Strength and Deflection Calculations of an Experimental Beam

Introduction

Our experimental beam is 5.5 in. wide by 9.5 in. deep and will have 1 - # 6 Grade 60 rebar centered 2.0 in. from the bottom face of the beam, or not. (By `or not' I mean that we may experiment with a beam or two that have no reinforcement.) It is 7 ft 0 in. long and will be placed (centered) over supports 6 ft 0 in. apart. We will load the beam with a single concentrated load at mid-span (or whatever other load point we/you agree upon). Our assumed concrete strength is 4000 psi; except we will not know beforehand what our concrete strength is. However, we will actually break some cylinders the day that we break the beams, so we will know the actual strength of the concrete (for the day we break). What this means is that we will estimate the concrete strengths, and, as needed, re-do some calcs based on actual strengths, or at least discuss how actual strength versus estimated strength affects our calculations and what we see in the experiment for actual strengths and predicted strengths of the beams.

Here is the information that we will want to calculate ...

... ω _{self weight} = the beam's weight (uniformly distributed load) ... and ...

... Δ _{self weight} = the deflection of the beam due to its weight.

Note: in general we will not measure this in the lab. Think of it this way - it will occur before we get the instrument up there to measure it.

... P _cr = the concentrated load P to (first) crack the beam (assuming it doesn't crack under its own weight).

... Δ _cr = the deflection of the beam due just before it cracks.

Note: if the beam is not reinforced the experiment is `over' once it cracks. The beam will be in two pieces on the floor. If the beam is reinforced, we'll just begin to see a small, hairline crack near the bottom of the beam probably directly under the concentrated load (mid-span).

Then we want to determine if it will `break' in ... shear, steel yield, or concrete crushing ...

... P_V or P_y or P_c ... whichever happens first ...

where

... P_V = the load P that `shears' the beam ...

... P_y = the load P that causes the steel to yield ...

... P_c = the load P that causes the `extreme fiber' of concrete to crush, in compression.

... and ...

... where `break' isn't necessarily the best choice of words ... as ...

... if the steel yields before the concrete shears or concrete crushes, we will be able to deform it (the steel and beam) further, yielding the steel more. The increased deformation of the beam will increase the stress in the concrete, and we can reach an `ultimate strength' state of steel yielded and concrete crushing ... P_ult.

P_ult = the load P that causes the steel to yield and the concrete to crush.

Along with the these additional `P's' we will want to calculate Δ`s.

... Δ _V = the deflection just before it shears ... (if the steel hasn't yielded first, or concrete crushed) ...

... Δ _y = the deflection just at (before) steel yields ... (if it hasn't sheared or concrete crushed first)

... Δ _c = the deflection just as (before) concrete crushes ... (if we ever reach this point).

Obviously if one event occurs, the others don't (except for P_y and on to P_ult).

In other words, if we calculate that the beam will shear before the steel yields or concrete crushes ... P_y and P_c have no real meaning (and neither do Δ _y and Δ _c ). And so on. If we determine that the steel will yield before the beam shears, then we would calculate P_y and Δ _y. P_V and Δ _V and P_c and Δ _c would have no real meaning. (And we wouldn't even calculate Δ _y and Δ _c).

And to make matters worse, at higher values of stress in the concrete, the concrete is non-linear, so our predictions will be `off' at high concrete stress.

With this in mind, or not, let's do the calcs ...

Self Weight

Our equation for self weight is ...

... ω _{self weight} = γ A_c ...

where

... γ = the specific weight of concrete, taken to be 150 pcf for normal weight concrete with average amounts of reinforcement, and ...

A_c = gross cross section area of the beam.

So,

... ω _{self weight} = 150 pcf (5.5/12 ft x 9.5 ft) = 54.4 lb/ft ... or ... dividing by 12 ... 4.5 lb/in.

... ω _{self weight} = 54.4 lb/ft = 4.5 lb/in.

