Stiffness Matrix Method: Symmetry vs Introducing a new node / joint

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The discussion focuses on solving a stiffness matrix problem using symmetry versus introducing an additional node where a load is applied. The original approach involved modifying the global stiffness matrix by doubling the entry in the bottom right corner and adjusting the force vector accordingly. However, this led to different results than expected. The user later identified an error in summing the stiffness matrices for the left and right halves, which was corrected by properly constructing the global stiffness matrix. This correction aligned the results with those presented in the referenced textbook.
Master1022
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This question is about comparing two methods in a stiffness matrix analysis question - using symmetry on half of the structure vs. introducing a new joint at the load
Hi,

In the question outlined in the images (apologies for the poor quality of the scans), the chosen solution has opted to use a symmetry argument and proceed from there.

ScannableDocumentStiffnessMatrix.jpg
StiffnessMatrix2.jpg


Question is from "Structures: theory and analysis" by Williams & Todd

My question is: How could we approach the same problem by introducing an extra node at the point where the load P acts?

Method: I believe that all we would need to do if we were considering an extra joint would be to effectively double the entry in the bottom right corner of the global stiffness matrix: (i.e.

K =<br /> \begin{pmatrix}<br /> \frac{4EI}{l} &amp; \frac{-6EI}{l^2} \\<br /> \frac{-6EI}{l^2} &amp; (\frac{24EI}{l^3} + k)\\<br /> \end{pmatrix}

and changing the force vector to: F = \begin{bmatrix} 0 \\ -P \\ \end{bmatrix}

I thought of doubling the bottom right entry of K as we would be adding it to itself again in another 2 x 2 global stiffness matrix and that would be the only parameter that we care about within K_{BC}

However, solving this yields different answers (letting EI = 1 and l = 1 for ease of typing):
\begin{bmatrix} \theta_{z_A} \\ u_{y_b} \\ \end{bmatrix} = \frac{-P}{15 + k} \begin{bmatrix} \frac{3}{2} \\ 1 \\ \end{bmatrix}.

Can anyone see where I have gone wrong in my thinking here?

Thanks in advance
 
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Update: I realized that I added up the stiffness matrices of the right and left half incorrectly.

The global stiffness matrix (for variables \theta_{z_A}, u_{y_B}, \theta_{z_C} respectively) should have been:
<br /> \begin{pmatrix}<br /> \frac{4EI}{l^3} &amp; \frac{-6EI}{l^2} &amp; 0 \\<br /> \frac{-6EI}{l^2} &amp; \frac{24EI}{l^3} + k &amp; \frac{6EI}{l^2} \\<br /> 0 &amp; \frac{6EI}{l^2} &amp; \frac{4EI}{l^3}<br /> \end{pmatrix}
Using this matrix leads to the same result as in the book.
 
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