Deformation of string from impact of an object with constant velocity

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Discussion Overview

The discussion revolves around the deformation of a spring and a beam structure when impacted by an object moving at a constant velocity. Participants explore the calculations involved in determining the deformation and the forces at play, considering both springs and beams under various conditions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant inquires about calculating the deformation of a spring when a car with constant velocity impacts it, questioning how to determine the force without acceleration.
  • Another participant suggests that upon impact, the car will experience acceleration due to the elastic force, and questions whether the maximum force resulting from this deformation is what the original poster seeks.
  • A participant proposes using the work-energy theorem, stating that the kinetic energy of the car is transferred to the spring, leading to a potential energy equation that can be solved for deformation.
  • There are inquiries about the deformation of a beam structure when impacted, with considerations for material properties and geometry affecting the outcome.
  • One participant mentions the equivalent stiffness of a simply supported beam under point loading, providing a formula that incorporates material and geometric factors.
  • There is a discussion about the assumptions involved in applying the energy equations, including the rigidity of the object and the elastic limits of the beam, as well as the potential energy changes if gravity is a factor.

Areas of Agreement / Disagreement

Participants express various viewpoints on the calculations and assumptions necessary for determining deformation in both springs and beams. There is no consensus on the specific methods or assumptions to be used, indicating that multiple competing views remain.

Contextual Notes

Limitations include assumptions about the rigidity of the impacting object, the elastic limits of the beam, and the applicability of Hooke's Law. The discussion also highlights the need to consider gravitational effects in certain scenarios.

andrestander
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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?
 
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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?
 
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
 
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?
 
andrestander said:
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).
 
so 0.5mv^2=0.5kx^2

Where k is the equivalent stiffness matrix for the beam?
 
andrestander said:
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.
 
PhanthomJay said:
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. :)
 

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