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Linear modelling solution from tudelft sOLUTION OF WEEKE 3
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Solution to practice problem Week 3 Simplifying the problem, I can represent it as follows: Let us consider element 1: There are two degrees of freedom at each node A and B in the global truss coordinate system. Lets call them 𝑢! and 𝑣! at node A and 𝑢! and 𝑣! at node B as shown in the figure below. The global coordinate system is shown as XY using the dashed lines.
Given the transformation matrix 𝜆 = 𝑙!" 𝑚!" 0 00 0 𝑙!" 𝑚!"
where, 𝑙!" = 𝑐𝑜𝑠𝜃 and 𝑚!" = 𝑠𝑖𝑛𝜃 and 𝜃 is the orientation to the global coordinate system., we can easily find the global stiffness matrix for element 1 as follows:
𝐾! = 𝜆 ! 𝑘! [𝜆]
where 𝑘! = !"!
1 −1−1 1
For element 1, !"!= 7.5𝑒5, and 𝜃 = 0°
Element 1
Element 2
Node A Node B
Node C
𝜃 = 36.869°
Element 1 Node A Node B
𝑢!
𝑣!
𝑢!
𝑣!
X
Y
∴ 𝜆 = 1 0 0 00 0 1 0
𝑢! 𝑣! 𝑢! 𝑣!
and, 𝐾! = 7.5𝑒51 0 −1 00 0 0 0−10
00
1 00 0
𝑢!𝑣!𝑢!𝑣!
Now, let us consider element 2 with degrees of freedom at each node as 𝑢! and 𝑣! at node C and 𝑢! and 𝑣! at node B and 𝜃 = 36.869°. Substituting in the same equation of 𝜆 and 𝐾, we get, 𝑢! 𝑣! 𝑢! 𝑣!
𝐾! = 6𝑒50.65 0.48 −0.65 −0.480.48 0.35 −0.48 −0.35−0.65−0.48
−0.48−0.35
0.65 0.480.48 0.35
𝑢!𝑣!𝑢!𝑣!
Now, the global stiffness matrix for the entire system should be assembled to lead to a 6x6 matrix as follows:
𝐾!"#$%" =
7.5𝑒5 0 −7.5𝑒50 0 0
−7.5𝑒5000
0000
11.4𝑒52.88𝑒5−3.9𝑒5−2.88𝑒5
0 0 00 0 0
2.88𝑒52.1𝑒5−2.88𝑒5−2.1𝑒5
−3.9𝑒5−2.88𝑒53.9𝑒52.88𝑒5
−2.88𝑒5−2.1𝑒52.88𝑒52.1𝑒5
The displacement vector will look like:
𝑢!𝑣!𝑢!𝑣!𝑢!𝑣!
=
00𝑢!𝑣!00
The force vector will look like:
𝐹!!𝐹!!𝐹!!𝐹!!𝐹!!𝐹!!
=
000
−10000
The nodes A and B are clamped so the corresponding displacements are zero, reducing the equilibrium equation to:
11.4𝑒5 2.88𝑒52.88𝑒5 2.1𝑒5
𝑢!𝑣! = 0
−100 So, that gets us to:
𝑢! = 1.8𝑒!! 𝑎𝑛𝑑 𝑣! = −7.3𝑒!! You can verify the reaction forces by yourself to check for accuracy.