김경종
494 lines
19 KiB
Markdown
494 lines
19 KiB
Markdown
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<details>
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<summary>text_image</summary>
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y
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t
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9
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3
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8
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4
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10
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s
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11
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12
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5
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6
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7
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2
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x
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</details>
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Figure 10–18 Cubic isoparametric element
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The cubic element in Figure 10–18 has four corner nodes and additional nodes taken to be at one-third and two-thirds of the length along each side. The shape functions of the cubic element (as derived in Reference [3]) are based on the incomplete quartic polynomial such that
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$$
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\begin{array}{l} x = a _ {1} + a _ {2} s + a _ {3} t + a _ {4} s ^ {2} + a _ {5} s t + a _ {6} t ^ {2} + a _ {7} s ^ {2} t + a _ {8} s t ^ {2} \\ + a _ {9} s ^ {3} + a _ {1 0} t ^ {3} + a _ {1 1} s ^ {3} t + a _ {1 2} s t ^ {3} \tag {10.6.34} \\ \end{array}
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$$
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with a similar polynomial for y. For the corner nodes ði ¼ 1; 2; 3; 4Þ,
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$$
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N _ {i} = \frac {1}{3 2} (1 + s s _ {i}) (1 + t t _ {i}) [ 9 (s ^ {2} + t ^ {2}) - 1 0 ] \tag {10.6.35}
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$$
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with $s _ { i }$ and $t _ { i }$ given by Eqs. (10.6.30). For the nodes on sides $s = \pm 1 \ ( i = 7 , 8 , 1 1 , 1 2 )$ ,
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$$
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N _ {i} = \frac {9}{3 2} (1 + s s _ {i}) (1 + 9 t t _ {i}) (1 - t ^ {2}) \tag {10.6.36}
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$$
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with $s _ { i } = \pm 1 \mathrm { a n d } t _ { i } = \pm \frac { 1 } { 3 } .$ For the nodes on sides $t = \pm 1 \ ( i = 5 , 6 , 9 , 1 0 )$ ,
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$$
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N _ {i} = \frac {9}{3 2} (1 + t t _ {i}) (1 + 9 s s _ {i}) (1 - s ^ {2}) \tag {10.6.37}
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$$
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with $t _ { i } = \pm 1$ and $s _ { i } = \pm { \frac { 1 } { 3 } } .$
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Having the shape functions for the quadratic element given by Eqs. (10.6.28) and (10.6.31) or for the cubic element given by Eqs. (10.6.35)–(10.6.37), we can again use Eq. (10.3.17) to obtain $\underline { { B } }$ and then Eq. (10.3.27) to set up $\underline { { k } }$ for numerical integration for the plane element. The cubic element requires a $3 \times 3$ rule (nine points) to evaluate the matrix $\underline { { k } }$ exactly. We then conclude that what is really desired is a library of shape functions that can be used in the general equations developed for sti¤- ness matrices, distributed load, and body force and can be applied not only to stress analysis but to nonstructural problems as well.
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Since in this discussion the element shape functions $N _ { i }$ relating x and y to nodal coordinates $x _ { i }$ and $y _ { i }$ are of the same form as the shape functions relating u and v to nodal displacements $u _ { i }$ and $v _ { i } ,$ this is said to be an isoparametric formulation. For instance, for the linear element $\begin{array} { r } { x = \sum _ { i = 1 } ^ { 4 } N _ { i } x _ { i } } \end{array}$ and the displacement function $u =$ $\textstyle \sum _ { i = 1 } ^ { 4 } N _ { i } u _ { i }$ ¼1 , use the same shape functions $N _ { i }$ given by Eq. (10.3.5). If instead the shape functions for the coordinates are of lower order (say, linear for x) than the
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shape functions used for displacements (say, quadratic for u), this is called a subparametric formulation.
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Finally, referring to Figure 10–18, note that an element can have a linear shape along, say, one edge (1–2), a quadratic along, say, two edges (2–3 and 1–4), and a cubic along the other edge (3–4). Hence, the simple linear element can be mixed with di¤erent higher-order elements in regions of a model where rapid stress variation is expected. The advantage of the use of higher-order elements is further illustrated in Reference [3].
