Question

Design a gearbox for lifting a weight of W - 1500 Kg with a pulley as shown in figure, the diameter of the pulley is the designer decision. The speed of lifting weight should be 40 meter/min. The design has to be compact with the least manufacturing cost. Pulley Gearbox Design Domain Motor Hex Coupler Input Shaft Output Shaft 3000 kg Shafts 1. You must also be able to draw the shear force and bending moment diagrams, as well as drawing the free-body diagrams and calculating the reactions at the supports. Since the shafts must be able to accommodate all the elements within the range of the supports, you must assign proper length dimensions for each of the shafts, considering the arrangement of the elements and allowing space for the width of each bearing. 2. From the bending moment diagrams obtained in task # 4, and after calculating the torque transmitted by each one of the shafts, you will select materials for the shafts and design the shafts. A factor of safety of 2.0 seems to be appropriate. Note that the diameter of the shaft will not be constant. There will be some keyways and/or shoulders on each of the shaft creating stress concentration, which need to be considered. 3. reliability requirements should all be considered. The life of the bearings should be at least 10$ cycles for this gearbox.

Design a gearbox for lifting a weight of W = 1500 Kg with a pulley as shown in figure, the diameter of the pulley is the designer decision. The speed of lifting weight should be 40 meter/min. The design has to be compact with the least manufacturing cost.

Shafts

1. You must also be able to draw the shear force and bending moment diagrams, as well as drawing the free-body diagrams and calculating the reactions at the supports. Since the shafts must be able to accommodate all the elements within the range of the supports, you must assign proper length dimensions for each of the shafts, considering the arrangement of the elements and allowing space for the width of each bearing.
2. From the bending moment diagrams obtained in task # 4, and after calculating the torque transmitted by each one of the shafts, you will select materials for the shafts and design the shafts. A factor of safety of 2.0 seems to be appropriate. Note that the diameter of the shaft will not be constant. There will be some keyways and/or shoulders on each of the shaft creating stress concentration, which need to be considered.

Answer 1

From the given load and the velocity, the required power can be obtained. The load in the figure is given as 3000kg while in the problem it is given as 1500kg. All calculations are done assuming the load to be 3000kg. Once we get the power as the product of load and velocity, we assume a radius for pulley. The radius has to be large so as to generate enough torque to raise the load easily. Also if the radius of pulley is large, it means that the angular velocity will be less for the pulley. The parameters such as angular velocity, torque can be obtained once a proper value of radius of pulley is assumed. The material choosen is AISI 1040 for the shaft. The factor of safety is taken as 2 and the shaft diameter is obtained. Since no information about the length of the shaft was given, so the bending moment can't be calculated. Shaft is designed purely on the basis of torsion. For 10^8 cycle, the bearing rated load is determined from which an appropriate bearing can be selected. The Free body diagram representing the torque load is presented.

Power Transmittel (= 40m/min 40 60 m/s = 2 m/s 3 او در سر v=Rw Radius of Palley of Pulley to be P= (3000x 9+81) * Assuming 6)-(393.33). R=333.33 ym Pe 19620w = 19.62 KW w = 2 radis For a factor of Safety of 2 T = S 1415 Material AISI 1040 N=19 RPM St = 415 Mia Tonque = W.R = 3000 x 9.81 x/333-333 (333.383) Torque = 9810 Nm = 207.5= mesmo portato No Information about length of Shaft is given, Bencing Moment is tartan Zeno assuming langth to be Swall. (207-5) = 16 (9810x1000) => d = 62-2119 m. Since 1 ιία Diameter of Qutput shaft)

10 cycles Revolutiow g Lt - 100 mi millions N = 19 KPM . 60 Nhp L می روم P = x + y f End ho => 4 = 87719-29€ (Life in Hoera.) (19860) Output F. B.D of. Shaft Gear Box End t=0 (Axial force on Bearing is zehe. ) &= 3000x9.81= 29430N = 29.43 KN ( Radial force on Bearing) x = 1. Pulley P=29.43 KN. 1013 1013 100 - L = (3/4) → (Assering Roller Jeanning) (22.42 e = 17162.9 + (Required Rating for Bearing)