김경종
855 lines
16 KiB
Markdown
855 lines
16 KiB
Markdown
<!-- source-page: 301 -->
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neutral axis to the top or bottom of the beam cross section, as used in the bending stress formula $\sigma = ( M c / I )$ .
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<details>
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<summary>text_image</summary>
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1000 lb
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4
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8
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12
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2000 lb
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3
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7
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11
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2000 lb
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2
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6
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10
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1
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1
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5
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9
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30 ft
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30 ft
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10 ft
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10 ft
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10 ft
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</details>
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Figure P5–22
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5.23–5.38 For the rigid frames or beams shown in Figures P5–23—P5–38, determine the displacements and rotations at the nodes, the element forces, and the reactions.
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<details>
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<summary>text_image</summary>
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3000 lb-ft
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1200 lb
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A = 10 in²
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I = 150 in⁴
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(for elements 3 and 4)
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2400 lb
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3
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③
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④
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300 in⁴
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A = 12 in²
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15 ft
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15 ft
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2
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①
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②
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2
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25 ft
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For cross members:
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I = 1.0 in⁴
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A = 2.0 in²
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E = 30 × 10⁶ psi
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(for all members)
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</details>
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Figure P5–23
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<details>
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<summary>text_image</summary>
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300 lb/ft
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25 ft
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5
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④
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6
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②
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⑥
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15 ft
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3
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③
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4
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①
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⑤
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1
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15 ft
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2
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E = 30 × 10⁶ psi
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I = 200 in⁴
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A = 15 in²
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</details>
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Figure P5–24
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<!-- source-page: 302 -->
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<details>
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<summary>text_image</summary>
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125 lb
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2
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25 lb
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1
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3
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4
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5
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6
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</details>
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(a)Design 1
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<details>
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<summary>text_image</summary>
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125 lb
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2
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25 lb
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1
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2
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3
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4
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5
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6
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7
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</details>
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(b）Design 2
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<details>
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<summary>text_image</summary>
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(10, 15)
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(30, 15)
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(10, 15)
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(32, 11.25)
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(38, 0)
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(0, 0)
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(20, 0)
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(32, 11.25)
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(38, 0)
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①
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②
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③
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④
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⑤
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⑥
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⑦
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⑧
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</details>
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(a)Design 1
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<details>
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<summary>text_image</summary>
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y
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1
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2 (10, 15)
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3
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4
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5
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6
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(30, 15)
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(32, 11.25)
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(38, 0)
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(20, 0)
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(38, 0)
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x
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①
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②
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③
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④
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⑤
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⑥
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⑦
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⑧
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⑨
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</details>
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(b）Design 2
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$$
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\begin{array}{l} \text { Case 1 } \\ E = 3 0 \times 1 0 ^ {6} \mathrm{psi} \\ \text { Case 2 } \\ E = 1 0 \times 1 0 ^ {6} \mathrm{psi} \\ A _ {1} = 0. 1 \text { in } ^ {2} \\ A _ {2} = A _ {3} = A _ {4} = A _ {5} = 0. 1 5 \mathrm{in} ^ {2} \\ A _ {6} = A _ {7} = A _ {8} = 0. 3 \text { in } ^ {2} \\ I _ {1} = 0. 0 1 \mathrm{in} ^ {4} \\ I _ {2} = I _ {3} = I _ {4} = I _ {5} = 0. 0 2 \text { in } ^ {4} \\ I _ {6} = I _ {7} = I _ {8} = 0. 1 \mathrm{in} ^ {4} \\ \end{array}
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$$
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Figure P5–25 Two bicycle frame models (coordinates shown in inches)
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<details>
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<summary>text_image</summary>
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1000 lb/ft
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3
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8 ft
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2
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I₂, A₂
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5
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8 ft
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6
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I₁, A₁
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12 ft
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I₁, A₁
