Download - DNEC
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- Presek 1 - ekvivalentne koncentrisane sile<br />
N 1 = N k cos(α 1 + β 1) ≅ N k = 1000 kN<br />
V 1 = N k sin(α 1 + β 1) = 1000 sin(4,004+2,862) = 119,5 kN<br />
- Presek 2 - ekvivalentna koncentrisana sila<br />
Komponenta usled preloma trase kabla<br />
V 2k = 2 x 1000 sinθ 2/2 = 2 x 1000 x sin(19,646/2) = 341,2 kN<br />
Komponenta usled preloma te`i{ne linije<br />
V 2TK = 2 x 1000 sinγ 2/2 = 2 x 1000 x sin(8,573/2) = 149,5 kN<br />
Ukupno<br />
V 2 = V 2k + V 2TL = 341,2 + 149,5 = 490,7 kN<br />
- Polje 2-4 - ekvivalentno podeljeno optere}enje<br />
Komponenta usled krivine kabla<br />
q k = 8N kf k/(L/2) 2 = 8 x 1000 x 1,40 /20 2 = 28 kN/m<br />
Komponenta usled krivine te`i{ne linije<br />
q TL = 8N kf TL/(L/2) 2 = 8 x 1000 x 0,50 /20 2 = 10 kN/m<br />
Ukupno<br />
q = q k + q TL = 28 + 10 = 38 kN/m<br />
Ekvivalentno optere}enje prikazano je na slici 2.23c.<br />
- Moment savijanja u sredini raspona - ekvivalentno optere}enja<br />
M 3 = -119,5 x 10,0 + 38,0x20 2 / 8 = 705kNm<br />
- Moment savijanja u sredini raspona - direktan prora~un<br />
M 3 = N k e = 1000 x (1,7 - 1,0) = 700 kNm<br />
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