김경종
678 lines
19 KiB
Markdown
678 lines
19 KiB
Markdown
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# Problems
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6.1 Sketch the variations of the shape functions $N _ { j }$ and $N _ { m } ,$ given by Eqs. (6.2.18), over the surface of the triangular element with nodes $i , j ,$ and m. Check that $N _ { i } + N _ { j } +$ $N _ { m } = 1$ anywhere on the element.
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6.2 For a simple three-noded triangular element, show explicitly that differentiation of Eq. (6.2.47) indeed results in Eq. (6.2.48); that is, substitute the expression for ½B and the plane stress condition for ½D into Eq. (6.2.47), and then differentiate $\pi _ { p }$ with respect to each nodal degree of freedom in Eq. (6.2.47) to obtain Eq. (6.2.48).
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6.3 Evaluate the stiffness matrix for the elements shown in Figure P6–3. The coordinates are in units of inches. Assume plane stress conditions. Let $E = 3 0 \times 1 0 ^ { 6 } \mathrm { p s i } , \nu = 0 . 2 5 \mathrm { } _ { \mathrm { \odot } }$ , and thickness t ¼ 1 in.
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<details>
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<summary>line</summary>
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| Point | x | y |
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|---|---|---|
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| (0,1) | 0 | 3 |
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| (0,-1) | 0 | 1 |
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| (2,0) | 2 | 0 |
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</details>
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<details>
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<summary>line</summary>
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| Point | x | y |
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|---|---|---|
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| 1 | 1.2 | 1 |
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| 2 | 2.4 | 0 |
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| 3 | 1.2 | 1 |
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</details>
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<details>
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<summary>line</summary>
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| x | y |
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|---|---|
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| 0 | 1 |
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| 2 | 0 |
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</details>
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Figure P6–3
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6.4 For the elements given in Problem 6.3, the nodal displacements are given as
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$$
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u _ {1} = 0. 0 \quad v _ {1} = 0. 0 0 2 5 \text { in. } \quad u _ {2} = 0. 0 0 1 2 \text { in. }
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$$
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$$
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v _ {2} = 0. 0 \quad u _ {3} = 0. 0 \quad v _ {3} = 0. 0 0 2 5 \text { in. }
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$$
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Determine the element stresses $\sigma _ { x } , \sigma _ { y } , \tau _ { x y } , \sigma _ { 1 }$ , and $\sigma _ { 2 }$ and the principal angle $\theta _ { p }$ . Use the values of E; n, and t given in Problem 6.3.
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6.5 Determine the von Mises stress for problem 6.4.
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6.6 Evaluate the stiffness matrix for the elements shown in Figure P6–6. The coordinates are given in units of millimeters. Assume plane stress conditions. Let $E = 2 1 0 ~ \mathrm { G P a }$ , $\nu = 0 . 2 5 ,$ and $t = 1 0$ mm.
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6.7 For the elements given in Problem 6.6, the nodal displacements are given as
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$$
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u _ {1} = 2. 0 \mathrm{mm} \quad v _ {1} = 1. 0 \mathrm{mm} \quad u _ {2} = 0. 5 \mathrm{mm}
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$$
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$$
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v _ {2} = 0. 0 \mathrm{mm} \quad u _ {3} = 3. 0 \mathrm{mm} \quad v _ {3} = 1. 0 \mathrm{mm}
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$$
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<!-- source-page: 362 -->
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Determine the element stresses $\sigma _ { x } , \sigma _ { y } , \tau _ { x y } , \sigma _ { 1 }$ , and $\sigma _ { 2 }$ and the principal angle $\theta _ { p }$ . Use the values of $E , \nu ,$ and t given in Problem 6.6.
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<details>
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<summary>radar</summary>
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| Vertex | X | Y |
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| 1 | 20 | 30 |
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| 2 | 80 | 30 |
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| 3 | 50 | 120 |
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</details>
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(a)
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<details>
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<summary>text_image</summary>
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y
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(15, 10)
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3
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1
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(10, 7.5)
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2
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(15, 5)
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x
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</details>
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(b)
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<details>
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<summary>radar</summary>
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| Vertex | x | y |
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|---|---|---|
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| 1 | 0 | 0 |
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| 2 | 10 | 0 |
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| 3 | 5 | 10 |
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| 4 | 0 | 0 |
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</details>
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Figure P6–6
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6.8 Determine the von Mises stress for problem 6.7
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6.9 For the plane strain elements shown in Figure P6–9, the nodal displacements are given as
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$$
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u _ {1} = 0. 0 0 1 \text { in. } \quad v _ {1} = 0. 0 0 5 \text { in. } \quad u _ {2} = 0. 0 0 1 \text { in. }
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$$
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$$
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v _ {2} = 0. 0 0 2 5 \text { in. } \quad u _ {3} = 0. 0 \text { in. } \quad v _ {3} = 0. 0 \text { in. }
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$$
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Determine the element stresses $\sigma _ { x } , \sigma _ { y } , \tau _ { x y } , \sigma _ { 1 }$ , and $\sigma _ { 2 }$ and the principal angle $\theta _ { p }$ . Let $E = 3 0 \times 1 0 ^ { 6 }$ psi and $\nu = 0 . 2 5$ , and use unit thickness for plane strain. All coordinates are in inches.
