公式

具体公式请查看这里。

1440 结果

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' 调整的弱轴弯矩设计值 Fby'

F_{by}' = F_{b 2} = F_{b} \cdot C_{D} \cdot C_{M} \cdot C_{L} \cdot C_{t} \cdot C_{fu} \cdot C_{F} \cdot C_{i} \cdot C_{r}

F_{by}' = F_{b 2} = 1, 100 psi \cdot 1.6 \cdot 1.0 \cdot 1.0 \cdot 1.0 \cdot 1.1 \cdot 1.0 \cdot 1.0 \cdot 1.0

F_{by}' = F_{b 2} = 1, 936 psi

(a) 开发长度

L_{dc} = (\frac{f_{y} \cdot ψ_{r}}{50 \cdot λ \cdot \sqrt{f'_{c}}}) \cdot d_{b} = (\frac{(60, 000 psi) \cdot (1.0)}{50 \cdot (1.0) \cdot \sqrt{4000 psi}}) \cdot (0.75 in .) = 14.23 in .

(b) 展开长度

< / p > < p > （ b ）

{}^{t + ∆ t}R - {}^{t + ∆ t}F^{0}

E_{Ed} = 1, 0 \cdot E_{EdX} \oplus 0, 3 \cdot E_{EdY} \oplus 0, 3 \cdot E_{EdZ}

E_{Ed} = 0, 3 \cdot E_{EdX} \oplus 1, 0 \cdot E_{EdY} \oplus 0, 3 \cdot E_{EdZ}

E_{Ed} = 0, 3 \cdot E_{EdX} \oplus 0, 3 \cdot E_{EdY} \oplus 1, 0 \cdot E_{EdZ}

B 点 – 轴力

P_{N} = (β_{1} cd) (0.85 F'_{c}) + A'_{s} F_{y} - A_{s} F_{y}

Bemessungssituation - GZG - Charakteristisch / Quasi-ständig

\sum_{j \geq 1} G_{k, j} (1 + k_{d e f}) + P + Q_{k, 1} (1 + ψ_{2, 1} k_{d e f}) + \sum_{i \geq 1} Q_{k, i} (Ψ_{0, i} + Ψ_{2, i} k_{d e f})

Bemessungssituation für Erdbeben nach EN 1990 Abschnitt 6.4.3.4

E_{d} = {ΣG}_{k, j} " + " {Σψ}_{2, i} Q_{k, i}

Biaxiales Flächenträgheitsmoment

I_{z y} = I_{y z} = - \int_{A} z y d A

C 点 – 弯矩

M_{N} = A_{s} F_{y} (d - \frac{a}{2})

C_{p} = \frac{P - P_{r e f}}{\frac{1}{2} ρ U_{r e f}^{2}}

D 点 – 纯受压

P_{N} = A_{s} F_{y}

Drucker-Prager 屈服准则

σ_{v} = \frac{m - 1}{2} (σ_{1} + σ_{2} + σ_{3}) + \frac{m + 1}{2} \sqrt{\frac{1}{2} [{(σ_{1} - σ_{2})}^{2} + {(σ_{2} - σ_{3})}^{2} + {(σ_{3} - σ_{1})}^{2}]}

m = \frac{σ_{c}}{σ_{t}}

E_{d, frequ} = E \{\sum_{j \geq 1} G_{k, j} " + " ψ_{1, 1} \cdot Q_{k, 1} " + " \sum_{i > 1} ψ_{2, i} \cdot Q_{k, i}\}

V_{pl, Rd} = \frac{A_{v} \cdot (f_{y} / \sqrt{3})}{γ_{M 0}}

EC5-分解位置刚度

ε (z) = \{\begin{matrix} ε_{x} \\ ε_{y} \\ γ_{x y} \end{matrix}\} = \{\begin{matrix} \frac{\partial u}{\partial x} \\ \frac{\partial v}{\partial y} \\ \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x} \end{matrix}\} + z \{\begin{matrix} \frac{\partial φ_{y}}{\partial x} \\ - \frac{\partial φ_{x}}{\partial y} \\ \frac{\partial φ_{y}}{\partial y} - \frac{\partial φ_{x}}{\partial x} \end{matrix}\}

Equation F.1-1 – AISC 360-22

C_{b} = \frac{12.5 M_{\max}}{2.5 M_{\max} + 3 M_{A} + 4 M_{B} + 3 M_{C}}

Euler-Bernoulli 微分方程

\frac{d^{2} u_{z}}{{dx}^{2}} = - \frac{M_{y}}{{EI}_{y}}

Flächenträgheitsmoment

I_{y} = \int_{A} z^{2} d A

I_{z} = \int_{A} y^{2} d A

\frac{{dβ}_{i}}{dt} = \dot{β_{i}} (θ, stupeň θ, β_{1}, . . . β_{n})

\to m_{0} = \frac{σ_{1} - σ_{0}}{ε_{1} - ε_{0}} = \frac{2 - 0}{0, 000645 - 0} = 3 100 kN / {cm}^{2}

f (σ_{ij}; ε_{ij}^{pl}) = F (σ_{ij} - α_{ij} (ε_{ij}^{pl})) - k plasticita kinematická

f (σ; ε_{ij}^{pl}; ε_{p}) = F (σ_{ij} - α_{ij} (ε_{ij}^{pl})) - k (ε_{p}) \leq 0 plasticita

G0减小的弯矩

μ_{Ed} = \frac{M_{{EdG}_{0}}}{A_{c} \cdot h \cdot f_{cd}}

GB 50005, 5.2.8

\frac{N_{c}}{b l_{b} K_{B} K_{Z c p}} \leq f_{c, 90}

GB 50017, 6.2.7-1

\frac{M_{x}}{φ_{d} \cdot W_{1 x} \cdot f} \leq 1.0

GB 50017, 6.2.7-3

λ_{n, b} = \sqrt{\frac{f_{y}}{σ_{c r}}}

GB 50017, 6.2.7-4

σ_{c r} = \frac{3.46 b_{1} t_{1}^{3} + h_{w} t_{w}^{3} (7.27 γ + 3.3) φ_{1}}{h_{w}^{2} (12 b_{1} \cdot t_{1} + 1.78 h_{w} t_{w})} E

GB 50017, 6.2.7-5

γ = \frac{b_{1}}{b_{w}} \sqrt{\frac{b_{1} t_{1}}{h_{w} t_{w}}}

GB 50017, 6.2.7-6

φ_{1} = \frac{1}{2} (\frac{5.436 γ h_{w}^{2}}{l^{2}} + \frac{l^{2}}{5.436 γ h_{w}^{2}})