PLEASE NOTE: there is an error at 42:57 ... this torque calculates to 72.02Nm, not 52.63Nm as stated in the video. I apologize for any confusion this may have caused!
My career as a mechanical engineer has gone down a bit of a rabbit hole these past five years. These fundamental videos you do are of great help to those of us who are trying to re-orient. Thank you.
So, first I thought I want to learn about torque wrenches. So I went and googled how a torque wrench works, watched a video about it and then I started wondering how someone calculates those values for bolts and here I am. I think I might be addicted to learning and it's so awesome to find lectures from RUclips. I may not need this information but at least I know little bit more about bolts now. Guess I should sometime finish my own studies. Thanks for uploading these!
I'm glad you found it interesting! If you want to learn about finding bolted joint stiffness constant for more complex joints, check this one out: ruclips.net/video/IpT4JzZ-4IE/видео.html Thanks for watching!
I found this to be one of the clearest lectures I have ever attended to. Kudos to the professor. He managed to clarify many of my doubts about the subject and make it way EASY to understand and apply in everyday situations and when I am actually busy designing something that has to work for real. I am of the opinion that making something EASY to understand is very difficult to do so I am happy to have found this channel. Greetings and please continue posting! Alfredo
This was perfect, I'm currently calculating bolt stress in a statically loaded member for an engineering team I'm a part of, and I haven't taken a machine design class yet, thank you.
Great lecture ,it's great to have access to those useful information from Vietnam. Hope you keep up with these amazing work and released more video like this
KN/ (MN/m) = mm ........................... 1MN = 10^-3 KN 1 m = 10^-3 mm ok 28.8 x 10^-3 / 554.6 x 10^-3 it is good, thanks to explained in metric units
Thank you! I'm making as much new content as I can while seeing to my other responsibilities. RUclips doesn't pay the bills, ha ha. In the meantime, here are some of my other playlists in case you haven't seen them yet and might be interested: ENGR122 (Statics & Engr Econ Intros): ruclips.net/p/PL1IHA35xY5H52IKu6TVfFW-BDqAt_aZyg ENGR220 (Statics & Mech of Mat): ruclips.net/p/PL1IHA35xY5H5sjfjibqn_XFFxk3-pFiaX MEMT203 (Dynamics): ruclips.net/p/PL1IHA35xY5H6G64khh8fcNkjVJDGMqrHo MEEN361 (Adv. Mech of Mat): ruclips.net/p/PL1IHA35xY5H5AJpRrM2lkF7Qu2WnbQLvS MEEN462 (Machine Design): ruclips.net/p/PL1IHA35xY5H5KqySx6n09jaJLUukbvJvB (MEEN 361 & 462 are taught from Shigley's Mechanical Engineering Design) Thanks for watching!
Amazing material, just quick doubt, where does the 0.625 comes from in the Torque formula where you add tan and sec with friction coefficients? or what does that constant means? Thanks and great job is so easy to understand the amazing world of fasteners
[56:00] Bolt stretches .05mm, plates compress .01mm, so the bolt must jut out & stretch .06mm past the end of the compressed plates. So add. (In detail, the stretch of the bolt would slightly increase the number of mm per turn?)
Thank you very much for this lecture, I loved the way you talked about your professor doctor LEE, it show's that you love what you do, where I can find a copy of the book you were using while solving the problem?
I use Shigley's Mechanical Engineering Design, 10th edition, but there are several other high quality texts out there as well (e.g. Norton). Thanks for watching!
Shigley's Mechanical Engineering Design 10th Edition. I am in this professors class in person, and have taken him for 3 other classes already. My last time taking him. Amazing professor, cool guy, very nice, but his homeworks are absolute ASS.
Just curious when Torque a bolt to its 73% of YS in addition to tension force , how we can take torsion force in account? How we can calculate combined load on bolt
At 59.40 you calculated that around 80 percent of additional loading on the bolt head is going into relaxing the member instead of stretching the bolt. Whilst it is shown that the member deforms about 5 times less than the bolt. Why is this? My intuition tells me that if the member is 5 times as stiff as the bolt most of the loading should go into stretching the bolt. Btw these videos are awesome.
