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marshall_mosty
12-28-2010, 10:23 PM
The CCR's are kinda fuzzy, since they are written around 3/8" bolts for 20lb plates...


If I'm going to mount 135# of ballast (rectangular plates), how many fastener's do I need to run and what size? I'm going to be drilling them and want to make sure they are properly secured. I'm thinking two grade 8 1/2" bolts.


Here are my nerdy engineering thoughts:

A 1/2" grade eight coarse thread bolt has a tensile breaking strength of 21,300. According to the Maximum Distortion Energy Theorem, to convert to shear breaking strength, multiply by .577 This gives 12,290 lbs per bolt, or 24,580 lbs total bolt strength.

A conservative approximation for a "worst case" crash with a concrete barrior at 150mph would be approx 80g (3 ft compression impact in 135 milliseconds from 150mph to zero, assuming linear deceleration), so the 135 lb plate stack will weigh 10,800 lbs during the impact.

Therefore, if I run the two bolts as mentioned above, I will have a factor of safety of approx 2.27.

Should be plenty, right?

GlennCMC70
12-28-2010, 10:41 PM
Do the AI rules address ballast mounting at all? As they would overrule the CCR's. And I know the 2011 rules are not out. How about the 2010?

marshall_mosty
12-28-2010, 11:42 PM
It just says, "...securely fastened and approved by NASA tech and safety officials..."

Last year I was able to use 2 7/16" bolts for two 45lb plates... but need to stack all three together for my current mounting setup...

So....

GlennCMC70
12-29-2010, 12:11 AM
The CCR says.....
15.20 Ballast
Unless superseded by class rules, all ballast shall be solid metal such as steel, lead, or
uranium, and consist of a minimum of five (5) pounds per piece. Each piece should be
bolted in place with through-bolts, fender washers, and a locking-nut / system (e.g. jamnuts,
Nylox, etc.). All bolts should be grade five (5). There should be at least one 3/8”
diameter bolt for every ten (10) pounds of weight (e.g. 20 pound block uses two bolts).

I find it odd that that NASA would require 15 3/8" bolts to hold 150lbs of ballast. They are spec'ing grade 5 though. In fact, they word it to lead you to think only grade 5 is allowed. I was thinking grade 8 would be better and possibly require fewer bolts.
Worth an email to NASA HQ.

GlennCMC70
12-29-2010, 12:20 AM
E-mail sent. I'll allow Todd a couple days to respond before I go over his head.

kbrewmr2
12-29-2010, 10:10 AM
the CCR version isn't a mistake, changed last year - luckily TT, PT, and ST chose to override it to something reasonable. Many other classes are choosing to do the same so I hope Todd sees the light and helps you guys out :)

marshall_mosty
12-29-2010, 10:15 AM
E-mail sent. I'll allow Todd a couple days to respond before I go over his head.
Oops... Since this was CCR's, I went to the "big guys" at Nationals...

GlennCMC70
12-29-2010, 10:26 AM
E-mail sent. I'll allow Todd a couple days to respond before I go over his head.
Oops... Since this was CCR's, I went to the "big guys" at Nationals...

Please use your chain of "command" when and if at all possible. In this case, myself and Rob. I know Todd is not the most available person, but give him a chance.

By all means, post up the responce you get.

cjlmlml
12-29-2010, 11:01 AM
http://wiki.answers.com/Q/Why_is_there_a_chain_of_command_in_business





I posted the above so everyone would know why there is a chain of command, you can thank me later Glen.

marshall_mosty
12-29-2010, 12:32 PM
E-mail sent. I'll allow Todd a couple days to respond before I go over his head.
Oops... Since this was CCR's, I went to the "big guys" at Nationals...

Please use your chain of "command" when and if at all possible. In this case, myself and Rob. I know Todd is not the most available person, but give him a chance.

By all means, post up the responce you get.Glenn,
Sorry, you mentioned NASA HQ... that made me believe I needed to talk to the California boys...

marshall_mosty
12-29-2010, 03:17 PM
Got a ruling from Jerry Kunzman...



