Deformation of string from impact of an object with constant velocity

[andrestander]

If a car with no acceleration, only velocity, hits my spring with a spring constant K

What will be the deformation in string? How do I calculate the force from impact while acceleration is equal to 0?

 

[nasu]

Once the car hits the spring, it will have an acceleration, produced by the elastic force.
The force will increase as the deformation of the spring increases and will reach some maximum value.
Is this maximum value that you want to calculate?

 

[nil1996]

use work energy theorem.

See the kinetic energy of your car will be transferred to the spring.This energy stored in the the spring is its potential energy denoted by 0.5kx².Here x is the deformation in spring.

so 0.5mv²=0.5kx²

m=mass of your car

from this you can calculate "x"
now
force exerted by the spring = -kx

 

[andrestander]

What if I have a simple beam structure and the object impacts in the middle of the beam? By how much the beam will deform and bend?

 

[PhanthomJay]

What if I have a simple beam structure and the object impacts in the middle of the beam? By how much the beam will deform and bend?

As a minimum, it depends on the material ( E = Young Modulus), and beam geometry (Area Moment of Inertia, I), and length between supports (L), impact speed, etc. Equivalent stiffness (k) of a simply supported beam subject to point loading at midpoint is 48EI/L^3, in units of force/length (N/m in SI).

 

[andrestander]

so 0.5mv^2=0.5kx^2

Where k is the equivalent stiffness matrix for the beam?

 

[PhanthomJay]

so 0.5mv^2=0.5kx^2

Where k is the equivalent stiffness matrix for the beam?

Yes, as I see it, but with a lot of assumptions, like the moving object is rigid and the beam remains within its elastic limit and obeys Hooke's Law when deforming. Also, if the moving object is is subject to gravity forces like for the case where it is moving straight down upon impact, then you must include its change in potential energy as well, that is, 0.5mv^2=0.5kx^2 -mgx.

 

[andrestander]

Yes, as I see it, but with a lot of assumptions, like the moving object is rigid and the beam remains within its elastic limit and obeys Hooke's Law when deforming. Also, if the moving object is is subject to gravity forces like for the case where it is moving straight down upon impact, then you must include its change in potential energy as well, that is, 0.5mv^2=0.5kx^2 -mgx.

Alright I'll do more searches and readings from there. Thanks a lot Jay. :)