Columns (Compression members):

Columns are element with height larger or equal three times the least dimension; primary
designed for compressive loads, and classified to short and Long (slender) columns.

Types of columns:
1- Tied columns: with separate ties at distance as allowable.

2- Spirally reinforced columns: effect of spiral indicated in figure.

3- Composite columns

1
Limits for reinforcement of compression members ( ACI-CODE 08):
1- Area of longitudinal reinforcement: For compression members shall be not less than
0.01Ag or more than 0.08Ag. Minimum number of longitudinal bars in compression
members shall be 4 for bars within rectangular or circular ties, 3 for bars within
triangular ties, and 6 for bars enclosed by spirals.
2- Tie reinforcement for compression members shall conform to the following:
3.1- Tie reinforcement for compression members shall conform to the following:
All bars shall be enclosed by lateral ties, at least No. 10 in size for longitudinal bars
No. 32 or smaller and at least No. 13 in size for No. 36, No. 43, No. 57, and bundled
longitudinal bars.
3.2- Vertical spacing of ties shall not exceed 16 longitudinal bar diameters, 48 tie bar
or wire diameters, or least dimension of the compression member. Ties shall be
arranged such that every corner and alternate longitudinal bar shall have lateral
support provided by the corner of a tie with an included angle of not more than 135
degrees and no bar shall be farther than 150 mm clear on each side along the tie from
such a laterally supported bar. Where longitudinal bars are located around the
perimeter of a circle, a complete circular tie shall be permitted.
3.3- Ties shall be located vertically not more than one-half a tie spacing above the top of
footing or slab in any story, and shall be spaced as provided herein to not more than
one-half a tie spacing below the lowest horizontal reinforcement in slab, drop panel,
or shear cap above.

3- Spirals: Spirals shall consist of evenly spaced continuous bar or wire of such size and so
assembled to permit handling and placing without distortion from designed dimensions. For
cast-in-place construction, size of spirals shall not be less than 10 mm diameter.
Clear spacing between spirals shall not exceed 75 mm, nor be less than 25 mm.
Volumetric spiral reinforcement ratio,  , shall be not less than the value given by
2

2   
2 
 = 0.45  2  1

Or  = 0.45  2  1
    

4  (   )
And the spacing 
=
    2
Anchorage of spiral reinforcement shall be provided by 1-1/2 extra turns of spiral bar or wire
at each end of a spiral unit. Spiral reinforcement shall be spliced, if needed, by any one of the
following methods:
(a) Lap splices not less than the larger of 300 mm and the length indicatedbelow:
(1) deformed uncoated bar or wire............... 48d
(2) plain uncoated bar or wire ...................... 72d
(3) epoxy-coated deformed bar or wire........ 72d

$[LDOVWUHQJWKij3Q :
for compression members shall not be taken greater than ij3QPD[ computed by ;
1- For nonprestressed members with spiral reinforcement ;
! = ", #$ = 0.85  "[0.85     (
 ) +   ]
2- For nonprestressed members with tie reinforcement;
! = ", #$ = 0.8  "[0.85     (
 ) +   ]

3
 Compression-controlled sections, as defined above the " value:
(a) Members with spiral reinforcement .................................0.75
(b) Other reinforced members ............................0.65
For sections in which the net tensile strain in the extreme tension steel at nomiQDOVWUHQJWKİt, is
between the limits for compression-controlled and tension-FRQWUROOHG VHFWLRQV ij shall be
permitted to be linearly increased from that for compression-FRQWUROOHGVHFWLRQVWRDVİt
increases from the compression controlled strain limit to 0.005.

$ORZHUij-factor is used for compression-controlled sections than is used for tension-controlled

sections because compression-controlled sections have less ductility, are more sensitive to
variations in concrete strength, and generally occur in members that support larger loaded areas
than members with tension-controlled sections. Members with spiral reinforcement are assigned
DKLJKHUij than tied columns because they have greater ductility or toughness.

Ex.1: For the two columns shown below compare the capacity of the columns if they specify
the minimum ACI-Code requirements. (f’c=25MPa, and fy = 414MPa)

For tied: ! = 0.8  "[0.85     (

 ) +   ]
4&20 2 4&20 2
! = 0.8  0.65 %0.85  25(450  450  )+  414'/1000= 2494.37 kN
4 4

For spiral:
&450 2 6&20 2 6&20 2
! = 0.85  0.75 %0.85  25(  )+  414'= 2627.36kN
4 4 4
The ratio = 2627.36/2494.37 = 1.05

Ex.2: Design a short tied column support service dead load of 1600kN and service live load of
900kN try   *- # = 28##    1.7% with f’c= 20.7MPa and fy =
350MPa.

Pu= 1.2 *1600 + 1.6* 900 = 3360 kN


= then Ast = 0.017 Ag

! = 0.8  "[0.85     (
 ) +   ]
3360  1000 = 0.8  0.65[0.85  20.7(
 0.017 
) + 0.017
 350]

4
Ag= 277963.45mm2 the - = 9277963.45 = 527.2## use 530mm

3360  1000 = 0.8  0.65[0.85  20.7(5302  ) +   350]

4570 .036
Ast= 4570.036mm2 no. of bars = = 7.4 use 8 bars
&14 2

8  &  142
 = = 0.0175 ;. > 1%  < 8%
530  530
For ties use 10mm bars with spacing  ? 16  28 = 448##
? 48  10 = 480##
? 530##

Use 10 mm ties @ 440mm c/c

Clear spacing for

530240210428
Sect. A= = 106## < 150##    @ 
3

530240210328
Sect. B= = 177## > 150##   @ .
2

Ex.3: Design a short circular spirally reinforced column support service dead load of 1100kN
and service live load of 950kN try   *- # = 25##    2% with f’c=
27.5 MPa and fy = 400MPa.

