PROJECT STRUCTURE CODES ELEMENT BS 5950 : PART 1 :2000 DESIGNED BY Reference Crane Gantry Girder REF CHECKED BY 26/09/2017 Calculations Output Based on the analysis critical member actions are as follows ● Ultimate Moment about major axis = 165.4 kNm ● Ultimate Shear through major axis = 217.7 kN ● Ultimate axial force = 20.7 kN ● Maximum deflection at service = 6.74 mm Girder Section = 356x171x67 356x171x67 Secion properties Flange width = B= 173.2 mm Outside height = D = 363.4 mm Flange thickness = T= 15.7 mm Web thickness = t= 9.1 mm Depth between fillets = d= 311.6 mm Table 9 Use S355 Steel; py = 355 N/mm 4 Ixx = 19460 cm Zxx = 1071 cm3 Sxx = 1211 cm3 A = 85.5 cm2 r y = 3.99 cm 4 J = 55.7 cm Iyy = 1362 cm4 r = 10.2 mm 2 Section classification Table 11 = (275/355)^0.5 = 0.88 Outstand element of compression flange Rolled section = b/T = 173.2/(2*15.7) = 5.51 <9 <9 ; Plastic Compression due to bending = b/T = 5.51 < 28 28 ; ; plastic Web of an an I d/t =311.6/9.1 = 34.2 < 80 80 plastic plastic Section is Plastic Section is Plastic Shear Capacity Clau 4.2.3. d/t = 34.2 < 70 70 ; No check required for Shear buckling Pv = 0.6*py*Av But Av = t*D = 9.1*363.4 Hence, Pv = 0.6*355*9.1*363.4/1000 =704.4 kN Ult Shear action = Fv = 217.7 kN < 704.4 kN; Section is adequate wrt shear capacity Clau 4.2.3. Adequate wrt shear Vertical Moment Capacity Fv/P Fv/Pv v = 217.7 217.7*1 *100 00/7 /704 04.4 .4 = 31% 31% < 60 60 %; %; Sub Subje ject cted ed to low low she shear ar Mc' = py *S = 355*1211*1000/1000000 = 429.8 kNm Check for irreversible deformation Mc' Mc' = 1.5 1.5*p *py* y*Z Z = 1.5* 1.5*35 355* 5*10 1071 71*1 *100 000/ 0/10 1000 0000 000 0 = 570 570 kNm kNm > 429. 429.8 8 kNm kNm Section Moment capacity = Mc = 429.8 kNm Ultima Ultimate te Moment Moment action action about about majo majorr axis axis = 165.4 165.4 kNm Section is adequate wrt to flexure < 429.8 429.8 kNm Adequate wrt flexure PROJECT STRUCTURE CODES ELEMENT BS 5950 : PART 1 :2000 DESIGNED BY Reference Clau 4.2.5.5. Crane Gantry Girder CHECKED BY Calculations REF 26/09/2017 Output Bolts holes No allowance is required as the bolts are in the compression flange at supports Lateral Torsional Buckling Table 13 Condition of restraint = Compression flange laterally unrestrained. Both flanges free to rotate on plan. Partial torsional restraint against rotation about longitudinal axis provided by connection of bottom flange to supports Loading condition : Normal assuming that rail are not on resilient pads Clau 4.3.5.2. Le for LTB = 1.0L LT + 2D LLT = L = 5.2 m Le (LTB)= 5.2 + 2*0.363 = 5.93 m Clau 4.3.6.2. Buckling Resistance Moment Mb = pb * Sx Clau 4.3.6.7 Equivalent Slenderness LT = u*v**sqrt(w) = Le/ry = 593/3.99 = 148.6 Annex B.2.3 hs = 363.4 - 15.7 = 347.7 mm x = 0.566*34.77x(85.5/55.7) = 34.77 cm = 24.38 0.5 2 -0.25 = 0.769 v = (1+0.05(148.6/24.38) ) = 1- 1362/19460 = 0.93 2 2 2 u = (4x1211 x0.93/(85.5 *34.77 )) Clau 4.3.6.9 0.25 = 0.886 w = 1.0 Equivalent Slenderness LT = u*v**sqrt(w) = 0.886*0.769*148.6*sqrt(1) = 101 Table 16 2 p = 137 N/mm Mb = pb * Sx = 137* 1211x1000/1000000 = 165.9 kNm Clau 4.3.6.2 