Chapter8 Column Design 2y3o6n

CE 134- Design of Reinforced Concrete Structures

Chapter 8 Design of Columns

Instructor: Richelle G. Zafra, PhD

Chapter 8 Outline 8.1

Introduction

8.2

Types of Columns

8.3

ACI/NS Code Requirements

8.4

Axially Loaded Short Columns

8.5

Short Columns under Combined Axial Load and Moment

2

Columns

Pier

Column

3

Axially Loaded Columns • Columns are defined as that carry loads chiefly in compression. • Columns with a ratio of height-to-least lateral dimension exceeding 3 are used primarily to axial compressive load. • Columns subjected to pure axial load rarely, if ever, exists.

4

Eccentically Loaded Columns Columns are subjected to some bending moment which may be caused by unbalanced floor loads on both exterior and interior columns.

Giesel Library Building, UC San Diego

5

Eccentically Loaded Columns Eccentric loads such as crane loads in industrial buildings also cause moment. Crane RC corbel

Industrial Building 6

Eccentically Loaded Columns Lateral loading due to earthquake

Olive View Hospital, 1971 San Fernando EQ

http://www.smate.wwu.edu/teched /geology/eq-CA-SanFernd.html

Collapsed parking structure, California State University, 1994 Northridge EQ

pubs.usgs.gov

7

Eccentically Loaded Columns Lateral loading due to wind

http://www.wbdg.org/resources/env_wind.php

Office Building Exterior and Curtain Wall, 2000 Forth Worth, USA Tornado

8

Types of RC Columns

ties

spirals

structural steel shape (c) composite column steel pipe

pitch (a) Tied column

(b) Spiral column

(d) Lally column 9

ACI/NS Code Requirements for Tied Columns 1. Min. dimension = 200 mm 2. Min. gross area, Ag = 200 mm x 300 mm col. = 60, 000 mm2 3. Min. main bars = 4 pcs – 16 mm ø 4. Min. clear bar cover = 40 mm (same as beams) 5. Lateral tie diameter = same method of determining stirrup diameter 6. Gross steel ratio: ρg = 1% min. to 8% max. (use only up to 4% when deg)

10

ACI/NS Code Requirements for Tied Columns (Cont.) 7. Main bar spacing, sm s > 1.5 main bar ø s > 40 mm s > 1.5 max size of coarse aggregate 8. Lateral tie spacing, st s = 16 main bar ø s = 48 lateral tie ø s = least col. dimension

ties

st sm

main bar

Smallest value 11

ACI/NS Code Requirements for Spiral Columns 1. Min. column diameter = 250 mm 2. Min. main bars = 6 pcs – 16 mm ø 3. Min. clear bar cover = same as tied column 4. Spiral bar diameter = same as ties 5. Gross steel ratio: ρg = same as tied column 6. Main bar spacing = same as tied column 12

ACI/NS Code Requirements for Spiral Columns (Cont.) 7. Min. spiral steel percentage  Ag  fc' Minimum ρs = 0.45 − 1  Ac  fy

(1)

where: Ag = gross cross-sectional area = π D 2 ; D = column diameter 4

Ac = core-concrete cross-sectional area

= π Dc2 ; Dc = concrete core diameter/ 4 outside diameter of spiral

Gross area Core concrete 13

ACI/NS Code Requirements for Spiral Columns (Cont.) 8. Actual/ required spiral steel ratio 40mm CC

~Dc Ac

D

Mean Dia, Ds

Core Dia Dc = D - 80

40mm CC

Core concrete Cover concrete

conc. core spiral

spiral pitch "s"

2

as = (pi/4)ds

14

Rebar cage

ACI/NS Code Requirements for Spiral Columns (Cont.) vol . of spiral in one (1) turn vol . of concrete core bounded asπDc 4as req ' d ρ s = = ; π 2 Dc s Dc s 4 where : as = area of one spiral bar req ' d ρ s =

s = spiral pitch

Note: To find the spiral pitch, equate req’d ρs to min ρs. 15

ACI/NS Code Requirements for Spiral Columns (Cont.) 9. Spiral pitch limits s > 40 mm (minimum) s < 75 mm (maximum) s < Dc / 6

