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9. What is the cross-section area of the I-beam in square mete = t = 1 cm 0.7 cm b= 10 cm h30 cm

9. What is the cross-section area of the I-beam in square mete = t = 1 cm 0.7 cm b= 10 cm h30 cm

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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### Cross-Section Area of an I-Beam #### Question: 9. **What is the cross-section area of the I-beam in square meters?** #### Diagram Description: An I-beam in the diagram consists of: - **Top flange width (tₜ): 1 cm** - **Web thickness (t_w): 0.7 cm** - **Total height (h): 30 cm** - **Flange width (b): 10 cm** The I-beam can be visualized as two horizontal rectangles connected by a vertical rectangle. #### Calculation Steps: 1. **Convert measurements to meters:** - \( t_{\text{top flange}} = 1 \text{ cm} = 0.01 \text{ m} \) - \( t_{\text{web}} = 0.7 \text{ cm} = 0.007 \text{ m} \) - \( h_{\text{total height}} = 30 \text{ cm} = 0.3 \text{ m} \) - \( b_{\text{flange width}} = 10 \text{ cm} = 0.1 \text{ m} \) 2. **Calculate the area of one flange:** \( A_{\text{flange}} = t_{t} \times b = 0.01 \text{ m} \times 0.1 \text{ m} = 0.001 \text{ m}^2 \) 3. **Calculate the area of the web:** \( A_{\text{web}} = t_{w} \times (h - 2 \times t_{t}) = 0.007 \text{ m} \times (0.3 \text{ m} - 2 \times 0.01 \text{ m}) = 0.007 \text{ m} \times 0.28 \text{ m} = 0.00196 \text{ m}^2 \) 4. **Calculate the total cross-sectional area:** \( A_{\text{total}} = 2 \times A_{\text{flange}} + A_{\text{web}} = 2 \times 0.001 \text{ m}^2 + 0.00196 \text{ m}^2 = 0.004 \text{ m}^
Transcribed Image Text:### Cross-Section Area of an I-Beam #### Question: 9. **What is the cross-section area of the I-beam in square meters?** #### Diagram Description: An I-beam in the diagram consists of: - **Top flange width (tₜ): 1 cm** - **Web thickness (t_w): 0.7 cm** - **Total height (h): 30 cm** - **Flange width (b): 10 cm** The I-beam can be visualized as two horizontal rectangles connected by a vertical rectangle. #### Calculation Steps: 1. **Convert measurements to meters:** - \( t_{\text{top flange}} = 1 \text{ cm} = 0.01 \text{ m} \) - \( t_{\text{web}} = 0.7 \text{ cm} = 0.007 \text{ m} \) - \( h_{\text{total height}} = 30 \text{ cm} = 0.3 \text{ m} \) - \( b_{\text{flange width}} = 10 \text{ cm} = 0.1 \text{ m} \) 2. **Calculate the area of one flange:** \( A_{\text{flange}} = t_{t} \times b = 0.01 \text{ m} \times 0.1 \text{ m} = 0.001 \text{ m}^2 \) 3. **Calculate the area of the web:** \( A_{\text{web}} = t_{w} \times (h - 2 \times t_{t}) = 0.007 \text{ m} \times (0.3 \text{ m} - 2 \times 0.01 \text{ m}) = 0.007 \text{ m} \times 0.28 \text{ m} = 0.00196 \text{ m}^2 \) 4. **Calculate the total cross-sectional area:** \( A_{\text{total}} = 2 \times A_{\text{flange}} + A_{\text{web}} = 2 \times 0.001 \text{ m}^2 + 0.00196 \text{ m}^2 = 0.004 \text{ m}^
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