Elastic and Inelastic
Column Buckling cont.
MORGAN STATE UNIVERSITY
SCHOOL OF ARCHITECTURE AND PLANNING
LECTURE IV
Dr. Jason E. Charalambides
Let's address an example of
a column that is braced
!
Bracing essentially
shortens what we address
as the “unbraced length”
"
"
What if the column is braced in
one direction and not braced in
another?
It should better be braced on the
weak axis otherwise the whole
concept is fallacious, i.e. the
weak axis is the governing axis
therefore the bracing does not
help.
2
Let's see it in an exercise
!
Given a W10x54 column of
A992 steel with pin
connection on top and
bottom, height of 15' and
bracing at mid-height on
weak axis, determine the
compressive load capacity
of the column.
3
Let's see it in an exercise
4
Braced Wide Flange Shape
subjected to axial loading
Problem Statement:
Determine the capacity in axial loading of the given W
shape. The element is pinned at top and bottom with no
intermediate bracing, therefore having an unbraced length
of 15ft in both directions.Use A992 steel
2
Area
Unbraced length on x axis:
Ag := 15.8in
Lux := 15ft
Unbraced length on y axis:
Luy := 7.5ft
radius of gyration y
radius of gyration x
ry := 2.56in
rx := 4.37in
K factor
K := 1
Young's Modulus of Elasticity
E := 29000ksi
Bolt diameter
d b := 0.875in
Yield Stress:
Ultimate Strength:
Fy := 50ksi
Fu := 65ksi
Factor of Safety phi
ϕ := 0.9
Solution:
1) Determining the governing slenderness ratio
λx :=
λy :=
K⋅ Lux
rx
ry
(
λx = 41.189931
 7.5ft⋅ 12 in 

ft 

= 35.156
λy = 35.156
4.37in
K⋅ Luy
r := min rx , ry
 15ft⋅ 12 in 

ft 

= 41.19
2.56in
)
governing radius of gyration
is not determinant
r = 2.56⋅ in
(
λ := max λx , λy
)
λ = 41.19
The above was already obvious but it was carried on just to "academicallly" justify the numbers
2) Calculating Euler's Buckling Stress
2
FE :=
2
π ⋅E
( λ)
3.14 ⋅ 29000 ksi
2
( 41.19 )
2
= 168.529 ⋅ ksi
FE = 168.7 ⋅ ksi
3) Determining if the buckling will be elastic or inelastic.

E

Fy
Buckling := if ( λ) ≤ 4.71

, "Inelastic" , "Elastic"
Buckling = "Inelastic"

Alternatively we can also follow the process below:
Fy
FE
= 0.296
 Fy 
Buckling := if 
 FE 

≤ 2.25 , "Inelastic" , "Elastic"

Buckling = "Inelastic"
governing unbraced length
is determinant
4) Calculating the Buckling Stress (Fcr) and the load capacity of the section:
 Fy 



 FE 

Fcr := 0.658   ⋅ Fy
 50ksi 



168.7ksi 
0.658 
 ⋅ 50ksi = 44.167 ksi
ΦPn := ϕ⋅ Ag ⋅ Fcr
0.9⋅ 15.8in ⋅ 44.167ksi = 628.055 kip
2
Solution Method 2: Using Table 4-22:
1) Determining the governing slenderness ratio
λx :=
K⋅ Lux
rx
 15ft⋅ 12 in 

ft 

= 41.19
4.37in
λx = 41.19
2) Using table we locate the KL/r value
corresponding to the Fy used for factrized critical
stress:
ΦFcr := 39.7ksi
Note: From our previous calculations:
Fcr = 44.167⋅ ksi
Therefore:
ϕ⋅ Fcr = 39.75 ⋅ ksi
3) Calculating the capacity of the element:
ΦPn := ϕ⋅ Ag ⋅ Fcr
ΦPn = 628.051 ⋅ kip
Fcr = 44.167⋅ ksi
ΦPn = 628.051 ⋅ kip
Solution Method 3:
Using Table 4-1 for
W shapes pp 4-12 to
4-23:
Oh you can surely do this too! Yes it
is KLx/rx but all you need to do is
multiply that by ry
λx ⋅ ry = 8.787 ft
Using the above length for the W10x54
we can interpolate the value of 628 kip
Efficiency in Axial vs
Flexural member design
!
!
Here we see four sections of about the
same cross sectional area. From left to
right we see better forms for axial design to
better forms for flexural design.
See moment of inertia for each.
5
Efficiency in Axial vs
Flexural member design
!
An efficient flexural member will have: d ≫ b f
!
An efficient axial member will have: d≈b f
r x ≫r y
rx
≈1.6 tο 3
ry
6
Taking Two Shapes as
Examples for Axial Loading
For the HSS shape @ Fy=42ksi, Lu=15':
!
Ix
=1
Iy
r x =r y=2.95
ΦP n=236 kip
For the W 14x30 @ Fy=50ksi, Lu=15':
Ix
=14.85
Iy
rx
≈5.5
ry
r y =1.49
ΦP n =not even listed iη Table 4−1, using Table 4−22≈133 kip
7
In Class Exercise
!
!
Given KLx=30', KLy=15' (column is
braced ad mid height), and Pu=1700
Kips, find the lightest W shape from
column tables.
We can start by referring to what
would be the most efficient W
shapes:
"
!
W8, W10, W12, & W14
Note: Since KLx≠KLy a designer has
to start with just an assumption.
"
Let's assume that the weak axis
controls:
8
In Class Exercise
!
Try W14
"
For KLy=15' → W14x159, From Table 4-1, → 1810k
"
Transform to KLx →
Strong axis controls
"
!
!
ft
rx
30
ft
ft
=1.6 →
=18.75 >15
ry
1.6
Checking for 18+ ft → Φpn<1700 kip NO GO
Try next heavier section, W14x176.
ft
rx
30
=1.6 →
=18.75 ft >15 ft
ry
1.6
"
Transform to KLx →
"
Strong axis controls
"
Checking for 18.75 ft → Φpn=1853 OK
Lightest W14 is the W14x176.
9
In Class Exercise
!
Try W12
"
For KLy=15' → W12x170, From Table 4-1, → 1790kip
"
Transform to KLx →
ft
rx
30
ft
ft
=1.78 →
=16.85 >15
ry
1.78
Strong axis controls again
"
!
Try next heavier section, W12x190.
"
!
Needless to try because already the W14x176 is lighter!
Try W10.
"
!
Checking for 16.85ft → Φpn<1688kip NO GO
For KLy=15ft → No section with Φpn≥1700kip
Lightest W section is the W14x176.
10