Chapter 5: Compression Members - Steel Design 4300:401 Chap05 - Compression Members.pdf · 5.3 5.3...

5.1

Chapter 5: Compression Members The following information is taken from “Unified Design of Steel Structures,” Second Edition, Louis F.

Geschwindner, 2012, Chapter 5.

5.1 Compression Members in Structures

A compression member is a structural element subjected to an axial force that

tends to push the ends of the member together.

• The column is the best known compression member in a building structure.

- Columns are usually considered to be vertical structural members that

support the horizontal elements of a roof or floor system.

- However, columns can be positioned in any orientation (e.g. diagonal and

horizontal compression elements in a truss).

• Other compression members include the chords of trusses, various bracing

members in floors and walls, and struts and posts.

- Struts are short vertical members subjected to compressive loads.

• Other members have compression elements: compression flanges of rolled

beams and built-up sections.

Compression members discussed in this chapter will be loaded only with concentric

axial forces.

• In real structures, additional load effects (e.g. load eccentricities, transverse

loads) acting on a compression member cause combined effects of axial force

and bending (e.g. beam-columns).

Chapter E of the Specification addresses members subjected to axial compression

resulting from forces applied concentrically at the centroidal axis.

5.2 Cross-Sectional Shapes for Compression Members

Compression members carry axial forces, so the primary cross-sectional property

of interest is the cross-sectional area.

• The basic relationship between force and stress in an axially loaded member is

given by the following equation.

f = P/A

• Theoretically, all cross sections with the same area will perform in the same

way.

• However, in real structures, other factors influence the strength of the

compression member, and the configuration of the area becomes important.

5.2

The typical compression member in a building is a column and the typical column is a

wide-flange shape.

• The wide-flange shape is not the most efficient section for a column.

- Round and square hollow structural steel sections and pipe sections are the

most efficient shapes.

• The wide-flange shape provides a compression member that can be easily

connected to other members of the structural system, such as beams and other

columns.

- The ease of the connections significantly influences the selection of the

wide-flange shape as an appropriate column cross-section.

There are a number of shapes that can be selected to resist a compression load in

a given structure.

• The number of possible options is limited by available sections, connection

methods, and the type of structure in which the section is to be used.

• Figure 5.2 (p. 114 of the textbook) shows various types of shapes used as

compression members.

- Many of these shapes are also used as tension members since the nature of

the loading in tension or compression is an axial load.

• Single-angles are used as bracing and as compression members in light trusses.

• Structural tees are used as the top chord compression members in roof trusses

because other truss members can be welded directly to the stems of the tees.

• Hollow structural sections (square, rectangular, or round HSS) and steel pipe

are common sections used for buildings, bridges, and other structures.

- Advantages of hollow structural sections include the following.

◦ Four-sided and round sections are easier to paint than open W, S, and M

sections.

◦ Hollow structural tubing does not have the problem of dirt collecting

between the flanges associated with open structural shapes.

- One disadvantage of hollow structural sections includes the following.

◦ The ends of the tube and pipe sections that are subject to corrosive

atmospheres may have to be sealed to protect the inaccessible inside

surfaces from corrosion.

• Built-up sections, a section comprised of two or more shapes and/or plates, may

be needed if the members are long and support very heavy loads.

• Composite columns, consisting of steel pipe and structural tubing filled with

concrete, or W shapes encased in concrete, are being used more frequently.

5.3

5.3 Compression Member Strength

A column’s ability to carry a load is greatly influenced by column slenderness and

column length.

• Very short, stout columns fail by crushing due to material failure.

- Failure occurs once the stress exceeds the elastic (yield point) limit of the

material.

• Long, slender columns fail by buckling – a function of the column’s dimensions

and its modulus of elasticity.

- Buckling is the sudden uncontrolled lateral displacement of a column at which

point no additional load can be supported.

- Failure occurs at a lower stress level than the column’s material strength

due to buckling (i.e. lateral instability).

Short columns

Short columns (a.k.a. stub columns) fail by crushing at high stress levels that

exceed the elastic limit of the column material.

The maximum compressive force for short columns is based on the following basic

stress equation.

Py = Fy Ag

where

Py = yield load (psi or ksi)

Fy = yield stress (psi or ksi)

Ag = gross cross-sectional area of the column (in2)

Euler Column

Long columns fail by buckling at stress levels that are below the elastic limit of the

column material.

Buckling in long, slender columns is due to the following effects.

• Eccentricities in loading.

• Irregularities in the column material.

Buckling can be avoided (theoretically) if the loads were applied absolutely axially,

the column material was totally homogeneous with no imperfections, and

construction was true and plumb.

5.4

A Swiss mathematician named Leonhard Euler (1707 – 1783) was the first to

investigate the buckling behavior of slender columns within the elastic limit of the

column’s material.

• Euler presented his analysis in 1759.

• Euler’s equation shows the relationship between the load that causes buckling of

a (pinned end) column and the material and stiffness properties of the column.

• Euler’s column model, usually called the perfect column or the pure column is

based on the following assumptions.

- The column ends are frictionless pins.

- The column is perfectly straight.

- The load is applied concentrically (i.e. along the centroidal axis)

- The material behaves elastically.

The critical buckling load (a.k.a. the Euler’s buckling load) can be determined by the

following equation.

Pcr = π2EImin/L2

where

Pcr = critical axial load that causes buckling in the column (pounds or kips)

E = modulus of elasticity of the column material (psi or ksi)

Imin = smallest moment of inertia of the column cross-section (in4)

(Most sections have Ix and Iy; angles have Ix, Iy and Iz.)

L = column length between pinned ends (inches)

• As the column length increases, the critical load rapidly decreases (since the

critical load is inversely proportional to L2), approaching zero as a limit.

Euler’s equation is valid only for long, slender columns that fail due to buckling.

• Euler’s equation contains no safety factors.

• Euler’s equation results in compressive stresses developed in columns that are

well below the elastic limit of the material.

Other Boundary Conditions

To generalize the buckling equation for end conditions other than that for pinned

ends, the column length L is replaced by the column effective length KL, where K is

the effective length factor.

• The general buckling equation takes the following form.

5.5

Pcr = π2EImin/(KL)2

End restraint affects the load-carrying capacity of columns.

• Columns with significant rotational or translational end restraint can support

considerably more load than those with little end restraint.

The effective length of a column (referred to as KL) is the distance between

inflection points, or points of zero moment in the column.

• The concept of effective length is simply a mathematical method of taking a

column, whatever its end bracing conditions, and replacing it with an equivalent

pinned-end braced column.

• The effective length factor K is used as a multiplier for converting the actual

column length to an effective column length based on end conditions.

The six most common theoretical end conditions and the associated theoretical and

recommended design values for K are shown in following table (ref. Table C-A-7.1,

AISC Commentary, Appendix 7, shown as Figure 5.6, p. 118 of the textbook).

5.6

a. Both ends fixed: Structure is adequately braced against lateral forces.

Theoretical: K = 0.5

Recommended design value: K = 0.65

Example: Steel column rigidly connected (welded) to large steel beams top and

bottom.

b. One end pinned and one end fixed: Structure is adequately braced against

lateral forces.



Example: Steel column rigidly connected to a concrete footing at the base and

attached to light-gauge roofing at the top.

c. Both ends fixed with some lateral translation.



Example: Steel column in a relatively flexible rigid frame structure.

d. Both ends pinned: Structure is adequately braced against lateral (wind and

earthquake) forces.



Examples: Steel column with simple clip angle connection top and bottom.

e. One end free and one end fixed: Lateral translation possible (develops eccentric

column load).



Examples: Water tank mounted on a simple pipe column; flagpole.

f. Base pinned, top fixed with some lateral translation.



Examples: Steel column with a rigid connection to a beam above and a simple pin

connection at the base.

5.7

Pinned end supports are often assumed for design purposes.

• Pinned ends are usually assumed even if the ends of steel columns are

restrained at the bottom (e.g. by being welded to base plates, which in turn are

anchor-bolted to concrete footings).

• Steel pipe columns generally have plates welded at each end, and then bolted to

other parts of the structure.

- Such restraints vary greatly and are difficult to evaluate.

• Designers rarely take advantage of the end restraint to increase the allowable

stress.

- Such design is conservative and therefore adds to the factor of safety of

the design.

In the case of fixed-end conditions, tests have indicated that the “theoretical” K

values are somewhat non-conservative when designing steel columns.

• Since true joint fixity is rarely possible, the AISC recommends the use of

recommended K-values (listed in Table C-A-7.1, AISC Commentary, Appendix 7,

shown as Figure 5.6, p. 118 of the textbook).

5.8

Example Problems – Theoretical Column Strength using the Euler Formula

Example

Given: W10 x 22 is used as a pin-connected column.

Steel: A992 (Fy = 50 ksi)

Find:

a) Column’s buckling load using the Euler formula for a 15’-long column.

b) Column’s buckling load using the Euler formula for an 8’-long column.

Solution

W10 x 22 (A = 6.49 in2, Iy = 11.4 in4)

a) 15’-long column

Effective length (pinned-pinned): KL = 1.0 (15.0) = 15.0’

Determine the Euler buckling load: Pcr = π2EImin/(KL)2

Pcr = π2 (29,000) 11.4/(15 x 12”/’)2 = 100.7 kips

Determine the load that would cause yielding.

Py = Fy As = 50 (6.49) = 324.5 kips

Since Pcr < Py, the theoretical column strength is 100.7 kips.

• The column would buckle before it reached its yield stress.

b) 8’-long column

Effective length (pinned-pinned): KL = 1.0 (8.0) = 8.0’

Determine the Euler buckling load: Pcr = π2EImin/(KL)2

Pcr = π2 (29,000) 11.4/(8 x 12”/’)2 = 354.0 kips

Determine the load that would cause yielding.

Py = Fy As = 50 (6.49) = 324.5 kips (same as before)

Since Pcr > Py, the theoretical column strength is 324.5 kips.

• The column would reach its yield stress before it buckled.

5.9

Example

Given: W10 x 22 with fixed supports.


Find: Column length that would theoretically cause the column to buckle elastically

before yielding.

Solution

W10 x 22 (A = 6.49 in2, Iy = 11.4 in4)

Determine the force that would cause the column to yield.

Py = Fy As = 50 (6.49) = 324.5 kips

Applicable equation: Pcr = π2EImin/(KL)2

Effective length (fixed-fixed): KL = 0.5 L

Determine the column length at which Py would also cause buckling.

Pcr = Py = 324.5 = π2 (29,000) 11.4/(0.5 L)2

324.5 = 13,051,565/L2

L2 = 13,051,565/324.5 = 40,221

L = 200.6” (16.7’)

The column length that would theoretically cause the column to buckle elastically

before yielding is 16.7’.

5.10

Real Column

Actual column strength observed by testing is less than that predicted by the

Euler buckling equation or by that expected by material yielding.

• A number of factors make it difficult to accurately predict column strength.

• The following three factors are the main factors that influence column

strength.

- Material inelasticity, resulting from built-in or residual stresses.

- Column initial out-of-straightness.

- End conditions, as previously discussed.

Residual stresses are the direct result of the manufacturing process.

• A major cause of residual stress is the uneven cooling of shapes after hot

rolling.

- For example, in a W-shape the outer tips of the flanges and the middle of

the web cool quickly, while the areas at the intersection of the flanges and

web cool more slowly.

• The magnitude of the residual stresses from uneven cooling varies from 10 to

30 ksi, depending on the shape.

• Welding can produce severe residual stresses that may approach the yield point

in the vicinity of the weld.

