= =

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.

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).

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:

At N/A σ=o: ÅÅ

ÅÅ

f = PA+M∗ y

I

y= IA∗ecc

Equilibrium

How equilibrium is established:

ÅÅ

f = PA+

(P∗ecc)∗ yI

ΣF=0 P=C−T

ΣM=P∗ecc

P∗ecc=C∗y1+T∗y2

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.

RC Column Analysis

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) 9.5.2.3 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:

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

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.)

In Class Example cont

In Class Example cont

In Class Example cont

In Class Example cont

In Class Example cont

In Class Example cont

θ/in M (k`) Comments

1.85E-05 47.56 just before cracking

41.77E-06 47.56 just after cracking

1.147E-04 130.62 fc=0.7*f`c

2.51E-04 286.26fs=fy (assumes linear

behavior)

In Class Example cont

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.

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)

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.

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.

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∗lur

≤34−12 [M 1

M 2 ]

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:

What is the Effect of Column Slenderness?

M c=δns∗M 2

eccmin=0.6+0.03∗h

δns=Cm

1−Pu

0.75∗Pc

≥1.0 Cm=0.6+0.4M 1

M 2≥0.4

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:

M magn=M u [ 1

1−PuPc ]

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

Cm=0.6+0.4M 1

M 2≥0.4

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∗l4

=90k '

Ec=57000√ f ' c=3122.02ksi I= b∗h3

12=3375inch

4

Pc=π2∗Ec∗I(k∗lu)

2 =2228.96kip M magn=M u [ 1

1−PuPc ]=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.

RC Tied Column Design Design 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 aboveare factorized. The column is in a braced frame with an unsupportedlength of 10 ft 6 inches:

Data:Pu 350kip:= Mu 110k':= Vu 14kip:= Φ .85:= fy 60ksi:= f'c 3ksi:= lu 10.5ft:= Clearcover 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%

ρt .015:=

Estimating the initial dimensios of the column:

The cross sectional area for a tied column is given by the following formula (variance of 22.4.2.2)

Ag_iniPu

.40 f'c fy ρt⋅+( ):= Ag_ini 224.36 in2

=

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

bini Ag_ini:= bini 14.98 in=

Given the fact that there are significant moments applied on this column, it would be wise to override the initial calculation that takesonly direct loads into account. Let's round it up about 10 -15% on each side:

btrial trunc bini 1.12⋅( )= btrial 16 in= b btrial:=

h b:=Ag b h⋅:= Ag 256 in2=

Determining the bar arrangement:

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

eccMuPu

:= ecc 3.77 in=

According to the figure indicated it will be more appropriate toapply re-bars on both sides of this column: ecc

h0.24=

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 for circular columns. Since this is abraced frame k is lesser or equal to1.0 and the ratio of M1 to M2 can vary between +/- 0.5. We can assume that k=1.0 andM1/M2=0.5. Therefore the above relations yield the following results:

r .3 h⋅:= k 1:=M1M2

0.5=klur

⋅ 26.25= 34 12M1M2

− 28=

Slendernesscondition if klur

⋅ 34 12M1M2

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

:=

Slendernesscondition "Neglect column slenderness"=

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 columndimension perpendicular to the bending axis: We shall assume that the ties are #3 rebars and the longitudinal bars are #7:

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

γb 2 Clearcover dtbar+

dlbar2

+

−

h:= γ 0.7109=

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

PuAg

1.367 ksi=Mu

Ag h⋅0.322 ksi=

For the above values the interaction diagram for gamma 0.6 gives a rho value of 0.014. The diagram for a gamma value of 0.75 givesa rho value of 0.013. We shall apply linear interpolation to compute the value of rho at gamma found. ACI code defines that rhoshould 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 0.014 0.014 0.013−( )γ .6−( )

