Discussion
A structural member supports loads, which produce at a particular point, a state of pure shear stress of 50 N/mm2. At what angles are the principal planes oriented with respect to the plane of pure shear?
Answer: B
The principal planes are at π/4 and 3π/4 to the plane of pure shear.
In pure shear (\(\sigma_{x} = \sigma_{y} = 0\), \(\tau_{xy} = 50\) N/mm2), the principal plane angle satisfies \(\tan 2\theta_{p} = \frac{2\tau_{xy}}{\sigma_{x} - \sigma_{y}}\), which is infinite. So \(2\theta_{p} = \pi/2\) or \(3\pi/2\), giving \(\theta_{p} = \pi/4\) and \(3\pi/4\), with principal stresses of +50 and −50 N/mm2.
This is the general rule that principal planes and planes of maximum shear are 45° apart. The other pairs are either not 45° from the shear plane or not 90° from each other, as two principal planes must be.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development