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1
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4
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12 ft
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E = 30 × 10⁶ psi
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I₁ = 300 in⁴
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I₂ = 600 in⁴
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A₁ = 15 in²
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A₂ = 30 in²
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</details>
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Figure P5–26
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<!-- source-page: 303 -->
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<details>
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<summary>text_image</summary>
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60 kN
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4 m
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4 m
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2
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②
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3
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③
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4
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①
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④
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6 m
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1
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5
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E = 210 GPa
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I = 1.0 × 10⁻⁴ m⁴
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A = 1.0 × 10⁻² m²
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</details>
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Figure P5-27
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<details>
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<summary>text_image</summary>
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10 m
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4
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4 m
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2
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4 m
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1
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300 kN/m
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5
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E = 210 GPa
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I = 0.5 × 10⁻⁴ m⁴
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A = 0.5 × 10⁻² m²
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</details>
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Figure P5-28
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<details>
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<summary>text_image</summary>
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E = 30 × 10⁶ psi
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I = 150 in⁴
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A = 10 in²
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1
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①
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2
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②
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16 kip
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16 kip
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4 kip
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3
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③
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4
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④
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5
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⑤
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6
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30 ft
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14 ft
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7 ft
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7 ft
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30 ft
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</details>
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Figure P5–29
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<!-- source-page: 304 -->
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$$
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\begin{array}{l} E = 3 0 \times 1 0 ^ {6} \mathrm{psi} \\ I = 2 0 0 \mathrm{in} ^ {4} \\ A = 1 2 \mathrm{in} ^ {2} \end{array}
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$$
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<details>
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<summary>text_image</summary>
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30 kip
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30 kip
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15 ft
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20 ft
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15 ft
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②
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3
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③
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4
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④
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2
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5
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10 ft
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①
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⑤
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30 ft
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1
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6
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50 ft
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</details>
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Figure P5–30
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<details>
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<summary>text_image</summary>
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E = 30 × 10⁶ psi
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I = 100 in⁴
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A = 8 in²
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①
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②
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③
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500 lb/ft
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10 ft
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15 ft
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3
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4
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1
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45°
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</details>
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Figure P5–31
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<details>
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<summary>text_image</summary>
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15 kN
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10 kN · m
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2
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6 m
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3
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①
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②
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③
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6 m
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1
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4
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</details>
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$$
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\begin{array}{l} E = 2 1 0 \mathrm{GPa} \\ I = 2 \times 1 0 ^ {- 4} \mathrm{m} ^ {4} \\ A = 2 \times 1 0 ^ {- 2} \mathrm{m} ^ {2} \\ \end{array}
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$$
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Figure P5–32
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<!-- source-page: 305 -->
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<details>
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<summary>text_image</summary>
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20 kN
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7
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60 kN·m
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I₁, A₁
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8
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I₃, A₃
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I₁, A₁
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I₃, A₃
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20 kN
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5
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6
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I₂, A₂
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I₂, A₂
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20 kN
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3
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I₁, A₁
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4
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I₂, A₂
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2
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1
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2
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10 m
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3 m
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3 m
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4 m
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</details>
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$$
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\begin{array}{l} E = 2 1 0 \mathrm{GPa} \\ I _ {1} = 2 \times 1 0 ^ {- 4} \mathrm{m} ^ {4} \\ A _ {1} = 2 \times 1 0 ^ {- 2} \mathrm{m} ^ {2} \\ I _ {2} = 1 \times 1 0 ^ {- 4} \mathrm{m} ^ {4} \\ A _ {2} = 1 \times 1 0 ^ {- 2} \mathrm{m} ^ {2} \\ I _ {3} = 0. 5 \times 1 0 ^ {- 4} \mathrm{m} ^ {4} \\ A _ {3} = 0. 5 \times 1 0 ^ {- 2} \mathrm{m} ^ {2} \\ \end{array}
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$$
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Figure P5-33
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<details>
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<summary>text_image</summary>
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7000 N/m
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1
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① 2 ② 3 ③ 4
