determined metal thickness needed

admin

Hello everyone. I've been welding for a few years now and always had the thickness of materials always told to me or i just really made it way over kill. How do you go about doing the calculations on determining the actual thickness needed. Like if i wanted to build trailing arms for my truck for example. I know i can cut them out on a plasma table out of some 1/4 but i feel like thatis way overkill. Thanks for the input.
Reply:Experience,study engineering or copy a proven design or all of the above. If you copy a proven design be sure to discover and copy the same alloys and heat treatment. Sometimes the shorter, quicker  and cheaper solution is what you've already described....overbuild until you have enough of the first one, experience. Observe lots of failures and try to understand causality. Some failures are not under built, just overly abused. Mostly have fun and learn from you failures and sucesses.---Meltedmetal
Reply:Originally Posted by MeltedmetalExperience,study engineering or copy a proven design or all of the above. If you copy a proven design be sure to discover and copy the same alloys and heat treatment. Sometimes the shorter, quicker  and cheaper solution is what you've already described....overbuild until you have enough of the first one, experience. Observe lots of failures and try to understand causality. Some failures are not under built, just overly abused. Mostly have fun and learn from you failures and sucesses.
Reply:Originally Posted by MeltedmetalObserve lots of failures and try to understand causality. Some failures are not under built, just overly abused.
Reply:Originally Posted by welndnfabHello everyone. I've been welding for a few years now and always had the thickness of materials always told to me or i just really made it way over kill. How do you go about doing the calculations on determining the actual thickness needed. Like if i wanted to build trailing arms for my truck for example. I know i can cut them out on a plasma table out of some 1/4 but i feel like thatis way overkill. Thanks for the input.
Reply:Educated guess (and experience) a lot of the time. Sometimes you want thicker material for rigidity more than for strength. Looking at commercially made items similar to what you're building and/or wanting yours the same strength or stronger.
Reply:There two basic waysFind trailer you like and is right size and just copy the metal size. The other way is to calculate the metal need for trailer Dave Originally Posted by welndnfabHello everyone. I've been welding for a few years now and always had the thickness of materials always told to me or i just really made it way over kill. How do you go about doing the calculations on determining the actual thickness needed. Like if i wanted to build trailing arms for my truck for example. I know i can cut them out on a plasma table out of some 1/4 but i feel like thatis way overkill. Thanks for the input.
Reply:Originally Posted by OscarWhile there are many methods to "determine" the metal thickness required, as meltedmetal mentioned, if you want to be able to apply and work out the proper equations to work out numerical calculations for stress/deflection, then that is exactly what needs to be done via engineering.  I hope you have a strong background in physics/analytical mechanics (IOW you need to know calculus, as a lot of stress/deflection models require integrating).Area Moment of Inertia or Moment of Inertia for an Area - also known as Second Moment of Area - I, is a property of shape that is used to predict deflection, bending and stress in beams. - Engineeringtoolbox.comObviously there is FEA analysis that can do all the stress/deflection analysis you need, and it surely has just a steep a learning curve.
Reply:"Design of Welded Structures" by Omer W. Blodgett is a book that is (or was) very commonly used by structural engineers.  It's a heavy book for its size and for its math.https://www.amazon.com/Design-Welded.../dp/9998474922https://www.jflf.org/SearchResults.asp?Cat=81
Reply:one lower iq method people left out is load to failure (destructive) testing, and it's less truthful cousin load testing.The final test for cranes and load bearing equipment is usually a "proof" test.  Overload to the intended safety factor and pray to the sweet metal gods up above, or down below (can't figure out which side metal work is on yet!).BE READY FOR PROOF TEST TO FAIL, be as overly safe as possible with live things first and equipment second.   Of course it's not always possible to do with every project etc [inset not so common sense disclaimers here].
