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26 KiB
The assemblage of the element matrices, Eqs. (13.4.54) and (13.4.56) and the other force matrices similar to Eq. (13.4.56), yields
1 0 0 \left[ \begin{array}{r r r r r} 1 & - 1 & 0 & 0 & 0 \\ - 1 & 2 & - 1 & 0 & 0 \\ 0 & - 1 & 2 & - 1 & 0 \\ 0 & 0 & - 1 & 2 & - 1 \\ 0 & 0 & 0 & - 1 & 1 \end{array} \right] \left\{ \begin{array}{l} t _ {1} \\ t _ {2} \\ t _ {3} \\ t _ {4} \\ t _ {5} \end{array} \right\} = \left\{ \begin{array}{c} F _ {1} + 5 0 \\ 1 0 0 \\ 1 0 0 \\ 1 0 0 \\ 5 0 \end{array} \right\} \tag {13.4.57}
Substituting the known temperature t _ { 1 } = 2 0 0 ^ { \circ } \mathrm { C } into Eq. (13.4.57), dividing both sides of Eq. (13.4.57) by 100, and transposing known terms to the right side, we have
\left[ \begin{array}{c c c c c} 1 & 0 & 0 & 0 & 0 \\ 0 & 2 & - 1 & 0 & 0 \\ 0 & - 1 & 2 & - 1 & 0 \\ 0 & 0 & - 1 & 2 & - 1 \\ 0 & 0 & 0 & - 1 & 1 \end{array} \right] \left\{ \begin{array}{l} t _ {1} \\ t _ {2} \\ t _ {3} \\ t _ {4} \\ t _ {5} \end{array} \right\} = \left\{ \begin{array}{c} 2 0 0 ^ {\circ} \mathrm{C} \\ 2 0 1 \\ 1 \\ 1 \\ 0. 5 \end{array} \right\} \tag {13.4.58}
The second through fifth equations of Eq. (13.4.58) can now be solved simultaneously to yield
t _ {2} = 2 0 3. 5 ^ {\circ} \mathrm{C} \quad t _ {3} = 2 0 6 ^ {\circ} \mathrm{C} \quad t _ {4} = 2 0 7. 5 ^ {\circ} \mathrm{C} \quad t _ {5} = 2 0 8 ^ {\circ} \mathrm{C} \tag {13.4.59}
Using the first of Eqs. (13.4.57) yields the rate of heat flow out the left end:
F _ {1} = 1 0 0 (t _ {1} - t _ {2}) - 5 0
F _ {1} = 1 0 0 (2 0 0 - 2 0 3. 5) - 5 0
F _ {1} = - 4 0 0 \mathrm{W}
The closed-form solution of the differential equation for conduction, Eq. (13.1.9), with the left-end boundary condition given by Eq. (13.1.10) and the right-end boundary condition given by Eq. (13.1.11), and with q _ { x } ^ { * } = 0 , is shown in Reference [2] to yield a parabolic temperature distribution through the wall. Evaluating the expression for the temperature function given in Reference [2] for values of x corresponding to the node points of the finite element model, we obtain
t _ {2} = 2 0 3. 5 ^ {\circ} \mathrm{C} \quad t _ {3} = 2 0 6 ^ {\circ} \mathrm{C} \quad t _ {4} = 2 0 7. 5 ^ {\circ} \mathrm{C} \quad t _ {5} = 2 0 8 ^ {\circ} \mathrm{C} \tag {13.4.60}
Figure 13–15 is a plot of the closed-form solution and the finite element solution for the temperature variation through the wall. The finite element nodal values and the closed-form values are equal, because the consistent equivalent force matrix has been used. (This was also discussed in Sections 3.10 and 3.11 for the axial bar subjected to distributed loading, and in Section 4.5 for the beam subjected to distributed loading.) However, recall that the finite element model predicts a linear temperature distribution within each element as indicated by the straight lines connecting the nodal temperature values in Figure 13–15.
