Understanding the State of Stress

The state of stress at a point in a body is described by normal stresses (\(\sigma_x\), \(\sigma_y\)) and shear stress (\(\tau_{xy}\)) in a given coordinate system. Principal stresses (\(\sigma_1\), \(\sigma_2\)) represent the maximum and minimum normal stresses at that point, which occur on planes where the shear stress is zero.

We are given the following information:

• Normal stress in x-direction, \(\sigma_x = 100 \text{ MPa}\)
• Normal stress in y-direction, \(\sigma_y = 200 \text{ MPa}\)
• One principal stress, \(\sigma_1 = 250 \text{ MPa}\)

We need to find the magnitude of the other principal stress (\(\sigma_2\)) and the shearing stress (\(\tau_{xy}\)).

Calculating the Other Principal Stress

A key concept in stress analysis is that certain quantities remain constant regardless of the orientation of the coordinate system. These are called stress invariants. One such invariant is the sum of the normal stresses, which is equal to the sum of the principal stresses:

\(\sigma_x + \sigma_y = \sigma_1 + \sigma_2\)

We can use this relationship to find the other principal stress, \(\sigma_2\).

Substitute the given values into the equation:

\(100 \text{ MPa} + 200 \text{ MPa} = 250 \text{ MPa} + \sigma_2\)

\(300 \text{ MPa} = 250 \text{ MPa} + \sigma_2\)

Now, solve for \(\sigma_2\):

\(\sigma_2 = 300 \text{ MPa} - 250 \text{ MPa}\)

\(\sigma_2 = 50 \text{ MPa}\)

So, the magnitude of the other principal stress is 50 MPa.

Determining the Shearing Stress

Another relationship connects the normal stresses, shear stress, and principal stresses. The product of the principal stresses is related to the normal and shear stresses by the formula:

\(\sigma_1 \sigma_2 = \sigma_x \sigma_y - \tau_{xy}^2\)

We know \(\sigma_x\), \(\sigma_y\), \(\sigma_1\), and we have just calculated \(\sigma_2\). We can use this to find \(\tau_{xy}\).

Substitute the known values:

\(250 \text{ MPa} \times 50 \text{ MPa} = (100 \text{ MPa} \times 200 \text{ MPa}) - \tau_{xy}^2\)

\(12500 \text{ MPa}^2 = 20000 \text{ MPa}^2 - \tau_{xy}^2\)

Rearrange the equation to solve for \(\tau_{xy}^2\):

\(\tau_{xy}^2 = 20000 \text{ MPa}^2 - 12500 \text{ MPa}^2\)

\(\tau_{xy}^2 = 7500 \text{ MPa}^2\)

Now, take the square root to find the magnitude of \(\tau_{xy}\):

\(\tau_{xy} = \sqrt{7500 \text{ MPa}^2}\)

\(\tau_{xy} = \sqrt{2500 \times 3} \text{ MPa}\)

\(\tau_{xy} = 50\sqrt{3} \text{ MPa}\)

The magnitude of the shearing stress is \(50\sqrt{3}\) MPa.

Summary of Results

Based on the calculations using stress invariants:

• The magnitude of the other principal stress is \(50 \text{ MPa}\).
• The magnitude of the shearing stress is \(50\sqrt{3} \text{ MPa}\).

The question asks for the magnitude of the other principal stress and shearing stress respectively. Therefore, the answer is 50 MPa and \(50\sqrt{3}\) MPa.