = = = lt

times times


2 2140 ( ) 140 024 232 (58 0552) 584



22 2

2

06 58 11 075 006 ( ) 1 1 075 006( ) 1 248

1 0552 2

p p

)

s s

l l m0

( ( m


2

2

0142 176 58 0142 142 0552 11 14 1 1 1 083

1 0552 4 2

p s )M

s p

l ( m0

( l mα



044 248 109 1 )

0 φ times = times = gt584 083 485 1




e

23

4S lesdot

=τ 12

8

101458 914112 10022 15

33639322 10 0144

M)a 2 M)aτ minus

minus

times times= = lt =

times times

1 101458

11440887

44 0

ξ = = =

2 2

324 552 9141128 8


424 55233639322

12 12

s3 (

cmtimes times

= = =

06 06 024 0144e




828 204 624 ) ) S 0 = minus = minus =

109 09 624

5111 11

)

0

times= = =

13



2 2 3

4

80 51 101 1 099

3000 3000 13248 10 15c

2

λ ξ

minus

minus




d

M M M m

ξ = = = times

3

4 6

51 10 0030421 15

13248 10 1218816 10t c

M)a 2 M)aσ minus

minus minus


times times

3

4 6

51 10 00304288 15

13248 10 1218816 10c c

M)a 2 M)aσ minus

minus minus


times times


03221 288

c sc

t c

(( m

σ

σ σ

times times= = =

+ +

21 0552023

288 21

t st

t c

(( m

σ

σ σ

times times= = =

+ +


300304 51 0064 10008

04 04

d an t

M t M

e e


2 012 2 032023 04

2 3 2 3

ct

(ae ( m


012 0552040 00642 2



123120 08 085

annec

3a a

cm

2 m m= = =



2 22 759 1518 123n nec

cm cm= times = gt =



3 35 3 24d d d cm+ + = 24

252695

d cm= =


alunecare

14



08 08 12 26 2496a

+ ad 0 = = times times = 035 035 552 26 502


2 2

mx 25 25 26 169

i+ d 0 = = times = 2 2 2 218 002 18 26 002 12 1505



3

min

008531 6

1505 10

an


times

Adoptăm 6 buloane


7 4 3 0026 4 0312e





2 20107-85



15




(Arce)



16







3 1 2 3

3 2 3

4 4

2 ( 2 ) 12 ( 2 ) ( 2) ( ) 2 12

2 (213 2 24) 24 12 (213 2 24) 24 (62 2) 1 (09 12) 2 24 86 12

10535 10

xp t t x p x 3 ( 3 ( ( E E 0p (

cm



= times1 1

3 3

( 2 ) ( 2) ( ) 2 ( 2) (213 2 24) 24 (62 2) 1 (09 12)

2 16 24 (62 2) 149 10


cm




129 38 34ix


12 5i

= 20 5

τ

=






2 2

1

(100 ) (100 )

i i

i

i e

0 0 ( (

τ

σ τ

+ le

part part

2 2

321 041017 1

100 10024( ) 20( )

60 60

+ = lt







x


8




1 1 069 024 045 x x t ( ( ( m= minus = minus =


1 1 1

3 2 3

4 4

2 ( 2 ) 12 ( 2 ) ( 2) ( ) 2 12

2 (213 2 24) 24 12 (213 2 24) 24 (45 2) (09 12) 2 24 69 12

503 10

x p t t x x 3 ( 3 ( ( E Ep 5p (

cm



1 11 1 1

3 3

( 2 ) ( 2) ( ) 2 ( 2) (213 2 24) 24 (45 2) (09 12)

2 24 24 (45 2) 1084 10

tx p t x t xS 3 ( ( E Ep 5p ( (

m

δ δ

minus




3 4

1 1 1

(2 ) 75 10 021 (2 503 10 ) 156i x ix x p




05 (05 ) 05 156 (05 156) 206 369

45 08 095 379

it i i i p d 2pta m

M)a


= times =




2 10 122 2 71 x cm= + =

Inaltimea sectiunii



2 0647 2 024 0167ix( m= minus times =



panou)


9



3 2 3

3 2

3 4 4

2 ( 2 ) 12( 2 ) ( 2) ( ) 12

2 (213 2 24) 24 12 (213 2 24) 24 (36 2) 1 (09 12)

