김경종
573 lines
23 KiB
Markdown
573 lines
23 KiB
Markdown
<!-- source-page: 431 -->
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<details>
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<summary>text_image</summary>
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z
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m
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i
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j
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θ
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r
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</details>
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Figure 9–1 Typical axisymmetric element ijm
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directions, respectively. Triangular torus elements are often used to idealize the axisymmetric system because they can be used to simulate complex surfaces and are simple to work with. For instance, the axisymmetric problem of a semi-infinite half-space loaded by a circular area (circular footing) shown in Figure 9–2(a), the domed pressure vessel shown in Figure 9–2(b), and the engine valve stem shown in Figure 9–2(c) can be solved using the axisymmetric element developed in this chapter.
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<details>
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<summary>text_image</summary>
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z, w
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Footing load
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r, u
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Soil mass
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</details>
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<details>
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<summary>text_image</summary>
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Load
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θ
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r
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</details>
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Plan view
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(a) soil mass
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<details>
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<summary>natural_image</summary>
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3D illustration of a capsule-shaped object with internal cross-section, enclosed by intersecting planes (no text or symbols)
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</details>
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(b) domed vessel
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<details>
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<summary>natural_image</summary>
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Cross-sectional illustration of a mechanical component with textured surface (no text or symbols)
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</details>
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(c) engine valve stem
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Figure 9-2 Examples of axisymmetric problems: (a) semi-infinite half-space (soil mass) modeled by axisymmetric elements, (b) a domed pressure vessel, and (c) an engine valve stem
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<details>
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<summary>text_image</summary>
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y
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C
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A
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B
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D
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u + \u03cfracu \u03cfracr dr
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d\u03c9
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r
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u
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\u03c9
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θ
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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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z
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C
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A
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D
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B
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dz
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E
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dr
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F
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r
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dθ
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θ
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x
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</details>
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(b)
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Figure 9–3 (a) Plane cross section of (b) axisymmetric element
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Because of symmetry about the z axis, the stresses are independent of the y coordinate. Therefore, all derivatives with respect to $\theta$ vanish, and the displacement component v (tangent to the y direction), the shear strains $\gamma _ { r \theta }$ and $\gamma _ { \theta z }$ , and the shear stresses $\tau _ { r \theta }$ and $\tau _ { \theta z }$ are all zero.
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Figure 9–3 shows an axisymmetric ring element and its cross section to represent the general state of strain for an axisymmetric problem. It is most convenient to express the displacements of an element ABCD in the plane of a cross section in cylindrical coordinates. We then let u and w denote the displacements in the radial and longitudinal directions, respectively. The side AB of the element is displaced an amount $u ,$ and side CD is then displaced an amount $\boldsymbol { u } + ( \boldsymbol { \hat { o } } \boldsymbol { u } / \boldsymbol { \hat { o } } \boldsymbol { r } )$ dr in the radial direction. The normal strain in the radial direction is then given by
