Couplings, Springs & Flywheels Feb 2018 Rev 2 Presentation-5

Continuation…COUPLINGS MACHINE DESIGN & SHOP PRACTICE “Satisfaction lies in the effort, not the attainment. Full effort is full victory.” - Mahatma Gandhi COUPLINGS Couplings • Are used to connect sections of shafts or to connect the shaft of a driving machine to the shaft of a driven machine. ANALYSIS OF COUPLINGS COUPLING ELEMENT FAILURE MODE STRESS FORMULA 𝟐𝐓 Shearing of key 𝐒𝐬 = 𝐰𝐋𝐃𝐬 KEY 𝟒𝐓 Compression of key 𝐒𝐜 = 𝐭𝐋𝐃𝐬 𝟐𝐓 Shearing of bolts 𝐒𝐬 = 𝛑 𝟐 𝟒 𝐛 𝐧𝐛 𝐃𝐁𝐂 𝐝 BOLTS Compression 𝟐𝐓 between bolt and 𝐒𝐜 = 𝐝𝐛 𝐭 𝐟 𝐧𝐛 𝐃𝐁𝐂 flange 𝟐𝐓 FLANGE Punching shear 𝐒𝐬 = 𝛑𝐭 𝐟 𝐃𝐡 COUPLINGS Problem 22 Two shafts are connected by a flange coupling. The coupling is secured by 6 bolts, 20 mm in diameter on a pitch circle diameter of 150 mm. If torque of 120 N-m is applied, find the shear stress in the bolts. A. 0.85 N/mm2 C. 0.85 Pa B. 0.85 kPa D. 0.95 Pa Ans: 𝟎. 𝟖𝟓 𝐌𝐏𝐚 COUPLINGS Problem 23 A flanged coupling has an outside diameter of 200 mm and connects two 40 mm shafts. There are four 16 mm bolts on a 140 mm bolt circle. The radial flanged thickness is 20 mm. If the torsional stress in the shaft is not to exceed 26 MPa, determine the shear stress in the bolts if uniformly distributed A. 1.2 MPa C. 4.3 MPa B. 2.9 MPa D. 5.8 MPa Ans: 𝟓. 𝟖 𝐌𝐏𝐚 COUPLINGS Problem 24 Two short shafts having identical diameters of 38.1 mm and rotating at 400 rpm are connected by a flange coupling having 4 bolts with a 100 mm bolt circle. The design shearing stress of the bolt is 12 N/mm2 and the design compressive stress of the flange is 15 N/mm2 . How thick should the flange be in mm? A. 11.51 mm C. 12.49 mm B. 13.60 mm D. 15.65 mm Ans: 𝟏𝟏. 𝟓𝟏 𝐦𝐦 COUPLINGS Problem 25 A flange coupling is to be designed, using 25-mm diameter bolts at a distance of 152 mm from the center of the shaft. Allowable shearing stress on the bolt is 103 MPa. If the shaft is to transmit 5,800 hp at a speed of 1,200 rpm, how many bolts are needed in the connection? A. 2 C. 4 B. 3 D. 5 Ans: 𝟓 𝐛𝐨𝐥𝐭𝐬 SPRINGS, FLYWHEELS MACHINE DESIGN & SHOP PRACTICE “Satisfaction lies in the effort, not the attainment. Full effort is full victory.” - Mahatma Gandhi SPRINGS • A spring is defined as an elastic body, whose function is to distort when loaded and to recover its original shape when the load is removed. • To cushion, absorb or control energy due to either shock or vibration; • To apply forces, as in brakes, clutches and spring- loaded valves; • To control motion by maintaining contact between two elements; • To measure forces; • To store energy. TERMS IN COMPRESSION SPRINGS • Solid Length • When the compression spring is compressed until the coils come in contact with each other. 