2 24 60 12 3592 10

ep t t r r 3 ( 3 ( ( E Ep 5p (

cm




Momentul static1 2

2 3 3

( 2 ) ( 2) ( ) 2 8 (213 2 24) 24 (36 2) 1 (09 12)

2 24 60 8 876 10

ep t r r S 3 ( ( E Ep 5p (

cm





90

( ) 6345 10 876 10 (3592 10 2 0024) 322

6 095 63

ep ep

p

4 S M)a

2 M)a


lt = =




( ) 6345 10 66 10 3592 10 2 024)

024 08 095 084

tep ep t 4 S (

2op n M)a

τ

γ




3 2 3 4 4

2 ( 2 ) 12 ( 2 ) ( 2) 2 (12 )

2 (213 2 24) 24 12 (213 2 24) 24 (833 2) 2 24 90 09 12 (12 1) 17268 10

e t t m m p Ep 5p E 3 ( 3 ( ( ( Ep 5p E

cm




(453 69 ) (453 69 06 09) 099 49

(213 2 24) 24 (833 2 24) 091

5

C

t i

β

β γ

γ




Calculam4 3 4 4 4

0

2 20

5 (384 ) 5 945 10 10 (384 10 17268 10 ) 0007

1 ( ) 0007 1 49 (09 10) 08 0012

0012 10 1 883 [ ]lim1 300 00033

n

m

l E e m

C ( l 5 m

l l


= = lt =

10




H

f i

x

q1 q2

1

2

c

x

f i


2 828 r s inc(

) l G G 5 = times + + =


0

1005 16 51 204

2 2



17136 09 154224 p p r

1 8 ( = times = times = 8568 09 77112



sticirclpilor0

03 14 08 51 17136 p

0 c t m


0 3 0225 14 04 51 8568



13173316 2 16 2

p cp p cp 9 1 1


= + = + =


2 2

1

17136 58( ) (154224 131733) 58 301272 2

p

p

9

M 1 x 9 0 m

ω times times


11



efortul de tăiere

1 152424 131733 17136 58 10146



1 2 828 ) 0 = = =


2

8568 58( ) (71112 131733) 58 265

2 2

cp

cp

9 M 1 x 9 0 m


2 77112 131733 8568 58 706



n

s(


0725 0387

5562

n

s( cm

+

= =


556161 16

345n 3ucati= = rArr


s 9 cm m= = =


4 224 552 13248 10 s


8 424 5523363932210

12 12

s x


= = =

2 26 324 552

1218816 106 6

s x


= = = times


80 1200289 0289 0552

s

l 9

i (λ


times times


c 2

1

M

e

d

e

le+=σ

2 2 3

4

1 1 80 828 100887

3000 3000 13248 10 15e c

2

λ ξ

minus

minus



1 30127

0340887

d

M M M m

ξ = = = times

3

4 6

828 10 03434 15

13248 10 1218816 10 c M) 2 M)aσ

minus

minus minus

times= + = lt =

times times

12




1le

sdotsdot+

sdotsdot

n

ce m

d

ce 21

M

2

ϕ ϕ


58836 120

0289 0289 024

l l

i 3λ = = = = lt

times times


2 2140 ( ) 140 024 232 (58 0552) 584



22 2

2

06 58 11 075 006 ( ) 1 1 075 006( ) 1 248

1 0552 2

p p

)

s s

l l m0

( ( m


2

2

0142 176 58 0142 142 0552 11 14 1 1 1 083

1 0552 4 2

p s )M

s p

l ( m0

( l mα



044 248 109 1 )

0 φ times = times = gt584 083 485 1




e

23

4S lesdot

=τ 12

8

101458 914112 10022 15

33639322 10 0144

M)a 2 M)aτ minus

minus

times times= = lt =

times times

1 101458

11440887

44 0

ξ = = =

2 2

324 552 9141128 8


424 55233639322

12 12

s3 (

cmtimes times

= = =

06 06 024 0144e




828 204 624 ) ) S 0 = minus = minus =

109 09 624

5111 11

)

