chapter - (c) new

7/27/2019 Chapter - (C) New

1/32

C-1

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 strutsor stanchions.

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:Stocky columns (short) fail by yielding of the material at

the cross section, but most columns fail by bucklingat

loads for less then yielding forces.

P

P

P

P

(a) (b)


2/32

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

x

L

y

x

y

Pcr Pcr

Pcr


3/32

C-3

From boundary conditions:

y = 0 @ x = 0, and

y = 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

cr

cr

rL

E

A

PF

---- Euler Buckling Critical Load

where: r = minor radius of gyration

The critical buckling load

is a function of the sectionproperties (A, L, r) and the

modulus of elasticity for

material, and is not a

function of the strength or

grade of the material.

Note:


4/32

C-4

Example C-1

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

Solut ion:

22

r

Lcr

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,

also note Fcr


5/32

Eccentric Loading; The Secant Formula

r

L

EA

P

r

ec

A

P eY

2

1sec1

2max

La carga no esta aplicada en el centroideEl miembro no es perfectamente recto


6/32

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 (Fpl), this is

called

Inelastic Buckling, constantly

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

in the inelastic zone.

Shanley (1947), resolved this

inconsistency.

Columnas largas: fallan por

inestabilidad elstica

Columnas intermedias: falla por

inestabilidad inelstica ( esfuerzo de

compresin es mayor que el esfuerzode fluencia)

Columnas cortas: No presentan

inestabilidad, la columna falla por

agotamiento de la resistencia del

material


7/32

C-6

Depending on (L/r) value the column buckling

strength was presented as shown by Shanley.

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.


8/32

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:

4CE

For,

EF

2

rKl

t2

cr2

rKL

2

cr

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:

1Perfectly straight column. (no crookedness).

2Load is concentric (no eccentricity).3Column is pinned on both ends.


9/32

C-8

AISC (Chapter E) of LRFD code stipulates:

Pu(factored load) cPn

where:

Pu= Sum of factored loads on column.

c = Resistance factor for compression = 0.90

Pn= Nominal compressive strength = FcrAg

Fcr= Critical buckling Stress. (E3 of LFRD)

a) for

3.4-E

EF

3.3-E0.877FF

0.44FFor4.71

r

Kl

3.2-EF0.658F

0.44FFor4.71r

Kl

2

rK L

2

e

ecr

yeF

E

ycr

yeF

E

y

eF

yF

y

b) for

where:


10/32

C-9

The above two equations of the LRFD

code can be illustrated as below:where:

E

F

r

Kl

y

c

The code further stipulates

that an upper value for column

should not exceed (200).

For higher slenderness ratio,

Equation (E-3.3) controls and

(Fy) has no effect on (Fcr).


11/32

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-2

Solution:

ksi30.56(96.77)

x2900

r

Kl

EFe

(0k)20096.772.48

240

r

Kl

2

2

2

2

max.

ksi21.99360.611

36x(0.658)F0.658F 1.178yF

F

cre

y

Kl =1 x 20 x 12 = 240 in

rmin= ry= 2.48

0.44 Fy= 0.44 x 36 =15.84 ksi

Fe 0.44 FyEqu. E-3.2

(controls)

cPn= 0.9 x Fcrx Ag= 0.9 x (21.99) x 21.8

= 433.44 kips (Answer)

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

cFcr= 19.75 ksi (by interpolation)

cPn= cFcrAg= 430.55 kips

(much faster)


12/32

C-11

For must profiles used as column, the

buckling of thin elements in the sectionmay 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 tothickness ratios (known as b/t ratio) as

shown here.

See AISC (B4)

(Page 16.1-14)

See also:

Part 1 on properties

of various sections.


13/32

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:pr (B4)

c) Slender Section:

> r (B4)

Certain strength reduction factors (Q) are introduced for slender

members. (AISC E7). This part is not required as most section

selected are compact.


14/32

C-13

Example C-3

Determine the design

compressive strength

(cPn) for W 12 x 65

column shown below,

(Fy= 50 ksi)?

