1. **Problem Statement:** Design the size of the foundation footing for a 12" x 12" column carrying a load of 100 kips with an eccentricity $e = 1.5'$. Given: - Column size: 12" x 12" - Load, $P = 100$ kips - Eccentricity, $e = 1.5'$ - Depth of footing, $D_f = 5'$ - Soil cohesion, $c' = 1200$ psf - Unit weight of soil, $\gamma = 120$ pcf - Angle of internal friction, $\phi' = 0$ 2. **Formula and Concepts:** Since $\phi' = 0$, the soil behaves as a purely cohesive soil. The ultimate bearing capacity $q_u$ for such soil is given by Terzaghi's bearing capacity formula simplified for $\phi' = 0$: $$q_u = c'N_c + q$$ where: - $N_c = 5.7$ (bearing capacity factor for $\phi' = 0$) - $q = \gamma D_f$ (overburden pressure at footing base) The allowable bearing capacity $q_{allow}$ is: $$q_{allow} = \frac{q_u}{FS}$$ Assuming a factor of safety $FS = 3$ (typical value). 3. **Calculate ultimate bearing capacity $q_u$:** $$q_u = c' N_c + \gamma D_f = 1200 \times 5.7 + 120 \times 5 = 6840 + 600 = 7440 \text{ psf}$$ 4. **Calculate allowable bearing capacity $q_{allow}$:** $$q_{allow} = \frac{7440}{3} = 2480 \text{ psf}$$ 5. **Determine the required footing area $A$ considering eccentricity:** The eccentricity reduces the effective area. The load is eccentric by $e = 1.5'$. The effective footing width $B$ must satisfy: $$A = B^2$$ The eccentricity reduces the effective width to: $$B_{eff} = B - 2e$$ The allowable soil pressure considering eccentricity is: $$q_{max} = \frac{P}{B_{eff} \times B} \leq q_{allow}$$ Rearranged: $$P \leq q_{allow} \times B (B - 2e)$$ 6. **Solve for $B$:** Let $x = B$ (in feet), then: $$100,000 = 2480 \times x (x - 3)$$ Simplify: $$2480 (x^2 - 3x) = 100,000$$ $$x^2 - 3x = \frac{100,000}{2480} \approx 40.32$$ $$x^2 - 3x - 40.32 = 0$$ 7. **Solve quadratic equation:** $$x = \frac{3 \pm \sqrt{9 + 4 \times 40.32}}{2} = \frac{3 \pm \sqrt{9 + 161.28}}{2} = \frac{3 \pm \sqrt{170.28}}{2}$$ $$\sqrt{170.28} \approx 13.05$$ So, $$x = \frac{3 + 13.05}{2} = 8.025 \text{ ft (positive root)}$$ 8. **Final footing size:** $$B = 8.025' \approx 8'$$ Therefore, the footing should be approximately $8' \times 8'$ to safely support the load with the given eccentricity and soil conditions. **Answer:** The foundation footing size is approximately **8 feet by 8 feet**.