Question

In a two-dimensional stress analysis, the state of stress at a point is shown in the figure. The values of length $PQ$, $QR$, and $RP$ are $4$, $3$, and $5$ units, respectively. The principal stresses are \underline{\hspace{3cm}. (round off to one decimal place)} \includegraphics[width=0.5\linewidth]{image8.png}

• A. $\sigma_x = 26.7$ MPa, $\sigma_y = 172.5$ MPa
• B. $\sigma_x = 54.0$ MPa, $\sigma_y = 128.5$ MPa
• C. $\sigma_x = 67.5$ MPa, $\sigma_y = 213.3$ MPa
• D. $\sigma_x = 16.0$ MPa, $\sigma_y = 138.5$ MPa

Hint:

For triangular stress elements, use unit normal/tangent on the inclined face and multiply by its length to get resultant force; then apply $\sum F_x=\sum F_y=0$ to recover the Cartesian stresses.

Answer 1

The Correct Option is C

Step 1: Geometry and unit vectors.
$PQ=4$, $QR=3$, $RP=5$ (a $3$–$4$–$5$ triangle).
For the slanted face $RP$, outward unit normal $\mathbf{n}=(4/5,\,3/5)$ and unit tangent $\mathbf{t}=(-3/5,\,4/5)$.

Step 2: Tractions on $RP$.
Given normal stress $\sigma_n=120$ MPa and shear stress $\tau=70$ MPa on $RP$.
Resultant force on $RP$ (unit thickness):
\[ \mathbf{F}_{RP}= (\sigma_n\,\mathbf{n}+\tau\,\mathbf{t})\times \text{area} = (120\,\mathbf{n}+70\,\mathbf{t})\times 5. \]
Compute components:
$120\times5\,\mathbf{n}=120(4,3)=(480,\,360)$, $70\times5\,\mathbf{t}=70(-3,4)=(-210,\,280)$.
Hence $\mathbf{F}_{RP}=(480-210,\,360+280)=(270,\,640)$.

Step 3: Forces on the other faces.
On $PQ$ (length $4$): normal stress $\sigma_x$ acts $\rightarrow$ force $\mathbf{F}_{PQ}=(-\sigma_x\cdot4,\,0)$.
On $QR$ (length $3$): normal stress $\sigma_y$ acts $\rightarrow$ force $\mathbf{F}_{QR}=(0,\,-\sigma_y\cdot3)$.

Step 4: Force equilibrium of the wedge.
$\mathbf{F}_{PQ}+\mathbf{F}_{QR}+\mathbf{F}_{RP}=\mathbf{0}$.
Thus, $-4\sigma_x+270=0 \Rightarrow \sigma_x=270/4=67.5\ \text{MPa}$.
And $-3\sigma_y+640=0 \Rightarrow \sigma_y=640/3=213.3\ \text{MPa}$.
\[ \boxed{\sigma_x=67.5\ \text{MPa}, \sigma_y=213.3\ \text{MPa}} \]