Lecture 45

Welcome to the next lecture in the course Basics of Mechanical Engineering I. I am Dr. Amardeep Singh along with me Professor Ramkumar is co-teaching this course. We have discussed about coupling and shafts in the previous lecture. I will talk about keys, nuts, bolts, screws and fasteners in this lecture. Predominantly what I am going to discuss here is the components or the small tools which are used to make joints. These are generally temporary joints. Keys are small metal component or small it could be rectangular, we will see that kinds of the keys, taper, rectangular. So, it is a metal component connects the rotating shaft with a stationary shaft. So, that the stationary shafts also starts rotating. So, nuts and bolts and screws are also there to fasten. To fasten means to connect or join the two different components which are in mechanical functioning of a system or a machine. So, we will talk about the keys and their function in shaft systems. Forces acting upon some key we will try to talk about, other keys we will just put it in the references you can discuss upon them. Bolts of uniform strength we will discuss, fastener and their types we will try to talk about the screws and we will try to talk about tensile and shear stress analysis. A very broad introduction to this would be given in the last few slides. Definition of keys, keys and their function in shaft system. Key is a mechanical component used to connect rotating machine elements to shaft. This is a kind of a key, this is a rectangular key, the cross section is rectangular and the ends are circular here. So, this is a rest for key, this is known as key rest, the key rests. Key prevents the two parts from rotating relative to each other which allow the torque transmission. So, key fits into the machining slots which are called keyways, this is key rest or this is also called keyway. So, in both the shafts keyways are there that is the gears, pulleys, couplings are connected to the rotating shaft. Its main purpose is to prevent relative rotation between the two parts that is what is the key made for. So, it is keyed joint, when the joint is made out of key it is known as a keyed joint. This is a temporary joint when the key is removed the joint could be removed. Purpose of keys, the purpose of key as I mentioned is majorly torque transmission. Keys helps to transmit torque between the shaft and the rotating component. Axial positioning is also the other function that it performs that is they can restrict the axial movement of the component of the shaft that is along the axis it would not move that is it would not rotate in this direction. However, in this direction the components could move that is along the radius they could move. Then comes a load distribution, load distribution that is a key distribute the torque load across the entire length of the key this complete length bears the torque. So, this reduce the stress concentration. So, lengthier the key is more distributed is the stress here. Let us try to talk about the types of keys. Keys are of various types, the following types of keys are important from the subject point of view that you are trying to learn here that is sunk keys, saddle keys, tangent keys, round keys and splines. These are rectangular keys, the cross section is rectangular. These are parallel keys, the cross section is rectangular, but the ends are parallel. These are skewer keys, here the cross section is also skewer and the ends are also completely straight. These are give head keys, I will try to talk about them when we will talk about the sunk keys. Spline keys, these are the splines you can see if the torque transmission is very heavy single key might not be able to bear the load that is the stress concentration is to be distributed into multiple keys. Then in place of keys we use splines, these are spline keys you can see 1, 2, 3, 4, 5, 6, 7, 8, 9, 9 or 10 splines are used here. Then we have saddle keys which are only to put the two components over each other, tangent keys are also there which are wedge shaped. These are sometimes put from the outer open surfaces, wood rough keys are there, then we have feather keys. Let us try to talk about the sunk keys only. These are called as sunk keys as the name suggests, these are submerged or put there inside the shaft completely, half of the portion is there in the shaft and half of that is there in the rotating element or that is it could be the gear or anything that you wish to rotate using the shaft. Here in sunk keys I provided half in the keyway of the shaft and half in the keyway of the hub or boss of the pulley, whether it is pulley, it is gear whatever you wish to rotate using this. The sunk keys are of following types, it could be rectangular sunk key, it could be a skewer sunk key, it could be parallel or it could be gib hat. This is a typical rectangular sunk key where width and thickness is given. Let us try to see each of the types of the sunk keys one by one. Rectangular sunk key, rectangular sunk key is shown in this figure, this is a rectangular sunk key where width and thickness is given. Here width is generally d by 4, d is diameter of the shaft and thickness is 2 w by 3 that is it is d by 6, width whatever it is given thickness is proportional to it. So, where diameter of the shaft is d, the key has a taper in general 1 in 100 on the top side only. So, this is 1 in 100 taper is also