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.

here are the solutions to Question 5.

QUESTION 5: STATICS

5.1

A simply supported beam is loaded as shown in FIGURE 1.

5.1.1 The reaction forces at supports B and D.

Step 1: Convert the distributed load into an equivalent point load. The distributed load is $6 \text{ kN/m}$ over $3.5 \text{ m}$. Equivalent point load $F_w = 6 kN/m \times 3.5 m = 21 \text{ kN}$. This load acts at the center of the distributed load, which is $1.5 m + \frac{3.5 m}{2} = 1.5 m + 1.75 m = 3.25 \text{ m}$ from point A.

Step 2: Apply the equilibrium equation for the sum of vertical forces. 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. $\sum F_y = 0$ $R_B + R_D - 18 kN - 21 kN - 40 kN - 22 kN = 0$ $R_B + R_D - 101 kN = 0$ $R_B + R_D = 101 kN \quad (1)$

Step 3: Apply the equilibrium equation for the sum of moments about support B. 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 $3.25 m - 1.5 m = 1.75 \text{ m}$. The distance from the $40 \text{ kN}$ load to B is $5 m - 1.5 m = 3.5 \text{ m}$. The distance from D to B is $6.5 m - 1.5 m = 5 \text{ m}$. The distance from the $22 \text{ kN}$ load to B is $9 m - 1.5 m = 7.5 \text{ m}$. $\sum M_B = 0$ $(-18 kN \times 1.5 m) + (21 kN \times 1.75 m) + (40 kN \times 3.5 m) - (R_D \times 5 m) + (22 kN \times 7.5 m) = 0$ $-27 + 36.75 + 140 - 5R_D + 165 = 0$ $314.75 - 5R_D = 0$ $5R_D = 314.75$ $R_D = \frac{314.75}{5} = 62.95 kN$

Step 4: Calculate $R_B$ using Equation (1). $R_B + 62.95 kN = 101 kN$ $R_B = 101 kN - 62.95 kN = 38.05 kN$

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

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 $R_B$: $V = -18 kN + 38.05 kN = 20.05 \text{ kN}$
• 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): $V_{5m, L} = 20.05 kN - (6 kN/m \times 3.5 m) = 20.05 kN - 21 kN = -0.95 \text{ kN}$
  • The shear force crosses zero at $x = 1.5 m + \frac{20.05 kN}{6 kN/m} \approx 1.5 m + 3.34 m = 4.84 \text{ m}$ 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: $V = -0.95 kN - 40 kN = -40.95 \text{ kN}$
• 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 $R_D$: $V = -40.95 kN + 62.95 kN = 22 \text{ kN}$
• 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: $V = 22 kN - 22 kN = 0 \text{ kN}$

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.

Step 1: Divide the lamina into simple geometric shapes and a hole. 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: $40 mm \times 80 \text{ mm}$
  • 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, to the right of Rectangle 1):
  • Dimensions: $(100 - 40) mm \times 40 mm = 60 mm \times 40 \text{ mm}$
  • Area $A_2 = 60 \times 40 = 2400 mm^2$
  • 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: $20 mm \times 20 \text{ mm}$
  • Area $A_3 = 20 \times 20 = 400 mm^2$ (This area will be subtracted)
  • Centroid $x_3 = 20 + \frac{20}{2} = 30 \text{ mm}$
  • Centroid $y_3 = 20 + \frac{20}{2} = 30 \text{ mm}$

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

Step 3: Calculate the x-coordinate of the centroid ($\bar{x}$). $\bar{x} = \frac{A_1 x_1 + A_2 x_2 - A_3 x_3}{A_{total}}$ $\bar{x} = \frac{(3200 mm^2)(20 mm) + (2400 mm^2)(70 mm) - (400 mm^2)(30 mm)}{5200 mm^2}$ $\bar{x} = \frac{64000 + 168000 - 12000}{5200} mm$ $\bar{x} = \frac{220000}{5200} mm = \frac{550}{13} mm \approx 42.31 mm$

Step 4: Calculate the y-coordinate of the centroid ($\bar{y}$). $\bar{y} = \frac{A_1 y_1 + A_2 y_2 - A_3 y_3}{A_{total}}$ $\bar{y} = \frac{(3200 mm^2)(40 mm) + (2400 mm^2)(20 mm) - (400 mm^2)(30 mm)}{5200 mm^2}$ $\bar{y} = \frac{128000 + 48000 - 12000}{5200} mm$ $\bar{y} = \frac{164000}{5200} mm = \frac{410}{13} mm \approx 31.54 mm$

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

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