16

Axially Loaded Short Columns Capacity Pu = φ Pn (max )

(2)

= φ (0.80Po ) ; φ = 0.65 for tied column = φ (0.85Po ) ; φ = 0.75 for spiral column Po = 0.85fc' (Ag − Ast ) + fy Ast

where:

[

(for analysis)

(3)

Ast = ρg Ag

]

Po = Ag 0.85fc' + ρg (fy − 0.85fc′ )

(for design) (4) 17

Axially Loaded Short Columns where:

φ = strength reduction factor Ag = gross area of section

Ast = total area of longitudinal reinforcement fc′ = specified compressive strength of concrete fy = specified yield strength of reinforcement 18

Example 1 A 500 mm x 500 mm tied column is reinforced with 8 - 28 mm ø bars. If f’c = 21 MPa and fy = 414 MPa, find the ultimate axial capacity of the column.

19

Example 2 A 450mm round spiral column is reinforced with 6-25mm Ф bars having fy=276MPa. Determine the ultimate strength if f’c=34MPa.

20

Example 3 Design a short axially loaded square tied column for a service dead load of 1080 KN and a service live load of 990 KN. The uned length is 2.60 m. Use f’c = 34.5 MPa, fy = 414 MPa, ρg = 2 %, 25 mm ø main bars, 10 mm ø ties, and 40 mm concrete cover.

21

Example 4 Design a round spiral column to an axial dead load of 800kN and an axial live load of 1350kN. Assume that 2% longitudinal steel is desired. Diameter of main bars is 25mm and the diameter of spiral ties is 10mm. Use f’c=27.6 MPa and fy=414 MPa.

22

Columns under Combined Axial Load and Moment

External Forces

Strain Diagram

Internal Forces on Columns

Equilibrium of external and internal axial forces requires that

Pn = 0.85fc' ab + As' fs' − As fs

(5)

23

Columns under Combined Axial Load and Moment (Cont.)

External Forces

Strain Diagram

Internal Forces on Columns

Moment about section centerline of internal forces must be equal and opposite the moment of external force Pn

Mn = Pne =

h ' 0.85fc ab

a h   ' 'h −  + As fs  − d ′  + As fs  d −   242    2 2 2 (6)

Columns under Combined Axial Load and Moment (Cont.) where: Pn e b h As′

= nominal axial capacity = eccentricity = width of section = height of section = compression steel area As = tension steel area d ′ = location of A’s from the compression face d = location of As from the compression face fs′ = stress in the compression steel fs = stress in the tension steel 25 a = β1c

Columns under Combined Axial Load and Moment (Cont.) If we know

Neutral axis (c, a) Strain condition (εs, ε’s) Stress condition (fs, f’s)

Can determine

Column Strength (Mn, Pn)

26

Columns under Combined Axial Load and Moment (Cont.) Strain in Tension Steel d −c ε s = ε cu (6) fs = ε s Es ≤ fy (7) c where: ε cu = 0.003 Strain Diagram

Strain in Compression Steel c − d′ ε s′ = ε cu (8) fs′ = ε s′ Es ≤ fy (9) c Concrete Stress Block a = β1 c ; c ≤ h (10)

Internal Forces on Columns

27

Jiravacharadet (2013)

Columns under Combined Axial Load and Moment (Cont.) External Force on Column Stress Diagram

C T

28

Interaction Diagram for Combined Bending and Axial Load Pn

Mn e= Pn

• For any eccentricity e, there is a unique pair of Pn and Mn .

