Beam deformation. FEM (solidmechanics)

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Payam30
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Homework Statement



I have this question. I need to know the diformation in P direction. Here is the question

Diagram of a rhombus-shaped frame with side length L, pinned to a support on the left, with a spring of stiffness k along the horizontal diagonal and a load P applied at the right vertex.


I use the symmetry and finally with a 1/4 modell I get the following:
Structural diagram of a symmetric frame: a spring of stiffness k/2 attached to a horizontal roller at the top, a diagonal rigid member pinned to a support at the lower left, and a load P/2 applied horizontally at the top joint.

which is correct according to the instruction. K is the stiffness of the spring. which with symmetry becomes k/2 and also P

now
I have this deformation according to my knowledge
Free-body diagram of a quarter-symmetry frame: a diagonal rigid bar pinned at a lower-left support, its upper end on a roller carrying a horizontal load P/2 to the right and a spring k/2 to a wall on the left; labels d1 (vertical) and d2 (horizontal) mark the displacements.


If I apply the formulas for beam deformation which I am allowed to do(in basic level) I got this:
Diagram of a two-node beam element in local coordinates x and z, with end forces and moments u1, f1x, w1, f1z, theta1, M1 at node 1 and u2, f2x, w2, f2z, theta2, M2 at node 2; labelled E, I, A and 2l (totallangd).


you see that deformation d1 in my third picture will have two componentes and it will not solve the problem. becouse :
Column vector of six entries: f1x, f1z, M1, f2x, f2z, M2, each with an overbar or tilde-style accent, in square brackets.

you see that i get problem writing f1x and f1z.. Can anybody tell me how to sett the solustion?

f1x is the force in node 1 at x direction and so on.
f2x will be p/2 cos45 and f2z will be -p/2sin45 and M2

Homework Equations


The Attempt at a Solution



What I wrote up there.
 
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Payam30: I would say your quarter model and displacements currently look good, except when you quarter the spring, I currently think its stiffness remains k, not 0.5*k. Also, I would probably number the displacements D5 and D7. I would currently say, use the frame member global stiffness matrix, instead of basic. I would probably number your three nodes node 1 for left end of spring, node 2 for lower end of frame member, and node 3 for upper end of frame member. I would probably number your members member 1 for the frame member, and member 2 for the spring.

Obtain the frame member global stiffness matrix for member 1, at a slope angle of alpha = 45 deg. Assemble its partitions into the structure stiffness matrix, uppercase K. The member global stiffness matrix for the spring is just k, which you add to K7,7.

In the structure stiffness matrix, note that you can cross out columns and rows 1, 2, 3, 4, 6, 8, and 9.

After you do all of this, I think you end up with the equations in the attached file. Check my work, to see if I made any mistake, because I did this very hurriedly.

After that, invert the stiffness matrix in the attached file, or solve the two equations by hand using simultaneous solution, to solve for displacements D5 and D7.

Do you have given numerical values for E, I, A, L, k, and P?
 
Attachments
  • Matrix equation: [0; 0.5·P] = (E/(2·L³)) times a 2x2 matrix with entries 12·I + A·L², 12·I − A·L², 12·I − A·L², and 12·I + A·L² + 2·k·L³/E, times the vector [D5; D7]. Dated nvn, 2013-02-21.
    fea-eqns01.png
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