• Residual stresses may be caused during fabrication when cambering is

performed by cold bending.

Initial column out-of-straightness significantly impacts column strength and is also

the result of the manufacturing process.

• Standard tolerances prescribed by the AISC Code of Standard Practice permit

an initial out-of-straightness of 1/1000 of the length between points of lateral

support – a seemingly small variation that still impacts column strength.

• The small lateral displacement of the column results in secondary moments

along the length of the column.

• When material inelasticity and out-of-straightness are combined, the Euler

equation cannot properly predict column behavior.

Curve plots of column test data have been developed on the basis of curve-fitting

in an attempt to develop a simple way to predict column behavior.

5.11

AISC Provisions

The AISC Specification provides one equation (based on the Euler equation) for

long columns with elastic buckling and an empirical equation for short and

intermediate columns.

• By using these equations a flexural buckling stress Fcr is determined for a

compression member.

• The flexural buckling stress Fcr is multiplied by the cross sectional area of the

member to obtain the nominal strength of the column.

The nominal compressive strength Pn and the LRFD design strength and ASD

allowable strength of a column member without slender elements are determined

from the following equations.

Pn = Fcr Ag AISC Equation E3-1

LRFD design compression strength (φc = 0.90): φc Pn = φc Fcr Ag

ASD allowable compression strength (Ωc = 1.67): Pn/Ωc = Fcr Ag/Ωc

The flexural buckling stress Fcr is determined from the following equations.

a) If KL/r ≤ 4.71(E/Fy)1/2 (or Fy/Fe ≤ 2.25),

then Fcr = [0.658 Fy/Fe] Fy AISC Equation E3-2

b) If KL/r > 4.71(E/Fy)1/2 (or Fy/Fe > 2.25),

then Fcr = 0.877 Fe AISC Equation E3-3

where

Fe = the elastic critical buckling stress (i.e. the Euler stress)

= π2E/(KL/r)2 AISC Equation E3-4

Calculations using the column equations are rarely made because the AISC Manual

provides computed values of critical stresses φcFcr and Fcr/Ωc in Table 4-22.

• Critical stress values are listed for slenderness ratios (i.e. KL/r) from 1 to 200

and for steels with Fy = 35, 36, 42, 46, and 50 ksi.

Tables 4-1 through 4-20 of the AISC Manual may be used to analyze and design

various column sections without the necessity of a trial-and-error procedure.

Earlier editions of the AISC Specification prescribed a maximum slenderness ratio

of KL/r = 200.

5.12

• The value was based on engineering judgment, practical economics, and the fact

that special care was required to prevent damage to very slender members

during fabrication, shipping, and assembly.

Recent editions of the AISC Specification no longer prescribe a specific maximum

slenderness ratio.

• Since many factors influence column strength a very slender column might

actually be acceptable.

• However, Section E2 of the AISC Commentary indicates that if KL/r > 200,

then the critical stress Fcr will be less than 6.3 ksi.

- By User Note, the 14th Edition of the Specification states, “For members

designed on the basis of compression, the effective slenderness ratio KL/r

preferably should not exceed 200.”

- The design aids for columns (e.g. Table 4-22) list do not provide critical

stress values for slenderness ratios in excess of 200.

• Based on these important practical considerations, it is prudent to select

compression members with slenderness ratios below 200.

5.13

Example Problems – Column Formulas

Example

Given: The column shown.

Steel: W12 x 72 (Fy = 50 ksi)

E = 29,000 ksi

Find: Determine the following.

a) Using the critical stress values in Table 4-22 of

the AISC Manual, determine the LRFD design

compression strength φcPn and the ASD

allowable compression strength Pn/Ωc.

b) Repeat the problem, using Table 4-1 of the AISC Manual.

c) Calculate φc Pn and Pn/Ωc using the column formulas from Section E3 of the

AISC Manual.

Solution

W12 x 72 (A = 21.1 in2, bf/2tf = 8.99, h/tw = 22.6, rx = 5.31”, ry = 3.04”)

K = 0.80 (Recommended design value, AISC Table C-A-7.1)

Check for slender elements using AISC Table B4.1a.

Flange (Case 1): λr = 0.56(E/Fy)1/2 = 0.56 (29,000/50)1/2 = 13.49 > bf/2tf = 8.99

The flange is not slender.

Web (Case 5): λr = 1.49(E/Fy)1/2 = 1.49 (29,000/50)1/2 = 35.9 > h/tw = 22.6

The web is not slender.

a) Using the critical stress values in Table 4-22 of the AISC Manual.

(KL/r)max = (KL/r)y = 0.80 (15) (12”/’)/3.04 = 47.37

By interpolating from Table 4-22 of the AISC Manual:

LRFD: φc Fcr = 38.19 ksi

ASD: Fcr/Ωc = 25.43 ksi

LRFD design compression strength (φc = 0.90):

φc Pn = (φc Fcr) Ag = 38.19 (21.1) = 805.8 kips

ASD allowable compression strength (Ωc = 1.67):

Pn/Ωc = (Fcr/Ωc) Ag = 25.43 (21.1) = 536.6 kips

5.14

b) Using Table 4-1 of the AISC Manual.

KL = 0.80 (15) = 12.0’

LRFD design compression strength: φc Pn = 806 kips

ASD allowable compression strength: Pn/Ωc = 536 kips

c) Using the column formulas from AISC Manual.

(KL/r)max = (KL/r)y = 0.80 (15) (12”/’)/3.04 = 47.37 (same as before)

Determine the flexural buckling stress Fcr.

Fe = π2E/(KL/r)2 = π2 (29,000)/(47.37)2 = 127.55 ksi

4.71 (E/Fy)1/2 = 4.71 (29,000/50)1/2 = 113.43 > KL/ry = 47.37

So, Fcr = [0.658 Fy/Fe] Fy = [0.658 50/127.55] 50 = 42.43 ksi


φc Pn = φc (Fcr Ag) = 0.90(42.43)(21.1) = 805.7 kips


Pn/Ωc = (Fcr Ag)/Ωc = 42.43 (21.1)/1.67 = 536.1 kips

5.15

Example

Given: An 18’ long HSS 16 x 16 x ½ column with pinned end supports.

Steel: Fy = 46 ksi

E = 29,000 ksi


a) Using the critical stress values in Table 4-22 of the AISC Manual, determine

the LRFD compression design strength φcPn and the ASD allowable

compression strength Pn/Ωc.

b) Repeat the problem, using Table 4-4 of the AISC Manual.

c) Calculate φcPn and Pn/Ωc using the column formulas from Section E3 of the

AISC Manual.

Solution

HSS 16 x 16 x ½ (A = 28.3 in2, b/t = 31.4, rx = ry = 6.31”)

K = 1.00 (Recommended design value, AISC Table C-A-7.1)


Walls (Case 6): λr = 1.40(E/Fy)1/2 = 1.40 (29,000/46)1/2 = 35.15 > b/t = 31.4

The walls are not slender.


(KL/r)max = (KL/r)y = 1.00 (18) (12”/’)/6.31 = 34.23









KL = 1.00 (18) = 18.0’



5.16


(KL/r)max = (KL/r) = 1.00 (18) (12”/’)/6.31 = 34.23 (same as before)


Fe = π2E/(KL/r)2 = π2 (29,000)/(34.23)2 = 244.28 ksi

4.71 (E/Fy)1/2 = 4.71 (29,000/46)1/2 = 118.26 > KL/r = 34.23

So, Fcr = [0.658 Fy/Fe] Fy = [0.658 46/244.28] 46 = 42.51 ksi




Pn/Ωc = (Fcr Ag)/Ωc = 42.51(28.3)/1.67 = 720.4 kips

5.17

Example

Given: A 32’ long W14 x 90 column, with the weak axis braced as shown.

Steel: Fy = 50 ksi

E = 29,000 ksi


a) Using the critical stress values in Table 4-22

of the AISC Manual, determine the LRFD

design compression strength φcPn and the

ASD allowable compression strength Pn/Ωc.

b) Repeat the problem, using Table 4-1 of the

AISC Manual.

c) Calculate φcPn and Pn/Ωc using the column

formulas from Section E3 of the AISC

Manual.

Solution

W14 x 90 (A = 26.5 in2, bf/2tf = 10.2, h/tw = 25.9, rx = 6.14”, ry = 3.70”)

K = 0.80 and 1.00 (Recommended design values, AISC Table C-A-7.1)


Flange (Case 1): λr = 0.56(E/Fy)1/2 = 0.56 (29,000/50)1/2 = 13.49 > bf/2tf = 10.2

The flange is not slender.

Web (Case 5): λr = 1.49(E/Fy)1/2 = 1.49 (29,000/50)1/2 = 35.9 > h/tw = 25.9

The web is not slender.


Determine the slenderness ratios:

(KL/r)x = 0.80 (32) (12”/’)/6.14 = 50.03 (controls)

(KL/r)y = 0.80 (12) (12”/’)/3.70 = 31.14

(KL/r)y = 1.00 (10) (12”/’)/3.70 = 32.43




5.18






KxLx = 0.80 (32) = 25.6’

KyLy = 0.80 (12) = 9.6’

KyLy = 1.00 (10) = 10.0’ (controls for weak axis)

Table 4-1 has been developed based the least radius of gyration ry. Due to

bracing, the strong axis (i.e. the x-axis) actually controls the design and as a

result a value of KxLx that is equivalent to KyLy needs to be determined in order

to use the table.

Thus, KxLx/rx = Equivalent KyLy/ry

Equivalent KyLy = KxLx (ry/rx) = 0.8 (32’) (3.70”/6.14”) = 15.43’

The equivalent KyLy = 15.43’ > actual KyLy = 10.0’; thus use KyLy = 15.43’.

From the column tables with KyLy = 15.43’, by interpolating:




(KL/r)max = (KL/r)x = 0.80 (32) (12”/’)/6.14 = 50.03 (from before)


Fe = π2E/(KL/r)2 = π2 (29,000)/(50.03)2 = 114.35 ksi

4.71 (E/Fy)1/2 = 4.71 (29,000/50)1/2 = 113.43 > KL/r = 50.03

So, Fcr = [0.658 Fy/Fe] Fy = [0.658 50/114.35] 50 = 41.64 ksi




Pn/Ωc = (Fcr Ag)/Ωc = 41.64(26.5)/1.67 = 660.8 kips

5.19

5.4 Additional Limit States for Compression

The two limit states for compression members presented previously include the

following.

1. Yielding

• The upper limit for column strength (i.e. FyAg) is reached only for the zero-

length column.

2. Flexural buckling (also called Euler buckling) is the primary type of buckling.

• Members are subject to flexure (bending) when they become unstable.

Loading of singly symmetric, unsymmetric, and certain doubly symmetric members

may be governed by the following limit states.

1. Torsional buckling or flexural-torsional buckling may occur in columns that

have certain cross-sectional configurations.

• These columns fail by twisting (torsion) or by a combination of torsional and

flexural buckling.

2. Local buckling occurs when some part or parts of the cross section of a column

are so thin that they buckle locally in compression before the other modes of

buckling can occur.

• Such columns are said to be a column with slender elements.

• The susceptibility of a column to local buckling is measured by the width-

thickness ratios of the parts of its cross section (e.g. thin flanges and webs

of W-shapes).

5.5 Length Effects

As previously discussed, the six most common theoretical end conditions and the

recommended design values for K are presented in Table C-A-7.1, AISC

Commentary, Appendix 7, and shown in Figure 5.17, p. 132 of the textbook.

• The theoretical K values were developed for columns with certain idealized

conditions of end restraint, which may be different from practical design

conditions.

