Technical Engineering question, a serious one

The roll pin is there just to keep it on the shaft.
The reason for the DD on the end only DD is easy to fabricate .

Getting this pedal in to where it goes, i.e. thru the floor pan, thru the frame, next to the exhaust and Trans is the reason it needs to come apart.

 

DD here
Not questioning the need for ease of assembly, but if you have to get splined weldments, etc. why not use splines on both ends, as far as I know your neighbors at Flaming River still make splined shafts to length.

 

Torque "in" equals torque "out" on any shaft.
12" long pedal so that will be standard pounds per foot.
Force and distance or distance/distance involves the ratio.

 

The attachment of the DD fitting into the plate....that's what appears to me to be the weakest link. You didn't specify how thick the plate is. The internal spline welded to the pedal is the other iffy part...

Note the commercially available pedal assembly has two critical welds, they are on "known" materials.

The factor of safety needs to be pretty high on something like this.

 

I could, and if splines are bigger better stronger I will.
It will involve a mock up, then precise measurements then depending on FR to get the splines in the right spot and when they feel like doing it.
I can control the DD type end all myself in house

 

No, I didn't specify but it would be 3/8 plate, like the pedal arm.
Yes the safety factor needs to be impeccable.
My welding capabilities are impeccable and up to the task.
In my idea here, there really can be no "guessing" or "should be".

Just for clarity:
The flat plate brake pedals (most) have a 1"-1/4 X 0.125 wall tube welded into them, they ride on bronze bushings that pivot on a 3/4" pin. That pin serves its duty in shear orientation. Usually the 3/4 pin has its primary attachment on a weldment bracket and there is a secondary bracket that braces off the pin's end creating a double shear.
I don't see the weld on the pedal bracket as critical, the welds on the pivot pin would be more so.

I'd trust this weld to hold the tube where it's at, but not much more. All that wels has to do is hold the tube, it's not involved with stopping the vehicle at all. If it should fail the pedal will wiggle around long before the pedal lost its connection to the master cylinder.

 

With the coupling welded to the pedal, my welds would be critical, the coupling connections to the shaft critical, and the shaft critical. The pivot tube for the shaft would be much less critical

 

cut a new pedal that will slip over the coupler and weld near / far side, using a longer shaft.

Or mimic the 39 clutch pedal (pined pedal and linkage arm floating in a fixed base.

 

The 55-59 chevy truck clutch pedal assembly also does pretty much the same thing that you're making...but it's a clutch pedal, not a brake pedal.

 

You might give Steve at Industrial Chassis a call. He makes some good products including brake pedals. He doesn't hang out here much anymore, but check FB, instagram and his website.

 

Yes, that is pretty similar.
I've seen quite a few clutch pedals hung on shafts that have splines or some type of DD end/flat spot stud and nut. But that's a clutch not a brake pedal.

Little back ground-
Here's what the other guy did


Since the truck is finished , if I come up for the floor penetration in the same place, that is out side the frame all will be good. Moving the brake pedal to a better spot inside the frame is going to bother the steering column

Moving the steering colum and poking holes in a finished truck really isn't a good option.
Under dash is crammed full of shit and I'm not allowed to clutter up the firewall.

 

And putting a skid plate under the vacuum booster would be a good idea....I knew someone who got hurt after the brakes went out after driving over a raised median, and the vacuum can came apart.

No matter what you do, it will be better than what you started with. I think you have a pretty good idea what it needs to be, when you're done.

 

I sorta figure whatever I do should be as strong as these the smaller of these 2 areas
1.00 X 3/8" that's pretty strong,,,, but isn't the the smallest area around the small hole,,,
That's 1/4" edge to hole in 3/8 plate. That area sees a different load

 

Hey, If you want to over engineer it just use a sector shaft out of an old steering box. And machine down the splined pitman for a weldable collar.
[think of the torsional loads parallel parking a non power steer "yank tank"]

In my old Corvette racing car, 100lbs pressure at the pedal [unboosted] would haul it up at 1G. Spikes in pedal pressure above that would lock up the brakes with racing slicks.
If you add a booster the loads at the pedal are reduced

 

I'm no engineer, but it seems like if this contraption will work, whatever you build will be fine.

