Steel Crane Girder Design example

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



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


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



Reference

Clau 4.5.3.1

DESIGNED 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

Steel Crane Girder Design example

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