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.

The distributed load is $6 \text{ kN/m}$ over $3.5 \text{ m}$.
Equivalent point load $F_w = 6 \text{ kN/m} \times 3.5 \text{ m} = 21 \text{ kN}$.
This load acts at the center of the distributed load, which is $1.5 \text{ m} + \frac{3.5 \text{ m}}{2} = 1.5 \text{ m} + 1.75 \text{ m} = 3.25 \text{ m}$ from point A.

Let $R_B$ and $R_D$ be the upward reaction forces at supports B and D, respectively. Upward forces are positive, downward forces are negative.

Taking moments about B (clockwise moments are positive):
The distance from A to B is $1.5 \text{ m}$.
The distance from the equivalent point load $F_w$ to B is $3.25 \text{ m} - 1.5 \text{ m} = 1.75 \text{ m}$.
The distance from the $40 \text{ kN}$ load to B is $5 \text{ m} - 1.5 \text{ m} = 3.5 \text{ m}$.
The distance from D to B is $6.5 \text{ m} - 1.5 \text{ m} = 5 \text{ m}$.
The distance from the $22 \text{ kN}$ load to B is $9 \text{ m} - 1.5 \text{ m} = 7.5 \text{ m}$.

The reaction forces are:
$R_B = \boxed{\text{38.05 kN}}$
$R_D = \boxed{\text{62.95 kN}}$

5.1.2 Draw the shear force diagram for the beam.

To draw the shear force diagram, we calculate the shear force at key points along the beam.

• At A (x = 0 m):
  • Just right of 18 kN load: $V_A = -18 \text{ kN}$
• From A to B (0 m < x < 1.5 m):
  • Shear force is constant: $V = -18 \text{ kN}$
• At B (x = 1.5 m):
  • Just left of $R_B$: $V = -18 \text{ kN}$
  • Just right of $R_B$: $V = -18 \text{ kN} + 38.05 \text{ kN} = 20.05 \text{ kN}$
• From B to 5 m (1.5 m < x < 5 m):
  • Distributed load of $6 \text{ kN/m}$ acts downwards. Shear force decreases linearly.
  • Shear force at 5 m (just left of 40 kN load): $V_{5m, L} = 20.05 \text{ kN} - (6 \text{ kN/m} \times 3.5 \text{ m}) = 20.05 \text{ kN} - 21 \text{ kN} = -0.95 \text{ kN}$
  • The shear force crosses zero at $x = 1.5 \text{ m} + \frac{20.05 \text{ kN}}{6 \text{ kN/m}} \approx 1.5 \text{ m} + 3.34 \text{ m} = 4.84 \text{ m}$ from A.
• At 5 m (x = 5 m):
  • Just left of 40 kN load: $V = -0.95 \text{ kN}$
  • Just right of 40 kN load: $V = -0.95 \text{ kN} - 40 \text{ kN} = -40.95 \text{ kN}$
• From 5 m to D (5 m < x < 6.5 m):
  • Shear force is constant: $V = -40.95 \text{ kN}$
• At D (x = 6.5 m):
  • Just left of $R_D$: $V = -40.95 \text{ kN}$
  • Just right of $R_D$: $V = -40.95 \text{ kN} + 62.95 \text{ kN} = 22 \text{ kN}$
• From D to E (6.5 m < x < 9 m):
  • Shear force is constant: $V = 22 \text{ kN}$
• At E (x = 9 m):
  • Just left of 22 kN load: $V = 22 \text{ kN}$
  • Just right of 22 kN load: $V = 22 \text{ kN} - 22 \text{ kN} = 0 \text{ kN}$

Shear Force Diagram Description:

• Start at $-18 \text{ kN}$ at point A.
• Draw a horizontal line from A to B (at $x=1.5 \text{ m}$) at $-18 \text{ kN}$.
• At B, draw a vertical line upwards to $20.05 \text{ kN}$.
• From B to $x=5 \text{ m}$, draw a linearly decreasing line from $20.05 \text{ kN}$ to $-0.95 \text{ kN}$. This line crosses the zero axis at approximately $x=4.84 \text{ m}$.
• At $x=5 \text{ m}$, draw a vertical line downwards to $-40.95 \text{ kN}$.
• Draw a horizontal line from $x=5 \text{ m}$ to D (at $x=6.5 \text{ m}$) at $-40.95 \text{ kN}$.
• At D, draw a vertical line upwards to $22 \text{ kN}$.
• Draw a horizontal line from D to E (at $x=9 \text{ m}$) at $22 \text{ kN}$.
• At E, draw a vertical line downwards to $0 \text{ kN}$.

5.2

Calculate the coordinates of the centroid of the lamina shown in FIGURE 2.

We can consider the lamina as two rectangles (Rectangle 1 and Rectangle 2) and subtract a rectangular hole (Hole 3).

• Rectangle 1 (Vertical part):
  • Dimensions: $40 \text{ mm} \times 80 \text{ mm}$
  • Area $A_1 = 40 \times 80 = 3200 \text{ mm}^2$
  • Centroid $x_1 = \frac{40}{2} = 20 \text{ mm}$
  • Centroid $y_1 = \frac{80}{2} = 40 \text{ mm}$
• Rectangle 2 (Horizontal part, to the right of Rectangle 1):
  • Dimensions: $(100 - 40) \text{ mm} \times 40 \text{ mm} = 60 \text{ mm} \times 40 \text{ mm}$
  • Area $A_2 = 60 \times 40 = 2400 \text{ mm}^2$
  • Centroid $x_2 = 40 + \frac{60}{2} = 40 + 30 = 70 \text{ mm}$
  • Centroid $y_2 = \frac{40}{2} = 20 \text{ mm}$
• Hole 3 (Rectangular hole):
  • Dimensions: $20 \text{ mm} \times 20 \text{ mm}$
  • Area $A_3 = 20 \times 20 = 400 \text{ mm}^2$ (This area will be subtracted)
  • Centroid $x_3 = 20 + \frac{20}{2} = 30 \text{ mm}$
  • Centroid $y_3 = 20 + \frac{20}{2} = 30 \text{ mm}$

The coordinates of the centroid of the lamina are $\boxed{(42.31 \text{ mm}, 31.54 \text{ mm})}$.