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.

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$^2$

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