AECT360 Lecture 33

Lecture 33 - Page 1 of 13

Lecture 33 – Steel Columns & Compression Members Similar to wood columns (or any column in general), steel columns are susceptible to buckling if not adequately laterally braced. The load-carrying capacity of compression members is inversely proportional to the slenderness ratio, which is defined as:

Slenderness ratio = r

kL < 200

where: k = modification factor based on column end conditions (see below)

L = unbraced length between lateral supports, inches r = radius of gyration

= AI

= from properties in textbook appendix


A graph of the load-carrying capacity of a column vs. the slenderness ratio is shown below: Column Shapes

The most efficient shape of a column section is square, circular or any symmetric shape such that one axis is not significantly weaker than another. Sections that are efficient as beams are NOT efficient as columns, and vice versa.

Col

umn

load

cap

acity

rkL

W12x65

6” dia. pipe

HSS 6x6x¼

Good column sections

W12x14

C6x8.2 channel

HSS 6x3x¼

Poor column sections


Allowable Loads in Columns and Compression Members

The allowable axial compressive load, Pallow, is determined as:

Pallow = FaAg

where: Pallow = allowable axial compressive load, kips Fa = allowable axial stress, KSI

=

( )

( ) ( )3

3

2

2

8/

8/3

35

2/1

cc

yc

CrkL

CrkL

FC

rkL

−+

⎥⎥⎦

⎤

⎢⎢⎣

⎡−

Cc = yFE22π

If kL/r > Cc, then:

Fa = ( )2

2

/2312

rkLEπ

Ag = gross cross-sectional area of member, in2

Fy = steel yield stress, KSI = 36 KSI for A36 steel = 50 KSI for A992 steel = 50 KSI for A572 steel = 46 KSI for A500 steel


Example 1 GIVEN: A W10x45 column using A992 steel is laterally braced at 14’-0” for both the X and Y axes. Use “k” = 1.0. REQUIRED: Determine the allowable axial compressive load, Pallow that can be applied to the column.

Step 1 – Determine the largest slenderness ratio kL/r: a) Check Weak Axis:

yrkL =

inftx

01.2)/"12'14)(0.1(

= 83.6

b) Check Strong axis:

xrkL =

inftx

33.4)/"12'14)(0.1(

= 38.8 Step 2 – Determine Cc:

Cc = yFE22π

= KSI

KSI50

)29000(2 2π

= 107

Use this value


Step 3 – Determine Fa:

Since Cc > kL/r then:

Fa =

( )

( ) ( )3

3

2

2

8/

8/3

35

2/1

cc

yc

CrkL

CrkL

FC

rkL

−+

⎥⎥⎦

⎤

⎢⎢⎣

⎡−

=

( )

( ) ( )3

3

2

2

)107(86.83

)107(86.833

35

50)107(2

6.831

−+

⎥⎦

⎤⎢⎣

⎡− KSI

= 18.28 KSI

Step 4 – Determine Pallow:

Pallow = FaAg = 18.28 KSI(13.3 in2) Pallow = 243 kips


Allowable Axial Stress Tables:

For convenience, the long calculations used to determine the allowable axial stress, Fa, have been summarized in the following tables (for Fy = 36

KSI and 50 KSI steels) based on MAXIMUM r

kL values:

Fy = 36 KSI


Fy = 50 KSI


Example 2 GIVEN: The W10x45 A992 column from Example 1. REQUIRED: Using the allowable axial stress table, determine Pallow.

Step 1 – Determine Fa @ kL/r = 83.6 from table above (Fy = 50 KSI):

Interpolating, Fa ≈ 18.30 KSI


Pallow = FaAg = 18.30 KSI(13.3 in2) Pallow = 243 kips (same answer as in Example 1)

Example 3 GIVEN: The free-standing 6” diameter standard weight ASTM A501 (Fy = 36 KSI) steel column supports a sign weighing 9000 lbs. For simplicity, disregard col. wt. REQUIRED: Determine if the column is adequate to carry the load.

Step 1 – Determine the maximum kL/r:

Since the pipe is symmetric about both axes, only need to check kL/r for either axes. From “k” table above, 1 since column is free-standing end conditions is type “e” → use k = 2.10

yrkL =

inftx

25.2)/"12'15)(10.2( = 168

Step 2 – Determine Fa @ kL/r = 168:

From table above (Fy = 36 KSI) → Fa = 5.29 KSI


Pallow = FaAg = 5.29 KSI(5.58 in2) = 29.5 kips Since Pallow = 29.5 kips > 9 kips → column is acceptable

15’-0” Ground

Sign


Column Load Tables

The simplest method of designing axially-loaded steel column is by use of the column load tables. Simply determine the effective length kL in terms of feet and select the column with a given allowable axial load that is greater than the actual load. The tables are based on buckling about the weak axis of the column section.




Example 4 GIVEN: A column has an axial load of 200 kips and a kL = 14’-0”. REQUIRED:

1) Design the lightest weight wide flange column using A992 steel (Fy = 50 KSI).

2) Design the lightest weight standard weight steel pipe column using A501 steel (Fy = 36 KSI).

Step 1 – Design lightest wide-flange shape using Fy = 50 KSI:

Using the column load tables above @ kL = 14’-0”: Possibilities:

Column Designation: Pallow: Weight (lb. per foot):

W8x40 217 kips 40 W10x39 207 kips 39

The lightest wide flange = W10x39

Step 2 – Design lightest standard weight pipe using Fy = 36 KSI:

Using the column load tables above @ kL = 14’-0”: Possibilities:

Column Designation: Pallow: Weight (lb. per foot): 10” Diameter 223 kips 40.48 12” Diameter 282 kips 49.56

The lightest pipe = 10” Diameter As can be seen in the example above, circular pipe columns have as much (or more) load-carrying capacity as an equal-weight Wide Flange shape of higher grade steel. Pipe columns are VERY EFFICIENT.


Column Base Plate Bolting Requirements

The Occupational Safety and Health Administration (OSHA) requires that a minimum of 4 anchor bolts be used to fasten column base plates to the concrete. This is in response to inadequate anchorage from the traditionally-used 2 anchor bolt requirement. For this requirement, W8, W6 and W4 wide flange shapes are very difficult to squeeze 4 anchor bolts and are therefore not recommended for columns anymore.

Column axial load

Base plate acts like a “footing” spreading out the load to the concrete footing

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AECT360 Lecture 33