Long Term Deflection in Concrete Beams UAA Civil Engineering CE 433 - Reinforced Concrete Design by Dr. Bart Quimby, P.E. Spring 2002 Long Term vs. Immediate Deflections
Immediate deflections are those deflections that are the immediate response to applied loads. There is virtually no delay between the application of the load and the deflection that results from the load application. Long term deflections are the result of either dry shrinkage or viscous flow (creep) under sustained loads. ACI 318 mixes the two effects together. Both are time dependent, with the creep component being the more dominate of the two terms. There are much more complex methods for determining the effects of resulting from creep and shrinkage than presented in ACI 318. See the text book discussion on these topics to get a better feel for what is happening. Shrinkage is significant issue, particularly in members that are unsymmetrical reinforced such as most RC beams. Without the reinforcing, the member would simply get shorter. Adding reinforcing steel that bonds to the concrete restrains shrinkage. With more restraint on one face than the other shrinkage occurs differently on each face and results in curvature that adds to that caused by bending. Adding more steel on the compression side tends to reduce the problem. Creep is viscous flow of the material under stress. The tension side gets longer and the compression side gets shorter. Similar to shrinkage, the presence of reinforcing steel (which does not measurably creep under load) restrains the deformation due to creep. The shortening of the compression side and lengthening of the tension side results in curvature that greatly magnifies deflections over time. Creep effects are most noticeable in the short term. The rate of creep decreases as with time. Figure R9.5.2.5 shows the typical creep vs time curve used in computing long term deflections by ACI 318. Sustained Loads
In determining creep deformations, deformations , the term "Sustained Loads" comes into play. Sustained loads are what drives the creep deformations. Two quantities are important here. The first is the MAGNITUDE of the sustained load and second is the DURATION of the sustained loads. The determination of these two quantities is left up to the engineer. The discussion in the textbook considers only the dead load as a sustained load and that it's duration is more than five years. This is a simplistic simplisti c approach. It is not uncommon for a portion of the live load to be sustained as well. For example, in an office occupancy, the desks, book shelves, file cabinets, copiers, etc. are all part of the live load, but they are sustained over a long period of time. Part of the live load is accounted for people coming in and out of the space and for temporary office equipment loadings. In this case, 50% of the design live load may be
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sustained for over 5 years, another 25% may be around for a cumulative total of say 2 years over the life of the structure and the remaining 25% being truly transient or rarely occurring. The creep effects of these portions on the load should be included in the long term deflection calculations as well. Other occupancies may have different estimates of sustained loading. There are no guides for determining the magnitudes or durations for any given design situation. These are left to the judgment of the engineer. ACI 318 Multiplier
ACI 318 9.5.2.5 presents a multiplier, λ, that should be used to estimate the effects of creep and shrinkage. The multiplier is a function of the creep curve and the amount of steel in the compression zone. The creep curve brings in the time effect. The reinforcement ratio accounts for presence of compression zone steel. Note that the addition of more compression zone steel has a significant effect in reducing the multiplier. When long term deflections are an issue (i.e. when they exceed the allowable), it may be a good idea to add more steel in the compression zone for that reason alone. The addition of compression zone steel does little to add to the flexural strength and is often omitted in the flexural capacity calculation (as has been stated several times before in the course). The long term deflection component is found by applying the multiplier to the immediate deflection due to the sustained load. The total deflection is the sum of the long term deflection component and the immediate deflections. An example is in order. We will continue the problem started in the notes for immediate deflection calculations (see those notes for the details) For this problem, we will assume that 50% of the live load is sustained for five years or more. An additional 25% of the live load is sustained for twelve months. We will need to compute two multipliers. For both cases the compression zone reinforcement ratio is zero since this beam has no compression steel. This means that λ equals ζ (zeta) in both cases. Lambda will equal 2.0 for the five year loads and 1.4 for the twelve month loads. The immediate deflections will also need to be computed for the various load stages. The spreadsheet results are shown in Table 1:
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Table 1 Immediate Deflections
load
Ma
(plf)
(ft-k)
DL Only
800
250
DL + .5 LL
1600
DL + .75 LL DL + LL
Stage
Mcr /Ma
Ie
defl
(in^4)
(in)
0.615
52108
0.599
500
0.307
47375
1.317
2000
625
0.246
47045
1.658
2400
750
0.205
46899
1.996
LL only
1.397
The graph in Figure 1 illustrates the deflection at various stages of loading. The five year and one year sustained loads/deflections are shown. The sustained deflections are multiplied by their respective λ values. Figure 1 Load vs Immediate Deflection
To check ACI 318 9.5.2.6 requirements "that part of the total deflection occurring after attachment of nonstructural elements" must be computed. This has several parts in our case. •
The total immediate live load only deflection. From the table above, this value is 1.996 in.