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7 m 7 m 7 m
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</details>
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$$
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E = 2 1 0 \mathrm{GPa}
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$$
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$$
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I = 1 \times 1 0 ^ {- 4} \mathrm{m} ^ {4}
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$$
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$$
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A = 1 \times 1 0 ^ {- 2} \mathrm{m} ^ {2}
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$$
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Figure P5-34
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<details>
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<summary>text_image</summary>
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2800 N · m
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4000 N → 9 → 2A₁, 3I₁ 10
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A₃, I₃ 4 m
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8000 N → 7 → 2A₁, 3I₁ 8
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A₃, I₃ 4 m
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8000 N → 5 → 2A₁, 3I₁ 6
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A₂, I₂ 4 m
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12,000 N → 3 → 2A₁, 3I₁ 4
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A₁, I₁ 6 m
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1 → 2
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10 m
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</details>
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$$
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E = 2 1 0 \mathrm{GPa}
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$$
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$$
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A _ {1} = 2. 0 \times 1 0 ^ {- 2} \mathrm{m} ^ {2}
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$$
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$$
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I _ {1} = 2. 0 \times 1 0 ^ {- 4} \mathrm{m} ^ {4}
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$$
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$$
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A _ {2} = 1. 5 \times 1 0 ^ {- 2} \mathrm{m} ^ {2}
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$$
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$$
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I _ {2} = 1. 5 \times 1 0 ^ {- 4} \mathrm{m} ^ {4}
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$$
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$$
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A _ {3} = 1. 0 \times 1 0 ^ {- 2} \mathrm{m} ^ {2}
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$$
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$$
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I _ {3} = 1. 0 \times 1 0 ^ {- 4} \mathrm{m} ^ {4}
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$$
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Figure P5-35
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$$
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A _ {1} = 2. 0 \times 1 0 ^ {- 2} \mathrm{m} ^ {2}
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$$
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$$
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I _ {1} = 2. 0 \times 1 0 ^ {- 4} \mathrm{m} ^ {4}
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$$
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$$
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A _ {2} = 1. 5 \times 1 0 ^ {- 2} \mathrm{m} ^ {2}
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$$
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$$
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I _ {2} = 1. 5 \times 1 0 ^ {- 4} \mathrm{m} ^ {4}
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$$
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$$
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A _ {3} = 1. 0 \times 1 0 ^ {- 2} \mathrm{m} ^ {2}
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$$
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$$
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I _ {3} = 1. 0 \times 1 0 ^ {- 4} \mathrm{m} ^ {4}
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$$
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<!-- source-page: 306 -->
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<details>
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<summary>text_image</summary>
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E = 210 GPa
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I = 1 × 10⁻⁴ m⁴
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A = 1 × 10⁻² m²
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10 kN
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10 kN
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4
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6
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20 kN
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13
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0.8m
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2.5 m
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2
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1 m
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2
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3
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5
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6
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7
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8
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9
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10
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11
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12
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13
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14
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1
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2
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3
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2 m
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2 m
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</details>
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Figure P5–36
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<details>
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<summary>text_image</summary>
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E = 210 GPa
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I = 4 × 10⁻⁴ m⁴
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A = 4 × 10⁻² m²
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18 kN 72 kN 72 kN
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6 m 4 m 3 m 5 m 6 m
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1 3 4 5 6 7
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10 m
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2 8
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10 m
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</details>
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Figure P5–37
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<details>
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<summary>text_image</summary>
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15 kN/m
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80 kN
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3.5 m
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E = 210 GPa
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I = 2.0 × 10⁻⁴ m⁴
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A = 1.0 × 10⁻² m²
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300 kN/m
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①
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②
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③
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④
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⑤
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⑥
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1
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3
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7
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8 m
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7 m
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7 m
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</details>
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Figure P5–38
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5.39 Consider the plane structure shown in Figure P5–39. First assume the structure to be a plane frame with rigid joints, and analyze using a frame element. Then assume the structure to be pin-jointed and analyze as a plane truss, using a truss element. If the structure is actually a truss, is it appropriate to model it as a rigid frame? How
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<!-- source-page: 307 -->
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<details>
|
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<summary>text_image</summary>
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10 kN
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7
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4 m
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8
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3 m
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20 kN
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5
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6
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3 m
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20 kN
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3
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4
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20 kN
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1
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2
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3
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4
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3 m
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1
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2
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</details>
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Figure P5–39
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can you model the truss using the frame (or beam) element? In other words, what idealization could you make in your model to use the beam element to approximate a truss?
