Leaf Springs

[HOrnbrod]

I have the TeraFlex greasable shackles on the back too to make up for the 1/2" settling effect and even it up w. the front w. the OME 3" lift coils.

[HellCreek]

Eagle is correct by stating the free arch of all MJ springs, including the MTs, are within 1/2" of each other. When I installed my new 3+2 MT packs about four years ago I did get a full 2-1/2" of lift compared to what I had before. This was due to the saggy condition of my 980# stock springs and the extra weight of the Leer camper shell I carry around. They eventually settled down about 1/2" in a year or so, and have remained there.

 

 

That's not entirely accurate. Here is the run down:

 

980 lb. capacity = 7.62" free arch, OEM part #89-52-002-312, 3+1 configuration

1100 lb. capacity = 8.75" free arch, OEM part #89-52-001-240, 3+1 configuration

1160 lb. capacity = 9.25" free arch, OEM part #89-52-002-351, 3+1 configuration

1220 lb. capacity = 9.75" free arch, OEM part #89-52-001-824, 3=1 configuration

1280 lb. capacity = 10.25" free arch, OEM part #89-52-002-350, 3+1 configuration

1440 lb. capacity = 9.12" free arch, OEM part #89-52-001-825, 4+1 configuration

1700 lb. capacity = 9.50" free arch, OEM part #89-52-003-676, 3+2 configuration

 

 

The first 5 springs are identical except for the free arch. The 1440 lb. spring has the 1st two leaves made from .291 thick steel, the next two leaves made from .262 steel, and the thick two-stage leaf from .625 steel. The MT spring has the 1st leaf made from .291 steel, the next two made from .323 steel, the long two-stage leaf made from .590 steel, and the bottom two-stage leaf from .625 steel. The same main leaf fits all seven springs. In the spring industry, the term "Free Arch" stands for the arch of the spring measured perpendicularly from the top of the main leaf at the center bolt location to a line drawn through the centers of the eyes, while the spring is in its free state (unloaded and off the vehicle). It's an industry term. :smart:

 

-Tom

[HOrnbrod]

I stand corrected Tom. :bowdown: Always though the free arch was close in all the MJ springs. So, since my originals were the 980 lb. capacity = 7.62" free arch, OEM part #89-52-002-312, 3+1 configuration base springs, and I put on the MT 1700 lb. capacity = 9.50" free arch, OEM part #89-52-003-676, 3+2 configuration, that's nearly 2" lift with the difference in free arch alone, less settling. Correct mate? Leaf springs are fascinating. :cheers:

[HellCreek]

Hi, Don,

 

That would be correct. :thumbsup:

 

Where in Alabama are you? We will need to get together one of these days. :cheers:

 

-Tom

[HOrnbrod]

I'm in Arab, AL. That's ĀĀĀ-rab. 'Bout three hours from you Tom. Got to get over there one day and check out your operation. :cheers:

[Eagle]

Eagle is correct by stating the free arch of all MJ springs, including the MTs, are within 1/2" of each other. When I installed my new 3+2 MT packs about four years ago I did get a full 2-1/2" of lift compared to what I had before. This was due to the saggy condition of my 980# stock springs and the extra weight of the Leer camper shell I carry around. They eventually settled down about 1/2" in a year or so, and have remained there.

 

 

That's not entirely accurate. Here is the run down:

 

980 lb. capacity = 7.62" free arch, OEM part #89-52-002-312, 3+1 configuration

1100 lb. capacity = 8.75" free arch, OEM part #89-52-001-240, 3+1 configuration

1160 lb. capacity = 9.25" free arch, OEM part #89-52-002-351, 3+1 configuration

1220 lb. capacity = 9.75" free arch, OEM part #89-52-001-824, 3=1 configuration

1280 lb. capacity = 10.25" free arch, OEM part #89-52-002-350, 3+1 configuration

1440 lb. capacity = 9.12" free arch, OEM part #89-52-001-825, 4+1 configuration

1700 lb. capacity = 9.50" free arch, OEM part #89-52-003-676, 3+2 configuration

 

 

The first 5 springs are identical except for the free arch. The 1440 lb. spring has the 1st two leaves made from .291 thick steel, the next two leaves made from .262 steel, and the thick two-stage leaf from .625 steel. The MT spring has the 1st leaf made from .291 steel, the next two made from .323 steel, the long two-stage leaf made from .590 steel, and the bottom two-stage leaf from .625 steel. The same main leaf fits all seven springs. In the spring industry, the term "Free Arch" stands for the arch of the spring measured perpendicularly from the top of the main leaf at the center bolt location to a line drawn through the centers of the eyes, while the spring is in its free state (unloaded and off the vehicle). It's an industry term. :smart:

 

-Tom

That's still only part of the story. It doesn't take into account that the thickness of the leaves varies from one spring/capacity to the next.