line
| x, m | T(x), °F (Finite element solution) | T(x), °F (Closed-form solution) |
|---|---|---|
| 0.00 | 200.0 | 200.0 |
| 0.25 | 203.5 | 203.5 |
| 0.50 | 206.0 | 206.0 |
| 0.75 | 207.5 | 207.5 |
| 1.00 | 208.0 | 208.0 |
Figure 13–15 Comparison of the finite element and closed-form solutions for Example 13.3
Example 13.4
The fin shown in Figure 13–16 is insulated on the perimeter. The left end has a constant temperature of 1 0 0 ^ { \circ } \mathrm { C } . \mathrm { A } positive heat flux of q = 5 0 0 0 \mathrm { W } / \mathrm { m } ^ { 2 } acts on the right end. Let K _ { x x } = 6 \mathrm { { W } } / ( \mathrm { { m } } { \cdot } ^ { \circ } \mathrm { { C } } ) and cross-sectional area A = 0 . 1 \mathrm { m } ^ { 2 } : Determine the temperatures at L/4, L/2, 3L/4, and L , where L = 0 . 4 \mathrm { m } .
text_image
T = 100°C → q = 5000 W/m² A = 0.1 m²
Figure 13–16 Insulated fin subjected to end heat flux
Using Eq. (13.4.22), with the second term set to zero as there is no heat transfer by convection from any surfaces due to the insulated perimeter and constant temperature on the left end and constant heat flux on the right end, we obtain
\begin{array}{l} \underline {{k}} ^ {(1)} = \underline {{k}} ^ {(2)} = \underline {{k}} ^ {(3)} = \frac {A K _ {x x}}{L} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \\ = \frac {(0 . 1 \mathrm{m} ^ {2}) (6 \mathrm{W} / (\mathrm{m} - ^ {\circ} \mathrm{C})}{0 . 1 \mathrm{m}} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] = \left[ \begin{array}{c c} 6 & - 6 \\ - 6 & 6 \end{array} \right] \mathrm{W} / ^ {\circ} \mathrm{C} \tag {13.4.61} \\ \end{array}
\underline {{k}} ^ {(4)} = \underline {{k}} ^ {(1)} \text { also }
\underline { { { f } } } ^ { ( 1 ) } = \underline { { { f } } } ^ { ( 2 ) } = \underline { { { f } } } ^ { ( 3 ) } = \left\{ { 0 \atop { 0 } } \right\} as Q ¼ 0 ðno internal heat sourceÞ and q ^ { * } = 0 ðno surface heat fluxÞ
\underline {{f}} ^ {(4)} = q A \left\{ \begin{array}{l} 0 \\ 1 \end{array} \right\} = (5 0 0 0 \mathrm{W} / \mathrm{m} ^ {2}) (0. 1 \mathrm{m} ^ {2}) \left\{ \begin{array}{l} 0 \\ 1 \end{array} \right\} = \left\{ \begin{array}{l} 0 \\ 5 0 0 \end{array} \right\} \mathrm{W} \tag {13.4.62}
Assembling the global stiffness matrix from Eq. (13.4.61), and the global force matrix from Eq. (13.4.62), we obtain the global equations as
\left[ \begin{array}{c c c c c} 6 & - 6 & 0 & 0 & 0 \\ & 1 2 & - 6 & 0 & 0 \\ & & 1 2 & - 6 & 0 \\ & & & 1 2 & - 6 \\ \text {Symmetry} & & & & 6 \end{array} \right] \left\{ \begin{array}{l} t _ {1} \\ t _ {2} \\ t _ {3} \\ t _ {4} \\ t _ {5} \end{array} \right\} = \left\{ \begin{array}{c} F _ {1 x} \\ 0 \\ 0 \\ 0 \\ 5 0 0 \end{array} \right\} \tag {13.4.63}
Now applying the boundary condition on temperature, we have
t _ {1} = 1 0 0 ^ {\circ} \mathrm{C} \tag {13.4.64}
Substituting Eq. (13.4.64) for t _ { 1 } into Eq. (13.4.63), we then solve the second through fourth equations (associated with the unknown temperatures t _ { 2 } - t _ { 5 } ) simultaneously to obtain
t _ {2} = 1 8 3. 3 3 ^ {\circ} \mathrm{C}, \quad t _ {3} = 2 6 6. 6 7 ^ {\circ} \mathrm{C}, \quad t _ {4} = 3 5 0 ^ {\circ} \mathrm{C}, \quad t _ {5} = 4 3 3. 3 3 ^ {\circ} \mathrm{C} \tag {13.4.65}
Substituting the nodal temperatures from Eq. (13.4.65) into the first of Eqs. (13.4.63), we obtain the nodal heat source at node 1 as
F _ {1 x} = 6 (1 0 0 ^ {\circ} \mathrm{C} - 1 8 3. 3 3 ^ {\circ} \mathrm{C}) = - 5 0 0 \mathrm{W} \tag {13.4.66}
The nodal heat source given by Eq. (13.4.66) has a negative value, which means the heat is leaving the left end. This source is the same as the source coming into the fin at the right end given by q A = ( 5 0 0 0 ) ( 0 . 1 ) = 5 0 0 \mathrm { W } . .
