Stability Of Structures: Additional Topics

26

Stability Of

Structures:

Additional Topics

P

=

π

_{E I}

L

Unified Column Buckling Formula

Effective length

P

=

π

_{E I}

L

Actual column length

Unified formula

for other end

conditions

Euler formula for

pinned-pinned

column

Effective Buckling Lengths For Several

End Condition Cases

P

pinned-pinned

(Euler column)

free-fixed

(cantilever)

pinned-fixed

fixed-fixed

L = L

L =2L

L

L

P

P

L

P

L =L/2

L =0.7 L

Slenderness Ratio

Given a column cross section with area A and minimum moment of

inertia I = I , its

radius of gyration

is defined as

The slenderness s is the ratio of the effective column length to

the radius of gyration:

This dimensionless ratio characterizes the

failure mode

of the

column, as described later.

r = I/A r = + I/A

√

eff min

s =

L

r

Critical Stress

P

=

π

E I

L

=

π

_{E A r}

L

=

π

_{E A}

s

Substituting I = A r in P =

π

E I / L and replacing L /r by s

yields

Dividing this axial load by A gives the axial stress at the critical load:

eff eff eff eff

σ

=

P

A

=

π

_E

s

Short vs Long Columns

Columns have two basic failure modes: yield and buckling.

They are classified according to which mode happen first:.

A long column (a.k.a. slender column)

buckles first

A short column (a.k.a. stout column)

yields first

One quick way to classify a given column is to compute the

stress at the critical load:

σ

=

π

E / s and compare it to the

yield stress

σ .

If

σ

is less than

σ

the column is long, since it will buckle first.

If

σ

exceeds

σ ,

yield will happens first and the column is short.

If

σ

=

σ ,

the column will simultaneously fail in both modes.

cr cr cr cr 2 2 Y Y Y

Failure Envelope Diagrams

To do column design quickly it is convenient to make use of

Failure Envelope Diagrams. These are constructed as follows.

Introduce two dimensionless ratios for the column material:

Divide both sides of the critical stress formula by

σ

and introduce

the foregoing ratios to get the

dimensionless

expression

This is graphed in the next slide for 3 materials:

structural steel

(E = 210 GPa,

σ

= 210 MPa, E = E/

σ

= 1000),

aluminum alloy

(E = 70 GPa,

σ

= 280 MPa, E = E/

σ

= 250) and

fir wood

(E =12.6 GPa,

σ

= 35 MPa in compression, E = E /

σ

= 360).

These curves delimit the so-called universal failure envelopes.

σ

=

σ

σ

,

E

=

E

σ

σ

=

π

¯E

s

Universal Slenderness Versus

Column Failure Diagram

1.2 1.0 0.8 0.6 0.4 0.2 0.0 0 50 100 150 200

σ

=

σ /σ

Short vs Long Columns: Example 1

A pinned-pinned streel column with E = 210 GPa and yield stress σ = 210 MPa has a pin-to-pin length of L = 5 m = 5000 mm, and a b x h solid rectangular cross section with b = 0.12 m = 120 mm and h = 0.08 m = 80 mm. Will the column fail first by yield or elastic buckling?

Solution by stress comparison. The critical Euler load is P = π E I / L since

L = L for the pinned-pinned case. The minimum second moment of inertia

is I = b h /12 because h < b. Replace and divide by A = b h to get σ = P /A = π H h /(12 L) = 44.8 N/mm = 44.8 MPa. Compare to yield: σ < σ = 210 MPa. Thus the column will fail first by buckling.

Solution by slenderness ratio. Alternatively, one can test the slenderness ratio:

s = L /r, in which L = L and r = I / A = h /12. A quick computation gives s = L 12 / h = 5000 12 /80 ~ 216, which is way into the long column range as

can be readily checked in the failure envelope disgram of the previous slide. eff Y cr cr cr cr eff eff Y

Short vs Long Columns: Example 2

A fixed-fixed streel column with E = 210 GPa and yield stress σ = 210 MPa of length L = 6 m has a solid circular cross section of unknown radius R. Find:

(1) the radius R in mm so the column fails simultaneously by yield and by elastic buckling, (2) the maximum load P that the designed column can support if the safety factor against both buckling and yield is 4.

Solution of (1). Equate σ = σ and solve for R. Details: L = L/2, I = (π/4) R , A = π R , r = I/A = R /4, σ = π E (R /4) = π E R /L = σ , whence R = σ / E L / π = L /(π E ) = 6000/99.34. Thus R = 60.4 mm.

Solution of (2). The cross section area of the designed column is A = π R = 11461 mm . The failure load is P = σ A = 2.407 ×10 N. Dividing by the safety factor of 4 gives

P = 6.02 × 10 N.

