Assignment2

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School

Concordia University *

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Course

6601

Subject

Mechanical Engineering

Date

Jan 9, 2024

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pdf

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Uploaded by v3nom227

MECH 6601 – Assignment 2 2 Question 1: Solution: The mean and standard deviation has been calculated using the formula below in MATLAB. % Assignment 2 - Elham R. Pathan % 40205903 % Given Data Points Xi = [648 570 540 552 595 580 575 533 564 630]; % Tensile strength in MPa % No. of Data Points N = length(Xi);

MECH 6601 – Assignment 2 3 % Calculating Mean for RTA and ETW mu_RTA = mean(Xi); mu_ETW = 0.8 * mu_RTA; % Calculating Standard Deviation sd = std(Xi); % A-basis and B-basis for RTA A_basis_RTA = mu_RTA - 2 * sd; B_basis_RTA = mu_RTA - 3 * sd; % A-basis and B-basis for ETW A_basis_ETW = mu_ETW - 2 * sd; B_basis_ETW = mu_ETW - 3 * sd; % Displaying results disp( 'RTA Properties:' ); disp([ 'Mean RTA: ' , num2str(mu_RTA), ' MPa' ]); disp([ 'A-basis RTA: ' , num2str(A_basis_RTA), ' MPa' ]); disp([ 'B-basis RTA: ' , num2str(B_basis_RTA), ' MPa' ]); disp( ' ' ); disp( 'ETW Properties:' ); disp([ 'Mean ETW: ' , num2str(mu_ETW), ' MPa' ]); disp([ 'A-basis ETW: ' , num2str(A_basis_ETW), ' MPa' ]); disp([ 'B-basis ETW: ' , num2str(B_basis_ETW), ' MPa' ]); disp( ' ' ); % Given Data Range x = [400:2:700]; % Calculating PDF and CDF for RTA for i = 1:length(x) z = (x(i) - mu_RTA) / sd; PDF_RTA = (1/sqrt(2*pi)) * exp(-z^2 / 2); % PDF function p_z_RTA = (1/sd) * PDF_RTA; PDF_results_RTA(i) = PDF_RTA; p_z_results_RTA(i) = p_z_RTA; CDF_RTA = normcdf(x(i), mu_RTA, sd); % CDF function CDF_results_RTA(i) = CDF_RTA; end % Plotting PDF and CDF for RTA figure(); subplot(2, 1, 1); plot(x, PDF_results_RTA); title( 'PDF for RTA Condition' ); xlabel( 'Tensile Strength (MPa)' ); ylabel( 'Probability Density' ); subplot(2, 1, 2); plot(x, CDF_results_RTA); title( 'CDF for RTA Condition' );

MECH 6601 – Assignment 2 4 xlabel( 'Tensile Strength (MPa)' ); ylabel( 'Cumulative Probability' ); % Calculating PDF and CDF for ETW for i = 1:length(x) z = (x(i) - mu_ETW) / sd; PDF_ETW = (1/sqrt(2*pi)) * exp(-z^2 / 2); % PDF function p_z_ETW = (1/sd) * PDF_ETW; PDF_results_ETW(i) = PDF_ETW; p_z_results_ETW(i) = p_z_ETW; CDF_ETW = normcdf(x(i), mu_ETW, sd); % CDF function CDF_results_ETW(i) = CDF_ETW; end % Plotting PDF and CDF for ETW figure(); subplot(2, 1, 1); plot(x, PDF_results_ETW); title( 'PDF for ETW Condition' ); xlabel( 'Tensile Strength (MPa)' ); ylabel( 'Probability Density' ); subplot(2, 1, 2); plot(x, CDF_results_ETW); title( 'CDF for ETW Condition' ); xlabel( 'Tensile Strength (MPa)' ); ylabel( 'Cumulative Probability' ); RTA Properties: Mean RTA: 578.7 MPa A-basis RTA: 504.7448 MPa B-basis RTA: 467.7671 MPa ETW Properties: Mean ETW: 462.96 MPa A-basis ETW: 389.0048 MPa B-basis ETW: 352.0271 MPa Published with MATLAB® R2023b

