Struggling with beam lever arm concept

[Joe117]

Studying for the SE and breaking my brain with:

Mn=fy*As(lever arm)

It makes sense when the forces are equal.
Meanwhile my brain tells me if the lever arm is equal, the distance from the neutral axis has to be equal (oversimplification - a balanced seesaw). However c/d is a range...

If the neutral axis moves depending on As, As*fy=0.85*f'c*a*b - wouldn't the force closest to the NA be higher which would break the forces being equal and Mn=fy*As(lever arm)? For a tensile failure, steel would stretch past 0.005 meaning the steel tension would have to be 0.005/0.003 times the concrete compression?

Or should I just hand-wavey thank the Whitney assumption and get on with life (not seriously, it'll just keep me up 15 min longer every night until it dawns on me, like my sleep tho, so please help)?

 

[BulbTheBuilder]

Think of it this way.
C(compression) = T(Tension)
C=Acf'c T=Asf[sub]y[/sub]

Moment about C or T =Mn
Mn=Tz or Mn=Cz

z is the lever arm. Distance from reference force (T or C) to the centroid other force (C or T)...Refer to attachement

If the neutral axis moves depending on As, As*fy=0.85*f'c*a*b - wouldn't the force closest to the NA be higher

Yes, for equilibrium in terms of Tz=Cz. Refer to attachment for additional info

For a tensile failure, steel would stretch past 0.005 meaning the steel tension would have to be 0.005/0.003 times the concrete compression?

The steel yields for maximum capacity as 0.005 (I think ACI says 0.004 but 0.005 is often used). The 0.005/0.003 ratio or equal triangles works from singly reinforced beams. For double reinforced beams you need to determine if the steel yields, so you know the right stress to use to determine your neutral axis

 

[Tomfh]

Neutral axis is simply the point of zero strain. It's not the pivot point of a see saw.

 

[Settingsun]

Not sure where the misunderstanding is but here are some possibilities.

The neutral axis moves so the internal stresses are in equilibrium with the external actions. If the external actions cause pure bending, the internal tension and compression components will be equal, resulting in no nett axial force.

The general case of calculating bending moment is (T*L_t) + (C*L_c). T = internal tension force; L_t = tension lever arm to neutral axis; similar notation for compression. Since T=C, this can be rewritten as T*(L_t + L_c) = T*L_tc ie tension force * distance between tension and compression resultants. This is exactly what you expect for a force couple. L_t and L_c don't need to be equal.

The equations you posted are for ultimate capacity of under-reinforced concrete sections. In that case, both the steel and concrete are beyond their linear ranges. You can't approach it as though the stress profile looks like the strain diagram. The steel has yielded so the tension force is As*Fy - the neutral axis location isn't in that equation, subject to the section being under-reinforced as assumed. The neutral axis only comes into the calculation of the concrete force, being the location necessary for the compression force in the concrete to match the steel yield force. When you assume a rectangular stress block, the neutral axis location is fairly easy to calculate. BUT, you don't need to calculate the neutral axis depth, because the compression stress block depth is the parameter you need to find the loaction of the compression resultant.

You might come across a formula like M = As * Fy * (0.85*d). This assumes the section is under-reinforced and the concrete stress block depth is 0.3*depth to reinforcement. This gives approximately the the shortest lever arm consistent with the under-reinforced assumption (ie 0.85*d) and only ~10% conservative at worst. Time saver for design.

 

[IDS]

I think it's worth looking more closely at the seesaw analogy.

In a seesaw the forces are (roughly) equal and in the same direction, so for equilibrium they have to be equal distance from the pivot point, so the moments are balanced and the nett moment is zero.

In a beam the forces are of equal magnitude, but in opposite directions, so the moments are additive and the nett moment is equal to the external applied moment. There is therefore no reason for the moments to be equal. They will be in a symmetrical beam of one material, but in a reinforced concrete beam, or an asymmetrical steel beam, the moments about the neutral axis will be different.

It might help to think about a steel T beam, rather than concrete. In that case I think it should be more obvious that centroids of the compressive and tensile zones cannot be equally distanced from the NA, if the forces are equal.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[Joe117]

Ah, not a seesaw... had a troublesome "moment" there.

Thank y'all for the help