Machine Elements: Design Project

Transcription

1 Machine Elements: Design Project Drag Chain Conveyor Matthew Gray, Michael McClain, Pete White, Kyle Gilliam, Alejandro Moncada and Matthew Gonzalez

2 Executive Summary The logistics of many industries use drag chain conveyors to move heavy loads short distances. The purpose of this project is to analytically define the characteristics and design parameters for our own drag chain conveyor system. The starting parameters used for our general design are listed below: Maximum load of 3000 pounds (treated as a 4 feet x 4 feet wooden pallet) Maximum speed of 35 feet per minute Load travel distance of 10 linear feet Conveyor frame is rigid Neglect weight of wooden pallet Chain idealized Drive Sprocket parameters are taken from manufacturer data The following components were the focus of our design: Pinion driven by the motor Main drive gear Shaft connecting main drive gear Bearings supporting the shaft Drag chain conveyors operate by rotating a shaft that drives two or more sprockets, which in turn, drive the chains. The chains are connected around a drive sprocket and a free rotating sprocket (an idler sprocket) at the opposite end of the conveyor. A load to be moved rests on top of the chains and as the chains rotate around the prescribed loop, the load is moved from one point to another. Our design approach was to work backwards from our starting parameters to determine the forces necessary for our conveyor to operate as prescribed. These forces were then used to analyze the key components of our conveyor. From our force analysis, we were able to determine that it should take the conveyor no longer than 17 seconds to move the 3000 pound load 10 feet. The total force required for our conveyor was calculated to be 776 pounds. Our sprocket design was based on our decision to use an ANSI No. 50 chain for the conveyor. With the properties of this chain, we were able to calculate the pitch diameter of the sprocket by using the American Chain Association Sprocket Design Guidelines. This diameter was then used in selecting an actual sprocket to be used in our design. Finally, we determined that a torque of 81 pound-foot on each sprocket is required to satisfy our initial parameters.

3 The initial parameters used for our gear design are listed below: 20 degree pressure angle Standard full depth gear 21 teeth for pinion 55 teeth for gear 2.62:1 gear ratio 3 inch pitch diameter for pinion These parameters were used to calculate the geometries of both the gear and pinion. After determining the geometries of the gear and pinion, a thorough stress analysis was conducted on the gear set. Notable calculations from this analysis include the rotational speed of the pinion and the number of cycles of both pinion and gear. The pinion was calculated to have a rotational speed of 45 revolutions per minute, while the number of cycles of the pinion and gear in a 5 year span were determined to be 27 million and 10 million respectively. Other calculations that comprised this stress analysis include: bending stress, surface stress, bending fatigue strength, and surface fatigue strength. After conducting the stress analysis of the gear set, the safety factors of the pinion and gear were determined. We then re-evaluated the design and re-iterated the process. The final step in our gear design was selecting the material of the gear set. Our choice of material is shown below: Gear Material (Table Norton) o AGMA Grade 2 (Premium Grade) o Class A5 o Carburized and case hardened o Brinell Hardness = 710 (corresponds to 64HRC) o Qv = 5 With this material, the final safety factors of the gear and pinion are as follows: Gear o Bending Safety Factor = 4.76 Pinion o Bending Safety Factor = 3.98 o Surface Fatigue Safety Factor = 2.38 Shaft Design Parameters: 1.75 inch diameter

4 2 3 inch overhang on the outside of each sprocket SAE/AISI 1010 Cold rolled steel As with the gear set, a stress analysis was conducted on the shaft. The values we calculated for the maximum stresses on the shaft are as follows: 7188 psi (bending stress), 219 psi (shear stress), and 1850 psi (torsional stress). After calculating the shaft s maximum stresses, we proceeded to determine its safety factor. With the shaft s material properties and maximum stresses, we used the maximum shear stress theory to get a safety factor of 5.4. In designing the bearings, we first had to determine the total force being subjected on each of the two bearings. The bearing closest to the drive gear was calculated to have a total force of 379 pounds, while the other bearing was calculated to have a total force of 85 pounds. Using the higher total force, we determined the dynamic load rating. The value of this rating was calculated to be 1296 pounds. Finally, using the dynamic load rating we were able to determine the safety factor of the bearing, which was calculated to be 15.

5 Report Outline I. Introduction & Overview A. Identification of Need B. Definition of Problem C. Drag Chain Conveyor System II. Analysis A. Nomenclature B. Preliminary Design C. Force Analysis D. Sprocket Design E. Gear Design 1. Gear and Pinion Stress Analysis a. Bending Stress Calculations b. Surface Stress Calculations c. Bending Fatigue Strengths d. Surface Fatigue Strengths 2. Safety Factor Calculation 3. Material Selection F. Shaft Design 1. Shaft Stress Analysis a. Bending Stress Calculation b. Shear Stress Calculation c. Torsional Stress Calculation 2. Safety Factor Calculations 3. Shaft Key Stress Calculations G. Bearing Design 1. Force Calculations 2. Dynamic Load Rating Calculation 3. Safety Factor Calculation III. Summary of Safety Factors IV. Conclusion V. References VI. Appendix VII. Engineering Drawings

