김경종
22 KiB
Step 3
Back-substitute the now known d into step 2 to obtain the actual nodal forces, \underline { { F } } ( = \underline { { K } } \underline { { d } } - \underline { { F } } _ { 0 } ) .
Hence, the thermal stress problem is solved in a manner similar to the distributed load problem discussed for beams and frames in Chapters 4 and 5. We will now solve the following examples to illustrate the general procedure.
Example 15.1
For the one-dimensional bar fixed at both ends and subjected to a uniform temperature rise T = 5 0 ^ { \circ } \mathrm { F } as shown in Figure 15–5, determine the reactions at the fixed ends and the axial stress in the bar. Let E = 3 0 \times 1 0 ^ { 6 } \ \mathrm { p s i } , A = 4 \ \mathrm { i n } ^ { 2 } , L = 4 \ \mathrm { f t } , and \alpha { = } 7 . 0 \times 1 0 ^ { - 6 } ~ \mathrm { ( i n . / i n . ) / ^ { \circ } F } .
Two elements will be sufficient to represent the bar because internal nodal displacements are not of importance here. To solve \underline { { F } } _ { 0 } = \underline { { K } } \underline { { d } } . , we must determine the global stiffness matrix for the bar. Hence, for each element, we have
\underline {{k}} ^ {(1)} = \frac {A E}{L / 2} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \frac {\mathrm{lb}}{\text { in. }} \quad \underline {{k}} ^ {(2)} = \frac {A E}{L / 2} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \frac {\mathrm{lb}}{\text { in. }} \tag {15.1.29}
where the numbers above the columns in the \underline { { k } } ^ { * } \mathbf { s } indicate the nodal displacements associated with each element.
Step 1
Using Eq. (15.1.20), the thermal force matrix for each element is given by
\underline {{f}} ^ {(1)} = \left\{ \begin{array}{c} - E \alpha T A \\ E \alpha T A \end{array} \right\} \quad \underline {{f}} ^ {(2)} = \left\{ \begin{array}{c} - E \alpha T A \\ E \alpha T A \end{array} \right\} \tag {15.1.30}
where these forces are considered to be equivalent nodal forces.
Step 2
Applying the direct stiffness method to Eqs. (15.1.29) and (15.1.30), we assemble the global equations as
\left\{ \begin{array}{c} F _ {1 x} - E \alpha T A \\ 0 \\ F _ {3 x} + E \alpha T A \end{array} \right\} = \frac {A E}{L / 2} \left[ \begin{array}{c c c} 1 & - 1 & 0 \\ - 1 & 1 + 1 & - 1 \\ 0 & - 1 & 1 \end{array} \right] \left\{ \begin{array}{c} d _ {1 x} \\ d _ {2 x} \\ d _ {3 x} \end{array} \right\} \tag {15.1.31}
T = 50^{\circ}\text{F} \]
1 \quad \textcircled{1} \quad 2 \cdot \quad \textcircled{2} \quad 3 \quad \rightarrow x
\[ L
Figure 15–5 Bar subjected to a uniform temperature rise
text_image
42,000 lb 1 ① 2 • ② 42,000 lb 3
Figure 15–6 Free-body diagram of the bar of Figure 15–5
Applying the boundary conditions d _ { 1 x } = 0 and d _ { 3 x } = 0 and solving the second of Eq. (15.1.31), we obtain
d _ {2 x} = 0 \tag {15.1.32}
Step 3
Back-substituting Eq. (15.1.30) into the global equation (Eq. (15.1.31)) (step 2) for the nodal forces, we obtain
\left\{ \begin{array}{l} F _ {1 x} \\ F _ {2 x} \\ F _ {3 x} \end{array} \right\} = \left\{ \begin{array}{l} 0 \\ 0 \\ 0 \end{array} \right\} - \left\{ \begin{array}{c} - E \alpha T A \\ 0 \\ E \alpha T A \end{array} \right\} = \left\{ \begin{array}{c} E \alpha T A \\ 0 \\ - E \alpha T A \end{array} \right\} \tag {15.1.33}
Using the numerical quantities for E , \alpha , T , and A in Eq. (15.1.33), we obtain
F _ {1 x} = 4 2, 0 0 0 \mathrm{lb} \quad F _ {2 x} = 0 \quad F _ {3 x} = - 4 2, 0 0 0 \mathrm{lb}
as shown in Figure 15–6. The stress in the bar is then
\sigma = \frac {4 2 , 0 0 0}{4} = 1 0, 5 0 0 \mathrm{psi} \quad (\text { compressive }) \tag {15.1.34}
Example 15.2
For the bar assemblage shown in Figure 15–7, determine the reactions at the fixed ends and the axial stress in each bar. Bar 1 is subjected to a temperature drop of 1 0 ^ { \circ } \mathrm { C } . . Let bar 1 be aluminum with E ¼ 70 GPa, \alpha = 2 3 \times 1 0 ^ { - 6 } (mm/mm)/ C, A = 1 2 \times 1 0 ^ { - 4 } ~ \mathrm { m } ^ { 2 } , and L = 2 m. Let bars 2 and 3 be brass with E = 1 0 0 \ \mathrm { G P a } . , \alpha = 2 0 \times 1 0 ^ { - 6 } (mm/mm)/ C, A = 6 \times 1 0 ^ { - 4 } \mathrm { m } ^ { 2 } , and L = 2 { \mathrm { m } } .
