DESIGN OF GANTRY GIRDER

Design a gantry girder to be used in an industrial building carrying a manually operated overhead traveling crane for the following data:Crane capacity - 200kNSelf-weight of the crane girder excluding trolley - 200kNSelf-weight of the trolley, electric motor, hook, etc. - 40kNApproximate minimum approach of the crane hook to the gantry girder - 1.20mWheel base - 3.5mc/c distance between gantry rails - 16mc/c distance between columns (span of gantry girder) - 8mSelf weight of rail section - 300N/mDiameter of crane wheels - 150mmSteel is of grade Fe 410. Design also the field welded connection if required. The support bracket connection need to designed.

SOLUTION :

For Fe410 grade of steel: fu = 410 Mpa fy = fyw = fyf = 410 MPa

For hand operated OT crane: Lateral loads = 5% of maximum static wheel load Longitudinal loads = 5% of weight of crab and weight lifted Maximum permissible deflection = L/500Partial safety factors m0 = 1.10ע mw = 1.50 (for site welds)ע

Load factor m1 = 1.50ע

ξ = ξw = ξf = √ (250/ fy) = √ (250/ 250) = 1.0

STEP 1: CALCULATION OF DESIGN FORCES

Maximum wheel load: Maximum concentrated load on crane = 200 + 40 = 240kN Maximum factored load on crane = 1.5 x 240 = 360kN The crane will carry the self-weight as a UDL = 200/16 = 12.5 kN/m Factored uniform load = 1.5 x 12.5 = 18.75kN/m

For maximum reaction on the gantry girder the loads are placed on the crane girder as shown in the figure below,

Taking moment about B,

RA x 16 = 360 x (16 - 1.2) + (18.75 x 16 x 16) / 2

RA = 483kN

RB = 177kN

The reaction from the crane girder is distributed equally on the two wheels at the end of the crane girder.

Therefore, Maximum wheel load on each = 483/2 = 241.5 kN wheel of the crane

STEP 2: CALCULATION OF MAXIMUM BM & SF:

Maximum Bending Moment:It consists of maximum moments caused by the moving wheel loads on the gantry girder and self weight of the gantry girder.

For Maximum Bending Moment, the wheel loads shall be placed as shown in figure.

The calculation of maximum bending moments due to wheel loads and self weight of gantry girder has been done separately because calculation of impact load and bending moment due to it involve live load only, Assume, Self Weight of the Gantry Girder as 2 kN/m. Total Dead Load, w = 2000 + 300 = 2300 N/m = 2.3 kN/m Factored Dead Load = 1.5 x 2.3 = 3.45kN/m

The position of one wheel load from the mid point of span = Wheel Base/4 = 3.5/4 = 0.875 mBending moment due to live load only,Taking moment about D,

Rc x 8 = 241.5 x (8 - 1.375) + 241.5 x 3.125

Rc = 294.33kN Taking moment about C,

RD x 8 = 241.5 x 1.375 + 241.5 x 4.875

RD = 188.67kN

Maximum Bending Moment due to Live Load = 188.67 x 3.125 = 589.6 kNmBending Moment due to Impact = 0.10 x 589.6 = 58.96 kNm

Total Bending Moment due to live load and impact loads = 589.6 + 58.96 = 648.56 kNmBending Moment due to Dead Load = wl2/8 = 3.45 x (82/8) = 27.6 kNmTherefore, Maximum B.M = 648.56 + 27.6 = 676.16 kNm = 676.16 x 106 N_mm

Maximum Shear Force: It consists of maximum shear due to moving wheel loads and self

weight of the girder.

For maximum shear force the wheel loads shall be placed as shown in figure, i.e one of the wheel loads should at the support.