Finally, remember that the most important advantage of the finite element method is that it enables us to approximate, with high confidence, more complicated problems, such as those with more then one thermal conductivity, for which closed-form solutions are difficult (if not impossible) to obtain. The automation of the finite element method through general computer programs makes the method extremely powerful.
13.5 Two-Dimensional Finite Element Formulation
Because many bodies can be modeled as two-dimensional heat-transfer problems, we now develop the equations for an element appropriate for these problems. Examples using this element then follow.
Step 1 Select Element Type
The three-noded triangular element with nodal temperatures shown in Figure 13–17 is the basic element for solution of the two-dimensional heat-transfer problem.
text_image
t_m m (x_m, y_m) t_i i (x_i, y_i) t_j j (x_j, y_j)
Figure 13–17 Basic triangular element with nodal temperatures
Step 2 Select a Temperature Function
The temperature function is given by
\{T \} = [ N _ {i} \quad N _ {j} \quad N _ {m} ] \left\{ \begin{array}{l} t _ {i} \\ t _ {j} \\ t _ {m} \end{array} \right\} \tag {13.5.1}
where t _ { i } , t _ { j } , and t _ { m } are the nodal temperatures, and the shape functions are again given by Eqs. (6.2.18); that is,
N _ {i} = \frac {1}{2 A} (\alpha_ {i} + \beta_ {i} x + \gamma_ {i} y) \tag {13.5.2}
with similar expressions for N _ { j } and N _ { m } . Here the a’s, \beta ^ { \circ } \mathbf { s } , and \gamma \mathbf { \bar { s } } are defined by Eqs. (6.2.10).
Unlike the CST element of Chapter 6 where there are two degrees of freedom per node (an x and a y displacement), in the heat transfer three-noded triangular element only a single scalar value (nodal temperature) is the primary unknown at each node, as shown by Eq. (13.5.1). This holds true for the three-dimensional elements as well, as shown in Section 13.7. Hence, the heat transfer problem is sometimes known as a scalar-valued boundary value problem.
