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Q: Strength of steel shafting with different radius ( Answered 5 out of 5 stars,   0 Comments )
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

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.


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.
Subject: Re: Strength of steel shafting with different radius
Answered By: krobert-ga on 25 Jan 2003 14:17 PST
Rated:5 out of 5 stars
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

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.



Deformation and Fracture Mechanics (4th Ed.), by Richard W. Hertzberg.
John Wiley (1996).

Mechanical Metallurgy (3rd Ed.), by George E. Dieter. McGraw Hill

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

Clarification of Answer by krobert-ga on 28 Jan 2003 08:00 PST

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

Best to you,


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:5 out of 5 stars
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!

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