ABE - Machine Design Handout PDF

0 5 /0 9 /2 0 2 4 Agricultural and Biosystems Engineering Licensure Exam Review on Machine Design Engr. Ma. Camille G. Acabal, MSc AE University of the Philippines Los Baños Design iterative decision-making process to formulate a plan resource...

0 5 /0 9 /2 0 2 4 Agricultural and Biosystems Engineering Licensure Exam Review on Machine Design Engr. Ma. Camille G. Acabal, MSc AE University of the Philippines Los Baños Design iterative decision-making process to formulate a plan resources are optimally converted into systems, processes or devices to solve a specific problem or to fulfill a specific need. 1  0 5 /0 9 /2 0 2 4 Design Identification of need Process Definition of problem Synthesis Analysis and optimization Evaluation Presentation Machine Design Machine A combination of linkages with definite motion, capable of performing useful work. 2 Machine Design Creation of plans for the machine to perform desired functions.  0 5 /0 9 /2 0 2 4 Steps in Machine Design 1. Define Problem- state the problem to be solved or the desired purpose of the machine 2. Select Mechanisms- possible mechanisms to achieve the desired motion or set of motions 3. Determine Forces- analyze forces acting on and energy transmitted by each machine element 4. Select Materials 5. Stress & Deflection- determine allowable stress and deflection values based on material and functional requirements 6. Design Elements- define size and shape to withstand applied loads 7. Modify Dimensions 8. Create Drawings Design Considerations Strength Noise Stiffness Safety Functionality Manufacturability Wear Marketability 3 Corrosion Corrosion Utility Surface Finish Thermal Properties Weight  0 5 /0 9 /2 0 2 4 Definition of Terms Constraints anything that limits the designer’s freedom of choice all the constraints should be clearly defined in the problem definition Optimization repetitive process of refining a set of criteria to achieve the best compromise. Static load does not change in magnitude or direction and gradually increases to a steady value Dynamic load changes in magnitude or direction or both with respect to time 4 Failure Criteria when a criterion crosses the acceptable limit (e.g. Stress or deflection of a particular element)  0 5 /0 9 /2 0 2 4 Factor of Safety reserve strength provided for any unexpected or unpredicted conditions depends on the effect of failure, type of load, accuracy in load calculations, material selected, desired reliability, service conditions, manufacturing quality, and cost Young’s Modulus or Modulus of Elasticity object's or substance's resistance to being deformed elastically when stress is applied to it Allowable Stresses strength of the material is divided by factor of safety to obtain allowable stress or design stress used to determine the dimensions of the components 𝑆𝑦 𝜎𝑎 = 𝑓𝑜𝑟 𝑑𝑢𝑐𝑡𝑖𝑙𝑒 