6 Introduction & Overview Identification of Need The logistics of many industries require the use of human operated forklifts to transport heavy loads. Implementation of automated conveyor systems has proven to be an exceptional solution to minimizing human interaction in moving heavy loads short distances. These conveyor systems are commonly used to move heavy pallets of finished products or raw materials from one area of a manufacturing facility to another. Definition of Problem The purpose of this project is to analytically define the characteristics and design parameters for a drag chain conveyor system. The criteria we chose for our design process came from our research on existing conveyor systems. The assumed maximum load of 3000 pounds was based on an average of maximum loads for these conveyor systems. In addition, these loads are typically placed on 4 feet 4 feet (l w) wooden pallets. For our design, we will also assume that the 3000 pounds maximum load takes the form of this pallet. We defined 10 linear feet as the distance that this load must be moved. This distance keeps the conveyor system small enough to be shipped as a finished product via motor freight. It also allows simple installation in a manufacturing facility with no need for additional setup or assembly. Again, based on average values found for existing systems, we chose our translational speed for the system to be 35 feet per minute. The frame is assumed rigid as it is a non-moving part. In addition, it is typically overdesigned compared to the moving elements of the conveyor system. The weight of the wooden pallet is taken to be negligible with respect to the 3000 pounds load, and any flexure of the pallet is ignored. The travel distance of 10 linear feet is measured from the center of the drive sprockets. Since the design of the chains is outside the scope of this project, all chains used are assumed to be idealized. This means that any flexure, stretching, or stress in the chain is ignored. The sprockets that drive the chains are taken directly from manufacturer s specifications to match the drive chains.

7 All our initial parameters are listed below: Maximum load of 3000 pounds (treated as a 4 feet x 4 feet wooden pallet) Maximum speed of 35 feet per minute Load travel distance of 10 linear feet Conveyor frame is rigid Neglect weight of wooden pallet Chain idealized Sprockets that drive chains are taken from manufacture s information The main elements to be designed are the pinion driven by the motor, the main drive gear, the shaft connecting the main drive gear, and the bearings supporting the shaft. Many assumptions and idealizations have been made in order to focus on the main elements required for this report: a shaft, gear and bearing. Drag Chain Conveyor System Drag chain conveyors operate by rotating a shaft that drives two or more sprockets, which in turn, drive the chains. The chains are connected around a drive sprocket and a free rotating sprocket (an idler sprocket) at the opposite end of the conveyor. See the Engineering Drawings section for a better illustration of the final design. These types of conveyors have many variations and are used in many different industries around the globe. A common application of these types of conveyors is in the area of material handling. A load to be moved rests on top of the chains and as the chains rotate around the prescribed loop, the load is moved from one point to another. Many different companies manufacture drag chain conveyors, however the design principles are similar industry wide. The parameters that do differ among designs are usually the component materials, load ratings, and speed of travel.

8 Analysis Nomenclature W p = distributed weight of load per side W c = distributed weight of chain per side W L = total distributed weight per side W T = total load (per side) applied to chain Q T = Normal Force F f = frictional force per side μ k = friction coeff. of UHMW and Steel V C = linear velocity of chain and load F C = force required per side to move load F T = force required (total)to move load P ds1 = pitch diameter of sprocket #1 P ds2 = pitch diameter of sprocket #2 T sp1 = torque at sprocket #1 T sp2 = torque at sprocket #2 ω sp = rotational speed of sprockets ω shaft = rotational speed of shaft P dp = diametral pitch of pinion r p = pitch radius of pinion P dg = diametral pitch of gear r g = pitch radius of gear N tp = number of teeth on pinion N tg = number of teeth on gear d p = pitch diamter of pinion d g = pitch diamter of gear P c = circular pitch P b = base pitch a = addendum b = dedendum D W = working depth D Wh = whole depth t ct = circular tooth thickness r f = fillet radius C bmin = minimum basic clearance W tlmin = minimum width of top land C sg = clearance W f = face width of tooth Z = length of action C = center distance m p = contact ratio of gear and pinion ω p = rotational speed of pinion ω g = rotational speed of gear N cyc p = number of cycles of pinion N cyc g = number of cycles of gear W t = tangential force on tooth T g = torque on gear K v = Dynamic Factor Q v = Quality Factor K m = Load Distribution Factor K B = Rim Thickness Factor K a = Application Factor K I = Idler Factor K S = Size Factor

9 J = Bending Strength Geometry Factor I = Surface Geometry Factor σ bp = bending stress on pinion σ bg = bending stress on gear σ cp = surface stress on pinion C v = Dyanmic Factor C m = Load Distribution Factor C a = Application Factor C S = Size Factor C p = Elastic Coefficient C f = Surface Finish Factor E p = Young s Modulus of Pinion E g = Young s Modulus of Gear v p = Poisson s ratio of Pinion v g = Poisson s ratio of gear S fbg = corrected AGMA bending strength for gear S fbp = corrected AGMA bending strength for pinion S fb = uncorrected AGMA bending strength K Lp = Life Factor for pinion K Lp = Life Factor for gear K T = Temperature Factor K R = Reliability Factor C H = Hardness Ratio Factor N bp = Safety Saftor for Bending on pinion N bg = Safety Saftor for Bending on gear N Cp = Safety Saftor for Surface Fatigue on pinion σ bmax = maximum bending stress on shaft M = maximum bending moment on shaft c = dist. from nuetral axis to outer edge of shaft I s = mass moment of inertia of shaft d s = diameter of shaft τ bmax = max shear bending stress V max = max shear force A shear = area under shear force τ torsion = max torsional stress T shaft = torque on shaft J s = polar moment of inertia of shaft τ max = largest principle shear stress N MSST = safety factor, Maximum Shear Theory N DE = safety factor, Distortion Energy Theory σ 1 = principal stress 1 σ 3 = principal stress 3 σ = von Mises effective stress S fcp = corrected surface fatique strength pinion S fcp = uncorrected surface fatique strength pinion C T = Temerpature Factor C R = Reliability Factor C L = Surface Life Factor

