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October 31, 2014

October 31, 2014

Posted by **sushil** on Thursday, July 25, 2013 at 12:25pm.

- solid mechanics -
**Elena**, Thursday, July 25, 2013 at 4:04pmnormal stress

σ = F/A = 250000/0.1•0.1 = 2.5•10⁷ N/m²

Tangential stress

τ = F /(A/cosα) = Fcosα/A =

=250000•0.5/0.1•0.1=1.25•10⁷ N/m²

- solid mechanics -
**MathMate**, Thursday, July 25, 2013 at 4:29pmIf I am not mistaken, tangential stress on a plane section at 60° to the longitudinal axis would be the shear stress, γ, and the normal stress on the same plane is the normal stress, σ.

σx=250 kN / (0.1²) m²

=25 MPa

σy=0

α=60°

Using the Mohr circle, we have

Tangential stress

=(σx-σy)/2 * sin(2*α)

=25/2 * sin(2*60°) MPa

=6.25√3 MPa

Normal stress

=((σx+σy)/2 + (σx-σy)/2 * cos(2*α)

= 12.5 + 12.5cos(2*60°)

=6.25 MPa

- solid mechanics -
**MathMate**, Thursday, July 25, 2013 at 4:56pmSorry, typo above:

shear stress should have read τ. γ would have been shear strain.

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