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙𝑠 𝑓𝑜𝑠 5 𝑆𝑢𝑡 𝜎𝑎 = 𝑓𝑜𝑟 𝑏𝑟𝑖𝑡𝑡𝑙𝑒 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙𝑠 𝑓𝑜𝑠 where Sy is yield strength Sut is ultimate tensile strength fos is factor of safety  0 5 /0 9 /2 0 2 4 Introduction to Stresses Definition Internal resistance to external force, force per unit area 𝑷 where 𝝈= 𝑨 P is external load A is area of the component Types Tensile Compressive Shear Bending Torsional Tensile Stress Occurs under pulling/stretching force, leading to elongation Tie rod in structures, connecting rod in engines where 𝑷 P is external load 𝝈𝒕 = ≤ 𝝈𝒂 6 𝑨 A is cross-sectional area of the component  0 5 /0 9 /2 0 2 4 Compressive Stress Occurs under compressive force, causing shortening Columns in buildings, pistons in engines 𝝈𝒄 = 𝑷 ≤ 𝝈 𝒂 𝑨 where P is external load A is cross-sectional area of the component Deformation tensile or compressive strain per unit length 𝒄𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒍𝒆𝒏𝒈𝒕𝒉, 𝝏𝒍 𝜺= 𝒐𝒓𝒊𝒈𝒊𝒏𝒂𝒍 𝒍𝒆𝒏𝒈𝒕𝒉, 𝒍 Hook's Law: stress is directly proportional to strain (within the elastic limit) 7 𝝈∞𝜺 𝒐𝒓 𝝈 = 𝑬𝜺 Where E is known as Young’s Modulus or Modulus of Elasticity (207 kN/mm2 for Carbon Steels, 100 kN/mm2 for Grey Cast Iron)  0 5 /0 9 /2 0 2 4 Shear Stress Occurs when layers slide past each other due to force Rivets and bolts, gear teeth 𝑷 where 𝝉= P is external load 𝑨 A is area parallel to the shearing force Shear Strain angular distortion of a material due to an applied shear stress ∆𝒙 where 𝜸= γ is shear strain in radians 𝒕 Δxis amount of sliding t is thickness of the material 8 Related to shear stress: 𝝉=𝜸∗𝑮  0 5 /0 9 /2 0 2 4 Bending Stress Occurs due to bending moment, tension on one side, compression on the other Distribution of bending stress is linear Beams in structures, crankshafts 𝑴𝒚 𝝈𝒃 = 𝑰 M is applied bending moment I is moment of inertia y is the distance of the fiber from the neutral axis Moment of Inertia for Common Cross-Sections Cross-Section I Parameters Rectangular 𝑏ℎ3 b: Width, h: Height 𝐼= 12 Circular 𝜋𝑟4 r: Radius 𝐼= 4 9 𝜋 Hollow Circular 𝐼 = (𝑅4-𝑟4) R: Outer radius, rrr: 4 Inner radius Triangular 𝑏ℎ3 b: Base, h: Height 𝐼= 36  0 5 /0 9 /2 0 2 4 Torsional Shear Stress - Occurs due to twisting force, causing shear stress over the cross-section - Drive shafts, helical gears where 𝑻𝒓 T is applied torque 𝝉𝑻 = r is radial distance 𝑱 J is polar moment of inertia Polar Moment of Shafts Cross-Section J Parameters Solid Circular Shaft 𝜋𝑟4 r: Radius 𝐽= 2 𝜋 Hollow Circular 𝐽 = (𝑅4-𝑟4) R: Outer radius, rrr: 2 Inner radius 10 Triangular 𝑏ℎ3 b: Base, h: Height 𝐽= 36  0 5 /0 9 /2 0 2 4 Bearing Stress Contact stress occurring in joints where a fastener presses against the material it is installed in Bolts, rivets, pins if n is the total number of rivets used, total projected area will become n x d x t 𝑃 where 𝜎𝑏 = P is applied force 𝐴 = 𝑑𝑡 A is contact/projected area Crushing Stress Compressive stress developed at the contact area between two bodies under load. Compressive loads on concrete blocks, or any compressive contact between two surfaces. 