Flexure of Reinforced Concrete Beams

1. Introduction

As already discussed we assign no tensile strength to structural concrete (except in footings and perhaps some other `foundation' elements). Therefore, in something bent (flexed) like a beam, where some of the material is (stretched) in tension, we need to add reinforcement. The bending strength, then, of a reinforced concrete beam is made up of concrete acting in compression in the `compression zone' of the beam and steel in tension in the tension zone. The concrete in the tension zone is given no no credit in the flexural capacity of the beam. It is assume to be `cracked'.

The role of concrete in the tension zone is thus:

1. to `develop' the reinforcement (transfer the tensile stresses to the reinforcement);
   
2. to `hold it in place'; and
   
3. to protect it (say against fire, weather, etc.).
   

As a result of having two different materials providing strength, and some of one of the materials providing no strength at all, and the somewhat unique features of concrete itself, we end up with formulas for bending that are quite different than those for, say, steel and wood. But, here goes ...

2. Flexural Strength

Ref: Ambrose Text, Pages 377 and on ...

I generally use the equivalent of substituting (his) Equation 13.3.4 into 13.3.7 to obtain the following.

R = r f_y (1 - 0.59 r f_y / f '_c),

where

R = Resistance Factor

r = reinforcement ratio = A_s / (b d)

where

A_s is the total amount of tension reinforcement

b = beam width

d = effective depth = distance from compression face to centroid of tension reinforcement

f_y = specified yield stress of the reinforcement

and

f '_c = concrete compressive strength.

(And I'm not the only one who does it this way.)

And, then,

φ Mn = φ b d² R

where ...

φ = strength reduction factor, flexurexure ... 0.90,

M_n is the nominal (`perfect worlflexuralural (moment) strength,

b and d already defined, and

R from above.

The first part of all this (the `R' part) is kind of cumbersome, so I generally pre pre-tabulated values for R as a function of r, f_y, f '_c.

3. Example

Let's calculate the factored moment strength of a reinforced concrete beam that is 8 in. wide, 12 in. deep, and reinforced with 2 - #4 Gr. 60 bars centered 2 in. from the bottom of the beam. Use 3000 psi concrete.

So,

b = 8 in.

d = 12 in. - 2 in. = 10 in.

A_s = 2 (0.196 in.²) = 0.392 in.²

and

r = reinforcement ratio = A_s / (b d) = (0.392 in.²) / (8 in. x 10 in.) = 0.0049

Now, R = r f_y (1 - 0.59 r f_y / f '_c) = 0.0049 (60,000 psi) [ 1 - 0.59 (0.0049)(60/3) ] = 277 psi.

Then,

φ M_n = φ b d² R = 0.9 (8 in.)(10 in.)² 277 psi = 199,400 lb-in.

Or, in lb-ft ... 199,400 lb-in. (1 ft / 12 in.) ...

φ M_n = 16,620 lb-ft.

Note: this is the strength of a section 8 x 10 with 2 - # 4 in the tension zone ... I haven't said how long the beam is, particulars of the load, whatever. The factoflexuralural strength is 16,620 lb-ft as long as the beam is loaded so that the tension zone is in tension!

4. Observations Regarding Flexural Strength

If you play around with the above equations for a while, you begin to observe several things.

1. Beam width `b' is not a big player. Yeah, the equation φ M<sub>n</sub> = φ b d<sup>2</sup> R makes it appear like the strength is proportional to b, but the `R' term has r in it that has b in the denominator so the nearly cancel out and the strength of the beam is only weakly a function of b. In terms of the `algebra' the b only enters in via the r in the second term in R. We will see later, however, that b is important in kind of a `back door' way in that it represents the `amount' of concrete and we need to make sure we have a safe `amount' of concrete. Actually, this will flesh out by making sure we have safe amounts of steel, and we often deal with amounts in terms of our familiar ratio ... steel area to concrete area.
   
2. Effective depth `d' is a big player. In the Mn equation it is to the second power. When we bring in the R we effectively bring a d into the denominator thus canceling one of the two in the numerator. But what we end up with is ... Mn is approximately proportional to d. Stated differently, if we need 20 % more strength in some application, then investigate increasing d by 20 % (placing the steel 20 % farther from the compressive face).
   
