1. **Problem Statement:** Determine the end shear forces and end moments for a beam subjected to a distributed load of 540 lb/ft along the hypotenuse of a right triangle with legs 4' (horizontal) and 6' (vertical). 2. **Step 1: Calculate the length of the hypotenuse** Use the Pythagorean theorem: $$L = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} \approx 7.211'$$ 3. **Step 2: Calculate the total load on the beam** The distributed load is 540 lb/ft along the hypotenuse, so total load $W$ is: $$W = 540 \times L = 540 \times 7.211 \approx 3899.94 \text{ lb}$$ 4. **Step 3: Determine the load components along horizontal and vertical directions** The load acts along the hypotenuse, so resolve it into horizontal and vertical components using the direction cosines: Horizontal component: $$W_x = W \times \frac{4}{L} = 3899.94 \times \frac{4}{7.211} \approx 2161.1 \text{ lb}$$ Vertical component: $$W_y = W \times \frac{6}{L} = 3899.94 \times \frac{6}{7.211} \approx 3241.7 \text{ lb}$$ 5. **Step 4: Calculate the location of the resultant load from the left end** The centroid of a triangular load is located at $\frac{1}{3}$ the length from the larger end (6' side). Since the load is uniform along the hypotenuse, the resultant acts at the midpoint: $$x_{res} = \frac{L}{2} = \frac{7.211}{2} = 3.6055'$$ 6. **Step 5: Calculate the reactions at the beam ends (assuming simply supported beam)** Sum of vertical forces: $$R_A + R_B = W_y = 3241.7$$ Taking moments about left end (A): $$R_B \times L = W_y \times x_{res}$$ $$R_B = \frac{W_y \times x_{res}}{L} = \frac{3241.7 \times 3.6055}{7.211} = 1620.85 \text{ lb}$$ Then, $$R_A = 3241.7 - 1620.85 = 1620.85 \text{ lb}$$ 7. **Step 6: Calculate the shear forces at the ends** Shear at left end $V_A = R_A = 1620.85$ lb upward. Shear at right end $V_B = R_B = 1620.85$ lb upward. 8. **Step 7: Calculate the bending moments at the ends** For a simply supported beam with a uniformly distributed load, end moments are zero: $$M_A = M_B = 0$$ **Final answers:** - Left end shear $V_A = 1620.85$ lb upward - Right end shear $V_B = 1620.85$ lb upward - End moments $M_A = M_B = 0$