COMPRESSION MEMBERSColumns are classified as being long, short or intermediate based on slenderness ratio KL/r. For each type of these column have different failure modes: 1- Long

Steel Design Misan University Fourth Year Engineering College Dr.Abbas Oda Dawood Civil Department

Lecture 3 ....... Page 1

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COMPRESSION MEMBERS

1. INTRODUCTION

Compression Members are Structural elements that are subjected to axial compressive

forces only, that is, the loads are applied along a longitudinal axis through the centroid of the

member.

The stress in the compression member cross-section can be calculated as

)1(A

Pf

where, f is assumed to be uniform over the entire cross-section. This ideal state is never

reached. The stress-state will be non-uniform due to:

1- Accidental eccentricity of loading with respect to the centroid

2- Member out-of –straightness (crookedness), or

3- Residual stresses in the member cross-section due to fabrication processes.

Accidental eccentricity and member out-of-straightness can cause bending moments in the

member. However, these are secondary and are usually ignored.

Residual stressed due to uneven cooling of standard sections after hot-rolling, and also

welding, can adversely affect the resistance of columns against buckling. As an example, in

an I-section, the outer tips of the flanges and the middle portion of the web cool more

quickly than the relatively thick portions at the intersection of the flanges and the web. The

result of this uneven cooling is that the areas cooled more quickly develop residual

compressive stresses, while the areas cooled more slowly develop residual tensile stresses.

The magnitude of the residual stresses can be as large as 10–15 ksi.

Bending moments cannot be neglected if they are acting on the member. Members with

axial compression and bending moment are called beam-columns.

The most common type of compression member in building and bridges is the column.

Column is a vertical member whose primary function is to support vertical loads. In many

cases these members are also resist bending, and in these cases the member is a beam-

column. Smaller compression members not classified as column are sometimes referred as

struts or posts. Thus Compression members are found as:



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1. Columns in buildings

2. Piers in bridges

3. Trusses ( Top chords)

4. Bracing members

Compression and tension members differ in the following ways:

1. Slender compression members can buckle.

2. In tension members, bolt holes reduce the effective cross-sectional area for carrying the

loads. In compression members, however, the bolts tend to fill the holes and the entire area

of the cross section is normally assumed to resist the loads.

In structural steel, the common shapes used for columns are wide flange shapes, round and

square hollow structural sections (HSS), and built-up sections. For truss members, double- or

single-angle shapes are used, as well as round and square HSS and WT-shapes

2. COLUMN BUCKLING

Consider the two axially loaded members shown in Figure below. In Figure, the column is

short enough that the failure mode is by compression crushing. This is called a short column.

For the longer column shown in Figure, the failure mode is buckling at the midspan of the

member. This is called a slender, or long, column. Intermediate columns fail by a

combination of buckling and yielding.



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)2()LK(

IEP

2

2

e

where,

Pe = Elastic critical buckling load (Euler load), lb.,

I = moment of inertia about axis of buckling

K = effective length factor based on end boundary conditions

L = Length of the column between brace points, in.

It should be noted that the critical buckling load given by Eq. (2) is independent of the

strength of the material (say, Fy, the yield stress).

Effective length factors are given in Table C-C 2.2 by the AISC manual as following.

Buckling occurs when a straight column subjected to

axial compression suddenly undergoes bending.

Consider a long slender compression member. If an

axial load P is applied and increased slowly, it will

ultimately reach a value Pe that will cause buckling of

the column. Pe is called the critical buckling load of the

column or Euler critical load.

The critical buckling load Pe for columns is

theoretically given by Equation (2)

KL is the distance between the points of zero

moment, or inflection points. The length KL is

known as the effective length of the column. The

dimensionless coefficient K is called the effective

length factor.



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It is convenient to rewrite Equation (2), and Knowing that I = Ar2as

)3()r/LK(

AE

)LK(

rAE

)LK(

IEP

2

2

2

22

2

2

e

Where

A = is the cross sectional area, r = is the radius of gyration with respect to the axis of buckling

KL/r = is the slenderness ratio and is the measure of a member's slenderness, with large

values corresponding to slender members.

If the critical load is divided by the cross-sectional area, the critical buckling stress or Euler

elastic critical buckling Fe is obtained:

)4()r/LK(

E

A

PF

2

2cr

e

If the column is not restricted to bend in a particular plane, it will buckle in a plane

perpendicular to the minor axis of the cross section. Hence, the moment of inertia and the

radius of gyration in equations 2 to 4 are with respect to the minor axis of the cross section

and the minimum moment of inertia and minimum radius of gyration of the cross section

should be used in these equations. Minor axis buckling usually governs for all doubly

symmetric cross-sections. However, for some cases, major (x) axis buckling can govern.

Major axis means axis about which it has greater moment of inertia (Ix > Iy)



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3. INELASTIC BUCKLING OF COLUMNS

Columns are classified as being long, short or intermediate based on slenderness ratio KL/r.

For each type of these column have different failure modes:

1- Long Columns: The Euler formula predicts very well the strength of long columns

where the axial stress remain below yielding stress. Such columns will buckle

elastically. This column fail by elastic buckling

2- Short Columns: For very short columns, the failure stress will equal yield stress and

no buckling will occur. The columns fall into this class are so short and have no

practical applications). This column fail by excessive yielding (compression crushing)

3- Intermediate Columns: For these columns some fiber will reach the yield stress and

some will not. The members will fail by both yielding and buckling, and their behavior

Example 1: Determine the buckling strength of a W 12 x 50 column. Its length is 20 ft. For minor axis buckling, it is pinned at both ends. For major buckling, is it pinned at one end and fixed at the other end.

Solution

According to Table C-C2.2 of the AISC Manual

- For pin-pin end conditions about the minor axis

Ky= 1.0 (recommended design value)

- For pin-fix end conditions about the major axis Kx

= 0.8 (recommended design value)

-According to the problem statement, the unsupported length for buckling about the major (x) axis = Lx = 20 ft. -The unsupported length for buckling about the minor (y) axis = Ly = 20 ft.

From AISC Manual, for W12 x 50: rx =5.18 in rx =1.96 in Ag = 14.6

1.3718.5

12*20*8.0

r

LK

x

xx

45.12296.1

12*20*0.1

r

LK

y

yy

Use larger Value (Minor (y) axis buckling governs)

45.122r

LK

r

LK

y

yy

kips7.278)45.122(

6.14*29000*

)r/LK(

AEP

axisyaboutbucklingforloadCritical

2

2

2

2

xe

Buckling strength of the column = Pe = 278.7 kips

major-axis buckling

Minor-axis buckling

The first step is visualize the problem

Cross section



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is said to be inelastic. Most columns fall into this range. This column fail by inelastic

buckling.

The variation of the critical stress fcr with the slenderness ratio KL/r is shown in Figure below:

Several other problems appear in the inelastic range.

- The member out-of-straightness has a significant influence on the buckling strength in the

inelastic region. It must be accounted for.

- The residual stresses in the member due to the fabrication process causes yielding in the

cross-section much before the uniform stress f reaches the yield stress Fy.

- The shape of the cross-section (W, C, etc.) also influences the buckling strength.

- In the inelastic range, the steel material can undergo strain hardening.

Euler formula Eq. (2) is valid only when the material everywhere in the cross-section is in the

elastic region. If the material goes inelastic then Equation (2) becomes useless and cannot be

used. Thus Euler equation cannot be used for short column due to its inelastic behavior.

