Math Problem Statement

Determine the location of the centroid of a beam with a cross-sectional area as shown in the figure, using centimeter units.

Solution

To determine the centroid of the given composite shape, let's analyze each part in the figure.

Steps to Solve

  1. Identify each section (A, B, and C):
    From the figure, it seems that section:

    • A is the horizontal rectangle at the top,
    • B is the vertical rectangle (central part),
    • C is the horizontal rectangle at the bottom.
  2. Calculate the area AiA_i for each section:

    • Section A: Length = 12 cm, Height = 2 cm. AA=12×2=24cm2A_A = 12 \times 2 = 24 \, \text{cm}^2
    • Section B: Length = 2 cm, Height = 12 cm. AB=2×12=24cm2A_B = 2 \times 12 = 24 \, \text{cm}^2
    • Section C: Length = 8 cm, Height = 2 cm. AC=8×2=16cm2A_C = 8 \times 2 = 16 \, \text{cm}^2
  3. Find the centroid coordinates xˉi\bar{x}_i and yˉi\bar{y}_i for each section:

    • Section A: The center is located at: xˉA=122=6cm,yˉA=12+1=13cm\bar{x}_A = \frac{12}{2} = 6 \, \text{cm}, \quad \bar{y}_A = 12 + 1 = 13 \, \text{cm}
    • Section B: The center is located at: xˉB=22=1cm,yˉB=122=6cm\bar{x}_B = \frac{2}{2} = 1 \, \text{cm}, \quad \bar{y}_B = \frac{12}{2} = 6 \, \text{cm}
    • Section C: The center is located at: xˉC=82=4cm,yˉC=1cm\bar{x}_C = \frac{8}{2} = 4 \, \text{cm}, \quad \bar{y}_C = 1 \, \text{cm}
  4. Calculate AixˉiA_i \cdot \bar{x}_i and AiyˉiA_i \cdot \bar{y}_i for each section:

    • Section A: AAxˉA=24×6=144A_A \cdot \bar{x}_A = 24 \times 6 = 144 AAyˉA=24×13=312A_A \cdot \bar{y}_A = 24 \times 13 = 312
    • Section B: ABxˉB=24×1=24A_B \cdot \bar{x}_B = 24 \times 1 = 24 AByˉB=24×6=144A_B \cdot \bar{y}_B = 24 \times 6 = 144
    • Section C: ACxˉC=16×4=64A_C \cdot \bar{x}_C = 16 \times 4 = 64 ACyˉC=16×1=16A_C \cdot \bar{y}_C = 16 \times 1 = 16
  5. Sum all areas and the products AixˉiA_i \cdot \bar{x}_i and AiyˉiA_i \cdot \bar{y}_i:

    • Total Area Ai=24+24+16=64cm2\sum A_i = 24 + 24 + 16 = 64 \, \text{cm}^2
    • Aixˉi=144+24+64=232\sum A_i \cdot \bar{x}_i = 144 + 24 + 64 = 232
    • Aiyˉi=312+144+16=472\sum A_i \cdot \bar{y}_i = 312 + 144 + 16 = 472
  6. Calculate the centroid coordinates xˉ\bar{x} and yˉ\bar{y}: xˉ=AixˉiAi=23264=3.625cm\bar{x} = \frac{\sum A_i \cdot \bar{x}_i}{\sum A_i} = \frac{232}{64} = 3.625 \, \text{cm} yˉ=AiyˉiAi=47264=7.375cm\bar{y} = \frac{\sum A_i \cdot \bar{y}_i}{\sum A_i} = \frac{472}{64} = 7.375 \, \text{cm}

Final Answer

The centroid of the composite shape is located at: (xˉ,yˉ)=(3.625cm,7.375cm)(\bar{x}, \bar{y}) = (3.625 \, \text{cm}, 7.375 \, \text{cm})

Would you like a detailed explanation of each calculation step, or do you have any questions?


Follow-up Questions

  1. What is the purpose of finding the centroid in this context?
  2. How would the centroid change if section dimensions were altered?
  3. What other methods can be used to find the centroid of complex shapes?
  4. Why is the centroid located at the calculated coordinates, and what does it imply about the shape’s balance?
  5. How can this method apply to 3-dimensional shapes?

Tip:

Remember, when calculating centroids for composite shapes, always break down the figure into simpler geometric parts.

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Math Problem Analysis

Mathematical Concepts

Centroid Calculation
Composite Areas
Statics

Formulas

Centroid X-coordinate: \( \bar{x} = \frac{\sum \bar{x}_i \cdot A_i}{\sum A_i} \)
Centroid Y-coordinate: \( \bar{y} = \frac{\sum \bar{y}_i \cdot A_i}{\sum A_i} \)

Theorems

Centroid of Composite Shapes

Suitable Grade Level

Grades 10-12