김경종
27 KiB
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(-1,1) 1 4 1 1 3 1 1 1 1 (-1,-1) (1,-1) t s
(a)
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Edge t = 1 y, v 4 Edge s = -1 1 (x₁, y₁) Edge t = -1 x, u s = 1/2 3 (x₃, y₃) u P(x, y) Edge s = 1 2 (x₂, y₂) Edge t = -1 t s y, v
(b)
Figure 10–4 (a) Linear square element in s-t coordinates and (b) square element mapped into quadrilateral in x-y coordinates whose size and shape are determined by the eight nodal coordinates x _ { 1 } , y _ { 1 } , \ldots , y _ { 4 }
complicated (higher-order) elements such as a quadratic plane element with three nodes along an edge, which can have straight or quadratic curved sides. Higherorder elements have additional nodes and use di¤erent shape functions as compared to the linear element, but the steps in the development of the sti¤ness matrices are the same. We will briefly discuss these elements after examining the linear plane element formulation.
Step 1 Select Element Type
First, the natural s-t coordinates are attached to the element, with the origin at the center of the element, as shown in Figure 10–4(a). The s and t axes need not be orthogonal, and neither has to be parallel to the x or y axis. The orientation of s-t coordinates is such that the four corner nodes and the edges of the quadrilateral are bounded by þ1 or 1. This orientation will later allow us to take advantage more fully of common numerical integration schemes.
We consider the quadrilateral to have eight degrees of freedom, u _ { 1 } , v _ { 1 } , \dotsc , u _ { 4 } . and v _ { 4 } associated with the global x and y directions. The element then has straight sides but is otherwise of arbitrary shape, as shown in Figure 10–4(b).
For the special case when the distorted element becomes a rectangular element with sides parallel to the global x-y coordinates (see Figure 10–3), the s-t coordinates can be related to the global element coordinates x and y by
x = x _ {c} + b s \quad y = y _ {c} + h t \tag {10.3.1}
where x _ { c } and y _ { c } are the global coordinates of the element centroid.
As we have shown for a rectangular element, the shape functions that define the displacements within the element are given by Eqs. (10.2.5). These same shape functions will now be used to map the square of Figure 10–4(a) in isoparametric coordinates s and t to the quadrilateral of Figure 10–4(b) in x and y coordinates whose size and shape are determined by the eight nodal coordinates x _ { 1 } , y _ { 1 } , \ldots , x _ { 4 } , and y _ { 4 } . That
is, letting
x = a _ {1} + a _ {2} s + a _ {3} t + a _ {4} s t \tag {10.3.2}
y = a _ {5} + a _ {6} s + a _ {7} t + a _ {8} s t
and solving for the \boldsymbol { a _ { i } } ^ { \prime } \mathbf { s } in terms of x _ { 1 } , x _ { 2 } , x _ { 3 } , x _ { 4 } , y _ { 1 } , y _ { 2 } , y _ { 3 } , and y _ { 4 } . , we establish a form similar to Eqs. (10.2.3) such that
\begin{array}{l} x = \frac {1}{4} [ (1 - s) (1 - t) x _ {1} + (1 + s) (1 - t) x _ {2} \\ \left. + (1 + s) (1 + t) x _ {3} + (1 - s) (1 + t) x _ {4} \right] \\ \end{array}
ð10:3:3Þ
y = \frac {1}{4} [ (1 - s) (1 - t) y _ {1} + (1 + s) (1 - t) y _ {2}
\left. + (1 + s) (1 + t) y _ {3} + (1 - s) (1 + t) y _ {4} \right]
\mathrm { o r } , in matrix form, we can express Eqs. (10.3.3) as
