Question

A thin flat plate is subjected to the following stresses: \[ \sigma_{xx} = 160 \, {MPa}; \, \sigma_{yy} = 40 \, {MPa}; \, \tau_{xy} = 80 \, {MPa}. \] Factor of safety is defined as the ratio of the yield stress to the applied stress. The yield stress of the material under uniaxial tensile load is 250 MPa. The factor of safety for the plate assuming that material failure is governed by the von Mises criterion is \_\_\_\_\_\_\_ (rounded off to two decimal places).
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Hint:

The von Mises stress is used to predict material failure when multiple stresses are acting on a material. The factor of safety is the ratio of the yield stress to the von Mises stress.

Answer 1

We are given the following:
\( \sigma_{xx} = 160 \, {MPa} \),
\( \sigma_{yy} = 40 \, {MPa} \),
\( \tau_{xy} = 80 \, {MPa} \),
Yield stress \( \sigma_y = 250 \, {MPa} \),
Material failure is governed by the von Mises criterion.
Step 1: The von Mises stress is given by the formula: \[ \sigma_v = \sqrt{\frac{\sigma_{xx}^2 + \sigma_{yy}^2}{2} - \sigma_{xx} \sigma_{yy} + 3 \tau_{xy}^2} \] Step 2: Substitute the given values: \[ \sigma_v = \sqrt{\frac{160^2 + 40^2}{2} - 160 \times 40 + 3 \times 80^2} \] Simplifying the terms: \[ \sigma_v = \sqrt{\frac{25600 + 1600}{2} - 6400 + 19200} = \sqrt{26400} = 162.49 \, {MPa} \] Step 3: The factor of safety is given by: \[ n = \frac{\sigma_y}{\sigma_v} = \frac{250}{162.49} \approx 1.25 \] Thus, the factor of safety for the plate is: \[ \boxed{1.25} \]