DEFLECTION CALCULATIONS (from Nilson And Nawy)

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DEFLECTION CALCULATIONS (from Nilson and Nawy)The deflection of a uniformly loaded flat plate, flat slab, or two-way slab supported bybeams on column lines can be calculated by an equivalent frame method that corresponds with the method for moment analysis. The definition of column and middlestrips, the longitudinal and transverse moment distribution coefficients, and many otherdetails are the same as for the moment analysis. Following the calculation of deflectionsby this means, they can be compared directly with limiting values like those of Table9.5(b) of ACI which are applicable to slabs as well as to beams.A slab region bounded by column centerlines is shown in Figure 1. While no columnline beams, drop panels, or column capitals are shown, the presence of any of theseintroduces no fundamental complication.The deflection calculation considers the deformation of such a typical region in onedirection at a time, alter which the contributions front each direction are added to obtainthe total deflection at any point of interest.From Figure 1a, the slab is considered to act as a broad, shallow beam of width equal tothe panel dimension ly, and having the span lx. Initially, the slab is considered to rest onunyielding support lines at x 0 and x lx.Because of variation of moment as well as flexural rigidity across the width of the slab,all unit strips in the X direction will not deform identically. Typically the slab curvaturein the middle-strip region will be less than that in the region of the column strips because themiddle-strip moments are less. The result is as shown in Figure 1a.Next the slab is analyzed for bending in the Y direction (see Figure 1b). Once again wecan see the effect of transverse variation of bending moment and flexural rigidity.We can see the actual deformed shape of the panel in Figure 1c. The mid-panel deflectionis the sum of the midspan deflection of the column strip in one direction and that of themiddle strip in the other direction: i.e.,Δ max Δ cx Δ my(1)Δ max Δ cy Δ mx(2)orIn calculations of the deformation of the slab panel in either direction, it is convenientfirst to assume that it deforms into a cylindrical surface, as it would if the bending moment atall sections were uniformly distributed across the panel width and if lateral bending of the1

panel were suppressed. We consider that the supports to be fully fixed against both rotationand vertical displacement at this stage. Thus, a reference deflection is computed:Δ f ,ref wl 4384 Ec I frame(3)where w is the load per foot along the span of length l and Iframe is the moment of inertia of thefull-width panel (Figure 3a) including the contribution of the column-line beam or droppanels and column capitals if present.The effect of the actual moment variation across the width of the panel and thevariation of stiffness due to beams, variable slab depth, etc., are accounted for bymultiplying the reference deflection by the ratio of M/E for the respective strips to that of thefull-width frame:Δ f ,col Δ f ,refM col Ec I frameM frame Ec I col(4)Δ f ,mid Δ f ,refM mid Ec I frameM frame Ec I mid(5)andThe subscripts relate the deflection Δ , the bending moment M, or the moment ofinertia I to the full-width frame, column strip, or middle strip, as shown in Figure 3a,b,and c respectively.Note that the moment ratios M mid / M frame and M col / M frame are identical to the lateralmoment-distribution factors for DDM (ACI 13.6.4.1-3).2

Figure 1. Basis of Equivalent Frame Method Deflection Analysis: (a) X-directionBending; (b) Y-Direction Bending; and (c) Combined Bending (From Nilson).3

Figure 2. Equivalent Frame Method Deflection Analysis: (a) Plate Panel Trasfered intoEquivalent Frames; (b) Profile of Deflected Shape at Centerline (From Nawy).4

Figure 3. Effective Cross Sections for Deflection Calculations; (a) Full-Width Frame; (b)Column Strip; (c) Middle Strips.5

The presence of drop panels or column capitals in the column strip of a flat slab floorrquires consideration of variation of moment of inertia in the span direction as shown inFigure 4 below.Figure 4. Flat Slab Span with Variable Moment of Inertia.Nilson and Walters (1975) suggested a weighted average moment of inertia be used insuch cases:I ave 2lcllIc 2 d Id s Islll moment of inertia of slab including both drop panel and capitalmoment of inertia of slab with drop panel onlymoment of inertia of slab alone(6)where:IcIdIs6

