ࡱ> wyv` bjbj r:&stttyyy8y${Mf|"LLLLLLL$AOhQRLtTTTLyMpppTtLpTLppNGX tZK | `AFyTH(LM0MHQQPZKQtZKL0"p_LLjMTTTT$iyyM Machine Design Important Topics Power screws Bolt design Bolt clamping loads Bolt shear loads Bolt groups Bolts in fatigue loading Gears Kinematics Forces Stresses Belts Welds Welds subjected to shear Welds subjected to torsion Welds subjected to bending Clutches/Brakes Disk type / Drum type Flywheels Spring design Helical Roller/ball bearing design Journal bearings Shaft design Against fatigue Whirling Lateral vibration Machine Design Recommended Text: Fundamentals of Machine Component Design by Juvinall and Marshek 3rd Edition Bolt Selections and Design You need Dimensions of standard threads (UNF/UNC) Strength specifications (grades) of bolts. Clamping forces     The bolt force is  EMBED Equation.3  Where Kb and KC are the bolt and the clamping material stiffness and Fi is the initial bolt tensioning. The clamping force is  EMBED Equation.3   Recommended initial tension Fi = 0.90 SpAt Where Sp is the proof strength and At is the tensile area of the bolt. Recommended tightening torque T = 0.20 Fid Where d is the nominal bolt size. Design of bolts in tension Fb = At Sp Where At is the tensile area. Problem #M1a : Given: Two plates are bolted with initial clamping force of 2250 lbs. The bolt stiffness is twice the clamping material stiffness. Find: External separating load that would reduce the clamping force to 225 lbs. Find the maximum bolt force at minimum clamping force. Answers: 6075 lbs and 6300 lbs Problem #M1b :Select a bolt that would withstand 6300 lbs nominal load in direct tension. Apply a factor of safety of 2.5. Use a bolt with SAE strength grade of 2 (which has a proof strength of 55 ksi). Answer: 10-UNC Bolts under shear loading Problem #M1c: A 1-12 UNF steel bolt of SAE grade 5 is under direct shear loading. The coefficient of friction between mating surfaces is 0.4. The bolt is tightened to its full proof strength. Tensile area is 0.663 in2 Proof strength is 85 kpsi, and yield strength is 92 kpsi What shear force would the friction carry? What shear load can the bolt withstand w/o yielding if the friction between clamped members is completely lost? Base the calculation on the thread root area. Approximate Answers: a) 22500 lbs, b)35300 lbs Bolt design under fatigue loading Analysis method: Root stress Goodman Criteria  SHAPE \* MERGEFORMAT  Example M2 Consider a UNF Grade 7 bolt with an initial tension of 90% of its proof load. The clamping material is 4 times stiffer than the bolt. If a cyclic load of 0 to 6150 lbs is applied, what would be the factor of safety guarding against eventual fatigue failure? The tensile area and proof strength are At = 1599 in2 and Sp=105 ksi. The initial tension is Fi = .9*(0.1599)(105) = 15.1 ksi The fluctuations in the bolt load is:  EMBED Equation.3  The alternating stress at the bolt root is:  EMBED Equation.3  Endurance limit  EMBED Equation.3  The fatigue strength is  EMBED Equation.3  The factor of safety is:  EMBED Equation.3  Design of Bolt Groups in Bending Assume bolted frame is rigid. Use geometry to determine bolt elongations. Assume load distribution proportional to elongations. Assume shear loads carried by friction.         Problem #M3 Consider the bracket shown above. Assume the bracket is rigid and the shear loads are carried by friction. The bracket is bolted by four bolts. The following is known: F=5400 lbs, L=40 inches, D=12 inches, d=4 inches. Find appropriate UNC bolt specifications for bolts of 120 Ksi proof strength using a factor of safety of 4. Answer: - 10UNC Design of Bolt Groups in Torsion Assume bolted frame is rigid. Use geometry to determine bolt distortion. Assume torque distribution proportional to distortions. Assume bracket rotates around the bolt group C.G. Exclude direct shear if its magnitude is small. Use the bolt shank area for analysis of stresses. Example #M4   The bolts are -13UNC. The distance between bolts is 1.25. The load is 2700 lbs and L=8. Find the shear stress on each bolt. Answer: 44250 psi Gear Geometry Kinematic model of a gear set        Terminology Diametral pitch (or just pitch) P : determines the size of the tooth. All standard meshing gears have the same pitch.  EMBED Equation.3  P is pitch, p is circular pitch and m is the module. I) Regular Gear Trains  EMBED Equation.3  N1 and N2 are the number of teeth in each gear, and n1 and n2 are the gear speed in rpm or similar units. Internal gears  EMBED Equation.3  II) Epicyclic (Planetary) Gear Trains    Epicyclic gear trains have two degrees of freedom They require two inputs.  