User:Eml4507.s13.team2.cc/R3

Honor Pledge
On our honor, we consulted the solution of FEAD F08 Team "Ramrod" HW4 located here. The following MATLAB codes are based upon those used to solve and plot a different truss system. The codes were rewritten to suit the particular plane structure and conditions of this problem and produce the desired results. Dependencies in the MATLAB code were obtained from a finite element companion site located here.

Given
A three-bar plane truss with identical members is presented, and is shown in the image below.

Element 2 is vertical, and all members make equal angles with each other. A force is applied at Node 1, while Nodes 2 and 3 have non-zero prescribed displacement degrees of freedom.



The following member properties are given:

$$ E = 206 \ GPa $$

$$ A = 1*10^{-4} \ m^2 $$

$$ L = 1 \ m $$

The following force and displacement DOFs are given:

$$ F = 20,000 \ N $$

$$ d_3= 2 \ cm $$

$$ d_4= -1 \ cm $$

$$ d_5= -3 \ cm $$

$$ d_6= 5 \ cm $$

$$ d_7= 0 \ cm $$

$$ d_8= 0 \ cm $$

Find
Perform the following:


 * Using MATLAB, solve for:
 * All unknown global displacement DOFs
 * All unknown force components (reactions)
 * All member forces


 * Plot the deformed truss shape superposed on the undeformed truss shape.


 * Verify the results with CALFEM.

Solving with MATLAB
In order to solve for the desired values, the following MATLAB code was used:

Output of MATLAB Code
The above code returns the following output:

k = 1.545e+007 -8.9201e+006 -1.545e+007  8.9201e+006 -8.9201e+006   5.15e+006  8.9201e+006   -5.15e+006 -1.545e+007 8.9201e+006   1.545e+007 -8.9201e+006 8.9201e+006  -5.15e+006 -8.9201e+006    5.15e+006 k = 7.7238e-026 1.2614e-009 -7.7238e-026 -1.2614e-009 1.2614e-009   2.06e+007 -1.2614e-009   -2.06e+007 -7.7238e-026 -1.2614e-009 7.7238e-026  1.2614e-009 -1.2614e-009  -2.06e+007  1.2614e-009    2.06e+007 k = 1.545e+007 8.9201e+006  -1.545e+007 -8.9201e+006 8.9201e+006   5.15e+006 -8.9201e+006   -5.15e+006 -1.545e+007 -8.9201e+006  1.545e+007  8.9201e+006 -8.9201e+006  -5.15e+006  8.9201e+006    5.15e+006 K = Columns 1 through 5 3.09e+007 1.8626e-009 -7.7238e-026 -1.2614e-009  -1.545e+007 1.8626e-009   3.09e+007 -1.2614e-009   -2.06e+007  8.9201e+006 -7.7238e-026 -1.2614e-009 7.7238e-026  1.2614e-009            0 -1.2614e-009  -2.06e+007  1.2614e-009    2.06e+007            0 -1.545e+007 8.9201e+006            0            0   1.545e+007 8.9201e+006  -5.15e+006            0            0 -8.9201e+006 -1.545e+007 -8.9201e+006           0            0            0 -8.9201e+006  -5.15e+006            0            0            0 Columns 6 through 8 8.9201e+006 -1.545e+007 -8.9201e+006 -5.15e+006 -8.9201e+006  -5.15e+006 0           0            0            0            0            0 -8.9201e+006            0            0 5.15e+006           0            0 0  1.545e+007  8.9201e+006 0 8.9201e+006    5.15e+006 R = 14142       14142            0            0            0            0            0            0 d = -0.028976    0.010785         0.02        -0.01        -0.03         0.05            0            0 rf = -2.6217e-011 -4.2816e+005 -3.6562e+005 2.1109e+005 3.5148e+005 2.0293e+005 forces = -4.2219e+005 -4.2816e+005 -4.0586e+005

The first, second, and third outputs k give the element stiffness matrices for element 1, 2, and 3, respectively.

K gives the global stiffness matrix.

R gives the RHS external force matrix.

d gives the global displacement DOFs.

rf gives the reaction force components.

forces gives the member forces.

Plot of Deformed and Undeformed Shape
To plot the figure, the following MATLAB code was used.

The resulting figure is shown below. Blue dashed lines represent the undeformed truss member shape, while red lines indicate the deformed shape. Axis units are in meters.



Honor Pledge
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions. The MATLAB code is based on that used in Problem 3.1 above, which contains references in its own honor pledge. The code was expanded to fit the plane truss and conditions of this particular problem.

Given
A plane truss structure is shown in the figure below with element numbers and encircled global node numbers.

The horizontal and vertical members have length $$ L $$, while inclined members have length $$ \sqrt{2}L $$.

Assume a Young's modulus of $$ E=100 \ GPa $$, cross sectional area $$ A = 1.0 \ cm^2 $$, and $$ L = 0.3 \ m $$.



