EXAMPLE 13 - KOCWelearning.kocw.net/KOCW/document/2015/hanyang/han... · • Practically, columns...

EXAMPLE 13.2

Copyright © 2011 Pearson Education South Asia Pte Ltd

A W150×24 steel column is 8 m long and is fixed at its ends as shown in Fig. 13–11a. Its load-carrying capacity is increased by bracing it about the y–y (weak) axis using struts that are assumed to be pin connected to its mid-height. Determine the load it can support so that the column does not buckle nor the material exceed the yield stress. Take Est = 200 GPa and σY = 410 MPa.

EXAMPLE 13.2 (cont)


• Effective length for buckling about the x–x and y–y axis is

• From the table in Appendix B,

• Applying Eq. 13–11,

Solutions

mm 2800m 8.22/87.0

mm 4000m 485.0

y

x

KLKL

46

46

mm 1083.1

mm 104.13

y

x

I

I

2 62

2 2

2 62

2 2

200 13.4 101653.2 kN (1)

4000

200 1.83 10460.8 kN (2)

2800

cr xx

cr yy

EIPKL

EIPKL

EXAMPLE 13.2 (cont)


• By comparison, buckling will occur about the y–y axis.

• The average compressive stress in the column is

• Since this stress is less than the yield stress, buckling will occur before the material yields.

• Thus,

• From Eq. 13–12 it can be seen that buckling will always occur about the column axis having the largest slenderness ratio, since it will give a small critical stress.

Solutions

32

460.8 10150.6 N/mm 150.6 MPa

3060cr

crPA

461 kN (Ans)crP

13.4 SECANT FORMULA


• For design of a column subjected to eccentric load, consider the moment-curvature equation

2 2 2

''

or '' where

EIV M P e vPv v eEI

13.4 SECANT FORMULA


• The solution is

• Since v = 0 at x = 0, so C2 = e

• Since v = 0 at x = L,

So,

2 2 2

''

or '' where

EIV M P e vPv v eEI

exCxCv cossin 21

21 cos 2sin and sin 2sin cos2 2 2L L LL L

2tan1

LeC

SECANT FORMULA (cont)


• Hence,

• And

• To find we require

1cossin

2tan xxLev

max 2 sec 2 1x Lv e L

crP

2sec L

EIPcr

22

L

EIPcr

2

2 crEIP

L



• Hence,

• And

• Note: A nonlinear relationship occurs between the load P and the defection v. As a result, the principle of superposition does not apply here.

1cossin

2tan xxLev

12sec2max Lev Lx



• Maximum moment occurs at the column’s midpoint, i.e.

• Hence the maximum stress is

max

max 2

sec2

Or, 1 sec2

P Mc P Pe c LA I A I

P ec L PA r r EA

2secor max

LPeMvePM



Design curves:• The above equation for maximum stress σmax is

transcendental, and cannot be solved explicitly. • Graphs to aid designer are available.

EXAMPLE 13.4


The W200 x 59 A-36 steel column shown in Fig. 13–17a is fixed at its base and braced at the top so that it is fixed from displacement, yet free to rotate about the y–y axis. Also, it can sway to the side in the y–z plane. Determine the maximum eccentric load the column can support before it either begins to buckle or the steel yields.

EXAMPLE 13.4 (cont)


• For y–y axis buckling, it is subjected to an axial load P.

• For x–x axis yielding, it is subjected to an axial load P and moment M.

Solutions

(Ans) kN 4.419N 419368

10143.1sec598.2110895.1

2sec1

36

2

x

xx

x

x

x

x

xY

PPP

EAP

rKL

rec

AP

kN 51362800

104.20102002

632

2

2

y

yycr KL

EIP

13.5 INELASTIC BUCKLING


• In engineering practice, columns are generally classified according to the type of stresses developed within the column at the time of failure.

• Long slender columns will become unstable when the compressive stress remain elastic. It is referred as elastic instability.

• Intermediate Columns fail due to inelastic instability, meaning that compressive stress at failure is greater than the material’s proportional limit (σpl).

• Short Columns do not become unstable; rather the material simply yields or fractures.

INELASTIC BUCKLING (cont)


• Consider a column having a slenderness ratio KL/r. The formula for critical stress depends on the value of the slenderness ratio.– When KL/r ≥ (KL/r)pl, apply the Euler’s formula.– When KL/r ≤ (KL/r)pl, apply the Engesser formula.

EXAMPLE 13.5


A solid rod has a diameter of 30 mm and is 600 mm long. It is made of a material that can be modeled by the stress–strain diagram shown in Fig. 13–19. If it is used as a pin-supported column, determine the critical load.

