Engineering Calculating stress and strain in complex loading scenario

[Bert2000]

Homework Statement

Stress and strain in complex loading scenario

Relevant Equations

Stress = F / A
Strain = change in length / original length

A bar is 100mm long and has a 20mm by 10mm cross section. It is subject the following complex loading a tensile load of 10,000N along its length

a compressive load of 100,000N on its 100mm by 20mm faces a tensile load of 100,000N on its 100mm by 10mm faces
Calculate the stress and strain on each axis
The Young's modulus, E = 200 GPa

The Poisson's ratio, v = 0.28
Stress = F / cross sectional area

Im ok with that.

Strain = change in length / original length which we don’t know.

 

[berkeman]

Welcome to PF.

What does your textbook say about how to handle multiple stresses on the same member in calculations like this?

 

[Lnewqban]

Welcome, Bert!
The material seems to be steel, which is isotropic.
You have the original unloaded dimensions on each axis.
For the each delta L, you will need to use both, the given Young's modulus and Poisson's ratio.
Consider that, for each axis, the final value of delta L is affected by the longitudinal and perpendicular loads.

Please, see:
https://en.m.wikipedia.org/wiki/Young's_modulus

https://en.m.wikipedia.org/wiki/Poisson's_ratio

 

[Chestermiller]

Homework Statement: Stress and strain in complex loading scenario
Relevant Equations: Stress = F / A
Strain = change in length / original length

A bar is 100mm long and has a 20mm by 10mm cross section. It is subject the following complex loading a tensile load of 10,000N along its length

a compressive load of 100,000N on its 100mm by 20mm faces a tensile load of 100,000N on its 100mm by 10mm faces
Calculate the stress and strain on each axis
The Young's modulus, E = 200 GPa

The Poisson's ratio, v = 0.28
Stress = F / cross sectional area

Im ok with that.

Strain = change in length / original length which we don’t know.

You forgot about the stress-strain relationship. For uni-axial loading, $$\sigma=E\epsilon$$ where $\sigma$ is the axial stress and $\epsilon$ is the axial strain. You also neglected to list the relationship between the axial strain, the transverse strain, and Poisson's ratio. What would you get if you only subjected the bar to the 10000 N axial load alone?