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$$
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\varepsilon_ {r} = \frac {\partial u}{\partial r} \tag {9.1.1a}
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$$
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In general, the strain in the tangential direction depends on the tangential displacement v and on the radial displacement u. However, for axisymmetric deformation behavior, recall that the tangential displacement v is equal to zero. Hence, the tangential strain is due only to the radial displacement. Having only radial displacement u, the new length of the arc $\widehat { A B }$ is $( r + u ) d \theta _ { : }$ , and the tangential strain is then given by
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$$
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\varepsilon_ {\theta} = \frac {(r + u) d \theta - r d \theta}{r d \theta} = \frac {u}{r} \tag {9.1.1b}
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$$
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Next, we consider the longitudinal element BDEF to obtain the longitudinal strain and the shear strain. In Figure 9–4, the element is shown to displace by amounts u and w in the radial and longitudinal directions at point E, and to displace additional amounts $( \hat { o } w / \hat { o } z )$ dz along line BE and $( \hat { o } u / \hat { o } r )$ dr along line EF. Furthermore, observing lines EF and BE, we see that point F moves upward an amount $( \hat { o } w / \hat { o } r )$ dr with respect to point E and point B moves to the right an amount $( \hat { o } u / \hat { o } z )$ dz with respect to point E. Again, from the basic definitions of normal and shear strain, we have the longitudinal normal strain given by
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$$
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\varepsilon_ {z} = \frac {\partial w}{\partial z} \tag {9.1.1c}
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$$
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and the shear strain in the r-z plane given by
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$$
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\gamma_ {r z} = \frac {\partial u}{\partial z} + \frac {\partial w}{\partial r} \tag {9.1.1d}
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$$
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<details>
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<summary>text_image</summary>
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z, w
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∂u/∂z dz
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w + ∂w/∂z dz
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dz
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w
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E
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u
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dr
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r, u
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B
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D
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F
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∂w/∂r dr
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u + ∂u/∂r dr
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</details>
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Figure 9–4 Displacement and rotations of lines of element in the r-z plane
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Summarizing the strain/displacement relationships of Eqs. (9.1.1a–d) in one equation for easier reference, we have
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$$
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\varepsilon_ {r} = \frac {\partial u}{\partial r} \quad \varepsilon_ {\theta} = \frac {u}{r} \quad \varepsilon_ {z} = \frac {\partial w}{\partial z} \quad \gamma_ {r z} = \frac {\partial u}{\partial z} + \frac {\partial w}{\partial r} \tag {9.1.1e}
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$$
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The isotropic stress/strain relationship, obtained by simplifying the general stress/strain relationships given in Appendix C, is
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$$
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\left\{ \begin{array}{l} \sigma_ {r} \\ \sigma_ {z} \\ \sigma_ {\theta} \\ \tau_ {r z} \end{array} \right\} = \frac {E}{(1 + \nu) (1 - 2 \nu)} \left[ \begin{array}{c c c c} 1 - \nu & \nu & \nu & 0 \\ \nu & 1 - \nu & \nu & 0 \\ \nu & \nu & 1 - \nu & 0 \\ 0 & 0 & 0 & \frac {1 - 2 \nu}{2} \end{array} \right] \left\{ \begin{array}{l} \varepsilon_ {r} \\ \varepsilon_ {z} \\ \varepsilon_ {\theta} \\ \gamma_ {r z} \end{array} \right\} \tag {9.1.2}
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$$
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The theoretical development follows that of the plane stress/strain problem given in Chapter 6.
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# Step 1 Select Element Type
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An axisymmetric solid is shown discretized in Figure 9–5(a), along with a typical triangular element. The element has three nodes with two degrees of freedom per node (that is, $u _ { i } ,$ wi at node i ). The stresses in the axisymmetric problem are shown in Figure 9–5(b).
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# Step 2 Select Displacement Functions
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The element displacement functions are taken to be
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$$
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u (r, z) = a _ {1} + a _ {2} r + a _ {3} z \tag {9.1.3}
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$$
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$$
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w (r, z) = a _ {4} + a _ {5} r + a _ {6} z
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$$
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so that we have the same linear displacement functions as used in the plane stress, constant-strain triangle. Again, the total number of $\boldsymbol { a } _ { i } { } ^ { \prime } \boldsymbol { \mathbf { s } }$ (six) introduced in the
