Section 4: Soil and Water Conservation Engineering
Soil Mechanics
Subtopic: Mohr's circle of stress
Question Statement:
A soil element in a shear testing device is subjected to principal stresses of $\sigma_1 = 320.0\text{ kPa}$ (major principal stress) and $\sigma_3 = 100.0\text{ kPa}$ (minor principal stress). The normal stress $\sigma_\theta$ acting on a plane inclined at $\theta = 60^\circ$ to the major principal plane in $\text{kPa}$ is ________ (answer in integer).
```
τ (Shear Stress)
| * (σ1+σ3)/2, R
| .-' | '-. Radius R = (σ1 - σ3)/2
| .' | '.
| / | \
----------+---(-------+-------)---------> σ (Normal Stress)
| σ3 C σ1
```
Answer Key:
155
Step-by-Step Solution & Explanation
**Method 1: Mohr's Circle Analytical Transformation**
1. Normal stress on a plane inclined at $\theta$ to the major principal plane:
$\sigma_\theta = \frac{\sigma_1 + \sigma_3}{2} + \frac{\sigma_1 - \sigma_3}{2} \cos(2\theta)$
2. Evaluating terms with $\theta = 60^\circ \implies 2\theta = 120^\circ$:
$\text{Center } C = \frac{320.0 + 100.0}{2} = \frac{420.0}{2} = 210.0\text{ kPa}$
$\text{Radius } R = \frac{320.0 - 100.0}{2} = \frac{220.0}{2} = 110.0\text{ kPa}$
$\cos(120^\circ) = -0.50$
$\sigma_\theta = 210.0 + 110.0 \times (-0.50) = 210.0 - 55.0 = 155.0\text{ kPa}$
**Method 2: Direction Cosine Projection**
$\sigma_\theta = \sigma_1 \cos^2(\theta) + \sigma_3 \sin^2(\theta)$
$\cos(60^\circ) = 0.50 \implies \cos^2(60^\circ) = 0.25$
$\sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.8660 \implies \sin^2(60^\circ) = 0.75$
$\sigma_\theta = 320.0 \times 0.25 + 100.0 \times 0.75 = 80.0 + 75.0 = 155.0\text{ kPa}$