Tolerance of The Hole : UL-LL = 30.5-29.5=1 Tolerance of The Shaft: UL-LL = 30.5-29.5=1 Fundamental Deviation of The Shaft = 0 ( Because its a shaft basis system) Fundamenta Deviation of the hole = -0.5 This is a transition fit, because it has a positive allowance in good case and negetive allowance in bad case.
Thank you for the video 😊. Answer for last problem: 1. Tolerance of hole and shaft = 30.5- 29.5= 1mm 3. Allowance (worst case)= 29.5-30.5= -1mm 4. Worst case tolerance= -1mm, Best case tolerance= 30.5-29.5 = 1mm Hence transition fit will be applicable. Now the question is how to decide the fundamental deviation... And hole or shaft basis system. Can you please help to resolve the problem? I watch the video twice but still couldn't able to understand.
1. Tolerance of Hole & Shaft = 1 2. Fundamental deviation of hole and shaft = 0 (Because shaft and hole maximum limit and minimum limit of size is same). 3. Worst case for assy = -1 (Interference fit ), Best fit for assy = 1 (Clearance fit) 4. Transition fit is having
How should we decide to go for either shaft or hole basis system and who should decide that parameter .since you are mentioning here it as shaft basis system in the problem
Great video! When designing a press fit assembly (brass pin into aluminum hole), after calculating the contact pressure of the MMC, what should be the max stress limit? Yield strength, Ultimate strength, other?
Your visual explanation is superb sir. Thanks a lot. It's very useful to understand the concept well. Thank you so much sir. Expecting more mechanical engineering subject videos sir.
Hello sir, In the example of Interference fit i think -0.25 will be the best case and -1.25 will be the worst case. Correct me if iam wrong. May be its some typing mistake. Thanks for sharing your knowledge.
What is the standard for choosing fits? And how do we know the standard chart is based on hole basis or shaft basis? Some chart contain zero line touching both shaft and hole at some point Confused please help
Tolerance of The Hole : UL-LL = 30.5-29.5=1 Tolerance of The Shaft: UL-LL = 30.5-29.5=1 Size of the shaft is 30 Fundamental Deviation of The Shaft = 0 ( Because its a shaft basis system) Fundamental Deviation of the hole = 0.5 Lower limit of size of hole = 30 -0.5 = 29.5 Upper limit of size of shaft = 30 +0.5= 30.5 Worst case for assy. = 29.5 - 30.5 = -1(-ve clearance) So this is interference fit. Upper limit of size of hole = 30 + 0.5 = 30.5 Lower limit of size of shaft = 30 - 0.5 = 29.5 Best case for assy. = 30.5 - 29.5 = 1 ( +ve clearance) So this is clearance fit.
Really a great video sir. I had one doubt in the last question. When you said it's a shaft basis system, does that mean that we have to ignore the UL of 30.5 and consider it to be equal to 30? Bcz only then the FD for hole will be 0.5.
1) Hole Tolerance = 1, Shaft Tolerance = 1 2) Fundamental deviation of Hole = -0.5, Fundamental deviation of Shaft = +0.5 3) Allowance for Hole = +1 (Best case for assembly), Allowance for Shaft = -1 (Worst case for assembly) 4) Transition fit
Your way of explaining is very good sir, And I have got all answers correct But, I am not convinced on fundamental deviation concept. And remaining concepts are excellent. Thanks for educating us😃
@@mechanicalvault in your last problem FD of shaft=0 But 30+-0.5 indicates both limits are away from zero line than how it's FD value is zero, Excuse me, if may I wrong. Thanks for your reply sir.
I would suggest to go through my another video H7/g6. You will get crystal clarity around FD with actual example and charts from ISO286 standard. FD is the location of of block from zero line. FD of either hole or shaft will be zero based upon if it is hole or shaft basis system. I am sure that after watching my latest video, you will get better clarity.
@@mechanicalvault Yes it was. One more request from my side to make video on geometric tolerance because there are lot of confusion in between circularity, cylindricity, run out & total run out. If possible please make separate video for each type of geometrical tolerance.
Standard IS 2102 explains general tolerances for form tolerances. But such tolerances should be further refined by tolerance stack up analysis and process capability study.
@18.38 (Interference fit)., is -0.25 the best case or worst case ?,i think -0.25 is less than -1.25 so it should be the best compared to -1.25.is it correct ?