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<details>
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<summary>text_image</summary>
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z, w
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Axis of
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symmetry
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(u_m, w_m)
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m(r_m, z_m)
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(u_j, w_j)
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j(r_j, z_j)
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i(r_i, z_i)
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(u_i, w_i)
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r, u
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</details>
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(a）Typical slice through an axisymmetric solid discretized into triangular elements
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<details>
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<summary>text_image</summary>
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z
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σz
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τrz
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σr
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r
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σθ
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</details>
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(b） Stresses in the axisymmetric problem
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Figure 9–5 Discretized axisymmetric solid
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displacement functions is the same as the total number of degrees of freedom for the element. The nodal displacements are
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$$
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\{d \} = \left\{ \begin{array}{l} \underline {{d}} _ {i} \\ \underline {{d}} _ {j} \\ \underline {{d}} _ {m} \end{array} \right\} = \left\{ \begin{array}{l} u _ {i} \\ w _ {i} \\ u _ {j} \\ w _ {j} \\ u _ {m} \\ w _ {m} \end{array} \right\} \tag {9.1.4}
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$$
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and u evaluated at node i is
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$$
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u (r _ {i}, z _ {i}) = u _ {i} = a _ {1} + a _ {2} r _ {i} + a _ {3} z _ {i} \tag {9.1.5}
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$$
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Using Eq. (9.1.3), the general displacement function is then expressed in matrix form as
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$$
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\{\psi \} = \left\{ \begin{array}{l} u \\ w \end{array} \right\} = \left\{ \begin{array}{l} a _ {1} + a _ {2} r + a _ {3} z \\ a _ {4} + a _ {5} r + a _ {6} z \end{array} \right\} = \left[ \begin{array}{c c c c c c} 1 & r & z & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & r & z \end{array} \right] \left\{ \begin{array}{l} a _ {1} \\ a _ {2} \\ a _ {3} \\ a _ {4} \\ a _ {5} \\ a _ {6} \end{array} \right\} \tag {9.1.6}
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$$
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Substituting the coordinates of the nodal points shown in Figure 9–5(a) into Eq . (9.1.6), we can solve for the $\boldsymbol { a _ { i } } ^ { \prime } \mathbf { s }$ in a manner similar to that in Section 6.2. The resulting expressions are
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$$
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\left\{ \begin{array}{l} a _ {1} \\ a _ {2} \\ a _ {3} \end{array} \right\} = \left[ \begin{array}{c c c} 1 & r _ {i} & z _ {i} \\ 1 & r _ {j} & z _ {j} \\ 1 & r _ {m} & z _ {m} \end{array} \right] ^ {- 1} \left\{ \begin{array}{l} u _ {i} \\ u _ {j} \\ u _ {m} \end{array} \right\} \tag {9.1.7}
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$$
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and $\left\{ \begin{array} { l } { a _ { 4 } } \\ { a _ { 5 } } \\ { a _ { 6 } } \end{array} \right\} = \left[ \begin{array} { l l l } { 1 } & { r _ { i } } & { z _ { i } } \\ { 1 } & { r _ { j } } & { z _ { j } } \\ { 1 } & { r _ { m } } & { z _ { m } } \end{array} \right] ^ { - 1 } \left\{ \begin{array} { l } { w _ { i } } \\ { w _ { j } } \\ { w _ { m } } \end{array} \right\}$ ð9:1:8Þ rm zm
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Performing the inversion operations in Eqs. (9.1.7) and (9.1.8), we have
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$$
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\left\{ \begin{array}{l} a _ {1} \\ a _ {2} \\ a _ {3} \end{array} \right\} = \frac {1}{2 A} \left[ \begin{array}{c c c} \alpha_ {i} & \alpha_ {j} & \alpha_ {m} \\ \beta_ {i} & \beta_ {j} & \beta_ {m} \\ \gamma_ {i} & \gamma_ {j} & \gamma_ {m} \end{array} \right] \left\{ \begin{array}{l} u _ {i} \\ u _ {j} \\ u _ {m} \end{array} \right\} \tag {9.1.9}
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$$
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and
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$$
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\left\{ \begin{array}{l} a _ {4} \\ a _ {5} \\ a _ {6} \end{array} \right\} = \frac {1}{2 A} \left[ \begin{array}{c c c} \alpha_ {i} & \alpha_ {j} & \alpha_ {m} \\ \beta_ {i} & \beta_ {j} & \beta_ {m} \\ \gamma_ {i} & \gamma_ {j} & \gamma_ {m} \end{array} \right] \left\{ \begin{array}{l} w _ {i} \\ w _ {j} \\ w _ {m} \end{array} \right\} \tag {9.1.10}
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$$
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where
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$$
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\alpha_ {i} = r _ {j} z _ {m} - z _ {j} r _ {m} \qquad \alpha_ {j} = r _ {m} z _ {i} - z _ {m} r _ {i} \qquad \alpha_ {m} = r _ {i} z _ {j} - z _ {i} r _ {j}
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$$
