A simply supported beam is loaded as shown in FIGURE 1. By referring to FIGURE 1 determine the following: The…

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A simply supported beam is loaded as shown in FIGURE 1. By referring to FIGURE 1 determine the following: The reaction forces at supports B and D. Draw the shear force diagram for the beam. Calculate the coordinates of the centroid of the lamina shown in FIGURE 2.

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Here are the solutions to Question 5: Question 5.1: Simply Supported Beam Step 1: Identify all forces and their positions. Let the origin be at point A. Point load at A: F_A = 18 kN (downwards) at x = 0 m. Uniformly Distributed Load (UDL): w = 6 kN/m from x = 0 m to x = 5 m (total length 5 m). The equivalent point load for the UDL is FUDL = 6 kN/m × 5 m = 30 kN, acting at the centroid of the UDL, which is x = 5/2 = 2.5 m. Support at B: Reaction R_B (upwards) at x = 1.5 m. Point load: F₄₀ = 40 kN (downwards) at x = 1.5 m + 3.5 m = 5 m. Support at D: Reaction R_D (upwards) at x = 5 m + 1.5 m = 6.5 m. Point load at E: F_E = 22 kN (downwards) at x = 6.5 m + 2.5 m = 9 m. 5.1.1 Determine the reaction forces at supports B and D. Step 2: Apply equilibrium equations. Sum of moments about B (∑ M_B = 0). Assume clockwise moments are positive. The distances are measured from B (x=1.5 m). F_A: 18 kN at x=0. Moment arm = 1.5 m. Moment = 18 × 1.5 (clockwise). FUDL: 30 kN at x=2.5 m. Moment arm = 2.5 − 1.5 = 1 m. Moment = 30 × 1 (clockwise). F₄₀: 40 kN at x=5 m. Moment arm = 5 − 1.5 = 3.5 m. Moment = 40 × 3.5 (clockwise). R_D: R_D at x=6.5 m. Moment arm = 6.5 − 1.5 = 5 m. Moment = R_D × 5 (counter-clockwise). F_E: 22 kN at x=9 m. Moment arm = 9 − 1.5 = 7.5 m. Moment = 22 × 7.5 (clockwise). ∑ M_B = (18 × 1.5) + (30 × 1) + (40 × 3.5) − (R_D × 5) + (22 × 7.5) = 0; 27 + 30 + 140 − 5R_D + 165 = 0; 362 − 5R_D = 0; 5R_D = 362; R_D = 362/5 = 72.4 kN; Sum of vertical forces (∑ F_y = 0). Assume upwards forces are positive. R_B + R_D − F_A − FUDL − F₄₀ − F_E = 0; R_B + 72.4 kN − 18 kN − 30 kN − 40 kN − 22 kN = 0; R_B + 72.4 kN − 110 kN = 0; R_B − 37.6 kN = 0; R_B = 37.6 kN The reaction forces are: R_B = 37.6 kN; R_D = 72.4 kN 5.1.2 Draw the shear force diagram for the beam. Step 3: Calculate shear force values at key points. At A (x=0): V_A = −18 kN (due to downward point load); Just before B (x=1.5^−): VB^− = −18 kN − (6 kN/m × 1.5 m) = −18 − 9 = −27 kN; Just after B (x=1.5^+): VB^+ = VB^− + R_B = −27 kN + 37.6 kN = 10.6 kN; Just before 40 kN load (x=5^−): The UDL continues from B to x=5 m. Length from B to x=5 m is 5 − 1.5 = 3.5 m. V40^− = VB^+ − (6 kN/m × 3.5 m) = 10.6 − 21 = −10.4 kN; Just after 40 kN load (x=5^+): V40^+ = V40^− − 40 kN = −10.4 − 40 = −50.4 kN; Just before D (x=6.5^−): No UDL between x=5 m and x=6.5 m. VD^− = −50.4 kN; Just after D (x=6.5^+): VD^+ = VD^− + R_D = −50.4 kN + 72.4 kN = 22 kN; Just before E (x=9^−): No loads between D and E. VE^− = 22 kN; At E (x=9): V_E = VE^− − 22 kN = 22 − 22 = 0 kN Step 4: Sketch the shear force diagram. The shear force diagram will show: A linear decrease from −18 kN at A to −27 kN at B. A sudden jump up by 37.6 kN at B, from −27 kN to 10.6 kN. A linear decrease from 10.6 kN at B to −10.4 kN at x=5 m (where the 40 kN load is applied). A sudden jump down by 40 kN at x=5 m, from −10.4 kN to −50.4 kN. A constant value of −50.4 kN from x=5 m to D. A sudden jump up by 72.4 kN at D, from −50.4 kN to 22 kN. A constant value of 22 kN from D to E. A sudden jump down by 22 kN at E, from 22 kN to 0 kN. ` Shear Force Diagram (SFD) (Values in kN) | 10.6|

A simply supported beam is loaded as shown in FIGURE 1. By referring to FIGURE 1 determine the following: The reaction forces at supports B and D. Draw the shear force diagram for the beam. Calculate the coordinates of the centroid of the lamina shown in FIGURE 2.

Question 5.1: Simply Supported Beam

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