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MECH 6601 – Assignment 2 5

MECH 6601 – Assignment 2 6 Question 2: Solution: % Given data from Question 1 Xi = [648 570 540 552 595 580 575 533 564 630]; % Fit Weibull distribution to the data pd = fitdist(Xi', 'Weibull' ); % Extract shape (beta) and scale (alpha) parameters beta = pd.ParameterValues(1); alpha = pd.ParameterValues(2); % Generate x values for PDF x_values = linspace(min(Xi), max(Xi), 100); % Calculate Weibull PDF using the fitted parameters weibull_pdf = pdf(pd, x_values); % Plot the histogram of the data figure; histogram(Xi, 'Normalization' , 'pdf' , 'BinMethod' , 'fd' , 'EdgeColor' , 'w' ); hold on ;

MECH 6601 – Assignment 2 7 % Plot the Weibull PDF curve plot(x_values, weibull_pdf, 'r' , 'LineWidth' , 2); % Display shape and scale parameters on the plot annotation( 'textbox' , [0.6, 0.8, 0.1, 0.1], 'String' , [ 'Shape (\beta): ' , num2str(beta)], 'Color' , 'b' , 'EdgeColor' , 'none' ); annotation( 'textbox' , [0.6, 0.75, 0.1, 0.1], 'String' , [ 'Scale (\alpha): ' , num2str(alpha)], 'Color' , 'b' , 'EdgeColor' , 'none' ); title( 'Weibull Probability Distribution Fit to Data' ); xlabel( 'Tensile Strength (MPa)' ); ylabel( 'Probability Density' ); legend( 'Data Histogram' , 'Weibull PDF Fit' ); % Display shape and scale parameters in the command window disp([ 'Shape parameter (beta): ' , num2str(beta)]); disp([ 'Scale parameter (alpha): ' , num2str(alpha)]); Shape parameter (beta): 596.0541 Scale parameter (alpha): 16.059 Published with MATLAB® R2023b Weibull Probability Distribution Fit to Data 520 540 560 580 600 620 640 660 Tensile Strength (MPa) 0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 Probability Density Data Histogram Weibull PDF Fit Shape ( - ): 596.0541 Scale ( , ): 16.059

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MECH 6601 – Assignment 2 8 1. Shape Parameter ( β ): • When β is less than 1, the failure rate decreases over time. This indicates a distribution with a decreasing hazard function. The tail of the distribution is heavy, and failures are more likely to occur early in the life cycle. • When β is equal to 1, the Weibull distribution becomes the exponential distribution, representing constant hazard over time. • When β is greater than 1, the failure rate increases over time. This indicates a distribution with an increasing hazard function. Failures are more likely to occur later in the life cycle. 2. Scale Parameter ( α ): • α determines the scale of the distribution. As α increases, the entire distribution is stretched horizontally. This means the time-to-failure values are higher, and failures are more likely to occur later in time. • As α decreases, the distribution is compressed horizontally, and failures are more likely to occur earlier in the life cycle. Question 3: Solution: Given: ࠵? ࠵? = ࠵?࠵? ࠵?࠵?࠵? ࠵? ࠵? = ࠵?࠵?࠵?࠵? × ࠵?࠵? "࠵? ࠵? ࠵? = −࠵?࠵?࠵? × ࠵?࠵? "࠵? ࠵? ࠵?࠵? = ࠵? ࠵? ࠵?(࠵? ࠵? − ࠵? ࠵? ) ࠵? ࠵?࠵? = ࠵?࠵? ࠵?(࠵?࠵?࠵?࠵? − (−࠵?࠵?࠵?)) × ࠵?࠵? "࠵? ࠵? ࠵?࠵? = ࠵?࠵?࠵?࠵? ࠵?࠵?࠵? = ࠵?. ࠵?࠵? ࠵?࠵?࠵?

