1. **Problem Statement:** Calculate the vertical deflection at point U on a simply supported beam with two downward point loads each of magnitude $\frac{P}{2}$ applied at points T and V. Given: - Total length $L = 17.20$ ft - Load $P = 140$ kip - Flexural rigidity $EI = 38600$ k-ft$^2$ - Beam divided into three equal segments of length $\frac{L}{3}$ - Point U is located at $\frac{L}{2}$ from the left support S 2. **Formula and Approach:** The deflection at a point due to point loads on a simply supported beam can be found using the principle of superposition and the standard beam deflection formulas. The deflection $V_x$ at a distance $x$ from the left support due to a point load $P$ at distance $a$ from the left support is given by: $$ V_x = \frac{P b x}{6 L EI} (L^2 - b^2 - x^2) $$ where $b = L - a$. 3. **Calculate distances:** - Length of each segment: $\frac{L}{3} = \frac{17.20}{3} = 5.7333$ ft - Positions: - $T$ at $\frac{L}{3} = 5.7333$ ft - $U$ at $\frac{L}{2} = 8.6$ ft - $V$ at $\frac{2L}{3} = 11.4667$ ft 4. **Calculate deflection at U due to load at T:** - Load magnitude: $\frac{P}{2} = 70$ kip - Distance from left support to load $a_T = 5.7333$ ft - Distance from load to right support $b_T = L - a_T = 17.20 - 5.7333 = 11.4667$ ft - Distance from left support to point U $x_U = 8.6$ ft Since $x_U > a_T$, use the formula for deflection at $x > a$: $$ V_{U,T} = \frac{P b_T a_T}{6 L EI} (L^2 - b_T^2 - a_T^2) $$ Calculate: - $L^2 = 17.20^2 = 295.84$ - $b_T^2 = 11.4667^2 = 131.48$ - $a_T^2 = 5.7333^2 = 32.87$ - $L^2 - b_T^2 - a_T^2 = 295.84 - 131.48 - 32.87 = 131.49$ Plug in values: $$ V_{U,T} = \frac{70 \times 11.4667 \times 5.7333}{6 \times 17.20 \times 38600} \times 131.49 $$ Calculate numerator: $$ 70 \times 11.4667 \times 5.7333 = 4600.5 $$ Calculate denominator: $$ 6 \times 17.20 \times 38600 = 3985920 $$ So: $$ V_{U,T} = \frac{4600.5}{3985920} \times 131.49 = 0.001154 \times 131.49 = 0.1517 \text{ ft} $$ Convert to inches: $$ 0.1517 \times 12 = 1.82 \text{ in} $$ 5. **Calculate deflection at U due to load at V:** - Load magnitude: $70$ kip - Distance from left support to load $a_V = 11.4667$ ft - Distance from load to right support $b_V = 5.7333$ ft - Distance from left support to point U $x_U = 8.6$ ft Since $x_U < a_V$, use the formula for deflection at $x < a$: $$ V_{U,V} = \frac{P b_V x_U}{6 L EI} (L^2 - b_V^2 - x_U^2) $$ Calculate: - $b_V^2 = 5.7333^2 = 32.87$ - $x_U^2 = 8.6^2 = 73.96$ - $L^2 - b_V^2 - x_U^2 = 295.84 - 32.87 - 73.96 = 189.01$ Plug in values: $$ V_{U,V} = \frac{70 \times 5.7333 \times 8.6}{6 \times 17.20 \times 38600} \times 189.01 $$ Calculate numerator: $$ 70 \times 5.7333 \times 8.6 = 3450.5 $$ Denominator is same as before: 3985920 So: $$ V_{U,V} = \frac{3450.5}{3985920} \times 189.01 = 0.000865 \times 189.01 = 0.1635 \text{ ft} $$ Convert to inches: $$ 0.1635 \times 12 = 1.96 \text{ in} $$ 6. **Total deflection at U:** $$ V_U = V_{U,T} + V_{U,V} = 1.82 + 1.96 = 3.78 \text{ in} $$ Since loads are downward, deflection is negative: $$ V_U = -3.78 \text{ in} $$ 7. **Compare with options:** Closest option is c. -3.93 in. **Final answer:** -3.93 in (option c)