Step 3 Define the Temperature Gradient=Temperature and Heat Flux=Temperature Gradient Relationships
We define the gradient matrix analogous to the strain matrix used in the stress analysis problem as
\{g \} = \left\{ \begin{array}{l} \frac {\partial T}{\partial x} \\ \frac {\partial T}{\partial y} \end{array} \right\} \tag {13.5.3}
Using Eq. (13.5.1) in Eq. (13.5.3), we have
\{g \} = \left[ \begin{array}{l l l} \frac {\partial N _ {i}}{\partial x} & \frac {\partial N _ {j}}{\partial x} & \frac {\partial N _ {m}}{\partial x} \\ \frac {\partial N _ {i}}{\partial y} & \frac {\partial N _ {j}}{\partial y} & \frac {\partial N _ {m}}{\partial y} \end{array} \right] \left\{ \begin{array}{l} t _ {i} \\ t _ {j} \\ t _ {m} \end{array} \right\} \tag {13.5.4}
The gradient matrix \{ g \} , written in compact matrix form analogously to the strain matrix \{ \varepsilon \} of the stress analysis problem, is given by
\{g \} = [ B ] \{t \} \tag {13.5.5}
where the ½B matrix is obtained by substituting the three equations suggested by Eq. (13.5.2) in the rectangular matrix on the right side of Eq. (13.5.4) as
[ B ] = \frac {1}{2 A} \left[ \begin{array}{c c c} \beta_ {i} & \beta_ {j} & \beta_ {m} \\ \gamma_ {i} & \gamma_ {j} & \gamma_ {m} \end{array} \right] \tag {13.5.6}
The heat flux/temperature gradient relationship is now
\left\{ \begin{array}{l} q _ {x} \\ q _ {y} \end{array} \right\} = - [ D ] \{g \} \tag {13.5.7}
where the material property matrix is
[ D ] = \left[ \begin{array}{c c} K _ {x x} & 0 \\ 0 & K _ {y y} \end{array} \right] \tag {13.5.8}
Step 4 Derive the Element Conduction Matrix and Equations
The element stiffness matrix from Eq. (13.4.17) is
[ k ] = \iint_ {V} [ B ] ^ {T} [ D ] [ B ] d V + \iint_ {S _ {3}} h [ N ] ^ {T} [ N ] d S \tag {13.5.9}
where [ k _ { c } ] = \iint _ { V } [ B ] ^ { T } [ D ] [ B ] d V
= \iiint_ {V} \frac {1}{4 A ^ {2}} \left[ \begin{array}{l l} \beta_ {i} & \gamma_ {i} \\ \beta_ {j} & \gamma_ {j} \\ \beta_ {m} & \gamma_ {m} \end{array} \right] \left[ \begin{array}{c c} K _ {x x} & 0 \\ 0 & K _ {y y} \end{array} \right] \left[ \begin{array}{l l l} \beta_ {i} & \beta_ {j} & \beta_ {m} \\ \gamma_ {i} & \gamma_ {j} & \gamma_ {m} \end{array} \right] d V \tag {13.5.10}
Assuming constant thickness in the element and noting that all terms of the integrand of Eq. (13.5.10) are constant, we have
[ k _ {c} ] = \iint_ {V} [ B ] ^ {T} [ D ] [ B ] d V = t A [ B ] ^ {T} [ D ] [ B ] \tag {13.5.11}
Equation (13.5.11) is the true conduction portion of the total stiffness matrix Eq. (13.5.9). The second integral of Eq. (13.5.9) (the convection portion of the total stiffness matrix) is defined by
[ k _ {h} ] = \iint_ {S _ {3}} h [ N ] ^ {T} [ N ] d S \tag {13.5.12}
We can explicitly multiply the matrices in Eq. (13.5.12) to obtain
\left[ k _ {h} \right] = h \iint_ {S _ {3}} \left[ \begin{array}{c c c} N _ {i} N _ {i} & N _ {i} N _ {j} & N _ {i} N _ {m} \\ N _ {j} N _ {i} & N _ {j} N _ {j} & N _ {j} N _ {m} \\ N _ {m} N _ {i} & N _ {m} N _ {j} & N _ {m} N _ {m} \end{array} \right] d S \tag {13.5.13}
To illustrate the use of Eq. (13.5.13), consider the side between nodes i and j of the triangular element to be subjected to convection (Figure 13–18). Then N _ { m } = 0 along side i \ – j , and we obtain
\left[ k _ {h} \right] = \frac {h L _ {i - j} t}{6} \left[ \begin{array}{l l l} 2 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 0 \end{array} \right] \tag {13.5.14}
where L _ { i - j } is the length of side i { - } j .
