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Old 12-10-2010, 10:15 PM   #61
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From Dieter, "Mechanical Metallurgy", page 281 :

Mechanical Metallurgy - Dieter (Si Edition)

"The modulus of elasticity is determined by the binding forces between atoms. Since these forces cannot be changed without changing the basic nature of the material, it follows that the modulus of elasticity is one of the most structure-insensitive of the mechanical properties. It is only slightly affected by allowing additions, heat treatment or cold work."

- Bart
How, or does, it matter with the differences between low carbon steel < .3% carbon and high carbon steel (which springs are typically made of) at .6 - .99% carbon?
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Old 12-10-2010, 10:17 PM   #62
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In the case of a tapered spring bar, we have a tapered cantilevered beam with a continuously varible cross section. Right?
This beam will:
-not deflect in a symmetrical arc
-not be relatively consistent in its tip's vertical (upward force applied at chain attachment/L-bracket, and force borne equally throughout the length of the beam) travel when equal and incremental load is applied (ie: 100 pounds of load deflects the tip considerably more than 1", the second 100 pounds applied moves the tip less than the first 100 pound application, but greater than the first beam above, and so on, up to the beam's yield point)
A constant spring rate can be expressed as follows:

Force = displacement * spring constant

Where displacement is how much the spring bar is deflected, and spring constant depends on size and length of the (steel) bar.

What is meant by a variable spring rate is where the spring constant isn't constant, but is also a function of X.

This doesn't happen for simple steel bars, regardless of shape, heat treatment, alloying materials, color or advertising budget.

If you want a variable spring rate, then you need to change the geometry of the spring or the amount of spring in play.

- Bart
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Old 12-10-2010, 10:19 PM   #63
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Hi, next we need to know how much pressure on the frame, from ball to spring bar perch, will it take before the frame buckles. Test this on 2" X 4", 2" X 5", or whatever the most common frame sizes are.
Exactly my concern.
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Old 12-10-2010, 10:20 PM   #64
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How, or does, it matter with the differences between low carbon steel < .3% carbon and high carbon steel (which springs are typically made of) at .6 - .99% carbon?
Modulus of Elasticity, Strength Properties of Metals - Iron and Steel - Engineers Edge

contains rather an exhaustive list.

- Bart
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Old 12-10-2010, 10:41 PM   #65
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A constant spring rate can be expressed as follows:

Force = displacement * spring constant

Where displacement is how much the spring bar is deflected, and spring constant depends on size and length of the (steel) bar.

What is meant by a variable spring rate is where the spring constant isn't constant, but is also a function of X.

This doesn't happen for simple steel bars, regardless of shape, heat treatment, alloying materials, color or advertising budget.

If you want a variable spring rate, then you need to change the geometry of the spring or the amount of spring in play.

- Bart
So, are you saying that two bars with the same rating (whatever that means), one which has uniform cross section through it's entire length will behave (deflect) the same as a tapered (continuously variable cross section) throughout it's operating range with the same loads applied? This makes my head spin!!!
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Old 12-10-2010, 10:50 PM   #66
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So, are you saying that two bars with the same rating (whatever that means), one which has uniform cross section through it's entire length will behave (deflect) the same as a tapered (continuously variable cross section) throughout it's operating range with the same loads applied? This makes my head spin!!!
The tapered bar will be more flexible; in other words, the spring constant will be lower. But doubling the amount of bend in each bar will always double the amount of force it exerts. Once that is no longer true, the bar will have taken a permanent set.

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Old 12-10-2010, 11:15 PM   #67
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The tapered bar will be more flexible; in other words, the spring constant will be lower. But doubling the amount of bend in each bar will always double the amount of force it exerts. Once that is no longer true, the bar will have taken a permanent set.

- Bart
So, if Andy only test to deflection at increasing loads we may not get many answers? (assuming no MFRs lie about their ratings) WAIT, I mean no MFRs marketing folks embellish in a positive fashion, as to the attributes of their particular product.

Seriously, though, I'm still struggling how force relates to flexibility when we imaging the motions going on back there during various geographic undulations.

There, I've used up all my big words with just a half a beer left. How's that for timing?
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Old 12-11-2010, 09:47 AM   #68
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The tapered bar will be more flexible; in other words, the spring constant will be lower. But doubling the amount of bend in each bar will always double the amount of force it exerts. Once that is no longer true, the bar will have taken a permanent set.

- Bart
Bart,
You're confusing me again with the "more flexible" statement. All I am interested in (I think) is tip travel at varying loads applied at the tip. When measuring from the ground to tip of spring bar and adding the additional tip travel, won't the graph look like the attached? X axis is load in pounds x 100, and the y axis is inches of vertical movement from an unloaded bar as load is applied.

Won't the graph look like the attached?

I'm beginning to wonder if we're not communicating again. I get that the two systems are going to have the same strength. I am not plotting strength, just tip deflection at various loads.

