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

• SenseAO
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.

#### SenseAO

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.

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?

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.

The 16mm corresponds to the bar

The length of the bar?

yes, uniformly across the bar

Figured it out.