Calculating reinforcing actual stress

[mjl23]

How do you calculate the actual stress in reinforcing under a given load? I have plenty of references showing how to get the capacity, but not the actual stress.

I know stress is P/A + M/S.

But let's say we have a reinforcing bar in the middle of a 12x12 beam/column.

For sake of this example what if P is zero and M is large.

I guess I'm confusing myself on the basics here with the situation where the reinforcing is at the c.g. of the section, which I presume is not the neutral axis.

If I missed some required information feel free to state some assumption, hopefully I can figure this out with a nudge in the right direction.

I'm doing this exercise in the first place because part of the code I'm working uses the actual reinforcing stress to calculate the required lap length.

 

[Qshake]

Sounds like you've never quite mastered or were never shown the strain compatibility concept. Check out some older concrete design books such as Wang and Salmon in the 80s.

Otherwise check out how a moment interaction diagram is developed as it is quite similar to what you want.

Regards,
Qshake

Eng-Tips Forums:Real Solutions for Real Problems Really Quick.

 

[DaveAtkins]

Stress is proportional to strain, in the elastic range (which is the only time you would calculate an actual stress in a reinforcing bar).

For bending, the further you are from the neutral axis, the larger the strain, and therefore, the larger the stress.

For axial loads, strain in the reinforcing bar must be the same as strain in the concrete.

Qshake's suggestion is good--any reinforced concrete textbook should cover this stuff.

DaveAtkins

 

[mjl23]

Thanks guys. I'll try looking again, but like you said I am missing something in the basic concept.
When you construct the interaction diagram you are dealing with strains/stresses assuming a stress/strain in either the concrete or steel (either the steel reached it's allowable limit or the concrete did). Really you start (at least I did when I made the interaction diagram) with assuming various values of k (various points for the neutral axis) and then go through all the compatibility equations to come up with the axial-moment capacity.

That approach does not seem to work for what I am talking about here. The concrete or steel would not be at one of the limits. Let's say I calculate the stress at the extreme fiber of the concrete using P/A + M/S to give the stress in the concrete at the face. How do I get the neutral axis corresponding to that point, so I can calculate the corresponding stress in the steel?

Like you said there's a basic point that I didn't imprint here.

 

[Tomfh]

If you read a reinforced concrete textbook it will explain it. It's difficult to answer in forum.

 

[shin25]

mjl23,

Your concept is completely right if the section is uncracked, but completely wrong if the section is cracked. The steel is about 9 times stiffer than the concrete. So the stress in steel will be about nine times than the stress in surrounding concrete.

If the section is uncracked, the stress at NA due to moment is zero, so the stress in the stess at this location should be n*P/A.

 

[mjl23]

This discussion should be equivalent for masonry or concrete. I'm posting my solution with some masonry symbols in case the solution is helpful to others or justifies scrutiny by others.

My solution to determine the actual stress in reinforcing steel due to an applied moment and axial force:
[tt]
t = depth of masonry beam

f_b = M/S - P/A = M (t/2) / I - P/A

[η] = E_s / E_m

[ρ] = A_s / (b d)

k = [2 ([ρ] [η]) + ([ρ] [η])²]^1/2 - ([ρ] [η])

f_s = [η] f_b [(1-k)/k]

[/tt]

I believe this addresses the comment of shin25 since [η] is in place to simulate the equivalent area of steel to masonry.

 

[Lion06]

that is using service load stresses, not ultimate.

 

[shin25]

mjl23,

The equations that you are writing for k and fs are for cracked sections. Where as equation for fb that you have written is for uncracked section.

To decide which equation is to use, you have to find out if the section is cracked or uncracked.

 

[mjl23]

Thanks for your help.

shin25
After reviewing an old Wang & Salmon book, I find where it is mentioned that Mc/I is applicable (only?) when using a transformed section, which I guess would be equivalent to the uncracked section.
Can you enlighten me and explain why the P/A + Mc/I is only applicable for an uncracked section? (something to do with not knowing I?)

Of the references I've been through the caveat is always mentioned "only applies to pure bending". Which is I guess why the earlier post suggested looking at deconstructing the development of the P-M diagram. I presume I could start with a given P-M and work backwards in the usual construction of the PM diagram.

When you mention needing to know if the section is cracked or not, are you just talking about checking that the concrete/masonry hasn't exceeded the allowable?

StructuralEIT
I may be missing your point. My question is about calculating the actual stress in the reinforcing, so I don't know why you would mention 'ultimate'.

 

[Lion06]

Mc/I assumes the entire section is acting.
In concrete, you assume that the concrete carries NO TENSILE STRESSES, only the reinforcement carries tension. This is not actually true until the concrete cracks, but the concrete will crack.

