Parallel Flange Channels

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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

Parallel Flange Channels (PFC)

Geometry Input

Section depth

h_{t}

430.00

mm

Section width

b_{t}

100.00

mm

Web thickness

t_{w}

11.00

mm

Flange thickness

t_{f}

19.00

mm

Root radius

r_{t}

15.00

mm

Material Properties

Young's modulus

E_{t}

210000

\frac{N}{{mm}^{2}}

Poisson's ratio

v_{t}

0.30000

Shear modulus

G_{t}

80769.2

\frac{N}{{mm}^{2}}

G_{t}

=

\frac{E_{t}}{(2 \cdot (1 + v_{t}))}

=

\frac{210000}{(2 \cdot (1 + 0.30000))}

Density

ρ_{t}

7850.00

\frac{kg}{m^{3}}

Section Properties

Depth of web

h_{w}

392.00

mm

h_{w}

=

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

=

430.00 - 2 \cdot 19.00

Depth between fillets

d_{t}

362.00

mm

d_{t}

=

h_{t} - 2 \cdot t_{f} - 2 \cdot r_{t}

=

430.00 - 2 \cdot 19.00 - 2 \cdot 15.00

Distance from major axis to flange centroid

h_{f}

205.50

mm

h_{f}

=

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

=

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

Cross-sectional area (1)

A_{I}

8112.00

{mm}^{2}

A_{I}

=

2 \cdot b_{t} \cdot t_{f} + h_{w} \cdot t_{w}

=

2 \cdot 100.00 \cdot 19.00 + 392.00 \cdot 11.00

Cross-sectional area (2)

A_{II}

96.57

{mm}^{2}

A_{II}

=

2 \cdot {r_{t}}^{2} \cdot (1 - \frac{π}{4})

=

2 \cdot {15.00}^{2} \cdot (1 - \frac{π}{4})

Cross-sectional area

A_{t}

8208.57

{mm}^{2}

A_{t}

=

A_{I} + A_{II}

=

8112.00 + 96.57

Distance to centroid in the z-z axis

C_{z}

215.00

mm

C_{z}

=

\frac{h_{t}}{2}

=

\frac{430.00}{2}

Distance to centroid in the y-y axis

C_{y}

26.35

mm

C_{y}

=

\frac{(2 \cdot {b_{t}}^{2} \cdot t_{f} + h_{w} \cdot {t_{w}}^{2})}{(2 \cdot b_{t} \cdot h_{t} - 2 \cdot h_{w} \cdot (b_{t} - t_{w}))}

=

\frac{(2 \cdot {100.00}^{2} \cdot 19.00 + 392.00 \cdot {11.00}^{2})}{(2 \cdot 100.00 \cdot 430.00 - 2 \cdot 392.00 \cdot (100.00 - 11.00))}

Mass per metre length

m_{m}

64.44

\frac{kg}{m}

m_{m}

=

\frac{A_{t} \cdot ρ_{t}}{1000000}

=

\frac{8208.57 \cdot 7850.00}{1000000}

Second moment of area about major axis (1)

I_{{yy}_{1}}

(2.158e+8)

{mm}^{4}

I_{{yy}_{1}}

=

\frac{b_{t} \cdot {h_{t}}^{3}}{12} - \frac{(b_{t} - t_{w}) \cdot {(h_{t} - 2 \cdot t_{f})}^{3}}{12}

=

\frac{100.00 \cdot {430.00}^{3}}{12} - \frac{(100.00 - 11.00) \cdot {(430.00 - 2 \cdot 19.00)}^{3}}{12}

Second moment of area about major axis (2)

I_{{yy}_{2}}

(3.584e+6)

{mm}^{4}

I_{{yy}_{2}}

=

2 \cdot {r_{t}}^{2} \cdot (1 - \frac{π}{4}) \cdot {(\frac{d_{t}}{2} + 0.7766 \cdot r_{t})}^{2}

=

2 \cdot {15.00}^{2} \cdot (1 - \frac{π}{4}) \cdot {(\frac{362.00}{2} + 0.7766 \cdot 15.00)}^{2}

Second moment of area about major axis

I_{yy}

(2.194e+8)

{mm}^{4}

I_{yy}

=

I_{{yy}_{1}} + I_{{yy}_{2}}

=

(2.158e+8) + (3.584e+6)

Second moment of area about minor axis (1)

I_{{zz}_{1}}

(7.210e+6)

{mm}^{4}

I_{{zz}_{1}}

=

\frac{{t_{w}}^{3} \cdot h_{w}}{12} + t_{w} \cdot h_{w} \cdot {(C_{y} - \frac{t_{w}}{2})}^{2} + \frac{2 \cdot {b_{t}}^{3} \cdot t_{f}}{12} + 2 \cdot b_{t} \cdot t_{f} \cdot {(C_{y} - \frac{b_{t}}{2})}^{2}

=

\frac{{11.00}^{3} \cdot 392.00}{12} + 11.00 \cdot 392.00 \cdot {(26.35 - \frac{11.00}{2})}^{2} + \frac{2 \cdot {100.00}^{3} \cdot 19.00}{12} + 2 \cdot 100.00 \cdot 19.00 \cdot {(26.35 - \frac{100.00}{2})}^{2}

