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P = 25 kN
RA = 25kN
1. Temat
a = 1500mm
2. Obliczenie reakcji podpór
Σ MiA = - P•a – P•2a + RB•3a = 0
RB = P = 25 kN
Σ Fiy = RA + RB - P - P = 0
RA = P = 25kN
3. Analiza kratownicy metodą analityczną
Na niebiesko zaznaczono wektory o nieznanych wartościach
Σ Fiy = RA – S1•sin60˚ = 0 → S2 = 28,86 kN
Σ Fix = S2•cos60˚ - S1 = 0 → S1 = 14,43 kN
RA = 25 kN
RB = 25 kN
S1 = 14,43 kN
S2 = 28,86 kN
S3 = 28,86 kN
Węzeł III
Σ Fiy = S2•sin60˚ - S3•sin60˚ = 0 → S3 = 28,86 kN
Σ Fix = -S2•cos60˚ - S3•cos60˚ + S4 = 0 → S4 = 28,86 kN
Węzeł II
Σ Fiy = -P + S5•sin60˚ + S3•sin60˚ = 0 → S5 = 0
Σ Fix = S3•cos60˚ + S1 – S6 = 0 → S6 = 28.86 kN
S4 = 28,86 kN
S5 = 0
S6 = 28,86 kN
S8 = 28,86 kN
Węzeł IV
Σ Fiy = S7•cos60˚ = 0 → S7 = 0
Σ Fix = -S4 + S8 = 0 → S8 = 28,86 kN
Węzeł VI
Σ Fix = -S9•cos60˚ + S11•cos60˚ - S8 = 0 → S11 = 28,86 kN
Σ Fiy = S9•sin60˚ + S11•sin60˚ = 0 → S9 = 28,86 kN
S7 = 0
S11 = 28,86 kN
S9...
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