© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 1
Contents :
Chapter 1 - Step1 : Determination Sway or Non-Sway ……………………………………………
1.1-Definition
Page 5
………………………………………………………………………………… Page 5
………………………………………………
Page 6
1.3-Methods of Determination Sway or Non-Sway …………………………………………...…
Page 7
1.3.1 - Method 1 …………………………………………………………………………...…
Page 7
1.3.2 - Method 2 ………………………………………………………………………...……
Page 7
1.3.3 - Method 3 ………………………………………………………………………......…
Page 9
1.2-First order analyses & Second order analyses
Chapter 2 - Step2 : Determination short or long (slender ) Column / Wall . ………….............…
Page 10
2.1 – In case Non-Sway column ………………………………………………………...……...…
Page 10
2.1.1 – ψ Factor …………………………………………………………………..…..…..…
Page 10
2.1.2 – effective length ratio ( k ) ………………………………………………………….…
Page 11
2.1.3 – The unsupported length (Lu ) ……………………………………………………...…
Page 12
2.1.4 – Radius of Gyration (r ) ………………………………………………………...….…
Page 13
2.1.5 – M1 / M2 ratio ……………………………………………………………...……..…
Page 13
2.2 – In case Sway column …………………………………………………...………...…………
Page 14
2.2.1 – ψ Factor …………………………………………………………………………….… Page 14
2.2.2 – effective length ratio ( k ) ……………………………………………………….....…
Page 14
2.2.3 – The unsupported length (Lu ) ……………………………………………………..…
Page 16
2.2.4 – Radius of Gyration (r ) ……………………………………………………...…….…
Page 16
Chapter 3 - Step3 : Moment Magnification for Non-sway column/wall …………………...…..…
Page 17
3.1 – The concept of the Moment Magnification for Non-sway column/wall ……………………
Page 17
3.2 – How to find Mmax Value………………………………………………………..………..…
Page 19
3.2.1 – 2nd order analysis method– Nonsway ………………………….……………...……
Page 19
3.2.2 – Moment magnification method – Nonsway………………………………………..…
Page 19
3.2.2.1 – Find Mmin………………………………………………………………….…
Page 19
3.2.2.2 – Find Cm factor……………………………………………………………...…
Page 19
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 2
3.2.2.3 – The critical buckling load Pc …………………………………...….....….…
Page 20
3.2.2.4 – Magnification factor dns ……………………………………………….….…
Page 20
3.2.2.5 - Magnified Moment Mc ………………………………………………….....…
Page 20
3.2.2.6 - Moments Magnification Ratio …………………………………………....…
Page 21
Chapter 4 - Step3 : Moment Magnification for sway column/wall …………………………….…
Page 22
4.1 – The concept of the Moment Magnification for sway column/wall ………………..………
Page 22
4.2 – How to find Mmax1 & Mmax2 values ………………………………………………..…..
Page 23
4.2.1 – Elastic 2nd order analysis method – Sway………………………………………..…
Page 23
4.2.2 – Inelastic 2nd order analysis method – Sway ……………………………………...…
Page 23
4.2.3 – Moment magnification method – Sway…………………………………………....…
Page 23
4.2.3.1 – Find SPu & SPc…………………………………………………………....…
Page 23
4.2.3.2 – Magnification factor ds ………………………………………………..……
Page 24
4.2.3.3 – Magnified Moments Mmax1 & Mmax2 ……………………………….……
Page 25
4.2.3.4 - Moments Magnification Ratio …………………………………………….…
Page 25
4.3 – check if there is buckling along the column in case Sway column……………………….…
Page 26
Chapter 5 - The Design Of The Column By Etabs ……………………………………………...…
Page 30
5.1 – Etabs doesn’t classify the column as Sway or Non-sway Automatically but
Users can do that manually …………………………………………………………….…
Page 30
5.1.1 – The Simplified procedure ……………………………………………………………
Page 30
5.1.2 - The code procedures …………………………………………………………...….…
Page 30
5.1.2.1 –Using Stability Index Q ( Method 2 ) ………………………………….….…
Page 30
5.1.2.2 –Using P-D effect ( Method 3 ) ……………………………..….…………..…
Page 34
5.2 – Etabs doesn’t classify the column as Short or Slender Automatically but
User can do that manually ………………………………………………………………..…
Page 37
5.2.1- column classification in Etabs ………………………………………………….…..…
Page 37
5.2.2- column classification by user………………………………………………………..…
Page 37
5.2.3 - k ( Effective Length Factor ) in Etabs ………………………………………….....…
Page 37
5.3 – The Methods and Formulas of Column ( Frame ) Design in Etabs …………………...……
Page 39
5.3.1- ACI code Methods which used by Etabs for the column (Frame ) Design ………..…
Page 39
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 3
5.3.1.1 – Non-Sway columns ……………………………………………………….…
Page 39
5.3.1.2 –Sway columns…………………………………………………………………
Page 39
5.3.2- The Etabs formulas of column (Frame ) Design…………………………………...….
Page 40
5.3.3- Comparison between Etabs formula and Code formulas ……………………..………
Page 40
5.3.3.1- Non-sway columns………………………………………………………….…
Page 40
5.3.3.2- Sway columns ……………………………………………………………....…
Page 41
5.3.3.2.1- ds Method……………………………………………………….….…
Page 41
5.3.3.2.2- P-D Method……………………………………………………………
Page 42
5.3.3.3- Conclusion and Summary ( The General Procedure ) ………………………...
Page 43
5.4 –The parameters of column Design in the Etabs with the correct modeling of columns…….
Page 43
5.4.1-Unbraced Length ( Clear Height ) Lu……………………………………………….…
Page 43
5.4.2- How Etabs calculates Lu , Cm , Mu’ …………………………………………………
Page 46
Example 5.4.2-a…………………………………………………..…………...………
Page 46
Example 5.4.2-b ………………………………………………………………..…..…
Page 51
5.4.3- the required consideration in case only gravity loads …………………………...……
Page 55
5.4.4- How to check the Code condition for the ratio of the second order
to first order moment in Etabs ………………………………………………………..….…
Page 56
5.4.5- What user has to do for all cases of the column situation ……………………….……
Page 57
Chapter 6 - Summary for how to adjust the inputs in Etabs to design the columns………….…
Page 61
Step 1 : Define the column Reinforcement as Design or check………………………………..…
Page 61
Step 2 : adjust the end length offset and check the column modeling ( before doing “ Run “ )…..
Page 61
Step 3 : activate P-D option…………………………………………………………………….…
Page 64
Step 4 : classify the columns Sway / Nonsway ………………………………………………..…
Page 65
Step 5 : Adjust Unbraced Length Ratio (After doing “Run” ) ………………………………....…
Page 66
Step 6 : Check The code condition , dns < 1.4 and dns M’/M < 1.4 …………………………….
Step 7 : Check the results and warnings ……………………………………………………….…
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 68
Page 69
Page 4
Chapter 1 - Step1 : Determination Sway or Non-Sway
1.1-Definition :
A column is considered as “Non-Sway “ for the considered direction if it is braced by bracing elements which
have such substantial lateral stiffness to resist the lateral deflections of the story along that direction , otherwise
it should be considered as “Sway “ .
( Bracing elements : shear walls or lateral bracing … ) PCA Notes on ACI 318-11 , Page 11-4
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 5
1.2-First order analyses & Second order analyses:
- First -order analyses, the deformed geometry of the structure is not included in the equations of equilibrium .
- second-order analyses, the deformed geometry of the structure is included in the equations of equilibrium so that
P-Delta effects are determined.

Elastic second-order analyses : The structure is assumed to remain elastic, but the effects of cracking and
creep are considered by using a reduced stiffness EI . ACI 318-14 - R6.7.1
Inelastic second-order analyses: consider material nonlinearity, member curvature and lateral drift, duration of
loads, shrinkage and creep …. ACI 318-14 - 6.8.1.1 , Refer to ACI Design Handbook ACI SP-17(14) - 3.5.8
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 6
1.3-Methods of Determination Sway or Non-Sway :
ACI 314-14 Provide three methods to determine if the column is considers as Sway Or Non-Sway :
1.3.1 - Method 1 :“ If bracing elements resisting lateral movement of a story have a total stiffness of at
least 12 times the gross lateral stiffness of the columns in the direction considered, it
shall be permitted to consider columns within the story to be braced against sidesway “ .
ACI 318-14 ( 6.2.4.4 )
Discussion :
1- bracing elements : (shear walls) , corewalls ,Special bracing System….
When do we consider the element as shear wall ?
ACI Design Handbook ACI SP-17(14)
2- do we have to study the wall as sway / nonsway ?
1.3.2 - Method 2 : if Q ( stability index ) doesn’t exceed 0.05 ( 5% ) the columns within the story to be
braced against sidesway . ACI 318-14 ( 6.6.4.3.b)
- SPu ,” is the sum all factored column and wall gravity loads ( not lateral load ) for load combination”
ACI Design Handbook ACI SP-17(14) – Columns Example 1 – page 363 .
Because there are many combinations ,
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 7
1)
2)
3)
4)
5)
6)
7)
1.4D
(Vert.)
1.2D+1.6L
(Vert.).
1.2D + 0.8W
(Lateral)
1.2D+0.5L +1.6W (Lateral)
1.2D+0.5L +1.0E
(Lateral)
0.9D+1.6W
(Lateral)
0.9D+1.0E
(Lateral)
* 1.6W can be reduced to 1.3W when wind load W is not reduced by directional factor
We consider the conservative value for SPu , to get max Q We have to consider max SPu from the ultimate
Lateral combinations , For that We consider SPu = 1.2 D +0.5 L
ACI Design Handbook ACI SP-17(14) – Columns Example 1 – page 363 .
-Vus ,Do : Vus is the factored lateral load in the combination which We considered to estimate SPu.
Do the Drift story due to Vus from first order analysis with the cracked sections .
ACI Design Handbook ACI SP-17(14) – 9.4.3.1 –Page 356
-Lc : story height measured center to-center of the joints .
Discussion :
1- What is the value of Do in case Torsional response ?
In Torsional response of the lateral-force resisting system due to eccentricity of the structural system
can increase second-order effects and should be considered ACI 318-14 ( R6.2.5 )
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 8
2- Which Vus We have to consider ?
The important value is the Ratio Do/ Vus
3- Lc value : some of stories could have two height or more , So which Lc do We have to consider ?
