ch 4 - ae design of columns

8/19/2019 CH 4 - AE Design of Columns

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CHAPTER 4

COLUMNS:

COMBINED AXIAL LOAD & BENDING

Abrham E.Sophonyas A.

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4.0 Introduction

A column is a vertical structural member supporting

axial compressive loads, with or without moments. The cross-sectional dimensions of a column are

generally considerably less than its height.

Columns support vertical loads from floors and roof and

transmit these loads to the foundations.

Columns may be classified based on the following

criteria:

a) On the basis of geometry; rectangular, square, circular,L-shaped, T-shaped, etc. depending on the structural

or architectural requirements

b) On the basis of composition; composite columns, in-

filled columns, etc. 2


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c) On the basis of lateral reinforcement; tied columns,

spiral columns.

d) On the basis of manner by which lateral stability isprovided to the structure as a whole; braced columns,

un-braced columns.

e) On the basis of sensitivity to second order effect due

to lateral displacements; sway columns, non-swaycolumns.

f) On the basis of degree of slenderness; short column,

slender column.g) On the basis of loading: axially loaded column,

columns under uni-axial bending, columns under

biaxial bending.

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4.0 Introduction

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4.0 Introduction The more general terms compression members and

members subjected to combined axial loads & bending

are used to refer to columns, walls, and members in

concrete trusses and frames.

These may be vertical, inclined, or horizontal.

A column is a special case of a compression memberthat is vertical.

Although the theory developed in this chapter applies to

columns in seismic regions, such columns require

special detailing to resist the shear forces and repeated

cyclic loading from the EQ.

In seismic regions the ties are heavier & more closely

spaced. 5


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4.1 Tied and Spiral Columns

Most of the columns in buildings in nonseismic regions

are tied columns.

Occasionally, when high strength and/or ductility are

required, the bars are placed in a circle, and the ties

are replaced by a bar bent into a helix or a spiral, with apitch from 35 to 85 mm.

Such a column, is called a spiral column. (Fig. 11-3)

The spiral acts to restrain the lateral expansion of the

column core under axial loads causing crushing and, indoing so, delays the failure of the core, making the

column more ductile.

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4.1.1 Behavior of Tied and Spiral Columns

Fig. 11-4a shows a portion of the core of a spiral columnenclosed by one and a half turns of a spiral.

Under a compressive load, the concrete in this column

shortens longitudinally under the stress f 1 and so, to

satisfy the Poisson’s ratio, it expands laterally.

This lateral expansion is especially pronounced at

stresses in excess of 70% of the cylinder strength.

In spiral column, the lateral expansion of the concreteinside the spiral (the core) is restrained by the spiral.

These stresses the spiral in tension (see fig 11.14).

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9

Fig 11-4 Triaxial stresses

in core of spiral column

f 1

f 2

f 2

f 2

f 1

f 2

f 1

f 2

f 2

f spf sp

s

f sp

f sp

Dc

f 1


f 2

f 2


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For equilibrium the concrete is subjected to lateral

compressive stresses, f 2.An element taken out of the core (see fig) is subjected

to triaxial compression which increases the strength of

concrete: f 1 = f c’+4.1f 2.

In a tied column in a non-seismic region, the ties are

spaced roughly the width of the column apart and thus

provide relatively little lateral restraint to the core.

Hence, normal ties have little effect on the strength ofthe core in a tied column.

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They do, however, act to reduce the unsupported length

of the longitudinal bars, thus reducing the danger ofbuckling of those bars as the bar stresses approach

yield.

Fig 11-5 presents load-deflection diagrams for a tied

column and a spiral column subjected to axial loads.

The initial parts of these diagrams are similar.

As the maximum load is reached, vertical cracks and

crushing develop in the concrete shell outside the tiesor spiral, and this concrete spalls off .

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When this occurs in a tied column, the capacity of the

core that remains is less than the load on the column. The concrete core is crushed, and the reinforcements

buckle outward b/n ties.