text_image
1 2 m 2 2 m 2 2 m 3 4 Rigid bar x
Figure 15–7 Bar assemblage for thermal stress analysis
We begin the solution by determining the stiffness matrices for each element.
Element 1
\underline {{k}} ^ {(1)} = \frac {(1 2 \times 1 0 ^ {- 4}) (7 0 \times 1 0 ^ {6})}{2} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] = 4 2, 0 0 0 \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \frac {\mathrm{kN}}{\mathrm{m}} \tag {15.1.35}
Elements 2 and 3
\underline {{k}} ^ {(2)} = \underline {{k}} ^ {(3)} = \frac {(6 \times 1 0 ^ {- 4}) (1 0 0 \times 1 0 ^ {6})}{2} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] = 3 0, 0 0 0 \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \frac {\mathrm{kN}}{\mathrm{m}} \tag {15.1.36}
Step 1
We obtain the element thermal force matrices by evaluating Eq. (15.1.20). First, evaluating -E\alpha TA for element 1, we have
- E \alpha T A = - (7 0 \times 1 0 ^ {6}) (2 3 \times 1 0 ^ {- 6}) (- 1 0) (1 2 \times 1 0 ^ {- 4}) = 1 9. 3 2 \mathrm{kN} \tag {15.1.37}
where the -10 term in Eq. (15.1.37) is due to the temperature drop in element 1. Using the result of Eq. (15.1.37) in Eq. (15.1.20), we obtain
\underline {{f}} ^ {(1)} = \left\{ \begin{array}{l} f _ {1 x} \\ f _ {2 x} \end{array} \right\} = \left\{ \begin{array}{c} 1 9. 3 2 \\ - 1 9. 3 2 \end{array} \right\} \mathrm{kN} \tag {15.1.38}
There is no temperature change in elements 2 and 3, and so
\underline {{f}} ^ {(2)} = \underline {{f}} ^ {(3)} = \left\{ \begin{array}{l} 0 \\ 0 \end{array} \right\} \tag {15.1.39}
Step 2
Assembling the global equations using Eqs. (15.1.35), (15.1.38), and (15.1.39), we obtain
1 0 0 0 \left[ \begin{array}{c c c c} 1 & 2 & 3 & 4 \\ 4 2 & - 4 2 & 0 & 0 \\ - 4 2 & 4 2 + 3 0 + 3 0 & - 3 0 & - 3 0 \\ 0 & - 3 0 & 3 0 & 0 \\ 0 & - 3 0 & 0 & 3 0 \end{array} \right] \left\{ \begin{array}{l} d _ {1 x} \\ d _ {2 x} \\ d _ {3 x} \\ d _ {4 x} \end{array} \right\} = \left\{ \begin{array}{c} F _ {1 x} + 1 9. 3 2 \\ - 1 9. 3 2 \\ F _ {3 x} \\ F _ {4 x} \end{array} \right\} \tag {15.1.40}
where the right-side thermal forces are considered to be equivalent nodal forces. Using the boundary conditions