Thanks! Here is the key confusion I think you are having: there is deformation of member and bolt that happen during joint pre-loading, and there is deformation that happens due to additional applied load. The bolt is less stiff, so it naturally deforms more during pre-loading (pre-load force is equal between bolt and member). However, when external additional load is applied, the amount that the bolt stretches is equal to the amount that the member relaxes (changes in length are equal in this case, not changes in force). Because the member is stiffer, that means that the change in how much force it carries is greater than the change in how much the bolt carries under additional applied loads. I hope this helps! Thanks for watching!
I already thanked you several times for your great work but every time I find another great solution to an engineering challenge I encounter at work I just can't hide my gratitude. Thank you. You the man! :) Can I please have the title of the handbook/s you are using in this video? And also the titles of any other good handbooks which you can recommend for learning stress analysis/machine design? Greetings from Poland! EDIT: I found the title in the previous comments below. Thank you
I've been looking for this breakdown for so long, thank you! Also, what stylus, tablet and drawing software are you using in your lectures? I'm keen to use something similar when I'm doing calcs for work to save me using pen and paper
I'm glad it helped! I use a Fujitsu T901, 16gb ram, 2nd gen i7, nvidia graphics. It is built with a wacom digitizer. That's the key, wacom makes the best digitizer technology IMO. I have a newer computer as well, an HP zbook X2, it also has a wacom digitizer. I use Microsoft Onenote 2007 since that was the last version of that software that allowed fully customizable toolbars. In case you haven't seen them yet and might be interested, here are some of my playlists: ENGR122 (Statics & Engr Econ Intros): ruclips.net/p/PL1IHA35xY5H52IKu6TVfFW-BDqAt_aZyg ENGR220 (Statics & Mech of Mat): ruclips.net/p/PL1IHA35xY5H5sjfjibqn_XFFxk3-pFiaX MEMT203 (Dynamics): ruclips.net/p/PL1IHA35xY5H6G64khh8fcNkjVJDGMqrHo MEEN361 (Adv. Mech of Mat): ruclips.net/p/PL1IHA35xY5H5AJpRrM2lkF7Qu2WnbQLvS MEEN462 (Machine Design): ruclips.net/p/PL1IHA35xY5H5KqySx6n09jaJLUukbvJvB (MEEN 361 & 462 are taught from Shigley's Mechanical Engineering Design) Thanks for watching!
Most sources are reluctant to quote any specific guidelines on factor of safety generally. This is partly due to the liability it could incur by doing so. I can tell you that NASA basically says that once the worst possible load is anticipated, there has to be a joint separation factor of at least 1. I don't give this as a prescription, just a reference point. Thanks for watching!
Thanks! I'm glad you liked it! In case you haven't seen them yet and might be interested, here are some of my playlists: ENGR122 (Statics & Engr Econ Intros): ruclips.net/p/PL1IHA35xY5H52IKu6TVfFW-BDqAt_aZyg ENGR220 (Statics & Mech of Mat): ruclips.net/p/PL1IHA35xY5H5sjfjibqn_XFFxk3-pFiaX MEMT203 (Dynamics): ruclips.net/p/PL1IHA35xY5H6G64khh8fcNkjVJDGMqrHo MEEN361 (Adv. Mech of Mat): ruclips.net/p/PL1IHA35xY5H5AJpRrM2lkF7Qu2WnbQLvS MEEN462 (Machine Design): ruclips.net/p/PL1IHA35xY5H5KqySx6n09jaJLUukbvJvB (MEEN 361 & 462 are taught from Shigley's Mechanical Engineering Design) Thanks for watching!
Dear Mr. TheBom_PE, - 46:50 here you are talking about the effects during my thesis. I made a test field to investigate 10.9 tZn bolts wether there is a possibillity to relate the angle of the nut to the achieved preload. The measurements starts at the hook's area. It is not possible to relate the angle to the preload. How can this be. E.g. One and the same angle has a variation of +/- 100 kN Thanks for you answer BR JW
Are any of your formulas only compatible with metric? I have been trying to dissect some of the problems and keep getting an unreasonably large result for material and bolt stretch. I have been keeping an eye on units so I don't think it is an issue of decimal translation.
All of the methods shown work for both SI and US systems. I would recommend carrying all of your units through every step to make sure you aren't doing one of your conversions improperly. Make sure you are using the correct units for elastic modulus, they are generally in GPa or Mpsi.