0

times= = =

13



2 2 3

4

80 51 101 1 099

3000 3000 13248 10 15c

2

λ ξ

minus

minus




d

M M M m

ξ = = = times

3

4 6

51 10 0030421 15

13248 10 1218816 10t c

M)a 2 M)aσ minus

minus minus


times times

3

4 6

51 10 00304288 15

13248 10 1218816 10c c

M)a 2 M)aσ minus

minus minus


times times


03221 288

c sc

t c

(( m

σ

σ σ

times times= = =

+ +

21 0552023

288 21

t st

t c

(( m

σ

σ σ

times times= = =

+ +


300304 51 0064 10008

04 04

d an t

M t M

e e


2 012 2 032023 04

2 3 2 3

ct

(ae ( m


012 0552040 00642 2



123120 08 085

annec

3a a

cm

2 m m= = =



2 22 759 1518 123n nec

cm cm= times = gt =



3 35 3 24d d d cm+ + = 24

252695

d cm= =


alunecare

14



08 08 12 26 2496a

+ ad 0 = = times times = 035 035 552 26 502


2 2

mx 25 25 26 169

i+ d 0 = = times = 2 2 2 218 002 18 26 002 12 1505



3

min

008531 6

1505 10

an


times

Adoptăm 6 buloane


7 4 3 0026 4 0312e





2 20107-85



15




(Arce)



16




1 1 069 024 045 x x t ( ( ( m= minus = minus =


1 1 1

3 2 3

4 4

2 ( 2 ) 12 ( 2 ) ( 2) ( ) 2 12

2 (213 2 24) 24 12 (213 2 24) 24 (45 2) (09 12) 2 24 69 12

503 10

x p t t x x 3 ( 3 ( ( E Ep 5p (

cm



1 11 1 1

3 3

( 2 ) ( 2) ( ) 2 ( 2) (213 2 24) 24 (45 2) (09 12)

2 24 24 (45 2) 1084 10

tx p t x t xS 3 ( ( E Ep 5p ( (

m

δ δ

minus




3 4

1 1 1

(2 ) 75 10 021 (2 503 10 ) 156i x ix x p




05 (05 ) 05 156 (05 156) 206 369

45 08 095 379

it i i i p d 2pta m

M)a


= times =




2 10 122 2 71 x cm= + =

Inaltimea sectiunii



2 0647 2 024 0167ix( m= minus times =



panou)


9



3 2 3

3 2

3 4 4

2 ( 2 ) 12( 2 ) ( 2) ( ) 12

2 (213 2 24) 24 12 (213 2 24) 24 (36 2) 1 (09 12)

2 24 60 12 3592 10

ep t t r r 3 ( 3 ( ( E Ep 5p (

cm




Momentul static1 2

2 3 3

( 2 ) ( 2) ( ) 2 8 (213 2 24) 24 (36 2) 1 (09 12)

2 24 60 8 876 10

ep t r r S 3 ( ( E Ep 5p (

cm





90

( ) 6345 10 876 10 (3592 10 2 0024) 322

6 095 63

ep ep

p

4 S M)a

2 M)a


lt = =




( ) 6345 10 66 10 3592 10 2 024)

024 08 095 084

tep ep t 4 S (

2op n M)a

τ

γ




3 2 3 4 4

2 ( 2 ) 12 ( 2 ) ( 2) 2 (12 )

