ContinuationโฆCOUPLINGS MACHINE DESIGN & SHOP PRACTICE โSatisfaction lies in the effort, not the attainment. Full effort is full victory.โ - Mahatma Gandhi COUPLINGS Couplings โข 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 ๐๐ Shearing of key ๐๐ฌ = ๐ฐ๐๐๐ฌ KEY ๐๐ Compression of key ๐๐ = ๐ญ๐๐๐ฌ ๐๐ Shearing of bolts ๐๐ฌ = ๐ ๐ ๐ ๐ ๐ง๐ ๐๐๐ ๐ BOLTS Compression ๐๐ between bolt and ๐๐ = ๐๐ ๐ญ ๐ ๐ง๐ ๐๐๐ flange ๐๐ FLANGE Punching shear ๐๐ฌ = ๐๐ญ ๐ ๐๐ก 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: ๐. ๐๐ ๐๐๐ 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: ๐. ๐ ๐๐๐ 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: ๐๐. ๐๐ ๐ฆ๐ฆ 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: ๐ ๐๐จ๐ฅ๐ญ๐ฌ SPRINGS, FLYWHEELS MACHINE DESIGN & SHOP PRACTICE โSatisfaction lies in the effort, not the attainment. Full effort is full victory.โ - Mahatma Gandhi SPRINGS โข 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. โข To cushion, absorb or control energy due to either shock or vibration; โข To apply forces, as in brakes, clutches and spring- loaded valves; โข To control motion by maintaining contact between two elements; โข To measure forces; โข To store energy. TERMS IN COMPRESSION SPRINGS โข Solid Length โข When the compression spring is compressed until the coils come in contact with each other. ๐๐ฌ๐จ๐ฅ๐ข๐๐ฌ = ๐ง๐ญ ๐๐ฐ โข Free Length โข The length of the spring in the free or unloaded condition. ๐๐๐ซ๐๐ = ๐ง๐ญ ๐๐ฐ + ๐ ๐ฆ๐๐ฑ + ๐ง๐ญ โ ๐ ๐ฑ ๐๐ฆ๐ฆ ๐๐๐ซ๐๐ = ๐ง๐ญ ๐๐ฐ + ๐ ๐ฆ๐๐ฑ + ๐. ๐๐๐ ๐ฆ๐๐ฑ TERMS IN COMPRESSION SPRINGS โข Spring index. The spring index is defined as the ratio of the mean diameter of the coil to the diameter of the wire. ๐๐ฆ๐๐๐ง ๐๐จ โ ๐๐ฐ ๐= = ๐๐ฐ ๐๐ฐ โข Spring rate. The spring rate (or stiffness or spring constant) is defined as the load required per unit deflection of the spring. ๐ ๐ ๐ค= = ๐ โ๐ฑ TERMS IN COMPRESSION SPRINGS โข Pitch. The pitch of the coil is defined as the axial distance between adjacent coils in uncompressed state. ๐๐๐ซ๐๐ ๐๐๐ซ๐๐ โ ๐๐ฌ๐จ๐ฅ๐ข๐ ๐ฉ= = + ๐๐ฐ ๐ง๐ญ โ ๐ ๐ง๐ญ TERMS IN COMPRESSION SPRINGS End Connections for Compression Helical Springs End type ๐ง๐ญ๐จ๐ญ๐๐ฅ ๐๐ฌ๐จ๐ฅ๐ข๐ ๐๐๐ซ๐๐ 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 ๐๐ ๐๐ฆ ๐๐ ๐ช ๐๐ฌ = ๐ ๐ฐ = ๐๐ฐ ๐ ๐๐๐๐ฐ ๐๐๐ฐ Where: ๐๐โ๐ ๐.