@@TheBomPE I have found my issue. I was mistakenly using proof load of the fastener as its modulus of elasticity. Since 85 ksi proof load is much lower than 29,000 ksi modulus, it was resulting in an absurd amount of bolt stretch. Thank you for uploading this video and helping people learn about bolted joints!
Sir thanks for this video as it is very interesting and knowledgeable. Can we safely say that np must be greater than 1, and nL & no must be greater than 3 in order to know that it is safe? thanks.
It needs to be greater than 1, but will seldom be as high as 3. The loading factor nL is often set at 3 or higher though. I generally avoid giving advice on how high Design factors "should" be, since there are so many real- world factors that should be considered when setting them. Thanks for watching!
I actually have a good many lectures presented in SI. I think it is important to understand that basically all units and unit systems are contrived by human beings, and are not transcendent entities. One of the best ways to learn deeply about this is to exercise the use of at least two systems of units while learning about engineering. Thanks for watching!
That´s very good lecture. Congratulations!!! I have one question: Is there any situation we should consider retorquing the bolt? Let´s say after a certain amount of hours of operation and when the bolts are submitted to transverse vibration.
Thank you! Many bolts are not installed to the levels of preload that are recommended in the text. Consider the case of a steel bolt installed into aluminum threads, for instance. It may be difficult to design such a connection that would be capable of stretching the bolt as much as we would want. Enough elastic deformation applied to the bolt and member at installation can hypothetically be sufficient to prevent loosening due to vibration. What is often done as another safety against vibration-induced loosening is to make washers or bolt heads that have some kind of teeth that dig into the mating surfaces. Other safeguards can include threadlocking compounds, cotter pins, safety wire, etc. It is often important to use these secondary safeguards in applications where vibration-induced loosening could cause mechanical failure that endangers life, health, or property. Thanks for watching!
First of all thank you for uploading these videos! They are really really helpful! I would like to know where could i find the book you use in your lectures. Thanks!
Great Learning here the video lecture as well as from the comments! Thank you for this. I would like to ask more. Do you have similar video/references about bolt preload where external moment force is involved in the bolt and the member?
First of All thanks for the amazing lecture, i am designing 4 bolts to hold a speed reducer so the force applied on them is due to moment and not pure tension in such case do we have to calculate the Km and Kb or we don't need them
You might check out this lecture: ruclips.net/video/yv26c_5H2rA/видео.html It deals with torsion applied to bolt groups, which is what it sounds to me like you are talking about. Thanks for watching!
thanks you so much for this useful lecture. i just have a question about equation Np = Sp* At/(C*P + Fi ) (1:01:11 ) My understanding, for the pre-load condition, Bolt is under tensile load and 2 metal plates is under compression load. Then if we apply external load in axial direction ( tensile load ), then 2 metal plates will be released the force. Mean that : force in bolt = Fi + 5kN and Clamping force in plates = Fi - 5kN. The above equation should not include C factor, it should present as : Np = Sp* At/( 5kN + Fi ) please correct me if i was wrong. thank you so much !
The C factor describes how much of the external load is used to stretch the bolt and how much is used to relax the clamped members. It is based on the relative stiffness of each part. I would refer you to the explanation in Shigley's Mechanical Engineering Design if you want to see the origins of that equation in greater detail. Thanks for watching!
Thanks for a great video, it was really helpful. I have one question in regards to the external load. The bolt is under axial loading from both ends shouldn't the axial load be doubled. I'll really appreciate your answer.
No, the axial force should not be considered twice. This is a case of static equilibrium. To illustrate, consider a 5 kg weight hanging from the ceiling, suspended by a (massless) rope. The resultant weight on the ceiling is 5 kg, which corresponds to the rope tension. Bringing it back to the worked example, if the tensile force on one end was 20 kN, and say 5 kN on the other end the bolt group would be subject to a tensile force of 5 kN, but the complete assembly would be subject to an acceleration equal to the ratio between the net (total) force (20-5=15 kN) and the mass of the assembly.
I'm confused by the units of Kb at timestamp 28:38 being MN/m. Isn't the answer 5,546,445,738,000 N/m or 5546 GN/m? Edit: Nvm, I realized 78mm^s is not 0.078m^2.I'll leave this here in the event anyone else comes along and makes the same mistake.