• In most cases, the recommended design value for K is slightly greater than the

theoretical value because actual column end conditions are unlikely to match the

theoretical assumptions.

When a column is part of a frame, the stiffness of the members framing into the

column impact the rotation that may occur at the column ends and a more detailed

analysis may be advantageous.

5.20

• In statically indeterminate structures, sidesway may occur due to the following

conditions.

- The frame deflects laterally due to the presence of lateral loads.

- Unsymmetrical vertical loads are applied to the frame.

- The frame is unsymmetrical.

• Sidesway also occurs in columns whose ends can move transversely when they

are loaded to the point that buckling occurs.

If frames are used with diagonal bracing or shear walls, then sidesway will be

prevented and some rotational restraint is provided at the ends of the column.

• In such instances, K values will fall somewhere between Case (a) (i.e. fixed-

fixed supports) and Case (d) (i.e. pinned-pinned support) of AISC Table C-A-7.1.

Section 7.3 of Appendix 7 of the AISC Specification (entitled “Alternative

Methods of Design for Stability”) states, “In braced frame systems, shear wall

systems, and other structural systems where lateral stability and resistance to

lateral loads does not rely on the flexural stiffness of columns, the effective

length factor, K, of members subject to compression shall be taken as 1.0, unless

rational analysis indicates that a lower value is appropriate.”

• A specification like K = 1.0 is often quite conservative and a more detailed

analysis may often result in some cost savings.

The true effective length of a column is a property of the entire structure.

• In many buildings, it is probable that the masonry walls used for elevator shafts

of stairways provide sufficient lateral support to prevent sidesway.

• On the contrary, light curtain walls provide little resistance to sidesway.

• Sidesway is present in tall buildings unless a bracing system or shear walls are

used.

• In some cases resistance to sidesway is provided by the lateral stiffness of the

frame alone.

Theoretical mathematical analyses may be used to determine effective lengths.

• Many such procedures are typically too lengthy and too difficult for most

designers.

• The usual procedure is to use AISC Table C-A-7.1 (cf. AISC Commentary to

Appendix 7), interpolating between the idealized values as the designer feels is

appropriate, or simply taking K = 1.0.

5.21

• The most common method for obtaining effective lengths is to employ the

alignment charts presented in the Commentary to Appendix 7.

- One chart was developed for columns braced against sidesway:

Figure C-A-7.1 entitled “Alignment chart – sidesway inhibited (braced

frame)”, and shown as Figure 5.19 (p. 134 of the textbook).

- One chart was developed for columns subject to sidesway: Figure C-A-7.2

entitled “Alignment chart - sidesway uninhibited (moment frame)”], and

shown as Figure 5.20 (p. 134 of the textbook).

- The use of these charts enables the designer to obtain good K values

without struggling through lengthy trial-and-error procedures.

The alignment charts were developed on the basis of a certain set of assumptions.

• A complete list of the assumptions is given in the Commentary to Appendix 7.

• Among these assumptions are the following.

1. The members are elastic, have constant cross sections, and are connected

with rigid joints.

2. All columns buckle simultaneously.

3. For columns in frames with sidesway inhibited, rotations at opposite ends of

the restraining beams are equal in magnitude and opposite in direction,

producing single curvature bending in the beam.

4. For columns in frames with sidesway uninhibited, rotations at opposite ends

of the restraining beams are equal in magnitude and direction, producing

double (reverse) curvature in the beam.

5. Axial compression forces in the girders are negligible.

The alignment charts are based on equations provided in the Commentary to

Appendix 7.

• The alignment chart for sidesway inhibited frames (shown in Figure 5.19) is

based on Equation C-A-7-1.

• The alignment chart for sidesway uninhibited frames (shown in Figure 5.20) is

based on Equation C-A-7-2.

To use the alignment charts, it is necessary to have preliminary sizes for the

girders and columns before the K factor can be determined for a column.

• Preliminary sizes for the columns and girders may be obtained using

approximate methods of analysis (e.g. portal method, or cantilever method).

• G values, a ratio of column stiffness ∑(Ic/Lc) to girder stiffness ∑(Ig/Lg), are

then determined for each end of the column.

5.22

The resistance to rotation furnished by the beams and girders meeting at one end

of a column is dependent on the rotational stiffnesses of those members.

• The rotational restraint at the end of a particular column is proportional to the

ratio of the sum of the column stiffnesses to the girder stiffnesses meeting at

that joint and located in the plane in which buckling of the column is being

considered (i.e. the value G).

G = ∑(Ic/Lc)/∑(Ig/Lg)

• For pinned connections of columns, G is theoretically infinite since the

resistance to rotation from the connection is nearly zero.

- Such is the case when a column is connected to a footing with a frictionless

hinge.

- Since such a connection is not actually frictionless, it is recommended that G

be made equal to 10.0 where such nonrigid supports are used.

• For rigid connections of columns to footings or to very stiff beams or girders,

the resistance to rotation from the beam (i.e. the beam’s stiffness) is very

high, and G theoretically approaches zero.

- From a practical standpoint, a value of 1.0 is recommended, because no

connection is perfectly rigid.

K factors for columns of a steel frame using the alignment charts may be

determined as follows.

1. Select the appropriate chart (sidesway inhibited or sidesway uninhibited).

2. Compute G at each end of the column and label the values GA and GB.

3. Draw a straight line on the chart between the GA and GB values, and read K

where the line hits the center K scale.

An initial design provides preliminary sizes for each of the members in the frame.

• After the effective lengths are determined, each column can be redesigned.

- Should the sizes change appreciably, new effective lengths can be

determined, and the column designs repeated.

The following equations provide approximate solutions that easily yield results as

accurate as those obtained by reading a value from the alignment charts.

• For a braced frame (sidesway inhibited):

K = [3 GA GB + 1.4(GA + GB) + 0.64] / [3 GA GB + 2(GA + GB) + 1.28]

• For an unbraced frame (sidesway uninhibited):

K = {[1.6 GA GB + 4(GA + GB) + 7.5] / (GA + GB + 7.5)}1/2

5.23

Example Problem – Frames Meeting Alignment Chart Assumptions

Example

Given: Frame shown.

Frame is not braced against sidesway.

Find: Effective length factor K for each column using the alignment chart from the

Commentary to Appendix 7 (Figure C-A-7.2), as shown in Figure 5.20 (p. 134 of

the textbook).

Solution

Determine the stiffness factors (Ix/L).

• In this example buckling of the columns is analyzed about the strong axis, so Ix

is used for the columns and Ix is used for the beams.

Member Shape Ix L Ix/L

AB W8 x 24 82.7 12 6.89

BC W8 x 24 82.7 10 8.27

DE W8 x 40 146 12 12.17

EF W8 x 40 146 10 14.6

GH W8 x 24 82.7 12 6.89

HI W8 x 24 82.7 10 8.27

BE W18 x 50 800 20 40.0

CF W16 x 36 448 20 22.4

EH W18 x 97 1750 30 58.33

FI W16 x 57 758 30 25.27

5.24

Determine G factors for each joint.

Joint ∑(Ic/Lc)/∑(Ig/Lg) G

A Pinned 10.0

B (6.89 + 8.27)/40.0 0.379

C 8.27/22.4 0.369

D Pinned 10.0

E (12.17 + 14.6)/(40.0 + 58.33) 0.272

F 14.6/(22.4 + 25.27) 0.306

G Pinned 10.0

H (6.89 + 8.27)/58.33 0.260

I 8.27/25.27 0.327

Column K factors from the alignment chart (Figure 7.2b).

Column GA GB K

AB 10.0 0.379 1.78

BC 0.379 0.369 1.12

DE 10.0 0.272 1.77

EF 0.272 0.306 1.11

GH 10.0 0.260 1.76

HI 0.260 0.327 1.11

Using the stiffness ratios GA and GB determined above, the K factor for column AB

is determined using Equation 5.19 (p. 135 of the textbook) as follows.

K = {[1.6 GA GB + 4(GA + GB) + 7.5]/(GA + GB + 7.5)}1/2

K = {[1.6(10.0)0.379 + 4(10.0 + 0.379) + 7.5]/(10.0 + 0.379 + 7.5)}1/2

= (55.08/17.879)1/2

K = 1.76 (which compares favorably with the value of 1.78 determined above)

Using the stiffness ratios GA and GB determined above, the K factor for column BC

is determined using Equation 5.19 (p. 135 of the textbook) as follows.

K = {[1.6 GA GB + 4(GA + GB) + 7.5]/(GA + GB + 7.5)}1/2

K = {[1.6(0.379)0.369 + 4(0.379 + 0.369) + 7.5]/(0.379 + 0.369 + 7.5)}1/2

= (10.716/8.248)1/2

K = 1.14 (which compares favorably with the value of 1.12 determined above)

5.25

In the previous example, buckling was examined about the strong axis (i.e. the x-

axis) for both the columns and beams.

• For most buildings, the values Kx and Ky should be examined separately.

- There may be different framing conditions in the two directions.

◦ Many multistory frames consist of rigid frames in one direction and

conventionally connected frames with sidesway bracing in the other.

◦ Points of lateral support may be different in the two directions.

The alignment chart of Figure 5.20 for frames with sidesway uninhibited always

indicates that K ≥ 1.0.

• Calculated K factors of 2.0 to 3.0 are common, and larger values are

occasionally obtained.

- To many designers, such large factors seem unreasonable.

• If seemingly high K factors are determined, the designer should carefully

review the G values used to enter the chart, as well as the basic assumptions

made in preparing the charts.

Effective Length for Inelastic Columns

It is important to remember that the alignment charts are based on the

assumptions of idealized conditions (including that the behavior is purely elastic);

however, these conditions seldom exist in real structures.

• Differing end conditions and residual stresses play a significant role in

determining column strength through inelastic behavior.

• When column buckling occurs in the inelastic range, the alignment charts usually

give conservative results.

For more accurate results, inelastic K-factors can be determined from the

alignment charts by multiplying the elastic stiffness ratios GA and GB by a

stiffness reduction factor, τb (i.e. replace EcIc with τbEcIc).

• The stiffness reduction factor τb is the ratio ET/E (i.e. the ratio of the

tangent modulus to the elastic modulus).

• AISC Manual Table 4-21 provides values for τb based on the required strength,

Pu/Ag (LRFD) and Pa/Ag (ASD).

- The use of Table 4-21 assumes that the column is loaded to its full available

strength.

- If the column is not loaded to its full available strength, then Table 4-21

provides a conservative assessment of the inelastic stiffness reduction

factor and the effective length.

5.26

5.6 Slender Elements in Compression

Elements that make up the column shape, such as the thin flanges or webs of a

column, can buckle locally in compression before the buckling strength of the whole

column is reached.

• When thin plates are used to carry compressive stresses, they are susceptible

to buckling about their weak axes due to the small moments of inertia about

those axes.

Section B4 of the AISC Specification provides limiting values for the width-

thickness ratios of the individual parts of shapes subject to axial compression.

• For members subject to axial compression, sections are classified as nonslender

element or slender-element sections.

• For a section to be considered as nonslender, its compression elements must

have width-thickness ratios equal to or less than the limiting values for λr

(Greek small letter “lambda”).

- These values are given for various elements in Table B4.1a of the AISC

Specification (and shown in part as Table 5.4, pp. 141 of the textbook).

• If the width-thickness ratio of any compression element exceeds λr, then the

section is a slender-element section.

Almost all of the W, M, and S shapes listed in the AISC Manual have no slender-

element sections for steels with a yield stress of 50 ksi.

• The shapes that do contain slender-element sections are so indicated by

footnote in the column tables of the AISC Manual (e.g. HSS 5 x 2½ x 1/8, in

Table 4-3).