Taking Moment about D,

Rc x 8 = (241.5 x 8) + (241.5 x 4.5)

Rc = 377.34 kN

Hence, Maximum Shear Force due to wheel loads = 377.34 kN

Lateral force transverse to the rails = 5% of weight of crab and weight lifted = 0.05 x 240 = 12kNFactored lateral force = 1.5 x 12 = 18 kN

STEP 3: CALCULATION OF LATERAL FORCES:

Maximum horizontal reaction due to lateral force by proportion at C, = (lateral force x Reaction at c due to vertical load)

(maximum wheel load due to vertical load) = (9 x 294.33) / 241.5 = 10.97 kNHorizontal reaction due to lateral force at D

= 18 - 10.97 = 7.03 kNMaximum bending moment due to lateral load by proportion = (9.0 / 241.5) x 589.6 = 21.97 kNmMaximum shear force due to lateral load by proportion

= (377.34 / 241.5) x 9.0 = 14.06 kN

Approximate depth of section = L / 12 = (8 x 103) / 12 = 666.66mm ≈ 600mm Approximate width of the flange = L / 30 = (8 x 103) / 30

= 266.66mm ≈ 300 mm

Approximate section modulus required, Zpz = 1.4 (Mz / fy )

= (1.4 x 676.16 x 106) / 250 = 3786.5 x 103 mm3

STEP 4: SELECTION OF PRELIMINARY TRIAL SECTION :

Let us try ISWB 600 @ 1311.6 N/m

with ISMC300 @ 351.2 N/m on its

top flange as shown in figure,

Properties Notation I-Section ISWB 600

Channel SectionISMC 300

Area A 17038 mm2 4564 mm2

Thickness of Flange tf 21.3 mm 13.6 mm

Thickness of Web tw 11.2 mm 7.6 mm

Width of Flange bf 250 mm 90 mm

Moment of Inertia Iz 106198.5 x 104 mm4 6362.6 x 104 mm4

Iy 4702.5 x 104 mm4 310.8 x 104 mm4

Depth of Section h 600 mm 300 mm

Radius at root Rl 17 mm

Cyy 23.6 mm

The distance of NA of built up section from the extreme fibre of compression flange, ỹ = (∑AY) / ∑A = {17038 X (300 +7.6)} + {4564 X 23.6} = 247.59 mm (4564 + 17038)Gross Moment of Inertia of the Built-up section Iz gross = Iz beam + Iz channel(The channel is placed over the I-section in such a manner that its yy-axis becomes the zz-axis. Therefore, in calculating Iz of the total section, Iy of the channel will become Iz of the channel.) Iz gross = [(106198.5 x 104) + (17038 x (307.6 - 247.59)2)] + [(310.8 x 104) + (4564 x (247.59 -23.6)2)]

= 135,543x104 mm4

STEP 5: MOMENT OF INERTIA OF GANTRY GIRDER :

Iy gross = Iy beam + Iy channel

= (4702.5 x 104) + (6362.6 x 104)

= 11,065.1x104 mm4

Zez = (Iy /y)

= (135,543 x 104) / (600 + 7.6 - 247.59) = 3764.98x103 mm3

Plastic Modulus of Section (Ignoring the Fillets)Equal area axis (Refer to Fig)

= 4564 + (250 X 21.3) + (ỹ1 t w)

= 250 X 21.3 + (600-2 X 21.3 ỹ1 ) X 11.2

Z = 74.95 mm(From Lower surface of top flange of I – Section)

Plastic section modulus of the section above equal axis,

Zpz1 = [300 X 7.6 X (74.95 + 21.3 + (7.6 / 2))]

+ [2 X (90-7.6)x13.6(74.95 + 21.3 – (90-7.6/2))] + [250 X 21.3 x (74.95 + (21.3/2))] + [74.95 X 11.2 X (74.95/2)]

Zpz1 = 838.775 X 103 mm3

Plastic section modulus of the section below equal axis,

Zpz2 = [250 X 21.3 X ( 600 - 21.3 - 74.95 - ( 21.3/2))]

+ [((600 – 2 X 21.3 – 74.95)2 / 2)]

