... 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.

[tex] P_{v} = \gamma z [/tex]

At the top:

[tex] P_{v} = (100 \frac{lb}{ft^3} )(0 ft) = 0 [/tex]

At the bottom:

[tex] P_{v} = (100 \frac{lb}{ft^3} )(19 ft) = 1900 \frac{lb}{ft^2} [/tex]

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).

[tex] P_{h} = P_{v} K_{a} - 2c \sqrt{K_{a}} + wK_{a} [/tex] (7)

In our case, we don't have a surcharge (w = 0) and c is 0, therefore the equation reduces to:

[tex] P_{h} = P_{v} K_{a} [/tex]

We proceed to calculate [itex] K_{a} [/itex] (coefficient of active pressure) by using Rankine's equation (8).

[tex] K_{a} = \tan ^{2} ( 45^{o} - \frac{\phi}{2}) [/tex] (8)

[tex] K_{a} = \tan ^{2} ( 45^{o} - \frac{30^{o}}{2}) [/tex]

we end up with:

[tex] K_{a} = 0.33 [/tex]

Therefore our equation for the horizontal pressure on this case is:

[tex] P_{h} = 0.33P_{v} [/tex]

Now we calculate the horizontal pressure at the top:

[tex] P_{h} = 0.33 (0) = 0 [/tex]

and the horizontal pressure at the bottom:

[tex] P_{h} = 0.33 (1900 \frac{lb}{ft^2} ) = 627 \frac{lb}{ft^2} [/tex]

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:

[tex] |\vec{R}| = \frac{1}{2} (627 \frac{lb}{ft^2}) (19 ft) (1 ft) [/tex]

[tex] |\vec{R}| = 5956.5 lb [/tex]

or

[tex] |\vec{R}| = 5.96 kip [/tex]

Now we need to calculate the position of our resultant (from the bottom), which from Statics is:

[tex] \bar{y} = \frac{1}{3} (19 ft) = 6.33 ft [/tex].

Solution:

[tex] |\vec{R}| = 5.96 kip [/tex]

[tex] \bar{y} = 6.33 ft [/tex]