EXAMPLE 13.5 (cont)


• The radius of gyration is

• The slenderness ratio is

• Engesser equation states that

• For elastic critical stress, thus .

• Inelastic buckling occurs since

Solutions

805.7

6001

rKL

mm 5.715

154/2

4

AIr

t

tcr E

rKLE 3

2

2

10542.1

GPa 150001.0

150E MPa 3.231cr

MPa 150 plcr

EXAMPLE 13.5 (cont)


• From the second line segment of the graph,

• Applying the value, we have

• Since this value falls within the limits of 150 MPa and 270 MPa, it is critical stress.

• The critical load on the rod is therefore

Solutions

MPa 1.1851012010542.1 33 cr

GPa 120001.0002.0

150270

tE

(Ans) kN 131015.01.185 2 AP crcr

13.6 ALLOWABLE STRESS FOR CONCENTRIC LOADING


• Practically, columns are not perfectly straight as assumed in theory. By performing experimental tests on a large number of axially loaded columns, the results may be plotted and a design formula developed by curve-fitting the mean of the data.

• Note: The experimental curve is similar to that determined from the secant formula. It accounts for the influence of an “accidental” eccentricity ratio on the column’s strength.

ALLOWABLE STRESS FOR CONCENTRIC LOADING (cont)


• For steel columns:

33

2

2

2

2

2

8//83352

1

2

21

cc

Yc

allow

Yc

cY

rKLrKLrKLrKLrKLrKL

Er

KL

rKLE

2

212 for 200

23allow

c

E KL KLr rKL r

for c

KL KLr r



• For aluminum columns:

rKL

rKL

rKL

rKLr

KL

allow

allow

allow

55 378125

5512 MPa 628.15.214

120 MPa 195

2



• For timber columns:

5026 MPa 3718

2611 MPa 0.26/

31125.8

110 MPa 25.8

2

2

dKL

dKL

dKLdKL

dKL

allow

allow

allow

EXAMPLE 13.6


An A-36 steel W250 x 149 member is used as a pin-supported column, Fig. 13–24. Using the AISC column design formulas, determine the largest load that it can safely support.

EXAMPLE 13.6 (cont)


• From Appendix B,

• Since k = 1 for both x and y axis buckling,

• From Eq. 13–22, we have

• Here , so

• The allowable load P on the column is therefore

Solutions

18.744.67

50001

rKL

mm 4.67 mm 117 mm 19000 2 yx rrA

crKLrKL //0

66.125250

1020022 322

Yc

Er

KL

2

2

3 3

12

110.85 MPa5 3 3 8 / / 8

Y

callow

c c

KL rKL r

KL r KL r KL r KL r

(Ans) kN 210619000

85110 ; PP.AP

allow

13.7 DESIGN OF COLUMNS FOR ECCENTRIC LOADING


• Method 1: Use available column formula

• Method 2: Use Interaction Formula

IMc

APmax where M = Pe such that σmax ≤ σallow

1a b

a ballow allow

where σa denotes the stress due to axial load, σb denotes the stress due to bending.

EXAMPLE 13.10


The column in Fig. 13–29 is made of aluminum alloy 2014-T6 and is used to support an eccentric load P. Determine the maximum magnitude of P that can be supported if the column is fixed at its base and free at its top. Use Eq. 13–30.

EXAMPLE 13.10 (cont)


• From Fig. 13–10b, K = 2.The largest slenderness ratio for the column is therefore

• By inspection, Eq. 13–26 must be used (277.1 > 55). Thus,

• The maximum compressive stress in the column is determined from the combination of axial load and bending.

• Assuming that this stress is uniform over the cross section, we require

Solutions

1.277

8040/408012/1

160023

r

KL

MPa 92.4

1.277378125

/378125

22 rKLallow

PI

cPeAP 00078125.0max

(Ans) kN 30.600078125.0.924 ;max PPallow

EXAMPLE 13.12


The timber column in Fig. 13–31 is made from two boards nailed together so that the cross section has the dimensions shown. If the column is fixed at its base and free at its top, use Eq. 13–30 to determine the eccentric load P that can be supported.

EXAMPLE 13.12 (cont)


• Since K = 2,

• Since 26 < KL/d < 50, the allowable axial stress is

• With σallow = σmax,

Solutions

MPa 324.2

403718

/3718

22 dKLallow

4060

12002

dKL

(Ans) kN 35.3

12060121)60)(80(

12060324.2 3

P

PPI

McAP

allow

EXAMPLE 13 - KOCWelearning.kocw.net/KOCW/document/2015/hanyang/han... · • Practically, columns...

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