To determine element displacements and forces, the transfer matrix is used. Calculating the internal reaction2.8 2.9 2.10 of Eq. (2.53) for drawing internal forces diagrams, the calculating trick is done (see Fig. 2.19).
where
Calculations of the internal reactions (internal forces correspond to Sign Convention 1) are presented in excerpt 2.13.
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In Fig. 2.19, minus and plus cross-sectional directions are shown. The sign of
the internal forces corresponds to Sign Convention 1 of the internal reactions.
To calculate an internal reaction for drawing internal forces diagrams, a calculating trick is done. As the sign of the initial parameters corresponds to Sign Convention 2, the internal reactions are calculated at cross-section (Eq. (2.53)).
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#------------- Element displacements and forces ---------- for i=1:NEARV krda=i; vF=zeros(6,12); EI=selem(i,13); % from topology EA=selem(i,14); % " " GAr=selem(i,15); % " " Li=lvarras(i,1); qx=qxZ(i,1); qz=qzZ(i,1); aLx=aLXx(i,1); Fz=FZz(i,1); Fx=FZx(i,1); xsamm=Li/Nmitmeks; % element is divided into 4 parts xx=0; AP=AlgPar(i,:)'; #--------- Transfer matrix equation -------- for ij=1:Nmitmeks+1 # 5 - displacements and forces at x=0.0 vvF=ylfhlin(1.0,xx,EA,GAr,EI); vvB=yzhqz(1.0,xx,qx,qz,EA,EI); vvFz=yzfzv(1.0,xx,aLx,Fx,Fz,EA,EI); Fvv(:,ij)=vvF*AP+vvB+vvFz; xx=xx+xsamm; endfor ... endfor #-------------------------------------------------------------
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