Thus trial and error solutions must be used for inelastic buckling, for this reason, most design

standards including the AISC Specifications contain empirical formulas for inelastic columns.

4. AISC SPECIFICATIONS FOR COLUMN STRENGTH

The AISC specifications account for the elastic and inelastic buckling of columns including all

issues (member crookedness, residual stresses, accidental eccentricity etc.) mentioned

above. the AISC specification defines the design compressive strength of a column as

follows:

Eq. (4)



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unc PP

9.0

AFP

c

)5(gcrn

Pu = Sum of the factored loads, kips,

Pn = Nominal compressive strength, kips,

Fcr = Flexural buckling stress (see below), ksi, and

Ag = Gross cross-sectional area of the column, in.2.

The flexural buckling stress, Fcr, according to AISC specifications is determined as follows:

)6(F)658.0(F

F44.0FORF

E71.4

r

LKWhen

yF/F

cr

yey

ey

)7(F877.0F

F44.0FORF

E71.4

r

LKWhen

ecr

yey

Equation (6) accounts for the case where inelastic buckling dominates the column behavior

because of the presence of residual stresses in the member, while equation (7) accounts for

elastic buckling in long or slender columns in which the 0.877 factor tries to account for

initial crookedness.

.

he AISC specification recommends limiting slenderness ratio for compression members such

that:

)8(200r

LK



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5. LOCAL STABILITY OF COLUMNS (LOCAL BUCKLING LIMIT STATE)

Local buckling of columns

Local buckling leads to a reduction in the strength of a compression member and prevents

the member from reaching its overall compression capacity. To avoid or prevent local

buckling, the AISC specification prescribes limits to the width-to-thickness ratios of the plate

components that make up the structural member. These limits are given in section B4 of the

AISC Manual.

Local buckling depends on the slenderness (width-to-thickness b/t ratio) of the plate

element and the yield stress (Fy) of the material. Each plate element must be stocky enough,

i.e., have a b/t ratio that prevents local buckling from governing the column strength.

The AISC specification B4 provides the slenderness (b/t) limits that the individual plate

elements must satisfy so that local buckling does not control. The AISC specification provides

two slenderness limits (λp and λr) for the local buckling of plate elements.

The AISC specifications for column strength

assume that column buckling is the governing limit

state. However, if the column section is made of

thin (slender) plate elements, then failure can

occur due to local buckling of the flanges or the

webs.

If local buckling of the individual plate elements

occurs, then the column may not be able to

develop its buckling strength. Therefore, the local

buckling limit state must be prevented from

controlling the column strength.

.

- If the slenderness ratio (b/t) of the plate element is greater than λr then it is slender. It will

locally buckle in the elastic range before reaching Fy

ElementSlendert/bIf r

- If the slenderness ratio (b/t) of the plate element is less than λr but greater than λp, then it

is non-compact. It will locally buckle immediately after reaching Fy

ElementcompactNont/bIf rp



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- If all the plate elements of a cross-section are compact, then the section is compact.

- If any one plate element is non-compact, then the cross-section is non-compact

- If any one plate element is slender, then the cross-section is slender.

When elements of a compression member that are slender a reduction is applied to the

flexural buckling stress, Fcr, in equations (6) and (7). For elements that are compact or non-

compact, equations (6) and (7) can be used directly.

Note that the slenderness limits (λp and λr) and the definition of plate slenderness (b/t) ratio

depend upon the boundary conditions for the plate.

- If the plate is supported along two edges parallel to the direction of compression force,

then it is a stiffened element. For example, the webs of W shapes

- If the plate is supported along only one edge parallel to the direction of the compression

force, then it is an unstiffened element. like, the flanges of W shapes.

If the slenderness ratio (b/t) of the plate

element is less than λp, then the element is

compact. It will locally buckle much after

reaching Fy

ElementCompactt/bIf p



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The slenderness limits λp and λr for various plate elements with different boundary

conditions are given in Table B4.1 of the AISC Spec.

The slenderness limits λr for various plate elements

The local buckling limit state can be prevented from controlling the column strength by using

sections that are non-compact. If all the elements of the cross-section have calculated

slenderness (b/t) ratio less than λr, then the local buckling limit state will not control.

Most wide flange shapes that are listed in the AISCM do not have slender elements. There

are, in fact, very few sections listed in the AISCM that have slender elements and these are

usually indicated by a footnote. However, some HSS (round and square), double-angle

shapes, and WT-shapes are made up of slender elements. The use of slender sections as

compression members is not efficient or economical; therefore, the its recommend to do

not use them in design practice.

yp

F

E12.1t/b



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6. AISC TABLES FOR COMPRESSION MEMBERS

The AISC manual contains many useful tables for analysis and design in order to simplify the

calculations. For compression members whose strength is governed by flexural buckling (no

local buckling), Table 4-22 of the Manual "design of Compression Members" can be used.

This table shows KL/r vs. φc Fcr for steels with various values of Fy.

Also tables of part 4 titled "column load tables" give the available strength of selected shape.

These tables shows KL vs. φc Pn for selected W, HP, single-angle, HSS, pipe, double-angle,

and composite shape.

7. ANALYSIS PROCEDURE FOR COMPRESSION MEMBERS

There are several methods available for the analysis of compression members and these are

discussed below. The first step is to determine the effective length, KL, and the slenderness

ratio, KL/r, for each axis of the column.

Method 1: Use equations (5) through (7), using the larger of x

xx

r

LKand

y

yy

r

LK.

Method 2: AISC Available Critical Stress Tables (AISCM, Table 4-22)

This gives the critical buckling stress, φc Fcr, as a function of r

LKfor various values of Fy.

For a given r

LK, determine φc Fcr from the table using the larger

x

xx

r

LKand

y

yy

r

LK.

(e.g., when 97r

LK and Fy = 36 ksi, AISCM, Table 4-22 gives φc Fcr = 19.7 ksi).

Knowing the critical buckling stress, the axial design capacity can be calculated from

φc Pn = φc Fcr Ag , where Ag is the gross cross-sectional area of the compression member.

Method 3: AISCM Available Compression Strength Tables (AICSM, Tables 4-1 through 4-12)

These tables give the design strength, φc Pn , of selected shapes for various effective lengths,

KL, and for selected values of Fy. Go to the appropriate table with KL, using the larger of

)r

r(

LK

y

x

xx and yy LK .



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Notes:

• Ensure that the slenderness ratio for the member is not greater than 200, that is,

200r

LK

• Check that local buckling will not occur.

• Use column load tables (i.e., the available compression strength tables) whenever possible.

(Note: Only a few selected sections are listed in these tables, but these are typically the most

commonly used for building construction.)

• Equations (5) through (7) can be used in all cases for column shapes that have no slender

elements.

Example 2: Calculate the design compressive strength of a W12 * 65 column, 20 ft. long, and pinned at both ends. Use ASTM A572 steel.

Solution

- Ky= Kx = 1.0 - Lx = Ly = 20 ft.

- Fy = 50 ksi

From AISC Manual, for W12 x 65: rx =5.28 in , rx =3.02 in and Ag = 19.1

45.4528.5

12*20*0.1

r

LK

x

xx

5.7902.3

12*20*0.1

r

LK

y

yy

OK2005.79r

LK

r

LK

y

yy

Check the slenderness criteria for compression member

in62/12bin12bf

in655.0t f

in39.0t w

in7.920.1*21.12kdh des

(Note: kdes, used for design, is smaller than kdet, used for detailing. The difference in these values is due to the variation in the fabrication processes.)

