____________________________________________________

ARCH5023/ARCH4343 - ARCHITECTURAL STRUCTURES II

CONCRETE STRUCTURES
THE UNIVERSITY OF OKLAHOMA - COLLEGE OF ARCHITECTURE
____________________________________________________


COLUMN LOAD

BUCKLING LOAD


Buckling of Columns:

E = modulus of elasticity

E = 57,000  x square root of  fc   p. 26 Leet

fc= 28 day compression strength of concrete

Imin = minimum moment of inertia


l 	= length of column between supports

k 	= effective length factor

r	= min. radius of gyration


Pc	= Euler Buckling Load

Pc = Critical Load


Pc = 3.14.. * 3.14.. * E * Imin
       ---------------------------------
   		  2
	     ( k * l)

3.14... = phi
fc = Buckling Stress

fc = ( 3.14..  * 3.14.. * E)
      ------------------------------
	           2
	(k * l/r)

but     
fc = < fy
concrete =< steel


COLUMN

Sample Problem

PROBLEM STATEMENT:
====================

What is the Euler buckling load (Critical Load) 

and the buckling stress  

for a reinforced concrete column, 

with a cross section of 10" x 12" 

and a length of 8'?

The column is fixed at one end 

and pin supported at the other end. 

Only 3,000 psi concrete is available. 

A

Given

b	=	10"
d	=	12"
l	=	8'-0"
fc	=	3,000 psi
one fixed end
one pin supported end
d = 12"
b = 10"
l = 8'- 0"
  = 96"

B

Graph


Graphic Representation:

==================

C

Asked



Pc = Euler buckling load
	= Critical Load
 

D

Solution


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

#1.  	Pc = 3.14.. * 3.14.. * E * I
              	       -----------------------------
			2
   		 (k * l)

#2. 	TT = phi = constant = 3.14....

#3.  	E = Modulus of Elasticity
	for fc < fy
                               ____
#4.  	E = 57,000 * V   fc'	 (lb/sqin) (Leet  26)
     
                      _______
#5.  	E = 57,000 *  V  3,000

#6.  	E = 57,000 * 54.77

#7. 	E = 3,122,018 lb/sqinch or psi
=================================



#8.	I = Moment of Inertia
	for common rectangular shape

	d = 12"

	
		  b = 10"	

#9. 	I = b * d * d *d
                   --------------------
		12

#10.	I = 10 * 12 * 12 * 12	
	     ------------------------    	
		   12

#11.	I = 10 * 1728
                   --------------
		12

#12.	I = 17,280
                    ----------
                      12

               	     4
#13.	 I = 1440 in 


#14.	 k = effective length factor
	

#15.	K- Table 7.9 p 299 Leet
	pin supported
	= rotation free translation free


	fixed end = rotation fixed & 								translation fixed
#16.	  k = 0.7

#17.	  l = 8'- 0"

#18.	  l = 8'-0 * 12"

#19.	  l = 96 "

#20.	  Pc = 3.14.. * 3.14.. * E * I
  	         --------------------------------
			    2
   		       (k * l)


 #21.	 Pc= 3.14 * 3.14 * 3,122,012 * 1440 
	       -------------------------------------------------	
				2
		          ( 0.7 * 96)

#22.	 Pc	= 9,850 kips
========================
	Euler Buckling Load
	= Critical Load

#23.	 Answer Graph
	Pc = 9,850 kips




#24.	  Buckling Stress

#25.	  fc = 3.14.. * 3.14.. * E   (lb/sq in)
	         --------------------------------------
	                                     2
		           (k * l/r )
                           	            
#26.	    r 	= radius of gyration

#27.	    r = square root out of     I/A

#28.	    I 	= Moment of Inertia

#29.	    A 	= Area

#30.	 Weaker Axis governs 

		d	


			   b
#31.	for rectangles

#32.	r = d/square root of 12

#33.	r = 12" / 3.464

#34.	r = 3.464  (just happens to be the same number)
             ========

Buckling Stress

#35.	 fc =  3.14.. * 3.14.. * E   (lb/sq in)
                       --------------------------------------
	                            2
	              	(k * l/r )
                       	 
#36.	 fc = 3.14 * 3.14 * 3,222,018
	        ---------------------------------- 

				   2

			(0.7 * 96)
			-------------

			  3.464


#37.  fc = 81.69 kips 
================




E

Answer








CONCRETE COLUMNS

Columns

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