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김경종
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Isotropy. An isotropic material has no preferred directions. Material properties are commonly expressed as a combination of two of the following: elastic modulus E, Poisson's ratio $\nu$ , and shear modulus G. In its upper triangle, [E] contains 12 zero coefficients and the 9 nonzero coefficients
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$$
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E _ {1 1} = E _ {2 2} = E _ {3 3} = (1 - \nu) c
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$$
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$$
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E _ {1 2} = E _ {1 3} = E _ {2 3} = \nu c \tag {1.7-3}
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$$
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$$
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E _ {4 4} = E _ {5 5} = E _ {6 6} = G
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$$
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where
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$$
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c = \frac {E}{(1 + \nu) (1 - 2 \nu)} \quad \text { and } \quad G = \frac {E}{2 (1 + \nu)}
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$$
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Because of the relation $E = 2(1 + \nu)G$ and the symmetry of [E], we see that [E] for an isotropic material contains only two independent coefficients.
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Plane strain is defined as a deformation state in which $w = 0$ everywhere and $u$ and $v$ are functions of $x$ and $y$ but not of $z$ . Thus, $\epsilon_z = \gamma_{yz} = \gamma_{zx} = 0$ . A typical slice of an underground tunnel that lies along the $z$ axis might deform in essentially plane strain conditions. The stress-strain relation $\{\sigma\} = [\mathrm{E}]\{\epsilon\}$ for isotropic and isothermal plane strain is
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$$
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\left\{ \begin{array}{l} \sigma_ {x} \\ \sigma_ {y} \\ \tau_ {x y} \end{array} \right\} = \frac {E}{(1 + \nu) (1 - 2 \nu)} \left[ \begin{array}{c c c c} 1 & - \nu & \nu & 0 \\ & \nu & 1 - \nu & 0 \\ & 0 & 0 & \frac {1 - 2 \nu}{2} \end{array} \right] \left\{ \begin{array}{l} \epsilon_ {x} \\ \epsilon_ {y} \\ \gamma_ {x y} \end{array} \right\} \tag {1.7-4}
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$$
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In Eq. 1.7-4, [E] is obtained by discarding rows and columns-3, -5, and 6 from the 6 by 6 matrix [E] of an isotropic solid.
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$\text{Stress } \sigma_z \text{ does not appear in Eq. 1.7-4, even though it is usually not zero. If needed, } \sigma_z \text{ can be obtained from the relation } \epsilon_z = 0 = (\sigma_z - \nu \sigma_y - \nu \sigma_x)/E \text{ after } \sigma_y \text{ and } \sigma_x \text{ are known.}$
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Plane stress is a condition that prevails in a flat plate in the xy plane, loaded only in its own plane and without z-direction restraint, so that $\sigma_{z} = \tau_{yz} = \tau_{zx} = 0$ . Then, for isotropic and isothermal conditions,
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$$
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\checkmark \left\{ \begin{array}{l} \epsilon_ {x} \\ \epsilon_ {y} \\ \gamma_ {x y} \end{array} \right\} = \frac {1}{E} \left[ \begin{array}{c c c} 1 & - \nu & 0 \\ - \nu & 1 & 0 \\ 0 & 0 & E / G \end{array} \right] \left\{ \begin{array}{l} \sigma_ {x} \\ \sigma_ {y} \\ \tau_ {x y} \end{array} \right\}
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$$
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or
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$$
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\left\{ \begin{array}{l} \sigma_ {x} \\ \sigma_ {y} \\ \tau_ {x y} \end{array} \right\} = \frac {E}{1 - \nu^ {2}} \left[ \begin{array}{c c c} 1 & \nu & 0 \\ \nu & 1 & 0 \\ 0 & 0 & \frac {1 - \nu}{2} \end{array} \right] \left\{ \begin{array}{l} \epsilon_ {x} \\ \epsilon_ {y} \\ \gamma_ {x y} \end{array} \right\} \tag {1.7-5}
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$$
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where $E / G = 2(1 + \nu)$ . The square matrices, including their scalar multipliers $1 / E$ and $E / (1 - \nu^2)$ , are, respectively, [C] and [E].
