Week 5 Challenge
1) AIM : To check whether the web of given plate girder is adequate or not for a given factored shear force.
INTRODUCTION :
Yield strength of plate = 250 Mpa.
Dimensions of plate girder
Overall Depth D = 1200mm.
Width of flange b = 300 mm.
Thickness of flange tf = 25 mm.
Thickness of web tw = 10 mm.
Factored shear force V = 1000 KN.
PROCEDURE :
Depth of web d = D - 2tf = 1200 - 2*25 = 1150 mm.
Shear area Av = dw*tw = 1150*10 = 11500mm2.
For the plate girder without stiffners, the resistance to shear buckling is checked when d/tw > 67 (From clause 8.4.2.1 - IS 800 : 2007)
d/tw = 1150/10 = 115 > 67ε
Shear resistance of the plate girder can be obtained by Simple post-critical method (From 8.4.2.2(a))
Kv = 5.35 (Since intermediate stiffners are not provided).
Assume poisson ratio of web u = 0.3
Elastic modulus = 2 * 10^5 N/mm2.
Elastic critical shear stress of the web Tcr,c = (Kv*pi^2*E)/(12*(1-u^2)*(d/tw)^2 = 73.12 N/mm2.
Non-dimensional web slenderness ratio for shear buckling stress l = sqrt((fyw/(sqrt(3)*Tcr,c)) = 1.4
Since lamda l = 1.4 > 1.2
Shear stress corresponding to web buckling Tb = (fyw)/((sqrt(3)*l^2) = 73.64 N/mm2.
Shear resistance Vcr = Av*Tb = 11500*73.64 = 846.86 KN < V = 1000KN.
Hence the web is not safe.
Result : The web of the given plate girder is not safe for the factored shear force of 1000KN.
2) AIM : To check the requirement of intermediate stiffners for the given plate girder to carry for a factored shear force of 2000KN.
INTRODUCTION :
Yield strength of plate = 250 Mpa.
Dimensions of plate girder
Overall Depth D = 1500mm.
Width of flange b = 500 mm.
Thickness of flange tf = 30 mm.
Thickness of web tw = 12 mm.
Factored shear force V = 2000 KN.
PROCEDURE :
Depth of web d = D - 2tf = 1500 - 2*30 = 1460 mm.
Shear area Av = dw*tw = 1460*12 = 17520mm2.
For the plate girder without stiffners, the resistance to shear buckling is checked when d/tw > 67 (From clause 8.4.2.1 - IS 800 : 2007)
d/tw = 1460/12 = 121.66 > 67ε
Shear resistance of the plate girder can be obtained by Simple post-critical method (From 8.4.2.2(a))
Kv = 5.35 (Since intermediate stiffners are not provided).
Assume poisson ratio of web u = 0.3
Elastic modulus = 2 * 10^5 N/mm2.
Elastic critical shear stress of the web Tcr,c = (Kv*pi^2*E)/(12*(1-u^2)*(d/tw)^2 = 65.33 N/mm2.
Non-dimensional web slenderness ratio for shear buckling stress l = sqrt((fyw/(sqrt(3)*Tcr,c)) = 1.49
Since lamda l = 1.49 > 1.2
Shear stress corresponding to web buckling Tb = (fyw)/((sqrt(3)*l^2) = 65.01 N/mm2.
Shear resistance Vcr = Av*Tb = 17520*65.01 = 1138.97KN < V = 2000KN.
Hence the web is not safe. So intermediate stiffners are required
For the plate girder with stiffners, the shear resistance is checked when d/tw > 67e*sqrt(Kv/5.35)
Assume c/d = 1. So Kv = 5.35 + 4.0/(c/d)^2 = 9.35, where 'c' is the spacing of stiffners
d/tw = 121.66
67e*sqrt(Kv/5.35) = 67*1*sqrt(9.35/5.35) = 88.57 < d/tw = 121.66. Hence resistance to shear buckling is verified.
Elastic critical shear stress of web Tcr,c = (Kv*pi^2*E)/(12*(1-u^2)*(d/tw)^2 = 114.17 N/mm2
Non dimensional web slenderness ratio l = sqrt(fyw/(sqrt(3)*Tcr,c) = 1.12 . So 0.8 < l < 1.2
So, the shear stress Tb = (1-0.8(l-0.8))(fyw/sqrt(3)) = 107.38 N/mm2
Shear resistance Vcr = Av*Tb = 17520*107.38 = 1881.29 KN < 2000KN. Hence not safe. So decrease spacing.
Assume c = 1000mm
c/d = 1000/1460 = 0.68 < 1.
Kv = 4 + 5.35/(c/d)^2 = 15.40
67e*sqrt(Kv/5.35) = 113.67 < d/tw = 121.66. Hence resistance to shear buckling check is required.