Pu= 1.2*1100+ 1.6* 950 = 2840 kN


= then Ast = 0.02 Ag

! = 0.8  "[0.85     (
 ) +   ]
2840  1000 = 0.8  0.65[0.85  27.5(
 0.02 
) + 0.02
 400]

4176705 .93
Ag= 176705.93 mm2 the  = A = 474.3## use 480 mm
&

5
 &  4802
2840  1000 = 0.8  0.65 B0.85  27.5(  ) +   400C
4
3270 .36
Ast = 3270.36 mm2 try bars with 30 mm diameter, then no. of bars = = 4.6 bars use
&15 2
minimum of 6 bars

6  &  152
 = = 0.0234 ;. > 1%  < 8%
&  4802
4
For spirals use bars with 10mm diameter then:

GH M OƍI ESDM MU.F

 = D. EF  M  N = D. EF R  NT = D. DNVW
GIJKL OPQ (ESDMED)M EDD

4  (    ) 4  79(400  10)


= 2 =
    0.0136  4002
= 56.6 ## ! 50## > 25  < 75## . ;.

Ex.4: Design a rectangular short tied column with h/b ratio of 1.5 , support service dead load of
300kN and service live load of 500kN try   *- # = 20##    2% with
f’c= 20.7MPa and fy = 400MPa.

Pu= 1.2 *300 + 1.6* 500 = 1160 kN


= then Ast = 0.02 Ag

! = 0.8  "[0.85     (
 ) +   ]
1160  1000 = 0.8  0.65[0.85  20.7(
 0.02 
) + 0.02
 400]

88371 .526
Ag= 88371.526mm2 = -   = 1.5 2 -  = A = 242.7## use 250mm
1.5

h= 1.5*250=375mm use 380mm Ag = 250*380mm2

1160  1000 = 0.8  0.65[0.85  20.7(250  380  ) +   400]

1807 .52
Ast = 1807.52mm2 no. of bars = = 5.7 use 6 bars
&10 2

6
 6  &  102
 = = 0.0198 ;. > 1%  < 8%
250  380
For ties use 10mm bars with spacing  ? 16  20 = 320##
? 48  10 = 480##
? 250##

Use 10 mm ties @ 250mm c/c

380240210320
Clear spacing for = = 110## <
2
150##    @ 

Plastic centroid: the point that the resultant of compressive loading should be applied through
and it coincide with the section center if symmetry exist.

Po = Cc + Cs = 0.85*f’c*b*h+As*fy+As’ *fy
-
X =   $ = Y  + (-   ) + 
2
Where:
Po= resultant of forces
Cc= concrete compressive strength
Cs= steel compressive strength.

Ex.5: Find the position of plastic centroid for the column section shown below, use f’c=30MPa
and fy= 400MPa. (As total = 10 ѫ 20mm )
Po= [0.85*30(300*600+200*200)+10*314*400]/1000

Po= 6866kN
Cc*Xc= 0.85*30[300*600*350+200*200*100]/106
= 1708.5kN.m
As*fyx=400[2*314*40+4*314*240+4*314*460]/106
= 361.73 kN.m

7
M= Cc*Xc +As*fyx= 1708.5+361.73=2070.23kN.m
Po*X= M then X= 2070.23/6866 = 0.302= 302mm

Analysis of rectangular eccentrically loaded columns:

:ΞΘϨΘδϧΕϻΎόϔϧϻ΍ϖϓ΍ϮΗϦϣϭ
   ƍ
Z = Z! and Z = Z! then  = 600 and ƍ = 600
   

:(ϥΪϠϟ΍ΰϛήϤϟ΍ϝϮΣϡϭΰόϟ΍άΧΎΑ)ϯϮϘϟ΍ϥί΍ϮΗϦϣϭ
\  = 0  = Y + Y  ^ then  = 0.85 . .  +   .   + . 


8
 -  - -
_X = 0 .  = Y    + Y     + ^(  )
2 2 2 2
-  - -
X = 0.85  . .     +   .   (   ) + . (  )
2 2 2 2
Ex.1; A short column 300*500mm supports load with eccentricity of e= 150 mm find nominal
section axial capacity and moment, use d’ =60mm, f’c=
25MPa and fy= 350MPa?

