notes_32_intro_matrix - Intro to Matrix Analysis ORIGIN:= 1...

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Intro to Matrix Analysis ORIGIN := 1 consider a 2-D structure consisting of four elements, linked at pinned joints with six nodes: Y 1 3 2 4 5 6 1 2 3 4 p Fx4 Fy4 X consider one of the elements, say element # 4 LINEAR ELASTIC behaviour and equilibrium => F = K ⋅∆ + f p + f ε 0 where F = forces at nodes (any direction => two at each node in X and Y) f p is the equivalent nodal force resulting in the same REACTION to the distributed pressure f ε 0 is the same for initial strain/stress in the element due to fit, temperature, etc. Fx and Fy are the external forces applied at node 4 p a distributed pressure on element #4 we attribute LINEAR ELASTIC behaviour to the structure and in turn to each of the elements: Y 4 4 p 5 6 X F = forces at each node for an example of the eqivalent nodal force consider the following uniformly loaded beam: p L has reaction forces note: p and R R1 = pL have opposite 2 directions in this example p L R1 R2 a force of p*L/2 in the same direction as p will create the same rection force hence L L R2 p*L/2 p*L/2 R1 f p = p 2 1
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U FeY henceforth we will assume that the nodal forces already account for the equivalent of distributed forces and that initial stress/strain = 0 therefore: . .. F = K ⋅∆ i.e. The nodal force is linearly proportional to the displacement of the nodes for the fourth element this is expressed as:. .. nod_el := 3 Fe 1 Ke 11 Ke 12 Ke 13 e 1 , , , , , , Fe Fe 2 Ke Ke 21 Ke 22 Ke 23 e e 2 Fe 3 Ke 31 Ke 32 Ke 33 e 3 , , , Ke = element stiffness matrix which for now we will assume can be determined by experiment or analysis similarly a matrix can be found such that:. .. σ e = Se ⋅∆ e Se = element stress matrix remember that F and in this two dimensional example each have two components X & Y FeX 1 1 U is X displacement 1 V 1 V is Y displacement FeX 2 U 2 and Ke is a 6 x 6 matrix of coefficients. Fe
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notes_32_intro_matrix - Intro to Matrix Analysis ORIGIN:= 1...

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