1.3.3 - Method 3 : if The increase in column end moments due to second-order effect doesn’t exceed 5%
of the first order end Moments , the columns to be braced against sidesway .
ACI 318-14 ( 6.6.4.3.a )
So , P-delta should be included in the analysis of The building :
Because there are many combinations , We consider the conservative value for P , For that We consider
SPu = 1.2 D +0.5 L
For example you can check the following example from CSI KNOWLEDGE BASE :
https://wiki.csiamerica.com/display/etabs/P-Delta+analysis+parameters
Discussion :
1- By this method , We can study the case of each Column & Wall , Different height , Torsional response
2- For vertical load combination , should we classify the column sway/nonsway ?
as mentioned above ,For combination (1) & (2) , P-delta effect should not be of concern , this means the
second analysis effect should not be of concern also and the column is Non-Sway .
anyway , We can check my method 3 to classify if the column is sway or non-sway even for vertical load
combination .
Note : The classification / sway or nonsway/ is related with the Vertical load , lateral stiffness .
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 9
Chapter 2 - Step2 : Determination short or long (slender ) Column / Wall .
2.1 – In case Non-Sway column :
For Non-Sway columns ,
If Slenderness Ratio
k.Lu / r < min of { 40 , 34 + 12 ( M1/M2) }
ACI 318-14 (6.2.5)
Slenderness effects shall be permitted to be neglected and the column is considered as Short column
otherwise it should be as Slender column .
Important Note : Before ACI 318-08 if ratio k.Lu/r > 100, slenderness effects cannot be accounted for
using moment magnification procedure. A more exact method must be used Like
software calculate the buckling along the column length .
but in ACI 318-14, ACI 318-11, ACI 318-08 this limit does not explicitly apply .
see ACI 318 -14 and Spcolumn Manual – Page 37
2.1.1 – ψ Factor :
ψ is the ratio of SEI / Lc of the columns to SEI / Lb of the beams in a plane at one end of the column,
ψ A and ψ B are the values of ψ at the upper end and the lower end of the column ,
ψ min = min of ( ψ A and ψ B )

I = is the gross moment of inertia Ig multiplied by the cracked section coefficients .
Lc , Lb = the lengths of the column and beams center to center

For a hinged end , ψ is very large. This happens in the case where SEI / Lb is very small (or zero)
relative to SEI / Lc at that end.
For a complete hinged end , ψ = ∞
For a complete Fixed end , ψ = 0
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 10
Discussion :
If there is no beam only flat slab , can We include the slab stiffness in ψ formula ?
Actually , For conservative procedure assume ψ = ∞ for slab without beams ,
but We can use the slab as equivalent beam with the following considerations:
The depth of the equivalent beam = The thickness of the slab
The width of the equivalent beam = L1/2+L2/2
The length of the equivalent beam = the span length C/C along the considered direction
The cracked section coefficients of the equivalent beam = The cracked section coefficients of the slab
ACI Design Handbook ACI SP-17(14) - Columns Example 1– Page 362
2.1.2 – effective length ratio ( k ) :
effective length ratio ( k ) is the distance between points of zero moment in the column divided be the clear
height .
For non-sway column , the value of k =kns must be between 0.5 and 1.0 ,
and the recommended value by ACI Code is 1.0 , Smaller values could be used by the methods below .
ACI 318-14 - R6.6.4.4.3
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 11
Method 1 : We find the value of the effective length factor, kns , from the following chart
(Jackson and Moreland Alignment Charts ) ACI 318-14 Fig. R6.2.5
Note : The Exact formula for Jackson and Moreland alignment chart is :
Spcolumn Manual – Page 36
Method 2 : As Per PCA Notes on ACI 318-11 Page 11-8 , the effective length factor may be taken as the
smaller value determined from the two equations to follow :
2.1.3 – The unsupported length (Lu ) :
The unsupported length (Lu ) , is the clear distance between floor slabs, beams , or other members
capable of providing lateral support in the direction being considered.
Where column capitals or haunches are present, Lu shall be measured to the lower extremity of the
capital or haunch in the plane considered.
ACI 318-11 - 10.10.1.1
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 12
2.1.4 – Radius of Gyration (r ) :
𝐼𝑔
r = √𝐴𝑔
: Ig = the gross moment of inertia (uncracked section )
Ag = The gross area of section
For rectangular section r = 0.3b or 0.3h depends on the considered direction
2.1.5 – M1 / M2 ratio :
M1 and M2 are the factored Moments at column ends .
M1 = min of { abs (M1 ) , abs ( M2 ) }
M2 = max of { abs (M1 ) , abs ( M2 ) }
where Ratio M1/M2 is considered as :
 Negative for single curvature ( M1,M2 are at the same side in the moment diagram )
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 13
 Positive for double curvature ( M1,M2 are in opposite sides in the moment diagram )
So , Ratio M1/M2 = - ̅̅̅̅
𝑀1 / ̅̅̅̅
𝑀2
: where ̅̅̅̅
𝑀1 , ̅̅̅̅
𝑀2 include the signs as per the moment diagram
Note1 : for nonsway columns , the end moments will be equal to the factored applied first order moment
( So P-Delta effects at the ends is not included )
Note2 : M1/M2 ratio is the minimum value from all load combinations .
Note3 :ACI codes before ACI 318 -14 had a contrary consideration for Ratio M1/M2 and Slenderness
condition was k.Lu / r < min of { 40 , 34 - 12 ( M1/M2) }
Note4 : If both moments are equal to zero , assume the Ratio M1 / M2 = -1
This corresponds to consideration Mmin instead of zero value for both moment with single
curvature as conservative procedure , And this also corresponds to Spcolumn assumption
Spcolumn Manual – Page 37
As per PCA Notes on ACI 318-11 Page 11-15 M1 / M2 = 0 , So k.Lu / r < 34
Assuming Ratio M1 / M2 = -1 is more conservative and corresponds with considering Cm =1 in
case M1=M2 =0 ( as per what will be explained later ) .
2.2 – In case Sway column :
For Sway columns ,
If Slenderness Ratio
k.Lu / r < 22
ACI 318-14 (6.2.5)
Slenderness effects shall be permitted to be neglected and the column is considered as Short column
otherwise it should be as Slender column .
Important Note : Before ACI 318-08 if ratio k.Lu/r > 100, slenderness effects cannot be accounted for
using moment magnification procedure. A more exact method must be used Like
software calculate the buckling due to displacement at the ends ( P-delta ) .
but in ACI 318-14, ACI 318-11, ACI 318-08 this limit does not explicitly apply .
see ACI 318 -14 and Spcolumn Manual – Page 33
2.2.1 – ψ Factor :
It has been explained in non-sway case 2.1.1
2.2.2 – effective length ratio ( k ) :
effective length ratio ( k ) is the distance between points of zero moment in the column divided be the clear
height .
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 14
For sway column , the value of k =ks must be between 1 and ∞ .
We find ks factor by the following Methods :
Method 1 : We find the value of the effective length factor, ks , from the following chart
(Jackson and Moreland Alignment Charts ) ACI 318-14 Fig. R6.2.5
Note : The Exact formula for Jackson and Moreland alignment chart is :
Spcolumn Manual – Page 32
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 15
Method 2 : As Per PCA Notes on ACI 318-11 Page 11-8 , the effective length factor may be determined
from the two equations to follow :
: ψavg = ( ψ A + ψ B ) / 2
2.2.3 – The unsupported length (Lu ) :
It has been explained in non-sway case 2.1.3
2.2.4 – Radius of Gyration (r ) :
It has been explained in non-sway case 2.2.4
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 16
Chapter 3 - Step3 : Moment Magnification for Non-sway column/wall
3.1 – The concept of the Moment Magnification for Non-sway column/wall :
Buckling for Sway or Non-Sway : is the sudden large deformation of the column due to Axial load .
The picture shows lateral deflection due to only an axial load , this deformation will cause additional moment .
Figures 3.1.a & 3.1.b show a column ,axially loaded by P and bent by end moments M1,M2 ( Where M2 is
the largest )
as per explained in 2.1.5
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 17
Due to only M1 & M2 (ignoring the effect of P with deflection ) , moment diagram will be changing from M1
to M2 values and the deflection will be do , Such Analysis is called first order analysis .
But , including the effect of P with deflection , the deflection Near the middle of the column (in case
Non-sway) will increase by d2 , and The axial load with the total deflection d=do+d2 ( including the first order
analysis ) will cause additional moment with maximum value P.d
The total moment diagram equals to first order moment diagram plus diagram due to P,
Such Analysis which include the effect of P.d is called Second order analysis .
When You look to the total Moment diagram in Figure 3.1.a , We will see that Mmax is located along the
column between the ends .
While in Figure 3.1.b , all value are still less than M2 , So Mmax = M2 .
Similarly , in case of double curvature ( shown in 3.1.c & 3.1.d )
Due to only M1 & M2 (ignoring the effect of P with deflection ) , moment diagram will be changing from
M1 to M2 values and the deflection will be do at two locations , and the deflection will increase due to
the effect of P with deflection by d2 at two locations up and down
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
.
Page 18
The axial load with the total deflection d=do+d2 ( including the first order analysis ) will cause additional
moment with two value P.d
.
In Figure 3.1.c Mmax is located along the column between the ends .
While in Figure 3.1.d , all value are still less than M2 , So Mmax = M2 .
3.2 – How to find Mmax Value
As per ACI 318-14 , Mmax Value could be found by two methods :
1) 2nd order analysis method – Nonsway
ACI 318-14 – 6.6.4.5
2) Moment magnification method – Nonsway . ACI 318-14 – R6.7.1.2
3.2.1 – 2nd order analysis method– Nonsway :
This method use the 2nd order analysis by software can analysis the buckling along the column
( for example : Etabs can do that using load case “ buckling “ ) .
In the modeling , columns may be subdivided using nodes along their length to evaluate slenderness
effects between the ends.
3.2.2 – Moment magnification method – Nonsway:
This method use The magnification factor dns to find Mmax as the following procedure ,
3.2.2.1 – Find Mmin :
Mmin = Pu ( 15+0.03 h ) : 15+0.03h in mm
Pu : the factored axial load (due to the considered load combination )
h : the section dimension (the diameter for circular sections) in the direction
being considered.
For the design , M2 should be not less than Mmin ACI 318-14 – 6.6.4.5.4
as per what will be explained later .
