Subject: Design and Drawing of Steel Structure

Topic: Compression Member

Difficulty: High

Design a built column composed of two channel sections placed back to back, carrying an axial factored load of 1600 KN. Effective length of column is 5.2 m. Design battened system. Draw neat sketches to show details. Use 4.6 grade 20mm dia. bolts for the connections.

Given

Fractured load = 1600 kN, Effective length = 5200 mm, 4.6 grade 20 mm diameter bolts.

To know: Design batten system, using two channels placing back to back.

For double channels slenderness ratio to be assumed between 40-80.

Therefore, assume allowable compression stress = 190 N/mm$^2$

C\S area required = $ \frac{1600 \times 10^3}{190} $ = 8421.05 mm$^2$

Let us try 215MC 300 at 351.2 N/m, properties of 15MC 300 are:

A = 4564 mm$^2$, I$_xx$ = 6362.6 x 10$^4$ mm$^4$

I$_yy$ = 310.8 x 10$^4$ mm$^4$, r$_xx$ = 118.1 mm, r$_yy$ = 26.1mm

c$_yy$ = 23.6 mm

Area provided = 2x4564=9128 mm$^2$

Effective slenderness ratio = $ \frac{KL}{r}|_c = 1.1 \times \frac{5200}{118.1} = 48.33 $

For effective slenderness ratio = 48.33, F$_y$ = 250 N/mm$^2$ and buckling class, the design compression stress = $ 198 - \frac{198-183}{50-40}(48.33-40) = 185.5 N/mm^2 $

Therefore, Design compression strength = $ F_c d A_c = 185.5 \times 9128 = 1693.24 kN \gt 1600 kN $

Therefore, safe.

Spacing of channels:

$ I_{xx} = I_{yy} \\ 2 \times 6362.6 \times 10^4 = 2[I_{yy} + Ah^2] \\ = 2[310.8 \times 10^4 + 4564(\frac{S}{2} + 23.6)^2] \\ 63.62 \times 10^6 = 310.8 \times 10^4 + 4564 (\frac{S}{2} + 23.6)^2 \\ 13259.42 = (\frac{S}{2} + 23.6)^2 \\ 115.15 = \frac{S}{2} + 23.6 \\ \frac{S}{2} = 91.55 \\ \therefore S = 183.099 \approx 185 mm $

Therefore provide 215MC 300 at 351.2 N/m at a spacing 185mm back to back.

Spacing of battens:

$ (\frac{C}{r_{min}})_{component} \ngtr 50 \hspace{0.20cm} \& \hspace{0.20cm} \ngtr 0.7 \lambda_e \\ \therefore \frac{C}{ry} \lt 0.7 \lambda_e \\ C = 0.7 \times 48.33 \times 26.1 = 882.98 mm $

Also,

$ \frac{C}{ry} \lt 50 \\ C = 50 \times 26.1 = 13.5 mm $

Therefore, provide spacing = 900 mm

Size of end battens:

Effective depth = S + 2 c$_yy$ = 185 + 2x23.6 = 232.2 mm

Overall depth = Effective depth + 2 x edge distance

(Provide 20 mm dia bolts, edge distance = 1.5 x 22 = 33 $\approx$ 35mm)

Therefore, overall depth = 232.2 + 2x35 = 302.2 $\approx$ 310 mm

Thickness of battens = $\frac{1}{50}(185+2 \times 50) = 5.7 \approx 6 mm$

Provide 6 mm thick batten.

Length of battens = 185 + 2 x 90 = 365 mm

Therefore, provide end battens plate of size 365x310x6 mm$^3$

Size of intermediate battens:

Effective depth = $\frac{3}{4}$ effective depth of end batten

= $\frac{3}{4} \times 232.2 = 174.15 mm$

Overall depth = 174.15 + 2x35 = 244.15 $\approx$ 250 mm

Therefore provide intermediate batten plate of size 365x250x6 mm$^3$

Design forces:

Transverse shear = vt = 2.5% of column load = $\frac{2.5}{100} \times 1600$ = 40 kN

Longitudinal shear = vb = $\frac{vtc}{NS} = \frac{40 \times 10^3 \times 900}{2 \times (185 + 2 \times 50)} = 63.157 kN $

$ Moment = M = \frac{vtc}{2N} = \frac{40 \times 10^3 \times .300}{2 \times 2} = 9 \times 10^6 N.mm $

Check for end battens:

Shear stress = $ \frac{vb}{A} = \frac{63.157 \times 10^3}{6 \times 310} = 33.95 N/mm^2 \lt \frac{Fy}{\sqrt{3} rm_1} = \frac{250}{\sqrt{3} \times 1.1} = 131.2 N/mm^2 $

Therefore, OK.

Bending stress = $ \frac{6M}{td^2} = \frac{6 \times 9 \times 10^6}{6 \times 310^2} = 93.65 N/mm^2 \lt \frac{Fy}{rm_1} = \frac{250}{1.1} = 227.27 N/mm^2 $

Therefore, Ok.

Check for intermediate battens:

Shear stress = $ \frac{vb}{A} = \frac{63.157 \times 10^3}{6 \times 250} = 42.10 N/mm^2 \lt 131.2 N/mm^2 $

Therefore, Ok.

Bending stress = $ \frac{6M}{td^2} = \frac{6 \times 9 \times 10^6}{6 \times 250^2} = 144 N/mm^2 \lt 227.27 N/mm^2 $

Therefore, Ok.

Connections:

Using 20mm dia., 4.6 grade bolt.

Strength of bolt in single shear = $ \frac{A_nbFvb}{\sqrt{3} r_{mb}} = \frac{0.78 \times \pi /4 \times 20^2 \times 400}{\sqrt{3} \times 1.25} = 45.26 kN $

Strength of bolt in bearing = $ \frac{2.5KbdtFv}{r_{mb}} = \frac{2.5 \times 1 \times 20 \times 6 \times 400}{1.25} = 96 kN $

Therefore, strength of bolt = 45.26 kN

Number of bolts required = $ \frac{ longitudinal \,\, shear}{strength \,\, of \,\, bolt} = \frac{63.157}{45.26} = 1.39 \approx 5 $

Increase the number of bolt to account for bending moment

Therefore, pitch = $ \frac{1}{4}(310- 2 \times 35) = 60 mm $ (For bolt in one row)

Force on each bolt due to longitudinal shear = $ \frac{63.157}{5} = 12.63 kN $

Force due to moment = $ \frac{M_r}{\sum r^2} = \frac{9 \times 10^6 \times 120}{2 \times 120^2 + 2 \times 60^2} = 30 kN $

Resultant force = $ \sqrt{12.63^2 + 30^2} \lt 45.26 kN $

Therefore, safe.

No. of battens plate on one face = $ \frac{length \,\, of \,\, column}{spacing \,\, of \,\, battens} + 1 = \frac{5200}{900} + 1 = 6.77 \approx 7 $

Revised centre to centre distance of battens = $ \frac{5200}{7.1} = 866.67mm $