10 Preliminary Design Our initial design was based on existing conveyor technology. After reviewing existing conveyor systems, we were able to make some justifiable preliminary design decisions. Many drag chain conveyors contain tensioning devices. These devices put an extremely high tension on the chains in order to minimize the flexure and keep the chain from riding on the metal frame. This aim at reducing metal on metal wear has its flaws. As a load is applied, the tension in the chain increases, and more force is transmitted to the sprockets that drive and guide the chain. This causes increased fatigue on the sprockets and connecting shafts. Another issue with tensioned conveyors is that when a load is applied on the chain, it may flex so much that there is contact between the chain and the conveyor s metal frame. This can cause extreme friction and premature failure of the frame and/or chain. We opted for a design that utilized Ultra High Molecular Weight (UHMW) Polyethylene wear blocks below the chains. These blocks have low coefficients of friction and wear extremely well. This decision reduces the need for extremely high tension in the chain, which in turn, reduces the overall fatigue on the system. The UHMW wear blocks can be easily replaced at regular intervals to prevent excessive wear on the chain and conveyor frame. In order to simplify the amount of analysis required, we chose a chain used by existing drag chain conveyors with sprockets that match the chain s properties. We will analyze the force acting on the chain and UHMW blocks to determine the amount of torque required at the shaft. The shaft diameter is determined by the sprockets that match the drive chains. However, all other design aspects of the shaft will be analyzed and a suitable material selection will be made. Finally, the drive gear and bearings will also be designed from scratch based on the forces required to transmit the load at the prescribed 35 feet per minute. Our approach to analyzing the assembly was to work backwards. From defining our load and desired travel speed, we calculated the force required to move the load. This force calculation was the foundation in designing the components of the conveyor.

11 Force Analysis The first step in our design was to use the given information to find the forces transmitted throughout the system. By analyzing the 3000 pound load acting downward on the chains and using the coefficient of kinetic friction of the UHMW block, we were able to determine the force needed in the chains to move the load at 35 feet per minute. The 3000 pound load is applied evenly between the 2 chains at 1500 pounds per chain. Each 1500 pound load is distributed over 4 feet of chain. The detailed calculations of our force analysis are shown below: 1500 lb W p = ( ) ( 1 ft lb ) = ft 12 in in (Eq. 1) W C = (4 lb ) ( 1 ft lb ) =. 333 ft 12 in in - Data from ANSI Chain # 50 (Eq. 2) W L = Load + Chain weight = lb lb lb +.33 = in in in (Eq. 3) W T = (31.58 lb in lb in ) (4ft (12 )) + (.33 ) (6 ft (12 )) = lb (Eq. 4) in 1 ft in 1 ft μ k = for UHMW Q T = W T = lb F f = μq T = (0.25) ( lb) = lb (Eq. 5) V c = (35 ft min in ) (1 ) (12 ) = 7 in min 60 s 1 ft s (Eq.6) 10 ft ( 12 in 1 ft 1 ) = 120 in ( = seconds to travel 10 feet )) (7 in s F = ma F C lbs = [3000 lb+(.33 lb in in ) (10ft) (12 1 ft )] 32.2 ft s 2 (a) (Eq. 7.a) F C lb = slugs (a) a = dv dt (Eq. 7.b) F C lb = slugs ( dv ) dt t (F V C lb)dt = 0 (94.41 slugs)dv (Eq. 7.c) s 0 7 in s (F C lb)dt = (94.41 slugs)dv (Eq. 7.d) 0 F C (17.14 s) lb(17.14 s) = slugs 7 in s (Eq. 7.e) F C = s slug in [( ) ( ft ) lb]= lbs (Eq. 7.f) s 12 in F T = lb per side 2 sides = lb (Eq. 8)

12 Sprocket Design Once the force to move the maximum load of 3000 pounds was calculated, we were able to use this information along with idealization of the chains to calculate the torque and angular velocity required at the sprockets and the shaft. The chains were idealized to have no stretch and it was assumed that any force on the chains was transmitted by the sprockets. We began with an ANSI No. 50 chain, which has a pitch of 5/8 inch. By using the properties of the ANSI No. 50 chain, the AGMA equations and the assumption that each sprocket has 40 teeth we were able to calculate a pitch diameter of the sprockets. The bore size was assumed to be 1.75 inches. This starting value would determine the diameter of the shaft and its feasibility is found later in the calculations. The calculation for finding the pitch diameter of a sprocket was found by using the following equation: P ds1 = P ds2 = chain pitch = ( 5 sin ( in) N ) sin( ) = in (Eq. 8) Based on this pitch diameter, the assumed number of teeth and bore size, we chose part #6236K744 from McMaster-Carr, a large material and component supplier. From this pitch diameter and our calculated total force, we were able to find the torque required at each sprocket to move the load at 35 feet per minute. T sp1 = T sp2 = F c ( P d 2 Thus, Tsp1 = Tsp2 = lb * ft in ) = ( lb) (7.966 ) = lb in ( 1 ft ) = lbft (Eq.9) 2 12 in T shaft = T sp1 + T sp2 = lbft (Eq.10) The rotational speed (angular velocity) of the sprockets is easily found through basic dynamic relationships. The chain must move at 35 feet per minute, and this speed corresponds with the tangential velocity of the sprockets at the pitch point or pitch diameter distance. We calculated the angular velocity of the sprocket and used the same value for the angular velocity of the shaft. Since both components are connected to each other, they share the same rotational speed. V c = ( P ds1 ) ω 2 sp ω sp = ( 2 35 ft in min in ) (12 ) ( 1 rev ) = RPM = 17 RPM (Eq.11) ft 2π rad Thus; ω sp = ω shaft = 17 rpm

