Subject: Design and Drawing of Steel Structure

Topic: Plate Girder

Difficulty: Medium

A welded plate girder of span 25 m is laterally supported throughout and required to carry superimposed udl 100 KN/m over the entire span. The girder will not have intermediate transverse stiffeners. Design the most economical section. Provide suitable curtailment of flange plates if necessary. Also design the welded connection between web &flange plate.

Given,

L = 25m

w = 100 kN/m (Superimposed)

1) Load calculation

Total factored load on girder = 1.5x100 = 150 kN/m

Maximum bending moment = $\frac{150 \times 25^2}{8}$ = 11,718.75 kN.m

Maximum shear force = $ \frac{150 \times 25}{2} $ = 1875 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{1171875 \times 10^6 \times 180}{250})^{0.33} = 1886.47 \approx 1900 mm $

Therefore, thickness of web = tw = $ (\frac{M_z}{FyK^2})^{0.33} = (\frac{1171875 \times 10^6}{250 \times 180^2})^{0.33} = 11.039 \approx 16mm $

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

Therefore, let us try web plate 1900x16 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{11718.75 \times 10^6 \times 1.1}{250 \times 1900} = 27,138.15 mm^2 $

Assume width of flange plate = 0.3d = 0.3x1900 = 570mm $\approx$ 600mm

Thickness o flange = $ \frac{27138.15}{600} $ = 45.23 $\approx$ 50mm

Let us try flange plate 600x50mm in size

4) Classification of flanges

Out-stand of flange, b = $ \frac{bf - tw}{2} = \frac{600-16}{2} = 292mm $

$ \frac{b}{tf} = \frac{292}{50} = 5.84 \lt 8.4 \varepsilon $

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{1900 \times 16 \times 250}{\sqrt{3} \times 1.1} = 3988.96 kN \gt 1875 kN $

Therefore, safe in shear

6) Moment capacity of flanges:

$ v = 1875 \lt 0.6 vd = 0.6 \times 3988.96 = 2393.379 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 600 \times 50}{2}(2000 - 50) \hspace{0.5cm} [D = 1900 + 2 \times 50 = 2000mm] \\ = 58.5 \times 10^6 mm^3 $

$ Md = \frac{1 \times 58.5 \times 10^6 \times 250}{1.1} = 13,295.45 kN.m \gt 11718.75 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-\mu^2)} (\frac{16}{1900})^2 =68.579 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 68.579}} = 1.45 \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 1.45^2} = 68.65 N/mm^2 $

Shear force corresponding to web buckling

$ v_{cr} = A_v \tau b \\ = d \times t\omega \times \tau b \\ = 1900 \times 16 \times 68.65 \\ = 2086.97 kN \gt 1875 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{600 \times 2000^3}{12} - \frac{(600-16)1900^3}{12} \\ = 4 \times 10^{11} - 3.33 \times 10^{11} = 6.619 \times 10^{10} mm^4 $

$ \bar{y} = \frac{D}{2} = \frac{2000}{2} = 1000mm $

AP = bFxtF = 600x50 = 30,000 mm$^2$

$ \omega = \frac{1875 \times 30,000 \times 1000}{2 \times 6.619 \times 10^{10}} = 0.42 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 0.42kN/mm $

Therefore Safe.