If you found this video useful, consider helping me upgrade the old tablet PC I use to create these videos! Thanks! www.gofundme.com/help-replace-my-2011-tablet-pc
Mathcad is sold by a company called PTC. It isn't free or open- source, unfortunately. We have a license for it at Louisiana Tech. I wish it was less expensive and more widely adopted!
Do you have a copy of the Shigley text I teach this course out of? If so, you might look at figure 8-2. Otherwise look at this link: en.wikipedia.org/wiki/ISO_metric_screw_thread The pitch diameter is the diameter where the "width" of one nut thread is the same as that of one bolt thread. Because of this, I believe it is reasonable to use pitch diameter as dm (the diameter where a distributed force over thread faces is lumped) in screw torque equations for threaded fastener thread profiles (as opposed to ACME or square profiles). Frankly, the other factors that lead me to use dp as dm rather than computing an actual dm for the profile is that it is easier to find, and it makes very little difference anyway.
I am actually in this professor's class right now at the college he teaches at! We use Shigley's Mechanical Engineering Design 10th edition, but the 11th edition is now standard. I recommend finding a pdf of the 10th edition and reading chapter 8.
I was wondering if the references you make to equations and tables etc. comes from a book. If yes, can you tell me the title and author of the book, please?
No, they add. As the member compresses, it relaxes some of the stress that would otherwise have developed in the bolt, because the bolt won't stretch as much. This is consistent with what our intuition would be: the less stiff either of the materials are (i.e. the bolt material and the member material), the more the nut will have to turn to generate the clamping forces.
@@TheBomPE for them to understand better, maybe its better to say it this way : the compression of the member takes away tension added to the bolt. so you need to add sigma (bolt,tension) and sigma (member,compression)
PLEASE NOTE: there is an error at 42:57 ... this torque calculates to 72.02Nm, not 52.63Nm as stated in the video. I apologize for any confusion this may have caused!
My career as a mechanical engineer has gone down a bit of a rabbit hole these past five years. These fundamental videos you do are of great help to those of us who are trying to re-orient. Thank you.
I'm glad the videos are helping! Good luck to you! Thanks for watching!
What do you mean gone down a rabbit hole? What happened?
So, first I thought I want to learn about torque wrenches. So I went and googled how a torque wrench works, watched a video about it and then I started wondering how someone calculates those values for bolts and here I am. I think I might be addicted to learning and it's so awesome to find lectures from RUclips. I may not need this information but at least I know little bit more about bolts now. Guess I should sometime finish my own studies.
Thanks for uploading these!
I'm glad you found it interesting! If you want to learn about finding bolted joint stiffness constant for more complex joints, check this one out: ruclips.net/video/IpT4JzZ-4IE/видео.html
Thanks for watching!
I found this to be one of the clearest lectures I have ever attended to. Kudos to the professor. He managed to clarify many of my doubts about the subject and make it way EASY to understand and apply in everyday situations and when I am actually busy designing something that has to work for real.
I am of the opinion that making something EASY to understand is very difficult to do so I am happy to have found this channel.
Greetings and please continue posting!
Alfredo
This was perfect, I'm currently calculating bolt stress in a statically loaded member for an engineering team I'm a part of, and I haven't taken a machine design class yet, thank you.
You're welcome! I'm glad it helped!
Which equation you used for stress calculation in bolt? I am also interetsed in stress of the bolt. Is there any help regarding it?
Thanks from Germany, absolutely understandable for a non native speaker.
I'm glad you liked it! Thanks for watching!
Great lecture ,it's great to have access to those useful information from Vietnam. Hope you keep up with these amazing work and released more video like this
Glad you found it helpful! Thanks for watching!
You are good at explaining things. Thank you for your time brother.
I'm glad I can help! Thanks for watching!
KN/ (MN/m) = mm ........................... 1MN = 10^-3 KN 1 m = 10^-3 mm ok 28.8 x 10^-3 / 554.6 x 10^-3 it is good, thanks to explained in metric units
Really appreciate the detailed discussion. It was super helpful to one of my projects.
I'm glad I could help! Thanks for watching!
Great seeing a SI unit problem solved! Congratulations again!
I'm glad you liked it! Thanks for the support!
This is really most helpful video technically for live projects also.
Thanks a lot for making videos on important topics
I'm glad you found it helpful! Thanks for watching!