2 (213 2 24) 24 12 (213 2 24) 24 (833 2) 2 24 90 09 12 (12 1) 17268 10

e t t m m p Ep 5p E 3 ( 3 ( ( ( Ep 5p E

cm




(453 69 ) (453 69 06 09) 099 49

(213 2 24) 24 (833 2 24) 091

5

C

t i

β

β γ

γ




Calculam4 3 4 4 4

0

2 20

5 (384 ) 5 945 10 10 (384 10 17268 10 ) 0007

1 ( ) 0007 1 49 (09 10) 08 0012

0012 10 1 883 [ ]lim1 300 00033

n

m

l E e m

C ( l 5 m

l l


= = lt =

10




H

f i

x

q1 q2

1

2

c

x

f i


2 828 r s inc(

) l G G 5 = times + + =


0

1005 16 51 204

2 2



17136 09 154224 p p r

1 8 ( = times = times = 8568 09 77112



sticirclpilor0

03 14 08 51 17136 p

0 c t m


0 3 0225 14 04 51 8568



13173316 2 16 2

p cp p cp 9 1 1


= + = + =


2 2

1

17136 58( ) (154224 131733) 58 301272 2

p

p

9

M 1 x 9 0 m

ω times times


11



efortul de tăiere

1 152424 131733 17136 58 10146



1 2 828 ) 0 = = =


2

8568 58( ) (71112 131733) 58 265

2 2

cp

cp

9 M 1 x 9 0 m


2 77112 131733 8568 58 706



n

s(


0725 0387

5562

n

s( cm

+

= =


556161 16

345n 3ucati= = rArr


s 9 cm m= = =


4 224 552 13248 10 s


8 424 5523363932210

12 12

s x


= = =

2 26 324 552

1218816 106 6

s x


= = = times


80 1200289 0289 0552

s

l 9

i (λ


times times


c 2

1

M

e

d

e

le+=σ

2 2 3

4

1 1 80 828 100887

3000 3000 13248 10 15e c

2

λ ξ

minus

minus



1 30127

0340887

d

M M M m

ξ = = = times

3

4 6

828 10 03434 15

13248 10 1218816 10 c M) 2 M)aσ

minus

minus minus

times= + = lt =

times times

12




1le

sdotsdot+

sdotsdot

n

ce m

d

ce 21

M

2

ϕ ϕ


58836 120

0289 0289 024

l l

i 3λ = = = = lt

times times


2 2140 ( ) 140 024 232 (58 0552) 584



22 2

2

06 58 11 075 006 ( ) 1 1 075 006( ) 1 248

1 0552 2

p p

)

s s

l l m0

( ( m


2

2

0142 176 58 0142 142 0552 11 14 1 1 1 083

1 0552 4 2

p s )M

s p

l ( m0

( l mα



044 248 109 1 )

0 φ times = times = gt584 083 485 1




e

23

4S lesdot

=τ 12

8

101458 914112 10022 15

33639322 10 0144

M)a 2 M)aτ minus

minus

times times= = lt =

times times

1 101458

11440887

44 0

ξ = = =

2 2

324 552 9141128 8


424 55233639322

12 12

s3 (

cmtimes times

= = =

06 06 024 0144e




828 204 624 ) ) S 0 = minus = minus =

109 09 624

5111 11

)

0

times= = =

13



2 2 3

4

80 51 101 1 099

3000 3000 13248 10 15c

2

λ ξ

minus

minus




d

M M M m

ξ = = = times

3

4 6

51 10 0030421 15

13248 10 1218816 10t c

M)a 2 M)aσ minus

minus minus


times times

3

4 6

51 10 00304288 15

13248 10 1218816 10c c

M)a 2 M)aσ minus

minus minus


times times


03221 288

c sc

t c

(( m

σ

σ σ

times times= = =

+ +

21 0552023

288 21

t st

t c

(( m

σ

σ σ

times times= = =

+ +


300304 51 0064 10008

04 04

d an t

M t M

e e


2 012 2 032023 04

2 3 2 3

ct

(ae ( m


012 0552040 00642 2



123120 08 085

annec

3a a

cm

2 m m= = =



2 22 759 1518 123n nec

cm cm= times = gt =



3 35 3 24d d d cm+ + = 24

252695

d cm= =


alunecare

14



08 08 12 26 2496a

+ ad 0 = = times times = 035 035 552 26 502


2 2

mx 25 25 26 169

i+ d 0 = = times = 2 2 2 218 002 18 26 002 12 1505



3

min

008531 6

1505 10

an


times

Adoptăm 6 buloane


7 4 3 0026 4 0312e





2 20107-85



15




(Arce)



16



3 2 3

3 2

3 4 4

2 ( 2 ) 12( 2 ) ( 2) ( ) 12

2 (213 2 24) 24 12 (213 2 24) 24 (36 2) 1 (09 12)

2 24 60 12 3592 10

ep t t r r 3 ( 3 ( ( E Ep 5p (

cm




Momentul static1 2

2 3 3

( 2 ) ( 2) ( ) 2 8 (213 2 24) 24 (36 2) 1 (09 12)

