1
$\begingroup$

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.

$\endgroup$
1
  • 2
    $\begingroup$ 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 $\endgroup$ Commented Dec 7, 2017 at 20:14

1 Answer 1

1
$\begingroup$

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: $$ \Delta L= \frac{F\cdot L_0}{A\cdot E} \qquad (5) $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.