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Also design curtailment of flange plates, end bearing stiffener and connection between flange and web using fillet welds.

Subject: Design and Drawing of Steel Structure

Topic: Plate Girder

Difficulty: Medium

A welded plate girder 20 m in span and laterally supported throughout is simply supported at the ends. It has to support a udl of 100 KN/m (factored) including self-weight. Design the central section of the plate girder without intermediate vertical stiffeners. Also design curtailment of flange plates, end bearing stiffener and connection between flange and web using fillet welds.

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Given,

L = 20m, wd = 100 kN/m (factored including self weight)

To know: Design of welded plate girder

1) Load calculation

Total factored load on girder = 100 kN/m

Maximum bending moment = $\frac{100 \times 20^2}{8}$ = 5000 kN.m

Maximum shear force = $ \frac{100 \times 20}{2} $ = 1000 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{5000 \times 10^6 \times 180}{250})^{0.33} = 1424.22 mm \approx 1450 mm $

Therefore, thickness of web = tw = $ (\frac{M_z}{FyK^2})^{0.33} = (\frac{5000 \times 10^6}{250 \times 180^2})^{0.33} = 8.33 \approx 12mm $

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

Therefore, let us try web plate 1450x12 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{5000 \times 10^6 \times 1.1}{250 \times 1450} = 15172.41 mm^2 $

Assume width of flange plate = 0.3d = 0.3x1450 = 435mm $\approx$ 440mm

Thickness of flange = $ \frac{15172.41}{440} $ = 34.48 $\approx$ 40mm

Let us try flange plate 440x40mm in size

4) Classification of flanges

Out-stand of flange, b = $ \frac{bf - tw}{2} = \frac{440-12}{2} = 214mm $

$ \frac{b}{tf} = \frac{214}{40} = 5.35 \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{1450 \times 12 \times 250}{\sqrt{3} \times 1.1} = 2283.15 kN \gt 1000 kN $

Therefore, safe in shear

6) Moment capacity of flanges:

$ v = 1000 \lt 0.6 vd = 0.6 \times 2283.15 = 1369.89 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 440 \times 40}{2}(1530 - 40) \hspace{0.5cm} [D = 1450 + 2 \times 40 = 1530mm] \\ = 26.22 \times 10^6 mm^3 $

$ Md = \frac{1 \times 26.22 \times 10^6 \times 250}{1.1} = 5959.09 kN.m \gt 5000 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{12}{1450})^2 = 66.23 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 66.23}} = 1.47 \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 1.47^2} = 66.795 N/mm^2 $

Shear force corresponding to web buckling

$ v_{cr} = A_v \tau b \\ = d \times t\omega \times \tau b \\ = 1450 \times 12 \times 66.795 \\ = 1162.22 kN \gt 1000 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{440 \times 1530^3}{12} - \frac{(440-12)1450^3}{12} \\ = 22590.198 \times 10^6 mm^4 $

$ \bar{y} = \frac{D}{2} = \frac{1530}{2} = 765mm $

AF = bFxtF = 440x40 = 17,600 mm$^2$

$ \omega = \frac{1000 \times 17600 \times 765}{2 \times 22590.198 \times 10^6} = 0.298 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 0.298 kN/mm $

Therefore Safe.

9) End bearing stiffeners

$ F_w = (b_1 + n_2) \frac{t\omega fy \omega}{rm_0} $

Assume b$_1$ = Bearing length = 0

n$_2$ = 2 x 40 x 2.5 = 200mm

$ F_w = (0 + 200) \times \frac{12 \times 250}{1.1} = 545.45 kN \lt 1000 kN $

Therefore, stiffeners are required

Let us try two-flat sections as stiffeners one on each side of web. Maximum available width = $\frac{bf - t\omega}{2} = \frac{440-12}{2} = 214 mm $

Let us provide 12 mm thick flat section.

Maximum permissible out-stand = 20 x 12 = 240 mm

Minimum permissible out-stand = 14 x 12 = 168 mm

Let us try flat section 180 x 12 mm in size

10) Check for buckling of stiffener:

Effective area of stiffener = (2x168x12) + (2x20x12x12) = 9792mm$^2$

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Moment of Inertia of stiffener $ = 2[\frac{168^3 \times 12}{12} + 12 \times 168 \times (\frac{168}{2} + \frac{12}{2})^2] = 42.14 \times 10^6 mm^4 $

Radius of gyration, $ r = \sqrt{\frac{1}{A}} = \sqrt{\frac{42.14 \times 10^6}{9792}} = 65.60 $

Slenderness ratio = $ \lambda = \frac{0.7 \times 1450}{65.60} = 15.47 $

For $ \lambda = 15.47, \,\, F_{cd} = 227 - \frac{227-224}{20-10}(15.47-10) = 225.36 N/mm^2 $

Buckling resistance pd = A x F$_{cd}$ = 9792 x 225.36 = 2206.71 kN $\gt$ 1000 kN

Therefore, Safe

11) Bearing capacity of stiffener

$ F_{psd} = \frac{Aa fya}{0.8 rm_0} \\ Aa = 2 \times 168 \times 12 = 4032 mm^2 $

Area of stiffener in contact with flange $ \frac{4032 \times 250}{0.8 \times 1.1} = 1145.45 kN \gt 1000 kN $

Therefore, safe

12) Torsional resistance provided by end bearing stiffener

$ I_s \geq 0.34 \alpha_s D^s \tau_c f \\ I_s = \frac{bd^3}{12} \times 2 = \frac{12 \times (168 \times 2)^3}{12} = 37.93 \times 10^6 mm^4 \\ I_y = \frac{2tfbf^3}{12} + \frac{df\omega^3}{12} \\ = \frac{2 \times 40 \times 440^3}{12} + \frac{1450 \times 12^3}{12} \\ = 567.89 \times 10^6 + 0.208 \times 10^6 \\ = 568.09 \times 10^6 mm^4 $

$ A = 2 \times 440 \times 40 + 1450 \times 12 = 52.6 \times 10^3 mm^2 $

$ r_y = \sqrt{\frac{I_y}{A}} = \sqrt{\frac{568.09 \times 10^6}{52.6 \times 10^3}} = 103.92 mm $

$ \lambda = \frac{LLT}{r_y} = \frac{20 \times 10^3}{103.92} = 192.45 \gt 100 $

For $ LLT \gt I_\omega \alpha_s = \frac{30}{\lambda^2} = \frac{30}{192.45} = 8.10 \times 10^{-30} \\ 0.34 \alpha_s D^3 \tau_c f = 0.34 \times 8.10 \times 10^{-4} \times (1450 + 2 \times 40)^3 \times 40 = 39.45 \times 10^6 mm^4 \\ \therefore I_s \ngtr 0.34 \alpha_s D^3 \tau_c f $

Therefore, Not safe

Therefore increase the size of stiffener

Therefore provide 180x12 mm in size

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