김경종
20 KiB
Problems
14.1 For the one-dimensional flow through the porous media shown in Figure P14–1, determine the potentials at one-third and two-thirds of the length. Also determine the velocities in each element. Let A = 0 . 2 \mathrm { m } ^ { 2 } .
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K_{xx}^{(1)} = 2 m/s K_{xx}^{(2)} = 4 m/s K_{xx}^{(3)} = 6 m/s p_1 = 10 m 1 ① 2 ② 3 ③ 4 p_4 = 0 m 1 m 1 m 1 m
Figure P14–1
14.2 For the one-dimensional flow through the porous medium shown in Figure P14–2 with fluid flux at the right end, determine the potentials at the third points. Also determine the velocities in each element. Let A = 2 ~ \mathrm { m } ^ { 2 } .
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Kxx = 1 m/s p1 = 10 m 3 m q* = 25 m/s
Figure P14–2
14.3 For the one-dimensional fluid flow through the stepped porous medium shown in Figure P14–3, determine the potentials at the junction of each area. Also determine the velocities in each element. Let K _ { x x } = 1 ~ \mathrm { i n . / s } .
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p₁ = 10 in.1 A₁ = 6 in² 2 • A₂ = 4 in² 3 • A₃ = 2 in² 4 • p₄ = 0 10 in. 10 in. 10 in.
Figure P14–3
14.4 For the one-dimensional fluid-flow problem (Figure P14–4) with velocity known at the right end, determine the velocities and the volumetric flow rates at nodes 1 and 2. Let K _ { x x } = 2 \mathrm { c m } / \mathrm { s } .
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1 • A₁ = 5 cm² 2 • A₂ = 3 cm² 3 • → v₃ = 2 cm/s 5 cm 5 cm
Figure P14–4
14.5 Derive the sti¤ness matrix, Eq. (14.2.15), using the first term on the right side of Eq. (13.4.17).
14.6 For the one-dimensional fluid-flow problem in Figure P14–6, determine the velocities and volumetric flow rates at nodes 2 and 3. Let \bar { K _ { x x } } = 1 0 ^ { - 1 } \mathrm { i n . / s } .
v _ {1} = 2 \text { in. / s } \boxed {1 \quad A _ {1} = 2 \text { in } ^ {2} \quad 2 \bullet \quad A _ {2} = 1 \text { in } ^ {2} \quad 3}
Figure P14–6
14.7 For the triangular element subjected to a fluid source shown in Figure P14–7, determine the amount of Q ^ { * } allocated to each node.
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(2, 7) Q* = 100 m²/s (2, 1) (4, 2) (9, 1) (All units meters)
Figure P14–7
14.8 For the triangular element subjected to the surface fluid source shown in Figure P14–8, determine the amount of fluid force at each node.
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q* = 5 in./s (2, 3) (2, 1) (4, 1) x All units inches)
Figure P14–8
14.9 For the two-dimensional fluid flow shown in Figure P14–9, determine the potentials at the center and right edge.
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p = 10 m 2 m 2 m q* = 25 × 10⁻⁵ m/s
Figure P14–9
14.10– Using a computer program, determine the potential distribution in the two-dimensional 14.15 bodies shown in Figures P14–10–P14–15.
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p = 500 m away from pump pumping rate = 5000 m²/day Pump Kₓₓ = Kᵧᵧ = 40 m/day
Figure P14–10
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p = 6 cm 2 1 2 cm 4 2 cm 6 3 5 8 1 cm 7 p = 3 cm ① ② ③ ④ ⑤ ⑥ Kₓₓ = Kᵧᵧ = 2 cm/s
Figure P14–11
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p = 100 m 200 m 400 m Pump Kxx = Ky = 100 m/day p = 100 m Pumping rate = 500 m²/day
Figure P14–12
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0.5 m/s 1 m 0.5 m 0.2 m 0.2 m
Figure P14–13
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0.5 m/s 1 m 0.5 m Ellipse |0.4 m| 0.2 m
Figure P14–14
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30 m Kxx = Ky = 50 m/day 22.5 m Kxx = Ky = 10 m/day
Figure P14–15
Introduction
In this chapter, we consider the problem of thermal stresses within a body. First, we will discuss the strain energy due to thermal stresses (stresses resulting from the constrained motion of a body or part of a body during a temperature change in the body).
The minimization of the thermal strain energy equation is shown to result in the thermal force matrix. We will then develop this thermal force matrix for the onedimensional bar element and the two-dimensional plane stress and plane strain elements.
We will outline the procedures for solving both one- and two-dimensional problems and then provide solutions of specific problems, including illustration of a computer program used to solve thermal stress problems for two-dimensional plane stress.