Zpz2 = 3929.202 X 103 mm3

Zpz = Zpz1 + Zpz2 = (838.775 X 103) + (3929.202 X 103) Zpz = 4767.977 X 103 mm3

Plastic section modulus of compression flange about yy – axis,

Zpfy = [((250 X 21.3 X 250) /4)] + [((2 X (300 – (13.6 X 2))2 X 7.6) / 8)]

+ [2 x (13.6 x 90 x ( 300 – 13.6) / 2)]

= 824.763 X 103 mm3

STEP 6: CLASSIFICATION OF SECTION:

Outstand of flange of I-Section,

b = bf/2 = 250/2 = 125 mm.

b/tf of flange of I-section

= 125/21.3 = 5.86 < 8.4 (8.4ε = 8.4 x 1 = 8.4)Outstand of flange of Channel Section,

b = bf - tw = 90 - 7.6 = 82.4 mm.

b/tf of flange of Channel section

= 82.4/13.6 = 6.05 < 8.4 (8.4 ε = 8.4 x 1 = 8.4)

d/tw of web of I-Section = ((h – 2tf) / tw)

= ((600 – 2 x 21.3) / 11.2) = 49.76 < 84 (84 ε = 84 x 1 = 84)

Hence the entire section is plastic, (βb = 1.0)

STEP 7: CHECK FOR MOMENT CAPACITY:

Local Moment capacity, Mdz = βb Zpz (fy / עm0 ) ≤ 1.2 x Ze (fy / עm0 )

Mdz = 1.0 x 4767.977 x 103 x (250 x 10-6 / 1.10)

= 1083.63 kNm ≤ 1.2 x 3764.98 x 103 x (250 x 10-6 / 1.10) = 1026.81 kNm

Hence, Moment capacity of the section,

Mdz = 1026.81 kNm > 676.16 kNm

which is safe.

Moment capacity of compression flange about y-axis,

Mdy,f = βb Zpyf (fy / עm0 ) ≤ 1.2 x Zey,f (fy / עm0 )

= 1.0 x 824.76 x 103 x (250 x 10-6 / 1.10) = 187.44 kNm ≤ 1.2 x 580.92 x 103 x (250 x 10-6 / 1.10) = 158.43 kNm

Hence moment capacity of flange,

Mdy,f = 158.43 kNm

Combined check for local moment capacity, (Mz / Mdz ) + (My,f / Mdy,f) ≤ 1.0 (676.16 / 1026.81) + (21.97 / 158.43) = 0.797 < 1.0 Which is safe.

The elastic lateral buckling moment,

Mcr = c1(π2EIYhf/2L2LT)[1+1/20((LLT/ry)/(hf/tf))2]0.5

Overall depth of the section,

hf = h = 600 + 7.6 = 607.6 mm

Note: hf is the centre to centre distance between flanges. In the

calculation, value of hf has been taken equal to overall depth of the

section to avoid cumbersome calculations. Further, this approximation will result in conservative design.

Effective length, LLT = 8 x 102 mm

Thickness of flange, tf = 21.3 + 7.6 = 28.9 mm

Radius of gyration, ry = √(IY/A)

= √((11065.1x104)/(17038+4564)) = 71.57mm

STEP 8: BUCKLING RESISTANCE IN BENDING CHECK:

The coefficient, c1 = 1.132 (From Appendix XIV, assuming uniform load condition) Mcr = [(1.132x2x2x105x11065.1x104x607.6) / (2x(8x103)2)]

x [1+1/20((8x103/71.57)/(607.6/28.9))2]0.5

= 1823.24 x 106 NmmNon dimensional slenderness ratio, λLTz = √(βb Zpz fy / Mcr)

= √((1 x 4783.594 x 103 x 250) / (1823.24 x 106))= 0.809

ΦLTz = 0.5 [1 + αLT (λLTz – 0.2) + λLTz2]αLT = 0.21 (assuming connection of channel with I-section flange by intermittent fillet welds. For uniform weld, αLT = 0.49)ΦLTz = 0.5 x [1+0.21 x (0.809 – 0.2) + 0.8092]

= 0.891

χLTz = [1/( ΦLTz + ΦLTz2 - λLTz

2)0.5]

= [1/(0.891 + 0.8912 – 0.8092)0.5] = 0.791

Design bending compressive stress,

fbd = χLTz (fy / עm0 )

= 0.791 (250/1.10) = 179.77 N/mm2

The design bending strength,

Mdz = βb Zpz fbd

= 1.0 x 4767.977 x 103 x 179.77 x 10-6

= 857.14 kNm > 676.16 kNm (which is alright)

Hence, the beam is safe in bending under vertical load.

