Why do you not shift the NA 7

[CorporalToe]

Hello I was doing a question in order to prep for the PE exam and I ran into this problem which I got wrong.

In this question since the cross section is being loaded axially plus a moment due to the eccentricity, that the neutral axis will shift from the centroid location. However in the solution they do not shift the NA why is this?

 

[rb1957]

NA = zero stress DUE to bending (not zero stress overall)

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[Eng16080]

NA = zero stress DUE to bending (not zero stress overall)

I don't think this is right, but maybe different definitions of neutral axis are being used. For concrete design, my understanding is aligned with the diagram above.

 

[Celt83]

I think this is all terminology semantics.
Here are my definitions:
Neutral Axis = Geometric Property where the first moment of area = 0 --> ∫ y dA = 0 or ∫ x dA = 0 (A Neutral Axis exists for any defined bending axis, the two formulas here are for the geometric x and y axis.)
Elastic Neutral Axis = Location of 0 strain of a fully elastic cross-section under combined actions
Cracked Neutral Axis = Location of 0 strain in a concrete cross-section after the extreme tension fiber of concrete has cracked
Plastic Neutral Axis (Steel) = Location where the cross-section is bisected such that C=T assuming the entire section is loaded in full yield stress in both compression and tension
Plastic Neutral Axis (Concrete) = Location of 0 strain when the extreme compression strain of concrete is at a code defined value (ACI 318 ec = 0.003)

 

[rb1957]

yeah, I'm not up on concrete design ...

but if those funny little circles are rebar ... won't that change the internal stresses ?

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[CorporalToe]

I just want to clear confusion, yes there are those circles but this question doesn’t mention rebar. I think they just used a bad image for the question but just do not worry about the rebar.

Also the thank you Celt83, you helped clear up my understanding.

 

[SirScole]

Hello all, I am a new poster here and a young EIT. I'm not entirely sure if my question should be a separate thread or not, but I believe it pertains to this topic so here goes. As I have been reading through this thread the top question that has been on my mind is when is it appropriate to use P/A-M/S vs P/A+M/S? In the PE handbook that the OP references it defines the normal stress as (+/-) Mc/I, but the usage of plus or minus isn't clear to me right now.

If at first you don't succeed, skydiving is not for you.

 

[CorporalToe]

It’s always P/A-Mc/I, you take M as positive or negative following the convention in the PE handbook, and c is positive or negative depending on if the fiber you are looking at is above or below the centroid.

In the reference handbook the formula for sigma should be sigma = -Mc/I, as long as you are following the sign convention above. If you get a negative sigma that means that Fiber is in compression. And if it is positive that means it is in tension.

 

[rb1957]

see that's where I differ ...

the stress equation is Mc/I ... +ve M and +ve c give +ve stress (by definition).

but in practice we're more interested in the compression stress from bending, so we write -Mc/I.

and then "c" ... using the furthest extreme fiber is correct for symmetric sections or sections with reversed bending,
but conservative for unsymmetric sections

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[SirScole]

The tension vs compression fiber makes sense. Thank you for that explanation CorporalToe

If at first you don't succeed, skydiving is not for you.