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# d References
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[1] Irons, B. M., ‘‘Engineering Applications of Numerical Integration in Sti¤ness Methods,’’ Journal of the American Institute of Aeronautics and Astronautics, Vol. 4, No. 11, pp. 2035– 2037, 1966.
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[2] Stroud, A. H., and Secrest, D., Gaussian Quadrature Formulas, Prentice-Hall, Englewood Cli¤s, NJ, 1966.
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[3] Ergatoudis, I., Irons, B. M., and Zienkiewicz, O. C., ‘‘Curved Isoparametric, Quadrilateral Elements for Finite Element Analysis,’’ International Journal of Solids and Structures, Vol. 4, pp. 31–42, 1968.
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[4] Zienkiewicz, O. C., The Finite Element Method, 3rd ed., McGraw-Hill, London, 1977.
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[5] Thomas, B. G., and Finney, R. L., Calculus and Analytic Geometry, Addison-Wesley, Reading, MA, 1984.
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[6] Gallagher, R., Finite Element Analysis Fundamentals, Prentice-Hall, Englewood Cli¤s, NJ, 1975.
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[7] Bathe, K. J., and Wilson, E. L., Numerical Methods in Finite Element Analysis, Prentice-Hall, Englewood Cli¤s, NJ, 1976.
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[8] Cook, R. D., Malkus, D. S., Plesha, M. E., and Witt, R. J., Concepts and Applications of Finite Element Analysis, 4th ed., Wiley, New York, 2002.
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[9] Bathe, Klaus-Jurgen, Finite Element Procedures in Engineering Analysis, Prentice-Hall, Englewood Cli¤s, New Jersey, 1982.
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# Problems
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10.1 For the three-noded linear strain bar with three coordinates of nodes $x _ { 1 } , x _ { 2 }$ , and x3, shown in Figure P10–1 in the global-coordinate system show that the Jacobian determinate is $\vert \underline { { J } } \vert = L / 2$ .
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<details>
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<summary>text_image</summary>
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1
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x₁
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L/2
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3
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L/2
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2
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x₃
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x₂
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</details>
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Figure P10–1
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10.2 For the two-noded one-dimensional isoparametric element shown in Figure P10–2 (a) and (b), with shape functions given by Eq. (10.1.5), determine (a) intrinsic coordinate s
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<details>
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<summary>text_image</summary>
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x₁ = 10 in.
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A
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xₐ = 14 in.
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x₂ = 20 in.
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</details>
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<details>
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<summary>text_image</summary>
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A
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x₁ = 5 in. xₐ = 7 in. x₂ = 10 in.
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</details>
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(b)
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Figure P10–2
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at point A and (b) shape functions $N _ { 1 }$ and $N _ { 2 }$ at point A. If the displacements at nodes one and two are respectively, $u _ { 1 } = 0 . 0 0 6$ in. and $u _ { 2 } = - 0 . 0 0 6$ in., determine (c) the value of the displacement at point A and (d) the strain in the element.
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10.3 Answer the same questions as posed in problem 10.2 with the data listed under the Figure P10–3.
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<details>
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<summary>text_image</summary>
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x₁ = 20 mm
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u₁ = 0.1 mm
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A
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xₐ = 40 mm
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x₂ = 60 mm
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u₂ = 0.2 mm
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</details>
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<summary>text_image</summary>
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x₁ = 10 mm
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u₁ = 0.05 mm
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A
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xₐ = 20 mm
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x₂ = 30 mm
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u₂ = 0.1 mm
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</details>
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(b)
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Figure P10–3
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10.4 For the four-noded bar element in Figure P10–4, show that the Jacobian determinate is $\vert \underline { { J } } \vert = L / 2$ . Also determine the shape functions $N _ { 1 } - N _ { 4 }$ and the strain/displacement matrix B. Assume $u = a _ { 1 } + a _ { 2 } s + a _ { 3 } s ^ { 2 } + a _ { 4 } s ^ { 3 }$ .
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<details>
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<summary>text_image</summary>
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-1
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-1/2
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→ s
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1/2
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1
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1
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2
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3
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4
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</details>
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Figure P10–4
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10.5 Using the three-noded bar element shown in Figure P10–5 (a) and (b), with shape functions given by Eq. (10.6.9), determine (a) the intrinsic coordinate s at point A and (b) the shape functions, $N _ { 1 } , \ N _ { 2 }$ , and $N _ { 3 }$ at A. For the displacements of the nodes shown in Figure P10–5, determine (c) the displacement at A and (d) the axial strain expression in the element.