3.2.2.2 – Find Cm factor :
Cm = 0.6 – 0.4 (M1/M2)
Eq 3.2.2.2
: M1/M2 has been explained in 2.1.5
Note1 : before ACI 318 – 08 , Cm factor had the value 0.4 as minimum limit , In codes
ACI 318–08 & ACI 318–11& ACI 318–14 this limit does not apply
Note2 : max Cm = 1
Note3 :
For column with transverse loading between the column Ends ,Cm = 1
ACI 318-14 – (6.6.4.5.3b)
Note4 : For M1=M2=0 , Cm = 1
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 19
Note5 : if M2 < Mmin , the value of Cm shall either be taken equal to 1.0, or shall be
computed by Eq 3.2.2.2 With the actual Moment values M1 & M2 .
PCA Notes on ACI 318-11 Page 11-13
3.2.2.3 – The critical buckling load Pc :
k = kns : have been explained in 2.1.2
Lu
: have been explained in 2.1.3
(EI)eff could be found for any one of the following equations :
Ig = the gross moment of inertia ( uncracked section )
I = the cracked moment of inertia calculated according to ACI - Table 6.6.3.1.1(b)
Ise = moment of inertia of reinforcement about centroidal axis of member cross
section
𝑃𝑢 𝑠𝑢𝑠
dns =
≤1
𝑃𝑢
Pu sus = the sustained factored gravity ( Dead & Live) loads
PCA Notes on ACI 318-11 Page 11-23
Pu
= the factored axial loads
Note : For simplification, it can be assumed thatdns = 0.6 ACI 318-14 – R6.6.4.4.4
3.2.2.4 – Magnification factor dns :
Note : Pu is not allowed to be greater than 0.75 Pc , Else the section should be revised .
3.2.2.5 - Magnified Moment Mc :
The factored moment used for design of columns and walls, Mc = Mmax , shall be the
first-order factored moment M2 amplified for the effects of member curvature :
Mc = dns . M2
Note1 : if M2 < Mmin , Mc = dns . Mmin
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 20
Note2 : dns ≥ 1 , means that Mmax will be between the column ends , see figures 3.1.a , 3.1.c
while dns < 1 , means that Mmax = M2 , for that reason dns should be considered (1) ,
see figures 3.1.b , 3.1.d
3.2.2.6 - Moments Magnification Ratio :
In Non-sway & Sway column the value of total magnified moment including second-order
analysis
cannot exceed 1.4 times the corresponding moment due to first order analysis, otherwise the
section should be revised ACI 318-14 - 6.2.6.
So the required conditions to accept the section in Non-sway column are :
a) If M2 ≥ Mmin → Mc ≤ 1.4 M2
→ dns . M2 ≤ 1.4 M2
→
dns ≤ 1.4
b) If M2 < Mmin → Mc ≤ 1.4 Mmin → dns . Mmin ≤ 1.4 Mmin → dns ≤ 1.4
Note : this limit was not exist before ACI 318-08
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 21
Chapter 4 - Step3 : Moment Magnification for sway column/wall
4.1 – The concept of the Moment Magnification for sway column/wall :
Figures 4.1 show a column , axially loaded by P And lateral load V and bent by end moments M1,M2 ( Where
M2 is the largest ).
To study the Sway columns, We divide the moments M1 & M2 to two types Mns & Ms :
Mns : the moment due to the load that cause no sidesway, calculated using a first-order elastic
(P-Delta effects at the ends is not included in the analysis )
For example : Moments due to Dead load , Live load .
Ms : the moment due to the load that cause sidesway, calculated using a first-order elastic
(P-Delta effects at the ends is not included in the analysis ) .
For example : Moments due to Wind load , Seismic load .
So , M1 = M1ns + M1s & M2 = M2ns + M2s
Due to only V & M1 & M2 (ignoring the effect of P with deflection ) the deflection will be Do , Such Analysis
is called first order analysis .
But , including the effect of P with deflection , the deflection at the end of the column will increase by D2 ,
and The axial load with the total deflection D =D o+D 2 ( including the first order analysis ) will cause
additional moments with maximum value P.D .
As per ACI code , The moment magnification due to the effect of P.D is considered applied on Ms moments
only .
The total moment diagram equals to first order moment plus P.D effect , Such Analysis which includes the
effect of P.D is called Second order analysis
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 22
4.2 – How to find Mmax1 & Mmax2 Values :
As per ACI 318-14 , Mmax Value could be found by three methods :
1) Elastic 2nd order analysis method – Sway
ACI 318-14 - 6.7
nd
2) Inelastic 2 order analysis method – Sway ACI 318-14 - 6.8
3) Moment magnification method – Sway .
ACI 318-14 - 6.6.4.6
4.2.1 – Elastic 2nd order analysis method – Sway:
This method use the 2nd order analysis by software can include the effect of P-D in the analysis .
( for example : Etabs can do that using P-D option ) .
By using P-D, software will get Directly the magnified Moments at the column Ends .
In this method , the cracked section coefficients should be used .
4.2.2 – Inelastic 2nd order analysis method – Sway:
This method is similar to Elastic 2nd order analysis method but many other consideration should
be considered , for example : nonlinearity properties of the material , the cracked section
coefficients should not be used .
4.2.3 – Moment magnification method – Sway:
This method use The magnification factor ds to find Mmax as the following procedure ,
4.2.3.1 – Find SPu & SPc:
SPu is the summation of all the factored vertical loads in a story ( due to the considered
load combination including the gravity and lateral loads )
SPc is the summation of critical buckling load for all sway-resisting columns & walls
in a story.
k = ks : have been explained in 2.2.2
Lu : have been explained in 2.1.3
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 23
(EI)eff could be found for any one of the following equations :
Ig = the gross moment of inertia ( uncracked section )
I = the cracked moment of inertia calculated according to ACI- Table 6.6.3.1.1(b)
Ise = moment of inertia of reinforcement about centroidal axis of member cross
section
ds =
𝑃𝑢 𝑠𝑢𝑠
𝑃𝑢
≤1
Pu sus = The sustained factored lateral loads
( for wind and seismic load Pu sus = 0 & ds = 0 , for earth pressure Pu sus ≠ 0)
Pu = The factored axial loads
4.2.3.2 – Magnification factor ds :
Magnification factor ds shall be calculated by (a) or (b) :
Q ( stability index ): has been explained in 1.3.2 , but here the value of SPu in Q
formula is as defined in 4.2.3.1 ( not only due to gravity loads )
But if ds >1.5 , this formula will be not applicable
Note1 : in codes only Before ACI 318 -08 ,There is limit ds ≤ 2.5 for Strength and stability of
the structure as a whole under factored gravity loads
Note2 : SPu is not allowed to be greater than 0.75 SPc , Else the Sections should be revised
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 24
4.2.3.3 – Magnified Moments Mmax1 & Mmax2 :
The factored moments used for design of columns and walls:
Mmax1 =Mns1 + ds . Ms1
Mmax2 =Mns2 + ds . Ms2
Note : by using 2nd order analysis methods – Sway :
Mmax = Mns + Ms’ : where Ms’ is the moment due to the lateral loads including
P-D .
So , Ms’ is equivalent to ds.Ms
4.2.3.4 - Moments Magnification Ratio :
In Non-sway & Sway column the value of total magnified moment including second-order effects
cannot exceed 1.4 times the corresponding moment due to first order effects , otherwise the section
should be revised ACI 318-14 - 6.2.6.
So the required conditions to accept the section in Sway column are :
Mmax1 ≤ 1.4 M1 → Mns1 + ds . Ms1 ≤ 1.4 M1
Mmax2 ≤ 1.4 M2 → Mns2 + ds . Ms2 ≤ 1.4 M2
Note1 : this limit was not exist before ACI 318-08
Note2 : For 2nd order analysis method , We do two analysis type :
M1=Mns1 + Ms1 and M2=Mns2 + Ms2 from first order analysis ( without P-D)
Mmax1 =Mns2 + Ms1’ and Mmax2= Mns2 + Ms2’ from second order analysis
( with P-D)
Then We check that :
Mmax1 ≤ 1.4 M1 → Mns1 + Ms1’ ≤ 1.4 M1
Mmax2 ≤ 1.4 M2 → Mns2 + Ms2’ ≤ 1.4 M2
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 25
4.3 – check if there is buckling along the column in case Sway column :
Second-order effects along the length of the column must be calculated for a sway or nonsway frame where
slenderness cannot be neglected.
ACI 318-14 - 6.6.4.6.4 Second-order effects shall be considered along the length of columns in sway frames.
It shall be permitted to account for these effects using 6.6.4.5 .
The magnified moment along the column length is generally less than the magnified end moment .
see figures 4.3.a
But this could not happen in all cases , in some cases the magnified moment along the column length will be
greater than the magnified end moment , see figures 4.3.b
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 26
The procedure to check and to find the magnified moment along the column in Sway column is the same In
non-sway column Chapter 3 , with some additional requirements :
1. Mmax1 & Mmax2 should be used instead of M1 & M2 in the calculation of Cm & Mc
Note : Mmax1 & Mmax2 should be compared to find the maximum and the minimum
(Mmax2 is not always the maximum in spite of M2 > M1 )
assume M1* is the min of { Mmax1 & Mmax2 } & M2* is the Max of { Mmax1 & Mmax2 }
2. In the calculation of Pc , use kns & dns ( as mentioned in chapter 3 )
3. Mc= dns . M2* : if M2* < Mmin → Mc= dns . Mmin
4. If dns > 1 , We have to check the condition M2nd-order ≤ 1.4 M1st-order :
if M2* ≥ Mmin → dns . M2* ≤ 1.4 M2
if M2* < Mmin → dns . Mmin ≤ 1.4 Mmin → dns ≤ 1.4
Note : According to the Codes before ACI 318 -08 , If the following condition is verified no need to
check if there is buckling along the column in case Sway column :
Starting from ACI 318 -08 , checking of the buckling along the column in case Sway column is
required Whatever the result of this condition .
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 27
Flowchart For Determining Column Slenderness Effect As Per ACI 318-14
Code
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 28
Flowchart For Determining Column Slenderness Effect As Per ACI 318-14
Code
Explained by Eng.Hassan Hammami
Step1 : Determination
Sway or Non-Sway
Step2 :
k.L/r < k.L/r
trohS
limit
telnelh
Sway
Step3 : Moment Magnification
for Sway
Non-Sway
Step3 : Moment Magnification
for Non-sway
To check if there is
buckling along the column
To be checked both in the
case of Sway or Non-Sway
To be design same as normal column
(for example : by interaction diagram )
but with 2nd-order moments
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 29
Chapter 5 - The Design Of The Column By Etabs
5.1 – Etabs doesn’t classify the column as Sway or Non-sway Automatically but Users can do that
manually :
As explained in chapter 1 , the first step to analyze and to Design the columns is to classify them as Sway or
Non-Sway
Etabs doesn’t classify the column as Sway or Non-sway , So User has to classify the column by himself
using method 1 or method 2 or method 3 ( see 1.3.1 – 1.3.2 – 1.3.3 in Chapter 1 ).