Deflection Due to Self Weight

The (immediate) deflection due to self weight is ... (uniformly distributed, simply supported beam) ...

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

where ...

E = 57,000 √f 'c psi = 57,000 √4000 psi = 57,000 (63.2 psi) = 3,600,000 psi.

Assuming the beam isn't cracked yet the Moment of Inertia, I, is taken to be that of simply a rectangular section ...

I = bh³ / 12 = 5.5 in. (9.5)³ / 12 = 393 in.⁴.

So,

Δ _{self weight} = (5/384) 4.5 lb/in. (6 x 12 in.)⁴ / (3,600,000 psi x 393 in.⁴) = 0.0011 in.

Δ _{self weight} = 0.0011 in.

Load P that Cracks the Beam, P_cr

Assuming the beam doesn't crack under its own weight, we will proceed to load it with a concentrated load at mid-span until it does. To determine the load P to crack it, we will come up with an expression for the bending moment load due to P, add in the moment due to the self weight of the beam, and set this expression equal to the moment strength of the beam just at cracking, M_cr.

First the load expression ...

The moment due to uniformly distributed loads on a simple beam (in this case the self weight) is ...

M = ω L² /8.

The moment due to a point load (P) at midspan, for a simple beam is ...

M = PL/4.

The expression for the total load is, thus,

M = ω _{self weight} L² /8 + PL/4.

In our case ...

M = 54.4 lb/ft (6 ft)² /8 + P (6 ft) / 4 = ...

M = 245 lb-ft + 1.5 P ft.

Now, for the moment strength at cracking ... we will set the stress under load, f ... where ...

... f = M/S ...

... equal to f_r the rupture strength (modulus of rupture) of the concrete, where

... f_r = 7.5 √f 'c psi ... (from the Code).

So, ...

... f_r= 7.5 √4000 psi = 474 psi.

S is our familiar section modulus I/c which is, for a rectangular section ...

S = bh²/6.

So, in our case ...

S = 5.5 in. (9.5 in.)² / 6 = 82.7 in.³.

So,

M_cr= f_r S = 474 psi x 82.7 in.³ = 39,200 lb-in. = 3266 lb-ft.

Now let's set load = strength and calculate P ...

M = 245 lb-ft + 1.5 P ft = M_cr = 3266 lb-ft ...

Solving for P ...

... 1.5 P ft = 3266 lb-ft - 245 lb-ft = 3021 lb-ft.

P = 3021 lb-ft / 1.5 ft = 2014 lb.

Call it ...

P_cr = 2014 lb.

The calculated / predicted load that will crack the beam is ... 2014 lb (call it 2010 lb).

If the beam has no reinforcement ... it will crack in half.

If the beam is reinforced ... a crack will appear but the reinforcement (hopefully / theoretically) will keep it together.

Deflection Just Before Cracking

The deflection of the beam just before it cracks (sag; how much it is `bent') will be made up of the deflection due to self weight, plus deflection due to the point load `P'.

We have already determined the deflection due to self weight ... 0.0011 in.

The deflection due to P is ... in terms of an equation ...

... Δ _P = PL³ / 48EI, in general.

For our situation, at P = Pcr = 2014 lb ...

... Δ _cr = P_cr L³ / 48 EI = 2014 lb (6 x 12 in.)³ / (48 x 3,600,000 psi x 393 in.⁴) =

... Δ _cr = 0.011 in.

Hmmm ... about 10 times as much (as that due to self weight).

Note: the total deflection is 0.011 + 0.0011 ... but, remember, the 0.0011 happened before we could measure it.

The Cracked Section

Once the beam cracks, it is either on the floor in two pieces, if not reinforced, or held together by reinforcement. For reinforced beams (provided they have enough reinforcement, another discussion), we assume that the section completely cracks ... meaning that no concrete carries tension, the tension is carried by the steel, and that compression is carried by the concrete above the neutral axis. We will handle the cracked section with `transformed section analysis', where we `transform' the steel into an equivalent amount of concrete, and this `equivalent' concrete we do let carry tension. The in-real-life fact that not all the beam is this cracked ... we will deal with later.

