p. 26 Leet
BEAM COLUMNS:
Columns
Members that carry both
compression force (axial load)
& moment
- pure beam action -moment governs
- pure column - axial load governs
COLUMN TYPES BY
length-to-thickness ratio:
Short Column:
short & stocky (h<10 Cross Sec.)
not based on absolute length
high bending stiffness
small secondary moments
def.: secondary moment <5%
primary moment = short column
Long or Slender Column:
less stocky
more flexible
P-delta effect:
deflection of the longitudinal axis
by primary moments create additional or secondary moments
equal to the product of the
axial force P and the
deflection of the center line
Primary moments:
member end moments or
moments due to transverse loads
TYPES OF COLUMNS:
Tied Columns
individual hoops
are used to position
the longitudinal steel
rectangular cross section typical
but can be round or L-shaped
95% of all nonseismic columns
min. 4 longitudinal bars
Spiral Columns
closely spaced continuous spirals
round cross section typical
min. 6 longitudinal bars
pitch 1 3/8" - 3 3/8"
for high strength
high ductility
high seismic forces
Core area:
area contained within the spiral
Design Guidelines for Columns:
A.) 8" - 10" min. width or min.
based on clearance requirements
B.) min. area of longitudinal
steel to be >= 1% and < 8% of the
cross-sectional area
C.) 5-6% practical to fit in column
and maintain spacing between bars
D.) Max. 4 bars can be bundled
E.) Spacing between bars
>=1.5 >= 1.5"
F.) Min. Concrete Coverage
p. 105 Leet Table 3.3
1.5" if not exposed
2" if exposed
applies to:
longitudinal rebar
steel ties
spirals
purpose of coverage:
fire protection
corrosion protection
G.) Min 4#4 bars / 5#4 for round c.
Buckling of Columns:
E = modulus of elasticity
E = 57,000
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 = * * E * Imin
2
( k * l)
fc = Buckling Stress
fc = ( * * E)
2
(k * l/r)
but fc = < fy
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"
pin
fixed
B. ASKED:
Pc = Euler buckling load
= Critical Load
Pc
C. SOLUTION:
Pc = * * E * I
2
(k * l)
= constant = 3.14....
E = Modulus of Elasticity
for fc < fy
E = 57,000 * fc' (lb/sqin) (Leet
26)
E = 57,000 * 3,000
E = 57,000 * 54.77
E = 3,122,018 lb/sqinch or psi
=========================
I = Moment of Inertia
for common rectangular shape
d = 12"
b
= 10"
I = b * d * d *d
12
I = 10 * 12 * 12 * 12
12
I = 10 * 1728
12
I = 17,280
12
4
I = 1440 in
k = effective length factor
Table 7.9 p 299 Leet
pin
supported
= rotation free translation free
fixed end = rotation fixed & translation fixed
k = 0.7
l = 8'- 0"
l = 8'-0 * 12"
l = 96 "
Pc = * * E * I
2
(k * l)
Pc= 3.14 * 3.14 * 3,122,012 * 1440
2
( 0.7 * 96)
Pc = 9,850 kips
==================
Euler Buckling Load
Critical Load
Answer Graph
Pc = 9,850 kips Buckling Stress
fc = * * E (lb/sq in)
2
(k * l )
r
r = radius of gyration
I
= Moment of Inertia
A = Area
Weaker Axis governs
d
b
for rectangles
r = 12"
3.464
r = 3.464
========
Buckling Stress
fc = * * E (lb/sq in)
2
(k * l )
r
fc = 3.14 * 3.14 * 3,222,018
2
(0.7 * 96 )
3.464
fc = 81.69 kips
================