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  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
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