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<!-- source-page: 363 -->
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<details>
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<summary>text_image</summary>
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(2, 5)
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3
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1
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2
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(2, 1)
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(4, 1)
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x
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(a)
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</details>
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<details>
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<summary>text_image</summary>
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y
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(4, 4)
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3
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1
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2
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(2, 2)
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(4, 2)
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x
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(b)
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</details>
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<details>
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<summary>line</summary>
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| Point | x | y |
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| (0, 0) | 0 | 0 |
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| (0, 2) | 2 | 2 |
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| (2, 0) | 2 | 0 |
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</details>
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<details>
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<summary>text_image</summary>
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y
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(4, 5)
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3
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1
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(2, 3)
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2
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(4, 1)
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x
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(d)
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</details>
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<details>
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<summary>text_image</summary>
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y
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(2.25, 3.25)
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(2, 3)
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3
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1
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2
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(3, 1)
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x
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(e)
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</details>
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<details>
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<summary>text_image</summary>
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y
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(1, 2)
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(0, 0)
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(2, 0)
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x
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(f)
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</details>
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Figure P6–9
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6.10 For the plane strain elements shown in Figure P6–10, the nodal displacements are given as
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$$
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u _ {1} = 0. 0 0 5 \mathrm{mm} \quad v _ {1} = 0. 0 0 2 \mathrm{mm} \quad u _ {2} = 0. 0 \mathrm{mm}
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$$
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$$
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v _ {2} = 0. 0 \mathrm{mm} \quad u _ {3} = 0. 0 0 5 \mathrm{mm} \quad v _ {3} = 0. 0 \mathrm{mm}
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$$
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Determine the element stresses $\sigma _ { x } , \sigma _ { y } , \tau _ { x y } , \sigma _ { 1 }$ , and $\sigma _ { 2 }$ and the principal angle $\theta _ { p }$ . Let E ¼ 70 GPa and n ¼ 0:3, and use unit thickness for plane strain. All coordinates are in millimeters.
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6.11 Determine the nodal forces for (a) a linearly varying pressure $p _ { x }$ on the edge of the triangular element shown in Figure P6–11(a); and (b) the quadratic varying pressure shown in Figure P6–11(b) by evaluating the surface integral given by Eq. (6.3.7). Assume the element thickness is equal to t.
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6.12 Determine the nodal forces for (a) the quadratic varying pressure loading shown in Figure P6–12(a) and the sinusoidal varying pressure loading shown in Figure P6–12(b)
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<!-- source-page: 364 -->
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<details>
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<summary>radar</summary>
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| Vertex | X | Y |
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| 1 | 10 | 10 |
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| 2 | 20 | 10 |
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| 3 | 10 | 25 |
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</details>
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(a)
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<details>
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<summary>text_image</summary>
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y
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(20, 20)
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3
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1
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(5, 5)
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2
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(20, 5)
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x
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</details>
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(b)
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<details>
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<summary>radar</summary>
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| Point | x | y |
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| 1 | 5 | 15 |
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| 2 | 25 | 5 |
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| 3 | 15 | 15 |
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</details>
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(c)
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<details>
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<summary>radar</summary>
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| Vertex | X | Y |
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| 1 | (5, 15) | |
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| 2 | (15, 5) | |
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| 3 | (25, 15) | |
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| 4 | (5, 15) | |
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</details>
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Figure P6–10
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<details>
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<summary>text_image</summary>
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y
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3
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p0
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L
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px
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a
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1
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2
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x
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</details>
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(a)
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<details>
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<summary>text_image</summary>
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p₀
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3
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L
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pₓ = p₀ y² / L²
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a
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1 2
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</details>
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(b)
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Figure P6–11
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<details>
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<summary>text_image</summary>
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y
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p(x)
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p2
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p3
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p1
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1
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2
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3
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L/2
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L/2
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x
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</details>
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(a)
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<details>
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<summary>text_image</summary>
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p = p₀ sin πx/L
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p₀
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1
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L
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2
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x
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</details>
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Figure P6–12
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<!-- source-page: 365 -->
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by the work equivalence method (use the surface integral expression given by Eq. (6.3.7)). Assume the element thickness to be t.