This occurs suddenly, w/o warning, in a brittle manner.

When the shell spalls off a spiral column, the column

does not fail immediately because the strength of the

cores has been enhanced by the triaxial stresses.

As a result, the column can undergo large deformations,eventually reaching a 2nd maximum load, when the

spirals yield and the column finally collapses.

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13

Fig 11-5



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Such a failure is much more ductile and gives warning

of impending failure, along with possible loadredistribution to other members

Fig 11-6 and 11-7 show tied and spiral columns,respectively, after an EQ. Both columns are in the same

building and have undergone the same deformations. The tied column has failed completely, while the spiral

column, although badly damaged, is still supporting aload.

The very minimal ties in Fig 11-6 were inadequate toconfine the core concrete.

Had the column been detailed according to ACI Section21.4, the column would have performed much better.

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Sway Frames vs. Nonsway Frames

A nonsway (braced) frame is one in which the lateralstability of the structure as a whole is provided by walls,

bracings, or buttresses, rigid enough to resist all lateral

forces in the direction under consideration.

A sway (unbraced) frame is one that depends on

moments in the columns to resist lateral loads and

lateral deflections. The applied lateral-load moment, Vl,

and the moment due to the vertical loads, ΣP∆ shall be

equilibrated by the sum of the moments at the top and

bottom of all the columns as shown in the figure below.

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Fig. Non-sway Frame / Braced

columnsFig. Sway Frame/ Un-braced

columns



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For design purpose, a given story in a frame canbe considered “non-sway” if horizontaldisplacements do not significantly reduce thevertical load carrying capacity of the structure.

In other words, a frame can be “non-sway” if theP-∆ moments due to lateral deflections are smallcompared with the first order moments due tolateral loads.

In sway frames, it is not possible to considercolumns independently as all columns in thatframe deflect laterally by the same amount.

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Columns are broadly categorized in to two as short

and slender columns.

Short columns are columns for which the strength is

governed by the strength of the materials and the

geometry of the cross section. In short columns, Second-order effects are negligible.

In these cases, it is not necessary to consider

slenderness effects and compression members can

be designed based on forces determined from first-

order analyses.

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Slender Columns vs. Short Columns


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When the unsupported length of the column is long,lateral deflections shall be so high that the moments

shall increase and weaken the column.

Such a column, whose axial load carrying capacity is

significantly reduced by moments resulting fromlateral deflections of the column, is referred to as a

slender column or sometimes as a long column.

“Significant reduction”, according to ACI, has been

taken to be any value greater than 5%.

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When slenderness effects cannot be neglected, the

design of compression members, restraining beamsand other supporting members shall be based on the

factored forces and moments from a second-order

analysis.

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Fig. Forces in slender column


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4.2 Strength of Axially Loaded Columns

When a symmetrical column is subjected to a

concentric axial load, P, longitudinal strains , developuniformly across the section as shown in Fig 11-8a.

Because the steel & concrete are bonded together, the

strains in the concrete & steel are equal.

For any given strain, it is possible to compute the

stresses in the concrete & steel using the stress-strain

curves for the two materials.

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Failure occurs when Po reaches amaximum:

Po = f cd(Ag – As,tot ) + f ydAs,tot


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4.3 Interaction Diagrams

Almost all compression members in concrete structures

are subjected to moments in addition to axial loads.

These may be due to misalignment of the load on the

column, as shown in Fig 11-9b, or may result from the

column resisting a portion of the unbalanced momentsat the ends of the beams supported by the columns (Fig

11-9c).

The distance e is referred to as the eccentricity of load.

These two cases are the same, because the eccentric

load can be replaced by an axial load P plus a moment

M=Pe about the centroid.

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Fig. 11-9 Load and moment

on column.


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The load P and moment M are calculated w.r.t. the

geometric centroidal axis because the moments andforces obtained from structural analysis are referred to

this axis.

