Primer Broj 1

BROJNI PRIMER -1 Potrebno je odrediti uticaje (sleganje w, nagib elastične linije θ, transverzalnu silu T i moment savijanja M) u tački A, B i nagib elastične linije ispod koncentrisane sile P, vrlo dugačkog nosača na deformabilnoj podlozi, prema opterećenju i dimenzijama na Slici 4.7. Temeljnu podlogu aproksimirati Vinklerovim modelom. Proračun izvršiti analitički, prema izrazima za beskonačni nosač. 0,4

P=1,3 MN

B

x

0,4

A

1,0

p=50,0 kN/m

k=30,0 MN/m3 3,0

2,0

B=1,5 4,0

1,0

L=10,0

Eb =21,0 GPa I=0,159 m4

z

Slika 4.7

Opterećenje između tačaka A i B beskonačnog nosača

Rešenje: Parametar krutosti sistema temeljni nosač – podloga (tlo), iznosi:

λ=

4

kB = 4Eb I

4

30,0 ⋅ 1,5 = 0,2409 m −1 4 ⋅ 21000 ⋅ 0,159

Proračun sleganja w, nagiba θ, transverzalne sile T i momenta savijanja M, u tački A : -

pomoćne veličine za proračun uticaja koncentrisanog opterećenja :

∆x = x − x P = 0 − 3,0 = −3 ,0 -

pomoćne veličine za proračun uticaja od jednako podeljenog opterećenja :

∆x L = x − x L = 0 − 5 ,0 = −5 ,0

∆x D = x − x D = 0 − 9 ,0 = −9 ,0

sgn(∆x L ) = sgn(− 5 ,0 ) = −1

sgn(∆x D ) = sgn(− 9 ,0 ) = −1

wA =

wA =

Pλ p [D(5λ ) − D(9λ )] A(3λ ) + 2 Bk 2 Bk

1300 ⋅ 0 ,2409 50 ,0 A(0 ,723) + [D(1,205 ) − D(2,169 )] = 2,48 ⋅ 10 −3 m 2 ⋅ 1,5 ⋅ 30000 2 ⋅ 1,5 ⋅ 30000

θA =

θA =

sgn(∆x ) = sgn(− 3,0 ) = −1

Pλ2 pλ [A(5λ ) − A(9λ )] B(3λ ) + Bk 2 Bk

1300 ⋅ 0 ,2409 2 50 ,0 ⋅ 02409 B (0 ,723) + [A(1,205 ) − A(2,169 )] = 0 ,586 ⋅ 10 −3 rad 1,5 ⋅ 30000 2 ⋅ 1,5 ⋅ 30000 MA =

P p C (3λ ) − 2 [B(5λ ) − B(9λ )] 4λ 4λ

MA =

1300 ,0 50 ,0 C (0 ,723 ) − [B(1,205 ) − B(2 ,169 )] = 18 ,028 kNm 0 ,964 0 ,232 P p [C (5λ ) − C (9λ )] D(3λ ) + 2 4λ

TA =

TA =

1300 ,0 50 ,0 D (0 ,723) + [C (1,205) − C (2 ,169 )] = 235 ,914 kN 2 0 ,964

Proračun sleganja w, nagiba θ, transverzalne sile T i momenta savijanja M, u tački B :

sgn(∆x ) = 1

∆x = 7 ,0

sgn(∆x L ) = 1

∆x D = 1,0

sgn(∆x D ) = 1

Pλ p [D(5λ ) − D(λ )] A(7 λ ) + 2 Bk 2 Bk

wB =

wB =

∆x L = 5 ,0

1300 ⋅ 0 ,2409 50 ,0 A(1,687 ) + [D(1,205) − D(0 ,241)] = 0 ,93 ⋅ 10 −3 m 2 ⋅ 1,5 ⋅ 30000 2 ⋅ 1,5 ⋅ 30000

θB = −

θB = −

1300 ⋅ 0 ,2409 2 50 ,0 ⋅ 02409 B(1,687 ) + [A(1,205) − A(0 ,241)] = −0 ,384 ⋅ 10 −3 rad 1,5 ⋅ 30000 2 ⋅ 1,5 ⋅ 30000 P p C (7 λ ) + 2 [B(5λ ) − B(λ )] 4λ 4λ

MB =

MA =

1300 ,0 50 ,0 C (1,687 ) + [B(1,205 ) − B(0 ,241)] = −257 ,049 kNm 0 ,964 0 ,232 T∞ = −

TB = −

Pλ2 pλ [A(5λ ) − A(λ )] B(7 λ ) + Bk 2 Bk

P p [C (5λ ) − C (λ )] D (7 λ ) + 2 4λ

1300 ,0 50 ,0 D(1,687 ) + [C (1,205 ) − C (0 ,241)] = −24 ,930 kN 2 0 ,964

Sleganje w i nagib elastične linije beskonačnog nosača θ ispod sile P :

w =

w =

1300 ⋅ 0 ,2409 50 ,0 1,00 + [D(0 ,482 ) − D(1,445 )] = 3,77 ⋅ 10 −3 m 2 ⋅ 1,5 ⋅ 30000 2 ⋅ 1,5 ⋅ 30000

θ=

θ=

Pλ p [D(2λ ) − D(6 λ )] A(0 ) + 2 Bk 2 Bk

Pλ 2 pλ [A(2λ ) − A(6 λ )] B(0 ) + Bk 2 Bk

1300 ⋅ 0 ,2409 2 50 ,0 ⋅ 02409 [A(0 ,482 ) − A(1,445 )] = 0 ,0765 ⋅ 10 −3 rad 0 ,0 + 1,5 ⋅ 30000 2 ⋅ 1,5 ⋅ 30000

Dijagram momenta savijanja na delu AB bekonačnog nosača (kNm) -400 -200 -208.17

0

-257.05

-123.29

18.03 -5.80

200 167.47

314.09

400 419.67

600 747.04

800

796.47

1,000 1,200 1,400

1337.66

1,600 0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0

9.0

10.0

Dijagram transverzalnih sila na delu AB beskonačnog nosača (kN) 800 674.15

700 600

508.96

500 360.50

400 300 200

235.91

100 -24.93

0 -100 -200

-141.46

-97.41

-75.98

8.0

9.0

-208.98

-300 -298.91

-400 -500

-457.26

-600 -625.85

-700 0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

10.0