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?
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Deflection in stepped shaft..... 5
- Thread starter Tagger
- Start date
- Thread starter
- #21
Oops, I meant to say 0.007" deflection with a Ø5.5" diameter section not 0.0007. But I calculated using case 3 in the Machinery's handbook a max deflection of 0.01659" for a shaft of uniform thickness of Ø4.438" at the same location you said desertfox (18.39157"
from the left end (x=0). Basically the load point is x=20.5 saying starting from left (x=0) to right (x=35.00), and the bearing supports are located at x=0 and x=35, the shaft configuration is 0<=x<=4.50" Ø=3.938", 4.50<=x<=31.5 Ø=4.438", 31.5<=x<=35 Ø=3.938", where W=11,036 lbs. at x=20.50". Thus for case 3: a=14.5, b=20.5, I=19.042, l=35.
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.
from the left end (x=0). Basically the load point is x=20.5 saying starting from left (x=0) to right (x=35.00), and the bearing supports are located at x=0 and x=35, the shaft configuration is 0<=x<=4.50" Ø=3.938", 4.50<=x<=31.5 Ø=4.438", 31.5<=x<=35 Ø=3.938", where W=11,036 lbs. at x=20.50". Thus for case 3: a=14.5, b=20.5, I=19.042, l=35.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.
- Thread starter
- #22
Using the program RKanik suggested I come up with increasing the Ø4.438 section to Ø4.938 will give a deflection of 0.0105", and based on a 35" long shaft this gives a 0.0003 in/in deflection. Since the customer will have to go to a greater QD bushing (Size M), and all this is theorectical, maybe going to the max bore of a size M QD bushing (Ø5.50" will work. I don't want them to purchase a new shaft, bushing, sheaves, and it still flexes. The customer can keep the same bearings, and center distances. Any thoughts?
will work. I don't want them to purchase a new shaft, bushing, sheaves, and it still flexes. The customer can keep the same bearings, and center distances. Any thoughts?Hi Tagger
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
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
- Thread starter
- #24
Hi desertfox, The customer brought this porblem to our attention, saying the shaft is flexing during operation. They wanted to know what needed to be done to eliminate the flexing. As to why the original design is now inadequate, I do not know. The belt pull can range from a minimum of 7360 pounds up to the 11,039 pounds. This could have been a problem since installation.
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?
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?
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1
- #25
I would look at the belt installation as a possible source of the deflection problem.
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.
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.
Hi Tagger
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"
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"
There is a program you can buy called Beam 2D which can do stresses and deflections in any shape of shaft. Since this program is basically for beam analysis, it does not take into account for torsion. However, I've been using this software for a while and it is well worth the price, which is about $200.
Hello,
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?
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?
I don't know if somebody is still looking at this thread but as I just have read it I find that some points have been either overlooked or not taken into account:
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
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
Hi sacem1
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
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
Not knowing the amount of flexing, two things can happen and both are bad. Depending on the amount the first thing to go is the bearing and the next is a shaft failure by fatigue. The failure is will probably be at the outboard of the pulley, usually at the step. The outboard bearing is probably getting a licking.
One thing I didn't mention is verify that all shaft are parallel, not matter what the bearing type.
One thing I didn't mention is verify that all shaft are parallel, not matter what the bearing type.
turboengineer
Mechanical
I'd start out with Roark too. How about rotordynamics? Have you address them as well? It's a pretty long and spindley shaft, so deflections during rotation could kill you if you pass through a critical speed. I use a program we have in house called Excel Rotor. It's results come pretty close to what I run into in the lab.
Turbo Engineer
Turbo Engineer
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