For an idealized homogeneous and elastic column with a

compressive strength, f cu, equal to its tensile strength,

f tu, failure would occur in compression when the

maximum stresses reached f cu, as given by:

Dividing both sides by f cu gives:

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f cu

f cu = 1

= f cu


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The maximum axial load the column can support occurs

when M = 0 and is Pmax = f cuA. Similarly, the maximum moment that can be supported

occurs when P = 0 and M is Mmax = f cuI/y.

Substituting Pmax and Mmax gives:

This equation is known as an interaction equation,

because it shows the interaction of, or relationship

between, P and M at failure. Points on the lines plotted in this figure represent

combinations of P and M corresponding to the

resistance of the section.28

Pmax Mmax = 1



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Fig. 11-10 Interaction diagram for an

elastic column, |f cu| = |f tu|. 29


A point inside the diagram, such as

E , represents a combination of Pand M that will not cause failure

Combinations of P and M falling onthe line or outside the line, such as

point F , will equal or exceed the

resistance of the section and hence

will cause failure


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4.4 Interaction Diagrams for Reinforced

Concrete Columns Since reinforced concrete is not elastic & has a tensile

strength that is lower than its compressive strength,

the general shape of the diagram resembles fig. 11.13

4.4.1 Strain Compatibility Solution

Interaction diagrams for columns are generallycomputed by assuming a series of strain distributions

at the ULS, each corresponding to a particular point on

the interaction diagram, and computing the

corresponding values of P and M.

Once enough such points have been computed, the

results are summarized in an interaction diagram.

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Assume strain distribution and select the location of the

neutral axis.

Compute the strain in each level of reinforcement from

the strain distribution.

Using this information, compute the size of thecompression stress block and the stress in each layer of

reinforcement.

Compute the forces in the concrete and the steel layers,by multiplying the stresses by the areas on which they

act.

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Concrete Columns

i i f i f d


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Finally, compute the axial force Pn by summing theindividual forces in the concrete and steel, and the

moment Mn by summing the moments of these forces

about the geometric centroid of the cross section.

These values of Pn and Mn represent one point on the

interaction diagram.

Other points on the interaction diagram can be

generated by selecting other values for the depth, c, tothe neutral axis from the extreme compression fiber.

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Concrete Columns


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33

10o/oo

(3/7)h

=2.0o/oo

A

B

10o/oo

=3.5o/oo


Concrete Columns


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Fig 11-13 Strain distributions

corresponding to points on

the interaction diagram

Moment resistance

A x i a l l o a d

r e s i s t a n c e

Onset of cracking

Balanced failure

Uniform compression


Concrete Columns


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In the actual design, interaction charts prepared foruniaxial bending can be used. The procedure involves:

Assume a cross section, d’ and evaluate d’/h to chooseappropriate chart

Compute:

Normal force ratio: = /ℎ Moment ratios:

= /ℎ

Enter the chart and pick ω (the mechanical steel ratio), ifthe coordinate (ν, μ) lies within the families of curves. Ifthe coordinate (ν, μ) lies outside the chart, the crosssection is small and a new trail need to be made.

Compute , = / Check Atot satisfies the maximum and minimum provisions Determine the distribution of bars in accordance with the

charts requirement

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f f d


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Draw the interaction diagram for the column crosssection. Use C-30 concrete and S-460 steel.

Show a minimum of 6 points on the interaction

diagram corresponding to1. Pure axial compression

2. Balanced failure

3. Zero tension (Onset of cracking)

4. Pure flexure5. A point b/n balanced failure and pure flexure

6. A point b/n pure axial compression and zero tension

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Concrete Columns

f f d


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Concrete Columns

i i f i f d


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• Solution

1. Pure axial compression

39

s2

s1

Cs2

Cs1

Ccc

= 2/oo

Cross section Strain distribution Stress resultant

(ULS)


