Mid-height deflection of a bar - Euler's Formula

Click For Summary
SUMMARY

The discussion focuses on calculating the mid-height deflection of a vertical aluminum bar, 1.0 m in length and with a cross-section of 12.5 mm x 4.8 mm, under axial loading using Euler's formula. The relevant equations include P = EIπ²/L² for critical load and I = bd³/12 for the moment of inertia. The calculated force is 79.6 N, but the deflection calculation yielded an unexpectedly large result. Participants clarified that the bending stress formula should be used instead, and noted that the section reaches full plasticity at a deflection of 45 mm.

PREREQUISITES
  • Understanding of Euler's formula for buckling analysis
  • Knowledge of bending stress calculations
  • Familiarity with the moment of inertia formula, I = bd³/12
  • Basic principles of material yield stress, specifically for aluminum
NEXT STEPS
  • Study the bending stress formula and its application in structural analysis
  • Learn about the implications of material yield strength in design
  • Research the differences between elastic and plastic deformation in materials
  • Explore advanced topics in structural stability and buckling analysis
USEFUL FOR

Engineering students, structural analysts, and professionals involved in materials science and mechanical design will benefit from this discussion, particularly those focused on buckling and deflection calculations in structural components.

SaRaH...
Messages
7
Reaction score
0

Homework Statement


I think I've got part of this question but it's multiple choice and nothing that I've got matches any of the options we were given. I'd really appreciate it if you could help me out.

A straight, vertical aluminium bar, 1.0-m in length and 12.5-mm x 4.8-mm in cross
section, is axially loaded until it buckles. Assuming Euler’s formula applies, determine
the mid-height deflection, in millimetres, of the vertical bar before the material attains
its plastic yield stress of 250-MPa.

Homework Equations



P = EIpi2/L2

I = bd3/12

deflection = PL/AE

The Attempt at a Solution



I = (12.5)(4.8)3/12 = 115.2mm4

P = (70*109)(115.2*10-12)*pi2/(1)2 = 79.6N

Then when I tried to get deflection it came out as a huge answer. I'm not certain if that's the right formula I'm using but it's the only one we've used in class so I don't know what else it could be.

The answers we were given were 57mm, 150mm, 31mm, 145mm, 378mm

Sarah
 
Last edited:
Physics news on Phys.org
SaRaH...: Hint 1: Your third relevant equation is inapplicable; you instead need the bending stress formula. Hint 2: Bending moment is force times distance. You correctly computed the force.

By the way, always leave a space between a numeric value and its following unit symbol. E.g., 79.59 N, not 79.59N. And, e.g., 250 MPa, not 250-MPa. See the international standard for writing units[/color] (ISO 31-0[/color]).
 
For what it's worth, I found that the section is FULLY plastic when deflection is 45 mm. So you could check whether extreme fibres first reach yield stress at 31 mm. I think the question is a bit ambiguous, but it's a well intentioned question.
 
pongo38: I am thinking your statement currently appears incorrect, unless I am misinterpreting. Would you be able to show how you obtained your answer? I currently did not find the question ambiguous.
 
Is th question asking for the deflection when the extreme fibres FIRST reach their yield stress, or when the whole section has gone plastic (about 50% more moment required)?
 
pongo38: Aluminum has no yield plateau. I interpreted the question as meaning it is asking for the deflection when only the extreme fibre reaches the tensile yield strength, Sty = 250 MPa.
 
Last edited: