14_440_407_Class Notes 3, 2009 [Compatibility Mode]

14_440_407_Class Notes 3, 2009 [Compatibility Mode] -...

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Plastic Deformation hear deformation Shear deformation The onset of plastic deformation in a crystalline solid coincides with e appearance of surface slip bands, which represent visible evidence the appearance of surface slip bands, which represent visible evidence for crystallographic shear deformation. -- shear plane is called “slip plane” -- shear direction is called “slip direction” -- combination shear plane / shear direction is called “slip system” Single versus multiple slip Shear deformation on one slip system is called single slip. -- commonly observed during plastic deformation of single crystal (one-grained) materials hear deformation on everal lip systems is called ultiple slip Shear deformation on several slip systems is called multiple slip . -- typically observed during plastic deformation of polycrystalline (many-grained) materials
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Figure 49. Slip in a zinc single Fig. 50. Slip lines on the surface of polycrystalline Cu that was electro- Operative slip systems crystal. pyy polished and deformed. 173x Shear deformation occurs in close-packed directions on close-packed planes in all crystalline solids.
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Solid Structure Slip Plane Slip direction Multiplicity Al fcc (111) [0-11] 12 Cu Fe Mo bcc (0-11) [111] g cp 001) 1 0] Mg Ti hcp (0001) [11-20] 3 chmid’sLaw Schmid s Law The value of the shear stress (acting on slip plane in slip direction) when plastic deformation is initiated is known as the critical resolved hear stress t CRSS which is a constant for a iven rystal The value shear stress , t CRSS, which is a constant for a given crystal. The value of τ crss is given by the expression ψ φ σ = τ cos cos where σ is the applied stress, Φ , is the angle between stress axis and slip direction, and Ψ is the angle between stress axis and slip plane normal. * Note that the normal stress acting on the slip plane has no influence on the slip process .
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Derivation F s = shear force on slip plane (F s = F cos Ф ) ea t o F n = tensile force normal to slip plane (F n = F cos ψ ) hear stress shear force/shear area Shear stress τ = shear force/shear area ψ = τ cos A F s cos ψ cos φ A F τ = ψ φ σ = τ cos cos
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Example A cubic crystal is 6 cm long x 2 cm 2 cross-section - 500 kg tensile load applied parallel to [100]. (a) Calculate the shear force on (1-10) in [110] (b) Calculate the shear stress on (1-10) in [110]. hear force Se a oc e F s = F cos Ф = 500 cos 45 ° 53 5 kg = 353.5 kg Shear stress τ = shear force/shear area Since A shear = A/cos ψ
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ψ = = τ cos A F A F s s ψ cos = = ) 707 . 0 ( 2 5 . 353 125 kg/cm 2
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aterial bserved Shear alculated Shear Material Observed Shear Strength (psi) Calculated Shear Strength (psi) Cu 32,000 1,120,000 Fe Ni SiC 42,000 69,000 250,000 1,805,000 1,750,000 4,690,000 Calculated shear strength overestimates actual strength by over an order of magnitude.
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14_440_407_Class Notes 3, 2009 [Compatibility Mode] -...

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