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•
•
The long term deflection due to the five year loads. This found by multiplying the DL + .5LL deflection by the λ for the five year plus loading. This equals 2.0*1.317 in = 2.635 in. The long term deflection due to the one year loads. Part of this deflection is already accounted for in the previous term, so here the additional creep is determined by multiplying the sustained load deformation not already accounted for by the λ for twelve months. The calculation is 1.4*(1.658-1.317) = 0.477 in.
The total long term deflection is the sum of the second two components: 2.635 + 0.477 = 3.112 in. That part of the total deflection occurring after attachment of nonstructural elements is the sum of all three components. The result is 4.509 in. For the beam in question, the limit on this deflection is (50')/240 = 2.50 in. This is much less than the computed 4.51". Something must be done to limit the long term deflections. The most likely way would be to add compression reinforcement. This will help in two ways. First it will increase I e by increasing the I cr. Secondly, it will reduce the long term effect multiplier lambda. Let’s try adding four #8 bars in the compression zone and see what happens. ρ' becomes 4(.79 in2)/(18 in)/32.625 in) = 0.00538. This reduces the five year λ from 2.0 to 1.58 and the one year λ from 1.4 to 1.10. Computing the new Icr requires the derivation of a new set of formulas for the doubly reinforced situation. Figure 2 Doubly Reinforced Section
Figure 2 shows the various parts of the problem. Note the (n-1) term in the computation for C s. This term is there so that the area of the bars is not double counted. The computation of C already includes the areas of the actual bars.
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As before, the equilibrium equation is : 0 = T - C s - C Note that f s = f c*(d-c)/c (as previously shown) and f s' = f c*(c-d')/c. Making these substitutions into the equilibrium equation results in: 0 = n*As*f c*(d-c)/c - (n-1)*A s'*f c*(c-d')/c - f c*c*b/2 2
0 = n*As*(d-c) - (n-1)*A s'*(c-d') - c *b/2 0 = (n*As*d + (n-1)*As'*d') - (n*As + (n-1)*As')*c - (b/2)*c
2
Solve this quadratic equation for c, the location of the neutral axis. In this case, c = 12.91 in. The cracked moment of inertia is then found about this axis using principles of statics. 3
2
2
Icr = [bc /3] + [(d-c) *n*As] + [(d'-c) *(n-1)*A s] Again we ignored the moment of interias of the bars about their own centriodal axis since it is 4 not significant. The resulting cracked moment of inertia for this problem is 57809 in . The recomputed the immediate deflections are given in Table 2. Table 2 Immediate Deflections in Doubly Reinforced Beam
load
Ma
(plf)
(ft-k)
DL Only
800
250
DL + .5 LL
1600
DL + .75 LL DL + LL
Stage
Mcr /Ma
Ie
defl
(in^4)
(in)
0.615
60638
0.515
500
0.307
58163
1.073
2000
625
0.246
57990
1.345
2400
750
0.205
57914
1.617
LL only
1.102
The live load only deflection is less than before and still well within the limiting value.
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The recomputed long term deflections are given in Table 3. Table 3 Long Term Deflections in Doubly Reinforced Beam
Stage
duration
λ
DL Only
5+ years
1.58
0.515
0.811
DL + .5 LL 5+ years
1.58
0.558
0.880
Increment Long Term
DL + .75 LL
1 year
1.10
0.272
0.300
DL + LL
zero years
0
0.271
0.000
Total Long Term Deflection
1.992
That part of the total deflection occurring after attachment of nonstructural elements is the sum of the LL only deflection and the total long term deflection. The result is 3.093 in. which is still greater than the permitted 2.50 inches. Further iterations show that seven #8 bars will get the resulting deflection below the 2.50 inch limit. Summary
There are several things that should be noted: • •
•
•
The long term deflections can be (and usually are) greater than the immediate deflections Adding compression steel can have a dramatic effect on long term deflection control even though it does not add significant flexural strength. It is still acceptable to ignore the presence of compression steel when doing flexural strength computations. Always remember that superposition does NOT work with RC deflection computations. You MUST consider load stages. Differential deflections are found by finding the differences in load stages. This is different than what you can do in other materials. The reason for this behavior is addition of cracking to the analysis. Remember that these calculations can be avoided by adhering to the span/depth ratios in ACI 318 table 9.5a. The problem with doing this is that the resulting members will probably have excessive depth.
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