𝐋𝐬𝐨𝐥𝐢𝐝𝐬 = 𝐧𝐭 𝐝𝐰 • Free Length • The length of the spring in the free or unloaded condition. 𝐋𝐟𝐫𝐞𝐞 = 𝐧𝐭 𝐝𝐰 + 𝛅𝐦𝐚𝐱 + 𝐧𝐭 − 𝟏 𝐱 𝟏𝐦𝐦 𝐋𝐟𝐫𝐞𝐞 = 𝐧𝐭 𝐝𝐰 + 𝛅𝐦𝐚𝐱 + 𝟎. 𝟏𝟓𝛅𝐦𝐚𝐱 TERMS IN COMPRESSION SPRINGS • Spring index. The spring index is defined as the ratio of the mean diameter of the coil to the diameter of the wire. 𝐃𝐦𝐞𝐚𝐧 𝐃𝐨 − 𝐝𝐰 𝐂= = 𝐝𝐰 𝐝𝐰 • Spring rate. The spring rate (or stiffness or spring constant) is defined as the load required per unit deflection of the spring. 𝐖 𝐅 𝐤= = 𝛅 ∆𝐱 TERMS IN COMPRESSION SPRINGS • Pitch. The pitch of the coil is defined as the axial distance between adjacent coils in uncompressed state. 𝐋𝐟𝐫𝐞𝐞 𝐋𝐟𝐫𝐞𝐞 − 𝐋𝐬𝐨𝐥𝐢𝐝 𝐩= = + 𝐝𝐰 𝐧𝐭 − 𝟏 𝐧𝐭 TERMS IN COMPRESSION SPRINGS End Connections for Compression Helical Springs End type 𝐧𝐭𝐨𝐭𝐚𝐥 𝐋𝐬𝐨𝐥𝐢𝐝 𝐋𝐟𝐫𝐞𝐞 Plain ends n (n+1) dw np + dw Ground ends n ndw np Squared ends n+2 (n+3)dw np + 3dw Squared & Ground ends n+2 (n+2)dw np + 2dw ANALYSIS OF COMPRESSION SPRINGS F F dw F Dm Dm F ANALYSIS OF COMPRESSION SPRINGS Maximum Torsional, Direct Shear Stress & Curvature of wire 𝟖𝐅𝐃𝐦 𝟖𝐅𝑪 𝐒𝐬 = 𝐊 𝐰 = 𝐊𝐰 𝟐 𝛑𝐝𝟑𝐰 𝛑𝐝𝐰 Where: 𝟒𝐂−𝟏 𝟎.𝟔𝟏𝟓 𝐊𝐰 = + 𝟒𝐂−𝟒 𝐂 𝐊𝐰 = 𝐊𝐒𝐊𝐂 Note: 𝐊 𝐬 − 𝐒𝐡𝐞𝐚𝐫 𝐬𝐭𝐫𝐞𝐬𝐬 𝐟𝐚𝐜𝐭𝐨𝐫 𝐊 𝒄 − 𝐬𝐭𝐫𝐞𝐬𝐬 𝐜𝐨𝐧𝐜𝐞𝐧𝐭𝐫𝐚𝐭𝐢𝐨𝐧 𝐟𝐚𝐜𝐭𝐨𝐫 𝐝𝐮𝐞 𝐭𝐨 𝐜𝐮𝐫𝐯𝐚𝐭𝐮𝐫𝐞 ANALYSIS OF COMPRESSION SPRINGS Deflection of Helical Springs of Circular wire 𝟖𝐅𝐃𝟑𝐦 𝐧 𝟖𝐅𝐂 𝟑 𝐧 𝛅= 𝟒 = 𝐆𝐝𝐰 𝐆𝐝𝐰 ANALYSIS OF COMPRESSION SPRINGS Energy Stored in a Helical Spring or Circular Wire 𝟏 𝟏 𝟏 𝟐 𝐄 = 𝐅𝛅 = 𝐅∆𝐱 = 𝐤 ∆𝐱 𝟐 𝟐 𝟐 Note : When a load (say P) falls on a spring through a height h, then the energy absorbed in a spring is given by: Work done by the Load P = Energy absorbed by the Spring 𝟏 𝟐 𝐏 𝐡 + 𝛅 = 𝐤𝛅 𝟐 ANALYSIS OF COMPRESSION SPRINGS Springs in Series 𝛅 = 𝛅𝟏 + 𝛅𝟐 𝟏 𝟏 𝟏 = + 𝐤 𝐤𝟏 𝐤𝟐 Springs in Parallel 𝐅 = 𝐅𝟏 + 𝐅𝟐 𝐤 = 𝐤𝟏 + 𝐤𝟐 SPRINGS • Leaf springs (also known as flat springs) are made out of flat plates. • The advantage of leaf spring over helical spring is that the ends of the spring may be guided along a definite path as it deflects to act as a structural member in addition to energy absorbing device. • Thus the leaf springs