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Resolved Shear Stress <ul><li>All working processes such as rolling, extrusion, forging etc; (stress system quite complex) cause plastic deformation by involving the processes of slip or twinning. (shear stress) </li></ul><ul><li>These stresses arises in most processes and test even the applied stress itself is not a pure shear stress. </li></ul>
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Illustration: <ul><li>A cylindrical single crystal of area A in a conventional tensile test under a uni axial load P. </li></ul><ul><li>The angle between normal to the slip plane and tensile axis is , </li></ul><ul><li>The angle which the slip direction makes with the tensile axis is . </li></ul><ul><li>The area of slip plane inclined at the angle will be A / cos , </li></ul><ul><li>The component of the axial load acting in the slip plane in the slip direction i.e shear stress. is P cos </li></ul><ul><li>Therefore the resolved shear stress R </li></ul><ul><li> R = P cos / A/ cos = P/ A cos cos </li></ul>
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<ul><li>It can be seen that the resolved shear stress has a maximum value when the slip plane is inclined at 45 o to the tensile axis i.e </li></ul><ul><li> R = 1/2 * P /A. </li></ul><ul><li>It becomes smaller for angles either greater than or less than 45 o . </li></ul><ul><li>If < 45 o ( slip plane nearly perpendicular to the tensile axis) , the applied stress has a greater tendency to pull the atoms apart than to slide them. When the slip plane becomes more nearly parallel to the tensile axis ( > 45 o ) the shear stress is again small because the area of the slip plane, A/ cos , is correspondingly large. </li></ul>
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<ul><li>It can be seen that the resolved shear stress has a maximum value when the slip plane is inclined at 45 o to the tensile axis i.e </li></ul><ul><li> R = 1/2 * P /A. </li></ul><ul><li>It becomes smaller for angles either greater than or less than 45 o . </li></ul><ul><li>If > 45 o ( slip plane nearly perpendicular to the tensile axis) , the applied stress has a greater tendency to pull the atoms apart than to slide them. When the slip plane becomes more nearly parallel to the tensile axis ( < 45 o ) the shear stress is again small because the area of the slip plane, A/ cos , is correspondingly large. </li></ul>
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Critical resolved shear stress <ul><li>Slip begins when the shearing stress on the slip plane in the slip direction reaches a threshold value called critical resolved shear stress. </li></ul><ul><li>This value is the yield stress of the single crystal in an ordinary stress strain curve. ( In practice it is very difficult to determine the stress at which the first slip bands are produced. </li></ul><ul><li>The value of the critical resolved shear stress depends chiefly on </li></ul><ul><li>composition and ii) temperature. </li></ul>
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M agnitude of the critical resolved shear stress <ul><li>i) In solid solution, critical resolved shear stress would be high. (Atoms of solute and solvent size different) </li></ul><ul><li>ii) Depends on the interaction of its population of dislocations with each other and with defects such as vacancies, interstitials, and impurity atoms. </li></ul><ul><li>On the basis of this reasoning, the critical resolved shear stress should decrease as the density of defects decreases, provided that the total number of imperfections is not zero. When the last dislocation is eliminated, the critical resolved shear stress should rise abruptly to the high value predicted for the shear strength of a perfect crystal. </li></ul>
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