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
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- #2
The basic cantilevered beam deflection formula, which you can find in any book on mechanics: y = F x L3 / (3 x E X I), where F is radial load, L is cantilevered length, E modulus of the elasticity of the material, and I is moment of inertia. Load F could be a cantilevered weight of the overhung load, a centrifugal force created by the end load unbalance, a hydraulic radial thrust of a centrifugal pump, or a combination of forces. These forces can be static and not changing direction (such as weight), or dynamic (such as rotating unbalance). Use superpostion to determine total deflection.
The Rayleigh and Dunkerley methods of calculating the first natural frequency of systems for dynamic deflection calculation.
The Rayleigh and Dunkerley methods of calculating the first natural frequency of systems for dynamic deflection calculation.
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EnglishMuffin
Mechanical
I suggest you get hold of a copy of Roark's "Formulas for Stress and Strain". In my 6th edition, three methods of solving your problem appear on p198, together with an example. Many books on elementary strength of materials also cover this topic.
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Luckily the shaft is symmetrical and it is easy to find the integration constants.
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.
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.
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- #10
Thanks stel8, The actual shaft dims are Ø3-15/16 for 7.5", the Ø4-7/16 for 27", then Ø3-15/16 for 6". Therefore I1 = 11.8 and I2 = 19.0. I don't know why I didn't use them in the first place.
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
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-
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- #11
Hi Tagger
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
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
For the new data:
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.
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.
Hi Tagger,
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
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
Yes DesertFox, I concur, slope underneath the load is not zero although pretty close to it. Remember that Stel8 did the calculation for the symmetrical case, that is, prior to the exact shaft dimensions as clarified by Tagger.
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
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
Hi Cockroach
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
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
Hi desertfox and all the others
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!!
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!!
Hi All,
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
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
- Thread starter
- #19
I just got back into the office, and would like to thank you guys for the help. I had a chance to look at the application. It was behind a safty guard so I had to "eyeball" the dims. I do know the CL/CL of the bearings is 35 inches. I believe it was 40.5 inches originally. The Ø4.438" x 27" long section is correct. The left bearing CL is 4.5" from the step-up, and the right bearing CL is 3.5" from the step-up. The piont of the 11,036 lb load is 11" from the right step-up shoulder near the right bearing. It si not symmetrical. The bearings are SKF SAF22522 pillow blocks. Using the program Rkanik suggested, I figured the deflection to be around 0.016. This is a rough figure since the "pull" is applied at a Dodge 8V Ø14.0x10 belt sheave with a J size QD bushing bored for Ø4.438 shaft. To get it down to a resonable deflection, I think increasing the size of the center section (27" to Ø5.5" will give about a 0.0007" deflection. The customer can keep the same bearings and locations, but will have to modify the sheaves for a larger QD bushing (say M size). I would like to analyze by hand. Can you use Castigliano's Method to solve this? I have textbook Fundametals of Machine Component Design 2nd Edition nad section 5.7 talks about a stepped shaft.
to Ø5.5" will give about a 0.0007" deflection. The customer can keep the same bearings and locations, but will have to modify the sheaves for a larger QD bushing (say M size). I would like to analyze by hand. Can you use Castigliano's Method to solve this? I have textbook Fundametals of Machine Component Design 2nd Edition nad section 5.7 talks about a stepped shaft.Hi Tagger
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
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
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