Reply:If the math is done right and use the right table it will not fail.I did metal work engineering for 30 years and everything is still standing today. Even in tornado world.Dave  Originally Posted by William McCormickMost of the failures today come from modern "math". They rarely use math to make something better they use it to cut into the ten-fold safety margin that has existed in buildings and cranes for over a hundred years. As far as software analysis it is a gamble as well. Experience is the only solution. Looking at modern builds is a bit dangerous, they are often built only to perform in a perfect world not in the actual world under the actual conditions it will face. Once you see a few devices or structures fail from the slightest deviation of "normal stress" you realize that the people who designed it are either on drugs or live in an isolated bubble away from the rest of the world or both. Experience is the best. Sincerely, William McCormick
Reply:Originally Posted by smithdoorIf the math is done right and use the right table it will not fail.I did metal work engineering for 30 years and everything is still standing today. Even in tornado world.Dave
Reply:Originally Posted by William McCormickThere is no way that math could be working when it is being done on structures that are engineeringly unsound or the math is done on methods that should not be allowed.
Reply:Originally Posted by IngenuityI think I will take my engineering theory from the likes of Galileo, Bernoulli and Euler, and not McCormick.Being a somewhat 'intelligent individual', I will state that is depends on the span (length) of the cantilever to the main span - and also on the floor framing type (eg steel beams, 2-way slabs etc). There is a 'sweet spot' where a cantilever will provide such a distribution, but commonly it does NOT.
Reply:Originally Posted by William McCormickAs soon as you create a cantilever, you double the weight on the wall supporting it. That is just the plain and simple truth of a lever.
Reply:Originally Posted by IngenuityWilliam:That is incorrect. You only 'double the weight' if the cantilever span equals the backspan. For any other ratio of backspan/cantilever the distribution of support reaction is as follows: This is first-year engineering statics - for some folks it was high school math/physics.Evidently, it appears to be misunderstood by you. The scale is NOT calibrated to read half the load placed upon it. The scale reads the tension in the cable - and assuming frictionless pulleys - is equal to the load at one end. If the load at each end was NOT equal the loading system (in this example) would move (displace) until it was in equilibrium. If you removed the left 25 lb pail (for example), and replaced it with a fixed support - like a clevis to secure the left end of the scale cable - and only have the 25 lb on the right end, the scale will read 25 lb. Action, reaction and equilibrium.
Reply:Originally Posted by IngenuityWilliam:That is incorrect. You only 'double the weight' if the cantilever span equals the backspan. For any other ratio of backspan/cantilever the distribution of support reaction is as follows: This is first-year engineering statics - for some folks it was high school math/physics.Evidently, it appears to be misunderstood by you. The scale is NOT calibrated to read half the load placed upon it. The scale reads the tension in the cable - and assuming frictionless pulleys - is equal to the load at one end. If the load at each end was NOT equal the loading system (in this example) would move (displace) until it was in equilibrium. If you removed the left 25 lb pail (for example), and replaced it with a fixed support - like a clevis to secure the left end of the scale cable - and only have the 25 lb on the right end, the scale will read 25 lb. Action, reaction and equilibrium.
Reply:Originally Posted by smithdoorI did metal work engineering for 30 years and everything is still standing today. Even in tornado world.Dave
Reply:Originally Posted by William McCormickYou have shown that if someone with a degree tells you something or publishes something that it must be true. If they make a table and put an engineering firm's name on it then that is gospel. As soon as you hang something over a fulcrum point, it doubles the weight of the payload on the fulcrum. But you have a table that shows otherwise. Look at any lever scenario and realize that as you hang something, a payload over a fulcrum point the amount that it is hungover for example say three feet, requires the point three feet on the other side of the fulcrum to create 25 pounds of counterforce to support the payload like a balance requires equal weight on both sides to balance it. This is basic engineering stuff that should be understood before you pick up a hammer or pry something with a screwdriver. So as I stated as soon as you hang something over a fulcrum point a wall supporting a cantilever for example the weight on the wall from the payload doubles. But do some more math until I am wrong. Haha. Sincerely, William McCormick
Reply:Originally Posted by William McCormickTake a look at the reality.Sincerely, William McCormick
Reply:Originally Posted by IngenuityI have a degree in structural engineering and have been in practice for over 30 years - if that disqualifies me in this discussion then so be it.But I shall minimize the math to avoid further confrontation.Let's use your fulcrum/lever example with say a crane beam with a 25 lb load on the LHS and a fulcrum such that the backspan:cantilever ratio is 3:1To balance the 25 lb applied load on the LHS we need a counterweight of 3 times = 75 lb on the RHS. Now the system is in equilibrium.So the total load that the fulcrum must now support is 25 lb + 75 lb = 100 lb.If the backspan:cantilever ratio is 2:1 then for a 25 lb applied load we would need 50 lb counterweight, and the total load to the fulcrum will be 75 lb.By adding a cantilever does indeed add load to the supporting wall/column (fulcrum), but its magnitude is dependent on the ratio of the backspan:cantilever.