\left\{ \begin{array}{l} x \\ y \end{array} \right\} = \left[ \begin{array}{c c c c c c c c} N _ {1} & 0 & N _ {2} & 0 & N _ {3} & 0 & N _ {4} & 0 \\ 0 & N _ {1} & 0 & N _ {2} & 0 & N _ {3} & 0 & N _ {4} \end{array} \right] \left\{ \begin{array}{l} x _ {1} \\ y _ {1} \\ x _ {2} \\ y _ {2} \\ x _ {3} \\ y _ {3} \\ x _ {4} \\ y _ {4} \end{array} \right\} \tag {10.3.4}
where the shape functions of Eq. (10.3.4) are now
N _ {1} = \frac {(1 - s) (1 - t)}{4} \quad N _ {2} = \frac {(1 + s) (1 - t)}{4} \tag {10.3.5}
N _ {3} = \frac {(1 + s) (1 + t)}{4} \quad N _ {4} = \frac {(1 - s) (1 + t)}{4}
The shape functions of Eqs. (10.3.5) are linear. These shape functions are seen to map the s and t coordinates of any point in the square element of Figure 1 0 { - } 4 ( \mathrm { a } ) to those x and y coordinates in the quadrilateral element of Figure 10–4(b). For instance, consider square element node 1 coordinates, where s = - 1 and t = - 1 . Using Eqs. (10.3.4) and (10.3.5), the left side of Eq. (10.3.4) becomes
x = x _ {1} \quad y = y _ {1} \tag {10.3.6}
Similarly, we can map the other local nodal coordinates at nodes 2, 3, and 4 such that the square element in s { - } t isoparametric coordinates is mapped into a quadrilateral element in global coordinates. Also observe the property that N _ { 1 } + N _ { 2 } + N _ { 3 } + N _ { 4 } = 1 for all values of s and t.
We further observe that the shape functions in Eq. (10.3.5) are again such that N _ { 1 } through N _ { 4 } have the properties that N _ { i } \ ( i = 1 , 2 , 3 , 4 ) is equal to one at node i and equal to zero at all other nodes. The physical shapes of N _ { i } as they vary over the element with natural coordinates are shown in Figure 10–5. For instance, N _ { 1 } represents the geometric shape for x _ { 1 } = 1 , y _ { 1 } = 1 , and x _ { 2 } , y _ { 2 } , x _ { 3 } , y _ { 3 } , x _ { 4 } , and y _ { 4 } all equal to zero.
Until this point in the discussion, we have always developed the element shape functions either by assuming some relationship between the natural and global
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N₁ t 1 4 3 1 s 2 1 1 1
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N₂ t 4 3 1 s 1 2 1 1 1
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N₃ t 4 3 1 s 1 2 1
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N₄ t 4 3 1 s 1 2 1 1 1
Figure 10–5 Variations of the shape functions over a linear square element
coordinates in terms of the generalized coordinates ( \boldsymbol { a _ { i } } ^ { \prime } \mathbf { s } ) as in Eqs. (10.3.2) or, similarly, by assuming a displacement function in terms of the \boldsymbol { a } _ { i } { \widetilde { \mathbf { s } } } . However, physical intuition can often guide us in directly expressing shape functions based on the following two criteria set forth in Section 3.2 and used on numerous occasions:
\sum_ {i = 1} ^ {n} N _ {i} = 1 \quad (i = 1, 2, \dots , n)
where n ¼ the number of shape functions corresponding to displacement shape functions N _ { i } , and N _ { i } = 1 at node i and N _ { i } = 0 at all nodes other than i. In addition, a third criterion is based on Lagrangian interpolation when displacement continuity is to be satisfied, or on Hermitian interpolation when additional slope continuity needs to be satisfied, as in the beam element of Chapter 4. (For a description of the use of Lagrangian and Hermitian interpolation to develop shape functions, consult References [4] and [6].)