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5.40 For the two-story, two-bay rigid frame shown, determine (1) the nodal displacement components and (2) the shear force and bending moments in each member. Let $E = \mathsf { 2 0 0 ~ G P a } , I = \overset { \cdot } { 2 } \times 1 0 ^ { - 4 } \mathrm { m } ^ { 4 }$ for each horizontal member and $I = 1 . 5 \times 1 0 ^ { - 4 } \mathrm { m } ^ { 4 }$ for each vertical member.
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<details>
|
||
<summary>text_image</summary>
|
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12 kN/m
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G
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H
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12 kN/m
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I
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5 m
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D
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E
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F
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5 m
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A
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B
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C
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10 m
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10 m
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</details>
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Figure P5–40
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5.41 For the two-story, three-bay rigid frame shown, determine (1) the nodal displacements and (2) the member end shear forces and bending moments. (3) Draw the shear force and bending moment diagrams for each member. Let E ¼ 200 GPa; $I = 1 . 2 9 \times 1 0 ^ { - 4 } \mathrm { m } ^ { 4 }$ for the beams and $I = 0 . 4 6 2 \times 1 0 ^ { - 4 } \mathrm { m } ^ { 4 }$ for the columns.
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<!-- source-page: 308 -->
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The properties for I correspond to a W $6 1 0 \times 1 5 5$ and a W $4 1 0 \times 1 1 4$ wide-flange section, respectively, in metric units.
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<details>
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<summary>text_image</summary>
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25 kN
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50 kN
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I J K L
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E F G H
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A B C D
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4 m
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6 m
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8 m 6 m 8 m
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</details>
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Figure P5–41
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5.42 For the rigid frame shown, determine (1) the nodal displacements and rotations and (2) the member shear forces and bending moments. Let $E = 2 0 0$ GPa, $\dot { I } = 0 . 7 9 5 \times 1 0 ^ { - 4 } \mathrm { m } ^ { 4 }$ for the horizontal members and $I = 0 . 3 1 6 \times 1 0 ^ { - 4 } \mathrm { m } ^ { 4 }$ for the vertical members. These I values correspond to a W $4 6 0 \times 1 5 8$ and a W $4 1 0 \times 8 5$ wide-flange section, respectively.
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<details>
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<summary>text_image</summary>
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20 kN → G H I
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40 kN → D E F
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A B C
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3 m
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3 m
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5 m 5 m
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</details>
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Figure P5–42
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5.43 For the rigid frame shown, determine (1) the nodal displacements and rotations and (2) the shear force and bending moments in each member. Let $E = 2 9 \times 1 0 6$ psi, $I = 3 1 0 0 ~ \mathrm { i n } . ^ { 4 }$ for the horizontal members and $I = 1 1 1 0 ~ \mathrm { i n } . ^ { 4 }$ for the vertical members. The I values correspond to a ${ \textsf { W } } 2 4 \times 1 0 4$ and a ${ \bf W } 1 6 \times 7 7 \mathrm { ~ \AA ~ }$ :
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<details>
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||
<summary>text_image</summary>
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||
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7.5 kip
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M N
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15 kip
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I J K L
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15 kip
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E F G H
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A B C D
|
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30 ft 20 ft 30 ft
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15 ft
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15 ft
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||
15 ft
|
||
</details>
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|
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Figure P5–43
|
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|
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<!-- source-page: 309 -->
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5.44 A structure is fabricated by welding together three lengths of I-shaped members as shown in Figure P5–44. The yield strength of the members is 36 ksi, $E = 2 9 \mathrm { e 6 p s i }$ , and Poisson’s ratio is 0.3. The members all have cross-section properties corresponding to a W18 by 76. That is, $A = 2 2 . 3 \mathrm { i n } ^ { 2 }$ , depth of section is $d = 1 8 . 2 1$ in., $I _ { x } = 1 3 3 0 \mathrm { i n } ^ { 4 }$ , $S _ { x } = 1 4 6 \mathrm { { i n } } ^ { 3 }$ , $I _ { \nu } = 1 5 2 \mathrm { i n } ^ { 4 }$ , and $S _ { \nu } = 2 7 . 6 \mathrm { i n } ^ { 3 }$ . Determine whether a load of $Q = 1 0 { , } 0 0 0$ lb downward is safe against general yielding of the material. The factor of safety against general yielding is to be 2.0. Also, determine the maximum vertical and horizontal deflections of the structure.