.75 .6−( )⋅−:= ρt 0.01326=

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 3.3947 in2

=

We can select six #7 bars, three on each face.Design the lap splices:

ld71.3dlbar fy

20 f'c= ld7 62.30344 in= ld7 5.19195 ft=

This splice is far too long, about half the length of the column. Consider using eight #6 bars

dlbar .75in:=8 .44⋅ in2 3.5 in2

=

ld61.3dlbar fy

25 f'c= ld6 42.72236 in= ld6 3.5602 ft=

So let's reevaluate our rho value:

nlbar 8:= Albar 0.44in2:=

As_fin nlbar Albar⋅:= ρAs_fin

Ag:= ρ 0.01375=As_fin 3.52 in2

=

Selecting the ties:

Based on ACI 318-14 sections 9.7.6.4.3, 23.6.3.1, and 25.7.2.1, the least of the following three conditions determines the spacingof the ties:

16 dlbar⋅ 12 in=

48 dtbar⋅ 18 in=

colleast_dim if b h< b, h, ( ):= colleast_dim 16 in=

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

θ 45deg:=

d b 1 Clearcover dtbar+dlbar

2+

⋅−:= d 13.75 in=

NuPu

tan θ( ):= Nu 350 kip=

ΦVc Φ 2⋅ f'c⋅ b⋅ d⋅ 1Nu( )

2000Ag+

⋅= ΦVc 34.5 kip=

Check ACI sections 318-14 sections 9.6.3.3, 9.7.6.2.2, 9.7.6.4.3, 10.6.2.2, 10.7.6.5.2, 15.4.2, 23.6.3.1, and 25.7.2.1 If Vc<Vu<Vc, the ACI code section ACI 314-11 7.10.5 governs.... The end result...#3 ties @ 12 inches o.c.

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.

COLUMN SLENDERNESSInserting Excel

Problem Statement:Determine whether or not a 16*20 inch section with 12#10 bars is adequate for Pu=1079k @ minimumeccentricity (Code clause 10.12.2) about the minor axis of the column. The Column height Lu=20.7ft

As_10 1.27in2:= f'c 5ksi:= fy 60ksi:= Lu 248.4in:= bw 16in:= h 20in:= Using inches and lb for consistency

Processing Data:DL 583000lbf:= DL_factored DL 1.4⋅:= DL_factored 816kip=

LL 154588lbf:= LL_factored LL 1.7⋅:= LL_factored 263kip= PD DL:=

Pu LL 1.7⋅ DL 1.4⋅+:= Pu 1079kip= PL LL:=

Assuming that M1=minM2 to cause compression on the same face such that M1/M2=1and k factor is forelastic connection on a multi-story building:k 0.7:=

kLu k Lu⋅:= kLu 173.88 in=Determining the M2min:M2min Pu 0.6in 0.03 bw⋅+( )( )⋅:= M2min 1165.32kip in⋅= OR M2min 97.11k'=

Note: As stated in the 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 wetreat 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 uniformmoment diagram:

Cm 0.6 0.4M1M2

+:= Cm 1=

Determining the modulus of elasticity of Concrete:

Ec 57000f'cpsi

psi⋅:= Ec 4030.509ksi=

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

βd 1.4PDPu⋅:= βd 0.756=

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

Igh bw

3⋅

12

:= Ig 6826.667 in4=

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":

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

Is As_10 n1 d12⋅ n2 d2

2⋅+( )⋅:= Is 324.414 in4=

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

EI0.2 Ec⋅ Ig⋅ Es Is⋅+( )

1 βd+( ):= EI 5.895 104× kip ft2⋅=

Calculating the (Pc) critical load:

Pcπ2 EI⋅

kLu( )2:= Pc 2771.234kip=

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

δnsCm

1Pu

0.75 Pc⋅−

:= δns 2.0796=

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

Mc δns M2⋅:= Mc 201.9517k'=Pu 1079kip=

Column designMoment vs Axial load

-500

0

500

1000

1500

2000

0 100 200 300 400 500 600

ΦMn (K)

ΦPn

(K)

Using the provided Excel file we can have the graph of this concrete column configuration. Based on ourdata, the Mc and Pu fall within the ΦPn/ΦMn curve. The column is approved.