Step 2 Select Displacement Functions
The displacement functions within an element are now similarly defined by the same shape functions as are used to define the element shape; that is,
\left\{ \begin{array}{l} u \\ v \end{array} \right\} = \left[ \begin{array}{c c c c c c c c} N _ {1} & 0 & N _ {2} & 0 & N _ {3} & 0 & N _ {4} & 0 \\ 0 & N _ {1} & 0 & N _ {2} & 0 & N _ {3} & 0 & N _ {4} \end{array} \right] \left\{ \begin{array}{l} u _ {1} \\ v _ {1} \\ u _ {2} \\ v _ {2} \\ u _ {3} \\ v _ {3} \\ u _ {4} \\ v _ {4} \end{array} \right\} \tag {10.3.7}
where u and v are displacements parallel to the global x and y coordinates, and the shape functions are given by Eqs. (10.3.5). The displacement of an interior point P located at ( x , y ) in the element of Figure 10–4(b) is described by u and v in Eq. (10.3.7).
Comparing Eqs. (10.2.6) and (10.3.7), we see similarities between the rectangular element with sides of lengths 2b and 2h (Figure 10–3) and the square element with sides of length 2. If we let b = 1 and h = 1 , the two sets of shape functions, Eqs. (10.2.5) and (10.3.5), are identical.
Step 3 Define the Strain= Displacement and Stress= Strain Relationships
We now want to formulate element matrix \underline { { B } } to evaluate \underline { { k } } . However, because it becomes tedious and di‰cult (if not impossible) to write the shape functions in terms of the x and y coordinates, as seen in Chapter 8, we will carry out the formulation in terms of the isoparametric coordinates s and t. This may appear tedious, but it is easier to use the s- and t-coordinate expressions than to attempt to use the x- and y-coordinate expressions. This approach also leads to a simple computer program formulation.
To construct an element sti¤ness matrix, we must determine the strains, which are defined in terms of the derivatives of the displacements with respect to the x and y coordinates. The displacements, however, are now functions of the s and t coordinates, as given by Eq. (10.3.7), with the shape functions given by Eqs. (10.3.5). Before, we could determine ( \partial f / \partial x ) and ( \partial f / \partial y ) , where, in general, f is a function representing the displacement functions u or v. However, u and v are now expressed in terms of s and t. Therefore, we need to apply the chain rule of di¤erentiation because it will not be possible to express s and t as functions of x and y directly. For f as a function of x and y , the chain rule yields
\frac {\partial f}{\partial s} = \frac {\partial f}{\partial x} \frac {\partial x}{\partial s} + \frac {\partial f}{\partial y} \frac {\partial y}{\partial s} \tag {10.3.8}
\frac {\partial f}{\partial t} = \frac {\partial f}{\partial x} \frac {\partial x}{\partial t} + \frac {\partial f}{\partial y} \frac {\partial y}{\partial t}
In Eq. (10.3.8), ( \hat { \sigma } f / \hat { \sigma } s ) , ( \hat { \sigma } f / \hat { \sigma } t ) , ( \hat { \sigma } x / \hat { \sigma } s ) , ( \hat { \sigma } y / \hat { \sigma } s ) , ( \hat { \sigma } x / \hat { \sigma } t ) , and ( \partial y / \partial t ) are all known using Eqs. (10.3.7) and (10.3.4). We still seek ( \partial f / \partial x ) and ( \partial f / \partial y ) . The strains can then be found; for example, \varepsilon _ { x } = ( \hat { o } u / \hat { o } x ) . Therefore, we solve Eqs. (10.3.8) for ( \partial f / \partial x ) and ( \partial f / \partial y ) using Cramer’s rule, which involves evaluation of determinants (Appendix B), as