there, so that the key is inserted easily here. So, this is a specific or typical rectangular key. Similarly, we have skewer sunk key, this is a skewer sunk key that is given here where the difference between rectangular sunk key and skewer sunk key is that width and thickness are equal in skewer sunk. So, here width is equal to thickness is equal to d by 4. In the rectangular sunk key the width was different than thickness where you can see thickness is d by 6, width is d by 4 or it could be some other proportion as well, but when we talk about a skewer key it is always the same that is width and thickness is same that is generally is equal to d by 4. Then comes the parallel sunk key, parallel sunk key may be of a rectangular or skewer section uniform in width and thickness throughout. Parallel sunk key is taper less and is used where pulley gear or other mating piece is required to slide along the shaft. So, I will put it here it is used when the pulley or you put it gear that is the other mating piece is required to slide along the shaft. So, this is why it is taper less which means for example, two shafts are there this is one shaft this is the other shaft generally a key would be there that would be fixed and this would be rotating for example, this is rotating this is a pulley that is rotating using this shaft. When it is a parallel sunk key this is allowed to move in this direction when it is allowed to move in this direction taper could not be allowed there is a taper less key would be put here for example, this button is a taper less key that is allowing the shaft to move in this direction along the axis. So, this is the purpose of using a parallel sunk key. Next comes the gib head key, gib head key it is a rectangular sunk key with head at one end that is known as gib head. So, this is a gib head key it is usually provided to facilitate removal of the key a gib head key it is used in it is shown in this figure this is a shaft this is a gib head key this allows you to hit here and helps you to remove the key using the gib head. Here generally width of the key is d by 4 and thickness of the key thickness is equal to 2 by 3 times of the width which is equal to d by 6 as we discussed about the rectangular key. So, these are the general proportions of the gib head key. So, this is gib head now let us try to see what are the forces which are acting on a key forces acting on a sunk key when a key is used in transmitting torque from a shaft to a rotor or hub the following 2 types of forces act on the key forces F 1 due to fit of the key in it is key way. As in a tight fitting straight key or in a tapered key driven in place these forces produce compressive stresses in the key. So, you can see this is a key that is fit and this is force F 1 that is because of the fit of the key. Then other force is F that you can see which are there along the thickness this is F 1 you can see along the width here this is F 1 this is F 1 this is F and this is F. So, F is the force due to the torque transmitted by the shaft this force is produced sharing and compressive or crushing stress in the key. So, it is very difficult to determine the force F 1 however, the force is F that is due to torque transmitted could be determined using experimentation or. So, that could be taken into account the distribution of forces along the length of the key is not uniform because the forces are concentrated near the torque input end. For example, this is rotating shaft and here we have a key this is a long key here the distribution of force would be different because this is a rotating end that is rotating here the torque would be more here the torque could be lesser. So, torque distribution is varying here that is what is mentioned here the torque distribution or the torque is concentrated at the input end that is what is mentioned here the forces are more concentrated at the torque input end. The force is acting on a key for a clockwise torque being transmitted from a shaft to a hub are shown figure here these are the gain forces that we looked at this is length of the key that is given and along the length as we said the forces concentration would be different along the torque input end. In designing a key forces due to fit of the key are neglected because it could not be determined and these are also very less in magnitude than the forces which are there due to the torque transmission. So, it is assumed that the distribution of the forces along the length of the key is uniform. So, as we can calculate or we can determine the strength of the key empirically and we try to take this assumption. Key connecting the shaft and hub is shown in figure let T is equal to torque transmitted by the shaft F is equal to tangential force acting at the circumference of the key this is F tangential force that is acting diameter of the shaft is D which is given here diameter length of the key is given L the width of the key is W the thickness of the key is T tau and sigma c are shear and crushing stresses for the material of the key. A little consideration will show that due to power transmitted by the shaft key may fail due to shearing or crushing that is why we have taken tau and sigma c considering shearing of the key the tangential shearing force acting at the circumference of the shaft is given like this F is equal to area resisting the shearing into shear stress that is length into width is the