Each direction of the building should be studied separately to classify the columns if Sway or Non-sway
5.1.1 – The Simplified procedure :
We can always consider all columns as slender sway columns including P-delta effect , this Procedure will
avoid all the following checks to classify the columns ( Sway / Non-Sway & short / Slender ) Specially if
We know the two following facts :
1. In Etabs , for the vertical reinforcement design of the column , the only difference between the Sway
and Non-Sway is to include P-Delta Effect - as will be explained later - which we usually include it
already .
2. Etabs considers all columns as Slender , this is a conservative consideration .
5.1.2 - The code procedures :
Although as per the code we can use Method 1 ( by check the ratio of walls to column to be less than 12 ) but we
will not explain how to use it in Etabs because it so conservative and the methods 2 and 3 are more accurate
and more used by the main references of the code .
5.1.2.1 –Using Stability Index Q ( Method 2 ) :
In most of the references , Method 2 ( Stability Index Q ) is the most used widely
if Q ( stability index ) doesn’t exceed 0.05 ( 5% ) the columns within the story to be braced against
sidesway .
ACI 318-14 ( 6.6.4.3.b)
- SPu , is the sum of all factored column and wall gravity loads for the considered Lateral load combination
ACI Design Handbook ACI SP-17(14) – Columns Example 1 – page 363
- Vus : the factored lateral load in the combination which We consider to estimate SPu .
ACI Design Handbook ACI SP-17(14) – 9.4.3.1 –Page 356
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 30
- Do : Drift story due to Vus from first order analysis ( No P-Delta) with the cracked sections .
ACI Design Handbook ACI SP-17(14) – 9.4.3.1 –Page 356
But because of the Torsional movement of the Plan in case lateral force with eccentricity, the
Drift will different for each point of the plan , So Do could be for the center of mass
“ the displacement plotted on the capacity curve should be for the center of mass”
The Seismic Design Handbook (Farzad Naeim ) – 15.3.8.1 –Page 773
Otherwise , we can use Method 3 which includes the torsional effect .
- Lc : story height measured center to-center of the joints .
Note : For any load case The lateral Story Stiffness K = Vus / Do is constant , So we can Re-write the
formula of the stability as follows :
Q = SPu / ( K . Lc )
How to Obtain Q in Etabs :
In this Method , P-delta should be inactive to obtain only the first order effect as per the code procedure
1. To get SPu :
a- Make combinations or load cases include only the part of the gravity loads of the lateral ultimate
load combinations :
For example ,
1.2D + 0.8W
→ 1.2D
1.2D+0.5L +1.6W
→ 1.2D+0.5L
1.2D+0.5L +1.0E
→ 1.2D+0.5L
0.9D+1.6W
→ 0.9D
0.9D+1.0E
→ 0.9D
So , three combination We have to use to find SPu : 1.2D , 1.2D+0.5L and
0.9D
b- go to Display Menu → Show Tables → analysis → Results → Structure Results → Story
Forces
c- Choose the combinations which you made in step a , and Choose the Results for the Bottom .
d- SPu equals to P for each story and each combination .
Important Note : For simplified and conservative procedure , We can make only the combination
1.2 D + 0.5 L in step (a) as maximum value , in this case we will find only one
value of SPu for each story .
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 31
2. To get Vus /Do = K :
Etabs show the lateral Story Stiffness ( K ) directly but it needs seismic load to be defined in the Model
Note1 : Etabs calculates ( K ) using Do as the average value of the max story drift and the min story
drift , which is approximately close to the drift of the center of mass
Note2 : if there are only gravity load , we can define the seismic load just to get (K) of the story
Note3 : in case many eccentricity cases Ex1 , Ex2 ,Ex3 for the same direction , (K min) should be
considered .
a- Go to Display Menu → Show Tables → analysis → Results → Structure Results → Story Stiffness
b- Get K value for each story and from column “ Stiffness “
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 32
3. To get Lc :
a- Go to Display Menu → Show Tables → Model → Structure Layout → Story Data
b- Get Lc value for each story and from column “ Height “
Note : In case there is difference in the height of the same story , it is recommended to use Method 3
or to use an average values based on the Engineer estimation .
4. To get The Stability index Q :
Find Q for each story and each direction using the formula :
Qx = SPu / ( Kx . Lc )
Qy = SPu / ( Ky . Lc )
If Q ≤ 0.05 → the Story is Non-Sway ( which means that all columns & walls inside it are
Non-sway )
If Q > 0.05 → the Story is Sway (which means that all columns & walls inside it are Sway )
You could use an excel sheet to make these calculations similarly to the Picture below .
All EHH spreadsheets could be obtained from Facebook Group “EHH Structural Spreadsheets”
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 33
5.1.2.2 –Using P-Delta effect ( Method 3 ) :
This method Determines the case of each column in each story if Sway or Non-sway and includes
the effect of the torsion in the Plan ( if exists ) .
In this method we have to compare the column end moments due to second-order effect with the
first order end Moments .
if The increase in column end moments due to second-order effect doesn’t exceed 5% of the first
order end Moments , the columns to be braced ( Non-Sway ) against sidesway .
ACI 318-14 ( 6.6.4.3.a )
So , P-delta should be active in the analysis of The building when the 2nd order effect is obtained .
Because there are many combinations , We consider the conservative value for SPu :
SPu = 1.2 D +0.5 L ( as explained in method 2 )
How to Obtain M2nd / M1st ratio in Etabs :
a- Make two copies of the Etabs Model :
one without P-delta , call it 1st order model
another one with P-delta ( 1.2 D +0.5L ) call it 2nd order model
b- Export the end moments of the both models separately ,do that for all columns and walls ( except
corewalls , no need ) due to the ultimate load combinations
call the end moments of 1st order Model as 1st order moments ( M 1st )
call the end moments of 2nd order Model as 2nd order moments (M 2nd)
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 34
c- Find the ratio M 2nd / M 1st for each column and wall
if M2nd / M1st > 1.05 → the column or walls is Sway else Non-Sway
All EHH spreadsheets could be obtained from Facebook Group “EHH Structural Spreadsheets”
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 35
Notes :
1- The Moments in Etabs are located at the End offset lengths ( the Ends of the clear height ) ,
So the End offset length option is recommended to be set before exporting the moments data .
(the effect of End offset length option will be explained later ) .
2- The Large or negative Ratios for small moments (less than 10 KN.m) could be ignored because at
most their direction is due to lateral force along the other direction .
3- The Ratios for Moments located not at the column ends can be ignored .
4- As mentioned in Chapter 1 ( 1.3.3 ) , For only vertical load combination , there is no difference
between the Procedure of the Sway or Non-Sway column because there is no lateral moment but the
lateral stability should be check (as will explained in chapter 6 )
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 36
5.2 – Etabs doesn’t classify the column as Short or Slender Automatically but User can do that manually
5.2.1- column classification in Etabs :
Etabs always Treats the columns as Slender columns , but this means Etabs will magnify the design moments for
all columns even for the short columns , it is a conservative procedure but it could give over-designed Section for
the short columns while actually it is not required in the code .
To classify each column as short or slender it is absolutely not easy even programmatically, because it is
required many data for each column as explained in Chapter 2 , such as :
- classification of the column if sway or non-sway
- the end moments of the column for each load combination ( for non-sway columns )
- the sizes and Height of the considered column .
- the size and Height of the columns above and below the considered column .
- the sizes , Length and locations of the beams above and below of the considered column in all directions
Wherefore, Etabs always considers the columns as Slender columns.
5.2.2- column classification by user :
User can classify separately each column Manually or by using softwares which can do that such as
EHH spreadsheets (EHH-Nonsway magnified moment V19.5.1 & EHH-sway magnified moment
V19.5.1 )
which are available to download in Facebook Group EHH Structural Spreadsheets
and explained in Youtube channel Eng.Hassan Hammami and in Facebook Page Eng.Hassan Hammami .
Knowing the classification of the column ( short / slender ) has high importance for those columns which are
Unsafe in the Etabs Results .
If user Finds column as short column, he can remove all magnifications of the moments in the design ( as will
be explained later ) by set dns = 1 and even by deactivation P-D also .
5.2.3 - k ( Effective Length Factor ) in Etabs :
In the frame design overwrites of the column in Etabs , there is k ( Effective Length Factor ) for both
direction Major and Minor .
As Explained in Chapter 2 , Chapter 3 and Chapter 4 there are two types of k factor :
1- kns used to :
a- classify the Non-Sway columns if short or Slender
b- calculate the value of Pc ns
2- ks used to :
a. classify the Sway columns if short or Slender
b. calculate the value of ∑Pc s
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 37
Now, The Question is : in Etabs there is only one “k” , so which Type is it ? kns or ks ? and what is the use of
it in Etabs ?
Actually , “k” in Etabs is kns and it is used only to calculate the value of Pc ns ( not to classify the column )
But ! where is ks in Etabs ? and how does Etabs Find ∑Pc s without ks ?
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 38
As Explained in this chapter , Etabs doesn’t classify the column as Short or Slender , So no need to use ks
for that , Further , the calculation of ∑Pc s is used to find ds ( Sway moment factor ) which used in Moment
magnification method ( 4.2.3 in Chapter 4 ) , Etabs always sets the value of ds equals to 1 , So it doesn’t
calculate ∑Pc s thus it doesn’t need ks for that also .
Note : user can calculate the value of ds manually or by using software such as EHH-sway magnified
moment V19.5.1 then he can change its value in Etabs , But it is so difficult because the value of ds
differs for each load combination . User has to remember that if he uses ds ,he has to deactivate P-D
to avoid double effect for second order analysis of the both Methods ( P-D Elastic 2nd order analysis
method and ds Moment magnification method )
5.3 – The Methods and Formulas of Column ( Frame ) Design in Etabs .