may carry lateral loads, brake torque, driving torque etc., in addition to shocks. SPRINGS ANALYSIS OF LEAF SPRINGS ANALYSIS OF LEAF SPRINGS Bending Stress on Leaf Springs 𝟔𝐅𝐋 𝐒𝐛 = 𝐧𝐬 𝐛𝐭 𝟐 Deflection of Leaf Springs Multiple Leaves (same Length) 𝟒𝐅𝐋𝟑 𝛅= 𝐧𝐬 𝐄𝐛𝐭 𝟑 Multiple Leaves (graduated Length) 𝟔𝐅𝐋𝟑 𝐒𝐛 𝐋𝟐 𝛅= 𝟑 = 𝐧𝐠 𝐄𝐛𝐭 𝐄𝐭 ANALYSIS OF LEAF SPRINGS Bending Stress on Leaf Springs Multiple Leaves (diff sizes) 𝟏𝟖𝐅𝐋 𝐒𝐬 = 𝐛𝐭 𝟐 𝟐𝐧𝐠 + 𝟑𝐧𝐟 Deflection of Leaf Springs Multiple Leaves (diff sizes) 𝟏𝟐𝐅𝐋𝟑 𝐒𝐬 = 𝐄𝐛𝐭 𝟑 𝟐𝐧𝐠 + 𝟑𝐧𝐟 SPRINGS Problem 1 A coiled spring with 1¾ in. outside diameter is required to work under load of 140 lb. Wire diameter used is 0.192 in., spring is to have seven active coils, and the ends are to be closed and ground. Determine the safe shearing stress and the unit deflection. Assume G equal to 12 million psi and mean radius to be 0.779 in. Ans: 𝟏. 𝟖𝟏𝟕 𝐢𝐧 SPRINGS Problem 2 A high alloy spring having squared and ground ends and has a total of 16 coils and modulus of elasticity in shear of 85 GPa. Compute the Wahl factor. The spring outside diameter is 9.66 cm and wire diameter is 0.65 cm. Ans: 𝟏. 𝟏𝟎 SPRINGS Problem 3 Compute the maximum deflection of a 22 coils helical spring having a load of 120 kgs. The spring is squared and ground ends with modulus of elasticity in shear of 80 GPa, outside diameter of 8 cm and wire diameter of 8 mm. Ans: 𝟐𝟏𝟒. 𝟓 𝐦𝐦 SPRINGS Problem 4 The load on a helical spring is 1600-lb and the corresponding deflection is to be 4-in. Rigidity modulus is 11 million psi and the maximum intensity of safe torsional stress is 60,000 psi. Design the spring for the total number of turns if the wire is circular in cross section with a diameter of 5/8 in. and a center line radius of 1 1/2 in. The spring is squared and ground ends. Ans: 𝟐𝟐 𝐜𝐨𝐢𝐥𝐬 SPRINGS Problem 5 All four compression coil spring support one load of 700 kg/mm. All four springs are arranged in parallel and rated same at 0.609 kg/mm. Compute the deflection in mm. Ans: 𝟐𝟖𝟕 𝐦𝐦 SPRINGS Problem 6 A Mercedes Benz gas engine valve spring is to have a mean diameter of 3.81 cm. The maximum load it will have to sustain is 45.45 kg with corresponding deflection of 1.27 cm. The spring is to be made of tempered steel wire. Since the material is to be subjected to repeated loading, and fatigue must be considered, a low working stress of 2280 kg/sq.cm. will be used. Determine the size of wire. Ans: 𝟎. 