TRANSFORMED SECTION ANALYSIS

Amount of steel ... A_s ... is that for 1 - # 6 ... 0.44 in.² ...

Equivalent amount of steel is ... n A_s,

where ...

n = E_s / E_c

where ...

E_s is the Modulus of Elasticity for Steel ... 29,000,000 psi,

and ...

E_c = 3,600,000 psi, for our 4,000 psi concrete (from before).

So,

... n = 29,000,000 / 3,600,000 = 8.06.

So, essentially, what we are saying, is that the steel acts like 8.06 times as much concrete ... but this concrete can carry tension.

... n A_s= 8.06 (0.44 in.²) = 3.54 in.².

So, in our transformed beam we have some amount of concrete above then neutral axis acting in compression, and 3.54 in.² of `equivalent' concrete acting in tension, where the steel is.

What gets wild is that we don't know where the neutral axis is. When the concrete `cracks' we lose section. We lose section in the tension zone (lower part of the beam), and thus the neutral axis moves up (to keep things in balance).

To find this new neutral axis, we will set up a `moment equation' that will `balance' the (moment of the area above the) neutral axis with the (moment of the) (transformed) area below.

Here is is ...

Area above ... y_cen b ...

where

... y_cen is so called because it is the distance (y) from the top of the beam down to the neutral axis, also called the centroid.

The moment of this area with respect to this new neutral axis is ...

... (y_cen /2)( y_cen b) ...

... where (y_cen /2) is simply the distance up to the `center' of this area in compression,

and,

... b = 5.5 in

The area of the steel-transformed-to-concrete is, already calculated ... 3.54 in.².

Its moment about the new neutral axis is ...

... (d - y_cen )(3.54 in.²),

where

... d is our 7.5 in. in this example.

So, now we can `balance' the (moments of the) areas

... (y_cen /2)( y_cen 5.5 in.) = (7.5 - y_cen)(3.54 in.²).

Simplifying a bit ...

... 2.75(y_cen)² = 7.5 - y_cen)(3.54 in.²) ...

Solving (by trial and error), we get ...

... y_cen = 2.53 in.

So, the neutral axis has moved up from about ½ of 9.5 or 4.75 in. from the top, ... to 2.53 in. from the top.

The section of the beam under compression is thus 2.53 in. deep (and 5.5 in. wide) ... and all the tension is assumed to be carried by the 3.54 in.² of transformed stuff 7.5 in. down from the top, or 7.5 - 2.53 = 4.97 in. below this new neutral axis.

The section strains (deforms) uniformly in both directions (up and down) from this new neutral axis, and in the `linear' (elastic) range of behavior, so do the stresses. And we can now calculate these stresses ...

... σ = M y / I ... at any location ...

... where σ is the stress ... M is the moment acting on the section at some `stage' of loading, ... y is the distance from the neutral axis ... and I is the Moment of Inertia.

First let's get I ...

This will be the I for the concrete with respect to the new neutral axis, plus the I for the n A_s with respect to the new neutral axis ...

I for the concrete ... the Moment of Inertial of a rectangle about its base ...

I = bh³/3 = (5.5 in.)(2.53 in.)³ / 3 = 29.7 in.4 ...

I for the n A_s ... is the area times the distance squared ...

I = (3.54 in.²)(4.97 in.)² = 87.4 in.4.

Adding these ... gives ... I = 29.7 + 87.4 = 117 in.⁴. ... we call it the `cracked' Moment of Inertia ... I_cr.

I_cr = 117 in.⁴ ...

Note that is way less than the I = 393 in.⁴ from before (cracking).

Now we can calculate the stress at any location at any stage of loading.

In particular, we want to look at extreme (`extreme fiber') stresses ... these will be at the `top' of the concrete ... y_c = 2.53 in. ...

... and ...