2 24 60 8 876 10

ep t r r S 3 ( ( E Ep 5p (

cm





90

( ) 6345 10 876 10 (3592 10 2 0024) 322

6 095 63

ep ep

p

4 S M)a

2 M)a


lt = =




( ) 6345 10 66 10 3592 10 2 024)

024 08 095 084

tep ep t 4 S (

2op n M)a

τ

γ




3 2 3 4 4

2 ( 2 ) 12 ( 2 ) ( 2) 2 (12 )

2 (213 2 24) 24 12 (213 2 24) 24 (833 2) 2 24 90 09 12 (12 1) 17268 10

e t t m m p Ep 5p E 3 ( 3 ( ( ( Ep 5p E

cm




(453 69 ) (453 69 06 09) 099 49

(213 2 24) 24 (833 2 24) 091

5

C

t i

β

β γ

γ




Calculam4 3 4 4 4

0

2 20

5 (384 ) 5 945 10 10 (384 10 17268 10 ) 0007

1 ( ) 0007 1 49 (09 10) 08 0012

0012 10 1 883 [ ]lim1 300 00033

n

m

l E e m

C ( l 5 m

l l


= = lt =

10




H

f i

x

q1 q2

1

2

c

x

f i


2 828 r s inc(

) l G G 5 = times + + =


0

1005 16 51 204

2 2



17136 09 154224 p p r

1 8 ( = times = times = 8568 09 77112



sticirclpilor0

03 14 08 51 17136 p

0 c t m


0 3 0225 14 04 51 8568



13173316 2 16 2

p cp p cp 9 1 1


= + = + =


2 2

1

17136 58( ) (154224 131733) 58 301272 2

p

p

9

M 1 x 9 0 m

ω times times


11



efortul de tăiere

1 152424 131733 17136 58 10146



1 2 828 ) 0 = = =


2

8568 58( ) (71112 131733) 58 265

2 2

cp

cp

9 M 1 x 9 0 m


2 77112 131733 8568 58 706



n

s(


0725 0387

5562

n

s( cm

+

= =


556161 16

345n 3ucati= = rArr


s 9 cm m= = =


4 224 552 13248 10 s


8 424 5523363932210

12 12

s x


= = =

2 26 324 552

1218816 106 6

s x


= = = times


80 1200289 0289 0552

s

l 9

i (λ


times times


c 2

1

M

e

d

e

le+=σ

2 2 3

4

1 1 80 828 100887

3000 3000 13248 10 15e c

2

λ ξ

minus

minus



1 30127

0340887

d

M M M m

ξ = = = times

3

4 6

828 10 03434 15

13248 10 1218816 10 c M) 2 M)aσ

minus

minus minus

times= + = lt =

times times

12




1le

sdotsdot+

sdotsdot

n

ce m

d

ce 21

M

2

ϕ ϕ


58836 120

0289 0289 024

l l

i 3λ = = = = lt

times times


2 2140 ( ) 140 024 232 (58 0552) 584



22 2

2

06 58 11 075 006 ( ) 1 1 075 006( ) 1 248

1 0552 2

p p

)

s s

l l m0

( ( m


2

2

0142 176 58 0142 142 0552 11 14 1 1 1 083

1 0552 4 2

p s )M

s p

l ( m0

( l mα



044 248 109 1 )

0 φ times = times = gt584 083 485 1




e

23

4S lesdot

=τ 12

8

101458 914112 10022 15

33639322 10 0144

M)a 2 M)aτ minus

minus

times times= = lt =

times times

1 101458

11440887

44 0

ξ = = =

2 2

324 552 9141128 8


424 55233639322

12 12

s3 (

cmtimes times

= = =

06 06 024 0144e




828 204 624 ) ) S 0 = minus = minus =

109 09 624

5111 11

)

