Deflection in stepped shaft.....
Deflection in stepped shaft.....
(OP)
Suppose you have a shaft starting from left to right with a Ø3.75" for 17" length, then it steps up to Ø4.5" long for 24" length then steps back down to Ø3.75" for 17" length. How do you take the equations and incorporate the differences in "I" the moment of inertia to determine the defection at the load?





RE: Deflection in stepped shaft.....
RE: Deflection in stepped shaft.....
The Rayleigh and Dunkerley methods of calculating the first natural frequency of systems for dynamic deflection calculation.
RE: Deflection in stepped shaft.....
where/how is this shaft supported?
RE: Deflection in stepped shaft.....
RE: Deflection in stepped shaft.....
RE: Deflection in stepped shaft.....
RE: Deflection in stepped shaft.....
Let I1 = PI*D1^4/64 = 3.14*3.75^4/64 = 9.7 inch4;
similarly I2 = 20.19 inch4.
The deflection curve is y = -M/(2EI)*x^2) + C1x +C2 for the small dia portion ( 0<= x <= 17 inch)and similarly with C3 and C4 for the large dia portion ( 17 <= x <= 29 inch)
From the left, at x = 0 deflection should be y = 0 which makes C2 = 0.
At x = 17 both the slope and deflection should be equal on the small dia and large dia joint. It gives the value of C1 and C4.
At x = 29 (middle of shaft) the slope is zero which gives C3 = 0.
Therefore the deflection at the middle of shaft is 8.1 *10^(-7) * P, where P is the applied concentrated load. The deflection at x = 17 is 7.8*10^(-7) * P, showing that the deflection is max at the middle of shaft where the load is applied.
If you need more details i would be glad to send them to you.
RE: Deflection in stepped shaft.....
RE: Deflection in stepped shaft.....
Basically since it is symmetrical you spit the shaft in half. My intergration skills are rusty so:
is the slope(T)= (-M/2EI)x^3/3 + C1x^2/2 + C2x? Sub I1 in there; and sub I2 using C3 and C4 respectively? The same can be said for the deflection equation?
Based on the updated dims of the shaft (P=11,036 lbs pulling up, I need to include the weight of the shaft which would be subtacted from P's influence). I understand the slope and deflection are the same at x=7.5. I understand solving for C2. To get C1 and C4 do you solve 2 equations (slope and delection) for 2 unknowns? How did you solve for C3? M is the moment at P ((P/2)*21") correct? I guess I need more details...Thanks
RE: Deflection in stepped shaft.....
I looked at this stepped shaft and have calculated a deflection based on your original shaft dimensions using a unity load as no load was given in the first post.
The method I used was the moment area method where you sketch out the bending moment diagram and then divide it by
the product of "modulus of elasticity for the material" and
its "second moment of area" for each section of the shaft.The beam was originally symmetrical so I plotted the above; over half the beam.Maximun deflection would occur in the centre of the larger diameter portion of the original shaft assuming that a point load was applied at that point.
Now there would be zero slope at maximum deflection of the beam so therefore I took this as the origin and calculated the "moment of area of the bending moment diagram" which gives directly the deflection between the centre of the beam and the supporting end.
I can do the same quite easily for the updated shaft dimensions however I would point out two things:-
1/ both the original shaft and new shaft do not satisfy
the length to depth ratio to class the shaft as a beam
The ratio of length to depth should be in the order of
20:1.
2/ The beam is not symmetrical if the smaller diameters
different lengths ie:- 7.5" and 6".
Finally can I assume the load you have given acts in the centre of the larger diameter portion of the shaft?
regards desertfox
RE: Deflection in stepped shaft.....
d=3.9375 inch for 7.5 inch
D=4.4375 inch for 27 inch, and
d=3.9375 inch for 6 inch.
Load applied at the middle of 27 inch portion or x=21 inch form the left.
CG is at x=20.3 inch for a shaft weight of 168 lb; load is 11036 lb, therefore the weight is only 1.5% and can be neglected.