13 Gear Design In designing our gear and pinion mesh, we referred heavily to Machine Design An Integrated Approach by Dr. Robert L. Norton. Our starting point was to define the pressure angle for the gear teeth. We chose a 20 degree pressure angle based on the fact that is the most widely used pressure angle in industry. Table 12-5 (Norton) shows the minimum number of pinion teeth to avoid interference between a full depth pinion and gear with a 20 degree pressure angle. We wanted to avoid interference so that we could eliminate the need to undercut the teeth. This saves time in drafting and manufacturing. We chose the range of 16 teeth minimum for the pinion to 101 teeth maximum for the gear; this range would give us more gear ratio options. We chose a standard full depth gear teeth because the equations for the geometry of this gear with a 20 degree pressure angle are readily available in Table 12-1 (Norton). Other standard tables have also been established for several factors, including the AGMA bending geometry factor J and I. We referred to these tables to narrow down our gear ratios. In the end, we chose the number of teeth to correspond with a pre-defined J and I values from the AGMA tables to further ease calculations. Because we also know that the motor will be at a lower torque and higher speed than the shaft requires, we estimated a required gear ratio of at least 2 to 1. Referring to the AGMA tables for bending and surface geometry factors and our teeth range to avoid interference, we were able to choose a desirable number of teeth for both the gear and pinion. We decided on 21 teeth for the pinion and 55 teeth for the gear, giving us a 2.62:1 gear ratio. We assumed a pinion pitch diameter of 3 inches. This would roughly make the gear pitch diameter very close to the pitch diameter of the sprockets. Once these factors were established we could now find the geometry for the gear and pinion. Gear and Pinion Dimensional Calculations One of the most useful properties for calculating the geometry of a gear is the diametral pitch. Once this diametral pitch is defined, all remaining gear and pinion geometry can be found based on this value. To find the diametral pitch, we used our assumed starting pinion pitch diameter and number of teeth on the pinion. P dp = N tp d p = 21 teeth 3 in = 7 teeth in (Eq.12) In order for gears to mesh properly, they must have the same pitch angle and diametral pitch. Using this information with our assumed 55 teeth on the gear, the pitch diameter of the gear can be calculated. P dg = 7 teeth in = N tg d g = 55 teeth d g d g = 55 teeth = in = 7.86 in (Eq.13) (7 teeth ) in

14 With the known pitch diameters and diametral pitches of both gear and pinion, we can now calculate the remaining dimensions of both the gear and pinion. For quick and simple calculations for the gear geometries, we created a Microsoft Excel spreadsheet (See appendix) that included all of the formulas for calculating gear geometry from Table 12-1 (Norton). These values apply to both the gear and pinion in order to ensure the gears mesh and transmit power properly. The calculated gear geometries are listed below: P c = in P b = in P d = 7 teeth/in a = in b = in t ct = in r f = in C bmin = in W tlmin = in C sg = in D W = in D Wh = in Another critical property needed for gear design is the face width. Face width (W f ) is roughly expressed as a range function of diametral pitch (8/Pd < W f < 16/Pd) (Norton). For our design, we used the following to find a starting face width: 8 < W P f < 16 d P d 8 7 teeth < W f < in 16 7 teeth in in tooth < W f < 2.29 in tooth (Eq. 14) We took the average of this range and defined our starting face width (W f ) of both gear and pinion to be inches. W f = in At this point, we have calculated all of the dimensions of the gear and pinion. We can now find the length of action of the gear mesh. Note that a represents the addendum length and is the same for both the gear and the pinion. Z = [(r p + a) 2 (r p cos ) 2 ] + [(r g + a) 2 (r g cos ) 2 ] Csin( ) (Eq.15) Z = [( ) 2 (1.5cos(20)) 2 ] + [( ) 2 (3.93cos (20)) 2 ] 5.43 sin(20) Thus, Z =.7037

15 With the length of action, we can calculate the contact ratio (mp) and ensure that it agrees with typical values for spur gears. According to Norton, the contact ratio is calculated using the length of action (Z), the diametral pitch (Pd), and the pressure angle (Ø). m p = p dz m πcos ( ) p = = 1.67 Thus; mp = 1.67 (Eq.16) πcos (20) This contact ratio falls within the acceptable range for spur gear sets of 1.4 to 2 (Norton). The gear contact ratio ensures that more than one tooth of the gear carries the load at any given time. Although true in theory, in reality, there will be times when one tooth carries the entire load. This occurs at the center of the mesh where the load is applied at a lower position on the tooth. This position is referred to as the highest point of single tooth contact (HPSTC) (Norton). We assumed this information when we chose our number of teeth based on the geometrical bending factors (J and I). This assumption will provide a more accurate stress analysis. Gear and Pinion Stress Analysis The first step in beginning a stress analysis is to gather the required values for the calculations. We must first find the number of cycles for the gear and pinion. We already found that the gear and shaft will spin at 17 revolutions per minute and we have the gear ratio, thus we can find the rotational speed of the pinion and in turn the number of cycles. After finding the rotational speed of the pinion, we will find the number of cycles for 5 years at 8 hour days and 250 working days in each year. We decided this was a reasonable period of working time for such a system. ω p = ω g gear ratio ω p = 17 rpm (2.62) = = 45 rpm (Eq.17) N cyc p = 45rpm ( 60min hr N cyc g = 17rpm ( 60min hr Thus; ωp = 45 rpm N cyc p =2.7x10 7 cycles in 5 years N cyc g =1.02x10 7 cycles in 5 years ) (8hr day yr ) (250 ) (5 day 1 yr ) (8hr day yr ) (250 ) (5 day 1 yr 1 ) = 2.7x107 cycles in 5 years (Eq.18) 1 ) = 1.02x107 cycles in 5 years (Eq.19) We assumed a gear quality value of Qv = 5, based on typical values found in Table 12-6 for mining conveyor systems (Norton). With this assumption and the data calculated so far we can now begin the calculation of various correction factors for the stress analysis of the gear and pinion.