0

times= = =

13



2 2 3

4

80 51 101 1 099

3000 3000 13248 10 15c

2

λ ξ

minus

minus




d

M M M m

ξ = = = times

3

4 6

51 10 0030421 15

13248 10 1218816 10t c

M)a 2 M)aσ minus

minus minus


times times

3

4 6

51 10 00304288 15

13248 10 1218816 10c c

M)a 2 M)aσ minus

minus minus


times times


03221 288

c sc

t c

(( m

σ

σ σ

times times= = =

+ +

21 0552023

288 21

t st

t c

(( m

σ

σ σ

times times= = =

+ +


300304 51 0064 10008

04 04

d an t

M t M

e e


2 012 2 032023 04

2 3 2 3

ct

(ae ( m


012 0552040 00642 2



123120 08 085

annec

3a a

cm

2 m m= = =



2 22 759 1518 123n nec

cm cm= times = gt =



3 35 3 24d d d cm+ + = 24

252695

d cm= =


alunecare

14



08 08 12 26 2496a

+ ad 0 = = times times = 035 035 552 26 502


2 2

mx 25 25 26 169

i+ d 0 = = times = 2 2 2 218 002 18 26 002 12 1505



3

min

008531 6

1505 10

an


times

Adoptăm 6 buloane


7 4 3 0026 4 0312e





2 20107-85



15




(Arce)



16




H

f i

x

q1 q2

1

2

c

x

f i


2 828 r s inc(

) l G G 5 = times + + =


0

1005 16 51 204

2 2



17136 09 154224 p p r

1 8 ( = times = times = 8568 09 77112



sticirclpilor0

03 14 08 51 17136 p

0 c t m


0 3 0225 14 04 51 8568



13173316 2 16 2

p cp p cp 9 1 1


= + = + =


2 2

1

17136 58( ) (154224 131733) 58 301272 2

p

p

9

M 1 x 9 0 m

ω times times


11



efortul de tăiere

1 152424 131733 17136 58 10146



1 2 828 ) 0 = = =


2

8568 58( ) (71112 131733) 58 265

2 2

cp

cp

9 M 1 x 9 0 m


2 77112 131733 8568 58 706



n

s(


0725 0387

5562

n

s( cm

+

= =


556161 16

345n 3ucati= = rArr


s 9 cm m= = =


4 224 552 13248 10 s


8 424 5523363932210

12 12

s x


= = =

2 26 324 552

1218816 106 6

s x


= = = times


80 1200289 0289 0552

s

l 9

i (λ


times times


c 2

1

M

e

d

e

le+=σ

2 2 3

4

1 1 80 828 100887

3000 3000 13248 10 15e c

2

λ ξ

minus

minus



1 30127

0340887

d

M M M m

ξ = = = times

3

4 6

828 10 03434 15

13248 10 1218816 10 c M) 2 M)aσ

minus

minus minus

times= + = lt =

times times

12




1le

sdotsdot+

sdotsdot

n

ce m

d

ce 21

M

2

ϕ ϕ


58836 120

0289 0289 024

l l

i 3λ = = = = lt

times times


2 2140 ( ) 140 024 232 (58 0552) 584



22 2

2

06 58 11 075 006 ( ) 1 1 075 006( ) 1 248

1 0552 2

p p

)

s s

l l m0

( ( m


2

2

0142 176 58 0142 142 0552 11 14 1 1 1 083

1 0552 4 2

p s )M

s p

l ( m0

( l mα



044 248 109 1 )

0 φ times = times = gt584 083 485 1




e

23

4S lesdot

=τ 12

8

101458 914112 10022 15

33639322 10 0144

M)a 2 M)aτ minus

minus

times times= = lt =

times times

1 101458

11440887

44 0

ξ = = =

2 2

324 552 9141128 8


424 55233639322

12 12

s3 (

cmtimes times

= = =

06 06 024 0144e




828 204 624 ) ) S 0 = minus = minus =

109 09 624

5111 11

)

