Immediate Deflection in Concrete Beams
UAA Civil Engineering
CE 433 - Reinforced Concrete Design
by Dr. Bart Quimby, P.E.
For prismatic beams consisting linear elastic materials, the elastic curve can be found by integrating moment equation twice with respect to the length variable. The resulting equation defines the displacement at any location along a span.
In most cases, we are interested in the magnitude and location of the MAXIMUM deflection. This is found by setting the first derivative of the the elastic curve (deflection) equation to zero and solving for the roots that fall within the span of the beam. These roots give the locations of the maximum or minimum deformations. The resulting roots can then be put into elastic curve equation to find the values of the maximum and minimum deformations. There are a number of sources that give these results for a variety of beam support and load conditions. (See your mechanics and structures texts. Also, any edition of the Manual of Steel Construction has an excellent set of beam tables.)
Note that deflections are a SERVICEABILITY condition. In other words, we want to know what the "actual" values are. Consequently we use SERVICE LOADS to compute deflections.
Adapting Elastic Equations for Use with RC Sections
Note that all the elastic deformation equations are function of E (modulus of elasticity) and I (moment of inertia). For RC members, we can assume a constant E IF deformations stay within the elastic region (the normal assumption for most materials) so there is no problem there. Ec varies with 28 day strength of concrete. The equation for computing Ec is found in ACI 318 8.5.1. The problem with RC members is the I.
Since there is cracking that occurs in the section there is a variable moment of inertia. In regions of cracking there is a reduced moment of inertia and in areas without cracking the the moment of inertia is much larger. Also, as the load increases, so does the cracking. THE MOMENT OF INERTIA CHANGES WITH LOAD!
This moment of inertia that changes with load makes our deflection equations non-linear with respect to loading and WE CANNOT USE SUPERPOSITION to determine load combinations.
For example: In any other material, we can compute the deflection due to just dead load and the deflection due to just live load. We can then find total load deflection simply by adding the two together. We can do this because both the dead and live load deflection calculations used the SAME I. In concrete we cannot do this because there is a different I for each load case and the parts are no longer additive. An example of this will be given later.
Moment of Inertia Calculations
The moment of inertia calculation for a composite section is covered quite well in most mechanics of materials texts. You should review this in your copy. The most common method for computing the section properties of a composite section is know as the TRANSFORMED SECTION METHOD.
To transform a section, the steel is transformed to an equivalent (i.e. same axial stiffness) area of concrete. This is done by multiplying the area, As, of each bar set by the modular ration n (n = Es/Ec). The transformed As is centered on the original As. With this transformed area, normal statics equations can be used to determine the location of the elastic neutral axis and the axis dependent properties such as moment of inertia of the transformed section. Most mechanics texts do a creditable job of showing how this is done.
The Uncracked Section
In the vast majority of the RC beams the moment of inertia for an uncracked section based on a transformed area is not much different than the moment of inertia of the gross section without transforming the steel. This is because the steel is a small portion of the overall cross section. This observation, along with the fact that the I from the non-transformed section is conservative, makes it appropriate to use the I without transforming the section whenever cracking is not present.
Cracks occur when the tensile bending stress in the concrete exceeds the concretes ability to resist it. A way to predict cracking is to determine the moment that causes cracking to occur and compare your actual moments to this cracking moment. The cracking moment is found by setting the elastic flexural stress equation (Mc/I) equal to the tensile stress capacity of the concrete, fr (aka modulus of rupture), then solving for M. This M is referred to as the cracking moment, Mcr. The equation can be found in ACI 318 188.8.131.52.
For a rectangular section, c = h/2 and I = bh3/12. fr, for normal weight concrete, is taken as 7.5 times the square root of the f'c. Not that some other codes that deal with concrete (the bridge design code for example) use different values for this. This particular value of fr is pretty conservative.
If the actual moment for the load stage under consideration is less than the moment that will cause cracking, then use the gross moment of inertia in the deflection equations. If the actual moment exceeds the cracking moment, then find the cracked moment of inertia for use in the deflection equations.
Finding the Cracked Moment of Inertia.
Since this computation is under elastic service conditions the stresses that are proportional to the strains. Whitney's wonderful constant stress block, which has been used all semester, is not applicable. This means that there is a linear stress distribution in the compression zone that goes from a value of zero at the neutral axis to a maximum value at the compression face. The strain in the concrete is no longer equal to 0.003 since this calculation is not looking at a failure condition.
