Technical A series of engineering questions and examples.

Im lolking for some guidance and wondering if I could get some help figuring a few things out and how they would relate to each other when combined.

I'd glady accept "a fish" here and would love a few fish but I need to learn a bit more about fishing.

Such as which concepts , their technical terms and calculation formulas that I'm trying to apply, then I could serch it with the proper terms and study it.



At first they are simple, but as the second part builds off the first and third off of the sum of those two, I am unsure.

So here goes - next post

 

Here is an 8" x 6" structure,
Assuming :
* the rotational pivot is fixed in shear
* the force generated is pushing against a fixed object
* the weight is the structure itself is Mute

Ill use 100 lbs of weight for example and I can easily extrapolate off of that.

A 100 lbs of mass hanging off of the 8" leg should generate 800 inch lbs of torque force at the pivot point. If that's correct, what's that "concept" or law called?

B Next that 100 lbs of mass generating that force should pass it thru the 6" leg and increase the force by 25%. If this is correct, what is the proper term for this law or concept

C there should be some load/force acting on the pivot in and angular direction- how is that calculated ?
For example-
Is it a combination of the 100 lbs and the force generated - being 225% of the weight ?
Is it a singular result of the input? - 100 lbs
Is it a cancelation or combination of some sort because its at an angle

What's this concept or law called?

 

I believe you'll increase the force by 33.3%...not 25%. DD

 

This is basically Newton's law you're describing here.

Here's round-about explanation of what you're getting at here...

 

This is correct. He'll end up with 133.33333333333333333333333333333lbs of force at the end of the 6' length.

 

I believe what you are looking for is statics.

 

Building off of the first example -
Keep in mind this is a side view and for now assume its as simple as its drawn.
Also lets assume the structure alone before any weight is added is in balance.

Here is the same 8" x 6" structure with the same 100 lbs of weight
What's added is a 48 long lever and an attachment point that's 3" in diameter

2A Ultimately in this example I'd like to know the amount of upwards force generated at the end of the 48" lever. I assume knowing the weight needed at the 48" end to achieve balance of 100 lbs on the 8" end could be extrapolated into the force being sought. What is this law or concept called ?


2B does the 1-1/2 radius (3" diameter) of the attachment points have about the true center have any bearing on calculations ? And again what would this law or concept be called?

2C I would imagine calculating the load under the pivot of a balanced structure would be a simple matter of adding the 3 weights, weight of the structure, weight on 8" end and weight on 48" end ?

 

Ok, I wasn't exactly sure how to apply the math.

8 is 33.3% more than 6
And
6 is 25% less than 8

 

Weird, huh? True scoop! DD

 

Yeah it's weird , but I got a pretty good lesson from my accountant on the finer workings of calculating money and percent. It's real easy to make 2+2 equal 3 or 5 when you start playing with it lol


Any way, as it pertains to this series of questions, what's the rule for applying that oddity to forces? I assure yo it will become more important to me and this thread as it progresses. It already has a big impact to the second part.

 

C) Resultant of forces in the horizontal and vertical directions - this is trigonometry; 133.33lb horiz. and 100lb vertical applied via Pythagorus: sqrt(133.33^2 + 100^2) = 166.66lb. The 100lb force is downwards and the 133.33lb force (as pushing against the lever) is to the left, so the angle would be the resultant of these two forces; arctan(100/133.33) = 36.87deg down and to the left, or, with 0deg @ three o'clock, 216.87deg. "Arctan" means "the angle whose tangent is", and is also written "tan^-1".

2B) No, the axis is the same.

2C) Torque is a moment, and moment is force times a lever arm - 100*8=800 - so to balance out, the opposed torque needs to be the same @800inch/lb; devide by the other lever arm length - 800/48=16.667lb. That seems light, but 8 goes into 48 6 times and 6*16.667 is 100. The vertical force on the pivot pin is the summation of all vertical forces, so that's 116.667lb.

 

As someone else mentioned, both examples are statics problems.

The working concept is that the sum of all the forces and moments (torques) will equal zero.

Looking at your 2nd diagram.......

Looking at the torques first, you've got a counterclockwise (ccw) torque of 800 lb-in and that has to be balanced by an equal clockwise torque. Since the cw torque's moment arm is 48", the force necessary to produce the opposing cw torque of 800 lb-in is found by 800 lb-in / 48 in = 16.67 lb

By setting up the division problem so it included the units as well as the values, you can check your work. The inches on top are cancelled by inches on the bottom, leaving your answer as lbs. This doesn't necessarily guarantee the answer is correct mathematically, but it does show the problem was set up such that it yields the correct units on the answer. If the units come out wrong, you can pretty well bet the answer is wrong and that will send you back to see where you've got the units messed up.