area that is resisting the shearing and tau is the shear stress that is L into W into tau is the force that is tangential shearing force that is here. Similarly, the torque transmitted could be calculated from this force F torque is equal to force into distance the distance is T by 2 force is given as L into W into tau. So, this becomes my torque L into W into tau into T by 2 considering the crushing of the key upper one was considering the shearing. Now, considering the crushing of the key tangential crushing force acting at the circumference of the shaft it is F would be area resisting the crushing into crushing stress crushing stress is sigma c area resisting the crushing here is L into T by T. You see in this figure the L into half of this thickness here the crushing is expected to happen or it is assumed to happen here. So, this T by 2 is taken as distance here. So, L into T by 2 into sigma c is here the torque transmitted by the shaft would be F into d by 2 that is L into T by 2 into sigma c into d by 2. Key to be designed to be equally strong in shearing and crushing we equate the equation 1 and 2, 2 is for crushing equation 1 was for the shearing. So, equating the shearing and crushing loads this becomes my relation while crossing out all the terms here we have gotten this relation where W by T that is ratio between width and the thickness of the key is equal to half of the ratio between the crushing and the shearing stress. So, this allows me to put this statement the permissible crushing stress of the usual key material is at least twice the permissible shearing stress. This is notable is at least twice what is that the permissible crushing stress for usual key is at least twice the permissible shearing stress. So, therefore, from the given equation 3 that is equation for the relation between the crushing stress and the shearing stress and the width and the thickness this is equation 3 that is here. When we have W is equal to T in other word the square key is equally strong in shearing and crushing. In order to find the length of the key to transmit full power to the shaft the shearing strength of the key is equal to the torsional strength of the shaft. So, in this way let us see. So, when we know that the shearing stress of the key that is the torque is equal to L into W into tau into d by 2 and torsional shear strength of the shaft that we discussed in the previous lectures that is pi by 16 into tau 1 into d cube. So, when we equate them from these two equations that is this is for key this is for shaft or I will put it here it is for key this is for shaft we get this relation. So, here tau 1 this is shear stress of shaft material and when we talk about square key where W is equal to d by 4 this could be put in this way L is equal to pi by a into tau 1 into d square by W into tau this gives me a relation 1.571 d into tau 1 by tau 2. So, when the material is same as that of shaft that is tau is equal to tau 1 L is equal to 1.571 d. So, let us go through a problem statement to understand this design a rectangular key for a shaft of 50 millimeter diameter the shearing and the crushing stresses are given here which are 42 mega Pascal and 70 mega Pascal respectively we have been given here this is 42 mega Pascal that is my shearing stress that is tau is equal to 42 mega Pascal which is equal to 42 Newton per millimeter square and I have sigma c is equal to 70 mega Pascal which is equal to 70 Newton per millimeter square to design this key only what we are given is d is 50 millimeter using all the relations for width for thickness let us design this key and length of key also to be determined. So, there are design tables where for a specific with diameter of a key while putting a material or so we can pick the width and thickness for 50 millimeter shaft dia from these design tables I have picked width as 16 millimeters and thickness as 10 millimeter. So, this is machine design only a brief introduction is given here hand books of machine design are there where for the different diameters of the shaft the key dimensions are given for a rectangular key width is 16 t is equal to 10. Now length of the key is to be determined L is equal to length of key which is unknown. So, I will do it in this way first I will design the key for shearing then we will design it for crushing you know the relation for shearing the t that is torque transmitted is equal to L into W into tau into the radius that is t by 2. So, this is equal to value of L is to be found W is equal to 16 tau is given as 42 and t by 2 would be 50 by 2 that is 25 this value is equal to 16800 into L Newton millimeter. So, and torsional shearing strength that is torque transmitted of the shaft this was for key. Now, we calculate the torque transmitted for the shaft this is equal to pi by 16 into tau of the shaft into d cube this is equal to pi by 16 into 42 into 50 cube this is equal to 1.03 into 10 raised to power 6 Newton millimeters. So, when I equate these two equations this is equation I would say 1 this is equation 2. So, when I say I equate 1 and 2 I would say equating 1 and 2 I got here length is equal to 1.03 into 10 raised to power 6 Newton raised to power 6 by 16800 this is equal to 61.31 millimeter this is the first answer when you consider the shearing stress only. Let us now consider the crushing stress which is given here as 70 mega Pascal for the crushing stress we know the torque