posted by sushil on .
a steel bar 10m long and 10cm*10cm in section .it is subjected to an axial pull of 250KN .determine the intensities of normal and tangential stresses on a plane section at 60 degree to the longitudinal axis .
σ = F/A = 250000/0.1•0.1 = 2.5•10⁷ N/m²
τ = F /(A/cosα) = Fcosα/A =
If 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
Using the Mohr circle, we have
=(σx-σy)/2 * sin(2*α)
=25/2 * sin(2*60°) MPa
=((σx+σy)/2 + (σx-σy)/2 * cos(2*α)
= 12.5 + 12.5cos(2*60°)
Sorry, typo above:
shear stress should have read τ. γ would have been shear strain.