lecture29 actual

7/31/2019 Lecture29 Actual

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Lecture 29 - Reinforced Columns

April 1, 2002

CVEN 444


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Lecture Goals

ColumnsLong Column Design


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Columns

Long with a relatively high slenderness ratiowhere lateral or shear walls are required

Long with a medium slenderness ration that will

cause a reduction in strength Short where the slenderness ratio is small

Slenderness ratio =

r

Klu


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Long ColumnsSlender Columns

Column with a significant reduction in

axial load capacity due to moments

resulting from lateral deflections of the

column (ACI Code: significant

reduction 5%)

Slender =

Column


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Long Columns

Less than 10 % of columns in braced or non-swayframes and less than half of columns in unbraced or

sway frames would be classified as slenderfollowing ACI Code Procedure.


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Effective Length

The effective length - Klu

lu

- It measures the clear distance between floors.

K - a factor, which represents the ratio of the distance

between points of zero moments in the columns


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H s K

beamsof/

columnsof/

u

u

lEI

lEI

YA v YBl cc h s top and bottom factors cact . Vi u khp Yl v hn (infinite) hoc l 10

v u c nh Y l zero hay 1


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H s KCc gi thit tng qut l

- Kt cu gm cc khung ch nht i xng

- The girder moment at a joint is distributed to columnsaccording to their relative stiffness

- All columns reach their critical loads at the same time


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General Formulation

Modulus of Elasticity

Reinforced Moment (ACI 10.11.1)c

c

5.1

c

57000

33

f

fwE

columnafor70.0

beamafor35.0

g

g

II

II


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General Formulation

Area

Moment of inertia shall be divided by (1 + bd)

with sustain lateral loads

gAA

bd = Max. factored sustain lateral loadMax. factored axial load


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K Factor

Use the Y values to

obtain the K factors

for the columns.


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Long Column

Lateral deflection -

increases moment

Eccentrically loaded pin-ended column.

M = P*( e + D )


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Long Column

OA - curve for end moment

OB - curve for maximumcolumn moment @ mid-height)

Eccentrically loaded pin-ended column.

Axial capacity is reduced from A

to B due to increase in maximummoment due to Ds (slendernesseffects)


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Long Columns

From ACI Sec. 12.10.2 , the slenderness effects maybe neglected if

k = effective length factor (function of end restraints)

Non-sway frames

Sway frames

2

1

ratiosslendernes

u 1234

M

M

r

kL

k

k

0.1

0.15.0


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Long Column - Slenderness Ratio

Slenderness Ratio forcolumns

Pinned-PinnedConnection

Fixed-Fixed

Connection

(a)

(b)


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Slenderness Ratio forcolumns

Fixed-PinnedConnection

Partial restrained

Connection

(c)

(d)


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Slenderness Ratio for columns in frames


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Slenderness Ratio for columns in frames


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Long Column

Unsupported height of column from top of

floor to bottom of beams or slab in floor

Radius of gyration

= 0.3* overall depth of rectangular columns= 0.25* overall depth of circular

columns

lu =

r =

A

I


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Long Column

singular curvature double curvature

Ratio of moments at two column ends, where

M2 > M1 (-1 to 1 range)

M1

/M2

=

0

2

1 M

M0

2

1 M

M


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Long Columns

M1/M2 = Ratio of moments at two column endswhere M2 > M1 (-1.0 to 1.0 range)

- single curvature

- double curvature

1.0kand

5.0

2

1

M

M

is typically conservative

(non-sway frames)

Note Code (10.12.2) M1/M2 -0.5 non-sway frames


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Long Column

From MacGregor

framessway-nonveconservatitypicallyis

0.1and

5.0

2

1

k

M

M

(non-sway frames)Note: Code 10.12.2 5.0

2

1 M

M

Possible range of = 22 to 40

r

ukl


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Design of Short Columns

Design a short square tied column to carry adesign load of Pu of 640 kips and a design

moment of 330 kip-ft. Place the reinforcing

uniformly in the four faces. Design the ties.

Use a 1.5 in. cover with an unsupported length

is 11 ft. and fc = 4000 psi and fy = 60 ksi.


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Problem Steps

Determine eccentricity.

Estimate column size

required base on axial load. Determine e/h and requiredfPn/Ag

Determine which chart touse.

Select steel sizes.

Design ties by ACI code

Design sketch


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Design Summary

lecture29 actual

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