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$$
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\beta_ {i} = z _ {j} - z _ {m} \quad \beta_ {j} = z _ {m} - z _ {i} \quad \beta_ {m} = z _ {i} - z _ {j} \tag {9.1.11}
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$$
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$$
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\gamma_ {i} = r _ {m} - r _ {j} \quad \gamma_ {j} = r _ {i} - r _ {m} \quad \gamma_ {m} = r _ {j} - r _ {i}
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$$
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We define the shape functions, similar to Eqs. (6.2.18), as
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$$
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N _ {i} = \frac {1}{2 A} \left(\alpha_ {i} + \beta_ {i} r + \gamma_ {i} z\right)
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$$
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$$
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N _ {j} = \frac {1}{2 A} \left(\alpha_ {j} + \beta_ {j} r + \gamma_ {j} z\right) \tag {9.1.12}
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$$
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$$
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N _ {m} = \frac {1}{2 A} (\alpha_ {m} + \beta_ {m} r + \gamma_ {m} z)
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$$
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Substituting Eqs. (9.1.7) and (9.1.8) into Eq. (9.1.6), along with the shape function Eqs. (9.1.12), we find that the general displacement function is
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$$
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\{\psi \} = \left\{ \begin{array}{l} u (r, z) \\ w (r, z) \end{array} \right\} = \left[ \begin{array}{c c c c c c} N _ {i} & 0 & N _ {j} & 0 & N _ {m} & 0 \\ 0 & N _ {i} & 0 & N _ {j} & 0 & N _ {m} \end{array} \right] \left\{ \begin{array}{l} u _ {i} \\ w _ {i} \\ u _ {j} \\ w _ {j} \\ u _ {m} \\ w _ {m} \end{array} \right\} \tag {9.1.13}
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$$
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or $\{ \psi \} = [ N ] \{ d \}$
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# Step 3 Define the Strain=Displacement and Stress=Strain Relationships
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When we use Eqs. (9.1.1) and (9.1.3), the strains become
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$$
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\{\varepsilon \} = \left\{ \begin{array}{c} a _ {2} \\ a _ {6} \\ \frac {a _ {1}}{r} + a _ {2} + \frac {a _ {3} z}{r} \\ a _ {3} + a _ {5} \end{array} \right\} \tag {9.1.15}
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$$
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Rewriting Eq. (9.1.15) with the $\boldsymbol { a _ { i } } ^ { \prime } \mathbf { s }$ as a separate column matrix, we have
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$$
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\left\{ \begin{array}{l} \varepsilon_ {r} \\ \varepsilon_ {z} \\ \varepsilon_ {\theta} \\ \gamma_ {r z} \end{array} \right\} = \left[ \begin{array}{c c c c c c} 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ \frac {1}{r} & 1 & \frac {z}{r} & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 \end{array} \right] \left\{ \begin{array}{l} a _ {1} \\ a _ {2} \\ a _ {3} \\ a _ {4} \\ a _ {5} \\ a _ {6} \end{array} \right\} \tag {9.1.16}
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$$
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Substituting Eqs. (9.1.7) and (9.1.8) into Eq. (9.1.16) and making use of Eq. (9.1.11), we obtain
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$$
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\{\varepsilon \} = \frac {1}{2 A} \left[ \begin{array}{c c c c c c} \beta_ {i} & 0 & \beta_ {j} & 0 & \beta_ {m} & 0 \\ 0 & \gamma_ {i} & 0 & \gamma_ {j} & 0 & \gamma_ {m} \\ \frac {\alpha_ {i}}{r} + \beta_ {i} + \frac {\gamma_ {i} z}{r} & 0 & \frac {\alpha_ {j}}{r} + \beta_ {j} + \frac {\gamma_ {j} z}{r} & 0 & \frac {\alpha_ {m}}{r} + \beta_ {m} + \frac {\gamma_ {m} z}{r} & 0 \\ \gamma_ {i} & \beta_ {i} & \gamma_ {j} & \beta_ {j} & \gamma_ {m} & \beta_ {m} \end{array} \right] \left\{ \begin{array}{l} u _ {i} \\ w _ {i} \\ u _ {j} \\ w _ {j} \\ u _ {m} \\ w _ {m} \end{array} \right\} \tag {9.1.17}
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$$
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or, rewriting Eq. (9.1.17) in simplified matrix form,
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$$
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\{\varepsilon \} = \left[ \begin{array}{l l l} \underline {{B}} _ {i} & \underline {{B}} _ {j} & \underline {{B}} _ {m} \end{array} \right] \left\{ \begin{array}{l} u _ {i} \\ w _ {i} \\ u _ {j} \\ w _ {j} \\ u _ {m} \\ w _ {m} \end{array} \right\} \tag {9.1.18}
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$$
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where $[ B _ { i } ] = \frac { 1 } { 2 A } \left[ \begin{array} { c c c } { \beta _ { i } } & { 0 } \\ { 0 } & { \gamma _ { i } } \\ { \frac { \alpha _ { i } } { r } + \beta _ { i } + \frac { \gamma _ { i } z } { r } } & { 0 } \\ { \gamma _ { i } } & { \beta _ { i } } \end{array} \right]$ ½B i ¼ 2 A ð9:1:19Þ
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Similarly, we obtain submatrices $\underline { { B } } _ { j }$ and $\underline { { B } } _ { m }$ by replacing the subscript i with j and then with m in Eq. (9.1.19). Rewriting Eq. (9.1.18) in compact matrix form, we have
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$$
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\{\varepsilon \} = [ B ] \{d \} \tag {9.1.20}
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$$
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where $\left[ B \right] = \left[ { \underline { { B } } } _ { i } \quad { \underline { { B } } } _ { j } \quad { \underline { { B } } } _ { m } \right]$ ð9:1:21Þ
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Note that $[ B ]$ is a function of the r and z coordinates. Therefore, in general, the strain $\varepsilon _ { \theta }$ will not be constant.