𝑷 11 𝝈𝒄𝒓 = where 𝑨 P is applied force A is contact area  0 5 /0 9 /2 0 2 4 Thermal Stress Occurs when a material or component expands or contracts due to temperature changes 𝜹𝒍 𝒍𝜶𝒕 𝑻𝒉𝒆𝒓𝒎𝒂𝒍 𝑺𝒕𝒓𝒂𝒊𝒏, 𝜺𝒕𝒉 = = = 𝜶𝒕 𝒍 𝒍 𝑻𝒉𝒆𝒓𝒎𝒂𝒍 𝑺𝒕𝒓𝒆𝒔𝒔, 𝝈𝒕𝒉 = 𝜺𝒕𝒉 𝑬 = 𝜶𝒕𝑬 Where l is original length of the member 𝛼 is coefficient of thermal expansion t is rise or fall of temperature KNUCKLE JOINT transfer loads in joints connect two rods subjected to axial tensile loads valve rod and eccentric rod connections link of a cycle chain 12 tie rod joints for roof trusses D= diameter of each rod (mm) D1 = enlarged diameter of each rod (mm) d = diameter of knuckle pin (mm) d0 = outside diameter of eye or fork (mm) d1 = diameter of pin head (mm) a = thickness of each eye of fork (mm) b = thickness of eye end of rod B (mm) http://mechanicalwalkins.com/ x = distance of the center of fork radius Rfrom the eye (mm)  0 5 /0 9 /2 0 2 4 Pin Analysis Shear Failure of Pin : Pin Analysis 13 Crushing Failure:  0 5 /0 9 /2 0 2 4 Pin Analysis Bending Failure of Pin: EYE ANALYSIS Tensile, crushing and shear stresses: 14  0 5 /0 9 /2 0 2 4 FORK ANALYSIS Tensile, crushing, and shear stresses: SHAFTS Transmit power and rotational motion. Provides the axis of rotation or oscillation of gears, pulleys, flywheels, cranks, sprockets Shaft Materials: 15 Hot-Rolled Plain Carbon Steel: Economical, commonly used Cold-Rolled Plain Carbon Steel: Better yield and endurance strength Alloy Steels: Used in severe loading/corrosive conditions, containing elements like Ni, Cr, Mo, and V  0 5 /0 9 /2 0 2 4 TYPES OF SHAFTS Axles: Stationary, supports rotating wheels/pulleys Transmission Shafts: Transmit power between sources and receivers Machine Shafts: Integral to the machine itself (crankshafts) Lineshafts/Mainshafts: Driven by prime movers. Countershafts/Jackshafts/Headshafts: Intermediate between a lineshaft and driven machine. Spindles: Short shafts used on machine Stress Analysis Torsional stress, bending stress, and combined loading 𝑻𝒓 𝑴𝒚 𝝉𝑻 = 𝝈𝒃 = 𝑱 𝑰 16  0 5 /0 9 /2 0 2 4 Torsional rigidity 𝑻𝑳 𝜽= ≤ 𝜽𝒂 𝑮𝑱 Where, T= Torque applied L = Length of the shaft J = Polar moment of inertia of the shaft about the axis of rotation G = Modulus of rigidity of the shaft material Sample Problem 1: What is the speed (rpm) of 63.42 mm (diameter) shaft transmitting 75 kW if stress is not to exceed 26 MPa? 𝜏= 26000 kPa 𝑻𝒓 𝑃 = 2𝜋𝑇𝑁 P= 75kW 𝝉𝑻 = r=0.03171 m 𝑱 75𝑘𝑊 = 2𝜋(1.302𝑘𝑁 − 𝑚)𝑁 17 J=𝜋𝑟4/2 2𝑇 𝑟 𝑠𝑒𝑐 26000𝑘𝑃𝑎 = 𝑁 = 9.168 𝑟𝑝𝑠 𝑥60 (𝜋𝑟4) 𝑚𝑖𝑛 2𝑇 𝑁 = 550 𝑟𝑝𝑚 26000𝑘𝑃𝑎 = 𝜋(0.031713) 𝑇 = 1.302𝑘𝑁 − 𝑚  0 5 /0 9 /2 0 2 4 Sample Problem 2: A steel shaft transmits 50hp at 1400rpm. If allowable stress is 500psi, find the diameter. 