3. The steel strenfyh f_y is a huge player. It shows up on R and we see that strength is proportional to R and R is strongly proportionalfyo f_y. Generally we will be using Gr. 60 reinforcement in our cast-in-place stuff. However, if someone tries to talk you into using Gr. 40 steel for your beam, then right off the bat you know it is only going to be 2/3 as strong (bar for bar). (Don't do it!)
   
4. Amount of steel (A_s or r). Amount of steel is a big player. It shows up strong in the equation for R. If you need double the strength, all things equal, then double the amount of steel. If all you have available is Gr. 40 steel for your beam designed for Gr. 60, then you could perhaps investigate putting in 3 - # 4 Gr. 40 ... (3/2 as much steel that is 2/3 as strong). Or, you could increase the strength by 50 % by using size # 5 bar over # 4 (0.31 in.² per bar vs. 0.20).
   
5. The strength of the concrete (f '_c) is a weak player when it comesflexurexure. But, in a real life project, you generally specify an economical (or at least practicf'c f'_c for, all the concrete in the superstructure, or, let's say, the beams, and don't go changing it around beam by beam, and so on.
   

5. Minimum Reinforcement

Generally we end up wflexuralural reinforcement comprising only a few percent of the beam volume. It's because steel is so strong that we can do this. But the Code sets lower limits. If we have too little reinforcement then there is the possibility of some situation that comes along and cracks the concrete, and if there is not enough reinforcement, the reinforcement will rupture immediately also. Not good. Not safe.

Section 10.5.1 of ACI 318 thus requires a minimum amount of reinforcement:

A_{s, min} = 3√f '_c b d / f_y ... But not less than, ... 200 b d / f_y..

Cast in terms of r ...

r _min = 3 √f '_c / f_y ... but not less than ... 200 / f_y.

Back to our example ...

r _min = 3√3000 / (60,000) = 0.0027 ... but not less than ... 200 / 60,000 = 0.0033 ...

So, r _min = 0.0033.

Our r in the example is 0.0049 ... GOOD!

6. Maximum Reinforcement

Steel makes our beams awesomely strong, but if we have too much steel, then, under load, the concrete can begin to crush (in the compression zone) before the steel yields. Steel yielding is much safer than concrete crushing (though both are considered `failure'). So, the Code also limits the maximum amount of reinforcement we can use. For members loaded primarily in flexure (little or no axial load) the Code demands that we limit the steel to an amount in which when the beam is loaded to its strength (concrete at verge of crushing) the steel will (already) be yielded 0.004 in./in. (we engineers call it `strain'). In terms of a ratio ...

r _{max, 0.004} = 0.364 β₁ f 'c / f_y ,

where,

β₁ = established by the Code and is 0.85 for f '_c of 4000 psi and less ... and decreased by 0.05 per thousand f '_c but in no case less than 0.65.

Back to our example,

r _{max, 0.004} = 0.364 β₁f '_c / f_y = 0.364 (0.85) 3000/ 60,000 = 0.0155 ( ... about 1-1/2 percent steel, based on effective depth d).

Our r in the example is 0.0049 ... GOOD!

Note that we have about one third of the maximum amount of steel. Our beam, if it fails, should fail `safely' ... a lot of steel yielding before the concrete crushes. It should, thus, exhibit ductile behavior during load-to-failure.

7. The `Balanced' Section

The so-called `Balanced' section is a section that has just the right amount of steel so that the steel will yield simultaneously with the concrete crushing (in the compression zone). One might argue that it would be the `perfect' use of the two (concrete and steel), or perhaps, the `most economical' use. And you would probably be right. Except that the Code trades a bit of perfection, and economy, for safety, and the strain limit thing described above will demand that the steel yield first (less than the balanced amount) by requiring that the beam has less than the balanced amount.

8. Conclusion

Now we have looflexurallexural strength. Our `look' has been limited to a rectangular beam loaded primarflexureflexure (no axial tension or compression). We can apply basically the same principles toflexurallexural members, with just a few changes (like one-way slabs, joists, etc.). Things don't get too ugly unless we add considerable axial loads.

Note that if we turn the example beam on its side, or upside down, we get way different strengths.

9. References

Reinforcement Development Length or Bond, Jeff Filler, Associated Content.

Simplified Engineering for Architects and Builders, Ambrose, J. TripenyTripthy, 10^th edition, John Wiley &HobokenHoboken, New Jersey.

Building Code Requirements for Structural Concrete, ACI 318, American Concrete Institute, P.O. BoxFarmingtonmington hills, Michigan, 48333.