\frac {\partial f}{\partial x} = \frac {\left| \begin{array}{l l} \frac {\partial f}{\partial s} & \frac {\partial y}{\partial s} \\ \frac {\partial f}{\partial t} & \frac {\partial y}{\partial t} \end{array} \right|}{\left| \begin{array}{l l} \frac {\partial x}{\partial s} & \frac {\partial y}{\partial s} \\ \frac {\partial x}{\partial t} & \frac {\partial y}{\partial t} \end{array} \right|} \quad \frac {\partial f}{\partial y} = \frac {\left| \begin{array}{l l} \frac {\partial x}{\partial s} & \frac {\partial f}{\partial s} \\ \frac {\partial x}{\partial t} & \frac {\partial f}{\partial t} \end{array} \right|}{\left| \begin{array}{l l} \frac {\partial x}{\partial s} & \frac {\partial y}{\partial s} \\ \frac {\partial x}{\partial t} & \frac {\partial y}{\partial t} \end{array} \right|} \tag {10.3.9}
where the determinant in the denominator is the determinant of the Jacobian matrix \underline { { J } } . Hence, the Jacobian matrix is given by
[ J ] = \left[ \begin{array}{l l} \frac {\partial x}{\partial s} & \frac {\partial y}{\partial s} \\ \frac {\partial x}{\partial t} & \frac {\partial y}{\partial t} \end{array} \right] \tag {10.3.10}
We now want to express the element strains as
\underline {{\varepsilon}} = \underline {{B}} \underline {{d}} \tag {10.3.11}
where \underline { { B } } must now be expressed as a function of s and t. We start with the usual relationship between strains and displacements given in matrix form as
\left\{ \begin{array}{l} \varepsilon_ {x} \\ \varepsilon_ {y} \\ \gamma_ {x y} \end{array} \right\} = \left[ \begin{array}{c c} \frac {\partial (\quad)}{\partial x} & 0 \\ 0 & \frac {\partial (\quad)}{\partial y} \\ \frac {\partial (\quad)}{\partial y} & \frac {\partial (\quad)}{\partial x} \end{array} \right] \left\{ \begin{array}{l} u \\ v \end{array} \right\} \tag {10.3.12}
where the rectangular matrix on the right side of Eq. (10.3.12) is an operator matrix; that is, \partial ( \mathbf { \xi } ) / \partial x and \partial ( { \bf \theta } ) / \partial y represent the partial derivatives of any variable we put inside the parentheses.
Using Eqs. (10.3.9) and evaluating the determinant in the numerators, we have
\frac {\partial (\quad)}{\partial x} = \frac {1}{| \underline {{{J}}} |} \left[ \frac {\partial y}{\partial t} \frac {\partial (\quad)}{\partial s} - \frac {\partial y}{\partial s} \frac {\partial (\quad)}{\partial t} \right] \tag {10.3.13}
\frac {\partial (\mathbf {\Phi})}{\partial y} = \frac {1}{| \underline {{J}} |} \left[ \frac {\partial x}{\partial s} \frac {\partial (\mathbf {\Phi})}{\partial t} - \frac {\partial x}{\partial t} \frac {\partial (\mathbf {\Phi})}{\partial s} \right]
where | \underline { J } | is the determinant of \underline { { J } } given by Eq. (10.3.10). Using Eq. (10.3.13) in Eq. (10.3.12) we obtain the strains expressed in terms of the natural coordinates (s-t) as
\left\{ \begin{array}{l} \varepsilon_ {x} \\ \varepsilon_ {y} \\ \gamma_ {x y} \end{array} \right\} = \frac {1}{| \underline {{{J}}} |} \left[ \begin{array}{c c} \frac {\partial y}{\partial t} \frac {\partial (\quad)}{\partial s} - \frac {\partial y}{\partial s} \frac {\partial (\quad)}{\partial t} & 0 \\ 0 & \frac {\partial x}{\partial s} \frac {\partial (\quad)}{\partial t} - \frac {\partial x}{\partial t} \frac {\partial (\quad)}{\partial s} \\ \frac {\partial x}{\partial s} \frac {\partial (\quad)}{\partial t} - \frac {\partial x}{\partial t} \frac {\partial (\quad)}{\partial s} & \frac {\partial y}{\partial t} \frac {\partial (\quad)}{\partial s} - \frac {\partial y}{\partial s} \frac {\partial (\quad)}{\partial t} \end{array} \right] \left\{ \begin{array}{l} u \\ v \end{array} \right\} \tag {10.3.14}
