Hot Formed Parallel Flange Channels

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Hot-formed Parallel Flange Channels (PFC) are C-shaped sections with parallel inner and outer flange surfaces, simplifying connections. Used for columns, lintels, and beams.

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Calculations

Description

Symbol Name

Value

Unit

Comment

Steel grade factor for the web

ε_{web}

0.92442

BS EN 1993-1-1 Table 5.2

ε_{web}

=

\sqrt{\frac{235}{f_{y,w}}}

=

\sqrt{\frac{235}{275.00}}

Steel grade factor for the flange

ε_{flange}

0.94170

BS EN 1993-1-1 Table 5.2

ε_{flange}

=

\sqrt{\frac{235}{f_{y,fl}}}

=

\sqrt{\frac{235}{265.00}}

isClassFour

false

BS EN 1993-1-1 Table 5.2

isClassFour

=

or (\frac{c_{w}}{t_{w}} > 124 \cdot ε_{web}, \frac{c_{f}}{t_{f}} > 14 \cdot ε_{flange})

=

or (32.91 > 124 \cdot 0.92442, 3.89 > 14 \cdot 0.94170)

isClassThree

false

isClassThree

=

{\begin{cases} 1 = 0 & if isClassFour \\ or (\frac{c_{w}}{t_{w}} > 83 \cdot ε_{web}, \frac{c_{f}}{t_{f}} > 10 \cdot ε_{flange}) & otherwise. \end{cases}

=

{\begin{cases} 1 = 0 & if false \\ or (32.91 > 83 \cdot 0.92442, 3.89 > 10 \cdot 0.94170) & otherwise. \end{cases}

isClassTwo

false

isClassTwo

=

{\begin{cases} 1 = 0 & if or (isClassThree, isClassFour) \\ or (\frac{c_{w}}{t_{w}} > 72 \cdot ε_{web}, \frac{c_{f}}{t_{f}} > 9 \cdot ε_{flange}) & otherwise. \end{cases}

=

{\begin{cases} 1 = 0 & if or (false, false) \\ or (32.91 > 72 \cdot 0.92442, 3.89 > 9 \cdot 0.94170) & otherwise. \end{cases}

isClassOne

true

isClassOne

=

{\begin{cases} 1 = 1 & if and (\frac{c_{w}}{t_{w}} \leq 72 \cdot ε_{web}, \frac{c_{f}}{t_{f}} \leq 9 \cdot ε_{flange}) \\ 1 = 0 & otherwise. \end{cases}

=

{\begin{cases} 1 = 1 & if and (32.91 \leq 72 \cdot 0.92442, 3.89 \leq 9 \cdot 0.94170) \\ 1 = 0 & otherwise. \end{cases}

Initial stress distribution for pure bending (Web)

ψ_{t,web}

-1.00

ψ_{t,web}

=

- 1

Buckling factor (web)

k_{sigma,web}

23.90

Plate slenderness of the web

λ_{p,web}

0.25641

λ_{p,web}

=

\frac{\frac{c_{w}}{t_{w}}}{(28.4 \cdot ε_{web} \cdot \sqrt{k_{sigma,web}})}

=

\frac{32.91}{(28.4 \cdot 0.92442 \cdot \sqrt{23.90})}

Reduction factor for the web

ρ_{web}

1.00

ρ_{web}

=

{\begin{cases} (λ_{p,web} - 0.055 \cdot (3 + ψ_{t,web})) \cdot (\frac{1}{{λ_{p,web}}^{2}}) & if λ_{p,web} > 0.5 + \sqrt{0.085 - 0.055 \cdot ψ_{t,web}} \\ 1 & otherwise. \end{cases}

=

{\begin{cases} (0.25641 - 0.055 \cdot (3 + -1.00)) \cdot (\frac{1}{{0.25641}^{2}}) & if 0.25641 > 0.5 + \sqrt{0.085 - 0.055 \cdot -1.00} \\ 1 & otherwise. \end{cases}

checkWeb

true

checkWeb

=

ρ_{web} \leq 1

=

1.00 \leq 1

Initial stress distribution for pure bending (Flange)

ψ_{t,flange}

1.00

Buckling factor (Flange)

k_{sigma,"flange"}

0.43000

Plate slenderness of the flange

λ_{p,flange}

0.22208

λ_{p,flange}

=

\frac{\frac{c_{f}}{t_{f}}}{(28.4 \cdot ε_{flange} \cdot \sqrt{k_{sigma,"flange"}})}