Lets go through the process of determining the location of the neutral axis and then finding the moment of inertia of the cracked section. Refer to the diagram above for a singly reinforce concrete section. (A similar derivation can be done for other situations.) When a flexural crack occurs, it begins at the tension face when the tensile capacity is exceeded and propagates upward until the concrete is in compression. Note that the section properties change as the crack progresses, causing the increased tensile stress that drives the crack upwards.
Equilibrium is achieved once the crack stops propagating. For the case of pure bending, C = T once equilibrium is reached.
Note that the concrete compression force, C, is found by integrating stress time dA over the area in compression. The stress is NOT constant, but a function of "y". In this case, we have a linearly varying stress on a constant width area, so the integration results in the area of the rectangle (c*b) times the AVERAGE stress (fc/2) over the area.
The tensile force, T, is simply the stress in the steel, fs, times that transformed area of steel, nAs.
Using similar triangles a relationship can be found between fs and fc in terms of c. The relationship is fs = fc*(d-c)/c. Creating the equilibrium equation and substituting for fs, you get:
0 = T- C = n*As*fs - fc*c*b/2
0 = n*As*fc*(d-c)/c - fc*c*b/2 <= Note that fc cancels out of the equation.
0 = (b/2)*c2 + (n*As)*c - (n*As*d)
This quadratic equation can then be solved for the location, c, of the neutral axis. The next step is to find the moment of inertia of the cracked section about the neutral axis. This done using the principles found in any statics book.
Icr = [moment of inertia of the concrete block about the N.A.] + [moment of inertia of the transformed steel about the N.A.]
Icr = [bc3/3] + [Io + (d-c)2*(n*As)] is approximately equal to [bc3/3] + [(d-c)2*(n*As)] since the moment of inertia of the transformed steel about it's own centroidal axis, Io, is small compared to everything else. Therefore use:
Icr = [bc3/3] + [(d-c)2*(n*As)]
Again... this result is ONLY VALID FOR A SINGLY REINFORCED RECTANGULAR SECTION.
Use a similar process to solve for any other situation. The textbook has some particularly good examples of different situations. In the fifth edition of the text, Figure 14.5.6 shows a typical continuous T-beam arrangement.
The Effective Moment of Inertia
The effective moment of inertia, Ie, is a moment of inertia that, when used with deflection equations developed for prismatic members, will yield approximately the same result as a more rigorous analysis that considers variable moment of inertia. This effective moment inertia will always be between the two extremes of the gross moment of inertia and the cracked moment of inertia, both of which have been discussed above.
In order to see what happens and the changing nature of the moment of inertia consider the following progression of loading. The situation shown is a simple span with a point load in the middle but the concept is the same for any other loading.
The following figure shows what happens relative to the moment diagram and beam cracking as the load is increased.
Load Stage #1 shows a load that creates a maximum moment that is less than the cracking moment. In this case no cracking occurs in the beam. The gross moment of inertia can be used to compute the deflection.
In Load Stage #2 the load P is increased until the maximum internal moment in the beam reaches the cracking moment. At this time the first flexural crack occurs in the beam. The crack propagates upward where ever the concrete is in tension. The moment of inertia of the beam right at the crack equals the cracked moment of inertia. The moment of inertia over the rest of the beam equals the gross moment of inertia. End effect is that the beam deflection exceeds that would be found using the gross moment of inertia, but not by much. So the effective moment of inertia is a little less than the gross moment of inertia.
As the load P is increased to the value at Load Stage #3 additional cracks form as the stress in the concrete due to bending exceeds the tensile stress, fr, of the concrete. The beam becomes more flexible as the more cracks form, meaning that the deflections become much larger than would be predicted by computing deflections using the gross moment of inertia. The effective moment of inertia continues to decrease as the load increases and more cracks form. This process illustrates why the effective moment of inertia changes with the applied moment.
Another interesting result is the fact that the cracks tend to become inclined as they form in the presence of both shear and bending forces. Recall that shear stress is zero at the top and bottom faces and maximum at the N.A. Bending stresses are maximum at the faces and zero at the N.A. Also bending stress is at its maximum when shear is zero. Using mechanics, the shear and normal stress resulting from bending can be determined at any point on the beam. These stresses can be represented on a small element as shown above. Knowing these stresses, the principle stresses and their orientation can be determined at the point in question. The cracks will be perpendicular to the principle tension stress. The combined shear and tension cause the principle stress to be inclined.
