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rod buckling calculation 2
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I've been looking in Machinery's Handbook and it gives me the calculations for a structural column (i.e. Euler's formula), but I don't know how to relate that to a cylinder. For example, what to substitute for your "k" and "l" values. The google search turned up nothing.
Calman
Calman
calcultr
I didn't use buckling analysis for a long time. But the info in Mchinery's handbook should be enough. You can also look at any machine design book such a Shigley, Norton. This is a very basic subject. What version of Macinery's handbook you have?
I didn't use buckling analysis for a long time. But the info in Mchinery's handbook should be enough. You can also look at any machine design book such a Shigley, Norton. This is a very basic subject. What version of Macinery's handbook you have?
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I have the 24th ed. It shows the formula P=(pi^2*I*E)/l^2.
I just don't know what to use for the effective length "l".
The cylinder is mounted by two pins (one at end of rod, the other at rear of cyl.) and the open center-center is 202".
Anyway,
I appreciate your help israelkk
I just don't know what to use for the effective length "l".
The cylinder is mounted by two pins (one at end of rod, the other at rear of cyl.) and the open center-center is 202".
Anyway,
I appreciate your help israelkk
The effective length will be the maximum piston rod length between the piston head (with the seals) and the support guide (with the seals) in the cylinder body. The worst case is when the piston rod is not extended meaning when its most length is still inside the cylinder and the piston starts its pushing. Your "problem" is to decide what type of "ending" you have (fixed, free etc). Here, a manufacturer technical info may help. To be on the safe side you can take the "both ends free" case.
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Since the cylinder is supported at the two extremes (rod end and rear end), I don't understand why the effective length wouldn't be the distance from pineye to pineye when the rod is fully extended.
Wouldn't the worst case be when the rod is fully extended out of the cylinder? That seems to be the most likely time for buckling failure to occur.
So let's say I used the "both ends free" case. What would that make my "k" factor?
Thanks
Wouldn't the worst case be when the rod is fully extended out of the cylinder? That seems to be the most likely time for buckling failure to occur.
So let's say I used the "both ends free" case. What would that make my "k" factor?
Thanks
In Euler formula there is no "K". Look at page 251 at the bottom of the table and use the "Both ends rounded" you can see that all you need is:
E - Modulus of elasticity
I - Rod section ineria = Pi*d^4/64
l - column maximum length
P=((Pi)^2*I*E)/(L^2)
E - Modulus of elasticity
I - Rod section ineria = Pi*d^4/64
l - column maximum length
P=((Pi)^2*I*E)/(L^2)
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hi there
eulers formula can be adapted for cylinders,
Fcr = pi2 x E X I/L2 ----- (eULERS )
sub. pi x d4/64 for I ( solid round shaft )
and 2.07 x 10*5 foe E (steel )
Fcr = pi*3 x d*4 x 2.07 x 10*5/l*2 x 64
then Fcr = d*4 x 10*5/l*2
then d*4 = Fcr x l*2/1 x 10*5
d = 4th root of Fcr x l*2/1 x 10*5
where Fcr = critical force in newtons
E = modulus of elasticity, N/mm*2
I = moment of inertia of rod mm*4
l = effective length
this gives rod dia for criticalforce, that is the force at which the rod may start to buckle, you must apply a safety factor to be sur that it doe's not collapse
d > 4th root of Fa x l*2 x sf/1 x 10*5
Fa is allowable force in newtons
sf is safety factor
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example.
cyl is 125mm dia p = 20Mpa extended rod 1000mm fixed at bottom free at end load factor 1,5 sf is 4
cyl force = 0,7854x(125)*2x20
= 2,45x10*5 N
Fa = 2,45x10*5x1,5
3,68x10*5 N
rod l = 1000x4
= 4000mm
d = 4th root 3,68x10*5x(4000)*2x4/1x10*5
d = 125mm dia rod
I hope this was of help to you.
greye.
eulers formula can be adapted for cylinders,
Fcr = pi2 x E X I/L2 ----- (eULERS )
sub. pi x d4/64 for I ( solid round shaft )
and 2.07 x 10*5 foe E (steel )
Fcr = pi*3 x d*4 x 2.07 x 10*5/l*2 x 64
then Fcr = d*4 x 10*5/l*2
then d*4 = Fcr x l*2/1 x 10*5
d = 4th root of Fcr x l*2/1 x 10*5
where Fcr = critical force in newtons
E = modulus of elasticity, N/mm*2
I = moment of inertia of rod mm*4
l = effective length
this gives rod dia for criticalforce, that is the force at which the rod may start to buckle, you must apply a safety factor to be sur that it doe's not collapse
d > 4th root of Fa x l*2 x sf/1 x 10*5
Fa is allowable force in newtons
sf is safety factor
-
example.
cyl is 125mm dia p = 20Mpa extended rod 1000mm fixed at bottom free at end load factor 1,5 sf is 4
cyl force = 0,7854x(125)*2x20
= 2,45x10*5 N
Fa = 2,45x10*5x1,5
3,68x10*5 N
rod l = 1000x4
= 4000mm
d = 4th root 3,68x10*5x(4000)*2x4/1x10*5
d = 125mm dia rod
I hope this was of help to you.
greye.
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