- The values listed in the column tables reflect the reduced design stresses

available for these slender-element sections.

The AISC Specification outlines provisions for members with slender elements.

• If the width-thickness limits for slender-element sections are exceeded for

members subject to axial compression, then refer to Section E7 of the AISC

Specification.

5.7 Column Design Tables

Tables 4-1 through 4-20 of the AISC Manual may be used to select various column

sections without the necessity of a trial-and-error procedure.

• These tables provide LRFD axial design strengths (φcPn) and ASD allowable

design strengths (Pn/Ωc) for various practical lengths of the steel sections

commonly used as columns.

5.27

• The AISC tables provide design strengths of columns with respect to the

weaker axis.

- Design strengths for W, HP, hollow structural sections (HSS), and pipe are

with respect to the y-axis.

◦ Standard steel pipe sections are labeled “Std” in the table.

◦ Extra strong pipe (XS) has thicker walls than standard steel pipe; it is

also heavier and more expensive than standard pipe.

◦ Double extra-strong pipe (XXS) has thicker walls and greater weight

than extra strong pipe.

- Design strengths for structural tees (i.e. WT) and double angles are with

respect to the axis indicated in the respective table.

- Design strengths for single angles are with respect to the z-axis.

To determine the design compression stress for a particular column, theoretically

both slenderness ratios (KL/r)x and (KL/r)y should be computed.

• For most steel sections used for columns, ry is much less than rx and only

(KL/r)y needs to be computed and used in the column analysis or design.

• If bracing of the weak axis is provided, both (KL/r)x and (KL/r)y must be

computed and the larger value used for determining the design stress φcFcr and

the allowable stress Fcr/Ωc for the member.

- Bracing may be accomplished by connecting braces or beams into the sides

of a column (e.g. such horizontal members, called girts).

- Steel columns may be built into masonry walls in such a manner that the

column is supported by the wall in the weaker direction.

5.28

Example Problems – AISC Column Design Tables

Example

Given: A column pinned at the top and bottom.

Service loads: PD = 130 kips, PL = 210 kips

KL = 10 feet

Find: a) Select the lightest W section (Fy = 50 ksi).

b) Select the lightest rectangular or square HSS section (Fy = 46 ksi).

c) Select the lightest round HSS section (Fy = 42 ksi).

d) Select the lightest pipe section (Fy = 35 ksi).

Solution

LRFD: Pu = 1.2 D + 1.6 L = 1.2 (130) + 1.6 (210) = 492 kips

ASD: Pa = D + L = 130 + 210 = 340 kips

a) Select the lightest W section (Fy = 50 ksi).

From AISC Manual Table 4-1

Possible selections (LRFD) Possible selections (ASD)

W8 x 48 (φcPn = 497 kips) W8 x 58 (Pn/Ωc = 403 kips)




Select W8 x 48 Select W10 x 49

b) Select the lightest rectangular or square HSS section (Fy = 46 ksi).

From AISC Manual Tables 4-3 and 4-4

Possible selections (LRFD)

HSS 9 x 7 x 5/8, Wt = 59.3 lb/ft (φcPn = 593 kips)







5.29











Select HSS 12 x 8 x 3/8 or HSS 10 x 10 x 3/8

Possible selections (ASD)

HSS 9 x 7 x 5/8, Wt = 59.3 lb/ft (Pn/Ωc = 395 kips)
















Select HSS 10 x 10 x 3/8

c) Select the lightest round HSS section (Fy = 42 ksi).



HSS 8.625 x 0.625, Wt = 53.5 lb/ft (φcPn = 498 kips)




5.30





Select HSS 16.000 x 0.312


HSS 10.000 x 0.625, Wt = 62.6 lb/ft (Pn/Ωc = 400 kips)







Select HSS 16.000 x 0.312

d) Select the lightest pipe section (Fy = 35 ksi).



XXS Pipe 8, Wt = 72.5 lb/ft (φcPn = 573 kips)

XS Pipe 12, Wt = 65.5 lb/ft (φcPn = 530 kips)

Select XS Pipe 12


XXS Pipe 8, Wt = 72.5 lb/ft (φcPn = 381 kips)

XS Pipe 12, Wt = 65.5 lb/ft (φcPn = 353 kips)

Select XS Pipe 12

5.31

The AISC Manual provides a method by which a section can be selected from its

tables when an axially loaded column is laterally restrained in its weak direction and

the unbraced lengths are different.

• For equal strengths about the x- and y-axes, the following relation must hold.

KxLx/rx = KyLy/ry

and KxLx = KyLy (rx/ry)

• The procedure is outlined as follows.

a. The designer enters the appropriate table with KyLy and selects a shape.

b. For that shape, the value rx/ry is taken from the table.

c. Multiply (rx/ry) by KyLy and compare with KxLx.

- If KyLy (rx/ry) > KxLx, then KyLy controls and the shape initially selected

is the correct one.

- If KyLy (rx/ry) < KxLx, then KxLx controls and the designer reenters the

table with a larger equivalent KyLy = KxLx/(rx/ry) to select the final

section.

5.32

Example Problem – Columns with Different Unbraced Lengths

Example

Given: A column with the following conditions.

Steel: Fy = 50 ksi


KxLx = 26’

KyLy = 13’

Find: Select the lightest W12 section using both the LRFD and ASD methods.

a) By trial-and-error, using AISC Table 4-22.

b) Using AISC Table 4-1.

Solution

a) By trial-and-error, using AISC Table 4-22.

LRFD

Pu = 1.2 D + 1.6 L = 1.2 (250) + 1.6 (400) = 940 kips

Assume KL/r = 50 and compute an estimated column area.

From AISC Table 4-22: φcFcr = 37.5 ksi

Required Ag = Pu/φcFcr = 940/37.5 = 25.1 in2

Select a trial section.

Try W12 x 87 (A = 25.6 in2, rx = 5.38”, ry = 3.07”)

Determine the critical slenderness ratio.

(KL/r)x = 26(12”/’)/5.38 = 57.99 (controls)

(KL/r)y = 13(12”/’)/3.07 = 50.81


φcPn = φcFcr Ag = 35.20 (25.6) = 901.1 kips < Pu = 940 kips NG

Try the next larger W12 section.

Try W12 x 96 (A = 28.2 in2, rx = 5.44”, ry = 3.09”)

5.33


(KL/r)x = 26(12”/’)/5.44 = 57.35 (controls)

(KL/r)y = 13(12”/’)/3.09 = 50.49


φcPn = φcFcr Ag = 35.40 (28.2) = 998.3 kips > Pu = 940 kips OK

Select W12 x 96

ASD

Pa = D + L = 250 + 400 = 650 kips


From AISC Table 4-22: Fcr/Ωc = 24.9 ksi

Required Ag = Pa/(Fcr/Ωc) = 650/24.9 = 26.1 in2


Try W12 x 96 (A = 28.2 in2, rx = 5.44”, ry = 3.09”)


(KL/r)x = 26.0 (12”/’)/5.44 = 57.35 (controls)

(KL/r)y = 13.0 (12”/’)/3.09 = 50.49

From AISC Table 4-22: Fcr/Ωc = 23.53

Pn/Ωc = FcrAg/Ωc = 23.53 (28.2) = 663.5 kips > Pa = 650 OK

Select W12 x 96

b) Using AISC Table 4-1.

LRFD

Pu = 1.2 D + 1.6 L = 1.2 (250) + 1.6 (400) = 940 kips

For KyLy = 13’: Try W12 x 87 (φcPn = 954 kips > 940 kips)

rx/ry = 1.75

Compare KyLy (rx/ry) with KxLx.

KyLy (rx/ry) = 13.0 (1.75) = 22.75’ < KxLx = 26’ Thus, KxLx controls.

Compute the equivalent KyLy.

Equivalent KyLy = KxLx/(rx/ry) = 26/1.75 = 14.86’

5.34

Reenter Table 4-1: Try W12 x 96

By interpolating, φcPn = 994.2 kips > Pu = 940 kips OK

Select W12 x 96

ASD

Pa = D + L = 250 + 400 = 650 kips

For KyLy = 13’: Try W12 x 96 (Pn/Ωc = 701 kips > 650 kips)

rx/ry = 1.76

Compare KyLy (rx/ry) with KxLx.

KyLy (rx/ry) = 13.0 (1.76) = 22.88’ < KxLx = 26’ Thus, KxLx controls.

Compute the equivalent KyLy.

Equivalent KyLy = KxLx/(rx/ry) = 26/1.76 = 14.77’

Reenter Table 4-1: Try W12 x 96

By interpolating, φcPn = 663.8 kips > Pa = 650 kips OK

Select W12 x 96

5.35

5.8 Torsional Buckling and Flexural Torsional Buckling

Up to this point, two limit states have been presented: flexural buckling and local

buckling.

• Two additional limit states for column behavior must also be considered:

torsional buckling and flexural-torsional buckling.

Axially loaded compression members can theoretically fail in four different ways.

1. Local buckling of the elements that form the cross section (e.g. flanges, webs).

2. Flexural buckling of the compression member.

- Flexural buckling (also called Euler buckling when elastic behavior occurs) is

the situation considered up to this point.

- The column design strengths given in the AISC column tables for W, M, S,

tube, and pipe sections are based on flexural buckling.

3. Torsional buckling of the compression member.

- Because torsional buckling can be very complex, it is desirable to prevent its

occurrence.

- Torsional buckling can be prevented by careful arrangement of the members

and by providing bracing to prevent lateral movement and twisting.

- If sufficient end supports and intermediate lateral bracing are provided,

flexural buckling will always control over torsional buckling.

4. Flexural-torsional buckling of the compression member.

Symmetrical members such as W sections are generally used as columns.

• The limit states of torsional buckling and flexural-torsional buckling are not

normally considered in the design of W-shape columns.

- Doubly symmetrical column members (such as W sections) are normally

subject only to local buckling, flexural buckling, and torsional buckling.

- The limit states of torsional buckling and flexural-torsional buckling

generally do not govern and when they do the critical load differs very little

from the strength determined from flexural buckling.

• If a torsional situation is encountered, it is best to use box sections or to make

box sections out of W sections by adding welded side plates.

Singly symmetric such as WT and non-symmetric shapes can fail through flexural,

torsional, or flexural-torsional buckling.

• Torsional problems can also be reduced by shortening the lengths of members

that are subject to torsion and through proper bracing of the column.

5.36

• Torsional problems can also be reduced by avoiding torsional loading.

- Torsion will not occur in such sections if the lines of action of the lateral

loads pass through the shear centers of the sections.

- The shear center is that point in the cross section of a member through

which the resultant of the transverse loads must pass so that no torsion will

occur.

- The figure at the right shows the

location of the shear center for

various shapes.

◦ The shear centers of commonly

used doubly symmetrical sections

occur at the centroids of these

sections.

◦ The shear center does not occur at

the centroid for sections such as

angles and channels.

The values given in the AISC column load tables for double-angle and structural

tee sections were computed for buckling about the weaker of the x- or y-axis and

for flexural-torsional buckling.

• For a singly symmetrical section such as a tee or double angle, Euler buckling

may occur about the x- or y-axis.

• For equal-leg single angles, Euler buckling may occur about the z-axis.

• For all these sections, flexural-torsional buckling is definitely a possibility and

may control.

- Flexural-torsional buckling will always control for unequal-leg single angle

columns.

Section E4 of the AISC Specification provides a method to determine the nominal

compressive strength Pn based on torsional or flexural-torsional buckling of

members without slender elements.