Concrete Columns

4 4 I i Di f R i f d


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Cont’d

Pu = Cs2 + Cs1 + Cc = s2As2 + s1As1 + f cdbh

yd = f yd/Es = 400/200000 = 0.002

reinforcement has yielded

Pu = f ydAs,tot/2 + f ydAs,tot/2 + f cdbh = f ydAs,tot + f cdbh

u = Pu / f cdbh = (f ydAs,tot )/ (f cdbh)+ (f cdbh)/(f cdbh)

= f yd/f cd + 1 = + 1 ; where is called the

mechanical reinforcement ratio and equal to= (6800/(400500))(400/13.6) = 1.0

u = 1 + 1 = 2.0

NB: νu= Relative design axial force (Pu/ (f cdAc)) 40


Concrete Columns

4 4 I i Di f R i f d


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Concrete Columns

4 4 Interaction Diagrams for Reinforced


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2. Balanced failure

– x/3.5 = d/(3.5+2)

x = (400/5.5)3.5 = 254.5454mm s2/(254.54-100)=3.5/254.54

s2 = 2.125 %0 > 2 %0 reinforcement has yielded

Cs2 = Ts1 = 3400400 = 1360000N 42

s2

ydTs1

cm=3.5 /oo

X Cs2Cc

= 2/oo

Cross section strain distribution stress resultant

(ULS)

cd


Concrete Columns

4 4 I t ti Di f R i f d


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•Cont’d

• cm > o and NA is within the section

c = kx(3cm – 2)/3cm

= (254.54/400)(33.5-2)/(33.5) = 0.5151 Cc = cf cdbd = 0.5113.6400400 = 1120967.7 N

• c = kx(cm(3cm-4)+2)/(2cm(3cm-2))

= (254.54/400)(3.5(3

3.5-4)+2)/(2

3.5(3

3.5-2))

= 0.2647 cd = 0.2647400 = 105.882 mm

Mu = Cc(h/2- cd) + Cs2(h/2-h’) + Ts1(h/2-h’)

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Concrete Columns



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– Cont’d

= 1120969.7(250-105.882) + 1360000(250-100) +1360000(250-100)

= 569551912.2 Nmm

• Pu= Cc+Ts1+Cs2 = Cc = 1120969.7

u = P/(f cdbh)

= 1120969.7/(13.6400500)

= 0.412u = Mu/(f cdbh2) = 569551912.2/(13.64005002)

= 0.419

NB: u = Mu/(f cdbh

2

) is called relative moment 44


Concrete Columns



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6. A point b/n pure axial compression and zero

tension

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Cs2

Cs1

Cc

= 2/oo

Cross Section Strain Distribution Stress Resultant

(ULS)

cm=3.0 /oo 1.0

C

e

b

(3/7)h

cba


Concrete Columns



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– Choose cm = 3 %o (strain profile passes also through C)

– Strain in the bottom concrete fiber cb from:

a = ((4/7)/(3/7))1 = 4/3 = 1.33

cb = 2 -1.33 = 0.667 %o

(entire cross section under compression as assumed) – Determine strain in reinforcement from:

– b/114.286 = 1/214.286 b = 0.533 %o

s2 = 2 + 0.533 = 2.533 %o

> 2 %o

reinforcement hasyielded and

– from e/185.714 = 1.33/285.714 e = 0.867 %o

s1 = 2-0.867 = 1.133 %o < 2 %o reinforcement has

not yielded 46


Concrete Columns



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– Cont’d

– cm > o and NA outside of the section

c = (1/189)(125+64cm-16cm2)

= (1/189)(125+64(3)-16(3)2) = 0.915344

Cc = cf cdbd = 0.91534413.6400400 = 1991788.4 N;

Cs2 = (As,tot/2)f yd = 3400400 = 1360000 N;

Cs1 = (As,tot/2)s1 = 3400(1.133/1000)200000

= 770666.67 N – c = 0.5-(40/7)(cm-2)

2/(125+64cm-16cm2)

