CVEN 2303 Truss Test - Eldon Liu [z5255871]

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Jan 9, 2024

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CVEN2303 Truss Lab Eldon Liu z5255871 04/08/2020 Introduction The stiffness method is one of several methods used to analyze forces and displacements in structures determinate or indeterminant. This assignment task involves the calculation of the axial forces of both a Warren Truss and a Basic Roof Truss using the stiffness method. The calculated theoretical results will then be compared with the calculated experimental results. Part 1.1: Calculate theoretical axial forces of the Warren and Roof Trusses. Figure 1: Code for Finding Kmatrix
Warren Truss The Warren Truss uses a left to right reference system with the structure containing 11 degrees of freedom. Figure 2: Reference System for Warren Truss The stiffness matrices for each member of the Warren Truss can be determined using the following code on MATLAB. Figure 3: Code for Finding the Stiffness Matrix for Warren Truss
The Resultant Stiffness Matrices for all members of the Warren Truss K1 (Member 1) [ ϴ = pi/6] K2 (Member 2) [ ϴ = 0] K3 (Member 3) [ ϴ = -pi/6] K4 (Member 4) [ ϴ = 0] K5 (Member 5) [ ϴ = pi/6] K6 (Member 6) [ ϴ = 0] K7 (Member 7) [ ϴ = -pi/6] K8 (Member 8) [ ϴ = 0] K9(Member 9) [ ϴ = pi/6] K10 (Member 10) [ ϴ = 0] K11 (Member 11) [ ϴ = -pi/6]
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Using the stiffness matrices found prior, the global matrix can be found. Figure 4: Code for Finding the Global Matrix Figure 5: Resultant Global Matrix The load vector for the Warren Truss has a -300N force acting on the eighth degree of freedom which makes the load vector P = [0;0;0;0;0;0;0; -300;0;0;0]. Thus, using the following code, the internal force matrix can be found for the Warren Truss. Figure 6: Code for Finding Internal Force Matrix.
The Resultant Internal Forces for all members of the Warren Truss N1 (Member 1) N2 (Member 2) N3 (Member 3) N4 (Member 4) N5 (Member 5) N6 (Member 6) N7 (Member 7) N8 (Member 8) N9 (Member 9) N10 (Member 10) N11 (Member 11)
The Theoretical Axial Forces of the Warren Truss can be calculated manually through the following method: Example Calculating member 1 of the Warren Truss N1 = √(100 ∗ 100) + (57.735 ∗ 57.735) N1 = 115.47 N Since N1 is known to be in compression, N1 = -115.47 N This Process is then repeated for all members of the truss. The resultant theoretical axial forces for the Warren Truss. Member of Warren Truss Calculated Theoretical Axial Force for Warren Truss 1 -115.47 2 57.74 3 115.47 4 -115.47 5 -115.47 6 173.21 7 115.47 8 -230.94 9 230.94 10 115.47 11 -230.94 Table 1: Calculated Theoretical Axial Forces for Warren Truss
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Roof Truss Like the Warren Truss, the Roof Truss uses a left to right reference system and also contains 11 degrees of freedom within its structure. Figure 7: Reference System for Roof Truss The stiffness matrices for each member of the Roof Truss can be determine using a code similar to the code used for the Warren Truss. Figure 8: Code for Finding the Stiffness Matrix for Roof Truss
The Resultant Stiffness Matrices for all members of the Roof Truss K1 (Member 1) [ ϴ = pi/6] K2 (Member 2) [ ϴ = 0] K3 (Member 3) [ ϴ = -pi/3] K4 (Member 4) [ ϴ = pi/6] K5 (Member 5) [ ϴ = pi/3] K6 (Member 6) [ ϴ = 0] K7 (Member 7) [ ϴ = -pi/3] K8 (Member 8) [ ϴ = -pi/6] K9(Member 9) [ ϴ = pi/3] K10 (Member 10) [ ϴ = 0] K11 (Member 11) [ ϴ = -pi/6]
The Global Matrix is then found using the previously found stiffness matrices. Figure 9: Resultant Global Matrix for Roof Truss Like the Warren Truss, the load vector for the Roof Truss has a -300N force acting on its eighth degree of freedom thus making the load vector for the Roof Truss the same as the load vector of the Warren Truss being P = [0;0;0;0;0;0;0; -300;0;0;0]. Then using the same code type of code, the internal force matrix for the Roof Truss can be found. Figure 10: Code for Finding Internal Force Matrix for Roof Truss
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The Resultant Internal forces for all members of the Roof Truss N1 (Member 1) N2 (Member 2) N3 (Member 3) N4 (Member 4) N5 (Member 5) N6 (Member 6) N7 (Member 7) N8 (Member 8) N9 (Member 9) N10 (Member 10) N11 (Member 11)
Similarly, the theoretical axial forces for the Roof Truss can be calculated manually using the same Method. Example Calculating member 1 of Roof Truss. N1 = √(100 ∗ 100) + (173.2051 ∗ 173.2051) N1 = 200.00 N Since N1 is known to be in compression, N1 = -200.00 N This Process is then repeated for all members of the truss. The resultant theoretical axial forces for the Roof Truss. Member of Roof Truss Calculated Theoretical Axial Force for Roof Truss 1 -200.00 2 173.21 3 0.00 4 -200.00 5 0.00 6 173.21 7 346.41 8 -400.00 9 0.00 10 346.41 11 -400.00 Table 2: Calculated Theoretical Axial Forces for Roof Truss
Part 1.2: Determine experimental axial forces of the Warren and Roof trusses The experimental axial forces for each of the Warren and Roof Trusses can be calculated using the given equation. N = 𝐸𝐴𝜀 Where 𝐸 is the elastic modulus, 𝐴 is the cross-sectional area and 𝜀 is the measured strain (x 10 6 ). Using the following code for each, their respective experimental axial forces can be found. Figure 11: Code for finding the Experimental Axial Forces for Warren Truss Figure 12: Code for finding the Experimental Axial Forces for Roof Truss The resultant experimental axial forces for the Warren Truss. The resultant experimental axial forces for the Roof Truss.
