C-1

Columns:Columns:

Are compression members which are subjected to concentric axial compressive forces. These are to be found in trusses and as a lateral bracing members in frame building. Short columns are sometimes referred to as to as “struts” or “stanchions”.

Beam-Columns:Beam-Columns:Are members subjected to combined axial compressive and bending stresses; These are found in single storey of multi-storey framed structures. These are treated independently in this course (chap. 12 in your text book).

Columns Theory:Columns Theory:

Stocky columns (short) fail by yielding of the material at the cross section, but most columns fail by “buckling” at loads for less then yielding forces.

P

P

P

P

(a) (b)

C-2

For “slender” columns, Euler (1759) predicted the critical buckling load (Pcr)

– also known as Euler Buckling Load as:

)1(2

2

CL

EIPcr

where: E = Young Modulus of Elasticity. I = Minor moment of Inertia.

L = Unbraced length of column.

Derivation of Euler Buckling Load:

0"

2

2

yEI

Py

EI

M

dx

yd

cr

Solution of this differential equation:y = A cos (cx) + B sin (cx)

where:

, A and B are constants.EI

Pc cr

xL

y

x

y

PcrPcr

C-3

From boundary conditions:y = 0 @ x = 0, andy = 0 @ x = L, we get (A = 0) and (B sin cL = 0)

if B ≠ 0, then cL = n where n = 0, 1, 2, 3 ………

cL =

)(

/

2

2

2

2

22

2

2

C

rL

EA

L

ArE

L

EIP

LEI

P

cr

cr

2

2

g

crcr

rL

Eπ

A

PF

---- Euler Buckling Critical Load where: r = minor radius of gyration

The critical buckling load

is a function of the section

properties (A, L, r) and

the modulus of elasticity

for material, and is not a

function of the strength or

grade of the material.

Note:

C-4

Example C-1Example C-1

Find the critical buckling load for W 12 x 50, supported in a pinned-pinned condition, and has an over-all length of 20 feet?

Solution:Solution:

2

2

rL

cr

EF

rmin = ry = 1.96 inch (properties of section).

ksiFcr 19

290002

96.11220

2

Note: The steel grade is not a factor affecting buckling.

Pcr = Fcr A = 19.1 x 14.7 = 280.8 kips

C-5

For short (stocky) columns; Equation (C-2) gives high values for (Fcr), sometimes greater then proportional limit, Engessor (1889)

proposed to use (Et) instead of (E) in Euler formula:

)3(2

2

CL

IEP tcr

where:

Et = Tangent Modulus of Elasticity

Et < E

When (Fcr) exceeds (FPR), this is called

“Inelastic Buckling”, constantly variable

(Et) need to be used to predict (Fcr)

in the inelastic zone.

Shanley (1947), resolved this inconsistency.

C-6

Depending on (L/r) value the column buckling

strength was presented as shown by Shanley.

Residual Stresses:-Residual Stresses:-

Due to uneven cooling of hot-rolled sections,residual stresses develop as seen here.

The presence of “residual stresses” in almost all hot-rolled sections further complicates the issue

of elastic buckling and leads towards inelastic buckling.

C-7

The Previous conditions are very

difficult to achieve in a realistic building

condition, especially the free rotation of pinned

ends. Thus an “effective slenderness factor” is

introduced to account for various end conditions:

Thus:

4

2

2

CE

Fr

KLcr 2

rKl

t2

cr

EπF or ,

where:

K = Effective length factor.(Kl) = Effective length.(Kl/r) = Effective slenderness ratio.

see commentary(C – C2.2) (page 16.1-240)

The Euler buckling formula (C-1) is based on:

1 – Perfectly straight column. (no crookedness).

2 – Load is concentric (no eccentricity).

3 – Column is pinned on both ends.

C-8

AISC (Chapter E) of LRFD code stipulates:Pu (factored load) c Pn

where:Pu = Sum of factored loads on column.

c = Resistance factor for compression = 0.90

Pn = Nominal compressive strength = Fcr Ag

Fcr = Critical buckling Stress. (E3 of LFRD)

a) for

3.4-E

EπF

3.3-E 0.877FF

0.44FF or 4.71r

Kl

3.2-E F0.658F

0.44FF or 4.71r

Kl

2

rKL

2

e

ecr

yeFE

ycr

yeFE

y

eFyF

y

b) for

where:

C-9

The above two equations of the LRFD code can beillustrated as below:where:

E

F

rπ

Klλ y

c

* The code further stipulates that an upper value for

for column should not exceed (200).