Assume c=300mm then a= 0.85*300=255mm

Fs=600(d-c)/c= 600(500-60-300)/300= 280MPa

T =280 *1323 /1000 = 345kN

Fs’= 600 (c-d’)/c= 600 (300-60)/300= 480 MPa > 350MPa

Cs’ = 350*1232 /1000 = 431.2kN

Cc = 0.85*25*255*300/1000 =1625.63kN

Pn= Cc+Cs’ –T = 1625.63431.2-345= 1711.83kN

-  - -
_X = 0 .  = Y    + Y      + ^   
2 2 2 2

X = 1625.63(0.25  0.128) + 431.2(0.25  0.06) + 345(0.44  0.25) = 345.8;`. #

E= 345.8/1711.83= 0.2m = 200mm more than 150 mm

Try c= 350mm then a= 298mm

Fs=600(d-c)/c= 600(500-60-350)/350= 154MPa

T =154 *1323 /1000 = 189.73kN

Fs’= 600 (c-d’)/c= 600 (350-60)/350= 350 MPa

Cs’ = 350*1232 /1000 = 431.2kN

Cc = 0.85*25*298*300/1000 =1899.75kN

Pn= Cc+Cs’ –T = 2141.22kN

-  - -
_X = 0 .  = Y    + Y     + ^   
2 2 2 2

9
X = 1899.75(0.25  0.15) + 431.2(0.25  0.06) + 189.73(0.44  0.25) = 307.95;`. #

e= 307.95/2414.22= 0.144m = 144mm less more than 150 mm try another value of c.

ΕϻΎΤϟ΍ϭ (e )Δϳΰϛήϣϼϟ΍ΔϤϴϗϰϠϋϚϟΫΪϤΘόϳϭΕϻΎΣΙϼΛϰϟ΍Ϟθϔϟ΍ϞϜηϰϠϋ΍ΩΎϤΘϋ΍ΓΪϤϋϻ΍ϒϨμΗ
-:ϲϫΔϴϟΎΘϟ΍
1- Balanced Failure ϥί΍ϮΘϤϟ΍Ϟθϔϟ΍:
The failure happened when the stresses in steel reach the value of maximum (fy) at the
same moment the strain in concrete reaches the maximum value of (Z! = 0.003)

0.003 600 ƍ

 = = then ƍ = 600 ? 
0.003+Z 600+ 

Y = 0.85 . .    = a  

 = 0.85  . .  +   .   + . 
-  - -
X = 0.85  . .     +   .   (   ) + . (  )
2 2 2 2

 = X/

2- Tension failure Ϊθϟ΍Ϟθϓ

3- Compression failureρΎϐπϧϻ΍Ϟθϓ

ςϠδϤϟ΍ϡΰόϟ΍ϭϞϤΤϟ΍ϰϠϋ΍ΩΎϤΘϋ΍ΔόϗϮΘϤϟ΍Ϟθϔϟ΍ϝΎϜη΍ϦϴΒϳϭΓΪϤϋϼϟϡϭΰόϟΎΑϯϮϘϟ΍ΔϗϼϋϊοϮϳϩΎϧΩ΍ϞϜθϟ΍ϥ΍

10
 :ΔϴϟΎΘϟ΍ΕϻΩΎόϤϟ΍ΐδΣ

11
Ex. 2: Find the balanced eccentricity for the column shown use f’c= 30MPa and Fy= 400MPa?

AS= As’ = 1847mm2

C b =600d /(600+fy) =600*440/(600+ 400) = 264mm
ab
f’s =600*(264 -60)/264 = 463 > 400MPa
T =Cs’= 400 *1847/1000 = 738.8kN
Cc = 0.85*30*224.4*300 /1000 = 1716.7kN
Pn b = Cc + Cs’-T = Cc = 1716.7kN
R R

Mn b = 1716(0.25-0.112) + 738.8(0.25-0.06) +738.8(0.44-

R R

0.25)= 517.65 kN.m

e b =517.65/1716.7 = 0.302m = 302 mm
R R

12
Ex.3: The column shown in figure below finds the axial strength and moment capacity, use
f’c=30MPa, fy= 400MPa and 4 bars with 28mm diameter if:

1- ex= 250mm
=0.85*264=
2- ey= 250mm
224.4mm

For c = 0.7 - d = 0.36

For c = 0.8 - d = 0.38

Then for c = 0.76 - d = 0.37

13
14
Ex.4: A short tied column with 400x550mm supports Pu=4000kN and Mu= 220kN.m .find area
of steel required if f’c =30MPa and fy= 400MPa and d’= 70mm use two layer one at each side.

4000  1000
d = = 0.93
0.65  30  400  550
For c = 0.7 - 
= 0.023
For c = 0.8 - 
= 0.021
Then for c = 0.75 - 
= 0.022

15
Ex.5: Design the rectangular column if it supports Pu= 2200kN and Mu= 660kN.m, using f’c=
30MPa and fy= 400MPa( assume 
= 0.03     ).

Try h= 600mm and d’= 75mm

6002  X! 
then c = = 0.75 = = 300## - = 0.5
600 ! -

For c = 0.7 - d = 0.46

For c = 0.8 - d = 0.50

Then for c = 0.75 - d = 0.48

 2200  1000
d = = = 0.48 e  = 392## !  = 400##
      - 0.65  30    600

As= 0.03 * 400* 600= 7200mm2

Use 8 ѫ 36mm (4 at eacg side) and
use ѫ 12mm for ties
Spacing ”  PP
”  PP
”PP

Ex. 6: A circular column with diameter of 400mm supports design load of 2800kN and design
moment of 135kN.m. Find area of steel and stirrups required for the section, use f’c= 30MPa,
d’= 60mm and fy= 400MPa.
X! 135  48
= = = 0.048# = 48## - = = 0.12
! 2800  400
&  4002

= = 125664 ##2
4
Try ѫ = 0.75 then

400260
and c= = 0.7
400
from the chart 
= 0.045 -  = 0.045  125664 = 5655##2
Use 8ѫ 30mm for stirrups use ѫ10mm dcore= 400 - 2*40= 320mm