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6.13 Determine the nodal displacements and the element stresses, including principal stresses, for the thin plate of Section 6.5 with a uniform shear load (instead of a tensile load) acting on the right edge, as shown in Figure P6–13. Use $E = 3 0 \times 1 0 ^ { 6 }$ psi, $\nu = 0 . 3 0 ,$ and t ¼ 1 in. (Hint: The ½K matrix derived in Section 6.5 and given by Eq. (6.5.22) can be used to solve the problem.)
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<details>
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<summary>text_image</summary>
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10 in.
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2
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3
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s = 1000 lb /in.
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1
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4
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20 in.
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</details>
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Figure P6–13
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6.14 Determine the nodal displacements and the element stresses, including principal stresses, due to the loads shown for the thin plates in Figure P6–14. Use E ¼ 210 GPa, $\nu = 0 . 3 0 ,$ and t ¼ 5 mm. Assume plane stress conditions apply. The recommended discretized plates are shown in the figures.
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6.15 Evaluate the body force matrix for the plates shown in Figures P6–14(a) and (c). Assume the weight density to be 77.1 kN/m3.
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6.16 Why is the triangular stiffness matrix derived in Section 6.2 called a constant strain triangle?
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6.17 How do the stresses vary within the constant strain triangle element?
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6.18 Can you use the plane stress or plane strain element to model the following:
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a. a flat slab floor of a building
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b. a wall subjected to wind loading (the wall acts as a shear wall)
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c. a tensile plate with a hole drilled through it
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d. an eyebar
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e. a soil mass subjected to a strip footing loading
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f. a wrench subjected to a force in the plane of the wrench
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g. a wrench subjected to twisting forces (the twisting forces act out of the plane of the wrench)
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h. a triangular plate connection with loads in the plane of the triangle
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i. a triangular plate connection with out-of-plane loads
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6.19 The plane stress element only allows for in-plane displacements, while the frame or beam element resists displacements and rotations. How can we combine the plane stress and beam elements and still insure compatibility?
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<!-- source-page: 366 -->
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<details>
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<summary>text_image</summary>
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4
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3
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250 mm
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5
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1
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2
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30°
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500 mm
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40 kN
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(a)
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</details>
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<details>
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<summary>text_image</summary>
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20 kN
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4
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3
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100 mm
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5
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100 mm
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20 kN
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(b)
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</details>
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<details>
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<summary>text_image</summary>
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400 mm
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4
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3
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5
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30 kN
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400 mm
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1
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2
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(c)
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</details>
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<details>
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<summary>text_image</summary>
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px = 10 MPa
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4
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3
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5
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400 mm
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1
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2
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400 mm
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(d)
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</details>
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Figure P6–14
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6.20 For the plane structures modeled by triangular elements shown in Figure P6–20, show that numbering in the direction that has fewer nodes, as in Figure P6–20(a) (as opposed to numbering in the direction that has more nodes), results in a reduced bandwidth. Illustrate this fact by filling in, with $X ^ { \mathfrak { s } } ,$ the occupied elements in K for each mesh, as was done in Appendix B.4. Compare the bandwidths for each case.
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<details>
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<summary>text_image</summary>
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7
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5
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3
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1
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8
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6
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4
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2
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(a)
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</details>
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<details>
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<summary>text_image</summary>
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4
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8
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3
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7
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2
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6
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1
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5
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(b)
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</details>
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Figure P6–20
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<!-- source-page: 367 -->
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6.21 Go through the detailed steps to evaluate Eq. (6.3.6).
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6.22 How would you treat a linearly varying thickness for a three-noded triangle?
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6.23 Compute the stiffness matrix of element 1 of the two-triangle element model of the rectangular plate in plane stress shown in Figure P6–23. Then use it to compute the stiffness matrix of element 2.
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<details>
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<summary>text_image</summary>
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y
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4
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3
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①
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②
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h
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1
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b
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2
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x
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</details>
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Figure P6–23
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<!-- source-page: 368 -->
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# Practical Considerations in Modeling; Interpreting Results; and Examples of Plane Stress=Strain Analysis
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# Introduction
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In this chapter, we will describe some modeling guidelines, including generally recommended mesh size, natural subdivisions modeling around concentrated loads, and more on use of symmetry and associated boundary conditions. This is followed by discussion of equilibrium, compatibility, and convergence of solution. We will then consider interpretation of stress results.
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Next, we introduce the concept of static condensation, which enables us to apply the concept of the basic constant-strain triangle stiffness matrix to a quadrilateral element. Thus, both three-sided and four-sided two-dimensional elements can be used in the finite element models of actual bodies.
