Step 1: Principal stresses.
For plane stress, rotation to principal planes yields \(\tau=0\); the normal stresses there are the principal stresses \(\sigma_1\) (major) and \(\sigma_2\) (minor) with \(\sigma_1 \ge \sigma_2\) algebraically.
\(\Rightarrow\) (A) is true.
By definition \(|\sigma_2| \le |\sigma_1|\) is guaranteed for ordered principal values (if both tensile or both compressive, the inequality is obvious; with opposite signs, \(\sigma_1\) is the algebraic maximum).
\(\Rightarrow\) (B) is true.
Step 2: Orientation of maximum shear planes.
Mohr's circle gives planes of maximum shear at \(45^\circ\) to the principal planes (i.e., \(2\theta=90^\circ\) on Mohr's circle).
\(\Rightarrow\) (C) is false (it is \(45^\circ\), not \(90^\circ\)).
Step 3: Normal stress on the max–shear planes.
On the two orthogonal planes of maximum shear, the normal stress equals the circle center \(\sigma_{\text{avg}}=\dfrac{\sigma_1+\sigma_2}{2}\) on both planes; hence they are equal to each other.
\(\Rightarrow\) (D) is true.
Final Answer:
\[
\boxed{(A),\ (B),\ (D)}
\]
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).
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).
