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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=29000000psiConcrete: E=3600000psi

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.

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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:

klur

⋅ 34 12M1M2

−≤

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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.

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Also we need to consider the following:ACI 10.12.3. defines Mc as the magnified moment and M2 the larger factored end moment of a no sway compression member:

In case our calculations provide minimal result we can apply the minimum eccentricity formula:

The moment magnifier δns is used to estimate the lateral deflection effect. It involves the code modificator Cm which is also given below:

emin 0.6 0.03 h⋅+

Mc δns M2⋅

δnsCm

1Pu

.75 Pc⋅−

1.0≥ Cm 0.6 0.4M1M2⋅+ 0.4≥

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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:

Mmagn Mu1

1PuPc

−

⋅:=

bñ~ãéäÉ

Mmagn 95.347 k'=Mmagn Mu1

1PuPc

−

⋅:=

Pc 2228.956 kip=Pcπ

2Ec⋅ I⋅

k lu⋅( )2:=

I 3375 in4=I

bw h3⋅

12:=

Using Euler's buckling load formula for secondary moment:

Ec 3122.02 ksi=Ec 57000f'cpsi

psi⋅:=

Calculating Young's modulus of concrete:

Mu 90 k'=MuPlat lu

4:=

Estimatiing primary moment: k 1:=h 1.25ft:=Vu 0kip:=Muini 0k' k':=Pu 125kip:=

bw 1ft:=Plat 20kip:=lu 18ft:=f'c 3ksi:=fy 60ksi:=

Calculate the primary moment due to a lateral 20k load and determine thtotal moment. bw=1ft, h=1.25ft, k=1.0 and lu=18ft

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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

Cm 0.6 0.4M1M2⋅+ 0.4≥

`lirjkp=Ó=pqo^fk=afpqof_rqflk=J=bñ~ãéäÉ

k 0.8:=

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:

PL LL:=Pu 1080 kip=Pu LL 1.6⋅ DL 1.2⋅+:=

PD DL:=LL_factored 264 kip=LL_factored LL 1.6⋅:=LL 165000lbf:=

DL_factored 816 kip=DL_factored DL 1.2⋅:=DL 680000lbf:=Processing Data:

Using inches and lb for consistencyh 20:=bw 16:=Lu 248.4in:=fy 60000:=f'c 5000:=As_10 1.27:=

Determine whether or not a 16*20 in section with 12#10 bars is adequate for Pu=1080k @ minimum eccentricity (Code clause 10.12.2) about the minor axis of the column. The Column height Lu=20.7ft

Problem Statement:

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Ec 4030.509 ksi=Ec 57000 f'cksi

1000

:=

Determining the modulus of elasticity of Concrete:

Cm 1=Cm 0.6 0.4M1M2

+:=

Determining the Cm (factor relating the actual moment diagram of a slender column to an equivalent uniform moment diagram:

Solution:

M1 M2min:=&M2 M2min:=

Note: As stated in the handed out code (ACI318 10.12.3.2), 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.

M2min 108 k'=ORM2min 1296 kip in⋅=M2min Pu 0.6 0.03 h⋅+( ) 1 in⋅( )⋅[ ]⋅:=

Determining the M2min:kLu 198.72 in=kLu k Lu⋅:=

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C l l ti th (P ) iti l l d

EI 1.026 107× kip in2⋅=EI0.2 Ec⋅ Ig⋅ Es Is⋅+( )

1 βd+( ):=

Calculating the EI (stiffness) - See ACI code 10.12.2 & 10.12.3:

Is 324.414 in4=Is As_10 n1 d12⋅ n2 d22⋅+( )⋅ in4( )⋅:=

Es 29000ksi:=n2 4:=n1 8:=d2 1.8333:=d1 5.5:=

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 inner bars shall be 11"/(3*2), or 1.833":

Ig 10666.667 in4=Igbw h3⋅

12

in4( )⋅:=

Determining the (Ig) gross moment of Inertia of the element along the minor axis (see problem statement):

βd 0.756=βd 1.2PDPu⋅:=

Calculating the (βd) ratio of maximum factored axial Dead Load to the total axial load:

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Calculating the (Pc) critical load:

Pcπ

2 EI⋅

kLu( )2:= Pc 2563.477 kip=

Calculating the δns (moment magnification factor - applied to frame columns that are braced against sidesway, reflecting effects of member curvature between ends):

δnsCm

1Pu

0.75 Pc⋅−

:= δns 2.2817=

Considering the ultimate moment (Mu) to be equivalent to the critical Moment (Mc) now we solve for the Mc:

Mc δns M2⋅:= Mc 246.4274 k'=Pu 1080 kip=

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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.

