Hoist Design Procedure for EOT crane…

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 trapezoidal
Selection of material : Mild steel
Load to lift : 2 tonne
Considering 50 % over loading.
So the design load = 2 tonne+50% of 2 tonne = 3tonne
Taking 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 mm
A = 2.75 C = 2.75 x 68 ≈ 187 mm
B = 1.31 C = 1.31 x 68 ≈ 89mm
D = 1.44 x C = 1.44 x 68 ≈ 98mm
E = 1.25C = 1.25 x 68 ≈ 85mm
F = C = 68mm
G = 35mm
G1 = M33, Pitch = 6mm (Coarse series)
H = 0.93 x C = 0.93 x 68 ≈ 63mm
J = 0.75 x C = 0.75 x 68 ≈ 51mm
K = 0.92 x C = 63mm
L = 0.7 x C = 0.7 x 68 ≈ 48mm
M = 0.6 x C = 0.6 x 68 ≈ 41mm
N = 1.2 x C = 82mm
P = 0.5 x C = 34mm ≈ 34mm
R = 0.5 x C = 0.5 x 68 ≈
U = 0.33 x C = 0.3 x 68≈20 mm
Checking for strength Area of the section = ½ x 63 x (41+8) = 1543.5 mm2
Centroid from ‘a’
= (.05 x 8 x 65) 63/3+(.5 x 41 x 63) x (2 x 63)/3
½ x 68 x (41+8)
= 38.571mm
= 38.6 mm= h2
So centroid from b = 63-38.6=24.4mm =h1
0 = 34 +24.4 = 58.4mm
r0 = A/(dA/u)
dA/u = [b2+r2/h (b1-b2)] ln r2/r1 – (b1-b2)
=28.65mm
r0 = A/(dA/u) = 1543.5 = 53.87  53.9 mm.
28.65
e= 0-r0 = 58.4 – 53.9 = 4.5mm

Moment
M = -P x 0
= -3 x 58.4
= – 175.2 (tonne x mm)
Stress due to bending is given by
b = M X 4
Ae r0-y
For point a
Y = -(e+h2)
= -(4.5+38.6)
= – 43.1 mm
For point b
Y = r0-r1
= 53.9 – 3.4
= 19.9 mm
Stress due to direct loading = P/A
= 3/1543.5
= 1.9436 x 10-3 Tonne/mm3
Stress due to curvature of ‘a’
ba = – (-175.2) x -43.1
1543.5 x 4.5 {53.99 – (-43.1)}
= 0.0112
So total tress at a
= – 0.0112 + 1.9436 x 10-3
= – 9.2642 x 10-3 Tonne/mm2
= – 9.2642 kg/mm2  -90.85 Mp
Stress due to curvature at b
bb = -(-175.2) x . 19.99 .
1543.5x 4.5 (53.9 – 19.9)
= 0.014763
So 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 MPa
Let the material be class 4 carbon steel ( 55C 8)
Ultimate tensil strength I 710MPa
Design strength = Ultimate tensil strength
Factor of safety
 = 710/4
 = 177.5 MPa
163.84 > 177.5
So design is safe
Determination of length of threaded portion
Pitch = 6mm
Nominal dia of thread = 33 mm (G1) = d
Considering the screw and thread are of single safest & square mean diameter of screw =
dm = d- (p/2)
= 33 – (6/2)
= 30
tan  = 1/dm = 6/( x 30)
tan  = tan-1 { 6/( x 30)} = 3.640
Let the co-efficient of fraction be 0.15
So  = tan  = 0.15
= 8.530
Torque required to resist the load

T = W x dm* tan ( + )
2
Where 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’)
So
T = 3050x 30 x tan (3.61+ 8.53)
2
=9866.42 Kgmm