From: jerry@drivenasa.com
To: marshallmosty@hotmail.com; jlindsey@drivenasa.com
Subject: RE: CCR Rule (15.20) Ballast
Date: Wed, 29 Dec 2010 12:05:46 -0800

Marshall,
Your thinking is sound and the setup you are asking for is acceptable.
Please print this email, in case a local inspector needs to see it.

Jerry Kunzman

Eecutive Director




My original email to Jerry and John...




From: marshall mosty [mailto:marshallmosty@hotmail.com]
Sent: Wednesday, December 29, 2010 6:01 AM
To: jerry@drivenasa.com; jlindsey@drivenasa.com
Subject: CCR Rule (15.20) Ballast


Jerry, John,
I'm not sure who to address this to and wanted an official ruling.

The CCR's are kinda fuzzy, since they are written around 3/8" bolts for 20lb plates...


If I'm going to mount 135# of ballast (rectangular plates), how many fastener's do I need to run and what size? I'm going to be drilling them and want to make sure they are properly secured. I'm thinking two grade 8, 1/2" dia bolts.


Here are my nerdy engineering thoughts:

A 1/2" grade eight coarse thread bolt has a tensile breaking strength of 21,300 lbs. According to the Maximum Distortion Energy Theorem, to convert to shear breaking strength, multiply by .577 This gives 12,290 lbs per bolt, or 24,580 lbs total bolt strength if two are used.

A conservative approximation for a "worst case" crash with a concrete barrior at 150mph would be approx 80g (3 ft compression impact in 135 milliseconds from 150mph to zero, assuming linear deceleration), so the 135 lb plate stack will weigh 10,800 lbs during the impact.

Therefore, if I run the two bolts as mentioned above, I will have a factor of safety of approx 2.27.

Should be plenty, right?

The AI series rules say, "...securely fastened and approved by NASA tech and safety officials..."

Last year I was able to use 2 7/16" dia bolts for two 45lb plates (90lbs total)... but need to stack all three together for my current mounting setup...

I will be finalizing my ballast setup this coming weekend and would like to be able to not have any issues at the first event.

-Marshall Mosty
AI #67
NASA TX

MikeP99Z
12-29-2010, 07:30 PM
Thanks Marshall.

You should forward the question/response to Balingit for the general tech file

MP

marshall_mosty
12-29-2010, 07:42 PM
Mike,
Great idea. Just sent the email.

-Marshall

Todd Covini
12-30-2010, 10:38 AM
I was looking for a page # in the Maximum Distortion Energy Theorem. :wink:

Glad you got the info you needed.

-=- Todd

marshall_mosty
12-30-2010, 11:23 AM
I was looking for a page # in the Maximum Distortion Energy Theorem. :wink:

Glad you got the info you needed.

-=- Todd

Todd,
I thought this was typical oil field stuff...

The Maximum distortion energy theory is also known as shear energy theory or von Mises-Hencky theory. This theory postulates that failure will occur when the distortion energy per unit volume due to the applied stresses in a part equals the distortion energy per unit volume at the yield point in uniaxial testing. The total elastic energy due to strain can be divided into two parts. One part causes change in volume, and the other part causes change in shape. Distortion energy is the amount of energy that is needed to change the shape.

The Von Mises yield criterion suggests that the yielding of materials begins when the second deviatoric stress invariant J2 reaches a critical value k. For this reason, it is sometimes called the J2-plasticity or J2 flow theory. It is part of a plasticity theory that applies best to ductile materials, such as metals. Prior to yield, material response is assumed to be elastic.

In materials science and engineering the von Mises yield criterion can be also formulated in terms of the von Mises stress or equivalent tensile stress, a scalar stress value that can be computed from the stress tensor. In this case, a material is said to start yielding when its von Mises stress reaches a critical value known as the yield strength. The von Mises stress is used to predict yielding of materials under any loading condition from results of simple uniaxial tensile tests. The von Mises stress satisfies the property that two stress states with equal distortion energy have equal von Mises stress.

