A flat-bottom grain bin has a square horizontal cross-section of side $B = 3.0\text{ m}$. Paddy grain having bulk density $\rho = 600\text{ kg/m}^3$ and internal angle of friction $\phi = 30^\circ$ is stored. According to Rankine's theory, the active lateral earth pressure coefficient is $k_a = \frac{1 - \sin\phi}{1 + \sin\phi}$. The total lateral thrust on one entire wall of height $H = 4.0\text{ m}$ under Rankine's active state in $\text{kN}$ is ________ (round off to two decimal places).
Answer Key:
47.09
Step-by-Step Solution & Explanation
Multi-Approach Solution:
**Method 1: Rankine Triangular Pressure Distribution**
1. Active lateral pressure coefficient:
$k_a = \frac{1 - \sin(30^\circ)}{1 + \sin(30^\circ)} = \frac{1 - 0.5}{1 + 0.5} = \frac{1}{3} \approx 0.3333$
2. Lateral pressure at base ($y = H = 4.0\text{ m}$):
$p_L = k_a \rho g H = \frac{1}{3} \times 600\text{ kg/m}^3 \times 9.81\text{ m/s}^2 \times 4.0\text{ m} = 7848\text{ Pa}$
3. Total thrust per unit width:
$P' = \frac{1}{2} p_L H = \frac{1}{2} \times 7848 \times 4.0 = 15696\text{ N/m}$
4. Total thrust on wall of width $B = 3.0\text{ m}$:
$P_{\text{total}} = P' \times B = 15696 \times 3.0 = 47088\text{ N} = 47.088\text{ kN} \approx 47.09\text{ kN}$
**Method 2: Direct Integration**
$P_{\text{total}} = \int_0^H k_a \rho g y B \, dy = k_a \rho g B \left[ \frac{H^2}{2} \right] = \frac{1}{3} (600)(9.81)(3.0) \frac{16}{2} = 47.088\text{ kN}$