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|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
90"
|
||
90"
|
||
Q
|
||
</details>
|
||
|
||
Figure P5–44
|
||
|
||
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|
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5.45 For the tapered beam shown in Figure P5–45, determine the maximum deflection using one, two, four, and eight elements. Calculate the moment of inertia at the midlength station for each element. Let $E = 3 0 \times 1 0 ^ { 6 }$ psi, $I _ { 0 } = 1 0 0 ~ \mathrm { i n } ^ { 4 }$ , and $L = 1 0 0$ in. Run cases where $n = 1 , 3 .$ , and 7. Use a beam element. The analytical solution for $n = 7$ is given by Reference [7]:
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|
||
$$
|
||
v _ {1} = \frac {P L ^ {3}}{4 9 E I _ {0}} (1 / 7 \ln 8 + 2. 5) = \frac {1}{1 7 . 5 5} \frac {P L ^ {3}}{E I _ {0}}
|
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$$
|
||
|
||
$$
|
||
\theta_ {1} = \frac {P L ^ {2}}{4 9 E I _ {0}} (\ln 8 - 7) = - \frac {1}{9 . 9 5} \frac {P L ^ {2}}{E I _ {0}}
|
||
$$
|
||
|
||
$$
|
||
I (x) = I _ {0} \left(1 + n \frac {x}{L}\right)
|
||
$$
|
||
|
||
where n ¼ arbitrary numerical factor and $I _ { 0 } = \mathrm { m o m e n t }$ of inertia of section at $x = 0$ .
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
P = 500 lb
|
||
1
|
||
A
|
||
2
|
||
x
|
||
w
|
||
h(x)
|
||
A-A
|
||
1
|
||
2
|
||
y
|
||
L
|
||
One-element approximation
|
||
</details>
|
||
|
||
Figure P5–45 Tapered cantilever beam
|
||
|
||
<!-- source-page: 310 -->
|
||
|
||
5.46 Derive the stiffness matrix for the nonprismatic torsion bar shown in Figure P5–46. The radius of the shaft is given by
|
||
|
||
$$
|
||
r = r _ {0} + (x / L) r _ {0}, \text { where } r _ {0} \text { is the radius at } x = 0.
|
||
$$
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
r₀工
|
||
x
|
||
L
|
||
</details>
|
||
|
||
Figure P5–46
|
||
|
||
5.47 Derive the total potential energy for the prismatic circular cross-section torsion bar shown in Figure P5–47. Also determine the equivalent nodal torques for the bar subjected to uniform torque per unit length (lb-in./in.). Let G be the shear modulus and J be the polar moment of inertia of the bar.
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
T (lb-in./in.)
|
||
1
|
||
2
|
||
L
|
||
</details>
|
||
|
||
Figure P5–47
|
||
|
||
5.48 For the grid shown in Figure P5–48, determine the nodal displacements and the local element forces. Let $E = 3 0 \times 1 0 ^ { 6 }$ psi, G ¼ 12 106 psi, $I = 2 0 0 \mathrm { i n } ^ { 4 }$ , and $J = 1 0 0 \mathrm { i n } ^ { 4 }$ for both elements.
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
5000 lb
|
||
10 ft
|
||
10 ft
|
||
x
|
||
y
|
||
z
|
||
1
|
||
2
|
||
3
|
||
</details>
|
||
|
||
Figure P5–48
|
||
|
||
5.49 Resolve Problem 5–48 with an additional nodal moment of 1000 k-in. applied about the x axis at node 2.
|
||
|
||
5.50–5.51 For the grids shown in Figures P5–50 and P5–51, determine the nodal displacements and the local element forces. Let E ¼ 210 GPa, G ¼ 84 GPa, $I = 2 \times 1 0 ^ { - 4 } \mathrm { ~ m } ^ { 4 }$ , $J = 1 \times 1 0 ^ { - 4 } ~ \mathrm { m } ^ { 4 }$ , and $A = 1 \times 1 0 ^ { - 2 } \mathrm { m } ^ { 2 }$ .
|