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plastic deformations / slip planes

plastic deformations / slip planes

plastic deformations / slip planes

(OP)
Hi guys, hope you're all on your game love

I was wondering why under tensile the shear plane is @45 degrees in regard to the applied load.
(is it exactly @45 degrees btw?)

Is this because:

a)the shear is at max under this orientation or
b)because the plane stacking is non-dense at this orientation for BCC & FCC crystals? or
c)because of the combination of a & b?

RE: plastic deformations / slip planes

321GO,

Read up on Mohr's circle.

--
JHG

RE: plastic deformations / slip planes

As has been said a Mohr's Circle shows that for a plane stress condition the maximum shear stress will occur at 45 degrees to the principal stress and for uniaxial tension this is always exactly 45 degrees.

Once we start to consider crystal structure and dislocations we then need to look at the critical resolved shear stresses and this is the component of stress resolved in the direction of slip.

From my increasingly dodgy memory it goes something like:

Resolved shear stress is given by τ = σ cos Φ cos λ (Schmid Factor ?) where σ is the magnitude of the applied tensile stress, Φ is the angle between the normal to the slip plane and the direction of the applied force and λ is the angle between the slip plane direction and the direction of the applied force. whereas, critical resolved shear stress value is given by τ =σ (cosΦ cosλ)max

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