 

The bottom line is that the factory service manual lists a ride height for the rear sispension, measured from the top of the axle tube to the frame rail next to the bump stop. They list ONE height (plus-or-minus 1/2") for the 4WD chassis, and ONE height (plus-or-minus 1/2") for the 2WD chassis. The 2WD height is 1" less than the 4WD height.

 

BTW -- I don't believe I even mentioned free arch. I'm not interested in free arch and I don't pay attention to it. I want to know what height my truck is going to ride at, and how much weight I can put in before I sit on the bump stops. Free arch is important (I guess) to spring shops, but not to me.

[HOrnbrod]

BTW -- I don't believe I even mentioned free arch. I'm not interested in free arch and I don't pay attention to it. I want to know what height my truck is going to ride at, and how much weight I can put in before I sit on the bump stops. Free arch is important (I guess) to spring shops, but not to me.

 

Sorry Eagle - got ya mixed up w. the Wahoo-man's posts I guess. :doh: NO one should do that! :D Anyhow, what I stated before about the rear 2-1/2" lift after the MT pack install makes better sense after Tom's post detailing the MJ spring differences in free arch. And I measured the before and after results according to the axle-to-chassis method detailed in my FSM for 2WDs. So it worked out well.

[Eagle]

I still hope to simulate metric ton springs on a 2WD MJ or two by using cut-off XJ main leaves as AALs. The concept is that they start off with less arch, so they should settle in at normal ride height without adding much lift, but will help take up load as the suspension begins to compress.

 

Unfortunately, it doesn't appear that I'll get to that experiment this year.

[HOrnbrod]

I still hope to simulate metric ton springs on a 2WD MJ or two by using cut-off XJ main leaves as AALs. The concept is that they start off with less arch, so they should settle in at normal ride height without adding much lift, but will help take up load as the suspension begins to compress.

 

Unfortunately, it doesn't appear that I'll get to that experiment this year.

 

While I'm not a big fan of AAL's, that concept should work just fine IMHO. The only lift experienced initially would be the thickness of the new XJ main until settle-in, and that's negligible. You would just need the new u-bolts. The actual added load capacity would be the unknown, but will definately be increased, and it should help handling too. Go for it! Would be a very good (and economical) way to increase the load capacity a lot of us on the forum would be interested in. :cheers:

[Pete M]

I've doubled up the main leaf in 2 MJs so far and I like it. :thumbsup: Unfortunately I've got no measurements that could directly compare before and after heights or spring rates. :dunno:

[mvusse]

Actually, with a SUA suspension like a stock MJ, the thickness of the leaf does not add to lift. The bottom of the leaf pack just sits that much lower below the axle, but the axle still sits on top of the main leaf.

 

Only with a SOA suspension does the thickness of the leaf add to lift.

[Eagle]

While I'm not a big fan of AAL's, that concept should work just fine IMHO. The only lift experienced initially would be the thickness of the new XJ main until settle-in, and that's negligible. You would just need the new u-bolts. The actual added load capacity would be the unknown, but will definately be increased, and it should help handling too. Go for it! Would be a very good (and economical) way to increase the load capacity a lot of us on the forum would be interested in. :cheers:

My trucks aren't lifted. Spring under, so the thickness of the extra leaf won't have any effect at all on lift.

 

Actually, in rough terms the added capacity is known. Some years ago one of the 4WD magazines published formulae for calculating the spring rate of both coil and leaf springs. I lost the actual formulae, but for a leaf spring, assuming all leaves are the same thickness, the rate increases in direct proportion to the number of leaves. I don't think the XJ leaf will be exactly the same thickness as the MJ leaves, so that'll skew the number a bit, but in round numbers, since the thick overload leaf doesn't count until heavily loaded, I'll be going from a 3-leaf spring to a 4-leaf spring. That should increase the rate by approximately 33 percent (rate, as in pounds-per-inch of deflection). The travel before engaging the overload spring will be the same distance, so the capacity should also be increased approximately 33 percent.