𝟔𝟑𝟓 𝐜𝐦 SPRINGS Problem 7 A body weighing 1000 lb falls 6 inches and strikes a 2000 lb per in spring. The deformation of the spring is: Ans: 𝟑 𝐢𝐧 SPRINGS Problem 8 Determine the width and thickness of the leaves of a six-leaf steel cantilever spring 13 in. long to carry a load of 375 lb with a deflection of 1¼ in. The maximum stress in stress in this spring is limited to 50,000 psi. Ans: 𝐭 = 𝟎. 𝟐𝟐𝟓 𝐢𝐧, 𝐰 = 𝟏. 𝟗𝟑 𝐢𝐧 FLYWHEELS • A flywheel used in machines serves as a reservoir which stores energy during the period when the supply of energy is more than the requirement and releases it during the period when the requirement of energy is more than supply FLYWHEELS Coefficient of Fluctuation of Speed • The ratio of the maximum fluctuation of speed to the mean speed. 𝐍𝟏 − 𝐍𝟐 𝟐 𝐍𝟏 − 𝐍𝟐 𝐂𝐟 = = 𝐍𝐦 𝐍𝟏 + 𝐍𝟐 𝛚𝟏 − 𝛚𝟐 𝟐 𝛚𝟏 − 𝛚𝟐 𝐂𝐟 = = 𝛚𝐦 𝛚𝟏 + 𝛚𝟐 𝐯𝟏 − 𝐯𝟐 𝟐 𝐯𝟏 − 𝐯𝟐 𝐂𝐟 = = Note: 𝐯𝐦 𝐯𝟏 + 𝐯𝟐 𝐍𝟏 + 𝐍𝟐 𝐍𝐦 = 𝟐 ANALYSIS OF FLYWHEELS ANALYSIS OF FLYWHEELS Energy Stored by the Flywheel (Kinetic Energy) 𝟏 𝟐 𝐊𝐄 = 𝐈𝛚𝐦 𝟐 ∆𝐊𝐄 = 𝐈𝛚𝟐𝐦 𝐂𝐟 = 𝐦𝐤 𝟐 𝛚𝟐𝐦 𝐂𝐟 = 𝐦𝐂𝐟 𝐯𝐦 𝟐 𝟏 𝟐 𝟐 𝟏 𝐖𝐅 𝟐 ∆𝐊𝐄 = 𝐦 𝐯𝟏 − 𝐯𝟐 = 𝐯𝟏 − 𝐯𝟐𝟐 𝟐 𝟐 𝐠 Where: WF = WArms + WHub + WRim 𝐖𝐑𝐢𝐦 = 𝛄𝐕𝐑𝐢𝐦 = 𝛄 𝛑𝐃𝐦 𝐭𝐛 ANALYSIS OF FLYWHEELS Energy needed to punch a hole Power Requirement 𝟏 𝐄 𝐄 = 𝐅𝐦 𝐭 𝐩 = 𝐅𝐭 𝐩 𝐏= 𝟐 𝐭𝐢𝐦𝐞 From: 𝐅 𝐅 𝐒𝐬𝐮 = = 𝐀𝐬 𝛑𝐝𝐡 𝐭 𝐩 𝟏 𝐄 = 𝐒𝐬𝐮 𝛑𝐝𝐡 𝐭 𝟐𝐩 𝟐 ANALYSIS OF FLYWHEELS Stresses induced in Flywheel arms Tensile Stress 𝟐 𝐒𝐭 = 𝛄𝐯𝐦 Bending Stress 𝐌 𝐓 𝐒𝐛 = = 𝐑−𝐫 𝐙 𝐑𝐧𝐚𝐫𝐦𝐬 𝐙 ANALYSIS OF FLYWHEELS From Machinery’s Handbook Pressure (Force) Required for Punching • General Equation Note: Steel/Iron – 50,000 psi 𝐏 = 𝑨𝒔 𝐒𝐬 = 𝛑𝐝𝐡 𝐭 𝐩 𝐒𝐬 Bronze – 40,000 psi Copper – 30, 000 psi • Steel Material Aluminum – 20, 000 psi Zinc – 10, 000 psi 𝐏 = 𝐝𝐡 𝐭 𝐩 × 𝟖𝟎 Tin/lead – 5000 psi • Brass Material Conversion: 1 ton = 2000 lbs 𝐏 = 𝐝𝐡 𝐭 𝐩 × 𝟔𝟓 Note: (For non- circular hole) 𝟏 𝐝𝐡 = 𝐏𝐡 𝟑 FLYWHEEL Problem 9 A flywheel weighing 457 kg has radius of 375 mm. How much energy in N -m does the flywheel loss from 3.0 rev/sec to 2.8 rev/sec? Ans: 𝟏𝟒𝟕𝟐 𝐍 − 𝐦 FLYWHEEL Problem 10 A flywheel weighing 6 kN has a mean diameter of 90 cm. The maximum speed of the flywheel is 7 rev/sec slowed down to 4 rev/sec during the shearing process. Determine the energy released by the flywheel. Ans: 𝟖𝟐𝟒𝟏 𝐤𝐠 − 𝐦 FLYWHEEL Problem 11 A flywheel rotates at 220 rpm slowed down to 65 % of its revolution during the three – fourth second punching portion of the cycle. Compute the angular acceleration of the flywheel in radian per second squared. Ans: −𝟏𝟎. 