Scissor mechanism: static analysis

[serbring]

Hi all,

I have this mechanism and I have to calcutate the spring applied by the spring to stand the force F.
I have done a simple analysis but unfortunately it's wrong since I get the system is never in equilibrium.

http://imageshack.us/photo/my-images/651/scissor2.jpg/

Would anybody help me to understand what's wrong in what I did?

Thanks

 

[Undoubtedly0]

Hi Serbring. Why do you say that the system is never in equilibrium? When the spring is contracted enough, it will exert a vertical force that balances the system.

 

[serbring]

Hi Serbring. Why do you say that the system is never in equilibrium? When the spring is contracted enough, it will exert a vertical force that balances the system.

Hi Undoubtedly0,

Look at the second picture. From the static analysis of the element 2, any horizontal force shoud be zero, thus for the element 1. Therefore the element 1 will not be in equilibrium since there is a resultant moment around the central joint always not equal to zero. Right?

 

[Undoubtedly0]

Hi Undoubtedly0,

Look at the second picture. From the static analysis of the element 2, any horizontal force shoud be zero, thus for the element 1. Therefore the element 1 will not be in equilibrium since there is a resultant moment around the central joint always not equal to zero. Right?

I think your free-body diagrams are missing a possibly non-zero horizontal force at the pin joint at the top left of element 2.

 

[serbring]

I think your free-body diagrams are missing a possibly non-zero horizontal force at the pin joint at the top left of element 2.

I think the force at that pin should be zero because at the element 3, no horizontal force is applied, isn't it?

 

[Studiot]

Should this be in the homework section?

I'd cheerfully help but cannot access the first picture.
Can you not simply post them here for all to see?

 

[serbring]

Should this be in the homework section?

I'd cheerfully help but cannot access the first picture.
Can you not simply post them here for all to see?

it's not an homework, therefore I posted it here.

I have attached the picture! Thanks!

 

[Studiot]

Well I'm have trouble reconciling your diagram with your force diagram.

As I read your mechanism two arms support a table top.
The arms are pin-joined in (at their centres?) to form an X between the table top and the ground.
One end of each arm is pinned to the table top and, vertically below to the ground at the right hand end.
The other end of each arm slides freely (NO friction) at the left hand end.
A compression spring joins the centre pin joint to the ground.

When force F is applied to the table top the left hand end sliders move further left, allowing the table top to settle downwards until the spring compression can support the load.

Is this a fair summary or way off beam?

 

[serbring]

Well I'm have trouble reconciling your diagram with your force diagram.

As I read your mechanism two arms support a table top.
The arms are pin-joined in (at their centres?) to form an X between the table top and the ground.
One end of each arm is pinned to the table top and, vertically below to the ground at the right hand end.
The other end of each arm slides freely (NO friction) at the left hand end.
A compression spring joins the centre pin joint to the ground.

When force F is applied to the table top the left hand end sliders move further left, allowing the table top to settle downwards until the spring compression can support the load.

Is this a fair summary or way off beam?

Hi Studiot, everything is right!

 

[Studiot]

You say this is not homework, so presumably the actual method of solution is not important.

This is a mechanism not an equilibrium structure so analysing its action by the methods of equilibrium structures will not work.

That is the mechanism moves until the applied forces are in equilibrium then it stops moving.
This is a different situation from a structure which deflects under the balance of internal forces and external loads that are always supposed to be in equilibrium.

When analysing a system it is always important to distinguish between a mechanism, where kinematics/dynamics are important and a structure where statics are important.

Remember also the geometry is important in both cases.


The simplest approach here is to note that there are only two forces doing any work. The applied force F and spring force.

Thus the work done by the downwards movement, h, of F = Strain energy stored in the spring, which by geometry compresses by an amount h.

Can you form and solve an equation to show this statement?

 

[serbring]

You say this is not homework, so presumably the actual method of solution is not important.

This is a mechanism not an equilibrium structure so analysing its action by the methods of equilibrium structures will not work.

That is the mechanism moves until the applied forces are in equilibrium then it stops moving.
This is a different situation from a structure which deflects under the balance of internal forces and external loads that are always supposed to be in equilibrium.

When analysing a system it is always important to distinguish between a mechanism, where kinematics/dynamics are important and a structure where statics are important.

Remember also the geometry is important in both cases.


The simplest approach here is to note that there are only two forces doing any work. The applied force F and spring force.

Thus the work done by the downwards movement, h, of F = Strain energy stored in the spring, which by geometry compresses by an amount h.

Can you form and solve an equation to show this statement?

Hi Studiot,
thanks for your reply. I didn't get to solve this mechanism in term of energy. So I'm able to solve this equation:

Fs*hs=F*h
hs=l*sin(teta)=h/2

where hs is the spring deflection and h is the force F displacement. But now I curious to know what I did wrong with the force analysis. Do you have any idea?

 

[Studiot]

Hello serbring,

First let me offer an apology.