* For higher slenderness ratio,

Equation (E-3.3) controls and

(Fy) has no effect on (Fcr).

C-10

Determine the design compressive strength (cPn) of W 14x74 with an untraced length of (20 ft), both ends are pinned, (A-36) steel is used?

Example C-2Example C-2

Solution:Solution:

ksi 30.56(96.77)

x2900π

rKl

EπFe

(0k) 20096.772.48

240

r

Kl

2

2

2

2

max.

ksi 21.99360.611

36 x (0.658)F 0.658F 1.178y

F

F

cre

y

Kl =1 x 20 x 12 = 240 inRmin = ry = 2.48

0.44 Fy = 0.44 x 36 =15.84 ksi

Fe ≥ 0.44 Fy Equ. E-3.2

(controls)

c Pn = 0.9 x Fcr x Ag = 0.9 x (21.99) x 21.8

= 433.44 kips (Answer)

Also from (table 4-22) LFRD Page 4-320

c Fcr = 19.75 ksi (by interpolation)

c Pn = c Fcr Ag = 430.55 kips

(much faster)

C-11

For must profiles used as column, the buckling of thin elements in the section may proceed the ever-all bucking of the member as a whole, this is called local bucking. To prevent local bucking from accruing prior to total buckling. AISC provides upper limits on width to thickness ratios (known as b/t ratio) as shown here.

See AISC (B4)(Page 16.1-14)

See also:Part 1 on propertiesof various sections.

C-12

Depending on their ( b/t ) ratios (referred to as ) ,sections are classified as:

a) Compact sections are those with flanges fully welded (connected) to their web and their:

p (AISC B4)

b) Non compact Sections:p r (B4)

c) Slender Section: > r (B4)

Certain strength reduction factors (Q) are introduced for slendermembers. (AISC E7). This part is not required as most sectionselected are compact.

C-13

Example C-3Example C-3Determine the design

compressive strength

(c Pn) for W 12 x 65

column shown below,

(Fy = 50 ksi)?

From properties: From properties:

Ag =19.1 inAg =19.1 in22

rrxx = 5.28 in = 5.28 in

rryy = 3.02 in = 3.02 inksi 40.22550 x 0.8045

F 0.658F

3.2) (E Equ. ksi) 22( F 0.44 F

ksi 96.2(54.55)

x29000π

rKl

EπF

31.795.28

12 x 8 x 1

r

LK

54.555.28

12 x 24 x 1

r

LK

y96.2

50

cr

ye

2

2

2

2

e

y

yy

x

xx

Solution:Solution:

c Pn = 0.9 x Fcr Ag = 0.9 x 40.225 x 19.1 = 691.5 kips

A) By direct LRFDA) By direct LRFD

C-14

B) From Table (4.22) LRFD

Evaluate = = 54.55

Enter table 4.22 (page 4 – 318 LRFD)

cFc = 36.235 ksi (by interpolation)

Pn = Fc x Ag = 692.0 kips

C) From (Table 4.1 LRFD)

maxr

Kl

ft 13.71.75

1x24

rrLK

(KL)

y

x

xxy

Enter table (4.1 ) page 4.17 LFRD with (KL)y = 13.7

Pn = 691.3 kips (by interpolation).

C-15

Design with Columns Load Table (4) LFRD:-Design with Columns Load Table (4) LFRD:-

A) Design with Column Load Table (4) LFRD: The selection of an economical rolled shape to resist a given compressive load is simple with the aid of the column load tables. Enter the table with the effective length and move horizontally until you find the desired design strength (or something slightly larger). In some cases, Usually the category of shape (W, WT, etc.) will have been decided upon in advance. Often the overall nominal dimensions will also be known because of architectural or other requirements. As pointed out earlier, all tabulated values correspond to a slenderness ratio of 200 or less. The tabulated unsymmetrical shapes – the structural tees and the single and double-angles – require special consideration and are covered later.