EMBED Equation.3  Epicyclic gear trains can always be solved by the following two relationships. Relative angular velocity formula:  EMBED Equation.3  Regular gear train formula with Arm stationary  EMBED Equation.3  Problem #M5: Gear kinematics The figure shows an epicyclic gear train. The number of teeth on each gear is as follows: N2=20 N5=16 N4=30 The input is Gear 2 and its speed is 250 rpm clockwise. Gear 6 is fixed. Determine the speed of the arm and the speed of Gear 4. The drawing is not to scale.  d5 + d6 = d2 + d4 and since all P are the same we get N5 + N6 = N2 + N4 and N6 = 34 teeth  EMBED Equation.3  Substituting for the number of teeth on each gear  EMBED Equation.3  Also  EMBED Equation.3  From above: n4=-357.1 rpm Kinematics of Automobile Differential Considering the Right Wheel, Left Wheel, the Ring Gear and the Drive Shaft as shown in the figure,  EMBED Equation.3  Gear Force Analysis  SHAPE \* MERGEFORMAT  Fn : Normal force Ft : Torque-producing tangential force Fr : Radial force. When n is in rpm and d is in inches:  EMBED Equation.3  and  EMBED Equation.3  Helical gears  SHAPE \* MERGEFORMAT  Geometric relationships:  EMBED Equation.3  Helical gear forces  EMBED Equation.3  Straight Bevel gears  SHAPE \* MERGEFORMAT  Bevel gear forces  EMBED Equation.3  These forces are for pinion and act through the tooth midpoint. Forces acting on the gear are the same but act on different directions. Worm Gear Kinematics  SHAPE \* MERGEFORMAT  The velocity ratio of a worm gear set is determined by the number of teeth in gear and worm thread (not the ratio of the pitch diameters).  EMBED Equation.3  Nw = Number of threads (single thread =1, double thread =2, etc) The worms lead is  EMBED Equation.3  The worms axial pitch pa is the same as the gears PR circular pitch p. The worms lead angle l is the same as the gear s helix angle y. Example For a speed reduction of 30 fold and a double threaded worm, what should be the number of teeth on a matching worm gear. Ng = (2) (30) = 60 teeth The geometric relation for finding worm lead angle  EMBED Equation.3  Worm Gear Forces The forces in a worm gear set when the worm is driving is Fgr = Fwr Fgt = Fwa Fga = Fwt  SHAPE \* MERGEFORMAT  Usually the Fwt is obtained from the motor hp and rpm as before. The other forces are:  EMBED Equation.3  The worm and gear radial forces are:  EMBED Equation.3  The worm gear set efficiency is:  EMBED Equation.3  Where f is the coefficient of friction. Condition for self-locking when worm is the driver  EMBED Equation.3  Bearing Reaction Forces  SHAPE \* MERGEFORMAT  Total thrust load on bearings is Fa For radial forces combine the radial and tangential forces into F:  EMBED Equation.3  Flat Belts Flat belts have two configurations: Open   EMBED Equation.3  Closed (Crossed)     EMBED Equation.3  Where C: Center-to-center distance D,d: Diameters of larger and smaller rims Slippage Relationship  EMBED Equation.3  q is in radians. Transmitted Hp is  EMBED Equation.3  Where F1 and F2 are in lbs and V is in ft/min. Initial Tension Belts are tensioned to a specified value of Fi. When the belt is not transmitting torque: F1=F2=Fi As the belt start transmitting power, F1 = Fi + DF F2 = Fi - DF The force imbalance continues until the slippage limit is reached. Problem M7 A 10 -wide flat belt is used with a driving pulley of diameter 16 and a driven pulley of rim diameter 36 in an open configuration. The center distance between the two pulleys is 15 feet. The friction coefficient between the belt and the pulley is 0.80. The belt speed is required to be 3600 ft/min. The belts are initially tensioned to 544 lbs. Determine the following. (answers are in parentheses) Belt engagement angle on the smaller pulley (3.03 radians). Force in belt in the tight side just before slippage. (1000 lbs). Maximum transmitted Hp. (99.4 hp) Formula for V-belts  EMBED Equation.3   where  EMBED Equation.3  m=Mass per unit length r=Pulley radius Disk Brakes and Clutches     Torque capacity under Uniform Wear condition per friction surface  EMBED Equation.3  Where f: Coefficient of friction pa: Maximum pressure on brake pad d,D: Inner and outer pad diameters Torque capacity under uniform pressure conditions per friction surface  EMBED Equation.3  Maximum clamping forces to develop full torque For Uniform Wear  EMBED Equation.3  For Uniform Pressure  EMBED Equation.3  Example M8 Given: A multi-plate disk clutch d=0.5 D=6 Pmax=100 psi Coefficient of friction=0.1 Power transmitted= 15 hp at 1500 rpm Find: Number of friction surfaces Answer: N=2 (uniform pressure) N=9 (uniform wear) Energy Dissipation in Clutches and Brakes The time it takes for two rotational inertia to reach the same speed after engagement through a clutch is:       EMBED Equation.3  where T: Common transmitted torque w: angular speed in rad/sec The total energy dissipated during clutching (braking) is:  EMBED Equation.3  If the answer is needed in BTU, divide the energy in in-lb by 9336. Problem M9 A brake with braking torque capacity of 230 ft-lb brings a rotational inertia I1 to rest from 1800 rpm in 8 seconds. Determine the rotational inertia. Also, determine the energy dissipated by the brake.    