I was too hasty in post#10 about the spring deflection. You are correct it is not equal to the downward movement of the force, h, but a fraction of this given by similar triangles. If the cross point is halfway along the arms it is h/2.

So the work done by the displacement of F = F*h.

The energy absorbed by the spring = 1/2 spring constant times (deflection)² = 0.5k(h/2)²

Please note this formulae if you do not know it.

equating

Fh = kh²/8

Fh - kh²/8 = 0

h(F-kh/8) = 0

either h= 0 or which we do not want

or

F- kh/8 = 0

h = 8F/k

 

[serbring]

Hello serbring,

First let me offer an apology.

I was too hasty in post#10 about the spring deflection. You are correct it is not equal to the downward movement of the force, h, but a fraction of this given by similar triangles. If the cross point is halfway along the arms it is h/2.

So the work done by the displacement of F = F*h.

The energy absorbed by the spring = 1/2 spring constant times (deflection)² = 0.5k(h/2)²

Please note this formulae if you do not know it.

equating

Fh = kh²/8

Fh - kh²/8 = 0

h(F-kh/8) = 0

either h= 0 or which we do not want

or

F- kh/8 = 0

h = 8F/k

Thanks Studiot for your kind reply! Don't worry about your previous reply, because it helped me a lot!

 

[serbring]

SSo I want to perform a dynamic calculation with a ground displacement excitation. the force F in this case is F=m*ah, where ah is the body acceleration.

Is it the right formula?

-m*ah*h+1/2*k(h/2-hg)=0

where hg is the ground displacement.

Unfortunately I don't get a linear equation so I think there is something wrong, right?

Thanks

 

[aortucre]

You have to look at the equilibrium of the joint between the two links.

 

[serbring]

You have to look at the equilibrium of the joint between the two links.

I'm sorry but I didn't get what I should do.

 

[serbring]

Could anyone help me, please?

 

[Studiot]

Perhaps if you gave a more detailed explanation of the dynamics?

What sort of ground excitation?

what is ah?

what is m?

 

[serbring]

Perhaps if you gave a more detailed explanation of the dynamics?

What sort of ground excitation?

what is ah?

what is m?

I'm sorry for the missing explanation:
m is the suspended body mass
ah is the suspended mass acceleration

thanks

 

[serbring]

up!

Any suggestion is appreciated

 

[serbring]

Could anybody help me, please?

 

[Studiot]

Look at the length of your posts (3 lines or so) and look at mine (15 lines or so).

Perhaps that has something to do with why I (at least) am having difficulty understanding what you are asking.

 

[6Stang7]

Maybe I'm not understanding you mechanism correctly, but if the spring only supports a compression/tension load and is always in the center of the two cross bars, then the load in the spring will be exactly the same as the vertical load applied at the top (assuming no friction). All horizontal components in the bars will cancel each other out. If the spring is fixed in place and ends up bending, then this is a different problem.

If you want to perform a dynamic calculation, you'll need to apply Newton's Law and write out the differential equation of motion.

Say you have a basic system like this:http://upload.wikimedia.org/wikipedia/commons/9/9f/Mass_spring.png

If the ground's movement with respect to time is y(t), then the differential equation of motion is:

m*x''(t)+k*[x(t)-y(t)]=0

Now, if the spring stays centered in this mechanism and only supports a compression/tension load, then this is a linear non-homogeneous differential equation and can easily be solved using the method of undetermined coefficients or Laplace transformations.

If the spring does not, then the spring (I think) will behave in a non-linear way, and solving the differential equation by hand might be hard/impossible. If this is a case, you could always use a numeric approximation to find a solution.

 

[serbring]

Look at the length of your posts (3 lines or so) and look at mine (15 lines or so).

Perhaps that has something to do with why I (at least) am having difficulty understanding what you are asking.

Thanks Studiot for your reply. I'm sorry for the not clearness of my post, I was convinced, it was clear. I'm trying to make my doubts more clear.
As I stated before I need to perform a dynamic analysis of the the scissor mechanism in the picture, so as 6Stang7 stated, I can use that formula:

m*x''(t)+k*[x(t)-y(t)]=0

where:

x''(t) is the mass displacement, in my case: h''
x(y)-y(t) is the spring deflection, in my case: hs-hg, where: hs is the displacement of the center point of the two cross links and hg is the ground displacement.
Due to the geometry construction: hs=h/2, where h is the top mass displacement.

I need the dynamic motion analysis of h in respect of hg, therefore:
m*h''(t)+k(h/2(t)-hg(t))=0 (1)

is it right?

In another way, starting from a work analysis, I get:
-m*h''(t)*h+1/2*k(h/2(t)-hg(t))^2=0 (2)

but I'm not able to simplify it. What I would like to undestand is:
a) is formula (1) right?
b) how could I get the motion equation from a work a analysis? What's wrong in what I did in formula (2)

May you be so kind to help me in that?

thank you in advance :)