Reply:Originally Posted by William McCormickI would get my money back from the school you went to. What you are not compensating for is what the boom adds to the equation. As soon as you pass the fulcrum point with a payload, the boom now adds the extra weight to the fulcrum to at least double it.  The payload now unnaturally has to try to lift the beam with an equal weight of the payload at an equal distance that it is extended over the fulcrum an equal distance away on the other side of the fulcrum. I figured this out while building boat davits and got my finger caught under a very light aluminum davit. I was stuck there for a long while I used the time to figure it out. Sincerely, William McCormick
Reply:What is a trailing arm?
Reply:Originally Posted by IngenuityOk, so math is not your strong suit. I intentionally made the beam 'weightless' to reduce the math - for your sake.Maybe you are a graphic person. Try this comparative graphic example:All three beams are in static equilibrium i.e. the sum of the forces in the vertical direction is zero, the some of the moments about any point along the beam is zero, and the sum of the forces in the horizontal direction is zero (which there are none in this example).If you cannot fathom this basic concept, then there is no hope.
Reply:Originally Posted by IngenuityWilliam:That is incorrect. You only 'double the weight' if the cantilever span equals the backspan. For any other ratio of backspan/cantilever the distribution of support reaction is as follows: This is first-year engineering statics - for some folks it was high school math/physics.Evidently, it appears to be misunderstood by you. The scale is NOT calibrated to read half the load placed upon it. The scale reads the tension in the cable - and assuming frictionless pulleys - is equal to the load at one end. If the load at each end was NOT equal the loading system (in this example) would move (displace) until it was in equilibrium. If you removed the left 25 lb pail (for example), and replaced it with a fixed support - like a clevis to secure the left end of the scale cable - and only have the 25 lb on the right end, the scale will read 25 lb. Action, reaction and equilibrium.Originally Posted by William McCormickThe scale in my graphic is showing that the scale is able to suspend 50 pounds against the force of gravity therefore it is under 50 pounds of force and registers half that force. Or I am missing something. Sincerely, William McCormick
Reply:Originally Posted by IngenuityAre you referring to this graphic you posted previously?If so, then yes, you are "missing something".There is only 25lb of applied load at each end, and the tension in the rope/cable is also only 25 lb - so the scale also reads 25 lb.Simple static free body diagram explains it. Cut a section through the horizontal rope and replace with unknown force, F:For static equilibrium: Sum of the forces in the vertical direction must equal zero:So:  Rv = 25 lb.....[Eqn 1]Sum of the forces in the horizontal direction must equal zero:So:  Rh = F.....[Eqn 2]But, we also know that (assuming prefect pulley (frictionless) and massless rope) that the total pulley support reaction Rtotal is at a 45o angle.So Rtotal = Rv / cos(45o).....[Eqn 3]Knowing Rtotal we can calculate Rh = Rtotal  * cos (45o).....[Eqn 4]Substituting Eqn 3 into Eqn 4 we have:Rh = Rv  /cos(45o) * cos (45o) = RvFrom Eqn 2:Rh = F = Rv = 25 lb==> Force in the scale = force in the horizontal rope/cable = 25 lb
Reply:You are missing the pulley reactions to complete the total system. There are 2 pulley reactions (at opposing 45o angles), and the sum of the vertical pulley components is 2 x 25 lb = 50 lb and that is equal to the magnitude of the 2 x 25 lb applied loads.Neglecting any pulley friction (and mass of the rope) - a rope with an applied load of F, will have such a force (F) over is entire length. A pulley only changes the direction of the force, NOT its magnitude.I can prove it to you with other some of my real-world work experiences, if you wish.Or if you do not believe me, Google it. It is the internet - so it has to be true :-)LINCOLNSquare Wave TIG 200Power MIG 210 MPRanger 330 MPXMILLERThunderbolt AC/DCHYPERTHERMPowermax 190CPowermax 45XPHOUGANHMD 130HMD 505FEIN/SLUGGER 14" COLD METAL CHOP SAW 9" COLD METAL HAND-HELD SAW
