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

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Donna Ellis
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1 Statics of the truss with force and temperature load - test problem Nr E := GPa - Young modulus for truss material - steel α t := - - thermal expanssion coeficient - steel D := cm - Cross section (pipe) diametr g := mm - Cross section (pipe) wall thicknes := π g ( D - g) - Cross sections area for elements = 7.9 cm

2 Some main parameters of the truss Ne := 9 - Number of elements Nn := - Number of nodes Nd := - Degree of freedom for D truss node Nq := Nd Nn - Number of equilibrium equations K onq, Nq Element cross section area := - Initiation of the global stiffness matrix with zero values A :=

3 Global coordinates of the nodes Element initial node (Ni) vector Element final node (Nj) - vector Temperature load for elements X := m Y := m. Ni := Nj := T := e :=.. Ne Loop for each element Truss graph is made for easy geometrical data verification Ø X Nie Y Ex Œ œ ( Nie ) e := Ey Œ e := Œ œ Ex, Ey - coordinates of the truss element nodes X Œ Nje Y Nje º ø œ ß Ø º ø œ ß

4 Ey Ey Ey Ey Ey Ey Ey 7 Ey Ey 9 7 Ex, Ex, Ex, Ex, Ex, Ex, Ex 7, Ex, Ex 9

5 Calculation of the stiffness matrix blocks (J) for each element Lx e := X Nje - X Nie Ly e := Y Nje - Y Nie L e := Lx e + ( Ly e ) Lx. = m Ly = m L.. 7. = m E A e J e := ( L e ) Ø Œ Œ º ( Lx e ) Lx e Ly e Lx e Ly e ( Ly e ) ø œ œ ß

6 J = J = J = m -7.. m m J = J = J =.. m.. m... m J 7 = J = J 9 =. 7. m m.. 9. m

7 Function MBL - Matrix Block Location, this function is used for global stiffness matrix agregation MBL ( A, B, r, c) := for i.. rows( B) - for j.. cols( B) - A r+ i, c+ j B + i, + j <----- r = row number, c =column number for B block location A

8 Agregation of the global stiffness matrix n e := Nd Ni e - k e := Nd Nj e - <--- n e = global number of degree of freedom of the initial node, <--- k e = global number of degree of freedom of the final node. In the agregation procedure, the user defined function - MBL is used K := e ( + MBL ( K o, J e, k e, k ) e - MBL ( K o, J e, n e, k ) e - MBL ( K o, J e, k e, n )) e MBL K o, J e, n e, n e K = m Global stiffness matrix K without boundary condition is singular K = K m. = <----- by the small errors in computer arithmetic, the value of the determinant can be different a bit from a zero

9 Global vektor of external forces - right hand side (RHS) vector Horizontal and vertical projection of the force acting in node (7) Fx := -7 sin ( deg) = -.9 Fy := -7 cos ( deg) = -... p := Fx Fy p =

10 - Nodal forces from temperature load in element "e" E A e Lx e t e := α t T e L e Ly e pt onq := Agregation of the thermal force vector pt pt := e ( - MBL ( pt o, t e, k e, )) MBL pt o, t e, n e, pt T =

11 Copy of the K matrix and p wector K o := K p o := p - pt Boundary conditions node No : degree of freedom s i s node No : degree of freedom s i s s := s := s := s := i :=.. Nq K os := K os := K os := K os, i, i, i, i := putting zero values in the K matrix rows K oi := K oi := K oi := K oi := putting zero values in the K matrix columns, s, s, s, s K os := K, s m os :=, s m K os := K, s m os, s m := putting on the diagonal of stifness matrix p os := p os := p os := p os := zero value for some rows in RHS vector

12 K o = p m o K o m =. - determinant of the modified stiffness matrix K o is always greather than, K o > Solving the system of linear equation: u := lsolve K o, p o u - vector of nodal displacements u T = mm The graph of the displaced truss allows to check the correctness of results scale := Ø u ( Nie -) Dx e Ex e scale Œ œ := + Dy Œ e := Ey e + scale u Nje - º ø œ ß Ø Œ Œ º u ( Nie ) u ( Nje ) ø œ œ ß

13 Ey Ey Ey Ey Ey Ey Ey 7 Dy Dy Dy Dy Dy Dy Dy 7 Dy Ex, Ex, Ex, Ex, Ex, Ex, Ex 7, Dx, Dx, Dx, Dx, Dx, Dx, Dx 7, Dx

14 Reaction of the supported nodes r := K u - p + pt r T = Calculation of the element internal forces N e := E A e ( L e ) Ø u Nje - u º - Nie - Lx e + ( u Nje - u Nie) Ly e ø ß - α t T e E A e N = N e e

15 Calculation of the normal stress in the truss elements σ e := N e A e σ = MPa σ e MPa e

ASSIGNMENT 1 INTRODUCTION TO CAD

Computer Aided Design(2161903) ASSIGNMENT 1 INTRODUCTION TO CAD Theory 1. Discuss the reasons for implementing a CAD system. 2. Define computer aided design. Compare computer aided design and conventional