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Axially symmetric solids require a 4 by 4 matrix [E]. This problem is discussed in Chapter 10.
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<!-- source-page: 42 -->
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Beam Bending. Consider again the beam of Fig. 1.5-2. Uniaxial stress prevails, so $\sigma_{x}=E\epsilon_{x}$ . Combining this equation with Eq. 1.5-7, we obtain $\sigma_{x}=-Ezw_{,xx}$ . Let the beam have cross-sectional area A and a constant modulus E. Multiplying by z dA and integrating over a cross section of depth t, we obtain
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$$
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\int_ {- t / 2} ^ {t / 2} \sigma_ {x} z d A = - E w _ {, x x} \int_ {- t / 2} ^ {t / 2} z ^ {2} d A \tag {1.7-6}
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$$
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The former integral is identified as bending moment M, positive when it bends the beam of Fig. 1.5-2 concave down. The latter integral is identified as the moment of inertia I of the cross-sectional area. Thus Eq. 1.7-6 becomes
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$$
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M = - E I w _ {, x x} \tag {1.7-7}
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$$
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This is a familiar expression from elementary mechanics of materials. It is the form of $\{\sigma\} = [E]\{\epsilon\}$ that applies to beam bending and is called a load–displacement relation. A similar expression, expanded to two dimensions, is used for plate bending (Chapter 11).
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Initial Stress and Strain. Thermal Effects. The term “initial stress” signifies a stress present before deformations are allowed. Effectively, it is a residual stress to be superposed on stress caused by deformation. With the addition of initial stresses $\{\sigma_{0}\}$ and initial strains $\{\epsilon_{0}\}$ , Eq. 1.7-2 becomes
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$$
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\{\boldsymbol {\sigma} \} = [ \mathrm{E} ] (\{\boldsymbol {\epsilon} \} - \{\boldsymbol {\epsilon} _ {0} \}) + \{\boldsymbol {\sigma} _ {0} \} \tag {1.7-8}
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$$
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As examples, $\{\epsilon_{0}\}$ might describe moisture-induced swelling and $\{\sigma_{0}\}$ might describe stresses produced by heating. Alternatively, both effects can be placed in $\{\epsilon_{0}\}$ , or $\{\epsilon_{0}\}$ and $\{\sigma_{0}\}$ can be viewed as alternative ways to express the same thing. For example, free expansion of an orthotropic material with principal axes xyz produces the initial strains
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$$
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\left\{\epsilon_ {0} \right\} = \left\lfloor \alpha_ {x} T \quad \alpha_ {y} T \quad \alpha_ {z} T \quad 0 \quad 0 \quad 0 \right] ^ {T} \tag {1.7-9}
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$$
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where $T$ is the temperature relative to an arbitrary reference temperature at which the body is free of stress, and the $\alpha$ 's are coefficients of thermal expansion in the principal material directions. Thus, to account for the effects of temperature change, we can substitute Eq. 1.7-9 into Eq. 1.7-8 and set $\{\sigma_0\} = \{0\}$ . Alternatively, we can substitute $\{\sigma_0\} = -[\mathbf{E}] \left[ \alpha_x T \quad \alpha_y T \quad \alpha_z T \quad 0 \quad 0 \quad 0 \right]^T$ into Eq. 1.7-8 and set $\{\epsilon_0\} = \{0\}$ . Stresses $\{\sigma\} = \{\sigma_0\}$ prevail when mechanical strains $\{\epsilon\}$ are prohibited. In the special case of isotropy we have, in three dimensions,
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$$
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\left\{\sigma_ {0} \right\} = - \frac {E \alpha T}{1 - 2 \nu} \left[ \begin{array}{l l l l l l} 1 & 1 & 1 & 0 & 0 & 0 \end{array} \right] ^ {T} \tag {1.7-10}
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$$
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<!-- source-page: 43 -->
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and, in plane stress,
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$$
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\left\{\epsilon_ {0} \right\} = \left\lfloor \alpha T \quad \alpha T \quad 0 \right] ^ {T} \quad \text {and} \quad \left\{\sigma_ {0} \right\} = - \frac {E \alpha T}{1 - \nu} \left\lfloor 1 \quad 1 \quad 0 \right] ^ {T} \tag {1.7-11}
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$$
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and, finally, in plane strain,
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$$
