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 HW Helper P: 2,277 ... And the examples, Part 2 1) We will assume the retaining wall is prismatic (constant cross section), and therefore will only analyze a foot of it. We start by calculating the effective vertical pressure. $$P_{v} = \gamma z$$ At the top: $$P_{v} = (100 \frac{lb}{ft^3} )(0 ft) = 0$$ At the bottom: $$P_{v} = (100 \frac{lb}{ft^3} )(19 ft) = 1900 \frac{lb}{ft^2}$$ The figure formed by the vertical effective pressure is a triangular prism. In cross section a triangle as shown on the picture in blue-green color. Now we calculate the horizontal pressure (active state) according to Rankine equation (7). $$P_{h} = P_{v} K_{a} - 2c \sqrt{K_{a}} + wK_{a}$$ (7) In our case, we don't have a surcharge (w = 0) and c is 0, therefore the equation reduces to: $$P_{h} = P_{v} K_{a}$$ We proceed to calculate $K_{a}$ (coefficient of active pressure) by using Rankine's equation (8). $$K_{a} = \tan ^{2} ( 45^{o} - \frac{\phi}{2})$$ (8) $$K_{a} = \tan ^{2} ( 45^{o} - \frac{30^{o}}{2})$$ we end up with: $$K_{a} = 0.33$$ Therefore our equation for the horizontal pressure on this case is: $$P_{h} = 0.33P_{v}$$ Now we calculate the horizontal pressure at the top: $$P_{h} = 0.33 (0) = 0$$ and the horizontal pressure at the bottom: $$P_{h} = 0.33 (1900 \frac{lb}{ft^2} ) = 627 \frac{lb}{ft^2}$$ The figure formed by the horizontal pressure is a triangular prism. In cross section a triangle as shown on the picture in magenta color. Now from our knowledge from Statics we calculate the resultant of the horizontal pressure triangle: $$|\vec{R}| = \frac{1}{2} (627 \frac{lb}{ft^2}) (19 ft) (1 ft)$$ $$|\vec{R}| = 5956.5 lb$$ or $$|\vec{R}| = 5.96 kip$$ Now we need to calculate the position of our resultant (from the bottom), which from Statics is: $$\bar{y} = \frac{1}{3} (19 ft) = 6.33 ft$$. Solution: $$|\vec{R}| = 5.96 kip$$ $$\bar{y} = 6.33 ft$$