Design Example 5 Welded joint - Steel-stainless.org

Job No. Sheet 4 of 8 Rev B102, Route de LimoursF-78471 St Rémy Lès Chevreuse CedexFranceTel : +33 (0)1 30 85 25 00Fax : +33 (0)1 30 52 75 38CALCULATION SHEETJob TitleSubjectClientECSCECSC Stainless Steel Valorisation ProjectDesign Example 5 Welded jointMade by IR Date Aug 2002Checked by FH/NB Date Oct 2002Revised by MEB Date April 2006For the present case of a plane fillet weld joint with right angle (equal leg) welds this lattercheck is not critical. However it may be so for partial penetration welds in bevelled joints.Instead of having to calculate the stresses (σ ⊥ , τ ⊥ and τ ⎥⎥ ) in the weld throat the followingdesign check expression may be used for y-z plane joints with right angle (equal leg)welds:2F2w, x+ 2F+ 2Fw,yFw,z2w, y+ 2F2w, z+ F2w, y⎛ fSinθCosθ≤ ⎜au⎝ βwγ2Cos θ +M22F w, z2⎞⎟⎠2Sin θ − 2FNote : The subscripts have been shortened: F w,x for F wx,Ed etc.w,xFw,ySinθ+ 2Fw,xFw,zCosθIn the above expression the angle θ is that between the y axis and the axis of the weld asshown in the following figure.attachedelementSection 1-1F w,xsupportattachedelement11F w,zzF w,yθyattachedelementFillet weld axis2θF w,zz2F w,yyF w,xattachedelementsupportSection 2-2The force components at the critical point of the weld are determined in theAppendix to this design example.1. Design using the simplified design shear strength approachThe design shear strength for the simplified design approach is:f u530f vw,d = =≈ 245N/mm 2 EN 1993-1-8,γ 3 1.0 × 1,25×3Eq. 4.4β wM2The value of the resultant induced force per unit length in a weld throat of 1mm is :2 2 2 2 2 2wx, Ed wy, Ed wz, Ed=F w,Ed = F + F + F = 243 + 747 + 966 1245 N/mmThe required throat size is therefore:Fw,Ed1245a ≥ = ≈ 5, 0 mmf 245vw,d138