Tính toán cột chịu nén lệch tâm xiên

Các thông số: \(N = 91.94 \, \text{kN}, \quad M_y = 1.63 \, \text{kNm}, \quad M_x = 24.25 \, \text{kNm}\)

Bê tông B20: \(R_b = 11.5 \, \text{MPa}, \quad E_b = 27000 \, \text{MPa}\)

Cốt thép CII: \(R_s = R_{sc} = 280 \, \text{MPa}, \quad E_s = 21000 \, \text{MPa}\)

\(C_x = 250 \, \text{mm}, \quad C_y = 300 \, \text{mm}, \quad L = 3300 \, \text{mm}, \quad a = 40 \, \text{mm}, \quad \mu_{\text{max}} = 3\%, \quad \mu_{\text{min}} = 1\%\)

Bố trí cốt thép theo chu vi.

Lời giải chi tiết:

Kiểm tra điều kiện: \(0.5 \leq \frac{C_x}{C_y} = \frac{250}{300} = 0.84 \leq 2\)

Lấy: \(\xi_R = \frac{X_R}{h_0} = \frac{0.8}{1 + \frac{\varepsilon_{s,el}}{\varepsilon_{b2}}} = 0.579\)

\(\Psi = 0.7 \Rightarrow L_{ox} = L_{oy} = 0.7 \times 3300 = 2310 \, \text{mm}\)

Độ lệch tâm ngẫu nhiên theo hai phương:

\(e_{\text{ax}} = \max \left( \frac{L}{600}, \frac{C_x}{30}, 10 \, \text{mm} \right) = \max \left( \frac{3300}{600}, \frac{250}{30}, 10 \, \text{mm} \right) = 10 \, \text{mm}\)

\(e_{\text{ay}} = \max \left( \frac{L}{600}, \frac{C_y}{30}, 10 \, \text{mm} \right) = \max \left( \frac{3300}{600}, \frac{300}{30}, 10 \, \text{mm} \right) = 10 \, \text{mm}\)

Độ mảnh theo hai phương:

\(\lambda_x = \frac{L_x}{0.288 C_x} = \frac{2310}{0.288 \times 250} = 32\)

\(\lambda_y = \frac{L_y}{0.288 C_y} = \frac{2310}{0.288 \times 300} = 26.7\)

\(\Rightarrow \lambda = \max (\lambda_x, \lambda_y) = \max (32, 26.7) = 32\)

Kết cấu siêu tĩnh:

\(e_{1x} = \frac{M_x}{N} = \frac{24.95}{91.34} = 0.264 \, \text{m} = 264 \, \text{mm} \Rightarrow e_{ox} = \max(e_{1x}, e_{ax}) = \max(264, 10) = 264 \, \text{mm}\)

\(e_{1y} = \frac{M_y}{N} = \frac{1.63}{91.34} = 0.018 \, \text{m} = 18 \, \text{mm} \Rightarrow e_{oy} = \max(e_{1y}, e_{ay}) = \max(18, 10) = 18 \, \text{mm}\)

Xét uốn dọc:

\(\lambda_x = 32 > 14 \Rightarrow I_x = \frac{C_y C_x^3}{12} = \frac{300 \times 250^3}{12} = 390625000 \, \text{mm}^4\)

\(N_x^{\text{cr}} = \frac{2.5 \, E_b \, I_x}{L_{\text{ox}}^2} = \frac{2.5 \times 27000 \times 390625000}{2310^2} = 4941284 \, \text{N} \Rightarrow \eta_x = \frac{1}{1 – \frac{N}{N_x^{\text{cr}}}} = \frac{1}{1 – \frac{91940}{4941284}} = 1.019\)

\(\lambda_y = 26.7 > 14 \Rightarrow I_y = \frac{C_x C_y^3}{12} = \frac{250 \times 300^3}{12} = 562500000 \, \text{mm}^4\)

\(N_y^{\text{cr}} = \frac{2.5 \, E_b \, I_y}{L_{\text{oy}}^2} = \frac{2.5 \times 27000 \times 562500000}{2310^2} = 7115449.5 \, \text{N} \Rightarrow \eta_y = \frac{1}{1 – \frac{N}{N_y^{\text{cr}}}} = \frac{1}{1 – \frac{91940}{7115449.5}} = 1.013\)

Moment gia tăng khi xét uốn dọc:

\(M_x^* = \eta_x M_x = 1.013 \times 24.25 = 24.57 \, \text{kNm}\)

\(M_y^* = \eta_y M_y = 1.019 \times 1.63 = 1.66 \, \text{kNm}\)

Mô hình tính toán:

\(\frac{M_x^*}{C_x} = \frac{24.57 \times 1000}{250} = 98.28 \, \text{kN} > \frac{M_y^*}{C_y} = \frac{1.66 \times 1000}{300} = 5.5 \, \text{kN}\)

\(\Rightarrow h = C_x = 250 \, \text{mm}; \quad b = C_y = 300 \, \text{mm}\)

\(\Rightarrow M_1 = M_x^* = 24.57 \, \text{kNm}; \quad M_2 = M_y^* = 1.66 \, \text{kNm}\)

\(\Rightarrow e_a = e_{\text{ax}} + 0.2 \, e_{\text{ay}} = 10 + 0.2 \times 10 = 12 \, \text{mm}\)

\(a = 40 \, \text{mm} \Rightarrow h_o = h – a = 250 – 40 = 210 \, \text{mm}; \quad z = h – 2a = 250 – 2 \times 40 = 170 \, \text{mm}\)

\(x_1 = \frac{N}{R_b b} = \frac{91.94 \times 1000}{11.5 \times 300} = 26.6 \, \text{mm} \leq h_o = 210 \, \text{mm} \Rightarrow m_o = 1 – \frac{0.6 \times 26.6}{210} = 0.924\)

Moment tương đương:

\(M = M_1 + m_o M_2 \frac{h}{b} = 24.57 + 0.924 \times 1.65 \times \frac{250}{300} = 25.84 \, \text{kNm}\)

\(e_1 = \frac{M}{N} = \frac{25.84 \times 1000}{91.94} = 281 \, \text{mm}\)

\(\Rightarrow e_o = \max(e_1 + e_a) = 281 \, \text{mm}\)

\(\Rightarrow e = e_o + 0.5h – a = 281 + 0.5 \times 250 – 40 = 366 \, \text{mm}\)

Trường hợp tính toán

\(\varepsilon = \frac{e_o}{h_o} = \frac{281}{210} = 1.34 > 0.3\)

\(x = 26.6 \, \text{mm} < \xi_R h_o = 0.579 \times 210 = 121.66 \, \text{mm}\)

⇒ TH3: Lệch tâm lớn

Diện tích cốt thép

Với k=0.4

\(A_{st} = \frac{N(e + 0.5x_1 – h_o)}{k R_s z} = \frac{91.94 \times 1000 (366 + 0.5 \times 26.6 – 210)}{0.4 \times 280 \times 170} = 817 \, \text{mm}^2\)

Đánh giá và xử lý kết quả

\(\mu_{\text{min}} = 0.5\% \leq \mu = \frac{A_{\text{st}} \times 100}{b h_o} = \frac{817 \times 100}{300 \times 210} = 1.3\% \leq \mu_{\text{max}} = 3\%\)

Bố trí cốt thép:  Chọn 4Φ18 bố trí 4 góc

\(A_c = \frac{4\pi d^2}{4} = 4\pi \left(\frac{18^2}{4}\right) = 1018 \, \text{mm}^2\)