Allowable stress in my project

  • #1
Ajl
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0

Main Question or Discussion Point

I'm doing a little project which came to a stop.

I calculated expected material stress to be ~10 MPa. (bending)

I know that my material (AlMgSi0,5) should carry well that stress, however I'm interested in max allowable stress for this material.
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Is the formula for max allowable stress correct?:
Sigma_max = R_p0,2 / Ni

Ni = 1,5 (common construction, no safety or health risks)
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Material properties:
Fatigue failure 140 - 155 MPa
Tensile strength 90 - 300 MPa
Yield strength 65 - 250 MPa

Since English is not my forte, which (if any) of these parameters should I use to determine R_p0,2?
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Wish you all a happy new year!

Thank you!
 

Answers and Replies

  • #2
256bits
Gold Member
3,048
1,076
Non-ferrous materials such as AL do not have an endurance limit , so a fatigue strength is listed as that which beyond a certain number of cycles ( in the millions usually ) of applied stress, the material is estimated to have a 100% chance of failure on the next cycle.
Is your load cycled, or repeated, duing the lifetime of your project?

( Ultimate ) tensile strength is that referred to as the axial stress applied for the material to rupture under tension.
Are you designing your project to be under a maximum stress just under failure of the project?

Yield stress is referrred to as the axial stress under which if relaxed, the material will return to its former shape, and above which the material will suffer some permanent plastic deformation.
Do you want your project to not ever deform?
 
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  • #3
Ajl
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Thank you for your extensive reply.

Estimated number of cycles will be low (~15600).

All I want to show(proof) with this calculation is, that the material deformation won't be present. (expected stress < allowable stress)

Wish you all the best in new year.
 

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