Week 5 Challenge
- Design of plate girder.
- Introduction - A plate girder (PG) is an I beam built up from plates using welding or bolting. They are deep flexure members used to carry loads that rolled beam cannot carry economically. Plate girders are more flexible and economical.
- Question no 1.
- Ans -
- Aim - Shear resistance of web problem, to find out Plate girder (PG) web thickness for given load without stiffener.
- Introduction - Based on load acting on the beam plate size and thickness are engineered and section is built.
- Deeper web will be weaker in shear and may cause web buckling. Thus enough thickness of web must be provided or thinner web with vertical or horizontal stiffners should be provided.The flanges resist all bending and the web resists all shear. These beams are divided into two types, shear resistant and partial tension field beams. The webs of shear resistant beams resist the shear load without buckling, and the webs of partial tension field beams buckle at less than the maximum beam load.
- Given -
- Plate girder (PG) of size 1200 x 300 x 25 x 10.
- Figure attached below.
- Check if the web is adequate to carry a factored shear force = 1000 kN.
- Yield strength of plate fy = 250 N/mm^2.
- Depth of section = D = 1200 mm.
- Width of flange plate = b = 300 mm.
- Thickness of flange plate = tf = 25 mm.
- Thickness of web = tw = 10 mm.
- stiffners are not provided use d/tw ≤ 200ɛ.
- Depth of web = d = 1200 - 50 = 1150 mm.
- d/tw = 1150/10 = 115 mm.
- ɛ = √250/fy.
- ɛ = √250/250.
- ɛ = 1.0.
- d/tw ≤ 200ɛ.
- 115 ≤ 200.
- Hence ok.
- Web connected to flange along both longitudinal edges.
- Check if the web is adequate to carry a factored shear force = 1000 kN.
- Vu = 1000 KN.
- V = 1000/1.5 = 666.67 KN.
- Where 1.5 is partial safety factor.
- τv = V/d x tw.
- τv = 666.67 x 10^3/1150 x 10.
- τv = 57.97 N/mm^2.
- Permissible τv = 0.4 x fy = 0.4 x 250 = 100 N/mm^2.
- Permissible τv = 100 N/mm^2.
- 57.97 < 100 N/mm^2.
- Hence safe.
- Web is adequate to carry factored shear force.
- If stiffeners are avoided then to find web thickness.
- d/tw ≤ 200ɛ.
- 1150/tw = 200 x 1.
- tw = 5.75 = 6.0 mm.
- To avoide stiffeners we get web thickness as 6.0 mm.
- Question no 2.
- Ans -
- Aim - To find specing and intermediate web stiffener required.
- Introduction - Stiffeners are secondary plates or sections which are attached to beam webs or flanges to stiffen them against out of plane deformations. Almost all main bridge beams will have stiffeners. Deep beams sometimes also have longitudinal web stiffeners.
- Intermediate web stiffner - Intermediate vertical and horizontal stiffeners not subjected to external loads shall be connected to web by rivets or welds so as to withstand a shearing force between each component of stiffener.
- Given -
- Plate girder (PG) of size 1500 x 500 x 30 x 12.
- Figure attached below.
- Check if the intermediate web stiffener is required to carry a factored shear force = 2000 kN.
- Yield strength of plate fy = 250 N/mm^2.
- Depth of section = D = 1500 mm.
- Width of flange plate = b = 500 mm.
- Thickness of flange plate = tf = 30 mm.
- Thickness of web = tw = 12 mm.
- Depth of web = d = 1500 - 60 = 1440 mm.
- d/tw = 1440/12 = 120 mm.
- ɛ = √250/fy.
- ɛ = √250/250.
- ɛ = 1.0.
- d/tw ≤ 200ɛ.
- 115 ≤ 200.
- Hence ok.
- Web connected to flange along both longitudinal edges.
- Check if the web is adequate to carry a factored shear force = 2000 kN.
- Vu = 2000 KN.
- V = 2000/1.5 = 1333.33 KN.
- Where 1.5 is partial safety factor.
- τv = V/d x tw.
- τv = 1333.33 x 10^3/1440 x 12.
- τv = 77.16 N/mm^2.
- Permissible τv = 0.4 x fy = 0.4 x 250 = 100 N/mm^2.
- Permissible τv = 100 N/mm^2.
- 77.16 < 100 N/mm^2.
- Hence safe.
- Web is adequate to carry factored shear force.
- Ratio of d/tw = 120 mm > 85.
- Vertical stiffeners are provided.
- Specing of stiffener at 1440/5 = 288 mm.
- Specing of intermediate web stiffener = 288 mm.
- Question no 3.
- Ans -
- Aim - To find web thickness required to avoid the stiffeners.
- Introduction - Stiffeners are secondary plates or sections which are attached to beam webs or flanges to stiffen them against out of plane deformations. Almost all main bridge beams will have stiffeners. Deep beams sometimes also have longitudinal web stiffeners.
- Intermediate web stiffner - Intermediate vertical and horizontal stiffeners not subjected to external loads shall be connected to web by rivets or welds so as to withstand a shearing force between each component of stiffener.
- Bearing stiffener - Provided to prevent crushing of web due to concentrated loads. They are placed in pairs.
- Load carrying stiffener - Provided to prevent local buckling of web due to concentrated load.
- Diagonal stiffener - Local strengthning of web which is under combined shear and bending.
- Torsional stiffener - Provided at support to restrain girders against torsional effects.
- Given -
- Plate girder (PG) of size 1500 x 500 x 30 x 12.
- Figure attached below.
