Hoist Desi gn Procedure for EOT crane…

Posted on July 29, 2008 by rajendra Parmanik

THE DESIGN

We are concerned with the design of the hoisting arrangement of 2 tonne capacity of EOT crane ,which will lift the load up to a distance of 8 meters.1.DESIGN OF HOOK

HOOK

hook-cross-section

LINK

SHEAVE

Selection of section : The section is trapezoidalSelection of material : Mild steelLoad to lift : 2 tonneConsidering 50 % over loading.So the design load = 2 tonne+50% of 2 tonne = 3tonneTaking the help of (IS 3815-1969) for selection of material for 8 dimensions of crane hook.In IS 3815-1969 the nearest selection for 3.3 tonne is 3.2 tonne.For load 3.2, proof load (P) is 6.4 tonne.So C = 26.73√P = 26.73 x √6.4= 67.62≈ 68 mmA = 2.75 C = 2.75 x 68 ≈ 187 mmB = 1.31 C = 1.31 x 68 ≈ 89mmD = 1.44 x C = 1.44 x 68 ≈ 98mmE = 1.25C = 1.25 x 68 ≈ 85mmF = C = 68mmG = 35mmG1 = M33, Pitch = 6mm (Coarse series)H = 0.93 x C = 0.93 x 68 ≈ 63mmJ = 0.75 x C = 0.75 x 68 ≈ 51mmK = 0.92 x C = 63mmL = 0.7 x C = 0.7 x 68 ≈ 48mmM = 0.6 x C = 0.6 x 68 ≈ 41mmN = 1.2 x C = 82mmP = 0.5 x C = 34mm ≈ 34mmR = 0.5 x C = 0.5 x 68 ≈U = 0.33 x C = 0.3 x 68≈20 mmChecking for strength Area of the section = ½ x 63 x (41+8) = 1543.5 mm2Centroid from ‘a’= (.05 x 8 x 65) 63/3+(.5 x 41 x 63) x (2 x 63)/3½ x 68 x (41+8)= 38.571mm= 38.6 mm= h2So centroid from b = 63-38.6=24.4mm =h10 = 34 +24.4 = 58.4mmr0 = A/(dA/u)dA/u = [b2+r2/h (b1-b2)] ln r2/r1 – (b1-b2)=28.65mmr0 = A/(dA/u) = 1543.5 = 53.87 53.9 mm.28.65e= 0-r0 = 58.4 – 53.9 = 4.5mm

MomentM = -P x 0= -3 x 58.4= – 175.2 (tonne x mm)Stress due to bending is given byb = M X 4Ae r0-y

For point aY = -(e+h2)= -(4.5+38.6)= – 43.1 mmFor point bY = r0-r1= 53.9 – 3.4= 19.9 mmStress due to direct loading = P/A= 3/1543.5= 1.9436 x 10-3 Tonne/mm3Stress due to curvature of ‘a’ba = – (-175.2) x -43.11543.5 x 4.5 {53.99 – (-43.1)}= 0.0112So total tress at a= – 0.0112 + 1.9436 x 10-3= – 9.2642 x 10-3 Tonne/mm2= – 9.2642 kg/mm2 -90.85 MpStress due to curvature at bbb = -(-175.2) x . 19.99 .1543.5x 4.5 (53.9 – 19.9)= 0.014763So total stress at b=bb + 1.9436 x10-3=0.014763 + 1.9436 x 10-3=0.0167 tonne/mm2= 16.7 Kg/mm2 163.84 MPaLet the material be class 4 carbon steel ( 55C 8)Ultimate tensil strength I 710MPaDesign strength = Ultimate tensil strengthFactor of safety = 710/4 = 177.5 MPa163.84 > 177.5So design is safeDetermination of length of threaded portionPitch = 6mmNominal dia of thread = 33 mm (G1) = dConsidering the screw and thread are of single safest & square mean diameter of screw =dm = d- (p/2)= 33 – (6/2)= 30tan = 1/dm = 6/( x 30)tan = tan-1 { 6/( x 30)} = 3.640Let the co-efficient of fraction be 0.15So = tan = 0.15= 8.530Torque required to resist the load

