The tensile testing was done on the three composite specimens (90°, and two 45°) were completed with a servo-hydraulic load frame with a wedge. The one in the lab was an MTS 647 Hydraulic Wedge Grip and an 810 Material Test System. The specimens had strain gages with a Wheatstone bridge to collect data such as time, distance, load, axial strain, and transverse strain in volts which was converted into pounds per force (lbf), pressure (psi), and etc. for data analysis. From the strain gages, evidence can support how and when the specimen material failed under the stress being applied to the specimen. The test would extend (stretch) the specimen at a constant rate which created tension on the material and caused the material to fail at or close
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The first factor that was required to find the Laminate Material Strength Properties was to plot Stress vs Strain (Figure 5 (a)) to find ultimate tensile strength (σUTS) and the maximum load (Pmax). Then to find Major Young’s Modulus (Ex), from the axial strain from the Axil/Longitudinal direction, was found by two points that contain a stress value and that are between 1000 µƐ and under 3000 µƐ Corrected Longitudinal Strain. Data point values are used in Tensile Chord Modulus of Elasticity equation to find Major Young’s Modulus. These values are listed Table 1 and the Major Young’s Modulus region is shown in Figure 5 (b). The final strength property found was Poisson’s Ratio (νxy) which uses Transverse Strain direction which is perpendicular to the axial/longitudinal direction. The same data points from the Major Young’s Modulus were applied but used the corrected Longitudinal and Transverse Strain values. The plot used to find Poisson’s Ratio is negative Transverse Strain vs Longitudinal Strain (Figure 5 (c)) and the equation is from the Chord Method (Figure 5 (d)). The plots and material strength properties show a brittle fracture due to the linear slope of the graphs. The next specimen tested was the First ±45° Glass/ Epoxy Laminate Composite …show more content…
The tensile test failure code was AGM [2] and the specimen fracture was at a 45° angle same as the fiber alignment (Image 3). The second ±45° specimen follows the same guidelines of what data is discovered of the first ±45° specimen. The Laminate Strength properties are in Table 4 and the In-Plane Shear properties are in Table 5. The plot of Stress vs Strain is in Figure 8 (a) found the Major Young’s Modulus (Ex) region is in Figure 8 (b). The Poisson’s Ratio region is plotted in Figure 8 (d). The Lamina Shear Stress vs Lamina Shear Strain plot (Figure 9, graph a) was used to calculate the Lamina Shear Modulus (Figure 9 (b)). The type of fracture of this specimen was a brittle
The first factor that was required to find the Laminate Material Strength Properties was to plot Stress vs Strain (Figure 5 (a)) to find ultimate tensile strength (σUTS) and the maximum load (Pmax). Then to find Major Young’s Modulus (Ex), from the axial strain from the Axil/Longitudinal direction, was found by two points that contain a stress value and that are between 1000 µƐ and under 3000 µƐ Corrected Longitudinal Strain. Data point values are used in Tensile Chord Modulus of Elasticity equation to find Major Young’s Modulus. These values are listed Table 1 and the Major Young’s Modulus region is shown in Figure 5 (b). The final strength property found was Poisson’s Ratio (νxy) which uses Transverse Strain direction which is perpendicular to the axial/longitudinal direction. The same data points from the Major Young’s Modulus were applied but used the corrected Longitudinal and Transverse Strain values. The plot used to find Poisson’s Ratio is negative Transverse Strain vs Longitudinal Strain (Figure 5 (c)) and the equation is from the Chord Method (Figure 5 (d)). The plots and material strength properties show a brittle fracture due to the linear slope of the graphs. The next specimen tested was the First ±45° Glass/ Epoxy Laminate Composite …show more content…
The tensile test failure code was AGM [2] and the specimen fracture was at a 45° angle same as the fiber alignment (Image 3). The second ±45° specimen follows the same guidelines of what data is discovered of the first ±45° specimen. The Laminate Strength properties are in Table 4 and the In-Plane Shear properties are in Table 5. The plot of Stress vs Strain is in Figure 8 (a) found the Major Young’s Modulus (Ex) region is in Figure 8 (b). The Poisson’s Ratio region is plotted in Figure 8 (d). The Lamina Shear Stress vs Lamina Shear Strain plot (Figure 9, graph a) was used to calculate the Lamina Shear Modulus (Figure 9 (b)). The type of fracture of this specimen was a brittle