Video 1

Now I will go through one worked out example on gantry girder. So in last lecture we have discussed the design steps of a gantry girder and I have given an example for solution. So, I hope many of you have done, now we will try to check the results means we will try to see whatever the actual results are coming and you try to correct or modify your results, if you have done something wrong.

Example:Design a simply supported gantry girder to carry electric overhead travelling crane, given:Span of gantry girder = 6 mSpan of crane girder = 15 mCrane capacity = 200 kNSelf weight of trolley, hook, electric motor etc. = 40 kNSelf weight of crane girder excluding trolley = 200 kNMinimum hook approach = 1.0 m

Distance between wheels = 3 mSelf weight of rails = 0.2 kN/mSolution:Maximum moment due to vertical forceWeight of trolley + lifted load = 40 + 200 = 240 kNSelf weight of crane girder = 200 kNFor maximum reaction on gantry girder, the moving load should be as close the gantry aspossible.Figure below shows the load position

RA=240 ×1415+2002=324 kNThis load is transferred to gantry girder, through two wheels, the wheel base being 3 m.So load on gantry girder from each wheel = 324/2 = 162 kNFactored wheel load = 162 × 1.5 = 243 kNMaximum moments due to moving loads occur under a wheel when the c.g. of wheel loadand the wheel are equidistant from the centre of girder. This is shown in figure:

RD=243×(3−1.5−0.75)+243×(3+0.75)6=182.25 kNMax. moment ME = 182.25 × 2.25 = 410 kN-mMoment due to impact = 0.25 × 410 = 102.5 kN-mAssume self weight of girder = 2 kN/mDead load due to self weight + rails = 2 + 0.2 = 2.2 kN/mFactored DL = 2.2 × 1.5 = 3.3 kN/mMoment due to DL = 3.3 × 62/8 = 14.85 kN-mFactored moment due to all vertical loads,Mz = 410 + 102.5 + 14.85 = 527.35 kN-m

Maximum moment due to lateral forceHorizontal force transferred to rails = 10% of weight of trolley plus load lifted = (10/100) ×(200 + 40) = 24 kNThis is distributed over 4 wheels.So, horizontal force on each wheel = 24/4 = 6 kNFactored horizontal force on each wheel = 1.5 × 6 = 9 kNFor maximum moment in gantry girder the position of loads is same as earlier except that it ishorizontal. Hence by proportioning we get,My= (9/243) × 410 = 15.18 kN-mShear forceFor maximum shear force on the girder, the trailing wheel should be just on the girder asshown in figure belowVertical shear due to wheel loads = 243 + (243 × 3)/6 = 364.5 kN

Vertical shear due to impact = 0.25 × 364.5 = 91.125 kNVertical shear due to self weight = (3.3 × 6)/2 = 9.9 kNTotal vertical shear = 364.5 + 91.125 + 9.9 = 465.52 kNBy proportioning lateral shear due to surge = (9/243) × 465.52 = 17.24 kN

Video 2

Preliminary SectionMinimum economic depth, L/12 = 6000/12 = 500 mmWidth of the compression flange may be taken as (1/40) to (1/30)th of the spanSo, flange width can be taken, L/40 = 6000/40 = 150 mm to L/30 = 6000/30 = 200 mmRequired Zp = 1.4 × M/fy = 1.4 × 527.35 × 106/250 = 2953.16 × 103 mm3Let us try a ISMB 550 with ISMC 250 on compression flange.

Let the distance of N. A. from the tension flange be ´ y ,Then, ´ y=13200× 275+3900 ×(550+7.2−23)13200+3900=334.11 mmIzz = 64900 × 104 + 13200 × (334.11-275)2 + 211 × 104 + 3900 × (550+7.2-23-334.11)2 =853.37 × 106 mm4Zez = 853.37 × 106 /334.11 = 2554.15 × 103 mm3For compression flange about y-y axis,I = 3880 × 104 + 1/12 × 19.3 × 1903 = 4984.15 × 104 mm4Zey for compression flange = 4983.15 × 104 / 125 = 398.73× 103 mm3Total area of section = 13200 + 3900 = 17100 mm2Let Plastic N.A. be at a distance Yp from tension flange. Then,(Yp – 19.3)×11.2 + 190 × 19.3 = 17100/2