16 Gear and Pinion Bending Stress Calculations The bending stress for a spur gear is defined by the AGMA Bending Stress Equation (Norton). The equation is as follows: σ b = ( W tp d W f J ) (K ak m K v ) K s K B K I (Eq.20) The Wt variable represents the tangential force on the tooth of the pinion and gear; other known values include the face width and diametral pitch. The remaining variables used in this equation are defined and justified below for both the gear and the pinion: Tangential Forces (Wt) Same for both pinion and gear This force is equal and opposite for both gear and pinion because they are in mesh. Because we already found the required torque on the shaft we can use this value and the pitch diameter of the gear to find the tangential force that acts on the gear and pinion teeth. W t = T g r g = lbft 3.93 in W t = lbs ( 12 in ft ) = lbs (Eq.21) Bending Strength Geometry Factor (J) Table 12-9 (Norton) Jg = 0.40 Jp = 0.34 Dynamic Factor (Kv) - Same for both pinion and gear For gears with a Qv factor less than or equal to 5 this factor is defined as: K v = V t in our case, the Vt term (pitch line velocity) is equal to 35 ft/min for both gears. K v = ft/min = (Eq.22) K v = Load Distribution Factor (Km) - Table 12-6 (Norton) - Same for both pinion and gear For face widths less than 2 inches, a Km factor is equal to 1.6. Km = 1.6 Rim Thickness Factor (KB = 1) - Same for both pinion and gear For solid-disk gears this value is always set to 1(Norton).

17 Idler Factor (KI = 1) - Same for both pinion and gear This value is always set to 1 for a non-idler gear (Norton). Application Factor (Ka) Table (Norton) - Same for both pinion and gear In our situation, we will be using an electric motor as the driving machine, which transmits a uniform load. Torque at start up may cause slight shock; therefore the decision was made to overdesign the component with a Ka value of Ka = 1.25 Size Factor (Ks) - Same for both pinion and gear In order to be conservative we chose a value of 1.25, typically this would be set to 1 unless we wanted to account for special situations (Norton) Ks = 1.25 Now that all of the necessary factors have been found, the bending stress for the gear and pinion teeth can be calculated. We employed Microsoft excel to speed these calculations as well as those calculations for the surface stresses. The obtained values for bending stress on the gear and pinion are found as: σ bp = ksi σ bg = ksi Gear and Pinion Surface Stress Calculations Because the gear teeth experience not only bending stresses but also surface stresses or pitting stresses, we must also take this into account for our design. The AGMA defines the pitting resistance formula as follows (Norton): σ c = C p ( W t W f Id p ) ( C ac m C v ) C s C f (Eq.23) This equation is only applied to the pinion because it experience more revolutions and thus will fail before the gear. The C variables used in this equation correspond to the K variables in the bending stress equation with exception of (Cp), and (Cf). The value of (d p ) is the pitch diameter of the pinion, for reasons mentioned above.

18 Surface Geometry Factor (I) The variable (I) represents the surface geometry factor, which takes into account the radii of curvature of the gear teeth and pressure angle. This value is readily available from AGMA 908- B89. I =.102 Elastic Coefficient (Cp) This value accounts for differences in tooth material and is found by (Norton): C p = ( 1 π( 1 v p 2 Ep )+(1 v g 2 Eg ) ) = ( 1 π( x10 6 )+( x10 6 ) ) = (Eq.24) Cp= Surface Finish Factor (Cf) This is usually set to 1 for gears made with standard manufacturing methods (Norton). Now that all of the factors have been defined, we may calculate the pitting resistance or surface stress of the pinion. σ cp = ( σ c = ksi ) ( ) (Eq.25) Now that the bending and surface stresses have been calculated, we may choose a gear and pinion material and determine the actual bending and surface strengths of that material. Once determined, a safety factor will be calculated and re-evaluation made if required. Bending Fatigue Strengths for Gear Materials In order to accurately design a gear set, one must ensure an acceptable safety factor. In order to check the safety of our design we must find the associated AGMA bending fatigue strengths associated with the gear materials. The AGMA publishes values; however they must be corrected to account for the situational use of a particular gear set. This corrected Bending fatigue strength(s fb ) is found by:

19 S fb = K L S K T K fb R (Eq.26) where S fb is the uncorrected bending fatigue strength given by AGMA. This is found for Grade 2 gear materials by: S fb = *HB HB 2 (Eq.27) Grade 2 implies a premium quality by the AGMA, ensuring a good homogenous microstructure and purity of the metal used to make the gear. HB designates the Brinell hardness of the metal to be used. We first decided on a steel that had HB = 300, this made our S fb = 47,095 psi. Other factors in this equation are defined and justified below: Life Factor (KL) This factor is based on the life required by the part. This will be different for the pinion and gear because they have a different number of cycles (N). An acceptable calculation of this factor for commercial uses is given by Figure as (See Norton): K L = (N) (Eq.28) K Lp = (2. 7x10 7 ) = K Lg = (1. 02x10 7 ) = Temperature Factor (KT =1) Same for both gear and pinion For temperatures less than 250 F, this value is set to 1(Norton). We are designing this for use at no more than this temperature. Reliability Factor (KR = 1.50) Same for both gear and pinion (Table 12-19)(Norton) We initially chose a reliability of 99.99%, which corresponds to a value of KR = Now that all factors have been identified we can calculate the actual bending strength of the gear and pinion. S fbg = K Lg S K T K fb = ,095 = ksi S R fb = ksi (Eq.26a) g S fbp = K Lp K T K R S fb = ,095 = ksi S fb p = ksi (Eq.26b)