Second moment of area about minor axis (2)

I_{{zz}_{2}}

13893.8

{mm}^{4}

I_{{zz}_{2}}

=

2 \cdot {r_{t}}^{2} \cdot (1 - \frac{π}{4}) \cdot {(C_{y} - t_{w} - 0.2234 \cdot r_{t})}^{2}

=

2 \cdot {15.00}^{2} \cdot (1 - \frac{π}{4}) \cdot {(26.35 - 11.00 - 0.2234 \cdot 15.00)}^{2}

Second moment of area about minor axis

I_{zz}

(7.224e+6)

{mm}^{4}

I_{zz}

=

I_{{zz}_{1}} + I_{{zz}_{2}}

=

(7.210e+6) + 13893.8

Web local buckling ratio

\frac{c_{w}}{t_{w}}

32.91

-

\frac{c_{w}}{t_{w}}

=

\frac{d_{t}}{t_{w}}

=

\frac{362.00}{11.00}

Flange local buckling ratio

\frac{c_{f}}{t_{f}}

3.89

-

\frac{c_{f}}{t_{f}}

=

\frac{(b_{t} - t_{w} - r_{t})}{(t_{f})}

=

\frac{(100.00 - 11.00 - 15.00)}{(19.00)}

Radius of gyration about major axis

K_{yy}

163.48

mm

K_{yy}

=

\sqrt{\frac{I_{yy}}{A_{t}}}

=

\sqrt{\frac{(2.194e+8)}{8208.57}}

Radius of gyration about minor axis

K_{zz}

29.67

mm

K_{zz}

=

\sqrt{\frac{I_{zz}}{A_{t}}}

=

\sqrt{\frac{(7.224e+6)}{8208.57}}

Elastic section modulus about the major axis

S_{yy}

(1.020e+6)

{mm}^{3}

S_{yy}

=

\frac{I_{yy}}{C_{z}}

=

\frac{(2.194e+8)}{215.00}

Elastic section modulus about the minor axis

S_{zz}

98079.6

{mm}^{3}

S_{zz}

=

\frac{I_{zz}}{(b_{t} - C_{y})}

=

\frac{(7.224e+6)}{(100.00 - 26.35)}

Plastic section modulus about major axis (1)

Z_{yy,1}

(1.203e+6)

{mm}^{3}

Z_{yy,1}

=

\frac{b_{t} \cdot {h_{t}}^{2}}{4} - \frac{(b_{t} - t_{w}) \cdot {h_{w}}^{2}}{4}

=

\frac{100.00 \cdot {430.00}^{2}}{4} - \frac{(100.00 - 11.00) \cdot {392.00}^{2}}{4}

Plastic section modulus about major axis (2)

Z_{yy,2}

18604.3

{mm}^{3}

Z_{yy,2}

=

2 \cdot {r_{t}}^{2} \cdot (1 - \frac{π}{4}) \cdot (\frac{d_{t}}{2} + 0.7766 \cdot r_{t})

=

2 \cdot {15.00}^{2} \cdot (1 - \frac{π}{4}) \cdot (\frac{362.00}{2} + 0.7766 \cdot 15.00)

Plastic section modulus about major axis

Z_{yy}

(1.222e+6)

{mm}^{3}

Z_{yy}

=

Z_{yy,1} + Z_{yy,2}

=

(1.203e+6) + 18604.3

Plastic neutral axis location - pna is inside the web

y_{pna,1}

9.54

mm

y_{pna,1}

=

\frac{A_{t}}{(2 \cdot h_{t})}

=

\frac{8208.57}{(2 \cdot 430.00)}

Plastic neutral axis location - pna is outside the web

y_{pna,2}

-8.01

mm

y_{pna,2}

=

t_{w} + \frac{(\frac{A_{t}}{2} - t_{w} \cdot h_{t} - 2 \cdot {r_{t}}^{2} \cdot (1 - \frac{π}{4}))}{(2 \cdot t_{f})}

=

11.00 + \frac{(\frac{8208.57}{2} - 11.00 \cdot 430.00 - 2 \cdot {15.00}^{2} \cdot (1 - \frac{π}{4}))}{(2 \cdot 19.00)}

Moment of Total Area about Rear

Z_{total}

215102

{mm}^{3}

Z_{total}

=

\frac{h_{t} \cdot {t_{w}}^{2}}{2} + t_{f} \cdot ({b_{t}}^{2} - {t_{w}}^{2}) + 2 \cdot {r_{t}}^{2} \cdot (1 - \frac{π}{4}) \cdot (t_{w} + 0.2234 \cdot r_{t})

=

\frac{430.00 \cdot {11.00}^{2}}{2} + 19.00 \cdot ({100.00}^{2} - {11.00}^{2}) + 2 \cdot {15.00}^{2} \cdot (1 - \frac{π}{4}) \cdot (11.00 + 0.2234 \cdot 15.00)

Plastic section modulus about minor axis (2)

