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ID number:739860
 
Author:
Evaluation:
Published: 08.11.2023.
Language: Latvian
Level: College/University
Literature: n/a
References: Not used
Time period viewed: 2016 - 2020 years
Extract

1. M, Q un N epīras konstruēšana
Balstu reakcijas noteikšana
W = 3·D - 2·L – Satb.
W = 3·3 - 2·(1+1) – (3+2) = 0
Momenta līdzsvara vienādojums pret locīklu C rāmja labajai daļai
∑M_C^(pa labi) = q_2⋅4·2+M_B-H_B⋅4=0
∑M_C^(pa labi) = 80+40-H_B⋅4=0
H_B=30 kN
Izmantojot spēka līdzsvara nosacījumu uz horizontālo asi ∑X=0 iegūstam
∑▒〖X=〗 P_2+H_B+H_A=0;
H_A=-50-30=-80 kN (mainām bultas virzienu un zīmi) H_A=80 kN
Lai atrastu momentu MA izmantojam līdzsvara vienādojumu pret locīklu n rāmja kreisajai daļai ∑▒〖M_n^"pa kreisi" =〗-q_2⋅4⋅2+H_A⋅4+M_A-P_2⋅"2"=0
∑▒〖M_n^"pa kreisi" =〗-80+320-100+M_A=0
M_A=-140 kN (mainām bultas virzienu un zīmi) M_A=140 kN
∑▒〖M_n^"pa labi" =〗 q_2⋅10⋅5-H_B⋅4+M_B-V_B⋅"6"=0
∑▒〖M_n^"pa labi" =〗 500-120+40-V_B⋅"6"=0
V_B=70 kN
Izmantojot spēka līdzsvara nosacījumu uz vertikālo asi ∑Y=0 iegūstam
∑▒〖Y=〗 V_A+V_B-q_2⋅14=0; V_A=70 kN
Pārbaude:
∑▒〖M_E^ =〗 H_A⋅2-M_A+V_A⋅3+M_B-V_B⋅"3 - " H_B⋅2=160-140+210+40-210-60=0…

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