Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

282 CHAPTER 8 Columns

Fig.8.18

Column section of Example 8.4.

The position of the shear center S is found using the method of Example 16.2; this gives
xS=−76.2mm.TheremainingsectionpropertiesarefoundbythemethodsspecifiedinExample8.3
andarelistednext:

A=600mm^2 Ixx=1.17× 106 mm^4 Iyy=0.67× 106 mm^4

I 0 =5.32× 106 mm^4 J=800mm^4 = 2488 × 106 mm^6

FromEq.(8.77)

PCR(yy)=4.63× 105 N PCR(xx)=8.08× 105 N PCR(θ)=1.97× 105 N

ExpandingEq.(8.79)

(P−PCR(xx))(P−PCR(θ ))I 0 /A−P^2 x^2 S=0(i)

RearrangingEq.(i)

P^2 ( 1 −AxS^2 /I 0 )−P(PCR(xx)+PCR(θ ))+PCR(xx)PCR(θ)=0(ii)

SubstitutingthevaluesoftheconstanttermsinEq.(ii),weobtain

P^2 −29.13× 105 P+46.14× 1010 =0(iii)

TherootsofEq.(iii)givetwovaluesofcriticalload,thelowestofwhichis

P=1.68× 105 N