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<details>
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<summary>text_image</summary>
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A (xA = 13 in.)
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x₁ = 10 in. x₃ = 15 in. x₂ = 20 in.
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u₁ = 0.006 in. u₃ = 0 u₂ = -0.006 in.
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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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A (x_A = 7 in.)
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x_1 = 0
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x_3 = 5 in.
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x_2 = 10 in.
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u_1 = 0
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u_3 = 0.001 in.
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u_2 = 0.003 in.
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</details>
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(b)
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Figure P10–5
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10.6 Using the three-noded bar element shown in Figure P10–5 (a) and (b), with shape functions given by Eq. (10.6.9), determine (a) the intrinsic coordinate s at point A and (b) the shape functions, $N _ { 1 } , N _ { 2 }$ , and $N _ { 3 }$ at point A. For the displacements of the nodes shown in Figure P10–6, determine (c) the displacement at A and (d) the axial strain expression in the element.
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<details>
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<summary>text_image</summary>
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A (xA = 1.5 mm)
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x1 = 0
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x3 = 1 mm
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x2 = 2 mm
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u1 = 0
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u3 = 0.001 mm
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u2 = 0.002 mm
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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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A (x_A = 2.5 mm)
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x_1 = 2 mm x_3 = 3 mm x_2 = 4 mm
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u_1 = -0.001 mm u_3 = 0 u_2 = 0.001 mm
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</details>
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(b)
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Figure P10–6
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10.7 For the bar subjected to the linearly varying axial line load shown in Figure P10–7, use the linear strain (three-noded element) with two elements in the model, to determine the nodal displacements and nodal stresses. Compare your answer with that in Figure 3–31 and Eqs. (3.11.6) and (3.11.7). Let $A = 2 \mathrm { i n } . ^ { 2 }$ and $E = 3 0 \times 1 0 ^ { 6 }$ psi. Hint: Use Eq. (10.6.22) for the element sti¤ness matrix.
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<details>
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<summary>text_image</summary>
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10 × lb/in.
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60 in.
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x
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</details>
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Figure P10–7
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10.8 Use the three-noded bar element and find the axial displacement at the end of the rod shown in Figure P10–8. Determine the stress at $x = 0 , x = L / 2$ and $x = L$ . Let $A = 2 \times 1 0 ^ { - 4 } \mathrm { ~ m } ^ { 2 } , E = 2 0 5 \mathrm { { G P a } } ,$ and $L = 4 ~ \mathrm { m }$ . Hint: use Eq. (10.6.22) for the element sti¤ness matrix.
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<details>
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<summary>text_image</summary>
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2 kN/m (uniform)
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L = 4 m
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</details>
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Figure P10–8
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10.9 Show that the sum $N _ { 1 } + N _ { 2 } + N _ { 3 } + N _ { 4 }$ is equal to 1 anywhere on a rectangular element, where $N _ { 1 }$ through $N _ { 4 }$ are defined by Eqs. (10.2.5).
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10.10 For the rectangular element of Figure 10–3 on page 450, the nodal displacements are given by
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$$
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u _ {1} = 0 \quad v _ {1} = 0 \quad u _ {2} = 0. 0 0 5 \text { in. }
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$$
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$$
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v _ {2} = 0. 0 0 2 5 \text { in. } \quad u _ {3} = 0. 0 0 2 5 \text { in. } \quad v _ {3} = - 0. 0 0 2 5 \text { in. }
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$$
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$$
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u _ {4} = 0 \quad v _ {4} = 0
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$$
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For b ¼ 2 in., h ¼ 1 in., $E = 3 0 \times 1 0 ^ { 6 }$ psi, and $\nu = 0 . 3 ,$ , determine the element strains and stresses at the centroid of the element and at the corner nodes.
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10.11 Derive jJj given by Eq. (10.3.22) for a four-noded isoparametric quadrilateral element.
