# Scissor mechanism: static analysis

1. Oct 11, 2012

### 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

2. Oct 11, 2012

### 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.

3. Oct 11, 2012

### serbring

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?

4. Oct 11, 2012

### Undoubtedly0

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.

5. Oct 12, 2012

### serbring

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

6. Oct 12, 2012

### 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?

7. Oct 12, 2012

### serbring

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

I have attached the picture! Thanks!

#### Attached Files:

• ###### scissor1.jpg
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8. Oct 12, 2012

### Studiot

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?

9. Oct 15, 2012

### serbring

Hi Studiot, everything is right!

10. Oct 15, 2012

### 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?

Last edited: Oct 15, 2012
11. Oct 17, 2012

### serbring

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?

12. Oct 17, 2012

### 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

13. Oct 18, 2012

### serbring

14. Oct 19, 2012

### 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

15. Oct 20, 2012

### aortucre

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

16. Oct 22, 2012

### serbring

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

17. Oct 25, 2012

### serbring

18. Oct 25, 2012

### Studiot

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

What sort of ground excitation?

what is ah?

what is m?

19. Oct 29, 2012

### serbring

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

thanks

20. Nov 5, 2012

### serbring

up!

Any suggestion is appreciated