Question

A shaft is subjected to pure torsional moment. The maximum shear stress developed in the shaft is 100 MPa. The yield and ultimate strengths of the shaft material in tension are 300 MPa and 450 MPa, respectively.
The factor of safety using maximum distortion energy (Von-Mises) theory is

• A. 1.732

Answer 1

The correct option is A 1.732
${\sigma }_1=100MPa$
${\sigma }_2=-100MPa$
As per Von-Mises yield theory:

$n^2\left[\right({\sigma }_1-{\sigma }_2{)}^2+({\sigma }_2-{\sigma }_3{)}^2+({\sigma }_1-{\sigma }_3{)}^2]\le 2{\sigma }_y^2$

In this case, ${\sigma }_3=0$

$2n^2[{\sigma }_1^2+{\sigma }_2^2-{\sigma }_1{\sigma }_2]\le 2{\sigma }_y^2$

$\Rightarrow n^2\left[\right(100{)}^2+(100{)}^2+(100{)}^2]\le (300{)}^2$

$\Rightarrow n^2=\frac{9}{3}\Rightarrow n\le \sqrt{3}$

$\therefore n=1.732$