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10.12 Show that for the quadrilateral element described in Section 10.3, ½J- can be expressed as
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$$
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[ J ] = \left[ \begin{array}{c c c c} N _ {1, s} & N _ {2, s} & N _ {3, s} & N _ {4, s} \\ N _ {1, t} & N _ {2, t} & N _ {3, t} & N _ {4, t} \end{array} \right] \left[ \begin{array}{c c} x _ {1} & y _ {1} \\ x _ {2} & y _ {2} \\ x _ {3} & y _ {3} \\ x _ {4} & y _ {4} \end{array} \right]
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$$
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10.13 Derive Eq. (10.3.18) with $B _ { i }$ given by Eq. (10.3.19) 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).
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10.14 Use Eq. (10.3.30) with $p _ { s } = 0$ and $p _ { t } = p$ (a constant) alongside 3-4 of the element shown in Figure 10–6 on page 460 to obtain the nodal forces.
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10.15 For the element shown in Figure P10–15, replace the distributed load with the energy equivalent nodal forces by evaluating a force matrix similar to Eq. (10.3.29). Let $h = 0 . 1$ in thick.
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10.16 Use Gaussian quadature with two and three Gauss points and Table 10–1 to evaluate the following integrals:
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(a) ðaÞ $\int _ { - 1 } ^ { 1 } \cos { \frac { s } { 2 } } d s$
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ðbÞ $\int _ { - 1 } ^ { 1 } s ^ { 2 } d s$
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ðcÞ $\int _ { - 1 } ^ { 1 } s ^ { 4 } d s$
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ðdÞ $\int _ { - 1 } ^ { 1 } { \frac { \cos s } { 1 - s ^ { 2 } } } d s$ s2
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ðeÞ $\int _ { - 1 } ^ { 1 } s ^ { 3 } d s$ -1
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ðf $\int _ { - 1 } ^ { 1 } s \cos s d s$
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ðgÞ $\int _ { - 1 } ^ { 1 } ( 4 ^ { s } - 2 s ) d s$
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Then use the Newton-Cotes quadrature with two and three sampling points and Table 10–2 to evaluate the same integrals.
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10.17 For the quadrilateral elements shown in Figure P10–17, write a computer program to evaluate the sti¤ness matrices using four-point Gaussian quadrature as outlined in Section 10.5. Let $E = 3 0 \times 1 0 ^ { 6 }$ psi and $\nu = 0 . 2 5$ . The global coordinates (in inches) are shown in the figures.
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Figure P10–15
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<details>
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<summary>text_image</summary>
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y
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(3, 4) 4 3
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(3, 2) 1 2 (5, 4)
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(5, 2) 2 (5, 4)
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</details>
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<details>
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<summary>text_image</summary>
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y
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3
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(5, 5)
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4
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(3, 4)
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(3, 2)
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1
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2
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(5, 2)
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x
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</details>
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(b)
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Figure P10–17
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10.18 For the quadrilateral elements shown in Figure P10–18, evaluate the sti¤ness matrices using four-point Gaussian quadrature as outlined in Section 10.5. Let $E = 2 1 0$ GPa and $\nu = 0 . 2 5$ . The global coordinates (in millimeters) are shown in the figures.
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10.19 Evaluate the matrix B for the quadratic quadrilateral element shown in Figure 10–16 on page 480 (Section 10.6).
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10.20 Evaluate the sti¤ness matrix for the four-noded bar of Problem 10.4 using three-point Gaussian quadrature.
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Figure P10–18
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10.21 For the rectangular element in Figure P10–21, with the nodal displacements given in Problem 10.10, determine the $\underline { { \sigma } }$ matrix at $s = 0 , t = 0$ using the isoparametric formulation described in Section 10.5. (Also see Example 10.5.)
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<details>
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<summary>text_image</summary>
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y
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(0, 2)
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4
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3
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(4, 2)
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t
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s
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(0, 0)
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1
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2
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x
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</details>
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Figure P10–21
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10.22 For the three-noded bar (Figure P10–1), what Gaussian quadrature rule (how many Gauss points) would you recommend to evaluate the sti¤ness matrix? Why?
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# Three-Dimensional Stress Analysis
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# Introduction
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In this chapter, we consider the three-dimensional, or solid, element. This element is useful for the stress analysis of general three-dimensional bodies that require more precise analysis than is possible through two-dimensional and/or axisymmetric analysis. Examples of three-dimensional problems are arch dams, thick-walled pressure vessels, and solid forging parts as used, for instance, in the heavy equipment and automotive industries. Figure 11–1 shows finite element models of some typical automobile parts. Also see Figure 1–7 for a model of a swing casting for a backhoe frame, Figure 1–9 for a model of a pelvis bone with an implant, and Figures 11–7 through 11–10 of a forging part, a foot pedal, a hollow pipe section, and an alternator bracket, respectively.