TOLERANCE FOR BOTH WILL BE 1, F.D FOR SHAFT WILL BE 0 AS UPPER LIMIT OF SHAFT WILL BE ZERO LINE SINCE IT IS SHAFT BASIS SYSTEM, ALLOWANCE WILL BE ALSO 0 AND IT IS PUSH TYPE OF TRANSITION FIT.
@@mechanicalvault sir can you share the FD diagram to be drawn because in example u have suggested tolerance of 0 to zero line but here it is+or - 5.hiw to draw it sir.
Hi Kevin! Thank you for your awesome content! As an example. I have a shaft that is connected to the bearing and the bearing to the housing. Is the housing the hole for the bearing and the bearing the "shaft" for the housing? And then the inside of the bearing would be the "hole" for the shaft and the shaft is the shaft itself? I mean, that is what this Hole and Shaft system is for, correct? Or for fir for bearings we would have to consider another type of engineering fit?
Yes, you understood it correct. You can imagine it like - Shaft and hole is like assembly between external and internal features of the parts. So, we can have different Engineering Fits between parts in the assembly or between su-assembly of parts and housing.
hello, we are not secondary school students. stop putting background music in your videos. which is very distracting while listeing to your instruction. although your videos are very informative and helpful, but background music has made it worse.
Ohh! Sorry that you are not secondary school student 😬 I had intended this video for secondary school students and that’s why I added background music. But all my recent videos are made up for grown up guys like you, so I have not added background music in them. SORRY GROWN UP MAN 👨
hi Answers: 1.Tolerance of hole and shaft = 1 2. Fundamental Deviation of Hole = 29.5 and for shaft 30.5 3.Allowance between Hole and shaft=0 4. it is Transition fit
Answer 1 & 4 are correct; Correct answers of Q2 is - FD of shaft is zero as it is shaft basis system; And FD of hole is 0.5; Correct answer for allowance = 29.5 (H) - 30.5(S) = -1.0 which is worse case for assy; I recommend you to go through Fundamental deviation diagram again and also through the example which I have explained in the video; But good try;
@@mechanicalvault Thanks for the confirmation. I get The 3rd answer but About FD , We mention FD as tolerance value or entire size value ? And you mentioned shaft size 30+/- 0.5 Bilateral tolerance, then how FD of shaft will be Zero? Can you please clarify? I took entire size value
I recommend you to watch video again especially understand hole and shaft basid system, fundamental deviation diagram and numerical example which I have explained. You are getting confused between upper, lower and fundamental deviation. If possible watch complete video, it will clarify your concept once and for all;
Tolerance of The Hole : UL-LL = 30.5-29.5=1
Tolerance of The Shaft: UL-LL = 30.5-29.5=1
Fundamental Deviation of The Shaft = 0 ( Because its a shaft basis system)
Fundamenta Deviation of the hole = -0.5
This is a transition fit, because it has a positive allowance in good case and negetive allowance in bad case.
Thank you for the video 😊.
Answer for last problem:
1. Tolerance of hole and shaft = 30.5- 29.5= 1mm
3. Allowance (worst case)= 29.5-30.5= -1mm
4. Worst case tolerance= -1mm,
Best case tolerance= 30.5-29.5 = 1mm
Hence transition fit will be applicable.
Now the question is how to decide the fundamental deviation... And hole or shaft basis system. Can you please help to resolve the problem? I watch the video twice but still couldn't able to understand.
1. Tolerance of Hole & Shaft = 1
2. Fundamental deviation of hole and shaft = 0 (Because shaft and hole maximum limit and minimum limit of size is same).
3. Worst case for assy = -1 (Interference fit ), Best fit for assy = 1 (Clearance fit)
4. Transition fit is having
That was so comprehensive. I found your explanations really easy to understand.
Glad that you liked it ✨️
How should we decide to go for either shaft or hole basis system and who should decide that parameter .since you are mentioning here it as shaft basis system in the problem
NOW ITS 500 COMPLETE SIR
THANKS TO EXPLAINE T
HIS CONCEPT OF FUNDAMENTAL DEVIATION
Thank you.
Hi
would like to ask how to determine the force needed to unpin/remove a shaft (according to the type of fitting) please?
Great video! When designing a press fit assembly (brass pin into aluminum hole), after calculating the contact pressure of the MMC, what should be the max stress limit? Yield strength, Ultimate strength, other?
Your visual explanation is superb sir. Thanks a lot. It's very useful to understand the concept well. Thank you so much sir. Expecting more mechanical engineering subject videos sir.