... at the steel-transformed-concrete ... y_s->c = 4.97 in. (see above).

So for any moment M, the extreme stresses in the concrete and steel-transformed-to-concrete are ...

Concrete ... call it f_c ... f_c = (2.53 in. / 117 in.4) M

Steel ... call it f_s->c ... f_s->c = (4.97 / 117) M ...

WAIT! ... we want to look at steel yielding, not ` steel-transformed-to-concrete' ... so we'll transform back. And we'll do so by multiplying by the `n' ... (since there is actually one-n^th as much actual area, it will feel n times as much stress).

So, f_s = 8.055 f_s->c ...

So ... f_s = 8.055 (4.97 / 117) M.

Now we are in a position to determine the moments that will bring the cracked section section up to the stress levels that crush concrete and / or yield steel ... and from those moments, the corresponding P values.

The P that Crushes Concrete

To bring the extreme fiber to f '_c = 4000 psi ...

... f_c = f '_c = 4000 psi = (2.53 in. / 117 in.⁴) M, ...

or ...

M = 4000 psi (117 in.⁴ ) / 2.53 in. = 185,000 lb-in. = 15,415 lb-ft.

( ... in equation form ... M_c = f '_c I_cr / y_c ...)

The load P that corresponds to this ... we can get from ... (remember)

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

So,

15,415 lb-ft = 245 lb-ft + 1.5 P ft ...

P = (15,415 - 245 ) / 1.5 = 10,113 lb ... round to 10,100 lb ... call it P_c.

Note: as mentioned, concrete is not linear at higher compressive stresses, so this calc is not exact. Further, the concrete is able to take on more load (not necessarily stress), if we can keep loading the steel. Right now it `appears' that if we doubled the steel, we would be limited at where we are at presently by the concrete. Interestingly, if we double the steel, we will be able to double the capacity of the beam, without having to increase the concrete strength. The non-linearity of the concrete will allow it to take on more load without increasing the `extreme' stress. But I'll save that for later.

The P that Yields the Steel

To determine the load that yields the steel ... we will let f_s = f_y = 60,000 psi ...

... 60,000 psi = (4.97 in. / 117 in.4 ) M (8.055) ...

or ...

... M = 60,000 psi (117 in.⁴ / 4.97 in.) / 8.055 ...

... M = 175,353 lb-in. = 14,612 lb-ft = M_y.

( ... in equation form ... M_y = f_y I_cr / y_s ... where y_s is the y_s->c = 4.97 in. above)

The corresponding load P ...

14,612 lb-ft = 245 lb-ft + 1.5 P ft ...

P = (14,612 lb-ft - 245 lb-ft ) / 1.5 ft = 9579 lb ... Call it P_y ... and round to 9600 lb.

Load to Shear the Beam

To determine P the load that shears the beam, we need to find the shear from the self weight, plus the shear due to P, and set it (the sum) equal to the shear strength of the beam.

For the loads ...

V = ω L /2 (from the uniform self weight) + P/2 (half of P goes each way) ...

So,

V = 54.4 lb/ft (6 ft / 2) + P / 2 = ...

V = 163 lb + P / 2 ...

The shear strength is given to be ...

V_c = 2 b d √f 'c psi ... (from the Code) ...

where,

... b = 5.5 in.,

... d = effective depth (top of beam down to center of bottom reinforcement) = 9.5 - 2 = 7.5 in..

So,

V_c = 2 (5.5 in.)(7.5 in.) √4000 psi = 5218 lb.

Set shear load = shear strength ... and solve for P ...

V = 163 lb + P / 2 = V_c = 5218 lb ...

... P/2 = 5218 lb - 163 lb = 5055 lb.

So,

P = P_V = 10,110 ... call it 10,100 lb.

Interesting that it is so close to the other values ... P_c and P_y.

Summarizing ...

P_c = 10,100 lb.

P_y = 9600 lb.

P_V = 10,100 lb.