𝟕𝟓𝟏 𝐫𝐚𝐝/𝐬 𝟐 FLYWHEEL Problem 12 A flywheel has a diameter of 1.5 m, and mass of 800 kg. What torque in N-m, is needed to produce an angular acceleration of 100 revolutions per minute, per second? Ans: 𝟐𝟑𝟓𝟔 FLYWHEEL Problem 13 A cast iron flywheel with a mean diameter of 36 inches changes speed from 300 rpm to 280 rpm while it gives up 8000 ft-lb of energy. What is the coefficient of fluctuation. Ans: 𝟎. 𝟎𝟔𝟗 FLYWHEEL Problem 14 What would be the weight of a flywheel in kg if the weight of the rim is 3 times the sum of the weight of the hub and arms. Given the outside diameter and inside diameter to be 24 in and 18 in respectively and the rim width is 4.5 in. (assume steel flywheel) Ans: 𝟏𝟓𝟐. 𝟗𝟓 𝐤𝐠 FLYWHEEL Problem 15 A flywheel has a mean diameter of 4 ft and is required to handle 2250 ft-lb of kinetic energy. The flywheel has a width of 8 in. Normal operating speed is 300 rpm and the coefficient of fluctuation is to be 0.05. Find the weight of the rim assuming that the arms and hub are equivalent to 10% of the rim weight. Ans: 𝟑𝟑𝟑. 𝟔𝟗 𝐥𝐛𝐬 FLYWHEEL Problem 16 A shearing machine requires 150 kg-m of energy to shear a steel sheet, and has a normal speed of 3 rev/sec, slowing down to 2.8 rev/sec during the shearing process. The flywheel of the machine has a mean diameter of 75 cm and weighs 0.0155 kg/cm3. The width of the rim is 30 cm. If the hub and arms of the flywheel account for 15% of its total weight, find the weight of the flywheel. Ans: 𝟒𝟓𝟕 𝐤𝐠 FLYWHEEL Problem 17 A punch punches a 1-in diameter hole in a steel plate ¾ inch thick every 10 sec. The actual punching takes 1 sec. The ultimate shear strength of the plate is 60,000 psi. The flywheel of the punch press has a mass moment of inertia of 500 in-lb-sec2 and rotates at a mean speed of 150 rpm. What is the horsepower required for the punch operation? Ans: 𝟖. 𝟎𝟒 𝐡𝐩 FLYWHEEL Problem 18 What pressure is required for punching a hole 3 inches in diameter through 1/4 inch steel stock? Ans: 𝟔𝟎 𝐭𝐨𝐧𝐬 FLYWHEEL Problem 19 What pressure is required for punching a 1.5 inch square hole in a 1/4 inch thick plate steel? Ans: 𝟒𝟎 𝐭𝐨𝐧𝐬 FLYWHEEL Problem 20 What pressure is required for punching a 1.5 inch by 2 inch rectangular hole in a 1/4 inch thick brass? Ans: 𝟑𝟕. 𝟗𝟐 𝐭𝐨𝐧𝐬