=

\frac{3.89}{(28.4 \cdot 0.94170 \cdot \sqrt{0.43000})}

Reduction factor for the flange

ρ_{flange}

1.00

ρ_{flange}

=

{\begin{cases} \frac{(λ_{p,flange} - 0.188)}{{λ_{p,flange}}^{2}} & if λ_{p,flange} > 0.748 \\ 1 & otherwise. \end{cases}

=

{\begin{cases} \frac{(0.22208 - 0.188)}{{0.22208}^{2}} & if 0.22208 > 0.748 \\ 1 & otherwise. \end{cases}

checkFlange

true

checkFlange

=

ρ_{flange} \leq 1

=

1.00 \leq 1

b_{eff,w}

181.00

mm

b_{eff,w}

=

ρ_{web} \cdot (\frac{d_{t}}{2})

=

1.00 \cdot (\frac{362.00}{2})

b_{eff,f}

74.00

mm

b_{eff,f}

=

ρ_{flange} \cdot (b_{t} - t_{w} - r_{t})

=

1.00 \cdot (100.00 - 11.00 - 15.00)

b_{lost,w}

mm

b_{lost,w}

=

\frac{d_{t}}{2} - b_{eff,w}

=

\frac{362.00}{2} - 181.00

b_{lost,f}

mm

b_{lost,f}

=

(b_{t} - t_{w} - r_{t}) - b_{eff,f}

=

(100.00 - 11.00 - 15.00) - 74.00

A_{lost,w}

{mm}^{2}

A_{lost,w}

=

b_{lost,w} \cdot t_{w}

=

0 \cdot 11.00

A_{lost,f}

{mm}^{2}

A_{lost,f}

=

b_{lost,f} \cdot t_{f}

=

0 \cdot 19.00

Effective area

A_{eff}

8208.57

{mm}^{2}

A_{eff}

=

{\begin{cases} A_{t} - A_{lost,w} - A_{lost,f} & if isClassFour \\ A_{t} & otherwise. \end{cases}

=

{\begin{cases} 8208.57 - 0 - 0 & if false \\ 8208.57 & otherwise. \end{cases}

Total reduction factor

Δ_{A,reduction}

1.00

Δ_{A,reduction}

=

\frac{A_{eff}}{A_{t}}

=

\frac{8208.57}{8208.57}

d_{f}

205.50

mm

d_{f}

=

\frac{h_{t}}{2} - \frac{t_{f}}{2}

=

\frac{430.00}{2} - \frac{19.00}{2}

d_{w}

107.50

mm

d_{w}

=

\frac{h_{t}}{4}

=

\frac{430.00}{4}

Neutral axis shift

e_{N}

mm

e_{N}

=

\frac{(A_{lost,w} \cdot d_{w} + A_{lost,f} \cdot d_{f})}{A_{eff}}

=

\frac{(0 \cdot 107.50 + 0 \cdot 205.50)}{8208.57}

I_{lost,w}

{mm}^{4}

I_{lost,w}

=

A_{lost,w} \cdot {d_{w}}^{2}

=

0 \cdot {107.50}^{2}

I_{lost,f}

{mm}^{4}

I_{lost,f}

=

A_{lost,f} \cdot {d_{f}}^{2}

=

0 \cdot {205.50}^{2}

I_{eff}

(2.194e+8)

{mm}^{4}

I_{eff}

=

{\begin{cases} I_{yy} - I_{lost,w} - I_{lost,f} - A_{eff} \cdot {e_{N}}^{2} & if isClassFour \\ I_{yy} & otherwise. \end{cases}

=

{\begin{cases} (2.194e+8) - 0 - 0 - 8208.57 \cdot 0^{2} & if false \\ (2.194e+8) & otherwise. \end{cases}

Effective section modulus (compression)

W_{eff,c}

(1.020e+6)

{mm}^{3}

W_{eff,c}

=

\frac{I_{eff}}{(\frac{h_{t}}{2} + e_{N})}

=

\frac{(2.194e+8)}{(\frac{430.00}{2} + 0)}

Effective section modulus (tension)

W_{eff,t}

(1.020e+6)

{mm}^{3}

W_{eff,t}

=

\frac{I_{eff}}{(\frac{h_{t}}{2} - e_{N})}

=

\frac{(2.194e+8)}{(\frac{430.00}{2} - 0)}

Effective section modulus

W_{eff}

(1.020e+6)

{mm}^{3}

W_{eff}

=

\min (W_{eff,c}, W_{eff,t})

=

\min ((1.020e+6), (1.020e+6))