Glad that you liked it.
Hello sir, In the example of Interference fit i think -0.25 will be the best case and -1.25 will be the worst case. Correct me if iam wrong. May be its some typing mistake.
Thanks for sharing your knowledge.
Great. I love your examinations. I have learnt some new in Ghana to also teach my students. Keep it up
That's great 👍
Superb video!
Glad that you liked it!
What is the standard for choosing fits?
And how do we know the standard chart is based on hole basis or shaft basis?
Some chart contain zero line touching both shaft and hole at some point
Confused please help
Tolerance of The Hole : UL-LL = 30.5-29.5=1
Tolerance of The Shaft: UL-LL = 30.5-29.5=1
Size of the shaft is 30
Fundamental Deviation of The Shaft = 0 ( Because its a shaft basis system)
Fundamental Deviation of the hole = 0.5
Lower limit of size of hole = 30 -0.5 = 29.5
Upper limit of size of shaft = 30 +0.5= 30.5
Worst case for assy. = 29.5 - 30.5 = -1(-ve clearance)
So this is interference fit.
Upper limit of size of hole = 30 + 0.5 = 30.5
Lower limit of size of shaft = 30 - 0.5 = 29.5
Best case for assy. = 30.5 - 29.5 = 1 ( +ve clearance)
So this is clearance fit.
What an pleasant explanation...We require more videos from this channel sir...
I am glad that you liked the video :-) Yes, lot more to come; keep watching and happy learning!!
Really a great video sir.
I had one doubt in the last question.
When you said it's a shaft basis system, does that mean that we have to ignore the UL of 30.5 and consider it to be equal to 30?
Bcz only then the FD for hole will be 0.5.
Great way of sharing knowledge,
Any course available on gd&t and tolerance stack up analysis?
Please reply
WhatsApp your name and email to 9881521371 for details
1) Hole Tolerance = 1, Shaft Tolerance = 1
2) Fundamental deviation of Hole = -0.5, Fundamental deviation of Shaft = +0.5
3) Allowance for Hole = +1 (Best case for assembly), Allowance for Shaft = -1 (Worst case for assembly)
4) Transition fit
Tolerance H&S=1
FDH=0.5
FDS=1
Allowance=-1
Interference fit
One doubht I have to refer FDD to solve
Your teaching method is fabulous❣️
Plz make gd&t series
And plz guide me from where i should learn plastic part design ..
Call Jagadish sir @8329029492 for plastic design course if you are interested.
Please share chart for every type of fit in hole basic system
Sir pls take lecture on plastic snap design calculations
Yes, very soon.
Superb explanation sir...
Expecting more videos
🙏🙏
it is transition fit.
tol of hole and shaft is 1 mm
allowance is +-1 mm,0
fd of hole -.5
fd of shaft +.5
correct me if wrong
fd of shaft is zero
Your way of explaining is very good sir,
And I have got all answers correct
But, I am not convinced on fundamental deviation concept.
And remaining concepts are excellent.
Thanks for educating us😃
What concept of fundamental deviation you are not agree about?
@@mechanicalvault in your last problem FD of shaft=0
But 30+-0.5 indicates both limits are away from zero line than how it's FD value is zero,
Excuse me, if may I wrong.
Thanks for your reply sir.
I would suggest to go through my another video H7/g6.
You will get crystal clarity around FD with actual example and charts from ISO286 standard.
FD is the location of of block from zero line.
FD of either hole or shaft will be zero based upon if it is hole or shaft basis system.
I am sure that after watching my latest video, you will get better clarity.
Thx for video
Thanks for fulfill my request on fits.
You are welcome, Kunar Sanu! Hope it was helpful to you:-)
@@mechanicalvault Yes it was. One more request from my side to make video on geometric tolerance because there are lot of confusion in between circularity, cylindricity, run out & total run out. If possible please make separate video for each type of geometrical tolerance.
Sir where to get values of straightness.flatness.parallism...?
Is there any chart for that according to asme
Standard IS 2102 explains general tolerances for form tolerances. But such tolerances should be further refined by tolerance stack up analysis and process capability study.
Good Work Sir !!!
Keep it up
Try to upload more videos....
Sure; lot more to come :-)
Dear sir can you take the lecture about surface roughness?
Yes, sure; in near future;
@18.38 (Interference fit)., is -0.25 the best case or worst case ?,i think -0.25 is less than -1.25 so it should be the best compared to -1.25.is it correct ?