For example, if the 48" had been stated as 4 ft and you'd used 4 ft in the bottom of the equation, you'd have gotten 200 as your answer. But the feet and inches wouldn't cancel, so the units of your answer would be lb-in/ft. This immediately alerts you to something being wrong since you know the answer should be lbs.

Sorry for the long explanation, but I saw this time after time in engineering school where people wouldn't use units trying to get stuff done faster, and end up with wrong answers that would've shown up immediately if they'd just included the units and made sure they cancelled such that the answer at least had the proper units.

Now to the forces.... they have to equal zero as well. Down force of 100# on the left plus down force of 16.67# on the right have to be balanced by an equal upward force for the sum to be zero. So, the upward force to balance the 116.67# downforce is also 116.67# applied upward thru the center of your 3" circle.

 

So what I'm gathering here from both of you guys is the 33.3% increase in example 1 does not transfer in any way to example 2. If this is true - where did it go by the conservation laws? Or am applying what i see incorrectly ?

 

Excellent answer by Ed in showing both the force and the resultant angle of that force in the first problem.

In practical real world application, if your pivot in the first problem was out on the end of a shaft, and you wanted to gusset the shaft to stiffen it, finding the angle as Ed showed would allow you to place a gusset at that angle where it would do the most good rather than just using the standard horizontal or vertical placement.

 

The forces are only a component; the moments are what is important and they are the same. In the first example, rotate the vertical lever CC to the horizontal, and now you have 133.33lb pushing downwards. Lengthen the lever arm from 6" to 48" and the weight required to create an equal opposing moment decreases to 16.67lb. So the moments have been conserved...

 

Trigonometry for Dummies is one of the cooler books I own .

PS, I made a mistake in post #11, the 0 should be @ three o'clock...

 

Yep. The 2nd problem is an example of where you have to reduce the drawing to its simplest terms. The 6" at the left could be 6 ft and nothing would change. Likewise, the triangular structure on the right could be a beam of the same length and nothing would change.

If the triangular structure was made of 2 small diameter rods and you wanted to know the forces in each rod, then you'd have to work that out in a manner similar to what Ed did on the first problem. At that point, the 6" vertical distance does come into play since it determines the angle of the lower leg of the triangle, and that influences the amount of force that has to be developed in that leg to counteract the vertical force of the 16.67#. To do this, you would assume the triangle is pinned at all 3 corners rather than fixed such as by welding. If it was welded at all 3 points both to itself and to the part its attached to, then it becomes statically indeterminate which means, in simple terms, the math gets a bunch more complicated and far past what can practically shown in a forum like this one.

 

Ok, I at least see why I'm having confusion then, that's good and a step closer to understand that confusion.

As you may be able to tell these are ladder bars for a rear end, the connections are bolted. "Statics," the study of loads and forces that generally do not move get dynamic real quick when the hammer gets dropped or traveling over roads an sane speeds.

 

Yep, that's it....good course, second year in engineering school, it's the foundation from which all knowledge flows.

 

You might want to start by understanding "free body diagrams"

They are very helpful for visualizing what's going on with forces, torques, moments, etc

 

31Vicky, If I can speak out of turn...in simplest terms: "Internal stress"

Go back for a moment to the roof truss again or better stated - you are working your questions THRU a structure - by doing so you will calculate several seemingly meaningless forces - by the conservation of energy they are still there - internally. Back to our truss - wall load is the same (think of that as an output force) the "snow load" our input force. Adding or removing internal supports will effect the loading (or internal stress) of the support system. Add a shit ton of (weightless) beams and your roof will get stiffer, but the wall load is the same. This can be a very tricky concept for some applications. Here is another way to look at it: Torque a nut to 100ft/#. Now use a 12" extension and torque it again to 100ft/#. The extension twists while you torque it....does the torque the nut "sees" remain the same? Now use a 12' long extension - again torque to 100ft/# - nut still see the same torque??? YES, it does. Some will argue it does not because it took energy to twist the extension, but they fail to remember the conservation of energy - equal and opposite reaction in this case. Yes we twisted the extension, but it was resisting this twist with the nut - the INPUT force the 100ft/# and the OUTPUT force the force the nut "saw" is the same regardless of the stress that the extension saw. Make sense?

