Gravity Dam Forces Sketch

Cheatsheet Content

### Problem Description This cheatsheet outlines the forces acting on a gravity dam constructed from bunker blocks. **Dam Dimensions:** * **Block size:** 2 ft (width) x 2 ft (depth) x 6 ft (length) * **Dam height:** 3 blocks high = 3 * 2 ft = 6 ft (assuming blocks are laid on their side, 2ft high) * **Dam base width:** 2 ft (width of a single block at the base) * **Dam length:** Assume a unit length (e.g., 6 ft, the length of one block) for 2D analysis. **Assumptions:** * Water level is at the top of the dam (full reservoir). * Neglect uplift pressure for simplicity in this initial sketch. * Concrete density for blocks: $\rho_{concrete} \approx 150 \text{ lb/ft}^3$. * Water density: $\rho_{water} \approx 62.4 \text{ lb/ft}^3$. * Coefficient of friction between dam base and foundation: $\mu \approx 0.6$. ### Dam Geometry A rectangular gravity dam, 6 ft high and 2 ft wide at the base, built from 2x2x6 ft blocks stacked 3 high (assuming 2ft height per block). #### Cross-Sectional View ``` Water Level | | H = 6 ft | +-----------------+ | | | | | | Dam Body | | +-----------------+ | | +-----------------+ | | +-----------------+ ### Forces on the Dam #### 1. Hydrostatic Pressure ($P_h$) The water exerts a triangular pressure distribution on the upstream face of the dam. * **Pressure at depth y:** $p(y) = \rho_{water} \cdot g \cdot y$ * **Maximum pressure at base:** $p_{max} = \rho_{water} \cdot g \cdot H$ * **Total horizontal force ($F_H$):** Resultant of the hydrostatic pressure, acting at H/3 from the base. $$F_H = \frac{1}{2} \cdot p_{max} \cdot H \cdot L_{unit}$$ $$F_H = \frac{1}{2} \cdot (\rho_{water} \cdot g \cdot H) \cdot H \cdot L_{unit}$$ Where $L_{unit}$ is the unit length of the dam (e.g., 1 ft or 6 ft for one block length). Assuming $L_{unit} = 6 \text{ ft}$ (length of one bunker block): $$F_H = \frac{1}{2} \cdot (62.4 \text{ lb/ft}^3) \cdot (6 \text{ ft})^2 \cdot (6 \text{ ft})$$ $$F_H = 6739.2 \text{ lb}$$ This force acts at $Y_{F_H} = H/3 = 6 \text{ ft} / 3 = 2 \text{ ft}$ from the base. #### 2. Weight of the Dam ($W$) The weight of the dam acts vertically downwards through its center of gravity. * **Volume of one block:** $2 \text{ ft} \times 2 \text{ ft} \times 6 \text{ ft} = 24 \text{ ft}^3$ * **Number of blocks (for unit length of 6 ft):** Since the dam is 2 ft wide and 6 ft long for this section, and blocks are 2x2x6, the base is one block (2 ft wide, 6 ft long). The dam is 3 blocks high, so it's 3 blocks stacked vertically. Total blocks = 3 blocks. * **Total volume of dam (for 6 ft length):** $3 \times 24 \text{ ft}^3 = 72 \text{ ft}^3$ * **Weight of dam ($W$):** $V_{dam} \cdot \rho_{concrete}$ $$W = 72 \text{ ft}^3 \cdot 150 \text{ lb/ft}^3 = 10800 \text{ lb}$$ This force acts at the center of gravity (CG) of the dam. For a rectangular dam, the CG is at $B/2$ from the heel (upstream toe) and $H/2$ from the base. So, $X_W = 2 \text{ ft} / 2 = 1 \text{ ft}$ from the heel. #### 3. Uplift Pressure ($U$) (Neglected for this initial sketch as per assumptions, but important in real design.) If considered, it acts upwards on the base of the dam. Distribution is often assumed trapezoidal, decreasing from maximum at the heel to minimum at the toe. #### 4. Frictional Resistance ($F_f$) This is the resisting force at the base of the dam that prevents sliding. * **Frictional force:** $F_f = \mu \cdot \sum \text{Vertical Forces}$ * **Vertical forces:** Primarily the weight of the dam ($W$). $$F_f = \mu \cdot W$$ $$F_f = 0.6 \cdot 10800 \text{ lb} = 6480 \text{ lb}$$ This force acts horizontally at the base, opposing the sliding tendency caused by $F_H$. #### 5. Reaction Force ($R_v$) The vertical reaction from the foundation, equal and opposite to the net vertical forces on the dam. For stability, $R_v = W - U$ (if uplift is considered). Here, $R_v = W$. ### Stability Analysis (Brief) #### 1. Sliding Stability * **Factor of Safety against Sliding (FOS_s):** Ratio of resisting forces to overturning forces. $$FOS_s = \frac{F_f}{F_H}$$ $$FOS_s = \frac{6480 \text{ lb}}{6739.2 \text{ lb}} \approx 0.96$$ A FOS_s less than 1.0 indicates the dam is likely to slide. Typically, FOS_s > 1.5 is required. This simple dam design is highly unstable against sliding. #### 2. Overturning Stability * **Overturning Moment ($M_O$):** Moment caused by $F_H$ about the toe (downstream edge) of the dam. $$M_O = F_H \cdot Y_{F_H} = 6739.2 \text{ lb} \cdot 2 \text{ ft} = 13478.4 \text{ lb-ft}$$ * **Resisting Moment ($M_R$):** Moment caused by $W$ about the toe. $$M_R = W \cdot X_W = 10800 \text{ lb} \cdot 1 \text{ ft} = 10800 \text{ lb-ft}$$ (Note: $X_W$ is distance from toe, not heel. For a rectangular dam, CG is at B/2 from the heel, so also B/2 from the toe). * **Factor of Safety against Overturning (FOS_o):** $$FOS_o = \frac{M_R}{M_O}$$ $$FOS_o = \frac{10800 \text{ lb-ft}}{13478.4 \text{ lb-ft}} \approx 0.80$$ A FOS_o less than 1.0 indicates the dam is likely to overturn. Typically, FOS_o > 2.0-3.0 is required. This dam design is highly unstable against overturning. #### 3. Eccentricity of Resultant Force The resultant force should fall within the middle third of the base to avoid tension at the heel/toe. For this dam, with a base of 2 ft, the middle third is from $2/3 \text{ ft}$ to $4/3 \text{ ft}$ from the heel. Given the severe instability, the resultant would fall outside this. ### Summary Sketch of Forces ``` Water Level (H = 6 ft) | | | | | | | | | +-------------------------+ | | | | | | Dam Body (6 ft high, 2 ft wide) F_H --> | | (6739.2 lb) | W (10800 lb) @ 2 ft from base | acting at CG (1 ft from heel/toe) | | | | | v +-------------------------+------+ | | | | | | | | | +-------------------------+------+ | | +-------------------------+ /|\ /|\ |_________________________| Heel (Upstream) Toe (Downstream)