• Part (a) is applicable specifically to double-angle and tee-shaped compression

members used as columns.

• Part (b) of the section presents a general method applicable to all shapes.

Using the AISC Specification, there are four steps involved to determine the

nominal compressive strength Pn based on flexural buckling, torsional buckling, or

flexural-torsional buckling of members without slender elements.

5.37

1. Determine the flexural buckling strength Fcr of the member for its x-axis using

AISC Equations E3-4, E3-2 or E3-3 as applicable, and the nominal compressive

strength Pn using Equation E3-1.

2. Determine the flexural buckling strength Fcr of the member for its y-axis using

AISC Equations E3-4, E3-2 or E3-3 as applicable, and the nominal compressive

strength Pn using Equation E3-1.

3. Determine the flexural torsional buckling strength Fcr of the member for its y-

axis using AISC Equations E4-2 through E4-11 as applicable, and the nominal

compressive strength Pn using Equation E4-1.

4. Select the smallest Pn value determined in the preceding three steps.

An additional factor in determining strength based on limit states of torsional

buckling and flexural-torsional buckling is the torsional effective length factor Kz.

• The Commentary recommends that conservatively Kz = 1.0 and provides several

other possibilities if greater accuracy is desired.

5.38

Example Problem - Flexural-Torsional Buckling of Compression Members

Example

Given: Column consisting of a WT10.5 x 66


KLx = 25 feet

KLy = KLz = 20 feet

The bracing is not adequate to resist torsion.

Find: The nominal compressive strength Pn.

Solution

WT10.5 x 66

A = 19.4 in2, tf = 1.04”, bf/2tf = 6.01, d/tw = 16.8, Ix = 181 in4, rx = 3.06”, Iy = 166 in4,

ry = 2.93”, y = 2.33”, J = 5.62 in4, G = 11,200 ksi

(Note: G is the shear modulus listed in Specification E4.)

Check for slender elements (AISC Table B4.1a)

Flange (Case 1): bf/2tf = 6.01 < 0.56(E/Fy)1/2 = 0.56(29,000/50)1/2 = 13.49

The flange is not a slender element.

Stem (Case 4): d/tw = 16.8 < 0.75(E/Fy)1/2 = 0.75(29,000/50)1/2 = 18.06

The stem is not a slender element.

a) Determine the flexural buckling strength Fcr for the x-axis (based on Section

E3 of the AISC Specification).

(KL/r)x = 25(12”/’)/3.06 = 98.04

Fex = π2E/(KL/r)x2 = π2 (29,000)/(98.04)2 = 29.78 AISC Equation E3-4

(KL/r)x = 98.04 < 4.71 (E/Fy)1/2

= 4.71 (29,000/50)1/2 = 113.43

Use Equation E3-2 for Fcrx.

Fcrx = (0.658 Fy/Fe) Fy = (0.658 50/29.78) 50 = 24.76 ksi AISC Equation E3-2

The nominal strength Pn for flexural buckling about the x-axis is

Pn = Fcrx Ag = 24.76 (19.4) = 480.3 kips AISC Equation E3-1

5.39

b) Determine the flexural buckling strength Fcr for the y-axis (based on Section

E3 of the AISC Specification).

(KL/r)y = 20(12”/’)/2.93 = 81.91

Fey = π2E/(KL/r)y2 = π2 (29,000)/(81.91)2 = 42.66 AISC Equation E3-4

(KL/r)y = 81.91 < 4.71 (E/Fy)1/2

= 4.71 (29,000/50)1/2 = 113.43

Use Equation E3-2 for Fcry.

Fcry = (0.658 Fy/Fe) Fy = (0.658 50/42.66) 50 = 30.61 ksi AISC Equation E3-2

The nominal strength Pn for flexural buckling about the y-axis is

Pn = Fcry Ag = 30.61 (19.4) = 593.8 kips AISC Equation E3-1

c) Determine the flexural torsional buckling strength Fcr of the member about the

y-axis (based on Section E4 of the AISC Specification).

Use Equation E4-2 for Fcr.

• The following values need to be determined: Fcry, Fcrz, and H.

Fcry = 30.61 ksi (previously determined)

Use Equation E4-3 to determine Fcrz.

xo = 0 Note: xo = x-coordinate of shear center with respect to the

centroid; the shear center of the tee is located at the intersection

of the stem and the flange.

yo = y - tf/2 = 2.33 – 1.04/2 = 1.81”

Note: yo = y-coordinate of shear center with respect to the centroid.

r o2 = xo

2 + yo2 + (Ix + Iy)/Ag AISC Equation E4-11

= 02 + 1.812 + (181 + 166)/19.4 = 21.16 in2

H = 1 – (xo2 + yo

2)/ r o2 AISC Equation E4-10

= 1 - (02 + 1.812)/21.16 = 1 - 0.155 = 0.845

Fcrz = (GJ/Ag r o2) = 11,200 (5.62)/19.4 (21.16) = 153.33 ksi AISC Eq. E4-3

For tee shaped compression members (ref. Commentary Table C-E4.1):

Fcr = [(Fcry + Fcrz)/2H] {1 – [1 – 4 Fcry Fcrz H/(Fcry + Fcrz)2]1/2 } AISC Eq. E4-2

= [(30.61 + 153.33)/2(0.845)] x

{1 – [1 – 4 (30.61)(153.33)(0.845)/(30.61 + 153.33)2]1/2 }

Fcr = 108.84 (0.271) = 29.50 ksi

5.40

The nominal strength Pn for flexural torsional buckling about the y-axis is

Pn = Fcr Ag = 29.50 (19.4) = 572.3 kips AISC Equation E4-1

d) The smallest of the Pn values determined in a), b), and c) above is the nominal

strength.

Flexural torsional buckling strength for the x-axis controls.

Thus, Pn = 480.3 kips

LRFD design strength (φ = 0.90): φcPn = 0.90 (480.3) = 432.3 kips

ASD allowable strength (Ωc = 1.67): Pn/Ωc = 480.3/1.67 = 287.6 kips

These values compare favorably with the values listed in AISC Table 4-7.

• Flexural buckling strength about the x-axis: Pn = 480.3 kips



From AISC Table 4-7 (KL = 25’): φcPn = 432.5 kips (by interpolation)

Pn/Ωc = 288.0 kips (by interpolation)

• Flexural torsional buckling strength about the y-axis: Pn = 572.3 kips



From AISC Table 4-7 (KL = 20’): φcPn = 515 kips

Pn/Ωc = 343 kips

5.41

5.9 Single-Angle Compression Members

The AISC has long been concerned about the problems involved in loading single-

angle compression members concentrically.

• Loads can be applied concentrically if the ends are milled and if the loads are

applied through bearing plates.

• In practice, single-angle columns are often used in such a manner that large

eccentricities of load applications are present.

- As a result, it is easy to under-design such members.

• Table 4-11 of the AISC Manual provides calculated strengths of concentrically

loaded single angles.

- Values are based on the slenderness ratio KL/rz.

Section E5 of the AISC Specification provides a special specification for the

design of single-angle compression members.

• For single angles with a width-thickness ratio b/t > 20, Section E4 shall be used.

- b = the angle leg length

- t = the angle thickness

• The effects of eccentricity on single angle members may be neglected when

evaluated as axially loaded compression members using one of the effective

slenderness ratios specified in Section E5(a) or E5(b), provided that:

1. Members are loaded at the ends in compression through the same one leg;

2. Members are attached by welding or by connections with a minimum of two

bolts; and

3. There are no intermediate transverse loads.

• Writers of the specification assume that connections to one leg of an angle

provide considerable resistance to bending about the y-axis of that angle (or

the axis that is perpendicular to the connected leg).

- The angle is assumed to bend and buckle about the x-axis of the member and

attention is given to the (L/r)x ratio.

To account for the eccentric loading of single angles, larger (L/r)x ratios for

various situations are provided by AISC Equations E5-1 to E5-4 and are used to

obtain the design stresses.

• Two cases are given for these provisions.

a) For equal-leg angles or unequal-leg angles connected through the longer leg

that are individual members or are web members of planar trusses with

5.42

adjacent web members attached to the same side of the gusset plate or

chord:

When L/rx ≤ 80: KL/r = 72 + 0.75(L/rx) AISC Equation E5-1

When L/rx > 80: KL/r = 32 + 1.25(L/rx) ≤ 200 AISC Equation E5-2

For unequal-leg angles with leg length ratios (bl/bs) less than 1.7 and

connected through the shorter leg, KL/r from Equations E5-1 and E5-2 shall

be increased by adding 4[(bl/bs)2 – 1], but KL/r of the members shall not be

less than 0.95 (L/rz).

where

bl = the length of the longer leg (inches)

bs = the length of the shorter leg (inches)

b) For equal-leg angles or unequal-leg angles connected through the longer leg

that are web members of box or space trusses with adjacent web members

attached to the same side of the gusset plate or chord:

When L/rx ≤ 75: KL/r = 60 + 0.8(L/rx) AISC Equation E5-3

When L/rx > 75: KL/r = 45 + (L/rx) ≤ 200 AISC Equation E5-4

For unequal-leg angles with leg length ratios (bl/bs) less than 1.7 and

connected through the shorter leg, KL/r from Equations E5-3 and E5-4 shall

be increased by adding 6[(bl/bs)2 – 1], but KL/r of the members shall not be

less than 0.82 (L/rz).

where

bl = the length of the longer leg (inches)

bs = the length of the shorter leg (inches)

Table 4-12 of the AISC Manual provides design values for angles eccentrically

loaded based on the following assumptions (ref. AISC Manual, p. 4-9).

• The long leg of the angle is assumed to be attached to a gusset plate with a

thickness of 1.5t.

• The tabulated values assume a load placed at the mid-width of the long leg of

the angle at a distance of 0.75t from the face of this leg (ref. Figure 4-4,

p. 4-9 of the AISC Manual).

• The effective length KL is assumed to be the same for all axes.

• The table considers the combined bending stresses at the heel and ends of

each leg of the angle produced by axial compression plus biaxial bending

moments about the principal axes using AISC Specification Equation H2-1.

5.43

Example Problem – Single-Angle Compression Member

Example

Given: A 10-foot long, L8 x 8 x ¾ with simple end connections used in a planar

truss. Other web members meeting at the ends of this member are

connected on the same side of the gusset plate.

Steel: A36

Find: Determine φcPn and Pn/Ωc.

Solution

L8 x 8 x ¾ (A = 11.5 in2, rx = 2.46”)

L/rx = 10(12”/’)/2.46 = 48.78 < 80

AISC Equation E5-1: KL/r = 72 + 0.75(L/rx) = 72 + 0.75(48.78) = 108.59

LRFD

From AISC Table 4-22 (KL/r = 108.59, Fy = 36 ksi): φcFcr = 17.38 ksi

φcPn = φcFcr A = 17.38 (11.5) = 199.9 kips

Compare with AISC Table 4-12 (KL = 10’): φcPn = 163 kips

ASD

From AISC Table 4-22 (KL/r = 108.59, Fy = 36 ksi): Fcr/Ωc = 11.58 ksi

Pn/Ωc = (Fcr/Ωc) A = 11.58 (11.5) = 133.2 kips

Compare with AISC Table 4-12: Pn/Ωc = 107 kips

5.44

5.10 Built-Up Members

Compression members may be constructed with two or more shapes built up into a

single member (ref. Figures 5.2h through n, p. 114 of the textbook).

• Built-up members may consist of parts in contact with each other (e.g. cover-

plated sections).

• Built-up members may consist of parts in near contact with each other (e.g.

pairs of angles separated by end connections or gusset plates).

