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PUSH FORCE CALCULATION
- Thread starter uppili11
- Start date
Any statics and dynamics book will tell almost all what you want about push pull calculations. Physics books have also very good chapters dealing with subject. Newton Laws apply. Friction plays an important role. Do not confuse with push pull ventilation systems available for commecial kitchens such as those availbable for restaurants.
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- #5
The temperature range is -40deg c to 170deg c.(automotive engine mounted application)
Have tried all available equation but not matching with the actual values.In fact the push force is checked at constant rate to avoid any impact load which could mislead.Do not know what is the missing thread .
desertfox:couldnot see any formula in your answer
Have tried all available equation but not matching with the actual values.In fact the push force is checked at constant rate to avoid any impact load which could mislead.Do not know what is the missing thread .
desertfox:couldnot see any formula in your answer
Hi again uppili
The formula that I gave you earlier were for obtaining the force required to make the assembly at room temperature.
I would point out that now you have added a temperature range, at the upper temperature of 170degrees C I calculate
that the brass collar will be loose on the shaft as you have
insufficient intereference between the steel and brass.
Brass expands 1.45 times that of steel for the same temperature rise.
exp coeff of steel= 11x10^-6 /K
exp coeff of brass= 16x10^6 / K
where K=Kelvin
You need to base your intereference between your brass and steel shaft on your maximum working temperature.
I can perphaps help you further if you require, but I will need more information about exactly what your trying to achieve.
regards desertfox
The formula that I gave you earlier were for obtaining the force required to make the assembly at room temperature.
I would point out that now you have added a temperature range, at the upper temperature of 170degrees C I calculate
that the brass collar will be loose on the shaft as you have
insufficient intereference between the steel and brass.
Brass expands 1.45 times that of steel for the same temperature rise.
exp coeff of steel= 11x10^-6 /K
exp coeff of brass= 16x10^6 / K
where K=Kelvin
You need to base your intereference between your brass and steel shaft on your maximum working temperature.
I can perphaps help you further if you require, but I will need more information about exactly what your trying to achieve.
regards desertfox
Hi shortstub
the coefficient of linear expansion that I have quoted are
per degree Kelvin which is what I have written, the only thing I have missed on the coefficient of brass was the minus on the 10^-6
coefficient of linear expansion of brass = 16*10^-6/K
the /k means 16*10^-6 divided by degrees Kelvin
the coefficient of linear expansion that I have quoted are
per degree Kelvin which is what I have written, the only thing I have missed on the coefficient of brass was the minus on the 10^-6
coefficient of linear expansion of brass = 16*10^-6/K
the /k means 16*10^-6 divided by degrees Kelvin
Hi uppili
Having re-checked my calculation it appears that there would still be some interference at 170 degrees C however
at this temperature the axial force required to move the brass collar along the shaft would be reduced.
regards
desertfox
Having re-checked my calculation it appears that there would still be some interference at 170 degrees C however
at this temperature the axial force required to move the brass collar along the shaft would be reduced.
regards
desertfox
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- #10
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- #11
Yes I have done the calculation with the assumption of coefficient of friction as 0.4 for steel and 0.15 for brass.
the push force as per calculation for brass with an interference of 16micron is 70kgs and 126 kgs for 29micron which matches with the actual.However for steel at 4 micron interference as per the calculation it is 102 kgs which in actual case is around 60kgs.any issue on the cOefficient of friction???
Regarding temp:the assy is done at room temperature and push force checking is also done at room temperature.But would like to calculate at the extreme working temperatures as mentioned above which is required to fulfill FMEA .
the push force as per calculation for brass with an interference of 16micron is 70kgs and 126 kgs for 29micron which matches with the actual.However for steel at 4 micron interference as per the calculation it is 102 kgs which in actual case is around 60kgs.any issue on the cOefficient of friction???
