Devin Hart Lab Stress Concentration

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MISSOURI WESTERN STATE UNIVERSITY COLLEGE OF PROFESSIONAL STUDIES DEPARTMENT OF ENGINEERING TECHNOLOGY CET 260-MECHANICS OF MATERIALS Lab 5: Stress Concentration Rinnie Treese Group Number 2 Purpose: Analyze how stress concentrations affect the ultimate strength of materials in tensile loads. Formulas: 1. Max Tensile Stress: 2. 3. Average Stress: Net area: a. b. c. Group A: Group B: Group C: - Dd (Figure 1.1) (Figure 1.2) (Figure 1.2) =k(38ksi) K= the stress concentration factor. (Figure 1.1
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  MISSOURI WESTERN STATE UNIVERSITY COLLEGE OF PROFESSIONAL STUDIES DEPARTMENT OF ENGINEERING TECHNOLOGY CET 260-MECHANICS OF MATERIALS Lab 5: Stress Concentration Rinnie TreeseGroup Number 2Purpose: Analyze how stress concentrations affect the ultimate strength of materials in tensile loads. Formulas: 1.   Max Tensile Stress:   =k(38ksi)2.   Average Stress:      3.   Net area:     a.   Group A:    - Dd (Figure 1.1)b.   Group B:    (Figure 1.2)c.   Group C:    (Figure 1.2)K= the stress concentration factor. (Figure 1.1, 1.2)P= the applied axial tensile load. (lb)    = the net cross-sectional area in the plane of the reduced cross section. (      Equipment/Materials: Metal specimen, Dial Calipers, Tinius Olson testing machine, Calculator, Figure 1.1, Figure 1.2 Procedure:   1.)   Obtain Aluminum Specimen a.   Group A: Center circular holeb.   Group B: Semicircular grooves  c.   Group C: Rectangular Grooves   2.)   Measure your specimen with the dial capilersa.   Determine     3.)   Using Figure 1.1 and 1.2a.   Determine stress concentration factor (k)4.)   Calculate theoretical max tensile force. The ultimate load is given in Appendix G as 38ksi.a.       5.)   Test the specimen in the Tinious Olson machine. Record data in tables. Compare all group’s data to yours.a.   Determine   from testing specimenb.   Record    c.   Print the graph for better evaluation of specimen’s failure point.    Group A: Center Circular Hole K-concentration factor: 2.35P-axial load (lb): 9,035    (    : .122     89.3 x         174,035     74,057Group B: Semicircular Grooves:K-concentration factor: 1.48P-axial load (lb): 8,800    (    :.092     56,240     141,565     95,652.2Group C: Rectangular Grooves:K-concentration factor: 4.2P-axial load (lb): 10,208    (    :.078     159,200     544,662     130,872Calculations:        D=.5035D=.154A=      K=     Discussion: 1.)   Compare the theoretical    you calculated to the    that developed in your specimenthroughout the test.a.)   Were they the same?b.)   Why was your group specimen different from the other groups?2.)   Compare      of your group’s specimen with the other groups.a.)   How did the different milling shapes effect the    ?b.)   Which specimen had the largest     Why?3.)   How important is it to understand stress concentrations in materials?a.)   In what situations do engineers need to consider concentration issues?4.)   Compare the   and the     a.)   What do you notice about the numbers?b.)   Why are they this way?  Conclusion:  The theoretical max was a lot higher than our actual max. This also happened when comparing our actualmax to the other groups. This could be because we had a hole in our specimen and each of the other groups hadgrooves cut out. Our max was a lot closer to Group B than Group C. Once again this could be because Group A andB had a circle and semi-circle. Group C had a rectangle so it completely changed the results. The specimen with thelargest max was the rectangle. It has a larger cross-sectional area. It is very important to understand stressconcentrations of materials because if stress is too high in one area a crack can be caused and eventually there willbe failure at that point. It is very important for engineers to understand this when designing any structure. Havingtoo much stress concentration in one area can be a safety hazard and could minimize the life of a structuredramatically. In each of the Groups the average was less than the max. This is probably to ensure that there will notbe too much stress on any point in an attempt to keep the structure from failing after long periods of time.  
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