= 0.5-(40/7(3-2)2/(125+643-1632) = 0.467

cd = 0.467400 = 186.788 mm 47


Concrete Columns



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– Cont’d

– Pu = Cc+Cs1+Cs2 = 4122455.1 N

– Mu = 1991788.4(250-186.788)+1360000(250-

100)-770666.67(250-100) = 214304927.8 Nmm

– u = Pu/(f cdbh) = 4122455.1/(13.6400500) =

1.516 – u = Mu/(f cdbh

2) = 214304927.8/(13.64005002)

= 0.1576

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Concrete Columns



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– 4. Pure flexure

– Start with cm/s1 = 3.5 /oo / 5 /oo and repeat until

u 0.00

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cm=3.5 /oo

X

Cs2Cc

Cross section strain distribution stress resultant

cd

s1=5.0 /oo


Concrete Columns

Ts1



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–Cont’d

– Use cm/s1 = 3.5 /oo / 6.227 /oo

– kx= 3.5/(3.5+6.227)=0.359823 x=143.9293mm

–

s2 = ((x-100)/x)3.5 = 1.06825 /oo

Cs2=3400(1.06825/1000)200000=726410N

– c = kx(3cm – 2)/3cm

= (143.9293/400)(33.5-2)/(33.5) = 0.291285

Cc = cf cdbd = 0.29128513.6400400 = 633837.1N

– Pu=Cc+Ts1+Cs2 = 633837.1-1360000+726410 = 247.1N

u = Pu/(f cdbh) = 247.1/(13.6400500) = 0.0050


Concrete Columns



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– c = kx(cm(3cm - 4) + 2)/(2cm(3cm - 2))

= (143.9293/400)(3.5(33.5-4)+2)/(23.5(33.5-2))

= 0.149674

cd = 0.149674400 = 59.8696 mm – Mu = Cc(h/2- cd) + Cs2(h/2-h’) + Ts1(h/2-h’)

= 633837.1(250-59.8696) + 726410(250-100) +

1360000(250-100)

= 433473111 Nmm

– u = Mu/(f cdbh2)

= 433473111 /(13.6 400 5002) = 0.31951


Concrete Columns

4 4 Interaction Diagrams for RC Columns


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0.2 0.4 0.6 0.8

4.4 Interaction Diagrams for RC Columns


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All columns are (in a strict sense) to be treated as

being subject to axial compression combined with

biaxial bending, as the design must account for

possible eccentricities in loading (emin at least) with

respect to both major and minor principal axes of thecolumn section.

Uniaxial loading is an idealized approximation which

can be made when the e/D ratio with respect to one

of the two principal axes can be considered to benegligible.

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4.5 Biaxially Loaded Columns


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Also, if the e/D ratios are negligible with respect to

both principal axes, conditions of axial loading may

be assumed, as a further approximation.

For a given cross section and reinforcing pattern, one

can draw an interaction diagram for axial load andbending about either axis.

These interaction diagrams form the two edges of an

interaction surface for axial load and bending about

2 axes.

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I i di f i l


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Interaction diagram for compression plus

bi-axial bending

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As shown in the figure above, the interaction diagram

involves a three-dimensional interaction surface foraxial load and bending about the two axes.

The calculation of each point on such a surface involvesa double iteration:

–

The strain gradient across the section is varied, and – The angle of the neutral axis is varied.

There are different methods for the design of Biaxiallyloaded columns:

– Strain compatibility method

– The equivalent eccentricity method

– Load contour method

– Bresler reciprocal load method

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Biaxial interaction diagrams calculated and prepared as

load contours or P-M diagrams drawn on planes of

constant angles relating the magnitudes of the biaxial

moments are more suitable for design (but difficult to

derive). Interaction diagrams are prepared as load contours for

biaxially loaded columns with different reinforcement

arrangement (4-corner reinforcement, 8-rebar

arrangement, uniformly distributed reinforcement on2-edges, uniformly distributed reinforcement on 4-

edges and so on.