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The stresses are given by
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$$
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\{\sigma \} = [ D ] [ B ] \{d \} \tag {9.1.22}
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$$
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<!-- source-page: 437 -->
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where ½D is given by the first matrix on the right side of Eq. (9.1.2). (As mentioned in Chapter 6, for $\nu = 0 . 5$ , a special formula must be used; see Reference [9].)
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# Step 4 Derive the Element Stiffness Matrix and Equations
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The stiffness matrix is
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$$
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[ k ] = \iint_ {V} [ B ] ^ {T} [ D ] [ B ] d V \tag {9.1.23}
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$$
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or
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$$
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[ k ] = 2 \pi \iint_ {A} [ B ] ^ {T} [ D ] [ B ] r d r d z \tag {9.1.24}
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$$
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after integrating along the circumferential boundary. The ½B matrix, Eq. (9.1.21), is a function of r and z. Therefore, ½k is a function of r and z and is of order $6 \times 6 .$ .
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We can evaluate Eq. (9.1.24) for ½k by one of three methods:
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1. Numerical integration (Gaussian quadrature) as discussed in Chapter 10.
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2. Explicit multiplication and term-by-term integration [1].
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3. Evaluate ½B for a centroidal point ðr; zÞ of the element
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$$
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r = \bar {r} = \frac {r _ {i} + r _ {j} + r _ {m}}{3} \quad z = \bar {z} = \frac {z _ {i} + z _ {j} + z _ {m}}{3} \tag {9.1.25}
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$$
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and define $[ B ( \bar { r } , \bar { z } ) ] = [ \bar { B } ]$ . Therefore, as a first approximation,
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$$
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[ k ] = 2 \pi \bar {r} A [ \bar {B} ] ^ {T} [ D ] [ \bar {B} ] \tag {9.1.26}
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$$
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If the triangular subdivisions are consistent with the final stress distribution (that is, small elements in regions of high stress gradients), then acceptable results can be obtained by method 3.
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# Distributed Body Forces
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Loads such as gravity (in the direction of the z axis) or centrifugal forces in rotating machine parts (in the direction of the r axis) are considered to be body forces (as shown in Figure 9–6). The body forces can be found by
|
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$$
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\left\{f _ {b} \right\} = 2 \pi \iint_ {A} [ N ] ^ {T} \left\{ \begin{array}{l} R _ {b} \\ Z _ {b} \end{array} \right\} r d r d z \tag {9.1.27}
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$$
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||
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||
<details>
|
||
<summary>text_image</summary>
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||
|
||
z
|
||
ω
|
||
Zb
|
||
Rh
|
||
r
|
||
</details>
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|
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Figure 9–6 Axisymmetric element with body forces per unit volume
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<!-- source-page: 438 -->
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where $R _ { b } = \omega ^ { 2 } \rho r$ for a machine part moving with a constant angular velocity o about the z axis, with material mass density $\rho$ and radial coordinate $r ,$ and where $Z _ { b }$ is the body force per unit volume due to the force of gravity.
