Wyliczenie reakcji.

∑Pix=0

R_Bx=0

∑M_b=0

q × 3l × 1,5l + 2ql × 4l + R_By × 6l +b4ql² = 0

R_Cy × 6l = - 4,5ql² - 8ql² - 4ql² /6l

R_Cy =-2,75ql

∑M_C=0

4ql² - 2ql × 2l + q × 3l ×7,5l – R_By × 6l =0

R_By × 6l = 4ql² – 4ql² × 22,5ql² / 6l

R_By = 3,75ql

Projektowanie przekroju poprzecznego belki.

A₁ = 16k²

A₂ = 12k²

A₃ = 8k²

A =36k²

S_y = 16k² × 9k + 12k² ×5k + 8k² × 1k = 212k³

$$z_{0} = \frac{S_{y}}{A} = \frac{{212k}^{3}}{{36k}^{2}} = 5,89k$$

$$I_{Y0} = \frac{8k \times \left( 2k \right)^{3}}{12} + \frac{2k \times \left( 6k \right)^{3}}{12} + \frac{4k \times \left( 2k \right)^{3}}{12} = 44k^{4}$$

$$W_{y}^{g} = \frac{{44k}^{4}}{4,11k} = 10,71k^{3}$$

$$W_{y}^{d} = \frac{{44k}^{4}}{5,89k} = 7,47k^{3}$$

σ_max ≤ R

$$\sigma_{\max} = \frac{M_{\max}}{W_{\min}}$$

M_max= 4,5ql² = 4,5×6kN × (4m)² = 432kNm

$$\sigma_{\max} = \frac{432kNm}{7,47k^{3}} \leq 175MPa$$

$$\frac{43200kNcm}{7,47k^{3}} \leq 17,5\frac{\text{kN}}{\text{cm}^{2}} \rightarrow k^{3} \geq \frac{43200}{131,075}\text{\ cm}$$

k ≥ 6,91cm

τ_max ≤ R_t

$$\tau_{\max} = \frac{T_{\max} \times S_{Y(x)}}{b \times I_{Y0}}$$

T_max =3ql = 3×6kN × 4m =72kN

$$\tau_{\max} = \frac{72kN \times \left( 8k \times 2k \times 3,11k + 2k \times 2,11k \times 1,055k \right)}{2k \times 44k^{4}} \leq 105MPa \rightarrow$$

$$\tau_{\max} = \ \frac{3903,2712kNk^{3}}{2k \times 44k^{4}} \leq 10,5\frac{\text{kN}}{\text{cm}^{2}}\ \rightarrow k^{3} \geq \frac{3903,2712}{2 \times 44 \times 10,5}\text{cm}$$

k ≥ 1, 62cm

Przyjęto k= 7cm

$$\sigma_{\max}^{g} = \frac{432kNm}{10,71 \times \left( 7cm \right)^{3}} = \frac{43200kNcm}{3673,53\text{cm}^{3}} = 11,76\frac{\text{kN}}{\text{cm}^{2}} = 117,6MPa$$

$$\sigma_{\max}^{d} = \frac{432kNm}{7,47 \times \left( 7cm \right)^{3}} = \frac{43200kNcm}{2562,21\text{cm}^{3}} = 16,86\frac{\text{kN}}{\text{cm}^{2}} = 168,6MPa$$

τ_max¹ = τ_max⁷ = 0

$$\tau_{\max}^{2} = \frac{72kN \times \left( 8k \times 2k \times 3,11k \right)}{8k \times 44k^{4}} = 0,21\frac{\text{kN}}{\text{cm}^{2}} = 2,1MPa$$

$$\tau_{\max}^{3} = \frac{72kN \times \left( 8k \times 2k \times 3,11k \right)}{2k \times 44k^{4}} = 0,83\frac{\text{kN}}{\text{cm}^{2}} = 8,3MPa$$

$$\tau_{\max}^{4} = \frac{72kN \times \left( 8k \times 2k \times 3,11k + 2k \times 2,11k \times 1,055k \right)}{2k \times 44k^{4}} = \frac{72kN \times \left( 49,76 + 4,4521 \right) \times \left( 7cm \right)^{3}}{88 \times \left( 7cm \right)^{5}} = \frac{72kN \times 54,2121}{88 \times \left( 7cm \right)^{2}} = \frac{3903,2712kN}{4312} = 0,91\frac{\text{kN}}{\text{cm}^{2}} = 9,1MPa$$

$$\tau_{\max}^{5} = \frac{72KN \times \left( 4k \times 2k \times 4,89 \right)}{2k \times 44k} = 0,65\frac{\text{kN}}{\text{cm}^{2}} = 6,5MPa$$

$$\tau_{\max}^{6} = \frac{72KN \times \left( 4k \times 2k \times 4,89 \right)}{4k \times 44k} = 0,33\frac{\text{kN}}{\text{cm}^{2}} = 3,3MPa$$

Wyznaczenie ugięcia i kąta obrotu w punkcie A.

φ_A = ?                                           W_A = ?

$$\varphi_{A} = \frac{1}{\text{EJy}} \times \left( \frac{1}{2} \times 3\text{ql}^{2} \times 4l \times \frac{7}{30} + 1,5\text{ql}^{2} \times 4l \times \frac{7}{20} + \frac{1}{2} \times 5,5\text{ql}^{2} \times 2l \times \frac{1}{5} - 4\text{ql}^{2} \times 2l \times \frac{3}{20} \right)$$

$$\varphi_{A} = \frac{1}{\text{EJy}} \times \left( \frac{7}{5} + \frac{21}{10} + \frac{11}{10} - \frac{6}{5} \right)\text{ql}^{3}$$

$$\varphi_{A} = \frac{3,4\text{ql}^{3}}{\text{EJy}}$$

$$W_{A} = \frac{1}{\text{EJy}} \times \left( \frac{1}{2} \times 3\text{ql}^{2} \times 4l \times \frac{13}{30} + 1,5\text{ql}^{2} \times 4l \times \frac{13}{20} + \frac{1}{2} \times 5,5\text{ql}^{2} \times 2l \times \frac{26}{30} - 4\text{ql}^{2} \times 2l \times \frac{13}{20} \right)$$

$$W_{A} = \frac{1}{\text{EJy}} \times \left( \frac{13}{5} + \frac{39}{10} + \frac{143}{30} - \frac{26}{5} \right)\text{ql}^{3}$$

$$W_{A} = \frac{6,1\text{ql}^{3}}{\text{EJy}}$$