5.3.1- ACI code Methods which used by Etabs for the column (Frame ) Design :
5.3.1.1 – Non-Sway columns :
As explained in chapter 3 , in the code there are two Methods to take account of the slenderness effect
For Non-Sway columns :
1) 2nd order analysis method – Nonsway
2) Moment magnification method – Nonsway ( dns )
2nd order analysis methods uses the 2nd order analysis by software can analyze the buckling along the
column , Even Etabs can do that using load case “ buckling “ but this method and buckling load case are
not used widely and still not workable to design all columns of the building ,in this method columns may
be subdivided using nodes along their length to evaluate slenderness effects between the ends.
To determine the magnified moment ( along the column ) , Etabs uses Moment magnification
method ( dns ) , this method supposes that no division is done between the column Ends .
5.3.1.2 –Sway columns :
As explained in chapter 4 , in the code there are Three Methods to take account of the slenderness effect
For Sway columns :
1) Elastic 2nd order analysis method ( P-D )
2) Inelastic 2nd order analysis method
3) Moment magnification method ( ds )
Etabs allows to use P-D Method and ds Method , but only one of them should be used at the same time .
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 39
5.3.2- The Etabs formulas of column (Frame ) Design :
Etabs use the interaction diagram to design the column so it determine the axial load Pu and the Moments in
each direction M2,M3 , then it check / design the column using the values of PMM (Pu,M2,M3) for each
combination comparing with the interaction diagram curves .
M2,M3 are the Magnified Moments due to the slenderness effect using the code formula .
The Magnified Moment formula which Etabs uses is M = dns ( Mns + ds Ms ) Concrete Frame Design
Manual ACI 318-14For ETABS - Provision No. 3.4.2.2
5.3.3- Comparison between Etabs formula and Code formulas :
User could ask : does Etabs formula deals with ACI code formula ? and can it be used for Sway and
Non-Sway column ?
To answer this question , Let’s Remember the formula of the code as explained in chapter 3 and chapter 4 :
5.3.3.1- Non-sway columns :
According to dns method ( which code and Etabs use) , the code formula for The Magnified Moment is :
Mc = dns M2
While the Etabs formula is
M = dns ( Mns + ds Ms )
Notice that symbol “M2” in the code is the ultimate Moment and it is same as symbol “Mu” in Etabs
formula Which Equals to Mns + Ms ( without magnification ) , And that symbol “Mc” in the code is the
magnified Moment and it is same as symbol “M” in Etabs formula , So we can re-write the code
formula using Etabs symbols :
M = dns Mu
The default value of in Etabs ds is 1 , and when we select “Non-Sway” in the frame design overwrites
Etabs will consider ds equals to 1 even user changes it to other value .
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 40
with ds = 1 , Etabs formula will become :
M = dns ( Mns + Ms )
M = dns Mu which is same as the code formula
So , To design the column as nonsway column you have to :
a. Select “ Nonsway” in the frame design overwrites or Keep ds=1 as the default value
b. No Need to activate P-D option .
5.3.3.2- Sway columns :
As explained in 5.3.1.2 , Etabs allows to use P-D Method and ds Method ( user has to use only one of
them )
5.3.3.2.1- ds Method :
According to ds Method which ( which code and Etabs use), the code formula for The Magnified
Moment is :
M = Mns +ds Ms
Then , to include the slenderness along the column , M should be multiplied by dns , So the final CODE
formula will be :
M = dns ( Mns + ds Ms ) which is same as the Etabs formula
So , To design the column as sway column by ds method you have to :
a. select “ Sway” in the frame design overwrites
b. calculate the value of ds and change it in the frame design overwrites .
c. deactivate P-D option .
Note : the Previous Procedure should be done for each load combination and for each story which is
almost impossible , because the value of ds differs for each load combination , where in the
formula of ds there is ∑Pu , ∑Pu is the sum of the axial loads of all vertical elements in the story
due to the ultimate load combinations including the axial loads due to the lateral forces which
differs for each load combination .
So , it is recommended to use P-D Method to design the sway columns .
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 41
5.3.3.2.2- P-D Method:
When P-D Method is used , the analysis will give the magnified moments at the column ends directly
Thus , ds Ms which used in the formula of ds method will be Replaced by Ms’ ( Magnified moment
including P-D effect )
So , the code formula M = dns ( Mns + ds Ms ) will become :
M = dns ( Mns + Ms’ )
In Etabs , When option P-D is active , all moments will be magnified due to including the additional
moment due to the vertical loads with the lateral drifts .
Using the default value of ds ( ds=1 ) the Etabs formula including P-D effect will become
M = dns ( Mns’ + Ms’ ) because both of gravity moments Mns and lateral moments Ms will be
affected by P-D , But P-D effect on the gravity moments Mns value is minor and could be
ignored so that we can consider Mns’ ≈ Mns , thus Etabs formula can be written as :
M = dns ( Mns + Ms’ ) which is same as the code formula
So , To design the column as sway column by P-D method you have to :
a. Active P-D option , using the maximum value of the gravity loads ( 1.2 Dead + 0.5 Live )
b. Keep the value of ds equals to 1 in the frame design overwrites .
c. Select the “ Sway” type in the frame design overwrites
( this step will not affect on the PMM result while ds = 1 , The effect is related to the shear
design of the column ) .
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 42
5.3.3.3- Conclusion and Summary ( The General Procedure ) :
For the columns Design , we can adopt one procedure for Both Sway and Non-sway columns to simplify
the work :
a. Active P-D option , using the maximum value of the gravity loads ( 1.2 Dead + 0.5 Live )
b. Keep the value of ds equals to 1 in the frame design overwrites .
c. Select the “ Non- sway “ or “ Sway” type in the frame design overwrites
( this step will not effect of the PMM result while ds = 1 , but The effect is related to the shear
design of the column ) .
We can call this Procedure ( The General Procedure )
M = dns Mu’
where Mu’ is the ultimate moment with P-D effect
5.4 –The parameters of column Design in the Etabs with the correct modeling of columns :
In this provision , We will explain the parameters of the column design in Etabs using the “ General
Procedure”
The formula of the magnified Moments is :
M = dns Mu’
If user set the value of End Length Offset , Unbraced Length Ratio With Correct Modeling , Etabs will
calculate dns and Mu’ correctly .
5.4.1 and 5.4.2 will explain how we set the values of End Length Offset and Unbraced Length Ratio to
obtain Lu and Cm and Mu’
Note : We will not use “ ks “ factor , because ( as explained in 5.2.3 ) ks used in the calculation of ds
while we will not use ds Method .
5.4.1-Unbraced Length ( Clear Hieght ) Lu :
The determination of the clear height Length and the location of its Ends ( which called also unbraced
Length “Lu”) has high importance because :
1- The Location of its Ends determines the values of the design moments Mu’ .
2- The values of the end moments are also included in Cm formula which affects on dns value
3- Lu is included in Pc formula .
After doing the analyze , Etabs calculates Lu and the unbraced length ratio for the both direction
( using the values of End Length offsets option ) and show it in the frame design overwrites .
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 43
unbraced length ratio (U.L.R) = Lu / Lc
Lu : unbraced length (clear height )
Lc : column length c/c
Lu determination as per Etabs :
As per ACI code , unbraced length is considered between the nearest surfaces at top and bot of the
column ( as shown in the picture below ) for each direction .
In Etabs use the values of the End Length offsets to determine Lu , Etabs considers the clear height
between the End Length offset at the top and the bottom of the column ( there are many cases will be
explained later by examples )
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 44
If we select Aromatic from connectivity Etabs considers the End Length offset as following :
 At the bottom ( End – I ) : Zero
 At the Top ( End – J ) : - Zero , if there is slab without beams or there is no slab and no beams
- The max depth of the beams , if there is beams .
If we select Define Length it considers the End Length offset to be same as what we insert for
End – I and End – J
Even Etabs calculates Lu using End length offset values , but user can change its values manually
in frame design overwrites.
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 45
5.4.2- How Etabs calculates Lu , Cm , Mu’ :
Let us see the following Example as General case then other special cases will
be explained .
Example 5.4.2-a :
In this example , we have a column C1 20X60 with Slabs and beams at the top
and the bottom . The Height of all stories is 4 m = 400cm
The End Moments about Minor axis ( axis 2 ) = bot 80 KN.m , top 60 KN.m
The End Moments about Major axis ( axis 3 )= bot 100 KN.m , top 150 KN.m
are shown in the Picture at the Right .
Find Lu , cm and Mu’ of the column and compare with the code requirements
.
Solution :
For column C1 20x60 there are slabs and beams
above and below the column , using Aromatic from
connectivity for End Length offset Etabs considers
the unbraced length from the bottom joint to the
deepest beam at top for both directions .
The Deeper beam at the top has 100cm depth , So :
Lu 2 = Lu 3 = Lu = 400-100 = 300 cm
Lc = 400 cm
U.L.R 2 = U.L.R 3 = U.L.R = Lu / Lc = 300/400 =
0.75
Note : Etabs always considers axis 2 as Minor axis and axis 3 as Major axis .
Etabs considers the top moments Mu’ of the columns at the end length offset ( 100cm )
M2 top = 30 , M3 top = 88
And uses the same moment in the formulas of Cm2 , Cm3
Cm2 = 0.6 – 0.4 (M1/M2) = 0.6 – 0.4 * 30/80 = 0.45
Cm3 = 0.6 – 0.4 (M1/M2) = 0.6 – 0.4 * 88/100 = 0.248
Note : Generally , the value of Cm and Mu’ are related to End Length offset only , but the value of Lu
is related to unbraced length Ratio .
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 46
ACI code considers the unbraced length ( Lu ) to be up to the deepest beam/slab but for each
direction separately . this means :
For Lu 2, the deepest beam is B20x100 , Lu 2 = 400-100 = 300 cm same as Etabs result
U.L.R 2 = 300/400 = 0.75 same as Etabs result
For Lu 3 , the deepest beam is B20x90 , Lu 3 = 400-90 = 310 cm greater than Etabs result
U.L.R 3 = 310/400 = 0.775 greater than Etabs result
On the other hand , ACI code considers also the moment at the deepest beam/slab but for each
direction separately to determine Cm2 , Cm3
the top moments Mu’ of the columns :
M2 top = 30 , M3 top = 94
Cm2 = 0.6 – 0.4 (M1/M2) = 0.6 – 0.4 * 30/80 = 0.45 same as Etabs result
Cm3 = 0.6 – 0.4 (M1/M2) = 0.6 – 0.4 * 94/100 = 0.224 less 9% than Etabs result
But to determine the Top Design Moments Mu’ for biaxial bending we have to use the moments
M2 top and M3 top at same height . There are two heights ( at 300cm and at 310 cm ) , the design
section should be studied at each height separately
At height 300 cm
At height 310 cm
M2 =30 KN.m
M3 =88 KN.m
M2 =35 KN.m
M3 =94 KN.m
Note1 : Column section is 20x60
Note2 : for direction 2 , We consider the moment up
to the beam face , so actually there is no moment
M2 =35 KN.m because it will be inside the beam and
the column section will include a part of the beam
section ( not just 20x60 ) which increases its capacity
ACI Code and all of its references do not talk about the next questions in ( Note 2 ) :
1. Where is the top design section for this case ?
2. if we consider it at height 310 cm What is the value of moment M2 we have to consider ?
