Answer according to n4 level

Here are the calculations for the beam problem. 7.1.1. Calculate the reactions at the supports. Step 1: Define the beam and loads. The beam is 8 m long. Support R_A is at 1 m from the left end (x=1 m). Support R_B is at 2 m from the right end, which is 8-2 = 6 m from the left end (x=6 m). Loads (all acting downwards): Point load P_1 = 450 kN at x=1 m. Concentrated load P_2 = 700 kN at x=6 m. Uniformly Distributed Load (UDL) w = 12 kN/m over the first 5 m from the right end. This means from x=8-5=3 m to x=8 m. The total force from the UDL is F_UDL = 12 kN/m × 5 m = 60 kN. This equivalent force acts at the centroid of the UDL, which is at x = 3 + (5)/(2) = 5.5 m from the left end. Concentrated load P_3 = 40 kN at x=8 m. Step 2: Apply the equilibrium equations. Sum of vertical forces F_y = 0 (taking upward forces as positive): R_A + R_B - P_1 - P_2 - F_UDL - P_3 = 0 R_A + R_B - 450 kN - 700 kN - 60 kN - 40 kN = 0 R_A + R_B = 1250 kN (Equation 1) Sum of moments about R_A (at x=1 m) M_A = 0 (taking clockwise moments as positive): P_1 is at R_A, so its moment arm is 0. R_B creates a counter-clockwise moment: -R_B × (6-1) = -5 R_B. P_2 creates a clockwise moment: +P_2 × (6-1) = +700 × 5 = 3500 kNm. F_UDL creates a clockwise moment: +F_UDL × (5.5-1) = +60 × 4.5 = 270 kNm. P_3 creates a clockwise moment: +P_3 × (8-1) = +40 × 7 = 280 kNm. -5 R_B + 3500 + 270 + 280 = 0 -5 R_B + 4050 = 0 5 R_B = 4050 R_B = (4050)/(5) = 810 kN Step 3: Calculate R_A using Equation 1. R_A + 810 kN = 1250 kN R_A = 1250 kN - 810 kN = 440 kN The reactions at the supports are: R_A = 440 kN R_B = 810 kN 7.1.2. Draw a fully labelled shear force diagram. We will calculate the shear force (V) at key points along the beam, starting from the left end (x=0). Upward forces are positive. At x=0 m: V_0 = 0 kN Just before x=1 m: V_1^- = 0 kN At x=1 m (after P_1 and R_A): V_1, after P_1 = V_1^- - P_1 = 0 - 450 = -450 kN V_1, after R_A = V_1, after P_1 + R_A = -450 + 440 = -10 kN From x=1 m to x=3 m: (No loads, shear is constant) V_x = -10 kN V_3^- = -10 kN From x=3 m to x=6 m: (UDL of 12 kN/m starts at x=3 m) V_x = V_3^- - w(x-3) = -10 - 12(x-3) V_3^+ = -10 kN V_6^- = -10 - 12(6-3) = -10 - 36 = -46 kN At x=6 m (after P_2 and R_B): V_6, after P_2 = V_6^- - P_2 = -46 - 700 = -746 kN V_6, after R_B = V_6, after P_2 + R_B = -746 + 810 = 64 kN From x=6 m to x=8 m: (UDL continues) V_x = V_6, after R_B - w(x-6) = 64 - 12(x-6) V_8^- = 64 - 12(8-6) = 64 - 24 = 40 kN At x=8 m (after P_3): V_8, after P_3 = V_8^- - P_3 = 40 - 40 = 0 kN Shear Force Diagram (SFD) values: V_0 = 0 kN V_1^- = 0 kN V_1, after P_1 = -450 kN V_1, after R_A = -10 kN V_3 = -10 kN V_6^- = -46 kN V_6, after P_2 = -746 kN V_6, after R_B = 64 kN V_8^- = 40 kN V_8, after P_3 = 0 kN ` Shear Force Diagram (SFD) 0 kN | | | +-------------------------------------------------------------------- (x=0) | | (x=1, before P1) | | -450 kN (at x=1, after P1) |-------------------------------------------------------------------- | | -10 kN (at x=1, after RA) +-------------------------------------------------------------------- (x=3) | | (x=6, before P2) | -46 kN |-------------------------------------------------------------------- | | -746 kN (at x=6, after P2) |-------------------------------------------------------------------- | | 64 kN (at x=6, after RB) +-------------------------------------------------------------------- (x=8, before P3) | | 40 kN |-------------------------------------------------------------------- | | 0 kN (at x=8, after P3) +-------------------------------------------------------------------- ` (Note: A graphical representation is required for a "fully labelled shear force diagram". Due to text-based limitations, the key values and their positions are listed above. The diagram would show a horizontal line from 0 to 1m, a drop to -450kN, a jump to -10kN, a horizontal line to 3m, a linear decrease to -46kN at 6m, a drop to -746kN, a jump to 64kN, a linear decrease to 40kN at 8m, and a final drop to 0kN.) 7.1.3. Draw a fully labelled bending moment diagram. We will calculate the bending moment (M) at key points along the beam. The change in bending moment between two points is the area under the shear force diagram between those points. Sagging moments are positive. At x=0 m (Left end): M_0 = 0 kNm At x=1 m (at P_1 and R_A): M_1 = M_0 + Area of SFD from 0 to 1 = 0 + (0 × 1) = 0 kNm At x=3 m (Start of UDL): M_3 = M_1 + Area of SFD from 1 to 3 = 0 + (-10 kN × (3-1) m) M_3 = 0 - 20 = -20 kNm At x=6 m (at P_2 and R_B): M_6 = M_3 + Area of SFD from 3 to 6 The SFD from 3 to 6 is a trapezoid with values -10 kN and -46 kN over a length of 3 m. Area = (-10 + (-46))/(2) × 3 = (-56)/(2) × 3 = -28 × 3 = -84 kNm M_6 = -20 - 84 = -104 kNm At x=8 m (Right end, at P_3): M_8 = M_6 + Area of SFD from 6 to 8 The SFD from 6 to 8 is a trapezoid with values 64 kN and 40 kN over a length of 2 m. Area = (64 + 40)/(2) × 2 = (104)/(2) × 2 = 52 × 2 = 104 kNm M_8 = -104 + 104 = 0 kNm Bending Moment Diagram (BMD) values: M_0 = 0 kNm M_1 = 0 kNm M_3 = -20 kNm M_6 = -104 kNm M_8 = 0 kNm ` Bending Moment Diagram (BMD) 0 kNm (at x=0) +-------------------------------------------------------------------- | | 0 kNm (at x=1) +-------------------------------------------------------------------- | | -20 kNm (at x=3) |-------------------------------------------------------------------- | | -104 kNm (at x=6) |-------------------------------------------------------------------- | | 0 kNm (at x=8) +-------------------------------------------------------------------- ` (Note: A graphical representation is required for a "fully labelled bending moment diagram". Due to text-based limitations, the key values and their positions are listed above. The diagram would show a horizontal line from 0 to 1m, then a linear decrease to -20kNm at 3m, then a parabolic curve decreasing to -104kNm at 6m, and finally a parabolic curve increasing to 0kNm at 8m.) Drop the next question.