springs in series or parallel?

[PietroG.]

Good morning to all,
I am preparing the examination of construction elements of the machines and I came across an exercise of fatigue verification of two helical springs of stiffness k1 and k2 and a flexional spring of rigidity kl, subject to the action of a circular cam that rotates around an eccentric point. the two helical springs are separated by an infinite stiffness plate.
I attach a sketch to make interpretation more clear.
But my problem lies in understanding the arrangement of the springs, and thus understanding how the action of the cam is distributed in the three springs.
Thinking about it, I thought that the two helical springs can be in series, and all in parallel with the flexional spring, but I am not too sure, as I have the bat that creates a little confusion: cry:.
I looked for other discussions on the site that could help me, but I couldn't find anything that would take away my doubt... so if anyone could help me out, I would be infinitely grateful!

 

[Stan9411]

I agree with you

the two helical springs are clearly in series: warn the same force but they will deform different quantities depending on their stiffness
their constant equivalent is keq = k1k2/(k1+k2);

for how the system is made, the arrow of the beam will correspond to the compression of the system of springs. If two springs are at equal deformation, they are in parallel.

the "rigidity of the beam" is 3ei/l^3

ktot = 3ei/l^3 + k1k2/(k1+k2)

 

[PietroG.]

I agree with you

the two helical springs are clearly in series: warn the same force but they will deform different quantities depending on their stiffness
their constant equivalent is keq = k1k2/(k1+k2);

for how the system is made, the arrow of the beam will correspond to the compression of the system of springs. If two springs are at equal deformation, they are in parallel.

the "rigidity of the beam" is 3ei/l^3

ktot = 3ei/l^3 + k1k2/(k1+k2)

Thank you very much ,

 

[meccanicamg]

to look at the system, the two helical springs are in series but according to me is also in series the flexional spring that you can well imagine as another helical spring. and that plate with infinite rigidity simply means that there is no residual deformation of other components.

If the eccentric pushes down 5 mm, it will charge the crossbow and the two springs so that everything is lowered by 5 mm. any of the three springs had a constant elastic tending to zero would collapse the package, so they can not be in parallel.
I don't see the parallel.

two compression springs in parallel are not put one above the other (series) but one next to the other (parallele).

 

[Mattymecc]

any of the three springs had a constant elastic tending to zero would collapse the package, so they can not be in parallel.
I don't see the parallel.

Why would he collapse? the system would oppose a lower force.
it would be less rigid as it would be less one of the two parallel springs, i.e. the elastic beam or the series of k1/k2 springs (whose rigidities are added).

 

[Stan9411]

any of the three springs had a constant elastic tending to zero would collapse the package, so they can not be in parallel.
I don't see the parallel.

Unfortunately it is the opposite.

we assume that none of the 3 springs is preloaded in the position shown in drawing.

if the helical springs had nothing stiffness, it would be the beam to govern the total deformation.. and vice versa. typical of parallel.

Nothing collapses.

 

[stevie]

the end of the flexional lamella and the group of the two helical springs (in series) are subject
to the same shift of the eccentric, therefore undergo the same shift, which is the situation of the springs in parallel.

the "infinite" rigidity plate simply indicates that one should not take into account the axial deformation of the plate, which would still be negligible.

 

[Fabietto]

even for me in parallel.
even if I arrived with a different reasoning:
the helical springs discharge their strength at a ground point along their axis,
the lamella at the point where it is attached to the wall,
therefore 2 different points, and therefore parallel. .
or is it a wrong reasoning?

 

[Stan9411]

Yes, less rigorously, but that's the same.

the juice is that if the springs were not there, the beam would absorb alone the deformation imposed by the cam, not if it falls on the ground.. therefore the beam is a subsystem that works in parallel to the springs.

 

[meccanicamg]

I can't see it yet. the two springs that have different rigidity are in series between them and do not suffer the same lowering between them....ahhh.... but the beam instead moves equal to the pack of the two springs....son of wood ...two series springs that parallel with the lamina. illuminated while writing.

 

[ceschi1959]

I agree with you

the two helical springs are clearly in series: warn the same force but they will deform different quantities depending on their stiffness
their constant equivalent is keq = k1k2/(k1+k2);

for how the system is made, the arrow of the beam will correspond to the compression of the system of springs. If two springs are at equal deformation, they are in parallel.

the "rigidity of the beam" is 3ei/l^3

ktot = 3ei/l^3 + k1k2/(k1+k2)

This report is the force that deforms the system, force that does not seem to have been taken into account in any other response.
f=ktot*f arrow that depends only on eccentricity.
as to whether it is parallel or serial it is intuitive to see that if it increases the rigidity of the springs in series it takes more strength to deform the beam, then the system behaves as said by @stan9411.

 

[Sfiligoy]

the end of the flexional lamella and the group of the two helical springs (in series) are subject
to the same shift of the eccentric, therefore undergo the same shift, which is the situation of the springs in parallel.

the "infinite" rigidity plate simply indicates that one should not take into account the axial deformation of the plate, which would still be negligible.

I agree with those who say that the two springs are in series between them and the lamina in parallel.
I try to say how I have reasoned, even if in the end it is similar to what is said by stevie: the end of the spring 1 is shortened by the same amount of the foil (stex elongation - parallel); spring 1 will react also on the rigid plate and consequently (or balance) we would have the same strength also on spring 2. (same force - series)

 

[PIETRO2002]

I try to say mine,

the lamina will be subjected to alternating bending, the load goes from zero, to maximum strength, when the eccentric has done the maximum run, the force will be given by the compression of the system of the two helical springs in series.

the wire of the helical springs will be subjected to alternating torque, given by the arrow set by the cam, each spring will yield according to the specific stiffness of the single spring.

the problem would divide it into two parts:

calculation of the system two helical springs in series and relative calculation of the single spring.

calculation of lamina.

 

[Stan9411]

the force will be given by the compression of the system of the two helical springs in series.

So if the foil is concrete and the pudding springs, to get the maximum arrow will be enough the 1-2 n necessary to deform the pudding?

lamina works in parallel to the springs. .

 

[PIETRO2002]

the constant equivalent of the system of the two springs in kequivalent series:

1/kequivalent = 1/kmolla1 + 1/kmolla2

 

[PIETRO2002]

So if the foil is concrete and the pudding springs, to get the maximum arrow will be enough the 1-2 n necessary to deform the pudding?

lamina works in parallel to the springs. .

hai ragione. . . .

 

[PIETRO2002]

then system of parallel springs composed of:

Lamina spring + system two helical springs?

 

[PIETRO2002]

then the kequivalent of the system:

kequivalent = k lamina spring + k equivalent system two elecoidal springs

Did I write a cable?

 

[Stan9411]

I wrote exactly the same formula as your post #2.

It seems to me that there is unanimity, however

 

[PietroG.]

You guys, thank you very much for the answers, have been very clear le