Regarding temp:the assy is done at room temperature and push force checking is also done at room temperature.But would like to calculate at the extreme working temperatures as mentioned above which is required to fulfill FMEA .
Further to my reply to desertfox:
You are of course, correct, when you stated that the unit of coefficient of expansion, Alpha, is 1/K. The fractional change of a material is related to Alpha times the change in temperature, or Delta(T). But, because a change in degrees K equals a change in degrees C, it is perfectly legitimate to use Delta(T) in deg C.
The following is an illustration for others interested in this thread. Therefore, please do not misconstrue it as demeaning or challenging your expertise:
Delta(L)/Lo = Alpha x Delta(T),
with Lo representing the original length, and Delta(T)the change of temperature, either in deg C or in deg K!
You are of course, correct, when you stated that the unit of coefficient of expansion, Alpha, is 1/K. The fractional change of a material is related to Alpha times the change in temperature, or Delta(T). But, because a change in degrees K equals a change in degrees C, it is perfectly legitimate to use Delta(T) in deg C.
The following is an illustration for others interested in this thread. Therefore, please do not misconstrue it as demeaning or challenging your expertise:
Delta(L)/Lo = Alpha x Delta(T),
with Lo representing the original length, and Delta(T)the change of temperature, either in deg C or in deg K!
Hi uppili
The references for sliding friction I have here are 0.15 to
0.25 for metal on metal, so while I would say you're close for the brass on steel, you appear to be a bit high for the friction coefficient steel on steel try using 0.24 to start.
Friction is a difficult thing to tie down and you won't always get a very accurate answer, at best these formula give an estimate of the force required and your error could easily be as much as you have seen before.
If you need real accurate figures then you need to run practical tests similar to what you're doing now however the calcs should put you in the right ball park.
For the calculation at 170 degrees C, I assumed that
both the brass and steel were free to expand and calculated the expansion for each material using the formula:-
x= dia x exp coeff x temp change
Once I had this information I calculated the new interference between steel and brass. Assuming the same coefficients etc as before I repeated the calculation for Pc
and found the new axial force required for the assembly based on this new interference.
If you can post all the diameters of the collars and shaft I
can calculate the figures separately and then we can compare
answers if you think that would help.
regards
desertfox
The references for sliding friction I have here are 0.15 to
0.25 for metal on metal, so while I would say you're close for the brass on steel, you appear to be a bit high for the friction coefficient steel on steel try using 0.24 to start.
Friction is a difficult thing to tie down and you won't always get a very accurate answer, at best these formula give an estimate of the force required and your error could easily be as much as you have seen before.
If you need real accurate figures then you need to run practical tests similar to what you're doing now however the calcs should put you in the right ball park.
For the calculation at 170 degrees C, I assumed that
both the brass and steel were free to expand and calculated the expansion for each material using the formula:-
x= dia x exp coeff x temp change
Once I had this information I calculated the new interference between steel and brass. Assuming the same coefficients etc as before I repeated the calculation for Pc
and found the new axial force required for the assembly based on this new interference.
If you can post all the diameters of the collars and shaft I
can calculate the figures separately and then we can compare
answers if you think that would help.
regards
desertfox
- Thread starter
- #16
here are the details.can provide any further info please.
THE DETAILS OF THE SHAFT & COLLARS AS FOLLOWS:
STAINLESS STEEL SHAFT:
DIAMETER: 4.001-4.004 mm
LENGTH : 89 mm
BRASS COLLAR:
OUTER DIAMETER: 6 mm
INNER DIAMETER: 3.975-3.985 mm
LENGTH : 2.45-2.55 mm
STAINLESS STEEL COLLAR:
OUTER DIAMETER: 8 mm
INNER DIAMETER: 3.992-3.997 mm
LENGTH : 1.975-2.025 mm
THE DETAILS OF THE SHAFT & COLLARS AS FOLLOWS:
STAINLESS STEEL SHAFT:
DIAMETER: 4.001-4.004 mm
LENGTH : 89 mm
BRASS COLLAR:
OUTER DIAMETER: 6 mm
INNER DIAMETER: 3.975-3.985 mm
LENGTH : 2.45-2.55 mm
STAINLESS STEEL COLLAR:
OUTER DIAMETER: 8 mm
INNER DIAMETER: 3.992-3.997 mm
LENGTH : 1.975-2.025 mm
Hi uppili
With the information you gave me in the last thread I have calculated axial forces for both the steel and brass collars as follows:-
brass collar= 75.6kg
steel collar= 59.79kg
For the brass collar I used a friction factor of 0.15 and for the steel 0.24.