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\left\{\boldsymbol {\epsilon} _ {0} \right\} = (1 + \nu) \left\lfloor \alpha T \quad \dot {\alpha} T \quad 0 \right] ^ {T} \quad \text { and } \quad \left\{\boldsymbol {\sigma} _ {0} \right\} = - \frac {E \alpha T}{1 - 2 \nu} \left\lfloor 1 \quad 1 \quad 0 \right] ^ {T} \tag {1.7-12}
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$$
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Temperature-dependent moduli can be accommodated by using the [E] appropriate to the temperature that prevails. A temperature-dependent expansion coefficient is more troublesome. By definition, $\alpha = \partial\epsilon/\partial T$ when deformation is unrestrained, so $\epsilon = \alpha T$ only if $\alpha$ is independent of T. Otherwise,
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$$
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\epsilon = \int_ {0} ^ {T} \alpha d T = \overline {{{\alpha}}} T, \quad \text { where } \quad \overline {{{\alpha}}} = \frac {1}{T} \int_ {0} ^ {T} \alpha d T \tag {1.7-13}
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$$
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Here $\bar{\alpha}$ is an average expansion coefficient, valid only over the temperature range from 0 to T.
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Remarks. For plane stress or plane strain conditions to prevail in the xy plane, the xy plane must be a plane of elastic symmetry. Thus, if the material is orthotropic, the z axis must be a principal material direction. If, in addition, the x and y axes are principal material directions, then $E_{13} = E_{31} = E_{23} = E_{32} = 0$ in the 3 by 3 matrix [E].
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Poisson's ratio $\nu$ is little affected by temperature. Modulus $E$ is affected more: for stainless steel $E$ decreases about $20\%$ when the temperature rises from $0^{\circ}$ to $450^{\circ}\mathrm{C}$ . Barring plastic flow, elastic properties are almost independent of stress. For example, an increase in hydrostatic pressure from 0 to 350 MPA increases the moduli of steel and aluminum $0.8\%$ and $2.6\%$ , respectively. Like $E$ , thermal expansion coefficient $\alpha$ is relatively insensitive to stress but may vary appreciably with temperature.
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When strain rates are high, as in wave propagation, modulus E is higher than its static value. The difference is appreciable for rubber-like materials but negligible for common metals unless strain rates are extreme, as in explosive forming processes.
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If a body is isotropic and linearly elastic, and its supports do not inhibit thermal expansion or contraction, then the body deforms but remains free of stress when T is a harmonic function; that is, when $\nabla^{2}T = 0$ . A special case of this is when T is a linear function of the coordinates, for which both isotropic and rectilinearly orthotropic bodies remain free of stress (but not curvilinearly orthotropic bodies, such as tree trunks).
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Capabilities of the finite element method far exceed the knowledge of material behavior on which an analysis must be based. If test data are lacking, as is often the case with anisotropic materials, we can only estimate the elastic constants. Even when the constants are known, anisotropy has an adverse effect on the accuracy of finite element solutions $[8.34]$ .
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<!-- source-page: 44 -->
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# 1.8 WARNING: THE COMPUTED ANSWER MAY·BE WRONG
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Users of finite element programs may be so impressed by the power of the method that its limitations are ignored. Whether computer-based or not, analytical methods rely on assumptions and on theory that is not universally applicable. The analyst may overlook or misjudge important aspects of physical behavior. There may be an error in the computer program. A large program has many options and many computational paths. Perhaps some paths have never before been exercised, were not anticipated by the program designers, and have never been checked. Far more likely causes of incorrect results are user errors, such as using an inappropriate program or supplying an appropriate program with the wrong data. A poor mesh may be used, an inappropriate element type may be chosen, yielding or buckling may be overlooked, support conditions may be misrepresented, and so on. Users must remember that a structure is not obliged to behave as a computer says it should, regardless of how expensive the program, how many digits are printed in the results, or how elegant the graphic display. Computer graphics has achieved such a level of polish and versatility as to inspire great trust in the underlying analysis, a trust that may be unwarranted. (One can now make mistakes with more confidence than ever before.)