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ll d d l


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The procedure for using interaction diagrams for

columns under biaxial bending involves:

Select cross section dimensions h and b and also h’

and b’

Calculate h’/h and b’/b and select suitable chart

Compute:

– Normal force ratio: = /ℎ – Moment ratios: = / ℎ and = /

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ll d d l


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Select suitable chart which satisfy and h’/h and b’/bratio:

Enter the chart to obtain ω

Compute

, = /

Check Atot satisfies the maximum and minimum

provisions

Determine the distribution of bars in accordance

with the chart’s requirement.

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60

Fig. SampleInteraction

diagram

Anal sis of col mns according to EBCS 2


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Analysis of columns according to EBCS 2

(short and slender)

Classification of Frames

A frame may be classified as non-sway for a given

load case if the critical load ratio for that load case

satisfies the criterion: ≤ 0.1

Where: Nsd is the design value of the total vertical load

Ncr is its critical value for failure in a sway mode


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In Beam-and-column type plane frames in buildingstructures with beams connecting each column at eachstory level may be classified as non-sway for a given

load case, when first-order theory is used, thehorizontal displacements in each story due to thedesign loads (both horizontal & vertical), plus the initialsway imperfection satisfy the following criteria.

≤ 0.1 Where: δ is the horizontal displacement at the top of the

story, relative to the bottom of the story

L is the story height

H is the total horizontal reaction at the bottom of the story

N is the total vertical reaction at the bottom of the story,


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The displacement δ in the above equation shall

be determined using stiffness values for beams

and columns corresponding to the ultimatelimit state.

All frames including sway frames shall also be

checked for adequate resistance to failure innon-sway modes.

D t i ti f t b kli L d N


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Determination of story buckling Load Ncr

Unless more accurate methods are used, the buckling load

of a story may be assumed to be equal to that of thesubstitute beam-column frame defined in Fig. and may bedetermined as:

=

Where: EIe is the effective stiffness of the substitute columndesigned using the equivalent reinforcement area.

Le is the effective length. It may be determined using the

stiffness properties of the gross concrete section for bothbeams and columns of the substitute frame (see Fig. 4.4-1c )

I li f t d t i ti th ff ti


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• In lieu of a more accurate determination, the effectivestiffness of a column EIe may be taken as:

= 0.2

Where: Ec = 1100f cd

Es is the modulus of elasticity of steel

Ic, Is, are the moments of inertia of the concrete andreinforcement sections, respectively, of the substitute

column, with respect to the centroid of the concretesection (see Fig. 4.4-1c).

• or alternatively

= (1 ) ≥ 0.4 • Where: Mb is the balanced moment capacity of the

substitute column


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1/rb is the curvature at balanced load & may be taken as

1 = 5

10−

The equivalent reinforcement areas, As, tot, in the substitute

column to be used for calculating Is and Mb may be obtained

by designing the substitute column at each floor level to

carry the story design axial load and amplified sway momentat the critical section.

Sl d R ti


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Slenderness Ratio

Slenderness ratio, λ of a column is defined as the ratio

of the effective length lei to the radius of gyration.NB: λ also provides a measure of the vulnerability to

failure of the column by elastic instability (buckling)

=

where: Le is the effective buckling length

i is the minimum radius of gyration.

The radius of gyration is equal to

= Where: I is the second moment of area of the section,

A is cross sectional area

Limits of Slenderness


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Limits of Slenderness The slenderness ratio of concrete columns shall not

exceed 140.

Second order moment in a column can be ignored if:

For sway frames, the greater of: ≤ 25

15/

= / For non-sway frames: ≤ 50 25 Where: M

1

and M2

are the first-order end moments, M2

being always positive and greater in magnitude than M1,

and M1 being positive if member is bent in single curvature

and negative if bent in double curvature

Effective Length of Columns


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Effective Length of Columns

Effective buckling length is the length between points

of inflection of columns and it is the length which iseffective against buckling.