Reply:Originally Posted by IngenuityAre you referring to this graphic you posted previously?If so, then yes, you are "missing something".There is only 25lb of applied load at each end, and the tension in the rope/cable is also only 25 lb - so the scale also reads 25 lb.Simple static free body diagram explains it. Cut a section through the horizontal rope and replace with unknown force, F:For static equilibrium: Sum of the forces in the vertical direction must equal zero:So:  Rv = 25 lb.....[Eqn 1]Sum of the forces in the horizontal direction must equal zero:So:  Rh = F.....[Eqn 2]But, we also know that (assuming prefect pulley (frictionless) and massless rope) that the total pulley support reaction Rtotal is at a 45o angle.So Rtotal = Rv / cos(45o).....[Eqn 3]Knowing Rtotal we can calculate Rh = Rtotal  * cos (45o).....[Eqn 4]Substituting Eqn 3 into Eqn 4 we have:Rh = Rv  /cos(45o) * cos (45o) = RvFrom Eqn 2:Rh = F = Rv = 25 lb==> Force in the scale = force in the horizontal rope/cable = 25 lb
Reply:Originally Posted by William McCormickDo you agree with this graphic? The Sprocket shaft fail is common.Sincerely, William McCormick
Reply:Originally Posted by IngenuityYou are missing the pulley reactions to complete the total system. There are 2 pulley reactions (at opposing 45o angles), and the sum of the vertical pulley components is 2 x 25 lb = 50 lb and that is equal to the magnitude of the 2 x 25 lb applied loads.Neglecting any pulley friction (and mass of the rope) - a rope with an applied load of F, will have such a force (F) over is entire length. A pulley only changes the direction of the force, NOT its magnitude.I can prove it to you with other some of my real-world work experiences, if you wish.Or if you do not believe me, Google it. It is the internet - so it has to be true :-)
Reply:Originally Posted by William McCormickYet the cable and scale are holding two 25 pound weights off the ground against 50 pounds of gravity.  That according to my schooling requires 50 pounds of force to do. Or perhaps at the gym when you hold two 25 pound weights off the ground you are only holding 25-pounds against the force of gravity in your opinion? A scale is calibrated to only recognize half the force upon it. A 25-pound weight in freefall represents 25 pounds of gravitational force, when you stop the 25-pound weight to weigh it, the scale applies a 25-pound counterforce to the object and records that it is applying a 25-pound counterforce to it.  Here we see Mr. Bill at the gym. Now physics requires that each one of Mr. Bills' arms must apply a little over 25 pounds to lift each 25-pound weight. After the weight is lifted it requires that a constant 25-pound force be maintained to hold the two weights off the ground. If Mr. Bill only used one arm to lift one weight applying 25 pounds of force he would be pulled over to the side of the weight machine unless he could use his feet, body weight, and stance to create a counterforce of 25 pounds. You just cannot hold two 25-pound weights off the floor against gravity without 50 pounds of force. What you are trying to justify is that at the docks they could have one crane cable that splits into two cables over two seperate pulleys, and lift two containers using the same energy as lifting one. Sincerely, William McCormick
Reply:Originally Posted by William McCormickMy graphic is depicting a dock crane lifting two 90,000 pound containers with one cable attached to two cables through a pulley system. According to you the two containers should only put 90,000 pounds of load on the dock crane.Sincerely, William McCormick
Reply:Originally Posted by InsanerideYou don't need stiñkingmath.