The reaction at left is 5313.6 lb and the bending moment formula is M = 5313.68*x which at x=21 is Mmax=111585 lb*in
From the left, when 0 <= x <=7.5, the slope1 = -(5313.6*x^2)/(2*EI1) +C1 and deflection,
y1 = -(5313.6*x^3)/(6*EI1) + C1*x + C2
When 7.5 <= x <= 21, the slope2 = - -(5313.6*x^2)/(2*EI2) +C3 and deflection, y2 = -(5313.6*x^3)/(6*EI2) + C3*x + C4
To determine the integration constants:
at x=0 y1 = 0, therefore C2 = 0
at x=21 assume slope2 =0, therefore C3 = 0
at x=7.5, both slope1 = slope2 and y1 = y2, solving for
C1 = 1.605*10^(-4) and C4 = 8.028*10^(-4)
Finally,
slope at x=0 is 1.605*10(-4) rads or 0.01 deg
deflection at x=7.5 is y1 = y2 = 0.0002 inch
deflection at load x= 21 is y2 = 0.0136 inch.
To be acceptable the deflection should be less than 0.0005 in/in for general machinery (40.5 inch shaft gives 0.020 inch) or if you have gears at load not to exceed 0.015 inch.
RE: Deflection in stepped shaft.....
I have calculated that the maximum deflection of the shaft
using the area moment method which is 0.02658216611"maximum and this deflection occurs at 20.0061" from the end with the 7.5" long reduced diameter.
As my calculation is a graphical method I did a check to see
what deflection the shaft would have if it was all the same diameter ie:- 4 7/16" dia
from machinery's handbook case 3 (simply supported beam load
at any point)
defl= W*a*v^3/(3*E*I*l)
v = point on beam of max deflection (20.4935")from 6"
reduced diameter end.
a = short span between load and end support = 19.5"
I = 19.03370627"
This gives a deflection of 0.02768" If the beam was the larger diameter throughout.
This shows my calculation to be incorrect at 0.02658216611"
as one would expect the stepped shaft to deflect more than this, however it appears to be in the order and e could say the shaft will deflect in the order of 0.030".
I think stel8 is incorrect in assuming that the slope of the beam is zero under the load and the machinery's handbook seems to confirm this.
In addition His deflection of the shaft in comparision with
a shaft of 4 7/16" throughout its length is out by a factor of 2.
regards desertfox
RE: Deflection in stepped shaft.....
The shaft is not symmetric about the load, you're 1 1/2 inches short on the RHS. This is why the CG is shifted left of the load.
I have run the shaft on SolidWorks but need to check it against my hand calculations since integration constants and theoretical assumptions are asked for. Tagger has mentioned he's using self aligning bearings, pillow block style. Any idea of model number? The length of supported shaft will give great detail on physical deflections!
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
RE: Deflection in stepped shaft.....
Thankyou for confirming my thoughts, sorry I have no idea
about the supports, I will be interested in your answer though.
I agree Tagger would have been correct with his assumption
of zero slope under the load with the original post.
My other concern about treating this as a beam was the length of beam to depth ratio which is less than 20:1.
Regards desertfox
RE: Deflection in stepped shaft.....
I checked your calculations and they are correct. I checked my calculatons and they are also correct. I humbly recognized that something is wrong if integration method (mine) does not agree with your grapphical method. I also agree that more likely your deflection estimate of about 0.030 is true.
This could be acceptable depending of the application. I found that for general machinery the acceptable deflection is between 0.0005 and 0.003 in/in of shaft (Mott, Machine elements in mech. design), which for this shaft of 40.5 inch is between 0.020 and 0.120 inch.
A Happy New Year!!
RE: Deflection in stepped shaft.....
Thank you for your last post and a happy new year to you
also.
regards desertfox
RE: Deflection in stepped shaft.....
Further to my earlier point, I have refined my graphical method and put it in a spreadsheet on Excel. In addition I also found a mathematical error, in that the maximum deflection occurs at 20.4939" from the 7.5" reduced diameter end and not as stated from the 6" long reduced diameter.
Having re-run my calculation with the corrections I have obtained a maximum deflection of 0.02851204736" at the 20.4939" dimension.
This answer now fits as it is clearly a larger deflection at the same point on the shaft when compared with a uniform diameter shaft of 4 7/16".
Regards
Desertfox
RE: Deflection in stepped shaft.....
RE: Deflection in stepped shaft.....
