Column Design. Columns Axial Load and Bending

Transcription

1 Column Design MORGAN STATE UNIVERSITY SCHOOL OF ARCHITECTURE AND PLANNING LECTURE VI Dr. Jason E. Charalambides = = Columns Axial Load and Bending We tend to have this image of columns that we envision as symmetrically shaped massive pilasters. They assume the loads applied upon them in a perfectly axial manner and transfer them straight down upon the foundations that support the structure. However, beside axial loads, moments are also assumed by columns, either through the application of eccentric loading conditions, or by the eccentricity of the forms.

2 How Bending is Applied to Columns Consider the effect of varying lengths of bays and the transferred moments from the beams. Then also consider the effect of lateral forces. How Moments are Produced Gravity and lateral loads generate moments on the structure. Again, asymmetrical structural forms or variation in loading patterns are the generators of bending moments. We will address: 1. Behavior of elastic homogeneous column, 2. Behavior of RC column, Uncracked, Cracked, Ultimate 1. Design of RC column. First we shall address slices of a column and then we will extrapolate to entire columns (short vs. long).

3 Elastic Homogenous Column Now the Neutral Axis and the geometric centroid are no longer at the same location. The location of the N.A. is dependent upon the eccentricity of the resultant of loads. óσ=(p/a)+(m*y/i) Locating the Neutral Axis Measure from the geometric centroid: f = P A + M ȳ I At N/A σ=o: ÅÅ y= I A ecc ÅÅ

4 Equilibrium How equilibrium is established: f = P A +(P ecc) y I ÅÅ ΣF =0 P=C T ΣM =P ecc P ecc=c y 1 +T y 2 RC Column Analysis For beams, service level behavior (cracking, deflections) is very important. Structural behavior tends to be controlled more by the ultimate behavior of the column, rather than service level behavior. So we place less emphasis on service level behavior of columns. Prior to Cracking: Use uncracked, transformed section. Concrete cracks at fr=7.5 f`c After Cracking: Use cracked, transformed section.

5 RC Column Analysis: Nomenclature fc is the computed compression flexural fiber stress at service loads fs is the calculated stress at reinforcement at service loads f's is the stress in compression reinforcement fy is the steel strength fr is the point where concrete cracks (modulus of rupture) Ec is the Young's modulus of elasticity of concrete which is given by the formula Ec=57000 f'c n is the ratio of elastic moduli of the two materials involved, in this case reinforcement steel and concrete In Class Example Construct the M-θ curve for the section shown. The section is subjected to a load at an eccentricity of 15 inches from its center. Construct the Eccentricity vs Stress curve according to the moment applied on this column:

6 In Class Example Construct the M-θ curve for the section shown. The section is subjected to a load at an eccentricity of 15 inches from its center. In Class Example cont

7 In Class Example cont In Class Example cont After Cracking: Locate NA by trial and error: Choose εc=>fc (reference only) Guess kd Calculate f`s, fs Calculate P & M See if calculated eccentricity (M/P) equals given ecc, Repeat as necessary. Note that the initial chosen fc is arbitrary. We compute P as a linear function of fc, and M as a linear function of fc. Then ecc=m/p is independent of fc.)

8 In Class Example cont In Class Example cont

9 In Class Example cont In Class Example cont

10 In Class Example cont In Class Example cont θ/in M (k`) Comments 1.85E just before cracking 41.77E just after cracking 1.147E fc=0.7*f`c 2.51E fs=fy (assumes linear behavior)

11 Columns Strain Distribution We see the interaction diagram can determine the capacity of stress that can be applied to a column. Let s use the Excel sheet provided for in class exercise. ΦPn (K) Column design Moment vs Axial load ΦMn (K') Columns Slenderness This is a general method that we can roughly apply in order to consider how the connections can effect the strength of the column. More on this will be addressed during the next lecture.

12 Columns Compare Materials Let s try to visualize the effect of strength and the geometry that would correspond to a design for a specific load. Let s take two very prominent materials: Steel: E=29,000,000psi Concrete: E=3,600,000psi It is obvious that in order to compensate the strength difference, we will address the geometric form, i.e. the cross sectional area. So a Steel column can be way more slender than a concrete column, just to bear the load.