http://www.kugelkomponents.com/zencart/index.php?main_page=index&cPath=4_5

 

If you're worried about your weld where the splined coupler welds to the pedal (photo in post 37) you could bevel the edges and weld it, then slide a sleeve over the welded area and weld the sleeve to the pedal and the coupler - that makes for a pretty stout attachment. Do whatever you can to eliminate play in the splines.
I understand why you want to use the D-D on one end but like DRE said, machine that D-D to fit precisely and pin it with a full shoulder aircraft bolt in a reamed hole as a little insurance. While I can't do the analysis to back it up, when you consider that it's all attached to power brakes, the gut/intuition is that it will be safe and work well.
A helluva lot better solution than what the previous guys did!

 

Totally different set of forces in play here--no splines etc involved and no torsional load on any shafts

Roo

 

Good grief!

 

I know right.
The rest of it came here just as bad.

 

If you know someone with a Bridgeport mill and a dividing head, putting a spline on the other end isn't that hard to do and the way to go. If not, nevermind...

 

Do a search for splines torque capacity. I found this.

Anthonyr (Mechanical)28 May 03 21:18
In the "Standard Handbook for Mechanical Engineers" by Marks, there is an equation for torque capacity of a spline. The equation is T=L*D*D*S/1.2732, where L is the spline length, D is the pitch diameter of the spline, and S is the allowable shear stress of the spline material. I believe this is for a spline shaft and not a hollow shaft. I am not sure which is the case you are looking at. I hope this helps.


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That should help with the spline strength. You would also need to check shaft torque. There are calculators on the internet to calc for a round shaft.


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I found the below: "Engineering Tool Box"is usually good.

If you search you can probably find a calculator that you can simply put your numbers into it.


When a shaft is subjected to a torque or twisting a shearing stress is produced in the shaft. The shear stress varies from zero in the axis to a maximum at the outside surface of the shaft.

The shear stress in a solid circular shaft in a given position can be expressed as:

τ = T r / J (1)

where

τ = shear stress (Pa, psi)

T = twisting moment (Nm, in lb)

r = distance from center to stressed surface in the given position (m, in)

J = Polar Moment of Inertia of Area (m4, in4)

Note

the "Polar Moment of Inertia of an Area" is a measure of a shaft's ability to resist torsion. The "Polar Moment of Inertia" is defined with respect to an axis perpendicular to the area considered. It is analogous to the "Area Moment of Inertia" - which characterizes a beam's ability to resist bending - required to predict deflection and stress in a beam.
"Polar Moment of Inertia of an Area" is also called "Polar Moment of Inertia", "Second Moment of Area", "Area Moment of Inertia", "Polar Moment of Area" or "Second Area Moment".

Polar Moment of Inertia vs. Area Moment of Inertia
"Polar Moment of Inertia" - a measure of a beam's ability to resist torsion - which is required to calculate the twist of a beam subjected to torque
"Area Moment of Inertia" - a property of shape that is used to predict deflection, bending and stress in beams
Circular Shaft and Maximum Moment or Torque
Maximum moment in a circular shaft can be expressed as:

Tmax = τmax J / R (2)

where

Tmax = maximum twisting moment (Nm, in lb)

τmax = maximum shear stress (Pa, psi)

R = radius of shaft (m, in)

Combining (2) and (3) for a solid shaft

Tmax = (π / 16) τmax D3 (2b)

Combining (2) and (3b) for a hollow shaft

Tmax = (π / 16) τmax (D4 - d4) / D (2c)

Circular Shaft and Polar Moment of Inertia
Polar Moment of Inertia of a circular solid shaft can be expressed as

J = π R4 / 2

= π (D / 2)4 / 2

= π D4 / 32 (3)

where

D = shaft outside diameter (m, in)

Polar Moment of Inertia of a circular hollow shaft can be expressed as

J = π (D4 - d4) / 32 (3b)

where

d = shaft inside diameter (m, in)

Diameter of a Solid Shaft
Diameter of a solid shaft can calculated by the formula

D = 1.72 (Tmax / τmax)1/3 (4)

Torsional Deflection of Shaft
Torsion deflection of shaft

The angular deflection of a torsion shaft can be expressed as

α = L T / (J G) (5)

where

α = angular shaft deflection (radians)

L = length of shaft (m, in)

G = Shear Modulus of Rigidity - or Modulus of Rigidity (Pa, psi)

The angular deflection of a torsion solid shaft can be expressed as

α = 32 L T / (G π D4) (5a)

The angular deflection of a torsion hollow shaft can be expressed as

α = 32 L T / (G π (D4- d4)) (5b)

The angle in degrees can be achieved by multiplying the angle θ in radians with 180 / π.