From properties:

Ag =19.1 in2rx= 5.28 in

ry= 3.02 inksi40.22550x0.8045

F0.658F

3.2)(EEqu.ksi)22(F0.44F

ksi96.2(54.55)x29000

r

KlEF

31.793.02

12x8x1

r

LK

54.555.28

12x24x1

r

LK

y

96.2

50

cr

ye

2

2

2

2

e

y

yy

x

xx

Solut ion:

cPn= 0.9 x FcrAg= 0.9 x 40.225 x 19.1 = 691.5 kips

A) By direct LRFD

(controls)


15/32

C-14

B) From Table (4.22) LRFD

Evaluate = = 54.55

Enter table 4.22 (page 4318 LRFD)

cFc= 36.235 ksi (by interpolation)

Pn= Fcx Ag = 692.0 kips

C) From (Table 4.1 LRFD)

maxr

Kl

ft13.7

1.75

1x24

r

r

LK(KL)

y

x

xxy

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

Pn= 691.3 kips (by interpolation).


16/32

C-15

Example C-4

Solut ion:

Find the maximum load capacity(Pn) of the W 14 x 53 (A-36)

column shown in figure ?

PP

25 ft.

15 ft.

10 ft.

A

C

A

B

C

x xx

x

x-axis Lx= 25 ft, kx= 0.8, rx= 5.89 in.

y-axisSection (AB) Ly= 15 ft, ky= 0.8, ry= 1.92 in.

Section (BC) Ly= 10 ft., ky= 1.0, ry= 1.92 in.

751.92

12150.8

r

Kl

415.98

12250.8

r

Kl

max

y

x

Enter table (4-22) , Fc= 24.1 ksi

Column capacity Pn= FcrAg= 24.1 x 15.6 = 376 kips

(controls)


17/32

C-16

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 unsymmetricalshapes the structural tees and the single and double-angles

require special consideration and are covered later.


18/32

C-17

EXAMPLE C - 5

A compression member is subjected to service loads of 165 kips dead loadand 535 kips live load. The member is 26 feet long and pinned in each end.

Use (A572Gr 50) steel and select a W14 shape.

SOLUTION 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 145

has design strength of 1230 kips.

ANSWER

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


19/32

C-18

EXAMPLE C - 6Select the lightest W-shape that can resists a factored compressive load Puof

190 kips. The effective length is 24 feet. Use ASTM A572 Grade 50 steel.

SOLUTION

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: W 8 58, cPn= 205 kips

W10: W10 49, cPn= 254 kips

W12: W12 53, cPn= 261 kips

W14: W14 61, cPn= 293 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.

ANSWER Use a W10 49.


20/32

C-19

Example C-7

Select the lightest W10 section made of

A 572-Gr50 steel to resist a factored load

of (600 kips) ?

Solut ion:

Assume weak axis (y-y) controls buckling:

Enter design tables of AISC (Section 4) with KyLy= 9 ft.

Select W 10 x 54 (capacity = 625 k > 600 k OK)

Check strong axis buckling strength:

Enter table for W10 x 54 with (KL)eq.= 10.53 ft.Capacity = 595.8 kips (by interpolation) N.G.

Select W10 x 60 capacity = 698 kips for KyLy= 9 ft.

capacity = 666 kips for (KL)eq.= 10.5 ft.


21/32

C-20

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. Ifthe strength is too small (unsafe) or too large (uneconomical), another trial must bemade. 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 Fcris the yield stress Fy.

2) From the requirement that cPnPu, let

cAgFcrPuand

3) Select a shape that satisfies this area requirement.

4) Compute Fcrand cPnfor 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 Fcrfound for the current trial shape as a value for Step 1

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

crc

u

F

P

gA


22/32

C-21

3.2.EEqu.LRFD(15.84)F0.44F

ksi22.9111.8

x29000

r

Kl

EF

ye

2

2

2

2

e

Example C-8

Select a W18 shape of A36 steel that can resist a factored load of 1054 kips.

The effective length KL is 26 feet.

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

Required2848

2490

1054in

F

PA

crc

ug .

)(.

Try W18 x 192:Ag= 56.4 in

2> 48.8in2

(OK)200111.82.79

26(12)

r

KL

min


23/32

C-22

(OK)200109.52.85

26(12)

r

KL

in62.83.in68.8 A

:234xW18Try

in62.830.9(18.64)

1054

F

PARequired

:192)xW18theforcomputedjustvalue(theksi18.64FTry

(N.G.)