20 Surface -Fatigue Strengths for Gear Materials We previously calculated the surface-fatigue stress for the pinion and now must calculate the actual strength of the pinion material in order to have a comparative analysis and formulate a safety factor. This calculation will not be done for the gear because it relies on the number of cycles of the gear or pinion being analyzed; in our case the pinion has significantly more cycles than the gear and thus will fail before the gear does. The AGMA publishes surface-fatigue strength values; however, like the bending strength values they must be corrected to account for operational situations. The AGMA surface- fatigue strength calculation is given by: S fcp = C LC H S C T C fcp (Eq.29) R where S fc is the uncorrected surface-fatigue strength given by AGMA. This is found for Grade 2 gear materials by: = HB. (Eq.30) S fc HB designates the Brinell hardness of the metal to be used. We first decided on a steel that had HB = 300, this made our S fc = psi. The factors CT and CR are identical in value to the K values with the same subscript. Other factors in this equation are defined and justified below: Surface-Life Factor (CL) The publish AGMA values for surface fatigue strength are based on 1 x10 7 cycles. So this factor allows for correction based on the actual number of cycles of the pinion (2.7 x10 7 ). From Table (Norton) and classifying our design for commercial use, we can calculate this value using the equation: C L = 1.448N (Eq.31) This makes our CL =.977 Hardness Ratio Factor (CH) This factor takes into account the difference in hardness of the gear and pinion teeth, if one exists. Because we are making both gear and pinion out of the same material with the same heat treatment, this value is set to CH = 1 (Norton page 728). Now that the factors have been identified and calculated, we may calculate the actual surfacefatigue strength of the pinion. S fcp = C LC H S C T C fcp = = = ksi (Eq.29a) R 1 1.5

21 Gear and Pinion Safety Factor Calculation Now that the bending and surface-fatigue stresses and strengths have been computed, we can calculate the safety factor for both the gear and pinion. These are basic calculations found by: N b = s fb σ b and N c = s fc σ c (Eq.32 and 33) The safety factors for the pinion are: N bp = ksi 16.6 ksi = N cp = 88.7 ksi ksi =. 754 The safety factor for the gear is: N bg = ksi ksi = It is clear from the safety factors that our pinion will fail due to surface-fatigue stresses. There are a number of values that we could change to increase the resistance to these stresses. These values include: the reliability, the material hardness, the number of cycles and face width. In order to speed the calculation process for changes we might make, we created a Microsoft Excel spreadsheet (See appendix) that did all calculations based on the input parameters. Gear and Pinion Material Selection and Final Values After much iteration, we decided to reduce the reliability of our gear and pinion from % to 99% and make the components from the following material, keeping the initial dimensions the same, including the face width. Gear Material (Table Norton) o AGMA Grade 2 (Premium Grade) o Class A5 Steel o Carburized and case hardened o Brinell Hardness = 710 (corresponds to 64HRC) o Qv = 5 Once this material was chosen, the calculations were easily repeated in the Excel spreadsheet and the safety factors are as follows: Gear o Bending Safety Factor (N bg ) = 4.76 Pinion o Bending Safety Factor (N bp ) = 3.98 o Surface Fatigue Safety Factor(N cp ) = 2.38 These safety factors are more than adequate for our design and from this point we moved on to designing the shaft and bearings.

22 Shaft Design for Conveyor System In order to design an acceptable shaft to transmit power from the gear to the sprockets some assumptions needed to be made. We determined earlier that the chain sprockets have an inside bore of 1.75 inches. To accommodate these sprockets we decided to choose a shaft that had an outside diameter of 1.75 inches. The length of the shaft was determined by close inspection of existing conveyor systems and standard dimensions of the pallets that the conveyor would move. The width of a wooden pallet is standardized at 48 inches. We decided on roughly a 2 to 3 inch overhang on the outside of each sprocket. This allows the main supports of the pallet to be directly over the chain. The remaining distances were found through iteration of the design and placement of the different components. We already had the dimensions of the sprockets and the width of their hubs; we also found the dimensions of the gear. The bearings dimensions were found from a standard manufacturer s catalog. This particular manufacturer had one pillow block bearing housing with a very wide range of dynamic load ratings available for the inserted bearing. So we used this housing s dimensions to finalize our locations and would choose our actual bearing based on the dynamic load rating later in our analysis. Referring to Norton and Mott we tried to adhere to common design guidelines for locating our power transmitting devices that would be attached to the shaft. Bearings were placed close to these power transmitting devices to minimize bending forces. The shaft material was initially chosen to be SAE/AISI 1010 Cold rolled steel. The final dimensions of the shaft and location of all of the power transmitting elements and bearings is shown in the appendix. Once we established the tentative location of all of the elements that will be attached to the shaft we were able to develop a loading diagram as well as the corresponding shear and moment diagrams. We used Autodesk Inventor to confirm our hand calculations and model the shaft and location of power transmitting components. Inventor has a very useful feature called Design Accelerator which allows the user to input the force and support bearing locations on the shaft. With this information and several dimensions required from the user, Inventor automatically calculates the shear and moment diagrams and maximum values for: Shear Stress, Bending Stress, Torsional Stress and maximum shaft deflection. One can even define the material to be used in creating the shaft, as well as the modulus of elasticity, Poisson s ratio and other metallurgical characteristics of the component. The plots of the maximum forces, moments and stresses generated by this software are shown in the appendix. We calculated all of these values by hand and compared them with the computer generated model. They agreed fairly well, discrepancies were attributed to rounding differences. The hand calculations for the bending, shear and torsional stresses are calculated below:

23 Maximum Bending Stress σ bmax = Mc I s = 32M πd s 3 = 32( lb ft) π(1.75 in) 3 (12 in ) = psi (Eq.34) ft Maximum Shear Stress τ bmax = V max A shear = lb π(1.75 in) 2 4 = psi (Eq.35) Maximum Torsional Stress τ torsion = T shaftc J s = 1.75 in ( lb ft)( ) 2 12 in.92 in 4 ft = psi (Eq.36) With this information and the assumption of SAE/AISI 1010 Cold rolled steel we were able to calculate several safety factors for bending, shear and torsional effects on the shaft. It is important to note that we assumed a constant loading condition; thus, eliminating any alternating or mean stresses. This idealization would prove to be feasible after we found our safety factors for bending, shear and torsional stresses to be relatively high. After the actual stresses were defined we used the Maximum Shear Stress Theory and Distortion Energy Theory to calculate the safety factor for the maximum shear stress and maximum bending stress respectively. The information related to the SAE 1010 Cold Rolled Steel is: o Sy = 44,000 psi o Sut = 53,000 psi o Se = 26,500 psi o Density =.284 lb/in 3 o Poisson s Ratio = 0.3 o Young s Modulus = 30x10 6 psi The calculations for the safety factors of the shaft under bending and shear are as follows: Maximum Shear Stress Theory for Static Failure τ max = ( σ x σ y 2 N MSST = 0.5S y τ max = ) 2 + τ 2 xy = ( psi 0 ) 2 + ( psi) 2 = psi ,000 psi psi 2 (Eq.37) = (Eq.38) Distortion Energy /Von Mises-Hencky Theory for Ductile Static Loading Failures σ 1 = σ x σ z 2 σ 3 = σ x+σ z 2 + τ max = τ max = psi psi psi = psi (Eq.39) psi = psi (Eq.40)

24 σ = σ σ 3 2 σ 1 σ 3 = ( psi) 2 + ( psi) 2 ( )( ) = psi (Eq.41) N DE = S y 44,000 psi σ = = 5.59 (Eq.42) psi Using the fact that the maximum shear theory is more conservative, our safety factor comes out to be This is more than reasonable for the level of precision required in this type of machine. Stress calculations on Shaft Keys We will be attaching the gear and sprockets to the shaft via profile parallel keys and keyways. The keys will be made from the same SAE 1010 Steel; however, the ultimate strength will be used to define the safety factor. This will be done such that the key will fracture and prevent the other components from spinning in the case of failure, protecting the remaining system from damage. From Table 10-2 (Norton) for a shaft having a diameter equal to 1.75 inches the suggested key width is inches. From Table 10-3 (Norton), the suggested height for a.375 inch by 1 inch long key is.437 inches. The length of our keys will be greater than or equal to the width of the element they are connecting to the shaft (i.e. gear width = 1.714, key length = 1.75 inch.). The force experienced by the shaft at the different locations along the shaft is equal to the torque at that location over the radius of the shaft at that location: F = T r F 1 = T r F 2 = T r = lbft 1.75 in 2 = lbft 1.75 in 2 = lbft 1.75 in 2 ( 12in ) = lbs (Eq.43) ft ( 12in ) = lbs (Eq.44) ft ( 12in ) = lbs (Eq.45) ft From these forces, the corresponding shear stresses can be computed for the keys at these locations. The forces on each sprocket are equal, thus simplifying the calculations. Once the actual shear stresses are calculated, the Von-Mises equivalent stresses can be computed and a safety factor developed. The width of the sprocket and key interaction is 1.25 inches, the width of the gear and key interaction is inches. All keys are inches in width. τ key@gear = τ key@sprockets = F lbs = A shear (0.375 in)(1.714 in) F lbs = A shear (0.375 in)(1.25 in) = psi (Eq.46) = psi (Eq.47) *Note that the shear stress on the key for the sprockets is equal.

25 σ = 3τ 2 xy = 3( psi) 2 = psi (Eq.48) σ key@sprockets = 3τ 2 xy = 3( psi) 2 = psi (Eq.49) *Note that the von-mises equivalent stresses on the key for the sprockets is equal. Calculation of the safety factors for the gear and sprocket keys. (Eq 6.18e pg 368 Norton) 1 N f = σ a Se +σ m where σ a = σ m (Eq.50) S ut 1 N gear = psi = psi psi 53000psi (Eq.51) 1 N sprockets = psi = psi psi 53000psi (Eq.52) Bearing Calculations and Design In order to find an acceptable bearing for our application we first needed to find the forces acting at the bearings. These forces are reactions to the radial and tangential components of the forces acting at the gear and sprockets on the shaft. The locations of the bearings, gear and sprockets is shown in the torque diagram in the appendix. We assumed no axial movement of the shaft because the shaft is locked in position at the bearings by clamp collars. Also note that the sprockets have no y-component of force because the only force experienced is that of the chain acting in the x direction. Calculations of the forces at the gear and sprockets are shown below: F gx = T g r g = lbft 7.86 in ( ) 2 ( 12 in ft ) = lbs (Positive x-direction) F gy = F gx tan( ) = lb tan(20 ) = lbs (Positive y-direction) F s1x = F gx T s1 r s1 = lbs F s1x = 0 F s2x = F s1x T s2 r s2 = lbs F s2x = lbft in in (12 ( ) ft lbft in in (12 ( ) ft 2 ) = lbs (Negative x-direction) ) = 6.58 lbs (Negative x-direction) It should be noted that the gear supplies a torque and the first sprocket removes roughly half of this torque to drive the chain. The remaining torque is transmitted to the second sprocket in order to drive the second chain. A minimal amount of force remains after the second sprocket. The bearing reaction forces are then found by taking the sum of the forces about the shaft and the