Using Eq. (10.3.7), we can express Eq. (10.3.14) in terms of the shape functions and global coordinates in compact matrix form as
\underline {{\varepsilon}} = \underline {{D}} ^ {\prime} \underline {{N}} \underline {{d}} \tag {10.3.15}
where \underline { { \boldsymbol { D } } } ^ { \prime } is an operator matrix given by
\underline {{{D}}} ^ {\prime} = \frac {1}{| \underline {{{J}}} |} \left[ \begin{array}{c c} \frac {\partial y}{\partial t} \frac {\partial (\quad)}{\partial s} - \frac {\partial y}{\partial s} \frac {\partial (\quad)}{\partial t} & 0 \\ 0 & \frac {\partial x}{\partial s} \frac {\partial (\quad)}{\partial t} - \frac {\partial x}{\partial t} \frac {\partial (\quad)}{\partial s} \\ \frac {\partial x}{\partial s} \frac {\partial (\quad)}{\partial t} - \frac {\partial x}{\partial t} \frac {\partial (\quad)}{\partial s} & \frac {\partial y}{\partial t} \frac {\partial (\quad)}{\partial s} - \frac {\partial y}{\partial s} \frac {\partial (\quad)}{\partial t} \end{array} \right] \tag {10.3.16}
and \underline { { N } } is the 2 \times 8 shape function matrix given as the first matrix on the right side of Eq. (10.3.7) and \underline { d } is the column matrix on the right side of Eq. (10.3.7).
Defining B as
\underline {{B}} = \underline {{D}} ^ {\prime} \quad \underline {{N}} \tag {10.3.17}
we have B expressed as a function of s and t and thus have the strains in terms of s and t. Here \underline { { B } } is of order 3 \times 8 . , as indicated in Eq. (10.3.17).
The explicit form of \underline { { B } } can be obtained by substituting Eq. (10.3.16) for \underline { { \boldsymbol { D } } } ^ { \prime } and Eqs. (10.3.5) for the shape functions into Eq. (10.3.17). The matrix multiplications yield
\underline {{{B}}} (s, t) = \frac {1}{| \underline {{{J}}} |} \left[ \begin{array}{l l l l} \underline {{{B}}} _ {1} & \underline {{{B}}} _ {2} & \underline {{{B}}} _ {3} & \underline {{{B}}} _ {4} \end{array} \right] \tag {10.3.18}
where the submatrices of \underline { { B } } are given by
\underline {{B}} _ {i} = \left[ \begin{array}{c c} a (N _ {i, s}) - b (N _ {i, t}) & 0 \\ 0 & c (N _ {i, t}) - d (N _ {i, s}) \\ c (N _ {i, t}) - d (N _ {i, s}) & a (N _ {i, s}) - b (N _ {i, t}) \end{array} \right] \tag {10.3.19}
Here i is a dummy variable equal to 1, 2, 3, and 4, and
\begin{array}{l} a = \frac {1}{4} \left[ y _ {1} (s - 1) + y _ {2} (- 1 - s) + y _ {3} (1 + s) + y _ {4} (1 - s) \right] \\ b = \frac {1}{4} \left[ y _ {1} (t - 1) + y _ {2} (1 - t) + y _ {3} (1 + t) + y _ {4} (- 1 - t) \right] \tag {10.3.20} \\ c = \frac {1}{4} \left[ x _ {1} (t - 1) + x _ {2} (1 - t) + x _ {3} (1 + t) + x _ {4} (- 1 - t) \right] \\ d = \frac {1}{4} \left[ x _ {1} (s - 1) + x _ {2} (- 1 - s) + x _ {3} (1 + s) + x _ {4} (1 - s) \right] \\ \end{array}
Using the shape functions defined by Eqs. (10.3.5), we have
N _ {1, s} = \frac {1}{4} (t - 1) \quad N _ {1, t} = \frac {1}{4} (s - 1) \quad (\text { and so on }) \tag {10.3.21}
