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Mechanical Properties: Review l0l0 A0A0 F  (stress)  (strain) Y nominal (engineering)true plastic – rearrange chemical bonds elastic – stretch chemical.

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Presentation on theme: "Mechanical Properties: Review l0l0 A0A0 F  (stress)  (strain) Y nominal (engineering)true plastic – rearrange chemical bonds elastic – stretch chemical."— Presentation transcript:

1 Mechanical Properties: Review l0l0 A0A0 F  (stress)  (strain) Y nominal (engineering)true plastic – rearrange chemical bonds elastic – stretch chemical bonds stress strain true ( ,  ) nominal (s, e)

2  (stress)  (strain) Y plastic elastic yy load Definitions unload 0.2 %  y = yield strength 0.2% offset yield strength  y = elastic limit strength – some value of stress

3 Definitions & Comparisons P. L. Mangonon “Principles of Materials Selection for Engineering Design” Prentice Hall, Upper Saddle River, NJ 1999. necking  UTS = (ultimate) tensile strength upper yield strength lower yield strength X fracture without yield (ionic/colavent bonding) X fracture strength ductility = strain at failure yy  = F/A i

4 Definitions  (stress)  (strain)  y = elastic limit ductility X resilience = energy to plastic deformationenergy/vol recall: Elastic region: ½ toughness = energy to failure

5 Shear stress from uniaxial tension A F force is normal to the plane normal stress A′A′ consider what happens on some other plane, A′  need to know area of A′ A′ = A/cos(  ) consider shear force parallel to A′ along an arbitrary direction F′F′  need to know magnitude of F′ F′ = Fcos(  ) F′ can be along any direction parallel to A′ (each corresponds to a different  ) Given  and  :

6 Final result:  max =  /2, in the orientation  =  =  /4

7 Solute Hardening

8 Precipitate Hardening coherent incoherent


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