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.
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QUESTION 5: STATICS 5.1.1 Calculate the reactions at the supports. Step 1: Define the forces and distances. The beam has supports at A and E. Let R_A and R_E be the upward reactions at A and E, respectively. Loads: Point load at A: 25 kN (downwards) Uniformly Distributed Load (UDL) from A to C: 10 kN/m over 5 m (total 10 × 5 = 50 kN acting at 2.5 m from A) Point load at C: 40 kN (downwards, at 5 m from A) Point load at E: 40 kN (downwards, at 7 m from A) Step 2: Apply the equilibrium equation for vertical forces ( F_y = 0). Assume upward forces are positive and downward forces are negative. R_A + R_E - 25 kN - (10 kN/m × 5 m) - 40 kN - 40 kN = 0 R_A + R_E - 25 - 50 - 40 - 40 = 0 R_A + R_E = 155 kN (1) Step 3: Apply the equilibrium equation for moments about point A ( M_A = 0). Assume clockwise moments are positive. (10 kN/m × 5 m) × (2.5 m) + (40 kN × 5 m) + (40 kN × 7 m) - (R_E × 7 m) = 0 (50 kN × 2.5 m) + (200 kNm) + (280 kNm) - 7 R_E = 0 125 kNm + 200 kNm + 280 kNm - 7 R_E = 0 605 kNm - 7 R_E = 0 7 R_E = 605 R_E = (605)/(7) kN ≈ 86.42857 kN Step 4: Substitute R_E into equation (1) to find R_A. R_A + 86.42857 kN = 155 kN R_A = 155 - 86.42857 kN R_A ≈ 68.57143 kN Rounding to three significant figures: R_A = 68.6 kN R_E = 86.4 kN 5.1.2 Calculate the bending moments at B, C and D. Step 1: Calculate the bending moment at B (at x = 1 m from A). Consider forces to the left of B. M_B = (R_A × 1 m) - (25 kN × 1 m) - (10 kN/m × 1 m × 0.5 m) M_B = (68.57143 × 1) - (25 × 1) - (10 × 0.5) M_B = 68.57143 - 25 - 5 = 38.57143 kNm Rounding to three significant figures: M_B = 38.6 kNm Step 2: Calculate the bending moment at C (at x = 5 m from A). Consider forces to the left of C. M_C = (R_A × 5 m) - (25 kN × 5 m) - (10 kN/m × 5 m × (5/2) m) M_C = (68.57143 × 5) - (25 × 5) - (50 × 2.5) M_C = 342.85715 - 125 - 125 = 92.85715 kNm Rounding to three significant figures: M_C = 92.9 kNm Step 3: Calculate the bending moment at D (at x = 6 m from A). Consider forces to the right of D for simplicity. M_D = (R_E × 1 m) - (40 kN × 1 m) M_D = (86.42857 × 1) - (40 × 1) M_D = 86.42857 - 40 = 46.42857 kNm Rounding to three significant figures: M_D = 46.4 kNm 5.1.3 Draw the SF and BM diagrams. Also, determine the magnitude of the maximum bending moment and its position. Shear Force (SF) Diagram: At A (just right): V_A = R_A - 25 kN = 68.57 - 25 = 43.57 kN. From A to C (UDL): Shear force decreases linearly. At B (just left/right): V_B = 43.57 - (10 × 1) = 33.57 kN. At C (just left): V_C,L = 43.57 - (10 × 5) = -6.43 kN. At C (just right): V_C,R = V_C,L - 40 kN = -6.43 - 40 = -46.43 kN. From C to E (no UDL): Shear force is constant. At D (just left/right): V_D = -46.43 kN. At E (just left): V_E,L = -46.43 kN. At E (just right): V_E,R = V_E,L + R_E - 40 kN = -46.43 + 86.43 - 40 = 0 kN. Bending Moment (BM) Diagram: At A: M_A = 0 kNm. From A to C (UDL): Bending moment is a parabolic curve. At B: M_B = 38.6 kNm. At C: M_C = 92.9 kNm. From C to E (no UDL): Bending moment is a linear curve. At D: M_D = 46.4 kNm. At E: M_E = 0 kNm. Maximum Bending Moment and its Position: The maximum bending moment occurs where the shear force is zero. The shear force changes sign between B and C. The shear force equation for 1 m < x < 5 m (from A) is V(x) = R_A - 25 - 10x = 68.57143 - 25 - 10x = 43.57143 - 10x. Set V(x) = 0: 43.57143 - 10x = 0 x = (43.57143)/(10) = 4.357143 m from A Now, calculate the bending moment at this position: M_max = (R_A × x) - (25 × x) - (10 × x × x/2) M_max = (68.57143 × 4.357143) - (25 × 4.357143) - (5 × (4.357143)^2) M_max = 298.775 - 108.928575 - 94.92495 ≈ 94.921 kNm Rounding to three significant figures: Magnitude of maximum bending moment: 94.9 kNm Position of maximum bending moment: 4.36 m from A 5.2 Calculate the centre of gravity of the body from the table on which it is placed. The body consists of a cylinder and a hemisphere (semi-ball) on top. Let the origin be at the base of the cylinder (on the table). Part 1: Cylinder Diameter = 50 mm, so radius r = 25 mm. Height h_c = 250 mm. Volume V_c = r^2 h_c = (25 mm)^2 (250 mm) = 156250 mm^3. Centroid from the base y_c = h_c / 2 = 250 mm / 2 = 125 mm. Part 2: Hemisphere Radius r = 25 mm. Volume V_h = (2)/(3) r^3 = (2)/(3) (25 mm)^3 = (2)/(3) (15625) = (31250)/(3) mm^3. Centroid from its flat base y_h' = (3)/(8)r = (3)/(8)(25 mm) = 9.375 mm. The flat base of the hemisphere is at the top of the cylinder, which is 250 mm from the table. Centroid from the table y_h = h_c + y_h' = 250 mm + 9.375 mm = 259.375 mm. Composite Body Centre of Gravity (Y): Y = (V_c y_c + V_h y_h)/(V_c + V_h) Y = ((156250 × 125) + (31250)/(3) × 259.375)156250 + (31250)/(3) Cancel from all terms: Y = ((156250 × 125) + (31250)/(3) × 259.375)156250 + (31250)/(3) Multiply numerator and denominator by 3 to simplify: Y = ((3 × 156250 × 125) + (31250 × 259.375))/((3 × 156250) + 31250) Y = (58593750 + 8105468.75)/(468750 + 31250) Y = (66699218.75)/(500000) Y ≈ 133.398 mm Rounding to three significant figures: 133 mm What's next?