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Draw the shear force diagram. Calculate the bending moments and draw the bending moment diagram. Calculate the position of the centroid from MO of the lamina shown in FIGURE 2 below.

Mathematics

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Answer

(*325.89 mm, 144.05 mm*)\text{(*325.89 mm, 144.05 mm*)}

Step-by-step solution

5 steps

  1. Divide the lamina into three rectangular sections and determine their dimensions, areas, and individual centroid coordinates relative to point O (the origin).

    • Rectangle 1 (Left):

      • Width (w1w_1) = 200 mm200 \text{ mm}
      • Height (h1h_1) = 400 mm400 \text{ mm}
      • Area (A1A_1) = w1×h1=200 mm×400 mm=80000 mm2w_1 \times h_1 = 200 \text{ mm} \times 400 \text{ mm} = 80000 \text{ mm}^2
      • Centroid x1=w12=2002=100 mmx_1 = \frac{w_1}{2} = \frac{200}{2} = 100 \text{ mm}
      • Centroid y1=h12=4002=200 mmy_1 = \frac{h_1}{2} = \frac{400}{2} = 200 \text{ mm}

    • Rectangle 2 (Middle):

      • Width (w2w_2) = 250 mm250 \text{ mm}
      • Height (h2h_2) = 90 mm90 \text{ mm}
      • Area (A2A_2) = w2×h2=250 mm×90 mm=22500 mm2w_2 \times h_2 = 250 \text{ mm} \times 90 \text{ mm} = 22500 \text{ mm}^2
      • Centroid x2=w1+w22=200+2502=200+125=325 mmx_2 = w_1 + \frac{w_2}{2} = 200 + \frac{250}{2} = 200 + 125 = 325 \text{ mm}
      • Centroid y2=h22=902=45 mmy_2 = \frac{h_2}{2} = \frac{90}{2} = 45 \text{ mm}

    • Rectangle 3 (Right):

      • Width (w3w_3) = 300 mm300 \text{ mm}
      • Height (h3h_3) = 220 mm220 \text{ mm}
      • Area (A3A_3) = w3×h3=300 mm×220 mm=66000 mm2w_3 \times h_3 = 300 \text{ mm} \times 220 \text{ mm} = 66000 \text{ mm}^2
      • Centroid x3=w1+w2+w32=200+250+3002=450+150=600 mmx_3 = w_1 + w_2 + \frac{w_3}{2} = 200 + 250 + \frac{300}{2} = 450 + 150 = 600 \text{ mm}
      • Centroid y3=h32=2202=110 mmy_3 = \frac{h_3}{2} = \frac{220}{2} = 110 \text{ mm}

  2. Calculate the total area of the lamina.

    Atotal=A1+A2+A3A_{\text{total}} = A_1 + A_2 + A_3 Atotal=80000 mm2+22500 mm2+66000 mm2=168500 mm2A_{\text{total}} = 80000 \text{ mm}^2 + 22500 \text{ mm}^2 + 66000 \text{ mm}^2 = 168500 \text{ mm}^2

  3. Calculate the sum of the moments of the areas about the y-axis (for xˉ\bar{x}).

    ∑Aixi=A1x1+A2x2+A3x3\sum A_i x_i = A_1 x_1 + A_2 x_2 + A_3 x_3 ∑Aixi=(80000 mm2×100 mm)+(22500 mm2×325 mm)+(66000 mm2×600 mm)\sum A_i x_i = (80000 \text{ mm}^2 \times 100 \text{ mm}) + (22500 \text{ mm}^2 \times 325 \text{ mm}) + (66000 \text{ mm}^2 \times 600 \text{ mm}) ∑Aixi=8000000 mm3+7312500 mm3+39600000 mm3=54912500 mm3\sum A_i x_i = 8000000 \text{ mm}^3 + 7312500 \text{ mm}^3 + 39600000 \text{ mm}^3 = 54912500 \text{ mm}^3

  4. Calculate the sum of the moments of the areas about the x-axis (for yˉ\bar{y}).

    ∑Aiyi=A1y1+A2y2+A3y3\sum A_i y_i = A_1 y_1 + A_2 y_2 + A_3 y_3 ∑Aiyi=(80000 mm2×200 mm)+(22500 mm2×45 mm)+(66000 mm2×110 mm)\sum A_i y_i = (80000 \text{ mm}^2 \times 200 \text{ mm}) + (22500 \text{ mm}^2 \times 45 \text{ mm}) + (66000 \text{ mm}^2 \times 110 \text{ mm}) ∑Aiyi=16000000 mm3+1012500 mm3+7260000 mm3=24272500 mm3\sum A_i y_i = 16000000 \text{ mm}^3 + 1012500 \text{ mm}^3 + 7260000 \text{ mm}^3 = 24272500 \text{ mm}^3

  5. Calculate the coordinates of the centroid (xˉ,yˉ)(\bar{x}, \bar{y}) from point O.

    xˉ=∑AixiAtotal=54912500 mm3168500 mm2≈325.89 mm\bar{x} = \frac{\sum A_i x_i}{A_{\text{total}}} = \frac{54912500 \text{ mm}^3}{168500 \text{ mm}^2} \approx 325.89 \text{ mm} yˉ=∑AiyiAtotal=24272500 mm3168500 mm2≈144.05 mm\bar{y} = \frac{\sum A_i y_i}{A_{\text{total}}} = \frac{24272500 \text{ mm}^3}{168500 \text{ mm}^2} \approx 144.05 \text{ mm}

    The position of the centroid from MO (which is the y-axis) is given by its coordinates (xˉ,yˉ)(\bar{x}, \bar{y}).
    The position of the centroid is (*325.89 mm, 144.05 mm*)\text{(*325.89 mm, 144.05 mm*)}.

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