26 sum of the moments about one of the bearings. The magnitude of the total bearing forces is shown below: R B1 = lbs R B2 = lbs From these forces, specifically the higher valued force at bearing one, we can calculate the required dynamic load rating of the bearings to be used. The bearing housing we assumed earlier in our design is manufactured by Rexnord Inc. We referred to their lengthy product catalog and found they had a step by step process for determining the dynamic load rating based on the materials and tolerances they use in their bearing design. The number of cycles for the shaft is based on the 17 rpm required to maintain a 35 feet per minute velocity of the pallet. We designed the gear to withstand 5 years of service, and the total number of cycles in this period was calculated previously to be 1.02 x 10 7 cycles. Using this value and the instructions from Rexnord, we were able to choose a bearing. The procedure provided by Rexnord is shown below: 1. Determine the number of hours the bearing is to be used over a 10 year period. a. hours = 8 hr 250 day 10yr = hrs day yr 2. Refer to Rexnord Table that correlates the number of hours in 10 years, and the RPM to the C/P ratio. Choose a corresponding C/P ratio. a. The lowest RPM that Rexnord has values for was 50 RPM, where we only required 17 RPM. We chose the corresponding C/P ratio for the 50 RPM b. The C/P ratio was shown to be Use the C/P ratio from the table and the actual applied load (P) to determine the required Dynamic Load Rating (C). C P = 3.42 C = 3.42 C = (379.15lb)(3.42) = lbs The dynamic load rating required is only lbs. Of the bearings available to fit in our housing, the lowest dynamic load rating was 20,200 lbs. We chose this double row roller ball bearing, though expensive, it would last must longer than all other components and require minimal maintenance. The part number and other information about this bearing are found in the appendix. We chose the method of calculating the bearing dynamic load rating from the manufacturer because the typical formula L 10 = ( C P )3 does not account for the bearing material makeup, hardness, revolutions per minute and manufacturing methods. The safety factor of our bearing is as follows: N B = C bearing = = (Eq.53) C required

27 Summary of Safety Factors and Allowable Stresses The design of this drag chain conveyor involves many complex parts that each must be analyzed with respect to the loads they are asked to transmit. Each component was analyzed and prescribed a safety factor. This safety factor is a simple ratio of the maximum stress allowed on that component before it fails to the actual stress it experiences. The higher the ratio the safer, or more overdesigned, that component will be. Below is a summary of the safety factors for each of our components. We list the lowest safety factor that was calculated for each component to remain consistent to a conservative design approach. Gear o Bending Safety Factor (N bg ) = 4.76 Pinion o Bending Safety Factor (N bp ) = 3.98 o Surface Fatigue Safety Factor(N cp ) = 2.38 Shaft o Shearing Safety Factor (N MSST ) = o Bending Safety Factor (N DE ) = 5.59 Bearing o Dynamic Load Safety Factor (N B ) = Shaft Keys o Safety Factor of Gear Key (N fkey ) = 2.95 o Safety Factor of Sprocket Keys (N fkeys ) = 4.30 It is evident from these values that the first component to fail will be the keys. This is consistent with the purpose of shaft keys. Keys are designed to lock a power transmitting device to the shaft and help transmit any rotational forces. If these forces become so much as to exceed the strength of the key, the key will fail prior to the shaft or transmitting device fails. This is optimal because the failure of the key keeps the main components of the system from experiencing forces that are over their design stresses. Keys are cheap to produce and provide a means of protection to the more expensive components.

28 Conclusion In conclusion, the drive shaft conveyor was designed to move a maximum load of 3000 pounds at a speed of 35 feet per minute. The root of all calculations was the force required to accomplish this task. Rather than defining our starting torque at the motor and working towards a feasible load that could be moved, we started with a typical load required in industry and worked backwards. This process proved successful, as we were able to design components specifically for the applied load. The lowest safety factor was 2.38 for the pinion under surface fatigue stresses. This is understandable because of the high number of cycles experienced by the pinion, relative to the remaining components. A safety factor of 2.38 is an acceptable conservative number in relation to the low speed operation of the conveyor. When the pinion fails according to this safety factor, it is due to pitting on the pinion teeth. Pitting is not a catastrophic failure and the component may still be used beyond this point for some time. As for the remaining components, the keys connecting the gear and sprockets would be the most likely to fail. This is a desirable result, as the keys act to protect the other components from overloading. The bearings were purposely overdesigned to accommodate a much longer life of the conveyor. The shaft had a very high safety factor in order to avoid bending stresses from repeated rotation under constant torque. Our initial assumption for the life expectancy of the system was 5 years of 250 days per year and 8 work hours per day. The calculated safety factors are all based on this assumption, with exception of the bearing which is rated for 10 years. Our final results show that the design should be successful under normal operating conditions implied during calculations.

29 References 1. Norton, Robert L., Machine Design: An Integrated Approach, 4 th edition, 2011, Chapters 5,6, Mott, Robert L., Machine Elements In Mechanical Design,1985, Chapters Taylor et al, Standard Handbook of Chains: Chains for Power Transmission and Material Handling, 2 nd edition, AGMA, Geometry Factors for Determining Pitting Resistance and Bending Strength of Spur, Helical and Herringbone Gear Teeth, AGMA Information Sheet , Rev B89, Rexnord, Link-belt, MB and Rexnord Bearings Product Catalog and Selection Guide, Rexnord Spherical Roller Bearings-Engineering Section, pp5-7, Kana Chain, ANSI Chain Designations 7. ASM Handbook, Characterization and Failure Analysis of Plastics, ASM International, 2003

30 APPENDIX Gear and Pinion Dimension Calculations in Microsoft Excel

31 Bending and Surface Stresses on Pinion and Gear Excel Screen Shot Bending and Surface Stresses on Pinion and Gear Excel Screen Shot

32 Shaft Torque Diagram Sprocket 1 Sprocket 2 Bearing 2 Gear Bearing 1 Shaft Shear Diagram Shaft Shear Stress Diagram

33 Shaft Bending Moment Diagram

34 Shaft Bending Stress Diagram Shaft Torsional Stress Diagram

35