The `event' that will happen `first' will be the lowest of these (P_c, P_y, and P_V) ... Controlling P = Py = 9579 lb. We expect the steel to yield before the concrete crushes or the concrete shears ... but not by much! Call it P = 9600 lb.

Ultimate Load

We can also calculate the `ultimate' load on the beam ... one that brings both concrete crushing and steel yield. To get this we will use our equation ...

M_n = bd² R ...

where

R = ρ f_y (1 - 0.59 ρ f_y / f '_c) ...

... ρ = A_s / (b d) = 0.44 in.² / (5.5 in. x 7.5 in.) = 0.01067

R = 0.01067 (60,000 psi) (1 - 0.59 (0.01067)(60,000)/4000) = 580 psi.

M_n = bd² R = 5.5 in. (7.5 in.)² 580 psi = 179,400 lb-in. = 14,950 lb-ft.

( ... in equation form ... M_ult = M_n = bd² ρ f_y (1 - 0.59 ρ f_y / f '_c) ...

This can be massaged into the following form ...

M_ult = bd² (A_s/bd)f_y [1- 0.59(A_s/bd)f_y/f '_c ]=

... = d (A_s f_y)[1 - .59(A_s /bd)f_y/f '_c ]

... = d T [ 1 - blah, blah ] ...)

where T is the tension force in the steel.

The `blah, blah' term is somewhat of a small number, so the 1 - blah, blah is a number a bit less than 1 ... so the Moment at steel yield is just a bit less than the tension force in the steel at yield, times the effective depth, d.)

The corresponding P is ...

M = 14,950 lb-ft = 245 lb-ft + 1.5 P ft ...

P = (14,950 - 245) lb-ft / 1.5 ft = 10,100 lb ... call it P_ult.

Interesting ... since P_y and P_c are pretty close ... So is P_y and P_ult. The beam will deform more ... not taking on much more load, and concrete starting to crush more and more.

And ... the dang thing is getting pretty close to shearing also ...

Summary (mini-summary):

P_cr = 2010 lb

P_c = 10,100 lb

P_y = 9600 lb

P_V = 10,100 lb

P_ult = 10,100 lb

So, we'll set the beam up on the supports ... and it will sag 0.0011 in. under its own weight.

Then we'll load it with a concentrated load at midspan. At a load of P = 2010 lb we'll anticipate the first crack. The crack will probably be in the bottom zone of the beam at mid-span under the concentrated load point - for this is where the moment in the beam is maximum. (This is, of course, with reinforcement in the beam. Without it, the beam will just break in half without warning.) The deflection of the beam just at (before) first crack will be 0.011 in. (plus 0.0011 in. from self weight ... but the self-weight deflection will occur before we get the dial gage up there, so all there will be to measure will be the deflection occurring after the immediate deflection due to self weight.)

Then we'll load the beam more ... more and more `P'. At a value of 9600 lb we will expect the steel to yield. This will be characterized by the cracks in the beam, particularly the one(s) under the load point, getting wider than hairline. We'll also see the beam taking on not much more load. We can keep `pumping' on the hydraulic cylinder that loads the beam, but all we will do will be deflect it more. Once the steel yields the deflections become `indeterminate', both in `elastic theory', and because, now, more and more `bent-ness' will be associated in essentially the same load P.

as we bend the beam more and more, we'll crush concrete, and, perhaps it will stay enough together to even shear (since PV was pretty close to P_y, P_c, and P_ult).

So, that said, let's calculate the deflection just before steel yield. That we can determine by elastic theory (at east approximately :) ... you'll see).

Deflection at Failure ( ... in this case ... Steel Yield)

EFFECTIVE MOMENT OF INERTIA

Okay, the beam will have deep vertical cracks near the mid-span of the beam, but as we move toward the supports, where the bending moments are less, the beam will be less cracked. In other words, even though we have gone through the trouble of calculating the Moment of Inertia for a (fully) cracked section, the beam as a whole is not that cracked. So ... the Code gives us an equation that deals with this ... I_e = I_eff ... the `Effective' Moment of Inertia ...

