19 buckling print

7/30/2019 19 Buckling Print

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20 Buckling Copyright G G Schierle, 2001-08 press Esc to end, for next, for previous slide 1

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Buckling

Buckling is a tendency of slender compression members

to bow out, which causes bending.

When the combined bending and compressive stress

exceeds the buckling capacity failure occurs.

Buckling effects all compression members, such as

columns, truss bars, bracing, etc

A Short columns fail in compression.

B Slender columns fail in buckling

P

P P

PA B C

C Free-body diagram of half column shows bending

moment M = P e (force times lever arm)

Buckling bends a column progressively.

Increasing offset e increases bending, which

in turn increases e further

which finally causes buckling failure

P

P P

PA B C

P

P P

PA B C

P

P P

PA B C


Euler formula, developed by the 18th century Swiss

Mathematician Euler defines critical buckling load as:

Pcr = 2 E I / L2Pcr= critical buckling load

E = Elastic modulus

I = Moment of inertia

L = Length (un-braced)

Since f=P/A, critical buckling stress Fcr=2EI/(AL2), or

Fcr = 2 E / (KL/r2)A = cross section area)

KL/r = slenderness ratior = radius of gyration, r =(I/A)1/2 I / A = r2

K = support factor defined as follows

Support type: Theoretical K Recommended K

A Pin supports: K = 1 K = 1

B One fixed support: K = 07 K = 0.8

C Both fixed supports: K = 0.5 K = 0.65

D Cantilever: K = 2 K= 2.1

E Moment frame K = 1 K= 1.2 -1.5


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fa/Fa + fb/Fb 1

fa = Actual axial stress

Fa= Allowable axial stress

fb = Actual bending stress

Fb = Allowable bending stress

Combined stress examples:

100%axial stress + 0%bending stress



0%axial stress +100%bending stress

Combined stress

Compression members may be subject to combined

axial and bending stress.

1 Wall subject to lateral load

2 Eccentric load causes bending moment M = P e

3 Column with eccentric beam support

4 Cantilever under gravity and lateral load5 Columns of moment frame under gravity load

6 Columns of moment frame under lateral load

7 Interaction triangle, visualizes interaction formula:


Kern,German for core denotes inner 1/3 of cross section

(rhomboid for rectangular, circular for round posts).

Load within the kern causes compressive stress only

Load outside the Kern adds bending stress fb=M/S (M=Pe)

Concentric load causes axial stress only fa= P/A

load inside Kern causes fb < fa

3 Load at Kern edge causes fb = fa

4 Laodoutside the Kern causes fb > fa(yields tensile bending stress)

Proof:

Triangular stress block has centroid @d/3

fa

=P/A

d


Arch and vau lt

Arches and vaults of un-reinforced brick or stone can only

resist compression and must be designed to avoid tension

Thus the funicular pressure line should be within the Kern

(Funicular line may be defined graphically)

1 Funicular arch under uniform load

2 Funicular line at Kern edge under asymmetrical load

3 Funicular line outside Kern under asymmetrical load

(would fail due to tensile stress)

A Compressive stress

B Bending stress compression avoids tensile stress

C Combined stress without tensile stressD Bending stress > compressive stress

E Combined stress includes tensile stress


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Wood buckling

Wood column slenderness is defined as L/d ratio

whered is the least thickness, unless the column

is braced by blocking or sheathing

For braced columns the largest L/d ratio governs

K is ignored since wood moment joints are difficult

d

d

d

d

1 The least dimension d typically defines L/d ratio

2 Square columns with equal d are most effective

3 Columns with blocking are governed by the

largestL/d ratio


Allowable buckling stress Fc:

Fc

= F*c

(CP)

F*c = Fc (CD) (CM) (Ct) X

Fc = Allowable compressive stress parallel to grain

CD= Duration factor (1=normal, 1.6=wind & seismic load)

CM= Moisture factor (1 for dry lumber)

Ct = Temperature factor (1for normal temperature)

