PKqTBHmimetypetext/x-wxmathmlPKqTQdBV55
format.txt
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PKqT;N;Ncontent.xml
kill(all);(%o0) doneequilibrium equations for the quarter(simplified) ladder frame chassiseqtrx : UA=0;eqtry : VD=0;eqtrz : -F+WA+WD;(%o1) UA=0(%o2) VD=0(%o3) WD+WA−FeqrOx : ThetaD + WD*b - F*c=0;(%o4) −F*c+WD*b+ThetaD=0eqrOy : PhiA - WA*a +F*a=0 ;(%o5) −WA*a+F*a+PhiA=0eqrOz : 0=0 ;(%o6) 0=0blindly solve the equilibrium equations in the parametric reactionslinsolve([ eqtrx , eqtry, eqtrz , eqrOx , eqrOy, eqrOz ],[UA,PhiA, WA, WD,VD,ThetaD]) ;solve: dependent equations eliminated: (6)(%o7) [UA=0,PhiA=−%r1*a,WA=F−%r1,WD=%r1,VD=0,ThetaD=F*c−%r1*b]there is one dependent equation, so I'm unable to solve the problem of obtaining all the reaction forces/moments by equilibrium alone.The quarter ladder frame is once statically indeterminate.The returned expression is parametric in a %r# parameter introduced by maxima.let's try removing the WA unknown reaction from the unknown listlinsolve([ eqtrx , eqtry, eqtrz , eqrOx , eqrOy, eqrOz ],[UA,PhiA, WD,VD,ThetaD]) , globalsolve=true;solve: dependent equations eliminated: (6)(%o8) [UA:0,PhiA:WA*a−F*a,WD:F−WA,VD:0,ThetaD:F*

c−b

+WA*b]the WA unknown reaction force has been removed fromthe unknowns list, thus obtaining a parametricrepresentation of the reaction forces and moments, with WA as a parameter (like F)to obtain the actual value of WA, the deflection in the z direction of the A point has to be evaluated andzeroed. Castigliano's Theorem fits for the goal.N : 0;Q1 : -VD;Q2 : -WD;(%o9) 0(%o10) 0(%o11) WA−Fequilibrium with respect to rotation along thelocal C3 axis linsolve( [ +Mt +ThetaD + 0*Q1 + 0*Q2 -e*WD +f*VD ] , Mt ) , globalsolve=true;(%o12) [Mt:

F−WA

*e−F*c+

F−WA

*b]equilibrium with respect to rotation along thelocal G1 and G2 axeslinsolve( [+M1+WD*s] , M1 ),globalsolve=true;(%o13) [M1:

WA−F

*s]linsolve( [+M2-VD*s] , M2),globalsolve=true;(%o14) [M2:0]Internal energy for the DB beamU : integrate((J22*M1^2+J11*M2^2+2*J12*M1*M2) /(2*E*(J11*J22-J12^2)) /alpha_flx+N^2/2/E/A /alpha_axl+( chi1 * Q1^2 + chi2 * Q2^2 + chi12 * Q1*Q2 ) /(2*G*A) /alpha_shr+Mt^2/2/G/Kt /alpha_trs,s,0,a),alpha_axl=1$The strain energy is now known as a quadratic function of F and WA aloneCastigliano's Theorem is now applied to obtain the z deflection at A; the WA skew-symm. reaction is tunedto make such a deflection vanish, and thus grandingthe material continuity at the A-B-L rigid bodyFILLME$the z deflection at the L point is evaluated for thequarter ladder frame structure, constrained according to two skew-symm planesd : diff(U , F)$The deflection of the whole chassis, mounted on theoriginal three supports, may be derived by observingthat its internal energy is four times its one-quartercounterpart.dw : diff(4*U , F)$such a results may be rationalized by superposingsuitable rigid body motions to the twice-mirrored deformed quarter ladder frame structure;such rigid body motions make the chassis comply with thenull z deflection condition a the three supports. the torsional stiffness of the chassis, measured in torque per unit twist angle, is derived from the dw deflectionasts : fullratsimp( (F * (2*c)^2) / ( dw )) ;such a quantity is independent on the actual axle trackdiff(ts,c);(%o19) 0The properties of a reference cross-section are introduced,in order to numerically evalate the resultsChannel section dimensions:_h : web heigth, measured at the thin wall midsurface_b : wing width, measured at the thin midsurface_t : web and wing wall thickness(e,f) : Gx,Gy position of the shear center, with respect to the center of gravity(m,n) : Ox,Oy position of the center of gravity, with respect to the web midpointupsilon: _b/_hepsilon: _t/_hChassis dimensions:a : half wheelbaseb : half the distance between the profilesc : half the axle trackdim : [ a=1200, b=400-m, c=600, _h=80 * _h_mult, upsilon=40/80, epsilon=6/80 * _t_mult / _h_mult],_t_mult=1,_h_mult=1;(%o20) [a=1200,b=400−m,c=600,_h=80,upsilon=12,epsilon=340]here, the "_h_mult" and the "_t_mult" multipliersare introduced to straighforwardly scale the cross section size and its thickness, for observingtheir influence on the torsional stiffness.They are provisionally set to unity.detailed section propertiessec :[ A = 2*epsilon*_h^2*upsilon+epsilon*_h^2, J11 =(epsilon*_h^4+6*epsilon*_h^4*upsilon)/12, J22 =(2*epsilon*_h^4*upsilon^3+epsilon*_h^4*upsilon^4)/(6*upsilon+3), J12 =0, Kt =(epsilon^3*_h^3*(_h+2*_h*upsilon))/3, chi1 =(15+78*upsilon+138*upsilon^2+96*upsilon^3+24*upsilon^4)/(20*upsilon^4+80*upsilon^3+80*upsilon^2), chi2 =(6+72*upsilon+300*upsilon^2+480*upsilon^3+240*upsilon^4)/(180*upsilon^2+60*upsilon+5), chi12 =0, e = (4*_h*upsilon^2+12*_h*upsilon^3)/(12*upsilon^2+8*upsilon+1), f = 0, m = (_h*upsilon^2)/(2*upsilon+1), n = 0],dim;(%o21) [A=960,J11=1024000,J22=160000,J12=0,Kt=11520,chi1=408125,chi2=125,chi12=0,e=25,f=0,m=10,n=0]material propertiesmat : [ G = 210000 / 2 / (1 +3/10), E = 210000, nu = 3/10 ];(%o22) [G=105000013,E=210000,nu=310]my specific value for the torsional stiffness, asa function of the various multipliersmyts : ev(ts, [dim, mat, sec]), infeval$a reference value, obtained for unit multipliersmytsref:ev(myts,alpha_flx=1,alpha_shr=1,alpha_trs=1);%,numer;(%o24) ts(%o25) tsI speculate the section bending and shear stiffness to vanish, i.e. the torsional stiffnessof the profile is left alone in elastically connecting the two rigid bodiesmyts/mytsref$ev(%, alpha_trs=1)$limit( %, alpha_flx,0)$limit( %, alpha_shr,0)$p:%;p,numer;(%o30) 1(%o31) 1the value shown is the fraction of residual stiffness remained.I suppose the torsional stiffness to vanish, while maintaining both the shear and the bending stiffness unchanged.myts/mytsref$ev(%, alpha_flx=1,alpha_shr=1)$notp:limit(%, alpha_trs,0);notp,numer;(%o34) 1(%o35) 1p and notp are complementaryp+notp;(%o36) 2I hypothesize the torsional stiffness to vanish, whilecase 1) maintaining the bending stiffness unaltered, with vanishing shear stiffnesscase 2) maintaining the shear stiffness unaltered, with vanishing bending stiffnessmyts/mytsref $ev(%, alpha_flx=1)$limit( %, alpha_trs,0)$limit( %, alpha_shr,0);(%o40) 1myts/mytsref $ev(%, alpha_shr=1)$limit( %, alpha_trs,0)$limit( %, alpha_flx,0);(%o44) 1no residual stiffness for the structure is obtained if the flexural and the shear stiffness of the profiles acts alone.by raising toward infinity the stiffness of one of the two contributions, the remaining compliance is isolated.myts/mytsref $ev(%, alpha_flx=1)$limit( %, alpha_trs,0)$q:limit( %, alpha_shr,infinity);(%o48) 1myts/mytsref $ev(%, alpha_shr=1)$limit( %, alpha_trs,0)$r:limit( %, alpha_flx,infinity);(%o52) 1the sum of the two isolated compliances returns the overall bending compliance1/q+1/r - 1/notp;(%o53) 1moreover, if only one amongst the flexural and the shear stiffness vanish no further stiffening is observed with respect to the torsion alonemyts/mytsref$ev( % , alpha_trs=1)$ev( % , alpha_flx=1)$limit( % , alpha_shr,0);%,numer;(%o57) 1(%o58) 1myts/mytsref$ev( % , alpha_trs=1)$ev(%, alpha_shr=1)$limit( % , alpha_flx,0);%,numer;(%o62) 1(%o63) 1We hence may rationalize the stiffness contributionsdue to flexure, shear and torsion as ___\\| trs.stiff | |\\|-----/\/\/\/\/\/\/\/\-----| | F*2*c\\| | |=======> action\\| | |\\| | | dw/2/c\\| flx.stiff. shr.stiff. | |-------| deflection\\|---/\/\/\/\----/\/\/\/\---| |\\| |___|The torsional stiffness of the profile acts **in parallel**with the combination **in series** of the flexural stiffness and of theshear stiffness of the same member. Pretty funny.PKqTBHmimetypePKqTQdBV55
5format.txtPKqT;N;Ncontent.xmlPKT