# A12_Design_for_buckling-columns_and_plates

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A12 - Design for Column and Plate Buckling 1 Design for Column and Plate Buckling The critical buckling load for a long slender column was previously obtained (see A10 and A11) by solving the governingdifferential equation of equilibrium and is given by: 22 cr   EI  P c L π  = where c is a constant depending upon the end conditions:clamped-free: c=0.25 pinned-pinned: c=1clamped-pinned: c=2clamped-clamped: c=4Equation can be written as a critical buckling stress, and can also be put in terms of a non-dimensional ratio called slenderness ratioas follows. The critical buckling stress is simply:  A12 - Design for Column and Plate Buckling 2 2 22 2 ( / ) cr cr   P EI E c c AL A L A I  π π σ  = = = The term (A/I) is related to the radius of gyration defined by/  I A  ρ  = (units of length)Equation becomes 22 2 (1/ ) cr   E c L π σ  ρ  = . So finally we write the  Euler critical buckling stress as: 22 (/)  E   E c L π σ  ρ  = The term/  L ρ  is non-dimensional and is known as the  slenderness ratio of the column.  A12 - Design for Column and Plate Buckling 3 When Euler's equation is compared to experimental results, itfound that the slenderness ratio must be large in order to obtainacceptable correlation. What is large will be considered shortly.For columns that have a cross-section such that the moments of inertia are different about the two axes, the minimum moment of inertia must be used. For example,suppose we have an aluminum W4x0.15cross-section. This is a cross-section thatis 4 deep and has a web that is 0.15 thick. The top and bottom caps are 0.23 thick and the shear web is 3.54 long. Wehave the following section properties: 244 1.965 , 5.62 , 1.04  xx yy  A in I in I in = = = Consequently, thecolumn will buckle so that bending occurs about the y-axis ( 4min 1.04  I in = ). x 4”0.23”0.23”3.54”0.15”  A12 - Design for Column and Plate Buckling 4 Example . Consider an aluminum column ( 6 10.410  E x psi = ) withthe cross-section above that is pinned on each end (c=1) andL=100 . The radius of gyration is min /0.727  I A  ρ  = = and theslenderness ratio is equal to/100 /0.727 137.6  L ρ  = = . The buckling stress becomes: 2 2 62 2 (10.4 10 )1 5,425( / ) (100 /0.727 )  E   E x psic psi L π π σ  ρ  = = = For a typical aluminum, we note that the yield stress is around 40,000  y psi σ  = (or greater). Hence, buckling will occur well before the yield stress is reached, and buckling for long, slender columns (large/  L ρ  ) is thus geometrically dominated, not materialyielding dominated.For very short columns (small/  L ρ  ), the column will not buckle but simply compress, and a simple /  P A σ  = model is sufficient.Failure will then be due to yielding of the material.

Nov 16, 2017

#### TensileTest

Nov 16, 2017
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