(1) σ (may be anisotropic) (2)

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1 T7S01 Session 1 Thermal Stresses (Linear Elasticity Only) Last Update: 18/5/14 Thermal stresses: formulation of e general elastic problem; Constrained and unconstrained cases; Uniform temperature gradient gives zero stress if unconstrained; Examples, e.g. temperature gradient rough a vessel wall; Anisotropy and inhomogeneity of α; Basics of heat transfer: Conduction equation; Convective boundary conditions, heat transfer coefficient; 1D temperature distribution, transient versus steady state Revision- What are e equations for e general elasticity problem? When ere are no ermal stresses, e ree sets of equations are (as discussed in session 7), Equilibrium: Hooke s Law: σ = b (1), j = Cklε kl 1 Compatibility: = ( u + u ) i σ (may be anisotropic) () ε i,j i, j j,i (3a) The last, e compatibility equations, can alternatively be written wiout reference to e displacements, i.e., as, ε xx,yy + ε yy,xx = ε xy,xy ; xy,xz + ε xz,xy = ε yz,xx + ε xx, yz ε ; etc. (3b) In is formulation, b is e applied load per unit volume, and ere will also be, in general, stresses and/or displacements prescribed as boundary conditions on e surface of e body. How does ermal loading arise? The reason why temperature distributions cause stress is at ey cause strain. That is, materials generally expand when eir temperature increases. The effect of is ermal expansion is to cause a purely ermal strain, ε, on an element of material which is raised in temperature uniformly by T. What is e coefficient of ermal expansion and is it a scalar? For small changes in temperature, e ermal strain will be proportional to e temperature change. The factor of proportionality is e coefficient of ermal expansion. Since strain is a nd rank tensor and temperature is a scalar, e most general coefficient of ermal expansion will be a nd rank tensor, α, ε = α T (4) In general, for anisotropic materials, e expansion may be different in e ree coordinate directions and ermal expansion might also cause shear strains. However, an isotropic material will expand isotropically, in which case e coefficient of ermal expansion reduces to a scalar, ε = α Tδ (5)

2 Here δ is e Kronecker delta ( δ = 1 if i= j, δ = 0 if i j ). For e examples below we shall assume isotropy, i.e., Equ.(5). There is a brief discussion of anisotropy later. Does ermal strain really vary linearly wi temperature? No. But is is addressed simply by accepting at e coefficient of expansion is temperature dependent. The temperature dependence is not dramatic. For example, a 6 CMn steel might have α ~ 1 10 / o C at 0 o 6 C, increasing to α ~ / o C at 350 o C (see R66). What are instantaneous and average coefficients of expansion? The instantaneous coefficient of expansion applies over a small range of temperature around e temperature, T, at which it is specified. It is defined by, instant dε instant α ( T ) = dε = α ( T )dt (6) dt However, we often want to know e total ermal strain due to a large temperature change, from T 1 tot. In e case where e lower temperature is 0 o C, is defines e average coefficient of expansion, o ( T )( T 0 C) average ε = α (7) From is it follows at e ermal strain increment due to a large temperature change from T 1 tot is, o average o ( T )( T C) α ( T )( T 0 C) average ε = α (8) The same ermal strain increment can be found in terms of e instantaneous coefficient of expansion, but is must be known at all temperatures fromt 1 tot and requires an integral to be carried out, T ( T ) ins tan t ε = α dt (9) T 1 R66 most often quotes average coefficients of expansion. However, R66 does give instantaneous values as well for some materials. Care is required when taking data from oer sources, since ey sometimes omit to tell you which definition is being used. WRONG! So, to find e stress we just plug e ermal strain into Hooke s Law? Why? What stress arises if a body is changed in temperature uniformly, and e body is not externally restrained from expansion? Assuming e material is homogeneous, is produces no stress. Why? Because e body is simply free to expand. But e body does expand, and so ere is a ermal strain namely e strain given by Equs.(4) or (5). There is strain, but no stress. This demonstrates e most important fact Thermal strain does not, of itself, give rise to stress

