Given:
L = 12 m, DL = 20 kN/m (excluding self wt), LL = 20 kN/m, 2 point load of 600 kN at 4m from each support, D = 1500mm restricted.
To know: Design of welded plate girder with end bearing stiffeners.
1) Load calculations:
Assume self wt of girder/m = $ \frac{Total \,\, load}{300} = \frac{(20+20) \times 12 + 2 \times 600}{300} = 5.6 $ kN/m
Total UDL = 20+20+5.6 = 45.6 kN/m
Fractured load, UDL/m = 45.6 x 1.5 = 68.4 kN/m
Point load = 1.5 x 600 = 900 kN
Maximum shear force = 1310.4 kN
Maximum bending moment at centre = 1310.4 x 6 - 68.4 x $\frac{6^2}{2}$ - 900 x 2 = 4831.2 kN.m
Maximum bending moment at point load = 1310.4 x 4 - 68.4 x $ \frac{4^2}{2} $ = 4694.4 kN.m
Therefore Maximum bending moment is at centre = 4831.2 kN.m
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
Therefore, depth of plate girder = d =
$
(\frac{M_2K}{fy} )^{0.33} = (\frac{4831.2 \times 10^6 \times 180}{250})^{0.33} = 1408.179 \approx 1420mm
$
Therefore, thickness of web t $\omega$ =
$
(\frac{M_2}{fyK^2})^{0.33} = (\frac{4831.2 \times 10^6}{250 \times 180^2})^{0.33} = 8.24 \approx 12mm
$
As intermediate transverse stiffeners are not provided therefore increase the thickness of web.
Therefore, let us try web plate 1420 x 12mm in size.
3) Design of flanges:
Assume that flanges carry the bending moment and shear by web
Therefore, area of flange required Af = $ \frac{M_2 \gamma m_0}{fyd} = \frac{4831.2 \times 10^6 \times 1.1}{250 \times 1420} = 14969.91 mm^2 $
Assume width of flange plate = 0.3d = 0.3 x 1420 = 426 $\approx$ 440mm
Therefore, thickness of flange = $\frac{14969.91}{440} = 34 \approx 40mm$
Let us try flange plate 440 x 40mm in size.
4) Classification of flanges:
Out-stand of flange,b = $\frac{bf-t \omega}{2} = \frac{440-12}{2} = 214mm$
Therefore, $\frac{b}{fF} = \frac{214}{40} = 5.35 \lt 8.4 $
Therefore, flanges are plastic
5) Shear capacity $v \lt vd$
$ vd = \frac{v_n}{\gamma m_0} $
Nominal plastic shear resistance, $ v_n = v_p = \frac{A_v f y \omega}{\sqrt{3}}$
For welded section, $ A_v = d t \omega $
$ V_n = \frac{d t\omega f y \omega}{\sqrt{3} \times 1.1} = \frac{1420 \times 12 \times 250}{\sqrt{3} \times 1.1} = 2235.92 kN \gt 1310.1 kN $
Hence safe in shear.
6) Moment capacity of flanges
v = 1310.4kN $\lt$ 0.6 vd = 0.6 x 2235.92 = 1341.55 kN
Therefore, case is low shear
$ M_d = \frac{Bb Z_{pz}fy}{\gamma m_0} $
B$_b$ = 1 for plastic section
$
Z_{pz} = \frac{A}{2}(\bar{y_1} + \bar{y_2}) = \frac{2bftf(D-tf)}{2} = \frac{2 \times 440 \times 40}{2} \times (1500 - 40) = 25.696 \times 10^6 mm^3
$
Moment capacity = $\frac{1 \times 25.696 \times 10^6 \times 250}{1.10} = 5840 kNm \gt 4831.2 kNm $
7) Shear buckling design by simple post-critical method.
Ter, $\tau$ = Elastic critical shear stress of the web = $\frac{K_v \pi^2E}{12(1-\mu^2)} (\frac{t \omega}{d})^2$
$ K_v = 5.35 $ when transverse stiffeners are provided only and supports.
