Statics of the truss with force and temperature load - test problem Nr 1

3Ey 1Ey 2Ey 3Ey 42Ey 5Ey 6Ey 7Ey 81Ey 9Ex 1 Ex 20 1 2 3 4 5 6 7 8, , Ex 3 , Ex 4 , Ex 5 , Ex 6 , Ex 7 , Ex 8 , Ex 9

Function MBL - Matrix Block Location, this function is used for global stiffness matrix agregationMBL ( A , B , r , c):=fori ˛ 0 .. rows( B) - 1forj ˛ 0 .. cols( B) - 1A r+ i , c+j ‹ B 1+ i , 1+j

Global vektor of external forces - right hand side (RHS) vectorHorizontal and vertical projection of the force acting in node 8 (7kN)Fx 4 := -7kN sin ( 25deg) = -2.958kN0Fy 4 := -7kN cos ( 25deg) = -6.344kN00001210.0000.000p :=0Fx 4Fy 4p =345670.0000.0000.0000.000-2.958kN000891011-6.3440.0000.0000.000Ł0ł120.000

- Nodal forces from temperature load in element "e"E A e Lx et e := α t T eL e ŁLy ełpT oNq := 0Agregation of the thermal force vector pTpT :=e( ( ) - MBL ( pT o , t e , k e , 1 ))MBL pT o , t e , n e , 1pT T =1 2 3 4 5 6 7 8 9 10 11 121 51.482 51.482 -51.482 -51.482 64.940 58.446 -64.940 -58.446 0.000 0.000 0.000 0.000kN

Copy of the K matrix and p wectorK o:= K p o := p - pTBoundary conditionsnode No 1: degree of freedom s1 i s2node No 6: degree of freedom s3 i s4s1 := 1s2 := 2s3 := 11s4 := 12i := 1 .. NqK os1 := 0K os2 := 0K os3 := 0K os4 0, i, i, i, i:= putting zero values in the K matrix rowsK oi := 0K oi := 0K oi := 0K oi := 0 putting zero values in the K matrix columns, s1, s2, s3, s4K os1 := 1 kNK, s1 m os2 := 1 kN, s2 mK os3 := 1 kNK, s3 m os4 1 kN, s4 m:= putting 1 on the diagonal of stifness matrixp os1 := 0p os2 := 0p os3 := 0p os4 := 0 zero value for some rows in RHS vector

4Ey 1Ey 2Ey 3Ey 43Ey 5Ey 6Ey 7Dy 12Dy 2Dy 3Dy 4Dy 51Dy 6Dy 7- 1 0 1 2 3 4 5 6 7 8 9 10 11Dy 8- 1Ex 1 , Ex 2 , Ex 3 , Ex 4 , Ex 5 , Ex 6 , Ex 7 , Dx 1 , Dx 2 , Dx 3 , Dx 4 , Dx 5 , Dx 6 , Dx 7 , Dx 8