16
 2

ƍ 4002 30
 = 0.45 R  1T = 0.45 f  1g = 0.016
  2  (320)2 400

4  (    ) 4  79(320  10)


= = = 59.8 ##
    2 0.016  3202

! 50## > 25  < 75## . ;

17
18
19
Ex.7: Check the adequency of the column shown below the loads Pu= 1200kN and the moments
are Mux= 90kN.m and Muy= 180kN.m, use f’c= 30MPa and fy= 400MPa.
Try ѫ = 0.65

Moment about y-axis :

X! 180
$ = = = 0.15# = 150##
! 1200

- = 0.3
-

Moment about x-axis :

X!$ 90 
 = = = 0.075# = 75## - = 0.25
! 1200 -
180260 
then c = = 0.6 from charts Kn = 0.67=
180 f’c bh

Pn= Pnx0= 0.67* 30* 300 *500 /1000 = 3015kN

For ex=0 and ey=0 Pno =[ 0.85*30(300*500-4928) +400*4928]/1000= 5670.54 kN

1 1 1 1
= +  then Pn = 1946.5kN
 2790 3015 5670 .54

Pu= 0.65*1946.5 = 1265.2 kN ”3X   5670.54 =2948.7 kN o.k.

Pu ”ѫ Pn o.k.

Ex.8: A rectangular tied column with 400*500mm

reinforced with 6 bars 0f 30mm diameter arranged as
shown below, calculate the loads ( Pu) that is applied
at the corner of the column, use f’c= 30MPa, and fy=
400MPa.

Moment about x-axis: e=200mm

20
e/h= 200/400=0.5
400260 6&30 2
then c = = 0.7 = = 0.0212
400 4005004

from charts Kn = 0.42=
f’c bh

Pn= Pnx0= 0.42* 30* 400 *500 /1000 = 2520 kN

500275
Moment about y-axis: e=250mm then e/h= 250/500=0.5 and c = = 0.7
500

4&30 2 
= = 0.01414 from charts Kn = 0.36=
4005004 f’c bh

Pn = Pnx0= 0.36* 30* 400 *500 /1000 = 2160 kN

Pno =[ 0.85*30(400*500-6*707) +400*6*707]/1000= 6688.63kN
1 1 1 1
= +  then Pn = 1408.2kN
 2520 2160 6688 .63

Pu= 0.65*1408.2 = 915.33kN ” Pu= 0.65* 0.8* 6688.63 =3478.08 kN o.k.

Ex.8: The circular spiral column with diameter of 500mm is reinforced by 8 bars of 25mm the
axial aplied load was Pu =2250kN , what moment would cause the column fail? Use f’c=
30MPa, and fy= 400MPa.
500275
c= = 0.7
500
 2250 1000
Kn = = = 0.507
"f’c Ag 0.7530&500 2 /4

8  &  252 /4
= = .02
&  5002 /4

From chart e/h= 0.25 then e= 0.25*500=125mm

Mu= 2250* 0.125 = 281.25 kN.m

Ex.9 : The circular spiral column with diameter of 500mm is reinforced by 8 bars of 25mm the
axial aplied load was Pu = 1200kN , the moments are Mux= 90kN.m and Muy= 180kN.m, use
f’c= 30MPa, d’= 75mm and fy= 400MPa. Check the adequency of the column shown below.

500275 X! 180

c= = 0.7 $ = = = 0.15# = 150##
500 ! 1200
21
  150
- = = 0.3
- 500

X!$ 90
 = = = 0.075# = 75##
! 1200

 !@ = o1502 + 752 = 167.7##

8&25 2 /4
= = .02
&500 2 /4


from charts Kn = 0.41=
f’c Ag

Pn = 0.41* 30* ʌ *5002 /(4*1000) = 2416.07 kN

Slender columns (long columns):

A column is said to be slender if its cross-sectional dimension are small compared with its
length (Height). Slender columns fail by buckling at the critical buckling ,compression Eular’s
load (Pc):
& 2 pq Pc t 2 EI t2E
 = (;r! ))2 Or buckling compression stress = = kLu 2
A A(kLu )2 ( )
r
Where:
{
K = A = K}~€ JO HPK}QJ JO Q‚L IKJ  LIQJ,
|

Lu= unsupported length of column ( clear span between

floors)
k= effective length factor ( measured between points of zero
moments)
;r!
= @   < 100

From the figure below it seen that the buckling stress (or load)
decreases rapidly with increasing slenderness
ratio for short column the value of buckling loads exceeds
the direct crushing strength
Pn = 0.85 f ƍ c Ac + Ast fy
;r! ;r!
When <  [ failur occurs by simple crushing
@#

regardless of concrete strength]

22
 ;r! ;r!
When >  [failure occurs by buckling]
@#

Effectve Length Measuerment:

Effective length measured between the points of zero moments or inflection points, it depends
on two factors :
a) Rrestrained against rotation .
b) Restrained against horizontal displacements.

