It's well explained and all the information is correct but omits some important information about group capacity. As it's poor practice to have single bolt connections in multi bolt connections for simply supported beams, beam flexure causes rotation of bolt holes which increases the load on the bolts. Although I don't believe this is directly covered by EN 1993-1 other literature does cover this such as The Steel Designer Manual 25.4.2 and SN017 where bolt shear capacity (or additional force applied) is reduced based based on bolt group geometry. It my be helpful to add this or create a separate video if you have not already. I ran some FE models and the effect is real, think it was something like 27% of vertical shear for a typical beam which tied in with SN017
Do you obtain total capacity by multiplying the shown result by number of bolts? Doesn't mater of their configuration? (bolts beside each other or one above the other)
Hi mate. Thanks for this very informative video. If I may ask, what code that refers your bolt min. distance and edge distance? Thanks By the way. Im one of your subrcirber now and Im keep on following your channel. Thanks mate
👏👏👏 Anyway of seeing what calcs you have planned ahead? I work for a leading consultancy in the UK.. and what you are doing, is taking the IstructE papers.... And doing a video tutorial version. I cannot stress how much these have been missing. Every engineering firm I've been at, there's always been (along with the code of conduct) a nessessity to share knowledge. However, to get that from the main engineering bodies, I have to have 3 memberships which is over £500 pa. On top of that, you need to pay a few hundred quid to do the day courses. It solely restricts the individuals to either pay and they may not be able to afford it. You remind me ALOT of the KHAN Academy. Nobody should be restricted access to this type of content. You really are a credit to the engineer field. How many are there of you?
Thank you very much for your lovely comment! We work very hard to help individuals in the engineering industry, we know how difficult engineering can be especially for those who are newly graduated or individuals who are just simply learning! We can't express our gratitude enough for comments like this! Its makes our day :) We are a team of 2, who are very passionate engineers! We put our videos together based solely on our own knowledge and experience. Combined we have over 30 years experience in structural engineering. If you have any suggestions or requests for tutorials please leave us a comment!
@@StructuralEngineerCalcs I see from your website you have structured your website in a similar fashion to that needed to cope with modern house jobs. Would it be better if I ping you an email through the site?
When you calculated the bearing capacity for the example in single shear, where did you get 410 for the UTS for S275? Is that the UTS for S275 for all plates of a thickness 10mm? If so are you able to link the website where I can find those figures from? Thanks
Assume a single hollow rectangular column a single boing through it. Dead load X is from top coming down into the bolt. The bolt is in one shear plane or two ? Plus the bolt Fub is highe then that of rectangular hollow section which Fub should be utilized. Lastly as the bolt is going through both walls of the rectangular section should we divide the load into half ? Because of two shear planes.
Hi! i was wondering if we have connection with multiple bolts of for example a resistance of 70 kN each (280 kN resistance in total). The force applied is 200 kN, is this then sufficient? Im asking this in relation to failure of one of the bolts for example. Should another safety factor be taken into account or should we for example design as if the full 200 kN is exerted to 1 bolt at an instant? I would love to hear from you what the best way is.
The tensile strength is calculated using the pitch diameter of the treads. When the length of thread engagement area exceeds the diameter of the bolt, longer engagement no longer affects the overall tensile strength as the diameter of the bolt is the weakest dimension.
So a 16mm bolt (M16) can take only around 6000-7000kg on shear load? I can't belive it... Also, by using 16mm diameter or full tensile area (so not on thread) of 201, the result is different: 61.4kN and 76.8kN. So why i can use the tensile area on the thread part but not on the untreaded one? I have a manual where it explain to use the tensile area, that for the not threaded part will be: 3.14*d*d / 4. For M16 is: 3.14*256 / 4 = 200.96 (so 201). There is also the different type for the thread part (with dn, dm, dres) that give around 157 (156,51 for M16). There is a document on fastenal website, that say for double shear a bolt "½-13 SAE J429 Grade 8", with strenght of 150000 psi (90000 psi with 60%) can allow 155xx kg, a value that is double from the calculated above (that is for M16 8.8 so 70k-75k psi on 60%!!). Also, if for the same size i use the same calculator, and converting the mm2 into sq-in, i get more than 21xxx kg for the M16, that is more realistic value for a M16 shear load. EDIT. some data is wrong in my comment, so read comment below for the correction.
Hi there! Thanks for reaching out with your question. The calculation of the shear load capacity of a bolt is dependent on many factors, including the material, the surface area, and the type of load. The manual you mentioned is using the tensile area of the untreaded part of the bolt to calculate its capacity, which is based on the formula of πd^2/4. However, the actual capacity of the bolt in shear load can be different, as it depends on other factors such as the type of thread, the surface condition, and the type of load. Regarding the Fastenal document, it's important to note that the shear load capacity of a bolt also depends on the type of load, such as double shear, as well as the standard used to calculate the strength. The document you mentioned is using the SAE J429 Grade 8 standard, which has a higher strength compared to the 8.8 standard used for the M16 bolt. That's why you are seeing a difference in the calculated values. I hope this clarifies your confusion. If you have any further questions, feel free to ask.
