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 Subject: Strength of steel shafting with different radius Category: Reference, Education and News > General Reference Asked by: 3rrotec-ga List Price: \$20.00 Posted: 25 Jan 2003 05:50 PST Expires: 24 Feb 2003 05:50 PST Question ID: 148349
 ```This question is a comparison of the strength of steel alloy (assume 4140 heat treated to 315 brn.). CASE 1: 5 inch o.d. material turned down 12 inches long to 2 inch diameter. Remainder of shaft length remains at 5 inch o.d. Shoulder between 2 inch and 5 inch has very little radius (if number needed use .016) Shaft is subjected to overhung, shock, rotating load. CASE 2: Same as case 1, but radius is .250. CASE 3 Same as case 1, but .500 radius. CASE 4 Same as case 1, but shoulder is a 45 degree bevel. The preceeding scenerio is hypothetical. You can probably see what I'm after and you may want to make some of your own assumptions. That's fine. I am a 50 year old experienced machinist with a strong mathematical background. I would like to be able to put some numbers (formulas) to this question. Thanks! 3rrotec``` Request for Question Clarification by krobert-ga on 25 Jan 2003 08:14 PST ```3rrotec-ga, This isn't really a question of strength, but fracture toughness. I assume that the failure point will be where the load is the highest (quite possibly the corner of the turned portion of the shaft). The simple answer is that the beveled shaft is going to be the strongest. The shaft with the 0.016 radius will be the weakest. Will you like an explanation of why this is the case? Of course this doesn't look at bending loads, but I'm going to wager that this is not what you are interested in. krobert-ga``` Clarification of Question by 3rrotec-ga on 25 Jan 2003 12:20 PST ```Dear Robert First off, am I trying to answer myself? My name is also Robert (Robert Williams). With two Roberts I bet we can figure this out (double trouble maybe!). Yes I would like an explanation why a right angle "funnels" or concentrates stresses. I agree with you on which case is strongest and weakest. My question is how much. Maybe a formula or rule of thumb giving percent decrease/increase in each case. For example if we test or assume our shaft will fail at 80,000 psi with a .016 radius, how much increase will we get with case 2, 3, & 4? Thanks Robert p.s. Are you left handed? I am.```
 ```A lefty huh? Not me... I've heard that you guys are trouble... Of course, a fellow Robert is an OK guy. Down to business. Basically, this "funneling" is what is referred to as "stress concentration". It is measured by a factor usually designated as K. K = Max_Stress / Nominal_Stress Lets look at the case where we don't have a bend (notch)... a straight shaft. In this case, the load is distributed approximately uniformly across the shaft, which is loaded at approximately the center and supported at the ends. Now lets introduce the notch. What happens? With the same loading, we have a discontinuity in the bending of the shaft, the notched portion of the shaft wants to bend further than the main portion of the shaft. BUT, the material is continuous (the ends are mechanically attached to the main shaft). The mechanical discontinuity is accompanied by a stress distribution discontinuity. The stress changes from two dimensional (through a cut section of shaft) to three dimensional (now there's torsion AND bending involved). The effect is to raise the stress between the two portions of the shaft at section of the discontinuity. Increasing the size of the radius between the two portions has the effect of spreading out this raised stress. According to the book in front of me (Dieter, no formulas... just graphs, it needs to be measured), if we have a narrowed shaft and the dimensions you gave me for Case 1, K = Off the graph (I couldn't even make a guess, it's above 3.5, and likely above 4). For Case 2, K = 1.5 (roughly). For Case 3, K = 1.4 (again, roughly). For Case 4, our scenario changed... now you need to be concerned about where your bevel meets the smaller portion of the shaft... that needs a radius too! Lesson: We dont need to give this shaft a radius of much more than 0.250. Why? It's related to the size of the smaller shaft (the ratio of the corner radius to the smaller shaft diameter). More radius surely helps, but not much (See the result for case 3 and compare to case 1). A final word... One thing I did not discuss was fatigue. With a rotating load like the one you described stress fatigue will definitely be a concern. This has to do with the formation of microcracks in the metal itself due to the alternating load. The principles are kind of the same because the stress experienced at the 90 degree bend will be a concentrated one. I hope this was what you were looking for, don't hesitate to ask for a question clarification if you need to. krobert-ga References: Deformation and Fracture Mechanics (4th Ed.), by Richard W. Hertzberg. John Wiley (1996). Mechanical Metallurgy (3rd Ed.), by George E. Dieter. McGraw Hill (1986).``` Request for Answer Clarification by 3rrotec-ga on 28 Jan 2003 05:52 PST ```Morning Robert, Thanks for your answer. Just clarify why a groove or shoulder throws a shaft into bending moment. I think I understand. If the stress along a shaft is concentrated by reducing o.d., can I make the assumption that all shafts with shoulders or different hardness or grain are subject to bending? Also please clarify the k factor. I know how to figure nominal strain But how do you put a number to maximun strain? Thanks Robert 3rrotec p.s. You will be hearing from me in the future. I like your answers. By the way, if the right side of your brain controls the left side of your body, then left handed people are the only ones in their right mind! Who's in trouble now? Thanks again Roteceng``` Clarification of Answer by krobert-ga on 28 Jan 2003 08:00 PST ```3rrotec-ga, A groove or other non-straight section of a shaft or any other piece of material will experience not-necessarily a bending moment, but a change from a 2 dimensional stress field (plane stress) to a 3 dimensional stress field. Think of it kind of like gravitational field lines. A star or other "non-uniformity" in the universe will have attractive field lines that "suck" things toward it. In our case, the non-uniformity in the shaft "sucks" stress toward it and concentrates it. This is really a problem for elasticity theory to explain (elasticity theory provides most of the math for the fracture theory). It has to do with mismatched stress fields. Think of a straight bar that has a 10,000 lb torsional load on it. Now take a similar bar that has only half the original bars cross-sectional area (not half the original bar's diameter... we're talking stress, which is a functin of area). Half the area means twice the stress. Now take the two bars and stack them along their axes. The area where they meet wants to have two different stress fields, but they are mechanically joined. Mother nature takes care of this nonuniformity by accomadating it with a 3 dimensional stress field which happens to be concentrated along the groove between the two sections of the bar. By "smoothing out" this groove, you are smoothing out the stress field. Your also giving it more material which is able to relieve the other material of some of the stress. As far as the different hardness or grain is concerned... what is happening in this situation is not really too different from the original case. Regardless of the hardness, a material has about the same elastic response (Young's or simply elastic modulus to those in the know). When the material fails though, you will see a big difference compared to the original case. This is because of the _plastic_ response of the material... which is a whole different question :) The k factor is just a empirical factor. For certain specific situations that can be handled by the elasticity theory I mentioned, you can figure out what this maximum stress is. Most elasticity theory nowadays is handled by numerical solutions to the elasticity equations... that's the finite element analysis I was talking about. For other relatively common situations, books of k-factors have been compiled (years ago) and can be used to get a rough approximation to the maximum stress that a groove or bend or other geometry will be experiencing. Best to you, krobert-ga P.S. to you: So, does your comment mean that rightys have never been in their right mind? That would explain a lot of the world today :-)```
 3rrotec-ga rated this answer: ```This has been the best "answer man" I have had the pleasure to question, even though he's in his left mind. I will specify him in the future 10 stars!```