𝑇𝑟 𝑃 = 2𝜋𝑇𝑁 𝜏𝑇 = 𝜏= 500 psi 𝐽 33000𝑙𝑏 − 𝑓𝑡/𝑚𝑖𝑛 P= 50 hp 50 ℎ𝑝 𝑥 = 2𝜋𝑇(1400𝑟𝑝𝑚) 1 ℎ𝑝 2𝑇 𝑟 N= 1400 rpm 12in 500 𝑙𝑏/𝑖𝑛2 = (𝜋𝑟 4 ) J=𝜋𝑟4/2 𝑇 = 187.575 lb − ft x ft (2250.91𝑙𝑏 − 𝑖𝑛)(2) 𝑇 = 2250.91 𝑙𝑏 − 𝑖𝑛 500 𝑙𝑏/𝑖𝑛2 = 𝜋(𝑟 3 ) r= 1.421 𝑖𝑛; d= 2.842 in Sample Problem 3: The shaft of the motor has a length of 20 times its diameter and has a maximum twist of 1 degree when twisted at 2kN-m torque. If the modulus of rigidity of the shaft is 80GPa, find the shaft diameter. 𝑇𝐿 𝜃= ≤ 𝜃𝑎 𝐺𝐽 𝐿= 20d 𝜋 1° 𝑥 𝜋 = (2 𝑘𝑁 − 𝑚) (20𝑑) 18 𝜃= 1° =180 180 (𝜋𝑑4) (80𝑥106 𝑘𝑃𝑎) 32 G= 80x106 kPa J=𝜋𝑑4/32 137077.8389 𝑑4 = 40𝑑 T=2kN-m 𝑑 = 6.633 𝑐𝑚  0 5 /0 9 /2 0 2 4 Sample Problem 4: A hollow shaft developed a torque of 5 kN-m and shaft stress is 40 MPa. If outside diameter of the shaft is 100 mm, determine the shaft inner diameter. 𝑇𝑟 𝜏𝑇 = 𝐽 𝜏= 40 x103 kPa T= 5 kN-m 𝜏 = 𝑇 𝑟 (32) 𝜋(𝑑𝑜4−𝑑𝑖4) do=0.1m J=𝜋(𝑑𝑜4-𝑑𝑖4)/32 (5 𝑘𝑁 − 𝑚) (0. 05𝑚)(32) 40000 𝑘𝑃𝑎 = 𝜋(0.14−𝑑𝑖4) 𝑑𝑖=0.07764 m or 7.7 cm Sample Problem 5: A hollow shaft has 100mm outside diameter and 80mm inside diameter is used to transmit 100 kW at 600 rpm. Determine the shaft stress. 𝑇𝑟 𝑃= 100 kW 𝑃 = 2𝜋𝑇𝑁 𝜏𝑇 = N= 600 rpm 𝐽 1 𝑚𝑖𝑛 do=0.1m 100 𝑘𝑊 = 2𝜋𝑇(600𝑟𝑝𝑚 𝑥 ) 1.592 𝑘𝑁 − 𝑚 (.05)(32) di=0.08m 60 𝑠 𝜏= 2𝜋(𝑑𝑜4−𝑑𝑖4) 19 J=𝜋(𝑑𝑜4-𝑑 𝑖4)/32 𝑇 = 1.592 𝑘𝑁 − 𝑚 (1.592 𝑘𝑁 − 𝑚) (0. 05𝑚)(32) 𝜏= 𝜋(0.14−0.084) 𝜏 = 13 7330 kPa  0 5 /0 9 /2 0 2 4 Sample Problem 6: What is the polar moment of inertia of a solid shaft that has a torque of 1.5 kN-m and a stress of 25 MPa? 𝑇𝑟 𝑇= 1.5 kN-m 𝜏𝑇 = 𝐽 𝜏 = 25 x103 kPa do=0.1m 𝑇𝑟(2) di=0.08m 𝜏= 𝜋𝑟4 J=𝜋𝑟4/2 (1.5 𝑘𝑁 − 𝑚) (𝑟)(2) 25000 kPa = 𝜋(𝑟4) r= 0.0337 m 4 J=𝜋𝑟 = 2.02x10-6 2 KEYS piece inserted in an axial direction between a shaft and hub of the mounted machine element such as pulley or gear transmit torque from shaft to hub of the mating element subjected to shearing and compressive stresses caused by the torque 20 Keyway a slot machined in the shaft or in the hub or both to accommodate the key reduces the strength of the shaft as it results in stress concentration  0 5 /0 9 /2 0 2 4 Key Materials ductile materials hardened and tempered steel of grades C30, C35, C40, C50 brass and stainless keys are used in corrosive environment factor of safety of 3 to 4 Types of Keys Sunk keys- ½ shaft+ ½ hub Saddle keys Tangent keys Round keys Splines SUNK KEYS Rectangular sunk key taper of about 1 in 100 is provided Square sunk key equal width and thickness Parallel sunk key no taper is provided Gib-head key rectangular sunk key 21 with a head parallel key made as an Feather key integral part of the shaft sunk key in the form of a Woodruff key semicircular disc of uniform thickness  0 5 /0 9 /2 0 2 4 Saddle keys - slot is only in the hub; used only for light applications flat saddle key, the bottom surface touching the shaft is flat; sits on the flat surface machined on the shaft. hollow