transmitted for the key. So, this is T is equal to L into T by 2 into the crushing stress into the radius that is T by 2 this is equal to L into 10 by 2 that is thickness is 10 into the crushing stress that is sigma c that is 70 and this into the d by 2 that is 50 by 2 this value is 8750 into L Newton millimeter this is also for key and this one was for shaft. Let me also try to put the relation for shaft for crushing and for the crushing also for the shaft relation is already given here. So, this relation I will use for the crushing stress as well this is for shaft that also is put here. So, we know this value tau is there for this becomes my equation 3 and we know here again equating 2 and 3 that is the shaft and the crushing stress that is there. So, we have gotten here L is equal to 1.03 into 10 raise to power 6 by 8750 this is equal to 117.7 millimeters which is equal to 120 millimeters. So, value or the length of the key that is taken from the design based upon the crushing stress is 120 mm and based upon the shearing stress is 61 mm. So, the final answer that we will take here is the longer value that is 120 mm. So, this is the final answer that we have taken the length of the key that will be designed or that is taken here is 120 mm. Let us now talk about the nuts and bolts nuts and bolts are type of threaded fasteners used for holding components together in mechanical structures the bolt provides the tensile force while the nut serves to create a clamping action. So, bolt is for tensile force and nut is for clamping action together they form a strong reliable connection which can be easily disassembled when needed. Bolt as a definition it is a cylindrical road with head on one end and threads on the other bolts typically pass through a hole in one or more components and are secured with a nut. Nut by definition it is a hexagonal or other shaped piece with an internal threads that pairs with the bolt to hold parts together. So, simply putting the definitions of nuts and bolts there are types of nuts and bolts hexagonal bolt hexagonal nut that you might have heard multiple times or might have used multiple times you have seen it multiple times you have designed that in your engineering drawing or so. Similar we will discuss here hexagonal bolt or square bolt eye bolt knock nut or so. So, let us see the types of bolts and nuts. Hexagonal bolt the most common type having a hexagonal head for easy wrenching this is a hexagonal bolt and this is hexagonal nut this is used in general purpose applications like machine assembly construction and automotive. Square bolt is similar to hexagonal bolt, but with a square head this is a square head that is here it is less common, but used for rustic or aesthetic applications. These are useful in applications where bolt needs to be held securely using pliers or a square socket. So, these are major two designs which are used other than that multiple places are there where eye bolt is used. A bolt with a loop or eye at one end is used to attach ropes or cables that these are all eye bolts here. So, you can see a loop here this is to attach a rope or we using a rod or so we try to lock it or unlock it. Typically used in lifting and rigging applications eye bolts are used. Similarly, we have a lock nut a type of a nut designed to resist loosening under vibration and torque does not get loose under vibration that is why we try to provide a sawn nut or a castle nut that is a kind of a lock nut this allow any loosening due to vibration. They can have internal mechanisms like a nylon insert or nylon nut or deformed threads or metal lock nut to increase friction between the bolt and nut. So, as you can see this is a deformed nut that is all metal which does not allow to rotate very easily or it could have a spring washer in between. So, these are the kind of a lock nut. Certain applications where nuts are used in mechanical structures that is bolts and nuts are essential in building frameworks and securing mechanical components like engine parts machines vehicle assemblies or so. These are widely used in construction for steel structures bridges and buildings where high strength reliable joints are necessary. Automotive and aerospace applications bolts and nuts are critical for holding engine components in aircraft structures or other high stress assemblies or so. Lifting and rigging eye bolts are used in systems where ropes or chains need to be attached securely for instance in automotive system for example, for the screw jack in automotive systems or so. When you use a screw jack it is a nut and bolt assembly which helps to lift your car up or bring it down it is nut and bolt assembly only. So, multiple applications where nuts and bolts are used for temporary joining and sometimes for the almost permanent joining. Next is bolts of uniform strength problem with standard bolts shock loading can cause stress concentration at threads leading to potential fractures at the weakest part of the world. So, uniform strength design is important that is turning down the shank reduce the shank diameter to match or to be slightly less than the core diameter of that threads that is d c. This redistributes stress making the bolts stronger and better at absorbing shock. Drilling an axial hole is another way to have the uniform strength that is we drill a hole through the bolt head to match the core of diameter of the threads. The