For W shape

OK48.1350

2900056.092.9

655.0

6

F

E56.0

t

b

y

OK88.3550

2900049.188.24

39.0

7.9

F

E49.1

t

h

yw

Determine the buckling stress, Fcr

ksi5.3150*)658.0(F)658.0(F

ksi3.45)5.79(

29000*

)r/LK(

EF

4.113F

E71.45.79

r

LK

4.11350

2900071.4

F

E71.4

3.45

50

yF

F

cr

2

2

2

2

e

y

y

e

y

.kips5411.19*5.31*9.0

AFP gcrcnc

Method 2 From AISCM, Table 4-22, φc Fcr could be obtained directly by entering the table with KL/r = 79.5and Fy= 50 ksi. By interpolation

ksi4.2835.28)5.7980(*7980

2.285.282.28Fcrc

A value of about φc Fcr = 28.4 is obtained, which confirms the calculation above.

Method 3 The design strength could be obtained directly from AISC, Table 4-1 (i.e., the column load tables).

44.11)02.3

28.5/(20*0.1)

r

r/(LK

y

xxx ,

20KLUseft2020*0.1LK yy .

Go to the table with KL = 20 ft. for W12 x 65

and obtain φc Pn = 541 ksi.



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Example 3: Determine the design strength for the axially loaded column of built up section shown. The length is 19 ft. and pinned at both ends. Use 50 ksi steel. Neglect local buckling.

Solution

- Ky= Kx = 1.0 - Lx = Ly = 19 ft.

- Fy = 50 ksi

From AISC Manual, for MC18 x 42.7: Ix =554 , Iy =14.3 and Ag = 12.6 m d =18,

Cofbackfromin877.0x

2

g in2.356.12*22/1*20A

in87.62.35

)2/18*6.12[225.0*2/1*20y

423

2x

in1721)25.087.6(*1012

5.0*20

)87.65.9(*6.12*2554*2I

43

2y in1554

12

5.0*20)877.6(*6.12*23.14*2I

in99.62.35

1721rx

in64.62.35

1554ry

62.3299.6

12*19*0.1

r

LK

x

34.3464.6

12*19*0.1

r

LK

y

OK20034.34r

LK

r

LK

y

Determine the buckling stress, Fcr

ksi87.4550*)658.0(F)658.0(F

ksi72.242)34.34(

29000*

)r/LK(

EF

4.113F

E71.434.34

r

LK

4.11350

2900071.4

F

E71.4

72.242

50

yF

F

cr

2

2

2

2

e

y

y

e

y

.kips2.14532.35*87.45*9.0

AFP gcrcnc

Method 2 From AISCM, Table 4-22, φc Fcr could be obtained directly by entering the table with KL/r = 34.34 and Fy= 50 ksi. By interpolation

ksi33.41)34.3435(*1

2.414.412.41Fcrc

A

Fcr = 45.92

Method 3 N/A HW. Determine the design strength of the simply supported 18 ft long column having section HSS 16*16*1/2. Use steel fy = 46 ksi. Using the three methods.



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8. DESIGN OF COLUMNS

In designing columns using the column load tables, follow these steps:

a. Calculate Pu (i.e., the factored load on the column).

b. Compute KL, using the larger of

)r

r(

LK

y

x

xx and yy LK .

c. Enter the column load tables (AISCM, Tables 4-1 through 4-21) with a KL value , and move

horizontally until the lightest column section is found with a design strength, φc Pn > the

factored load, Pu.

Example 4: Select a W14 column of ASTM A572, grade 50 steel, 14 ft. long, pinned at both ends, and subjected to the following service loads: PD = 160 kips and PL =330 kips Solution - Ky= Kx = 1.0 - Lx = Ly = 14 ft.

- Fy = 50 ksi

kips720330*6.1160*2.1

P6.1P2.1P LDu

ft1414*0.1KLKLLLtoDue yyx

From the column load tables in part 4 of the AISCM, find the W14 tables. Enter these tables with KL = 14 ft. and find the

lightest W14 with φc Pn > 720 kips

We obtain a W14 * 82 with φc Pn =774 kips > Pu = 720

kips.

Check slenderness ratio

OK2007.6748.2

12*14*0.1

r

LK

y

Check local buckling

The shape W 14 * 82 is not slender (there is no footnote in the dimensions and properties indicate that it is), so local buckling does not have to be investigated.

Use W14 * 82

Example 5: Select the lightest W-shape column for a factored compression load, Pu=194 kips, and a column length, L=24 ft. Use ASTM A572, grade 50 steel and assume that the column is pinned at both ends. Solution - Ky= Kx = 1.0 - Lx = Ly = 24 ft.

- Fy = 50 ksi kips194Pu

ft2424*0.1KLKLLLtoDue yyx

Enter these tables with KL = 24 ft. and find the

lightest W shapes with φc Pn > 194kips

Selected Size φc Pn , kips

W 8 * ?? W 8 * 58 205

W 10 * ?? W 10 * 49 254

W 12 * ?? W 12 * 53 261

W 14 * ?? W 14 * 61 293

Check slenderness ratio

OK2004.11354.2

12*24*0.1

r

LK

y


The shape W 10*49 is not slender (there is no footnote in the dimensions and properties indicate that it is), so local buckling does not have to be investigated.

Use W10 * 49



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Example 6: Select an ASTM A572, grade 50 steel W-shape column to resist a factored compression load, Pu = 780 kips. The unbraced lengths are Lx = 25 ft. and Ly = 12.5 ft. and the column is pinned at each end. Solution - Ky= Kx = 1.0 - Lx = 25 ft - Ly = 12.5 ft.

- Fy = 50 ksi kips780Pu

ft5.125.12*0.1LK y

Assume that the weak axis governs

ft5.12KLKL y

Enter the table with KL = 12.5 ft. and Pu = 7801 ksi.

Selected Size φc Pn , kips

W 8 * ?? N/A

W 10 * ?? W 10 * 88 918.5

W 12 * ?? W 12 * 79 874.5

W 14 * ?? W 14 * 82 828

Use W 12*79 the lightest section From AISC Manual, for W12 x 79: rx =5.34 in , rx =3.05 in rx /rx =1.75 (from AISC Tabel 4-1)

12WnewSelectand

tablesenterreand3.14LKUse

controlsLK

LKft3.14)75.1/(25*0.1)r

r/(LK

yy

xx

yyy

xxx

Also W 12*79 is the lightest section

KL φc Pn , kips

14.0 836

14.3 ??

15.0 809

ksi9.827)3.1415(*1415

809836809Pnc

For W 12*79 φc Pn =827.9 > Pu = 780 OK Check slenderness ratio

OK2002.4905.3

12*5.12*0.1

r

LK

y

OK2002.5634.5

12*25*0.1

r

LK

x



Use W12*79 .



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9. DESIGN AND ANALYSIS OF COLUMNS IN FRAMES

Braced Versus Unbraced Frames

Braced frames exist in buildings where the lateral loads are resisted by diagonal bracing or

shearwalls as shown in Figure below. The beams and girders in braced frames are usually

connected to the columns with simple shear connections and thus there is very little

moment restraint at these connections. The ends of columns in braced frames are assumed

to have no appreciable relative lateral sway; therefore, the term nonsway or sidesway-

inhibited is used to describe these frames. The effective length factor for columns in braced

frames is taken as 1.0.

In unbraced or moment frames, the lateral loads are resisted through bending of the beams,

girders, and columns, and thus the girder-to-column and beam-to-column connections are

designed as moment connections. The ends of columns in unbraced frames undergo

relatively appreciable sidesway; therefore, the term sway or sway-uninhibited is used to

describe these frames. The effective length of columns in moment frames is usually greater

than 1.0.

