Hot Formed Rectangular Hollow Sections

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Hot-formed Rectangular Hollow Sections are rectangular profiles formed at elevated temperatures, providing lower residual stresses. Suitable for columns, beams, and trusses.

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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}

566.21

kN

BS EN 1993-1-1, 6.2.4

N_{c,Rd}

=

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

=

\frac{1594.96 \cdot 355.00}{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}

18577.4

N

N_{cr,y}

=

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

=

\frac{π^{2} \cdot 210000 \cdot (1.757e+6)}{{14000.0}^{2}}

Non-dimensional slenderness

λ_{yy}

5.52

BS EN 1993-1-1, 6.3.1.3

λ_{yy}

=

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

=

{\begin{cases} \sqrt{\frac{1594.96 \cdot 355.00}{18577.4} \cdot 1.00} & if false \\ \sqrt{\frac{1594.96 \cdot 355.00}{18577.4}} & otherwise. \end{cases}

Imperfection factor

α_{buckling,y}

0.21000

BS EN 1993-1-1, Table 6.2

α_{buckling,y}

=

{\begin{cases} 0.13 & if SteelGrade = 460 \\ 0.21 & otherwise. \end{cases}

=

{\begin{cases} 0.13 & if 355.00 = 460 \\ 0.21 & otherwise. \end{cases}

An intermediate auxiliary factor

Φ_{yy}

16.30

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 + 0.21000 \cdot (5.52 - 0.2) + {5.52}^{2})

Buckling reduction factor

χ_{yy,1}

0.03161

BS EN 1993-1-1, (6.49)

χ_{yy,1}

=

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

=

\frac{1}{(16.30 + \sqrt{{16.30}^{2} - {5.52}^{2}})}

χ_{yy}

0.03161

χ_{yy}

=

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

=

{\begin{cases} 1 & if 0.03161 > 1 \\ 0.03161 & otherwise. \end{cases}

N_{b,y,Rd}

17.90

kN

N_{b,y,Rd}

=

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

=

\frac{0.03161 \cdot 1594.96 \cdot 355.00}{1.00} \cdot (\frac{1}{10^{3}})

Buckling resistance under uniform compression (Z axis)

L_{cr,z}

14000.0

mm

N_{cr,z}

6152.96

N

N_{cr,z}

=

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

=

\frac{π^{2} \cdot 210000 \cdot 581863}{{14000.0}^{2}}

λ_{zz}

9.59

λ_{zz}

=

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

=

{\begin{cases} \sqrt{\frac{1594.96 \cdot 355.00}{6152.96} \cdot 1.00} & if false \\ \sqrt{\frac{1594.96 \cdot 355.00}{6152.96}} & otherwise. \end{cases}

α_{buckling,z}

0.21000

α_{buckling,z}

=

{\begin{cases} 0.13 & if SteelGrade = 460 \\ 0.21 & otherwise. \end{cases}

=

{\begin{cases} 0.13 & if 355.00 = 460 \\ 0.21 & otherwise. \end{cases}

Φ_{zz}

47.50

Φ_{zz}

=

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

=

0.5 \cdot (1 + 0.21000 \cdot (9.59 - 0.2) + {9.59}^{2})

χ_{zz,1}

0.01064

χ_{zz,1}

=

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

=

\frac{1}{(47.50 + \sqrt{{47.50}^{2} - {9.59}^{2}})}

χ_{zz}

0.01064

χ_{zz}

=

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

=

{\begin{cases} 1 & if 0.01064 > 1 \\ 0.01064 & otherwise. \end{cases}

N_{b,z,Rd}

6.02

kN

N_{b,z,Rd}

=

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

=

\frac{0.01064 \cdot 1594.96 \cdot 355.00}{1.00} \cdot (\frac{1}{10^{3}})

Buckling resistance under uniform compression (Torsion)

L_{cr,t}

14000.0

mm

N_{cr,t}

(8.708e+7)

N

N_{cr,t}

=

\frac{1}{({K_{yy}}^{2} + {K_{zz}}^{2})} \cdot (\frac{π^{2} \cdot E_{t} \cdot 0}{{L_{cr,t}}^{2}} + G_{t} \cdot I_{t})

=

\frac{1}{({33.19}^{2} + {19.10}^{2})} \cdot (\frac{π^{2} \cdot 210000 \cdot 0}{{14000.0}^{2}} + 80769.2 \cdot (1.581e+6))

λ_{torsion}

0.08064

λ_{torsion}

=

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

=

\sqrt{\frac{1594.96 \cdot 355.00}{(8.708e+7)}}

Φ_{torsion}

0.49072

Φ_{torsion}

=

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

=

0.5 \cdot (1 + 0.21000 \cdot (0.08064 - 0.2) + {0.08064}^{2})

χ_{torsion,1}

1.03

χ_{torsion,1}

=

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

=

\frac{1}{(0.49072 + \sqrt{{0.49072}^{2} - {0.08064}^{2}})}

χ_{torsion}

1.00

χ_{torsion}

=

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

=

{\begin{cases} 1 & if 1.03 > 1 \\ 1.03 & otherwise. \end{cases}

N_{b,t,Rd}

566.21

kN

N_{b,t,Rd}

=

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

=

\frac{1.00 \cdot 1594.96 \cdot 355.00}{1.00} \cdot (\frac{1}{10^{3}})