 

It the design of the ladder bars remains a statics problem for the most part. You'd use dynamics to determine the peak forces on the bars based on available torque, traction, weight of the vehicle, and acceleration. And then the peak forces you determine thru that would be applied to the statics problem to determine the reactions and size the materials.

Ladder bars often have bracing between the upper and lower tubes in a triangular pattern for a couple reasons.

First, the triangle shape is strong as compared to bracing that creates a parallelogram. And the resulting structure becomes a truss which is far stronger and stiffer than its individual components, while remaining lightweight as compared to most other structures of equivalent strength.

And second, you could size the tubes to withstand the tensile and compressive forces, but it would still be likely that the lower tube by itself would fail by buckling. Under acceleration the lower tube is loaded in compression as it resists the attempt of the axle housing to rotate in a pinion upward direction. And its further loaded in compression by the tractive force that's transmitted thru the axle housing and onto the tubes to push the car forward. Even though the lower tube may appear to have sufficient strength to handle these loads, the small ratio of the tube's diameter to its length (known as slenderness ratio) would still make buckling likely. By trussing the bars you convert the long tubes into a bunch of short tubes whose ratio of diameter to length becomes much larger and therefore much more resistant to buckling.

When you look at the various forces, its fairly easy to understand why I say the lower tube would be more likely to buckle. The lower tube is loaded in compression by two separate forces as explained above. But the upper tube is loaded in compression by the tractive force that pushes the car forward, but that is somewhat offset by the fact that the upper tube's force to resist axle housing rotation is either non-existant if the tube is in line with the housing, or a tension force if the upper tube is above the axle centerline.

If you intend to get further into this, these sorts of calculations would be covered in most textbooks under column loading, buckling, and slenderness ratios.

 

Question, how does a Snap-On torque stick adapter work on my impact gun?

Here is the kit

http://store.snapon.com/Torque-Stic...rd-Duty-Master-1-2-drive-10-pcs--P646363.aspx

 

Never seen them before - or at least never stopped to think what they are doing. But since they are used with an impact - I would think they LIMIT torque by deflecting - sounds contradictory to what I said earlier, but hear me out. Previously we are talking about applying a constant input force with a torque wrench - resulting in a static equation. These are entering in the dynamic realm. The stick twists and never achieves equilibrium thus reducing the torque they transmit - now take these and use them with a HAND torque wrench and we are back to 100ft/# in - 100ft/# out.

EDIT: Think of those sticks as torsion springs - the spring will bend (deflect) allowing only so much rotational force to be transmitted before they start to twist - the impact wrench isn't a constant force like you applying a torque wrench but a series of small hits. That series of small hits adds up to the same OUTPUT (rotation?) as the fixed torque wrench. I would suspect they are spring steel (guessing of course). Would love to hear from the experts to see how close I got.

 

I do intend to and have ordered a few books today.

What I'm trying to explore (and hopefully without hurting myself too much) is the 16% forward transfer of load of that suspension design. I do realize that this is just one component of many involved, but what intrigues me the most is the effect that has on sprung weight , and at the very least making sure I do not make the ratio of sprung to unsprung worse, Hopefully much better.

Very interesting that the 33.3% gain from example 1 doesn't cary to into example 2. Not knowing how to factor in the 33.3% gain was keeping the train of thought parked.

 

Two young engineering students were making their way across the A&M campus one day. These two young men were life long friends from childhood. As they were approaching each other from different directions, one noticed that the other was on a very nice new bicycle. He rushed over to him and said "MAN, that's a nice bike, where'd you get that?". The fella on the bike says "Man....you'll never believe the story on how I got this bike. Just yesterday I was walking across this same courtyard and the most beautiful woman I have ever seen in my life approached me on this bike. She got off the bike, tossed it to the ground and pressed her body against mine and said 'Take anything you want....', so I took the bike." The friend was awestruck at the story and paused for a few seconds to take it all in and he said "Ya know, that was probably a good choice, I would've taken the bike also because I highly doubt her clothes woulda fit either one of us."

 

(Engineers often miss the obvious)

 

I'm surprised that an engineering student would say something like that; those clothes could certainly be altered to fit.

 

Yes indeed, however then he would have to red-line the drawings. And we all know how much engineers hate to do that shit, especially the young ones.

 

Yeah I was thinking the same thing. All the excessive energy is returned to the impact gun. Those sticks are pretty accurate (I've checked them). I've never tried one without an impact gun tho. I think if it was loaded with a torque wrench, it might be a little scary if it slipped off