d _ {1 x} = 0 \quad d _ {3 x} = 0 \quad d _ {4 x} = 0 \tag {15.1.41}
we obtain, from the second equation of Eq. (15.1.40),
1 0 0 0 (1 0 2) d _ {2 x} = - 1 9. 3 2
Solving for d _ { 2 x } , we obtain
d _ {2 x} = - 1. 8 9 \times 1 0 ^ {- 4} \mathrm{m} \tag {15.1.42}
Step 3
Back-substituting Eq. (15.1.42) into the global equation for the nodal forces, \underline { { F } } = \underline { { K } } \underline { { d } } - \underline { { F } } _ { 0 } , we have
\left\{ \begin{array}{l} F _ {1 x} \\ F _ {2 x} \\ F _ {3 x} \\ F _ {4 x} \end{array} \right\} = 1 0 0 0 \left[ \begin{array}{r r r r} 4 2 & - 4 2 & 0 & 0 \\ - 4 2 & 1 0 2 & - 3 0 & - 3 0 \\ 0 & - 3 0 & 3 0 & 0 \\ 0 & - 3 0 & 0 & 3 0 \end{array} \right] \left\{ \begin{array}{c} 0 \\ - 1. 8 9 \times 1 0 ^ {- 4} \\ 0 \\ 0 \end{array} \right\} - \left\{ \begin{array}{c} 1 9. 3 2 \\ - 1 9. 3 2 \\ 0 \\ 0 \end{array} \right\} \tag {15.1.43}
Simplifying Eq. (15.1.43), we obtain
F _ {1 x} = - 1 1. 3 8 \mathrm{kN}
F _ {2 x} = 0. 0 \mathrm{kN} \tag {15.1.44}
F _ {3 x} = 5. 6 9 \mathrm{kN}
F _ {4 x} = 5. 6 9 \mathrm{kN}
A free-body diagram of the bar assemblage is shown in Figure 15–8. The stresses in each bar are then
\sigma^ {(1)} = \frac {1 1 . 3 8}{1 2 \times 1 0 ^ {- 4}} = 9. 4 8 \times 1 0 ^ {3} \mathrm{kN} / \mathrm{m} ^ {2} \quad (9. 4 8 \mathrm{MPa}) \tag {15.1.45}
\sigma^ {(2)} = \sigma^ {(3)} = \frac {5 . 6 9}{6 \times 1 0 ^ {- 4}} = 9. 4 8 \times 1 0 ^ {3} \mathrm{kN} / \mathrm{m} ^ {2} \quad (9. 4 8 \mathrm{MPa})
text_image
1 11.38 kN 2 2 3 5.69 kN 2 4 5.69 kN
Figure 15–8 Free-body diagram of the bar assemblage of Figure 15–7
For the plane truss shown in Figure 15–9, determine the displacements at node 1 and the axial stresses in each bar. Bar 1 is subjected to a temperature rise of 7 5 ^ { \circ } \mathrm { F } . Let { \cal E } = 3 0 \times 1 0 ^ { 6 } \mathrm { p s i } , \alpha = 7 \times 1 0 ^ { - 6 } ( \mathrm { i n . / i n . } ) / \mathrm { ° F } . , and A = 2 ~ \mathrm { i n } ^ { 2 } for both bar elements.
text_image
y 1 8 ft ① ② 2 x 3 6 ft
Figure 15–9 Plane truss for thermal stress analysis
First, using Eq. (3.4.23), we determine the stiffness matrices for each element.