@@StructuralEngineerCalcs Thanks for the answer. I know that there are many factor that can make a difference, but here i was talking that ½-13 SAE J429 Grade 8, that's around 12.5mm and have 150000 psi, was much stronger than M16 8.8 (16mm and around 120000 psi) where instead have close to same performance BUT with 3.5mm less in size diameter. I've also found (maybe) the cause of my doubt. In your video you added safety factor that fastenal didn't use. Also in my previous comment i've calculated with double shear, that's why i've wrote around 20T, so single shear here is around 10T but without safety factor. If you use the same formula of the fastenal website the result is close the same (but you need to add safety factor). Now everything is much clear, many thanks :D
@@Peppins You're welcome! I'm glad I could help clarify things for you. It's great to hear that you were able to understand the difference and found the cause of your doubt. Yes, it's important to keep in mind that the safety factor can greatly impact the results, especially when calculating the strength of a bolt. Thank you for taking the time to explain your findings and I'm happy I could assist you.
@@StructuralEngineerCalcs I've also found how to calculate directly the kN or the kg by using different number for the Ym2. For example, calcuating with safety factor of 25% so 1,25 in the video, you have the N. If you divide by 1250 instead you will have the kN. If you divide for 12,258 you will have the kg. :D
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Very useful! worth noting the 0.6 shear factor can be different, in Eurocodes it can be 0.5 also!
Thanks for the video. How do you account for bending stress on the bolt if the connecting plates have more spacing between them than this simple case?
Thanks for the explanation, In the case of the middle plate will thicker than the plates of the extremity?
excellent tutorials sir calculations made easy
Great work sir . Great respect to you sir
Great video! Could you tell me the source of the Bearing capacity equations? A citation would be great!
Google mate
An excellent video tutorial !!
Thank you so much !!
Nice brother 🎉🎉🎉
Thanks for it!
Any particular standard we can refer to?
Great video and so helpful, I wonder if the partial safety factor can vary its value in the formula?
Nice Video
It's well explained and all the information is correct but omits some important information about group capacity. As it's poor practice to have single bolt connections in multi bolt connections for simply supported beams, beam flexure causes rotation of bolt holes which increases the load on the bolts. Although I don't believe this is directly covered by EN 1993-1 other literature does cover this such as The Steel Designer Manual 25.4.2 and SN017 where bolt shear capacity (or additional force applied) is reduced based based on bolt group geometry. It my be helpful to add this or create a separate video if you have not already. I ran some FE models and the effect is real, think it was something like 27% of vertical shear for a typical beam which tied in with SN017
Do you obtain total capacity by multiplying the shown result by number of bolts? Doesn't mater of their configuration? (bolts beside each other or one above the other)
Thanks for the video🙏
Great video! thanks
Hi mate. Thanks for this very informative video. If I may ask, what code that refers your bolt min. distance and edge distance? Thanks
By the way. Im one of your subrcirber now and Im keep on following your channel. Thanks mate
splendid
👏👏👏
Anyway of seeing what calcs you have planned ahead?
I work for a leading consultancy in the UK.. and what you are doing, is taking the IstructE papers.... And doing a video tutorial version.
I cannot stress how much these have been missing.
Every engineering firm I've been at, there's always been (along with the code of conduct) a nessessity to share knowledge. However, to get that from the main engineering bodies, I have to have 3 memberships which is over £500 pa. On top of that, you need to pay a few hundred quid to do the day courses.
It solely restricts the individuals to either pay and they may not be able to afford it.
You remind me ALOT of the KHAN Academy. Nobody should be restricted access to this type of content. You really are a credit to the engineer field.
How many are there of you?
Thank you very much for your lovely comment!
We work very hard to help individuals in the engineering industry, we know how difficult engineering can be especially for those who are newly graduated or individuals who are just simply learning!
We can't express our gratitude enough for comments like this! Its makes our day :)
We are a team of 2, who are very passionate engineers!
We put our videos together based solely on our own knowledge and experience. Combined we have over 30 years experience in structural engineering.
If you have any suggestions or requests for tutorials please leave us a comment!
@@StructuralEngineerCalcs I see from your website you have structured your website in a similar fashion to that needed to cope with modern house jobs. Would it be better if I ping you an email through the site?
@@markr8604 You are welcome anytime
is the calculation based on any standard or reference? like ISO, or DIN, JIS?
When you calculated the bearing capacity for the example in single shear, where did you get 410 for the UTS for S275? Is that the UTS for S275 for all plates of a thickness 10mm? If so are you able to link the website where I can find those figures from? Thanks
UTS is not dependent on the plate thickness. Its a material property
Which Is Code you used for Stadard formula of Shearing and Bearing ?
Eurocode 3-1-8
Can you give me load calculation of bolt fastener for basin counter to hold basin
Assume a single hollow rectangular column a single boing through it.
Dead load X is from top coming down into the bolt.
The bolt is in one shear plane or two ?
Plus the bolt Fub is highe then that of rectangular hollow section which Fub should be utilized.
Lastly as the bolt is going through both walls of the rectangular section should we divide the load into half ? Because of two shear planes.
Hi! i was wondering if we have connection with multiple bolts of for example a resistance of 70 kN each (280 kN resistance in total). The force applied is 200 kN, is this then sufficient? Im asking this in relation to failure of one of the bolts for example. Should another safety factor be taken into account or should we for example design as if the full 200 kN is exerted to 1 bolt at an instant? I would love to hear from you what the best way is.
this example applied to 4 bolts
For the shear capacity you have considered Fub of 8.8 grade and for bearing capacity you have considered fub of S275 ???
one for bolt， one for plate
would like to know the reference of ur calculating method.
Where is the tensile area actually referring to? If I have a longer thread engagement, would it change the tensile area?
The tensile strength is calculated using the pitch diameter of the treads. When the length of thread engagement area exceeds the diameter of the bolt, longer engagement no longer affects the overall tensile strength as the diameter of the bolt is the weakest dimension.
May I know what is the value of 0.6 stands for?
How about shear, axial and bending combined?
Do you have a source specifying the shear strength is 60% of the ultimate tensile strength?
Refer machine design Book by Author RS khurmi
How about when plate thickness isn't equal, what changes ?
Where did you find all the factors used?
He's explain bolt connection in Eurocode 3-1-8
How did you come up with the 157?
Tensile Area of the M16 Bolt (as per the table on the right side of the screen)
So a 16mm bolt (M16) can take only around 6000-7000kg on shear load? I can't belive it... Also, by using 16mm diameter or full tensile area (so not on thread) of 201, the result is different: 61.4kN and 76.8kN. So why i can use the tensile area on the thread part but not on the untreaded one?
I have a manual where it explain to use the tensile area, that for the not threaded part will be: 3.14*d*d / 4. For M16 is: 3.14*256 / 4 = 200.96 (so 201). There is also the different type for the thread part (with dn, dm, dres) that give around 157 (156,51 for M16).
There is a document on fastenal website, that say for double shear a bolt "½-13 SAE J429 Grade 8", with strenght of 150000 psi (90000 psi with 60%) can allow 155xx kg, a value that is double from the calculated above (that is for M16 8.8 so 70k-75k psi on 60%!!). Also, if for the same size i use the same calculator, and converting the mm2 into sq-in, i get more than 21xxx kg for the M16, that is more realistic value for a M16 shear load.
EDIT. some data is wrong in my comment, so read comment below for the correction.
Hi there! Thanks for reaching out with your question. The calculation of the shear load capacity of a bolt is dependent on many factors, including the material, the surface area, and the type of load. The manual you mentioned is using the tensile area of the untreaded part of the bolt to calculate its capacity, which is based on the formula of πd^2/4. However, the actual capacity of the bolt in shear load can be different, as it depends on other factors such as the type of thread, the surface condition, and the type of load. Regarding the Fastenal document, it's important to note that the shear load capacity of a bolt also depends on the type of load, such as double shear, as well as the standard used to calculate the strength. The document you mentioned is using the SAE J429 Grade 8 standard, which has a higher strength compared to the 8.8 standard used for the M16 bolt. That's why you are seeing a difference in the calculated values. I hope this clarifies your confusion. If you have any further questions, feel free to ask.
@@StructuralEngineerCalcs Thanks for the answer. I know that there are many factor that can make a difference, but here i was talking that ½-13 SAE J429 Grade 8, that's around 12.5mm and have 150000 psi, was much stronger than M16 8.8 (16mm and around 120000 psi) where instead have close to same performance BUT with 3.5mm less in size diameter.
I've also found (maybe) the cause of my doubt. In your video you added safety factor that fastenal didn't use. Also in my previous comment i've calculated with double shear, that's why i've wrote around 20T, so single shear here is around 10T but without safety factor.
If you use the same formula of the fastenal website the result is close the same (but you need to add safety factor).
Now everything is much clear, many thanks :D
@@Peppins You're welcome! I'm glad I could help clarify things for you. It's great to hear that you were able to understand the difference and found the cause of your doubt. Yes, it's important to keep in mind that the safety factor can greatly impact the results, especially when calculating the strength of a bolt. Thank you for taking the time to explain your findings and I'm happy I could assist you.
@@StructuralEngineerCalcs I've also found how to calculate directly the kN or the kg by using different number for the Ym2. For example, calcuating with safety factor of 25% so 1,25 in the video, you have the N. If you divide by 1250 instead you will have the kN. If you divide for 12,258 you will have the kg. :D
which code is it BS CODE