Find

 * Develop a MATLAB program to determine the deflections and element forces for the following three load cases:
 * Case A: $$ F_{x13}=F_{x14}=10,000 \ N $$
 * Case B: $$ F_{y13}=F_{y14}=10,000 \ N $$
 * Case C: $$ F_{x13}=10,000 \ and \ F_{x14}=-10,000 \ N $$


 * Plot the deformed shape


 * Use the average tip deflections obtained to compute the equivalent section properties
 * $$ (EA)_{eq} $$
 * $$ (EI)_{eq} $$
 * $$ (GA)_{eq} $$

Load Analysis with MATLAB
In order to solve for the desired values, the following MATLAB code was used. Lines 47-50 were altered as needed to fit each load condition.

Load Case A
For Case A, the MATLAB code returns the following deflections in d and member forces in forces:

d = 0 -4.1321e-019 0           0       0.0003      -0.0003       0.0003      -0.0003       0.0006      -0.0006       0.0006      -0.0006       0.0009      -0.0009       0.0009      -0.0009       0.0012      -0.0012       0.0012      -0.0012       0.0015      -0.0015       0.0015      -0.0015       0.0018      -0.0018       0.0018      -0.0018 forces = 10000       10000        10000        10000        10000        10000        10000        10000        10000        10000        10000        10000  1.3774e-011 1.4456e-011 1.807e-011 2.1684e-011 2.1684e-011 7.228e-012 0 -2.0585e-011 -2.2999e-011 -2.9388e-011 -4.0888e-011 -1.5333e-011 5.111e-012

Load Case B
For Case B, the MATLAB code returns the following deflections in d and member forces in forces:

d = 0      0.0006            0            0        0.003    0.0064971      -0.0036    0.0058971       0.0054     0.018394      -0.0066     0.017794       0.0072     0.035091       -0.009     0.034491       0.0084     0.055388      -0.0108     0.054788        0.009     0.078085       -0.012     0.077485        0.009      0.10168      -0.0126      0.10138 forces = 1e+005 80000       60000        40000        20000 -1.1565e-010 -1.2e+005 -1e+005 -80000      -60000       -40000       -20000       -20000       -20000       -20000       -20000       -20000       -20000       -10000        28284        28284        28284        28284        28284        28284

Load Case C
For Case C, the MATLAB code returns the following deflections in d and member forces in forces:

d = 0  2.628e-018 0           0       0.0003       0.0003      -0.0003       0.0003       0.0006       0.0012      -0.0006       0.0012       0.0009       0.0027      -0.0009       0.0027       0.0012       0.0048      -0.0012       0.0048       0.0015       0.0075      -0.0015       0.0075       0.0018       0.0108      -0.0018       0.0108 forces = 10000       10000        10000        10000        10000        10000       -10000       -10000       -10000       -10000       -10000       -10000 -8.7601e-011 -8.8543e-011 -1.0119e-010 -1.1565e-010 -1.1565e-010 -2.8912e-011 0 1.2295e-010 1.2777e-010 1.38e-010 1.6355e-010 8.1776e-011 -6.1332e-011

Equivalent Cross-Sectional Properties
The desired cross sectional values are included in the following relations:

Axial rigidity from the axial deflection of a beam due to an axial force F:
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$$  \displaystyle u_{tip}=\frac{Fl}{(EA)_{eq}} $$      (6.1)
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Flexural and shear rigidity from the transverse deflection due to a transverse force F at the tip:
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$$  \displaystyle v_{tip}=\frac{Fl^3}{3(EI)_{eq}}+\frac{Fl}{(GA)_{eq}} $$      (6.2)
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Flexural rigidity from the transverse deflection due to an end couple C:
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$$  \displaystyle v_{tip}=\frac{Cl^2}{2(EI)_{eq}} $$      (6.3)
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First, the results from Load Case A are used to solve for $$ (EA)_{eq} $$ using Eq. 6.1. The average axial deflection at the tip was 0.0018 meters. The beam length is 1.8 meters, and the total applied axial load was 20,000 newtons.


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$$  \displaystyle 0.0018=\frac{20000*1.8}{(EA)_{eq}} $$      (6.4)
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$$  \displaystyle (EA)_{eq}=20*10^6 \ N $$ (6.5)
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Next, the results from Load Case C are used to solve for $$ (EI)_{eq} $$ using Eq. 6.3. The average transverse deflection at the tip was 0.0108 meters. The beam length is 1.8 meters, and the magnitude of the end couple C was 10,000 newtons.


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$$  \displaystyle .0108=\frac{10000*(1.8)^2}{2(EI)_{eq}} $$      (6.6)
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$$  \displaystyle (EI)_{eq}=1500 \ kN $$ (6.7)
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Finally, the results from Load Case B and the flexural rigidity above are used to solve for $$ (GA)_{eq} $$ using Eq. 6.2. The average transverse deflection at the tip was 0.10153 meters. The beam length is 1.8 meters, and the total applied transverse load was 20,000 newtons.


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$$  \displaystyle 0.10153=\frac{20000(1.8)^3}{3(1500000)}+\frac{20000*1.8}{(GA)_{eq}} $$      (6.8)
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$$  \displaystyle (GA)_{eq} \approx 476 \ kN $$ (6.9)
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