• Built-up members may consist of parts that are spread well apart (e.g. plates

with pairs of channels, or four angles).

Built-up Columns Whose Components are in Contact with Each Other

Several requirements concerning built-up columns are presented in Section E6.2 of

the AISC Specification.

• When built-up columns consist of different components that are in contact with

each other and are bearing on base plates or milled surfaces, the components

must be connected at the ends with bolts or welds.

- If welds are used, the weld lengths must at least equal the maximum width

of the member.

- If bolts are used, the bolts may not be spaced longitudinally more than four

bolt diameters on center.

◦ The connection must extend a distance at least equal to 1½ times the

maximum width of the member.

The design strength of a built-up column is reduced if the spacing of the

connectors is such that one of the components of the column can buckle before the

whole column buckles.

When the component of a built-up column consists of an outside plate, Section E6.2

of the AISC Specification provides specific maximum spacing for fastening.

• If intermittent welds are used along the edges of the components, then the

maximum spacing of the welds may not be greater than the thickness of the

thinner plate times 0.75 (E/Fy)1/2, nor greater than 12”.

• If bolts are used along the edges of the components, then the maximum spacing

of the bolts may not be greater than the thickness of the thinner plate times

0.75 (E/Fy)1/2, nor greater than 12”.

• If the fasteners are staggered, then the maximum spacing along each gage line

shall not exceed the thickness of the thinner plate times 1.12 (E/Fy)1/2 nor 18”.

5.45

If compression members consisting of two or more shapes are used, these

members must meet the following requirement.

• The shapes must be connected at intervals such that the effective slenderness

ratio Ka/ri of each of the component shapes between the connectors is not

larger than ¾ times the governing or controlling slenderness ratio of the whole

built-up member.

where

a = distance between connectors

ri = the least radius of gyration of an individual component of the column

• The end connections must be made using welds or using slip-critical bolts with

Class A or B faying surfaces (ref. AISC Specification Section J3.8).

- Faying surface is that surface of an object which comes in contact with

another object to which it is fastened (e.g. plates, angle irons, etc).

◦ Class A faying surface is an unpainted clean mill scale steel surface or a

surface with Class A coatings on blast-cleaned steel and hot dipped

galvanized and roughened surfaces.

◦ Class B faying surface is an unpainted blast-cleaned steel surface or

surface with Class B coatings on blast-cleaned steel.

• The design strength of compression members built-up from two or more shapes

in contact with each other is determined with the usual applicable AISC

Equations E3-2, E3-3, and E3-4, with one exception.

- If the column tends to buckle in such a manner that relative deformations in

the different parts cause shear forces in the connectors between the parts,

it is necessary to modify the KL/r value for that axis of buckling (per AISC

Specification Section E6).

a. For intermediate connectors that are snug-tight bolted:

(KL/r)m = [(KL/r)o2 + (a/ri)2]1/2 AISC Equation E6-1

b. For intermediate connectors that are welded or are connected by means

of pretensioned bolts, as required for slip-critical joints:

i) When a/ri ≤ 40

(KL/r)m = (KL/r)o AISC Equation E6-2a

ii) When a/ri > 40

(KL/r)m = [(KL/r)o2 + (Kia/ri)2]1/2 AISC Equation E6-2b

where

(KL/r)m = modified slenderness ratio of the built-up member

5.46

(KL/r)o = slenderness ratio of the built-up member acting as a unit

in the buckling direction being considered

Ki = 0.50 for angles back-to-back

= 0.75 for channels back-to-back

= 0.86 for all other cases

a = distance between connectors (inches)

ri = minimum radius of gyration of individual component (inches)

5.47

Example Problem – Built-Up Columns

Example

Given: Column section shown consisting of a

W12 x 120 plus cover plates.

Steel: Fy = 50 ksi


KL = 14 feet

Design cover plates to be snug-tight bolted at 6” spacings.

Find: Design the column.

Solution:

W12 x 120 (A = 35.2 in2, d = 13.1”, bf = 12.3”, Ix = 1070 in4, Iy = 345 in4)

LRFD

Pu = 1.2 D + 1.6 L = 1.2 (750) + 1.6 (1000) = 2500 kips

Assume KL/r = 50 and determine the area of the plates.


Required A = 2500/37.50 = 66.67 in2

Area of the plates: 66.67 – 35.2 = 31.47 in2 (15.74 in2 each)

Try plates: PL 1 x 16 for each flange

Compute the area for the built-up section.

A = 35.2 + 2(1) (16) = 67.2 in2

Determine the critical slenderness ratio of the built-up section.

Ix = 1070 + 2 (16) (13.1/2 + 1.0/2)2 = 2660 in4

rx = (Ix/A)1/2 = (2660/67.2)1/2 = 6.29”

(KL/r)x = 14(12”/’)/6.29 = 26.71

Iy = 345 + 2 (1) (16)3/12 = 1027.7 in4

ry = (Iy/A)1/2 = (1027.7/67.2)1/2 = 3.91”

(KL/r)y = 14(12”/’)/3.91 = 42.97 (controls)

5.48

Compute the modified slenderness ratio for buckling about the x-axis of the built-

up section since this buckling mode produces shear forces in the connectors

between the individual shapes.

Slenderness ratio for each plate:

ri = (I/A)1/2 = {[16(1)3/12]/16(1)}1/2 = 0.289”

a = 6” (the distance between connectors)

a/ri = 6/0.289 = 20.76

The modified slenderness ratio for intermediate connectors that are snug-tight

bolted is computed as follows:

(KL/r)m = [(KL/r)o2 + (a/ri)2] 1/2 AISC Equation E6-1

= [(26.71)2 + (20.76)2]1/2 = 33.83 < (KL/r)y = 42.97

(KL/r)m does not control

Check the slenderness ratio of the plates.

a/ri = 20.76 < ¾ (KL/r)y = 0.75(42.97) = 32.23 OK

From AISC Table 4-22 (Fy = 50 ksi): φcFcr = 39.31 ksi for KL/r = 42.97

φcPn = φcFcr A = 39.31 (67.2) = 2641.6 kips > 2500 kips OK

ASD

Pa = D + L = 750 + 1000 = 1750 kips

Assume KL/r = 50 and determine the area of the plates.

From AISC Table 4-22: Fcr/Ωc = 24.9 ksi

Required A = 1750/24.9 = 70.28 in2

Area of the plates: 70.28 – 35.2 = 35.08 in2 (17.54 in2 each)

Try plates: PL 1 x 16 for each flange

Compute the area for the built-up section.

A = 35.2 + 2(1) (16) = 67.2 in2

Determine the critical slenderness ratio of the built-up section.

Ix = 1070 + 2 (16) (13.1/2 + 1.0/2)2 = 2660 in4

rx = (Ix/A)1/2 = (2660/67.2)1/2 = 6.29”

(KL/r)x = 14(12”/’)/6.29 = 26.71

Iy = 345 + 2 (1) (16)3/12 = 1027.7 in4

ry = (Iy/A)1/2 = (1027.7/67.2)1/2 = 3.91”

(KL/r)y = 14(12”/’)/3.91 = 42.97 (controls)

5.49

Compute the modified slenderness ratio for buckling about the x-axis of the built-

up section since this buckling mode produces shear forces in the connectors

between the individual shapes.

Slenderness ratio for each plate:

ri = (I/A)1/2 = {[16(1)3/12]/16(1)}1/2 = 0.289”

a = 6” (the distance between connectors)

a/ri = 6/0.289 = 20.76

The modified slenderness ratio for intermediate connectors that are snug-tight

bolted is computed as follows:

(KL/r)m = [(KL/r)o2 + (a/ri)2] 1/2 AISC Equation E6-1

= [(26.71)2 + (20.76)2]1/2 = 33.83 < (KL/r)y = 42.97

(KL/r)m does not control

Check the slenderness ratio of the plates.

a/ri = 20.76 < ¾ (KL/r)y = 0.75(42.97) = 32.23 OK

From AISC Table 4-22 (Fy = 50 ksi): Fcr/Ωc = 26.20 ksi for KL/r = 42.97

Pn/Ωc = (Fcr/Ωc) A = 26.20 (67.2) = 1760.6 kips > 1750 kips OK

5.50

Built-up Columns with Components Not in Contact with Each Other

The parts of such members need to be connected or laced together across their

open sides.

• The open sides of compression members that are built up from plates or shapes

may be connected together with continuous cover plates featuring perforated

holes for access purposes, or they may be connected together with lacing and

tie plates.

- The perforated cover plates and the lacing hold the various parts parallel

and the correct distance apart, and equalize the stress distribution between

the various parts.

- Each part tends to buckle individually unless they are tied together to act as

a unit in supporting the load.

Continuous Cover Plates Perforated with Access Holes

Section E6.2 of the AISC Specification prescribes the dimensional requirements if

continuous cover plates perforated with access holes are used to tie the members

together.

1) The plates must comply with the limiting width-thickness ratios specified for

compression elements in Section B4 of the AISC Specification.

2) The ratio of the access hole length (in the direction of stress) to the hole

width may not exceed 2.

3) The clear distance between holes in the direction of stress may not be less

than the transverse distance between the nearest lines of connecting fasteners

or weld.

4) The periphery of the holes at all points must have a radius no less than 1½”.

Stress concentrations and secondary bending stresses are usually neglected in the

cover plates, but lateral shearing forces must be checked.

The unsupported width of the cover plates at access holes is assumed to

contribute to the design strength of the member if the conditions as to sizes and

width-thickness ratios described in Section E6.2 of the AISC Specification are

met (as stated above).

Perforated cover plates for built-up members are attractive to many designers

because of the following advantages.

1. They are easily fabricated with modern gas cutting methods.

5.51

2. Some specifications permit the net areas of the cover plates to be included in

the effective section of the main members, provided that the holes are made in

accordance with the AISC requirements.

3. Painting of the members is simplified, compared with painting of ordinary lacing

bars.

Lacing and Tie Plates

As an alternative to perforated cover plates, lacing with tie plates is permitted for

built-up members.

Dimensions and details for lacing with tie plates in built-up members are controlled

by the requirements of Section E6 of the AISC Specification.

• Tie plates shall be as near the ends as practical.

• End tie plates shall have a length (i.e. the distance along the member) of not

less than the distance between the lines of fasteners or welds connecting them

to the components of the members.

• Intermediate tie plates shall have a length not less than one-half the distance

between the lines of fasteners or welds connecting them to the components of

the member.

• In bolted construction, tie plates shall have a width (between the members) of

not less than the distance between the lines of fasteners plus adequate edge

distance.

• In welded construction, tie plates shall have a width (between the members) of

the distance between the lines of welds.

• Tie plates shall have a thickness at least equal to 1/50 of the distance between

lines of welds or fasteners connecting them to the segments of the members.

• In welded construction, the welding on each line connecting a tie plate shall

total not less than one-third of the length of the plate.

• In bolted construction, the spacing in the direction of stress in tie plates shall

be not more than six bolt diameters and the tie plates shall be connected to

each segment by at least three fasteners.

• The inclination of lacing bars to the axis of the member shall preferably be not

less than 60° for single lacing and 45° for double lacing.

• When the distance between the lines of weld or fasteners is more than 15”, the

lacing shall preferably be double or made of angles.

5.52

Additional details for lacing in built-up members are also controlled by the

requirements of Section E6 of the AISC Specification.

• Lacing may consist of flat bars, angles, channels, or other rolled sections.

• Lacing pieces must be spaced so that the individual parts being connected will

not have L/r values of the member between the connections greater than

three-fourths times the governing slenderness ratio for the whole built-up

member.

