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- AO 3 xy 3 z 3 is the solving system, in which the effective degrees of freedom are mutuallyorthogonal and the force quantities can be computed independently from the active loads.For this cross-section the sectorial area at the wall mid-lines is zero, without lateral supports,for the arm h s (s) of a parallel to s unit vector i ( s) around the pole A is zero for all s:h s (s) ≡ 0 , ~ ω () s ≡ 0 (7)This fulfils at the non-compressible axes the condition ~ ω ( s ) ~1= ω ( s2 ) = 0 . For the solvingsystem AO 3 xy 3 z 3 the pole is still at the corner A and sectorial area is zero:a y3 − a y0 = 0 , a z3 − a z0 = 0 , ω 3 (s) ≡ 0 (8)The guided origin of the solving system is placed at the intersectional point O 3 of thestraight line between the non-compressible axes with the normal drawn to this line throughthe centroid. In this system the cross-sectional integrals are expressed below. The subscriptzero ( 0 ) refers to the centroidal system AOxyz, parallel to the solving system.6 5 3 3 5 6ta ( + 4ab+ 6ab + 4ab + b)I v3 = I v0 =2 212( a + b )( a+b)I w0 = ta 2 b 2( a + b )2 212( a + b )I w3 = I w0 + Ah 2 /4 = ta 2 b 2( a + b )2 2+ ta 2 b 2( a + b )2 212( a + b ) 4( a + b )s= ta 2 b 2( a + b )2 23( a + b )= ta 2 b 2( a + b )23cI vw3 = I vw0 = tab ( a 3 − b 3 )2, S w3 = − Ah/2 = − tab ( a + b ), S v3 = 0 (12),(13),(14)12c2cThe model of the quarter of the box beam as a guided Vlasov beam on elastic foundationThe functioning of the upper left corner part of the box beam of figure 1 with its effectivedegrees of freedom can be modelled as a guided Vlasov beam in figure 3. The active degreesof freedom of the cross-section are the deflection of some pole A 5 of the axis w 5 andthe angle of twist . In the reality the centre B of the box beam, in figure 3 the point A 4 , hasno deflections due torsion or distortion, but in the analogy with the Vlasov BEF (beam onelastic foundation) it has a motion v 4 (x) normal to the guided neutral line w 4 (P 1 P 2 ). Indirection w it is fixed both in the model and the reality. The corner A, the concentrationaxis of the loads q D and q T , has deflections as if the part beam with rigid-in-plane crosssectionwere supported according to figure 3.If the shear deformation γ * (x) were neglected at the profile line like usually is supposedfor the usual sectorial areaγ xs (x,s) = 0 , (15)it would not be possible for a beam of figure 3 to twist at all, except if the parallel to P 1 P 2plane support passed through corner A of the cross-section. That is to say it is not possibleto draw any usual sectorial area ω 5 (s) − ω 5 (s 01 ) around any pole axis A 5 in the supportplane w 5 so that it were zero in the same time at both non-compressible axes:u(x,s 01 ) = u(x,s 02 ) = 0 (16)(9)(10)(11)
q D (x)ba y5 − a y0v 1 w ~ w z z3 ,z 5v,v 3 ,F vq T (x)~v P 1z 1 ,w 1A ~ z αw 5Oa O 3 ,O 5y, y 5~ yπ/2β d 1v 4 A 5v 5P 2B = A 4p 2Figure 3 Analogous model for a quarter segment of a box beam under torsion and distortionThe spring acting in direction v 4 is due to the plate bending of the walls. q D (x) is adistorting and q T (x) a twisting load. There influences are connected by the spring.In this case equation (15) must be replaced by an equation taking into account the sheardeformation from constant shear stress flowγ xs (x,s) = ∂ * *uxs ( , ) ∂υs( xs , )*τzs( xs , ) Ts( x)+ = γzs( xs , ) = = (17)∂s∂xG Gt()sT* ( x) is the constant shear stress flowsThe warping gives now rise to axial displacement distributionsu(x,s) = − ∫[ θ T*u′ ( xh )s5( s)−( x ) sGt() s]ds = θ u´(x)[ω 5 (s) − ω 5 (s 01 )] + T * x s( ) (tG s − s 01 ) (18)s01Equation (16) is now valid, ifωT* 5( s02) − ω5( s01)s( x) =− Gtθu′( x) , (19)s02 − s01if the wall thickness t(s) ≡ t is constant. This shear stress flow results now in a torque correspondingthe StVenant torque of closed beams:
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Page 3: Our basic element is an angle cross
Page 7 and 8: surface or volume loadings:∂σx(
Page 9: References:/1/ Koivula Risto: Avopr

ANALYSIS OF DISTORTION OF A THIN-WALLED ... - students.tut.fi

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