# I Difference between Young's Modulus and spring constant?

Tags:
1. Sep 30, 2016

### Metals

I have read that Young's Modulus, like spring constant, is a measure of stiffness (how hard it is to deform a material). Though apparently, Young's Modulus is a way of doing so that applies only to the material and not its shape, where a spring constant value depends on the dimensions of the material. Is Young's Modulus meant to be the same as spring constant but for comparison of materials disregarding dimensions?

If this is true, then why is it that stress and strain incorporate cross-sectional area and length? If it has nothing to do with dimensions and only compares the materials' stiffness, why bring in those values?

Finally, I just want to confirm that spring constant doesn't literally apply only to springs, right? It does apply to any material with ductility/malleability, doesn't it?

2. Sep 30, 2016

### lychette

You are on the right track.
Spring constant (stiffness) relates to an object.
Young's modulus relates (stiffness) to a material

3. Sep 30, 2016

### JBA

At their base the two do relate in that they both represent displacement vs applied force. For Young's Modulus, stress is proportional to the applied force (as one factor) and strain is the amount of displacement in the material due to that amount of stress/force. In a physical application sense they are directly related in that the spring factor displacement is related to the amount of tensile or torsional strain (depending upon the type of spring) in the springs material.

4. Sep 30, 2016

### Staff: Mentor

Here's an example which might help. Consider a rod of length l and cross sectional area A. A force F is applied to the ends of the rod, and the rod stretches to a new length $l+\Delta l$. The stress in the rod is $\sigma = F/A$ and the strain in the rod is $\epsilon = \frac{\Delta l}{l}$. From Hooke's law of elasticity
$$\sigma=E\epsilon$$where E is Young's modulus. So,$$\frac{F}{A}=E\frac{\Delta l}{l}$$or, equivalently,
$$F=\frac{EA}{l}\Delta l$$So, the spring constant for the rod is $$k=\frac{EA}{l}$$
Of course, in terms of material behavior, $\sigma=E\epsilon$ is fundamental and $F=k\Delta l$ is not.

5. Oct 1, 2016

### Metals

My first question was confirmed, thank you, but the second and third are yet to be answered. It's especially the second that confuses me. If Young's Modulus is not related to the dimensions of a material, why does it incorporate them (A in stress and l in strain) whilst spring constant does not?

6. Oct 1, 2016

### Staff: Mentor

You have it backwards. The Young's modulus is a physical property of the material (independent of the shape and size of the material). I doesn't incorporate any dimensions of the material. It is the spring constant that incorporates the dimensions of the material: $$k=\frac{A}{l}E$$

7. Oct 2, 2016

### CWatters

Young modulus has units because it is a ratio of two quantities measured in different units (so they don't cancel). Not because the material has dimensions.