15.1 Formulation of the Thermal Stress Problem and Examples
In addition to the strains associated with the displacement functions due to mechanical loading, there may be other strains within a body due to temperature variations, swelling (moisture differential), or other causes. We will concern ourselves only with the strains due to temperature variation, eT , and will consider both one- and two-dimensional problems.
Temperature changes in a structure can result in large stresses if not considered properly in design. In bridges, improper constraint of beams and slabs can result in large compressive stresses and resulting buckling failures due to temperature changes. In statically indeterminate trusses, members subjected to large temperature changes can result in stresses induced in members of the truss. Similarly, machine parts constrained from expanding or contracting may have large stresses induced in them due to temperature changes. Composite members made of two or more different materials
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α₁ α₂
Figure 15–1 Composite member composed of two materials with different coefficients of thermal expansion
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A L B
(a)
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L δT A B
Figure 15–2 (a) Unconstrained member, and (b) same member subjected to uniform temperature increase
may experience large stresses due to temperature change if they are not thermally compatible; that is, if the materials have large differences in their coefficients of thermal expansion, stresses may be induced even under free expansion (Figure 15–1)
When a member undergoes a temperature change the member attempts to change dimensions. For an unconstrained member AB (Figure 15–2) undergoing uniform change in temperature T, the change in the length L is given by
\delta_ {T} = \alpha T L \tag {15.1.1}
where a is called the coefficient of thermal expansion and T is the change in temperature. The coefficient a is a mechanical property of the material having units of 1/ F (where F is degrees Fahrenheit) in the USCS of units or 1/ C (where C is degrees Celsius) in the SI system. In Eq. (15.1.1), \delta _ { T } is considered to be positive when expansion occurs and negative when contraction occurs. Typical values of a are: for structural steel \alpha = 6 . 5 \times 1 0 ^ { - 6 } / ^ { \circ } \mathrm { F } ( 1 2 \times 1 0 ^ { - 6 } ) / ^ { \circ } \mathrm { C } and for aluminum alloys \alpha = 1 3 \times 106/ F ð23 106Þ/ C.
Based on the definition of normal strain, we can determine the strain due to a uniform temperature change. For the bar subjected to a uniform temperature change T (Figure 15–2), the strain is the change in a dimension due to a temperature change divided by the original dimension. Considering the axial direction, we then have
\varepsilon_ {T} = \alpha T \tag {15.1.2}
Since the bar in Figure 15–2 is free to expand, that is, it is not constrained by other members or supports, the bar will not have any stress in it. In general, for statically determinate structures, a uniform temperature change in one or more members does not result in stress in any of the members. That is, the structure will be stress-free. For statically indeterminate structures, a uniform temperature change in one or more members of the structure usually results in stress \sigma _ { T } in one or more members.
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σ σ = E(ε - ε_T) E 1 ε₀ = ε_T ε
Figure 15–3 Linear stress=strain law with initial thermal strain
We can have strain due to temperature change \varepsilon _ { T } without stress due to temperature change, and we can have \sigma _ { T } without any actual change in member lengths or without strains.
We will now consider the one-dimensional thermal stress problem. The linear stress/strain diagram with initial (thermal) strain \left( \varepsilon _ { 0 } = \varepsilon _ { T } \right) is shown in Figure 15–3.
For the one-dimensional problem, we have, from Figure 15–3,
\varepsilon_ {x} = \frac {\sigma_ {x}}{E} + \varepsilon_ {T} \tag {15.1.3}
If, in general, we let 1 / E = \underline { { D } } ^ { - 1 } , then in general matrix form Eq. (15.1.3) can be written as
\underline {{\varepsilon}} = [ D ] ^ {- 1} \underline {{\sigma}} + \underline {{\varepsilon}} _ {T} \tag {15.1.4}
From Eq. (15.1.4), we solve for s as
\underline {{{\sigma}}} = \underline {{{D}}} (\underline {{{\varepsilon}}} - \underline {{{\varepsilon}}} _ {T}) \tag {15.1.5}
The strain energy per unit volume (called strain energy density) is the area under the \sigma - \varepsilon diagram in Figure 15–3 and is given by
u _ {0} = \frac {1}{2} \underline {{{\sigma}}} (\underline {{{\varepsilon}}} - \underline {{{\varepsilon}}} _ {T}) \tag {15.1.6}
Using Eq. (15.1.5) in Eq. (15.1.6), we have
u _ {0} = \frac {1}{2} (\underline {{\varepsilon}} - \underline {{\varepsilon}} _ {T}) ^ {T} \underline {{D}} (\underline {{\varepsilon}} - \underline {{\varepsilon}} _ {T}) \tag {15.1.7}
where, in general, the transpose is needed on the strain matrix to multiply the matrices properly.
