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

Here's a step-by-step solution for the given problems.

5.1.1 The reaction forces at supports B and D.

First, we need to determine the equivalent point load for the uniformly distributed load (UDL). The UDL is $6 \text{ kN/m}$ over a length of $1.5 m + 3.5 m = 5 \text{ m}$. The total force from the UDL is $F_{UDL} = 6 kN/m \times 5 m = 30 \text{ kN}$. This force acts at the centroid of the UDL, which is $5 m / 2 = 2.5 \text{ m}$ from point A.

Let's define the positions of all forces and supports from point A (left end of the beam):

• Point A: $x = 0 \text{ m}$ (18 kN downward force)
• Support B: $x = 1.5 \text{ m}$ (Reaction $R_B$ upward)
• UDL equivalent force: $x = 2.5 \text{ m}$ (30 kN downward force)
• Point C: $x = 1.5 m + 3.5 m = 5 \text{ m}$ (40 kN downward force)
• Support D: $x = 5 m + 1.5 m = 6.5 \text{ m}$ (Reaction $R_D$ upward)
• Point E: $x = 6.5 m + 2.5 m = 9 \text{ m}$ (22 kN downward force)

Step 1: Apply the equilibrium equation for the sum of vertical forces. Assume upward forces are positive. $\sum F_y = 0$ $R_B + R_D - 18 kN - 30 kN - 40 kN - 22 kN = 0$ $R_B + R_D - 110 kN = 0$ $R_B + R_D = 110 kN \quad (1)$

Step 2: Apply the equilibrium equation for the sum of moments about support B. Assume clockwise moments are positive. $\sum M_B = 0$ The moment arms are calculated relative to point B ($x=1.5 \text{ m}$). $(18 kN \times 1.5 m) + (30 kN \times (2.5 m - 1.5 m)) + (40 kN \times (5 m - 1.5 m)) - (R_D \times (6.5 m - 1.5 m)) + (22 kN \times (9 m - 1.5 m)) = 0$ $(18 \times 1.5) + (30 \times 1) + (40 \times 3.5) - (R_D \times 5) + (22 \times 7.5) = 0$ $27 + 30 + 140 - 5R_D + 165 = 0$ $362 - 5R_D = 0$ $5R_D = 362$ $R_D = \frac{362}{5} = 72.4 kN$

Step 3: Substitute the value of $R_D$ into Equation (1) to find $R_B$. $R_B + 72.4 kN = 110 kN$ $R_B = 110 kN - 72.4 kN$ $R_B = 37.6 kN$

The reaction forces are:

• $R_B = \text{37.6 kN}$
• $R_D = \text{72.4 kN}$

5.1.2 Draw the shear force diagram for the beam.

We will calculate the shear force at key points along the beam, starting from the left (point A).

• At A ($x=0 \text{ m}$): $V_A = -18 \text{ kN}$ (downward force)

• Just before B ($x=1.5^- \text{ m}$): The UDL acts from $x=0$ to $x=5 \text{ m}$. Shear force due to UDL up to B = $6 kN/m \times 1.5 m = 9 \text{ kN}$. $V_{B^-} = -18 kN - 9 kN = -27 \text{ kN}$

• Just after B ($x=1.5^+ \text{ m}$): $V_{B^+} = V_{B^-} + R_B = -27 kN + 37.6 kN = 10.6 \text{ kN}$

• Just before C ($x=5^- \text{ m}$): Shear force from B to C due to UDL = $6 kN/m \times (5 m - 1.5 m) = 6 kN/m \times 3.5 m = 21 \text{ kN}$. $V_{C^-} = V_{B^+} - 21 kN = 10.6 kN - 21 kN = -10.4 \text{ kN}$

• Just after C ($x=5^+ \text{ m}$): $V_{C^+} = V_{C^-} - 40 kN = -10.4 kN - 40 kN = -50.4 \text{ kN}$

• Just before D ($x=6.5^- \text{ m}$): There are no distributed loads or point loads between C and D. $V_{D^-} = V_{C^+} = -50.4 \text{ kN}$

• Just after D ($x=6.5^+ \text{ m}$): $V_{D^+} = V_{D^-} + R_D = -50.4 kN + 72.4 kN = 22 \text{ kN}$

• At E ($x=9 \text{ m}$): There are no distributed loads or point loads between D and E. $V_E = V_{D^+} - 22 kN = 22 kN - 22 kN = 0 \text{ kN}$ (This confirms the calculations are correct as the shear force closes to zero at the end of the beam).