Braced Frames Unbraced Frame



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EFFECTIVE LENGTH OF COLUMNS IN FRAMES

The effective length of a column is defined as KL, where K is usually determined by one of

two methods:

1. AISCM, Table C-C2.2: This table, is used for isolated columns that are not part of a

continuous frame. In Table C-C2.2, a through c represent columns in braced frames,

while d through f represent columns in unbraced frames.

If a compression member is supported differently with respect to each of its principle

axes. Thus slenderness ratio must be computed for both minor and major axes and

the larger value would be used for determination of the axial compressive strength. If

a W-shape is used as a column and is braced by horizontal members in two

perpendicular directions at the top and at the mid-height the column is braced but

only in one direction the KL for each case is:

Example 7: A W 12 x 58, 24 ft long column is pinned at both ends and braced in the weak direction at the points shown in Figure. A992 steel is used. Determine the available compressive strength

Solution

From AISC Manual, for W12 x 58: rx =5.28 in , rx =2.51 in , Ag = 17

45.4528.5

12*20*0.1

r

LK

x

xx

25.3851.2

12*8*0.1

r

LK

y

yy

Use larger Value 5.4545.45r

LK

r

LK

x

xx

From AISCM, Table 4-22, with KL/r = 45.5 and Fy=

50 ksi. ksi65.38Fcrc

.kips65717*65.38

AFP gcrcnc



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So far, it is looked at the buckling strength of individual columns. These columns had various

boundary conditions at the ends, but they were not connected to other members with

moment (fix) connections.

2. Nomographs or alignment charts (AISCM, Tables C-C2.3 and C-C2.4)—The alignment

charts use the actual restraints at the girder-to-column connections to determine the

effective length factor, K. They provide more accurate values for the effective length factor

than AISCM, Table C-C2.2 , but the process of obtaining these values is more tedious than

the first method, and the alignment charts can only be used if the initial sizes of the columns

and girders are known.

However, when these individual columns are part of a

frame, their ends are connected to other members (beams

etc.). The effective length factor K will depend on the

relative rigidity (stiffness) of the members connected at the

ends.

Consider the rigid unbraced frame shown. If Table C-C2.2 is

used for this frame, the lower- story columns are best

approximated by condition (f) with K=2.0. For column AB a

value of K=1.2 corresponding to condition (c).



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The effective length factor for columns in frames must be calculated as follows:

1- Determine whether the column is part of a braced frame or an unbraced

(moment resisting) frame.

- If the column is part of a braced frame then its effective length factor 0 < K ≤ 1

- If the column is part of an unbraced frame then 1 < K ≤ ∞

2- Determine the relative rigidity factor G for both ends of the column

- G is defined as the ratio of the summation of the rigidity (EI/L) of all columns coming

together at an end to the summation of the rigidity (EI/L) of all beams coming together at

the same end.

1GSupportFixedFor

10GSupportPinnedFor)9(

L

EI

L

EI

G

b

b

c

c

3- Determine the effective length factor K for the column using the calculated value of

G at both ends, i.e., GA and GB and the appropriate alignment chart. For columns in

braced (sidesway inhibited) frames use Figure C-C2.3 of the AISC manual in which 0 < K

≤ 1 . For columns in unbraced (sidesway uninhibited) frames use Figure C-C2.4 of the

AISC manual in which 1 < K ≤ ∞ . The procedure for calculating G is the same for both

cases.

Note: The effective length factor obtained in this manner is with respect to the axis of

bending, which is the axis perpendicular to the plane of the frame. A separate analysis

must be made for buckling about the other axis. Normally the beam-to-column

connections in this direction will not transmit moment; sidesway is prevented by bracing

; and K can be taken as 1.0.

INELASTIC STIFFNESS REDUCTION FACTOR, τa

The inelastic stiffness reduction factor is incorporated in to the AISC column design method

through the use of Table 4-21 (pp. 4-317)of the AISC manual. Table 4-21 gives the stiffness

reduction factor (τa) as a function of the yield stress Fy and the stress Pu/Ag in the column,

where Pu is factored design load.

)10(G*G ainelastic



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Example 6: A rigid unbraced frame is shown in Figure. All members are oriented so that bending is about the strong axis. Lateral support at each joint by simply connected bracing in the direction perpendicular to the frame. Determine the effective length factors with respect to each axis for member AB and find the design strength for it. The service dead load is 35.5 kips and the service live load is 142 kips. A992 steel is used. Solution

- It is an unbraced (sidesway uninhibited) frame.

- Ky = 1.0 - Lx = Ly = 12 ft

- Fy = 50 ksi - Kx = depends on boundary conditions,

which involve restraints due to beams and columns connected to the ends of column AB. Thus Need to calculate Kx using alignment charts.

From AISC Manual, for beam W12 x 14, Ix =88.6 in4 , for beam W14 x 22, Ix =199 in4 for column W10 x 33, Ix =171 in4, Ag =9.71 in2

rx =4.19 in , ry =1.94 in

52.1352.9

25.14

18/6.8820/6.88

12/171

L/I

L/IG

bb

cc

A

36.101.21

5.28

18/19920/199

)12/171(2

L/I

L/IG

bb

cc

B

Enter the alignment chart for unbraced frames with GA

and GB and read Kx= 1.45 (based on elastic behavior)

Determine whether the column behavior is elastic or inelastic

usedbecanKinelasticF

E71.4

r

LK

4.11350

2900071.4

F

E71.4

83.4919.4

12*12*45.1

r

LK

inelasticy

y

x

x

kips8.269142*6.15.35*2.1

P6.1P2.1P LDu

Find stiffness reduction factor τa

Pu/Ag = 269.8/9.71 = 27.79

Enter Table 4-21 of AISCM with Fy = 50 ksi and Pu/Ag = 27.79 ksi and read the stiffness reduction

factor τa pp.4-317:

8105.0)79.2728(*1

804.0835.0804.0a

23.152.1*8105.0G*G elasticAainelasticA

1.136.1*8105.0G*G elasticBainelasticB

from alignment chart with GAinelastic and GBinelastic and

read Kx= 1.35 (based on inelastic behavior)

The design strength could be obtained directly from AISC, Table 4-1 pp.4-20

5.7)94.1

19.4/(12*35.1)

r

r/(LK

y

xxx ,

12KLUseft1212*0.1LK yy .

Go to the table with KL = 12 ft. for W10 x 33

and obtain φc Pn = 292 ksi.

OKPP unc

OK200r

LK





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Example 6: Calculate the effective length factor and the design strength for a W10 x 60 column AB made from 50 ksi steel in the unbraced frame shown. Column AB has a design factor load Pu=450 kips. The columns are oriented such that major axis bending occurs in the plane of the frame. The columns are braced continuously along the length for out-of-plane buckling. Assume that the same column section is used for the story above. Solution


- Ly = 0 - Ky = has no meaning because out-of-plane

buckling is not possible. - Lx = 15 ft

- Fy = 50 ksi - Kx = depends on boundary conditions,


From AISC Manual, for beam W14 x 74, Ix =796 in4 for column W10 x 60, Ix =341 in4, Ag =17.6 in2

rx =4.39 in , ry =2.57 in

609.0002.7

2625.4

20/79618/796

15/34112/341

L/I

L/IG

bb

cc

A

)portsuppinned(10GB


and GB and read Kx= 1.8



E71.4

r

LK

4.11350

2900071.4

F

E71.4

8.7339.4

12*15*8.1

r

LK

inelasticy

y

x

x


Pu/Ag = 450/17.6 = 25.57


factor τa (pp.4-317):

8746.0)57.2526(*1

863.089.0863.0a


10GB

from alignment chart with GAinelastic and GB and read

Kx= 1.75

The design strength could be obtained directly from AISC, Table 4-1 pp.4.19

37.15)57.2

39.4/(15*75.1)

r

r/(LK

y

xxx ,

37.15KLUse0LK yy .