where the comma followed by the variable s or t indicates di¤erentiation with respect to that variable; that is, N _ { 1 , s } \equiv \hat { o } N _ { 1 } / \hat { o } s , and so on. The determinant | \underline { J } | is a polynomial in s and t and is tedious to evaluate even for the simplest case of the linear plane element. However, using Eq. (10.3.10) for ½J- and Eqs. (10.3.3) for x and y , we can
evaluate | \underline { J } | as
| \underline {{{J}}} | = \frac {1}{8} \{X _ {c} \} ^ {T} \left[ \begin{array}{c c c c} 0 & 1 - t & t - s & s - 1 \\ t - 1 & 0 & s + 1 & - s - t \\ s - t & - s - 1 & 0 & t + 1 \\ 1 - s & s + t & - t - 1 & 0 \end{array} \right] \{Y _ {c} \} \tag {10.3.22}
where \left\{ X _ { c } \right\} ^ { T } = \left[ x _ { 1 } \quad x _ { 2 } \quad x _ { 3 } \quad x _ { 4 } \right] ð10:3:23Þ
and \left\{ Y _ { c } \right\} = \left\{ \begin{array} { c } { { y _ { 1 } } } \\ { { y _ { 2 } } } \\ { { y _ { 3 } } } \\ { { y _ { 4 } } } \end{array} \right\} ð10:3:24Þ
We observe that \left| \underline { J } \right| is a function of s and t and the known global coordinates x _ { 1 } , x _ { 2 } , \ldots , y _ { 4 } . Hence, \underline { { B } } is a function of s and t in both the numerator and the denominator [because of | \underline { { J } } | given by Eq. (10.3.22)] and of the known global coordinates x _ { 1 } through y _ { 4 } .
The stress/strain relationship is again \underline { { \sigma } } = \underline { { D } } \underline { { B } } \underline { { d } } _ { \mathrm { ~ } } , where because the \underline { { B } } matrix is a function of s and t , so also is the stress matrix \underline { { \sigma } } . .
Step 4 Derive the Element Stiffness Matrix and Equations
We now want to express the sti¤ness matrix in terms of s \ – t coordinates. For an element with a constant thickness h , we have
[ k ] = \iint_ {A} [ B ] ^ {T} [ D ] [ B ] h d x d y \tag {10.3.25}
However, \underline { { B } } is now a function of s and t , as seen by Eqs. (10.3.18)–(10.3.20), and so we must integrate with respect to s and t. Once again, to transform the variables and the region from x and y to s and t , we must have a standard procedure that involves the determinant of \underline { { J } } . . This general type of transformation [4, 5] is given by
\iint_ {A} f (x, y) d x d y = \iint_ {A} f (s, t) | \underline {{{J}}} | d s d t \tag {10.3.26}
where the inclusion of | \underline { J } | in the integrand on the right side of Eq. (10.3.26) results from a theorem of integral calculus (see Reference [ 5 ] for the complete proof of this theorem). Using Eq. (10.3.26) in Eq. (10.3.25), we obtain
[ k ] = \int_ {- 1} ^ {1} \int_ {- 1} ^ {1} [ B ] ^ {T} [ D ] [ B ] h | \underline {{J}} | d s d t \tag {10.3.27}
The | \underline { J } | and \underline { { B } } are such as to result in complicated expressions within the integral of Eq. (10.3.27), and so the integration to determine the element sti¤ness matrix is usually done numerically. A method for numerically integrating Eq. (10.3.27) is given in Section 10.4. The sti¤ness matrix in Eq. (10.3.27) is of the order 8 \times 8 .
Body Forces
The element body-force matrix will now be determined from
\left\{f _ {b} \right\} = \int_ {- 1} ^ {1} \int_ {- 1} ^ {1} \left[ N \right] ^ {T} \left\{X \right\} h | \underline {{{I}}} | d s d t \tag {10.3.28}
Like the sti¤ness matrix, the body-force matrix in Eq. (10.3.28) has to be evaluated by numerical integration.
