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5.2. Ritterova metodaP1IP2B1,5 bbAI4ar7P1IS6R8r6S7r8AR6IS8R7OČITANI KRAKOVI SILA:r6 = 3,54 mr7 = 21,24 mr8 = 3,25 m31
∑ MA = 0- P1⋅a - P2⋅2a + B⋅4a = 0B = 0,25 (P1⋅ + P2⋅2) = 0,25 ( 110 + 200 ) = 77,5 kN∑ MB = 0- A ⋅ 4a + P1⋅3a + P2⋅2a = 0A = 0,25 (P1⋅3 - P2⋅2) = 0,25 ( 330 + 200 ) = 132, 5 kN∑ MR6 = 0- A ⋅ a + S6⋅r6 = 0S6 = 0,282 ⋅ ( 2,4 ⋅ 132,5 ) = 89,8 kN∑ MR7 = 0- A ⋅ 28,77 + P1 ⋅ 26, 40 + S7 ⋅r7 = 0S7 = 0,047 ⋅ ( 28,77 ⋅ 132,5 - 26,40 ⋅ 110 ) = 42,68 kN∑ MR8 = 0- A ⋅ 2a + P1⋅a + S8⋅r8 = 0S8 = 0,31 ⋅ ( 4,8 ⋅ 132,5 – 2,4 ⋅ 110 ) = 114,46 kN32
Page 1 and 2: Sveučilište u ZagrebuGrađevinski
Page 3 and 4: 1. UVODU ovom ćemo radu obraditi s
Page 5 and 6: nš = 2nč.Zaključujemo kako je za
Page 7 and 8: 3. GRAFIČKE METODE ODREĐIVANJA SI
Page 9 and 10: 3.3. Maxwell Cremonin plan silaOvaj
Page 11 and 12: P1P2P1A2112314563784P394zBS7S5S9S6S
Page 13 and 14: ABSlika 12.Kao i kod postupka u koj
Page 15 and 16: ABSlika 15.Također treba razmotrit
Page 17 and 18: 2 7P13 P3P284nč=10nš=17ne=3a.AA16
Page 19 and 20: P3P113S1S2S1P457P5a.AzS2656 BRlKC12
Page 21 and 22: Sile u vertikalama nad ležajevima
Page 23 and 24: 2 S231S2S4S44S5S1S4S5S3IS1S4S3AHAS5
Page 25 and 26: AB'B'C'a) Cb)BCACA,B,CC'C''c)A,B,CC
Page 27 and 28: 4.3. Pomaci grednih rešetkastih no
Page 29 and 30: 5. PRIMJERP1P21,5 bb4aP1=110 kNP2=1
Page 31: OČITANE SILE U ŠTAPOVIMA:ŠTAP DU
Page 35 and 36: 5.4. Williotov plan pomaka∆l i =
Page 37 and 38: 6. ZAKLJUČAKNakon rješavanja zada

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