公式

具体公式请查看这里。

1871 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{(60000 psi) \cdot (1.0)}{50 \cdot (1.0) \cdot \sqrt{4000 psi}}) \cdot (0.75 in) = 14.23 in

(b) 展开长度

L_{dc} = 0.0003 \cdot f_{y} \cdot ψ_{r} \cdot d_{b} = 0.0003 \cdot (60000 psi) \cdot (1.0) \cdot (0.75 in) = 13.5 in

1. 迭代

{}^{t + Δ}R -^{t + Δt} F^{0}

1. 双向偏心心宽（菱形）

|e_{y}| \leq - \frac{\frac{w_{y}}{6}}{\frac{w_{x}}{6}} \cdot |e_{x}| + \frac{w_{y}}{6}

100/30 规则

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}

2. 双向偏心截面宽度(椭圆)

e_{y} \leq \frac{w_{y}}{3} \cdot \sqrt{1 - {(\frac{e_{x}}{\frac{w_{x}}{3}})}^{2}}

AISC 焊接极限应力

\emptyset R_{n} = \emptyset \cdot β \cdot F_{nw} \cdot k_{ds}

\emptyset R_{n} = 0.75 \cdot 0.912 \cdot (0.60 \cdot 70 ksi) \cdot 1.0

\emptyset R_{n} = 28.7 ksi

AISC 焊接极限应力

\emptyset R_{n} = \emptyset \cdot F_{nw} \cdot k_{ds}

\emptyset R_{n} = 0.75 \cdot (0.60 \cdot 70 ksi) \cdot 1.40

\emptyset R_{n} = 44.1 ksi

Ansatzfunktion für die Durchbiegung

w (x) = e^{λ x}

A_s1计算书

A_{s_{1}} = \frac{ω_{1} \cdot b \cdot d \cdot f_{cd} + N_{Ed}}{σ_{sd}} = \frac{0.1265 \cdot 100 \cdot 17 \cdot 1.13 + 0}{45.37} = 5.36 {cm}^{2} / m

Aufstellen der richtigen Differentialgleichung der Durchbiegung

λ^{2} + α^{2} = 0 | - α^{2}

λ^{2} = - α^{2} | \sqrt{}

λ_{1} = - i α

λ_{2} = i α

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}

Berechnung von C_D,B

C_{D, B} = \frac{E}{4 \cdot (1 - ν^{2})} \cdot \frac{1}{\frac{\bar{h}}{t_{w}^{3}} c_{1} \cdot \frac{b}{t_{f}^{3}}} =

= \frac{21000}{4 \cdot (1 - 0 . 3^{2})} \cdot \frac{1}{\frac{(25 - 1.25)}{{0.75}^{3}} \frac{0.5 \cdot 26}{{1.25}^{3}}} = 91.64 kNm / m

Biaxiales Flächenträgheitsmoment

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

BSt 500 S (B) 的最大钢筋应力 σ_s,max

σ_{s,max} = 0.8 \cdot f_{yk} = 0.8 \cdot 500 = 400 {N/mm}^{2}

C 点 – 弯矩

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

CFL 编号

CFL = Δt \cdot \frac{u_{max}}{{Δx}_{min}}

Cp 值

C_{p} = \frac{P - P_{ref}}{\frac{1}{2} {ρU}_{ref}^{2}}

D 点 – 纯受压

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

Der Querkontraktionszahl (Poissonzahl)

G = \frac{E}{2 \cdot (1 + μ)}

Die Bauwerksantwort

S_{p} = \sqrt{\sum_{i = 1}^{N} {(S_{i} - S_{m})}^{2}}

Dimensionslose Variable u+

u^{+} = \frac{U}{u_{τ}}

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}}

EC 2 | 公式

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}\}

EC3 | 公式

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}\}

EN 1991-1-2 (3.1) 中的净热通量

{\overset{\cdot}{h}}_{net} = {\overset{\cdot}{h}}_{net, c} {\overset{\cdot}{h}}_{net, r}