QQuestionEngineering
QuestionEngineering
Question:
. In a 2 -D stress system compressive stresses of magnitudes 100 MPa and 150 MPa act
in two perpendicular directions. Shear stresses on these planes have magnitude of 80
MPa. Use Mohr's circle to find,
(i) Principal stresses and their planes
(ii) Maximum shears stress and their planes and
(iii) Normal and shear stresses on a plane inclined at 450 to 150 MPa stress.
13 days agoReport content
Answer
Full Solution Locked
Sign in to view the complete step-by-step solution and unlock all study resources.
Step 1: Identify the given stresses and sketch the stress situation.
- Shear stress of magnitude 80 MPa acting on the x-y plane (denoted as $\tau_{xy} = \tau_{yx} = 80$ MPa)
We are given a 2 -D stress system with the following stress components:
Step 2: Construct Mohr's circle using the given stresses.
$C = \frac{\sigma_{xx} + \sigma_{yy}}{2} = \frac{-100 - 150}{2} = -125$ MPa
To construct Mohr's circle, we need to find two points on the circle, which are given by the stresses on the two perpendicular planes. Now, we can calculate the radius of Mohr's circle: The center of Mohr's circle is given by:
Final Answer
(i) Principal stresses and their planes: - Maximum Principal Stress: $\sigma_{1} = - 14.15$ MPa, acting on a plane perpendicular to a plane inclined at 45 degrees to the x-axis. - Minimum Principal Stress: $\sigma_{2} = - 235.85$ MPa, acting on a plane perpendicular to a plane inclined at 45 degrees to the y-axis. (ii) Maximum shear stress and their planes: - Maximum Shear Stress: $\tau_{max} = 110.85$ MPa, acting on planes inclined at 45 degrees to the principal stress planes. (iii) Normal and shear stresses on a plane inclined at 45 degrees to the 150 MPa stress: - Normal Stress: $\sigma = - 125$ MPa - Shear Stress: $\tau = - 80$ MPa
Need Help with Homework?
Stuck on a difficult problem? We've got you covered:
- Post your question or upload an image
- Get instant step-by-step solutions
- Learn from our AI and community of students