03- Columns Under Eccentric Loading

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M.Shedid 1

1The British University inEgypt

Marwan ShedidPh.D., P.Eng.

Reinforced Concrete Design (2)14CIVL11I

Lecture III

Columns

Under eccentric loading

Reinforced Concrete (2)

2The British University inEgypt

Marwan ShedidPh.D., P.Eng.

Reinforced Concrete Design (2)14CIVL11I

- Analysis under eccentric compression- Construction of Interaction Diagrams- Design using Interaction Diagrams- Approximate method for large eccentricities- Design under eccentric tension

Lecture contents

M.Shedid 2

3The British University inEgypt

Marwan ShedidPh.D., P.Eng.

Reinforced Concrete Design (2)14CIVL11I

Analysis under eccentric Axial loading

Usually moment is represented by axial load times eccentricity,i.e. M = P x e

4The British University inEgypt

Marwan ShedidPh.D., P.Eng.

Reinforced Concrete Design (2)14CIVL11I

Resultant forces acting at centroid

Moment about plastic centroid

Analysis under eccentric Axial loading (cont.)

M.Shedid 3

5The British University inEgypt

Marwan ShedidPh.D., P.Eng.

Reinforced Concrete Design (2)14CIVL11I

Plastic centroid

Analysis under eccentric Axial loading (cont.)

6The British University inEgypt

Marwan ShedidPh.D., P.Eng.

Reinforced Concrete Design (2)14CIVL11I

An Interaction Diagram is a Graph Representing the RelationshipBetween Axial Load and Moment Capacities of a RC cross section( Failure Envelope) or

( Design Envelope)

Concrete crushesbefore tensionsteel yields

Tension steelyields beforeconcrete crushes

Analysis under eccentric Axial loading (cont.)

M.Shedid 4

7The British University inEgypt

Marwan ShedidPh.D., P.Eng.

Reinforced Concrete Design (2)14CIVL11I

How to construct ID• Select a c value

(essentially you areselecting a strain profile).

• Calculate the stress in thesteel components

• Calculate the forces in thesteel and concrete, Cc, Cs andTs.

• Determine Pr value.• Compute the Mr about the

centroid.• Compute eccentricity,

e = Mr / Pr.

Analysis under eccentric Axial loading (cont.)

8The British University inEgypt

Marwan ShedidPh.D., P.Eng.

Reinforced Concrete Design (2)14CIVL11I

Interaction DiagramAnalysis under eccentric Axial loading (cont.)

M.Shedid 5

9The British University inEgypt

Marwan ShedidPh.D., P.Eng.

Reinforced Concrete Design (2)14CIVL11I

Analysis under eccentric Axial loading (cont.)

First, start with concentric load, that is e = 0:

For concentric load, maximumstrain for concrete is 0.003

Now increase eccentricity until

tensile face strain = 0, and

compressive face strain = 0.003

c = h

0.003

0.003

a

10The British University inEgypt

Marwan ShedidPh.D., P.Eng.

Reinforced Concrete Design (2)14CIVL11I

Analysis under eccentric Axial loading (cont.)

Next, increase eccentricity until tensile face rebar reaches yield:

0.003

Next, double, triple, quadruple,… steel strainand determine M and N

0.003

M.Shedid 6

11The British University inEgypt

Marwan ShedidPh.D., P.Eng.

Reinforced Concrete Design (2)14CIVL11I

Analysis under eccentric Axial loading (cont.)

Example

Draw Interaction Diagram for the column shown

fcu = 30 MPa

fy = 400 MPa

12The British University inEgypt

Marwan ShedidPh.D., P.Eng.

Reinforced Concrete Design (2)14CIVL11I

Example (cont.)

Point 1 (Pure compression)

Point 2 (Compression and minimum Moment)

Point 5 (Pure Tension)

M.Shedid 7

13The British University inEgypt

Marwan ShedidPh.D., P.Eng.

Reinforced Concrete Design (2)14CIVL11I

Example (cont.)

Point 3

Balanced condition

14The British University inEgypt

Marwan ShedidPh.D., P.Eng.