Elastic critical shear stress Tcr,c = 188.044 N/mm2
Non dimensional web slenderness ratio l = 0.88 . So 0.8 < l < 1.2
Shear stress Tb = 135.09 N/mm2
Shear resistance Vcr = Av*Tb = 17520*135.09 = 2366.77 KN > V = 2000KN. Hence safe.
Result :
1) Intermediate stiffners are required.
2) Spacing of Intermediate stiffners = 1000mm.
3) The spacing of stiffners may vary in economic point of view.
3) AIM : To check whether the web of given plate girder is adequate or not for a given factored shear force.
INTRODUCTION :
Yield strength of plate = 250 Mpa.
Dimensions of plate girder
Overall Depth D = 1500mm.
Width of flange b = 500 mm.
Thickness of flange tf = 30 mm.
Factored shear force V = 2000 KN.
PROCEDURE :
Depth of web d = D - 2tf = 1500 - 2*30 = 1460 mm.
Assume thickness of web tw = 16mm
d/tw = 91.25 > 67. Hence shear bucling check is required
Kv = 5.35 (Since intermediate stiffners are not provided).
Assume poisson ratio of web u = 0.3
Elastic modulus = 2 * 10^5 N/mm2.
Elastic critical shear stress of the web Tcr,c = (Kv*pi^2*E)/(12*(1-u^2)*(d/tw)^2 = 116.14 N/mm2.
Non-dimensional web slenderness ratio for shear buckling stress l = sqrt((fyw/(sqrt(3)*Tcr,c)) = 1.11. So 0.8 < l < 1.2
So, the shear stress Tb = (1-0.8(l-0.8))(fyw/sqrt(3)) = 107.38 N/mm2
Shear resistance Vcr = Av*Tb = 1460*16*107.38 = 2508.39 KN > 2000KN. Hence safe.
Result :
Hence thickness of web required is 16mm to avoid stiffners for the given web plate.
4) AIM : To find out the dimensions of the web and flange for the plate girder for a given load.
INTRODUCTION :
Span of beam l= 15m
UDL on beam = 100KN/m
PROCEDURE :
Step 1 : Loading
Assume self weight of plate girder = W/200 = 100*15/200 = 7.5 KN/m.
Total factored load w = 1.5(100+7.5) = 161.25 KN/m.
Step 2 : S.F & B.M
Maximum shear force V = wl/2 = 1209.375
Maximum bending moment M = wl^2/8 = 4535 KNm
Step 3 : Proportioning of web
Assume d = l/15 = 15/15 = 1m = 1000mm
Consider 'tw' is the thickness of web
k = d/tw = 1/tw
Thickness of web tw = (M/(fy*k^2))^0.33 `⇒`tw = 18 mm (approximately)
`Area of web Av = 1000*18 = 18000mm2`
Step 4 : Proportioning of flange
For optimum area of flange,
Af = (M/(fy*d))-Av/6 = 15140mm2.
Width of flange bf = 0.3*d = 300mm
thickness of flange = Af/b = 15140/300 = 50mm
bf/tf = 300/50 = 6 mm < 8.4. Hence plastic.
Step 5 : Web thickness requirement
d/tw = 1000/18= 55.55 < 67. So decrease thickness of web
Consider tw = 14mm
Hence d/tw = 1000/14 = 71.42 > 67.
Consider plate girder with transverse stiffners
Assume c/d = 1.4 ⇒ c = 1400mm, where 'c' is the spacing of stiffners.
Depth of web d = 1000 mm
3*d = 3000mm.
Hence 3d > c > d
d/tw = 71.42 < 200.
Hence web thickness is okay.
Step 6 : Shear resistance
Shear area Av = 18000mm2
d/tw = 71.42, c/d = 1.4 ⇒ Shear stress corresponding to web buckling Tb = 144.3 N/mm2. (From Table 7.1 of N.Subramanium text book)
Shear strength Vd = Av*Tb = 18000*144.3 = 2597 KN > V = 1209.375. Hence we can reduce thickness of web.
Step 7 : Bending strength
0.6*Vd = 0.6*2597 = 1558 > V = 1209.375 ( Hence low shear).
So bending strength Md = βb*Zp*fy/γm0
βb = 1.0 Since section is plastic.
γm0 = 1.1
Width of flange bf = 300mm.
Thickness of flange tf = 50mm.
Depth of web dw = 1000mm.
Thickness of web tw = 14mm.
Zp = bf*tf(D-tf) + tw(D-2tf)^2/4 = 19250000 mm3
⇒ Md = 1*19250000*250/10^6 = 4812.5 KNm > M = 4535 KNm.
Hence flange is okay.
Result :
Width of flange bf = 300mm.
Thickness of flange tf = 50mm.
Depth of web dw = 1000mm.
Thickness of web tw = 14mm.
Spacing of stiffners c = 1400 mm.
The thickness of web may be reduced as the shear strength of the assumed web is very high.
Flange curtailment may be required as the moment is not same throughout the section.