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The finite element method is a most versatile tool, but not the best analytical tool for every problem. It is foolish, but not unheard of, to use three-dimensional finite elements to compute stresses obtainable by the flexure formula $\sigma = Mc/I$ . In other cases experiment may be the most appropriate method, especially if experiment is needed anyway to obtain data needed for analysis (data such as material properties, the effective stiffness of joints, damping properties, the time history of loads, etc.). If an analysis is to be done by numerical methods, finite elements are not the only choice. For example, finite difference methods are effective for shells of revolution, and boundary elements are effective for some problems with boundaries at infinity.
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Powerful computer programs cannot be used without training. Their results cannot be trusted if users have no knowledge of their internal workings and little understanding of the physical theories on which they are based. An error caused by misunderstanding or oversight is not correctible by mesh refinement or by use of a more powerful computer. Some authorities have suggested that users be “qualified,” somewhat in the manner of practitioners having to be licensed before engaging in a profession in which the potential for damage to the public is substantial. Although the finite element method can make a good engineer better, it can make a poor engineer more dangerous.
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In years past, when analysis was done by hand, the analyst was required to invent a mathematical model before undertaking its analysis. Invention of a good model required sound physical understanding of the problem. Understanding can now be replaced by activation of a computer program. Having had little need to sharpen intuitions by devising simple models, the computer user may lack the physical understanding needed to prepare a good model and to check computed results. Or, what the user perceives as understanding may instead be familiarity with previous computer output.
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Computed results must in some way be judged or compared with expectations. Alternative results, useful for comparison, might be obtained from a different computer program that relies on a different analytical basis, from a simplified
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<!-- source-page: 45 -->
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model amenable to hand calculation, from the behavior of similar structures already built, and from experiment. Experiment may be expensive and has its own pitfalls, but is desirable if the analytical process is pushed beyond previous experience and established practice.
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The overall message of this discussion is that a competent analyst must have sound engineering judgment and experience, and that doubts raised in the course of an analysis should be taken seriously.
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# PROBLEMS
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# Section 1.1
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1.1 In Fig. 1.1-1a, let cross-sectional area A vary linearly from $3A_{0}$ at x = 0 to $A_{0}$ at $x = L_{T}$ . Model the bar by uniform elements. Let the cross-sectional area of each element be that of the actual bar at the x coordinate of the element midpoint. Assume that elastic modulus E is constant. Solve for the displacement of load P in the following ways, and compute the percentage error of each result. The exact answer is $PL_{T} \ln 3/2EA_{0}$ .
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(a) Use a single uniform element. Let $L_{1} = L_{T}$ (and $A = 2A_{0}$ ).
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(b) Use two uniform elements. Let $L_{1} = L_{2} = L_{T} / 2$ .
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(c) Use three uniform elements. Let $L_{1} = L_{2} = L_{3} = L_{T} / 3$ .
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(d) Use four uniform elements, each of length $L_{T} / 4$ .
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1.2 Derive the “exact answer” stated in Problem 1.1.
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1.3 The bar shown has a uniform circular cross section of radius $a$ . It carries only torsional loads. Develop the 2 by 2 element stiffness matrix in terms of $a$ , $L$ , and shear modulus $G$ .
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<details>
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<summary>text_image</summary>
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θ₁,T₁
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L
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S₂
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θ₂,T₂
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</details>
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Problem 1.3
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1.4 Let $\phi$ be a function that is interpolated over an element, where $\phi$ is a function of $x, y$ , element dimensions, and nodal values of $\phi$ . For each element shown, write this expression for $\phi$ .
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(a) Consider an element of length $L$ . Let $\phi$ vary linearly with axial coordinate $x$ .
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(b) Consider the triangular element (see Eq. 1.1-1).
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(c) Consider the rectangular element (see Eq. 1.1-2).