For our beam ...

I _e = I_g (Mcr/Ma)³ + Icr [1 - (Mcr/Ma)³] ...

where

Ig is our `gross' (un-cracked) moment of inertia ... in our case 393 in.4.

Mcr = the `cracking moment' ... in our case ... 3266 lb-ft (from above).

Ma is the applied load at any stage of loading ... in this case, M_y (steel yield) = 14,612 lb-ft ...

So, ...

I_e = 393 (3266 / 14,612)³ + 117 [ 1 - (3266 / 14,612)³ ] = 393 (0.2235)³ + 117 [ 1 - (0.2235)³ ] =

... I_e= 393 (0.011) + 117 [ 1- 0.011 ] = 4 + 116 = 120 in.⁴.

Wow ... this shows that it's (the whole beam) is probably pretty cracked ... I_e pretty close to I_g.

The deflection of the beam will be made up of the deflection due to self weight, and the `deflection due to P'. Since P cracks the beam, the Moment of Inertia lessens, even in its resistance of the self weight. So, the deflection we anticipate seeing will be ...

Δ _{y, apparent} = Δ _{self weight, cracked beam} + Δ _{due to P} - Δ _{immediate due to self weight} ...

Here goes ...

Δ _{y, apparent} = (5/384) ω _{s w} L⁴ / E I_e + P L³ / 48 E I_e - 0.0011 in.

Now let me show you something: the equation for the deflection due to self weight is exactly the same as before, except that we are using a new I, in the denominator. The I is less so the deflection will be more. Let's take the old I out, and put the new one in ... giving us ...

... for the first term ... 0.0011 in. (393 in.⁴ / 120 in.⁴) = 0.004 in.

For the second term we are doing the same calc as for P_cr, except that now we have P = P_y = 9600 lb, and I = I_e ... so ...

... for the second term ... 0.011 (9600 / 2010) (393 / 120) = 0.172 in.

So ...

Δ _{y, apparent} = 0.004 in. + 0.172 in. - 0.001 in. = 0.175 in.

Steel will yield at an apparent deflection of 0.175 in. (about 1/6^th of an inch), at a load of about 9600 lb ... (and then wildly beyond that).

If you don't like the ratio-ing thing I just did above ... fine ... go back to the full equation(s) and plug and chug ...

Δ _{y, apparent} = (5/384) ω _{s w} L⁴ / E I_e + P L³ / 48 E I_e - 0.0011 in.

Δ _{y, apparent} = ...

... (5/384) 4.5 lb/in (72 in.)⁴ /(3,600,000 psi x 120 in.⁴) + 9600 lb (72 in.)³ / (48 x 3,600,000 x 120 in.⁴) - 0.001 in. = ... 0.175 in.

Yeah!

Conclusion

Okay, this is how we `predict' our beam will perform ...

Deflection under own weight ... 0.0011 in. (but it will occur before we can measure it) ...

P_cr = 2014 lb ... the concentrated load (additional to self weight) that will (first) crack the beam ...

Deflection just before it cracks ... 0.011 in. (only coincidental that it is 10x that of self weight) ...

The beam will fail in steel yielding at P_y = 9600 lb. But, this value is not a lot less than the loads that will (theoretically) crush concrete, and shear the beam; so, it will be interesting. And, note: if we end up using lower strength concrete, the concrete may indeed shear, or crush, before steel yields.

The deflection just before steel yield ... 0.175 in.

We anticipate that the beam will fail `safely' (steel yielding ... beam becoming more and more deformed, but staying together).

The ultimate load on the beam is calculated to be P ult = 10,100 lb, where we should see steel yielded and concrete crushing.

The deflection associated with the ultimate load ... we can't really calculate ... but it will be large. (Actually, we probably could calculate it, but not with the tools we have used so far.)

References

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

Design of Concrete Structures, Christian Meyer, Prentice Hall, 1996.