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2.3 PRIMERI<br />

2.3.1 Primer 1<br />

Sra~unati i nacrtati ekvivalentno optere}enje usled prethodnog naprezanja za nosa~ sistema<br />

proste grede i datu trasu rezultante kablova, slika 2.22a. U preseku 2 tangenta parabole<br />

je horizontalna. Pretpostavlja se da je u pitanju 'plitka trasa' kabla, odnosno da je sila<br />

prethodnog naprezanja konstanta du` nosa~a i jednaka sili prethodnog naprezanja N k.<br />

Podaci: L=3000cm; N k = 2860 kN.<br />

20<br />

N 1<br />

N k Nk<br />

a)<br />

M 1<br />

b)<br />

α 1<br />

L/4=750<br />

L/2=1500 L/4=750 L/4=750<br />

- Paraboli~ni deo 1-2 - ekvivalentno podeljeno optere}enje<br />

tgα 1 = 2(70+20)/1500 = 0,120 α 1 = 6,843 0<br />

f = (70+20)/4 = 22,5 cm<br />

q k = 8N kf/L 2 = 8 x 2860 x 0,225/15 2 = 22,88 kN/m<br />

- Presek 1- ekvivalentne koncentrisane sile<br />

N 1 ≅ N k = 2860 kN<br />

V 1 = N k sinα 1 = 340,8 kN<br />

M 1 = N k e = 2860 x 0,20 = 572 kN<br />

- Presek 3 - ekvivalentna koncentrisana sila<br />

tgα 3 = 90/750 = 0,120<br />

sinα 3/2 = 0,05968<br />

V 3 ≅ 2 N k sinα 3/2 = 2 x 2860 x 0,05968 = 341.4 kN<br />

f=22,5<br />

1 2 3 4<br />

V 1<br />

- Presek 4 - ekvivalentne koncentrisane sile<br />

α 4 = α 3<br />

N 4 ≅ N k = 2860 kN<br />

T 4 = N k sinα4 = 341,4kN<br />

q<br />

L=3000<br />

1 2 3 4<br />

L/2=1500 L/4=750 L/4=750<br />

Slika 2.22 Nosa~, trasa 'rezultuje}eg kabla' i ekvivalentno optere}enje<br />

2- 19<br />

TL<br />

V 3<br />

α 3<br />

α 4<br />

20<br />

70<br />

V 4<br />

M 4<br />

N 4

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