Hello friends here in this video we will see a problem on Design of key and for that purpose here we have a question design a Rectangular key for a shaft of 50 mm diameter the shearing and crushing stresses are 42 mega Pascal and 70 mega Pascal for key material respectively this is the question which is based on the design of key whatever is given here I will write that in the form of data first so let us get started in data first of all it is clear from reading this that it is a rectangular key the key is of rectangular cross-section I’ll draw it this key it has width W its thickness would be P just by reading that it is a rectangular key I have drawn the cross section next design a rectangular key for a shaft of 50 mm diameter so the diameter of shaft is given that is small D is equal to 50 mm the shearing and crushing stresses are 42 mega Pascal is the shearing stress so tau is 42 MPA and crushing stress is 70 MPA so Sigma CR that is crushing stress is 70 mega Pascal now for these data available we have to design the key see by designing the key we mean that we should know how much is the length of the key then we should know how much should be the width of key and the last part that is the thickness what is the thickness of the key so these three dimensions we need to find out so let us get these solution I will say that in the solution part since the diameter of shaft is 50mm so I will see that therefore let small L be equal to the length of shaft or I can say let it be equal to even the length of hub here I am assuming this and therefore let it be L is equal to 1.25 times of D that is the diameter of shaft so here I can directly get the length of key which is therefore small L will be equal to 1.25 into diameter of shaft is 50 so from this I will get my answer L comes out to be 62 point five mm so I will say that therefore the length of shaft length of hub key length may also be taken as same so I will say that therefore length of key is equal to 62 point five millimeters now after getting the length of key we need to find out weight and thickness see whenever key is there key has to fail it will fail in two possible ways first by shearing second by crushing so we are seeing the first kind of failure for the key and that is considering sheering failure of key so considering the shearing failure I will draw the diagram just to explain it here we have a hub this is the key and here we have this shaft this portion which I am sectioning it is the hub portion also called as the sleeve also called as the bearing or boss now this is the hub here is the shaft and by shearing we mean that considering shearing failure since we know that shaft is a rotating member shaft will try to rotate but hub is stationary at first it is stationary so what may happen when the shaft will try to rotate in clockwise direction hub will try to opposite that is it will try to rotate it in anti-clockwise direction because of that what may happen when the shaft is rotating clockwise one part of the key on that the force is acting towards right this is called as a tangential force it is acting towards right now here as I have told when the shaft is rotating clockwise a tangential force is acting towards right and as the hub is stationary it offers anti-clockwise motion so that is it tries to oppose the motion of the shaft so here I have another tangential force now because of the effect of these two sliding forces what can happen here I have taken this key separately one tangential force is acting towards the right other is acting towards the left one is because of the shaft other is because of the hub so because of this what can happen the sliding of the key takes place and it would be shearing breaking into two halves and how it will look like that I draw see here this is the braking area or the failing area means the shaft it would this key would be sheering like this it would be breaking into two halves so this is the shearing area when it is failing by shear shearing area capital a that will be equal to this is the width of the key as we can see here this is the width and here we have length of key so the failing area is within two lengths so now I can say that considering sharing failure of key therefore torque transmitted by key that will be P is equal to see torque is basically tangential force multiplied by the radius so therefore this T now the tangential force if I want to find out that will be force is stress into area here we are having shear stress so tangential forces area which is w into L that is the shearing area into the stress which is shear stress and radius is nothing but the radius of the shaft up till here the radius of the shot so that is a diameter of shaft by 2 so this is I can call it as equation number one one equation of torque but we know another equation therefore that torque transmitted by the shaft is PI by 16 D cube into tau so I will call this as equation number two this is from strength criteria and when we equate equation 1 and 2 as it on to the next page therefore equating 1 & 2 we get now if I simplify this I will get this as W into L into tau into D by 2 is equal to PI by 16 d cube into tau that is I have equated equations 1 & 2 from both the sides tau will get cancelled shear stress your diameter gets cancelled so here we have d square so finally we are left with I want to width of key lengths is already known so W will be equal to here I have 2 pi this would be 2 pi and then after this 2 gets multiplied here so it is 2 pi and d square so d square upon 16 this L will come here into the denominator so here if I write the values therefore 2 pi into diameter is 50 square upon 16 into length of the key is sixty two point five so from this I am getting width of key as fifteen point seven zero or I’ll round it off to 16mm so this is my second answer that as I have found out the width of key previously we had calculated the length of key this was our first answer answer number one so now we are left with only the thickness of key and for that I’ll write down considering crushing failure of key so when I consider crushing failure Sikhi how can it crush if I draw the diagram now since half of the portion of the key is in the hub the remaining half is in the shaft so both because of them there would be compressive force one compressive force would be there from hub the other compressive force would be there from shaft and because of that this key it can crush form small powders here so now if I take the crushing area since t is the thickness and crushing area I can take it as and this is the length of the key we know that it is L so therefore crushing area capital a it will be T by 2 that is approximately it is taken as half the thickness multiplied by length so finally I say that therefore torque transmitted by key now the formula would be T is equal to again it is tangential force into the radius so therefore P is equal to tangential forces stress into area area is T by 2 into L stress since here I am considering crushing so the stress would be crushing stress so this much is the tangential force radius is nothing but the radius of shaft it is d by 2 now this I will say that it is equation number 3 for us I will say that also torque transmitted by the shaft is 5 by 16 d cube into tau I will keep this as equation number 4 so therefore equating equation number 3 and 4 so I’ll equate them I will get it as therefore P by 2 into L into Sigma C R that is crushing stress into D by 2 is equal to PI by 16 B cube into tau so from this if I simplify I will get it as therefore P will be on one side here I have pi by 16 next D and D will get cancelled from both the sides where 1b gets cancelled it becomes d square so it is PI by 16 into d square multiplied by tau next here we have this 2 which would get multiplied on to the other side so therefore here we have another 2 which would get multiplied so the next thing is therefore this is this 2 and 2 will go into the numerator and in the denominator I have the length into Sigma C R so I will go on putting the values T is equal to PI by 16 diameter is 50 square shear stress is given in the question shear stress is 42 into 2 into 2 divided by length is 62 point five into Sigma C R is given as 70 so from this I will get P 18 point 8 5 or I can say that as 20mm so this is our third answer so as we have seen in this video we have designed the key and it was considered that when we were designing this key first it it was taken in shear and then it was taken in crushing that is considering shearing and crushing failure we have designed the key that is we have calculated the length width and thickness of key I hope this problem is understood