The shear force diagram will be drawn based on these values.

• From A to B, the shear force decreases linearly from $-18 \text{ kN}$ to $-27 \text{ kN}$.
• At B, it jumps up to $10.6 \text{ kN}$.
• From B to C, it decreases linearly from $10.6 \text{ kN}$ to $-10.4 \text{ kN}$.
• At C, it jumps down to $-50.4 \text{ kN}$.
• From C to D, it remains constant at $-50.4 \text{ kN}$.
• At D, it jumps up to $22 \text{ kN}$.
• From D to E, it remains constant at $22 \text{ kN}$ and then drops to $0 \text{ kN}$ at E.

The point where the shear force is zero between B and C can be found by similar triangles or by setting the shear force equation to zero. Let $x'$ be the distance from B where shear force is zero. $10.6 - 6x' = 0 \implies x' = \frac{10.6}{6} \approx 1.767 \text{ m}$ from B. This corresponds to $x = 1.5 + 1.767 = 3.267 \text{ m}$ from A.

The shear force diagram is as follows:

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

We will divide the lamina into simpler geometric shapes and use the principle of composite areas. Let's divide the shape into three rectangles:

• Rectangle 1 (Vertical part on left):

  • Width = $40 \text{ mm}$
  • Height = $80 \text{ mm}$
  • Area $A_1 = 40 \times 80 = 3200 mm^2$
  • Centroid $x_1 = \frac{40}{2} = 20 \text{ mm}$
  • Centroid $y_1 = \frac{80}{2} = 40 \text{ mm}$

• Rectangle 2 (Horizontal part on right):

  • Width = $100 mm - 40 mm = 60 \text{ mm}$
  • Height = $40 \text{ mm}$
  • Area $A_2 = 60 \times 40 = 2400 mm^2$
  • Centroid $x_2 = 40 mm + \frac{60}{2} = 40 + 30 = 70 \text{ mm}$
  • Centroid $y_2 = \frac{40}{2} = 20 \text{ mm}$

• Rectangle 3 (Hole):

  • This is a cutout, so its area will be negative.
  • Width = $20 \text{ mm}$
  • Height = $20 \text{ mm}$
  • Area $A_3 = -(20 \times 20) = -400 mm^2$
  • Centroid $x_3 = 20 mm + \frac{20}{2} = 20 + 10 = 30 \text{ mm}$
  • Centroid $y_3 = 20 mm + \frac{20}{2} = 20 + 10 = 30 \text{ mm}$

Step 1: Calculate the total area $A_{total}$. $A_{total} = A_1 + A_2 + A_3$ $A_{total} = 3200 mm^2 + 2400 mm^2 - 400 mm^2$ $A_{total} = 5200 mm^2$

Step 2: Calculate the x-coordinate of the centroid ($\bar{x}$). $\bar{x} = \frac{A_1 x_1 + A_2 x_2 + A_3 x_3}{A_{total}}$ $\bar{x} = \frac{(3200 \times 20) + (2400 \times 70) + (-400 \times 30)}{5200}$ $\bar{x} = \frac{64000 + 168000 - 12000}{5200}$ $\bar{x} = \frac{220000}{5200}$ $\bar{x} = \frac{2200}{52} = \frac{550}{13} \approx 42.31 mm$

Step 3: Calculate the y-coordinate of the centroid ($\bar{y}$). $\bar{y} = \frac{A_1 y_1 + A_2 y_2 + A_3 y_3}{A_{total}}$ $\bar{y} = \frac{(3200 \times 40) + (2400 \times 20) + (-400 \times 30)}{5200}$ $\bar{y} = \frac{128000 + 48000 - 12000}{5200}$ $\bar{y} = \frac{164000}{5200}$ $\bar{y} = \frac{1640}{52} = \frac{410}{13} \approx 31.54 mm$

The coordinates of the centroid are: $(\bar{x, \bar{y}) = (42.31 mm, 31.54 mm)}$

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