Go to the table with KL = 15.37 ft. for W10 x 60

and obtain

kips545)37.1516(*1

528555528Pnc

OKPP unc

OK200r

LK





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Example 6: Design Column AB of the frame shown for a design load of 500 kips. Assume that the column is oriented in such a way that major axis bending occurs in the plane of the frame. Assume that the columns are braced at each story level for out-of-plane buckling. Assume that the same column section is used for the stories above and below. Solution


- Ky = 1.0 - Lx = Ly = 12 ft

- Fy = 50 ksi

- Pu = 500 kips - Kx = depends on boundary conditions,


ft1212*0.1LK y

Assume that the weak axis governs

ft12KLKL y

Enter the table on page 4-18 with KL = 12 ft. and Pu = 500 ksi. Select section W 12*53

kips500kips547Pnc

From AISC Manual, for beam W14 x 68, Ix =722 in4 for column W12 x 53, Ix =425 in4, Ag =15.6 in2

rx =5.23 in , ry =2.48 in

02.120/72218/722

12/42510/425

L/I

L/IG

bb

cc

A

836.020/72218/722

15/42512/425

L/I

L/IG

bb

cc

B


and GB and read Kx= 1.3



E71.4

r

LK

4.11350

2900071.4

F

E71.4

79.3523.5

12*12*3.1

r

LK

inelasticy

y

x

x


Pu/Ag = 500/15.6 = 32.05


factor τa (pp.4-317):

658.0)05.3233(*1

62.066.062.0a


55.0836.0*658.0G*G elasticBainelasticB

from alignment chart with GAinelastic and GB and read

Kx= 1.2

OKLK83.6)48.2/23.5/(12*2.1)r

r/(LK yy

y

xxx

OK200r

LK





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10. DESIGN OF SINGLY SYMMETRIC CROSS-SECTIONS

(FLEXURAL-TORSIONAL BUCKLING)

When an axially loaded compression member becomes unstable overall (that is, not locally

unstable), it can buckle in one of three ways, as shown in Figure below.

1. Flexural buckling: We have considered this type of buckling up to now. It is a deflection

caused by bending, or flexure, about the axis corresponding to the largest slenderness ratio.

Compression members with any type of cross-sectional configuration can fail in this way.

2. Torsional buckling: This type of failure is caused by twisting about the longitudinal axis of

the member. It can occur only with doubly symmetrical cross sections with very slender

cross-sectional elements. Standard hot-rolled shapes are not susceptible to torsional

buckling.

3. Flexural-torsional buckling: This type of failure is caused by a combination of flexural

buckling and torsional buckling. The member bends and twists simultaneously. This type of

failure can occur only with unsymmetrical cross sections, both those with one axis of

symmetry—such as channels, structural tees, double-angle shapes, and equal-leg single

angles— and those wi th no axis of symmetry, such as unequal-leg single angles.

- So far, we have been talking about doubly

symmetric wide-flange (I-shaped) sections

and channel sections. These rolled shapes

always fail by flexural buckling.

- Singly symmetric (Tees and double angle)

sections fail either by flexural buckling

about the axis of non-symmetry or by

flexural-torsional buckling about the axis of

symmetry and the longitudinal axis.

Flexural buckling will occur about the x-axis

Flexural-torsional buckling will occur about the y and z-axis

Smaller of the two will govern the design strength



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The AISC specification for flexural-torsional buckling is given by Spec. E4. The design strength

can be computed as

gcrcnc AFP

2crzcry

crzcrycrzcrycr

)FF(

HFF411

H2

FFF

2og

crzrA

JGF

Where:

cryF critical stress for buckling about the y-axis

G = shear modulus (ksi) = 11,200 ksi for structural steel

J = torsional constant (equal to the polar moment of inertia only for circular cross sections)

2or = polar radius of gyration about shear center (in.)

For double angle the constants 2or and H are given in AISC Manual Table 1-15 pp1-100,

while J are double the values given for single angles.

For T shape only J is given in Table 1-8 pp1-48 while the constants 2or and H are computed

as following:

0xo (Because the shear center of a tee is located at the intersection of the centerlines of the

flange and the stem)

2

tyy f

o

g

yx2o

2o

2o

A

IIyxr

2o

2o

2o

r

yx1H

In the above equations, y is defined as the axis of symmetry (regardless of the orientation of

the member), and flexural-torsional buckling will take place only about this axis (flexural

buckling about this axis will not occur). The x-axis (the axis of no symmetry) is subject only to

flexural buckling. Therefore, for singly symmetrical shapes, there are two possibilities for

the strength: either flexural-torsional buckling about the axis of symmetry or flexural

buckling about the axis of no symmetry. To determine which one controls, compute the

strength corresponding to each axis and use the smaller value.



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Example 6: Compute the compressive strength

of a WT12×81 of A992 steel. The effective length with respect to the x-axis is 25 feet 6 inches, the effective length with respect to the y-axis is 20 feet, and the effective length with respect to the z-axis is 20 feet. Solution Because this shape is a WT, both flexural buckling and flexural- torsional must be checked.

- Kx Lx = 25.5 ft - Ky Ly = 20 ft.

- Fy = 50 ksi - G=11200 ksi

From AISC Manual, for WT12 x 81: rx =3.5 in , ry =3.05 in and Ag = 23.9 , , tf =1.22 in in70.2y , Ix =293 , Iy =221, J=9.22

Step I: Flexural buckling strength about axis of no symmetry, namely x-axis

43.875.3

12*5.25

r

LK

x

xx

ksi44.37)43.87(

29000*

)r/LK(

EF

2

2

2

2

e

ksi59.2850*)658.0(F)658.0(F

4.113F

E71.45.79

r

LK

4.11350

2900071.4

F

E71.4

44.37

50

yF

F

cr

y

y

e

y

.kips3.6839.23*59.28*9.0

AFP gcrcnc

Step II : Compute the flexural-torsional buckling strength about the the axis of symmetry, namely y-axis:

- Compute Fcry

69.7805.3

12*20

r

LK

y

yy

ksi22.46)69.78(

29000*

)r/LK(

E

)r/LK(

EF

2

2

2yyy

2

2

2

e

ksi79.3150*)658.0(F)658.0(F

F

E71.4

r

LK

4.11350

2900071.4

F

E71.4

22.46

50

yF

F

cry

yy

y

e

y

Because the shear center of a tee is located at the intersection of the centerlines of the flange and the stem:

0xo

in09.22

22.170.2

2

tyy f

o

87.259.23

22129309.20

A

IIyxr 2

g

yx2o

2o

2o

8312.087.25

09.201

r

yx1H

2

2o

2o

2o

ksi16787.25*9.23

22.9*11200

rA

JGF

2og

crz

ksi8.19816779.31FF crzcry

63.30)8.198(

8312.0*167*79.31*411

8312.0*2

8.198

)FF(

HFF411

H2

FFF

2

2crzcry

crzcrycrzcrycr

ksi3.683.kips1.7329.23*63.30*9.0

AFP gcrcnc

The flexural buckling strength controls, and the nominal strength is 683.3 kips.