text_image
m i h j
Figure 13–18 Heat loss by convection from side i-j
The evaluation of the force matrix integrals in Eq. (13.4.16) is as follows:
\left\{f _ {Q} \right\} = \iiint_ {V} Q [ N ] ^ {T} d V = Q \iiint_ {V} [ N ] ^ {T} d V \tag {13.5.15}
for constant heat source Q. Thus it can be shown (left to your discretion) that this integral is equal to
\{f _ {Q} \} = \frac {Q V}{3} \left\{ \begin{array}{l} 1 \\ 1 \\ 1 \end{array} \right\} \tag {13.5.16}
where V = A t is the volume of the element. Equation (13.5.16) indicates that heat is generated by the body in three equal parts to the nodes (like body forces in the elasticity problem). The second force matrix in Eq. (13.4.16) is
\left\{f _ {q} \right\} = \iint_ {S _ {2}} q ^ {*} [ N ] ^ {T} d S = \iint_ {S _ {2}} q ^ {*} \left\{ \begin{array}{l} N _ {i} \\ N _ {j} \\ N _ {m} \end{array} \right\} d S \tag {13.5.17}
This reduces to
\frac {q ^ {*} L _ {i - j} t}{2} \left\{ \begin{array}{l} 1 \\ 1 \\ 0 \end{array} \right\} \quad \text { on side } i - j \tag {13.5.18}
\frac {q ^ {*} L _ {j - m} t}{2} \left\{ \begin{array}{l} 0 \\ 1 \\ 1 \end{array} \right\} \quad \text { on side } j - m \tag {13.5.19}
\frac {q ^ {*} L _ {m - i} t}{2} \left\{ \begin{array}{l} 1 \\ 0 \\ 1 \end{array} \right\} \quad \text { on side } m - i \tag {13.5.20}
where L _ { i - j } , L _ { j - m } . , and L _ { m - i } are the lengths of the sides of the element, and q ^ { * } is assumed constant over each edge. The integral \int \int _ { S _ { 3 } } h { { T } _ { \infty } } [ N ] ^ { T } dS can be found in a manner similar to Eq. (13.5.17) by simply replacing \check { q } ^ { * } with h T _ { \infty } in Eqs. (13.5.18)–(13.5.20).
Steps 5–7
Steps 5–7 are identical to those described in Section 13.4.
To illustrate the use of the equations presented in Section 13.5, we will now solve some two-dimensional heat-transfer problems.
For the two-dimensional body shown in Figure 13–19, determine the temperature distribution. The temperature at the left side of the body is maintained at 1 0 0 ^ { \circ } \mathrm { F } . The edges on the top and bottom of the body are insulated. There is heat convection from the right side with convection coefficient h = 2 0 \mathrm { \ B t u } / ( \mathrm { h } \mathrm { - f t } ^ { 2 } \mathrm { - } ^ { \circ } \mathrm { F } ) . The freestream temperature is T _ { \infty } = 5 0 ^ { \circ } \mathrm { F } . The coefficients of thermal conductivity are K _ { x x } = K _ { y y } = 2 5 \ \mathrm { B t u } / ( \mathrm { h – f t – ^ { \circ } F } ) . The dimensions are shown in the figure. Assume the thickness to be 1 ft.
text_image
T = 100°F 2 ft h = 20 Tx = 50°F 2 ft
Figure 13–19 Two-dimensional body subjected to temperature variation and convection
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y 4 3 2 ft ③ ② 5 ④ 1 2 ft x 2
Figure 13–20 Discretized two-dimensional body of Figure 13–19
The finite element discretization is shown in Figure 13–20. We will use four triangular elements of equal size for simplicity of the longhand solution. There will be convective heat loss only over the right side of the body because the other faces are insulated. We now calculate the element stiffness matrices using Eq. (13.5.11) applied for all elements and using Eq. (13.5.14) applied for element 4 only, because convection is occurring only across one edge of element 4.
