hw_3_solutions

# hw_3_solutions - 1 You have been hired by the quality...

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1. You have been hired by the quality control section of Stanley and you have been tasked with determining the optimal material for the manufacture of hammers. As you know, hammers are constantly subjected to deformation during their usage and you have determined that testing the fatigue of materials is the best way about it. You have carried out fatigue tests on two materials and have established the plot below. a. What do quality control engineers do? b. What are the maximum and minimum stress levels that have been subjected by the two materials? c. Assuming the materials in question have shown failure at the indicated spots, which of the two would you choose for the manufacture of a hammerhead? Solution : The maximum and minimum stresses are 510 MPa and 490MPa, respectively; the stress modulates between these two extremes following a sinusoidal trend. Based on the chart above, material A would break sooner than material B and should be avoided. 485 490 495 500 505 510 515 0 0.5 1 1.5 2 2.5 3 3.5 Stress (kPa) time in hours Fatigue A B

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2. You are to design a single-poled stool made of aluminum to be used by the general public. What must be its minimum diameter to withstand the weight of an average person? What would the diameter be for gold? Solution: To withstand a person’s weight, the stool must’s yield stress must be greater than the stress a person would exert, or: ߪ > ߪ ௣௘௥௦௢௡ We can find the stress a person would exert from the weight of the person and the area of the stool: ߪ ௣௘௥௦௢௡ = ܹ݁݅݃ℎݐ ܣݎ݁ܽ = ݉ܽݏݏ ௣௘௥௦௢௡ ݔ݃ ߨݎ Assume the average person weighs around 160lb (72.5kg). Since the maximum stress that a person should exert is the yield strength, then equate the two and: ߪ ௬௜௘௟ௗ = 72.5݇݃ݔ9.81 ݉ ݏ ߨݎ Solving for r: ݎ = 72.5݇݃ ݔ 9.81 ݉ ݏ ߨߪ ௬௜௘௟ௗ The yield strength for aluminum is 15,000,000Pa and for gold it is 210,000,000Pa. Thus, we find that the minimum radii for the construction of the chair are: ݎ ௚௢௟ௗ = 72.5݇݃ ݔ 9.81 ݉ ݏ ߨ210000000ܲܽ = 0.00104݉ = 1.04݉݉ ݎ ௔௟௨௠௜௡௨௠ = 72.5݇݃ ݔ 9.81 ݉ ݏ ߨ15000000ܲܽ = 0.0039݉ = 3.9݉݉
3. The following was taken for an aluminum alloy at 480°C that has a cube shape of side 2.3m and a constant compressive stress of 2.75MPa. Determine: a. The steady state creep rate. b. The length of the aluminum slab at rupture. Solution: Part a Steady state creep is the slope of the strain over time in the linear region, or: ܵݐ݁ܽ݀ݕ ܵݐܽݐ݁ ܥݎ݁݌ = ∆ߝ ∆ݐ = 0.9 − 0.55 20 − 10 = 0.045 ݏ ିଵ 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 5 10 15 20 25 30 35 40 Strain Time (sec) Creep Rate

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Part b At rupture, the strain is 1.75, we use the definition of strain to find the final dimension: ߝ = ݈ − ݈ ݈ = ݈ − 2.3݉ 2.3݉ ݈ = 6.33݉
4.

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