T = W x dm* tan ( + )2Where w is the weight of load is 3 tonne and the load of the hook itself.The maximum weight of the hook is 50kg (from the use of the soft ware ‘Pro-Engineer’)SoT = 3050x 30 x tan (3.61+ 8.53)2=9866.42 Kgmm

Stress induced in the screwDirect tensible stress (allowable or design) =4w/ d02d0 = core diameter of the screw.dc = d-p = 33-6 = 27 mm1 = 4x 3050 272= 5.326 kg/mm2 52.24MR

Torssional shear stress = 16T = 16×9866.42 de3 x 273= 2.5529 Kg/mm2 = 25MP

Maximum shear stress in the screwmax = ½ (2 + 4 2)= ½ √ (52.242 + 4 x 252)= max = 36.15Mpa

Height of the nut

a)Considering bearing action between the thread in engagement.Let ‘n’ is no of thread in engagement with screw.Considering bearing action between nut & screw.Let the permissible bearing pressure =pi= 6 MR.We knowPi = 4W (d2-dc2) x nSo 6 = 4×3050 x 9.8(332-272) x n

N = 4 x 3050 x 9.8(332-272) x 6 1.27 x 9.8 12.5So the height of the nut is = 2 x 12.5 = 25mm.b) Considering shear failure of thread across rootShear stress induced = . W .dc(0.5xP) xn

= . 3050 x 9.8 . x 27 x (0.5 x 6) x n

= 117.46n= 0.5 x 177.5 = 117.46n n= 117.46177.5 x0.5= 1.32 = 2So height is n x p = 2 x 6 =12mmTacking the highest value 25mm

Design of pin which will carry the dead load & the load of hook.We have to determine the dimension of ‘t’.n = M x YI= M x (24)I 2Where “I” is moment of inertial about bh3 t x 243 x 212 12Maximum bending moment for = M = ¼ x W x LLet L = 70 mmSo M = ¼ x 3050 x 70 = 2 x 26687.5 Kgmm.= 2 x 261.715 x 103 N-mm = 2x 261.715 x 103 x 241 t x 243 2177.5 = 1363 x 2tt = 1363 x 2 = 7.8 x 2 15.2 MM177.5 16 mmTaking 20 m for additional safetyDiameter of the projected portion

The projected position is undergo only shearing failure.Design shear stress = 0.5 x design tensive stress.Force action on each side i.e. projected portion is3050/2 =1525 kg 14.95 x 103 15 KNSo the minimum value of the height of the projected portion is 15mm, taking n= 40mm for screwing arrangement. = 15 x 103/4 x dk2= 177.5 x 0.5 = 15 x 103(/4) dk2= dk2 = 15 x 103 x 4177.5 x 0.5 x = dk2 = 15 x 103 x 4177.5 x 0.5 x

= 215.19= 14.66 15mmTaking 20 mm for additional safety purpose.So that it can be turned to thread in size ‘m20’.Height of the projected portionP * x * dk = w/2Where p = bearing pressure or crushing stress.Let p = 210 MPxAllow crushing stress =210/4 =210/4 x x x 20 = 3050 x 9.82x = 3050 x 9.8 x 52 x 20 x 210= 3.56 mm x 4= 14.024 =15 mmSo the minimum value of the height of the projected portion is 15 mm, taking x = 40 mm forscrewing arrangement.

Design of link and the cover plate

coverplate

Thickness of link and cover plate should not be minimum.

Let the material be (55C8).a)Hole fopr placing the pin which will carry the hook will be ‘dk’ i.e. 20 mm.zchecking for the failure of link & cover plate combindely(as they are of same material and undergone same condition of failure).