saddle key has a concave surface at the bottom to match the circular surface of the shaft. flat saddle key- less slip, more power t Tangent keys - used to transmit high torque in one direction Round keys - into holes drilled partly in the shaft and partly in the hub -used for low torque transmission 22 Spline keys -made as an integral part of the shaft -high torque transmission e.g. in automobile transmission - permit the axial movement.  0 5 /0 9 /2 0 2 4 design of sunk keys b- width h- height l- length P-force P’ – resisting force that keeps the key in position d- diameter of shaft design of sunk keys Shear stress Crushing stress 𝑃 𝑃 𝜏= 𝜎𝑐𝑟 = 𝑏𝑙 ℎ 𝑙 23 2 Rectangular Key: b = d / 4 and h = d / 6 Square Key: b = h = d /4  0 5 /0 9 /2 0 2 4 SLEEVE hollow cylinder fitted on the ends of input and output shaft made of cast iron (factor of safety of 6-8) Torsional shear stress Standard proportions: Outer diameter of the sleeve, D = 2d + 13 Length of the sleeve, L = 3.5d where d is the diameter of the shaft COUPLINGS, CLUTCHES, SPLINE Couplings used to connect two rotating shafts to transmit torque from one to the other 24  0 5 /0 9 /2 0 2 4 TYPES: Rigid couplings -shafts are colinear; in a fixed angular relation with respect to each other 1. Clamp shaft- split and bolted sleeve coupling, proportioned to clamp firmly on the shafts 2. Flange -permanent installations; heavy loads and large sizes; vertical drives (agitators) 3. Sleeve/ Muff- hollow cylinder fitted on the ends of input and output shaft; made of cast iron (factor of safety of 6-8) Outer diameter of the sleeve, D = 2d + 13 Length of the sleeve, L = 3.5d where d is the diameter of the shaft Flexible couplings -misaligned laterally or angularly; absorb impacts 1. Oldham (double slider)-eliminates the need for large clearances and noisy backlash 2. Rubber-bushed- self-lubricated bronze bushings, rubber-cushioned in the opposite flange; electrical insulation 3. Gear-type coupling- have integral external gear teeth that mesh with internal teeth in the casing 25 4. Roller chain flexible coupling- two opposing hubs with integral sprockets over which a double roller chain is fitted 5. Rubber-flexible coupling- torque is transmitted through rubber in compression; quietness is desired 6. Universal joint -larger values of misalignment than can be tolerated by the other types of flexible couplings.  0 5 /0 9 /2 0 2 4 COUPLINGS, CLUTCHES, SPLINE Clutches couplings which permit the disengagement of the coupled shafts during rotation Types: Friction– reduce coupling shock and torque by slipping Centrifugal- produces torque by centrifugal force of weights pressing against the driving or frictionally driven member Conical friction- frustum of a cone, fitted to shaft by feather key Positive clutches- transmit torque without slip, jaw clutches (square jaw or spiral jaw); slow-moving shafts