3 But ermal strains do sometimes cause stress, don t ey? Yes, of course. Here is one way in which ermal stresses can arise. Heat a bar to a higher temperature, but ensure at it cannot expand by restraining its ends. Since e expansion is now zero, e total (longitudinal) strain is zero. But we know at e free ermal strain would have been α T if e bar had not been restrained. So e el restraint has caused an additional, equal and opposite, strain of ε = α T to arise. This is e elastic strain which corresponds to stress via Hooke s Law. The ermal stress is us σ = Eε el = Eα T, compressive because e expansion has been prevented. The total strain is e sum of e ermal and elastic strains, ε = ε + ε. For a case of complete restraint of e ermal expansion, like is example, e total strain is zero. It is only e elastic part of e strain which relates to stress via Hooke s Law total el It is e total strain which is observable as e change in e dimensions of e body. It is e total strain which is related to e displacements in e usual manner. How is e general elastic ermal stress problem formulated? The stresses must still be in equilibrium, whatever e nature of e loading, so Equ.(1) still holds. Similarly, Hooke s Law, Equ.(), still holds provided at by strain we mean e elastic strain, not e total strain or e ermal strain. The most significant change is in regard to compatibility. It is e total strain which must be compatible wi e displacements of e material. Finally, we also have an extra equation namely Equ.(4) or (5) which specifies e ermal strain. The complete set of equations for a ermal stress problem is us, Equilibrium: Hooke s Law: σ = b (10), j el Cklε kl total 1 Compatibility: = ( + u ) i σ = (may be anisotropic) (11) ε i, j u i, j j, i (1) Thermal Expansion: ε = α T or ε = α Tδ (13) Plus: total el ε = ε + ε (14) Note at oer loads may be applied as well as ermal loading. The compatibility total total total equations, (1), may also be expressed in e alternative form ε xx, yy + ε yy, xx = ε xy, xy, ε + +, etc. total total total total xy, xz ε xz, xy = ε yz, xx ε xx, yz

4 WRONG! So, a body has to be externally restrained to produce ermal stress? A body which is completely free can also give rise to ermal stress but e ermal strains must be non-uniform. Thermal stresses can erefore arise in isotropic materials due to non-uniform temperature distributions. Alternatively, a uniform temperature change can result in ermal stresses if e coefficient of ermal expansion varies across e body (i.e., e body is inhomogeneous). Bo of ese situations cause non-uniform ermal strains, and hence ermal stresses (in general). But not all non-uniform temperature distributions will give non-zero ermal stresses. But doesn t a temperature gradient cause ermal stress? Not necessarily, no. This is a common, but serious, misunderstanding. It is quite common for analysts to assume at e ermal stresses in a structure are proportional to e temperature gradient in some region of e body. Actually, is can sometimes be a reasonable approximation if e temperature gradient exists only over a portion of e body, and separates two regions of uniform, but different, temperature. But if e body is unrestrained and has a temperature gradient which is uniform across e whole body, and assuming a homogeneous material, en e ermal stresses are zero! If a body of homogeneous material is unrestrained and its temperature varies linearly wi e Cartesian x,y,z coordinates, en e ermal stresses are zero. Is is simple to prove? Yes. Or, at least, it is simple so long as we take it on trust at e solution to e set of equations (10-14) is unique. Assume at e stresses are indeed zero. This means at e elastic strains are zero, total and hence at e total strains equal e ermal strains, i.e., ε = α T. So we have already ensured at equations (10), (11), (13) and (14) are obeyed. The only ing left is compatibility. Now if we use e alternative form of e compatibility total total total total total total total relations, i.e., ε xx, yy + ε yy, xx = ε xy, xy, ε xy, xz + ε xz, xy = ε yz, xx + ε xx, yz, etc., we see at all terms are second order derivatives in e Cartesian coordinates. Since, by assumption, e expansion coefficients, α, are uniform, and e temperature change, T, varies linearly wi e Cartesian coordinates, it follows at all ese terms are zero. Hence compatibility is respected. So all e required equations are obeyed, and hence e stresses are indeed zero. Note at e proof applies even for anisotropic materials, provided at e structure is homogeneous. If e temperature was linear in polar coordinates, would e ermal stresses be zero? No, not in general. This is because when e compatibility equations are expressed in polar coordinates, ere are terms involving e first derivatives as well as e second derivatives so e above proof breaks down.

