Consider the loaded beam in FIGURE 1 below and answer the following questions. Calculate the reactions at the supports. Calculate bending moments at B, C and D. Draw the SF and the BM diagrams. Also, determine the magnitude of the maximum bending moment and its position.

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: $25 \text{ kN}$ (downwards)
• Uniformly Distributed Load (UDL) from A to C: $10 \text{ kN/m}$ over $5 \text{ m}$ (total $10 \times 5 = 50 \text{ kN}$ acting at $2.5 \text{ m}$ from A)
• Point load at C: $40 \text{ kN}$ (downwards, at $5 \text{ m}$ from A)
• Point load at E: $40 \text{ kN}$ (downwards, at $7 \text{ m}$ from A)

Assume upward forces are positive and downward forces are negative.

$$R_A + R_E - 25 \text{ kN} - (10 \text{ kN/m} \times 5 \text{ m}) - 40 \text{ kN} - 40 \text{ kN} = 0$$ $$R_A + R_E - 25 - 50 - 40 - 40 = 0$$ $$R_A + R_E = 155 \text{ kN} \quad (1)$$

Assume clockwise moments are positive.

$$(10 \text{ kN/m} \times 5 \text{ m}) \times (2.5 \text{ m}) + (40 \text{ kN} \times 5 \text{ m}) + (40 \text{ kN} \times 7 \text{ m}) - (R_E \times 7 \text{ m}) = 0$$ $$(50 \text{ kN} \times 2.5 \text{ m}) + (200 \text{ kNm}) + (280 \text{ kNm}) - 7 R_E = 0$$ $$125 \text{ kNm} + 200 \text{ kNm} + 280 \text{ kNm} - 7 R_E = 0$$ $$605 \text{ kNm} - 7 R_E = 0$$ $$7 R_E = 605$$ $$R_E = \frac{605}{7} \text{ kN} \approx 86.42857 \text{ kN}$$

$$R_A + 86.42857 \text{ kN} = 155 \text{ kN}$$ $$R_A = 155 - 86.42857 \text{ kN}$$ $$R_A \approx 68.57143 \text{ kN}$$

Rounding to three significant figures:

$$R_A = \boxed{\text{68.6 kN}}$$ $$R_E = \boxed{\text{86.4 kN}}$$

5.1.2 Calculate the bending moments at B, C and D.

Consider forces to the left of B.

$$M_B = (R_A \times 1 \text{ m}) - (25 \text{ kN} \times 1 \text{ m}) - (10 \text{ kN/m} \times 1 \text{ m} \times 0.5 \text{ m})$$ $$M_B = (68.57143 \times 1) - (25 \times 1) - (10 \times 0.5)$$ $$M_B = 68.57143 - 25 - 5 = 38.57143 \text{ kNm}$$

Rounding to three significant figures:

$$M_B = \boxed{\text{38.6 kNm}}$$

Consider forces to the left of C.

$$M_C = (R_A \times 5 \text{ m}) - (25 \text{ kN} \times 5 \text{ m}) - (10 \text{ kN/m} \times 5 \text{ m} \times (5/2) \text{ m})$$ $$M_C = (68.57143 \times 5) - (25 \times 5) - (50 \times 2.5)$$ $$M_C = 342.85715 - 125 - 125 = 92.85715 \text{ kNm}$$

Rounding to three significant figures:

$$M_C = \boxed{\text{92.9 kNm}}$$

Consider forces to the right of D for simplicity.

$$M_D = (R_E \times 1 \text{ m}) - (40 \text{ kN} \times 1 \text{ m})$$ $$M_D = (86.42857 \times 1) - (40 \times 1)$$ $$M_D = 86.42857 - 40 = 46.42857 \text{ kNm}$$

Rounding to three significant figures:

$$M_D = \boxed{\text{46.4 kNm}}$$

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): $V_A = R_A - 25 \text{ kN} = 68.57 - 25 = 43.57 \text{ kN}$.
• From A to C (UDL): Shear force decreases linearly.
  • At B (just left/right): $V_B = 43.57 - (10 \times 1) = 33.57 \text{ kN}$.
  • At C (just left): $V_{C,L} = 43.57 - (10 \times 5) = -6.43 \text{ kN}$.
• At C (just right): $V_{C,R} = V_{C,L} - 40 \text{ kN} = -6.43 - 40 = -46.43 \text{ kN}$.
• From C to E (no UDL): Shear force is constant.
  • At D (just left/right): $V_D = -46.43 \text{ kN}$.
  • At E (just left): $V_{E,L} = -46.43 \text{ kN}$.
• At E (just right): $V_{E,R} = V_{E,L} + R_E - 40 \text{ kN} = -46.43 + 86.43 - 40 = 0 \text{ kN}$.

