What Is the Equivalent Linear Spring Constant at Steady Deflection?

In summary, the force-deflection relation of a steel helical spring used in an engine is given by \(F(x) = 200x + 50x^2 + 10x^3\), where \(F\) is measured in pounds and \(x\) in inches. If the spring undergoes a steady deflection of \(0.5\) in during operation, the equivalent linear spring constant is \(253.75\) lb/in. This can be found by solving the cubic equation for \(x\) and using Hooke's Law, \(F = kx\).
  • #1
Dustinsfl
2,281
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The force-deflection relation of a steel helical spring used in an engine is found experimentally as \(F(x) = 200x + 50x^2 + 10x^3\), where the force \((F)\) and deflection \((x)\) are measured in pounds and inches, respectively. If the spring undergoes a steady deflection of \(0.5\) in during the operation of the engine, determine the equivalent linear spring constant of the spring at its steady deflection.

If I knew \(F\), I could simply solve the cubic for the real \(x\) and then use the Newtons 2nd law \(F = k_{eq}x\).

In this problem, I don't have \(F\) instead I have the steady deflection. How do I solve this type of problem?

The answer is \(253.75\) lb per in
 
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  • #2
dwsmith said:
The force-deflection relation of a steel helical spring used in an engine is found experimentally as \(F(x) = 200x + 50x^2 + 10x^3\), where the force \((F)\) and deflection \((x)\) are measured in pounds and inches, respectively. If the spring undergoes a steady deflection of \(0.5\) in during the operation of the engine, determine the equivalent linear spring constant of the spring at its steady deflection.

If I knew \(F\), I could simply solve the cubic for the real \(x\) and then use the Newtons 2nd law \(F = k_{eq}x\).

In this problem, I don't have \(F\) instead I have the steady deflection. How do I solve this type of problem?

The answer is \(253.75\) lb per in
What do you mean by "steady deflection?" It sounds like the spring is at a particular point and doesn't move?

Also, F = kx is Hooke's Law, not Newton's.

-Dan
 
  • #3
topsquark said:
What do you mean by "steady deflection?" It sounds like the spring is at a particular point and doesn't move?

Also, F = kx is Hooke's Law, not Newton's.

-Dan

That is just the terminology the book uses.
 

FAQ: What Is the Equivalent Linear Spring Constant at Steady Deflection?

What is an equivalent spring constant?

An equivalent spring constant is a measure of the stiffness of a spring. It represents the amount of force required to stretch a spring by a given distance, and is typically measured in units of newtons per meter (N/m).

How is the equivalent spring constant calculated?

The equivalent spring constant is calculated by dividing the applied force by the resulting change in length of the spring. This is known as Hooke's law, which states that the force applied to a spring is directly proportional to the amount of stretch or compression of the spring.

What factors affect the equivalent spring constant?

The equivalent spring constant can be affected by several factors, including the material of the spring, its dimensions, and its shape. It can also be affected by the temperature and the presence of other forces acting on the spring.

Why is the equivalent spring constant important?

The equivalent spring constant is important because it is a fundamental property of a spring and is used to predict the behavior of a spring in different situations. It is also used in various engineering and scientific applications, such as in the design of suspension systems and measuring the elasticity of materials.

Can the equivalent spring constant change?

Yes, the equivalent spring constant can change depending on the factors mentioned above. For example, if the dimensions of the spring are altered, its equivalent spring constant will also change. Additionally, the equivalent spring constant can change if the material of the spring undergoes a phase change, such as melting or freezing.

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