Non-sway side sway

First: horizontal restrained (braced against side sway)

The structurs built with (steel bracing, shear walls around stairs or lifts etc), it depends on the
end supports and its rotation. Fig. a) shows hinge-hinge ends free end rotaion k=1 ,Fig. b) fixed-
fixed ends (full restrained case) k=0.5 and Fig. c) partailly restrained in structrue which depends
of stiffness of columns ratio to stiffness of beams (degree of restrained)
EI
’ 
Lc columns
‘= EI attached - 0.5 < ; < 1
’ 
Lc beams

Second : free restrained (side sway exist)

There is horizontal movement in the ends of column assume that one end move and the other
restrained. Fig. d) shows fixed free end then k = 2 , Fig. e) no rotation for both ends k=1 and
Fig. f) partially restrained in structures - 1 < ; < “

Note: From above Pc value for braced column is larger than Pc for unbraced column and
depending on degree of restrained k value could be calculated from the chart below.

23
Slenderness effects:
ACI-Code 318 allow to neglect slenderness effect and design as short column if :
a) Braced columns (non- sway)
”•– ˜N
neglect slenderness if ? VE  NM ? ED
— ˜M

24
 where : M 1 (smaller) and M 2 (larger) end moments
˜N ˜N
™ D. F the sign of is (+) for single curvature and (-) for double curvature.
˜M ˜M
— = D. Vš  ›!   

!@ @!#
( *-  -  -  #@ #  
)
— = D. MFœ   !@ @!# ( D is the diamerter)

b) Un-braced columns (sway)

”•–
neglect slenderness if ? MM
—

Neglect side sway:

ACI-Code 318 allow to neglect side sway effect if there are walls or any bracing as well as to:
1- Increase in moments due to sway case less than 5% from moment in linear calculation.
2- Neglect side sway if the stability index (Q) less or equal (0.05):

’Pu žo
= ? 0.05
Vu Lc
Where:
’Pu = sum of axial load in the floor (kN).
Vu= sum of shear force in floor (kN).
žo = horizontal displacement ( difference between displacement at the ends due to horizontal
loads) (m)
Lc = length of column center to center between floors (m).

ACI-Code Slenderness effects:

ACI code for magnify the moments for long (slender) columns depends on increasing the
moments by multyplying the moment by factor larger than one. This increase came from axial
load multyplied by displacement due to rotation . The amount of increase depends on
direction and magnitude of main moments at the ends, if the structure is braced or
unbraced to side sway and the slenderness ratio. For single curvature shows increase in
moment at mid-span if the end moments are equal. For double curvature case the moments have
same sign, the moments become larger near the ends.

25
 Increase in moment

due to single curvature

Seco
Secondary
ondary
mome
m nt
moment

Main
moments

Increase in moment
due to double
curvature

Moment magnification for Braced frames

Slender columns shall be designed using the factored axial load obtained from conventional
frame analysis and magnified moment defined for non-sway frames
X =   X2
Where:
X2 = value of larger factored end moments. (M 2 min = Pu (15+0.03h) in N.mm)
  = moment magnification factor
Y#
  = ! ™1
1
0.75
Pu= design axial load from analysis (kN)
& 2 pq
Pc = buckling load for column  = (;r! ))2
Cm = factor curvature effect without transverse loads between supports

26
 X1
Y# = 0.6 + 0.4 ™ 0.4
X2
¡¢£¤¤¥— ˜N
(sign of is (+) for single curvature and (-) for double curvature) (for other cases Cm
¤£—¦¥— ˜M
=1.0) . To calculate Pc it is required:
EI
’ 
Lc columns
1- Calculate k for the column ‘ = EI (for hing support=’WDNH ‘ = 10 and for
’ 
Lc beams
fixed support =0 take ‘ = 1)
p = 4700o   and Ig moment of interia of section without reinforcement.
Beams I b = 0.35I g , Lc for beams are taken span cneter to center.
Columns I c = 0.7I g , Lc for columns are taken span cneter to center.
Walls no cracks I w = 0.7I g ,
Walls with cracks I w = 0.35I g ,
Slabs without beams I s = 0.25I g
Area for beams, columns and slabs is taken Ag without reinforcements, Ig moment of interia
for hall section. For (T) or (I) beams use Ig = 2I web .
0.4 p q

2- Find EI for columns in Pc equation pq = or more exact and if steel is known

1+a
and exist
o. 2Ec Ig + Es. Ise
EI =
1 + §d
Es = 200000MPa for steel and Ise moment of interia of reinforcement about centroidal axis of
cross section.
§d = ratio of maximum factored sustained axial dead load to maximum total factored axial load
( effect of creep)
’PuD.L. sus
§d =
PuDL + PuLL

Ex.1) A rectangular tied braced column is reinforced with 6-28mm bars as shown in figure
below check the adeqancy of the column for the followning gravity loads . Service dead load=
900kN, service live loads = 800kN both are at eccentricity of 40mm. Service dead load
moments=34kN.m, service live load moments =25kN.m, F’c =27.5MPa, fy = 400MPa, Lu
=5m ,‘ = ‘# = 1.0 ? Sol:

Pu = 1.2 (900)+1.6 ( 800) = 1080+1280 =2360 kN

Mu = 1.2* 34+ 1.6* 25 + 2360*0.04 = 175.2 kN.m

Check slenderness ‘ = ‘ = 1.0    @!#

27
k= 0.77 from chart , r= 0.3* h= 0.3* 500=150mm
kLu 0.775000
= = 25.7 < 100
r 150
X1
34  12 = 34  12(+1) = 22
X2

kLu
= 25.7 > 22 the column is slender
r

single curvature opposite direction moments

X1
Y# = 0.6 + 0.4 ™ 0.4
X2

Cm =0.6 + 0.4 (1) =1

’Pu D .L . sus 1.4900 1260
§d = = = = 0.534
Pu DL +Pu LL 2360 2360

5003
0.4 p q
0.4  350  12   4700927.5
pq = =
1 + a (1 + 0.534)  109
= 23431.26 ;`. #2
4
&   &  284 &  282
q = 6 R +  2 T = 6 R + (186)2 T = 6(0 + 21302561) = 127815370.4 ##4
64 64 4

