# The compressive stress in a material

I am trying to work out the compressive stress in a material.

I have a strut diameter of 30mm. Length is 2.5m. Load 0f 30kN. Modulus of elasticity for the material is 160GPa.

I need to find the amount it compresses by.

I have came up with the answer: 1.130973355x10^10m

By using Stress/Strain.

I am stuck on the 160GPa. I have wrote it in the equation as: 160x10^9.

But not sure if this is correct. And it should just be 160.

And not sure if I have calculated it correctly.

Can anyone help or give advice?

Thanks.

• Show the formula that you used, and all the numbers that you put into the equation at every stage - 160x10^9 Pa works, but only if you also used 30x10^3 N, and 0.003m – Jonathan R Swift Dec 7 '17 at 20:14

For a one-dimensional linear-elastic case you can use Hooke's Law: $$\varepsilon=\frac{\Delta L}{L_0} \qquad (1)$$ where $L_0$ … initial length of the member, $\Delta L$ … change in length (The convention that I use is, that tensile stresses cause positive strains and compressive stresses cause negative strains.) Assuming a uniform distribution of stress: $$\sigma=\frac{F}{A} \qquad (2)$$ where $\sigma$ … normal stress, $F$ … Force acting on member, $A$ … cross-sectional area of member Now, Young's Modulus is defined as the slope of the stress-strain diagram in the linear-elastic region: $$\sigma=E\cdot \varepsilon \qquad (3)$$
By combining equations (2) and (3) you get the resulting strain: $$\varepsilon =\frac{F}{A\cdot E} \qquad (4)$$ and by inserting equation (1) and solving for $\Delta L$ you get: $$\Delta L= \frac{F\cdot L_0}{A\cdot E} \qquad (5)$$