I ran your latest shaft dimensions through my spreadsheet
using the moment area method and I calculate a figure of
0.008733" occuring at a length of 18.39157416" from the
left hand support.I also calculated the shaft deflection at
the same point for a uniform shaft of dia 4.4375" using the machinery's handbook which worked out at 0.00713".
I think Castigliano's method could be used to solve this
however I am not sure how exactly one would begin without studying my books a little more as its been a while since I used the theorem.
regards desertfox
RE: Deflection in stepped shaft.....
I agree v=18.391574, so for a uniform shaft Ø of 4.438" the deflection I determined was 0.01659. I got TK Solver 4 here at work and will install it. I haven't used it since school years ago, so I will have to freshen up on it. I guess I can set up a program to solve a stepped shaft. Has any on already set up a program?
Cockroach, do you have Comos set up in conjunction with your Solidworks. We have here Inventor 7, haven't installed 8 yet. I doubt that Inventor can calculate deflection of a stepped shaft, I will probably need Comos or something similar installed with it.
RE: Deflection in stepped shaft.....
RE: Deflection in stepped shaft.....
Your correct, my last calculations are wrong your stepped shaft of 35" long would according to my moment area method would be 0.01736823732" at 18.39157416" and a uniform shaft
of 4.4375" would be 0.0172136369" at the same point using the case 3 of the machinery's handbook.Increasing the centre portion of the shaft to 5.5" diameter according to my moment area method gives a deflection of 0.00765259479" at the same location as previous and this also shows that my method and the program that Rkanik suggested are in the correct ball park.
This shaft arrangement from your posts obviously exists and you are looking at replacing it, what I am not sure about is why your worried about the deflection, what as or will change in service that as prompted the question of deflection and why can you not just replace it with a shaft to the current design?
regards desertfox
RE: Deflection in stepped shaft.....
What spreadsheet program are you using? Excel? I would like to set up a program using the Moment-Area method. Do you have the textbook Mechanics of Materials by Hibbeler, or Fundamentals of Machine Component Design by Juvinall/Marshek? Marshek has a BASIC program, and I can't figure out how to tanslate it to Excel. How did you set yours up?
RE: Deflection in stepped shaft.....
We have quite a few large Buffalo Fans that are driven by 5V belt drives and after about 35 years started breaking shafts. The initial thought was that the bearings ,sheave, had worn causing allowing a slight deflection of the shafts. No failure analysis was done on the parts. The bearings were changed out and new shafts installed. 2 of the same shafts broke after a very short time. Metallurgical inspection of the last two failures revealed that the failure mode was multi-origin fatigue.
Investigation after the second failure revealed that the belts had been changed out due a "flopping noise". I caught the installation of the next set of belts prior to startup. The jackscrew adjustment was all the way out on the driver, in other words the belts had been tightened about 1 1/2" beyond the correct tension setting. The problem was that the crew had never changed the belts on these fans and were told to get the noise out.
We immediately adjusted the remaining belts in question to proper tension.
The high tension on the belts was actually bending the shaft.
RE: Deflection in stepped shaft.....
Sorry for the delay in my response I was looking at Castigliano's theorem for your problem and I think I have
a result!
Firstly I used excel to do the moment area method but only for your particular shaft, all I did was split the shaft up into discrete lengths and worked out the bending moment at each length by way of formula, then knowing where the section of the shaft changes I calculated and listed the ((bending moment)/EI) for all lengths. Now where the max deflection occurs ie 18.391 etc I used values of
((bending moment)/EI) at this point and also at the point where shaft changes section to calculate the deflection,although this latter part was done by hand.
Now using Castigliano's I calculated a deflection of 0.017368" at the 18.391" position which seems to tally with the moment area method and that software program and here is my method:-
(1/(2EI))* intergral of [M^2] for the 4.5"
long reduced
diameter
so intergrate M^2= (R1*x)^2 = R1^2 * x^3/3
calculate strain energy for small diameter of shaft:-
(1/(2EI))* R1^2 * x^3/3
"I"= second moment of area of smaller diameter on shaft
limits for the intergral are x=0 to x=4.5"
R1= left hand end reaction = 4572.057143lbs.