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Considering the body force at node i, we have
|
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|
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$$
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\left\{f _ {b i} \right\} = 2 \pi \iint_ {A} \left[ N _ {i} \right] ^ {T} \left\{ \begin{array}{l} R _ {b} \\ Z _ {b} \end{array} \right\} r d r d z \tag {9.1.28}
|
||
$$
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where ½Ni T ¼ Ni 00 Ni $\left[ N _ { i } \right] ^ { T } = \left[ { N _ { i } \begin{array} { c c } { N _ { i } } & { 0 } \\ { 0 } & { N _ { i } } \end{array} } \right]$ ð9:1:29Þ
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Multiplying and integrating in Eq. (9.1.28), we obtain
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$$
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\{f _ {b i} \} = \frac {2 \pi}{3} \left\{ \begin{array}{l} \overline {{R}} _ {b} \\ Z _ {b} \end{array} \right\} A \bar {r} \tag {9.1.30}
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$$
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where the origin of the coordinates has been taken as the centroid of the element, and $\overline { { R } } _ { b }$ is the radially directed body force per unit volume evaluated at the centroid of the element. The body forces at nodes j and m are identical to those given by Eq. (9.1.30) for node i. Hence, for an element, we have
|
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|
||
$$
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\left\{f _ {b} \right\} = \frac {2 \pi \bar {r} A}{3} \left\{ \begin{array}{l} \bar {R} _ {b} \\ Z _ {b} \\ \bar {R} _ {b} \\ Z _ {b} \\ \bar {R} _ {b} \\ Z _ {b} \end{array} \right\} \tag {9.1.31}
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||
$$
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where $\overline { { R } } _ { b } = \omega ^ { 2 } \rho \bar { r }$ ð9:1:32Þ
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|
||
Equation (9.1.31) is a first approximation to the radially directed body force distribution.
|
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|
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# Surface Forces
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||
|
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Surface forces can be found by
|
||
|
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$$
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||
\{f _ {s} \} = \iint_ {S} \left[ N _ {s} \right] ^ {T} \{T \} d S \tag {9.1.33}
|
||
$$
|
||
|
||
where again $[ N _ { s } ]$ denotes the shape function matrix evaluated along the surface where the surface traction acts.
|
||
|
||
For radial and axial pressures $p _ { r }$ and $p _ { z }$ , respectively, we have
|
||
|
||
$$
|
||
\left\{f _ {s} \right\} = \iint_ {S} \left[ N _ {s} \right] ^ {T} \left\{ \begin{array}{l} p _ {r} \\ p _ {z} \end{array} \right\} d S \tag {9.1.34}
|
||
$$
|
||
|
||
For example, along the vertical face jm of an element, let uniform loads $p _ { r }$ and $p _ { z }$ be applied, as shown in Figure 9–7 along surface $r = r _ { j }$ . We can use Eq. (9.1.34)
|
||
|
||
<!-- source-page: 439 -->
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
m
|
||
i
|
||
j
|
||
p_r
|
||
p_z
|
||
</details>
|
||
|
||
Figure 9–7 Axisymmetric element with surface forces
|
||
|
||
written for each node separately. For instance, for node $j ,$ substituting $N _ { j }$ from Eqs. (9.1.12) into Eq. (9.1.34), we have
|
||
|
||
$$
|
||
\{f _ {s j} \} = \int_ {z _ {j}} ^ {z _ {m}} \frac {1}{2 A} \left[ \begin{array}{c c} \alpha_ {j} + \beta_ {j} r + \gamma_ {j} z & 0 \\ 0 & \alpha_ {j} + \beta_ {j} r + \gamma_ {j} z \end{array} \right] \left| \left\{ \begin{array}{l} p _ {r} \\ p _ {z} \end{array} \right\} 2 \pi r _ {j} d z \right. \tag {9.1.35}
|
||
$$
|
||
|
||
evaluated at $r = r _ { j } , \ z = z$
|
||
|
||
Performing the integration of Eq. (9.1.35) explicitly, along with similar evaluations for $f _ { s i }$ and $f _ { s m } ,$ we obtain the total distribution of surface force to nodes $i , j ,$ and m as
|
||
|
||
$$
|
||
\left\{f _ {s} \right\} = \frac {2 \pi r _ {j} \left(z _ {m} - z _ {j}\right)}{2} \left\{ \begin{array}{l} 0 \\ 0 \\ p _ {r} \\ p _ {z} \\ p _ {r} \\ p _ {z} \end{array} \right\} \tag {9.1.36}
|
||
$$
|
||
|
||
# Steps 5–7
|
||
|
||
Steps 5–7, which involve assembling the total stiffness matrix, total force matrix, and total set of equations; solving for the nodal degrees of freedom; and calculating the element stresses, are analogous to those of Chapter 6 for the CST element, except the stresses are not constant in each element. They are usually determined by one of two methods that we use to determine the LST element stresses. Either we determine the centroidal element stresses, or we determine the nodal stresses for the element and then average them. The latter method has been shown to be more accurate in some cases [2].