Z_{zz,2,temp}

26320.2

{mm}^{3}

Z_{zz,2,temp}

=

\frac{h_{t} \cdot {t_{w}}^{2}}{2} + 2 \cdot {r_{t}}^{2} \cdot (1 - \frac{π}{4}) \cdot (t_{w} + 0.2234 \cdot r_{t}) + t_{f} \cdot ({y_{pna,2}}^{2} - {t_{w}}^{2})

=

\frac{430.00 \cdot {11.00}^{2}}{2} + 2 \cdot {15.00}^{2} \cdot (1 - \frac{π}{4}) \cdot (11.00 + 0.2234 \cdot 15.00) + 19.00 \cdot ({-8.01}^{2} - {11.00}^{2})

Plastic section modulus about minor axis

Z_{zz}

175927

{mm}^{3}

Z_{zz}

=

{\begin{cases} Z_{total} - h_{t} \cdot {y_{pna,1}}^{2} & if 0.5 \cdot A_{t} \leq h_{t} \cdot t_{w} \\ Z_{total} - 2 \cdot Z_{zz,2,temp} & otherwise. \end{cases}

=

{\begin{cases} 215102 - 430.00 \cdot {9.54}^{2} & if 0.5 \cdot 8208.57 \leq 430.00 \cdot 11.00 \\ 215102 - 2 \cdot 26320.2 & otherwise. \end{cases}

Warping constant

I_{w}

(2.177e+11)

{mm}^{6}

I_{w}

=

\frac{{(h_{t} - t_{f})}^{2}}{4} \cdot (I_{zz} - (A_{t} \cdot {(C_{y} - \frac{t_{w}}{2})}^{2} \cdot (\frac{{(h_{t} - t_{f})}^{2} \cdot A_{t}}{(4 \cdot I_{yy})} - 1)))

=

\frac{{(430.00 - 19.00)}^{2}}{4} \cdot ((7.224e+6) - (8208.57 \cdot {(26.35 - \frac{11.00}{2})}^{2} \cdot (\frac{{(430.00 - 19.00)}^{2} \cdot 8208.57}{(4 \cdot (2.194e+8))} - 1)))

Geometric coefficient

α_{1}

0.09208

-

α_{1}

=

\frac{0.2621 \cdot t_{w}}{t_{f}} - 0.0908 + \frac{0.1231 \cdot r_{t}}{t_{f}} - \frac{0.0752 \cdot r_{t} \cdot t_{w}}{{t_{f}}^{2}} - \frac{0.0945 \cdot {t_{w}}^{2}}{{t_{f}}^{2}}

=

\frac{0.2621 \cdot 11.00}{19.00} - 0.0908 + \frac{0.1231 \cdot 15.00}{19.00} - \frac{0.0752 \cdot 15.00 \cdot 11.00}{{19.00}^{2}} - \frac{0.0945 \cdot {11.00}^{2}}{{19.00}^{2}}

Geometric coefficient

D_{1}

23.22

mm

D_{1}

=

2 \cdot ((3 \cdot r_{t} + t_{w} + t_{f}) - \sqrt{2 \cdot (2 \cdot r_{t} + t_{w}) \cdot (2 \cdot r_{t} + t_{f})})

=

2 \cdot ((3 \cdot 15.00 + 11.00 + 19.00) - \sqrt{2 \cdot (2 \cdot 15.00 + 11.00) \cdot (2 \cdot 15.00 + 19.00)})

Torsional constant

I_{T}

630028

{mm}^{4}

I_{T}

=

\frac{2}{3} \cdot b_{t} \cdot {t_{f}}^{3} + \frac{1}{3} \cdot (h_{t} - 2 \cdot t_{f}) \cdot {t_{w}}^{3} + 2 \cdot α_{1} \cdot {D_{1}}^{4} - 0.42 \cdot {t_{f}}^{4}

=

\frac{2}{3} \cdot 100.00 \cdot {19.00}^{3} + \frac{1}{3} \cdot (430.00 - 2 \cdot 19.00) \cdot {11.00}^{3} + 2 \cdot 0.09208 \cdot {23.22}^{4} - 0.42 \cdot {19.00}^{4}

Torsional index

X

22.44

-

X

=

\sqrt{\frac{(π^{2} \cdot E_{t} \cdot A_{t} \cdot I_{w})}{(20 \cdot G_{t} \cdot I_{T} \cdot I_{zz})}}

=

\sqrt{\frac{(π^{2} \cdot 210000 \cdot 8208.57 \cdot (2.177e+11))}{(20 \cdot 80769.2 \cdot 630028 \cdot (7.224e+6))}}

Geometric coefficient

g

0.98340

-

g

=

\sqrt{1 - \frac{I_{zz}}{I_{yy}}}

=

\sqrt{1 - \frac{(7.224e+6)}{(2.194e+8)}}

Buckling parameter

U

0.91836

-

U

=

{(\frac{Z_{yy} \cdot g}{A_{t}})}^{0.5} \cdot {(\frac{I_{zz}}{I_{w}})}^{0.25}

=

{(\frac{(1.222e+6) \cdot 0.98340}{8208.57})}^{0.5} \cdot {(\frac{(7.224e+6)}{(2.177e+11)})}^{0.25}