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(a)
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<details>
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<summary>text_image</summary>
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y
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3 ← a →
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b
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1 2 x
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</details>
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(b)
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<details>
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<summary>text_image</summary>
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y
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a
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4
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3
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b
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1
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2
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x
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</details>
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(c)
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Problem 1.4
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<!-- source-page: 46 -->
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1.5 The quadrilateral element shown might be used in the finite element model of Fig. 1.1-2b. Imagine that its $x$ -direction displacement field is $u = a_1 + a_2x + a_3y + a_4xy$ , where the $a_l$ are constants. How does $u$ vary with $x$ or $y$ along each side? Do you think this element will be compatible with its neighbors?
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<details>
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<summary>text_image</summary>
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y
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4 3
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45°
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1 2 x
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</details>
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Problem 1.5
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# Section 1.2
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1.6 Following the procedure used in Eqs. 1.2-10 to Eq. 1.2-14, derive the stiffness coefficients in columns 2, 3, and 4 of the stiffness matrix of a uniform beam element.
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1.7 The bar shown can have both axial and bending deformations. Regard the bar as a single element with the four translational and two rotational d.o.f. shown. Without calculation, determine the algebraic sign of each coefficient in the 6 by 6 element stiffness matrix [k] (or enter zero for a null coefficient). Assume that displacements and rotations are small. Suggestion: Sketch each of six separate deformation states. Show the forces and moments needed to produce these states while satisfying static equilibrium. Finally, compare the directions of these loads with the assumed positive directions (which are those of the six nodal d.o.f.).
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<details>
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<summary>text_image</summary>
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d₁
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d₂
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d₃
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d₆
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d₄
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d₅
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</details>
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Problem 1.7
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<details>
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<summary>text_image</summary>
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d₃
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d₁
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d₂
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d₆
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d₅
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d₄
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45°
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</details>
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Problem 1.8
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1.8 Repeat the instructions of Problem 1.7, but with reference to the inclined bar shown. Assume that the bar has an axial stiffness much greater than its bending stiffness. Also assume that displacements and rotations are small.
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1.9 Repeat the instructions of Problem 1.7, but with reference to the bent bar shown. Suggestion: Assume that $[k]$ is symmetric. Also, start with column 2 of $[k]$ , then proceed to columns 1, 3, 4, 5, and 6.
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<!-- source-page: 47 -->
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<details>
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<summary>text_image</summary>
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d₁
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d₂
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d₃
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d₄
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d₅
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d₆
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</details>
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Problem 1.9
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# Section 1.3
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1.10 The structure shown consists of a rigid, weightless bar and two linear springs of stiffnesses $k_{1}$ and $k_{2}$ . Only small vertical displacements are permitted. The stiffness matrix [K] of this structure is 2 by 2 but can have various forms depending on the choice of d.o.f. Write [K] for each of the following choices of d.o.f.
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(a) Displacements $v_{1}$ at $x = 0$ and $v_{2}$ at $x = L$ (shown in the second sketch).
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(b) Displacements $v_{1}$ at $x = 0$ and $v_{A}$ at $x = L / 2$ .
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(c) Displacements $v_{2}$ at $x = L$ and $v_{B}$ at $x = 2L$ .
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(d) Displacement $v_{1}$ at $x = 0$ and a small rotation $\theta$ about $x = 0$ .
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(e) Displacement $v_{B}$ at $x = 2L$ and a small rotation $\theta$ about $x = 2L$ .
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<details>
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<summary>text_image</summary>
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y,v
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L
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L
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A
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B
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x
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k₁
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k₂
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</details>
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<details>
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<summary>text_image</summary>
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Sketch for part (a)
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v₁
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v₂
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k₁
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k₂
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</details>
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Problem 1.10
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1.11 The angled bar is rigid and weightless. With its two linear springs it forms a structure whose stiffness matrix [K] is 2 by 2. Various forms of [K] are
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<details>
|
||||
<summary>text_image</summary>
|
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|
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k₂
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k₁
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a
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||||
b
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||||
</details>
|
||||
|
||||
|
||||
|
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<details>
|
||||
<summary>text_image</summary>
|
||||
|
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v₁
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v₂
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||||
</details>
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|
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(a)
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|
||||
|
||||
<details>
|
||||
<summary>text_image</summary>
|
||||
|
||||
θ₁
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v₁
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||||
</details>
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||||
|
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(b)
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||||
|
||||
<details>
|
||||
<summary>text_image</summary>
|
||||
|
||||
a/2
|
||||
v₁ v₃
|
||||
</details>
|
||||
|
||||
(c)
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Problem 1.11
|
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|
||||
<!-- source-page: 48 -->
|
||||
|
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possible for various choices of d.o.f. Write [K] for each of the choices (a), (b), and (c) shown in the sketch. Displacements are small in each case.