The total strain energy is then
U = \int_ {V} u _ {0} d V \tag {15.1.8}
Substituting Eq. (15.1.7) into Eq. (15.1.8), we obtain
U = \int_ {V} \frac {1}{2} \left(\underline {{\varepsilon}} - \underline {{\varepsilon}} _ {T}\right) ^ {T} \underline {{D}} \left(\underline {{\varepsilon}} - \underline {{\varepsilon}} _ {T}\right) d V \tag {15.1.9}
Now, using \underline { { \varepsilon } } = \underline { { B } } \underline { { d } } in Eq. (15.1.9), we obtain
U = \frac {1}{2} \int_ {V} \left(\underline {{B}} \underline {{d}} - \underline {{\varepsilon}} _ {T}\right) ^ {T} \underline {{D}} \left(\underline {{B}} \underline {{d}} - \underline {{\varepsilon}} _ {T}\right) d V \tag {15.1.10}
Simplifying Eq. (15.1.10) yields
U = \frac {1}{2} \int_ {V} \left(\underline {{d}} ^ {T} \underline {{B}} ^ {T} \underline {{D}} \underline {{B}} \underline {{d}} - \underline {{d}} ^ {T} \underline {{B}} ^ {T} \underline {{D}} \underline {{\varepsilon}} _ {T} - \underline {{\varepsilon}} _ {T} ^ {T} \underline {{D}} \underline {{B}} \underline {{d}} + \underline {{\varepsilon}} _ {T} ^ {T} \underline {{D}} \underline {{\varepsilon}} _ {T}\right) d V \tag {15.1.11}
The first term in Eq. (15.1.11) is the usual strain energy due to stress produced from mechanical loading—that is,
U _ {L} = \frac {1}{2} \int_ {V} \underline {{d}} ^ {T} \underline {{B}} ^ {T} \underline {{D}} \underline {{B}} \underline {{d}} d V \tag {15.1.12}
Terms 2 and 3 in Eq. (15.1.11) are identical and can be written together as
U _ {T} = \int_ {V} \underline {{d}} ^ {T} \underline {{B}} ^ {T} \underline {{D}} \underline {{\varepsilon}} _ {T} d V \tag {15.1.13}
The last (fourth) term in Eq. (15.1.11) is a constant and drops out when we apply the principle of minimum potential energy by setting
\frac {\partial U}{\partial \underline {{d}}} = 0 \tag {15.1.14}
Therefore, letting U = U _ { L } + U _ { T } and substituting Eqs. (15.1.12) and (15.1.13) into Eq. (15.1.14), we obtain two contributions as
\frac {\partial U _ {L}}{\partial \underline {{d}}} = \int_ {V} \underline {{B}} ^ {T} \underline {{D}} \underline {{B}} d V \underline {{d}} \tag {15.1.15}
and qUTqd ¼ ð V BTDeT dV ¼ f fTg \frac { \partial U _ { T } } { \partial \underline { { d } } } = \int _ { V } \underline { { B } } ^ { T } \underline { { D } } \underline { { \varepsilon } } _ { T } d V = \{ f _ { T } \} ð15:1:16Þ
We recognize the integral term in Eq. (15.1.15) that multiplies by the displacement matrix \underline { d } as the general form of the element stiffness matrix \underline { { k } } , , whereas Eq. (15.1.16) is the load or force vector due to temperature change in the element.
We will now consider the one-dimensional thermal stress problem. We define the thermal strain matrix for the one-dimensional bar made of isotropic material with coefficient of thermal expansion \alpha , and subjected to a uniform temperature rise T , as
\{\varepsilon_ {T} \} = \{\varepsilon_ {x T} \} = \{\alpha T \} \tag {15.1.17}
where the units on a are typically ( \mathrm { i n . / i n . } ) / ^ { \circ } \mathrm { F } or ( { \mathrm { m m / m m } } ) / { } ^ { \circ } { \mathrm { C } } .
For the simple one-dimensional bar (with a node at each end), we substitute Eq. (15.1.17) into Eq. (15.1.16) to obtain the thermal force matrix as
\left\{f _ {T} \right\} = A \int_ {0} ^ {L} [ B ] ^ {T} [ D ] \left\{\alpha T \right\} d x \tag {15.1.18}
Recall that for the one-dimensional case, from Eqs. (3.10.15) and (3.10.13), we have
[ D ] = [ E ] \quad [ B ] = \left[ - \frac {1}{L} \quad \frac {1}{L} \right] \tag {15.1.19}
Figure 15–4 Differential two-dimensional element (a) before and (b) after being subjected to uniform temperature change for an anisotropic material
Substituting Eqs. (15.1.19) into Eq. (15.1.18) and simplifying, we obtain the thermal force matrix as
\left\{f _ {T} \right\} = \left\{ \begin{array}{c} f _ {T 1} \\ f _ {T 2} \end{array} \right\} = \left\{ \begin{array}{c} - E \alpha T A \\ E \alpha T A \end{array} \right\} \tag {15.1.20}
For the two-dimensional thermal stress problem, there will be two normal strains, \varepsilon _ { x T } and \varepsilon _ { y T } along with a shear strain \gamma _ { x y T } due to the change in temperature because of the different mechanical properties (such as E _ { x } \neq E _ { y } ) in the x and y directions for the anisotropic material (See Figure 15–4). The thermal strain matrix for an anisotropic material is then
\{\varepsilon_ {T} \} = \left\{ \begin{array}{l} \varepsilon_ {x T} \\ \varepsilon_ {y T} \\ \gamma_ {x y T} \end{array} \right\} \tag {15.1.21}
For the case of plane stress in an isotropic material with coefficient of thermal expansion a subjected to a temperature rise T, the thermal strain matrix is
\{\varepsilon_ {T} \} = \left\{ \begin{array}{c} \alpha T \\ \alpha T \\ 0 \end{array} \right\} \tag {15.1.22}
No shear strains are caused by a change in temperature of isotropic materials, only expansion or contraction.