Or, Yp = 455.28 mmZpz = (190 × 19.3) ×(455.28-11.2/2)+(455.28-19.3)2/2 ×11.2+(550-455.28-19.3)2/2×11.2+19.3 ×190 ×(550-455.28-19.3/2)+3900 ×(550-455.28+7.2-22.3)= 3367.74 × 103 mm3For top flange,Zpy = 1/4 ×19.3 ×1902+1/4(250 - 2 ×14.1)2 ×7.8+2 ×80 × 14.1 ×(125-14.1/2)= 536.203 × 103 mm3Section classificationb/t of flange of ISMB 550 = (190-11.2)/(2 ×19.3) = 4.63 < 9.4d/t of web of ISMB 550 = (550-2 ×(19.3+18))/11.2 =42.44 < 84And b/t of flange of channel = (80-7.2)/14.1 = 5.16 < 9.4Hence the section is plastic.Check for local moment capacityLocal moment capacity for bending in vertical plane:Mdz = fyZp/1.1 = 250 × 3367.74 × 103 /1.1 = 765.31 kN-m1.2Zezfy/1.1 = 1.2× 2554.15 × 103×250/1.1 = 696.58 kN-mSo, Mdz = 696.58 kN-mFor top flange,Mdy = 250 × 536.203 × 103 /1.1 = 121.86 kN-m1.2Zeyfy/1.1 = 1.2× 332.21× 103 × 250/1.1 = 90.6 kN-mSo for top flange Mdy = 90.6 kN-mCheck for combined local capacity527.35/ 696.58 + 15.18/90.6 = 0.92 < 1Check for buckling resistanceMd = bZpfbdFor plastic section b = 1f cr, b=1.1π2E(rLy¿)2 √1+ 20 1 (KL rthfy )2LLT = 6000 mm, E = 2 × 105 N/mm2hf = 550 – (19.3/2) + (14.1/2) = 547.4 mmIy= 1830 ×104 + 3880 × 104 = 5710 ×104 mm4A = 13200 + 3900 = 17100 mm2ry= (Iy/A)1/2 = (5710 × 104/17100)1/2 = 57.78 mmf cr, b=1.1π2×2×105(57.78 6000 )2 √1+20 1 (57.78 547.4 6000 19.3 )2=260.23 N/mm2 λ¿=√260.23 250 =0.98φ¿=0.5[1+0.21×(0.98−0.2)+0.98 2]=1.062X¿=1{1.062+(1.0622−0.98 2)0.5}=0.68 N/mm2 fbd = 0.733×250/1.1 = 164.11Mdz = 1.0×166.6× 3367.74 × 103 = 531.06 kN-mSince lateral force is also acting, the beam must be checked for bi-axial bending.So, Mdy = 250/1.1 × (1830 × 104 + 3880 × 104 )/125 = 103.81 kN-mHence,527.35561.06+15.18103.81=1.086The section is unsafe against torsional buckling.Check for shearVz = 465.52 kNShear capacity =Av f yw√3×1.1 =550×11.2×250√3× 1.1 =808.29 kN > 465.52 kNNow, 0.6 × 808.29 = 484.974 kNSo, it is a case of low shear.Check for Web bucklingAssuming b1 = 150 mmn1 = 225 + 7.6 = 232.6 mmd = 550 - 2(19.3+18) = 475.4 mm, t = 11.2 mmλ=2.42 dt=2.42 × 475.411.2=102.72For buckling class a, from Table 9(a), Is 800: 2007Fcd = 128.6 N/mm2Buckling resistance = (150+232.6)×11.2×128.6 = 551.66 kN > 465.52 kNCheck for deflection at working load(i) Vertical deflectionServiceability vertical wheel load excluding impact = 162 kNDeflection at mid-spanδ=Wl3[(43aL)−(a L33)]/(6 EI)Where, a = (L - c)/2 = (6000-3000)/2 = 1500 mmCombined Izz = 842.7 × 106 mm4δ=162× 103×60003[(3 4××1500 3000)−(1500 60003 3)]/(6×2×105×853.37× 10 6)= 12.27 mm(ii) Horizontal deflectionI = (Iz)ch + IF = 4983.15 × 104 mm4δ=6×103 × 60003[(3 4× ×1500 3000)−(1500 60003 3)]/(6× 2× 105× 4983.15 × 10 4)= 7.788 mmDeflection limit, L/750 = 6000/750 = 8 mmSo the section is unsafe in vertical direction.Design for weldThe required shear capacity of the weld, q=VA ´y/I zV = 465.52 kNA = 3900 mm2 (Area of the channel section)´ y = (550 - 334.11 + 14.1/2) = 222.94 mmIzz = 853.37 × 106 mm4q = 465.52 ×103×3900×222.94 /853.37×103 = 474.3 N/mmSo, (0.7s×410)/( √3 ×1.5) = 474.3or, s = 4.29 mmSo use 5 mm fillet weld for the connection.So, this is all about the gantry girder, if you go through some example given in the book thenit will be much more clear.