Element 1
Choosing x^ from node 2 to node 1, \theta = 9 0 ^ { \circ } , and so cos \theta = 0 , sin \theta = 1 , and
\underline {{k}} ^ {(1)} = \frac {(2) (3 0 \times 1 0 ^ {6})}{(8 \times 1 2)} \left[ \begin{array}{c c c c} 0 & 0 & 0 & 0 \\ & 1 & 0 & - 1 \\ & & 0 & 0 \\ \text { Symmetry } & & & 1 \end{array} \right] \frac {\text { lb }}{\text { in. }} \tag {15.1.46}
Element 2
Choosing x^ from node 3 to node 1, \theta = 1 8 0 ^ { \circ } - 5 3 . 1 3 ^ { \circ } = 1 2 6 . 8 7 ^ { \circ } , and so cos \theta = - 0 . 6 , sin \theta = 0 . 8 , and
\underline {{k}} ^ {(2)} = \frac {(2) (3 0 \times 1 0 ^ {6})}{(1 0 \times 1 2)} \left[ \begin{array}{c c c c} 0. 3 6 & - 0. 4 8 & - 0. 3 6 & 0. 4 8 \\ & 0. 6 4 & 0. 4 8 & - 0. 6 4 \\ & & 0. 3 6 & - 0. 4 8 \\ \text { Symmetry } & & & 0. 6 4 \end{array} \right] \frac {\text { lb }}{\text { in. }} \tag {15.1.47}
Step 1
We obtain the element thermal force matrices by evaluating Eq. (15.1.20) as follows:
- E \alpha T A = - (3 0 \times 1 0 ^ {6}) (7 \times 1 0 ^ {- 6}) (7 5) (2) = - 3 1, 5 0 0 \mathrm{lb} \tag {15.1.48}
Using the result of Eq. (15.1.48) for element 1, we then have the local thermal force matrix as
\underline {{f}} ^ {(1)} = \left\{ \begin{array}{l} \hat {f} _ {2 x} \\ \hat {f} _ {1 x} \end{array} \right\} = \left\{ \begin{array}{c} - 3 1, 5 0 0 \\ 3 1, 5 0 0 \end{array} \right\} \text { lb } \tag {15.1.49}
There is no temperature change in element 2, so
\underline {{f}} ^ {(2)} = \left\{ \begin{array}{l} \hat {f} _ {3 x} \\ \hat {f} _ {1 x} \end{array} \right\} = \left\{ \begin{array}{l} 0 \\ 0 \end{array} \right\} \tag {15.1.50}
Recall that by Eq. ( 3 . 4 . 1 6 ) , \underline { { { \hat { f } } } } = \underline { { { T f } } } . Since we have shown that \underline { { T } } ^ { - 1 } = \underline { { T } } ^ { T } , we can obtain the global forces by premultiplying Eq. (3.4.16) by \underline { { T } } ^ { T } to obtain the element nodal forces in the global reference frame as
\underline {{f}} = \underline {{T}} ^ {T} \underline {{\hat {f}}} \tag {15.1.51}
Using Eq. (15.1.51), the element 1 global nodal forces are then
\left\{ \begin{array}{l} f _ {2 x} \\ f _ {2 y} \\ f _ {1 x} \\ f _ {1 y} \end{array} \right\} = \left[ \begin{array}{c c c c} C & - S & 0 & 0 \\ S & C & 0 & 0 \\ 0 & 0 & C & - S \\ 0 & 0 & S & C \end{array} \right] \left\{ \begin{array}{l} \hat {f} _ {2 x} \\ \hat {f} _ {2 y} \\ \hat {f} _ {1 x} \\ \hat {f} _ {1 y} \end{array} \right\} \tag {15.1.52}
where the order of terms in Eq. (15.1.52) is due to the choice of the x^ axis from node 2 to node 1 and where T, given by Eq. (3.4.15), has been used.
Substituting the numerical quantities C = 0 and S = 1 (consistent with x^ for element 1), and \bar { \hat { f } } _ { 1 x } = 3 1 , 5 0 0 , \hat { f } _ { 1 y } = 0 , \hat { f } _ { 2 x } = - 3 1 , 5 0 0 , and \hat { f } _ { 2 y } = 0 into Eq. (15.1.52), we obtain
f _ {2 x} = 0 \quad f _ {2 y} = - 3 1, 5 0 0 \mathrm{lb} \quad f _ {1 x} = 0 \quad f _ {1 y} = 3 1, 5 0 0 \mathrm{lb} \tag {15.1.53}
These element forces are now the only equivalent global nodal forces, because element 2 is not subjected to a change in temperature.