Element 1
The coordinates of the element 1 nodes are x _ { 1 } = 0 , y _ { 1 } = 0 , x _ { 2 } = 2 , y _ { 2 } = 0 , x _ { 5 } = 1 , and y _ { 5 } = 1 . Using these coordinates and Eqs. (7.2.10), we obtain
\beta_ {1} = 0 - 1 = - 1 \quad \beta_ {2} = 1 - 0 = 1 \quad \beta_ {5} = 0 - 0 = 0 \tag {13.5.21}
\gamma_ {1} = 1 - 2 = - 1 \quad \gamma_ {2} = 0 - 1 = - 1 \quad \gamma_ {5} = 2 - 0 = 2
Using Eqs. (13.5.21) in Eq. (13.5.11), we have
\left[ k _ {c} ^ {(1)} \right] = \frac {1 (1)}{2 (2)} \left[ \begin{array}{c c} - 1 & - 1 \\ 1 & - 1 \\ 0 & 2 \end{array} \right] \left[ \begin{array}{c c} 2 5 & 0 \\ 0 & 2 5 \end{array} \right] \left[ \begin{array}{c c c} - 1 & 1 & 0 \\ - 1 & - 1 & 2 \end{array} \right] \tag {13.5.22}
Simplifying Eq. (13.5.22), we obtain
\left[ k _ {c} ^ {(1)} \right] = \left[ \begin{array}{c c c} 1 & 2 & 5 \\ 1 2. 5 & 0 & - 1 2. 5 \\ 0 & 1 2. 5 & - 1 2. 5 \\ - 1 2. 5 & - 1 2. 5 & 2 5 \end{array} \right] \mathrm{Btu} / (\mathrm{h} \cdot {} ^ {\circ} \mathrm{F}) \tag {13.5.23}
where the numbers above the columns indicate the node numbers associated with the matrix.
Element 2
The coordinates of the element 2 nodes are x_{1} = 0 , y_{1} = 0 , x_{5} = 1 , y_{5} = 1 , x_{4} = 0 , and y_{4} = 2 . Using these coordinates, we obtain
\beta_ {1} = 1 - 2 = - 1 \quad \beta_ {5} = 2 - 0 = 2 \quad \beta_ {4} = 0 - 1 = - 1 \tag {13.5.24}
\gamma_ {1} = 0 - 1 = - 1 \quad \gamma_ {5} = 0 - 0 = 0 \quad \gamma_ {4} = 1 - 0 = 1
Using Eqs. (13.5.24) in Eq. (13.5.11), we have
\left[ k _ {c} ^ {(2)} \right] = \frac {1}{4} \left[ \begin{array}{c c} - 1 & - 1 \\ 2 & 0 \\ - 1 & 1 \end{array} \right] \left[ \begin{array}{c c} 2 5 & 0 \\ 0 & 2 5 \end{array} \right] \left[ \begin{array}{c c c} - 1 & 2 & - 1 \\ - 1 & 0 & 1 \end{array} \right] \tag {13.5.25}
Simplifying Eq. (13.5.25), we obtain
\left[ k _ {c} ^ {(2)} \right] = \left[ \begin{array}{c c c} 1 & 5 & 4 \\ 1 2. 5 & - 1 2. 5 & 0 \\ - 1 2. 5 & 2 5 & - 1 2. 5 \\ 0 & - 1 2. 5 & 1 2. 5 \end{array} \right] \mathrm{Btu} / (\mathrm{h} \cdot {} ^ {\circ} \mathrm{F}) \tag {13.5.26}
Element 3
The coordinates of the element 3 nodes are x_4 = 0 , y_4 = 2 , x_5 = 1 , y_5 = 1 , x_3 = 2 , and y_3 = 2 . Using these coordinates, we obtain
\beta_ {4} = 1 - 2 = - 1 \quad \beta_ {5} = 2 - 2 = 0 \quad \beta_ {3} = 2 - 1 = 1 \tag {13.5.27}
\gamma_ {4} = 2 - 1 = 1 \quad \gamma_ {5} = 0 - 2 = - 2 \quad \gamma_ {3} = 1 - 0 = 1
Using Eqs. (13.5.27) in Eq. (13.5.11), we obtain
\left[ k _ {c} ^ {(3)} \right] = \left[ \begin{array}{c c c} 4 & 5 & 3 \\ 1 2. 5 & - 1 2. 5 & 0 \\ - 1 2. 5 & 2 5 & - 1 2. 5 \\ 0 & - 1 2. 5 & 1 2. 5 \end{array} \right] \mathrm{Btu} / (\mathrm{h} \cdot {} ^ {\circ} \mathrm{F}) \tag {13.5.28}
Element 4
The coordinates of the element 4 nodes are x_{2} = 2 , y_{2} = 0 , x_{3} = 2 , y_{3} = 2 , x_{5} = 1 , and y_{5} = 1 . Using these coordinates, we obtain
