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 π’! and π£! at node A and π’! and π£! 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 π = π!" π!" 0 00 0 π!" π!"
where, π!" = πππ π and π!" = π πππ and π is the orientation to the global coordinate system., we can easily find the global stiffness matrix for element 1 as follows:
πΎ! = π ! π! [π]
where π! = !"!
1 β1β1 1
For element 1, !"!= 7.5π5, and π = 0Β°
Element 1
Element 2
Node A Node B
Node C
π = 36.869Β°
Element 1 Node A Node B
π’!
π£!
π’!
π£!
X
Y
β΄ π = 1 0 0 00 0 1 0
π’! π£! π’! π£!
and, πΎ! = 7.5π51 0 β1 00 0 0 0β10
00
1 00 0
π’!π£!π’!π£!
Now, let us consider element 2 with degrees of freedom at each node as π’! and π£! at node C and π’! and π£! at node B and π = 36.869Β°. Substituting in the same equation of π and πΎ, we get, π’! π£! π’! π£!
πΎ! = 6π50.65 0.48 β0.65 β0.480.48 0.35 β0.48 β0.35β0.65β0.48
β0.48β0.35
0.65 0.480.48 0.35
π’!π£!π’!π£!
Now, the global stiffness matrix for the entire system should be assembled to lead to a 6x6 matrix as follows:
πΎ!"#$%" =
7.5π5 0 β7.5π50 0 0
β7.5π5000
0000
11.4π52.88π5β3.9π5β2.88π5
0 0 00 0 0
2.88π52.1π5β2.88π5β2.1π5
β3.9π5β2.88π53.9π52.88π5
β2.88π5β2.1π52.88π52.1π5
The displacement vector will look like:
π’!π£!π’!π£!π’!π£!
=
00π’!π£!00
The force vector will look like:
πΉ!!πΉ!!πΉ!!πΉ!!πΉ!!πΉ!!
=
000
β10000
The nodes A and B are clamped so the corresponding displacements are zero, reducing the equilibrium equation to:
11.4π5 2.88π52.88π5 2.1π5
π’!π£! = 0
β100 So, that gets us to:
π’! = 1.8π!! πππ π£! = β7.3π!! You can verify the reaction forces by yourself to check for accuracy.
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