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We then show some computer program results. A computer program facilitates the solution of complex, large-number-of-degrees-of-freedom plane stress/plane strain problems that generally cannot be solved longhand because of the larger number of equations involved. Also, problems for which longhand solutions do not exist (such as those involving complex geometries and complex loads or where unrealistic, often gross, assumptions were previously made to simplify the problem to allow it to be described via a classical differential equation approach) can now be solved with a higher degree of confidence in the results by using the finite element approach (with its resulting system of algebraic equations).
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# 7.1 Finite Element Modeling
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We will now discuss various concepts that should be considered when modeling any problem for solution by the finite element method.
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<!-- source-page: 369 -->
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# General Considerations
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Finite element modeling is partly an art guided by visualizing physical interactions taking place within the body. One appears to acquire good modeling techniques through experience and by working with experienced people. General-purpose programs provide some guidelines for specific types of problems [12, 15]. In subsequent parts of this section, some significant concepts that should be considered are described.
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In modeling, the user is first confronted with the sometimes difficult task of understanding the physical behavior taking place and understanding the physical behavior of the various elements available for use. Choosing the proper type of element or elements to match as closely as possible the physical behavior of the problem is one of the numerous decisions that must be made by the user. Understanding the boundary conditions imposed on the problem can, at times, be a difficult task. Also, it is often difficult to determine the kinds of loads that must be applied to a body and their magnitudes and locations. Again, working with more experienced users and searching the literature can help overcome these difficulties.
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# Aspect Ratio and Element Shapes
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The aspect ratio is defined as the ratio of the longest dimension to the shortest dimension of a quadrilateral element. In many cases, as the aspect ratio increases, the inaccuracy of the solution increases. To illustrate this point, Figure 7–1(a) shows five different finite element models used to analyze a beam subjected to bending. The element used here is the rectangular one described in Section 10.2. Figure 7–1(b) is a plot of the resulting error in the displacement at point A of the beam versus the aspect ratio. Table 7–1 reports a comparison of results for the displacements at points A and B for the five models, and the exact solution [2].
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There are exceptions for which aspect ratios approaching 50 still produce satisfactory results; for example, if the stress gradient is close to zero at some location of the actual problem, then large aspect ratios at that location still produce reasonable results.
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In general, an element yields best results if its shape is compact and regular. Although different elements have different sensitivities to shape distortions, try to maintain (1) aspect ratios low as in Figure 7–1, cases 1 and 2, and (2) corner angles of quadrilaterals near 90. Figure 7–2 shows elements with poor shapes that tend to promote poor results. If few of these poor element shapes exist in a model, then usually only results near these elements are poor. In the Algor program [12], when a X 170 in Figure 7–2(c), the program automatically divides the quadrilateral into two triangles.
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# Use of Symmetry
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The appropriate use of symmetry\* will often expedite the modeling of a problem. Use of symmetry allows us to consider a reduced problem instead of the actual problem.
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<details>
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<summary>text_image</summary>
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y, v
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24 in.
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8 in.
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x, u
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48 in.
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40,000-lb
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total shear
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load
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Parabolic
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load distribution
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</details>
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$$
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E = 3 0 \times 1 0 ^ {6} \mathrm{psi}
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$$
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$$
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v = 0. 3
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$$
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$$
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t = 1. 0 \text { in. }
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$$
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<details>
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<summary>text_image</summary>
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12 in. × ½ in.
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elements (typical)
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</details>
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(5) AR = 24
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<details>
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<summary>text_image</summary>
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6 in. × 1 in.
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elements (typical)
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</details>
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(4) AR =6
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<details>
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<summary>text_image</summary>
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4.8 in. × 1½ in.
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elements (typical)
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||
</details>
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(3)AR=3.6
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<details>
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||
<summary>text_image</summary>
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3 in. × 2 in.
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elements (typical)
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||
</details>
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(2) AR = 1.5
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<details>
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||
<summary>text_image</summary>
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||
2.4 in. × 2½/3 in.
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elements (typical)
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(1) AP = 1.1
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</details>
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(1)AR=1.1
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Figure 7–1 (a) Beam with loading; effects of the aspect ratio (AR) illustrated by five cases with different aspect ratios
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Thus, we can use a finer subdivision of elements with less labor and computer costs. For another discussion on the use of symmetry, see Reference [3].
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Figures 7–3—7–5 illustrate the use of symmetry in modeling (1) a soil mass subjected to foundation loading, (2) a uniaxially loaded member with a fillet, and (3) a plate with a hole subjected to internal pressure. Note that at the plane of symmetry the displacement in the direction perpendicular to the plane must be equal to zero. This is modeled by the rollers at nodes 2–6 in Figure 7–3, where the plane of symmetry is the vertical plane passing through nodes 1–6, perpendicular to the plane of the model. In Figures $7 { - } 4 ( \mathrm { a } )$ and 7–5(a), there are two planes of symmetry. Thus, we need model only one-fourth of the actual members, as shown in Figures 7–4(b) and 7–5(b).
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