Very nice videos! I'm currently taking Machine Design II and Shigley is also our reference textbook so these videos are very helpful and relevant! ♥️
Glad I could help! Thanks for watching!
Thank you for providing this very thorough example. I worked the specs along with your video. Thanks again!
I'm glad you found it useful! Thanks for watching!
Hi Bom, Nice lecture and content. Please keep going! Thanks
Thank you! I'm making as much new content as I can while seeing to my other responsibilities. RUclips doesn't pay the bills, ha ha. In the meantime, here are some of my other playlists in case you haven't seen them yet and might be interested:
ENGR122 (Statics & Engr Econ Intros): ruclips.net/p/PL1IHA35xY5H52IKu6TVfFW-BDqAt_aZyg
ENGR220 (Statics & Mech of Mat): ruclips.net/p/PL1IHA35xY5H5sjfjibqn_XFFxk3-pFiaX
MEMT203 (Dynamics): ruclips.net/p/PL1IHA35xY5H6G64khh8fcNkjVJDGMqrHo
MEEN361 (Adv. Mech of Mat): ruclips.net/p/PL1IHA35xY5H5AJpRrM2lkF7Qu2WnbQLvS
MEEN462 (Machine Design): ruclips.net/p/PL1IHA35xY5H5KqySx6n09jaJLUukbvJvB
(MEEN 361 & 462 are taught from Shigley's Mechanical Engineering Design)
Thanks for watching!
Amazing material, just quick doubt, where does the 0.625 comes from in the Torque formula where you add tan and sec with friction coefficients? or what does that constant means? Thanks and great job is so easy to understand the amazing world of fasteners
Love your work mate. Valuable and generous, much appreciated.
Glad you like it! Thanks for watching!
[56:00] Bolt stretches .05mm, plates compress .01mm, so the bolt must jut out & stretch .06mm past the end of the compressed plates. So add. (In detail, the stretch of the bolt would slightly increase the number of mm per turn?)
- Bolt Class/Grade : 9.8 means :
9 x 100 = 900 MPa ------> S_ut
900 x 0.8 = 720 MPa --------> S_y
Thank you very much for this lecture, I loved the way you talked about your professor doctor LEE, it show's that you love what you do,
where I can find a copy of the book you were using while solving the problem?
Wonderfully clear lecture, even for an EE. What book are you using for this class. I would like to buy it.
I use Shigley's Mechanical Engineering Design, 10th edition, but there are several other high quality texts out there as well (e.g. Norton). Thanks for watching!
@@TheBomPE aside from Norton and Shigley, what are the other high quality machine design textbooks? I'm interested to know. Thank you
What book are you using during the lecture? It is possible to follow the same book?
Shigley's Mechanical Engineering Design 10th Edition. I am in this professors class in person, and have taken him for 3 other classes already. My last time taking him. Amazing professor, cool guy, very nice, but his homeworks are absolute ASS.
Just curious when Torque a bolt to its 73% of YS in addition to tension force , how we can take torsion force in account? How we can calculate combined load on bolt
At 59.40 you calculated that around 80 percent of additional loading on the bolt head is going into relaxing the member instead of stretching the bolt. Whilst it is shown that the member deforms about 5 times less than the bolt. Why is this? My intuition tells me that if the member is 5 times as stiff as the bolt most of the loading should go into stretching the bolt.
Btw these videos are awesome.
Thanks! Here is the key confusion I think you are having: there is deformation of member and bolt that happen during joint pre-loading, and there is deformation that happens due to additional applied load. The bolt is less stiff, so it naturally deforms more during pre-loading (pre-load force is equal between bolt and member). However, when external additional load is applied, the amount that the bolt stretches is equal to the amount that the member relaxes (changes in length are equal in this case, not changes in force). Because the member is stiffer, that means that the change in how much force it carries is greater than the change in how much the bolt carries under additional applied loads. I hope this helps! Thanks for watching!
Bolts Pretensioned to specified torque Eliminates stress reversal and resulting fatigue cracking in fasteners .
Bolts stay in constant tension .
I already thanked you several times for your great work but every time I find another great solution to an engineering challenge I encounter at work I just can't hide my gratitude. Thank you. You the man! :)
Can I please have the title of the handbook/s you are using in this video? And also the titles of any other good handbooks which you can recommend for learning stress analysis/machine design?
Greetings from Poland!