\beta_ {2} = 2 - 1 = 1 \quad \beta_ {3} = 1 - 0 = 1 \quad \beta_ {5} = 0 - 2 = - 2 \tag {13.5.29}
\gamma_ {2} = 1 - 2 = - 1 \quad \gamma_ {3} = 2 - 1 = 1 \quad \gamma_ {5} = 2 - 2 = 0
Using Eqs. (13.5.29) in Eq. (13.5.11), we obtain
\left[ k _ {c} ^ {(4)} \right] = \left[ \begin{array}{c c c} 2 & 3 & 5 \\ 1 2. 5 & 0 & - 1 2. 5 \\ 0 & 1 2. 5 & - 1 2. 5 \\ - 1 2. 5 & - 1 2. 5 & 2 5 \end{array} \right] \mathrm{Btu} / (\mathrm{h} \cdot {} ^ {\circ} \mathrm{F}) \tag {13.5.30}
For element 4, we have a convection contribution to the total stiffness matrix because side 2–3 is exposed to the free-stream temperature. Using Eq. (13.5.14) with i ¼ 2 and j ¼ 3, we obtain
[ k _ {h} ^ {(4)} ] = \frac {(2 0) (2) (1)}{6} \left[ \begin{array}{l l l} 2 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 0 \end{array} \right] \tag {13.5.31}
Simplifying Eq. (13.5.31) yields
\left[ k _ {h} ^ {(4)} \right] = \left[ \begin{array}{c c c} 2 & 3 & 5 \\ 1 3. 3 & 6. 6 7 & 0 \\ 6. 6 7 & 1 3. 3 & 0 \\ 0 & 0 & 0 \end{array} \right] \mathrm{Btu} / (\mathrm{h} \cdot {} ^ {\circ} \mathrm{F}) \tag {13.5.32}
Adding Eqs. (13.5.30) and (13.5.32), we obtain the element 4 total stiffness matrix as
[ k ^ {(4)} ] = \left[ \begin{array}{c c c} 2 & 3 & 5 \\ 2 5. 8 3 & 6. 6 7 & - 1 2. 5 \\ 6. 6 7 & 2 5. 8 3 & - 1 2. 5 \\ - 1 2. 5 & - 1 2. 5 & 2 5 \end{array} \right] \mathrm{Btu} / (\mathrm{h} \cdot {} ^ {\circ} \mathrm{F}) \tag {13.5.33}
Superimposing the stiffness matrices given by Eqs. (13.5.23), (13.5.26), (13.5.28), and (13.5.33), we obtain the total stiffness matrix for the body as
\underline {{K}} = \left[ \begin{array}{c c c c c} 2 5 & 0 & 0 & 0 & - 2 5 \\ 0 & 3 8. 3 3 & 6. 6 7 & 0 & - 2 5 \\ 0 & 6. 6 7 & 3 8. 3 3 & 0 & - 2 5 \\ 0 & 0 & 0 & 2 5 & - 2 5 \\ - 2 5 & - 2 5 & - 2 5 & - 2 5 & 1 0 0 \end{array} \right] \mathrm{Btu} / (\mathrm{h} - ^ {\circ} \mathrm{F}) \tag {13.5.34}
Next, we determine the element force matrices by using Eqs. (13.5.18)–(13.5.20) with q ^ { * } replaced by h T _ { \infty } . Because Q = 0 , q ^ { * } = 0 , and we have convective heat transfer only from side 2–3, element 4 is the only one that contributes nodal forces. Hence,
\{f ^ {(4)} \} = \left\{ \begin{array}{l} f _ {2} \\ f _ {3} \\ f _ {5} \end{array} \right\} = \frac {h T _ {\infty} L _ {2 - 3} t}{2} \left\{ \begin{array}{l} 1 \\ 1 \\ 0 \end{array} \right\} \tag {13.5.35}
Substituting the appropriate numerical values into Eq. (13.5.35) yields
\{f ^ {(4)} \} = \frac {(2 0) (5 0) (2) (1)}{2} \left\{ \begin{array}{l} 1 \\ 1 \\ 0 \end{array} \right\} = \left\{ \begin{array}{l} 1 0 0 0 \\ 1 0 0 0 \\ 0 \end{array} \right\} \frac {\mathrm{Btu}}{\mathrm{h}} \tag {13.5.36}
Using Eqs. (13.5.34) and (13.5.36), we find that the total assembled system of equations is