Step 2
Assembling the global equations using Eqs. (15.1.46), (15.1.47), and (15.1.53), we obtain
0. 5 0 \times 1 0 ^ {6} \left[ \begin{array}{c c c c c c} 0. 3 6 & - 0. 4 8 & 0 & 0 & 0 & 0 \\ & 1. 8 9 & 0 & - 1. 2 5 & 0 & 0 \\ & & 0 & 0 & 0 & 0 \\ & & & 1. 2 5 & 0 & 0 \\ & & & & 0. 3 6 & - 0. 4 8 \\ \text {Symmetry} & & & & & 0. 6 4 \end{array} \right] \left\{ \begin{array}{l} d _ {1 x} \\ d _ {1 y} \\ d _ {2 x} \\ d _ {2 y} \\ d _ {3 x} \\ d _ {3 y} \end{array} \right\} = \left\{ \begin{array}{c} F _ {1 x} + 0 \\ 3 1, 5 0 0 \\ F _ {2 x} + 0 \\ - 3 1, 5 0 0 + F _ {2 y} \\ F _ {3 x} + 0 \\ F _ {3 y} + 0 \end{array} \right\} \tag {15.1.54}
The boundary conditions are given by
d _ {1 x} = 0 \quad d _ {2 x} = 0 \quad d _ {2 y} = 0 \quad d _ {3 x} = 0 \quad d _ {3 y} = 0 \tag {15.1.55}
Using the boundary condition Eqs. (15.1.55) and the second equation of Eq. (15.1.54), we obtain
(0. 9 4 5 \times 1 0 ^ {6}) d _ {1 y} = 3 1, 5 0 0
or d _ { 1 y } = 0 . 0 3 3 3 \mathrm { i n } . ð15:1:56Þ
Step 3
We now illustrate the procedure used to obtain the local element forces in local coordinates; that is, the local element forces are
\underline {{{\hat {f}}}} = \underline {{{\hat {k}}}} \underline {{{\hat {d}}}} - \underline {{{\hat {f}}}} _ {0} \tag {15.1.57}
We determine the actual local element nodal forces by using the relationship \underline { { \hat { d } } } = \underline { { T } } ^ { * } \underline { { d } } . the usual bar element \underline { { \hat { k } } } matrix [Eq. (3.1.14)], the transformation matrix \underline { { T } } ^ { * } \left[ \mathrm { E q } \right. . (3.4.8)], and the calculated displacements and initial thermal forces applicable for the element under consideration. Substituting the numerical quantities for element 1, from Eq. (15.1.57), we have
\left\{ \begin{array}{l} \hat {f} _ {2 x} \\ \hat {f} _ {1 x} \end{array} \right\} = \frac {2 (3 0 \times 1 0 ^ {6})}{8 \times 1 2} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \left[ \begin{array}{c c c c} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right] \left\{ \begin{array}{l} d _ {2 x} = 0 \\ d _ {2 y} = 0 \\ d _ {1 x} = 0 \\ d _ {1 y} = 0. 0 3 3 3 \end{array} \right\} - \left\{ \begin{array}{c} - 3 1, 5 0 0 \\ 3 1, 5 0 0 \end{array} \right\} \tag {15.1.58}
Simplifying Eq. (15.1.58), we obtain
\hat {f} _ {2 x} = 1 0, 7 0 0 \mathrm{lb} \quad \hat {f} _ {1 x} = - 1 0, 7 0 0 \mathrm{lb} \tag {15.1.59}
Dividing the local element force \hat { f } _ { 1 x } (which is the far-end force consistent with the convention used in Section 3.5) by the cross-sectional area, we obtain the stress as
\sigma^ {(1)} = \frac {- 1 0 , 7 0 0}{2} = - 5 3 5 0 \mathrm{psi} \tag {15.1.60}
Similarly, for element 2, we have
\left\{ \begin{array}{l} \hat {f} _ {3 x} \\ \hat {f} _ {1 x} \end{array} \right\} = \frac {2 \left(3 0 \times 1 0 ^ {6}\right)}{1 0 \times 1 2} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \left[ \begin{array}{c c c c} - 0. 6 & 0. 8 & 0 & 0 \\ 0 & 0 & - 0. 6 & 0. 8 \end{array} \right] \left\{ \begin{array}{l} 0 \\ 0 \\ 0 \\ 0. 0 3 3 3 \end{array} \right\} \tag {15.1.61}