EDIT: I found the title in the previous comments below. Thank you
You are very kind! Thanks for the encouragement! The other machine design book I really like is Norton.
I've been looking for this breakdown for so long, thank you! Also, what stylus, tablet and drawing software are you using in your lectures? I'm keen to use something similar when I'm doing calcs for work to save me using pen and paper
I'm glad it helped! I use a Fujitsu T901, 16gb ram, 2nd gen i7, nvidia graphics. It is built with a wacom digitizer. That's the key, wacom makes the best digitizer technology IMO. I have a newer computer as well, an HP zbook X2, it also has a wacom digitizer. I use Microsoft Onenote 2007 since that was the last version of that software that allowed fully customizable toolbars.
In case you haven't seen them yet and might be interested, here are some of my playlists:
ENGR122 (Statics & Engr Econ Intros): ruclips.net/p/PL1IHA35xY5H52IKu6TVfFW-BDqAt_aZyg
ENGR220 (Statics & Mech of Mat): ruclips.net/p/PL1IHA35xY5H5sjfjibqn_XFFxk3-pFiaX
MEMT203 (Dynamics): ruclips.net/p/PL1IHA35xY5H6G64khh8fcNkjVJDGMqrHo
MEEN361 (Adv. Mech of Mat): ruclips.net/p/PL1IHA35xY5H5AJpRrM2lkF7Qu2WnbQLvS
MEEN462 (Machine Design): ruclips.net/p/PL1IHA35xY5H5KqySx6n09jaJLUukbvJvB
(MEEN 361 & 462 are taught from Shigley's Mechanical Engineering Design)
Thanks for watching!
18:41
Great job .Thanks for your efforts
Thanks for watching!
Excellent video..need to know, how much will be typical factor of safety against joint seperation to be considered.
Most sources are reluctant to quote any specific guidelines on factor of safety generally. This is partly due to the liability it could incur by doing so. I can tell you that NASA basically says that once the worst possible load is anticipated, there has to be a joint separation factor of at least 1. I don't give this as a prescription, just a reference point. Thanks for watching!
Great lecture. Enjoyed every minute of it
Thanks! I'm glad you liked it! In case you haven't seen them yet and might be interested, here are some of my playlists:
ENGR122 (Statics & Engr Econ Intros): ruclips.net/p/PL1IHA35xY5H52IKu6TVfFW-BDqAt_aZyg
ENGR220 (Statics & Mech of Mat): ruclips.net/p/PL1IHA35xY5H5sjfjibqn_XFFxk3-pFiaX
MEMT203 (Dynamics): ruclips.net/p/PL1IHA35xY5H6G64khh8fcNkjVJDGMqrHo
MEEN361 (Adv. Mech of Mat): ruclips.net/p/PL1IHA35xY5H5AJpRrM2lkF7Qu2WnbQLvS
MEEN462 (Machine Design): ruclips.net/p/PL1IHA35xY5H5KqySx6n09jaJLUukbvJvB
(MEEN 361 & 462 are taught from Shigley's Mechanical Engineering Design)
Thanks for watching!
Hi thank you for the the video, but do you have any other video for max stress calculation in the bolts
What would be the maximum tightening torque recommended for M 48 x 5 Hex Head bolt ; 300 mm long ; 8.8 Gr.
Dear Mr. TheBom_PE, - 46:50 here you are talking about the effects during my thesis.
I made a test field to investigate 10.9 tZn bolts wether there is a possibillity to relate the angle of the nut to the achieved preload. The measurements starts at the hook's area. It is not possible to relate the angle to the preload. How can this be. E.g. One and the same angle has a variation of +/- 100 kN
Thanks for you answer
BR JW
Sir do you have factor of safety to check if the number of bolts is already sufficient? Thanks. God bless.
Are any of your formulas only compatible with metric? I have been trying to dissect some of the problems and keep getting an unreasonably large result for material and bolt stretch. I have been keeping an eye on units so I don't think it is an issue of decimal translation.
All of the methods shown work for both SI and US systems. I would recommend carrying all of your units through every step to make sure you aren't doing one of your conversions improperly. Make sure you are using the correct units for elastic modulus, they are generally in GPa or Mpsi.
@@TheBomPE I have found my issue. I was mistakenly using proof load of the fastener as its modulus of elasticity. Since 85 ksi proof load is much lower than 29,000 ksi modulus, it was resulting in an absurd amount of bolt stretch. Thank you for uploading this video and helping people learn about bolted joints!