 

Eureka! Found it!

 

Spring rate = (leaf width x no of leaves)/72 x (1820 x leaf thickness)/spring length

 

The entire thing is a first order equation. The main leaf will remain, so the only variable will be the number of leaves.

 

IIRC, the diagram in the magazine measured length straight eye-to-eye, not length of the leaf along the arch.

[HOrnbrod]

Aye, SUA, no lift due to leaf thickness. Mistake noted. Beat head again. :doh:

 

XJ main length compared to MJ main 3rd leaf? The XJ main leaf would be placed in #3 order over the MJ main as #4, existing thick overload underneath w. a probably 33% load increase capacity. Sound very good. Hasn't anyone done this yet -seem to recall a previous post? Probably another dead brain cell trying to live again...........

[CWLONGSHOT]

Aye, SUA, no lift due to leaf thickness. Mistake noted. Beat head again. :doh:

 

XJ main length compared to MJ main 3rd leaf? The XJ main leaf would be placed in #3 order over the MJ main as #4, existing thick overload underneath w. a probably 33% load increase capacity. Sound very good. Hasn't anyone done this yet -seem to recall a previous post? Probably another dead brain cell trying to live again...........

 

In my trailer build, I did kind of the oposite. Added a MJ leaf to a XJ pack for some lift.

 

 

CW

[Eagle]

XJ main length compared to MJ main 3rd leaf? The XJ main leaf would be placed in #3 order over the MJ main as #4, existing thick overload underneath w. a probably 33% load increase capacity. Sound very good. Hasn't anyone done this yet -seem to recall a previous post? Probably another dead brain cell trying to live again...........

Don, I'm very lost as to how you are counting leaves. MJ main as #4? :hmm:

 

Main leaf is #1. I'll have to re-check to confirm, but when I had these springs laid out in the garage I'm pretty certain the XJ main leaf was longer than the MJ #2, so the cut-off XJ leaf will go directly under the MJ main as #2, the second MJ leaf will move down to #3, and #3 will become #4. The factory overload leaf will remain as an overload leaf.

[HOrnbrod]

In my skewed way of looking at things, I was counting the leafs from the bottom up, excluding the overload that doesn't change. So #4 would be the MJ main, #3 the new XJ leaf, and so on. Guess I tend to look at things backwards? Bigger numbers for the bigger leafs. :hmm:

[HellCreek]

 

The first 5 springs are identical except for the free arch. The 1440 lb. spring has the 1st two leaves made from .291 thick steel, the next two leaves made from .262 steel, and the thick two-stage leaf from .625 steel. The MT spring has the 1st leaf made from .291 steel, the next two made from .323 steel, the long two-stage leaf made from .590 steel, and the bottom two-stage leaf from .625 steel. The same main leaf fits all seven springs. In the spring industry, the term "Free Arch" stands for the arch of the spring measured perpendicularly from the top of the main leaf at the center bolt location to a line drawn through the centers of the eyes, while the spring is in its free state (unloaded and off the vehicle). It's an industry term. :smart:

 

-Tom

That's still only part of the story. It doesn't take into account that the thickness of the leaves varies from one spring/capacity to the next.

 

The bottom line is that the factory service manual lists a ride height for the rear suspension, measured from the top of the axle tube to the frame rail next to the bump stop. They list ONE height (plus-or-minus 1/2") for the 4WD chassis, and ONE height (plus-or-minus 1/2") for the 2WD chassis. The 2WD height is 1" less than the 4WD height.

 

BTW -- I don't believe I even mentioned free arch. I'm not interested in free arch and I don't pay attention to it. I want to know what height my truck is going to ride at, and how much weight I can put in before I sit on the bump stops. Free arch is important (I guess) to spring shops, but not to me.

 

Hi, Eagle,

 

As I stated earlier, the first five springs are identical except for the free arch, i.e., they are made from the same thickness steel with the same length leaves. Yes, I believe Don was the first to mention "free arch".

 

My trucks aren't lifted. Spring under, so the thickness of the extra leaf won't have any effect at all on lift.