500 3
o.2Ec Ig +Es .Ise 0.2350 4700 927.5+200000 127815370 .4
12
EI = = (1+0.534)10 9
= 28379.95 kN.m2
1+§d

& 2 pq ª 2 28379 .95

 = (;r! ))2 = (0.775)2
= 18896.87 ;`

«¬ N
  = ­€ = MVWD = 1.199 ™ 1
N N
D.UF­I D.UF18896 .87

Mmin = Pu*e min = 2360*(15+0.03*500)/1000= 70.8 kN.m

Mc=    X2 = 1.199  175.2 = 210.06 ;`. #
Design values:
Pu=2360kN , Mc =213.64 kN.m
210.06  1000
= = 89##
2360
 89
= = 0.178 > 0.1 . ;
- 500

28
 28
5002(40+10+ )
2
c= = 0.75
500
! 2360  1000
; = = = 0.754
"  - 0.65  27.5  350  500
For c = 0.7  = 0.018
c = 0.8  =0.017
c = 0.75  =0.0175
As = 0.0175*350*500 = 3062.5 mm2
As provided 6-28 As= 3694mm2 more than required. The column is adequate.Ties -10mm
spaecing ? 16*28= 448mm
?48*10= 480mm
?350mm then adequate.

Ex.2: The column with (300*375mm) and single curvature about y-axis shown
below in a building braced against side sway,and ultimate loads shwon in figure
(the service dead load of 133kN). Find steel area required and check the slenderness
about bending axis (y-axis) use Lu=4.8m and k=0.85 f’c = 30 MPa fy = 400MPa,

Try as short column:

3752(60)
c= = 0.68 use c = 0.7 then
375
! 490  1000
; = = = 0.223
"  - 0.65  30  300  375
X! 117  1000000
® = = = 0.142
"  -2 0.65  30  300  3752
Then from charts find  = 0.013 which is less than maximum percent
29
 Check slenderness:
kLu 0.854800
= = 35.4 < 100
r 0.3375
X1 115
34  12 = 34  12   = 22.5 < 35.4 then column is slender
X2 117

single curvature opposite direction moments

X1 112
Y# = 0.6 + 0.4 = 0.6 + 0.4 0.893 ™ 0.4
X2 117
’PuD.L. sus 1.4  133
§d = = = 0.38
PuDL + PuLL 490
3753
0.4 p q
0.4  300  12 
 4700930
pq = = = 9837.2 ;`. #2
1 + a (1 + 0.38)  109

& 2 pq ª 2 9837.2
 = (;r! ))2 = (0.834.8)2
= 5832 ;`

Y# 0.983
  = = = 1.107 ™ 1
! 490
1 1
0.75 0.75  5832
Mmin = Pu*e min = 490*(15+0.03*370)/1000= 12.86 kN.m
Mc=    X2 = 1.107  117 = 128.7 ;`. #

Design values:
Pu = 490kN , Mc = 128.7 kN.m
X! 128.7  1000000
® = = = 0.16
"  -2 0.65  30  300  3752

For c = 0.7  = 0.015

As = 0.015* 300* 375= 1688 mm2
As provided 6-20 As= 1886mm2 more than required.
Ties -10mm spacing
? 16*20= 320mm
?48*10= 480mm
?300mm then adequate. Use ѫ 10mm @300mm c/c

30
Ex.3: The figure below indicates the plan and side veiw of reinforced concrete
building with 10 stories.The ground floor has clear hight of 6.4m the other with
3.4m, the size of beams 500x600mm, External columns 500x500mm, Interior
columns 600x600mm and f’c=30MPa and fy= 400 MPa, Design the columns A3
and C3 of ground floor the shear wall used for bracing and side sway pevent, and
a = 0.8 . Check slenderness in the north-south N-S direction, the columns are
under double curvature, the axial loads and bending moments are:

31
 For column C3 try d’= 75mm
6002(75)
c= = 0.75
600
! 8950  1000
; = = = 1.28
"  - 0.65  30  6002
X! 16  1000000
® = = = 0.003
"  -2 0.65  30  6003
Then from charts find  = 0.052 Check slenderness use k=1:
kLu 16400
= = 35.6 < 100
r 0.3600
X1 8.4
34  12 = 34  12   = 40.3 > 40 ! 40 *-- > 35.6 - -  @!#
X2 16

For column A3 try d’= 75mm

5002(75)
c= = 0.7
500
! 4870  1000
; = = = 1.0
"  - 0.65  30  5002
X! 158  1000000
® = = = 0.065
"  -2 0.65  30  5003
Then from charts find  = 0.028
kLu 16400
Check slenderness use k=1: = = 42.7 < 100
r 0.3500
X1 81
34  12 = 34  12   = 40.14 > 40 ! 40 *-- < 42.7 - @ @!#
X2 158