The shape W 10 * 60 is not slender (there is no footnote in the dimensions and properties indicate that it is), so local buckling does not have to be



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11. BUILT-UP COMPRESSION MEMBERS

11.1: Built-up Members with Components not in contact

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

See Example 3 of lectures pp.13.

The open sides of compression members that are built up from plates and shapes may be

connected together with continuous cover plates with perforated holes for access purpose,

or they may be connected together with lacing and tie plates. The purpose of the perforated

cover plates and the lacing or latticework, are to hold the various parts parallel and the

correct distance apart, and to equalize the stress distribution between various parts.

11.2: Built-Up Members With Components in Contact

Which consist either from cover-plates and standard rolled shapes or Composed of Rolled

Shapes as shown in figure below.

The most common built-up shape is one that is composed of rolled shapes, namely, the

double-angle shape. Double-angle sections are very popular as compression members in

trusses and bracing members in frames. These sections consist of two angles placed back-to-

back and connected together using bolts or welds.



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If the planks are unconnected, they will slip along the surface of contact when loaded and

will function as two separate beams. When connected by bolts (or any other fasteners, such

as nails), the two planks will behave as a unit, and the resistance to slip will be provided by

shear in the bolts. This behavior takes place in the double-angle shape when bending about

its y-axis. If the plank beam is oriented so that bending takes place about its other axis (the

b-axis), then both planks bend in exactly the same manner, and there is no slippage and

hence no shear. This behavior is analogous to bending about the x-axis of the double-angle

shape. When the fasteners are subjected to shear, a modified slenderness ratio larger than

the actual value may be required.



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If the column tend 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.

AISC E6 considers two categories of intermediate connectors: (1) snug-tight bolts and (2)

welds or fully-tensioned bolts (slip-critical). Snug-tight bolts are that are tightened until all

piles of a connections are in firm contact with each other. This usually means the tightness

obtained by the full manual effort of a worker. Slip-critical bolts are tightened much more

firmly than are snug-tight bolts. They are tightened until their bodies have very high tensile

stresses. Such bolts clamp the fastened parts of a connection so tightly together between

the bolt and nut heads that loads are resisted by friction and slippage is nil.

According to AISC Specification, a modified (KL/r)m must be calculated for the built up

section for buckling about axis in which the connectors are subjected to shearing forces (like

y-axis of double angle section or x-axis of W-shape with cover palte)

When the connectors are snug-tight bolts, the modified slenderness ratio is:

2

i

2

om r

a

r

LK

r

LK

If cover-plated column tends to buckle about y-axis, the connectors

between W shape and the plates are not subjected to any

calculated load. If it tend to buckle about x-axis, the connectors are

subjected to shearing forces. The flanges of the W shape and the

cover plates will have different stresses and thus different

deformation. The result will be shear in the connection between

these parts and (KL/r)x will have to be modified.

snug-tight bolts fully-tensioned bolts (slip-critical)



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Where : (KL/r)o = original unmodified slenderness ratio, a = spacing of the connectors and

ri= smallest radius of gyration of the component

When the connectors are fully-tensioned bolts or welds, the modified slenderness:

- omi r

LK

r

LK40

r

aIf

-

2

i

i2

omi r

aK

r

LK

r

LK40

r

aIf

- Where :Ki = 0.5 for angles back-to-back

= 0.75 for channnels back-to-back = 0.86 for all other cases

The column load tables for double angles are based on the use of welds or fully

tightened bolts.

To ensure that the built-up member acts as a unit, AISC E6.2 requires that the

slenderness of an individual component be no greater than three-fourths of the

slenderness of the built-up member; that is,

r

LK

4

3

r

aK

i

Where : Ka / ri = effective slenderness ratio of the component

KL/r = maximum slenderness ratio of the built-up member

11.2.1 Design Procedure

1- Assume b and d

2- Assume rx and ry (use approximate values)

3- Find KL/r and find crcF from Table 4-22

4- Find required area crc

urequired

F

PA

5- Select suitable sections with requiredupbuilt AA

6- For select shape check unc PP

7- Check local buckling



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Example 7: Compute the available strength of the compression member consist of W12*120 and cover plate PL 1*16 snug-tight bolted at 6 in spacings to the W section as shown in Figure.. The effective length KL is 14 feet. A992 steel is used.

Solution

- Kx Lx = Ky Ly = 14 ft

- Fy = 50 ksi From AISC Manual, for W12*120: Ix =1070 Iy =345 , rx =5.51 in , ry =3.13 in and Ag = 35.3

2in3.67)1*16*23.35A

42

x in26602

11.13*)1*16*21070I

in29.63.67

2660

A

Ir xx

71.2629.6

12*14

r

LK

x

43

y in5.102712

16*1*2345I

in91.33.67

7.1027

A

Ir xy

97.4291.3

12*14

r

LK

y

For x-axis use modified slenderness ratio (KL/r)m

For snug-tight bolted, the modified slenderness ratio is:

2

i

2

om r

a

r

LK

r

LK

71.26r

LK

r

LK

xo

For W12*120 , rx =5.51 in , ry =3.13 in

in289.016*1

12/1*16

A

Ir

3

plate

51.5and13.3in289.0rr platei

76.20289.0

6

r

a

i

r

LK

4

3

r

aKCheck

i

OK23.3297.42*75.0r

LK

4

376.20

289.0

6

r

aK

i

83.3376.2071.26r

a

r

LK

r

LK 22

2

i

2

om

97.42r

LK

r

LK

83.33r

LK97.42

r

LK

y

my

ksi0.155)97.42(

29000*

)r/LK(

EF

2

2

2

2

e

ksi69.4350*)658.0(F)658.0(F

F

E71.497.42

r

LK

4..11350

2900071.4

F

E71.4

155

50

yF

F

cr

y

y

e

y

.kips26463.67*69.43*9.0

AFP gcrcnc


PL 1*16



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Example 7: Compute the available strength of the compression member shown in Figure. Two angles, 5* 3*1/2, are oriented with the long legs back-to-back (2L5* 3 *1⁄2 LLBB) and separated by 3⁄8 inch. The effective length KL is 16 feet, and there are three fully tightened intermediate connectors. A36 steel is used.

Solution Because this shape is a double angle, both flexural buckling and flexural- torsional must be checked.


- Fy = 36 ksi - G=11200 ksi

From AISC Manual, for 2L5* 3 *1⁄2 LLBB: rx =1.58

in , ry =1.24 in and Ag = 7.51 , ,H =0.646 , 2or =2.51.

For single angle J=0.322, Ag = 3.75, rz =0.642 in

Step I: Flexural buckling strength about axis of no symmetry, namely x-axis

5.12158.1

12*16

r

LK

x

xx

ksi39.19)5.121(

29000*

)r/LK(

EF

2

2

2

2

e

ksi55.1636*)658.0(F)658.0(F

134F

E71.45.121

r

LK

13436

2900071.4

F

E71.4

39.19

36

yF

F

cr

y

y

e

y

.kips1.12451.7*55.16*9.0

AFP gcrcnc

Step II : Compute the flexural-torsional buckling strength about y-axis use the modified slenderness ratio:

8.15424.1

12*16

r

LK

r

LK

yo

double angles are based on the use of fully tightened bolts.