The greater the effective length, the more likely the

column is to be buckle.

Le = kL

where k is the effective length factor (i.e., the ratio of

effective length to the unsupported length whose value

depends on the degrees of rotational and translation

restraints at the column ends.

For idealized boundary conditions:


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y


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The rotational restraint at a column end in a building

frame is governed by the flexural stiffnesses of the

members framing into it, relative to the flexuralstiffness of the column itself.

Hence, it is possible to arrive at measures of the

‘degree of fixity’ at column ends, and thereby arrive at

a more realistic estimate of the effective length ratio of

a column than the estimates (for idealized boundary

conditions).

These involve use of alignment charts or approximateequations.

The alignment chart (Nomograph): for members


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The alignment chart (Nomograph): for members

that are parts of a framework

Approximate equations


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Approximate equations

The following approximate equations can be used

provided that the values of α1 and α2 don’t exceed 10(see EBCS 2).

Non-sway mode

= 0.4 0.8 ≥ 0.7 In Sway mode

=

.+

+

+.

.++ ≥ 1.15

Or Conservatively,

1 0 8 ≥ 1 15


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Where α1 and α2 are as defined above and αm is

defined as:

= 2 Note that: for flats slab construction, an equivalent

beam shall be taken as having the width and

thickness of the slab forming the column strip.

The effect of end restrainet is quantified by the two

end restrain factors α1 and α2

= / /


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Where Ecm is modulus of elasticity of concrete

Lcol is column height

Lb is span of the beam

Icol, Ib are moment of inertia of the column and beam

respectively

β is factor taking in to account the condition of

restraint of the beam at the opposite end

= 1.0 opposite end elastically or rigidly restrained

= 0.5 opposite end free to rotate

= 0 for cantilever beam


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Note that: if the end of the column is fixed, theoretically,α = 0, but an α value of 1 is recommended for use.

On the other hand, if the end of the member is pinned,the theoretical value of α is infinity, but an α value of 10is recommended for use.

The rationale behind the foregoing recommendations isthat no support in reality can be truly fixed or pinned.

Design of columns, EBSC-2 1995


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General:

The internal forces and moments may generally be

determined by elastic global analysis using either first

order theory or second order theory.

First-order theory, using the initial geometry of the

structure, may be used in the following cases – Non-sway frames

– Braced frames

– Design methods which make indirect allowances for

second-order effects.

Second-order theory, taking into account the influence

of the deformation of the structure, may be used in all

cases.

Design of Non-sway Frames


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Design of Non sway Frames

Individual non-sway compression members shall be

considered to be isolated elements and be designedaccordingly.

Design of Isolated Columns

For buildings, a design method may be used which

assumes the compression members to be isolated.

The additional eccentricity induced in the column by

its deflection is then calculated as a function of

slenderness ratio and curvature at the critical section

Total Eccentricity


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Total Eccentricity

The total eccentricity to be used for the design of

columns of constant cross-section at the critical

section is given by:

= Where: ee is equivalent constant first-order

eccentricity of the design axial load

ea is the additional eccentricity allowance

for imperfections.

For isolated columns: = 300 ≥ 20

e2 is the second-order eccentricity

First order equivalent eccentricity


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q y

i. For first-order eccentricity e0 is equal at both ends of

a column = ii. For first-order moments varying linearly along the

length, the equivalent eccentricity is the higher of the

following two values: = 0.6 0.4

= 0.4 Where: e01 and e02 are the first-order eccentricities at the

ends, e02 being positive and greater in magnitude than e01.

e01 is positive if the column bents in single curvature and

negative if the column bends in double curvature.