X = Other adjustment factors for size, etc. per NDS

CP= Buckling stability defined by Ylinenformula:

2* * *

cE c cE c cE cP

1+F /F 1+F /F F /FC = - -

2c 2c c

c = 0.80 for sawn lumber

0.85 for round timber

0.90 for glulam posts

FcE= KcE E/(L/d)2 (Euler buckling stress)

KcE= 0.300 for visually graded lumber

0.384 for MEL (Machine Evaluated Lumber)

0.418 for glulam & MSR (Machine Stress Rated)

E = Elastic modulus

Wood buckling

Allowable bucklinggraph

Horizontal axis: slenderness L/d

Vertical axis: allowable stress Fc

Allowable buckling load:

P = A Fc

A = cross section area


Example Douglas fir-larch # 1, Fc = 1000 psi

E = 1,600,000 psi, sawn lumber c = 0.8;

visually graded, KcE= 0.3, 4x6 post, L = 8

F*c = Fc (no adjustments needed)

Find allowable load P

Slenderness L/d = 8(12)/3.5 L/d = 27.4

FcE= KcEE /(L/d)2

FcE= 0.3 (1,600,000)/27.42 FcE= 639 psi

FcE/F*c = 639/1000 FcE/F*c = 0.639

Allowable stress

Fc = Fc CP = 1000 (0.524) Fc = 524 psi

Allowable load

P = FcA = 524 (3.5x5.5) / 1000 P = 10 k

Using graph

At Fc=1000 psi, L/d = 27.4 Fc= 520 psi

Allowable load

P=FcA=520 (3.5x5.5)/1000 P = 10 k

Confirming result with good accuracy

8.0

639.0

6.1

639.01

6.1

639.012

+

+=

PC Cp= 0.524


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Interpolation exampleAssume:

8x12 post nominal, 7.25x11.25 actual

F*c=1150 psi, E =1,600,000 psi

L = 12, braced about weak axis at mid-height

Weak axis slenderness

L/d = (12/2) x 12 / 7.25 L/d = 10

Strong axis slenderness

L/d = 12 x 12 / 11.25 L/d = 13

Strong axis slenderness governs 13 > 10

At F*c = 1000 and L/d = 13 Fc= 920 psi

At F*c = 1200 and L/d = 13 Fc=1060 psi

Use conservative 920 psi buckling stress,

or interpolate between 920 and 1060 psi

Allowable stress = X

X/1150 = 1060/1200

Allowable stress

Fc = 1150x1060/1200 Fc=1016 psi

Allowable load

P= FcA=1016x7.25x11.25/1000 P = 82.9 k


Load graph for square posts

Design example

Post: P = 40 K, L=10un-braced length

Find nearest curve above 10 and 40 k

Use 8x8 post

Anal ysis exampl e

Find allowable load for 12x12 post, L= 24

Find intersection of 24 with 12x12 curve

Allowable load P = 64 K

Anal ysis exampl e

Find allowable load for 6x6 post, 20' long, braced at 8

Find intersection of 208=12 with 6x6 curve

Allowable load P = 15 k

Metric design example

Design a 4 m post for P = 200 k

Find curve above intersection of 4 m and 200 kN

Use 10x10 post

Metric analysis example

Find allowable load for a 8x8 post of 6 m long

Find intersection of 6 m with 8x8 curve

Allowable load P = 90 kN


Other posts 1 Glulamposts are effective to resist combined axial and

bending stress by adjusting the strength of laminations

to reflect stress distribution over the cross section.

2 Round poles are very effective for concentric load, since

buckling resistance is uniform in all direction.

But poles are least effective for eccentric load, since

material is focused around the centroid with minimal

lever arm to resist bending stress.

Round poles are designed like square posts, with

diameter adjusted for equal cross section area, i. e.,

d2 = 2/4, ord = 0.886 d = square post thickness

= pole diameter3 Spaced columns for posts and truss bars are designed

for critical L/d ratio (L1/d1, or L2/d2)

19 buckling print

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