5 To express is in physical terms consider a cylinder. If e temperature varies linear wi e radial coordinate, r, is will clearly cause hoop stresses because e inner and outer surfaces will want to expand by differing amounts. In summary, when do we expect non-zero ermal stresses? Thermal stresses will generally be non-zero when, [1] The body is unrestrained and homogeneous but subject to a temperature distribution which deviates from being linear wrt e Cartesian coordinates, or, [] The body is unrestrained but has an inhomogeneous coefficient of expansion and is subject to an arbitrary temperature change (possibly uniform), or, [3] A homogeneous body is restrained in some way and is subject to an arbitrary temperature change (possibly uniform). Example: Flat Plate Restrained From Bending (but free to expand) wi Uniform Temperature Gradient Through-Thickness, Temperature Difference T Between Surfaces The temperatures on e surfaces are ± T / wi respect to e middle. (Overall expansion is not restrained so we re not concerned about e mean temperature change). So e ermal strains on e surfaces are ±α T / in all ree directions (assuming an isotropic and homogeneous material). The plate is fully restrained from el bending, so ere is no net bending strain. Hence ε = ε = m α T /, where e terms relate to bending strains. The in-plane directions, x and y, are equivalent, so surface stresses σ σ = σ. The x = y normal stress at e free surface is zero, σ z= 0. Hooke s Law for e in-plane strains is us, That is, e in-plane surfaces stresses are, el ( 1 ν ) σ = E T / Eε = Eε = σ νσ = m α x y Eα T σ = m (15) 1 ( ν ) Note at e stress is tensile on e side where e temperature is lowest ( T / wrt e middle). This is an equi-biaxial stress in all in-plane directions of e plate. Example: Thermal Shock of a Cylindrical Vessel / Pipe A rapid change in temperature of e fluid which is in contact wi e wall of e pipe causes a very in layer of e surface of e pipe wall to cool by T. It is assumed at, transiently, e bulk of e wall ickness remains unchanged in temperature. Consequently e bulk of e wall effectively fully restrains e in layer whose temperature has dropped. The same analysis as above en gives, Eα T σ = ( 1 ν ) The minus sign indicates at cooling causes tension. Note at e stress in is case is double at derived above for a uniform temperature gradient. This stress is also equi-biaxial (i.e. bo hoop and axial). (16)

6 Can we safely assume at cases of interest to us are isotropic? To be honest, not really, no. Nevereless, analyses are usually based on an isotropic (i.e., scalar) α. This is OK for parent material in most cases, if grain sizes are small and eir orientation is random. Then isotropic behaviour will prevail, on average, on structural scales. (What might happen at grain boundaries is anoer, raer interesting, question). The most obviously anisotropic situation is for weldments. Grains are not randomly oriented in weld metal. The cooling process, proceeding from e fusion boundary inwards towards e weld centre, leads to a preferred spatial direction. For example, e weld metal grains at e fusion boundary are often columnar and oriented perpendicular to e fusion face. Weld metal can have significantly different expansion coefficients parallel to, and transverse to, ese columnar grains. Most often if weld expansion is measured at all it will be measured on a bulk weldment specimen in directions longitudinal and transverse to e welding direction. This is not e same ing as e anisotropic expansion of e individual weld grains, being bo a bulk measurement and not aligned wi e individual grain axes. The HAZ can also be affected for similar reasons, ough probably to a lesser extent. (Actually Young s modulus can be quite different in weld metal compared wi parent, which is a furer reason why ermal stresses in weld material may differ from ose calculated for parent). Since weldments are e feature of principal concern in most assessments, is issue might be significant. However, if e ermal stresses arise due to large scale temperature gradients, e weldment anisotropy might cause only a minor blip in e stress field. On e oer hand, if e ermal stresses arise due to temperature gradients over a spatial scale similar to e size of e weld, en e weld anisotropy may result in significant deviations from e stresses derived assuming isotropy. When is e ermal expansion coefficient inhomogeneous? To some degree, even if anisotropy is ignored, e mean value of α will vary between weld and parent material, so a degree of inhomogeneity is common, ough usually ignored. The really important cases of inhomogeneous α are transition joints. That is, welds joining a ferritic material to an austenitic material. The former typically have an α in e range 1 14 x 10-6 / o C, whereas e latter is more like x 10-6 / o C. This alpha mismatch over e fusion face creates large, but very localised, ermal stresses. R5 advises on e magnitude of alpha mismatch stress to assume (variously E α T or 1.5E α T, depending upon e context).

7 Heat Transfer This is really a different SQEP area, so only a few general observations will be made. What process governs e temperature distribution in a solid? The temperature distribution in a solid is controlled by e conduction of heat (since convection and radiation are not relevant wiin a solid). What is e definition of ermal conductivity Heat flows from hot to cold: e steeper e temperature gradient, e faster e flux of heat. The ermal conductivity is e factor of proportionality between e two, Q = K T (3D) (17a) dt Q = K (1D) (17b) dx where Q is e heat flux (energy rough unit area per second), and K is e ermal conductivity, and T is e temperature at position r (or x). What is e definition of specific heat The specific heat, C, is e energy required to raise e temperature of one kg of p material by 1 o C. If e material density is ρ (in units kg/m 3 ) e energy required to raise e temperature of 1m 3 of material by 1 o C is us C. What is e heat balance equation? Assume at a certain region of a body contains no heat sources or sinks. If e T temperature is increasing at a rate, en e energy per unit volume of material at t T at point is increasing at e rate C p ρ. But e heat flowing out of is volume t dq per second equals e divergence of e heat flux, i.e., Q, or in 1D. In e dx absence of sources and sinks, e conservation of energy us requires at a net outward flux of energy must be balanced by a reduction in e heat energy wiin e region, and hence a reduction of its temperature. Thus, e heat balance equation is, ρ p T Q = C pρ (3D) (18a) t dq T = C pρ (1D) (18b) dx t What equation governs e temperature distribution in a solid? The heat conduction equation follows by substituting (17a) or (17b) into (18a) or (18b), giving, T κ T = (3D) (19a) t T T κ = (1D) (19b) x t