diameter of the hole d e can be calculated as d is equal to d o minus d c where d c is diameter of the threads and d o is our outside diameter. So, this ensure that uniform strength is there that is the shank area equal to the thread root area. So, here to put it in more detail for example, this is a hexagonal bolt and this is a shank and this is the threaded portion here only the threads are there this is our shank and this is thread. Here what will you have when we talk about the thread here or you say this is axis this is other part here this is an outside diameter that is nominal diameter. This diameter is our outside diameter d o which is nominal diameter and this smaller diameter that is the minimum diameter of the internal of the thread is d c that is core diameter. So, the hole diameter that is there if we drill a hole here there could be hole that could be drilled here. So, here diameter of this hole is d this hole diameter is equal to square root of the differences of the squares of the nominal diameter and the core diameter of a bolt. So, this is a bolt here that is a thread if I try to put in that color once again for example, there could be shank here for the bolt and this could be bolt head here. So, this is how this is designed to make it more clear let me show you a numerical where this is taken from a hand book from a m 48 bolt designs m is the diameter 48 that is m is notation m is generally for metric thread metric thread 48 size the pitch is 5 millimeter and the major or nominal diameter here is d is 48 and you can see the effective pitch diameter or nut of the bolt is d p and bolt and nut diameters that is core diameters are given for the bolt and nut which is 41.7 and 5 and 42.5 and 7 for the bolt and nut respectively that is d c depth of the bolt is given 3.067 stress area it is in millimeter square is 1465. So, determine the diameter of the hole that must be drilled in a m 48 that is metric 48 bolt such that the bolt becomes of a uniform strength design dimensions of screw threads bolts and nuts according to IS 4218 part 3 1976 is given here. So, to solve this we will just use the relation that is given in the previous slide that is we will calculate the diameter of the hole from this relationship here how do we put that we know that in this slide it is given the nominal diameter that is d 0 is equal to 48 millimeters and the core diameter they are asking about the bolt and for the bolt it is given this is for bolt the core diameter given is d c is equal to 41.795 millimeters. So, for the bolts of uniform strength for uniform strength we will say d is equal to under root of d o square minus d c square which is equal to 48 square minus 41.795 square under root and this value is 23.64 millimeters. So, this was a brief introduction to bolts and nuts let us try to go through fasteners and their types fastener are hardware devices used to join or secure two or more components together these are again joining devices then these include bolts screws nuts rivets washers retaining rings and. So, about bolts screws nuts we will keep discussing types of fasteners fasteners are rivets washers and retainer rings these are rivets which are there you can see these are washers that helps you to lock two bolts or nuts and these are retainer rings. So, that is the things or the bolts or nuts that you are put in a machine component that those are retained there types of fasteners first is rivets solid rivets solid rivets because the name itself tells solid these are permanent fasteners which are used in applications like aircraft and bridges they are deformed at one end to secure the joint. So, this is a permanent rivet that is there it is deformed at one end. So, it is a kind of a permanent it is cannot be removed easily once it is put in any two mating surfaces then we have blind rivets these are used when access to one side of the joint is not possible through only one side only this rivet is put inside and joint is made wherever joint is to be made this rivet gets in there two mating surfaces could be there on the other side access is not available. The rivet expands and locks in place when mandrel is pulled mandrel is used to lock in these rivets which are blind rivets and when mandrel is pulled the rivet expands the other side of rivet this side also expands and it locks in place. So, that the rivet is tightened then is tubular rivets. So, these are tubular rivets the kind of a tube which are hollow from inside these are used where the joint needs to be very strong and they often are used in leather work and electronics. So, these are tubular rivets which are there. Next come washers, washers are there to secure the spacing between nuts and bolts between screws even before the rivets also washers could be put. So, there are multiple kinds of washers flat washers, lock washers, spring washers. Flat washers distribute the load of a fastener to prevent damage to the surface and reduce friction lock washer is there as I have mentioned in the previous slides as well these have serrations here. So, that the load is distributed here and these prevent fasteners from loosening due to vibrations they include split tooth and serrated types. So, these are tooth washers it could be serrations it could be split washers also spring washers this is a spring washer that provide a spring effect to the connection to the nut on the bolt to absorb shock and maintain