Mode of failureTearing of cover plate & link at the edge.Crushing of cover plate and link.Breaking at the lowest cross section.i).Considering tearing of cover plate & link at the edge

Experiments from the riveted joints have shown that if the distance between the centre of rivetand the edge is 1.5 times the diameter of the rivet.The element will not undergo the failure of tearing at the edge.The same condition is also applicable in our case.But for more safety reasons taking the distance between the centre of the projected element and the edge of the cover plate & linkis 2 times the dia of the projected element.So Z = 2 x dk= 2 x 20 = 40 mmConsidering tensile failure at the lower cross sectionSo σd = w/2A177.5 = 3050 x 9.8/22 x x x 30

x = 3050 x 9.8/22 x 30 x 177.5

= 28/2 mm ≈ 1.4 mm

Considering crushing failureForce = σcrushing(d) x projected area3050 x 9.8 = σcrushing(d) x (20 x x)FOS= 3050 x 9.8 x 4 = 14.24 mm ≈ 15 mm2 x 210 x 20So taking maximum of x i.e 15 mm, So x = 15 mm

cover-plate-auto-cad

AutoCAD drawing Of Cover plate

auto_link Auto CAD

AutoCAD Drawing Of Link

Design of shaft carrying the pulley

shaft to carry pully

Pulley

The weight of each cover plate is 2.5 kg.Weights of each link weigh 2 kg.So weight of 2 covers plate & 2 linkIs 2 x 2.5 + 2 x 2 = 9 kg.So the total weight which the shafts carry is 3050 + 9 = 3059 ≈ 3060Each Side subjected to a load of 3060/2 = 1530 kg.The shaft is only subjected toCrushing failure (at the cover plate and link).Shear failure.Crushing failure (at the pulley)Considering shear failure = f/a = allowable shear stress= 0.5 x 177.5 = 0 .5 x 177.5 = 22.2 MRFOS 4Stress induced = 1530 x 9.8 = 150042π/4 x 302 π/4 x 302= 21.22 MPaAs induced stress is less than that allowable stress, the design is safe

Considering crushing failure at the cover plateCrushing stress = Force/Protected areaAllowable stress = 240/4 = 60MPa

Induced Crushing stress = ForceProjected area= 3060 x 9.870 x 40As allowable induced stress 60.86 KN. The selection is feasible.

Design of Sheave

sheave

a = 40mmb = 30mmc = 7mmd = 18mme = 1mml = 10mmr = 12mmr4 = 8mmh = 25mmr1 = 4mmr2 = 3mmr3 = 12mmMaterial is caste ironLet dia of the sheave = 20 x d= 20 x 14= 240mmThe reference is made from Rudenko,N.’Materials Handling Equipment ‘,Mir Publishers, Moscow(1969).P .86,Table 16.

Drum designDrum grove size

Referring to Redenko,N “Materials Handing Equipment”,Mir publishers,Moscow (1969),P.No.90 table 17.Considering standard groove of drum, for, diameter of wire 15 mm as it is nearest to 14mm.Drum diameter = sheeve diameter = 240 mm.r 1 = 0.53 x d (d ≈15) = 9mms1 = 1.15d = 17mmC1 = 0.25 d = 5mmNo. of turn on each side of drumZ = (hi/πd) + 2 =WhereH = Lifting Height.I = Ratio of the pulley system.D = Drum diameter ≈ 45 x 14 = 630So Z = ( 8.00 x 2) + 2Π x63Z = 10Full lenth of drum for one rope.L = (2HI + 7)Si (I = 2 assumed)πD=(2 x 800 x 2 + 7)1.5=34.75≈35 cmThe drum is made up of IS grade = SG 80/2With stress = 480Mpa

W = 0.02 x 630 = 10= 22.6≈ 23 mmOutside dia of the drum Do = D + 6d= 630 + 6 x 14 = 714mmInside dia of the drum = Di = D – 2W= 630 – 2 x 25= 584 mmChecking of strengthBending Stress in drum σbend = 8WLDπ(D4 – Di4)= 8 x 6200 x 9.8 x 35 x 630Π x (6304 – 5844)= 0.0828MP

Maximun TorqueTmax = W (D + d)2= 6200 x 9.8 (630 + 14)2= 19.564 x 106Maximum Shear Stress = 16 Tmax Dπ(D4 – Di4)= 16 x 19.56 x 106 x 630π x (6304 – 5844)≈ 1.523 MPxDirect Compressive Stress= W = σcwSi= 6200 x 9.823 x 17= 155.4MPx

Maximum Stress in the Drumσ = √σbend2 + σc2 + 4max2= (0.08282 +155.42 = 4 x 1.5232)1/2

σ < 480 MPx, S the Design is safe.