8 K where e ermal diffusivity is κ=. ρ Boundary Conditions for e Heat Conduction Equation The following types of boundary condition are all important, C p [1] Specified, constant, boundary temperature; [] Specified, but varying, boundary temperature as a given function of time; [3] Specified heat flux into/out of a surface, as a given function of time; [4] Adiabatic boundary (i.e., insulated): This is when e heat flux at e surface is specified to be zero at all times. As a consequence of Equ.(17), is is equivalent to requiring e gradient of e temperature normal to e surface to be zero. [In mas speak, [3] and [4] are Neumann boundary conditions, whereas [1] and [] are Dirichlet boundary conditions]. [5] Convective boundary condition: In is case we do not know in advance eier e temperature or e heat flux at e boundary. Instead e temperature of a fluid in contact wi e boundary is specified. The heat flux into e surface of e solid is en specified via a heat transfer coefficient, h, as Q = h( T f T s ), where T f is e fluid temperature, and Ts is e solid temperature at its surface. Discussion of how to estimate e heat transfer coefficient, h, would take us firmly into e ermal analysis SQEP area. Seek advice! [6] Radiative boundary condition: A surface will radiate heat into its surroundings in accord wi Stefan s four-power law. Equally, e surface will absorb radiative heat from its surroundings. Thus, e net heat flux into e surface is 4 4 Q = εσ ( T env T s ) where σ is Stefan s constant and ε is an emissivity/absorptivity factor. T env is e temperature of e surroundings. The temperatures in is expression must be absolute. In tru, a radiative boundary condition is more complicated an is because e radiation may originate from anywhere wiin line of sight of e surface and is will generally involve regions of widely varying temperature. The resulting integrated effect is best left to experts. Units? 1 1o 1 K= Jm s C ; C 1o 1 p = Jkg C ; What is meant by penetration dep? 1 κ = m s 1o 1 h= Jm s C Exact analytic solutions to e heat conduction equation tend to be awkward, generally involving e sum of an infinite series. But a good intuitive feel for e conduction behaviour during a ermal transient can be gained using e concept of penetration dep. This is how far heat has penetrated from e surface in a given time. Be warned, is is an approximate concept only. Since e only material property which enters e heat conduction equation, (19), is 1 e ermal diffusivity, κ, and since its dimensions are κ = m s, dimensional analysis implies at e penetration dep must be roughly of order d ~ κt at time t. A more careful study of a range of accurate solutions shows at a better approximation is, d 3κt (0)

9 What is meant by transit time? If we consider a slab of material of a given ickness, w, e transit time,τ, is e time taken for heat to cross from one surface to e oer. In oer words, it is e time when e penetration dep equals e ickness. Hence, () gives, w τ= (1) 3κ Note at transit time is also just an approximate concept. When are transient ermal stresses significant? Consider a vessel or pipe wi an initially uniform temperature but which contains a flowing fluid whose temperature undergoes a change. This will in general produce a temperature variation rough e wall of e vessel, and hence a ermal stress. But if e fluid temperature changes temperature sufficiently slowly e conduction of heat rough e wall of e vessel may be able to maintain e vessel's material temperature close to uniform. In is case e ermal stresses will be small. This situation is more likely to occur for materials wi high ermal conductivity because e temperature gradient depends upon e competition between e rate of heat flowing into e surface from e fluid and e rate of heat conduction away from e surface. Consequently e ermal stresses will be small if K is large or h is small. A dimensionless quantity which gives a feel for wheer ermal transient stresses are likely to be large or small is erefore e Biot number, defined by, hl Bi= () K where L is a characteristic dimension, often taken as e ratio of e volume of e body to e surface area in contact wi e fluid. For vessels it can simply be taken to be e wall ickness. If Bi << 1 en ermal transient stresses are expected to be small. Conversely if Bi >> 1 en ermal transient stresses might be large (ough, even en, not necessarily - as we see below). This can be seen from e following diagram, representing e temperatures during an up-shock, fluid T f T 1 vessel wall L T The rate of flow of heat from e fluid into e surface per unit area is ( T ) whereas e rate of flow of heat by conduction across e wall ickness is K ( T1 T )/ L. Equating ese two heat flow rates gives, h f, T 1