Interference for the brass collar was 16 microns and for the steel 4 microns.
For the contact diameter I used the average of the two diameters ie:-
for the brass collar and shaft
Dc= 4.001+3.985/(2)=3.993
for the steel collar and shaft
Dc= 4.001+3.997/(2)=3.999
Finally the longest length for each collar was used in finding the axial force.
Now for calculating at the temperature 170 degrees C I assumed that both the brass and steel could expand freely and this would not only change the interference but also the contact diameter Dc and the o.d of the collar.
The steel collar and shaft showed virtually no change in push force at the higher temperature being 59.68kg as compared with 59.79kg at room temp.The brass collar however showed a reduction in push force.Using the shaft dia of 4.001 and collar i.d. of 3.985 before allowing for expansion due to temp rise, which put the diameters up to
steel shaft =4.00760165mm
collar i.d.(brass) =3.994564mm
interference=0.01303765mm
Dc=4.00760165+3.994564/(2) = 4.001082825mm
using a friction factor of 0.15 as before
push force for brass collar now 61.59kg as compared with 75.6kg at room temp.
Obviously these figures can change significantly with the friction coefficient which without a series of practical tests cannot be defined accurately,also the friction factors here are for dry sliding friction which can be dramatically reduced if any oil or grease or any other lubricant comes into contact with the mating parts.
I have worked these figures based on minimum interference and therefore if the forces above are way above what you need for the design then you can take comfort in the fact you have a good safety margin and a good platform to work from.
hope this helps and good luck
regards desertfox
With the information you gave me in the last thread I have calculated axial forces for both the steel and brass collars as follows:-
brass collar= 75.6kg
steel collar= 59.79kg
For the brass collar I used a friction factor of 0.15 and for the steel 0.24.
Interference for the brass collar was 16 microns and for the steel 4 microns.
For the contact diameter I used the average of the two diameters ie:-
for the brass collar and shaft
Dc= 4.001+3.985/(2)=3.993
for the steel collar and shaft
Dc= 4.001+3.997/(2)=3.999
Finally the longest length for each collar was used in finding the axial force.
Now for calculating at the temperature 170 degrees C I assumed that both the brass and steel could expand freely and this would not only change the interference but also the contact diameter Dc and the o.d of the collar.
The steel collar and shaft showed virtually no change in push force at the higher temperature being 59.68kg as compared with 59.79kg at room temp.The brass collar however showed a reduction in push force.Using the shaft dia of 4.001 and collar i.d. of 3.985 before allowing for expansion due to temp rise, which put the diameters up to
steel shaft =4.00760165mm
collar i.d.(brass) =3.994564mm
interference=0.01303765mm
Dc=4.00760165+3.994564/(2) = 4.001082825mm
using a friction factor of 0.15 as before
push force for brass collar now 61.59kg as compared with 75.6kg at room temp.
Obviously these figures can change significantly with the friction coefficient which without a series of practical tests cannot be defined accurately,also the friction factors here are for dry sliding friction which can be dramatically reduced if any oil or grease or any other lubricant comes into contact with the mating parts.
I have worked these figures based on minimum interference and therefore if the forces above are way above what you need for the design then you can take comfort in the fact you have a good safety margin and a good platform to work from.
hope this helps and good luck
regards desertfox
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