(Mn, Pn)

• Plotting a series of (Mn, Pn) pairs corresponding to different e will result in an interaction diagram.

emin

Po e=0

Direct axial failure

Compression e range failure

eb

Tension failure range

e=∞

Mo

Mn

Column Interaction Diagram 29

Interaction Diagram for Combined Bending and Axial Load Pn

• Any combination of loading that falls inside the curve is satisfactory

emin

Po e=0

Direct axial failure Compression failure range

eb

Tension failure range

e=∞

Mo

• However, any combination falling outside the curve represents failure.

Mn

Column Interaction Diagram 30

Balanced Failure, eb • Concrete reaches the strain limit εcu at the same time that the tensile steel reaches the yield strain εy • Dividing point between compression failure (small eccentricities) and tension failure (large eccentricities) Strain Diagram

c = cb = d

ε cu

ε cu + ε y

(11)

where: ε cu = 0.003 a = ab = β1 cb

(12) 31

Balanced Failure, eb Strain Diagram

Mb = eb Pb

(15)

fs′ = ε s′ Es ≤ fy

where:

Pb = 0.85 fc' ab b + As' fs' − As fy Mb =

h ' 0.85 fc ab b 

cb − d ′ ε s′ = ε cu cb (13)

ab  h   ' 'h −  + As fs  − d ′  + As fy  d −  (14) 2  32   2 2  2

Short Columns Under Combined Axial Load and Moment Pn

emin = 0.10h (Tied)

Po

emin = 0.05D (Spiral)

B. Actual Eccentricity

Mu Mn e= or Pu Pn

e=0

A. Minimum Eccentricity

(Mn, Pn)

e e=∞

Mo

Mn

Column Interaction Diagram 33

Behavior at Failure: Columns under Combined P and M Pn

Region I: Negligible Moment

e=0

Po

e < emin Direct axial failure (Region I)

eb e=∞

Mo

Capacity: See axially loaded column Mn

Column Interaction Diagram

34

Behavior at Failure: Columns under Combined P and M Pn

Region II: Proportioned Axial Load and Moment

Po e=0

emin < e < eb Compression failure range (Region II)

eb e=∞

Mo

Capacity: Pn > Pb Mn

Column Interaction Diagram

35

Approximate Capacity Formulas (Region II) 1. From straight line relation on interaction curve

Mn Po = Po − (Po − Pb ) Pn = Po  e M b 1+  − 1 Pb  eb

(16)

36

Approximate Capacity Formulas (Region II) 2. Whitney’s Formula b h fc ' As ' fy Pn = + 3he e +1 + 0 .5 2 d − d' d

Ag fc ' 9 .6 D e

(17)

As fy Pn = + (18) 3e + 1.18 +1 2 Ds (0.8D + 0.67D s )

Tied Column

Spiral Column

37

Behavior at Failure: Columns under Combined P and M Pn

Region III: Proportioned Axial Load and Moment

Po e=0

e > eb Capacity: Pn < Pb

eb Tension failure range (Region III)

e=∞

Mo

Mn

Column Interaction Diagram

38

Approximate Capacity Formulas (Region III) Approximate Whitney’s Capacity Formulas (Tied Columns) 2  e'  e'  d '  e'   Pn = 0.85fc ' bd 1 − ρ − + 1 −  + 2 ρ (m − 1)(1 − ) +  d d  d   d   (19) As ; As = bars in tension where: ρ = bd fy m= ' 0.85fc h ' e = e + − d' 39 2

Approximate Capacity Formulas (Region III) Approximate Whitney’s Capacity Formulas (Spiral Columns) 2 ρ mDs   0.85e g  − 0.38  + Pn = 0.85fc ' D 2   2.5D    D

 e 0 . 85   − − 0.38   D  (20)

40

Bending is about y-axis of the column.

3-28mmφ

• ultimate axial capacity at balanced condition; • load eccentricity at balanced condition; • ultimate axial capacity if e = 200 mm.