Stress induced in the screw
Direct tensible stress (allowable or design)
 =4w/ d02
d0 = core diameter of the screw.
dc = d-p = 33-6 = 27 mm
1 = 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 screw
max = 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 know
Pi = 4W
 (d2-dc2) x n
So 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
 12.5
So the height of the nut is = 2 x 12.5 = 25mm.
b) Considering shear failure of thread across root
Shear stress induced
 = . W .
dc(0.5xP) xn

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

= 117.46
n
= 0.5 x 177.5 = 117.46
n n
= 117.46
177.5 x0.5
= 1.32 = 2
So height is n x p = 2 x 6 =12mm
Tacking 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 Y
I
= M x (24)
I 2
Where “I” is moment of inertial about bh3 t x 243 x 2
12 12
Maximum bending moment for = M = ¼ x W x L
Let L = 70 mm
So 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 24
1 t x 243 2
177.5 = 1363 x 2
t
t = 1363 x 2 = 7.8 x 2  15.2 MM
177.5
 16 mm
Taking 20 m for additional safety
Diameter 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 is
3050/2 =1525 kg  14.95 x 103
 15 KN
So 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 4
177.5 x 0.5 x 
= dk2 = 15 x 103 x 4
177.5 x 0.5 x 
= 215.19
= 14.66
 15mm
Taking 20 mm for additional safety purpose.
So that it can be turned to thread in size ‘m20’.
Height of the projected portion
P * x * dk = w/2
Where p = bearing pressure or crushing stress.
Let p = 210 MPx
Allow crushing stress =210/4 =
210/4 x x x 20 = 3050 x 9.8
2
x = 3050 x 9.8 x 5
2 x 20 x 210
= 3.56 mm x 4
= 14.024 =15 mm
So the minimum value of the height of the projected portion is 15 mm, taking x = 40 mm for
screwing 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 failure
Tearing 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 rivet
and 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 mm
Considering tensile failure at the lower cross section
So σd = w/2
A
177.5 = 3050 x 9.8/2
2 x x x 30

x = 3050 x 9.8/2
2 x 30 x 177.5

= 28/2 mm ≈ 1.4 mm

Considering crushing failure
Force = σcrushing(d) x projected area
3050 x 9.8 = σcrushing(d) x (20 x x)
FOS
= 3050 x 9.8 x 4 = 14.24 mm ≈ 15 mm
2 x 210 x 20
So 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 link
Is 2 x 2.5 + 2 x 2 = 9 kg.
So the total weight which the shafts carry is 3050 + 9 = 3059 ≈ 3060
Each Side subjected to a load of 3060/2 = 1530 kg.
The shaft is only subjected to
Crushing 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 MR
FOS 4
Stress induced = 1530 x 9.8 = 150042
π/4 x 302 π/4 x 302
= 21.22 MPa
As induced stress is less than that allowable stress, the design is safe

Considering crushing failure at the cover plate
Crushing stress = Force/Protected area
Allowable stress = 240/4 = 60MPa

Induced Crushing stress = Force
Projected area
= 3060 x 9.8
70 x 40
As allowable induced stress 60.86 KN. The selection is feasible.

Design of Sheave

sheave

a = 40mm
b = 30mm
c = 7mm
d = 18mm
e = 1mm
l = 10mm
r = 12mm
r4 = 8mm
h = 25mm
r1 = 4mm
r2 = 3mm
r3 = 12mm
Material is caste iron
Let dia of the sheave = 20 x d
= 20 x 14
= 240mm
The reference is made from Rudenko,N.’Materials Handling Equipment ‘,Mir Publishers, Moscow(1969).
P .86,Table 16.