13 What is the Effect of Column Slenderness? Imagine the effect of purely axial load applied in this element. What do you think will happen? Even if there is no shear or moment applied, do you believe that it will crush from the axial load? The uniformity and homogeneity of the material should be challenged. Even with prefabricated materials that are made under the strictest of regulations, we can expect some slight abnormalities. Those will render the element asymmetrical and stronger on one direction versus another. The formula that determines a column to be slender or not is the following: k l u r 34 12[ M 1 M 2 ] What is the Effect of Column Slenderness? Also we need to consider the following: ACI defines Mc as the magnified moment and M2 the larger factored end moment of a no sway compression member: M c =δ ns M 2 In case our calculations provide minimal result we can apply the minimum eccentricity formula: ecc min = h The moment magnifier δns is used to estimate the lateral deflection effect. It involves the code modificator Cm which is also given below: C m δ ns = 1 P u 0.75 P c 1.0 C m = M 1 M 2 0.4

14 How Bending is Applied To Columns Once a slight deflection takes place on an axially loaded element, there is more eccentricity generated, which in turn produces a second generation moment, which will result in further deflection, one more round of moment and deflection and so on and so forth, until equilibrium is reached. Looping this process to analyze the deflection and the applied moment over and over may be extraordinarily tedious and the result will not vary tremendously once two or three cycles are reached. Timoshenko resolves this process by multiplying the primary moment by the following formula, which can give us a result that is precise enough for us: u[ 1 M magn ] =M 1 P u P c What About Double Curvature and Other Scenarios? What happens in the case of double curvature with equal but reverse moments, or in the case where we have no moment on one end? In the first scenario we have moment and deflection equal to zero and in the second, we have a deflection that is about half of what the amplification factor provides, and a very large moment. Therefore, the code addresses the issue by the use of the modification factor Cm which can vary between 0.4 and 1.0 that is to be used for braced frames without transverse loads. For other cases the value to be taken is 1.0 C m = M 1 M 2 0.4

15 In Class Example We use McCormac and Nelson example on page 332. Calculate the primary moment due to a lateral load of 20k when the column is subjected to 125k axial. Bw= 12, h=15, k=1.0, lu=18ft. M u = P l 4 =90k ' E c =57000 f ' c = ksi P c = π 2 u[ E c I = kip M magn =M (k l u ) 2 I= b h =3375inch 1 1 P u P c ]=95.35 k ' Let s Consider the Design of a Column Subjected to Moment, Shear, and Axial Load Design a tied column cross section to support an axial load of 300 kips, a moment load of 110kip feet, and a shear load of 14kips. All of the above are factorized. The column is in a braced frame with an unsupported length of 10 ft 6 inches.

16 RC Tied Column Design Design a tied column cross section to support an axial load of 350 kips, a moment load of 110kip feet, and a shear load of 14kips. All of the above are factorized. The column is in a braced frame with an unsupported length of 10 ft 6 inches: Data: P u := 350kip M u := 110k' V u := 14kip Φ :=.85 f y := 60ksi f' c := 3ksi l u := 10.5ft Clear cover := 1.5in Estimatiing an initial ratio of steel for tied columns ρt: Based on the most efficient ratio that would be between 1% and 2%, we select an initial ratio of 1.5% ρ t :=.015 Estimating the initial dimensios of the column: The cross sectional area for a tied column is given by the following formula (variance of ) P u A g_ini :=.40( f' c + f y ρ t ) A g_ini = in 2 Given the option that we may design a square base column...we estimate an initial base value: b ini := A g_ini b ini = in Given the fact that there are significant moments applied on this column, it would be wise to override the initial calculation that takes only direct loads into account. Let's round it up about 10-15% on each side: ( ) b trial = trunc b ini 1.12 b trial = 16 in b := b trial A g := bh A g = 256 in 2 Determining the bar arrangement: h := b To determine the preferable bar arrangement we compute the ratio of eccentricity to the height "h" of the column: Note: This is "h" in cross section, not the actual column height. M u ecc := ecc = 3.77 in P u According to the figure indicated it will be more appropriate to apply re-bars on both sides of this column: Column slenderness can be neglected if: l u M 1 k r M 2 ecc h = 0.24 According to ACI code , the radius of gyration of rectangular columns is 0.3h and.25d for circular columns. Since this is a braced frame k is lesser or equal to1.0 and the ratio of M1 to M2 can vary between +/ We can assume that k=1.0 and M1/M2=0.5. Therefore the above relations yield the following results: l u M 1 k = = 28 r M 2 r :=.3 h k := 1 M 1 = 0.5 M 2

17 l u Slenderness condition := if k r M 1 M 2 Slenderness condition = "Neglect column slenderness", "Neglect column slenderness", "Design slender column" Computing the "γ" ratio : At this point we need to compute the value gamma (γ) which is the ratio of distance of centroids of outer rows of bars and column dimension perpendicular to the bending axis: We shall assume that the ties are #3 rebars and the longitudinal bars are #7: γ := d lbar b 2 Clear cover + d tbar + 2 h d lbar :=.875in d tbar :=.375in γ = We need to point out that the assumption we make about the #7 rebars may prove imprecise, in which case we shall need to reiterate this process. Given the gamma value above we will refer to the ACI interaction diagrams (or use our own system!!!) to define again the ratio of steel ρt. P u A g = ksi M u A g h = ksi For the above values the interaction diagram for gamma 0.6 gives a rho value of The diagram for a gamma value of 0.75 gives a rho value of We shall apply linear interpolation to compute the value of rho at gamma found. ACI code defines that rho should lie between 0.01 and 0.08.

18 ρt is the ratio of total reonforcement area divided by the cross sectional area of a column ( γ.6) ρ t := ( ) ρ (.75.6) t = ( ) ( >, "High value", "OK" ) Rho lo_condition := if ρ t <.01, "Low value", "OK" Rho lo_condition = "OK" Rho hi_condition := if ρ t.08 Rho hi_condition = "OK" Selecting reinforcement: A s := ρ t A g A s = in 2 We can select six #7 bars, three on each face. Design the lap splices: l d7 = 1.3d lbar f y 20 f' c l d7 = in l d7 = ft This splice is far too long, about half the length of the column. Consider using eight #6 bars l d6 = 1.3d lbar f y 25 f' c d lbar :=.75in l d6 = in l d6 = ft 8.44in 2 = 3.5 in 2 So let's reevaluate our rho value: n lbar := 8 A lbar := 0.44in 2 A s_fin A s_fin := n lbar A lbar ρ := ρ = A s_fin = 3.52 in 2 A g Selecting the ties: Based on ACI sections , , and , the least of the following three conditions determines the spacing of the ties: 16 d lbar = 12 in 48 d tbar = 18 in col least_dim := if ( b < h, b, h) col least_dim = 16 in We need to make reference to the subject we addressed on principal stresses to visualize the effect of the following formula. The factor "Nu" represents an axial tension force resulting from the compression. The angle theta " θ" to be used is the critical 45 degrees. θ := 45deg d lbar d := b 1 Clear cover + d tbar + d = in 2 P u N u := N tan( θ) u = 350 kip ( ) N u ΦV c = Φ 2 f' c b d 1 + ΦV 2000A c = 34.5 kip g Check ACI sections sections , , , , , , , and If Vc<Vu<Vc, the ACI code section ACI governs.... The end result...#3 12 inches o.c.

19 pmfo^i=obfkclo`bjbkq `lk`obqb=`lirjk=abpfdk Problem Statement: Select a cross section for a spirally reinforced column section to support the loads indicated below, using f'c of 5 ksi and grade 60 steel reinforcement. Try to use ρg of 4%. Processing Data: Factorizing Dead and Live load: Dead load is multiplied by a factor of 1.2 and Live load by a factor of 1.6: PD := 620kip PD_factored := PD 1.2 PD_factored = 744 kip PL := 328kip PL_factored := PL 1.6 PL_factored = kip Pu := PD_factored + PL_factored MD := 0k' MD_factored := MD 1.2 MD_factored = 0 k' Pu = kip ML := 80k' ML_factored := ML 1.6 ML_factored = 128 k' Mu := MD_factored + ML_factored f'c := 5ksi fy := 60ksi Solution: Φ.75 Mu = 128 k' := ACI : For spiral ϕ=.75, for tied ϕ=.65 Also, the first factor changes from.85 to.80 for tied. The length value that is the result of the division of the applied moment by the applied axial load is the eccentricity "ecc" of the column. Almost always, a compression member may assume a moment either because the axial load is not perfectly centered on the column, or because the column will resist portion of the unbalanced moments at the ends of the beams it supports. In order not to confuse the term "e" with the base of the natural logarithm, we can use the term ecc. Mu ecc := ecc = in Pu Eccentricity should not be higher than 10%, so we can set a minimum diameter of one foot For reinforcement of 3% the letter "ρ" that signifies density is used ρg = Ast Ag ρg_max.08 ρg_min.01 ρg :=.04 The reason we use the ρg is to determine the area of the column. The ACI code gives the following formula for non prestressed members w/spiral reinforcements (ACI ) maxφpn := Pu Inversing data: maxφpn = 0.85 Φ[ 0.85 f'c ( Ag) + ρg Ag ( fy 0.85 f'c) ] Ag := maxφpn Φ 0.85 [ 0.85 f'c + ρg ( fy 0.85 f'c) ] Ag = in 2

20 Since the column shall be circular, we can use the formula of the circle's area to estimate an approximate radius and round it. r := Ag π r = in Rounding.. r := 10in Therefore: Ag π r 2 := Ag = in 2 Using the following formulas now we can determine the steel reinforcement maxφpn = 0.85 Φ[ 0.85f'c ( Ag As) + fy As] OR maxφpn = 0.85 Φ[ 0.85 f'c ( Ag) + ρg Ag ( fy 0.85 f'c) ] However, none of the above formulas can be inverted for us to use in terms of As which is what we are trying to solve for. Therefore, we can use a system called "solve block" with an initial guess and allow a series of iterations to take place until a solution is found. maxφpn := Pu Guess values: As := ρg Ag As = in 2 This seems like a good starting point. Given maxφpn 0.85Φ [ 0.85f'c ( Ag As) + fy As] = (ACI ) Ast_value := Find( As) As := Ast_value As = in 2 For rebars smaller than #9 a formula can be used to define the As: Try 12 #9 rebars. The formula will yield slightly imprecise result As should be 12 sq. inches BarSize1 := 11 n1 := 8 BarSize2 := 14 n2 := 0 2 BarSize1 BarSize2 As := n1 + n π in2 As = in 2 As := 12.48in 2 Verifying... maxφpn := 0.85Φ [ 0.85f'c ( Ag As) + fy As] maxφpn = kip ρg := As Ag ρg = = This is a very good result. We are only very slightly above our ρg ratio is as we initially aimed for.

21 COLUMN SLENDERNESS Inserting Excel Problem Statement: Determine whether or not a 16*20 inch section with 12#10 bars is adequate for minimum eccentricity (Code clause ) about the minor axis of the column. The Column height Lu=20.7ft A s_10 := 1.27in 2 f'c := 5ksi fy := 60ksi Lu := 248.4in b w := 16in h := 20in Using inches and lb for consistency Processing Data: DL := lbf DL_factored := DL 1.4 DL_factored = 816 kip LL := lbf LL_factored := LL 1.7 LL_factored = 263 kip PD := DL Pu := LL DL 1.4 Pu = 1079 kip PL := LL Assuming that M1=minM2 to cause compression on the same face such that M1/M2=1and k factor is for elastic connection on a multi-story building: k := 0.7 klu := k Lu klu = in Determining the M2min: M 2min := Pu 0.6in b w M 2min = kip in OR M 2min = 97.11k' ( ( )) Note: As stated in the code (ACI ), the units of 0.6 and (h) (or c1) are taken in inches. Also note that in this case we treat as h (or c1) the short side of the column because we solve for the max moment on the weakest side. The result can be written in any format the user prefers. As k' are defined above, the k' option is provided. M 2 := M 2min & M 1 := M 2min Solution: Determining the Cm (factor relating the actual moment diagram of a slender column to an equivalent uniform moment diagram: C m M 1 := + C m = 1 M 2 Determining the modulus of elasticity of Concrete: E c := f'c psi psi E c = ksi Calculating the (βd) ratio of maximum factored axial Dead Load to the total axial load: β d := 1.4 PD β d = Pu Determining the (Ig) gross moment of Inertia of the element along the minor axis (see problem statement): 3 hb w I g := I g = in 4 12 Solving for the moment of Inertia of the steel rebars using the Ad^2 formula. There shall be 2.5" cover, allowing 11" along the short axis, so the distance of the outer bars shall be 5.5" and the distance of the

22 inner bars shall be 11"/(3*2), or 1.833": d 1 := 5.5in d 2 := in n 1 := 8 n 2 := 4 E s := 29000ksi ( ) 2 2 I s := A s_10 n 1 d 1 + n 2 d 2 I s = in 4 Calculating the EI (stiffness) - See ACI code & : ( 0.2 E c I g + E s I s ) EI := ( 1 + β d ) EI = kip ft 2 Calculating the (Pc) critical load: π 2 EI Pc := ( klu) 2 Pc = kip Calculating the δns (moment magnification factor - applied to frame columns that are braced against sidesway, reflecting effects of member curvature between ends): δ ns C m := δ ns = Pu Pc Considering the ultimate moment (Mu) to be equivalent to the critical Moment (Mc) now we solve for the Mc: M c := δ ns M 2 M c = k' Pu = 1079 kip Column design Moment vs Axial load ΦPn (K) ΦMn (K) Using the provided Excel file we can have the graph of this concrete column configuration. Based on our data, the Mc and Pu fall within the ΦPn/ΦMn curve. The column is approved.