$ \mu = 0.3 $
$
\tau = \frac{5.35 \times \pi^2 \times 2 \times 10^5}{12(1-0.3^2)}(\frac{12}{1420})^2 = 69.06 N/mm^2
$
$ \lambda_\omega $ = Non-dimensional web slenderness ratio for shear buckling stress,
$ \lambda_\omega = \sqrt{\frac{fy \omega}{\sqrt{3} \tau}} = \sqrt{\frac{250}{\sqrt{3} \times 69.06}} = 1.44 \gt 1.2
$
$ \therefore \tau_b $ = 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 1.44^2} = 69.60 N/mm^2 $
Shear force corresponding to web buckling,
$ V_{er} = A_v \tau_b= d \times t \omega \times \tau_b = 1420 \times 12 \times 69.60 = 1186.10 kN \lt 1310.4kN$ which is safe.
Therefore revised the web thickness
Let us try 14mm thickness for web,
$
\tau_{er} = \frac{5.35 \times \pi^2 \times 2 \times 10^5}{12(1-0.3^2)}(\frac{14}{1420})^2 = 94 N/mm^2 \\
\lambda_\omega = \sqrt{\frac{fy \omega}{\sqrt{3} \times \tau}} = \sqrt{\frac{250}{\sqrt{3} \times 94}} = 1.234 \gt 1.2 \\
\tau_b = \frac{250}{\sqrt{3} \times 1.234^2} = 94.78 N/mm^2 \\
V_{er} = 1420 \times 14 \times 94.78 = 1884.36 kN \gt 1310.4kN \\
\therefore 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{V_df\bar{y}}{2I_z} $
$
I_z = \frac{bfD^3}{12} - \frac{(bf-t\omega)d^3}{12} = \frac{440 \times 1500^3}{12} - \frac{(440-14)1420^3}{12} = 22103.27 \times 10^6 mm^4
$
$
\bar{y} = \frac{D}{2} = \frac{1500}{2} = 750 mm \\
Af = bf \times tf = 440 \times 40 = 17600 mm^2
$
$
\therefore \omega = \frac{1310.4 \times 17600 \times 750}{2 \times 22103.27 \ times 10^6} = 0.39 kN/mm
$
Provide size of weld = 7mm
K$_s$ = 0.7 x 7 = 4.9 mm
Strength of weld per unit length
$ f\omega d = \frac{4.9 \times 250}{\sqrt{3} \times 1.50} = 0.471 kN/mm \gt 0.39 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 1310.4 kN
$
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 = (2x180x12) + (2x20x12x12) = 10080mm$^2$
Moment of Inertia of stiffener
$
= 2[\frac{180^3 \times 12}{12} + 12 \times 180 \times (\frac{180}{2} + \frac{12}{2})^2] = 51.47 \times 10^6 mm^4
$
Radius of gyration,
$
r = \sqrt{\frac{1}{A}} = \sqrt{\frac{51.47 \times 10^6}{10080}} = 71.46
$
Slenderness ratio = $ \lambda = \frac{0.7 \times 1420}{71.46} = 13.90 $
For $ \lambda = 13.90, \,\, F_{cd} = 227 - \frac{227-224}{20-10}(13.90-10) = 225.83 N/mm^2 $
Buckling resistance pd = A x F$_{cd}$ = 10080 x 225.83 = 2276.36 kN $\gt$ 1310.4 kN
Therefore, Safe
11) Bearing capacity of stiffener
$
F_{psd} = \frac{Aa fya}{0.8 rm_0} \\
Aa = 2 \times 180 \times 12 = 4320 mm^2
$
Area of stiffener in contact with flange $ \frac{4320 \times 250}{0.8 \times 1.1} = 1227.27 kN \ngtr 1310.4 kN $
Therefore, Not safe
Therefore, Revised stiffener plate 200 from 180mm
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 (200 \times 2)^3}{12} = 64 \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{1420 \times 12^3}{12} \\
= 567.85 \times 10^6 + 204.48 \times 10^3 \\
= 568.05 \times 10^6 mm^4
$
$
A = 2 \times 440 \times 40 + 1420 \times 12 = 52.24 \times 10^3 mm^2
$
$
r_y = \sqrt{\frac{I_y}{A}} = \sqrt{\frac{568.05 \times 10^6}{52.24 \times 10^3}} = 104.27 mm
$
$
\lambda = \frac{LLT}{r_y} = \frac{12 \times 10^3}{104.27} = 115.07 \gt 100
$
For $ LLT \gt I_\omega \alpha_s = \frac{30}{\lambda^2} = \frac{30}{115.072} = 2.26 \times 10^{-30} \\
0.34 \alpha_s D^3 \tau_c f = 0.34 \times 2.26 \times 10^{-3} \times (1420 + 2 \times 40)^3 \times 40 = 103.37 \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 210x18 mm in size
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