Deflections of an Uncracked Reinforced Concrete Beam

Background

Now we want to predict (calculate) how much our beam will `deflect' (bend, bow, sag) in the lab. Recall that we are working with (in this example) a beam that is 11 in. wide, 5 in. deep, with 2 - # 3 Gr. 60 bars centered 2 in. from the bottom face (Strength of a Reinforced Concrete Beam). The beam is 7'-4" long and we will place it on supports 6 ft apart, and load it with a concentrated load at midspan (or where ever else we decide to load it).

Deflection Due to Dead Loads and Self Weight

With reinforced concrete there are `two kinds' of loads we need to deal with in regard to deflections. One kind is `sustained' loads, and the other is `transient' (or not sustained). The most common sustained load is Dead load and / or self weight. By sustained we mean `for a long time'. And with regard to sustained loads, there are two kinds of deflection; there is the immediate deflection due to sustained loads, and then creep. Creep is time-dependent additional deflection. It is plastic - in the sense that once it occurs, it stays. So, when we load a beam with sustained loads, it will bend, flex, sag, ... and, in time, bend, flex, sag some more. In the lab we are going to be dealing with short time frames, so we will not worry about creep; only in more permanent situations (such as real buildings) will we worry about about creep. Long term deflections we will probably reserve for another `conversation'.

Let us assume that our beam is not cracked. (This will have more meaning as we continue.) The only Dead load on the beam will be the beam's weight. As we place our beam up on the supports, and let go, it will `sag' under its own weight. Let's calculate this sag (deflection).

Aside from the fact that the beam hangs a bit beyond the supports at both ends, it's essentially a uniformly distributed, simply supported beam. For a beam with constant E and I, we have an equation for the sag (deflection) at mid-span (see, for example, your Ambrose Text, p. 94):

Δ = (5/384) ω L⁴ / E I ... or ... (5/384) W L³ / E I ... where W = ω L.

The above is for a beam without overhangs. For now we will neglect the effect of the overhangs, but simply acknowledge in the backs of our minds that the overhangs will tend to `reverse bend' the beam a bit (reduce the sag).

Gross Moment of Inertia

Again, assuming the beam is not cracked, we will assume, for now, that all concrete acts to resist bending (yes, resists flexural tension in addition to flexural compression). And, for now, with the concrete working throughout, we will assume that the effect of the steel is ... well, at this point, small. (The steel won't really `kick in' in a meaningful way until the concrete cracks.) So, in terms of a structural section, it is all concrete, and it is rectangular. The Moment of Inertia, I, for such a section is, ...

... I = (1/12) b h³ ... and we are going to call this the `Gross MOI' ... I_g.

In our case, then,

... I_g = (1/12) 11 in. (5 in.)³ = 114.6 in.⁴.

I could include the steel, using the so-called `transformed section' approach, but it changes the answer by so little, that for now (and ever) it's not worth it.

Modulus of Elasticity

Let's say we have a concrete compressive strength of ... 3500 psi.

The MOE for concrete (E or E_c) is given by ...

... E_c = 57,000 √f '_c psi ...

The above equation shows the 28-day compressive strength. We may not (and probably will not) break our beams at `exactly' 28 days ... so the number we will put in to our equation will be the `strength for the day' of the experiment. And we'll break some test cylinders to come up with that number.

But, for now, let's say, ... 3500 psi.

So,

... E_c = 57,000 √3500 psi = 3,372,000 psi

We earlier determined the self weight of the beam to be 57 plf.

So,

... Δ _{self weight, immediate} = (5/384) ω L⁴ / E I = ...

... = (5/384) (57/12 lb/in.) (6 x 12 in.)⁴ / (114.6 in.⁴ x 3,372,000 psi) = ...

... = 0.004 in.

What?!!!

Yes, ... four one-thousandths of an inch.

If we include the `stiffness' provided by the rebar, the deflection will be a percent or two less. And if we include the effect of the overhangs, the deflection will be a few (more) percent less. What is a few percent less than 0.004? ... Probably still about 0.004. Small!

Not only is it small ... it will occur before we can measure it. It will occur before we can get the dial gage up there to measure defections.

So, in our lab, the defections we measure will be additional to this immediate dead load deflection, which we can calculate, but not measure. (I guess we could measure it ... we'd have to get our beam up on the supports, and supported fully also between the supports, get the dial gage in place, and then remove the support between the supports, ... and measure the sag as the support between the supports is removed.)

Cracking Moment

Before we go on, let's make sure our section was not cracked. In other words, assuming it didn't get cracked by dropping it while moving it into the lab from the loading dock, it won't be cracked unless we have loaded it up to the `cracking moment', M _cr. The cracking moment is given to be (from the ACI Code),

... M_cr = f_r I_g / y_t ...

where

... f_r is the Modulus of Rupture for the concrete ... = 7.5 √f '_c psi ...

