- #1

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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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The cross sectional area is pi(0.008m)^2 = 2.01x10^-4In summary, the normal axial stress in the steel bar with a diameter of 16mm and resisting a force of 25kN is calculated by dividing the force by the cross-sectional area of the bar. The cross-sectional area can be found using the formula pi(r)^2 where r is the radius (half of the diameter). Therefore, the stress is equal to 25kN divided by 2.01x10^-4 sqm, which is approximately 1.24x10^8 MPa.

- #1

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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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- #2

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Start by looking up the definition of stress.

- #3

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Yes i have and i get the picture in my head. But i just can't place where the 16mm goes.

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Perhaps you're simply missing the implication of the use of the word

- #5

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F/A?

Force being 25kn and the area being 16mmx16mm?

Force being 25kn and the area being 16mmx16mm?

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Good. What precisely does A stand for? It's an area, but the area of what?

- #7

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The cross sectional area of the beam?

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am i somewhat right?

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What does the 16-mm given correspond to? The problem statement implies what the shape is.

- #11

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The 16mm corresponds to the bar

- #12

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The length of the bar?

- #13

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yes, uniformly across the bar

- #14

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Figured it out.

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