Surface Forces
The surface-force matrix, say, along edge t ¼ 1 (Figure 10–6) with overall length L, is
\left\{f _ {s} \right\} = \int_ {- 1} ^ {1} \left[ N _ {s} \right] ^ {T} \left\{T \right\} h \frac {L}{2} d s \tag {10.3.29}
\text { or } \quad \left\{ \begin{array}{l} f _ {s 3 s} \\ f _ {s 3 t} \\ f _ {s 4 s} \\ f _ {s 4 t} \end{array} \right\} = \int_ {- 1} ^ {1} \left[ \begin{array}{c c c c} N _ {3} & 0 & N _ {4} & 0 \\ 0 & N _ {3} & 0 & N _ {4} \end{array} \right] ^ {T} \Bigg | _ {\text { evaluated along } t = 1} \left\{ \begin{array}{l} p _ {s} \\ p _ {t} \end{array} \right\} h \frac {L}{2} d s \tag {10.3.30}
because N _ { 1 } = 0 and N _ { 2 } = 0 along edge t ¼ 1, and hence, no nodal forces exist at nodes 1 and 2. For the case of uniform (constant) p _ { s } and p _ { t } along edge t = 1 , the total surface-force matrix is
\left\{f _ {s} \right\} = h \frac {L}{2} \left[ \begin{array}{l l l l l l l l} 0 & 0 & 0 & 0 & p _ {s} & p _ {t} & p _ {s} & p _ {t} \end{array} \right] ^ {T} \tag {10.3.31}
Surface forces along other edges can be obtained similar to Eq. (10.3.30) by merely using the proper shape functions associated with the edge where the tractions are applied.
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(-1, 1) p_s 4 3 t p_t (1, 1) s 1 2 (-1, -1) (1, -1)
Figure 10–6 Surface traction: p _ { s } and p _ { t } acting at edge t = 1
For the four-noded linear plane element shown in Figure 10–7 with a uniform surface traction along side 2–3, evaluate the force matrix by using the energy equivalent nodal forces obtained from the integral similar to Eq. (10.3.29). Let the thickness of the element be h = 0 . 1 in.
line
| Point | x | y |
|---|---|---|
| 1 | 0 | 0 |
| 2 | 8 | 0 |
| 3 | 5 | 4 |
| 4 | 0 | 4 |
Figure 10–7 Element subjected to uniform surface traction
Using Eq. (10.3.29), we have
\{f _ {s} \} = \int_ {- 1} ^ {1} \left[ N _ {s} \right] ^ {T} \{T \} h \frac {L}{2} d t \tag {10.3.32}
With length of side 2–3 given by
L = \sqrt {(5 - 8) ^ {2} + (4 - 0) ^ {2}} = \sqrt {9 + 1 6} = 5 \tag {10.3.33}
Shape functions N _ { 2 } and N _ { 3 } must be used, as we are evaluating the surface traction along side 2 - 3 \ : ( \mathrm { a t } \ : s = 1 ) . Therefore, Eq. (10.3.33) becomes
\{f _ {s} \} = \int_ {- 1} ^ {1} \left[ N _ {s} \right] ^ {T} \{T \} h \frac {L}{2} d t = \int_ {- 1} ^ {1} \left[ \begin{array}{c c c c} N _ {2} & 0 & N _ {3} & 0 \\ 0 & N _ {2} & 0 & N _ {3} \end{array} \right] ^ {T} \left\{ \begin{array}{l} p _ {s} \\ p _ {t} \end{array} \right\} h \frac {L}{2} d t \tag {10.3.34}
evaluated along s ¼ 1
