Subject: Design and Drawing of Steel Structure

Topic: Plate Girder

Difficulty: Medium

Design the central section of 25 m long plate girder subjected to a factored load of 60 KN/m inclusive of self-weight. Provide suitable curtailment of flange plate. Also design suitable welded connection between web and flange plates

Given,

L = 25m, w = 60 kN/m including self-weight

To know, Design of welded plate girder.

1) Load calculation

Total factored load on girder = 60 kN/m

Maximum bending moment = $\frac{60 \times 25^2}{8}$ = 4687.5 kN.m

Maximum shear force = $ \frac{60 \times 25}{2} $ = 750 kN

2) Design of web

Section classification

Serviceability requirements

$ \frac{d}{t \omega} \leq 200 \varepsilon $

flange buckling requirements, $ \frac{d}{t \omega} \leq 345 \varepsilon f^2 $

Assume $ K = \frac{d}{t \omega} $ = web slenderness ratio = 180

Depth of plate girder = d = $ (\frac{M_zK}{Fy})^{0.33} = (\frac{4687.5\times 10^6 \times 180}{250})^{0.33} = 1394.2 \approx 1400 mm $

Therefore, thickness of web = tw = $ (\frac{M_z}{FyK^2})^{0.33} = (\frac{4687.5 \times 10^6}{250 \times 180^2})^{0.33} = 8.15 \approx 10mm $

As intermediate transverse stiffeners are not provided therefore increase the thickness of web.

Therefore, let us try web plate 1400x10 mm in size.

3) Design of flanges:

Assume that flanges carry the bending moment and shear by web

Area of flange AP = $ \frac{M_z rm_0}{Fyd} = \frac{4687.5 \times 10^6 \times 1.1}{250 \times 1400} = 14732.14 mm^2 $

Assume width of flange plate = 0.3d = 0.3x1400 = 420mm

Thickness of flange = $ \frac{14732.14}{420} $ = 35.097 $\approx$ 40mm

Let us try flange plate 420x40mm in size.

4) Classification of flanges

Out-stand of flange, b = $ \frac{bf - tw}{2} = \frac{400-10}{2} = 195mm $

$ \frac{b}{tf} = \frac{195}{40} = 4.875 \lt 8.4 $

Therefore, flanges are plastic.

5) Shear capacity: v \leq vd

$ vd = \frac{v_n}{rm_0} $

Nominal plastic shear resistance, $ v_n = v_p = \frac{A_vfy \omega}{\sqrt{3}} $

For welded section, $A_v = dt \omega$

$ v_n = \frac{dt \omega Fy\omega}{\sqrt{3} \times 1.1} = \frac{1400 \times 10 \times 250}{\sqrt{3} \times 1.1} = 1837.02 kN \gt 750 kN $

Therefore, safe in shear

6) Moment capacity of flanges:

$ v = 750 kN \lt 0.6 vd = 0.6 \times 1837.02 = 1102.2 kN \\ $

Therefore, Case is low shear

$ M_d = \frac{BbZ_{pz} Fy}{rm_0} $

Bb = 1 for plastic section.

$ Z_{pz} = \frac{A}{2} (\bar{y_1} + \bar{y_2}) \\ = \frac{2bFtF}{2} (D - \lambda F) \\ = \frac{2 \times 420 \times 40}{2}(1480 - 40) \hspace{0.5cm} [D = 1400 + 2 \times 40 = 1480mm] \\ = 24.19 \times 10^6 mm^3 $

$ Md = \frac{1 \times 24.19 \times 10^6 \times 250}{1.1} = 5498.18 kN.m \gt 4687.5 kN.m $

Therefore, safe.

7) Shear buckling design by simple post-critical method

$\tau$ = Elastic critical shear stress of the web

$ \tau = \frac{K_v \pi^2 E}{12(1-\mu^2)}(\frac{t \omega}{d})^2 $

$K_v$ = 5.35, when stiffeners (transverse) are provided only at supports.

$\mu$ = 0.3

$ \tau = \frac{5.35 \times \pi^2 \times 2 \times 10^5}{12(1-0.3^2)} (\frac{10}{1400})^2 = 49.34 N/mm^2 $

Non-dimensional web slenderness ratio for shear buckling stress.

$ \lambda_\omega = \sqrt{\frac{Fy \omega}{\sqrt{3} \tau}} \\ = \sqrt{\frac{250}{\sqrt{3} \times 49.34}} = 1.73 \gt 1.2 $

Therefore shear stress corresponding to buckling (when $ \lambda_\omega \gt 1.2 $)

$ \tau b = \frac{Fy \omega}{\sqrt{3} \lambda_\omega^2} \\ = \frac{250}{\sqrt{3} \times 1.71^2} = 19.36 N/mm^2 $

Shear force corresponding to web buckling

$ v_{cr} = A_v \tau b \\ = d \times t\omega \times \tau b \\ = 1400 \times 10 \times 49.36 \\ = 691 kN \lt 750 kN $

This is unsafe.

Therefore, revised the web thickness

Let us try 12mm thickness for web.

$ \tau_{cr} = \frac{5.35 \times \pi^2 \times 2 \times 10^5}{12(1-0.3^2)}(\frac{12}{1400})^2 = 71.05 N/mm^2 $

$ \lambda_\omega = \sqrt{\frac{Fy \omega}{\sqrt{3} \tau}} \\ = \sqrt{\frac{250}{\sqrt{3} \times 73.05}} = 1.42 \gt 1.2 $

$ \tau b = \frac{Fy \omega}{\sqrt{3} \lambda_\omega^2} \\ = \frac{250}{\sqrt{3} \times 1.42^2} = 71.58 N/mm^2 $

$ v_{cr} = A_v \tau b \\ = d \times t\omega \times \tau b \\ = 1400 \times 12 \times 71.58 \\ = 1202.54 kN \gt 750 kN $

This is safe.

8) Flange to web connection

There will be two weld lengths along the span for each flange to web connection

$\omega$ = Horizontal shear between flange to web

= $\frac{VAFbar{y}}{2 I_z}$

$ I_z = \frac{bf D^3}{12} - \frac{(bF - t\omega)d^3}{12} \\ = \frac{420 \times 1480^3}{12} - \frac{(420-12)1400^3}{12} \\ = 20166.72 \times 10^{10} mm^4 $

$ \bar{y} = \frac{D}{2} = \frac{1480}{2} = 740mm $

AF = bFxtF = 420x40 = 16800 mm$^2$

$ \omega = \frac{750 \times 16800 \times 740}{2 \times 20166.72 \times 10^6} = 0.231 kN/mm $

Provide size of weld = 7mm

$K_s$ = 0.7x7 = 4.9mm

Strength of weld per unit length

$ F_{wd} = \frac{4.9 \times 250}{\sqrt{3} \times 1.50} = 0.471 kN/mm \gt 0.231 kN/mm $

Therefore Safe.