Bending Moment (BM) Diagram:

• At A: $M_A = 0 \text{ kNm}$.
• From A to C (UDL): Bending moment is a parabolic curve.
  • At B: $M_B = 38.6 \text{ kNm}$.
  • At C: $M_C = 92.9 \text{ kNm}$.
• From C to E (no UDL): Bending moment is a linear curve.
  • At D: $M_D = 46.4 \text{ kNm}$.
• At E: $M_E = 0 \text{ kNm}$.

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 $1 \text{ m} < x < 5 \text{ m}$ (from A) is $V(x) = R_A - 25 - 10x = 68.57143 - 25 - 10x = 43.57143 - 10x$.
Set $V(x) = 0$:

$$43.57143 - 10x = 0$$ $$x = \frac{43.57143}{10} = 4.357143 \text{ m} \text{ from A}$$

Now, calculate the bending moment at this position:

$$M_{max} = (R_A \times x) - (25 \times x) - (10 \times x \times x/2)$$ $$M_{max} = (68.57143 \times 4.357143) - (25 \times 4.357143) - (5 \times (4.357143)^2)$$ $$M_{max} = 298.775 - 108.928575 - 94.92495 \approx 94.921 \text{ kNm}$$

Rounding to three significant figures:
Magnitude of maximum bending moment: $\boxed{\text{94.9 kNm}}$
Position of maximum bending moment: $\boxed{\text{4.36 m from A}}$

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 = $50 \text{ mm}$, so radius $r = 25 \text{ mm}$.
• Height $h_c = 250 \text{ mm}$.
• Volume $V_c = \pi r^2 h_c = \pi (25 \text{ mm})^2 (250 \text{ mm}) = 156250 \pi \text{ mm}^3$.
• Centroid from the base $y_c = h_c / 2 = 250 \text{ mm} / 2 = 125 \text{ mm}$.

Part 2: Hemisphere

• Radius $r = 25 \text{ mm}$.
• Volume $V_h = \frac{2}{3}\pi r^3 = \frac{2}{3}\pi (25 \text{ mm})^3 = \frac{2}{3}\pi (15625) = \frac{31250}{3}\pi \text{ mm}^3$.
• Centroid from its flat base $y_h' = \frac{3}{8}r = \frac{3}{8}(25 \text{ mm}) = 9.375 \text{ mm}$.
• The flat base of the hemisphere is at the top of the cylinder, which is $250 \text{ mm}$ from the table.
• Centroid from the table $y_h = h_c + y_h' = 250 \text{ mm} + 9.375 \text{ mm} = 259.375 \text{ mm}$.

Composite Body Centre of Gravity ($\bar{Y}$):

$$\bar{Y} = \frac{V_c y_c + V_h y_h}{V_c + V_h}$$ $$\bar{Y} = \frac{(156250 \pi \times 125) + (\frac{31250}{3}\pi \times 259.375)}{156250 \pi + \frac{31250}{3}\pi}$$

Cancel $\pi$ from all terms:

$$\bar{Y} = \frac{(156250 \times 125) + (\frac{31250}{3} \times 259.375)}{156250 + \frac{31250}{3}}$$

Multiply numerator and denominator by 3 to simplify:

$$\bar{Y} = \frac{(3 \times 156250 \times 125) + (31250 \times 259.375)}{(3 \times 156250) + 31250}$$ $$\bar{Y} = \frac{58593750 + 8105468.75}{468750 + 31250}$$ $$\bar{Y} = \frac{66699218.75}{500000}$$ $$\bar{Y} \approx 133.398 \text{ mm}$$

Rounding to three significant figures:

$$\boxed{\text{133 mm}}$$