- #1

- 85

- 0

## Main Question or Discussion Point

imagine a solid metal shaft.

If we apply a torque, there is shear stress in the material. and we get [tex] \tau_{xy} [/tex] from [tex] \frac {\tau}{r} = \frac {T}{J}[/tex]

If we applied direct tension, a tensile force, we get a [tex] \sigma_x [/tex] value from [tex] \sigma_x = \frac {F}{A}[/tex]

If we apply a bending moment, we get a bending stress of value [tex] \sigma_B [/tex] and bending moment, M = [tex] \frac {I \sigma_B }{y} [/tex]

can anyone confirm the fact that only shear (tau) is created by torque, sigma(x) by direct tension and sigma(y) is created by bending moment in every case.

i have a problem which has torque only and shows the principle stresses to be:

[tex] \sigma_1 = \frac {\sigma_x + \sigma_y}{2} + \frac {1}{2} \sqrt{ (\sigma_x - \sigma_y)^2 + 4 {\tau_{xy}}^2 }[/tex]

reduces to this [tex] \sigma_1 = \tau_{xy}[/tex] (and then minus for sigma(2) ) i.e. [tex] \sigma_2 = -\tau_{xy}[/tex]

and when we work out the maximum shear stress from TRESCA, its zero ?

If we apply a torque, there is shear stress in the material. and we get [tex] \tau_{xy} [/tex] from [tex] \frac {\tau}{r} = \frac {T}{J}[/tex]

If we applied direct tension, a tensile force, we get a [tex] \sigma_x [/tex] value from [tex] \sigma_x = \frac {F}{A}[/tex]

If we apply a bending moment, we get a bending stress of value [tex] \sigma_B [/tex] and bending moment, M = [tex] \frac {I \sigma_B }{y} [/tex]

can anyone confirm the fact that only shear (tau) is created by torque, sigma(x) by direct tension and sigma(y) is created by bending moment in every case.

i have a problem which has torque only and shows the principle stresses to be:

[tex] \sigma_1 = \frac {\sigma_x + \sigma_y}{2} + \frac {1}{2} \sqrt{ (\sigma_x - \sigma_y)^2 + 4 {\tau_{xy}}^2 }[/tex]

reduces to this [tex] \sigma_1 = \tau_{xy}[/tex] (and then minus for sigma(2) ) i.e. [tex] \sigma_2 = -\tau_{xy}[/tex]

and when we work out the maximum shear stress from TRESCA, its zero ?