Analysis of a doubly reinforced beam under special conditions!!! No modification is possible 1

[Marc Rogue]

Hello,

I am looking at a doubly reinforced beam 8"x18" with 2#8 bars top and bot witg 2.5" cover everywhere meaning d 15.5". when i do the strain analysis i get a strain in the compression steel of .0018 which isn't quite yielded yet, based on hooks law we have .0018X29000 for the stress in the comp steel but when i sum forces to find my new a for the whitney stress block i get a value of .8 which is above the comp steel. My gut is telling me this way sum of forces equals 0 which is fine but what about the n/a, if my new C is above the top steel that would mean the top steel is in tension and the neutral axis has moved up soo much that it no longer makes sense to me. So the question is how do we calculate the capacity of such beam?

 

[TLHS]

If the top steel is below the concrete compression block but is too high to take as tension steel, just neglect it for design of bending in that direction.

 

[Ingenuity]

Maybe your strain analysis is incorrect.

For 4 ksi concrete, I calc a NA of 3.01", and the top steel is only at <15 ksi of compression and M_n = 109.3 k.ft

Also Robert Hooke's name has an "e" - he was no Captain Hook!

 

[JoshPlumSE]

This is a simple first principles problem. But, I remember it's something that a lot of people struggled with back in college.

1) First Iteration: Assume your "compression steel" is in compression. But, when you're done solving you realized it would be in tension.

2) Second Iteration: You assume the "compression steel force strain as a value based on it's distance from the neutral axis.
Strain in extreme compression fiber of the section = E_c = 0.003 (ACI 22.2.2.1) which is at a distance of d_c from the neutral axis. Strain in the "tension" steel is then Es = d_s *Ec/ d_c.

Check to see if both layers of steel yield. If not, calculate the stress in the bars.

3) Plug these numbers back into your force balance equation. And, re-solve for "a". Compare it to the "a" value from your last iteration.

 

[Marc Rogue]

Hi guys,
Thanks for all your input, I got home and put some pen to paper, this is what I came up with, with the added top steel the section only gain a 3.5% capacity.
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[Ingenuity]

...put some pen to paper, this is what I came up with, with the added top steel the section only gain a 3.5% capacity.

Yep, for 3 ksi concrete, I get M_n = 107.4 k.ft (with top steel), and 103.6 k.ft (when excluding top steel).

 

[Settingsun]

the added top steel the section only gain a 3.5% capacity.

That's generally the order of magnitude for under-reinforced sections. You'd only bother calculating it if you have software that does it automatically, or you're struggling to get something over the line.

Compression reinforcement only has substantial value for strength when the section would be over-reinforced without it. It also reduces creep (and therefore long-term deflection), but deeper beams are usually a cheaper if you have space available.

 

[Marc Rogue]

Thank all for your input, also wouldn't it be necessary to calculate the capacity for negative moment(usually higher than positive) at a beam column joint cast monolithically.