\left[ \begin{array}{c c c c c} 2 5 & 0 & 0 & 0 & - 2 5 \\ 0 & 3 8. 3 3 & 6. 6 7 & 0 & - 2 5 \\ 0 & 6. 6 7 & 3 8. 3 3 & 0 & - 2 5 \\ 0 & 0 & 0 & 2 5 & - 2 5 \\ - 2 5 & - 2 5 & - 2 5 & - 2 5 & 1 0 0 \end{array} \right] \left\{ \begin{array}{l} t _ {1} \\ t _ {2} \\ t _ {3} \\ t _ {4} \\ t _ {5} \end{array} \right\} = \left\{ \begin{array}{l} F _ {1} \\ 1 0 0 0 \\ 1 0 0 0 \\ F _ {4} \\ 0 \end{array} \right\} \tag {13.5.37}
We have known nodal temperature boundary conditions of t _ { 1 } = 1 0 0 ^ { \circ } \mathrm { F } and t _ { 4 } = 1 0 0 ^ { \circ } \mathrm { F } . We again modify the stiffness and force matrices as follows:
\left[ \begin{array}{c c c c c} 1 & 0 & 0 & 0 & 0 \\ 0 & 3 8. 3 3 & 6. 6 7 & 0 & - 2 5 \\ 0 & 6. 6 7 & 3 8. 3 3 & 0 & - 2 5 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & - 2 5 & - 2 5 & 0 & 1 0 0 \end{array} \right] \left\{ \begin{array}{l} t _ {1} \\ t _ {2} \\ t _ {3} \\ t _ {4} \\ t _ {5} \end{array} \right\} = \left\{ \begin{array}{l} 1 0 0 \\ 1 0 0 0 \\ 1 0 0 0 \\ 1 0 0 \\ 5 0 0 0 \end{array} \right\} \tag {13.5.38}
The terms in the first and fourth rows and columns corresponding to the known temperature conditions t _ { 1 } = 1 0 0 ^ { \circ } \mathrm { F } and t _ { 4 } = 1 0 0 ^ { \circ } \mathrm { F } have been set equal to zero except for the main diagonal, which has been set equal to one, and the first and fourth rows of the force matrix have been set equal to the known nodal temperatures. Also, the term ( - 2 5 ) ( 1 0 0 ^ { \circ } \mathbf { F } ) + ( - 2 5 ) \times ( 1 0 0 ^ { \circ } \mathbf { F } ) = - 5 0 0 0 on the left side of the fifth equation of Eq. (13.5.37) has been transposed to the right side in the fifth row ( \mathrm { a s } + 5 0 0 0 ) o f Eq. (13.5.38). The second, third and fifth equations of Eq. (13.5.38), corresponding to the rows of unknown nodal temperatures, can now be solved in the usual manner. The resulting solution is given by
t _ {2} = 6 9. 3 3 ^ {\circ} \mathrm{F} \quad t _ {3} = 6 9. 3 3 ^ {\circ} \mathrm{F} \quad t _ {5} = 8 4. 6 2 ^ {\circ} \mathrm{F} \tag {13.5.39}
Example 13.6
For the two-dimensional body shown in Figure 13–21, determine the temperature distribution. The temperature of the top side of the body is maintained at 1 0 0 ^ { \circ } \mathrm { C } . The body is insulated on the other edges. A uniform heat source of Q = 1 0 0 0 { \mathrm { W } } / { \mathrm { m } } ^ { 3 } acts over the whole plate, as shown in the figure. Assume a constant thickness of 1 m. Let K _ { x x } = K _ { y y } = 2 5 \ : \mathrm { W / ( m \cdot ^ { \circ } C ) } .
We need consider only the left half of the body, because we have a vertical plane of symmetry passing through the body 2 m from both the left and right edges. This vertical plane can be considered to be an insulated boundary. The finite element model is shown in Figure 13–22.
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T = 100°C 2 m 4 m
Figure 13–21 Two-dimensional body subjected to a heat source