105494318.64x56.4x0.9FA0.9P

ksi18.64

36x0.532x360.658F0.658F

min

2g

2

crc

ug

cr

kkipscrgnc

yF

F

cr

2

22.9

36

e

y


24/32

C-23

234xW18aUseAnswer

(OK)42.236

25313.8th

(OK)15.836

952.8

2t

b

:cheackedbemustratios

thicknes-widththesotables,loadcolumntheinnotisshapeThis

(OK)1054118519.15x68.8x0.9FA0.9P

ksi19.1536x0.53236x0.658F0.658F

3.2)-(Equ.ELFRDUse0.44FF

23.87ksi

109.5

29000EF

w

f

f

kkipscrgnc

y

F

F

cr

ye

2

2

2

r

Kl

2

e

23.87

36

e

y


25/32

C-24

The effective length factor (K) was introduced in page (C-7) for six ideal conditions,

these are not encountered in practical field conditions. LRFD commentary provides

both real conditions and standard ideal conditions (C-C2.2) (page 16.1-239 to 242)

Braced Frames:

No lateral movement is allowed

(0.5 < K < 1.0) (sideway prevented)

Unb raced Frames:

Lateral movement possible

(1.0 < K < 20.0) (sideway allowed)

a) Diagonal

bracing

b) Shear Walls

(masonry,

reinforcement concrete

or steel plate)


26/32

C-25

gg

ccB

gg

ccA

/LI

/LIG

/LI

/LIG

* For fixed footing G = 1.0

* For pinned support G = 10.0

where

A is top of column

where

B is bottom of column


27/32

C-26

In the rigid frame shown below, Determine Kxfor columns(AB) & (BC). Knowing that all columns webs are in the plane.

Column (AB):

Joint (A):

0.94

169.21586

1830/181350/201070/12833/12

/LI

/LIG

gg

ccA

Example C 9:-

Solut ion:

A

B

C

W24 x 55

W24 x 55

W24 x 68

W24 x 68W12x120

W12x120

W12x96

12'

15'

12'

18'20'20'


28/32

C-27

For joint B,:-

0.95169.2160.5

169.21070/151070/12

/LI/LIG

gg

cc

From the alignment chart for sideways uninhibited, with GA= 0.94 and GB= 0.95,

Kx= 1.3 for column AB.

Column (BC):

For joint B, as before,

G = 0.95

For joint C, at a pin connection the situation is analogous to that of a very

stiff column attached to infinitely flexible girdersthat is, girders of zero

stiffness. The ratio of column stiffness to girder stiffness would therefore

be infinite for a perfectly frictionless hinge. This end condition is only beapproximated in practice, so the discussion accompanying the alignment

chart recommends that G be taken as 10.0.

From the alignment chart with GA= 0.95 and GB= 10.0, Kx= 1.85 for column BC.


29/32

C-28

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.

Flexural buckling. We have considered this type of buckling up to now. It

is a deflection caused by bending, or flexure, about this axis corresponding

to the largest slenderness ratio (Figure a). This is usually the minor

principal axis the one with the smallest radius of gyration. Compression

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

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 (Figure b).

Standard hot-rolled shapes are not susceptible to torsional buckling, but

members built up from thin plate elements may be and should beinvestigated. The cruciform shape shown is particularly vulnerable to this

type of buckling. This shape can be fabricated from plates as shown in the

figure, or built up from four angles placed back to back.

1.

2.


30/32

C-29

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

of flexural buckling and torsional buckling. The member bends and twists

simultaneously (Figure c). This type of failure can occur only with

unsymmetrical cross sections, both those with one axis of symmetrysuch as

channels, structural tees, double-angle shapes and equal-leg single angles

and those with no axis of symmetry, such as unequal-leg single angles.

3.

The AISC Specification requires

an analysis of torsional or flexural-

torsional buckling when appropriate.

Section E3 of the Specification covers

double-angle and tee-shaped members,

and Appendix E3 provides a more

general approach that can be used for

any unsymmetrical shape.


31/32

When length exceeds requirements for

a single section, built-up compression

section are used as shown below:

The code provides details for built-upsection under LRFD EG.

C-30


32/32

Calculate the capacity of the built-up column shown below.

Lx= Ly= 25 ft, Kx= 1.6, Ky= 1.0 Fy= 42 ksi ?

Solut ion:-

82.903.62

12251.0

r

lK

(controls)108.844.41

12251.6rlK

in3.6222.70

298.0

A

Ir

in4.4122.70

440.5

A

Ir

in22.7047.352A

in298)8(0.953)7.35(57.252I

in440.50.25)(58x1102I

y

yy

x

xx

yy

y

xxx

2

g

43

21

1212

yy

42

21

xx

C-31

Example C 10 :-

From table 4.22 page 4.321

cFcr = 18.3 ksi

Design Nominal Strength = cFcr Ag=18.3 x 22.70

= 415.4 kips.

chapter - (c) new

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