• Lacing shall be proportioned to provide a shearing strength normal to the axis

of the member equal to 2% of the available compressive strength of the

member.

• The AISC column formulas are used to design the lacing in the usual manner.

- Slenderness ratios are limited to 140 for single lacing and 200 for double

lacing.

5.53

Example Problem – Built-Up Columns with Components Not in Contact with Each

Other

Example

Given: Column shown consisting of a pair of

12” standard channels.

Steel: Fy = 50 ksi (channels)

Fy = 36 ksi (flat bar for lacing)


Bolt diameter: 3/4”

Find: Select the channels. Consider both

LRFD and ASD. Design the bolted single

lacing.

Solution

LRFD

Pu = 1.2 D + 1.6 L = 1.2 (100) + 1.6 (300) = 600 kips


From AISC Table 4-22 (Fy = 50 ksi): φcFcr = 37.50 ksi

Required A = 600/37.50 = 16.00 in2


Try 2 - C12 x 30 (For each channel: A = 8.81 in2, Ix = 162.0 in4, Iy = 5.12 in4,

x = 0.674”, ry = 0.762”, workable gage = 1.75”)


Ix = 2 (162.0) = 324.0 in4

rx = (Ix/A)1/2 = [324.0/2(8.81)]1/2 = 4.29”

(KL/r)x = 1(20)(12”/’)/4.29 = 55.94 (controls)

Iy = 2 [5.12 + 8.81 (6.00 – 0.674)2] = 510.0 in4

ry = (Iy/A)1/2 = [510.0/2(8.81)]1/2 = 5.38”

(KL/r)y = 1(20)(12”/’)/5.38 = 44.61

From AISC Table 4-22 (Fy = 50 ksi): φcFcr = 35.82 ksi for (KL/r)x = 55.94

φcPn = φcFcr A = 35.82 (2)(8.81) = 631.1 kips > Pu = 600 kips OK

5.54

ASD

Pa = D + L = 100 + 300 = 400 kips


From AISC Table 4-22 (Fy = 50 ksi): Fcr/Ωc = 24.90 ksi

Required A = 400/24.90 = 16.06 in2


Try 2 - C12 x 30 (For each channel: A = 8.81 in2, Ix = 162.0 in4, Iy = 5.12 in4,

x = 0.674”, ry = 0.762”, workable gage = 1.75”)


Ix = 2 (162.0) = 324.0 in4

rx = (Ix/A)1/2 = [324.0/2(8.81)]1/2 = 4.29”

(KL/r)x = 1(20)(12”/’)/4.29 = 55.94 (controls)

Iy = 2 [5.12 + 8.81 (6.00 – 0.674)2] = 510.0 in4

ry = (Iy/A)1/2 = [510.0/2(8.81)]1/2 = 5.38”

(KL/r)y = 1(20)(12”/’)/5.38 = 44.61

From AISC Table 4-22 (Fy = 50 ksi): Fcr/Ωc = 23.81 ksi for (KL/r)x = 55.94

Pn/Ωc = Fcr/Ωc A = 23.81 (2)(8.81) = 419.5 kips > Pa = 400 kips OK

Check the width-thickness ratios of the channels.

C12 x 30 (d = 12.0”, bf = 3.17”, tf = 0.501”, tw = 0.510”, k = 1.125”)

Flanges

AISC Table B4.1a (Case 1):

Check b/t ≤ 0.56 (E/Fy)1/2

b/t = bf/tf = 3.17/0.501 = 6.33 < 0.56(29,000/50)1/2 = 13.49 OK

Webs

AISC Table B4.1a (Case 5):

Check h/tw ≤ 1.49 (E/Fy)1/2

h/tw = (d – 2k)/tw = [12.0 – 2(1.125)]/0.510 = 19.12

< 1.49(29,000/50)1/2 = 35.88 OK

5.55

Design Bolted Single Lacing

Assume that lacing bars (flat bars) are inclined at 60° with the axis of the

channels.

Determine the length of the column channels between connections.

Distance between connections = 12 – 2g = 12.0 – 2(1.75) = 8.50”

where

g = workable gage (listed in AISC Table 1-5)

Length = (8.5”/cos 30°) = 9.8”

(Note: Since the lacing forms an equilateral triangle, the hypotenuse of the 30-60-90 triangle is

also equal to the length of channel between connections.)

Check the slenderness ratio of the column channels between the lacing connections.

Channel L/r < ¾ (governing KL/r for the built-up member)

Channel L/r = 9.8/0.762 = 12.86 < ¾ (55.94) = 41.95 OK

LRFD

Vu = 2% of the design compression strength of the built-up member

Design compression strength: φcPn = φcFcr A = 631.1 kips (previously determined)

Vu = 0.02 φcPn = 0.02 (631.1) = 12.62 kips

Determine the shear force on each of the

two planes of the single lacing.

½ Vu = 0.5 (12.62) = 6.31 kips

Determine the force in each lacing bar.

(9.8/8.5) (6.31) = 7.28 kips

(Note: The forces in the bars and the bar dimensions are geometrically similar.)

5.56

Determine the geometric properties of a flat bar.

Imin = bt3/12 and A = bt

Thus, rmin = [(bt3/12)/bt]1/2 = 0.289 t

Design the lacing bar.

Assume L/r = 140

L/rmin = 9.8/0.289 t = 140 t = 0.242” (Try ¼” flat bar.)

Calculate the actual slenderness ratio for a ¼” lacing bar.

L/r = L/0.289 t = 9.8/0.289 (0.25) = 135.6 < 140 OK

Determine the lacing bar width.

From AISC Table 4-22 (Fy = 36 ksi): φcFcr = 12.28 ksi for KL/r = 135.6

Required A = Force in each bar/φcFcr = 7.28/12.28 = 0.593 in2

Required width = A/t = 0.593/0.25 = 2.37”

Minimum edge distance for ¾” bolt = 1¼” AISC Table J3.4

Minimum width (for ¾” bolts) = 2(1¼”) = 2.50” > 2.37” (Use 2.50”)

Determine the minimum lacing bar length:

Length = 9.8 + 2(1¼”) = 12.3” (Say 14”)

Use ¼” x 2½” x 14” bars for the lacing.

ASD

Va = 2% of the design compression strength of the built-up member

Design compression strength: Pn/Ωc = (Fcr/Ωc) A

= 419.5 kips (previously determined)

Va = 0.02 Pn/Ωc = 0.02 (419.5) = 8.39 kips

Determine the shear force on each of the two planes of the single lacing.

½ Va = 0.5 (8.39) = 4.20 kips

Determine the force in each lacing bar.

(9.8/8.5) (4.20) = 4.84 kips

(Note: The forces in the bars and the bar dimensions are geometrically similar.)

Determine the geometric properties of a flat bar.

Imin = bt3/12 and A = bt

Thus, rmin = [(bt3/12)/bt]1/2 = 0.289 t

5.57

Design the lacing bar.

Assume L/r = 140

L/rmin = 9.8/0.289 t = 140 t = 0.242” (Try ¼” flat bar.)

Calculate the actual slenderness ratio for a ¼” lacing bar.

L/r = L/0.289 t = 9.8/0.289 (0.25) = 135.6 < 140 OK

Determine the lacing bar width.

From AISC Table 4-22 (Fy = 36 ksi): Fcr/Ωc = 8.18 ksi for KL/r = 135.6

Required A = Force in each bar/(Fcr/Ωc) = 4.84/8.18 = 0.592 in2

Required width = A/t = 0.592/0.25 = 2.37”

Minimum edge distance for ¾” bolt = 1¼” AISC Table J3.4

Minimum width (for ¾” bolts) = 2(1¼”) = 2.50” > 2.37” (Use 2.50”)

Determine the minimum lacing bar length:

Length = 9.8 + 2(1¼”) = 12.3” (Say 14”)

Use ¼” x 2½” x 14” bars for the lacing.

Design the end tie plates

Minimum length = distance between fasteners = 8.5”

Minimum t = (1/50) x distance between fasteners

= (1/50) 8.5 = 0.17” (Use 3/16”)

Minimum width (for ¾” bolts) = 8.5 + 2(1¼”) = 11” (Use 12”)

Use 3/16” x 8½” x 12” end tie plates.

5.58

5.11 Column Base Plates

When a steel column is supported by a footing, it is necessary for the column load

to be spread over a sufficient area to keep the footing from being overstressed.

• Loads from steel columns are transferred through a steel base plate to a fairly

large area of the footing below.

Methods for connecting base plates to columns include the following.

• Base plates for steel columns can be welded directly to the columns.

• Base plates for steel columns may be fastened to the columns by means of some

type of bolted or welded angles.

• A minimum of four anchor bolts are used to anchor the base plate to the

footing.

Practical considerations for the design and installation of base plates include the

following.

• The lengths and widths of column base plates are usually selected in multiples

of inches.

• The thicknesses of base plates are usually selected in multiples of 1/8” up to

1.25”, and in multiples of 1/4” for plates 1.25” and thicker.

• To make certain that column loads are spread uniformly over the base plate, it

is essential to have good contact between the column and the base plate.

- Surface preparation of these plates is governed by the AISC Specification

Section M2.8.

• At least one hole should be provided near the center of large area base plates

for placing grout.

- These holes permit more even placement of grout under the plates, which

will tend to prevent air pockets.

The basis for designing column base plates is outlined as follows.

• The column is assumed to apply a total load to the base plate equal to Pu (for

LRFD) or Pa (for ASD).

• The load is assumed to be transmitted uniformly through the plate to the

footing below with a pressure equal to Pu/A or Pa/A where A is the area of the

base plate.

5.59

• The footing will push back with an equal pressure.

- This pressure will tend to curl

up the cantilevered parts of

the base plate outside of the

column.

- This pressure will also tend to

push up the base plate between

the flanges of the column.

• The AISC Manual (ref. p. 14-5)

suggests that the maximum

moments in a base plate occur at

distances 0.80bf and 0.95d apart.

- The bending moment can be

calculated at each of these

sections, and the larger value is

used to determine the required

plate thickness.

- This method of analysis is

approximate, because the

actual plate stresses are

caused by a combination of

bending in two directions.

Plate Area

The design strength of the concrete in bearing beneath the base plate must at

least equal the load to be carried.

• When the base plate covers the entire area of concrete, the nominal bearing

strength of the concrete (Pp) is

Pp = 0.85 fc’ A1 AISC Equation J8-1

where

fc’ = the 28-day compression strength of the concrete

A1 = the area of the base plate (Note: The minimum A1 = bf d.)

= B x N

and, B = 0.80 bf + 2 n

N = 0.95 d + 2 m

5.60

• When the base plate covers less than the entire area of concrete, the nominal

bearing strength of the concrete (Pp) can then be determined by the following

equation.

Pp = (0.85 fc’ A1) (A2/A1)1/2 ≤ 1.7 fc’ A1 AISC Equation J8-2

• If the plate does not cover the full area of the concrete support, the concrete

under the plate, surrounded by concrete outside, will be somewhat stronger.

- For this situation AISC Specification permits the nominal strength 0.85fc’A1

to be increased by multiplying it by (A2/A1)1/2.

where

A2 = the maximum area of the portion of the supporting concrete that is

geometrically similar to and concentric with the loaded area

The quantity (A2/A1)1/2 is limited to a maximum value of 2.0.

The plate area can be determined based on the following equations.