Reinforced Concrete Design (2)14CIVL11I

Example (cont.)

Point 4

Tension failure zone

M.Shedid 8

15The British University inEgypt

Marwan ShedidPh.D., P.Eng.

Reinforced Concrete Design (2)14CIVL11I

Example (cont.)

Interaction diagram

M= 700 kN.m and P=1000 kN

M=200kN.m and P=2000kN

P= 2000 kN and e = 100 mm

16The British University inEgypt

Marwan ShedidPh.D., P.Eng.

Reinforced Concrete Design (2)14CIVL11I

Design using I-D

Each design chart (dimensionless interaction diagrams) includesseveral curves. Each curve represents a different ρt value fordifferent fy and f’c

In design charts:

x- axis represents Mr/(fcubt2),y-axis represents Pr /(fcubt)

To select a proper design chart, check the;- Steel and concrete grade- g value- Arrangement of steel

In the final design stage objective is to select the cross section dimensions and tofind the steel area and arrangement. Therefore engineer makes assumption aboutarrangement of steel and enters the charts.

M.Shedid 9

17The British University inEgypt

Marwan ShedidPh.D., P.Eng.

Reinforced Concrete Design (2)14CIVL11I

Design using I-D (cont.)

18The British University inEgypt

Marwan ShedidPh.D., P.Eng.

Reinforced Concrete Design (2)14CIVL11I

Design using I-D (cont.)

Design a rectangular column 250x650 to carry:

PUL= 1700 kN, MUL= 300 kN.m

fcu= 30 MPa, fy = 360 MPa

M.Shedid 10

19The British University inEgypt

Marwan ShedidPh.D., P.Eng.

Reinforced Concrete Design (2)14CIVL11I

Design using I-D (cont.)

20The British University inEgypt

Marwan ShedidPh.D., P.Eng.

Reinforced Concrete Design (2)14CIVL11I

Design using I-D (cont.)

M.Shedid 11

21The British University inEgypt

Marwan ShedidPh.D., P.Eng.

Reinforced Concrete Design (2)14CIVL11I

Design using I-D (cont.)

22The British University inEgypt

Marwan ShedidPh.D., P.Eng.

Reinforced Concrete Design (2)14CIVL11I

Design for large eccentricitySummary

M.Shedid 12

23The British University inEgypt

Marwan ShedidPh.D., P.Eng.

Reinforced Concrete Design (2)14CIVL11I

Design for large eccentricity (cont.)

Compression failure Tension failure

I.D. for α = 0.4 to 1.0 Economical to use α < 0.4

24The British University inEgypt

Marwan ShedidPh.D., P.Eng.

Reinforced Concrete Design (2)14CIVL11I

Design for large eccentricity (cont.)

Tension failure

Calculate C1 and determine J

M.Shedid 13

25The British University inEgypt

Marwan ShedidPh.D., P.Eng.

Reinforced Concrete Design (2)14CIVL11I

Example 1

Design a rectangular section using the following information

MUL = 300 kN.m, NUL=900 kN, b = 250 mm

2Φ20 2Φ12 2Φ20

26The British University inEgypt

Marwan ShedidPh.D., P.Eng.

Reinforced Concrete Design (2)14CIVL11I

Example 2

Design a rectangular section using the following information

MUL = 500 kN.m, NUL=750 kN, b = 300 mm

M.Shedid 14

27The British University inEgypt

Marwan ShedidPh.D., P.Eng.

Reinforced Concrete Design (2)14CIVL11I

Design eccentric tension

Small eccentricity Large eccentricity

28The British University inEgypt

Marwan ShedidPh.D., P.Eng.

Reinforced Concrete Design (2)14CIVL11I

Design eccentric tension (cont.)

Small eccentricity

M.Shedid 15

29The British University inEgypt

Marwan ShedidPh.D., P.Eng.

Reinforced Concrete Design (2)14CIVL11I

Design eccentric tension (cont.)

Large eccentricity (approx. method)

Calculate C1 and determine J

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