3. what is the column section at height 310 ?
4. is the value of M3 is really 94 KN.m ? and the torsional inertia of B20x100 will not affect on the
M3 diagram above height 300 cm ?
Actually , these questions need many academic studies to answer because There are many variables
control the location of the top design section , one of these variable is the relative size of the column and
the connected beams and the locations of these beams , Let us see the three cases below and to think how
to determine the top design section for each case :
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 47
Case 1 : The beams are located along both direction at four sides and their size are
big than column size :
In this case we can expect that the failure could happen only at the bottom
of the deeper beam B1 (as hatched ) because of the high stiffness of the
joint above this level and the torsional inertia of B20x100 will affect on the
M3 diagram above height 300 cm ( as shown in the figure at the right )
Case 2 : The beams are located along both direction , but B1 is small than the
column size and it is not continues to the other side , this situation of B1
reduces its effect while B2 is huge , in this case we can expect that the
failure could happen the bottom of beam B2 .
Case 3 : The beams are located along both direction , B1 is small than the column size and it is not
continues to the other side , but also B2 is very small than the column size . so the effect of both
beams is small that the huge size of the column , in this case it is recommended to consider the
design section at the column end .
Based on the above , Engineer can use his experience to determine the location of the top section because
there is no exact way to determine the location of top design section , We can call the way which uses the
Engineer experience ( The estimated Method )
But also we need to find a conservative way if we want to simplify the design procedure which We can
call it ( The conservative Method ) .
For this example , we will use the both ways .
The estimated Method :
Our case could be considered similar to case 1 , because the column size is 20x60 while the beams sizes
are 20x90 at direction 2 and 20x100 at direction 3 , So we can estimate the top design section to be at
300cm ( same as we consider already using Aromatic from connectivity for End Length offset ) .
using this value of Automatic End Length offset will make the top Moments Mu’
M2 =30 KN.m , M3 =88 KN.m ✓
and same moments will be used to determine Cm
Cm2 = 0.45 , Cm3 = 0.248 ✓
and Lu = 300 cm for both directions ✘
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 48
As explained above , we accepted the estimated location thus the values of the moments M2, M3 and
the value of Cm2,Cm3 which related with the moments values but the value of Lu should be modified
to be matched with the Code Requirements .
Lu2 (Etabs ) = Lu2 (Code ) = 300 cm ✓
But Lu3 ( Etabs ) should be 310 cm as per code value , We can set Lu3 by changing U.L.R3
Lu3 = 310 cm, Lc= 400 → U.L.R3 = 310/400 = 0.775 ✓
Select the column and go to frame design overwrites option then change the value of unbraced length
ratio (Major ) to 0.775
To check your work , you have to see the effective length in the design Details , the values of Lu should
be:
Lu2 = 3000 mm , Lu3 = 3100 mm ✓
The conservative Method :
We can consider the top design section to be at the top of the
column and beams ( at the top End – J ) or at the bottom of the
slab ( if there is slab ) .
In this example , assume there is slab with thickness 240 mm .
Select the column and go to end length offset and select define
length , then make the value for End – J to be 240 mm ( slab
thickness )
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 49
The top Moments Mu’ :
M2 =53 KN.m , M3 =135 KN.m ✓
and same moments will be used to determine Cm
Cm2 = 0.335 , Cm3 = 0.304 ✓
and Lu = 376 cm for both directions ✘
Also , the value of Lu should be modified to be matched with the Code Requirements .
Lu2 (Code ) = 300 cm → U.L.R2 ( Etabs ) = 300/400 = 0.75 ✓
Lu3 (Code ) = 310 cm → U.L.R3 ( Etabs ) = 310/400 = 0.775 ✓
Select the column and go to frame design overwrites option then change the value of unbraced length
ratio (Major ) to 0.775 and change the value of unbraced length ratio (Minor ) to 0.75
To check your work , you have to see the effective length in the design Details , the values of Lu should
be:
Lu2 = 3000 mm , Lu3 = 3100 mm ✓
Note : The comparison between The estimated Method and The conservative Method shows that :
direction
2
direction
3
M2(conservative) is 176% of M2(estimated )
Cm2(conservative) is 74% of Cm2(estimated )
So the final magnified M2 (conservative) will be 130 % of final
magnified M2(estimated )
M3(conservative) is 153% of M3(estimated )
Cm3(conservative) is 123% of Cm3(estimated )
So the final magnified M3 (conservative) will be 191 % of final
magnified M3(estimated )
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 50
Example 5.4.2-b :
Using same data and results of the Previous example and the new Data in this example :
Pu = 1400 KN , Pu dead = 840 KN , Fc’ = 32 MPa
Find dns and the final Magnified moments M for both direction at the top of the column using the
estimated and the conservative Methods
Solution :
Etabs uses same Code formula of dns :

Cm : we have explained it , and here a general comparison for Cm properties between Etabs and ACI
code :
Cm
M1/M2 sign
transverse loading
M1=M2=0
Cm For Response
Spectrum load case


ACI Code
Cm = 0.6 – 0.4 (M1/M2) ≤ 1
Negative if Mu,top and Mu,bot have
same Sign
Positive if Mu,top and Mu,bot has
opposite Sign
For column with transverse loading
between the column Ends , Cm = 1
For M1=M2=0 , Cm = 1
The code is not clear about the
calculation of Cm from the moments
due to the Response Spectrum load
case .
Etabs
Same
Same
Same
Will be explained later in an example
Same
Etabs uses min negative value at an
end with the max positive value at the
other end to determine M1,M2
Will be explained later
Pu = 1400 KN ( the ultimate axial load )
Euler buckling load , it should be calculated for each direction .
a- As per ACI code (EI)eff could be found for any one of the following three equations but
Etabs uses only formula a , where Ig is the gross inertia ( without considering the
applied stiffness modifiers )
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 51
dns =
𝑃𝑢 𝑠𝑢𝑠
𝑃𝑢
≤1
As per ACI code Pu sus is the sustained axial load ( dead load with A part of live load , usually
DL+0.25 LL ) But Etabs considers only the dead load as sustained load , This consideration
reduces the value of dns and dns considers , this Procedure could be accepted because of other
used conservative values and Etabs may suppose that no part of the live load could be considered
as sustained in the project .
So Pu sus ( Etabs ) = Pu dead = 840 KN
dns = 840 / 1400 = 0.6
Note : if Pu > 0.75 Pc → The section should be revised and Etabs shows warning O/S #5
b- k factor : as explained in 5.2.3 , k ( effective
length ratio ) in dns formula is kns ( which its
value is equals or less than 1 ) , The default
value of k in Etabs for both direction is 1 ( the
maximum and conservative value ) but user
can change it for each direction after he
calculate it manually (see 2.1.2 ) to enhance/
reduce the design .
in this example we will keep the default values
k2 = k3 = 1

Mmin : If the ultimate moment Mu’ is less than Mmin , Mmin should be used in the formula
of the final magnitude moment M = dns Mmin , Etabs uses same Code formula of Mmin
Mmin = Pu ( 15+0.03 h ) : 15+0.03h in mm , where h in the dimension of the column
perpendicular to the considered axis .
but Etabs doesn’t use Mmin for the both direction at the same time .
a- if M2 and M3 are less than M2min and M3min respectively , Etabs use M2 , M3min as
design moments .
b- if M2 > M2min and M3 < M3min , Etabs use M2 , M3min as design moments .
c- if M2 < M2min and M3 > M3min , Etabs use M2 , M3 as design moments .
Etabs always call direction 3 as Major even it is not actually , So user has to make axis 3 as
Major axis in the column define to be matched with the Etabs consideration .
✓
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
✘
Page 52
Now , We will use the examination above to calculate the magnified moment for the estimated and
the conservative Method .
The Estimated Method :
Direction 2
Cm2 = 0.45 ( from the Previous example )
Pu = 1400 KN
Direction 3
Cm3 = 0.248 ( from the Previous example )
Pu = 1400 KN
Ec = 4700 √ f’c = 4700 √ 32 = 26587.2 MPa
= 2.659 x 1010 Pa
Ec = 4700 √ f’c = 4700 √ 32 = 26587.2 MPa
= 2.659 x 1010 Pa
Ig2 = 60 x 203 / 12 = 40,000 cm4 = 4 x 10-4 m4
Ig3 = 20 x 603 / 12 = 360,000 cm4 = 36 x 10-4 m4
dns = 0.6
dns = 0.6
EIeff 2 = 0.4 Ec Ig / ( 1+dns )
= 0.4 (2.659 x 1010) (4 x 10-4 ) / ( 1 + 0.6 )
= 2.569 x 106 N. m2
EIeff 3 = 0.4 Ec Ig / ( 1+dns )
= 0.4 (2.659 x 1010) (36 x 10-4 ) / ( 1 + 0.6 )
= 23.121 x 106 N. m2
Lu2 = 300 cm = 3 m ( from the Previous example )
Lu3 = 310 cm = 3.1 m ( from the Previous example )
k2 = 1 ( default value )
k3 = 1 ( default value )
Pc2 = p2 * EIeff2 / ( k2 Lu2 )2
= p2 * 2.569 x 106 / ( 1 x 3 )2
= 2,817,224 N = 2817 KN
Pc3 = p2 * EIeff3 / ( k3 Lu3 )2
= p2 * 23.121 x 106 / ( 1 x 3.1 )2
= 23,745,590 N = 23,746 KN
Pu < 0.75 x min ( Pc2, Pc3 ) OK
dns2 = Cm2 / ( 1 - Pu / 0.75 Pc2)
dns3 = Cm3 / ( 1 - Pu / 0.75 Pc3)
= 0.45 / ( 1 – 1400 / ( 0.75 * 2817 ) )
= 0. 248 / ( 1 – 1400 / ( 0.75 * 23,746 ) )
= 1.33
= 0.27 < 1 → consider dns3=1
As per ACI code ( see chapter 3 and 4 ) dns should be less than 1.4
dns2 = 1.33 OK , dns3 = 1 OK
Mu2’ =30 KN.m ( from the Previous example )
Mu3’ =88 KN.m ( from the Previous example )
M2min = Pu ( 15+0.03 h )
M3min = Pu ( 15+0.03 h )
=1400*(15+0.03*200)
=1400*(15+0.03*600)
= 29,400 KN.mm = 29.4 KN.m
= 46,200 KN.mm = 46.2 KN.m
Mu2’ > M2min , use Mu2’
Mu3’ > M3min , use Mu3’
Design M2 = dns2 Mu2’
Design M3 = dns3 Mu3’
= 1.33 x 30 = 39.9 KN.m
= 1 x 88 = 88 KN.m
The Conservative Method :
Direction 2
Direction 3
The difference between the estimated and the Conservative Methods is only in Cm and M’ values ,other
values are the same .