|
||
|
||
# Example 9.1
|
||
|
||
For the element of an axisymmetric body rotating with a constant angular velocity o ¼ 100 rev/min as shown in Figure 9–8, evaluate the approximate body force matrix. Include the weight of the material, where the weight density $\rho _ { w }$ is 0.283 lb/in3. The coordinates of the element (in inches) are shown in the figure.
|
||
|
||
We need to evaluate Eq. (9.1.31) to obtain the approximate body force matrix. Therefore, the body forces per unit volume evaluated at the centroid of the element are
|
||
|
||
$$
|
||
Z _ {b} = 0. 2 8 3 \mathrm{lb} / \mathrm{in} ^ {3}
|
||
$$
|
||
|
||
<!-- source-page: 440 -->
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
(2, 3)
|
||
3
|
||
ω
|
||
r̄ = 2.333 in.
|
||
C
|
||
1
|
||
(2, 2)
|
||
2
|
||
(3, 2)
|
||
Axis of symmetry
|
||
</details>
|
||
|
||
Figure 9–8 Axisymmetric element subjected to angular velocity
|
||
|
||
and by Eq. (9.1.32), we have
|
||
|
||
$$
|
||
\bar {R} _ {b} = \omega^ {2} \rho \bar {r} = \left[ \left(1 0 0 \frac {\mathrm{rev}}{\mathrm{min}}\right) \left(2 \pi \frac {\mathrm{rad}}{\mathrm{rev}}\right) \left(\frac {1 \mathrm{min}}{6 0 \mathrm{s}}\right) \right] ^ {2} \frac {(0 . 2 8 3 \mathrm{lb} / \mathrm{in} ^ {3})}{(3 2 . 2 \times 1 2) \mathrm{in} . / \mathrm{s} ^ {2}} (2. 3 3 3 \mathrm{in}.)
|
||
$$
|
||
|
||
$$
|
||
\bar {R} _ {b} = 0. 1 8 7 \mathrm{lb/in} ^ {3}
|
||
$$
|
||
|
||
$$
|
||
\frac {2 \pi \bar {r} A}{3} = \frac {2 \pi (2 . 3 3 3) (0 . 5)}{3} = 2. 4 4 \mathrm{in} ^ {3}
|
||
$$
|
||
|
||
$$
|
||
f _ {b 1 r} = (2. 4 4) (0. 1 8 7) = 0. 4 5 7 \mathrm{lb}
|
||
$$
|
||
|
||
$$
|
||
f _ {b 1 z} = - (2. 4 4) (0. 2 8 3) = - 0. 6 9 1 \mathrm{lb} \quad (\text { downward })
|
||
$$
|
||
|
||
Because we are using the first approximation Eq. (9.1.31), all r-directed nodal body forces are equal, and all z-directed body forces are equal. Therefore,
|
||
|
||
$$
|
||
f _ {b 2 r} = 0. 4 5 7 \mathrm{lb} \quad f _ {b 2 z} = - 0. 6 9 1 \mathrm{lb}
|
||
$$
|
||
|
||
$$
|
||
f _ {b 3 r} = 0. 4 5 7 \mathrm{lb} \quad f _ {b 3 z} = - 0. 6 9 1 \mathrm{lb}
|
||
$$
|
||
|
||
# 9.2 Solution of an Axisymmetric Pressure Vessel
|
||
|
||
To illustrate the use of the equations developed in Section 9.1, we will now solve an axisymmetric stress problem.
|
||
|
||
# Example 9.2
|
||
|
||
For the long, thick-walled cylinder under internal pressure p equal to 1 psi shown in Figure 9–9, determine the displacements and stresses.
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
p
|
||
</details>
|
||
|
||
Figure 9–9 Thick-walled cylinder subjected to internal pressure
|