MECH 6601 – Assignment 2 9 Question 4: Solution: The short beam shear (SBS) test conducted according to ASTM D2344 is a three-point bending test used to measure the interlaminar shear strength of composite materials. The recommended limits for specimen dimensions and test span are crucial for obtaining accurate and meaningful results. The failure modes in such tests can be influenced by specimen dimensions, and understanding these factors is essential for interpreting the test outcomes. Specimen Dimensions: The ASTM D2344 standard provides guidelines for the dimensions of the specimen. For a typical 15- to 20-ply carbon/epoxy laminate, recommended dimensions include a span length (L) of 1 cm (0.4 in), width (b) of 0.64 cm (0.25 in), and thickness (h) of 1.9 to 2.5 mm (0.075 to 0.100 in). These dimensions are chosen to ensure that the test primarily induces interlaminar shear failure rather than flexural failure. [1] Failure Modes: Interlaminar Shear Failure: In an ideal short beam shear test, failure should occur due to interlaminar shear stresses at the midplane. This failure mode is characterized by delamination between the plies. Flexural Failure: If the beam is too long compared to its depth, flexural failure may occur at the outer plies of the beam. This type of failure is undesirable as it does not provide an accurate measure of interlaminar shear strength. The span-to-depth ratio (L/h) must be controlled to avoid flexural failure. Impact of Specimen Dimensions: Span-to-Depth Ratio (L/h): The ratio of the span length to the depth of the specimen is critical. If the span is too long compared to the depth, flexural failure may dominate. To ensure interlaminar shear failure, the span-to-depth ratio must satisfy specific relationships, as mentioned in the ASTM D2344 standard. Width (b) and Thickness (h): The width and thickness of the specimen also influence the stress distribution and can impact the failure mode. Specimens that are too thin might experience local compressive failure near the loaded points, affecting the results. Material Considerations : The SBS test may be problematic for materials with a low flexural to interlaminar shear strength ratio, such as Kevlar, textile, and carbon/carbon composites.

MECH 6601 – Assignment 2 10 Modifications to the test method, such as using a short sandwich beam (SSB) test, may be necessary for accurate results. In summary, adherence to recommended specimen dimensions is crucial to ensure that the SBS test induces interlaminar shear failure and provides meaningful results. Deviations from these dimensions can lead to flexural failure or other undesirable outcomes, impacting the accuracy of interlaminar shear strength measurements. REFERENCE : “Engineering Mechanics of Composite Materials”, I.N. Daniel and O. Ishai, Oxford, 2nd Edition, 2006.[1] Question 5: Solution: Stress Relaxation: Definition: Stress relaxation is the gradual decrease in stress within a material under a constant strain or deformation. In other words, when a material is subjected to a constant deformation, the stress within the material decreases with time. Testing Method: To evaluate stress relaxation, Dynamic Mechanical Analysis (DMA) is a common test technique to characterize the mechanical properties of materials, including polymers like epoxy which involves applying a constant strain to a material and measuring the resulting stress over time. The material is typically deformed to a certain strain level and then held at that strain while the stress is monitored. The stress values are recorded over a specified period, and the rate of stress decrease provides information about the material's stress relaxation behavior.

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MECH 6601 – Assignment 2 11 Creep: Definition: Creep is the gradual deformation of a material over time under a constant applied stress. In other words, when a material is subjected to a constant stress, it continues to deform over time. Testing Method: To evaluate creep, Dynamic Mechanical Analysis (DMA) is a common test technique to characterize the mechanical properties of materials, including polymers like epoxy which involves applying a constant stress to a material and measuring the resulting strain over time. The material is typically loaded with a constant stress and the resulting strain is measured as a function of time. Creep tests often involve keeping the material under stress for an extended period to observe its long-term deformation behavior. Key Differences: Deformation Condition: Stress Relaxation: Constant strain, decreasing stress. Creep: Constant stress, increasing strain. Testing Approach: Stress Relaxation: Apply a constant strain, measure stress over time. Creep: Apply a constant stress, measure strain over time. Material Response: Stress Relaxation: Focus on how stress decreases over time under constant deformation. Creep: Focus on how deformation increases over time under constant stress. Reference: Professor Slides (moodle).