Week 5 Challenge
Q.1 AIM AND INTODUCTION : DESIGN OF PLATE GIRDER TO DETERMINE PG WEB THICKNESS FOR GIVEN LOAD WITHOUT STIFFNERS
PROCEDURE: DATA GIVEN 1) plate girder size 1200x300x25x10
2) Factored shear force 1000 KN
3)YIELD STRENGTH fy 250 N/mm2
Shear resistance of web Vd= Vn/`gammamo` = AV Fyw/`gammamo`x `sqrt3`= 1200 x 10 x250/1.1x sqrt3
=1574.6 KN > 1000 KN
RESULT: Hence section is adequate .NO stiffner required.
Q.2 AIM AND INTODUCTION : DESIGN OF PLATE GIRDER TO DETERMINE IF INTERMEDIATE STIFFNERS ARE REQUIRED FOR GIVEN FACTORED LOAD IF REQUIRED SPACING OF STIFFNER
PROCEDURE: DATA GIVEN 1) plate girder size 1500 x 500 x 30 x12
2) Factored shear force 2000 KN
3)YIELD STRENGTH fy 250 N/mm2
d/tw = 1500/12 = 125
Hence resistance to shear buckling should be verified.Considering simple post critical method.
`taucr = Kv x pi2 x E / 12( 1 - mu2) (d/tw)2`= 5.35 x pi2 x 2x 10^5
12(1-0.3^2)(d/tw)2
d/tw > 67 , Kv = 5.35
= 5.35 x pi2 x 2x10^5/12(1- 0.3^2) x( 1500/12)^2
= 61.89N/mm2
`gammaw= sqrtfyw/sqrt3 x 61.9`= 144.34/61.9 = 1.527
since it is more than 1.2
`taub = fyw/sqrt3xlamda w2` = 250/sqrt3 x(1.527)2=144.33/(1.527)2=61.89
Vcr = dxtwxtaub=(1500x12)x61.89
=1114KN
Vd= Vcr/gammamo=1114/1.1=1012.7KN<2000KN
Hence intermediate stiffners are to be provided to improve the buckling strength of the slender web.
Q3. to determine the web thickness reqd to avoid the stiffner`
Two design requirements for minimum web thickness with no stiffners
`nbsp; d/tw < 200epsilo` (for serviceability)
`nbsp; d/tw < 345 epsilon` ( to avoid flange buckling`
if the web is made thick .
d/tw < 67epsilon (these requirements are met)
Thus minimum web thickness should be tw > 1500 = 22.3 mm = 23 mm
67x1
RESULT: hence web thickness of 23 mm to provided without stiffners.
Q.4 AIM AND INTODUCTION : DESIGN OF PLATE GIRDER TO DETERMINE THE MINI DETPTH & WEB THK OF BEAM SPAN 15M AND UDL 100KN/m considered simply supported also calculate the flange depth
considering the section as plastic and check the section is adequate to carry the moment.
PROCEDURE: DATA GIVEN 1)span of the beam = 15 m
2) U.D.L = 100 KN/m
3)YIELD STRENGTH fy 250 N/mm2
`nbsp; MOMENT at mid span = 100 x (15)2/8= 2812.5 KN-m
let us assume web plate 1200 x 12 mm size.Assuming b.m is carried by flanges.
`nbsp; Af = Mz x gammamo `/ (fy x D ) = 2813 x10^6 x1.1 = 10314.3 mm2
250 x 1200
`nbsp; limiting 8.4epsio` for the plastic section the approx flange thk
flange thk tf = sqrt 10314/2x8.4= 24.77mm= 25mm
`nbsp; breadth of flange bf = 10314/25= 412.6 mm =420mm
d/tw= 1200/12=100> 67epsilon
moment capacity check
The flanges are plastic .
`nbsp; Moment of resistance = betabxZpxfy/gammamo`
Considering that flanges only resist the bending moment Zp = 2 x 420 x25(1250-25)/2=12.86 x 10^6mm3
design moment capacity Md = 1 x 12.86 x 10^6 x250 = 2922.73 kn-m >2813 kn-m
1.1x10^6
RESULT: Hence the section is safe
Q.5 AIM AND INTODUCTION : DESIGN OF PLATE GIRDER TO DETERMINE LOAD CARRYING STIFFNER
PROCEDURE: DATA GIVEN 1) Beam size ISMB 500
2) load 1000 KN
3)YIELD STRENGTH fy 250 N/mm2
For ISMB 500 tw = 10.2 Av= 180 x 17.2
Factored Load = 1000 KN
Shear capacity of web of ISMB 500 = Av x fyw /``gammamo x sqrt3
= 180 x17.2 x 250 = 406.24 KN
1.1 x sqrt3x 1000
Load to be resisted by the stiffner =1000 - 406.24 =593.75 KN
Assume the thickness of the stiffner plate as 6mm
The maximum allowable breadth (outstand) of stiffner =20 x tqxepsilon= 20x6( fy =250mpa )
=120mm
RESULT: take 120 x 6 mm stiffner