 

[40818]

I will throw in my halfpenny about steel reinforced concrete analysis:

Concrete is a brittle and very weak in tension, and so a pure concrete beam has virtually zero bending strength as cracking will occur at the extreme tension side. The general way to overcome this is to embedd stell rods in the tension fibres of a concrete beam. When the concrete is cast around a steel rod, the setting of the concrete shrinks and grips the steel rods. As the coefficients of linear expansion of concrete and steel are nearly equal, only negligible stresses are induced through thermal loading.
The bending of an ordinary reinforced concrete beam may be treated on the basis of transformed sections.

if you can visualise a rectangular concrete beam of breadth "b" and height from the top concrete fibre to the centre of the steel reinforcing "h". The beam is bent so that tensile stresses occur in the lower fibres. the total area of cross section of the steel rods is "A". Compressive stresses are set up in the concrete above the neutral axis. The concrete below the neutral axis is assumed to crack and become innefectual, so the tenile stresses are carried in the bars.
Call "m" the ratio of youngs modulus of steel "Es", to youngs modulus of concrete "Ec", so that m=(Es/Ec)
If the area of steel is transformed to concrete, its equivalent area is "mA" the depth of the neutral axis "Ixx" beloe the extreme upper fibres is "n". the equivalent concrete area "mA" on the tension side of the beam is concentrated aproximately at the depth "h".
The neutral axis of the beam occurs at the centroid of the equivalent concrete beam so that:

(bn) x (1/2n) = (mA(h-n))

So that "n" is the root of the queadratic equation:

(1/2bn^2)+(mAn)-(mAh)=0

And so the relevent root is :

n=(mA/b)(sqrt(1+(2bh/mA)-1))

The second moment of area of the equivalent concrete beam is:

Ixx = (1/3bn^3)+((mA(h-n)^2)

The max compressive stress in the upper fibres is:
"M" is the bending moment

Comp Stress = (Mn/Ixx)

The tensile stress in the steel is:

Tens Stress = (M(h-n)/Ixx) x (Es/Ec) = (mM(h-n)/Ixx)

This is just the bending component, and you have to add/subtract the effects of any endloads within the section as normal, i.e P/A

 

[Lion06]

I have a method different from that of 40818 that I will post as soon as I get a chance today.

 

[shin25]

mjl23,

Per code, the concrete section will be uncracked if the stress in the concrete due to external forces is less than 7.5*sqrt(f'c). So, if the stress is less than this then you can use P/A+M/S, which considers moment of inertia of gross concrete section. On the other hand, if the stress is more than the cracking stress, then you need to use cracked moment of inertia and related equations.

 

[Lion06]

shin-
where does ACI address this value? This is addressed in PCI, but not in ACI (to my knowledge).

 

[shin25]

Per ACI 18.3.3, the section is defined uncracked if the stress is less than 7.5*sqrt(f'c).

 

[Lion06]

that is pre-stressed, I am not sure that you are allowed to use that same criteria for mild-reinforced members.
Does anyone else know for sure?

 

[civilperson]

5* SQ.RT(f'c), Plain concrete as (22-2) ACI 318-05

 

[shin25]

Also look under the following articles for 7.5*sqrt(f'c)-

ACI-9.5.2.3
AASHTO standard-8.15.2
AASHTO LRFD-5.4.2.6 (use the lower value).

 

[mjl23]

What if it's masonry?

I couldn't find anything similar in ACI 530-05. I think a previous edition of ACI 530 may have used f_s in the derivation of the bar development length, but the 05 version does not.

The Florida Building Code appears to still reference what I suspect to be the older ACI 530 formula:

2107.2.3 ACI 530/ASCE 5/TMS 402, Section 2.1.10.6.1.1, lap splices. 
The minimum length of lap splices for reinforcing bars in tension or compression, l[sub]ld[/sub] , shall be calculated by Equation 21-2, but shall not be less than 15 inches (380 mm). 
l[sub]ld[/sub] = 0.002 d[sub]b[/sub] f[sub]s[/sub]     (Equation 21-2) 
but not less than 12 inches (305 mm). In no case shall the length of the lapped splice be less than 40 bar diameters. 
where: 
d[sub]b[/sub] = Diameter of reinforcement, inches (mm). 
f[sub]s[/sub] = Computed stress in reinforcement due to design loads, psi (MPa). 
In regions of moment where the design tensile stresses in the reinforcement are greater than 80 percent of the allowable steel tension stress F[sub]s[/sub] , the lap length of splices shall be increased not less than 50 percent of the minimum required length. Other equivalent means of stress transfer to accomplish the same 50-percent increase shall be permitted to be used.

Note development length per ACI 530-05 is:

l[sub]ld[/sub] = 0.13 d[sub]b[/sub][sup]2[/sup] f[sub]y[/sub] gamma / [K sqrt(f'm)]

So, much simpler to calculate per ACI 530-05.