The ACI 318 Method for computing the Effective Moment of Inertia, Ie
The ACI 318 for approximating the effective moment of inertia, Ie, is found in ACI 318 184.108.40.206. You should turn to this section while reading this section. Equation 9-7 is a function of a number of constants that have been discussed above. The variable in the equation is Ma, the actual maximum moment in the load stage under consideration. Whenever Ma < Mcr, the Ie = Ig. Once Ma exceeds Mcr, Ie varies from Ig to nearly Icr. You should take the time to graph out this function to get a feel for its behavior.
The resulting Ie can be used in the elastic deflection equations to approximate the actual deflections. Remember that a different Ie must be computed for each load stage. An example follows below.
The discussion above involved a simply supported beam with one zone of cracking. If a beam has negative moments, a similar analysis can be preformed to compute the cracked moment of inertia in the negative moment zones. These will be different than the positive moment zones because the compression face is now on the opposite side and different steel is in place to handle the negative bending. ACI 318 220.127.116.11 discusses how to compute the Ie for the span when there is continuity that causes both positive and negative bending.
Given: A simply supported roof beam with a span of 50 ft has the cross section shown. The beam is subjected to a dead load of 800 lb/ft and a live load of 1,600 lb/ft. The 28 day strength of the concrete is 4,000 psi. Take the effective depth, d, to be 32.5 inches. The beam is not attached to any non-structural elements likely to damaged by large deflections.
Wanted: Determine the Dead Load Only, the Live Load Only, and Total Load deflections.
From the equation in ACI 318 8.5.1, Ec = 3,600,000 psi
The modular ratio, n, = Es/Ec = 8.0 (Note that n is usually computed to the nearest 0.5)
Ig = bh3/12 = 69984 in4.
From ACI 318 18.104.22.168, fr = 474 psi. yt = h/2 for a rectangular section, which leads to a cracking moment of 154 ft-k using ACI 318 equation 9-8.
We will also need to find the location of the neutral axis for the cracked case. To do this we will need to find "c" using the following quadratic equation:
9*c2 + 88.48*c - 2876 = 0.
The relevant root is c = 13.62 in.
Now we can compute the cracked moment of inertia.
Icr = (18*(13.62)3)/3 + 88.48*(32.5-13.62)2 = 46700 in4
In the following table the moment is taken as w*L2/8 and the deflection as (5*w*L4)/(384*E*I). Both are applicable to a uniformly loaded simply supported beam.
Note that both calculations were done at LOAD STAGES. The complete load for each stage is included in the calculation. This must be because of nature of the Ie calculation.
The Live Load Only deflection is taken as the difference between the two load stages. LL Only deflection = 1.996 in - 0.599 in = 1.397 in.
The graph shown here illustrates the load vs deflection curved for the given beam. Various stages and their deflections are shown. Note that the graph is nonlinear.
The following calculation is for demonstration purposes only!
Don't get tempted to try and compute LL Only by computing the deflection with only the LL present. This incorrect computation would look as follows:
Note that the resulting deflection (1.317 in) is less than the correct live load only deflection of 1.397 in. They are different because there is no load stage where only the live load is present. Dead load is always present and will show up in all load stages. The moments used in computing Ie are actual moments that the beam will see, not portions of moments.
How Much Deflection is Too Much?
The next logical question, after computing the deflection, is: Is this deflection okay or not? The answer to the question is: It depends...
Since this is a serviceability issue and not a strength issue, excess deflection is not likely to be a safety issue and is not well covered in the codes. How much deflection is to be permitted depends on the use of the structure. The amount of deflection may be controlled by occupant comfort levels or the effect of the deflections on supported components.
ACI 318 makes an effort to give a minimum set of deflection limits in section 22.214.171.124 and the associated table 9.5b. This is by no means an exhaustive list. Care should be taken to meet not only these requirements, but those of your client as well. Other deflection limits are often found in the building codes and other structural design references. For now, we will only worry about the ACI 318 requirements.
ACI 318 table 9.5b looks principally at the deflections that result from loads placed on the structure after all the dead loads are installed. Hence the ability to correctly compute live load only deflections is important. The table also has two conditions where the long term deflection effects of creep and shrinkage must be considered. These long term deflections are covered in the next lecture.
The limit state associated with deflection comes down to the following design inequality:
actual deflections <= allowable deflections.
A number of basic ideas were presented above. Among them are: