User:Eml4500.f08.gravy.mmm/HW4

21
Closing the loop, infinitesimal displacement, eigenvalues problems, zero eigenvalues, rigid body mechanisms.

HW: Plot the eigenvectors corresponding to zero eigenvalue of 2. bar truss system. - Brian
Eval pb. $$\ \mathbf{kv} = \boldsymbol{\lambda}  \mathbf{v}\,$$

Let {u1,u2,u3,u4} be the pure eigenvalues corresponding to four zero eigenvalues.

$$\ \mathbf{k}_{6x6} \mathbf{u}_i =0 \cdot \mathbf{u}_i \,$$, i=1,2,3,4

where $$ 0 \cdot u_i $$ is the zero vector of size 6x1.

Linear combination of {u_i, i=1,2,3,4}

$$ \sum_{i=1}^4 \alpha_i u_i =: \mathbf{W}_{6x1}$$

=: means equal by definition and $$\alpha$$ is a real number so it is 1x1 and u_1 is 6x1.

W is also an eigenvector corresponding to a zero eigenvalue:

$$\ \mathbf{KW}= \mathbf{K}( \sum_{i=1}^4 \boldsymbol{\alpha}_i \mathbf{W}_i ) =  \sum_{i=1}^4 \boldsymbol{\alpha}_i (\mathbf{Ku}_i) \,$$ where $$(\mathbf{Ku}_i)$$ is a zero matriz of size 6x6. $$\ =\mathbf{0}_{6x1} = 0 \cdot \mathbf{W}_{6x1} \,$$

HW: matlab- Sean


$$\ \mathbf{kv} = \boldsymbol{\lambda}  \,$$

Plot evect. evct. corresp. zero eval for case a use matlab.

a=b=1, E=2,  A=3

22
Justication of assembly of element stiffness matrix $$k^{(e)} e=1, number of elements$$ into global stiffness matrix K

consider example of 2 bar truss from prevous lecture.

Recall element force displacement relation

$$\ k^{(e)}_{4x4}d^{(e)}_{4x1}=f^{(e)}_{4x1} \,$$

from prevous lecture: Euler cut principal method 2 (equilbrium of global node 2) Euler cut principal method

from prevous report: Free body digram of element 1 and element 2 element degress of freedom d(e) Trusses, matrix method

from prevous lecture: for node 2, identify global degrees of freedom to element degrees of freedom for both element 1 and 2. Global_F-D_Relationship



$$\ \sum F_x= 0-f_3^{(1)}- f_1^{(2)}= 0 \,$$  Equation (1)

$$ \sum F_y= P-f_4^{(1)}- f_2^{(2)}= 0$$  Equation (2)

Next use element Force Displacement relation $$\mathbf{k}^{(e)} \mathbf{d}^{(e)}=\mathbf{f}^{(e)}$$

23
Equation 1 becomes $$\ f_3^{(1)} + f_1^{(2)}=0 \,$$

Equation 2 becomes $$\ f_4^{(1)} + f_2^{(2)}=P \,$$

So substituting for the fs in equation 1 gives

$$\ \underbrace{[ k_{31}^{(1)} d_{1}^{(1)} + k_{32}^{(1)} d_{2}^{(1)} + k_{33}^{(1)} d_{3}^{(1)} + k_{34}^{(1)} d_{4}^{(1)} ]}_{f_3^{(1)}} + \underbrace{[k_{11}^{(2)} d_{1}^{(2)} + k_{12}^{(2)} d_{2}^{(2)} + k_{13}^{(2)} d_{3}^{(2)}+ k_{14}^{(2)} d_{4}^{(2)}]}_{f_1^{(2)}}=0 \,$$

Replacing the local ds with the global ds as explained Global_F-D_Relationship You end up with

$$[ k_{31}^{(1)} d_{1} + k_{32}^{(1)} d_{2} + k_{33}^{(1)} d_{3} + k_{34}^{(1)} d_{4} ]

+ [k_{11}^{(2)} d_{3} + k_{12}^{(2)} d_{4} + k_{13}^{(2)} d_{5}+ k_{14}^{(2)} d_{6}  ]=0 $$

Which is the 3rd row of the K matrix in Kd=F

Hw. Derive details of Eq 2 on 23.1 and the assembly of row 4. - Patrick
The assembly of K(e) e=1 to the number of elements, into the global stiffness matrix K can be represented as

$$\ K_{nxn}=Ak^{(e)}_{mxm}\,$$

Where A is the Assembly operator, n is the total number of global degrees of freedom, m is the number of element degrees of freedom and m is much less than n. Which is the principal of virtual work insert link here

Elimination of rows corresponding to the bountary conitions lets you obtain $$\mathbf{\bar{q}}_{2x2}$$

Remeber that:

$$\ \mathbf{q}^{(e)}_{2x1} = \mathbf{T}^{(e)}_{2x4} d^{(e)} _{4x1} \,$$

$$\ k^{(e)} = \mathbf{T}^{(e)T} \hat{k}^{(e)}  \mathbf{T}^{(e)} \,$$

Deriving FEM of Partial differential Equations

Force Displacement relation of a bar or spring: Kd=F implies that Kd-F=0  Equation 3

which is equivalent to w(Kd-F)=0 for all w  Equation 4

Equation 4 is a "weak form" which is Principal of virtual work

Prove:

A) Equation 3 implies equation 4 which is a trivial conclusion

B) Equation 4 implies equation 3 which is not a trivial conclusion

Since equation 4 is valid for all w, first select w=1 and equation 4 becomes 1(Kd-F)=0 which is equation 3