Practice Question: Active Earth Pressure on Cantilever Retaining Wall

---

---

Solution

Answer: D) 472 kip

Calculate the Coulumb coefficient of active earth pressure:

\[K_a = \frac{\cos^2 (\phi - \theta)}{\cos^2 \theta \cos (\theta + \delta) \left[ 1 + \sqrt{\left( \frac{\sin (\phi + \delta) \sin (\phi - \alpha)}{\cos (\theta + \delta) \cos (\theta - \alpha)} \right)} \right]^2}\]\(Wall\ face\ angle=\theta = 0 \degree\)

\(\alpha = 15 \degree\)

\(\delta = 0\)

\[K_a = \frac{\cos^2 (30)}{\cos^2 (0) \cos (0 + 19) \left[ 1 + \sqrt{\left( \frac{\sin (30 + 19) \sin (30 - 15)}{\cos (0 + 19) \cos (0 - 15)} \right)} \right]^2}\]\[K_a = \frac{\cos^2 (30)}{\cos(19)\left[ 1 + \sqrt{\left( \frac{\sin (49) \sin (15)}{\cos(19)\cos (- 15)} \right) }\right]^2}\]

\[K_a = 0.37\]

Calculate the horizontal component of the coefficient:

\[K_{ah}\ =K_{a}\ast\cos{19 \degree}\ =\ 0.37 \ast \cos{19°} =\ 0.35\]

Calculate the horizontal component of the resultant force:

\[R_{ah}=\frac{1}{2}\gamma\ H^2K_{ah}L=\ \frac{1}{2}\ast120\ pcf\ast\left(30\ ft\right)^2\ast0.35\ast(25\ ft)=472,500\ lb\]

 \[R_{ah}=472.5\ kip\]
 

---