Simplifying, Eq. (15.1.61), we obtain
\hat {f} _ {3 x} = - 1 3, 3 1 0 \mathrm{lb} \quad \hat {f} _ {1 x} = 1 3, 3 1 0 \mathrm{lb} \tag {15.1.62}
where no initial thermal forces were present for element 2 because the element was not subjected to a temperature change. Dividing the far-end force \hat { f } _ { 1 x } by the
cross-sectional area results in
\sigma^ {(2)} = 6 6 6 0 \mathrm{psi} \tag {15.1.63}
For two- and three-dimensional stress problems, this direct division of force by cross-sectional area is not permissible. Hence, the total stress due to both applied loading and temperature change must be determined by
\underline {{\sigma}} = \underline {{\sigma}} _ {L} - \underline {{\sigma}} _ {T} \tag {15.1.64}
We now illustrate Eq. (15.1.64) for bar element 1 of the truss of Example 15.3. For the bar, \sigma _ { L } can be obtained using Eq. (3.5.6), and \sigma _ { T } is obtained from
\sigma_ {T} = \underline {{{D}}} \underline {{{\varepsilon}}} _ {T} = E \alpha T \tag {15.1.65}
because \underline { { \boldsymbol { D } } } = E and \varepsilon _ { T } = \alpha T for the bar element. The stress in bar element 1 is then determined to be
\sigma^ {(1)} = \frac {E}{L} \left[ - C - S C S \right] \left\{ \begin{array}{l} d _ {2 x} \\ d _ {2 y} \\ d _ {1 x} \\ d _ {1 y} \end{array} \right\} - E \alpha T \tag {15.1.66}
Substituting the numerical quantities for element 1 into Eq. (15.1.66), we obtain
\sigma^ {(1)} = \frac {3 0 \times 1 0 ^ {6}}{8 \times 1 2} [ 0 - 1 0 1 ] \left\{ \begin{array}{l} 0 \\ 0 \\ 0 \\ 0. 0 3 3 3 \end{array} \right\} - (3 0 \times 1 0 ^ {6}) (7 \times 1 0 ^ {- 6}) (7 5) \tag {15.1.67}
or \sigma ^ { ( 1 ) } = - 5 3 5 0 \mathrm { \ p s i } ð15:1:68Þ 9
We will now illustrate the solutions of two plane thermal stress problems.
Example 15.4
For the plane stress element shown in Figure 15–10, determine the element equations. The element has a 2 0 0 0 { - } 1 \mathrm { b } / \mathrm { i n } ^ { 2 } pressure acting perpendicular to side j-m and is subjected to a 3 0 ^ { \circ } \mathrm { F } temperature rise.
Recall that the stiffness matrix is given by [Eq. (6.2.52) or (6.4.1)]
[ k ] = [ B ] ^ {T} [ D ] [ B ] t A \tag {15.1.69}
and
\beta_ {i} = y _ {j} - y _ {m} = - 3 \quad \gamma_ {i} = x _ {m} - x _ {j} = - 1
\beta_ {j} = y _ {m} - y _ {i} = 3 \quad \gamma_ {j} = x _ {i} - x _ {m} = - 1 \tag {15.1.70}
\beta_ {m} = y _ {i} - y _ {j} = 0 \quad \gamma_ {m} = x _ {j} - x _ {i} = 2
and
A = \frac {(3) (2)}{2} = 3 \mathrm{in} ^ {2}
text_image
y -1 in. m 3 in. 2000 lb/in² i 2 in. j x
t = 1 \text { in. }
E = 3 0 \times 1 0 ^ {6} \mathrm{psi}
\alpha = 7 \times 1 0 ^ {- 6} (\text { in. } / \text { in. }) / ^ {\circ} \mathrm{F}
\nu = 0. 2 5
Figure 15–10 Plane stress element subjected to mechanical loading and a temperature change
