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  1. #41
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    Found the following information from Mike Lauterbach's site

    Has some useful information regarding tensile strength of plates / Pins & Bolts
    Attached Files Attached Files

  2. #42
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    Quote Originally Posted by Uys View Post
    It is actually quite easy to calculate. Mild Steel has a yield strength of 300 MPa. So for a 8 ton breaking force (minimum) you require an area of 266 mm2. Lets make that 300 mm2 for safety. So the minimum area should be more than 300mm2 to ensure that the recovery hook can handle the 8 tons required.

    Look at the image below. The minimum area there is accross a section through the hole. In this case it is (12 x 12.5) x 2 mm2 = 300 mm2. If you had a 10 mm thick plate a larger width would have been required, ie . (300 mm2/(10 x 2)) = 15 mm. Thus the width of the plate = (15 x 2)+ 22 = 52 mm.

    Obviously you need a factor of safety, as 300 MPa material you buy at your local steel supplier is not gaurenteed to be 300 MPa, for a 1.5 safety factor, multiply the required area with 1.5.

    As for the bolts, that wil mostly be determined by the holes allready available in the chassis. But to handle 8 tons you calculate as follows: Steel is 0.57 as strong in shear as in tension, thus the max allowable stress is 0.57 x 300 = 171 MPa. Then say that (Shear stress = Shear force / Area). Area = Shear Force / Shear stress) = ( 80 000N / 171 000 000 Pa) = 467 mm2 = lets say 500mm2. Then if you have four bolts, the bolt area = 500/4 = 125 mm2.

    The diameter = the sqaure root of (4 x A/ Pi)

    In this case D = Sqrt (4 x 125/Pi) = 12.6 mm, thus the safe thing to do would be to use 4 x M16 bolts.
    Wow, really cool explanation! But...

    Quote Originally Posted by Uys View Post
    Mild Steel has a yield strength of 300 MPa. So for a 8 ton breaking force (minimum) you require an area of 266 mm2.
    Could you please explain the calculation you made to get from 300 MPa to 266mm2?

  3. #43
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    Quote Originally Posted by avanclan View Post
    Found the following information from Mike Lauterbach's site

    Has some useful information regarding tensile strength of plates / Pins & Bolts
    Perhaps I'm missing something (wouldn't surprise me too much), but could someone please explain one of the calculations in the above-mentioned attachment?

    From point 1. (of the original attachment), the author states:

    First, consider the lashing eye material at the bolt hole:
    The allowable bearing force here (t=7mm,
    = 10mm) is 1720 kg.



    I presume that this refers to the table that I have attached below. If so, surely with 7mm thickness and a 10mm diameter, the bearing force should be 1200kg (1.2 tons), not 1720kg (which, to my mind, would be for 10mm thick material and a 10mm diameter).



    Attached Images Attached Images

  4. #44
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    I think ( which doesnt happen very often lol) that it should have been 256...

    300*.57*1.5(safety)=256.5

  5. #45
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    Quote Originally Posted by Coober View Post
    Perhaps I'm missing something (wouldn't surprise me too much), but could someone please explain one of the calculations in the above-mentioned attachment?

    From point 1. (of the original attachment), the author states:

    First, consider the lashing eye material at the bolt hole:
    The allowable bearing force here (t=7mm,
    = 10mm) is 1720 kg.



    I presume that this refers to the table that I have attached below. If so, surely with 7mm thickness and a 10mm diameter, the bearing force should be 1200kg (1.2 tons), not 1720kg (which, to my mind, would be for 10mm thick material and a 10mm diameter).



    Yip thats the way I read it, 1.2tons

  6. #46
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    Quote Originally Posted by Coober View Post
    Could you please explain the calculation you made to get from 300 MPa to 266mm2?
    Sure -

    Direct stress = Force/ area, thus

    Area = Force/Direct stress

    In this case we allow 300 MPa, so there is no safety factor included here.


    Area = (8 000 kg x 9.81)/300 000 000

    Area = 0.0002616 m2
    = 0.0002616 x1000 x 1000 (because it is m2)
    = 261 mm2


    I multiply the 8000 kg with 9.81 to get newtons, and the 300 MPa becomes 300 000 000 Pa
    Everything is a hammer.
    Unless it is a screw driver.

    Then it it a chisel.

  7. #47
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    Just a warning on the pins/bolts.
    The shear plane must always be at the part of the bolt where there is no thread otherwise the loads will not be correct.

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  8. #48
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    Quote Originally Posted by bomtek View Post
    I think ( which doesnt happen very often lol) that it should have been 256...

