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 hookcrosssection 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 38151969) for selection of material for 8 dimensions of crane hook.In IS 38151969 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 = 6338.6=24.4mm =h1
0 = 34 +24.4 = 58.4mm r0 = A/(dA/u) dA/u = [b2+r2/h (b1b2)] ln r2/r1
–
(b1b2) =28.65mm r0 = A/(dA/u) = 1543.5 = 53.87
53.9 mm. 28.65 e=
0r0 = 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 r0y For point a Y = (e+h2) = (4.5+38.6) =
–
43.1 mm
For point b Y = r0r1 = 53.9
–
3.4 = 19.9 mm Stress due to direct loading = P/A = 3/1543.5 = 1.9436 x 103 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 103 =
–
9.2642 x 103 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 x103 =0.014763 + 1.9436 x 103 =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
= tan1 { 6/(
x 30)} = 3.640 Let the coefficient 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 = dp = 336 = 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
(d2dc2) x n So 6 = 4×3050 x 9.8
(332272) x n N = 4 x 3050 x 9.8
(332272) 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