8.2 Inelastic Buckling 261
Also,Pisappliedthroughthecentroidofeachendsectionadistanceefromnnsothat
∫d^1
0
σx(y 1 +e)dA+
∫d^2
0
σv(y 2 −e)dA=−Pv (8.10)
FromFig.8.8(b),
σx=
σ 1
d 1
y 1 σv=
σ 2
d 2
y 2 (8.11)
Theanglebetweentwoclose,initiallyparallel,sectionsofthecolumnisequaltothechangeinslope
d^2 v/dz^2 ofthecolumnbetweenthetwosections.This,inturn,mustbeequaltotheangleδφinthestrain
diagramofFig.8.8(c).Hence,
d^2 v
dz^2
=
σ 1
Ed 1
=
σ 2
Etd 2
(8.12)
andEq.(8.9)becomes,fromEqs.(8.11)and(8.12)
E
d^2 v
dz^2
∫d^1
0
y 1 dA−Et
d^2 v
dz^2
∫d^2
0
y 2 dA= 0 (8.13)
Further,inasimilarmanner,fromEq.(8.10)
d^2 v
dz^2
⎛
⎝E
∫d^1
0
y^21 dA+Et
∫d^2
0
y^22 dA
⎞
⎠+ed
(^2) v
dz^2
⎛
⎝E
∫d^1
0
y 1 dA−Et
∫d^2
0
y 2 dA
⎞
⎠=−Pv (8.14)
Thesecondtermontheleft-handsideofEq.(8.14)iszerofromEq.(8.13).Therefore,wehave
d^2 v
dz^2
(EI 1 +EtI 2 )=−Pv (8.15)
inwhich
I 1 =
∫d^1
0
y^21 dA and I 2 =
∫d^2
0
y^22 dA
the second moments of area aboutnnof the convex and concave sides of the column, respectively.
Putting
ErI=EI 1 +EtI 2
or
Er=E
I 1
I
+Et