48spaces4

12*16aconnectorsofspacingThe

624.0rr,r

LK

4

3

r

aKCheck zi

i

OK1.1168.154*75.0r

LK

4

377.74

624.0

48

r

aK

z

- compute the modified slenderness ratio (KL/r)m

Ki = 0.5 for angles back-to-back

4077.74624.0

48

r

a

i

2

i

i2

om r

aK

r

LK

r

LK

38.37642.0

48*5.0

r

aK

i

i

2.15938.378.154r

aK

r

LK

r

LK 22

2

i

i2

om

- Compute Fcry

2.159r

LK

r

LK

my

ksi29.11)2.159(

29000*

)r/LK(

E

)r/LK(

EF

2

2

2m

2

2

2

e

ksi901.936*877.0F877.0F

134F

E71.4

r

LK

13436

2900071.4

F

E71.4

ycry

yy

y

ksi6.15251.2*51.7

)322.0*2(*11200

rA

JGF

22og

crz

ksi5.1626.152901.9FF crzcry

ksi599.9)5.162(

646.0*6.152*832.9*411

646.0*2

5.162

)FF(

HFF411

H2

FFF

2

2crzcry

crzcrycrzcrycr

.kips8.6451.7*599.9*9.0

AFP gcrcnc

The flexural-torsional buckling strength controls, and the nominal strength is 64.8 kips.



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Example 8: Design a built up column consist of W12and cover plate, snug-tight bolted at 6 in spacings to the W section as shown in Figure.. The effective length KL is 14 feet. A992 steel is used.

Solution


- Fy = 50 ksi - Assume d =14

- Assume b=15

kips25001000*6.1750*2.1Pu

in36.316*21.0b21.0r

in6.514*4.0d4.0r

y

x

kips600300*6.1100*2.1Pu

OK2005036.3

12*14

r

LK

r

LK

y

From Table 4-22: ksi5.37Fcrc

67.665.37

2500

F

PA

crc

urequired

69.152/37.31A37.313.3567.66A2

3.35A120*12WTry

plateplate

gp

OK69.1516A16*1PLTry

From AISC Manual, for W12*120: Ix =1070 Iy =345 , rx =5.51 in , ry =3.13 in and Ag = 35.3 Try PL 16*1

2in3.67)1*16*23.35A

42

x in26602

11.13*)1*16*21070I

in29.63.67

2660

A

Ir xx

71.2629.6

12*14

r

LK

x

43

y in5.102712

16*1*2345I

in91.33.67

7.1027

A

Ir xy

97.4291.3

12*14

r

LK

y

For x-axis use modified slenderness ratio (KL/r)m

For snug-tight bolted, the modified slenderness ratio is:

2

i

2

om r

a

r

LK

r

LK

71.26r

LK

r

LK

xo

For W12*120 , rx =5.51 in , ry =3.13 in

in289.016*1

12/1*16

A

Ir

3

plate

51.5and13.3in289.0rr platei

76.20289.0

6

r

a

i

r

LK

4

3

r

aKCheck

i

OK23.3297.42*75.0r

LK

4

376.20

289.0

6

r

aK

i

83.3376.2071.26r

a

r

LK

r

LK 22

2

i

2

om

97.42r

LK

r

LK

83.33r

LK97.42

r

LK

y

my

ksi0.155)97.42(

29000*

)r/LK(

EF

2

2

2

2

e

ksi69.4350*)658.0(F)658.0(F

F

E71.497.42

r

LK

4..11350

2900071.4

F

E71.4

155

50

yF

F

cr

y

y

e

y

OK2500P.kips26463.67*69.43*9.0

AFP

u

gcrcnc

Use W12*120 with one cover plate 1*16 each flange


k1000P

k750P

L

D



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11.3 Design of Laced Columns

11.3.1 Design of Lacing

The lacing either single lacing or double lacing as shown in figure below. In additional to

lacing, it is necessary to have tie plate (also called stay plate or batten plate) as near the

ends of the member as possible, and at intermediate points if lacing is interrupted.

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

single lacing made with angles should preferably be used if the distance between connection

lines is greater than 15 in. The AISC column formulas are used to design the lacing in the

usual manner.

)5045(45LacingDoubleFor

)6560(60LacingSingleFor

ooo

ooo

LacingDoubleUse"15wIf

LacingSingleUse"15wIf

7.0KandLacingDouble200

0.1KandLacingSingle140r

LK

c

c

memberupbuiltmemberconnectedmin

o

r

LK

4

3

r

K

boltsofcetandisedgeimummin*2bc

boltsofcetandisedgeimummin*2LlacingoflengthMinmum c

ncu P*02.0V

sin2

VP u

c

Single Lacing Double Lacing

Lacing is assumed to be subjected to a shearing force normal to the member, equal

to not less than 2% of the compression design strength φc Pn of the member.

o

w

cLcL

o

w

2

Vu

cP

cP



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11.3.2 Tie Plates

Its required to use tie plate at ends and mid-span of column or if lacing is interrupted.

50

wtthicknessMinmum plate

2

wLplatespanmidFor

wLplateEndFor

LplatetieofLengthMinmum

plate

plate

plate

boltsofdistaceedge*2wwMinmum platetie

- Connections

- At least three fasteners per side with spacing, s= (3db-6db)

- If welding is used : Length of welding plateL

11.3.3 Design Of Laced Built Up Column

8- Assume b and d

9- Assume rx and ry (use approximate values)

10- Find KL/r and find crcF from Table 4-22

11- Find required area crc

urequired

F

PA

12- Select suitable sections with requiredupbuilt AA

13- For select shape check unc PP

14- Check local buckling

15- Design of lacing and tie plate

w

plateLs



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Example 9: A select 2C12 column with lacing. The effective length is KL= 20 ft. and pinned at both ends. Use 50 ksi steel for channels and 36 ksi steel for lacing. Neglect local buckling. Also design bolted lacing and tie plates. Assume 3/4 in bolts.

Solution

- KLy= KLx = 20 ft

in920*45.0b45.0r

in2.720*36.0d36.0r

y

x

kips600300*6.1100*2.1Pu

OK20033.332.7

12*20

r

LK

r

LK

x

From Table 4-22:

ksi53.41)33.3334(*1

4.416.414.41Fcrc

44.1453.41

600

F

PA

crc

urequired

OK44.1462.1781.8*2A30*12C2Try upbuilt

From AISC Manual, for C12 x 30:

Ix =162 , Iy =5.12 and Ag = 8.81 , d =12 rx =4.29 , ry =0.762

Cofbackfromin674.0x 2

g in62.1781.8*2A

4x in324162*2I

42y in510)326.5(*81.8*212.5*2I

in29.462.17

324rx , in38.5

62.17

510ry

94.5529.4

12*20

r

LK

x

, 61.44

38.5

12*20

r

LK

y

OK20094.55r

LK

r

LK

x

From AISCM, Table 4-22, enter the table with KL/r = 55.94 and Fy= 50 ksi

ksi82.35)94.5556(*1

8.351.368.35Fcrc

.

