A Grade 5 bolt,3/8-16 in Single Shear, would take 5500 lbs to shear at the threads, according to this page: https://site.alliedbolt.com/files/ShearStrength2.pdf So, with the amount of force required to break that bolt in Double Shear, I'm pretty sure that something else would bend or break first. Same goes for the plates.
Pull all 4 springs and take them to a spring shop and have them re-heat treated to a softer spring rate. I had a "T" roadster with this same kind of setup and I had to run about 20 psi in the tires. try lower tire pressure first,it's free and might do the job. Also replace 4 nuts on each side that support your "Friction Shocks" yes you have shocks they are adjustable by tightening the center bolt in their big end
Yes, pivots should be bushed, that is fundamental mechanical design. And the bushing should not load onto the threaded portion of the bolt, because the threads will ruin the bushing.
I used to work for a spring manufacturer, hand making prototypes and small batches, including aircraft and the orbiting Space Station. Yes, really. Springs are Tempered to a certain temperature for a reason. Having them "re-heat treated to a softer spring rate." risks shortening their cycle life. If the OP wants softer springs, the leaves need to be some combination of thinner, narrower, and longer. Did someone ever get away with heating their springs to make them softer ? I am sure they did. Doesn't make it a good idea.
Hydraulics can be engineered to work so much better. High and low speed dampening for example. Can't do that with a friction shock.
Finally! Some one with understanding of this suspension setup, that has the degrees to back it up. With any bolts in double shear, the breaking point doubles over the same bolt in a single shear. This chassis does not look like it was thrown together by someone that didn't know what they were doing. There may be some less then perfect things done or that may have been changed from the original design (the brake hose is classic), but over all, without disassembly for inspection, it looks pretty sound. If you hit a big enough pot hole to provide enough force on the bolts to break them, almost any front suspension under our cars would need to be seriously inspected. That would be a pot hole that would wake your butt up for sure!
I’m always curious when threads on 1/4 elipitcal springs/suspension come up because that is what I’ll be using on my speedster build which is a copy of Richard Scaldwell’s GN JAP which uses this setup. He and his daughter I believe have been racing his car for many years and seem to do just fine. I agree that because it is not widely used here in NA, there are some misconceptions as to how they function and handle.
Not exactly. Bolt strength is not per bolt. It is measured in pounds-per-square-inch. A 3/8" bolt is ~0.442 square-inches. Single shear yield on a Grade 5 3/8" bolt has a at-thread single-shear strength of 2466.36 pounds of force.
When calculating area of a Circle, you have to Square the Radius, not the Diameter. You squared the Diameter. #Highschoolmath 3/8" dia = .375" Radius = .1875" r² = .0351 π ≈ 3.14159265 ≈ 3.14 Circle area = π × r² = π × 0.0352 inch² ≈ 0.11045 in² https://www.omnicalculator.com/math/area-of-a-circle 3/8-16 Minor dia ~.314" Radius = .157" r² =.0246π ≈ 3.14159265 ≈ 3.14 π ≈ 3.14159265 ≈ 3.14 Circle area = π × r² = π × 0.0247 inch² ≈ 0.07744 in² https://www.fastenal.com/fast/services-and-solutions/engineering-calculators/load-calculator# Last place I worked at, I had to build a test fixture to prove one of my designs. Engineering couldn't do the math, "because we don't have the software". My math said a 5/16" bolt was strong enough to pass the test; it needed to support a 5000 lb load in bending without complete failure. They guessed that a 5/8" bolt wasn't strong enough. We stopped the test around 9000 lbs. The bolt only had a slight bend.
Grade 5 bolts: Proof Load: 85,000 PSI Yield Strength: 92,000 PSI Tensile strength: 120,000 PSI https://boltdepot.com/Fastener-Information/Materials-and-Grades/Bolt-Grade-Chart Yield strength of the Minor Dia, for a Grade 5 bolt 3/8-16 = 92,000PSI x 0.07744in² = 7124.5 lbs. No idea how you got 2466.36lbs.
This one ? https://site.alliedbolt.com/files/ShearStrength2.pdf Top of that page, it says: "The industry standard for determining shear strengths of fasteners is to take 60% of the minimum Ultimate Tensile Strength of the fastener for single shear joints and 120% for double shear joints." While 5580 x .442 does = 2466, the logic is wrong. 5580 lbs in the chart is the answer, for Single Shear, based on the Minor Dia of the bolt and the strength of the material. The material strength is a constant, shown at the top. 120,000 PSI. The chart shows the strength, based on area, then multiplied by 60%. 120,000 x .0775 = 9300 x 60% = 5580 lbs. As shown above. This application is in Double Shear, so the chart numbers need to be doubled. 120,000 x .0775 = 9,300 x 120% = 11,160 lbs. You multiplied the answer given in the chart, by another area, to get a smaller answer. Strength x one Area, x 60%, then x another Area, does not give the correct answer.
I definitely won’t try to add much, but I don’t prefer to rely on nylocs to stay in place when not fully torqued. I realize the OP mentioned using a jam nut, but maybe a better alternative would be drill the bolt for a cotter pin and using a castle nut? Wouldn’t be any less visually appealing than a jam nut.
And if it is assembled correctly, the shear point should be at the shank of the bolt rather then at the threads. All the bolt threads should be outside of the brackets. Using the bolt shank shear numbers raises the shear point even higher. Drilling the hole in a 3/8" diameter bolt for a cotter key will weaken the end of the bolt allowing the drilled end break off more easily. If the end retaining the cotter key is broken off, the nut can easily loosen. The NY lock nuts are designed for this purpose and are suppose to be equal in holding force as having a nut with a jam nut locked against it. As long as there isn't enough heat present to break down the nylon in the NY lock nut, it is efficient enough for this application. The NY lock nut should be replaced every time it is removed though.
I guess I see your point. But I wasn’t thinking of that portion being filled breaking off, there isn’t any force on it. I was figuring the castle nut and a cotter pin pretty much ensures a non-torqued nut to stay put. All good!
That is a 1923 or earlier Dodge Brothers chassis with aftermarket cross spring setup replacing the two quarter eliptical upper rear springs. Lower half eliptical springs are original. I've had two of those setups and cars. Similar kinda thing to Hassler shock springs. Dave
Just wanted to clear up the error in this statement. I'm the old HAMB Metallurgist. Heat treating or not, *does not* change the spring rate. Spring rate is a function of elastic modulus, which does not change based on strength. What *does* change with heat treatment or not is the yield strength, which determines how much the spring can deflect and return back to original shape. This is why springs are heat treated, to enable higher deflection without distorting. Permant distortion is officially called plastic deformation. This is also why heating springs to lower a car is such a bad idea. Now back to the original programming.
I’d like to see more pictures of the chassis overall. The PO had some interesting ideas on hot rod suspension. You’ve made the change I would have suggested. If it seems to work, run it and keep an eye on it. I did quarter elliptic springs at all four corners of my ‘27 T using four cheap swap meet Model A front springs cut just past the center. six leaves in front and four in the rear. It rode just fine. In the front, I used the spring as the lower locator instead of a four bar knowing that it would change the caster slightly as it cycled. It doesn’t move enough to matter. I found a way to get a better angle on the front shocks and the tie rod is a mock up - these are early pictures while figuring out the radiator mounting. Model A rails with modified front and rear.
Spring rate is a function of the spring design. Removing a leaf or two will make a lower rate. Or longer spring, thinner leaf thickness, or narrower spring. Rate is determined by the design (and material, steel vs alum for example), not strength. Strength determines how much deflection before permanent set takes place.
The Jalopy Journal
