I. OBLICZENIE REAKCJI PODPOROWYCH

1. ∑M_C^L = 0

-23kN/m × 24m × 1/2 × 1/3 × 24m + V_B × 19m = 0

V_B × 19m = 2208kNm/:19m

V_B = $\frac{\mathbf{2208}}{\mathbf{19}}$kN = 116.211kN

2. ∑Piy_E^P = 0

-2kN + V_G = 0

V_G = 2kN

3. ∑Piy_E^L = 0

V_D – 23kN/m × 24m × $\frac{1}{2}$ + $\frac{2208}{19}$kN = 0

V_D = $\frac{\mathbf{3036}}{\mathbf{19}}$kN = 159.789kN

4. ∑M_C^P = 0

13kNm - $\frac{3036}{19}$kN × 5m + 2kN × 18m – 2kN × 37m + M_G = 0

M_G = $\frac{\mathbf{15655}}{\mathbf{19}}$kNm = 823.947kNm

Spr.

∑M_G = 0

$\frac{15655}{19}$kNm – 2kN × 19m + $\frac{3036}{19}$kN × 32m + 13kNm – (23kN/m × 24m × ½)×((1/3×24)+37) + + $\frac{2208}{19}$kN × 56m = 0

0 = 0

L = P

I. OBLICZENIE SIŁ WEWNĘTRZNYCH

x₁ є (0m – 5m)

$\frac{q*}{x} = \frac{23kN/m}{24m}$ => q* = $\frac{23kN/m\ \times \text{\ x}}{24m}$

T(x) = -q* × x ×1/2 = - $\frac{\mathbf{23}\mathbf{\text{kN}}\mathbf{/}\mathbf{m}\mathbf{\ }\mathbf{\times}\mathbf{\ }\mathbf{x}}{\mathbf{24}\mathbf{m}}$ × x × 1/2

T(0) = 0kN T(5) = - 11.979kN

M(x) = -q* × x ×1/2 × 1/3 × x = - $\frac{\mathbf{23}\mathbf{\text{kN}}\mathbf{/}\mathbf{m}\mathbf{\ \times \ }\mathbf{x}}{\mathbf{24}\mathbf{m}}$ × x² × 1/6

M(0) = 0kNm M(5) = -19.965kNm

x₂ є (0m – 19m)

$\frac{q*}{(5 + x)} = \frac{23\text{kN}/m}{24m}$ => q* = $\frac{23\text{kN}/m\ \times \ (5 + x)}{24m}$

T(x) = -q* × (5+x) × 1/2 + $\frac{\mathbf{2208}}{\mathbf{19}}\mathbf{\text{kN}}$ = - $\frac{\mathbf{23}\mathbf{\text{kN}}\mathbf{/}\mathbf{m}\mathbf{\ }\mathbf{\times}\mathbf{\ (5 +}\mathbf{x}\mathbf{)}}{\mathbf{24}\mathbf{m}}$ × (5+x) × 1/2 + $\frac{\mathbf{2208}}{\mathbf{19}}\mathbf{\text{kN}}$

T(0) = 104.23kN T(19) = -159.79kN

M = -q* × (5+x) × 1/2 × 1/3 × (x+5) + $\frac{\mathbf{2208}}{\mathbf{19}}\mathbf{\text{kN}}$ × x = - $\frac{\mathbf{23}\mathbf{\text{kN}}\mathbf{/}\mathbf{m}\mathbf{\ }\mathbf{\times}\mathbf{\ (5 +}\mathbf{x}\mathbf{)}}{\mathbf{24}\mathbf{m}}\mathbf{\ }$× (5+x) × 1/2 × 1/3 × (x+5) + $\frac{\mathbf{2208}}{\mathbf{19}}\mathbf{\text{kN}}$ × x

M(0) = -19.965kNm M(19) = 0kNm

Extremum

- $\frac{23\text{kN}/m\ \times \ (5 + x)}{24m}$ × (5+x) × 1/2 + $\frac{2208}{19}\text{kN}$ = 0 x=7,48m

M( 7,48 ) = 558,792kNm

x₃ є (0m – 19m)

T(x) = -2kN = T(2) = T(19)

M(x) = - $\frac{\mathbf{15655}}{\mathbf{19}}$kNm + 2kN × x

M(0) = - 823.947kNm M(19) = - 785,947kNm

x₄ є (0m – 5m)

T(x) = 2kN – 2kN = 0kN = T(0) = T(5)

M(x) = - $\frac{\mathbf{15655}}{\mathbf{19}}$kNm + 2kN × (x + 19m) – 2kN × x

M(0) = - 785,947kNm M(5) = -785,947kNm

x₅ є (0m – 8m)

T(x) = 2kN – 2kN = 0kN = T(0) = T(8)

M(x) = - $\frac{\mathbf{15655}}{\mathbf{19}}$kNm + 2kN × (x + 24m) – 2kN × (x + 5m)

M(0) = -785,947kNm M(8) = -785,947kNm

x₆ є (0m – 5m)

T(x) = 2kN – 2kN - = $\frac{\mathbf{3036}}{\mathbf{19}}$kN = -159.789kN = T(0) = T(5)

M(x) = - $\frac{\mathbf{15655}}{\mathbf{19}}$kNm + 2kN × (x + 32m) – 2kN × (x + 13m) – 13kNm + $\frac{\mathbf{3036}}{\mathbf{19}}$kN × x

M(0) = -798,947kNm M(5) = 0kNm