sir standARD KEY CROSS SECTIONS ALREADY AVAILABLE so why need to the finding width and thickness

Sir can you explain me the reason behind the length of the shaft = 1.25d ?

great lectures

super great work sir,I will be coming often to this channel until I complete my degree

sir for rectangular key w = d/4 and t = d/6 as per textbook of machine design by r.s. khurmi…..what is your point of view about this ??? waiting for your answer

hllo sir calculate ve naal kro plz answer na likho sidda plz kyu ke mere answer aunda ni thunda wale numerical da plz vo batna plz vasse lecture bht vdia. good luck

thanks for your help.

I wish you were muslim.

great work sir….for making our problem easy and helpful….

Length direct hi 1.25l kyu li h

Design a key for fixing a gear to a shaft dia 25 mm KW of power at 720 rpm is transmitted from the shaft to the gear .The key is made up of steel 50C4.the yeild strength in compression can be assumed to be equal to the yield strength in tension

Why not avoid keys in the first place! They are the weakest point in the design! Why not make a coupling system or like a shaft connects to a hub type of design ?

alot of confusions why did you take shear stress for shaft and key same while it is not stated in problem and crushing area that you have taken is that of only one face why? while when the crushing force will be applied both the side faces will come into contact plz clear my concept as soon as possible

I found on some books L=1.5d

For the doubt of why length = 1.25d, this is because, it is a condition that the key cannot be greater than 1.25 times the diameter, so we assumed it to be equal to that and followed with the calcs.

Sir ..volume problem

Thank you sirji

very important question for torsion Anna university exam

Excellent Sir

Sir please jo bhi videos design of machine Ko aapne chupa ke Rakha hai use bhi open karo samjhta nhi hai ek dekhne ke bad dusra AAP chupa lie ho to doubt hota hai. I hope ki use bhi public karoge.

In my text book ans of W=6.91

and l = 142mm

Tech max book

Can you believe that someone has accused me of murdering with one key? I am not that stupid. Why? The Definition!

A steel shaft has a diameter of 25 mm. The shaft rotates at a speed of 600 r.p.m. and transmits 30 kW

through a gear. The tensile and yield strength of the material of shaft are 650 MPa and 353 MPa

respectively. Taking a factor of safety 3, select a suitable key for the gear. Assume that the key and

shaft are made of the same mater

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Length of key =1.25d not shaft length

Plz tell the reason for each step

Sir the crushing stress is applied on total thickness of the key but why we consider it as t/2.

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Please speak loudly 🙏🙏🙏🙏🙏

wrong length and thickness is determined, As t=10mm & l=117.7

refer RS KHURMI

He is only focused in solving question no extra information given by him…. 🤨