The shape functions for the four-noded linear plane element are taken from Eq. (10.3.5) as
N _ {2} = \frac {(1 + s) (1 - t)}{4} = \frac {s - t - s t + 1}{4} \quad N _ {3} = \frac {(1 + s) (1 + t)}{4} = \frac {s + t + s t + 1}{4} \tag {10.3.35}
The surface traction matrix is given by
\{T \} = \left\{ \begin{array}{l} p _ {s} \\ p _ {t} \end{array} \right\} = \left\{ \begin{array}{c} 2 0 0 0 \\ 0 \end{array} \right\} \tag {10.3.36}
Substituting Eq. (10.3.33) for L and Eq. (10.3.36) for the surface traction matrix and the thickness h ¼ 0:1 inch into Eq. (10.3.32), we obtain
\{f _ {s} \} = \int_ {- 1} ^ {1} \left[ N _ {s} \right] ^ {T} \{T \} h \frac {L}{2} d t = \int_ {- 1} ^ {1} \left[ \begin{array}{c c} N _ {2} & 0 \\ 0 & N _ {2} \\ N _ {3} & 0 \\ 0 & N _ {3} \end{array} \right] \left\{ \begin{array}{c} 2 0 0 0 \\ 0 \end{array} \right\} 0. 1 \frac {5}{2} d t \tag {10.3.37}
evaluated along s ¼ 1
Simplifying Eq. (10.3.37), we obtain
\{f _ {s} \} = 0. 2 5 \int_ {- 1} ^ {1} \left[ \begin{array}{c} 2 0 0 0 N _ {2} \\ 0 \\ 2 0 0 0 N _ {3} \\ 0 \end{array} \right] d t = 5 0 0 \int_ {- 1} ^ {1} \left[ \begin{array}{c} N _ {2} \\ 0 \\ N _ {3} \\ 0 \end{array} \right] d t \tag {10.3.38}
evaluated along s ¼ 1
Substituting the shape functions from Eq. (10.3.35) into Eq. (10.3.38), we have
\left\{f _ {s} \right\} = 5 0 0 \int_ {- 1} ^ {1} \left[ \begin{array}{c} \frac {s - t - s t + 1}{4} \\ 0 \\ \frac {s + t + s t + 1}{4} \\ 0 \end{array} \right] d t \tag {10.3.39}
evaluated along s ¼ 1
Upon substituting s ¼ 1 into the integrand in Eq. (10.3.39) and performing the explicit integration in Eq. (10.3.40), we obtain
\left\{f _ {s} \right\} = 5 0 0 \int_ {- 1} ^ {1} \left[ \begin{array}{c} \frac {2 - 2 t}{4} \\ 0 \\ \frac {2 t + 2}{4} \\ 0 \end{array} \right] d t = 5 0 0 \left[ \begin{array}{c} 0. 5 0 t - \frac {t ^ {2}}{4} \\ 0 \\ 0. 5 0 t + \frac {t ^ {2}}{4} \\ 0 \end{array} \right] _ {- 1} ^ {1} \tag {10.3.40}
Evaluating the resulting integration expression for each limit, we obtain the final expression for the surface traction matrix as
\{f _ {s} \} = 5 0 0 \left[ \begin{array}{c} 0. 5 0 - 0. 2 5 \\ 0 \\ 0. 5 0 + 0. 2 5 \\ 0 \end{array} \right] - 5 0 0 \left[ \begin{array}{c} - 0. 5 0 - 0. 2 5 \\ 0 \\ - 0. 5 0 + 0. 2 5 \\ 0 \end{array} \right] = 5 0 0 \left[ \begin{array}{c} 1 \\ 0 \\ 1 \\ 0 \end{array} \right] l b \tag {10.3.41}
Or in explicit form the surface tractions at nodes 2 and 3 are
\left\{ \begin{array}{l} f _ {s 2 s} \\ f _ {s 2 t} \\ f _ {s 3 s} \\ f _ {s 3 t} \end{array} \right\} = \left[ \begin{array}{c} 5 0 0 \\ 0 \\ 5 0 0 \\ 0 \end{array} \right] l b \tag {10.3.42}