We also need to correct for the rotations of the equivalent frame at the supports, whichuntil now we have assumed to be fully fixed. If the ends of the columns are consideredfixed at the floor above and below, the rotation of column at the floor is:θ M netK ec(7)whereθM netK ec angle change, radiansdifference in floor moments to left and right of columnstiffness of equivalent column.Once we know the rotation, the associated mid-span deflection of the Equivalent Framecan be calculated. The midspan deflection of of a member experiencing an end rotationof θ radian having the far end fixed is:Δθ θl8(8)Therefore, the total deflection at mid-span of the column strip or middle strip is the sumof three parts:Δ col Δ f ,col Δθ l Δθ r(9)Δ mid Δ f ,mid Δθ l Δθ r(10)andwhere the subscripts l and r refer to the left and right ends of the span respectively.7

Example Problem (From Nilson’s Book).Find the deflections at the center of typical exterior panel of the two-way slab floorsystem designed before (shown below), due to dead load and live loads. The live loadmay be considered a short-term load and will be distributed uniformly over all panels.The floor will support non-structural elements that are likely to be damaged by largedeflections. Take Ec 3,600 ksi.8

9

First calculating the deflection of the floor in the short-span (N-S) direction of the panel(l2 25 ft), from the Equation (3):Δ f ,ref wl 4(88 /12)(25)(20 12) 4 0.016 in384 Ec I frame 384(3, 600, 000)(27,900)(11)Note that we used the centerline span distance, although we used clear span for momentanalysis. From the moment analysis in the short-span direction, we found that 68% of themoment both negative and positive sections was taken by the column strip and 32% bythe middle strips. Therefore, from Equation (4) and(5) we have:Δ f ,col Δ f ,refM col Ec I frame27,900 0.016 0.68 0.014 in21, 000M frame Ec I colandM mid Ec I frame27,900 0.016 0.32 0.028 in5150M frame Ec I midFor the panel under consideration, which is fully continuous over both supports in theshort direction, we can assume that support reactions are negligible; and therefore,Δ f ,mid Δ f ,refΔθ l Δθ r 0 inTherefore,Δ col Δ f ,col Δθ l Δθ r 0.014 inandΔ mid Δ f ,mid Δθ l Δθ r 0.028 inNow calculating the deformation in the long direction (E-W):Δ f ,ref wl 4(88 /12)(20)(25 12) 4 0.033 in384 Ec I frame 384(3, 600, 000)(25,800)10

From the moment analysis it was found that the column strip would take 93% of theexterior negative moment, 81% of the positive moment, and 81% of the interior negativemoment. Therefore, the average lateral distribution factor for the column strip is: 93 81 1 81 0.84 2 2or 84%, while the middle strips are assigned 16%, therefore,Δ f ,col Δ f ,refM col Ec I frame25,800 0.033 0.84 0.0034 in21, 000M frame Ec I colandM mid Ec I frame25,800 0.033 0.16 0.040 in3430M frame Ec I midWe cannot ignore the rotation at the exterior column. The full static moment due to deadload is:Δ f ,mid Δ f ,ref1M 0 0.088 20 252 137.5 ft-kips8We found that 16% of the static moment, or 22 ft-kips should be assigned to the exteriorsupport section. The resulting (assuming that from Equivalent Frame analysis we havethe equivalent column stiffness as 169E.M net22, 000 12 0.00043 radK ec 169 3, 600, 000From Equation (8) we have:θ Δθ θl8 0.00043 (25 12) 0.016 in8Therefore,Δ col Δ f ,col Δθ l Δθ r 0.034 0.016 0.050 inand11

Δ mid Δ f ,mid Δθ l Δθ r 0.040 0.016 0.056 inThe short-term mid-span deflection due to self-weight isΔ max 0.05 0.028 0.078 inThe long term deflection due to dead load isΔ long term 3.0 0.078 0.234 inThe short term live load deflection isloadΔ livelong term 1250.078 0.111 in88The ACI code limiting value for the present case is found to be 1/480 times the span, orΔ limit 20 12 0.500 in480based on the sum of the long-time delfectin due to sustained load and the immediatedeflection due to live load. The sum of these deflection components in the present case isΔ max 0.234 0.111 0.345 inwhich is less than the permissible value of 0.500 inches.12

The deflection of a uniformly loaded flat plate, flat slab, or two-way slab supported by beams on column lines can be calculated by an equivalent frame method that cor-responds with the method for moment analysis. The definition of column and middle strips, the longitudinal and transverse moment distribution coefficients, and many other

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