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Results of Axial Forces Member of Truss Warren Truss Roof Truss Theoretical Axial Force (N) Measured Strain (x10 6 ) Experimental Axial Force (N) Theoretical Axial Force (N) Measured Strain (x10 6 ) Experimental Axial Force (N) 1 -115.47 -183 -114.38 -200.00 -341 -213.13 2 57.74 102 63.75 173.21 283 176.88 3 115.47 221 138.13 0.00 17 10.63 4 -115.47 -182 -113.75 -200.00 -336 -210.00 5 -115.47 -185 -115.63 0.00 -5 -3.13 6 173.21 260 162.50 173.21 277 173.13 7 115.47 193 120.63 346.41 578 361.25 8 -230.94 -384 -240.00 -400.00 -660 -412.50 9 230.94 387 241.88 0.00 26 16.25 10 115.47 207 129.38 346.41 543 339.38 11 -230.94 -423 -264.38 -400.00 -668 -417.50 Table 3: Results of Axial Forces Error Percentages Member of Truss Warren Truss Roof Truss Theoretical Axial Force (N) Experimental Axial Force (N) Percentage Error (%) Theoretical Axial Force (N) Experimental Axial Force (N) Percentage Error (%) 1 -115.47 -114.38 0.96 -200.00 -213.13 6.16 2 57.74 63.75 9.44 173.21 176.88 2.07 3 115.47 138.13 16.40 0.00 10.63 100.00 4 -115.47 -113.75 1.51 -200.00 -210.00 4.76 5 -115.47 -115.63 0.13 0.00 -3.13 100.00 6 173.21 162.50 6.59 173.21 173.13 0.05 7 115.47 120.63 4.27 346.41 361.25 4.11 8 -230.94 -240.00 3.77 -400.00 -412.50 3.03 9 230.94 241.88 4.52 0.00 16.25 100.00 10 115.47 129.38 10.75 346.41 339.38 2.07 11 -230.94 -264.38 12.65 -400.00 -417.50 4.19 Table 4: Error Percentages Error is determined using the following equation.
Part 1.3: Results Discussion The provided experimental results were obtained from reading the measurements from the strain gauges after applying the load. Then the experimental axial forces were calculated by multiplying the strain readings by the elastic modulus and the cross-sectional areas of the truss members. The theoretical results were calculated from using the stiffness method. For the Warren Truss, majority of errors were within 10% with the highest value of error being 16.40%. For the Roof Truss, majority of errors were within 5%, excluding zero force members, the highest value of error for members within the Roof Truss was 6.16%. Although the methodology of the lab was valid, the lab had several errors and unaccounted-for variables. From the results, the errors appear to be random errors as the resultant errors that appear above and below the theoretical axial force seems to have no consistency and form no particular pattern. Multiple factors such as temperature, physical defects and friction in the trusses may have contributed to some of these errors. Temperature may be a contributing factor towards the errors as heating causes the trusses to expand whilst cooling causes the trusses to contract. This phenomenon varies the length of the truss which also changes the strain of the truss which will also go on to change the value of the axial force due to it being directly related to the strain value. Physical defects produced either during manufacturing or it’s usage over time will also affect the experimental axial force as the cross-sectional areas of the trusses will not be consistent throughout the entire structure due to the defections. Friction is another factor that may have possibly generated errors. The joints connecting members of the truss may not be ideal and could have some unaccounted-for forces whether it be friction or other forces such as shear force or bending moments. Following the methodology, the lab had also followed a few assumptions. One such assumption is that the structure follows Young’s Modulus throughout the entir ety of its loading. This assumption does not account for potential physical deformations during the loading of the material. Other assumptions refer back to the gauge and its readings. The lab had assumed that the gauges were set to zero and that there is no difference between all the gauges in that they were all functional and calibrated. In the possibility that gauges were faulty or that they were not calibrated, the experimental results would differ.