C-16

EXAMPLE C - 4EXAMPLE C - 4

A compression member is subjected to service loads of 165 kips dead load and 535 kips live load. The member is 26 feet long and pinned in each end. Use (A572 – Gr 50) steel and select a W14 shape.

SOLUTIONSOLUTION Calculate the factored load:

Pu = 1.2D + 1.6L = 1.2(165) + 1.6(535) = 1054 kips

Required design strength cPn = 1054 kips

From the column load table for KL = 26 ft, a W14 176

has design strength of 1150 kips.

ANSWERANSWER

Use a W14 145, But practically W14 132 is OK.

C-17

EXAMPLE C - 5EXAMPLE C - 5Select the lightest W-shape that can resists a factored compressive load Pu of 190 kips. The effective length is 24 feet. Use ASTM A572 Grade 50 steel.

SOLUTIONSOLUTION

The appropriate strategy here is to fined the lightest shape for each nominal size and then choose the lightest overall. The choices are as follows.

W4, W5 and W6: None of the tabulated shape will work.W8: W8 58, cPn = 194 kipsW10: W10 49, cPn = 239 kipsW12: W12 53, cPn = 247 kipsW14: W14 61, cPn = 276 kips

Note that the load capacity is not proportional to the weight (or cross-sectional area). Although the W8 58 has the smallest design strength of the four choices, it is the second heaviest.

ANSWERANSWER Use a W10 49.

C-18

B) Design for sections not from Column Load Tables:

For shapes not in the column load tables, a trial-and-error approach must be used. The general procedure is to assume a shape and then compute its design strength. If the strength is too small (unsafe) or too large (uneconomical), another trial must be made. A systematic approach to making the trial selection is as follows.

1) Assume a value for the critical buckling stress Fcr. Examination of AISC Equations

E3-2 and E3-3 shows that the theoretically maximum value of Fcr is the yield stress Fy.

2) From the requirement that cPn Pu, let

cAgFcr Pu and

3) Select a shape that satisfies this area requirement.4) Compute Fcr and cPn for the trial shape.

5) Revise if necessary. If the design strength is very close to the required value,the next tabulated size can be tried. Otherwise, repeat the entire procedure,using the value of Fcr found for the current trial shape as a value for Step 1

6) Check local stability (check width-thickness ratios). Revise if necessary.

crc

u

FφP

gA

C-19

3.2 . E Equ. LRFD (15.84) F 0.44F

ksi 22.9111.8

x29000π

rKl

EπF

ye

2

2

2

2

e

Example C-6Example C-6Select a W18 shape of A36 steel that can resist a factored load of 1054 kips.The effective length KL is 26 feet.

Solution:Solution:Try Fcr = 24 ksi (two-thirds of Fy):

Required2848

2490

1054in

F

PA

crc

ug .

)(.

Try W18 x 192:Ag = 56.4 in2 > 48.8in2

(OK) 200111.82.79

26(12)

r

KL

min

C-20

(OK) ..

)(

r

KL

in 62.83 .in . A

:234 x Try W18

in .).(.

ARequired

:192) x W18the for computed just value (the ksi .FTry

(N.G.)

105494318.64 x 56.4 x 0.9F A0.9P

ksi 18.64

36 x 0.532x360.658F 0.658F

min

2g

2g

cr

kkipscrgnc

yF

F

cr

22.9

36

e

y

2005109852

1226

868

8362641890

1054

6418

2

crc

u

F

P

C-21

234 x W18a Use Answer

(OK) 42.236

25313.8

t

h

(OK) 15.836

952.8

2t

b

:cheacked bemust ratios

thickness- widththe so tables, load column the innot is shape This

(OK) 1054118519.15 x 68.8 x 0.9F A0.9Pφ

ksi 19.15 36 x 0.53236 x0.658F 0.658F

3.2)-(Equ.E LFRD Use 0.44FF

23.87ksi109.5

29000πEπF

w

f

f

kkipscrgnc

yF

F

cr

ye

2

2

2

rKl

2

e

23.87

36

e

y