Clau 4.11.3 Mx b/mLT mLT = 1.0 Mx = 165.4 < 165.9 / 1.0; Adequate wrt to Lateral torsional buckling Adequate w rt to lateral torsional buckling PROJECT STRUCTURE ELEMENT CODES BS 5950 : PART 1 :2000 Reference DESIGNED BY Crane Gantry Girder CHECKED BY REF 26/09/2017 Calculations Output Horizontal Moment Capacity of Top flange 2 Plastic modulus of top flange S f = TB /4 2 = 15.7*173.2 /4 3 = 117743 mm 2 3 Elastic modulus of top flange =Z f = TB /6 = 78495.3 mm Mc = py*Sf = 355*117743 = 41.79 kNm Mc irreversible def = 1.5*py*Zf = 1.5*355*78495.3 = 41.79 kNm =Mc Based on crane specification horizontal force at top of the flange = 2.6 kN Design horizontal force = 2.6*1.4 = F h = 3.64 kN Maximum horizontal moment to flange = M yf = from analysis = 3.78 kNm Myf < Mc Satisfied for most onerous arrangement of horizontal load Top flange is adequate for hor.loads Combined Effect Clau 4.8.3.2 For simplicity mx, my = 1.0 165.4/429.8 +3.78/41.79 = 0.475 < 1 ; OK Local Compression under the wheels Clau 4.11.4 xR = 2(Hr+T) = 2( 50 + 15.7) = 131.4 mm 50x50 railing Factored wheel loads = 1.25*1.4*30.63 + 1.25*1.6*37.49 = 128.58 kN local compressive stress on web = 128.58x1000/(128.58*9.1) = 107.53 N/mm2 < py = 355 N/mm2 Adequate wrt to local compression under wheels Clau 4.5.2.1 Web Bearing Considering that bolt connection restraint both rotation & lateral movement relative to the both flange and web P bw = (b1 + nk)tpw b1= 250 mm ; bearing plate width n = 2 for a end of a girder; Conservatively reduce capacity) k = T +r = 15.7 + 10.2 = 25.9 mm pyw = 355 N/mm2 P w = (250+2*25.9)9.1*355/1000 '= 975 kN Maximum Reaction at support = R = 242 kN < 975 kN ; Web bearing is adequate the supports Adequate wrt local compression under wheels PROJECT STRUCTURE CODES ELEMENT BS 5950 : PART 1 :2000 Reference Clau 4.5.3.1 DESIGNED BY Crane Gantry Girder CHECKED BY Calculations REF 26/09/2017 Output Web Buckling At ends of the girder, ae = Bplate width/2 = 250/2 = 125 mm < 0.7d = 0.7*311.6 = 211 mm So Px = ((125+0.7*311.6)/(1.4*311.6))*((25*0.88*9.1/(sqrt(250+2*25.9)*311.6))) * 975 Px = 500.6 kN Maximum Reaction at support = R =242 kN < 500.6 kN ;Safe against web buckling the supports Safe against Web failures Deflection Maximum permissble deflection = Span / 600 = 5200/600 = 8.66 mm Service deflection 6.74 mm < 8.66 mm Deflection is satisfied Section 356x171x67 with S355 grade steel girder can withstand the given forces Deflections are within limits
Structural steelwork design to BS 5950. The coverage is extensive and includes beams, purlins, crane girdcrs (excluding plate girders), trusses, columns, connections and bracing. A chapter is also devoted to comIx)site construction. The purpose of this section of the book is to set out in detail the design of individual structural elements. The author's preferred course of action w:uld be as follows: a) Design the crane stop for 1 00 of the rated speed with crane empty, for a deceleration rate of 16 ft/sec 2. b) Design the bracing and the foundation for the above crane stop force or 20$ of the total maximum wheel loads, which ever is greater.