Question

A thin cylindrical tube closed at ends is subjected to internal pressure. A torque is also applied to the tube. The principal stresses $p_1$ and $p_2$ developed are $80.0$ units and $20.0$ units respectively. If the yield stress is $240$ units, then what is the factor of safety according to maximum shear stress theory?

• A. $3.00$
• B. $4.00$
• C. $5.00$
• D. $6.00$

Answer 1

The correct option is B $4.00$

Maximum shear stress due to applied loading,
${\tau }_{max}=\frac{p_1-p_2}{2}=\frac{80-20}{2}=30$ units
Maximum shear stress in material at yield stress under uniaxial tension
${\tau }_y=\frac{f_y}{2}=\frac{240}{2}=120units$
$Factorofsafety=\frac{{\tau }_y}{{\tau }_{max}}=\frac{120}{30}=4.0$