32
 EI
’ 
Lc columns
Check k exactly : ‘ = EI
’ 
Lc beams

0.7  5004
q 12
# = = 0.548  106 ##4
r 6650
0.7  5004
q 12
 = = 0.935  106 ##4
r 3900
0.35  600  5003
q 12
= = 0.26  106 ##4
r 8400
EI
’ 
Lc columns 0.54810 6 +0.93510 6
‘ = EI = = 5.7
’  0.2610 6
Lc beams

‘# = 1.0 $  # #
kLu 0.846400
k = 0.84 and : = = 35.9 <40
r 0.3500
X1 81
34  12 = 34  12   = 27.85 < 35.9  
@ ! ¯!  - @ @!#
X2 158

X1 81
Y# = 0.6 + 0.4 = 0.6 + 0.4 = 0.8 ™ 0.4 DQGȕG WKHQ
X2 138

5004
0.4 
0.4 p q
 12   4700930
pq = = = 29795 ;`. #2
1 + a (1 + 0.8)  109

& 2 pq ª 2 29795
 = (;r! ))2 = (0.846.4)2
= 10175 ;`

Y# 0.8
  = = = 2.21 ™ 1
! 4870
1 1
0.75 0.75  10175
M2 min= 4870(15+0.03*500)/1000=146.1 kN.m < M2=158kN.m
Mc=    X2 = 2.21  158 = 349.2 ;`. #
Design values:
5002(75)
c= = 0.7
500

33
 ! 4870  1000
; = = = 1.0
"  - 0.65  30  5002

X! 349.2  1000000
® = = = 0.14
"  -2 0.65  30  5003
For c = 0.7  = 0.042
As = 0.042* 500*500= 10500 mm2
As provided 12 ѫ 36 As = 12215 mm2 more than required.
Ties -12mm spacing
? 16*36= 576mm
? 48*12= 576mm
? 500mm then adequate. Use ѫ 12mm @500mm c/c

34
Moment magnification for un-brasecd frames
For compression members (columns) not braced against sidesway effect of slenderness may be
neglected when:
kLu
? 22
r
The moments M1 and M2 at the ends of an individual compression member shall be taken as:
X1 = X1 +   X1

X2 = X2 +   X2
Where:
X1 = factored end moments on a compression members at end at which X1 acts, due to loads
that cause no appreciable side sway.
X2 = factored end moments on a compression members at end at which X2 acts, due to loads
that cause no appreciable side sway.
X1 = factored end moments on a compression members at end at which X1 acts, due to loads
that cause appreciable side sway.
X2 = factored end moments on a compression members at end at which X1 acts, due to loads
that cause appreciable side sway.
  = moment magnification factor for frames not braced against sidesway to reflect drift
results from lateral (wind , earth pressure and gravity loads… etc.).

There are three methods to calculate moment magnification factor   X:

First- from nonlinear analysis of moments depending on moment of inertia :

Beams I b = 0.35I g , Lc for beams are taken span center to center.

Columns I c = 0.7I g , Lc for columns are taken span center to center.
Walls no cracks I w = 0.7I g ,
Walls with cracks I w = 0.35I g ,
Slabs without beams I s = 0.25I g
For (T) or (I) beams use Ig = 2I web .
Second: calculate from
1
  = ™ 1 and should be   ? 1.5 if not use the first or third method.
1
1
Third : the same   = ’ ! ™1
1
0.75’ 

(Note if   ? 2.5 the frame must be stiffened to reduce   and not exceed 2.5 )
’! = sum of axil vertical factored loads for all columns in the floor under consideration.
’ = sum of all buckling loads for all columns in the floor under consideration.

35
 & 2 pq
 = (;r! ))2 *-  1 ? ; < “

0.4 p q

Find EI for columns in Pc equation pq = or more exact and if steel is known and
1+a
o.2Ec Ig +Es .Ise
exist EI =
1+§d

Es = 200000MPa for steel and Ise moment of interia of reinforcement about centroidal axis of
cross section.

§d = ratio of maximum factored sustained lateral dead load to maximum total factored latera
load ( effect of creep)

total sustained lateral (horizontal) loads in the floor

§d =
total lateral (horizontal) shear force in the floor

for wind load and earthquake loads ±² = D because are generally at short duration but
for other cases like earth pressure ±² ³ D £´² ¡–µ¶¥·¸¥² ¹´ ¹´¥ ¡º²¥.

If an individual compression member has:

Lu 35
>
rPu
A
Ag. fc
Then the maximum moments will be in a point between the column ends and not at the ends
and the maximum moment will be larger than the magnified moment by 5% . then replay
moment calculation as it is braced against side sway then magnify M1 and M2 as:

X1 = X1 +   X1
X2 = X2 +   X2
Thus the moments will be magnify two times first as non sway and second sway case with
moments equal that gained from the first magnify as:
Mc = ¼ns M2 = ¼ns ( M2ns + ¼s M2s )

The ultimate loads calulation with lateral should be calculated as:

U = 1.2D + 1.6L
U =1.2D + 1.6L+ 0.8W
U = 0.9D + 1.6W
U = 1.2D + 1.6L + 1.6H

Where; D= dead loads , L = live loads, W = wind loads, H = lateral soil pressure.

