CONCRETE STRUCTURES


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


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.

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"


Graphic Representation: ==================


Pc = Euler buckling load = Critical Load


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
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