COUPLINGS, CLUTCHES, SPLINE Splines used for the transmission of power from a shaft to hub or vice versa Types: Square splines- employed in multiple-spline fittings having 4, 6, 10, or 16 splines Involute splines- multiple keys in the general form of internal and 26 external gear teeth; prevent relative rotation of cylindrically fitted machine parts Recommended design practice: 𝟏𝟎𝟎 𝟑 Where 𝑷𝒏 = 𝑺𝒇𝑷𝒓( )𝟒 Pn is nominal power at 100rpm, Watts 𝑵𝒓 Sf is service factor Pr is required power, Watts Nr is required rpm  0 5 /0 9 /2 0 2 4 BELTS when there is larger center distance between the shafts simple, less noisy, have low initial and maintenance cost; absorbs shock loads and damping vibrations Power is transmitted from the driver pulley to the belt and from the belt to the driven pulley by friction Slippage- when the friction between a belt and the pulleys it drives is insufficient to transmit the required power or motion DESIGN OF BELTS Types: Flat belts- used to transmit moderate amount of power between shafts less than 6m apart V-belts- sed along with pulleys having similar cross-section as that of the belt 27 Ropes- used to transmit large power between shafts more than 10m apart Sheaves- grooved pulleys, used with the ropes  0 5 /0 9 /2 0 2 4 DESIGN OF BELTS Types of belts drives: DESIGN OF BELTS 𝑵𝟐 𝒅𝟏 𝑽𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝑹𝒂𝒕𝒊𝒐 = = 𝑵𝟏 𝒅𝟐 𝑵𝟐 𝒅𝟏 + 𝒕 𝑽𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝑹𝒂𝒕𝒊𝒐 = = 𝑵𝟏 𝒅𝟐 + 𝒕 𝑵𝟐 𝒅𝟏 + 𝒕 𝒔 28 𝑽𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝑹𝒂𝒕𝒊𝒐 = = (𝟏 − ) 𝑵𝟏 𝒅𝟐 + 𝒕 𝟏𝟎𝟎 Where N1, d1 = speed (rpm) and diameter of driving pulley N2, d2 = speed (rpm) and diameter of driven pulley t = thickness of belt s=slip  0 5 /0 9 /2 0 2 4 DESIGN OF BELTS Length of the Belt 𝟐 𝑳 = 𝟐𝑪 + 𝝅 𝒅 + 𝑫 + (𝒅−𝑫) (open belt drive) 𝟐 𝟒𝑪 𝟐 𝑳 = 𝟐𝑪 + 𝝅 𝒅 + 𝑫 + (𝒅+𝑫) (cross belt drive) 𝟐 𝟒𝑪 Where d & D are diameters of smaller and larger pulleys respectively C as the center distance between the axes of the pulleys DESIGN OF BELTS Angle of Wrap - smaller (𝛼s) and larger pulleys (𝛼l) 𝑫−𝒅 𝜶𝒔 = 𝟏𝟖𝟎° − 𝟐𝒔𝒊𝒏−𝟏 (𝒐𝒑𝒆𝒏 𝒃𝒆𝒍𝒕) 𝟐𝑪 29 𝑫−𝒅 𝜶𝒍 = 𝟏𝟖𝟎° + 𝟐𝒔𝒊𝒏−𝟏 (𝒐𝒑𝒆𝒏 𝒃𝒆𝒍𝒕) 𝟐𝑪 𝑫+𝒅 𝜶𝒔 = 𝜶𝒍 = 𝟏𝟖𝟎° + 𝟐𝒔𝒊𝒏−𝟏 (𝒄𝒓𝒐𝒔𝒔 𝒃𝒆𝒍𝒕) 𝟐𝑪  0 5 /0 9 /2 0 2 4 DESIGN OF BELTS Belt Tension due to tangential force Tension in tight side, T1 = Ti + dT Tension in slack side, T2 = Ti - dT 𝑻𝟏 + 𝑻𝟐 𝑻𝒊 = 𝟐 𝑻𝟏 = 𝒆𝝁𝜽 (𝒇𝒍𝒂𝒕 𝒃𝒆𝒍𝒕 𝒅𝒓𝒊𝒗𝒆) μ= coefficient of friction between belt & pulley 𝑻𝟐 Θ= angle of contact 𝑻𝟏 β= half of the groove angle of v-belt = 𝒆𝝁𝜽 𝒄𝒐𝒔𝒆𝒄𝜷 (𝒄𝒓𝒐𝒔𝒔 𝒃𝒆𝒍𝒕 𝒅𝒓𝒊𝒗𝒆) 𝑻𝟐 DESIGN OF BELTS Belt Tension due to centrifugal force Centrifugal tension Tc = mv2 m = mass of the belt per unit length v = velocity of the belt in m/s 30 Tmax = T1 + TC If b and t are width & thickness of a flat belt: 𝑻𝒎𝒂𝒙 𝝈𝒎𝒂𝒙 = 𝒃𝒕  0 5 /0 9 /2 0 2 4 Sample Problem 7: 5 kN force tangent to 1 foot diameter pulley is mounted on a 2 inches shaft. Determine the torsional deflection if G= 83 x106 kPa. 