... I_g is the gross section Moment of Inertia, already discussed,

and,

... y_t is the distance from the neutral axis to the extreme fiber in tension.

Since we are at this stage `dealing with' a rectangular section, y_t = h/2 = 5 in./2 = 2.5 in.

And ... f_r= 7.5 √3500 psi = 444 psi.

So,

... M_cr = 444 psi 114.6 in.⁴ / 2.5 in. = 20,350 lb-in. = ...

... M_cr = 1696 lb-ft

In an earlier lesson (here) we found the expression for the load on our reinforced beam ...

M _total = M _{sw, simple span} - M _{sw overhang} + M _{applied ω} + M _{from P} ...

M _total = 258 lb-ft - 12.7 lb-ft + 4.5 ω ft² + 1.5 P ft.

The expression took into account the overhangs, and allowed for a uniformly applied load. If we take out the uniform applied load, we get ...

M _total = 245 lb-ft + 1.5 P ft,

where, recall,

P = the concentrated applied load at mid-span.

If we have not yet applied the concentrated load at mid-span, then P = 0 and ...

M _total = 258 lb-ft.

Since the moment due to self weight is less than the cracking moment (1696 lb-ft), the beam is assumed to be un-cracked.

Uncracked Stiffness

Now let's apply a concentrated load P, at mid-span, but keep it small, for the time being.

In general, for a beam of constant E and I ... under concentrated load P, the deflection is ... (see Ambrose text, p. 94 again) ...

... Δ = P L³ / 48 E I

As long as P doesn't crack the beam,

... Δ _{P, uncracked} = P (6 x 12 in.)³ / (48 x 3,372,000 psi x 114.6 in.⁴) ...

... Δ _{P, uncracked} = P (0.000020 in. / lb).

Let's put in a load of, say, 500 lb.

... Δ _{P = 500 lb, uncracked} = 500 lb (0.000020 in./lb) = 0.010 in.

We could say that this `adds' to the 0.004 in. (that we couldn't measure) due to its self weight. But we can measure the 0.010 in., with our dial gage. It is still a pretty small deflection.

Now, here is the cool part. Let's talk in terms of `stiffness' (k).

It takes 500 lb to deflect the beam 0.01 in. ...

Stiffness is the amount of load required to make a certain deflection ...

In this case, then, ... k = k _{P = 500, uncracked} = 500 lb / 0.010 in. = 50,000 lb/in.

Note, I used 500 lb ... but we could have used 100 lb, or even 1000 lb, or 10,000 lb ... to get the (number) k = 50,000 lb/in.

We need to be a bit careful, though, since, in reality, too great of values of P will indeed crack the beam.

Note: k = 50,000 lb/in. ... is a big number. It takes 50,000 lb to bend the beam 1.0 in. Actually, it will start cracking first. Let's find out when it starts cracking ...

The Load That Cracks The Beam

Okay, the beam didn't crack under its weight, but there is some load P that we could apply, mid-span, that will eventually cause it to crack.

We can get this by going back to our general load equation (above and from earlier lesson, here) ...

M _total = 245 lb-ft + 1.5 P ft.

And now let's set M _total equal to M_cr ...

M _total = 245 lb-ft + 1.5 P = M_cr = 1696 lb-ft.

P_cr = 967 lb.

Let's find the corresponding deflection (just before it cracks) ... call it ... ... Δ _{cr ...}

We can get this number a number of ways ...

First ... if 50,000 lb will bend the beam 1.0 in., then how much will 967 lb? ...

Δ _cr = 967 lb / (50,000 lb per in.) = ...

Δ _cr = 0.019 in.

In equation form ... k = P / Δ ... or ... Δ = P / k.

Or we could go back to the ... Δ = P L³ / 48 E I ... and start from scratch.

Note a couple things: 1) this whole business of ... 0.01 inch for 500 lb, 0.019 in. for 967 lb, 1 in. for 50,000 lb ... assumes the beam is `linear elastic' ... deflection is proportional to load. And, in real life, the behavior of the beam will be linear ... until it cracks! ... and 2) this is small deflection, and a relatively small load ... so be careful as we load up the beam ... load it slowly, or you'll miss the excitement of the first crack!

(And, shudder the thought, if the beam has no reinforcement in it - be REALLY careful - before you know it it will be cracked IN HALF - and on the ground.)

Well, ... enough for now. In the next piece (here) we will look at the `cracked' Moment of Inertia, `cracked' stiffness, and so on.

References

Simplified Engineering for Architects and Builders, Ambrose, J. and P. Tripeny, 10^th edition, John Wiley & Sons, Hoboken, New Jersey.

Building Code Requirements for Structural Concrete, ACI 318, American Concrete Institute, P.O. Box 9094, Farmington hills, Michigan, 48333.

Strength of a Reinforced Concrete Beam, Jeff Filler, Associated Content.

Cracked Section Neutral Axis and Moment of Inertia, Jeff Filler, Associated Content.