What am I missing with the "flexibility" phrase?
Attached Files
File Type: xls spring bar.xls (31.0 KB, 80 views)
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Old 12-11-2010, 09:50 AM   #69
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Oh, yeah, don't get hung up on the values....totally bogus numbers, for illustrative purposes only.
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Old 12-11-2010, 11:39 PM   #70
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A simple spring bar will produce force in proportion to the displacement from the rest position:

F = kx,

where F is in lbs, x is in inches, and k, the spring constant, is in lbs/inch.

The value of k will depend on the geometry of the bar. Different thickness of bar will produce different slopes, but in each case the line will pass through zero, and the plot of force vs displacement will be a straight line so long as the bar isn't bent so far that it takes a permanent set.

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Old 12-12-2010, 07:00 AM   #71
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f = kx,

where f is in lbs, x is in inches, and k, the spring constant, is in lbs/inch.

the value of k will depend on the geometry of the bar. Different thickness of bar will produce different slopes,
- bart
BINGO!!!!! By gosh, I think you've got it!
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Old 12-12-2010, 08:06 AM   #72
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Today, we started a research program on load equalizing hitches.

We are measuring the bend in the load equalizing bars, at 100 pound increments, with different brands as well as square and round bars.

The range of the tests start at a low level and up to and including a 20 to 25 percent overload.

We will publish the results as soon as the tests are completed, hopefully by the end of this year.

Andy
Looking forward to it!

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Old 12-12-2010, 05:47 PM   #73
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Andy's work ties in precisely with my Load Distribution Hitch Calculator at:

http://www.airforums.com/forums/f464...sis-19236.html

At post 9, markdoane points out that what is required to complement the calculator is precisely the work Andy is doing.
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Old 12-12-2010, 08:31 PM   #74
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Nick,

Since then I have measured a couple of sets of bars to determine the spring rates. The bars I measured were a set of Reese 600# bars, and a set of Draw-Tite 800# bars.

The Reese bars developed a spring deflection rate of 166 lbs/inch, and the Draw-Tite bars measured 183 lbs/inch. Both sets were 1.125 inches wide.

If we assume a 2" bend for optimal deflection, the 600 lb bars measured 664 lbs, and the 800 lb bars measured 732 lbs. I think that is a reasonable variation, considering the bars were the same width.
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Old 12-12-2010, 09:29 PM   #75
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Nick,

Since then I have measured a couple of sets of bars to determine the spring rates. The bars I measured were a set of Reese 600# bars, and a set of Draw-Tite 800# bars.

The Reese bars developed a spring deflection rate of 166 lbs/inch, and the Draw-Tite bars measured 183 lbs/inch. Both sets were 1.125 inches wide.

If we assume a 2" bend for optimal deflection, the 600 lb bars measured 664 lbs, and the 800 lb bars measured 732 lbs. I think that is a reasonable variation, considering the bars were the same width.
Mark, are you assuming the second inch of travel is double the load of the first inch, or linear, or did you actually measure the load at two inches of travel? And, I have to assume these were both sets of trunion type bars?

Did you measure any round bars of simular advertised weight?
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Old 12-12-2010, 09:35 PM   #76
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I think he's assuming linear. 2" X 166 lbs X 2 bars= 664 lbs?????
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Old 12-12-2010, 09:38 PM   #77
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I think he's assuming linear. 2" X 166 lbs X 2 bars= 664 lbs?????
Yes, I did the math also, but personally, I don't think the bars load linear.
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Old 12-12-2010, 09:46 PM   #78
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Yes, I did the math also, but personally, I don't think the bars load linear.
Yeah, I'm still having trouble reconciling the above posts between Bart and me. My head has started hurting trying to figger it out. I'm waiting till the numbers roll in. I'm also trying to figure out how we can make a spreadsheet to understand how deflection and associated force rises as the angle between the TV and trailer frame changes.... ie, the steep approach angle thing.
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Old 12-12-2010, 10:52 PM   #79
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I'm also trying to figure out how we can make a spreadsheet to understand how deflection and associated force rises as the angle between the TV and trailer frame changes.... ie, the steep approach angle thing.
If we assume bars have length L and spring constant k, and we
denote the truck beginning to climb a hill as a positive angle theta, we
can say (assuming small angles of deflection, blah blah):

force due to hill = L * sin(theta) * k

or for a 24" bar, 10% grade (5.7 degree slope) and 500 lb/inch spring constant:

24 * sin(5.7) * 500 = 1190 lbs....

So climbing a steep hill effectively bends the bars a bit more than 2" extra....

Those w/ 1000 lb bars can wince now...

This is why I feel a more constant force hitch would be a good thing for
one's trailer if one needs a WD hitch... and why those headed off road may wish to relax their WD bars somewhat...

- Bart
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Old 12-13-2010, 05:24 AM   #80
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Yes, I did the math also, but personally, I don't think the bars load linear.
That is correct. The bars were loaded at 260 lbs and 420 lbs. The spring rate increases as the load increases.

Here are the spring rates at 260 lbs load and 420 lbs load.

600# Reese bar
166 lbs/inch at 260 lb load
179 lbs/inch at 420 lb load

800# Draw-Tite bar
173 lbs/inch at 260 lb load
195 lbs/inch at 420 lb load

The spring rates are the average for two bars at each treatment. All were trunnion bars.
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