36
Ex. 4: The figure below indicates the side veiw of reinforced concrete building not braced
against sidesway dimension of beams are (1200x300mm), interior columns (450x450mm),
exterior columns 400x400mm clear height Lu= 3.9m, f’c= 30MPa , fy=400MPa, and horizontal
displacement due to total shear wind loads (250 kN) at third floor = 20mm. Design the column
C, the service loads and moments from linear analysis for third floor are :

loads Columns A3,F3 Columns B3, E3 Columns D3, C3

Deadload Pd (kN) 500 1000 1000
Live load P L (kN) 400 750 750
Wind load Pw (kN) ½125 ½75 ½25
Wind shear (kN) 25 50 50
Dead load moment 3
M2d (kN.m)
live load moment 150
M2 L (kN.m)
Wind load moment ½115
M2w (kN.m)
Dead load moment -3
M1d (kN.m)
live load moment 135
M1 L (kN.m)
Wind load moment ½95
M1w (kN.m)

37
Vu= 1.6*250 =400kN
¨o= 1.6* 20 = 32mm

Pu = 1.2* 500 + 1.0 *400 = 1000kN for A3, F3 columns

Pu= 1.2* 1000+ 1.0* 750 = 1950kN for B3, C3, D3, E3 columns

_ ! = 2  1000 + 4  1950 = 9800;`

\ ! ¾ 9800  32
= = = 0.19 > 0.05 !    #
¿! r 400  4200
0.7  4504
q 12
 = = 570  103 ##4
r 4200
0.35  1200  3003
q 12
= = 263  103 ##4
r 7200
EI
’ 
Lc columns 2  570
‘ = = = 2.17 = ‘#
EI 2  263
’ 
Lc beams
kLu 1.643900 X1
k = 1.64 and : = = 47.4 > 34  12 X2 = 34  12(1) = 22
r 0.3450

For column C3 case of loading:

Load cases Design load M2ns (due to gravity M2s (due to wind
(kN) loads only) (kN.m) loads) (kN.m)
Wu=1.2D + 1.6L 2400 243.6 0
Wu=1.2D + 0.8W 1200 3.6 92
Wu= 1.2D +1.6W +1.0L 1950 153.6 184
Wu= 0.9D +1.6W 900 2.7 184

38
39
40
Ex.5: Design a typical exterior column (col.D4) of the third multi-stroy building . The stability
index for this story Q= 0.36 f’c=27.5MPa and fy=400MPa. All columns are 500x500mm and
height center to center of floor 5.8m, all the beams are 300*600mm.

41
 Loading Exterior columns Interior columns Notes
Factored axial load 2600 3500 for a assume factored
(kN) dead load is 50% of total
factored loads
Factored non-sway M top= 100 M top= 60
moment Mns Mbottom= 170 Mbottom= 100
(kN.m)
Factored sway M top= 70 M top= 100 Assume all lateral loads
moment Ms (kN.m) Mbottom= 80 Mbottom= 100 are live loads a=0

Sol; (‘ ¯@! $ . @!#)

500 3 600 3
q
500 q
300
( )@!# = 12
= 900  103 ##3 , ( )# = 12
= 1200  103 ##3
r 5800 r 4500

0.7(900 + 900)
Àext. col. = Àtop = Àbottom = =3
0.35(1200)

since Q = 0.36 the story is not braced against sidesway k(ext.) unbraced=1.82

0.7(900 + 900)
Àint. col. = Àtop = Àbottom = = 1.5
0.35(1200 + 1200)
then k(int.) unbraced = 1.45

check slenderness col. D4

;@! 1.82  5200
= = 63 > 22  < 100 @ @.
0.3  500
0.44700 927.5 500500 3
pq = (1+0)10 9
 = 51348;`. #2
12

All lateral loads are live load

t 2 51348
for ext. columns Pcunbraced = = 5658 kN
(1.825.2)2

t 2 51348
for int. columns Pcunbraced = = 8914 kN
(1.455.2)2

1 1
  = = = 1.563 > 1.5  $. @.  . ;.
1 10.36

42
 1
  = ’ ! ™ 1.0    > 2.5 -    # or increase number
1
0.75’ 

of column or use shear wall.

’! = 16  2600 + 16  3500 = 97600 ;`

’ = 16  5658 + 16  891 = 233152 ;`

1
  = 97600 = 2.263 > 1.0 and < 2.5
1
0.75(233152 )

M2=M2ns +   M2s = 170+2.263*80 = 351kN.m

M1= M1ns +   X1 = 100+ 2.263* 70 =258.4kN.m

r! 5200
= = 34.67
0.3500

35 35 r!
= = 56.91 > = 34.67 -    X =    X2
! 2600  103
A A
 
27.5  500  500
r! 35
If >
!
A  

then take M2 new and magnified by   as braced col. and use a for axial load.
3511000
Design Pu =2600 kN M2 = 351kN.m  = = 135##
2600

 135 2600 1000

= = 0.27 K = = 0.58
- 500 0.6527.5500500

5002(40+10+11)
assume using bars with 22mm and ties 10mm; c = = 0.756
500

c = 0.7 -  = 0.015

c = 0.8 -  = 0.012

c = 0.75 -  = 0.0135 As = 0.0135*500*500 = 3375mm2

3375
No. of bars = (22)2
= 8.88 ! 10   Â 22## As= 10*380 = 3800mm2
&
4

For ties use bars with 10mm

43
Spacing ? 16  22 = 352##

? 10  48 = 480##
? 500## then use 10mm @350mm c/c
500(240+522+210)
Clear spacing in side the stirrups= = 72.5## < 150##
4

Use additional to provide for corner bars

Or the other order

44
45