Solution hints: Convert rpm to rad/sec: w1 = 188 rad/sec Note that w2=0 Find the ratio (I1I2/I1+I2) using time and torque=>9.79 Note that I2 is infinitely large => I1=9.79 slugs-ft Find energy from equation=>173000 ft-lb Springs Coverage: Helical compression springs in static loading Terminology: d: Wire diameter D: Mean coil diameter C: Spring index (D/d) Nt: Total # of coils N: Number of active coils p: Coil pitch Lf: Free length = N*p Ls: Solid length La: Assembled length Lm: Minimum working length End detail and number of active coils: Spring Rate of Helical Springs (compression/extension)  EMBED Equation.3  where : N is the number of active coils Plain ends: N=Nt Plain and ground ends: N=Nt-1 Square ends: N=Nt-2 Square and ground ends: N=Nt-2 G: shear modulus = E/2(1+n) G=11.5*106 psi for steels Shear stress in helical springs for static loading  EMBED Equation.3  where  EMBED Equation.3  and C is the spring index. Shear strength in springs  EMBED Equation.3  Ferrous without presetting  EMBED Equation.3  Ferrous with presetting Solid Lengths Ls=(Nt+1)d with plain ends Ls=(Nt)d with ground ends Spring Surge Frequency  EMBED Equation.3  Where g is the gravitational acceleration and Wa is the weight of the active coils:  EMBED Equation.3  with g being the specific gravity of spring material. For steel springs when d and D are in inches:  EMBED Equation.3  Example M10 Consider a helical compression spring with the following information (not all are necessarily needed): Ends: Squared and ground Spring is not preset Material: Music wire (steel) with Sut=283 ksi d=.055 inches and D=0.48 inches Lf=1.36 inches and Nt=10 Find the following. Answers are given in parentheses. Spring constant, K (14.87 lb/in) Length at minimum working load of 5 lbs (1.02) Length at maximum load of 10 lbs (0.69) Solid length (0.55) Load corresponding to solid length (12.04 lbs) Clash allowance (0.137) Shear stress at solid length (77676 psi) Surge frequency of the spring (415 Hz) Design of Welds Welds in parallel loading and transverse loading  SHAPE \* MERGEFORMAT  Weld Geometry      Analysis Convention Critical stresses are due to shear stresses in throat area of the weld in both parallel and transverse loading. For convex welds, t=0.707w is used. Yield strength of weld rods used in analysis is 12 ksi smaller than their nominal minimum yield strength. Analysis Methodology Under combined loading, different stresses per unit leg length are calculated and combined as vectors. Stresses based on weld leg (w) Direct tension/compression:  EMBED Equation.3  Direct shear:  EMBED Equation.3  Bending:  EMBED Equation.3  Torsion:  EMBED Equation.3  Formulas for Aw, Sw, and Jw are attached for different weld shapes. Problem M11a -Welds subject to direct shear Two steel plates welded and are under a direct shear load P. 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The weld material is E60 with a yield strength of 60 ksi nominal.  SHAPE \* MERGEFORMAT  Solution (of M11a) From Table: Aw = 2d = 6  EMBED Equation.3  The design strength of the weld material in shear is: Sys=.58 Sy = .58(60-12) = 48*.58 = 27.84 ksi Using a factor of safety of 2, the allowable shear stress is: Sys,a = 27.84/2 = 13.92 ksi Equating stress and strength .6284F = 13920 ( F=22150 lbs Problem M11b Welds subject to torsion A round steel bar is welded to a rigid surface with a fillet weld all around. The bars outer diameter is 4.5. Determine the critical shear stresses in the weld when the bar is subjected to a 20,000 lb-in pure torque.  EMBED Equation.3   EMBED Equation.3  Problem M11c Welds subject to bending Solve the previous problem with a bending moment of 35000 lb-in acting on the welds instead of the torsion load.  EMBED Equation.3   EMBED Equation.3  Problem M11d Welds subject to combined loads If the design shear strength (Sys) in the weld is 27800 psi, what is the factor of safety against yielding when both stresses in previous two problems are acting on the bar.  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