Southwell Plot Configuration

v

v

P

experimental data points

~

slope = P_cr

Exterimental Data Recorded For Pinned-Pinned

Column, Fall 2010 Column Buckling Lab Demo

(Converted from Excel spreadsheets to TeX table format)

e= 1.5 mm (1 notch) e = 3 mm (2 notches) e = 4.5 mm (3 notches) e = 6 mm (4 notches) Offset=1.7 mm† Offset = 4 mm Offset = 4.5 mm Offset = 7.5 mm TLoad(N)‡ Def(mm) TLoad(N) Def(mm) TLoad(N) Def(mm) TLoad(N) Def(mm)

4 2.1 4 5 3 5 3 8 8 2.8 8 7 6 6 6 9 12 3.5 12 8 9 7 9 10 16 4.2 16 9 12 8 12 11 20 5.6 20 11 15 9.5 15 13 24 7.5 24 15 18 11.5 18 15 28 11.2 28 21.5 21 13.5 21 18 30 14.5 30 25.5 24 16.5 24 21 32 19.8 32 34 27 22.5 27 28 34 30.0 34 48 30 31.5 30 39 36 53.1 35 61 33 48.5 33 59 34 59

† Offsets are chosen by trial and error so lower left portion of the S-plots look reasonable. ‡ TLoad means “tray load”. Actual load on tested columns is (4/3)× tray load.

Mathematica Script To Produce Southwell Plots

For Data of Previous Slide: Pinned-Pinned Column

<<Graphics`MultipleListPlot`;

(* Southwell plots for Pinned-Pinned column - Fall 2010 lab *)

offs1=1.7; offs2=4; offs3=4.5; offs4=7.5;

PPdata1={{4,2.1},{8,2.8},{12,3.5},{16,4.2},{20,5.6},{24,7.5},{28,11.2}, {30,14.5},{32,19.8},{34,30},{36,53.1}}; PPdata2={{4,5},{8,7},{12,8},{16,9},{20,11},{24,15},{28,21.5}, {30,25.5},{32,34},{34,48},{35,51}}; PPdata3={{3,5},{6,6},{9,7},{12,8},{15,9.5},{18,11.5},{21,13.5}, {24,16.5},{27,22.5},{30,31.5},{33,48.5},{34,59}}; PPdata4={{3,8},{6,9},{9,10},{12,11},{15,13},{18,15},{21,18}, {24,21},{27,28},{30,39},{33,59}}; PPSouth1=Table[N[{(PPdata1[[i,2]]-offs1)/((4/3)*PPdata1[[i,1]]), PPdata1[[i,2]]-offs1}],{i,1,Length[PPdata1]}]; PPSouth2=Table[N[{(PPdata2[[i,2]]-offs2)/((4/3)*PPdata2[[i,1]]), PPdata2[[i,2]]-offs2}],{i,1,Length[PPdata2]}]; PPSouth3=Table[N[{(PPdata3[[i,2]]-offs3)/((4/3)*PPdata3[[i,1]]), PPdata3[[i,2]]-offs3}],{i,1,Length[PPdata3]}]; PPSouth4=Table[N[{(PPdata4[[i,2]]-offs4)/((4/3)*PPdata4[[i,1]]), PPdata4[[i,2]]-offs4}],{i,1,Length[PPdata4]}]; MultipleListPlot[PPSouth1,PPSouth2,PPSouth3,PPSouth4, PlotJoined->True,Frame->True];

Southwell Plot for Pinned-Pinned Case

Data Recorded in Fall 2010 Lab For 4 Eccentricities

0 0.2 0.4 0.6 0.8 1 1.2 0 10 20 30 40 50 Deflection in mm

Deflection over column axial load in mm/N

"Eyeballed" Fit

Deflection over column axial load in mm/N

Analytical Buckling Load For

Pinned-Pinned Case (Euler Column)

Critical load of pinned-pinned test column (Euler column)

Pcr=50.542 Nw Ptray=37.90 Nw

Em=190000*Nw/mm^2; L=600*mm; t=1.67*mm; w=25*mm; Izz=w*t^3/12; Pcr=N[Pi^2]*Em*Izz/L^2; Ptray=Pcr*3/4; Print["Pcr=",Pcr," Ptray=",Ptray];

Comparison Of Southwell Plot Critical Load

Predictions Versus Analytical Values - Fall 2010

P_crtest 60 1.28− 0.12 51.7 N P_crt est 60 0.61− 0.06 109.1 N P_crtest 60 0.76− 0.05 83.3 N Pinned-pinned (Euler) column:

Very good agreement with analytical result of 50.5 N

Pinned-fixed column:

Moderately good agreement with analytical result of 103.4 N

Pinned-restrained column:

Mediocre agreement with analytical result of 73.6 N

Probable reason for discrepancy in the last case: torsional spring model doesn't do a good job of capturing the rigid angle bracket at column bottom end. The presence of this bracket may increase the equivalent torsional stiffness significantly.

ITL Column Buckling Test Module