1. **State the problem:** We need to draw the shear force and bending moment diagrams for a beam with the following loads and supports: - Point load 5 kips downward at A (x=0 ft) - Point load 15 kips upward at B (x=5 ft) - Uniform distributed load (UDL) 2 kips/ft from B to E (x=5 ft to x=28 ft) - Point load 10 kips downward at E (x=28 ft) - A clockwise moment of 25 kip-ft applied between B and D (at x=13 ft) - Supports at B and D 2. **Identify beam segments and reactions:** - Total beam length: 5 + 8 + 12 + 3 = 28 ft - Supports at B (x=5 ft) and D (x=25 ft) 3. **Calculate reactions at supports B and D:** Let reaction forces be $R_B$ and $R_D$ (vertical). Sum of vertical forces = 0: $$R_B + R_D - 5 + 15 - 10 - (2 \times 23) = 0$$ The distributed load length is from B (5 ft) to E (28 ft), length = 23 ft, total load = $2 \times 23 = 46$ kips downward. So: $$R_B + R_D - 5 + 15 - 10 - 46 = 0 \Rightarrow R_B + R_D - 46 = 0 \Rightarrow R_B + R_D = 46$$ Sum moments about B = 0 (taking clockwise positive): - Moment due to 5 kip at A (5 ft left of B): $5 \times 5 = 25$ kip-ft clockwise - Moment due to 10 kip at E (28 - 5 = 23 ft right of B): $10 \times 23 = 230$ kip-ft counterclockwise - Moment due to distributed load (46 kips) acts at midpoint of 23 ft span from B, i.e., at 11.5 ft from B: $46 \times 11.5 = 529$ kip-ft counterclockwise - Moment applied at 13 ft from B: 25 kip-ft clockwise - Moment due to reaction at D (at 20 ft from B): $R_D \times 20$ kip-ft counterclockwise Sum moments about B: $$25 - 230 - 529 + 25 + R_D \times 20 = 0$$ Simplify: $$25 + 25 - 230 - 529 + 20 R_D = 0$$ $$50 - 759 + 20 R_D = 0$$ $$20 R_D = 709$$ $$R_D = \frac{709}{20} = 35.45 \text{ kips}$$ From $R_B + R_D = 46$: $$R_B = 46 - 35.45 = 10.55 \text{ kips}$$ 4. **Shear force diagram (V):** - Start at A (x=0): Shear = $-5$ kips (downward load) - Jump up by $R_B = 10.55$ kips at B (x=5 ft): Shear = $-5 + 10.55 = 5.55$ kips - From B to E (x=5 to 28 ft), shear decreases linearly due to UDL of 2 kips/ft: Shear at E just before load = $5.55 - 2 \times 23 = 5.55 - 46 = -40.45$ kips - Jump down by 10 kips at E (x=28 ft): Shear = $-40.45 - 10 = -50.45$ kips - Jump up by $R_D = 35.45$ kips at D (x=25 ft) (note order: D is before E, so at x=25 ft shear jumps up by 35.45 kips) 5. **Moment diagram (M):** - Moment at A = 0 - Moment increases/decreases by area under shear diagram - Moment jump of 25 kip-ft clockwise at x=13 ft Calculate moments at key points: - At B (x=5 ft): $$M_B = M_A + \int_0^5 V dx = 0 + (-5) \times 5 = -25 \text{ kip-ft}$$ - Between B and 13 ft (8 ft segment): shear decreases linearly from 5.55 kips to value at 13 ft: Shear at 13 ft: $$V_{13} = 5.55 - 2 \times (13 - 5) = 5.55 - 16 = -10.45 \text{ kips}$$ Moment at 13 ft: $$M_{13} = M_B + \text{area under shear from 5 to 13}$$ Area under shear from 5 to 13 is trapezoid: $$\frac{(5.55 + (-10.45))}{2} \times 8 = \frac{-4.9}{2} \times 8 = -19.6 \text{ kip-ft}$$ So: $$M_{13} = -25 - 19.6 = -44.6 \text{ kip-ft}$$ Add moment jump of 25 kip-ft clockwise (negative moment): $$M_{13^+} = -44.6 - 25 = -69.6 \text{ kip-ft}$$ - Between 13 ft and D (x=25 ft): shear continues decreasing linearly: Shear at 25 ft: $$V_{25} = V_{13} - 2 \times (25 - 13) = -10.45 - 24 = -34.45 \text{ kips}$$ Area under shear from 13 to 25: $$\frac{(-10.45 + (-34.45))}{2} \times 12 = \frac{-44.9}{2} \times 12 = -269.4 \text{ kip-ft}$$ Moment at D: $$M_D = M_{13^+} + (-269.4) = -69.6 - 269.4 = -339 \text{ kip-ft}$$ - Between D and E (3 ft): shear jumps up by $R_D = 35.45$ kips at D, so shear at D just right is: $$V_{25^+} = -34.45 + 35.45 = 1 \text{ kip}$$ Shear at E just before load is 1 kip - 2 kips/ft over 3 ft: $$V_E = 1 - 2 \times 3 = 1 - 6 = -5 \text{ kips}$$ Area under shear from 25 to 28: $$\frac{(1 + (-5))}{2} \times 3 = \frac{-4}{2} \times 3 = -6 \text{ kip-ft}$$ Moment at E: $$M_E = M_D + (-6) = -339 - 6 = -345 \text{ kip-ft}$$ 6. **Summary:** - Reactions: $R_B = 10.55$ kips upward, $R_D = 35.45$ kips upward - Shear diagram starts at -5 kips at A, jumps to 5.55 kips at B, decreases linearly to -40.45 kips at E, with jumps at D and E - Moment diagram starts at 0 at A, negative moments throughout, with a jump of -25 kip-ft at 13 ft This completes the shear and moment diagram analysis.