Since lateral forces are also acting, the beam must be checked for bi-axial bending. The bending strength about y-axis will be provided by the top flange only as the lateral loads are applied there only.

Mdy =Zyt (fy/γm0)

Zyt = section modulus of top flange about yy-axis. = (4702.5x104/2 +6362.6x104)/(300/2) = 580.92 x 103 mm3

(Assuming the moment of inertia of top flange to be half of the moment of inertia of I-section) Mdy = 580.92 x 103 x (250/1.1) x (10-6) = 132.02kNm. (Mz/Mdz) + (My/Mdy) < 1.0 = (676.16/859.946) + (21.97/132.02) = 0.952 < 1.0 Hence, the section is safe.

STEP 9:

Maximum shear force due to wheel load = 377.34 kNImpact load = 0.1 x 377.34 = 37.73 kNDesign shear force = 377.34 + 37.73 = 415.05 kN

Shear capacity = Av fyw / (√3 x γmo)

= [((600x11.2) x 250x10-3) / (√3x1.1) ] = 881.77 kN > 415.07kN, which is safe.

Maximum shear, V = 415.07 kN < 529 kN

(i.e. 0.6*Vd = 0.6 x 881.77 = 529kN)

Since V < 0.6*Vd, the case is of low shear.

No reduction will be therefore there in the moment capacity.

STEP 10: CHECK FOR SHEAR CAPACITY :

We should check for buckling under the wheel load. Buckling Resistance = (b1+n1)tw*fcd Where, b1 = bearing length=150mm(diameter of wheel) n1 = (600/2) +2x7.6 = 315.2 mmSlenderness ratio of the web, λw = 2.45 (d1/tw) = 2.45(600 -2x(21.3 + 17))/11.2 = 114.49For λw=114.49, fy=250N/mm2 and buckling curve c, the design compressive stress from Table 7.6,

fcd = 89.64 N/mm2 Therefore, Buckling Resistance = (150 + 315.2) x 7.6 x 89.64 x 10-3 = 316.9 kN > 241.5kN (which is safe)

STEP 11: WEB-BUCKING CHECK:

δ = wL3x(3a/4L –a3/L3) 6EI

W = Maximum Static Wheel Load = (241.5 / 1.5) = 161 kN a = (L - c) /2 = [(8x103) – (3.5x103)] / 2 = 2.25 x 103 mm Therefore, Vertical deflection = (161x103)x(8x103)3x{(3x2.25x103)/(4x8x103)-(2.25x103)3/(8x103)3)

[6 x (2x105) x 135,543x104] = 9.56 mmPermissible maximum deflection, = L / 500 = 8000 / 500 = 16 mm > 9.56 mm, which is safe.

STEP 12: DEFLECTION CHECK:

The required shear capacity of the weld,

q = Vaỹ / Iz

V = 377.34 kN A = 4564 mm2 (area above the section) ỹ = (247.59 -7.6) = 239.9 mm

Iz = 135,543x104 mm4

Therefore, q = [(377.34 x 103) x 4564 x 239.9] / [135,543x104] = 304.82 N/mmLet us provide 3 mm weld to connect channel with flange of I-section. Strength of the weld provided, = (0.7x3x410)/(√3x1.50) = 331.39 N/mm > 304.82 N/mmHence, provide 3 mm size intermittent fillet welds for making the connection.

STEP 13: DESIGN OF CONNECTIONS :

DRAWINGS & DETAILING