Reply:Originally Posted by IngenuityWell, I disagree. Whilst simple tasks of calculating lengths of weld is not multivariable calculus, it is math...and like it or not, you use it every day.
Reply:In a LINCOLN welding book, I read about structural design, no math. Simply put, you design a prototype , analyze failure and improve design. Math not required. Farmer way. No offense to farmers. For fancy talkers: iterate design after failure, reiterate as needed. Hire Mathematicians for numbers and statistics for better results and profit. Bookkeepers and secretaries job.
Reply:Originally Posted by InsanerideWell now mister smarty pants, I do statistics in my head every single day so yes, I use math everyday. As a matter of fact for your information, at one point in time I used todo trigonometry in my head not only because I could but because I could and it saved time. So there
Reply:Originally Posted by InsanerideWacky Mac.  Loved your Mr Bill explanation. Saw him in post #17 also.You are a pillar of welding world. You don't need stiñkingmath. Keep up with your good work.Edit: is it about tension or compression at the fulcrum tho? Whenever cantilevered.
Reply:Originally Posted by IngenuityNo, according to me (and science, physics, engineering, the real world, etc. etc.) - and the explanation below - the single upper cable attached to the dock crane will be acted upon by 180,000 lbf of force i.e. sum of the force in each of two container cables as the two are connected to the one dock crane cable.Free body diagram of the upper pulleys and the 2-to-1 cable connection:
Reply:are the listed values weights against gravity or purely tension against fixed points?I think the confusion lies in jumping between the two, force vs weight.Last edited by SlowBlues; 3 Weeks Ago at 04:41 PM.
Reply:A cantilever by definition has to resist the forces levered against it, which depend on forces against and length of lever.  Every lever has a point of fulcrum, if lever length and forces on either end remain the same the force on the fulcrum point will not change if moved, only the ratios on either end.  essentially the same as a cable split into two under tension with fixed points, gear/belt reduction, etc.  all the same idea.Most often the best way to simplify it is to look at both tension AND compression in each member.  The "wall" of the cantilever doesn't need to support twice the weight, but it does need to resist two forces (compression and tension) in opposite directions (one side of base of wall pulling up, one side pushing down).Last edited by SlowBlues; 3 Weeks Ago at 04:43 PM.
Reply:Originally Posted by William McCormickWell if you agree that there is 180,000 pounds created by the two suspended weights then the two cables holding them are also under 180,000 pounds and the single cable holding both cables is being subjected to 360,000 pounds. Just like a hanging scale a cable holding a 25 pound weight off the ground against the force gravity and motionless is under 50 pounds of stress. The scale and the cable salesmen just sells them like they do to make it seem simpler for the new guy.
Reply:All i know is, if you decide not to tighten the cap on an argon cylinder because it is cocked and on " Tight enough", then proceed to carry the cylinder by holding the cap....then you observe one of Newton's laws of motion.  An object (cap) in motion will remain in motion until it meets an opposing force(chin) of equal or greater magnitude.I also know that lifting blocks have reduced capacity when lifting vertically. Since the mass being lifted and the lifting force both act on the block.Perhaps there is confusion of what is actually happening? I'on kno.I still don't know what a trailing arm is.
Reply:[/B][/QUOTE] Originally Posted by SlowBluesare the listed values weights against gravity or purely tension against fixed points?I think the confusion lies in jumping between the two, force vs weight.