๐๐๐ ๐๐ฐ = + ๐๐โ๐ ๐ ๐๐ฐ = ๐๐๐๐ Note: ๐ ๐ฌ โ ๐๐ก๐๐๐ซ ๐ฌ๐ญ๐ซ๐๐ฌ๐ฌ ๐๐๐๐ญ๐จ๐ซ ๐ ๐ โ ๐ฌ๐ญ๐ซ๐๐ฌ๐ฌ ๐๐จ๐ง๐๐๐ง๐ญ๐ซ๐๐ญ๐ข๐จ๐ง ๐๐๐๐ญ๐จ๐ซ ๐๐ฎ๐ ๐ญ๐จ ๐๐ฎ๐ซ๐ฏ๐๐ญ๐ฎ๐ซ๐ ANALYSIS OF COMPRESSION SPRINGS Deflection of Helical Springs of Circular wire ๐๐ ๐๐๐ฆ ๐ง ๐๐ ๐ ๐ ๐ง ๐ = ๐ = ๐๐๐ฐ ๐๐๐ฐ ANALYSIS OF COMPRESSION SPRINGS Energy Stored in a Helical Spring or Circular Wire ๐ ๐ ๐ ๐ ๐ = ๐ ๐ = ๐ โ๐ฑ = ๐ค โ๐ฑ ๐ ๐ ๐ 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 ๐ ๐ ๐ ๐ก + ๐ = ๐ค๐ ๐ ANALYSIS OF COMPRESSION SPRINGS Springs in Series ๐ = ๐ ๐ + ๐ ๐ ๐ ๐ ๐ = + ๐ค ๐ค๐ ๐ค๐ Springs in Parallel ๐ = ๐ ๐ + ๐ ๐ ๐ค = ๐ค๐ + ๐ค๐ SPRINGS โข Leaf springs (also known as flat springs) are made out of flat plates. โข 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. โข 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 ๐๐ ๐ ๐๐ = ๐ง๐ฌ ๐๐ญ ๐ Deflection of Leaf Springs Multiple Leaves (same Length) ๐๐ ๐๐ ๐ = ๐ง๐ฌ ๐๐๐ญ ๐ Multiple Leaves (graduated Length) ๐๐ ๐๐ ๐๐ ๐๐ ๐ = ๐ = ๐ง๐ ๐๐๐ญ ๐๐ญ ANALYSIS OF LEAF SPRINGS Bending Stress on Leaf Springs Multiple Leaves (diff sizes) ๐๐๐ ๐ ๐๐ฌ = ๐๐ญ ๐ ๐๐ง๐ + ๐๐ง๐ Deflection of Leaf Springs Multiple Leaves (diff sizes) ๐๐๐ ๐๐ ๐๐ฌ = ๐๐๐ญ ๐ ๐๐ง๐ + ๐๐ง๐ SPRINGS Problem 1 A coiled spring with 1ยพ 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: ๐. ๐๐๐ ๐ข๐ง 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: ๐. ๐๐ 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: ๐๐๐. ๐ ๐ฆ๐ฆ 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: ๐๐ ๐๐จ๐ข๐ฅ๐ฌ 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: ๐๐๐ ๐ฆ๐ฆ 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: ๐. ๐๐๐ ๐๐ฆ 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: ๐ ๐ข๐ง 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 1ยผ in. The maximum stress in stress in this spring is limited to 50,000 psi. Ans: ๐ญ = ๐. ๐๐๐ ๐ข๐ง, ๐ฐ = ๐. ๐๐ ๐ข๐ง FLYWHEELS โข 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 โข The ratio of the maximum fluctuation of speed to the mean speed. ๐๐ โ ๐๐ ๐ ๐๐ โ ๐๐ ๐๐ = = ๐๐ฆ ๐๐ + ๐๐ ๐๐ โ ๐๐ ๐ ๐๐ โ ๐๐ ๐๐ = = ๐๐ฆ ๐๐ + ๐๐ ๐ฏ๐ โ ๐ฏ๐ ๐ ๐ฏ๐ โ ๐ฏ๐ ๐๐ = = Note: ๐ฏ๐ฆ ๐ฏ๐ + ๐ฏ๐ ๐๐ + ๐๐ ๐๐ฆ = ๐ ANALYSIS OF FLYWHEELS ANALYSIS