Because the von Mises yield criterion is independent of the first stress invariant, it is applicable for the analysis of plastic deformation for ductile materials such as metals, as the onset of yield for these materials does not depend on the hydrostatic component of the stress tensor.

Although formulated by Maxwell in 1865, it is generally attributed to Richard Edler von Mises (1913). Tytus Maksymilian Huber (1904), in a paper in Polish, anticipated to some extent this criterion. This criterion is also referred to as the Maxwell–Huber–Hencky–von Mises theory.

Mathematically the yield function for the von Mises condition is expressed as:
http://sphotos.ak.fbcdn.net/hphotos-ak-snc4/hs242.snc4/39427_1377206289576_1814430887_715723_5532710_n.jp g


An alternative form is:
http://sphotos.ak.fbcdn.net/hphotos-ak-ash1/hs742.ash1/163418_1377206209574_1814430887_715722_5386649_n.j pg


where k can be shown to be the yield stress of the material in pure shear. As it will become evident later in the article, at the onset of yielding, the magnitude of the shear yield stress in pure shear is √3 times lower than the tensile yield stress in the case of simple tension. Thus, we have
http://sphotos.ak.fbcdn.net/hphotos-ak-snc6/hs011.snc6/166146_1377206369578_1814430887_715724_7127582_n.

Furthermore, if we define the von Mises stress as http://sphotos.ak.fbcdn.net/hphotos-ak-snc6/hs032.snc6/166229_1377206449580_1814430887_715725_6388662_n.j pg, the von Mises yield criterion can be expressed as:
http://sphotos.ak.fbcdn.net/hphotos-ak-ash1/hs740.ash1/163238_1377206569583_1814430887_715726_7999172_n.j pg

Substituting J2 in terms of the principal stresses into the von Mises criterion equation we have
http://sphotos.ak.fbcdn.net/hphotos-ak-snc6/hs048.snc6/167823_1377206649585_1814430887_715727_1785451_n.j pg

or
http://sphotos.ak.fbcdn.net/hphotos-ak-snc6/hs070.snc6/168066_1377206729587_1814430887_715728_6396713_n.j pg

or as a function of the stress tensor components
http://sphotos.ak.fbcdn.net/hphotos-ak-snc6/hs074.snc6/168435_1377206809589_1814430887_715729_2991623_n.j pg

This equation defines the yield surface as a circular cylinder (See Figure Below) whose yield curve, or intersection with the deviatoric plane, is a circle with radius http://sphotos.ak.fbcdn.net/hphotos-ak-ash1/hs752.ash1/164357_1377206889591_1814430887_715730_7914388_n.j pg, or http://sphotos.ak.fbcdn.net/hphotos-ak-snc4/hs1396.snc4/164893_1377206969593_1814430887_715731_8377073_n.j pg. This implies that the yield condition is independent of hydrostatic stresses.

http://sphotos.ak.fbcdn.net/hphotos-ak-snc4/hs1369.snc4/164133_1377206089571_1814430887_715721_7753186_n.j pg

ShadowBolt
12-30-2010, 12:03 PM
I was looking for a page # in the Maximum Distortion Energy Theorem. :wink:

Glad you got the info you needed.

-=- Todd

Todd,
I thought this was typical oil field stuff...

The Maximum distortion energy theory is also known as shear energy theory or von Mises-Hencky theory. This theory postulates that failure will occur when the distortion energy per unit volume due to the applied stresses in a part equals the distortion energy per unit volume at the yield point in uniaxial testing. The total elastic energy due to strain can be divided into two parts. One part causes change in volume, and the other part causes change in shape. Distortion energy is the amount of energy that is needed to change the shape.

The Von Mises yield criterion suggests that the yielding of materials begins when the second deviatoric stress invariant J2 reaches a critical value k. For this reason, it is sometimes called the J2-plasticity or J2 flow theory. It is part of a plasticity theory that applies best to ductile materials, such as metals. Prior to yield, material response is assumed to be elastic.