𝑇𝐿 𝜃= 𝐺𝐽 F=5 kN 𝜃 0.762 𝑘𝑁 − 𝑚 = D= 12 𝑖𝑛 = 0.3048 𝑚 𝐿 𝜋(0.0254)4 d= 2 in= 0.0508 m 83𝑥106𝑘𝑃𝑎 𝑥 2 G= 83x106 kPa 𝜃 𝑟𝑎𝑑 180 L is not given = 0.0104 𝑥 = 0.805° 𝐿 𝑚 𝜋 J=𝜋𝑟4/2 0.3048𝑚 𝑇 = 𝐹𝑥𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 5 𝑘𝑁 𝑥 = 0.762 kN − m 2 Sample Problem 8: A 12 cm pulley turning at 600 rpm is driving a 20 cm pulley by means of belt. If total belt slip is 5%, determine the speed of driving gear. 𝑵𝟐 𝒅𝟏 𝑽𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝑹𝒂𝒕𝒊𝒐 = = (𝟏 − 𝐬) 𝑵𝟏 𝒅𝟐 𝑁2 12 𝑐𝑚 𝑁1 = 600 𝑟𝑝𝑚 = 31 d1= 12 cm 600 𝑟𝑝𝑚 20 𝑐𝑚 d2= 20 cm 𝑁2 = 360 𝑟𝑝𝑚 𝑥 (1 − 0.05) Slip, S= 5% 𝑁2 = 342 𝑟𝑝𝑚  0 5 /0 9 /2 0 2 4 Sample Problem 9: The torque transmitted in a belt connected to 300 mm diameter pulley is 4 kN-m. Find the power driving pulley if belt speed is 20 m/sec. 𝑷 = 𝟐𝝅𝑻𝑵 V= 20 𝑚𝑠 d= 0.3 m 𝑃 = 2𝜋(4𝑘𝑁 − 𝑚)(21.22 𝑟𝑝𝑠) T= 4 kN-m 𝑃 = 533.32 𝑘𝑊 𝑉 = 2𝜋𝑟𝑁 20 m/s = 𝜋(0.3 𝑚)𝑁 𝑁 = 21.22 𝑟𝑝𝑠 Sample Problem 10: Pulleys connected with belt have diameters of 10 cm and 30 cm. If center distance is 50 cm, determine the angle of contact of smaller pulley. −𝟏 𝑫−𝒅 𝜶𝒔 = 𝟏𝟖𝟎° − 𝟐𝒔𝒊𝒏 (𝒐𝒑𝒆𝒏 𝒃𝒆𝒍𝒕) 𝟐𝑪 𝑫−𝒅 𝐶 = 50 𝑐𝑚 𝜶𝒍 = 𝟏𝟖𝟎° + 𝟐𝒔𝒊𝒏−𝟏 (𝒐𝒑𝒆𝒏 𝒃𝒆𝒍𝒕) 𝟐𝑪 32 d= 10 cm D= 30 cm 30 − 10 𝛼𝑠 = 180° − 2𝑠𝑖𝑛−1 = 156.93° 2(50  0 5 /0 9 /2 0 2 4 Sample Problem 11: A compressor is driven by an 18 kW motor by means of V-belt. The service factor is 1.4 and the corrected horsepower capacity of V-belt is 3.5 hp. Determine number of belts needed. 𝑃𝑎 = 𝑆𝐹 (𝑃𝑚) 𝑃𝑎 = 1.4 (18 𝑘𝑊) Pm= 18 𝑘𝑊 𝑃𝑎 = 25.2 𝑘𝑊 SF=1.4 Pb= 3.5 hp/belt 𝑃𝑎 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑏𝑒𝑙𝑡𝑠 = 𝑃𝑏 25.2 𝑘𝑊 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑏𝑒𝑙𝑡𝑠 = = 10 (3.5 hp/belt )(0.746 𝑘𝑊/ℎ𝑝) Sample Problem 12: A pulley 600 mm in diameter transmits 50 kW at 600 rpm by means of belt. Determine the effective belt pull in kN. 𝑃 = 2𝜋𝑇𝑁 𝑇 = 𝐹(𝑟) 50 𝑘𝑊 = 2𝜋(𝑇)(10 𝑟𝑝𝑠) 0.7958 𝑘𝑁 − 𝑚 = 𝐹(0.3𝑚) 33 𝐹 = 2.653 𝑘𝑁 𝑇 = 0.7958 𝑘𝑁 − 𝑚 d=0.6 m P= 50 kW N= 600 rpm x1min/60 sec = 10rps  0 5 /0 9 /2 0 2 4 Sample Problem 13: A pulley has an effective belt pull of 3 kN and an angle of belt contact of 160 degrees. The working stress of belt is 2 MPa. Determine the thickness of belt to be used if the width of belt is 350 mm and coefficient of friction is 0.32. 𝑻 𝒎𝒂𝒙 𝝈𝒎𝒂𝒙 = 𝑻𝟏 𝒃𝒕 = 𝒆𝝁𝜽 = 𝟐. 𝟒𝟒𝟒 𝑻𝟐 5.083 𝑘𝑁 F= 3kN= T1-T 2 2000𝑘𝑃𝑎 = 𝜃 = 160° = 2.793 𝑟𝑎𝑑 (0.35 𝑚)𝑡 𝐹 = 𝑇1 − 𝑇2 𝜎𝑡 =2000 kPa b=0.35 m 𝑇1 𝑡 = 7.26 𝑚𝑚 𝜇 = 0.32 3 𝑘𝑁 = 𝑇1 − 2.44 𝑇1 = 5.083 𝑘𝑁 Sample Problem 14: A pulley has a belt pull of 2.5 KN. If 20 hp motor is used to drive the pulley, determine the belt speed. 𝑷 = 𝑭𝒙 𝒗 F= 2.5 kN 14.92 𝑘𝑊 = 2.5 𝑘𝑁 (𝑣) 34 𝑃𝑚 = 20 ℎ𝑝 = 14.92 𝑘𝑊 𝑣 = 5.968 𝑚/𝑠  0 5 /0 9 /2 0 2 4 Sample Problem 15: An 8 in diameter pulley turning at 600 rpm is belt connected to a 14in diameter pulley. If there is a 4% slip, find the speed of 14 in pulley. 