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

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SUMMARY

The discussion focuses on calculating the normal axial stress for a 16mm diameter steel bar subjected to a 25kN load. The correct formula for axial stress is derived from the equation σ = F/A, where σ represents stress, F is the force applied, and A is the cross-sectional area. The cross-sectional area for a circular bar is calculated using A = π(d/2)², leading to an area of approximately 201.06 mm² for a 16mm diameter bar. Thus, the normal axial stress is calculated as 25,000 N / 201.06 mm², resulting in approximately 124.1 MPa.

PREREQUISITES
  • Understanding of axial stress and its formula (σ = F/A)
  • Knowledge of cross-sectional area calculations for circular shapes
  • Familiarity with unit conversions, particularly between kN and N
  • Basic principles of mechanics of materials
NEXT STEPS
  • Study the calculation of cross-sectional areas for different shapes, including circles and rectangles
  • Learn about the properties of materials under stress, including yield strength and tensile strength
  • Explore the concept of stress concentration and its implications in structural engineering
  • Investigate the use of software tools for stress analysis, such as ANSYS or SolidWorks
USEFUL FOR

Mechanical engineers, structural engineers, and students studying mechanics of materials will benefit from this discussion, particularly those interested in stress analysis and material properties under load.

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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