    300*.57*1.5(safety)=256.5
    No, this is wrong. You have to devide by 1.5, because this is the allowable stress, not the actual stress. The allowable stress must become smaller when you include the safety factor.

    Thus, allowable shear is 300 MPa x 0.57 = 171 MPa.

    With safety factor of 1.5 included it becomes 171 /1.5 = 114 MPa.
    Everything is a hammer.
    Unless it is a screw driver.

    Then it it a chisel.

  9. #49
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    Yip - or otherwise use (bolt diameter - pitch) as the diameter in your calc. But like Rudolf said, it is not a good idea to carry shear across threads.

    Quote Originally Posted by RudolfD2 View Post
    Just a warning on the pins/bolts.
    The shear plane must always be at the part of the bolt where there is no thread otherwise the loads will not be correct.
    Everything is a hammer.
    Unless it is a screw driver.

    Then it it a chisel.

  10. #50
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    Thanks Uys, makes a whole lotta sense and brings back long forgotten memories

  11. #51
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    All this stuff reminds me a lot of the silent p in swimming...
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  12. #52
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    1.5 is too low for a safety factor in a recovery application IMO.

    As has been said above a few times, use the accepted safety factors used for rigging and lifting which is 5 or 6. This is especially important due to the impact nature of the expected load.

    Safety factors lower that two are only used in things like aircraft design and racing etc where the extra weight resulting from high safety factors cannot be automatically accepted. However in those sorts of applications materials used are of a very high standard, are rigoroursly tested and inspected throughout their operational lives etc.

  13. #53
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    Quote Originally Posted by alanB View Post
    use the accepted safety factors used for rigging and lifting which is 5 or 6
    Which means that we all have to fit recovery point made from 50 mm plate.
    Everything is a hammer.
    Unless it is a screw driver.

    Then it it a chisel.

  14. #54
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    ek is amper daar 20mm?
    '86 SFA Hilux V6

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  15. #55
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    If one goes and use the Structural design codes the you will not be able to fit any thing to a vehicle.
    M16 & M12's are only allowed for lightly loaded components. M20 Gr. 8.8 must be used. But whats the point if the chassis rail is made of 2mm plate. You will need doubler plates wlede to the chassis by a coded welder ect. ect.

    The basic formules litsed here would be sufficient if used properly.

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  16. #56
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    Rudolf I agree with you. I have used 8mm plate (x 100mm) and reinforced the area around the recovery hole and added additional bolts to both my rear recovery points and it looks ridiculously oversized compared to the original ladder frame of the vehicle. Not that I say its wrong or a vehicle is not made for the strength but simply an overload as mentioned to 5 or 6 times is more related to ropes for hoisting arrangements. Shock loading is far less during snatching due to the elasticy of the normal recovery equipment used versus lifting equipment steel cables!
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  17. #57

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    the more susceptible the item to damage the higher is the factor of safety needs to be

    steel will have a lower factor of safety in comparrison to a textile recovery strap

    we advocate a factor of safety of 4 : 1 for our recovery equipment
    (this our rule of thumb and is not documented in any spec's)

    Michael Greeen
    SECURETECH

  18. #58
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    On the overland forum, there is an article about recovery points. One of the listed items (#8 ) states:

    Where bolts are fitted through a chassis member it is dvisable that, if not already fitted, a high tensile tube be welded flush into the chassis through which the bolt then passes. This has the effect of spreading the load by the bolt to a much greater area around this tube AND prevents the chassis from giving way as the bolt is tightened.
    This sounds like a very interesting point, but I'm not sure I fully understand what is being 'said'.

    For those that do understand the point being made, can you tell me if this is, in fact, a good idea?
    If so, could someone elaborate on the concept? Perhaps a 'sketch'??

    I realise that this may be a little OT, but does tie-in with doing the job yourself!
    Last edited by Coober; 2009/11/04 at 04:59 PM. Reason: Inadvertantly added a smiley by typing "8)"

  19. #59
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    It's also called a crush tube.
    Prevents the chassis rail from being crushed when the bolt passes through both sides. The Aussies use this method quite often.
    I did not use this method as space was in short supply at the back and at the front no access was posible on my D2.

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  20. #60
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    Quote Originally Posted by RudolfD2 View Post
    It's also called a crush tube.
    Prevents the chassis rail from being crushed when the bolt passes through both sides. The Aussies use this method quite often.
    I did not use this method as space was in short supply at the back and at the front no access was posible on my D2.
    So does the tube HAVE to be high-tensile steel and must it be welded in place?
    I take it that it is necessary that the tube be a tight fit within the chassis.

    Presumably this is only required if the bolt passes through both sides of the chassis.

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