Excellent explanation on simple but very important topic sir!!! Bravo 👌Keep inspiring sir. 👍
Happy learning; Keep watching;
very usefull thanks alot
Glad that you liked it!
TOLERANCE FOR BOTH WILL BE 1, F.D FOR SHAFT WILL BE 0 AS UPPER LIMIT OF SHAFT WILL BE ZERO LINE SINCE IT IS SHAFT BASIS SYSTEM, ALLOWANCE WILL BE ALSO 0 AND IT IS PUSH TYPE OF TRANSITION FIT.
Tolerance oh hole = 1;
Tolerance of shaft = 1
It's shaft basis system:
F.D. oh shaft = 0;
F.D. oh hole = 0.5
Worst case clearances:
A1 = 30.5(H) - 29.5(S) = 1 Clearance
A2 = 29.5(H) - 30.5(S) = -1 interference
Allowance = -1 (Worst case of A1 & A2)
Since we get both clearance & interference, it's a transition fit;
How are you getting 1 as tolerance , it should be 0.5 right @@mechanicalvault
@@mechanicalvault sir can you share the FD diagram to be drawn because in example u have suggested tolerance of 0 to zero line but here it is+or - 5.hiw to draw it sir.
I also have same doubt
Very good explanation
Thank you Nilesh! Watch my other videos as well exclusively made for Mechanical Design Engineers.
@@mechanicalvault sir can you give information about which of limit pair should be use to achieve specific fit.
Sir , why the worst case formula is different in clearance fit and interference fit
Hi Kevin! Thank you for your awesome content!
As an example. I have a shaft that is connected to the bearing and the bearing to the housing. Is the housing the hole for the bearing and the bearing the "shaft" for the housing? And then the inside of the bearing would be the "hole" for the shaft and the shaft is the shaft itself?
I mean, that is what this Hole and Shaft system is for, correct? Or for fir for bearings we would have to consider another type of engineering fit?
Yes, you understood it correct.
You can imagine it like - Shaft and hole is like assembly between external and internal features of the parts.
So, we can have different Engineering Fits between parts in the assembly or between su-assembly of parts and housing.
Awesome! Many thanks for your clarification. @@mechanicalvault
Hi sir i want to learn stackup analysis
WhatsApp to 9881521371 for demo videos of the course.
Great work
🙏🙏
Nice and useful information about engineering fits :-)
Thank you Anita!!
Excellent work sir....keep it up...we are waiting for next videos.
Thank you Yogesh!
@@mechanicalvault sir Please make detail video on advance gd&t.
14:33 Isn't the best case 2.25?
I got the same results 2.25
1.SHAFT:
UL= 30
LL = 29.5
T= 0.5
HOLE:
UL= 30.5
LL= 29.5
T= 1
2. F.D for shaft is 0
F.D for hole is 0.5
3. Allowance = 29.5-30= -0.5
4. Transition fit
hello, we are not secondary school students. stop putting background music in your videos. which is very distracting while listeing to your instruction. although your videos are very informative and helpful, but background music has made it worse.
Ohh! Sorry that you are not secondary school student 😬
I had intended this video for secondary school students and that’s why I added background music.
But all my recent videos are made up for grown up guys like you, so I have not added background music in them.
SORRY GROWN UP MAN 👨
hi
Answers:
1.Tolerance of hole and shaft = 1
2. Fundamental Deviation of Hole = 29.5 and for shaft 30.5
3.Allowance between Hole and shaft=0
4. it is Transition fit
Answer 1 & 4 are correct;
Correct answers of Q2 is - FD of shaft is zero as it is shaft basis system;
And FD of hole is 0.5;
Correct answer for allowance = 29.5 (H) - 30.5(S) = -1.0 which is worse case for assy; I recommend you to go through Fundamental deviation diagram again and also through the example which I have explained in the video;
But good try;
@@mechanicalvault
Thanks for the confirmation.
I get The 3rd answer but
About FD ,
We mention FD as tolerance value or entire size value ?
And you mentioned shaft size 30+/- 0.5
Bilateral tolerance, then how FD of shaft will be Zero?
Can you please clarify?
I took entire size value
I recommend you to watch video again especially understand hole and shaft basid system, fundamental deviation diagram and numerical example which I have explained. You are getting confused between upper, lower and fundamental deviation. If possible watch complete video, it will clarify your concept once and for all;
What would have been the results if the deviation was a hole basic system