OF FLYWHEELS Energy Stored by the Flywheel (Kinetic Energy) ๐ ๐ ๐๐ = ๐๐๐ฆ ๐ โ๐๐ = ๐๐๐๐ฆ ๐๐ = ๐ฆ๐ค ๐ ๐๐๐ฆ ๐๐ = ๐ฆ๐๐ ๐ฏ๐ฆ ๐ ๐ ๐ ๐ ๐ ๐๐ ๐ โ๐๐ = ๐ฆ ๐ฏ๐ โ ๐ฏ๐ = ๐ฏ๐ โ ๐ฏ๐๐ ๐ ๐ ๐ Where: WF = WArms + WHub + WRim ๐๐๐ข๐ฆ = ๐๐๐๐ข๐ฆ = ๐ ๐๐๐ฆ ๐ญ๐ ANALYSIS OF FLYWHEELS Energy needed to punch a hole Power Requirement ๐ ๐ ๐ = ๐ ๐ฆ ๐ญ ๐ฉ = ๐ ๐ญ ๐ฉ ๐= ๐ ๐ญ๐ข๐ฆ๐ From: ๐ ๐ ๐๐ฌ๐ฎ = = ๐๐ฌ ๐๐๐ก ๐ญ ๐ฉ ๐ ๐ = ๐๐ฌ๐ฎ ๐๐๐ก ๐ญ ๐๐ฉ ๐ ANALYSIS OF FLYWHEELS Stresses induced in Flywheel arms Tensile Stress ๐ ๐๐ญ = ๐๐ฏ๐ฆ Bending Stress ๐ ๐ ๐๐ = = ๐โ๐ซ ๐ ๐๐ง๐๐ซ๐ฆ๐ฌ ๐ ANALYSIS OF FLYWHEELS From Machineryโs Handbook Pressure (Force) Required for Punching โข General Equation Note: Steel/Iron โ 50,000 psi ๐ = ๐จ๐ ๐๐ฌ = ๐๐๐ก ๐ญ ๐ฉ ๐๐ฌ Bronze โ 40,000 psi Copper โ 30, 000 psi โข Steel Material Aluminum โ 20, 000 psi Zinc โ 10, 000 psi ๐ = ๐๐ก ๐ญ ๐ฉ ร ๐๐ Tin/lead โ 5000 psi โข Brass Material Conversion: 1 ton = 2000 lbs ๐ = ๐๐ก ๐ญ ๐ฉ ร ๐๐ Note: (For non- circular hole) ๐ ๐๐ก = ๐๐ก ๐ 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: ๐๐๐๐ ๐ โ ๐ฆ 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: ๐๐๐๐ ๐ค๐ โ ๐ฆ FLYWHEEL Problem 11 A flywheel rotates at 220 rpm slowed down to 65 % of its revolution during the three โ fourth second punching portion of the cycle. Compute the angular acceleration of the flywheel in radian per second squared. Ans: โ๐๐. ๐๐๐ ๐ซ๐๐/๐ฌ ๐ 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: ๐๐๐๐ 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: ๐. ๐๐๐ 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: ๐๐๐. ๐๐ ๐ค๐ 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: ๐๐๐. ๐๐ ๐ฅ๐๐ฌ 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: ๐๐๐ ๐ค๐ FLYWHEEL Problem 17 A punch punches a 1-in diameter hole in a steel plate ยพ 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: ๐. ๐๐ ๐ก๐ฉ FLYWHEEL Problem 18 What pressure is required for punching a hole 3 inches in diameter through 1/4 inch steel stock? Ans: ๐๐ ๐ญ๐จ๐ง๐ฌ FLYWHEEL Problem 19 What pressure is required for punching a 1.5 inch square hole in a 1/4 inch thick plate steel? Ans: ๐๐ ๐ญ๐จ๐ง๐ฌ 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: ๐๐. ๐๐ ๐ญ๐จ๐ง๐ฌ
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