Solid shaft (π replaced)

αdegrees ≈ 584 L T / (G D4) (6a)

Hollow shaft (π replaced)

αdegrees ≈ 584 L T / (G (D4- d4) (6b)

Torsion Resisting Moments from Shafts of Various Cross Sections
Shaft Cross Section Area Maximum Torsional
Resisting Moment
- Tmax -
(Nm, in lb) Nomenclature
Torsion - solid cylinder shaft Solid Cylinder Shaft
(π / 16) τmax (2 r)3

= (π / 16) τmax D3

Torsion - Hollow Cylinder Shaft Hollow Cylinder Shaft
(π / 16) τmax ((2 R)4 - (2 r)4) / (2 R)

= (π / 16) τmax (D4 - d4) / D

Torsion - Ellipse Shaft Ellipse Shaft (π / 16) τmax b2 h h = "height" of shaft
b = "width" of shaft
h > b
Torsion - Rectangle Shaft Rectangle Shaft (2 / 9) τmax b2 h h > b
Torsion - Square Shaft Square Shaft (2 / 9) τmax H3
Torsion - Triangle Shaft Triangle Shaft (1 / 20) τmax b3 b = length of triangle side
Torsion - Hexagon Shaft Hexagon Shaft (b3 / 1.09) τmax
b = length of hexagon side
Example - Shear Stress and Angular Deflection in a Solid Cylinder
A moment of 1000 Nm is acting on a solid cylinder shaft with diameter 50 mm (0.05 m) and length 1 m. The shaft is made in steel with modulus of rigidity 79 GPa (79 109 Pa).

Maximum shear stress can be calculated as

τmax = T r / J

= T (D / 2) / (π D4 / 32)

= (1000 Nm) ((0.05 m) / 2) / (π (0.05 m)4 / 32)

= 40764331 Pa

= 40.8 MPa

The angular deflection of the shaft can be calculated as

θ = L T / (J G)

= L T / ((π D4 / 32) G)

= (1 m) (1000 Nm) / ((π (0.05 m)4 / 32) (79 109 Pa))

= 0.021 (radians)

= 1.2 o

Example - Shear Stress and Angular Deflection in a Hollow Cylinder
A moment of 1000 Nm is acting on a hollow cylinder shaft with outer diameter 50 mm (0.05 m), inner diameter 30 mm (0.03 m) and length 1 m. The shaft is made in steel with modulus of rigidity 79 GPa (79 109 Pa).

Maximum shear stress can be calculated as

τmax = T r / J

= T (D / 2) / (π (D4 - d4) / 32)

= (1000 Nm) ((0.05 m) / 2) / (π ((0.05 m)4 - (0.03 m)4) / 32)

= 46.8 MPa

The angular deflection of the shaft can be calculated as

θ = L T / (J G)

= L T / ((π D4 / 32) G)

= (1 m) (1000 Nm) / ((π ((0.05 m)4 - (0.03 m)4) / 32) (79 109 Pa))

= 0.023 radian)

= 1.4 o

Example - Required Shaft Diameter to Transmit Power
Electric motor

A 15 kW electric motor shall be used to transmit power through a connected solid shaft. The motor and the shaft rotates with 2000 rpm. The maximum allowable shear stress - τmax - in the shaft is 100 MPa.

The connection between power and torque can be expressed

P = 0.105 nrpm T (7)

where

P = power (W)

nrpm = speed of shaft (rpm)

Re-arranged and with values - the torque can be calculated

T = (15 103 W) / (0.105 (2000 rpm))

= 71 Nm

Minimum diameter of the shaft can be calculated with eq. 4

D = 1.72 ((71 Nm) / (100 106 Pa))1/3

= 0.0153 m

= 15.3 mm

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Lastly, you should check deflection (twist). This will create a spongy feeling if it deflects too much

Also note that deflection is purely a function of geometry (diameter, length). And the material modulus of elasticity. Going to a higher strength steel does not improve deflection. Higher strength allows for more twist before permanent bending. But does not change the twist per unit of torque.