Fastening Of Rope With The DrumFor 14 mm diaLocating DimensionPitch of screw = 53 mmT = 43 mmScrew sizeLo = 18 mmL = 50 mm

PlateC = 7 mmNo. of fastenings = 1

Selection of MotorW = 3100 kg x 9.8 m/s2V = 0.2m/secSo power required = 3100 x 9.8 x 0.2= 6080.123Nm/sec= 6.08 KwattTaking 11 Kwatt 3 phase induction motor (flange type) of 11 KW (nearest to 6.08KW)Frame No. 132 MFlange designation F265BThe speed is 1000 RPMBy using gear box the speed can be reduced to 300 RPM

GearBox CalculationGearbox ratio = (Input RPM x π x rope drum dia)(speed x no. of falls /2)= ( 980 x π x .714) = 183.54(3.00 x 4 )

Brake calculationRequired Brake Torque = 1.5 x 716.2 x mech H.P.Motor RPM= 1.5 x 716.2 x 50.12300= 53.84 NmDesign of wormset Power = P = 6.08 KwattRPM of Worm = Nw = V x 60πd= 0.2 x 60π x 0.63RPM of Worm Gear NG = 300 RPMSo R = NGNw= 3006Let Ф = 14.5oLet the centre distance = C = 300 mm =0.3mPitch circle diameter of the wormDw = C.8750 = 3Pc3.48Dw = 3.8750 = 0.10020m3.48

≈ 100mmDw = 3Pc = 3Pa = 3 x π x ma

So ma = Dw/3π = 100.2/3π= 10.63 mm ≈ 11mm

Pitch circle dia of the gear Dg = 2C – Dw = 2x 300 – 100 = 500 mmVelocity ratio = Ng/Nw = 50Ng/Nw = Dg /(ma x Nw)50 = Dg/(11 x Nw)Dg = 50 x 11 x Nw

Taking Ng = 1 , because Dw is closer to the calculated.So, Dg = 550Dw = 2C – Dg= 2 x 300 -500= 50mmFace width of the gearb = 0.73 x Dw= 0.73 x 50= 36.5= 40 mmStatic strength of Bronzeσd = 90MPaIn worm drive irrespective of materials of worm and worm ger,the gear is weak.So design should be based on gear.

Tangential load on the gear :Ft = σ π mn y b= (σd x Cv)π mn y bVelocity factor Cv = 6 .6 + VgVg = π x Dg x Ng60= 8.69Cv = 6 = 0.416 + 8.69Form factorY = 0.124 – 0.684Ng= 0.11132Tan λ = m x NwDw

= 11 x 150λ = tan-1 (.22) = 12.4o

Nominal Module = m (ma) x cos λ= 11 x cos 12.4

= 10.743 mmFt = 90 x .41 x π x 1o.743 x .11132 x 40= 5545.47 N= 5.55 KNPower Capacity = P1 = Ft x Vg= 5.55 x 103 x 8.69= 48.19 K wattWhich is greater than capacity so the design is safe.Powe capacity of drive from wear point of view

P2 = Dg x b x W x Vg= 550 x 40 x W x 8.64W = .550 = Material combination factor.For worm and worm gear made up of hardened steel and phosphor bronzeSo, P2 = (550 / 1000) x 40 x .55 x 8.64= 104.4 KwattPower capacityof the drive, from the heat dissipation point of view is given by :P3 = 3650 x C17R + 5= 3650 x 300 1750 + 5= 361.99K wattSo The safe power capacity is minimum i.e P1 = 48.19 Kw

CONCLUSION : The design of the hoist of EOT crane is done Numerically .We can implement the design practically in industries for various lifting jobs.(For More Information Regarding This Topic Please Give Your Email id