Therefore, substituting the results of Eqs. (15.1.70) into Eq. (6.2.34) for ½B , we obtain
[ B ] = \frac {1}{6} \left[ \begin{array}{r r r r r r} - 3 & 0 & 3 & 0 & 0 & 0 \\ 0 & - 1 & 0 & - 1 & 0 & 2 \\ - 1 & - 3 & - 1 & 3 & 2 & 0 \end{array} \right] \tag {15.1.71}
Assuming plane stress conditions to be valid, we have
[ D ] = \frac {E}{1 - v ^ {2}} \left[ \begin{array}{c c c} 1 & v & 0 \\ v & 1 & 0 \\ 0 & 0 & \frac {1 - v}{2} \end{array} \right] = \frac {3 0 \times 1 0 ^ {6}}{1 - (0 . 2 5) ^ {2}} \left[ \begin{array}{c c c} 1 & 0. 2 5 & 0 \\ 0. 2 5 & 1 & 0 \\ 0 & 0 & 0. 3 7 5 \end{array} \right]
= (4 \times 1 0 ^ {6}) \left[ \begin{array}{l l l} 8 & 2 & 0 \\ 2 & 8 & 0 \\ 0 & 0 & 3 \end{array} \right] \text { psi } \tag {15.1.72}
[ B ] ^ {T} [ D ] = \frac {1}{6} \left[ \begin{array}{r r r} - 3 & 0 & - 1 \\ 0 & - 1 & - 3 \\ 3 & 0 & - 1 \\ 0 & - 1 & 3 \\ 0 & 0 & 2 \\ 0 & 2 & 0 \end{array} \right] (4 \times 1 0 ^ {6}) \left[ \begin{array}{r r r} 8 & 2 & 0 \\ 2 & 8 & 0 \\ 0 & 0 & 3 \end{array} \right] \tag {15.1.73}
Simplifying Eq. (15.1.73), we obtain
[ B ] ^ {T} [ D ] = \frac {4 \times 1 0 ^ {6}}{6} \left[ \begin{array}{r r r} - 2 4 & - 6 & - 3 \\ - 2 & - 8 & - 9 \\ 2 4 & 6 & - 3 \\ - 2 & - 8 & 9 \\ 0 & 0 & 6 \\ 4 & 1 6 & 0 \end{array} \right] \tag {15.1.74}
Therefore, substituting the results of Eqs. (15.1.71) and (15.1.74) into Eq. (15.1.69) yields the element stiffness matrix as
[ k ] = (1 \text { in. }) \frac {(3 \text { in } ^ {2})}{6} \frac {4 \times 1 0 ^ {6}}{6} \left[ \begin{array}{r r r} - 2 4 & - 6 & - 3 \\ - 2 & - 8 & - 9 \\ 2 4 & 6 & - 3 \\ - 2 & - 8 & 9 \\ 0 & 0 & 6 \\ 4 & 1 6 & 0 \end{array} \right] \left[ \begin{array}{r r r r r r} - 3 & 0 & 3 & 0 & 0 & 0 \\ 0 & - 1 & 0 & - 1 & 0 & 2 \\ - 1 & - 3 & - 1 & 3 & 2 & 0 \end{array} \right] \tag {15.1.75}
Simplifying Eq. (15.1.75), we have the element stiffness matrix as
[ k ] = \frac {1 \times 1 0 ^ {6}}{3} \left[ \begin{array}{r r r r r r} 7 5 & 1 5 & - 6 9 & - 3 & - 6 & - 1 2 \\ 1 5 & 3 5 & 3 & - 1 9 & - 1 8 & - 1 6 \\ - 6 9 & 3 & 7 5 & - 1 5 & - 6 & 1 2 \\ - 3 & - 1 9 & - 1 5 & 3 5 & 1 8 & - 1 6 \\ - 6 & - 1 8 & - 6 & 1 8 & 1 2 & 0 \\ - 1 2 & - 1 6 & 1 2 & - 1 6 & 0 & 3 2 \end{array} \right] \frac {\mathrm{lb}}{\text {in.}} \tag {15.1.76}
Using Eq. (15.1.25), the thermal force matrix is given by
\left\{f _ {T} \right\} = \frac {\alpha E t T}{2 (1 - v)} \left\{ \begin{array}{l} \beta_ {i} \\ \gamma_ {i} \\ \beta_ {j} \\ \gamma_ {j} \\ \beta_ {m} \\ \gamma_ {m} \end{array} \right\} = \frac {(7 \times 1 0 ^ {- 6}) (3 0 \times 1 0 ^ {6}) (1) (3 0)}{2 (1 - 0 . 2 5)} \left\{ \begin{array}{c} - 3 \\ - 1 \\ 3 \\ - 1 \\ 0 \\ 2 \end{array} \right\} = 4 2 0 0 \left\{ \begin{array}{c} - 3 \\ - 1 \\ 3 \\ - 1 \\ 0 \\ 2 \end{array} \right\}
\text { or } \quad \left\{f _ {T} \right\} = \left\{ \begin{array}{c} - 1 2, 6 0 0 \\ - 4 2 0 0 \\ 1 2, 6 0 0 \\ - 4 2 0 0 \\ 0 \\ 8 4 0 0 \end{array} \right\} \text { lb } \tag {15.1.77}