10 Bi hl T T 1 = = (3) K Tf T1 From (3) we see at e temperature difference across e wall, T1 T, will be small if Bi << 1. Conversely, e temperature difference across e wall, T1 T, will be many times larger an e fluid/metal surface temperature difference, T f T1, if Bi >> 1. The latter condition does not necessarily imply large ermal stresses, however, because e rate of change of fluid temperature might be sufficiently slow at T f T1 remains tiny at all times. We conclude e following, If Bi << 1 e ermal transient stresses will be small. If dt f / dt is sufficiently slow en e ermal transient stresses will be small. If Bi >> 1 and dt f / dt is sufficiently fast en e ermal transient stresses will be large. Wheer e fluid rate-of-change-of-temperature is slow or fast may be judged dt f L from how much its temperature changes in one transit time. Thus if is dt 3κ just a few degrees or less en e rate is slow and ermal stresses are likely to be small. Example Approximate Solution for Conduction Only Consider a slab of material of ickness w which is originally isoermal (at datum zero temperature). Suppose at one surface, at x = 0, is raised suddenly to a temperature T 0 and held fixed at is temperature. The oer surface (at x = w) is insulated (adiabatic). The gradient of e temperature distribution must erefore be zero at x = w. Hence we assume for simplicity at e temperature distribution is roughly parabolic, of e form, ( ) x T x, t = MAX 1 ( T0 + T1 ) T1, 0 () w where e MAX part ensures at if e function on e RHS is negative en it is replaced by 0. Equ.() obeys e boundary conditions on bo x = 0 and x = w. (But it is an approximation only, because it does not exactly obey e heat conduction equation). The term T 1 is a function of time, and has yet to be determined. We find it by requiring at () is consistent wi e penetration dep given by (0). Hence, we require, 3κt w ( T + T ) T = (3) giving, for 0 t τ : T T 1 ( ) = 0 3κ t t = t w τ (4)

11 Using (4) wiin () gives our approximate solution. w For times later an e transit time, t>, e temperature of e insulated face of 3κ e slab starts to increase and (4) is not appropriate. However, we can appeal to anoer physical approximation. Retaining e form () we can evaluate e average temperature, and hence e rate of increase of ermal energy in e body. This can be equated to e rate at which energy is flowing into e body, which is given by (17b) using e temperature gradient at x=0 from (). This gives, For t > τ : (noting at T 1 < 0 for T 1 T ( t) 0 t = exp 1 1 τ t > τ, and e temperature of e insulate face is T1 ). This example illustrates how rough estimates of transient temperature distributions can be found using physically motivated approximations, and wiout solving e strict conduction equation. I don t vouch for e accuracy of (,4,5), but ese solutions do have roughly e right qualitative form as shown by e Figure below. A prediction from is approximate solution is at e temperature across e whole slab becomes a uniform T 0, to a tolerance of %, after 5 times e transit time, i.e., 5w after e time t uniform =. I d be interested to know if is is accurate. 3κ (5)

12 Figure Approximate solution, Equs.(,4,5), for e temperature distribution rough a slab of material whose surface undergoes a step change in temperature at time zero: Temperature normalised by e surface temperature change (T/T 0 ) versus normalised position (x/w). Each curve corresponds to a different time, as shown in e legend normalised by e transit time t= t=0.001 t=0.01 t=0.05 t=0.1 t=0. t=0.3 t=0.4 t=0.5 t=0.6 t=0.7 t=0.8 t=0.9 t=1 t=1.1 t=1. t=1.4 t=1.7 t= t=.5 t=3 t=3.5 t=4 t=5 t=6 What is a steady state temperature distribution? The steady state temperature distribution is simply what e temperature distribution tends to after a long time. If e boundary conditions continue to change indefinitely, for example if we apply a sinusoidally varying surface temperature, en ere will be no steady distribution. The simplest example of a steady state distribution occurs when all e boundary conditions are prescribed, constant temperatures. A qualitative picture of what e steady state temperature distribution will look like can be constructed as follows, Draw lines of heat flux, flowing from e hot boundaries to e cool boundaries; The isoerms (lines of constant temperature) are en drawn by following e rule at ey are everywhere perpendicular to e lines of heat flux. For a 1D problem, e steady state is a linear distribution (i.e., a uniform temperature gradient).