Drum design
Drum 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) = 9mm
s1 = 1.15d = 17mm
C1 = 0.25 d = 5mm
No. of turn on each side of drum
Z = (hi/πd) + 2 =
Where
H = Lifting Height.
I = Ratio of the pulley system.
D = Drum diameter ≈ 45 x 14 = 630
So Z = ( 8.00 x 2) + 2
Π x63
Z = 10
Full 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 cm
The drum is made up of IS grade = SG 80/2
With stress = 480Mpa
W = 0.02 x 630 = 10
= 22.6
≈ 23 mm
Outside dia of the drum Do = D + 6d
= 630 + 6 x 14 = 714mm
Inside dia of the drum = Di = D – 2W
= 630 – 2 x 25
= 584 mm
Checking of strength
Bending Stress in drum
σbend = 8WLD
π(D4 – Di4)
= 8 x 6200 x 9.8 x 35 x 630
Π x (6304 – 5844)
= 0.0828MP

Maximun Torque
Tmax = W (D + d)
2
= 6200 x 9.8 (630 + 14)
2
= 19.564 x 106
Maximum Shear Stress
 = 16 Tmax D
π(D4 – Di4)
= 16 x 19.56 x 106 x 630
π x (6304 – 5844)
≈ 1.523 MPx
Direct Compressive Stress
= W = σc
wSi
= 6200 x 9.8
23 x 17
= 155.4MPx

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

σ < 480 MPx, S the Design is safe.

Fastening Of Rope With The Drum
For 14 mm dia
Locating Dimension
Pitch of screw = 53 mm
T = 43 mm
Screw size
Lo = 18 mm
L = 50 mm

Plate
C = 7 mm
No. of fastenings = 1

Selection of Motor
W = 3100 kg x 9.8 m/s2
V = 0.2m/sec
So power required = 3100 x 9.8 x 0.2
= 6080.123Nm/sec
= 6.08 Kwatt
Taking 11 Kwatt 3 phase induction motor (flange type) of 11 KW (nearest to 6.08KW)
Frame No. 132 M
Flange designation F265B
The speed is 1000 RPM
By using gear box the speed can be reduced to 300 RPM

GearBox Calculation
Gearbox 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 calculation
Required Brake Torque = 1.5 x 716.2 x mech H.P.
Motor RPM
= 1.5 x 716.2 x 50.12
300
= 53.84 Nm
Design of wormset
Power = P = 6.08 Kwatt
RPM of Worm = Nw = V x 60
πd
= 0.2 x 60
π x 0.63
RPM of Worm Gear NG = 300 RPM
So R = NG
Nw
= 300
6
Let Ф = 14.5o
Let the centre distance = C = 300 mm =0.3m
Pitch circle diameter of the worm
Dw = C.8750 = 3Pc
3.48
Dw = 3.8750 = 0.10020m
3.48

≈ 100mm
Dw = 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 mm
Velocity ratio = Ng/Nw = 50
Ng/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 = 550
Dw = 2C – Dg
= 2 x 300 -500
= 50mm
Face width of the gear
b = 0.73 x Dw
= 0.73 x 50
= 36.5
= 40 mm
Static strength of Bronze
σd = 90MPa
In 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 b
Velocity factor Cv = 6 .
6 + Vg
Vg = π x Dg x Ng
60
= 8.69
Cv = 6 = 0.41
6 + 8.69
Form factor
Y = 0.124 – 0.684
Ng
= 0.11132
Tan λ = m x Nw
Dw

= 11 x 1
50
λ = tan-1 (.22) = 12.4o

Nominal Module = m (ma) x cos λ
= 11 x cos 12.4
= 10.743 mm
Ft = 90 x .41 x π x 1o.743 x .11132 x 40
= 5545.47 N
= 5.55 KN
Power Capacity = P1 = Ft x Vg
= 5.55 x 103 x 8.69
= 48.19 K watt
Which 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 8.64
W = .550 = Material combination factor.
For worm and worm gear made up of hardened steel and phosphor bronze
So, P2 = (550 / 1000) x 40 x .55 x 8.64
= 104.4 Kwatt
Power capacityof the drive, from the heat dissipation point of view is given by :
P3 = 3650 x C17
R + 5
= 3650 x 300 17
50 + 5
= 361.99K watt
So 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.
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