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1.12 The structure shown consists of linear springs whose stiffnesses are $k_{1}, k_{2}, k_{3}$ , and $k_{4}$ . Only horizontal displacements are allowed.
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|
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(a) In matrix form, write the three equilibrium equations of the structure. The d.o.f. are $u_{1}, u_{2}$ , and $u_{3}$ .
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|
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(b) Let $k_{1} = k_{2} = k_{3} = k_{4} = k$ and $F_{1} = F_{2} = 0$ . Determine $u_{1}, u_{2}$ , and $u_{3}$ in terms of $k$ and $F_{3}$ .
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|
||||
|
||||
|
||||
<details>
|
||||
<summary>text_image</summary>
|
||||
|
||||
u₁,F₁
|
||||
u₂,F₂
|
||||
u₃,F₃
|
||||
k₁
|
||||
k₂
|
||||
k₃
|
||||
k₄
|
||||
</details>
|
||||
|
||||
Problem 1.12
|
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|
||||
1.13 The structure shown consists of rigid bars $AB$ and $CD$ and linear springs that connect $A$ to $C$ and $B$ to $D$ . Only horizontal motions and small rotations of the bars are permitted. In terms of $k_1, k_2$ , and $a$ , determine the 4 by 4 stiffness matrix that operates on $[u_1 \theta_1 u_2 \theta_2]^T$ to yield $[F_1 M_1 F_2 M_2]^T$ . Suggestion: Since rotations are small, $u_A = u_1 - a\theta_1$ , with similar relations for $u_B, u_C$ , and $u_D$ .
|
||||
|
||||
|
||||
|
||||
<details>
|
||||
<summary>text_image</summary>
|
||||
|
||||
A
|
||||
k₁
|
||||
C
|
||||
1
|
||||
θ₁,M₁
|
||||
a
|
||||
θ₂,M₂
|
||||
u₁,F₁
|
||||
2
|
||||
u₂,F₂
|
||||
B
|
||||
k₂
|
||||
D
|
||||
</details>
|
||||
|
||||
Problem 1.13
|
||||
|
||||
1.14 The uniform, linearly elastic bar shown is fixed at both ends. Force $P$ is applied at node 2. Use the finite element method to compute nodal displacements and stresses in each element in terms of $P, L, A$ , and $E$ . Compare these results with exact values.
|
||||
|
||||
1.15 The uniform, linearly elastic bar shown carries a uniformly distributed load $q_0$ . Assume that $q_0$ produces the respective nodal loads $q_0L / 2$ , $q_0L$ , and $q_0L / 2$ .
|
||||
|
||||
(a) Compute nodal displacements by the finite element method and compare them with the exact values, which are $u_{2} = 3q_{0}L^{2} / 2AE$ and $u_{3} = 2q_{0}L^{2} / AE$ .
|
||||
|
||||
(b) Using the procedure of Eqs. 1.3-6, compute the axial stress in each element. On a single set of axes, plot these stresses as well as the actual stress distribution along the bar. Do the results suggest a rule regarding where stresses should be calculated in a finite element?
|
||||
|
||||
<!-- source-page: 49 -->
|
||||
|
||||
|
||||
|
||||
<details>
|
||||
<summary>text_image</summary>
|
||||
|
||||
P 2 3
|
||||
1
|
||||
L L L
|
||||
4
|
||||
</details>
|
||||
|
||||
Problem 1.14
|
||||
|
||||
|
||||
|
||||
<details>
|
||||
<summary>text_image</summary>
|
||||
|
||||
1
|
||||
q₀
|
||||
2
|
||||
3
|
||||
L
|
||||
L
|
||||
</details>
|
||||
|
||||
Problem 1.15
|
||||
|
||||
# Section 1.6
|
||||
|
||||
1.16 (a) The sketch shows a plane differential element similar to that in Fig. 1.6-1a but in polar coordinates. The stresses are $\sigma_r$ (radial), $\sigma_\theta$ (circumferential), and $\tau_{r\theta}$ (shear). Derive the differential equations of equilibrium in polar coordinates.
