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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

Partial Factors

The partial factor for the resistance of cross-sections

γ_{M0}

1.00

The partial factor for the resistance of cross-sections

γ_{M1}

1.00

Design resistance under uniform compression

N_{c,Rd}

2175.27

kN

BS EN 1993-1-1, 6.2.4

N_{c,Rd}

=

\frac{A_{eff} \cdot f_{y,section}}{γ_{M0}} \cdot (\frac{1}{10^{3}})

=

\frac{8208.57 \cdot 265.00}{1.00} \cdot (\frac{1}{10^{3}})

Buckling resistance under uniform compression (Y axis)

Buckling length

L_{cr,y}

14000.0

mm

N_{cr,y}

(2.320e+6)

N

N_{cr,y}

=

\frac{π^{2} \cdot E_{t} \cdot I_{yy}}{{L_{cr,y}}^{2}}

=

\frac{π^{2} \cdot 210000 \cdot (2.194e+8)}{{14000.0}^{2}}

Non-dimensional slenderness

λ_{yy}

0.96832

BS EN 1993-1-1, 6.3.1.3

λ_{yy}

=

{\begin{cases} \sqrt{\frac{A_{eff} \cdot f_{y,section}}{N_{cr,y}} \cdot Δ_{A,reduction}} & if isClassFour \\ \sqrt{\frac{A_{eff} \cdot f_{y,section}}{N_{cr,y}}} & otherwise. \end{cases}

=

{\begin{cases} \sqrt{\frac{8208.57 \cdot 265.00}{(2.320e+6)} \cdot 1.00} & if false \\ \sqrt{\frac{8208.57 \cdot 265.00}{(2.320e+6)}} & otherwise. \end{cases}

Imperfection factor

α_{buckling,y}

0.49000

BS EN 1993-1-1, Table 6.2

An intermediate auxiliary factor

Φ_{yy}

1.16

BS EN 1993-1-1, (6.49)

Φ_{yy}

=

0.5 \cdot (1 + α_{buckling,y} \cdot (λ_{yy} - 0.2) + {λ_{yy}}^{2})

=

0.5 \cdot (1 + 0.49000 \cdot (0.96832 - 0.2) + {0.96832}^{2})

Buckling reduction factor

χ_{yy,1}

0.55853

BS EN 1993-1-1, (6.49)

χ_{yy,1}

=

\frac{1}{(Φ_{yy} + \sqrt{{Φ_{yy}}^{2} - {λ_{yy}}^{2}})}

=

\frac{1}{(1.16 + \sqrt{{1.16}^{2} - {0.96832}^{2}})}

χ_{yy}

0.55853

χ_{yy}

=

{\begin{cases} 1 & if χ_{yy,1} > 1 \\ χ_{yy,1} & otherwise. \end{cases}

=

{\begin{cases} 1 & if 0.55853 > 1 \\ 0.55853 & otherwise. \end{cases}

N_{b,y,Rd}

1214.96

kN

N_{b,y,Rd}

=

\frac{χ_{yy} \cdot A_{eff} \cdot f_{y,section}}{γ_{M1}} \cdot (\frac{1}{10^{3}})

=

\frac{0.55853 \cdot 8208.57 \cdot 265.00}{1.00} \cdot (\frac{1}{10^{3}})

Buckling resistance under uniform compression (Z axis)

L_{cr,z}

14000.0

mm

N_{cr,z}

76390.6

N

N_{cr,z}

=

\frac{π^{2} \cdot E_{t} \cdot I_{zz}}{{L_{cr,z}}^{2}}

=

\frac{π^{2} \cdot 210000 \cdot (7.224e+6)}{{14000.0}^{2}}

λ_{zz}

5.34

λ_{zz}

=

{\begin{cases} \sqrt{\frac{A_{eff} \cdot f_{y,section}}{N_{cr,z}} \cdot Δ_{A,reduction}} & if isClassFour \\ \sqrt{\frac{A_{eff} \cdot f_{y,section}}{N_{cr,z}}} & otherwise. \end{cases}

