03- Columns Under Eccentric Loading

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)




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

Lecture contents

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Analysis under eccentric Axial loading

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




Resultant forces acting at centroid

Moment about plastic centroid

Analysis under eccentric Axial loading (cont.)

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Plastic centroid





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


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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.





Interaction DiagramAnalysis under eccentric Axial loading (cont.)

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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





Next, increase eccentricity until tensile face rebar reaches yield:

0.003

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

0.003

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Example

Draw Interaction Diagram for the column shown

fcu = 30 MPa

fy = 400 MPa




Example (cont.)

Point 1 (Pure compression)

Point 2 (Compression and minimum Moment)

Point 5 (Pure Tension)

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Example (cont.)

Point 3

Balanced condition




Example (cont.)

Point 4

Tension failure zone

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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




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.

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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

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Design for large eccentricitySummary

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Design for large eccentricity (cont.)

Compression failure Tension failure

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




Design for large eccentricity (cont.)

Tension failure

Calculate C1 and determine J

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Example 1

Design a rectangular section using the following information

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

2Φ20 2Φ12 2Φ20




Example 2

Design a rectangular section using the following information

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

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Design eccentric tension

Small eccentricity Large eccentricity




Design eccentric tension (cont.)

Small eccentricity

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Design eccentric tension (cont.)

Large eccentricity (approx. method)

Calculate C1 and determine J

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03- Columns Under Eccentric Loading