For the case of plane strain in an isotropic material, the thermal strain matrix is
\left\{\varepsilon_ {T} \right\} = (1 + \nu) \left\{ \begin{array}{c} \alpha T \\ \alpha T \\ 0 \end{array} \right\} \tag {15.1.23}
For a constant-thickness ðtÞ, constant-strain triangular element, Eq. (15.1.14) can be simplified to
\left\{f _ {T} \right\} = [ B ] ^ {T} [ D ] \left\{\varepsilon_ {T} \right\} t A \tag {15.1.24}
The forces in Eq. (15.1.24) are contributed to the nodes of an element in an unequal manner and require precise evaluation. It can be shown that substituting Eq. (6.1.8) for ½D , Eq. (6.2.34) for ½B , and Eq. (15.1.22) for \left\{ \varepsilon _ { T } \right\} for a plane stress condition
into Eq. (15.1.24) reveals the constant-strain triangular element thermal force matrix to be
\left\{f _ {T} \right\} = \left\{ \begin{array}{c} f _ {T i x} \\ f _ {T i y} \\ \vdots \\ f _ {T m y} \end{array} \right\} = \frac {\alpha E t T}{2 (1 - \nu)} \left\{ \begin{array}{c} \beta_ {i} \\ \gamma_ {i} \\ \beta_ {j} \\ \gamma_ {j} \\ \beta_ {m} \\ \gamma_ {m} \end{array} \right\} \tag {15.1.25}
where the \beta \mathbf { \hat { s } } and \gamma \smash { \vdots } s are defined by Eqs. (6.2.10).
For the case of an axisymmetric triangular element of isotropic material subjected to uniform temperature change, the thermal strain matrix is
\left\{\varepsilon_ {T} \right\} = \left\{ \begin{array}{l} \varepsilon_ {\tau T} \\ \varepsilon_ {z T} \\ \varepsilon_ {\theta T} \\ \gamma_ {r z T} \end{array} \right\} = \left\{ \begin{array}{l} \alpha T \\ \alpha T \\ \alpha T \\ 0 \end{array} \right\} \tag {15.1.26}
The thermal force matrix for the three-noded triangular element is obtained by substituting the B from Eq. (9.1.19) and Eq. (9.1.21) into the following:
\underline {{f}} _ {T} = 2 \pi \int_ {A} \underline {{\varepsilon}} ^ {T} \underline {{D}} \varepsilon_ {\underline {{T}}} r d A \tag {15.1.27}
For the element stiffness matrix evaluated at the centroid (-r, -z), Eq. (15.1.25) becomes
\underline {{f}} _ {T} = 2 \pi \bar {r} A \overline {{B}} ^ {T} \underline {{D}} \underline {{\varepsilon}} _ {T} \tag {15.1.28}
where \underline { { \overline { { B } } } } is given by Eq. (9.2.3), A is the surface area of the element which can be found in general from Eq. (6.2.8) when the coordinates of the element are known and D is given by Eq. (9.2.6).
We will now describe the solution procedure for both one- and two-dimensional thermal stress problems.
Step 1
Evaluate the thermal force matrix, such as Eq. (15.1.20) or Eq. (15.1.25). Then treat this force matrix as an equivalent (or initial) force matrix \underline { { F } } _ { 0 } analogous to that obtained when we replace a distributed load acting on an element by equivalent nodal forces (Chapters 4 and 5 and Appendix D).
Step 2
Apply \underline { { F } } = \underline { { K } } \underline { { d } } - \underline { { F } } _ { 0 } , where if only thermal loading is considered, we solve \underline { { F } } _ { 0 } = \underline { { K } } \underline { { d } } for the nodal displacements. Recall that when we formulate the set of simultaneous equations, \underline { { F } } represents the applied nodal forces, which here are assumed to be zero.