Lecture 4 6pp - 1 Mechanical Properties of Metals and...

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Unformatted text preview: 1 Mechanical Properties of Metals and Alloys: DeFnitions and Units • Stress $ = F/A, MPa = 10 6 N/m 2 • Strain % = & L/L o (engineering strain) • Elastic Modulus or Young’s Modulus E = $ / % when % << 1 • Poisson Ratio ’ = - % L / % A ! L ! A 2 Mechanical Properties of Metals and Alloys: DeFnitions and Units • Shear Stress ! = F/A, MPa = 10 6 N/m 2 • Shear Strain " = tan # • Shear Modulus G = ! / " when " << 1 G = E 2(1 + " ) for linear elastic materials A F F F F ! 3 Young’s Modulus • Measure of resistance against elastic deformation • Direction-dependent – For iron, average E = 205 GPa; – E(111) = 280 GPa; E(100) = 125 GPa 4 Yield Strength • Measure of the stress level at which permanent or plastic deformation begins • 0.2% offset yield strength de¡ned for metallic systems Material Yield Strength (MPa) Annealed 1100 Al alloys 35 Annealed 304 SS 200 Aged 7075 Al alloys 500 Ti-6Al-4V 800 17-7PH SS 1200 5 A “Typical” Stress-Strain Curve 0.2% strain Stress $ y $ u Strain Ultimate tensile strength Yield strength True stress/strain Eng. stress/strain 6 Plastic Strain " tot = " el + " pl = # E + " pl total strain Stress Strain elastic strain plastic strain 1 2 3 7 Hardness • Rockwell • Vickers • Knoop • ….. 8 Vickers Hardness • (4-face) Pyramidal diamond indentor • Indent at load force F • Measured area A total = A proj / cos 68 o • H V = F / A total Indentor ( # = 136 o ) Projected indented area 9 Plastic Deformation • When stressed beyond elastic limit, metals often develop slip bands • Slip band due to sliding of crystallographic planes 10 Slip Bands Side view Top view 200 μ 11 Critical Shear Stress • Expect shear stress to initiate permanent or plastic deformation to be about 1/4 G • More precise analysis gives about 1/10 G • Experiment: 1/100 to 1/1000 G ~0.5 R 2R 12 Critical Shear Stress Why is the actual shear strength so much smaller?...
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Lecture 4 6pp - 1 Mechanical Properties of Metals and...

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