 

Actually, in rough terms the added capacity is known. Some years ago one of the 4WD magazines published formulae for calculating the spring rate of both coil and leaf springs. I lost the actual formulae, but for a leaf spring, assuming all leaves are the same thickness, the rate increases in direct proportion to the number of leaves. I don't think the XJ leaf will be exactly the same thickness as the MJ leaves, so that'll skew the number a bit, but in round numbers, since the thick overload leaf doesn't count until heavily loaded, I'll be going from a 3-leaf spring to a 4-leaf spring. That should increase the rate by approximately 33 percent (rate, as in pounds-per-inch of deflection). The travel before engaging the overload spring will be the same distance, so the capacity should also be increased approximately 33 percent.

 

Eureka! Found it!

 

Spring rate = (leaf width x no of leaves)/72 x (1820 x leaf thickness)/spring length

 

The entire thing is a first order equation. The main leaf will remain, so the only variable will be the number of leaves.

 

IIRC, the diagram in the magazine measured length straight eye-to-eye, not length of the leaf along the arch.

 

As you pointed out, that formula only holds true for springs made from the same thickness of steel throughout. A 4 leaf spring made with, say, a .323 main leaf, a .291 #2 & 3 leaf, and a .262 #4 leaf will have a totally different rate than a spring with all four leaves made from .291 steel. Also, this formula does not take into account the variations in the lengths of the leaves below the main leaf. For example, a spring made with leaves 4 inches progressively shorter will have a vastly different rate than a spring made with leaves 8 inches progressively shorter, i.e., 4 inches shorter on each end of each leaf. This formula is only helpful in comparing apples to apples, so to speak. We have not found any formula adequate in determining an accurate spring rate, since there are too many variables which are not accounted for. If anyone has a formula for calculating an accurate spring rate, taking into account any and all variables, I would love to see it, but I'm afraid it would be about as long and complicated as this thread.

 

-Tom

[Eagle]

As you pointed out, that formula only holds true for springs made from the same thickness of steel throughout. A 4 leaf spring made with, say, a .323 main leaf, a .291 #2 & 3 leaf, and a .262 #4 leaf will have a totally different rate than a spring with all four leaves made from .291 steel. Also, this formula does not take into account the variations in the lengths of the leaves below the main leaf. For example, a spring made with leaves 4 inches progressively shorter will have a vastly different rate than a spring made with leaves 8 inches progressively shorter, i.e., 4 inches shorter on each end of each leaf. This formula is only helpful in comparing apples to apples, so to speak. We have not found any formula adequate in determining an accurate spring rate, since there are too many variables which are not accounted for. If anyone has a formula for calculating an accurate spring rate, taking into account any and all variables, I would love to see it, but I'm afraid it would be about as long and complicated as this thread.

 

-Tom

Length of the leaves below the main leaf doesn't make any difference. A leaf spring is nothing but a simple beam, upside down, that's working within the elastic limit of the steel. If you look at the bending moment diagram for a simple beam, you'll see that it's a wedge, with zero moment at the ends and maximum moment at the center (or the point where the point load is applied, if off-center like on an MJ rear axle). Progressively shortening the leaves only results in not wasting steel where there isn't any need to have it there because it wouldn't be doing anything, anyway.

 

You can use the formula to calculate the rate of a spring with different thickness leaves, but not easily. For a three-leaf spring, for example, you would have to run a separate calculation for each leaf (using the same eye-to-eye measurement as the length), then add up the resultant numbers to get the total rate of the spring.

 

I tried to Google up a beam diagram for a simple beam with a point load, but everything I found is in .PDF format and has pages of diagrams, that would be more confusing than helpful to those who don't understand what all the different diagrams illustrate.

[HellCreek]

Length of the leaves below the main leaf doesn't make any difference. A leaf spring is nothing but a simple beam, upside down, that's working within the elastic limit of the steel. If you look at the bending moment diagram for a simple beam, you'll see that it's a wedge, with zero moment at the ends and maximum moment at the center (or the point where the point load is applied, if off-center like on an MJ rear axle). Progressively shortening the leaves only results in not wasting steel where there isn't any need to have it there because it wouldn't be doing anything, anyway.

 

You can use the formula to calculate the rate of a spring with different thickness leaves, but not easily. For a three-leaf spring, for example, you would have to run a separate calculation for each leaf (using the same eye-to-eye measurement as the length), then add up the resultant numbers to get the total rate of the spring.

 

I tried to Google up a beam diagram for a simple beam with a point load, but everything I found is in .PDF format and has pages of diagrams, that would be more confusing than helpful to those who don't understand what all the different diagrams illustrate.