As 300 mm

For the column shown with f’c = 28 MPa and fy = 414 MPa, determine:

y

e As’ 3-28mmφ

Example 3

62.5 187.5 187.5 500 mm Top View

Pn x 62.5

e Pn

Elevation View

41

Biaxial Bending • Axial compression is accompanied by simultaneous bending about both principal axes of the section. • Such is the case of corner columns of tier buildings • Beams and girders frame into the columns in both directions and transfer their end moments into the columns in two perpendicular planes

nees-anchor.ceas.uwm.edu

nisee.berkeley.edu

42

Strength Interaction Diagram for Biaxial Bending Uniaxial bending about Y axis

Uniaxial bending about X axis

Biaxial bending

Column Interaction Diagram

43

Reciprocal Load Method • A simple, approximate developed by Bresler. • Acceptably accurate provided Pn ≥ 0.10P0

for

design design

method purposes

44

Reciprocal Load Method Bresler’s reciprocal load equation is given by 1 1 1 1 = + − Pn Pnx 0 Pny 0 P0

(21)

where: Pn = approximate value of nominal load in biaxial bending with eccentricities ex and ey Pny 0 = nominal load when only eccentricity ex is present (ey = 0) Pnx 0 = nominal load when only eccentricity ey is present (ex = 0) 45 Po = load for concentrically loaded column

Example 4 y 300 mm

Using Bresler’s equation, determine the strength of the column shown given the biaxial capacities Pnx = 1880 KN, Pny = 1000 KN, f’c = 21 MPa, fy = 414 MPa.

8-20mmφ

x

200 mm 200 mm

46

Use of Column Interaction Diagram for Design • The preceding lectures have clearly shown that the analysis and design of columns with eccentricities, using static equations, is very tedious and complicated. • Consequently, designers resort almost completely to tables, computers, or diagrams (e. g. column interaction diagram) for their column calculations. • Interaction diagrams are useful for studying strength of columns with varying proportions of loads and moment. 47

ρg

engmechanics.blogspot.com

Use of Column Interaction Diagram for Design

e/h

φ Pn Ag (ksi)

φ Pn e Ag

×

h

=

φMn Ag h

(ksi)

48

How to Use Column Interaction Diagram for Design In order to correctly use the column interaction diagram, it is necessary to compute the value of γ γ =

h γh

γh h

where: γh = center to center distance of bars on each side of the column h = column depth

b

49

Use of Column Interaction Diagram Note: • Both γh and h should be taken in the direction of bending. • In using the column interaction diagram, be sure that the column picture at the upper right of the diagram being used agrees with the column being considered. • For example, are there bars on two faces of the column or on all four faces? 50

Example 5

75

4-25mmφ

4-25mmφ

e 400 mm

Calculate the nominal axial capacity (in kN) of the column shown if the eccentricity of the load is 200 mm. Use f’c = 21 MPa and fy = 414 MPa. Use the interaction diagram.

Pn

450 75 600 mm

51

Column Interaction Diagram (Rectangular Section)

1.2 ksi 52

500 mm

8-22mmφ

375

e

62.5

Calculate the nominal axial load (in kN) that can be applied in the column at an eccentricity of 200 mm. Use f’c = 28 MPa Pn and fy = 414 MPa. Use the interaction diagram.

62.5

Example 6

53

Column Interaction Diagram (Circular Section)

1.0 ksi 54

Example 7 e 500 mm

Calculate the nominal axial load (in kN) that can be applied in the column at an eccentricity of 400 mm. Use f’c = 21 MPa and fy = 414 MPa. Use the interaction diagram.

16-28mmφ

70

Pn

70 360 500 mm 55

Column Interaction Diagram (Rectangular Section)

0.65 ksi 56

Column Interaction Diagram (Rectangular Section)

0.82 ksi 57

References Jiravacharadet, M. Lecture Notes in Reinforced Concrete Columns. School of Civil Engineering, Suranaree University of Technology, Thailand. Accessed June 10, 2013. Nilson, A. H. 1997. Design of Concrete Structures. 12th Edition. McGraw-Hill: Singapore. McCormac, J. C. and Nelson, J.K. 2005. Design of Reinforced Concrete. 6th Edition. John Wiley & Sons, Inc: New Jersey. 58