0

times= = =

13



2 2 3

4

80 51 101 1 099

3000 3000 13248 10 15c

2

λ ξ

minus

minus




d

M M M m

ξ = = = times

3

4 6

51 10 0030421 15

13248 10 1218816 10t c

M)a 2 M)aσ minus

minus minus


times times

3

4 6

51 10 00304288 15

13248 10 1218816 10c c

M)a 2 M)aσ minus

minus minus


times times


03221 288

c sc

t c

(( m

σ

σ σ

times times= = =

+ +

21 0552023

288 21

t st

t c

(( m

σ

σ σ

times times= = =

+ +


300304 51 0064 10008

04 04

d an t

M t M

e e


2 012 2 032023 04

2 3 2 3

ct

(ae ( m


012 0552040 00642 2



123120 08 085

annec

3a a

cm

2 m m= = =



2 22 759 1518 123n nec

cm cm= times = gt =



3 35 3 24d d d cm+ + = 24

252695

d cm= =


alunecare

14



08 08 12 26 2496a

+ ad 0 = = times times = 035 035 552 26 502


2 2

mx 25 25 26 169

i+ d 0 = = times = 2 2 2 218 002 18 26 002 12 1505



3

min

008531 6

1505 10

an


times

Adoptăm 6 buloane


7 4 3 0026 4 0312e





2 20107-85



15




(Arce)



16



efortul de tăiere

1 152424 131733 17136 58 10146



1 2 828 ) 0 = = =


2

8568 58( ) (71112 131733) 58 265

2 2

cp

cp

9 M 1 x 9 0 m


2 77112 131733 8568 58 706



n

s(


0725 0387

5562

n

s( cm

+

= =


556161 16

345n 3ucati= = rArr


s 9 cm m= = =


4 224 552 13248 10 s


8 424 5523363932210

12 12

s x


= = =

2 26 324 552

1218816 106 6

s x


= = = times


80 1200289 0289 0552

s

l 9

i (λ


times times


c 2

1

M

e

d

e

le+=σ

2 2 3

4

1 1 80 828 100887

3000 3000 13248 10 15e c

2

λ ξ

minus

minus



1 30127

0340887

d

M M M m

ξ = = = times

3

4 6

828 10 03434 15

13248 10 1218816 10 c M) 2 M)aσ

minus

minus minus

times= + = lt =

times times

12




1le

sdotsdot+

sdotsdot

n

ce m

d

ce 21

M

2

ϕ ϕ


58836 120

0289 0289 024

l l

i 3λ = = = = lt

times times


2 2140 ( ) 140 024 232 (58 0552) 584



22 2

2

06 58 11 075 006 ( ) 1 1 075 006( ) 1 248

1 0552 2

p p

)

s s

l l m0

( ( m


2

2

0142 176 58 0142 142 0552 11 14 1 1 1 083

1 0552 4 2

p s )M

s p

l ( m0

( l mα



044 248 109 1 )

0 φ times = times = gt584 083 485 1




e

23

4S lesdot

=τ 12

8

101458 914112 10022 15

33639322 10 0144

M)a 2 M)aτ minus

minus

times times= = lt =

times times

1 101458

11440887

44 0

ξ = = =

2 2

324 552 9141128 8


424 55233639322

12 12

s3 (

cmtimes times

= = =

06 06 024 0144e




828 204 624 ) ) S 0 = minus = minus =

109 09 624

5111 11

)

0

times= = =

13



2 2 3

4

80 51 101 1 099

3000 3000 13248 10 15c

2

λ ξ

minus

minus




d

M M M m

ξ = = = times

3

4 6

51 10 0030421 15

13248 10 1218816 10t c

M)a 2 M)aσ minus

minus minus


times times

3

4 6

51 10 00304288 15

13248 10 1218816 10c c

M)a 2 M)aσ minus

minus minus


times times


03221 288

c sc

t c

(( m

σ

σ σ

times times= = =

+ +

21 0552023

288 21

t st

t c

(( m

σ

σ σ

times times= = =

+ +


300304 51 0064 10008

04 04

d an t

M t M

e e


2 012 2 032023 04

2 3 2 3

ct

(ae ( m


012 0552040 00642 2



123120 08 085

annec

3a a

cm

2 m m= = =



2 22 759 1518 123n nec

cm cm= times = gt =



3 35 3 24d d d cm+ + = 24

252695

d cm= =


alunecare

14



08 08 12 26 2496a

+ ad 0 = = times times = 035 035 552 26 502


2 2

mx 25 25 26 169

i+ d 0 = = times = 2 2 2 218 002 18 26 002 12 1505



3

min

008531 6

1505 10

an


times

Adoptăm 6 buloane


7 4 3 0026 4 0312e





2 20107-85



15




(Arce)



