The principal stresses at a certain point in a strained material are 150 MPa and 48 Mpa both tensile. Find normal and tangential stresses on a plane inclined at 20 degree with the major principal plane.
You will find the equations you need at
http://en.wikipedia.org/wiki/Mohr's_circle
To find the normal and tangential stresses on a plane inclined at an angle with the major principal plane, you can use the transformation equations for stress.
The transformation equations for stress relate the normal and tangential stresses on a plane inclined at an angle \(\theta\) with the major principal plane to the principal stresses.
The equations are:
\[
\begin{align*}
\sigma_n &= \frac{1}{2}(\sigma_1 + \sigma_2) + \frac{1}{2}(\sigma_1 - \sigma_2)\cos(2\theta) \\
\tau_t &= \frac{1}{2}(\sigma_1 - \sigma_2)\sin(2\theta)
\end{align*}
\]
where \(\sigma_n\) is the normal stress on the inclined plane and \(\tau_t\) is the tangential or shear stress on the plane. \(\sigma_1\) and \(\sigma_2\) are the principal stresses.
In this case, the major principal stress is 150 MPa, and the minor principal stress is 48 MPa. The angle of inclination \(\theta\) is given as 20 degrees.
Now, we can substitute the given values into the transformation equations to find the normal and tangential stresses on the inclined plane:
\[
\begin{align*}
\sigma_n &= \frac{1}{2}(150 + 48) + \frac{1}{2}(150 - 48)\cos(2\times 20^\circ) \\
\tau_t &= \frac{1}{2}(150 - 48)\sin(2\times 20^\circ)
\end{align*}
\]
Calculating \(\sigma_n\) and \(\tau_t\) using a calculator or software, we get:
\[
\begin{align*}
\sigma_n &\approx 107.19 \text{ MPa (tensile)} \\
\tau_t &\approx 46.74 \text{ MPa}
\end{align*}
\]
Therefore, the normal stress (\(\sigma_n\)) on the inclined plane is approximately 107.19 MPa (tensile), and the tangential stress (\(\tau_t\)) is approximately 46.74 MPa.