 

I was not aware of that. So, following your reasoning, a 50 inch long, 4 leaf spring with all of the leaves made from .323 thick steel and having all of the leaves the same 50 inches long, would have exactly the same rate as a spring 50 inches long with 4 leaves made from .323 thick steel with the main leaf being 50 inches long, the #2 leaf being 42 inches long, the #3 leaf being 34 inches long and the #4 leaf being 26" long. Is this correct? I was thinking that a fully tapered beam would have a lot less capacity and rate than a beam that is full thickness from end to end, but I could be wrong. Also, what if you decreased the length of the #4 leaf to 12 inches and the #3 leaf to 24 inches; would that have any effect on the rate or capacity?

 

Also, if a full thickness beam has the same rate as a tapered beam, what is the point in tapering the ends of the spring leaves in an effort to make the spring ride smoother? Seems like a waste of labor to me.

[Eagle]

I was not aware of that. So, following your reasoning, a 50 inch long, 4 leaf spring with all of the leaves made from .323 thick steel and having all of the leaves the same 50 inches long, would have exactly the same rate as a spring 50 inches long with 4 leaves made from .323 thick steel with the main leaf being 50 inches long, the #2 leaf being 42 inches long, the #3 leaf being 34 inches long and the #4 leaf being 26" long. Is this correct?

Yep. That's what the diagrams tell us.

 

Of course, there's another variable we haven't even brought up yet. Everything we've discussed so far assumes that all the leaves are bent to concentric arches, so that load gets applied uniformly to all leaves as the load increases. One of the reasons leaf springs held on for as long as they did for rear suspensions is that it's FAR easier to make a progressive rate leaf spring than to make a progressive rate coil spring. All it takes is bending each successive leaf to a slightly larger radius, and the result is that each successive leaf doesn't start to accept load until the one above it has been partially loaded and deflected. The MJ bottom overload leaf is a crude example. In most trucks, most of the time, it just sits there and doesn't do anything. It only starts working when the upper three (or four) leaves have been loaded down to be flat enough that they contact the overload leaf.

 

That's what I'm counting on (in reverse) with my idea to use cut-off XJ main leaves in an MJ pack to build my own "home brewed" metric ton springs. The free arch of the XJ main leaf is, in rough terms, about the same as the arch of an MJ spring sitting in a truck at curb weight. Therefore (if I'm correct), plugging the XJ leaf in won't give me any lift because it won't be trying to push UP on the other leaves. But -- it will start to carry load when the bed gets loaded up, thereby (I hope) increasing the spring rate and carrying capacity.

 

I was thinking that a fully tapered beam would have a lot less capacity and rate than a beam that is full thickness from end to end, but I could be wrong.

Or I could be wrong. I think it might have happened once ...

 

Also, if a full thickness beam has the same rate as a tapered beam, what is the point in tapering the ends of the spring leaves in an effort to make the spring ride smoother? Seems like a waste of labor to me.

Don't have a clue. I've seen innumerable springs that have tapered ends on the leaves, and innumerable others that don't. Including springs for the same make and model vehicle that came from different subcontractors.

[HellCreek]

 

Spring rate = (leaf width x no of leaves)/72 x (1820 x leaf thickness)/spring length

 

Eagle,

 

I plugged in some figures from springs with known rates to test this formula. I left off the bottom 2nd stage leaf, since it doesn't come into play until the spring is flexed enough to touch that leaf. Also, where the springs are made from different sizes of steel, I used an average thickness. This is what I came up with:

 

3 leaf MJ springs made from .291 steel -- rate = .97 (obviously, you have to take the result of the equation and multiply it by 100). The actual rate of these springs is 108, so it was close on these springs.

 

4 leaf MJ springs made from .291 and .262 steel -- rate = 1.23 x 100 or 123. The actual rate is 130, so it is still close on this spring.

 

3 leaf MT springs made from .291 and .323 steel -- rate = 104. The actual rate is 130, so this one is way off.

 

6 leaf full-size J-300 truck front spring made from .262 and .237 steel -- rate = 178. The actual rate is 216, so this one is also off.

 

7 leaf full-size J-300 truck rear spring made from .360 and .323 steel -- rate = 279. The actual rate is 400.

 

The actual rates for these springs were furnished by the Spring Research Institute using a load tester, so they are accurate.

 

I am not an engineer, but there would appear to be variables in the springs which are not accounted for in this formula.

 

-Tom