Column designMoment vs Axial load

-600

-400

-200

0

200

400

600

800

1000

1200

1400

0 50 100 150 200 250 300 350

ΦMn (K')

ΦPn

(K)

pmfo^i=obfkclo`bjbkq`lk`obqb=`lirjk=abpfdkProblem Statement:

Select a cross section for a spirally reinforced column section to support the loads indicated below, usingf'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 524.8 kip= Pu PD_factored PL_factored+:=

Pu 1268.8 kip=MD 0k':= MD_factored MD 1.2⋅:= MD_factored 0 k'=

ML 80k':= ML_factored ML 1.6⋅:= ML_factored 128 k'= Mu MD_factored ML_factored+:=

Mu 128 k'=f'c 5ksi:= fy 60ksi:=

Φ .75:= ϕ ϕACI 318 - 10.3.6: For spiral =.75, for tied =.65Also, the first factor changes from .85 to .80 for tied.

Solution:The length value that is the result of the division of the applied moment by the applied axial load is theeccentricity "ecc" of the column. Almost always, a compression member may assume a moment either because the axial load is notperfectly centered on the column, or because the column will resist portion of the unbalanced momentsat the ends of the beams it supports. In order not to confuse the term "e" with the base of the naturallogarithm, we can use the term ecc.

Eccentricity should not be higher than 10%, sowe can set a minimum diameter of one foot ecc

MuPu

:= ecc 1.211 in=

For reinforcement of 3% the letter "ρ" that signifies density is used

ρg_max .08≤ ρg_min .01≥ ρg .04:=ρg

AstAg

=

The reason we use the ρg is to determine the area of the column. The ACI code gives the following formula for non prestressedmembers w/spiral reinforcements (ACI 318 10.3.6.1)

maxΦPn Pu:= maxΦPn 0.85 Φ 0.85 f'c⋅ Ag( )⋅ ρg Ag⋅ fy 0.85 f'c⋅−( )⋅+[ ]⋅=Inversing data:

AgmaxΦPn

Φ 0.85⋅ 0.85 f'c⋅ ρg fy 0.85 f'c⋅−( )⋅+[ ]⋅:= Ag 307.141 in2=

Since the column shall be circular, we can use the formula of thecircle's area to estimate an approximate radius and round it.

rAgπ

:= r 9.888 in=

Rounding.. r 10in:=Therefore: Ag π r2:= Ag 314.159 in2=

Using the following formulas now we can determine the steelreinforcement

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 whichis 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 12.566 in2= This seems like a good starting point.

Given maxΦPn 0.85Φ 0.85f'c Ag As−( )⋅ fy As⋅+[ ]⋅= (ACI 10.3.5.1)

Ast_value Find As( ):=

As Ast_value:= As 11.751 in2=

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:=1.56 8⋅ 12.48=

AsBarSize1

16

2

n1⋅BarSize2

16

2

n2⋅+

π in2⋅:= As 11.879 in2=

As 12.48in2:=Verifying.....

maxΦPn 0.85Φ 0.85f'c Ag As−( )⋅ fy As⋅+[ ]⋅:= maxΦPn 1294.722 kip=

ρgAsAg

:=ρg 0.0397=

This is a very good result. We are only very slightly above our ρg ratio is as we initially aimed for.

o`=qfba=`lirjk=aÉëáÖåDesign a tied column cross section to support an axial load of 350 kips, amoment load of 110kip feet, and a shear load of 14kips. All of the above arefactorized. The column is in a braced frame with an unsupported length of10 ft 6 inches.

fy 60ksi:= f'c 4ksi:= lu 10.5ft:= Clearcover 1.5in:=

Pu 300kip:= Mu 110k':= Vu 14kip:=ϕ .65:=

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%

ρt .015:=

Estimating the initial dimensios of the column:

The cross sectional area for a tied column is given by the following formula: ACI 318: 10.3.6.2

Ag_ini5 Pu⋅

2.21 f'c⋅ 2.6 fy⋅ ρt⋅+ 2.21 f'c⋅ ρt⋅−:= Ag_ini 135.8 in2

=

Given the option that we may design a square base column.....we estimate an initial base value:

bini Ag_ini:= bini 11.65 in=

Given the fact that there are significant moments applied on this column, it would be wise tooverride the initial calculation that takes only direct loads into account. Let's round it up about10 -15% on each side:

btrial truncbiniin

1.25⋅

in:= btrial 14 in= b btrial:=

h b:=Ag b h⋅:= Ag 196 in2=

Determining the bar arrangement:

To determine the preferable bar arrangementwe compute the ratio of eccentricity to theheight "h" of the column: Note: This is "h" incross section, not the actual column height.