=

{\begin{cases} \sqrt{\frac{8208.57 \cdot 265.00}{76390.6} \cdot 1.00} & if false \\ \sqrt{\frac{8208.57 \cdot 265.00}{76390.6}} & otherwise. \end{cases}

α_{buckling,z}

0.49000

Φ_{zz}

16.00

Φ_{zz}

=

0.5 \cdot (1 + α_{buckling,z} \cdot (λ_{zz} - 0.2) + {λ_{zz}}^{2})

=

0.5 \cdot (1 + 0.49000 \cdot (5.34 - 0.2) + {5.34}^{2})

χ_{zz,1}

0.03218

χ_{zz,1}

=

\frac{1}{(Φ_{zz} + \sqrt{{Φ_{zz}}^{2} - {λ_{zz}}^{2}})}

=

\frac{1}{(16.00 + \sqrt{{16.00}^{2} - {5.34}^{2}})}

χ_{zz}

0.03218

χ_{zz}

=

{\begin{cases} 1 & if χ_{zz,1} > 1 \\ χ_{zz,1} & otherwise. \end{cases}

=

{\begin{cases} 1 & if 0.03218 > 1 \\ 0.03218 & otherwise. \end{cases}

N_{b,z,Rd}

70.00

kN

N_{b,z,Rd}

=

\frac{χ_{zz} \cdot A_{eff} \cdot f_{y,section}}{γ_{M1}} \cdot (\frac{1}{10^{3}})

=

\frac{0.03218 \cdot 8208.57 \cdot 265.00}{1.00} \cdot (\frac{1}{10^{3}})

Buckling resistance under uniform compression (Torsion)

L_{cr,t}

14000.0

mm

N_{cr,t}

(1.927e+6)

N

N_{cr,t}

=

\frac{1}{({K_{yy}}^{2} + {K_{zz}}^{2})} \cdot (\frac{π^{2} \cdot E_{t} \cdot I_{w}}{{L_{cr,t}}^{2}} + G_{t} \cdot I_{T})

=

\frac{1}{({163.48}^{2} + {29.67}^{2})} \cdot (\frac{π^{2} \cdot 210000 \cdot (2.177e+11)}{{14000.0}^{2}} + 80769.2 \cdot 630028)

λ_{torsion}

1.06

λ_{torsion}

=

\sqrt{\frac{A_{eff} \cdot f_{y,section}}{N_{cr,t}}}

=

\sqrt{\frac{8208.57 \cdot 265.00}{(1.927e+6)}}

Φ_{torsion}

1.28

Φ_{torsion}

=

0.5 \cdot (1 + α_{buckling,z} \cdot (λ_{torsion} - 0.2) + {λ_{torsion}}^{2})

=

0.5 \cdot (1 + 0.49000 \cdot (1.06 - 0.2) + {1.06}^{2})

χ_{torsion,1}

0.50452

χ_{torsion,1}

=

\frac{1}{(Φ_{torsion} + \sqrt{{Φ_{torsion}}^{2} - {λ_{torsion}}^{2}})}

=

\frac{1}{(1.28 + \sqrt{{1.28}^{2} - {1.06}^{2}})}

χ_{torsion}

0.50452

χ_{torsion}

=

{\begin{cases} 1 & if χ_{torsion,1} > 1 \\ χ_{torsion,1} & otherwise. \end{cases}

=

{\begin{cases} 1 & if 0.50452 > 1 \\ 0.50452 & otherwise. \end{cases}

N_{b,t,Rd}

1097.47

kN

N_{b,t,Rd}

=

\frac{χ_{torsion} \cdot A_{eff} \cdot f_{y,section}}{γ_{M1}} \cdot (\frac{1}{10^{3}})

=

\frac{0.50452 \cdot 8208.57 \cdot 265.00}{1.00} \cdot (\frac{1}{10^{3}})