COLUMNS

SHORT COLUMNS LONG COLUMNSFail by crushing, cracking Fail by buckling

4.71y

KL Er F

4.71y

KL Er F

= Slenderness Ratio

r = Least radius of gyration, inches

E = Modulus of elasticity, ksi

Fy = Yield strength of steel, ksi

KLr

SHORT COLUMNSLONG COLUMNS

SLENDERNESS RATIO

CRITICALSTRESS, ksi (Fcr)

Fe = Euler’s Buckling Stress, ksi2

2( )

E

KL r

KLr

Pn = Nominal Strength (Failure Strength) of Column = Fcr. Ag, kips

Design strength of column = 0.9 x Pn

FOR LRFDDESIGN STRENTH DESIGN LOAD Pn Pu

C

C = Resistance Factor for Compression, 0.9

Fixed FreePin

FIXED

FIXED

FIXED

PINPIN

PIN

K = 0.65 K = 1.0 K = 0.8

RECOMMENDED VALUES OF EFFECTIVE LENGTH FACTOR , K

POINT OF INFLECTION

K = 2.1

FIXED

FREE TO ROTATE & TRANSLATE

HSS 10x6x ½ASTM A500,Grade B steel(Fy = 46ksi)

FIXED

FIXED

K = 0.65

PROBLEM # 1

Determine the available strength of the compression member shown in Figure 1, in each of the following ways:a. Use AISC Equation E3-2 or E3-3. Compute both the design strength for LRFD and

the allowable strength for ASD.b. Use table 4-22 from Part 4 of the Manual. Compute both the design strength for

LRFD and the allowable strength for ASD.

Figure 1

3.57xr

2.39yr

PROBLEM # 1

1) Find Slenderness Ratio

KL 0.65(15x12)48.95 200

r 2.39 OK

2) Find Euler’s Buckling Stress

2 2

22eE x29,000

F 119.5ksi48.95KL

r

3) Find if Short or Long Column

29,0004.71 4.71 118.3

46

y

E

F

KL48.95 118.3

r SHORT COLUMN

4) Find Fcr

0.658 .

Fy

FeyF Fcr

46119.5

cr 0.658 .46

39.15ksi

F

HSS 10x6x ½ASTM A500,Grade B steel(Fy = 46ksi)

FIXED

K = 0.65

FIXED

Figure 1

PROBLEM # 1

5) Design Strength of Column = 0.9 x Fcr x Ag = 0.9 x 39.15 x 13.5 = 475.7 kips

KL = 0.65 x 15 = 9.75’

Effective length

9' 486kips9.75' 474kips10' 470kips

*By interpolation = 486 -

= 474 kips

486 470x0.75

1

Page # 4-31

Page # 4-319

KLr

crF

49 35.2

Load = 35.2 x 13.5 = 475.2 kips

C

PROBLEM # 1

For the conditions shown in Figure 1, use LRFD and

a. Select a W12 of A992 steel.b. Select a steel pipe.c. Select a square HSS.d. Select a rectangular HSS.

W12 PIPE, RECTANGULAR HSS, SQUARE HSS

Pu = 1.2D + 1.6L = 1.2 x 90 + 1.6 x 260 = 524 kips

KL = 0.65 x 15.33’ = 9.97’

D = 90kL= 260k

Figure 1

PROBLEM # 2

WEB

FLANGES

A 992STEEL

PROBLEM # 2

W12

Page # 4-18

COLUMN DESIGNATION

WEIGHT(lbs/ft)

LOAD CAPACITY Pn (kips)

STRENGTH REQUIRE (Pu)

W12 53 590 524 kips

C

PROBLEM # 2

W12

PROBLEM # 2

Page # 4-64

ROUND HSS

Page # 4-65

PROBLEM # 2

ROUND HSS

PROBLEM # 2

Page # 4-65

ROUND HSS

Page # 4-66

PROBLEM # 2

ROUND HSS

COLUMN DESIGNATION

WEIGHT(lbs/ft)

LOAD CAPACITY Pn (kips)

STRENGTH REQUIRE (Pu)

HSS 16.000x0.375 62.6 630 524 kips

HSS 16.000x0.312 52.3 528 524 kips

HSS 14.000x0.375 54.6 545 524 kips

HSS 10.750x0.500 54.8 531 524 kips

C

PROBLEM # 2

ROUND HSS

PROBLEM # 2

SQUARE HSS

Page # 4-48

PROBLEM # 2

SQUARE HSS

Page # 4-49

PROBLEM # 2

SQUARE HSS

Page # 4-51