Cm2 = 0. 335 ( from the Previous example )
Cm3 = 0.304 ( from the Previous example )
dns2 = Cm2 / ( 1 - Pu / 0.75 Pc2)
dns3 = Cm3 / ( 1 - Pu / 0.75 Pc3)
= 0.335 / ( 1 – 1400 / ( 0.75 * 2817 ) )
= 0. 304 / ( 1 – 1400 / ( 0.75 * 23,746 ) )
= 0.99 < 1 → consider dns2=1
= 0.33 < 1 → consider dns3=1
dns2 = 1 OK , dns3 = 1 OK
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 53
Direction 2
Mu2’ =53 KN.m ( from the Previous example )
Mu2’ > M2min , use Mu2’
Design M2 = dns2 Mu2’
= 1x 53 = 53 KN.m
Direction 3
Mu3’ =135 KN.m ( from the Previous example )
Mu3’ > M3min , use Mu3’
Design M3 = dns3 Mu3’
= 1 x 135 = 135 KN.m
Notes :
1- The Design Moment for the conservative Method is much more than the Estimated Method , So it is
recommended to think about the actual behavior to use the Estimated Method .
2- As required in this example , the results are for the Top section , Etabs does same Procedure to Design the
column section at the Top and Bottom and at the Middle ( user can increase the number of the design
sections using output Station option .
3- in sway columns , the condition is not only dns ≯ 1.4 , the original condition is :
Magnified M / Mu ≯ 1.4 → dns * Mu’ / Mu ≯ 1.4 .
If dns = 1 → Mu’ / Mu ≯ 1.4 , where Mu’ include P-D effect
As Per James MacGregor (Consulting Members In ACI Code ) In his book Reinforced concrete
Mechanics & design 6E Page 583 , He seems that he allows to accept dns more than 1.4
( up to 1.75 to 2), actually he said : “ if dns exceed 1.75 or 2.0 a larger cross section should be
selected “ , he said that in example for Non-sway column :
Page 583
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 54
While in other example for Sway column Page 622 , He checked only the condition Mu’ / Mu ≯ 1.4
and ignored the condition dns * Mu’ / Mu ≯ 1.4
Page 622
We can conclude that MacGregor interests in verifying the condition dns ≯ 1.4 for Non-sway
column and Mu’ / Mu ≯ 1.4 for Sway Column more than dns * Mu’ / Mu ≯ 1.4 .
We will Talk later about checking the Code conditions .
5.4.3- the required consideration in case only gravity loads :
Actually , for gravity loads only there is no effect on the design moment whether Non-sway or sway,
because there is only gravity moments Mns and no lateral moments Ms
For nonsway column : M = dns Mu = dns ( Mns + Ms ) = dns Mns
For Sway column : M = dns (Mns + ds Ms )
= dns Mns
However , the Sideway Stability should be checked .
In This Case , stability index Q is used check To the sideway stability , and because there is no Lateral
loads we apply an arbitrary lateral load Vu and we see its story drift Do Then calculate Q using Vu
and Do, Notice that we can use the lateral stiffness K which explained in 5.1.2.1 to calculate Q .
∑Pu is due to the gravity load combination ( Max of 1.4 DL , 1.2DL+1.6LL )
If Q > 0.25 → Unstable → increase the lateral stiffness of the story
See ACI 318 – 14 - R6.2.6 and MacGregor book Reinforced concrete Mechanics & design 6E
EXAMPLE 12-3 Page 615.
If Q < 0.25 → Stable → Treat with the column as Non-sway column
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 55
5.4.4- How to check the Code condition for the ratio of the second order to first order moment in Etabs :
Based on what already explained , the code condition for stability is :
dns ( M’ / Mu ) < 1.4
-
The Ratio M’ / Mu we can get it in Etabs using Procedure and EHH spread sheets which explained in 5.1.2
for column classification .
-
dns values are not shown in Etabs up Result Table to Version 17.0.1, and it is so difficult to get dns value
from the design detail report for each column each combination .
Even the Newer Etabs versions will show dns values as Table ,We need to calculate the
Ratio dns ( M’ / Mu ) to compare with 1.4 .
Now , The more important is to check M’ / Mu ( as explained in 5.1.2 ) , Then to check dns which
shows in the design detail report .
In case EHH spreadsheet has been made in later to calculate the value of dns and the
Ratio dns ( M’ / Mu ) it will be informed in the following accounts :
Facebook Page : Eng.Hassan Hammami
Facebook Group : EHH Structural Spreadsheets
Youtube : Eng.Hassan Hammami.
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 56
5.4.5- What user has to do for all cases of the column situation :
As explain before , To get best result for column design , user has to Estimate the location of the top
design section then to set the value of unbraced length ratio and the end length offset and to Model
the column correctly .
We will discuss how to do that for the different cases of the column .
The General Case :
- End Length Offset :
Estimate the location of top design section .
* if it is at the bottom of deepest beam
Keep End Length Offset as default ( automatic )
* if it is at the bottom of beam B2x90
change End Length Offset at End – J to be 900 mm
* if it is at the soffit of slab
change End Length Offset at End – J to be 240 mm
(assuming slab thickness = 240 mm )
* if it is at the Top Joint (conservative )
change End Length Offset at End – J to be 0
Note : if the beams has same depth around the column , you can
keep End Length Offset as default ( automatic )
- Unbraced Length Ratio (U.L.R) :
U.L.R2 = (400-100 ) / 400 =0.75
U.L.R3 = (400-90 ) / 400 =0.775
Etabs will calculate it automatically if End Length Offset is Auto
User has to change it manually
Note : if the beams has same depth around the column , Etabs will calculate U.L.R correctly
automatically
Special Case 1 , No beams above the column :
- End Length Offset :
Keep it as default ( automatic ) , but you can consider the slab
thickness ( End – J = slab thickness ) to make the top design section
at the soffit of slab .
- Unbraced Length Ratio (U.L.R) :
Etabs will calculate it automatically
-
Modeling comment :
Be attentive to if there is no diaphragm assigned to the slab , Etabs
may not detected the slab thus it will extend the column height , So
it is always recommended to assign diaphragm to the slabs
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 57
Special Case 2 , No Slab above the column but there are beams
around the column at the top :
Same as the General case
Special Case 3 , Column extends Double height ( two stories ) and
there is beams along one direction at the middle
story level .
In this case , user has to set the parameter for both columns C1 , C2
, Where C1 and C2 are the two parts of one column
- End Length Offset :
For C1 :
Keep it as default ( automatic ) .
Etabs will consider the end length offset to be 900 mm for
direction 3 only and he will ignore it for direction 2 where it
will consider double height for direction 2 only .
For C2 :
Same as the General case
- Unbraced Length Ratio (U.L.R) :
For C1 :
U.L.R2 = (400+360-100 ) / 400 =1.65
Etabs will calculate it automatically if End Length Offset
is Auto at the Top of C2
U.L.R3 = (400-90 ) / 400 =0.775
Etabs will calculate it automatically
For C2 :
U.L.R2 = (400+360-100 ) / 360 =1.83
Etabs will calculate it automatically if End Length Offset
is Auto at the Top of C2
U.L.R3 = (360-90 ) / 360 = 0.75
User has to change it manually
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 58
Special Case 4 , No Slab above the column but there is beam in one
direction and inclined beam in other direction .
In this case these is no beam along direction 3 but there is beams makes
angle with the main directions 2 and 3 .
The Question in this case is : does Etabs consider this beam as support
for direction 2 or direction 3 ?
Actually , it depends on the value of the angle , but generally we cannot
depends on Etabs for this consideration because sometimes it gives
illogical values for Lu .
So , user has to decide if this beam is support to which direction then he
set the values of end length offset unbraced length ratio .
For the Example in the picture , if the angle with axis 2 is small we can
consider B20x70 as support about axis 3 only and this case will be
similar to Special Case 3 , but if the angle with axis 3 is small we can
consider B20x70 as support about axis 2 only and this case will be
similar to Special Case 2
Special Case 5 , Column extends Double height ( two stories ) without beams at
the middle story level .
In this case , Etabs considers Cm=1 ( max value )
automatically for both direction this at most will
give over design .
To consider correct values for Cm , user has to draw
the column as one Part ( as shown in the right
picture) then this case will be Treated as one column
( as explained before )
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 59
Special Case 6 , There is beam/beams connected to the column between its Ends
In this case , Etabs considers Cm=1 ( max value )
automatically , this value is correct as mentioned
in 3.2.2.2 - Note 3 .
But user has to divide the column at the
intersection point to two column ( as shown in
the right picture) .
then treat with the both column same as Special
Case 2 or 3 according to beams number and
direction .
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 60
Chapter 6 - Summary for how to adjust the inputs in Etabs to design the columns
Step 1 : Define the column Reinforcement as Design or check :
In Check option , Etabs will check the Reinforcement using the distribution and the size of the bars
In Design option , Etabs will Design the Required Reinforcement area using the distribution of the bars
Step 2 : adjust the end length offset and check the column modeling ( before doing “ Run “ ):
If the depths of the beams at the Top of the column are not different , Etabs will calculate
the unbraced length ratio automatically .
For simplified and fast work , Make the End Length offset for all columns to be Zero as very
conservative Procedure , Then if you face undesirable results or you want more accurate results , you
can use the following Procedures as per the case of each column :
The General Case :
Estimate the location of top design section .