Glad I could help!
Sir thanks for this video as it is very interesting and knowledgeable. Can we safely say that np must be greater than 1, and nL & no must be greater than 3 in order to know that it is safe? thanks.
It needs to be greater than 1, but will seldom be as high as 3. The loading factor nL is often set at 3 or higher though. I generally avoid giving advice on how high Design factors "should" be, since there are so many real- world factors that should be considered when setting them. Thanks for watching!
Thanks for using metric units.
I actually have a good many lectures presented in SI. I think it is important to understand that basically all units and unit systems are contrived by human beings, and are not transcendent entities. One of the best ways to learn deeply about this is to exercise the use of at least two systems of units while learning about engineering. Thanks for watching!
Please mention book name you use in this lacture
may i please know the book that provides these equations?
Thnkyeou
I'm referencing Shigley's Mechanical Engineering Design, 10th edition. Thanks for watching!
Hi! Very interesting and understandable explanation, thanks for that. Only one question: from where does 0,625 come from (at 35:45)?
Does that extra twist during torquing reduce the final axial stretch of the fastener or is it negligible, if using a torque spec for installation
since some of the torque you are measuring is going into torsional stress rather than tension
That´s very good lecture. Congratulations!!! I have one question: Is there any situation we should consider retorquing the bolt? Let´s say after a certain amount of hours of operation and when the bolts are submitted to transverse vibration.
Thank you! Many bolts are not installed to the levels of preload that are recommended in the text. Consider the case of a steel bolt installed into aluminum threads, for instance. It may be difficult to design such a connection that would be capable of stretching the bolt as much as we would want. Enough elastic deformation applied to the bolt and member at installation can hypothetically be sufficient to prevent loosening due to vibration. What is often done as another safety against vibration-induced loosening is to make washers or bolt heads that have some kind of teeth that dig into the mating surfaces. Other safeguards can include threadlocking compounds, cotter pins, safety wire, etc. It is often important to use these secondary safeguards in applications where vibration-induced loosening could cause mechanical failure that endangers life, health, or property. Thanks for watching!
@@TheBomPE Thanks
Sir, from which book you are refering the page numbers?
I teach this course from Shigley's Mechanical Engineering Design, 10th edition. Thanks for watching!
First of all thank you for uploading these videos! They are really really helpful! I would like to know where could i find the book you use in your lectures. Thanks!
I'm glad you are finding my videos helpful! I use Shigley's Mechanical Engineering Design, 10th edition. Thanks for watching!
@@TheBomPE Appreciate your answer! Thanks!
Sir, thanks for this video. in 35:50 where 0.625fc is included in the formula, I just wan to ask about the value of fc. thanks
fc is the coefficient of friction for the reaction collar. Thanks for watching!
Great Learning here the video lecture as well as from the comments! Thank you for this.
I would like to ask more. Do you have similar video/references about bolt preload where external moment force is involved in the bolt and the member?
First of All thanks for the amazing lecture, i am designing 4 bolts to hold a speed reducer so the force applied on them is due to moment and not pure tension in such case do we have to calculate the Km and Kb or we don't need them
You might check out this lecture: ruclips.net/video/yv26c_5H2rA/видео.html
It deals with torsion applied to bolt groups, which is what it sounds to me like you are talking about. Thanks for watching!
thanks you so much for this useful lecture.
i just have a question about equation Np = Sp* At/(C*P + Fi ) (1:01:11 )
My understanding, for the pre-load condition, Bolt is under tensile load and 2 metal plates is under compression load.
Then if we apply external load in axial direction ( tensile load ), then 2 metal plates will be released the force.
Mean that : force in bolt = Fi + 5kN
and Clamping force in plates = Fi - 5kN.
The above equation should not include C factor, it should present as : Np = Sp* At/( 5kN + Fi )
please correct me if i was wrong.
thank you so much !
The C factor describes how much of the external load is used to stretch the bolt and how much is used to relax the clamped members. It is based on the relative stiffness of each part. I would refer you to the explanation in Shigley's Mechanical Engineering Design if you want to see the origins of that equation in greater detail. Thanks for watching!
Book you used for this?? Kindly tell me
I use Shigley's Mechanical Engineering Design, 10th ed. Thanks for watching!
