Week 5 Challenge
1. Plate girder size : 1200 x 300 x 25 x 10 ; fy = 250Mpa ; Shear force = 1000kN
= d/tw> 67 = 1150/10 = 115>67 hence safe.
Minimum thickness requirement of web-
d/tw< 200ε
1150/10 = 115< 200ε , ok
Shear Capacity check - Vd = Av x fyw/`root`3 x `gamma`mw
= 1200 x 10 x 250/ 1.732 x 1.1
= 1509kN
=1509>1000kN, Safe inshear.
- Shear buckling strength Vn = Vcr
d/tw = 1150/10 = 115 > 67
Tcr = 5.35 x (3.14)^2 x 2 x 10^5 / [12 (1-0.3^2)(1150/10)^2] = 73.05
= 1.405 > 1.2
= 73.12 N/mm2
Vn = Vcr = Av x Tb = 1150 x 10 x 73.12 = 840.88 kN < 1000 kN, Hence, Intermediate stiffeners are required to improve buckling strength of the slender web.
2) The factored design shear force, V, in a beam due to external actions shall satisfy V< Vd,
Where, Vd = design strength = Vn / ymo
Where, ymo = partial safety factor against shear failure (cl 5.4.1) = 1.1
Vd = Av x Fyw / ymo 'Sqrt' (3) = d x tw x Fyw / ymo 'sqrt' (3)
Vd = 1460 x 12 x 250 / 1.1 sqrt(3) = 2299 kN > factored shear force 2000 kN
Hence, adequate.
- Shear buckling strength by simple post-critical method Vn = Vcr
Since, d/tw = 1460/12 = 121.6 > 67
Tcr = 5.35 x (3.14)^2 x 2 x 10^5 / [12 (1-0.3^2)(1460/12)^2] = 65.26
= 1.48 > 1.2
= 65.89 N/mm2
Vn = Vcr = Av x Tb = 1460 x 12 x 65.89 = 1154 kN < 2000 kN
Result: Hence, Intermediate stiffeners are required to improve buckling strength of the slender web.
- Spacing of Stiffener
Providing a spacing of 2 m for intermediate transverse stiffeners
c/d = 2000/1460 = 1.36 < Sqrt 2
Hence minimum Is = 1.5 d^3 tw^3/ c^2 = 1.5 x 1460^3 x 12^3/2000^2 = 20,16,664 mm4
Trying intermediate stiffener of size 120 x 20 on each side <20 tq
Is = 1/12 x 20 x (120+12+120)^3 – 1/12 x 20 x 12^3 = 266,68,800 mm4 ;Is required
Hence OK.
3) If stiffeners are to be avoided, d/tw <= 67
That is, tw >= d/67 = 1460 / 67 = 21.8
Hence web thickness required to avoid the stiffener is 22 mm.
4) a) Factored load (UDL) = 1.5 x 100 = 150 kN/m
self weight of plate girder = wl /200 = 100 * 15 *15/400 = 56.25
self weight of girder /m length = 56.25/15 =3.75kN/m
Factored self weight = 1.5 * 3.75 = 5.625kN/m
Total factored load = 150 +5.625 = 155.625kN/m
b) Max bending moment = wl * l /8 = 155.625 x 15 x 15/8 = 4376.95kNm
Max shear force Vmax = wl/2 = 155.625 * 15/2 = 1167.18 kN
c) optimum depth of web plate d = (Mk/fy)^0.33 ; k = d/tw
let us assume k = d/tw = 180
dw = ( 4376.95*10^6 x 180 /250)^0.33 = 1363.032mm
= 1500mm
Thickness of web tw = (M/k^2*fy)^0.33
= 7.97mm ~10mm
Assume web plate = 1500 x 10 mm
Design of flange Af = Mx'gamma' m0/fy*dw = 4376.95*10^6*1.1/250*1400 = 13756.12mm2
b/tf <= 8.4 ; b/2tf <= 8.4
b= 16.8tf; Af = bf x tf ; 13756.12 = 16.8 x tf x tf ; tf = 30mm
width of flange = Af/tf = 137656.12/30 =458.53mm = 500mm
Vumax < 0.6Vd
Vd = Avfyw/ ('root' 3x 'gamma' m0) = 1*1500*10*250/1.732*1.1 = 1968.29kN
0.6Vd = 1180.97 kN > Vumax= 1167.18kN
Md = {Bb Zp fy / 'gamma'm0 } <= Ze fy/ 'gamma' m0
Zp = 2 x bf x tf x (D-tf)/2 = 2 * 500 * 30 * (1500 +30)/2 = 22.95 x 10^6 mm3
Md = 22.95 x 10^6 * 250 /1.1 = 5215.9 x 10^6 Nmm
Md> Mmax