Reply:Originally Posted by IngenuityNo, the upper cable has 180,000 lb, and each of the two lower cables has 90,000 lb - so combined, the two lower cables total 180,000lb. The sum of the vertical forces at the connection between the 2 lower cables to the 1 upper cable must = zero.No, the scale does NOT measure half the force, it is measuring the only force that is applied to the system - 25 lb - that is in static equilibrium.==> So you are not believing my math, and my graphics was not intuitive to you, so lets try this:EXPERIMENTAL PROOF:  The foundation upon which scientific theory is proven. I used my DILLON tension gauge and a setup that mimics your 'scrap yard scale' pictorial:First, measured the mass of the 2 x 25 lb weights to verify that they are indeed 50 lb total - measured in the vertical direction, using the DILLON gauge.Photographic proof: A little hard to read in this photo, but it states 50 lbf. But I have other photos I can share if you need further proof. And the weights are stamped with 25lb - and they have been checked from the USPS scale because I use them to calibrate my bicycle power meter.Next, I applied 2 x 25lb masses to each end of a horizontal beam with end pulleys attached, connected via ropes (different color ropes to show that there is no 'sleight-of-hand'), with a centrally placed DILLON tension gauge (the same one I used to verify the 50 lb weights).Photographic proof:Then read the DILLON tension gauge:Photographic proof:And low and behold, the gauge read 50 lbf.Exactly what static equilibrium states.So, all is good, the earth still spins on its axis, the sun will shine tomorrow, William McCormack is a believer in science, and I do NOT need to get a refund on my engineering degree. Life is good!
Reply:Here is another way to look at it. Now if I am correct, then the scales record half of the total force they are under when supporting an object. I am not suggesting chaos and madness, I am suggesting that most do not understand how a scale is actually calibrated. In the graphic below each hanging scale is correctly reporting the weight of the object hanging underneath it. Yet you can see the dilemma if the ceiling has a force of fifty pounds of downward force upon it. The hanging scale only reports the gravities effect on the weight, not the counterforce holding the scale-up. Sincerely, William McCormickIf I wasn't so.....crazy, I wouldn't try to act normal, and you would be afraid.
Reply:I have been played by a fool. I did some 'research' on the infamous 'William McCormick' and he/she/it is notorious on this site (and several others, namely, Miller, Hobart etc) on the insanity of this very subject (and several others).This subject - along with the very same graphics that McCormick uses above - were the cause of his ban back in 2010: https://weldingweb.com/vbb/threads/4...igure-this-outYou can believe what ever you wish, but rejecting/disputing scientific fact is denialism. Surely, after more than a decade of your 'crusade' on this subject, you must stop and think "maybe I am wrong".Because you are wrong, William McCormick. Sincerely, wrong.LINCOLNSquare Wave TIG 200Power MIG 210 MPRanger 330 MPXMILLERThunderbolt AC/DCHYPERTHERMPowermax 190CPowermax 45XPHOUGANHMD 130HMD 505FEIN/SLUGGER 14" COLD METAL CHOP SAW 9" COLD METAL HAND-HELD SAW
Reply:Originally Posted by IngenuityI have been played by a fool. I did some 'research' on the infamous 'William McCormick' and he/she/it is notorious on this site (and several others, namely, Miller, Hobart etc) on the insanity of this very subject (and several others).This subject - along with the very same graphics that McCormick uses above - were the cause of his ban back in 2010: https://weldingweb.com/vbb/threads/4...igure-this-outYou can believe what ever you wish, but rejecting/disputing scientific fact is denialism. Surely, after more than a decade of your 'crusade' on this subject, you must stop and think "maybe I am wrong".Because you are wrong, William McCormick. Sincerely, wrong.
Reply:Originally Posted by IngenuityI have been played by a fool. I did some 'research' on the infamous 'William McCormick' and he/she/it is notorious on this site (and several others, namely, Miller, Hobart etc) on the insanity of this very subject (and several others).This subject - along with the very same graphics that McCormick uses above - were the cause of his ban back in 2010: https://weldingweb.com/vbb/threads/4...igure-this-outYou can believe what ever you wish, but rejecting/disputing scientific fact is denialism. Surely, after more than a decade of your 'crusade' on this subject, you must stop and think "maybe I am wrong".Because you are wrong, William McCormick. Sincerely, wrong.