Lect10 - Columns Interaction Diagrams

1 ` =f ~ =. a ~ ~ . rkfsbopfqv=lc=tfp`lkpfk=pqlrq `liibdb=lc=p`fbk`bI=qb`eklildvI=bkdfkbbo fkdI=^ka=j^qebj^qf`p ib`qrob=u a K=g~ =bK=` ~ ~ ~ . ` = =. ` ~ =j~ ~ . 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=29000000psi Concrete: E=3600000psi It is obvious that in order to compensate the strength difference, we will address the geometric form, the cross sectional area. So a Steel column can be way more slender than a concrete column, just to bear the load. t ~ =f =q =b =l =. ` =p \. Imagine the effect of purely axial load applied in this element.

2 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: lu M1 . k 34 12 . r M2 . ` = =. p . 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.

3 T ~ = = =b =l =. ` =p \. 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: Mc ns M 2. In case our calculations provide minimal result we can apply the minimum eccentricity formula: e min + h The moment magnifier ns is used to estimate the lateral deflection effect. It involves the code modificator Cm which is also given below: Cm M1. ns C + . m Pu M2. 1 .75 P. c .. e =_ =f =^ =l =. ` . 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.

4 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: M magn := M u . 1. Pu 1 . Pc . b ~ . Calculate the primary moment due to a lateral 20k load and determine th total moment. bw=1ft, h= , k= and lu=18ft fy := 60ksi f'c := 3ksi lu := 18ft Plat := 20kip b w := 1ft Pu := 125kip M uini := 0k' k' Vu := 0kip h := k := 1. Estimatiing primary moment: Plat lu M u := M u = 90 k'.

5 4. Calculating Young's modulus of concrete: f'c Ec := 57000 psi Ec = ksi psi Using Euler's buckling load formula for secondary moment : 3. b w h 4. I := I = 3375 in 12. 2. Ec I. Pc := Pc = kip ( k lu )2. M magn := M u . 1. M magn = k'. Pu 1 . Pc . t ~ =^ =a =` ~ =. ^ =l =p ~ \. 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 and that is to be used for braced frames without transverse loads.

6 For other cases the value to be taken is M1. Cm + M2. `lirjkp= =. pqo^fk=afpqof_rqflk=J=. b ~ . Problem Statement: Determine whether or not a 16*20 in section with 12#10 bars is adequate for Pu=1080k @ minimum eccentricity (Code clause ) about the minor axis of the column. The Column height Lu= As_10 := f'c := 5000 fy := 60000 Lu := b w := 16 h := 20 Using inches and lb for consistency Processing Data: DL := 680000lbf DL_factored := DL DL_factored = 816 kip LL := 165000lbf LL_factored := LL LL_factored = 264 kip PD := DL. Pu := LL + DL Pu = 1080 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 := b ~ = K.

7 KLu := k Lu kLu = in Determining the M2min: M2min := Pu [ ( + h) ( 1 in) ] M2min = 1296 kip in OR M2min = 108 k'. Note: As stated in the handed out code (ACI318 ), the units of 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. M2 := M2min & M1 := M2min Solution: Determining the Cm (factor relating the actual moment diagram of a slender column to an equivalent uniform moment diagram: Cm := + . M1 Cm = 1. M2 . Determining the modulus of elasticity of Concrete: ksi.)

8 Ec := 57000 f'c Ec = ksi 1000 . b ~ = K. Calculating the ( d) ratio of maximum factored axial Dead Load to the total axial load: PD. d := d = Pu Determining the (Ig) gross moment of Inertia of the element along the minor axis (see problem statement): bw h3 . Ig := ( ). in 4. Ig = in 4. 12 . Solving for the moment of Inertia of the steel rebars using the Ad^2 formula. There shall be " cover, allowing 11" along the short axis, so the distance of the outer bars shall be " and the distance of the inner bars shall be 11"/(3*2), or ": d1 := d2 := n1 := 8 n2 := 4 Es := 29000ksi ( 2. Is := As_10 n1 d1 + n2 d2 in 2 ) ( 4) Is = in 4. Calculating the EI (stiffness) - See ACI code & : ( Ec Ig + Es Is).

9 EI := 7 2. ( 1 + d) EI = 10 kip in C l l ti th (P ) iti ll d b ~ = K. Calculating the (Pc) critical load: 2. EI. Pc := Pc = kip 2. ( kLu). Calculating the ns (moment magnification factor - applied to frame Columns that are braced against sidesway, reflecting effects of member curvature between ends): Cm ns := ns = Pu 1 . Pc Considering the ultimate moment (Mu) to be equivalent to the critical Moment (Mc) now we solve for the Mc: Mc := ns M2 Mc = k'. Pu = 1080 kip `lirjkp= =. pqo^fk=afpqof_rqflk We see the Interaction Column design Moment vs Axial load diagram can 1400. determine the 1200. capacity of stress 1000. that can be 800. applied to a 600. Pn (K).

10 Column. 400. 200. Let's use the 0. Excel sheet -200. 0 50 100 150 200 250 300 350. provided for in -400. class exercise. -600. Mn (K'). 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 g of 4%. Processing Data: Factorizing Dead and Live load: Dead load is multiplied by a factor of and Live load by a factor of : PD := 620 kip PD_factored := PD PD_factored = 744 kip PL := 328 kip PL_factored := PL PL_factored = kip Pu := PD_factored + PL_factored Pu = kip MD := 0 k' MD_factored := MD MD_factored = 0 k'.