See picture at right :
* if it is at the bottom of deepest beam ( see Example 5.4.2-a to
know the estimation concept )
Keep End Length Offset as default ( automatic )
* if it is at the bottom of beam B2x90
change End Length Offset at End – J to be 900 mm
* if it is at the soffit of slab
change End Length Offset at End – J to be 240 mm
(assuming slab thickness = 240 mm )
* if it is at the Top Joint (conservative )
change End Length Offset at End – J to be 0
Note : if the beams has same depth around the column , you can
keep End Length Offset as default ( automatic )
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 61
Special Case 1 , No beams above the column :
Keep it as default ( automatic ) , but you can consider the slab
thickness ( End – J = slab thickness ) to make the top design section
at the soffit of slab .
-
Modeling comment :
it is always recommended to assign diaphragm to the slabs to
avoid any problem . otherwise Etabs could not detect the slab and
considers the column as double height
Special Case 2 , No Slab above the column but there are beams around
the column at the top :
Same as the General case
Special Case 3 , Column extends Double height ( two stories ) and
there is beams along one direction at the middle story
level .
In this case , user has to set the parameter for both columns C1 , C2
, Where C1 and C2 are the two parts of one column
For C1 :
Keep it as default ( automatic ) .
For C2 :
Same as the General case
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 62
Special Case 4 , No Slab above the column but there is beam in one
direction and inclined beam in other direction .
if the angle with axis 2 is small we can consider B20x70 as support
about axis 3 only and this case will be similar to Special Case 3 , but
if the angle with axis 3 is small we can consider B20x70 as support
about axis 2 only and this case will be similar to Special Case 2
Special Case 5 , Column extends Double height ( two stories ) without
beams at the middle story level .
draw the column as one Part ( as shown in the right
picture) then this case will be Treated as one column
( as explained before )
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 63
Special Case 6 , There is beam/beams connected to the column between its Ends
divide the column at the intersection point to two
column ( as shown in the right picture) .
then treat with the both column same as Special
Case 2 or 3 according to beams number and
direction .
Step 3 : activate P-D option .
Activate P-D option , using the vertical load with the scale factor 1.2DL + 0.5 LL
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 64
Step 4 : classify the columns Sway / Nonsway :
There are two methods M2nd / M1st method and Q method , using M2nd / M1st method is better
because we will need the value of M2nd / M1st ratio later to check the lateral stability in step 6 .
M 2nd / M 1st method :
a- Make two copies of the Etabs Model , one without P-D , call it 1st order model and another one
with P-D ( 1.2 D +0.5L ) call it 2nd order model
b- Export the end moments of the both models separately , do that for all columns and walls ( except
corewalls , no need ) due to the lateral ultimate load combinations , call the end moments of 1st order
Model as 1st order moments ( M 1st ) , and call the end moments of 2nd order Model as 2nd order
moments (M 2nd)
c- Find the Ratio of the end moments M 2nd / M 1st for each column and wall directions (EHH
spreadsheet can help you ) :
if M2nd / M1st < 1.05 → the column or walls is Non-Sway
if 1.05 < M2nd / M1st < 1.4 → the column or walls is Sway
if M2nd / M1st > 1.4 → The story Needs to increase lateral stiffness ( see the notes below )
Note1 : The Large or negative Ratios for small moments (less than 10 KN.m) could be ignored
because at most their direction is due to lateral force along the other direction .
Note2 : The Ratios for Moments located not at the column ends can be ignored .
Note3 : in case the ratio of a column is more than 1.4 for a load combination includes vertical loads
not 1.2DL+0.5LL , you can use the actual vertical load of the combination in P-D option .
Ex : if the ratio due to 0.9 DL + 1 E is more than 1.4 for a column , you can change the loads
in P-D option to be 0.9DL just to check this column .
Q method :
a-
Deactivate P-D
b- Go to Display Menu → Show Tables → analysis → Results → Structure Results → Story Forces
Choose the ultimate combinations then Choose the Results for the Bottom
(SPu equals to P for each story )
c- Go to Display Menu → Show Tables → analysis → Results → Structure Results → Story Stiffness
Get K value for each story and from column “ Stiffness “
Note : if Etabs doesn’t show K value , define any seismic forces for both directions .
d- Go to Display Menu → Show Tables → Model → Structure Layout → Story Data
Get Lc value for each story and from column “ Height “
e- Calculate Q for the gravity combinations for each story for both directions using formula
Q = SPu / ( K . Lc ) (EHH spreadsheet can help you ) :
If Q < 0.05 → Non-Sway
If Q > 0.05 → Sway
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 65
Special case ( Gravity loads only ) : Check only the lateral stability .
In case there is only Gravity loads , it is no need and not possible to apply M2nd / M1st method to classify
the columns as sway or nonsway because there are no lateral loads , actually in this case we need only to
check the Sideway stability and the column will be designed as Non-Sway .
Follow same Q method which explained above , but SPu will be for the ultimate gravity combinations
and the results of Q should be check as below :
If Q < 0.25 → Stable
If Q > 0.25 → The story is sideway unstable , Need to increase the story lateral stiffness
Step 5 : Adjust Unbraced Length Ratio (After doing “Run” ) :
If the depths of the beams at the Top of the column are not different , Etabs will calculate the
unbraced length ratio automatically
If you set End length offset = 0 , unbraced length ratio automatically as maximum height (conservative )
Otherwise , you have to adjust the value of unbraced length ratio manually , see the following example :
The General Case :
U.L.R2 = (400-100 ) / 400 =0.75 Etabs will calculate it
automatically if End Length Offset is Auto
U.L.R3 = (400-90 ) / 400 =0.775
User has to change it manually
Note : if the beams has same depth around the column , Etabs
will calculate U.L.R correctly automatically
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 66
Special Case 1 , No beams above the column :
Etabs will calculate it automatically
Special Case 2 , No Slab above the column but there are beams
around the column at the top :
Same as the General case
Special Case 3 , Column extends Double height ( two stories ) and
there is beams along one direction at the middle
story level .
In this case , user has to set the parameter for both columns C1 , C2
, Where C1 and C2 are the two parts of one column
For C1 :
U.L.R2 = (400+360-100 ) / 400 =1.65
Etabs will calculate it
automatically if End
Length Offset is Auto
at the Top of C2
U.L.R3 = (400-90 ) / 400 =0.775
Etabs will calculate it
automatically
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 67
For C2 :
U.L.R2 = (400+360-100 ) / 360 =1.83
Etabs will calculate it
automatically if End
Length Offset is Auto at
the Top of C2
U.L.R3 = (360-90 ) / 360 = 0.75
User has to change it manually
Special Case 4 , No Slab above the column but there is beam in one
direction and inclined beam in other direction .
For the Example in the picture , if the angle with axis 2 is small we can
consider B20x70 as support about axis 3 only and this case will be
similar to Special Case 3 , but if the angle with axis 3 is small we can
consider B20x70 as support about axis 2 only and this case will be
similar to Special Case 2
Step 6 : Check The code condition , dns < 1.4 and dns M’/M < 1.4 :


If the column is nonsway ( from Step 4 results ) : check that dns < 1.4 in the design detail
If the column is sway ( from Step 4 results ) : check that dns M’/M < 1.4 in the design detail
The values of M’/M are obtained from step 4 using M2nd / M1st method ( therefore we need
M2nd / M1st method anyway )
M’ : the ultimate moment with P-D , M : the ultimate moment without P-D
It is so difficult to find and to check dns for each column and each load combination because there it
doesn’t show in the result table , in the future EHH spreadsheet could be made to find dns values for
nonsway column and dns M’/M values for sway column as table .
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 68
In case the conditions are not achieved , you can use the following solutions :
a. If the column is short ( EHH spreadsheet could be used to classify the column ) ,
you can ignore these conditions .
b. Be sure that you set the parameters as explained in the previous steps
c. Calculate and insert the actual value of k ( EHH spreadsheet could be used )
d. Decrease the axial load if available .
e. Increase the column size ( specially the width )
f. Reduce the height if available .
g. Increase f’c
Step 7 : Check the results and warnings :
Warning O/S #5 :
This warning means Pu > 0.75 Pc
The Solutions :
a.
b.
c.
d.
e.
f.
Be sure that you set the parameters as explained in the previous steps
Calculate and insert the actual value of k ( EHH spreadsheet could be used )
Decrease the axial load if available .
Increase the column size ( specially the width )
Reduce the height if available .
Increase f’c
Warning O/S #2 :
This warning means that the required reinforcement ratio exceeds 6% for special sway or
8% for other cases This warning shows in case the column reinforcement is defined for
Design . At splice area should be checked of the half of these ratio 3% and 4% respectively
The Solutions :
a- Be sure that you set the parameters as explained in the previous steps
b- Calculate and insert the actual value of k if dns > 1 ( EHH spreadsheet could
be used )
c- If the column is short ( EHH spreadsheet could be used to classify the column )
, you can design the column with inserting the value of dns = 1 in Frame Design
defgh-
Overwrite and with deactivation P-D
If the column is Slender But Nonsway ( from Step 5 results ) , you can design
the column with deactivation P-D
Increase the column size ( specially the width )
Try to change the distribution of the bars in the column section
Decrease the axial load and the moments if available .
Increase f’c
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 69
Warning O/S #35 :
This warning means that the capacity ratio D/C ( ratio of required reinforcement to the used
reinforcement ) exceeds 1 , This warning shows in case the column reinforcement is defined
for Check , Note that user has to check the if the applied reinforcement exceeds the max
reinforcement ratio .
The Solutions :
a- Be sure that you set the parameters as explained in the previous steps
b- Calculate and insert the actual value of k if dns > 1 ( EHH spreadsheet could be used )
c- If the column is short ( EHH spreadsheet could be used to classify the column ) , you
can design the column with inserting the value of dns = 1 in Frame Design Overwrite
defghi-
and with deactivation P-D
If the column is Slender But Nonsway (from Step 5 results ) , you can design the
column with deactivation P-D
Increase the size and/or the number of the bars in the column section
Try to change the distribution of the bars in the column section
Increase the column size ( specially the width )
Decrease the axial load and the moments if available .
Increase f’c
© copyright Eng .Hassan Hammami All copyrights reserved
This note is allowed only for personal use not for Commercial use
Facebook : Eng.Hassan Hammami , Youtube : Eng.Hassan Hammami
Page 70