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iii. For different eccentiries at the ends, (2) above, the

critical end section shall be checked for first order

moments:

=

Second order eccentricity


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Second order eccentricity

i. The second-order eccentricity e2 of an isolated

column may be obtained as

=

10 (1 ) Where: Le is the effective buckling length of the column

k1= λ/20 - 0.75 for 15≤ λ ≤ 35

k1= 1.0 for λ >35

l/r is the curvature at the critical section.

ii. The curvature is approximated by:1 =

5 10

−

Wh d i h ff i l di i i h l f


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Where: d is the effective column dimension in the plane of

buckling

k2 =Md /Mb Md is the design moment at the critical section including

second-order effects

Mb is the balanced moment capacity of the column.

iii. The appropriate value of k2 may be found iteratively

taking an initial value corresponding to first-order

actions.

Reinforcement Detail


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Reinforcement Detail

Size

The minimum lateral dimension of a column shall beat least 150mm

Longitudinal Reinforcement

The minimum number of bars to be used in columnsshall be:

Four in rectangular columns (One at each corner)

Six in circular columns

The minimum bar diameter shall be 12 mm.

The minimum area of steel shall be 0.008 Ac for

ductility.

Th i f t l h ll b 0 08 A i l di


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The maximum area of steel shall be 0.08 Ac, including

laps.

Lateral Reinforcement Minimum bar size shall be ¼ times the largest

compression bar, but not less than 6mm.

Center to center spacing shall not exceed:

12 times minimum diameter of longitudinal bars

Least dimension of column

300 mm.

Spiral or circular ties may be used for longitudinal

bars located around the perimeter of a circle. The

pitch of spirals shall not exceed 100 mm.

Design of Sway Frames


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Design of Sway Frames

Nonsway moments, Mns, are the factored end

moments on a column due to loads that cause

no appreciable sidesway as computed by a

first-order elastic frame analysis. They result

from gravity loads.

Sway moments, Ms, are the factored end

moments on a column due to loads whichcause appreciable sidesway, calculated by a

first-order elastic frame analysis.

First-order vs. Second-order Frame


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First order vs. Second order Frame

Analyses:

A first-order frame analysis is one in which the effect oflateral deflections on bending moments, axial forces,

and lateral deflections is ignored.

The resulting moments and deflections are linearlyrelated to the loads.

When slenderness effects cannot be neglected, the

design of compression members, restraining beams and

other supporting members shall be based on thefactored forces and moments from a second-order

analysis.

First order vs Second order Frame


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A second-order frame analysis, is one which considersthe effects of deflections on moments, and so on.

The resulting moments and deflections include the

effects of slenderness and hence are nonlinear withrespect to the load.

The stiffness, EI, used in the analysis should represent

the stage immediately prior to yielding of the flexural

reinforcement since the moments are directly affectedby the lateral deflections.

First-order vs. Second-order Frame

Analyses:

Methods of Second Order Analysis


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y

a) Iterative P-∆ Analysis:

When a frame is displaced by an amount ∆ due to

lateral loads, additional moments shall be induced

due to the vertical loads (ΣP∆).

This shall be loaded as sway force (ΣP∆/l) and thestructure shall be re-analyzed until convergence is

achieved.

The values can be assumed to converge when the

change in deflection between two consecutiveanalyses is less than 2.5 percent . (ACI)

89


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Fig. 4 Determination of sway forces

90

b) Di t P ∆ l i


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b) Direct P-∆ analysis

It describes the iterative P-∆ analysis mathematicallyas an infinite series.

An estimate of final deflections is obtained directly

from the first-order deflections (from the sum of the

terms in the series).

Thus the second-order moments are δsMs Where: and

)()(1 0

0

Vl P u

0.11

1

Q s

vl

P Q u 0

)(

91

c) Amplified Sway Moments Method:


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) p y

First-order sway moments are multiplied by a sway

moment magnifier, δs, to obtain the second ordermoments.

M1 = M1ns + δsM1s

M2 = M

2ns + δ

sM

2s

Where: (According to ACI)

(According to EBCS) Where: Pu, NSd is the design value of the total vertical load

Pc, Ncr is its critical value for failure in a sway mode

1)75.0(1

1

cu s

P P

cr Sd

s

N N 1

1

ch 4 - ae design of columns

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