Now for the next step, here we need the strain energy stored
in the larger diameter between 4.5" and 18.391..." from the left hand end, so subtract 4.5" from 18.391..." to give length of larger diameter in which the strain energy is stored ie:- 18.391..."- 4.5 " = 13.8915...."
now the bending moment in this larger diameter within the above length is:-
M= R1(4.5"+ x) where x = 0 and 13.8915"
therefore M^2 = R1^2*(4.5" + x)^2
and (1/(2*E*I))* intergral R1^2*(4.5 + x)^2
(1/(2*E*I))* R1^2*(20.25*x + (9*x^2/2)+ (x^3/3)
again calcute the strain energy, remeber that "I" here should be for the larger diameter of the shaft.
Sum the strain energy for both parts of the shaft and then equate the strain energy to:-
0.5*R1*Delta= strain energy
therefore Delta (deflection)=2*strain energy/(R1)
This according to my calculations is as stated earlier:-
Delta = 0.017368"
RE: Deflection in stepped shaft.....
RE: Deflection in stepped shaft.....
I have a problem similar to the one described by Tagger. I have a cantilever beam which is axially symmetrical but diameter is not constant throughout. Some part of it is tapered and some part is stepped. Load is at the tip.
I need to find the highest stress and a deflection at certain point in the beam. I have some idea on how to do it but would like to know whether my concept is right.
For stress, can i just use the bending moment equation Mc/I to find the stress at the smallest diameter? M would be force X (distance from tip to the desired position), c would be diameter/2 and I would be the moment of inertia at the particular point of interest.
For deflection, using castiliagno's theorem, i use a spreadsheet to compute the deflection at 1mm interval using (Mm/EI)dx. (Length of my cantilever is about 1m long.)
Is that the correct concept? What will be the estimated accuracy that I will get?
RE: Deflection in stepped shaft.....
Assuming your cantilever as the correct length to depth ratio to class it as a beam I see no problem with your concept.
hope this helps
regards desertfox
RE: Deflection in stepped shaft.....
A) If you have a wide pulley installed on a shaft and it is fitted to the shaft with some type of mechanical fitting to attach it quite solidly then no deflection will occur in the inside of the pulley and that will make the calcs to change quite drastically, I'm not involved in those calcs but would like to see them done by some of you guys.
B) the main problem for a deflection on a pulley shaft is excesive belt tensioning, and that is usually done to eliminate belt slip over the smaller size pulley (but the tension affects both pulleys via the belts), to correct the slip it is better practice to increase the size of both pulley in a way to not change the transmission rate of speed but enough to eliminate the slip with out excesive belt tension.
C) in fan operations many times the main culprit of both the noise and the vibration which finally cause the failure of the shaft not by shear but instead by fatigue caused by minimal shaft bending done thousands of times a day, is the inbalance of the rotor caused by several factors, grime deposited on some rotor blades, or the same that has been washed away in only a part of the rotor, a balancing weight added in manufacture that has torn loose, or ultimatly a piece of rotor that has corroded and unbalanced the whole rotor.
Your calcs are right but maybe eliminating the cause of the deflection will ease the shaft fatigue better.
Cheers,
SACEM1
RE: Deflection in stepped shaft.....
The shaft deflection above is based on point loads, which in practice can never be achieved ie:- the beam is simply supported but in reality the beam is actually supported in bearings which give support somewhere between simply supported and built in ends. The load created by a wide pulley may well be better modelled as a uniformly distributed load but to model it as that would be an assumption in itself. Therefore from Tagger's original problem data and to err completely on the safe side by considering the shaft as simply supported and under deflection from point loads I considered this would give the worst case of deflection possible. At the end of the day
I agree at best this deflection would only be an estimate and not an accurate figure that one might see in practice.
regards desertfox
RE: Deflection in stepped shaft.....
You are right besides nobody can be sure if the fit between pulley is done all the way thru the pulley section or if its only on one point.
We must always take the worst case scenario as you stated
Cheers
SACEM1
RE: Deflection in stepped shaft.....
I'm curious what problem the "flexing" is believed to be causing.
RE: Deflection in stepped shaft.....
One thing I didn't mention is verify that all shaft are parallel, not matter what the bearing type.
RE: Deflection in stepped shaft.....
Turbo Engineer