tension. So, these are spring washers which are mentioned here. Then comes retaining rings the word retaining itself tells that these are the rings which are used to fasten parts such as bearings to prevent them from coming off. So, these are put in place and the part does not come off. So, that is why we put it there. So, these are ears of the rings and this is gap these rings helps to lock the two mating parts I would call mating parts to the parts which are to be joined this could be internal this could be external depending upon the where do you put the rings if these rings are here this could be called as external when the ears are outside when the ears are inside this is internal retaining rings this is a kind of a C style ring. So, internal retaining rings are there which fits inside a bore or groove and hold components in place this is a internal ring external rings are there which fit outside or around a shaft or other external parts to keep them in place these are external rings. Snap rings are there types of retaining ring that can be easily installed and removed simply snapping is putting in taking out putting in taking out easy installation in and taking out then it is known as snap ring. So, these are there retaining rings. Applications of the fasteners fasteners have multiple applications in automotive construction electronics and aerospace industries. In automotive industry bolts and nuts are used to assemble engine components and chassis. Rivets are used to join structural steel beams in construction. In electronics washers are used to prevent damage to delicate components and to sometimes provide spring actions to ensure secure connections. In aerospace industry blind rivets are used where excess is limited and high strength is required when excess is there from the one direction or one side only for example, these are blind rivets where excess to the other side of the surface is not possible. So, blind rivets are used here. Next comes screws stress concentration in screws. Screws experience stress concentrations particularly at the root of the thread due to the reduced cross sectional area this can lead to higher local stresses compared to the rest of the screw. So, as I mentioned in the bolts as well. So, this is crest and rough this could be axis. So, here at the internal surface area the stress concentration is high. Load distribution in threads, load distribution when we talk about the load applied to a screw is distributed along the threads where it creates both tensile and shear stresses the distribution is not uniform and can vary depending on the screws design and the load applied. It depends what kind of load is applied it also depends what kind of screw are we using and what kind of the thread configuration are we using, what pitch are we using depending upon the load, the size, the diameter, the pitch, the kind of the screw, the kind of the thread everything is decided accordingly this is machine design. Then again let us recall the tensile stress that is sigma t that we have discussed in the previous lectures. Tensile stress in a screw is calculated by dividing the applied load by the cross section area of the screw. So, tensile stress is simply force per unit area or load per unit area where F is applied load it is F by A, it is F by A where F is applied load and A is cross section area of the screw. Here cross section area A is equal to pi D square by 4 where D is diameter of the screw. Shear stress could also be calculated it is shear force per unit shear area. The shear stress in a screw is calculated based upon the applied shear force and shear area tau is equal to F shear by A shear, F shear is shear force, A shear is shear area. For a screw with diameter D the area of shear could be calculated again as pi D square by 4. This approximation assumes that the shear force is uniformly distributed over the screws cross section. So, with this lecture I am resting I will just recapitulate the things that I have covered. Let us try to see the question that could help you to ponder over the topics that we have discussed in this lecture. First question is what are the primary functions of a shaft in a mechanical system? Differentiate between solid and hollow shafts, which one is preferable for weight reduction and why? What is the purpose of a key in a shaft and assembly? How would you calculate the stress on a key when subjected to a given torque? What is the difference between a bolt and a screw? Explain the significance of thread pitch and lead in threaded fasteners? How do you calculate clamping force in a bolted joint? What are the different types of nuts and their applications? Describe various failure modes of fasteners that is fatigue failure, shear failure, grip failure this you can discuss and try to understand that. What are locking devices for bolts and nuts? Give examples of common locking devices methods. So, common locking methods that I have discussed certain fasteners could be used to lock as well. Why is material selection important when designing fasteners for high temperature applications? There are certain questions which I have repeated from the previous lecture as well, but you can go through them and try to understand the concept of keys, nuts, bolts, screws, fasteners in detail. And we will keep discussing about different components and the systems in the machine design in the coming lectures. Thank you.