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The tetrahedron is the basic three-dimensional element, and it is used in the development of the shape functions, stiffness matrix, and force matrices in terms of a global coordinate system. We follow this development with the isoparametric formulation of the stiffness matrix for the hexahedron, or brick element. Finally, we will provide some typical three-dimensional applications.
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In the last section of this chapter, we show some three-dimensional problems solved using a computer program.
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# 11.1 Three-Dimensional Stress and Strain
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We begin by considering the three-dimensional infinitesimal element in Cartesian coordinates with dimensions dx; dy, and dz and normal and shear stresses as shown in Figure 11–2. This element conveniently represents the state of stress on three mutually perpendicular planes of a body in a state of three-dimensional stress. As usual, normal stresses are perpendicular to the faces of the element and are represented by
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<details>
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<summary>natural_image</summary>
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3D rendered mechanical component with a star-like cutout and grid pattern (no text or symbols)
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</details>
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<details>
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<summary>natural_image</summary>
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3D rendered mechanical component with visible internal cavities and embedded debris (no text or symbols)
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</details>
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Figure 11–1 (a) wheel rim; (b) engine block. ((a) Courtesy of Mark Blair; (b) courtesy of Mark Guard.)
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$\sigma _ { x } , \sigma _ { y } .$ , and $\sigma _ { z }$ . Shear stresses act in the faces ( planes) of the element and are represented by $\tau _ { x y } , \tau _ { y z } , \tau _ { z x } ,$ and so on.
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From moment equilibrium of the element, we show in Appendix C that
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$$
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\tau_ {x y} = \tau_ {y x} \quad \tau_ {y z} = \tau_ {z y} \quad \tau_ {z x} = \tau_ {x z}
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$$
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Hence, there are only three independent shear stresses, along with the three normal stresses.
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<details>
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<summary>text_image</summary>
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y, v
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σy
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τyx
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dy
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τyz
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τxy
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τxz
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σx
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x, u
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σz
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τzx
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dz
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z, w
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dx
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</details>
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Figure 11–2 Three-dimensional stresses on an element
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The element strain/displacement relationships are obtained in Appendix C. They are repeated here, for convenience, as
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$$
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\varepsilon_ {x} = \frac {\partial u}{\partial x} \quad \varepsilon_ {y} = \frac {\partial v}{\partial y} \quad \varepsilon_ {z} = \frac {\partial w}{\partial z} \tag {11.1.1}
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$$
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where $u , v ,$ and w are the displacements associated with the $x , y ,$ and z directions. The shear strains g are now given by
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$$
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\gamma_ {x y} = \frac {\partial u}{\partial y} + \frac {\partial v}{\partial x} = \gamma_ {y x}
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$$
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$$
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\gamma_ {y z} = \frac {\partial v}{\partial z} + \frac {\partial w}{\partial y} = \gamma_ {z y} \tag {11.1.2}
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$$
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$$
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\gamma_ {z x} = \frac {\partial w}{\partial x} + \frac {\partial u}{\partial z} = \gamma_ {x z}
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$$
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where, as for shear stresses, only three independent shear strains exist.
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We again represent the stresses and strains by column matrices as
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$$
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\{\sigma \} = \left\{ \begin{array}{l} \sigma_ {x} \\ \sigma_ {y} \\ \sigma_ {z} \\ \tau_ {x y} \\ \tau_ {y z} \\ \tau_ {z x} \end{array} \right\} \quad \{\varepsilon \} = \left\{ \begin{array}{l} \varepsilon_ {x} \\ \varepsilon_ {y} \\ \varepsilon_ {z} \\ \gamma_ {x y} \\ \gamma_ {y z} \\ \gamma_ {z x} \end{array} \right\} \tag {11.1.3}
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$$
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The stress/strain relationships for an isotropic material are again given by
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$$
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\{\sigma \} = [ D ] \{\varepsilon \} \tag {11.1.4}
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$$
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