16




1le

sdotsdot+

sdotsdot

n

ce m

d

ce 21

M

2

ϕ ϕ


58836 120

0289 0289 024

l l

i 3λ = = = = lt

times times


2 2140 ( ) 140 024 232 (58 0552) 584



22 2

2

06 58 11 075 006 ( ) 1 1 075 006( ) 1 248

1 0552 2

p p

)

s s

l l m0

( ( m


2

2

0142 176 58 0142 142 0552 11 14 1 1 1 083

1 0552 4 2

p s )M

s p

l ( m0

( l mα



044 248 109 1 )

0 φ times = times = gt584 083 485 1




e

23

4S lesdot

=τ 12

8

101458 914112 10022 15

33639322 10 0144

M)a 2 M)aτ minus

minus

times times= = lt =

times times

1 101458

11440887

44 0

ξ = = =

2 2

324 552 9141128 8


424 55233639322

12 12

s3 (

cmtimes times

= = =

06 06 024 0144e




828 204 624 ) ) S 0 = minus = minus =

109 09 624

5111 11

)

0

times= = =

13



2 2 3

4

80 51 101 1 099

3000 3000 13248 10 15c

2

λ ξ

minus

minus




d

M M M m

ξ = = = times

3

4 6

51 10 0030421 15

13248 10 1218816 10t c

M)a 2 M)aσ minus

minus minus


times times

3

4 6

51 10 00304288 15

13248 10 1218816 10c c

M)a 2 M)aσ minus

minus minus


times times


03221 288

c sc

t c

(( m

σ

σ σ

times times= = =

+ +

21 0552023

288 21

t st

t c

(( m

σ

σ σ

times times= = =

+ +


300304 51 0064 10008

04 04

d an t

M t M

e e


2 012 2 032023 04

2 3 2 3

ct

(ae ( m


012 0552040 00642 2



123120 08 085

annec

3a a

cm

2 m m= = =



2 22 759 1518 123n nec

cm cm= times = gt =



3 35 3 24d d d cm+ + = 24

252695

d cm= =


alunecare

14



08 08 12 26 2496a

+ ad 0 = = times times = 035 035 552 26 502


2 2

mx 25 25 26 169

i+ d 0 = = times = 2 2 2 218 002 18 26 002 12 1505



3

min

008531 6

1505 10

an


times

Adoptăm 6 buloane


7 4 3 0026 4 0312e





2 20107-85



15




(Arce)



16



2 2 3

4

80 51 101 1 099

3000 3000 13248 10 15c

2

λ ξ

minus

minus




d

M M M m

ξ = = = times

3

4 6

51 10 0030421 15

13248 10 1218816 10t c

M)a 2 M)aσ minus

minus minus


times times

3

4 6

51 10 00304288 15

13248 10 1218816 10c c

M)a 2 M)aσ minus

minus minus


times times


03221 288

c sc

t c

(( m

σ

σ σ

times times= = =

+ +

21 0552023

288 21

t st

t c

(( m

σ

σ σ

times times= = =

+ +


300304 51 0064 10008

04 04

d an t

M t M

e e


2 012 2 032023 04

2 3 2 3

ct

(ae ( m


012 0552040 00642 2



123120 08 085

annec

3a a

cm

2 m m= = =



2 22 759 1518 123n nec

cm cm= times = gt =



3 35 3 24d d d cm+ + = 24

252695

d cm= =


alunecare

14



08 08 12 26 2496a

+ ad 0 = = times times = 035 035 552 26 502


2 2

mx 25 25 26 169

i+ d 0 = = times = 2 2 2 218 002 18 26 002 12 1505



3

min

008531 6

1505 10

an


times

Adoptăm 6 buloane


7 4 3 0026 4 0312e





2 20107-85



15




(Arce)



16



08 08 12 26 2496a

+ ad 0 = = times times = 035 035 552 26 502


2 2

mx 25 25 26 169

i+ d 0 = = times = 2 2 2 218 002 18 26 002 12 1505



3

min

008531 6

1505 10

an


times

Adoptăm 6 buloane


7 4 3 0026 4 0312e





2 20107-85



15




(Arce)



16




(Arce)



16

lemn vedrasco

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