|
||||
|
||||
(b) Similarly, use cylindrical coordinates to derive equilibrium equations analogous to the first set of Eqs. 1.6-4.
|
||||
|
||||
|
||||
|
||||
<details>
|
||||
<summary>text_image</summary>
|
||||
|
||||
Fθ
|
||||
Fτ
|
||||
τrθ
|
||||
σr
|
||||
dθ
|
||||
τrθ
|
||||
σθ
|
||||
dr
|
||||
</details>
|
||||
|
||||
Problem 1.16
|
||||
|
||||
1.17 Imagine that stresses in the $xy$ plane are given as $\sigma_x = -6a_1x^2$ , $\sigma_y = 12a_1x^2$ , and $\tau_{xy} = 12a_1y^2$ , where $a_1$ is a constant.
|
||||
|
||||
(a) Consider the square region $0 \leqslant x \leqslant b$ , $0 \leqslant y \leqslant b$ . Write expressions for tractions $\Phi_x$ and $\Phi_y$ on each side of this square, in terms of $x, y, b$ , and $a_1$ .
|
||||
(b) If body forces are zero, is the given stress field in fact possible? Explain.
|
||||
|
||||
1.18 In a certain approximation method, stresses in a plane region are assumed to have the forms
|
||||
|
||||
$$
|
||||
\sigma_ {x} = a _ {1} + a _ {2} x + a _ {3} y, \quad \sigma_ {y} = a _ {4} + a _ {5} x + a _ {6} y, \quad \tau_ {x y} = a _ {7} + a _ {8} x + a _ {9} y
|
||||
$$
|
||||
|
||||
where each $a_{i}$ is a constant. Let all body forces vanish. What must be the relation among the $a_{i}$ if equilibrium is to be satisfied?
|
||||
|
||||
1.19 Determine whether or not the following stress field is a valid solution of a plane elasticity problem: $\sigma_x = 3a_1x^2y$ , $\sigma_y = a_1y^3$ , and $\tau_{xy} = -3a_1xy^2$ , where $a_1$ is a constant. The body is isotropic and linearly elastic and body forces are zero.
|
||||
|
||||
# Section 1.7
|
||||
|
||||
1.20 Judging by Eq. 1.7-4, what property does a material display if $\nu = 0.5$ ?
|
||||
1.21 Combine the latter form of Eqs. 1.7-5, the strain-displacement relations, and the equilibrium equations with $F_{x} = F_{y} = 0$ , and show that
|
||||
|
||||
<!-- source-page: 50 -->
|
||||
|
||||
$$
|
||||
u _ {, x x} + u _ {, y y} = (1 + \nu) (u _ {, y y} - v _ {, x y}) / 2
|
||||
$$
|
||||
|
||||
This equation, and its companion (obtained by interchange of u with v and x with y), are known as the equilibrium equations expressed in terms of displacements.
|
||||
|
||||
1.22 Let displacements in a plane stress problem be given by
|
||||
|
||||
$$
|
||||
\begin{array}{l} u = a _ {1} + a _ {2} x + a _ {3} y + a _ {4} x ^ {2} + a _ {5} x y + a _ {6} y ^ {2} \\ v = a _ {7} + a _ {8} x + a _ {9} y + a _ {1 0} x ^ {2} + a _ {1 1} x y + a _ {1 2} y ^ {2} \\ \end{array}
|
||||
$$
|
||||
|
||||
where each $a_{i}$ is a constant. Let all body forces vanish. What relation among the $a_{i}$ is needed to satisfy equilibrium? The medium is isotropic.
|
||||
|
||||
1.23 Start with Eqs. 1.7-8 and 1.7-9, and derive the plane strain form of $\{\epsilon_0\}$ (Eq. 1.7-12).
|
||||
|
||||
[Unreadable]
|
||||
Reference in New Issue
Block a user