Machine Design Notes PDF

# Classification of Machine Design - **Adaptive Design:** This type of design requires special knowledge or skill. Designers only make minor modifications to existing designs of products. - **Development Design:** This type of design requires scientific training and design ability. Designers start...

# Classification of Machine Design - **Adaptive Design:** This type of design requires special knowledge or skill. Designers only make minor modifications to existing designs of products. - **Development Design:** This type of design requires scientific training and design ability. Designers start with the existing design to modify it and arrive at a final product design. - **New Design:** This type of design requires research, technical ability, and creative thinking. There are several methods for creating new designs: - Rational design - Industrial design - System design - Element design - Computer-aided design # Factors Governing the Design of Machine Elements - **Strength:** The machine component should not fail under external forces acting on it. It should have sufficient strength to avoid failure due to yielding, which is called strength. - **Rigidity:** The machine component should be rigid enough so that it will not deflect or bend due to forces or moments acting on it. A transmission shaft is designed based on lateral rigidity and torsional rigidity. - **Wear Resistance:** A machine component should be wear resistant because wear reduces the accuracy of the machine tool as well as its life cycle. Surface hardening increases the wear resistance of the machine component. - **Safety:** The shape and dimension of the machine part should ensure safety for the operator of the machine. - **Minimum Dimension and Weight:** A machine part should have the minimum possible dimension and weight, which will reduce material cost. - **Conformance to Standards:** Machine parts should conform to national and international standards covering dimensions, profiles, and materials. - **Minimum Cycle Cost:** The total cost that is paid for purchasing, operating, and maintaining the machine for its lifespan should be minimum. # Design Procedure of Machine Elements The general procedure of design is as follows: 1. **Recognition of Need:** First of all, recognize the need or purpose for which the machine is to be designed. 2. **Mechanism:** Select a possible mechanism or group of mechanisms that will give the desired motion. 3. **Analysis of Forces:** Find the forces acting on each member of the machine and the energy transmitted by each member. 4. **Material Selection:** Select the material best suited for each member of the machine. 5. **Design of Elements:** Find the size of each member of the machine by considering the forces acting on it and the permissible stresses for the material used. 6. **Modification:** Modify the sizes of members to reduce overall cost. 7. **Detailed Drawing:** Draw the detailed drawing of the components and assembly of the machine with complete specifications for the manufacturing process. # Advantages of Welded Joints - The welded joint structure is usually lighter than other structures. - Welded joints provide maximum efficiency, which is not possible in the case of other joints. - Alterations or additions can be easily made to existing structures. - In welded connections, the tension member cannot weaken. - The welded joint has a great strength. - The welded joint provides very rigid joints. - It is possible to weld any part of the structure at any point. - The process of welding takes less time than other joints. # Disadvantages of Welded Joints - Requires highly skilled labor and supervision. - Inspecting welded work is more difficult than other work. - Welded joints can develop cracks over time. - Welded joints are harder to remove, replace, or repair than fasteners. # Failure of Riveted Joints - **Tearing of the Plate at an Edge:** The joint may fail due to tearing of the plate at an edge. This can be avoided by keeping m >1.5d / l, where m = margin or margin pitch, l = diameter of rivet, d = pitch length. - **Tearing of the Plate Across Rows of Rivets:** The plate can tear across rows of rivets due to tensile stresses in the main plate or cover plate. The resistance offered by the plate against tearing is known as tearing resistance or tearing strength. - **Shearing of the Rivet:** The resistance offered by the rivet to being sheared is known as shearing resistance or shearing strength. - **Crushing of the Rivet to Plates:** The resistance offered by the rivet to be crushed is known as crushing resistance or crushing strength. # Assumptions in Designing of Boiler Joints - The load on the joints is equally sheared by all rivets. - The tensile stress is evenly distributed over the section being the rivets. - The shearing stress in all rivets is uniform. - The crushing stress is uniform. - There is no bending stress in the rivet. - The friction between the rivet and plate is neglected. - The hole into which the rivet is driven does not weaken the member. - The rivet fills in the hole after it is driven. # Determining the Diameter of a Solid Shaft A solid shaft transmits 1 kW at 240 rpm. Determine the diameter of the shaft if the maximum torque transmitted exceeds the mean torque by 20%. Take the maximum allowable shear stress as 60 MPa. **Given Data:** - Speed of shaft N = 240 rpm - Power P = 1 kW = 1000 W - τ = 60 MPa = 60 N/mm² **Calculations:** - Let d = diameter of the solid shaft - T_max exceeds the mean torque by 20% - Therefore, T_max = 1.2 T_mean - We know P = 2πNT_mean/60 - T_mean = (P * 60) / (2πN) - T_mean = (1000 * 60) / (2 * π * 240) = 39788.73 N-mm = 39.78873 N-m - T_max = 1.2 * 39.78873 N-m = 47.746476 N-m - We also know that T = (π/16) * τ * d^3 - T_max = (π/16) * τ * d^3 - d^3 = (16 * T_max) / (π * τ) - d = ³√((16 * T_max) / (π * τ)) - d = ³√((16 * 47.746476 N-m) / (π * 60 N/mm²)) - d = 73.85 mm ≈ 74 mm # Determining the Maximum Shear Stress in a Helical Spring Consider a compression helical spring made of alloy steel with the following specifications: - Mean diameter of coil (D) = 50 mm - Diameter of wire (d) = 6 mm - Number of active coils (n) = 20 - Axial load (W) = 500 N Calculate the maximum shear stress to which the spring material is subjected. **Given data:** - D = 50 mm - d = 6 mm - n = 20 - W = 500 N **Calculations:** - Spring index (C) = D/d = 50/6 = 8.33 - Shear stress factor (k_s) = 1 + 1/4C = 1 + 1/(4 * 8.33) = 1.03 - Maximum shear stress (τ_max) = k_s * (8WD) / (πd³) = 1.03 * (8 * 500 N * 50 mm) / (π * (6 mm)³) = 534.76 N/mm² = 534.76 MPa.