5 1(Compression Members)

7/27/2019 5 1(Compression Members)

1/30

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 struts orstanchions.

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 buckling at

loads for less then yielding forces.

P

P

P

P

(a) (b)


2/30

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


3/30

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:


4/30

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.

Pcr= FcrA = 19.1 x 14.7 = 280.8 kips


5/30C-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.


6/30

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.


7/30

C-7

The Previous conditions are very

difficult to achieve in a realistic building

condition, especially the free rotation of pinnedends. Thus an effective slenderness factor is

introduced to account for various end

conditions:

Thus:

4

2

2

CE

F

rKL

cr 2

rK l

t

2

cr

EFor,

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.


8/30

C-7


9/30

C-7


10/30

C-7


11/30

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 = 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:


12/30

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

for column should not exceed (200).

* For higher slenderness ratio,

Equation (E-3.3) controls and

(Fy) has no effect on (Fcr).


13/30

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 Fy Equ. E-3.2

(controls)

c Pn = 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

c Fcr= 19.75 ksi (by interpolation)

c Pn = c FcrAg = 430.55 kips

(much faster)


14/30

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 properties

of various sections.


15/30

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.


16/30

C-13

Example C-3

Determine the design

compressive strength

(c Pn) 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

31.795.28

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:

c Pn = 0.9 x FcrAg = 0.9 x 40.225 x 19.1 = 691.5 kips

A) By direct LRFD


17/30

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)

maxrKl

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).


18/30

C-15

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.


19/30

C-16

EXAMPLE C - 4

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 (A572 Gr 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 176

has design strength of 1150 kips.

ANSWER

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


20/30

C-17

EXAMPLE 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.

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: W8 58, cPn = 194 kips

W10: W10 49, cPn = 239 kips

W12: W12 53, cPn = 247 kips

W14: 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.

ANSWER Use a W10 49.


21/30

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. 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 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 Fcrand 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 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


22/30

C-19

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

ksi22.9111.8

x29000

r

Kl

EF

ye

2

2

2

2

e

Example C-6

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


23/30

C-20

(OK)..

)(

r

KL

in62.83.in.A

:234xTry W18

in.).(.

ARequired

:192)xW18theforcomputedjustvalue(theksi.FTry

(N.G.)

105494318.64x56.4x0.9FA0.9P

ksi18.64

36x0.532x360.658F0.658F

min

2g

2g

cr

kkipscrgnc

yF

F

cr

22.9

36

e

y

2005109852

1226

868

8362641890

1054

6418

2

crc

u

F

P


24/30

C-21

234xW18aUseAnswer

(OK)42.236

25313.8t

h

(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

kkips

crgnc

y

F

F

cr

ye

2

2

2

r

Kl

2

e

23.87

36

e

y

COMPRESSIVE STRENGTH FOR TORSIONAL AND


25/30

C-21

COMPRESSIVE STRENGTH FOR TORSIONAL AND

FLEXURAL-TORSIONAL BUCKLING OF MEMBERS

WITHOUT SLENDER ELEMENTS

This section applies to singly symmetric and unsymmetric members, and certaindoubly symmetric members, such as cruciform or built-up columns with compact

and noncompact sections

For double-angle and tee-shaped compression members:

where Fcry is taken as Fcr from Equation E3-2 or E3-3, for flexural buckling

about the y-axis of symmetry and KL/r= KL/ry, and

COMPRESSIVE STRENGTH FOR TORSIONAL AND FLEXURAL


26/30

C-21

COMPRESSIVE STRENGTH FOR TORSIONAL AND FLEXURAL-

TORSIONAL BUCKLING OF MEMBERS WITHOUT SLENDER ELEMENTS

COMPRESSIVE STRENGTH FOR TORSIONAL AND FLEXURAL


27/30

C-21

COMPRESSIVE STRENGTH FOR TORSIONAL AND FLEXURAL-

TORSIONAL BUCKLING OF MEMBERS WITHOUT SLENDER ELEMENTS


28/30

C-21

Single angle compression membera) 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 adjacent web

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

For unequal-leg angles with leg

length ratios 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 themembers shall not be less than

0.95L/rz .

where

L = length of member between work points at truss chord centerlines,

in. (mm)

bl = longer leg of angle, in. (mm)bs = shorter leg of angle, in. (mm)

rx = radius of gyration about geometric axis parallel to connected leg,

in. (mm)

rz = radius of gyration for the minor principal axis, in. (mm)

Si l l i b


29/30

C-21

Single angle compression member

(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 membersattached to the same side of the gusset plate or chord:



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 member shall not be less

than 0.82L/rz ,

where


in. (mm)



in. (mm)


Si l l i b


30/30

C-21

Single angle compression member

(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 membersattached to the same side of the gusset plate or chord:



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 member shall not be less

than 0.82L/rz ,

where


in. (mm)



in. (mm)


5 1(Compression Members)

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