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I found this. Engineers edge is also a good site.


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You might find something useful by looking at sprint car torsion bars and arms. They come in a variety of diameters, but you'd likely have to have one cut down to the length you need and have the splines cut on one end. A short section of torsion bar would likely handle your worst-case torque scenario.

 

I presume the double D shaft is only a double D on the end of a round shaft that passes through your frame mounting bracket? I would have some concerns with a double D shaft against the frame bracket bushing, damaging the bushings as it turned under pressure.

I believe your frame mounting bracket will be your weak point in your design. You will need to be sure it can not deflect where it is attached to the frame.

I've seen a lot of brake pedal pivot brackets that were no where near as strong as the steering components you are using at the pivot point. I would probably want to keep the arms as short as possible to reduce the stress levels. Gene

 

100 lb of pedal force on a 12" lever arm would give you 1200 in-lb of torque on your 3/4" shaft.

The shear stress due to torque would be 4640 psi. This stress will vary directly with pedal force. IOW, 200 lb of pedal force would give you a shear stress of 9280 psi. Note that its necessary to use in-lb rather than ft-lb for your answers to work out in psi rather than pounds per square foot.

That 3/4-36 spline is likely a serration type spline. IOW, its cross section is a series of V's as opposed to a more coarse spline like you'd find on a driveshaft which will have an involute cross section. This spline is undoubtedly rolled rather than cut. At the stress levels involved, you could safely assume the spline has no significant effect on strength as compared to the 3/4" smooth section.

That D-D section is a whole nuther thang. To get a true value for its polar moment of inertia would require some calculus rather than simple math. However, you'd be safe to approximate it as a rectangular shaft as that would give you a smaller area and therefore a somewhat higher value for shear stress than what the shaft would actually see.

The dimensions of the rectangle would be the width across the flat and the height is what's remaining after the flats are cut. Go to http://www.kmp.tul.cz/system/files/hearn_krouc_nekruh_tenkost_profilu.pdf for the formulas to calculate the shear stress on a rectangular shaft.

For the formula, the longer side of your rectangle is d and the shorter side is b. Scroll down to the next page and you'll find table 5.1 which gives values for k1 and k2 based on the ratio of d to b. At that point you can plug the values for d,b,k1, and k2 into the formula to get the max shear stress. Remember, torque must be expressed in in-lb and both d and b in inches to get the correct answer.

Once you have this stress value, we're in a much better position to evaluate the adequacy of the shaft as designed since the D-D section will definitely have significantly higher stress than the rest of the shaft.

As I understand it, the pedal will attach at the outboard side of the frame. The shaft will pass thru the frame and will be cantilevered from that point out to where it aligns with the master cylinder. If that's correct, you may also need to look at the resultant bending force on the shaft, and in particular, the load that's put on the oilite bushings you mentioned. Because oilite is an open sintered structure, it doesn't have a lot of compressive strength. You may find you need to go to 660 bronze for the bushings to get the necessary compressive strength. Calculating the loads on the bushings would require the cantilevered length and the length of the bushings and the spacing between them.

 

First, your design is much like my '57 Chevy STOCK clutch pedal. A D-D pressed into the pedal arm and a splined end with a clamp holding the through firewall rod. I highlighted stock because the shaft diameter is 3/4 as is yours. I think the shaft is up to the task.

My suggestion is to use two set screws on the splined end at 90* rotation of each other. You should assemble the splined end before installing into the car and drill a small hole in the splined shaft at the location of the set screws. It only needs to be deep enough to go to the bottom of the spline. This will allow you to tighten the assembly to prevent back off and allow future disassembly. Some type of thread lock should also be used.

For the opposite end I would not use a roll pin. Instead, you could tap a hole in the shaft end and use a washer/bolt to hold the "swing arm" in place. You could weld a D-D washer on the shaft as a stop for the back of the "swing arm".

 

Thanks for the detailed explanation, metlmunchr.