Question

The state of 2-D stress at a point is given by the following stresses
$\left[\begin{array}{cc}120& 40\\ 40& 80\end{array}\right]MPa$
If a specimen is designed by the maximum shear stress theory applied FOS is . ($F_y=210MPa)$

• A. 2.9
• B. 1.45
• C. 3.84
• D. 4.21

Answer 1

The correct option is B 1.45

According to maximum shear stress theory:

$({\tau }_{max}{)}_{abs}\le \frac{F_y}{2\times FOS}$

Principal stresses:

${\sigma }_{1/2}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}^2}$

Here: ${\sigma }_x=120MPa,{\sigma }_y=80MPa,{\tau }_{xy}=40MPa$

$So,{\sigma }_{1/2}=\frac{120+80}{2}\pm \sqrt{{\left(\frac{120-80}{2}\right)}^2+{40}^2}$

$=100\pm 44.72$

${\sigma }_1=144.72MPa$

${\sigma }_2=55.28MPa$

$\text{Here}{\sigma }_3\text{is not given, so assume}{\sigma }_3=0$

${\tau }_{max}=max\{\left|\frac{{\sigma }_1-{\sigma }_2}{2}\right|,\left|\frac{{\sigma }_2-{\sigma }_3}{2}\right|,\left|\frac{{\sigma }_1-{\sigma }_3}{2}\right|\}$

$=\frac{144.72-0}{2}\Rightarrow {\tau }_{max}=72.36MPa$

$\because {\tau }_{max}\le \frac{F_y}{2\times FOS}\Rightarrow FOS=\frac{210}{72.36\times 2}=1.45$