In materials science and engineering the von Mises yield criterion can be also formulated in terms of the von Mises stress or equivalent tensile stress, a scalar stress value that can be computed from the stress tensor. In this case, a material is said to start yielding when its von Mises stress reaches a critical value known as the yield strength. The von Mises stress is used to predict yielding of materials under any loading condition from results of simple uniaxial tensile tests. The von Mises stress satisfies the property that two stress states with equal distortion energy have equal von Mises stress.

Because the von Mises yield criterion is independent of the first stress invariant, it is applicable for the analysis of plastic deformation for ductile materials such as metals, as the onset of yield for these materials does not depend on the hydrostatic component of the stress tensor.

Although formulated by Maxwell in 1865, it is generally attributed to Richard Edler von Mises (1913). Tytus Maksymilian Huber (1904), in a paper in Polish, anticipated to some extent this criterion. This criterion is also referred to as the Maxwell–Huber–Hencky–von Mises theory.

Mathematically the yield function for the von Mises condition is expressed as:
http://sphotos.ak.fbcdn.net/hphotos-ak-snc4/hs242.snc4/39427_1377206289576_1814430887_715723_5532710_n.jp g


An alternative form is:
http://sphotos.ak.fbcdn.net/hphotos-ak-ash1/hs742.ash1/163418_1377206209574_1814430887_715722_5386649_n.j pg


where k can be shown to be the yield stress of the material in pure shear. As it will become evident later in the article, at the onset of yielding, the magnitude of the shear yield stress in pure shear is √3 times lower than the tensile yield stress in the case of simple tension. Thus, we have
http://sphotos.ak.fbcdn.net/hphotos-ak-snc6/hs011.snc6/166146_1377206369578_1814430887_715724_7127582_n.

Furthermore, if we define the von Mises stress as http://sphotos.ak.fbcdn.net/hphotos-ak-snc6/hs032.snc6/166229_1377206449580_1814430887_715725_6388662_n.j pg, the von Mises yield criterion can be expressed as:
http://sphotos.ak.fbcdn.net/hphotos-ak-ash1/hs740.ash1/163238_1377206569583_1814430887_715726_7999172_n.j pg

Substituting J2 in terms of the principal stresses into the von Mises criterion equation we have
http://sphotos.ak.fbcdn.net/hphotos-ak-snc6/hs048.snc6/167823_1377206649585_1814430887_715727_1785451_n.j pg

or
http://sphotos.ak.fbcdn.net/hphotos-ak-snc6/hs070.snc6/168066_1377206729587_1814430887_715728_6396713_n.j pg

or as a function of the stress tensor components
http://sphotos.ak.fbcdn.net/hphotos-ak-snc6/hs074.snc6/168435_1377206809589_1814430887_715729_2991623_n.j pg

This equation defines the yield surface as a circular cylinder (See Figure Below) whose yield curve, or intersection with the deviatoric plane, is a circle with radius http://sphotos.ak.fbcdn.net/hphotos-ak-ash1/hs752.ash1/164357_1377206889591_1814430887_715730_7914388_n.j pg, or http://sphotos.ak.fbcdn.net/hphotos-ak-snc4/hs1396.snc4/164893_1377206969593_1814430887_715731_8377073_n.j pg. This implies that the yield condition is independent of hydrostatic stresses.

http://sphotos.ak.fbcdn.net/hphotos-ak-snc4/hs1369.snc4/164133_1377206089571_1814430887_715721_7753186_n.j pg

If BP had know all of this that well in the Gulf would not have blown out.


JJ

jdlingle
12-30-2010, 03:17 PM
:shock: :shock: :shock: :shock: :shock: :shock:

I don't think they taught that in my school. :)

BryanL
12-30-2010, 05:27 PM
[quote="marshall_mosty"]
A conservative approximation for a "worst case" crash with a concrete barrior at 150mph would be approx 80g (3 ft compression impact in 135 milliseconds from 150mph to zero, assuming linear deceleration), so the 135 lb plate stack will weigh 10,800 lbs during the impact.

quote]

So what part of the car can you put 2 half inch holes that will hold 10,800 lbs during an impact?