𝑵𝟐 𝒅𝟏 𝑽𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝑹𝒂𝒕𝒊𝒐 = = (𝟏 − 𝐬) 𝑵𝟏 𝒅𝟐 𝑁2 8 𝑖𝑛 = 𝑥 (1 − 0.04) 600 𝑟𝑝𝑚 14 𝑖𝑛 𝑁2 = 360 𝑟𝑝𝑚 𝑥 (1 − 0.04) 𝑁2 = 329. 14 𝑟𝑝𝑚 Sample Problem 16: A 25.4 cm pulley is belt-driven with a net torque of 339 N-m. The ratio of tension in the tight side of the belt is 4:1. What is the maximum tension in the belt? 𝑇1 25.4 =4 𝑇 = 𝐹𝑥 𝑟 𝐹 = 𝑇1 − 𝑇2 𝑇2 339 𝑁𝑚 = 𝐹𝑥 𝑟 35 𝑇1 = (889.764 𝑁)(4) 339 𝑁𝑚 = (0.127 𝑚)(𝑇1 − 𝑇2) 𝑇1 = 3559.0551 𝑁 𝑇2 = 889.764 𝑁 T2 T1  0 5 /0 9 /2 0 2 4 DESIGN OF GEARS used for the transmission of motion and power from one shaft to another by progressive engagement of teeth can transmit very large power can be operated at very low speeds very high efficiency require lesser space in comparison to the belt and chain drives TYPES OF GEARS Spur gears have teeth parallel to the axis of rotation and are used for parallel shafts simplest type of gears impose radial loads on the shafts 36 Helical gears used for parallel shafts teeth are inclined to the axis of rotation less noisy in comparison to the spur gears develop axial thrust loads in addition to the radial load used to transmit motion between nonparallel shafts  0 5 /0 9 /2 0 2 4 Bevel Gears have the shape of a truncated cone can have straight or spiral teeth. Miter bevel gears- equal numbers of driver and driven gear teeth and operate at axes with right angles Straight bevel gears- teeth are straight but the sides are tapered Worm Gears form of a threaded screw which engages with a wheel have very high reduction ratio. GEAR TERMINOLOGY 37  0 5 /0 9 /2 0 2 4 the line of intersection of the length of the arc of the pitch the pitch cylinder by a circle between two consecutive plane perpendicular to corresponding profiles which is the axis of the gear measured at the large end of the tooth; 𝑝=𝜋𝑑/𝑁 where N is number of teeth; d is pitch diameter the radial distance the circle that between the addendum bounds the outer circle and the pitch circle; ends of the teeth a= mwhere mis module (d/N) 38  0 5 /0 9 /2 0 2 4 theradial distance GEAR TERMINOLO GY between the dedendum circle and the pitch circle; b = 1.25m GEAR TERMINOLOGY the length of the arc of the pitch circle between two consecutive corresponding profiles 39 the amount by which the dedendum in a given gear exceeds the addendum of its meshing gear  0 5 /0 9 /2 0 2 4 Gear ratio ratio between the number of teeth of the driver and the driven gear 𝒕𝟐 𝑮𝒆𝒂𝒓 𝒓𝒂𝒕𝒊𝒐 = 𝒕𝟏 Spur Gears If torque to be transmitted, T is known, tangential component of force can be calculated as, 𝑻 𝑭= 𝒓 DESIGN OF ROLLER CHAINS AND SPROCKETS Bore Caliper Diameter 40 Bottom Diameter Pitch Diameter Outside Diameter Maximum hub and Groove Diameter https://www.usarollerchain.com/  0 5 /0 9 /2 0 2 4 Types: Center distance between sprockets, C (General Rule) should not be less than 1.5 times the diameter of the larger sprocket 30P