Force in reinforcing steel of underutilized beam. 1

[StrEng007]

Scenario: Concrete beam, single layer of bottom reinforcing, Applied Mu = 0.75Mn

Question: Does anyone have a reference for how to calculate the theoretical force in the reinforcing steel of this beam, when subjected to 75% of it's nominal moment capacity?

Let's assume, when fully loaded to it's nominal moment capacity, the steel strain is et=0.005 and concrete strain ec=0.003.

 

[Celt83]

Its an iterative process of adjusting the peak compressive strain and neutral axis depth. You'll need to use a more accurate parabolic stress strain relationship, Eurocode EC2 has one and PCA Notes on ACI 318 has one.

Being a beam finding the compression block force and moment isn't to bad, can use Gauss Integration or break up the stress block into several rectangular strips.

As an initial step you check the elastic transformed section and see if you reach tensile rupture stress of the concrete, if not you can do a normal elastic stress analysis using M*y/I, where y is the distancd from the steel to the transformed centroid and I is for the transformed section, n=Es/Ec





My Personal Open Source Structural Applications:

https://github.com/buddyd16/Structural-Engineering

Open Source Structural GitHub Group:

https://github.com/open-struct-engineer

 

[WARose]

Most concrete texts do this in the introductory sections where the neutral axis is shown how to be calculated then the elastic stresses are determined. I don't know if you have Jack McCormac's text, but he has such a section. (I've got the 3rd edition.)

Also, if you have a masonry text, you'll see working stress design methods to determine it.

 

[IDS]

Assuming the section is under-reinforced, at 75% of design capacity the steel will not be yielded, and the concrete strain will be nowhere near 0.003.

It would therefore be reasonable and conservative to assume linear behaviour in the steel and concrete.

You can find the depth of the neutral axis assuming some arbitrary strain in the concrete (either by iteration, or with a closed form solution), than factor the strain to give the required bending moment, and find the stress in the steel.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[Settingsun]

The earlier replies are right but, if close enough is good enough, 0.75 * nominal yield stress is the one-second answer.

 

[RPMG]

Couple things to note:
[li]The theoretical rebar stress is at service state, not LRFD strength values (x1.5)[/li]
[li]Steel yields at 60 ksi / 29000 ksi = 0.00207 strain[/li]
[li]M = T * jd, where j may not follow the whitmore section assumption[/li]
[li]If you are below the yield stress, this is working stress method, which is similar to CMU ASD design[/li]

 

[r13]

Let's assume, when fully loaded to it's nominal moment capacity, the steel strain is et=0.005 and concrete strain ec=0.003.

The strains provided by OP indicates the stresses are at the limited/ultimate state, thus, agree with steveh49 stated above.

 

[IDS]

The earlier replies are right but, if close enough is good enough, 0.75 * nominal yield stress is the one-second answer.

Well it would be a reasonable approximation, but the 5 second answer (typing the section details into an RC elastic design spreadsheet) would give a much better approximation.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[rapt]

It would depend on the beam, but one I just checked came in at fs = 290MPa at service with Ms = .75Mu and fsy = 500.

So about 58%!

 

[IDS]

It would depend on the beam, but one I just checked came in at fs = 290MPa at service with Ms = .75Mu and fsy = 500.

So about 58%!

That's strange. Using the Australian code, I get 413 MPa at 0.75M_u, or 351 MPa at 0.75Phi.M_u, using the Australian reduction factor of Phi = 0.85.

For the steel at 0.75fy, assuming a triangular stress block, the lever arm will be less, so the moment should be less than 0.75M_u.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[rapt]

IDS,

It would be the phiMu = M* version in AS terminology!

I missed that reinforcement had been added for crack control. That brought it back to 336MPa so 67.2%.

Main point being, it is not 75% as Phi affects the ratio.

 

[IDS]

Main point being, it is not 75% as Phi affects the ratio.

But the change in lever arm affects the ratio in the other direction, so with a phi of 0.9 the stress does come out pretty close to 0.75fy.

But I agree that it is worth spending the extra 4 seconds to calculate it.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[Settingsun]

So we have 82.6% from Doug and 67.2% from Rapt, average 74.9%.

But I think you're talking about different things. The ACI code takes the design capacity as phi.Mn whereas we are at 75% of Mn (Doug's calc) not 75% of phi.Mn (Rapt's calc, I think).

I offered 75% in case it's not very critical (hopefully it isn't; see below). It's not a 5 second task when you're on a forum asking how to do it. Also, while I realise we need to base calculations on something, we gloss over the approximations in:

- Actual load (75% of Mn is an estimate).
- Concrete strength (usually >f'c by unknown amount)
- Stress-strain relationship
- Steel modulus.

Edit: My 5-second calculation gave 74%, for my particular trial section and stress-strain curve (40 MPa, EC2 parabolic-linear, AS3600 limit on concrete stress of 0.9*40MPa).

 

[IDS]

Steve - could you post your cross section details.

I used:
1000 mm wide
300 mm deep
5 no. 20 mm bars top and bottom with 50 mm cover
Esteel = 200000 MPa
Econc = 30000 MPa
40 MPa concrete with linear stress/strain (max concrete stress was < 20 MPa, so I think linear is reasonable).




Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[Settingsun]

I used:

600x400
40 MPa
5*34mm bars to get the same k_u as in the first post. d=540mm.
No top reo.
E_s = 200,000 MPa
Concrete stress gets up to 32 MPa


For yours, I get M_n (aka M_u) = 192 kNm, and 412 MPa (=0.824*F_yield) for M=0.75*Mu

 

[rapt]

Agreed if we use .75 Mn, it is about 400MPa.