A light, horizontal beam, ABCDE, with A on the left-hand side is 26 m long. It is supported at two points, A and D. A point load of 20 kN is at B, 12 m from A. A point load of 30 is at C, 6 m from B. A uniform distributed load of 8 kN/m is between A and B. D is 4 m from C.

Question

Asked on June 28, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-06-28T20:25:26.006Z

Here are the solutions to your statics problem: 2.1 Make a neat, labelled diagram of the beam as described above. A light, horizontal beam ABCDE, 26 m long, with A on the left-hand side. Support at A (0 m) with reaction R_A. Uniformly Distributed Load (UDL) of 8 kN/m from A to B (0 m to 12 m). Point load of 20 kN downwards at B (12 m from A). Point load of 30 kN downwards at C (6 m from B, so 12+6=18 m from A). Support at D (4 m from C, so 18+4=22 m from A) with reaction R_D. End of beam at E (26 m from A). ` <------------------ 26 m ------------------> A UDL=8kN/m B C D E |-----------------------|-------|-------|-------| ^ | | ^ RA 20kN 30kN RD 0m 12m 18m 22m 26m ` 2.2 Calculate the reactions of the supports at points A and D and test your answers. Step 1: Convert the UDL to an equivalent point load. The UDL is 8 kN/m over 12 m (from A to B). Equivalent UDL force F_UDL = 8 kN/m × 12 m = 96 kN. This force acts at the midpoint of A-B, which is 12 m / 2 = 6 m from A. Step 2: Apply the equilibrium equation for moments about point A ( M_A = 0). Taking clockwise moments as positive: (96 kN × 6 m) + (20 kN × 12 m) + (30 kN × 18 m) - (R_D × 22 m) = 0 576 kN · m + 240 kN · m + 540 kN · m - 22 R_D = 0 1356 kN · m - 22 R_D = 0 22 R_D = 1356 kN · m R_D = (1356)/(22) ≈ 61.64 kN Step 3: Apply the equilibrium equation for vertical forces ( F_y = 0). Taking upward forces as positive: R_A + R_D - 96 kN - 20 kN - 30 kN = 0 R_A + 61.64 kN - 146 kN = 0 R_A - 84.36 kN = 0 R_A = 84.36 kN Step 4: Test the answers by taking moments about point D ( M_D = 0). (R_A × 22 m) - (96 kN × (22-6) m) - (20 kN × (22-12) m) - (30 kN × (22-18) m) = 0 (84.36 × 22) - (96 × 16) - (20 × 10) - (30 × 4) = 0 1856.00 - 1536 - 200 - 120 = 0 1856.00 - 1856 = 0 The reactions are correct. The reactions are: R_A = 84.36 kN R_D = 61.64 kN 2.3 Draw a shear-force diagram and show ALL the main values on the diagram. At A (x=0 m): Shear force V_A = R_A = 84.36 kN. From A to B (0 m to 12 m): Shear force decreases linearly due to UDL. At B (just left, x=12 m): V_B,left = R_A - (8 kN/m × 12 m) = 84.36 - 96 = -11.64 kN. At B (x=12 m): Shear force drops due to 20 kN point load. At B (just right): V_B,right = -11.64 - 20 = -31.64 kN. From B to C (12 m to 18 m): Shear force is constant. At C (just left, x=18 m): V_C,left = -31.64 kN. At C (x=18 m): Shear force drops due to 30 kN point load. At C (just right): V_C,right = -31.64 - 30 = -61.64 kN. From C to D (18 m to 22 m): Shear force is constant. At D (just left, x=22 m): V_D,left = -61.64 kN. At D (x=22 m): Shear force jumps up due to R_D. At D (just right): V_D,right = -61.64 + R_D = -61.64 + 61.64 = 0 kN. From D to E (22 m to 26 m): Shear force is constant. At E (x=26 m): V_E = 0 kN. The shear force diagram starts at 84.36 kN at A, decreases linearly to -11.64 kN at B, drops to -31.64 kN, remains constant to C, drops to -61.64 kN, remains constant to D, jumps to 0 kN, and remains 0 kN to E. The point of zero shear force occurs between A and B where R_A - 8x = 0 84.36 - 8x = 0 x = 10.55 m from A. 2.4 Calculate the bending moments at B and C. Bending Moment at A (M_A): M_A = 0 kN · m (support at end of beam). Bending Moment at B (M_B, at x=12 m): Consider forces to the left of B: R_A and the UDL. M_B = (R_A × 12 m) - (8 kN/m × 12 m × 12 m2) M_B = (84.36 kN × 12 m) - (96 kN × 6 m) M_B = 1012.32 kN · m - 576 kN · m M_B = 436.32 kN · m Bending Moment at C (M_C, at x=18 m): Consider forces to the left of C: R_A, UDL, and 20 kN point load at B. M_C = (R_A × 18 m) - (96 kN × (18-6) m) - (20 kN × (18-12) m) M_C = (84.36 kN × 18 m) - (96 kN × 12 m) - (20 kN × 6 m) M_C = 1518.48 kN · m - 1152 kN · m - 120 kN · m M_C = 246.48 kN · m 2.5 Draw a neat bending-moment diagram and show ALL the main values on the diagram. Bending Moment at A (M_A): 0 kN · m. Maximum Bending Moment (between A and B): Occurs at x = 10.55 m (where shear force is zero). M_max = (R_A × 10.55) - (8 × 10.55 × (10.55)/(2)) M_max = (84.36 × 10.55) - (4 × 10.55^2) M_max = 889.70 - 445.21 = 444.49 kN · m Bending Moment at B (M_B): 436.32 kN · m. Bending Moment at C (M_C): 246.48 kN · m. Bending Moment at D (M_D, at x=22 m): M_D = (R_A × 22) - (96 × (22-6)) - (20 × (22-12)) - (30 × (22-18)) M_D = (84.36 × 22) - (96 × 16) - (20 × 10) - (30 × 4) M_D = 1856.00 - 1536 - 200 - 120 = 0 kN · m Bending Moment at E (M_E): 0 kN · m (free end). The bending moment diagram starts at 0 kN · m at A, curves parabolically upwards to a maximum of 444.49 kN · m at x=10.55 m, then decreases to 436.32 kN · m at B. From B to C, it decreases linearly to 246.48 kN · m. From C to D, it decreases linearly to 0 kN · m. From D to E, it remains 0 kN · m. 3 done, 2 left today. You're making progress.