- Check if the intermediate web stiffener is required to carry a factored shear force = 2000 kN.
- Yield strength of plate fy = 250 N/mm^2.
- Depth of section = D = 1500 mm.
- Width of flange plate = b = 500 mm.
- Thickness of flange plate = tf = 30 mm.
- Thickness of web = tw = 12 mm.
- Depth of web = d = 1500 - 60 = 1440 mm.
- d/tw = 1440/12 = 120 mm.
- ɛ = √250/fy.
- ɛ = √250/250.
- ɛ = 1.0.
- d/tw ≤ 200ɛ.
- 115 ≤ 200.
- Hence ok.
- Web connected to flange along both longitudinal edges.
- Check if the web is adequate to carry a factored shear force = 2000 kN.
- Vu = 2000 KN.
- V = 2000/1.5 = 1333.33 KN.
- Where 1.5 is partial safety factor.
- τv = V/d x tw.
- τv = 1333.33 x 10^3/1440 x 12.
- τv = 77.16 N/mm^2.
- Permissible τv = 0.4 x fy = 0.4 x 250 = 100 N/mm^2.
- Permissible τv = 100 N/mm^2.
- 77.16 < 100 N/mm^2.
- Hence safe.
- Web is adequate to carry factored shear force.
- If stiffeners are avoided then to find web thickness.
- d/tw ≤ 200ɛ.
- 1440/tw = 200 x 1.
- tw = 7.2 mm.
- To avoide stiffeners we get web thickness as 7.2 mm.
- Question no 4.
- Ans -
- Aim - To find the minimum depth & web thickness requirement of a beam, calculate flange depth & thickness.
- Introduction - Cross section of girder must be praportioned such that the following functional are met.
- Adequate Zp to carry bending moment.
- Adequate Iz to improve vertical stiffness to satisfy vertical deflection.
- Low value of Lb/r1 to improve lateral stiffness to prevent lateral torsional buckling.
- Adequate web area to carry shear force d/tw and c/d ratio to improve post buckling strenght of web.
- Given -
- Span L = 15 m.
- UDL = 100 kN/m.
- Figure attached below.
- Design of web -
- It is assumed that the girder is laterally supported.
- Let us assume fy = 250 N/mm^2.
- σbc = 0.66fy = 0.66 x 250 = 165 N/mm^2.
- Let us assume thickness of web plate = 10 mm.
- The economical depth of girder is given by
- d = 1.1 √M/σbc x tw.
- M = wL^2/8.
- M = 100 x 15^2/8.
- M = 2813 KNm. = 2813 x 10^6 Nmm.
- d = 1.1 √2813 x 10^6/165 x 10.
- d = 1436 mm.
- Atual depth selected should be slightly lower.
- Hence Keep d = 1400 mm.
- Depth of section d = 1400 mm.
- Let us check for shear to adequate carry the moment.
- V = wL/2 = 100 x 15/2.
- V = 750 KN.
- τv = V/d x tw.
- τv = 750 x 10^3/1400 x 10.
- τv = 53.57 N/mm^2.
- Permissible τv = 0.4 x fy = 0.4 x 250 = 100 N/mm^2.
- Permissible τv = 100 N/mm^2.
- 53.57 < 100 N/mm^2.
- Hence safe.
- Web is adequate to carry moment of the section.
- Section is plastic.
- Width of flange plate as 0.3 times depth of section.
- Width of flange = bf = 0.3 x 1400 = 420 mm.
- Flange width to thickness ratio limited to 24.
- bf/tf = 24.
- 420/tf = 24.
- tf = 17.5 mm = 18 mm.
- Minimum depth of beam = d = 1400 mm.
- Web thickness = tw = 10 mm.
- Width of flange = bf = 420 mm.
- Thickness of flange = tf = 18 mm.
- Size of plate girder (PG) = 1400 x 420 x 18 x 10.
- Question no. 5
- Ans -
- Aim - Design of load carrying stiffener that is design a itermediate stiffener.
- Introduction - Stiffeners are secondary plates or sections which are attached to beam webs or flanges to stiffen them against out of plane deformations. Almost all main bridge beams will have stiffeners. Deep beams sometimes also have longitudinal web stiffeners.
- Intermediate stiffner - Intermediate vertical and horizontal stiffeners not subjected to external loads shall be connected to web by rivets or welds so as to withstand a shearing force between each component of stiffener.
- These are provided to avoid diagonal buckling of web. The vertical stiffeners also help in maintaining the 90 degree angle between the flanges and web.
- Given -
- Beam ISMB 500.
- Load = 1000 KN.
- From Steel table given that -
- Cross section area = A = 11074 mm^2.
- Weight = w = 86.9 Kg/m.
- Depth of section = D = 500.
- Width of flange = bf = 180 mm.
- Thickness of flange = tf = 17.2 mm.
- thickness of web = tw = 10.2 mm.
- Clear depth of web = d1 = 500 - 2 x 17.2 = 465.6 mm.
- Twice clear distance from nuetral axis to toe of compreesion face flange.
- d2 = 2 x (232.8) = 465.6 mm.
- d1 = d2 = 465.6 mm.
- Ratio d1/tw = 465.6/10.2 = 45.64 = d2/tw.
- Since d1/tw > 85 vertical stiffeners are reuired.
- Since d2/tw > 250 horizontal stiffeners are required.
- From above we are not to provide stiffener.
- So that Beam ISMB 500 are used as to avoid stiffener.
- If d/tw ≤ 67ε. Then section is classified as beam.
- Web connected to flange along one longitudinal edge only (Near and far side).