Thanks for a great video, it was really helpful. I have one question in regards to the external load. The bolt is under axial loading from both ends shouldn't the axial load be doubled. I'll really appreciate your answer.
No, the axial force should not be considered twice. This is a case of static equilibrium. To illustrate, consider a 5 kg weight hanging from the ceiling, suspended by a (massless) rope. The resultant weight on the ceiling is 5 kg, which corresponds to the rope tension. Bringing it back to the worked example, if the tensile force on one end was 20 kN, and say 5 kN on the other end the bolt group would be subject to a tensile force of 5 kN, but the complete assembly would be subject to an acceleration equal to the ratio between the net (total) force (20-5=15 kN) and the mass of the assembly.
I'm confused by the units of Kb at timestamp 28:38 being MN/m. Isn't the answer 5,546,445,738,000 N/m or 5546 GN/m?
Edit: Nvm, I realized 78mm^s is not 0.078m^2.I'll leave this here in the event anyone else comes along and makes the same mistake.
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Very nice Video! Thank you!
I'm glad you enjoyed it! Thanks for watching!
At 51：53， Why l=24mm.,not12mm? Sorry, I can't understand.
Dr. Micheal, where do I download Mathcad for training purposes?
Mathcad is sold by a company called PTC. It isn't free or open- source, unfortunately. We have a license for it at Louisiana Tech. I wish it was less expensive and more widely adopted!
@@TheBomPE I will propose my employer to purchase licence for us:) Thanks.
Sir what book did you used here?
I use Shigley's Mechanical Engineering Design, 10th ed. Thanks for watching!
Whay book’s pages is being referred to?
I use Shigley's Mechanical Engineering Design, 10th edition. Thanks for watching!
Thanks!
Thank you so much!
thanks alot for your lecture , however can you put the references that you are using in the description of the video , tables ..etc
You're welcome! All of the references I make in this video are in Shigley's Mechanical Engineering Design, 10th edition. Thanks for watching!
If dm is the same as mean diameter, why is that equal to the pitch diameter dp?
Do you have a copy of the Shigley text I teach this course out of? If so, you might look at figure 8-2. Otherwise look at this link: en.wikipedia.org/wiki/ISO_metric_screw_thread
The pitch diameter is the diameter where the "width" of one nut thread is the same as that of one bolt thread. Because of this, I believe it is reasonable to use pitch diameter as dm (the diameter where a distributed force over thread faces is lumped) in screw torque equations for threaded fastener thread profiles (as opposed to ACME or square profiles). Frankly, the other factors that lead me to use dp as dm rather than computing an actual dm for the profile is that it is easier to find, and it makes very little difference anyway.
What is a good engineering book on this topic?
I teach this from Shigley's Mechanical Engineering Design, 10th edition. There are likely other good sources as well. Thanks for watching!
What is the reference text book
Shigley's Mechanical Engineering Design, 10th ed. (US version)
what books do you recommend me to learn this theory? I work in Europe and with DIN and Eurocodes. Thanks a lot.
I am actually in this professor's class right now at the college he teaches at! We use Shigley's Mechanical Engineering Design 10th edition, but the 11th edition is now standard. I recommend finding a pdf of the 10th edition and reading chapter 8.
@@kebha6308 thanks a lot
Which book you referred to in this video?
Shigley's Mechanical Engineering Design, 10th ed.
Can you recommend any comprehensive book on Valve Design Engineering?
I was wondering if the references you make to equations and tables etc. comes from a book. If yes, can you tell me the title and author of the book, please?
I use Shigley's Mechanical Engineering Design, 10th edition. Thanks for watching!
Make more vedios
@54:45 should it be a subtraction since the bolt in under tension and the member under compression
No, they add. As the member compresses, it relaxes some of the stress that would otherwise have developed in the bolt, because the bolt won't stretch as much. This is consistent with what our intuition would be: the less stiff either of the materials are (i.e. the bolt material and the member material), the more the nut will have to turn to generate the clamping forces.
@@TheBomPE
for them to understand better, maybe its better to say it this way :
the compression of the member takes away tension added to the bolt.
so you need to add
sigma (bolt,tension)
and
sigma (member,compression)
Great
Glad you liked it!
Good
Glad you liked it! Thanks for watching!