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ARCH5023/ARCH4343 - ARCHITECTURAL STRUCTURES II

CONCRETE STRUCTURES
THE UNIVERSITY OF OKLAHOMA - COLLEGE OF ARCHITECTURE
____________________________________________________


DESIGN

OF

LONG TIED COLUMNS


Long Tied Square Column

Sample Problem


PROBLEM STATEMENT:

Design a slender column.

Given is a tied column,

which is a member of a braced frame.

The factored moments shown

produce bending about the y-y axis.

Moments about the x-x axis are zero.

lu = 16 feet

Pu = 250 kips top

Pu = 250 kips bottom

Mu = 60 ft kips top

Mu = 10 ft kips bottom

The effective length factor

k equals 0.9 with regards

to bending about the y-y axis, and

k equals 0.85 with respect

to bending about the x-x axis.

Determine the longitudinal reinforcement required

to support the factored loads.

Only 3000 psi concrete is available.

Use Grade 60 rebar.

The ratio of

factored axial dead load to

factored axial total load is

0.6

12" x 14" cross section pre-selected


Pu = 250 kips

Mu = 60 ft kips

lu = 16'

14"

Pu = 250 kips

Mu = 10 ft kips

12"

14"


A

Given


A. GIVEN:

lu = 16 feet

Pu = 250 kips top

Pu = 250 kips bottom

Mu = 60 ft kips top

Mu = 10 ft kips bottom

k = 0.9 y-y axis

k = 0.85 x-x axis

fc = 3,000 psi = 3 kips/sqin

fy = 60,000 psi = 60 kips/sqin

d = 0.6

ratio of area of reinforcement

terminated to total area of reinforcement

12" x 14" cross section pre-selected

x = 14" = h

y = 12"

y = 12"

x = 14"


B

Asked


B. ASKED:

longitudinal reinforcement required

number of bars

bar sizes

ties

14"

?

?

12" ?




C

Graph


C. GRAPH:
=========

D

Solution


Step (#) by Step (#) Solution
=======================

D. SOLUTION:

#1.

The strength of the column will be checked independently for

bending about both principal axes.


#2.

Bending about the y-y axis

Figure 7.9 page 299 Leet

does not apply

0.9 was previously calculated for a

braced frame

= k * lu

r

k = 0.9

<.p> #3.

r = 0.3 * y



#4.

= k * lu

r

= 0.9 * 16 ft * 12

0.3 * 14 in

= 41.14

=======

= 34 - 12 M1t top = given

M2b bottom = given

34 - (12 -10)

60

34 - (12 * (-)0.1666)

34 - (-2)

36

==


36 < 41.14

The column is long

consider secondary moments

Bending about the y-y Axis

= k * lu

r

k = 0.85 given

r = 0.3 * x

= k * lu

r

= 0.85 * 16 * 12

0.3 *12

= 45.3

=====

34 - 12 M1

M2

34 - 12 0

M2

34 - 0

= 34

====

34 < 45.3

The column is long

consider secondary moment

Analysis for bending about the y-y axis

Design for:

M1b = -10 ft kips

M2b = 60 ft kips


E

Answer





CONCRETE COLUMNS

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Sweet's System


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Prof. Dr. Hermann Gruenwald
(mail comments to: HGRUENWALD@ou.edu)
College of Architecture
The University of Oklahoma
©Dr. Gruenwald 1996, 1997