Stability Of Structures: Additional Topics

Transcription

1 26 Stability Of Structures: Additional Topics ASEN 3112 Lecture 26 Slide 1

2 Unified Column Buckling Formula Euler formula for pinned-pinned column P cr = π 2 EI L 2 Actual column length Unified formula for other end conditions P cr = π 2 EI L 2 ef f Effective length ASEN 3112 Lecture 26 Slide 2

3 Effective Buckling Lengths For Several End Condition Cases ASEN Structures P P P P L L eff =0.7 L L eff = L L =2L L L eff L =L/2 eff fictitious continuation about fixed end pinned-pinned (Euler column) free-fixed (cantilever) pinned-fixed fixed-fixed ASEN 3112 Lecture 26 Slide 3

4 Slenderness Ratio ASEN Structures Given a column cross section with area A and minimum moment of inertia I = I, its radius of gyration is defined as min 2 r = I/A r = + I/A The slenderness s is the ratio of the effective column length to the radius of gyration: s = L r eff This dimensionless ratio characterizes the failure mode of the column, as described later. ASEN 3112 Lecture 26 Slide 4

5 Critical Stress ASEN Structures Substituting I = A r in P = π E I / L and replacing L eff /r by s yields cr eff P cr = π 2 EI L 2 = π 2 EAr 2 L 2 eff eff = π 2 EA s 2 Dividing this axial load by A gives the axial stress at the critical load: σ cr cr = P A = π 2 E s 2 ASEN 3112 Lecture 26 Slide 5

6 Short vs Long Columns ASEN Structures 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 2 E / s 2 cr and compare it to the yield stress σ. If σ cr is less than σ Y the column is long, since it will buckle first. If σ cr exceeds σ, yield will happens first and the column is short. If σ = σ, the column will simultaneously fail in both modes. cr Y Y ASEN 3112 Lecture 26 Slide 6

7 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: - σ def = σ cr, E def = E cr σ Y σ Y Divide both sides of the critical stress formula by σ Y and introduce the foregoing ratios to get the dimensionless expression σ- cr = π 2 Ē This is graphed in the next slide for 3 materials: structural steel - (E = 210 GPa, σ = 210 MPa, E = E/ σ = 1000), aluminum alloy Y - (E = 70 GPa, σ Y = 280 MPa, E = E/ σ Y = 250) and fir wood - (E =12.6 GPa, σ Y = 35 MPa in compression, E = E / σ Y = 360). These curves delimit the so-called universal failure envelopes. s 2 - ASEN 3112 Lecture 26 Slide 7

8 Universal Slenderness Versus Column Failure Diagram σ cr = σ cr /σ Y Failure by yield (short columns) Aluminum Failure by buckling (long columns) Fir Wood Steel Slenderness ratio s = L /r eff ASEN 3112 Lecture 26 Slide 8

9 Short vs Long Columns: Example 1 A pinned-pinned streel column with E = 210 GPa and yield stress σ Y = 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? 2 2 Solution by stress comparison. The critical Euler load is P cr = π E I / L since L eff = L for the pinned-pinned case. The minimum second moment of inertia 3 is I = b h /12 because h < b. Replace and divide by A = b h to get σ cr = P cr /A = 2 3 π H h /(12 L) = 44.8 N/mm = 44.8 MPa. Compare to yield: σ cr < σ Y = 210 MPa. Thus the column will fail first by buckling. Solution by slenderness ratio. Alternatively, one can test the slenderness ratio: 2 2 s = L eff /r, in which L eff = L and r = I / A = h /12. A quick computation gives s = L 12 / h = /80 ~ 216, which is way into the long column range as can be readily checked in the failure envelope disgram of the previous slide. ASEN 3112 Lecture 26 Slide 9

10 Short vs Long Columns: Example 2 ASEN Structures A fixed-fixed streel column with E = 210 GPa and yield stress σ Y = 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 max that the designed column can support if the safety factor against both buckling and yield is 4. Solution of (1). Equate σ cr = σ Y and solve for R. Details: L eff = L/2, I = (π/4) R, A = π R, r = I/A = R /4, σ cr = π E (R /4) = π E R /L = σ Y, whence R = σ Y / E L / π = L /(π E ) = 6000/ Thus R = 60.4 mm. 4 2 max 2 2 Solution of (2). The cross section area of the designed column is A = π R = mm. 12 The failure load is P cr = σ Y A = N. Dividing by the safety factor of 4 gives 11 P = N. ASEN 3112 Lecture 26 Slide 10

11 Southwell Plot Configuration v m experimental data points ~ slope = P cr v m P ASEN 3112 Lecture 26 Slide 11

12 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) 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. ASEN 3112 Lecture 26 Slide 12

13 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]; ASEN 3112 Lecture 26 Slide 13

14 Southwell Plot for Pinned-Pinned Case Data Recorded in Fall 2010 Lab For 4 Eccentricities 50 Deflection in mm Deflection over column axial load in mm/n ASEN 3112 Lecture 26 Slide 14

15 "Eyeballed" Fit ASEN Structures 50 Deflection in mm "Eyeballed" best-fit Deflection over column axial load in mm/n ASEN 3112 Lecture 26 Slide 15

16 Analytical Buckling Load For Pinned-Pinned Case (Euler Column) ASEN Structures Critical load of pinned-pinned test column (Euler column) 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]; Pcr= Nw Ptray=37.90 Nw ASEN 3112 Lecture 26 Slide 16

17 Comparison Of Southwell Plot Critical Load Predictions Versus Analytical Values - Fall 2010 Pinned-pinned (Euler) column: P test cr P test cr Very good agreement with analytical result of 50.5 N Pinned-fixed column: P test cr N N Moderately good agreement with analytical result of N Pinned-restrained column: Mediocre agreement with analytical result of 73.6 N 83.3 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. ASEN 3112 Lecture 26 Slide 17

18 ITL Column Buckling Test Module ASEN Structures 3L L counterweight load arm knife edge beam-column specimen of high-strength steel ruler stop load tray beam clamps frame restraint beam ASEN 3112 Lecture 26 Slide 18