Accepted Answer · 2026-06-28T20:25:26.006Z

Here are the solutions to your statics problem: 2.1 Make a neat, labelled diagram of the beam as described above. A light, horizontal beam ABCDE, 26 m long, with A on the left-hand side. Support at A (0 m) with reaction R_A. Uniformly Distributed Load (UDL) of 8 kN/m from A to B (0 m to 12 m). Point load of 20 kN downwards at B (12 m from A). Point load of 30 kN downwards at C (6 m from B, so 12+6=18 m from A). Support at D (4 m from C, so 18+4=22 m from A) with reaction R_D. End of beam at E (26 m from A). ` <------------------ 26 m ------------------> A UDL=8kN/m B C D E |-----------------------|-------|-------|-------| ^ | | ^ RA 20kN 30kN RD 0m 12m 18m 22m 26m ` 2.2 Calculate the reactions of the supports at points A and D and test your answers. Step 1: Convert the UDL to an equivalent point load. The UDL is 8 kN/m over 12 m (from A to B). Equivalent UDL force F_UDL = 8 kN/m × 12 m = 96 kN. This force acts at the midpoint of A-B, which is 12 m / 2 = 6 m from A. Step 2: Apply the equilibrium equation for moments about point A ( M_A = 0). Taking clockwise moments as positive: (96 kN × 6 m) + (20 kN × 12 m) + (30 kN × 18 m) - (R_D × 22 m) = 0 576 kN · m + 240 kN · m + 540 kN · m - 22 R_D = 0 1356 kN · m - 22 R_D = 0 22 R_D = 1356 kN · m R_D = (1356)/(22) ≈ 61.64 kN Step 3: Apply the equilibrium equation for vertical forces ( F_y = 0). Taking upward forces as positive: R_A + R_D - 96 kN - 20 kN - 30 kN = 0 R_A + 61.64 kN - 146 kN = 0 R_A - 84.36 kN = 0 R_A = 84.36 kN Step 4: Test the answers by taking moments about point D ( M_D = 0). (R_A × 22 m) - (96 kN × (22-6) m) - (20 kN × (22-12) m) - (30 kN × (22-18) m) = 0 (84.36 × 22) - (96 × 16) - (20 × 10) - (30 × 4) = 0 1856.00 - 1536 - 200 - 120 = 0 1856.00 - 1856 = 0 The reactions are correct. The reactions are: R_A = 84.36 kN R_D = 61.64 kN 2.3 Draw a shear-force diagram and show ALL the main values on the diagram. At A (x=0 m): Shear force V_A = R_A = 84.36 kN. From A to B (0 m to 12 m): Shear force decreases linearly due to UDL. At B (just left, x=12 m): V_B,left = R_A - (8 kN/m × 12 m) = 84.36 - 96 = -11.64 kN. At B (x=12 m): Shear force drops due to 20 kN point load. At B (just right): V_B,right = -11.64 - 20 = -31.64 kN. From B to C (12 m to 18 m): Shear force is constant. At C (just left, x=18 m): V_C,left = -31.64 kN. At C (x=18 m): Shear force drops due to 30 kN point load. At C (just right): V_C,right = -31.64 - 30 = -61.64 kN. From C to D (18 m to 22 m): Shear force is constant. At D (just left, x=22 m): V_D,left = -61.64 kN. At D (x=22 m): Shear force jumps up due to R_D. At D (just right): V_D,right = -61.64 + R_D = -61.64 + 61.64 = 0 kN. From D to E (22 m to 26 m): Shear force is constant. At E (x=26 m): V_E = 0 kN. The shear force diagram starts at 84.36 kN at A, decreases linearly to -11.64 kN at B, drops to -31.64 kN, remains constant to C, drops to -61.64 kN, remains constant to D, jumps to 0 kN, and remains 0 kN to E. The point of zero shear force occurs between A and B where R_A - 8x = 0 84.36 - 8x = 0 x = 10.55 m from A. 2.4 Calculate the bending moments at B and C. Bending Moment at A (M_A): M_A = 0 kN · m (support at end of beam). Bending Moment at B (M_B, at x=12 m): Consider forces to the left of B: R_A and the UDL. M_B = (R_A × 12 m) - (8 kN/m × 12 m × 12 m2) M_B = (84.36 kN × 12 m) - (96 kN × 6 m) M_B = 1012.32 kN · m - 576 kN · m M_B = 436.32 kN · m Bending Moment at C (M_C, at x=18 m): Consider forces to the left of C: R_A, UDL, and 20 kN point load at B. M_C = (R_A × 18 m) - (96 kN × (18-6) m) - (20 kN × (18-12) m) M_C = (84.36 kN × 18 m) - (96 kN × 12 m) - (20 kN × 6 m) M_C = 1518.48 kN · m - 1152 kN · m - 120 kN · m M_C = 246.48 kN · m 2.5 Draw a neat bending-moment diagram and show ALL the main values on the diagram. Bending Moment at A (M_A): 0 kN · m. Maximum Bending Moment (between A and B): Occurs at x = 10.55 m (where shear force is zero). M_max = (R_A × 10.55) - (8 × 10.55 × (10.55)/(2)) M_max = (84.36 × 10.55) - (4 × 10.55^2) M_max = 889.70 - 445.21 = 444.49 kN · m Bending Moment at B (M_B): 436.32 kN · m. Bending Moment at C (M_C): 246.48 kN · m. Bending Moment at D (M_D, at x=22 m): M_D = (R_A × 22) - (96 × (22-6)) - (20 × (22-12)) - (30 × (22-18)) M_D = (84.36 × 22) - (96 × 16) - (20 × 10) - (30 × 4) M_D = 1856.00 - 1536 - 200 - 120 = 0 kN · m Bending Moment at E (M_E): 0 kN · m (free end). The bending moment diagram starts at 0 kN · m at A, curves parabolically upwards to a maximum of 444.49 kN · m at x=10.55 m, then decreases to 436.32 kN · m at B. From B to C, it decreases linearly to 246.48 kN · m. From C to D, it decreases linearly to 0 kN · m. From D to E, it remains 0 kN · m. 3 done, 2 left today. You're making progress.

A light, horizontal beam, ABCDE, with A on the left-hand side is 26 m long. It is supported at two points, A and D. A point load of 20 kN is at B, 12 m from A. A point load of 30 is at C, 6 m from B. A uniform distributed load of 8 kN/m is between A and B. D is 4 m from C.

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A light, horizontal beam, ABCDE, with A on the left-hand side is 26 m long. It is supported at two points, A and D. A point load of 20 kN is at B, 12 m from A. A point load of 30 is at C, 6 m from B. A uniform distributed load of 8 kN/m is between A and B. D is 4 m from C.

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