Question

Ideal Column with Pin Supports Learning Goal: To use the formula for the critical load, i.e., the Euler buckling load, for pin-supported columns to calculate various parameters of columns. Ideally, a column that is perfectly straight and has an axial load applied exactly at the centroid of its cross section will not yield until the internal normal stress reaches the yield stress of the material. Real-world columns, however, are subject to small asymmetries, whether due to irregularities of shape or just lateral vibrations, When the axial load is large enough relative to the length of the column, such asymmetries cause the column to become unstable and buckle. It bends and continues bending until it fails. The load required to cause buckling depends on how the column is attached to its supports. The maximum or critical load for a column of length L that is pin supported at both ends is # EI Per A factor of safety is used to account for uncertainties in material properties and

Answer 1

Note : For Applying Euler's buckling Load formula a column must fail through buckling only ( ie not by crushig or combination of crushing and buckling.

Solution :-

PEI Critical Load (Per) L2

Note : Above critical load formula is only applicable for column supported by hing on both ends. In other cases we use L_e ( effective length ) in place of L .

Part A)

Given data : L = 2000 mm , breadth = 48 mm , depth = 89 mm , E = 2*10⁵ N/mm² , Yield Stress = 250 MPa

Inertia of a Rectangular section (Imin) depth * breadth 12

Inertia of a Rectangular section (Imin) 89 * 483 12 820224mm

$\mathbf{Critical \,Load \,(P_{cr})=\frac{\pi^2(2*10^5)(820224)}{2000^2}=404764.2 \,N}$

$\mathbf{Critical \,Load \,(P_{cr})=404.7642 \,KN}$

$\mathbf{Yield \,Load \,For \,This \,Rectangular \,Column \,(F)=\sigma_y*(Area \,of \,cross-section)}$

$\mathbf{Yield \,Load \,For \,This \,Rectangular \,Column \,(F)=250*(48*89)=1068000\,N}$

$\mathbf{F=1068\,KN}$

$\mathbf{P_{cr}<F .....(ie \,column \,fails \,by \,buckling \,first\,)}$

Hence Max Load Supported By Column is Pcr (ie 404.7642 KN)

Part B)

Given data : L = ? , d = 10cm = 100 mm , P_cr = 760 KN = 760 *1000 N , E = 2*10⁵ N/mm²

Assumed that column does not fail in yielding .

$\mathbf{Inertia\, of \,a \,Circular\,section \,\,(I_{min})=\frac{\pi*d^4}{64}}$

$\mathbf{Inertia\, of \,a \,Rectangular \,section \,\,(I_{min})=\frac{\pi*100^4}{64}=4908738.521\,mm^4}$

PEI Critical Load (Per) L2

$\mathbf{760*1000=\frac{\pi^2(2*10^5)(4908738.521)}{L^2}\Rightarrow L=3570.6149\,mm}$

$\mathbf{ L=3570.6149\,mm=3.5706149\,m}$

Hence , Max Height Of the column can be 3.5706149 m

Part C)

Given FS = 2.5 , L = 2000 mm , P_safe = 960*1000 N ,

Assumed that column does not fail in yielding .

$\mathbf{Inertia\, of \,a \,Circular\,section \,\,(I_{min})=\frac{\pi*d^4}{64}}$

PEI Critical Load (Per) L2

$\mathbf{safe \,Load \,(P_{safe})=\frac{P_{cr}}{FS}=\frac{\pi^2EI}{L^2*FS}}$

$\mathbf{960*1000=\frac{\pi^2(2*10^5)(\pi*d^4/64)}{2000^2*2.5}}$

$\mathbf{d=132.82516\,mm=13.282516 \,cm}$

So, Min dia requred to resist this load with a factor of safety of 2.5 is 13.282516 cm

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