.kips63162.17*82.35PAFP ncgcrcnc

Use 2C12*30

Design of lacing

w < 15 in , thus use single lacing , bf =3.17 Minimum edge distance for 3/4" bolts =1.25 in

"5.8wUsein16.825.1*22*17.312w o60Let

in8.960sin

5.8Lc

in8.9)60cos8.9(*2o

memberupbuiltmemberconnectedmin

o

r

LK

4

3

r

KCheck

OK96.4194.55*4

39.12

762.0

8.9*0.1

ncu P*02.0V kips62.12631*02.0Vu

kips28.760sin*2

62.12

sin2

VP u

c

For lacing bars

t289.012

t

tb

12/tb

A

IrtbA,

12

tbI

3

ccc

3c

c

barflatin25.0tTry

242.0t140t289.0

8.9*0.1140

r

LK

c

c

13625.0*289.0

8.9*0.1

r

LK

actualc

c

From AISCM, Table 4-22, with KL/r = 136 and Fy= 36 ksi. ksi2.12Fcrc

2

crc

urequired in597.0

2.12

28.7

F

PA

"5.2bUse

OK5.225.1*2cetandisedge.minCheck

5.2bUse39.225.0

597.0bin597.025.0*b cc

2c

"14LUse3.1225.1*281.9LMinimum cc

Use 1/4*2.5 *14" bars, Fy=36 Design of End tie plate

"16/3t17.050

5.8

50

wtthickness.Min plateplate

"5.8LwLplatetieofLengthMinmum plateplate

"12Use1125.1*25.8

boltsofdistaceedge*2wwMinmum plate

Use 3/16*8.5 *12" End tie plate



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12. Design of Column Base Plates

Column base plates are shop welded to the bottom of a column to provide bearing for the

column usually and to help in transferring the column axial loads to the concrete pier or

footing. They help prevent crushing of the concrete underneath the column and also help to

provide temporary support to the column during steel erection by allowing the column (in

combination with anchor bolts) to act temporarily as a vertical cantilever.

The thickness of base plates varies from 1⁄2 in. to 6 in. and they are more commonly

available in ASTM A36 steel. The base plate is usually larger than the column size (depending

on the shape of the column) by as much as 3 to 4 in. all around to provide room for the

placement of the anchor bolt holes outside of the column footprint.

The design strength of concrete in bearing from ACI 318 is given as

1cc1

21ccpc Af7.1

A

AA)f85.0(P and

2A

A1

1

2

where

A1 = Base plate area = B *N

B = Width of base plate,

N = Length of base plate,

A2= Area of concrete supporting base plate .

The column base plate often bears on a layer of

3 ⁄4-in. to 1.5-in. nonshrink grout that provides

a uniform bearing surface. The compressive

strength of the grout should be at least equal to

the compressive strength of the concrete used

in the pier or footing; however, a grout

compressive.

cf Compressive strength of the

concrete pier or footing, ksi, and

,65.0c Strength reduction factor for

concrete in bearing (ACI 318).



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1- Determination of Plate Dimensions B and N

1

2cc

u1

pcu

A

A)f85.0(

PA

PP

and d*bAAA fminmin,1

For economical design to keep the plate thickness to a minimum the values of m and n are

roughly equal. The condition n = m can be approached if:

1AN

)b8.0d95.0(5.0 f

N

AB 1

2- Determination of Plate Thickness t

NBF9.0

P2t

y

u

0.1assumeand)n,n,m(max

n,n,m Cantilever lengths of the base plate beyond the edges of the critical area of the

column.

For a square HSS column,

2

bNm

2

bBn

4

bn

For a W shape column,

2

d95.0Nm

2

b80.0Bn f

fbd4

1n



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Example 10: A W10 × 49 is used as a column and is supported by a concrete pier as shown in Figure. The top surface of the pier is 18 inches by 18 inches. Design an A36 base plate for a column dead load of 98 kips and a live load of 145 kips.

The concrete strength is cf 3000 psi.

Solution

From AISC Manual, for W10 x 49:

bf =10 , d =10

kips6.349145*6.198*2.1Pu

22

1

1

22

1

1

1

1

1

2cc

u1

in3.137324

92.210A

A

324

92.210)A(

A

324

92.210A

A

18*18)3*85.0(65.0

6.349

A

A)f85.0(

PA

OK3.137A10010*10d*bA 1fmin

b- Dimensions of Plate B and N

75.0)10*8.010*95.0(5.0

)b8.0d95.0(5.0 f

"13Say47.1275.03.137AN 1

"13BSay56.1013

3.137

N

AB 1

Check the bearing strength of the concrete

OK6.34986.38713*13

18*18*)13*13(*)3*85.0(*65.0

A

AA)f85.0(P

1

21ccpc

c- Thickness of plate t

in5.22

10*80.013

2

b80.0Bn f

in75.12

10*95.013

2

d95.0Nm

5.210*104

1bd

4

1n f

in5.2)0.1*5.2,5.2,75.1(max)n,n,m(max

"1tSay,893.013*13*36*9.0

6.349*25.2

NBF9.0

P2t

y

u

Use PL 1*13*13

Example 11: Design a base plate of A36 steel for

W12 × 65 that supports the loads of PD=200k and PL=300k. The concrete has a compressive

strength is cf 3 ksi and the footing has the

dimensions 9 ft * 9 ft .

Solution:

From AISC Manual, for W12 x 65: bf =12 , d =12.1

kips720300*6.1200*2.1Pu 2

2 in11664)12*9(*)12*9(A (108"*108")

The area of supporting concrete A2 is much greater than the base plate area, Thus

0.2A

A

1

2

21

1

1

2cc

u1

in2.217A

2*)3*85.0(65.0

720A

A

A)f85.0(

PA

OK2.217A2.1451.12*12d*bA 1fmin

a- Dimensions of Plate B and N

947.0)12*8.01.12*95.0(5.0

)b8.0d95.0(5.0 f

"16Say7.15947.02.217AN 1

"16BSay6.1316

2.217

N

AB 1


OK720k6.8482*)16*16(*)3*85.0(*65.0

A

AA)f85.0(P

1

21ccpc



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Example 11, cont

Thickness of plate t

in2.32

12*80.016

2

b80.0Bn f

in25.22

1.12*95.016

2

d95.0Nm

01.312*1.124

1bd

4

1n f

in2.3)0.1*01.3,25.2,2.3(max)n,n,m(max

"5.1tSay

33.1)16*16(*36*9.0

720*22.3

NBF9.0

P2t

y

u

Use PL 1.5*16*16

Example 12: A base plate is to be designed for a

W12 × 152 that supports the loads of PD=200k and PL=450k. Select an A36 plate to cover the entire area of the 3 ksi concrete pedestal underneath.

Solution:

From AISC Manual, for W12 x 152: bf =12.5 , d =13.7

kips960450*6.1200*2.1Pu

A2 = A1, Thus

0.1A

A

1

2

21

1

1

2cc

u1

in2.579A

1*)3*85.0(65.0

960A

A

A)f85.0(

PA

OK2.579A2.11715.12*7.13d*bA 1fmin

d- Dimensions of Plate B and N

51.1)5.12*8.07.13*95.0(5.0

)b8.0d95.0(5.0 f

"26Say6.2551.12.579AN 1

"23BSay3.2226

2.579

N

AB 1


OK960k2.9911*)26*23(*)3*85.0(*65.0

A

AA)f85.0(P

1

21ccpc

Thickness of plate t

in5.62

5.12*80.023

2

b80.0Bn f

in49.62

7.13*95.026

2

d95.0Nm

27.35.12*7.134

1bd

4

1n f

in5.6)0.1*27.3,49.6,5.6(max)n,n,m(max

"5.1tSay

05.2)23*26(*36*9.0

960*25.6

NBF9.0

P2t

y

u

Use PL 2 1/8*26*23 A36 base plate with 23*26

concrete pedestal 3 ksi strength

Documents

COMPRESSION MEMBERSColumns are classified as being long, short or intermediate based on slenderness ratio KL/r. For each type of these column have different failure modes: 1- Long