• For LRFD (φc = 0.65):

Pu = φc Pp = φc (0.85 fc’ A1) (A2/A1)1/2

A1 = Pu/[φc (0.85 fc’)(A2/A1)1/2]

• For ASD (Ωc = 2.31):

Pa = Pp/Ωc = (0.85 fc’ A1) (A2/A1)1/2/Ωc

A1 = Pa Ωc/[(0.85 fc’) (A2/A1)1/2]

After the controlling value of A1 is determined, the plate dimensions B and N are

selected to the nearest 1” or 2” so that the values of m and n are approximately

equal.

• From a practical standpoint, designers will often use square base plates (i.e.

B = N) with anchor bolts arranged in a square pattern.

B = N = (A1)1/2

• Designers may use a rectangular base plate to keep the plate thickness to a

minimum.

- The values for “m” and “n” are set approximately equal so that the cantilever

moments in the two directions are approximately the same.

A1 = area of plate = B x N

Plate Thickness

To determine the required plate thickness, t, moments are taken in two directions

as though the plate were cantilevered out by the dimensions “m” and “n.”

5.61

• The moments in the two directions (assuming a 1” width of plate) are

determined by the following equations.

For LRFD: (Pu/BN) m (m/2) = Pum2/2BN or (Pu/BN) n (n/2) = Pun2/2BN

For ASD: (Pa/BN) m (m/2) = Pam2/2BN or (Pa/BN) n (n/2) = Pan2/2BN

Because of the possibility of a lightly loaded plate (with correspondingly small

areas and small thicknesses), a modified method is used to determine the required

plate thickness.

• The thickness of the plate is determined by the largest of m, n, or λn’.

- This largest value is ℓ, where ℓ = max (m, n, or λn’).

where

m = (N – 0.95 d)/2 Equation 14-2

n = (B - 0.80 bf)/2 Equation 14-3

n’ = ¼ (d bf)1/2 Equation 14-4

λ = 1.0 (conservative for all cases)

• Letting the largest value of m, n, or λn’ be referred to as ℓ, the largest moment

in the plate will equal

For LRFD: Pu ℓ2/2BN

For ASD: Pa ℓ2/2BN

• The required plate thickness can be determined from the following equations.

For LRFD: tmin = ℓ (2 Pu/0.9 Fy B N)1/2 Equation 14-7a

For ASD: tmin = ℓ (3.33 Pa/Fy B N)1/2 Equation 14-7b

5.62

Example Problems – Column Base Plates

Example

Given: Base plate for W12 x 65 column.

Column steel: Fy = 50 ksi

Base plate steel: A36 (Fy = 36 ksi)


Concrete compression strength: fc’ = 3 ksi

Footing dimensions: 9’ x 9’

Find: Design the column base plate.

Solution

W12 x 65 (d = 12.1”, bf = 12.0”)

LRFD

Pu = 1.2 D + 1.6 L = 1.2 (200) + 1.6 (300) = 720 kips

Determine the base plate area (φc = 0.65).

• The area of the supporting concrete is much greater than the base plate area,

such that (A2/A1)1/2 = 2.0.

A1 = Pu/[φc (0.85 fc’)(A2/A1)1/2] = 720/[0.65(0.85)(3.0)(2.0)] = 217.2 in2

• The base plate must be at least as large as the column.

bf d = 12.0 (12.1) = 145.2 in2 < A1 = 217.2 in2 OK

Determine the base plate dimensions.

• Simplify the base plate dimensions by making it square.

B = N = (217.2)1/2 = 14.74” (Say 15” x 15”)

Check the bearing strength of the concrete (φc = 0.65).

φc Pp = φc 0.85 fc’ A1 (A2/A1)1/2

= 0.65(0.85)(3.0)(15 x 15)(2.0) = 745.9 kips > Pu = 720 kips OK

Compute the required base plate thickness.

m = (N – 0.95d)/2 = [15 – 0.95(12.1)]/2 = 1.753”

n = (B - 0.80bf)/2 = [15 – 0.80(12.0)]/2 = 2.700”

5.63

n’ = ¼ (d bf)1/2 = ¼ [12.1 (12.0)]1/2 = 3.012”

ℓ = max (m, n, or λn’) = 3.012”

tmin = ℓ (2 Pu/0.9 Fy B N)1/2

= 3.012 [2(720)/0.90(36.0)(15)(15)]1/2 = 1.34” (Say 1.50”)

Use PL 1½” x 15” x 15” (A36 steel)

Alternative rectangular base plate design

• Optimize the base plate dimensions to make m and n approximately equal.

N x B = 217.2 = (0.95 d + 2Δ) (0.80 bf + 2Δ)

217.2 = [0.95(12.1) + 2Δ] [0.80(12.0) + 2Δ] = (11.495 + 2Δ)(9.60 + 2Δ)

217.2 = 110.352 + 42.19Δ + 4Δ2

4Δ2 + 42.19Δ – 106.85 = 0

• Using the quadratic equation, Δ = 2.110”

N = 0.95 d + 2 Δ = 0.95(12.1) + 2(2.110) = 15.72” (Say 16”)

B = 0.80 bf + 2 Δ = 0.80(12.0) + 2(2.110) = 13.82” (Say 14”)


φc Pp = φc 0.85 fc’ A1 (A2/A1)1/2

= 0.65(0.85)(3.0)(16 x 14)(2.0) = 742.6 kips > Pu = 720 kips OK


m = (N – 0.95d)/2 = [16 – 0.95(12.1)]/2 = 2.253”

n = (B - 0.80bf)/2 = [14 – 0.80(12.0)]/2 = 2.200”

n’ = ¼ (d bf)1/2 = ¼ [12.1 (12.0)]1/2 = 3.012”

ℓ = max (m, n, or λn’) = 3.012”

tmin = ℓ (2 Pu/0.9 Fy B N)1/2

= 3.012 [2(720)/0.90(36.0)(16)(14)]1/2 = 1.34” (Say 1.50”)

Use PL 1½” x 16” x 14” (A36 steel)

5.64

ASD

Pa = D + L = 200 + 300 = 500 kips

Determine the base plate area (Ωc = 2.31).


such that (A2/A1)1/2 = 2.0.

A1 = Pa Ωc/[(0.85 fc’) (A2/A1)1/2] = 500 (2.31)/[0.85(3.0)(2.0)] = 226.5 in2


bf d = 12.0 (12.1) = 145.2 in2 < 226.5 in2 OK

Determine the base plate dimensions.

• Simplify the plate by making it square.

B = N = (226.5)1/2 = 15.05” (Say 16” x 16”)

Check the bearing strength of the concrete (Ωc = 2.31).

Pp/Ωc = (1/Ωc) 0.85 fc’ A1 (A2/A1)1/2

= (1/2.31)(0.85)(3.0)(16 x 16)(2.0) = 565.2 kips > Pa = 500 kips OK


m = (N – 0.95d)/2 = [16 – 0.95(12.1)]/2 = 2.252”

n = (B - 0.80bf)/2 = [16 – 0.80(12.0)]/2 = 3.200”

n’ = ¼ (d bf)1/2 = ¼ [12.1 (12.0)]1/2 = 3.012”

ℓ = max (m, n, or λn’) = 3.20”

tmin = ℓ (3.33 Pa/Fy B N)1/2

= 3.20 [3.33(500)/(36.0)(16)(16)]1/2 = 1.36” (Say 1.50”)

Use PL 1½” x 16” x 16” (A36 steel)

Alternative rectangular base plate design

Optimize the base plate dimensions to make m and n approximately equal.

N x B = 245.1 = (0.95 d + 2Δ) (0.80 bf + 2Δ)

226.5 = [0.95(12.1) + 2Δ] [0.80(12.0) + 2Δ] = (11.495 + 2Δ)(9.60 + 2Δ)

226.5 = 110.352 + 42.19Δ + 4Δ2

4Δ2 + 42.19Δ – 116.15 = 0

• Using the quadratic equation, Δ = 2.266”

N = 0.95 d + 2 Δ = 0.95(12.1) + 2(2.266) = 16.03” (Say 17”)

B = 0.80 bf + 2 Δ = 0.80(12.0) + 2(2.266) = 14.13” (Say 15”)

5.65


Pp/Ωc = (1/Ωc) 0.85 fc’ A1 (A2/A1)1/2

= (1/2.31)(0.85)(3.0)(17 x 15)(2.0) = 563.0 kips > Pa = 500 kips OK


m = (N – 0.95d)/2 = [17 – 0.95(12.1)]/2 = 2.753”

n = (B - 0.80bf)/2 = [15 – 0.80(12.0)]/2 = 2.700”

n’ = ¼ (d bf)1/2 = ¼ [12.1 (12.0)]1/2 = 3.012”

ℓ = max (m, n, or λn’) = 3.012”

tmin = ℓ (3.33 Pa/Fy B N)1/2

= 3.012 [3.33(500)/(36.0)(17)(15)]1/2 = 1.283” (Say 1.50”)

Use PL 1½” x 17” x 15” (A36 steel)

5.66

Example

Given: Base plate for HSS 10 x 10 x 5/16 column.

Column steel: Fy = 46 ksi

Base plate steel: A36 (Fy = 36 ksi, Fu = 58 ksi)


Concrete compression strength: fc’ = 4 ksi

Spread footing dimensions: 9’ x 9’

Find: Design the column base plate.

Solution

LRFD

Pu = 1.2 D + 1.6 L = 1.2 (100) + 1.6 (150) = 360 kips

Determine the base plate area (φc = 0.65).


such that (A2/A1)1/2 = 2.0.

A1 = Pu/[φc (0.85 fc’)(A2/A1)1/2] = 360/[0.65(0.85)(4.0)(2.0)] = 81.45 in2


bf d = 10.0 (10.0) = 100.0 in2 > 81.45 in2

• Try a square base plate extending 4” from the face of the column in each

direction.

B = N = 18” (Say 18” x 18”)


φc Pp = φc 0.85 fc’ A1 (A2/A1)1/2

= 0.65(0.85)(4.0)(18 x 18)(2.0) = 1432.1 kips > 360 kips OK


m = n = (N – 0.95d)/2 = [18 – 0.95(10.0)]/2 = 4.25”

n’ = ¼ (d bf)1/2 = ¼ [10 (10)]1/2 = 2.50”

ℓ = max (m, n, or λn’) = 4.25”

tmin = ℓ (2 Pu/0.9 Fy B N)1/2

= 4.25 [2(360)/0.90(36.0)(18)(18)]1/2 = 1.11” (Say 1.25”)

Use PL 1¼” x 18” x 18” (A36 steel)

5.67

ASD

Pa = D + L = 100 + 150 = 250 kips

Determine the base plate area (Ωc = 2.31).


such that (A2/A1)1/2 = 2.0.

A1 = Pa Ωc/[(0.85 fc’) (A2/A1)1/2] = 250 (2.31)/[0.85(4.0)(2.0)] = 84.9 in2


bf d = 10.0 (10.0) = 100.0 in2 > 84.9 in2

• Try a square base plate extending 4” from the face of the column in each

direction.

B = N = 18” (Say 18” x 18”)


Pp/Ωc = (1/Ωc) 0.85 fc’ A1 (A2/A1)1/2

= (1/2.31)(0.85)(4.0)(18 x 18)(2.0) = 953.8 kips > 250 kips OK


m = n = (N – 0.95d)/2 = [18 – 0.95(10.0)]/2 = 4.25”

n’ = ¼ (d bf)1/2 = ¼ [10 (10)]1/2 = 2.50”

ℓ = max (m, n, or λn’) = 4.25”

tmin = ℓ (3.33 Pa/Fy B N)1/2

= 4.25 [3.33(250)/(36.0)(18)(18)]1/2 = 1.14” (Say 1.25”)

Use PL 1¼” x 18” x 18” (A36 steel)

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