Calculating Axial Stress for a 16mm Steel Bar Under 25kN Load

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Homework Help Overview

The discussion revolves around calculating the normal axial stress in a 16mm diameter steel bar subjected to a 25kN load. Participants are exploring the relationship between force, area, and stress in the context of material mechanics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the formula for stress and the significance of the bar's diameter in the calculations. There are questions about the correct interpretation of the area involved in the stress calculation and the shape of the cross-section.

Discussion Status

The discussion is active, with participants providing guidance on the definitions and implications of the terms used. There is a focus on clarifying the area calculation and its dependence on the shape of the cross-section. Some participants are exploring different interpretations of the problem setup.

Contextual Notes

There is a lack of explicit information regarding the shape of the cross-section, which has led to some confusion in the calculations. Participants are navigating through assumptions about the bar's geometry and the implications for the area used in the stress formula.

SenseAO
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A steel bar that is 16mm in diameter is resisting a force of 25kN. What is the normal axial stress in the bar(MPa)


Since MPa = N sqmm would it be 25kn/(0.016m x 0.016m)

Im more after the process on how to get the answer than the answer itself.
 
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Start by looking up the definition of stress.
 
Yes i have and i get the picture in my head. But i just can't place where the 16mm goes.
 
What quantities goes into calculating the stress?

Perhaps you're simply missing the implication of the use of the word diameter.
 
F/A?

Force being 25kn and the area being 16mmx16mm?
 
Good. What precisely does A stand for? It's an area, but the area of what?
 
The cross sectional area of the beam?
 
Right. The cross-sectional area depends on the shape of the cross section. A=0.016m x 0.016m would work if the cross section were square, but is that the case here?
 
Does not specify. So than we would just do 25kn/0.016m = to give us 1562.5kNm converting this to Nmm would give us 1562 x 10^6 Nmm

am i somewhat right?
 
  • #10
No, that's not correct. For one thing, when you divide, the units divide as well, so you end up with kN/m, not kN m.

What does the 16-mm given correspond to? The problem statement implies what the shape is.
 
  • #11
The 16mm corresponds to the bar
 
  • #12
The length of the bar?
 
  • #13
yes, uniformly across the bar
 
  • #14
Figured it out.
 

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