eccMuPu

:= ecc 4.4 in=

According to the figure indicated it willbe more appropriate to apply re-bars onboth sides of this column:

ecch

0.31=

Column slenderness can be neglected if:

klur

⋅ 34 12M1M2

⋅−≤

According to ACI code 10.11.2, the radius of gyration of rectangular columns is 0.3h and .25D forcircular columns. Since this is a braced frame k is lesser or equal to1.0 and the ratio of M1 to M2can vary between +/- 0.5. We can assume that k=1.0 and M1/M2=0.5. Therefore the aboverelations yield the following results:

r .3 h⋅:= k 1:=M1M2

0.5=klur

⋅ 30= 34 12M1M2

− 28=

Slendernesscond if klur

⋅ 34 12M1M2

⋅−≤ "Neglect column slenderness", "Design slender column",

:=

Slendernesscond "Design slender column"=

Ib h3

⋅

12:= I 3201.33 in4

=

Ec 57000f'cpsi

psi⋅:= Ec 3605 ksi=

Pcπ

2 Ec⋅ I⋅

k lu⋅( )2:= Pc 7174.55 kip=

Mmagn Mu1

1PuPc

−

⋅:= Mmagn 114.8 k'=

Computing the "γ" ratio :

At this point we need to compute the value gamma (γ) which is the ratio of distance of centroids ofouter rows of bars and column dimension perpendicular to the bending axis: We shall assume thatthe ties are #3 rebars and the longitudinal bars are #7:

dlbar .875in:= dtbar .375in:=

γ

h 2 Clearcover dtbar+dlbar

2+

−

h:=

γ 0.6696=

We need to point out that the assumption we make about the #7 rebars may prove imprecise, inwhich case we shall need to reiterate this process. Given the gamma value above we will refer tothe ACI interaction diagrams (or use our own system!!!) to define again the ratio of steel ρt.

Puϕ f'c⋅ Ag⋅

0.589=

Mmagnϕ f'c⋅ Ag⋅ h⋅

0.193=

For the above values the interaction diagram for gamma 0.6 gives a rho value of 0.03. The diagramfor a gamma value of 0.7 gives a rho value of 0.025. We shall apply linear interpolation to computethe value of rho at gamma found. ACI code defines that rho should lie between 0.01 and 0.08.

ρt is the ratio of total reonforcement area divided by the cross sectional area of a column

ρt 0.03 0.03 0.025−( )γ .6−( ).7 .6−( )

⋅−:= ρt 0.0265=

Rholo_condition if ρt .01< "Low value", "OK", ( ):= Rholo_condition "OK"=

Rhohi_condition if ρt .08> "High value", "OK", ( ):= Rhohi_condition "OK"=

Selecting reinforcement:As ρt Ag⋅:= As 5.1975 in2

= As

.44in211.8=

We can select twelve 6 bars, six on each face but let's verifythat spacing will be approved:.

nlbar 6:= dlbar .75in:= Albar 0.44in2:=

sb 2 Clearcover⋅− 2dtbar− nlbar dlbar⋅−( )

nlbar 1−:= s 1.15 in=

Spacing if s max dlbar 1in, ( )< "Redesign", "OK", ( ):= Spacing "OK"=

Design the lap splices:

Bar Designation Number

Weight per foot (lbf) Diameter db Area As Perimeter

3 0.376 0.375 0.11 1.1784 0.668 0.500 0.20 1.5715 1.043 0.625 0.31 1.9636 1.502 0.750 0.44 2.3567 2.044 0.875 0.60 2.7498 2.670 1.000 0.79 3.1429 3.400 1.128 1.00 3.54410 4.318 1.270 1.27 3.99011 5.304 1.410 1.56 4.43014 7.650 1.693 2.25 5.31918 13.600 2.257 4.00 7.091

ld6

1.3dlbarfypsi

25f'cpsi

:= ld6 36.99865 in=

ld6 3.08322 ft=

So let's reevaluate our rho value:

As_fin 2nlbar Albar⋅:= As_fin 5.28 in2=

ρAs_fin

Ag:= ρ 0.02694=

Selecting the ties:

Based on ACI 7.10.5.2, the least of the following three conditions determines the spacing of theties:colleast_dim if b h< b, h, ( ):= colleast_dim 14 in= 48 dtbar⋅ 18 in= 16 dlbar⋅ 12 in=

TieSpacing min colleast_dim 48 dtbar⋅, 16 dlbar⋅, ( ):= TieSpacing 12 in=

We need to make reference to the subject we addressed on principal stresses to visualize the effectof the following formula. The factor "Nu" represents an axial tension force resulting from thecompression. The angle theta "θ" to be used is the critical 45 degrees.

θ 45deg:=ϕ .85:=

d b 2 Clearcover dtbar+dlbar

2+

⋅−:= d 9.5 in=

NuPu

tan θ( ):= Nu 300 kip=

ΦVc ϕ 2⋅f'cpsi

⋅ psi⋅ b⋅ d⋅ 1

Nulbf

2000Ag

in2

+

⋅:= ΦVc 25.2 kip=

Check ACI sections 7.10.5, 11.4.5.1 and 11.4.6.3.If Vc<Vu<Vc, the ACI code section 7.10.5 governs.The end result...#3 ties @ 12 inches o.c.

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Reading: Required: Furlong, Chapt 7.3 through 7.6 incl.Recommended: McCormac & Nelson, Chapter 11 (pp. 317-333) for this week’s lectures.