Design Of Reinforced Concrete Structures II Columns Before discussion of sway and non-sway columns we have to know some important concepts. What is the difference between Building frame system and Moment-resisting frame system? Building frame system: A structural system with an essentially complete space frame providing support for gravity loads. Beams and columns are connected and the system is incapable of resisting any lateral loads. Resistance to lateral load is provided by shear walls or braced frames. Moment-resisting frame system: A structural system with an essentially complete space frame providing support for gravity loads. Moment-resisting frames provide resistance to lateral load primarily by flexural action of members.)UBC1629.6)

FRAME

Beam

Column

Flexural

Compression

Shear

Eccentrricity

Slenderness

1

Joint

shear

Design Of Reinforced Concrete Structures II Columns

Types of Columns

COLUMNS

By reinforcing method

By lateral displacement ability

By slenderness

Tied

Sway

Long

Spiral

Non-sway

Short

Composite

2

Design Of Reinforced Concrete Structures II Columns 

Check for sway or non sway for the story If Q 

P  u

Vu l c

o

 0.05 Non-Sway

Where Stability index

Q

P

Total Vertical Load in the story

0

First order relative deflection between the top and the bottom of the story.

u



Vu

Story Shear

Lc

Center to center length

Slenderness Check for each column

A. Find the k factor. B. Check 

Find k Calculate  for each joint in the columns

Ec I c  l c   E I  lb b b Where

lc

Length of column center-to-center of the joints

lb

Length of beam center-to-center of the joints

Ec

Modulus of elasticity of column concrete

Eb

Modulus of elasticity of beam concrete

Ic

Moment of inertia of column cross section about an axis perpendicular to the Plane of buckling being considered. 3

Design Of Reinforced Concrete Structures II Columns Ib

Moment of inertia of beam cross section about an axis perpendicular to the

plane of buckling being considered.

Note :

 

4

Design Of Reinforced Concrete Structures II Columns 

Check k

Effective length factor

lu

Unsupported length of member

r

Radius of gyration associated with axis about which bending occurs For Rectangular r = 0.30 h, and for Circular r = 0.25 h h = column dimension in the direction of bending.

For Non - sway Frames

M1

Smaller factored end moment on column

M2

Larger factored end moment on column Positive when single curvature Negative when double curvature

For Sway Frames

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Design Of Reinforced Concrete Structures II Columns Example (1) For the frame shown in the figure, check whether column AB is adequate for resisting the following service loads applied on member BC : Dead Load including own weight = 6 t/m. Live Load = 2.5 t/m , and a concentrated Earthquake load at joint C = 4 ton “ acting to the left ”. Note : Consider D + L + E combination only. Related Data : Frame members are 40 X 100 cm in cross section, and story drift for this case of loading considered is equal to 5.0 cm.

 For D+L+E Combination: U = 1.2 DL +LL + E … from code. 1) Check If Sway or Non-Sway : Q=



Wu=1.2 6+2.5=9.7 t/m Pu = Wu L = 9.7

15 =145.5 ton. 6

Design Of Reinforced Concrete Structures II Columns o

= 5.0 cm = 0.05 m.

u

= 1 4=4 ton.

Lc = 6 m. Q=

=0.303 > 0.05→Sway

2) Check If Long or Short : 

…. Short, or >22 and ≤ 100 … long and the Moment magnification method can be used, or > 100 , so the P- analysis should be used. 

Calc. of K :.

=

. A =  … pin supportB =

5.

From alignment chart for Sway – Column. K = 3.375.

. =

= 61.875 > 22, and ≤ 100 → Long column and Moment

Magnification Method can be used.

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Design Of Reinforced Concrete Structures II Columns 3) Calculations and Design : “ Column is Long And Sway ” o Mmax = Mu = Mns + δs Ms . Mns = 0.0 “Pin supported column”. Ms = 1 4 6=24 t.m o δs = = 145.5 ton “calculated before”. o

=

“Euler Critical Buckling Load”

=

=

EI =

3.37

1011 … “note

= 0.0 because

sustained shear is 0.0”. 10-3 = 964.31 ton

=

= =2

= 2

δs =

964.31 = 1928.62 ton. “Due to symmetry”.

= 1.13 > 1.0 ….. OK. …. “it is better to be larger but close to 1 ”

 Mmax = Mu = 0.0 + 1.13 For

24 = 26.7 t.m .

on the column, take M about point B >>>> Pu on the column AB = 74.35 ton.

After calculating

& Mu , we can use interaction diagrams for design.

For Design : “ VI note : units of loads to be used for interaction diag. is Kg , cm ”

** From Interaction Diagram with: = 0.9 ,

.

req is less than min= 0.01 ….. so use min = 0.01. 8

Design Of Reinforced Concrete Structures II Columns Ast. req. = 

Ag = 0.01 100 40 = 40 cm2.

Ast. ava. = 1025 = 49.1 cm2 > Ast.)req = 40 cm2 …. OK For Stirrups Check : bar = 25 mm , stirrup = 10 mm. Smax = smaller of :

16 bar = 16 2.5 = 40 cm. 48 stirrup = 48 1 = 48 cm. Smaller dimension of cross section = 40 cm.

Sava. = 20 cm … which is smaller than the smallest spacing required above …… OK. »

Column AB is adequate for resisting the loads applied on member BC

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